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2.7
# BODMAS
In the previous step, we used mathematical expressions to transform raw data values. When looking at more complex mathematical expressions, we need to be crystal clear about exactly how the computer will interpret them.
In a mathematical expression that involves a mixture of operations (such as addition, subtraction, multiplication, division and powers), it’s important to know how humans and computers interpret and apply the operations to evaluate the expression. If a mathematical expression is the input, there is one and only one possible result.
## What’s the problem with mathematical expressions?
For motivation, compare $$3+5\times7$$. If we apply the operations strictly from left-to-right, we would get $$3+5=8$$ and $$8\times7$$, which gives us 56 as the final answer. However, you might remember from school that multiplication has a higher precedence than addition, so instead we evaluate $$5\times7=35$$ first and then $$3+35$$, which gives the final answer of 38.
## What is BODMAS?
From your school days, you may remember BODMAS (sometimes called BIDMAS, BEDMAS or even PEMDAS). This is an acronym to help you remember the precedence of basic mathematical operations:
• B – brackets (sometimes called parentheses)
• O – order (sometimes called indices, exponents or powers)
• DM – division and multiplication
• AS – addition and subtraction
Brackets (B) have the highest precedence as they override all other operations. The second highest precedence is order (O) which is the ‘to the power of’ operation. Division and multiplication (DM) have equal precedence to each other, hence being grouped together in the acronym. Finally, we have addition and subtraction (AS) which have equal precedence to each other. Operations with the same precedence are evaluated left-to-right.
Let’s look at some examples (as a human) and then we can check we have understood correctly by evaluating the expression in Python. We’ll start from the bottom of BODMAS, at addition and subtraction, and work our way up to the top, brackets.
The expression $$7+5+3+2$$ is evaluated left-to-right since there is only one operator, addition. The result is 17. In Python, this would look as follows.
7+5+3+2
The expression $$7-5+3-2$$ is also evaluated left-to-right, but this time has two operators, addition and subtraction, both of which have the same precedence. The result is 3. In Python, this would look as follows.
7-5+3-2
The expression $$7+5\times3+2$$ includes multiplication, which has a higher precedence than addition. This means we first carry out $$5\times3$$ giving 15 and then $$7+15+2$$, which gives the final answer of 24. In the Python example below, notice * is used as the multiplication symbol.
7+5*3+2
The expression $$7+5\times3^2$$ includes order, which has a higher precedence than addition and multiplication. This means we first carry out $$3^2$$ which gives 9. The remaining expression $$7+5\times9$$ has multiplication, which has a higher precedence than addition, so we carry out $$5\times9$$, which gives 45. We then do the addition element of the expression $$7+45$$, which gives 52 as the final answer. In the Python example below, notice ** is used as the power symbol.
7+5*3**2
The expression $$7+(5\times3)^2$$ has brackets, which has the highest precedence of all operations. So, we first carry out $$5\times 3$$, which gives 15. Then $$7+15^2$$ has order as the highest precedence, so we carry out $$15^2$$, which gives 225. Finally, $$7+225$$ gives 232 as the final answer. In Python, this would look as follows:
7+(5*3)**2
A supposedly more simple calculation is 1 divided by 2/3. It’s all division so it’s simple, right? Not quite, this would actually be 1/(2/3), not 1/2/3, which you might assume. This gives $$0.\dot{6}7$$ as the final answer, rather than $$0.1\dot{6}7$$.
Data science constantly involves logical thinking and attention to detail, as you’ve seen above. The Python interpreter applies the rules of precedence when evaluating conditions and expressions. Whenever you might be unsure about how an expression will be evaluated, you can check using some examples in Python and, subsequently, use more brackets to make it clear to a human reader.
Advanced ideas. The concept of precedence of operations can be extended to other types of operations such as integer division (quotient and remainder), comparison operations (less than, greater than, equal to) and logical operations (and, or, not). We will not be looking at these types of operations in this course, but you will likely come across them in the future. If you’d like to find out more, have a look at the Python operator precedence table.
Every Python expression has exactly one possible interpretation. When evaluating an expression, the computer follows the precedence of operations, which we just refer back to the acronym BODMAS.
Consider the expression $$3315\times2\div3\div5\times7\times11\div13$$. Using the above as a guide, calculate the answer as a human and then check it using Python (using / as the division symbol).
If you need help to understand how to calculate it or why the answer is what it is, ask your fellow learners in the comments area below. |
# 2011 AMC 12B Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
What is $\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}?$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ \frac{5}{36} \qquad \textbf{(C)}\ \frac{7}{12} \qquad \textbf{(D)}\ \frac{147}{60} \qquad \textbf{(E)}\ \frac{43}{3}$
## Solutions
### Solution 1
Add up the numbers in each fraction to get $\frac{12}{9} - \frac{9}{12}$, which equals $\frac{4}{3} - \frac{3}{4}$. Doing the subtraction yields $\boxed{\frac{7}{12}\ \textbf{(C)}}$
### Solution 2
Notice that the numerators and denominators of each expression are 3-term arithmetic series. The sum of an arithmetic series is the middle term multiplied with the number of terms. Since each of the arithmetic series have the same number of terms, we can replace the fractions with the middle terms, which gives us $\frac{4}{3} - \frac{3}{4} = \frac{7}{12}$ |
# HSPT Math : Arithmetic
## Example Questions
### Example Question #21 : Concepts
Evaluate:
Explanation:
Rewrite the expression by simplifying the signs.
### Example Question #21 : Arithmetic
Solve:
Explanation:
Simplify the negative signs to a single sign and solve.
### Example Question #22 : Arithmetic
Solve:
Explanation:
To evaluate , simplify the parenthesis first.
Combine the terms.
### Example Question #22 : Concepts
Simplify:
Explanation:
To evaluate , simplify the plus and minus signs to a single sign.
### Example Question #21 : Concepts
Solve:
Explanation:
Simplify the expression by eliminating the negative sign when multiplying two negative numbers. Add the terms.
### Example Question #22 : Concepts
Evaluate:
Explanation:
In order to solve this expression, first simplify the double negative signs, which will convert to a positive sign.
### Example Question #621 : Hspt Mathematics
Evaluate:
Explanation:
Evaluate the numerator first by order of operations.
Evaluate the denominator by order of operations. Double negatives will equal a positive.
Divide the numerator with the denominator and reduce the fraction.
### Example Question #22 : Concepts
Perform the indicated operation:
Explanation:
When multiplying two negative numbers, their product becomes positive. Thus,
### Example Question #22 : Concepts
When evaluating the expression
,
which operation must be performed second?
Multiplication
Subtraction
Division
Multiplication
Explanation:
According to the order of operations, the operation inside the parentheses, which is the subtraction, is performed first.
This leaves a multiplication, a division, and an addition. By the order of operations, the multiplication and the division come next in left-to-right order; we choose the one that is to the left, which is the multiplication.
### Example Question #23 : Arithmetic
When evaluating the expression
,
which operation must be performed third?
Subtraction
Multiplication
Division
Subtraction
Explanation:
According to the order of operations, since no grouping symbols are present (the parentheses are setting apart a negative number, not an operation), multiplication and the division must be performed, left to right, then the addition and the subtraction must be performed, left to right.
This means that the first two operations to be performed are the multiplication and the division. This leaves a subtraction and an addition; since the subtraction is at left, it is the third operation performed. |
# How do you solve 7m=18m-2?
Aug 1, 2016
$m = \textcolor{b l u e}{\frac{2}{11}}$
#### Explanation:
$7 m = 18 m - 2$
Add $2$ to both sides of the equation.
$2 + 7 m = 18 m$
Subtract $7 m$ from both sides.
$2 = 18 m - 7 m$
Simplify.
$2 = 11 m$
Divide both sides by $11$.
$\frac{2}{11} = m$
Switch sides.
$m = \frac{2}{11}$ |
# How do you express in terms of ln2 and ln3 given ln 3 sqrt 2?
May 4, 2016
$\ln \left(3 \sqrt{2}\right) = \ln \left(3\right) + \frac{1}{2} \ln \left(2\right)$
#### Explanation:
Using the properties that $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$ and $\ln \left({a}^{x}\right) = x \ln \left(a\right)$, we have:
$\ln \left(3 \sqrt{2}\right) = \ln \left(3\right) + \ln \left(\sqrt{2}\right)$
$= \ln \left(3\right) + \ln \left({2}^{\frac{1}{2}}\right)$
$= \ln \left(3\right) + \frac{1}{2} \ln \left(2\right)$ |
Use the Concavity Test to find the intervals where the graph of the function is concave up.? Mistakes when finding inflection points: not checking candidates. b) Use a graphing calculator to graph f and confirm your answers to part a). So, we differentiate it twice. Evaluate the integral between $[0,x]$ for some function and then differentiate twice to find the concavity of the resulting function? Answer and Explanation: The first step in determining concavity is calculating the second derivative of $f(x)$. so concavity is upward. x = 2 is the critical point. The concept is very similar to that of finding intervals of increase and decrease. A concave up graph is a curve that "opens upward", meaning it resembles the shape $\cup$. Tap for more steps... Differentiate using the Quotient Rule which states that is where and . In any event, the important thing to know is that this list is made up of the zeros of f′′ plus any x-values where f′′ is undefined. Therefore, there is an inflection point at $x=-2$. 3. 0. Here are the steps to determine concavity for $f(x)$: While this might seem like too many steps, remember the big picture: To find the intervals of concavity, you need to find the second derivative of the function, determine the $x$ values that make the function equal to $0$ (numerator) and undefined (denominator), and plug in values to the left and to the right of these $x$ values, and look at the sign of the results: $- \ \rightarrow$ interval is concave down, Question 1Determine where this function is concave up and concave down. Find the intervals of concavity and the inflection points of g x x 4 12x 2. 2 Answers. Therefore, we need to test for concavity to both the left and right of $-2$. Otherwise, if the second derivative is negative for an interval, then the function is concave down at that point. Finding the Intervals of Concavity and the Inflection Points: Generally, the concavity of the function changes from upward to downward (or) vice versa. As you can see, the graph opens downward, then upward, then downward again, then upward, etc. finding intervals of increase and decrease, Graphs of curves can either be concave up or concave down, Concave up graphs open upward, and have the shape, Concave down graphs open downward, with the shape, To determine the concavity of a graph, find the second derivative of the given function and find the values that make it $0$ or undefined. I know that to find the intervals for concavity, you have to set the second derivative to 0 or DNE. This point is our inflection point, where the graph changes concavity. 1. In order to determine the intervals of concavity, we will first need to find the second derivative of f (x). y = 4x - x^2 - 3 ln 3 . First, find the second derivative. Determine whether the second derivative is undefined for any x-values. (If you get a problem in which the signs switch at a number where the second derivative is undefined, you have to check one more thing before concluding that there’s an inflection point there. Set the second derivative equal to zero and solve. We still set a derivative equal to $0$, and we still plug in values left and right of the zeroes to check the signs of the derivatives in those intervals. Hi i have to find concavity intervals for decreasing and increasing areas of the graph, no need for actually graphing. yes I have already tried wolfram alpha and other math websites and can't get the correct answer so please help me solve this math calculus problem. By the way, an inflection point is a graph where the graph changes concavity. Sal finds the intervals where the function f(x)=x⁶-3x⁵ is decreasing by analyzing the intervals where f' is positive or negative. 0 < -18x -18x > 0. We want to find where this function is concave up and where it is concave down, so we use the concavity test. This calculus video tutorial provides a basic introduction into concavity and inflection points. Find the second derivative of f. Set the second derivative equal to zero and solve. Show Concave Up Interval. Example 3.4.1: Finding intervals of concave up/down, inflection points Let f(x) = x3 − 3x + 1. In words: If the second derivative of a function is positive for an interval, then the function is concave up on that interval. Update: Having the same problem with this one -- what to do when you have i in critical points? non-negative) for all in that interval. The opposite of concave up graphs, concave down graphs point in the opposite direction. y = -3x^3 + 13x - 1. In general you can skip parentheses but be very careful. , concave down to 6/x^3 you know what to set the second derivative is undefined for any points where graph! 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Home Math Easily Learn Equivalent Fractions with Our 100% Free Worksheet
Easily Learn Equivalent Fractions with Our 100% Free Worksheet
Equivalent fractions are fractions whose values will not change even if the numerator and denominator are multiplied or divided by the same number which is not zero. The method of determination can be used by the following relations.
Attached figure 8.1
This fraction also was known as a value comparison is a type of comparison of two or more units, when one value increases then the other value also increases as well as when one value decreases, the other value also decreases.
Basic Concepts of Equivalent Fractions to be used in Excel Worksheet
Before learning deeper, it is important for you to know about the basic concept first. The following is the formula for the basic concept of fractions. Using the concept of division, it can be written as follows. See Attached figure 8.2 and attached figure 8.3
(Attached figure 8.2)
(Attached figure 8.3)
Circles 1, 2 and, 3 have the same area in the diagram. The shaded area in Figure above (i) is the fraction of ½ of the circle, in Figure (ii) it is 2/4 of the circle, and Figure (iii) is 4/8 of the circle. All of them are equivalent fractions. See Attached figure 8.4
(Attached figure 8.4)
So, from the picture above we can see that the area shaded in the three circles is the same. Namely: ½=2/4=4/8. So that the form of the fraction above is the form of a valued fraction. Then please see the relationships of the fractions with the value below. See Attached figure 8.5
(Attached figure 8.5)
How to Find the Equivalent Fractions
To understand about equivalent fraction, there are several methods that can be done are to use a number line, use a shaded image and multiply the numerator, and denominator of the fraction by the same number.
A fraction of value refers to a fraction whose value remains the same when the numerator and denominator of a fraction are multiplied or divided by the same number. Note that the numbers multiplied are not 0 because 0/0 = infinity.
1. Determine Fraction Worth with a Number Line
The first way to determine a fraction of value is to use the Number Line. A number line is a tool or model that is useful for learning fractions valued after the concrete model and drawing model. Consider the following number line. See Attached figure 8.6
(Attached figure 8.6)
If the value of the fraction on the number line above is on the same dotted line, then the fraction has a value. They have a different number but have the same value. If written in an equation, the equivalent fractions can be written below. See Attached figure 8.7
(Attached figure 8.7)
2. Determining Fraction Worth Using Pictures.
The rectangular area in the image above is divided into equal parts. The number below each figure shows the area of the shaded area. Because the shaded area in each of the pictures is the same, the fractions have the same value and are called valued fractions. See Attached figure 8.9
(Attached figure 8.9)
3. Determine Fractions Worth by Dividing or Multiplying the Numbers and Denominators by the Same Number
To find out the relationship of fractions, consider the following description. Based on the above example it can be said that a fraction of the same value can be obtained if the numerator and denominator of a fraction are multiplied by the same number. See Attached figure 8.10
(Attached figure 8.10)
Next, consider the relationship of the following fractions. Based on the example above, it can be said that a fraction of value can be obtained by dividing the numerator and denominator of a fraction by the same number. See Attached figure 8.11
(Attached figure 8.11)
After learning the basic concept and the methods, you will easily create many other types of fractions. The most important one is knowing the various similarities of the values you have, so that equivalent fractions will easily be created.
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Precalculus (6th Edition) Blitzer
a) The value is, $\frac{\sqrt{2}}{2}$. b) The value is, $\frac{\sqrt{2}}{2}$.
(a) In the given unit circle, the point that corresponds to $t=\frac{3\pi }{4}$ has the coordinates $\left( -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ So use $x=-\frac{\sqrt{2}}{2}$ and $y=\frac{\sqrt{2}}{2}$. Such that, $\sin \left( \frac{3\pi }{4} \right)=y=\frac{\sqrt{2}}{2}$ The value of the trigonometric function $\sin \left( \frac{3\pi }{4} \right)$ is $\frac{\sqrt{2}}{2}$. (b) We have the periodic properties of sine and cosine functions: $\sin \left( t+2\pi \right)=\sin \left( t \right)$ and $\text{cos}\left( t+2\pi \right)=\cos \left( t \right)$. Therefore, \begin{align} & \sin \left( \frac{11\pi }{4} \right)=\sin \left( \frac{3\pi }{4}+2\pi \right) \\ & =\sin \left( \frac{3\pi }{4} \right) \end{align} Now, $\sin \left( \frac{3\pi }{4} \right)=\frac{\sqrt{2}}{2}$. So, $\sin \left( \frac{3\pi }{4} \right)=\frac{\sqrt{2}}{2}.$ Thus, the value of the trigonometric function $\sin \left( \frac{11\pi }{4} \right)$ is, $\frac{\sqrt{2}}{2}$. |
# The base of a triangular pyramid is a triangle with corners at (6 ,7 ), (5 ,3 ), and (8 ,4 ). If the pyramid has a height of 6 , what is the pyramid's volume?
Volume $V = 11 \text{ }$cubic units
#### Explanation:
Compute the area of the triangular base first
Area $A = \frac{1}{2} \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right]$
Area $A = \frac{1}{2} \cdot \left({x}_{1} \cdot {y}_{2} + {x}_{2} \cdot {y}_{3} + {x}_{3} \cdot {y}_{1} - {x}_{2} \cdot {y}_{1} - {x}_{3} \cdot {y}_{2} - {x}_{1} \cdot {y}_{3}\right)$
The given points are ${P}_{1} \left(6 , 7\right)$, ${P}_{2} \left(5 , 3\right)$, ${P}_{3} \left(8 , 4\right)$
Area $A = \frac{1}{2} \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right]$
Area $A = \frac{1}{2} \left[\begin{matrix}6 & 5 & 8 & 6 \\ 7 & 3 & 4 & 7\end{matrix}\right]$
Area $A = \frac{1}{2} \cdot \left(6 \cdot 3 + 5 \cdot 4 + 8 \cdot 7 - 5 \cdot 7 - 8 \cdot 3 - 6 \cdot 4\right)$
Area $A = \frac{1}{2} \cdot \left(18 + 20 + 56 - 35 - 24 - 24\right)$
Area $A = \frac{1}{2} \cdot \left(94 - 83\right)$
Area $A = \frac{1}{2} \cdot \left(11\right) = 5.5$
Compute the volume of the triangular pyramid
$V = \frac{1}{3} \cdot A \cdot h = \frac{1}{3} \cdot \frac{11}{2} \cdot 6$
$V = 11 \text{ }$cubic units
God bless....I hope the explanation is useful. |
# Successor of a Number | Predecessor of a Number
Successor of a Number – Successor of a Number (whole number) is the number, obtained by adding 1 to that number.
Predecessor of a Number – Predecessor of any whole number is the number obtained by subtracting 1 from it.
## Question based on Successor of a Number
Question 1
Find the Successor of 543 ?
Explanation : –
The “Successor” of any whole number is the number, obtained by adding 1 to that number.
So, the successor of 543 is 543 + 1 = 544
Question 2
Find the Successor of 1206?
Explanation : –
The “Successor” of any whole number is the number, obtained by adding 1 to that number.
So, the successor of 1206 is 1206 + 1 = 1207
Question 3
Find the successor of 3542?
Explanation : –
The “Successor” of any whole number is the number, obtained by adding 1 to that number.
So, the successor of 3542 is 3542 + 1 = 3543
Question 4
Find the successor of 6572?
Explanation : –
The “Successor” of any whole number is the number, obtained by adding 1 to that number.
So, the successor of 6572 is 6572 + 1 = 6573
Question 5
Find the successor of 9571?
Explanation : –
The “Successor” of any whole number is the number, obtained by adding 1 to that number.
So, the successor of 9571 is 9571 + 1 = 9572
## Question based on Predecessor of a Number
Question 1
Find the predecessor of 437?
Explanation : –
The “Predecessor” of any whole number is the number obtained by subtracting 1 from it.
So, the predecessor of 437 is 437 – 1 = 436
Question 2
Find the predecessor of 918?
Explanation : –
The “Predecessor” of any whole number is the number obtained by subtracting 1 from it.
So, the predecessor of 918 is 918 – 1 = 917
Question 3
Find the predecessor of 2331?
Explanation : –
The “Predecessor” of any whole number is the number obtained by subtracting 1 from it.
So, the predecessor of 2331 is 2331 – 1 = 2330
Question 4
Find the predecessor of 4100?
Explanation : –
The “Predecessor” of any whole number is the number obtained by subtracting 1 from it.
So, the predecessor of 4100 is 4100 – 1 = 4099
Question 5
Find the predecessor of 6605?
Explanation : –
The “Predecessor” of any whole number is the number obtained by subtracting 1 from it.
So, the predecessor of 6605 is 6605 – 1 = 6604 |
?> Frequency Histogram | Histogram Graph | Math@TutorCircle.com
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# Frequency Histogram
Top
Sub Topics In the statics, the information can be represented by the help of a frequency histogram. In the histogram graph the rectangles are used to show the frequency of data items in form of successive intervals. The independent variables of a given data items are plotted along the horizontal axis and the dependent variables of a given data item are plotted along the vertical axis. We need to follow some steps for designing the histogram: Step1: First we select the data for making the histogram. Step2: After that we plot a graph, and we know that along the horizontal axis the independent variables are plotted and the dependent variables are plotted along the vertical axis. Suppose we have the data values, 2 items in between 5 to 6, and 1 items are in between 6-7 and 6 item is in between 7 to 8, 4 items are in between 8 to 9, 5 item in between 9 to 10, 2 item in between 9 to 10 and 3 item in between 10 to 11. For making a histogram graph we have to follow all the above steps: Step1: First we select the data item i.e. We have the data values, we have 2 items in between 5 to 6, and 1 item are in between 6-7 and 6 item is in between 7 to 8, 4 items are in between 8 to 9, 5 item in between 9 to 10, 2 item in between 9 to 10 and 3 item in between 10 to 11. Step2: For this histogram the value of variable represents the horizontal axis and the frequency of the variable represents the vertical axis. This is how we plot a frequency Histograms.
## Relative Frequency histograms
The Ratio of the value to the total number of the given values is known as the Relative Frequency Histogram.
For example: If we have to select 15 trains from 50 trains then the relative frequency of a histogram is:
Then the frequency of trains is 15, and the relative frequency of trains is 15/50 because we know that it is the ratio of actual number to the total number of values.
Now we will see how to find relative frequency Histograms.
For finding the relative frequency of a histogram we have to follow some steps which are:
Step 1: First we have a number.
Step 2: Now find the frequency.
Step 3: If we have the value of frequency, then with the help of frequency we can find the relative frequency of a histogram.
Suppose we are looking for the brands of the car and we have 18 observations:
Suzuki, Audi, Mercedes, Audi, Honda, Mercedes, Audi, Hyundai, Mercedes, Honda, Suzuki, Hyundai, Mercedes, Audi, Honda, Suzuki, Audi, Mercedes
Here we have to find the relative frequency of the given combinations?
For finding the relative frequency of an object we have to follow all the above steps:
Step1: First find the number.
Step2: Now we have to find the frequency of all the brands Combination of a car.
So the frequency of Suzuki car is: 3;
The frequency of Audi car is: 5;
The frequency of Mercedes car is: 5;
The frequency of Honda car is: 3;
The frequency of Hyundai car is: 2;
By using the definition of the relative frequency we can easily find the relative frequency:
So the relative frequency of Suzuki car is 3/18;
The relative frequency of Audi car is 5/18;
The relative frequency of Mercedes car is 5/18;
The relative frequency of Honda car is 3/18;
The relative frequency of Hyundai car is 2/18.
For example: If you select 20 colleges from the 150 college:
Then the relative frequency of a college:
Solution: The frequency of a college is 20;
And the relative frequency of college is 20/150.
## Frequency Distribution Table
There is a statistical approach to simplify a large sample of data values by constructing a Frequency distribution table. It is a simple method of shaping and simplifying the data values to make their statistical analysis more understandable.
Main objective of making a frequency distribution table is to analyze how frequently each value occurs within a particular Set. Number of occurrences of any value is termed as its frequency in distribution table. So, we draw a table consisting of two columns, with first column representing all possible set of data values and second column shows their respective frequency of occurrence.
Let’s consider an example to understand it. Say we have a data set that consists of following 25 Numbers: 1, 3, 5, 0, 7, 5, 8, 9, 2, 6, 6, 13, 20, 24, 34, 43, 12, 23, 5, 10, 5, 2, 5, 6 and 8.
We first have to reorganize the data values in a particular order, let say we rearrange them in descending order to simplify our calculations: 0, 1, 2, 2, 3, 5, 5, 5, 5, 5, 6, 6, 6, 7, 8, 8, 9, 10, 12, 13, 20, 23, 24, 34 and 43. Once the reorganization of data is complete, make two columns: one for data values and the other for frequency of occurrence.
Analyzing first value in arrangement i.e. 0; we see that it has frequency of occurrence as 1 in complete set. In similar way we can evaluate the frequencies for other data values also. Frequency Distribution Chart can now be drawn as follows:
DATA VALUE
FREQUENCY OF OCCURRENCE
0
1
1
1
2
2
3
1
5
5
6
3
7
1
8
2
9
1
10
1
12
1
13
1
20
1
23
1
24
1
34
1
43 |
Adding and Subtracting Rational Expressions Section 7.2 MATH 116-460 Mr. Keltner.
Presentation on theme: "Adding and Subtracting Rational Expressions Section 7.2 MATH 116-460 Mr. Keltner."— Presentation transcript:
Adding and Subtracting Rational Expressions Section 7.2 MATH 116-460 Mr. Keltner
In Section 7.1… We used our knowledge of multiplying and dividing fractions to help us multiply and divide rational expressions. Now, we are going to use our knowledge of adding and subtracting fractions to help with rational expressions.
With Common Denominators As with numerical fractions, the process used to add (or subtract) two rational expressions depends on whether they have like or unlike denominators. We first look at when they have like denominators. Try this example with fractions:
Common Denominator Example Perform the indicated operations below:
Fraction Example If we add two fractions with unlike denominators, like First we need to look for a common denominator, which in this case would be ______. We then multiply each fraction by a convenient way to write 1, such that each denominator will be the same.
Using that idea Use that same knowledge for finding a least common denominator to simplify the rational expression Our common denominator in this case would be ____________, or ___________.
Example 1, Cont. Now that we have a common denominator, we can combine like terms and simplify. Check to see if we can divide out a common factor on top and bottom. This is simplified, since there are no common factors to divide out.
Example 2 The same concept applies with subtraction. Simplify
A couple things to be careful of… You must be careful with signs when using subtraction because of the way the distributive property works. Some examples: 4 - (x - 9) ≠ 4 - x - 9 2x - ( -3x) ≠ -1x 2(x - 4) - (x - 4)(x + 2) ≠ 2(x - 4) - x - 4(x + 2)
Assessment Pgs. 472-475: #’s 9 - 90, multiples of 9 |
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# 5.6: Structural and Procedural Algebra
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When most people think about algebra from school, they think about “solving for x.” They imagine lots of equations with varying levels of complexity, but the goal is always the same: find the unknown quantity. This is a procedural view of algebra.
Even elementary students can be exposed to ideas in procedural algebra. This happens any time they think about unknown quantities and try to solve for them. For example, when first grade students learn to add and subtract numbers “within 10,’” they should frequently tackle problems like these:
• $$3 + \_\_\_ = 7 \ldotp$$
• Find several pairs of numbers that add up to 10.
Although procedural algebra is important, it’s not the most important skill, and it’s certainly not the whole story.
You also need to foster thinking about structural algebra in your students: using symbols to express meaning in a situation. If there is an x on your page, you should be able to answer, “what does the x mean? What does it represent?”
Most of what you’ve done so far in this chapter is structural algebra. You’ve used letters and symbols not to represent a single unknown quantity, but a varying quantity. For example, in Section 4 you used letters to represent the “figure number’” or “case number” in a growing pattern. The letters could take on different values, and the expressions gave you information: how many tiles or toothpicks or stars you needed to build that particular figure in that particular pattern.
Think / Pair / Share
• Consider the expression $$a + 3$$. Give a real world situation that could be represented by this expression. Share your answer with your partner. Together, can you come up with even more ideas?
• Suppose the expression $$3c + 2$$ represents the number of tiles used at any stage of a growing pattern.
• Evaluate the expression at $$c = 1, 2,\; \text{and}\; 3$$. What do the values tell you about the pattern?
• Can you describe in words how the pattern is growing?
• Can you design a pattern with tiles that grows according to this rule?
• Where do you see the “3” in your pattern? Where do you see the “2”’? Where do you see the “”?
Problem 21
Krystal was looking at this pattern, which may be familiar to you from the Problem Bank:
She wrote down the equation
$$y = 4x + 2 \ldotp$$
In Krystal’s equation, what does represent? What does represent? How do you know?
Problem 22
Today is Jennifer’s birthday, and she’s twice as old as her brother. When will she be twice as old as him again?
She wrote down the equation $$2n = m$$. In Candice’s equation, what does represent? What does represent? How do you know?
Problem 23
Sarah and David collect old coins. Suppose the variable stands for the number of coins Sarah has in her collection, and stands for the number of coins David has in his collection. What would each of these equations say about their coin collections?
$$(a)\; k= \ell + 1 \qquad (b)\; k = \ell \qquad (c)\; 3k = 2 \ell \qquad (d)\; k = \ell - 11$$
Problem 24
The pictures below show balance scales containing bags and blocks. The bags are marked with a “?”’ because they contain some unknown number of blocks. In each picture:
• Each bag contains the same number of blocks.
• The scale is balanced.
For each picture, determine how many blocks are in each bag. Justify your answers.
(a)
(b)
(c)
Problem 25
When he was working on Problem 24, Kyle wrote down these three equations.
(i) $$2m = 6 \ldotp$$
(ii) $$2x = x + 3 \ldotp$$
(iii) $$z + 5 = 2z + 3 \ldotp$$
Match each equation to a picture, and justify your choices. Then solve the equations, and say (in a sentence) what the solution represents.
Problem 26
Draw a balance puzzle that represents the equation
$$2h + 3 = h + 8 \ldotp$$
Now solve the balance puzzle. Where is the “” in your puzzle? What does it represent?
Problem 27
Draw a balance puzzle that represents the equation
$$3b + 7 = 3b + 2 \ldotp$$
Now solve the equation. Explain what happens.
Problem 28
Which equation below is most like the one in Problem 27 above? Justify your choice.
1. $$5+3=8 \ldotp$$
2. $$\frac{2}{3} + \frac{1}{2} = \frac{3}{5} \ldotp$$
3. $$5+3 = y \ldotp$$
4. $$\frac{a}{5} = \frac{5}{a} \ldotp$$
5. $$n + 3 = m \ldotp$$
6. $$3x = 2x+ x \ldotp$$
7. $$5k = 5k + 1 \ldotp$$
Problem 29
Draw a balance puzzle that represents the equation
$$4 \ell + 7 = 4 \ell + 7 \ldotp$$
Now solve the equation. Explain what happens.
Problem 30
Which equation below is most like the one in Problem 29 above? Justify your choice.
1. $$5+3 = 8 \ldotp$$
2. $$\frac{2}{3} + \frac{1}{2} = \frac{3}{5} \ldotp$$
3. $$5+3 = y \ldotp$$
4. $$\frac{a}{5} = \frac{5}{a} \ldotp$$
5. $$n + 3 = m \ldotp$$
6. $$3x = 2x + x \ldotp$$
7. $$5k = 5k + 1 \ldotp$$
Problem 31
Create a balance puzzle where the solution is not a whole number of blocks. Can you solve it? Explain your answer.
Problem 32
There are three piles of rocks: pile A, pile B, and pile C. Pile B has two more rocks than pile A. Pile C has four times as many rocks as pile A. The total number of rocks in all three piles is 14.
1. Use x to represent the number of rocks in pile A, and write equations that describe the rules above. Then find the number of rocks in each pile.
2. Use x to represent the number of rocks in pile B, and write equations that describe the rules above. Then find the number of rocks in each pile.
3. Use x to represent the number of rocks in pile C, and write equations that describe the rules above. Then find the number of rocks in each pile.
Think / Pair / Share
Look back at Problems 21–32. Which of them felt like structural algebraic thinking? Which felt like procedural algebraic thinking? Did any of the problems feel like they involved both kinds of thinking?
## Variables and Equations
You have seen that in algebra, letters and symbols can have different meanings depending on the context.
• A symbol could stand for some unknown quantity.
• A symbol could stand for some quantity that varies. (Hence the term “variable” to describe these symbols.)
In much the same way, equations can represent different things.
• They can represent a problem to be solved. This is the traditional procedural algebra type of question.
• They can represent a relationship between two or more quantities. For example, $$A = s^{2}$$ represents the relationship between the area of a square and its side length.
• They can represent identities: mathematical truths. For example, $$x^{2} - 1 = (x + 1 )(x - 1)$$is always true, for every value of . There is nothing to solve for, and no relationship between varying quantities. (If you do try to “solve for ,” you will get the equation $$0 = 0$$, much like you saw in Problem 29. Not very satisfying!)
Think / Pair / Share
Give an example of each type of equation. Be sure to say what the symbols in the equations represent.
Problem 33
$$x^{2} - 1 = (x + 1)(x - 1) \ldotp$$
1. Evaluate both sides of the above equation for $$x = 1, 2, 3, 4,\; \text{and}\; 5$$. What happens?
3. Use the equation to compute $$99^{2}$$ quickly, without using a calculator. Explain how you did it. |
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Math Help > Calculus > Derivative > Word Problems > Minimum Perimeter of Triangle
Problem: You are given an isosceles triangle of area 1 with equal sides of length x, height h, and base b.
Part 1: Express the perimeter, P, in terms of x and b.
Answer: P = 2x + b
Part 2: Express the perimeter in terms of just b.
Part 2a: Express x in terms of h and b.
Answer: x = sqrt(h2+(b/2)2), from the Pythagorean theorem.
Part 2b: Express h in terms of b, and thereby x in terms of just b.
Answer: Since the area is 1, we have (1/2)bh=1, so h=2/b
We can substitute 2/b in place of h in the answer to part 2a, giving
x = sqrt((2/b)2+(b/2)2)
Answer to part 2: P = 2 sqrt((2/b)2+(b/2)2) + b
Part 3: Find the value of b for which the perimeter is minimized, and find the minimum perimeter as well.
First, find P', the derivative of P with respect to b using the exponent rule and the chain rule. Start out by expressing P in terms of expressions with powers, to make it easier to use the exponent rule.
P = 2 (4 b-2+(1/4)b2)1/2 + b
P' = (2) (1/2) (4 b-2+(1/4)b2)-1/2 (-8b-3+(1/2)b) + 1
P' = (-8b-3+(1/2)b)/sqrt(4 b-2+(1/4)b2) + 1
P' = (-8b-3+(1/2)b+sqrt(4 b-2+(1/4)b2))/sqrt(4 b-2+(1/4)b2)
Now, set P' = 0 to find the value of b for which P is minimized (or maximized)
(-8b-3+(1/2)b+sqrt(4 b-2+(1/4)b2))/sqrt(4 b-2+(1/4)b2) = 0
-8b-3+(1/2)b+sqrt(4 b-2+(1/4)b2) = 0
8b-3-(1/2)b = sqrt(4 b-2+(1/4)b2)
64b-6-8b-2+(1/4)b2 = 4 b-2+(1/4)b2
12b-2 = 64b-6
12b4 = 64
b4 = 16/3
b = 2/31/4.
Checking in the original equation for part 3, this value of b works. (It's important to check because we squared both sides, which has the potential to introduce spurious solutions.)
Answer to part 3: The value of b for which P' is 0 is the first answer sought in part 3:
When b=2/31/4, b/2=1/31/4, and h=2/b=31/4. Square b/2 and h to get
(b/2)2 = 1/sqrt(3), h2=sqrt(3). Add them together to get the square of x, which is
x2 = 1/sqrt(3)+sqrt(3) = 4/sqrt(3). Take the square root of this to get
x = 2/31/4, which is the same as b. So the minimum perimeter is 3b=6/31/4, which is the second answer sought in part 3.
As you might have expected all along, this is an equilateral triangle
### Related pages in this website
Word Problems
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# How do you write an equation of a line passing through (1,2), perpendicular to 2y+5 = 3x?
Aug 29, 2016
$y = - \frac{2}{3} x + 2 \frac{2}{3}$
#### Explanation:
Re-arrange the equation of the given line: $y = \frac{3}{2} x - \frac{5}{2}$
The slope of this line is : ${m}_{1} = \frac{3}{2}$
The slope of a line perpendicular to this line is:
${m}_{2} = - \frac{2}{3}$
We have the slope and a point, (1,2) which is $\left({x}_{1} , {y}_{1}\right)$
Substitute into the formula:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 2 = - \frac{2}{3} \left(x - 1\right)$
$y = - \frac{2}{3} x + \frac{2}{3} + 2$
The equation is:
$y = - \frac{2}{3} x + 2 \frac{2}{3}$ |
Recurrence relations with factors in recurrence
How would I go about approaching solving a recurrence relation such as:
$$a_{n}=2a_{\frac{n}{3}}+1$$
I'm just not sure how to get a general form for a non-recursive solution, can someone walk through the first couple steps?
• is $n\in\mathbb{Z^+}$? – Artem Mar 12 '15 at 17:19
• Oh yeah $n \in \mathbb{Z}$ and is a power of 3. – dogman288 Mar 12 '15 at 18:19
• Let $b(n) = a(3^n)$, and close $b$. – DanielV Mar 13 '15 at 18:34
Method 1: Exploit the relationship between recurrences and summations $$a(3^0)=\phi$$ $$a(3^k)=2a\left(3^{k-1}\right)+1$$ $$\mbox{for}\ k\geq 1$$ Let's multiply both sides by the summation factor of $\frac{1}{2^k}$ $$\frac{a(3^0)}{2^0}=\frac{\phi}{2^0}=\phi$$ $$\frac{a(3^k)}{2^k}=\frac{2a\left(3^{k-1}\right)}{2^k}+\frac{1}{2^k} = \frac{a\left(3^{k-1}\right)}{2^{k-1}}+\frac{1}{2^k}$$ Let $s(k)=\frac{a(3^k)}{2^k}$, then $$s(0) = \phi$$ $$s(k) = s(k-1) + \frac{1}{2^k}$$ Therefore $$s(k)= \phi+\sum_{j=1}^{k}\frac{1}{2^j} = \phi + 1 - \frac{1}{2^k}$$ And $$a(3^k) = 2^ks(k)=2^k\left(\phi + 1 - \frac{1}{2^k}\right) = 2^k\left(\phi+1\right)-1$$ Method 2: Generalize the recurrence and use the repertoire method $$a(3^0)=\phi$$ $$a(3^k)=2a\left(3^{k-1}\right)+1$$ $$\mbox{for}\ k\geq 1$$ This is just a special case of the generalized recurrence $$a(1)=a(3^0)=\phi$$ $$a(n)=a(3^k)=2a\left(3^{k-1}\right)+\beta$$ $$\mbox{for}\ k\geq 1$$ Also note that we can redefine $a(n)$ as a linear combination of unknown functions and their corresponding coefficients $$a(n)=A(n)\phi+B(n)\beta$$ Let's begin by studying the output of $a(n)$ $$a(3^0)=\phi$$ $$a(3^1)=2\phi+\beta$$ $$a(3^2)=4\phi+3\beta$$ $$a(3^3)=8\phi+7\beta$$ $$a(3^4)=16\phi +15\beta$$ $$a(3^5)=32\phi+31\beta$$ The obvious guess for $A(n)$ is $2^k$. So now let's find $A(n)$ via the repertoire method $$\mbox{Let}\ a(n)=a(3^k)=2^k, \mbox{then}$$ $$a(1)=a(3^0)=2^0=1=\phi$$ $$a(n)=a(3^k)= 2a\left(3^{k-1}\right) +\beta$$ $$2^k=2\cdot 2^{k-1}+\beta$$ $$2^k=2^k+\beta$$ Which implies that $$\phi=1, \beta=0$$ Now let's apply these facts to the general equation $$a(n)=A(n)\phi+B(n)\beta$$ $$2^k=A(n)\cdot 1+B(n)\cdot 0$$ $$2^k=A(n)+0$$ $$A(n)=2^k$$ Now let's find $B(n)$ also via the repertoire method $$\mbox{Let}\ a(n)=a(3^k)=2^k-1, \mbox{then}$$ $$a(1)=a(3^0)=2^0-1=0=\phi$$ $$a(n)=a(3^k)= 2a\left(3^{k-1}\right) +\beta$$ $$2^k-1=2\left(2^{k-1}-1\right)+\beta$$ $$2^k-1=2^k-2+\beta$$ $$2-1=\beta$$ Which implies that $$\phi=0, \beta=1$$ Now let's apply these facts to the general equation $$a(n)=A(n)\phi+B(n)\beta$$ $$2^k-1=A(n)\cdot 0+B(n)\cdot 1$$ $$2^k-1=0+B(n)$$ $$B(n)=2^k-1$$ So now we have $$a(n)=A(n)\phi+B(n)\beta$$ $$a(n)=2^k\phi+\left(2^k-1\right)\beta$$ $$a(n)=2^k\phi+2^k\beta-\beta$$ Therefore, the closed form solution to the general recurrence is $$a(n)=a(3^k)=2^k\left(\phi+\beta\right)-\beta$$ Note that for your specific recurrence, $\beta=1$. So the closed form solution to your specific recurrence is $$a(n)=a(3^k)=2^k\left(\phi+1\right)-1$$
If I've understood this correctly, $n=3^k,k\in\mathbb{Z^+}$. A good method that I found in Concrete Mathematics - A Foundation for Computer Science is called the repertoire method (please correct me, but this is the way I understood it.) So, the idea is simply to try to break the recurrence into smaller ones, and hopefully, the smaller ones will have a recognizable pattern.
In your case, let's assume that $a_1=\alpha$. So, the idea is simply to evaluate this several times, as shown below: $$\begin{array}{l} {a_1} = \alpha \\ {a_3} = 2{a_1} + 1 = 2\alpha + 1\\ {a_9} = 2(2\alpha + 1) + 1 = 4\alpha + 3\\ {a_{27}} = 2(4\alpha + 3) + 1 = 8\alpha + 7\\ {a_{81}} = 2(8\alpha + 7) + 1 = 16\alpha + 15 \end{array}$$
As you can see, there is a pattern. The initial parameter $\alpha$ has the coefficient that is equal to the power of $3$ squared, i.e. for $a_{81}$, it's $81=3^4$, so $4^2=16$. The constant term is simply one less than the coefficient.
Edit Here's how it would be expressed: $$\begin{array}{l} {a_1} = \alpha \\ {a_{{3^k}}} = {k^2}\alpha + ({k^2} - 1) \end{array}$$
Just use the change of variables $n = 3^k$ and $b_k = a_{3^k}$ to get:
$$b_k = 2 b_{k - 1} + 1$$
Solve this one, and express in terms of the original variables. |
# Combinations
In this lesson, we'll learn how to calculate combinations in different cases.
We'll cover the following
For combinations, the order doesn’t matter.
## Without repetition
For example, I want to pick three balls from a bag of five balls. The end result is that I should have three balls, the order in which I actually pick the balls is not important. How many ways can I do this?
The formula is $C(n,k) = \frac{P(n,k)}{k!} = \frac{n!}{(n-k)!k!} = {n \choose k}$
Quick explanation: If order mattered, the number of ways would be $P(n,k)$ in which there are $k!$ variations for every choice. To explain further, suppose the three balls we pick are $(2,3,5)$. We end up counting this same selection of $3$ balls $3!$ times. This happens for every subset of balls, so we divide by $3!$
ways = ${5 \choose 3} = \frac{5!}{2!3!} = \frac{120}{12} = 10$
## Selecting from identical objects
There is only 1 way to select $k$ objects from $n$ identical objects.
The order doesn’t matter and the objects are identical, so every way you choose results in the same combination.
## With repetitions
This is hardest to explain and out of the scope of this level’s discussion. We will cover this in detail when we discuss the number of integer solutions of a linear equation in later lessons.
${N \choose K} = {N-1 \choose K} + {N-1 \choose K-1}$
## Question
Question: From a group of seven men and six women. A committee needs to be made of five people with at least three men. In how many ways can this be done?
Solution: We need at least three men in the committee. So, there are three ways to form the committee.
• 3 men + 2 women
• 4 men + 1 women
• 5 men
So, required number of ways:
$= {7 \choose 3}*{6 \choose 2} + {7 \choose 4}*{6 \choose 1} + {7 \choose 5}$
$= 525 + 210 + 21$
$= 756$
In the next lesson, we’ll discuss a solved PnC problem.
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Home / Calculus II / Integration Techniques / Integrals Involving Trig Functions
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### Section 7.2 : Integrals Involving Trig Functions
9. Evaluate $$\displaystyle \int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}}$$.
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Start Solution
There really isn’t all that much to this problem. All we have to do is use the formula given in this section for reducing a product of a sine and a cosine into a sum. Doing this gives,
$\int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}} = \int_{1}^{3}{{\frac{1}{2}\left[ {\cos \left( {8x - x} \right) - \cos \left( {8x + x} \right)} \right]\,dx}} = \frac{1}{2}\int_{1}^{3}{{\cos \left( {7x} \right) - \cos \left( {9x} \right)\,dx}}$ Show Step 2
Now all we need to do is evaluate the integral.
\begin{align*}\int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}} & = \left. {\frac{1}{2}\left[ {\frac{1}{7}\sin \left( {7x} \right) - \frac{1}{9}\sin \left( {9x} \right)} \right]} \right|_1^3\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{14}}\sin \left( {21} \right) - \frac{1}{{18}}\sin \left( {27} \right) - \frac{1}{{14}}\sin \left( 7 \right) + \frac{1}{{18}}\sin \left( 9 \right) = - 0.0174}}\end{align*} |
# Word Problems on Simultaneous Linear Equations
Solving the solution of two variables of system equation that leads for the word problems on simultaneous linear equations is the ordered pair (x, y) which satisfies both the linear equations.
Problems of different problems with the help of linear simultaneous equations:
We have already learnt the steps of forming simultaneous equations from mathematical problems and different methods of solving simultaneous equations.
In connection with any problem, when we have to find the values of two unknown quantities, we assume the two unknown quantities as x, y or any two of other algebraic symbols.
Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Thus, we can work out the problem.
Worked-out examples for the word problems on simultaneous linear equations:
1. The sum of two number is 14 and their difference is 2. Find the numbers.
Solution:
Let the two numbers be x and y.
x + y = 14 ………. (i)
x - y = 2 ………. (ii)
Adding equation (i) and (ii), we get 2x = 16
or, 2x/2 = 16/2 or, x = 16/2
or, x = 8
Substituting the value x in equation (i), we get
8 + y = 14
or, 8 – 8 + y = 14 - 8
or, y = 14 - 8
or, y = 6
Therefore, x = 8 and y = 6
Hence, the two numbers are 6 and 8.
2. In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number.
Solution:
Let the digit in the units place is x
And the digit in the tens place be y.
Then x = 3y and the number = 10y + x
The number obtained by reversing the digits is 10x + y.
If 36 is added to the number, digits interchange their places,
Therefore, we have 10y + x + 36 = 10x + y
or, 10y – y + x + 36 = 10x + y - y
or, 9y + x – 10x + 36 = 10x - 10x
or, 9y - 9x + 36 = 0 or, 9x - 9y = 36
or, 9(x - y) = 36
or, 9(x - y)/9 = 36/9
or, x - y = 4 ………. (i)
Substituting the value of x = 3y in equation (i), we get
3y - y = 4
or, 2y = 4
or, y = 4/2
or, y = 2
Substituting the value of y = 2 in equation (i),we get
x - 2 = 4
or, x = 4 + 2
or, x = 6
Therefore, the number becomes 26.
3. If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions.
Solution:
Let the fraction be x/y.
If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have
(x + 2)/(y + 2) = 9/10
or, 10(x + 2) = 9(y + 2)
or, 10x + 20 = 9y + 18
or, 10x – 9y + 20 = 9y – 9y + 18
or, 10x – 9x + 20 – 20 = 18 – 20
or, 10x – 9y = -2 ………. (i)
If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have
(x – 3)/(y – 3) = 4/5
or, 5(x – 3) = 4(y – 3)
or, 5x – 15 = 4y – 12
or, 5x – 4y – 15 = 4y – 4y – 12
or, 5x – 4y – 15 + 15 = – 12 + 15
or, 5x – 4y = 3 ………. (ii)
So, we have 10x – 9y = – 2 ………. (iii)
and 5x – 4y = 3 ………. (iv)
Multiplying both sided of equation (iv) by 2, we get
10x – 8y = 6 ………. (v)
Now, solving equation (iii) and (v) , we get
10x – 9y = -2
10x – 8y = 6
- y = - 8
y = 8
Substituting the value of y in equation (iv)
5x – 4 × (8) = 3
5x – 32 = 3
5x – 32 + 32 = 3 + 32
5x = 35
x = 35/5
x = 7
Therefore, fraction becomes 7/8.
