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Finding Slope 2009
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Finding Slope 2009
1. 1. Objective find the slope of a line given a line on a graph. The students will be able to: find the slope of a line given 2 points on a graph. find the slope of a line given 1 point and the slope of the graph.
3. 3. What does the 7% mean? 7% is the slope of the road. It means the road drops 7 feet vertically for every 100 feet horizontally. So, what is slope??? 7 feet 100 feet Slope is the steepness of a line. 7%
4. 4. Slope can be expressed differently as: A line has a positive slope if it slopes uphill from left to right . A line has a negative slope if it slopes downhill from left to right .
5. 5. When given the graph, it is easier to apply “ rise over run ” . Determine the slope of the line given a line on a graph.
6. 6. Determine the slope of the line. Start with the lower point and count how much you rise and run to get to the other point! 6 3 run 3 6 = = rise Notice the slope is positive! the line slopes upward and to the right 1 2
7. 7. Find the slope of the line that passes through the points (-2, -2) and (4, 1). When given points, it is easier to use the formula. y 2 is the y coordinate of the 2 nd ordered pair (y 2 = 1) y 1 is the y coordinate of the 1 st ordered pair (y 1 = -2)
8. 8. Did you notice that both previous examples were the same problem written differently? (-2, -2) and (4, 1) Which method do you think is easiest? You can do the problems either way! 6 3
9. 9. Find the slope of the line that goes through the points (-1, 2) and (1, 4). a. -1 b. 1 c. 2 d. 4
10. 10. Determine the slope of the line shown. a. -3 b. 3 c. -3/5 d. 3/5
11. 11. Determine the slope of the line using any two points on the line. The line is decreasing (slope is negative). 5 -2.5 Choose two points on the graph. Apply rise over run.
12. 12. What is the slope of a vertical line? <ul><li>Notice the line doesn’t run! </li></ul>All vertical lines have an undefined slope.
13. 13. What is the slope of a horizontal line? <ul><li>Notice the line doesn’t rise! </li></ul>All horizontal lines have a slope = 0.
14. 14. Remember the word “VUXHOY” V ertical lines U ndefined slope X = number; this is the equation of the line. H orizontal lines O - zero is the slope Y = number; this is the equation of the line.
15. 15. Draw a line through the point (2,0) that has a slope of 3. <ul><li>1. Plot the ordered pair (2, 0). </li></ul>1 3 2. From (2, 0), apply rise over run (write 3 as a fraction). 3. Plot a point at this location. <ul><li>Draw a straight line through </li></ul><ul><li>the points. </li></ul>
16. 16. The slope of a line that goes through points (r, 6) and (4, 2) is 4. Find r. To solve this, plug the given information into the formula continued
17. 17. We must solve for r (x 1 ), simplify and write as a proportion. x (4 – r ) multiply each side by the common denominator x (4 – r ) 16 – 4 r = -4 continued
18. 18. Simplify and solve the equation. The ordered pairs are (5, 6) and (4, 2) |
Sections:
# Subtraction Test #2
In this section, we continue to review subtraction. Subtraction is the act of taking away. We previously learned to identify the parts of a subtraction problem such as: minuend, subtrahend, and difference.
Example 1: Identify the parts of the subtraction problem: 7 - 4 = 3
7 (Minuend) This is the staring amount or whole amount we are taking away from
4 (Subtrahend) This is the amount to be subtracted away or taken away
3 (Difference) This is the amount that is left after 4 is taken away from 7
In easier examples, subtraction can be done mentally. What happens when we need to perform subtraction with: 2 digit numbers, 3 digit numbers, etc...This is where we begin subtracting using a vertical format. This is known as vertical subtraction. We can use this operation when we need to subtract multi-digit whole numbers. Vertical subtraction breaks the subtraction of larger numbers up into a series of smaller manageable steps. We begin by aligning our numbers by place value with the minuend on the top and the subtrahend on the bottom. We then subtract in the ones’ place followed by the tens, hundreds,…etc. We will also review subtracting with regrouping. This is also known as subtracting with borrowing. This occurs when we are subtracting in a column and the top number is smaller than the bottom number. |
# Second derivative of $x^{2}$
The calculator will find the second derivative of $x^{2}$, with steps shown.
Related calculators: Derivative Calculator, Logarithmic Differentiation Calculator
Leave empty for autodetection.
Leave empty, if you don't need the derivative at a specific point.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Find $\frac{d^{2}}{dx^{2}} \left(x^{2}\right)$.
### Find the first derivative $\frac{d}{dx} \left(x^{2}\right)$
Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 2$:
$${\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = {\color{red}\left(2 x\right)}$$
Thus, $\frac{d}{dx} \left(x^{2}\right) = 2 x$.
### Next, $\frac{d^{2}}{dx^{2}} \left(x^{2}\right) = \frac{d}{dx} \left(2 x\right)$
Apply the constant multiple rule $\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$ with $c = 2$ and $f{\left(x \right)} = x$:
$${\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$
Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:
$$2 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 2 {\color{red}\left(1\right)}$$
Thus, $\frac{d}{dx} \left(2 x\right) = 2$.
Therefore, $\frac{d^{2}}{dx^{2}} \left(x^{2}\right) = 2$.
$\frac{d^{2}}{dx^{2}} \left(x^{2}\right) = 2$A |
Analytics Check
# The math behind exploding dice rolls
2021/09/19
Exploding dice is a mechanic that appears in various tabletop games. It works like this: a dice is rolled, if it lands on its maximum values, it “explodes” — the dice is then rolled again, adding the new value to the max value of the dice. If dice again lands on its maximum, the dice explodes a second time. This process continues and may lead to chains of explosions that push the roll total far beyond what is normally expected.
This mechanic appears in a number of tabletop games, with Savage Worlds likely being the most popular system within the RPG space that uses it. In this post, we dig into the infinite nature of exploding dice and explore the math and expected outcomes from this fun mechanic.
## Exploding probabilities
Suppose we roll an $n$ sided exploding dice. The probability of rolling a non-exploded value, that is between $1$ and $n-1$ is no different than a normal dice roll. Each face below $n$ appears with equal probability $1/n$. If instead the dice explodes — which happens with the remaining probability $1/n$ — we add $n$ to the outcome of a new exploding dice roll.
Let’s look at an example when we are rolling a 6-sided dice, so $n = 6$. Any result between 1 and 5 appears with probability $1/6$. How about the probability of rolling a 7? Rolling a 7 requires rolling a 6 on the first roll, triggering the explosion, and then rolling a 1 on the second:
$P(X = 7) = P(X = 6)P(X=1) = \frac{1}{6^2}$
Any value of the exploded dice roll can be written like this — as powers of $1/n$. Suppose we want to know the probability of rolling a 25 on an exploding d6. Well that’s just rolling 6 four times in a row followed by a 1, so $p = 1/6^5$. In general, for an $n$ sided dice, the probability of rolling any given value is:
$\boxed{ P(X = x) = \begin{cases} \left(\frac{1}{n}\right)^{\lfloor{x/n}\rfloor + 1} & x \textrm{ not a multiple of }n \\ 0 & \textrm{otherwise} \end{cases} }$
Where $\lfloor{x/n}\rfloor$ represents the rounded down result of dividing $x$ by $n$. The probability of rolling a multiple of $n$ is 0, because every time a multiple of $n$ would show up, it triggers a dice explosion. The below graph shows the outcome probabilities when rolling an exploding d6.
Within a given number of explosions, each outcome is equally likely. As the number of explosions increases, the probability decays rapidly due to the rarity of rolling many 6s in a row. This particular graph cuts off at 18, but in reality it continues to infinity with ever decreasing probabilities.
The probability of at least $k$ explosions on an $n$ sided dice is $P(\textrm{at least k explosions}) = n^{-k}$. Continuing with the d6 example, we have:
\begin{aligned} P(\textrm{at least 1 explosion on d6}) &\approx &17\% \\ P(\textrm{at least 2 explosions on d6}) &\approx &3\% \\ P(\textrm{at least 3 explosions on d6}) &\approx &0.5\% \\ P(\textrm{at least 4 explosions on d6}) &\approx &0.08\% \\ \end{aligned}
While one or two explosions is fairly common, more beyond that are quite rare, with a less than 1-in-1000 chance that you roll at least 4 explosions on a d6.
## Explosive expectations
Another way to understand the effect of the exploding dice mechanic is to look at how it changes the average or expected outcome of a roll. In a previous post, we showed that the expectation of a normal $n$ sided dice roll is $(n + 1) / 2$. It is clear that exploding dice would have a higher average, but by how much?
We’ll calculate this by appealing to something called the law of total expectation. It allows us to write the expectation of an exploding dice roll as follows:
$E(X) = E(X | \textrm{explode}) P(\textrm{explode}) + E(X | \textrm{not explode}) P(\textrm{not explode})$
By breaking up our expectation into two distinct cases — exploding and not exploding — we simplify our approach. First, the probability statements are are easily addressed by our previous work:
$P(\textrm{explode}) = \frac{1}{n}$
$P(\textrm{not explode}) = \frac{n-1}{n}$
Next, we consider the expected value of a roll with no explosion. Since each outcome below the maximum value is equally likely, this the same as the expected result of an $n-1$ sided roll. Plugging in to $(n + 1) / 2$, we get:
$E(X | \textrm{not explode}) = \frac{n}{2}$
The last term, which is the expected value of an exploded dice is less clear, but we can get around it with a little trick. Recall that once a dice explodes, you can view it as a new exploding dice roll with $n$ added to the final result. We can express this as:
$E(X | \textrm{explode}) = n + E(X)$
While it may seem odd to write the original thing that we were trying to solve for — $E(X)$ — on the right hand side, it’s perfectly valid to do and will help us later. Now that we have expressions for all four terms in our original expectation equation, we can plug in to get:
$E(X) = \left(n + E(X)\right) \frac{1}{n} + \frac{n}{2}\frac{n - 1}{n}$
The above simplifies to:
$E(X) = 1 + \frac{E(X)}{n} + \frac{n - 1}{2}$
We take advantage of our trick by subtracting $E(X)/n$ from both sides to gather all the $E(X)$ terms on the right and solve to get:
$\boxed{ E(X) = \frac{n}{n-1} \frac{n + 1}{2} }$
Careful inspection of our final result shows that the expectation of a normal dice roll appears on in our result, $(n + 1)/2$. A succinct way to summarize the effects of the exploding dice mechanic is that it increases the expected result by a factor of $n / (n - 1)$.
Revisiting our $n=6$ case, we plug in to see that $E(X) = 4.2$. A notable increase over the normal average of 3.5. The below graph summarizes the differences over the standard dice from d4 up to d20.
While the effect is of a similar magnitude across different types, it is most significant for low-faced dice. The factor of increase, $n / (n - 1)$, quickly approaches 1 as $n$ increases, but with few sides it is relatively large. For a d4, the increase from an average roll of 2.5 to one of 3.33, represents a 33% increase. On the other hand, the d20’s 10.5 to 11.05 increase is a meager 5%. The graph below shows percent increase of the exploding mechanic across the standard dice types:
Despite explosions having a consistent increase the expected outcome, as a percent difference, the effect is attenuated for high-sided dice. Such a difference suggests some reasoning behind the prevalence of exploding dice in d6 systems, but less so in those with a greater d20 focus.
Regardless of those differences, exploding dice is a fun mechanic that leads to some interesting math and meaningfully alters outcomes in systems that employ it. |
# Using the 1 in 60 Rule for navigation
Have you ever wondered why we talk about our boat speed and wind speed in knots and not kilometers or miles per hour as we do on land?
It all has to do with the number sixty. 60 is a very important number for navigators. In one degree of latitude, for example, we have 60 minutes. In 60 minutes we have sixty seconds.
If a boat is moving over the ground at one knot, and does so for the next sixty hours, it will have traveled sixty nautical miles.
Positions are expressed in degrees, minutes and seconds, each component having sixty units (sixty degree, sixty minutes and sixty seconds.) When working with paper charts this relationship is more tangible because there are scales and instruments specific to the task of navigating, but when we use computerised chart plotting and GPS all of this is dealt with by the machine and not our heads.
So let me introduce you to the 1 in 60 rule. The 1 in 60 rule adheres to the fact that with a right-angle triangle, and for argument's sake lets say the longest side measures 60 'units,' then the length of the shortest side will be the same in units as the degrees of the angle opposite. Yeah right. I learn by doing, so here's how to apply the rule in your navigation.
Example 1:
When sailing a boat, the wind sometimes (OK, most times) hampers you from being able to set a direct course for your waypoint. Say you are 5 degrees low of your intended mark and that your intended mark is 30 miles off in the distance. If I keep going for 30 miles 5 degrees low of the mark, the question is "how far will I be from the intended mark once I've run 30 miles?"
Use this two step simple formula to find the answer.
So let's use our formulas above to solve a real world problem. You have 30 miles to run to the waypoint and you are pointing 5 degrees low of that mark. How far will I be from the waypoint holding this course?
Example 2:
Let's use another example. You have 45 miles to run and are 15 degrees off your intended track.
Get it? Nifty eh! |
# Exponential Equations (Indicial Equations)
Home > Algebra The equation $a^x=y$ is an example of a general exponent equation (indicial equation) and $2^x = 32$ is an example of a more specific exponential equation (indicial equation). To solve one of these equations it is necessary to write both sides of the equation with the same base if the unknown is [...]
# Algebraic Factorisation with Exponents (Indices)
Home > Algebra $\textit{Factorisation}$ We first look for $\textit{common factors}$ and then for other forms such as $\textit{perfect squares}$, $\textit{difference of two squares}$, etc. Example 1 Factorise $2^{n+4} + 2^{n+1}$. Show Solution \begin{align} \displaystyle &= 2^{n+1} \times 2^{3} + 2^{n+1} \\ &= 2^{n+1}(2^{3} + 1) \\ &= 2^{n+1} \times 9 \\ \end{align} Example [...]
# Multiplication using Exponents (Indices)
Home > Algebra If we wish to calculate $5^4 \times 5^3$, we could write in factor form to get: \begin{align} \displaystyle 5^4 \times 5^3 &= (5 \times 5 \times 5 \times 5) \times (5 \times 5 \times 5) \\ &= 5^7 \\ \end{align} Example 1 Simplify $7^2 \times 7^3$ after first writing in [...]
# Proof by Contradiction
$\textbf{Introduction to Proof by Contradiction}$ The basic idea of $\textit{Proof by Contradiction}$ is to assume that the statement that we want to prove is $\textit{false}$, and then show this assumption leads to nonsense. We then conclude that it was wrong to assume the statement was $\textit{false}$, so the statement must be $\textit{true}$. As an example […]
# The Sign of One
Make the following numbers using FOUR 1s using any mathematics operators and/or symbols, such as $\dfrac{x}{y}$, $\sqrt{x}$, decimal dots, $+$, $-$, $\times$, $\div$, $($ $)$, etc by the Sign of One. Click the numbers below to see the answers. The first one is done for you.
# Number Curiosity
Are these numbers beautiful? \begin{align} \displaystyle 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 &= 98 \\ 123 \times 8 + 3 &= 987 \\ 1234 \times 8 + 4 &= 9876 \\ 12345 \times 8 + 5 &= 98765 \\ 123456 \times 8 + 6 &= 987654 \\ 1234567 […]
# Nicomachus Theorem
$\textbf{Nicomachus}$ discovered “Nicomachus Theorem” interesting number patterns involving cubes and sums of odd numbers. Nicomachus was born in Roman Syria (now, Jerash, Jordan) around 100 AD. He wrote in Greek was a Pythagorean. $\textbf{Nicomachus Theorem: Cubes and Sums of Odd numbers}$ \begin{eqnarray*} 1 &=& 1^3 \\ 3 + 5 &=& 8 = 2^3 \\ 7 […]
# 10 Deadly Common Algebra Mistakes
Share0 Share +10 Tweet0 Common Patterns of Algebra Mistakes Students often make Common Algebra Mistakes due to confusions such as expand and simplify rules, fractions, indices and equations which that lead the students to the wrong answer. Also these error patterns are very basic and quite easily rectified. Check yourself whether you make similar mistakes, […]
# Absolute Value Inequalities
Share0 Share +10 Tweet0 Absolute Value Inequalities are usually proved by the absolute value of a certain value is greater than or equal to it. The square of the value is equal to the square of its absolute value. Proof of Absolute Value Inequalities Prove $|a| + |b| \ge |a+b|$. |
# Tricks and Problems on SET Theory
#### SET THEORY
Set theory has its own notations and symbols that can seem unusual for many. In this tutorial, we look at some solved examples to understand how set theory works and the kind of problems it can be used to solve.
#### Definition
A set is a collection of objects.
It is usually represented in flower braces.
For example:
Set of natural numbers = {1,2,3,…..}
Set of whole numbers = {0,1,2,3,…..}
Each object is called an element of the set.
The set that contains all the elements of a given collection is called the universal set and is represented by the symbol ‘µ’, pronounced as ‘mu’.
For two sets A and B,
n(A ᴜ B) is the number of elements present in either of the sets A or B.
n(A ∩ B) is the number of elements present in both the sets A and B.
n(AᴜB) = n(A) + (n(B) – n(A∩B)
For three sets A, B and C,
n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)
The intersection of sets is only those elements common to all sets. Let’s call our sets A, B, and C. If ‘∩ = intersection‘ and ‘ᴜ = union‘, the need-to-know formulas are:
P(A U B U C) = P(A) + P(B) + P(C) – P(A B) – P(A C) – P(B C) + P(A B C)
To find the number of people in exactly one set:
P(A) + P(B) + P(C) – 2P(A B) – 2P(A C) – 2P(B C) + 3P(A B C)
To find the number of people in exactly two sets:
P(A B) + P(A C) + P(B C) – 3P(A B C)
To find the number of people in exactly three sets:
P(A B C)
To find the number of people in two or more sets:
P(A B) + P(A C) + P(B C) – 2P(A B C)
To find the number of people in at least one set:
P(A) + P(B) + P(C) – P(A B) – P(A C) – P(B C) + 2 P(A B ∩ C)
Correct! Wrong!
Correct! Wrong!
Correct! Wrong!
Correct! Wrong!
#### Let A = {0, 1, 3, 5}, B = {5, 6, 1, 3, 9} and C = {0, 1, 2, 3, 9, 13}. Then, (A&B) OR C :
Correct! Wrong!
set theory
Great job...
Good
Better Luck Next Time |
# What is a Matrix?
This lesson introduces the matrix - the rectangular array at the heart of matrix algebra. Matrix algebra is used quite a bit in advanced statistics, largely because it provides two benefits.
• Compact notation for describing sets of data and sets of equations.
• Efficient methods for manipulating sets of data and solving sets of equations.
## Matrix Definition
A matrix is a rectangular array of numbers arranged in rows and columns. The array of numbers below is an example of a matrix.
21 62 33 93 44 95 66 13 77 38 79 33
The number of rows and columns that a matrix has is called its dimension or its order. By convention, rows are listed first; and columns, second. Thus, we would say that the dimension (or order) of the above matrix is 3 x 4, meaning that it has 3 rows and 4 columns.
Numbers that appear in the rows and columns of a matrix are called elements of the matrix. In the above matrix, the element in the first column of the first row is 21; the element in the second column of the first row is 62; and so on.
## Matrix Notation
Statisticians use symbols to identify matrix elements and matrices.
• Matrix elements. Consider the matrix below, in which matrix elements are represented entirely by symbols.
A11 A12 A13 A14 A21 A22 A23 A24
By convention, first subscript refers to the row number; and the second subscript, to the column number. Thus, the first element in the first row is represented by A11. The second element in the first row is represented by A12. And so on, until we reach the fourth element in the second row, which is represented by A24.
• Matrices. There are several ways to represent a matrix symbolically. The simplest is to use a boldface letter, such as A, B, or C. Thus, A might represent a 2 x 4 matrix, as illustrated below.
A =
11 62 33 93 44 95 66 13
Another approach for representing matrix A is:
A = [ Aij ] where i = 1, 2 and j = 1, 2, 3, 4
This notation indicates that A is a matrix with 2 rows and 4 columns. The actual elements of the array are not displayed; they are represented by the symbol Aij.
Other matrix notation will be introduced as needed. For a description of all the matrix notation used in this tutorial, see the Matrix Notation Appendix.
## Matrix Equality
To understand matrix algebra, we need to understand matrix equality. Two matrices are equal if all three of the following conditions are met:
• Each matrix has the same number of rows.
• Each matrix has the same number of columns.
• Corresponding elements within each matrix are equal.
Consider the three matrices shown below.
A =
111 x y 444
B =
111 222 333 444
C =
l m n o p q
If A = B, we know that x = 222 and y = 333; since corresponding elements of equal matrices are also equal. And we know that matrix C is not equal to A or B, because C has more columns than A or B.
## Test Your Understanding
Problem 1
The notation below describes two matrices - matrix A and matrix B.
A = [ Aij ]
where i = 1, 2, 3 and j = 1, 2
B =
111 222 333 444 555 666 777 888
Which of the following statements about A and B are true?
I. Matrix A has 5 elements.
II. The dimension of matrix B is 4 x 2.
III. In matrix B, element B21 is equal to 222.
(A) I only
(B) II only
(C) III only
(D) All of the above
(E) None of the above
Solution
The correct answer is (E).
• Matrix A has 3 rows and 2 columns; that is, 3 rows, each with 2 elements. This adds up to 6 elements, altogether - not 5.
• The dimension of matrix B is 2 x 4 - not 4 x 2. That is, matrix B has 2 rows and 4 columns - not 4 rows and 2 columns.
• And, finally, element B21 refers to the first element in the second row of matrix B, which is equal to 555 - not 222. |
Measuring Rotation
Explore trig ratios of angles greater than 90 degrees
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Introduction to Angles of Rotation, Coterminal Angles, and Reference Angles
In which quadrant does the terminal side of the angle \begin{align*}-500^{\text{o}}\end{align*} lie and what is the reference angle for this angle?
Angles of Rotation
Angles of rotation are formed in the coordinate plane between the positive \begin{align*}x\end{align*}-axis (initial side) and a ray (terminal side). Positive angle measures represent a counterclockwise rotation while negative angles indicate a clockwise rotation.
Since the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} axes are perpendicular, each axis then represents an increment of ninety degrees of rotation. The diagrams below show a variety of angles formed by rotating a ray through the quadrants of the coordinate plane.
An angle of rotation can be described infinitely many ways. It can be described by a positive or negative angle of rotation or by making multiple full circle rotations through \begin{align*}360^{\text{o}}\end{align*}. The example below illustrates this concept.
For the angle \begin{align*}525^{\text{o}}\end{align*}, an entire \begin{align*}360^{\text{o}}\end{align*} rotation is made and then we keep going another \begin{align*}165^{\text{o}}\end{align*} to \begin{align*}525^{\text{o}}\end{align*}. Therefore, the resulting angle is equivalent to \begin{align*}525^\text{o} - 360^\text{o}\end{align*}, or \begin{align*}165^\text{o}\end{align*}. In other words, the terminal side is in the same location as the terminal side for a \begin{align*}165^\text{o}\end{align*} angle. If we subtract \begin{align*}360^\text{o}\end{align*} again, we get a negative angle, \begin{align*}-195^\text{o}\end{align*}. Since they all share the same terminal side, they are called coterminal angles.
Let's determine two coterminal angles to \begin{align*}837^\text{o}\end{align*}, one positive and one negative.
To find coterminal angles we simply add or subtract \begin{align*}360^\text{o}\end{align*} multiple times to get the angles we desire. \begin{align*}837^\text{o} - 360^\text{o} = 477^\text{o}\end{align*}, so we have a positive coterminal angle. Now we can subtract \begin{align*}360^\text{o}\end{align*} again to get \begin{align*}477^\text{o} - 360^\text{o}=117^\text{o}\end{align*}.
Reference Angle
A reference angle is the acute angle between the terminal side of an angle and the \begin{align*}x\end{align*} – axis. The diagram below shows the reference angles for terminal sides of angles in each of the four quadrants.
Note: A reference angle is never determined by the angle between the terminal side and the \begin{align*}y\end{align*} – axis. This is a common error for students, especially when the terminal side appears to be closer to the \begin{align*}y\end{align*} – axis than the \begin{align*}x\end{align*} – axis.
Now, let's determine the quadrant in which \begin{align*}-745^\text{o}\end{align*} lies and hence determine the reference angle.
Since our angle is more than one rotation, we need to add \begin{align*}360^\text{o}\end{align*} until we get an angle whose absolute value is less than \begin{align*}360^\text{o}\end{align*}: \begin{align*}-745^\text{o} + 360^\text{o} = -385^\text{o}\end{align*}, again \begin{align*}-385^\text{o} + 360^\text{o} = -25^\text{o}\end{align*}.
Now we can plot the angle and determine the reference angle:
Note that the reference angle is positive \begin{align*}25^\text{o}\end{align*}. All reference angles will be positive as they are acute angles (between \begin{align*}0^\text{o}\end{align*} and \begin{align*}90^\text{o}\end{align*}).
Finally, let's give two coterminal angles to \begin{align*}595^\circ\end{align*}, one positive and one negative, and find the reference angle.
To find the coterminal angles we can add/subtract \begin{align*}360^\circ\end{align*}. In this case, our angle is greater than \begin{align*}360^\circ\end{align*} so it makes sense to subtract \begin{align*}360^\circ\end{align*} to get a positive coterminal angle: \begin{align*}595^\circ - 360^\circ = 235^\circ\end{align*}. Now subtract again to get a negative angle: \begin{align*}235^\circ - 360^\circ = -125^\circ\end{align*}.
By plotting any of these angles we can see that the terminal side lies in the third quadrant as shown.
Since the terminal side lies in the third quadrant, we need to find the angle between \begin{align*}180^\circ\end{align*} and \begin{align*}235^\circ\end{align*}, so \begin{align*}235^\circ - 180^\circ = 55^\circ\end{align*}.
Examples
Example 1
Earlier, you were asked to find the reference angle of \begin{align*}-500^\text{o}\end{align*} and find the quadrant in which the terminal side lies.
Since our angle is more than one rotation, we need to add \begin{align*}360^\text{o}\end{align*} until we get an angle whose absolute value is less than \begin{align*}360^\text{o}\end{align*}: \begin{align*}-500^\text{o} + 360^\text{o} = -200^\text{o}\end{align*}.
If we plot this angle we see that it is \begin{align*}-200^\text{o}\end{align*} clockwise from the origin or \begin{align*}160^\text{o}\end{align*} counterclockwise. \begin{align*}160^\text{o}\end{align*} lies in the second quadrant.
Now determine the reference angle: \begin{align*}180^\text{o} - 160^\text{o} = 20^\text{o}\end{align*}.
Example 2
Find two coterminal angles to \begin{align*}138^\text{o}\end{align*}, one positive and one negative.
\begin{align*}138^\text{o} + 360^\text{o} = 498^\text{o}\end{align*} and \begin{align*}138^\text{o} - 360^\text{o} = -222^\text{o}\end{align*}
Example 3
Find the reference angle for \begin{align*}895^\text{o}\end{align*}.
\begin{align*}895^\text{o} - 360^\text{o} = 535^\text{o}, 535^\text{o} - 360^\text{o} = 175^\text{o}\end{align*}. The terminal side lies in the second quadrant, so we need to determine the angle between \begin{align*}175^\text{o}\end{align*} and \begin{align*}180^\text{o}\end{align*}, which is \begin{align*}5^\text{o}\end{align*}.
Example 4
Find the reference angle for \begin{align*}343^\text{o}\end{align*}.
\begin{align*}343^\text{o}\end{align*} is in the fourth quadrant so we need to find the angle between \begin{align*}343^\text{o}\end{align*} and \begin{align*}360^\text{o}\end{align*} which is \begin{align*}17^\text{o}\end{align*}.
Review
Find two coterminal angles to each angle measure, one positive and one negative.
1. \begin{align*}-98^\text{o}\end{align*}
2. \begin{align*}475^\text{o}\end{align*}
3. \begin{align*}-210^\text{o}\end{align*}
4. \begin{align*}47^\text{o}\end{align*}
5. \begin{align*}-1022^\text{o}\end{align*}
6. \begin{align*}354^\text{o}\end{align*}
7. \begin{align*}-7^\text{o}\end{align*}
Determine the quadrant in which the terminal side lies and find the reference angle for each of the following angles.
1. \begin{align*}102^\text{o}\end{align*}
2. \begin{align*}-400^\text{o}\end{align*}
3. \begin{align*}1307^\text{o}\end{align*}
4. \begin{align*}-820^\text{o}\end{align*}
5. \begin{align*}304^\text{o}\end{align*}
6. \begin{align*}251^\text{o}\end{align*}
7. \begin{align*}-348^\text{o}\end{align*}
8. Explain why the reference angle for an angle between \begin{align*}0^\text{o}\end{align*} and \begin{align*}90^\text{o}\end{align*} is equal to itself.
To see the Review answers, open this PDF file and look for section 13.5.
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
This is the write-up for Assingnment 2, problem # 7 and #8.
By Doug Westmoreland
This write-up is an exploration of the curves and using Graphing Calculator 2.7.1.
Let's first take a look at how the "a" value in effects the graph. The following are graphs with the "a" value of .1, .5, 1, 2, 5, 10.
a=.1 purple graph : a=.5 red graph : a=1 blue graph : a=2 green graph : a=5 lt.blue graph : a=10 yellow graph
We can see from these examples that the "a" value is definitely either narrowing or expanding the graph depending upon the value of a. As the value of "a" gets larger the graph gets narrower. As the value of "a" decreases to zero the graph expands toward the x-axis.
Question: What do you think negative values of "a" will do to the graph? Click here for the answer.
Now let's take a look at how the value of "d" effects the graph of . The following are graphs with the "d" value of -10, -5, -3, -2, -1, 0, 2, 3, 5, 10.
a=-10 (left red) , a=-5 (left purple) , a=-3 (left blue) , a=-2 green , a=-1 lt.blue , a=0 yellow , a=2 black
a=3 (right purple) , a=5 (right blue) , a=10 (right blue)
We can see the the "d" value effects the graph of in the graphs above by shifting the graph of
either right or left depending upon the value of "d." Click here for a better presentation of the translation of the graph.
SUMMARY
The value of "a" in dialates the graph. If |a| is greater than 1, then the graph will get narrower toward the y-axis. If |a| is between 0 and 1, then the graph will expand toward the x-axis.
The value of "d" in translates the graph of . If "d" is a positive value, the graph will shift to the right "d" units. If the "d" value is negative, then the graph will shift to the left |d| units.
Return |
offered the focus and directrix the a parabola , how do we discover the equation the the parabola?
If we consider only parabolas that open upwards or downwards, then the directrix will certainly be a horizontal line the the type y = c .
let ( a , b ) be the focus and also let y = c it is in the directrix. Permit ( x 0 , y 0 ) be any suggest on the parabola.
any point, ( x 0 , y 0 ) ~ above the parabola satisfies the an interpretation of parabola, for this reason there are two ranges to calculate:
Distance between the allude on the parabola come the emphasis Distance between the suggest on the parabola come the directrix
To uncover the equation of the parabola, equate these 2 expressions and solve because that y 0 .
discover the equation that the parabola in the instance above.
Distance in between the point ( x 0 , y 0 ) and ( a , b ) :
( x 0 − a ) 2 + ( y 0 − b ) 2
street between point ( x 0 , y 0 ) and the line y = c :
| y 0 − c |
(Here, the distance in between the allude and horizontal heat is distinction of their y -coordinates.)
Equate the 2 expressions.
( x 0 − a ) 2 + ( y 0 − b ) 2 = | y 0 − c |
Square both sides.
( x 0 − a ) 2 + ( y 0 − b ) 2 = ( y 0 − c ) 2
increase the expression in y 0 ~ above both sides and also simplify.
( x 0 − a ) 2 + b 2 − c 2 = 2 ( b − c ) y 0
This equation in ( x 0 , y 0 ) is true for all other values top top the parabola and hence we deserve to rewrite with ( x , y ) .
Therefore, the equation of the parabola with emphasis ( a , b ) and also directrix y = c is
( x − a ) 2 + b 2 − c 2 = 2 ( b − c ) y
You are watching: Using a directrix of y = −3 and a focus of (2, 1), what quadratic function is created?
Example:
If the emphasis of a parabola is ( 2 , 5 ) and the directrix is y = 3 , uncover the equation of the parabola.
allow ( x 0 , y 0 ) be any allude on the parabola. Uncover the distance between ( x 0 , y 0 ) and also the focus. Then uncover the distance in between ( x 0 , y 0 ) and directrix. Equate these 2 distance equations and also the streamlined equation in x 0 and y 0 is equation of the parabola.
The distance in between ( x 0 , y 0 ) and ( 2 , 5 ) is ( x 0 − 2 ) 2 + ( y 0 − 5 ) 2
The distance in between ( x 0 , y 0 ) and also the directrix, y = 3 is
| y 0 − 3 | .
Equate the 2 distance expressions and square on both sides.
See more: Darius Rucker Wedding Songs, The 10 Best Darius Rucker Songs
( x 0 − 2 ) 2 + ( y 0 − 5 ) 2 = | y 0 − 3 |
( x 0 − 2 ) 2 + ( y 0 − 5 ) 2 = ( y 0 − 3 ) 2
Simplify and bring every terms to one side:
x 0 2 − 4 x 0 − 4 y 0 + 20 = 0
create the equation v y 0 on one side:
y 0 = x 0 2 4 − x 0 + 5
This equation in ( x 0 , y 0 ) is true for all other values top top the parabola and hence we have the right to rewrite through ( x , y ) .
So, the equation the the parabola with emphasis ( 2 , 5 ) and directrix is y = 3 is |
# 42 Times Table
Greetings, dear student! Today, we embark on an exciting mathematical journey through the 42 times table. Brace yourself for a thrilling adventure into the world of multiplication!
Let's start exploring the magnificent 42 times table, shall we? Prepare yourself for some mind-blowing results!
First things first, let's begin with the easiest multiplication in any times table: 42 times 1. Now, when we multiply any number by 1, it's like a gentle tap on the shoulder—nothing really changes. So, 42 times 1 equals... you guessed it... 42! It's as simple as that!
Now, let's spice things up a bit. What happens when we multiply 42 by 2? Well, we're doubling the original number, which means we're taking 42 and adding another 42 to it. So, 42 times 2 equals 84. Double the fun, double the result!
Moving along, let's try something a little more adventurous. How about 42 times 5? To find the answer, we can think of it as multiplying by 10 and then dividing by 2.
When we multiply 42 by 10, we get 420. And then, we divide that by 2, which gives us 210. So, 42 times 5 equals 210. Exciting, isn't it?
Let's keep pushing our mathematical boundaries. How about 42 times 9? We can use a similar approach here.
We multiply 42 by 10 to get 420 and then subtract 42 from that result to account for the missing 1. That leaves us with 378. Therefore, 42 times 9 equals 378. Marvelous!
But wait, there's more! Let's take on a slightly bigger challenge. What happens when we multiply 42 by 13? Well, we can break it down into smaller steps. First, we multiply 42 by 10 to get 420.
Then, we multiply 42 by 3 to get 126. Lastly, we add those two results together: 420 plus 126 equals 546. So, 42 times 13 equals 546. Incredible!
Now, my eager learner, it's your turn to explore the 42 times table further. See if you can uncover more exciting patterns and surprising results. Remember, the more you practice, the more confident you'll become!
If you have any questions or need assistance with any other mathematical endeavors, feel free to ask. I'm here to guide you on your mathematical journey. Happy multiplying!
## Forty-two Multiplication Table
Read, Repeat and Learn Forty-two times table and Check yourself by giving a test below
## Table of 42
42 Times table Test
## How much is 42 multiplied by other numbers?
@2024 PrintableMultiplicationTable.net |
# 8-2 Basics of Hypothesis Testing
## Presentation on theme: "8-2 Basics of Hypothesis Testing"— Presentation transcript:
8-2 Basics of Hypothesis Testing
This section presents individual components of a hypothesis test. We should know and understand the following: How to identify the null hypothesis and alternative hypothesis from a given claim, and how to express both in symbolic form How to calculate the value of the test statistic, given a claim and sample data How to choose the sampling distribution that is relevant How to identify the P-value or identify the critical value(s) How to state the conclusion about a claim in simple and nontechnical terms
Definitions A hypothesis is a claim or statement about a property of a population. A hypothesis test is a procedure for testing a claim about a property of a population.
Rare Event Rule for Inferential Statistics
If, under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption is probably not correct.
Null Hypothesis The null hypothesis (denoted by H0) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value. We test the null hypothesis directly in the sense that we assume it is true and reach a conclusion to either reject H0 or fail to reject H0.
Alternative Hypothesis
The alternative hypothesis (denoted by H1 or HA) is the statement that the parameter has a value that somehow differs from the null hypothesis. The symbolic form of the alternative hypothesis must use one of these symbols: <, >, ≠.
If you are conducting a study and want to use a hypothesis test to support your claim, the claim must be worded so that it becomes the alternative hypothesis.
Steps 1, 2, 3 Identifying H0 and H1
Example Assume that 100 babies are born to 100 couples treated with the XSORT method of gender selection that is claimed to make girls more likely. We observe 58 girls in 100 babies. Write the hypotheses to test the claim the “with the XSORT method, the proportion of girls is greater than the 50% that occurs without any treatment”.
Step 4 Select the Significance Level α
Significance Level The significance level (denoted by α) is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true (making the mistake of rejecting the null hypothesis when it is true). This is the same α introduced in Section 7-2. Common choices for α are 0.05, 0.01, and 0.10.
Step 5 Identify the Test Statistic and Determine its Sampling Distribution
Test Statistic The test statistic is a value used in making a decision about the null hypothesis, and is found by converting the sample statistic to a score with the assumption that the null hypothesis is true.
Step 6 Find the Value of the Test Statistic, Then Find Either the P-Value or the Critical Value(s)
First transform the relevant sample statistic to a standardized score called the test statistic. Then find the P-Value or the critical value(s).
Example Let’s again consider the claim that the XSORT method of gender selection increases the likelihood of having a baby girl. Preliminary results from a test of the XSORT method of gender selection involved 100 couples who gave birth to 58 girls and 42 boys. Use the given claim and the preliminary results to calculate the value of the test statistic. Use the format of the test statistic given above, so that a normal distribution is used to approximate a binomial distribution.
Example - Continued The claim that the XSORT method of gender selection increases the likelihood of having a baby girl results in the following null and alternative hypotheses: We work under the assumption that the null hypothesis is true with p = 0.5. The sample proportion of 58 girls in 10 births results in:
Example – Convert to the Test Statistic
We know from previous chapters that a z score of 1.60 is not “unusual”. At first glance, 58 girls in 100 births does not seem to support the claim that the XSORT method increases the likelihood a having a girl (more than a 50% chance).
Types of Hypothesis Tests: Two-tailed, Left-tailed, Right-tailed
The tails in a distribution are the extreme regions bounded by critical values. Determinations of P-values and critical values are affected by whether a critical region is in two tails, the left tail, or the right tail. It, therefore, becomes important to correctly ,characterize a hypothesis test as two-tailed, left-tailed, or right-tailed.
α is divided equally between the two tails of the critical region
Two-tailed Test α is divided equally between the two tails of the critical region
Left-tailed Test All α in the left tail
Right-tailed Test All α in the right tail
P-Value The P-value (or probability value) is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. Critical region in the left tail: P-value = area to the left of the test statistic Critical region in the right tail: P-value = area to the right of the test statistic Critical region in two tails: P-value = twice the area in the tail beyond the test statistic
P-Value The null hypothesis is rejected if the P-value is very small, such as 0.05 or less.
Example The claim that the XSORT method of gender selection increases the likelihood of having a baby girl results in the following null and alternative hypotheses: The test statistic was :
Example The test statistic of z = 1.60 has an area of to its right, so a right-tailed test with test statistic z = 1.60 has a P-value of
Procedure for Finding P-Values
Critical Region The critical region (or rejection region) is the set of all values of the test statistic that cause us to reject the null hypothesis. For example, see the red-shaded region in the previous figures.
Critical Value A critical value is any value that separates the critical region (where we reject the null hypothesis) from the values of the test statistic that do not lead to rejection of the null hypothesis. The critical values depend on the nature of the null hypothesis, the sampling distribution that applies, and the significance level α.
Example For the XSORT birth hypothesis test, the critical value and critical region for an α = 0.05 test are shown below:
Caution Don’t confuse a P-value with a proportion p.
Know this distinction: P-value = probability of getting a test statistic at least as extreme as the one representing sample data p = population proportion
Step 7 : Make a Decision: Reject H0 or Fail to Reject H0
The methodologies depend on if you are using the P-Value method or the critical value method.
Decision Criterion P-value Method: Using the significance level α:
If P-value ≤ α, reject H0. If P-value > α, fail to reject H0.
Decision Criterion Critical Value Method:
If the test statistic falls within the critical region, reject H0. If the test statistic does not fall within the critical region, fail to reject H0.
Example For the XSORT baby gender test, the test had a test statistic of z = 1.60 and a P-Value of We tested: Using the P-Value method, we would fail to reject the null at the α = 0.05 level. Using the critical value method, we would fail to reject the null because the test statistic of z = 1.60 does not fall in the rejection region. (You will come to the same decision using either method.)
Step 8 : Restate the Decision Using Simple and Nontechnical Terms
State a final conclusion that addresses the original claim with wording that can be understood by those without knowledge of statistical procedures.
Example For the XSORT baby gender test, there was not sufficient evidence to support the claim that the XSORT method is effective in increasing the probability that a baby girl will be born.
Wording of Final Conclusion
Caution Never conclude a hypothesis test with a statement of “reject the null hypothesis” or “fail to reject the null hypothesis.” Always make sense of the conclusion with a statement that uses simple nontechnical wording that addresses the original claim.
Accept Versus Fail to Reject
Some texts use “accept the null hypothesis.” We are not proving the null hypothesis. Fail to reject says more correctly that the available evidence is not strong enough to warrant rejection of the null hypothesis.
Type I Error A Type I error is the mistake of rejecting the null hypothesis when it is actually true. The symbol α is used to represent the probability of a type I error. page 398 of Elementary Statistics, 10th Edition
Type II Error A Type II error is the mistake of failing to reject the null hypothesis when it is actually false. The symbol β (beta) is used to represent the probability of a type II error. Page 398 of Elementary Statistics, 10th Edition
Type I and Type II Errors
Example Assume that we are conducting a hypothesis test of the claim that a method of gender selection increases the likelihood of a baby girl, so that the probability of a baby girls is p > 0.5. Here are the null and alternative hypotheses: a) Identify a type I error. b) Identify a type II error.
Example - Continued a) A type I error is the mistake of rejecting a true null hypothesis: We conclude the probability of having a girl is greater than 50%, when in reality, it is not. Our data misled us. b) A type II error is the mistake of failing to reject the null hypothesis when it is false: There is no evidence to conclude the probability of having a girl is greater than 50% (our data misled us), but in reality, the probability is greater than 50%.
Controlling Type I and Type II Errors
For any fixed α, an increase in the sample size n will cause a decrease in β For any fixed sample size n, a decrease in α will cause an increase in β. Conversely, an increase in α will cause a decrease in β. To decrease both α and β, increase the sample size.
Part 2: Beyond the Basics of Hypothesis Testing: The Power of a Test
Definition The power of a hypothesis test is the probability 1 – β of rejecting a false null hypothesis. The value of the power is computed by using a particular significance level α and a particular value of the population parameter that is an alternative to the value assumed true in the null hypothesis.
Power and the Design of Experiments
Just as 0.05 is a common choice for a significance level, a power of at least 0.80 is a common requirement for determining that a hypothesis test is effective. (Some statisticians argue that the power should be higher, such as 0.85 or 0.90.) When designing an experiment, we might consider how much of a difference between the claimed value of a parameter and its true value is an important amount of difference. When designing an experiment, a goal of having a power value of at least 0.80 can often be used to determine the minimum required sample size.
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## Part 1: Understanding the Basics of Circles
Hello from MKSprep, the trusted SAT preparation center in Putalisadak, Kathmandu, Nepal. We are now ready to take a circular journey in our SAT math series! This first part of our in-depth exploration into Circles will set the foundation for understanding this key geometric shape.
### Introduction to Circles
In mathematics, a circle is a shape consisting of all points in a plane at a given distance from a certain point, known as the circle’s center. The radius is the distance between any point on the circle and its center.
### Diameter, Circumference, and Area
Here are some fundamental concepts related to circles:
• Diameter: The longest line can be drawn in a circle, passing through the center. The diameter is twice the radius of the circle.
• Circumference: The circumference is the boundary or the distance around the circle. For any circle, the circumference is more than three times the diameter. This relationship is expressed by the formula C = πd or C = 2πr.
• Area: The area of a circle is the number of square units that can fit inside it. It is calculated as A = πr².
Understanding these basic concepts is crucial for solving circle-related problems in the SAT math section.
Stay tuned for the next part, where we delve into the world of arcs and sectors. At MKSprep, we are committed to making your SAT preparation as smooth as possible!
## Part 2: Arcs and Sectors in Circles
Welcome back to the SAT preparation series by MKSprep, your trusted SAT preparation center located in Putalisadak, Kathmandu, Nepal. In this second part on Circles, we dive deeper into two integral components – Arcs and Sectors.
### Arcs in Circles
An arc in a circle is a portion of the circumference. The length of an arc is proportional to the degree measure of its central angle. In other words, if you have a 60° angle at the center of the circle, the length of the arc will be 1/6 of the total circumference. This relationship is expressed as Arc length = (θ/360°) × 2πr, where θ is the degree measure of the central angle.
### Sectors in Circles
A sector of a circle is the region enclosed by two radii and their intercepted arc. It resembles a slice of pie. Similar to arcs, the area of a sector is proportional to the degree measure of its central angle. Therefore, if a circle is divided into a sector by a 60° angle, the area of the sector will be 1/6 of the total area of the circle. This relationship is expressed as Sector area = (θ/360°) × πr², where θ is the degree measure of the central angle.
Understanding these concepts will help you efficiently tackle SAT math questions related to arcs and sectors.
Stay tuned for our next lesson, where we will explore the properties of chords in circles. At MKSprep, we are committed to equipping you with the knowledge and skills needed for SAT success!
## Part 3: Chords in Circles
Greetings once more from MKSprep, your preferred SAT preparation center located in Putalisadak, Kathmandu, Nepal. Continuing with our deep dive into Circles, in this third part, we will discuss an essential concept: Chords.
### What is a Chord?
In the context of a circle, a chord is a straight-line segment that connects two points on the circle’s circumference. The longest possible chord of a circle runs through the circle’s center and is known as the diameter.
### Properties of Chords
Here are some key properties of chords:
• Equal Chords: In the same circle or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.
• Perpendicular to the Chord: A radius or diameter that is perpendicular to a chord bisects the chord (divides it into two equal parts) and the arc it subtends (the arc that it cuts off).
• Chords Equidistant from the Center: In the same circle or congruent circles, two chords are equidistant from the center if and only if they are congruent.
Comprehending these properties is crucial for solving SAT math questions that involve chords in circles.
In our next installment, we will examine the concept of tangents to a circle. At MKSprep, we are dedicated to ensuring your SAT preparation is thorough and fruitful!
## Part 4: Tangents to Circles
Welcome back to our SAT preparation series provided by MKSprep, your leading SAT preparation center in Putalisadak, Kathmandu, Nepal. In this fourth part of our deep dive into Circles, we’re going to explore the fascinating world of tangents.
### What is a Tangent?
A tangent to a circle is a straight line that touches the circle at exactly one point, known as the point of tangency. An important property to remember is that a radius drawn to the point of tangency is always perpendicular to the tangent line.
### Properties of Tangents
Here are some significant properties of tangents to remember:
• Tangent Segments from a Point: From a point outside a circle, the two tangent segments to the circle are congruent. This means they have the same length.
• Angles between Tangents and Radii: The angle between the radius of a circle and a tangent line drawn to that circle from the endpoint of the radius is always a right angle (90 degrees).
• Angles formed by Tangents: The angle formed by two tangents, two secants, or a secant and a tangent drawn from a point outside the circle is half the difference of the measures of the intercepted arcs.
Understanding these properties will significantly assist you in successfully navigating SAT math questions involving tangents to circles.
Join us for the next lesson, where we’ll delve into the intriguing topic of inscribed angles in circles. At MKSprep, we’re committed to providing you with the most effective SAT preparation possible!
## Part 5: Inscribed Angles in Circles
Hello again from MKSprep, your SAT preparation center situated in the heart of Putalisadak, Kathmandu, Nepal. As we continue our exploration of Circles, this fifth part introduces a key concept: Inscribed Angles.
### What is an Inscribed Angle?
An inscribed angle is formed by two chords in a circle with a common endpoint. This common endpoint forms the vertex of the inscribed angle. The other two endpoints define what is known as an intercepted arc on the circle’s circumference.
### Properties of Inscribed Angles
Here are the essential properties of inscribed angles:
• Measure of an Inscribed Angle: The measure of an inscribed angle is half the measure of its intercepted arc.
• Inscribed Angles on the Same Arc: All inscribed angles that intercept the same arc are equal.
• Inscribed Angle and a Diameter: An inscribed angle that intercepts a semicircle (where the arc is a diameter) is always a right angle (90 degrees).
Grasping these properties will greatly assist you in solving SAT math problems involving inscribed angles in circles.
Stay tuned for our next section, where we will dive into the topic of sector and arc length in circles. At MKSprep, we are here to ensure that your SAT preparation journey is comprehensive and rewarding!
## Part 6: Sectors and Arc Lengths in Circles
Welcome once again to our SAT preparation series offered by MKSprep, your trustworthy SAT preparation center in Putalisadak, Kathmandu, Nepal. We’re discussing sectors and arc lengths in this sixth part of our circle exploration.
### What are Sectors and Arc Lengths?
A sector of a circle is a section of the circle enclosed by two radii and their intercepted arc. The length of an arc is simply the distance along the path of the circle from one point to another.
### How to Find Sectors and Arc Lengths?
Here’s how you calculate these two key components:
• Sector Area: To find the area of a sector, you need to know the measure of the central angle that subtends it. The formula for the sector area is (central angle/360) * π * r², where r is the radius of the circle.
• Arc Length: The arc length is found using a similar formula as the sector area. The formula is (central angle/360) * 2πr, again where r is the radius of the circle.
Understanding these principles will be incredibly beneficial in tackling SAT math questions concerning sectors and arc lengths in circles.
Join us in the next segment, where we’ll investigate the compelling topic of chords in circles. MKSprep is dedicated to providing you with the highest standard of SAT preparation!
## Part 7: Chords in Circles
Welcome back to MKSprep, your dedicated SAT preparation center in Putalisadak, Kathmandu, Nepal. In this seventh part of our circle series, we are focusing on the concept of chords.
### What are Chords?
A chord is a straight-line segment that connects two points on the circumference of a circle. The longest possible chord in a circle is its diameter.
### Significant Properties of Chords:
Let’s explore some of the important properties of chords:
• Equal Chords: Equal chords of a circle subtend equal angles at the circle’s center. Conversely, if the angles subtended by two chords at the center of the circle are equal, the chords are also equal.
• Perpendicular from the Center: If a perpendicular is drawn from the center of the circle to a chord, it bisects the chord. That means it divides the chord into two equal parts.
• Chords Equidistant from Center: Chords that are equidistant from the center of the circle are equal in length.
Knowing these chord properties can greatly assist you in answering SAT math questions related to chords in circles.
Join us in our final installment of this series, where we’ll be dealing with tangents and their properties. MKSprep is committed to making your SAT preparation journey as comprehensive as possible!
## Part 8: Tangents to Circles
Congratulations on reaching the final part of our SAT preparation series on circles offered by MKSprep, your reliable SAT preparation center in Putalisadak, Kathmandu, Nepal. In this concluding segment, we’re studying the concept of tangents to circles.
### What is a Tangent?
A tangent is a line that touches a circle at exactly one point, known as the point of tangency. No matter where the tangent line touches the circle, it is always perpendicular to the radius drawn to the point of tangency.
### Key Properties of Tangents:
Below are some significant properties of tangents:
• Tangent Segment: A tangent segment is a line segment whose endpoints are the point of tangency and a point on the tangent line.
• Tangent Segments from an External Point: Those segments are congruent if two tangent segments are drawn to a circle from an external point.
Understanding these properties will be immensely helpful in answering SAT math questions related to tangents to circles.
With the conclusion of this part, we’ve covered all the fundamental elements of circles that you’ll encounter in the SAT. Thank you for choosing MKSprep as your companion in your SAT preparation journey! |
## May 24, 2020
### VECTORS – 2 (MASTER ADVANCED PHYSICS)
(FOR ALL TOP EXAMS)
VECTORS – 2
Few Basic Concepts in Vector Algebra
We have seen that a vector is inherently a quantity that has both magnitude and direction. We cannot add vectors like normal addition. We need to consider both magnitude and direction and find ways to get the final effect or what is called a resultant.
Before we go into that, let us understand a few basic concepts that are needed to find resultants.
Negative vector –
A negative vector is simply the same vector with the same magnitude but exactly the opposite direction.
Like and unlike vectors –
Like vectors are those that may differ in magnitude but have the same direction.
Unlike vectors have different magnitude but have directions exactly opposite to each other.
Equality of vectors –
If two vectors have BOTH same magnitude and the same direction, they are called equal vectors.
Unit vector –
A vector having unit magnitude, means one, but the same direction as a given vector, is called UNIT VECTOR of the given vector.
So, we if we have a vector of 5 units and a direction of 60 degrees from horizontal, then the unit vector is simply ONE unit at 60 degrees from horizontal.
It is convenient to have 3 unit vectors in 3 directions, x, y, and z as shown.
Then we have a special symbol ,
I cap, j cap, k cap for unit vectors along x, y, and z axes respectively.
Multiplication of a vector by a real number -
We simply multiply the real number with the magnitude of the vector and the direction remains the same.
For example, if we have a vector with 5 units magnitude and 60 degrees from the horizontal as direction. Multiplication of 3 with the number is simply
3 multiplied by 5. So, we have 15 units of magnitude and the same 60 degrees direction. |
# 4.3 Calculating the Present Value
## LEARNING OBJECTIVES
• Calculate present value for compound interest.
The principal of the loan or investment is called the present value ($PV$). The present value is the amount of money borrowed for a loan or the amount of money invested for an investment at the start of the term. The present value is the amount at some earlier point in time than when the future value is known, and so excludes the future interest.
## The Present Value Formula
By solving for $PV$ in the future value formula from the previous section, the present value for compound interest is
$\displaystyle{PV=FV \times (1+i)^{-n}}$
where
• $PV$ is the present value. The present value is the starting amount upon which compound interest is calculated.
• $FV$ is the future value. The future value includes the principal plus all of the interest accumulated over the term.
• $i$ is the periodic interest rate. The periodic interest rate is the interest rate per compounding period: $i=\frac{j}{m}$ where $j$ is the nominal interest rate and $m$ is the compounding frequency.
• $n$ is the total number of compounding periods over the term. $n=m \times t$ where $m$ is the compounding frequency and $t$ is the length of the term in years.
Castillo’s Warehouse will need to purchase a new forklift for its warehouse operations three years from now, when its new warehouse facility becomes operational. If the price of the new forklift is $38,000 and Castillo’s can invest its money at 7.25% compounded monthly, how much money should it put aside today to achieve its goal? Solution: The timeline for the investment is shown below. Step 1: The given information is $\begin{eqnarray*} FV & = & \38,000 \\ j & = & 7.25\% \\ m & = & 12 \\ t & = & 3 \mbox{ years} \end{eqnarray*}$ Step 2: Calculate the periodic interest rate. $\begin{eqnarray*} i & = & \frac{j}{m} \\ & = & \frac{7.25\%}{12} \\ & = & 0.60416...\% \\ & = & 0.0060416...\end{eqnarray*}$ Step 3: Calculate the total number of compoundings. $\begin{eqnarray*} n & = & m \times t \\ & = & 12 \times 3 \\ & = & 36 \end{eqnarray*}$ Step 4: Calculate the present value. $\begin{eqnarray*} PV & = & FV \times (1+i)^{-n} \\ & = & 38,000 \times (1+0.0060416...)^{-36} \\ & = & \30,592.06 \end{eqnarray*}$ If Castillo’s Warehouse places$30,592.06 into the investment, it will earn enough interest to grow to $38,000 three years from now to purchase the forklift. ## Using a Financial Calculator As in the previous section, a financial calculator can be used to solve for the present value in compound interest problems. You use the financial calculator in the same way as described previously, but the only difference is that the unknown quantity is PV instead of FV. You must still load the other six variables into the calculator and apply the cash flow sign convention carefully. ## USING THE TI BAII PLUS CALCULATOR TO FIND THE PRESENT VALUE FOR COMPOUND INTEREST The time value of money buttons are located in the TVM row (the third row from the top) of the calculator. The five buttons located on the third row of the calculator are five of the seven variables required for time value of money calculations. This row’s buttons are different in colour from the rest of the buttons on the keypad. The other two variables are in a secondary menu above the I/Y key and are accessed by pressing 2nd I/Y. Altogether, there are seven variables required to complete time value of money calculations. Note that P/Y and C/Y are not main button keys in the TVM row. The P/Y and C/Y variables are located in the secondary function accessed by pressing 2nd I/Y. Variable Meaning N Total number of compounding periods. This is the same value as $n$ in the future value/present value formulas: $N=\mbox{time in years} \times \mbox{compounding frequency}$ I/Y Interest rate per year (i.e. the nominal interest rate). The interest rate is entered in percent form (without the % sign). For example, 5% is entered as 5. PV Present value or principal. PMT Periodic annuity payment. For compound interest only calculations, PMT=0. (Note: in later chapters you will learn about annuities where PMT will not be 0.) FV Future value or maturity value. P/Y Payment frequency for annuity payment. For compound interest only calculations, P/Y is set to the same value as C/Y. (Note: in later chapters you will learn about annuities where P/Y will be set to the frequency of the payments.) C/Y Compounding frequency. This is the value of $m$. To enter values into the calculator: • For the main button keys in the TVM row (i.e. N, I/Y, PV, PMT, FV), enter the number first and then press the corresponding button. • For example, to enter N=34, enter 34 on the calculator and then press N. • For P/Y and C/Y, press 2nd I/Y. At the P/Y screen, enter the value for P/Y and then press ENTER. Press the down arrow to access the C/Y screen. At the C/Y screen, enter the value for C/Y and then press ENTER. Press 2nd QUIT (the CPT button) to exit the menu. • For example, to enter P/Y=4 and C/Y=4, press 2nd I/Y. At the P/Y screen, enter 4 and press ENTER. Press the down arrow. At the C/Y screen, enter 4 and press ENTER. Press 2nd QUIT to exit. After all of the known quantities are loaded into the calculator, press CPT and then PV to solve for the present value. Compound Interest (Present and Future Values) by Joshua Emmanuel [6:56] (transcript available). ## EXAMPLE Castillo’s Warehouse will need to purchase a new forklift for its warehouse operations three years from now, when its new warehouse facility becomes operational. If the price of the new forklift is$38,000 and Castillo’s can invest its money at 7.25% compounded monthly, how much money should it put aside today to achieve its goal?
