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# Fruity Brain Teaser? Math Problem With Apples, Bananas, Coconuts Is Harder Than It Looks
A tricky brain teaser filled with pictures of apples, bananas, and coconuts has recently been stumping quite a few people on Facebook.
At first glance, it may seem as if the answers are easy giveaways — especially when you focus on the very first line.
When you look at an equation that features three apples which equal 30, which was analyzed on The Problem Site, chances are that the average Algebra student would quickly fill in the blanks with the number “10.”
10 + 10 + 10 = 30
Keeping in mind that each individual apple has the value of 10, everything else apparently seems to fall into place.
The second line claims that two bunches of bananas along with an apple equals 18, which is why it’s easy to think that the two bananas equal 8 (which means that each individual bundle has a value of 4).
10 + 4 + 4 = 18
The third line claims that a bundle of bananas (with an assumed value of 4) minus coconuts equals 2. Therefore, it is easy to assume that the coconut has an individual value of 2.
4 – 2 = 2
As of right now, then, here’s what you may have so far when it comes to the individual value of each fruit:
• Apples = 10 each
• Bananas = 4 each
• Coconuts = 2 each
Therefore, solving the final line of this confusing brain teaser may seem a little too easy.
1 coconut (2) + 1 apple (10) + 1 banana (4) = 16
### THE ONE FACTOR THAT CHANGES EVERYTHING
If 16 was your final answer, you might honestly think that you are correct. If so, you are not the only one…that was unfortunately mistaken!
There is one particular factor to this confusing brain teaser that you more than likely overlooked the first time around. What is it? Pay close attention to the bananas.
How many individual bananas are found in each bundle? Each bundle has four bananas. That is, of course, until you get to the bananas on the final line of this fruity brain teaser.
Within that particular banana bundle, there are only three bananas.
That’s not the only fruit that had a drastic makeover at some point throughout the progression of this brain teaser.
Take a quick look at the coconuts on each line. On the third line of the brain teaser, the coconut was split into two halves. Both halves were seen in the picture, right?
However, when you move down to the last line of the puzzle, there is only one half of the coconut that is pictured.
How does this change the math of this fruity brain teaser? A bundle of four bananas has a value of 4, which means that each individual banana has a value of 1. Therefore, a bundle of 3 bananas represents a value of 3, not 4.
The same principle is applied to the coconut. If a full coconut has a value of 2, then one-half of that coconut only has a value of 1.
With this in mind, the math of the final line of this fruity brain teaser looks like this:
“1 + 10 + 3 = 14”
Even though most people that shared their responses to the fruity brain teaser seemed to agree with 16, there were some that paid attention to the detail of the missing fruit and guessed the correct answer.
There are quite a few different brain teasers and puzzles that have been circulating online, especially on Facebook. As previously reported by the Inquisitr, there are popular puzzles based on such topics as rock bands from the ’90s as well as a picture puzzle with hidden words.
[Image Credit: Dollar Photo Club]
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## MAP Recommended Practice
### Course: MAP Recommended Practice>Unit 26
Lesson 1: Volume with unit cubes
# Volume intro
Finding volume, length, and area all involve measuring dimensions in order to quantify a certain aspect of an object or space. Although each property requires different calculations, they all rely on the same fundamental principle: taking measurements and using those measurements to derive a numerical value. When finding volume, you are measuring the amount of space that an object occupies. Created by Sal Khan.
## Want to join the conversation?
• What is the difference between capacity and volume?
• Volume is the amount of space taken up by an object, while capacity is the measure of an object's ability to hold a substance, like a solid, a liquid or a gas. ... Volume is measured in cubic units, while capacity can be measured in almost every other unit, including liters, gallons, pounds, etc.
• How do you calculate the volume of an irregular shape?
• You can calculate the volume of an irregular shape - it just takes a few more steps:
First, break the irregular shape into smaller, regular shapes. (You should know how to calculate the volume of each of these smaller shapes.)
Next - one at a time - calculate the volume of each of the smaller shapes.
Finally, add up the volumes of all the smaller shapes. This gives you the total volume of the irregular shape.
Hope this helps!
• a 4-d is when a figure is 3-d holographic
• Human beings are literally incapable of seeing a 4D shape in its most true form. Consider:
If I had a sphere and we could only see in 2D, we'd just see a circle, not the whole sphere.