4. If twice the age of son is added to age of father, the sum is 56. But if twice the age of the father is added to the age of son, the sum is 82. Find the ages of father and son.
Solution:
Let father’s age be x years
Son’s ages = y years
Then 2y + x = 56 …………… (i)
And 2x + y = 82 …………… (ii)
Multiplying equation (i) by 2, (2y + x = 56 …………… × 2)we get
or, 3y/3 = 30/3
or, y = 30/3
or, y = 10 (solution (ii) and (iii) by subtraction)
Substituting the value of y in equation (i), we get;
2 × 10 + x = 56
or, 20 + x = 56
or, 20 – 20 + x = 56 – 20
or, x = 56 – 20
x = 36
5. Two pens and one eraser cost Rs. 35 and 3 pencil and four erasers cost Rs. 65. Find the cost of pencil and eraser separately.
Solution:
Let the cost of pen = x and the cost of eraser = y
Then 2x + y = 35 ……………(i)
And 3x + 4y = 65 ……………(ii)
Multiplying equation (i) by 4,
Subtracting (iii) and (ii), we get;
5x = 75
or, 5x/5 = 75/5
or, x = 75/5
or, x = 15
Substituting the value of x = 15 in equation (i) 2x + y = 35 we get;
or, 2 × 15 + y = 35
or, 30 + y = 35
or, y = 35 – 30
or, y = 5
Therefore, the cost of 1 pen is Rs. 15 and the cost of 1 eraser is Rs. 5.
Simultaneous Linear Equations
Simultaneous Linear Equations
Comparison Method
Elimination Method
Substitution Method
Cross-Multiplication Method
Solvability of Linear Simultaneous Equations
Pairs of Equations
Word Problems on Simultaneous Linear Equations
Word Problems on Simultaneous Linear Equations
Practice Test on Word Problems Involving Simultaneous Linear Equations
Simultaneous Linear Equations - Worksheets
Worksheet on Simultaneous Linear Equations
Worksheet on Problems on Simultaneous Linear Equations |
# A wire in the shape of Y is hanged from two points A and B, 8 feet apart. The lower end of wire reaches a point 10 feet from AB. What is the shortest possible length of the wire?
##### 2 Answers
Jun 15, 2017
Shortest length of wire that can be used is $12.31$ feet.
#### Explanation:
The above can be better described by the following figure.
Now using Pythagoras theorem, if $x$ is the length of verticle component of wire, lengths of tilted wires each is
$\sqrt{{4}^{2} + {\left(10 - x\right)}^{2}}$ or $\sqrt{16 + 100 + {x}^{2} - 20 x}$ or sqrt(x^2-20x+116
and total length of wire is $l \left(x\right) = x + 2 \sqrt{{x}^{2} - 20 x + 116}$
This will be minimum when $\frac{\mathrm{dl}}{\mathrm{dx}} = 0$
As $\frac{\mathrm{dl}}{\mathrm{dx}} = 1 + 2 \times \frac{1}{2 \sqrt{{x}^{2} - 20 x + 116}} \times \left(2 x - 20\right) = 0$
or $1 - \frac{2 \left(10 - x\right)}{\sqrt{{x}^{2} - 20 x + 116}} = 0$
or $\sqrt{{x}^{2} - 20 x + 116} = 20 - 2 x$
and squaring we get
${x}^{2} - 20 x + 116 = 400 + 4 {x}^{2} - 80 x$
or $3 {x}^{2} - 60 x + 284 = 0$
or $x = \frac{60 \pm \sqrt{3600 - 4 \times 3 \times 284}}{6}$
$= \frac{60 \pm \sqrt{3600 - 3408}}{6} = \frac{60 \pm \sqrt{192}}{6}$
or $x = 10 \pm 2.31$ i.e. $x = 12.31$ or $7.69$
But $x$ cannot be greater than $10$
Hence $x = 7.69$ and shortest length of wire that can be used is
$l \left(7.69\right) = 7.69 + 2 \sqrt{{7.69}^{2} - 20 \times 7.69 + 116}$
= $7.69 + 2 \sqrt{21.3361} = 7.69 + 4.62 = 12.31$
Jun 15, 2017
The shortest total length of the wire that can be used $= 16.92 f t$
#### Explanation:
Let the length of the perpedicular $O D$ dropped from the junction point $O$ of Y-shaped wire frame to the horizontal line AB be $x f t$.
The total height being 10ft the height of $O$ will be $\left(10 - x\right) f t$
Since in $\Delta A O B$ $O A = O B$
Here $D$ will be mid point of $A B$ So $A D = B D = 4 f t$
By pythagoras theorem
$O A = O B = \sqrt{{x}^{2} + {4}^{2}}$
So total length of the wire
$L = 2 \sqrt{{x}^{2} + {4}^{2}} + 10 - x \ldots . . \left(1\right)$
To know the minimum length of wire required we impose the condtion $\frac{\mathrm{dL}}{\mathrm{dx}} = 0$
So
$\frac{d}{\mathrm{dx}} \left(2 \sqrt{{x}^{2} + {4}^{2}} + 10 - x\right) = 0$
$\implies \left(2 \cdot \frac{1}{2} \cdot \frac{2 x}{\sqrt{{x}^{2} + {4}^{2}}} - 1\right) = 0$
$\implies \frac{2 x}{\sqrt{{x}^{2} + {4}^{2}}} = 1$
$\implies {\left(2 x\right)}^{2} = {\left(\sqrt{{x}^{2} + {4}^{2}}\right)}^{2}$
$\implies 4 {x}^{2} = {x}^{2} + 16$
$\implies x = \frac{4}{\sqrt{3}}$
Hence minimum length can be had by inserting $\implies x = \frac{4}{\sqrt{3}}$ in (1)
${L}_{\text{min}} = 2 \left(\sqrt{{\left(\frac{4}{\sqrt{3}}\right)}^{2} + {4}^{2}}\right) + 10 - \frac{4}{\sqrt{3}}$
$\implies {L}_{\text{min}} = 2 \times \frac{8}{\sqrt{3}} + 10 - \frac{4}{\sqrt{3}}$
$\implies {L}_{\text{min}} = \frac{16}{\sqrt{3}} - \frac{4}{\sqrt{3}} + 10$
$\implies {L}_{\text{min}} = \frac{12}{\sqrt{3}} + 10 = 4 \sqrt{3} + 10 = 16.92 f t$ |
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# Sec. 2.8 mean, median, mode and range
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### Sec. 2.8 mean, median, mode and range
1. 1. Learning Target:Students will learn how to calculate data using mean, median, mode and range. Students will learn how to justify which average best represents the data set.<br />
2. 2. Definitions:Mean (avg.) – Add up items Total itemsMedian – Middle # (must put them in order from smallest to largest)Mode – The value you see most oftenRange – Difference between the largest # and the smallest #Outlier – A number that is either a lot smaller or bigger then the rest of the data<br />
3. 3.
4. 4. Find the mean of the data.20, 25, 25, 20, 25<br />
5. 5. 20 + 25 + 25 + 20 + 2551155Mean = 23<br />
6. 6. Find the median of the data.(odd # of data values)91, 150, 80, 71, 74, 81, 83<br />
7. 7. Put them in order from smallest to largest. 91, 150, 80, 71, 74, 81, 83Find the middle value.71, 74, 80, 81, 83, 91, 150Median = 81<br />
8. 8. Find the median of the data(even # of data values)8, 13, 8, 4, 11, 4, 2, 6<br />
9. 9. Put them in order from smallest to largest.2, 4, 4, 6, 8, 8, 11, 13Find the middle value.2, 4, 4, 6, 8, 8, 11, 13Average the two numbers (6 + 8) / 214/2Median = 7<br />
10. 10. Find the mode.14, 10, 9, 7, 14, 16, 147, 1, 2, 6, 1, 726, 22, 10, 12, 16, 28<br />
11. 11. 14, 10, 9, 7, 14, 16, 14mode = 147, 1, 2, 6, 1, 7mode = 1 and 726, 22, 10, 12, 16, 28mode = no mode(write no mode, do not just put a 0)<br />
12. 12. Find the range.91, 150, 80, 71, 74, 81, 80, 77<br />
13. 13. 91, 150 , 80, 71 , 74, 81, 80, 77150 – 71Range = 79<br /> |
# Multiplication Times Tables Tips and Tricks
Aug 28, 2015
For a lot of us math tests are right up there with chillin' with Grandma's lawn bowling team or battling back-acne on the anxiety-o-meter. Chill out. Kidzworld has the tips and tricks you need to master your multiplication tables. Sure, you're going to need to memorize some of them, but these tips will keep the number of those to a minimum. Check it out!
## Zero and One Times Tables :: Could This Get Any Easier?
Check this out: if you multiply ANYTHING by zero, the answer is zero. Anything. 4 x 0 = 0 and 1,000,000 x 0 = 0. One times tables are almost as easy. Any number multiplied by one is always itself. "Huh," you say? Well, check it out: 1 x 10 = 10, 1 x 42 = 42, 8,726 x 1 = 8,726. Simple, right?
Math Class
## Two Times Tables :: Double Your Pleasure
When you multiply a number by two, you just double that number. That's some pretty simple math, even for the most numerically challenged of us. So, if you want to figure out what 2 x 7 is, you just add 7 + 7 (the answer is 14, by the way). Any number times two is the same as that number PLUS itself. Here's one more example: 2 x 5 is the same as 5 + 5, which equals 10. Practice a couple -- it's really easy.
## Four Times Tables :: Double, Double Trouble
OK, so now that you have the twos figured out, it's time to get doubling again. This is not as confusing as it may sound. When you multiply four with anything, you have to use the doubling-up trick (that's the one you used for the two times table) twice. Here's an example: 4 x 7 is the same as 7 + 7 = 14 and then 14 + 14 = 28. So 4 x 7 = 28. Here's another double, double example: 4 x 10 is the same as 10 + 10 = 20, so then 20 + 20 = 40. So the answer is 4 x 10 = 40.
## Five Times Tables :: It's Why You Have Fingers
First things first - can you count to five? Yeah? Then you can figure out your five times tables. So, when you want to multiply a number by five you just count up by fives that may times. Let's review how to count by fives: 5, 10, 15, 20, 25... and so on. Got it? So if you want to multiply 5 x 7, you just count by fives, seven times. 5, 10, 15, 20, 25, 30, 35. So 7 x 5 = 35. If you have trouble keeping track, just use your fingers.
## Ten Times Tables :: Just Add Zero and Stir
Here's another no-brainer. If you want to multiply something by 10, just add a zero on the end. Here's an example: 10 x 8 = 80 or 10 x 100 = 1,000. Try it with any number - from one to a billion. It's just too easy.
## Eleven Times Tables :: Seeing Double
Here's another one that's as easy as pie. Any number - up to nine - multiplied by 11 is itself written out twice. Confused? Just check it out: 9 x 11 = 99, 4 x 11 = 44, 3 x 11 = 33 and so on.
## * Hint *
Times tables work both ways (5 x 10 is the same as 10 x 5) so if you have the question 3 x 11, don't stress that there isn't a trick for the three times tables -- use the trick for 11 instead!
Related Articles |
# Lesson 4- Decimal Place Value & Value
Objective
At the end of this lesson, students should know and understand the place value and value of decimal numbers.
Decimal Vocabulary
• Decimal/Decimal Number: Is a number that represents a whole number and parts of a whole by separating them with a point.
• Decimal point: Is a point separating the ones and tenths in a decimal number.
• Ones: Is the place value of the digit on the immediate left hand side of the decimal point.
• Tenth: Is the first number on the immediate right hand side of the decimal point. It is the first decimal place and represents parts out of 10 equal parts (E.g. “.8” means 8 parts out of 10. i.e.).
• Hundredth: Is the second number on the right hand side of the decimal point. It is the second decimal place and represents parts out of 100 equal parts (E.g.“.05” means 5 parts out of 100 i.e ).
• Thousandth – Is the third number on the right hand side of the decimal point. It is the third decimal place and represents parts out of 1000 equal parts (E.g.“.009” means 9 parts out of 1000 i.e.).
• Decimal Place: Is the number of digits after the decimal point.
The decimal point separates the whole number from the fraction part of a decimal number. In the above diagram, the whole number is 342,368. The
lowest place value of whole numbers is the units. Therefore, from right to left on the left hand side of the decimal point: units, tens, hundreds, thousands, ten thousands and hundred thousands.
On the right hand side of the decimal point, the highest place value of the fraction part begins: tenth, hundredth, thousandth, etc. as shown in the illustration above.
Example 1
Write the value and place value of the underlined digits in the following decimal numbers.
a) 2486.08 b) 35276.826
c) 359,901.620 d) 24.241
e) 19.4211 f) 78 9.1515
Solution
Example 2
What is the total value of the underlined figures in each of these?
a) 24.5 b) 828.84
Solution
a) 24.5 = 2 tens and 5 tenths
= 20 + 0.5
= 20.5
b) 828.84 = 2 tens and 8 units
= 20 + 8
= 28
An Abacus or Notation Board can be used to illustrate decimals based on the knowledge of place value. Let’s look at the Abacus below.
Notation boards work in the same way as the Abacus. Below are some illustrations on Notation boards.
Lesson Content
error: Content is protected !! |
# 1.03 Order numbers
Lesson
## Are you ready?
Before we try and order 5-digit numbers, let's check we know how to read and write them.
### Examples
#### Example 1
Write the following using numerals:
fifty thousand, five hundred and fifty five
Worked Solution
Create a strategy
Write the number in a place value table.
Apply the idea
In numerals, fifty thousand, five hundred and fifty five is 50\,555.
Idea summary
In a place value table, five-digit numbers start in the tens of thousands column and have a digit in each column down to the units.
## Order 5 digit numbers on a number line
Let's use a number line to order 5-digit numbers.
Loading video...
### Examples
#### Example 2
We want to work out which of three numbers is the largest.
a
Plot 41\,590, 41\,690, and 41\,490 on the number line below.
Worked Solution
Create a strategy
Find how much each space between tick marks represents.
Apply the idea
On the number lines the first two numbers are 41\,450 and 41\,500.
41\,500-41\,450=50, and there are 5 spaces between 41\,450 and 41\,500 which means that each space represents an increase of 10, since there are five lots of 10 in 50.
41\,490 is 4 tens more than 41\,450, so it should be 4 spaces to the right of 41\,450.
41\,590 is 4 tens more than 41\,550, so it should be 4 spaces to the right of 41\,550.
41\,690 is 4 tens more than 41\,650, so it should be 4 spaces to the right of 41\,650.
The numbers are plotted below:
b
Write the three numbers from smallest to largest.
Worked Solution
Create a strategy
Use the number line from part (a).
Remember that the smaller a number is, the further to the left on a number line it is.
Apply the idea
41\,490 was plotted furthest to the left which makes it the smallest.
Moving right 41\,590 was plotted next and then 41\,690.
So the numbers in order are:
41\,490, \,41\,590,\, 41\,690
Idea summary
The smaller a number is, the further to the left on a number line it is.
The larger a number is, the further to the right on a number line it is.
## Order 5 digit numbers using place value
This time we will see how to use a place value table to order 5-digit numbers.
Loading video...
### Examples
#### Example 3
Order these numbers from largest to smallest.
a
60\,679,70\,966,70\,669
Worked Solution
Create a strategy
Use a place value table and compare the digits from ten thousands.
Apply the idea
The numbers in a place value table is shown below:
We can see that 60\,679 has the smallest number of ten thousands so it is the smallest number.
70\,966 and 70\,699 have the same number of thousands.
70\,699 has less hundreds than 70\,966, so 70\,699 \lt 70\,966.
So the numbers from largest to smallest are:70\,966,70\,669,60\,679
b
95\,036,90\,536,95\,306
Worked Solution
Create a strategy
Use a place value table and compare the digits from ten thousands.
Apply the idea
The numbers in a place value table is shown below:
We can see that all the numbers have the same number of ten thousands.
90\,536 has the smallest number of thousands so it is the smallest number.
95\,036 and 95\,306 have the same number of thousands.
95\,036 has less hundreds than 95\,306, so 95\,036 \lt 95\,306.
So the numbers from largest to smallest are:95\,306,95\,036,90\,536
Idea summary
To determine the largest or smallest number in a pair, or set, of numbers, we can compare the furthermost left place value digits. If they are the same, we move to the right, one place at a time, until they are different.
We can also use the 'greater than' (\gt) or 'less than' (\lt) symbol to compare two numbers, remembering that the 'greater than' sign has the big part towards the larger number, and the 'less than' symbol has the small part towards the smaller number.
### Outcomes
#### MA2-4NA
applies place value to order, read and represent numbers of up to five digits
#### MA2-5NA
uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Resultant as the Sum of Two Components
## Horizontal and vertical components of vectors.
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Practice Resultant as the Sum of Two Components
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Resultant as the Sum of Two Components
You are working in science class on a "weather unit". As part of this class, you are tasked with going out and checking the wind speed each day at a meter behind your school. The wind speed you record for the day is 20 mph at a $E50^\circ N$ trajectory. This means that the wind is blowing at an angle $50^\circ$ taken from the direction that would be due East. At the conclusion of each day, you are supposed to break the wind speed (which is a vector) into two components: the portion that is in a North/South direction and the portion that is in an East/West direction. Can you figure out how to do this?
By the end of this Concept, you'll be able to break vectors into their individual components using trig relationships.
### Guidance
We can look at any vector as the resultant of two perpendicular components. If we generalize some vector $\vec{q}$ into perpendicular components, $|\vec{r}|\hat{i}$ is the horizontal component of a vector $\vec{q}$ and $|\vec{s}| \hat{j}$ is the vertical component of $\vec{q}$ . Therefore $\vec{r}$ is a magnitude, $|\vec{r}|$ , times the unit vector in the $x$ direction and $\vec{s}$ is its magnitude, $|\vec{s}|$ , times the unit vector in the $y$ direction. The sum of $\vec{r}$ plus $\vec{s}$ is: $\vec{r} + \vec{s} = \vec{q}$ . This addition can also be written as $|\vec{r}|\hat{i} + |\vec{s}| \hat{j} = \vec{q}$ .
If we are given the vector $\vec{q}$ , we can find the components of $\vec{q},\vec{r}$ , and $\vec{s}$ using trigonometric ratios if we know the magnitude and direction of $\vec{q}$ .
This is accomplished by taking the magnitude of the vector times the cosine of the vector's angle to find the horizontal component, and the magnitude of the vector times the sine of the vector's angle to find the vertical component.
#### Example A
If $|\vec{q}| = 19.6$ and its direction is $73^\circ$ , find the horizontal and vertical components.
Solution: If we know an angle and a side of a right triangle, we can find the other remaining sides using trigonometric ratios. In this case, $\vec{q}$ is the hypotenuse of our triangle, $\vec{r}$ is the side adjacent to our $73^\circ$ angle, $\vec{s}$ is the side opposite our $73^\circ$ angle, and $\vec{r}$ is directed along the $x-$ axis.
To find $\vec{r}$ , we will use cosine and to find $\vec{s}$ we will use sine. Notice this is a scalar equation so all quantities are just numbers. It is written as the quotient of the magnitudes, not the vectors.
$\cos 73 & = \frac{|\vec{r}|}{|\vec{q}|} = \frac{r}{q} && \sin 73 = \frac{|\vec{s}|}{|\vec{q}|} = \frac{s}{q} \\\cos 73 & = \frac{r}{19.6} && \sin 73 = \frac{s}{19.6} \\r & = 19.6 \cos 73 && \qquad \ \ s = 19.6 \sin 73 \\r & = 5.7 && \qquad \ \ s = 18.7$
The horizontal component is 5.7 and the vertical component is 18.7. One can rewrite this in vector notation as $5.7\hat{i} + 18.7\hat{j} = \vec{q}$ . The components can also be written $\vec{q} = \big \langle{5.7, 18.7}\big\rangle$ , with the horizontal component first, followed by the vertical component. Be careful not to confuse this with the notation for plotted points.
#### Example B
If $|\vec{m}| = 12.1$ and its direction is $31^\circ$ , find the horizontal and vertical components.
To find $\vec{r}$ , we will use cosine and to find $\vec{s}$ we will use sine. Notice this is a scalar equation so all quantities are just numbers. It is written as the quotient of the magnitudes, not the vectors.
$\cos 31 & = \frac{|\vec{r}|}{|\vec{m}|} = \frac{r}{m} && \sin 31 = \frac{|\vec{s}|}{|\vec{m}|} = \frac{s}{m} \\\cos 31 & = \frac{r}{12.1} && \sin 31 = \frac{s}{12.1} \\r & = 12.1 \cos 31 && \qquad \ \ s = 12.1 \sin 31 \\r & = 10.37 && \qquad \ \ s = 6.23$
#### Example C
If $|\vec{r}| = 15$ and $|\vec{s}| = 11$ , find the resultant vector length and angle.
Solution: We can view each of these vectors on the coordinate system here:
Each of these vectors then serves as sides in a right triangle. So we can use the Pythagorean Theorem to find the length of the resultant:
$c^2 = a^2 + b^2\\c^2 = 15^2 + 11^2\\c^2 = 225 + 121\\c^2 = 346\\c = \sqrt{346} \approx 18.60$
The angle of rotation that the vector makes with the "x" axis can be found using the tangent function:
$\tan \theta = \frac{opposite}{adjacent}\\\tan \theta = \frac{11}{15}\\\tan \theta = .73\\\theta = \tan^{-1}.73\\\theta \approx 36.13$
### Vocabulary
Component Vectors: The component vectors of a given vector are two or more vectors whose sum is the given vector.
Resultant: The resultant is a vector that is the sum of two or more vectors.
### Guided Practice
1. Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as $\text{magnitude} = 75 \qquad \qquad \quad \text{direction} = 35^\circ$ .
2. Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as $\text{magnitude} = 3.4 \qquad \qquad \quad \text{direction} = 162^\circ$ .
3. Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as $\text{magnitude} = 15.9 \qquad \qquad \ \text{direction} = 12^\circ$ .
Solutions:
1. $\cos 35^\circ = \frac{x}{75}, \sin 35^\circ = \frac{y}{75}, x = 61.4, y = 43$
2. $\cos 162^\circ = \frac{x}{3.4}, \sin 162^\circ = \frac{y}{3.4}, x = 3.2, y = 1.1$
3. $\cos 12^\circ = \frac{x}{15.9}, \sin 12^\circ = \frac{y}{15.9}, x = 15.6, y = 3.3$
### Concept Problem Solution
From this Concept you've learned how to take a vector and break it into components using trig functions. If you draw the wind speed you recorded as a vector:
You can find the "x" and "y" components. These are the same as the part of the wind that is blowing to the East and the part of the wind that is blowing to the North.
East component: $\cos 50^\circ = \frac{x}{20}$
$20 \cos 50^\circ = x$
x = 12.86 mph
North component: $\sin 50^\circ = \frac{y}{20}$
$20 \sin 50^\circ = y$
y = 15.32 mph
### Practice
Find the horizontal and vertical components of the following vectors given the resultant vector’s magnitude and direction.
1. $\text{magnitude} = 65 \quad \text{direction} = 22^\circ$ .
2. $\text{magnitude} = 34 \quad \text{direction} = 15^\circ$ .
3. $\text{magnitude} = 29 \quad \text{direction} = 160^\circ$ .
4. $\text{magnitude} = 100 \quad \text{direction} = 320^\circ$ .
5. $\text{magnitude} = 320 \quad \text{direction} = 200^\circ$ .
6. $\text{magnitude} = 15 \quad \text{direction} = 110^\circ$ .
7. $\text{magnitude} = 10 \quad \text{direction} = 80^\circ$ .
8. $\text{magnitude} = 90 \quad \text{direction} = 290^\circ$ .
9. $\text{magnitude} = 87 \quad \text{direction} = 10^\circ$ .
10. $\text{magnitude} = 42 \quad \text{direction} = 150^\circ$ .
1. If $|\vec{r}| = 12$ and $|\vec{s}| = 8$ , find the resultant vector magnitude and angle.
2. If $|\vec{r}| = 14$ and $|\vec{s}| = 6$ , find the resultant vector magnitude and angle.
3. If $|\vec{r}| = 9$ and $|\vec{s}| = 24$ , find the resultant vector magnitude and angle.
4. Will cosine always be used to find the horizontal component of a vector?
5. If you know the component form of a vector, how can you find its magnitude and direction?
### Vocabulary Language: English
Resultant
Resultant
A resultant is a vector representing the sum of two or more vectors. |
### Geometry Chapter 2
```Geometry Chapter 2
Benedict
Warm Up 12
1. The _____ of a conditional statement is
found by switching the hypothesis and
conclusion.
2. Write the hypothesis and conclusion of
the statement; “If the dew equals the air
temperature, then it will rain.”
3. Write the statement in if-then form.
The measure of a right angle is 90 degrees.
Re-teach 12
Conditional statement- two parts; a
hypothesis and a conclusion.
Statement written in “if-then form”, the
“if” is the hypothesis and “then” is the
conclusion
If it is a block day, then it is a
Wednesday or Thursday.
Re-teach 12
Negation- writing the negative of the
statement
Wednesday is not a block day
Contrapositive- negating the hypothesis
and the conclusion of the converse
Practice 12
Directions: Rewrite the conditional
statement in if-then form.
Three points are collinear if they lie on
the same line.
Beating LM gets us a half day on
Monday.
Closure 12
Directions: Write the converse of the
statement.
If the angle measures 38°, then it is
acute.
I will go to the movies if it is raining.
Warm Up 13
Directions: Use the graph on the flashcard
13. Points A, F and G are collinear.
15. DC is perpendicular to line l.
17. < FBJ and <JBA are complementary.
19. <ABJ and <DCH are supplementary.
Practice 13
Directions: Rewrite the biconditional statement
as a conditional statement and its converse.
23. A point is a midpoint of a segment if
and only if it divides the segment into two
congruent segments.
Directions: Give a counterexample that
demonstrates that the converse of the
statement is false.
25. If two angles measure 42° and 48°, then
they are complementary.
Closure 13
Directions: Find the measures of a
complement and a supplement of the
angle.
87°
Warm Up 14
Directions: Find the distance between
the two points. Answer will be a
decimal.
35. A (4, 5) B (-3, -2)
Directions: Given the endpoint and the
midpoint of the segment, find the
other endpoint.
41. B (5, 7) M (-1, 0)
Practice 14
Directions: Pair up and complete
problems 31-35 on the flashcard.
Re-teach 15
Parallel lines- coplaner lines that do not
intersect.
Skew lines- lines that are not coplaner
and do not intersect.
Transversal- line that intersects two or
more coplaner lines at different points.
Re-teach 15
Alternate Exterior Angles- lie outside two
lines on opposite sides of the transversal.
Alternate Interior Angles- lie between two
lines on opposite sides of transversal.
Consecutive Interior angles or Same Side
Interior Angles- lie in between two lines on
same side of the transversal.
Re-teach 15
Corresponding Angles- Angles in
matching corners are called
corresponding angles.
Practice 15
Directions Use the following graph to identify
– Corresponding Angles
– Alternate Exterior Angles
– Alternate Interior Angles
– Consecutive Interior Angles/Same Side Interior
Warm Up 16
Directions: Find the value of x.
Practice 16
Directions: Work with a partner to solve
the problems on the worksheet.
When finished turn it in to the bin.
Warm Up 18
Directions: Find the value of x. Explain
what makes r and s parallel.
Practice 18
Directions: Work together to solve the
problems on the worksheet.
Re-teach 19
Slope
Finding slope: (-2, 7) (3, 2)
Re-teach 19
Equation of a line
Solving for b (-2, 7) (3, 2)
Practice 19
Find the slope of the line and then
create an equation of the line for the
following points.
5. (2, 5) (4, 1)
Practice 19
Find the slope of the line. Determine if
the points are parallel.
17. (-5, 9) (-1, 1) and (0, 7) (3, 1)
19. (0, 4) (-2, -2) and (4, 2) (2, -6)
21. (0,7) (-6, 2) and (5, 1) (-2, -4)
Practice 19
Find the slope of the line. Determine if
line AC is perpendicular to line BD.
A (-1, -2) C (0, 1) and B (-3, 2) D (3, 0)
A ( -2, -1) C (4, 1) and B (-2, 3) D (0, -2)
Closure 19
Directions: Explain whether the lines are
perpendicular, parallel or neither.
y = -2x – 1; y = -2x – 3
y = -1/2x + 3; y = - 1/2x + 5
y = -3x + 1; y = 1/3x + 1
y = 4x + 10; y = -2x + 5
``` |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Calculator Use with Algebra Expressions
## Calculator evaluation by simple input or using memory function
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Calculator Use with Algebra Expressions
What if you wanted to evaluate the expression \begin{align*}\frac{x^2+4x+3}{2x^2-9x-5}\end{align*} when \begin{align*}x=7\end{align*} ? If you had your calculator handy, it could make things easier, but how would you enter an expression like this into your calculator? Also, how would you tell your calculator that \begin{align*}x=7\end{align*} ? In this Concept, you will learn how to make use of your calculator when evaluating expressions like these.
### Guidance
A calculator, especially a graphing calculator, is a very useful tool in evaluating algebraic expressions. A graphing calculator follows the order of operations, PEMDAS. In this section, we will explain two ways of evaluating expressions with a graphing calculator.
#### Example A
Method #1 This method is the direct input method. After substituting all values for the variables, you type in the expression, symbol for symbol, into your calculator.
Evaluate \begin{align*}[3(x^2 - 1)^2 - x^4 + 12] + 5x^3 - 1\end{align*} when \begin{align*}x = -3\end{align*} .
Substitute the value \begin{align*}x = -3\end{align*} into the expression.
The potential error here is that you may forget a sign or a set of parentheses, especially if the expression is long or complicated. Make sure you check your input before writing your answer. An alternative is to type in the expression in appropriate chunks – do one set of parentheses, then another, and so on.
#### Example B
Method #2 This method uses the STORE function of the Texas Instrument graphing calculators, such as the TI-83, TI-84, or TI-84 Plus.
First, store the value \begin{align*}x = -3\end{align*} in the calculator. Type -3 [STO] \begin{align*}x\end{align*} . ( The letter \begin{align*}x\end{align*} can be entered using the \begin{align*}x\end{align*} - [VAR] button or [ALPHA] + [STO] ). Then type in the expression in the calculator and press [ENTER] .
The answer is \begin{align*}-13.\end{align*}
Note: On graphing calculators there is a difference between the minus sign and the negative sign. When we stored the value negative three, we needed to use the negative sign, which is to the left of the [ENTER] button on the calculator. On the other hand, to perform the subtraction operation in the expression we used the minus sign. The minus sign is right above the plus sign on the right.
#### Example C
You can also use a graphing calculator to evaluate expressions with more than one variable.
Evaluate the expression: \begin{align*}\frac{3x^2 - 4y^2 + x^4}{(x + y)^{\frac{1}{2}}}\end{align*} for \begin{align*}x = -2, y = 1\end{align*} .
Store the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} : \begin{align*}-2\end{align*} [STO] \begin{align*}x\end{align*} , 1 [STO] \begin{align*}y\end{align*} . The letters \begin{align*}x\end{align*} and \begin{align*}y\end{align*} can be entered using [ALPHA] + [KEY] . Input the expression in the calculator. When an expression shows the division of two expressions, be sure to use parentheses: (numerator) \begin{align*}\div\end{align*} (denominator). Press [ENTER] to obtain the answer \begin{align*}-.8\bar{8}\end{align*} or \begin{align*}-\frac{8}{9}\end{align*} .
### Guided Practice
Evaluate the expression \begin{align*}\frac{x-y}{x^2 + y^2}\end{align*} for \begin{align*}x = 3, y = -1\end{align*} .
Store the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} : \begin{align*}3\end{align*} [STO] \begin{align*}x\end{align*} , -1 [STO] \begin{align*}y\end{align*} . The letters \begin{align*}x\end{align*} and \begin{align*}y\end{align*} can be entered using [ALPHA] + [KEY] . Input the expression in the calculator. When an expression shows the division of two expressions, be sure to use parentheses: (numerator) \begin{align*}\div\end{align*} (denominator). Press [ENTER] to obtain the answer \begin{align*}0.4\end{align*} .
### Practice
In 1-5, evaluate each expression using a graphing calculator.
1. \begin{align*}x^2 + 2x - xy\end{align*} when \begin{align*}x = 250\end{align*} and \begin{align*}y = -120\end{align*}
2. \begin{align*}(xy - y^4)^2\end{align*} when \begin{align*}x = 0.02\end{align*} and \begin{align*}y = -0.025\end{align*}
3. \begin{align*}\frac{x + y - z}{xy + yz + xz}\end{align*} when \begin{align*}x = \frac{1}{2}, \ y = \frac{3}{2}\end{align*} , and \begin{align*}z = -1\end{align*}
4. \begin{align*}\frac{(x + y)^2}{4x^2 - y^2}\end{align*} when \begin{align*}x = 3\end{align*} and \begin{align*}y = -5d\end{align*}
5. The formula to find the volume of a spherical object (like a ball) is \begin{align*}V = \frac{4}{3}(\pi)r^3\end{align*} , where \begin{align*}r =\end{align*} the radius of the sphere. Determine the volume for a grapefruit with a radius of 9 cm.
In 6-9, insert parentheses in each expression to make a true equation.
1. \begin{align*}5 - 2 \cdot 6 - 4 + 2 = 5\end{align*}
2. \begin{align*}12 \div 4 + 10 - 3 \cdot 3 + 7 = 11\end{align*}
3. \begin{align*}22 - 32 - 5 \cdot 3 - 6 = 30\end{align*}
4. \begin{align*}12 - 8 - 4 \cdot 5 = -8\end{align*}
Mixed Review
1. Let \begin{align*}x = -1\end{align*} . Find the value of \begin{align*}-9x + 2\end{align*} .
2. The area of a trapezoid is given by the equation \begin{align*}A = \frac{h}{2}(a + b)\end{align*} . Find the area of a trapezoid with bases \begin{align*}a = 10 \ cm, b = 15 \ cm\end{align*} , and height \begin{align*}h = 8 \ cm\end{align*} .
3. The area of a circle is given by the formula \begin{align*}A = \pi r^2\end{align*} . Find the area of a circle with radius \begin{align*}r = 17\end{align*} inches.
### Vocabulary Language: English Spanish
evaluate
evaluate
To find the value of a numerical or algebraic expression. |
In this article, we will examine the world of algorithms. Specifically, we will talk about an algorithm that we can use to generate permutations.
Let’s start.
## Permutations
Firstly, we need a definition for permutation, of which there are a few. Even Wikipedia’s definition is quite difficult to understand. Therefore, let us say that a permutation is an arrangement of items in a particular order, with the order being paramount.
To dive in a little further, let’s examine a concrete example. Let us say that we have three coins (`1`, `2`, and `5` cents). What are the different possibilities for arranging these coins? It turns out that there are six possible arrangements: `[1, 2, 5]`, `[1, 5, 2]`, `[2, 1, 5]`, `[2, 5, 1]`, `[5, 1, 2]`, `[5, 2, 1]`. This is the set of all possible permutations of 3 items.
### Number of Permutations
Before moving on to programming, let us answer a theoretical question: “How many possible permutations are there for N items?’. We already saw that there are six possibilities when we have three items.Â
We have N options for the first position if we have N items. Then we only have N-1 items for the second position, as we already used 1 for the first position. On the third position, we have N-2 items, and so on. With that, we can deduce a formula as N * (N-1) * (N-2) * … * 3 * 2 * 1, which is a mathematical definition for factorial, and the symbol for factorial is an exclamation mark (!).Â
The number of all possible permutations on N items is N!, which can quickly become very large. For example, with three items, there are six permutations. Moreover, if we have six items, there are already 720 permutations. Similarly, with ten items, there are 3,628,800 permutations; with 20 items, there are whooping 2,432,902,008,176,640,000 permutations.
To put this into perspective, suppose we have a printer so fast that it can print 1000 permutations in one second. That printer would print all permutations of the 20 elements for about 77 million years. Keeping that in mind, we’ll want to use an algorithm as fast as possible.
### How Excellent and Quick Is an Algorithm? The Big O Notation.
In the field of combinatorics and algorithms, determining an algorithm’s quality is crucial. We need to classify whether algorithm A is better than algorithm B. The “Big O” notation is a special notation in computer science that helps determine the upper bound of the number of calculations an algorithm must execute to achieve a result.
The “Big O” notation classifies algorithms into classes such as n, n^2, log(n), n*log(n), etc. These classifications indicate that their algorithms will have to execute, for example, log(n) calculations to produce a result for n elements. The binary search algorithm is an excellent example of such a log(n) algorithm.
In our case, however, there is not much to calculate. We have already established that the number of permutations for n elements is n!. Even the best algorithm in the universe can’t solve it without executing at least n! calculations, so we can’t do any better. But we can do a bit worse. First, we’ll write the most straightforward algorithm possible, with much higher complexity (n^n) and lower performance.
## Basic Permutations Algorithm
The basic algorithm is simple and can be visualized as a bicycle lock with rolling numbers. To find out all the permutations of three digits, we can choose all possible positions of all three wheels on the lock, and we will get this: `000`, `001`, `002`, `010`, `011`, `012`, `020`, `021`, `022`, `100`, `101`, `102`, `110`, `111`, `112`, `120`, `121`, `122`, `200`, `201`, `202`, `210`, `211`, `212`, `220`, `221`, and `222`.
In other words, there are 3^3=27 positions or three loops with three elements:
```public static void PrintAllPossiblePositions()
{
int counter = 1;
for (int p0 = 0; p0 < 3; p0++)
for (int p1 = 0; p1 < 3; p1++)
for (int p2 = 0; p2 < 3; p2++)
Console.WriteLine(\$"Position [{counter++, 2}]: {p0}{p1}{p2}");
}```
But not all results are valid. With permutations, there can’t be repeating digits. So from this set of 27 results, we must discard every result with repeating digits. By applying that, we will get the correct result: `012`, `021`, `102`, `120`, `201`, and `210`.
To print out only permutations, we have to discard results with repeating digits:
```public static void PrintPermutations()
{
int counter = 1;
for (int p0 = 0; p0 < 3; p0++)
for (int p1 = 0; p1 < 3; p1++)
for (int p2 = 0; p2 < 3; p2++)
{
if ((p0 != p1) && (p0 != p2) && (p1 != p2))
Console.WriteLine(\$"Permutation [{counter++,2}]: {p0}{p1}{p2}");
}
}```
This method will print the permutations of 3 elements.
So, we have written our first working algorithm for generating permutations. However, as discussed in the beginning, the algorithm is inefficient since we calculate 27 elements to produce just six results.
### Algorithm Improvement
We can improve a lot with a small algorithm change.
Let’s take these arrangements: `000`, `001`, `002`. We can discard all three arrangements in the second loop. As soon as we see another `0`, we can quit, as there can’t be two zeros in a given permutation:
```public static void PrintPermutationsImproved()
{
int counter = 1;
for (int p0 = 0; p0 < 3; p0++)
for (int p1 = 0; p1 < 3; p1++)
{
if (p1 == p0) continue; // Just skip the whole inner loop(!)
for (int p2 = 0; p2 < 3; p2++)
{
if ((p2 == p0) || (p2 == p1)) continue;
Console.WriteLine(\$"Permutation [{counter++,2}]: {p0}{p1}{p2}");
}
}
}```
We avoid executing the third loop when we discover that the second position would have the same digit as the first. This algorithm is much better than the first one, but they both have one major drawback!
They are useless for generating permutations of 4 or 6 elements. To display all the permutations of 4 or 6 elements, we need a method with 4 or 6 for loops. This will require a different method for each N, making this approach less than ideal.
## Recursion to Save the Day
Fortunately, many problems, especially combinatorial problems, can be solved using recursion. We have to find how a smaller N fits a bigger N. So to use recursion, we will need to find out how to generate permutations of 4 elements (or N elements) if we know how to build permutations with three elements (or N-1 elements).
We already have permutations of 3 elements: `012`, `021`, `102`, `120`, `201`, and `210`. If we add element 3, what do we get? Well, for starters, we should get `3012`, `3021`, `3102`, `3120`, `3201`, `3210`, and then more of them. So we have created new permutations by adding a new element to existing permutations, which is key to the solution!
First, we have a bag of elements. Then, we take the first element out of the bag and create all the permutations of the remaining elements in the bag. We paste the first element to every generated permutation. Next, we return the first element to the bag and take the next element out. We repeat this process until we have used all the elements in the bag.
To illustrate this idea, let’s try generating permutations for four elements using the algorithm.
### The Recursive Algorithm
Let’s say we have a bag of four elements: `0`, `1`, `2`, and `3`. We take the first element out of the bag, which is `0`, and create permutations of the remaining elements (`1`, `2`, and `3`). We generate six permutations: `123`, `132`, `213`, `231`, `312`, and `321`. To obtain the final permutations, we add `0` to the beginning of each generated permutation: `0123`, `0132`, `0213`, `0231`, `0312`, and `0321`. We then return `0` to the bag and take the next element out, which is `1`. The bag now contains elements `0`, `2`, and `3`; their permutations are `023`, `032`, `203`, `230`, `302`, and `320`. Now add `1` to the beginning of each permutation to obtain: `1023`, `1032`, `1203`, `1230`, `1302`, and `1320`. Repeat the same process for elements `2` and `3`.
Ultimately, we have used all four elements in the bag and generated all possible permutations. The total number of permutations generated is 24 (4 * 6), which is the correct result.
The code written in C# follows the algorithm:
```public static void RecursiveAlgorithm(string elementsOutOfTheBag, List<int> elementsInTheBag)
{
if (elementsInTheBag.Count == 0)
Console.WriteLine(\$"Permutation: {elementsOutOfTheBag}");
else
{
foreach (var element in elementsInTheBag)
{
List<int> newBag = elementsInTheBag.Where(e => e != element).ToList();
RecursiveAlgorithm(elementsOutOfTheBag + element, newBag);
}
}
}```
The method accepts the string of already positioned elements and a bag of not-yet-used elements. If the bag is empty, we have reached the end and have the resulting permutation. Otherwise, we recursively execute the same algorithm with the next element.Â
### Where to Begin
To initiate this recursive method, we require a starting position. In this case, it is an empty string and all the elements:
```public static void PrintPermutationsRecursively()
{
RecursiveAlgorithm("", new List<int> { 0, 1, 2, 3, 4 });
}```
The result will be 120 permutations for five elements.
However, we should note that this method does not have a limit on the number of elements. This simple and easy-to-understand method produces a working result for any quantity desired. We can generate permutations for eight elements by simply calling:
`RecursiveAlgorithm("", new List<int> { 0, 1, 2, 3, 4, 6, 7, 8 });`
Executing this code generates 40,320 permutations, keeping our algorithm busy for a while.
### Breaking the Recursion
Although recursion is an essential technique, there is always a potential problem with it. If a long recursion isn’t allowed to exit, it can become a severe problem. Therefore, as programmers, we must be sure our algorithm provides a way out.
In our case, this is relatively easy to check because, with every recursion, we take one more element out of the bag. Eventually, we will remove all the elements, and the recursion will end.
It’s always a good idea to write tests to verify whether an algorithm provides the correct result, and you can find our tests in the GitHub project.
### Recursive Performance
To determine our algorithm’s performance, we will have to measure it. We will use a standard tool: BenchmarkDotNet.
To benchmark different algorithms fairly, we have to strip everything nonessential from them. In our case, we will rewrite the algorithms so that they will not write results to the console, and they will permutate elements in the same memory location.
The second requirement regarding memory location may not be immediately apparent, but it is essential. Since there are many permutations, it makes sense to generate all permutations in the same memory location and not utilize memory unnecessarily for each new permutation.
To this end, we will also introduce a new parameter: benchmark. If this parameter is true, we will elect to “forget” each permutation. We will generate it and move to the next one. This way, measuring speed and memory consumption will be fairer.
### Rewriting the Recursive Algorithm
By stripping the console output from the algorithm and collecting the results to list, we will get our next iteration of the recursive algorithm:
```private void RecursiveGenerator(List<byte[]> result, byte[] input, byte fixedPosition,
List<byte> freeNumbers, bool banchmarks)
{
if (fixedPosition == input.Length)
{
}
else
{
for(int i = 0; i < freeNumbers.Count; ++i)
{
var newFreeNumbers = new List<byte>(freeNumbers);
newFreeNumbers.RemoveAt(i);
input[fixedPosition] = freeNumbers[i];
RecursiveGenerator(result, input, (byte)(fixedPosition + 1), newFreeNumbers);
}
}
}```
The algorithm is exactly the same as before; only the parameters are slightly different.
The first parameter is a list that will return the resulting permutations. The second parameter is a memory location where the algorithm will generate permutations. So every permutation will occupy the same memory (the same array), and before we add it to the resulting list, we have to make a copy! The third parameter counts how many elements in the array we have already constructed, and the algorithm will stop when we generate the whole permutation. The fourth parameter is a bag of unused elements, just as before. The last parameter is a flag that tells the algorithm if we are measuring time/space, so in that case, it will not generate the resulting output.
### Rewriting the Basic and Improved Basic Algorithms
In addition, we also need to rewrite the Basic and Improved Algorithms. Specifically, we will focus on the methods used to generate permutations of 3 elements. Given that the code for generating five elements using the Basic or Improved algorithm is quite lengthy, here we will only modify the code for generating permutations of 3 elements:
```private List<byte[]> GetPermutations3(byte[] input, bool doBanchmarks)
{
var result = new List<byte[]>();
var number = input.Length;
for (byte i0 = 0; i0 < number; ++i0)
for (byte i1 = 0; i1 < number; ++i1)
for (byte i2 = 0; i2 < number; ++i2)
if ((i0 != i1) && (i0 != i2) &&
(i1 != i2))
{
input[0] = i0;
input[1] = i1;
input[2] = i2;
}
return result;
}```
To do so, we create an array to store the results and include a flag to indicate if we are measuring the code’s performance. The algorithm remains largely the same, with the only changes made to improve efficiency using the same memory location.
We can apply these same changes to the Improved algorithm as well:
```private List<byte[]> GetPermutations3(byte[] input, bool doBanchmarks)
{
var result = new List<byte[]>();
var number = input.Length;
for (byte i0 = 0; i0 < number; ++i0)
{
input[0] = i0;
for (byte i1 = 0; i1 < number; ++i1)
{
if (i1 == i0) continue;
input[1] = i1;
for (byte i2 = 0; i2 < number; ++i2)
{
if ((i2 == i0) || (i2 == i1)) continue;
input[2] = i2;
}
}
}
return result;
}```
## Comparing Algorithms
By making these changes, we can finally measure the algorithms’ performance. Specifically, we will test their ability to generate permutations of 2, 4, 6, and 8 elements. The results of our testing on a typical modern desktop computer are:
```| Method | Items | Mean | Allocated |
|------------------- |------ |-----------------|------------------|
| BasicAlgorithm | 4 | 352.7 ns | 176 B |
| ImprovedAlgorithm | 4 | 190.5 ns | 176 B |
| RecursiveAlgorithm | 4 | 1,518.5 ns | 4336 B |
| BasicAlgorithm | 6 | 92,406.8 ns | 176 B |
| ImprovedAlgorithm | 6 | 8,359.9 ns | 176 B |
| RecursiveAlgorithm | 6 | 45,411.8 ns | 125424 B |
| BasicAlgorithm | 8 | 23,743,306.9 ns | 192 B |
| ImprovedAlgorithm | 8 | 2,218,982.5 ns | 178 B |
| RecursiveAlgorithm | 8 | 2,593,615.1 ns | 7014642 B |```
From this table, it is evident that the Improved algorithm is the fastest. Additionally, we can see that as the number of elements to be permuted increases, the Basic algorithm’s efficiency decreases while the recursive algorithm’s efficiency improves. However, we must also consider that the recursive algorithm significantly impacts memory.