Solution:
The timeline for the investment is shown below.
N $12 \times 3=36$ PV ? FV $38,000$ PMT $0$ I/Y $7.25$ P/Y $12$ C/Y $12$
$\displaystyle{PV=\30,592.06}$
## TRY IT
A debt of $37,000 is owed 21 months from today. If prevailing interest rates are 6.55% compounded quarterly, what amount should the creditor be willing to accept today? Click to see Solution N $4 \times 1.75=7$ PV ? FV $-37,000$ PMT $0$ I/Y $6.55$ P/Y $4$ C/Y $4$ $\displaystyle{PV=\33,023.56}$ ## Present Value Calculations with Variable Changes Addressing variable changes in present value calculations follows the same techniques as future value calculations discussed in the previous section. You must break the timeline into separate time segments, each of which involves its own calculations. Solving for the unknown PV at the left of the timeline means you must start at the right of the timeline. You must work from right to left, one time segment at a time using the formula for PV each time. Note that the present value for one time segment becomes the future value for the next time segment to the left. 1. Read and understand the problem. Identify the future value. Draw a timeline broken into separate time segments at the point of any change. For each time segment, identify any principal changes, the nominal interest rate, the compounding frequency, and the length of the time segment in years. 2. Starting with the future value in the last time segment (starting on the right), solve for the present value. 3. Let the present value calculated in the previous step become the future value for the next segment to the left. If the principal changes, adjust the future value accordingly. 4. Calculate the present value of the next time segment. 5. Repeat the previous steps until you obtain the final present value from the leftmost time segment. ### NOTE To use your calculator efficiently in working through multiple time segments, follow a procedure similar to that for future value. 1. Load the calculator with all known compound interest variables for the last time segment (on the right). 2. Compute the present value at the beginning of the segment. 3. With the answer still on your display, adjust the principal if needed, change the cash flow sign by pressing the $\pm$ key, and then store the unrounded number back into the future value button by pressing FV. Change the N, I/Y, and C/Y as required for the next segment. 4. Return to step 2 for each time segment until you have completed all time segments. ## EXAMPLE Sebastien needs to have$9,200 saved up three years from now. The investment he is considering pays 7% compounded semi-annually, 8% compounded quarterly, and 9% compounded monthly in successive years. To achieve his goal, how much money does he need to place into the investment today?
Solution:
The timeline shows today through to the future value three years from now
Step 1: Calculate the present value at the start of the last segment on the right.
N $12 \times 1=12$ PV ? FV $9,200$ PMT $0$ I/Y $9$ P/Y $12$ C/Y $12$
$\displaystyle{PV_1=\8,410.991...}$
Step 2: Calculate the present value at the start of the second segment on the right. The present value from the first step becomes the future value for the second step: $PV_1=\8,410.991...=FV_2$.
N $4 \times 1=4$ PV ? FV $8,410.991...$ PMT $0$ I/Y $8$ P/Y $4$ C/Y $4$
$\displaystyle{PV_2=\7,770.4555...}$
Step 3: Calculate the present value at the end of the third segment on the right. The present value from the second step becomes the future value for the third step: $PV_2=\7,770.455...=FV_3$.
N $2 \times 1=2$ PV ? FV $7,770.455..$ PMT $0$ I/Y $7$ P/Y $2$ C/Y $2$
$\displaystyle{PV_3=\7,253.80}$
Sebastien needs to place $7,253.80 into the investment today to have$9,200 three years from now.
For the first 4.5 years, a loan was charged interest at 4.5% compounded semi-annually. For the next 4 years, the rate was 3.25% compounded annually. If the maturity value was $45,839.05 at the end of the 8.5 years, what was the principal of the loan? Click to see Solution N $4$ $9$ PV $\textcolor{blue}{40,334.378...}$ $\textcolor{blue}{33,014.56}$ FV $-45,839.05$ $-40,334.378...$ PMT $0$ $0$ I/Y $3.25$ $4.5$ P/Y $1$ $2$ C/Y $1$ $2$ $\displaystyle{PV=\33,014.56}$ ## Exercises 1. A loan is repaid with$14,000. If the loan was taken out 14 years ago at 9% compounded semi-annually, how much money was borrowed? How much interest was paid on the loan?
PV=$4,081.99, I=$9,9180.01
2. In 9 years and 3 months, you want to have $97,000 in your savings account. How much money must you invest today if the savings account earns 6% compounded monthly? Click to see Answer$55,762.07
3. Eight and a half years ago, Tom took out a loan. The interest rate on the loan was 4.5% compounded semi-annually for the first four and half years and 3.25% compounded annually for the last four years. Tom repaid the loan today with a payment of $45,839.05. How much money did Tom borrow? How much interest did Tom pay? Click to see Answer PV=$33,014.56, I=$12,824.49 4. George wants to invest some money today. In 6.5 years, George wants to have$7,223.83 in his investment. The investment earns 8.05% compounded semi-annually for the first 2 years and 6 months, then 7.95% compounded quarterly for 1 year and 3 months, and then 7.8% compounded monthly for 2 years and 9 months. How much money does George need to invest?
$4,340 5. Dovetail Industries needs to save$1,000,000 for new production machinery that it expects will be needed six years from today. If money can earn 8.35% compounded monthly, how much money should Dovetail invest today?
$606,976.63 6. A debt of$37,000 is owed 21 months from today. If prevailing interest rates are 6.55% compounded quarterly, what amount should the creditor be willing to accept today?
$33,023.56 7. Rene wants to invest a lump sum of money today to make a$35,000 down payment on a new home in five years. If he can place his money in an investment that will earn 4.53% compounded quarterly in the first two years followed by 4.76% compounded monthly for the remaining years, how much money does he need to invest today?
$27,736.24 8. In August 2004, Google Inc. made its initial stock offering. The value of the shares grew to$531.99 by July 2011. What was the original value of a share in August 2004 if the stock has grown at a rate of 26.8104% compounded monthly?
$85 9. If a three-year and seven-month investment earned$8,879.17 of interest at 3.95% compounded monthly, what amount was originally placed into the investment?
$58,499.97 10. A lottery ticket advertises a$1 million prize. However, the fine print indicates that the winning amount will be paid out on the following schedule: $250,000 today,$250,000 one year from now, and $100,000 per year thereafter. If money can earn 9% compounded annually, what is the value of the prize today? Click to see Answer$836,206.54
11. Your company is selling some real estate and has received three potential offers:
• Option A: $520,000 today plus$500,000 one year from now.
• Option B: $200,000 today,$250,000 six months from now, and $600,000 15 months from now. • Option C:$70,000 today, $200,000 in six months, then four quarterly payments of$200,000 starting six months later.
Prevailing interest rates are expected to be 6.75% compounded semi-annually in the next year, followed by 6.85% compounded quarterly afterwards. Rank the three offers from best to worst based on their values today.
Option B is best with PV=$993,846.61; Option C is second best with PV=$993,391.13; Option A is third best with PV=\$987,884.82
9.3: Determining the Present Value” from Business Math: A Step-by-Step Handbook (2021B) by J. Olivier and Lyryx Learning Inc. through a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License unless otherwise noted. |
This section looks at a number of the key statistical graphs required for GCSE Mathematics.
It will also show how you can find averages from a given graph.
Being able to read and then relate the information which a graph shows is an important skill in real-life
Cumulative Frequency Graph
If we are given a table containing continuous data, we can find a running total of the frequency. This is called Cumulative Frequency.
Height (m) Frequency Cumulative Frequency In context 7 7 7 people less than 1.4m 8 15 15 people less than 1.6m 22 37 37 people less than 1.8m 3 40 40 people less than 2.0m
We can therefore plot a graph:
· The x-axis (horizontal) will be Height(m)
· The y-axis will (vertical) will be Cumulative Frequency
· We plot the numbers in red
· The graph starts at 1.2m, because this was the lowest value
We can now use this graph to estimate a number of key values
In this case n=40 (the total number of people
Median: On a cumulative frequency graph we find value
We can read this off the graph to get 1.64m
Lower Quartile: On a cumulative frequency graph we find value
We can read this off the graph to get 1.48m
Upper Quartile: On a cumulative frequency graph we find value
We can read this off the graph to get 1.73m
From this we can deduce the IQR = UQ - LQ = 1.73m-1.48m=0.25m
Histogram – the A/A* graph!
Suppose you are given a table of continuous data (see below).
Given a class eg. the class width is
In the table below, you can see how the class widths change.
In this case, we will need to construct a histogram to represent the data.
In a histogram, the AREA of the bars equals the FREQUENCY
To achieve this, we need to calculate a value called the FREQUENCY DENSITY
Height (m) Frequency class-width Frequency density 5 0.2 25 12 0.3 40 15 0.3 50 2 0.1 20
We now draw a graph with:
· Height(m) on the x-axis
· Frequency density on the y-axis
The next section considers how to read graphs to find an average
Finding the mean and median from a frequency graph
A frequency graph to show the frequency of scores in a test
This graph can be turned into a frequency table
Mark Midpoint Frequency 5 8 15 12 25 11 35 3 TOTAL 34 |
# How do you factor the expression 4x^4-6x^2+2?
May 12, 2018
$4 {x}^{4} - 6 {x}^{2} + 2 = \left(x - 1\right) \left(x + 1\right) \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right)$
$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = 4 \left(x - 1\right) \left(x + 1\right) \left(x - \frac{\sqrt{2}}{2}\right) \left(x + \frac{\sqrt{2}}{2}\right)$
#### Explanation:
Given:
$4 {x}^{4} - 6 {x}^{2} + 2$
Note that the sum of the coefficients is zero, i.e. $4 - 6 + 2 = 0$.
So we can tell that $x = \pm 1$ are zeros and $\left(x - 1\right) \left(x + 1\right) = {x}^{2} - 1$ is a factor.
Also all of the terms are divisible by $2$, so we could separate that out as a factor, but given what we find, it's probably better to leave it in there until the end:
$4 {x}^{4} - 6 {x}^{2} + 2 = \left({x}^{2} - 1\right) \left(4 {x}^{2} - 2\right)$
$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = \left({x}^{2} - {1}^{2}\right) \left({\left(2 x\right)}^{2} - {\left(\sqrt{2}\right)}^{2}\right)$
$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = \left(x - 1\right) \left(x + 1\right) \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right)$
$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = 4 \left(x - 1\right) \left(x + 1\right) \left(x - \frac{\sqrt{2}}{2}\right) \left(x + \frac{\sqrt{2}}{2}\right)$ |
# SEBA CLASS 10 MATHS CHAPTER 5 MCQ ARITHMETIC PROGRESSIONS
3 Questions of 1 mark each will come from SEBA Class 10 Maths Chapter 5 MCQ (Arithmetic Progressions). Arithmetic Progressions is an important chapter not only in Class 10 but it has immense application afterwards.
So, we have prepared 10 important MCQ Questions from SEBA Class 10 Maths Chapter 5 MCQ Arithmetic Progressions. Before that, we will provide a detailed syllabus and topics of Arithmetic Progressions prescribed by SEBA for Class 10. Moreover, we have provided mark distribution of SEBA Class 10 Chapter 5 Arithmetic Progressions.
## SYLLABUS OF SEBA CLASS 10 MATHS CHAPTER 5 MCQ (ARITHMETIC PROGRESSIONS)
First let us know the syllabus of SEBA Class 10 Maths Chapter 5 – Arithmetic Progressions.
• Motivation for studying Arithmetic Progression
• Derivation of the nth term and sum of the first n terms of A.P.
• Application of A.P in solving daily life problems.
## TOPICS COVERED UNDER SEBA CLASS 10 MATHS CHAPTER 5 MCQ (ARITHMETIC PROGRESSIONS)
The SCERT Maths textbook for Class 10 has the following topics in Chapter 5 – Arithmetic Progressions.
• 5.1 Introduction
• 5.2 Arithmetic Progressions
• 5.3 nth Term of an AP
• 5.4 Sum of First n Terms of an AP
There are 4 exercises covering the above topics including optional exercise. The solution of Chapter 5 exercises can be found in this link.
## SEBA CLASS 10 MATHS CHAPTER 5 MCQ (ARITHMETIC PROGRESSIONS)
An arithmetic progression (AP) is a sequence of numbers in which the difference between a number and the next one (i.e consecutive numbers) is always the same.
The 1st term or number of AP is donated by a and the number of terms by n. This difference or common difference is usually denoted by the letter d.
For example, the sequence 1, 3, 5, 7, 9 is an arithmetic progression with a common difference of 2. The 1st term is 1 and the number of terms is 5.
The nth term of an arithmetic progression with first term a and common difference d is given by the formula:
an = a + (n-1)d
The sum of n terms of an arithmetic progression is given by:
Sn = n/2 * (2a + (n-1)d)
If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP
is given by :
Sn = n/2 * (a + l)
Some of the important properties of arithmetic progressions are:
• The product of the first and last term of an arithmetic progression is given by:
• a * l = a*(a + (n-1)d)
• The sum of an arithmetic progression is infinite if the common difference is a fraction.
## MARKS DISTRIBUTION OF CHAPTER 5 MCQ (ARITHMETIC PROGRESSIONS)
Exactly 50% of marks in SEBA Class 10 HSLC Examination is assigned for MCQ and Very Short Answer (VSA) Questions.
Four (5) types of questions or question format will be asked from Class 10 Maths Chapter 5 Arithmetic Progressions as listed below.
• Multiple Choice Questions (MCQ) – 1 Marks each
• Very Short Answer (VSA) – 1 Marks each
• Short Answer Type Questions (SA-II) – 3 Marks each
These questions of this chapter will be based on knowledge and understanding. 3 MCQ of 1 marks each of SEBA Class 10 Maths Chapter 5 MCQ (Arithmetic Progressions) will be asked in SEBA HSLC Examination.
The detail marks distribution is shown in the table below.
## SEBA CLASS 10 MATHS CHAPTER 5 MCQ (ARITHMETIC PROGRESSIONS)
1. What is the common difference in the arithmetic sequence 3, 7, 11, 15, 19?
a) 2
b) 3
c) 4
d) 5
1. In the arithmetic sequence 2, 5, 8, 11, 14, … what is the 30th term?
a) 87
b) 88
c) 89
d) 90
1. In the arithmetic sequence 2, 5, 8, 11, 14, … what is the sum of the first 20 terms?
a) 380
b) 385
c) 390
d) 395
1. In the arithmetic sequence -4, -1, 2, 5, 8, … what is the 50th term?
a) 197
b) 198
c) 199
d) 200
1. The 4th term from the end of the AP: –11, –8, –5, …, 49 is
a) 37
b) 40
c) 43
d) 58
1. If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th
term will be
a) 7
b) 11
c) 18
d) 0
1. Two APs have the same common difference. The first term of one of these is
–1 and that of the other is – 8. Then the difference between their 4th terms is
a) -1
b) -8
c) 7
d) -9
1. In an AP if a = –7.2, d = 3.6, an = 7.2, then n is
a)1
b) 3
c) 4
d) 5
1. In an AP if a = 1, an= 20 and Sn = 399, then n is
a)19
b) 21
c) 38
d) 42
1. The list of numbers – 10, – 6, – 2, 2,… is
(a) an AP with d = – 16
(b) an AP with d = 4
(c) an AP with d = – 4
(d) not an AP
## SEBA CLASS 10 MATHS MCQ PDF
The SEBA Class 10 Maths Chapter 3 MCQ PDF is given below
More chapters will be added in this list. So, keep visiting www.guwahatilive.com Education Section
### What are arithmetic progressions Class 10?
An arithmetic progression (AP) is a sequence of numbers in which the difference between a number and the next one (i.e consecutive numbers) is always the same.
The 1st term or number of AP is donated by a and the number of terms by n. This difference or common difference is usually denoted by the letter d.
### What is arithmetic progression Class 10 with an example?
An arithmetic progression (AP) is a sequence of numbers in which the difference between a number and the next one (i.e consecutive numbers) is always the same.
The 1st term or number of AP is donated by a and the number of terms by n. This difference or common difference is usually denoted by the letter d.
For example, the sequence 1, 3, 5, 7, 9 is an arithmetic progression with a common difference of 2. The 1st term is 1 and the number of terms is 5.
### Is arithmetic progression in syllabus class 10?
Yes, arithmetic progression in syllabus class 10.
3 Questions of 1 mark each will come from SEBA Class 10 Maths Chapter 5 MCQ (Arithmetic Progressions). Arithmetic Progressions is an important chapter not only in Class 10 but it has immense application afterwards.
## CONCLUSION
We have prepared 10 important MCQ Questions from SEBA Class 10 Maths Chapter 5 Arithmetic Progressions. We have also provided detailed syllabus and topics of Chapter 5 Arithmetic Progressions prescribed by SEBA for Class 10. Moreover, we have also provided mark distribution of SEBA Class 10 Chapter 5. |
• Multiplying and Dividing Integers Temperature Ups and Downs
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• Multiplying and Dividing Integers Temperature Ups and Downs
```Multiplying and Dividing Integers
ACTIVITY 10
Temperature Ups and Downs
Lesson 10-1 Multiplying Integers
My Notes
Learning Targets:
Multiply integers.
Solve real-world problems by multiplying integers.
•
•
SUGGESTED LEARNING STRATEGIES: Marking the Text,
Visualization, Predict and Confirm, Create Representations,
Look for a Pattern
In science class, Mariah is learning about temperature. Mariah has
decided to use the Celsius scale to investigate the three states of water.
• When the temperature of water is below 0°C, it is a solid called ice.
• From 0°C to 100°C, it is a liquid called water.
• Above 100°C, it is a gas called steam.
Mariah starts with a container of water with a temperature of 0°.
That is the temperature at which ice changes to water. To study changes
in the water’s state, she increases the temperature at a constant rate for
10 minutes until it begins to steam.
1. What is the increase in temperature?
CONNECT TO SCIENCE
Substances can exist in three
different forms:
• solid,
• liquid, and
• gas.
Each form is called a state. Heating
and cooling causes a substance to
change from one state to another.
Heating causes ice, the solid form
of water, to become water, the
liquid form, and finally to become
steam, the gas form.
2. What is the rate at which the temperature increased? Write the
3. Would it be more appropriate to represent this rate of increase as a
positive integer or as a negative integer? Explain your reasoning.
Mariah drew this number line to represent the minute-by-minute changes
in temperature:
0 10 20 30 40 50 60 70 80 90 100
4. Let + represent the rate you found in Item 2. Use
represent the total change in temperature.
+
to
Activity 10 • Multiplying and Dividing Integers
117
Lesson 10-1
Multiplying Integers
ACTIVITY 10
continued
My Notes
Next, Mariah places the container in the laboratory freezer. Over the next
20 minutes, she lowers the temperature at a steady rate until the water
begins to freeze.
5. What is the rate of decrease in temperature? Write the answer in
degrees per minute.
6. Would it be more appropriate to represent this rate of decrease as a
positive integer or a negative integer? Explain your reasoning.
7. Mark the number line to represent the minute-by-minute changes in
temperature:
0 10 20 30 40 50 60 70 80 90 100
–
8. Let
represent the rate that you found in Item 5. Use
represent the total change in temperature.
–
to
In Items 4 and 8, you used triangles to represent multiplication.
–
represents a temperature increase of 10 degrees.
represents a temperature decrease of 5 degrees.
So, 10
+
= +100 and 20
–
= −100.
9. a. If – represents −8, what does
represent?
b. If
–
represents −5, what does
c. If
+
represents +3, what does
d. If + represents +9, what does
represent?
118
Unit 2 • Integers
–
–
–
–
–
–
+
+
+
+
+
+
–
–
represent?
+
+
represent?
+
+
+
Lesson 10-1
Multiplying Integers
ACTIVITY 10
continued
10. Write the number sentence represented by each diagram. Each
counter stands for 10 or −10. The first one has been done for you.
a. + + + +
b. − − −
c.
+ +
+ + + +
+ +
+ + + +
f. − −
2 × 40 = 80
−−−
+ +
d. − − − − −
e.
My Notes
−−−−−
+ + + +
−−
−−
+ + + +
−−
11. Use + and – to represent each number sentence.
a. 2 × 50 = 100
b. 3 × (−30) = −90
12. Look at your results for Items 10 and 11. Answer the following
questions:
a. What was the sign of the product when you multiplied two
positive integers?
b. What was the sign of the product when you multiplied a positive
integer and a negative integer?
c. Use these observations to write rules to find the sign of the
product of integers.
Activity 10 • Multiplying and Dividing Integers
119
Lesson 10-1
Multiplying Integers
ACTIVITY 10
continued
My Notes
13. You can use patterns to find the sign of the product of two negative
integers.
a. Fill in ONLY the first four squares in the table below.
×
3
2
1
0
−1
−2
−3
−3
b. Make use of structure. Describe the pattern in the four squares
you filled in.
c. Fill in the last three squares by continuing this pattern.
d. Use the same procedure you used in parts a–c to complete the
multiplication tables below.
×
3
2
1
0
−1
−2
−3
3
2
1
0
−1
−2
−3
3
2
1
0
−1
−2
−3
−5
×
−8
×
−11
14. Complete the table at the right
showing the sign of the
product of integers. Write a
positive sign or negative sign
in each box.
×
positive
negative
positive
negative
15. Make use of structure. In parts a–c, state whether the product is
positive or negative.
a. the product of two negative integers
b. the product of three negative integers
c. the product of four negative integers
d. State a rule for finding the sign of the product of an even number
of negative integers.
e. State a rule for finding the sign of the product of an odd number
of negative integers.
120
Unit 2 • Integers
e. Use your results in parts a and d to write a rule to find the sign of
the product of two negative integers.
Lesson 10-1
Multiplying Integers
ACTIVITY 10
continued
My Notes
16. Find each product.
a. 3(−9)
c. −10(−2)
b. 7 ⋅ 6
d. 5(−8)
17. Complete.
a. 3 × (−4) = −3 × _______ b. 2 × 5 = −2 × _______
c. −6 × 7 = 6 × _______
d. −9 × (−9) = 9 × _______
18. Copy the number line below and use it to show the product of 2
and (−4).
–8 –7 –6 –5 –4 –3 –2 –1
0
1
2
3
4
5
6
7
8
2(−4) = _______
19. Which property justifies this equation? −5 × 12 = 12 × −5?
LESSON 10-1 PRACTICE
20. At 4 p.m. the temperature in Clarksville was 0°C. Over the next
8 hours the temperature fell at a rate of 3 degrees per hour.
a. Write the rate that the temperature fell as an integer.
b. Determine the temperature at midnight.
21. a. Blake borrowed \$25 from his brother. Write an integer to express
the \$25 that Blake owes his brother.
b. Blake borrowed \$25 from his brother a total of 7 times. Write an
integer that represents the amount that Blake owes his brother.
22. Find the 11th number in the pattern below:
0, −7, −14, −21,…
23. A submarine on the surface of the ocean descended at a rate of 7 feet
per second for 2 minutes. Then it ascended at a rate of 4 feet per
second for 3 minutes, Finally, it descended at a rate of 9 feet per
second for 5 minutes. What was the final elevation of the submarine?
(The elevation of the ocean surface is 0 feet.)
Activity 10 • Multiplying and Dividing Integers
121
```
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# Do you have Direct Proportion questions?
If you’re looking for the answers to your ‘direct proportion’ questions, then you’ve come to the right blog! The concept of direct proportion is a relatively simple one β as one value increases, the other increases at the same rate. Even though this seems relatively simple, it can be applied to a lot of different situations and the types of questions you can get on a GCSE paper will vary a lot.
Letβs consider two variables, π₯ and π¦, that are directly proportional. If π₯ is doubled, then π¦ is also doubled. If π₯ is halved, then π¦ is halved. In fact, if we multiply or divide one of the variables by any number, the other variable will be changed in the same way.
We can plot the two variables on a set of axes. Whenever the two variables are directly proportional, the graph will be a straight line that passes through the point (0, 0). We can use this relationship to find the value of one variable when we know the value of the other. One way to do this is to first find the value of π₯ when π¦ = 1 or of π¦ when π₯ = 1. This is called the unitary method.
Example 1
3 pens and 5 exercise books cost Β£7.36.
2 exercise books cost Β£2.50.
How much will 10 pens cost?
Since 2 exercise books cost Β£2.50, we can divide this by 2 to find the cost of 1 exercise book.
2 exercise books = Β£2.50
1 exercise book = 2.50 Γ· 2
= Β£1.25
Now we can multiply the cost of 1 exercise book by 5 to find the cost of 5 exercise books.
5 exercise books = 1.25 Γ 5
= Β£6.25
Now we can subtract this from the total cost to find the cost of 3 pens and repeat the process to find the cost of 10 pens.
3 pens = 7.36 β 6.25
= Β£1.11
1 pen = 1.11 Γ· 3
= Β£0.37
10 pens = 0.37 Γ 10
= Β£3.70
Example 2
Here is a list of ingredients for making 12 cookies.
a. How much flour will Pedra need to make 24 cookies?
b. Louisa only has 10 eggs. She has plenty of other ingredients. How many cookies can she make?
250g flour
200g chocolate chips
125g butter
125g sugar
3 eggs
a. Since 24 is a multiple of 12, we donβt need to find the amount of flour to make one cookie. Instead, we can find the amount of flour needed by doubling the amount of flour needed for 12.
250 Γ 2 = 500g
If you prefer, you could still find the amount of flour for 1 cookie and multiply it by 24 β itβs just not the most efficient method.
b. For this part, itβs easier to find the number of cookies that we can make using 1 egg. |
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Transcript
```Name: ___________________________________
Date: _________________
TRIGONOMETRIC IDENTITIES – DAY #2
ALGEBRA 2 WITH TRIGONOMETRY
In this lesson we will develop and use identities that relate the sine and cosine of an angle to that of twice its
measure. These identities, which are useful in calculus and equation solving, are known as the Double Angle
Identities.
THE DOUBLE ANGLE IDENTITIES
Exercise #1: Using the sum identities shown below, derive the identity for (a) sin 2A and (b) the first
identity for cos 2A .
(a) sin A B sin A cos B sin B cos A
(b) cos A B cos A cos B sin A sin B
The other two identities for cos 2A are found by substituting the Pythagorean Identity into the first identity
for cos 2A . These derivations will be left for the homework.
Exercise #2: Verify the identity for sin 2A using A 30 and 2 A 60 .
Exercise #3: For an angle it is known that 90 180 . If cos 3
(a) sin 2
5
then find:
(b) cos 2
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #8 – TRIGONOMETRIC ALGEBRA – LESSON #5
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
THE DOUBLE ANGLE IDENTITIES
Exercise #4: For the angle A it is known that 180 A 270 and sin A 11
(a) sin 2A
6
. Find the following:
(b) cos 2A
Exercise #5: Which of the following is equivalent to 2sin 50 cos50 ?
(1) sin 25
(3) cos100
(2) sin100
(4) cos 25
Exercise #6: Which of the following is not equal to cos80 ?
(1) 1 sin 2 80
(3) 2cos2 160 1
(2) 1 2sin 2 40
(4) cos2 40 sin 2 40
Exercise #7: If cos 0.28 then cos 2 is closest to
(1) 0.56
(3) 0.68
(2) 0.14
(4) 0.84
Exercise #8: The value of cos130 is equal to
(1) 1 sin 2 130
(3) 2cos2 260 1
(2) 1 2sin 2 65
(4) 2cos 65
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #8 – TRIGONOMETRIC ALGEBRA – LESSON #5
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
Name: ___________________________________
Date: _________________
TRIGONOMETRIC IDENTITIES – DAY #2
ALGEBRA 2 WITH TRIGONOMETRY - HOMEWORK
THE DOUBLE ANGLE IDENTITIES
SKILLS
1. Which of the following is equal to the expression 2cos2 60 1 ?
(1) cos30
(3) cos120
(2) sin 30
(4) sin120
2. Which of the following is equal to sin10 ?
(1) 1 2sin 2 20
(3) 2sin 5 cos5
(2) 2sin 20 cos 20
(4) cos2 5 sin 2 5
3. Which of the following is not equal to cos50 ?
(1) 1 sin 2 50
(3) cos310
(2) 2cos2 100 1
(4) 1 2sin 2 25
4. If cos 1
(1) 7
(2) 5
9
9
3
then cos 2 ?
(3) 2
3
(4) 3
5. If 90 A 180 and cos 2 A 5
(1) 3
(2)
(3)
8
6
2
(4) 5
3
3
8
then sin A ?
4
7
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #8 – TRIGONOMETRIC ALGEBRA – LESSON #5
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #8 – TRIGONOMETRIC ALGEBRA – LESSON #5
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
THE DOUBLE ANGLE IDENTITIES
6. For an angle it is known that 90 180 and cos 5
(a) sin 2
13
. Find in simplest form:
(b) cos 2
7. For an angle it is known that cos 2
4
and 270 360 . Find in simplest form:
(b) cos 2
(a) sin 2
REASONING
8. We saw in the lesson where the first identity for cos 2A comes from. By combining this identity with the
Pythagorean Identity, cos2 A sin 2 A 1 the other two can be found.
(a) By
substituting
the Pythagorean Identity
into the first identity for
, derive the second identity for
.
(b) By
substituting
the Pythagorean Identity
into the first identity for
, derive the third identity for
ALGEBRA 2 WITH TRIGONOMETRY, UNIT #8 – TRIGONOMETRIC ALGEBRA – LESSON #5
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2009
.
```
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# Lesson 10: Inference for One Mean: Sigma Known (Confidence Interval)
(Redirected from Lesson 10)
These optional videos discuss the contents of this lesson.
## 1 Lesson Outcomes
By the end of this lesson, you should be able to:
• Confidence Intervals for a single mean with σ known:
• Calculate and interpret a confidence interval for a population mean given a confidence level.
• Explain how the margin of error changes with the sample size and the level of confidence.
• Identify a point estimate and margin of error for the confidence interval.
• Show the appropriate connections between the numerical and graphical summaries that support this confidence interval.
• Check the requirements the confidence interval.
• Calculate a desired sample size given a level of confidence and margin of error.
## 2 Political Polls
During an election in the United States, many polls are conducted to determine the attitudes of likely voters. Poll results are usually reported as percentages. For example, a poll might state that 49% favor the Republican candidate and 51% favor the Democratic candidate.
Polls always include a margin of error. The margin of error gives an estimate of the variability in the responses. A common value for the margin of error in political polls is 3%.
When we incorporate the margin of error, we estimate that the true proportion of people who favor the Republican candidate is 49% ± 3%, or in other words between 46% and 52%. For the Democratic candidate, we get 48% to 54%. There is a lot of overlap in these numbers. In this case, the political race is too close to know who might win.
In this reading, we will explore the margin of error and its role in estimating a parameter.
## 3 Background
### 3.1 Point Estimators
We have learned about several statistics. Remember, a statistic is any number computed based on data. The sample statistics we have discussed are used to estimate population parameters.
Sample Statistic Population Parameter
Mean $\bar x$ $\mu$
Standard Deviation $s$ $\sigma$
Variance $s^2$ $\sigma^2$
$\vdots$ $\vdots$ $\vdots$
The statistics above are called point estimators because they are just one number (one point on a number line) that is used to estimate a parameter. Parameters are generally unknown values. Think about the mean. If $\mu$ is unknown, how do we know if $\bar x$ is close to it?
The short answer is that we will never know for sure if $\bar x$ is close to $\mu$. This does not mean that we are helpless. The laws of probability and the normal distribution provide a way for us to create a range of plausible values for a parameter (e.g. $\mu$) based on a statistic (e.g. $\bar x$).
### 3.2 Interval Estimators
A point estimator gives one specific value as an estimate of a parameter. An interval estimator is a range of plausible values for a parameter. We can create an interval estimate by starting with a point estimate and adding or subtracting the margin of error.
In the political poll mentioned above, the point estimate for the support of the Republican candidate is 49%. By adding and subtracting the margin of error, we get the interval estimate: 46% to 52%.
A confidence interval is a commonly used interval estimator. In this reading, we will explore how to create a confidence interval for the mean when $\sigma$ is known.
### 3.3 The Margin of Error
#### 3.3.1 Properties of Bell-shaped Curves
1. Fill in the blank in the following sentence.
"The 68-95-99.7% rule only applies for distributions that are _________."
The 68-95-99.7% rule only applies for distributions that are bell-shaped.
In the past, some students have answered that the data must be normally distributed. Actually, the 68-95-99.7% rule will work for any distribution that is mound-shaped and symmetrical. As long as the data are unimodal and not skewed left or skewed right, this rule works well.
2. Approximately what percentage of data from a bell-shaped distribution will lie within two standard deviations of the mean?
Using the 68-95-99.7% rule, about 95% of the observations will lie within two standard deviations of the mean.
#### 3.3.2 The Distribution of the Sample Mean
We learned in the reading Distribution of Sample Means & The Central Limit Theorem about the characteristics of the sample mean, $\bar x$. Specifically, if the population from which the data are drawn is (i) approximately normal or (ii) if the sample size is large, then $\bar x$ will be approximately normally distributed. Furthermore, if the original population has mean $\mu$ and standard deviation $\sigma$, then the sampling distribution of $\bar x$ will have mean $\mu$ and standard deviation $\sigma/\sqrt{n}$.
So, if either condition (i) or (ii) is met, then we can consider the sample mean $\bar x$ as a normal random variable with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$. According to the 68-95-99.7% rule for symmetric bell-shaped distributions, about 95% of the time, the sample mean ($\bar x$) will lie within two standard deviations of the population mean ($\mu$).
This is an important concept. Make sure that you understand the logic above before you continue reading.
#### 3.3.3 How Far is $\bar x$ from $\mu$, or in other words, How Far is $\mu$ from $\bar x$?
Assuming that $\bar x$ is approximately normally distributed, about 95% of the time, it will be within two standard deviations of $\mu$.
Remember, the standard deviation of $\bar x$ is $\frac{\sigma}{\sqrt{n}}$. For the variable $\bar x$, two standard deviations would be equal to $2 \frac{\sigma}{\sqrt{n}}$.
If we collect a random sample from a population and $\bar x$ is normally distributed, then about 95% of the time the sample mean $\bar x$ will be less than $2 \frac{\sigma}{\sqrt{n}}$ units away from the population mean $\mu$. Notice that this is true, whether or not we know $\mu$.
We can express this as a probability statement:
$$P\left( \mu - 2 \frac{\sigma}{\sqrt{n}} < \bar X < \mu + 2 \frac{\sigma}{\sqrt{n}} \right) \approx 0.95$$
Here is the magic: If $\bar x$ is within 2 standard deviations of $\mu$, then $\mu$ is within 2 standard deviations of $\bar x$. It may seem silly to state it, but this is very important.
Starting with $$\mu - 2 \frac{\sigma}{\sqrt{n}} < \bar X < \mu + 2 \frac{\sigma}{\sqrt{n}}$$ We subtract $\mu$ from all parts of the inequality: $$\mu - 2 \frac{\sigma}{\sqrt{n}} - \mu < \bar X - \mu < \mu + 2 \frac{\sigma}{\sqrt{n}} - \mu$$ Which reduces to $$- 2 \frac{\sigma}{\sqrt{n}} < \bar X - \mu < 2 \frac{\sigma}{\sqrt{n}}$$ Now, subtract $\bar X$ from all the terms: $$- 2 \frac{\sigma}{\sqrt{n}} - \bar X < \bar X - \mu -\bar X < 2 \frac{\sigma}{\sqrt{n}} -\bar X$$ And this simplifies to $$- 2 \frac{\sigma}{\sqrt{n}}-\bar X < -\mu < 2 \frac{\sigma}{\sqrt{n}}-\bar X$$ Multiplying all the terms by $-1$, we get $$2 \frac{\sigma}{\sqrt{n}}+\bar X > \mu > -2 \frac{\sigma}{\sqrt{n}}+\bar X$$ Rewriting this reversing the order of the terms, we get $$\bar X - 2 \frac{\sigma}{\sqrt{n}} < \mu < \bar X + 2 \frac{\sigma}{\sqrt{n}}$$
So, the statement $$P\left( \mu - 2 \frac{\sigma}{\sqrt{n}} < \bar X < \mu + 2 \frac{\sigma}{\sqrt{n}} \right) \approx 0.95$$ is equivalent to the statement $$P\left( \bar X - 2 \frac{\sigma}{\sqrt{n}} < \mu < \bar X + 2 \frac{\sigma}{\sqrt{n}} \right) \approx 0.95$$
## 4 Putting It All Together: Confidence Intervals for $\mu$ when $\sigma$ is Known
Notice that if we know $\sigma$, then with approximately 95% confidence, $\mu$ will be between the following two values: $$\left( \bar X - 2 \frac{\sigma}{\sqrt{n}}, ~ \bar X + 2 \frac{\sigma}{\sqrt{n}} \right)$$ This equation gives a confidence interval for $\mu$.
A confidence interval is actually a point estimate $\left( \bar x \right)$ plus or minus the margin of error $\left( 2 \frac{\sigma}{\sqrt{n}} \right)$.
We use the letter $m$ to denote the margin of error: $$m = 2 \frac{\sigma}{\sqrt{n}}$$ Using this definition for $m$, our confidence interval can be written as $$( \bar x - m, ~ \bar x + m )$$ or $$\bar x \pm m$$ This pattern will be repeated throughout the course as we create confidence intervals of the form: \left( \begin{align} point~ ~ ~ & & margin~ & & point~ ~ ~ & & margin~ \\ estimate & ~ ~- & of~error & ~ ~, & estimate & ~ ~+ & of~error \end{align} \right) or $$(\text{point estimate}) \pm (\text{margin of error})$$
### 4.1 A Little More Precision
#### 4.1.1 What $z$ Corresponds to 95% of the Area?
The 68-95-99.7% Rule for bell-shaped distribution is just a quick approximation. This is useful for estimation, but more precision is usually required.
3. For a standard normal distribution, between what two $z$-scores will 95% of the data fall? In other words, find the values $-z$ and $z$ such that the area between them is equal to 0.95. Use the Normal Probability Applet.
To find the value of $z^*$, given a specific confidence level, do the following:
• Open the applet at Normal Probability Applet
• Shade the area between the two red lines (the middle portion of the graph) only and enter the decimal value for the desired percentage (e.g. 0.95 for 95%) in the Area box.
• The value of $z^*$ will be displayed below. Two numbers will be given along the horizontal axis. The only difference in these numbers is their sign: one is positive and one is negative. The positive value is the desired $z^*$.
95% of the area under the standard normal curve is between $z=-1.96$ and $z=1.96$
For normally distributed data, 95% of the observations will fall within 1.96 standard deviations of the mean.
The actual formula for a 95% confidence interval for $\mu$ (when $\sigma$ is known) is: $$\left( \bar x - 1.96 \frac{\sigma}{\sqrt{n}}, \bar x + 1.96 \frac{\sigma}{\sqrt{n}} \right)$$ Please notice that the number 2 was used as an approximation for the actual value of 1.96. When computing a 95% confidence interval for a mean with $\sigma$ known, please use 1.96 in the equation.
#### 4.1.2 Worked Example: Rolling a Die 25 Times
We will compute a 95% confidence interval for the mean of $n=25$ rolls of a fair die. A die was rolled 25 times. The values rolled were: $$3 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 3 ~ ~ ~ 4 ~ ~ ~ 2 ~ ~ ~ 2 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 5$$ The mean of these values is $\bar x = 3.04$. It is a fact that for the outcome of a six-sided die, $\sigma=\sqrt{\frac{35}{12}} \approx 1.7078$.
Applying the formula for a 95% confidence interval, $$\left( \bar x - z^* \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^* \frac{\sigma}{\sqrt{n}} \right)$$ we get: $$\left( 3.04 - 1.96 \frac{1.7078}{\sqrt{25}}, ~ 3.04 + 1.96 \frac{1.7078}{\sqrt{25}} \right)$$ or $$(2.37, ~ 3.71)$$
#### 4.1.3 What about Other Confidence Levels?
Most of the time, researchers report 95% confidence intervals. The number 95% is called the confidence level. Sometimes it is desirable to use a level of confidence that is different than 95%. In that case, we need to use a number besides 1.96 in the calculations.
Suppose we want to create a 90% confidence interval. What value would we put in the blanks in the confidence interval formula below? $$\left( \bar x - \underline{~ ~ ~ ? ~ ~ ~} \frac{\sigma}{\sqrt{n}}, ~ \bar x + \underline{~ ~ ~ ? ~ ~ ~} \frac{\sigma}{\sqrt{n}} \right)$$ Another way to state this question is to ask, between what two $z$-scores will 90% of the data in a standard normal distribution fall? We need to find the values of $-z$ and $z$ such that the area between them is equal to 0.90. Again, we use the Normal Probability Applet.
So, for a 90% confidence interval, the value of $z*$ would be 1.645. Note: We use an asterisk ($*$) to indicate that $z$ was not computed from data, but was determined based on a chosen confidence level, 90%.
The formula for a 90% confidence interval for a mean when $\sigma$ is known is: $$\left( \bar x - 1.645 \frac{\sigma}{\sqrt{n}}, ~ \bar x + 1.645 \frac{\sigma}{\sqrt{n}} \right)$$
When $\sigma$ is known, we use $z^* = 1.96$ to create 95% confidence intervals, and we use $z^* = 1.645$ to create 90% confidence intervals.
#### 4.1.4 Formula for the Confidence Interval for $\mu$ ($\sigma$ Known)
With this notation, the confidence interval formula generalizes to: $$\left( \bar x - z^* \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^* \frac{\sigma}{\sqrt{n}} \right)$$ where $z^*$ is determined by the level of confidence. If you want a 90% confidence interval, then $z^*$ is the number of standard deviations on either side of the mean that you must go to capture 90% of the data.
4. If $\bar x$ follows a normal distribution and $\sigma$ is known, what is the equation for a 99% confidence interval for the true population mean?
For a 99% confidence level, $z^* = 2.576$. The equation for the 99% confidence interval becomes:
$\left( \bar x - 2.576 \frac{\sigma}{\sqrt{n}}, ~ \bar x + 2.576 \frac{\sigma}{\sqrt{n}} \right)$
5. Apply the equation from the previous problem to find a 99% confidence interval for the mean value rolled on the die. Use the data given above. For convenience, they are reproduced here:
$3 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 3 ~ ~ ~ 4 ~ ~ ~ 2 ~ ~ ~ 2 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 5$
If we were making a 99% confidence interval, we follow the same procedure as for a 95% confidence interval, only we use $z^*=2.576$. The 99% confidence interval for the true mean is:
$\left( 3.04 - 2.576 \frac{1.7078}{\sqrt{25}}, ~ 3.04 + 2.576 \frac{1.7078}{\sqrt{25}} \right)$
which simplifies to:
$(2.16, ~ 3.92)$
The most commonly used values for $z^*$ are:
Confidence
Level
$z^*$
90% 1.645
95% 1.960
99% 2.576
Any other values that you need can be determined using the Normal Probability Applet.
### 4.2 Factors Affecting the Width of the Confidence Interval
Recall that the formula to compute the confidence interval for a mean, where $\sigma$ is known is: $$\left( \bar x - z^* \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^* \frac{\sigma}{\sqrt{n}} \right)$$
6. What would happen to the confidence interval if the sample size $n$ was increased, but the other values were still the same?
Since the sample size $n$ is in the denominator, if $n$ is increased, $\sqrt{n}$ would increase, and the fraction $\frac{\sigma}{\sqrt{n}}$ would decrease, which would lead to a smaller margin of error. In other words, if the sample size is increased, the width of a confidence interval will decrease--it will become narrower.
7. What would happen to the confidence interval if the confidence level were increased, say from 95% to 99%?
Increasing the confidence level will lead to larger values of $z^*$. If $z^*$ is increased, then the margin of error $z^* \frac{\sigma}{\sqrt{n}}$ would increase. This would lead to a wider confidence interval.
### 4.3 Interpretation of Confidence Intervals
How do we interpret confidence intervals? What do they really mean?
(Image source: Flickr/ICMA Photos)
Consider a coin with two sides: one called "heads" and the other called "tails". Imagine that you flipped this coin, but you have not looked at it yet. What is the probability that the coin shows heads?
Strangely enough, the answer is, it depends! If the head is facing up, then the probability that the coin shows heads is 1. If the head is facing down, then the probability that it shows heads is 0.
The coin has been tossed. There is no randomness left in the process. So, either the head is facing up or it is facing down. So, the probability that the coin shows heads is either 1 or 0. (We just don't know which.) The fact that we do not know the outcome does not change it or make it random.
Before we toss the coin, the probability that the coin will show heads is $\frac{1}{2}=0.5$. After we toss the coin, the probability that we get heads is 1 or 0.
Transferring this reasoning to confidence intervals, we get a similar result.
Once we have collected data on something, there is no randomness in the system. Any confidence interval that is created using that data will either contain the true parameter ($\mu$) or it will not. After collecting data, the probability that a specific confidence interval will contain $\mu$ is either 1 or 0.
The correct interpretation of a 95% confidence interval is to say, "We are 95% confident that the true mean lies within the lower and upper bounds of the confidence interval."
Consider the 95% confidence interval for the true mean of 25 rolls of a fair die. We found the 95% confidence interval to be: $(2.37, 3.71)$. When we interpret this confidence interval, we say, "We are 95% confident that the true mean is between 2.37 and 3.71."
The word, "confident" implies that if we repeated this process many, many times, 95% of the confidence intervals we would get would contain the true mean $\mu$. It does not imply anything about whether or not one specific confidence interval will contain the true mean.
We do not say that "there is a 95% probability (or chance) that the true mean is between 2.37 and 3.71." The probability that the true mean $\mu$ is between 2.37 and 3.71 is either 1 or 0.
### 4.4 Requirements
There are three requirements that need to be checked when computing a confidence interval for a mean with $\sigma$ known:
1. A simple random sample was drawn from the population
2. $\bar x$ is normally distributed
3. $\sigma$ is assumed to be known
The requirement of normality is satisfied if (a) the raw data are normally distributed or (b) the sample size is large. This procedure is robust to moderate departures from normality. Even if the requirement that $\bar x$ is normally distributed is not satisfied perfectly, it is usually okay to conduct the test.
In practice, we never really know $\sigma$. This procedure is primarily used to help you understand the idea of confidence intervals. When $\sigma$ is unknown, we use a slightly different computation.
### 4.5 Example: Costs of CABG Surgery
(Photo credit: Vanderbilt Photo/Neil Brake)
This reading focuses on an important aspect of designing a study: determining the sample size.
It is important for health care administrators to know the mean hospital costs for patients who have coronary artery bypass graft (CABG) surgery. A large hospital is planning a study to determine their mean costs for patients who have CABG surgery.
A study will be conducted in which the charts of patients who had CABG surgery will be sampled, and their hospital costs will be recorded. For budgetary reasons, the hospital administrators do not want to collect a sample that is too large. However, if the sample size is not large enough, the confidence interval will be too wide to be useful as a planning tool.
After a discussion among the senior administration, they have determined that they want to estimate the mean hospital costs of CABG surgery within $2000 (i.e., plus or minus$2000.) In other words, they want the confidence interval for the true mean to have a margin of error of $2000 dollars. Recall the equation for the confidence interval is: $$\left( \bar x - z^* \frac{\sigma}{\sqrt{n}}, \bar x + z^* \frac{\sigma}{\sqrt{n}} \right)$$ The part of the equation that is added to and subtracted from$\bar x$is called the margin of error. We will denote the margin of error by the letter$m$. $$m=z^* \frac{\sigma}{\sqrt{n}}$$ To use this formula, the parameter$\sigma$must be given to us. It is the true standard deviation of the data you are observing. If you do not know$\sigma$, you can estimate it using the standard deviation reported in a previous study or by conducting a pilot study. A study published by another hospital reported that the standard deviation of the costs for CABG surgery was$28,705 .
In the following questions, you will compute the margin of error, $m$, for a future study of the hospital costs for CABG surgery. The hospital administrators want to use a 95% level of confidence. Assume the standard deviation can be estimated to be $\sigma = \$28,705$. Answer the following questions: 8. If the hospital administrators want to be 95% confident in the results, what should the value of$z^*$be?$ \begin{align} z^*=1.96 \end{align} $9. If the hospital collects a sample of$n=100$patients' costs, what would the margin of error be?$ \begin{array}{1cl} m &= z^* \frac{\sigma}{\sqrt{n}} \\ &= 1.96 \cdot \frac{28705}{\sqrt{100}} \\ &= \$5626.18 \end{array}$
10. If the hospital collects a sample of $n=1000$ patients' costs, what would the margin of error be?
$\begin{array}{1cl} m &= z^* \frac{\sigma}{\sqrt{n}} \\ &= 1.96 \cdot \frac{28705}{\sqrt{1000}} \\ &= \$1779.15 \end{array} $11. Choose any other value for the sample size,$n$. Find the margin of error for your sample size. Answers will vary. 12. Repeat Question 11 until you find the sample size that will yield a margin of error as close to$2000 as possible, without going over.
$n = 792$
## 5 Sample Size Calculations
The process you followed in Questions 8 - 12 is effective, but tedious. There must be an easier way! The trick is to solve the margin of error equation $m = z^* \frac{\sigma}{\sqrt{n}}$ for $n$.
Starting with $$m = z^* \frac{\sigma}{\sqrt{n}}$$ Multiply both sides of the equation by $\sqrt{n}$: $$m \cdot \sqrt{n} = z^* \frac{\sigma}{\sqrt{n}} \cdot \sqrt{n}$$ Which reduces to: $$m \cdot \sqrt{n} = z^* \cdot \sigma$$ Divide both sides of the equation by $m$. Cancelling the $m$'s on the left hand side: $$\sqrt{n} = \frac{z^* \cdot \sigma}{m}$$ Squaring both sides: $$\left( \sqrt{n} \right)^2 = \left( \frac{z^* \cdot \sigma}{m} \right)^2$$ Which simplifies to the desired result: $$n = \left( \frac{z^* \cdot \sigma}{m} \right)^2$$
This gives us the sample size formula, which tells the number of observations required in order to obtain a specified margin of error: $$n = \left( \frac{z^*\sigma}{m} \right)^2$$
Once the level of confidence is selected, $z^*$ is automatically determined. In practice, the most common level of confidence is 95%, which means that $z^*$ would equal 1.96.
If you need a reminder on finding the value of $z^*$, click here.
13. In Question 12, you used the process of guess-and-check to find the sample size. For this question, use the sample size formula to compute the sample size required to estimate the mean cost of CABG surgery, $\mu$, within $2000 with 95% confidence. Recall that in a previous study, the standard deviation was found to be$\sigma = \$28,705$ .
\begin{align} n &= \left( \frac{z^*\sigma}{m} \right)^2 \\ &= \left( \frac{1.96 \cdot 28705}{2000} \right)^2 \\ &= 791.3 \end{align}
### 5.1 Rounding Up
In Question 13, you computed the sample size required to get a margin of error of $2000. Notice that the result was 791.3, which is not a whole number. What does that mean? Does that suggest that you will survey only a fraction of the last patient's costs? Of course not! When doing sample size calculations, if the answer is not a whole number, you always round up to the next highest whole number. This will allow you to get a sample size that is large enough to attain your desired margin of error. If you want to estimate the mean cost of CABG surgery within$2000 with 95% confidence, you will need to survey the files of 792 patients.
When doing sample size calculations, you always round up.
14. Answer this question without doing any computations. If the hospital administrators wanted to estimate the mean cost of CABG surgery with a margin of error of $1000 at the 95% confidence level, should the sample size be larger or smaller than$n=792$? In question 5, we determined that the sample size required to have a margin of error of$2000 is 792. The margin of error is in the denominator of the sample size formula. If we want to reduce the margin of error, we need to increase the sample size. The sample size should be larger than $n=792$.
15. Find the sample size required to estimate the mean cost of CABG surgery with a margin of error of $1000 and with a 95% confidence level.$ \begin{array}{1cl} n &= \left( \frac{z^*\sigma}{m} \right)^2 \\ &= \left( \frac{1.96 \cdot 28705}{1000} \right)^2 \\ &= 3165.4 \end{array} $We always round up in sample size calculations, so the required sample size is$ n=3166$. 16. What happened to the sample size required when the margin of error is cut in half from$2000 to $1000? If we divide the two sample sizes, we get:$ \frac{3166}{792} \approx 4 $In order to reduce the margin of error by half, we need to quadruple the sample size. 17. Remember the sample size required to have a margin of error of$2000 at the 95% level of confidence is $n=792$. If we wanted to estimate the mean cost with the same margin of error at the 99% level of confidence, would the sample size be larger or smaller?
The value of $z^*$ would increase from 1.96 to 2.576, which would increase the required sample size. |
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# A Mathematical Model of Motion
CHAPTER 5 PHYSICS
## 5.1 Graphing Motion in One Dimension
Interpret graphs of position versus time for a moving object to determine the velocity of the object Describe in words the information presented in graphs and draw graphs from descriptions of motion Write equations that describe the position of an object moving at constant velocity
Parts of a Graph
X-axis
Y-axis
All
## 5.1 Position vs. Time
The x-axis is always time The y-axis is always position The slope of the line indicates the velocity of the object. Slope = (y2-y1)/(x2-x1)
d1-d0/t1-t0 d/t
Position vs. Time
20 18 16 14
Position (m)
12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 Time (s)
Uniform Motion
Uniform
motion is defined as equal displacements occurring during successive equal time periods (sometimes called constant velocity) Straight lines on position-time graphs mean uniform motion.
Given below is a diagram of a ball rolling along a table. Strobe pictures reveal the position of the object at regular intervals of time, in this case, once each 0.1 seconds.
Notice that the ball covers an equal distance between flashes. Let's assume this distance equals 20 cm and display the ball's behavior on a graph plotting its xposition versus time.
The slope of this line would equal 20 cm divided by 0.1 sec or 200 cm/sec. This represents the ball's average velocity as it moves across the table. Since the ball is moving in a positive direction its velocity is positive. That is, the ball's velocity is a vector quantity possessing both magnitude (200 cm/sec) and direction (positive).
## Steepness of slope on PositionTime graph
Slope
is related to velocity Steep slope = higher velocity Shallow slope = less velocity
## Different Position. Vs. Time graphs
Position vs. Time
20
Position (m)
Uniform Motion
Accelerated Motion
15 10 5 0 1 2 3 4 5 6 7 8 9 10 Time (s)
## Constant positive velocity (zero acceleration)
Position vs. Time
25
Position (m)
## Increasing positive velocity (positive acceleration)
20 15 10 5 0 1 2 3 4 5 6 7 8 9 10 Time (s)
## Different Position. Vs. Time
Changing slope means changing velocity!!!!!!
## Increasing negative slope = ??
X
B A
t
C A Starts at home (origin) and goes forward slowly B Not moving (position remains constant as time progresses) C Turns around and goes in the other direction quickly, passing up home
During which intervals was he traveling in a positive direction? During which intervals was he traveling in a negative direction? During which interval was he resting in a negative location? During which interval was he resting in a positive location? During which two intervals did he travel at the same speed?
A) 0 to 2 sec B) 2 to 5 sec C) 5 to 6 sec D)6 to 7 sec E) 7 to 9 sec F)9 to 11 sec
x
C
Graphing w/ Acceleration
t
A Start from rest south of home; increase speed gradually B Pass home; gradually slow to a stop (still moving north)
D Continue heading south; gradually slow to a stop near the starting point
## You try it..
Using the Position-time graph given to you, write a one or two paragraph story that describes the motion illustrated. You need to describe the specific motion for each of the steps (a-f) You will be graded upon your ability to correctly describe the motion for each step.
Tangent Lines
t
Positive
Negative Zero
Positive
Negative Zero
Steep
Gentle Flat
Fast
Slow Zero
## Increasing & Decreasing
t
Increasing Decreasing
On a position vs. time graph: Increasing means moving forward (positive direction).
## Decreasing means moving backwards (negative direction).
Concavity
t
On a position vs. time graph: Concave up means positive acceleration. Concave down means negative acceleration.
x Q P S R
Special Points
t
Inflection Pt.
Peak or Valley Time Axis Intercept
P, R
Q P, S
Change of concavity, change of acceleration Turning point, change of positive velocity to negative Times when you are at home, or at origin
## 5.2 Graphing Velocity in One Dimension
Determine, from a graph of velocity versus time, the velocity of an object at a specific time Interpret a v-t graph to find the time at which an object has a specific velocity Calculate the displacement of an object from the area under a v-t graph
5.2
## Velocity vs. Time
Velocity vs. Time
20 18 16 14
Velcoity (m/s)
X-axis is the time Y-axis is the velocity Slope of the line = the acceleration
12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 Time (s)
## Different Velocity-time graphs
Velocity vs. Time
20
Velocity (m/s)
15 10 5 0 1 2 3 4 5 6 7 8 9 10 Time (s)
## Velocity vs. Time
25
Velocity (m/s)
20 15 10 5 0 1 2 3 4 5 6 7 8 9 10 Time (s)
## Velocity vs. Time
Horizontal lines = constant velocity Sloped line = changing velocity
Steeper
## Acceleration vs. Time
Time is on the x-axis Acceleration is on the y-axis Shows how acceleration changes over a period of time. Often a horizontal line.
Acceleration vs. Time
12
10
Acceleration (m/s^2)
0 1 2 3 4 5 6 7 8 9 10 Time (s)
All 3 Graphs
t
a t
Real life
Note how the v graph is pointy and the a graph skips. In real life, the blue points would be smooth curves and the orange segments would be connected. In our class, however, well only deal with constant acceleration. v
a t
## Leftward Velocity with Rightward Acceleration
Graph Practice
Try making all three graphs for the following scenario: 1. Newberry starts out north of home. At time zero hes driving a cement mixer south very fast at a constant speed. 2. He accidentally runs over an innocent moose crossing the road, so he slows to a stop to check on the poor moose. 3. He pauses for a while until he determines the moose is squashed flat and deader than a doornail. 4. Fleeing the scene of the crime, Newberry takes off again in the same direction, speeding up quickly. 5. When his conscience gets the better of him, he slows, turns around, and returns to the crash site.
## Area Underneath v-t Graph
If you calculate the area underneath a v-t graph, you would multiply height X width. Because height is actually velocity and width is actually time, area underneath the graph is equal to Velocity X time or Vt
## Remember that Velocity = d t Rearranging, we get d = velocity X t
So.the area underneath a velocity-time graph is equal to the displacement during that time period.
positive area
Area
t
negative area
Note that, here, the areas are about equal, so even though a significant distance may have been covered, the displacement is about zero, meaning the stopping point was near the starting point. The position graph shows this as well. x
## Velocity vs. Time
The area under a velocity time graph indicates the displacement during that time period. Remember that the slope of the line indicates the acceleration. The smaller the time units the more instantaneous is the acceleration at that particular time. If velocity is not uniform, or is changing, the acceleration will be changing, and cannot be determined for an instant, so you can only find average acceleration
5.3 Acceleration
Determine from the curves on a velocitytime graph both the constant and instantaneous acceleration Determine the sign of acceleration using a v-t graph and a motion diagram Calculate the velocity and the displacement of an object undergoing constant acceleration
5.3 Acceleration
Like speed or velocity, acceleration is a rate of change, defined as the rate of change of velocity Average Acceleration = change in velocity Elapsed time
V 1 V 0 a t
Units of acceleration?
