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1.
Introduction
1.1.
Problem Description🗒️
1.2.
Sample Examples😄
2.
Approach🤖
2.1.
Algorithm👽
2.2.
C++ Code⌨️
2.3.
JAVA Code⌨️
2.3.1.
Complexity Analysis
3.
3.1.
What is a Binary Tree?
3.2.
What is a BST?
3.3.
How can we calculate the shortest distance between two nodes in BST?
4.
Conclusion
Last Updated: Mar 27, 2024
Medium
# Calculate the Shortest Distance Between Two Nodes in BST
Manish Kumar
1 upvote
## Introduction
Hi Ninja! In this article, we will learn to solve the problem of calculating the shortest distance between two nodes in BST. We will use the property of BST that the values of all the nodes in the left subtree are less than the current node's value and the values of nodes in the right subtree are greater than the current node's value.
I hope you will enjoy and have fun learning a medium-level DSA problem on binary search tree♟ "Calculate the shortest distance between two nodes in BST."
### Problem Description🗒️
We have been given a binary search tree and two values. We have to find the shortest distance between two nodes having these values. Assume that both values exist in the given BST.
### Sample Examples😄
Input 1:
``Root of above tree, x = 7, y = 12``
Output:
``4``
Explanation:
Distance between 7 and 12 in above given BST is 4.
Input 2:
``Root of above tree, x = 6, y = 15``
Output:
``2``
Explanation:
Distance between 6 and 15 in above given BST is 2.
## Approach🤖
In the case of Binary Search Trees, we can find the distance between two nodes faster as compared to normal binary trees. In this approach, we will use BST’s properties.
We know that BST follows these properties:
1. For a node, all the nodes in the left subtree have values less than the current node's value.
2. For a node, all the nodes in the right subtree have values greater than the current node's value.
3. The left and right subtree should also be binary search trees and follow all three properties.
### Algorithm👽
We will start from the root, and for every node, we do the following:
• If both the given values are smaller than the current node, we move to the left subtree of the current node.
• If both the given values are greater than the current node, we move to the right subtree of the current node.
• If one of the given values is smaller and the other value is greater than the current node, then the current node is the Lowest Common Ancestor (LCA) of the two nodes with the given value.
• We find the distances of the current node from these two given values, and the sum of the distances will be our answer.
### C++ Code⌨️
``````//Program to calculate the
//shortest distance between two nodes in BST
#include <bits/stdc++.h>
using namespace std;
struct Node {
struct Node* left;
struct Node* right;
int value;
};
struct Node* newNode(int value)
{
struct Node* ptr = new Node;
ptr->value = value;
ptr->left = ptr->right = NULL;
return ptr;
}
// Insert function in BST
struct Node* insert(struct Node* root, int value)
{
if (!root)
root = newNode(value);
else if (root->value > value)
root->left = insert(root->left, value);
else if (root->value < value)
root->right = insert(root->right, value);
return root;
}
// This function returns distance of x from root.
int rootDistance(struct Node* root, int x)
{
if (root->value == x)
return 0;
else if (root->value > x)
return 1 + rootDistance(root->left, x);
return 1 + rootDistance(root->right, x);
}
// This returns the minimum distance between x and y.
// assuming that x and y exist in BST.
int distance(struct Node* root, int x, int y)
{
if (!root)
return 0;
// Both values lie on left
if (root->value > x && root->value > y)
return distance(root->left, x, y);
// Both values lie in the right
if (root->value < x && root->value < y)
return distance(root->right, x, y);
// Lie in opposite directions
// Root is the LCA of two nodes
if (root->value >= x && root->value <= y)
return rootDistance(root, x) +
rootDistance(root, y);
}
//function to check x smaller than y
int findDistance(Node *root, int x, int y)
{
if (x > y)
swap(x, y);
return distance(root, x, y);
}
// main
int main()
{
struct Node* root = NULL;
root = insert(root, 10);
insert(root, 5);
insert(root, 15);
insert(root, 2);
insert(root, 7);
insert(root, 12);
insert(root, 20);
insert(root, 17);
insert(root, 25);
int x = 7, y = 12;
cout << findDistance(root, 7, 12);
return 0;
}``````
Output:
``4``
### JAVA Code⌨️
``````//Program to calculate the
//shortest distance between two nodes in BST
class Ninja {
static class Node {
Node left, right;
int value;
}
static Node newNode(int value)
{
Node ptr = new Node();
ptr.value = value;
ptr.left = null;
ptr.right = null;
return ptr;
}
// Insert function in BST
static Node insert(Node root, int value)
{
if (root == null)
root = newNode(value);
else if (root.value > value)
root.left = insert(root.left, value);
else if (root.value < value)
root.right = insert(root.right, value);
return root;
}
// This function returns the distance of x from the root.
static int distanceFromRoot(Node root, int x)
{
if (root.value == x)
return 0;
else if (root.value > x)
return 1 + distanceFromRoot(root.left, x);
return 1 + distanceFromRoot(root.right, x);
}
// This returns the minimum distance between x and y.
// assuming that x and y exist in BST.
static int distance(Node root, int x, int y)
{
if (root == null)
return 0;
// Both values lie in left
if (root.value > x && root.value > y)
return distance(root.left, x, y);
// Both values lie in right
if (root.value < x && root.value < y) // same path
return distance(root.right, x, y);
// Lie in opposite directions
// Root is the LCA of two nodes
if (root.value >= x && root.value <= y)
return distanceFromRoot(root, x) + distanceFromRoot(root, y);
return 0;
}
//function to check x smaller than y
static int findDistance(Node root, int x, int y)
{
int temp = 0;
if (x > y)
{
temp = x;
x = y;
y = temp;
}
return distance(root, x, y);
}
// Driver code
public static void main(String[] args)
{
Node root = null;
root = insert(root, 10);
insert(root, 5);
insert(root, 15);
insert(root, 2);
insert(root, 7);
insert(root, 12);
insert(root, 20);
insert(root, 17);
insert(root, 25);
int x = 7, y = 12;
System.out.println(findDistance(root, 7, 12));
}
}``````
Output:
``4``
#### Complexity Analysis
Time Complexity: O(H), where H is the height of the given binary search tree. We find the distance of nodes from the LCA and in BST, the maximum number of nodes we have to visit is the tree's height.
Space Complexity: O(H), where H is the height of the BST and O(H) space is required for the call stack.
### What is a Binary Tree?
As the name suggests, Binary means two, i.e., a tree node is allowed to have a maximum of two children.
### What is a BST?
BST (Binary Search Tree) is a special type of node-based binary tree data structure. It follows properties like:
1. For a node, all the nodes in the left subtree have values less than the current node's value.
2. For a node, all the nodes in the right subtree have values greater than the current node's value.
3. The left and right subtree should also be binary search trees and follow all three properties.
### How can we calculate the shortest distance between two nodes in BST?
In the case of Binary Search Trees, we can find the distance between two nodes faster as compared to normal binary trees. In this approach, we will use BST’s properties.
We know that BST follows these properties:
For a node, all the nodes in the left subtree have values less than the current node's value.
For a node, all the nodes in the right subtree have values greater than the current node's value.
The left and right subtree should also be binary search trees and follow all three properties.
## Conclusion
In this blog, we learned how to solve the popular interview question to calculate the shortest distance between two nodes in BST. I hope you enjoyed reading our blog on ‘Calculate the shortest distance between two nodes in BST’.
Check out this problem - Connect Nodes At Same Level
Refer to our Guided Path on Coding Ninjas Studio to upskill ourselves in Data Structures and AlgorithmsCompetitive ProgrammingJavaScriptSystem DesignMachine learning, and many more! But suppose we have just started our learning process and are looking for questions from tech giants like Amazon, Microsoft, Uber, etc. In that case, we must look at the problemsinterview experiences, and interview bundle for placement preparations.
Nevertheless, we may consider our paid courses to give our career an edge over others!
Do upvote this blog if you find it helpful and engaging!
Happy Learning!!
Live masterclass
|
# Lesson 16
Features of Trigonometric Graphs (Part 2)
## 16.1: Which One Doesn't Belong: Graph Periods (5 minutes)
### Warm-up
This warm-up prompts students to compare four trigonometric functions. It gives students a reason to use language precisely (MP6) and gives the opportunity to hear how they use terminology and talk about characteristics of the items in comparison to one another.
### Launch
Arrange students in groups of 2–4. Display the graphs for all to see. Give students 1 minute of quiet think time and then time to share their thinking with their small group. In their small groups, ask each student to share their reasoning why a particular item does not belong, and together find at least one reason each item doesn't belong.
### Student Facing
Which one doesn't belong?
### Activity Synthesis
Ask each group to share one reason why a particular item does not belong. Record and display the responses for all to see. After each response, ask the class if they agree or disagree. Since there is no single correct answer to the question of which one does not belong, attend to students’ explanations and ensure the reasons given are correct.
During the discussion, ask students to explain the meaning of any terminology they use, such as scale factor, horizontal translation, or period. Also, press students on unsubstantiated claims.
## 16.2: Any Period (15 minutes)
### Activity
This activity continues to develop the idea of period for a trigonometric function both graphically and from the point of view of expressions and equations. In the previous lesson, students first saw a trigonometric function whose period was scaled. The function $$\cos(2\theta)$$ has period $$\pi$$ because the expression $$2\theta$$ goes through all values between 0 and $$2\pi$$ when $$\theta$$ goes from 0 to $$\pi$$. For the same reason, $$\sin(5\theta)$$ has period $$\frac{2\pi}{5}$$. From a graphical point of view, decreasing the period means that the wave is compressed horizontally.
The new content in this activity is that students begin to look at functions defined by expressions like $$\cos(\pi \theta)$$. The impact of the coefficient $$\pi$$ is again to change the period, here from $$2\pi$$ to 2, since $$\frac{2\pi}{\pi} = 2$$. Finding trigonometric functions whose period is a whole or rational number is important in the applications that students study for the remainder of this unit. When they model real world phenomena with trigonometric functions, the period of those phenomena is usually a rational number.
### Student Facing
1. For each graph of a trigonometric function, identify the period.
2. Here are some trigonometric functions. Find the period of each function.
function period
$$y=\cos(\theta)$$
$$y=\cos(3\theta)$$
$$y=\sin(6\theta)$$
$$y=\sin(10\theta)$$
$$y=\cos\left(\frac{1}{3}\theta\right)$$
3. What is the period of the function $$y=\cos(\pi \theta)$$? Explain your reasoning.
4. Identify a possible equation for a trigonometric function with this graph.
### Anticipated Misconceptions
As students calculate the period for the 5 functions listed, some may take the coefficient of $$\theta$$ as the period. For example, they may think that the period of $$y=\cos(10\theta)$$ is 10. Ask, “What value does $$\theta$$ need to be in order for $$10\theta$$ to be 0? To be $$2\pi$$?” (0 and $$\frac{2\pi}{10}$$, respectively.) “What does that tell you about the period of $$\cos(10\theta)$$?” (That it repeats every $$\frac{2\pi}{10}$$, so that's the period of $$y=\cos(10\theta)$$.)
### Activity Synthesis
Focus student attention on the two ways of understanding the period of a trigonometric function that appear in this activity: interpreting a graph and interpreting an expression. For the graphical interpretation, ask students questions like:
• "How can you find the period of a trigonometric function from its graph?" (The period is how long it takes the function to complete one full "wave" or cycle of output values.)
• "What is the period of the trigonometric function from the last problem? Explain how you know." (1, because the graph completes one cycle when the input increases by 1.)
For interpreting expressions, ask students questions like:
• “When $$\theta = 0$$, what value does $$2\pi \theta$$ take? What about $$\cos(2\pi \theta)$$?” (0 and 1.)
• “When $$\theta = 1$$ what value does $$2\pi \theta$$ take? What about $$\cos(2\pi \theta)$$?” ($$2\pi$$ and 1.)
• “How long does it take the graph of $$y=\cos(2\pi \theta)$$ to complete one full period? Explain how you know.” (1 because whenever $$\theta$$ increases by 1, $$2\pi \theta$$ goes through all values between 0 and $$2\pi$$.)
Highlight that the function $$y = \cos(2\pi \theta)$$ defines the graph in the last question because the period is 1. The function $$y=\cos(4\pi \theta)$$ defines the third graph of the first problem as its period is $$\frac{1}{2}$$.
Reading, Writing, Speaking: MLR3 Clarify, Critique, Correct. Before students share how they calculated the period for the functions in the table, present an incorrect statement that represents an incomplete understanding of the coefficient of $$\theta$$. For example, “The period of $$y=\cos(10 \theta)$$ is 10 because you look at the coefficient to find it.” Ask students to identify the error, critique the reasoning, and write a correct explanation. As students discuss with a partner, listen for students who identify and clarify the ambiguous language in the statement. For example, the author could improve their statement by stating that the period is determined by finding the value that makes $$10 \theta$$ equal $$2\pi$$. This helps students evaluate, and improve upon, the written mathematical arguments of others, as they clarify how to find the period of a function written as an equation.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness
Representation: Internalize Comprehension. Use color and annotations to illustrate student thinking. As students share their strategies for interpreting trigonometric functions presented graphically and as functions, scribe their thinking on a visual display. Highlight connections students make between the graph and the corresponding function.
Supports accessibility for: Visual-spatial processing; Conceptual processing
## 16.3: Around the World’s Largest Ferris Wheel (15 minutes)
### Activity
In this activity, students return to the context of a Ferris wheel with their new understanding of how to transform a periodic function by changing the midline, amplitude, and period. The purpose of this activity is for students to interpret a sine function in context and then make a sketch of the function.
Monitor for students using clear reasoning about the expression for $$f$$ as they identify the diameter, time per revolution, and times the passenger seat is at its lowest point. For example, students may make and solve equations like $$\frac{2\pi t}{30}=2\pi$$ in order to determine the time $$t$$ for one full revolution.
A note on the use of $$t$$ as a variable: Until this activity, the input for trigonometric functions has typically been $$\theta$$. In a previous course students used $$\theta$$ to represent angles in different geometric shapes, so its use throughout this unit was meant to remind students of the connection between cosine and sine and the unit circle. Now, however, students are transitioning to working with functions modeling contexts where the input isn't an angle, so it makes sense to choose a variable representative of the input value. Since $$f$$ takes an input of time and gives and output of height, the variables $$t$$ and $$h$$ are reasonable choices. If time allows and students wonder why $$t$$ is used instead of $$\theta$$, invite students to suggest why they think the input variable is $$t$$ and highlight the reasons listed here for the change.
### Launch
Tell students that they will be examining the position of a seat on a very large Ferris wheel. The Ferris wheel moves slowly enough that passengers can get on and off while the Ferris wheel continues its motion.
It time allows, recommend students try making a sketch of $$f$$ before they graph the function using technology.
Speaking, Reading: MLR5 Co-Craft Questions. Use this routine to help students interpret a sine function in context of a Ferris wheel. Display the image and Student Task Statement, leaving out the questions. Ask students to write down possible mathematical questions that could be asked about the situation. Invite students to compare their questions before revealing the activity’s questions. Listen for and amplify any questions involving features of the Ferris wheel, the height of passenger seat $$f(t)$$, or the graph of the function $$f$$. This will help students create the language of mathematical questions before feeling pressure to produce solutions.
Design Principle(s): Maximize meta-awareness; Support sense-making
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer interactions. Invite students to talk about their ideas with a partner before writing them down. Display sentence frames to support students when they explain their strategy. For example, “ Why did you . . . ?”, “How do you know . . . ?”, “First, I _____ because. . .”, “Then/next, I . . . .”
Supports accessibility for: Language; Social-emotional skills
### Student Facing
The world’s tallest Ferris wheel is in Las Vegas. The height $$h$$ in feet of one of the passenger seats on the Ferris wheel can be modeled by the function $$f(t) = 275+ 260 \sin\left(\frac{2\pi t}{30}\right)$$ where time $$t$$ is measured in minutes after 8:00 a.m.
1. What is the diameter of the Ferris wheel? Explain how you know.
2. How long does it take the Ferris wheel to make a complete revolution? Explain how you know.
3. Give at least three different times when the passenger seat modeled by $$f$$ is at its lowest point. Explain how you know.
4. Sketch a graph of the height of the seat on the Ferris wheel for at least two full revolutions.
### Student Facing
#### Are you ready for more?
Here is a graph of a wave where the amplitude is not constant but rather decreases over time. Write an equation which could match this graph.
### Anticipated Misconceptions
If students are unsure how to interpret the expression inside the sine function, ask them to consider what value they would use for $$t$$ if they want the input expression to be 0, $$\frac{\pi}{2}$$, $$\pi$$, $$\frac{3\pi}{2}$$, or $$2\pi$$.
### Activity Synthesis
Select previously identified students to share their reasoning about the first three questions. Highlight reasoning such as:
• 260 feet is the amplitude. The value of $$260\sin(\frac{2\pi t}{30})$$ will go between -260 and +260 for different inputs. This means the diameter of the wheel is 520 feet.
• The expression $$\frac{2\pi t}{30}$$ gives the period. When $$t$$ goes from 0 to 30, this expression goes from 0 to $$2\pi$$ so the Ferris wheel makes one full revolution in 30 minutes meaning its period is 30 minutes.
• The function has its lowest value when $$260\sin(\frac{2\pi t}{30})=\text-260$$. This is when $$\sin(\frac{2\pi t}{30})=\text-1$$, which happens when the input to sine is $$\frac{3\pi}{2}$$. We can calculate the value of $$t$$ by solving $$\frac{2\pi t}{30}=\frac{3\pi}{2}$$, which is when $$t=22.5$$.
• Since the period of the wheel is 30 minutes, then 22.5 minutes after 8 a.m. and every 30 minutes after that passenger seat is at its lowest point.
As students share, highlight how each of these features can be seen on a displayed graph of $$y=f(t)$$.
## Lesson Synthesis
### Lesson Synthesis
Arrange students in groups of 2. The purpose of this discussion is for students to compare graphs two at a time where one graph is a transformation of the other. Display each of the given pairs one at a time. Give students brief quiet think time and then time to share their answer with a partner before selecting students to share their responses.
• “How does the graph of $$f(\theta) = \sin(\theta-5) + 2$$ compare to the graph of $$g(\theta) = \sin(\theta)$$?” (It is translated up two units and right 5 units.)
• “How does the graph of $$f(\theta) = 2\sin(\theta)-5$$ compare to the graph of $$g(\theta) = \sin(\theta)$$?” (It has twice the amplitude and is translated down 5 units. Or, it has a midline at -5, a maximum at -3, and a minimum at -7.)
• “How does the graph of $$f(\theta) =5 \sin(2\theta)$$ compare to the graph of $$g(\theta) = \sin(\theta)$$?” (It has 5 times the amplitude and half the period.)
• “How does the graph of $$f(\theta) = 5\sin(2\pi \theta)$$ compare to the graph of $$g(\theta) = \sin(\theta)$$?” (It has 5 times the amplitude and a period of 1 instead of $$2\pi$$.)
If time allows, invite students to come up with their own pairs and challenge their partners to identify the transformations.
## Student Lesson Summary
### Student Facing
Here is a point $$P$$ on a wheel.
Imagine the height $$h$$ of $$P$$ in feet relative to the center of the wheel after $$t$$ seconds is given by the equation $$h= \sin(2\pi t)$$. When $$t = 1$$, that is after 1 second, the wheel will be back where it started. It will return to its starting position every second.
What if $$h= \sin(4\pi t)$$ was the equation representing the height of $$P$$ instead? What does this equation tell us about how long it takes the wheel to complete a full revolution? In this case, when $$t = 1$$ the wheel has made 2 complete revolutions, so it makes one complete revolution in 0.5 seconds. Here are the graphs of these two functions.
Notice that the midline is the $$t$$-axis for each function and the amplitude is 1. The only difference is the period, how long it takes each function to complete one full revolution.
Now let's consider the height of a point on a wheel with a radius of 11 inches, the center of the wheel 11 inches off the ground, and completing two revolutions per second.
Notice that the midline and amplitude are 11. An equation defining this graph is $$h = 11\sin\left(4\pi t\right)+11$$ where $$h$$ is the height, in inches, of the point on the wheel after $$t$$ seconds.
|
# Logarithms change of base
In this section ask-math explains you the logarithms change of base formula.
Any calculator gives us the value of logarithm either base 10 or base 'e'. But if the common and natural logarithm has base other than 10 or 'e' then we use logarithms change of base formula.
Change of base : Let us now learn how to convert from base 'b' to any other base 'a' by proving that for any positive real numbers 'r' and 'b', b ≠ 1.
Proof : Let N = log b r
b
N = r ( By definition of logarithm)
Taking log to the base a on both sides, we get
∴ N log
a b = log a r
Note that we can use any base in place of a.
## Examples on logarithms change of base
Example 1 : Find log 9 3 .
Solution: Consider the common base as 3.
log
9 3 = (log 3 3)/(log 3 9)
= (log
3 3) / (log 3 (3) 2 )
= (log
3 3) / 2 log 3 3
= 1/2.
Example 2 : Given log 2 16 = 4. Find log 16 2 .
Solution:
Here we have , b = 16, r = 2 and a = 2
log
16 2 = (log 2 2)/(log 2 16)
= (log
2 2) / (log 2 (2) 4 )
= (log
2 2) / 4 log 2 2
= 1/4.
Solve each equation using logarithms. Round to the nearest ten-thousandth.
1) 2
x = 3
Solution : 2 x = 3
log
2 3 = x
log
2 3 = (log 3)/(log 2) (consider the base as 10, if not mentioned)
= 0.4771 / 0.3010 (using calculator)
∴ x = 1.5850
2) 8 + 10
x = 1008
Solution :
8 + 10
x = 1008
10
x = 1008 - 8
10
x = 1000
log
10 1000 = x
log
10 1000 = (log 1000)/(log 10)
= (log 10
3 )/(log 10)
= 3 log 10/ log 10
∴ x = 3
Use the change of base formula to evaluate each.
1) log
2 9 = log 9/ log 2 = 0.9542/0.3010 = 3.17
2) log
4 8 = log 8/ log 4 = 0.9031/0.6021 = 1.499 = 1.5
3) log
3 50 = log 50/ log 3 = 1.6989/.4771 = 3.5608
4) log
4.6 12.5 = log 12.5/ log 4.6 = 1.0969/0.6627 = 1.655
|
The current approach to learning math in Singapore primary schools is through the math model method. Under this approach, students are required to draw math models that illustrate the questions. For lower primary students, modeling might seem unnecessary but the skills being mastered will come in handy at the upper primary level where problem sums becoming more challenging. To give your lower primary child a head start, this article will show you some techniques for mathematical modeling.
### 1. Background on math model method
This methodology was created by a Singaporean teacher called Hector Chee. Due to its practicality, the method was soon taught in all schools, starting from Primary One. This new method presented a challenge to parents who might have been taught using algebra or other math methods. As a result, many could not help their children to develop the right math model techniques for the different kinds of math problem sums.
### Main concepts in Singapore math model method
In the math model method, there are basically 2 concepts that form the foundation for all further iterations.
### The part-whole concept
In this concept, the child starts with understanding the relationship between parts. Once understood, they can represent these relationships using rectangular blocks to model math questions.
Let?s look at one example to better understand how the part-whole concept works. In the image below, the child starts off learning how to add the individual balls to understand a simple addition question of 3 +2 =5. At this stage, it is important to use real images such as the balls to let the child connect the dots.
### Singapore Math Model Method: Part-Whole Concept
Once the child has understood the above, we can take out the balls and uses blocks as representations. Below is how the image can be drawn. Once the kid accepts the blocks as representations, he or she will be in a good position to understand further abstraction.
### Singapore Math Model Method: Part-Whole Concept
After the above, we can go one step further to visualize the question in even more abstract terms. Here, we don?t need the individual blocks. Instead, just a visual distinction between 3 and 2 is enough to represent the relationship between the blocks.
### Singapore Math Model Method: Part-Whole Concept
In summary, this technique uses the relationship of the parts to let children learn about the whole.
" />
The current approach to learning math in Singapore primary schools is through the math model method. Under this approach, students are required to draw math models that illustrate the questions. For lower primary students, modeling might seem unnecessary but the skills being mastered will come in handy at the upper primary level where problem sums becoming more challenging. To give your lower primary child a head start, this article will show you some techniques for mathematical modeling.
### 1. Background on math model method
This methodology was created by a Singaporean teacher called Hector Chee. Due to its practicality, the method was soon taught in all schools, starting from Primary One. This new method presented a challenge to parents who might have been taught using algebra or other math methods. As a result, many could not help their children to develop the right math model techniques for the different kinds of math problem sums.
### Main concepts in Singapore math model method
In the math model method, there are basically 2 concepts that form the foundation for all further iterations.
### The part-whole concept
In this concept, the child starts with understanding the relationship between parts. Once understood, they can represent these relationships using rectangular blocks to model math questions.
Let?s look at one example to better understand how the part-whole concept works. In the image below, the child starts off learning how to add the individual balls to understand a simple addition question of 3 +2 =5. At this stage, it is important to use real images such as the balls to let the child connect the dots.
### Singapore Math Model Method: Part-Whole Concept
Once the child has understood the above, we can take out the balls and uses blocks as representations. Below is how the image can be drawn. Once the kid accepts the blocks as representations, he or she will be in a good position to understand further abstraction.
### Singapore Math Model Method: Part-Whole Concept
After the above, we can go one step further to visualize the question in even more abstract terms. Here, we don?t need the individual blocks. Instead, just a visual distinction between 3 and 2 is enough to represent the relationship between the blocks.
### Singapore Math Model Method: Part-Whole Concept
In summary, this technique uses the relationship of the parts to let children learn about the whole.
" />
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# 9.7 Graph Quadratic Functions Using Transformations
The topics covered in this section are:
## 9.7.1 Graph Quadratic Functions of the form $f(x)=x^{2}+k$
In the last section, we learned how to graph quadratic functions using their properties. Another method involves starting with the basic graph of $f(x)=x^{2}$ and ‘moving’ it according to information given in the function equation. We call this graphing quadratic functions using transformations.
In the first example, we will graph the quadratic function $f(x)=x^{2}$ by plotting points. Then we will see what effect adding a constant, $k$, to the equation will have on the graph of the new function $f(x)=x^{2}+k$.
#### Example 1
Graph $f(x)=x^{2}, g(x)=x^{2}+2,$ and $h(x)=x^{2}-2$ on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.
Solution
Plotting points will help us see the effect of the constants on the basic $f(x)=x^{2}$ graph. We fill in the chart for all three functions.
The $g(x)$ values are two more than the $f(x)$ values. Also, the $h(x)$ values are two less than the $f(x)$ values. Now we will graph all three functions on the same rectangular coordinate system.
The graph of $g(x)=x^{2}+2$ is the same as the graph of $f(x)=x^{2}$ but shifted up $2$ units.
The graph of $h(x)=x^{2}+2$ is the same as the graph of $f(x)=x^{2}$ but shifted down $2$ units.
The last example shows us that to graph a quadratic function of the form $f(x)=x^{2}+k$, we take the basic parabola graph of $f(x)=x^{2}$ and vertically shift it up $(k>0)$ or shift it down $(k<0)$.
This transformation is called a vertical shift.
### GRAPH A QUADRATIC FUNCTION OF THE FORM $f(x)=x^{2}$ USING A VERTICAL SHIFT
The graph of $f(x)=x^{2}+k$ shifts the graph of $f(x)=x^{2}$ vertically $k$ units.
• If $k>0$, shift the parabola vertically up $k$ units.
• If $k<0$, shift the parabola vertically down $|k|$ units.
Now that we have seen the effect of the constant, $k$, it is easy to graph functions of the form $f(x)=x^{2}+k$. We just start with the basic parabola of $f(x)=x^{2}$ and then shift it up or down.
It may be helpful to practice sketching $f(x)=x^{2}$ quickly. We know the values and can sketch the graph from there.
Once we know this parabola, it will be easy to apply the transformations. The next example will require a vertical shift.
#### Example 2
Graph $f(x)=x^{2}-3$ using a vertical shift.
Solution
## 9.7.2 Graph Quadratic Functions of the form $f(x)=(x-h)^{2}$
In the first example, we graphed the quadratic function $f(x)=x^{2}$ by plotting points and then saw the effect of adding a constant $k$ to the function had on the resulting graph of the new function $f(x)=x^{2}+k$.
We will now explore the effect of subtracting a constant, $h$, from $x$ has on the resulting graph of the new function $f(x)=(x-h)^{2}$.
#### Example 3
Graph $f(x)=x^{2}, g(x)=(x-1)^{2},$ and $h(x)=(x+1)^{2}$ on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.
Solution
Plotting points will help us see the effect of the constants on the basic $f(x)=x^{2}$ graph. We fill in the chart for all three functions.
The $g(x)$ values and the $h(x)$ values share the common numbers $0,1,4,9,$ and $16,$ but are shifted.
The last example shows us that to graph a quadratic function of the form $f(x)=(x-h)^{2}$, we take the basic parabola graph of $f(x)=x^{2}$ and shift it left $(h>0)$ or shift it right $(h<0)$.
This transformation is called a horizontal shift.
### GRAPH A QUADRATIC FUNCTION OF THE FORM $f(x)=(x-h)^{2}$ USING A HORIZONTAL SHIFT
The graph of $f(x)=(x-h)^{2}$ shifts the graph of $f(x)=x^{2}$ horizontally $h$ units.
• If $h>0$, shift the parabola horizontally right $h$ units.
• If $h<0$, shift the parabola horizontally left $|h|$ units.
Now that we have seen the effect of the constant, $h$, it is easy to graph functions of the form $f(x)=(x-h)^{2}$. We just start with the basic parabola of $f(x)=x^{2}$ and then shift it left or right.
The next example will require a horizontal shift.
#### Example 4
Graph $f(x)=(x-6)^{2}$ using a horizontal shift.
Solution
Now that we know the effect of the constants h and k, we will graph a quadratic function of the form $f(x)=(x-h)^{2} +k$ by first drawing the basic parabola and then making a horizontal shift followed by a vertical shift. We could do the vertical shift followed by the horizontal shift, but most students prefer the horizontal shift followed by the vertical.
#### Example 5
Graph $f(x)=(x+1)^{2}-2$ using transformations.
Solution
This function will involve two transformations and we need a plan.
Let’s first identify the constants $h,k$.
$f(x)=(x+1)^{2}-2$
$f(x)=(x-h)^{2}+k$
$f(x)=(x-(-1))^{2}+(-2)$
$h=-1$ and $k=-2$
The $h$ constant gives us a horizontal shift and the k gives us a vertical shift.
We first draw the graph of $f(x)=x^{2}$ on the grid.
## 9.7.3 Graph Quadratic Functions of the Form $f(x)=ax^{2}$
So far we graphed the quadratic function $f(x)=x^{2}$ and then saw the effect of including a constant $h$ or $k$ in the equation had on the resulting graph of the new function. We will now explore the effect of the coefficient a on the resulting graph of the new function $f(x)=ax^{2}$.
If we graph these functions, we can see the effect of the constant $a$, assuming $a>0$.
To graph a function with constant a it is easiest to choose a few points on $f(x)=x^{2}$ and multiply the $y$-values by $a$.
### GRAPH OF A QUADRATIC FUNCTION OF THE FORM $f(x)=ax^{2}$
The coefficient $a$ in the function $f(x)=ax^{2}$ affects the graph of $f(x)=x^{2}$ by stretching or compressing it.
• If $0<|a|<1$, the graph of $f(x)=ax^{2}$ will be “wider” than the graph of $f(x)=x^{2}$.
• If $|a| >1$, the graph of $f(x)=ax^{2}$ will be “skinnier” than the graph of $f(x)=x^{2}$.
#### Example 6
Graph $f(x)=3x^{2}$.
Solution
We will graph the functions $f(x)=x^{2}$ and $g(x)=3x^{2}$ on the same grid. We will choose a few points on $f(x)=x^{2}$ and the multiply the $y$-values by $3$ to get the points for $g(x)=3x^{2}$.
## 9.7.4 Graph Quadratic Functions Using Transformations
We have learned how the constants $a,h,$ and $k$ in the functions, $f(x)=x^{2}, f(x)=(x-h)^{2},$ and $f(x)=ax^{2}$ affect their graphs. We can now put this together and graph quadratic functions $f(x)=ax^{2}+bx+c$ by first putting them into the form $f(x)=a(x-h)^{2}+k$ by completing the square. This form is sometimes known as the vertex form or standard form.
We must be careful to both add and subtract the number to the SAME side of the function to complete the square. We cannot add the number to both sides as we did when we completed the square with quadratic equations.
When we complete the square in a function with a coefficient of $x^{2}$ that is not one, we have to factor that coefficient from just the x-terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the $x$-terms.
Once we get the constant we want to complete the square, we must remember to multiply it by that coefficient before we then subtract it.
#### Example 7
Rewrite $f(x)=-3x^{2}-6x-1$ in the $f(x)=a(x-h)^{2}+k$ form by completing the square.
Solution
Once we put the function into the $f(x)=(x-h)^{2}+k$ form, we can then use the transformations as we did in the last few problems. The next example will show us how to do this.
#### Example 8
Graph $f(x)=x^{2}+6x+5$ by using transofmraions.
Solution
Step 1. Rewrite the function in $f(x)=a(x-h)^{2}+k$ vertex form by completing the square.
Step 2. Graph the function using transformations.
Looking at the $h,k,$ values, we see the graph will take the graph of $f(x)=x^{2}$ and shift it to the left $3$ units and down $4$ units.
We first draw the graph of $f(x)=x^{2}$ on the grid.
We list the steps to take to graph a quadratic function using transformations here.
### HOW TO: Graph a quadratic function using transformations.
1. Rewrite the function in $f(x)=a(x-h)^{2}+k$ form by completing the square.
2. Graph the function using transformations.
#### Example 9
Graph $f(x)=-2x^{2}-4x+2$ by using transformations.
Solution
Step 1. Rewrite the function in $f(x)=a(x-h)^{2}+k$ vertex form by completing the square.
Step 2. Graph the function using transformations.
We first draw the graph of $f(x)=x^{2}$ on the grid.
Now that we have completed the square to put a quadratic function into $f(x)=a(x-h)^{2}+k$ form, we can also use this technique to graph the function using its properties as in the previous section.
If we look back at the last few examples, we see that the vertex is related to the constants $h$ and $k$.
In each case, the vertex is $(h,k)$. Also the axis of symmetry is the line $x=h$.
We rewrite our steps for graphing a quadratic function using properties for when the function is in $f(x)=a(x-h)^{2}+k$ form.
### HOW TO: Graph a quadratic function in the form $f(x)=a(x-h)^{2}+k$ using properties.
1. Rewrite the function in $f(x)=a(x-h)^{2}+k$ form.
2. Determine whether the parabola opens upward, $a>0$, or downward, $a<0$.
3. Find the axis of symmetry, $x=h$
4. Find the vertex, $(h,k)$.
5. Find the $y$-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
6. Find the $x$-intercepts.
7. Graph the parabola.
#### Example 10
• Rewrite $f(x)=2x^{2}+4x+5$ in $f(x)=a(x-h)^{2}+5$ form
• Graph the function using properties.
Solution
## 9.7.5 Find a Quadratic Function from its Graph
So far we have started with a function and then found its graph.
Now we are going to reverse the process. Starting with the graph, we will find the function.
#### Example 11
Determine the quadratic function whose graph is shown.
Solution
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# Calculating the Subfactorial
## Difficulty: ★ ★ ★ ★
Calculating a factorial is a common programming exercise, often coupled with introducing the concept of recursion. Mathematically, the factorial has a twin: the subfactorial. Its calculation can be a fun programming challenge, which is the subject of this month’s Exercise.
As a review, a factorial is a value written as n!. The value represents integer n multiplied sequentially. For example, 5! is 1×2×3×4×5, which equals 120. This value is the number of permutations possible with five items.
Figure 1 illustrates 3! or three factorial. It shows the number of ways to arrange three objects, which is what 3! represents.
Figure 1. The value 3! shows how many ways to arrange three objects. 3! = 6 different arrangements.
The subfactorial works along similar lines. It’s written with the exclamation point before the value: subfactorial three is !3. The subfactorial limits the arrangements so that no objects can be in the same position as in the original. Value !3 calls out all non-unique arrangements for three objects. An example is shown in Figure 2.
Figure 2. Three subfactorial (!3) shows that only two arrangements are possible when compared with the original. !3 = 2.
The best explanation of the subfactorial process, also called derangement, is found in this video from Andy Math. Do take time to watch it, as it describes a formula you can use to calculate subfactorials. The formula is illustrated in Figure 3.
Figure 3. The equation for calculating a subfactorial.
Your task for this month’s Exercise is to write code that outputs values for subfactorials !0 through !13. You can use the equation from Figure 3 as your inspiration, which is what I did. You can also check out the Derangement Wikipedia page, which offers more formulas and mathematical mumbojumbo.
Here is output from my solution, the list of subfactorial values for !0 through !13:
```!0 = 1 !1 = 0 !2 = 1 !3 = 2 !4 = 9 !5 = 44 !6 = 265 !7 = 1854 !8 = 14833 !9 = 133496 !10 = 1334961 !11 = 14684570 !12 = 176214841 !13 = 2290792932```
As with factorials, these subfactorials are known values, though you can’t just write a solution that outputs the results shown above. No, you must code a solution in C.
Try this Exercise on your own before you check out my solution.
## 3 thoughts on “Calculating the Subfactorial”
1. ” . . . you can’t just write a solution that outputs the results shown above”. Why not? That’s the sort of thing people on Instagram do! Possibly not the world’s greatest source of programming expertise though.
Is Andy Math his real name? 🙂
I don’t understand the meaning of !0. The result implies that if you have zero squares, circles and triangles (for example) there is 1 unique way of rearranging them which obviously doesn’t make sense. I’d have assumed !0 was undefined.
According to Wikipedia the easy way is n!/e (e being Euler’s number) but it gets increasingly inaccurate as n gets larger. Maybe you need e to gazillions of decimal places.
Anyway, this is my solution. I pre-calculated the factorials in an array. I might have gone a bit over the top with doubles but I didn’t want to risk losing any precision. They might not be entirely necessary but I couldn’t be bothered to try downsizing.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int main()
{
long factorials[14];
double subfactorial;
double interim;
// const float e = 2.71828; // inaccurate from !10
const float e = 2.7182818284590452353602874713527; // inaccurate from !11
factorials[0] = 1;
factorials[1] = 1;
// pre-calculate factorials
for(int i = 2; i <=13; i++)
{
factorials[i] = factorials[i-1] * i;
}
// the n!/e method
for(int n = 1; n <= 13; n++)
{
subfactorial = round(factorials[n]/e);
printf(“%d\t%.0lf\n”, n, subfactorial);
}
puts(“\n———————\n”);
// the complicated method
for(double n = 0; n <= 13; n++)
{
interim = 0;
for(double k = 0; k <= n; k++)
{
interim += ( pow(-1.0, k) / factorials[(int)k] );
}
subfactorial = factorials[(int)n] * interim;
printf(“%d\t%.0lf\n”, (int)n, subfactorial);
}
return EXIT_SUCCESS;
}
2. The explanation behind !0 is that when you have zero object there is only one way to arrange them: Not at all. I’ve seen other proofs as well, though this is the one I remember.
Thanks for the solution!
3. I messed that up. I declared e as a float but should have used double. I changed it and got correct results up to !13. (Possibly higher but I didn’t check.)
BUT the n!/e method gives !0 as 0. I still don’t understand because the example with 3 shapes in fig. 2 shows that the subfactorial is the number of ADDITIONAL ways of arranging items therefore both !0 and !1 should be 0. (Assuming !0 is valid anyway which I’m still dubious about.)
Actually I’m also not sure why 0! = 1. Probably just due to mathematicians being a bit mad.
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# Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions – Exercise 5.1
### (i). The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
Solution:
Initially fare = 0
After completing 1 km, fare = 15
After completing 2 km, fare = 15 + 8 = 23
After completing 3 km, fare = 23 + 8 = 31
And so on
So you can write fare in series as
15, 23, 31, 39,……… and so on
Here you can see that the first term is 15 and the difference between any two-term is 8. So, this series is in arithmetic progression.
Note: A series is said to be in arithmetic progression if any term can be found out by adding a fixed number to the previous term. The first term is denoted as and the fixed number is known as a difference which is denoted by d.
In the above series a = 15 and d = 8
### (ii). The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Solution:
Initial volume of air in cylinder = πR2H
Amount of air removed by vacuum pump = πR2H/4
Remaining air = 3 * πR2H/4
Again amount of air removed by vacuum pump = 3 * πR2H/16
Remaining air = 3 * πR2H/4 – 3 * πR2H/16 = 9 * πR2H/16
and so on
The series can be written as: πR2H, 3 * πR2H/4, 9*πR2H/16,……… and so on
You can see here is no common difference between any two-term. So, this series is not arithmetic progression.
### (iii). The cost of digging a well after every meter of digging, when it costs Rs. 150 for the first meter and rises by Rs. 50 for each subsequent meter.
Solution:
Cost of digging well for first meter is 150 rupees
Cost of digging well for two meters is 150 + 50 = 200 rupees
Cost of digging well for three meters is 200 + 50 = 250 rupees
And so on
The series for cost of digging a well look like 150, 200, 250, 300,……… and so on
Here you can see that after the first term, any term can be found out by adding 50 to the previous term. Hence, the above series is in arithmetic progression with the first term 150 and the common difference is 50.
### (iv). The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Solution:
Formula for compound interest is given by P(1 + r/100)n where P is the principal amount, r is the rate of interest and n is the time duration.
So, the series can be written as: 10000(1 + 8/100), 10000(1 + 8/100)2,10000(1 + 8/100)3,……… and so on
As time duration increases, compound interest will increase exponentially. Hence, this can’t form a series in arithmetic series because there is no fixed common difference between any two terms.
### (i). a = 10, d = 10
Solution:
Given:
a1 = a = 10
d = 10
Now we find the remaining terms:
Second term(a2) = a1 + d = 10 + 10 = 20
Third term(a3) = a2 + d = 20 + 10 = 30
Fourth term(a4) = a3 + d = 30 + 10 = 40
and so on.
So, four terms of this A.P. will be as follows
10, 20, 30, 40,….
### (ii) a = -2, d = 0
Solution:
Given:
a1 = a = -2
d = 0
Now we find the remaining terms:
Second term(a2) = a1 + d = -2 + 0 = -2
Third term(a3) = a2 + d = -2 + 0 = -2
Fourth term(a4) = a3 + d = -2 + 0 = -2
and so on.
So, four terms of this A.P. will be as follows
-2, -2, -2, -2,….
### (iii). a = 4, d = -3
Solution:
Given:
a1 = a = 4
d = -3
Now we find the remaining terms:
Second term(a2) = a1 + d = 4 + (-3) = 1
Third term(a3) = a2 + d = 1 + (-3) = -2
Fourth term(a4) = a3 + d = -2 + (-3) = -5
and so on.
So, four terms of this A.P. will be as follows
4, 1, -2, -5,….
### (iv). a = -1, d = 1/2
Solution:
Given:
a1 = a = -1
d = 1/2
Now we find the remaining terms:
Second term(a2) = a1 + d = -1 + 1/2 = -1/2
Third term(a3) = a2 + d = (-1/2) + (1/2) = 0
Fourth term(a4) = a3 + d = 0 + (1/2) = 1/2
and so on.
So, four terms of this A.P. will be as follows
-1, -1/2, 0, 1/2,….
### (v). a = -1.25, d = -0.25
Solution:
Given:
a1 = a = -1.25
d = -0.25
Now we find the remaining terms:
Second term(a2) = a1 + d = -1.25 – 0.25 = – 1.50
Third term(a3) = a2 + d = -1.50 – 0.25 = -1.75
Fourth term(a4) = a3 + d = -1.75 – 0.25 = -2.00
and so on.
So, four terms of this A.P. will be as follows
-1.25, -1.50, -1.75, -2.00,….
### (i). 3, 1, -1, -3,….
Solution:
From the above A.P., first term (a) = 3
Common difference (d) = second term – first term
= 1 – 3 = -2
### (ii). -5, -1, 3, 7,….
Solution:
From the above A.P., first term (a) = -5
Common difference (d) = second term – first term
= -1 – (-5)
= -1 + 5 = 4
### (iii). 1/3, 5/3, 9/3, 13/3,….
Solution:
From the above A.P., first term (a) = 1
Common difference (d) = second term – first term
= 5/3 – 1/3
= 4/3
### (iv). 0.6, 1.7, 2.8, 3.9,….
Solution:
From the above A.P., first term (a) = 0.6
Common difference (d) = second term – first term
= 1.7 – 0.6
= 1.1
### (i). 2, 4, 8, 16,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 4 – 2 = 2
d2 = a3 + a2 = 8 – 4 = 4
So d1 ≠ d2
Hence, this series doesn’t form an AP because there is no fixed common difference.
### (ii). 2, 5/2, 3, 7/2,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 5/2 – 2 = 1/2
d2 = a3 + a2 = 3 – 5/2 = 1/2
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is 1/2.
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = 7/2 + 1/2 = 4
Sixth term(a6) = a5 + d = (4) + (1/2) = 9/2
Seventh term(a7) = a6 + d = 9/2 + (1/2) = 5
So, the next three terms are: 4, 9/2, 5
### (iii). -1.2, -3.2, -5.2, -7.2,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = -3.2 – (-1.2) = -2.0
d2 = a3 + a2 = -5.2 – (-3.2) = -2.0
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is -2.0
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = (-7.2) + (-2.0) = -9.2
Sixth term(a6) = a5 + d = (-9.2) + (-2.0) = -11.2
Seventh term(a7) = a6 + d = (-11.2) + (-2.0) = -13.2
So, the next three terms are: -9.2, -11.2, -13.2
### (iv). -10, -6, -2, 2,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = -6 – (-10) = 4
d2 = a3 + a2 = -2 – (-6) = 4
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is 4
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = 2 + 4 = 6
Sixth term(a6) = a5 + d = 6 + 4 = 10
Seventh term(a7) = a6 + d = 10 + 4 = 14
So, the next three terms are 6, 10, 14
### (v). 3, 3+√2, 3+2√2, 3+3√2,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 3 + √2 – 3 = √2
d2 = a3 + a2 = 3 + 2√2 – (3 + √2) = √2
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is √2
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = (3 + 3√2) + √2 = 3 + 4√2
Sixth term(a6) = a5 + d = (3 + 4√2) + √2 = 3 + 5√2
Seventh term(a7) = a6 + d = (3 + 5√2) + √2 = 3 + 6√2
So, the next three terms are 3+4√2, 3+5√2, 3+6√2
### (vi). 0.2, 0.22, 0.222, 0.2222,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 0.22 – 0.2 = 0.02
d2 = a3 + a2 = 0.222 – 0.22 = 0.002
So d1 ≠ d2
Hence, this series doesn’t form an AP because there is no fixed common difference.
### (vii). 0, -4, -8, -12,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = -4 – 0 = -4
d2 = a3 + a2 = -8 – (-4) = -4
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is -4
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = (-12) + (-4) = -16
Sixth term(a6) = a5 + d = (-16) + (-4) = -20
Seventh term(a7) = a6 + d = (-20) + (-4) = -24
So, the next three terms are -16, -20, -24
### (viii). -1/2, -1/2, -1/2, -1/2,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = -1/2 – (-1/2) = 0
d2 = a3 + a2 = -1/2 – (-1/2) = 0
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is 0
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = (-1/2) + 0 = -1/2
Sixth term(a6) = a5 + d = (-1/2) + 0 = -1/2
Seventh term(a7) = a6 + d = (-1/2) + 0 = -1/2
So, the next three terms are -1/2, -1/2, -1/2
### (ix). 1, 3, 9, 27,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 3 – 1 = 2
d2 = a3 + a2 = 9 – 3 = 6
So d1 ≠ d2
Hence, this series doesn’t form an AP because there is no fixed common difference.
### (x). a, 2a, 3a, 4a,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 2a – a = a
d2 = a3 + a2 = 3a – 2a = a
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is a
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = 4a + a = 5a
Sixth term(a6) = a5 + d = 5a + a = 6a
Seventh term(a7) = a6 + d = 6a + a = 7a
So, the next three terms are 5a, 6a, 7a
### (xi). a, a2, a3, a4,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = a2 – a
d2 = a3 + a2 = a3 – a2
So d1 ≠ d2
Hence, this series doesn’t form an AP because there is no fixed common difference.
### (xii). √2, √8, √18, √32,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 2√2 – √2 = √2
d2 = a3 + a2 = 3√2 – 2√2 = √2
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is √2
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = √32 + √2 = 5√2
Sixth term(a6) = a5 + d = 5√2 + √2 = 6√2
Seventh term(a7) = a6 + d = 6√2 + √2 = 7√2
So, the next three terms are 5√2, 6√2, 7√2
### (xiii). √3, √6, √9, √12,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = √6 – √3 = √3(√2 – 1)
d2 = a3 + a2 = √9 – √6 = √3(√3 – √2)
So d1 ≠ d2
Hence, this series doesn’t form an AP because there is no fixed common difference.
### (xiv). 12, 32, 52, 72,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 9 – 1 = 8
d2 = a3 + a2 = 25 – 9 = 16
So d1 ≠ d2
Hence, this series doesn’t form an AP because there is no fixed common difference.
### (xv). 12, 52, 72, 73,….
Solution:
First we check the given series is AP or not by finding common difference:
d1 = a2 + a1 = 25 – 1 = 24
d2 = a3 + a2 = 49 – 25 = 24
So d1 = d2
Hence, the above series is in arithmetic progression and the common difference is 24
Now we find the remaining terms of the A.P.:
Fifth term(a5) = a4 + d = 73 + 24 = 97
Sixth term(a6) = a5 + d = 97 + 24 = 121
Seventh term(a7) = a6 + d = 121 + 24 = 145
So, the next three terms are 97, 121, 145
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# The plans for a shed call for a rectangular floor with a perimeter of 198 ft. The length is The two times the width. How do you find the length and width?
Jan 7, 2017
See full explanation for how to find the length and width below in the explanation:
#### Explanation:
The formula for a perimeter is:
$p = 2 w + 2 l$
Where
$p$ is the perimeter
$w$ is the width
$l$ is the length.
And, we know from the problem:
$l = 2 w$
We can then substitute $2 w$ for $l$ in the equation for the perimeter to give:
$p = 2 w + \left(2 \times 2 w\right)$
$p = 2 w + 4 w$
$p = 6 w$
We also know the perimeter will be 198 ft so we can substitute this for $p$ and solve for $w$, the width:
$198 = 6 w$
$\frac{198}{\textcolor{red}{6}} = \frac{6 w}{\textcolor{red}{6}}$
$33 = w$ or $w = 33$
And because we know:
$l = 2 w$ we can substitute $33$ for $w$ and solve for $l$:
$l = 2 \times 33$
$l = 66$
The length of the floor will be 66 ft and the width will be 33 ft.
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The Radius of Convergence of a Power Series Examples 2
# The Radius of Convergence of a Power Series Examples 2
Recall from The Radius of Convergence of a Power Series page that we can calculate the radius of convergence of a power series using the ratio test, that is if $\lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = L$, then the radius of convergence is $R = \frac{1}{L}$. If $L = 0$ then the radius of convergence $R = \infty$ and if $L = \infty$ then the radius of convergence $R = 0$.
We will now look at some more examples of determining the radius of convergence of a given power series.
## Example 1
Determine the radius of convergence of the power series $\sum_{n=1}^{\infty} \frac{x^n}{4^n \ln n}$.
Applying the ratio test and we get that:
(1)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{4^n \ln (n) }{4^{n+1} \ln (n + 1)} \biggr \rvert = \lim_{n \to \infty} \frac{\ln (n)}{4 \ln (n + 1)} \end{align}
Let $y = \frac{\ln (x)}{4 \ln (x + 1)}$. Then applying L'Hospital's rule we get that:
(2)
\begin{align} \quad \lim_{x \to \infty} \frac{\ln (x)}{4 \ln (x + 1)} \overset H = \lim_{x \to \infty} \frac{\frac{1}{x}}{4 \frac{1}{x +1}} = \lim_{x \to \infty} \frac{x + 1}{4x} = \frac{1}{4} \end{align}
Therefore the radius of convergence is $4$.
## Example 2
Determine the radius of convergence of the power series $\sum_{n=1}^{\infty} \frac{(3x - 2)^n}{n3^n}$.
We must first rewrite the series above as follows:
(3)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{(3x - 2)^n}{n3^n} = \sum_{n=1}^{\infty} \frac{\left (3 \left (x - \frac{2}{3} \right ) \right )^n}{n3^n} = \sum_{n=1}^{\infty} \frac{3^n \left (x - \frac{2}{3} \right )^n}{n3^n} = \sum_{n=1}^{\infty} \frac{\left (x - \frac{2}{3} \right )^n}{n} \end{align}
Now applying the ratio test and we get that:
(4)
\begin{align} \quad \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{n}{n+1} = 1 \end{align}
Therefore the radius of convergence is $1$.
## Example 3
Determine the radii of convergence of the power series $\sum_{n=1}^{\infty} \frac{(ax + b)^n}{c^n}$. where $a, b, c \in \mathbb{R}$ and $a, c \neq 0$.**
We will first rewrite the series above as follows:
(5)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{(ax + b)^n}{c^n} = \sum_{n=1}^{\infty} \frac{\left (a \left (x + \frac{b}{a} \right ) \right )^n}{c^n} = \frac{a^n \left ( x + \frac{b}{a} \right )^n}{c^n} \end{align}
Therefore applying the ratio test we get that:
(6)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a^{n+1}c^n}{a^{n}c^{n+1}} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{a}{c} \biggr \rvert \end{align}
Therefore the radius of convergence is $\biggr \rvert \frac{c}{a} \biggr \rvert$.
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Calculate square root with pencil and paper
Introduction
In this article the ancient art of calculating a square root, using pencil and paper, is rediscovered.
The reader should know the addition and multiplication tables by heart .