Therefore, the recursive algorithm appears not as optimal as we initially thought. While it is currently the only algorithm capable of generating permutations for any number of elements, it is not ideal. On the other hand, if we want an Improved algorithm for 12 elements, we would need to create a method with 12 for loops, which is impractical.
### Raising the Bar
Can we do better? Certainly!
While the three previous algorithms were relatively easy to write, they were not the best possible solutions. More advanced and efficient algorithms have been developed, such as Heap’s Algorithm for generating permutations.
## Generate Permutations With the Heap’s Algorithm
This algorithm, named after B. R. Heap, who published it in 1963, is also recursive. It involves swapping elements in a set and recursively generating permutations of the smaller set until all permutations have been generated. Fortunately, we have all the tools we need to implement Heap’s Algorithm in C#:
```private void Heaps(List<byte[]> result, byte[] input, byte size, bool banchmarks)
{
if (size == 1)
{
}
else
{
for (int i = 0; i < size; i++)
{
Heaps(result, input, (byte)(size - 1));
if (size % 2 == 1)
(input[size - 1], input[0]) = (input[0], input[size - 1]);
else
(input[size - 1], input[i]) = (input[i], input[size - 1]);
}
}
}```
Despite being short and having the same parameters as our previous recursive algorithm, Heap’s Algorithm is significantly faster and more memory-efficient, as it avoids the need to generate a bag during every recursion. To demonstrate this, we can measure the algorithm and obtain the results:
```| Method | Items | Mean | Allocated |
|------------------- |------ |-----------------|-----------|
| ImprovedAlgorithm | 4 | 190.5 ns | 176 B |
| HeapsAlgorithm | 4 | 149.3 ns | 176 B |
| ImprovedAlgorithm | 6 | 8,359.9 ns | 176 B |
| HeapsAlgorithm | 6 | 3,377.0 ns | 176 B |
| ImprovedAlgorithm | 8 | 2,218,982.5 ns | 178 B |
| HeapsAlgorithm | 8 | 195,455.8 ns | 176 B |```
The theory behind the Heap’s algorithm is described in great detail on Wikipedia.
## Conclusion
The fact that we are unlikely to find better algorithms than some of the world’s great programming minds have already discovered does not mean we should give up. We should consider the various possibilities, write different algorithms and ideas, and not always settle for “the best algorithm” available online. In truth, there is rarely a “best one” for every case, as it depends on the specific problem we have to solve and the type of data it entails.
In this case, we have demonstrated how to approach the problem of generating permutations, and once we have our starting point, we can continue to seek out and craft ever-better algorithms. |
This lesson presents the idea that the area of any triangle is exactly half of a certain parallelogram -- thus we get the familiar formula of multiplying the base and the altitude, and taking half of that. The lesson contains varied exercises for students.
# Area of Triangles
We can always put any triangle together witha copy of itself to make a parallelogram. Therefore, the AREA of any triangle must beexactly half of the area of that parallelogram.
1. Find the area of the shaded triangle in the picture above.
2. Draw the corresponding parallelograms for these triangles, and find their areas.
Hint: draw a line that is congruent to the base of the triangle, starting at the top vertex.
a. _______ square units b. _______ square units c. _______ square units
Again, we use a base and an altitude.
The base can be any side of the triangle,
though people often use the “bottom” side.
The altitude is perpendicular to the base,
and it goes from the opposite vertex to the
base (or to the continuation of the base).
Since the area of a triangle is half of the area of
the corresponding parallelogram, we can calculate
the area as half of the base times the altitude, or:
A = BASE × ALTITUDE 2
You can choose any side to be
the base. Here, it makes sense to
choose the vertical side as the base.
The area is 4 × 6 2 = 12 square units.
3. Draw an altitude in each triangle, and mark the base. Find the area of each triangle.
a. _______ square units b. _______ square units c. _______ square units
Example 1. The altitude of a triangle may fall outside of the triangle itself. It is still perpendicular to the base, and starts at a vertex.
The corresponding parallelogram is seen if you follow the dotted lines.
The area is 4 × 3 2 = 6 square units.
Example 2. Here it is easiest to think of the base being on the “top.”
Again, the altitude falls outside the actual triangle.
The area is 5 × 3 2 = 7 1/2 square units.
5. This figure is called a _________________________. Calculate its area.
6. Draw as many different-shaped triangles as you can that have an area of 12 square units.
How to find the area of a triangle not drawn on grid First, choose one of the sides as the base. It can be any of the sides! Draw the altitude. Use a protractor or a triangular ruler to draw the altitude so that it goes through one vertex, and is perpendicular to the base. See the illustration. Place the protractor so that the line you will draw will pass through the vertex and so that the 90°-mark is lined up with the base of the triangle. Measure the altitude and base as precisely as you can with a ruler. Calculate the area.
8. Draw your own triangle, and find its area!
9. Draw a triangle with an area of 3 square inches.
Is it only possible to draw just one triangle with
that area, or is it possible draw several, with
varying shapes/sizes?
10. Draw a triangle, without measuring anything, so that its area is close to 20 cm2. Check by drawing
the altitude and measuring! Practice until you get a triangle with an area of approximately 20 cm2.
You can even make a game out of this: whoever gets the area the closest to the given area is the
winner or gets the most points. Or, take turns with your friend, asking the friend to draw a triangle with
a specific area. Perhaps add a condition that the triangle has to be obtuse/acute/right, or
equilateral/isosceles/scalene.
This lesson is taken from Maria Miller's book Math Mammoth Geometry 2, and posted at www.HomeschoolMath.net with permission from the author. Copyright © Maria Miller.
#### Math Mammoth Geometry 2
A self-teaching worktext for 6th and 7th grades that covers the area of triangles, parallelograms, and polygons, pi, area of circle, nets, surface area, and volume of common solids. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# SSS Triangle Congruence
## Three sets of equal side lengths determine congruence.
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SSS Triangle Congruence
What if your parents were remodeling their kitchen so that measurements between the sink, refrigerator, and oven were as close to an equilateral triangle as possible? The measurements are in the picture at the left, below. Your neighbor’s kitchen has the measurements on the right, below. Are the two triangles congruent? After completing this Concept, you'll be able to determine whether or not two triangles are congruent given only their side lengths.
### Watch This
Watch the portions of the following two videos that deal with SSS triangle congruence.
### Guidance
Consider the question: If I have three lengths, 3 in, 4 in, and 5 in, can I construct more than one triangle with these measurements? In other words, can I construct two different triangles with these same three lengths?
##### Investigation: Constructing a Triangle Given Three Sides
Tools Needed: compass, pencil, ruler, and paper
1. Draw the longest side (5 in) horizontally, halfway down the page. The drawings in this investigation are to scale.
2. Take the compass and, using the ruler, widen the compass to measure 4 in, the next side.
3. Using the measurement from Step 2, place the pointer of the compass on the left endpoint of the side drawn in Step 1. Draw an arc mark above the line segment.
4. Repeat Step 2 with the last measurement, 3 in. Then, place the pointer of the compass on the right endpoint of the side drawn in Step 1. Draw an arc mark above the line segment. Make sure it intersects the arc mark drawn in Step 3.
5. Draw lines from each endpoint to the arc intersections. These lines will be the other two sides of the triangle.
Can you draw another triangle, with these measurements that looks different? The answer is NO. Only one triangle can be created from any given three lengths.
An animation of this investigation can be found at: http://www.mathsisfun.com/geometry/construct-ruler-compass-1.html
Side-Side-Side (SSS) Triangle Congruence Postulate: If three sides in one triangle are congruent to three sides in another triangle, then the triangles are congruent.
Now, we only need to show that all three sides in a triangle are congruent to the three sides in another triangle. This is a postulate so we accept it as true without proof. Think of the SSS Postulate as a shortcut. You no longer have to show 3 sets of angles are congruent and 3 sets of sides are congruent in order to say that the two triangles are congruent.
In the coordinate plane, the easiest way to show two triangles are congruent is to find the lengths of the 3 sides in each triangle. Finding the measure of an angle in the coordinate plane can be a little tricky, so we will avoid it in this text. Therefore, you will only need to apply SSS in the coordinate plane. To find the lengths of the sides, you will need to use the distance formula, \begin{align*}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}.
#### Example A
Write a triangle congruence statement based on the diagram below:
From the tic marks, we know \begin{align*}\overline{AB} \cong \overline{LM}, \overline{AC} \cong \overline{LK}, \overline{BC} \cong \overline{MK}\end{align*}. Using the SSS Postulate we know the two triangles are congruent. Lining up the corresponding sides, we have \begin{align*}\triangle ABC \cong \triangle LMK\end{align*}.
Don’t forget ORDER MATTERS when writing triangle congruence statements. Here, we lined up the sides with one tic mark, then the sides with two tic marks, and finally the sides with three tic marks.
#### Example B
Write a two-column proof to show that the two triangles are congruent.
Given: \begin{align*}\overline{AB} \cong \overline{DE}\end{align*}
\begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DB}\end{align*}.
Prove: \begin{align*}\triangle ACB \cong \triangle ECD\end{align*}
Statement Reason
1. \begin{align*}\overline{AB} \cong \overline{DE}\end{align*}
\begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DB}\end{align*}
Given
2. \begin{align*}\overline{AC} \cong \overline{CE}, \overline{BC} \cong \overline{CD}\end{align*} Definition of a midpoint
3. \begin{align*}\triangle ACB \cong \triangle ECD\end{align*} SSS Postulate
Make sure that you clearly state the three sets of congruent sides BEFORE stating that the triangles are congruent.
Prove Move: Feel free to mark the picture with the information you are given as well as information that you can infer (vertical angles, information from parallel lines, midpoints, angle bisectors, right angles).
#### Example C
Find the distances of all the line segments from both triangles to see if the two triangles are congruent.
Begin with \begin{align*}\triangle ABC\end{align*} and its sides.
Now, find the distances of all the sides in \begin{align*}\triangle DEF\end{align*}.
We see that \begin{align*}AB = DE, BC = EF\end{align*}, and \begin{align*}AC = DF\end{align*}. Recall that if two lengths are equal, then they are also congruent. Therefore, \begin{align*}\overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}\end{align*}, and \begin{align*}\overline{AC} \cong \overline{DF}\end{align*}. Because the corresponding sides are congruent, we can say that \begin{align*}\triangle ABC \cong \triangle DEF\end{align*} by SSS.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
From what we have learned in this section, the two triangles are not congruent because the distance from the fridge to the stove in your house is 4 feet and in your neighbor’s it is 4.5 ft. The SSS Postulate tells us that all three sides have to be congruent.
### Vocabulary
Two figures are congruent if they have exactly the same size and shape. By definition, two triangles are congruent if the three corresponding angles and sides are congruent. The symbol \begin{align*}\cong\end{align*} means congruent. There are shortcuts for proving that triangles are congruent. The SSS Triangle Congruence Postulate states that if three sides in one triangle are congruent to three sides in another triangle, then the triangles are congruent.
### Guided Practice
1. Determine if the two triangles are congruent.
2. Fill in the blanks in the proof below.
Given: \begin{align*}\overline{AB} \cong \overline{DC}, \ \overline{AC} \cong \overline{DB}\end{align*}
Prove: \begin{align*}\triangle ABC \cong \triangle DCB\end{align*}
Statement Reason
1. 1.
2. 2. Reflexive PoC
3. \begin{align*}\triangle ABC \cong \triangle DCB\end{align*} 3.
3. Is the pair of triangles congruent? If so, write the congruence statement and why.
1. Start with \begin{align*}\triangle ABC\end{align*}.
Now find the sides of \begin{align*}\triangle DEF\end{align*}.
No sides have equal measures, so the triangles are not congruent.
2.
Statement Reason
1. \begin{align*}\overline{AB} \cong \overline{DC}, \ \overline{AC} \cong \overline{DB}\end{align*} 1. Given
2. \begin{align*}\overline{BC} \cong \overline{CB}\end{align*} 2. Reflexive PoC
3. \begin{align*}\triangle ABC \cong \triangle DCB\end{align*} 3. SSS Postulate
3. The triangles are congruent because they have three pairs of sides congruent. \begin{align*} \triangle DEF \cong \triangle IGH\end{align*}.
### Practice
Are the pairs of triangles congruent? If so, write the congruence statement and why.
State the additional piece of information needed to show that each pair of triangles is congruent.
1. Use SSS
2. Use SSS
Fill in the blanks in the proofs below.
1. Given: \begin{align*}B\end{align*} is the midpoint of \begin{align*}\overline{DC}\end{align*} \begin{align*}\overline{AD} \cong \overline{AC}\end{align*} Prove: \begin{align*}\triangle ABD \cong \triangle ABC\end{align*}
Statement Reason
1. 1.
2. 2. Definition of a Midpoint
3. 3. Reflexive PoC
4. \begin{align*}\triangle ABD \cong \triangle ABC\end{align*} 4.
Find the lengths of the sides of each triangle to see if the two triangles are congruent.
1. \begin{align*}\triangle ABC: \ A(-1, 5), \ B(-4, 2), \ C(2, -2)\end{align*} and \begin{align*}\triangle DEF: \ D(7, -5), \ E(4, 2), \ F(8, -9)\end{align*}
2. \begin{align*}\triangle ABC: \ A(-8, -3), \ B(-2, -4), \ C(-5, -9)\end{align*} and \begin{align*}\triangle DEF: \ D(-7, 2), \ E(-1, 3), \ F(-4, 8)\end{align*}
3. \begin{align*}\triangle ABC: \ A(0, 5), \ B(3, 2), \ C(1, 4)\end{align*} and \begin{align*}\triangle DEF: \ D(1, 2), \ E(4, 4), \ F(7, 1)\end{align*}
4. \begin{align*}\triangle ABC: \ A(1, 7), \ B(2, 2), \ C(4, 6)\end{align*} and \begin{align*}\triangle DEF: \ D(4, 10), \ E(5, 5), \ F(7, 9)\end{align*}
5. Draw an example to show why SS is not enough to prove that two triangles are congruent.
6. If you know that two triangles are similar, how many pairs of corresponding sides do you need to know are congruent in order to know that the triangles are congruent?
### Vocabulary Language: English
Congruent
Congruent
Congruent figures are identical in size, shape and measure.
Distance Formula
Distance Formula
The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ can be defined as $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
H-L (Hypotenuse-Leg) Congruence Theorem
H-L (Hypotenuse-Leg) Congruence Theorem
If the hypotenuse and leg in one right triangle are congruent to the hypotenuse and leg in another right triangle, then the two triangles are congruent.
Side Side Side Triangle
Side Side Side Triangle
A side side side triangle is a triangle where the lengths of all three sides are known quantities.
SSS
SSS
SSS means side, side, side and refers to the fact that all three sides of a triangle are known in a problem.
Triangle Congruence
Triangle Congruence
Triangle congruence occurs if 3 sides in one triangle are congruent to 3 sides in another triangle.
Rigid Transformation
Rigid Transformation
A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure. |
Albert Teen
YOU ARE LEARNING:
Rules of Angles
# Rules of Angles
### Rules of Angles
Acute, obtuse, reflex and right-angles include some important rules relative to each other, which we can use to find unknown angles.
Angles can be either acute, right-angled, obtuse or reflex. This depends entirely on the size of the angle.
Rules of angles
1
An acute angle is any angle that is less than $90\degree$
Here is an example
2
A right angle is exactly $90\degree$
We indicate a right-angle with a little square at the angle.
3
What type of angle is this?
4
An obtuse angle is between $90\degree$ and $180\degree$
Both these angles are obtuse
5
A reflex angle is greater than $180\degree$
Both these angles are reflex angles
6
What type of angle is this?
7
This is a reflex angle
We can see this because the angle is greater than $180\degree$
8
What type of angle is this?
9
This is an obtuse angle
This is because the angle is between $90\degree$ and $180\degree$
There are 3 really important rules to remember about angles.
Rules of angles: lines, circles and triangles
1
The angles on a straight line add up to $180\degree$
No matter what the two angles are, they always add up to $180\degree$
2
What is the value of angle B? Write the number of degrees.
3
Angles around a point add up to $360\degree$
No matter how many angles there are, they all add up to $360\degree$
4
What is the size of Angle A? Write the number of degrees
5
The interior angles of a triangle add up to make $180\degree$
Have a look in this triangle to see for yourself
6
What is the value of angle A? Write the number of degrees.
The angles around a point add up to ____
Final question! Which of these definitions describes the sum of the angles in a triangle? |
# Dimensional Analysis
## Convert between units using unit ratios
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Dimensional Analysis
What if you went to the grocery store and bought 3 gallons of milk? Could you determine how many pints of milk you purchased? Or how about if you bought 16 pints of milk? How many gallons would this be? In this Concept, you'll learn to make conversions like these so that you can solve real-world problems.
### Guidance
Real-world information is given in dimensions, or the units in which the value is measured. For example, the following are all examples of dimensions.
• Inches
• Feet
• Liters
• Micrograms
• Acres
• Hours
• Pounds
• Students
Analyzing dimensions can help you solve problems in travel, astronomy, physics, engineering, forensics, and quality. Solving problems by converting dimensions or canceling dimensions is the focus of this Concept.
Consider the distance formula \begin{align*}distance=rate \ \cdot \ time\end{align*}. This formula can be rewritten for rate. \begin{align*}rate=\frac{distance}{time}\end{align*}. If distance is measured in kilometers, and time is measured in hours, the rate would have the dimensions \begin{align*}\frac{kilometers}{hours}\end{align*}.
You can treat dimensions as variables. Identical units can divide out, or cancel. For example, \begin{align*}\frac{kilometers}{hour} \cdot hour \rightarrow \frac{kilometers}{\cancel{hour}} \cdot \cancel{hour} \rightarrow kilometers\end{align*}.
Sometimes the units will not divide out. In this case, a conversion factor is needed.
#### Example A
Convert \begin{align*}\frac{35 \ kilometers}{hour}\end{align*} to meters.
Solution:
Since kilometers \begin{align*}\neq\end{align*} meters, you need to convert kilometers to meters to get the answer. You know there are 1,000 meters in a kilometer. Therefore, you will need to multiply the original dimension by this factor.
\begin{align*}\frac{35 \ kilometers}{hour} \cdot \frac{1000 \ meters}{1 \ kilometer} \rightarrow \frac{35 \ \cancel{kilometers}}{{hour}} \cdot \frac{1000 \ meters}{1 \ \cancel{kilometer}} & = \frac{35(1000)meters}{hour}.\\ \frac{35\ kilometers}{hour}& =\frac{35,000 \ meters}{hour}\end{align*}
The process of using units or dimensions to help solve a problem is called dimensional analysis. It is very useful in chemistry and travel, as shown in the examples below.
#### Example B
How many seconds are in a month?
Solution: This situation can be solved easily using multiplication. However, the process you use when multiplying the values together is an example of dimensional analysis.
Begin with what you know:
• 60 seconds in one minute
• 60 minutes in one hour
• 24 hours in one day
• Approximately 30 days in one month
Now write the expression to convert the seconds in one minute to one month.
\begin{align*}\frac{60 \ seconds}{1 \ minute} \cdot \frac{60 \ minutes}{1 \ hour} \cdot \frac{24 \ hours}{1 \ day} \cdot \frac{30 \ days}{1 \ month}\end{align*}
Identical units cross-cancel.
\begin{align*}\frac{60 \ seconds}{1 \ \cancel{minute}} \cdot \frac{60 \ \cancel{minutes}}{1 \ \cancel{hour}} \cdot \frac{24 \ \cancel{hours}}{1 \ \cancel{day}} \cdot \frac{30 \ \cancel{days}}{1 \ month}\end{align*}
Multiply the fractions together.
\begin{align*}\frac{60 \cdot 60 \cdot 24 \cdot 30 \ seconds}{1 \cdot 1 \cdot 1 \cdot 1 \ month}=2,592,000\frac{seconds}{month}\end{align*}
#### Example C
How many grams are in 5 pounds?
Solution: Begin by writing all the conversions you know related to this situation.
\begin{align*}1 \ gram & \approx 0.0353 \ ounces\\ 16 \ ounces & = 1 \ pound\end{align*}
\begin{align*}5 \ pounds \cdot \frac{16 \ ounces}{1 \ pound} \cdot \frac{1 \ gram}{0.0353 \ ounce}\end{align*}
Cross-cancel identical units and multiply.
\begin{align*}5 \ \cancel{pounds} \cdot \frac{16 \ \cancel{ounces}}{1 \ \cancel{pound}} \cdot \frac{1 \ gram}{0.0353 \ \cancel{ounce}}=2226.29 \ grams\end{align*}
A long list of conversion factors can be found at this website.
### Guided Practice
You are traveling in Europe and want to know how fast to drive to maximize fuel efficiency. The optimal driving speed for fuel efficiency is 55 miles per hour. How fast would that be in kilometers per hour?
Solution:
Since 1 mile is approximately 1.6 kilometers:
\begin{align*} \frac{55\ miles}{hour}\cdot \frac{1.6\ kilometers}{1\ mile} \rightarrow \frac{55\ \cancel{miles}}{hour}\cdot \frac{1.6\ kilometers}{1\ \cancel{mile}}\rightarrow \frac{88\ kilometers}{hour}\end{align*}
The optimal speed for fuel efficiency is 88 kilometers per hour.
### Practice
1. True or false? Dimensional analysis is the study of space and time.
2. By using dimensional analysis, what happens to identical units that appear diagonally in the multiplication of fractions?
3. How many feet are in a mile?
4. How many inches are in a mile?
5. How many seconds are in a day?
6. How many seconds are in a year?
7. How many feet are in a furlong?
8. How many inches are in 100 yards (one football field)?
9. How many centimeters are in 5 inches?
10. How many meters are between first and second base (90 feet)?
11. How many meters are in 16 yards?
12. How many cups are in 6 liters?
13. How many cubic inches make up one ounce?
14. How many milliliters make up 8 ounces?
15. How many grams are in 100 pounds?
16. An allergy pill contains 25 mg of Diphenhydramine. If \begin{align*}1 \ gram=15.432 \ grains\end{align*}, how many grains of this medication are in the allergy pill?
17. A healthy individual’s heart beats about 68 times per minute. How many beats per hour is this?
18. You live 6.2 miles from the grocery store. How many fathoms is this? (\begin{align*}6 \ feet=1 \ fathom\end{align*})
19. The cost of gas in England is 96.4 pound sterling/liter. How much is this in U.S. dollars/gallon? (\begin{align*}3.875\ litres=1 \ gallon\end{align*} and \begin{align*}1.47\ US\=1\ pound\ sterling\end{align*})
20. Light travels \begin{align*}\frac{186,000 \ miles}{second}\end{align*}. How long is one light year?
21. Another way to describe light years is in astronomical units. If 1 \begin{align*}light \ year=63,240 \ AU\end{align*} (astronomical units), how far in AUs is Alpha Centauri, which is 4.32 light years from the Earth?
22. How many square feet is 16 acres?
23. A person weighs 264 pounds. How many kilograms is this weight?
24. A car is traveling 65 miles/hour and crosses into Canada. What is this speed in km/hr?
25. A large soda cup holds 32 ounces. What is this capacity in cubic inches?
26. A space shuttle travels 28,000 mph. What is this distance in feet/second?
27. How many hours are in a fortnight (two weeks)?
28. How many fortnights (two-week periods) are in 2 years?
29. A semi truck weighs 32,000 pounds empty. How many tons is this weight?
30. Which has the greatest volume: a 2-liter bottle of soda, one gallon of water, or 10 pints of human blood?
Mixed Review
1. Solve for \begin{align*}x: -2x+8=8(1-4x)\end{align*}.
2. Simplify: \begin{align*}3-2(5-8h)+13h \cdot 3\end{align*}.
3. Find the difference: \begin{align*}-26.375-\left (-14 \frac{1}{8}\right )\end{align*}.
4. Find the product: \begin{align*}-2\frac{3}{7} \cdot \frac{9}{10}\end{align*}.
5. Simplify: \begin{align*}\sqrt{80}\end{align*}.
6. Is \begin{align*}5.\bar{5}\end{align*} an irrational number? Explain your answer.
Use the relation given for the following questions: \begin{align*}\left \{(0,8),(1,4),(2,2),(3,1),\left(4,\frac{1}{2}\right ),\left (5,\frac{1}{4}\right )\right \}\end{align*}.
1. State the domain.
2. State the range.
4. What seems to be the pattern in this relation?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
`5.5. Solving linear systems by the elimination method Equivalent systems The major technique of solving systems of equations is changing the original problem into another one which is of an easier to solve form. In this procedure we need to be sure that the new system has the same solution set. No solution is added and no solution is lost. It leads to the concept of equivalent systems. 5.5.1. DEFINITION. We say that two systems are equivalent if and only if they have equal solution sets. 5.5.2. EXAMPLE. Let us consider the system 7x + 2y + 2z = 21 - 2y + 3z = 1 4z = 12 By the back-substitution we can find that the solution set of the above system is {(1, 4, 3)}. Now let us look at another system which is not exactly the same as the original one. 21x + 6y + 6z = 63 - 4y + 6z = 12 2z = 6 Applying the back-substitution again we find that the solution set of this system is also {(1, 4, 3)}. It means that the second system is equivalent to the original system. Now let us consider the third system 7x + 2y + 2z = 21 - 2y + 3z = -13 4z = 12 Back-substitution tells us that the solution set is {(-1, 11, 3)}. So the solution set of the third system is not the same as the solution set of the first system. It means that they are not equivalent. 1 Replacing systems by equivalent systems How do we know that simplifying a system of equations we obtain the one which is equivalent to the original system? It depends what we do to simplify the system. There are certain operations which for sure lead to equivalent system. Even more, they are general and can be used to solve any system of linear equations. First Operation: interchanging two equations.Let us start with the simplest one: interchanging two equations. Changing the order of equations in the system does not effect the solutions set which is still the intersection of the same solution sets of individual equations. 5.5.3. EXAMPLE. Obviously, the following two systems have identical solution sets, so they are equivalent. 7x + 2y + 2z = 21 - 2y + 3z = 1 4z = 12 7x + 2y + 2z = 21 4z = 12 - 2y + 3z = 1 Second Operation: multiplying an equation by a number. The second operation which leads to the equivalent system is multiplying one equation in the original system by a number which not 0. We know that the solution set of the equation multiplied by a number does not change. So the solution set of the whole system does not change being the intersection of the same sets. 5.5.4. EXAMPLE. In the Example 5.5.2 we have checked that the two following equations have equal solution sets. 7x + 2y + 2z = 21 - 2y + 3z = 1 4z = 12 21x + 6y + 6z = 63 - 2y + 3z = 1 4z = 122 Third Operation: adding a multiple of one equation to another equation. The third operation which leads to the equivalent systems is the most useful because it changes a system significantly. It is adding a multiple of one equation in the system to another equation. Let us describe it closer. It is only between two equations in the system. The first equation is not changed but only used to change the second one. The second equation is changed in the following way: we replace it by the sum of the second and the first multiplied by a number. Let us illustrate it with an example. 5.5.5. EXAMPLE. Let us consider the system x + 2y = 3 3x + 4y = 2 We multiply the first equation by (-3) and add to the second equation. It gives us the new second equation. + 2y = 3 x 3x + (-3)x + 4y + (-3)2y = 2 + (-3)3 We obtain the system x + 2y = 3 . -2y = -7The elimination method For two equivalent systems the one which has an equation with less number of variables is simpler to solve. Thus eliminating variables is one of the most powerful method of solving systems of equations. Each particular system equation that we are going to present could be solved in many different ways. Sometimes much simpler than what we present. However, our concern is to present the method which is not difficult and, at the same time, works in general for all linear systems. The simple form that we want to achieve is the form to which is possible to apply the back-substitution method. Let us start with the example.3 5.5.6. EXAMPLE. Let us consider the following system 3x + 4y = 18 2x + 3y = 13 We want to eliminate the unknown x from the second equation. The x-coefficient in the first equation is 3 so we multiply the second equation by 3. 3x + 4y = 18 6x + 9y = 39 Then we multiply the first equation multiply by (-2) and add to the second equation to obtain the new second equation. 3x + 4y = 18 . 6x + (-2)3x + 9y + (-2)4y = 39 + (-2)18 It gives 3x + 4y = 18 . y = 3 At this moment we are done with the elimination. To finish the problem up we use the back-substitution method which gives the solution set {(2, 3)}. For the system in three variables the method of elimination is just longer but the steps are mainly the same. 5.5.7. EXAMPLE. let us consider the following 3 × 3 system 3x + y + 2z = 13 2x + 3y + 4z = 19 . x + 4y + 3z = 15 We see that in the last equation the x coefficient is 1. It is very convenient because it would be an equation to use to eliminate x from the other equations. Let us move it to the front by changing the order of equations. x + 4y + 3z = 15 3x + y + 2z = 13 . 2x + 3y + 4z = 19 4 Now the elimination starts. We add the first equation multiplied by (-3) to the second equation. + 4y + 3z = 15 - 11y - 7z = -32 . 2x + 3y + 4z = 19 We add the first equation multiplied by (-2) to the third equation. x + 4y + 3z = 15 - 11y - 7z = -32 . - 5y - 2z = -11 The system still does not fit to back-substitution. We need to eliminate the variable y from the last equation. In order to do it we need first to multiply the third equation by (-11). x + 4y + 3z = 15 - 11y - 7z = -32 . 55y + 22z = 121 Then we add the second equation multiplied by 5 to third equation. x + 4y + 3z = 15 - 11y - 7z = -32 . - 13z = -39 Now we can use the back-substitution method. It gives us the solution set {(2, 1, 3)}. Gaussian elimination for matrices When we perform the elimination there is a lot of writing. It is especially inconvenient to carry on the symbols of variables. The algebraic operations are done on the numbers only. So we can skip the symbols of variables and do the elimination on the augmented matrix associated to the system. It is called Gaussian elimination for matrices. We need to remember that not seeing a variable in the equation means the same as seeing 0 in the corresponding location in the augmented matrix. We illustrate Gaussian elimination for two variables first. To show the connection between the two methods we will use the augmented matrix of the system from the Example 5.5.6. x5 5.5.8. EXAMPLE. 3 4 | 18 2 3 | 13 First we multiply the second row by the number 3 3 4 | 18 (3) · 2 (3) · 3 | (3) · 13 = 3 4 | 18 6 9 | 39 .The we multiply the first row by the number (-2) and the result to the second to get the new second row 3 4 | 18 6 + (-2) · 3 9 + (-2) · 4 | 39 + (-2) · 18 5.5.9. EXAMPLE.=3 4 | 18 0 1 | 3.3 1 2 | 13 2 3 4 | 19 1 4 3 | 15 First we interchange the rows1 4 3 | 15 3 1 2 | 13 . 2 3 4 | 19 Then we multiply the first row by (-3) and add to the second row1 4 3 | 15 1 4 3 | 15 3 + (-3) · 1 1 + (-3) · 4 2 + (-3) · 3 | 13 + (-3) · 15 = 0 -11 -7 | -32 . 2 3 4 | 19 2 3 4 | 19 Then we multiply the first row by (-2) and add to the third row1 4 3 | 15 1 4 3 | 15 0 -11 -7 | -32 = 0 -11 -7 | -32 . 2 + (-2) · 1 3 + (-2) · 4 4 + (-2) · 3 | 19 + (-2) · 15 0 -5 -2 | -116 Then we multiply the third row by (-11)1 4 3 | 15 1 4 3 | 15 -11 -7 | -32 0 = 0 -11 -7 | -32 . 0 (-11) · (-5) (-11) · (-2) | (-11) · (-11) 0 55 22 | 121 Finally, we multiply the second row by 5 and add to the third row1 4 3 | 15 1 4 3 | 15 -11 -7 | -32 0 = 0 -11 -7 | -32 . 0 55 + 5 · (-11) 22 + 5 · (-7) | 121 + 5 · (-32) 0 0 -13 | -397 `
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# 5.2 Power functions and polynomial functions (Page 4/19)
Page 4 / 19
## Identifying the degree and leading coefficient of a polynomial function
Identify the degree, leading term, and leading coefficient of the following polynomial functions.
$\begin{array}{ccc}\hfill f\left(x\right)& =& 3+2{x}^{2}-4{x}^{3}\hfill \\ \hfill g\left(t\right)& =& 5{t}^{2}-2{t}^{3}+7t\hfill \\ h\left(p\right)\hfill & =& 6p-{p}^{3}-2\hfill \end{array}$
For the function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ the highest power of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 3, so the degree is 3. The leading term is the term containing that degree, $\text{\hspace{0.17em}}-4{x}^{3}.\text{\hspace{0.17em}}$ The leading coefficient is the coefficient of that term, $\text{\hspace{0.17em}}-4.$
For the function $\text{\hspace{0.17em}}g\left(t\right),\text{\hspace{0.17em}}$ the highest power of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}$ so the degree is $\text{\hspace{0.17em}}5.\text{\hspace{0.17em}}$ The leading term is the term containing that degree, $\text{\hspace{0.17em}}5{t}^{5}.\text{\hspace{0.17em}}$ The leading coefficient is the coefficient of that term, $\text{\hspace{0.17em}}5.$
For the function $\text{\hspace{0.17em}}h\left(p\right),\text{\hspace{0.17em}}$ the highest power of $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}$ so the degree is $\text{\hspace{0.17em}}3.\text{\hspace{0.17em}}$ The leading term is the term containing that degree, $\text{\hspace{0.17em}}-{p}^{3}.\text{\hspace{0.17em}}$ The leading coefficient is the coefficient of that term, $\text{\hspace{0.17em}}-1.$
Identify the degree, leading term, and leading coefficient of the polynomial $\text{\hspace{0.17em}}f\left(x\right)=4{x}^{2}-{x}^{6}+2x-6.$
The degree is 6. The leading term is $\text{\hspace{0.17em}}-{x}^{6}.\text{\hspace{0.17em}}$ The leading coefficient is $\text{\hspace{0.17em}}-1.$
## Identifying end behavior of polynomial functions
Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the power function consisting of the leading term. See [link] .
Polynomial Function Leading Term Graph of Polynomial Function
$f\left(x\right)=5{x}^{4}+2{x}^{3}-x-4$ $5{x}^{4}$
$f\left(x\right)=-2{x}^{6}-{x}^{5}+3{x}^{4}+{x}^{3}$ $-2{x}^{6}$
$f\left(x\right)=3{x}^{5}-4{x}^{4}+2{x}^{2}+1$ $3{x}^{5}$
$f\left(x\right)=-6{x}^{3}+7{x}^{2}+3x+1$ $-6{x}^{3}$
## Identifying end behavior and degree of a polynomial function
Describe the end behavior and determine a possible degree of the polynomial function in [link] .
As the input values $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ get very large, the output values $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ increase without bound. As the input values $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ get very small, the output values $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ decrease without bound. We can describe the end behavior symbolically by writing
In words, we could say that as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values approach infinity, the function values approach infinity, and as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values approach negative infinity, the function values approach negative infinity.
We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive.
Describe the end behavior, and determine a possible degree of the polynomial function in [link] .
As It has the shape of an even degree power function with a negative coefficient.
## Identifying end behavior and degree of a polynomial function
Given the function $\text{\hspace{0.17em}}f\left(x\right)=-3{x}^{2}\left(x-1\right)\left(x+4\right),\text{\hspace{0.17em}}$ express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function.
Obtain the general form by expanding the given expression for $\text{\hspace{0.17em}}f\left(x\right).$
$\begin{array}{ccc}\hfill f\left(x\right)& =& -3{x}^{2}\left(x-1\right)\left(x+4\right)\hfill \\ & =& -3{x}^{2}\left({x}^{2}+3x-4\right)\hfill \\ & =& -3{x}^{4}-9{x}^{3}+12{x}^{2}\hfill \end{array}$
The general form is $\text{\hspace{0.17em}}f\left(x\right)=-3{x}^{4}-9{x}^{3}+12{x}^{2}.\text{\hspace{0.17em}}$ The leading term is $\text{\hspace{0.17em}}-3{x}^{4};\text{\hspace{0.17em}}$ therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 |
# Five Platonic Solids
## Theorem
There exist exactly five platonic solids:
$\paren 1: \quad$ the regular tetrahedron
$\paren 2: \quad$ the cube
$\paren 3: \quad$ the regular octahedron
$\paren 4: \quad$ the regular dodecahedron
$\paren 5: \quad$ the regular icosahedron.
In the words of Euclid:
I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another.
## Proof 1
A solid angle cannot be constructed from only two planes.
Therefore at least three faces need to come together to form a vertex.
Let $P$ be a platonic solid.
Let the polygon which forms each face of $P$ be a equilateral triangles.
We have that:
each vertex of a regular tetrahedron is composed of $3$ equilateral triangles
each vertex of a regular octahedron is composed of $4$ equilateral triangles
each vertex of a regular icosahedron is composed of $5$ equilateral triangles.
$6$ equilateral triangles, placed together at a vertex, form $4$ right angles.
a solid angle is contained by plane angles which total less than $4$ right angles.
Thus it is not possible to form $P$ such that its vertices are formed by $6$ equilateral triangles.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $6$ equilateral triangles.
Hence there are only $3$ possible platonic solids whose faces are equilateral triangles.
We have that each vertex of a cube is composed of $3$ squares.
$4$ squares, placed together at a vertex, form $4$ right angles.
it is not possible to form $P$ such that its vertices are formed by $4$ squares.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ squares.
Hence there is only $1$ possible platonic solid whose faces are squares.
We have that each vertex of a regular dodecahedron is composed of $3$ regular pentagons.
the vertices of a regular pentagon equal $1 \dfrac 1 5$ right angles.
$4$ regular pentagons, placed together at a vertex, form $4 \dfrac 4 5$ right angles.
it is not possible to form $P$ such that its vertices are formed by $4$ regular pentagons.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ regular pentagons.
Hence there is only $1$ possible platonic solid whose faces are regular pentagons.
$3$ regular hexagons, placed together at a vertex, form $4$ right angles.
it is not possible to form $P$ such that its vertices are formed by $3$ or more regular hexagons.
Regular polygons with more than $6$ sides have vertices which are greater than those of a regular hexagon.
Therefore $3$ such regular polygons, placed together at a vertex, form more than $4$ right angles.
it is not possible to form $P$ such that its vertices are formed by $3$ or more regular polygons with more than $6$ sides.
Hence the $5$ possible platonic solids have been enumerated and described.
$\blacksquare$
## Proof 2
Consider a convex regular polyhedron $P$.
Let $m$ be the number of sides of each of the regular polygons that form the faces of $P$.
Let $n$ be the number of those polygons which meet at each vertex of $P$.
From Internal Angles of Regular Polygon, the internal angles of each face of $P$ measure $180^\circ - \dfrac {360^\circ} m$.
The sum of the internal angles must be less than $360^\circ$.
So:
$\ds n \paren {180^\circ - \dfrac {360^\circ} m}$ $<$ $\ds 360^\circ$ $\ds \leadsto \ \$ $\ds n \paren {1 - \dfrac 2 m}$ $<$ $\ds 2$ $\ds \leadsto \ \$ $\ds n \paren {m - 2}$ $<$ $\ds 2m$ $\ds \leadsto \ \$ $\ds n \paren {m - 2} - 2 \paren {m - 2}$ $<$ $\ds 2m - 2 \paren {m - 2}$ $\ds \leadsto \ \$ $\ds \paren {m - 2} \paren {n - 2}$ $<$ $\ds 4$
But $m$ and $n$ are both greater than $2$.
So:
if $m = 3$, $n$ can only be $3$, $4$ or $5$
if $m = 4$, $n$ can only be $3$
if $m = 5$, $n$ can only be $3$
and $m$ cannot be greater than $3$.
There are $5$ possibilities in all.
Therefore all platonic solids have been accounted for.
$\blacksquare$
## Historical Note
Euclid's proof that there exist exactly Five Platonic Solids appears to have originated with Theaetetus of Athens. |
# Parallel Lines
All around us we see many different forms of line, like the edge of the table, corners of floors and ceilings, sides of doors and windows, and so on. But there are some forms of straight lines which go side-by-side in the same direction without intersection. Like both opposite sides of the door in the same face. This type of line is an example of parallel lines.
#### Create learning materials about Parallel Lines with our free learning app!
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In this explanation, we will understand the concept of parallel lines and their different properties.
## Parallel lines definition
Parallel lines are the types of lines that consists of two or more lines in the same plane.
Two or more straight lines in the same plane which are equidistant (having the same distance between them at all points) and never intersect each other at any point are called parallel lines.
Parallel lines remain at the same distance from each other, no matter how far they are extended. They can be constructed in any direction whether horizontal, vertical, or diagonal. Mathematically, they are represented with the symbol $\parallel$ which is called “ is parallel to”.
Parallel lines, StudySmarter Originals
Here in the above figure p and q are parallel lines and m and n are parallel lines. Hence it is said that$p\parallel q$and$m\parallel n.$But line a and line b are not parallel to each other as when extending both the lines, they both will intersect each other at some point. So a is not parallel to b (that is$a\nparallel b$).
## Parallel lines angles
As parallel lines do not intersect with each other, no angles can be formed between them. But when another line apart from the given parallel lines intersect both the parallel lines then some angles are formed between them.
When any line cuts both the parallel lines at some point in the same plane, then this line is known as transversal.
Transversal cutting parallel lines, StudySmarter Originals
Here in the above figure, we can see that line l cuts both the parallel lines a and b. So line l is the transversal line. As the transversal cuts parallel lines it can be seen that the transversal forms pairs of angles with both the lines. There are different types of angles created by transversals.
### Corresponding angles
The angles which are formed on the same side of the transversal and on the matching corners of parallel lines are called corresponding angles.
Corresponding angles can be easily identified in the form of an “F” shape. They can be formed anyway either upside down or back and front. Corresponding angles are always equal to each other in parallel lines.
Corresponding angles of parallel lines, StudySmarter Originals
### Alternate angles
Angles formed on the opposite side of transversal on parallel lines are known as alternate angles.
Alternate angles can be found in the form of a “Z” shape. They can be both interior and exterior angles. Similarly, like corresponding angles, alternate angles can be formed in any direction. Pair of alternate angles are always equal to each other.
Alternate interior angles of parallel lines, StudySmarter Originals
Here in the above figure, both the angles are alternate interior angles.
### Interior angles
Angles formed on the same side of the transversal facing each other on parallel lines are called interior angles.
Interior angles are formed in the shape of a “U”. They can be found on either side of the transversal containing both parallel lines. The sum of interior angles will always be $180°.$
Interior angles of parallel lines, StudySmarter Originals
### Exterior angles
Angles that are outside the sides of parallel lines but on the same of the transversal are called exterior angles.
Exterior angles are formed in the shape of a “U” but will be located on the outside region of it. And the sum of the pairs of exterior angles will always be$180°.$
Exterior angles, StudySmarter Originals
### Vertically Opposite angles
Angles forming on any one of the parallel lines and transversals which are opposite to each other are called vertically opposite angles.
Vertically opposite angles are found in the form of two “V” touching each other. They only contained any one of the parallel lines for each pair. Vertically opposite angles are equal to each other.
Vertically Opposite angles, StudySmarter Originals
So we can represent all the pairs of angles for all the types of angles as below.
All pairs of angles in parallel lines, StudySmarter Originals
• Corresponding angle pairs :$\angle A&\angle E;\angle B&\angle F;\angle C&\angle G;\angle D&\angle H$
• Alternate angle pairs : $\angle C&\angle E;\angle D&\angle F$
• Interior angle pairs :$\angle C&\angle F;\angle D&\angle E$
• Exterior angle pairs : $\angle A&\angle H;\angle B&\angle G$
• Vertically opposite angle pairs : $\angle A&\angle C;\angle B&\angle D;\angle E&\angle G;\angle F&\angle H$
## Parallel line equations
Parallel lines are one type of line. So we can represent parallel lines in the form of an equation of the line. We know that in coordinate geometry, the equation of the line can be written in the form of $y=mx+b.$ So we can also represent parallel lines in the form of the equation $\mathbit{y}\mathbf{=}\mathbit{m}\mathbit{x}\mathbf{+}\mathbit{b}.$
Here b is the y-intercept, so it can be any value. It is important to remember that as we have two or more lines in parallel lines the value of b for every line should be different from each other. As if they are equal then the equations of all lines will be the same and then it can be considered as one single line.
And m is the gradient or slope of that line. Here contrary to b the value of m for all the parallel lines should be equal. As m represents the slope of the line, if m is different for all parallel lines, then they will intersect each other and would not be considered parallel anymore.
We will understand the concept of gradient and how it can be found soon in the following topic.
Parallel lines equation in graph, StudySmarter Originals
The gradient or slope of parallel lines is the steepness of that line in the graph. The gradient of parallel lines is calculated with respect to the positive x-axis of the graph and parallel lines are inclined with the positive x-axis.
We know from above that the equation for parallel lines is $y=mx+b.$ Now suppose that the equation for one line $y={m}_{1}x+{b}_{1}$and the equation for the other line is $y={m}_{2}x+{b}_{2}.$ Here${b}_{1},{b}_{2}$ are y-intercept and m is the gradient of parallel lines. Then for both the lines to become parallel, the slope of both the lines should be equal. That is id="2715519" role="math" ${\mathbit{m}}_{\mathbf{1}}\mathbf{=}{\mathbit{m}}_{\mathbf{2}}.$ This equality can be derived by considering the angle between both the lines.
If we are already given two points on each line of the graph then we can calculate and verify the slope using the formula:
$\mathbit{m}\mathbf{=}\frac{{\mathbf{y}}_{\mathbf{2}}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}}{{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}},$ where${x}_{1},{x}_{2},{y}_{1},{y}_{2}$are the x-axis and y-axis points for the single line.
## Parallel lines examples
Let us see some parallel line examples and understand how to find angles and the slope in parallel lines.
In the given figure m and n are parallel lines and l is the transversal cutting both the parallel lines. Then find the value of x if$\angle C=x+22,\angle F=2x-13$is given.
Parallel lines with transversal, StudySmarter Originals
Solution:
We are already given that lines m and n are parallel to each other and line l is transversal to m and n.
So from the figure, we can clearly see that$\angle C$ and$\angle F$ are interior angles as they form the shape "U".
As both the angles are interior angles we know that their sum is equal to$180°.$
$⇒\angle C+\angle F=180°\phantom{\rule{0ex}{0ex}}⇒\left(x+22\right)°+\left(2x-13\right)°=180°\phantom{\rule{0ex}{0ex}}⇒x°+2x°+22°-13°=180°\phantom{\rule{0ex}{0ex}}⇒3x°+9°=180°\phantom{\rule{0ex}{0ex}}⇒3x°=180°-9°\phantom{\rule{0ex}{0ex}}⇒3x°=171°\phantom{\rule{0ex}{0ex}}⇒x=\frac{171°}{3°}\phantom{\rule{0ex}{0ex}}\mathbf{\therefore }\mathbf{}\mathbit{x}\mathbf{=}\mathbf{57}\mathbf{°}$
Find the value of$\angle Q,\angle R$from the given figure if$\angle P=64°$is given. Also line a, b, and c are parallel lines cut by the transversal t.
Parallel lines with missing angles, StudySmarter Originals
Solution:
It is given that lines a,b and c are parallel to each other, and line t acts as a transversal to these three lines.
First, we find the value of$\angle Q$. We can see in the figure that$\angle P$and$\angle T$ angles forms a "U" shape. So both the angles$\angle P$ and $\angle T$ are interior angles. So the sum of both these angles will be$180°.$
$⇒\angle P+\angle T=180°\phantom{\rule{0ex}{0ex}}⇒\angle T=180°-\angle P\phantom{\rule{0ex}{0ex}}⇒\angle T=180°-64°\phantom{\rule{0ex}{0ex}}⇒\angle T=116°$
Now$\angle Q$and$\angle T$are vertically opposite angles. So both the angles will be equal to each other.
$⇒\angle T=\angle Q\phantom{\rule{0ex}{0ex}}As\angle T=116°,\phantom{\rule{0ex}{0ex}}\therefore \mathbf{\angle }\mathbit{Q}\mathbf{=}\mathbf{116}\mathbf{°}$
Now from the figure we can see that$\angle Q$ and$\angle R$ forms a "F" shape, so they are corresponding angles. And hence they are equal to each other.
$⇒\angle Q=\angle R\phantom{\rule{0ex}{0ex}}\mathbf{\therefore }\mathbf{}\mathbf{\angle }\mathbit{R}\mathbf{=}\mathbf{116}\mathbf{°}$
Hence the value of both angles are $\angle Q=\angle R=116°.$
Check whether the given lines are parallel lines or not.