It's the same idea with 4D VS 3D.
• what is volume
• In math, volume is the amount of space in a certain 3D object. For instance, a fish tank has 3 feet in length, 1 foot in width and two feet in height. To find the volume, you multiply length times width times height, which is 3x1x2, which equals six. So the volume of the fish tank is 6 cubic feet. Volume is also how loud a sound is. Look on your TV remote. There is a volume control button on it.
• how do you calculate the volume of an irregular shape?
From,
Charlotte :D
• Break the irregular shape into smaller shapes, then calculate the volumes of those. When you are done, add all the volumes together to find the total volume
• Hi everyone!💖
Hope y'all are all having a great day ✨
Just want to say, don't worry you're an AMAZING person, and I hope your life is going well ✨
And if you're having a bad day. Its okay, things will get better for you! 🥺❤
• You are a good person you make people happy
• How did human beings find out one line is longer than the other?
• they Measure
• Is it okay if i am going ahead of my grade
• Yes! Absolutely!
• why do we have to watch the videos?
(1 vote)
• because are teachers force us:3
• This is not a question I just wanted to thank Kahn for helping me when I did not understand something! (:
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Ordering Decimals Unit 12 > Lesson 4 of 12
Example 1: The Glosser Family drove to a gasoline station in their neighborhood. The station has three gas pumps, each marked in price per gallon. Which pump has the lowest price per gallon? Which pump has the highest price per gallon?
\$1.79, \$1.96, \$1.61
Analysis: We know that 1.79 < 1.96 and that 1.79 > 1.61. Writing one decimal beneath the other in order, we get:
1 . 6 1 least 1 . 7 9 1 . 9 6 greatest
Answer: The pump marked \$1.61 has the lowest price per gallon. The pump marked \$1.96 has the highest price per gallon.
In the example above, we ordered three decimal numbers from least to greatest by comparing them two at a time. Let's look at some more examples.
Example 2: Order these decimals from least to greatest: 0.5629, 0.5621, 0.6521
Let's examine these decimals in our place-value-chart.
0 . 5 6 2 9 0 . 5 6 2 1 0 . 6 5 2 1
Now let's order these decimals from least to greatest without our place-value chart. We will do this by comparing two decimals at a time.
0 . 5 6 2 1 0 . 5 6 2 9 0 . 6 5 2 1
Answer: Ordering these decimals from least to greatest we get: 0.5621, 0.5629, 0.6521.
In the examples above, the decimals in each problem had the same number of digits. Thus, they lined up nicely, one beneath the other. Let's look at some examples in which the decimals presented have a different number of decimal digits.
Example 3: Order these decimals from least to greatest: 6.01, 0.601, 6.1
Let's start by writing one decimal beneath the other in their original order. Note that these three decimals have a different number of decimal digits.
6 . 0 1 0 0 . 6 0 1 6 . 1 0 0
Next, examine each decimal, writing one or more zeros to the right of the last digit, so that all decimals have the same number of decimal digits.
6 . 0 1 0 0 . 6 0 1 6 . 1 0 0
Now we can compare two decimals at a time.
6 . 0 1 0 0 . 6 0 1 6 . 1 0 0
From least to greatest, we get: 0.6010, 6.010, 6.100.
Answer: Ordering these decimals from least to greatest we get: 0.601, 6.01, 6.1.
Sometimes it is helpful to place a number in a circle to the right of each decimal you are trying to order. This is done in Example 4.
Example 4: Order these decimals from least to greatest: 3.87, 3.0875, 3.87502, 3.807
We have been asked to order four decimal numbers. Let's start by writing one decimal beneath the other in their original order.
3 . 8 7 0 0 0 3 . 0 8 7 5 0 3 . 8 7 5 0 2 3 . 8 0 7 0 0
Next, examine each decimal, writing one or more zeros to the right of the last digit, so that all decimals have the same number of decimal digits.
3 . 8 7 0 0 0 3 . 0 8 7 5 0 3 . 8 7 5 0 2 3 . 8 0 7 0 0
Now we can compare two decimals at a time. We will write a number in a circle next to each decimal to denote its order.