After the examples an explanation follows why this method is correct.
Notation: the square root of 100 is written as SQRT(100).
Examples
1.
We calculate SQRT(4096)
step 1:
split the number (rigth to left) in groups of 2 digits, result 40 | 96
step 2:
find the digit which square is closest (below or equal) to 40,
this is the 6 , because 6 * 6 = 36
step 3:
subtract 36 from 40 and shift next 2 digits in place
``` SQRT(40 96) = 6
36
---
4 96
```
The temparary answer is 6. The remainder is 496
Step 4:
multiply temporary answer by 2, so 2 * 6 = 12
Step 5:
write 12 as 12? * ?
step 6:
find digit ? for value closest (below) or equal to remainder.
This digit is 4 , because 124 * 4 = 496
step 7:
subtract and add digit 4:
``` W(40 96) = 64
36
---
4 96
4 96
------
0
```
SQRT(4086) = 64, because 64 *64 = 4096
2.
Calculating SQRT(1522756)
- split number in groups of 2 digits, add 0 when number of digits is odd
01 | 52 | 27 | 56
- find first square = 1
``` SQRT(01 52 27 56) = 1
1
---
0 52
```
- multiply by 2 : 2 * 1 = 2
- write 2? * ?
- ? = 2
- 22 * 2 = 44
``` SQRT(01 52 27 56) = 12
1
---
0 52
44
---
8 27
```
- multiply temporary answer by 2 : 2 * 12 = 24
- write 24? * ?
- ? = 3, because 243 * 3 = 729
``` SQRT(01 52 27 56) = 123
1
---
0 52
44
---
8 27
7 29
------
98 56
```
- multiply temporary answer by 2 : 2 * 123 = 246
- write 246? * ?
- find ? = 4 ( 3 too small, 5 too large)
- 2464 * 4 = 9856
``` SQRT(01 52 27 56) = 1234
1
---
0 52
44
---
8 27
7 29
------
98 56
98 56
-----
0
```
3.
Calculate SQRT(5).
Because 5 is no square, the answer will be an approximation.
- write 05| . 00 | 00 | 00 ................
- first square = 2
``` SQRT(05) = 2.
4
---
1 00
```
place decimal point in answer, but do not use in calculations.
- double answer : 2 * 2 = 4
- write 4? * ?
- ? = 2
- 42 * 2 = 84
``` SQRT(05) = 2.2
4
---
1 00
84
---
16 00
```
- double answer : 2 * 22 = 44
- write 44? * ?
- find ? = 3
- 443 * 3 = 1329
- subtract
``` SQRT(05) = 2.23
4
---
1 00
84
---
16 00
13 29
-----
2 71 00
```
- double answer : 2 * 223 = 446
- write as 446? * ?
- ? = 6
- 4466 * 6 = 26796
``` SQRT(05) = 2.236
4
---
1 00
84
---
16 00
13 29
-----
2 71 00
2 67 96
-------
3 04
```
- double answer : 2 * 2236 = 4472
- write as 4472? * ?
- ? = 0, so only shift down 2 zero digits
``` SQRT(05) = 2.2360
4
---
1 00
84
---
16 00
13 29
-----
2 71 00
2 67 96
-------
3 04 00
```
- reapeat steps before for more digits in answer.
Why this works
Core of the method is the product
(a + b)2 = a2 + 2ab + b2
For clarity this notation is introduced:
if a number consists of digits a and b then
[ab] = 10a + b
so
[abc] = 100a + 10b + c
[23b] = 230 + b
[(2*3)b] = [6b] = 60 + b
Example : SQRT(4096) again.
Let the answer be [ab] = 10a + b so
4096 = [ab]2 = (10a + b)2 = 100a2 + 2*10*ab + b2 = 100a2 + (20a + b)b
We have to find digits a and b.
Write number as 40.96
62 = 36 < 40 and 72 = 49 > 40,
a = 6
[6b]2 = 36 00 + 2*10*6b + b2 = 4096
120b +b2 = (120 + b)*b = [12b]*b
[12b]*b = 4096 - 36 00 = 496
b = 4
but this is exactly what we did before, finding the value of ? in 12? * ? .
Only, b is used instead of ?
Another example
SQRT(01 52 27 56)
Write as SQRT( 01 . 52 27 56) and we observe
a = 1
write number as 1 52.27 56
[1b]2 = 1 00 + 2*10*b + b2 = 152 ,27 56
20b + b2 = (20 + b)*b = [2b]*b
[2b]*b = 1 52, 27 56 - 100 = 52, 27 56
b = 2
The temporary answer [12].
Write original number as 152 27. 56
Multiply temporary answer by 10, result is [12b]
So far, number 1 52 27 . 56 - [120]2 is the remainder:
1 5227, 56 - 14400 = 827, 56
And this is the tric:
regard [12] as the new value of a and find value of b in [12b]
Following instructions above, we find b = 3, so the new temporary answer is [123]
Call [123] the new value of a.
Again shift decimal point 2 places right and multiply temporary answer by 10
1 52 27 56 - 12 302 = 9856
Calculated value b = 4 and remainder is 0.
The general explanation
Say the root of an 8 digit number xx xx xx xx must be calculated.
step action 1 write number as xx . xx xx xx = a . ------.2 2 calculate a (temporary answer) and subtract : remainder = number - a2 3 multiply remainder by 100, temporary answer by 10 4 calculate b in [ab] 5 calculate new remainder = remainder - [(2a)b]*b 6 new a = [ab] 7 repeat steps 3 .. 7
Do not worry about tehe decimal point : it is automatically right.
Two digits of the number produce one digit of the answer.
The purpose of the decimal point was only to facilitate the explanation.
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# 2.7 Piecewise Functions
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### 2.7 Piecewise Functions
1. 1. 2.7 Piecewise Functions
2. 2. What is a Piecewise Function?• A function that combines pieces of different equations.• Each piece is for a different domain (set of x values).• Example:
3. 3. Why Are They Important?• In real life, lots of problems are modeled by piecewise functions.• Examples: ▫ Finding shipping costs ▫ Income taxes ▫ Ordering t-shirts
4. 4. Examples:
5. 5. Writing Piecewise Functions• We know how to graph, now go backwards!• First, find the domains (where the graph is cut)• Then, find the slopes and y-intercepts.• Fill in the equation for each domain.• Example: ___ x + ___ , if x ______ ___ x + ___ , if x ______
6. 6. Example: ___ x + ___ , if x ______ ___ x + ___ , if x ______
7. 7. Your Turn! ___ x + ___ , if x ______ ___ x + ___ , if x ______
8. 8. Evaluating from a Graph• Move left/right to the x you need, then move up/down to find y.• Example:• Evaluate f(x) for the function shown when:• x = -3• x = -1•x=2
9. 9. Your Turn!• Evaluate f(x) for the function shown when:•x = -1•x = 1•x = 2•x = 4
10. 10. Evaluating Piecewise Functions• The domain tells you which equation to use.• Evaluate f(x) when:a)x = 0b)x = 2c)x = 4
11. 11. Your Turn!• Evaluate f(x) when:•x = 0•x = 3•x = 6
12. 12. Step FunctionsEach piece of the function is a flat,horizontal line.
1. #### A particular slide catching your eye?
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# Implicit differentiation and its use in derivatives
Implicit differentiation is one of the types of derivatives used widely in differentiation calculus is a sort of derivative in which the derivative of the equation must be determined. Differentiation calculus is a type of calculus that is mainly used to find the rate of change of the functions. In calculus, it is commonly used to find the derivatives of an algebraic expression. In this post, we’ll look at what implicit differentiation is and how to identify it using a variety of examples.
###### What is Implicit Differentiation in calculus?
Before starting the introduction of implicit differentiation, we must have a sound knowledge about implicit function as in implicit differential we have to calculate the derivative of implicit functions. The function becomes an implicit function when the dependent variable is not explicitly isolated on either side of the equation.
A function defined in terms of both dependent and independent variables, such as,
X2 + 3Y = 0
is an implicit function. An explicit function, on the other hand, is expressed in terms of an independent variable, such as y = f(x). An implicit function can be easily differentiated without rearranging the function and instead distinguishing each word in the case of differentiation. Because y is a function of x, we’ll use the chain rule, as well as the product and quotient rules.
The technique of obtaining the derivative of an implicit function is known as implicit differentiation. Explicit and implicit functions are the two types of functions. The dependent variable “y” is on one of the sides of the equation in an explicit function of the type y = f(x). However, having ‘y’ on one side of the equation is not necessarily required. Consider the following functions, for example:
• X3 + 3Y = 5
• xy2 + cos(xy) = 0
Even though ‘y’ is not one of the sides of the equation in the first case, we can still solve it to write it as y = 1/3 (5 – x3), and it is an explicit function. However, in the second situation, we cannot readily solve the equation for ‘y,’ and this sort of function is known as an implicit function. Implicit differentiation calculator can be used for the accurate results for this type of derivative.
###### Method to solve this type of differentiation
We can’t start with dy/dx in the process of implicit differentiation since an implicit function isn’t of the type y = f(x), but rather f (x, y) = 0. It’s important to be aware of derivative rules like the power rule, addition rule, product rule, exponent rule, quotient rule, chain rule, and so on. The step-by-step method to find the implicit differentiation is mentioned below.
• In the given equation, first, take the derivative of the given equation with respect to x, differentiate each term of f (x, y) = 0.
• Sort the terms that have dy/dx on one side and the ones that don’t have dy/dx on the other.
• Make dy/dx a function of either x or y, or both.
###### How to evaluate the problems of implicit differentiation?
The steps for performing implicit differentiation have been demonstrated. Is there a certain formula we came across along the way? No!! There is no specific formula for implicit differentiation; instead, to obtain the implicit derivative, we follow the processes mentioned above.
Important Points to Remember About Implicit Differentiation:
• When the function is of the form f(x, y) = 0, implicit differentiation is the process of determining dy/dx.
• Simply differentiate on both sides and solve for dy/dx to discover the implicit derivative dy/dx. However, whenever we are distinguishing y, we should write dy/dx.
• In the process of implicit differentiation, all derivative formulas and techniques must be applied as well.
To understand this concept let’s take some examples.
###### Example 1
Find the implicit differentiation of x3 + y2 = 9
Solution
Step 1: write the given function.
x3 + y2 = 9
Step 2: Apply d/dx on both sides of the given equation.
d/dx (x3 + y2) = d/dx (9)
Step 3: Apply the rule of the differentiation.
d/dx (x3) +d/dx (y2) = d/dx (9)
Step 4: Solve the derivative.
3x2 + 2y dy/dx = 0
Step 5: Arrange the above equation to get dy/dx.
3x2 + 2y dy/dx = 0
2y dy/dx = -3x2
dy/dx = -3x2/2y
Thus, the implicit differentiation of the given function is dy/dx = -3x2/2y.
###### Example 2
Find the implicit differentiation of 4x2 + 2y2 = 6y
Solution
Step 1: write the given function.
4x2 + 2y2 = 6y
Step 2: Apply d/dx on both sides of the given equation.
d/dx (4x2 + 2y2) = d/dx (6y)
Step 3: Apply the rule of the differentiation.
d/dx (4x2) +d/dx (2y2) = d/dx (6y)
Step 4: Solve the derivative.
8x + 4y dy/dx = 6dy/dx
Step 5: Arrange the above equation to get dy/dx.
8x + 4y dy/dx = 6dy/dx
4y dy/dx = 6 dy/dx – 8x
so 4y dy/dx – 6 dy/dx = – 8x
(4y -6) dy/dx = -8x
dy/dx = -8x / (4y – 6)
so dy/dx = -8x / 2(2y – 3)
dy/dx = -4x / (2y – 3)
Thus, the implicit differentiation of the given function is dy/dx = -4x / (2y – 3). You can also find the antiderivative or integral of a function using antiderivative calculator.
###### Example 3
Find the implicit differentiation of x2 + y2 = 7y2 + 7x
Solution
Step 1: Write the given function.
x2 + y2 = 7y2 + 7x
Step 2: Apply d/dx on both sides of the given equation.
d/dx (x2 + y2) = d/dx (7y2 + 7x)
Step 3: Apply the rule of the differentiation.
d/dx (x2) +d/dx (y2) = d/dx (7y2) + dy/dx (7x)
Step 4: Solve the derivative.
2x + 2y dy/dx = 14y dy/dx +7
Step 5: Arrange the above equation to get dy/dx.
2x + 2y dy/dx = 14y dy/dx +7
2y dy/dx = 14y dy/dx +7 – 2x
so 2y dy/dx – 14y dy/dx = 7 – 2x
(2y -14y) dy/dx = 7 – 2x
dy/dx = (7 – 2x) / (2y -14y)
Thus, the implicit differentiation of the given function is dy/dx = (7 – 2x) / (2y -14y).
###### Summary
A function defined in terms of both dependent and independent variables, such as X2 + 3Y = 0, is an implicit function. Implicit differentiation is a process of finding dy/dx of the given implicit function equation such as f (x, y) = 0. Implicit differentiation has no specific formula to solve the problems rather it has some steps to solve the problems of implicit differentiation.
By following those steps we can easily solve any problem related to this type of differentiation.
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# PLAYING WITH NUMBERS 249
CHAPTER
## Playing with Numbers
16
16.1 Introduction
You have studied various types of numbers such as natural numbers, whole numbers,
integers and rational numbers. You have also studied a number of interesting properties
about them. In Class VI, we explored finding factors and multiples and the relationships
among them.
In this chapter, we will explore numbers in more detail. These ideas help in justifying
tests of divisibility.
## 16.2 Numbers in General Form
Let us take the number 52 and write it as
52 = 50 + 2 = 10 × 5 + 2
Similarly, the number 37 can be written as
37 = 10 × 3 + 7
In general, any two digit number ab made of digits a and b can be written as
ab = 10 × a + b = 10a + b Here ab does not
mean a × b!
What about ba? ba = 10 × b + a = 10b + a
Let us now take number 351. This is a three digit number. It can also be written as
351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1
Similarly 497 = 100 × 4 + 10 × 9 + 1 × 7
In general, a 3-digit number abc made up of digits a, b and c is written as
abc = 100 × a + 10 × b + 1 × c
= 100a + 10b + c
In the same way,
cab = 100c + 10a + b
bca = 100b + 10c + a and so on.
250 MATHEMATICS
TRY THESE
1. Write the following numbers in generalised form.
(i) 25 (ii) 73 (iii) 129 (iv) 302
2. Write the following in the usual form.
(i) 10 × 5 + 6 (ii) 100 × 7 + 10 × 1 + 8 (iii) 100 × a + 10 × c + b
## 16.3 Games with Numbers
(i) Reversing the digits – two digit number
Minakshi asks Sundaram to think of a 2-digit number, and then to do whatever she asks
him to do, to that number. Their conversation is shown in the following figure. Study the
It so happens that Sundaram chose the number 49. So, he got the reversed number
94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this
number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi
PLAYING WITH NUMBERS 251
TRY THESE
Check what the result would have been if Sundaram had chosen the numbers shown
below.
1. 27 2. 39 3. 64 4. 17
## Now, let us see if we can explain Minakshi’s “trick”.
Suppose Sundaram chooses the number ab, which is a short form for the 2-digit
number 10a + b. On reversing the digits, he gets the number ba = 10b + a. When he adds
the two numbers he gets:
(10a + b) + (10b + a) = 11a + 11b
= 11 (a + b).
So, the sum is always a multiple of 11, just as Minakshi had claimed.
Observe here that if we divide the sum by 11, the quotient is a + b, which is exactly the
sum of the digits of chosen number ab.
You may check the same by taking any other two digit number.
The game between Minakshi and Sundaram continues!
Minakshi: Think of another 2-digit number, but don’t tell me what it is.
Sundaram: Alright.
Minakshi: Now reverse the digits of the number, and subtract the smaller number from
the larger one.
Sundaram: I have done the subtraction. What next?
Minakshi: Now divide your answer by 9. I claim that there will be no remainder!
Sundaram: Yes, you are right. There is indeed no remainder! But this time I think I know
how you are so sure of this!
In fact, Sundaram had thought of 29. So his calculations were: first he got
the number 92; then he got 92 – 29 = 63; and finally he did (63 ÷ 9) and got 7 as
quotient, with no remainder.
TRY THESE
Check what the result would have been if Sundaram had chosen the numbers shown
below.
1. 17 2. 21 3. 96 4. 37
Let us see how Sundaram explains Minakshi’s second “trick”. (Now he feels confident
of doing so!)
Suppose he chooses the 2-digit number ab = 10a + b. After reversing the digits, he
gets the number ba = 10b + a. Now Minakshi tells him to do a subtraction, the
smaller number from the larger one.
• If the tens digit is larger than the ones digit (that is, a > b), he does:
(10a + b) – (10b + a) = 10a + b – 10b – a
= 9a – 9b = 9(a – b).
252 MATHEMATICS
• If the ones digit is larger than the tens digit (that is, b > a), he does:
(10b + a) – (10a + b) = 9(b – a).
• And, of course, if a = b, he gets 0.
In each case, the resulting number is divisible by 9. So, the remainder is 0. Observe
here that if we divide the resulting number (obtained by subtraction), the quotient is
a – b or b – a according as a > b or a < b. You may check the same by taking any
other two digit numbers.
(ii) Reversing the digits – three digit number.
Now it is Sundaram’s turn to play some tricks!
Sundaram: Think of a 3-digit number, but don’t tell me what it is.
Minakshi: Alright.
Sundaram: Now make a new number by putting the digits in reverse order, and subtract
the smaller number from the larger one.
Minakshi: Alright, I have done the subtraction. What next?
Sundaram: Divide your answer by 99. I am sure that there will be no remainder!
In fact, Minakshi chose the 3-digit number 349. So she got:
• Reversed number: 943; • Difference: 943 – 349 = 594;
• Division: 594 ÷ 99 = 6, with no remainder.
TRY THESE
Check what the result would have been if Minakshi had chosen the numbers shown
below. In each case keep a record of the quotient obtained at the end.
1. 132 2. 469 3. 737 4. 901
## Let us see how this trick works.
Let the 3-digit number chosen by Minakshi be abc = 100a + 10b + c.
After reversing the order of the digits, she gets the number cba = 100c + 10b + a. On
subtraction:
• If a > c, then the difference between the numbers is
(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a
= 99a – 99c = 99(a – c).
• If c > a, then the difference between the numbers is
(100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).
• And, of course, if a = c, the difference is 0.
In each case, the resulting number is divisible by 99. So the remainder is 0. Observe
that quotient is a – c or c – a. You may check the same by taking other 3-digit numbers.
(iii) Forming three-digit numbers with given three-digits.
Now it is Minakshi’s turn once more.
PLAYING WITH NUMBERS 253
## Minakshi: Think of any 3-digit number.
Sundaram: Alright, I have done so.
Minakshi: Now use this number to form two more 3-digit numbers, like this: if the
number you chose is abc, then
• ‘the first number is cab (i.e., with the ones digit shifted to the “left end” of
the number);
• the other number is bca (i.e., with the hundreds digit shifted to the “right
end” of the number).
Now add them up. Divide the resulting number by 37. I claim that there will
be no remainder.
Sundaram: Yes. You are right!
In fact, Sundaram had thought of the 3-digit number 237. After doing what Minakshi had
asked, he got the numbers 723 and 372. So he did:
2 3 7
+ 7 2 3 Form all possible 3-digit numbers using all the digits 2, 3 and
7 and find their sum. Check whether the sum is divisible by
+ 3 7 2 37! Is it true for the sum of all the numbers formed by the
digits a, b and c of the number abc?
1 3 3 2
Then he divided the resulting number 1332 by 37:
1332 ÷ 37 = 36, with no remainder.
TRY THESE
Check what the result would have been if Sundaram had chosen the numbers
shown below.
1. 417 2. 632 3. 117 4. 937
## Will this trick always work?
Let us see. abc = 100a + 10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
abc + cab + bca = 111(a + b + c)
= 37 × 3(a + b + c), which is divisible by 37
## 16.4 Letters for Digits
Here we have puzzles in which letters take the place of digits in an arithmetic ‘sum’, and
the problem is to find out which letter represents which digit; so it is like cracking a code.
Here we stick to problems of addition and multiplication.
254 MATHEMATICS
## Here are two rules we follow while doing such puzzles.
1. Each letter in the puzzle must stand for just one digit. Each digit must be
represented by just one letter.
2. The first digit of a number cannot be zero. Thus, we write the number “sixty
three” as 63, and not as 063, or 0063.
A rule that we would like to follow is that the puzzle must have just one answer.
Example 1: Find Q in the addition.
3 1 Q
+ 1 Q 3
5 0 1
Solution:
There is just one letter Q whose value we have to find.
Study the addition in the ones column: from Q + 3, we get ‘1’, that is, a number whose
ones digit is 1.
For this to happen, the digit Q should be 8. So the puzzle can be solved as shown below.
3 1 8
+ 1 8 3
5 0 1
That is, Q = 8
Example 2: Find A and B in the addition.
A
+ A
+ A
B A
Solution: This has two letters A and B whose values are to be found.
Study the addition in the ones column: the sum of three A’s is a number whose ones digit
is A. Therefore, the sum of two A’s must be a number whose ones digit is 0.
This happens only for A = 0 and A = 5.
If A = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0 too. We do not want this
(as it makes A = B, and then the tens digit of BA too becomes 0), so we reject this
possibility. So, A = 5.
Therefore, the puzzle is solved as shown below.
5
+ 5
+ 5
That is, A = 5 and B = 1. 1 5
PLAYING WITH NUMBERS 255
## Example 3: Find the digits A and B.
BA
× B 3
5 7 A
Solution:
This also has two letters A and B whose values are to be found.
Since the ones digit of 3 × A is A, it must be that A = 0 or A = 5.
Now look at B. If B = 1, then BA × B3 would at most be equal to 19 × 19; that is,
it would at most be equal to 361. But the product here is 57A, which is more than 500. So
we cannot have B = 1.
If B = 3, then BA × B3 would be more than 30 × 30; that is, more than 900. But 57A
is less than 600. So, B can not be equal to 3.
Putting these two facts together, we see that B = 2 only. So the multiplication is either
20 × 23, or 25 × 23.
The first possibility fails, since 20 × 23 = 460. But, the 2 5
second one works out correctly, since 25 × 23 = 575. × 2 3
So the answer is A = 5, B = 2. 5 7 5
DO THIS
Write a 2-digit number ab and the number obtained by reversing its digits i.e., ba. Find
their sum. Let the sum be a 3-digit number dad
i.e., ab + ba = dad
(10a + b) + (10b + a) = dad
The sum a + b can not exceed 18 (Why?).
Is dad a multiple of 11?
Write all the 3-digit numbers which are multiples of 11 upto 198.
Find the values of a and d.
EXERCISE 16.1
Find the values of the letters in each of the following and give reasons for the steps involved.
1. 3 A 2. 4 A 3. 1 A
+ 2 5 + 9 8 × A
B 2 C B 3 9 A
256 MATHEMATICS
4. A B 5. A B 6. A B
+ 3 7 × 3 × 5
6 A C A B C A B
2 A B
7. A B 8. A 1 9.
+ A B 1
× 6 + 1 B
B 1 8
B B B B 0
10. 1 2 A
+ 6 A B
A 0 9
## 16.5 Tests of Divisibility
In Class VI, you learnt how to check divisibility by the following divisors.
10, 5, 2, 3, 6, 4, 8, 9, 11.
You would have found the tests easy to do, but you may have wondered at the same
time why they work. Now, in this chapter, we shall go into the “why” aspect of the above.
16.5.1 Divisibility by 10
This is certainly the easiest test of all! We first look at some multiples of 10.
10, 20, 30, 40, 50, 60, ... ,
and then at some non-multiples of 10.
13, 27, 32, 48, 55, 69,
From these lists we see that if the ones digit of a number is 0, then the number is a
multiple of 10; and if the ones digit is not 0, then the number is not a multiple of 10. So, we
get a test of divisibility by 10.
Of course, we must not stop with just stating the test; we must also explain why it
“works”. That is not hard to do; we only need to remember the rules of place value.
Take the number. ... cba; this is a short form for
... + 100c + 10b + a
Here a is the one’s digit, b is the ten’s digit, c is the hundred’s digit, and so on. The
dots are there to say that there may be more digits to the left of c.
Since 10, 100, ... are divisible by 10, so are 10b, 100c, ... . And as for the number a
is concerned, it must be a divisible by 10 if the given number is divisible by 10. This is
possible only when a = 0.
Hence, a number is divisible by 10 when its one’s digit is 0.
16.5.2 Divisibility by 5
Look at the multiples of 5.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50,
PLAYING WITH NUMBERS 257
We see that the one’s digits are alternately 5 and 0, and no other digit ever
appears in this list.
So, we get our test of divisibility by 5.
If the ones digit of a number is 0 or 5, then it is divisible by 5.
Let us explain this rule. Any number ... cba can be written as:
... + 100c + 10b + a
Since 10, 100 are divisible by 10 so are 10b, 100c, ... which in turn, are divisible
by 5 because 10 = 2 × 5. As far as number a is concerned it must be divisible by 5 if the
number is divisible by 5. So a has to be either 0 or 5.
TRY THESE
(The first one has been done for you.)
1. If the division N ÷ 5 leaves a remainder of 3, what might be the ones digit of N?
(The one’s digit, when divided by 5, must leave a remainder of 3. So the one’s digit
must be either 3 or 8.)
2. If the division N ÷ 5 leaves a remainder of 1, what might be the one’s digit of N?
3. If the division N ÷ 5 leaves a remainder of 4, what might be the one’s digit of N?
16.5.3 Divisibility by 2
Here are the even numbers.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, ... ,
and here are the odd numbers.
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... ,
We see that a natural number is even if its one’s digit is
2, 4, 6, 8 or 0
A number is odd if its one’s digit is
1, 3, 5, 7 or 9
Recall the test of divisibility by 2 learnt in Class VI, which is as follows.
If the one’s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
The explanation for this is as follows.
Any number cba can be written as 100c + 10b + a
First two terms namely 100c, 10b are divisible by 2 because 100 and 10 are divisible
by 2. So far as a is concerned, it must be divisible by 2 if the given number is divisible by
2. This is possible only when a = 0, 2, 4, 6 or 8.
TRY THESE
(The first one has been done for you.)
1. If the division N ÷ 2 leaves a remainder of 1, what might be the one’s digit of N?
(N is odd; so its one’s digit is odd. Therefore, the one’s digit must be 1, 3, 5, 7 or 9.)
258 MATHEMATICS
2. If the division N ÷ 2 leaves no remainder (i.e., zero remainder), what might be the
one’s digit of N?
3. Suppose that the division N ÷ 5 leaves a remainder of 4, and the division N ÷ 2
leaves a remainder of 1. What must be the one’s digit of N?
## 16.5.4 Divisibility by 9 and 3
Look carefully at the three tests of divisibility found till now, for checking division by
10, 5 and 2. We see something common to them: they use only the one’s digit of the
given number; they do not bother about the ‘rest’ of the digits. Thus, divisibility
is decided just by the one’s digit. 10, 5, 2 are divisors of 10, which is the key
number in our place value.
But for checking divisibility by 9, this will not work. Let us take some number say 3573.
Its expanded form is: 3 × 1000 + 5 × 100 + 7 × 10 + 3
This is equal to 3 × (999 + 1) + 5 × (99 + 1) + 7 × (9 + 1) + 3
= 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 3) ... (1)
We see that the number 3573 will be divisible by 9 or 3 if (3 + 5 + 7 + 3) is divisible
by 9 or 3.
We see that 3 + 5 + 7 + 3 = 18 is divisible by 9 and also by 3. Therefore, the number
3573 is divisible by both 9 and 3.
Now, let us consider the number 3576. As above, we get
3576 = 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 6) ... (2)
Since (3 + 5 + 7 + 6) i.e., 21 is not divisible by 9 but is divisible by 3,
therefore 3576 is not divisible by 9. However 3576 is divisible by 3. Hence,
(i) A number N is divisible by 9 if the sum of its digits is divisible by 9. Otherwise it is
not divisible by 9.
(ii) A number N is divisible by 3 if the sum of its digits is divisible by 3. Otherwise it is
not divisible by 3.
If the number is ‘cba’, then, 100c + 10b + a = 99c + 9b + (a + b + c)
= 9(11c + b) + (a + b + c)
divisible by 3 and 9
## Hence, divisibility by 9 (or 3) is possible if a + b + c is divisible by 9 (or 3).
Example 4: Check the divisibility of 21436587 by 9.
Solution: The sum of the digits of 21436587 is 2 + 1 + 4 + 3 + 6 + 5 + 8 + 7 = 36.
This number is divisible by 9 (for 36 ÷ 9 = 4). We conclude that 21436587 is divisible by 9.
We can double-check:
21436587
= 2381843 (the division is exact).
9
PLAYING WITH NUMBERS 259
## Example 5: Check the divisibility of 152875 by 9.
Solution: The sum of the digits of 152875 is 1 + 5 + 2 + 8 + 7 + 5 = 28. This number
is not divisible by 9. We conclude that 152875 is not divisible by 9.
TRY THESE
Check the divisibility of the following numbers by 9.
1. 108 2. 616 3. 294 4. 432 5. 927
Example 6: If the three digit number 24x is divisible by 9, what is the value of x?
Solution: Since 24x is divisible by 9, sum of it’s digits, i.e., 2 + 4 + x should be
divisible by 9, i.e., 6 + x should be divisible by 9.
This is possible when 6 + x = 9 or 18, ....
But, since x is a digit, therefore, 6 + x = 9, i.e., x = 3.
## THINK, DISCUSS AND WRITE
1. You have seen that a number 450 is divisible by 10. It is also divisible by 2 and 5
which are factors of 10. Similarly, a number 135 is divisible 9. It is also divisible
by 3 which is a factor of 9.
Can you say that if a number is divisible by any number m, then it will also be
divisible by each of the factors of m?
2. (i) Write a 3-digit number abc as 100a + 10b + c
= 99a + 11b + (a – b + c)
= 11(9a + b) + (a – b + c)
If the number abc is divisible by 11, then what can you say about
(a – b + c)?
Is it necessary that (a + c – b) should be divisible by 11?
(ii) Write a 4-digit number abcd as 1000a + 100b + 10c + d
= (1001a + 99b + 11c) – (a – b + c – d)
= 11(91a + 9b + c) + [(b + d) – (a + c)]
If the number abcd is divisible by 11, then what can you say about
[(b + d) – (a + c)]?
(iii) From (i) and (ii) above, can you say that a number will be divisible by 11 if
the difference between the sum of digits at its odd places and that of digits at
the even places is divisible by 11?
## Example 7: Check the divisibility of 2146587 by 3.
Solution: The sum of the digits of 2146587 is 2 + 1 + 4 + 6 + 5 + 8 + 7 = 33. This
number is divisible by 3 (for 33 ÷ 3 = 11). We conclude that 2146587 is divisible by 3.
260 MATHEMATICS
## Example 8: Check the divisibility of 15287 by 3.
Solution: The sum of the digits of 15287 is 1 + 5 + 2 + 8 + 7 = 23. This number is not
divisible by 3. We conclude that 15287 too is not divisible by 3.
TRY THESE
Check the divisibility of the following numbers by 3.
1. 108 2. 616 3. 294 4. 432 5. 927
EXERCISE 16.2
1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
You will find that there are two answers for the last problem. Why is this so?
3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one
of these numbers: 0, 3, 6, 9, 12, 15, 18, ... . But since x is a digit, it can only be that
6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of
four different values.)
4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
## WHAT HAVE WE DISCUSSED?
1. Numbers can be written in general form. Thus, a two digit number ab will be written as
ab = 10a + b.
2. The general form of numbers are helpful in solving puzzles or number games.
3. The reasons for the divisibility of numbers by 10, 5, 2, 9 or 3 can be given when numbers are
written in general form.
|
# Computing Derivatives: Part 2 (Explaining Calculus #8)
Most recently in the series on calculus, we did some overview on some “precalculus” topics that we’d need for later calculus discussions. Having now done this, we move on to several examples of more ‘complicated’ rules that derivatives follow. We will then investigate some more difficult specific functions and their derivatives. Finally, at the end, I will leave a list of “homework problems” for any of my readers who want practice.
The Product Rule
We have already tackled to to take derivatives of functions like $f(x) + g(x)$. We handled addition, one of the main operations of arithmetic. By doing addition, we were able to handle subtraction as well. But what about multiplication? That would be the natural next step, just as in school multiplication is the next natural step after learning about addition and subtraction. Our purpose now is to lay out the so-called product rule, which enables us to take derivatives of functions like $f(x) g(x)$.
Fact: For any two functions $f(x)$ and $g(x)$ that have derivatives,
$\dfrac{d}{dx}[f(x) g(x)] = f^\prime(x) g(x) + f(x) g^\prime(x).$
Proof: This proof uses a slick trick of “adding zero” to an expression. First, by the definition of derivatives,
$\dfrac{d}{dx}[f(x) g(x)] = \lim\limits_{h \to 0} \dfrac{f(x+h) g(x+h) - f(x) g(x)}{h}.$
Now, we want to be clever. Our clever move will be the fact that $f(x) g(x+h) - f(x) g(x+h) = 0$. While this looks very strange to point out something so obvious, this fact means that
$\lim\limits_{h \to 0} \dfrac{f(x+h) g(x+h) - f(x) g(x)}{h} = \lim\limits_{h \to 0} \dfrac{f(x+h) g(x+h) - f(x) g(x+h) + f(x) g(x+h) - f(x) g(x)}{h}.$
The first to pieces of this numerator have a common factor, as do the second two. Therefore, this messy expression can be simplified a little bit as
$\lim\limits_{h \to 0} \dfrac{g(x+h)(f(x+h) - f(x))}{h} + \lim\limits_{h \to 0} \dfrac{f(x)(g(x+h) - g(x))}{h}.$
Since $\lim\limits_{h \to 0} g(x+h) = g(x)$, the first of these simplifies as
$\lim\limits_{h \to 0} \dfrac{g(x+h)(f(x+h) - f(x))}{h} = g(x) \lim\limits_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = g(x) f^\prime(x)$.
The second piece, using the exact same process, is equal to $f(x) g^\prime(x)$. When everything is put back together, $f(x) g(x)$ has derivative $f^\prime(x) g(x) + f(x) g^\prime(x)$.
The Chain Rule
You’d think that the next thing we would do is division. And in fact, there is a way I could do division next. But it will actually be easier to do something else first. I want to do functions inside of other functions next, like $f(g(x))$, which just means “copy-paste the expression $g(x)$ wherever there would have been an $x$. This is called the chain rule, because the functions sort of link together like a chain, inextricably connected.
Fact: For any two functions $f(x), g(x)$ for which $f(g(x))$ makes sense, we have $\dfrac{d}{dx}[f(g(x))] = f^\prime(g(x)) * g^\prime(x).$ In different notation, if $y$ is a function of $u$, which itself is a function of $x$, then
$\dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{dy}{dx}.$
In an earlier post, I made a comment about a similarity between this derivative notation and genuine fractions. The chain rule is the central such similarity – when written in the dy-over-dx style, it looks as if the du’s are cancelling out, like they would if these were actual fractions.
Proof: This one is actually quite tricky compared to the others. For this reason, I won’t actually do a totally correct proof. Instead, I’m going to do a proof that “usually” works. For those who want a great challenge, try to figure out where this proof goes wrong. (If you end up wanting to know, the Wikipedia article on the chain rule will actually tell you. If you don’t care, then reading this proof will give you the right idea.)
The definition of the derivative tells us that
$\dfrac{d}{dx}[f(g(x))] = \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{h}.$
We now make a clever step of multiplying top and bottom by $g(x+h) - g(x)$.
$\dfrac{d}{dx}[f(g(x))] = \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \dfrac{g(x+h) - g(x)}{h}.$
We can even split this up into two limits multiplied together.
$\dfrac{d}{dx}[f(g(x))] = \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \lim\limits_{h \to 0} \dfrac{g(x+h) - g(x)}{h} = g^\prime(x) \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)}.$
Since $g(x)$ will be a continuous function (it has a derivative so it must be continuous), $g(x+h) \to g(x)$ as $h \to 0$. This enables us to treat $g(x+h)$ as if it were something like $g(x) + h$. There is a more careful way to write that down, but since we are being informal I won’t do that (and no, this isn’t the real problem in the proof… that already happened earlier). This will mean that
$\lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} = \lim\limits_{h \to 0} \dfrac{f(g(x) + h) - f(g(x))}{(g(x) + h) - g(x)} = \lim\limits_{h \to 0} \dfrac{f(y+h) - f(y)}{h},$
where $y = g(x)$ is used to make the formulas easier to follow. This leads directly to the value $f^\prime(g(x))$ for this limit. When we combine all of our work, we find that $\dfrac{d}{dx}[f(g(x))] = f^\prime(g(x)) g^\prime(x)$.
Fact: If the functions $f(x), g(x)$ have derivatives, then the derivative of $\dfrac{f(x)}{g(x)}$ is $\dfrac{g(x) f^\prime(x) - f(x) g^\prime(x)}{g(x)^2}$.
Proof: The first thing we do is to view the division as a multiplication by
$\dfrac{f(x)}{g(x)} = f(x) \cdot \dfrac{1}{g(x)}.$
Secondly, we want to use the chain rule for $\dfrac{1}{g(x)}$. To do this, we will use the function $h(x) = \dfrac{1}{x}$. Then $\dfrac{1}{g(x)} = h(g(x))$. Therefore,
$\dfrac{f(x)}{g(x)} = f(x) \cdot h(g(x)).$
We can then use the product rule to begin this derivative:
$\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = f(x) \cdot \dfrac{d}{dx}[ h(g(x)) ] + h(g(x)) f^\prime(x).$
From the chain rule, we can simplify the derivative of $h(g(x))$ to arrive at
$\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = f(x) \cdot [h^\prime(g(x)) \cdot g^\prime(x)] + h(g(x)) f^\prime(x).$
Now, since $h(x) = \dfrac{1}{x} = x^{-1}$, the rule for taking derivatives of powers of $x$ proven in an earlier post tells us that $h^\prime(x) = - x^{-2} = \dfrac{-1}{x^2}$. Therefore,
$\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = f(x) \cdot \bigg(\dfrac{-1}{g(x)^2} \cdot g^\prime(x) \bigg) + \dfrac{1}{g(x)} \cdot f^\prime(x),$
and we can simplify this as
$\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = \dfrac{- f(x) g^\prime(x)}{g(x)^2} + \dfrac{f^\prime(x) g(x)}{g(x)^2} = \dfrac{g(x) f^\prime(x) - f(x) g^\prime(x)}{g(x)^2}.$
This is the original formula we wanted to prove. So, our proof is now done.
More Special Functions
We now move on from these general principles to some more specific examples. In the previous post on computing derivatives, we built up all the tools to compute the derivative of any polynomial expression, and in fact any expression involving $x$ raised to constant powers. Now, we move on to trigonometry, exponentials, and logarithms. These derivatives are more difficult to discover, but the importance of these functions requires that any complete study of calculus should include their derivatives. Also, a thorough study of these functions and their derivatives will provide a very complete picture of how to calculate derivatives using the limit definition.
Fact: The derivatives of $\sin{x}$ and $\cos{x}$ are
$\dfrac{d}{dx}[\sin{x}] = \cos{x} \text{ and } \dfrac{d}{dx}[\cos{x}] = -\sin{x}.$
Proof: The definition of the derivative tells us that
$\dfrac{d}{dx}[ \sin{x} ] = \lim\limits_{h \to 0} \dfrac{\sin{(x+h)} - \sin{x}}{h}.$
In a previous post in the series, where I defined the function $\sin{x}$, I also gave the following rule for computing the value of $\sin{(x+h)}$:
$\sin{(x+h)} = \sin{x} \cos{h} + \cos{x} \sin{h}.$
Therefore,
$\dfrac{d}{dx}[ \sin{x} ] = \lim\limits_{h \to 0} \dfrac{\sin{(x+h)} - \sin{x}}{h} = \lim\limits_{h \to 0} \dfrac{(\sin{x} \cos{h} + \cos{x} \sin{h}) - \sin{x}}{h}.$
We can now split this limit up into two pieces – one for $\sin{x}$ and one for $\cos{x}$.
$\lim\limits_{h \to 0} \dfrac{(\sin{x} \cos{h} + \cos{x} \sin{h}) - \sin{x}}{h} = \lim\limits_{h \to 0} \dfrac{\sin{x}(\cos{h} - 1)}{h} + \lim\limits_{h \to 0} \dfrac{\cos{x} \sin{h}}{h}.$
There are some facts we have to know in order to continue, these are that $\lim\limits_{h \to 0} \dfrac{\sin{h}}{h} = 1$ and $\lim\limits_{h \to 0} \dfrac{\cos{h} -1}{h} = 0$. The proofs of these are a bit tricky, and so I don’t want to go off on a tangent (math pun!) talking about those here. I will add an appendix to the end of this post in which I talk about how to find these limits.
Moving on, once we know the values of these limits, we know that
$\lim\limits_{h \to 0} \dfrac{\sin{x}(\cos{h} - 1)}{h} = \sin{x} \cdot \lim\limits_{h \to 0} \dfrac{\cos{h} - 1}{h} = 0$
and
$\lim\limits_{h \to 0} \dfrac{\cos{x} \sin{h}}{h} = \cos{x} \cdot \lim\limits_{h \to 0} \dfrac{\sin{h}}{h} = \cos{x}.$
Therefore, putting together all the steps we’ve laid out,
$\dfrac{d}{dx}[\sin{x}] = \cos{x}.$
This completes the first half of the proof. I will leave the proof about the derivative of $\cos{x}$ as practice for any of my curious readers, only providing a few guiding hints. The proof begins the same way. Instead of using the special rule for $\sin{(x+h)}$, you need to use the rule for $\cos{(x+h)}$ that I gave in the same post in which I gave the rule for $\sin{(x+h)}$. After this rule is used, you should be able to finish the proof by following the same ideas I use here.
This completes our discussion of
Fact: The derivative of $f(x) = b^x$ is $f^\prime(x) = b^x * \log{b}$.
Proof: The first thing we will do is to “compress” every value of $b$ into the very special number $e$. Remember that, earlier in the series, we defined the “natural” exponential function $e^x$ and “natural” logarithm function $\log{x}$. Also, remember that $e^{\log{x}} = x$. Using $x = b$, we conclude that $b = e^{\log{b}}$ and therefore
$b^x = (e^{\log{b}})^x = e^{x * \log{b}}.$
Now, let’s call $f(x) = e^x$ and $g(x) = x * \log{b}$. Then $b^x = f(g(x))$. Using the chain rule, we then know that
$\dfrac{d}{dx}[b^x] = \dfrac{d}{dx}[ f(g(x)) ] = f^\prime(g(x)) * g^\prime(x) = \log{b} * f^\prime(x*\log{b}).$
What we have done here is to express the derivative of $b^x$ in terms of the derivative of $e^x$. This means that we now only need to know how to find the derivative of this most special function. Using the definition of the derivative, along with some basic rules of exponents, we have
$\dfrac{d}{dx}[e^x] = \lim\limits_{h \to 0} \dfrac{e^{x+h} - e^x}{h} = \lim\limits_{h \to 0} \dfrac{e^x(e^h - 1)}{h} = e^x \lim\limits_{h \to 0} \dfrac{e^h-1}{h}.$
The proof here is a bit detailed, but the value of $\lim\limits_{h \to 0} \dfrac{e^h - 1}{h}$ is 1. I will delay this proof to the appendix. But, the key fact here is that $e^x$ is so special because it is its own derivative. That is, if $f(x) = e^x$, then $f^\prime(x) = f(x) = e^x$! Using all of these facts, we conclude that
$\dfrac{d}{dx}[b^x] = e^{x * \log{b}} * \log{b} = b^x * \log{b}.$
This completes our discussion of the derivative of exponential functions.
Fact: The derivative of the function $f(x) = \log_b{x}$ is $f^\prime(x) = \dfrac{1}{x * \log{b}}$.
Proof: There is a way to do this similar to the previous proof about $b^x$. Instead of doing that, I want to be clever. Since the functions $b^x$ and $\log_b{x}$ are so closely related, shouldn’t their derivatives be really closely related too? We should think so. That just makes sense. Then maybe we can find a way to take advantage of our formula for the derivative of $b^x$ to find the derivative of $\log_b{x}$. Can you find the way? Take a moment and think about it.
The way I’ll do this is to transform the equation $f(x) = \log_b{x}$ into the equation $b^{f(x)} = b^{\log_b{x}}$. The second step is to notice that $b^{\log_b{x}} = x$ because of the relationship between $b^x$ and $\log_b{x}$. Therefore, $b^{f(x)} = x$. The other clever observation we have to make is that if two things are equal, their derivatives must also be equal. This means that $\dfrac{d}{dx}[ b^{f(x)} ] = \dfrac{d}{dx}[x]$. This is so clever because we can find these two derivatives by different methods. The right-hand side is much easier, the derivative of $x$ is just 1. The left-hand side can be found by using the fact we already found about the derivative of exponentials along with the chain rule:
$\dfrac{d}{dx}[b^{f(x)}] = (b^{f(x)}*\log{b}) * f^\prime(x) = (x*\log{b})*f^\prime(x)$
since $b^{f(x)} = x$ was already known to us. By combining these two derivative computations, we conclude that
$(x*\log{b})*f^\prime(x) = 1,$
and all we need to do is divide both sides of the equation by $x*\log{b}$ to find our answer.
Notice that from this fact, we can also deduce that $\dfrac{d}{dx}[\log{x}] = \dfrac{1}{x}$. I will leave this to my reader. (Hint: why is $\log{e} = 1$?)
This basically completes the list of important derivatives. But for the sake of learning, I want to compute one more that uses a clever trick much like the trick I used in the previous proof that enables us to find the derivative of a function that is “more complicated” than any of those we have previously discussed. The reason I want to show this is to demonstrate how we can take advantage of the rules we have already learned to find derivatives of more complex functions.
Fact: The derivative of $f(x) = x^x$ is $f^\prime(x) = x^x(\log{x} + 1)$.
Proof: The function $x^x$ isn’t actually susceptible to any derivative rule we’ve used to far. This is where we make a clever move to bring this function into the realm we know how to deal with by taking a logarithm. By the normal rules of logarithms, $\log{f(x)} = \log{x^x} = x \log{x}$. The derivative of the right hand side can be done using the product rule:
$\dfrac{d}{dx}[ x \log{x} ] = x \dfrac{d}{dx}[\log{x}] + \log{x} \dfrac{d}{dx}[x] = \dfrac{x}{x} + \log{x} = \log{x} + 1.$
One the other hand, the left hand side can be done using the chain rule:
$\dfrac{d}{dx}[\log{f(x)}] = \dfrac{1}{f(x)} * f^\prime(x) = \dfrac{f^\prime(x)}{x^x}.$
Since the left and right hand sides are equal, we conclude that
$\dfrac{f^\prime(x)}{x^x} = \log{x}+1 \implies f^\prime(x) = x^x(\log{x} + 1).$
Conclusion
We have now done everything we need to do with the admittedly tedious work of computing special kinds of derivatives. If you’ve made it this far, well done. Things get less tedious and more conceptual from here. Think about how tedious it was to learn how to add and multiply small numbers. Despite this tediousness, the effort is clearly worthwhile because of how useful addition and multiplication are for solving real-world problems. What we’ve done is “learned out times tables” for derivatives. Now that we’ve finished our “tables,” we can move on to grander topics. We can stop looking at individual trees and see the forest. Beginning with the next post, this is what we shall do. In the meantime, if you want some more practice, I’ve left some practice problems and extra material in an appendix that my more curious leaders could look at.
Homework/Practice
Problem 1: Find the derivatives of $x^2 e^x$ and of $\dfrac{x^2+2}{x^3 - x}$.
Problem 2: Work out the details of the proof that the derivative of $\cos{x}$ is $- \sin{x}$. Also, prove that the derivative of $\tan{x} = \dfrac{\sin{x}}{\cos{x}}$ is $\dfrac{1}{\cos^2{x}}$. Go on to find the derivatives of $\cot{x}, \sec{x}$, and $\csc{x}$.
Problem 3: Find the derivatives of $e^{e^x}$ and of $\log{(\log{x})}$. (Hint: Use the chain rule for both!)
Appendix
In this appendix, we evaluate the various limits we have delayed in the proofs given above.
Fact: $\lim\limits_{h \to 0} \dfrac{\sin{h}}{h}$ and $\lim\limits_{h \to 0} \dfrac{\cos{h} - 1}{h}.$
Proof: The proof for the second limit is similar to the first, so we won’t consider it here. We only do the first limit. To do so, we consider the following diagram:
We need to establish some initial values of the geometry here. The angle at $A$ is defined to be $x$. The side $AB$ has length 1, which means immediately that $AE$ has length $\cos{x}$ and side $CE$ has length $\sin{x}$. (I’m not currently very good at uploading labelled diagrams like this… sketch it on a piece of paper with labelled sides if you need that visual).
We first notice the triangle $\Delta ABC$. The area of this triangle is one half its base times its height. Its height is $\sin{x}$, and its base is 1. Therefore,
$\text{Area}(\Delta ABC) = \dfrac{1}{2} \sin{x}.$
Before we carry on, why is the length of $BD$ equal to $\tan{x}$? This is because of “similar triangles.” Remember from geometry that two triangles are similar if their angles have the same degrees. The two triangles $\Delta ACE$ and $ABD$ are similar. The key fact about similar triangles is that any fractions of side lengths are always the same. This means that
$\dfrac{\text{Length}(AE)}{\text{Length}(CE)} = \dfrac{\text{Length}(AB)}{\text{Length}(BD)}.$
The length of $AE$ is just $\cos{x}$. The length of $AB$ is 1, and the length of $CE$ is $\sin{x}$. Therefore,
$\dfrac{\cos{x}}{\sin{x}} = \dfrac{1}{\text{Length}(BD)}.$
Isolating $\text{Length}(BD) = \dfrac{\sin{x}}{\cos{x}} = \tan{x}$. So, the length of $BD$ truly is equal to $\tan{x}$. This quickly tells us (as with the smaller triangle) that $\text{Area}(\Delta ABD) = \dfrac{1}{2} \tan{x}$. One more thing we want to observe. Instead of the triangle $\Delta ABC$, look at the “pizza slice” $ABC$. It definitely has more area than the triangle $\Delta ABC$ and less area that the larger triangle $\Delta ABD$. Therefore, by putting together our earlier calculations,
$\text{Area}(\Delta ABC) \leq \text{Area}(ABC) \leq \text{Area}(\Delta ABD)$
$\implies \dfrac{1}{2} \sin{x} \leq \text{Area}(ABC) \leq \dfrac{1}{2} \tan{x}.$
The way angles are defined, the “pizza slice” $ABC$ has area $\dfrac{x}{2\pi}$ of a full circle. The area of the full unit circle is $\pi$, so the area of the pizza slice is $\dfrac{1}{2} x$. Therefore,
$\dfrac{1}{2} \sin{x} \leq \dfrac{1}{2} x \leq \dfrac{1}{2} \tan{x} \implies \sin{x} \leq x \leq \tan{x}$.
If we divide both sides by $\sin{x}$, then since $\dfrac{\tan{x}}{\sin{x}} = \dfrac{1}{\cos{x}}$,
$1 \leq \dfrac{x}{\sin{x}} \leq \dfrac{1}{\cos{x}}.$
If we “flip” all these fractions and reverse the inequalities, this means that
$1 \geq \dfrac{\sin{x}}{x} \geq \cos{x}.$
By taking limits on all parts of this equation,
$\lim\limits_{x \to 0} 1 \geq \lim\limits_{x \to 0} \dfrac{\sin{x}}{x} \geq \lim\limits_{x \to 0} \cos{x}.$
Notice now that the third of these limits is just $\cos{0}$, which if we look at the triangle construction earlier is just 1, the length of $AB$. Therefore,
$1 \leq \lim\limits_{x \to 0} \dfrac{\sin{x}}{x} \leq 1.$
This technique is often called the Sandwich Theorem because we stuck the expression we wanted in between two other expressions. Since, of course, the only number between 1 and 1 is 1, the limit we wanted to find must be 1.
Fact: $\lim\limits_{h \to 0} \dfrac{e^h - 1}{h} = 1$.
Proof: This is done by very cleverly using the way the number $e$ is defined. Remember that
$e = \lim\limits_{x \to \infty} \bigg( 1 + \dfrac{1}{x} \bigg)^x.$
The clever move we can make here is to define $h = \dfrac{1}{x}$ and rewrite this same limit in terms of $h$. Notice that if $x \to \infty$, then $h \to 0$. (Technically this is only a limit “from the right”. We would first have to prove that this limit actually exists. This isn’t terribly difficult to do, but it would require a lot of additional writing. So, I won’t do this. If you look at a graph of the function $\dfrac{e^x-1}{x}$, you can convince yourself that this limit does exist.)
By interchanging variables in this way,
$e = \lim\limits_{h \to 0} (1 + h)^{1/h}.$
Using this as a substitution for $e$ inside the limit we want to actually compute,
$\lim\limits_{h \to 0} \dfrac{e^h-1}{h} = \lim\limits_{h \to 0} \dfrac{((1+h)^{1/h})^h - 1}{h} = \lim\limits_{h \to 0} \dfrac{(1+h) - 1}{h} = \lim\limits_{h \to 0} \dfrac{h}{h} = 1.$
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# Mathematics 1010 online
## Working with Polynomials
Polynomials occur ubiquitously in applications. A number of techniques have been developed for working with them and exploiting their special structure. In this class you will learn how to evaluate, factor, and divide polynomials.