## Rearrangement of the equation
V 1 V 0 a t
at v1 v0
v0 at v1
v1 v0 at
Finally
v1 v0 at
This equation is to be used to find (final) velocity of an accelerating object. You can use it if there is or is not a beginning velocity
## Remember that displacement under constant velocity was
d = vt
With
or
d1 = d0 + vt
acceleration, there is no one single instantaneous v to use, but we could use an average velocity of an accelerating object.
Average velocity of an accelerating object would simply be of sum of beginning and ending velocities
## Average velocity of an accelerating object
V = (v0 + v1)
So.
d1 d 0 vt d1 d 0 1 / 2(v1 v0)t
Key equation
d1 d 0 1 / 2(v1 v 0)t
## Some other equations
d1 d 0 v0t 1/ 2at
This equation is to be used to find final position when there is an initial velocity, but velocity at time t1 is not known.
v1 v 0 d1 d 0 2a
2
## Finding final velocity if no time is known
v1 v0 2a(d 1 d 0)
2 2
## The equations of importance
v1 v0 at
d d 0 1 / 2(v1 v0)t
V 1 V 0 a t
d d 0 v0t 1/ 2at
2
v1 v 0 d1 d 0 2a v1 v0 2a(d 1 d 0)
2 2 2
## Practical Application Velocity/Position/Time equations
Calculation of arrival times/schedules of aircraft/trains (including vectors) GPS technology (arrival time of signal/distance to satellite) Military targeting/delivery Calculation of Mass movement (glaciers/faults) Ultrasound (speed of sound) (babies/concrete/metals) Sonar (Sound Navigation and Ranging) Auto accident reconstruction Explosives (rate of burn/expansion rates/timing with det. cord)
## 5.4 Free Fall
Recognize the meaning of the acceleration due to gravity Define the magnitude of the acceleration due to gravity as a positive quantity and determine the sign of the acceleration relative to the chosen coordinate system Use the motion equations to solve problems involving freely falling objects
Freefall
Defined
No
## Acceleration Due to Gravity
Galileo Galilei recognized about 400 years ago that, to understand the motion of falling objects, the effects of air or water would have to be ignored. As a result, we will investigate falling, but only as a result of one force, gravity.
## Galileo Galilei 1564-1642
Galileos Ramps
Because gravity causes objects to move very fast, and because the timekeepers available to Galileo were limited, Galileo used ramps with moveable bells to slow down falling objects for accurate timing.
Galileos Ramps
Galileos Ramps
## To keep accurate time, Galileo used a water clock.
For the measurement of time, he employed a large vessel of water placed in an elevated position; to the bottom of this vessel was soldered a pipe of small diameter giving a thin jet of water, which he collected in a small glass during the time of each descent... the water thus collected was weighed, after each descent, on a very accurate balance; the difference and ratios of these weights gave us the differences and ratios of the times...
## Displacements of Falling Objects
Looking at his results, Galileo realized that a falling ( or rolling downhill) object would have displacements that increased as a function of the square of the time, or t2 Another way to look at it, the velocity of an object increased as a function of the square of time, multiplied by some constant.
Galileo
also found that all objects, no matter what slope of ramp he rolled them down, and as long as the ramps were all the same height, would reach the bottom with the same velocity.
Galileos Finding
Galileo
found that, neglecting friction, all freely falling objects have the same acceleration.
## Hippo & Ping Pong Ball
In a vacuum, all bodies fall at the same rate.
If a hippo and a ping pong ball were dropped from a helicopter in a vacuum (assuming the copter could fly without air), theyd land at the same time.
## Proving Galileo Correct
Galileo could not produce a vacuum to prove his ideas. That came later with the invention of a vacuum machine, and the demonstration with a guinea feather and gold coin dropped in a vacuum.
9.8
2 m/s
## Acceleration Due to Gravity
Because
this value is an acceleration value, it can be used to calculate displacements or velocities using the acceleration equations learned earlier. Just substitute g for the a
Example problem
## A brick is dropped from a high building.
What
is its velocity after 4.0 sec.? How far does the brick fall during this time?
## The Churchs opposition to new thought
Church leaders of the time held the same views as Aristotle, the great philosopher. Aristotle thought that objects of different mass would fall at different ratesmakes sense huh?????? All objects have their natural position. Rocks fall faster than feathers, so it only made sense (to him) Galileo, in attempting to convince church leaders that the natural position view was incorrect, considered two rocks of different mass.
## Falling Rock Conundrum
Galileo presented this in his book Dialogue Concerning the Two Chief World Systems(1632) as a discussion between Simplicio (as played by a church leader) and Salviati (as played by Galileo) Two rocks of different masses are dropped Massive rock falls faster???
Rocks continued
Now consider the two rocks held together by a length of string. If the smaller rock were to fall slower, it would impede the rate at which both rocks would fall. But the two rocks together would actually have more mass, and should therefore fall faster. A conundrum????? The previously held views could not have been correct.
Galileo had data which proved Aristotle and the church wrong, but church leaders were hardly moved in their position that all objects have their correct position in the world Furthermore, the use of Simplicio (or simpleton) as the head of the church in his dialog, was a direct insult to the church leaders themselves. |
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# Conversion of Customary Units of Measurement
## Convert between customary units for length, mass and volume.
%
Progress
Practice Conversion of Customary Units of Measurement
Progress
%
Conversion of Customary Units of Measurement
Remember the tables that Mr. Potter suggested? Take a look at this dilemma.
Tyrone has his first measurement done when he meets Mr. Potter in the auditorium. He has written the measurement of the first table on the paper.
The first table is $8' \times 4'$ .
When Tyrone arrives at the auditorium, Mr. Potter has another table all clean and set up for Tyrone to check out.
“I think this one is larger than the other one,” Mr. Potter says. “It measures $96'' \times 30''$ .”
Tyrone looks at the table. He doesn’t think this one looks larger, but he can’t be sure.
On his paper he writes.
Table $2 = 96'' \times 30''$
To figure out which table is larger, Tyrone will need to convert customary units of measure. Then he will need to compare them.
This Concept will teach you all that you need to know about how to do this. When finished, you will know which table is larger and so will Tyrone.
### Guidance
Imagine that you are cooking with a recipe that calls for 13 tablespoons of whipping cream. Since you are cooking for a large banquet you need to make 4 times what the recipe makes. So you are multiplying all of the ingredient quantities by 4 and combining them in a very large bowl. You realize that this requires you to use 52 tablespoons of whipping cream. To measure out 1 tablespoon 52 times will take forever!
Don’t worry, though.
You can convert tablespoons to a larger unit of measurement like cups and be able to measure out the whipping cream in larger quantities.
Look at the chart of customary units again.
Customary Units of Length
$& \text{inch} \ (in)\\& \text{foot} \ (ft) && 12 \ in.\\& \text{yard} \ (yd) && 3 \ ft.\\& \text{Mile} \ (mi) && 5,280 \ ft.$
Customary Units of Mass
$& \text{ounce} \ (oz)\\& \text{pound} \ (lb) && 16 \ oz.$
Customary Units of Volume
$& \text{ounce} \ (oz)\\& \text{cup} \ (c) && 8 \ oz.\\& \text{pint} \ (pt) && 16 \ oz.\\& \text{quart} \ (qt) && 32 \ oz.\\& \text{gallon} \ (gal) && 4 \ qt.$
Customary Units of Volume Used in Cooking
$& \text{teaspoon} \ (tsp)\\& \text{tablespoon} \ (tbsp) && 3 \ tsp.\\& \text{cup} \ (c) && 16 \ tbsp.$
Do you notice a relationship between the various units of volume? If we have 8 ounces of a liquid, we have 1 cup of it. If we have 16 ounces of a liquid, we have 1 pint of it, or 2 cups of it. If we have 2 pints of a liquid, we have 1 quart, or 4 cups, or 32 ounces of it.
We just looked at volume relationships by going from ounces to quarts, let’s look at a larger quantity and break it down to smaller parts. If we have 1 cup, how many teaspoons do we have? 1 cup is 16 tablespoons and 1 tablespoon is 3 teaspoons, so 1 cup is $16 \cdot 3$ teaspoons. 1 cup is 48 teaspoons.
Once we know how the different units of measurement relate to each other it is easy to convert among them . Don’t forget that 12 inches is equal to 1 foot. As you work more with measurements, it will be helpful for you to learn many of these relationships by heart.
Convert 374 inches into feet.
There are 12 inches in 1 foot, so to go from inches to feet we divide 374 by 12.
Our answer is $31 \frac{1}{6}$ feet.
Notice this rule again.
If you go from a smaller unit to a larger unit, we divide. If we go from a larger unit to a smaller unit, we multiply.
Dividing and multiplying any units always has to do with factors and multiples of different numbers.
Studying measurement is particularly helpful because you learn to estimate more accurately what things around you measure. You catch a bass that weighs about 10 pounds, or your book bag filled with books weighs about 25 pounds. Maybe your good friend is about $5 \frac{1}{2}$ feet tall or your bicycle is 4 feet long.
Once you start measuring things, you can compare different quantities. Is the distance from your house to school farther than the distance from your house to the grocery store? Does your math book weigh more than your history book?
To be accurate in comparing and ordering with measurements, it is essential that you are comparing using the same unit. So, if you compare pounds and ounces, you should convert ounces to pounds and then compare pounds to pounds. Or, you can convert pounds to ounces and compare ounces to ounces.
Compare $4 \frac{1}{2}$ lbs ___ 74 oz.
First, notice that we have two different units of measurement. To accurately compare these two quantities, we need to make them both into the same unit. We can do this by multiplying. Multiply the pounds by 16 to get ounces.
$4 \frac{1}{2} \cdot 16 = 72 \ oz.$
Our answer is that $4 \frac{1}{2} \ lbs < 74 \ oz$ .
Compare 62 ft. ___ 744 in.
First, we need to convert the units so that they are both the same. We can do this by converting inches to feet. The inches measurement is so large, that it is difficult to get an idea the exact size. Divide inches by 12 to get feet.
$744 \div 12 = 62 \ feet$
Our answer is that 62 ft. = 744 inches.
You can also use this information when ordering units of measurement from least to greatest and from greatest to least.
Now it's time for you to practice. Convert these measurements into quarts.
#### Example A
82 pints
Solution: 41 quarts
#### Example B
80 ounces
Solution:2.5 quarts
#### Example C
$8 \frac{1}{2}$ gallons
Solution:34 quarts
Here is the original problem once again.
Tyrone has his first measurement done when he meets Mr. Potter in the auditorium. He has written the measurement of the first table on the paper.
The first table is $8' \times 4'$ .
When Tyrone arrives at the auditorium, Mr. Potter has another table all clean and set up for Tyrone to check out.
“I think this one is larger than the other one,” Mr. Potter says. “It measures $96'' \times 30''$ .”
Tyrone looks at the table. He doesn’t think this one looks larger, but he can’t be sure.
On his paper he writes.
Table $2 = 96'' \times 30''$
To figure out which table is larger, Tyrone will need to convert customary units of measure. Then he will need to compare them.
To figure out which table is larger, Tyrone needs to convert both tables to the same unit of measure. One has been measured in inches and one has been measured in feet. Tyrone will convert both tables to inches.
He takes his measurements from table one.
$8' \times 4'$
There are 12 inches in 1 foot, so if he multiplies each by 12 they will be converted to inches.
$8 \times 12 = 96''$
The length of both tables is the same. Let’s check out the width.
$4 \times 12 = 48''$
The first table is $96'' \times 48''$ .
The second table is $96'' \times 30''$ .
Tyrone shows his math to Mr. Potter. The first table that the students already have is the larger of the two tables. Tyrone thanks Mr. Potter, but decides to stick with the first table.
### Guided Practice
Here is one for you to try on your own.
Henrietta is having her 8 best friends over for a luncheon. She wants to prepare salads on which she uses exactly 7 tablespoons of Romano cheese. If she is preparing 8 salads, how many cups of Romano cheese does Henrietta require?
We know that each salad requires 7 tablespoons and that Henrietta is making 8 salads. To find out the total amount of Romano cheese she needs in tablespoons, we multiply 7 by 8 to get 56 tablespoons. Now we need to convert 56 tablespoons to cups. There are 16 tablespoons in a cup, so we need to divide 56 by 16.
$56 \div 16 = 3 \frac{1}{2}$
Henrietta will need $3 \frac{1}{2}$ cups of Romano cheese.
### Explore More
Directions: Convert the following measurements into yards. You may round to the nearest tenth when necessary.
1. 195 inches
2. 0.2 miles
3. 88 feet
4. 90 feet
5. 900 feet
Directions: Convert the following measurements into pounds.
6. 2,104 ounces
7. 96 ounces
8. 3 tons
9. 15 tons
Directions: Convert the following measurements into pints.
10. 102 quarts
11. 57 ounces
12. 9.5 gallons
13. 4 quarts
14. 18 quarts
15. 67 gallons
### Vocabulary Language: English
Customary System
Customary System
The customary system is the measurement system commonly used in the United States, including: feet, inches, pounds, cups, gallons, etc.
Metric System
Metric System
The metric system is a system of measurement commonly used outside of the United States. It contains units such as meters, liters, and grams, all in multiples of ten.
Proportion
Proportion
A proportion is an equation that shows two equivalent ratios.
Ratio
Ratio
A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”. |
# Chapter 3 Descriptive Statistics Numerical Methods I Measures
• Slides: 8
Chapter 3 Descriptive Statistics: Numerical Methods
I. Measures of Location Quantitative data is described with measures of location and measures of dispersion. Measures of location are also called measures of central tendency. Examples include the mean, median, and mode.
A. Mean • Often referred to as the “average”, the mean is calculated as: • Sample mean • Population mean
B. Trimmed Mean • Extremely high or low values will affect the mean and possibly lead to mistaken conclusions. • The trimmed mean would drop the upper and lower 5% of the data. This removes extreme values.
C. Median • This is the value in the middle of the data when the data are rank ordered from smallest to largest. Ex. 1, 2, 3, 4, 5. The median is 3. Ex. 2, 4, 6, 8, 10, 12 The median is the mean of the middle two values: (6+8)/2=7
D. Mode • The mode is the value in a sample that appears with the greatest frequency. Ex. 1, 2, 3, 4, 4, 4, 5 The mode is 4. Ex. 1, 2, 2, 3, 3, 4 There are two modes, 2 and 3. Statisticians rarely report more than two modes.
E. Percentiles • The pth percentile is a value such that at least p% of the items take this value or less and at least (100 -p)% of the items take this value or more. • A good example of percentiles is how SAT scores are reported. When you’re told you are in the 90 th percentile it means only 10% of the scores were above yours.
How to calculate percentiles. 1. Rank order the data. 2. Compute an index: i=(p/100)*n 3. a. If i isn’t an integer, round up. The next integer greater than i is the position of the pth percentile. b. If i is an integer, the pth percentile is the average for the data values in positions i and (i+1). |
# Biased coin probability example: Bayes' rule and the law of total probability
My textbook gives the following example:
You have one fair coin, and one biased coin which lands Heads with probability $$3/4$$. You pick one of the coins at random and flip it three times. It lands Heads all three times. Given this information, what is the probability that the coin you picked is the fair one?
Solution:
Let $$A$$ be the event that the chosen coin lands Heads three times and $$F$$ be the event that we picked the fair coin. We are interested in $$P(F \mid A)$$, but it is easier to find $$P(A \mid F)$$ and $$P(A \mid F^c)$$ since it helps to know which coin we have; this suggests using Bayes' rule and the law of total probability. Doing so, we have
\begin{align} P(F \mid A) &= \dfrac{P(A \mid F)P(F)}{P(A)} \\ &= \dfrac{P(A \mid F) P(F)}{P(A \mid F) P(F) + P(A \mid F^c) P(F^c)} \\ &= \dfrac{(1/2)^3 \cdot 1/2}{(1/2)^3 \cdot 1/2 + (3/4)^3 \cdot 1/2} \\ &\approx 0.23 \end{align}
Bayes' rule is as follows:
$$P(A \mid B) = \dfrac{P(B \mid A) P(A)}{P(B)}$$
And the law of total probability is as follows:
Let $$A_1, \dots, A_n$$ be a partition of the sample space $$S$$, with $$P(A_i) > 0$$ for all $$i$$. Then
$$P(B) = \sum_{i}^n P(B \mid A_i) P(A_i)$$
What I don't see is how the authors got $$P(A) = P(A \mid F) P(F) + P(A \mid F^c) P(F^c)$$?
I would appreciate it if people could please take he time to clarify this.
• That's just the Law of Total Probability.
– lulu
Commented Aug 5, 2019 at 19:58
• @lulu hmm, how? Commented Aug 5, 2019 at 20:00
• Well, $P(A\,|\,\,B)\times P(B)=P(A\cap B)$ and $F,F^c$ are mutually exclusive and exhaustive so...
– lulu
Commented Aug 5, 2019 at 20:05
• @lulu But I don't see where the complements are coming from? Commented Aug 5, 2019 at 20:07
• What do you mean? For any event $F$ the two events $F, F^c$ are mutually exclusive and exhaustive, hence the law of total probability can be applied.
– lulu
Commented Aug 5, 2019 at 20:08
lulu is right: In your formula for the law of total probability take $$A_1 = F$$ and $$A_2 = F^c$$. Then $$A_1,A_2$$ is a partition of the sample space and $$P(A) = \sum_{i=1}^2 P(A \mid A_i) P(A_i) = P(A \mid F) P(F) + P(A \mid F^c) P(F^c)$$ is exactly that formula with the sum written out. |
# Conditional Probability: Definition & Examples
Statistics Definitions > Conditional Probability
Conditional probability is the probability of one event occurring with some relationship to one or more other events. For example:
• Event A is that it is raining outside, and it has a 0.3 (30%) chance of raining today.
• Event B is that you will need to go outside, and that has a probability of 0.5 (50%).
A conditional probability would look at these two events in relationship with one another, such as the probability that it is both raining and you will need to go outside.
The formula for conditional probability is:
P(B|A) = P(A and B) / P(A)
which you can also rewrite as:
P(B|A) = P(A∩B) / P(A)
## Conditional Probability Formula Examples
Watch the video, or read the examples below:
## Example 1
In a group of 100 sports car buyers, 40 bought alarm systems, 30 purchased bucket seats, and 20 purchased an alarm system and bucket seats. If a car buyer chosen at random bought an alarm system, what is the probability they also bought bucket seats?
Step 1: Figure out P(A). It’s given in the question as 40%, or 0.4.
Step 2: Figure out P(A∩B). This is the intersection of A and B: both happening together. It’s given in the question 20 out of 100 buyers, or 0.2.
P(B|A) = P(A∩B) / P(A) = 0.2 / 0.4 = 0.5.
The probability that a buyer bought bucket seats, given that they purchased an alarm system, is 50%.
Venn diagram showing that 20 out of 40 alarm buyers purchased bucket seats.
## Example 2:
This question uses the following contingency table:
What is the probability a randomly selected person is male, given that they own a pet?
Step 1: Repopulate the formula with new variables so that it makes sense for the question (optional, but it helps to clarify what you’re looking for). I’m going to say M is for male and PO stands for pet owner, so the formula becomes:
P(M|PO) = P(M∩PO) / P(PO)
Step 2: Figure out P(M∩PO) from the table. The intersection of male/pets (the intersection on the table of these two factors) is 0.41.
Step 3: Figure out P(PO) from the table. From the total column, 86% (0.86) of respondents had a pet.
Step 4: Insert your values into the formula:
P(M|PO) = P(M∩PO) / P(M) = 0.41 / 0.86 = 0.477, or 47.7%.
Why do we care about conditional probability? Events in life rarely have simple probability. Think about the probability of getting rainfall.
## Conditional Probability in Real Life
Conditional probability is used in many areas, including finance, insurance and politics. For example, the re-election of a president depends upon the voting preference of voters and perhaps the success of television advertising — even the probability of the opponent making gaffes during debates!
The weatherman might state that your area has a probability of rain of 40 percent. However, this fact is conditional on many things, such as the probability of…
• …a cold front coming to your area.
• …rain clouds forming.
• …another front pushing the rain clouds away.
We say that the conditional probability of rain occurring depends on all the above events.
## Where does the Conditional Probability Formula Come From?
The formula for conditional probability is derived from the probability multiplication rule, P(A and B) = P(A)*P(B|A). You may also see this rule as P(A∪B). The Union symbol (∪) means “and”, as in event A happening and event B happening.
Step by step, here’s how to derive the conditional probability equation from the multiplication rule:
Step 1: Write out the multiplication rule:
P(A and B) = P(A)*P(B|A)
Step 2: Divide both sides of the equation by P(A):
P(A and B) / P(A) = P(A)*P(B|A) / / P(A)
Step 3: Cancel P(A) on the right side of the equation:
P(A and B) / P(A) = P(B|A)
Step 4: Rewrite the equation:
P(B|A) = P(A and B) / P(A)
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# 50 Special Segments In Triangles Answer Key
## Introduction
In geometry, triangles are one of the most fundamental shapes. They have three sides and three angles, and their properties and characteristics have been extensively studied. One interesting aspect of triangles is the existence of special segments within them. These segments are derived from the vertices or sides of the triangle, and they have unique properties that make them worth exploring. In this article, we will delve into these special segments in triangles and provide an answer key to help you understand and solve related problems.
## Altitude
An altitude is a line segment drawn from a vertex of the triangle perpendicular to the opposite side or its extension. It can be extended outside the triangle or lie entirely within it. The altitude has several important properties:
• It intersects the opposite side at a right angle.
• If the triangle is acute, all three altitudes intersect at a point called the orthocenter.
• If the triangle is obtuse, the orthocenter lies outside the triangle.
• The length of the altitude can be calculated using the formula: altitude = (2 * area) / base.
## Median
A median is a line segment drawn from a vertex of the triangle to the midpoint of the opposite side. Each triangle has three medians, and they have the following properties:
• All three medians intersect at a point called the centroid.
• The centroid divides each median into two segments, with the length of the segment from the centroid to the vertex being twice as long as the length of the segment from the centroid to the midpoint.
• The centroid divides the triangle's area into six smaller triangles, each with equal area.
## Angle Bisector
An angle bisector is a line segment that divides an angle into two congruent angles. Each triangle has three angle bisectors, and they have the following properties:
• All three angle bisectors intersect at a point called the incenter.
• The incenter is equidistant from the triangle's three sides.
• The incenter is the center of the circle inscribed within the triangle.
• The length of the angle bisector can be calculated using the formula: angle bisector = (2 * side1 * side2 * cos(angle/2)) / (side1 + side2).
## Perpendicular Bisector
A perpendicular bisector is a line segment that divides a side of the triangle into two equal parts and is perpendicular to that side. Each triangle has three perpendicular bisectors, and they have the following properties:
• All three perpendicular bisectors intersect at a point called the circumcenter.
• The circumcenter is equidistant from the triangle's three vertices.
• The circumcenter is the center of the circle circumscribed around the triangle.
• The length of the perpendicular bisector can be calculated using the formula: perpendicular bisector = (side1 * side2 * cos(angle)) / (side1 + side2).
## Centroid
The centroid is the point of concurrency of the medians in a triangle. It is also the center of mass or balance point of the triangle. The centroid has the following properties:
• The centroid divides each median into two segments, with the length of the segment from the centroid to the vertex being twice as long as the length of the segment from the centroid to the midpoint.
• The centroid is the center of gravity of the triangle, meaning that if the triangle were a solid object, it would balance perfectly on the centroid.
• The centroid divides the triangle's area into six smaller triangles, each with equal area.
## Orthocenter
The orthocenter is the point of concurrency of the altitudes in a triangle. It is the point where the altitudes intersect. The orthocenter has the following properties:
• If the triangle is acute, all three altitudes intersect at the orthocenter.
• If the triangle is obtuse, the orthocenter lies outside the triangle.
• The orthocenter is equidistant from the triangle's three sides.
## Incenter
The incenter is the point of concurrency of the angle bisectors in a triangle. It is the point where the angle bisectors intersect. The incenter has the following properties:
• The incenter is equidistant from the triangle's three sides.
• The incenter is the center of the circle inscribed within the triangle.
• The incenter is the center of gravity of the triangle, meaning that if the triangle were a solid object, it would balance perfectly on the incenter.
## Circumcenter
The circumcenter is the point of concurrency of the perpendicular bisectors in a triangle. It is the point where the perpendicular bisectors intersect. The circumcenter has the following properties:
• The circumcenter is equidistant from the triangle's three vertices.
• The circumcenter is the center of the circle circumscribed around the triangle.
• The circumcenter is the center of gravity of the triangle, meaning that if the triangle were a solid object, it would balance perfectly on the circumcenter.
## Area of a Triangle
The area of a triangle can be calculated using different formulas, depending on the given information. Some common formulas include:
• Heron's formula: area = sqrt(s * (s - side1) * (s - side2) * (s - side3)), where s is the semiperimeter of the triangle (s = (side1 + side2 + side3) / 2).
• Using the base and height: area = (base * height) / 2.
• Using the lengths of two sides and the included angle: area = (side1 * side2 * sin(angle)) / 2.
## Similar Triangles
Similar triangles are triangles that have the same shape but may differ in size. They have proportional sides and congruent angles. Some properties of similar triangles include:
• The corresponding angles of similar triangles are congruent.
• The corresponding sides of similar triangles are proportional.
• If two pairs of corresponding angles are congruent, the triangles are similar.
• If two pairs of corresponding sides are proportional, the triangles are similar.
• Similar triangles have the same shape but may differ in size.
## Pythagorean Theorem
The Pythagorean theorem is a fundamental relationship between the sides of a right triangle. It states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. The theorem can be written as:
a^2 + b^2 = c^2
where a and b are the lengths of the legs of the triangle, and c is the length of the hypotenuse.
The Pythagorean theorem is widely used in geometry and has numerous applications in real-world problems.
## Triangle Inequality Theorem
The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In other words:
side1 + side2 > side3
side2 + side3 > side1
side1 + side3 > side2
This theorem is a fundamental property of triangles and is used to determine if a given set of side lengths can form a valid triangle.
## Conclusion
Special segments in triangles play a crucial role in understanding the properties and characteristics of these geometric shapes. The altitude, median, angle bisector, and perpendicular bisector each have unique properties that can be used to solve various triangle-related problems. Additionally, the centroid, orthocenter, incenter, and circumcenter are points of concurrency that have significant implications in triangle geometry. Understanding these special segments and their properties is essential for mastering the study of triangles and their applications in mathematics and real-world situations.
With the answer key provided in this article, you now have a comprehensive resource to help you solve problems related to special segments in triangles. Whether you're a student learning geometry or someone who wants to deepen their understanding of triangles, this information will serve as a valuable reference. So go ahead, explore the world of special segments in triangles, and unlock new possibilities in your mathematical journey. |
Question #15fe4
Sep 19, 2017
$n = 6$
Explanation:
Step one: collect like-terms
$7 n + 3 = 45$
7n + (2+1) = 45
Step two: : subtract both sides by 3
$7 n = 42$
7n + 3 -3 = 45 *-3
7n =
*42
Step three : divide both sides by 7
Ans is $n = 6$
Sep 19, 2017
$n = 6$
Explanation:
Combine like terms:
$7 n + \left(2 + 1\right) = 45$
$7 n + 3 = 45$
Isolate the variable:
$7 n + 3 - 3 = 45 - 3$
$7 n = 42$
$\frac{7 n}{7} = \frac{42}{7}$
$n = 6$ |
### Mode, Median and Mean
```MODE
Mode = the item that appears most often
- the most common number
2, 3, 6, 4, 2, 2, 3, 6, 5
this
group
fruit
group
of of
numbers
In In
this
transport
survey
APPLES
areMODE.
the MODE.
2 is
the
MODE.
CARS
are
the
There
more
apples
2 appears
more
than
any
There
areare
cars
than
than
anyvehicle
other fruit.
other
number
any
other
Click mouse
MODE - continued
Click mouse
What is the mode in each of the following? :Click the mouse when you want to see each answer
10, 11, 10, 8, 5, 8, 10, 8, 6, 7, 8, 8, 9, 12
8
20,
20, 20,
20, 22,
22, 25,
25, 40,
40, 13,
13, 20,
20, 21,
21, 22,
22, 20,
20, 40,
40,
20
2, 4, 8, 2, 3, 3, 3, 2, 4, 4, 2, 2, 2, 9, 5, 2, 4
2
102, 105, 120, 102, 103, 104, 166, 124, 132
102
65, 61, 67, 66, 67, 66, 61, 66, 64, 66, 67, 65
66
Median
The value of the middle number when they are arranged in order
10, 5, 2, 1, 8, 7, 6
1
2
First, put the numbers into order:5
6
7
8
10
middle
Now find the number that is in the middle
6 is the MEDIAN
Click mouse
Median
The value of the middle number when they are arranged in order
4, 19, 3, 22, 15, 1, 9 First, put the numbers into order:1
3
4
9
15
19
22
middle
Now find the number that is in the middle
9 is the MEDIAN
Click mouse
Median - continued
Click mouse
Find the MEDIAN in the following:Click the mouse to see the answers
4, 9, 5, 2, 3, 10, 1
1
2
3
4
5
9
10
16
22
median
22, 16, 5, 29, 32, 10, 1
2, 6, 15, 24, 32, 10, 9
1
2
5
6
10
median
9
10
median
15
29
32
24
32
Now that you’ve got the idea of MEDIAN
there’s just one more thing to remember!
If there are an EVEN amount of numbers you
can’t find the middle straight away……
4
6
7
9
What’s the
middle
number?
Click mouse for more
12
15
To find the median ADD the
two middle numbers together,
then DIVIDE by 2.
7 + 9 = 16 16÷ 2 = 8
So, the MEDIAN is 8
Click mouse
Let’s find the median of this group of numbers
6, 5, 5, 7, 1, 4, 10, 7, 8
FIRST, put the numbers into ORDER
1
4
5
5
6
7
7
8
10
Make sure you’ve
got the right
amount – there
are 9 numbers in
this group
MEDIAN
NOW, find the middle
Click mouse for more
The MEDIAN is 6
Click mouse
Let’s find the median of this group of numbers
Click mouse for more
6, 5, 5, 7, 1, 4, 10, 7,
FIRST, put the numbers into ORDER
1
4
5
5
6
7
7 10
Make sure you’ve
got the right
amount – there
are 8 numbers in
this group
What is the MEDIAN? Add the middle two numbers:
5 + 6 = 11
NOW, find the middle
Divide by 2
The MEDIAN is 5 . 5
11 ÷ 2 = 5 . 5 (or 5 ½)
Try these: click the mouse when you are ready to see the answer
Find the MEDIAN
Find the middle
5, 9, 5, 3, 10, 5, 8, 6, 5, 10, 12 3 5 5 5 5 6 8 9 10 10 12
Make sure you’ve got 11
Put into
order
median
Find the middle
5, 9, 7, 2, 1, 4, 9, 6, 11, 10, 42
Put into
order
1 2 4 5 6 7 9 9 10 11 42
Make sure you’ve got 11
median Click mouse
Mean
Mean is another word for AVERAGE
It’s easy to work out
First, add all the numbers together:Then, divide by the number of items (the number of ‘numbers’!)
6
4
5
6
2
8
4
1
2
3
4
5
6
7
6 + 4 + 5 + 6 + 2 + 8 + 4 = 35
35 ÷ 7 = 5
5 is the MEAN
Click mouse
Mean
Try this one – find the MEAN
First, add all the numbers together:Then, divide by the number of items (the number of ‘numbers’!)
8
1
4
20
8
2
8
4
2
2
3
4
5
6
7
8
8 + 4 + 20 + 8 + 2 + 8 + 4 + 2 = 56
56 ÷ 8 = 7
7 is the MEAN
Click mouse
Click the mouse for help / answers
Find the MEAN of these numbers:10, 12, 3, 21, 4, 8, 12, 5, 9, 6
1 2 3 4 5 6 7 8 9 10
10 + 12 + 3 + 21 + 4 + 8 + 12 + 5 + 9 + 6 = 90
15, 22, 12, 9, 6, 10, 3
1 2 3 4 5 6 7
15 + 22 + 12 + 9 + 6 + 10 + 3 = 77
90 ÷ 10 = 9
The MEAN is 9
77 ÷ 7 = 11
The MEAN is 11
5, 7, 10, 3, 12, 6, 7, 19, 5, 1, 13
1 2 3 4 5 6 7 8 9 10 11
88 ÷ 11 = 8
5 + 7 + 10 + 3 + 12 + 6 + 7 + 19 + 5 + 1 + 13 = 88
The MEAN is 8
Click mouse
Questions on mode, median and mean
Find the MODE and MEDIAN shoe size of 15
children whose shoe sizes are:-
4
2
3
4
4
5
3
6
3
7
5
4
4
3
1
Don’t forget – put
the numbers in
ascending order
1 2 3 3 3 3 4 4 4 4 4 5 5 6 7
Check you have 15 numbers
The MODE is 4
The MEDIAN is 4
4 is the middle number
4 occurs most often
Click
Click mouse
mouse for
for help
Questions on mode, median and mean
Find the Mean shoe size of 15 children whose
shoe sizes are:-
4
2
3
4
4
5
3
7
3
8
all the shoe sizes
together
5
4
4
3
1
Divide by the number of
children
4 + 2 + 3 + 4 + 4 + 5 + 3 + 7 + 3 + 8 + 5 + 4 + 4 + 3 + 1 = 60
60 ÷ 15 = 4
The MEAN shoe
size is 4
Click
Click mouse
mouse for
for help
Questions on mode, median and mean
Find the MODE, MEDIAN and MEAN of the following data:
17 0131 11
0
3 69 77 910100 11
0 1133 17
6
MODE is 0
MEDIAN
is 7
Put in order; check
you have 11 numbers
MEAN: add all the numbers together
divide by number of items (11)
0 + 0 + 1 + 3 + 6 + 7 + 9 + 10 + 11 + 13 + 17 = 77
77 ÷ 11 = 7
The MEAN is 7
Click mouse
Click mouse for help / answer
Questions on mode, median and mean
Find the MODE, MEDIAN and MEAN of the following data:
10
14 10
15 10
11 11
12
10 12
11 14
10 15
10 15
10
10
10
11 11
11
12
14
15
15
MODE is 10
MEDIAN
is 11
Put in order; check
you have 9 numbers
MEAN: add all the numbers together
divide by number of items (9)
10 + 10 + 10 + 11 + 11 + 12+ 14 + 15 + 15 = 108
108 ÷ 9 = 12
The MEAN is 12
Click mouse
Click mouse for help / answer
Questions on mode, median and mean
Find the MODE, MEDIAN and MEAN of the following data:
117
6 671783810
131313
131381717
13 313
10
1011
61111
7
Put in order; check
you have 10 numbers
MODE is 13
MEAN: add all the numbers together
MEDIAN is
(8 + 10)÷ 2 = 9
divide by number of items (10)
1 + 3 + 6 + 7 + 8 + 10 + 11 + 13 + 13 + 17 = 90
90 ÷ 10 = 9
The MEAN is 9
Click mouse
Click mouse for help / answer
The number of absences for each pupil in a class of 25 is
shown in the table below.
0
18
0
3
1
Absences
1
0
0
3
2
2
0
1
1
2
5
20
2
24
2
5
3
4
0
3
a. Calculate the MEAN number of absences per pupil
b. Find the MEDIAN number of absences per pupil
Click mouse for help on next page
a. Calculate the MEAN number of absences per pupil
Absences
0
1
0
0
5
18
3
2
2
3
0
0
1
1
4
3
2
5
20
0
1
2
24
2
3
First, add together all the absences
Then, divide by the number of
pupils (25)
0 + 1 + 0 + 0 + 5 + 18 + 3 + 2 + 2 + 3 + 0 + 0 + 1 + 1 + 4 + 3 + 2 + 5 + 20 + 0 + 1 + 2 + 24 + 2 + 3= 102
It’s really important to include the 0s because
you must count them in your total number of
pupils
The MEAN number
of absences per
pupil is 4·08
102 ÷ 25 = 4.08
Click mouse for next page
b. Find the MEDIAN number of absences per pupil
First, put all the absences in order
Absences
0
1
0
0
5
18
3
2
2
3
0
0
1
1
4
3
2
5
20
0
1
2
24
2
3
Then, check that you have 25
numbers!
Now, find the middle number
0 0 0 0 0 0 1 1 1 1 2 2 2 2 2 3 3 3 3 4 5 5 18 20 24
median
The MEDIAN number of
absences is 2
END
``` |
# The device shown is used to straighten the frames 1
The device shown is used to straighten the frames of wrecked autos. Determine the tension of each segment of the chain, i.e., AB and BC, if the force which the hydraulic cylinder DB exerts on point B is 3.50 kN, as shown.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
#### Solution:
Let us first draw our free body diagram. The angles have been calculated in the diagram, but steps to calculating those angles will be shown below.
To figure out the angles, we will use simple trigonometry.
The green angle is found by:
$\text{tan}^{-1}(\frac{400}{450})=41.6^0$
The brown angle is found by:
$\text{tan}^{-1}(\frac{250}{450})=29^0$
We will assume forces going $\rightarrow^+$ to be positive and $\uparrow+$ to be positive.
First, we will write equations of equilibrium for the y-axis forces as this will give us a solution to $F_{BC}$.
$\sum \text{F}_\text{y}=0$
$3.5\text{cos}\,(41.6^0)-F_{BC}\text{cos}\,(29^0)=0$
(Solve for $F_{BC}$)
$F_{BC}=2.99$ kN
Now, we will write equations of equilibrium for the x-axis forces.
$\sum \text{F}_\text{x}=0$
$3.5\text{sin}\,(41.6^0)+2.99\text{sin}\,(29^0)-F_{AB}=0$
(Substituted the value of $F_{BC}$ we found earlier. Solve for $F_{AB}$.)
$F_{AB}=3.77$ kN |
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# Chapter 27
## Volumes of common solids
27.1 Introduction
The volume of any solid is a measure of the space occupied by the solid. Volume is measured in cubic units such as mm3 , cm3 and m3 . This chapter deals with nding volumes of common solids; in engineering it is often important to be able to calculate volume or capacity to estimate, say, the amount of liquid, such as water, oil or petrol, in different shaped containers. A prism is a solid with a constant cross-section and with two ends parallel. The shape of the end is used to describe the prism. For example, there are rectangular prisms (called cuboids), triangular prisms and circular prisms (called cylinders). On completing this chapter you will be able to calculate the volumes and surface areas of rectangular and other prisms, cylinders, pyramids, cones and spheres, together with frusta of pyramids and cones. Volumes of similar shapes are also considered.
h l b
Figure 27.1
A cube is a square prism. If all the sides of a cube are x then Volume = x 3 and surface area = 6x 2 Problem 1. A cuboid has dimensions of 12 cm by 4 cm by 3 cm. Determine (a) its volume and (b) its total surface area The cuboid is similar to that in Figure 27.1, with l = 12 cm, b = 4 cm and h = 3 cm.
## 27.2 Volumes and surface areas of common shapes
27.2.1 Cuboids or rectangular prisms
A cuboid is a solid gure bounded by six rectangular faces; all angles are right angles and opposite faces are equal. A typical cuboid is shown in Figure 27.1 with length l, breadth b and height h. Volume of cuboid = l b h and surface area = 2bh + 2hl + 2lb = 2(bh + hl + lb)
DOI: 10.1016/B978-1-85617-697-2.00027-2
(a)
## Volume of cuboid = l b h = 12 4 3 = 144 cm3
(b) Surface area = 2(bh + hl + lb) = 2(4 3 + 3 12 + 12 4) = 2(12 + 36 + 48) = 2 96 = 192 cm2 Problem 2. An oil tank is the shape of a cube, each edge being of length 1.5 m. Determine (a) the maximum capacity of the tank in m3 and litres and (b) its total surface area ignoring input and output orices
## Volumes of common solids
(a) Volume of oil tank = volume of cube = 1.5 m 1.5 m 1.5 m = 1.53 m3 = 3.375 m3 1 m3 = 100 cm 100 cm 100 cm = 106 cm3 . Hence, volume of tank = 3.375 106 cm3 1 litre = 1000 cm3, hence oil tank capacity 3.375 106 litres = 3375 litres = 1000 (b) Surface area of one side = 1.5 m 1.5 m = 2.25 m2 . A cube has six identical sides, hence total surface area of oil tank = 6 2.25 = 13.5 m2 Problem 3. A water tank is the shape of a rectangular prism having length 2 m, breadth 75 cm and height 500 mm. Determine the capacity of the tank in (a) m3 (b) cm3 (c) litres Capacity means volume; when dealing with liquids, the word capacity is usually used. The water tank is similar in shape to that in Figure 27.1, with l = 2 m, b = 75 cm and h = 500 mm. (a) Capacity of water tank = l b h. To use this formula, all dimensions must be in the same units. Thus, l = 2 m, b = 0.75 m and h = 0.5 m (since 1 m = 100 cm = 1000 mm). Hence, capacity of tank = 2 0.75 0.5 = 0.75 m3 (b) 1 m3 = 1 m 1 m 1 m = 100 cm 100 cm 100 cm i.e., 1 m3 = 1 000 000 =106 cm3 . Hence, capacity = 0.75 m = 0.75 10 cm
3 6 3
241
Figure 27.2
Volume = r 2 h Curved surface area = 2rh Total surface area = 2rh + 2r 2 Total surface area means the curved surface area plus the area of the two circular ends. Problem 4. A solid cylinder has a base diameter of 12 cm and a perpendicular height of 20 cm. Calculate (a) the volume and (b) the total surface area (a) Volume = r 2 h = 12 2
2
20
= 720 = 2262 cm3 (b) Total surface area = 2rh + 2r 2 = (2 6 20) + (2 62 ) = 240 + 72 = 312 = 980 cm2 Problem 5. A copper pipe has the dimensions shown in Figure 27.3. Calculate the volume of copper in the pipe, in cubic metres.
2.5 m
= 750 000 cm3 (c) 1 litre = 1000 cm3 . Hence, 750 000 cm3 = 750,000 = 750 litres 1000
25 cm 12 cm
27.2.2 Cylinders
A cylinder is a circular prism. A cylinder of radius r and height h is shown in Figure 27.2.
Figure 27.3
## 242 Basic Engineering Mathematics
Outer diameter, D = 25 cm = 0.25 m and inner diameter, d = 12 cm = 0.12 m. Area of cross-section of copper D 2 d 2 (0.25)2 (0.12)2 = = 4 4 4 4 = 0.0491 0.0113 = 0.0378 m2 Hence, volume of copper = (cross-sectional area) length of pipe = 0.0378 2.5 = 0.0945 m
3
16 mm
12 mm
40 mm
Figure 27.5
## 27.2.3 More prisms
A right-angled triangular prism is shown in Figure 27.4 with dimensions b, h and l.
The solid shown in Figure 27.5 is a triangular prism. The volume V of any prism is given by V = Ah, where A is the cross-sectional area and h is the perpendicular height. Hence, volume = 1 16 12 40 = 3840 mm3 2 = 3.840 cm3 (since 1 cm3 = 1000 mm3 ) Problem 7. Calculate the volume of the right-angled triangular prism shown in Figure 27.6. Also, determine its total surface area
I
Figure 27.4
40 cm
1 Volume = bhl 2 and surface area = area of each end + area of three sides Notice that the volume is given by the area of the end (i.e. area of triangle = 1 bh) multiplied by the length l. 2 In fact, the volume of any shaped prism is given by the area of an end multiplied by the length. Problem 6. Determine the volume (in cm3 ) of the shape shown in Figure 27.5
Figure 27.6
6 cm
B 8 cm
## Volumes of common solids
Total surface area = area of each end + area of three sides. In triangle ABC, from which, AC2 = AB2 + BC2 AC = AB2 + BC 2 = 62 + 82 = 10 cm Hence, total surface area 1 =2 bh + (AC 40) + (BC 40) + (AB 40) 2 = (8 6) + (10 40) + (8 40) + (6 40) = 48 + 400 + 320 + 240 i.e. total surface area = 1008 cm2 Problem 8. Calculate the volume and total surface area of the solid prism shown in Figure 27.7 5.
11 cm 4 cm
243
Now try the following Practice Exercise Practice Exercise 105 Volumes and surface areas of common shapes (answers on page 351) 1. 2. 3. 4. Change a volume of 1 200 000 cm3 to cubic metres. Change a volume of 5000 mm3 to cubic centimetres. A metal cube has a surface area of 24 cm2 . Determine its volume. A rectangular block of wood has dimensions of 40 mm by 12 mm by 8 mm. Determine (a) its volume, in cubic millimetres (b) its total surface area in square millimetres. Determine the capacity, in litres, of a sh tank measuring 90 cm by 60 cm by 1.8 m, given 1 litre = 1000 cm3 . A rectangular block of metal has dimensions of 40 mm by 25 mm by 15 mm. Determine its volume in cm3 . Find also its mass if the metal has a density of 9 g/cm3 . Determine the maximum capacity, in litres, of a sh tank measuring 50 cm by 40 cm by 2.5 m(1 litre = 1000 cm3). Determine how many cubic metres of concrete are required for a 120 m long path, 150 mm wide and 80 mm deep. A cylinder has a diameter 30 mm and height 50 mm. Calculate (a) its volume in cubic centimetres, correct to 1 decimal place (b) the total surface area in square centimetres, correct to 1 decimal place. 10. Find (a) the volume and (b) the total surface area of a right-angled triangular prism of length 80 cm and whose triangular end has a base of 12 cm and perpendicular height 5 cm. 11. A steel ingot whose volume is 2 m2 is rolled out into a plate which is 30 mm thick and 1.80 m wide. Calculate the length of the plate in metres.
6.
15 cm
7.
8.
5 cm 5 cm 5 cm
Figure 27.7
9.
The solid shown in Figure 27.7 is a trapezoidal prism. Volume of prism = cross-sectional area height 1 = (11 + 5)4 15 = 32 15 2 = 480 cm3 Surface area of prism = sum of two trapeziums + 4 rectangles = (2 32) + (5 15) + (11 15) + 2(5 15) = 64 + 75 + 165 + 150 = 454 cm2
## 244 Basic Engineering Mathematics
12. The volume of a cylinder is 75 cm3 . If its height is 9.0 cm, nd its radius. 13. Calculate the volume of a metal tube whose outside diameter is 8 cm and whose inside diameter is 6 cm, if the length of the tube is 4 m. 14. The volume of a cylinder is 400 cm3 . If its radius is 5.20 cm, nd its height. Also determine its curved surface area. 15. A cylinder is cast from a rectangular piece of alloy 5 cm by 7 cm by 12 cm. If the length of the cylinder is to be 60 cm, nd its diameter. 16. Find the volume and the total surface area of a regular hexagonal bar of metal of length 3 m if each side of the hexagon is 6 cm. 17. A block of lead 1.5 m by 90 cm by 750 mm is hammered out to make a square sheet 15 mm thick. Determine the dimensions of the square sheet, correct to the nearest centimetre. 18. How long will it take a tap dripping at a rate of 800 mm3/s to ll a 3-litre can? 19. A cylinder is cast from a rectangular piece of alloy 5.20 cm by 6.50 cm by 19.33 cm. If the height of the cylinder is to be 52.0 cm, determine its diameter, correct to the nearest centimetre. 20. How much concrete is required for the construction of the path shown in Figure 27.8, if the path is 12 cm thick?
8.5 m 2m
27.2.4 Pyramids
Volume of any pyramid 1 = area of base perpendicular height 3 A square-based pyramid is shown in Figure 27.9 with base dimension x by x and the perpendicular height of the pyramid h. For the square-base pyramid shown, 1 volume = x 2 h 3
x x
Figure 27.9
Problem 9. A square pyramid has a perpendicular height of 16 cm. If a side of the base is 6 cm, determine the volume of a pyramid Volume of pyramid = 1 area of base perpendicular height 3 1 = (6 6) 16 3 = 192 cm3 Problem 10. Determine the volume and the total surface area of the square pyramid shown in Figure 27.10 if its perpendicular height is 12 cm. Volume of pyramid
1.2 m 3.1 m
2.4 m
Figure 27.8
## Volumes of common solids
A
245
from which,
h=
3 540 = 27 cm 6 10
## i.e. perpendicular height of pyramid = 27 cm
27.2.5 Cones
A cone is a circular-based pyramid. A cone of base radius r and perpendicular height h is shown in Figure 27.11.
B 5 cm C D 5 cm E
Volume =
## 1 area of base perpendicular height 3
Figure 27.10
The total surface area consists of a square base and 4 equal triangles. Area of triangle ADE = 1 base perpendicular height 2 1 = 5 AC 2
Figure 27.11
The length AC may be calculated using Pythagoras theorem on triangle ABC, where AB = 12 cm and BC = 1 5 = 2.5 cm. 2 AC = Hence, area of triangle ADE = 1 5 12.26 = 30.65 cm2 2 = 147.6 cm
2
AB2 + BC 2 =
i.e.
## Total surface area of pyramid = (5 5) + 4(30.65)
Problem 12. Calculate the volume, in cubic centimetres, of a cone of radius 30 mm and perpendicular height 80 mm 1 1 Volume of cone = r 2 h = 302 80 3 3 = 75398.2236 . . . mm3 1 cm = 10 mm and 1 cm3 = 10 mm 10 mm 10 mm = 103 mm3, or 1 mm3 = 103 cm3 Hence, 75398.2236 . . . mm3 = 75398.2236 . . . 103 cm3 i.e., volume = 75.40 cm3
Problem 11. A rectangular prism of metal having dimensions of 5 cm by 6 cm by 18 cm is melted down and recast into a pyramid having a rectangular base measuring 6 cm by 10 cm. Calculate the perpendicular height of the pyramid, assuming no waste of metal Volume of rectangular prism = 5 6 18 = 540 cm3 Volume of pyramid 1 = area of base perpendicular height 3 1 540 = (6 10) h 3
Hence,
## 246 Basic Engineering Mathematics
Alternatively, from the question, r = 30 mm = 3 cm and h = 80 mm = 8 cm. Hence, 1 1 volume = r 2 h = 32 8 = 75.40 cm3 3 3 Problem 13. Determine the volume and total surface area of a cone of radius 5 cm and perpendicular height 12 cm The cone is shown in Figure 27.12. Problem 14. Find the volume and surface area of a sphere of diameter 10 cm Since diameter = 10 cm, radius, r = 5 cm. 4 4 Volume of sphere = r 3 = 53 3 3 = 523.6 cm3 Surface area of sphere = 4r 2 = 4 52 = 314.2 cm2
l h 12 cm
Problem 15. The surface area of a sphere is 201.1 cm2 . Find the diameter of the sphere and hence its volume Surface area of sphere = 4r 2 . Hence, 201.1 cm2 = 4 r 2 , from which and 201.1 = 16.0 4 radius, r = 16.0 = 4.0 cm r2 =
5 cm
Figure 27.12
1 1 Volume of cone = r 2 h = 52 12 3 3 = 314.2 cm3 Total surface area = curved surface area + area of base = rl + r 2 From Figure 27.12, slant height l may be calculated using Pythagoras theorem: l= 122 + 52 = 13 cm
from which, diameter = 2 r = 2 4.0 = 8.0 cm 4 4 Volume of sphere = r 3 = (4.0)3 3 3 = 268.1 cm3 Now try the following Practice Exercise Practice Exercise 106 Volumes and surface areas of common shapes (answers on page 351) 1. If a cone has a diameter of 80 mm and a perpendicular height of 120 mm, calculate its volume in cm3 and its curved surface area. A square pyramid has a perpendicular height of 4 cm. If a side of the base is 2.4 cm long, nd the volume and total surface area of the pyramid.
## Hence, total surface area = ( 5 13) + ( 52) = 282.7 cm2 .
27.2.6 Spheres
For the sphere shown in Figure 27.13: 4 Volume = r 3 3 and surface area = 4r 2
2.
Figure 27.13
## Volumes of common solids
Cylinder
r
247
3.
A sphere has a diameter of 6 cm. Determine its volume and surface area. If the volume of a sphere is 566 cm3 , nd its radius. A pyramid having a square base has a perpendicular height of 25 cm and a volume of 75 cm3. Determine, in centimetres, the length of each side of the base. A cone has a base diameter of 16 mm and a perpendicular height of 40 mm. Find its volume correct to the nearest cubic millimetre.
4.
## Volume = r 2 h Total surface area = 2rh + 2r 2
5.
Triangular prism
6.
7.
Determine (a) the volume and (b) the surface area of a sphere of radius 40 mm. The volume of a sphere is 325 cm3 . Determine its diameter. Given the radius of the earth is 6380 km, calculate, in engineering notation (a) its surface area in km2 . (b) its volume in km3 .
1 Volume = bhl 2
8.
## Surface area = area of each end + area of three sides
b
Pyramid
h
9.
1 Ah 3 Total surface area = sum of areas of triangles forming sides + area of base Volume =
10. An ingot whose volume is 1.5 m3 is to be made into ball bearings whose radii are 8.0 cm. How many bearings will be produced from the ingot, assuming 5% wastage?
Cone 1 Volume = r 2 h 3
r
l h
## 27.3 Summary of volumes and surface areas of common solids
A summary of volumes and surface areas of regular solids is shown in Table 27.1. Table 27.1 Volumes and surface areas of regular solids Rectangular prism (or cuboid)
## 27.4 More complex volumes and surface areas
Here are some worked problems involving more complex and composite solids.
## Volume = l b h Surface area = 2(bh + hl + lb)
l
Problem 16. A wooden section is shown in Figure 27.14. Find (a) its volume in m3 and (b) its total surface area
## 248 Basic Engineering Mathematics
F
r5
8 cm
15.0 cm
12 cm
Figure 27.14
15.0 cm
0 cm 15. 0 cm 15.
3m
(a) The section of wood is a prism whose end comprises a rectangle and a semicircle. Since the radius of the semicircle is 8 cm, the diameter is 16 cm. Hence, the rectangle has dimensions 12 cm by 16 cm. 1 Area of end = (12 16) + 82 = 292.5 cm2 2 Volume of wooden section = area of end perpendicular height = 292.5 300 = 87 750 cm3 = 87750 3 m , since 1 m3 = 106 cm3 106
3
C D
cm
E G
H .40 cm 5
3.60
Figure 27.15
Using Pythagoras theorem on triangle BEF gives B F 2 = E B2 + E F 2 from which EF = = Volume of pyramid 1 = (area of base)(perpendicular height) 3 1 = (3.60 5.40)(14.64) = 94.87 cm3 3 Area of triangle ADF (which equals triangle BCF) = 1(AD)(FG), where G is the midpoint of AD. 2 Using Pythagoras theorem on triangle FGA gives FG = 15.02 1.802 = 14.89 cm BF 2 EB 2 15.02 3.2452 = 14.64 cm
= 0.08775 m
(b) The total surface area comprises the two ends (each of area 292.5 cm2 ), three rectangles and a curved surface (which is half a cylinder). Hence, total surface area = (2 292.5) + 2(12 300) 1 + (16 300) + (2 8 300) 2 = 585 + 7200 + 4800 + 2400 = 20 125 cm2 or 2.0125 m2
Problem 17. A pyramid has a rectangular base 3.60 cm by 5.40 cm. Determine the volume and total surface area of the pyramid if each of its sloping edges is 15.0 cm The pyramid is shown in Figure 27.15. To calculate the volume of the pyramid, the perpendicular height EF is required. Diagonal BD is calculated using Pythagoras theorem, i.e. Hence, BD = 3.602 + 5.402 = 6.490 cm 1 6.490 EB = BD = = 3.245 cm 2 2
1 Hence, area of triangleADF = (3.60)(14.89) 2 = 26.80cm2 Similarly, if H is the mid-point of AB, FH = 15.02 2.702 = 14.75cm
Hence, area of triangle ABF (which equals triangle 1 CDF) = (5.40)(14.75) = 39.83 cm2 2
## Volumes of common solids
Total surface area of pyramid = 2(26.80) + 2(39.83) + (3.60)(5.40) = 53.60 + 79.66 + 19.44 = 152.7 cm2 Problem 18. Calculate the volume and total surface area of a hemisphere of diameter 5.0 cm 1 Volume of hemisphere = (volume of sphere) 2 2 2 5.0 = r 3 = 3 3 2 = 32.7 cm3 Total surface area = curved surface area + area of circle 1 = (surface area of sphere) + r 2 2 1 = (4r 2 ) + r 2 2 = 2r 2 + r 2 = 3r 2 = 3 5.0 2
2 3
249
Problem 20. A rivet consists of a cylindrical head, of diameter 1 cm and depth 2 mm, and a shaft of diameter 2 mm and length 1.5 cm. Determine the volume of metal in 2000 such rivets 1 cm = 0.5 cm and 2 height of cylindrical head = 2 mm = 0.2 cm. Radius of cylindrical head = Hence, volume of cylindrical head = r 2 h = (0.5)2 (0.2) = 0.1571 cm3 Volume of cylindrical shaft = r 2 h = 0.2 2
2
## (1.5) = 0.0471 cm3
Total volume of 1 rivet= 0.1571 + 0.0471 = 0.2042 cm3 Volume of metal in 2000 such rivets = 2000 0.2042 = 408.4 cm3 Problem 21. A solid metal cylinder of radius 6 cm and height 15 cm is melted down and recast into a shape comprising a hemisphere surmounted by a cone. Assuming that 8% of the metal is wasted in the process, determine the height of the conical portion if its diameter is to be 12 cm Volume of cylinder = r 2 h = 62 15 = 540 cm3
= 58.9 cm2 Problem 19. A rectangular piece of metal having dimensions 4 cm by 3 cm by 12 cm is melted down and recast into a pyramid having a rectangular base measuring 2.5 cm by 5 cm. Calculate the perpendicular height of the pyramid Volume of rectangular prism of metal = 4 3 12 = 144 cm Volume of pyramid 1 = (area of base)(perpendicular height) 3 Assuming no waste of metal, 1 144 = (2.5 5)(height) 3
3
If 8% of metal is lost then 92% of 540 gives the volume of the new shape, shown in Figure 27.16.
144 3 2.5 5
Figure 27.16
12 cm
= 34.56 cm
## 250 Basic Engineering Mathematics
Hence, the volume of (hemisphere + cone) = 0.92 540 cm3 i.e. 1 2 4 3 1 r + r 2 h = 0.92 540 3 3 (c) Area of circle = r 2 or hence, 0.1122 = from which, d = d 2 4 4 0.1122 = 0.3780 cm d 2 4
Dividing throughout by gives 2 3 1 2 r + r h = 0.92 540 3 3 Since the diameter of the new shape is to be 12 cm, radius r = 6 cm, hence 2 3 1 2 (6) + (6) h = 0.92 540 3 3 144 + 12h = 496.8
i.e. diameter of cross-section is 3.780 mm. Problem 23. A boiler consists of a cylindrical section of length 8 m and diameter 6 m, on one end of which is surmounted a hemispherical section of diameter 6 m and on the other end a conical section of height 4 m and base diameter 6 m. Calculate the volume of the boiler and the total surface area The boiler is shown in Figure 27.17.