3 . 8 7 0 0 0 3 . 0 8 7 5 0 3 . 8 7 5 0 2 3 . 8 0 7 0 0
From least to greatest, we get: 3.08750, 3.80700, 3.87000, 3.87502
Answer: Ordering these decimals from least to greatest we get: 3.0875, 3.807, 3.87, 3.87502
Example 5: Order these decimals from least to greatest: 5.364, 6.0364, 5.36, 5.00364, 5.40364
We have been asked to order five decimal numbers. Let's start by writing one decimal beneath the other in their original order. Next, examine each decimal, writing one or more zeros to the right of the last digit, as needed.
5 . 3 6 4 0 0 6 . 0 3 6 4 0 5 . 3 6 0 0 0 5 . 0 0 3 6 4 5 . 4 0 3 6 4
Now we can compare two decimals at a time. We will write a number in a circle next to each decimal to denote its order.
5 . 3 6 4 0 0 6 . 0 3 6 4 0 5 . 3 6 0 0 0 5 . 0 0 3 6 4 5 . 4 0 3 6 4
From least to greatest, we get: 5.00364, 5.36000, 5.36400, 5.40364, 6.03640
Answer: Ordering these decimals from least to greatest we get: 5.00364, 5.36, 5.364, 5.40364, 6.0364
Let's look at some non-routine problems that involve comparing and ordering decimals.
Example 6: Write 3 decimals between 4.35 and 4.36 in order from least to greatest.
Analysis: We need to write a zero in the thousandths place for each of the given numbers.
4 . 3 5 0 4 . 3 6 0
Now we must find 3 numbers that are between the given numbers. We can use any digit between 1 and 9.
4 . 3 5 0 4 . 3 5 1 4 . 3 5 2 4 . 3 5 3 4 . 3 5 4 4 . 3 5 5 4 . 3 5 6 4 . 3 5 7 4 . 3 5 8 4 . 3 5 9 4 . 3 6 0
The question asks us to write 3 decimals between 4.35 and 4.36 in order from least to greatest. We can choose any 3 numbers in red from above as long as they are in order from least to greatest. Accordingly, our answers will vary. Below are some sample answers.
Sample Answer 1: 4.351, 4.352, 4.353
Sample Answer 2: 4.354, 4.356, 4.358
Example 7: Write 3 decimals between 7.418 and 7.419 in order from least to greatest.
Analysis: We need to write a zero in the ten-thousandths place for each of the given numbers.
7 . 4 1 8 0 7 . 4 1 9 0
Now we must find 3 numbers that are between the given numbers. We can use any digit between 1 and 9.
7 . 4 1 8 0 7 . 4 1 8 1 7 . 4 1 8 2 7 . 4 1 8 3 7 . 4 1 8 4 7 . 4 1 8 5 7 . 4 1 8 6 7 . 4 1 8 7 7 . 4 1 8 8 7 . 4 1 8 9 7 . 4 1 9 0
The question asks us to write 3 decimals between 7.418 and 7.419 in order from least to greatest. We can choose any 3 numbers in red from above as long as they are in order from least to greatest. Accordingly, our answers will vary. Below are some sample answers.
Sample Answer 1: 7.4182, 7.4183, 7.4184
Sample Answer 2: 7.4185, 7.4187, 7.4189
Example 8: Write the smallest possible decimal between zero and one that uses the digits 5, 0, 4, 1, 9, and 6 exactly once. Answer: .014569. Note that a leading zero was not used here.
Example 9: Write the greatest possible decimal between zero and one that uses the digits 9, 0, 2, 7, 3 and 5 exactly once. Answer: .985320 (without a leading zero) or 0.98532 (with a leading zero). Note that these two decimals are equivalent.
Summary: When ordering decimals, first write one decimal beneath the other in their original order. Then compare them two at a time. When ordering four or more decimals, it is helpful to write a number in a circle next to each to order them.