### Evaluating a polynomial
Consider a polynomial such as
It is clear, for example, that Suppose you want to know . You could evaluate those powers of in , multiply with the coefficients, and add the individual terms to get
It turns out that this is a clumsy procedure. The evaluation becomes much simpler if we rewrite as
(You should take a moment to convince yourself that and are in fact equivalent.) Evaluating in the form is much easier since we don't have to compute powers. In our special case we obtain:
We get the same answer, of course, but the computation is simpler and it involves fewer basic operations. The advantages of this form are more pronounced for polynomials of high degree. The above technique works in general and is described in the literature as synthetic division (for reasons discussed below), nested multiplication (for reasons that are obvious when considering ), or Horner's Scheme (after the English Algebraist William George Horner, 1786-1837). It's a good rule of thumb that if something is known under several names it is usually powerful or otherwise important.
You can use synthetic division to evaluate a polynomial in your head, or using a calculator (storing just one number, the value of ). The usual way to do it on paper is to construct a table which has the coefficients of the polynomial in the first row, and the intermediate results of the Calculation in the second and third rows:
Each entry in the third row is the sum of the entries above it. In the first column there is a blank in the second row that you can think of as zero. Each entry in the second row is obtained by multiplying the entry to the lower left of it with the number at which we evaluate the polynomial (in this case 7). Usually we take note of that number by writing it to the left of the array in the first row:
The final result of the evaluation (in this case ) is given (and underlined) in the lower right corner of the array.
The process is called synthetic division because it can be thought of as dividing our polynomial by with remainder. The quotient is a polynomial whose coefficients are the other entries in the last row. In this case,
It's a little involved but a good exercise to see that this works in general. Send me an e-mail or talk with me if you are interested in more information.
### Factoring a Polynomial
We write a polynomial in standard form as a sum of monomials. In factored form a polynomial is written as a product of lower degree polynomials. It is completely factored if all factors are of as low a degree as possible.
For example, you can easily check that our polynomial satisfies
The significance of having a polynomial in factored form is that it makes it easier to solve equations of the form . A solution of that equation is called a root or zero of . A product is zero if and only if one of the factors is zero, so to find the roots of we need to look only at the roots of the individual factors. It is clear from that the roots of our particular polynomial are , , and . This is not at all obvious by looking at the original definition of .
Factoring is emphasized in a class like this as a means of solving polynomial equations. It works beautifully if you are able to find the factors. However, the usual flow is often the other way, if you really want factors you find them by first finding the roots of a polynomial.
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# Find the HCF and LCM of 8/9, 10/27 and 16/81.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
To find: HCF and LCM of 8/9, 10/27 and 16/81.
To determine the HCF of fractional numbers, first we have to find the HCF of numerators and then LCM of denominators. So, the HCF of 8/9, 10/27 and 16/81 can be calculated in the following way.
HCF of (a/b, c/d, e/f) = HCF of (a, c, d) / LCM of (b, d, f)
HCF of (8/9, 10/27, 16/81) = HCF of (8, 10, 16) / LCM of (9, 27, 81)
HCF of (8/9, 10/27, 16/81) = 2 / 81
Now, to find the LCM of 8/9, 10/27 and 16/81 we have to find the LCM of numerators and then HCF of denominators.
LCM of (a/b, c/d, e/f) = LCM of (a, c, d) / HCF of (b, d, f)
LCM of (8/9, 10/27, 16/81) = LCM of (8, 10, 16) / HCF of (9, 27, 81)
LCM of (8/9, 10/27, 16/81) = 80 / 9
Hence, HCF of (8/9, 10/27, 16/81) is 2/81 and LCM of (8/9, 10/27, 16/81) is 80/9.
Table of content
## HCF and LCM of 8/9, 10/27 and 16/81
As shared above, the HCF and LCM of 8/9, 10/27 and 16/81 are 2/81 and 80/9 respectively. The HCF and LCM can be calculated by following the step by step process explained above. Further, it is important to have knowledge of HCF and LCM to solve this question.
• HCF is an abbreviation for Highest Common Factor. The highest (or greatest) common factor of two or more given numbers is the HCF. It is also referred to as the Greatest Common Divisor (GCD).
• LCM is an abbreviation for Lowest Common Multiple. The smallest (or least or lowest) of two or more given numbers’ common multiples is the LCM.
Summary:
## Find the HCF and LCM of 8/9, 10/27 and 16/81.
The HCF and LCM of (8/9, 10/27, 16/81) is 2/81 and 80/9 respectively. To find the HCF of the given fractions, calculate the HCF of numerators and then LCM of denominators. Similarly to find the LCM of the given numbers, first we have to determine LCM of numerators and then HCF of denominators.
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# Word Problems - Ratio & Proportion
Related Topics: More lessons for Singapore Math
In these lessons, we will learn how to solve ratio and proportion word problems using models or block diagrams. (Singapore Math).
Word problem on finding out the ratio between 2 items given their quantities explained using models
Example:
300 children participated in a painting contest. 120 of them were boys
a) Find the ratio of the number of boys to the number of girls
b) Find the ratio of the number of girls to the number of boys
c) Find the ratio of the number of boys to the total number of children
d) the ratio of the number of girls to the total number of children.
How to use models to find out the ratio between the heights of two objects and what ratio really means?
Example:
Last year Ben was 4 feet tall and Jenny was 3 feet tall. Both children have grown by 1 foot since last year. Find the ratio of Ben's present height to Jenny's present height.
A ratio word problem to explain how to find out the quantity of one item given the quantity of a second item and their ratio
Example:
Mia has 20 black marbles and 15 red marbles. Alex has 8 black marbles and some red marbles. The ratio of the number of black marbles to the number of red marbles that Mia has is the same as the ratio of the black marbles to the red marbles that Alex has. Find the number of red marbles that Alex has.
A word problem on ratio of intermediate difficulty level explained using the visual method of model drawing
Example:
Mac has 200 white and black horses on his farm.The ratio of the number of black horses to the number of white horses is 2:3. Find the number of white horses on Mac's farm.
A ratio word problem of intermediate difficulty level
Finding out ratio between 2 things given their quantities. Involves unit conversion. Explained using visual drawings called models.
Example:
Rita is 12 years 6 months old. Jet is 4 years 2 months younger than Rita. Find the ratio of Jet's age to Rita's age.
How to solve a word problem solved using 2 different approaches? 1) unit method of model drawing and 2) ratio.
Example:
The following ingredients are used to make strawberry shake for four children. 8 cups milk, 2 cups strawberries, 4 cups ice-cubes.
What are the quantities of each ingredient that you will need to make strawberry shake for 7 children? A word problem to show the relation between ratios and fractions using Singapore's model method of problem solving
Example:
Town A has 2/5 of the population of Town B. What is the ratio of the population of Town A to the population of Town B?
A ratio word problem of intermediate level involving 3 objects and their quantities
Solved using models.
Example:
A piece of fabric is colored green, white and pink in the ratio 3:2:1.
If 45 cm of the fabric is green, what is the length of the fabric that is colored white?
A word problem based on the principles of ratio and measurement (length) Explained and solved using the visual method of model drawing.
Example:
160m of fence is used to border a 50-m long rectangular field.
Find the ratio of the length to the width (breadth) of this rectangular field.
An intermediate level ratio and measurement word problem involving unit conversion and explained using the model method or solving mathematics problems
Example:
Alice has 250 ml of soda and 1 liter of water. What is the ratio of the volume of soda to the volume of water that Alic has?
A word problem to reinforce the basic principle of equivalent ratio
Example:
A fruit seller makes fruit baskets with apples, oranges and mangoes in the ratio 3:5:2.
If he packs a total of 90 mangoes with 6 mangoes in each basket, how many pieces of fruit does he pack in total?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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8.12: Congruent Figures
Difficulty Level: At Grade Created by: CK-12
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Mrs. Gilman brought a small group of students over to look at this tile floor in the hallway of the art museum.
“You see, there is even math in the floor,” she said smiling. Mrs. Gilman is one of those teachers who loves to point out every place where math can be found.
“Okay, I get it,” Jesse started. “I see the squares.”
“There is a lot more math than just squares,” Mrs. Gilman said walking away with a huge smile on her face.
“She frustrates me sometimes,” Kara whispered staring at the floor. “Where is the math besides the squares?”
“I think she is talking about the size of the squares,” Hannah chimed in. “See there are two different sizes.”
“Actually there are three different sizes and there could be more that I haven’t found yet,” Jesse said.
“Remember when we learned about comparing shapes that are alike and aren’t alike, it has to do with proportions or something like that,” Hannah chimed in again.
All three students stopped talking and began looking at the floor again.
“Oh yeah, congruent and similar figures, but which are which?”Kara asked.
What are congruent figures? This Concept will teach you all about congruent figures. When you are all finished with this Concept, you will have a chance to study the floor again and see if you can find the congruent figures.
Guidance
The word congruent means exactly the same. Sometimes, you will see this symbol \begin{align*}\cong\end{align*}.
In this Concept, we are going to use the word congruent to compare figures.
Congruent figures have exactly the same size and shape. They have congruent sides and congruent angles. Here are some pairs of congruent figures.
Compare the figures in each pair. They are exactly the same! If you’re not sure, imagine that you could cut out one figure and place it on top of the other. If they match exactly, they are congruent.
How can we recognize congruence?
We test for congruency by comparing each side and angle of two figures to see if all aspects of both are the same. If the sides are the same length and the angles are equal, the figures are congruent. Each side and angle of one figure corresponds to a side or angle in the other. We call these corresponding parts. For instance, the top point of one triangle corresponds to the top point of the other triangle in a congruent pair.
It is not always easy to see the corresponding parts of two figures. One figure may be rotated differently so that the corresponding parts appear to be in different places. If you’re not sure, trace one figure and place it on top of the other to see if you can make them match. Let’s see if we can recognize some congruent figures.
Which pair of figures below is congruent?
Let’s analyze one pair at a time to see if we can find any corresponding angles and sides that are congruent.
The figures in the first pair appear to be the same shape, if we rotate one 180\begin{align*}180^\circ\end{align*} so they both point up. Now we can see all of the corresponding parts, such as the angle at the top and the two long sides on the left and right. This is not enough to go on, however. We need to compare the measures of the angles and the lengths of the sides. If any one set of corresponding parts doesn’t match, the figures cannot be congruent.
We only know the measure of one angle in the first two figures. We can compare these angles if they are corresponding parts. They are, because if we rotate one figure these angles are in the same place at the top of each figure. Now compare their measures. The angle in the first figure is 45\begin{align*}45^\circ\end{align*}. The corresponding angle in the second figure is 55\begin{align*}55^\circ\end{align*}. Because the angles are different, these two figures are not congruent. Let’s look at the next pair.
The two triangles in the second pair seem to have corresponding parts: a long base and a wide angle at the top. We need to know whether any of these corresponding parts are congruent, however. We know the measure of the top angle in each figure: it is 110\begin{align*}110^\circ\end{align*} in both. These figures might be congruent, but we need to see if the sides are congruent to be sure (as we said, similar figures also have congruent angles, but their sides are different lengths). We know the measure of each triangle’s base: one is 2 inches and the other is 4 inches. These sides are not congruent, so the triangles are not congruent. Remember, every side and every angle must be the same in order for figures to be congruent.
That leaves the last pair. Can you find the corresponding parts? If we rotate the second figure 180\begin{align*}180^\circ\end{align*}, we have two shapes that look like the letter L\begin{align*}L\end{align*}. Now compare the corresponding sides. The bottom side of each is 8 cm, the long left side of each is 8 cm, two sides are 6 cm, and two sides are 2 cm. All of the angles in both figures are 90\begin{align*}90^\circ\end{align*}. Because every side and angle in one figure corresponds to a congruent side and angle in the second, these two figures are congruent.
What about angle measures of congruent figures?
We know that congruent figures have exactly the same angles and sides. That means we can use the information we have about one figure in a pair of congruent figures to find the measure of a corresponding angle or side in the other figure. Let’s see how this works. Take a look at the congruent figures below.
We have been told these two parallelograms are congruent.
Can you find the corresponding parts?
If not, trace one parallelogram and place it on top of the other. Rotate it until the parts correspond.
Which sides and angles correspond?
We can see that side AB\begin{align*}AB\end{align*} corresponds to side PQ\begin{align*}PQ\end{align*}. Because they are congruent, we write.
ABPQ.\begin{align*}AB \cong PQ.\end{align*}
What other sides are congruent? Let’s write them out.
ABBCADDCPQQRPSSR\begin{align*}AB &\cong PQ\\ BC &\cong QR\\ AD &\cong PS\\ DC &\cong SR\end{align*}
We can also write down the corresponding angles, which we know must be congruent because the figures are congruent.
ABPQDSCR\begin{align*}\angle A &\cong \angle P && \angle D \cong \angle S\\ \angle B &\cong \angle Q && \angle C \cong \angle R\end{align*}
Now that we understand all of the corresponding relationships in the two figures, we can use what we know about one figure to find the measure of a side or angle in the second figure.
Can we find the length of side AB\begin{align*}AB\end{align*}?
We do not know the length of AB\begin{align*}AB\end{align*}. However, we do know that it is congruent to PQ\begin{align*}PQ\end{align*}, so if we can find the length of PQ\begin{align*}PQ\end{align*} then it will be the same for AB.PQ\begin{align*}AB. PQ\end{align*} is 7 centimeters. Therefore AB\begin{align*}AB\end{align*} must also be 7 centimeters long.
Now let’s look at the angles. Can we find the measure of C\begin{align*}\angle C\end{align*}?
It corresponds to R\begin{align*}\angle R\end{align*}, but we do not know the measure of R\begin{align*}\angle R\end{align*} either. Well, we do know the measures of two of the angles in the first parallelogram: 70\begin{align*}70^\circ\end{align*} and 110\begin{align*}110^\circ\end{align*}. If we had three, we could subtract from 360\begin{align*}360^\circ\end{align*} to find the fourth, because all quadrilaterals have angles that add up to 360\begin{align*}360^\circ\end{align*}. We do not know the measure of B\begin{align*}\angle B\end{align*}, but this time we do know the measure of its corresponding angle, Q\begin{align*} \angle Q\end{align*}. These two angles are congruent, so we know that B\begin{align*}\angle B\end{align*} must measure 70\begin{align*}70^\circ\end{align*}. Now we know three of the angles in the first figure, so we can subtract to find the measure of C\begin{align*}\angle C\end{align*}.
360(70+110+70)360250110=C=C=C\begin{align*}360 - (70 + 110 + 70) &= \angle C\\ 360 - 250 &= \angle C\\ 110^\circ &= \angle C\end{align*}
We were able to combine the given information from both figures because we knew that they were congruent.
Yes and the more you work on puzzles like this one the easier they will become.
Now it's time for you to try a few on your own.
Example A
True or false. Congruent figures have the same number of sides and angles.
Solution: True
Example B
True or false. Congruent figures can have one pair of angles with the same measure, but not all angles have the same measure.
Solution: False
Example C
True or false. Congruent figures can be different sizes as long as the angle measures are the same.
Solution: False
Now back to the congruent figures at the art museum.
Mrs. Gilman brought a small group of students over to look at this tile floor in the hallway of the art museum.
“You see, there is even math in the floor,” she said smiling. Mrs. Gilman is one of those teachers who loves to point out every place where math can be found.
“Okay, I get it,” Jesse started. “I see the squares.”
“There is a lot more math than just squares,” Mrs. Gilman said walking away with a huge smile on her face.
“She frustrates me sometimes,” Kara whispered staring at the floor. “Where is the math besides the squares?”
“I think she is talking about the size of the squares,” Hannah chimed in. “See there are two different sizes.”
“Actually there are three different sizes and there could be more that I haven’t found yet,” Jesse said.
“Remember when we learned about comparing shapes that are alike and aren’t alike, it has to do with proportions or something like that,” Hannah chimed in again.
All three students stopped talking and began looking at the floor again.
“Oh yeah, congruent and similar figures, but which are which?”Kara asked.
The congruent figures are exactly the same. We can say that the small dark brown squares are congruent because they are just like each other. They have the same side lengths. What is one other pair of congruent squares?
Make a note of the congruent figures that you can find and then share your findings with a friend. Compare answers and continue with the Concept.
Vocabulary
Here are the vocabulary words in this Concept.
Congruent
having exactly the same shape and size. All side lengths and angle measures are the same.
Guided Practice
Here is one for you to try on your own.
What is the measure of M\begin{align*}\angle M\end{align*}?
We can use reasoning to figure out this problem. First, we know that the two triangles are congruent. We also know two of the three angle measures of the triangles.
Let's write an equation.
95+35+x=180\begin{align*}95 + 35 + x = 180\end{align*}
Now we can solve for the missing angle measure.
130+x=180\begin{align*}130 + x = 180\end{align*}
x=50\begin{align*}x = 50\end{align*}
The measure of the missing angle is 50\begin{align*}50^\circ\end{align*}.
Video Review
Here is a video for review.
Practice
Directions: Name the corresponding parts to those given below.
1. R\begin{align*}\angle R\end{align*}
2. MN\begin{align*}MN\end{align*}
3. O\begin{align*}\angle O\end{align*}
Directions: Use the relationships between congruent figures to find the measure of g\begin{align*}g\end{align*}. Show your work.
4.
Directions: Use the relationships between congruent figures to find the measure of T\begin{align*}\angle T\end{align*}. Show your work.
5.
Directions: Answer each of the following questions.
6. Triangles ABC\begin{align*}ABC\end{align*} and DEF\begin{align*}DEF\end{align*} are congruent. If the measure of angle A\begin{align*}A\end{align*} is 58\begin{align*}58^{\circ}\end{align*}, what is the measure of angle D\begin{align*}D\end{align*} if it corresponds to angle A\begin{align*}A\end{align*}?
7. True or false. Congruent figures are exactly the same in every way.
Directions: Identify the given triangles as visually congruent or not.
8.
9.
10.
11.
12.
Directions: Answer each of the following questions.
13. Triangles ABC\begin{align*}ABC\end{align*} and DEF\begin{align*}DEF\end{align*} are congruent. Does this mean that their angle measures are the same? Why?
14.Define Congruent.
15. True or false. If two figures are congruent, then they have the same length sides but not the same angle measures.
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# Internal angles
Find the internal angles of the triangle ABC if the angle at the vertex C is twice the angle at the B and the angle at the vertex B is 4 degrees smaller than the angle at the vertex A.
Result
A = 48
B = 44
C = 88
#### Solution:
A+B+C=180
C = 2B
B = A-4
A+B+C = 180
2B-C = 0
A-B = 4
A = 48
B = 44
C = 88
Calculated by our linear equations calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Do you have a system of equations and looking for calculator system of linear equations? See also our trigonometric triangle calculator.
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# Revolutions to Radians – Formulas and Examples
Radians are a way of measuring angles. Radians are mainly used when we want to perform advanced mathematical operations such as differential or integral calculus. This is because the radian has a relationship to the radius of the circle. On the other hand, revolutions are a way of considering a complete turn around a circle. This means that one revolution is equal to 2π. Therefore, to convert revolutions to radians, we multiply the number of revolutions by 2π.
Here, we will apply the process of transforming revolutions to radians by solving some practice problems.
##### TRIGONOMETRY
Relevant for
Learning to transform from revolutions to radians with examples.
See examples
##### TRIGONOMETRY
Relevant for
Learning to transform from revolutions to radians with examples.
See examples
## How to convert from revolutions to radians?
To convert from revolutions to radians, we have to multiply the number of revolutions by 2π and we will get the angle in radians that corresponds to the given number of revolutions. Therefore, we have the following formula:
where x represents the number of revolutions and y is the answer in radians.
This formula is derived considering a circle. If we go around a full circle, we have an angle of 2π radians. Also, by definition, one revolution equals one complete turn of the circle. This means that we can form the relation 1 rev = 2π.
Therefore, having any number of revolutions, we simply have to multiply by 2π to find the equivalent radians.
The following practice examples are solved using the formula for the transformation of revolutions to radians given above. Try solving the problems yourself before looking at the answer.
### EXAMPLE 1
If we have 3 revolutions, how many radians do we have?
We substitute the value given in the transformation formula, to obtain:
$latex (x\text{ rev})\times 2\pi=y \text{ rad}$
$latex (3\text{ rev})\times 2\pi=6 \pi\text{ rad}$
Therefore, 3 revolutions equal 6π radians.
### EXAMPLE 2
Using $latex x = 6$ which is the number of revolutions, we have:
$latex (x\text{ rev})\times 2\pi=y \text{ rad}$
$latex (6\text{ rev})\times 2\pi=12 \pi\text{ rad}$
Therefore, 6 revolutions equal 12π radians.
### EXAMPLE 3
How many radians is equal to 2.5 revolutions?
In this case, we have a fractional number, but the formula to use is the same. We use the value in the formula and we have:
$latex (x\text{ rev})\times 2\pi=y \text{ rad}$
$latex (2.5\text{ rev})\times 2\pi=5 \pi\text{ rad}$
Therefore, 2.5 revolutions is equal to 5π radians.
### EXAMPLE 4
If we have 3.8 revolutions, how many radians do we have?
We use the value $latex x = 3.8$ in the transformation formula and solve:
$latex (x\text{ rev})\times 2\pi=y \text{ rad}$
$latex (3.8\text{ rev})\times 2\pi=7.6 \pi\text{ rad}$
### EXAMPLE 5
How many radians is equal to 6.5 revolutions?
We can use the formula with the value $latex x = 6.5$ to obtain:
$latex (x\text{ rev})\times 2\pi=y \text{ rad}$
$latex (6.5\text{ rev})\times 2\pi=13 \pi\text{ rad}$
Therefore, 6.5 revolutions equal 13π radians.
## Revolutions to radians – Practice problems
Practice using the revolutions to radians transformation formula by solving the following problems. Select an answer and check it to see if you got the correct answer.
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# How do you differentiate xe^(x^2+y^2)?
Mar 31, 2015
To differentiate a two-variable function, you need to build the gradient. The gradient is a vector with as many coordinates as the variables the function depends on.
Each coordinate of the vector is a derivative with respect to one of the variables. So, in the two-variables case, you need to calculate the derivatives with respect to $x$ and $y$, and then put them together in a vector.
Since deriving with respect to a variable means to consider the other as a constant, it's easier to derivate your function if it's expressed in the form
$x {e}^{{x}^{2} + {y}^{2}} = x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}}$
So, deriving with respect to $x$, and using the product rule $\left(f g\right) ' = f ' g + f g '$ where $f \left(x\right) = x$ and $g \left(x\right) = {e}^{{x}^{2}}$, we get
$\frac{d}{\mathrm{dx}} x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}} = {e}^{{y}^{2}} \left({e}^{{x}^{2}} + x \cdot {e}^{{x}^{2}} \cdot 2 x\right) =$
${e}^{{x}^{2}} \cdot {e}^{{y}^{2}} \left(1 + 2 {x}^{2}\right) = {e}^{{x}^{2} + {y}^{2}} \left(1 + 2 {x}^{2}\right)$
Where the derivative of ${e}^{{x}^{2}}$ has been calculated using the chain rule, which states that $f \left(g \left(x\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$, where $f \left(x\right) = {e}^{x}$, and $g \left(x\right) = {x}^{2}$.
The derivative with respect to $y$ is easier, since the only factor to differentiate is ${e}^{{y}^{2}}$, while the others depend only on $x$ and are thus to be considered as constant. So, we have
$\frac{d}{\mathrm{dy}} x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}} = x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}} \cdot 2 y =$
$2 x y {e}^{{x}^{2} + {y}^{2}}$
The gradient is thus the vector
$\left({e}^{{x}^{2} + {y}^{2}} \left(1 + 2 {x}^{2}\right) , 2 x y \setminus {e}^{{x}^{2} + {y}^{2}}\right)$
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# Question Video: Comparing Expressions Involving Multiplication and Division of Decimal Numbers Mathematics
Calculate 947 ÷ 0.89 ＿ (947 ÷ 8.9) × 0.01 using <, =, or >.
02:47
### Video Transcript
Complete nine hundred and forty-seven divided by nought point eight nine is something to nine hundred and forty-seven divided by eight point nine times nought point nought one using the less than sign, the equal to sign, or the greater than sign.
So we’ve got two different expressions and we’ve got to decide which one is bigger than the other. Now if I can rearrange the expression on the right-hand side so that it contains exactly the same expression on the left-hand side but multiplied or divided by something, that will help me to decide which is the bigger expression.
Well in fact to make my life easier, before I start I’m going to reexpress the left-hand side: nine hundred and forty-seven divided by nought point eight nine can be rewritten as nine hundred and forty-seven over nought point eight nine.
So now let’s do a similar thing to the right-hand side: nine hundred and forty-seven divided by eight point nine can be rewritten as nine hundred and forty-seven over eight point nine, and of course that’s got to be multiplied by nought point nought one.
Now nought point nought one has got a one in the hundredths column, so that means one hundredth. So I’m going to write this as, instead of nought point nought one, I’m gonna write it as one hundredth, one over a hundred.
Now these two terms here look quite a lot like each other. We’ve got nine hundred and forty-seven over zero point eight nine, and we’ve got nine hundred and forty-seven over eight point nine. Now I’m gonna reexpress eight point nine as nought point eight nine times ten; if I multiply nought point eight nine by ten, I get eight point nine.
And because I’m multiplying two fractions together on this right-hand side over here, it doesn’t matter whether I do the multiplying by ten over here or if I move it to over here, and that’s what I’m gonna do.
And now I’ve got nine hundred and forty-seven over nought point eight nine here, and I’ve got nine hundred and forty-seven over nought point eight nine here. But in this case, I’m multiplying it by one over a hundred times ten, or one over a thousand.
So we’ve got the same expression or the same term here and here, but on the right-hand side, we’re gonna take that expression and multiply it by a thousandth or divide by a thousand; we’re gonna make it smaller. So that’s gonna be the smaller number, and this one is gonna be the bigger number.
So we can write our sign this way round: nine hundred and forty-seven divided by nought point eight nine is greater than nine hundred and forty-seven divided by eight point nine times nought point nought one. So there we have our answer.
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# Change of Variables in Multiple Integrals
## Introduction
Substitution (or change of variables) is a powerful technique for evaluating integrals in single variable calculus. An equivalent transformation is available for dealing with multiple integrals. The idea is to replace the original variables of integration by the new set of variables. This way the integrand is changed as well as the bounds for integration. If we are lucky enough to find a convenient change of variables we can significantly simplify the integrand or the bounds.
## Change of variables formula
Two dimensional pictures are the easiest to draw so we will start with functions of two variables. Our first task is to get familiar with transformations of two dimensional regions.
### Transformations in $$\mathbb R^2$$
Assume that $$S$$ is a region in $$\mathbb R^2$$. We want to study the ways in which this region can be transformed to another region $$T$$.
This is easiest to explain by considering an example. Let $$S=[0,2]\times[0,2]$$. Consider the functions $$u:S\to \mathbb R$$ and $$v:S\to\mathbb R$$ defined in the following way: \begin{eqnarray*} u(x,y)&=&x+2y\\ v(x,y)&=&x-y.\end{eqnarray*} To every point $$(x,y)\in S$$ ($$S$$ is painted blue in the diagram on the left) we can assign a new green point with coordinates $$(u(x,y),v(x,y))$$. This way we obtain a green region $$T$$.
The mapping $$(x,y)\mapsto (u(x,y),v(x,y))$$ is one-to-one and onto, hence a bijection (you may want to review these terms in the section Functions). We can also write the inverse transformation, that maps each point $$(u,v)\in T$$ to the point $$(x(u,v),y(u,v))$$ in the following way: \begin{eqnarray*} x(u,v)&=&\frac{u+2v}3 \\ y(u,v)&=&\frac{u-v}3. \end{eqnarray*}
### Change of variables in double integrals
Assume that $$S\subseteq \mathbb R^2$$ is a region in the plane. Let $$T\subseteq\mathbb R^2$$ be another region and assume that there are continuously differentiable functions $$X:T\to\mathbb R$$ and $$Y:T\to\mathbb R$$, such that the mapping $$\Phi(u,v)= (X(u,v),Y(u,v))$$ is a bijection between $$T$$ and $$S$$.
The Jacobian of the mapping $$\Phi$$ is defined as $\frac{\partial(X,Y)}{\partial(u,v)}=\det\left|\begin{array}{cc} \frac{\partial X}{\partial u}& \frac{\partial X}{\partial v}\\ \frac{\partial Y}{\partial u}& \frac{\partial Y}{\partial v}\end{array}\right|.$
Theorem (Change of variables in double integrals)
Assume that $$S$$ and $$T$$ are domains in $$\mathbb R^2$$ and that there are two continuously differentiable functions $$X,Y:T\to \mathbb R$$ such that $$\Phi: T\to S$$ defined by $$\Phi(u,v)=(X(u,v),Y(u,v))$$ is a bijection whose Jacobian is never $$0$$. For each continuous bounded $$f:S\to\mathbb R$$ the following equality holds: $\iint_S f(x,y)\,dxdy=\iint_T f(X(u,v),Y(u,v)) \cdot \left|\frac{\partial(X,Y)}{\partial(u,v)}\right|\,dudv.$
Example 1.
Using the substitution $$u=2x+3y$$, $$v=x-3y$$, find the value of the integral $\iint_D e^{2x+3y}\cdot \cos(x-3y)\,dxdy,$ where $$D$$ is the region bounded by the parallelogram with vertices $$(0,0)$$, $$\left(1,\frac13\right)$$, $$\left(\frac43,\frac19\right)$$, and $$\left(\frac13,-\frac29\right)$$.
### Change of variables in triple integrals
Assume that $$S, T\subseteq \mathbb R^3$$ are two regions in space. Assume that there are continuously differentiable functions $$X:T\to\mathbb R$$, $$Y:T\to\mathbb R$$, and $$Z:T\to\mathbb R$$, such that the mapping $$\Phi:T\to S$$ defined as $$\Phi(u,v,w)= (X(u,v,w),Y(u,v,w),Z(u,v,w))$$ is a bijection.
The Jacobian of the mapping $$\Phi$$ is defined as $\frac{\partial(X,Y,Z)}{\partial(u,v,w)}=\det\left|\begin{array}{ccc} \frac{\partial X}{\partial u}& \frac{\partial X}{\partial v}& \frac{\partial X}{\partial w}\\ \frac{\partial Y}{\partial u}& \frac{\partial Y}{\partial v}& \frac{\partial Y}{\partial w} \\ \frac{\partial Z}{\partial u}& \frac{\partial Z}{\partial v}& \frac{\partial Z}{\partial w}\end{array}\right|.$
Theorem (Change of variables in triple integrals)
Assume that $$S$$ and $$T$$ are domains in $$\mathbb R^2$$ and that there are three continuously differentiable functions $$X,Y,Z:T\to \mathbb R$$ such that $$\Phi: T\to S$$ defined by $$\Phi(u,v,w)=(X(u,v,w),Y(u,v,w),Z(u,v,w))$$ is a bijection whose Jacobian is never $$0$$. For each continuous bounded function $$f:S\to\mathbb R$$ the following equality holds: $\iiint_S f(x,y,z)\,dxdydz=\iiint_T f(X(u,v,w),Y(u,v,w),Z(u,v,w)) \cdot \left|\frac{\partial(X,Y,Z)}{\partial(u,v,w)}\right|\,dudvdw.$
## Polar, cylindrical, and spherical substitutions
We will now study very important substitutions that are used to simplify integrations over circular, spherical, cylindrical, and elliptical domains. One of them is applicable to double integral and is called polar change of variables and the other two, cylindrical and spherical, are used in triple integralds.
### Polar substitution
The following change of variables is called the polar substitution: \begin{eqnarray*} x&=&r\cos\theta\\ y&=&r\sin \theta. \end{eqnarray*} The Jacobian for the polar substitution is equal to: $\frac{\partial(x,y)}{\partial(r,\theta)}=\det\left|\begin{array}{cc} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta\end{array}\right|=r\cos^2\theta+r\sin^2\theta=r.$
The variables $$r$$ and $$\theta$$ have the geometric meaning in the $$xy$$-coordinate system. The distance between $$(x,y)$$ and the origin is precisely $$r$$, while $$\theta$$ is the angles between the $$x$$-axis and the line connecting $$(x,y)$$ with $$(0,0)$$.
Example 2.
Evaluate the integral $\iint_D \cos\left(x^2+y^2\right)\,dxdy,$ where $$D$$ is the disc of radius $$3$$ centered at the origin.
When dealing with ellipses it is very common to use the modified polar substitution. If the equation of the ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, the following substitution is used to describe its interior: \begin{eqnarray*} x&=&ar\cos\theta\\ y&=&br\sin\theta\\ 0\leq&r&\leq 1\\ 0\leq&\theta&\leq 2\pi. \end{eqnarray*}
Example 3.
Let $$a$$ and $$b$$ be two positive real numbers. Find the area of the region enclosed by the ellipse with the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$.
### Cylindrical substitution
In cylindrical substitution the original variables $$(x,y,z)$$ are replaced by $$(r,\theta, z)$$ using the following equations: \begin{eqnarray*} x&=&r\cos\theta\\ y&=&r\sin\theta\\ z&=&z. \end{eqnarray*} Again we can find the Jacobian by calculating the appropriate determinant: $\frac{\partial(x,y,z)}{\partial(r,\theta,z)}=\det\left|\begin{array}{ccc} \cos\theta &-r\sin\theta &0\\ \sin\theta&r\cos\theta&0\\ 0&0&1\end{array}\right|=r .$
Example 4.
Determine the value of the integral $\iiint_D e^{x^2+y^2}\,dV$ where $$D$$ is the the region in bounded by the planes $$y=0$$, $$z=0$$, $$y=x$$, and the paraboloid $$z=4-x^2-y^2$$.
### Spherical substitution
Spherical substitution means replacing the original variables $$(x,y,z)$$ by the variables $$(\rho,\theta, \phi)$$, where $$\rho$$ is the distance of the points $$(x,y,z)$$ from the origin $$(0,0,0)$$; $$\theta$$ is the angle that the line connecting $$(0,0,0)$$ and $$(x,y,0)$$ forms with the $$x$$-axis, and $$\phi$$ is the angle between the $$z$$-axis and the line connecting $$(x,y,z)$$ with $$(0,0,0)$$. Mathematically, the equations are: \begin{eqnarray*} x&=&\rho\cos\theta\sin\phi\\ y&=&\rho\sin\theta\sin\phi\\ z&=&\rho\cos\phi. \end{eqnarray*} We can find the Jacobian by calculating the appropriate determinant: \begin{eqnarray*} \frac{\partial(x,y,z)}{\partial(\rho,\theta,\phi)}&=&\det\left|\begin{array}{ccc} \cos\theta\sin\phi &-\rho\sin\theta\sin\phi &\rho\cos\theta\cos\phi\\ \sin\theta\sin\phi&\rho\cos\theta\sin\phi&\rho\sin\theta\cos\phi\\ \cos\phi&0&-\rho\sin\phi\end{array}\right| \\ &=&-\rho^2\cos^2\theta\sin^3\phi-\rho^2\sin^2\theta\sin\phi\cos^2\phi-\rho^2\cos^2\theta\sin\phi\cos^2\phi-\rho^2\sin^2\theta\sin^3\phi \\&=&-\rho^2\sin^3\phi-\rho^2\sin\phi\cos^2\phi=-\rho^2\sin\phi .\end{eqnarray*} Since in evaluation of the integral we are using the absolute value of the Jacobian, and $$\phi\in\left(0,\frac{\pi}2\right)$$ it is sufficient and more convenient to remember that $\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\phi)}\right|=\rho^2\sin\phi.$
Example 5.
Determine the value of the integral $\iiint_D e^{\sqrt{x^2+y^2+z^2}}\,dV$ where $$D$$ is the the region in bounded by the planes $$y=0$$, $$z=0$$, $$y=x$$, and the sphere $$x^2+y^2+z^2=9$$.
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# What is the Common Difference of an A.P. in Which A21 – A7 = 84? - CBSE Class 10 - Mathematics
#### Question
What is the common difference of an A.P. in which a21 – a7 = 84?
#### Solution
Let a be the first term and d be the common difference of AP.
We know that
an = a +(n − 1)d
∴ a21 = a +(21 − 1)d = a + 20d
and a7=a + (7 − 1)d = a + 6d
Given:
a21 − a7 = 84
∴ (a + 20d) − (a + 6d) = 84
⇒ a + 20d − a − 6d = 84
⇒14d = 84
⇒ d = 6
Thus, the common difference of the AP is 6
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [6]
Solution What is the Common Difference of an A.P. in Which A21 – A7 = 84? Concept: Arithmetic Progression.
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## Problem 2 Solution
Round the following numbers as specified
a.
1000.34532 rounded to 6 digits
b.
12314.643 rounded to 3 digits
c.
0.00006574 rounded to 6 decimal places
d.
10.0029245 rounded to 2 decimal places
```(a) 1000.35 (b) 12300 or 12.3 thousand (c) 0.000066 (d) 10.00
```
The key ideas: remember the rule, and the definition of "digit" or "decimal place"
if the digit is 0, 1, 2, 3 or 4 round down
if the digit is 5, 6, 7, 8 or 9 round up
### Solution Steps for Part (a):
Round 1000.34532 to 6 digits (not decimal places)
The 6th digit is the 4 in the hundredth's place, and the next digit is 5 so we round up, and the answer is 1000.35
### Solution Steps for Part (b):
Round 12314.643 to 3 digits (not decimal places)
The 3rd digit is the 3 in the hundreds place, and the next digit is 1 so we round down, and the answer is 12300 or 12.3 thousand
This may be considered by some to be a trick question, so we'll present two different answers and let you decide which is more appropriate. If we are only allowed to use 3 digits in the answer, then we need to use some other way of expressing the answer: 12.3 thousand has only 3 digits. In the next section you will learn about significant digits for which some scientists use "digits" as shorthand notation. They would likely tell you that 12300 would also be an acceptable answer, since the last two digits (zeros) are just "place holders." If you ever have any doubt as to what the problem calls for, be sure you ask your instructor.
### Solution Steps for Part (c):
Round 0.00006574 to 6 decimal places (not digits)
The 6th decimal place is the 5 in the millionth's place, and the next digit is 7 so we round up, and the answer is 0.000066
### Solution Steps for Part (d):
Round 10.0029245 to 2 decimal places (not digits)
The 2nd decimal place is the 0 in the hundredth's place, and the next digit is 2 so we round down, and the answer is 10.00
Try another problem like this one.
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Transcript
```Answer Key
Name: ____________________________________
Date: __________________
RATIONAL EXPONENTS
COMMON CORE ALGEBRA II
When you first learned about exponents, they were always positive integers, and just represented repeated
multiplication. And then we had to go and introduce negative exponents, which really just represent repeated
division. Today we will introduce rational (or fractional) exponents and extend your exponential knowledge
that much further.
Exercise #1: Recall the Product Property of Exponents and use it to rewrite each of the following as a
simplified exponential expression. There is no need to find a final numerical value.
(a) 23
(b) 52
4
(c) 37
5
0
4
2
2 2
(d)
2 3 4
5 2 5
37 0
422 2
212
5 10
30
4 8
We will now use the Product Property to extend our understanding of exponents to include unit fraction
exponents (those of the form 1 where n is a positive integer).
n
1
Exercise #2: Consider the expression 16 2 .
1
(a) Apply the Product Property to simplify
(b) You can now say that 16 2 is equivalent to
what more familiar quantity?
16 . What other number squared yields
1
2
2
16?
16
1 2
2
161 16
16
1
16 4
2
We can also square 4 to get 16 (and -4).
This is remarkable! An exponent of 1 is equivalent to a square root of a number!!!
2
Exercise #3: Test the equivalence of the 1
exponent to the square root by using your calculator to evaluate
2
each of the following. Be careful in how you enter each expression.
(a) 25
1
2
1
(b) 81 2
5
(c) 100
9
1
2
10
We can extend this now to all levels of roots, that is square roots, cubic roots, fourth roots, etcetera.
UNIT FRACTION EXPONENTS
For n given as a positive integer:
b
1
n
nb
COMMON CORE ALGEBRA II, UNIT #4 – EXPONENTIAL AND LOGARITHMIC FUNCTIONS – LESSON #2
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2015
Exercise #4: Rewrite each of the following using roots instead of fractional exponents. Then, if necessary,
evaluate using your calculator to guess and check to find the roots (don't use the generic root function). Check
(a) 125
1
(b) 16
3
1
(c) 9
4
3
125
16
3 53
4 24
5
2
1
4
(d) 32
2
1
9 2
1
1
32 5
1
9
1
3
5
1
1
1
5
5
32
1
2
5
1
2
We can now combine traditional integer powers with unit fractions in order interpret any exponent that is a
rational number, i.e. the ratio of two integers. The next exercise will illustrate the thinking. Remember, we
want our exponent properties to be consistent with the structure of the expression.
3
Exercise #5: Let's think about the expression 4 2 .
(b) Fill in the missing blank and then evaluate this
expression:
(a) Fill in the missing blank and then evaluate this
expression:
4
3
2
1
2
4
3
2
43
64
1
1
4
2
3
2
3
4
3
2
2
4
64 8
1
3
2
4
3
23 8
2
(d) Evaluate 27 3 without your calculator. Show
(c) Verify both (a) and (b) using your calculator.
Done on calculator.
27
1
3
2
3
27
2
32 9
RATIONAL EXPONENT CONNECTION TO ROOTS
For the rational number m
n
we define b
m
n
to be:
n
b m or
b
n
m
.
Exercise #6: Evaluate each of the following exponential expressions involving rational exponents without the
(a) 16
3
(b) 25
4
16
1
4
23 8
3
4
16
3
3
(c) 8
2
25
1
2
3
53 125
25
2
3
3
1
8
2
3
8
1 1
22 4
COMMON CORE ALGEBRA II, UNIT #4 – EXPONENTIAL AND LOGARITHMIC FUNCTIONS – LESSON #2
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2015
1
1
3
2
1
8
3
2
Name: ____________________________________
Date: __________________
RATIONAL EXPONENTS
COMMON CORE ALGEBRA II HOMEWORK
FLUENCY
1. Rewrite the following as equivalent roots and then evaluate as many as possible without your calculator.
(a) 36
1
(b) 27
2
36
6
(e) 625
1
1
(c) 32
3
3 27
3
(f) 49
4
1
(d) 100
5
5 32
2
1
(g) 81
2
49
7
1
1
100
1
1
2
100
(h) 343
4
2
1
1
10
3
1
4 625
5
1
1
81 4
1
4
3 343
7
81
1
3
2. Evaluate each of the following by considering the root and power indicated by the exponent. Do as many as
(a) 8
2
3
8
1
(b) 4
2
3
8
3
3
2
4
2
22 4
(e) 4
5
(f) 128
4
5
2
1
4
1
2
5
(c) 16
3
2
4
3
3
4
1
1
25 32
7
128
3
3
23 8
16
3
(h) 243
4
625
625
3
4
81
4
1
4
1
5
4
81
5
35 243
4
3
4
1
4
1 1
23 8
7
1
16
3
(g) 625
128
5
1
7
1
5
(d) 81
4
23 8
2
1
1
3
3
5 3 125
3
5
5
243
243
5
1
3
3
33 27
3
3. Given the function f x 5 x 4 2 , which of the following represents its y-intercept?
(1) 40
(2) 20
(3) 4
(4) 30
f 0 5 0 4
5 4
3
2
5
3
2
4
3
5 8 40
COMMON CORE ALGEBRA II, UNIT #4 – EXPONENTIAL AND LOGARITHMIC FUNCTIONS – LESSON #2
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2015
(1)
4. Which of the following is equivalent to x
1
(1) x
2
(3)
(2) x
(4)
1
2
?
1
x
x
3x
2
1
(2)
3
x
2
2
1
x
1
1
2
x
(3)
1
2x
5. Written without fractional or negative exponents, x
(1)
1
3
2
is equal to
1
(3)
x3
x
1
x
(4)
3
2
1
1
3/ 2
x
x3
1
2
1
x3
(3)
3
6. Which of the following is not equivalent to 16 2 ?
4096
(1)
(3) 64
16
(2) 8
3
3
2
163
1
2
163 4096 64
3
(2)
(4) 16
REASONING
7. Marlene claims that the square root of a cube root is a sixth root? Is she correct? To start, try rewriting the
expression below in terms of fractional exponents. Then apply the Product Property of Exponents.
3
a
3
a a
1
1
3
2
1 1
1
a3 2 a6 6 a
Marlene is correct. The square root of a cube root is a sixth root.
8. We should know that 3 8 2 . To see how this is equivalent to 8
do this, we can rewrite the equation as:
2
3 n
2
3
n
21
23 n 21
3
2 we can solve the equation 8n 2 . To
21
How can we now use this equation to see that 8
8n 2
1
1
3
2?
Because the bases are equal we
can now set the exponents equal
3n 1
1
n
3
COMMON CORE ALGEBRA II, UNIT #4 – EXPONENTIAL AND LOGARITHMIC FUNCTIONS – LESSON #2
eMATHINSTRUCTION, RED HOOK, NY 12571, © 2015
```
Related documents
|
6
Multiplication What happens if you hang more than one weight on the same number? Try balancing these examples. (1) 3×2=____
(2) 4×3=____
4
×3
×2
Division
9 < =>
Look at the following example balances. How many weights are needed in the “x?” space to balance the arms? (1) 4÷2=____ Find the correct number of weights to put on the number 2 spot on the right hand side.
×?
(2) 20÷5=____ What is the correct number of weights to put on the number 5 spot on the right hand side of the balance.
×2
×?
#1026 21 PCS
4+
8> LEARN THROUGH PLAY
3
Product Description
Sure, we all know 2+2=4, but how did you learn it? Was it guess work? Do you still have to use your fingers and toes?
Add the following weights to the left and right arm as shown, then try to balance it! (1) 5+3=____
(2) 4+5=_____
(3) 2+____=9
(4) ____+4=7
The Number Equalizer Balance is a great tool for educators to share the physical intuition of numbers and equalities with their students. The balance can be used to solve simple math questions, and can also be used for students beginning algebra. The Number Equalizer Balance is a simple T-shape arm balance with a set of weights. Simply hang weights on a number on one side, and balance it by placing numbers on the other side.
Tips and Tricks Assembly Steps
Displaying Number Relationships: equal to (=), greater than (>), less than (<). Count the fruits below and hang the same number on the left and right arms of the balance. Which side does the balance lean toward? Compare the numbers and discuss with your classmates.
Subtraction
Set up your balance as shown in the examples below. Try taking one or two weights off to balance the arms. (1) 6 _ ____=2
(2) 10 _ ____=7
(3) 19 _ ___=15
(4) 20 _ ____=6
Example: (1) Hang a weight on the left arm of the balance at the number 2 position; now hang a weight under the right arm of the balance at the number 3 position. Which is bigger, 2 or 3? (1) 2<3
(2) 6>4
< (3) 5=5
> (3) 4____7
=
__
|
# Chapter 8
## 8.1 Introduction
In our daily life, there are many occasions when we compare two quantities. Suppose we are comparing heights of Heena and Amir. We find that
1. Heena is two times taller than Amir.
Or
2. Amir’s height is of Heena’s height.
Consider another example, where 20 marbles are divided between Rita and Amit such that Rita has 12 marbles and Amit has 8 marbles. We say,
1. Rita has times the marbles that Amit has. Or
2. Amit has part of what Rita has.
Heena Amir
150 cm 75cm
Yet another example is where we compare speeds of a Cheetah and a man.
The speed of a Cheetah is 6 times the speed of a Man.
Or
The speed of a Man is of the speed of the Cheetah.
Speed of Cheetah 20 km per hour
Speed of Man 120 km per hour
Do you remember comparisons like this? In Class VI, we have learnt to make comparisons by saying how many times one quantity is of the other. Here, we see that it can also be inverted and written as what part one quantity is of the other.
In the given cases, we write the ratio of the heights as :
Heena’s height : Amir’s height is 150 : 75 or 2 : 1.
Can you now write the ratios for the other comparisons?
These are relative comparisons and could be same for two different situations.
If Heena’s height was 150 cm and Amir’s was 100 cm, then the ratio of their heights would be,
Heena’s height : Amir’s height = 150 : 100 = or 3 : 2.
This is same as the ratio for Rita’s to Amit’s share of marbles.
Thus, we see that the ratio for two different comparisons may be the same. Remember that to compare two quantities, the units must be the same.
A ratio has no units.
Example 1 Find the ratio of 3 km to 300 m.
Solution First convert both the distances to the same unit.
So, 3 km = 3 × 1000 m = 3000 m.
Thus, the required ratio, 3 km : 300 m is 3000 : 300 = 10 : 1.
## 8.2 Equivalent Ratios
Different ratios can also be compared with each other to know whether they are equivalent or not. To do this, we need to write the ratios in the form of fractions and then compare them by converting them to like fractions. If these like fractions are equal, we say the given ratios are equivalent.
Example 2 Are the ratios 1:2 and 2:3 equivalent?
Solution To check this, we need to know whether .
We have, ;
We find that , which means that .
Therefore, the ratio 1:2 is not equivalent to the ratio 2:3.
Use of such comparisons can be seen by the following example.
Example 3 Following is the performance of a cricket team in the matches it played:
Year Wins Losses
Last year 8
This year 4
In which year was the record better?
How can you say so?
Solution Last year, Wins: Losses = 8 : 2 = 4 : 1
This year, Wins: Losses = 4 : 2 = 2 : 1
Obviously, 4 : 1 > 2 : 1 (In fractional form, )
Hence, we can say that the team performed better last year.
In Class VI, we have also seen the importance of equivalent ratios. The ratios which are equivalent are said to be in proportion. Let us recall the use of proportions.
## Keeping things in proportion and getting solutions
Aruna made a sketch of the building she lives in and drew sketch of her mother standing beside the building.
Mona said, “There seems to be something wrong with the drawing”
Can you say what is wrong? How can you say this?
In this case, the ratio of heights in the drawing should be the same as the ratio of actual heights. That is
= .
Only then would these be in proportion. Often when proportions are maintained, the drawing seems pleasing to the eye.
Another example where proportions are used is in the making of national flags.
Do you know that the flags are always made in a fixed ratio of length to its breadth? These may be different for different countries but are mostly around 1.5 : 1 or 1.7 : 1.
We can take an approximate value of this ratio as 3 : 2. Even the Indian post card is around the same ratio.
Now, can you say whether a card with length 4.5 cm and breadth 3.0 cm is near to this ratio. That is we need to ask, is 4.5 : 3.0 equivalent to 3 : 2?
We note that
Hence, we see that 4.5 : 3.0 is equivalent to 3 : 2.
We see a wide use of such proportions in real life. Can you think of some more situations?
We have also learnt a method in the earlier classes known as Unitary Method in which we first find the value of one unit and then the value of the required number of units.
Let us see how both the above methods help us to achieve the same thing.
Example 4 A map is given with a scale of 2 cm = 1000 km. What is the actual distance between the two places in kms, if the distance in the map is 2.5 cm?
## Solution
Arun has solved it by equating ratios to make proportions and then by solving the equation. Meera has first found the distance that corresponds to 1 cm and then used that to find what 2.5 cm would correspond to. She used the unitary method.
Let us solve some more examples using the unitary method.
Example 5 6 bowls cost 90. What would be the cost of 10 such bowls?
Solution Cost of 6 bowls is 90.
Therefore, cost of 1 bowl =
Hence, cost of 10 bowls = × 10 = 150
Example 6 The car that I own can go 150 km with 25 litres of petrol. How far can it go with 30 litres of petrol?
Solution With 25 litres of petrol, the car goes 150 km.
With 1 litre the car will go km.
Hence, with 30 litres of petrol it would go km = 180 km
In this method, we first found the value for one unit or the unit rate. This is done by the comparison of two different properties. For example, when you compare total cost to number of items, we get cost per item or if you take distance travelled to time taken, we get distance per unit time.
Thus, you can see that we often use per to mean for each.
For example, km per hour, children per teacher etc., denote unit rates.
# Think, Discuss and Write
An ant can carry 50 times its weight. If a person can do the same, how much would you be able to carry?
# Exercise 8.1
1. Find the ratio of:
(a) ` 5 to 50 paise (b) 15 kg to 210 g
(c) 9 m to 27 cm (d) 30 days to 36 hours
2. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
3. Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs.
Area of Rajasthan = 3 lakh km2 and area of UP = 2 lakh km2.
(i) How many people are there per km2 in both these States?
(ii) Which State is less populated?
## 8.3 Percentage – Another Way of Comparing Quantities
Anita’s Report Rita’s Report
Total 320/400 Total 300/360
Percentage: 80 Percentage: 83.3
Anita said that she has done better as she got 320 marks whereas Rita got only 300. Do you agree with her? Who do you think has done better?
Mansi told them that they cannot decide who has done better by just comparing the total marks obtained because the maximum marks out of which they got the marks are not the same.
She said why don’t you see the Percentages given in your report cards?
Anita’s Percentage was 80 and Rita’s was 83.3. So, this shows Rita has done better.
Do you agree?
Percentages are numerators of fractions with denominator 100 and have been used in comparing results. Let us try to understand in detail about it.
## 8.3.1 Meaning of Percentage
Per cent is derived from Latin word ‘per centum’ meaning ‘per hundred’.
Per cent is represented by the symbol % and means hundredths too. That is 1% means 1 out of hundred or one hundredth. It can be written as: 1% = = 0.01
To understand this, let us consider the following example.
Rina made a table top of 100 different coloured tiles. She counted yellow, green, red and blue tiles separately and filled the table below. Can you help her complete the table?
# Try These
1. Find the Percentage of children of different heights for the following data.
2. A shop has the following number of shoe pairs of different sizes.
Size 2 : 20 Size 3 : 30 Size 4 : 28
Size 5 : 14 Size 6 : 8
Write this information in tabular form as done earlier and find the Percentage of each shoe size available in the shop.
## Percentages when total is not hundred
In all these examples, the total number of items add up to 100. For example, Rina had 100 tiles in all, there were 100 children and 100 shoe pairs. How do we calculate Percentage of an item if the total number of items do not add up to 100? In such cases, we need to convert the fraction to an equivalent fraction with denominator 100. Consider the following example. You have a necklace with twenty beads in two colours.