## i.e. height of conical portion, h= 496.8 144 = 29.4 cm 12
P 6m 8m A 4m C Q 3m R B
Problem 22. A block of copper having a mass of 50 kg is drawn out to make 500 m of wire of uniform cross-section. Given that the density of copper is 8.91 g/cm3 , calculate (a) the volume of copper, (b) the cross-sectional area of the wire and (c) the diameter of the cross-section of the wire (a) A density of 8.91 g/cm3 means that 8.91 g of copper has a volume of 1 cm3 , or 1 g of copper has a volume of (1 8.91) cm3 . Density = from which mass volume mass volume = density
Figure 27.17
Volume of hemisphere,
2 P = r 3 3 2 = 33 = 18 m3 3 = 72 m3
Hence, 50 kg, i.e. 50 000 g, has a 50000 3 mass = cm = 5612 cm3 volume = density 8.91 (b) Volume of wire = area of circular cross-section length of wire. Hence, 5612 cm3 = area (500 100 cm) from which, area = 5612 cm2 500 100
Volume of cylinder,
Q = r 2 h = 32 8
Volume of cone,
1 1 R = r 2 h = 32 4 3 3 = 12 m3
Total volume of boiler = 18 + 72 + 12 = 102 = 320.4 m3 1 Surface area of hemisphere, P = (4r 2 ) 2 = 2 32 = 18m2
= 0.1122 cm2
## Volumes of common solids
Curved surface area of cylinder, Q = 2rh = 2 38 = 48 m2 The slant height of the cone, l, is obtained by Pythagoras theorem on triangle ABC, i.e. l= 42 + 32 = 5
251
(iv) a 2.5 cm by 2.5 cm square pyramid of perpendicular height 5.0 cm. (v) a 4.0 cm by 6.0 cm rectangular pyramid of perpendicular height 12.0 cm. (vi) a 4.2 cm by 4.2 cm square pyramid whose sloping edges are each 15.0 cm (vii) a pyramid having an octagonal base of side 5.0 cm and perpendicular height 20 cm. 7. A metal sphere weighing 24 kg is melted down and recast into a solid cone of base radius 8.0 cm. If the density of the metal is 8000 kg/m3 determine (a) the diameter of the metal sphere. (b) the perpendicular height of the cone, assuming that 15% of the metal is lost in the process. 8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 16.0 cm and the side of the base is 3.0 cm. A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere is 2.5 m and the slant height of the cone is 4.0 m. Determine the volume and surface area of the buoy.
Curved surface area of cone, R = rl = 3 5 = 15 m2 Total surface area of boiler= 18 + 48 + 15 = 81 = 254.5 m2 Now try the following Practice Exercise Practice Exercise 107 More complex volumes and surface areas (answers on page 351) 1. 2. 3. Find the total surface area of a hemisphere of diameter 50 mm. Find (a) the volume and (b) the total surface area of a hemisphere of diameter 6 cm. Determine the mass of a hemispherical copper container whose external and internal radii are 12 cm and 10 cm, assuming that 1 cm3 of copper weighs 8.9 g. A metal plumb bob comprises a hemisphere surmounted by a cone. If the diameter of the hemisphere and cone are each 4 cm and the total length is 5 cm, nd its total volume. A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 mand the cylindrical portion has a height of 3.5 m with a diameter of 15 m. Calculate the surface area of material needed to make the marquee assuming 12% of the material is wasted in the process. Determine (a) the volume and (b) the total surface area of the following solids. (i) a cone of radius 8.0 cm and perpendicular height 10 cm. (ii) a sphere of diameter 7.0 cm. (iii) a hemisphere of radius 3.0 cm.
9.
4.
10. A petrol container is in the form of a central cylindrical portion 5.0 m long with a hemispherical section surmounted on each end. If the diameters of the hemisphere and cylinder are both 1.2 m, determine the capacity of the tank in litres (1litre = 1000 cm3 ). 11. Figure 27.18 shows a metal rod section. Determine its volume and total surface area.
5.
6.
1.00 m
Figure 27.18
## 252 Basic Engineering Mathematics
12. Find the volume (in cm3) of the die-casting shown in Figure 27.19. The dimensions are in millimetres.
## 27.5 Volumes and surface areas of frusta of pyramids and cones
The frustum of a pyramid or cone is the portion remaining when a part containing the vertex is cut off by a plane parallel to the base. The volume of a frustum of a pyramid or cone is given by the volume of the whole pyramid or cone minus the volume of the small pyramid or cone cut off. The surface area of the sides of a frustum of a pyramid or cone is given by the surface area of the whole pyramid or cone minus the surface area of the small pyramid or cone cut off. This gives the lateral surface area of the frustum. If the total surface area of the frustum is required then the surface area of the two parallel ends are added to the lateral surface area. There is an alternative method for nding the volume and surface area of a frustum of a cone. With reference to Figure 27.21,
r I
60
10
25
50
Figure 27.19
13. The cross-section of part of a circular ventilation shaft is shown in Figure 27.20, ends AB and CD being open. Calculate (a) the volume of the air, correct to the nearest litre, contained in the part of the system shown, neglecting the sheet metal thickness (given 1litre = 1000 cm3 ). (b) the cross-sectional area of the sheet metal used to make the system, in square metres. (c) the cost of the sheet metal if the material costs 11.50 per square metre, assuming that 25% extra metal is required due to wastage.
2m A 500 mm B 1.5 m
Figure 27.21
1 Volume = h(R 2 + Rr + r 2 ) 3 Curved surface area = l(R + r) Total surface area = l(R + r) + r 2 + R 2 Problem 24. Determine the volume of a frustum of a cone if the diameter of the ends are 6.0 cm and 4.0 cm and its perpendicular height is 3.6 cm (i) Method 1 A section through the vertex of a complete cone is shown in Figure 27.22. AP DR = Using similar triangles, DP BR Hence, 3.6 AP = 2.0 1.0 AP = (2.0)(3.6) 1.0 = 7.2 cm
1.5 m
C 800 mm
from which
Figure 27.20
## Volumes of common solids
Problem 25. Find the total surface area of the frustum of the cone in Problem 24.
253
(i) Method 1 Curved surface area of frustum = curved surface area of large cone curved surface area of small cone cut off. From Figure 27.22, using Pythagoras theorem, AB2 = AQ2 + BQ2
4.0 cm D 2.0 cm 3.6 cm E P
B 1.0 cm
R Q
C
## = (BQ)(AB) = (3.0)(11.21) = 105.65 cm2
3.0 cm 6.0 cm
and curved surface area of small cone = (DP)(AD) = (2.0)(7.47) = 46.94 cm2 Hence, curved surface area of frustum = 105.65 46.94 = 58.71 cm2 Total surface area of frustum = curved surface area + area of two circular ends = 58.71 + (2.0)2 + (3.0)2 = 58.71 + 12.57 + 28.27 = 99.6 cm2 (ii) Method 2 From page 252, total surface area of frustum = l(R + r) + r 2 + R 2 where l = BD = 11.21 7.47 = 3.74 cm, R = 3.0 cm and r = 2.0 cm. Hence, total surface area of frustum = (3.74)(3.0 + 2.0) + (2.0)2 + (3.0)2 = 99.6 cm2
Figure 27.22
The height of the large cone = 3.6 + 7.2 = 10.8 cm Volume of frustum of cone = volume of large cone volume of small cone cut off 1 1 = (3.0)2 (10.8) (2.0)2 (7.2) 3 3 = 101.79 30.16 = 71.6 cm3 (ii) Method 2 From above, volume of the frustum of a cone 1 = h(R 2 + Rr + r 2 ) 3 R = 3.0 cm, r = 2.0 cm and h = 3.6 cm Hence, volume of frustum 1 = (3.6) (3.0)2 + (3.0)(2.0) + (2.0)2 3 1 = (3.6)(19.0) = 71.6 cm3 3
where
## 254 Basic Engineering Mathematics
Problem 26. A storage hopper is in the shape of a frustum of a pyramid. Determine its volume if the ends of the frustum are squares of sides 8.0 m and 4.6 m, respectively, and the perpendicular height between its ends is 3.6 m The frustum is shown shaded in Figure 27.23(a) as part of a complete pyramid. A section perpendicular to the base through the vertex is shown in Figure 27.23(b).
Figure 27.24
C
## 4.6 m R 4.6 m P O U 8.0 m T 8.0 m Q S
4.6 cm G 2.3 m A 8.0 m (a) H 1.7 m 2.3 m (b) 2.3 m F E 4.0 m D 3.6 m
B
8.0 m
4.6 cm
QT =
OQ 2 + OT 2 =
## 3.62 + 1.72 = 3.98 m
Area of trapezium PRSU 1 = (4.6 + 8.0)(3.98) = 25.07 m2 2 Lateral surface area of hopper = 4(25.07) = 100.3 m2
Figure 27.23
## BH CG = BG AH (2.3)(3.6) = 4.87 m 1.7
BH AH
Problem 28. A lampshade is in the shape of a frustum of a cone. The vertical height of the shade is 25.0 cm and the diameters of the ends are 20.0 cm and 10.0 cm, respectively. Determine the area of the material needed to form the lampshade, correct to 3 signicant gures The curved surface area of a frustum of a cone = l(R + r) from page 252. Since the diameters of the ends of the frustum are 20.0 cm and 10.0 cm, from Figure 27.25,
r 5 5.0 cm
Height of complete pyramid = 3.6 + 4.87 = 8.47 m 1 Volume of large pyramid = (8.0)2 (8.47) 3 = 180.69 m3 1 Volume of small pyramid cut off = (4.6)2 (4.87) 3 = 34.35 m3 Hence, volume of storage hopper = 180.69 34.35
h 5 25.0 cm
= 146.3 m3 Problem 27. Determine the lateral surface area of the storage hopper in Problem 26 The lateral surface area of the storage hopper consists of four equal trapeziums. From Figure 27.24, 1 Area of trapezium PRSU = (PR + SU )(QT ) 2
Figure 27.25
5.0 cm R 5 10.0 cm
## Volumes of common solids
r = 5.0 cm, R = 10.0 cm and l= 25.02 + 5.02 = 25.50 cm If 40% of space is occupied then volume of air space = 0.6 10928 = 6557 m3 Now try the following Practice Exercise Practice Exercise 108 Volumes and surface areas of frusta of pyramids and cones (answers on page 352) 1. The radii of the faces of a frustum of a cone are 2.0 cm and 4.0 cm and the thickness of the frustum is 5.0 cm. Determine its volume and total surface area. 2. A frustum of a pyramid has square ends, the squares having sides 9.0 cm and 5.0 cm, respectively. Calculate the volume and total surface area of the frustum if the perpendicular distance between its ends is 8.0 cm. 3. A cooling tower is in the form of a frustum of a cone. The base has a diameter of 32.0 m, the top has a diameter of 14.0 m and the vertical height is 24.0 m. Calculate the volume of the tower and the curved surface area. 4. A loudspeaker diaphragm is in the form of a frustum of a cone. If the end diameters are 28.0 cm and 6.00 cm and the vertical distance between the ends is 30.0 cm, nd the area of material needed to cover the curved surface of the speaker.
2
255
from Pythagoras theorem. Hence, curved surface area = (25.50)(10.0 + 5.0) = 1201.7 cm2 i.e., the area of material needed to form the lampshade is 1200 cm2 , correct to 3 signicant gures. Problem 29. A cooling tower is in the form of a cylinder surmounted by a frustum of a cone, as shown in Figure 27.26. Determine the volume of air space in the tower if 40% of the space is used for pipes and other structures
12.0 m
25.0 m
Figure 27.26
## Volume of cylindrical portion = r 2 h = 25.0 2 (12.0)
12.0 m
30.0 m
= 5890 m3 1 Volume of frustum of cone = h(R 2 + Rr + r 2 ) 3 h = 30.0 12.0 = 18.0 m, where R = 25.0 2 = 12.5 m and r = 12.0 2 = 6.0 m.
5. A rectangular prism of metal having dimensions 4.3 cm by 7.2 cm by 12.4 cm is melted down and recast into a frustum of a square pyramid, 10% of the metal being lost in the process. If the ends of the frustum are squares of side 3 cm and 8 cm respectively, nd the thickness of the frustum. 6. Determine the volume and total surface area of a bucket consisting of an inverted frustum of a cone, of slant height 36.0 cm and end diameters 55.0 cm and 35.0 cm. 7. A cylindrical tank of diameter 2.0 m and perpendicular height 3.0 m is to be replaced by a tank of the same capacity but in the form of a frustum of a cone. If the diameters of the ends of the frustum are 1.0 m and 2.0 m, respectively, determine the vertical height required.
Hence, volume of frustum of cone 1 = (18.0) (12.5)2 + (12.5)(6.0) + (6.0)2 3 = 5038 m3 Total volume of cooling tower = 5890 + 5038 = 10 928 m3
## 256 Basic Engineering Mathematics
27.6 Volumes of similar shapes
Problem 30. A car has a mass of 1000 kg. A model of the car is made to a scale of 1 to 50. Determine the mass of the model if the car and its model are made of the same material Volume of model = Volume of car 1 50
3
Figure 27.27 shows two cubes, one of which has sides three times as long as those of the other.
since the volume of similar bodies are proportional to the cube of corresponding dimensions. Mass = density volume and, since both car and model are made of the same material,
3x x x x (a) 3x (b) 3x
3
1 50
## 1000 = 0.008 kg or 8 g 503
Figure 27.27
Now try the following Practice Exercise Practice Exercise 109 Volumes of similar shapes (answers on page 352) 1. The diameter of two spherical bearings are in the ratio 2:5. What is the ratio of their volumes? 2. An engineering component has a mass of 400 g. If each of its dimensions are reduced by 30%, determine its new mass.
Volume of Figure 27.27(a) = (x)(x)(x) = x 3 Volume of Figure 27.27(b) = (3x)(3x)(3x) = 27x 3 Hence, Figure 27.27(b) has a volume (3)3 , i.e. 27, times the volume of Figure 27.27(a). Summarizing, the volumes of similar bodies are proportional to the cubes of corresponding linear dimensions. |
# Mathematics of a price war
In my last post I looked at two companies competing over a natural monopoly. We saw that once they started competing it would not then make sense for either company to drop out. This is why price wars tend to continue once they have started.
Here is the maths behind the problem:
To keep things simple let’s assume that the two companies are only going to compete for two years.
They can drop out before they even enter the market, after a year or after two years.
Let’s assume that the second company drops out at the start with probability x and after the first year with probability y. This leaves the probability of him dropping out after two years as 1-x-y.
We will say that the cost to each company of competing is represented by ‘c’ each year and the profit they can make in a monopoly is ‘p’.
Let’s work out the profit that the first company will make at the end of each year.
It profit at the start is zero because they haven’t yet entered the market.
After the end of the first year it has either incurred a cost c if the other company hasn’t dropped out, which has a probability of 1-x, or it has a profit of p if the other company dropped out at the start, which has a probability of x.
In other words its expected profit after a year = xp – c(1-x)
After two years its expected profit = 2xp + (p-c)y – 2c(1-x-y)
This is because 2p is two years’ worth of profit and he gets this if the other company dropped out at the start, which happens with probability x. This gives us the 2xp bit.
If the other company drops out after a year, which happens with a probability y, then he will make a profit p in the second year, but have incurred a cost c in the first year. Overall he makes (p-c) with a probability y. This gives us the (p-c)y part of the equation.
The last bit of the equation is the two years of cost (2c) multiplied by the probability that the other company waits until the final year to drop out (1-x-y)
We now know the expected profit for the first company at the end of each year based on the probability of second company dropping out at each point. The second company will set the probabilities so that first company’s profits are the same whatever strategy it chooses.
This means that the expected profit of dropping out at the start must equal the expected profit of dropping out at the end of year 1, so xp – c(1-x) = 0
The other two equations must be equal as well so:
xp – c(1-x) = 2xp + (p-c)y – 2c(1-x-y)
This is only true if y=0.
y is the probability that the second company drops out after a year. y is 0 so there is no chance that the company will drop out half way through the game. The game is symmetrical so the same applies to the first company. They either drop out at the start or keep going to the end.
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# How do you integrate int x^3sqrt(1+x^2) using integration by parts?
Nov 8, 2016
$\frac{{\left(1 + {x}^{2}\right)}^{\frac{3}{2}} \left(3 {x}^{2} - 2\right)}{15} + C$
#### Explanation:
$I = \int {x}^{3} \sqrt{1 + {x}^{2}} \mathrm{dx}$
Before trying any integration by parts, we should attempt more basic substitutions to simplify the integral. Let $u = 1 + {x}^{2}$. This implies that $\mathrm{du} = 2 x \mathrm{dx}$ and ${x}^{2} = u - 1$.
Rearranging the integral:
$I = \frac{1}{2} \int {x}^{2} \sqrt{1 + {x}^{2}} \left(2 x \mathrm{dx}\right) = \frac{1}{2} \int \left(u - 1\right) \sqrt{u} \mathrm{du}$
So, we see that integration by parts won't be necessary at all!
$I = \frac{1}{2} \int \left({u}^{\frac{3}{2}} - {u}^{\frac{1}{2}}\right) \mathrm{du} = \frac{1}{2} \left({u}^{\frac{5}{2}} / \left(\frac{5}{2}\right) - {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right)$
$\textcolor{w h i t e}{I} = \frac{1}{2} \left(\frac{2}{5} {u}^{\frac{5}{2}} - \frac{2}{3} {u}^{\frac{3}{2}}\right) = {u}^{\frac{5}{2}} / 5 - {u}^{\frac{3}{2}} / 3 = \frac{3 {u}^{\frac{5}{2}} - 5 {u}^{\frac{3}{2}}}{15}$
$\textcolor{w h i t e}{I} = \frac{{u}^{\frac{3}{2}} \left(3 u - 5\right)}{15} = \frac{{\left(1 + {x}^{2}\right)}^{\frac{3}{2}} \left(3 {x}^{2} - 2\right)}{15} + C$ |
# Challenge 12a: Flipping Coins
1. Dec 27, 2013
### Office_Shredder
Staff Emeritus
A man begins flipping coins according to the following rules: He starts with a single coin that he flips. Every time he flips a heads, he removes the coin from the game and then puts out two more coins to add to his stack of coins to flip, every time he flips a tails he simply removes the coin from the game.
For example, if he flips a heads on his first flip, he now has two coins. He flips both, if he gets two tails then he has run out of coins. On the other hand if he flipped two heads he would now have four coins to flip.
Once he begins flipping, what is the probability that he ever runs out of coins?
For a harder challenge check out Challenge 12b.
2. Dec 27, 2013
### Staff: Mentor
So heads means +1, and tail means -1?
3. Dec 27, 2013
### D H
Staff Emeritus
That's how I take it, Borek -- except presumably it's game over once he runs out of coins. If that's the case, the probability is 1 that he runs out of coins eventually.
The person will only run out of coins on odd numbered flips. It helps to look at just those odd numbered flips. The probability that the person is still in the game (hasn't run out of coins) after flip $2n+1$ is $\binom{2n+1}{n+1}\,2^{-(2n+1)}$. This goes to zero as n→∞. In particular, it asymptotically approaches $1/\sqrt{n\pi}$.
4. Dec 27, 2013
### PeroK
Here's an alternative solution.
$P(2) = \frac14 + \frac12 P(2) + \frac14 P(4)$
And, in general:
$P(2n) = \frac14 P(2n-2) + \frac12 P(2n) + \frac14 P(2n+2)$
Hence:
$P(2n+2) = 2P(2n) - P(2n-2)$
By induction:
$P(2n) = nP(2) - (n-1)$
Hence:
$P(2) = \frac1n P(2n) + \frac{n-1}{n}$
Letting n → ∞ gives:
$P(2) = 1$
So, the probability of losing eventually if you start with 2 coins is 1. And, therefore, you're bound to lose if you start with 1 coin.
Last edited: Dec 27, 2013
5. Dec 27, 2013
### Office_Shredder
Staff Emeritus
Nice inductive solution Perok!
6. Dec 27, 2013
### PeroK
In fact, as a corollary, starting from:
$P(2n) = nP(2) - (n-1)$
$P(2n) = n - (n-1) = 1$
So, it doesn't matter how many coins you start with, you're bound to lose eventually! Which, I guess, is intuitive if you know enough about probability.
7. Dec 27, 2013
### D H
Staff Emeritus
Exactly. This is the gambler's ruin problem. The house presumably has an infinite supply of pennies in this challenge.
8. Dec 28, 2013
### PeroK
Just for completeness, using the brilliant observation by mfb on the harder problem that:
$P(2) = P(1)^2$
(Losing with two coins is equivalent to losing two independent games with one coin each.)
Then:
$P(1) = \frac12 + \frac12 P(2) = \frac12 + \frac12 P(1)^2$
$∴ \ P(1)^2 - 2P(1) + 1 = 0 \ ∴ \ (P(1) - 1)^2 = 0 \ ∴ \ P(1) = 1$
9. Jan 1, 2014
### ssd
If consecutive coins in action have the P(H)=1 then the game never ends. So the answer is not unique unless a sequence of probabilities for the coins in action is defined. For P(H)=p (fixed and 0<p<1) the problem is a famous college problem as indicated by others. Of course p=0 and 1 are trivial cases.
Last edited: Jan 1, 2014
10. Jan 1, 2014
### ssd
I note that, if for all coins P(H)=0.6 the the game never ends with probability = 1- (1-0.6)/0.6= 1-2/3= 1/3.
In fact P(H)≤ 0.5 => definite end of game. Else, there is a positive chance of a never ending game.
Last edited: Jan 1, 2014
11. Jan 3, 2014
### The_Duck
Let $p$ be the probability that he eventually runs out of coins. There are two ways for this to happen. He can get tails on the first flip--this happens with probability $\frac{1}{2}$. Alternatively with probability $\frac{1}{2}$ he can get heads on the first flip. If he gets heads on the first flip he essentially plays the same game twice, and the probability of running out of coins in *both* of these games is $p^2$. So we have
$p = \frac{1}{2} + \frac{1}{2}p^2$
The solution to this equation is $p=1$. |
Chapters
We write algebraic fractions as we write simple fractions with a numerator and denominator. The difference between a normal fraction and algebraic fraction is that the normal fraction usually contains numbers in numerator or denominator, while algebraic fractions have algebraic expressions in numerator and denominator. Algebraic fractions are basically the rational expressions which can be written in the following fractional form:
where
We simplify algebraic fractions in the same way as we simplify the normal fractions, i.e. numerator and denominator are divided by a common factor. The techniques for adding and subtracting algebraic fractions is also the same as the normal fractions. In this article, we will see how to simplify the fractions by adding, subtracting, multiplying and dividing them.
We take L.C.D to add two or more algebraic fractions when we have addition and subtraction sign between them. L.C.D stands for the least common denominator and the process of finding L.C.D is the same as the process of finding L.C.M of the fractions. L.C.D shows the smallest number divisible by the denominator of each fraction in the question. L.C.M is the least common multiple of two numbers.
The following examples will make this concept more clear.
### Solution
To add the above two fractions, take the L.C.D of 5 and 3.
By multiplying 3 and 5 with the terms inside the brackets:
Now, combine the like terms and add or subtract them depending on their signs:
Since the above fraction cannot be simplified further, so is the correct answer.
### Solution
the above algebraic expression is different than the first example because it contains algebraic expressions both in the numerator and the denominator. Take L.C.D of the expressions. We know that .
Use the FOIL method of multiplication to multiply the binomial terms in the numerator:
Combine the like terms together and apply their respective operations of addition or subtraction like this:
Since, the fraction cannot be simplified further, so it is the final answer.
### Solution
Take the L.C.D of the expressions and write the resultant expression as:
The terms in the numerators will be multiplied using the FOIL method:
You can also use the FOIL method to multiply the terms in the denominator and write the resultant expression as:
Since the above expression cannot be simplified further, so it is the final answer.
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## 2. Subtracting Algebraic Fractions
We subtract algebraic fractions in the same way as we add them. The only difference is when we apply the operation of subtraction after multiplication. The signs of all the factors inside the brackets are reversed because of the minus sign outside the bracket. Let us explain this concept through following examples
### Example 1
Subtract fractions
### Solution
Take the L.C.D to write the expression as:
Since , so we will write the expression as:
Use the FOIL method to multiply the terms in the numerator:
It is better to add or subtract the like terms together inside the parentheses before opening the brackets in the numerator:
Open the bracket in the numerator and reverse the signs of the terms which were inside the parenthesis:
The above fraction cannot be simplified further, hence it is the final answer.
Simplify
### Solution
Take L.C.D of the denominators and write the expression as:
Use FOIL method to multiply the expressions:
Multiply the terms inside the brackets by variables and constant in the numerator:
Simplify
### Solution
Taking L.C.D, we will write the above expression as:
Use FOIL method to multiply the numerator:
## 3. Multiplying Algebraic Fractions
Multiplying algebraic fractions is very simple. You just have to multiply the numerator with the numerator and denominator with the denominator. Let us explain the concept through the following examples.
### Solution
Multiply the numerators and the denominators together to simplify the above algebraic expression:
Factoring we will get the factors :
is the common factor both in the numerator and denominator, hence we will cancel out this factor and write the final answer as:
### Solution
Multiplying the numerators and the denominators together will give the following expression:
Remember that while writing the final answer simplifying fractions is necessary. Simplifying means reducing fractions to the lowest possible fraction. You can see that in the above expression, is the common factor both in the numerator and the denominator. Hence, we will write the expressions as:
Cancel out in the numerator and the denominator to write the above expression as a single fraction like this:
### Solution
Multiplying the numerators and the denominators together will give us the following expression:
You can see that 5 is the common term in the numerator and the denominator:
will be cancelled out to write the final expression as:
## 4. Dividing Algebraic Fractions
We divide algebraic fractions using the same process as we follow while dividing fractions, i.e. we take reciprocal of the second fraction. Let us explain the concept through the following examples.
### Solution
Take the reciprocal of the second fraction:
Multiply the numerators and denominators together:
### Solution
Take the reciprocal of the second fraction:
Factor the terms in the numerator and the denominator as much as possible:
Write the resulting expression by cancelling out and :
can be written as , hence the above expression can be simplified further as:
in the numerator and denominator will be cancelled out:
Hence, -1 is the simplest form of the above question.
### Example 3
Simplify fractions
### Solution
To divide, we will take the reciprocal of the second fractional expression. We will rewrite the above expression as:
Factor the expressions in the numerator and the denominator as much as possible:
Factoring of different fractions will yield following factors:
=
Write the expressions in factored form like this:
Cross multiply the two algebraic expressions to cancel out the common factors:
Hence, is in its simplest form.
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# How To Find Angles When The Lengths Of The Sides Of A Triangle Are Known
## Video: How To Find Angles When The Lengths Of The Sides Of A Triangle Are Known
The values of the angles lying at the vertices of the triangle, and the lengths of the sides forming these vertices, are related to each other by certain ratios. These ratios are most often expressed in terms of trigonometric functions - mainly in terms of sine and cosine. Knowing the lengths of all sides of the figure is enough to restore the values of all three angles using these functions.
## Instructions
### Step 1
Use the cosine theorem to calculate the magnitude of any of the angles of an arbitrary triangle. It says that the square of the length of any side (for example, A) is equal to the sum of the squares of the lengths of the other two sides (B and C), from which the product of their own lengths and the cosine of the angle (α) lying in the vertex they form is subtracted. This means that you can express the cosine in terms of the side lengths: cos (α) = (B² + C²-A²) / (2 * A * B). To get the value of this angle in degrees, apply the inverse cosine function to the resulting expression - the inverse cosine: α = arccos ((B² + C²-A²) / (2 * A * B)). In this way, you will calculate the value of one of the angles - in this case, the one that lies opposite side A.
### Step 2
To calculate the two remaining angles, you can use the same formula, swapping the lengths of the known sides in it. But a simpler expression with fewer mathematical operations can be obtained using another postulate from the field of trigonometry - the theorem of sines. She claims that the ratio of the length of any side to the sine of the opposite angle in a triangle is equal. This means that you can express, for example, the sine of the angle β opposite side B in terms of the length of side C and the already calculated angle α. Multiply the length of B by the sine α, and divide the result by the length of C: sin (β) = B * sin (α) / C. The value of this angle in degrees, as in the previous step, calculate using the inverse trigonometric function - this time the arcsine: β = arcsin (B * sin (α) / C).
### Step 3
The value of the remaining angle (γ) can be calculated using any of the formulas obtained in the previous steps, by swapping the lengths of the sides in them. But it is easier to use one more theorem - about the sum of angles in a triangle. She claims that this sum is always 180 °. Since two of the three angles are already known to you, simply subtract their values from 180 ° to get the value of the third: γ = 180 ° -α-β. |
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##### Presentation Transcript
1. Please close your laptops and turn off and put away your cell phones, and get out your note-taking materials. Today’s daily homework quiz will be given at the end of class.
2. This week’s schedule: • Today: Lecture on Section 4.5A • HW 4.5A: 16 word problems, due tomorrow • Tomorrow: Lecture on Section 4.5B • 16 more word problems due Wed. (HW 4.5B) • Wednesday: Review for Test 2 • Practice Test 2 due Thursday • Thursday:Take Test 2 (125 points) Next week’s schedule: Spring Break!!
3. Steps in Solving Problems Involving Systems of Two Linear Equations in Two Variables: • Understand the problem. • Read and reread the problem. • Choose two variables to represent the two unknowns. • Translate the problem into two equations. • Solve the system of equations. • Interpret the results. • Check proposed solution in the problem. • State your conclusion. • Check proposed solution in the problem !!!!!
4. Example One number is 4 more than twice the second number. Their total is 25. Find the numbers. 1. UNDERSTAND Read and reread the problem. Since we are looking for two numbers, we let x = first number y = second number continued
5. One number is 4 more than twice the second number. continued 2. TRANSLATE x = 2y + 4 Their total is 25. x + y = 25 continued
6. continued 3. SOLVE We are solving the system x = 2y +4 x + y = 25 Using the substitution method, we substitute the solution for x from the first equation into the second equation. x + y = 25 (2y +4) + y = 25 Replace x with 4 + 2y. 3y + 4 = 25 Simplify. 3y = 21 Subtract 4 from both sides. y = 7 Divide both sides by 3. continued
7. continued Now we substitute 7 for y into the first equation. x= 4 + 2y= 4 + 2(7) = 4 + 14 = 18 4. INTERPRET Check:Substitute x = 18 and y = 7 into both of the equations. First equation: x = 4 + 2y 18 = 4 + 2(7) True Second equation: x + y = 25 18 + 7 = 25 True State: The two numbers are 18 and 7.
8. Example Hilton University Drama club sold 311 tickets for a play. Student tickets cost 50 cents each; non-student tickets cost \$1.50. If the total receipts were \$385.50, find how many tickets of each type were sold. 1. UNDERSTAND Read and reread the problem. Since we are looking for two numbers, we let s = the number of student tickets n = the number of non-student tickets continued
9. Hilton University Drama club sold 311 tickets for a play. total receipts were \$385.50 Admission for students continued 2. TRANSLATE s + n = 311 Admission for non-students Total receipts = + 1.50n 0.50s 385.50 continued
10. continued 3. SOLVE We are solving the system s + n = 311 0.50s + 1.50n = 385.50 Since the equations are written in standard form (and we might like to get rid of the decimals anyway), we’ll solve by the addition/elimination method. (Substitution could be used instead, if you prefer to do it that way.) Question: If we wanted to eliminate s, what would we multiply the 0.50s in the second equation by to make it become -1s? Answer: Multiply the second equation by –2. s + n = 311 s + n = 311 –s – 3n = –771 –2(0.50s + 1.50n) = –2(385.50) –2n = –460 n = 230 continued
11. continued Now we substitute 230 for n into the first equation to solve for s. s + n = 311 s + 230 = 311 s = 81 4. INTERPRET Check: Substitute s = 81 and n = 230 into both of the equations. s + n = 311First Equation 81+ 230 = 311 True 0.50s + 1.50n = 385.50Second Equation 0.50(81) + 1.50(230) = 385.50 40.50 + 345 = 385.50 True State: There were 81 student tickets and 230 non student tickets sold.
12. How could we set this problem up using two variables? How would you check these answers? x = ? Ounces of 16% solution y = ? Ounces of 8 % solution Equation 1? x + y = 32 Equation 2? 0.16 ● x + 0.08 ● y = 0.11●32
13. You may now OPEN your LAPTOPS and begin working on the homework assignment until it’s time to take the quiz on HW 4.3. The assignment on this material (HW 4.5A) Is due at the start of the next class session.
14. You will have access to the online calculator on your laptop during this quiz. No other calculator may be used. Please open Quiz 4.3. • IMPORTANT NOTE: If you have time left after you finish the problems on this quiz, use it to check your answers before you submit the quiz! • Remember to turn in your answer sheetto the TA when the quiz time is up. |
# How do you simplify 3sqrt(96k)+2sqrt180?
Jul 23, 2018
See a solution process below:
#### Explanation:
First rewrite each term as:
$3 \sqrt{16 \cdot 6 k} + 2 \sqrt{36 \cdot 5}$
$\sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}}$
$3 \sqrt{\textcolor{red}{16} \cdot \textcolor{b l u e}{6 k}} + 2 \sqrt{\textcolor{red}{36} \cdot \textcolor{b l u e}{5}} \implies$
$3 \sqrt{\textcolor{red}{16} \cdot \textcolor{b l u e}{6 k}} + 2 \sqrt{\textcolor{red}{36} \cdot \textcolor{b l u e}{5}} \implies$
$3 \sqrt{\textcolor{red}{16}} \sqrt{\textcolor{b l u e}{6 k}} + 2 \sqrt{\textcolor{red}{36}} \sqrt{\textcolor{b l u e}{5}} \implies$
$\left(3 \cdot 4\right) \sqrt{\textcolor{b l u e}{6 k}} + \left(2 \cdot 6\right) \sqrt{\textcolor{b l u e}{5}} \implies$
$12 \sqrt{6 k} + 12 \sqrt{5}$
Now, facture out the common term:
$\textcolor{red}{12} \sqrt{6 k} + \textcolor{red}{12} \sqrt{5} \implies$
$\textcolor{red}{12} \left(\sqrt{6 k} + \sqrt{5}\right)$ |
# Prime Factors
Est. Class Sessions: 2
### Part 1: Finding Prime Factors
If necessary, review the terms that are involved in the tasks in this lesson. For example, ask:
• If I write this number sentence, “36 = 4 × 9,” which of the numbers are factors and which is the product? (36 is the product and 4 and 9 are factors.)
• What does it mean to “write 36 as a product of three factors”? (Write the same kind of number sentence, but multiply three numbers instead of two. For example, 2 × 2 × 9 = 36.)
Find Factors of Composite Numbers. Have students work with a partner on this task:
• Find as many different ways as you can to write 36 as a product of three or more factors.
Possible responses include:
36 = 3 × 4 × 3
36 = 3 × 2 × 6
36 = 9 × 2 × 2
36 = 2 × 2 × 3 × 3
Note that the last example is a number sentence that renames 36 as a product of primes. If some students find factors that are all prime numbers, ask them to tell how they found their answers.
• Can the factors in your number sentence be factored into smaller numbers?
For example, in the sentence 36 = 9 × 2 × 2, the 9 can be factored into 3 × 3. The sentence then becomes
36 = 3 × 3 × 2 × 2.
Ask students to work with the same partner to complete the following:
• Write 24 as a product of at least three factors. (Possible responses include
2 × 3 × 4, 2 × 2 × 6, 2 × 2 × 2 × 3.)
• Try to write 24 as a product of factors that are all prime numbers (that cannot be factored any further).
(2 × 2 × 2 × 3 = 24)
Have students share their number sentences with the class.
• Are the number sentences the same? (The correct number sentence is 24 = 2 × 2 × 2 × 3, though the factors need not be in that order.)
• Does it matter if the order of the factors is not the same? (No, because of the turn-around rule or the commutative property.)
• How can you tell that all the factors in these number sentences are prime numbers? (Students may say that the factors cannot be factored any further, or that each factor cannot be divided by any number other than one and the number itself.)
Ask students to read the vignette in the Finding Prime Factors section of the Student Guide. After reading Nicholas's explanation of his strategy, ask students to compare Nicholas's strategy to their own.
Make Factor Trees. Show the class how to use a factor tree to organize the work in a search for prime factors. Draw the factor tree that Mrs. Dewey drew on the board as shown in the Student Guide. Have a student show how each step of the factor tree matches with each of Nicholas's steps. Then ask a student to explain each step of Nila's factor tree. Ask another student to finish the factor tree that John started in Question 1. See Figure 1.
The Prime Factors pages in the Student Guide give examples of factor trees for 24, starting with 3 × 8, 6 × 4, and 2 × 12. For additional class practice, work together to make factor trees for 30 and 40.
When students are ready, ask them to answer Questions 2–3, which provide additional practice with factor trees. Students can discuss their solutions with partners.
Use Check-In: Question 3 on the Prime Factors page in the Student Guide to assess students' abilities to find the prime factorization of a number [E6].
In Question 4, students distinguish between finding all the factors of a number and finding the prime factorization of a number. Clarify for students the meaning of each. See Content Note.
“Finding all the factors of a number” refers to naming the numbers found in all of the unique factor pairs of a number. For example, the factors of 18 are 1, 2, 3, 6, 9, and 18, since 18 can be written as 1 × 18, 2 × 9, or 3 × 6 (and the reverse of each).
The prime factorization of a number, on the other hand, is the number written as a product of prime numbers. If the order of the factors is not considered, there is only one way to write a number as a product of primes. For example, the prime factorization of 18 is 2 × 3 × 3.
The prime factorization of a prime number is just the number itself. For example, the prime factorization of 7 is 7.
(It is not 1 × 7, since by definition, 1 is not a prime number.)
Use Exponents. Students practice using exponents to write prime factorizations of numbers in Questions 5 and 6.
Assign the Factors and Primes section of the Homework page in the Student Guide after Part 1.
+
+
+
A factor tree for 24 for Question 1 |
# NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4
Ncert Solutions For Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 Practical Geometry Exercise 14.4 Ncert Solutions For Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 Chapter 14 Practical Geometry Class 6 Chapter 14 Maths Chapter 14 Class 6 Maths Maths Class 6 Chapter 14 Maths Chapter 14 Class 6 Ncert Maths Class 6 Chapter 14 Ncert Solutions For Class 6 Maths Chapter 14 Practical Geometry Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 Class 6 Maths Ncert Solutions
## NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4
Question 1.
Draw any line segment AB¯¯¯¯¯¯¯¯ . Mark any point M on it. Through M draw a perpendicular to AB¯¯¯¯¯¯¯¯. (use ruler and compasses).
Solution :
Step 1. Given a point M on any line AB¯¯¯¯¯¯¯¯.
Step 2. With M as centre and a convenient radius, construct a part circle (arc) intersecting the line segment AB¯¯¯¯¯¯¯¯ at two points C and D.
Step 3. With C and D as centres and a radius greater than CM, construct two arcs which cut each other at N.
Step 4. Join MN¯¯¯¯¯¯¯¯¯¯. Then MN¯¯¯¯¯¯¯¯¯¯ is perpendicular to AB¯¯¯¯¯¯¯¯ at M, i.e., MN¯¯¯¯¯¯¯¯¯¯ AB¯¯¯¯¯¯¯¯.
Question 2.
Draw any’line segment PQ¯¯¯¯¯¯¯¯. Take any point R not on it. Through R draw a perpendicular to PQ¯¯¯¯¯¯¯¯. (use ruler and set-square).
Solution :
Step 1. Let PQ¯¯¯¯¯¯¯¯ be the given line segment and R be a point not on it.
Step 2. Place a set-square on PQ¯¯¯¯¯¯¯¯ such that one arm of the right angle aligns along PQ¯¯¯¯¯¯¯¯.
Step 3. Place a ruler along the edge opposite of the right angle.
Step 4. Hold the ruler fixed. Slide the set-square along the ruler all the point R touches the arm of the set-square.
Step 5. Join RS along the edge through R, meeting PQ¯¯¯¯¯¯¯¯ at S. Now RS¯¯¯¯¯¯¯ PQ¯¯¯¯¯¯¯¯.
Question 3.
Draw a line l and a point X on it. Through X, draw a line segment XY¯¯¯¯¯¯¯¯ perpendicular to l. Now draw a perpendicular to XY at Y. (use ruler and compasses)
Solution :
Step 1. Given a point X on a line l.
Step 2. With X as centre and a convenient radius, construct a part circle (arc) intersecting the line l at two points A and B.
Step 3. With A and B as centres and a radius greater than AX, construct two arcs which cut each other at Y.
Step 4. Join XY¯¯¯¯¯¯¯¯. Then XY¯¯¯¯¯¯¯¯ is perpendicular to l atX, i.e., XY¯¯¯¯¯¯¯¯ l.
error: |
## Intermediate Algebra (12th Edition)
$r=\left\{ -1-3\sqrt{2},-1+3\sqrt{2} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $(r-3)(r+5)=2 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} r(r)+r(5)-3(r)-3(5)=2 \\\\ r^2+5r-3r-15=2 \\\\ r^2+2r-15=2 .\end{array} Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} r^2+2r-15-2=0 \\\\ r^2+2r-17=0 .\end{array} The quadratic equation above has $a= 1 , b= 2 , c= -17 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} r=\dfrac{-2\pm\sqrt{2^2-4(1)(-17)}}{2(1)} \\\\ r=\dfrac{-2\pm\sqrt{4+68}}{2} \\\\ r=\dfrac{-2\pm\sqrt{72}}{2} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} r=\dfrac{-2\pm\sqrt{36\cdot2}}{2} \\\\ r=\dfrac{-2\pm\sqrt{(6)^2\cdot2}}{2} \\\\ r=\dfrac{-2\pm6\sqrt{2}}{2} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} r=\dfrac{\cancel2(-1)\pm\cancel2(3)\sqrt{2}}{\cancel2(1)} \\\\ r=-1\pm3\sqrt{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} r=-1-3\sqrt{2} \\\\\text{OR}\\\\ r=-1+3\sqrt{2} .\end{array} Hence, $r=\left\{ -1-3\sqrt{2},-1+3\sqrt{2} \right\} .$ |
# How do you simplify (1/16)^(-3/4)?
${\left(\frac{1}{16}\right)}^{- \frac{3}{4}} = 8$
#### Explanation:
We can use the following rules to figure this out:
${x}^{- 1} = \frac{1}{x}$
${\left({x}^{a}\right)}^{b} = {x}^{a b}$
We can now start working. Let's first use the second rule:
${\left(\frac{1}{16}\right)}^{- \frac{3}{4}}$
${16}^{\frac{3}{4}}$
Notice that $16 = {2}^{4}$
${\left({2}^{4}\right)}^{\frac{3}{4}}$
${2}^{4 \times \frac{3}{4}}$
${2}^{3} = 8$ |
# Knowing Our Numbers
## Ex 1.2 Question 1.
A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Solution-
Number of tickets sold on the first day = 1094
Number of tickets sold on the second day = 1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the final day = 2751
∴ Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7,707
## Ex 1.2 Question 2.
Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Solution-
Shekhar wishes to complete = 10,000 runs.
Shekhar has so far scored = 6980 runs
∴ Total number of runs needed by him = 10,000 – 6980 = 3020 runs
## Ex 1.2 Question 3.
In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Solution-
Number of votes secured by the successful candidate = 5,77,500
Number of votes secured by his nearest rival = 3,48,700
∴Margin of votes to win the election = 5,77,500 – 3,48,700 = 2,28,800
## Ex 1.2 Question 4.
Kirti bookstore sold books worth ₹2,85,891 in the first week of June and books worth ₹4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Solution-
Books sold in the first week of June = ₹2,85,891
Books sold in the second week of the June = ₹4,00,768
∴ Total books sold
= ₹2,85,891 + ₹4,00,768 = ₹6,86,659
In the second week of the month, the sale of books was greater.
The difference between the sale of books
= ₹4,00,768 – ₹2,85,891 = ₹1,14,877
Hence, in the second week, the sale of books was more by ₹1,14,877.
## Ex 1.2 Question 5.
Find the difference between the greatest and the least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once.
Solution-
Given digits = 6, 2, 7, 4, 3
Greatest number = 76432
Least number = 23467
∴ Difference = 76432 – 23467 = 52,965
## Ex 1.2 Question 6.
A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Solution-
The number of screws manufactured in a day = 2,825.
Number of screws manufactured in month of January (31 days) = 31 x 2825 = 87,575
## Ex 1.2 Question 7.
A merchant had ₹78,592 with her. She placed an order for purchasing 40 radio sets at ₹1200 each. How much money will remain with her after the purchase?
Solution-
Number of radio sets = 40
Cost of one radio set = ₹1200
∴Cost of 40 radio sets = ₹1200 x 40 = ₹48,000
Amount of money with the merchant = ₹78,592
money spent by her =₹48,000
Remaining money with the merchant = ₹78,592 – ₹48000 = ₹30,592
## Ex 1.2 Question 8.
A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Solution-
The student has multiplied 7236 by 65 instead of multiplying by 56.
Difference between the two multiplications = (65 – 56) x 7236 = 9 x 7236 = 65124
## Ex 1.2 Question 9.
To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Solution-
Total length of the cloth = 40 m = 40 x 100 cm = 4000 cm.
Cloth needed to stitch a shirt = 2 m 15 cm = 2 x 100 + 15 cm = 215 cm
Therefore, the number of shirts stitched = 4000/215
So, the number of shirts stitched = 18
Remaining cloth = 130 cm = 1 m 30 cm
## Ex 1.2 Question 10.
Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solution-
Weight of one box = 4 kg 500 g = 4 x 1000 + 500 = 4500 g
Maximum load can be loaded in van = 800 kg = 800 x 1000 = 800000 g
Number of boxes =800000/4500
Therefore, 177 boxes can be loaded in the van.
## Ex 1.2 Question 11.
The distance between the school and the house of a student is 1 km 875 m. Every day she walks both ways. Find the total distance covered by her in six days.
Solution-
Distance between school and house = 1 km 875 m = (1000 + 875) m = 1875 m.
Total ditance covered in one day = 2 x 1875 = 3750 m
Distance travelled in 6 days = 3750 m x 6 – 22500 m = 22 km 500 m.
Hence, the total distance covered in six days = 22 km 500 m
## Ex 1.2 Question 12.
A vessel has 4 liters and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Solution-
Quantity of curd in a vessel = 4 liters 500 mL = (4 x 1000 + 500) mL = 4500 mL.
Capacity of 1 glass = 25 mL
Number of glasses can be filled =4500/25
∴The number of glasses can be filled =180
### UKPSC Forest Guard Exam 09 April 2023 – Answer Key
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# Single Phase Voltage Calculation | Matlab
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Here, we will find phase voltages VAN, VBN, and VCN, shown in the following figure, using Matlab.
By applying KVL, we come up with the following three equations:
$\begin{matrix} 110\angle {{0}^{o}}=(1+j1){{I}_{1}}+(5+j12){{I}_{1}} & \cdots & (1) \\ 110\angle -{{120}^{o}}=(1-j2){{I}_{2}}+(3+j4){{I}_{2}} & \cdots & (2) \\ 110\angle {{120}^{o}}=(1-j0.5){{I}_{3}}+(5-j12){{I}_{3}} & \cdots & (3) \\\end{matrix}$
After simplifying above equations, we have
$\begin{matrix} 110\angle {{0}^{o}}=(6+j13){{I}_{1}} & \cdots & (4) \\ 110\angle -{{120}^{o}}=(4+j2){{I}_{2}} & \cdots & (5) \\ 110\angle {{120}^{o}}=(6-j12.5){{I}_{3}} & \cdots & (6) \\\end{matrix}$
Let’s put the above three equations in matrix form,
$\left[ \begin{matrix} 6+j13 & 0 & 0 \\ 0 & 4+j2 & 0 \\ 0 & 0 & 6-j12.5 \\\end{matrix} \right]\left[ \begin{matrix} {{I}_{1}} \\ {{I}_{2}} \\ {{I}_{3}} \\\end{matrix} \right]=\left[ \begin{matrix} 110\angle {{0}^{o}} \\ 110\angle -{{120}^{o}} \\ 110\angle {{120}^{o}} \\\end{matrix} \right]$
Now, we can write the above matrix as:
$\left[ Z \right]\left[ I \right]=\left[ V \right]$
From above, we can easily calculate unknown currents using:
$I=inv(Z)*V$
And for the phase voltages:
\begin{matrix} \begin{align} & {{V}_{AN}}=(5+j12){{I}_{1}} \\ & {{V}_{BN}}=(3+j4){{I}_{2}} \\ & {{V}_{CN}}=(5-j12){{I}_{3}} \\\end{align} & \cdots & (7) \\\end{matrix}
Now, it’s time to write Matlab code to find out the phase voltages using the above formulas.
## Measure phase voltages in three-phase system using Matlab
clear all;close all;clc
% Phase Voltages Calculation using Matlab
Z = [6-13*j 0 0;
0 4+2*j 0; % Z-Matrix (Impedance Matrix) from text
0 0 6-12.5*j];
c1=110; %Angle is 0 degree here so we simply ommit it
c2 = 110*exp(j*pi*(-120/180)); % Voltages expressed in phasor form (V=Vm*exp(j*theta))
c3 = 110*exp(j*pi*(120/180)); % Angles are converted into radians (=degrees*pi/180)
V = [c1; c2; c3]; % Voltage Vector [V] mentioned in the text
I = inv(Z)*V; % Calculate unknown Loop currents
%% Phase Voltages Calcualtion
V_an = (5+12*j)*I(1);
V_bn = (3+4*j)*I(2); % Calcualting Phase Voltages using equation (7) in text
V_cn = (5-12*j)*I(3);
% Magnitude and Angle Calculation for each Phase Voltage
V_an_abs = abs(V_an);
V_an_ang = angle(V_an)*180/pi;
V_bn_abs = abs(V_bn);
V_bn_ang = angle(V_bn)*180/pi;
V_cn_abs = abs(V_cn);
V_cn_ang = angle(V_cn)*180/pi;
%% Print out Results
fprintf('Phase Voltage Van \n Magnitude: %f \n Angle in degree: %f \n', V_an_abs, V_an_ang)
fprintf('Phase Voltage Vbn \n Magnitude: %f \n Angle in degree: %f \n', V_bn_abs, V_bn_ang)
fprintf('Phase Voltage Vcn \n Magnitude: %f \n Angle in degree: %f \n', V_cn_abs, V_cn_ang)
Results
Phase Voltage Van
Magnitude: 99.875532
Angle in degree: 132.604994
Phase Voltage Vbn
Magnitude: 122.983739
Angle in degree: -93.434949
Phase Voltage Vcn
Magnitude: 103.134238
Angle in degree: 116.978859
Did you find apk for android? You can find new Free Android Games and apps. |
# Lesson 12
Completing the Square (Part 1)
### Problem 1
Add the number that would make the expression a perfect square. Next, write an equivalent expression in factored form.
1. $$x^2 - 6x$$
2. $$x^2 + 2x$$
3. $$x^2 + 14x$$
4. $$x^2 - 4x$$
5. $$x^2 + 24x$$
### Solution
For access, consult one of our IM Certified Partners.
### Problem 2
Mai is solving the equation $$x^2 + 12x = 13$$. She writes:
\displaystyle \begin{align} x^2 + 12x &= 13\\ (x + 6)^2 &= 49\\ x &= 1 \text { or } x = \text- 13\\ \end{align}\\
Jada looks at Mai’s work and is confused. She doesn’t see how Mai got her answer.
Complete Mai’s missing steps to help Jada see how Mai solved the equation.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 3
Match each equation to an equivalent equation with a perfect square on one side.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 4
Solve each equation by completing the square.
$$x^2-6x+5=12$$
$$x^2-2x=8$$
$$11=x^2+4x-1$$
$$x^2-18x+60=\text-21$$
### Solution
For access, consult one of our IM Certified Partners.
### Problem 5
Rewrite each expression in standard form.
1. $$(x+3)(x-3)$$
2. $$(7+x)(x-7)$$
3. $$(2x-5)(2x+5)$$
4. $$(x+\frac18)(x-\frac18)$$
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 7, Lesson 8.)
### Problem 6
To find the product $$203 \boldcdot 197$$ without a calculator, Priya wrote $$(200+3)(200-3)$$. Very quickly, and without writing anything else, she arrived at 39,991. Explain how writing the two factors as a sum and a difference may have helped Priya.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 7, Lesson 8.)
### Problem 7
A basketball is dropped from the roof of a building and its height in feet is modeled by the function $$h$$.
Here is a graph representing $$h$$.
A:
When $$t=0$$ the height is 0 feet.
B:
The basketball falls at a constant speed.
C:
The expression that defines $$h$$ is linear.
D:
The expression that defines $$h$$ is quadratic.
E:
When $$t=0$$ the ball is about 50 feet above the ground.
F:
The basketball lands on the ground about 1.75 seconds after it is dropped.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 6, Lesson 5.)
### Problem 8
A group of students are guessing the number of paper clips in a small box.
The guesses and the guessing errors are plotted on a coordinate plane.
What is the actual number of paper clips in the box?
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 13.) |
Premium
# The Mean, Median, and Mode Card Game
### What You Need:
• Pencils
• White paper
• Playing cards
• Calculator (optional)
### What You Do:
Review the definitions of these key terms with your child:
Mean Is the average of all of the numbers in a sample. Add up all of the numbers in a set and divide by the total number of items to calculate a mean.
Median Is the middle number in a series of numbers that's ordered from least to greatest. If there's an even number of items in the data set, the median can be calculated by averaging the two middle numbers.
Mode Is the number that appears the most times in the data set.
Once your kid grasps the differences between these vocabulary words, gather 4 players to tackle this card game and practise each of these maths concepts. Using only the Ace through 10 cards, deal out 7 cards to each player. Have everyone arrange their cards in sequential order, with every Ace representing the number 1. Then, depending upon which game you want to play, follow the directions below:
1. Finding the Mean Game. Instruct everyone playing to find the total value of the numbers on their cards. Each player should then divide their total by 7 (the total number of items in the set) and round to the nearest whole number to find the mean. For example, if the cards in your hand are 2, 2, 3, 6, 7, 7, 9, then the sum of those digits is 38. Divide the sum by 7 to get 5 as your answer. Your answer represents the number of points you receive in each round. Provide scratch paper and pencils to help players find the answer to every division problem, or use a calculator to speed up the process.
2. Finding the Median Game. In this game, players get a number of points that match the median card in their hands. For example, the point value for the hand above would be 6, since 6 is the value of the median card in that set.
3. Finding the Mode Game. Just like in the games above, the number of points in the game rounds here is reflected by the mode in each hand of cards. If there isn't a mode (a number appearing more than once), then that player scores a 0 for that round. In the situation where there are multiple modes, such as in the hand above, the player receives a number of points that matches the sum of the modes. For example, the mode for the hand above would be 9, since 7 and 2 are both modes.
In each game, the winner is the first person to score a total of 21 points. This number can be lowered for kids who have shorter attention spans.
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# HOW TO GRAPH RATIONAL FUNCTIONS 2
## About the topic "How to graph rational functions 2"
"How to graph rational functions 2" is the question having had by almost all the students who study math in high schools.
Even though students can get this stuff on internet, they do not understand exactly what has been explained.
To make the students to understand "How to graph rational functions 2", we have explained the above mentioned stuff step by step.
## Rational function - Example
Before learning "How to graph the rational functions", first you have to be knowing the following stuff.
1. Vertical Asymptote
2. Horizontal Asymptote
3. Slant Asymptote
4. Hole
To know more about the above mentioned stuff, please click the topics given above.
If you had already learned the above mentioned stuff, then you are ready to learn the stuff, "How to graph rational functions 2".
Now let us take an example and understand graphing rational functions.
Example :
Graph the rational function given below.
Solution :
Step 1:
First, we have to find hole, if any.
To find hole of the rational function, we have to see whether there is any common factor found at both numerator and denominator.
In our problem, clearly there is no common factor found at both numerator and denominator.So, there is no hole.
Step 2:
Now, we have to find vertical asymptote, if any.
Most of the rational function will have vertical asymptote.
To find vertical asymptote, we have to make the denominator equal to zero.
When we do so, x² - 9 = 0 ====> (x+3)(x-3) = 0
x + 3 = 0 or x - 3 = 0
x = -3 or x = 3
So, the vertical asymptotes are x = -3 and x = 3
Step 3 :
Now we have to find horizontal asymptote, if any.
In the rational function given above, the highest exponent of the numerator is 1 and denominator is 2.
If the highest exponent of the numerator is less than the highest exponent of the denominator.
So, the horizontal asymptote is y = o or x-axis
Step 4 :
Now we have to find slant asymptote, if any.
Since there is horizontal asymptote, there is no slant asymptote.
Step 5 :
In the given rational function, now we have to plug some random values for "x" and find the corresponding values of "y".
We have already known that the vertical asymptotes are x = -3 and x = 3.
Now, we have to take some random values for x in the following intervals.
x<-3, -3<x<3, x>3 but not x = -3 & x = 3.
(Because, x = -3 and x = 3 are vertical asymptotes) |
# optimization problem please help!!!
Printable View
• May 29th 2008, 11:29 PM
eawolbert
optimization problem please help!!!
A cylinder shaped can needs to be constructed to hold 300 cubic centimeters of soup. The material for the sides of the can costs 0.04 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.08 cents per square centimeter. Find the dimensions for the can that will minimize production cost.
Helpful information:
h : height of can, r : radius of can
Volume of a cylinder: V=πr2h
Area of the sides: A=2πrh
Area of the top/bottom: A=πr2
• May 29th 2008, 11:49 PM
earboth
Quote:
Originally Posted by eawolbert
A cylinder shaped can needs to be constructed to hold 300 cubic centimeters of soup. The material for the sides of the can costs 0.04 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.08 cents per square centimeter. Find the dimensions for the can that will minimize production cost.
Helpful information:
h : height of can, r : radius of can
Volume of a cylinder: V=πr2h
Area of the sides: A=2πrh
Area of the top/bottom: A=πr2
You have all the information needed to solve this question:
$V=300=\pi r^2 h~\implies~\boxed{h=\frac{300}{\pi r^2}}$
$A=2 \pi r^2 + 2 \pi r h$
Let C denote the costs:
$C= 0.08 \cdot 2 \pi r^2 + 0.04 \cdot 2 \pi r h$
Substitute h by term calculated above and you'll get a function dependent on r:
$C(r)= 0.08 \cdot 2 \pi r^2 + 0.04 \cdot 2 \pi r \cdot \frac{300}{\pi r^2}$
$C(r)= 0.16 \cdot \pi r^2 + \frac{24}{ r}$
Calculate the first derivative and solve C'(r) = 0 for r
I've got $r = \sqrt[3]{\frac{24}{0.32 \cdot \pi}}\approx 2.879$
Plug in this value into the equation of h.
• May 30th 2008, 12:58 AM
eawolbert
thanks a lot really helped me.(Nod) |
# Area of Triangle and Quadrilateral
Perimeter and Area of Geometrical figures
Rectangle:
The perimeter of rectangle = 2(l + b).