### Exercises
1. Which of the following is the smallest decimal number? 0.4981 0.52 0.4891 0.6 RESULTS BOX:
2. Which of the following is the largest decimal number? 0.0231 0.231 0.03 0.2 RESULTS BOX:
3. Which of the following choices lists these decimals in order from least to greatest: 0.910, 0.091, 0.9? 0.9, 0.091, 0.910 0.910, 0.9, 0.091 0.091, 0.9, 0.910 None of the Above RESULTS BOX:
4. Which of the following choices lists these decimals in order from least to greatest: 3.45, 3.0459, 3.5, 3.4059? 3.0459, 3.4059, 3.45, 3.5 3.4059, 3.5, 3.0459, 3.45 3.5, 3.45, 3.4059, 3.0459 None of the above. RESULTS BOX:
5. Which of the following choices lists these decimals in order from least to greatest: 7.102, 7.0102, 7.012, 7.00102, 7.102021? 7.102021, 7.00102, 7.012, 7.0102, 7.102 7.00102, 7.0102, 7.012, 7.102, 7.102021 7.102021, 7.102, 7.012, 7.0102, 7.00102 None of the above. RESULTS BOX:
Lessons on Decimals, Part I Introduction Reading and Writing Decimals Comparing Decimals Ordering Decimals Estimating Decimal Sums Adding Decimals Estimating Decimal Differences Subtracting Decimals Solving Decimal Word Problems Practice Exercises Challenge Exercises Solutions
Related Activities Interactive Puzzles Printable Worksheets Percent Goodies Game
Need More Practice? Try our Decimal Worksheet Generator Try our Money Worksheet Generator Try our Place Value Worksheet Generator
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# How do you find (dy)/(dx) given y+y^3=x^2?
Feb 24, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{1 + 3 {y}^{2}}$
#### Explanation:
When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.
However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.
When this is done in situ it is known as implicit differentiation.
We have:
$y + {y}^{3} = {x}^{2}$
Differentiate wrt $x$:
$\setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 + 3 {y}^{2}\right) = 2 x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{1 + 3 {y}^{2}}$
There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us
(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))
So Let $F \left(x , y\right) = y + {y}^{3} - {x}^{2}$; Then;
$\frac{\partial F}{\partial x} = - 2 x \setminus \setminus \setminus$ and $\setminus \setminus \setminus \frac{\partial F}{\partial y} = 1 + 3 {y}^{2}$
And so:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{- 2 x}{1 + 3 {y}^{2}} = \frac{2 x}{1 + 3 {y}^{2}}$, as before.
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# Subtraction of Decimals
## What is Decimal Subtraction?
Last updated date: 19th Mar 2023
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Total views: 61.2k
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Views today: 1.41k
Decimal subtraction is like the standard deduction of entire numbers. We simply have to put the decimal numbers, one beneath the other, as per their place values alongside the decimal point. At times the two given numbers are unique. If necessary, we change the given decimal portions to like decimal divisions to make the deduction simpler. Two decimals are called like decimal portions on the off chance that they have an equivalent number of decimal spots.
## How to Subtract Decimals or Point Subtraction
Taking away decimals with pulling together is equivalent to the refocusing that is finished in entire numbers. The accompanying advances make sense of the way to deduct decimals by refocusing. This is also termed as point subtraction.
Stage 1: After changing the numbers to like decimals, and putting the more modest number beneath the more noteworthy number, guarantee that the decimal focuses are adjusted and the digits are set in the comparing place esteem section.
Stage 2: Subtract the digits in every section exclusively, beginning from the right and move to the left.
Stage 3: If the number in the upper line is more modest than the number in the lower column, we acquire 1 from the digit to one side and add 10 to the ongoing digit. This step is known as refocusing.
Stage 4: Since 1 has been acquired, this number has decreased by 1. Change the number as needed and take it away.
For instance, let us take away the given numbers.
Model: Subtract 0.84 from 0.93.
Arrangement: Taking away decimals with refocusing
Subtraction of Decimal
In the above numbers 0.93 and 0.84, the digit 3 is more modest than 4 in the hundredths section, so we really want to acquire 1 from 9 to do the deduction. We want to diminish 1 from 9 since it has given 1 to the number to one side. Along these lines, 9 will become 8, and 3 will become 13. Presently, we take away 13 - 4 = 9 and 8 - 8 = 0.
At last, we place a decimal direct in the response to get 0.09.
## Deducting Decimals with the Same Number of Decimal Places
Deducting decimals with a similar number of decimal spots is basic. We simply have to do the straightforward course of deduction and afterward place the decimal focuses as per the given numbers.
## Rules for Subtracting Decimals
There are sure principles and steps which ought to be recalled while deducting decimals.
Following are the subtraction rules:
• Change the given decimal parts to like decimals.
• Compose the more modest number beneath the more prominent number so that the digits are put in the relating places and the decimal focuses are put in a similar vertical line.