Anwar found the Percentage of red beads like this
Hence, out of 100, the number of red beads is
(out of hundred) = 40%
Asha does it like this
= 40%
We see that these three methods can be used to find the Percentage when the total does not add to give 100. In the method shown in the table, we multiply the fraction by . This does not change the value of the fraction. Subsequently, only 100 remains in the denominator.
Anwar has used the unitary method. Asha has multiplied by to get 100 in the denominator. You can use whichever method you find suitable. May be, you can make your own method too.
The method used by Anwar can work for all ratios. Can the method used by Asha also work for all ratios? Anwar says Asha’s method can be used only if you can find a natural number which on multiplication with the denominator gives 100. Since denominator was 20, she could multiply it by 5 to get 100. If the denominator was 6, she would not have been able to use this method. Do you agree?
# Try These
1. A collection of 10 chips with different colours is given .
Fill the table and find the percentage of chips of each colour.
2. Mala has a collection of bangles. She has 20 gold bangles and 10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as done in the above example?
# Think, Discuss and Write
1. Look at the examples below and in each of them, discuss which is better for comparison.
In the atmosphere, 1 g of air contains:
.78 g Nitrogen 78% Nitrogen
.21 g Oxygen or 21% Oxygen
.01 g Other gas 1% Other gas
2. A shirt has:
Cotton 60% Cotton
Polyster or 40% Polyster
## 8.3.2 Converting Fractional Numbers to Percentage
Fractional numbers can have different denominator. To compare fractional numbers, we need a common denominator and we have seen that it is more convenient to compare if our denominator is 100. That is, we are converting the fractions to Percentages. Let us try converting different fractional numbers to Percentages.
Example 7 Write as per cent.
Solution We have,
=
Example 8 Out of 25 children in a class, 15 are girls. What is the percentage of girls?
Solution Out of 25 children, there are 15 girls.
Therefore, percentage of girls = ×100 = 60. There are 60% girls in the class.
Example 9 Convert to per cent.
Solution We have,
From these examples, we find that the percentages related to proper fractions are less than 100 whereas percentages related to improper fractions are more than 100.
# Think, Discuss and Write
(i) Can you eat 50% of a cake? Can you eat 100% of a cake?
Can you eat 150% of a cake?
(ii) Can a price of an item go up by 50%? Can a price of an item go up by 100%?
Can a price of an item go up by 150%?
## 8.3.3 Converting Decimals to Percentage
We have seen how fractions can be converted to per cents. Let us now find how decimals can be converted to per cents.
Example 10 Convert the given decimals to per cents:
(a) 0.75 (b) 0.09 (c) 0.2
Solution
(a) 0.75 = 0.75 × 100 % (b) 0.09 = = 9 %
= × 100 % = 75%
(c) 0.2 = × 100% = 20 %
Try These
1. Convert the following to per cents:
(a) (b) 3.5 (c) ( d) ( e) 0.05
2. (i) Out of 32 students, 8 are absent. What per cent of the students are absent?
(ii) There are 25 radios, 16 of them are out of order. What per cent of radios are out of order?
(iii) A shop has 500 items, out of which 5 are defective. What per cent are defective?
(iv) There are 120 voters, 90 of them voted yes. What per cent voted yes?
## 8.3.4 Converting Percentages to Fractions or Decimals
Make some more such examples and solve them.
We have so far converted fractions and decimals to percentages. We can also do the reverse. That is, given per cents, we can convert them to decimals or fractions. Look at the table, observe and complete it:
## Parts always add to give a whole
In the examples for coloured tiles, for the heights of children and for gases in the air, we find that when we add the Percentages we get 100. All the parts that form the whole when added together gives the whole or 100%. So, if we are given one part, we can always find out the other part. Suppose, 30% of a given number of students are boys.
This means that if there were 100 students, 30 out of them would be boys and the remaining would be girls.
Then girls would obviously be (100 – 30)% = 70%.
# Try These
1. 35% + _______% = 100%, 64% + 20% +________ % = 100%
45% = 100% – _________ %, 70% = ______% – 30%
2. If 65% of students in a class have a bicycle, what per cent of the student do not have bicycles?
3. We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what per cent are mangoes?
# Think, Discuss and Write
Consider the expenditure made on a dress
20% on embroidery, 50% on cloth, 30% on stitching.
Can you think of more such examples?
## 8.3.5 Fun with Estimation
Percentages help us to estimate the parts of an area.
Solution We first find the fraction of the figure that is shaded. From this fraction, the percentage of the shaded part can be found.
You will find that half of the figure is shaded. And,
Thus, 50 % of the figure is shaded.
# Try These
What per cent of these figures are shaded?
Tangram
## 8.4.1 Interpreting Percentages
We saw how percentages were helpful in comparison. We have also learnt to convert fractional numbers and decimals to percentages. Now, we shall learn how percentages can be used in real life. For this, we start with interpreting the following statements:
5% of the income is saved by Ravi. — 20% of Meera’s dresses are blue in colour.
— Rekha gets 10% on every book sold by her.
What can you infer from each of these statements?
By 5% we mean 5 parts out of 100 or we write it as . It means Ravi is saving
` 5 out of every ` 100 that he earns. In the same way, interpret the rest of the statements given above.
## 8.4.2 Converting Percentages to “How Many”
Consider the following examples:
Example 12 A survey of 40 children showed that 25% liked playing football. How many children liked playing football?
Solution Here, the total number of children are 40. Out of these, 25% like playing football. Meena and Arun used the following methods to find the number. You can choose either method.
Arun does it like this
Out of 100, 25 like playing football
So out of 40, number of children who like
playing football = = 10
Meena does it like this
25% of 40 =
= 10
Hence, 10 children out of 40 like playing football.
# Try These
1. Find:
(a) 50% of 164 (b) 75% of 12 (c) % of 64
2. 8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain.
Example 13 Rahul bought a sweater and saved 200 when a discount of 25% was given. What was the price of the sweater before the discount?
Solution Rahul has saved 200 when price of sweater is reduced by 25%. This means that 25% reduction in price is the amount saved by Rahul. Let us see how Mohan and Abdul have found the original cost of the sweater.
Mohan’s solution
25% of the original price = 200
Let the price (in ) be P
So, 25% of P = 200 or
or, or P = 200 × 4
Therefore, P = 800
Abdul’s solution
25 is saved for every 100
Amount for which 200 is saved
= ` 800
Thus both obtained the original price of sweater as rs800.
# Try These
1. 9 is 25% of what number? 2. 75% of what number is 15?
# Exercise 8.2
1. Convert the given fractional numbers to per cents.
(a) (b) (c) (d)
2. Convert the given decimal fractions to per cents.
(a) 0.65 (b) 2.1 (c) 0.02 (d) 12.35
3. Estimate what part of the figures is coloured and hence find the per cent which is coloured.
(i)
(ii)
(iii)
4. Find:
(a) 15% of 250 (b) 1% of 1 hour (c) 20% of ` 2500 (d) 75% of 1 kg
5. Find the whole quantity if
(a) 5% of it is 600. (b) 12% of it is ` 1080. (c) 40% of it is 500 km.
(d) 70% of it is 14 minutes. (e) 8% of it is 40 litres.
6. Convert given per cents to decimal fractions and also to fractions in simplest forms:
(a) 25% (b) 150% (c) 20% (d) 5%
7. In a city, 30% are females, 40% are males and remaining are children. What per cent are children?
8. Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
9. Meeta saves ` 4000 from her salary. If this is 10% of her salary. What is her salary?
10. A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
## 8.4.3 Ratios to Percents
Sometimes, parts are given to us in the form of ratios and we need to convert those to percentages. Consider the following example:
Example 14 Reena’s mother said, to make idlis, you must take two parts rice and one part urad dal. What percentage of such a mixture would be rice and what percentage would be urad dal?
Solution In terms of ratio we would write this as Rice : Urad dal = 2 : 1.
Now, 2 + 1=3 is the total of all parts. This means part is rice and part is urad dal.
Then, percentage of rice would be .
Percentage of urad dal would be .
Example 15 If ` 250 is to be divided amongst Ravi, Raju and Roy, so that Ravi gets two parts, Raju three parts and Roy five parts. How much money will each get? What will it be in percentages?
Solution The parts which the three boys are getting can be written in terms of ratios as 2 : 3 : 5. Total of the parts is 2 + 3 + 5 = 10.
Amounts received by each Percentages of money for each
× ` 250 = ` 50 Ravi gets
= ` 75 Raju gets
= ` 125 Roy gets
# Try These
1. Divide 15 sweets between Manu and Sonu so that they get 20 % and 80 % of them respectively.
2. If angles of a triangle are in the ratio 2 : 3 : 4. Find the value of each angle.
## 8.4.4 Increase or Decrease as Per cent
There are times when we need to know the increase or decrease in a certain quantity as percentage. For example, if the population of a state increased from 5,50,000 to 6,05,000. Then the increase in population can be understood better if we say, the population increased by 10 %.
How do we convert the increase or decrease in a quantity as a percentage of the initial amount? Consider the following example.
Example 16 A school team won 6 games this year against 4 games won last year. What is the per cent increase?
Solution The increase in the number of wins (or amount of change) = 6 – 4 = 2.
Percentage increase = × 100
= = × 100 = 50
Example 17 The number of illiterate persons in a country decreased from 150 lakhs to 100 lakhs in 10 years. What is the percentage of decrease?
Solution Original amount = the number of illiterate persons initially = 150 lakhs.
Amount of change = decrease in the number of illiterate persons = 150 – 100 = 50 lakhs
Therefore, the percentage of decrease
= × 100 =
# Try Theses
1. Find Percentage of increase or decrease:
– Price of shirt decreased from 280 to 210.
– Marks in a test increased from 20 to 30.
2. My mother says, in her childhood petrol was 1 a litre. It is 52 per litre today. By what Percentage has the price gone up?
## 8.5 Prices Related to an Item or Buying and Selling
The buying price of any item is known as its cost price. It is written in short as CP.
The price at which you sell is known as the selling price or in short SP.
What would you say is better, to you sell the item at a lower price, same price or higher price than your buying price? You can decide whether the sale was profitable or not depending on the CP and SP. If CP < SP then you made a profit = SP – CP.
If CP = SP then you are in a no profit no loss situation.
If CP > SP then you have a loss = CP – SP.
Let us try to interpret the statements related to prices of items.
A toy bought for72 is sold at 80.
A T-shirt bought for 120 is sold at 100.
A cycle bought for 800 is sold for 940.
Let us consider the first statement.
The buying price (or CP) is 72 and the selling price (or SP) is 80. This means SP is more than CP. Hence profit made = SP – CP = 80 – 72 = 8
Now try interpreting the remaining statements in a similar way.
## 8.5.1 Profit or Loss as a Percentage
The profit or loss can be converted to a percentage. It is always calculated on the CP.
For the above examples, we can find the profit % or loss %.
Let us consider the example related to the toy. We have CP = 72, SP = 80,
Profit =
8. To find the percentage of profit, Neha and Shekhar have used the following methods.
Thus, the profit is 8 and
profit Per cent is .
Similarly you can find the loss per cent in the second situation. Here,
CP = 120, SP = 100.
Therefore, Loss = 120 – 100 = 20
Try the last case.
Now we see that given any two out of the three quantities related to prices that is, CP, SP, amount of Profit or Loss or their percentage, we can find the rest.
Example 18 The cost of a flower vase is 120. If the shopkeeper sells it at a loss of 10%, find the price at which it is sold.
Solution We are given that CP = 120 and Loss per cent = 10. We have to find the SP.
Thus, by both methods we get the SP as 108.
Example 19 Selling price of a toy car is 540. If the profit made by shopkeeper is 20%, what is the cost price of this toy?
Solution We are given that SP =540 and the Profit = 20%. We need to find the CP.
Thus, by both methods, the cost price is 450.
# Try These
1. A shopkeeper bought a chair for 375 and sold it for 400. Find the gain Percentage.
2. Cost of an item is 50. It was sold with a profit of 12%. Find the selling price.
3. An article was sold for 250 with a profit of 5%. What was its cost price?
4. An item was sold for 540 at a loss of 5%. What was its cost price?
## 8.6 Charge Given on Borrowed Money or Simple Interest
Sohini said that they were going to buy a new scooter. Mohan asked her whether they had the money to buy it. Sohini said her father was going to take a loan from a bank. The money you borrow is known as sum borrowed or principal.
This money would be used by the borrower for some time before it is returned. For keeping this money for some time the borrower has to pay some extra money to the bank. This is known as Interest.
You can find the amount you have to pay at the end of the year by adding the sum borrowed and the interest. That is, Amount = Principal + Interest.
Interest is generally given in per cent for a period of one year. It is written as say 10% per year or per annum or in short as 10% p.a. (per annum).
10% p.a. means on every 100 borrowed, 10 is the interest you have to pay for one year. Let us take an example and see how this works.
## Example 20
Anita takes a loan of 5,000 at 15% per year as rate of interest. Find the interest she has to pay at the end of one year.
## Solution
The sum borrowed = 5,000, Rate of interest = 15% per year.
This means if 100 is borrowed, she has to pay 15 as interest for one year. If she has borrowed 5,000, then the interest she has to pay for one year
= ` 750
So, at the end of the year she has to give an amount of 5,000 + 750 = 5,750.
We can write a general relation to find interest for one year. Take P as the principal or sum and R % as Rate per cent per annum.
Now on every 100 borrowed, the interest paid is R
Therefore, on P borrowed, the interest paid for one year would be = .
## 8.6.1 Interest for Multiple Years
Try These
If the amount is borrowed for more than one year the interest is calculated for the period the money is kept for. For example, if Anita returns the money at the end of two years and the rate of interest is the same then she would have to pay twice the interest i.e., 750 for the first year and 750 for the second. This way of calculating interest where principal is not changed is known as simple interest. As the number of years increase the interest also increases. For 100 borrowed for 3 years at 18%, the interest to be paid at the end of 3 years is 18 + 18 + 18 = 3 × 18 = 54.
We can find the general form for simple interest for more than one year.
We know that on a principal of P at R% rate of interest per year, the interest paid for one year is . Therefore, interest I paid for T years would be
And amount you have to pay at the end of T years is A = P + I
## Try These
1. 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one year.
2. 3,500 is given at 7% p.a. rate of interest. Find the interest which will be received at the end of two years.
3. 6,050 is borrowed at 6.5% rate of interest p.a.. Find the interest and the amount to be paid at the end of 3 years.
4. 7,000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the amount to be paid at the end of the second year.
Just as in the case of prices related to items, if you are given any two of the three quantities in the relation , you could find the remaining quantity.
## Example 21
If Manohar pays an interest of 750 for 2 years on a sum of
` 4,500, find the rate of interest.
# Try These
1. You have 2,400 in your account and the interest rate is 5%. After how many years would you earn 240 as interest.
2. On a certain sum the interest paid after 3 years is 450 at 5% rate of interest per annum. Find the sum.
# Exercise 8.3
1. Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.
(a) Gardening shears bought for 250 and sold for 325.
(b) A refrigerater bought for 12,000 and sold at 13,500.
(c) A cupboard bought for 2,500 and sold at 3,000.
(d) A skirt bought for 250 and sold at 150.
2. Convert each part of the ratio to percentage:
(a) 3 : 1 (b) 2 : 3 : 5 (c) 1:4 (d) 1 : 2 : 5
3. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
4. Arun bought a car for 3,50,000. The next year, the price went upto
3,70,000. What was the Percentage of price increase?
5. I buy a T.V. for ` 10,000 and sell it at a profit of 20%. How much money do I get for it?
6. Juhi sells a washing machine for 13,500. She loses 20% in the bargain. What was the price at which she bought it?
7. (i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?
8. Amina buys a book for 275 and sells it at a loss of 15%. How much does she sell it for?
9. Find the amount to be paid at the end of 3 years in each case:
(a) Principal = 1,200 at 12% p.a. (b) Principal = 7,500 at 5% p.a.
10. What rate gives 280 as interest on a sum of 56,000 in 2 years?
11. If Meena gives an interest of 45 for one year at 9% rate p.a.. What is the sum she has borrowed?
## What have We Discussed?
1. We are often required to compare two quantities in our daily life. They may be heights, weights, salaries, marks etc.
2. While comparing heights of two persons with heights150 cm and 75 cm, we write it as the ratio 150 : 75 or 2 : 1.
3. Two ratios can be compared by converting them to like fractions. If the two fractions are equal, we say the two given ratios are equivalent.
4. If two ratios are equivalent then the four quantities are said to be in proportion. For example, the ratios 8 : 2 and 16 : 4 are equivalent therefore 8, 2, 16 and 4 are in proportion.
5. A way of comparing quantities is percentage. Percentages are numerators of fractions with denominator 100. Per cent means per hundred.
For example 82% marks means 82 marks out of hundred.
6. Fractions can be converted to percentages and vice-versa.
For example, whereas, 75% =
7. Decimals too can be converted to percentages and vice-versa.
For example, 0.25 = 0.25 × 100% = = 25%
8. Percentages are widely used in our daily life,
(a) We have learnt to find exact number when a certain per cent of the total quantity is given.
(b) When parts of a quantity are given to us as ratios, we have seen how to convert them to percentages.
(c) The increase or decrease in a certain quantity can also be expressed as percentage.
(d) The profit or loss incurred in a certain transaction can be expressed in terms of percentages.
(e) While computing interest on an amount borrowed, the rate of interest is given in terms of per cents. For example, ` 800 borrowed for 3 years at 12% per annum.
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# A soccer player kicks a ball from the ground to a maximum height of 12 m. The high point in the trajectory of the ball occurs at a distance of 18 m from the kicker. On the downward path, another player heads the ball at a height of 2.2 m from the ground.
Write a quadratic function that models the situation.
my ans -
y = a (x+2.2)2 + 12
y = a (0 +2.2)2 + 12
12 = a(2.2)2
a = 12/(2.2)2
a= 2.47..
## y(0) = 0 and the vertex is at x=18 (halfway to the other root) and the parabola opens downward, so
y = ax(36-x)
the vertex is at (18,12) so
18a*(18) = 12
x = 12/18^2 = 2/3*18 = 2/54
y = 2/54 x(36-x)
## To derive the quadratic function that models the situation, we can use the general form of a quadratic equation, which is y = ax^2 + bx + c.
Given the information in the problem, we know that the maximum height of the ball is 12 m, which occurs at a distance of 18 m from the kicker. This means that the vertex of the parabolic trajectory is at the point (18, 12).
To find the value of the quadratic coefficient, a, we can substitute the vertex coordinates into the equation:
12 = a(18)^2 + b(18) + c
Since the ball starts at ground level (y = 0) and reaches a maximum height of 12 m, we also know that the y-intercept is 0. We can substitute this information into the equation as well:
0 = a(0)^2 + b(0) + c
Now we have a system of equations:
1) 12 = a(18)^2 + b(18) + c
2) 0 = c
From equation 2, we can conclude that c = 0.
Substituting this value into equation 1, we have:
12 = a(18)^2 + b(18)
Simplifying further:
12 = 324a + 18b
Now we have a linear equation with two variables (a and b). Since we need another equation to solve the system, we can use the given information that another player heads the ball at a height of 2.2 m from the ground. This means that when y = 2.2, there is some corresponding value of x.
Using the vertex form of a parabolic equation, we know that the x-coordinate of the vertex (h) is given by:
h = -b/2a
In this case, the x-coordinate of the vertex (h) is 18. Substituting this value into the equation:
18 = -b/2a
We can rearrange the equation as:
b = -36a
Now we have two equations:
1) 12 = 324a + 18b
2) b = -36a
Substituting equation 2 into equation 1:
12 = 324a + 18(-36a)
Simplifying:
12 = 324a - 648a
Combining like terms:
12 + 648a = 324a
648a - 324a = 12
324a = 12
a = 12/324
a = 0.037... (approximately)
Now that we have the value of a, we can substitute it back into equation 2 to find b:
b = -36(0.037...)
b = -1.33... (approximately)
Finally, substituting the values of a and b into the general quadratic equation form, we have:
y = 0.037...x^2 - 1.33...x
Therefore, the quadratic function that models the situation is:
y = 0.037...x^2 - 1.33...x
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Inequalities in one triangle
In a triangle, we can order the lengths of the legs from shortest to longest if we know what the three angles of the triangle are. There is a nice rule that says, in a triangle, the shortest leg is opposite the smallest angle, the longest leg is opposite the largest angle, and the middle-length leg is opposite the middle-sized angle.
If you want to see an original proof of this law that is relatively easy to understand, you can check out Proposition 18 and 19 from Book I of Euclid’s Elements. Born thousands of years ago, Euclid was the “original geometer” and he discovered most of the basic properties of geometry that we still use to this day.
Let’s use the rule to help us order the sides of triangles from shortest to longest.
Example: Order the sides of the triangle from shortest to longest.
SOLUTION: By our rule, we know that the shortest side must be $$\overline {EF}$$, since it is opposite the smallest angle. Next, we know that the largest side must be $$\overline {EG}$$, since it is opposite the largest angle. Finally, the middle side will be $$\overline {FG}$$.
Then the sides of the triangle, from shortest to longest, are $$\overline {EF}$$, $$\overline {FG}$$, $$\overline {EG}$$.
EXAMPLE: Order the sides of the triangle from shortest to longest.
SOLUTION: By our rule, we know that the shortest side is $$\overline {UW}$$, since it is opposite the smallest angle. Similarly, we know that the largest side must be $$\overline {VW}$$, since it is opposite the largest angle. Then the middle sized side will be $$\overline {UV}$$.
Then the sides of the triangle, in order from shortest to longest, are $$\overline {UW}$$, $$\overline {UV}$$, $$\overline {VW}$$.
2088 x
Name the largest and smallest angle in each triangle.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
1782 x
Order the sides of each triangle from shortest to longest.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
1612 x
Order the angles in each triangle from smallest to largest.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
Geometry
Circles
Congruent Triangles
Constructions
Parallel Lines and the Coordinate Plane
Properties of Triangles
Algebra and Pre-Algebra
Beginning Algebra
Beginning Trigonometry
Equations
Exponents
Factoring
Linear Equations and Inequalities
Percents
Polynomials
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## College Algebra (11th Edition)
The parent function is $f(x)=(\frac{1}{3})^x$ (with green) the given function is $g(x)=-(\frac{1}{3})^{-x}$ (with blue). The parent function can be graphed by calculating a few coordinates and connecting them with a smooth curve: $f(-2)=(\frac{1}{3})^{-2}=9$ $f(-1)=(\frac{1}{3})^{-1}=3$ $f(0)=(\frac{1}{3})^0=1$ $f(1)=(\frac{1}{3})^1=\frac{1}{3}$ $f(2)=(\frac{1}{3})^2=\frac{1}{9}$ For every corresponding x-value the following equation is true: $-f(-x)=g(x)$ This means that the graph is reflected across both the x-axis and the y-axis. First, the reflection across the y-axis. We only consider the g(x) function as $g'(x)=\frac{1}{3}^{-x}$ For every corresponding x-value the following equation is true: $f(x)=g'(-x)$ This means that the graph is reflected across the y-axis. Because when f(x)=g'(x), the g'(x) function acts like the f(x). For example if $f(-1)=3$ in the original $f(x)$, this will be equal to $g'(1)=f(-1)=3$. Here,$f(-1)=g'(1)$ also, $f(-2)=g'(2)$ We can see that here, each point in the parent function was reflected across the y-axis. Second, the reflection across the x-axis. Now, the following equation is true: $-g'(x)=g(x)$ Every $g'(x)$ value will be multiplied by (-1). For example if $g'(1)=3$, this will be translated as $g(1)=-g'(1)=-3$. Also, $g(2)=-g'(2)=-9$ We can see that here, every $g'(x)$ is multiplied by (-1), therefore g'(x) is reflected across the x-axis.)
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# How do you find the variance of the sum of a random variable?
## How do you find the variance of the sum of a random variable?
We can also find the variance of Y based on our discussion in Section 5.3. In particular, we saw that the variance of a sum of two random variables is Var(X1+X2)=Var(X1)+Var(X2)+2Cov(X1,X2).
What is the variance of sum of two random variables?
The variance of the sum of two or more random variables is equal to the sum of each of their variances only when the random variables are independent. Rule 1. The covariance of two constants, c and k, is zero.
Is the sum of independent variables independent?
Independent random variables This means that the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances (i.e., the square of the standard deviation is the sum of the squares of the standard deviations).
### How do you find the variance of an independent variable?
For independent random variables X and Y, the variance of their sum or difference is the sum of their variances: Variances are added for both the sum and difference of two independent random variables because the variation in each variable contributes to the variation in each case.
How do you find the variance of two independent variables?
If you have two independent random variables, then: E(X/Y) = E(X)·E(1/Y).
What is the variance of the random variable?
In words, the variance of a random variable is the average of the squared deviations of the random variable from its mean (expected value). Notice that the variance of a random variable will result in a number with units squared, but the standard deviation will have the same units as the random variable.
## Are sum and product of random variables independent?
Two random variables X and Y are independent if all events of the form “X ≤ x” and “Y ≤ y” are independent events. The expected value of the sum of several random variables is equal to the sum of their expectations, e.g., E[X+Y] = E[X]+ E[Y] .
How do you find the sum of variance?
How to Calculate Variance
1. Find the mean of the data set. Add all data values and divide by the sample size n.
2. Find the squared difference from the mean for each data value. Subtract the mean from each data value and square the result.
3. Find the sum of all the squared differences.
4. Calculate the variance.
How do I find the variance?
We can combine variances as long as it’s reasonable to assume that the variables are independent. Here’s a few important facts about combining variances: Make sure that the variables are independent or that it’s reasonable to assume independence, before combining variances.
How do I test if two random variables are independent?
You can tell if two random variables are independent by looking at their individual probabilities. If those probabilities don’t change when the events meet, then those variables are independent. Another way of saying this is that if the two variables are correlated, then they are not independent.
What is the formula for a random variable?
1. If X is a random variable, then V(aX+b) = a2V(X), where a and b are constants.
## How do you calculate the expected value of a random?
For most simple events, you’ll use either the Expected Value formula of a Binomial Random Variable or the Expected Value formula for Multiple Events. The formula for the Expected Value for a binomial random variable is: P(x) * X. X is the number of trials and P(x) is the probability of success.
Are X and Y independent?
Thus, X and Y are not independent, or in other words, X and Y are dependent. This should make sense given the definition of X and Y. The winnings earned depend on the number of heads obtained. So the probabilities assigned to the values of Y will be affected by the values of X.
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# How do you find the derivative of y = x^(cos x)?
Jan 16, 2016
For a function that involves a variable base to a variable exponent, we usually need some form of logarithmic differentiation.
#### Explanation:
$y = {x}^{\cos} x$
Method 1
$\ln y = \ln \left({x}^{\cos} x\right) = \cos x \ln x$
Now differentiate implicitly:
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = - \sin x \ln x + \cos \frac{x}{x}$
$\mathrm{dy} = y \left(\cos \frac{x}{x} - \sin x \ln x\right) = {x}^{\cos} x \left(\cos \frac{x}{x} - \sin x \ln x\right)$
Method 2
$y = {e}^{\ln} \left({x}^{\cos} x\right) = {e}^{\cos x \ln x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{\cos x \ln x} \frac{d}{\mathrm{dx}} \left(\cos x \ln x\right)$ (we just did this derivative above)
$= {e}^{\cos x \ln x} \left(\cos \frac{x}{x} - \sin x \ln x\right)$
$= {x}^{\cos} x \left(\cos \frac{x}{x} - \sin x \ln x\right)$
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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2
# NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 6 Lines and Angles Ex 6.2.
## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2
Ex 6.2 Class 9 Maths Question 1.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Solution:
We have, AB || CD and CD || EF [Given]
AB || EF
x = z [Alternate interior angles] ….(1)
Again, AB || CD
x + y = 180° [Co-interior angles]
z + y = 180° … (2) [By (1)]
But y : z = 3 : 7
Let y = 3k and z = 7k
7k + 3k = 180° [By (2)]
10k = 180°
k = 18°
y = 3 × 18° = 54° and z = 7 × 18° = 126°
x = 126° [From (1)]
Ex 6.2 Class 9 Maths Question 2.
In figure, if AB || CD, EF CD and GED = 126°, find AGE, GEF and FGE.
Solution:
We have AB || CD and GE is a transversal.
AGE = GED [Alternate interior angles]
But
GED = 126° [Given]
∴ ∠AGE = 126°
Also,
GEF + FED = GED
or
GEF + 90° = 126° [ EF CD (given)]
GEF = 126° – 90° = 36°
Now, AB || CD and GE is a transversal.
FGE + GED = 180° [Co-interior angles]
or
FGE + 126° = 180°
or FGE = 180° – 126° = 54°
Thus,
AGE = 126°, GEF = 36° and FGE = 54°.
Ex 6.2 Class 9 Maths Question 3.
In the given figure, if PQ || ST,
PQR = 110° and RST = 130°, find QRS.
[Hint: Draw a line parallel to ST through point R.]
Solution:
Given, PQ || ST,
PQR = 110° and RST = 130°
Draw a line AB parallel to ST through R.
Now, ST || AB and SR is a transversal.
So,
RST + SRB = 180° [Since, sum of the co-interior angles is 180°]
130° + SRB = 180°
SRB = 180° – 130°
SRB = 50° …(i)
Since, PQ || ST and AB || ST, so PQ || AB and then QR is a transversal.
So,
PQR + QRA = 180° [Since, sum of the co-interior angles is 180°]
110° + QRA = 180°
QRA = 180° – 110°
QRA = 70° ...(ii)
Now, ARB is a line.
QRA + QRS + SRB = 180° [Angles on a line]
70° + QRS + 50° = 180°
120° + QRS = 180°
QRS = 180° – 120°
QRS = 60°
Ex 6.2 Class 9 Maths Question 4.
In the given figure, if AB||CD,
APQ = 50° and PRD = 127°, find x and y.
Solution:
Given, APQ = 50° and PRD = 127°
In the given figure, AB || CD and PQ is a transversal.
APQ = PQR [Alternate interior angles]
x = 50°
Also, AB || CD and PR is a transversal.
So,
APR = PRD [Alternate interior angles]
50° + y = 127° [ APR = APQ + QPR = 50° + y]
y = 127° – 50°
y = 77°
Hence, x = 50° and y = 77°.
Ex 6.2 Class 9 Maths Question 5.
In the given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
Solution:
Draw BE
PQ and CF RS.
BE || CF
Also,
a = b …(i) [ angle of incidence = angle of reflection]
and x = …(ii) [ angle of incidence = angle of reflection]
Since, BE || CF and BC is a transversal.
b = [Alternate interior angles]
2b = 2x [Multiplying by 2 on both sides]
b + b = x + x
a + b = x + y [from eqs. (i) and (ii)]
ABC = DCB
which are alternate interior angles.
Hence, AB || CD. Hence proved.
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
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Q:
# You have a standard deck of 52 cards (i.e., 4 aces, 4 twos, 4 threes, …, 4 tens, 4 jacks, 4 queens, and 4 kings) that contains 4 suits (hearts, clubs, spades, and diamonds). We draw one card from the deck. What is the probability that the card is NEITHER a face card (jack, king, or queen) NOR a heart? 27/52
Accepted Solution
A:
Answer:30/52 or 0.5769 or 57.69%Step-by-step explanation:In a standard deck of 52 cards, the number of face cards (F) and the number of hearts (H) is given by:$$F=4+4+4 =12\\H=\frac{52}{4}=13$$Out of all hearts, three of them are face cards (jack, king, and queen). Therefore, the probability of a card being EITHER a face card or a heart is:$$P(F \cup H) = P(F) +P(H) - P(F \cap H) \\P(F \cup H)=\frac{12+13-3}{52} =\frac{22}{52}$$Therefore, the probability of card being NEITHER a face card NOR a heart is:$$P=1-P(F \cup H) \\P=1-\frac{22}{52}=\frac{30}{52}\\\\P=0.5769\ or\ 57.69\%$$
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# Operations with Fractions using Pattern Blocks
## Overview and Objective
In this lesson, students practice four operations with fractions using pattern blocks by expressing one block in terms of the other.
## Warm-Up
The traditional pattern blocks are under the Polygon tiles of the Polypad. If students are already familiar with the blocks, you may want to change their colors to the original one be selecting Alternate Tile Colours in the settings
Start by inserting four pattern blocks into a blank canvas; the equilateral triangle, rhombus, trapezium, and hexagon. Ask students how many triangles, rhombuses, or trapeziums can be used to cover the hexagon?
Share this canvas with the students and let them express each block in terms of one another.
In the first part, where the triangle is assumed to be 1 unit, it is easy to express the others since we can use whole numbers to describe them. In the second section, where the rhombus is 1, students still can easily find the hexagon. Since the triangle is half of the rhombus, they might also find $1/2$ by intuition. Remind them to write any fraction in the answer boxes as a/b format with no space in between.
Let them work on the possible ways to express the trapezium in terms of the rhombus. Remind them to make their thinking visible on the canvas using the blocks.
In the third section, where the trapezium is set to be 1, the hexagon and the triangle can still be found easily. Give students some time to work on the rhombus. There are many ways to find its value. Here are some possible expressions.
In the last section, where the hexagon is valued as 1, students can express the other blocks using unit fractions.
## Main Activity
Remind students that they may use the given block(s) to cover the asked shapes to help them work towards the solution. They may also change the transparency of the blocks when overlapping them. Here are some examples:
Share some student work with the class. Invite students to share which approaches they used to find the answers. There are several methods to find the answer; some might only consider the proportional relations between the given and asked fraction models. Some students might use different blocks as unit fractions to build up the given models.
## Closure
To close the lesson, share this canvas with students and ask them to create fraction models using the four pattern blocks. At the end of the lesson, let them share their models with the class. Finally, compare and comment on different models that represent the same fractions.
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# 5.3: Multiply Special Products
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# The Pythagorean Theorem: Detailed Explanation
Contributed by:
This pdf covers the basics of the Pythagoras theorem and how to find an unknown side by using this theorem with examples.
1. Section 2.1: The Pythagorean Theorem
The Pythagorean Theorem is a formula that gives a relationship between the sides of a right
triangle The Pythagorean Theorem only applies to RIGHT triangles. A RIGHT triangle is a
triangle with a 90 degree angle. The 90 degree angle in a right triangle is often depicted with a
a c Pythagorean Theorem: a2 + b2 = c2
b
NOTE: The side “c” is always the side opposite the right angle. Side “c” is called the
The sides adjacent to the right angle are named “a” and “b”.
There is no choice when selecting a side to be named “c”, but the sides you choose to call “a”
and “b” are interchangeable.
Example: Name each side of the triangle below. That is name one side a, one b and the
hypotenuse c.
The c is the side opposite the right angle. There is no flexibility when it comes to choosing the
side we name c.
The sides a, b are adjacent to the right angle and we have flexibility when naming them.
Each of these is a correct answer.
Answer 1: a = 3, b = 4 and c = 5
Answer 2: a = 4, b = 3, and c = 5
2. Example: Find the length of the missing side. Round to 2 decimals. Make sure to use the
Step 1: Name each side. (It isn’t necessary to use the units (cm) until I write my answer.)
The side with the question mark is opposite the right angle. It needs to be called c.
The 2 and the 4 are the sides that need to be labeled a and b. I have a bit of flexibility.
It would be correct to call:
a = 2 and b = 4 or a = 4 and b = 2
I will pick a = 2 and b = 4
Step 2: put the values into the a2 + b2 = c2 formula and do the algebra to solve for the missing
22 + 42 = c2
4 + 16 = c2
20 = c2
√20 = c
Example: Find the length of the missing side. Make sure to use proper units in your answer.
Step 1: Name each side. (It isn’t necessary to use the units (cm) until I write my answer.)
The side opposite the right angle is called c. c = 20
The side with the x and the 12 are the “a” and “b” for this problem. I will call a = 12 and
b=x
Step 2: Put the values into the a2 + b2 = c2 formula and do the algebra to solve for the missing
122 + x2 = 202
144 + x2 = 400
-144 -144
x2 = 256
x = √256
3. Example: Find the length of the missing side (round your answer to 2 decimals). Be sure to
Step 1: Name each side.
The side opposite the right angle is called c. c = 6.2
The x and the 4.9 are the sides for “a” and “b” in this problem. I will call a = 4.9 and
b=x
Step 2: put the values into the a2 + b2 = c2 formula and do the algebra to solve for the missing
4.92 + x2 = 6.22
24.01 + x2 = 38.44
-24.01 -24.01
x2 = 14.43
𝑥 = √14.43
x = 3.798
4. Example: Find x. Be sure to include units in your answer.
This is a Pythagorean Theorem problem in disguise. I can break this rectangle up into these two
Now I just need to solve for x. Each of the triangles has the same three sides and will give the
same answer regardless of the triangle I use when I solve for x.
Step 1: Name each side. I will use the left triangle.
The side opposite the right angle is called c. c = 10
The x and the 8 are the “a” and “b” for this problem. I will call a = 8 and b = x
Step 2: put the values into the a2 + b2 = c2 formula and do the algebra to solve for the missing
82 + x2 = 102
64 + x2 = 100
-64 -64
x2 = 36
𝑥 = √36
Answer: x = 6 cm (I put my units back in at the end. I don’t do my computations with the units
in place as they don’t affect my algebra.)
5. Example: Find the height of the triangle. That is, solve for h (round your answer to 2 decimal
This is also a Pythagorean Theorem problem in disguise. I can break this triangle up into these
two triangles.
Now I just need to solve for h. I will use triangle on the right as it has all three sides labeled.
Step 1: Name each side.
The side opposite the right angle is called c.
The h and the 4 are the “a” and “b” for this problem. I will call a = 4 and b = h
Step 2: Put the values into the a2 + b2 = c2 formula and do the algebra to solve for the missing
42 + h2 = 82
16 + h2 = 64
-16 -16
h2 = 48
ℎ = √48
h = 6.928
Answer: h = 6.93 ft (I put units in my answer as each number in the problem has units.)
6. Homework #1-20: Use the Pythagorean Theorem to find the missing length. Round to two
decimal places when necessary. Make sure to use units in your answer when the lengths are
given with units.
1) 2)
3) 4)
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# How do you divide decimals by decimals?
Several methods of determining decimal point location are discussed below. It is assumed you are already familiar with integer long division and can continue it to as many decimal places as required. (You can do 9/8 = 1.125, for example.) Express each as an integer divided by a power of 10. Compute the value of the fraction in the usual way. Example 12.34/2.683 = (1234/100)/(2683/1000) = 1234/2683*1000/100 = 12340/2683 (perform this division to get your answer) = 4.5993... Note that this is equivalent to adding enough zeros to the right of the shorter decimal so the numbers have the same number of digits to the right of the decimal point. Then remove the decimal points and treat the numbers as integers. 12.34 becomes 12.340 so it has 3 decimal fraction digits like 2.683 does. Now we treat the problem as 12340/2683 _____ It can also be convenient to use powers of 10 to rewrite the numbers to the form x.xxx...*10^yy then you only have to know whether the quotient is larger or smaller than 1. In the above example, this would look like (1.234*10)/(2.683) The quotient of 1.234/2.683 is smaller than 1 (because the denominator is larger), so the result will be .45993...*10 = 4.5993... The result of the division performed on these 1-digit numbers will always be between .1000...
and 9.999... Those of us who learned multiplication and division using a slide rule made much use of this method.
You can also pay attention to the "multiply the divisor by the quotient digit" part of the operation. This will tell you how big the multiplier has to be to make the product line up for subtraction. Example .0073/12 We know that 72/12 = 6. In order to have the correct product to subtract from .0073, we realize the multiplier must be .0006. Successive quotient digits will go to the right of that. Once the decimal position is determined, division proceeds as with integers.
thanked the writer.
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5th Grade Math Home > Teacher Resources > Lesson Plans > The Literature Connection
# The Literature Connection
### Concepts Covered:
Tables, Fractions, Decimals, Percents, Ratio, Proportion, Probability
MA.3.A.2.2
MA.3.A.6.2
MA.4.A.2.3
MA.4.A.6.3
MA.6.A.2.2
MA.6.A.5.1
MA.7.P.7.2
### Materials for each group:
10 sets of 10 different colored snap cubes
1 box of markers
1 paper lunch bag with candy
10 sticky labels
assorted packs of sticky notes
1 calculator
grid/chart paper
### Student Arrangement:
Students should work in groups of 3 or 4.
### Procedure:
#### Day 1: Table
1. Read Cucumber Soup by Vickie Leigh Krudwig to students.
2. Have students make representations for the number of each kind of bug found in the book. (For example, 10 ants can be represented with 10 blue snap cubes and 2 praying mantises can be represented with 2 red snap cubes etc…)
3. Label each group of bugs with stickers.
Discussion:
How did you construct and organize your models of the insects?
Why did you arrange them that way?
What are other methods we could use to organize this information?
#### Day 2: Multiples
1. Using chart paper, create data tables to organize the information.
2. Share/Explain tables with class, use chart paper.
Discussion:
How many bugs are there in all?
What would happen to the number of bugs if 2 or more cucumbers fell on the anthill?
How many of each bug would be needed to move the cucumbers? (Fill in table)
What patterns do you see?
What kinds of numbers are these?
#### Day 3: Fractions
1. Direct each student to pick a bug and write its name on a sticky note. Each student in the group must pick a different bug.
2. On your sticky note, write what part of the whole bug group your bug is. (For example, in the book there are 9 mosquitoes and 55 bugs all together, so the relationship is the fraction 9/55.)
3. Students share answers in their groups explaining how and why they wrote their answer the way they did.
Discussion:
Show each bug’s fractional representation of whole bug group on chart paper.
Discuss need to simplify some of the fractions.
1. Working in pairs, the students should complete a data table on grid paper.
2. Demonstrate how to convert decimals to percentages
3. Complete the data table with the matching decimal and percent. (Students work in pairs to complete their data tables.)
4. Ask students how fractions, decimals, and percent are related.
5. Ask students how fractions, decimals, and percents help them understand data collected.
#### Day 4: Ratio Part to Whole
1. Lead discussion about the need to compare data, and ways fractions, decimals, percents can be used to help make those comparisons.
2. Another way to compare data.
Ex. There are 9 mosquitoes for every group of 55 bugs. State this comparison as 9 to 55. Explain this is called a ratio. It can be written in several ways. In words, it can be written as "9 to 55". We also have a symbol to replace the word "to." Ask if anyone already knows this symbol. Show an example of a ratio using the colon (9:55).
3. Add another column to the data table.
4. On the sticky note, write the ratio for the bug group chosen earlier. The ratio should be written in all 3 ways: word, with colon, and as fraction.
5. Share with your group.
6. Fill in the ratio of bugs in each separate group to the total bugs.
7. Have students add data to data table.
Discussion:
What about the fractions we simplified?
What happened to their ratio?
What about the equivalent fractions we calculated earlier?
Will their ratio be any different than the ones we just completed? Why?
#### Day 5: Ratio Part to Part
1. Discuss possibility of comparing a group of bugs to another group of bugs rather than a group of bugs to the whole group of bugs.
2. Pick two bug groups and predict what the ratio would be.
3. Write the bug group names and the three forms of the ratio on the back of the grid paper.
4. Share answers with class on chart.
#### Day 6: Probability
Discussion:
Which bug would a hungry robin eat if he swooped down on the mixed group of bugs?
Why do you think that?
Is it guaranteed that the bird will really catch that bug?
Discuss terms used when talking about probability.
Discuss probability and its mathematical representation
Lead the group to discover the fractional representation, decimal, and percent form - the probability of catching a mosquito is 9/55, or about 0.16, or about 16%.
Probability Activity/Game
Directions:
Pull apart snap cubes and place inside paper bag (Just the right number of cubes for each bug group).
Predict bug/cube will be pulled from the bag.
T
ell the group the probability of getting that bug.
Students who successfully predict which bug is pulled may choose one piece of candy
Return all cubes to the bag after each try.
Variation 1:
This time, pick two bugs that you predict will be chosen
Pick one cube from the bag.
Variation 2:
Predict which bug is likely to be chosen
Pick two cubes from the bag.
#### Discussion:
How do probability and ratio tie into fractions, decimals, and percent?
5th Grade Math Home > Teacher Resources > Lesson Plans > The Literature Connection
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These notes are taken from the resource book and were originally written by Dr Erwin. I will be editing and adding to them throughout. Most mistakes within them can thus be presumed to be mine rather than Dr Erwin’s.
In this section we are going to develop a new set of methods to solve a type of problem we are relatively familiar with. We will find a way to translate between methods we know well, but which turn out not to be very efficient, methods which are graphically very intuitive, but not very calculationally useful, and methods which are computationally extremely powerful, but appear rather abstract compared with the other two ways of looking at these problems. These three methods which we will utilise in detail in the coming sections are shown in the following diagram:
As we go through I will try and show how we can go between these apparently different formalisms. To start with we won’t use matrices, but as we come on to them we will first think that they are horrible and abstract, and then realise that the machinery we develop for them is incredibly powerful.
Let’s start with a very simple example of a question which will be the type that we will eventually want to answer using matrices.
Suppose that we want to find all values of $x$ and $y$ for which
$x + y =3$
$2x-y = 4$
(to be clear: we wish to find the values of $x$ and $y$ that satisfy both of these equations simultaneously). This is a system of linear equations. Solving this system is easy: Rewrite the first equation as $y=3-x$ and substitute this into the second equation to get $2x-(3-x)=4.$ From this, it follows that $x=\frac{7}{3}$ and so $y=3-\frac{7}{3}=\frac{2}{3}.$
Here we have treated the system purely algebraically, but let’s see what we’ve done graphically. Each of these equations is a constraint on the points in the ${\mathcal R}^2$ plane – this simply corresponds in this case to two lines. Which look like this:
What we mean by having solved the equations is to have converted them into two new equations, one which involves only x, and one which involves only y (note that we can’t think of the former as a function, because it’s not single valued). These two new equations $x=\frac{7}{3}$ and $y=3-\frac{7}{3}=\frac{2}{3}$ correspond to two new lines which we plot on the same graph:
In some way we’ve combined the two equations together to get two new equations which correspond to horizontal and vertical lines, where before we had intersecting slanted lines. The solution corresponds to the point of intersection of the two original lines, and of course to the intersection of the two new lines.
In two dimensions this is relatively easy to see, and indeed in three dimensions as we will see too, but in higher dimensions it is harder to visualise. However, we should still keep in mind what is happening geometrically in higher dimensions.
In principle, we can apply the same method as above (eliminating the variables) to any system of linear equations. However, if we are asked to solve a system like
$u-v+3w+x-5y+2z=3$
$9u-3v-w+2x-y+12z=-8$
$17u+w-21x-y+11z=35$
$2u+2v+3w+x-6y-3z=0$
$4u-2v+w-54y+2z=-71$
then things are going to get messy, fast, unless we have a systematic method. In this section of the course, we shall develop some systematic methods for solving systems of linear equations.
Systems of linear equations
Definition: An equation of the form $a_1x_1+a_2x_2+ \cdots + a_nx_n = b$ (where $x_1,x_2,\ldots,x_n$ are variables and $a_1,a_2,\ldots,a_n,b$ are real numbers) is called a linear equation in the variables $x_1,x_2,\ldots,x_n$.
Each of the following is either a linear equation or can be rewritten as one:
$3x-5y=7$
$y = 2-6x$
$9x+y-2z=3$
$x_1 + 2x_2 = 5x_3 - 12$
$10x_1 - \frac{3}{2} x_7 + x_{21} = 0$
$2x_1 + 3x_2 +4x_3 -2 x_5 + 9x_6 +11x_8 = \pi$
Each of the following is not a linear equation:
$x^2+y = 4$
$2xyz = 5$
$\sin (x_1) - 3x_2 + 4x_3 = -11$
$e^{x_1} \ln (x_2 + x_3) - x_4^3 = 2$
Suppose we are given several linear equations and asked to determine which values of the variables satisfy all of them simultaneously. To be specific, suppose that we are given $m$ linear equations in the $n$ variables $x_1,x_2,\ldots,x_n$:
$a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1$
$a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2$
$\cdots$
$a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = b_m$
This is called a system of linear equations. If $c_1,c_2,\ldots,c_n$ are real numbers for which
$a_{11}c_1 + a_{12}c_2 + \cdots + a_{1n}c_n = b_1$
$a_{21}c_1 + a_{22}c_2 + \cdots + a_{2n}c_n = b_2$
$\cdots$
$a_{m1}c_1 + a_{m2}c_2 + \cdots + a_{mn}c_n = b_m$
(i.e., if $(x_1,x_2,\ldots,x_n) = (c_1,c_2,\ldots,c_n)$ satisfies every one of these linear equations), then the vector $(c_1,c_2,\ldots,c_n)$ is a solution of this system of linear equations.
Consider the system of linear equations
$x + 2y -z =5$
$-x+3y+z = 0$
$2x - y -2z = 5$
$1(2) + 2(1) -(-1) =5$
$-1(2)+3(1)+(-1) = 0$
$2(2) -(1) -2(-1) = 5$
the vector $(x,y,z) = (2,1,-1)$ is a solution of this system. Note that here we are moving away from the notation for a vector as $\left$ and interchangeably using the coordinate and the vector which goes from the origin to that coordinate.
Consider the system of linear equations
$x+y=1$
$x+y=-1$
From the first equation, we have $y=1-x$. Substituting $y=1-x$ into the second equation, we get $x+(1-x)=-1$ which after simplification becomes $1=-1,$ which is not true. This system of linear equations therefore has no solution. We can see the reason for this when we plot these equations and see that they look like
They don’t intersect anywhere and thus they have no solution.
We can see from the example above that some systems of linear equations have no solution. And, while we found a solution to the system of linear equations in the first example, we do not know whether $(2,1,-1)$ is the only solution or whether there are others. We shall therefore ask two, related, questions:
Given a system of linear equations:
1) Does the system have at least one solution?
2) If it does, how do we find all the solutions?
When we are asked to solve a system of linear equations, we must either show that the system has no solution, or we must find all the solutions of the system. The set of all such vectors is called the solution of that system.
How clear is this post?
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# Video: Evaluating Negative Single Term Mixed Number Expressions with Negative Integer Exponents
Which of the following is equal to (−3 1/2)⁻²? [A] 4/49 [B] −3 1/4 [C] 49/4 [D] 14/−4 [E] 6 1/2
01:30
### Video Transcript
Which of the following is equal to negative three and a half to the power of negative two? Is it (A) four over 49, (B) negative three and one-quarter, (C) 49 over four, (D) 14 over negative four, or (E) six and one-half?
In this question, we have a mixed number that’s being raised to a negative power. Before we can evaluate this, we need to change our mixed number into an improper fraction. We know that three and a half is equal to seven over two. We achieve this by multiplying the integer by the denominator — that’s three times two, which is six — and then adding the numerator, which is seven. The denominator stays the same. So three and a half is equal to seven over two. This means negative three and a half is equal to negative seven over two. And so negative three and a half to the power of negative two is negative seven over two to the power of negative two.
Next, we recall what we do when we have a negative exponent. We know that 𝑥 to the power of negative 𝑎 is one over 𝑥 to the power of 𝑎. A negative exponent means we need to take the reciprocal. The reciprocal of negative seven over two is negative two over seven. So negative seven over two to the power of negative two is negative two over seven squared. Then, to square our fraction, we simply square the numerator and separately square the denominator. We also know that a negative number squared is positive. So we get two squared, which is four, over seven squared, which is 49.
And we see that negative three and a half to the power of negative two is equal to four over 49. That’s (A).
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## Intro to Fractions
### Transcript
Intro to Fractions. I realize that fractions are not necessarily everyone's favorite topic. I hope to convince you during the course of the rest of this module that you can actually do fractions. You can make sense of this topic. Let's start at the beginning.
Here's a number line with a few fractions. Of course most of the fractions fall between the integers. Notice that fractions can be positive or negative. Also, there are an infinite number of fractions between any two integers. So between, say, one and two, there's an infinity of fractions. We won't spend a lot of time on this idea, but it's just an important perspective to have on some of the things the test will ask about fractions.
Notice that any negative fraction can be written with the negative sign in a few different places. For example, we could have the negative sign out in front of the fraction, negative one fifth. We could also write the negative in front of the one, so negative one divided by 5. We could also write 1 divided by negative 5.
So, all three of those are perfectly equivalent and interchangeable. And we could switch back and forth freely between them. Sometimes people get really locked up, they think that, that negative sign has to sit in one place and can't move. Well no, all three of these are equivalent, you're free to move that negative sign around.
So, the, the test will actually expect you to have that fluency. So let's talk about some terms. Suppose we have the fraction 3 over 16, what do we call the place where the three is sitting and where the 16 is sitting? The top part, the upstairs of a fraction is the numerator. This fraction has a numerator of 3.
The bottom part, the downstairs, of a fraction is the denominator. This fraction has a denominator of 16. Two ways of thinking about a fraction, well first of all we can think of it as division. Two sevenths means two is divided by sevens, so there's an actual arithmetic operation occurring.
A completely different way is pieces of pie. If a pie, a hypothetical pie is cut into seven equal pieces, then two sevenths means two of those seven pieces. It would mean this much of the circle. So notice that the first one, you could call it an arithmetic way of thinking about fractions.
The second one is more a visualization. And it's actually very important to cultivate both of these, because these employ different sides of your brain. And if every time you look at a fraction and you think both of the division as well as the diagram, then you're going to be using both sides of your brain, and you're gonna be understanding fractions much more deeply.
For fractions such as 35 over 5 it's useful to think about the fraction as division because that division we can actually perform. 35 divided by 5 equals 7. Notice that 35 over 5, that, in every way, is an equivalent way to write the number 7. When we write 35 over 5, we are writing the number seven in kind of a hidden form.
This leads to the idea of equivalent fractions. Two fractions are equivalent if they have the same numerical value even though they have different numerators and different denominators. So, two thirds equals ten fifteenths. Those fractions are equal. We are totally allowed to put an equal sign between them, which of course is a very powerful statement.