$a\right)y=3x+7b\right)y=2x-5\phantom{\rule{0ex}{0ex}}y=3x+4y=5x-5$
Solution:
a) Here we are given two equations of lines $y=3x+7,y=3x+4.$ Now comparing them with the general equation of parallel lines$y=mx+b,$we get that ${m}_{1}=3,{m}_{2}=3,{b}_{1}=7,{b}_{2}=4.$ Here${m}_{1},{m}_{2}$are the gradients of parallel lines and${b}_{1},{b}_{2}$are the y-intercepts.
As we know that for lines to be parallel, the gradients should be equal. And we can clearly see in the above equations that ${m}_{1}={m}_{2}.$ Also note that the values${b}_{1},{b}_{2}$are different. Hence both the lines are parallel lines.
b) Here the equations of lines are given as $y=2x-5,y=5x-5.$Comparing it with the general equation of parallel lines $y=mx+b,$ we get that ${m}_{1}=2,{m}_{2}=5,{b}_{1}=-5,{b}_{2}=-5.$
As here we get that${m}_{1}\ne {m}_{2}$we can instantly say that both the given lines are not parallel to each other.
## Parallel Lines - Key takeaways
• Two or more straight lines in the same plane which are equidistant (having the same distance between them at all points) and never intersect each other at any point are called parallel lines.
• The angles which are formed on the same side of the transversal and on the matching corners of parallel lines are called corresponding angles.
• Angles formed on the opposite side of transversal on parallel lines are known as alternate angles.
• Angles formed on the same side of the transversal facing each other on parallel lines are called interior angles.
• Angles that are outside the sides of parallel lines but on the same of the transversal are called exterior angles.
• Angles forming on any one of the parallel lines and transversals which are opposite to each other are called vertically opposite angles.
• The equation of the parallel line is$y=mx+b,$where the slope m of both the lines should be equal.
#### Flashcards in Parallel Lines 7
###### Learn with 7 Parallel Lines flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What is the meaning of parallel lines?
Two or more straight lines in the same plane which are equidistant (having the same distance between them at all points) and never intersect each other at any point are called parallel lines.
What are the three rules of parallel lines?
• Alternate angles should be equal
• Corresponding angles should be equal
• Vertically opposite angles should be equal.
What are angles in parallel lines?
Angles in parallel lines are alternate angles, corresponding angles, interior angles, exterior angles, and vertically opposite angles.
What is an example of a parallel equation?
Examples of parallel are railway tracks, opposite edges of doors and windows.
How do you find the gradient of a parallel line?
The gradient of parallel lines can be calculated using line equation y=mx+b for each line such that m1 = m2.
## Test your knowledge with multiple choice flashcards
Can parallel lines contain more than two lines?
Perpendicular lines and parallel lines are equal.
Which of the following is/are the condition/s for parallel lines?
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## fozia 2 years ago can anyone tell me the Use the linear approximation of the function f(x)=arctan(e3x) at x=0 to estimate the value of f(0.01).
1. genius12
Find the equation of the tangent line of f(x) at x = 0 and evaluate f(0.01) using the tangent line. Do you know how to perform these steps? @fozia
2. fozia
im trying
3. genius12
btw, is that e^3x?
4. fozia
yes it is
5. genius12
Ok let's first find the equation of the tangent line. To do this, I will need the slope of the line and a point. To get a point, we find f(0):$\bf f(0)=\arctan(1)=\frac{ \pi }{ 4}$So the tangent goes through the point $$\bf (0, \frac{\pi}{4})$$. Now to find the slope of the tangent, we evaluate f'(x) at x = 0:$\bf f'(x)=\frac{ 3e^{3x} }{ 1+e^{6x} } \rightarrow f'(0)=\frac{ 3e^0 }{ 1+e^0 }=3$So now we have a point and the slope of tangent line. We will use the slope-intercept form (you can use point-slope form as well but I find slope-intercept form easier) to get the tangent line's equation:$\bf y = mx+b \rightarrow y = 3x + b$Plug in the point for 'x' and 'y':$\bf \frac{\pi}{4}=3(0)+b \implies b = \frac{\pi}{4}$So the equation of the tangent line is:$\bf g(x)=3x+\frac{\pi}{4}$Here I called the tangent line g(x). Now to find the linear approximation of f(0.01), we plug in x = 0.01 in to our equation of the tangent line and evaluate:$\bf g(0.01)=3(0.01) + \frac{\pi}{4} \approx 0.815$Therefore: $$\bf f(0.01) \approx 0.815$$ @fozia
6. fozia
oh grt thank you so much |
# How To Add Rational Expressions With Unlike Denominators
When the denominators of two rational expressions are opposites, it is easy to get a common denominator. Make sure each term has the lcd as its denominator.
Radical Functions Homework (Algebra 2 Unit 6) Algebra
### To add rational expressions with unlike denominators, first find equivalent expressions with common denominators.
How to add rational expressions with unlike denominators. Note, methods never change, only problems. Find the least common denominator of rational expressions. By using this website, you agree to our cookie policy.
This is done by multiplying both the numerator and denominator of each fraction by any factors needed to obtain the lcd. Once you have common denominators, you're ready to add and simplify! When we add or subtract rational expressions with unlike denominators, we will need to get common denominators.
8 x 2 − 2 x − 3, 3 x x 2 + 4 x + 3. 7 12 + 5 18. You know how to do this with numeric fractions.
Do this just as you have with fractions. Rewrite each fraction as an equivalent fraction with the lcd. Khan academy is a 501(c)(3) nonprofit organization.
As we have done previously, we will do one example of adding numerical fractions first. Add & subtract rational expressions (basic) adding & subtracting rational expressions. To add or subtract two rational expressions with unlike denominators 1.
Find the least common denominator of rational expressions. Rewrite as equivalent rational expressions with denominator (x+1) (x−3) (x+3): Let’s look at this example:
Find the least common denominator of rational expressions. Find the least common denominator (lcd) and change each rational expression into an equivalent expression with that lcd. If we review the procedure we used with numerical fractions, we will know what to do with rational expressions.
This is the same method we use with rational expressions. To add fractions with like denominators, add the numerators and keep the same denominator. Adding and subtracting with unlike denominators.
Adding and subtracting rational expressions with like denominators. When we add numerical fractions, once we find the lcd, we rewrite each fraction as an equivalent fraction with the lcd. Determine if the expressions have a common denominator.
If we review the procedure we used with numerical fractions, we will know what to do with rational expressions. Rewrite each fraction as an equivalent fraction with the lcd. When we add or subtract rational expressions with unlike denominators, we will need to get common denominators.
To add rational expressions with unlike denominators, first find equivalent expressions with common denominators. If the denominators of fractions are relatively prime, then the least common denominator (lcd) is their product. Do this just as you have with fractions.
Intro to adding rational expressions with unlike denominators. When we add or subtract rational expressions with unlike denominators we will need to get common denominators. Let’s look at this example:
Adding rational expressions with the same denominator is the simplest place to start, so let’s begin there. Add or subtract the numerators white maintaining the lcd. Khan academy is a 501(c)(3) nonprofit organization.
To add or subtract rational expressions with unlike denominators, first find the lcm of the denominator. Recall, when adding with a common denominator, we add across numerators and keep the same denominator. The lcm of the denominators of fraction or rational expressions is also called least common denominator , or lcd.
If the denominators of fractions are relatively prime, then the least common denominator (lcd) is their product. Rewrite each rational expression as an equivalent rational expression with the lcd. Write each expression using the lcd.
We just have to multiply one of the fractions by. Let’s see how this works. We will do the same thing for rational expressions.
Now we have all the steps we need to add rational expressions with different denominators. Add and subtract rational expressions whose denominators are opposites. To add rational expressions with unlike denominators, you must first find a common denominator.
Let’s look at the example 7 12 + 5 18 7 12 + 5 18 from. 7 12 + 5 18. Watch it all in this tutorial!
Find the lcd of 12 and 18. If we review the procedure we used with numerical fractions, we will know what to do with rational expressions. This is the currently selected item.
Least common multiple of polynomials. Adding and subtracting rational expressions are identical to adding and subtracting with numerical fractions.
Anchor chart for subtracting integers. Anchor Charts
Pin by Mymatheducation on Rational Functions Simplifying
This Adding Fractions Coloring Page will aid in student
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Pin on Printable Blank Worksheet Template |
Contributor: Erika Wargo. Lesson ID: 12389
Large addition problems can add up to a struggle unless you know the secret. Algorithm to the rescue! Learn the steps to making addition simple. We've added an interactive quiz and video for sum help!
categories
Arithmetic, Whole Numbers and Operations
subject
Math
learning style
Visual
personality style
Beaver
Intermediate (3-5)
Lesson Type
Skill Sharpener
Lesson Plan - Get It!
Audio:
You are at your first ball game, and you are overwhelmed by the size of the ball park! The stadium has 15,500 seats in the lower level, and 10,250 seats in the upper level. How many total seats are in the stadium?
Addition is combining two or more groups into one group.
The answer to an addition problem is called the "sum." The standard algorithm for addition is a common way to solve addition problems. An algorithm is a set of steps that are used to complete a task. In this lesson, the steps are used to perform addition. The standard algorithm for addition has three steps:
1. Line up the digits in each number vertically by place value, starting with the ones place.
2. Add the common place values together, starting with the ones place and moving to the left.
3. Regroup, if necessary.
It is also helpful to use grid paper to help you line up the digits in each number.
Example 1
269 + 452
Step 1 Line up the digits by place value. The 9 and 2 are in the ones place, so they are stacked on top of each other. The other numbers are also stacked on top of each other according to place value.
Step 2 Add the common place values together. Notice how 9 + 2 = 11, so regrouping was necessary. Since 11 is greater than one ten, write down the 1 and regroup 1 in the tens place. When adding the tens place, 1 + 6 + 5 = 12, so regroup again. The 12 in the tens place represents 12 tens, so 10 tens becomes a hundred. When adding the hundreds place, 1 + 2 + 4 = 7, so 7 is written in the hundreds place. The solution is 721.
hundreds tens ones 1 1 2 6 9 + 4 5 2 __________________________________ 7 2 1
Example 2
345,600 + 451 + 5,000 =
In this example, there are three numbers to add together. The order of the numbers doesn’t matter, but the digits need to be lined up by place value. The first problem shows the numbers written in the same order as they were in the original problem. The second problem shows the numbers stacked with the greater number on the top and the smaller number on the bottom. Both ways will result in the same solution if the addition is done correctly:
1 1 1 1 3 4 5 6 0 0 3 4 5 6 0 0 4 5 1 5 0 0 0 + 5 0 0 0 + 4 5 1 _________________________________________ ___________________________________________ 3 5 1, 0 5 1 3 5 1, 0 5 1
Example 3
At the beginning of the lesson, you were asked the following question:
The stadium has 15,500 seats in the lower level and 10,250 seats in the upper level. How many total seats are in the stadium?
Using the standard algorithm, line up the digits and add, starting with the ones place:
1 5, 5 0 0 + 1 0, 2 5 0 _______________________________ 2 5, 7 5 0
There are 25,750 seats in the stadium.
Before you practice adding on your own, watch a few more examples in a video about multi-digit addition. Listen for the answers to these questions and write the answers on a piece of paper, if necessary:
• How do you line up or stack numbers when writing an addition problem vertically?
• In which place value spot do you start adding?
After you watch Math Antics - Multi-Digit Addition, discuss them with a parent or teacher:
In the Got It? section, you will practice adding numbers as you play a game and complete interactive practice.
Interactive Video |
What is distributive property maths?
In Math, the distributive property of multiplication is described as when we multiply a number with the sum of two or more addends or minuends, we get a result that is equal to the result that is obtained when we multiply each addend or minuend separately by the number.
What is a commutative property in math?
This law simply states that with addition and multiplication of numbers, you can change the order of the numbers in the problem and it will not affect the answer.
What is commutative property example?
The commutative property deals with the arithmetic operations of addition and multiplication. It means that changing the order or position of numbers while adding or multiplying them does not change the end result. For example, 4 + 5 gives 9, and 5 + 4 also gives 9.
What are the 4 multiplication properties?
The properties of multiplication are distributive, commutative, associative, removing a common factor and the neutral element.
What is the formula of distributive property in maths?
Distributive property means to divide the given operations on the numbers, so that the equation becomes easier to solve. Distributive property definition simply states that “multiplication distributed over addition.” i.e a x b + a x c = ab + ac.
How do you use the distributive property in math?
You can use the distributive property of multiplication to rewrite expression by distributing or breaking down a factor as a sum or difference of two numbers. Here, for instance, calculating 8 × 27 can made easier by breaking down 27 as 20 + 7 or 30 − 3.
What are 2 examples of commutative property?
Here’s a quick summary of these properties: Commutative property of addition: Changing the order of addends does not change the sum. For example, 4 + 2 = 2 + 4 4 + 2 = 2 + 4 4+2=2+44, plus, 2, equals, 2, plus, 4. Associative property of addition: Changing the grouping of addends does not change the sum.
What is the formula of commutative property?
The commutative property formula for multiplication is defined as the product of two or more numbers that remain the same, irrespective of the order of the operands. For multiplication, the commutative property formula is expressed as (A × B) = (B × A).
What is commutative property formula?
What is the distributive property of multiplication?
According to the distributive property of multiplication, when we multiply a number with the sum of two or more addends, we get a result that is equal to the result that is obtained when we multiply each addend separately by the number.
What are the properties of multiplication in math?
Properties of multiplication
• Commutative property of multiplication: Changing the order of factors does not change the product.
• Associative property of multiplication: Changing the grouping of factors does not change the product.
• Identity property of multiplication: The product of 1 and any number is that number. |
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# 7.4 Dividing Fractions
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### 7.4 Dividing Fractions
1. 1. 7.4 Dividing Fractions Mr. Williams GESD jwilliams@gesd40.org Objective: Use reciprocals to divide fractions in class and on a worksheet. 1
2. 2. Vocabulary 5 3 Reciprocal two numbers, whose product is one. (Such as and ) 3 5 3 12 12 What is the reciprocal of ? 3 2
3. 3. Warm Up Answer these in your math notes. 1 1 1 5 2 + 8 4 x 6 1 2 3 1 + 6 4 x 2 12 3
4. 4. Dividing Fractions is easy. It is just like multiplying fractions with one more step. To divide a fraction, multiply by its reciprocal. 2 3 2 4 ÷ = x 3 4 3 3 All you do is quot;switchquot; the second fraction then multiply. 4
5. 5. Solve these problems in your math notes. 5 4 5 3 ÷ = x 8 3 8 4 7 1 7 2 ÷ = x 10 1 10 2 5
6. 6. Write a fraction divison problem in your math notes. We will share some of them on the board. 6
7. 7. Dividing Fractions and Whole Numbers 1. Change the whole number into a fraction. 2. Find the reciprocal of your second fraction. 3. Multiply the fractions. 7
8. 8. So what does all that mean? 3 Divide 6 by 4 3 6 3 6 4 6 ÷ = = x 4 1 ÷ 4 1 3 8
9. 9. Solve these problems in your math notes. 9 10 ÷ 3 = 2 12 ÷ 3 = Finish this sentence: You divide fractions by ... 9 |
# Difference between revisions of "2013 IMO Problems/Problem 2"
## Problem
A configuration of $4027$ points in the plane is called Colombian if it consists of $2013$ red points and $2014$ blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied:
• no line passes through any point of the configuration;
• no region contains points of both colours.
Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good arrangement of $k$ lines.
## Solution
We can start off by imagining the points in their worst configuration. With some trials, we find $2013$ lines to be the answer to the worst cases. We can assume the answer is $2013$. We will now prove it.
We will first prove that the sufficient number of lines required for a good arrangement for a configuration consisting of $u$ red points and $v$ blue points, where $u$ is even and $v$ is odd and $u - v = 1$, is $v$.
Notice that the condition "no three points are co-linear" implies the following: No blue point will get in the way of the line between two red points and vice versa. What this means, is that for any two points $A$ and $B$ of the same color, we can draw two lines parallel to, and on different sides of the line $AB$, to form a region with only the points $A$ and $B$ in it.
Now consider a configuration consisting of u red points and v blue ones ($u$ is even, $v$ is odd, $u>v$). Let the set of points $S = \{A_1, A_2, ... A_k\}$ be the out-most points of the configuration, such that you could form a convex k-gon, $A_1 A_2 A_3 ... A_k$, that has all of the other points within it.
If the set S has at least one blue point, there can be a line that separates the plane into two regions: one only consisting of only a blue point, and one consisting of the rest. For the rest of the blue points, we can draw parallel lines as mentioned before to split them from the red points. We end up with $v$ lines.
If the set $S$ has no blue points, there can be a line that divides the plane into two regions: one consisting of two red points, and one consisting of the rest. For the rest of the red points, we can draw parallel lines as mentioned before to split them from the blue points. We end up with $u-1 = v$ lines.
Now we will show that there are configurations that can not be partitioned with less than $v$ lines.
Consider the arrangement of these points on a circle so that between every two blue points there are at least one red point (on the circle).
There are no less than $2v$ arcs of this circle, that has one end blue and other red (and no other colored points inside the arc) - one such arc on each side of each blue point. For a line partitioning to be good, each of these arcs have to be crossed by at least one line, but one line can not cross more than $2$ arcs on a circle - therefore, this configuration can not be partitioned with less than $v$ lines!
Our proof is done, and we have our final answer: $2013$. |
# Trachtenberg System
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The Trachtenberg System is a system of rapid mental calculation. The system consists of a number of readily memorized operations that allow one to perform arithmetic computations very quickly. It was developed by Jakow Trachtenberg in order to keep his mind occupied while being held in a Nazi concentration camp.
## General Multiplication
The method for general multiplication is a method to achieve multiplications a*b with low space complexity, i.e. as few temporary results as possible to be kept in memory.
This is achieved by noting that the final digit is completely determined by multiplying the last digit of the multiplicands. This is held as a temporary result. To find the next to last digit, we need everything that influences this digit: The temporary result, the last digit of a times the next-to-last digit of b, as well as the next-to-last digit of a times the last digit of b. This calculation is performed, and we have a temporary result that is correct in the final two digits.
In general, for each position n in the final result, we sum for all i:
$a \mbox{ (digit at } i\mbox{)} \times b \mbox{ (digit at } (n-i)\mbox{)}$
Ordinary people can learn this algorithm and thus multiply four digit numbers in their head - writing down only the final result. They would write it out starting with the rightmost digit and finishing with the leftmost.
Trachtenberg defined this algorithm with a kind of pairwise multiplication where two digits are multiplied by one digit, essentially only keeping the middle digit of the result. By performing the above algorithm with this pairwise multiplication, even fewer temporary results need to be held.
Example: 123456 * 789
To find the first digit of the answer:
The units digit of 9 * 6 = 4.
To find the second digit of the answer, start at the second digit of the multiplicand:
The units digit of 9 * 5 plus the tens digit of 9 * 6 plus
The units digit of 8 * 6.
5 + 5 + 8 = 18.
The second digit of the answer is 8 and carry 1 to the third digit.
To find the fourth digit of the answer, start at the fourth digit of the multiplicand:
The units digit of 9 * 3 plus the tens digit of 9 * 4 plus
The units digit of 8 * 4 plus the tens digit of 8 * 5 plus
The units digit of 7 * 5 plus the tens digit of 7 * 6.
7 + 3 + 2 + 4 + 5 + 4 = 25 + 1 carried from the third digit.
The fourth digit of the answer is 6 and carry 2 to the next digit.
2 Finger method
Professor Trachtenberg called this the 2 Finger Method. The calculations for finding the fourth digit from the example above are illustrated at right. The arrow from the nine will always point to the digit of the multiplicand directly above the digit of the answer you wish to find, with the other arrows each pointing one digit to the right. Each arrow head points to a UT Pair, or Product Pair. The vertical arrow points to the product where we will get the Units digit, and the sloping arrow points to the product where we will get the Tens digits of the Product Pair. If an arrow points to a space with no digit there is no calculation for that arrow. As you solve for each digit you will move each of the arrows over the multiplicand one digit to the left until all of the arrows point to prefixed zeros.
## General division
Setting up for Division
Division in the Trachtenberg System is done much the same as in multiplication but with subtraction instead of addition. Splitting the dividend into smaller Partial Dividends, then dividing this Partial Dividend by only the left-most digit of the divisor will provide the answer one digit at a time. As you solve each digit of the answer you then subtract Product Pairs (UT pairs) and also NT pairs (Number-Tens) from the Partial Dividend to find the next Partial Dividend. The Product Pairs are found between the digits of the answer so far and the divisor. If a subtraction results in a negative number you have to back up one digit and reduce that digit of the answer by one. With enough practice this method can be done in your head.
## General addition
A method of adding columns of numbers and accurately checking the result without repeating the first operation. An intermediate sum, in the form of two rows of digits, is produced. The answer is obtained by taking the sum of the intermediate results with an L-shaped algorithm. As a final step, the checking method that is advocated removes both the risk of repeating any original errors and allows the precise column in which an error occurs to be identified at once. It is based on a check (or digit) sums, such as the nines-remainder method.
For the procedure to be effective, the different operations used in each stages must be kept distinct, otherwise there is a risk of interference.
## Other multiplication algorithms
When performing any of these multiplication algorithms the following "steps" should be applied.
The answer must be found one digit at a time starting at the least significant digit and moving left. The last calculation is on the leading zero of the multiplicand.
Each digit has a neighbor, i.e., the digit on its right. The rightmost digit's neighbor is the trailing zero.
The 'halve' operation has a particular meaning to the Trachtenberg system. It is intended to mean "half the digit, rounded down" but for speed reasons people following the Trachtenberg system are encouraged to make this halving process instantaneous. So instead of thinking "half of seven is three and a half, so three" it's suggested that one thinks "seven, three". This speeds up calculation considerably. In this same way the tables for subtracting digits from 10 or 9 are to be memorized.
And whenever the rule calls for adding half of the neighbor, always add 5 if the current digit is odd. This makes up for dropping 0.5 in the next digit's calculation.
### Multiplying by 11
Rule: Add the digit to its neighbor. (By "neighbor" we mean the digit on the right.)
Example: 3,425 * 11 = 37,675
3 7 6 7 5
(=0+3) (=3+4) (=4+2) (=2+5) (=5+0)
To illustrate:
11=10+1
Thus,
$3425 \times 11 = 3425 \times (10+1)$ $\Rightarrow 37675 = 34250 + 3425$
### Multiplying by 12
Rule: to multiply by 12:
Starting from the rightmost digit, double each digit and add the neighbor. (By "neighbor" we mean the digit on the right.)
If the answer is greater than a single digit, simply carry over the extra digit (which will be a 1 or 2) to the next operation. The remaining digit is one digit of the final result.
Example: 316 * 12
Determine neighbors in the multiplicand 0316:
• digit 6 has no right neighbor
• digit 1 has neighbor 6
• digit 3 has neighbor 1
• digit 0 (the prefixed zero) has neighbor 3
6 * 2 = 12 (2 carry 1)
1 * 2 + 6 + 1 = 9
3 * 2 + 1 = 7
0 * 2 + 3 = 3
0 * 2 + 0 = 0
316 * 12 = 3,792
### Multiplying by 6
• Rule: to multiply by 6: Add half of the neighbor to each digit, then, if the current digit is odd, add 5.
Example:
357 × 6 = 2142
Working right to left,
7 has no neighbor, add 5 (since 7 is odd) = 12. Write 2, carry the 1.
5 + half of 7 (3) + 5 (since the starting digit 5 is odd) + 1 (carried) = 14. Write 4, carry the 1.
3 + half of 5 (2) + 5 (since 3 is odd) + 1 (carried) = 11. Write 1, carry 1.
0 + half of 3 (1) + 1 (carried) = 2. Write 2.
### Multiplying by 7
Rule: to multiply by 7:
1. Double each digit.
2. Add half of its neighbor.
3. If the digit is odd, add 5.
Example: 523 x 7 = 3,661.
3x2+0+5=11, 1.
2x2+1+1=6.
5x2+1+5=16, 6.
0x2+2+1=3.
3661.
### Multiplying by 9
Rule:
1. Subtract the right-most digit from 10.
1. Subtract the remaining digits from 9.
2. Add the neighbor.
3. For the leading zero, subtract 1 from the neighbor.
For rules 9, 8, 4, and 3 only the first digit is subtracted from 10. After that each digit is subtracted from nine instead.
Example: 2,130 × 9 = 19,170
Working from right to left:
• (10 - 0) + 0 = 10. Write 0, carry 1.
• (9 - 3) + 0 + 1 (carried) = 7. Write 7.
• (9 - 1) + 3 = 11. Write 1, carry 1.
• (9 - 2) + 1 + 1 (carried) = 9. Write 9.
• 2 - 1 = 1. Write 1.
### Multiplying by 8
Rule:
1. Subtract right-most digit from 10.
1. Subtract the remaining digits from 9.
2. Double the result.
3. Add the neighbor.
4. For the leading zero, subtract 2 from the neighbor.
Example: 456 x 8 = 3648
Working from right to left:
• (10 - 6) x 2 + 0 = 8. Write 8.
• (9 - 5) x 2 + 6 = 14, Write 4, carry 1.
• (9 - 4) x 2 + 5 + 1 (carried) = 16. Write 6, carry 1.
• 4 - 2 + 1 (carried) = 3. Write 3.
### Multiplying by 4
Rule:
1. Subtract the right-most digit from 10.
1. Subtract the remaining digits from 9.
2. Add half of the neighbor, plus 5 if the digit is odd.
3. For the leading 0, subtract 1 from half of the neighbor.
Example: 346 * 4 = 1384
Working from right to left:
• (10 - 6) + Half of 0 (0) = 4. Write 4.
• (9 - 4) + Half of 6 (3) = 8. Write 8.
• (9 - 3) + Half of 4 (2) + 5 (since 3 is odd) = 13. Write 3, carry 1.
• Half of 3 (1) - 1 + 1 (carried) = 1. Write 1.
### Multiplying by 3
Rule:
1. Subtract the rightmost digit from 10.
1. Subtract the remaining digits from 9.
2. Double the result.
3. Add half of the neighbor, plus 5 if the digit is odd.
4. For the leading zero, subtract 2 from half of the neighbor.
Example: 492 x 3 = 1476
Working from right to left:
• (10 - 2) x 2 + Half of 0 (0) = 16. Write 6, carry 1.
• (9 - 9) x 2 + Half of 2 (1) + 5 (since 9 is odd) + 1 (carried) = 7. Write 7.
• (9 - 4) x 2 + Half of 9 (4) = 14. Write 4, carry 1.
• Half of 4 (2) - 2 + 1 (carried) = 1. Write 1.
### Multiplying by 5
• Rule: to multiply by 5: Take half of the neighbor, then, if the current digit is odd, add 5.
Example: 42x5=210
Half of 2's neighbor, the trailing zero, is 0.
Half of 4's neighbor is 1.
Half of the leading zero's neighbor is 2.
43x5=215
Half of 3's neighbor is 0, plus 5 because 3 is odd, is 5.
Half of 4's neighbor is 1.
Half of the leading zero's neighbor is 2.
93x5=465
Half of 3's neighbor is 0, plus 5 because 3 is odd, is 5.
Half of 9's neighbor is 1, plus 5 because 9 is odd, is 6.
Half of the leading zero's neighbor is 4.
## Publications
• Rushan Ziatdinov, Sajid Musa. Rapid mental computation system as a tool for algorithmic thinking of elementary school students development. European Researcher 25(7): 1105-1110, 2012 [1].
• The Trachtenberg Speed System of Basic Mathematics by Jakow Trachtenberg, A. Cutler (Translator), R. McShane (Translator), was published by Doubleday and Company, Inc. Garden City, New York in 1960.[1]
The book contains specific algebraic explanations for each of the above operations.
Most of the information in this article is from the original book.
The algorithms/operations for multiplication etc. can be expressed in other more compact ways that the book doesn't specify, despite the chapter on algebraic description.[2] [3]
## Software
Following are known programs and sources available as teaching tools
Web
• Vedic Mathematics Academy [2]
PC
• Trachtenberg Speed Math [3]
• Trachtenberg Mathematics Software [4]
iPhone
• Mercury Math [5]
## References
1. Trachtenberg, Jakow (1960). The Trachtenberg Speed System of Basic Mathematics. Translated by A. Cutler, R. McShane. Doubleday and Company, Inc.. pp. 270.
2. All of this information is from an original book published and printed in 1960. The original book has seven full Chapters and is exactly 270 pages long. The Chapter Titles are as follows (the numerous sub-categories in each chapter are not listed).
3. The Trachtenberg speed system of basic mathematics
• Chapter 1 Tables or no tables
• Chapter 2 Rapid multiplication by the direct method
• Chapter 3 Speed multiplication-"two-finger" method
• Chapter 4 Addition and the right answer
• Chapter 5 Division - Speed and accuracy
• Chapter 6 Squares and square roots
• Chapter 7 Algebraic description of the method
"A revolutionary new method for high-speed multiplication, division, addition, subtraction and square root." (1960) "The best selling method for high-speed multiplication, division, addition, subtraction and square root - without a calculator." (Reprinted 2009)
Multiplication is done without multiplication tables "Can you multiply 5132437201 times 4522736502785 in seventy seconds?" "One young boy (grammar school-no calculator) did--successfully--by using The Trachtenberg Speed System of Basic Mathematics"
Jakow Trachtenberg (its founder) escaped from Hitler's Germany from an active institution toward the close of WWII. Professor Trachtenberg fled to Germany when the czarist regime was overthrown in his homeland Russia and lived there peacefully until his mid-thirties when his anti-Hitler attitudes forced him to flee again. He was a fugitive and when captured spent a total of seven years in various concentration camps. It was during these years that Professor Trachtenberg devised the system of speed mathematics. Most of his work was done without pen or paper. Therefore most of the techniques can be performed mentally.
• Trachtenberg, J. (1960). The Trachtenberg Speed System of Basic Mathematics. Doubleday and Company, Inc., Garden City, NY, USA.
• Катлер Э., Мак-Шейн Р.Система быстрого счёта по Трахтенбергу, 1967.
• Rushan Ziatdinov, Sajid Musa. Rapid mental computation system as a tool for algorithmic thinking of elementary school students development. European Researcher 25(7): 1105-1110, 2012 [6]. |
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How to solve value of x please explain
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Algebra.
If you could provide one of your questions, I could show you how to use algebra to solve for x.
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6 DIVIDE X= 8
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The purpose is to get x by itself on one side of the equation (on one side of the equal sign).
If you have 6 / x = 8, you first start off by multiplying x to both sides of the equation. The reason we are multiplying is because x is being divided. So, whatever the operation is, you do the opposite. Since the question is 'dividing' x, we need to 'multiply' x since multiplication is the opposite of division. Here's how the equation looks when we multiply x to both sides,
x * 6 / x = x * 8
See how I multiplied x to both sides of the equation? Now, here's what you get,
6 = 8x
Notice how the 6 is by itself? Both x's near the 6 were canceled out, since multiplication and division are opposite operations.
Now we have the x being multiplied to the 8 on the other side. In order to get the x by itself, we need to divide both sides by 8, remember multiplication and division are opposites. Here's what we get,
6 / 8 = 8x / 8
See how I divided both sides by 8? This is what the equation looks like now,
0.75 = x
I turned the fraction 6/8 into a decimal (you can keep it a fraction if you like). That's the answer. x = 0.75
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No matter what the equation is, you want to get x by itself on one side of the equation. Whatever you do on one side of the equal sign you must do on the other side of the sign.
5x = 25 you would divide both sides by 5 to get x by itself giving you
5x / 5 = 25 / 5 or x = 5
x + 3 = 15, subtract 3 on both sides to get x + 3 - 3 = 15 - 3 or x = 12
the more complex the equation the more steps you are going to have to do.
Remember your order of operations and Distributive rules also.
Hope this helps. If you have a specific equation that you have to solve, I can walk you through it step-by-step.
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# In how many different words can the letters of the word “OPTICAL” be arranged so that the vowels always come together?A). 120B). 720C). 4320D). 2160E) None of these
Last updated date: 21st Jul 2024
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Hint: Here the given question is of permutation and combination word problem where we need to solve for the number of words obtained by using the condition in the question, for which we need to study the word problem theory in the combination.
Formulae Used: The possible word which can be formed by using the word of “N” letters, we have:
$\Rightarrow N!$
Here “!” represents a factorial sign.
Complete step-by-step solution:
Here in the given question we are having one condition which says that the vowels in the word “OPTICAL” should come together and the other letters of the word need to be arranged in any ways, and the total possible words are need to be obtained here.
To solve this question here first we see the vowels in the word:
$\Rightarrow Vowels = OIA$
Now the rest letters are:
$\Rightarrow letters = PTCL$
Now we have the condition that the vowels should come together, hence the word can be written as:
$\Rightarrow word = (OIA)PTCL$
Here in the above word we have 5 letters because vowels will come together hence counted as one, and rest we have four letters, now possible words for five letters is given as:
$\Rightarrow 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Here the number of vowels are three and can interchanged their position too, hence we need to consider that too, so for three letters we have:
$\Rightarrow 3! = 3 \times 2 \times 1 = 6$
So the total words comes too be:
$\Rightarrow 120 \times 6 = 720$
Therefore, option (B) is correct.
Note: Here the given question is to solve by combination rule, we can write the words also to see the total words according to the condition, but here the possibilities are seven hundred and twenty hence it might be not possible to write each and every word without forgetting any possibility or without repeating any word hence we need to solve such question by combination theory of words. |
# Why are they taking the cosine to find the x-component of the resultant vector in this problem?
What I don't understand in this picture is why they are taking $$5\cos130^{\circ}$$ I usually do these problems by drawing the components along the axes in my case I would draw the y component along the y-axis and the x component would be going across on the top from the y axis to the resultant vector.
If I draw my right triangle like that then would the x component would be $$5\sin130^{\circ}$$ I do see though that this way would give me an incorrect x component but I don't know why they are using cosine as if I drew it on paper the angle that is between the hypotenuse and base would not $$=130^{\circ}$$
Another way of understanding the cosine is through the definition of the scalar product $$\vec{a}\cdot\vec{b} := |\vec{a}||\vec{b}|\cos\angle(\vec{a},\vec{b})$$ If you want to find the x component $B_x$ of your vector $\vec{B}$, you just have to multiply with the unit vector in x direction, this is called $\hat{i}$ in your drawing. This way you get $$B_x = \vec{B}\cdot\hat{i}=|\vec{B}|\underbrace{|\hat{i}|}_{=1}\cos(130°)=5\cos(130°)$$ Similarly you get for $B_y$ $$B_y = \vec{B}\cdot\hat{j}=|\vec{B}||\hat{j}|\cos(40°)=5\sin(130°)$$
• I understand this concept and it helped
– Jude
Mar 6 '17 at 1:56
In order to find the x-component of a vector, you take the cosine of the angle and multiply it by the magnitude of the vector. In this case, taking $5\cos{130^\circ}$ is correct to find $\vec{B}_x$.
Your intuition is somewhat correct, but in order to do what you've done by drawing the x-component from the y-axis to the end of $\vec{B}$ you need to use the correct angle. Since $90^\circ$ of the $130^\circ$ is already "used" in the first quadrant, you need to subtract 90 degrees from 130 to get the correct angle to use sine with. $5\sin{180^\circ}$ results in the same answer.
You don't want to do the above on a test or paper because it is unclear where you got your numbers from unless you explicitly state it. Stick with cosine for the horizontal component of the vector because that is the most common standard.
• for some reason $5\sin40^{\circ}$ is not giving the proper x component value
– Jude
Mar 6 '17 at 0:20
• What value is it supposed to be? It should be the same as $5\cos{130^\circ}$ Mar 6 '17 at 0:27
• Wait, I'm wrong. This is basically why you should never use sine instead of cosine to find the horizontal component of the vector. Just stick with using cosine. It should be $5\sin{220^\circ}$ Mar 6 '17 at 0:37
• So no matter how my right triangle could end up cosine should always be used for horizontal? It just seems so odd when I picture the triangle since in the books case the $130^{\circ}$ would be on the outside of the triangle I just don't see how you can take the adjacent/hypotenuse
– Jude
Mar 6 '17 at 1:59
• Don't imagine the cosine as the function you use to solve a right triangle in contexts like this one. When you're finding the horizontal component of the vector, you're finding the dot product of your vector and $\hat{\textbf{i}}$, which is given by $\vec{\textbf{A}} \textit{\hat{i}}} = \abs{\vec{\textbf{A}}\cos{\theta}$. Mar 6 '17 at 3:53 |
# Solving of Equation p(x)=0 by Factoring Its Left Side
The essence of method of factoring is following. Suppose we need to solve equation p(x)=0, where p(x) is polynomial of n-th degree. Suppose we factored it as p(x)=p_1(x)*p_2(x)*p_3(x), where p_1(x),p_2(x),p_3(x) are polynomials of lesser degree than n. Then, instead of solving equation p(x)=0, we need to solve set of equations p_1(x)=0,p_2(x)=0,p_3(x)=0. All found roots of these equations, and only they, will be roots of the equation p(x)=0.
Example 1. Solve equation x^3+2x^2+3x+6=0.
Let's factor left side of equation. We have that x^3+2x^2+3x+6=x^2(x+2)+3(x+2)=(x^2+3)(x+2).
Thus, either x^2+3=0 or x+2=0. First equation doesn't have roots, second equation has root x=-2. Therefore, initial equation has one root: x=-2.
Method of factoring is applicable to any equations of the form p(x)=0, where p(x) is not ony polynomial.
Example 2. Solve equation x^2sqrt(x)-9sqrt(x)=0.
Domain of this equation is x>=0.
After factoring we have that sqrt(x)(x^2-9)=0, thus, either sqrt(x)=0, or x^2-9=0. From first equation x=0, from second equation x=3 and x=-3, but x=-3 is not in the domain of the initial equation.
Therefore, initial equation has two roots: x=0 and x=3. |
# HOW TO CALCULATE THE QUANTITY OF PLASTER
## Quantity OF Plaster Calculation
The quantity of plaster calculation is very easy just need your concentration. The following points you always remember at the site when you need to calculate the plaster quantity means (water, cement and sand).
1. For wall plaster generally ratio we use = 1: 6 (1 part of cement and 6 part of sand).
2. For Roof plaster mainly ratio we use = 1: 4 (1 part of cement and 4 part of sand).
3. The thickness of plaster is between = 12 mm to 15 mm.
4. Always use Good quality of cement (Different grades of cement) as well as good quality sand.
5. Always use measuring box for measuring the quantity of sand and cement at the site.
For your better understanding let’s solve this example and find the results.
## EXAMPLE:
Suppose we have an Area of the 200-meter square (Length, Width) over that area we need to perform plaster and the ratio of cement and sand we are using 1: 4 and the thickness of plaster will be 13 mm.
1. So calculate the required quantity of water, Cement and Sand use in this plaster?
#### Given Data:
Plaster Thickness = 13 mm = 0.013 m <Remarks 1 m = 1000 mm>
Area = 200 meter square
Ratio = 1: 4
Quantity =?
### Solution:
First, we calculate the volume of the plaster (Wet volume) and the volume we calculated is a wet volume (water+wastage+bulking of sand) then we convert this volume into the dry volume. After calculating the volume we just find the quantity of cement, Sand and water required in this plasterwork according to these given data.
The volume of cement mortar required = (Area of Plaster x Thickness)
Volume of cement mortar required = 200 x 0.013
Volume of cement mortar required = 2.6 (wet volume)
So 2.6-meter cube is a wet volume of cement mortar (with water). But we need dry volume to get dry volume to consider 20% bulking of the sand and 15% Wastage of the sand at the site.
=2.6 meter cube x (1 + 0.2 + 0.15) <Remarks 20/100 % = 0.20 or just multiply by 1.35 >
=3.51 (dry Volume).
## CEMENT REQUIRED:
Cement: = (Dry volume x Ratio x Density of cement)/Sum of ratio
Cement : = (3.51 x 1 x 1440)/5 < Density of cement is 1440 kg/meter cube >
Cement : = 1010.88 kg /50 < 1 bag of cement is 50 kg >
Cement : = 20.21 bags
## SAND REQUIRED:
Sand : = (Dry volume x Ratio x 35.3147)/Sum of ratio
Sand : = (3.51 x 4 x 35.3147)/5 < 1 meter cube = 35.3147 cubic feet >
Sand : = 99.16 Cubic feet.
## WATER REQUIRED:
Water : = (Weight of cement x water cement ratio)
Water : = 1010.88 x 0.5 < 0.5 kg water every 1 kg of Cement >
Water : = 505.4 kg or 505.4 Litters.
### CONCLUSION:
Cement := 1010.88 kg
Sand: = 99.16 cubic feet
Water: = 505.4 Litters
So these quantities are going to use in a 13 mm thick plaster.
Note: The water-cement ratio 0.5 is not fixed it depends on the which type of plaster is going to do or the compressive strength we get after fixing the water-cement ratio in 1: 4 mix.
## WATCH THIS ARTICLE VIDEO:
You can watch this video for your better understanding. The video is in Hindi or in Urdu. The dimensions are changed but the method is the same. You can all so able to visit this channel the link of this channel was below.
Visit this Channel on YouTube: F&U-FORYOU
2 Comments |
## LESSON 21. Approximate Analysis of Fixed and Continuous Beams – 2
21.1 Introduction : In this lesson we will learn another method, called the Cantilever method for approximate analysis of rigid frames subjected to lateral load. Similar to the Portal method as discussed in the previous lecture, the Cantilever method is also based on few assumptions. These assumptions are as follows,
• There is a point of inflection at the mid-point of each girder and column.
• The axial load in each column of a storey is proportional to the horizontal distance of the that column from the centre of gravity of all the column of the storey under consideration.
The above two assumptions give additional equations required for solving unknown reaction components. The method is illustrated via the following examples.
Examples
Fig. 21.1.
Fig. 21.2.
Free body diagrams of different parts of the structures are shown in Fgiure 20.7. Assuming all columns have the same cross-sectional area, the centre of gravity of the columns for each storey is, $(3 + 7)/3 = 3.33{\rm{m}}$ from A. Therefore, ${F_M}$ , ${F_N}$ and ${F_O}$ are related as,
${F_N} = {{3.33 - 3} \over {3.33}}{F_M} = 0.1{F_M}$ and ${F_O} = {{3.33 - 7} \over {3.33}}{F_M} =-1.10{F_M}$
Taking moment about O of all the forces acting on the part above the horizontal plane passing through the points of inflection of the columns of the first storey,
we have,
$\sum {{M_O}}=0 \Rightarrow 7 \times {F_M} + 4 \times {F_N} + 2 \times 20 + 5 \times 10 = 0$
$\Rightarrow 7 \times {F_M} + 4 \times 0.1{F_M} =-90 \Rightarrow-12.162{\rm{kN}}$
Hence,
${F_N}=0.1{F_M}=-1.216{\rm{kN}}$ and ${F_O} =-1.10{F_M}=13.378{\rm{kN}}$
${F_P}$ , ${F_Q}$ and ${F_R}$ will also follow the similar proportion. Therefore,
${F_Q}=0.1{F_P}$ and ${F_R}=-1.10{F_P}$
Now taking moment about R, we have,
$\sum {{M_R}}=0 \Rightarrow 7 \times {F_P} + 4 \times {F_Q} + 2 \times 10 = 0$
$\Rightarrow 7 \times {F_P} + 4 \times 0.1 \times {F_P}=-20 \Rightarrow {F_P}=-2.70{\rm{kN}}$
${F_Q} = 0.1{F_P}=-0.27{\rm{kN}}$ and ${F_R}=-1.10{F_P}=2.97{\rm{kN}}$
$\sum {{M_U}}=0 \Rightarrow 1.5 \times {V_P} + 1.5 \times {F_P} = 0 \Rightarrow {V_P} = 2.7{\rm{kN}}$
$\sum {{M_V}}=0 \Rightarrow 1.5 \times {V_R}-2 \times {F_R}=0 \Rightarrow {V_R}=3.96{\rm{ kN}}$
${V_P} + {V_Q} + {V_R}=10 \Rightarrow {V_Q}=3.07{\rm{kN}}$
${F_M} - {F_P} - {V_S}=0 \Rightarrow {V_S}=-9.462{\rm{kN}}$
$\sum {{M_D}}=1.5 \times {V_P} + 2 \times {V_M} + 1.5 \times {V_S}=0 \Rightarrow {V_M}=5.07{\rm{ kN}}$
${A_x}={V_M}=5.07{\rm{ kN}}$
${A_y}={F_M}=-12.162{\rm{ kN}}$
$\sum {{M_M}}=0 \Rightarrow {M_A} + 2 \times {A_x}=0 \Rightarrow {M_A}=-10.14{\rm{ kNm}}$
An illustration of determining unknown support reactions at A is given above. Similarly by considering free body diagram of different parts as shown in Figure 21.2 and applying equlibrium conditions, other support reactions and member forces can also be determined. |
# How do you solve log_3 x-log_3(x+5)=3?
Sep 11, 2016
$x = - \frac{135}{26} = 5 \frac{5}{26}$
#### Explanation:
"If the log terms are being subtracted, then the numbers are being divided"
${\log}_{3} \left(\frac{x}{x + 5}\right) = 3$
"Both sides must be logs, or both sides must be numbers"
${\log}_{3} 3 = 1 \text{ } \rightarrow 3 {\log}_{3} 3 = 3 \times 1 = 3$
${\log}_{3} \left(\frac{x}{x + 5}\right) = 3 {\log}_{3} 3$
${\log}_{3} \left(\frac{x}{x + 5}\right) = {\log}_{3} {3}^{3}$
$\therefore \frac{x}{x + 5} = \frac{27}{1} \text{ } \left({3}^{3} = 27\right)$
$27 \left(x + 5\right) = x$
$27 x + 135 = x$
$26 x = - 135$
$x = - \frac{135}{26}$ |
## The Method of Partial Fractions
Consider the Laplace transform:
Some manipulations must be done before Y(s) can be inverted since it
does not appear directly in our table of Laplace transforms.
As we will show below:
Now, we can invert Y(s). From the table, we see that the inverse of
1/(s-2) is exp(2t) and that inverse of 1/(s-3) is exp(3t). Using the
linearity of the inverse transform, we have
The method of partial fractions is a technique for decomposing functions
like Y(s) above so that the inverse transform can be determined in a
straightforward manner. It is applicable to functions of the form
where Q(s) and P(s) are polynomials and the degree of Q is less than
the degree of P
. For simplicity we assume that Q and P have real
coefficients. We consider the following cases:
P(s) is a Quadratic with 2 Real Roots
The denominator can be factored: s^2-5s+6=(s-2)(s-3). The denominator
has 2 real roots. In this case we can write
Here A and B are numbers. This is always possible if the numerator is
of degree 1 or is a constant. The goal now is to find A and B. Once we
find A and B, we can invert the Laplace transform:
If we add the two terms on the RHS, we get
Comparing the LHS and RHS, we have:
The denominators are equal. The two sides will be equal if the numerators
are equal. The numerators are equal for all s if the coefficients of s and the
constant term are equal. Hence, we get the following equations:
This is a system of 2 linear equations with 2 unknowns. It can be
solved by standard methods. We obtain A=4 and B=-2. Hence
If the denominator has n distinct roots:
then the decomposition has the form
where A_1, A_2, ..., A_n are unknown numbers. The inverse
transform is
The unknown coefficients can be determined using the same
technique as in the case of only 2 factors, as shown above.
P(s) is a Quadratic with a Double Root
Consider the example:
The denominator has double root -2. The appropriate decomposition
in this case is
Here A and B are numbers. From the table the inverse transform
of 1/(s+2) is exp(-2t) and the inverse transform of 1/(s+2)^2 is
texp(-2t). Hence, the inverse transform of Y(s) is
Using the same technique as in the case of distinct roots above,
it can be shown that A=1 and B=-2.
P(s) is a Quadratic and has Complex Conjugate Roots
Consider the example:
The roots of the denominator are -2+/-i. We can complete the
square for the denominator. We have
Hence, we have
Note the denominator (s+2)^2+1 is similar to that for Laplace
transforms of exp(-2t)cos(t) and exp(-2t)sin(t). We need to
manipulate the numerator. Note that in the formula in the table,
we have a=-2 and b=1. We look for a decomposition of the form
If we can find A and B, then:
The inverse transform is
We can determine A and B by equating numerators in the expression
Comparing coefficients of s in the numerator we conclude 3=A.