• Then, at that point, take away the numbers similarly as on account of entire numbers.
• After the deduction, place the decimal point in accordance with the other decimal places. By applying the above stated subtraction rules, we can easily get our answer.
## How to Subtract Decimals with Whole Numbers
To deduct decimal numbers from entire numbers, we place a decimal point in the entire number as per the quantity of digits (after the decimal) in the other number. Then, at that point, we add the expected number of zeros, so both the numbers have an equivalent length.
For instance, let us deduct 0.999 from 6.
Arrangement: Since there are three digits after the decimal in 0.999, we will put a decimal after 6 and add three zeros after it.
Subtracting From Whole Numbers
## Solved Example
Let us deduct the given numbers:
Model: Subtract: 0.56 - 0.42.
Arrangement: Since these are like decimals, we will put them all together and do the customary deduction.
0.56 - 0.42 = 0.14
## Summary
This article discussed decimals and subtractions in decimals. It is equivalent to refocusing that is completed in whole numbers to remove decimals by pulling them together. The supplementary developments clarify how to subtract decimals by focusing. The numbers are first cushioned with zero contingent on the most extreme digits present after the decimal for any of the numbers. The numbers are fixed up upward alongside one another. Finally, deduct the decimal numbers like whole numbers and spot the decimal point as needs are.
## FAQs on Subtraction of Decimals
1. How do you deduct decimals within 1, and how do you deduct decimals with different decimal places?
For deducting decimals inside 1, we follow the very methodology that we do on account of the other decimal numbers. Decimals inside 1 imply that the given numbers are under 1.
At the point when the numbers have different numbers of decimal spots, they are named as dissimilar to decimal parts. For this situation, we count the quantity of digits after the decimal point in both the numbers and recognize the higher one in these. Then, at that point, we add the necessary number of zeros to the more modest decimal number to come to a similar length as the other number. After this, we take away the decimals numbers.
2. What is Subtraction in Maths?
Subtraction is an operation used to find the difference among numbers. When you've got a set of objects and also you get rid of a few objects from it, the group becomes smaller. For example, you purchased nine cupcakes at your celebration and your buddies ate 7 cupcakes. Now you're left with 2 cupcakes. This may be written in the form of a subtraction expression: 9 - 7 = 2 and is studied as "9 minus seven equals ". When we subtract 7 from 9, (9 - 7) we get 2. Here, we performed the subtraction operation on two numbers 9 and seven to get the difference of two.
3. How do we solve problems involving the addition and subtraction of decimals?
Identify the important numbers and keywords in the problem that will indicate the operation(s) you will perform. Stack the values to perform the operation. Then add or subtract the decimal values by lining up the decimal point and using zeros as placeholders if needed to get your final answer.
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University Physics Volume 1
# 10.6Torque
## Learning Objectives
By the end of this section, you will be able to:
• Describe how the magnitude of a torque depends on the magnitude of the lever arm and the angle the force vector makes with the lever arm
• Determine the sign (positive or negative) of a torque using the right-hand rule
• Calculate individual torques about a common axis and sum them to find the net torque
An important quantity for describing the dynamics of a rotating rigid body is torque. We see the application of torque in many ways in our world. We all have an intuition about torque, as when we use a large wrench to unscrew a stubborn bolt. Torque is at work in unseen ways, as when we press on the accelerator in a car, causing the engine to put additional torque on the drive train. Or every time we move our bodies from a standing position, we apply a torque to our limbs. In this section, we define torque and make an argument for the equation for calculating torque for a rigid body with fixed-axis rotation.
## Defining Torque
So far we have defined many variables that are rotational equivalents to their translational counterparts. Let’s consider what the counterpart to force must be. Since forces change the translational motion of objects, the rotational counterpart must be related to changing the rotational motion of an object about an axis. We call this rotational counterpart torque.
In everyday life, we rotate objects about an axis all the time, so intuitively we already know much about torque. Consider, for example, how we rotate a door to open it. First, we know that a door opens slowly if we push too close to its hinges; it is more efficient to rotate a door open if we push far from the hinges. Second, we know that we should push perpendicular to the plane of the door; if we push parallel to the plane of the door, we are not able to rotate it. Third, the larger the force, the more effective it is in opening the door; the harder you push, the more rapidly the door opens. The first point implies that the farther the force is applied from the axis of rotation, the greater the angular acceleration; the second implies that the effectiveness depends on the angle at which the force is applied; the third implies that the magnitude of the force must also be part of the equation. Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point. Figure 10.31 shows counterclockwise rotations.