We're saying that those two have exactly the same, mathematical value, even though they look very different. In fact, one way to see what's going on here, we can write the 10 and the 15 as products. 2 times 5 over 3 times 5. In other words, the numerator and the denominator of the original fraction, both got multiplied by the same factor.
In fact we can expand that idea. If we are given one fraction, we can always find an equivalent fraction by multiplying the numerator, and the denominator by the same factor. So for example, we start out with three eighths we could multiply the numerator and the denominator by 4. This would give us twelve over thirty two an equivalent fraction.
Essentially, we're multiplying the fraction by 4 over 4 which equals 1. As we can multiple both numerator and denominator by the same factor, we can also factor out the same positive integer from both the numerator and the denominator. So for example, suppose we have 6 over 42, well we could multiply. We can express both of those as products, 6 times 1 over 6 times 7.
And then, we can get, we can factor out those six, and cancel them, and get down to 1/7ths. The sixes cancel. Notice that canceling is a form of division. That's very important. Canceling is a form of division.
Some folks have a have misconceptions about cancelling. Cancelling the common number in the numerator denominator does not merely quote go away. A lot of people have this sloppy way of talking about it. The sixes go away. That's not a very healthy way to think about this, it leads to mistakes.
For example, suppose we had 8 over 40, we wanted to simplify it. We'll clearly, we can divide, we can cancel the eights. So 40 divided by 8 is 5. That gives us the denominator. Well people were stuck on the, the concept of eights go away. Well then they wonder what happens in the numerator.
What if the eight goes away, what's left in that numerator? Instead of thinking at of it in terms of the eights go away, we cancel the eights. That is to say we divide by the eights, and 8 divided by 8 is 1. The eights don't go away. They divide, 8 over 8 to equal 1. This is a crucially important idea for anyone who previously has been using the go away understanding of cancelling.
The equivalent fraction with the lowest possible integer value of numerator and denominator is the fraction written in lowest terms. So for example, we start up with 72 over 96, both of which are divisible by 6. We can divide that to 12 over 16. Both of that are divisible by 4. We can divide down to 3 over 4.
We cannot divide any further. That fraction is in lowest terms. Once the fraction is in lowest terms we cannot simplify any further. Lowest terms is also known as simplest form. Make a habit of always, always, always writing all your fractions in simplest form.
Well why? First of all it will make all your calculations much easier because you'll be dealing with smaller numbers. On the GRE multiple choice, the answer choice listed will almost always be in simplest form. So you'll have to simplify the simplest form in order to find something that matches the given answer choice.
On the numerical entry, you can enter a non simplified fraction, as long as it fits in the box. Still it's a good idea to practice simplifying though, because simplifying along the way may make your calculation easier so there it still will be helpful. Practice writing these factions in lowest terms. Pause the video here and then we'll talk about these.
. Okay, here are the answers. The last big fraction topic for this video is how to handle fractions that are greater than one. If we have a fraction greater than one, we have a choice about its form. We can write the fraction as either an improper fraction, that is to say a fraction in which the numerator is larger than the denominator- That's what it means to be an improper fraction, the numerator is larger than the denominator.
Or, we can write it as a mixed numeral, that is to say, something that is part integer and part fraction. So for example, we could say five thirds equals one and two thirds. Five thirds is an improper fraction. One and two thirds is the mixed numeral of the same numerical value. And that's why we can put an equal sign between them.
Notice when a mixed numeral is written correctly the fraction part is always a fraction less than one. Any part greater than one is put into the integer. Also notice that by convention we write the integer and the fraction parts of a mixed numeral right next to each other, but the understood operation between them is addition.
So when we write, four and three fifths, what we actually mean, we don't have to write it, but what that actually means is 4 plus three fifths. This makes it easy to change from a mixed numeral to improper fraction form. Write the mixed numeral as addition and then multiply the integer by c over c, where c is the denominator of the fraction part. So, for example, we'll start with four and three fifths.
We'll write that as 4 plus three fifths. Now we're going to multiple that 4 by 5 over 5, and of course we're allowed to multiply by 5 over 5 because that equals 1. And that will give us 20 over 5, plus 3 over 5, and those add to 23 over 5. That is the improper fraction form that is equal to the mixed numeral four and three fifths.
Sometimes a test problem will give all the answers in either mixed numeral or improper fractions form. So, either one could appear on the test. Notice that if you have a choice, improper fractions are almost always more efficient for calculations of all sort, than are mixed numerals. Unless the problem forces you to use mixed numerals, you'll be better off changing everything to improper fractions.
In summary, we talked about the basic fraction terms and how to think about fraction, the division way of looking at it versus the pie method. We talked about the important idea of equivalent fractions. We talked about how cancelling is a form of division, it doesn't mean go away, it means we're actually performing division. We talked about the importance of learning how to put fractions in lowest terms, also known as simplest form.
And we talked a little bit about the idea of mixed numerals and improper fractions
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# Thread: Partial Fraction Decomposition Problem
1. ## Partial Fraction Decomposition Problem
I'm trying to figure out this problem:
At the moment, I've got it down to this:
$3x-4=A(x-1)+B(x-1)^2$
To find $B$, I let $x=1$, which also gets rid of $A$. Any ideas?
2. Originally Posted by BeSweeet
I'm trying to figure out this problem:
At the moment, I've got it down to this:
$3x-4=A(x-1)+B(x-1)^2$
To find $B$, I let $x=1$, which also gets rid of $A$. Any ideas?
Hi BeSweet,
$\frac{3x-4}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$
There must be a partial fraction present for each power of $(x - c)^n$ less than or equal to n. We can find B, but not A using Heaviside's method. Only the constant corresponding to $(x-c)^n$ can be found using this method.
First, multiply the original proper fraction by $(x-c)^2$ and evaluate the result at x = c to get the value of the constant.
$\frac{3x-4}{(x-1)^2}(x-1)^2=3x-4 \Rightarrow B=3(1)-4=-1$
$\boxed{B=-1}$
$\frac{3x-4}{(x-1)^2}=\frac{A}{x-1}+\frac{-1}{(x-1)^2}$
To find the other constant, multiply both sides by the least common denominator to clear the fractions:
$3x-4=A(x-1)-1$
$3x-4=Ax + (-1-A)$
The polynomial on the left of the equal sign must have the same coefficient and constant as the polynomial on the right of the equal sign.
$\boxed{A=3}$
You could also say $-4=(-1-A)$
You still get $\boxed{A=3}$
The decomposition is:
$\boxed{\frac{3x-4}{(x-1)^2}=\frac{3}{x-1}+\frac{-1}{(x-1)^2}}$
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Question
# Compute the gradient of the following functions and evaluate it at the given point P. g(x,y)=x^{2}-4x^{2}y-8xy^{2}. P(-1,2)
Derivatives
Compute the gradient of the following functions and evaluate it at the given point P.
$$\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{{2}}}-{4}{x}^{{{2}}}{y}-{8}{x}{y}^{{{2}}}.{P}{\left(-{1},{2}\right)}$$
2021-05-08
Step 1
The function is given as, $$\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{{2}}}-{4}{x}^{{{2}}}{y}-{8}{x}{y}^{{{2}}}$$
To find the gradient of the given function at (−1 , 2).
The gradient is just the vector of partial derivatives, the partial derivatives will be as,
$$\displaystyle{\frac{{\partial{g}}}{{\partial{x}}}}={2}{x}-{8}{x}{y}-{8}{y}^{{{2}}}$$ and
$$\displaystyle{\frac{{\partial{g}}}{{\partial{y}}}}=-{4}{x}^{{{2}}}-{16}{x}{y}$$
At point (−1,2),
$$\displaystyle{\frac{{\partial{g}}}{{\partial{x}}}}={2}{\left(-{1}\right)}-{8}{\left(-{1}\right)}{\left({2}\right)}-{8}{\left({2}\right)}^{{{2}}}$$
=-2+16-32
=-18
Step 2
and
$$\displaystyle{\frac{{\partial{g}}}{{\partial{y}}}}=-{4}{\left(-{1}^{{{2}}}\right)}-{16}{\left(-{1}\right)}{\left({2}\right)}$$
=-4+32
=28
Since, the gradient is represented in vector form as, $$\displaystyle{\frac{{\partial{g}}}{{\partial{x}}}}{i}+{\frac{{\partial{g}}}{{\partial{y}}}}{j}$$,
gradient $$\displaystyle={\frac{{\partial{g}}}{{\partial{x}}}}{i}+{\frac{{\partial{g}}}{{\partial{y}}}}{j}$$
=-18i+28j
Hence, the gradient for the given function is −18i+28j.
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# Basic structure
Assumptions
We demonstrate how structured derivations works through some examples. Along the way we add useful features to the examples by using the structured derivation method.
The assignment is to solve equation $$3x+7=15-2x$$. We can manipulate the equation by adding, subtracting, multiplying or dividing both sides of equation with same terms. The goal is to get the unknown variable (for example $$x$$) on one side of the equation and its value on the other side.
You might have used notation like this:
$$3x+7=15-2x$$ | $$-7$$
$$3x=8-2x$$ | $$+2x$$
$$5x=8$$ | / $$5$$
$$x=\frac{8}{5}$$
Is this familiar? If we write the same example as structured derivation it would look like following:
$$\bullet$$ $$3x + 7 = 15 -2x$$ $$\Leftrightarrow$$ {Subtract 7 from both sides of the equation} $$3x=15-2x-7$$ $$\Leftrightarrow$$ {Add $$2x$$ to both sides of the equation} $$3x + 2x = 15 -7$$ $$\Leftrightarrow$$ {Calculations} $$5x=8$$ $$\Leftrightarrow$$ {Divide both sides of the equation by 5} $$x=\frac{8}{5}$$
This solution includes the same steps as the first version, but in this version we have more text which explains the solution and some new symbols.
# What are the basic elements of structured derivation?
The beginning of the derivation is marked by a bullet ($$\bullet$$), which is followed by steps. Every step includes three parts: terms, relation and motivation.
$$3x = 15 - 2x + 7$$ Term $$\Leftrightarrow$$ {Add $$2x$$ to both sides of the equation} Relation and motivation $$3x + 2x = 15 - 7$$ Term
Terms are mathematical expressions. The relation (in this case an equivalence $$\Leftrightarrow$$) tells how the terms are related to each other. The motivation explains (or justifies) why this relation holds between the first term and the second one.
Each term and motivation is written on its own line so that there is enough room for proper explanations and even longer terms. Derivations are written in two columns to make them easier to read, write and understand. Relations (here $$\Leftrightarrow$$) are written in the first column. Terms and motivations are written in the second column.
So the step above tells that expression $$3x=15-2x+7$$ is equivalent ($$\Leftrightarrow$$) to the expression $$3x+2x=15-7$$ since we have added the same expression to both sides of the first term. The whole solution is constructed by adding more steps with a motivation for each step.
Assumptions
The contents of the website reflect the authors' views. The Managing Authority of the Central Baltic INTERREG IVA Programme cannot be held liable for the information published by the project partners.
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Re: GAMSAT Physics Translational Motion
Scalars and Vectors
Knowing the difference between vectors and scalars will be advantageous when solving physics problems in the GAMSAT. A vector is a physical quantity that has both direction and magnitude; a scalar is a physical quantity that has magnitude but no direction.
Arrows can be used to reveal the direction of the vector. The length of the arrow will reveal the magnitude of the vector.
When adding vectors, the head of the first vector is placed to the tail of the second vector. An arrow is drawn from the tail of the first to the head of the second. The resulting arrow represents the vector sum of the other two vectors.
For the subtraction of vectors, place the heads of the two vectors together and draw an arrow from the tail of the first to the tail of the second. The resulting arrow represents the vector difference between the two vectors.
Resolution of Vectors and Trigonometric Functions
A vector can be divided into two components. These two components are perpendicular to each other. A vector is resolved into its scalar components. The lengths of the component vectors are found through Pythagoras Theorem and Trigonometric Ratios.
This skill is required in projectile motion problems in the GAMSAT. It is essential that students are able to understand and apply Pythagoras Theorem and Trigonometric Ratios.
Distance and Displacement
Distance and displacement are respectively scalar and vector components. Displacement is distance with an added dimension. This added dimension is direction.
Simply put, we can say that distance refers to how much ground an object has covered and displacement is the object’s overall change in position.
To test your understanding of this distinction, consider the motion depicted in the diagram below. A lady walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.
Even though the lady has walked a total distance of 12 meters, her displacement is 0 meters. During the course of her motion, she has “covered 12 meters of ground” (distance = 12 m). Yet when she is finished walking, she is not ‘out of place. In other words, there is no displacement for her motion (displacement = 0 m). Displacement, being a vector quantity, must give attention to direction. The 4 meters east cancels the 4 meters west; and the 2 meters south cancels the 2 meters north. Vector quantities such as displacement are direction aware. Scalar quantities such as distance are ignorant of direction. In determining the overall distance traveled by the lady, the various directions of motion can be ignored.
Speed and velocity
Speed and velocity are respectively scalar and vector components. Velocity is speed with the added dimension of direction. The units for velocity are expressed in length divided by time – e.g. meters/sec (m/s); feet/sec; miles/hour.
The following two equations must be memorised for the GAMSAT.
Speed = distance/time
Velocity = displacement/time
Acceleration
Acceleration is the rate of change in velocity. Acceleration is a vector, meaning that it has a direction and a magnitude. An object is accelerating if it is changing its velocity. The units for acceleration are expressed as velocity divided by time such as meters/sec2.
When an object experiences negative acceleration it is termed deceleration.
The formula for acceleration must be memorised for the GAMSAT.
Acceleration = rate of change of velocity/time.
Uniformly Accelerated Motion
Uniformly accelerated motion is motion that involves constant acceleration. Constant acceleration means that both the magnitude and direction of acceleration must remain constant. A particle that exhibits uniform accelerated motion will always accelerate at a constant rate. When this particle is accelerating on a linear path, there are 4 variables that can describe its motion. These are:
Time (t) – scalar
Displacement (x) – vector
Velocity (v) – vector
Acceleration (a) – vector
Three basic equations can be used to derive the values of these variables. These equations are usually provided on the GAMSAT stimulus material, but they should be memorised for speed and efficiency. Manipulation of these equations will become second nature through repetition and practice.
The linear motion equations are as follows:
x = xo + vot + 1/2at2
v = vo + at
v2 = vo2 + 2ax
When deciding which equation to use, choose the one for which you know the value of all but one of the variables.
Remember: constant acceleration and linear motion are required in order to use these equations.
Another important concept in the GAMSAT is average velocity. When an object experiences uniformly accelerated motion the average velocity can be determined using the following formula:
Vavg = ½(v + vo)
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# 11.3 Systems of nonlinear equations and inequalities: two variables (Page 4/9)
Page 4 / 9
## Graphing a system of nonlinear inequalities
Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region .
Given a system of nonlinear inequalities, sketch a graph.
1. Find the intersection points by solving the corresponding system of nonlinear equations.
2. Graph the nonlinear equations.
3. Find the shaded regions of each inequality.
4. Identify the feasible region as the intersection of the shaded regions of each inequality or the set of points common to each inequality.
## Graphing a system of inequalities
Graph the given system of inequalities.
$\begin{array}{r}\hfill {x}^{2}-y\le 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill 2{x}^{2}+y\le 12\end{array}$
These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ will be eliminated. Then we solve for $\text{\hspace{0.17em}}x.$
Substitute the x -values into one of the equations and solve for $\text{\hspace{0.17em}}y.$
$\begin{array}{r}\hfill {x}^{2}-y=0\\ \hfill {\left(2\right)}^{2}-y=0\\ \hfill 4-y=0\\ \hfill y=4\\ \hfill \\ \hfill {\left(-2\right)}^{2}-y=0\\ \hfill 4-y=0\\ \hfill y=4\end{array}$
The two points of intersection are $\text{\hspace{0.17em}}\left(2,4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-2,4\right).\text{\hspace{0.17em}}$ Notice that the equations can be rewritten as follows.
$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}-y\le 0\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}\le y\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\ge {x}^{2}\hfill \\ \hfill \\ 2{x}^{2}+y\le 12\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\le -2{x}^{2}+12\hfill \end{array}$
Graph each inequality. See [link] . The feasible region is the region between the two equations bounded by $\text{\hspace{0.17em}}2{x}^{2}+y\le 12\text{\hspace{0.17em}}$ on the top and $\text{\hspace{0.17em}}{x}^{2}-y\le 0\text{\hspace{0.17em}}$ on the bottom.
Graph the given system of inequalities.
Shade the area bounded by the two curves, above the quadratic and below the line.
Access these online resources for additional instruction and practice with nonlinear equations.
## Key concepts
• There are three possible types of solutions to a system of equations representing a line and a parabola: (1) no solution, the line does not intersect the parabola; (2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See [link] .
• There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one solution, the line is tangent to the parabola; (3) two solutions, the line intersects the circle in two points. See [link] .
• There are five possible types of solutions to the system of nonlinear equations representing an ellipse and a circle:
(1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points. See [link] .
• An inequality is graphed in much the same way as an equation, except for>or<, we draw a dashed line and shade the region containing the solution set. See [link] .
• Inequalities are solved the same way as equalities, but solutions to systems of inequalities must satisfy both inequalities. See [link] .
stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter
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# Velocity-definition, types, practice problems, FAQs
Let’s assume your friend lives 10 km away from your home. The road between your home and your friend’s home is straight. If someone asks you, will you be able to reach his house? The answer is no. Why so? You require information about the direction in which you need to travel. Such quantities which are meaningful only when the magnitude and direction both are given are called vector quantities. Velocity is one such quantity. Velocity is defined as the displacement per unit time. In this article, we will explore more about velocity and its types.
## What is velocity?
The change in position vector or the displacement of a body divided by the time taken is called velocity. In the following diagram, the position vector of the body changes from point A to B. Displacement is defined as the shortest possible distance between two points.
Motion between two points A and B
Imagine yourself cycling in a circular path of radius r as shown in figure.
Distance between points A and B =r
Displacement between points A and B =2r
It is a vector quantity. Velocity carries a SI unit of ms-1.
Velocity of an object is said to be positive if the body moves in the positive direction, and negative if it moves in the negative direction.
## Velocity vs speed
Though both of them have the same units, yet they are totally different. Speed is scalar, whereas velocity is a vector.
## Average velocity
The total displacement covered by a body divided by the total time taken is called average velocity(vavg). In the following displacement time (s-t) graph, the average velocity in the time interval ab is equal to the slope of the graph between a and b.
vavg=Total displacementTotal time=st
s indicates the displacement undergone in time t.
## Instantaneous velocity
The velocity of an object(v) at a particular instant of time is called “instantaneous velocity vinst ”. If ds indicates displacement in time dt, then
vinst =dsdt; dsdt indicates the derivative of displacement with respect to time at a given point, P in our case.
The graphical meaning of dsdt at point P is the slope of the graph at point P i.e the slope of the tangent drawn on the s-t curve at point P. Direction of instantaneous velocity at an instant is the same as the direction of average velocity for a straight line motion if the body does not reverse its direction of motion. Direction of instantaneous velocity at an instant may be opposite to the direction of average velocity for a straight line motion if the body reverses its direction of motion.
## Uniform and non-uniform motion
A motion is said to be uniform if the body covers equal distance in equal intervals of time. For eg. if a body covers 2 m of distance each second, the body is said to be in uniform motion. Whereas a motion is said to be non-uniform if the body covers unequal distances in equal intervals of time. A body in non-uniform motion is said to have acceleration.
## Relative velocity
Motion is never absolute; it is always “relative”. It can only be defined wrt an observer. So, relative velocity is defined as the velocity of one body with respect to another.
The relative velocity of A wrt B can be written as vAB = vA- vB
## Practice problems
1) A car A moves along east with a velocity of 20 ms-1. Another car B moves west with a speed of 40 ms-1. The relative velocity of A wrt B has magnitude
(a)20 ms-1 (b)10 ms-1 (c)60 ms-1 (d) 25 ms-1
Soln) c
Given vA=20i;vB=-40i
The relative velocity of A wrt B; vAB=vA-vB=20i-(-40i)=60i
Hence, the velocity is 60 ms-1
2) A police van is moving on a highway with a speed of 30 kmhr-1 fires a bullet at a thief car which is moving away in the same direction with a speed of 190 kmhr-1. If the muzzle speed of the bullet is 150 ms-1, find the speed of the bullet with respect to the thief's car.
Ans) Muzzle velocity of the bullet vBP=vB-vP=150 185=540 kmhr-1
Velocity of thief’s car, vT=190 kmhr-1
Since bullet is fired from the police car, the velocity of bullet, vB=540+30=570 kmhr-1
Velocity of the bullet with respect to the thief , vBT= vB- vT= 570-190=380 kmhr-1
3)A particle moves along the x-axis. At a time t, the displacement is given by x=40+12t-t3. It’s x coordinate when it comes to rest ?
(a)24 m (b)40 m (c)56 m (d)16 m
Ans) c
When it comes to rest, velocity=0
v=dxdt=ddt(40+12t-t3)=12-3t2=0t=2 s
∴x(at t=2)=40+12(2)-(23)=56 m
4) A car travels with a velocity vu from X to Y. It returns from Y to X with a velocity vd. What is the average velocity of the car during this entire trip?
Solution)
Average velocity =displacementTime taken
Since the displacement for the entire trip is zero, average velocity is zero.
## FAQs
Q.What is velocity?
Ans)Velocity is defined as the change in displacement divided by the change in time.
Q.Is velocity a scalar or vector quantity?
Ans)Velocity is a vector quantity. It has direction.
Q.What is the difference between velocity and speed?
Ans) Velocity is total displacement divided by the total time taken. Speed is the total distance divided by the total time taken. Speed is a scalar quantity whereas velocity is a vector quantity.
Q.Is average velocity greater than average speed?
Ans) Average speed =Total distance Time taken
Average velocity=Total displacement Time taken
Since distance is greater than or equal to the magnitude of displacement, average speed is greater than or equal to average velocity.
## Related Topics:
Relative Motion-Definition, Solved examples, FAQs Non-uniform accelerated motion - Definition, graphs, Solved examples, FAQs Derivation of equations by Graphical method - 3 equations, Solved examples, FAQs Motion under gravity - sign convention, Solved examples, FAQs Graphs- velocity-time, acceleration time, interconversion, Solved examples, FAQs Graphs- position-time, velocity-time, interconversion, Solved examples, FAQs
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Difference between revisions of "2015 AIME I Problems/Problem 10"
Problem
Let $f(x)$ be a third-degree polynomial with real coefficients satisfying $$|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.$$ Find $|f(0)|$.
Solution
Let $f(x)$ = $ax^3+bx^2+cx+d$. Since $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing $12$ and $-12$, it is easy to see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$; otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that $f(1)=12$, and $f(2)=-12$. This provides the following system of equations. $$a + b + c + d = 12$$ $$8a + 4b + 2c + d = -12$$ $$27a + 9b + 3c + d = -12$$ $$125a + 25b + 5c + d = 12$$ $$216a + 36b + 6c + d = 12$$ $$343a + 49b + 7c + d = -12$$ Using any four of these functions as a system of equations yields $d = |f(0)| = \boxed{072}$
Solution 2 (fast)
By drawing the function, and similar to Solution 1, WLOG let $f(1)=f(5)=f(6)=12$. Then, $f(2)=f(3)=f(7)$. Set $g(x)+12=f(x)$. Then the roots of $g(x)$ are $1,5,6$. So, $g(x)=a(x-1)(x-5)(x-6)$. Plug in $x=2$ to find a. We know $$-24=-12-12=f(2)-12=g(2)=a(1)(-3)(-4)=12a$$. So, $a=-2$. Thus, $f(x)=g(x)+12=-2(x-1)(x-5)(x-6)+12$, and then $|f(0)|=60+12=\boxed{072}$.
Solution 3
Without loss of generality, let $f(1) = 12$. (If $f(1) = -12$, then take $-f(x)$ as the polynomial, which leaves $|f(0)|$ unchanged.) Because $f$ is third-degree, write $$f(x) - 12 = a(x - 1)(x - b)(x - c)$$ $$f(x) + 12 = a(x - d)(x - e)(x - f)$$ where $\{b, c, d, e, f \}$ clearly must be a permutation of $\{2, 3, 5, 6, 7\}$ from the given condition. Thus $b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.$ However, subtracting the two equations gives $-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]$, so comparing $x^2$ coefficients gives $1 + b + c = d + e + f$ and thus both values equal to $\dfrac{24}{2} = 12$. As a result, $\{b, c \} = \{5, 6 \}$. As a result, $-24 = a (12)$ and so $a = -2$. Now, we easily deduce that $f(0) = (-2) \cdot (-1) \cdot (-5) \cdot (-6) + 12 = 72,$ and so removing the without loss of generality gives $|f(0)| = \boxed{072}$, which is our answer.
Solution 4
The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$. Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$. Hence $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}$. Note that $g(x)=(f(x)+12)(f(x)-12)$. Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\neq 0$ (else $f$ is not cubic) where $\{q,r,s,t,u,v\}$ is the same as the set $\{1,2,3,5,6,7\}$. Subtracting the second equation from the first, expanding, and collecting like terms, we have that $$24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)$$ which must hold for all $x$. Since $a\neq 0$ we have that (1) $t+u+v=q+r+s$, (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$. Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\geq 12$ with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be $\{2,3,7\}$ and $\{1,5,6\}$. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that $|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2$. Since $a$ is the leading coefficient of $f(x)$, the leading coefficient of $(f(x))^2$ is $a^2=|a|^2=2^2=4$. Thus the leading coefficient of $g(x)$ is 4, i.e. $k=4$. Then from earlier, $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72$ so that the answer is $\boxed{072}$.
Solution 5
Express $f(x)$ in terms of powers of $(x-4)$: $$f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d$$ By the same argument as in the first Solution, we see that $f(x)$ is an odd function about the line $x=4$, so its coefficients $b$ and $d$ are 0. From there it is relatively simple to solve $f(2)=f(3)=-12$ (as in the above solution, but with a smaller system of equations): $$a(1)^3 + c(1) = -12$$ $$a(2)^3 + c(2) = -12$$ $a=2$ and $c=-14$ $$|f(0)| = |2(-4)^3 - 14(-4)| = \boxed{072}$$
Solution 6 (Finite differences)
Because a cubic must come in a "wave form" with two "humps" (Called points of inflections) , we can see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$. By symmetry, $f(4)=0$. Now, WLOG let $f(1)=12$, and $f(2)=f(3)=-12$. Then, we can use finite differences to get that the third (constant) difference is $12$, and therefore $f(0)=12+(24+(24+12))=\boxed{072}$.
Solution 7 (Like solution 1 without annoying systems)
We can rewrite our function as two different cubics, $f(x)=k(x-a)(x-b)(x-c)\pm12=k(x-d)(x-e)(x-f)\mp12$. Note that $k$ is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that $a+b+c=d+e+f$ and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing $a=1$, $b=5$, $c=6$, $d=2$, $e=3$, $f=7$ yields: $$kx^3-12kx^2+41kx-30k\pm12=kx^3-12kx^2+41kx-42k\mp12$$ For the constant terms to have a difference of 24 ($|\pm12-\mp12|$), we must have $k=\pm2$, so the constant term of our polynomial is $\pm72$, the absolute value of which is $\boxed{072}$. -- Solution by eiis1000
Solution 8(godspeed)
Trivially, assume the function can be written as $f(x) = (x-4)(g(x))$, where $g(x)$ is quadratic. Then we set up the following equations using the coefficients of $g(x)$.
$a + b + c = 4$
$4a + 2b + c = -6$
$9a + 3b + c = -12$
Subtract the first equation from the second and the second from the third to get a system of two variables. Solve and plug it in to get $g(x) = 2x^2-16x+18$.$|f(0)| = |18*-4| = | -72| = \boxed{072}$
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# Calculate: f(x)=(6)/(sqrt(4-3x))
## Expression: $f\left( x \right)=\frac{ 6 }{ \sqrt{ 4-3x } }$
Take the derivative of both sides
$f '\left( x \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 6 }{ \sqrt{ 4-3x } } \right)$
Use differentiation rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ a }{ f } \right)=-a \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( f \right) }{ {f}^{2} }$
$f '\left( x \right)=-6 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sqrt{ 4-3x } \right) }{ {\sqrt{ 4-3x }}^{2} }$
Use the chain rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( f\left( g \right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}g} \left( f\left( g \right) \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( g \right)$, where $g=4-3x$, to find the derivative
$f '\left( x \right)=-6 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}g} \left( \sqrt{ g } \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 4-3x \right) }{ {\sqrt{ 4-3x }}^{2} }$
Find the derivative
$f '\left( x \right)=-6 \times \frac{ \frac{ 1 }{ 2\sqrt{ g } } \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 4-3x \right) }{ {\sqrt{ 4-3x }}^{2} }$
Find the derivative of the sum or difference
$f '\left( x \right)=-6 \times \frac{ \frac{ 1 }{ 2\sqrt{ g } } \times \left( -3 \right) }{ {\sqrt{ 4-3x }}^{2} }$
Substitute back $g=4-3x$
$f '\left( x \right)=-6 \times \frac{ \frac{ 1 }{ 2\sqrt{ 4-3x } } \times \left( -3 \right) }{ {\sqrt{ 4-3x }}^{2} }$
Simplify the expression
$f '\left( x \right)=\frac{ 9 }{ \sqrt{ 4-3x }\left( 4-3x \right) }$
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2-basic-integration-how-mathbff
# Interactive video lesson plan for: ❤︎² Basic Integration... How? (mathbff)
#### Activity overview:
For my video on how to integrate using U-SUBSTITUTION, jump to: https://youtu.be/kTXnhwQKuvU
For my video on how to do INTEGRATION BY PARTS, jump to: https://youtu.be/pqKs3zcqJwU
1) POWER RULE: If you're integrating a polynomial, or just a power of x, you can use the Power Rule on each x-term in the polynomial. The Power Rule says that for a term that's just a power of x, such as x^3, you can integrate by raising the power by 1 AND dividing by that number (the new number you got by increasing the power by 1). For example, the integral of x^3 would be (x^4)/4. If there was a constant multiplied in front of the x-power, you can keep the constant and integrate the rest of the term (Constant Multiple Rule). For example, the integral of 6x^2 would be 6 times (x^3)/3, which simplifies to 2x^3. You can repeat these steps to integrate every term in the polynomial and string them together for the full integral. You can also keep the addition and subtraction between the terms (because of the Sum and Difference Rules). NOTE: The anti-derivative of a constant (just a number) is always that number times x (the Constant Rule), so the integral of 1 is 1*x, or x. At the end, it's very important to remember to add a constant of integration, to include a "+ c" at the end of your answer, when it is an indefinite integral (integral with no limits). ANOTHER NOTE: The Power Rule only works when the power is not -1.
2) NEGATIVE & FRACTIONAL POWERS, and RADICALS: For negative powers, you can still use the Power Rule, as long as the power is not -1. For fractional powers, you can also use the Power Rule. When you increase a fractional power by 1, you will have to simplify the power (before integrating) by getting a common denominator. For roots, like the square root of x, it's best to rewrite the radical as a fractional power, and then use the Power Rule.
3) IF THE X POWER IS -1 in the integrand, either written as x^(-1) or 1/x, you have to use a special rule, the LOG RULE, that you can find on a Table of Integrals. It says that the anti-derivative of x^(-1), or (1/x) is the natural log of the absolute value of x, plus c.
4) MORE ALGEBRA: Sometimes you may need to do a bit more algebra before integrating, so that the integrand is in a form that fits a basic integration rule, like the Power Rule. If you're integrating a rational expression, you can sometimes separate the numerator, or break the fraction into separate fractions, simplify each term, and then integrate with the Power Rule. If you have a product of x expressions, you can multiply it out, or distribute the factors so that you have just a polynomial (and can use the Power Rule).
5) TRIG & EXPONENTIAL: You can find a lot of trigonometric (and exponential) integral rules in the Table of Integrals. If you don't see it in the table, you may need to use a trig identity first, to rewrite the integrand into a form that you can integrate.
Tagged under: integrate,integration,integral,calculus,antiderivative,basic integration,rules,integration rules,power rule,indefinite integral,trig,sin,trigonometric,indentities,radicals,root,exponential,differential,dx,derivative,differentiate,khan,patrickjmt,log rule,natural log,fractional power,rational power,constant rule,constant multiple rule,sum difference,table integrals,formula,polynomial,algebra,-, , ,introduction,intro,function,find,problem
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# Simplification and Expansion of Simple Fractions
🏆Practice reduce and expand simple fractions
## Simplification and Expansiono f Simple Fractions
### To amplify fractions
We will perform the same multiplication operation on the numerator and the denominator: the value of the fraction will be preserved.
You can expand as many times as you want and by any number.
### To simplify fractions:
We will perform the same division operation on the numerator and the denominator: the value of the fraction will be preserved.
This can only be done with a number that is completely divisible by both the numerator and the denominator.
It is possible to simplify only until reaching a fraction in which it is not possible to find a number that divides without remainder both in the numerator and the denominator.
## Test yourself on reduce and expand simple fractions!
Simplify the following fraction by a factor of 4:
$$\frac{4}{8}=$$
## Simplify and expand simple fractions
Simplifying and amplifying fractions is an easy and enjoyable topic that will accompany you in almost all exercises with fractions.
Simplifying and amplifying a fraction is actually a multiplication or division operation that is performed on the fraction so that the real value of the fraction does not change and simply looks different.
Expansion: a multiplication operation (which is performed on both the numerator and the denominator).
Simplification: a division operation (which is performed on both the numerator and the denominator).
### Expansion of Fractions
#### A daily example
Think of a pizza with $8$ slices: a normal family-sized pizza sold at pizzerias.
You call to place the order and ask the seller to put olives on half of the pizza.
You will receive $1 \over 2$ pizza with olives.
But, what if you told the seller, please put olives on $4$ triangles of the pizza?
Let's see:
Even then you would get half a pizza with olives.
That is to say: $4 \over 8$ -> four slices out of $8$ is equal to $1 \over 2$
Now let's see this in the exercise
We expand the fraction $1 \over 2$ by $4$
Solution:
We multiply by $2$ both in the numerator and the denominator and we get:
Note the sign that is usually used to expand.
If we multiply both the numerator and the denominator by the same number, the value of the fraction will not change and the fractions will be equal, exactly as in the pizza.
Important note: we can expand the fraction as many times as we want and by any number we want and still the value will not change!
For example, even if instead of being expanded by $4$ we would expand by $2$, we would get:
which is also equal to $4 \over 8$
#### More exercises of another type
Complete the missing number
$\frac{2}{8}=\frac{4}{□}$
Solution:
We see that $2$ in the numerator becomes $4$. That is, it is multiplied by $2$. Therefore, we also multiply the denominator $8$ by the number $2$ to reach the correct result.
After all, we learned that fractions will be equal only if we perform the operation on both the numerator and the denominator.
We obtain:
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### Now we move on to the simplification of fractions
Simplifying fractions is a division operation that is performed on both the numerator and the denominator and preserves the value of the fraction.
It is possible to simplify only by a number that is divisible without remainder in both the numerator and the denominator.
When can we not simplify a fraction?
When the numerator and the denominator are not completely divisible by the same number.
#### Exercise
Simplify the fraction $6 \over 12$ by $6$.
Solution:
We divide both the numerator and the denominator by $6$ and we get:
$\frac{1}{2}=\frac{6}{12}$
Extra section:
We simplify the fraction $6 \over 12$ by the number you choose (except for $6$)
Solution:
We must choose a number that divides both the numerator and the denominator without remainder.
We are asked: What number can you divide $6$ and $12$ by?
The answer can be: $2$ or $3$. For example
We choose: $3$ and simplify:
$\frac{2}{4}=\frac{1}{2}=\frac{6}{12}$
#### Another exercise
We simplify the fraction: $3 \over 5$
Solution:
This exercise has no solution. This fraction cannot be simplified further since there is no number that is divisible by $3$ and also by $5$ without a remainder.
#### Another exercise
Complete the missing number
$\frac{8}{10}=\frac{4}{□}$
Solution:
We see that $8$ in the numerator becomes $4$. That is, it is divided by $2$. Therefore, we also divide in the denominator $10$ by the number $2$ to arrive at the correct result since we learned that fractions will be equal only if we perform the operation both in the numerator and in the denominator.
We obtain:
Note: Expansion does not mean that the fraction gets bigger and simplification does not mean that the fraction gets smaller!
## Examples and exercises with solutions for simplifying and expanding simple fractions
### Exercise #1
Simplify the following fraction by a factor of 4:
$\frac{4}{8}=$
### Step-by-Step Solution
Let's simplify as follows, we'll divide both the numerator by 4 and the denominator by 4:
$\frac{4:4}{8:4}=\frac{1}{2}$
$\frac{1}{2}$
### Exercise #2
Simplify the following fraction by a factor of 1:
$\frac{3}{10}=$
### Step-by-Step Solution
We will reduce in the following way, divide the numerator by 1 and the denominator by 1:
$\frac{3:1}{10:1}=\frac{3}{10}$
$\frac{3}{10}$
### Exercise #3
Simplify the following fraction:
$\frac{2}{10}=$
### Step-by-Step Solution
Let's simplify as follows, we'll divide both the numerator by 2 and the denominator by 2:
$\frac{2:2}{10:2}=\frac{1}{5}$
$\frac{1}{5}$
### Exercise #4
Simplify the following fraction:
$\frac{4}{16}=$
### Step-by-Step Solution
We will reduce in the following way, divide the numerator by 4 and the denominator by 4:
$\frac{4:4}{16:4}=\frac{1}{4}$
$\frac{1}{4}$
### Exercise #5
Simplify the following fraction by a factor of 3:
$\frac{3}{6}=$
### Step-by-Step Solution
We will reduce as follows, divide the numerator by 3 and the denominator by 3:
$\frac{3:3}{6:3}=\frac{1}{2}$
$\frac{1}{2}$
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# How do you solve 6x^2 - x -3 =0?
Nov 22, 2015
$x \approx - 0.63 , \textcolor{w h i t e}{\times} x \approx 0.79$
#### Explanation:
It is a matter of practice so that you get used to spotting the factors.
The only 2 factors you are going to get for 3 is 1 and 3. This is because 3 is a prime number. The thing is that we have -3 so one has to be positive and the other negative.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Factors of 3 are only {1,3} because 3 is a prime number
Factors of 6 are {1,6} , {2,3}
Looking for -1 in $- x$
These factors do not work so revert to the standard solution formula:
$a {x}^{2} + b x + c = 0$
where
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$a = 6$
$b = - 1$
$c = - 3$
$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(6\right) \left(- 3\right)}}{2 \left(6\right)}$
$\textcolor{g r e e n}{\text{The use of brackets round negative numbers reduce}}$
$\textcolor{g r e e n}{\text{the chance of making a mistake!}}$
$x = \frac{1 \pm \sqrt{1 + 72}}{12}$
$x = \frac{1}{12} \pm \frac{\sqrt{73}}{12}$
$x \approx 0.08 \dot{3} \pm 0.71$
$x \approx - 0.63 , x \approx 0.79$
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# 2004 AIME II Problems/Problem 11
## Problem
A right circular cone has a base with radius $600$ and height $200\sqrt{7}.$ A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is $125$, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}.$ Find the least distance that the fly could have crawled.
## Solution
The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$-axis and the angle $\theta$ going counterclockwise. The circumference of the base is $C=1200\pi$. The sector's radius (cone's sweep) is $R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{360000+280000}=\sqrt{640000}=800$. Setting $\theta R=C\implies 800\theta=1200\pi\implies\theta=\frac{3\pi}{2}$.
If the starting point $A$ is on the positive $x$-axis at $(125,0)$ then we can take the end point $B$ on $\theta$'s bisector at $\frac{3\pi}{4}$ radians along the $y=-x$ line in the second quadrant. Using the distance from the vertex puts $B$ at $(-375,-375)$. Thus the shortest distance for the fly to travel is along segment $AB$ in the sector, which gives a distance $\sqrt{(-375-125)^2+(-375-0)^2}=125\sqrt{4^2+3^2}=\boxed{625}$.
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# Prove that if the altitude and median of a triangle form equal angles with sides then the triangle is right.
Problem statement:
Prove that if the altitude and median drawn from the same vertex of a nonisosceles triangle lie inside the triangle and form equal angles with its sides, then this is a right triangle.
After many attempts, I came up with this: let $ABC$ be the triangle where $CH$ is the altitude, $CM$ is the median, and name the angles $ACH = MCB = \theta$, $CAH = \alpha$, $CBM = \beta$ and $HCM = \gamma$. Using the sine theorem in the triangle $CMB$ we get $$\frac{MB}{\sin(\theta)} = \frac{MC}{\sin(\beta)},$$ while using the sine theorem in the triangle $ACM$ we get $$\frac{MC}{\sin(\alpha)} = \frac{MA}{\sin(\theta + \gamma)}.$$ Combining these equations I found $$\sin(\alpha) \sin(\theta) = \sin(\beta) \sin(\theta + \gamma).$$ Applying the identity $$\sin(a)\sin(b) = \frac{\cos(a-b) - \cos(a+b)}{2}$$ I concluded $$\cos(\alpha - \theta) = \cos(\theta + \gamma - \beta).$$ Since all angles are within $[0, \pi]$ range I concluded $$\alpha - \theta = \theta + \gamma - \beta,$$ thus $2 \theta + \gamma = \alpha + \beta$. From the original triangle we know $$2 \theta + \gamma + \alpha + \beta = \pi,$$ and combining the equations proves the triangle is right.
Is this correct? Is there a synthetic way to do it?
• Here's a late response, using synthetic geometry; I only recently noticed this interesting post. May 8, 2018 at 16:48
It can also be proved using geometry only.
1] Through $$M$$, draw $$MX \parallel BC$$ cutting $$AC$$ at $$X$$. From that, we have
• (1.1) $$\beta = \beta’$$; and
• (1.2) $$AX = XC$$.
2] Through $$X$$, draw $$XYZ \parallel AB$$ cutting $$CH$$ at $$Y$$ and $$CM$$ at $$Z$$. From that, we have
• (2.1) $$XY$$ is the perpendicular bisector of $$CH$$; and
• (2.2) $$Z$$ is the midpoint of $$CM$$.
(2.1) + (1.1) implies $$\alpha’ = \alpha = \beta = \beta’$$. Since $$H$$ and $$M$$ are distinct ($$\triangle ABC$$ is not isosceles), the inscribed angle theorem means $$XHMC$$ is a cyclic quadrilateral. This also gives us that $$\angle MXC = \angle MHC = 90^\circ$$.
Then since $$MX \parallel BC$$, also $$\angle ACB = 90^\circ$$ as required.
• Could you explain why $CM$ is the diameter of the dotted circle? Why is $\angle HXC$ a right angle? Why is $\angle ACB$ a right angle?
– robjohn
Jan 5, 2021 at 22:32
• @robjohn CH is the given as the altitude of $\triangle ABC$. This means $\angle CHM = 90^0$. By converse of angle in semi-circle, CM must be the diameter of the circle passing through CHM.
– Mick
Jan 6, 2021 at 12:46
• Okay, I agree that $CM$ is the diameter of the circle (if this explanation were included in your answer, it would be a lot clearer), but it would seem that $\angle MXC$ would be a right angle, not $\angle HXC$ (it would help if $X$ were actually shown on the circle). It would be helpful to mention the role that the angle between $\alpha$ and $\beta$ plays.
– robjohn
Jan 6, 2021 at 13:21
• @robjohn The point that I want to make is ‘initially we don’t know whether X is one of the con-cyclic point of the dotted circle’. That is why it was intentionally not drawn on the circle. You are right about the typo of $\angle HXC$. Fixed.
– Mick
Jan 6, 2021 at 16:57
• (+1) I have added a section to my answer clarifying what seemed confusing to me.
– robjohn
Jan 7, 2021 at 21:30
Let $\triangle ABC$ be a right triangle in the semi-circle with diameter $AB$. Then with altitude $CD$ and median $CM$, since $\triangle CMB$ is isosceles$$\angle \beta=\angle \gamma$$But since by Euclid [Elements, VI, 8]$$\triangle ACD\sim\triangle CBD$$with$$\angle\alpha=\angle\gamma$$therefore$$\angle\alpha=\angle\beta$$Hence in a right triangle, the altitude and median from the vertex of the right angle make equal angles with the adjacent sides.
Now we must prove the converse: If the altitude and median from the vertex of a triangle make equal angles with the adjacent sides, the angle at the vertex is right.
First let the angle at $C$ be obtuse, and suppose, in the figure below, that altitude $CD$ and median $CM$ make $$\angle\alpha=\angle\beta$$
Construct a semi-circle on $AB$ as diameter, extend $AC$ to $E$ on the circumference, join $EB$, $EM$, and drop altitude $EF$.
Then, as shown above, since $\triangle ABE$ is right$$\angle\delta=\angle\epsilon$$And since$$EF\parallel CD$$then$$\angle\delta=\angle\alpha$$and it follows that$$\angle\epsilon=\angle\beta$$making$$\triangle EGB\sim\triangle CGM$$and $EBMC$ therefore a cyclic quadrilateral.
But since $\angle CEB$ is right, $CEBM$ is cyclic only if opposite $\angle CMB$ is also right, that is if $D$ coincides with $M$, making $\triangle ACB$ isosceles, as in the figure below.
But this contradicts the given condition that the triangle is not isosceles.
Therefore, (in the second figure) $EBMC$ is not a cyclic quadrilateral, $\triangle EGB$ and $\triangle CGM$ are not similar, and$$\angle\epsilon\ne\angle\beta$$
Therefore, in obtuse $\triangle ACB$$\angle\alpha\ne\angle\beta$$ And by the same reductio argument it can be shown that$\angle\alpha\ne\angle\beta$when the angle at$C\$ is acute.
If the altitude and median from the vertex of a triangle make equal angles with the adjacent sides, the angle at the vertex is right.
Trigonometric Approach
We are given that $$\angle MAC=\angle BAP$$ and since $$\triangle BPA$$ is a right triangle, $$\angle BAP=\frac\pi2-B$$. Therefore, $$\angle MAC=\frac\pi2-B\tag1$$ and, since $$\angle MAC+\angle BAM=A$$, we have $$\angle BAM=A+B-\frac\pi2\tag2$$ Since $$M$$ is the bisector of $$\overline{BC}$$, the areas of $$\triangle BAM$$ and $$\triangle MAC$$ are equal. Thus, \begin{align} \overbrace{\frac12cm\sin\left(A+B-\frac\pi2\right)}^{\left|\triangle BAM\right|} &=\overbrace{\frac12bm\sin\left(\frac\pi2-B\right)}^{\left|\triangle MAC\right|}\tag{3a}\\ c\cos(C)&=b\cos(B)\tag{3b}\\[3pt] c\sin(B)&=b\sin(C)\tag{3c}\\[3pt] \sin(2B)&=\sin(2C)\tag{3d} \end{align} Explanation:
$$\text{(3b)}$$: apply $$A+B-\frac\pi2=\frac\pi2-C$$ and multiply by $$\frac2m$$
$$\text{(3c)}$$: Law of Sines
$$\text{(3d)}$$: cross multiply $$\text{(3b)}$$ and$$\text{(3c)}$$ and multiply by $$\frac2{bc}$$
Since $$\triangle ABC$$ is not isosceles, $$\text{(3d)}$$ implies that $$2B=\pi-2C$$; which, since $$A=\pi-B-C$$, means $$\bbox[5px,border:2px solid #C0A000]{A=\frac\pi2}\tag4$$
Geometric Approach
This is an approach similar, but slightly different, to that in Mick's answer.
Start with the diagram above. We are given that $$\angle MAC$$ is equal to $$\angle BAP$$, which is $$\frac\pi2-B$$ since $$\angle APB$$ is a right angle.
Add $$N$$ and $$R$$, the midpoints of $$\overline{AC}$$ and $$\overline{AB}$$ respectively. Then add the segments from $$R$$ to $$N$$, $$M$$, and $$P$$.
Add the circle with $$\overline{AM}$$ as its diameter. Since $$\angle APM$$ is a right angle, $$P$$ sits on this circle. Since $$\overline{RN}$$ is parallel to $$\overline{BC}$$, $$\overline{RN}$$ is also perpendicular to $$\overline{AP}$$.
Since $$\triangle ARN$$ is similar to $$\triangle ABC$$ with a linear ratio of $$\frac12$$, $$Q$$, the intersection of $$\overline{AP}$$ and $$\overline{RN}$$, bisects $$\overline{AP}$$. Thus, $$\angle RPQ$$ is also $$\frac\pi2-B$$.
Since $$\overline{RM}$$ is parallel to $$\overline{AC}$$, $$\angle RMA$$ is equal to $$\angle MAC$$, which equals $$\frac\pi2-B$$.
The locus of points at which $$\overline{AR}$$ subtends an angle of $$\frac\pi2-B$$ is the circle which contains $$A$$, $$R$$, $$P$$, and $$M$$. This is the same circle that was added above since it contains $$A$$, $$P$$, and $$M$$. Thus, $$\angle ARM$$ is a right angle, and since $$\overline{RM}$$ is parallel to $$\overline{AC}$$, so is $$A$$.
• Great answer! Did you keep this question in your backlog and then came back to it? It's been a while. Jan 6, 2021 at 20:16
• Thanks! Actually, it showed up on the main list. Once I realized how old the question was, I looked for the edit that put it onto the main list. It turns out it was an automatic bump by the system.
– robjohn
Jan 6, 2021 at 21:35
Extend median $$\overline{AM}$$ of $$\triangle ABC$$ to meet the circumcircle at $$M'$$, and let $$D$$ be the foot of the altitude from $$A$$. (Note that $$D$$ and $$M$$ are distinct —and, thus, $$\overline{AM'}\not\perp\overline{BC}$$— because we assume $$\triangle ABC$$ is non-isosceles.)
Since $$\angle M' \cong\angle B$$ (by the Inscribed Angle Theorem), we conclude $$\angle ACM'\cong\angle ADB=90^\circ$$, so that $$\overline{AM'}$$ is a diameter. Now, a diameter can only bisect a non-perpendicular chord at its own midpoint (aka, the circumcenter); consequently, $$M$$ is that circumcenter, $$\overline{BC}$$ is a diameter, and $$\angle BAC$$ is a right angle. $$\square$$
• Amazing synthetic answer. Thanks for your contribution! Jan 10, 2021 at 16:29
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25 is a perfect square number. 25 is likewise the amount of the first five odd natural numbers (1+ 3 + 5 + 7+ 9). Let’s consider the multiples that 25. If you think about the number line, friend will need to jump every 25 places to get the multiples of 25.
First five multiples of 25: 25, 50, 75, 100, 125Prime administrate of 25: 5 × 5
1.You are watching: 25+25+25 What space the Multiples that 25? 2. First 20 Multiples that 25 3. Important Notes 4. FAQs on Multiples that 25
To understand the principle of detect multiples, let us take a couple of more examples.
25 × 1= 25 25 × 2 = 50 25 × 3 = 75 25 × 4 = 100 25 × 5 = 125 25 × 6 = 150 25 × 7 = 175 25 × 8 = 200 25 × 9 = 225 25 × 10 = 250 25 × 11 = 275 25 × 12 = 300 25 × 13 = 325 25 × 14 = 350 25 × 15 = 375 25 × 16 = 400 25 × 17 = 425 25 × 18 = 450 25 × 19 = 475 25 × 20 = 500
You can obtain the multiples of 25 by multiplying 25 through the integers.You can achieve the nth lot of of 25 by multiplying n with 25.Multiples the 50 and 100 space all multiples that 25.
Oliver has 2050 notebooks. Deserve to he group them in a stack of 25 notebooks equally?Amber has actually to uncover the multiples of 15. Is 25 one of its multiples?
Example 1: Stuart have the right to bake 25 various kinds the bread. Every week, that distributes about 3000 loaves of bread to various restaurants. Deserve to he division them into teams of 25?
Solution:
He distributes about 3000 loaves of breads.
_____ groups of 25 = 3000 loaves
_____× 25 = 3000
This means 3000 ÷ 25 = 120 groups
120 × 25 = 3000
Therefore, 3000 loaves can be separated into 120 teams of 25 lump each.
Example 2: Andy has to recognize whether the number of weeks in a year can be group as multiples that 25? If not, then uncover out how many weeks should be diminished or added so the they have the right to be grouped as multiples of 25?
Solution:
There are 52 mainly in a year.
To determine whether the weeks in a year can be group as multiples that 25, Andy will usage the complying with steps:
52 ÷ 25 = 2, Remainder = 2 weeks
There space 2 extra main in a year as result of which they can not be grouped together multiples that 25.
Math will no longer be a tough subject, particularly when you know the ideas through visualizations.
### How carry out you uncover the multiples the 25?
The multiples the 25 are acquired by multiplying 25 through other numbers. For example, 25 x 15 = 375. 375 is a multiple of 25.
### Is 25 a multiple of 2 or 5? just how do friend know?
25 is a lot of of 5 as the is divisible through 5. 25 is no a lot of of 2 as it is no divisible by 2.
### Are multiples that 25 also multiples the 1000?
Multiples the 25 room 25, 50, 75, 100, .....1000, 1025,....1975, 2000,....
Multiples the 1000 are 1000, 2000, 3000, 4000, 5000 and also so on.
All the multiples of 1000 space multiples that 25, however all the multiples the 25 space not multiples that 1000.
### What is the 15th lot of of 25?
The 15multiple that 25 is 15 × 25 = 375.
See more: How To Dry Fleshlight : Here"S How To Keep It Fresh And Feeling Great
### How perform you create multiples the 25?
To get the multiples the 25, you have to multiply 25 by any kind of integer and the result obtained will certainly be the multiples the 25.
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# Percent Increase and Percent Decrease Problems
Percent Increase and Percent Decrease Problems. Percent Decrease. A sofa regularly sells for \$700.00. Today it is on sale for \$630. What is the percent decrease of the price of the sofa?. Percent Decrease. A sofa regularly sells for \$700.00.