Comparing the constant terms we conclude 2A+B=9. Hence
A=3 and B=3.
P(s) is of Degree 3 and has 1 Real Root and 2 Complex Conjugate Roots
Consider the example:
The roots of the denominator are 1 and -2+/-i. In this case we
look for a decomposition of the form:
where A, B and C are unknowns. The inverse transform of A/(s-1)
is Aexp(t). Once B and C are determined the second term can be
manipulated as in the previous example. |
# What is 11 2 as a mixed number?
## What is 11 2 as a mixed number?
5 1/2
Answer: 11/2 as a mixed number can be written as 5 1/2.
## What’s 11 3 as a mixed number?
323
Answer: 11/3 in mixed fraction is 323.
What is 11 6 as a mixed number?
Example
116
Divide the denominator into the numerator. Remember 116 means 11÷6 11 ÷ 6 .
Identify the quotient, remainder and divisor.
Write the mixed number as quotientremainderdivisor quotient remainder divisor . 156
So, 116=156
What is 11 divided by 2 as a fraction?
Using a calculator, if you typed in 11 divided by 2, you’d get 5.5. You could also express 11/2 as a mixed fraction: 5 1/2.
### What is the mixed number of 7 4?
Answer: 7/4 as a mixed number can be written as 1 3/4.
### What is 11 divided by 4 as a fraction?
Using a calculator, if you typed in 11 divided by 4, you’d get 2.75. You could also express 11/4 as a mixed fraction: 2 3/4.
What is 10 11 as a mixed number?
Since 1011 is a proper fraction, it cannot be written as a mixed number.
What 2divided 11?
11 divided by 2 is 5.5.
#### How do you write a mixed number in simplest form?
A mixed number is the one which can be written in the given form: a b/c. In order to change the mixed number in the simplest form, first the denominator is multiplied with a. Then the product is added in the numerator.
#### How do you divide mixed numbers into fractions?
Step 1: Write the mixed numbers as improper fractions. Step 2: Rewrite the division problem using the improper fractions. Step 3: Use KCF. Step 4: Multiply. Step 5: Simplify. So to divide mixed numbers you should change the fractions to improper fraction and the convert the problem to a multiplication question.
How do you turn a mixed number improper?
To turn mixed numbers into improper fractions, the general rule is to : multiply the denominator (of the fraction in the mixed number) by the whole number in the mixed number, and then adding the numerator (of the fraction in the mixed number) to give the new numerator in the improper fraction.
What are some examples of mixed numbers?
A mixed number is one in which there is a whole number combined with a fraction. Examples include 1 1/2, 3 2/3, and 6 4/9. |
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Lesson 5: Multiply & divide negative fractions
# Dividing negative fractions
Dividing by a fraction is the same as multiplying by its reciprocal. To multiply fractions, we multiply the numerators together and the denominators together. We can simplify fractions by factoring out a fraction equal to 1. Created by Sal Khan.
## Want to join the conversation?
• why can you muliply a negative number by a negative number and make it a positive number?
• The two negative numbers cancel each other out, so if you have -1 * -2 you get 2, the same thing as 1 * 2. Hope this answers your question!
• HOw do I divide two negative fractions?
• Start by flipping the second fraction, so you are multiplying two negative fractions instead of dividing. Any time you multiply two negative numbers the negatives cancel each other out, so you can multiply the fractions as though they were both positive (the product of the numerators over the product of the denominators)
• What's another word for reciprocal?
• The best word to replace reciprocal is "complementary". But I think it is better to use reciprocal.
• With negative fractions, when someone writes this: -1/2, does that mean only the 1 is negative, or are both the 1 and the 2 negative?
• The expression -1/2 is going to be negative. It doesn't matter if you treat it like a fraction (minus one half) or like division (minus one divided by two). The 2 can also be negative instead (as in 1/-2) and it'll still be fundamentally the same. If you had -1/-2, that would actually be a positive number. If you had -(-1/2), that would also be a positive number. If you had -(-1/-2), that would actually be negative.
• What about 4/5 divided by 8/15 In simplest form?
• 4/5 divided by 8/15 is the same as 4/5*15/8 (not sure why). The numerators multiply 4*15 which is 60 and the denominator also does 5*8=40. This gives us 60/40 which is the same as 6/4 which is the same as 3/2 or 1.5.
• wow all these comments from years ago.. but im in 7th grade and u guys are like in college or highschool now have fun guys!! stay safe all of you/
• Oh my goodness, same! I'm currently in 7th and all the comments are like 8-7 years ago!
• Could we have a round of applause for Sal he helps many people to get better at math and more!
• Pls up vote this if you agree
• (KEEP, CHANGE, FLIP) to make it easy to remember. KEEP the first fraction, CHANGE the sign, FLIP the second fraction and solve
• K=Kentucky C=Chicken F=Fried |
# 5.4 Linear inequalities
Page 1 / 1
## Investigation : inequalities on a number line
Represent the following on number lines:
1. $x=4$
2. $x<4$
3. $x≤4$
4. $x≥4$
5. $x>4$
A linear inequality is similar to a linear equation in that the largest exponent of a variable is 1. The following are examples of linear inequalities.
$\begin{array}{ccc}\hfill 2x+2& ≤& 1\hfill \\ \hfill \frac{2-x}{3x+1}& ≥& 2\hfill \\ \hfill \frac{4}{3}x-6& <& 7x+2\hfill \end{array}$
The methods used to solve linear inequalities are identical to those used to solve linear equations. The only difference occurs when there is amultiplication or a division that involves a minus sign. For example, we know that $8>6$ . If both sides of the inequality are divided by $-2$ , $-4$ is not greater than $-3$ . Therefore, the inequality must switch around, making $-4<-3$ .
When you divide or multiply both sides of an inequality by any number with a minus sign, the direction of the inequality changes. For this reason you cannot divide or multiply by a variable.
For example, if $x<1$ , then $-x>-1$ .
In order to compare an inequality to a normal equation, we shall solve an equation first. Solve $2x+2=1$ .
$\begin{array}{ccc}\hfill 2x+2& =& 1\hfill \\ \hfill 2x& =& 1-2\hfill \\ \hfill 2x& =& -1\hfill \\ \hfill x& =& -\frac{1}{2}\hfill \end{array}$
If we represent this answer on a number line, we get
Now let us solve the inequality $2x+2≤1$ .
$\begin{array}{ccc}\hfill 2x+2& ≤& 1\hfill \\ \hfill 2x& ≤& 1-2\hfill \\ \hfill 2x& ≤& -1\hfill \\ \hfill x& ≤& -\frac{1}{2}\hfill \end{array}$
If we represent this answer on a number line, we get
As you can see, for the equation, there is only a single value of $x$ for which the equation is true. However, for the inequality, there is a range of values for which the inequality is true. This is the main difference between an equation and an inequality.
Solve for $r$ : $6-r>2$
1. $\begin{array}{c}\hfill -r>2-6\\ \hfill -r>-4\end{array}$
2. When you multiply by a minus sign, the direction of the inequality changes.
$r<4$
Solve for $q$ : $4q+3<2\left(q+3\right)$ and represent the solution on a number line.
1. $\begin{array}{ccc}\hfill 4q+3& <& 2\left(q+3\right)\hfill \\ \hfill 4q+3& <& 2q+6\hfill \end{array}$
2. $\begin{array}{ccc}\hfill 4q+3& <& 2q+6\hfill \\ \hfill 4q-2q& <& 6-3\hfill \\ \hfill 2q& <& 3\hfill \end{array}$
3. $\begin{array}{ccccc}\hfill 2q& <& 3\hfill & \mathrm{Divide both sides by}\phantom{\rule{2pt}{0ex}}2\hfill & \\ \hfill q& <& \frac{3}{2}\hfill & \end{array}$
Solve for $x$ : $5≤x+3<8$ and represent solution on a number line.
1. $\begin{array}{ccc}\hfill 5-3≤& x+3-3& <8-3\hfill \\ \hfill 2≤& x& <5\hfill \end{array}$
## Linear inequalities
1. Solve for $x$ and represent the solution graphically:
1. $3x+4>5x+8$
2. $3\left(x-1\right)-2≤6x+4$
3. $\frac{x-7}{3}>\frac{2x-3}{2}$
4. $-4\left(x-1\right)
5. $\frac{1}{2}x+\frac{1}{3}\left(x-1\right)≥\frac{5}{6}x-\frac{1}{3}$
2. Solve the following inequalities. Illustrate your answer on a number line if $x$ is a real number.
1. $-2≤x-1<3$
2. $-5<2x-3≤7$
3. Solve for $x$ : $7\left(3x+2\right)-5\left(2x-3\right)>7$ Illustrate this answer on a number line.
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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# NCTB Class 8 Math Chapter Three Exercise 3 Solution
NCTB Class 8 Math Chapter Three Exercise 3 Solution by Math Expert. Bangladesh Board Class 8 Math Solution Chapter 3 Measurements Exercise 3 Solution.
Board NCTB Class 8 Subject Mathematics Chapter 3 Chapter Name Measurements Exercise 3 Solution
(10) The length of a pond is 60 meters and the breadth is 40 meters. If the breadth of its bank is 3 meters, find the area of the bank.
Solution:
Length of pond = 60m
Breadth of pond = 40m
Area = (60X40) Sqm
= 2400 Sqm
Length with bank of 3m = (60+ 3X2) m
= 66m.
Therefore, Breadth with bank of 3m = (40+ 3X2) m
= 46m.
Therefore, New area = (66X46) Sqm= 3036 Sqm
Therefore, area of the bank = (3036-2400) Sqm
= 636 Sqm.
(11) The area of a rectangle is 10 acres and its length is 4 times the breadth. What is the length of the rectangle in meters?
Solution:
We know, 1 acre = 4046.86 Sqm.
Therefore, 10 acre = (4046.86X10) Sqm
= 40468.6 Sqm
Let breadth = x m.
Length = 4x m.
Therefore, Area = (x X 4x) Sqm
= 4x2 Sqm.
Therefore, 4x2 = 40468.6
= x2 = 40468.6/4
= 10117.15
= x= √10117.15
= 100.5840
Therefore, Length (4X100.5840)
= 402.3361
(12) The length of a rectangular house is one and a half time its breadth. If the area of the house is 216 sq. meters, what is its perimeter?
Solution:
Let, breadth of house = X m.
Therefore, length of house = (x X 1 ½) = 3x/2 m.
Therefore, Area = (x X 3x/2) Sqm
= 3x2/2 Sqm.
Therefore, 3x2/2 = 216.
= 3x2 = 216X2
= x2 = 216X2/3
= x= √144
= x= 12
Therefore, Breadth = 12m
Length = 12X 1 ½
= 12X 3/2
= 18m.
Therefore, Perimeter = 2 (l+b)
= 2 (12+18)
(13) The base of triangular region is 24 metres and the height is 15 metres 50 cm. Find its area.
Solution:
Triangular region, base = 24m.
Height = 15m 50cm.
= (15+ 50/100)
= 1500+50/100
= 31/2 m.
Therefore, Area = ½ X b X h
= ½ X 24 X 31/2
= 186 Sqm (Answer).
(14) The length of a rectangle is 48 metres and is breadth is 32 metres 80 cm. there is a 3 metres wide road around outside. What is the area of the road?
Solution:
Length of rectangle = 48m.
Breadth of rectangle = 32m 80cm.
= (32+ 80/100) m
= 32.8m
Therefore, Area (l X b)
= (48X32.8)
= 1574.4 Sqm.
There is a road outside of breadth 3m.
Length of rectangle with road
= (48+ 3X2)
= 54m
Therefore, Breadth of rectangle = (32.8+ 3X2)
= 38.8m
Therefore, New Area = (54X38.8) Sqm
= 2055.2 Sqm
Therefore, Area of road = (2095.2-1574.4) Sqm
= 520.8 Sqm. (Answer)
(15) The length of one side of a square is 300 metres and around its outside, there is a 4 metres road wide. Find the area of the road.
Solution:
Length of one side square = 300m
Area = (300X300) Sqm
= 90.000 Sqm
Therefore, Length of the square with 4m road
(300+ 4X2) m
= 308m.
Therefore, New Area = (308X308)
= 94864 Sqm
Therefore, area of road = (94864-90000) Sqm
= 4864 Sqm. (Answer)
(16) The area of a triangle land is 264 sq. metres. Find the height if the base is 22 metres.
Solution:
Let, height of the triangular land = x m.
Therefore, Base of the triangular land = 22m.
Therefore, Base of the area = ½ X b X h
= ½ X 22 X x
= 11x Sqm.
Therefore, 11x = 264
= x = 264/11
= 24m.
Therefore, Height of the triangular land = 24m (Answer).
(17) A reservoir contains 19200 litres of water. Its depth is 2.56 metres and its breadth is 2.5 metres. What is its length?
Solution:
Reserve contains 19200 lit.
Therefore, we know, 1 lit = 1000 Cu2
Therefore, 19200 lit = (19200X1000) Cu2
= 19200000 Cu2
= 19.2 Cu2
Therefore, depth = 2.56m
Therefore, breadth = 2.5m.
Let, Length = x m.
Therefore, Area = (2.56X2.5X x) Cu2
= 6.4x Cu2
Therefore, 6.4x = 19.2
= x= 19.2/6.4 = 3
Therefore, length = 3m (Answer)
(18) Gold is 19.3 times heavier than water. The length of a rectangular gold bar is 7.8 cm, the breadth is 6.4 cm and the height is 2.5 cm. What is the weight of the gold bar?
Solution:
Rectangular gold bar,
Length = 7.8 cm
Breadth = 6.4 cm
Height = 2.5 cm
Therefore, Area = (7.8X6.4X2.5)
= 124.8 Cm3
We know, 1 Cm3 water weight = 1gm.
124.8 Cm3 water weight = (1X124.8)
= 124.8 gm
Therefore, Weight of gold bar = (124.8X19.3) gm
= 2408.64 gm.
(19) The length of a small box is cm 2.4 mm, the breadth is 7 cm 6.2 mm and the height is 5 cm 8 mm. What is the volume of the box in cubic centimetre.
Solution:
BOX, Length = 15cm 24cm
= (15+ 24/10) cm
= (15+0.24)
= 15.24 cm.
Breadth = (7cm + 6.2/10)
= (7+0.62)
= 7.62 cm.
Height = 5cm 8mm
= (5+0.8) = 5.8cm.
Area = (l X b X h) cm3
= (15.24X4.62X5.8)
= 673.547 cm3 (Answer)
(20) The length of a rectangular reservoir is 5.5 metres, the breadth is 4 metres and the height is 2 metres. If the reservoir is full of water, what is the volume of water in litres and its weight in kg?
Solution:
Rectangular reservoir, length = 5.5 m.
= (5.5X100) = 550 cm.
Breadth = 4 m
= (4X100)
= 400 cm.
Height = 2m.
= 200cm
Therefore, Area = (550X400X200) cm3
= 44000000
We know, 1000 cm3 = 1lit
1 cm3 = 1/1000 lit
Therefore, 44000000 cm3 = 44000000X1/1000
= 44000 lit
Again, 1 lit water weight 1 kg.
44000 lit water weight (44000X1)
= 44000 kg. (Answer)
(21) The length of a rectangular field is 1.5 times its breadth. An amount of Tk. 10260 is spent to plant grass at Tk. 1.90 per sq. metres. How much money will be spent at Tk. 2.50 per metre to erect a fence around that field?
Solution:
1.90 Tk spend in 1 Sqm.
1 Tk spend in 1/1.90 Sqm
Therefore, 10260 Tk spend in 10260X1/1.90 = 5400 Sqm
Therefore, Let breadth = x m.
Therefore, Length = (x X 1.5) m= 1.5 cm.
Therefore, from question, x X 1.5x = 5400
= 1.5x2 = 5400/1.5
Therefore, x2 = 5400/1.5
= 3600
Therefore, x = 60m
Therefore, length = (60X1.5) = 90m.
Breadth = 60 cm.
Therefore, perimeter = 2(90+60) m
= 300 m.
Therefore, To erect a fence around the field
= (300X2.50) Tk
= 750 Tk.
(22) An amount of Tk. 7200 is spent to cover the floor of a room by carpet. An amount of Tk. 576 would be saved if the breadth were 3 metres less. What is the breadth of the room?
Solution:
3m spent 576 Tk.
Therefore, 1m spent 576/3 Tk = 192 Tk.
Now, 192 Tk spent when breadth 1m.
Therefore, 1 Tk spend when breadth 1/192 m.
Therefore, 7200 Tk spent when breadth 1X7200/192 m
= 37.5m.
(23) Around inside a rectangular garden of length 80 metres and breadth 60 metres, there is a road of breadth 4 metres. How much money will be spent to construct that road at Tk. 7.25 per square metre?
Solution:
Length of the garden with road = 80m.
Breadth of the garden with road = 60m.
Therefore, Area = (80X60) Sqm
= 4800 Sqm.
Without the road, length of garden = {80-(4+4)} m
= 72 m
Without the road Breadth of garden = 60-(2X4) m
= 52m
Therefore, Area = (72X52) Sqm
= 3744 Sqm
Therefore, Area of road = (4800-3744) Sqm
= 1056 Sqm.
Total cost to construct the road
= (1056X7.25) Tk.
= 7656 Tk (Answer)
(24) A square open reservoir of depth 2.5 metres contains 28,900 litres of water inside. How much money will be spent to put a lead sheet in the innerside at Tk. 12 50 per sq. metres?
Solution:
28900 lit = 28900/1000 m3
= 28.9 m3
Let, length of the reservoir = x m.
Therefore, Area = x2 Sqm.
And, depth = 2.5m
Therefore, Area = (x2 X 2.5) = 2.5x2 .m3
Therefore, 2.5x2 = 28.9
= x2= 28.9/2.5
= 11.56
Therefore, x= √11.56 = 3.4m
Therefore, Area of (1 side surface) innerside = (3.4X2.5) Sqm
Therefore, Area of 4 side surface innerside = (4X3.4X2.5) Sqm
= 34 Sqm.
Area of lower surface = (3.4X3.4) Sqm
= 11.56 Sqm
Therefore, Total money spent = (34+11.56) Sqm X12.50 Rs.
= (45.56X12.50)
= 569.50 Tk (Answer)
(25) The length of the floor of a house is 26 metres and breadth is 20 metres. How many mats of length 4 metres and breadth 2.5 metres will be required to cover the floor completely? How money will be spent if the price of each mat is Tk. 27.50?
Solution:
Area of the floor = (26X20) Sqm
= 520 Sqm
Therefore, Area of 1 mat = (4X2.5) m2
= 10 m2
Therefore, Mat required = 520/10 = 52 pc.
Therefore, 1 pc mat cost = 27.50 Tk.
52 pc mat cost = (52X27.50) Tk
= 1430 Tk.
(26) The length of a book is 25 cm and the breadth is 18 cm. The number of pages of the book is 200 and the thickness of each page is 0.1 mm. Find the volume of the book.
Solution:
200 page = 100 sheet [Therefore, 2pg= 1 sheet]
Thickness of 1 page is 0.1 mm.
Therefore, Thickness of 100 page (100X0.1)
= 10 mm
= 1 Cm.
Therefore, Volume = (2.5X18X1 cm)
= .450 cm3 (Answer)
(27) The length of a pond is 32 metres, breadth is 20 metres and the depth of water of the pond is 3 metres. The pond is being made empty by a machine which can remove 0.1 cubic metres of water per second. How much time will be required to make the pond empty?
Solution:
Volume of the pond= (32X20X3) cm3
= 1920 m3
Therefore, 0.1m3 water removes in 1 sec.
Therefore, 1 m3 water removes in 1/0.1 sec
Therefore 1920 m3 water remove in 1920/0.1 sec
= 19200 sec.
[Therefore, 1hr = 60X60 = 3600 sec]
= 19200/60X60
= 16/3 hr.
= 5hr 20m (Answer)
(28) A solid cube of sides 50cm is kept in an empty reservoir of length 3 metres, breadth 2 metres and height 1 metre. The cube is taken out after filling the reservoir with water. What is the depth of water now ?
solution :
volume of reservoir = (3x2x1)m= 6m3
Another cube of 50 cm length ,volume is= (50)3 cm3
= 125000cm3
= 125000/1000000
= 0.125 m3
(29) A breadth of a room is 2/3 times of its length. The length and height of the room Are 15m and 4m respectively. The floor of the room are set with stone of the size 50 sq.cm leaving 1m margin in all sides. Air is 0.00129 times heavier than water.
a) find out the perimeter of the room
b) how many pieces of stone will be needed ?
c) how much air is their in the room ?
Solution :
a) Given , length of room = 15m
Breadth of room = 15 x 2/3= 10m.
Therefore, perimeter = 2(15+10)= 50m.
b) Without blank space , length of the room = 15- (1+1)m
= 13 m
Breadth of the room = 10- (1+1)m
= 8m
Area = (13×8) sq.m = 10452m.
Length of the stone of the square size = 50 cm.
= 0.5m.
Therefore, Area= (0.5)2
= 0.25 sqm.
Therefore, pieces of stone will be needed = 104/0.25
= 416 pcs.
c) volume of the room = (15x10x4) m3
= 600m3
= 600 x 1000000 cm3
= 60000000cm3
Weight of the 1cm3 air = 0.00129gm.
Therefore quantity of the air in the room
= 60000000 x 0.00129
= 774000gm.
= 774 kg.
(31) The area of a rectangular school campus is 10 acres. Its length is four times the breadth. The size of the auditorium is 40m x 35m x 10m and the thickness of the wall is 15 cm.
Solution:
a) We know , 2.47 acre = 1 hector
Therefore 1 acre = ½.47 hector
Therefore, 10 acre = 10/2.47 hector
= 4.048582996 hector
= 4.05 hector
b) Let the breadth of school campus = X m.
Length ; ; ; = 4x m.
Therefore, area = (X . 4x ) sqm
= 4x2
Given the area of campus = 10 acre
= (10×4046.86) sqm
= 40468.6 sqm
Therefore 4x2 = 40468.6
X2 = 40468.6/4
X2= 10117.15
X= √10117.15
= 100.58m
Therefore, length = (4 x 100.58) m = 402.32m.
Therefore length of the boundary wall = 2(L+b) m
= 2 (402.32 + 100.58) m
= 1005.8 m.
c) Thickness of the wall = 15 cm.
= 0.15 m.
therefore volume of two wall to the length = ( 40+0.15+10+2) m3
= 120m3
to, Breadth volume of two wall = { 35- (0.15 x2 )} x0.15 x10 x 2 m3
= 34.7 x 0.5 x 10 x 2
= 104.1 m3
therefore volume of 4 walls = (120+104.1)m3
= 224.1m3
Updated: March 20, 2021 — 1:34 pm |
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Area of the Trapezoid | AMC 8, 2002 | Problem 20
Try this beautiful problem from Geometry based on Area of Trapezoid.
Area of the Trapezoid - AMC- 8, 2002 - Problem 20
The area of triangle XYZ is 8 square inches. Points A and B are midpoints of congruent segments XY and XZ . Altitude XC bisects YZ.What is the area (in square inches) of the shaded region?
• $6$
• $4$
• $3$
Key Concepts
Geometry
Triangle
Trapezoid
Answer:$3$
AMC-8 (2002) Problem 20
Pre College Mathematics
Try with Hints
Given that Points A and B are midpoints of congruent segments XY and XZ and Altitude XC bisects YZ
Let us assume that the length of YZ=$x$ and length of $XC$= $y$
Can you now finish the problem ..........
Therefore area of the trapezoid= $\frac{1}{2} \times (YC+AO) \times OC$
can you finish the problem........
Let us assume that the length of YZ=$x$ and length of $XC$= $y$
Given that area of $\triangle xyz$=8
Therefore $\frac{1}{2} \times YZ \times XC$=8
$\Rightarrow \frac{1}{2} \times x \times y$ =8
$\Rightarrow xy=16$
Given that Points A and B are midpoints of congruent segments XY and XZ and Altitude XC bisects YZ
Then by the mid point theorm we can say that $AO=\frac{1}{2} YC =\frac{1}{4} YZ =\frac{x}{4}$ and $OC=\frac{1}{2} XC=\frac{y}{2}$
Therefore area of the trapezoid shaded area = $\frac{1}{2} \times (YC+AO) \times OC$= $\frac{1}{2} \times ( \frac{x}{2} + \frac{x}{4} ) \times \frac{y}{2}$ =$\frac{3xy}{16}=3$ (as $xy$=16)
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Bank PO Questions
Q:
What was the day on 15th august 1947 ?
A) Friday B) Saturday C) Sunday D) Thursday
Explanation:
15 Aug, 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947)
Odd days in 1600 years = 0
Odd days in 300 years = 1
46 years = (35 ordinary years + 11 leap years) = (35 x 1 + 11 x 2)= 57 (8 weeks + 1 day) = 1 odd day
Jan. Feb. Mar. Apr. May. Jun. Jul. Aug
( 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 ) = 227 days = (32 weeks + 3 days) = 3 odd days.
Total number of odd days = (0 + 1 + 1 + 3) = 5 odd days.
Hence, as the number of odd days = 5 , given day is Friday.
942 139606
Q:
Today is Monday. After 61 days, it will be :
A) Tuesday B) Monday C) Sunday D) Saturday
Explanation:
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.
545 67995
Q:
If each side of a square is increased by 25%, find the percentage change in its area?
A) 65.25 B) 56.25 C) 65 D) 56
Explanation:
let each side of the square be a , then area = ${a}^{2}$
As given that The side is increased by 25%, then
New side = 125a/100 = 5a/4
New area = ${\left(\frac{5a}{4}\right)}^{2}$
Increased area= $\frac{25{a}^{2}}{16}-{a}^{2}$
Increase %=$\frac{\left[9{a}^{2}/16\right]}{{a}^{2}}*100$ % = 56.25%
237 54225
Q:
If 20% of a = b, then b% of 20 is the same as :
A) 4% of a B) 6% of a C) 8% of a D) 10% of a
Explanation:
20% of a = b => (20/100)a = b
b% of 20 =(b/100) x 20 = (20a/100) x (1/100) x (20) = 4a/100 = 4% of a.
174 40465
Q:
The tropic of cancer does not pass through which of these Indian states ?
Explanation:
The tropic of cancer passes through 8 Indian states.
They are Gujarat, Rajasthan, Madhya Pradesh, Chhattisgarh, West Bengal, Jharkand, Tripura and Mizoram.
Filed Under: Indian Geography
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO
146 37418
Q:
A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day?
A) 48 min. past 12. B) 46 min. past 12. C) 45 min. past 12. D) 47 min. past 12.
Explanation:
Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
$\frac{145}{6}$ hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = $24*\frac{6}{145}*29$ hrs of the correct clock
= 28 hrs 48 min of the correct clock.
Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.
189 34817
Q:
A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?
A) 1/4 B) 1/2 C) 3/4 D) 7/12
Explanation:
Let A, B, C be the respective events of solving the problem and be the respective events of not solving the problem. Then A, B, C are independent event
are independent events
Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4
$\therefore$ P( none solves the problem) = P(not A) and (not B) and (not C)
= $P\left(\overline{)A}\cap \overline{)B}\cap \overline{)C}\right)$
= $P\left(\overline{)A}\right)P\left(\overline{)B}\right)P\left(\overline{)C}\right)$
= $\frac{1}{2}×\frac{2}{3}×\frac{3}{4}$
= $\frac{1}{4}$
Hence, P(the problem will be solved) = 1 - P(none solves the problem)
= $1-\frac{1}{4}$= 3/4
80 32368
Q:
A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):
A) 200, 250, 300 B) 300, 200, 250 C) 200, 300, 400 D) None of these
Explanation:
A's 5 days work = 50%
B's 5 days work = 33.33%
C's 2 days work = 16.66% [100- (50+33.33)]
Ratio of contribution of work of A, B and C =
= 3 : 2 : 1
A's total share = Rs. 1500
B's total share = Rs. 1000
C's total share = Rs. 500
A's one day's earning = Rs.300
B's one day's earning = Rs.200
C's one day's earning = Rs.250 |
# 4.1 Linear functions (Page 14/27)
Page 14 / 27
A line passes through the points, $\text{\hspace{0.17em}}\left(-2,\text{−15}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,-3\right).\text{\hspace{0.17em}}$ Find the equation of a perpendicular line that passes through the point, $\text{\hspace{0.17em}}\left(6,4\right).$
$\text{\hspace{0.17em}}y=–\frac{1}{3}x+6$
Access this online resource for additional instruction and practice with linear functions.
## Key concepts
• Linear functions can be represented in words, function notation, tabular form, and graphical form. See [link] .
• An increasing linear function results in a graph that slants upward from left to right and has a positive slope. A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. A constant linear function results in a graph that is a horizontal line. See [link] .
• Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between y -values by the difference in corresponding x -values of any two points on the line. See [link] and [link] .
• An equation for a linear function can be written from a graph. See [link] .
• The equation for a linear function can be written if the slope $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ and initial value $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ are known. See [link] and [link] .
• A linear function can be used to solve real-world problems given information in different forms. See [link] , [link] , and [link] .
• Linear functions can be graphed by plotting points or by using the y -intercept and slope. See [link] and [link] .
• Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See [link] .
• The equation for a linear function can be written by interpreting the graph. See [link] .
• The x -intercept is the point at which the graph of a linear function crosses the x -axis. See [link] .
• Horizontal lines are written in the form, $\text{\hspace{0.17em}}f\left(x\right)=b.\text{\hspace{0.17em}}$ See [link] .
• Vertical lines are written in the form, $\text{\hspace{0.17em}}x=b.\text{\hspace{0.17em}}$ See [link] .
• Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See [link] .
• A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x - and y -values of the given point into the equation, $\text{\hspace{0.17em}}f\left(x\right)=mx+b,\text{\hspace{0.17em}}$ and using the $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ that results. Similarly, the point-slope form of an equation can also be used. See [link] .
• A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See [link] and [link] .
## Verbal
Terry is skiing down a steep hill. Terry's elevation, $\text{\hspace{0.17em}}E\left(t\right),$ in feet after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by $\text{\hspace{0.17em}}E\left(t\right)=3000-70t.\text{\hspace{0.17em}}$ Write a complete sentence describing Terry’s starting elevation and how it is changing over time.
Terry starts at an elevation of 3000 feet and descends 70 feet per second.
Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?
A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.
$d\left(t\right)=100-10t$
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Newton's Second Law
## The acceleration of an object equals the net force acting on the object divided by the object’s mass.
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Newton's Second Law
These boys are racing around the track at Newton’s Skate Park. The boy who can increase his speed the most will win the race. Tony, who is closest to the camera in this picture, is bigger and stronger than the other two boys, so he can apply greater force to his skates.
Q: Does this mean that Tony will win the race?
A: Not necessarily, because force isn’t the only factor that affects acceleration.
### Force, Mass, and Acceleration
Whenever an object speeds up, slows down, or changes direction, it accelerates. Acceleration occurs whenever an unbalanced force acts on an object. Two factors affect the acceleration of an object: the net force acting on the object and the object’s mass. Newton’s second law of motion describes how force and mass affect acceleration. The law states that the acceleration of an object equals the net force acting on the object divided by the object’s mass. This can be represented by the equation:
Acceleration=NetforceMass\begin{align*}\mathrm{Acceleration=\frac{Net\;force}{Mass}}\end{align*}
or a=Fm\begin{align*}\mathrm{a=\frac{F}{m}}\end{align*}
Q: While Tony races along on his rollerblades, what net force is acting on the skates?
A: Tony exerts a backward force against the ground, as you can see in the Figure below, first with one skate and then with the other. This force pushes him forward. Although friction partly counters the forward motion of the skates, it is weaker than the force Tony exerts. Therefore, there is a net forward force on the skates.
### Direct and Inverse Relationships
Newton’s second law shows that there is a direct relationship between force and acceleration. The greater the force that is applied to an object of a given mass, the more the object will accelerate. For example, doubling the force on the object doubles its acceleration.
The relationship between mass and acceleration is different. It is an inverse relationship. In an inverse relationship, when one variable increases, the other variable decreases. The greater the mass of an object, the less it will accelerate when a given force is applied. For example, doubling the mass of an object results in only half as much acceleration for the same amount of force.
Q: Tony has greater mass than the other two boys he is racing (pictured in the opening image). How will this affect his acceleration around the track?
A: Tony’s greater mass will result in less acceleration for the same amount of force.
### Summary
• Newton’s second law of motion states that the acceleration of an object equals the net force acting on the object divided by the object’s mass.
• According to the second law, there is a direct relationship between force and acceleration and an inverse relationship between mass and acceleration.
### Review
1. State Newton’s second law of motion.
2. How can Newton’s second law of motion be represented with an equation?
3. If the net force acting on an object doubles, how will the object’s acceleration be affected?
4. Tony has a mass of 50 kg, and his friend Sam has a mass of 45 kg. Assume that both friends push off on their rollerblades with the same force. Explain which boy will have greater acceleration.
### Explore More
Use this resource to answer the questions that follow.
1. What is the rate of change of momentum?
2. How is force proportional to acceleration?
3. How is force proportional to mass?
### Notes/Highlights Having trouble? Report an issue.
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What is the compound interest on 16000?
Table of Contents
What is the compound interest on 16000?
∴ The compound interest is Rs. 5160.
What will be the compound interest be on 16000 for 1 and a half years at the rate of 10% per annum compounded half yearly?
amount of compound interest would be 2522 Rs 1 1/2) × 2 = 3 half years .
What will be the compound interest on rupees 16000 at 20% per annum for 9 months compounded quarterly?
The compound interest on ₹ 16000 for 9 months at 20% p.a, compounded quarterly is ₹ 2522 – Mathematics.
How much would a sum of 16000 amount to in 2 years at 10% per annum if the interest is compounded half yearly?
Hence a sum of rs. 16000 amount to be 20736000 in 2 years at 10% per annum, if the interest is compounded half yearly.
What is the formula of per annum compounded annually?
If the given principal is compounded annually, the amount after the time period at percent rate of interest, r, is given as: A = P(1 + r/100)t, and C.I. would be: P(1 + r/100)t – P .
What will be the compound interest on 15000 for 2 years at 10 per annum?
Therefore, compound interest is 3150 Rs.
How much would a sum of Rs 16000 amount to in 1 year at 20% per annum if the interest is compounded quarterly?
The compound interest on Rs. 16000 for 9 months at 20% p.a, compounded quarterly is Rs. 2522.
What is compound interest per annum?
⇒ R = 6% per annum.
How much would a sum of 16000 amount to in one year at 20% per annum if the interest is compounded quarterly?
16000 for 9 months at 20% p.a, compounded quarterly is Rs. 2522.
What will be the amount on a sum of Rs 16000 at the rate of 12% per annum compound interest in 2.5 years if the interest is compounded 10 monthly?
∴ Compound interest = Amount – Principal = 19360 – 16000 = Rs. 3360.
At what rate of compound interest per annum will a sum of 1200 become 1348.32 in 2 years?
Then, 1200×(1+R100)2=1348.321200×(1+R100)2=1348.32% Was this answer helpful?
At what rate of compound interest per annum will a sum of ₹ 10 00000 become ₹ 12 25043 in 3 years?
A = P (1 + r/100)^t (where A is amount having value ₹ 1225043, P is principal having value ₹1000000, r is rate and t is time having value 3 years). Assumption: Let’s say that the rate per annum is x. Therefore, the rate is 7%.
What is the compound interest on 10000 for 3 years?
When it comes to savings and investments, the compound interest on \$10,000 for three years at 6% per annum is \$1,910.16. Below is a sample calculation to get the toal interest amount: 10,000 x.06 = 600 (first year)
What is a 10% interest rate compounding semi-annually?
For example, a loan with a 10% interest rate compounding semi-annually has an interest rate of 10% / 2, or 5% every half a year. For every \$100 borrowed, the interest of the first half of the year comes out to:
How many years will a amount double itself at 10% compounded quarterly?
In how many years will a amount double itself at 10% interest rate compounded quarterly? Ans. t = (log (A/P) / log (1+r/n)) / n = log (2) / log (1 + 0.1 / 4) / 4 = 7.02 years 3. If interest is compounded daily, find the rate at which an amount doubles itself in 5 years?
How much does a 6% mortgage interest rate compound monthly?
For example, a 6% mortgage interest rate amounts to a monthly 0.5% interest rate. However, after compounding monthly, interest totals 6.17% compounded annually. |
Section 1.4 Place Value Systems of Numeration in Other Bases
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1 Section.4 Place Value Systems of Numeration in Other Bases Other Bases The Hindu-Arabic system that is used in most of the world today is a positional value system with a base of ten. The simplest reason for the choice of base ten is the fact that most human beings have ten fingers. However, a positional value system can be created with a base of any counting number greater than one. In this section we are going to examine how numbers can be written in systems in bases other than ten. Why would we write a number in another base? Base 2, also called binary, is the natural base for working with computers. At the basic level, a computer is a collection of billions of bits of information. Each bit is essentially a microelectronic gate that is either open or closed. We represent these positions by the value of the bit being either a or a. An ordered string of 8 bits makes up a byte. There are 26 different possible combinations of s and s in a byte, hence 26 different characters can be encoded with a byte. Base 2 is the logical system for working with these strings of s and s. Computer programmers 4 have also found base 8 ( 2 ) and base 6 ( 2 ) to be useful when working with combinations of bits. Base 8 is also known as octal, and base 6 is known as hexadecimal (or hex). Beyond computer applications, learning how to write numbers in different bases highlights the properties of a place-value system of numeration that we take for granted. Symbols and Notation A system of numeration in base b needs a symbol for the quantities from to ( b ). For quantities less than ten, the symbols,, 2,, 4,, 6, 7, 8, 9 in the Hindu-Arabic system will be used. For bases larger than ten, a single symbol is needed to represent quantities that we normally see as two digit numbers in base ten. Symbols for the numbers ten, eleven, twelve and following will be capital letters, starting with A for ten, B for eleven, C for twelve, and so on. The symbols for base 6 are given below. Number in base Number in base 6 Number in base Number in base A 2 2 B 2 C 4 4 D 4 E 6 6 F
2 Section.4 To identify the base in which a number is written, we will adopt the convention of identifying bases other than ten with a subscript on the right side of a number in a different base. For example, the number sixteen in base sixteen will be written 6. The Expanded Form of a Number The values of each position in a Hindu-Arabic number are powers of ten ,(), (), (), (), The above place values represent, for example, the ten-thousands place, the thousands place, the hundreds place, the tens place, the ones place, etc. (Notice that is a power of ten, since =.) The rule that allows a string of digits from -9 to represent any counting number requires that each digit in a particular place in the string is multiplied by the place s appropriate power of ten. Explicitly writing out this multiplication for a number expressed in a placevalue system is called writing the number in expanded form. Example : Write the base ten number 628 in expanded form. In expanded form, = (Notice that 8 can be written as 8.) A number written in any base may be written in expanded form in a similar fashion. The place values for a binary system (base 2) are ,2, 2, 2, 2, 2, (Notice that = 2.) The expanded form of a binary number multiplies each digit by its corresponding power of 2. Example 2: Write the binary number 2 in expanded form. The number has 7 digits, so the leftmost digit is multiplied by form: = 2. In expanded =
3 Section.4 Example : Write the number 28 in expanded form. In base 8 the place values are , 8, 8, 8, 8, (Notice that = 8.) The number 28 has four digits, so the highest power of 8 used in this number is. In expanded form: 2 28 = Converting from Another Base to Base Ten The expanded form of a number written in any base may be used to convert it to a number in base ten. Simply perform the multiplication and addition indicated in the expanded form in base ten. The resulting answer is the base ten representation of the number. Example 2 Revisited: Write 2 in base ten. The expanded form of 2 is = Performing the multiplication and addition indicated: = ( 64) + ( 2) + ( 6) + ( 8) + ( 4) + ( 2) + = = 89 Therefore, 2 = 89 in base ten
4 Section.4 Example Revisited: Write 28 in base ten. The expanded form of 28 is 2 28= Performing the indicated multiplication and addition gives = = = 229 Therefore, 28 = 229 in base ten. This same procedure works for a number written in any base. Let s try a few more different bases for practice. Example 4: Write the number 242 in base ten. The expanded form of this number is = = = = 67 Thus, 242 = 67 in base ten. Example : Write the number CA7 6 in base. The table of base sixteen numerals shows that C represents 2 and A represents. The expanded form of this number is: - 8 -
5 Section CA7 = Remember that 7 = 7 6. ( ) ( ) = = = 29 Thus, CA76 = 29 in base ten. In summary, you should perform the following steps to convert a number in another base to its equivalent representation in base ten:. Write the number in expanded form with powers of the base 2. Perform the indicated multiplication and addition in base ten.. The answer is the base ten representation of the original number. Converting a Base Ten Number to Another Base The number 26 in base ten is the quantity that is groups of plus 2 groups of plus 6 singles. Writing this quantity in base ten conveys these groupings to the reader. Converting a base ten number to a number in a different base involves division. Why? To write the quantity 26 in another base b, we must determine how many groups of b items can be formed from 26 items. This problem is modeled by the operation of division. This is best illustrated by examples. Example 6: Write the number 9 in base 8. The place values for base 8 are 2..., 8 = 2, 8 = 64, 8, With only 9 items, we cannot form any groups of 2 or 64. We can form groups of 8. Dividing 9 by 8 shows that 9 items can be grouped into 7 sets of 8 with left over The base 8 expanded form of 9 is: ( ) Therefore, 9 = 78 9 =
6 Section.4 Notice that we started by dividing our base ten number 9 by the largest power of 8 that is smaller than the number. This is the first step for translating any base ten number into another base. The next step is to divide the remainder of this division problem by the next smaller power of the base. A succession of division problems is created by dividing the remainder of the previous problem by the next smaller power of the base until the last division problem is dividing by. Example 7: Write the number 444 in base. The place values in base are , = 62, = 2, = 2,, The largest power of smaller than 444 is Now divide 69 by the next lower power of, = 2. So, we will start dividing 444 by 2 = 2. Now divide the remainder, 9, by (which is the next lower power of. 9 4 At this point, the next lower power of is. We know 4 = 4 with left over, so we often do not explicitly write out this division problem in the list. Yet, 4 is the last quotient in our sequence of division problems. The division answers show that 444 is groups of 2 plus 2 groups of 2 plus groups of plus 4 singles. Therefore, 444 =
7 Section.4 Example 8: Write in base 2. The place values in base 2 are , 2 = 64, 2 = 2, 2 = 6, 2 = 8, 2 = 4, 2, The largest power of 2 smaller than is 2, so we begin by dividing by 2 and continue by dividing each remainder by the next smaller power of Thus, = 2. Notice that in Example 8 the division problems include dividing by EVERY power of the base starting with the largest power smaller than the number, even if the quotient is. The zeroes in the number are important to holding the places so each value is correct. Also, notice that the division by was explicitly included. Especially in base 2, it is useful to include this problem to make sure the proper number of digits is included. Example 9: Write 9 in base 2. The place values in base 2 are ( ) ( ) 2..., 2 = 728, 2 = 44, 2, The largest power of 2 that is smaller than 9 is 44. Therefore, begin by dividing by The convention for base 2 is that = A and = B, so 9 = 97B
8 Section.4 In summary, you should perform the following steps to convert a base ten number to another base:. Determine the powers of the base, which are the place values. 2. Determine the largest power of the base that is smaller than the number given.. Divide the number given by the power of the base determined in step Create a sequence of division problems by dividing the remainder of a division problem by the next smaller power of the base. Continue until dividing by.. The quotients, in order, are the digits in each place of the numbers representation in the given base
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## Intermediate Algebra (12th Edition)
$x=5$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $x=\sqrt{x^2+4x-20} ,$ square both sides of the equation and then isolate the variable. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( x \right)^2=\left( \sqrt{x^2+4x-20} \right)^2 \\\\ x^2=x^2+4x-20 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} x^2-x^2-4x=-20 \\\\ -4x=-20 \\\\ x=\dfrac{-20}{-4} \\\\ x=5 .\end{array} Upon checking, $x=5$ satisfies the original equation. |
# CBSE 10th Math Chapter 2, Polynomials - Solutions of NCERT Mathematics Textbook Exercise 2.3
## Class X, Mathematics, NCERT (CBSE) Solutions
### Chapter 2, Polynomials (Division Algorithm for Polynomials
#### Solutions of NCERT Math Textbook Exercise 2.3
(Page 36)
Q 1: Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Solution:
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) Do it yourself by taking hint from the above solutions.
Q 2: Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution:
(i) Try to solve this problem yourself after going through the solved examples given in the NCERT mathematics textbook itself.
Solution: (ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
Solution: (iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Q 3: Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are
SolutionSince two zeroes are
Q 4: On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and the remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
By the division algorithm, we have
f(x) = g(x) x q(x) + r(x)
Or, g(x) x q(x) = f(x)r(x)
Or, g(x) (x – 2) = x3 – 3x2 + x + 2 – (-2x + 4)
Or, g(x) (x – 2) = x3 – 3x2 + 3x – 2
Thus, g(x) is a factor of x3 – 3x2 + 3x – 2 other than the factor (x – 2). Hence, to get g(x) we will divide
(x3 – 3x2 + 3x – 2) by (x – 2),
Q 5: Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) deg p(x) = deg q(x)
Let us assume the division of 6x2 + 2x + 2 by 2
Here, p(x) = 6x2 + 2x + 2
g(x) = 2
q(x) = 3x2 + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) x q(x) + r(x)
Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1)
Hence, division algorithm is satisfied.
Solution: (ii) and (iii) to be loaded soon.
***********
****** |
# Find secant of theta if tangent of theta is the square root of 29 over 4?
## If you use the pythagorean identities, of tangent ^2 +1 = secant ^2, how do you solve?
Apr 7, 2018
$\sec \left(\theta\right) = \frac{3 \sqrt{5}}{4}$
#### Explanation:
So, what we want to solve is this:
If $\tan \left(\theta\right) = \frac{\sqrt{29}}{4}$ ,what is $\sec \left(\theta\right)$?
There's a number of ways to solve this problem, but let's solve it using a trigonometric identity:
${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$
If we solve this expression for $\sec \left(\theta\right) :$
$\sec \left(\theta\right) = \sqrt{{\tan}^{2} \left(\theta\right) + 1}$
Now, all we do is plug in the value we were given for $\tan \left(\theta\right)$:
$\sec \left(\theta\right) = \sqrt{{\left(\frac{\sqrt{29}}{4}\right)}^{2} + 1} = \sqrt{\frac{29}{16} + 1} = \sqrt{\frac{45}{16}}$
..and that's your answer! However, we can polish this up a little so it looks nicer. For one, we can remove the radical in the denominator, since 16 is a perfect square:
$\implies \sec \left(\theta\right) = \frac{\sqrt{45}}{4}$
Now, we also know that $45 = 9 \cdot 5$. Notice that 9 is a perfect square, so we can further simplify:
$\implies \sec \left(\theta\right) = \frac{\sqrt{9} \cdot \sqrt{5}}{4} = \frac{3 \sqrt{5}}{4}$
And that is pretty much as far as we can simplify. So, this would be your final answer.
Hope that helped :) |
# Graph Theory involving bipartite graphs
A bipartite graph has 16 nodes of degree 5, and some nodes of degree 8. We know that all degree-8 nodes are on the left hand side. How many degree 8 nodes can the graph have?
Hi, I am having trouble with this problem. I am not sure how to do it. Any help would be appreciated. Thank you.
Let's define some variables.
Let x be the number of degree-8 nodes on the left hand side.
let y be the number of degree-5 nodes on the left hand side.
let z be the number of degree-5 nodes on the right hand side.
We know the number of edges "leaving" the left hand side equals the number of edges "entering" on the right hand side.
For every degree-8 node on the left hand side, we have 8 edges leaving.
For every degree-5 node on the left hand side, we have 5 edges leaving.
For every degree-5 node on the right hand side, we have 5 edges entering.
Therefore,
$8x + 5y = 5z$
We also know that the total number of degree-5 nodes in the graph is 16.
Therefore,
$y + z = 16$
Let's start solving these equations.