Figure 10.31 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) A counterclockwise torque is produced by a force $F→F→$ acting at a distance r from the hinges (the pivot point). (b) A smaller counterclockwise torque is produced when a smaller force $F→′F→′$ acts at the same distance r from the hinges. (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) A smaller counterclockwise torque is produced by the same magnitude force as (a) acting at the same distance as (a) but at an angle $θθ$ that is less than $90°90°$.
Now let’s consider how to define torques in the general three-dimensional case.
## Torque
When a force $F→F→$ is applied to a point P whose position is $r→r→$ relative to O (Figure 10.32), the torque $τ→τ→$ around O is
$τ→=r→×F→.τ→=r→×F→.$
10.22
Figure 10.32 The torque is perpendicular to the plane defined by $r→andF→r→andF→$ and its direction is determined by the right-hand rule.
From the definition of the cross product, the torque $τ→τ→$ is perpendicular to the plane containing $r→andF→r→andF→$ and has magnitude
$|τ→|=|r→×F→|=rFsinθ,|τ→|=|r→×F→|=rFsinθ,$
where $θθ$ is the angle between the vectors $r→r→$ and $F→F→$. The SI unit of torque is newtons times meters, usually written as $N·mN·m$. The quantity $r⊥=rsinθr⊥=rsinθ$ is the perpendicular distance from O to the line determined by the vector $F→F→$ and is called the lever arm. Note that the greater the lever arm, the greater the magnitude of the torque. In terms of the lever arm, the magnitude of the torque is
$|τ→|=r⊥F.|τ→|=r⊥F.$
10.23
The cross product $r→×F→r→×F→$ also tells us the sign of the torque. In Figure 10.32, the cross product $r→×F→r→×F→$ is along the positive z-axis, which by convention is a positive torque. If $r→×F→r→×F→$ is along the negative z-axis, this produces a negative torque.
If we consider a disk that is free to rotate about an axis through the center, as shown in Figure 10.33, we can see how the angle between the radius $r→r→$ and the force $F→F→$ affects the magnitude of the torque. If the angle is zero, the torque is zero; if the angle is $90°90°$, the torque is maximum. The torque in Figure 10.33 is positive because the direction of the torque by the right-hand rule is out of the page along the positive z-axis. The disk rotates counterclockwise due to the torque, in the same direction as a positive angular acceleration.
Figure 10.33 A disk is free to rotate about its axis through the center. The magnitude of the torque on the disk is $rFsinθrFsinθ$.When $θ=0°θ=0°$, the torque is zero and the disk does not rotate. When $θ=90°θ=90°$, the torque is maximum and the disk rotates with maximum angular acceleration.
Any number of torques can be calculated about a given axis. The individual torques add to produce a net torque about the axis. When the appropriate sign (positive or negative) is assigned to the magnitudes of individual torques about a specified axis, the net torque about the axis is the sum of the individual torques:
$τ→net=∑iτ→i.τ→net=∑iτ→i.$
10.24
## Calculating Net Torque for Rigid Bodies on a Fixed Axis
In the following examples, we calculate the torque both abstractly and as applied to a rigid body.
We first introduce a problem-solving strategy.
## Problem-Solving Strategy
### Finding Net Torque
1. Choose a coordinate system with the pivot point or axis of rotation as the origin of the selected coordinate system.
2. Determine the angle between the lever arm $r→r→$ and the force vector.
3. Take the cross product of $r→andF→r→andF→$ to determine if the torque is positive or negative about the pivot point or axis.
4. Evaluate the magnitude of the torque using $r⊥Fr⊥F$.
5. Assign the appropriate sign, positive or negative, to the magnitude.
6. Sum the torques to find the net torque.
## Example 10.14
### Calculating Torque
Four forces are shown in Figure 10.34 at particular locations and orientations with respect to a given xy-coordinate system. Find the torque due to each force about the origin, then use your results to find the net torque about the origin.
Figure 10.34 Four forces producing torques.