Area of rectangle = l × b; (l and b are the length and breadth of rectangle)
Diagonal of rectangle = $\sqrt {{{\rm{l}}^2}{\rm{\: }} + {\rm{\: }}{{\rm{b}}^2}}$
Square:
Perimeter of square = 4 × S.
Area of square = S × S.
Diagonal of square = S√2; (S is the side of square)
Triangle:
Perimeter of triangle = (a + b + c); (a, b, c are 3 sides of a triangle)
Area of triangle = $\sqrt {{\rm{s}}\left( {{\rm{s\: }} - {\rm{\: a}}} \right)\left( {{\rm{s\: }} - {\rm{\: b}}} \right)\left( {{\rm{s\: }} - {\rm{\: c}}} \right)} ;$ (s is the semi-perimeter of triangle)
S = $\frac{1}{2}$ (a + b + c)
Area of triangle = $frac{1}{2}$× b × h; (b base , h height)
Area of an equilateral triangle = $\frac{{{{\rm{a}}^2}\sqrt 3 }}{4};$ (a is the side of triangle)
Parallelogram:
Perimeter of parallelogram = 2 (sum of adjacent sides)
Area of parallelogram = base × height
Rhombus:
Area of rhombus = base × height
Area of rhombus = $frac{1}{2}$× length of one diagonal × length of other diagonal
Perimeter of rhombus = 4 × side
Trapezium:
Area of trapezium =$frac{1}{2}$(sum of parallel sides) × (perpendicular distance between them)
= $frac{1}{2}$☐(p1 + p2) × h (p1, pare two parallel sides)
Circle:
Circumference of circle = 2πr
= πd
Where, π = 3.14 or π = 22/7
r is the radius of circle
d is the diameter of circle
Area of circle = πr2
Area of ring = Area of outer circle - Area of inner circle.
Theorems
Parallelograms standing on the same base and between the same parallel lines are equal in area.
Given: ☐PQRS and ☐AQRB are standing on the same base and between the same paralleled lines
TO prove: ☐PQRS = ☐AQRB
S. NO Statement Reason 1. 2. 3. 4. 5. 6. In Δ PQAand ΔSRB ∠ PAQ=∠ AQR =∠ SRB(A) ∠ APQ=∠ SRQ =∠ RSB(A) AQ= BR (S) Δ PQA≃ΔSRB Δ PQA+ AQRS=AQRS +ΔSRB ☐PQRS = ☐AQRB Being an alternate angleand opposite angle of parallelogram Being an alternate angleand opposite angle of parallelogram Opposite side of the parallelogram BYA.A. Saxiom Adding common part From 5
The area of the triangle is equal to the area of the parallelograms, both standing on the same base and between the sane parallel lines
Given:Δ TQR and☐PQRS are standing on the same base and between the same parallel linesPT and RQ
TO prove: Δ TQR == $frac{1}{2}$☐ PQRS
Construction:join QU
S.No Statement Reason 1. 2. 3. 4. RQUTis a parallelogram Δ TQR =$frac{1}{2}$☐TRQU ☐TRQU = ☐QRSP Δ TQR = $frac{1}{2}$☐TRQU By constructionQU || TR The diagonal bisects the parallelogram. Parallelogram standing on the same base and between the same parallel lines are equal in area From 2 and 3
Examples 1
ABCD and PQRD are the two parallelograms
Prove that:
☐ABCD =☐PQRD
Given: ABCD and PQRD are parallelogram
To prove:☐ABCD = ☐PQRD
S.NO Statement Reason 1. 2. 3. Δ DPC=$frac{1}{2}$☐ ABCD Δ DPC=$frac{1}{2}$☐PQRD ☐ ABCD = ☐PQRD 1. Triangle and the parallelogram standing on the same base and between the same parallel lines. 2. Triangle and the parallelogram standing on the same base and between the same parallel lines. From2 and 3
Examples 2
In the given figure, ABCD is the parallelogram X is the any point within it .Prove that he sum of the area of XCD and XCB is equal to half of the area of ABCD .
Given: ABCD is a parallelogram X is at any point in centre.
To prove:Δ XCD + Δ XAB =$\frac{1}{2}{\rm{\: }}$parallelogram ABCD
Construction: Draw MN || AB
S.NO Statement Reason 1. 2. 3. 4. MNCD is a parallelogram ABNM is a parallelogram Δ XCD =$frac{1}{2}$MNCD Δ XAB= $frac{1}{2}$ABNM AB ||MNand AB ||CD AB ||MNand AB ||CD Triangle and the parallelogram standing on the same base and between same parallel lines Triangle and the parallelogram standing on the same base and between same parallel lines
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# How to graph a parabola y=1/4x^2?
May 12, 2015
1. The curve will be curved along the axis which has less power.
Well,if you are concerned with this question, firstly you can write it as $4 y = {x}^{2}$,
Now as i said earlier the curve will be bent towards that axis which has less power irrespective of the coefficient of x and y .
About this question, you can see that y has less power so the curve will be bent towards y axis.the graph will be as follows,
graph{x^2/4 [-10, 10, -5, 5]}
Here's another method you could use.
You can put different values of x keeping y constant and note the points that you will get; similarly, you can find values of y keeping x constant.
You can plot these points on the graph and the same graph would be in front of you.
Jun 23, 2016
Choose x-values, work out the y-values and plot points.
Join them with a smooth curve to obtain the graph.
#### Explanation:
The first thing is to have some points to plot to be able to find out what the graph will look like.
Draw a table and select some suitable x values.
$x : - 4 , - 3 , - 2 , - 1 , - 1 , \text{ " 2," " 3," } 4$
For each x-value, substitute into the equation given and calculate the y-values,
$y : 4 , \text{ "2.25," " 1," " 0.25," " 0," " 0.25, " " 1, " " 2.25," } 4$
Notice that the $y$-values are all positive and are symmetrical on either side of $0$
Now plotting the points on a set of axes will give the graph of the parabola.
graph{y=1/4x^2 [-9.42, 10.58, -1.64, 8.36]} |
# Fermat's factorization method
Fermat's factorization method
Fermat's factorization method is based on the representation of an odd integer as the difference of two squares: :$N = a^2 - b^2.$ That difference is algebraically factorable as $\left(a+b\right)\left(a-b\right)$; if neither factor equals one, it is a proper factorization of "N".
Each odd number has such a representation. Indeed, if $N=cd$ is a factorization of "N", then :$N = \left[\left(c+d\right)/2\right] ^2 - \left[\left(c-d\right)/2\right] ^2.$ Since "N" is odd, then "c" and "d" are also odd, so those halves are integers. (A multiple of four is also a difference of squares: let "c" and "d" be even.)
In its simplest form, Fermat's method might be even slower than trial division (worst case). Nonetheless, the combination of trial division and Fermat's is more effective than either.
The basic method
One tries various values of "a", hoping that $a^2-N = b^2$ is a square.
:FermatFactor(N): // N should be odd::A ← ceil(sqrt(N))::Bsq ← A*A - N::while Bsq isn't a square::::A ← A + 1:::Bsq ← A*A - N // equivalently: Bsq ← Bsq + 2*A + 1::endwhile::return A - sqrt(Bsq) // or A + sqrt(Bsq)
For example, to factor $N=5959$, one computes
A: 78 79 80 Bsq: 125 282 441
The third try produces a square. $A=80$, $B=21$, and the factors are $A-B = 59$, and $A+B = 101$.
Suppose N has more than two prime factors. That procedure first finds the factorization with the least values of "a" and "b". That is, $a+b$ is the smallest factor ≥ the square-root of "N". And so $a-b = N/\left(a+b\right)$ is the largest factor ≤ root-"N". If the procedure finds $N=1*N$, that shows that "N" is prime.
For $N=cd$, let "c" be the largest subroot factor. $a = \left(c+d\right)/2$, so the number of steps is approximately $\left(c+d\right)/2 - sqrt N = \left(sqrt d - sqrt c\right)^2 / 2 = \left(sqrt N - c\right)^2 / 2c$.
If "N" is prime (so that $c=1$), one needs $O\left(N\right)$ steps! This is a bad way to prove primality. But if "N" has a factor close to its square-root, the method works quickly. More precisely, if c differs less than $\left\{left\left(4N ight\right)\right\}^\left\{1/4\right\}$ from $sqrt N$ the method requires only one step. Note, that this is independent of the size of N.
Fermat's and trial division
Let's try to factor the prime number N=2345678917, but also compute B and A-B throughout. Going up from $sqrt\left\{N\right\}$, we can tabulate:
A: 48433 48434 48435 48436 B2: 76572 173439 270308 367179 B: 276.7 416.5 519.9 605.9 A-B: 48156.3 48017.5 47915.1 47830.1
In practice, one wouldn't bother with that last row, until "B" is an integer. But observe that if "N" had a subroot factor above $A-B=47830.1$, Fermat's method would have found it already.
Trial division would normally try up to 48432; but after only four Fermat steps, we need only divide up to 47830, to find a factor or prove primality.
This all suggests a combined factoring method. Choose some bound $c > sqrt\left\{N\right\}$; use Fermat for factors between $sqrt\left\{N\right\}$ and $c$. This gives a bound for trial division which is $c - sqrt\left\{c^2 - N\right\}$. In the above example, with $c = 48436$ the bound for trial division is 47830. A reasonable choice could be $c = 55000$ giving a bound of 28937.
In this regard, Fermat's method gives diminishing returns. One would surely stop before this point:
A: 60001 60002 B2: 1.25444e+09 1.25456e+09 B: 35418.1 35419.8 A-B: 24582.9 24582.2
ieve improvement
One needn't compute all the square-roots of $a^2-N$, nor even examine all the values for $a$. Examine the tableau for $N=2345678917$:
A: 48433 48434 48435 48436 Bsq: 76572 173439 270308 367179 B: 276.7 416.5 519.9 605.9
One can quickly tell that none of these values of Bsq are squares. Squares end with 0, 1, 4, 5, 9, or 16 modulo 20. The values repeat with each increase of $a$ by 10. For this example $a^2-N$ produces 3, 4, 7, 8, 12, and 19 modulo 20 for these values. It is apparent that only the 4 from this list can be a square. Thus, $a^2$ must be 1 mod 20, which means that $a$ is 1 or 9 mod 10; it will produce a Bsq which ends in 4 mod 20, and if Bsq is a square, $b$ will end in 2 or 8 mod 10.
This can be performed with any modulus. Using the same $N=2345678917$,
modulo 16: Squares are 0, 1, 4, or 9 N mod 16 is 5 so $a^2$ can only be 9 and $a$ must be 3 or 5 modulo 16 modulo 9: Squares are 0, 1, 4, or 7 N mod 9 is 7 so $a^2$ can only be 7 and $a$ must be 4 or 5 modulo 9
One generally chooses a power of a different prime for each modulus.
Given a sequence of "a"-values (start, end, and step) and a modulus, one can proceed thus:
:FermatSieve(N, Astart, Aend, Astep, Modulus)::A ← Astart::do Modulus times::::Bsq ← A*A - N:::if Bsq is a square, modulo Modulus:::::FermatSieve(N, A, Aend, Astep * Modulus, NextModulus):::endif:::A ← A + Astep::enddo
But one stops the recursion, when few "a"-values remain; that is, when (Aend-Astart)/Astep is small. Also, because "a"'s step-size is constant, one can compute successive Bsq's with additions.
Multiplier improvement
Fermat's method works best when there is a factor near the square-root of "N". Perhaps one can arrange for that to happen.
If one knows the approximate ratio of two factors ($d/c$), then one can pick a rational number $v/u$ near that value. $Nuv = cv * du$, and the factors are roughly equal: Fermat's, applied to "Nuv", will find them quickly. Then $gcd\left(N,cv\right)=c$ and $gcd\left(N,du\right)=d$. (Unless "c" divides "u" or "d" divides "v".)
Generally, one does not know the ratio, but one can try various $u/v$ values, and try to factor each resulting "Nuv". R. Lehman devised a systematic way to do this, so that Fermat's plus trial-division can factor N in $O\left(N^\left\{1/3\right\}\right)$ time.See R. Lehman, "Factoring Large Integers", "Mathematics of Computation", 28:637-646, 1974.
Other improvements
The fundamental ideas of Fermat's factorization method are the basis of the quadratic sieve and general number field sieve, the best-known algorithms for factoring "worst-case" large semiprimes. The primary improvement that quadratic sieve makes over Fermat's factorization method is that instead of simply finding a square in the sequence of "a"2−"n", it finds a subset of elements of this sequence whose "product" is a square, and it does this in a highly efficient manner. The end result is the same: a difference of square mod "n" that, if nontrivial, can be used to factor "n".
See also J. McKee, " [http://www.ams.org/mcom/1999-68-228/S0025-5718-99-01133-3/home.html Speeding Fermat's factoring method] ", "Mathematics of Computation", 68:1729-1737 (1999).
* [http://www.patrickkonsor.com/code/ Java Implementation Of Fermat's method and trial division]
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### Look at other dictionaries:
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• Méthode De Factorisation De Fermat — En arithmétique modulaire, la méthode de factorisation de Fermat est un algorithme de décomposition en produit de facteurs premiers. Sommaire 1 Intuition 2 Méthode de base 2.1 Analyse … Wikipédia en Français
• Méthode de factorisation de fermat — En arithmétique modulaire, la méthode de factorisation de Fermat est un algorithme de décomposition en produit de facteurs premiers. Sommaire 1 Intuition 2 Méthode de base 2.1 Analyse … Wikipédia en Français |
# What is the general solution of the differential equation (d^2y)/(dx^2) + 3dy/dx + 2y = 2sinx?
Jul 21, 2017
$y \left(x\right) = A {e}^{- x} + B {e}^{- 2 x} - \frac{3}{5} \cos x + \frac{1}{5} \sin x$
#### Explanation:
We have:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 3 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y = 2 \sin x$ ..... [A]
This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, ${y}_{c}$ of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, ${y}_{p}$ of the non-homogeneous equation.
Complementary Function
The homogeneous equation associated with [A] is
$y ' ' + 3 y ' + 2 y = 0$
And it's associated Auxiliary equation is:
${m}^{2} + 3 m + 2 = 0$
$\left(m + 1\right) \left(m + 2\right) = 0$
Which has two real and distinct solutions $m = - 1 , - 2$
Thus the solution of the homogeneous equation is:
${y}_{c} = A {e}^{- 1 x} + B {e}^{- 2 x}$
$\setminus \setminus \setminus = A {e}^{- x} + B {e}^{- 2 x}$
Particular Solution
With this particular equation [A], a probably solution is of the form:
$y = a \cos \left(x\right) + b \sin \left(x\right)$
Where $a$ and $b$ are constants to be determined by substitution
Let us assume the above solution works, in which case be differentiating wrt $x$ we have:
$y ' \setminus \setminus = - a \sin \left(x\right) + b \cos \left(x\right)$
$y ' ' = - a \cos \left(x\right) - b \sin \left(x\right)$
Substituting into the initial Differential Equation $\left[A\right]$ we get:
$- a \cos \left(x\right) - b \sin \left(x\right) + 3 \left\{- a \sin \left(x\right) + b \cos \left(x\right)\right\} + 2 \left\{a \cos \left(x\right) + b \sin \left(x\right)\right\} = 2 \sin \left(x\right)$
$- a \cos \left(x\right) - b \sin \left(x\right) - 3 a \sin \left(x\right) + 3 b \cos \left(x\right) + 2 a \cos \left(x\right) + 2 b \sin \left(x\right) = 2 \sin \left(x\right)$
Equating coefficients of $\cos \left(x\right)$ and $\sin \left(x\right)$ we get:
$\cos \left(x\right) : - a + 3 b + 2 a = 0 \implies a + 3 b = 0$
$\sin \left(x\right) : - b - 3 a + 2 b = 2 \setminus \implies b - 3 a = 2$
Solving simultaneously we get:
$a = - \frac{3}{5}$ and $b = \frac{1}{5}$
And so we form the Particular solution:
${y}_{p} = - \frac{3}{5} \cos x + \frac{1}{5} \sin x$
General Solution
Which then leads to the GS of [A}
$y \left(x\right) = {y}_{c} + {y}_{p}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = A {e}^{- x} + B {e}^{- 2 x} - \frac{3}{5} \cos x + \frac{1}{5} \sin x$ |
# Thornton & Marion, Classical Dynamics, 5th Edition
## Chapter 1. Matrices, Vectors, and Vector Calculus
### Problem 02. Trigonometric properties of direction cosines
#### Solution
For Equation (1.10): $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ Proof)
Using Figure 1-4 (a) in the book, let the length of a point P on the line $(\alpha, \beta, \gamma)$ from the origin be $l$, then
$l^2\cos^2\alpha + l^2\cos^2\beta + l^2\cos^2\gamma = l^2$
Therefore, it becomes
$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
For Equation (1.11): $\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma' = \cos\theta$
In Figure 1-4 (b) in the book, let the length of a point P on the line $(\alpha, \beta, \gamma)$ from the origin be $l$ and the length of a point P’ on the line $(\alpha', \beta', \gamma')$ be $l'$.
$l = \sqrt{l^2\cos^2\alpha + l^2\cos^2\beta + l^2\cos^2\gamma}$
$l' = \sqrt{l'^2\cos^2\alpha' + l'^2\cos^2\beta' + l'^2\cos^2\gamma'}$
Using the law of cosines,
$\overline{PP'}^2 = l^2 + l'^2 - 2 l l' \cos\theta$
Since $\begin{array}{rcl} \overline{PP'}^2 & = & (l\cos\alpha - l'\cos\alpha')^2 + (l\cos\beta - l'\cos\beta')^2 + (l\cos\gamma - l'\cos\gamma')^2 \\ & = & l^2(\cos^2\alpha + \cos^2\beta + \cos^2\gamma) + l'^2(\cos^2\alpha' + \cos^2\beta' + \cos^2\gamma') - 2 l l' (\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma') \\ & = & l^2 + l'^2 - 2 l l' (\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma') \end{array}$ the equation becomes
$l^2 + l'^2 - 2 l l' (\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma') = l^2 + l'^2 - 2 l l' \cos\theta$
Therefore,
$\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma' = \cos\theta$
#### Reference
https://en.wikipedia.org/wiki/Law_of_cosines |
# Time and Work Tricks – 1
0
15
Time and Work is yet another easy topic and almost all the questions are predictable. Please go through the following solved examples and I am sure that 90% questions would be similar to these examples. Most probably I will complete Time and Work in 3 parts.
Note: In the complete Time and Work series, Efficiency would mean “Work Done in 1 day”, and efficiency has been denoted by small letters, e.g. “a” means “Efficiency of A”.
1. 1)
Let the total work be 8 units (because 8 is the LCM of 4 and 8)
Efficiency of x (Work done by x in 1 hour) = 8/4 = 2 units
Efficiency of y (Work done by y in 1 hour) = 8/8 = 1 unit
Work done by (x + y) in 1 hour = 3 units
3 units work in completed in 1 hour. Hence 8 units work will be completed in 8/3 hours or 160 minutes.
1. 2)
Let total work be 60 units (LCM of 10 and 12)
Raj completes the work in 12 days. Hence efficiency or per day work of Raj = 60/12 = 5 units
Raj and Ram take 10 days to complete the work, hence their efficiency = 60/10 = 6 units
Now Efficiency of Ram = (Efficiency of Raj and Ram) – (Efficiency of Raj) = 6 – 5 = 1 unit
That means Raj completes 1 unit of work per day. So to perform 60 units of work, he will take 60 days.
1. 3)
Let total work = 120 units
Efficiency of A + B = 120/15 = 8 units
Efficiency of B + C = 120/12 = 10 units
Efficiency of C + A = 120/10 = 12 units
Adding all the above 3 equations –
2 * (A + B + C) = 30
Efficiency of (A + B + C) = 15 units
Efficiency of B + C = 120/12 = 10 units
Hence Efficiency of A = Efficiency of (A + B + C) – Efficiency of (B + C) = 15 – 10 = 5 units
Hence time taken by A to do 120 units of work = 120/5 = 24 days
1. 4)
Let the total work be 16 units.
Efficiency of first pipe = 16/4 = +4 units
Efficiency of second pipe = 16/16 = -1 units [negative sign because this pipe is emptying the tank]
When both the pipes are opened together, their efficiency = (+4) + (-1) = +3 units [The positive sign indicates that when both the pipes are opened together, their net result will fill the tank]
3 units of work is done in 1 hour
16 units of work is done in 16/3 hours
Note : In questions where one pipe is emptying the cistern while another is filling it, you must put a positive or negative sign before the efficiency. But in questions where both the pipes are emptying the cistern or both the pipes are filling the cistern, you can take the efficiency of both the pipes as positive.
1. 5)
Let the total work be 15 units.
Efficiency of first pipe = 15/3 = +5 units
Efficiency of second pipe = 15/3.75 = +4 units
Efficiency of third pipe = 15/1 = -15 units
Efficiency of all the three pipes = 5 + 4 – 15 = -6 units
If all the pipes are opened, it will take 15/6 or 5/2 hours to empty the cistern, but the cistern is already half empty, hence only 5/4 hours are required to empty it.
1. 6)
Let the total work = 60 units
Efficiency of A = 60/20 = +3 units
Efficiency of B = 60/30 = -2 units
Now total work to be performed is 60 units. When 57 units work is complete, A will take 1 more minute to add 3 units and hence will make it a total of 60 units.
Hence time taken to fill the tank = Time taken to perform 57 units of work + 1 minute
Now A and B are opened alternatively. That means for the first minute only A is opened, for the second minute A is closed and B is opened, then for third minute again B is closed and A is opened and so on.
So for each 2 minutes cycle, work done = Efficiency of A + Efficiency of B = +3 + (-2) = 1 unit
1 unit work is done in 2 minutes, so 57 units work is done in 114 minutes
Time taken to fill the tank = 114 + 1 = 115 minutes
Explanation : We have to perform a total of 60 units of work. For the 1st minute – A adds 3 units of work, but in the 2nd minute, B adds (-2) units of work and hence makes total work for 2 minutes = (+3) + (-2) = 1 unit. So effectively in 2 minutes, we are just adding 1 unit of work. Hence in 4 minutes, 2 units of work will be performed and in 6 minutes 3 units of work will be performed. Same sequence will continue till 57 units. As soon as 57 units of work is done (in 114 minutes), it will be A’s turn to do the work. A will add 3 units of work(in 1 minute) and hence take the total work from 57 units to 60 units. B won’t be needed any more.
1. 7)
Let the total work be 240 units.
40 men complete the work in 6 months. Hence 10 men can complete the work in 6*4 = 24 months. Hence, Efficiency of 10 men = 240/24 = 10 units
60 women complete the work in 6 months. Hence 10 women can complete the work in 6*6 = 36 months. Hence, Efficiency of 10 women = 240/36 = 20/3 units
80 boys complete the work in 6 months. Hence 10 boys can complete the work in 6*8 = 48 months. Hence, Efficiency of 10 boys = 240/48 = 5 units
Efficiency of 10 men + Efficiency of 10 women + Efficiency of 10 boys = 10 + 20/3 + 5 = 65/3 units
So, 10 men, 10 women and 10 boys complete 65/3 units of work in 1 month. To complete 120 units(half of the work), they will take = 120 * 3/65 = 72/13 months
1. 8)
Let the total work be 112 units and the efficiency of 1 man and 1 woman be m and w respectively
2m + w = 112/14 = 8
4w + 2m = 112/8 = 14
Solve the equations and you will get w = 2 and m = 3
Hence the wage of woman = 2/3 * 180 = Rs. 120
1. 9)
A and C complete 19/23 of the work. Hence B does 4/23 of the work
Amount paid to B = 4/23 * 575 = Rs. 100
1. 10)
Let A and B complete the work in x days
Then A will complete the work in (x + 8) days and B will complete the work in (x + 4.5) days. Now,
1/(x + 8) + 1/(x + 4.5) = 1/x
Solve the equation and you will get x = 6 hours
1. 11)
The question is same the previous one.
Let A, B and C take ‘x’ days to do the job. Then,
A takes (x + 6) days, B takes (x + 1) days and C taken 2x days
1/(x + 6) + 1/(x + 1) + 1/2x = 1/x
1/(x + 6) + 1/(x + 1) = 1/x – 1/2x
1/(x + 6) + 1/(x + 1) = 1/2x … (1)
Solve it and you will get x = 2/3
From equation (1) you can see that A and B take 2x days to complete the work |
## Vector Addition: 6 + 8 = ?
Most of us are accustomed to the following form of mathematics:
6 + 8 = 14.
Yet, we are extremely uneasy about this form of mathematics:
6 + 8 = 10
and
6 + 8 = 2
and
6 + 8 = 5.
When we become students of physics and approach the task of adding vector quantities, we soon become aware of the fact that the addition of two vector quantities with magnitudes of 6 and 8 will not necessarily result in an answer of 14. The rules for adding vector quantities are different than the rules for adding two quantities arithmetically. Thus, vectors with magnitudes of 6 + 8 will not necessarily sum to 14.
Vectors are quantities which include a direction. As such, the addition of two or more vectors must take into account that the quantities being added have a directional characteristic. There are a number of methods for carrying out the addition of two (or more) vectors. The most common method is the head-to-tail method of vector addition. Using such a method, the first vector is drawn to scale in the appropriate direction. The second vector is then drawn such that its tail is positioned at the head (vector arrow) of the first vector. The sum of two such vectors is then represented by a third vector which stretches from the tail of the first vector to the head of the second vector. This third vector is known as the resultant - it is the result of adding the two vectors. The resultant is the vector sum of the two individual vectors. Of course, the actual magnitude and direction of the resultant is dependent upon the direction which the two individual vectors have.
This principle of the head-to-tail addition of vectors is illustrated in the animation below. In each frame of the animation, a vector with magnitude of 6 (in green) is added to a vector with magnitude of 8 (in blue). The resultant is depicted by a black vector which stretches from the tail of the first vector (8 units) to the head of the second vector (6 units).
As can be seen from this animation, 8 + 6 could be equal to 14, but only if the two vectors are directed in the same direction. All that can be said for certain is that 8 + 6 can add up to a vector with a maximum magnitude of 14 and a minimum magnitude of 2. The maximum is obtained when the two vectors are directed in the same direction. The minimum s obtained when the two vectors are directed in the opposite direction.
For more information on physical descriptions of motion, visit The Physics Classroom Tutorial. Detailed information is available there on the following topics:
Vectors and Direction
Vector Addition
Resultants
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Chapter 5: Exponential and Logarithmic Functions 5.4: Common and Natural Logarithmic Functions
# Chapter 5: Exponential and Logarithmic Functions 5.4: Common and Natural Logarithmic Functions
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## Chapter 5: Exponential and Logarithmic Functions 5.4: Common and Natural Logarithmic Functions
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##### Presentation Transcript
1. Chapter 5: Exponential and Logarithmic Functions5.4: Common and Natural Logarithmic Functions Essential Question: What is the relationship between a logarithm and an exponent?
2. 5.4: Common and Natural Logarithmic Functions • You’ve ran across a multitude of inverses in mathematics so far... • Additive Inverses: 3 & -3 • Multiplicative Inverses: 2 & ½ • Inverse of powers: x4 & or x¼ • But what do you do when the exponent is unknown? For example, how would you solve 3x = 28, other than guess & check? • Welcome to logs…
3. 5.4: Common and Natural Logarithmic Functions • Logs • There are three types of commonly used logs • Common logarithms (base 10) • Natural logarithms (base e) • Binary logarithms (base 2) • We’re only going to concentrate on the first two types of logarithms, the 3rd is used primarily in computer science. • Want to take a guess as to why I used the words “base” above?
4. 5.4: Common and Natural Logarithmic Functions • Common logarithms • The functions f(x) = 10x and g(x) = log x are inverse functions • log v = u if and only if 10u = v • All logs can be thought of as a way to solve for an exponent • Log base answer = exponent x x 10 = 2 10 = 2 x log 10
5. 5.4: Common and Natural Logarithmic Functions • Common logarithms • Scientific/graphing calculators have the logarithmic tables built in, on our TI-86s, the “log” button is below the graph key. • To find the log of 29, simply type “log 29”, and you will be returned the answer 1.4624. • That means, 101.4624 = 29 • Though the calculator will give you logs to a bunch of places, round your answers to 4 decimal places
6. 5.4: Common and Natural Logarithmic Functions • Evaluating Common Logarithms • Without using a calculator, find the following • log 1000 • log 1 • log • log (-3) If log 1000 = x, then 10x = 1000. Because 103 = 1000, log 1000 = 3 If log 1 = x, then 10x = 1 Because 100 = 1, log 1 = 0 If log (-3) = x, then 10x = (-3) Because there is no real number exponent of 10 to get -3 (or any negative number, for that matter), log(-3) is undefined
7. 5.4: Common and Natural Logarithmic Functions • Using Equivalent Statements (log) • Solve each by using equivalent statements (and calculator, if necessary) • log x = 2 • 10x = 29 • Remember • Log base answer = exponent log x = 2 → 102 = x → 100 = x 10x = 29 → log 29 = x → 1.4624 = x
8. 5.4: Common and Natural Logarithmic Functions • Natural logarithms • (or Captain’s Log, star date 2.71828182846…) • Common logarithms are used when the base is 10. • Another regular base is used with exponents, that being the irrational constant e. • For natural logarithms, we use “ln” instead of “log”. The ln key is located beneath the log key on your calculator.
9. 5.4: Common and Natural Logarithmic Functions • Evaluating Natural Logarithms • Use a calculator to find each value. • ln 0.15 • ln 0.15 = -1.8971, which means e-1.8971 = 0.15 • ln 186 • ln 186 = 5.2257, which means e5.2257 = 186 • ln (-5) • Undefined, as it’s not possible for a positive number (e) to somehow yield a negative number.
10. 5.4: Common and Natural Logarithmic Functions • Using Equivalent Statements (ln) • Solve each by using equivalent statements (and calculator, if necessary) • ln x = 4 • ex = 5 • Remember • Log base answer = exponent ln x = 4 → e4 = x → 54.5982 = x ex = 5 → ln 5 = x → 1.6094 = x
11. 5.4: Common and Natural Logarithmic Functions • Assignment • Page 361, 2 – 36 (even problems) • Even problems are done exactly like the odd problems, which are in the back of the book)
12. Chapter 5: Exponential and Logarithmic Functions5.4: Common and Natural Logarithmic FunctionsDay 2 Essential Question: What is the relationship between a logarithm and an exponent?
13. 5.4: Common and Natural Logarithmic Functions • Graphs of Logarithmic Functions
14. 5.4: Common and Natural Logarithmic Functions • Transforming Logarithmic Functions • Same as before… • Changes next to the x affect the graph horizontally and opposite as would be expected • Changes away from the x affect the graph vertically and as expected • Example • Describe the transformation from the graph of g(x) = log x to the graph of h(x) = 2 log (x – 3). Give the domain and range. • Vertical stretch by a factor of 2 • Horizontal shift to the right 3 units • Domain: The domain of a log function is all positive real numbers (x > 0). Shifting three units right means the new domain is x > 3. • Range: The range of a log function is all real numbers. That doesn’t change by transforming the graph.
15. 5.4: Common and Natural Logarithmic Functions • Transforming Logarithmic Functions • Example #2 • Describe the transformation from the graph of g(x) = ln x to the graph of h(x) = ln (2 – x) - 3. Give the domain and range. • x is supposed to come first, so h(x) should be rewritten as h(x) = ln [-(x – 2)] - 3 • Horizontal reflection • Horizontal shift to the right 2 units • Vertical shift down 3 units • Domain: The domain of a log function is all positive real numbers (x > 0). The horizontal reflection flips the sign, and shifting two units right means the new domain is x < 2. • Range: The range of a log function is all real numbers. That doesn’t change by transforming the graph.
16. 5.4: Common and Natural Logarithmic Functions • Assignment • Page 361, 37 – 48 (all problems) • Problems 37 – 40 only ask to find the domain, but you may need to figure out the translation first. • Even problems are done exactly like the odd problems, which are in the back of the book) |
# IBPS RRB Quantitative Aptitude Quiz Day 3
5 Steps - 3 Clicks
# IBPS RRB Quantitative Aptitude Quiz Day 3
### Introduction
What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.
A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.
The article IBPS RRB Quantitative Aptitude Quiz Day 3 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. IBPS RRB has released IBPS RRB Officer 2019 Official Notification for Officer Scale(I, II, and II). Quantitative Aptitude plays major role to qualify examination. The article IBPS RRB Quantitative Aptitude Quiz Day 3 will assist the students to know the expected questions from Quantitative Aptitude.
### Quiz
Directions(1-5): LCM and HCF Problems
1. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
A. 123
B. 127
C. 235
D. 305
Explanation: Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.
2. Which of the following has the most number of divisors?
A. 99
B. 101
C. 176
D. 182
Explanation: 99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
3. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
A. 28
B. 32
C. 40
D. 64
Explanation: Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
The numbers are 16 and 24.
∴ Hence, required sum = (16 + 24) = 40.
4. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
A. $$\frac{55}{601}$$
B. $$\frac{601}{55}$$
C. $$\frac{11}{120}$$
D. $$\frac{120}{11}$$
Explanation: Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
∴ The required sum = $$\frac{1}{a}$$ + $$\frac{1}{b}$$
= $$\frac{a + b}{ab}$$
= $$\frac{55}{600}$$
= $$\frac{11}{120}$$
5. Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
A. 75
B. 81
C. 85
D. 89
Explanation: Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number = $$\frac{551}{29}$$ = 19, Third number = $$\frac{1073}{29}$$ = 37
∴ Required sum = (19 + 29 + 37) = 85.
Directions(1-5): Simple Interest
1. A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:
A. 5%
B. 8%
C. 12%
D. 15%
Explanation: S.I. for 3 years = Rs. (12005 – 9800) = Rs. 2205.
S.I. for 5 years = Rs. [$$\frac{2205}{3}$$ x 5]= Rs. 3675
∴ Principal = Rs. (9800 – 3675) = Rs. 6125.
Hence, rate = $$\frac{100 × 3675}{6125 × 5}$$ = 12%
2. A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6% p.a for 2 years. Find his gain in the transaction per year.
A. Rs. 112.50
B. Rs. 125
C. Rs. 225
D. Rs. 167.50
Explanation: Gain in 2 years = [ [5000 x $$\frac{25}{4}$$ x $$\frac{2}{100}$$] – [$$\frac{5000 × 4 × 2}{100}$$ ]]
= Rs. (625 – 400)
= Rs. 225.
∴ Gain in 1 year = Rs. $$\frac{225}{2}$$ = Rs. 112.50
3. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?
A. 3%
B. 4%
C. 5%
D. 6%
Explanation: TS.I. = Rs. (15500 – 12500) = Rs. 3000.
Rate = $$\frac{100 × 3000 }{12500 × 4}$$ = 60
4. A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
A. Rs. 2000
B. Rs. 10,000
C. Rs. 15,000
D. Rs. 20,000
Explanation: Principal = Rs. $$\frac{100 × 5400 }{12 × 3}$$
= Rs. 15000.
5. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
A. Rs. 4462.50
B. Rs. 8032.50
C. Rs. 8900
D. Rs. 8925
Explanation: Principal
= Rs.$$\frac{100 × 4016.25 }{9 × 5}$$
= Rs.$$\frac{401625 }{45}$$
= = Rs. 8925
Directions(1-5): Averages
1. How many marks did Tarun secure in English?
I. The average mark obtained by Tarun in four subjects including English is 240.
II. The total marks obtained by him in English and Mathematics together are 170.
III. The total marks obtained by him in Mathematics and Science together are 180.
A. I and II only
B. II and III only
C. I and III only
D. All I, II and III
Explanation: I gives, total marks in 4 subjects = (60 x 4) = 240.
II gives, E + M = 170
III gives, M + S = 180.
2. In a cricket team, the average age of eleven players in 28 years. What is the age of the captain?
I. The captain is eleven years older than the youngest player.
II. The average age of 10 players, other than the captain is 27.3 years.
III. Leaving aside the captain and the youngest player, the average ages of three groups of three players each are 25 years, 28 years and 30 years respectively.
A. Any two of the three
B. All I, II and III
C. II only or I and III only
D. II and III only
Explanation: Total age of 11 players = (28 x 11) years = 308 years.
I. C = Y + 11 C – Y = 11 …. (i)
II. Total age of 10 players (excluding captain) = (27.3 x 10) years = 273 years.
Age of captain = (308 – 273) years = 35 years.
Thus, C = 35. …. (ii)
From (i) and (ii), we get Y = 24
III. Total age of 9 players = [ (25 x 3) + (28 x 3) + (30 x 3)] years = 249 years.
C + Y = (308 – 249) = 59 …. (iii)
From (i) and (iii), we get C = 35.
Thus, II alone gives the answer.
Also, I and III together give the answer.
3. The average age of P, Q, R, and S is 30 years. How old is R?
I. The sum of ages of P and R is 60 years.
II. S is 10 years younger than R.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
Explanation: P + Q + R + S = (30 x 4) P + Q + R + S = 120 …. (i)
I. P + R = 60 …. (ii)
II. S = (R – 10) …. (iii)
From (i), (ii) and (iii), we cannot find R.
4. How many candidates were interviewed every day by the panel A out of the three panels A, B and C?
I. The three panels on average interview 15 candidates every day.
II. Out of a total of 45 candidates interviewed everyday by the three panels, the number of candidates interviewed by panel A is more by 2 than the candidates interviewed by panel c and is more by 1 than the candidates interviewed by panel B.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
Explanation: I. Total candidates interviewed by 3 panels = (15 x 3) = 45.
II. Let x candidates be interviewed by C.
Number of candidates interviewed by A = (x + 2).
Number of candidates interviewed by B = (x + 1).
x + (x + 2) + (x + 1) = 45
3x = 42
x = 14
5. What is the average age of children in the class?
I. The age of the teacher is as many years as the number of children.
II. The average age is increased by 1 year if the teacher’s age is also included.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
Explanation: Let there be x children.
I gives, age of teacher = x years.
II gives, average age of (x + 1) persons = (x + 1) years.
Teacher’s age = (x + 1) (x + 1) – $${x}^{2}$$ = ($${x}^{2}$$ + 1 + 2x) – $${x}^{2}$$ = (1 + 2x)
Thus, teacher’s age cannot be obtained.
### Exams
Competitive Exams – College Entrance Exams
PG GATE 2020 ATMA 2019
Competitive Exams – Recent Job Notifications
Category
Banking SSC Railway
Defence Police Insurance
### SP Quiz
Competitive Exams – Practice Sets
Category Quiz
Quant Aptitude Spotting Errors
Profit and Loss
Reasoning Ability Puzzles
Insurance Awareness Insurance Awareness
### GK
General Knowledge for Competitive Examinations
Topic Name of the Article
GK – World Multinational Companies Headquarters
Grammy Awards 2019
GK – India Niti Aayog Objectives for India 2022- 2023
NITI Aayog
GK – Abbreviations Insurance Domain Abbreviations
GK – Banking & Insurance African Countries Capitals Currencies
Asian Countries Capitals Currencies
GK – Science & Technology Indian Research Institutes
Ama Ghare LED Scheme |
Home | | Statistics 11th std | Measurement Scales
# Measurement Scales
There are four types of data or measurements scales called nominal, ordinal, interval and ratio.
Measurement Scales
There are four types of data or measurements scales called nominal, ordinal, interval and ratio. These measurement scale is made by Stanley Stevens.
## Nominal scales:
Nominal measurement is used to label a variable without any ordered value. For example, we can ask in a questionnaire ‘What is your gender? The answer is male or female. Here gender is a nominal variable and we associate a value 1 for male and 2 for a female.’
They are numerical for name sake only. For example, the numbers 1,2,3,4 may be used to denote a person being single, married, widowed or divorced respectively. These numbers do not share any of the properties of numbers we deal with in day to life. We cannot say 4 > 1 or 2 < 3 or 1+3 = 4 etc. The order of listings in the categories is irrelevant here. Any statistical analysis carried out with the ordering or with arithmetic operations is meaningless.
## Ordinal scales:
These data share some properties of numbers of arithmetic but not all properties.
For example, we can classify the cars as small, medium and big depending on the size.
In the ordinal scales, the order of the values is important but the differences between each one is unknown. Look at the example below.
How did you feel yesterday after our trip toVedanthangal? The answers would be:
(1) Very unhappy (2) Unhappy (3) Okay (4) Happy (5) Very happy
In each case, we know that number 5 is better than number 4 or number 3, but we don’t know how much better it is. For example, is the difference between “Okay” and “Unhappy” the same as the difference between “Very Happy” and “Happy?” In fact we cannot say anything.
Similarly, a medical practitioner can say the condition of a patient in the hospital as good, fair, serious and critical and assign numbers 1 for good, 2 for fair, 3 for serious and 4 for critical. The level of seriousness can be from 1 to 4 leading to 1 < 2 or 2 < 3 or 3 < 4. However, the value here just indicates the level of seriousness of the patient only.
## Interval scales:
In an interval scale one can also carryout numerical differences but not the multiplication and division. In other words, an interval variable has the numerical distances between any two numbers. For example, suppose we are given the following temperature readings in Fahrenheit: 60c , 65c , 88c , 105c , 115c , and 120c. It can be written that 105c> 88cor 60c < 65c which means that 105cis warmer than 88c and that 60c is colder than 65c. It can also be written that 65c– 60c = 120c - 115c because equal temperature differences are equal conveying the same amount of heat needed to increase the temperature from an object from 60c to 65c and from 115c to 125c. However it does not mean that an object with temperature 120c is twice as hot as an object with temperature 60c, though 120c divided by 60c is 2. The reason is converting the temperature in to Celsius we have:
In the above equations, it is clear from the left hand side that 120cF is twice of 60cF while the right hand side says 48.87cC is more than three times of 15.57cC . The reason for the difficulty is that the Fahrenheit and Celsius scales have artificial origins namely zeros(freezing point of centigrade measure is 0o C and the freezing point of Farenheit is 32o F) and there is no such thing as ‘no temperature.’
## Ratio scales:
Ratio scales are important when it comes to measurement scales because they tell us about the order, they tell us the exact value between units, and they also have an absolute zero–which allows for a wide range of both descriptive and inferential statistics to be applied. Good examples of ratio variables include height and weight. Ratio scales provide a wealth of possibilities when it comes to statistical analysis. These variables can be meaningfully added, subtracted, multiplied, divided). Central tendency can be measured by mean, median, or mode; Measures of dispersion, such as standard deviation and coefficient of variation can also be calculated from ratio scales.
In summary, nominal variables are used to “name,” or label a series of values. Ordinal scales provide good information about the order of choices, such as in a customer satisfaction survey. Interval scales give us the order of values plus the ability to quantify the difference between each one. Finally, Ratio scales give us the ultimate–order, interval values, plus the ability to calculate ratios since a “true zero” can be defined. The distinction made here among nominal, ordinal, interval and ratio data are very much important as these concepts used in computers for solving statistical problems using statistical packages like SPSS, SAS, R, STATA etc.,
Tags : Statistics , 11th Statistics : Chapter 1 : Scope of Statistics and Types of Data
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th Statistics : Chapter 1 : Scope of Statistics and Types of Data : Measurement Scales | Statistics |
### Theory:
Theorem
If $$p(x)$$ is a polynomial of degree $$n \geq 1$$and ‘$$a$$’ is any real number then
(i) $$p(a) = 0$$ implies $$(x - a)$$ is a factor of $$p(x)$$.
(ii) $$(x-a)$$ is a factor of $$p(x)$$ implies $$p(a) = 0$$.
Proof:
Let $$p(x)$$ be the dividend and $$(x-a)$$ be the divisor.
Division algorithm
Given any two integers $$a$$ and $$b$$ with $$a > 0$$, there exists a unique interger $$q$$ and $$r$$ such that $$b = qa + r$$ with $$0 \leq r < a$$.
By division algorithm we have, $$p(x) = q(x)(x-a) + p(a)$$ where $$q(x)$$ is the quotient and $$p(a)$$ is the remainder.
(i) Assume $$p(a) = 0$$.
To prove: $$(x - a)$$ is a factor of $$p(x)$$.
Substitute $$p(a) = 0$$ in $$p(x) = q(x)(x-a) + p(a)$$.
Then, $$p(x) = q(x)(x-a)$$
This implies, $$(x - a)$$ is a factor of $$p(x)$$.
(ii) Assume $$(x-a)$$ is a factor of $$p(x)$$.
To prove: $$p(a) = 0$$.
By assumption, we have $$p(x) = q(x)(x-a)$$, where $$q(x)$$ is some polynomial.
Substitute $$(x = a)$$ in $$p(x) = q(x)(x-a)$$.
Then, $$p(a) = q(a)(a-a)$$
$$= q(a)(0)$$
$$= 0$$
Hence, the proof.
Example:
Check whether $$x - 5$$ is the factor of $$x^{3} + x^{2} + 2x - 3$$.
Given:
The polynomial $$p(x) = x^{3} + x^{2} + 2x - 3$$.
To check:
If $$x - 5$$ is the factor of $$p(x)$$.
Solution:
Step 1: Find the zero of the polynomial $$x = a$$.
Equate $$x - 5$$ to zero and solve for $$x$$.
$$x$$ $$-$$ $$5$$ $$=$$ $$0$$
$$x$$ $$=$$ $$5$$
Step 2: Check if $$p(5) = 0$$.
By the factor theorem, for $$x - 5$$ to be a factor of $$p(x)$$ it must satisfy $$p(5) = 0$$.
Substitute $$x = 5$$ in $$p(x) = x^{3} + x^{2} + 2x - 3$$.
$$p(5) = 5^{3} + 5^{2} + 2(5) - 3$$
$$= 125 + 25 + 10 - 3$$
$$= 157$$
Here, $$p(5) \neq 0$$.
This implies that, $$x - 5$$ is not the factor of $$x^{3} + x^{2} + 2x - 3$$. |
# What is the slope of any line perpendicular to the line passing through (3,13) and (-8,17)?
May 16, 2017
write the equation in the form y=mx + b using the points (3,13) and (-8,17)
Find the slope $\frac{13 - 17}{3 + 8} = - \frac{4}{11}$
Then find the y-intercept, plug in one of the points for (x,y)
$13 = \left(- \frac{4}{11}\right) \cdot \left(3\right) + b$
Simplify
$13 = - \frac{12}{11} + b$
Solve for b, add $\frac{12}{11}$ to both sides to isolate b
$b = 14 \frac{1}{11}$
Then you get the equation
$y = - \frac{4}{11} x + 14 \frac{1}{11}$
To find a PERPENDICULAR equation
The the slope of the perpendicular equation is
Opposite Reciprocal of the original equation
So the original equation had a slope of $- \frac{4}{11}$
Find the opposite reciprocal of that slope to find the slope of the perpendicular equation
The new slope is: $\frac{11}{4}$
Then find b, by plugging in a given point so either (3,13) or (-8,17)
$17 = \left(\frac{11}{4}\right) \cdot \left(- 8\right) + b$
Simplify
$17 = - 22 + b$
Add 22 to both sides to isolate b
$b = 39$
The Perpendicular Equation is: $y = \frac{11}{4} x + 39$ |
# How do you factor ax – bx – 4a + 4b?
Mar 11, 2018
$\left(x - 4\right) \left(a - b\right)$
#### Explanation:
$a x - b x - 4 a + 4 b$
$\implies x \left(a - b\right) - 4 \left(a - b\right)$
$\implies \left(x - 4\right) \left(a - b\right)$
Mar 11, 2018
$\left(a - b\right) \left(x - 4\right)$
#### Explanation:
$a x - b x - 4 a + 4 b$
Regroup the terms
$a x - 4 a + 4 b - b x$
$\left(a x - 4 a\right) + \left(4 b - b x\right)$
$\left(a x - 4 a\right) - \left(b x - 4 b\right)$
Factorise $a$ and $b$,
$a \left(x - 4\right) - b \left(x - 4\right)$
Factorise $\left(x - 4\right)$,
$\left(a - b\right) \left(x - 4\right)$
There you have it! |
# Codesmith
Vector Rotation for Dummies
Bilgin Esme June 25, 2014 Coding
Maybe the title is not very appropriate, because the vector rotation operations are not so simple in nature. You should take a couple of engineering courses to completely understand it.
The bad news is; as a game developer, you should know your math quite well. Now, relax! You're in better hands you hoped for.
The Rotation in Standard Coordinate System and in Ours - Tap/Click to Enlarge
First of all, lets start with the coordinate system presented on the image above. On the left, you see the 2-D standard coordinate system. And on the right, we have the coordinate system used in computers in general. The only difference is that, the y-coordinate origin is on the top of the screen, not on the bottom.
As for the rotation, a positive Θ angle is counter-clockwise. And in our coordinate system of computer graphics, a positive Θ angle is simply clockwise.
The reason is that: the angle is defined by the arctangent of the division y/x. And for counter-clockwise angles yield positive arctangent values. This is only for your information, you can skip it if you don't bother with trigonometry for now.
### What is a Vector?
The definition of a vector changes insanely depending in which discipline you're in. We can define simply as:
A Vector is a quantity having elements of more than one dimensions. For example, a point on 2 dimensional space with x and y coordinates is simply a vector.
In this article, from now on, you can regard a vector as a point in 2-dimensional space. But remember that a vector can be an entity even with even thousand dimensions.
Our task is to obtain a rotated version of this vector according to the origin with an angle Θ. The rotation task is done via a transformation matrix such as:
Don't be intimidated with this matrix. It's much easier than you're afraid of. If you know your rotation angle Θ, everything will be solved simply. This is the standard rotation transformation matrix for a 2D vector. Using the non-standard coordinate system doesn't mean that we should use the modified versions of these matrices. Just keep in mind that, everything is same, only the y values are visually upside-down (not negative).
If you wonder, how this rotation matrix is derived, you can refer to this article. In fact, it's simply putting together all the trigonometric relations between the two coordinates (x and y) of the two vectors (initial and rotated).
### Remembering the Matrix Multiplication
For those, who're not very familiar with the matrix operations, here's the matrix multiplication (the multiplication of two matrices) simply explained. This is the only matrix operation we'll need for vector rotation.
You cannot multiply all matrices with each other, the dimensions should fit. The number of columns of the first matrix, should be equal to the number of rows of the second matrix. In our example, we have a 2x2 transformation matrix, and a 2x1 vector. Which is OK to multiply. After this multiplication, we'll get a 2x1 vector. And remember that, matrix dimensions are displayed as rows x columns
The method is simple, you multiply the row of the matrix, with the column of the vector.
Now, lets use this method on our rotation matrix:
Here, on the left-hand side of the equation, x' and y' are the coordinates of the rotated vector. The first thing you see on the right-hand side of the equation is our beloved rotation matrix; and the rightmost entity is the original vector with x and y coordinates.
The equations that yields the rotated vector (x', y' ) are as follows:
### An Example
Lets go on with a simple example. As you see on the figure below, let's find the rotated value of the vector (564, 128) with an angle of 20o, with respect to the coordinate origin.
A Simple Vector Rotation Example - Tap/Click to Enlarge
The angle Θ is 20o; so cosΘ is 0.94 and sinΘ is 0.34 approximately. Now let's put all these values into our matrix operation.
From this matrix operation, we have two equations as I explained before. From now on, all the work is simple arithmetics.
Finally, ve obtain the rotated vector as (486.64, 312.08).
As you noticed, the point (564, 128) is in fact (564, -128) according to the standard coordinate system. But we're using the non-standard coordinate system of computer graphics.
### The Code
After all these mind-boggling matrices and equations, the code to rotate a vector is ridiculously simple. This is why most of the developers just prefer using the code, without learning the science beneath. This approach works most of the time, but does not help to create your own geometrical tricks, or to solve complicated problems.
If you don't understand the math beneath fully, you can't solve even the derivative problems; such as rotating a point around another point.
``` ```
public static Vector2 GetRotatedVector (Vector2 pos, float theta)
{
float cosTheta = (float) Math.Cos(theta);
float sinTheta = (float) Math.Sin(theta);
float x = pos.X * cosTheta - pos.Y * sinTheta;
float y = pos.X * sinTheta + pos.Y * cosTheta;
return new Vector2(x, y);
}
```
```
This code is written in C# and based on Windows XNA Game framework. Besides coding for PC, XBox and Windows Phone; you can write iOS and Android games by using Xamarin Mono. Even if you use other coding languages and platforms, more or less, everything's nearly the same.
### A Final Word
Learning trigonometry and linear algebra helps you to achieve wonders in your game. It may be hard at the beginning, but it will pay off eventually. Vector rotation is the first door opening to this magical world of wonders.
As further study, I recommend you to cover affine transformations, such as shearing, reflecting and projection; which you can start learning via Wikipedia: Transformation matrix.
Good luck with the magical tricks that you imagine to implement. |
# Square inside a triangle area
##### 2019-12-13 11:30
Feb 24, 2017 Inscribed Square. The perimeter of square S is 40. Square T is inscribed in square S. If we were to calculate the area of the triangles, we would see that this area is smaller than when T has side lengths of 5sqrt(2). And that makes sense, since T is larger and takes up a greater portion of the area of S. Now, we want to draw square TIntuition for why the area of a triangle is A21bh. Whoa! You made a rectangle that's twice as big as the triangle! The area of the rectangle is bh4520 square units, so the area of the triangle is 21bh square units. Key intuition: A triangle is half as big as the rectangle that surrounds it, which is why the area of a triangle is onehalf base times height. square inside a triangle area
2 Answers. Thus, the area of the triangle is half the area of the square. In fact, the top vertex of the triangle could be anywhere on the straightline containing the top edge of the square, even beyond the edges of the square, and the triangle would still have base, height, and area. In the figure below, for instance, triangles,
Square feet are widely used to measure area in the United States and a few other countries. While an area defined by a triangle can be calculated in a number of ways, the Heron's Theorem (formula) allows you a straightforward computation of the triangles area. Given the sides of the triangle, find the length x of the side of the square. Solution to Problem: The sum of the areas of triangles BEC and BEA is equal to the total area of the right trianglesquare inside a triangle area Nov 18, 2014 Where alpha is the small angle in the 345 triangle (about 36. 87 deg). Solving, we get theta is approx. 14. 0362 degrees. From this the side length of the square is found to be a 4cos(theta) 3. in close agreement with your result. and the area is about 15. 0588
Rating: 4.86 / Views: 777
## Square inside a triangle area free
Squares Inscribed in a Right Triangle. 5. Suppose in the right triangle ABC the square of side length s inscribed in the right angle has an area of 441 and the square of side lenght x inscribed along the hypotenuse has an area of 440. square inside a triangle area Understanding the Square footage: The area of a triangle is the space contained within its 3 sides. To find out the area of a triangle, we need to know the length of its three sides. The sides should be measured in feet (ft) for square footage calculations and if needed, converted to inches (in), yards (yd), centimetres (cm), millimetres (mm) and metres (m). Aug 13, 2016 Well, the triangle sides are going to be x over 3, x over 3, and x over 3 as an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now it's easy to figure out an expression for the area of the square So if you add the areas of these three shapes, you get the area of the large triangle: Since we know that the area of a triangle is half the base times the height and the area of a square is the side squared, this makes: ()(4)(3) ()(x)(3 x) ()(4 x)(x) x2. |
## What are the roots of an equation?
The roots of a function are the x-intercepts. By definition, the y-coordinate of points lying on the x-axis is zero. Therefore, to find the roots of a quadratic function, we set f (x) = 0, and solve the equation, ax2 + bx + c = 0.
## What is the formula for equal roots?
For an equation ax2+bx+c = 0, b2-4ac is called the discriminant and helps in determining the nature of the roots of a quadratic equation. If b2-4ac > 0, the roots are real and distinct. If b2-4ac = 0, the roots are real and equal. If b2-4ac < 0, the roots are not real (they are complex).
## How do you find the roots of an equation by factoring?
Since the roots of a function are the points at which y = 0, we can find the roots of y = ax2 + bx + c = 0 by factoring ax2 + bx + c = 0 and solving for x. We can also find the roots of y = ax2 + bx + c = 0 using the quadratic formula, and we can find the number of roots using the discriminant.
## Are roots and zeros the same?
A zero is of a function. A root is of an equation. But, when the equation only has numbers and one variable, the ONLY appropriate term is roots. However, when looking at just a polynomial (no equation) then either term is appropriate, because they both imply making the polynomial equal to zero first.
## What is root math example?
A square root of a number is a value that, when multiplied by itself, gives the number. Example: 4 × 4 = 16, so a square root of 16 is 4. Note that (−4) × (−4) = 16 too, so −4 is also a square root of 16. The symbol is √ which always means the positive square root. Example: √36 = 6 (because 6 x 6 = 36)
## What is the sum of roots?
The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient. The product of the roots of a quadratic equation is equal to the constant term (the third term), The roots will be represented as r1 and r2.
## What are two equal roots?
When solving a quadratic equation for which the discriminant is zero (ie. b2−4ac=0) we say there are “two real and equal roots”.
## Are there two distinct real roots?
It use it to ‘discriminate’ between the roots (or solutions) of a quadratic equation. If the discriminant is greater than zero, this means that the quadratic equation has two real, distinct (different) roots. If the discriminant is equal to zero, this means that the quadratic equation has two real, identical roots.
## Is Root 2 a polynomial?
The square root of 2 is a number. A polynomial is an expression in a variably, x, consisting of sums of multiples of non-negative powers of x. Root 2 is not a polynomial .
## How do you find all roots of a polynomial?
The Rational Zeros Theorem states: If P(x) is a polynomial with integer coefficients and if is a zero of P(x) (P( ) = 0), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x). We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial.
## What is the rational root theorem equation?
The theorem states that each rational solution x = pq, written in lowest terms so that p and q are relatively prime, satisfies: p is an integer factor of the constant term a, and.
## How do you solve system of equations?
Here’s how it goes:Step 1: Solve one of the equations for one of the variables. Let’s solve the first equation for y: Step 2: Substitute that equation into the other equation, and solve for x. Step 3: Substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.
### Releated
#### Law.Of conservation of momentum equation
What is the formula for the law of conservation of momentum? It’s hard to imagine how momentum could still be conserved, but it is. The equation for conservation of momentum looks like this: total momentum before = total momentum after. pbefore = pafter. (m1v1 + m2v2)before = (m1v1 + m2v2)after. What is law of conservation […]
#### Equation of a trend line
How do you calculate a trend line? Lesson SummaryStep 1: Complete each column of the table.Step 2: Calculate the slope (m) of your trend line by dividing the total for Column 3 by the total for Column 4.Step 3: Calculate the y-intercept (b) of your trend line using the average of the slope from Step […] |
# Important Questions for Class 10 Maths Chapter 15 Probability
Important Questions for Class 10 Maths Chapter 15 Probability
### Probability Class 10 Important Questions Very Short Answer (1 Mark)
Question 1.
In a family of 3 children calculate the probability of having at least one boy. (2014OD)
Solution:
S = {bbb, bbg, ggb, ggg}
Atleast 1 boy = {bbb, bbg, ggb}
∴ P(atleast 1 boy) = $\frac{3}{4}$
Question 2.