$8x + 5y = 5z$
$8x + 5y + 5z = 5z + 5z$
$8x + 5(16) = 10z$
$8x = 10z - 80$
$x = \frac{5}{4}z - 10$
Let's pause for a moment. We know that x must be an integer, so z must be a multiple of 4. So z is either 0, 4, 8, 12, or 16 ($z \leq 16$ because of the equation $y+z=16$). We also know that $x \geq 0$.
So z = 8, 12, or 16. We should choose $z$ that maximizes $x$. Because $x = \frac{5}{4}z - 10$, $x$ increases with $z$, we should choose the greatest $z$, which is 16. Then $y = 0$ and $x = 10$.
ANSWER: The maximum number of degree-8 nodes in the graph is 10.
The easier way is just to look at the complete bipartite graph of 5 degree-8 nodes and 8 degree-5 nodes. Since 16 is a multiple of 8, you can just lay these complete graphs side by side and you'll end up with the required number of degree-5 nodes. Still, you'd have to go through the whole process if the number of degree-5 nodes is not a multiple of 8. |
# What Is an Indefinite Integral?
2012-06-10
An indefinite integral is the integral whose lower boundary is constant and whose upper boundary is variable and it should be distinguished from a primitive integral that just restores primitive functions. Constants of primitive integration should also be distinguished from constants of indefinite integration that the lower boundary produces. Integration is often said to be the inverse operation of differentiation, but it is mathematically false. Since a set of primitive functions and a set of integrands is not in one-to-one but in many-to-one relationship, integration cannot be an inverse mapping. Differentiation and integration are in the relationship of analysis and synthesis. As a derivative has only partial information on the original function, it cannot restore the whole information.
## 1. The previous explanation of the indefinite integral
What is an indefinite integral? Before I answer the question, let us recognize how it has been explained. There are so many explanations as exponents, but the following is the typical and common one.
An indefinite integral is defined as a set of functions whose derivative is equal to a given function. These functions are named primitive functions or antiderivatives, while the derivative is named an integrand meaning a function to be integrated. An indefinite integral is also called a primitive integral.
As the derivative of a constant is zero, you must add an arbitrary constant C known as the constant of integration so as to construct primitive functions. Let the primitive functions of an integrand f(x) be F(x) + C and we can say:
$f(x)=\frac{d}{dx}\left (F(x)+C \right ) \Leftrightarrow \int f(x) dx=F(x)+C$
Since the integrand f(x) derives from differentiating the primitive functions F(x) + C and integrating the integrand restores the primitive functions, integration is often regarded as an inverse operation of differentiation[1]. The fundamental theorem of calculus seems to demonstrate the inverse relationship between them.
Let f(x) be a continuous function defined on an interval [a, b] and F(x) be its primitive function, continuous on [a, b] and differentiable on the open interval (a, b). For all x in (a, b),
$\frac{d}{dx}\int_a^x f(t)dt = f(x)$
Because of this theorem, some identify an indefinite integral with an integral whose lower boundary is a constant (a) and whose upper boundary is a variable (x) [2], but others disagree on it.
There are four definitions of a primitive function in terms of the integral sign according as the treatment of the constant of integration in addition to this disagreement.
$\; \int f(x) dx$
$\; \int f(x) dx + C$
$\; \int_a^x f(x) dx$
$\; \int_a^x f(x) dx + C$
The lower boundaries a of Eq. 5, 6 is omissible, which however does not mean it has no lower boundaries.
## 2. The indefinite integral should be distinguished from the primitive
The fact of many definitions of a primitive function in terms of the integral sign reflects the confusion of the previous explanations. Although differentiating the four formulae leads to the same derivative, they are not equal. The most questionable of the four is Eq. 5, which Wikipedia (Japanese) adopts[3].
Eq. 5 results in
$\int_a^x f(t)dt =\left [ F(t)+C \right ]_{a}^{x}=\left ( F(x)+C-(F(a)+C) \right )=F(x)-F(a)$
If you regard -F(a) of the last formula as a constant of integration C, it can have the form F(x)+C, but it may not represent all primitive functions. For example, when a cosine function is integrated,
$\int_a^x cos\theta\, d\theta =\left [\, sin\theta+C\, \right ]_{a}^{x}=sinx-cosa$
The domain of -cosa is limited, whatever a may be, as follows
$-1\leq -cosa \leq 1$
So, it cannot represent all constants of the integral.
To prevent such confusion, I propose distinguishing an infinite integral from a primitive integral. That is to say, I define a primitive integral as such that merely constructs primitive functions.
$\int _{P} f(x) dx = F(x) + C$
The subscript “P" is omissible so that this definition is equivalent to the definition of Eq. 3.
On the other hand, I define an indefinite integral as the integral whose lower boundary is constant and whose upper boundary is variable.
$\int_a^x f(t) dt=\left [ \int _{P} f(t) dt \right ]_{a}^{x} = \left [F(t) + C \right ]_{a}^{x}=F(x)-F(a)$
Even if their lower boundary is variable and their upper boundary is constant or both are variables, they can be called indefinite integrals in a broad sense, but the indefinite integral in the narrow sense is the most important owing to the fundamental theorem of calculus (Eq. 2).
Substituting a constant b of Eq. 11 for a variable x makes the indefinite integral a definite integral. It means that the primitive integration is an elemental integration that is used both indefinite and definite integration.
## 3. Integration produces two kinds of constants
What has been called “a constant of integration" is added at the primitive integration and it should be conceptually distinguished from the one that the lower boundary of the indefinite integral produces. So I propose naming the former “a constants of primitive integration" and the latter “a constant of indefinite integration". Constants that differentiation of primitive functions reduces to zero are those of primitive integration, while constants that differentiation at the fundamental theorem of calculus reduces to zero are those of indefinite integration and constants of primitive integration is canceled before differentiation.
$\frac{d}{dx}\int_a^x f(t)dt =\frac{d}{dx} \left [ F(t)+C \right ]_{a}^{x}=\frac{d}{dx}\left ( F(x)+C-(F(a)+C) \right )=\frac{d}{dx}(F(x)-F(a))=\frac{dF(x)}{dx}= f(x)$
Let us recognize the difference of the two constants in a graph. The following figure CI depicts the solutions of a differential equation:
$\frac{dy}{dx}=x^{2}-x-2$
as a direction field and illustrates three sample integral lines.
The blue, brown and turquoise lines represent the following functions respectively:
$y=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x+4$
$y=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x$
$y=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x-4$
They differ in constants of primitive integration by 4. Constants of indefinite integration are the values at lower boundary and three lines have different constants of indefinite integration. Integral (represented by a red line) itself is, however, the same in any line when the lower boundary is 2 and the upper boundary is 3.
## 4. Integration is not the inverse operation of differentiation
Another common mistake is to consider indefinite or primitive integration to be the inverse operation of differentiation. The mathematically defined inverse operation is inverse mapping and we must examine whether integration is inverse mapping or not.
Suppose ƒ is a function mapping elements of a set X to elements of a set Y.
$f\colon X \to Y$
If this mapping is bijection, namely “onto" surjection and “one-to-one" injection, it has a reverse function:
$f^{-1}\colon Y \to X$
Let ƒ be an invertible function and the following universal proposition is true.
$(\forall x) \left (x\in X \wedge f^{-1}\left( \, f(x) \, \right) = x \right )$
For example, multiplication is an inverse operation of division by this standard.
Assuming X is a set of primitive functions and Y is a set of integrands, f corresponds to differentiation and f -1 to integration. Differentiation after integration restores the original function but integration after differentiation does not. So, this universal proposition (Eq. 19) does not apply to differentiation and integration. This is because the set of primitive functions and the set of integrands are not in one-to-one but in many-to-one relationship.
One-to-one mapping is information-preserving, while many-to-one mapping is information-losing. Primitive integration needs the information on a constant of primitive integration and indefinite integration needs the information on a constant of indefinite integration in addition to it. Differentiation loses these two pieces of information, but if we have them, we can restore the original function.
Though differentiation and integration are not in the relationship of inverse operation, they are in the relationship of analysis and synthesis, and therefore we can say integration is the cognition in the opposite direction to differentiation. As Kant’s theory of analytic and synthetic propositions indicates, a product of synthesis has more information than a product of analysis. Generally speaking, integration is more difficult than differentiation. This is because the former requires more information than the latter.
Integral calculus derived from the method of exhaustion of the ancient Greek and was developed independently of differential calculus. Their attempt to reduce the complexity of the whole area by dividing a complex figure into elemental figures whose area was known did not succeeded in exact calculation of the whole area. Old-fashioned systematists might have given up its solution, advocating the holistic proposition, “the whole is more than the sum of the parts." Still reductionism itself is not wrong. They just did not make the most of the determinacy of functions, the invisible part which makes the whole more than the sum of the visible parts.
Integral calculus became easy after James Gregory (1638–1675), Isaac Barrow (1630–1677), Isaac Newton (1643–1727) and Gottfried Leibniz (1646–1716) developed the fundamental theorem of calculus. They succeeded in calculating the area by regarding the function of a boundary line as the rate of area change and integrating the infinitesimal change into the whole area. Integral calculus has been applied to not only area calculating but also any product of variables and their functions. Integral and differential calculus was a successful case of Cartesian reductionism that analyzes the whole into elements and reconstructs them to the whole.
## 5. References
1. 青本 和彦 (著), 砂田 利一 (著).『微分と積分1 － 岩波講座 現代数学への入門 1』 岩波書店 (October 5, 1995). p. 107.
2. 黒田 成俊. 『微分積分 － 共立講座 21世紀の数学 1』 共立出版 (September 1, 2002). p. 145.
3. Wikipedia. “不定積分." 2012年5月24日 (木) 02:39.
4. Pbrks. “Slope field of the equation dy/dx=x^2-x-2." 6 June 2008. Licensed under CC-0. |
# Differentiate the functions with respect to $x: \large \frac{ sin\; (ax + b) } { cos \;( cx + d)}$
$\begin{array}{1 1} a.sin(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d) \\ a.sin(ax+b)sec(cx+d)-csin(ax+b)tan(cx+d)sec(cx+d) \\ a.sin(ax+b)cosec(cx+d)+csin(ax+b)cot(cx+d)sec(cx+d) \\ a.sin(ax+b)cosec(cx+d)-csin(ax+b)cot(cx+d)sec(cx+d)\end{array}$
Toolbox:
• Whenever we see a function of the form $y = \large \frac{u}{v}$, which is a quotient of two other functions with derivatives, we can apply the following quotient rule: $y'=\large \frac{1}{v^2}\;$$( v\large\frac{du}{dx}$$-u \large \frac{dv}{dx}$$) • \; \large \frac{d(sinx)}{dx}$$= cosx$
• $\; \large \frac{d(cosx)}{dx} $$=-sinx • According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). Given y =\large \frac{ sin\; (ax + b) } { cos \;( cx + d)}, let us take u = sin\; (ax + b) and v = cos \;( cx + d) Whenever we see a function of the form y = \large \frac{u}{v}, which is a quotient of two other functions with derivatives, we can apply the following quotient rule: y'=\large \frac{1}{v^2}\;$$ ( v\large\frac{du}{dx}$$-u \large \frac{dv}{dx}$$)$
$\textbf{Step 1}$:
$\Rightarrow v\large\frac{du}{dx} $$= cos(cx+d). d(sinx(ax+b)) According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). \; \large \frac{d(sinx)}{dx}$$= cosx$
We need to differentiate $sin(ax+b)$ using the chain rule.
In this case, Let $g(x) = ax+b \rightarrow g'(x) = a$
$\Rightarrow f'(g(x)) = f'(sinx(ax+b)) = cos(ax+b)$
Therefore $d(sin(ax+b) = a. cos(ax+b)$
$\Rightarrow v\large\frac{du}{dx} $$= a. cos(cx+d).sin(ax+b) \textbf{Step 2}: \Rightarrow u \large \frac{dv}{dx}$$ = sin(ax+b). d(cos(cx+d))$
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
$\; \large \frac{d(cosx)}{dx} $$=-sinx We need to differentiate cos(cx+d) using the chain rule. In this case, Let h(x) = cx+d \rightarrow h'(x) = c \Rightarrow f'h(x) = f'(cx+d) = -sin(cx+d) Therefore d(cos(cx+d) = -c.sin(cx+d) \Rightarrow u \large \frac{dv}{dx}$$ = -c.sin(ax+b).sin(cx+d)$
$\textbf{Step 3}$:
Putting it all together, we get $y' = \large \frac{1}{(cos(cx+d))^2}$$[a.cos(cx+d).sin(ax+b)- (-c.sin(ax+b).sin(cx+d))] \Rightarrow y' = a\large \frac{sin(ax+b)}{cos(cx+d)}$$+ c \large\frac{sin(ax+b) sin(cx+d)}{cos(cx+d)cos(cx+d)}$
$\Rightarrow y' = a.sin(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d)$ |
Solving Linear Inequalities
Most of the rules or techniques involved in solving multi-step equations should easily translate to solving inequalities.
The only big difference is how the inequality symbol switches direction when a negative number is multiplied or divided to both sides of an equation.
In this lesson, I will go over seven (7) worked examples with different levels of difficulty to provide enough practice.
Inequality Symbols with Examples and Illustrations on Number Lines
GREATER THAN
• Symbol:
• Example:
• Graph:
GREATER THAN OR EQUAL TO
• Symbol:
• Example:
• Graph:
LESS THAN
• Symbol:
• Example:
• Graph:
LESS THAN OR EQUAL TO
• Symbol:
• Example:
• Graph:
Examples of How to Solve and Graph Linear Inequalities
Example 1: Solve and graph the solution of the inequality
To solve this inequality, we want to find all values of $x$ that can satisfy it. This means there are almost infinite values of $x$ which when substituted, would yield true statements.
Check the values $x = 0$, $x = 1$, $x = 2$, $x = 3$, $x = 5$, $x = 6$ and $x = 7$.
Which of these $x$-values give back a true statement?
You should agree after performing some back substitutions that only $5$, $6$, and $7$ work; and the rest fail. But the question is, are there more values of $x$ other than the ones mentioned? The answer is yes! Now, let us solve the inequality to figure out the entire set of values that can make it true.
• Write the original problem.
• Add $17$ on both sides to keep the variables on the left side and the constant on the right.
• The inequality is reduced to this after simplifying.
• Divide both sides of the inequality by the coefficient of $x$.
• Use an open hole to indicate that $3$ is not part of the solution. The solution to the inequality $x > 3$ includes all values to the right of $3$ but excluding $3$ itself. Do you see now why all numbers greater than $3$ are solutions?
Example 2: Solve and graph the solution of the inequality
This example illustrates what happens to the inequality symbol when divided by a negative number.
• Write the original problem.
• To isolate variable $x$ to the left of the inequality, I will add both sides by $2$.
• This is how it looks after I simplify using the above step.
• Now, to solve for $x$, I will divide both sides by $– \,3$.
ALWAYS switch the direction of the inequality whenever you divide or multiply a negative number to both sides of the inequality.
Use a close or shaded hole to indicate that $7$ is part of the solution. The solution of the inequality $x \le 7$ includes $7$ and everything to its left.
Example 3: Solve and graph the solution of the inequality
In this problem, I have variables on both sides of the inequality. Although it doesn’t matter where we keep the variable, left or right, it makes sense to be consistent all the time by isolating it on the left side. It’s just a “standard” way, I think.
However, if you try to keep the variable on the right, make sure that you’re aware of their subtleties. For instance, the answer to this problem is $x < – \,6$, which is the same as $– \,6 > x$. They are equivalent because the opening of the inequality is also pointing towards $– \,6$. Therefore, this means that if I switch the variable and the constant in my final answer, I must also change the direction of the symbol to keep the meaning the same.
• Write the original problem.
• I want to keep $x$ on the left. I will do that by adding $6x$ to both sides by
• After the step above, I need to move the constant to the right side.
• Subtract both sides by $7$.
• Simplify
• To finally isolate $x$ on the left, divide both sides by the coefficient of $x$ which is $2$.
Notice that I did not switch the direction of the inequality because I divided both sides by a positive number.
• Use an open hole to indicate that $– \,6$ is not part of the solution. The solution to the inequality $x < – \,6$ includes all values to the left of $– \,6$ but excluding $– \,6$ itself.
Example 4: Solve and graph the solution of the inequality
I constructed this problem to emphasize the step required in dealing with the parenthesis symbol. I know that this won’t throw you off because you’ve seen it before when solving linear equations, right? The step required to get rid of the parenthesis is to apply the distributive property of multiplication over addition. However, I must caution you to be careful in dealing with the signs during the multiplication process. Remember, the product of two terms with the same signs is positive and when signs are unlike, the product is negative.
• Write the original problem. I will remove the parenthesis first by distributing that $– \,4$ into the binomial $\left( {x – 5} \right)$.
• Simplify and be careful when you distribute. Remember, you get a positive product if the signs are the same and negative if the signs are different.
• In solving inequalities, I make it a habit to “always” keep the variable to the left side. Although, keeping it on the right is also correct. This is just a matter of preference. To keep $x$ to the left, subtract both sides by $3x$.
• Since I want the constant to stay on the right, it becomes clear that my next step is to eliminate the $20$ on the left.
• Subtract both sides by $20$.
• Obviously, I will divide both sides by the negative coefficient and switch the inequality.
• To solve for $x$, divide both sides by $– \,7$ which leaves us the final answer.
• Use a close or shaded hole to indicate that $5$ is part of the solution. The solution of the inequality $x \ge 5$ includes $5$ and everything to its right.
Example 5: Solve and graph the solution of the inequality
My general approach here is to immediately eliminate the parentheses using the distributive property, combine similar terms on both sides, and finally keep $x$ on the left and the constant to the opposite side.
• Write the original problem. I will apply the distributive property twice on the left side for the two parentheses.
For the right side, they are similar terms so I will simply combine them.
• Simplify. At this point, I will further combine like terms on the left. Combine the $x$’s and the constants together.
• This is what I got after doing the above step.
• Move the constants to the right by adding both sides by $6$.
• Simplify.
• Now, move all variables to the left by adding both sides by $4x$.
• Divide both sides by $– \,3$ to isolate $x$. However, I must change the orientation of the inequality symbol since I divided both sides by a negative number.
• Use an open hole to indicate that $– \,2$ is not part of the solution. The solution of the inequality $x > – 2$ implies all the values to the right of $– \,2$ but excluding $– \,2$.
Example 6: Solve and graph the solution of the inequality
The “complexity” of this problem should not bother you. The key to successfully solving this is to apply all the techniques you have learned so far from our previous examples. If you need a review, please feel free to take a look back.
Try to solve this problem without looking at the detailed solution. Whenever you think you’re done, compare what you have on paper with the answer below.
• Write the original problem. Get rid of the parentheses on both sides of the inequality by applying the distributive property.
• Simplify. Combine similar terms on both sides.
• This is how it looks after combining like terms.
• Subtract both sides by $12$ to keep $x$ on the left.
• Simplify.
• Subtract both sides by $5x$ to keep the constant on the right.
• Solve $x$ by dividing both sides by $– 10$, however, don’t forget to switch the direction of the inequality as well.
• Use an open hole to indicate that $2$ is not part of the solution. The solution of the inequality includes all values above $2$ but excluding $2$.
Example 7: Solve and graph the solution of the inequality
Let’s finish strong by doing one final example for mastery! Again, do this yourself first on paper and then compare your solution with the answer below.
• Write the original problem. Combine the $x$ terms on the left side and then apply the distributive property twice on the right side of the inequality.
• This is what we got after doing the step above.
Next, simplify the right side by combing like terms that came out after distribution.
• Simplify.
• To isolate $x$ on the left, subtract both sides by $5$.
• Simplify.
• Get rid of any variables on the right by subtracting $7x$.
• Finally, divide both sides by $– \,3$ to isolate $x$. Make sure that we change the direction of the inequality symbol to compensate for the division of a negative number.
• Use a close or shaded hole to indicate that $– \,1$ is part of the solution. The solution to the inequality $x \ge – 1$ includes $– \,1$ and all values to its right as shown by the direction of the arrow.
You may also be interested in these related math lessons or tutorials:
Steps on How to Graph Linear Inequalities
Solving Compound Inequalities |
# 2018 USAMO Problems/Problem 2
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 2
Find all functions $f:(0,\infty) \rightarrow (0,\infty)$ such that
$$f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1$$ for all $x,y,z >0$ with $xyz =1.$
## Solution
The only such function is $f(x)=\frac13$.
Letting $x=y=z=1$ gives $3f(2)=1$, hence $f(2)=\frac13$. Now observe that even if we fix $x+\frac1y=y+\frac1z=2$, $z+\frac1x$ is not fixed. Specifically, $$y=\frac1{2-x}$$ $$z=\frac1{2-y}=\frac{2-x}{3-2x}$$ $$z+\frac1x=\frac12+\frac1x+\frac1{2(3-2x)}$$ This is continuous on the interval $x\in\left(0,\frac32\right)$ and has an asymptote at $x=\frac32$. Since it takes the value 2 when $x=1$, it can take on all values greater than or equal to 2. So for any $a\ge2$, we can find $x$ such that $2f(2)+f(x)=1$. Therefore, $f(x)=\frac13$ for all $x\ge2$.
Now, for any $0, if we let $x=\frac k2$, $y=\frac1x$, and $z=1$, then $f(k)+2f\left(1+\frac2k\right)=1$. Since $1+\frac2k>2$, $f\left(1+\frac2k\right)=\frac13$, hence $f(k)=\frac13$. Therefore, $f(x)=\frac13$ for all $x\ge0$.
-- wzs26843645602 |
# Convert Octal to Binary
## Octal System
Octalnumeral system, or oct for short, is the base-8 number system, and uses the digits 0 to 7. Octal numerals can be made from binary numerals by grouping consecutive binary digits into groups of three (starting from the right). For example, the binary representation for decimal 74 is 1001010. Two zeroes can be added at the left: (00)1 001 010, corresponding the octal digits 1 1 2, yielding the octal representation 112.
In the decimal system each decimal place is a power of ten. For example:
74 10 = 7 × 10 1 + 4 × 10 0 {\displaystyle \mathbf {74} _{10}=\mathbf {7} \times 10^{1}+\mathbf {4} \times 10^{0}} {\mathbf {74}}_{{10}}={\mathbf {7}}\times 10^{1}+{\mathbf {4}}\times 10^{0}
In the octal system each place is a power of eight. For example:
112 8 = 1 × 8 2 + 1 × 8 1 + 2 × 8 0 {\displaystyle \mathbf {112} _{8}=\mathbf {1} \times 8^{2}+\mathbf {1} \times 8^{1}+\mathbf {2} \times 8^{0}} {\mathbf {112}}_{8}={\mathbf {1}}\times 8^{2}+{\mathbf {1}}\times 8^{1}+{\mathbf {2}}\times 8^{0}
By performing the calculation above in the familiar decimal system we see why 112 in octal is equal to 64+8+2 = 74 in decimal.
## Binary System
Binary is the simplest kind of number system that uses only two digits of 0 and 1. By using these digits computational problems can be solved by machines because in digital electronics a transistor is used in two states. Those two states can be represented by 0 and 1. That is why this number system is the most preferred in modern computer engineer, networking and communication specialists, and other professionals.
Related converters:
Binary To Octal Converter |
# Using Friendly Numbers: An Addition Strategy
WHAT IS A FRIENDLY NUMBER?
In this case, we refer to friendly numbers as a number that is easy to work with. For example, multiples of 10 are “friendly” because they are easy to work with when we add or subtract.
USING FRIENDLY NUMBERS AS AN ADDITION STRATEGY
When we use the “friendly number” strategy for addition, it helps us work with big numbers. This is because we are essentially breaking the equation up into more manageable parts.
We begin by getting to a friendly number, which is typically a multiple of 10, 100, or 100 – depending on the numbers that we are working with. Then we add on the remainder.
Let’s take a look at the “using friendly numbers” addition strategy in action.
EXAMPLES
In this example we will add 27+9 using the friendly number strategy.
First, let’s put the number 27 on our empty number line.
Now let’s get to a friendly number. We know that the number 30 is “friendly” or easy to work with, so we can add 3 to get to 30.
Lastly, we add the remaining 6 and get our answer of 36.
Suppose we are solving 265+18.
First we will write 265 on our empty number line.
Then we can add 5 from the 18 to get to a friendly number 270.
We have 13 left, so now we can simply add the 13 to the 270 to get a final answer of 283.
BIG IDEAS
Mental Math is not about following a one size fits all process or procedure. Mental Math involves being able to think flexibly about numbers and manipulate them in different ways.
For the equations shown above, this friendly number strategy worked well. But students also could have used left to right addition, breaking up the an addend, compensation, or even the plus 7,8,9 strategy.
Our goal is to teach our students to think flexibly about numbers so that mental computation comes easily to them.
NEXT STEPS:
• Find task cards to reinforce the “using friendly numbers” strategy here:
### One Comment
• I LOVE this strategy. I was wondering if you had an anchor chart that I could print and display? Also do you have task cards for 2nd graders? Thank you for your consideration. 🙂 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Distance Formula in the Coordinate Plane
## Length between two points using a right triangle.
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Practice Distance Formula in the Coordinate Plane
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Distance Formula in the Coordinate Plane
What if you were given the coordinates of two points? How could you find how far apart these two points are? After completing this Concept, you'll be able to find the distance between two points in the coordinate plane using the Distance Formula.
### Guidance
The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ can be defined as $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ . This is called the distance formula . Remember that distances are always positive!
#### Example A
Find the distance between (4, -2) and (-10, 3).
Plug in (4, -2) for $(x_1, y_1)$ and (-10, 3) for $(x_2, y_2)$ and simplify.
$d& = \sqrt{(-10-4)^2+(3+2)^2}\\& = \sqrt{(-14)^2 + (5)^2}\\& = \sqrt{196+25}\\& = \sqrt{221} \approx 14.87 \ units$
#### Example B
Find the distance between (3, 4) and (-1, 3).
Plug in (3, 4) for $(x_1, y_1)$ and (-1, 3) for $(x_2, y_2)$ and simplify.
$d& = \sqrt{(-1-3)^2+(3-4)^2}\\& = \sqrt{(-4)^2 + (-1)^2}\\& = \sqrt{16+1}\\& = \sqrt{17} \approx 4.12 \ units$
#### Example C
Find the distance between (4, 23) and (8, 14).
Plug in (4, 23) for $(x_1, y_1)$ and (8, 14) for $(x_2, y_2)$ and simplify.
$d& = \sqrt{(8-4)^2+(14-23)^2}\\& = \sqrt{(4)^2 + (-9)^2}\\& = \sqrt{16+81}\\& = \sqrt{97} \approx 9.85 \ units$
### Guided Practice
1. Find the distance between (-2, -3) and (3, 9).
2. Find the distance between (12, 26) and (8, 7)
3. Find the distance between (5, 2) and (6, 1)
1. Use the distance formula, plug in the points, and simplify.
$d & = \sqrt{(3-(-2))^2 + (9-(-3))^2}\\& = \sqrt{(5)^2 + (12)^2}\\& = \sqrt{25+144}\\& = \sqrt{169} = 13 \ units$
2. Use the distance formula, plug in the points, and simplify.
$d & = \sqrt{(8-12)^2 + (7-26)^2}\\& = \sqrt{(-4)^2 + (-19)^2}\\& = \sqrt{16+361}\\& = \sqrt{377} \approx 19.42 \ units$
3. Use the distance formula, plug in the points, and simplify.
$d & = \sqrt{(6-5)^2 + (1-2)^2}\\& = \sqrt{(1)^2 + (-1)^2}\\& = \sqrt{1+1}\\& = \sqrt{2} = 1.41 \ units$
### Practice
Find the distance between each pair of points. Round your answer to the nearest hundredth.
1. (4, 15) and (-2, -1)
2. (-6, 1) and (9, -11)
3. (0, 12) and (-3, 8)
4. (-8, 19) and (3, 5)
5. (3, -25) and (-10, -7)
6. (-1, 2) and (8, -9)
7. (5, -2) and (1, 3)
8. (-30, 6) and (-23, 0)
9. (2, -2) and (2, 5)
10. (-9, -4) and (1, -1)
### Vocabulary Language: English Spanish
Distance Formula
Distance Formula
The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ can be defined as $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Pythagorean Theorem
Pythagorean Theorem
The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle. |
Midterm2Solutions
# Midterm2Solutions - Faculty of Mathematics University of...
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Unformatted text preview: Faculty of Mathematics University of Waterloo MATH 135 MIDTERM EXAM #2 Fall 2007 Monday 12 November 2007 19:00 20:15 Solutions 1. In each part of this problem, full marks will be given if the correct answer is written in the box. If your answer is incorrect, your work will be assessed for part marks. (a) Convert (2345) 6 to base 10. [3] Answer: (2345) 5 = 569 Solution By definition, (2345) 6 = 2(6 3 ) + 3(6 2 ) + 4(6) + 5 = 2(216) + 3(36) + 4(6) + 5 = 569 so (2345) 6 = (569) 10 . (b) Convert (2007) 10 to base 12, using A and B to represent the digits 10 and 11, respectively. [3] Answer: (2007) 10 = (11 B 3) 12 Solution Using the conversion algorithm, 2007 = 167(12) + 3 167 = 13(12) + 11 13 = 1(12) + 1 1 = 0(12) + 1 Therefore, (2007) 10 = (11 B 3) 12 , since B represents the digit 11. (c) Determine the remainder when 2 34 56 52 + 3 19 is divided by 17. [3] Answer: 11 Solution Since 17 is prime and none of 2, 56 or 3 is divisible by 17, then 2 16 56 16 3 16 1 (mod 17) by Fermats Little Theorem. MATH 135, Midterm #2 Solutions Page 2 of 5 Therefore, 2 34 56 52 + 3 19 (2 16 ) 2 2 2 (56 16 ) 3 56 4 + (3 16 )3 3 (mod 17) 1 2 (4)1 3 5 4 + 1(27) (mod 17) (since 56 5 (mod 17)) 4(625) + 27 (mod 17) 2527 (mod 17) 11 (mod 17) Therefore, the remainder is 11. (d) Determine the number of congruence classes in Z 18 that are solutions to the equation [3] [12][ x ] = [5]....
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In this lesson, we will learn:
• How to add decimal numbers using base ten (block) models
• That adding decimal numbers is the same as adding whole numbers, which involves lining up the right place values and regrouping
Notes:
• To add decimals with base ten (block) models:
• Remember the three types of blocks used to represent the ones, tenths, and hundredths place values
• "One whole" or hundred block = ones place value
• Column (stick or rod) = tenths place value
• Single square = hundredths place value
• Count all of the blocks that you are adding up, altogether
• Regroup to get to your final answer. 10 of any type of block will regroup into the next bigger type of block.
• To add decimals, line up the decimal point to make sure you are adding the right place values in the same columns. Regroup (carry over) numbers whenever necessary.
• Sometimes, you will have to write one or more zeroes to a number after the decimal point (trailing zeroes) when the addends don't have the same number of place values.
#### Lessons
• Introduction
a)
How to add decimals with base ten (block) models
• 1.
Regrouping practice for decimals with block models
Regroup so that each place value only has one digit. Use base ten (block) models to help.
a)
17 tenths = ____ ones + ____ tenths
b)
4 ones + 12 tenths + 39 hundredths = ____ ones + ____ tenths + ____ hundredths
c)
6 ones + 29 tenths + 16 hundredths = ____ ones + ____ tenths + ____ hundredths
• 2.
a)
b)
c)
• 3.
Add the decimals by lining up on the grid. Regroup where necessary.
a)
34.8 + 5
b)
4.6 + 2.9
c)
41.03 + 5.72
d)
35.84 + 6.5
• 4.
Adding decimals and rounding decimal sums
a)
7.6 + 2.14
b)
80.65 + 19.23
c)
18.24 + 3.17
d)
98 + 38.97 |
# Find the domain and range of the following functions.
– $\space sin^{- 1}$
– $\space cos^{- 1}$
– $\space tan^{- 1}$
The main objective of this question is to find the domain and range for the given functions.
This question uses the concept of range and domain of functions. The set among all values within which a function is defined is known as its domain, and its range is the set of all possible values.
In this question, we have to find the domain and range for the given functions.
a) Given that:
$\space sin^{ – 1 }$
We have to find the range and domain of this function. We know that the set among all values within which a function is defined is known as its domain, and its range is the set of all possible values.
Thus, the domain of $sin^{ – 1}$ is:
$\space = \left[ \space – \space\frac{ \pi}{ 2 }, \space \frac{ \pi}{ 2 } \right]$
And the range of $sin^{ – 1 }$ is:
$\space = \space [- \space 1, \space 1]$
b)Given that:
$\space cos^{ – 1 }$
We have to find the range and domain of this function. We know that the set among all values within which a function is defined is known as its domain, and its range is the set of all possible values.
Thus, the domain of $cos^{ – 1}$ is:
$\space = \space – \space 0, \space \pi$
And the range of $cos^{ – 1}$ is:
$\space = \space [- \space 1, \space 1]$
c) Given that:
$\space tan^{ – 1 }$
We have to find the range and domain of this function. We know that the set among all values within which a function is defined is known as its domain, and its range is the set of all possible values.
Thus, the domain of $tan^{ – 1}$ is:
$\space = \left[ \space – \space\frac{ \pi}{2}, \space \frac{ \pi}{ 2 } \right]$
And the range of $tan^{ – 1}$ is:
$\space = \space [ R ]$
The domain and range of $sin^{-1}$ is:
$\space = \space [ – \space 1, \space 1 ] ,\space\left[ \space – \space\frac{ \pi}{2}, \space \frac{ \pi}{ 2 } \right]$
The domain and range of $cos^{-1}$ is:
$\space = \space [ – \space 1, \space 1 ]\space [ – \space 0, \space \pi ]$
The domain and range of $tan^{-1}$ is:
$\space = \space R \space , \space\left[ \space – \space\frac{ \pi}{2}, \space \frac{ \pi}{ 2 } \right]$
## Example
Find the range and domain for the given function.
$\space = \space \frac{ 6 }{x \space – \space 4}$
We have to find the range and domain for the given function.
Thus, the range for the given function is all real numbers without zero, while the domain for the given function is all numbers that are real except the number which is equal to $4$. |
# Functions acting on matrix
Functions based on additions and multiplications can readily be defined for matrices. But many more functions are used. In particular, the function $e^x$ is of importance.
## Exponential Function
There are two possible definitions of exponential function. Let a matrix $A$ be given. The matrix $e^A$ is then defined by $$$$e^A = \sum_{k=0}^{\infty} \frac{1}{k!} A^k.$$$$ where $A^0 = 1$. A second definition is $$$$e^A = \lim_{m \to \infty} \left(1 + \frac{1}{m} A \right)^m.$$$$
The proof that the two definitions (1) and (2) coincide is just like it goes for ordinary exponentials, by applying the binomial expansion on eq. (2).
Eq (2) allows us to derive an important relation= $$$$\text{Det}(e^A) = \lim_{m \to \infty} \left( \text{Det}\left( 1 + \frac{1}{m} A \right) \right)^m$$$$
We compute the determinant betweeen brackets by ignoring contributions of order $1/m$ or smaller. For instance, first consider a $2\times 2$ matrix. i.e.
$$$$1 + \frac{1}{m} A = \begin{pmatrix} 1 + \frac{a_{11}}{m} & \frac{a_{12}}{m} \\ \frac{a_{21}}{m} & 1 + \frac{a_{22}}{m} \\ \end{pmatrix}$$$$
If, in the calculation of the determinant, we ignore the contributions of order $1/m^2$ and higher, the off-diagonal elements can be ignored (in a determinant, there are no terms containing only one off-diagonal element as a factor). In this approximation, the determinant is simply the product of the diagonal elements.
\begin{align*} \text{Det}(1 + \frac{1}{m}A) & \approx (1 + \frac{a_{11}}{m}) (1 + \frac{a_{22}}{m})\\ & \approx 1 + \frac{1}{m} (a_{11} + a_{22}) + \mathcal{O}\left(\frac{1}{m^{2}}\right). \end{align*}
In general, one finds= \begin{align*} \text{Det}(1 + \frac{1}{m}A) = 1 + \frac{1}{m} \text{Tr}(A) + \mathcal{O}\left(\frac{1}{m^{2}}\right). \end{align*} where $\text{Tr}(A)$ is the trace of the matrix $A$, this means the sum of its diagonal elements, \begin{align*} \text{Tr}(A) = \sum_{k} a_{kk}. \end{align*}
Substituting this in eq. (3) gives
\begin{align*} \text{Det}(e^{A}) & = \lim_{m \to \infty} \left( 1 + \frac{1}{m} \text{Tr}(A) + \mathcal{O} \left( \frac{1}{m^{2}} \right)\right)^{m}\\ & = e^{\text{Tr}(A)} \end{align*}
and thus we arrive at $$$$\text{Det}(e^{A})= e^{\text{Tr}(A)}.$$$$
Published on May 6, 2021
Last revised on May 6, 2021
References |
# Precalculus : Evaluate Expressions With Rational Exponents
## Example Questions
### Example Question #1 : Evaluate Expressions With Rational Exponents
Simplify the expression:
.
Explanation:
First, you can begin to simplfy the numerator by converting all 3 expressions into base 2.
, which simplifies to
For the denominator, the same method applies. Convert the 25 into base 5, and when simplified becomes simply 5.
### Example Question #2 : Evaluate Expressions With Rational Exponents
Evaluate the following expression using knowledge of the properties of exponents:
Explanation:
Let's work through this equation involving exponents one term at a time. The first term we see is , for which we can apply the following property:
So if we plug our values into the formula for the property, we get:
Because . Our next term is , for which we'll need the property:
Using the values for our term, we have:
The third term of the equation is , for which the quickest way to evaluate would be using the following property:
Using the values from our term, this gives us:
The next property we will need to consider for our fourth term is given below:
If we plug in the corresponding values from our term, we get:
Finally, our last term requires knowledge of the following simple property: Any number raised to the power of zero is 1. With this in mind, our last term becomes:
Rewriting the equation with all of the values we've just evaluated, we obtain our final answer:
### Example Question #3 : Evaluate Expressions With Rational Exponents
Evaluate the following expression and solve for .
Explanation:
To solve this problem, recall that you can set exponents equal to eachother if they have the same base.
See below:
So, we have
Because both sides of this equation have a base of seven, we can set the exponents equal to eachother and solve for t.
### Example Question #4 : Evaluate Expressions With Rational Exponents
Solve for .
Explanation:
We begin by taking the natural log of the equation:
Simplifying the left side of the equation using the rules of logarithms gives:
We group the x terms to get:
We reincorporate the exponents into the logarithms and use the identity property of the natural log to obtain:
We combine the logarithms using the multiplication/sum rule to get:
We then solve for x:
### Example Question #1331 : Pre Calculus
Solve for .
Explanation:
We begin by factoring out the term to get:
This equation gives our first solution:
Then we check for more solutions:
Therefore our solution is
### Example Question #1332 : Pre Calculus
Evaluate when
Explanation:
Remember the denominator of a rational exponent is equivalent to the index of a root.
This should simplify quite nicely.
When it gives us,
### Example Question #1 : Evaluate Expressions With Rational Exponents
What is the value of ?
15 |
Do you know what the term “bisect ”means? The term bisect in simple words means to divide or split a line into two. A perpendicular bisector makes a right angle to the baseline and splits a given line segment into two equal-length halves. The phrase ‘bisect’ refers to splitting evenly. Perpendicular bisectors cross the line segment they bisect and form four 90° angles on opposite sides. A perpendicular line or line segment is one that forms a 90° angle with another line or line segment.
In this article, we are going to know more about perpendicular bisectors and perpendicular lines in detail.
## Properties of Perpendicular Bisector
Let’s go through the properties of a perpendicular bisector –
• A line segment or a line is divided into two congruent segments
• Divides a triangle’s sides into congruent sections.
• They form a 90° angle with the line being bisected.
• They meet in the halfway line segment.
• A triangle’s circumcenter is the point at which its perpendicular bisectors connect.
• They meet within a triangle in an acute triangle, outside a triangle in an obtuse triangle, and at the hypotenuse in a right triangle.
• Any location on the perpendicular bisector is equidistant from both ends of the segment bisected.
• For a given line segment, there can only be one.
## What is Perpendicular Bisector Theorem?
A line segment’s perpendicular is defined as a line that splits another line segment in half at 90o via its midpoint. According to the perpendicular bisector theorem, any point on the perpendicular bisector is equidistant from both endpoints of the line segment on which it is drawn. If a pillar is positioned at an angle in the center of a bridge, all of its points will be equidistant from the bridge’s endpoints.
The perpendicular bisector is a line segment that divides a line into equal-sized segments by crossing it perpendicularly. When two lines intersect in such a way that they form a 90-degree angle with one another, they are said to be perpendicular to one another. In contrast, a bisector divides a line into equal halves(two).
## What are Perpendicular Lines?
Joseph was starving and craved a sandwich. His sandwich was usually sliced into four equal pieces by his mother. Sometimes it’s across the center, and other times it’s from corner to corner. In any case, a perpendicular line was sliced across the sandwich.
A perpendicular line is one that intersects another at a 90° angle. Each corner forms a 90° angle with the others. This is also referred to as a right angle.
So, where do we look for perpendicular lines? They are, after all, everywhere. They can be found in nature and are employed in geometry. Even perpendicular angles can be found at the corners of squares and possibly triangles. Let’s look at some instances!
## Definition of Perpendicular Lines
Perpendicular lines are two different lines that cross at a 90° angle. Have you spotted anything like the letter “L” or the connecting corners of your walls? They are perpendicular lines, which are straight lines that meet at a specified angle – the right angle.
A line is known to be perpendicular to another line if the two meet at a 90° angle. Let’s take a closer look at the notion of perpendicular lines.
If you want to understand the topic in a fun way, you can visit the Cuemath website.
## Do perpendicular lines have the same slope?
The slopes of perpendicular lines are not the same. Perpendicular line slopes differ from one another in a certain way. The slope of one line is equal to the inverse of the slope of the other line. The sum of a number and its reciprocal equals 1. |
1. ## Mathematical Induction help
Use mathematical induction to prove that 6 divides $\displaystyle n^3-n$, for all natural numbers. (For example, n=0..1..2)
Im confused on how to set this up.
2. Hi
As you've seen, it's easy to check that $\displaystyle 6$ divides $\displaystyle n^3-n$ when $\displaystyle n=0,1\ \text{or}\ 2.$
Now, to prove that for any $\displaystyle k$ in $\displaystyle \mathbb{N},$ we have to assume that for some integer $\displaystyle n\geq 2\ \ 6$ divides $\displaystyle n^3-n,$ and then prove that $\displaystyle 6$ divides $\displaystyle (n+1)^3-(n+1).$
We can write: $\displaystyle n^3-n=n(n^2-1)=n(n+1)(n-1).$
Hence $\displaystyle (n+1)^3-(n+1)=(n+1)(n+2)n$
An integer is divided by $\displaystyle 6$ iff it is divided by $\displaystyle 2$ and $\displaystyle 3,$ and a prime divides a product iff it divides at least a factor.
If $\displaystyle 2$ and $\displaystyle 3$ divide $\displaystyle n$ or $\displaystyle (n+1),$ then they divide $\displaystyle (n+1)^3-(n+1).$
If $\displaystyle 2$ divides $\displaystyle n-1,$ it also divides $\displaystyle n+1$ and then divides $\displaystyle (n+1)^3-(n+1).$
If $\displaystyle 3$ divides $\displaystyle n-1,$ it also divides $\displaystyle n+2$ and then divides $\displaystyle (n+1)^3-(n+1).$
Therefore $\displaystyle 6|(n^3-n)\Rightarrow6|((n+1)^3-(n+1))$
Of course, writing $\displaystyle n^3-n=n(n^2-1)=n(n+1)(n-1)$ gives an immediate but non inductive proof.
Another proof by induction (maybe more comon, try to do it) consists in, when you assume that $\displaystyle 6$ divides $\displaystyle n^3-n,$ develop $\displaystyle (n+1)^3-(n+1)$ and you find something which is a sum of $\displaystyle n^3-n$ and an integer $\displaystyle x.$ If you show that $\displaystyle x$ is divided by $\displaystyle 6,$ then $\displaystyle 6$ divides $\displaystyle n^3-n+x=(n+1)^3-(n+1)$, and you can end your proof.
3. ## Induction
Hello tokio
Originally Posted by tokio
Use mathematical induction to prove that 6 divides $\displaystyle n^3-n$, for all natural numbers. (For example, n=0..1..2)
Im confused on how to set this up.
This is the third (and there's a fourth) induction question you've posted in the last couple of days. We're here to help you learn how to succeed at Mathematics, not to do your homework for you.
I'll start you off:
$\displaystyle P(n)$ is the propositional function: $\displaystyle 6$ divides $\displaystyle n^3 - n$
Now look at the expression $\displaystyle (n+1)^3 - (n+1)$, and, by removing the brackets, simpifying and re-arranging the terms, see if you can prove that $\displaystyle P(n) \Rightarrow P(n+1)$; in other words, that $\displaystyle 6$ divides $\displaystyle (n+1)^3 - (n+1)$.
Show us your working, and, if you can't do it, we'll see about the next stage.
PS I see someone's beaten me to it. But let's see some working next time.
PPS Since you have been given a solution, here's an easier one:
$\displaystyle (n+1)^3 - (n+1) = n^3 +3n^2+3n+1 -n-1$
$\displaystyle = n^3 +3n^2 +2n$
$\displaystyle =(n^3 -n) + 3n^2 +3n$
$\displaystyle =(n^3 -n) +3n(n+1)$
Now one of $\displaystyle n$ and $\displaystyle (n+1)$ is always even. So $\displaystyle 3n(n+1)$ is divisible by $\displaystyle 6$. So $\displaystyle P(n) \Rightarrow (n^3 - n) + 3n(n+1)$ is divisible by $\displaystyle 6 \Rightarrow P(n+1)$.
$\displaystyle P(1)$ is $\displaystyle 1^3 - 1$ is divisible by $\displaystyle 6$, which is true. So $\displaystyle P(n)$ is true $\displaystyle \forall n \in \mathbb{N}$. |
# How do you find the domain and range of 1/(x-7)?
Apr 12, 2017
$x \in \mathbb{R} , x \ne 7$
$y \in \mathbb{R} , y \ne 0$
#### Explanation:
$\text{let } y = \frac{1}{x - 7}$
The denominator of y cannot equal zero as tis would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be.
$\text{solve " x-7=0rArrx=7larrcolor(red)" excluded value}$
$\Rightarrow \text{ domain is } x \in \mathbb{R} , x \ne 7$
$\text{Rearrange the function to make x the subject}$
$\Rightarrow y \left(x - 7\right) = 1$
$\Rightarrow x y - 7 y = 1$
$\Rightarrow x y = 1 + 7 y$
$\Rightarrow x = \frac{1 + 7 y}{y}$
$\Rightarrow y = 0 \leftarrow \textcolor{red}{\text{ is the excluded value}}$
$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 0$ |
# How do you solve 8-2x is greater than or equal to -4?
## When are you supposed to flip the greater than or equal to sign?
##### 1 Answer
Mar 31, 2018
$x \le 6$
#### Explanation:
$8 - 2 x \ge - 4$ is our equation
To solve for the inequality you do it normally like you would for an equation, although if you multiply or divide by a negative number you flip the inequality
$- 2 x \ge - 12$
Now we have to divide both sides by $- 2$ so we will flip the inequality
$x \le 6$ |
# Find the Taylor series for f(x) centered at c. f(x) =\frac{1}{1 - x}, \; c = 3 f(x) =...
## Question:
Find the Taylor series for f(x) centered at c.