### Strategy
This problem requires calculating torque. All known quantities––forces with directions and lever arms––are given in the figure. The goal is to find each individual torque and the net torque by summing the individual torques. Be careful to assign the correct sign to each torque by using the cross product of $r→r→$ and the force vector $F→F→$.
### Solution
Use $|τ→|=r⊥F=rFsinθ|τ→|=r⊥F=rFsinθ$ to find the magnitude and $τ→=r→×F→τ→=r→×F→$ to determine the sign of the torque.
The torque from force 40 N in the first quadrant is given by $(4)(40)sin90°=160N·m(4)(40)sin90°=160N·m$.
The cross product of $r→r→$ and $F→F→$ is out of the page, positive.
The torque from force 20 N in the third quadrant is given by$−(3)(20)sin90°=−60N·m−(3)(20)sin90°=−60N·m$.
The cross product of $r→r→$ and $F→F→$ is into the page, so it is negative.
The torque from force 30 N in the third quadrant is given by $(5)(30)sin53°=120N·m(5)(30)sin53°=120N·m$.
The cross product of $r→r→$ and $F→F→$ is out of the page, positive.
The torque from force 20 N in the second quadrant is given by $(1)(20)sin30°=10N·m(1)(20)sin30°=10N·m$.
The cross product of $r→r→$ and $F→F→$ is out of the page.
The net torque is therefore $τnet=∑i|τi|=160−60+120+10=230N·m.τnet=∑i|τi|=160−60+120+10=230N·m.$
### Significance
Note that each force that acts in the counterclockwise direction has a positive torque, whereas each force that acts in the clockwise direction has a negative torque. The torque is greater when the distance, force, or perpendicular components are greater.
## Example 10.15
### Calculating Torque on a rigid body
Figure 10.35 shows several forces acting at different locations and angles on a flywheel. We have $|F→1|=20N,|F→1|=20N,$ $|F→2|=30N|F→2|=30N$, $|F→3|=30N|F→3|=30N$, and $r=0.5mr=0.5m$. Find the net torque on the flywheel about an axis through the center.
Figure 10.35 Three forces acting on a flywheel.
### Strategy
We calculate each torque individually, using the cross product, and determine the sign of the torque. Then we sum the torques to find the net torque.
### Solution
We start with $F→1F→1$. If we look at Figure 10.35, we see that $F→1F→1$ makes an angle of $90°+60°90°+60°$ with the radius vector $r→r→$. Taking the cross product, we see that it is out of the page and so is positive. We also see this from calculating its magnitude:
$|τ→1|=rF1sin150°=0.5m(20N)(0.5)=5.0N·m.|τ→1|=rF1sin150°=0.5m(20N)(0.5)=5.0N·m.$
Next we look at $F→2F→2$. The angle between $F→2F→2$ and $r→r→$ is $90°90°$ and the cross product is into the page so the torque is negative. Its value is
$|τ→2|=−rF2sin90°=−0.5m(30N)=−15.0N·m.|τ→2|=−rF2sin90°=−0.5m(30N)=−15.0N·m.$
When we evaluate the torque due to $F→3F→3$, we see that the angle it makes with $r→r→$ is zero so $r→×F→3=0.r→×F→3=0.$ Therefore, $F→3F→3$ does not produce any torque on the flywheel.
We evaluate the sum of the torques:
$τnet=∑i|τi|=5−15=−10N·m.τnet=∑i|τi|=5−15=−10N·m.$
### Significance
The axis of rotation is at the center of mass of the flywheel. Since the flywheel is on a fixed axis, it is not free to translate. If it were on a frictionless surface and not fixed in place, $F→3F→3$ would cause the flywheel to translate, as well as $F→1F→1$. Its motion would be a combination of translation and rotation.
A large ocean-going ship runs aground near the coastline, similar to the fate of the Costa Concordia, and lies at an angle as shown below. Salvage crews must apply a torque to right the ship in order to float the vessel for transport. A force of $5.0×105N5.0×105N$ acting at point A must be applied to right the ship. What is the torque about the point of contact of the ship with the ground (Figure 10.36)?
Figure 10.36 A ship runs aground and tilts, requiring torque to be applied to return the vessel to an upright position.