A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant. (2015D)
Solution:
Total English alphabets = 26
Number of consonants = 21
∴ P (letter is a consonant) = $\frac{21}{26}$
Question 3.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen. (2016OD)
Solution:
S = 52
P(neither a red card nor a queen)
= 1 – P(red card or a queen)
Question 4.
A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number which is a perfect square. (2013OD)
Solution:
Total number of cards = 50 – 6 + 1 = 45
Perfect square numbers are 9, 16, 25, 36, 49, i.e.,
5 numbers
∴ P(a prefect square) = $\frac{5}{45}=\frac{1}{9}$
Question 5.
Cards marked with number 3, 4, 5, …, 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number. (2016D)
Solution:
Total no. of cards = 50 – 3 + 1 = 48
Perfect square number cards are 4, 9, 16, 25, 36, 49 i.e., 6 cards
∴ P(perfect square number) = $\frac{6}{48}=\frac{1}{8}$
Question 6.
A number is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1? (2017D)
Solution:
(-3)2 = 9; (-2)2 = 4; (-1)2 = 1; (0)2 = 0
(1)2 = 1; (2)2 = 4; (3)22 = 9
∴ P(Sq. of nos. of ≤ 1) = $\frac{3}{7}$
Question 7.
If two different dice are rolled together, calculate the probability of getting an even number on both dice. (2014D)
Solution:
Two dice can be thrown as 6 × 6 = 36 ways
Even numbers on both the dice can be obtained as
∴ P(even numbers) = $\frac{9}{36}=\frac{1}{4}$
Question 8.
Two different dices are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6. (2015OD)
Solution:
Total outcomes = 6= 62 = 36
Possible outcomes having the product of the two numbers on the top of the dice as 6 are
(3 × 2, 2 × 3, 6 × 1, 1 × 6), i.e, 4
P(Product of two nos. is 6) = $\frac{4}{36}=\frac{1}{9}$
Question 9.
The probability of selecting a rotten apple ran domly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap? (2017OD)
Solution:
∴ No. of rotten apples = 900 × 0.18 = 162
### Probability Class 10 Important Questions Short Answer-I (2 Marks)
Question 10.
A coin is tossed two times. Find the probability of getting both heads or both tails. (2011D)
Solution:
S = {HH, HT, TH, TT) = 4
P (both heads or both tails)
= P (both heads) + P (both tails)
= $\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
Question 11.
In a simultaneous toss of two coins, find the probability of getting:
Solution:
The sample space is given by
S = {HH, HT, TH, TT}
Total events n(S) = 4
(i) exactly one head = {HT, TH} = 2
P(exactly one head) = $\frac{2}{4}=\frac{1}{2}$
(ii) atmost one head = {HT, TH, TT} = 3
P(atmost one head) = $\frac{3}{4}$
Question 12.
Two coins are tossed simultaneously. Find the probability of getting at least one head. (2013OD)
Solution:
S = {HH, HT, TH, TT}, i.e., 4
∴ P (atleast one head) = $\frac{3}{4}$
Question 13.
Three coins are tossed simultaneously. Find the probability of getting exactly two heads. (2013OD)
Solution:
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8
P(exactly two heads) = $\frac{3}{8}$
Question 14.
Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail. (2014D)
Solution:
S = {HH, HT, TH, TT}, i.e., 3
P(at least one tail) = $\frac{3}{4}$
Question 15.
A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king. (2011OD)
Solution:
P (neither an ace nor a king)
= 1 – P (either an ace or a king)
= 1 – [P (an ace) + P (a king)]
Question 16.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen. (2013D)
Solution:
P (neither a king nor a queen)
= 1 – P (king or queen)
Question 17.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a jack nor an ace. (2013D)
Solution:
Total number of cards = 52
Numbers of jacks = 4
Numbers of aces = 4
Card is neither a jack nor an ace
= 52 – 4 – 4 = 44
∴ Required probability = $\frac{44}{52}=\frac{11}{13}$
Question 18.
Two dice are thrown simultaneously. Find the probability of getting a doublet. (2013OD)
Solution:
Two dice can be thrown as 6 x 6 = 36 ways
“a doublet” can be obtained by (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), i.e., 6 ways
P(a doublet) = $\frac{6}{36}=\frac{1}{6}$
Question 19.
A dice is tossed once. Find the probability of getting an even number or a multiple of 3. (2013D)
Solution:
S = {1, 2, 3, 4, 5, 6} = 6
‘an even number or multiple of 3’ are 2, 3, 4, 6, i.e., 4 numbers
∴ Required probability = $\frac{4}{6}=\frac{2}{3}$
Question 20.
Two different dice are tossed together. Find the probability. (2014OD, 2015D)
(i) that the number on each die is even.
(ii) that the sum of numbers appearing on the two dice is 5.
Solution:
Two dice can be thrown as 6 × 6 = 36 ways
(i) The probability of number on each die is even are
### Probability Class 10 Important Questions Short Answer-II (3 Marks)
Question 21.
A coin is tossed two times. Find the probability of getting at least one head. (2011D)
Solution:
S = {HH, HT, TH, TT}
Total number of ways = 4
Atleast one head = {HH, HT, TH}, i.e., 3 ways
∴ P (atleast one head) = $\frac{3}{4}$
Question 22.
A coin is tossed two times. Find the probability of getting not more than one head. (2011D)
Solution:
S = {HH, HT, TH, TT} = 4
Favourable cases are HT, TH, TT i.e., 3 cases
∴ P (not more than 1 head) = $\frac{3}{4}$
Question 23.
Three distinct coins are tossed together. Find the probability of getting (2015 D)
Solution:
Total number of possible outcomes = 21 = 23 = 8
(HHH, TIT, HHT, THH, THT, HTH, TTH, HTT)
(i) Possible outcomes of at least two heads = 4
(HHT, THH, HHH, HTH)
∴ P(at least two heads) = $\frac{4}{8}=\frac{1}{2}$
(ii) Possible outcomes of at most two heads = 7
(HHT, TTT, THH, THT, HTH, TTH, HTT)
∴ P(at most two heads) = $\frac{7}{8}$
Question 24.
Three different coins are tossed together. Find the probability of getting
(iii) at least two tails. (2016OD)
Solution:
S {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of ways = 8
= HHT, HTH, THH, i.e., 3 ways
∴ P (exactly two heads) = $\frac{3}{8}$
= HHT, HTH, THH, HHH i.e., 4 ways
∴ P (atleast two heads) = $\frac{4}{8}=\frac{1}{2}$
(iii) Atleast two tails
= TTH, THT, HTT,TTT i.e., 4 ways
∴ P (atleast two tails) = $\frac{4}{8}=\frac{1}{2}$
Question 25.
A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5. (2011OD)
Solution:
Total number of tickets = 40
‘A multiple of 5’ are
5, 10, 15, 20, … 40, i.e., 8 tickets
∴ P (A multiple of 5) = $\frac{8}{10}=\frac{1}{5}$
Question 26.
A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. (2012 D)
Solution:
Multiples of 3 & 4 are 12, 24, 36, 48, i.e., 4 nos.
P(a multiple of 3 and 4) = $\frac{4}{50}=\frac{2}{5}$
Question 27.
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is $\frac{1}{4}$ . The probability of selecting a blue ball at random from the same jar is $\frac{1}{3}$ . If the jar contains 10 orange balls, find the total number of balls in the jar. (2015OD)
Solution:
Question 28.
A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is $\frac{3}{10}$ and that of a black ball is $\frac{2}{5}$, then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag. (2015OD)
Solution:
Let W, B and R denote the White, Black and Red balls respectively,
We know, Total probability = 1
Question 29.
A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. (2017OD)
Solution:
Let the number of black balls = x
the number of white balls = 15
∴ Total number of balls = x + 15
∴ P(black ball) = 3P(White balls)
Question 30.
In a single throw of a pair of different dice, what is the probability of getting
(i) a prime number on each dice?
(ii) a total of 9 or 11? (2016D)
Solution:
Two dice can be thrown in 6 × 6 = 36 ways
(i) “a prime number on each dice” can be obtained as (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5), i.e., 9 ways.
∴ P(a prime no. on each dice) = $\frac{9}{36}=\frac{1}{4}$
(ii) “a total of 9 or 11” can be obtained as (3, 6),(6, 3),(4, 5),(5, 4) (5, 6),(6, 5).
Total ‘9’ Total ’11’ i.e., 6 ways
∴ P(a total of 9 or 11) = $\frac{6}{36}=\frac{1}{6}$
Question 31.
A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that:
(i) Ramesh will buy the selected shirt?
(ii) ‘Kewal will buy the selected shirt? (2016D)
Solution:
No. of good shirts = 88
= P(good shirts) = $\frac{88}{100}=\frac{22}{25}$
(ii) No. of shirts without major defect = 96
= P(shirts without major defect)
= $\frac{88+8}{100}=\frac{96}{100}=\frac{24}{25}$
Question 32.
A card is drawn at random from a well shuffled pack of 52 cards. Find the probability of getting: (2012 OD)
(i) a red king.
(ii) a queen or a jack.
Solution:
Cards in a pack = 52
Number of kings = 4
Number of red kings = 2
Question 33.
All red face cards are removed from a pack of playing cards. The remaining cards were well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (2015D)
(i) a red card
(ii) a face card
(iii) a card of clubs.
Solution:
Number of red face cards removed = 6
∴ Remaining cards = 52 – 6 = 46
Hence, Total no. of outcomes = 46
(i) Possible outcomes of red cards = 26 – 6 = 20
∴ P(a red card) = $\frac{20}{46}=\frac{10}{23}$
(ii) Possible outcomes of face cards = 6
∴ P(a face card) = $\frac{6}{46}=\frac{3}{23}$
(iii) Possible outcomes of card of clubs = 13
∴ P(a card of clubs) = $\frac{13}{46}$
Question 34.
From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:
(i) a black King
(ii) a card of red colour
(iii) a card of black colour (2016OD)
Solution:
Total cards in the pack = 52
Cards removed
= 2(Jacks) + 2(Queens) + 2(Kings) = 6
∴ Remaining cards = 52 – 6 = 46
(i) Number of black Kings = 2
∴ P(a black King) = $\frac{2}{46}=\frac{1}{23}$
(ii) Total red cards in the pack = 26
Red cards removed = 6
Remaining red cards = 26 – 6 = 20
∴ P(a card of red colour) = $\frac{20}{46}=\frac{10}{23}$
(iii) Total black cards in the pack = 26
∴ P(a card of black colour) = $\frac{26}{46}=\frac{13}{23}$
Question 35.
There are 100 cards in a bag on which num bers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card
(i) is divisible by 9 and is a perfect square
(ii) is a prime number greater than 80. (2016OD)
Solution:
Total cards = 100
(i) Numbers which are “Divisible by 9 and perfect squares” are 9, 36, 81, i.e., 3 .
∴ P(Divisible by 9 & perfect square) = $\frac{3}{100}$
(ii) Numbers which are “prime numbers greater than 80” are 83, 89, 97, i.e., 3
∴ P(Prime nos. > 80) = $\frac{3}{100}$
Question 36.
Two different dice are rolled together. Find the probability of getting: (2015 D)
(i) the sum of numbers on two dice to be 5.
(ii) even numbers on both dice.
Solution:
Total possible outcomes = 6n = 62 = 36
(i) The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) when the sum of numbers on two dice is 5, i.e., 4
∴ Required Probability, P(E) = $\frac{4}{36}=\frac{1}{9}$
(ii) The possible outcomes are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) for even numbers on both dice; 9
∴ Required Probability, P(E) = $\frac{9}{36}=\frac{1}{4}$
Question 37.
Two different dice are thrown together. Find the probability of:
(i) getting a number greater than 3 on each die
(ii) getting a total of 6 or 7 of the numbers on two dice (2016D)
Solution:
Two dice can be thrown in 6 x 6 = 36 ways
(i) “getting a number > 3 on each die” can be obtained as (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), i.e., 9 ways.
P(number > 3 on each die) = $\frac{9}{36}=\frac{1}{4}$
(ii) “a total of 6 or 7 can be obtained as (2, 4),(4, 2),(3, 3), (1, 6),(6, 1),(2, 5) (1, 5),(5, 1) (5, 2),(3, 4),(4, 3), i.e., 11 ways
Total 6 Total “7′
P(a total of 6 or 7) = $\frac{11}{36}$
Question 38.
Two different dice are thrown together. Find the probability that the numbers obtained.
(i) have a sum less than 6
(ii) have a product less than 16
(iii) is a doublet of odd numbers. (2017D)
Solution:
Two dice can be thrown in (6 × 6) = 36 ways
(i) Sum less than ‘6’ are (1, 1) (1, 2) (1, 3) (1, 4), (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (4, 1) i.e., 10 ways
∴ P(sum < 6) = $\frac{10}{36}=\frac{5}{18}$
(ii) Product less than ’16’ are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2,5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3,5)(4, 1) (4, 2) (4, 3) (5, 1) (5, 2) (5,3) (6, 1) (6, 2) i.e., 25 ways
∴ P(product less than 16) = $\frac{25}{36}$
(iii) (1, 1) (3, 3) (5, 5) i.e., 3 ways
∴ P(doublet of odd nos.) = $\frac{3}{36}=\frac{1}{12}$
### Probability Class 10 Important Questions Long Answer (4 Marks)
Question 39.
A survey has been done on 100 people out of which 20 use bicycles, 50 use motorbikes and 30 use cars to travel from one place to another. Find the probability of persons who use bicycles, motorbikes and cars respectively? (2011D)
Solution:
Question 40.
A box contains 35 blue, 25 white and 40 red ‘marbles. If a marble is drawn at random from the box, find the probability that the drawn marble is (i) white (ii) not blue (iii) neither white nor blue. (2012D)
Solution:
No. of blue marbles = 35
No. of white marbles = 25
No. of red marbles = 40
Total no. of marbles = 35 + 25 + 40 = 100
Question 41.
A box contains 70 cards numbered from 1 to 70. If one card is drawn at random from the box, find the probability that it bears
(i) a perfect square number.
(ii) a number divisible by 2 and 3. (2012OD)
Solution:
(i) Perfect squares upto 70 are
12, 22, …, 82 = 8
∴ P(a perfect square) = $\frac{8}{70}=\frac{4}{35}$
(ii) Numbers divisible by 2 and 3 are:
6, 12, 18, …, 66, i.e., 11 nos.
∴ P(a no. divisible by 2 and 3) = $\frac{11}{70}$
Question 42.
A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (2013D)
(i) extremely patient
(ii) extremely kind or honest.
Solution:
Extremely patient = 3
Extremely honest = 6
Extremely kind = 12 – 3 – 6 = 3
Question 43.
A box contains cards numbered 3, 5, 7, 9, …, 35, 37. A card is drawn at random from the box. Find the probability that the number on the drawn card is a prime number. (2013OD)
Solution:
Total number of cards = 18
Prime numbers are: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, i.e., 11
∴ P(Prime number) = $\frac{11}{180}$
Question 44.
Find the probability that a leap year selected at random, will contain 53 Mondays. (2013OD)
Solution:
In a leap year, total number of days = 366
∴ 366 days
= 52 complete weeks + 2 extra days
Thus, a leap year always has 52 Mondays and extra 2 days.
Extra 2 days can be,
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Let E be the event that a leap year has 53 Mondays.
∴ E = {Sun and Mon, Mon and Tues}
∴ P(E) = $\frac{2}{7}$
Question 45.
A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is:
(i) an odd number
(ii) a multiple of 5
(iii) a perfect square
(iv) an even prime number (2014D)
Solution:
Total number of cards = 49
(i) Odd numbers are 1, 3, 5, …., 49, i.e., 25
∴ P(an odd number) = $\frac{25}{49}$
(ii) ‘A multiple of 5’ numbers are 5, 10, 15, ……., 45, i.e., 9
∴ P(a multiple of 5) = $\frac{9}{49}$
(iii) “A perfect square” numbers are 1, 4, 9, …….., 49, i.e., 7
∴ P(a perfect square number) = $\frac{7}{49}=\frac{1}{7}$
(iv) “An even prime number” is 2, i.e., only one number
∴ P(an even prime number) = $\frac{1}{49}$
Question 46.
A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (2015D)
(i) divisible by 2 or 3,
(ii) a prime number.
Solution:
(i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15, 20 = 13
Total outcomes = 20
Possible outcomes = 13
∴ P(divisible by 2 or 3) = $\frac{13}{20}$
(ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8
Total Outcomes = 20
Possible outcomes = 8
∴ P(a prime number) = $\frac{8}{20}=\frac{2}{5}$
Question 47.
A bag contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is: (2015D)
(i) divisible by 3 or 5
(ii) a perfect square number.
Solution:
Total number of outcomes = 25
(i) Possible outcomes of numbers divisible by 3 or 5 in numbers 1 to 25 are (3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25) = 12
∴P(No. divisible by 3 or 5) = $\frac{12}{25}$
(ii) Possible outcomes of numbers which are a perfect square = 5, i.e., (1, 4, 9, 16, 25)
∴ P(a perfect square no.) = $\frac{5}{25}=\frac{1}{5}$
Question 48.
A game of chance consists of spinning an arrow on a 3 circular board, divided into / 4 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, …, 8 (Figure), which are equally likely outcomes. What is the probability that the arrow will point at
(i) an odd number
(ii) a number greater than 3
(iii) a number less than 9. (2016 D)
Solution:
Total numbers = 8
(i) “Odd numbers” are 1, 3, 5, 7, i.e., 4
∴ P(an odd number) = $\frac{4}{8}=\frac{1}{2}$
(ii) “nos. greater than 3” are 4, 5, 6, 7, 8, i.e., 5
∴ P(a number > 3) = $\frac{5}{8}$
(iii) “numbers less than 9” are 1, 2, 3, …8 i.e., 8
∴ P(a number < 9) = $\frac{8}{8}$ = 1
Question 49.
A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16. (2016OD)
Solution:
X can be any one of 1, 2, 3, and 4 i.e., 4 ways
Y can be any one of 1, 4, 9, and 16 i.e., 4 ways
Total no. of cases of XY = 4 × 4 = 16 ways
No. of cases, where product is less than 16 (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4), (4,1) i.e., 8 ways
∴ P (product x & y less then 16) = $\frac{8}{16}=\frac{1}{2}$
Question 50.
The probability of guessing the correct answer to a certain question is $\frac{x}{12}$ . If the probability of guessing the wrong answer is $\frac{3}{4}$ , find x. If a student copies the answer, then its probability is $\frac{2}{6}$ . If he doesn’t copy the answer, then the probability is $\frac{2y}{3}$ . Find the value of y. (2011OD)
Solution:
P (guessing) + P (guessing wrong) = 1
Question 51.
A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16. (2016OD)
Solution:
x can be any one of 1, 4, 9, or 16, i.e., 4 ways
y can be any one of 1, 2, 3, or 4, i.e., 4 ways
Total number of cases of xy = 4 × 4 = 16 ways
Number of cases, where product is more than 16
(9, 2), (9, 3), (9, 4), (16, 2), (16, 3), (16, 4), i.e., 6 ways
∴ Required probability = $\frac{6}{16}=\frac{3}{8}$
Question 52.
A game consists of tossing a coin 3 times and noting its outcome each time. Hanif wins if he gets three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. (2011D)
Solution:
The possible outcomes on tossing a coin 3 times are,
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT) = 8
Outcomes when Hanif wins = {HHH, TTT} = 2
∴ P (Hanif wins) = $\frac{2}{8}=\frac{1}{4}$
∴ P (Hanif will lose) = 1 – $\frac{1}{4}=\frac{3}{4}$
Question 53.
Two dice are rolled once. Find the probability of getting such numbers on the two dice, whose product is 12. (2011OD)
Solution:
Two dice can be thrown in 6 × 6 = 36 ways
‘Product is 12′ can be obtained as (2, 6), (6, 2), (3, 4), (4, 3), i.e., in 4 ways
∴ P(Product is 12) = $\frac{4}{36}=\frac{1}{9}$
Question 54.
A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be
(i) a blue card
(ii) not a yellow card
(iii) neither yellow nor a blue card. (2012D)
Solution:
No. of red cards = 100
No. of yellow cards = 200
No. of blue cards = 50
Total no. of cards = 100 + 200 + 50 = 350
Question 55.
Two different dice are thrown together. Find the probability that the numbers obtained have (2017OD)
(i) even sum, and
(ii) even product.
Solution:
Two dice can be thrown as 6 × 6 = 36 ways.
Question 56.
A 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) the queen of diamonds. (2012 D)
Solution:
(i) Total cards in a deck = 52
Total no. of kings = 4
Total no. of red kings = 2
Question 57.
All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is
(i) a black face card.
(ii) a red card. (2012OD)
Solution:
Total no. of cards = 52
No. of cards removed = (4 + 4 + 4) = 12
Remaining cards = 40
Question 58.
Red kings and black aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is
(i) a black face card.
(ii) a red card. (2012OD)
Solution:
Total nos. of cards = 52
Cards removed = 2 + 2 = 4
Remaining cards = 52 – 4 = 48
Question 59.
All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a:
(i) face card
(ii) red card
(iii) black card
(iv) king (2014D)
Solution:
Total number of cards = 52
Black face cards = 6
Remaining cards = 52 – 6 = 46
Question 60.
Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is: (2014D)
(i) an odd number
(ii) a perfect square number
(iii) divisible by 5
(iv) a prime number less than 20
Solution:
Total number of cards = 60 – 11 + 1 = 50
(i) Odd nos, are 11, 13, 15, 17, …. 59 = 25 no.
∴ P(an odd number) = $\frac{25}{50}=\frac{1}{2}$
(ii) Perfect square numbers are 16, 25, 36, 49 = 4 numbers
∴ P(a perfect square no.) = $\frac{4}{50}=\frac{2}{25}$
(iii) “Divisible by 5” numbers are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, = 10 numbers
∴ P(divisible by 5) = $\frac{10}{50}=\frac{2}{25}$
(iv) Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8 numbers
∴ P(a prime no. less than 20) = $\frac{8}{50}=\frac{4}{25}$
Question 61.
Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is (2014OD)
(i) a king
(ii) of red colour
(iii) a face card
(iv) a queen
Solution:
Number of red queens = 2
Number of black jacks = 2
Remaining cards = 52 – 2 – 2 = 48
Question 62.
All the red face cards are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is (2014OD)
(i) of red colour
(ii) a queen
(iii) an ace
(iv) a face card
Solution:
Total number of cards = 52
Red face cards = 6
Remaining cards = 52 – 6 = 46
Question 63.
Five cards, the ten, jack, queen, king and ace of diamonds, are well shuffled with their faces downwards. One card is then picked up at random. (2014OD)
(a) What is the probability that the drawn card is the queen?
(b) If the queen is drawn and put aside, and a second card is drawn, find the probability that the second card is (i) an ace (ii) a queen.
Solution:
(a) Total events = 5; P(queen) = $\frac{1y}{5}$
(b) Now total events = 4
(i) P (an ace) = $\frac{1}{4}$
(ii) P (a queen) = $\frac{0}{4}$ = 0 …[As there is no queen left
Question 64.
A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is (2015OD)
(i) a card of spade or an ace.
(ii) a black king.
(iii) neither a jack nor a king.
(iv) either a king or a queen.
Solution:
Total no. of outcomes = 52 |
## Slot -1 – Quantitative Aptitude – Arithmetic – Averages – In an apartment complex, the number
In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?
a) 27
b) 25
c) 26
d) 28
Solution:
Let there are n people whose ages are below 51 years. So
Sum of ages of all people = 38×(30+n)
The average age of the people whose ages are below 51 years will be maximum if average age of people aged 51 years or above will be 51 years.
So Sum of ages of these n people = 38(30+n)-51×30=38n-390
So required average age = (38n-390)/n=38-390/n
This will be maximum when n will be maximum and maximum possible value of n = 39
So maximum possible average age of the people whose ages are below 51 years = 38 –390/39=28
## Online Coaching Course for CAT Exam Preparation
a) 750+ Videos covering entire CAT syllabus
b) 2 Live Classes (online) every week for doubt clarification
c) Study Material & PDFs for practice and understanding
d) 10 Mock Tests in the latest pattern
e) Previous Year Questions solved on video
## Slot -1 – Quantitative Aptitude – Arithmetic – Averages – A CAT aspirant appears for a certain number
A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is?
Solution: let total number test taken by him = n and average score = x
So average score of last (n-10) test = x+1
Sum of scores of last n-10 test = (n-10)*(x+1) = nx – ( 10x – n + 10)
So total score of first 10 test = 10x – n + 10
20*10 = 10x – n + 10
10x – n = 190 —————1)
average score of first (n-10) test = x-1
Sum of scores of first n-10 test = (n-10)*(x-1) = nx – ( n +10x – 10)
So total score of first 10 test = ( n +10x – 10)
30*10 = ( n +10x – 10)
n + 10x = 310 —————2)
by solving eq 1 ) & eq 2)
n = 60, x = 25
so total number of test taken by him = 60
## Online Coaching Course for CAT Exam Preparation
a) 750+ Videos covering entire CAT syllabus
b) 2 Live Classes (online) every week for doubt clarification
c) Study Material & PDFs for practice and understanding
d) 10 Mock Tests in the latest pattern
e) Previous Year Questions solved on video |
# A use differentiation to find a power series representation for fx 1 7 x2
## Differentiation
Differentiation is a process of finding a power series representation for a given function. This process is useful in many mathematical and physical problems. In this article, we will discuss the process of differentiation and how it can be used to find a power series representation for a given function.
### Find the derivative of the function
The derivative of a function at a point is the limit of the ratio of the change in function values to the change in x-values as the latter approaches 0. If this limit exists, we say that the function is differentiable at this point, and we denote it by f'(x).
### Use differentiation to find a power series representation for the function
Differentiation is a process that allows us to find a power series representation for a given function. It is a crucial tool in mathematics, and allows us to find things like Taylor series and Maclaurin series. To differentiate a function, we take the derivative of each term in the function, with respect to the variable that the function is written in.
## Integration
Differentiation is the process of finding a function’s rate of change at a given point. It is the inverse of integration, which is the process of finding the area under a curve. Differentiation can be used to find a power series representation for a function. In this section, we’ll learn how to use differentiation to find a power series representation for a function.
### Find the integral of the function
To find the integral of a function, we must first find its power series representation. To do this, we use differentiation. We take the derivative of the function, and then integrate it. This gives us the power series representation for the original function.
### Use integration to find a power series representation for the function
If you’re looking for a power series representation of a function, one approach is to use integration. The basic idea is to start with the Taylor series for some function, and then to use integration to “massage” it into the form of the power series that you’re looking for.
Suppose that we want to find a power series representation for the function f(x) = (1 + x)^5. One way to do this is to start with the Taylor series for f(x), which is given by:
f(x) = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5
Now we can use integration to rearrange this into the form of a power series. First, we’ll need to introduce some new variables. Let y = f(x), and let z = xy. Then we have:
dy/dx = 5 + 30x + 40x^2 + 20x^3 + x^4
dz/dx = y+ xdy/dx
Now we can integrate both sides with respect to x:
∫ dy/dx dx = ∫ (5+30x+40x^2+20x^3+x^4) dx
y = 5x+15x^2+20x^3+12 x4+C1
∫ dz/dx dx = ∫ (y+ xdy/dx) dx
z = ∫ y dx + ∫ x dy/dx dx
z = yx- 5xx- 15xx2- 20xx3- 12xx4+ C1* C2 / 2 3 4\
=yx- 1/2* 5* x – 1/3* 15* xx – 1 20* xx – 12 xx C12 C22
/ 2 \ / 3 \ / 4 \5位数电话号码查询</p><br /><h2>Conclusion</h2><br /><p>
Differentiating both sides of the given equation, we get:
$$f'(x) = \frac{1}{7x^2 – 1}$$
Now, we can use differentiation to find a power series representation for $f(x)$.
Consider the Maclaurin series for $f(x)$:
$$f(x) = \sum_{n=0}^{\infty} f^{(n)}(a) \frac{(x-a)^n}{n!}$$
Evaluating at $a=0$, we get:
$$f(x) = \sum_{n=0}^{\infty} f^{(n)}(0) \frac{x^n}{n!}$$
Now, using our expression for $f'(x)$, we can write:
$$f(x) = \sum_{n=0}^{\infty} \frac{1}{7^{2n}}\frac{(-1)^n x^{2n}}{n!}$$ |
# The Monty Hall Problem: A Simple Visual Explanation
In an old game show titled Let’s Make a Deal, host Monty Hall presented contestants with three doors.
One of the doors contained a prize while the other two did not.
Monty would ask the contestant to choose which door they believed contained the prize.
After the contestant selected a door, Monty would then open one door that did not contain the prize.
The prize remained in one of the two unopened doors.
Monty would then ask the contestant if they would like to switch doors.
While it might be hard to believe, it turns out that switching doors at this point in the game actually gives you a higher probability of winning.
Keep reading to find out why!
## The Monty Hall Problem Explained Visually
To illustrate why switching doors gives you a higher probability of winning, consider the following scenarios where you pick door 1 first.
Scenario 1: You pick door 1 and the prize is actually behind door 1.
In this case, Monty will open either door 2 or 3 and show you that nothing is behind one of those doors. If you stay with door 1, you win.
Scenario 2: You pick door 1 and the prize is actually behind door 2.
In this case, Monty must open door 3 and show you that nothing is behind it. If you stay with door 1, you lose.
Scenario 3: You pick door 1 and the prize is actually behind door 3.
In this case, Monty must open door 2 and show you that nothing is behind it. If you stay with door 1, you lose.
These are all of the possible outcomes if you pick door 1. Notice that if you stay with door 1, you only win one-third of the time. But if you switch, you win two-thirds of the time.
The following table summarizes all of the possible scenarios in this game show along with the outcomes associated with staying and switching:
From the table we can see that you win 33% of the time when you stay, but you win 66% of the time when you switch.
Thus, switching doors increases the probability that you will win the prize.
Although this doesn’t seem to make sense intuitively, the math doesn’t lie. |
# 0.3.e,ine,det.
50 %
50 %
Education
Published on March 19, 2014
Author: m2699
Source: slideshare.net
Equations, Inequations and Determinants
Equations An equation is a mathematical statement that has two expressions separated by an equal sign. The expression on the left side of the equal sign has the same value as the expression on the right side. One or both of the expressions may contain variables. Solving an equation means manipulating the expressions and finding the value of the variables. An example might be : x=4+8 to solve this equation we would add 4 and 8 and find that x = 12. A mathematical expression can have a variable as part of the expression. If x=3, the expression 7x + 4 becomes 7 * 3 + 4 which is equal to 21 + 4 or 25. To evaluate an expression with a variable, simply substitute the value of the variable into the expression and simplify. A mathematical expression can have variables as part of the expression. If x=3, and y=5, the expression 7x + y - 4 becomes 7 * 3 +5 - 4 which is equal to 21 + 5 - 4 or 22. To evaluate an expression with two or more variables, substitute the value of the variables into the expression and simplify.
Examples 1. Solve: 7x - 7 = 42 Solution: 7x - 7 = 42 7x - 7 + 7 = 42 + 7 7x = 49 (7x) / 7 = 49 / 7 x = 7 The variable needs to be isolated. To undo subtracting 7, add 7 to both sides. Adding 7 hasn't isolated the variable, so we need to continue. To undo multiplying by 7, divide both sides by 7. 2. Solve: 5(x + 2) = 25 Solution: 5(x + 2) = 25 [5(x + 2]/5 = 25/5 x + 2 = 5 x + 2 -2 = 5 -2 x = 3 The variable needs to be isolated. To undo multiplying by 5, divide by 5 on both sides. Dividing by 5 hasn't isolated the variable, so we need to continue. To undo adding 2, subtract 2, from both sides.
Equations with Multi-variables Solve for 3x + 2y = 3 and x = 3y -10 . Solution: Replace x in the first equation with its equivalent, (3y - 10) from the second equation. 3x + 2y = 3 (Top equation. 3(3y - 10) + 2y = 3 Replaced x with (3y - 10). 9y - 30 + 2y = 3 Multiplied out. 11y = 33 Simplified. y = 3 Divide each side by 11 to get answer.) Now that y has a value, you can plug that value in either equation and find a value for x. Because the second equation has already been solved for x, it will be easier to plug 3 in for y in that equation. x = 3(3) - 10 x = 9 - 10 x = -1 . The solution is the ordered pair (- 1,3).
Inequalities Math problems containing <, >, <=, and >= are called inequalities. A solution to any inequality is any number that makes the inequality true. On many occasions, you will be asked to show the solution to an inequality by graphing it on a number line. 1. Addition Principle for Inequalities - If a > b then a + c > b + c. Example: 1. Solve: x + 3 > 6 Solution: Using the Addition Principle, add -3 to each side of the inequality. x + 3 - 3 > 6 - 3 After simplification, x > 3. 2. Multiplication Principle for Inequalities - If a >b and c is positive, then ac > bc. If a > b and c is negative, then ac < bc (notice the sign was reversed). Example: 2. Solve: -4x < .8 Solution: Using the Multiplication Principle, multiply both sides of the inequality by -.25. Then reverse the signs. -.25(-4x) > -.25(.8) x > -.2 Absolute value becomes even more complicated when dealing with equations. However, there is a theorem that tells us how to deal with equations with absolute value and complicated inequalities. 1. If X is any expression, and b any positive number, and |X| = b it is the same as |X| = b or |X|=-b. 2. If X is any expression, and b any positive number, and |X| < b it is the same as -b < X < b. 3. If X is any expression, and b any positive number, and |X| > b it is the same as X < -b, X > b. Example: 3. Solve: |5x - 4| = 11 Solution: Use the theorem stated above to rewrite the equation. |X| = b X = 5x - 4 and b = 11 5x - 4 = 11, 5x - 4 = -11 Solve each equation using the Addition Principle and the Multiplication Principle. 5x = 15, 5x = -7 x = 3, x = -(7/5)
Determinants With each square matrix corresponds just one number. This number is called the determinant of the matrix. The determinant of a matrix A is denoted det(A) or |A|. Now we'll define this correspondence. Determinant of a 1 x 1 matrix The determinant of the matrix is the element itself. Ex: det([-7]) = -7 Row and columns of the determinant If we say the i-th row of a determinant we mean the i-th row of the matrix corresponding with this determinant. If we say the i-th column of a determinant we mean the i-th column of the matrix corresponding with this determinant.
Quick References: Determinant of a 1x1 matrix The determinant of a 1x1 matrix is the element itself. Determinant of a 2x2 matrix |a b| |c d| = ad – cb Determinant of a 3x3 matrix The Sarrus rule : |a b c| |d e f| |g h i| = aei + bfg + cdh - ceg - afh – bdi Cofactor of ai,j The cofactor Ai,j is independent of the elements of the i-th row and the elements of the j-th column. The value of Ai,j = (-1)i+j.(the determinant of the sub-matrix of A, obtained from A by crossing out the i-th row and the j-th column. Example : Take the 3 x 3 matrix A = [4 5 7] [1 2 3] [2 5 6] We calculate the cofactor corresponding with the element a1,3 = 7. We delete the first row and the third column. The cofactor A1,3 = (-1)4.(5 - 4) = 1 Determinant of a nxn matrix : Choose a fixed row value i. The determinant can be calculated emanating from the i-th row. |A| = Ai,1 . ai,1 + Ai,2 . ai,2 + Ai,3 . ai,3 + ... Ai,n . ai,n Ai, j is the cofactor of ai,j. We say we have unfold the determinant following row i. Example : Take the 3 x 3 matrix A = [4 5 7] [1 2 3] [2 5 6] We unfold the determinant following row 1. The three cofactors are -3 , 0 , 1. |A| = 4.(-3) + 5.0 + 7.1 = -5. We unfold the determinant following column 3. The three cofactors are 1, -10, 3. |A| = 7.1 - 3.10 + 6.3 = -5
Properties A matrix A and its transpose have the same determinant. If we swap two columns in A, |A| changes sign. If we swap two rows in A, |A| changes sign. If we multiply a row in A by a real number r, |A| changes in r.|A| If we multiply a column in A by a real number r, |A| changes in r.|A| If a row of a determinant only consists of zeros, the determinant is 0. If a column of a determinant only consists of zeros, the determinant is 0. If a determinant has two equal rows or two equal columns, the determinant is 0. If a determinant has two proportional rows or two proportional columns, the determinant is 0. |a b c| |a b' c| |a b+b' c| |d e f|+|d e' f| = |d e+e' f| |g h i| |g h' i| |g h+h' i| A determinant does not change if we add a multiple of a row to another row. The same rule holds for columns The determinant of the identity matrix is 1. The determinant of a diagonal matrix is the product of the diagonal elements. |A|.|B| = |A.B|
Cramer’s Rule A system of n linear equations in n unknowns is called a Cramer system if and only if the matrix formed by the coefficients is regular. There is a special method to solve such a system. This method is called Cramer's rule. We'll prove the rule for a system of 3 equations in 3 unknowns, but the rule is universal. Take, / ax + by + cz = d | a'x+ b'y + c'z = d‘ (1) a"x+ b"y + c"z = d" |a b c | with |N| = |a' b' c'| |a" b" c"| Then we have x.|N| = |xa b c | |xa' b' c'| |xa" b" c"| and using the properties of determinants |xa +by +cz b c | x.|N| = |xa'+b'y+c'z b' c'| |xa"+b"y+c"z b" c"| and appealing to (1) |d b c | x.|N| = |d' b' c'| |d" b" c"| Thus, x =|d b c | |d' b' c'| / |N| (2) |d" b" c"| Similarly, y =|a d c | |a' d' c‘|/|N| (3) |a" d" c"| z = |a b d | |a' b' d'| / |N| (4) |a" b" d"| The formulas (2), (3), (4) constitute Cramer's rule. It can be proved that this solution is the only solution of (1). Example: Solve the system in x, y and z . This system has parameter p. / p x - y + z = 0 | 6 x + y - 2z = 2 px - 2y - z = 1 We find : |N| = -4p - 18 We assume that p is not -4.5. Men find : x.|N| = -5 y.|N| = -2p + 6 z.|N| = 3p + 6 So, x = 5/(4p+18) ; y = (p- 3)/(2p+9) ; z = (3p+6)/(-4p-18)
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# Balbharati solutions for Textbook for SSC Class 10 Mathematics 2 chapter 4 - Geometric Constructions [Latest edition]
Textbook page
## Chapter 4: Geometric Constructions
Practice set 4.1Practice set 4.2Problem set 4
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 2 Chapter 4 Geometric Constructions Exercise Practice set 4.1 [Page 96]
Practice set 4.1 | Q 1 | Page 96
∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that $\frac{BC}{MN} = \frac{5}{4} .$
Practice set 4.1 | Q 1 | Page 96
Δ ABC ~ Δ LMN. In Δ ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. If MN = 4.8 cm, then construct Δ ABC and Δ LMN.
Practice set 4.1 | Q 2 | Page 96
∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that $\frac{PQ}{LT} = \frac{3}{4} .$
Practice set 4.1 | Q 3 | Page 96
∆RST ~ ∆XYZ. In ∆RST, RS = 4.5 cm, ∠RST = 40°, ST = 5.7 cm Construct ∆RST and ∆XYZ, such that $\frac{RS}{XY} = \frac{3}{5} .$
Practice set 4.1 | Q 4 | Page 96
∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm. $\frac{AM}{AH} = \frac{7}{5} .$Construct ∆AHE.
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 2 Chapter 4 Geometric Constructions Exercise Practice set 4.2 [Pages 98 - 99]
Practice set 4.2 | Q 1 | Page 98
Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.
Practice set 4.2 | Q 2 | Page 98
Draw a circle of radius 2.7 cm. Draw a tangent to the circle at any point on it.
Practice set 4.2 | Q 3 | Page 98
Draw a circle of radius 3.6 cm. Draw a tangent to the circle at any point on it without using the centre.
Practice set 4.2 | Q 4 | Page 98
Draw a circle of radius 3.3 cm Draw a chord PQ of length 6.6 cm. Draw tangents to the circle at points P and Q. Write your obeservation about the tangents.
Practice set 4.2 | Q 5 | Page 99
Draw a circle with radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. construct tangents at point M and N to the circle.
Practice set 4.2 | Q 6 | Page 99
Draw a circle with centre P and radius 3.4 cm. Take point Q at a distance 5.5 cm from the centre. Construct tangents to the circle from point Q.
Practice set 4.2 | Q 7 | Page 99
Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.
#### Balbharati solutions for Textbook for SSC Class 10 Mathematics 2 Chapter 4 Geometric Constructions Exercise Problem set 4 [Page 99]
Problem set 4 | Q 1.1 | Page 99
Select the correct alternative for the following question.
The number of tangents that can be drawn to a circle at a point on the circle is ............... .
• 3
• 2
• 1
• 0
Problem set 4 | Q 1.2 | Page 99
Select the correct alternative for the following question.
The maximum number of tangents that can be drawn to a circle from a point out side it is .............. .
• 2
• 1
• one and only one
• 0
Problem set 4 | Q 1.3 | Page 99
Select the correct alternative for the following question.
(3) If ∆ABC ~ ∆PQR and (AB)/(PQ) = 7/5, then ...............
• ∆ABC is bigger.
• ∆PQR is bigger.
• Both triangles will be equal.
• Can not be decided.
Problem set 4 | Q 2 | Page 99
Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.
Problem set 4 | Q 3 | Page 99
Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.
Problem set 4 | Q 4 | Page 99
Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.
Problem set 4 | Q 5 | Page 99
Draw a circle with centre P. Draw an arc AB of 100° measure. Draw tangents to the circle at point A and point B.
Problem set 4 | Q 6 | Page 99
Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E-F-A and FA = 4.1 cm. Draw tangents to the circle from point A.
Problem set 4 | Q 7 | Page 99
∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1cm, ∠B = 40°, BC = 4.8 cm,$\frac{AC}{LN} = \frac{4}{7}$. Construct ∆ABC and ∆LBN.
Problem set 4 | Q 8 | Page 99
Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If $\frac{YZ}{YQ} = \frac{6}{5},$ then construct ∆XYZ similar to ∆PYQ.
## Chapter 4: Geometric Constructions
Practice set 4.1Practice set 4.2Problem set 4
## Balbharati solutions for Textbook for SSC Class 10 Mathematics 2 chapter 4 - Geometric Constructions
Balbharati solutions for Textbook for SSC Class 10 Mathematics 2 chapter 4 (Geometric Constructions) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Textbook for SSC Class 10 Mathematics 2 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Balbharati textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Textbook for SSC Class 10 Mathematics 2 chapter 4 Geometric Constructions are To Construct Tangents to a Circle from a Point Outside the Circle., Construction of Triangle If the Base, Angle Opposite to It and Either Median Altitude is Given, Construction of Tangent Without Using Centre, Construction of Tangents to a Circle, Construction of Tangent to the Circle from the Point on the Circle, Basic Geometric Constructions, Division of a Line Segment.
Using Balbharati Class 10th Board Exam solutions Geometric Constructions exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Balbharati Solutions are important questions that can be asked in the final exam. Maximum students of Maharashtra State Board Class 10th Board Exam prefer Balbharati Textbook Solutions to score more in exam.
Get the free view of chapter 4 Geometric Constructions Class 10th Board Exam extra questions for Textbook for SSC Class 10 Mathematics 2 and can use Shaalaa.com to keep it handy for your exam preparation
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# What Is Standard Notation In Math latest 2023
You are searching about What Is Standard Notation In Math, today we will share with you article about What Is Standard Notation In Math was compiled and edited by our team from many sources on the internet. Hope this article on the topic What Is Standard Notation In Math is useful to you.
Page Contents
## Rhythm in Music
Quarter note = 1 beat
A quarter note is all black with a rod on it that goes up or down. I call it our “stepping” note because the note just steps and moves. With 4/4 time, you would have a measure of 4 quarter notes because a quarter note gets 1 count. Remember that music and math go hand in hand.
White = 2 beats
A half note is all white with a stem on it. When you played this kind of note, you paused, as if you were coming to a yellow light. You would play a half note by counting 1 & 2 &. With 4/4 time, you would have 2 half notes in a bar to play because 2 + 2 = 4. Don’t forget that music and math go hand in hand!
If you’re not sure what a bar is in written music, always remember that notes placed between barlines are a bar. Bar lines divide music into bars. When watching a piece of music, go to the very end of a song and you will find a double bar line there.
Repeat dots (repeat sign)
The colon at the end of the song is a repeat sign, meaning to play the song from the beginning. Sometimes my students ask, “Oh, do I have to do this?” The answer is “yes” because the composer intended to play this section of music again! Another way of looking at it is that you won’t have any more pages to turn. Just replay the first page or maybe it will just be a few lines repeated. The sign of repetition is indeed a very precious sign.
Whole rating = 4 counts
A whole note is also a white note, but it doesn’t have a stem on it. I call it our donut note. Some of my students call it a “hamburger” note! When you see a whole note, you have to hold down the note and count, 1 & 2 & 3 & 4 &. It’s like coming to a red light and coming to a complete stop.
Dotted half note = 3 beats
A dot after a note adds half the value of the note. So you now have 2 + 1 = 3 beats. You hold down the note and count, 1 & 2 & 3 &.
Now, what’s cool is that each type of musical note has a corresponding rest. A rest sign means “do not play” like a rest area. Sometimes that means getting your hand ready in position and going up, but still don’t play a note, rest! Examples of some silences are:
rest shift
It looks like a “Z” and a quarter rest is worth 1 count.
half rest
It looks like a black top hat. A half-rest gets 2 beats or beats of silence. A semi-rest is found above the third line.
Everything else
Looks like an upside-down black top hat. Everything else hangs below the fourth line. He gets 4 counts or beats of silence.
Each bar in 4/4 has notes and rests totaling up to 4 beats.
Time signature
The time signature is made up of the two numbers written at the start of a piece. The top number indicates the number of beats in each measure. The lower number indicates which type of note gets a beat.
2 means two beats in each measure
4 means that the quarter note receives a time or a count.
3 means three beats in each measure.
4 means quarter note gets a beat.
4 means four beats in each bar
4 means quarter note gets a beat.
Practice Guidelines
1. Clap and count the beat out loud.
2. Play and say the note names out loud.
3. Play and count the beat out loud.
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## What Is Significant Figures In Math latest 2023
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In this video, we are going to divide radical expressions. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.
One simple example is:
$\dfrac{\sqrt{25}}{\sqrt{9}}$
By simplifying the numerator and the denominator, we now have $\dfrac{5}{3}$
A slightly more difficult problem would be:
$\dfrac{\sqrt{50}}{\sqrt{18}}$
Just like the other problem, first find perfect square factors for each expression
$\dfrac{\sqrt{25}\times\sqrt{2}}{\sqrt{9}\times\sqrt{2}}$
Simplify each expression
$\dfrac{5\sqrt{2}}{3\sqrt{2}}$
And the final answer is $\dfrac{5}{3}$
If the expression has a radical in the denominator and a rational number in the numerator, then it is necessary to rationalize the fraction.
For example:
$\dfrac{3}{\sqrt{10}}$
Multiple the denominator to both the numerator and the denominator. Like terms cancel each other out so the final answer is:
$\dfrac{3\times\sqrt{10}}{\sqrt{10}\times\sqrt{10}}$
$\dfrac{3\sqrt{10}}{10}$
## Examples of Dividing Radical Expressions
### Example 1
$\dfrac{3}{8\sqrt{2}}$
Since the denominator has a radical, we have to rationalize the fraction.
Multiply the numerator and denominator by $\sqrt{2}$
$\dfrac{3}{8\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}$
Now, we have
$\dfrac{3\sqrt{2}}{8(2)}$
$\dfrac{3\sqrt{2}}{16}$
### Example 2
$\dfrac{3xy^2}{\sqrt{25x^3y}}$
Simplify first the terms inside the radical
$\dfrac{3y}{\sqrt{25x^2}}$
Then $\sqrt{25}$ is $5$ and $\sqrt{x^2}$ is $x$
Now, we have
$\dfrac{\sqrt{3y}}{5x}$
## Video-Lesson Transcript
Let’s go over how to divide radical expressions.
Let’s say we have
$\dfrac{\sqrt{25}}{\sqrt{9}}$
$\dfrac{5}{3}$
And there’s not much to do.
But what if we have
$\dfrac{\sqrt{50}}{\sqrt{18}}$
$= \dfrac{\sqrt{25} \sqrt{2}}{\sqrt{9} \sqrt {2}}$
$= \dfrac{5 \sqrt{2}}{3 \sqrt{2}}$
And just like if we have
$\dfrac{5x}{3x}$
$x$ will be cancelled out.
In this case, $\sqrt{2}$ will be cancelled out.
$= \dfrac{5}{3}$
Of course, not all is going to work out like this.
Let’s look at another example.
$\dfrac{\sqrt{27}}{\sqrt{30}}$
Now, let’s break this down.
$= \dfrac{\sqrt{9} \sqrt{3}}{\sqrt{30}}$
$= \dfrac{3 \sqrt{3}}{\sqrt{30}}$
This is how far we can go as far as the radicals are concerned.
But this is not our final answer.
Because we don’t want a radical in the denominator.
But even more so, we can reduce this.
Just like this example:
$\dfrac{\sqrt{15}}{\sqrt{5}}$
$= \dfrac{\sqrt{3}}{\sqrt{1}}$
$= \sqrt{3}$
Another way to look at that is:
$\dfrac{\sqrt{15}}{\sqrt{5}}$
$= \dfrac{\sqrt{5} \sqrt{3}}{\sqrt{5}}$
$\sqrt{5}$ cancels out and we’re left with
$= \sqrt{3}$
Now, let’s apply this in solving. Let’s go back
$= \dfrac{3 \sqrt{3}}{\sqrt{30}}$
We can break this down even more into
$= \dfrac{3 \sqrt{3}}{\sqrt{3} \sqrt{10}}$
Here, $= \sqrt{3}$ is going to cancel out.
And we have
$= \dfrac{3}{\sqrt{10}}$
Now, we want to rationalize the denominator.
The trick in getting rid of $\sqrt{10}$ is to multiply the numerator and denominator by the same number.
$= \dfrac{3 \times \sqrt{10}}{\sqrt{10} \times \sqrt{10}}$
$= \dfrac{3 \sqrt{10}}{10}$
Now, let’s look at some more.
$= \dfrac{ \sqrt{45}}{\sqrt{15}}$
Let’s break this down
$= \dfrac{\sqrt{9} \sqrt{5}}{\sqrt{15}}$
$= \dfrac{3 \sqrt{5}}{\sqrt{15}}$
Then let’s reduce $5$ and $15$
We’ll have
$= \dfrac{3 \sqrt{1}}{\sqrt{3}}$
$= \dfrac{3}{\sqrt{3}}$
Now, we don’t want a radical in the denominator.
We have to rationalize it. So let’s multiply the numerator and denominator by $\sqrt{3}$.
$= \dfrac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$
$= \dfrac{3 \sqrt{3}}{3}$
Which we cancel and we’re left with our final answer
$= \sqrt{3}$
This is a $100\%$ correct way to solve it.
Any way you solve for the correct answer is a good way to do it.
Let me show you another way.
We can reduce this right off the bat.
$= \dfrac{ \sqrt{45}}{\sqrt{15}}$
$15$ goes into $45$, so this can be reduced to
$= \dfrac{ \sqrt{3}}{\sqrt{1}}$
$\sqrt{3}$
Two different options to solve the same problem. We have the same answer both ways. |
using the difference formula
• Nov 6th 2009, 09:28 PM
maryanna91
using the difference formula
So Im supposed to use the difference formula to figure out cos(2x). This is what I have so far:
cos(2x)=cos(3x-x)
cos(3x-x)=(cos3x)(cosx)+(sin3x)(sinx)
Im just not sure how to finish solving! (my math homework is done online so it's supposed to be as simplified as possible).
• Nov 6th 2009, 09:33 PM
Gusbob
Are you sure you can use only the difference formula? Most people derive it with the sum formula using cos(2x) = cos(x+x)
• Nov 6th 2009, 09:36 PM
maryanna91
yep we're supposed to use the difference formula.
• Nov 6th 2009, 09:49 PM
Gusbob
In that case you will need these identities:
$\displaystyle \sin(3x) = 3\sin(x) - 4\sin^3(x)$
$\displaystyle \cos(3x) = 4\cos^3(x) - 3\cos(x)$
• Nov 6th 2009, 10:05 PM
maryanna91
Hmmm...tried plugging them in and I'm still still stuck (Headbang). Would you mind showing me what you'd do after they've been plugged in?
• Nov 6th 2009, 10:56 PM
Gusbob
$\displaystyle \cos(3x)\cos(x) + \sin(3x)\sin(x)$ = $\displaystyle [4\cos^3(x) - 3\cos(x)] \cos(x) + [3\sin(x) - 4\sin^3(x)] \sin(x)$
= $\displaystyle 4\cos^4(x) - 3\cos^2(x) + 3\sin^2(x) - 4\sin^4(x)$
= $\displaystyle 4[\cos^4(x) - \sin^4(x)] - 3[ \cos^2(x) - \sin^2(x)]$
Now notice that $\displaystyle \cos^4(x) - \sin^4(x)$ can be written as a difference of two squares to $\displaystyle (\cos^2(x) + \sin^2(x) ) (\cos^2(x) - \sin^2(x))$.
Do you see where to go from here?
• Nov 6th 2009, 11:16 PM
maryanna91
yep got it!!! thanks sooo much for all the help! |
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Unformatted text preview: Chapter Five 5.1 Random variable: A variable whose value is determined by the outcome of a random experiment is called a random variable. An example of this is the income of a randomly selected family. Discrete random variable: A random variable whose values are countable is called a discrete random variable. An example of this is the number of cars in a parking lot at any particular time. Continuous random variable: A random variable that can assume any value in one or more intervals is called a continuous random variable. An example of this is the time taken by a person to travel by car from New York City to Boston. 5.3 a. a discrete random variable b. a continuous random variable c. a continuous random variable d. a discrete random variable e. a discrete random variable f. a continuous random variable 5.5 The number of cars x that stop at the Texaco station is a discrete random variable because the values of x are countable: 0, 1, 2, 3, 4, 5 and 6. 5.7 The two characteristics of the probability distribution of a discrete random variable x are: 1. The probability that x assumes any single value lies in the range 0 to 1, that is, 1 ) ( ≤ ≤ x P . 2. The sum of the probabilities of all values of x for an experiment is equal to 1, that is: 1 ) ( = ∑ x P . 5.9 a. P ( x = 1) = .17 b. P ( x ≤ 1) = P (0) + P (1) = .03 + .17 + = .20 c. P ( x ≥ 3) = P (3) + P (4) + P (5) = .31 + .15 + .12 = .58 d. P (0 ≤ x ≤ 2) = P (0) + P (1) + P (2) = .03 + .17 + .22 = .42 e. P ( x < 3) = P (0) + P (1) + P (2) = .03 + .17 + .22 = .42 f. P ( x > 3) = P (4) + P (5) = .15 + .12 = .27 g. P (2 ≤ x ≤ 4) = P (2) + P (3) + P (4) = .22 + .31 + .15 = .68 5.11 a. b. i. P (exactly 3) = P (3) = .25 ii. P (at least 4) = P ( x ≥ 4) = P (4) + P (5) + P (6) = .14 + .07 + .03 = .24 71 10 20 30 1 2 3 4 5 Number of Systems Installed P(x) iii. P (less than 3) = P ( x< 3) = P (0) + P (1) + P (2) = .10 + .18 + .23 = .51 iv. P (2 to 5) = P (2) + P (3) + P (4) + P (5) = .23 + .25 + .14 + .07 = .69 5.13 a. x P ( x ) 1 8 / 80 = .10 2 20 / 80 = .25 3 24 / 80 = .30 4 16 / 80 = .20 5 12 / 80 = .15 b. The probabilities listed in the table of part a are approximate because they are obtained from a sample of 80 days. c. i. P (x = 3) = .30 ii. P ( x ≥ 3) = P (3) + P (4) + P (5) = .30 + .20 + .15 = .65 iii. P (2 ≤ ≤ x 4) = P (2) + P (3) + P (4) = .25 + .30 + .20 = .75 iv. P ( x ≤ 4) = P (2) + P (3) + P (4) = .10 + .25 + .30 = .65 5.15 Let Y = owns a cell phone and N = does not own a cell phone. Then P ( Y ) = .64 and P ( N ) = 1 – .64 = .36 Let x be the number of adults in a sample of two who own a cell phone. The following table lists the probability distribution of x . Note that x = 0 if neither adult owns a cell phone, x = 1 if one adult owns a cell phone and the other does not, and x = 2 if both adults own a cell phone. The probabilities are written in the table using the tree diagram above. The probability that x = 1 is obtained by adding the probabilities of YN and NY...
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# Which is the decomposed form for the expression x^2 + 4x + 3?
neela | Student
It is required to decompose x^2+4x+3.
We shal treat the given expression of secon degree to consist of 2 factors like x+a and x+b. Then the product of the x+a and x+b must be the given expression. So.
(x+a)(x+b) = x^2+4x+3. So,
X^2+(a+b)x+ab = x^2+4x+3. Comparing the coefficients of x^2, x and constant terms on both sides, we get:
1/1 = 4/(a+b) = 3/(ab). Therefore,
a+b =4 and
ab =3.
Solving these inequality, you get, (We do it by guess here without prolonging the procedure): a = 3 and b =1.
Threfore, x^2+4x+3 = (x+3)(x+1)
krishna-agrawala | Student
I believe, by decomposed for of the expression you mean a form in which the expression is represented as multiple of two or more constituent factors.
The given expression is:
E = x^2 + 4x + 3
We can find its factors by suitably modifying the expression in stages without changing its value. For the given expression it can be done as follows:
E = x^2 + 4x + 3
= x^2 + 3x + x + 3 = (x^2 + 3x) + (x + 3)
= x*(x + 3) + 1*(x + 3)
= (x + 1)*(x + 3)
Thus factors of the given expression are (x + 1) and (x + 3).
giorgiana1976 | Student
We’ll consider the expression as a second grade polynomial.
A polynomial could be written as a product of linear factors, according to his roots. For this reason, first of all, we’ll try to find out the roots of the expression.
x^2+4x+3=0
The coefficients of the polynomial x^2+4x+3=0 are a=1, b=4, c=3
x1=[-b + sqrt (b^2-4*a*c)]/2*a
x1=[-4 + sqrt (16-4*1*3)]/2*1
x1= (-4+sqrt4)/2
x1=(-4+2)/2
x1=-1
From Viete first relation between roots and coefficients, we’ll have:
x1+x2=-(b/a)
-1 + x2=-4 => x2=-4+1
x2=-3
After finding the roots, we could write the expression as a product of linear factors.
x^2+4x+3= (x-x1)*(x-x2)
x^2+4x+3= (x+1)*(x+3) |
# Question Video: Finding the Total Surface Area of a Right Cone Mathematics • 8th Grade
Find the total surface area of the right cone approximated to the nearest two decimal places.
03:05
### Video Transcript
Find the total surface area of the right cone approximated to the nearest two decimal places.