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## Percent Increase and Percent Decrease Problems
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### Presentation Transcript
1. Percent Increase and Percent Decrease Problems
2. Percent Decrease A sofa regularly sells for \$700.00. Today it is on sale for \$630. What is the percent decrease of the price of the sofa?
3. Percent Decrease A sofa regularly sells for \$700.00. Today it is on sale for \$630. What is the percent decrease of the price of the sofa? First, find the amount of the decrease. 700.00 – 630.00 = 70.00
4. Percent Decrease A sofa regularly sells for \$700.00. Today it is on sale for \$630. What is the percent decrease of the price of the sofa? The amount of the decrease is 70.00. If you like making an equation to solve, The question is “70.00 is what percent of 700”?
5. Percent Decrease A sofa regularly sells for \$700.00. Today it is on sale for \$630. What is the percent decrease of the price of the sofa? “70.00 is what percent of 700”? 70 = x(700) divide by 700 x = 1/10 or .10 or 10%
6. Percent Decrease • “70.00 is what percent of 700”? • If you prefer using the percent proportion: • P = P that’s what we don’t know • B = 700 “of 700” • A = 70
7. Percent Decrease
8. When asked to calculate the percent increase or decrease: Figure out the amount of the increased or decrease Divide by the original amount If it went up, it’s an increase. If it went down, it’s a decrease.
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# PRESENTATION 9 Ratios and Proportions
## Presentation on theme: "PRESENTATION 9 Ratios and Proportions"— Presentation transcript:
PRESENTATION 9 Ratios and Proportions
RATIOS A ratio is the comparison of two like quantities
The terms of a ratio must be compared in the order in which they are given Terms must be expressed in the same units The first term is the numerator of a fraction, and the second term is the denominator A ratio should be expressed in lowest fractional terms
RATIOS Ratios are expressed in two ways:
With a colon between the terms, such as 4 : 9 This is read as “4 to 9” With a division sign separating the two numbers, such as 4 ÷ 9 or
RATIOS Example: Express 5 to 15 as a ratio in lowest terms
Write the ratio as a fraction and reduce The ratio is 1 : 3
RATIOS Example: Express 10 to as a ratio in lowest terms Divide
The ratio is 12 : 1
PROPORTIONS A proportion is an expression that states the equality of two ratios Proportions are expressed in two ways As 3 : 4 = 6 : 8, which is read as “3 is to 4 as 6 is to 8” As , which is the equation form
PROPORTIONS A proportion consists of four terms
The first and fourth terms are called extremes The second and third terms are called means In the proportion 3 : 4 = 6 : 8, 3 and 8 are the extremes and 4 and 6 are the means The product of the means equals the product of the extremes (if the terms are cross-multiplied, their products are equal)
PROPORTIONS Example: Solve the proportion below for F:
Cross multiply: F = 6.2(9.8) Divide both sides by 21.7: Therefore F = 2.8
DIRECT PROPORTIONS Two quantities are directly proportional if a change in one produces a change in the other in the same direction When setting up a direct proportion in fractional form: Numerator of the first ratio must correspond to the numerator of the second ratio Denominator of the first ratio must correspond to the denominator of the second ratio
DIRECT PROPORTIONS Example: A machine produces 280 pieces in 3.5 hours. How long does it take to produce 720 pieces? Analyze: An increase in the number of pieces produced (from 280 to 720) requires an increase in time Time increases as production increases; therefore, the proportion is direct
DIRECT PROPORTIONS Set up the proportion and let t represent the time required to produce 720 pieces The numerator of the first ratio corresponds to the numerator of the second ratio (280 pieces to 3.5 hours) The denominator of the first ratio corresponds to the denominator of the second ratio (720 pieces to t)
DIRECT PROPORTIONS Solve for t:
It will take 9 hours to produce 720 pieces
INVERSE PROPORTIONS Two quantities are inversely or indirectly proportional if a change in one produces a change in the other in the opposite direction Two quantities are inversely proportional if An increase in one produces a decrease in the other A decrease in one produces an increase in the other
INVERSE PROPORTIONS When setting up an inverse proportion in fractional form: The numerator of the first ratio must correspond to the denominator of the second ratio The denominator of the first ratio must correspond to the numerator of the second ratio
INVERSE PROPORTIONS Example: Five identical machines produce the same parts at the same rate. The 5 machines complete the required number of parts in 1.8 hours. How many hours does it take 3 machines to produce the same number of parts? Analyze: A decrease in the number of machines (from 5 to 3) requires an increase in time Time increases as the number of machines decrease and this is an inverse proportion
INVERSE PROPORTIONS Let x represent the time required by 3 machines to produce the parts The numerator of the first ratio corresponds to the denominator of the second ratio; 5 machines corresponds to 1.8 hours The denominator of the first ratio corresponds to the numerator of the second ratio; 3 machines corresponds to x
INVERSE PROPORTIONS Solve for x: It will take 3 hours
PRACTICAL PROBLEMS A piece of lumber 2.8 meters long weighs 24.5 kilograms A piece 0.8 meters long is cut from the 2.8-meter length Determine the weight of the 0.8- meter piece
PRACTICAL PROBLEMS Analyze: Since the weight of 0.8 meters is less than the total weight of the piece of lumber, this is a direct proportion Set up the proportion and let x represent the weight of the 0.8-meter piece
PRACTICAL PROBLEMS Solve for x: The piece of lumber weighs 7 kilograms
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# Calculus 1 : How to find acceleration
## Example Questions
### Example Question #11 : How To Find Acceleration
A car is moving at a constant speed of miles per hour. What is the acceleration after hours?
Explanation:
The car is moving at a constant speed of 40 miles per hour, so the velocity function is:
The derivative of the velocity function is the acceleration function.
The acceleration at any particular time is zero.
### Example Question #11 : How To Find Acceleration
The position of an object, in meters, is given by the following equation:
Find the acceleration of the object.
Explanation:
Velocity is the derivative of position, and acceleration is the derivative of velocity, so acceleration is the second derivative of position. With that in mind, all we have to do to find the acceleration of the object is take the derivative of the equation for its position twice.
### Example Question #13 : How To Find Acceleration
The velocity of an object is given by the following equation:
Find the equation for the acceleration of the object.
Explanation:
Acceleration is the derivative of velocity, so in order to find the equation for the object's acceleration, we must take the derivative of the equation for its velocity:
We will use the power rule to find the derivative which states:
### Example Question #14 : How To Find Acceleration
The acceleration of an object is given by the folowing indefinite integral:
If , find the acceleration of the object at seconds.
Explanation:
In order to find a(3), our first step is to evaluate the integral in the equation for acceleration:
Now we use the initial acceleration, a(0)=0.1, to solve for the constant C:
So if C=0.1, then our final equation for acceleration is as follows, which we can then plug t=3 into to find the acceleration of the object after 3 seconds, a(3):
### Example Question #2 : Derivatives
The position of an object is described by the following equation:
Find the acceleration of the object at second.
Explanation:
Acceleration is the second derivative of position, so we must first find the second derivative of the equation for position:
Now we can plug in t=1 to find the acceleration of the object after 1 second:
### Example Question #15 : How To Find Acceleration
The velocity of an object is given by the following equation:
Find the acceleration of the object at seconds.
Explanation:
Acceleration is the derivative of velocity, so we must take the derivative of the given equation to find an equation for acceleration:
Now we can plug in t=2 to find the acceleration of the object at 2 seconds:
### Example Question #16 : How To Find Acceleration
The position vector of an object moving in a plane is given by . Find the object's acceleration when .
Explanation:
To find the acceleration, we must differentiate the position vector twice. Differentiating the position vector once gives the velocity vector:
Differentiating the second time gives acceleration:
Using 3 as the value for gives
### Example Question #17 : How To Find Acceleration
If the position of a particle is: , What is the acceleration at ?
Explanation:
To find the acceleration equation of the particle, we can differentiate the position equation twice.
Differentiating once gives the velocity equation:
Differentiating again give the acceleration equation:
Using 3 as the value for ,
### Example Question #18 : How To Find Acceleration
What is the acceleration at time when the acceleration is given by is time in seconds.
Explanation:
To find the acceleration at time 4 seconds, we simply subsitute in 4 seconds into the acceleration function.
### Example Question #19 : How To Find Acceleration
The velocity function is .
What is the acceleration function?
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# Solving Equations Algebraically
#### Scatter Plot
Contents: This page corresponds to § 2.4 (p. 200) of the text.
Suggested Problems from Text:
p. 212 #7, 8, 11, 15, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, 60, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97
### Equations Involving Fractional Expressions or Absolute Values
A quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0.
### Factoring
This approach to solving equations is based on the fact that if the product of two quantities is zero, then at least one of the quantities must be zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, see the review section P.3 (p.26) of the text.
Example 1.
2x2 - 5x - 12 = 0.
(2x + 3)(x - 4) = 0.
2x + 3 = 0 or x - 4 = 0.
x = -3/2, or x = 4.
### Square Root Principle
If x2 = k, then x = ± sqrt(k).
Example 2.
x2 - 9 = 0.
x2 = 9.
x = 3, or x = -3.
Example 3.
Example 4.
x2 + 7 = 0.
x2 = -7.
x = ± .
Note that = = , so the solutions are
x = ± , two complex numbers.
### Completing the Square
The idea behind completing the square is to rewrite the equation in a form that allows us to apply the square root principle.
Example 5.
x2 +6x - 1 = 0.
x2 +6x = 1.
x2 +6x + 9 = 1 + 9.
The 9 added to both sides came from squaring half the coefficient of x, (6/2)2 = 9. The reason for choosing this value is that now the left hand side of the equation is the square of a binomial (two term polynomial). That is why this procedure is called completing the square. [ The interested reader can see that this is true by considering (x + a)2 = x2 + 2ax + a2. To get "a" one need only divide the x-coefficient by 2. Thus, to complete the square for x2 + 2ax, one has to add a2.]
(x + 3)2 = 10.
Now we may apply the square root principle and then solve for x.
x = -3 ± sqrt(10).
Example 6.
2x2 + 6x - 5 = 0.
2x2 + 6x = 5.
The method of completing the square demonstrated in the previous example only works if the leading coefficient (coefficient of x2) is 1. In this example the leading coefficient is 2, but we can change that by dividing both sides of the equation by 2.
x2 + 3x = 5/2.
Now that the leading coefficient is 1, we take the coefficient of x, which is now 3, divide it by 2 and square, (3/2)2 = 9/4. This is the constant that we add to both sides to complete the square.
x2 + 3x + 9/4 = 5/2 + 9/4.
The left hand side is the square of (x + 3/2). [ Verify this!]
(x + 3/2)2 = 19/4.
Now we use the square root principle and solve for x.
x + 3/2 = ± sqrt(19/4) = ± sqrt(19)/2.
x = -3/2 ± sqrt(19)/2 = (-3 ± sqrt(19))/2
So far we have discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right form to apply the square root principle may be rearranged and solved by factoring as we see in the next example.
Example 7.
x2 = 16.
x2 - 16 = 0.
(x + 4)(x - 4) = 0.
x = -4, or x = 4.
In some cases the equation can be solved by factoring, but the factorization is not obvious.
The method of completing the square will always work, even if the solutions are complex numbers, in which case we will take the square root of a negative number. Furthermore, the steps necessary to complete the square are always the same, so they can be applied to the general quadratic equation
ax2 + bx + c = 0.
The result of completing the square on this general equation is a formula for the solutions of the equation called the Quadratic Formula.
The solutions for the equation ax2 + bx + c = 0 are
We are saying that completing the square always works, and we have completed the square in the general case, where we have a,b, and c instead of numbers. So, to find the solutions for any quadratic equation, we write it in the standard form to find the values of a, b, and c, then substitute these values into the Quadratic Formula.
One consequence is that you never have to complete the square to find the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so you still need to know how to do it!
Example 8.
2x2 + 6x - 5 = 0.
In this case, a = 2, b = 6, c = -5. Substituting these values in the Quadratic Formula yields
Notice that we solved this equation earlier by completing the square.
Note: There are two real solutions. In terms of graphs, there are two intercepts for the graph of the function f(x) = 2x2 + 6x - 5.
Example 9.
4x2 + 4x + 1 = 0
In this example a = 4, b = 4, and c = 1.
• There is only one solution. In terms of graphs, this means there is only one x-intercept.
• The solution simplified so that there is no square root involved. This means that the equation could have been solved by factoring. (All quadratic equations can be solved by factoring! What I mean is it could have been solved easily by factoring.)
4x2 + 4x + 1 = 0.
(2x + 1)2 = 0.
x = -1/2.
Example 10.
x2 + x + 1 = 0
a = 1, b = 1, c = 1
Note: There are no real solutions. In terms of graphs, there are no intercepts for the graph of the function f(x) = x2 + x + 1. Thus, the solutions are complex because the graph of y = x2 + x + 1 has no x-intercepts.
The expression under the radical in the Quadratic Formula, b2 - 4ac, is called the discriminant of the equation. The last three examples illustrate the three possibilities for quadratic equations.
1. Discriminant > 0. Two real solutions.
2. Discriminant = 0. One real solution.
3. Discriminant < 0. Two complex solutions.
Notes on checking solutions
None of the techniques introduced so far in this section can introduce extraneous solutions. (See example 3 from the Linear Equations and Modeling section.) However, it is still a good idea to check your solutions, because it is very easy to make careless errors while solving equations.
The algebraic method, which consists of substituting the number back into the equation and checking that the resulting statement is true, works well when the solution is "simple", but it is not very practical when the solution involves a radical.
For instance, in our next to last example, 4x2 + 4x + 1 = 0, we found one solution x = -1/2.
The algebraic check looks like
4(-1/2)2 +4(-1/2) + 1 = 0.
4(1/4) - 2 + 1 = 0.
1 - 2 + 1 = 0.
0 = 0. The solution checks.
In the example before that, 2x2 + 6x - 5 = 0, we found two real solutions, x = (-3 ± sqrt(19))/2. It is certainly possible to check this algebraically, but it is not very easy. In this case either a graphical check, or using a calculator for the algebraic check are faster.
First, find decimal approximations for the two proposed solutions.
(-3 + sqrt(19))/2 = 0.679449.
(-3 - sqrt(19))/2 = -3.679449.
Now use a graphing utility to graph y = 2x2 + 6x - 5, and trace the graph to find approximately where the x-intercepts are. If they are close to the values above, then you can be pretty sure you have the correct solutions. You can also insert the approximate solution into the equation to see if both sides of the equation give approximately the same values. However, you still need to be careful in your claim that your solution is correct, since it is not the exact solution.
Note that if you had started with the equation 2x2 + 6x - 5 = 0 and gone directly to the graphing utility to solve it, then you would not get the exact solutions, because they are irrational. However, having found (algebraically) two numbers that you think are solutions, if the graphing utility shows that intercepts are very close to the numbers you found, then you are probably right!
Exercise 1:
(a) 3x2 -5x - 2 = 0. Answer
(b) (x + 1)2 = 3. Answer
(c) x2 = 3x + 2. Answer
Equations with radicals can often be simplified by raising to the appropriate power, squaring if the radical is a square root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked.
If there is only one radical in the equation, then before raising to a power, you should arrange to have the radical term by itself on one side of the equation.
Example 11.
Now that we have isolated the radical term on the right side, we square both sides and solve the resulting equation for x.
Check:
x = 0
When we substitute x = 0 into the original equation we get the statement 0 = 2, which is not true!
So, x = 0 is not a solution.
x = 3
When we substitute x = 3 into the original equation, we get the statement 3 = 3. This is true, so x = 3 is a solution.
Solution: x = 3.
Note: The solution is the x-coordinate of the intersection point of the graphs of y = x and y = sqrt(x+1)+1.
Look at what would have happened if we had squared both sides of the equation before isolating the radical term.
This is worse than what we started with!
If there is more than one radical term in the equation, then in general, we cannot eliminate all radicals by raising to a power one time. However, we can decrease the number of radical terms by raising to a power.
If the equation involves more than one radical term, then we still want to isolate one radical on one side and raise to a power. Then we repeat that process.
Example 12.
Now square both sides of the equation.
This equation has only one radical term, so we have made progress! Now isolate the radical term and then square both sides again.
Check:
Substituting x = 5/4 into the original equation yields
sqrt(9/4) + sqrt(1/4) = 2.
3/2 + 1/2 = 2.
This statement is true, so x = 5/4 is a solution.
Note on checking solutions:
The algebraic check was easy to do in this case. However, the graphical check has the advantage of showing that there are no solutions that we have not found, at least within the scope of the viewing rectangle. The solution is the x-coordinate of the intersection point of the graphs of y = 2 and y = sqrt(x+1)+sqrt(x-1).
Exercise 2:
Solve the equation sqrt(x+2) + 2 = 2x. Answer
## Polynomial Equations of Higher Degree
We have seen that any degree two polynomial equation (quadratic equation) in one variable can be solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a problem, then either the polynomial is of a special form which allows us to factor it, or we must approximate the solutions with a graphing utility.
### Zero Constant
One common special case is where there is no constant term. In this case we may factor out one or more powers of x to begin the problem.
Example 13.
2x3 + 3x2 -5x = 0.
x (2x2 + 3x -5) = 0.
Now we have a product of x and a quadratic polynomial equal to 0, so we have two simpler equations.
x = 0, or 2x2 + 3x -5 = 0.
The first equation is trivial to solve. x = 0 is the only solution. The second equation may be solved by factoring. Note: If we were unable to factor the quadratic in the second equation, then we could have resorted to using the Quadratic Formula. [Verify that you get the same results as below.]
x = 0, or (2x + 5)(x - 1) = 0.
So there are three solutions: x = 0, x = -5/2, x = 1.
Note: The solution is found from the intercepts of the graphs of f(x) = 2x3 + 3x2 -5x.
### Factor by Grouping
Example 14.
x3 -2x2 -9x +18 = 0.
The coefficient of x2 is -2 times that of x3, and the same relationship exists between the coefficients of the third and fourth terms. Group terms one and two, and also terms three and four.
x2 (x - 2) - 9 (x - 2) = 0.
These groups share the common factor (x - 2), so we can factor the left hand side of the equation.
(x - 2)(x2 - 9) = 0.
Whenever we find a product equal to zero, we obtain two simpler equations.
x - 2 = 0, or x2 - 9 = 0.
x = 2, or (x + 3)(x - 3) = 0.
So, there are three solutions, x = 2, x = -3, x = 3.
Note: These solutions are found from the intercepts of the graph of f(x) = x3 -2x2 -9x +18.
Example 15.
x4 - x2 - 12 = 0.
This polynomial is not quadratic, it has degree four. However, it can be thought of as quadratic in x2.
(x2) 2 -(x2) - 12 = 0.
z2 - z - 12 = 0 This is a quadratic equation in z.
(z - 4)(z + 3) = 0.
z = 4 or z = -3.
We are not done, because we need to find values of x that make the original equation true. Now replace z by x2 and solve the resulting equations.
x2 = 4.
x = 2, x = -2.
x2 = -3.
x = i , or x = -i.
So there are four solutions, two real and two complex.
Note: These solutions are found from the intercepts of the graph of f(x) = x4 - x2 - 12.
A graph of f(x) = x4 - x2 - 12 and a zoom showing its local extrema.
Exercise 3:
Solve the equation x4 - 5x2 + 4 = 0. Answer
## Equations Involving Fractional Expressions or Absolute Values
Example 16.
The least common denominator is x(x + 2), so we multiply both sides by this product.
Checking is necessary because we multiplied both sides by a variable expression. Using a graphing utility we see that both of these solutions check. The solution is the x-coordinate of the intersection point of the graphs of y = 1 and y = 2/x-1/(x+2).
Example 17.
5 | x - 1 | = x + 11.
The key to solving an equation with absolute values is to remember that the quantity inside the absolute value bars could be positive or negative. We will have two separate equations representing the different possibilities, and all solutions must be checked.
Case 1. Suppose x - 1 >= 0. Then | x - 1 | = x - 1, so we have the equation
5(x - 1) = x + 11.
5x - 5 = x + 11.
4x = 16.
x = 4, and this solution checks because 5*3 = 4 + 11.
Case 2. Suppose x - 1 < 0. Then x - 1 is negative, so | x - 1 | = -(x - 1). This point often confuses students, because it looks as if we are saying that the absolute value of an expression is negative, but we are not. The expression (x - 1) is already negative, so -(x - 1) is positive.
Now our equation becomes
-5(x - 1) = x + 11.
-5x + 5 = x + 11.
-6x = 6.
x = -1, and this solution checks because 5*2 = -1 + 11.
If you use the Java Grapher to check graphically, note that abs() is absolute value, so you would graph
5*abs(x - 1) - x - 11 and look at x-intercepts, or you can find the solution as the x-coordinates of the intersection points of the graphs of y = x+11 and y = 5*abs(x-1).
Exercise 4:
(b) Solve the equation | x - 2 | = 2 - x/3 Answer
|
# Wilson's Theorem/Necessary Condition
## Theorem
Let $p$ be a prime number.
Then:
$\paren {p - 1}! \equiv -1 \pmod p$
## Proof 1
If $p = 2$ the result is obvious.
Therefore we assume that $p$ is an odd prime.
Let $p$ be prime.
Consider $n \in \Z, 1 \le n < p$.
As $p$ is prime, $n \perp p$.
From Law of Inverses (Modulo Arithmetic), we have:
$\exists n' \in \Z, 1 \le n' < p: n n' \equiv 1 \pmod p$
By Solution of Linear Congruence, for each $n$ there is exactly one such $n'$, and $\paren {n'}' = n$.
So, provided $n \ne n'$, we can pair any given $n$ from $1$ to $p$ with another $n'$ from $1$ to $p$.
We are then left with the numbers such that $n = n'$.
Then we have $n^2 \equiv 1 \pmod p$.
Consider $n^2 - 1 = \paren {n + 1} \paren {n - 1}$ from Difference of Two Squares.
So either $n + 1 \divides p$ or $n - 1 \divides p$.
Observe that these cases do not occur simultaneously, as their difference is $2$, and $p$ is an odd prime.
$p - 1 \equiv -1 \pmod p$
Hence $n = 1$ or $n = p - 1$.
So, we have that $\paren {p - 1}!$ consists of numbers multiplied together as follows:
in pairs whose product is congruent to $1 \pmod p$
the numbers $1$ and $p - 1$.
The product of all these numbers is therefore congruent to $1 \times 1 \times \cdots \times 1 \times p - 1 \pmod p$ by modulo multiplication.
$\paren {p - 1}! \equiv -1 \pmod p$
$\blacksquare$
## Proof 2
If $p = 2$ the result is obvious.
Therefore we assume that $p$ is an odd prime.
Consider $p$ points on the circumference of a circle $C$ dividing it into $p$ equal arcs.
By joining these points in a cycle, we can create polygons, some of them stellated.
From Number of Different n-gons that can be Inscribed in Circle, the number of different such polygons is $\dfrac {\paren {p - 1}!} 2$.
When you rotate these polygons through an angle of $\dfrac {2 \pi} p$, exactly $\dfrac {p - 1} 2$ are unaltered.
These are the regular $p$-gons and regular stellated $p$-gons.
That there are $\dfrac {p - 1} 2$ of them follows from Number of Regular Stellated Odd n-gons.
The remaining $\dfrac {\paren {p - 1}!} 2 - \dfrac {p - 1} 2$ polygons can be partitioned into sets of $p$ elements: those which can be obtained from each other by rotation through multiples of $\dfrac {2 \pi} p$.
The total number of such sets is then:
$\dfrac {\paren {p - 1}! - \paren {p - 1} } {2 p}$
Thus we have that:
$2 p \divides \paren {\paren {p - 1}! - \paren {p - 1} }$
That is:
$p \divides \paren {\paren {p - 1}! + 1}$
$\blacksquare$
## Proof 3
Let $p$ be prime.
Consider $\struct {\Z_p, +, \times}$, the ring of integers modulo $m$.
From Ring of Integers Modulo Prime is Field, $\struct {\Z_p, +, \times}$ is a field.
Hence, apart from $\eqclass 0 p$, all elements of $\struct {\Z_p, +, \times}$ are units
As $\struct {\Z_p, +, \times}$ is a field, it is also by definition an integral domain, we can apply:
From Product of Units of Integral Domain with Finite Number of Units, the product of all elements of $\struct {\Z_p, +, \times}$ is $-1$.
But the product of all elements of $\struct {\Z_p, +, \times}$ is $\paren {p - 1}!$
The result follows.
$\blacksquare$
## Also known as
Some sources refer to Wilson's Theorem as the Wilson-Lagrange theorem, after Joseph Louis Lagrange, who proved it.
## Source of Name
This entry was named for John Wilson.
## Historical Note
The proof of Wilson's Theorem was attributed to John Wilson by Edward Waring in his $1770$ edition of Meditationes Algebraicae.
It was first stated by Ibn al-Haytham ("Alhazen").
It appears also to have been known to Gottfried Leibniz in $1682$ or $1683$ (accounts differ).
It was in fact finally proved by Lagrange in $1793$.
|
# Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 782: 34
$4\mathbf{w}-3\mathbf{v}=5\mathbf{i}-45\mathbf{j}$
#### Work Step by Step
Here, \begin{align} & {{a}_{1}}=-3 \\ & {{a}_{2}}=-1 \\ & {{b}_{1}}=7 \\ & {{b}_{2}}=-6 \end{align} \begin{align} & {{k}_{1}}=3 \\ & {{k}_{2}}=4 \\ \end{align} Therefore, the value of $3\mathbf{v},4\mathbf{w}$ is, \begin{align} & 3\mathbf{v}=\left[ 3\times \left( -3 \right) \right]\mathbf{i}+\left[ 3\times 7 \right]\mathbf{j} \\ & =-9\mathbf{i}+21\mathbf{j} \end{align} \begin{align} & 4\mathbf{w}=\left[ 4\times \left( -1 \right) \right]\mathbf{i}+\left[ 4\times \left( -6 \right) \right]\mathbf{j} \\ & =-4\mathbf{i}-24\mathbf{j} \end{align} Subtract $3\mathbf{v}$ from $4\mathbf{w}$ to get, \begin{align} & 4\mathbf{w}-3\mathbf{v}=\left[ -4-\left( -9 \right) \right]\mathbf{i}+\left[ -24-\left( 21 \right) \right]\mathbf{j} \\ & =5\mathbf{i}-45\mathbf{j} \end{align}
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# 8.5: Equal Costs
Difficulty Level: At Grade Created by: CK-12
Carla and Dan went to the store. Carla bought 3 notebooks and a $2 pen. Dan bought 2 notebooks and an$8 stapler. All notebooks cost the same. They spent the same amount of money. Can you figure out the cost of one notebook? In this section, we will learn how to use equations to solve problems having to do with equal costs.
### Equal Costs 6
In order to answer questions about equal costs like the one above, we can use the problem solving steps to help.
• First, describe what you know. What did each person buy?
• Second, identify what your job is. In these problems, your job will be to solve for the price of an item.
• Third, make a plan. In these problems, your plan should be to write an expression for what each person spent. Then, set those expressions equal to each other since each person spent the same amount. Finally, solve the equation.
• Fourth, solve. Implement your plan.
• Fifth, check. Substitute your answer into the original problem and make sure it works.
#### Finding the Cost
1. Al bought 2 sandwiches. Bob bought one sandwich and a 4 large soda. All sandwiches cost the same. They spent the same amount of money. What is the cost of one sandwich? • Use \begin{align*}b\end{align*} to stand for the cost of one sandwich. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}b\end{align*}. • Show your work. We can use problem solving steps to help us with this problem. \begin{align*}& \mathbf{Describe:} && \text{Al:} \ 2 \ \text{sandwiches}\\ &&& \text{Bob:} \ 1 \ \text{sandwich and a} \ \4 \ \text{large soda}\\ &&& 2 \ \text{sandwiches} \ \text{cost the same as} \ 1 \ \text{sandwich and} \ \4\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one sandwich.}\\ \\ & \mathbf{Plan:} && \text{Use} \ b \ \text{to stand for the cost of one sandwich.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ b.\\ \\ & \mathbf{Solve:} && \text{Al's cost:} \ 2b\\ &&& \text{Bob's cost:} \ b+4.\\ &&& \text{The costs are the same so} \ 2b=b+4. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2b=b+4\\ &&& \text{Subtract} \ b \ \text{from each side.} \qquad \qquad 2b-b=b+4-b\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ b=4\\ &&& \text{One sandwich is} \ \4\\ \\ & \mathbf{Check:} && \text{Al:} \ 2 \ \text{sandwiches is} \ 2 \times \4, \ \text{or} \ \8.\\ &&& \text{Bob:} \ 1 \ \text{sandwich and a} \ \4 \ \text{large soda is} \ 1 \times \4+\4, \ \text{or} \ \8.\\ &&& \8= \8\end{align*} 2. Camilla bought 4 small sodas and a2 cookie. Darla bought 3 small sodas and a 5 dessert. All small sodas cost the same. They spent the same amount of money. What is the cost of one small soda? • Use \begin{align*}c\end{align*} to stand for the cost of one small soda. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}c\end{align*}. • Show your work. We can use problem solving steps to help us with this problem. \begin{align*}& \mathbf{Describe:} && \text{Camilla:} \ 4 \ \text{small sodas and a } \ \2 \ \text{cookie}\\ &&& \text{Dana:} \ 3 \ \text{small sodas and a } \ \5 \ \text{dessert}\\ &&& 4 \ \text{small sodas and } \ \2 \ \text{costs the same as} \ 3 \ \text{notebooks and} \ \5\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one small soda.}\\ \\ & \mathbf{Plan:} && \text{Use} \ c \ \text{to stand for the cost of one small soda.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ c.\\ \\ & \mathbf{Solve:} && \text{Camilla's cost:} \ 4c+2\\ &&& \text{Dana's cost:} \ 3c+5.\\ &&& \text{The costs are the same so} \ 4c+2=3c+5. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4c+2=3c+5\\ &&& \text{Subtract} \ 3c \ \text{from each side.} \qquad \qquad 4c+2-3c=3c+5-3c\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ c+2=5\\ &&& \text{Subtract} \ 2 \ \text{from each side.} \qquad \qquad \ \ c=3\\ &&& \text{One small soda is} \ \3\\ \\ & \mathbf{Check:} && \text{Camilla:} \ 4 \ \text{small sodas and a} \ \2 \ \text{cookie is} \ 4 \times \3+\2, \ \text{or} \ \14.\\ &&& \text{Dana:} \ 3 \ \text{small sodas and a} \ \5 \ \text{dessert is} \ 3 \times \3+\5, \ \text{or} \ \14.\\ &&& \14= \14\end{align*} 3. Erin bought 6 muffins Fred bought 4 muffins and a3 drink. All muffins cost the same. They spent the same amount of money. What is the cost of one muffin?
• Use \begin{align*}d\end{align*} to stand for the cost of one muffin.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}d\end{align*}.
• Show your work.
We can use problem solving steps to help us with this problem.
\begin{align*}& \mathbf{Describe:} && \text{Erin:} \ 6 \ \text{muffins}\\ &&& \text{Fred:} \ 4 \ \text{muffins and a} \ \3 \ \text{drink}\\ &&& 6 \ \text{muffins} \ \text{costs the same as} \ 4 \ \text{notebooks and} \ \3\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one muffin.}\\ \\ & \mathbf{Plan:} && \text{Use} \ d \ \text{to stand for the cost of one muffin.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ d.\\ \\ & \mathbf{Solve:} && \text{Erin's cost:} \ 6d\\ &&& \text{Fred's cost:} \ 4d+3.\\ &&& \text{The costs are the same so} \ 6d=4d+3. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 6d=4d+3\\ &&& \text{Subtract} \ 4d \ \text{from each side.} \qquad \qquad 6d-4d=4d+3-4d\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 2d=3\\ &&& \text{Divide each side by} \ 2. \qquad \qquad \ \ d=1.5\\ &&& \text{One muffin is} \ \1.50\\ \\ & \mathbf{Check:} && \text{Erin:} \ 6 \ \text{muffins is} \ 6 \times \1.50, \ \text{or} \ \9.\\ &&& \text{Fred:} \ 4 \ \text{muffins and a} \ \3 \ \text{drink is} \ 4 \times \1.50+\3, \ \text{or} \ \9.\\ &&& \9= \9\end{align*}
#### Earlier Problem Revisited
Remember the problem about Carla and Dan? Carla bought 3 notebooks and a $2 pen. Dan bought 2 notebooks and an$8 stapler. All notebooks cost the same. They spent the same amount of money. What is the cost of one notebook?
We can use problem solving steps to help us with this problem.
\begin{align*}& \mathbf{Describe:} && \text{Carla:} \ 3 \ \text{notebooks and a} \ \2 \ \text{pen}\\ &&& \text{Dan:} \ 2 \ \text{notebooks and an} \ \8 \ \text{stapler}\\ &&& 3 \ \text{notebooks and} \ \2 \ \text{costs the same as} \ 2 \ \text{notebooks and} \ \8\\ \\ & \mathbf{My \ job:} && \text{Figure out the cost of one notebook.}\\ \\ & \mathbf{Plan:} && \text{Use} \ a \ \text{to stand for the cost of one notebook.}\\ &&& \text{Write an expression showing the money spent by each person.}\\ &&& \text{Since they spent the same amount of money, the two expressions are equal.}\\ &&& \text{Write the equation.}\\ &&& \text{Solve for} \ a.\\ \\ & \mathbf{Solve:} && \text{Carla's cost:} \ 3a+2\\ &&& \text{Dan's cost:} \ 2a+8.\\ &&& \text{The costs are the same so} \ 3a+2=2a+8. \ \text{Solve the equation.}\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3a+2=2a+8\\ &&& \text{Subtract} \ 2a \ \text{from each side.} \qquad \qquad 3a+2-2a=2a+8-2a\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ a+2=8\\ &&& \text{Subtract} \ 2 \ \text{from each side.} \qquad \qquad \ \ a+2-2=8-2\\ &&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ a=6\\ &&& \text{One notebook is} \ \6.\\ \\ & \mathbf{Check:} && \text{Carla:} \ 3 \ \text{notebooks and} \ \2 \ \text{pen is} \ 3 \times \6+\2, \ \text{or} \ \20.\\ &&& \text{Dan:} \ 2 \ \text{notebooks and} \ \8 \ \text{stapler is} \ 2 \times \6+\8, \ \text{or} \ \20.\\ &&& \20 = \20\end{align*}
### Vocabulary
In math, an expression is a phrase that can contain numbers, operations, and variables without an equal's sign. An equation is a statement that two expressions are equal. An equation is two expressions combined with an equals sign. To solve an equation means to figure out the value(s) for the variable(s) that make the equation true.
### Examples
#### Example 1
Gary bought 5 CDs and a $5 CD case. Helen bought 3 CDs and a set of$29 earphones. All CDs cost the same. They spent the same amount of money. What is the cost of one CD?
• Use \begin{align*}f\end{align*} to stand for the cost of one CD.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}f\end{align*}.
• Show your work.
One CD costs 12. Here is the equation you should have written and the steps to solve: \begin{align*}5f + 5 &= 3f + 29\\ 5f + 5 - 3f &= 3f + 29 - 3f\\ 2f + 5 &= 29\\ 2f + 5 - 5 &= 29 - 5\\ 2f &= 24\\ f &= 12\end{align*} #### Example 2 Ina bought 4 movie tickets and a2 soda. Jen bought 2 movie tickets, a $4 bag of popcorn, a$3 drinks, and a 5 bag of candy. All movie tickets cost the same. They spent the same amount of money. What is the cost of one movie ticket? • Use \begin{align*}g\end{align*} to stand for the cost of one movie ticket. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}g\end{align*}. • Show your work. One movie ticket costs5. Here is the equation you should have written and the steps to solve:
\begin{align*}4g + 2 &= 2g + 4 + 3 + 5\\ 4g + 2 &= 2g + 12\\ 4g + 2 - 2g &= 2g + 12 - 2g\\ 2g + 2 &= 12\\ 2g + 2 - 2 &= 12 - 2\\ 2g &= 10\\ g &= 5\end{align*}
#### Example 3
Ken bought 10 pounds of apples. Larry bought 5 pounds of apples and a 10 jar of honey. Each pound of apples cost the same. They spent the same amount of money. What is the cost of one pound of apples? • Use \begin{align*}j\end{align*} to stand for the cost of one pound of apples. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}j\end{align*}. • Show your work. One pound of apples costs2. Here is the equation you should have written and the steps to solve:
\begin{align*}10j &= 5j + 10\\ 10j - 5j &= 5j + 10 - 5j\\ 5j &= 10\\ j &= 2\end{align*}
### Review
1. Hal bought 2 bagels and a large hot chocolate for $2.50. Jon bought 4 bagels and a$1.00 cream cheese. All bagels cost the same. They spent the same amount of money. What is the cost of one bagel?
• Use \begin{align*}a\end{align*} to stand for the cost of one bagel.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}a\end{align*}.
• Show your work.
2. Kaelyn bought 5 pads of paper and a $1.50 box of binder clips. Lexa bought 2 pads of paper and a$6.00 box of pens. All pads of paper cost the same. They spent the same amount of money. What is the cost of one pad of paper?
• Use \begin{align*}b\end{align*} to stand for the cost of one pad of paper.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}b\end{align*}.
• Show your work.
3. Mary bought 4 decks of cards and a $2.00 crossword puzzle book. Nina bought 2 decks of cards and 2 boxes of dominoes for$3.50 each. All card decks cost the same. They spent the same amount of money. What is the cost of one deck of cards?
• Use \begin{align*}c\end{align*} to stand for the cost of one deck of cards.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}c\end{align*}.
• Show your work.
4. Mark bought 5 sandwiches and a $2 bag of chips. Dave bought 3 sandwiches and a$10 pie. All sandwiches cost the same and they spent the same amount of money. What is the cost of one sandwich?
• Use \begin{align*}d\end{align*} to stand for the cost of one sandwich.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}d\end{align*}.
• Show your work.
5. Jess bought 2 boxes of paper clips and a $1 notepad. John bought one box of paper clips and a$1.75 pen. They spent the same amount of money. What is the cost of one box of paper clips?
• Use \begin{align*}e\end{align*} to stand for the cost of one box of paper clips.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}e\end{align*}.
• Show your work.
6. Sarah bought 7 binders and a 4.50 pack of pens. Ben bought 8 binders. They spent the same amount of money. What is the cost of one binder? • Use \begin{align*}f\end{align*} to stand for the cost of one binder. • Write an equation to represent the costs of the two people. • Solve for the value of \begin{align*}f\end{align*}. • Show your work. 7. Bob bought 2 t-shirts and a22.75 pair of pants. Jeff bought 3 t-shirts and a \$12.50 hat. All t-shirts cost the same. They spent the same amount of money. What is the cost of one t-shirt?
• Use \begin{align*}g\end{align*} to stand for the cost of one t-shirt.
• Write an equation to represent the costs of the two people.
• Solve for the value of \begin{align*}g\end{align*}.
• Show your work.
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### Cubes
How many faces can you see when you arrange these three cubes in different ways?
### Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Bracelets
Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads?
# Cycling Squares
##### Age 7 to 11 Challenge Level:
William from Tattingstone School said that he tried four times before he came up with a solution to Cycling Squares:
We're glad you didn't give up, William. Beth and Pheobe from Exminster CP School and Skye and Molly from Breckland Middle School also used trial and error (or trial and improvement as I like to call it!). Beth and Phoebe said it tested their skills of perseverance.
Matha, also from Tattingstone, said that she started with the number $2$ and then added $14$ because it was the only number she could have added to make a square number. She goes on to say:
I then added $11$ to $14$ to get $25$ which is another square number and carried on like this.
Martha sent in a drawing of her circle which is the same as William's answer, just written in a different way:
.
Dominic from Stonehill took a logical approach:
I listed the numbers and wrote down their pairs to make square numbers.
Some of the numbers only had two possible combinations to make square numbers, so I started with one of these. I put in $2$ first and put $14$ and $34$ on either side of it.
From there, I put $11$ on the other side of the $14$ as these were the only combinations, and so on.
Some numbers had four or even five possible combinations but by trial and error I moved these around until all the combinations made a square number.
Brandon, Antonia and Oliver from Mayhill Junior used a similar method to Dominic. Emilie and Bethany from Alverstoke Junior School said:
To solve the problem we first worked out the highest square number was $64$.
We then worked out all the square numbers from $4$ to $64$.
We then worked out which pairs of numbers made square numbers and we used trial and error to solve the problem.
Very well done to you all. Here are some comments we received in 2015 from pupils at Queen Edith, Cambridge having completed the challenge.
Edward:- "I worked it out by using trial and improvement and it took us a couple of times to figure it out. I started with one pair of numbers and went both ways around the circle"
James:- "I worked out what numbers added together which would make a square number. Next we put two adjoining numbers to make a square number and we used trial and improvement to work out which number came next"
Lavinia:-" We worked together and persevered. If we got it wrong, we helped each other to make the answer right. Cooperating we completed the challenge even though it was hard!
Abigail :-"Clare and I chose the biggest number that was available in the range and made the first possible number that we could think of: 64 (30 + 34) We found out putting 34 first did not work so we put 30 first. We worked our way around the circle clockwise using trial and improvement to complete the challenge!
Lauren :-"Even though it was tricky we still persevered and tried our best. I had to use trial and improvement to make the circle but the best thing was to have fun!
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# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.2
• Difficulty Level : Easy
• Last Updated : 21 Dec, 2020
### Compute the mean number of calls per interval.
Solution:
Let the assumed mean(A) be =3 (Generally we choose the middle element to be the assumed mean, but it’s not mandatory),
hence, the table is,
hence, mean of the calls =
=
=
Therefore, mean number of calls per interval is 3.54
### Question 2: Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.
Solution:
Let the assumed mean (A) be = 2
hence, the table is,
Mean number of head per toss =
=
= 2.47
Therefore, mean number of head per toss is 2.47
### Calculate the average number of branches per plant.
Solution:
Let the assumed mean(A) be = 4
hence, the table is,
Average Number of branches per plant =
=
= 3.615
Therefore, mean number of branches per plant is 3.615
### Find the average number of children per family.
Solution:
Let the assumed mean(A) be = 2
Hence, the table is,
Average number of children per family =
=
= 2.35 (approximately)
Therefore, the average number of children per family is 2.35 (approximately)
### Find the average number of marks.
Solution:
Let the assume mean (A) be = 25
hence, the table is,
Average number of marks =
=
= 26.08 (approximately)
Therefore, average number of marks is 26.08 (approximately)
### Find the mean number of students absent per day.
Solution:
Let mean assumed mean (A) be = 3
Mean number of students absent per day =
=
= 3.525
Therefore, the mean number of students absent per day is 3.525
### Find the average number of misprints per page.
Solution:
Let the assumed mean (A) be = 2
Average number of misprints per day =
=
= 0.73
Therefore, the average number of misprints per day is 0.73
### Find the average number of misprints per page.
Solution:
Let the assumed mean (A) = 2
Average no of accidents per day workers =
=
= 0.83
Therefore, average no of accidents per day workers 0.83
### Question 9: Find the mean from the following frequency distribution of marks at a test in statistics:
Solution:
Let the assumed mean (A) be = 25
Mean =
=
= 22.075
Therefore, the mean is 22.075
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# Using the definition of a limit, prove that $\lim_{n \rightarrow \infty} \frac{n^2+3n}{n^3-3} = 0$
Using the definition of a limit, prove that $$\lim_{n \rightarrow \infty} \frac{n^2+3n}{n^3-3} = 0$$
I know how i should start: I want to prove that given $\epsilon > 0$, there $\exists N \in \mathbb{N}$ such that $\forall n \ge N$
$$\left |\frac{n^2+3n}{n^3-3} - 0 \right | < \epsilon$$
but from here how do I proceed? I feel like i have to get rid of $3n, -3$ from but clearly $$\left |\frac{n^2+3n}{n^3-3} \right | <\frac{n^2}{n^3-3}$$this is not true.
This is not so much of an answer as a general technique.
What we do in this case, is to divide top and bottom by $n^3$: $$\dfrac{\frac{1}{n} + \frac{3}{n^2}}{1-\frac{3}{n^3}}$$ Suppose we want this to be less than a given $\epsilon>0$. We know that $\frac{1}{n}$ can be made as small as we like. First, we split this into two parts: $$\dfrac{\frac{1}{n}}{1-\frac{3}{n^3}} + \dfrac{\frac{3}{n^2}}{1-\frac{3}{n^3}}$$
The first thing we know is that for large enough $n$, say $n>N$, $\frac{3}{n^3} < \frac{3}{n^2} < \frac{1}{n}$. We will use this fact.
Let $\delta >0$ be so small that $\frac{\delta}{1-\delta} < \frac{\epsilon}{2}$. Now, let $n$ be so large that $\frac{1}{n} < \delta$, and $n>N$.
Now, note that $\frac{3}{n^3} < \frac{3}{n^2} < \frac{1}{n} < \delta$. Furthermore, $1- \frac{3}{n^3} > 1 - \frac{3}{n^2} > 1-\delta$.
Thus, $$\dfrac{\frac{1}{n}}{1-\frac{3}{n^3}} + \dfrac{\frac{3}{n^2}}{1-\frac{3}{n^3}} < \frac{\delta}{1+\delta} + \frac{\delta}{1+\delta} < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon$$
For large enough $n$. Hence, the limit is zero.
I could have had a shorter answer, but you see that using this technique we have reduced powers of $n$ to this one $\delta$ term, and just bounded that $\delta$ term by itself, bounding all powers of $n$ at once.
• I am hoping that you have understood the above process. As an exercise, or just for fun, try the limit $\displaystyle\lim_{n \to \infty} \frac{n^4+7n^6+35n+42}{12n^3+34n^2+278n^6 + 1728}$. The answer is just the coefficients of the largest powers, $\frac{7}{278}$. If you can answer this, then you will be able to appreciate this technique. – астон вілла олоф мэллбэрг Sep 28 '16 at 0:40
• yeah I think I get an idea from your proof. but my professor never proved something using $\delta$ yet. so I was wondering if you could prove this without using $\delta$ – Allie Sep 28 '16 at 0:48
• Of course. See, $\delta$ is just a "proxy name" for $\frac{1}{n}$. In place of $\delta$, I could have just taken $\frac{1}{n}$ in that expression, and then I will directly get a bound for $\frac{1}{n}$ instead of $\delta$. Just for ease of reading, I decided to split this into two parts, one having $\delta$ and one having $\frac{1}{n}$.Hence, $\delta$ can be avoided easily, – астон вілла олоф мэллбэрг Sep 28 '16 at 1:41
Another way to show $\lim_{n \rightarrow \infty} \frac{n^2+3n}{n^3-3} = 0$.
Let $f(n) =\frac{n^2+3n}{n^3-3}$.
For $n \ge 3$, $n^2+3n < 2n^2$.
Similarly, for $n \ge 3$, $n^3-3 = \frac12 n^3 + \frac12 n^3 - 3 \gt \frac12 n^3$ for $\frac12 n^3 > 3$, which is certainly true for $n \ge 2$.
Therefore, for $n \ge 3$, $f(n) =\frac{n^2+3n}{n^3-3} < \frac{2n^2}{\frac12 n^3} =\frac{4}{n}$.
You should now be able to easily show that $\lim_{n \to \infty} f(n) = 0$.
Note that this is not the best upper bound for $f(n)$, but it is enough to show what you want.
The generalizations are left to you.
How about this： when $$n>2$$, $$0<\frac{n^2+3n}{n^3-3}<\frac{4n^2}{n^3-n^2}=\frac{4}{n-1}$$
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## How do you write the equation of a line in general form?
General form of a line The general form ax+by+c=0 is one of the many different forms you can write linear functions in. Other ones include the slope intercept form y=mx+b or slope-point form.
## How do you write a function for a line?
The equation of a linear function is expressed as: y = mx + b where m is the slope of the line or how steep it is, b represents the y-intercept or where the graph crosses the y-axis and x and y represent points on the graph.
## How many ways are there to write an equation of a line?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.
## Is general and standard form the same?
General form will typically be in the form “y=mx+b”. M is the slope of the graph, x is the unknown, and b is the y-Intercept. this means that the product of a number and x added to b will equal y. Standard form will always be “x+y= a number value.” so, let’s get some practice.
## What is the general form of a line?
The “General Form” of the equation of a straight line is: Ax + By + C = 0. A or B can be zero, but not both at the same time. The General Form is not always the most useful form, and you may prefer to use: The Slope-Intercept Form of the equation of a straight line: y = mx + b.
## What’s a standard form?
Standard form is a way of writing down very large or very small numbers easily. So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.
## How do you write an equation of a line in slope intercept form?
To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope. This is the value of m in the equation. Next, find the coordinates of the y-intercept–this should be of the form (0, b).
## What two things do you need to write the equation of a line?
from 2 points on the line, you can calculate the slope and then, using one of the points, can determine the equation. (x1,y1) is one point on the line. (x2,y2) is the other point on the line. once you know the slope, you can use either the point slope form or the y intercept form to generate the equation of the line.
## How do you find standard form?
Definitions: Standard Form: the standard form of a line is in the form Ax + By = C where A is a positive integer, and B, and C are integers. The standard form of a line is just another way of writing the equation of a line.
## What is standard form for slope?
Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. A, B, C are integers (positive or negative whole numbers) No fractions nor decimals in standard form.
## How do I find the slope of the line?
The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .
### Releated
#### Find the real solutions of the equation
What are real solutions of an equation? How To Solve Them? The “solutions” to the Quadratic Equation are where it is equal to zero. There are usually 2 solutions (as shown in this graph). Just plug in the values of a, b and c, and do the calculations. What does it mean to find all […]
#### Write an equation for the polynomial graphed below
What is the formula for a polynomial function? A polynomial is a function of the form f(x) = anxn + an−1xn−1 + + a2x2 + a1x + a0 . The degree of a polynomial is the highest power of x in its expression. What are examples of polynomial functions? Basic knowledge of polynomial functions Polynomial […]
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# How do you find the midpoint of the segment with the endpoints (2/3,9/2) and (1/3,11/2)?
Jun 6, 2017
See a solution process below:
#### Explanation:
The formula to find the mid-point of a line segment give the two end points is:
$M = \left(\frac{\textcolor{red}{{x}_{1}} + \textcolor{b l u e}{{x}_{2}}}{2} , \frac{\textcolor{red}{{y}_{1}} + \textcolor{b l u e}{{y}_{2}}}{2}\right)$
Where $M$ is the midpoint and the given points are:
$\left(\textcolor{red}{{x}_{1}} , \textcolor{red}{{y}_{1}}\right)$ and $\left(\textcolor{b l u e}{{x}_{2} , {y}_{2}}\right)$
Substituting the values from the points in the problem gives:
$M = \left(\frac{\textcolor{red}{\frac{2}{3}} + \textcolor{b l u e}{\frac{1}{3}}}{2} , \frac{\textcolor{red}{\frac{9}{2}} + \textcolor{b l u e}{\frac{11}{2}}}{2}\right)$
$M = \left(\frac{\left(\frac{\textcolor{red}{2} + \textcolor{b l u e}{1}}{3}\right)}{2} , \frac{\frac{\textcolor{red}{9} + \textcolor{b l u e}{11}}{2}}{2}\right)$
$M = \left(\frac{\frac{3}{3}}{2} , \frac{\frac{20}{2}}{2}\right)$
$M = \left(\frac{1}{2} , \frac{10}{2}\right)$
$M = \left(\frac{1}{2} , 5\right)$
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# Two-step equations
It is valid:
• if , then :
(1)
• if , then :
(2)
• if and
(3)
Most equations require more than one step to find the solution. In this lesson we are going to cover two-step equations.
Two-step equations are equations that can be solved in two steps. These equations can be written in form of:
where is the unknown variable, is the coefficient and and are numbers. The coefficient can not be or .
When solving this type of equation we should follow this order of operations:
2. multiplication
3. division
Note that we don’t have to follow this order, but it makes the calculation a lot easier.
Let’s get started with some examples, one of each type of the two-step equation.
This example can be solved in following steps:
We should subtract from both sides first.
Now we have one step equation:
Note that in this example we could have done division first and then subtraction, but it would make the solving of the equation more difficult since we would have to deal with fractions and it is easier to deal with regular numbers.
The example above can be solved in steps similar to the previous example:
First we need to add 8 to both sides. The resulting equation is:
In this step we only need to divide both sides with 4.
The result is .
Look at this example that you need to divide to get the unknown variable :
In this step we are going to multiply the equation by and get the result.
The result is .
All other two step equations are solving like those two examples.
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# Supplementary Angles
## Two angles that add to 180 degrees.
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Practice Supplementary Angles
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Supplementary Angles
### Supplementary Angles
Two angles are supplementary if they add up to 180\begin{align*}180^\circ\end{align*}. Supplementary angles do not have to be congruent or adjacent.
What if you were given two angles of unknown size and were told they are supplementary? How would you determine their angle measures?
### Examples
#### Example 1
Find the measure of an angle that is supplementary to ABC\begin{align*}\angle ABC\end{align*} if mABC\begin{align*}m\angle ABC\end{align*} is 118\begin{align*}118^\circ\end{align*}.
180118=62\begin{align*}180^\circ-118^\circ=62^\circ\end{align*}.
#### Example 2
Find the measure of an angle that is supplementary to ABC\begin{align*}\angle ABC\end{align*} if mABC\begin{align*}m\angle ABC\end{align*} is 32\begin{align*}32^\circ\end{align*}.
18032=148\begin{align*}180^\circ-32^\circ=148^\circ\end{align*}.
#### Example 3
The two angles below are supplementary. If mMNO=78\begin{align*}m\angle MNO = 78^\circ\end{align*} what is mPQR\begin{align*}m\angle PQR\end{align*}?
Set up an equation. However, instead of equaling 90\begin{align*}90^\circ\end{align*}, now the sum is 180\begin{align*}180^\circ\end{align*}.
78+mPQRmPQR=180=102\begin{align*}78^\circ + m\angle PQR & = 180^\circ\\ m\angle PQR & = 102^\circ\end{align*}
#### Example 4
What are the measures of two congruent, supplementary angles?