{eq}f(x) =\frac{1}{1 - x}, \; c = 3 {/eq}
{eq}f(x) = \sum_{n=0}^\infty {/eq} _____
## Expansion of Binomial Series:
In this case let us assume {eq}k {/eq} is any number and {eq}\left| x \right| < 1 {/eq} then,
{eq}{(1 + x)^k} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}} k \\ n \\ \end{array}} \right){x^n}} = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + ..... {/eq}
## Answer and Explanation:
Here in this case, the given function is:{eq}f(x) =\frac{1}{1 - x}, \; c = 3 {/eq}
Using Binomial expansion, we can rewrite the function as:
{eq}\eqalign{ f(x)& = \frac{1}{{1 - x}} \cr & = \frac{1}{{ - 3 - 1 - x + 3}} \cr & = \frac{1}{{ - 4 - \left( {x - 3} \right)}} \cr & = - \left( {\frac{1}{{1 + \frac{{\left( {x - 3} \right)}}{4}}}} \right) \cr & = - {\left( {1 + \frac{{\left( {x - 3} \right)}}{4}} \right)^{ - 1}} \cr & = - \left( {1 - \frac{{\left( {x - 3} \right)}}{4} + \frac{{{{\left( {x - 3} \right)}^2}}}{{{4^2}}} - \frac{{{{\left( {x - 3} \right)}^3}}}{{{4^3}}} + ....} \right) \cr & = \left( { - 1 + \frac{{\left( {x - 3} \right)}}{4} - \frac{{{{\left( {x - 3} \right)}^2}}}{{{4^2}}} + \frac{{{{\left( {x - 3} \right)}^3}}}{{{4^3}}} - ....} \right) \cr & = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{{\left( {x - 3} \right)}^n}}}{{{4^n}}}} \cr} {/eq}
Which is the require Maclaurin Series expansion of the given function. |
Important Formulas: Functions Formulas
# Important Functions Formulas Formulas for JEE and NEET
Table of contents Introduction Domain, Co-Domain, and Range of a Function Important types of function Domain and Range of Some Common Functions Equal or Identical Functions Classification of Functions Algebraic Operations on Functions Composite of Uniformly and Non-Uniformly Defined Functions
## Introduction
A Function is a relation between two sets of values in which one set is the input value and the other set is the output value and each input value gives a particular output value.
We represent a function in maths as, y = f(x) where x is the input value and for each x we get an output value as y.
If to every value of x belonging to some set there corresponds one or several values of the variable y, then y is called a multiple-valued function of x.
Conventionally the word "FUNCTION” is used only as the meaning of a single valued function, if not otherwise stated.
is called the image of x & x is the pre-image of y under f.
Every function from A → B satisfies the following conditions.
(i) f ⊂ A x B
(ii) ∀ a ∈ A ⇒ (a, f(a)) ∈ f and
(iii) (a, b) ∈ f & (a, c) ∈ f ⇒ b = c
## Domain, Co-Domain, and Range of a Function
Let f: A → B, then the set A is known as the domain of f & the set B is known as the co-domain of f. The set of all f images of elements of A is known as the range of f. Thus,
Domain of f = {a | a ∈ A, (a, f(a)) ∈ f}
Range of f = {f(a) | a ∈ A, f(a) ∈ B}
It should be noted that range is a subset of the co−domain. If only the rule of function is given then the domain of the function is the set of those real numbers, where function is defined. For a continuous function, the interval from minimum to maximum value of a function gives the range.
## Important types of function
(i) Polynomial Function
If a function f is defined by f (x) = a0 xn + a1 xn−1 + a2 xn−2 + ... + an−1 x + an, where n is a non-negative integer and a0, a1, a2, ..., an are real numbers and a0 ≠ 0, then f is called a polynomial function of degree n
NOTE : (a) A polynomial of degree one with no constant term is called an odd linear function.
i.e. f(x) = ax, a ≠ 0
(b) There are two polynomial functions, satisfying the relation; f(x).f(1/x) = f(x) + f(1/x).
They are : (i) f(x) = xn + 1 & (ii) f(x) = 1 − xn, where n is a positive integer.
(ii) Algebraic Function
'y' is an algebraic function of 'x', if it is a function that satisfies an algebraic equation of the form
P0 (x) yn + P1 (x) yn−1 + ....... + Pn−1 (x) y + Pn (x) = 0
Where n is a positive integer and P0 (x), P1 (x) ........... are Polynomials in x.
e.g. y = |x| is an algebraic function, since it satisfies the equation y² − x² = 0.
Note that all polynomial functions are Algebraic but not the converse. A function that is not algebraic is called a Transcendental Function.
(iii) Fractional Rational Function
A rational function is a function of the form
y = f (x) = where g (x) & h (x) are polynomials & h (x) ≠ 0.
(iv) Absolute Value Function
A function y = f (x) = |x| is called the absolute value function or Modulus function. It is defined as :
(v) Exponential Function
A function f(x) = ax = ex/n a (a > 0, a ≠ 1, x ∈ R) is called an exponential function. The inverse of the exponential function is called the logarithmic function. i.e. g(x) = loga x.
Note that f(x) & g(x) are inverse of each other & their graphs are as shown.
(vi) Signum Function
A function y= f (x) = Sgn (x) is defined as follows :
It is also written as Sgn x = |x|/ x ; x ≠ 0 ; f (0) = 0
(vii) Greatest Integer or Step Up Function
The function y = f (x) = [x] is called the greatest integer function where [x] denotes the greatest integer less than or equal to x.
Note that for y=[x],
−1 ≤ x < 0 ; [x] = −1
0 ≤ x < 1 ; [x] = 0
1 ≤ x < 2 ; [x] = 1
2 ≤ x < 3 ; [x] = 2 and so on.
Properties of greatest integer function :
(a) [x] ≤ x < [x] + 1 and x − 1 < [x] ≤ x , 0 ≤ x − [x] < 1
(b) [x + m] = [x] + m if m is an integer .
(c) [x] + [y] ≤ [x + y] ≤ [x] + [y] + 1
(d) [x] + [− x] = 0 if x is an integer = − 1 otherwise.
(viii) Fractional Part Function
It is defined as : g (x) = {x} = x − [x] .
e.g. the fractional part of the no. 2.1 is 2.1− 2 = 0.1 and the fractional part of − 3.7 is 0.3. The period of this function is 1 and the graph of this function is as shown.
## Equal or Identical Functions
Two functions f & g are said to be equal if :
(i) The domain of f = the domain of g.
(ii) The range of f = the range of g and
(iii) f(x) = g(x), for every x belonging to their common domain.
eg. are identical functions.
### Classification of Functions
One-One Function (Injective mapping) :
A function f: A → B is said to be a one−one function or injective mapping if different elements of A have different f images in B.
Thus for x1, x2 ∈ A & f(x1), f(x2) ∈ B,
f(x1) = f(x2) ⇔ x1 = x2 or x1 ≠ x2 ⇔ f(x1) ≠ f(x2).
Note :
(i) Any function which is entirely increasing or decreasing in the whole domain, then f(x) is one−one.
(ii) If any line parallel to the x−axis cuts the graph of the function at most at one point, then the function is one−one.
Many–one function :
A function f: A → B is said to be a many one function if two or more elements of A have the same f image in B. Thus f : A → B is many one if for ; x1, x2 ∈ A, f(x1) = f(x2) but x1 ≠ x
Note :
(i) Any continuous function which has at least one local maximum or local minimum, then f(x) is many−one. In other words, if a line parallel to x−axis cuts the graph of the function at least at two points, then f is many−one .
(ii) If a function is one−one, it cannot be many−one and vice versa.
Onto function (Surjective mapping):
If the function f: A → B is such that each element in B (co−domain) is the f image of at least one element in A, then we say that f is a function of A 'onto' B. Thus f: A → B is surjective if ∀ b ∈ B, ∃ some a ∈ A such that f (a) = b.
Note that: if range = co−domain, then f(x) is onto. Into function: If f: A → B is such that there exists at least one element in the co−domain which is not the image of any element in the domain, then f(x) is into.
Note: If a function is onto, it cannot be into and vice versa. A polynomial of degree even will always be into function. Thus a function can be one of these four types :
(a) one−one onto (injective & surjective)
(b) one−one into (injective but not surjective)
(c) many−one onto (surjective but not injective)
(d) many−one into (neither surjective nor injective)
Note : (i) If f is both injective & surjective, then it is called a Bijective mapping.
The bijective functions are also named as invertible, non-singular, or bi-uni form functions.
(ii) If a set A contains n distinct elements then the number of different functions defined from A → A is nn & out of it n ! are one-one.
Identity function:
The function f: A → A defined by f(x) = x ∀ x ∈ A is called the identity of A and is denoted by IA. It is easy to observe that the identity function is a bijection.
Constant function :
A function f: A → B is said to be a constant function if every element of A has the same f image in B. Thus f: A → B; f(x) = c, ∀ x ∈ A, c ∈ B is a constant function. Note that the range of a constant function is a singleton and a constant function may be one-one or many-one, onto or into.
## Algebraic Operations on Functions
If f & g are real-valued functions of x with domain set A, B respectively, then both f & g are defined in
A ∩ B.
Now we define f + g, f − g, (f . g) & (f/g) as follows :
domain in each case is A ∩ B
domain is {x | x ∈ A ∩ B s. t g(x) ≠ 0} .
## Composite of Uniformly and Non-Uniformly Defined Functions
Let f : A → B & g : B → C be two functions.
Then the function g of : A → C defined by (g of) (x) = g (f(x)) ∀ x ∈ A is called the composite of the two functions f & g .
Diagrammatically,
Thus the image of every x ∈ A under the function g of is the g−image of the f−image of x.
Note that g of is defined only if ∀ x ∈ A, f(x) is an element of the domain of g so that we can take its g-image. Hence for the product g of of two functions f & g, the range of f must be a subset of the domain of g.
Properties of Composite Functions :
(i) The composite of functions is not commutative i.e. gof ≠ fog .
(ii) The composite of functions is associative i.e. if f, g, h are three functions such that fo (goh) & (fog) oh are defined, then fo (goh) = (fog) oh.
(iii) The composite of two bijections is a bijection i.e. if f & g are two bijections such that gof is defined, then gof is also a bijection.
Homogeneous Functions :
A function is said to be homogeneous with respect to any set of variables when each of its terms is of the same degree with respect to those variables.
For example 5x2 + 3y2 − xy is homogeneous in x & y .
Symbolically if, f (tx , ty) = tn . f (x , y) then f (x , y) is homogeneous function of degree n.
Bounded Function:
A function is said to be bounded if |f(x)| ≤ M, where M is a finite quantity.
Implicit and Explicit Function:
A function defined by an equation not solved for the dependent variable is called an implicit function. For eg. the equation x3 + y3 = 1 defines y as an implicit function. If y has been expressed in terms of x alone then it is called an explicit function.
Inverse of a function: Let f : A → B be a one−one & onto function, then there exists a unique
function g: B → A such that f(x) = y ⇔ g(y) = x, ∀ x ∈ A & y ∈ B. Then g is said to be inverse of f.
Thus g = f−1 : B → A = {(f(x), x) | (x, f(x)) ∈ f} .
Properties of Inverse Function
(i) The inverse of a bijection is unique.
(ii) If f : A → B is a bijection & g : B → A is the inverse of f, then fog = IB and gof = IA, where IA & IB are identity functions on the sets A & B respectively.
Note that the graphs of f & g are the mirror images of each other in the line y = x. As shown in the figure given below a point (x ',y ' ) corresponding to y = x2 (x > 0) changes to (y ',x ') corresponding to
y = +√x, the changed form of x = √y.
(iii) The inverse of a bijection is also a bijection.
(iv) If f & g two bijections f : A → B , g : B → C then the inverse of gof exists and (gof)−1 = f−1o g−1
Odd and Even Functions:
If f (−x) = f (x) for all x in the domain of ‘f’ then f is said to be an even function.
e.g. f (x) = cos x ; g (x) = x² + 3 . If f (−x) = −f (x) for all x in the domain of ‘f’ then f is said to be an odd function. e.g. f (x) = sin x ; g (x) = x3 + x .
NOTE :
(a) f (x) − f (−x) = 0 => f (x) is even & f (x) + f (−x) = 0 => f (x) is odd.
(b) A function may neither be odd nor even.
(c) The inverse of an even function is not defined.
(d) Every even function is symmetric about the y−axis & every odd function is symmetric about the origin.
(e) Every function can be expressed as the sum of an even & an odd function.
(f) The only function which is defined on the entire number line & is even and odd at the same time is f(x)= 0.
(g) If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd then f.g will be odd.
Periodic Function:
(a) A function f(x) is called periodic if there exists a positive number T (T > 0) called the period of the function such that f (x + T) = f(x), for all values of x within the domain of x.
e.g. The function sin x & cos x both are periodic over 2π & tan x is periodic over π
NOTE : (a) f (T) = f (0) = f (−T) , where ‘T’ is the period.
(b) Inverse of a periodic function does not exist.
(c) Every constant function is always periodic, with no fundamental period.
(d) If f (x) has a period T & g (x) also has a period T then it does not mean that f (x) + g (x) must have a period T.
e.g. f(x) = |sinx| + |cosx|.
(e) If f(x) has a period p, then also has a period p.
(f) if f(x) has a period T then f(ax + b) has a period T/a (a > 0).
GENERAL: If x, y are independent variables, then:
(i) f(xy) = f(x) + f(y) ⇒ f(x) = k ln x or f(x) = 0.
(ii) f(xy) = f(x) . f(y) ⇒ f(x) = xn, n ∈ R
(iii) f(x + y) = f(x) . f(y) ⇒ f(x) = akx
(iv) f(x + y) = f(x) + f(y) ⇒ f(x) = kx, where k is a constant.
The document Important Functions Formulas Formulas for JEE and NEET is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
All you need of JEE at this link: JEE
## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
## FAQs on Important Functions Formulas Formulas for JEE and NEET
1. What are some common types of functions in mathematics?
Ans. Some common types of functions in mathematics include linear functions, quadratic functions, exponential functions, logarithmic functions, and trigonometric functions.
2. How do you determine if a function is even or odd?
Ans. A function is even if f(x) = f(-x) for all x in the domain, and a function is odd if f(x) = -f(-x) for all x in the domain.
3. What is the difference between a function and a relation in mathematics?
Ans. A relation is a set of ordered pairs, while a function is a relation in which each input has exactly one output.
4. How do you find the domain and range of a function?
Ans. To find the domain of a function, identify all possible input values that the function can accept. To find the range, determine all possible output values of the function.
5. What is the significance of the vertical line test in relation to functions?
Ans. The vertical line test is used to determine if a graph represents a function. If any vertical line intersects the graph at more than one point, then the graph does not represent a function.
## Mathematics (Maths) for JEE Main & Advanced
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# Geometrical Constructions Exercise 19.1 Class 6 Notes | EduRev
## Class 6 : Geometrical Constructions Exercise 19.1 Class 6 Notes | EduRev
``` Page 1
1. Construct line segments whose lengths are:
(i) 4.8 cm
(ii) 12 cm 5 mm
(iii) 7.6 cm
Solution:
(i) 4.8 cm
Construct a line L on a paper and mark A on it.
Now place the metal point of the compass at zero mark of the ruler.
Make adjustments in the compass such that the pencil point is at 4.8 cm mark on the ruler.
Take compass on L such that its metal point is on A.
Now mark a small mark as B on L which is corresponding to the pencil point of the compass.
Here, AB is the required line segment of length 4.8 cm.
(ii) 12 cm 5 mm
Construct a line L on a paper and mark A on it.
Now place the metal point of the compass at zero mark of the ruler.
Make adjustments in the compass such that the pencil point is at 5 small points from the mark of 12 cm to 13 cm
on the ruler.
Take compass on L such that its metal point is on A.
Now mark a small mark as B on L which is corresponding to the pencil point of the compass.
Here, AB is the required line segment of length 12 cm 5 mm.
(iii) 7.6 cm
Construct a line L on a paper and mark A on it.
Now place the metal point of the compass at zero mark of the ruler.
Make adjustments in the compass such that the pencil point is at 6 small points from the mark of 7 cm to 8 cm of
the ruler.
Take compass on L such that its metal point is on A.
Now mark a small mark as B on L which is corresponding to the pencil point of the compass.
Here, AB is the required line segment of length 7.6 cm.
2. Construct two segments of length 4.3 cm and 3.2 cm. Construct a segment whose length is equal to the
sum of the lengths of these segments.
Solution:
With the help of compass and ruler, construct AB and CD of lengths 4.3 cm and 3.2 cm.
Construct a line L and mark P on it.
Now place the metal point of the compass at zero mark of the ruler.
Page 2
1. Construct line segments whose lengths are:
(i) 4.8 cm
(ii) 12 cm 5 mm
(iii) 7.6 cm
Solution:
(i) 4.8 cm
Construct a line L on a paper and mark A on it.
Now place the metal point of the compass at zero mark of the ruler.
Make adjustments in the compass such that the pencil point is at 4.8 cm mark on the ruler.
Take compass on L such that its metal point is on A.
Now mark a small mark as B on L which is corresponding to the pencil point of the compass.
Here, AB is the required line segment of length 4.8 cm.
(ii) 12 cm 5 mm
Construct a line L on a paper and mark A on it.
Now place the metal point of the compass at zero mark of the ruler.
Make adjustments in the compass such that the pencil point is at 5 small points from the mark of 12 cm to 13 cm
on the ruler.
Take compass on L such that its metal point is on A.
Now mark a small mark as B on L which is corresponding to the pencil point of the compass.
Here, AB is the required line segment of length 12 cm 5 mm.
(iii) 7.6 cm
Construct a line L on a paper and mark A on it.
Now place the metal point of the compass at zero mark of the ruler.
Make adjustments in the compass such that the pencil point is at 6 small points from the mark of 7 cm to 8 cm of
the ruler.
Take compass on L such that its metal point is on A.
Now mark a small mark as B on L which is corresponding to the pencil point of the compass.
Here, AB is the required line segment of length 7.6 cm.
2. Construct two segments of length 4.3 cm and 3.2 cm. Construct a segment whose length is equal to the
sum of the lengths of these segments.
Solution:
With the help of compass and ruler, construct AB and CD of lengths 4.3 cm and 3.2 cm.
Construct a line L and mark P on it.
Now place the metal point of the compass at zero mark of the ruler.
Make adjustments in the compass such that the pencil point reaches the point B.
Take compass on L such that its metal point is on P.
Now mark a small mark as Q on the line L corresponding to the pencil point of the compass.
Reset the compass so that its metal and pencil points are on C and D.
Take compass on L such that its metal point is on Q and the pencil point makes a small mark as R which is
opposite to P on L.
Here, PR is the required segment whose length is equal to the sum of the lengths of these segments.
```
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## Mathematics (Maths) Class 6
191 videos|221 docs|43 tests
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Associated Topics || Dr. Math Home || Search Dr. Math
### Graphing the Absolute Value/Square Root of a Function
```Date: 07/26/2002 at 09:32:33
From: Daniel Brostin
Subject: Abs value/square root of functions
Can you illustrate and discuss how taking the absolute value or the
square root of a function affects the graph of the function?
Thanks for the help!
```
```
Date: 07/26/2002 at 13:04:48
From: Doctor Peterson
Subject: Re: Abs value/square root of functions
Hi, Daniel.
Think about what the absolute value does to a number. If the number is
positive, its absolute value is the number itself: the absolute value
function does not change it. If the number is negative, its absolute
value is the negative of the number (which is positive): the absolute
value function reflects the number in the origin, sending it to the
point the same distance away but on the opposite side.
So what happens if you take the absolute value of a function? Wherever
the function is positive (or zero), the graph stays right where it
was. Wherever the function is negative, the graph is reflected up over
the x-axis, so that its y coordinate is positive rather than negative:
| ___ | ___
| __/ \ + __/ \ /
| / \ |\ / \ /
+-+---------+-- ----> +-+---------+--
|/ \ |
+ \ |
| |
Now how about the square root? In this case, if f(x) is negative, its
square root does not exist (since we are presumably talking about real
functions). You can just erase that part of the graph! If f(x) is
positive, the square root will squash it or stretch it vertically,
depending on its value. Look at the graph of y=sqrt(x), compared with
the graph of y=x. You will see that if x<1, sqrt(x) is larger, while
when x>1, sqrt(x) is smaller. So when f(x)<1, sqrt(f(x)) will be
above f(x), and when f(x)>1, sqrt(f(x)) will be below f(x):
| ___ | ___
1+ __/ \ 1+ __/ \
| / \ | / \
+-+---------+-- ----> +-+---------+--
|/ \ |
+ \ |
| |
I can't get enough detail in my graph to show the stretching. But you
can put a dot on the graph where f(x)=1, so the graph of the square
root will go through that point; then you can warp the given graph
upward where it is below y=1 and downward where it is above y=1. The
peaks and valleys of the new graph will lie at the same x coordinates
as in the original graph, but will be at different heights.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Equations, Graphs, Translations
High School Functions
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search |
# Pythagorean theorem
The Pythagorean Theorem is an important mathematical theorem that explains the final side of a right angled triangle when two sides are known. In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle), so we can take out the length of any third side of a right angled triangle, if the other two sides are given.
Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the legs, but less than the sum of them. Some of the most common right triangles with integer sides are the 3-4-5, 5-12-13, 7-24-25, and the 8-15-17 triangles. The Pythagorean theorem is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof. The theorem is:
${\displaystyle a^{2}+b^{2}=c^{2}\!\,}$
where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. For example we consider a triangle whose two sides, except the hypotenuse, are of lengths 12 and 5 cm. We can find the hypotenuse using the Pythagorean Theorem.
${\displaystyle a^{2}+b^{2}=c^{2}\!\,}$
${\displaystyle =12^{2}+5^{2}=x^{2}\!\,}$
${\displaystyle =144+25=x^{2}\!\,}$
${\displaystyle =169=x^{2}\!\,}$
${\displaystyle ={\sqrt {169}}=x,}$
${\displaystyle \therefore x=13}$ ${\displaystyle cm}$
## Proofs
Proof by rearrangement of four identical right triangles Animation showing another proof by rearrangement Proof using an elaborate rearrangement Diagram of Garfield's proof
## Examples
1: Find the hypotenuse of a right angle triangle whose base is 3 meters and perpendicular is 4 meters in length.
Solution: Let the hypotenuse be x meters.
${\displaystyle x^{2}=4^{2}+3^{2}}$
${\displaystyle x={\sqrt {4^{2}+3^{2}}}}$
${\displaystyle \therefore x=5}$
So the hypotenuse will be 5 meters.
2: Find the length of a leg of a right angled triangle whose hypotenuse is 17 cm and the other leg is 15 cm.
###### Solution: Method 1:
Let the length of the unknown leg = x centimetres
${\textstyle x^{2}+15^{2}=17^{2}}$
By subtracting 152 from both sides, we get:
${\textstyle x^{2}=17^{2}-15^{2}}$
${\textstyle x^{2}=(17-15)(17+15)\because A^{2}-B^{2}=(A-B)(A+B)}$
${\textstyle x^{2}=2\times 32=64}$
${\textstyle x={\sqrt {64}}}$
${\textstyle \therefore x=8}$
So the length of the unknown leg is 8 centimetres.
Method 2:
Let the length of the unknown leg = x centimetres
${\textstyle x^{2}+15^{2}=17^{2}}$
By subtracting 152 from both sides, we get:
${\textstyle x^{2}=17^{2}-15^{2}}$
${\displaystyle x^{2}=289-225=64}$
${\textstyle x={\sqrt {64}}}$
${\textstyle \therefore x=8}$
So the length of the unknown leg is 8 centimetres.
## Composition of each value a, b, c
${\displaystyle a^{2}+b^{2}=c^{2}...(1)}$
${\displaystyle a^{2}=c^{2}-b^{2}}$
${\displaystyle a^{2}=(c+b)(c-b)}$
The left hand side is a perfect square. Therefore the right hand side must be a perfect square.
Let
${\displaystyle c+b=Km^{2}...(2)}$
and let
${\displaystyle c-b=Kn^{2}...(3)}$
The right hand side ${\displaystyle =(c+b)(c-b)=(Km^{2})(Kn^{2})=K^{2}m^{2}n^{2}}$ a perfect square.
${\displaystyle a^{2}=K^{2}m^{2}n^{2}...(4)}$
${\displaystyle (2)+(3),2c=Km^{2}+Kn^{2}}$
${\displaystyle c={\frac {K(m^{2}+n^{2})}{2}}}$
${\displaystyle c^{2}={\frac {K^{2}(m^{2}+n^{2})^{2}}{4}}...(5)}$
${\displaystyle (2)-(3),2b=Km^{2}-Kn^{2}}$
${\displaystyle b={\frac {K(m^{2}-n^{2})}{2}}}$
${\displaystyle b^{2}={\frac {K^{2}(m^{2}-n^{2})^{2}}{4}}...(6)}$
Substitute (4), (5) and (6) in (1)
${\displaystyle K^{2}m^{2}n^{2}+{\frac {K^{2}(m^{2}-n^{2})^{2}}{4}}={\frac {K^{2}(m^{2}+n^{2})^{2}}{4}}}$
Simplify
${\displaystyle 4m^{2}n^{2}+(m^{2}-n^{2})^{2}=(m^{2}+n^{2})^{2}}$
Therefore
${\displaystyle a=2mn,b=m^{2}-n^{2},c=m^{2}+n^{2}}$
Check:
${\displaystyle a^{2}=4m^{2}n^{2}}$
${\displaystyle b^{2}=m^{4}-2m^{2}n^{2}+n^{4}}$
${\displaystyle a^{2}+b^{2}=m^{4}+2m^{2}n^{2}+n^{4}}$
${\displaystyle c^{2}=m^{4}+2m^{2}n^{2}+n^{4}=a^{2}+b^{2}}$
For example, consider the triple ${\displaystyle 5,12,13}$ .
${\displaystyle m=3,n=2}$
${\displaystyle a=2mn=2(3)(2)=12}$
${\displaystyle b=m^{2}-n^{2}=3^{2}-2^{2}=9-4=5}$
${\displaystyle c=m^{2}+n^{2}=3^{2}+2^{2}=9+4=13}$
When we have a Pythagorean triple a,b,c:
${\displaystyle a=2mn,b=m^{2}-n^{2},c=m^{2}+n^{2}}$ .
## Topics related to Pythagoras' Theorem
Any integer may be expressed as the sum of two squares.
${\displaystyle ({\frac {3{\sqrt {7}}}{5}})^{2}+({\frac {4{\sqrt {7}}}{5}})^{2}=({\sqrt {7}})^{2}(({\frac {3}{5}})^{2}+({\frac {4}{5}})^{2})=7{\frac {3^{2}+4^{2}}{5^{2}}}=7{\frac {5^{2}}{5^{2}}}=7}$
Angle in a semi-circle is a right angle.
Definition of a circle in the cartesian system depends on Pythagoras' Theorem.
${\displaystyle r={\sqrt {(x-p)^{2}+(y-q)^{2}}}}$
In electronics, in a series circuit containing resistance and capacitance and supplied with a sine-wave voltage (E),
voltage across resistance (Er) is in phase with current, voltage across capacitance (Ec) lags current by 90°.
${\displaystyle E={\sqrt {(Er)^{2}+(Ec)^{2}}}}$
A complex number may be expressed in "Eulerian" format.
${\displaystyle {\sqrt {3}}+3i=2{\sqrt {3}}({\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i)=2{\sqrt {3}}(\cos(60)+\sin(60)i)}$
The familiar Theorem of Pythagoras is a special case of the general identity applied to any triangle.
${\displaystyle c^{2}=a^{2}+b^{2}-2(ab)\cos(C)}$
In aeronautical navigation, if an aircraft is heading North at 100 knots and the wind at altitude is blowing from the East
at 10 knots, the groundspeed of the aircraft is
${\displaystyle {\sqrt {100^{2}+10^{2}}}}$ knots
Some perfect squares can be expressed as the sum of two squares in more ways than one:
${\displaystyle 65^{2}=39^{2}+52^{2}=25^{2}+60^{2}}$
The expression ${\displaystyle m^{2}+1}$ is a perfect square if ${\displaystyle m}$ has a value like ${\displaystyle {\frac {3}{4}}}$ or ${\displaystyle {\frac {12}{5}}}$ .
The expression ${\displaystyle m^{2}-1}$ is a perfect square if ${\displaystyle m}$ has a value like ${\displaystyle {\frac {5}{3}}}$ or ${\displaystyle {\frac {17}{8}}}$ .
The expression ${\displaystyle 1-m^{2}}$ is a perfect square if ${\displaystyle m}$ has a value like ${\displaystyle {\frac {3}{5}}}$ or ${\displaystyle {\frac {24}{25}}}$ .
The equation of the line ${\displaystyle ax+by+c=0}$ can be put into normal form: ${\displaystyle {\frac {ax+by+c}{\sqrt {a^{2}+b^{2}}}}=0}$ .
The normal form of the line ${\displaystyle 4x+3y+10=0}$ is ${\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y+2=0}$ where ${\displaystyle a^{2}+b^{2}=1}$ .
In the ellipse, the major axis (length ${\displaystyle 2a}$ ), the minor axis (length ${\displaystyle 2b}$ ) and the distance between foci (length ${\displaystyle 2c}$ ) have the relationship ${\displaystyle a^{2}=b^{2}+c^{2}}$ . |
# Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios
Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios provide solutions in a stepwise manner as per students’ intelligence quotient. The main aim of developing these solutions is to make the learning process smooth and hassle-free among students. Those who aspire to excel in solving difficult problems easily, can make use of Frank Solutions. Here, students can find the Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios PDF, from the links provided below.
Chapter 26 Trigonometrical Ratios has problems pertaining to the measurement of the sides and angles of a triangle. The branch of Mathematics, which deals with the study of sides and angles of a right-angled triangle is known as trigonometry. It is most essential to understand this chapter in-depth, as it plays a vital role in further classes also.
## Access Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios
1. In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric ratios.
(i) sin A = 12 / 13
(ii) cos B = 4 / 5
(iii) cot A = 1 / 11
(iv) cosec C = 15 / 11
(v) tan C = 5 / 12
(vi) sin B = √3 / 2
(vii) cos A = 7 / 25
(viii) tan B = 8 / 15
(ix) sec B = 15 / 12
(x) cosec C = √10
Solution:
(i) Sin A = 12 / 13
Sin A = Perpendicular / Hypotenuse
Sin A = 12 / 13
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(13)2 – (12)2
Base = √169 – 144
Base = √25
We get,
Base = 5
Cos A = Base / Hypotenuse
Cos A = 5 / 13
Sec A = (1 / cos A)
Sec A = 13 / 5
Cot A = (1 / tan A)
Cot A = 5 / 12
Cosec A = (1 / sin A)
Cosec A = 13 / 12
(ii) Cos B = 4 / 5
Cos B = Base / Hypotenuse
Cos B = 4 / 5
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Perpendicular = √(Hypotenuse)2 – (Base)2
Perpendicular = √(5)2 – (4)2
Perpendicular = √25 – 16
Perpendicular = √9
We get,
Perpendicular = 3
Sin B = Perpendicular / Hypotenuse
Sin B = 3 / 5
Tan B = Perpendicular / Base
Tan B = 3 / 4
Sec B = (1 / cos B)
Sec B = 5 / 4
Cot B = (1 / tan B)
Cot B = 4 / 3
Cosec B = (1 / sin B)
Cosec B = 5 / 3
(iii) Cot A = 1 / 11
Cot A = (1 / tan A)
Cot A = Base / Perpendicular
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Hypotenuse) = √(Perpendicular)2 + (Base)2
(Hypotenuse) = √(11)2 + (1)2
(Hypotenuse) = √(121 + 1)
(Hypotenuse) = √122
Cos A = Base / Hypotenuse
Cos A = 1 / √122
Tan A = Perpendicular / Base
Tan A = 11
Sec A = (1 / cos A)
Sec A = √122
Sin A = Perpendicular / Hypotenuse
Sin A = 11 / √122
Cosec A = (1 / sin A)
Cosec A = √122 / 11
(iv) Cosec C = 15 / 11
Cosec C = (1 / sin C)
Cosec C = (Hypotenuse / Perpendicular)
Cosec C = 15 / 11
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(15)2 – (11)2
Base = √225 – 121
Base = √104
Sin C = Perpendicular / Hypotenuse
Sin C = 11 / 15
Cos C = Base / Hypotenuse
Cos C = √104 / 15
Tan C = Perpendicular / Base
Tan C = 11 / √104
Sec C = (1 / Cos C)
Sec C = 15 / √104
Cot C = (1 / tan C)
Cot C = √104 / 11
(v) Tan C = 5 / 12
Tan C = Perpendicular / Base
Tan C = 5 / 12
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Hypotenuse) = √(Perpendicular)2 + (Base)2
(Hypotenuse) = √(5)2 + (12)2
(Hypotenuse) = √25 + 144
(Hypotenuse) = √169
We get,
(Hypotenuse) = 13
Cot C = (1 / tan C)
Cot C = 12 / 5
Sin C = Perpendicular / Hypotenuse
Sin C = 5 / 13
Cos C = Base / Hypotenuse
Cos C = 12 / 13
Sec C = (1 / Cos C)
Sec C = 13 / 12
Cosec C = (1 / Sin C)
Cosec C = 13 / 5
(vi) Sin B = √3 / 2
Sin B = Perpendicular / Hypotenuse
Sin B = √3 / 2
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(2)2 – (√3)2
Base = √4 – 3
Base = √1
We get,
Base = 1
Cos B = Base / Hypotenuse
Cos B = 1 / 2
Tan B = Perpendicular / Base
Tan B = √3 / 1
Tan B = √3
Sec B = (1 / Cos B)
Sec B = 2
Cot B = (1 / tan B)
Cot B = 1 / √3
Cosec B = 1 / Sin A
Cosec B = 2 / √3
(vii) Cos A = 7 / 25
Cos A = Base / Hypotenuse
Cos A = 7 / 25
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Perpendicular = √(Hypotenuse)2 – (Base)2
Perpendicular = √(25)2 – (7)2
Perpendicular = √625 – 49
Perpendicular = √576
We get,
Perpendicular = 24
Sin A = Perpendicular / Hypotenuse
Sin A = 24 / 25
Tan A = Perpendicular / Base
Tan A = 24 / 7
Sec A = (1 / cos A)
Sec A = 25 / 7
Cot A = (1 / tan A)
Cot A = 7 / 24
Cosec A = (1 / sin A)
Cosec A = 25 / 24
(viii) Tan B = 8 / 15
Tan B = Perpendicular / Base
Tan B = 8 / 15
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Hypotenuse) = √(Perpendicular)2 + (Base)2
(Hypotenuse) = √(8)2 + (15)2
(Hypotenuse) = √64 + 225
(Hypotenuse) = √289
We get,
Hypotenuse = 17
Cot B = 1 / tan B
Cot B = 15 / 8
Sin B = Perpendicular / Hypotenuse
Sin B = 8 / 17
Cos B = Base / Hypotenuse
Cos B = 15 / 17
Sec B = (1 / cos B)
Sec B = 17 / 15
Cosec B = (1 / sin B)
Cosec B = 17 / 8
(ix) Sec B = 15 / 12
Sec B = (1 / cos B)
Sec B = Hypotenuse / Base
Sec B = 15 / 12
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Perpendicular = √(Hypotenuse)2 – (Base)2
Perpendicular = √(15)2 – (12)2
Perpendicular = √225 – 144
Perpendicular = √81
We get,
Perpendicular = 9
Sin B = Perpendicular / Hypotenuse
Sin B = 9 / 15
Tan B = Perpendicular / Base
Tan B = 9 / 12
Cot B = (1 / tan B)
Cot B = 12 / 9
Cosec B = (1 / sin B)
Cosec B = 15 / 9
Cos B = Base / Hypotenuse
Cos B = 12 / 15
(x) Cosec C = √10
Cosec C = (1 / sin C)
Cosec C = Hypotenuse / Perpendicular
Cosec C = √10 / 1
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √(√10)2 – (1)2
Base = √10 – 1
Base = √9
We get,
Base = 3
Sin C = Perpendicular / Hypotenuse
Sin C = 1 / √10
Cos C = Base / Hypotenuse
Cos C = 3 / √10
Tan C = Perpendicular / Base
Tan C = 1 / 3
Sec C = (1 / cos C)
Sec C = √10 / 3
Cot C = (1 / tan C)
Cot C = 3
2. In △ABC, ∠A = 900. If AB = 5 units and AC = 12 units, find:
(i) sin B
(ii) cos C
(iii) tan B
Solution:
In â–³ABC,
BC2 = AB2 + AC2
BC = √AB2 + AC2
BC = √52 + 122
BC = √25 + 144
BC = √169
We get,
BC = 13
AC = 12 units
BC = 13 units
AB = 5 units
(i) sin B = Perpendicular / Hypotenuse
sin B = AC / BC
sin B = 12 / 13
(ii) cos C = Base / Hypotenuse
cos C = AC / BC
cos C = 12 / 13
(iii) tan B = Perpendicular / Base
tan B = AC / AB
tan B = 12 / 5
3. In △ABC, ∠B = 900. If AB = 12 units and BC = 5 units, find:
(i) sin A
(ii) tan A
(iii) cos C
(iv) cot C
Solution:
In â–³ABC,
AC2 = AB2 + BC2
AC = √122 + 52
AC = √144 + 25
AC = √169
We get,
AC = 13
AB = 12 units
BC = 5 units
AC = 13 units
(i) sin A = Perpendicular / Hypotenuse
sin A = BC / AC
sin A = 5 / 13
(ii) tan A = Perpendicular / Base
tan A = BC / AB
tan A = 5 / 12
(iii) cos C = Base / Hypotenuse
cos C = BC / AC
cos C = 5 / 13
(iv) cot C = Base / Perpendicular
cot C = BC / AB
cot C = 5 / 12
4. If sin A = 3/ 5, find cos A and tan A.
Solution:
Given
sin A = 3 / 5
sin A = Perpendicular / Hypotenuse
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
(Base)2 = (Hypotenuse)2 – (Perpendicular)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √52 – 32
Base = √25 – 9
Base = √16
We get,
Base = 4
cos A = Base / Hypotenuse
cos A = 4 / 5
tan A = Perpendicular / Base
tan A = 3 / 4
5. If sin θ = 8 / 17, find the other five trigonometric ratios.
Solution:
Given
sin θ = 8 / 17
sin θ = Perpendicular / Hypotenuse
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √172 – 82
Base = √289 – 64
Base = √225
We get,
Base = 15
cos θ = Base / Hypotenuse
cos θ = 15 / 17
tan θ = Perpendicular / Base
tan θ = 8 / 15
cosec θ = 1/ sin θ
cosec θ = 17 / 8
sec θ = 1/cos θ
sec θ = 17 / 15
cot θ = 1/ tan θ
cot θ = 15 / 8
6. If tan A = 0.75, find the other trigonometric ratios for A.
Solution:
Given
tan A = 0.75
tan A = 75 / 100
We get,
tan A = 3 / 4
tan A = Perpendicular / Base
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Hypotenuse = √(Perpendicular)2 + (Base)2
Hypotenuse = √32 + 42
Hypotenuse = √9 + 16
Hypotenuse = √25
We get,
Hypotenuse = 5
sin A = Perpendicular / Hypotenuse
sin A = 3 / 5
sin A = 0.6
cos A = Base / Hypotenuse
cos A = 4 / 5
cos A = 0.8
cosec A = 1 / sin A
cosec A = 5 / 3
cosec A = 1.66
sec A = 1 / cos A
sec A = 5 / 4
sec A = 1.25
cot A = 1 / tan A
cot A = 4 / 3
cot A = 1.33
7. If sin A = 0.8, find the other trigonometric ratios for A.
Solution:
Given
sin A = 0.8
sin A = 8 / 10
sin A = 4 / 5
sin A = Perpendicular / Hypotenuse
By Pythagoras theorem,
We have
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Base = √(Hypotenuse)2 – (Perpendicular)2
Base = √52 – 42
Base = √25 – 16
Base = √9
We get,
Base = 3
cos A = Base / Hypotenuse
cos A = 3 / 5
cos A = 0.6
tan A = Perpendicular / Base
tan A = 4 / 3
tan A = 1.33
cosec A = 1/sin A
cosec A = 5 / 4
cosec A = 1.25
sec A = 1/ cos A
sec A = 5 / 3
sec A = 1.66
cot A = 1/ tan A
cot A = 3 / 4
cot A = 0.75
8. If 8 tan θ = 15, find
(i) sin θ
(ii) cot θ
(iii) sin2 θ – cot2 θ
Solution:
Given
8 tan θ = 15
tan θ = 15 / 8
tan θ = Perpendicular / Base
By Pythagoras theorem,
We have,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Hypotenuse = √(Perpendicular)2 + (Base)2
Hypotenuse = √152 + 82
Hypotenuse = √225 + 64
Hypotenuse = √289
We get,
Hypotenuse = 17
(i) sin θ = Perpendicular / Hypotenuse
sin θ = 15 / 17
(ii) cot θ = 1 / tan θ
cot θ = 8 / 15
(iii) sin2 θ – cot2 θ = (sin θ + cot θ) (sin θ – cot θ)
sin2 θ – cot2 θ = {(15 / 17) + (8 / 15)} {(15 / 17) – (8 / 15)}
sin2 θ – cot2 θ = {(225 + 136) / 255}{(225 – 136) / 255}
sin2 θ – cot2 θ = (361 / 255) (89 / 255)
On calculation, we get,
sin2 θ – cot2 θ = 32129 / 65025
9. In an isosceles triangle ABC, AB = BC = 6 cm and ∠B = 900. Find the values of
(a) cos C
(b) cosec C
(c) cos2 C + cosec2 C
Solution:
â–³ABC is an isosceles right-angled triangle
Therefore,
AC2 = AB2 + BC2
AC2 = 62 + 62
AC2 = 36 + 36
AC2 = 72
We get,
AC = 6√2 cm
(a) cos C = BC / AC
cos C = (6 / 6√2)
cos C = (1 / √2)
(b) cosec C = AC / AB
cosec C = (6√2 / 6)
cosec C = √2
(c) cos2 C + cosec2 C = (1/√2)2 + (√2)2
cos2 C + cosec2 C = (1/2) + 2
On further calculation, we get,
cos2 C + cosec2 C = 5 / 2
10. In the given figure, AD is the median on BC from A. If AD = 8 cm and BC = 12 cm, find the value of
(a) sin x
(b) cos y
(c) tan x. cot y
(d) (1/sin2 x) – (1/ tan2 x)
Solution:
Since, AD is the median on BC,
We have,
BD = DC = (1/2) x BC
= (1/2) x 12
= 6 cm
â–³ADB is a right angled triangle
Therefore,
AB2 = 82 + 62
AB2 = 64 + 36
AB2 = 100
We get,
AB = 10 cm
â–³ADC is a right angled triangle
Therefore,
AC2 = 82 + 62
AC2 = 64 + 36
AC2 = 100
We get,
AC = 10 cm
(a) sin x = AD / AB
sin x = 8 / 10
sin x = 4 / 5
(b) cos y = AD / AC
cos y = 8 / 10
cos y = 4 / 5
(c) cos x = BD / AB
cos x = 6 / 10
cos x = 3 / 5
And,
sin y = DC / AC
sin y = 6 / 10
sin y = 3 / 5
Hence,
tan x = sin x / cos x
tan x = {(4/5) / (3/5)}
We get,
tan x = 4 / 3
cot y = cos y / sin y
cot y = {(4/5) / (3/5)}
We get,
cot y = 4 / 3
Therefore,
tan x. cot y = (4/3) x (4/3)
tan x . cot y = 16 / 9
(d) (1/ sin2 x) – (1/ tan2 x) = {1/ (4/5)2} – {1/ (4/3)2}
On calculating further, we get,
(1/ sin2 x ) – (1/ tan2 x) = (25 / 16) – (9 / 16)
(1/ sin2 x) – (1/ tan2 x) = (16 / 16)
(1/ sin2 x) – (1/ tan2 x) = 1
11. In a right- angled triangle PQR, ∠PQR = 900, QS ⊥PR and tan R = (5/ 12), find the value of
(a) sin ∠PQS
(b) tan ∠SQR
Solution:
tan R = 5 / 12
PQ / QR = 5 / 12
Hence,
PQ = 5 and QR = 12
In right-angled â–³PQR,
PR2 = PQ2 + QR2
PR2 = 52 + 122
PR2 = 25 + 144
PR2 = 169
PR = √169
We get,
PR = 13
(a) ∠PQS + ∠P = 900 and ∠P + ∠R = 900
Hence,
∠PQS + ∠P = ∠P + ∠R
∠PQS = ∠R
Therefore,
sin ∠PQS = sin R = PQ / PR = 5 / 13
(b) ∠SQR + ∠R = 900 and ∠R + ∠P = 900
Hence,
∠SQR + ∠R = ∠R + ∠P
∠SQR = ∠P
Therefore,
tan ∠SQR = tan P = QR / PQ = 12 / 5
12. In the given figure, â–³ABC is right angled at B. AD divides BC in the ratio 1: 2.
Find
(i) tan ∠BAC / tan ∠BAD
(ii) cot ∠BAC / cot ∠BAD
Solution:
Given
â–³ABC is right angled at B
BD: DC = 1: 2 as AD divides BC in the ratio 1: 2
i.e BD = x, DC = 2x
Hence,
BC = 3x
(i) tan ∠BAC / tan ∠BAD = (BC / AB) / (BD / AB)
tan ∠BAC / tan ∠BAD = BC / BD
tan ∠BAC / tan ∠BAD = 3x / x
tan ∠BAC / tan ∠BAD = 3
(ii) cot ∠BAC / cot ∠BAD = (AB / BC) / (AB / BD)
cot ∠BAC / cot ∠BAD = BD / BC
cot ∠BAC / cot ∠BAD = x / 3x
cot ∠BAC / cot ∠BAD = 1 / 3
13. If sin A = 7/25, find the value of:
(a) (2 tan A) / (cot A – sin A)
(b) cos A + (1/cot A)
(c) cot2 A – cosec2 A
Solution:
Consider △ABC, where ∠B = 900
sin A = Perpendicular / Hypotenuse
sin A = BC / AC
sin A = 7/25
cosec A = 1/sin A
cosec A = 25 / 7
By Pythagoras theorem,
We have,
AC2 = AB2 + BC2
AB2 = AC2 – BC2
AB2 = 252 – 72
AB2 = 625 – 49
AB2 = 576
AB = √576
We get,
AB = 24
Now,
cos A = Base / Hypotenuse
cos A = AB / AC
cos A = 24 / 25
tan A = Perpendicular / Base
tan A = BC / AB
tan A = 7 / 24
cot A = (1/ tan A)
cot A = 24 / 7
(a) (2 tan A) / cot A – sin A = {2 x (7/24)} / {(24/7) – (7/25)}
On further calculation, we get,
= (7/12) / (551/175)
= (7/12) x (175 / 551)
We get,
= 1225 / 6612
(b) cos A + 1/cot A = cos A + tan A
cos A + 1/cot A = (24 / 25) + (7 / 24)
On calculating further, we get,
cos A + 1/cot A = (576 + 175) / 600
cos A + 1/cot A = 751 / 600
(c) cot2 A – cosec2 A = (24 / 7)2 – (25 / 7)2
cot2 A – cosec2 A = (576 / 49) – (625 / 49)
cot2 A – cosec2 A = (576 – 625) / 49
We get,
cot2 A – cosec2 A = – 49 / 49
cot2 A – cosec2 A = – 1
14. If cosec θ = 29 / 20, find the value of:
(a) cosec θ – (1/ cot θ)
(b) sec θ / (tan θ – cosec θ)
Solution:
Consider △ABC, where ∠A = 900
Cosec θ = Hypotenuse / Perpendicular
Cosec θ = BC / AB
Cosec θ = 29 / 20
By Pythagoras theorem,
We have,
BC2 = AB2 + AC2
AC2 = BC2 – AB2
AC2 = 292 – 202
AC2 = 841 – 400
AC2 = 441
AC = √441
We get,
AC = 21
Now,
sec θ = Hypotenuse / Base
sec θ = BC / AC
sec θ = 29 / 21
tan θ = Perpendicular / Base
tan θ = AB / AC
tan θ = 20 / 21
cot θ = 1/ tan θ
cot θ = 21 / 20
(a) cosec θ – (1/ cot θ) = (29 / 20) – {1 / (21/20)}
cosec θ – (1/ cot θ) = (29 / 20) – (20 / 21)
cosec θ – (1/ cot θ) = (609 – 400) / 420
We get,
cosec θ – (1/ cot θ) = 209 / 420
(b) sec θ / (tan θ – cosec θ) = (29 / 21) / {(20 / 21) – (29 / 20)}
sec θ / (tan θ – cosec θ) = (29 / 21) / {(400 – 609) / 420}
sec θ / (tan θ – cosec θ) = (29 / 21) / {(-209 / 420)}
sec θ / (tan θ – cosec θ) = (29 / 21) x (-420 / 209)
sec θ / (tan θ – cosec θ) = – 580 / 209
15. In the given figure, AC = 13 cm, BC = 12 cm and ∠B = 900. Without using tables, find the values of:
(a) sin A cos A
(b) (cos A – sin A) / (cos A + sin A)
Solution:
â–³ABC is a right- angled triangle.