Weβre told on the diagram that the height of the cone is 14.5 centimetres. And its slant height is 16.5 centimetres. The radius is currently unknown. We can calculate the length of the radius by using Pythagorasβs theorem. This states that π squared plus π squared is equal to π squared, where π is the longest side of a right triangle, known as the hypotenuse. Substituting in our values gives us π squared plus 14.5 squared is equal to 16.5 squared. Subtracting 14.5 squared from both sides gives us π squared is equal to 16.5 squared minus 14.5 squared. Square-rooting both sides of this equation gives us π is equal to 16.5 squared minus 14.5 squared. This is equal to the square root of 62.
For accuracy, we will leave our answer in this form at present. We were asked to calculate the total surface area of the cone. A cone has two surfaces, a curved surface and a base. Therefore, the total surface area is equal to the area of the curved surface plus the area of the base. The area of the curved or lateral surface is equal to πππ. We multiply π by the radius by the slant height. As the base is a circle, we work out the area of the base by multiplying π by the radius squared. Substituting in our values for the radius and slant height gives us π multiplied by the square root of 62 multiplied by 16.5 plus π multiplied by the square root of 62 squared.
The square root of 62 squared is just equal to 62. As we need to calculate this to two decimal places and not in terms of π, we can type this calculation into our calculator. This gives us an answer of 602.93801 and so on. The eight in the thousandths column is the deciding number. When this digit is greater than or equal to five, we round up. The total surface area of the cone to two decimal places is 602.94 square centimetres. Any surface area will be measured in square units. |
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# Arithmetic Mean Formula
Last updated date: 29th Nov 2023
Total views: 311.7k
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## Arithmetic Mean
According to statistics, the arithmetic mean is the ratio of all observations in a data set to the total number of observations. For instance, the average rainfall of a place and the average salary of an organization's employees can be used. A lot of time, we hear statements like "the average family income is ₹15,000 or there is about 1000 mm of rain every month in a certain area.". Average is also referred to as Arithmetic Mean.
### What Does Arithmetic Mean?
In mathematics, the arithmetic mean is usually referred to as the mean or as an average. An arithmetic mean is calculated by adding all the numbers in a set of data and then dividing by the number of items included in the set. As a result, the middle number is the arithmetic mean for equally distributed numbers. As an added bonus, there are numerous ways to calculate the arithmetic mean, which is based on the amount of data and the distribution of that data.
Let's take a look at an example where the arithmetic mean is used. Since 6 + 8 + 10 = 24, and 24 divided by 3 (there are three numbers) is 8, the mean of the numbers 6, 8, 10 is 8. During a particular month, the arithmetic mean continues to be used to calculate the average closing price of a stock. In the example, suppose there are 24 trading days in a month. Where would we find the mean? The arithmetic mean can be calculated by adding all the prices together and dividing by 24.
### Calculation of Arithmetic Mean
The arithmetic mean is derived using the mathematical formula below.
Where,
A- arithmetic mean or average
n- number of items or terms being averaged
x1- value of every single item in the list of numbers being averaged
A= $\frac {1} {n}$ × n i=0 xi
Here is a more understandable version of the arithmetic mean formula.
A= $\frac {S} {N}$
Where,
A- Arithmetic mean or average.
n- number of items or terms being averaged.
S- The sum total of the numbers in the set of interest being averaged.
### Arithmetic Mean for Ungrouped Data
Using the formula below, we calculate the arithmetic mean:
Mean x̄ = Sum of all observations / Number of observations
Example: Calculate the arithmetic mean of the first six odd, natural numbers.
The first six odd, natural numbers are: 1, 3, 5, 7, 9, 11.
The formula $\frac {(1+3+5+7+9+11)} {6}$ = $\frac {36} {6}$ = 6.
As a result, the arithmetic mean is 6.
### Arithmetic Mean for Grouped Data
To determine the arithmetic mean of grouped data, three methods are available (Direct method, Short-cut method, and Step-deviation method). According to the numerical values of xi and fi, the method to be used must be chosen. The sum of all data inputs is xi, and the sum of their frequencies is fi. ∑ represents the summation. Direct methods work when xi and fi are sufficiently small. When the values are large, we use the assumed arithmetic mean method or the step deviation method.
## FAQs on Arithmetic Mean Formula
1. What is Meant by the Arithmetic Mean?
The arithmetic mean is the easiest and extensively used measure of a mean, or what we say average. It simply requires taking the sum of a group of numbers, then dividing that sum total by the count of the numbers used in the series.
For example, take the numbers 9, 21, 46, and 68. The sum is 144. The arithmetic mean is 144 divided by 4 (144/4), or 36.
2. How Many Types of Mean are There?
There are several other types of means that are being widely used such as the geometric mean and harmonic mean. These types of means come into play in certain situations like investing, stock trading analysis and finance. Another example includes the trimmed mean, used when computing economic data such as CPI and CPE.
3. How the Arithmetic Mean Works?
The arithmetic mean surprisingly reserves its position in finance as well. For example, mean earnings approximates essentially are arithmetic mean. Say you want to find out the average earnings expectation of the 21 analysts covering a specific stock. Simply add up all the approximates and divide by 21 to obtain the arithmetic mean.
The same holds true when we seek to find out a stock’s average closing price during a specific month. Say there are 22 trading days in the month. Simply consider all the prices, add them up, and divide by 22 to obtain the average or arithmetic mean.
4. What are the Advantages of Arithmetic Meaning?
In addition to statistics and mathematics, the arithmetic mean is used in experimental science, economics, sociology, and numerous other disciplines. The following are some of the main advantages of using an arithmetic mean.
• Since the formula for finding the arithmetic mean is rigid, the result remains the same. Unlike the median, it isn't influenced by how a value is positioned in the data set.
• It considers every value in the data set.
• Mathematical means are quite easy to find; even a person with no financial or math skills can calculate them.
• In addition, it provides useful results even with large number of groupings, making it useful for measuring central tendency.
• Contrary to mode and median, it can be further treated algebraically. From the mean of an individual series, for example, it is possible to obtain the mean of two or more series.
5. What are the Disadvantages of Arithmetic Mean?
In the next section, we will look at a few disadvantages/demerits of using the arithmetic mean.
• Data sets with extreme values have the greatest tendency to affect the arithmetic mean.
• Neither visual inspection nor graphic representation can locate the arithmetic mean.
• There is no way to use it to display qualitative data such as honesty, milkshake flavor favorites, most popular products, etc.
• If one observation is missing or lost, the arithmetic mean cannot be found.
6. What do Arithmetic mean tips and tricks?
• Out of the three methods to find the arithmetic mean, the direct method is preferred when there are fewer classes and values of a smaller magnitude.
• A step deviation will work best in scenarios in which we have a grouped frequency distribution, where the width is constant for every interval, and we have a large number of intervals in the frequency distribution.
7. How do you calculate the Arithmetic Mean?
According to statistics, the arithmetic mean is the sum of all observations divided by the number of observations. As an example, if the data set consists of five observations, you can calculate the arithmetic mean by adding all the five observations and dividing them by 5.
8. What are the Uses of Arithmetic Mean?
In mathematics, the arithmetic mean represents the central tendency. By considering all the observations, we can determine the center of the frequency distribution. |
Given the matrices
Question:
Given the matrices
$A=\left[\begin{array}{rrr}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right], B=\left[\begin{array}{rrr}9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6\end{array}\right]$ and $C=\left[\begin{array}{rrr}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$
Verify that (A + B) + C = A + (B + C).
Solution:
Here,
LHS $=(A+B)+C$
$=\left(\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6\end{array}\right]\right)+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$
$=\left(\left[\begin{array}{cc}2+9 & 1+7 & 1-1 \\ 3+3 & -1+5 & 0+4 \\ 0+2 & 2+1 & 4+6\end{array}\right]\right)+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$
$=\left[\begin{array}{ccc}11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10\end{array}\right]+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]$
$=\left[\begin{array}{cc}11+2 & 8-4 & 0+3 \\ 6+1 & 4-1 & 4+0 \\ 2+9 & 3+4 & 10+5\end{array}\right]$
$=\left[\begin{array}{ccc}13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15\end{array}\right]$
$\mathrm{RHS}=A+(B+C)$
$=\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left(\left[\begin{array}{ccc}9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6\end{array}\right]+\left[\begin{array}{ccc}2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5\end{array}\right]\right)$
$=\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left(\left[\begin{array}{cc}9+2 & 7-4 & -1+3 \\ 3+1 & 5-1 & 4+0 \\ 2+9 & 1+4 & 6+5\end{array}\right]\right)$
$=\left[\begin{array}{ccc}2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11\end{array}\right]$
$=\left[\begin{array}{ccc}2+11 & 1+3 & 1+2 \\ 3+4 & -1+4 & 0+4 \\ 0+11 & 2+5 & 4+11\end{array}\right]$
$=\left[\begin{array}{ccc}13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15\end{array}\right]$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
Hence proved. |
# Find the bounderies given by the curve x = 4 sqrt(y) and y = 4 sqrt(x). Show your solution.
## Question:
Find the bounderies given by the curve {eq}x = 4 \sqrt y {/eq} and {eq}y = 4 \sqrt x {/eq}. Show your solution.
## Area Between Functions With Integration
In 2D calculus, the definite integral has many applications, for example, we can find the area between functions, we need the integration limits and solve the integral. Remember, the bigger area minus the smaller area.
The area is:
The functions are:
{eq}\displaystyle \ f(x) = 4\sqrt{x} \\ {/eq}
and
{eq}\displaystyle x= 4\sqrt{y} \; \Rightarrow \; \frac{x^2}{16}=y \; \Rightarrow \; \ g(x)= \frac{x^2}{16} \\ {/eq}
Integration limits at interceptions {eq}x {/eq} values.
Matching functions:
{eq}\displaystyle \frac{x^2}{16}= 4\sqrt{x} \\ \displaystyle \frac{x^2}{16}- 4\sqrt{x}=0 \; \Leftrightarrow \; x=0 \; \text{or} \;x=16 \\ {/eq}
Terefore, the area is:
{eq}\displaystyle A \; = \; \displaystyle \int_{0}^{16} \ f(x) - \ g(x) \;dx \\ \displaystyle A \; = \; \displaystyle \int_{0}^{16} \ 4 \sqrt{x} - \frac{x^2}{16} \;dx \\ \displaystyle A \; = \; 8/3\,{x}^{3/2}-1/48\,{x}^{3} \bigg|_{0}^{16 } \\ \displaystyle A \; = \; 8/3\,{(16)}^{3/2}-1/48\,{(16)}^{3} - \left( 8/3\,{(0)}^{3/2}-1/48\,{(0)}^{3} \right) \\ \displaystyle A \; = \; \frac{256}{3} \; \text{or} \; A \; = \; 85.3333333333333 \; \; \text{ squared units } \; \\ {/eq} |
What Is 1 Percent of 500 + Solution with Free Steps?
The 1 percent of 500 is equal to 5. It can be easily calculated by dividing 1 by 100 and multiplying the answer with 500 to get 5.
The easiest way to get this answer is by solving a simple mathematical problem of percentage. You need to find 1% of 500Â Â for some sale or real-life problem. Divide 1Â by 100, multiply the answer with 500Â , and get the 1% of 500Â Â value in seconds.
This article will explain the full process of finding any percentage value from any given quantity or number with easy and simple steps.
What Is 1 percent of 500?
The 1 percent of 500 is 5.
The percentage can be understood with a simple explanation. Take 500, and divide it into 100 equal parts. The 1Â number of parts from the total 100 parts is called 1 percent, which is 5Â in this example.
How To Calculate 1 percent of 500?
You can find 1 percent of 500Â by some simple mathematical steps explained below.
Step 1
Firstly, depict 1 percent of 500 as a fractional multiple as shown below:
1% x 500
Step 2
The percentage sign % means percent, equivalent to the fraction of 1/100.
Substituting this value in the above formula:
= (1/100) x 500
Step 3
Using the algebraic simplification process, we can arithmetically manipulate the above equation as follows:
= (1 x 500) / 100
= 500 / 100
= 5
This percentage can be represented on a pie chart for visualization. Let us suppose that the whole pie chart represents the 500Â value. Now, we find 1 percent of 500, which is 5. The area occupied by the 5Â value will represent the 1Â percent of the total 500Â value. The remaining region of the pie chart will represent 99 percent of the total 500Â value. The 100% of 500 will cover the whole pie chart as 500Â is the total value.
Any given number or quantity can be represented in percentages to better understand the total quantity. The percentage can be considered a quantity that divides any number into hundred equal parts for better representation of large numbers and understanding.
Percentage scaling or normalization is a very simple and convenient method of representing numbers in relative terms. Such notations find wide application in many industrial sectors where the relative proportions are used. |
# Distance Constraint
The next equality constraint we will derive is the distance constraint. A distance constraint can be used to join two bodies at a fixed distance. It can also be used as a spring between two bodies.
Problem Definition
It’s probably good to start with a good definition of what we are trying to accomplish.
We want to take two or more bodies and constrain their motion in some way. For instance, say we want two bodies to only be able to rotate about a common point (Revolute Joint). The most common application are constraints between pairs of bodies. Because we have constrained the motion of the bodies, we must find the correct velocities, so that constraints are satisfied otherwise the integrator would allow the bodies to move forward along their current paths. To do this we need to create equations that allow us to solve for the velocities.
What follows is the derivation of the equations needed to solve for a Distance constraint.
Process Overview
Let’s review the process:
1. Create a position constraint equation.
2. Perform the derivative with respect to time to obtain the velocity constraint.
3. Isolate the velocity.
Using these steps we can ensure that we get the correct velocity constraint. After isolating the velocity we inspect the equation to find J, the Jacobian.
Most constraint solvers today solve on the velocity level. Earlier work solved on the acceleration level.
Once the Jacobian is found we use that to compute the K matrix. The K matrix is the A in the Ax = b general form equation.
Position Constraint
So the first step is to write out an equation that describes the constraint. A Distance Joint should allow the two bodies to move and rotate freely, but should keep them at a certain distance from one another. In other words:
which says that the current distance, and the desired distance, must be equal to zero.
The Derivative
The next step after defining the position constraint is to perform the derivative with respect to time. This will yield us the velocity constraint.
The velocity constraint can be found/identified directly, however its encouraged that a position constraint be created first and a derivative be performed to ensure that the velocity constraint is correct.
Another reason to write out the position constraint is because it can be useful during whats called the position correction step; the step to correct position errors (drift).
First if we write out how is computed we obtain:
Where and are the anchor points on the respective bodies.
We can rewrite this equation changing the magnitude to:
Since the squared magnitude of a vector is the dot product of the vector and itself.
Now we perform the derivative:
First let:
so
by the chain rule:
by the chain rule again:
Since the dot product is cumulative and distributive over addition:
If we clean up a bit:
Now we know that:
The derivative of a fixed length vector under a rotation frame is the cross product of the angular velocity with that fixed length vector.
So
We can also let:
Giving us:
Isolate The Velocities
The next step involves isolating the velocities and identifying the Jacobian. This may be confusing at first because there are two velocity variables. In fact, there are actually four, the linear and angular velocities of both bodies. To isolate the velocities we will need to employ some identities and matrix math.
The linear velocities are already isolated so we can ignore those for now. The angular velocities on the other hand have a pesky cross product. In 3D, we can use the identity that a cross product of two vectors is the same as the multiplication by a skew symmetric matrix and the other vector; see here. For 2D, we can do something similar by examining the cross product itself:
Remember that the angular velocity in 2D is a scalar.
Removing the cross products using the process above yields:
Now if we employ some matrix multiplication we can separate the velocities from the known coefficients:
Now, by inspection, we obtain the Jacobian:
Compute The K Matrix
Lastly, to solve the constraint we need to compute the values for A (I use the name K) and b:
See the “Equality Constraints” post for the derivation of the A matrix and b vector.
The b vector is fairly straight forward to compute. Therefore I’ll skip that and compute the K matrix symbolically:
Multiplying left to right the first two matrices we obtain:
Multiplying left to right again:
Multiplying left to right again:
Finally distributing the last term and pulling out scalar values we get:
Remember the inertia tensor in 2D is a scalar, therefore we can pull it out to the front of the multiplications.
Since n is normalized:
We get:
Then if we perform the matrix multiplication in the other terms:
We obtain (remember the cross product in 2D is a scalar):
Plug the values of the K matrix and b vector into your linear equation solver and you will get the impulse required to satisfy the constraint.
Note here that if you are using an iterative solver that the K matrix does not change over iterations and as such can be computed once each time step.
Another interesting thing to note is that the K matrix will always be a square matrix with a size equal to the number of degrees of freedom (DOF) removed. This is a good way to check that the derivation was performed correctly.
## 7 thoughts on “Distance Constraint”
• roehit says:
Hi William,
You have mentioned that the size of the K matrix will be equal to the no. of degrees of freedom removed.
This makes the ‘K’ matrix in the distance constraint a 1×1(a scalar).
And with ‘J’ having a ‘nT’ (transpose of n) term before the 3×12 matrix, that makes ‘b’ a scalar too.
Does this mean ‘x’ (lambda) will be a scalar too? And if so, how will this lambda be used to calculate the impulses to be applied on both the bodies?
• William says:
Yes, you are exactly right, lambda will be a scalar in this case. Once lambda (the scalar impulse) is found, we can use the equation from the Equality Constraints post to find the change in velocity:
Now if we plug in the mass matrix, JT and lambda we get 4 equations to find the change in velocity for the two bodies (note in 2D the mass matrix entries are scalars and as you pointed out lambda is a scalar):
As a side note, the above explains how to apply the impulse to the bodies. The impulse is simply (in this case):
William
• roehit says:
Hi William,
Are Vf and Vi velocities about the pivot or about the COM?
If they are about the pivot, shouldn’t Vf be 0 after every frame(for Pin Constraint of a body with the world[very heavy static object])?
Thanks
• William says:
vf and vi are the linear velocities of the bodies at their center of mass.
The velocity of the pivot point would be:
This is seen in the derivation of the distance constraint. However, the formulation we have setup here is to find the velocities at the center of mass for the bodies.
William
• roehit says:
Hi William,
Thanks for the replies.
I am getting the right responses from the bodies when i’ve implemented the distance constraint, but there seems to be a constant drift arising out of the numerical integration errors. I’ve noticed this error for the point-to-point constraint also.
How do we solve/cancel out these errors and ensure close-to-perfect behavior with the constraints?
• William says:
Exactly. In the Equality Constraints post I briefly touch on the drift problem. Basically as you pointed out, the velocity constraints can easily be satisfied, however, because our integration techniques are approximations, we will always get drift in the position constraints. The generally accepted solution to this problem is to use some sort of post stabilization method.
Unfortunately I don’t have a post for this yet but I can offer a few details. If we take our original position constraint that we defined:
and we solve this for C we get the position constraint error. We can feed this error back into the velocity constraint by moving/rotating the bodies by this error (using the effective mass to apply the position correction appropriately):
Then we can apply this position correction by:
This will give the change in position of the Center of Mass and the change in Rotation of both bodies. Because we changed the position and rotation, upon the next iteration of the simulation, the velocity constraint will see these changes (that’s why we say “the position error gets feed back into the velocity constraint”).
Naturally, each constraint will have a different position constraint equation. The one above is specifically for the Distance Constraint. But they all follow the same process, solve the position constraint for C (which could be a vector instead of a scalar for some constraints) and then scale this by the “effective mass” (the K matrix). Then apply the result to the bodies.
William
• NK says:
You say it can be used to model a spring – how?
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# Dividing Polynomials VOCABULARY
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1 - Dividing Polynomials TEKS FOCUS TEKS ()(C) Determine the quotient of a polynomial of degree three and degree four when divided by a polynomial of degree one and of degree two. TEKS ()(A) Apply mathematics to problems arising in everyday life, society, and the workplace. Additional TEKS ()(G) VOCABULARY Synthetic division Synthetic division is a process for dividing a polynomial by a linear expression x - a. You list the standard-form coefficients (including zeros) of the polynomial, omitting all variables and exponents. You use a for the divisor and add instead of subtract throughout the process. Apply use knowledge or information for a specific purpose, such as solving a problem ESSENTIAL UNDERSTANDING You can divide polynomials using steps that are similar to the long-division steps that you use to divide whole numbers. Key Concept The Division Algorithm for Polynomials You can divide polynomial P (x) by polynomial D (x) to get polynomial quotient Q (x) and polynomial remainder R (x). The result is P (x) = D (x)q (x) + R (x). Q (x) D (x))p (x) # ## R (x) If R (x) =, then P (x) = D (x)q (x) and D (x) and Q (x) are factors of P (x). To use long division, P (x) and D (x) should be in standard form with zero coefficients where appropriate. The process stops when the degree of the remainder, R (x), is less than the degree of the divisor, D (x). Theorem The Remainder Theorem If you divide a polynomial P (x) of degree n Ú by x - a, then the remainder is P (a). PearsonTEXAS.com
2 Problem Using Polynomial Long Division Use polynomial long division to divide x + x x by x. What is the quotient and remainder? How can you check your result? Show that (divisor)(quotient) + remainder = dividend. x x - )x + x - x - x - x x - Divide: x x = x. Multiply: x(x - ) = x - x. Subtract to get x. Bring down -. Repeat the process of dividing, multiplying, and subtracting. x + x - )x + x - x - x - x x - x - - The quotient is x + with remainder -. Check (x - )(x + ) - = (x + x - x - ) - = x + x - x - Divide: x x = Multiply: (x - ) = x -. Subtract to get -. You can say: x, R. Multiply (x - )(x + ). Simplify. Problem TEKS Process Standard ()(G) Using Polynomial Long Division to Check Factors A Use polynomial long division to divide P(x) = x x + x + by x +. Is x + a factor of P(x)? x - x + x + x + )x - x + x + x + x + x + x -x + x + x -x + x - x x + x + x + x + x - Include x terms. The degree of the remainder is less than the degree of the divisor. Stop! The remainder is not zero. x + does not divide x - x + x + evenly and is not a factor of P(x). continued on next page Lesson - Dividing Polynomials
3 Problem continued Can you use the Factor Theorem to help answer this question? Yes; recall that if P(a) =, then x - a is a factor of P(x). B Is x a factor of P (x) = x? If it is, write P (x) as a product of two factors. Step Use the Factor Theorem to determine if x - is a factor of x -. P () = - = - = Since P () =, x - is a factor of P (x). Step Use polynomial long division to find the other factor. x + x + x + x - )x + x + x + x - x - x x + x x - x x + x x - x x - x - P (x) = (x - )(x + x + x + ) Problem Using Synthetic Division To divide by x + what number do you use for the synthetic divisor? x + = x - (-), so use -. Use synthetic division to divide x x + x x by x +. What is the quotient and remainder? Step Reverse the sign of +. Write the coefficients of the polynomial. - ; Step Multiply the coefficient by the divisor. Add to the next coefficient. - ; Step Bring down the first coefficient. - ; Step Continue multiplying and adding through the last coefficient. - ; The quotient is x - x + x -, R. PearsonTEXAS.com
4 Problem TEKS Process Standard ()(A) How can you use the picture to help solve the problem? The picture gives the width of the box. Remember for a rectangular prism, V = / * w * h. Using Synthetic Division to Solve a Problem Crafts The polynomial x + x x expresses the volume, in cubic inches, of the shadow box shown. A What are the dimensions of the box? (Hint: The length is greater than the height (or depth).) - ; x + x - = (x - )(x + ) So, x + x - x - = (x + )(x + x - ) = (x + )(x - )(x + ) The length, width, and height (or depth) of the box are (x + ) in., (x + ) in., and (x - ) in., respectively. B If the width of the box is in., what are the other two dimensions? The width of the box is x +. So if x + =, then x =. Substitute for x to find the length and height (or depth). Length: x + = + = in. Height: x - = - = in. x + Is there a way to find P() without substituting? Use synthetic division. P() is the remainder. Problem Evaluating a Polynomial Given that P (x) = x x x +, what is P ()? By the Remainder Theorem, P () is the remainder when you divide P (x) by x -. P () =. ; Lesson - Dividing Polynomials
5 ONLINE H O M E W O R K PRACTICE and APPLICATION EXERCISES Scan page for a Virtual Nerd tutorial video. Divide using long division. Check your answers. For additional support when completing your homework, go to PearsonTEXAS.com.. x + x -, (x + ). x + x - x +, (x - ). x - x - x -, (x - ). x + x - x -, (x - ) Divide.. x + x + x +, (x + ). x + x + x +, (x + ). x + x - x -, (x - ). x - x + x + x -, (x - ) Determine whether each binomial is a factor of x + x + x.. x +. x +. x +. x - Divide using synthetic division.. x + x - x -, (x - ). x - x + x -, (x - ). x - x - x +, (x + ). x - x - x -, (x - ). x +, (x - ). x + x + x -, (x + ). x +, (x + ). x - x -, (x - ) Use synthetic division and the given factor to completely factor each polynomial function.. y = x + x - x - ; (x + ). y = x - x - x + ; (x + ). Apply Mathematics ()(A) The volume, in cubic inches, of the decorative box shown can be expressed as the product of the lengths of its sides as V (x) = x + x - x. What linear expressions with integer coefficients represent the length and height of the box? x Use synthetic division and the Remainder Theorem to find P (a).. P (x) = x + x - x - ; a =-. P (x) = x + x + x; a =-. P (x) = x - x + x - ; a =. P (x) = x + x + x; a =-. P (x) = x - x + x + ; a =. P (x) = x - x + x + ; a =. P (x) = x + x - x - ; a =. P (x) = x + x + x - ; a =- PearsonTEXAS.com
6 . Select Techniques to Solve Problems ()(C) Your friend multiplies x + by a quadratic polynomial and gets the result x - x - x +. The teacher says that everything is correct except for the constant term. Find the quadratic polynomial that your friend used. What is the correct result of multiplication?. Display Mathematical Ideas ()(G) A student used synthetic division to divide x - x - x by x +. Describe and correct the error shown.. Connect Mathematical Ideas ()(F) When a polynomial is divided by (x - ), the quotient is x + x + with remainder. Find the polynomial.. Apply Mathematics ()(A) The expression (x + x + x + ) represents the volume of a square pyramid. The expression x + represents the height of the pyramid. What expression represents the side length of the base? (Hint: The formula for the volume of a pyramid is V = Bh.). Analyze Mathematical Relationships ()(F) Divide. Look for patterns in your answers. a. x -, (x - ) b. x -, (x - ) c. x -, (x - ) d. Using the patterns, factor x -.. Select Tools to Solve Problems ()(C) The remainder from the division of the polynomial x + ax + ax + by x + is. Find a.. Use synthetic division to find (x + ), (x - i).. Display Mathematical Ideas ()(G) Suppose, -, and are zeros of a cubic polynomial function f (x). What is the sign of f () # f ()? (Hint: Sketch the graph; consider all possibilities.) TEXAS Test Practice. What is the remainder when x - x + is divided by x +? A. B. C. D.. What is the least degree of a polynomial that has a zero of multiplicity at, a zero of multiplicity at, and a zero of multiplicity at? F. G. H. J.. The equation y =.x relates your weight on the Moon y to your weight on Earth x in pounds. If Al weighs lb on Earth, what would he weigh on the Moon? A.. lb B.. lb C. lb D.. lb. The formula for the area of a circle is A = pr. Solve the equation for r. If the area of a circle is. cm, what is the radius? Use. for p. Lesson - Dividing Polynomials
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# Difference between revisions of "2017 AMC 8 Problems/Problem 25"
## Problem
In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $\overarc{TR}$ and $\overarc{SR}$ are each one-sixth of a circle with radius 2. What is the area of the region shown?
$[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("U", (2,3.464), N); label("S", (1,1.732), W); label("T", (3,1.732), E); label("R", (2,0), S);[/asy]$
$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$
## Solution
Let the centers of the circles containing arcs $\overarc{SR}$ and $\overarc{TR}$ be $X$ and $Y$, respectively. Extend $\overline{US}$ and $\overline{UT}$ to $X$ and $Y$, and connect point $X$ with point $Y$. $[asy] unitsize(1 cm); pair U,S,T,R,X,Y; U =(2,3.464); S=(1,1.732); T=(3,1.732); R=(2,0); X=(0,0); Y=(4,0); draw(U--S); draw(S--U--T); draw(S--X--Y--T,red); draw(arc(X,R,S),red); draw(arc(Y,T,R),red); label("U",U, N); label("S", S, W); label("T", T, E); label("R", R, S); label("X",X, W); label("Y", Y, E); [/asy]$ We can clearly see that $\triangle UXY$ is an equilateral triangle, because the problem states that $m\angle TUS = 60^\circ$. We can figure out that $m\angle SXR= 60^\circ$ and $m\angle TYR = 60^\circ$ because they are $\frac{1}{6}$ of a circle. The area of the figure is equal to $[\triangle UXY]$ minus the combined area of the $2$ sectors of the circles(in red). Using the area formula for an equilateral triangle, $\frac{a^2\sqrt{3}}{4},$ where $a$ is the side length of the equilateral triangle, $[\triangle UXY]$ is $\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.$ The combined area of the $2$ sectors is $2\cdot\frac16\cdot\pi r^2$, which is $\frac 13\pi \cdot 2^2 = \frac{4\pi}{3}.$ Thus, our final answer is $\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$
## Video Solutions
https://youtu.be/wc5rGulTTR8 - Happytwin
https://youtu.be/uvwLT5xBNdU ~DSA_Catachu |
# APPLICATIONS. are symmetric, but. are not.
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1 CHAPTER III APPLICATIONS Real Symmetric Matrices The most common matrices we meet in applications are symmetric, that is, they are square matrices which are equal to their transposes In symbols, A t = A Examples are symmetric, but are not,, 4 Symmetric matrices are in many ways much simpler to deal with than general matrices First, as we noted previously, it is not generally true that the roots of the characteristic equation of a matrix are necessarily real numbers, even if the matrix has only real entries However, if A is a symmetric matrix with real entries, then the roots of its characteristic equation are all real Example The characteristic equations of and are λ = and λ += respectively Notice the dramatic effect of a simple change of sign The reason for the reality of the roots (for a real symmetric matrix) is a bit subtle, and we will come back to it later sections The second important property of real symmetric matrices is that they are always diagonalizable, that is, there is always a basis for R n consisting of eigenvectors for the matrix
2 III APPLICATIONS Example We previously found a basis for R consisting of eigenvectors for the symmetric matrix A = The eigenvalues are λ =,λ =, and the basis of eigenvectors is { v =, v = } If you look carefully, you will note that the vectors v and v not only form a basis, but they are perpendicular to one another, ie, v v =( ) + () = The perpendicularity of the eigenvectors is no accident It is always the case for a symmetric matrix by the following reasoning First, recall that the dot product of two column vectors u and v in R n can be written as a row by column product u v = u t v =[u u u n ] v v v n = n i= u i v i Suppose now that Au = λu and Av = µv, ie, u and v are eigenvectors for A with corresponding eigenvalues λ and µ Assume λ µ Then () u (Av) =u t (Av)=u t (µv)=µ(u t v)=µ(u v) On the other hand, () (Au) v =(Au) t v=(λu) t v=λ(u t v)=λ(u v) However, since the matrix is symmetric, A t = A, and (Au) t v =(u t A t )v=(u t A)v=u t (Av) The first of these expressions is what was calculated in () and the last was calculated in (), so the two are equal, ie, µ(u v) =λ(u v) If u v, we can cancel the common factor to conclude that µ = λ, which is contrary to our assumption, so it must be true that u v =, ie, u v We summarize this as follows Eigenvectors for a real symmetric matrix which belong to different eigenvalues are necessarily perpendicular This fact has important consequences Assume first that the eigenvalues of A are distinct and that it is real and symmetric Then not only is there a basis consisting of eigenvectors, but the basis elements are also mutually perpendicular
3 REAL SYMMETRIC MATRICES This is reminiscent of the familiar situation in R and R, where coordinate axes are almost always assumed to be mutually perpendicular For arbitrary matrices, we may have to face the prospect of using skew axes, but the above remark tells us we can avoid this possibility in the symmetric case In two or three dimensions, we usually require our basis vectors to be unit vectors There is no problem with that here Namely, if u is not a unit vector, we can always obtain a unit vector by dividing u by its length u Moreover, if u is an eigenvector for A with eigenvalue λ, then any nonzero multiple of u is also such an eigenvector, in particular, the unit vector u u is Example, revisited The eigenvectors v and v both have length So we replace them by the corresponding unit vectors [ ] v = v = [ which also constitute a basis for R There is some special terminology which is commonly used in linear algebra for the familiar concepts discussed above Two vectors are said to be orthogonal if they are perpendicular A unit vector is said to be normalized The idea is that if we started with a non-unit vector, we would produce an equivalent unit vector by dividing it by its length The latter process is called normalization Finally, a basis for R n consisting of mutually perpendicular unit vectors is called an orthonormal basis Exercises for Section 4 (a) Find a basis of eigenvectors for A = 4 (b) Check that the basis vectors are orthogonal, and normalize them to yield an orthonormal basis (a) Find a basis of eigenvectors for A = 8 (b) Are the basis vectors orthogonal to one another? If not what might be the problem? (a) Find a basis of eigenvectors for A = (b) Check that the basis vectors are orthogonal, and normalize them to yield an orthonormal basis 4 Let A = 4 4 Find an orthonormal basis of eigenvectors Let A be a symmetric n n matrix, and let P be any n n matrix Show that P t AP is also symmetric ]
4 4 III APPLICATIONS Repeated Eigenvalues, The Gram Schmidt Process We now consider the case in which one or more eigenvalues of a real symmetric matrix A is a repeated root of the characteristic equation It turns out that we can still find an orthonormal basis of eigenvectors, but it is a bit more complicated Example Consider The characteristic equation is det λ λ λ A = = ( + λ)(( + λ) ) ( λ ) + ( + + λ) = ( + λ)(λ +λ)+(λ+) = (λ +λ 4)= Using the method suggested in Chapter, we may find the roots of this equation by trying the factors of the constant term The roots are λ =, which has multiplicity, and λ =, which has multiplicity For λ =, we need to reduce A I = The general solution is v = v,v =v with v free A basic eigenvector is v = but we should normalize this by dividing it by v = This gives u = For λ =, the situation is more complicated Reduce A +I =
5 REPEATED EIGENVALUES, THE GRAM SCHMIDT PROCESS which yields the general solution v = v v with v,v free This gives basic eigenvectors v =, v = Note that, as the general theory predicts, v is perpendicular to both v and v (The eigenvalues are different) Unfortunately, v and v are not perpendicular to each other However, with a little effort, this is easy to remedy All we have to do is pick another basis for the subspace spanned by {v, v } The eigenvectors with eigenvalue are exactly the non-zero vectors in this subspace, so any basis will do as well Hence, we arrange to pick a basis consisting of mutually perpendicular vectors It is easy to construct the new basis Indeed we need only replace one of the two vectors Keep v, and let v = v cv where c is chosen so that v v = v v cv v =, ie, take c = v v v v (See the diagram to get some idea of the geometry behind this calculation) v v We have v v v = v v v = v v = = We should also normalize this basis by choosing u = v v =, u = v v = Putting this all together, we see that u =, u =, u = form an orthonormal basis for R consisting of eigenvectors for A
6 6 III APPLICATIONS The Gram Schmidt Process In Example, we used a special case of a more general algorithm in order to construct an orthonormal basis of eigenvectors The algorithm, called the Gram Schmidt Process works as follows Suppose {v, v,,v k } is a linearly independent set spanning a certain subspace W of R n We construct an orthonormal basis for W as follows Let v = v v = v v v v v v v = v v v v v v v v v v v k v k v k v j = v k v j v j v j j= It is not hard to see that each new v k before it For example, is perpendicular to those constructed v v = v v v v v v v v v v v v v v However, we may suppose that we already know that v v = (from the previous stage of the construction), so the above becomes v v = v v v v = The same argument works at each stage It is also not hard to see that at each stage, replacing v j by v j in {v, v,,v j,v j } does not change the subspace spanned by the set Hence, for j = k, we conclude that {v, v,,v k } is a basis for W consisting of mutually perpendicular vectors Finally, to complete the process simply divide each v j by its length u j = v j v j Then {u,,u k } is an orthonormal basis for W
7 REPEATED EIGENVALUES, THE GRAM SCHMIDT PROCESS 7 Example Consider the subspace of R 4 spanned by v =, v =, v = Then Normalizing, we get v = v = = v = 9 u = u = u = 4 = = 4 = The Principal Axis Theorem The Principal Axis Theorem asserts that the process outlined above for finding mutually perpendicular eigenvectors always works If A is a real symmetric n n matrix, there is always an orthonormal basis for R n consisting of eigenvectors for A Here is a summary of the method If the roots of the characteristic equation are all different, then all we need to do is find an eigenvector for each eigenvalue and if necessary normalize it by dividing by its length If there are repeated roots, then it will usually be necessary to apply the Gram Schmidt process to the set of basic eigenvectors obtained for each repeated eigenvalue 4
8 8 III APPLICATIONS Exercises for Section Apply the Gram Schmidt Process to each of the following sets of vectors (a), (b),, Find an orthonormal basis of eigenvectors for A = Find an orthonormal basis of eigenvectors for A = Hint: is an eigenvalue 4 Let {v, v, v } be a linearly independent set Suppose {v, v, v } is the set obtained (before normalizing) by the Gram-Schmidt Process (a) Explain why v is not zero (b) Explain why v is not zero The generalization of this to an arbitrary linearly independent set is one reason the Gram-Schmidt Process works The vectors produced by that process are mutually perpendicular provided they are non-zero, and so they form a linearly independent set Since they are in the subspace W spanned by the original set of vectors and there are just enough of them, they must form a basis a basis for W Change of Coordinates As we have noted previously, it is probably a good idea to use a special basis like an orthonormal basis of eigenvectors Any problem associated with the matrix A is likely to take a particularly simple form when expressed relative to such a basis To study this in greater detail, we need to talk a bit more about changes of coordinates Although the theory is quite general, we shall concentrate on some simple examples In R n, the entries in a column vector x may be thought of as the coordinates x,x,,x n of the vector with respect to the standard basis To simplify the algebra, let s concentrate on one specific n, sayn= In that case, we may make the usual identifications e = i, e = j, e = k for the elements of the standard basis Suppose {v, v, v } is another basis The coordinates of x with respect to the new basis call them x,x,x are defined by the relation () x = v x + v x + v x =[v v v ] x x =[v v v ]x x
9 CHANGE OF COORDINATES 9 One way to view this relation is as a system of equations in which the old coordinates x = x x x are given, and we want to solve for the new coordinates x = x x x The coefficient matrix of this system P =[v v v ] is called the change of basis matrix It s columns are the old coordinates of the new basis vectors The relation () may be rewritten () x = P x and it may also be interpreted as expressing the old coordinates of a vector in terms of its new coordinates This seems backwards, but it is easy to turn it around Since the columns of P are linearly independent, P is invertible and we may write instead () x = P x where we express the new coordinates in terms of the old coordinates Example Suppose in R we pick a new set of coordinate axes by rotating each of the old axes through angle θ in the counterclockwise direction Call the old coordinates (x,x ) and the new coordinates (x,x ) According to the above discussion, the columns of the change of basis matrix P come from the old coordinates of the new basis vectors, ie, of unit vectors along the new axes From the diagram, these are [ cos θ sin θ ] sin θ cos θ
10 III APPLICATIONS x x x θ θ x Hence, [ x x ] [ cos θ sin θ x = sin θ cos θ The change of basis matrix is easy to invert in this case (Use the special rule which applies to matrices) cos θ sin θ cos θ sin θ cos θ sin θ = sin θ cos θ cos θ + sin = θ sin θ cos θ sin θ cos θ (You could also have obtained this by using the matrix for rotation through angle θ) Hence, we may express the new coordinates in terms of the old coordinates through the relation x x = [ cos θ sin θ sin θ cos θ x ][ x For example, suppose θ = π/6 The new coordinates of the point with original coordinates (, 6) are given by [ ] x / / = = / / 6 x x ] ] [ ] + + So with respect to the rotated axes, the coordinates are ( +, ) Orthogonal Matrices You may have noticed that the matrix P obtained in Example has the property P = P t This is no accident It is a consequence of the fact that its columns are mutually perpendicular unit vectors Indeed, The columns of an n n matrix form an orthonormal basis for R n if and only if its inverse is its transpose An n n real matrix with this property is called orthogonal
11 CHANGE OF COORDINATES Example Let [ P = 4 4 ] The columns of P are u = [ ] 4, u = 4, and it is easy to check that these are mutually perpendicular unit vectors in R To see that P = P t, it suffices to show that P t P = [ 4 4 ][ 4 ] = 4 Of course, it is easy to see that this true by direct calculation, but it may be more informative to write it out as follows [ P t (u ) P = t (u ) t ] [ u u ]= u u u u u u u u where the entries in the product are exhibited as row by column dot products The off diagonal entries are zero because the vectors are perpendicular, and the diagonal entries are ones because the vectors are unit vectors The argument for n n matrices is exactly the same except that there are more entries Note The terminology is very confusing The definition of an orthogonal matrix requires that the columns be mutually perpendicular and also that they be unit vectors Unfortunately, the terminology reminds us of the former condition but not of the latter condition It would have been better if such matrices had been named orthonormal matrices rather than orthogonal matrices, but that is not how it happened, and we don t have the option of changing the terminology at this late date The Principal Axis Theorem Again As we have seen, given a real symmetric n n matrix A, the Principal Axis Theorem assures us that we can always find an orthonormal basis {v, v,,v n } for R n consisting of eigenvectors for A Let P =[v v v n ]
12 III APPLICATIONS be the corresponding change of basis matrix As in Chapter II, Section, we have λ Av = v λ =[v v v n ] Av =v λ =[v v λ v n ] Av n =v n λ n =[v v v n ] where some eigenvalues λ j for different eigenvectors might be repeated These equations can be written in a single matrix equation λ λ A [ v v v n ]=[v v v n ] λ n or AP = PD where D is a diagonal matrix with eigenvalues (possibly repeated) on the diagonal This may also be written (4) P AP = D Since we have insisted that the basic eigenvectors form an orthonormal basis, the change of basis matrix P is orthogonal, and we have P = P t Hence, (4) can be written in the alternate form () P t AP = D with P orthogonal 7 4 Example Let A = The characteristic equation of A turns out 4 7 to be λ 6 =, so the eigenvalues are λ = ± Calculation shows that an orthonormal basis of eigenvectors is formed by [ ] u = 4 4 for λ = and u = for λ = Hence, we may take P to be the orthogonal matrix [ 4 ] The reader should check in this ] case that P t AP = [ [ ][ 4 ] = 4 λ n
13 CHANGE OF COORDINATES Appendix A Proof of the Principal Axis Theorem The following section outlines how the Principal Axis Theorem is proved for the very few of you who may insist on seeing it It is not necessary for what follows In view of the previous discussions, we can establish the Principal Axis Theorem by showing that there is an orthogonal n n matrix P such that (6) AP = PD or equivalently P t AP = D where D is a diagonal matrix with the eigenvalues of A (possibly repeated) on its diagonal The method is to proceed by induction on n If n = there really isn t anything to prove (Take P = [ ]) Suppose the theorem has been proved for (n ) (n ) matrices Let u be a unit eigenvector for A with eigenvalue λ Consider the subspace W consisting of all vectors perpendicular to u It is not hard to see that W is an n dimensional subspace Choose (by the Gram Schmidt Process) an orthonormal basis {w, w,w n } for W Then {u, w,,w n } is an orthonormal basis for R n, and Au = u λ =[u w w n ] } {{ } P This gives the first column of AP, and we want to say something about its remaining columns Aw, Aw,,Aw n To this end, note that if w is any vector in W, then Aw is also a vector in W For, we have u (Aw) =(u ) t Aw=(u ) t A t w=(au ) t w=λ (u ) t w)=λ (u w)=, which is to say, Aw is perpendicular to u if w is perpendicular to u It follows that each Aw j is a linear combination just of w, w,,w n, ie, Aw j =[u w w n ] where denotes some unspecified entry Putting this all together, we see that AP = P λ A } {{ } A λ
14 4 III APPLICATIONS where A is an (n ) (n ) matrix P is orthogonal (since its columns form an orthonormal basis) so P t AP = A, and it is not hard to derive from this the fact that A is symmetric Because of the structure of A, this implies that A is symmetric Hence, by induction we may assume there is an (n ) (n ) orthogonal matrix P such that A P = P D with D diagonal It follows that A λ P = A P } {{ } P = λ A P = P = λ P D λ D } {{ } } {{ } P D Note that P is orthogonal and D is diagonal Thus, AP P }{{ = P } A P = P P D } {{ } P P or AP = PD = P D However, a product of orthogonal matrices is orthogonal see the Exercises so P is orthogonal as required This completes the proof There is one subtle point involved in the above proof We have to know that a real symmetric n n matrix has at least one real eigenvalue This follows from the fact, alluded to earlier, that the roots of the characteristic equation for such a matrix are necessarily real Since the equation does have a root, that root is the desired eigenvalue Exercises for Section Find the change of basis matrix for a rotation through (a) degrees in the counterclockwise direction and (b) degrees in the clockwise direction Let P (θ) be the matrix for rotation of axes through θ Show that P ( θ) = P(θ) t =P(θ)
15 4 CLASSIFICATION OF CONICS AND QUADRICS An inclined plane makes an angle of degrees with the horizontal Change to a coordinate system with x axis parallel to the inclined plane and x axis perpendicular to it Use the change of variables formula derived in the section to find the components of the gravitational acceleration vector gj in the new coordinate system Compare this with what you would get by direct geometric reasoning 4 Let A = Find a orthogonal matrix P such that P t AP is diagonal What are the diagonal entries? Let A = 4 4 Find a orthogonal matrix P such that P t AP is diagonal What are the diagonal entries? 6 Show that the product of two orthogonal matrices is orthogonal How about the inverse of an orthogonal matrix? 7 The columns of an orthogonal matrix are mutually perpendicular unit vectors Is the same thing true of the rows? Explain 4 Classification of Conics and Quadrics The Principal Axis Theorem derives its name from its relation to classifying conics, quadric surfaces, and their higher dimensional analogues The general quadratic equation ax + bxy + cy + dx + ey = f (with enough of the coefficients non-zero) defines a curve in the plane Such a curve is generally an ellipse, a hyperbola, a parabola, all of which are called conics, or two lines, which is considered a degenerate case (See the Appendix to this section for a review) Examples x + y 4 = x y = x xy +y = x +xy y +x y= If the linear terms are not present (d = e = and f ), we call the curve a central conic It turns out to be an ellipse or hyperbola (but its axes of symmetry may not be the coordinate axes) or a pair of lines in the degenerate case Parabolas can t be obtained this way
16 6 III APPLICATIONS In this section, we shall show how to use linear algebra to classify such central conics and higher dimensional analogues such as quadric surfaces in R Once you understand the central case, it is fairly easy to reduce the general case to that (You just use completion of squares to get rid of the linear terms in the same way that you identify a circle with center not at the origin from its equation) In order to apply the linear algebra we have studied, we adopt a more systematic notation, using subscripted variables x,x instead of x, y Consider the central conic defined by the equation f(x) =a x +a x x + a x = C (The reason for the will be clear shortly) It is more useful to express the function f as follows f(x) =(x a + x a )x +(x a + x a )x = x (a x + a x )+x (a x + a x ), where we have introduced a = a The above expression may also be written in matrix form f(x) = x j a jk x k = x t Ax j,k= where A is the symmetric matrix of coefficients Note what has happened to the coefficients The coefficients of the squares appear on the diagonal of A, while the coefficient of the cross term bx x is divided into two equal parts Half of it appears as b in the, position (corresponding to the product x x ) while the other half appears as b in the, position (corresponding to the product x x which of course equals x x ) So it is clear why the matrix is symmetric This may be generalized to n> in a rather obvious manner Let A be a real symmetric n n matrix, and define f(x) = n j,k= For n = this may be written explicitly x j a jk x k = x t Ax f(x) =(x a + x a + x a )x +(x a + x a + x a )x +(x a + x a + x a )x = a x + a x + a x +a x x +a x x +a x x The rule for forming the matrix A from the equation for f(x) is the same as in the case The coefficients of the squares are put on the diagonal The
17 4 CLASSIFICATION OF CONICS AND QUADRICS 7 coefficient of a cross term involving x i x j is split in half, with one half put in the i, j position, and the other half is put in the j, i position The level set defined by f(x) =C is called a central hyperquadric It should be visualized as an n dimensional curved object in R n For n = it will be an ellipsoid or a hyperboloid (of one or two sheets) or perhaps a degenerate quadric like a cone (As in the case of conics, we must also allow linear terms to encompass paraboloids) If the above descriptions are accurate, we should expect the locus of the equation f(x) =Cto have certain axes of symmetry which we shall call its principal axes It turns out that these axes are determined by an orthonormal basis of eigenvectors for the coefficient matrix A To see this, suppose {u, u,,u n } is such a basis and P =[u u u n ] is the corresponding orthogonal matrix By the Principal Axis Theorem, P t AP = D is diagonal with the eigenvalues, λ,λ,,λ n,ofa appearing on the diagonal Make the change of coordinates x = P x where x represents the old coordinates and x represents the new coordinates Then f(x) =x t Ax=(Px ) t A(Px )=(x ) t P t AP x =(x ) t Dx Since D is diagonal, the quadratic expression on the right has no cross terms, ie λ λ (x ) t Dx =[x x x n] λ n =λ (x ) +λ (x ) + +λ n (x n) In the new coordinates, the equation takes the form λ (x ) + λ (x ) + +λ n (x n) =C and its graph is usually quite easy to describe Example We shall investigate the conic f(x, y) =x +4xy + y = First rewrite the equation x [ x y] = y (Note how the 4 was split into two symmetrically placed s) Next, find the eigenvalues of the coefficient matrix by solving λ det =( λ) 4=λ λ = λ This equation is easy to factor, and the roots are λ =,λ= For λ =, to find the eigenvectors, we need to solve v = v x x x n
18 8 III APPLICATIONS Reduction of the coefficient matrix yields with the general solution v = v, v free A basic normalized eigenvector is u = For λ =, a similar calculation (which you should make) yields the basic normalized eigenvector u = (Note that u u as expected) From this we can form the corresponding orthogonal matrix P and make the change of coordinates x x = P y y, and, according to the above analysis, the equation of the conic in the new coordinate system is (x ) (y ) = It is clear that this is a hyperbola with principal axes pointing along the new axes y y x x Example Consider the quadric surface defined by x + x + x x x = We take f(x) =x +x +x x x =[x x x ] x x x
19 4 CLASSIFICATION OF CONICS AND QUADRICS 9 Note how the coefficient in x x was split into two equal parts, a inthe,-position and a in the, -position The coefficients of the other cross terms were zero As usual, the coefficients of the squares were put on the diagonal The characteristic equation of the coefficient matrix is det λ λ =( λ) ( λ) = (λ )(λ )λ = λ Thus, the eigenvalues are λ =,, For λ =, reduce to obtain v = v,v = with v free Thus, v = is a basic eigenvector for λ =, and u = is a basic unit eigenvector Similarly, for λ = reduce which yields v = v = with v free Thus a basic unit eigenvector for λ =is u = Finally, for λ =, reduce This yields v = x,v = with v free Thus, a basic unit eigenvector for λ =is u =
20 III APPLICATIONS The corresponding orthogonal change of basis matrix is P =[u u u ]= Moreover, putting x = P x, we can express the equation of the quadric surface in the new coordinate system () x +x +x =x +x = Thus it is easy to see what this quadric surface is: an elliptical cylinder perpendicular to the x,x plane (This is one of the degenerate cases) The three principal axes in this case are the two axes of the ellipse in the x,x plane and the x axis, which is the central axis of the cylinder Tilted cylinder relative to the new axes The new axes are labelled Representing the graph in the new coordinates makes it easy to understand its geometry Suppose, for example, that we want to find the points on the graph which are closest to the origin These are the points at which the x -axis intersects the surface These are the points with new coordinates x = ±,x =x = If you want the coordinates of these points in the original coordinate system, use the change of coordinates formula x = P x Thus, the old coordinates of the minimum point with new coordinates (/,, ) are given by =
21 4 CLASSIFICATION OF CONICS AND QUADRICS Appendix A review of conics and quadrics You are probably familiar with certain graphs when they arise in standard configurations In two dimensions, the central conics have equations of the form ± x a ± y b = If both signs are +, the conic is an ellipse If one sign is + and one is, then the conic is a hyperbola The + goes with the axis which crosses the hyperbola Some examples are sketched below Ellipse Hyperbola Two signs result in an empty graph, ie, there are no points satisfying the equation Parabolas arise from equations of the form y = px with p Parabolas For x = py, the parabola opens along the positive or negative y-axis There are also some degenerate cases For example, x a y b = defines two lines which intersect at the origin In three dimensions, the central quadrics have equations of the form ± x a ± y b ± z c =
23 CLASSIFICATION OF CONICS AND QUADRICS Its graph is a double cone with elliptical cross sections Another would be ± x a ± y b = with at least one + sign Its graph is a cylinder perpendicular to the x, y-plane The cross sections are ellipses or hyperbolas, depending on the combination of signs Cone Cylinder Exercises for Section 4 Find the principal axes and classify the central conic x + xy + y = Identify the conic defined by x +4xy + y = 4 Find its principal axes, and find the points closest and furthest (if any) from the origin Identify the conic defined by x +7xy +y = Find its principal axes, and find the points closest and furthest (if any) from the origin 4 Find the principal axes and classify the central quadric defined by x y + z 4xy 4yz = (Optional) Classify the surface defined by x +y +z +xy +yz z = Hint: This is not a central quadric To classify it, first apply the methods of the section to the quadratic expression x +y +z +xy+yz to find a new coordinate system in which this expression has the form λ x + λ y + λ z Use the change of coordinates formula to express z in terms of x,y, and z and then complete squares to eliminate all linear terms At this point, it should be clear what the surface is
24 4 III APPLICATIONS Conics and the Method of Lagrange Multipliers There is another approach to finding the principal axes of a conic, quadric, or hyperquadric Consider for an example an ellipse in R centered at the origin One of the principal axes intersects the conic in the two points at greatest distance from the origin, and the other intersects it in the two points at least distance from the origin Similarly, two of the three principal axes of a central ellipsoid in R may be obtained in this way Thus, if we didn t know about eigenvalues and eigenvectors, we might try to find the principal axes by maximizing (or minimizing) the function giving the distance to the origin subject to the quadratic equation defining the conic or quadric In other words, we need to minimize a function given a constraint among the variables Such problems are solved by the method of Lagrange multipliers, which you learned in your multidimensional calculus course Here is a review of the method Suppose we want to maximize (minimize) the real valued function f(x) = f(x,x,,x n ) subject to the constraint g(x) = g(x,x,,x )=c For n =, this has a simple geometric interpretation The locus of the equation g(x,x )=cis a level curve of the function g, and we want to maximize (minimize) the function f on that curve Similarly, for n =, the level set g(x,x x )=cis a surface in R, and we want to maximize (minimize) f on that surface g( ) = c x g( ) = c x n = Level curve in the plane n = Level surface in space Examples Maximize f(x, y) =x +y on the ellipse g(x, y) =x +4y = (This is easy if you draw the picture) Minimize f(x, y, z) = x +xy + y + xz 4z on the sphere g(x, y, z) = x +y +z = Minimize f(x, y, z, t) =x +y +z t on the hypersphere g(x, y, z, t) = x +y +z +t = We shall concentrate on the case of n = variables, but the reasoning for any n is similar We want to maximize (or minimize) f(x) on a level surface g(x) =c in R, where as usual we abbreviate x =(x,x,x ) At any point x on the level surface at which such an extreme value is obtained, we must have () f(x) =λ g(x) for some scalar λ
25 CONICS AND THE METHOD OF LAGRANGE MULTIPLIERS f is parallel to g maximum point f other point () is a necessary condition which must hold at the relevant points (It doesn t by itself guarantee that there is a maximum or a minimum at the point There could be no extreme value at all at the point) In deriving this condition, we assume implicitly that the level surface is smooth and has a well defined normal vector g, and that the function f is also smooth If these conditions are violated at some point, that point could also be a candidate for a maximum or minimum Taking components, we obtain scalar equations for the 4 variables x,x,x,λ We would not expect, even in the best of circumstances to get a unique solution from this, but the defining equation for the level surface g(x) =c provides a 4th equation We still won t generally get a unique solution, but we will usually get at most a finite number of possible solutions Each of these can be examined further to see if f attains a maximum (or minimum) at that point in the level set Notice that the variable λ plays an auxiliary role since we really only want the coordinates of the point x (In some applications, λ has some significance beyond that) λ is called a Lagrange multiplier The method of Lagrange multipliers often leads to a set of equations which is difficult to solve However, in the case of quadratic functions f, there is a typical pattern which emerges Example Suppose we want to minimize the function f(x, y) =x +4xy + y on the circle x + y = For this problem n =, and the level set is a curve Take g(x, y) =x +y Then f = x +4y, 4x +y, g= x, y, and f = λ g yields the equations to which we add x +4y =λ(x) 4x +y =λ(y) x + y = g
26 6 III APPLICATIONS After canceling a common factor of, the first two equations may be written in matrix form x x =λ y y which says that x y is an eigenvector for the eigenvalue λ, and the equation x + y = says it is a unit eigenvector You should know how to solve such problems, and we leave it to you to make the required calculations (See also Example in the previous section where we made these calculations in another context) The eigenvalues are λ = and λ = For λ =, a basic unit eigenvector is u = [ and every other eigenvector is of the form cu The latter will be a unit vector if and only c =, ie, c = ± We conclude that λ = yields two solutions of the Lagrange multiplier problem: (/, / ) and ( /, / ) At each of these points f(x, y) =x +4xy + y = For λ =, we obtain the basic unit eigenvector u = [ and a similar analysis (which you should do) yields the two points: (/, / ) and ( /, / ) At each of these points f(x, y) =x +4xy + y = ], ], Min Max Max Min Hence, the function attains its maximum value at the first two points and its minimum value at the second two
27 CONICS AND THE METHOD OF LAGRANGE MULTIPLIERS 7 Example Suppose we want to minimize the function g(x, y) =x +y (which is the square of the distance to the origin) on the conic f(x, y) =x +4xy + y = Note that this is basically the same as the previous example except that the roles of the two functions are reversed The Lagrange multiplier condition g = λ f is the same as the condition f =(/λ) g provided λ (λ in this case since otherwise g =, which yields x = y = However, (, ) is not a point on the conic) We just solved that problem and found eigenvalues /λ = or /λ = In this case, we don t need unit eigenvectors, so to avoid square roots we choose basic eigenvectors v = and corresponding respectively to λ = and λ = The endpoint of v does not lie on the conic, but any other eigenvector for λ = is of the form cv, so all we need to do is adjust c so that the point satisfies the equation f(x, y) =x +4xy +y = Substituting (x, y) =(c, c) yields 6c =orc=±/ 6 Thus, we obtain the two points (/ 6, / 6) and ( / 6, / 6) For λ =, substituting (x, y) = ( c, c) in the equation yields c = which has no solutions Thus, the only candidates for a minimum (or maximum) are the first pair of points: (/ 6, / 6) and ( / 6, / 6) A simple calculation shows these are both / units from the origin, but without further analysis, we can t tell if this is the maximum, the minimum, or neither However, it is not hard to classify this conic see the previous section and discover that it is a hyperbola Hence, the two points are minimum points The Rayleigh-Ritz Method Example above is typical of a certain class of Lagrange multiplier problems Let A be a real symmetric n n matrix, and consider the problem of maximizing (minimizing) the quadratic function f(x) = x t Axsubject to the constraint g(x) = x = This is called the Rayleigh Ritz problem For n =orn=, the level set x = is a circle or sphere, and for n>, it is called a hypersphere Alternately, we could reverse the roles of the functions f and g, ie, we could try to maximize (minimize) the square of the distance to the origin g(x) = x on the level set f(x) = Because the Lagrange multiplier condition in either case asserts that the two gradients f and g are parallel, these two problems are very closely related The latter problem finding the points on a conic, quadric, or hyperquadric furthest from (closest to) the origin is easier to visualize, but the former problem maximizing or minimizing the quadratic function f on the hypersphere x = is easier to compute with Let s go about applying the Lagrange Multiplier method to the Rayleigh Ritz problem The components of g are easy: g =x i, i =,,n x i The calculation of f is harder First write n n f(x) = x j ( a jk x k ) j= k=
28 8 III APPLICATIONS and then carefully apply the product rule together with a jk = a kj The result is f x i = n a ij x j j= i =,,,n (Work this out explicitly in the cases n = and n = if you don t believe it) Thus, the Lagrange multiplier condition f = λ g yields the equations n a ij x j = λ(x i ) j= i =,,,n which may be rewritten in matrix form (after canceling the s) () Ax = λx To this we must add the equation of the level set g(x) = x = Thus, any potential solution x is a unit eigenvector for the matrix A with eigenvalue λ Note also that for such a unit eigenvector, we have f(x) =x t Ax=x t (λx)=λx t x=λ x =λ Thus the eigenvalue is the extreme value of the quadratic function at the point on the (hyper)sphere given by the unit eigenvector The upshot of this discussion is that for a real symmetric matrix A, the Rayleigh Ritz problem is equivalent to the problem of finding an orthonormal basis of eigenvectors for A The Rayleigh Ritz method may be used to show that the characteristic equation of a real symmetric matrix only has real eigenvalues This was an issue left unresolved in our earlier discussions Here is an outline of the argument The hypersphere g(x) = x = is a closed bounded set in R n for any n It follows from a basic theorem in analysis that any continuous function, in particular the quadratic function f(x), must attain both maximum and minimum values on the hypersphere Hence, the Lagrange multiplier problem always has solutions, which by the above algebra amounts to the assertion that the real symmetric matrix A must have at least one eigenvalue This suggests a general procedure for showing that all the eigenvalues are real First find the largest eigenvalue by maximizing the quadratic function f(x) on the set x = Let x = u be the corresponding eigenvector Change coordinates by choosing an orthonormal basis starting with u Then the additional basis elements will span the subspace perpendicular to u and we may obtain a lower dimensional quadratic function by restricting f to that subspace We can now repeat the process to find the next smaller real eigenvalue Continuing in this way, we will obtain an orthonormal basis of eigenvectors for A and each of the corresponding eigenvalues will be real
29 6 NORMAL MODES 9 Exercises for Section Find the maximum and minimum values of the function f(x, y) =x +y given the constraint x + xy + y = Find the maximum and/or minimum value of f(x, y, z) =x y +z 4xy 4yz subject to x + y + z = (Optional) The derivation of the Lagrange multiplier condition f = λ g assumes that the g, so there is a well defined tangent plane at the potential maximum or minimum point However, a maximum or minimum could occur at a point where g =, so all such points should also be checked (Similarly, either f or g might fail to be smooth at a maximum or minimum point) With these remarks in mind, find where f(x, y, z) =x +y +z attains its minimum value subject to the constraint g(x, y, z) =x +y z = 4 Consider as in Example the problem of maximizing f(x, y) =x +4xy + y given the constraint x + y = This is equivalent to maximizing F (x, y) =xy on the circle x +y = (Why?) Draw a diagram showing the circle and selected level curves F (x, y) =cof the function F Can you see why F (x, y) attains its maximum at (/, / ) and ( /, / ) without using any calculus? Hint: consider how the level curves of F intersect the circle and decide from that where F is increasing, and where it is decreasing on the circle 6 Normal Modes Eigenvalues and eigenvectors are an essential tool in solving systems of linear differential equations We leave an extended treatment of this subject for a course in differential equations, but it is instructive to consider an interesting class of vibration problems that have many important scientific and engineering applications We start with some elementary physics you may have encountered in a physics class Imagine an experiment in which a small car is placed on a track and connected to a wall though a stiff spring With the spring in its rest position, the car will just sit there forever, but if the car is pulled away from the wall a small distance and then released, it will oscillate back and forth about its rest position If we assume the track is so well greased that we can ignore friction, this oscillation will in principle continue forever k m We want to describe this situation symbolically Let x denote the displacement of the car from equilibrium, and suppose the car has mass m Hooke s Law tells x
30 4 III APPLICATIONS us that there is a restoring force of the form F = kx where k is a constant called the spring constant Newton s second law relating force and acceleration tells us () m d x dt = kx This is also commonly written d x dt + k x = You may have learned how to solve m this differential equation in a previous course, but in this particular case, it is not really necessary From the physical characteristics of the solution, we can pretty much guess what it should look like () x = A cos(ωt) where A is the amplitude of the oscillation and ω is determined by the frequency or rapidity of the oscillation It is usually called the angular frequency and it is related to the actual frequency f by the equation ω =πf A is determined by the size of the initial displacement It gives the maximum displacement attained as the car oscillates ω however is determined by the spring constant To see how, just substitute () in () We get m( ω A cos(ωt)) = ka cos(ωt) which after canceling common factors yields mω = k k or ω = m The above discussion is a bit simplified We could not only have initially displaced the car from rest, but we could also have given it an initial shove or velocity In that case, the maximal displacement would be shifted in time The way to describe this symbolically is x = A cos(ωt + δ) where δ is called the phase shift This complication does not change the basic character of the problem since it is usually the fundamental vibration of the system that we are interested in, and that turns out to be the same if we include a possible phase shift We now want to generalize this to more than one mass connected by several springs This may seem a bit bizarre, but it is just a model for situations commonly met in scientific applications For example, in chemistry, one often needs to determine the basic vibrations of a complex molecule The molecule consists of atoms connected by interatomic forces As a first approximation, we may treat the atoms as point masses and the forces between them as linear restoring forces from equilibrium positions Thus the mass-spring model may tell us something useful about real problems
31 6 NORMAL MODES 4 Example Consider the the configuration of masses and springs indicated below, where m is the common mass of the two particles and k is the common spring constant of the three springs k m k m k x x Look at the first mass When it is displaced a distance x to the right from equilibrium, it will be acted upon by two forces Extension of the spring on the left will pull it back with force kx At the same time, the spring in the middle will push or pull it depending on whether it is compressed or stretched If x is the displacement of the second mass from equilibrium, the change in length of the second spring will be x x, so the force on the first mass will be k(x x ) This yields a total force of kx k(x x )= kx + kx A similar analysis works for the second mass differential equations Thus, we obtain the system of m d x dt = kx + kx m d x dt = kx kx The system may also be rewritten in matrix form () m d x dt = [ k k k k ] x where x = Note that the matrix on the right is a symmetric matrix This is always the case in such problems It is an indirect consequence of Newton s third law which asserts that the forces exerted by two masses on each other must be equal and opposite To solve this, we look for solutions of the form [ x x ] (4) x = v cos(ωt) x = v cos(ωt) In such a solution, the two particles oscillate with the same frequency but with possibly different amplitudes v and v Such a solution is called a normal mode General motions of the system can be quite a bit more complicated First of all, we have to worry about possible phase shifts More important, we also have to
32 4 III APPLICATIONS allow for linear combinations of the normal modes in which there is a mixture of different frequencies In this way the situation is similar to that of a musical instrument which may produce a complex sound which can be analyzed in terms of basic frequencies or harmonics We leave such complications for another course Here we content ourselves at doing the first step, which is to find the fundamental oscillations or normal modes (4) may be rewritten in matrix form () x = v cos(ωt) where ω and v are to be determined Then d x dt Hence, putting () in () yields = ω v cos(ωt) [ m( ω k k v cos(ωt)) = k k ] vcos(ωt) Now factor out the common scalar factor cos(ωt) to obtain ω k k mv = v k k Note that the amplitude v is a vector in this case, so we cannot cancel it as we did in the case of a single particle The above equation may now be rewritten v = ω m k v This is a trifle messy, but if we put abbreviate λ = ω m k for the scalar on the right, we can write it v = λv This equation should look familiar It says that v is an eigenvector for the matrix on the left, and that λ = ω m is the corresponding eigenvalue However, k we know how to solve such problems First we find the eigenvalues by solving the characteristic equation For each eigenvalue, we can find the corresponding λm frequency ω from ω = Next, for each eigenvalue, we can determine basic k eigenvectors as before In this example, the characteristic equation is λ det =( λ) λ =λ +4λ+4 =λ +4λ+ =(λ+ )(λ +)=
33 6 NORMAL MODES 4 Hence, the roots are λ = (ω= k/m) and λ = (ω= k/m) For λ = (ω= k/m), finding the eigenvectors results in reducing the matrix + = + Hence, the solution is v = v with v free A basic solution vector for the subspace of solutions is v = The corresponding normal mode has the form x = cos( k/m t) Note that x (t) =x (t) for all t, so the two particles move together in tandem with the same angular frequency k/m This behavior of the particles is a consequence of the fact that the components of the basic vector v are equal Similarly, for λ = (ω= k/m), we have + = + The solution is v = v with v free, and a basic solution vector for the system is The corresponding normal mode is is x = v = cos( k/m t) Note that x (t) = x (t) for all t, so the two masses move opposite to one another k with the same amplitude and angular frequency m Note that in the above example, we could have determined the two vectors v and v by inspection As noted, the first corresponds to motion in which the particles move in tandem and the spring between them experiences no net change in length The second corresponds to motion in which the particles move back and forth equal amounts in opposite directions but with the same frequency This would have simplified the problem quite a lot For, if you know an eigenvector of a matrix, it is fairly simple to find the corresponding eigenvalue, and hence the angular frequency In fact, it is often true that careful consideration of the physical arrangement of the particles, with particular attention to any symmetries that may be present, may suggest possible normal modes with little or no calculation
34 44 III APPLICATIONS Relation to the Principal Axis Theorem As noted above normal mode problems typically result in systems of the form d x (7) dt = Ax where A is a real symmetric matrix (In the case that all the particles have the same mass, A = K, where K is a symmetric matrix of spring constants If the m masses are different, the situation is a bit more complicated, but the problem may still be restated in the above form) If P is a matrix with columns the elements of a basis of eigenvectors for A, then we saw earlier that AP = PD where D is a diagonal matrix with the eigenvalues on the diagonal Assume we make the change of coordinates x = P x Then d P x dt = AP x P d x = AP x dt d x dt = P AP x = Dx However, since D is diagonal, this last equation may be written as n scalar equations d x j dt = λ jx j j =,,,n In the original coordinates, the motions of the particles are coupled since the motion of each particle may affect the motion of the other particles In the new coordinate system, these motions are decoupled The new coordinates are called normal coordinates Each x j may be thought of as the displacement of one of n fictitious particles, each of which oscillates independently of the others in one of n mutually perpendicular directions The physical significance in terms of the original particles of each normal coordinate is a but murky, but they presumably represent underlying structure of some importance Example, revisited d x dt = k m x A basis of eigenvectors for the coefficient matrix is as before { } v =, v =
35 6 NORMAL MODES 4 If we divide the vectors by their lengths, we obtain the orthonormal basis { }, This in turn leads to the change of coordinates matrix [ ] P = x x x π /4 π /4 x If you look carefully, you will see this represents a rotation of the original x,x - axes through an angle π/4 However, this has nothing to do with the original geometry of the problem x and x stand for displacements of two different particles along the same one dimensional axis The x,x plane is a fictitious configuration space in which a single point represents a pair of particles It is not absolutely clear what a rotation of axes means for this plane, but the new normal coordinates x,x obtained thereby give us a formalism in which the normal modes appear as decoupled oscillations Exercises for Section 6 Determine the normal modes, including frequencies and relative motions for the system m d x dt = k(x x )= kx + kx m d x dt = k(x x )+k(x x )=kx kx + kx m d x dt = k(x x )=kx kx
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Associated Topics || Dr. Math Home || Search Dr. Math
```Date: 12/07/2003 at 20:55:16
From: Kristina
I don't understand on how to solve radical equations. Example:
__
\/5x = 1.5
I always seem to lose a step, or skip a step.