Supplementary angles add up to 180\begin{align*}180^\circ\end{align*}. Congruent angles have the same measure. So, 180÷2=90\begin{align*}180^\circ \div 2 = 90^\circ\end{align*}, which means two congruent, supplementary angles are right angles, or 90\begin{align*}90^\circ\end{align*}.
#### Example 5
Find the measure of an angle that is a supplementary to MRS\begin{align*}\angle MRS\end{align*} if mMRS\begin{align*} m\angle MRS\end{align*} is 70\begin{align*} 70^\circ\end{align*}.
Because supplementary angles have to add up to 180\begin{align*}180^\circ\end{align*}, the other angle must be 18070=110\begin{align*}180^\circ-70^\circ=110^\circ\end{align*}.
### Review
Find the measure of an angle that is supplementary to ABC\begin{align*}\angle ABC\end{align*} if mABC\begin{align*}m\angle ABC\end{align*} is:
1. 114\begin{align*}114^\circ\end{align*}
2. 11\begin{align*}11^\circ\end{align*}
3. 91\begin{align*}91^\circ\end{align*}
4. 84\begin{align*}84^\circ\end{align*}
5. 57\begin{align*}57^\circ\end{align*}
6. x\begin{align*}x^\circ\end{align*}
7. (x+y)\begin{align*}(x+y)^\circ\end{align*}
Use the diagram below for exercises 8-9. Note that NK¯¯¯¯¯¯¯¯¯IL\begin{align*}\overline{NK} \perp \overleftrightarrow{IL}\end{align*}.
1. Name two supplementary angles.
1. If mINJ=63\begin{align*}m\angle INJ = 63^\circ\end{align*}, find mJNL\begin{align*}m\angle JNL\end{align*}.
For exercise 10, determine if the statement is true or false.
1. Supplementary angles add up to 180\begin{align*}180^\circ\end{align*}.
For 11-12, find the value of x\begin{align*}x\end{align*}.
To see the Review answers, open this PDF file and look for section 1.8.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Supplementary angles
Supplementary angles are two angles whose sum is 180 degrees.
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CIRCLES: ARCS and SECTORS
Central Angles - Circumference Arc
Circular Clock Face
In 2 minutes, the time on this clock will be 15 minutes to 4 o'clock. Another way to say this time is a quarter to 4. We call 15 minutes "a quarter of an hour" because the minute hand rotates ¼ of the way around the circular face in that time. We know that 360° is a full circle, and we also know that a full circle is an hour or 60 minutes. So 90° or ¼ of 360°, corresponds to 15 minutes which is ¼ of 60 minutes.
The length of the circumference arc, traced by the minute hand as it moves through a 90° angle, is exactly ¼ the length of the clockface circumference. If we know the length of the minute hand (radius), we can find the length of the arc it traces. It will measure ¼ the length of the circumference.
The length of a sector arc is proportional to the sector angle.
We find the length of an arc created by a central angle of a circle with this proportion
Remember that circumference is so we have to know the diameter or radius of the circle in order to find its circumference. If the minute hand on the clock is 1.5 cm. long, we find the distance its tip travels in 15 minutes with this proportion:
since 90° is ¼ of 360°, the arc is ¼ the circumference or
Example:
Find the length of the arc traced by a 3 cm. minute hand in 24 minutes.
Solution:
The circumference of the face is:
24 minutes is of an hour or 360°,
so the arc length =
The ratio of arc length to circumference
equals the ratio of the central angle to 360°.
Sometimes, we're given the length of the arc and we have to find the central angle.
Example:
A circle has radius = 11 inches. How big is the central angle that cuts off an arc of 23.4 inches?
Solution:
If we know the ratio of Arc to Circumference, we know the ratio of central angle to 360°. We first find the circumference:
Now we set up the proportion. We'll use x to represent the measure of the central angle.
.
Central Angles - Sector Area
In the same way that the size of a sector arc is proportional to the central angle that subtends it, the ratio of sector area to circle area equals the ratio of central angle to 360°.
Example
We find the area of a sector created by a central angle of a circle with this proportion
Example:
A circle has radius = 11 inches. Find the area of a sector created by a central angle of 20°.
Solution:
We find the circle's area:
Now we set up the proportion. We'll use x to represent the sector area.
In some questions, we know the area, so we need to find the central angle. We use the proportion statement but the variable now is in the angle position instead of the sector area spot.
Example
The area of a sector is one-fifteenth the area of the whole circle. How big is the central angle?
Solution
Since the sector area is 1/15 the circle's area, the angle must be 1/15 of 360°.
.
Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.
Practice Exercise
Make a diagram if you're stuck. It really helps!
1)
a) How long is the arc formed by a 37° central angle in a circle with radius = 3.6 inches.
b) What size central angle subtends an arc of 3.77 cm in a circle with circumference = 18.85 cm?
c) Angle AOB = 65°. AO is a radius = 7.3 cm. What is the area of sector AOB?
2) An arc of 27.3 cm is cut off by a 45.6° central angle in a circle.
a) What is the circumference of the circle?
b) What is the radius of the circle?
c) What is the area of the disc?
d) What is the area of the sector?
.
Solutions
1)
a) Perimeter = 22.62 inches so
The arc is 2.32 inches long.
b) What size central angle subtends an arc of 3.77 cm in a circle with circumference = 18.85 cm?
The angle is 72 °.
c) Angle AOB = 65°. AO is a radius = 7.3 cm. What is the area of sector AOB?
The area of a circle with r = 7.3 is 167.42 cm².
The area of the sector is 30.23 cm².
2)
a) We use a proportion since C = arc cut off by 360°
So,
b) What is the radius of the circle?
Since
c) What is the area of the circle?
d) What is the area of the sector?
So,
.
(all content © MathRoom Learning Service; 2004 - ).
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# Question Video: Finding the Length of a Rectangle given a Relation between Its Dimensions Mathematics • 8th Grade
ππππ is a rectangle where ππ = 9π₯ β 8 and ππ = 8π₯ + 1. Find ππ.
02:23
### Video Transcript
ππππ is a rectangle where ππ is equal to nine π₯ minus eight and ππ is equal to eight π₯ plus one. Find ππ.
So weβve been told that ππππ is a rectangle and asked to find ππ which is the length of one of the sides of the rectangle. Weβve been given an expression for its length in terms of the variable lowercase π₯. It is equal to eight π₯ plus one. Weβve also been given an expression for the opposite side of the rectangle ππ, which is equal to nine π₯ minus eight.
In order to find the length of the side ππ, we need to find the value of the variable lowercase π₯. Letβs think about how we can use the information weβve been given to do this. Well, a key fact about rectangles is that opposite sides are equal in length. For our rectangle, this means that ππ and ππ are equal in length.
But also more usefully, the sides ππ and ππ are equal in length. As weβve been given expressions for the lengths of these two sides, we can equate them, giving the equation nine π₯ minus eight is equal to eight π₯ plus one. We can now solve this equation in order to find the value of the variable π₯. First we add eight to each side of the equation, giving nine π₯ is equal to eight π₯ plus nine.
Next, I want to group all of the π₯ terms on the same side of the equation. So Iβll subtract eight π₯ from each side. This gives π₯ is equal to nine. So weβve solved the equation and found the value of π₯. Remember, in this question, weβre asked to find the length of the side ππ. ππ is equal to eight π₯ plus one. So we need to substitute the value that we have calculated four π₯ into this expression.
This gives eight multiplied by nine plus one. Eight multiplied by nine is 72, and adding one gives 73. We havenβt been given any units to use in this question, so the length of ππ is 73.
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# 2.3 - Interpretations of Probability
2.3 - Interpretations of Probability
## Interpretations of Probability
Before discussing the specific rules of probability, it's important to recognize there are three main interpretations of probability.
Classical Interpretation of Probability
The probability that event E occurs is denoted by P(E). When all outcomes are equally likely, then:
$P(E) = \frac{number\ of\ outcomes\ in\ E}{number\ of\ possible\ outcomes}$
Subjective Probability
Subjective probability reflects personal belief which involves personal judgment, information, intuition, etc.
For example, what is P (you will get an A in a certain course)? Each student may have a different answer to the question.
Relative Frequency Concept of Probability (Empirical Approach)
If a particular outcome happens over a large number of events then the percentage of that outcome is close to the true probability.
For example, if we flip the given coin 10,000 times and observe 4555 heads and 5445 tails, then for that coin, P(H)=0.4555.
$P(E) \approx \frac{number\ of\ outcomes\ in\ E}{number\ of\ possible\ outcomes}$
## Example 2-5
1. Find the probability that exactly one head appears in two flips of a fair coin.
The possible outcomes are listed as: {(H, H), (H, T), (T, H), (T, T)}.
Note that the four outcomes are of equal probability since the coin is fair.
$P(getting\ exactly\ one\ H\ in\ two\ flips\ of\ a\ fair\ coin)=P\{(H,T),(T,H)\}=\frac{2}{4}=\frac{1}{2}$
2. Find the probability that two heads appear in two flips of a fair coin.
$P(getting\ two\ H\ in\ two\ flips\ of\ a\ fair\ coin)=P\{(H,H)\}=\frac{1}{4}$
3. Find the probability that the sum of two faces is greater than or equal to 10 when one rolls a pair of fair dice.
The possible outcomes of the experiment are:
{ (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) }
If S denotes the sum of the points in the two faces:
\begin {align} P(S\ greater\ than\ or\ equal\ to\ 10)& =P(S=10)+P(S=11)+P(S=12)\\ & = P\{(4,6),(5,5),(6,4)\}+P\{(5,6),(6,5)\}+P\{(6,6)\} \\ & = \frac{3}{36}+\frac{2}{36}+\frac{1}{36} \\ & =\frac{1}{6} \\ \end {align}
## Try It! Rolling Dice
Directions: Try each of these problems and click the buttons to reveal the correct answers.
Again using the example where we flip two fair six-sided dice, find the following:
• $A={(3, 5)}$
• $B=\text {a 4 is rolled on the first die}$
• $C=\text {a 5 is rolled on the second die}$
• $D=\text {the sum of the dice is 7}$
• $E={(7, 4)}$
1. $P(B\cap D)$ and $P(B\cup D)$.
$P(B\cap D)=\frac{1}{36}$ and $P(B\cup D)=\frac{11}{36}$.
2. $P(D\cap C)$ and $P(D\cup C)$.
$P(D\cap C)=\frac{1}{36}$ and $P(D\cup C)=\frac{11}{36}$
3. $P(A\cap D)$ and $P(A\cup D)$
$P(A\cap D)=0$ and $P(A\cup D)=\frac{7}{36}$
4. $P(B\cap C)$ and $P(B\cup C)$
$P(B\cap C)=\frac{1}{36}$ and $P(B\cup C)=\frac{11}{36}$
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility
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## Sunday, October 27, 2013
### SV #4: Unit I Concept 2: Graphing Logarithmic Equations
This problem goes over how to graph logarithmic equations and how to find key points such as x and y-intercepts. We also identify the asymptote, domain, and range. This type of graph is very different than exponential ones.
What does the viewer need to pay special attention to in order to understand?
Make sure you are correctly able to find your h because it will be crucial to finding your asymptote. In these types of graph we only care about h and k. Our asymptote equation will be x = h. P.S. h will be the opposite of what you see in the equation. Be sure to use the change of base formula when solving for your y-intercept and to correctly exponentiate for your x-intercept.
Happy solving!
## Thursday, October 24, 2013
### SP #3: Unit I Concept 1: Finding Parts & Graphing Exponential Functions
Let's start off by breaking off the steps to solve this problem!
So you first find the parts of the equation such as a, h, b, and k as in the first box. You then find your asymptote because your equation for this is y=k (remember YAK). You then proceed to find x and y intercepts (hint, you won't be able to find one of these *nudge nudge*) After this you can plot in the equation into your graphing calculator as is but remember to put parantheses around (x-1). Your domain and range should be especially easy to find, domain anyway. Your range depends on the location of the asymptote.
Ok so now:
This problem covers our favorite exponential functions and how to solve for their key parts. We use horizontal asymptotes for exponential graphs. So this is like intertwining two concepts (the graphing functions and exponential stuff)
What do you need to pay special attention to in order to understand?
The exponentual YaK died! Don't forget that there will be no x-intercept because a is positive and it must go above the asymptote of y = 2. Range will depend on the location of the asymptote and direction the graph goes off in. Other than that don't forget to pay attention to what you input in your calculator! And do not make simple mistakes as I do and think h will be -1 instead of +1. OH and do not forget that when (1/2) is raised to -1, it will be become because you flip the fraction to get its reciprocal.
## Tuesday, October 15, 2013
### SV #3: Unit H Concept 7: Finding logs when given approximations
This problem is about finding logs when given approximations of course. The clues are given and the equation we are aiming to solve for. We need to break down the problem and expand our log in order to use the clues given to us. You then substitute in your clues. These problems demonstrate our ability to solve these without a calculator (for the most part because sometimes you do need help to break down the terms)
What does the viewer need to pay special attention to in order to understand?
Please be sure to remember log base 8 of 8 will equal to one. Do not forget how the quotient and power property will play a role in expansion and solving our log. Please please please take into careful consideration what these properties will do when you write out the expanded form. Also be sure to use the correct clues when you end up with your final form with your approximations!
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# Balbharati Solutions Class 6 Mathematics HCF LCM
## Balbharati Solutions Class 6 Mathematics HCF LCM
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Balbharati Class 6 Mathematics Book, Practice Set 23, 24 and 25, HCF LCM. Here students can easily find step by step solutions of all the problems for HCF LCM. Also our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get HCF LCM solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.
HCF LCM all Question Solutions :
Practice Set 23 :
Question no – (1)
Solution :
(1) Given, numbers are 12, 16
Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 16 = 1, 2, 4, 8, 16
Common Factors = 1, 2, 4
(2) Given, numbers are 21, 24
Factors of 21 = 1, 3, 7, 21
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Common Factors of both number is 1 and 3
(3) Given, numbers are 25, 30.
Factors of 25 = 1, 5, 25
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
Common Factors of both number is 1 and 5.
(4) Given, numbers are 24, 25.
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 25 = 1, 5, 25
Common Factor = 1
(5) Given, numbers are 56, 72.
Factors of 56 is = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 72 is = 1, 2, 3, 4, 6, 8, 9, 18, 24, 36, 72
Common Factors of both number are 1, 2, 4, 8.
Practice Set 24 :
Question no – (1)
Solution :
(1) To find factor of 45 and 30.
Factors of 45 = 1, 3, 5, 9, 15, 45
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
Common factors = 1, 3, 5, 15
Thus, HCF ( 45 and 30) = 15
(2) To find factor of 16 and 48
Factors of 16 = 1,2, 4, 8, 16
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Common factors = 1, 2, 4, 8, 16
Thus, the HCF of 16 and 48 is 16
(3) To find factor of 39 and 25.
Factors of 39 is = 1, 3,13, 39
Factors of 25 is = 1, 5, 25
Common factor of both the number is 1
Thus, the HCF of 39 and 25 is = 1
(4) To find factor of 49 and 56
Factors of 49 = 1, 7 , 49
Factors of 56 = 1, 2, 4, 7, 8, 14, 28 , 56
Common factors of both number is 1, 7
Thus, the HCF of 49 and 56 is 1, 7
(5) To find factor of 120, 144.
Factors of 120 is = 1 , 2, 3, 4, 5, 12 , 24 , 30, 60, 120
Factors of 144 is = 1, 2, 3, 4, 12, 24, 76, 144
Common factors of both the numbers are = 1, 2, 3, 4, 12, 24
Thus, the HCF of 120 and 144 is = 24
(6) To find factor of 81 and 99
Factors of 81 = 1, 3, 9, 27, 81
Factors of 99 = 1, 3, 9, 11, 33, 99
Common factors of both the numbers = 1,3, 9
Thus, HCF ( 81,99) = 9
(7) To find factor of 24, 36
Factors of 24 is = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 is = 1, 2,3, 4, 6, 9, 12, 18, 36
Common factors of the both number is = 1, 2, 3, 4 , 6 , 12
Hence, the HCF of 24 and 36 = 12
(8) To find factor of 25 and 75
Factors of 25 is = 1, 5, 25
Factors of 75 is = 1, 5, 15, 25, 75
Common factors for both the number is = 1, 5, 25
Hence, the HCF of 25 and 75 is = 25
(9) To find factor of 48, 54
Factors of 48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 54 : 1, 2,3, 6, 9, 18, 27, 54
Common factors of both the number are 1, 2, 3, 6,
∴ The HCF of 48 and 54 is 6.
(10) To find factor of 150, 225
Factors of 150 is = 1,2, 3, 5, 6, 10, 15, 25, 30, 75, 150
Factors of 225 is = 1, 3, 5, 15, 25, 75, 225
Common factors of both the number is = 1, 3, 5, 15, 25, 75
Thus, the HCF of 150 and 225 is = 75.
Question no – (2)
Solution :
We required to find HCF of 18 and 15
Factors of 18 is = 1, 2, 3, 9, 18
Factors of 15 is = 1, 3, 5, 15
Common factors of both the number is = 1,3
Thus, the HCF of 18 and 15 = 3
Hence, the maximum length of each bed is 3 meters.
Practice Set 25:
Question no – (1)
Solution :
(1) 9 and 15.
9 time table = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
15 time table = 15, 30, 45, 60, 75
Since, lowest common multiple is 45
∴ The LCM of 9 and 15 = 45
(2) 2, 3 and 5.
2 time table is = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
3 time table is = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
5 time table is = 5, 10, 15, 20, 25, 30, 35, 40
Since, lowest common multiple is 30
Hence, The LCM of 2,3 and 5 is 30.
(3) 12 and 28
12 time table = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
28 time table = 28, 56, 84, 112, 140
Since, lowest common multiple is 84
The LCM of 12 and 28 = 84.
(4) 15, 20
15 time table is = 15, 30, 45, 60, 75, 90, 105, 120, 135, 150
20 time table is = 20, 40, 60, 80
Since, lowest common multiple is 60
Therefore, the LCM of 15 and 20 = 60
(5) 8 and 11.
8 time table is = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
11 time table is = 11, 22, 33, 44, 55, 66, 77, 88, 99
Since, the lowest common multiple is 88
The LCM of 8 and 11 is = 88.
Question no – (2)
Solution :
(1) We required to find LCM of 20 and 25.
20 time table is = 20,40,60,80,100,120,140
25 time table is = 25, 50, 75, 100, 125
Since, The lowest common multiple is 100
Therefore, the LCM of 20 and 25 = 100
(2) We required to find LCM of 16, 24, and 40.
16 time table is = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240
24 time table is = 24, 48, 72, 96, 120, 144, 168, 192, 216, 240
40 time table is = 40, 80, 120, 160, 200, 240
Since, lowest common multiple is 240.
Hence, the LCM of 16, 24 and 40 = 240
(4) We required to find LCM of 60, 120 and 24
60 time table is = 60, 120, 180
120 time table is = 120, 240,360
24 time table is = 24, 48, 72, 96,120, 144
Since, lowest common multiple is 120.
Therefore, the LCM of 60, 120 and 24 is = 120
(5) Given the fractions 13/45 and 22/ 75
We required to find LCM of 45 and 75.
45 time table is = 45, 90, 135, 180, 225, 270, 315
75 time table is = 75, 150, 225, 300
Since, the lowest common multiple is 225.
Thus, the LCM of 45 and 75 = 225
Therefore,
13 × 5/45 × 5 = 65/225 ….. (Eqn.1)
22×3/75×3 =66 /225 …. (Eqn. 2)
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# Magic Egg Tangram
## Overview and Objective
The goal of this activity is to enhance students’ creative thinking in mathematics. In this activity, students will use geometric constructions to build the pieces of the magic egg tangram. Then, they will use the tangram pieces to form different shapes. At the end of the lesson, students will make their own creations to express their geometric imagination using this extraordinary tangram puzzle.
## Warm-Up
The most well-known tangram is the square tangram. According to an ancient Chinese legend, an old wise man was tasked with constructing a window at the Chinese emperor's palace. He wrapped a square glass in the silk, but upon arriving at the palace he dropped the pane, and the square glass had broken into seven geometric pieces. He attempted to collect the pieces back, but he came up with a different shape every time he tried. In the end, the emperor became more pleased with the infinite amount of combinations and interesting shapes.
Soon the puzzle was reincarnated with paper, wood, and other materials and spread across the world. When re-creating a tangram puzzle, one must construct the pieces carefully by measuring and calculating the angles and side ratios.
• Diagonals of the square create two big triangles.
• Connecting the midpoints of the sides of the square and the triangles create the other shapes.
• The big triangles have $1/4$ of the square area. The purple one has $1/8$ the smallest two triangles have $1/16$ , and so on.
## Main Activity
Now, Introduce the Magic Egg on Polypad. The Magic Egg consists of nine pieces. You can flip, slide and turn them to create various patterns. The most popular are birds.
Invite students to drag all the pieces of the egg tangram to a blank canvas and create the egg form. Then, discuss the pieces they can identify by using measuring tools like a ruler and protractor. They might identify some of the pieces - isosceles right triangles, quarter circles, and small $1/8$ circles.
Challenge them to come up with different ways to construct the egg tangram pieces and perhaps maybe use some Polypad's shortcuts. For example, here is one of the possible construction methods.
### Construction of Egg Tangram
• Ask students to try to construct a circle that overlaps with the outer curve of the egg. You may want to label the endpoints of the widest line of the egg as A and B. Clarify with the students that the first step of the construction is drawing two intersecting circles (like a Venn Diagram) with centers A and B and the radius |AB|.
• Discuss how the vertical axis can be constructed by using the fact that the vertical halves of the egg are identical. Thus, the vertical axis is the axis of symmetry for the egg. After the students come up with the idea of a perpendicular bisector, you can remind them to select the AB line segment to see the construction options at the bottom center toolbar.
• Tell the students to construct and label three intersection points by using the point tool as C, D, E.
The upper part of the egg started to appear., now we can have a close look at the two large triangles and the quarter circles at the bottom. After getting students' comments, you can clarify that another circle is needed with center C.
• Construct a circle with center C and radius |CB|. Construct point F, and G the point of intersection of the new circle and the perpendicular line.
Students can always check their progress and think about the next steps by dragging the original egg tangram on their drawing. They can also adjust the transparency of the tangram's colors to see the lines and circles better.
Discuss possible next steps.
Only the very top of the egg remained without outlined. Therefore, we need a smaller circle at the top to create the 1/8 circle pieces.
• Use the ruler the draw a segment passing through B and F and another segment from A and F. You can label the new intersection points as H and I.
• Draw a circle with the center F and radius FH.
• At the last step of the construction, we need a congruent circle with center G. To do that; we can repeat the last two steps for the bottom half of the egg tangram.
• To create the small white triangle piece, we can draw another circle with center C and radius |CK|. Finally, the points of intersection of the circle and segment AB can be labeled as M and N to help construct the last piece.
Students can drag the tangram back to check each piece.
## Closure
At the end of the lesson, discuss that the Tangram puzzles have been around for over 2000 years. Therefore, it was essential for each piece to be constructed with a compass and ruler, which are also the only available technologies then.
To close the lesson, share this canvas with the students. Invite them to try solving the puzzles. Additionally, you may encourage them to make some of their own creations with the egg tangram pieces. Click here to read more about students making puzzles for their classmates to solve.
Here are the solutions to the birds above.
## Support and Extension
For students ready for additional extension in this lesson, consider showing the heart tangram and ask them to re-create the tangram pieces using the Polypad tools.
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Example.Let A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 6 & 8 \\ -1 & 2 & -3\end{array}\right] and B=\left[\begin{array}{ccc}2 & 3 & 4 \\ -4 & 6 & -8 \\ 1 & 2 & 5\end{array}\right] Then(iii)\begin{aligned}A B &=\left[\begin{array}{ccc}2-8+3 & 3+12+6 & 4-16+15 \\8-24+8 & 12+36+16 & 16-48+40 \\-2-8-3 & -3+12-6 & -4-16-15\end{array}\right] \\&=\left[\begin{array}{ccc}-3 & 21 & 3 \\-8 & 64 & 8 \\-13 & 3 & -35\end{array}\right]\end{aligned}
17. Let \mathrm{A}=\left[\begin{array}{ccc}1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2\end{array}\right] \mathrm{B}=\left[\begin{array}{ccc}2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2\end{array}\right] and C=\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 5 & 8 \\ 2 & 7 & 6\end{array}\right] Show that :(V) \mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}^{-7}=\pm=
Evaluate the following determinants using their properties.8. \left|\begin{array}{lll}x & y & z \\ z & x & y \\ y & z & x\end{array}\right|
18. Prove the following identities :$\left\{\left[\begin{array}{lll}1 & \omega & \omega^{2} \\\omega & \omega^{2} & 1 \\\omega^{2} & 1 & \omega\end{array}\right]+\left[\begin{array}{lll}\omega & \omega^{2} & 1 \\\omega^{2} & 1 & \omega \\\omega & \omega^{2} & 1\end{array}\right]\right\}\left[\begin{array}{l}1 \\\omega \\\omega^{2}\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$where \omega is a complex cube root of unity.
1. Use matrices if possible to solve the following systems of linear equations by:(i) the matrix inversion method (ii) the Cramers rule.(vii)$\begin{array}{l}2 x-2 y=4 \\-5 x-2 y=-10\end{array}$
8. Let A=\left[\begin{array}{ccc}-2 & 1 & 0 \\ -1 & 4 & 3 \\ 0 & 8 & 5\end{array}\right] and X=\left[\begin{array}{ccc}2 & 1 & -1 \\ -3 & 2 & -4 \\ 5 & 4 & 0\end{array}\right] Find:(ii) \lambda I-A where \lambda is any scalar and I is a unit matrix of order 3 .
Evaluate the following determinants using their properties.7. \left|\begin{array}{ccc}a+b+2 c & b & a \\ c & b & b+c+2 a \\ c & c+a+2 b & a\end{array}\right|
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# Resistors in Parallel
## Many resistors branch from a single point. The resistance of the circuit is the sum of the inverse of each resistor.
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Resistors in Parallel
#### Resistors in Parallel
All resistors are connected together at both ends. Recall the river analogy: current is analogous to a river of water, but instead of water flowing, charge does. With resistors in parallel, there are many rivers (i.e. the main river branches off into many other rivers), so all resistors receive different amounts of current. But since they are all connected to the same point at both ends they all receive the same voltage.
Key Equations
1Rtotal=1R1+1R2+1R3+\begin{align*}\frac{1} {R_{total}} = \frac{1} {R_1} + \frac{1}{R_2} + \frac{1} {R_3} + \ldots\end{align*}
#### Example
A circuit is wired up with 2 resistors in parallel.
Both resistors are directly connected to the power supply, so both have the same 20V across them. But they are on different ‘rivers’ so they have different current flowing through them. Lets go through the same questions and answers as with the circuit in series.
What is the total resistance of the circuit?
The total resistance is 1Rtotal=1R1+1R2=190Ω+110Ω=190Ω+990Ω=1090Ω\begin{align*}\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{90\Omega}+\frac{1}{10\Omega}=\frac{1}{90\Omega}+\frac{9}{90\Omega}=\frac{10}{90\Omega}\end{align*} thus, Rtotal=90Ω10=9Ω\begin{align*}R_{total}=\frac{90\Omega}{10}=9\Omega\end{align*}
\begin{align*}^*\end{align*}Note: Total resistance for a circuit in parallel will always be smaller than smallest resistor in the circuit.
What is the total current coming out of the power supply?
Use Ohm’s Law (V=IR)\begin{align*}(V=IR)\end{align*} but solve for current (I=V/R)\begin{align*}(I=V/R)\end{align*}.
Itotal=VtotalRtotal=20V9Ω=2.2A\begin{align*}I_{total}=\frac{V_{total}}{R_{total}}=\frac{20V}{9\Omega}=2.2A\end{align*}
How much power does the power supply dissipate?
P=IV\begin{align*}P=IV\end{align*}, so the total power equals the total voltage multiplied by the total current. Thus, Ptotal=ItotalVtotal=(2.2A)(20V)=44.4W\begin{align*}P_{total}=I_{total}V_{total}=(2.2A)(20V)=44.4W\end{align*}. So the Power Supply outputs 44W (i.e. 44 Joules of energy per second).
How much power is each resistor dissipating?
Each resistor has different current across it, but the same voltage. So, using Ohm’s law, convert the power formula into a form that does not depend on current. P=IV=(VR)V=V2R\begin{align*}P=IV=\left(\frac{V}{R}\right) V=\frac{V^2}{R}\end{align*} Substituted I=V/R\begin{align*}I=V/R\end{align*} into the power formula. P90Ω=V290ΩR90Ω=(20V)290Ω=4.4W;P10Ω=V210ΩR10Ω=(20V)210Ω=40W\begin{align*}P_{90\Omega}=\frac{V^2_{90\Omega}}{R_{90\Omega}}=\frac{(20V)^2}{90\Omega}=4.4W; P_{10\Omega}=\frac{V^2_{10\Omega}}{R_{10}\Omega}=\frac{(20V)^2}{10\Omega}=40W\end{align*}
\begin{align*}^*\end{align*}Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should–Energy is always conserved.
How much current is flowing through each resistor?
Use Ohm’s law to calculate the current for each resistor.
I90Ω=V90ΩR90Ω=20V90Ω=0.22AI10Ω=V10ΩR10Ω=20V10Ω=2.0A\begin{align*}I_{90\Omega}=\frac{V_{90\Omega}}{R_{90\Omega}}=\frac{20V}{90\Omega}=0.22A \qquad I_{10\Omega}=\frac{V_{10\Omega}}{R_{10\Omega}}=\frac{20V}{10\Omega}=2.0A\end{align*}
Notice that the 10Ω\begin{align*}10\Omega\end{align*} resistor has the most current going through it. It has the least resistance to electricity so this makes sense.
\begin{align*}^*\end{align*}Note: If you add up the currents of the individual ‘rivers’ you get the total current of the of the circuit, as you should.
### Review
1. Three 82 Ω\begin{align*}82\ \Omega\end{align*} resistors and one 12 Ω\begin{align*}12\ \Omega\end{align*} resistor are wired in parallel with a 9V\begin{align*}9\;\mathrm{V}\end{align*}battery.
1. Draw the schematic diagram.
2. What is the total resistance of the circuit?
2. What does the ammeter read and which resistor is dissipating the most power?
3. Given three resistors, 200 Ω,300 Ω\begin{align*}200\ \Omega, 300\ \Omega\end{align*} and 600 Ω\begin{align*}600\ \Omega\end{align*} and a 120V\begin{align*}120\;\mathrm{V}\end{align*}power source connect them in a way to heat a container of water as rapidly as possible.
1. Show the circuit diagram
2. How many joules of heat are developed after 5 minutes?
1. b. 8.3W\begin{align*}8.3 \;\mathrm{W}\end{align*}
2. 0.8A\begin{align*}0.8\mathrm{A}\end{align*} and the 50 Ω\begin{align*}50 \ \Omega\end{align*} on the left
3. b. 43200J\begin{align*}43200\mathrm{J}\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
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# Difference between revisions of "2016 AIME II Problems/Problem 5"
## Problem
Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \geq 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$. Find $p$.
## Solution 1
Note that by counting the area in 2 ways, the first altitude is $\dfrac{ab}{c}$. By similar triangles, the common ratio is $\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$. Multiplying by the denominator and expanding, the equation becomes $\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a$. Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$, so $7ab = 6bc+6b^2$ and $7a=6b+6c$. Checking for Pythagorean triples gives $13,84,$ and $85$, so $p=13+84+85=\boxed{182}$
Solution modified/fixed from Shaddoll's solution.
## Solution 2
We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.
First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\geq2$. The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)$. Thus,
$p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$.
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)$. (1)
Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even,
$p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$.
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)$. (2)
Adding (1) and (2) together, we find that
$6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB$.
Setting $a=C_{0}B$, $b=C_{0}A$, and $c=AB$, we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\boxed{182}$.
Solution by brightaz
## Solution 3
$[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("A", A, SE); label("B", B, NW); label("C_0", C0, SW); label("C_1", C1, NE); label("C_2", C2, W); label("C_3", C3, NE); label("C_4", C4, W); label("a", (B+C0)/2, W); label("b", (A+C0)/2, S); label("c", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]$
Let $a = BC_0$, $b = AC_0$, and $c = AB$. Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$.
The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$, shown in red in the figure below. Since each of the triangles $\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\frac{a+c}{b}$. In other words, we have that $a\left(\frac{a+c}{b}\right) = 6p$.
Guessing and checking Pythagorean triples reveals that $a = 84$, $b=13$, $c = 85$, and $p = a + b + c = \boxed{182}$ satisfies this equation.
$[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("A", A, SE); label("B", B, NW); label("C_0", C0, SW); label("a", (B+C0)/2, W); label("b", (A+C0)/2, S); label("c", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); draw(C0--B, blue); [/asy]$
## Solution 4
This solution proceeds from $\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)$. Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$. Thus $a = 168n^2, b = 26n^2, c = 170n^2$. Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = \boxed{182}$.
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0
# What two numbers add up to make seven and multiply together to make three?
Updated: 9/18/2023
Wiki User
14y ago
This is equivalent to solving the simultaneous equation set {x, y: x + y = 7, xy = 3}.
By rearranging and substituting, we find that x = 7 - y, and that (7 - y)y = 3.
We can expand and simply to gain -y2 + 7y - 3 = 0, a relatively simple equation called a quadratic, and one which has a precise formula for its solution.
Using this formula, we find that y = 0.459 and y = 6.541. Note that because these add to 7, we will end up with the same answers for x, i.e. for y = 0.459, x = 6.541 and vice versa.
Regardless of the way you choose to then label the numbers, the solution is (0.459, 6.541). These add to 7 and multiply to 3.
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14y ago
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Q: What two numbers add up to make seven and multiply together to make three?
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Related questions
### How would you find the LCM of three prime numbers?
Multiply them together.
### What three numbers do you multiply together to get 210?
They can be: 6*5*7 = 210
### What two numbers multiply to 27?
Nine times three equals twenty-seven.
33
### What three numbers can you multiply together to make 245?
7 × 7 × 5 = 245
### What does associative properties of multiplication mean?
The associative property of multiplication states that for any three numbers a, b and c, (a * b) * c = a * (b * c) and so we can write either as a * b * c without ambiguity. ie, when multiplying three numbers together, you can multiply the first two together and then multiply the result of that by the third, or multiply the second two numbers together and multiply that result by the first, and you will get the same answer.
### What numbers multiply together to get -3?
minus one and three -1 X 3 = -3
### What is the smallest composite number that has three different prime factors?
Just multiply the first three prime numbers together.
### What 3 numbers multiple equals 54?
Three numbers that multiply together to equal 54 are 1x2x27, 1x3x18, 1x6x9, and 2x3x9.
### What three consecutive numbers multiply together to give you 990?
The numbers are 9, 10 and 11.
2, 4, 27
### What three numbers multiply together to make 120?
4.93, i just answered my own question :L
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# Simplification III – Simplification Questions solved using Unit Digit Method for IBPS PO Exam
In this post, we will discuss Unit Digit Method to solve Simplification Questions for IBPS PO Exam
Do huge simplification questions scare you in IBPS PO Exam? Do you often tend to skip these questions due to large calculations? Here we are with a simple yet smart method which will help you to solve simplification questions in IBPS PO Exam within 10 to 15 seconds.
In our previous post Simplification I- Simplification Tricks to solve Simplification Problems and Simplification II- Simplification Questions for IBPS PO and SSC CGL Exam we have solved few Simplification Questions using the smart method.In this post, we will discuss the unit digit method which will help you to find the answers through smart ways.
Unit digit Method
Unit digit method is a method in which we find the solution by taking the unit digits of the numbers and try to find the unit digit of the answer which will help us eliminate options in IBPS PO Exam. Unit digit method works only with Addition, Subtraction and Multiplication. We cannot find the solution through unit digit method for Division.
Example 8: Simplification Questions for IBPS PO and SSC CGL Exams
Question:
(12 x 19) + (13 x 8) = (15 x 14) - ?
1) 124 2) 122 3)126 4)128 5) None of these
Solution:
Step 1:
(12 x 9) = 9 x 2 = 8
(13 x 8) = 3 x 8 = 4
(15 x 14) = 0
Step 2:
Now as we know the unit digits of the answer, we can balance the equation to find the unit digit of the unknown value-
8 + 4 = 0 - ?
The unit digit of the right-hand side equation is 2
2 = 0 - ?
? = 2
Step 3:
As we now know the unit digit of the unknown value we can start eliminating the options one by one.
Option 1 is not the correct answer as it doesn’t end with 2.
Option 2 is the suitable answer as it ends with 2.
Option 3 is not the correct answer as it doesn’t end with 2.
Option 4 is not the correct answer as it doesn’t end with 2.
Option 5 none of these can be the correct answer as we don’t know the whole number.
Step 4:
After cutting down the options we now have Option 2 and Option 5 which can be the suitable answers. There is 50% chance of them being correct. In exams such as IBPS PO, it is important that we save our time. So, we are supposed to make a calculated guess. You can either choose the option 2 or option 5 whichever option you feel to be correct. In IBPS PO Exam, we need to take the risks, as we have limited time. But, here we select option 2 and move on. The choice is up to you to solve the whole problem and spend a minute in it or solve it in few seconds and move on by taking the risk.
Therefore, the answer is option 2: 122
Example 9: Simplification Questions for IBPS PO and SSC CGL Exams
Question:
?² + 65² = 160² - 90² - 7191
1) 75 2)77 3)79 4) 81 5)None of these.
Solution:
Step 1:
This question appears to be tough but it can be solved in 5 to 10 seconds through the unit digit method in IBPS PO Exam.
Now finding the square root is a tough task,
So just see the digit of the unit place and perform the operations as follow,
?² + 5 = 0 - 0 -1 ( as we need to borrow a digit from the tens place)
?² + 5 = 9
?² = 9 – 5
?² = 4
Step 2:
As we know that unit place of the unknown value we can start eliminating the options.
Option 1 is not the correct answer as it doesn’t end with 4.
Option 2 is not the correct answer as it doesn’t end with 4.
Option 3 is not the correct answer as it doesn’t end with 4.
Option 4 is not the correct answer as it doesn’t end with 4.
Option 5 none of these is the correct answer as other option doesn’t satisfy the condition.
Therefore, the correct answer is option 4: None of these.
Example 10: Simplification Questions for IBPS PO and SSC CGL Exams
Question:
398 x ? x 7 = 47362
1) 15 2)13 3)17 4)19 5) None of these
Solution:
Step 1:
By using the unit digit method, we can get the unit digit of the unknown value which will help you to eliminate the options.
First, multiply the unit digits-
By multiplying 8 x 7 we get the unit digit as 6.
Now-
6 x ? = 2 ( as L.H.S = R.H.S)
Step 2:
Multiply 6 with the unit digit of the options to get the unit digit as 2.
For example, by multiplying 6 with option 1 of the unit digit i.e. 5 we get 0. So, option 1 is not eligible to be the answer, we eliminate it.
Likewise,
6 x 3 = 8 ; option 2 eliminated
6 x 7 = 2 ; option 3 satisfies our requirement
6 x 9 = 4 ; option 4 eliminated.
Option 5; none of these can be our answer too.
Step 3:
Now that we know option 2 and option 5 can be our answer. The option which highly seems to be correct can be marked. Though there are 50% chances of it being wrong, In IBPS PO and SSC CGL we cannot afford to waste time, just mark the answer and move on instead of solving the entire sum. Again the choice is yours.
Therefore, the answer is Option 3: 17
Example 11: Simplification Questions for IBPS PO and SSC CGL Exams
Question:
1)181/331 2)164/441 3)155/246 5)161/241
Solution:
This question is time-consuming. It takes around two and half minute to solve this question. The smart way to solve this simplification question is to leave the question. Move on to the next question and come back to this question at the end when the time is left.
Do write down in the comment section, how this post helped you to solve these Simplification Question for IBPS PO Exam and Save Time.
Stay tuned for more Simplification Questions.
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# Math Assignment Help With Roman Numerals
## Chapter 6. Roman Numerals
6.1 Introduction: Roman numerals are number system of ancient Rome represented in terms of alphabets. It does not include zero. They are commonly used in numbered lists, clock, pages preceding the main body of a book, months of the year, and even for naming successive political leaders or children like Elizabeth II.
6.2 Symbols
I – Stands for One
V - Stands for Five
X- Stands for Ten
L- Stands for Fifty
C- Stands for One Hundred
D- Stands for Five hundred
M- Stands for One Thousand
### 6.3 How to read and write Roman numerals?
Roman numerals are written by iterating the symbols.
I = 1
II = 2
III = 3
IV = 4
V = 5
VI = 6
VII = 7
VIII = 8
IX = 9
X = 10
Numerals above X (10) are XI, XII, XIII, XIV, XV, XVI, XVII, XVIII, XIX and XX is 20.
Then XXX = 30,
XXXI = 31
and so on.
XL = 40, XLI = 41 etc.
LX = 60, LXX = 70, LXXX = 80, XC = 90
### 6.3 How to write Large numbers?
Let's try to create a large number in Roman numerals. Let's write 1947 in roman.
Start from the left:
1000 = M
900 = CM
40 = XL
7 = VII
Now put them together MCMXLVII
### 6.4 Subtracting rules:
IV means 1 from 5, while VI means 5+1. Therefore a digit could be both positive and negative depending on whether it is to the right or left of the higher digit.
4 = IV (5 - 1)
9 = IX (10-1)
19 = XIX (10+10-1)
40 = XL (50-10)
90 = XC (100-10)
Various rules can be used for subtraction with Roman Numerals.
(a) Only subtract powers of ten (I, X, or C, but not V or L).
(b) Only subtract one numeral from another.
(c) Do not subtract a numeral from another that is more than 10 times greater.
### Email Based Assignment Help in Roman Numerals
To Schedule a Roman Numerals tutoring session
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# What Is Inverse Function In Math?
## How do you find the inverse of a function?
Finding the Inverse of a Function
1. First, replace f(x) with y.
2. Replace every x with a y and replace every y with an x.
3. Solve the equation from Step 2 for y.
4. Replace y with f−1(x) f − 1 ( x ).
5. Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.
## Why do we find the inverse of a function?
Inverses. A function normally tells you what y is if you know what x is. The inverse of a function will tell you what x had to be to get that value of y. The domain of f is the range of f 1 and the range of f is the domain of f 1.
## What is the inverse of 12?
The multiplicative inverse of 12 is 1/12.
## Do all functions have an inverse?
Not all functions have inverse functions. Those that do are called invertible. For a function f: X → Y to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that f(x) = y.
## How do you solve inverse variations?
An inverse variation can be represented by the equation xy=k or y=kx. That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0. Suppose y varies inversely as x such that xy=3 or y=3x.
You might be interested: Quick Answer: How To Learn Math Quickly?
## What is the relationship between a function and its inverse?
For a function that is defined to be the set of all ordered pairs (x, y), the inverse of the function is the set of all ordered pairs (y, x). The domain of the function becomes the range of the inverse of the function. The range of the function becomes the domain of the inverse of the function.
## What does inverse mean?
In mathematics, the word inverse refers to the opposite of another operation. Let us look at some examples to understand the meaning of inverse. Example 1: So, subtraction is the opposite of addition. Hence, addition and subtraction are opposite operations.
## What’s the inverse of 3?
The multiplicative inverse of 3 is 1/3.
## What is the property of inverse?
The purpose of the inverse property of addition is to get a result of zero. The purpose of the inverse property of multiplication is to get a result of 1. We use inverse properties to solve equations. Inverse Property of Addition says that any number added to its opposite will equal zero.
## How do you know if a function has an inverse algebraically?
To find the inverse of a function using algebra ( if the inverse exists), set the function equal to y. Then, swap x and y and solve for y in terms of x.
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# The centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a
The centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and Questions Navigation Menu preliminary estimate of the proportion who smoke of .26.
a) How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02?(to the nearest whole number) Use 95% confidence.
b) Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population (to 4 decimals)?
c) What is the 95% confidence interval for the proportion of smokers in the population?(to 4 decimals)?
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Step 1
Given:
Margin of error (E) = 0.02
Confidence level = 95%
Population proportion (p) = 0.26
Step 2
(a)
The sample size could be calculated using the following formula:
$n=p\left(1-p\right){\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}$
Step 3
The sample size at 95% confidence level can be calculated as:
The value of $\frac{{Z}_{\alpha }}{2}$ corresponding to 95% confidence level is 1.96
$n=p\left(1-p\right){\left(\frac{{Z}_{\frac{\alpha }{2}}}{E}\right)}^{2}$
$=0.26\left(1-0.26\right){\left(\frac{1.96}{0.02}\right)}^{2}$
$=0.1924×9604$
=1847.810
$\approx 1848$
Step 4
(b)
Here, X = 520 and n = 1848.
The point estimate can be calculated as:
$\stackrel{^}{p}=\frac{X}{n}$
$=\frac{520}{1848}$
=0.2814
Step 5
(c)
The 95% confidence interval for the proportion of smokers in the population can be calculated as:
$CI=\stackrel{^}{p}±{Z}_{\frac{\alpha }{2}}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$=0.2814±1.96×\sqrt{\frac{0.2814\left(1-0.2814\right)}{1848}}$
$=0.2814±0.0205$
=(0.2609,0.3019)
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# Dinosaur Sensory Table
I rolled out the sensory table today. Thank you, Mr. Surber, for loading it on the U-haul truck and bringing it to our temporary classroom! I filled the table with a new item and thought it would be fun to have the students guess what that item was….
Beans! 50 pounds of beans! What better to dig in for our dinosaur unit, then beans! Today, the students dug in the beans to uncover double tens frames.
They counted the number of squares that were colored in and found the corresponding number on the worksheet.
# Terrific Ten Frame
In math this week, students have been exploring the ten frame. A ten frame is simply a graphic way for students to see numbers.
This ten frame shows the number eight.
We started the lesson with some fun activities on the Smart Board where students had to build numbers less than ten.
Then students had to look at a ten frame and identify the number of dogs in the frame.
As an extension, I asked students if there were five dogs in the ten frame, how many empty spots were there? Five of course! This paved the way for our next activity that I like to call Spill the Beans. Each student received a cup with ten beans that were white on one side and orange on the other. They were to shake up the cup and spill out the beans, sorting them into groups according to color. Students then recorded how many orange beans they had and how many white beans they had. They colored in the number of orange beans on the ten frame. Finally, students wrote and solved the addition equation.
So what exactly does a ten frame teach? Ohhhh….so many concepts!
First, students learn number sense. It is through this number exploration that students learn the value of each number. For example, if numbers don’t fill one row, then the number is less than five. If the numbers fill more than the first row, the number is larger than five. Number sense is key to students’ success in advanced math.
Second, students look at numbers as sums including five. Students make the numbers to 10 and write them as composites of 5 and another number: i.e., 8= 5 + 3.
Third, students look at numbers in the context of ten. In other words, by using a ten frame, they figure out how many are needed to add to 6 to make ten. This will later help students decompose addition greater than ten: i.e., 8 plus 8 is 8 plus 2 plus 6, or 16.
In short, a ten frame can be used to build number sense, help students gain “mental math” fluency, and to better understand how to use the math strategies of “composing and decomposing” numbers, to complete operations over places (i.e., from tens to hundreds, or thousands to hundreds.) Go ten frame! Wait until you see the math centers I have planned for the students!
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# I would like to draw an isosceles triangle with the vertex measuring 75 degrees. Whats is the measure of each base angle?
Nov 21, 2015
52.5˚
#### Explanation:
In an isosceles triangle, there are at least $2$ congruent angles—the base angles.
Here's what we know:
The vertex angle is 75˚.
The sum of the angles in a triangle is 180˚.
The base angles are congruent.
If we call the measure of each base angle $x$, we can write the following equation relating the degree measures of all three angles in the triangle:
$x + x + 75 = 180$
Therefore,
$2 x + 75 = 180$
$2 x = 105$
$x = 52.5$
So, the vertex angle of this triangle is 75˚ and each base angle is 52.5˚.
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# Points, Lines, Planes and Angles Mathematics Notes | EduRev
## Mathematics : Points, Lines, Planes and Angles Mathematics Notes | EduRev
The document Points, Lines, Planes and Angles Mathematics Notes | EduRev is a part of the Mathematics Course Additional Topics for IIT JAM Mathematics.
All you need of Mathematics at this link: Mathematics
An introduction to geometry
A point in geometry is a location. It has no size i.e. no width, no length and no depth. A point is shown by a dot.
A line is defined as a line of points that extends infinitely in two directions. It has one dimension, length. Points that are on the same line are called collinear points.
A line is defined by two points and is written as shown below with an arrowhead.
Two lines that meet in a point are called intersecting lines.
A part of a line that has defined endpoints is called a line segment. A line segment as the segment between A and B above is written as:
A plane extends infinitely in two dimensions. It has no thickness. An example of a plane is a coordinate plane. A plane is named by three points in the plane that are not on the same line. Here below we see the plane ABC.
A space extends infinitely in all directions and is a set of all points in three dimensions. You can think of a space as the inside of a box.
Measure line segments
The length of a line segment can be measured (unlike a line) because it has two endpoints. As we have learnt previously the line segment can be written as
While the length or the measure is simply written AB. The length could either be determined in Metric units (e.g. millimeters, centimeters or meters) or Customary units (e.g. inches or foot).
Two lines could have the same measure but still not be identical.
AB and CD have the exact same measure and are said to be congruent and is noted as
This is read as the line AB is congruent to the line CD.
Finding distances and midpoints
If we want to find the distance between two points on a number line we use the distance formula:
AB = |b − a| or |a − b|
Example
Point A is on the coordinate 4 and point B is on the coordinate -1.
AB = |4−(−1)|=|4+1|=|5|= 5
If we want to find the distance between two points in a coordinate plane we use a different formula that is based on the Pythagorean Theorem where (x1,y1) and (x2,y2) are the coordinates and d marks the distance:
The point that is exactly in the middle between two points is called the midpoint and is found by using one of the two following equations.
Method 1: For a number line with the coordinates a and b as endpoints:
Method 2: If we are working in a coordinate plane where the endpoints has the coordinates (x1,y1) and (x2,y2) then the midpoint coordinates is found by using the following formula:
Measure and classify an angle
A line that has one defined endpoint is called a ray and extends endlessly in one direction. A ray is named after the endpoint and another point on the ray e.g.
The angle that is formed between two rays with the same endpoint is measured in degrees. The point is called the vertex
The vertex is written as
In algebra we used the coordinate plane to graph and solve equations. You can plot lines, line segments, rays and angles in a coordinate plane.
In the coordinate plane above we have two rays
That form an angle with the vertex in point B.
You can use the coordinate plane to measure the length of a line segment. Point B is at (-2, -2) and C (1. -2). The distance between the two points is 1 - (-2) = 3 units.
Angles can be either straight, right, acute or obtuse.
An angle is a fraction of a circle where the whole circle is 360°. A straight angle is the same as half the circle and is 180° whereas a right angle is a quarter of a circle and is 90°.
You measure the size of an angle with a protractor.
Two angles with the same measure are called congruent angles. Congruent angles are denoted as
Or could be shown by an arc on the figure to indicate which angles that are congruent.
Two angles whose measures together are 180° are called supplementary e.g. two right angles are supplementary since 90° + 90° = 180°.
Two angles whose measures together are 90° are called complementary.
m∠A +m∠B = 180
m∠C + m∠D = 90
Polygons
A polygon is a closed figure where the sides are all line segments. Each side must intersect exactly two others sides but only at their endpoints. The sides must be noncollinear and have a common endpoint.
A polygon is usually named after how many sides it has, a polygon with n-sides is called a n-gon. E.g. the building which houses United States Department of Defense is called pentagon since it has 5 sides.
Sides Polygon 3 triangle 4 quadrilateral 5 pentagon 6 hexagon 7 heptagon 9 nonagon 10 decagon
A regular polygon is a polygon in which all sides are congruent and all the angles are congruent.
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# Lesson Travel and Distance problems
Algebra -> Algebra -> Customizable Word Problem Solvers -> Travel -> Lesson Travel and Distance problems Log On
Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Word Problems: Travel and Distance Solvers Lessons Answers archive Quiz In Depth
This Lesson (Travel and Distance problems) was created by by ikleyn(4) : View Source, Show
## Travel and Distance problems
In this lesson simple typical word problems on Travel and Distance are presented to show the approach and the methodology of their solutions.
### Problem 1. Two objects moving toward each other
Two cars entered an Interstate highway at the same time and traveled toward each other
(see Figure 1). The initial distance between cars was 390 miles.
First car was going 70 miles per hour, the second car was going 60 miles per hour.
How long will it take for two cars to pass each other?
What distance each car will travel before passing?
Figure 1. Two objects moving toward each other
Solution
We will solve the problem by reducing it to the simple linear equation.
Let us denote as the unknown time in hours for two cars to pass each other.
Since the rate of the first car is 70 miles per hour, the distance it travels for time is equal to miles.
Since the rate of the second car is 60 miles per hour, the distance it travels for time is equal to miles.
At the moment two cars pass each other, the sum of these distances is equal to the initial distance between two cars, i.e. 390 miles.
This gives us an equation
.
Simplify the left side of the equation using the distributive law:
.
So, our equation takes the form
.
Divide both sides of the equation by 130. You get
.
Thus, it will take 3 hours for two cars to pass each other.
Now, you can check the solution.
The distance that the first car will travel for 3 hours is equal to miles.
The distance that the second car will travel for 3 hours is equal to miles.
Since , the solution is correct.
Answer. It will take 3 hours for two cars to pass each other.
First car will travel 210 miles, and the second car will travel 180 miles before passing.
Note. For those who prefer more physical thinking, the following alternative solution may seem more suitable.
The value of miles per hour is the rate of decreasing the distance between two cars, so the time before passing is equal to hours. Based on this, the distance traveled by the first car for this time is equal to miles, and the distance traveled by the second car for this time is equal to miles.