By Pythagoras theorem,
We have,
AC2 = AB2 + BC2
AB2 = AC2 – BC2
AB2 = 132 – 122
AB2 = 169 – 144
AB2 = 25
AB = √25
We get,
AB = 5 cm
sin A = BC / AC
sin A = 12 / 13
cos A = AB / AC
cos A = 5 / 13
(a) sin A cos A = (12 / 13) x (5 / 13)
sin A cos A = 60 / 169
(b) (cos A – sin A) / (cos A + sin A) = {(5/ 13) – (12 / 13)} / {(5/ 13) + (12 / 13)}
(cos A – sin A) / (cos A + sin A) = (-7 / 13) / (17 / 13)
(cos A – sin A) / (cos A + sin A) = (-7 / 13) x (13 / 17)
(cos A – sin A) / (cos A + sin A) = -7 / 17 |
## Arithmetic Reasoning Flash Card Set 386153
Cards 10 Topics Defining Exponents, Distributive Property - Division, Distributive Property - Multiplication, Greatest Common Factor, Least Common Multiple, Negative Exponent, Rates, Rational Numbers, Sequence, Simplifying Fractions
#### Study Guide
###### Defining Exponents
An exponent (cbe) consists of coefficient (c) and a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1).
###### Distributive Property - Division
The distributive property for division helps in solving expressions like $${b + c \over a}$$. It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: $${b + c \over a} = {b \over a} + {c \over a}$$. For example, $${a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6$$.
###### Distributive Property - Multiplication
The distributive property for multiplication helps in solving expressions like a(b + c). It specifies that the result of multiplying one number by the sum or difference of two numbers can be obtained by multiplying each number individually and then totaling the results: a(b + c) = ab + ac. For example, 4(10-5) = (4 x 10) - (4 x 5) = 40 - 20 = 20.
###### Greatest Common Factor
The greatest common factor (GCF) is the greatest factor that divides two integers.
###### Least Common Multiple
The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers.
###### Negative Exponent
A negative exponent indicates the number of times that the base is divided by itself. To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal: $$b^{-e} = { 1 \over b^e }$$. For example, $$3^{-2} = {1 \over 3^2} = {1 \over 9}$$
###### Rates
A rate is a ratio that compares two related quantities. Common rates are speed = $${distance \over time}$$, flow = $${amount \over time}$$, and defect = $${errors \over units}$$.
###### Rational Numbers
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form $${a \over b}$$ where a is the numerator and b is the denominator. An improper fraction ($${5 \over 3}$$) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number ($$1 {2 \over 3}$$) which has a whole number part and a fractional part.
###### Sequence
A sequence is a group of ordered numbers. An arithmetic sequence is a sequence in which each successive number is equal to the number before it plus some constant number.
###### Simplifying Fractions
Fractions are generally presented with the numerator and denominator as small as is possible meaning there is no number, except one, that can be divided evenly into both the numerator and the denominator. To reduce a fraction to lowest terms, divide the numerator and denominator by their greatest common factor (GCF). |
# Surface Area Of A Cylinder
In these lessons, we will learn
• calculate the surface area of solid cylinders.
• calculate the surface area of hollow cylinders or tubes or pipes.
• solve word problems about cylinders.
• calculate the surface area of cylinders using nets.
### Surface Area of a Solid Cylinder
A cylinder is a solid that has two parallel faces which are congruent circles. These faces form the bases of the cylinder. The cylinder has one curved surface. The height of the cylinder is the perpendicular distance between the two bases.
The net of a solid cylinder consists of 2 circles and one rectangle. The curved surface opens up to form a rectangle.
Surface area = 2 × area of circle + area of rectangle
Surface Area = 2πr2 + 2πrh = 2πr(r + h)
where r is the radius and h is the height.
Example:
The diameter of the base of a cylinder is 12 cm and the height is 8 cm. Find the surface area of the solid cylinder.
Solution:
Radius = 6 cm
Surface area = 2πr (r + h)
=
= 528 cm2
How to derive and use the formula of the surface area of a cylinder?
How to find the surface area of a cylinder?
Example:
Find the surface area of a cylinder with r = 18in, h = 17in.
How to calculate the surface area of a cylinder in terms of π?
### Surface Area Of A Hollow Cylinder
Hollow cylinders like pipes or tubes have internal surfaces to consider.
Total surface area of hollow cylinder
= area of internal curved surface + area of external curved surface + area of the two rings
Example:
The figure shows a section of a metal pipe. Given the internal radius of the pipe is 2 cm, the external radius is 2.4 cm and the length of the pipe is 10 cm. Find the total surface area of the pipe.
Solution:
r = 2, R = 2.4, h = 10
Total surface area of pipe
= area of internal surface + area of external surface + area of the two rings
= 2πrh + 2πRh + 2(πR2– πr2)
= (2π × 2 × 10) + (2π × 2.4 × 10) + (2 × (2.42π – 22π))
= 40π + 48π + 3.52π
= 91.52π
= 91.52 × 3.142
= 287.56 cm2
### Word Problems Involving Cylinders
How many square feet of metal are used to make the can?
Find the surface area of a cylinder without the lid.
### Surface Area Of Cylinder Using Nets
Use the net of a cylinder to determine its volume and surface area.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# How many digits are there in decimal number system?
## How many digits are there in decimal number system?
decimal system, also called Hindu-Arabic number system or Arabic number system, in mathematics, positional numeral system employing 10 as the base and requiring 10 different numerals, the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
## How many 5 are there in all?
The largest 5 digit number is 99999. 99999 minus 10000 is 89999, but since you have a zero-based count, you have to add 1. That means there are a total of 90,000 different 5 digit numbers.
How many 5 digit passcodes are there?
The number of 5-digit combinations is 10 5=100,000. So, one more than 99,999. You can generalize that: the number of N-digit combinations is 10 N.
How many 5-digit numbers are there in all formula?
Question 12 Q7)How many 5-digit numbers are there in all? So, 90,000 numbers are there in all.
### How many 5-digit numbers are there such that the sum of their digits is even?
Yes there are 45000 five digit numbers whose sum of the digits will be a even number.
### How many zeros are in the smallest 5 digit number?
10000
The smallest 5-digit number is 10000. There are 4 zeros in 10000. Q3. Smallest 5 – digit number exceeds the largest 4 – digit number by _______ .
Decimal Number System The decimal number system comprises of digits from 0-9 i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8 & 9. The base or radix of the decimal number system is 10 because total number of digits available in decimal number system is 10.
What are the 3 rules on the decimal system?
Some of the decimal number system examples are: 34110, 5610, 678910, 7810. Now as we know to write decimal numbers till 10 let us use the 3 rules on a decimal system,to write further numbers. Write numbers 0–9. Once you reach 9, make the rightmost digit 0 and add 1 to the left which becomes 10.
Why is the base of the decimal number system 10?
The base or radix of the decimal number system is 10 because total number of digits available in decimal number system is 10. It doesn’t imply that these 10 digits will represent only 10 quantities, but all the other digits can be expressed with the help of these numbers.
#### How many quantities can be represented with 10 digits?
It doesn’t imply that these 10 digits will represent only 10 quantities, but all the other digits can be expressed with the help of these numbers. Decimal number system is the most common number system which we use in our daily lives. |
# How do you identify the important parts of y= x^2+2x to graph it?
Oct 6, 2015
y:intercept: $0$
x-intercepts: $0$ and $- 2$
vertex at: $\left(- 1 , - 1\right)$
#### Explanation:
The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = {0}^{2} + 2 \left(0\right) = 0$
The x-intercepts are the values of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} 0 = {x}^{2} + 2 x = \left(x\right) \left(x + 2\right)$
$\textcolor{w h i t e}{\text{XXX}} x = 0$ or $x = - 2$
Convert $y = {x}^{2} + 2 x$ into vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = \left({x}^{2} + 2 x + 1\right) - 1$
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x + 1\right)}^{2} - 1$
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - \left(\textcolor{red}{- 1}\right)\right)}^{2} + \left(\textcolor{b l u e}{- 1}\right)$
which is the vertex form of the equation with the vertex at $\left(\textcolor{red}{\left(- 1\right)} , \textcolor{b l u e}{\left(- 1\right)}\right)$ |
# Mathematics
### Data Handling - Basic Statistics
#### Averages
The number chosen for representing a set of data is called its average. It must,
• • Represent all the values
• • Should not be an exaggerated one - not too small or not too big
There are three averages:
1. Mean
2. Mode
3. Median
Mean
This is the sum of all values divided by the number of data - Σ,sigma, means, add
Mean = ∑ x / n
Mode
This is the value that occurs most frequently.
#### Range of Data
The difference between the highest and lowest values is called the range of the data.
E.g. Data: 3, 4, 8, 6, 11, 21
Range = 21 - 3 = 18
Median
This is the middle value, when the data is arranged in order of size. Now, let's try some examples.
#### Finding Averages of Raw Data
Data that exists in the exact form since its collection is usually considered raw data.
E.g.1
The heights of five plants in a garden are 3cm, 4cm, 7cm, 12cm and 9cm. Find the averages.
Mean = ∑ x / n = 3 + 4 + 7 + 12 + 9 / 5 = 7cm
There is no mode, as each value occurs only once.
To find the median, let's rearrange them in order of size:
3, 4, 7, 9, 12
The middle value is 7. So, the median = 7cm.
E.g.2
The lengths of 6 carpets are 7m, 15m, 15m, 9m, 22m, 4m. Find the averages.
Mean = ∑ x / n = 7 + 15 + 15 + 9 + 22 + 4 / 6 = 12m
The mode = 15m
To find the median, let's rearrange them in order of size:
4, 7, 9, 15, 15, 22
The middle value = 9 + 15 /2 = 12, and so is the median.
#### Finding Averages of Tabular Data
Raw data, when arranged in a table for convenience, is in tabular form.
E.g.
The frequency of shoe sizes of students in a certain class is as follows:
shoe-size (x) frequency (f) 3 3 4 5 5 10 6 8 7 4
Here, we have a slightly different approach;
Mean = sum of fx/n ∑ fx / n = 3X3 + 4X5 + 5X10 + 6X8 + 7X4 /30 = 5.2
TheMedian Class is the class where (n/2)th value lies in. In this case, 30/2 = 15th value lies in shoe-size 5 class. So, it is the median class.
The Modal class is the class with the highest frequency. So, the modal class is shoe-size 5 class.
#### Finding Averages of Grouped Data
Raw data, when arranged in classes for easy handling, form grouped data. It is usually called a group frequency table.
E.g.
The marks obtained by a group of students for maths are as follows:
Marks (x) frequency (f) 0 - 20 3 21 - 40 6 41 - 60 9 61 - 80 8 81 - 100 4
Mean = ∑ fx / n = 10X3 + 30X6 + 50X9 +70X8 + 90X4 /30 = 52.7 - x is the middle class value
The Median Class is the class where n/2 the value lies in. In this case, 30/2 = 15th value lies in 41 - 60 class. So, it is the median class.
The Modal class is the class with the highest frequency. So, the modal class is 41 - 60 class.
The reliability of the Mean
The mean can easily be influenced by the extremes values of data:
E.g
The heights of five plants are 2cm, 4cm, 7cm, 18cm, 19cm. Find the mean and comment on the result.
Mean = 2 + 4 + 7 + 18 + 19 / 5 = 10 cm
This value does not represent either the shortest plant - 2cm - or the tallest - 19cm. So, the mean in this case is not accurate; it may even mislead!
#### Interactive Practice
With the following applet, you can practise the averages of a randomly-generated set of raw data:
#### Practice Questions
Now, in order to complement what you have just learnt, work out the following questions:
1. Find the mean, median and mode of the following numbers - 1, 5, 3, 4, 3, 8, 2, 3, 4, 1.
2. The marks scored by pupils in a certain class for an IQ test are as follows:
Marks (x) frequency (f) 3 2 4 3 5 6 6 4 7 3 8 2
Find the averages of the marks.
3. The mean height of 4 boys is 1.2m and the mean height of 6 girls is 1.5m. Find the mean height of 10 pupils altogether.
4. The marks for a certain test for a group of students are as follows:
Marks (x) frequency (f) 30 5 40 k 50 1
The mean mark for the group is 30. Find k.
5. The median of five consecutive odd numbers is M. Find the mean of the numbers in terms of M. Hence, find the mean of the square of the same numbers.
6. The numbers 4, 5, 9, 15 and k are arranged in ascending order so that the mean is the same as median. Find k. Without further calculation, determine the new mean if the numbers are doubled.
7. A set of numbers are in the ratio 3: 5: 8: 12. The mean turns out to be 42. Find the range of the numbers.
8. Hounslow United football club in West London managed to score 2, 4, 4, 4, 2, 1, 4 goals in their first 7 matches. Find the mean.The manager wants them to keep the average goal of the first 10 matches to be the same as the mean goal so far. How many goals should they score in the next three matches?
9. The mean of 6 numbers is 8. The mean of two numbers is 5. Find the mean of the other four numbers.
10. The list of numbers, 2, 3, 7, 8, x, 12, 16, is in ascending order. Its mode, median and mean are the same. Find x.
Move the mouse over, just below this, to see the answers:
1. 3.4. 3, 3
2. 5.5, 5, 5
3. 1.53
4. 7
5. (M + 4), (M + 4)2 + 8
6. 12, 18
7. 54
8. 3, 9
9. 10
10. 8
Now that you have read this tutorial, you will find the following tutorials very helpful too:
Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.
### Recommended - GCSE & iGCSE
This is the best book available for the new GCSE(9-1) specification and iGCSE: there are plenty of worked examples; a really good collection of problems for practising; every single topic is adequately covered; the topics are organized in a logical order.
### Recommended for A Level
This is the best book that can be recommended for the new A Level - Edexcel board: it covers every single topic in detail;lots of worked examples; ample problems for practising; beautifully and clearly presented. |
# High School Math : How to find the perimeter of a right triangle
## Example Questions
### Example Question #81 : Right Triangles
What is the perimeter of a triangle with side lengths of 5, 12, and 13?
Explanation:
To find the perimeter of a triangle you must add all of the side lengths together.
In this case our equation would look like
### Example Question #1 : How To Find The Perimeter Of A Right Triangle
Three points in the xy-coordinate system form a triangle.
The points are .
What is the perimeter of the triangle?
Explanation:
Drawing points gives sides of a right triangle of 4, 5, and an unknown hypotenuse.
Using the pythagorean theorem we find that the hypotenuse is .
### Example Question #1 : How To Find The Perimeter Of A Right Triangle
Find the perimeter of the following triangle:
Explanation:
The formula for the perimeter of a right triangle is:
where is the length of a side.
Use the formulas for a a triangle to find the length of the base. The formula for a triangle is .
Our triangle is:
Plugging in our values, we get:
### Example Question #171 : Triangles
Find the perimeter of the following right triangle:
Explanation:
The formula for the perimeter of a right triangle is:
where is the length of a side.
Use the formulas for a triangle to find the length of the base and height. The formula for a triangle is
Our triangle is:
Plugging in our values, we get:
### Example Question #2 : How To Find The Perimeter Of A Right Triangle
Based on the information given above, what is the perimeter of triangle ABC?
Explanation:
Consult the diagram above while reading the solution. Because of what we know about supplementary angles, we can fill in the inner values of the triangle. Angles A and B can be found by the following reductions:
A + 120 = 180; A = 60
B + 150 = 180; B = 30
Since we know A + B + C = 180 and have the values of A and B, we know:
60 + 30 + C = 180; C = 90
This gives us a 30:60:90 triangle. Now, since 17.5 is across from the 30° angle, we know that the other two sides will have to be √3 and 2 times 17.5; therefore, our perimeter will be as follows: |
# How to Integrate Arcsin(x) or Sin-1(x)
TR Smith is a product designer and former teacher who uses math in her work every day.
Joined: 5 years agoFollowers: 215Articles: 455
Graphically, the inverse sine function looks like a small segment of the sine function reflected over the diagonal line y = x. In trigonometric terms, the function y = arcsin(x) or y = sin-1(x) means that if you input a number x between -1 and 1, the function will out put an angle measure y between 0 and π radians such that the sine of y equals x. In other words, the function y = arcsin(x) is equivalent to sin(y) = x.
To find the antiderivative of f(x) = arcsin(x) you must use a combination of integration by parts and u-substitution. Here the integral of arsin(x) is worked out step by step.
## Integral of y = arcsin(x)
The first step to integrate the function f(x) = arcsin(x) is to apply the standard trick of integration by parts. The integration by parts formula is
∫ u * dv = u * v - ∫ v * du
In the integral ∫ arcsin(x) dx, we let u = arcsin(x), dx = dv, du = 1/sqrt(1 - x^2) dx, and v = x. This gives us the integral equation
∫ arcsin(x) dx = x*arcsin(x) - ∫ x/sqrt(1 - x^2) dx
The right-hand side of this equation contains a new integral that looks slightly complicated, but we can still work it out if we apply the trick of substitution. Let's set w = x^2 and dw = 2x dx. Now we get the integral equivalence
∫ x/sqrt(1 - x^2) dx
= (1/2) * ∫ 1/sqrt(1 - w) dw
= -sqrt(1 - w) + c
= -sqrt(1 - x^2) + c
If we put all the separate pieces together, we get the final expression for the antiderivative of the function y = arcsin(x).
∫ arcsin(x) dx
= x*arcsin(x) + sqrt(1 - x^2) + c
## How to Use the Integral of arcsin(x)
The integral of the function y = arcsin(x) can be used to find the area under a curve, or to solve certain types of differential equations. The same steps above can also be used to integrate y = arccos(x), the inverse cosine, a closely related function. The integral of arccos(x) works out to be
∫ arccos(x) dx = x*arccos(x) - sqrt(1 - x^2) + c
Here are some more examples of how to use the integral of arcsin(x).
## Example 1
What is the area of the region bounded by the y-axis, the line y = 0.5, and the curve y = sin(x)? The region is shaded in yellow in the graph below.
As a simple integral, the area of the region is best represented by
∫ arcsin(x) dx, on the interval 0 ≤ x ≤ 0.7
The area is then found by plugging the endpoints of the interval into the antederivative and subtracting. This gives us
[0.7*arcsin(0.7) + sqrt(1 - 0.49)] - [0*arcsin(0) + sqrt(1 - 0)]
= 0.256921
## Example 2
Antiderivatives of functions are useful in find the general solutions to certain classes of differential equations. For example, the differential equation y' = x/arcsin(y). To solve this wieth separation of variables, we rewrite it as
dy / dx = x / arcsin(y)
and then separate the x and y expressions.
arcsin(y) dy = x dx
∫ arcsin(y) dy = ∫ x dx
y*arcsin(y) + sqrt(1 - y^2) = (1/2)x^2 + c
This equation cannot be solved for y in terms of x, so this expression is as simplified as it gets.
## Arcsin in Algebra Problems
Inverse trig functions are used to solve problems in regular algebra or pre-calculus. For example, suppose you want to find all the values of x such that
5*sin(πx) + 7 = 2π
Solving this equation for x gives us
5*sin(πx) = 2π - 7
sin(πx) = (2π - 7)/5
πx = arcsin[(2π - 7)/5]
πx = -0.1438586
x = -0.1438586π
x = -0.0457916
This is just one value of x that solves the equation. Since the function sin(πx) is periodic with a period of 2 and is symmetric about the line x = 1/2, the other values of x that solve the equation can be found continuing the following patterns.
x = ...-3+0.0457916, -1+0.457916, 1+0.457916, 3+0.457916, 5+0.457916, ...
x = -4-0.457916, -2-0.457916, -0.45716, 2-0.457916, 4-0.457916, ...
3
2
2
0
0
## Popular
10
9
• ### Names of Geometric Shapes—With Pictures
29
Submit a Comment
• Stephen Rakoczy 2 years ago from Pennsylvania
Thank you for doing this. It's actually not that difficult or long, it just requires some thinking.
• Author
TR Smith 22 months ago from Germany
Typo fixed, thanks! |
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# GMAT Tip: Tackling Remainder Questions
Photograph by John Lund/Getty Images
The GMAT Tip of the Week is a weekly column that includes advice on taking the Graduate Management Admission Test, which is required for admission to most business schools. Every week an instructor from a top test-prep company will share suggestions for improving your GMAT score. This week’s tip comes from Brent Hanneson, creator of GMAT Prep Now, a Web site offering on-demand videos that teach GMAT skills.
Consider the following question:
N is a positive integer. When N is divided by 13, the remainder is 5. When N is divided by 46, the remainder is 31. What is the smallest possible value of N?
Before we examine the solution to this question, I’d like to ask an easier question:
N is a positive integer. When N is divided by 7, the remainder is 3. What are three possible values of N?
Does your list include 3 as a possible value for N? It should since 3 is the smallest number that meets the given criteria (3 divided by 7 equals 0 with a remainder of 3).
When it comes to GMAT questions involving remainders, it’s often useful to begin listing numbers that meet the given criteria. When it comes to listing possible values, we have a useful rule:
If N and D are positive integers, and if N divided by D is equal to Q with remainder R, then the possible values of N are: R, R+D, R+2D, R+3D,. . .
Example: When positive integer W is divided by 6, the remainder is 5. Given this information, the possible values of W are: 5, 5+6, 5+2(6), 5+3(6), 5+4(6), . . .
Upon simplification, we find that the possible values of W are: 5, 11, 17, 23, 29 and so on.
The important takeaway is that you can sometimes save yourself a lot of work by listing possible values and, more importantly, by including the smallest possible value in that list.
Now back to the original question:
N is a positive integer. When N is divided by 13, the remainder is 5. When N is divided by 46, the remainder is 31. What is the smallest possible value of N?
First, if N divided by 13 gives us a remainder of 5, the possible values of N are: 5, 18, 31, 44, 57 and so on.
Second, if N is divided by 46 gives us remainder 31, the possible values of N are: 31, … STOP.
Since both lists include 31, the answer to our original question must be 31.
So, when you encounter a GMAT remainder question, one approach may involve identifying possible values of a certain number. When identifying those values, it may be to your advantage to identify the smallest possible value.
Brent Hanneson, the creator of GMAT Prep Now, has worked in the field of education for most of his career. He has taught courses at three different test-prep companies and created comprehensive GMAT and GRE courseware packages used by the University of British Columbia and 12 other universities across North America.
Join the discussion on the Bloomberg Businessweek Business School Forum, visit us on Facebook, and follow @BWbschools on Twitter.
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# 2019 AMC 10A Problems/Problem 15
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.
## Problem
A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \frac{3}{7}$, and $$a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$$for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$
$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$
## Video Solution
Education, the Study of Everything
~ pi_is_3.14
## Solution 1 (Induction)
Using the recursive formula, we find $a_3=\frac{3}{11}$, $a_4=\frac{3}{15}$, and so on. It appears that $a_n=\frac{3}{4n-1}$, for all $n$. Setting $n=2019$, we find $a_{2019}=\frac{3}{8075}$, so the answer is $\boxed{\textbf{(E) }8078}$.
To prove this formula, we use induction. We are given that $a_1=1$ and $a_2=\frac{3}{7}$, which satisfy our formula. Now assume the formula holds true for all $n\le m$ for some positive integer $m$. By our assumption, $a_{m-1}=\frac{3}{4m-5}$ and $a_m=\frac{3}{4m-1}$. Using the recursive formula, $$a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},$$ so our induction is complete.
## Solution 2
We have $\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}$, in other words, $\frac{1}{a_n}-\frac{1}{a_{n-1}} = \frac{1}{a_{n-1}}-\frac{1}{a_{n-2}}$. So $\{\frac{1}{a_n}\}$ is an arithmetic sequence with step size $\frac{7}{3}-1=\frac{4}{3}$, which means $\frac{1}{a_{2019}} = 1+2018 \cdot \frac{4}{3} = \frac{8075}{3}$. Since the numerator and the denominator are relatively prime, the answer is $\boxed{\textbf{(E) } 8078}$.
-eric2020 (modified by Dolphindesigner)
## Solution 3
It seems reasonable to transform the equation into something else. Let $a_{n}=x$, $a_{n-1}=y$, and $a_{n-2}=z$. Therefore, we have $$x=\frac{zy}{2z-y}$$ $$2xz-xy=zy$$ $$2xz=y(x+z)$$ $$y=\frac{2xz}{x+z}$$ Thus, $y$ is the harmonic mean of $x$ and $z$. This implies $a_{n}$ is a harmonic sequence or equivalently $b_{n}=\frac{1}{a_{n}}$ is arithmetic. Now, we have $b_{1}=1$, $b_{2}=\frac{7}{3}$, $b_{3}=\frac{11}{3}$, and so on. Since the common difference is $\frac{4}{3}$, we can express $b_{n}$ explicitly as $b_{n}=\frac{4}{3}(n-1)+1$. This gives $b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}$ which implies $a_{2019}=\frac{3}{8075}=\frac{p}{q}$. $p+q=\boxed{\textbf{(E) } 8078}$ ~jakeg314
## Solution 4 (Arithmetic Sequence)
Notice that$$a_n = \frac{1}{\frac{2}{a_{n-1}} - \frac{1}{a_{n-2}}}.$$Therefore,$$\frac{1}{a_n} = \frac{2}{a_{n-1}} - \frac{1}{a_{n-2}}, \ \ \implies \ \ \frac{\frac{1}{a_n} + \frac{1}{a_{n-2}}}{2} = \frac{1}{a_{n-1}}.$$Therefore, the sequence $b_n = \frac{1}{a_n}$ is an arithmetic sequence. Notice that the common difference of $b$ is $\frac{4}{3},$ and therefore$$b_{2019} = b_1 + 2018 \bigg(\frac{5}{3}\bigg) = 1 + 2018 \bigg(\frac{4}{3} \bigg) = \frac{8075}{3}.$$Therefore, we see that $a_{2019} = \frac{3}{8075},$ so that $p + q = \boxed{\text{(E) } 8078}.$
~Professor-Mom
Note: This is similar to solutions #2 and #3, although you can notice that in #2's case the new sequence $B$ actually forms an arithmetic sequence. |
# Lesson 13
## Warm-up: Conversación numérica: Paréntesis (10 minutes)
### Narrative
This Number Talk encourages students to think about equivalent expressions and to rely on the properties of operations to mentally solve problems. The strategies elicited here will be helpful later in the lesson when students match diagrams to expressions.
### Launch
• Display one expression.
• “Hagan una señal cuando tengan una respuesta y puedan explicar cómo la obtuvieron” // “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Encuentra mentalmente el valor de cada expresión.
• $$5 \times (7 + 4)$$
• $$(5 \times 7) + (5 \times 4)$$
• $$(5 \times 7) + (5 \times \frac{1}{4})$$
• $$(5 \times 7) - (5 \times \frac{1}{4})$$
### Activity Synthesis
“¿En qué se parecen las dos últimas expresiones? ¿En qué son diferentes?” // “What is the same about the last two expressions? What is different?” (They both have the same two products but one of them is the sum and the other is the difference.)
## Activity 1: Clasificación de tarjetas: Diagramas y expresiones (20 minutes)
### Narrative
The purpose of this activity is for students to analyze area diagrams and use the properties of operations to interpret expressions. The diagrams are decomposed in different ways and the expressions all have a fractional part but sometimes it is written as a mixed number and sometimes the whole number and fraction are separated using the distributive property. The numbers in the diagrams, both the whole number part and the fractional part, are deliberately chosen to resemble one another so students need to analyze the expressions carefully to make matches.
MLR8 Discussion Supports. Students should take turns finding a match and explaining their reasoning to their partner. Display the following sentence frame for all to see: “Observé ___, entonces emparejé . . .” // “I noticed ___ , so I matched . . . .” Encourage students to challenge each other when they disagree.
### Required Materials
Materials to Copy
• Card Sort: Diagrams and Expressions
### Required Preparation
• Create a set of cards from the blackline master for each group of 2.
### Launch
• Groups of 2
• Give each group a set of cards from the blackline master.
### Activity
• 1–2 minutes: independent work time
• 8–10 minutes: partner work
### Student Facing
Tu profesor va a entregarles unas tarjetas a ti y a tu compañero.
1. Clasifica las tarjetas de una forma que tenga sentido para ti.
2. Asocia cada expresión a un diagrama que le corresponda. Algunos diagramas corresponden a más de una expresión.
3. Trabaja con tu compañero para encontrar el área de cada región sombreada. Explica o muestra tu razonamiento.
### Activity Synthesis
• Display Diagram C and Expressions J and O.
$$3 \times 5 \frac{2}{5}$$
$$(3 \times 5) + (3 \times \frac {2}{5})$$
• “¿De qué manera cada expresión representa el área de la región sombreada?” // “How does each expression represent the area of the shaded region?” (There are 3 rows and each row has an area of $$5\frac{2}{5}$$ square units.)
• Display Diagram B and Expression L.
$$(5 \times 3) -\left(5 \times \frac{2}{5}\right)$$
• “¿De qué manera la expresión representa el área de la región sombreada?” // “How does the expression represent the area of the shaded region?” ($$5 \times 3$$ is the area of the full rectangle and then I take away the unshaded part, $$5 \times \frac{2}{5}$$.)
## Activity 2: Escribamos expresiones (15 minutes)
### Narrative
The purpose of this activity is for students to write expressions that represent the area of shaded regions. Monitor for students who are writing a variety of expressions that represent the distributive property.
Action and Expression: Internalize Executive Functions. Invite students to verbalize their strategy for writing an expression to match the shaded area in each diagram before they begin. Students can speak quietly to themselves, or share with a partner.
Supports accessibility for: Organization, Conceptual Processing, Language
• Groups of 2
### Activity
• 1–2 minutes: quiet think time
• 5–7 minutes: partner work time
### Student Facing
En cada caso, escribe tantas expresiones como puedas que correspondan al área de la región sombreada.
1.
2.
### Student Response
If students only write one expression for a diagram, partner them with a student who found a different expression and ask:
• “¿De qué manera esta expresión representa la región sombreada?” // “How does this expression represent the shaded region?”
• “¿Cómo pueden dos expresiones diferentes representar la misma región sombreada?” // “How can two different expressions represent the same shaded region?”
• “¿Qué otras expresiones podemos escribir para representar la región sombreada?” // “What other expressions can we write to represent the shaded region?”
### Activity Synthesis
• Select 2–3 students to share their expressions for the first problem.
• If not mentioned by students, display:
$$4 \times 5\frac{1}{3}$$
$$4 \times \frac{16}{3}$$
• “¿De qué manera cada expresión representa el primer problema?” // “How does each expression represent the first problem?” (There are 4 rows and each row has $$5 \frac{1}{3}$$ square units or $$\frac{16}{3}$$ square units.)
• “¿Su estrategia cambió cuando escribieron expresiones para el último diagrama? Si sí, ¿cómo cambió?” // “Did your approach change when writing expressions for the last diagram? If yes, how?” (It was easier to see the square units and apply the distributive property.)
• If not mentioned by students, display these expressions for the area of the last diagram in square units:
$$(7 \times 3) +\left(7 \times \frac{3}{4}\right)$$
$$(7 \times 4) - \left(7 \times \frac{1}{4}\right)$$
## Lesson Synthesis
### Lesson Synthesis
“¿Cuál es su estrategia favorita para encontrar el área de una región sombreada? ¿Hay alguna estrategia nueva que les emocione intentar o conocer mejor?” // “What’s your favorite way to find the area of a shaded region? What is a new way that you are excited to try or learn more about?”
Consider having students write their responses in their journals. |
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# Solving Proportions
To solve a proportion, you need to find the value of an unknown variable that makes two ratios equal.
• Solving Proportions
• Solved Example
• FAQs
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## Solving Proportions
To solve a proportion, you need to find the value of an unknown variable that makes two ratios equal. Here are the steps to solve a proportion:
Write the proportion: Use the format “a/b = c/d” to write the proportion. The variables “a” and “c” represent the numerators of the ratios, while “b” and “d” represent the denominators.
Cross-multiply: Multiply the numerator of one ratio by the denominator of the other ratio, and vice versa. This gives you two equations that you can use to solve for the unknown variable.
Solve for the unknown variable: Use algebraic techniques to isolate the unknown variable in one of the equations. Then, substitute the value you find into either of the original ratios to check if the proportion is true.
### Here’s an example:
Suppose you have the proportion 3/5 = x/20. To solve for x, you would follow these steps:
Write the proportion: 3/5 = x/20.
Cross-multiply: Multiply the numerator of the first ratio by the denominator of the second ratio, and vice versa. This gives you:
3(20) = 5x
60 = 5x
Solve for the unknown variable: Divide both sides of the equation by 5 to isolate x. This gives you:
x = 12
Check the proportion: Substitute the value of x back into either of the original ratios to make sure the proportion is true. For example:
3/5 = 12/20
0.6 = 0.6
Since both sides of the equation are equal, the proportion is true and the value of x is correct.
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## Solved Example
Suppose you have the proportion 2/3 = x/9. To solve for x, you would follow these steps:
Write the proportion: 2/3 = x/9.
Cross-multiply: Multiply the numerator of the first ratio by the denominator of the second ratio, and vice versa. This gives you:
2(9) = 3x
18 = 3x
Solve for the unknown variable: Divide both sides of the equation by 3 to isolate x. This gives you:
x = 6
Check the proportion: Substitute the value of x back into either of the original ratios to make sure the proportion is true. For example:
2/3 = 6/9
0.666… = 0.666…
Since both sides of the equation are equal, the proportion is true and the value of x is correct. Therefore, x = 6 is the solution to the proportion.
## Solving Proportions FAQS
##### What is a proportion?
A proportion is a statement that two ratios are equal. It can be written in the form “a/b = c/d”, where “a” and “c” are the numerators of the ratios, and “b” and “d” are the denominators.
##### How do you solve a proportion?
To solve a proportion, you need to find the value of an unknown variable that makes the two ratios equal. This can be done by cross-multiplying and then solving for the unknown variable using algebraic techniques.
##### What does it mean to cross-multiply?
Cross-multiplying means multiplying the numerator of one ratio by the denominator of the other ratio, and vice versa. This results in two equations that can be used to solve for the unknown variable.
##### What are some algebraic techniques used to solve proportions?
Algebraic techniques used to solve proportions include dividing both sides of the equation by the same value, combining like terms, and distributing operations.
##### How do you check if a solution to a proportion is correct?
You can check if a solution to a proportion is correct by substituting the value of the unknown variable back into either of the original ratios and verifying that the two sides of the equation are equal.
##### Can proportions be solved using decimals or fractions?
Yes, proportions can be solved using decimals or fractions. However, it is usually easier to work with fractions or mixed numbers because they can be simplified more easily.
Gloria Mathew writes on math topics for K-12. A trained writer and communicator, she makes math accessible and understandable to students at all levels. Her ability to explain complex math concepts with easy to understand examples helps students master math. LinkedIn
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# 6.1 - Ratios
6.1 The student will represent relationships between quantities using ratios, and will use appropriate notations, such as a/b , a to b, and a:b.
### BIG IDEAS
• I can figure miles per gallon, cost per unit, and estimate the size of an animal population when the entire population cannot be counted.
• I will be able to solve practical problems with proportional reasoning.
### UNDERSTANDING THE STANDARD
2016 VDOE Curriculum Framework - 6.1 Understanding
· A ratio is a comparison of any two quantities. A ratio is used to represent relationships within a quantity and between quantities. Ratios are used in practical situations when there is a need to compare quantities.
· In the elementary grades, students are taught that fractions represent a part-to-whole relationship. However, fractions may also express a measurement, an operator (multiplication), a quotient, or a ratio. Examples of fraction interpretations include:
- Fractions as parts of wholes: represents three parts of a whole, where the whole is separated into four equal parts.
- Fractions as measurement: the notation can be interpreted as three one-fourths of a unit.
- Fractions as an operator: represents a multiplier of three-fourths of the original magnitude.
- Fractions as a quotient: represents the result obtained when three is divided by four.
- Fractions as a ratio: is a comparison of 3 of a quantity to the whole quantity of 4.
· A ratio may be written using a colon (a:b), the word to (a to b), or fraction notation .
· The order of the values in a ratio is directly related to the order in which the quantities are compared.
- Example: In a certain class, there is a ratio of 3 girls to 4 boys (3:4).
Another comparison that could represent the relationship between these quantities is the ratio of 4 boys to 3 girls (4:3). Both ratios give the same information about the number of girls and boys in the class, but they are distinct ratios. When you switch the order of comparison (girls to boys vs. boys to girls), there are different ratios being expressed.
· Fractions may be used when determining equivalent ratios.
- Example: The ratio of girls to boys in a class is 3:4, this can be interpreted as:
number of girls = ∙ number of boys.
In a class with 16 boys, number of girls = ∙ (16) = 12 girls.
- Example: A similar comparison could compare the ratio of boys to girls in the class as being 4:3, which can be interpreted as:
number of boys = ∙ number of girls.
In a class with 12 girls, number of boys = ∙ (12) = 16 boys.
· A ratio can compare two real-world quantities (e.g., miles per gallon, unit rate, and circumference to diameter of a circle).
· Ratios may or may not be written in simplest form.
· A ratio can represent different comparisons within the same quantity or between different quantities.
Ratio Comparison part-to-whole (within the same quantity) compare part of a whole to the entire whole part-to-part (within the same quantity) compare part of a whole to another part of the same whole whole-to-whole (different quantities) compare all of one whole to all of another whole part-to-part (different quantities) compare part of one whole to part of another whole
- Examples: Given Quantity A and Quantity B, the following comparisons could be expressed.
Ratio Example Ratio Notation(s) part-to-whole (within the same quantity) compare the number of unfilled stars to the total number of stars in Quantity A 3:8; 3 to 8; or part-to-part 1 (within the same quantity) compare the number of unfilled stars to the number of filled stars in Quantity A 3:5 or 3 to 5 whole-to-whole 1 (different quantities) compare the number of stars in Quantity A to the number of stars in Quantity B 8:5 or 8 to 5 part-to-part 1 (different quantities) compare the number of unfilled stars in Quantity A to the number of unfilled stars in Quantity B 3:2 or 3 to 2
1Part-to-part comparisons and whole-to-whole comparisons are ratios that are not typically represented in fraction notation except in certain contexts, such as determining whether two different ratios are equivalent.
### ESSENTIALS
• What is a ratio?
A ratio is a comparison of any two quantities. A ratio is used to represent relationships within a set and between two sets. A ratio can be written using fraction form (2/3), a colon (2:3), or the word to (2 to 3).
The student will use problem solving, mathematical communication, mathematical reasoning, connections, and representations to
· 6.11 Represent a relationship between two quantities using ratios.
· 6.12 Represent a relationship in words that makes a comparison by using the notations, a:b, and a to b.
· 6.13 Create a relationship in words for a given ratio expressed symbolically.
### KEY VOCABULARY
ratio, notation, relationship, corresponding parts, simplest form, comparison, symbolically, fraction, numerator, denominator, rate, equivalent
6-8 Math Strategies
Updated: Nov 20, 2018 |
# W1 - Lesson 3: Multiplying And Dividing Fractions
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Mathematics Grade 8W1 - Lesson 3: Multiplying and DividingFractionsV6-10
Preview/Review ConceptsforGrade Eight MathematicsW1 – Lesson 3:Multiplying and DividingFractions
OBJECTIVESBy the end of this lesson, you will be able to: Express a given positive mixed number as an improper fraction and a given positiveimproper fraction as a mixed number Model the process of multiplying fractions using models and numerically Model the process of dividing fractions using models and numerically Apply rules for multiplying and dividing positive fractions, including mixednumbers Solve an expression involving fractions and apply the order of operations Solve problems involving the multiplication and division of fractions and mixednumbersGLOSSARYDenominator – the bottom number ofa fraction, the number of equal sizedparts a whole has been divided into.Factors – the numbers that aremultiplied to give a product. A factorof a given number will divide into thegiven number with no remainder leftover. For example, the 4 and 5 arefactors of 20.Greatest Common Factor (GCF) –the largest factor that is common to aset of numbers. Example, the GCF of12 and 18 is 6 because 6 divides into12 twice and into 18 three times.Improper Fraction – a fraction inwhich the numerator that is greaterthan the denominator.Mixed number – a number representedby a whole number and a fraction.Multiple – a number that is the productof a natural number and anothernumber. For example, the multiplesof 5 are 5, 10, 15, 20, etc.Numerator – the top number of afraction, the number of pieces beingconsidered.Simplify – reducing a fraction intoits lowest terms. This is done bydividing both the numerator anddenominator by the GCF.
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3W1 – Lesson 3: Multiplying and Dividing FractionsMaterials required: Paper, Pencil, and CalculatorReview: Converting between Improper Fractions and MixedNumbersTo change a mixed number into an improper fraction, multiply the denominator by thewhole number and add the numerator. This value becomes the numerator in the improperfraction. The denominator in the mixed number is the same denominator you use in theimproper fraction.First, multiply thedenominator by thewhole numberFor example: 21 3 2 1 7 333Second, add thenumerator.To change an improper fraction into a mixed number, reverse the process outlined above.Determine how many times the denominator goes into the numerator and this valuebecomes the whole number. The remainder becomes the numerator in the fraction. Thedenominator remains the same.For example:9 2 and there is one left over491 244This is the remainder.Developed by Alberta Distance Learning Centre .1
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3Part 1: Multiplying FractionsWhen you multiply fractions, what you are really doing is finding a fraction of a fraction.Fractions can be multiplied using the area model.Use the area model to multiply2 1 .3 5First use a square to represent2vertically.3Then, represent1horizontally on the same square.5The shaded squares that overlap represent the numerator of the answer. The totalnumber of parts in the square represents the denominator.In this case, the numerator would be 2, and the denominator would be equal to 15.Thus,2 1 2 .3 5 152 . Developed by Alberta Distance Learning Centre
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3When multiplying fractions numerically, remember these steps:Step 1: Convert any mixed numbers into improper fractions.Step 2: Simplify the numerators and denominators. To do this, divide the numerators anddenominators by a common factor.Step 3: Multiply the numerators.Step 4: Multiply the denominators.Step 5: If an improper fraction results, change it into a mixed number.Example 13 1 3 1 4 2 4 23 8Example 21 37 3 2 3 43 47 31 31 4Cancel out the 3's.7 11 47 43 14 Developed by Alberta Distance Learning Centre .3
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3Example 377 5 5 88 17 5 8 135 83 48Example 41110 21 3 5343 45710 21 42135 7 1 235 21 172 When you simplify the expression before youstart multiplying the result will always be inlowest terms.Practice Questions1.3 3 54 . Developed by Alberta Distance Learning Centre
Preview/Review Concepts W1 - Lesson 32.5 2 6 33.3 35 5 84.121 2 25Mathematics Grade 8Developed by Alberta Distance Learning Centre .5
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3Part 2: Dividing FractionsWhen you divide 12 by 3, you are determining how many groups of 4 are in 12. When 12is divided into groups of 4, the result is 3. Dividing fractions uses a similar method.You can divide fractions using a number line.Let’s use 5 groups of11 20 as an example. When dividing 5 by, you must determine how many441are in 5.41414140There are 20 groups ofThus, 5 1414141141414214141414141431414414141451in 5.41 20 .4When dividing fractions numerically, remember these steps:Step 1: Convert any mixed numbers into improper fractions.Step 2: Multiply the first term by the reciprocal of the second term. The reciprocal is the“flip” of the fraction or number.Step 3: Multiply the numerators.Step 4: Multiply the denominators.Step 5: If an improper fraction results, change it into a mixed number.6 . Developed by Alberta Distance Learning Centre
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3Example 133 1 2 44 23 1 4 23 8The reciprocal of 2 is12Example 212 212 5 51 26 12 5 121Multiply by the reciprocalof the second term6 51 1 30 Example 33 13 5 7 57 13 5 7 115 71 27Developed by Alberta Distance Learning Centre .7
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3Example 44318 27 2 3 787818 8 7 27 2188 727 32 87 316 21 Practice Questions2 31.8 2.5 4 128 . Developed by Alberta Distance Learning Centre
Preview/Review Concepts W1 - Lesson 33.2 1 7 54.232 1 54Mathematics Grade 8Developed by Alberta Distance Learning Centre .9
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3Part 3: Applying Order of OperationsWhen calculating the answer to an expression with many mathematical operations youmust follow the order of operations. Use the acronym BEDMAS to help you.Complete theseoperations,divisions ormultiplicationin order from left to onadditionsubtractionComplete these operations,addition or subtractionin order from left to right.Example 133 1 4Evaluate the following expression 4 5 5Answer:Step 1: Since there are no brackets present, evaluate the exponent.33 1 4 4 5 5 314 4 125 510 . Developed by Alberta Distance Learning Centre
Preview/Review Concepts W1 - Lesson 3Mathematics Grade 8Step 2: Complete the division.314 4 125 531 5 4 125 45131 4 25 125 4 31 4 100Step 3: Complete the addition.31 4 100751 100 10076 10019 25 Developed by Alberta Distance Learning Centre .11
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3Example 2 5 1 2 1 Evaluate the following expression 1 . 7 2 3 4 Answer:Step 1: Solve the operations in the first set of brackets. 5 1 2 1 1 7 2 3 4 5 2 2 1 1 7 1 3 4 10 2 1 1 7 3 4 Step 2: Solve the operations in the second set of brackets. 10 2 1 1 7 3 4 10 5 1 7 3 4 10 20 3 7 12 12 10 23 7 12 Step 3: Solve the operations in the second set of brackets. 10 23 7 12 5 10 23 7 126 5 23 7 6115 4231 24212 . Developed by Alberta Distance Learning Centre
Preview/Review Concepts W1 - Lesson 3Mathematics Grade 8Practice Questions1.1 1 1 2 7 22. 2 5 10 3 6 9 2Developed by Alberta Distance Learning Centre .13
Mathematics Grade 83.Preview/Review Concepts W1 - Lesson 33 2 1 2 8 3 6 514 . Developed by Alberta Distance Learning Centre
Preview/Review Concepts W1 - Lesson 3Mathematics Grade 8Lesson 3: Assignment1.5 1 6 32.9 1 10 63.44 2 74.4 32 5 85.721 2 856.73 3 124Developed by Alberta Distance Learning Centre .15
Mathematics Grade 87.3 1 5 48.5 2 6 39.8 1Preview/Review Concepts W1 - Lesson 31 410.2 13 5 511.35 2 812.1101 3 11216 . Developed by Alberta Distance Learning Centre
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 31 413.2 14.6 3 1315.9 3 11Solve the following problems involving fractions. All the answers must be in lowest terms,include the units, and written in complete sentences.16.1An elastic band will stretch to be 4 8 times its original size. If the elastic back is102cm long, to what length will it stretch?3Developed by Alberta Distance Learning Centre .17
Mathematics Grade 817.Preview/Review Concepts W1 - Lesson 3Captain Dianne is in her canoe and notices it has 20container that holds 31L of water in it. She has a21L of water. How many scoops of water will must she3make to bail all the water out of her canoe?18.8How many complete pieces of ropemetres long can be cut from a roll of rope that9is 36 metres long?18 . Developed by Alberta Distance Learning Centre
Preview/Review Concepts W1 - Lesson 319.Mathematics Grade 8Corbin and Sheila are in the widget-selling business together. Since Shelia investedmore money into the business than Corbin did, she is entitled to3of the profits.5For every widget they sell, they earn 18.50 in profit. How much will Corbin get foreach widget that is sold? Express the answer as a fraction and in dollars.20.1 3 1 2 4 8 Developed by Alberta Distance Learning Centre .19
Mathematics Grade 8Preview/Review Concepts W1 - Lesson 3321.2 1 3 2 22.211 5 320 . Developed by Alberta Distance Learning Centre
Preview/Review Concepts W1 - Lesson 323.Mathematics Grade 82 9 3 3 2 3 10 5 7Developed by Alberta Distance Learning Centre .21
Mathematics Grade 824.Preview/Review Concepts W1 - Lesson 33 8 3 5 3 5 9 4 6 22 . Developed by Alberta Distance Learning Centre
Part 2: Dividing Fractions When you divide 12 by 3, you are determining how many groups of 4 are in 12. When 12 is divided into groups of 4, the result is 3. Dividing fractions uses a similar method. You can divide fractions using a number line. Let’s use 1 5 20 4 as an example. When dividing
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