```
```
Date: 12/08/2003 at 12:31:11
From: Doctor Riz
Hi Kristina -
Thanks for writing Dr. Math. Radical equations are one of the trickier
things in algebra because they have some subtle points that need to be
The general rule for solving radical equations is to isolate the
radical on one side of the equation, then raise both sides to whatever
power you need to undo or cancel the radical.
For example, if you have a square root, you will square both sides of
the equation. If you have a cube root, you will cube both sides.
Here's an example:
__
\/x = 5
Square both sides, and get x = 25.
So in your problem, you have:
__
\/5x = 1.5
Squaring both sides gives 5x = 2.25, and then you just need to divide
by 5 to find x.
Now, here are the two main subtleties that come into play on radical
equations. First, when you square both sides, you have to square the
entire side, not piece by piece. Suppose you have:
_
\/x = x + 2
When you square both sides, you can't just square the x and the 2
separately on the right. You have to take the whole side and square
it:
_
\/x = x + 2
x = (x + 2)^2
x = x^2 + 4x + 4
Many students make the mistake of just squaring each piece and getting
x^2 + 4 on the right rather than x^2 + 4x + 4.
The second and more common subtlety involves what happens when you
square both sides. Here's an untrue statement:
-2 = 2
But when I square both sides, that untrue statement becomes:
4 = 4
which is clearly true! This means that sometimes when you square both
sides of an equation you wind up with answers that don't really check
in the original equation. Therefore, you ALWAYS have to check your
answer and make sure it works. Here's an example:
_
\/x = -3
Squaring both sides gives x = 9, but when we check that we have:
_
\/9 = -3
Is that a true statement? Remember that the square root of 9 could be
3 or -3, but we agree that the way the radical is written will tell us
which one we mean:
_
\/9 = 3
_
-\/9 = -3
So in this example, the answer of 9 does not work because it gives us
3 = -3
when we check it. That means there is no possible solution to the
given equation.
Confusing? Sometimes, yes. The things to remember as you work on
1) Isolate the radical and raise to a power to get rid of it.
2) When you raise to that power, raise both sides of the equation.
You can't do it term-by-term. Slap ( ) around the other side
of the equation and put the exponent outside the ( ) to remind
you to raise the whole side to that power.
work or be off by a negative, in which case you have to throw
it out.
Hope that helps. Good luck, and write back if you need more help on
this.
- Doctor Riz, The Math Forum
http://mathforum.org/dr.math/
```
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High School Square & Cube Roots
Middle School Square Roots
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# Basic Description
Logarithms are considered the inverse or opposite operations to exponents, just as subtraction is the inverse to addition or square rooting is the inverse to squaring.
For example, suppose we have the exponential expression $2^3$ , which we know will equal 8. Now, suppose we want to do the inverse operation and go from the value 8 to the exponent 3 with a base of 2. We could do the inverse operation by using logarithms and write $log_{2}(8) = 3$ , which is read "logarithm of 8 base 2 is equal to 3".
In order words, if we have an exponential equation: $base^x = n\,$
we can write an equivalent logarithmic equation: $log_{base}(n) = x\,$
What about exponential equations such as $10^x = 932$? It might seem harder to solve for x in this case because there is no that will give us the value of 932 with a base of 10. However, if we simply rewrite the equation as an logarithmic equation $x = log_{10}(932)$ , we can find quite easily with a calculator that x is about 2.969.
To look at some more examples of switching between exponential and logarithmic equations:
Exponential Equation Logarithmic Equation
$7^3 = 343 \,$
$6 = log_{3}(729)\,$
$4 = log_{10}(10000)\,$
$e^7 \approx 1096.6 \,$
$3.5 \approx log_{6}(529.1)\,$
$22^x = 57643 \,$
# A More Mathematical Description
Definition of a Logarithm
$y = log_a(x)\,$ if and only if $x = a^y\,$, where b > 1 and x > 0
In words: The logarithm of a value at a given base is the power (exponent) that the base must be raised to produce the value.
## Bases
As seen from the definition above, the base of a logarithm affects how a logarithm is evaluated. Bases can be any positive number except for 1, and the logarithms of a value can be found at different bases using a change of base formula, as shown in the section below the Common Bases chart.
### Common Bases
There are three main bases that are most frequently used:
Base Exponential Representation Logarithmic Representation Notes Example
Base 10 $y = 10^x\,$ $x = log_{10}(y)\,$ can be written simply as $log(x)\,$
also called Common Logarithms
$100 = 10^x\,$
$x = log(100)\,$
where x = 2
Base 2 $y = 2^x\,$ $x = log_2(y)\,$ basis for the Binary System $16 = 2^x\,$
$x = log_{2}(16)\,$
where x = 4
Base $y = e^x\,$ $x = log_{e}(y)\,$ can be written simply as $ln(y)\,$
also called Natural Logarithms
where $ln(e) = 1\,$
$25 = e^x\,$
$x = ln(25)\,$
where $x \approx 3.22$
### Changing Bases
To go from a logarithm of base k to a logarithm of base a, we use the formula:
$\frac{log_k(x)}{log_k(a)} = log_a(x) \,$
$x = a^n \,$
$log_k(x) = log_k(a^n) \,$ Taking the $log_k$ of both sides
$log_k(x) = nlog_k(a) \,$ Using the exponential property of logarithms
$\frac{log_k(x)}{log_k(a)} = n \,$Dividing by $log_k(a)$
$\frac{log_k(x)}{log_k(a)} = log_a(x) \,$From a logarithm's definition $x = a^n$ and thus $n = log_a(x)$
For example, let us take the , which we are given, and find the , which we do not know.
We would designate k = 10, a = 2, and x = 14.
$log_a(x)=\frac{log_k(x)}{log_k(a)}$
$log_{2}(14)=\frac{log_{(10)}(14)}{log_{(10)}(12)}$
$log_2(14) \approx \frac{1.146}{0.301}$
$log_2(14) \approx 3.807$
## Graphing a Logarithmic Function
The image on the left is a graph of a basic logarithmic function: $y = log(x)\,$. A logarithmic function, such as the one used to create the featured image, takes the basic form $y = log_b(x)\,$, where b is fixed while y and x are variables. In addition, there is an vertical at x = 0, so that logarithmic functions are undefined when x is less than or equal to 0. However, non-real logarithms for negative x values can be found using complex logarithms with complex numbers.
The second image compares the graphical representations of the most common logarithmic functions: $y = log(x)\,$, $y = log_2(x)\,$, and $y = log_e(x)\,$.
## Basic Properties of Logarithms
Logarithms possess various properties and identities including the following:
Identities
Multiplication $log(x * y) = log(x) + log(y) \,$
Division $log\left(\frac{x}{y}\right) = log(x) - log(y) \,$
Exponentiation $log(x^n) = nlog(x)\,$
Integration $\int log(x) = xlog(x) - x + constant\,$
Differentiation $\frac{d}{dx}(log(x)) = \frac{1}{x} \,$
Other properties
$log(1) = 0\,$
$log(0) = undefined\,$
$log_a(a^x) = x\,$
$a^{log_a(x)} = x \,$
$log_a(b) = \frac{1}{log_b(a)}\,$ |
# What is the limit of (x +sqrt((x^2)+(3x))) as x goes to infinity?
Oct 6, 2015
${\lim}_{x \rightarrow \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = \infty$
#### Explanation:
$\left(x + \sqrt{{x}^{2} + 3 x}\right) = x + \sqrt{{x}^{2} \left(1 + \frac{3}{x}\right)}$ for $x \ne 0$
= x +sqrt(x^2)sqrt(1+3/x)) for $x \ne 0$
= x +absx sqrt(1+3/x)) for $x \ne 0$
Now, as $x \rightarrow \infty$, we have
$\left\mid x \right\mid \rightarrow \infty$ and
$\sqrt{1 + \frac{3}{x}} \rightarrow \sqrt{1 + 0} = 1$
So,
${\lim}_{x \rightarrow \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = {\lim}_{x \rightarrow \infty} \left(x + \left\mid x \right\mid \sqrt{1 + \frac{3}{x}}\right) = \infty$
Bonus
As $x$ decreases without bound, the limit is quite different.
As $x \rightarrow - \infty$, we get indeterminate form: $- \infty + \infty$.
We think of $\left(x + \sqrt{{x}^{2} + 3 x}\right)$ as a ratio (over $1$) and eliminate the square root from the numerator.
$\left(x + \sqrt{{x}^{2} + 3 x}\right) \cdot \frac{x - \sqrt{{x}^{2} + 3 x}}{x - \sqrt{{x}^{2} + 3 x}} = \frac{- 3 x}{x - \sqrt{{x}^{2} + 3 x}}$
$= \frac{- 3 x}{x - \left\mid x \right\mid \sqrt{1 + \frac{3}{x}}}$.
For $x < 0$, we have $\left\mid x \right\mid = - x$, so we have:
${\lim}_{x \rightarrow - \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = {\lim}_{x \rightarrow - \infty} \frac{- 3 x}{x - \left\mid x \right\mid \sqrt{1 + \frac{3}{x}}}$
$= {\lim}_{x \rightarrow - \infty} \frac{- 3 x}{x + x \sqrt{1 + \frac{3}{x}}}$
$= {\lim}_{x \rightarrow - \infty} \frac{- 3}{1 + \sqrt{1 + \frac{3}{x}}}$
$= - \frac{3}{2}$.
Oct 6, 2015
${\lim}_{x \to + \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = + \infty$
${\lim}_{x \to - \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = - \frac{3}{2}$
#### Explanation:
$\sqrt{{x}^{2} + 3 x} \ge 0$ for all $x \ge 0$
So
${\lim}_{x \to + \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) \ge {\lim}_{x \to + \infty} x = + \infty$
Much more interesting is the case ${\lim}_{x \to - \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right)$
First notice that ${\lim}_{t \to \infty} \left(t - \sqrt{{t}^{2} - C}\right) = 0$ for any constant $C > 0$
To see this: Given $\epsilon > 0$
${\left(t - \epsilon\right)}^{2} = {t}^{2} - 2 t \epsilon + {\epsilon}^{2}$
So if $t > \frac{C + {\epsilon}^{2}}{2 \epsilon}$, then
${\left(t - \epsilon\right)}^{2} = {t}^{2} - 2 t \epsilon + {\epsilon}^{2} < {t}^{2} - \left(C + {\epsilon}^{2}\right) + {\epsilon}^{2} = {t}^{2} - C$
So
$t - \epsilon < \sqrt{{t}^{2} - C} < \sqrt{{t}^{2}} = t$
Let $t = - \left(x + \frac{3}{2}\right)$ and $C = \frac{9}{4}$
Then:
$0 = {\lim}_{t \to \infty} \left(\sqrt{{t}^{2} - C} - t\right)$
$= {\lim}_{x \to - \infty} \left(\sqrt{{\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{4}} + x + \frac{3}{2}\right)$
$= {\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x + \frac{9}{4} - \frac{9}{4}} + x + \frac{3}{2}\right)$
$= {\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x} + x + \frac{3}{2}\right)$
$= {\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x} + x\right) + \frac{3}{2}$
Hence:
${\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x} + x\right) = - \frac{3}{2}$ |
### Learning Outcomes
Identify rational number from a list of numbersIdentify irrational numbers from a list of numbers
In this chapter, we’ll make sure your an abilities are firmly set. We’ll take one more look at the kinds of numbers we have operated with in every previous chapters. We’ll job-related with properties of numbers that will help you enhance your number sense. And we’ll exercise using them in methods that we’ll use as soon as we solve equations and complete other actions in algebra.
You are watching: Is 1/8 an irrational number
We have already described numbers together counting numbers, entirety numbers, and also integers. Carry out you remember what the distinction is amongst these species of numbers?
counting numbers 1,2,3,4dots whole numbers 0,1,2,3,4dots integers dots -3,-2,-1,0,1,2,3,4dots
### Rational Numbers
What form of numbers would certainly you gain if you started with all the integers and then had all the fractions? The numbers you would have form the collection of reasonable numbers. A rational number is a number that have the right to be composed as a ratio of 2 integers.
### Rational Numbers
A reasonable number is a number that deserve to be composed in the form fracpq, wherein p and also q space integers and also q e o.
All fractions, both positive and also negative, space rational numbers. A few examples are
frac45,-frac78,frac134, extand-frac203
Each numerator and also each denominator is one integer.
We have to look at every the numbers we have used therefore far and verify that they space rational. The an interpretation of rational number tells us that all fractions are rational. We will currently look in ~ the counting numbers, entirety numbers, integers, and decimals come make certain they space rational.Are integers reasonable numbers? To decision if an essence is a reasonable number, we try to compose it as a ratio of two integers. An easy method to do this is to create it as a fraction with denominator one.
3=frac31-8=frac-810=frac01
Since any kind of integer deserve to be created as the ratio of 2 integers, every integers room rational numbers. Remember the all the count numbers and also all the whole numbers are additionally integers, and also so they, too, are rational.
What around decimals? are they rational? Let’s look in ~ a few to watch if we have the right to write every of them together the proportion of two integers. We’ve already seen the integers are rational numbers. The essence -8 could be created as the decimal -8.0. So, clearly, some decimals room rational.
Think around the decimal 7.3. Deserve to we compose it together a proportion of two integers? due to the fact that 7.3 means 7frac310, we deserve to write it together an not correct fraction, frac7310. So 7.3 is the ratio of the integers 73 and 10. It is a rational number.
In general, any type of decimal that ends after ~ a variety of digits such together 7.3 or -1.2684 is a reasonable number. We can use the ar value the the last digit as the denominator when writing the decimal together a fraction.
### example
Write each together the ratio of two integers:
1. -15
2. 6.81
3. -3frac67
Solution:
1. -15 Write the integer as a portion with denominator 1. frac-151
2. 6.81 Write the decimal together a combined number. 6frac81100 Then transform it come an improper fraction. frac681100
3. -3frac67 Convert the mixed number come an not correct fraction. -frac277
### try it
Let’s look in ~ the decimal kind of the number we recognize are rational. We have seen that every creature is a reasonable number, due to the fact that a=fraca1 for any integer, a. We deserve to also adjust any integer come a decimal by including a decimal suggest and a zero.
Integer -2,-1,0,1,2,3
Decimal -2.0,-1.0,0.0,1.0,2.0,3.0These decimal number stop.
We have additionally seen that every portion is a rational number. Look at the decimal kind of the fountain we just considered.
Ratio the Integers frac45,frac78,frac134,frac203
Decimal develops 0.8,-0.875,3.25,-6.666ldots,-6.overline66These decimal either stop or repeat.
What do these examples tell you? Every reasonable number deserve to be created both as a ratio of integers and as a decimal that either stop or repeats. The table listed below shows the numbers us looked at expressed as a proportion of integers and as a decimal.
Rational Numbers
FractionsIntegers
Numberfrac45,-frac78,frac134,frac-203-2,-1,0,1,2,3
Ratio the Integerfrac45,frac-78,frac134,frac-203frac-21,frac-11,frac01,frac11,frac21,frac31
Decimal number0.8,-0.875,3.25,-6.overline6-2.0,-1.0,0.0,1.0,2.0,3.0
### Irrational Numbers
Are there any decimals that carry out not stop or repeat? Yes. The number pi (the Greek letter pi, pronounce ‘pie’), i m sorry is an extremely important in relenten circles, has actually a decimal form that does not protect against or repeat.
pi = ext3.141592654…….Similarly, the decimal depictions of square roots of number that room not perfect squares never stop and also never repeat. Because that example,
sqrt5= ext2.236067978…..A decimal the does no stop and does no repeat cannot be created as the proportion of integers. We contact this kind of number an irrational number.
### Irrational Number
An irrational number is a number that cannot be created as the ratio of 2 integers. That is decimal form does not stop and does no repeat.
Let’s summary a technique we have the right to use to determine whether a number is reasonable or irrational.If the decimal kind of a number
stops or repeats, the number is rational.does not stop and does not repeat, the number is irrational.
### example
Identify each of the adhering to as rational or irrational:1. 0.58overline32. 0.4753. 3.605551275dots
Show Solution
Solution:1. 0.58overline3The bar above the 3 suggests that that repeats. Therefore, 0.58overline3 is a repeating decimal, and also is because of this a rational number.
2. 0.475This decimal stops after the 5, so the is a reasonable number.
See more: How Many Is A Yoke Of Oxen, Wednesday Economic History
3. 3.605551275dotsThe ellipsis (dots) method that this number does not stop. Over there is no repeating pattern of digits. Since the number doesn’t stop and also doesn’t repeat, that is irrational. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Graphs Using Slope-Intercept Form
## Use the y-intercept and the 'rise over run' to graph a line
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MCC9 - 12.F.IF.7a GraphLinear Functions (in Slope-Intercept Form)
Standard
MCC9-12.F.IF.7a Graph linear functions and show intercepts.
Suppose the linear function f(x) = -0.25x + 10 represents the amount of money you have left to play video games, where f(x) is the amount of money you have left and x is the number of video games that you have played so far. Do you know how to graph this function? What would be the slope and intercept of the graph? In this Concept, you'll learn how to graph linear functions like this one by finding the graph's slope and intercept.
### Guidance
You can see that the notations are interchangeable. This means you can substitute the notation for and use all the concepts you have learned on linear equations.
#### Example A
\begin{align*}\text{Graph} \ f(x)& =\frac{1}{3}x+1.\\ \text{Replace} \ f(x)& = \text{with} \ y=.\\ y& =\frac{1}{3} x+1\end{align*}
This equation is in slope-intercept form. You can now graph the function by graphing the intercept and then using the slope as a set of directions to find your second coordinate.
### Try This
To get a better understanding of what happens when you change the slope or the \begin{align*}y-\end{align*}intercept of a linear equation, try playing with the Java applet at http://standards.nctm.org/document/eexamples/chap7/7.5/index.htm.
Identify Slope and \begin{align*}y-\end{align*}intercept
So far, we’ve been writing a lot of our equations in slope-intercept form—that is, we’ve been writing them in the form \begin{align*}y = mx + b\end{align*}, where \begin{align*}m\end{align*} and \begin{align*}b\end{align*} are both constants. It just so happens that \begin{align*}m\end{align*} is the slope and the point \begin{align*}(0, b)\end{align*} is the \begin{align*}y-\end{align*}intercept of the graph of the equation, which gives us enough information to draw the graph quickly.
#### Example A
Identify the slope and \begin{align*}y-\end{align*}intercept of the following equations.
a) \begin{align*}y = 3x + 2\end{align*}
b) \begin{align*}y = 0.5x - 3\end{align*}
c) \begin{align*}y = -7x\end{align*}
d) \begin{align*}y = -4\end{align*}
Solution
a) Comparing , we can see that \begin{align*}m = 3\end{align*} and \begin{align*}b = 2\end{align*}. So \begin{align*}y = 3x + 2\end{align*} has a slope of 3 and a \begin{align*}y-\end{align*}intercept of (0, 2).
b) has a slope of 0.5 and a \begin{align*}y-\end{align*}intercept of (0, -3).
Notice that the intercept is negative. The \begin{align*}b-\end{align*}term includes the sign of the operator (plus or minus) in front of the number—for example, \begin{align*}y = 0.5x - 3\end{align*} is identical to \begin{align*}y = 0.5x + (-3)\end{align*}, and that means that \begin{align*}b\end{align*} is -3, not just 3.
c) At first glance, this equation doesn’t look like it’s in slope-intercept form. But we can rewrite it as \begin{align*}y = -7x + 0\end{align*}, and that means it has a slope of -7 and a \begin{align*}y-\end{align*}intercept of (0, 0). Notice that the slope is negative and the line passes through the origin.
d) We can rewrite this one as \begin{align*}y = 0x - 4\end{align*}, giving us a slope of 0 and a \begin{align*}y-\end{align*}intercept of (0, -4). This is a horizontal line.
Graph an Equation in Slope-Intercept Form
Once we know the slope and intercept of a line, it’s easy to graph it.
Ali is trying to work out a trick that his friend showed him. His friend started by asking him to think of a number, then double it, then add five to the result. Ali has written down a rule to describe the first part of the trick. He is using the letter \begin{align*}x\end{align*} to stand for the number he thought of and the letter \begin{align*}y\end{align*} to represent the final result of applying the rule. He wrote his rule in the form of an equation: \begin{align*}y = 2x + 5\end{align*}.
Help him visualize what is going on by graphing the function that this rule describes.
In that example, we constructed a table of values, and used that table to plot some points to create our graph.
We also saw another way to graph this equation. Just by looking at the equation, we could see that the \begin{align*}y-\end{align*}intercept was (0, 5), so we could start by plotting that point. Then we could also see that the slope was 2, so we could find another point on the graph by going up 2 units and right one unit. The graph would then be the line between those two points.
Here’s another problem where we can use the same method.
#### Example C
Graph the following function: \begin{align*}y=-3x+5\end{align*}
Solution
To graph the function without making a table, follow these steps:
1. Identify the \begin{align*}y-\end{align*}intercept: \begin{align*}b = 5\end{align*}
2. Plot the intercept: (0, 5)
3. Identify the slope: \begin{align*}m = -3\end{align*}. (This is equal to \begin{align*}\frac {-3}{1}\end{align*}, so the rise is -3 and the run is 1.)
4. Move down 3 units and right one unit to find another point on the line: (1, 2)
5. Draw the line through the points (0, 5) and (1, 2).
Notice that to graph this equation based on its slope, we had to find the rise and run—and it was easiest to do that when the slope was expressed as a fraction. That’s true in general: to graph a line with a particular slope, it’s easiest to first express the slope as a fraction in simplest form, and then read off the numerator and the denominator of the fraction to get the rise and run of the graph.
#### Example D
Find integer values for the rise and run of the following slopes, then graph lines with corresponding slopes.
a) \begin{align*}m=3\end{align*}
b) \begin{align*}m=-2\end{align*}
Solution
a)
b)
### Vocabulary
• A common form of a line (linear equation) is slope-intercept form: \begin{align*}y=mx+b\end{align*}, where \begin{align*}m\end{align*} is the slope and the point \begin{align*}(0, b)\end{align*} is the \begin{align*}y-\end{align*}intercept
• Graphing a line in slope-intercept form is a matter of first plotting the \begin{align*}y-\end{align*}intercept \begin{align*}(0, b)\end{align*}, then finding a second point based on the slope, and using those two points to graph the line.
### Guided Practice
Find integer values for the rise and run of the following slopes, then graph lines with corresponding slopes.
a) \begin{align*}m=0.75\end{align*}
b) \begin{align*}m=-0.375\end{align*}
Solution:
a)
b)
### Practice
Identify the slope and \begin{align*}y-\end{align*}intercept for the following equations.
1. \begin{align*}y=2x+5\end{align*}
2. \begin{align*}y=-0.2x+7\end{align*}
3. \begin{align*}y=x\end{align*}
4. \begin{align*}y=3.75\end{align*}
Identify the slope of the following lines.
Identify the slope and \begin{align*}y-\end{align*}intercept for the following functions.
For 7-10, plot the following functions on a graph.
1. \begin{align*}y=2x+5\end{align*}
2. \begin{align*}y=-0.2x+7\end{align*}
3. \begin{align*}y=x\end{align*}
4. \begin{align*}y=3.75\end{align*}
1. Which two of the following lines are parallel?
1. \begin{align*}y=2x+5\end{align*}
2. \begin{align*}y=-0.2x+7\end{align*}
3. \begin{align*}y=x\end{align*}
4. \begin{align*}y=3.75\end{align*}
5. \begin{align*}y= -\frac{1}{5}x-11\end{align*}
6. \begin{align*}y=-5x+5\end{align*}
7. \begin{align*}y=-3x+11\end{align*}
8. \begin{align*}y=3x+3.5\end{align*}
2. What is the \begin{align*}y-\end{align*}intercept of the line passing through (1, -4) and (3, 2)?
3. What is the \begin{align*}y-\end{align*}intercept of the line with slope -2 that passes through (3, 1)?
4. Line \begin{align*}A\end{align*} passes through the points (2, 6) and (-4, 3). Line \begin{align*}B\end{align*} passes through the point (3, 2.5), and is parallel to line \begin{align*}A\end{align*}
1. Write an equation for line \begin{align*}A\end{align*} in slope-intercept form.
2. Write an equation for line \begin{align*}B\end{align*} in slope-intercept form.
5. Line \begin{align*}C\end{align*} passes through the points (2, 5) and (1, 3.5). Line \begin{align*}D\end{align*} is parallel to line \begin{align*}C\end{align*}, and passes through the point (2, 6). Name another point on line \begin{align*}D\end{align*}. (Hint: you can do this without graphing or finding an equation for either line.)
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
# Proof the recurrence relation of order statistics using induction
I read order statistics from the book The Design and Analysis of Computer Algorithms", by Aho, Hopcroft, Ullman, Addison-Wesley
As per the algorithm the recurrence relation given
T(n)<= cn for n<=49 .....1
T(n)<= T(n/5) + T(3n/4) + cn for n>=50 .....2
How can I prove it by induction T(n)<= 20cn?
I have two more questions
For the 1st equation why they define n<=49 and for the 2nd equation n>=50
Thank you.
## 1 Answer
The first equation is the base case for the recurrence. The second equation represents the recurrence formula, on everything that isn't a base case.
Now, you can prove by induction this statemen, simply by using complete induction as follows:
The base case is obviously correct. For the induction hypothesis, assume correctness for any $$k.
Now, for $$T(n)$$, we can use the recurrence formula to get $$T(n)=T\left(\frac{n}{5}\right) + T\left(\frac{3n}{4}\right) + cn$$ And notice that $$\frac{n}{5},\frac{3n}{4}, and hence we can use the induction hypothesis on them. Doing so, will yield: $$T(n)\le \frac{20cn}{5}+\frac{20c\cdot 3n}{4}+cn=4cn+15cn+cn=20cn$$ And hence we also proved that $$T(n)\le 20cn$$ and with that we completed the induction proof.
• The induction hypothesis is P(k )holds for an arbitrary integer k , need to proof that P (k + 1) is also true. I can not understand where do you use induction hypothesis? Additionally, T(n)<= 20cn/5+20c.3n/4+cn then, how could you make it equal to(=) 4cn+15cn+cn. Commented Oct 5, 2021 at 23:52
• Complete induction assumes $P(k)$ for all $k<n$, and you will have to prove $P(n)$. For your other doubt, the equality is between $20cn/5+20c\cdot3n/4+cn$ and $4cn+15cn+cn$, its just writing a simpler form for the same value Commented Oct 6, 2021 at 7:59
• Thanks. However as per my last question I would like to know that how could you convert inequality to equality? Commented Oct 6, 2021 at 15:11
• Why would you want equality? The inequality is between $T(n)$ and $20cn$ Commented Oct 6, 2021 at 16:17 |
# Proportional Relationships
Proportional relationships form the very fabric of our mathematical understanding, weaving connections between quantities in a way that reflects balance and harmony. At the heart of these relationships lies the concept that two or more quantities vary in such a manner that their ratios remain constant. This fundamental principle finds applications across various fields, from everyday scenarios to complex mathematical models, making it an indispensable tool for understanding and solving problems. In this exploration, we will delve into the essence of proportional relationships, their properties, real-world applications, and the role they play in shaping our mathematical comprehension.
## Understanding Proportional Relationships:
At its core, a proportional relationship signifies that two quantities change in size in a consistent manner. Mathematically, if two variables, say x and y, are in proportion, their ratio, $$\frac{x}{y}$$, remains constant. This fundamental idea can be expressed as an equation: $$y = kx$$, where $$k$$ is the constant of proportionality. In this equation, as $$x$$ changes, $$y$$ changes in a way that maintains the same ratio throughout.
## Properties of Proportional Relationships:
### 1. Direct Variation:
One of the most straightforward types of proportional relationships is direct variation. In this scenario, as one quantity increases (or decreases), the other quantity also increases (or decreases) in a linear fashion. The constant of proportionality in this case is simply the slope of the line that represents the relationship on a graph.
### 2. Inverse Variation:
Inverse variation occurs when one quantity increases as the other decreases, and vice versa. The classic example of this is the relationship between speed and time, where speed is inversely proportional to the time taken to cover a certain distance. The product of the two variables remains constant.
### 3. Nonlinear Proportional Relationships:
Proportional relationships need not always be linear. Nonlinear proportional relationships exist when the ratio of two variables remains constant, but the relationship is expressed through a nonlinear equation. An example is the relationship between the radius and the area of a circle.
## Real-World Applications:
Proportional relationships find applications in numerous real-world scenarios, ranging from simple everyday situations to complex scientific and economic models.
### 1. Finance:
In the world of finance, proportional relationships play a pivotal role. Interest rates, for instance, often follow a proportional relationship with the amount of money borrowed or invested. The more significant the principal amount, the higher the interest accrued.
### 2. Physics:
Proportional relationships are abundant in the field of physics. Newton’s Law of Universal Gravitation, for example, expresses a proportional relationship between the masses of two objects and the force of gravity between them. The law can be mathematically represented as $$F = \frac{G \cdot m_1 \cdot m_2}{r^2}$$, where $$F$$ is the force, $$G$$ is the gravitational constant, $$m_1$$ and $$m_2$$ are the masses, and $$r$$ is the distance between the centers of the masses.
### 3. Cooking:
relationships even find their way into the kitchen. Recipes are often based on proportions, where the quantities of ingredients are adjusted in relation to one another. For example, a recipe might call for a ratio of 2 parts flour to 1 part sugar, ensuring the right balance of flavors and textures.
### 4. Scaling:
relationships are crucial in scaling objects or images. Architects, engineers, and artists use proportional reasoning to scale down or enlarge structures, ensuring that the relationships between dimensions remain constant.
## Educational Significance:
Understanding relationships is a foundational skill in mathematics education. It provides students with a versatile tool for solving a wide range of problems, fostering critical thinking and analytical skills. Moreover, the concept serves as a bridge between arithmetic and algebra, laying the groundwork for more advanced mathematical concepts.
### Conclusion:
In conclusion, relationships are an integral part of our mathematical landscape, connecting diverse areas of study and enriching our understanding of the world. From the simplicity of direct and inverse variations to the intricacies of nonlinear relationships, this concept serves as a powerful tool for problem-solving and modeling in various disciplines. As we continue to explore and apply proportional relationships, we uncover the symphony of order that underlies the seemingly complex tapestry of our mathematical reality.
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# How do you factor 4m^4 - 37m^2 + 9?
Nov 13, 2016
$\left(4 {m}^{2} - 1\right) \left({m}^{2} - 9\right)$
#### Explanation:
Factoring this polynomial gives:
$\left(4 {m}^{2} - 1\right) \left({m}^{2} - 9\right)$
Nov 18, 2016
$4 {m}^{4} - 37 {m}^{2} + 9 = \left(2 m - 1\right) \left(2 m + 1\right) \left(m - 3\right) \left(m + 3\right)$
#### Explanation:
The difference of squares identity can be written:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
We will use this a couple of times, but first note that:
$4 {m}^{4} - 37 {m}^{2} + 9 = 4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9$
So we can treat this quartic as a quadratic in ${m}^{2}$
Factor using an AC method:
Look for a pair of factors of $A C = 4 \cdot 9 = 36$ with sum $B = 37$
The pair $36 , 1$ works.
Use this pair to split the middle term and factor by grouping:
$4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9 = 4 {\left({m}^{2}\right)}^{2} - 36 \left({m}^{2}\right) - \left({m}^{2}\right) + 9$
$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left(4 {\left({m}^{2}\right)}^{2} - 36 \left({m}^{2}\right)\right) - \left(\left({m}^{2}\right) - 9\right)$
$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = 4 {m}^{2} \left({m}^{2} - 9\right) - 1 \left({m}^{2} - 9\right)$
$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left(4 {m}^{2} - 1\right) \left({m}^{2} - 9\right)$
$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left({\left(2 m\right)}^{2} - {1}^{2}\right) \left({m}^{2} - {3}^{2}\right)$
$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left(2 m - 1\right) \left(2 m + 1\right) \left(m - 3\right) \left(m + 3\right)$ |
# 5.4.1 - Errors
5.4.1 - Errors
## Committing an Error
Every time we make a decision and come to a conclusion, we must keep in mind that our decision is based on probability. Therefore, it is possible that we made a mistake.
Consider the example of the previous Lesson on whether the majority of Penn State students like the cold. In that example, we took a random sample of 500 Penn State students and found that 278 like the cold. We rejected the null hypothesis, at a significance level of 5% with a p-value of 0.006.
Type I Error
Rejecting $$H_0$$ when $$H_0$$ is really true, denoted by $$\alpha$$ ("alpha") and commonly set at .05
$$\alpha=P(Type\;I\;error)$$
The significance level of 5% means that we have a 5% chance of committing a Type I error. That is, we have 5% chance that we rejected a true null hypothesis.
Type II Error
Failing to reject $$H_0$$ when $$H_0$$ is really false, denoted by $$\beta$$ ("beta")
$$\beta=P(Type\;II\;error)$$
If we failed to reject a null hypothesis, then we could have committed a Type II error. This means that we could have failed to reject a false null hypothesis.
Decision Reality
$$H_0$$ is true $$H_0$$ is false Probability Level
Reject $$H_0$$, (conclude $$H_a$$) Type I error Correct decision P is LESS than .05 that the null is true Small probabilities (less than .05) lead to rejecting the null
Fail to reject $$H_0$$ Correct decision Type II error P is GREATER than .05 that the null is true Large probability (greater than .05) lead to not rejecting the null
## How Important are the Conditions of a Test?
In our six steps in hypothesis testing, one of them is to verify the conditions. If the conditions are not satisfied, we cannot, however, make a decision or state a conclusion. The conclusion is based on probability theory.
If the conditions are not satisfied, there are other methods to help us make a conclusion. The conclusion, however, may be based on other parameters, such as the median. There are other tests (called nonparametric) that can be used.
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# Definition
The rank of a matrix is the number of non-zero rows in its RREF.
MathJax TeX Test Page $$\begin{bmatrix}1 & 2 & 5 & 10 \\ 3 & 4 & 8 & 10 \\ 3 & 4 & 8 & 10\end{bmatrix} \to \begin{bmatrix}1 & 2 & 5 & 10 \\ 0 & -2 & -7 & -20 \\ 0 & 0 & 0 &0\end{bmatrix} \to \begin{bmatrix}\boxed{1} & 0 & -2 & -10 \\ 0 & \boxed{1} & 3.5 & 10 \\ 0 & 0 & 0 &0\end{bmatrix}$$ In this example, the rank of this matrix is 2. As you can see if the rank equals the number of non-zero rows, it equals the number of pivots. The pivots cannot be greater than the number of rows or columns. $$\text{Rank } \le \min(m,n) \text{ (m = rows, n = columns)}$$
# Injective
MathJax TeX Test Page Injective means a function is one-to-one. That is to say $$f(x) \text{ is injective iff } \forall x,y \text{ if } f(x) = f(y) \to x = y$$ One theorem is T(x) is injective iff T(x) = 0 is unique. $$## Theorem: Injective means Tx = 0 is trivial MathJax TeX Test Page One theorem is T(x) is injective iff T(x) = 0 has only the trivial solution.$$\text{ Assume T(x) is injective}T(0) = 0 \text{ (by definition}) \text{ If }T(x) = T(0), x = 0 \text{So, } T(x) = 0 \text{ only has the trivial solution.}$$Now, we prove it the other way. Let's prove the contrapositive. That is to say, instead of proving (T(x) = 0 trivial) \to (T(x) injective), we prove (T(x) not injective) \to (T(x) = 0 non-trivial)$$\text{Assume } T(x) = 0 \text{ is not injective.}\exists u,v, u \neq v \text{ s.t. } Tu = Tv = bTu - Tv = b - b = 0 \to T(u-v) = 0T(u-v) = T(0) = 0T(x) = 0 \text{ is not unique}$$Why is this important? Because we can now use another important theorem. ## Theorem: Ax = 0 is trivial means columns of A are linearly independent MathJax TeX Test Page Let's dissect what Ax = 0 is trivial means. This means$$A\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ ... \\x_n\end{bmatrix} = \vec{0}_mx_1\vec{a_1} + x_2\vec{a_2} + ... + x_n\vec{a_n} = \vec{0}_m$$Saying that this equals 0 iff x_1 ... x_n = 0 is the same as saying a_1 ... a_n are linearly independent. ## Theorem: If the columns of A are L.I., then the number of pivots equals n. MathJax TeX Test Page Row-reduction is a linear transformation, because it can be expressed as a matrix. All matrix products are linear transformations. So, we can say B = M*A$$Ax = 0 \to MAx = M*0 \to Bx = 0$$Observe that the values that satisfy x will not change. That is to say, the columns of B are also linearly independent. This means none of the columns are 0. This means there is a non-zero value in every column. If there are n columns with a 1, with each one on a different row, then there are n rows with a 1, meaning there are n pivots. By definition, this means there are n non-zero rows. So, \boxed{\text{rank} = n} # Surjective MathJax TeX Test Page A linear transformation means that \forall b \in \mathbb{R}^m, \exists x \text{ s.t. } Tx = b. Alternatively, you can say the columns of T span \mathbb{R}^m. ## Theorem: A is surjective means that there m pivots. MathJax TeX Test Page If A is surjective, then there exists a solution for every Ax = b. We can write this as an augmented matrix.$$\begin{bmatrix} A_1 & b_1 \\ A_2 & b_2 \\ A_3 & b_3 \\ ...\\ A_m & b_m\end{bmatrix}$$Assume it doesn't have m pivots, then there exists a row that is all 0.$$\begin{bmatrix} A_1 & b_1 \\ A_2 & b_2 \\ A_3 & b_3 \\ ...\\ \vec{0} & b_m - c_1b_1 + ... + c_{m-1}b_{m-1}\end{bmatrix} In order for this equation to be consistent, $b_m - c_1b_1 + ... + c_{m-1}b_{m-1} = 0$. However, we specified that this works for any $b_1 ... b_m$. Therefore, there is a contradiction.
We conclude that A has m pivots.
# Row Space
A row space is the span of the rows of a matrix. The row space of two row-equivalent matrices is the same. Why? because to row reduce you just add linear combinations of rows. The basis of the row space is equal to the non-zero rows in the row-reduced matrix, which equals the dimension of the row space.
# Column Space
The column space is the span of the columns of A. In other words, it’s the image of a matrix A. I will prove that the dimension of the column space equals the dimension of the row space which equals the rank.
# Null Space
The null space is the set of all x such that Ax = 0. I will later prove that the Rank + Dim (Null Space) = n (columns in A). |
# Question Video: Applying the Right-Angled Triangle Altitude Theorem and the Pythagorean Theorem Mathematics
From the figure, determine the length of the line π΅π·. If necessary, round your answer to the nearest hundredth.
02:26
### Video Transcript
From the figure, determine the length of the line π΅π·. If necessary, round your answer to the nearest hundredth.
Weβre asked in this example to find the length of the line π΅π·, which we can see is the perpendicular projection to π· on the hypotenuse of the right triangle π΄π΅πΆ from the right angle at π΅. To find π΅π·, our first step will be to use the Pythagorean theorem to find the length of the hypotenuse π΄πΆ of triangle π΄π΅πΆ. We can then use this in one part of the right triangle altitude theorem to find the side length πΆπ· and then use this value in the Pythagorean theorem in right triangle πΆπ·π΅ to find the side length we want, which is π΅π·.
So, letβs start with the Pythagorean theorem applied to triangle π΄π΅πΆ to find the length of π΄πΆ. We have π΄πΆ squared equal to πΆπ΅ squared plus π΄π΅ squared. And substituting our two known side lengths, this gives 15 squared plus eight squared on the right-hand side. This evaluates to 289. And taking the positive square root on both sides, positive since lengths are positive, we have π΄πΆ equals the square root of 289, which is 17 centimeters.
So, now marking this on the diagram and clearing some space, next we can use the right triangle altitude theorem to find side length πΆπ·. We know that πΆπ΅ is 15 centimeters. And weβve just found that π΄πΆ is 17 centimeters. And so we have 15 squared equals πΆπ· multiplied by 17. Now, dividing through by 17 and evaluating 15 squared, we have πΆπ· equal to 225 over 17.
Next, applying the Pythagorean theorem to triangle πΆπ·π΅, we have π΅π· squared equals π΅πΆ squared minus πΆπ· squared. Thatβs 15 squared minus 225 over 17 all squared. This is 49.826 and so on. And taking the square root on both sides, we have π΅π· equal to 7.058 and so on. Finally, rounding to the nearest hundredth, which is to two decimal places, we have that π΅π· is equal to 7.06 centimeters. |
# In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line.
Question:
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are 10 potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and he continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
Solution:
Distance covered by the competitor to pick and drop the first potato = 2 × 5 m = 10 m
Distance covered by the competitor to pick and drop the second potato = 2 × (5 + 3) m = 2 × 8 m = 16 m
Distance covered by the competitor to pick and drop the third potato = 2 × (5 + 3 + 3) m = 2 × 11 m = 22 m and so on.
∴ Total distance covered by the competitor = 10 m + 16 m + 22 m + ... up to 10 terms
This is an arithmetic series.
Here, = 10, d = 16 − 10 = 6 and n = 10
Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get
$S_{10}=\frac{10}{2}[2 \times 10+(10-1) \times 6]$
$=5 \times(20+54)$
$=5 \times 74$
$=370$
Hence, the total distance the competitor has to run is 370 m. |
Video: GCSE Mathematics Foundation Tier Pack 5 β’ Paper 3 β’ Question 28
GCSE Mathematics Foundation Tier Pack 5 β’ Paper 3 β’ Question 28
03:02
Video Transcript
π΄π΅πΆ is an isosceles triangle, with π΄π΅ equal to π΄πΆ. πΆπ·πΈπΉ is a rectangle. Angle πΆπ΄π΅ is equal to 42 degrees. Angle πΉπΆπ΄ is equal to 63 degrees. Angle π΄π΅π· is equal to 90 degrees. Show that triangle πΆπ΅π· is isosceles. You must give a reason for each stage of your working.
These questions can look really scary, but I like to go through and add the angles I do know or can work out to the diagram. The answer will usually follow fairly quickly. We mustnβt, however, forget to give a reason each time we work out an angle.
Firstly, weβre given that π΄π΅ is equal to π΄πΆ in an isosceles triangle π΄π΅πΆ. This means that the angles π΄π΅πΆ and π΄πΆπ΅ as marked must be equal. We can work out the value of these two angles by subtracting 42 from 180, since angles in a triangle sum to 180 degrees. 180 minus 42 is 138, so the two angles have a total measure of 138 degrees.
Since these two angles are equal, we can divide 138 by two to work out the measure of each of the angles. 138 divided by two is 69, so angles π΄πΆπ΅ and π΄π΅πΆ are both 69 degrees. There are two angles we can now work out. The first is this angle down here. This is called angle πΆπ΅π·.
We are told that angle π΄π΅π· is equal to 90 degrees. We can therefore find the measure of angle πΆπ΅π· by subtracting what we just worked out from 90. 90 minus 69 is 21, so πΆπ΅π· is equal to 21 degrees.
And we can use a similar process here for angle π΅πΆπ·. This time, we use the fact that angles around a point add to 360 degrees. So we can subtract the given angles of 63 and 90 plus the one we just worked out, 69, from 360. Thatβs 138, so angle π΅πΆπ· is 138 degrees.
Notice we now have two angles out of a possible three in triangle πΆπ΅π·. We can calculate the measure of angle πΆπ·π΅ by subtracting the two that we know from 180 degrees. 180 minus 138 plus 21 is 21 degrees.
Now letβs refer back to the question. We were trying to prove that the triangle πΆπ΅π· is isosceles; that is to say, two of its angles are equal. We have just shown that angles πΆπ·π΅ and πΆπ΅π· are both 21 degrees. So we have shown that triangle πΆπ΅π· must be isosceles as required. |
# Arithmetic and Geometric Means
## Presentation on theme: "Arithmetic and Geometric Means"— Presentation transcript:
Arithmetic and Geometric Means
OBJ: • Find arithmetic and geometric means
Arithmetic means are the terms between two given terms of an arithmetic progression or sequence.
For example, three arithmetic means between 2 and 18 in the progression below are 6, 10, and 14 since 2, 6, 10, 14, 18, is an arithmetic progression. 2, 6, 10, 14, 18, . . .
As shown in the example below, you can find any specified number of arithmetic means between two given numbers. EX: Find two arithmetic means between 29 and 8. 29, ____, ____, 8 an = a1 + (n – 1) d 8 = d -21 = 3d -7 = d 29, 22, 15, 8
As shown in the example below, you can find any specified number of arithmetic means between two given numbers. EX: Find the five arithmetic means between 30 and 21. 30,__,__,__,__,__, 21 an = a1 + (n – 1) d 21 = d -9 = 6d -1.5 = d 30, 28.5, 27, 25.5, 24, 22.5,21
As shown in the example below, you can find any specified number of arithmetic means between two given numbers. EX: Find the one arithmetic mean between 5 and 17. 5, ____, 17 an = a1 + (n – 1) d 17 = 5 + 2d 12 = 2d 6 = d 5, 11, 17
Since this is the same as the average of 5 and 17, it easier to use the formula: x + y. 2 which is called the arithmetic mean of the real numbers x and y. EX: Find the arithmetic mean of -8 and 22. 2 14 7
Find the real number solution.
125 r = -4 5
Geometric means are the terms between two given terms of a geometric progression or sequence.
For example, four geometric means between 3 and 96 in the progression below are 6, 12, 24, and 48 since 3, 6, 12, 24, 48, 96, is a geometric progression. 3, 6, 12, 24, 48, and
As shown in the example below, you can find any specified number of geometric means between two given numbers. EX: Find the two real geometric means between –3 and -3, ____, ____, 24 8 l = a •rn – 1 24 = -3 •r 3 -64 = r 3 -4 = r
As shown in the example below, you can find any specified number of geometric means between two given numbers. EX: Find three geometric means between 32 and 2. 32, ____, ____, ____, 2 l = a •rn – 1 2 = 32 •r4 1 = r4 16 ± 1 2
As shown in the example below, you can find any specified number of geometric means between two given numbers. EX: Find one geometric mean between 5 and 10 5, ____, 10 l = a •rn – 1 10 = 5 •r2 2 = r2 ±2
The geometric mean (mean proportional) of the real numbers x and y (xy > 0) is
xy or – xy . EX: Find the positive geometric mean of 4 and 8.
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