Surely, the results are the same.
### Problem 2. Two objects moving in the same direction
Two cars entered an Interstate highway at the same time at different locations and traveled
in the same direction as shown in Figure 2. The initial distance between cars was 30 miles.
First car was going 70 miles per hour, the second car was going 60 miles per hour.
How long will it take for the first car to catch the second one?
What distance will each car travel before the first car catches the second one?
Figure 2. Two objects moving in the same direction
Solution
I will show you first how to solve the problem by reducing it to the simple linear equation.
Let us denote as the unknown time in hours for the first car to catch the second one.
Since the rate of the first car is 70 miles per hour, the distance it travels for time is equal to miles.
Since the rate of the second car is 60 miles per hour, the distance it travels for time is equal to miles.
At the moment when the first car catches the second one, the difference of these distances is equal to the initial distance between two cars, i.e. 30 miles.
This gives us an equation
.
Simplify the left side of the equation using the distributive law:
.
So, our equation takes a form
.
Divide both sides of the equation by 10. You get
.
Thus, it takes 3 hours for the first car to catch the second one.
You can check the solution.
The distance first car will travel for 3 hours is equal to miles.
The distance the second car will travels for 3 hours is equal to miles.
Since , the solution is correct.
Answer. It will take 3 hours for the first car to catch the second one.
First car will travel 210 miles, and the second car will travel 180 miles before the first car catches the second one.
Note. Again, there is alternative physical solution.
The value of miles per hour is the rate of decreasing the distance between two cars in this case, so the time before passing is equal to hours. Based on this, the distance traveled by the first car for this time is equal to miles, and the distance traveled by the second car for this time is equal to miles.
The results are the same.
### Problem 3. Two objects moving toward each other
Two cars entered an Interstate highway at the same time and traveled toward each other
(see Figure 3). The initial distance between cars was 390 miles.
The speed of the first car was in 10 miles per hour greater than that of the second car.
It took 3 hours for two cars to pass each other.
What was the speed of each car? What distance each car traveled before they pass each other?
Figure 3. Two objects moving toward each other
Solution
The situation is similar to that of the Problem 1, but the time to get passing is given and the speed is under the question in this case.
Again, we will solve the problem by reducing it to the simple linear equation.
Let us denote as the unknown speed (in miles per hour) of the first car.
Then the distance the first car traveled for 3 hours is equal to miles.
Since the rate of the first car is in 10 miles per hour greater than that of the second car, the rate of the second car is equal to miles per hour.
Then the distance the second car traveled for 3 hours is equal to miles.
At the moment two cars pass each other, the sum of these distances is equal to the initial distance between two cars, i.e. 390 miles.
This gives us an equation
.
Simplify the left side of the equation by opening brackets and collecting like terms:
.
So, our equation takes the form
.
Transfer the constant term from the left side to the right and collect like terms. You get
.
Divide both sides by 6. You get
.
Thus, the speed of the first car was equal to 70 miles per hour.
Hence, the speed of the second car was equal to miles per hour.
The distance first car traveled for 3 hours was equal to miles.
The distance the second car traveled for 3 hours was equal to miles.
Since , the solution is correct.
Answer. The speed of the first car was 70 miles per hour, the speed of the second car was 60 miles per hour.
First car traveled 210 miles, and the second car traveled 180 miles before they pass each other.
This lesson has been accessed 11305 times.
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# STUDY GUIDE FOR PROBLEM # 15
STUDY GUIDE FOR PROBLEM # 15
By Anthony Marc
15) It is claimed that the average annual per person spending on prescription drugs is \$410. If a survey of 65 randomly selected people indicated an average spending of \$425 with a standard deviation of \$45, do we reject the claim that the average is \$410? Use a 5% level of significance.
• The first thing to do is construct a hypothesis test with a null and alternative hypothesis to test whether the claim will be greater, less than or equal to the average of what is already stated. The tails of the test refer to the area in the normal curve which corresponds to the alternative hypothesis (the region where we you reject the null hypothesis). The claim states that the average per person annually spending on prescription drugs is \$410.
• Ho (The null hypothesis) would be that the average spent would be equal to \$410 and Ha (the alternative hypothesis) would also not be equal to \$410. Both hypotheses are considered not equal due to the fact that average is NOT more or less than \$410 but is exactly that number \$410 being spent.
• Itās a two tailed test
• The level of significance is the amount on the normal curve whose maximum probability determines whether or not to reject the claim from the mean to the rejection area.
• The level of significance is 5% which comes out to 5/100= .0500. Since the normal distribution curve is two tailed the level of significance has to be divided by two to determine the rejection area.
• .0500/2= .0250 for both ends of the curve. Looking on the table the critical value comes out to z = -1.96.
Ā
Ā SOLUTION:
• The problem would be done as: ,Ā Z= 425-410/45 and the square root of 65
• What made it easier for me to solve this problem was do it step by step starting off with the square root of n= 65, which comes out to 8.062257748.
• Then divide that sum by the standard deviation of 45, so you would divide 45/8.062257748 which comes to the answer of 5.581563057.
• Then subtract the average by the sample mean (425-410) and divide (5.581563057) which come to the answer of Z= 2.687419249.
• So as a result we REJECT the claim.
The claim is rejected due to the fact that the Z score is higher than the critical value which determines whether or not it falls within the range of values whether the claim is acceptable. So people annually spend more than the averageĀ \$410 per year on prescription drugs.
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## Give a whole number, give the number of its multiples and click on OK to display its multiples
Find the multiples of an integer inside an interval of two integers
Find all the factors of a whole number
Retrieve prime numbers inside an interval of two whole numbers
## What are the multiples of 31?
The ten first multiples of 31 greater than zero are: 31 ; 62 ; 93 ; 124 ; 155 ; 186 ; 217 ; 248 ; 279 ; 310.
In arithmetic a positive integer is a multiple of 31 when it can be divided by 31; this means that its division by 31 gives an integer quotient and zero remainder (if an integer is a multiple of 31 then 31 is its divisor). If a whole number is a multiple of 31 it exists another whole number we can multiply by 31 to get it.
Example 1:
155 divided by 31 gives an integer quotient equal to 5 and zero remainder, the multiplication of 31 by 5 returns 155, then 155 is a multiple of 31. 5 and 31 are called factors or divisors of 155 and 155 is their common multiple.
Example 2:
If we divide 62 by 31 the quotient is equal to 2 and the remainder is nul. By multiplying 31 by 2 we get 62 again, so 62 is a multiple of 31 (so we can say 62 is divisible by 31 or 31 divides 62).
Example 3:
By dividing 60 by 31 we get 1 as quotient and a remainder equal to 29, since the remainder is not equal to zero then 60 is not a multiple of 31 (31 doesn't divide 60 or 60 is not divisible by 31). we can't find a positive integer whose the multiplication by 31 gives 60.
Remark:
31 is a prime number (opposite of a composite number), if the decomposition into product of prime factors of a whole number contains 31 so that number is a multiple of 31.
In maths the whole numbers have an infinite number of multiples, 31 is an odd number, if a number is an odd number its multiples are at the same time even numbers and odd numbers, here are some positive integers that are multiples of 31 (between 1 to 10000):
Multiples of 31 up to 1 and less than 2000:
31 ; 62 ; 93 ; 124 ; 155 ; 186 ; 217 ; 248 ; 279 ; 310 ; 341 ; 372 ; 403 ; 434 ; 465 ; 496 ; 527 ; 558 ; 589 ; 620 ; 651 ; 682 ; 713 ; 744 ; 775 ; 806 ; 837 ; 868 ; 899 ; 930 ; 961 ; 992 ; 1023 ; 1054 ; 1085 ; 1116 ; 1147 ; 1178 ; 1209 ; 1240 ; 1271 ; 1302 ; 1333 ; 1364 ; 1395 ; 1426 ; 1457 ; 1488 ; 1519 ; 1550 ; 1581 ; 1612 ; 1643 ; 1674 ; 1705 ; 1736 ; 1767 ; 1798 ; 1829 ; 1860 ; 1891 ; 1922 ; 1953 ; 1984
Multiples of 31 greater than 2000 and less than 4000:
2015 ; 2046 ; 2077 ; 2108 ; 2139 ; 2170 ; 2201 ; 2232 ; 2263 ; 2294 ; 2325 ; 2356 ; 2387 ; 2418 ; 2449 ; 2480 ; 2511 ; 2542 ; 2573 ; 2604 ; 2635 ; 2666 ; 2697 ; 2728 ; 2759 ; 2790 ; 2821 ; 2852 ; 2883 ; 2914 ; 2945 ; 2976 ; 3007 ; 3038 ; 3069 ; 3100 ; 3131 ; 3162 ; 3193 ; 3224 ; 3255 ; 3286 ; 3317 ; 3348 ; 3379 ; 3410 ; 3441 ; 3472 ; 3503 ; 3534 ; 3565 ; 3596 ; 3627 ; 3658 ; 3689 ; 3720 ; 3751 ; 3782 ; 3813 ; 3844 ; 3875 ; 3906 ; 3937 ; 3968 ; 3999
Multiples of 31 between 4000 and 6000:
4030 ; 4061 ; 4092 ; 4123 ; 4154 ; 4185 ; 4216 ; 4247 ; 4278 ; 4309 ; 4340 ; 4371 ; 4402 ; 4433 ; 4464 ; 4495 ; 4526 ; 4557 ; 4588 ; 4619 ; 4650 ; 4681 ; 4712 ; 4743 ; 4774 ; 4805 ; 4836 ; 4867 ; 4898 ; 4929 ; 4960 ; 4991 ; 5022 ; 5053 ; 5084 ; 5115 ; 5146 ; 5177 ; 5208 ; 5239 ; 5270 ; 5301 ; 5332 ; 5363 ; 5394 ; 5425 ; 5456 ; 5487 ; 5518 ; 5549 ; 5580 ; 5611 ; 5642 ; 5673 ; 5704 ; 5735 ; 5766 ; 5797 ; 5828 ; 5859 ; 5890 ; 5921 ; 5952 ; 5983
Multiples of 31 up to 6000 and less than 8000:
6014 ; 6045 ; 6076 ; 6107 ; 6138 ; 6169 ; 6200 ; 6231 ; 6262 ; 6293 ; 6324 ; 6355 ; 6386 ; 6417 ; 6448 ; 6479 ; 6510 ; 6541 ; 6572 ; 6603 ; 6634 ; 6665 ; 6696 ; 6727 ; 6758 ; 6789 ; 6820 ; 6851 ; 6882 ; 6913 ; 6944 ; 6975 ; 7006 ; 7037 ; 7068 ; 7099 ; 7130 ; 7161 ; 7192 ; 7223 ; 7254 ; 7285 ; 7316 ; 7347 ; 7378 ; 7409 ; 7440 ; 7471 ; 7502 ; 7533 ; 7564 ; 7595 ; 7626 ; 7657 ; 7688 ; 7719 ; 7750 ; 7781 ; 7812 ; 7843 ; 7874 ; 7905 ; 7936 ; 7967 ; 7998
The multiples of 31 up to 8000 and less than 10000:
8029 ; 8060 ; 8091 ; 8122 ; 8153 ; 8184 ; 8215 ; 8246 ; 8277 ; 8308 ; 8339 ; 8370 ; 8401 ; 8432 ; 8463 ; 8494 ; 8525 ; 8556 ; 8587 ; 8618 ; 8649 ; 8680 ; 8711 ; 8742 ; 8773 ; 8804 ; 8835 ; 8866 ; 8897 ; 8928 ; 8959 ; 8990 ; 9021 ; 9052 ; 9083 ; 9114 ; 9145 ; 9176 ; 9207 ; 9238 ; 9269 ; 9300 ; 9331 ; 9362 ; 9393 ; 9424 ; 9455 ; 9486 ; 9517 ; 9548 ; 9579 ; 9610 ; 9641 ; 9672 ; 9703 ; 9734 ; 9765 ; 9796 ; 9827 ; 9858 ; 9889 ; 9920 ; 9951 ; 9982
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Home > Grade 7 (page 4)
# Grade 7
## Maximizing Rectangular Prism Volume
Directions: Using the whole number 1 through 9, at most one time each, fill in the boxes to list the dimensions of a rectangular prism with the greatest volume. Source: Robert Kaplinsky
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## Maximizing Rectangular Prism Surface Area
Directions: Using the whole number 1 through 9, at most one time each, list the dimensions of a rectangular prism with the greatest surface area. Source: Robert Kaplinsky
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## Converting Between Fractions and Decimals
Directions: Using the numbers 0 through 9, at most one time each, fill in each of the boxes so that the fraction equals the decimal. Source: Robert Kaplinsky
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## Undefined Quotient with Fraction Division
Directions: Give at least two different examples where the quotient is undefined by filling in the boxes with whole numbers 0 through 9, using each number at most once for each example. Source: Daniel Luevanos
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## Product of Distributive Property
Directions: Decide if 30x – 12 could be a result of using the distributive property. If it is, find the possible combinations of factors whose product would be 30x – 12 (using integer coefficients and constants). Source: adapted from Nathan Charlton
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## Multiply to Make -64
Directions: Find three numbers whose product is -64. You may use integers from -10 to 10. You may not use the same absolute value twice. Find all possible combinations. Source: Nathan Charlton and Daniel Martinez
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## Subtracting Decimals (Middle School)
Directions: Make the smallest (or largest) difference by filling in the boxes using the whole numbers 1-9 no more than one time each. Note: This problem’s difficulty can be adjusted by altering the number of digits (boxes), picking smallest or largest, or by picking either a positive, negative, or both. Source: Robert Kaplinsky
Read More »
## Dividing Two-Digit Numbers (Middle School)
Directions: Make the smallest (or largest) quotient by filling in the boxes using the whole numbers 1-9 no more than one time each Note: This problem’s difficulty can be adjusted by altering the number of digits (boxes), picking smallest or largest, or by picking either a positive, negative, or both. Source: Robert Kaplinsky
Read More »
## Adding Two-Digit Numbers (Middle School)
Directions: Make the smallest (or largest) sum by filling in the boxes using the whole numbers 1-9 no more than one time each Note: This problem’s difficulty can be adjusted by altering the number of digits (boxes), picking smallest or largest, or by picking either a positive, negative, or both. Source: Robert Kaplinsky
Read More »
## Subtracting Two-Digit Numbers (Middle School)
Directions: Make the smallest (or largest) difference by filling in the boxes using the whole numbers 1-9 no more than one time each Note: This problem’s difficulty can be adjusted by altering the number of digits (boxes), picking smallest or largest, or by picking either a positive, negative, or both. Source: Robert Kaplinsky
Read More »
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“A Box Of Cubes: How Many Can Fit?”
You may have thought that a box can only fit a certain number of cubes, but you would be wrong. With a little bit of creativity, you can fit an infinite number of cubes into a box.
How many cubes are in the box
It’s a common question, one that we’ve all asked ourselves at one point or another: how many cubes are in the box? The answer, of course, is dependent on the size of the box. A small box might contain a few dozen cubes, while a large box could hold hundreds or even thousands.
But what if we’re talking about an average-sized box? How many cubes can we expect to find inside?
To get an estimate, we can start by looking at the volume of a cube. A cube has six faces, each with an area of s^2 (where s is the length of a side). The total volume of the cube is therefore 6s^2.
Now, let’s say our box has a length of l, a width of w, and a height of h. The internal volume of the box is then lwh. We can convert this to cubic units by dividing by 1 l = 1 dm3 (or 1 m3). This gives us a volume of l/dm3 * w/dm3 * h/dm3 = lwh/dm3.
We can now compare the two volumes. For our cube, we have 6s^2/dm3. For our box, we have lwh/dm3. If we set these equal to each other and solve for s, we find that s = (lwh/6)^(1/2). In other words, the length of a side of our cube is equal to the square root of the product of the length, width, and height of our box, divided by six.
So how many cubes are in our box? We can calculate this by taking the number of cubic units in our box and dividing by the number of cubic units in a single cube. That is, we take lwh/dm3 and divide by (l/dm3)(w/dm3)(h/dm3) = (lwh/6)^(1/2)/dm3 = (lwh/6)^(1/2)/s^3.
This gives us a final answer of (lwh/6)^(1/2)/s^3 = (lwh/6)^(1/2)/[(lwh/6)^(1/2)]^3 = 6^(1/2)/6 = 1/6. In other words, there is one cube in our box for every six unit cubes that could fit inside it.
Of course, this assumes that our box is filled completely with cubes. In reality, there will likely be some empty space between them. But even so, this gives us a good starting point for estimating how many cubes are in our box.
How many more cubes does Stephanie need to put in the box
Stephanie is trying to figure out how many more cubes she needs to put in the box. She knows that the box is 1 foot long, 1 foot wide, and 1 foot tall. She also knows that each cube is 1 inch long, 1 inch wide, and 1 inch tall. So far, she has put 25 cubes in the box. How many more does she need to put in the box?
To figure this out, Stephanie will need to do some math. She will need to calculate the volume of the box and the volume of the cubes. The volume of the box is 1 foot times 1 foot times 1 foot, which equals 1 cubic foot. The volume of the cubes is 25 cubic inches. To find out how many more cubes Stephanie needs to put in the box, she will need to divide the volume of the box by the volume of the cubes. One cubic foot divided by 25 cubic inches equals 4. So, Stephanie needs to put 4 more cubes in the box.
What is the volume of the box
When it comes to calculating the volume of a box, there are a few different methods that can be used. The most common method is to use the formula V = l x w x h, where l is the length, w is the width and h is the height of the box. However, this formula only works for rectangular boxes. If you have a box that is not rectangular, you will need to use a different formula.
To calculate the volume of a rectangular box, you will need to measure the length, width and height of the box. Once you have these measurements, you can plug them into the formula V = l x w x h. For example, if your box has a length of 10 inches, a width of 5 inches and a height of 2 inches, the volume would be 10 x 5 x 2, or 100 cubic inches.
If you have a box that is not rectangular, you can still use the V = l x w x h formula, but you will need to find the length, width and height of each individual side first. To do this, you will need to measure the length and width of the top and bottom of the box, as well as the height of each side. Once you have these measurements, you can plug them into the formula V = l x w x h.
For example, let’s say you have a cylindrical box with a diameter of 10 inches and a height of 5 inches. To find the length and width of the top and bottom of the box, you would need to measure the diameter (10 inches) and divide it by 2 (5 inches). The length and width of each side would then be 5 inches. To find the height of each side, you would need to measure the height of the cylinder (5 inches).
Once you have these measurements, you can plug them into the formula V = l x w x h. In this example, the volume would be 5 x 5 x 5, or 125 cubic inches.
There are other formulas that can be used to calculate the volume of a box, but V = l x w x h is the most common.
How many cubes would fit in the box if it were twice as big
Assuming that we are talking about a rectangular box, the answer is simple. If the length, width, and height of the box are all doubled, then the volume of the box is also doubled. This means that twice as many cubes would fit into the box.
To visualize this, imagine that you have a square box. If you double the length and width of the box, then you have effectively created a new square box that is twice the size of the original. The area of the new square is four times the area of the original square because 2 x 2 = 4. This means that four times as many small squares could fit into the new square.
The same idea applies to a rectangular box. If you double the length, width, and height of the box, then you have effectively created a new rectangular box that is twice the size of the original. The volume of the new rectangular box is eight times the volume of the original rectangular box because 2 x 2 x 2 = 8. This means that eight times as many cubes could fit into the new rectangular box.
So, if you have a box that is 12 inches long, 12 inches wide, and 12 inches tall, and you double the length, width, and height to 24 inches each, then you have created a new box that is twice the size of the original. The new box has a volume of 24 x 24 x 24 = 5184 cubic inches. Since each cube is 1 cubic inch, this means that 5184 cubes would fit into the larger box.
How long will it take Stephanie to put 30 cubes in the box
It’s a question that has stumped Stephanie for weeks now. How long will it take her to put 30 cubes in the box? She has tried various methods, but so far nothing has worked.
The first method she tried was the ‘brute force’ approach. She simply picked up each cube and placed it in the box one at a time. This was extremely tedious and time-consuming, and after an hour she had only managed to put 10 cubes in the box.
The second method she tried was to group the cubes into smaller piles and then place them in the box all at once. This saved some time, but she still ended up with 20 cubes in the box after two hours.
The third method was similar to the second, except she used a funnel to pour the cubes into the box. This reduced the number of cubes that fell out of the pile, but she still couldn’t get all 30 cubes into the box.
Stephanie is starting to think that maybe the answer to this question is ‘it depends’. It depends on how you approach the problem, and it also depends on luck. Maybe she just needs to keep trying different methods until she finally gets all 30 cubes into the box.
What is the weight of the box with 30 cubes in it
If you’re asking what the weight of the box is with 30 cubes in it, then the answer is quite simple. The weight of the box is going to be determined by the weight of the 30 cubes inside of it. However, if you’re asking what the weight of the box itself is without taking into account the weight of the 30 cubes, then that’s a different story.
The weight of an empty box will always be lighter than a box that is filled with something, regardless of what that something may be. This is because there is less material in an empty box than there is in a box that contains objects. The more objects that are inside of a box, the heavier the box will be.
So, if you’re asking about the weight of an empty box, the answer is that it depends on the size and material of the box. A small cardboard box will weigh less than a large metal box, for example. But if you’re asking about the weight of a box that contains 30 cubes, then the answer is that it will be heavier than an empty box of the same size and material.
What is the density of the cubes
The density of the cubes is a measure of how much mass is contained in a given volume. The unit for density is typically grams per cubic centimeter (g/cm3). The density of the cubes can be affected by a number of factors, including the type of material from which they are made, the size of the cubes, and the amount of air or other material that is present in the space between the cubes.
In general, denser objects have more mass per unit volume than less dense objects. This means that if you were to take two identical objects and one was made of a more dense material, it would weigh more than the less dense object. If you took two identical objects and one was twice as large as the other, the larger object would have half the density of the smaller object.
The density of the cubes can also be affected by the presence of air or other materials in the space between them. For example, if the cubes are made of a material that is not very dense, such as styrofoam, and there is a lot of space between them, they will not weigh very much. Conversely, if the cubes are made of a very dense material, such as lead, and there is no space between them, they will weigh more than if there was air present.
In conclusion, the density of the cubes is a measure of how much mass is contained in a given volume and can be affected by a number of factors including the type of material from which they are made, the size of the cubes, and the amount of air or other material that is present in the space between them.
What material are the cubes made of
The cubes are made of a variety of materials, including wood, plastic, metal, and glass. Each material has its own unique properties that make it suited for different purposes. For example, wood is a durable material that is often used in construction, while metal is a strong material that is often used in manufacturing. Glass is a transparent material that is often used in windows and other applications where light needs to be able to pass through.
Are the cubes all the same size
No, the cubes are not all the same size. In fact, they come in a variety of sizes, from small to large. And while some people may think that all cubes are created equal, others know that there can be a big difference in size between different types of cubes.
So, why are the cubes not all the same size? Well, it all has to do with the manufacturing process. The size of the cube is determined by the mold that is used to create it. And depending on the manufacturer, they may use different sized molds to create their cubes.
But it’s not just the size of the mold that determines the size of the cube. The type of material that is used to make the cube also plays a role. For example, if a cube is made out of plastic, it will be smaller than a cube made out of wood.
So, if you’re ever wondering why the cubes you see in stores are not all the same size, now you know. It all has to do with the manufacturing process and the materials that are used to make them.
How many faces does each cube have
There are six faces on each cube.
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# Learning Objectives
### Learning Objectives
By the end of this section, you will be able to do the following:
• Explain the law of conservation of energy in terms of kinetic and potential energy
• Perform calculations related to kinetic and potential energy. Apply the law of conservation of energy
law of conservation of energy
# Mechanical Energy and Conservation of Energy
### Mechanical Energy and Conservation of Energy
We saw earlier that mechanical energy can be either potential or kinetic. In this section we will see how energy is transformed from one of these forms to the other. We will also see that, in a closed system, the sum of these forms of energy remains constant.
Quite a bit of potential energy is gained by a roller coaster car and its passengers when they are raised to the top of the first hill. Remember that the potential part of the term means that energy has been stored and can be used at another time. You will see that this stored energy can either be used to do work or can be transformed into kinetic energy. For example, when an object that has gravitational potential energy falls, its energy is converted to kinetic energy. Remember that both work and energy are expressed in joules.
Refer back to Figure 9.3. The amount of work required to raise the TV from point A to point B is equal to the amount of gravitational potential energy the TV gains from its height above the ground. This is generally true for any object raised above the ground. If all the work done on an object is used to raise the object above the ground, the amount work equals the object’s gain in gravitational potential energy. However, note that because of the work done by friction, these energy–work transformations are never perfect. Friction causes the loss of some useful energy. In the discussions to follow, we will use the approximation that transformations are frictionless.
Now, let’s look at the roller coaster in Figure 9.6. Work was done on the roller coaster to get it to the top of the first rise; at this point, the roller coaster has gravitational potential energy. It is moving slowly, so it also has a small amount of kinetic energy. As the car descends the first slope, its PE is converted to KE. At the low point much of the original PE has been transformed to KE, and speed is at a maximum. As the car moves up the next slope, some of the KE is transformed back into PE and the car slows down.
Figure 9.6 During this roller coaster ride, there are conversions between potential and kinetic energy.
### Virtual Physics
#### Energy Skate Park Basics
This simulation shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KE and PE by clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation.
Figure 9.7
Grasp Check
The bar graphs show how KE and PE are transformed back and forth. Which statement best explains what happens to the mechanical energy of the system as speed is increasing?
1. The mechanical energy of the system increases, provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing.
2. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing.
3. The mechanical energy of the system increases provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing.
4. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing.
On an actual roller coaster, there are many ups and downs, and each of these is accompanied by transitions between kinetic and potential energy. Assume that no energy is lost to friction. At any point in the ride, the total mechanical energy is the same, and it is equal to the energy the car had at the top of the first rise. This is a result of the law of conservation of energy, which says that, in a closed system, total energy is conserved—that is, it is constant. Using subscripts 1 and 2 to represent initial and final energy, this law is expressed as
$K E 1 +P E 1 =K E 2 +P E 2 . K E 1 +P E 1 =K E 2 +P E 2 .$
Either side equals the total mechanical energy. The phrase in a closed system means we are assuming no energy is lost to the surroundings due to friction and air resistance. If we are making calculations on dense falling objects, this is a good assumption. For the roller coaster, this assumption introduces some inaccuracy to the calculation.
# Calculations involving Mechanical Energy and Conservation of Energy
### Tips For Success
When calculating work or energy, use units of meters for distance, newtons for force, kilograms for mass, and seconds for time. This will assure that the result is expressed in joules.
### Watch Physics
#### Conservation of Energy
This video discusses conversion of PE to KE and conservation of energy. The scenario is very similar to the roller coaster and the skate park. It is also a good explanation of the energy changes studied in the snap lab.
Grasp Check
Did you expect the speed at the bottom of the slope to be the same as when the object fell straight down? Which statement best explains why this is not exactly the case in real-life situations?
1. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction.
2. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction.
3. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy.
4. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy.
### Worked Example
#### Applying the Law of Conservation of Energy
A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fallen 10 m?
### Strategy
Choose the equation.
9.6$K E 1 +P E 1 =K E 2 +P E 2 K E 1 +P E 1 =K E 2 +P E 2$
9.7
9.8$1 2 m v 1 2 +mg h 1 = 1 2 m v 2 2 +mg h 2 1 2 m v 1 2 +mg h 1 = 1 2 m v 2 2 +mg h 2$
List the knowns.
m = 10 kg, v1 = 0, g = 9.80
9.9$m s 2 , m s 2 ,$
h1 = 20 m, h2 = 10 m
Identify the unknowns.
KE2 and PE2
Substitute the known values into the equation and solve for the unknown variables.
Solution
9.10
9.11
Discussion
Alternatively, conservation of energy equation could be solved for v2 and KE2 could be calculated. Note that m could also be eliminated.
### Tips For Success
Note that we can solve many problems involving conversion between KE and PE without knowing the mass of the object in question. This is because kinetic and potential energy are both proportional to the mass of the object. In a situation where KE = PE, we know that mgh = (1/2)mv2.
Dividing both sides by m and rearranging, we have the relationship
2gh = v2.
# Practice Problems
### Practice Problems
A child slides down a playground slide. If the slide is 3 m high and the child weighs 300 N, how much potential energy does the child have at the top of the slide? (Round g to )
1. 0 J
2. 100 J
3. 300 J
4. 900 J
A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to the ground, converting all of its PE to kinetic energy. What is the velocity of the apple just before it hits the ground?
1. 0 m/s
2. 2 m/s
3. 10 m/s
4. 50 m/s
### Snap Lab
#### Converting Potential Energy to Kinetic Energy
In this activity, you will calculate the potential energy of an object and predict the object’s speed when all that potential energy has been converted to kinetic energy. You will then check your prediction.
You will be dropping objects from a height. Be sure to stay a safe distance from the edge. Don’t lean over the railing too far. Make sure that you do not drop objects into an area where people or vehicles pass by. Make sure that dropping objects will not cause damage.
You will need the following:
Materials for each pair of students:
• Four marbles (or similar small, dense objects)
• Stopwatch
Materials for class:
• Metric measuring tape long enough to measure the chosen height
• A scale
Instructions
Procedure
1. Work with a partner. Find and record the mass of four small, dense objects per group.
2. Choose a location where the objects can be safely dropped from a height of at least 15 meters. A bridge over water with a safe pedestrian walkway will work well.
3. Measure the distance the object will fall.
4. Calculate the potential energy of the object before you drop it using PE = mgh = (9.80)mh.
5. Predict the kinetic energy and velocity of the object when it lands using PE = KE and so,
6. One partner drops the object while the other measures the time it takes to fall.
7. Take turns being the dropper and the timer until you have made four measurements.
8. Average your drop multiplied by and calculate the velocity of the object when it landed using v = at = gt = (9.80)t.
Grasp Check
Galileo’s experiments proved that, contrary to popular belief, heavy objects do not fall faster than light objects. How do the equations you used support this fact?
1. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance.
2. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance.
3. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance.
4. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance.
Exercise 3
Describe the transformation between forms of mechanical energy that is happening to a falling skydiver before his parachute opens.
1. Kinetic energy is being transformed into potential energy.
2. Potential energy is being transformed into kinetic energy.
3. Work is being transformed into kinetic energy.
4. Kinetic energy is being transformed into work.
Exercise 4
True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy.
1. True
2. False
Exercise 5
Identify equivalent terms for stored energy and energy of motion.
1. Stored energy is potential energy, and energy of motion is kinetic energy.
2. Energy of motion is potential energy, and stored energy is kinetic energy.
3. Stored energy is the potential as well as the kinetic energy of the system.
4. Energy of motion is the potential as well as the kinetic energy of the system.
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## A Multiplication Based Logic Puzzle
### 760 Not a Misprint: One side is 84, the other 48. What are the other sides?
• 760 is a composite number.
• Prime factorization: 760 = 2 x 2 x 2 x 5 x 19, which can be written 760 = (2^3) x 5 x 19
• The exponents in the prime factorization are 3, 1, and 1. Adding one to each and multiplying we get (3 + 1)(1 + 1)(1 + 1) = 4 x 2 x 2 = 16. Therefore 760 has exactly 16 factors.
• Factors of 760: 1, 2, 4, 5, 8, 10, 19, 20, 38, 40, 76, 95, 152, 190, 380, 760
• Factor pairs: 760 = 1 x 760, 2 x 380, 4 x 190, 5 x 152, 8 x 95, 10 x 76, 19 x 40, or 20 x 38
• Taking the factor pair with the largest square number factor, we get √760 = (√4)(√190) = 2√190 ≈ 27.5680975.
There is no misprint in this puzzle. One side really is 84 while the other side really is 48. Can you find the other sides?
Print the puzzles or type the solution on this excel file: 10 Factors 2016-02-04
—————————————
Here’s more about the number 760:
760 is the sum of consecutive numbers two different ways. (Two of its factor pairs show up in those ways.):
• 150 + 151 + 152 + 153 + 154 = 760; that’s 5 consecutive numbers.
• 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 = 760; that’s 19 consecutive numbers.
760 is the hypotenuse of a Pythagorean triple so that 456² + 608² = 760²
760 is the sum of three squares: 20² + 18² + 6² = 760.
760 is palindrome 1A1 in Base 23 because 1(23²) + 10(23) + 1(1) = 760.
Wikipedia informs us that 760 is the 23rd centered triangular number because (3⋅22² + 3⋅22 + 2)/2 = 760.
—————————————
Here’s the same puzzle with out the lists of triples:
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# The Ultimate Guide To The Bar Model: How To Teach It And Use It In Elementary School
Teaching the bar model in elementary school is essential if you want students to do well in their reasoning and problem solving skills. Here’s your step-by-step guide on how to teach the bar model using math mastery lessons from Pre-K and Kindergarten right up to 5th grade level questions.
First we will look at the four operations and a progression of bar model representations that can be applied across all grades in elementary school.
Then we will look at more complex bar model examples, including how to apply bar models to other concepts such as fraction and equations.
Fun Math Games and Activities Packs for Kindergarten to 5th Grade
Individual packs for K-5 containing fun math games and activities to do with your students
### What is a bar model in math?
What is a bar model? In math a bar model, also known as a strip diagram, is a pictorial representation of a problem or concept where bars or boxes are used to represent the known and unknown quantities. Bar models are most often used to solve number problems with the four operations – addition and subtraction, multiplication and division.
In word problems, bar models help children visualize the problem and decide which operations to use.
The bar model is the pictorial stage in the concrete pictorial abstract (CPA) approach to learning and is central to math mastery. Bar models will not, however, do the calculations for the students; they simply make it easier for learners to work out which equation must be done to solve the problem.
#### What is the bar model in math?
The bar model is used by drawing out a bar to represent the known and unknown quantities in a given problem. Encouraging a child to use the bar model in a problem can help them to understand conceptually what math operation is required from the problem, and how each part combines to make the whole.
An understanding of bar models is essential for teaching word problems, especially those using four operations (addition, subtraction, multiplication, division), fractions and algebra. Here the bar model shows us that if 2 rectangles out of 3 rectangles are green, then two thirds of 12 must be 8.
#### Why use the bar model method in math?
The bar model is used in Singapore and Asian Math textbooks and is an essential part of the mastery math approach used by schools at all stages.
By using the bar method to visualize problems, students are able to tackle any kind of number problem or complex word problem.
Because bar models only require pencils and paper, they are highly versatile and can be very useful for tests.
The use of bar models are also useful for learning at an early stage, from showing number bonds to ten or partitioning numbers as part of your place value work.
Once a child is secure in their use of the bar model for the four operations and can conceptualize its versatility, they can start to use it to visualize many other math topics and problems, such as statistics and data handling.
#### The Singapore Bar Model Method
‘The Singapore Math Model’ is another name for the bar model method. Despite this name however the Singapore bar model (like most of the math mastery approach) is based heavily on the work of Bruner, Dienes and Bishop about the best way to help children learn: teaching for mastery.
Bar models act as a ‘bridge’ between the concrete, pictorial and abstract (CPA in math); once children are secure with using pictorial versions of their concrete materials, they can progress to using bars as visual representations.
Bars are a more abstract way of representing amounts, making the transition to using wholly abstract numbers significantly less difficult.
#### The Part Whole Method
Also known as the ‘part part whole’ method or the comparison model, this kind of bar model uses rectangles to represent the known and unknown quantities as parts of a whole. This is an excellent method to help students represent the very common ‘missing number’ problems.
This can be done in two ways:
• As discrete parts to a whole – each unit in the problem has its own individual box, similar to using Numicon cubes.
• As continuous parts to a whole – units are grouped into one box for each amount in the problem e.g. in 26 + 52, 26 would have one long bar, not 26 smaller rectangles joined together.
When using the part-whole method, proportionality is key; all the bars must be roughly proportional to each other e.g. 6 should be about twice the length of 3.
Part-whole models are generally used to visually represent the four operations, fractions, measurement, algebra and ratios, but can be applied to many more topics in the common core (as long as they’re relevant!).
Read more: What Is The Part Whole Model?
### How to draw a bar model
There are a few steps involved in drawing a bar model and using it to solve a problem:
2. Circle the important information
3. Determine the variables: who? what?
4. Make a plan for solving the problem: what operation needs to be used?
5. Draw the unit bars based on the information
6. Re-read the problem to make sure that the bar models match the information given
7. Complete the problem using the determined operation.
### Bar models: Pre-K to third grade
Students in Pre-K and Kindergarten will routinely come across expressions such as 4+3.
Often, these expressions will be presented as word problems: Aliya has 4 oranges. Alfie has 3 oranges. How many oranges are there altogether? To help children fully understand later stages of the bar model, it is crucial they begin with concrete representations.
There are 2 models that can be used to represent addition:
Once they are used to the format and able to represent word problems with models in this way themselves (assigning ‘labels’ verbally), the next stage is to replace the ‘real’ objects with objects that represent what is being discussed (in this case, we replace the ‘real’ oranges with button counters):
The next stage is to move away from the concrete to the pictorial. As with all the stages, when students are ready for the next stage is a judgment call that is best decided upon within your school.
A general rule of thumb would be that towards the end of kindergarten or start of first grade, students should be able to understand and represent simple addition and subtraction problems pictorially and assign written labels in a bar model.
The penultimate stage is to represent each object as part of a bar, in preparation for the final stage:
The final stage moves away from the 1:1 representation. Each quantity is represented approximately as a rectangular bar:
As mentioned before, it is a judgment call for your school to make, but if you want students to use the bar model to support them at the end of semester tests towards the later stage of their early years, they are going to need to have had a fair amount of experience in this final stage.
#### Bar model subtraction
The same concrete to pictorial stages can be applied to subtraction. However, whereas with addition it is really down to the student’s preference as to which of the 2 bar representations to use, with subtraction the teacher can nudge students to one or the other.
The reason? One represents a ‘part-part-whole’ model, the other a ‘find the difference’ model. Each will be more suited to different word problems and different learners. Let’s examine those at the final stage of the bar model:
Part-part-whole
Austin has 18 lego bricks. He used 15 pieces to build a small car. How many pieces does he have left?
Equation: 18 – 15 =
Find the difference
Austin has 18 lego bricks. Lionel has 3 lego bricks. How many more lego bricks does Austin have than Lionel?
Equation: 18 – 3 =
#### Bar model multiplication
Bar model multiplication starts with the same ‘real’ and ‘representative counters’ stages as addition and subtraction. Then it moves to its final stage, drawing rectangular bars to represent each group:
#### Bar model division
Due to the complexity of division, it is recommended to keep grouping and sharing until the final stage of the bar model is understood. Then word problems such as the 2 below can be introduced:
Sharing
Grace has 27 lollipops. She wants to share them amongst 9 friends by putting them in party bags. How many lollipops will go into each party bag?
Grouping
Grace has 27 lollipops for her party friends. She wants each friend to have 3 lollipops. How many friends can she invite to her party?
#### Bar models in assessments
The key question at any grade level is, “what do we know?” By training students to ask this when presented with word problems themselves, they quickly become independent at drawing bar models.
For example, in the problem: Egg boxes can hold 6 eggs. We need to fill 7 boxes. How many eggs will we need?
We know that there will be 7 egg boxes, so we know we can draw 7 rectangular bars. We know that each box holds 6 eggs, so we can write ‘6 eggs’ or ‘6’ in each of those 7 rectangular bars. We know we need to find the amount of eggs we have altogether. We can see we will need to use repeated addition or multiplication to solve the problem.
#### Bar model for the four operations word problems
Let’s ramp up the difficulty a little. In a sample assessment, students are asked:
A bag of 5 lemons costs $1. A bag of 4 oranges costs$1.80. How much more does one orange cost than one lemon?
Students could represent this problem in the below bar model, simply by asking and answering ‘what do we know?’
From here it should be straightforward for the students to ‘see’ or visualize their next step. Namely, dividing $1.80 by 4 and$1 by 5. Some students will not need the bar model to represent the next stage, but if they do, they would solve and then allocate the cost onto the model:
Then those students that needed this stage, should be able to see that to answer the question, they need to solve 45¢ – 20¢. With the answer of 25¢.
#### Bar model for word problems with fractions
Here’s another example from tests involving fractions and how it can be solved using a fraction bar model
On Saturday Lara read two fifths of her book. On Sunday, she read the other 90 pages to finish the book. How many pages are there in Lara’s book? If we create our bar model for what we know:
Students will then see that they can divide 90 by 3:
As fractions are ‘equal parts’ – a concept they should be familiar with from third grade – they know that the other 2 fifths (Saturday’s reading) will be 30 pages each:
Then they can solve 30 x 5 = 150
### Bar models: Middle school and beyond
#### Equations with the bar model
There are lots of other areas where bar models can assist student’s understanding such as ratio, percentages and equations. In this final example, we look at how an equation can be demystified using the comparison model:
2a + 7 = a + 11
Let’s draw what we know in a comparison model, as we know both sides of the equation will equal the same total:
The bars showing 7 and 11 could have been a lot smaller or larger as we don’t know their relative value to ‘a’ at this stage. However, it is crucial that the ‘a’ appearing first in both bars is understood to be equal (even if it is only approximately equal when drawn freehand in the bar). This allows the student to ‘see’ that to work out the second ‘a’ in the top bar, they can solve 11-7.
So if that ‘a’ is 4, then both the other ‘a’s will also be 4. So each side of the equation will total 15. The below model shows all sections completed. This is not necessary for the students to do, the representation is merely useful until they can see the steps necessary to solve whatever they are faced with:
Where next?
Now you’re persuaded (we hope!) that the bar model in math is going to revolutionize problem solving across your school from early years onwards.
If you don’t believe me, use this problem to grab their attention:
With or without bar models? Which is easier? My guess is your staff will be hooked!
More on the bar model and math mastery techniques
Do you have students who need extra support in math?
Give your students more opportunities to consolidate learning and practice skills through personalized math tutoring with their own dedicated online math tutor.
Each student receives differentiated instruction designed to close their individual learning gaps, and scaffolded learning ensures every student learns at the right pace. Lessons are aligned with your state’s standards and assessments, plus you’ll receive regular reports every step of the way.
Personalized one-on-one math tutoring programs are available for:
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## Mathematical Background
A complex number z is a linear combination of two real numbers x and y together with a symbol i defined via the property that i2 = −1 as follows:
z = x + i y, where i is a symbol such that i2 = −1 and x, y are real numbers
The variable x is called the real part of z, or Re(z) = x, while y is called the imaginary part, or Im(z) = y. The constant i is the imaginary unit, defined by the property that i2 = −1 (which makes it non-real or imaginary--you can not find it on the usual number line). Informally you can think of i as the square root of −1, i.e. i = = (−1)1/2.
Please note that--contrary to what you might think--the imaginary part of a complex number is a real number (confusing, isn't it). The real part is also a real number, but that "sounds" right. In other words, if z = 2 + 3 i, then Re(z) = 2 and Im(z) = 3.
Complex numbers can be added, subtracted, multiplied, and divided using rules familiar from the real numbers together with that fact that i2 = −1. For example:
(1 + 2i) + (3 − 4i) = (1 + 3) + (2 − 4)i = 4 - 2i = 2(2 − i)
or
(1 − 2i) + (3 + 4i)(5 − 6i) = (1 − 2i) + (3 · 5 − 3 · 6i + 5 · 4i − 4 · i2) = (1 − 2i) + (15 − 18i + 20i + 24) = (1 − 2i) + (39 + 2i) = 40 + 0i = 40
or
Since real and imaginary parts of a complex number are independent of each other, the space of all complex numbers is two-dimensional. It can be visualized as a plane with a horizontal (the real) and vertical (the imaginary) axis. For example, the complex numbers
A = 1 + i, B = −1 − 2 i, and C = 2 i
are as shown in Figure 1.
Figure 1: The complex plane with a few points marked
But then a function that takes complex numbers as input and produces (possibly) complex values as output is a map from two-dimensional space (the domain) to another two-dimensional space (the range).
#### Example
Let's consider the (complex) square function f(z) = z2. We can easily generate a table of input/output values, as shown below. You can see, for example, that the function f maps −1 to 1, 1 + i to 2i and so on. We can visualize this mapping property by drawing the domain in one coordinate system, say on the left, and the range in a second coordinate system next to the domain. Then we plot the points in the domain and their corresponding images in the range (see Figure 2).
Figure 2. Domain and Range with points and images
z f(z) = z2
z = 1 f(1) = 12 = 1
z = −1 f(−1) = (−1)2 = 1
z = i f(i) = i2 = −1
z = −i f(−i) = (−i)2 = (−1) i (−1) i = −1
z = 1 + i f(1 + i) = (1 + i) (1 + i) = 1 + 2i − 1 = 2i
z = 1 − i f(1 − 1) = (1 − i) (1 − i) = 1 − 2i − 1 = −2i
To go one step further, we draw complete curves--not only single points--in the oordinate system representing the domain and see what the map does to this collection of points by drawing their images in another coordinate system on the right.
#### Example
Let's again consider the (complex) square function f(z) = z2. This time we want to see what that function does to a horizontal (say) line going through the point z = i.
A horizontal line passing through the point z = i has as imaginary part (or height) always 1, whereas the real part is arbitrary, since the line stretches from left to right infinitely long. Therefore, a (parametric) equation of such a line is given by l(t) = t + i, where t is a real number. To find the image of this line, we substitute its equation into the function. We get:
f(l(t)) = f(t + i) = (t + i)2 = (t + i) (t + i) = t2 − 1 + 2 i t
But t2 − 1 + 2 i t is equivalent to the parametric curve (t2 − 1, 2t), splitting up the complex result into its real and imaginary parts. But then we have:
x = t2 − 1 and y = 2t
Solving the second equation for t we get y / 2 = t. We can substitute that into the first equation to get x = (1 / 4) y2 − 1. We recognize that as the equation of a parabola going left to right, opening to the right, with vertex at (−1, 0), and with y-intercepts (0, −2) and (0, 2). Figure 3 shows the result:
Figure 3. f(z) = z2 maps the horizontal line through z = i to a left-to-right parabola with vertex at w = −1
This is exactly what ZMap does: you define a function as well as curves in the domain and ZMap will show you the image of the curves in the range. Below is the ZMap program, showing again what f(z) = z2 will do to the horizontal line z = l(t) = t + i. For more information on how to use ZMap, please see the quick guide or features page, or click on the Help icon of the ZMap program below.
The ZMap program for f(z) = z2
This idea of drawing domain and range separately next to each other to visualize the graph of a complex function is useful but not entirely enlightening. To understand the limitations of this approach, ask yourself how the "graph" of a regular real function, such as for example f(x) = x2 would look if we drew (real) domain and (real) range next to each other?
We would draw one real line as the domain, and next it another real line as the range. As x moves from −∞ to ∞, covering the entire domain line, the range would go from +∞ (since negative squared is positive) down to 0 and then back to +∞. The end result, show in Figure 4, would look nothing like the usual parabola we normally visualize with the function f(x) = x2.
Figure 4. Visualizing f(x) = x2 using two number lines
Only when we combine domain and range axis at right angles into our usual Cartesian coordinate system do we see how x and f(x) move together to give our true parabola graph. Thus, to get a perfect picture of a graph of a complex function we really should combine the domain coordinate system at right angles with the range coordinate system instead of drawing them next to each other. That, however, would require a 4-dimensional coordinate system. Mathematically, the graph of a complex function is indeed a 4-dimensional object, but since we can not visualize 4-dimensional space we can not use this approach as visualization help.
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# Area and Perimeter
## Area and Perimeter
What can you say about these two shapes?
What is the area of each one? What is the perimeter of each one?
What can you say about the shapes below?
You can print out a set of shapes and cut them into separate cards. These cards have the coloured background.
Can you draw a shape in which the area is numerically equal to its perimeter? And another?
Can you draw a shape in which the perimeter is numerically twice the area?
Can you draw a shape in which the area is numerically twice the perimeter?
Can you make the area of your shape go up but the perimeter go down?
Can you make the perimeter of your shape go up but the area go down?
Can you draw some shapes that have the same area but different perimeters?
Can you draw some shapes that have the same perimeter but different areas?
### Why do this problem?
This problem offers opportunities for children to consolidate their understanding of area and perimeter. The exploratory nature of the task means that learners will be grappling with the two concepts at the same time rather than tackling them independently which might usually be the case. The activity is likely to require persistence and a 'tinkering' or trial and improvement approach.
### Possible approach
It is essential to have squared paper available (preferably $1$ cm squared) and the shapes printed out and cut into eight separate cards. These cards can be downloaded here in black and white, and here with a coloured background. It might also be helpful to have post-it notes so that pupils could attach details of area and perimeter onto each card, rather than continually having to re-calculate them.
You could start with the whole group looking at the two shapes given at the beginning of the problem and invite learners to talk about anything they notice. (These are also two of the shapes given on the set of cards.) If area and/or perimeter doesn't come up naturally, you could ask direct questions to shift their attention to these concepts.
Pairs could then explore the shapes on the remaining cards and you can challenge them with the specific questions given in the problem itself. Copies of these for printing out can be downloaded here. It is important that you stress we are looking at numerically equal values. The area and perimeter cannot be equal because they are measured in different units.
When you gather the whole group together again, invite them to share not just solutions (i.e. shapes that fit the criteria), but their methods for creating the shapes. Did they establish any 'principles' that helped them? What actions could they perform on a shape without changing its perimeter? For example, what happens to the area and perimeter if you take a 'corner square' off a shape? What happens if you take an 'edge square' off a shape? What happens if you take a 'middle square' out of a shape? It may be that some children notice that for a given perimeter, a square gives the maximum possible area.
### Key questions
How will you find out the perimeter?
How will you find out the area?
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What’s the Current Job Market for square root of sum of squares Professionals Like?
square root of sum of squares is the sum of all the squares of each of the numbers in the square. This is the most commonly used formula to find the square root of a sum of numbers. For example, the square root of 10 is the sum of the squares of all 10s.
This is exactly the same thing as taking the cube root of the sum of the cubes in the square. It’s just that the cube is often the most useful number to use for multiplication.
Square roots and cubes are often the same thing, but the cube is often used for multiplication too. The cube is a very useful number for multiplying, yet it is not the most useful number for finding the square root of. To find the cube of a number, you multiply it by its cube. For example, the cube of 50 is the square root of the sum of the squares of all the cubes in the 2×2 square (50 x 50).
A common technique for learning how to use a digital calculator is to read the number and look at the answer. The calculator does the math for you. If the answer is negative, then it is a mistake. If the answer is positive, then it is a good thing.
This is the reason why square root (or log) is a useful technique for learning math. The log is a very important concept for any number system which has to do with a property of the logarithm of a number. For example, the logarithm of 3 is 3 (or 1) times the log of 3.
The word “log” refers to a series of symbols, but its meaning can be more precise than that of “log.” A logarithm is a series of digits, not a series of letters.
The logarithm is a logarithmic number, which means it goes up by a power of 10. This is important for the way we use the logarithm to figure out how many factors we need to multiply a number to figure out the product. One important property of a logarithm is that it is an even number. That means that, if we multiply two logarithms, we get the same number.
The square root of a number is the number we multiply a number by by its square root. This is another important property of a logarithm because it means that we multiply the logarithm by its own logarithm, rather than by its reciprocal. This is useful for figuring out how many times we need to multiply a number because we are multiplying by its logarithm.
So, what is the square root of a sum of squares? Well, take the sum of all the squares in that sum. We can then take the square root of that sum without dividing by its reciprocal because we know that the square root of a number is the same as a power of two.
We can also use this to multiply number by number to find the square root of a number: 2^0 = 1.2 = 1. How about 10^2? Well, 10^2 = 10^2 = 100, so it has the same square root as 10 because we know that if you multiply a number x times the square root of that number, you get the square root of the product of the two numbers.
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# Search by Topic
#### Resources tagged with Fibonacci sequence similar to Equal Temperament:
Filter by: Content type:
Stage:
Challenge level:
### There are 17 results
Broad Topics > Sequences, Functions and Graphs > Fibonacci sequence
### Leonardo of Pisa and the Golden Rectangle
##### Stage: 2, 3 and 4
Leonardo who?! Well, Leonardo is better known as Fibonacci and this article will tell you some of fascinating things about his famous sequence.
### LOGO Challenge - Circles as Bugs
##### Stage: 3 and 4 Challenge Level:
Here are some circle bugs to try to replicate with some elegant programming, plus some sequences generated elegantly in LOGO.
### Gnomon Dimensions
##### Stage: 4 Challenge Level:
These gnomons appear to have more than a passing connection with the Fibonacci sequence. This problem ask you to investigate some of these connections.
### Fibonacci Surprises
##### Stage: 3 Challenge Level:
Play around with the Fibonacci sequence and discover some surprising results!
### Building Gnomons
##### Stage: 4 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.
### Continued Fractions I
##### Stage: 4 and 5
An article introducing continued fractions with some simple puzzles for the reader.
### The Golden Ratio, Fibonacci Numbers and Continued Fractions.
##### Stage: 4
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
### Got a Strategy for Last Biscuit?
##### Stage: 3 and 4 Challenge Level:
Can you beat the computer in the challenging strategy game?
### Fibonacci's Three Wishes 2
##### Stage: 2 and 3
Second of two articles about Fibonacci, written for students.
##### Stage: 3 and 4 Challenge Level:
Using logo to investigate spirals
### Cellular
##### Stage: 2, 3 and 4 Challenge Level:
Cellular is an animation that helps you make geometric sequences composed of square cells.
### Fibs
##### Stage: 3 Challenge Level:
The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?
### First Forward Into Logo 11: Sequences
##### Stage: 3, 4 and 5 Challenge Level:
This part introduces the use of Logo for number work. Learn how to use Logo to generate sequences of numbers.
### Whirling Fibonacci Squares
##### Stage: 3 and 4
Draw whirling squares and see how Fibonacci sequences and golden rectangles are connected.
### Stringing it Out
##### Stage: 4 Challenge Level:
Explore the transformations and comment on what you find.
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