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# 6.7 Integer exponents and scientific notation
Page 1 / 10
By the end of this section, you will be able to:
• Use the definition of a negative exponent
• Simplify expressions with integer exponents
• Convert from decimal notation to scientific notation
• Convert scientific notation to decimal form
• Multiply and divide using scientific notation
Before you get started, take this readiness quiz.
1. What is the place value of the $6$ in the number $64,891$ ?
If you missed this problem, review [link] .
2. Name the decimal: $0.0012.$
If you missed this problem, review [link] .
3. Subtract: $5-\left(-3\right).$
If you missed this problem, review [link] .
## Use the definition of a negative exponent
We saw that the Quotient Property for Exponents introduced earlier in this chapter, has two forms depending on whether the exponent is larger in the numerator or the denominator.
## Quotient property for exponents
If $a$ is a real number, $a\ne 0$ , and $m\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}n$ are whole numbers, then
$\begin{array}{c}\frac{{a}^{m}}{{a}^{n}}={a}^{m-n},m>n\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}},n>m\hfill \end{array}$
What if we just subtract exponents regardless of which is larger?
Let’s consider $\frac{{x}^{2}}{{x}^{5}}$ .
We subtract the exponent in the denominator from the exponent in the numerator.
$\begin{array}{c}\hfill \frac{{x}^{2}}{{x}^{5}}\hfill \\ \hfill {x}^{2-5}\hfill \\ \hfill {x}^{-3}\hfill \end{array}$
We can also simplify $\frac{{x}^{2}}{{x}^{5}}$ by dividing out common factors:
This implies that ${x}^{-3}=\frac{1}{{x}^{3}}$ and it leads us to the definition of a negative exponent .
## Negative exponent
If $n$ is an integer and $a\ne 0$ , then ${a}^{\text{−}n}=\frac{1}{{a}^{n}}$ .
The negative exponent tells us we can re-write the expression by taking the reciprocal of the base and then changing the sign of the exponent.
Any expression that has negative exponents is not considered to be in simplest form. We will use the definition of a negative exponent and other properties of exponents to write the expression with only positive exponents.
For example, if after simplifying an expression we end up with the expression ${x}^{-3}$ , we will take one more step and write $\frac{1}{{x}^{3}}$ . The answer is considered to be in simplest form when it has only positive exponents.
Simplify: ${4}^{-2}$ ${10}^{-3}.$
## Solution
1. $\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{4}^{-2}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{4}^{2}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{16}\hfill \end{array}$
2. $\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{10}^{-3}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{10}^{3}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{1000}\hfill \end{array}$
Simplify: ${2}^{-3}$ ${10}^{-7}.$
$\frac{1}{8}$ $\frac{1}{{10}^{7}}$
Simplify: ${3}^{-2}$ ${10}^{-4}.$
$\frac{1}{9}$ $\frac{1}{10,000}$
In [link] we raised an integer to a negative exponent. What happens when we raise a fraction to a negative exponent? We’ll start by looking at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{1}{{a}^{\text{−}n}}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{\frac{1}{{a}^{n}}}\hfill \\ \text{Simplify the complex fraction.}\hfill & & & \phantom{\rule{4em}{0ex}}1·\frac{{a}^{n}}{1}\hfill \\ \text{Multiply.}\hfill & & & \phantom{\rule{4em}{0ex}}{a}^{n}\hfill \end{array}$
This leads to the Property of Negative Exponents.
## Property of negative exponents
If $n$ is an integer and $a\ne 0$ , then $\frac{1}{{a}^{\text{−}n}}={a}^{n}$ .
Simplify: $\frac{1}{{y}^{-4}}$ $\frac{1}{{3}^{-2}}.$
## Solution
1. $\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{1}{{y}^{-4}}\hfill \\ \text{Use the property of a negative exponent,}\phantom{\rule{0.2em}{0ex}}\frac{1}{{a}^{\text{−}n}}={a}^{n}.\hfill & & & \phantom{\rule{4em}{0ex}}{y}^{4}\hfill \end{array}$
2. $\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{1}{{3}^{-2}}\hfill \\ \text{Use the property of a negative exponent,}\phantom{\rule{0.2em}{0ex}}\frac{1}{{a}^{\text{−}n}}={a}^{n}.\hfill & & & \phantom{\rule{4em}{0ex}}{3}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}9\hfill \end{array}$
Simplify: $\frac{1}{{p}^{-8}}$ $\frac{1}{{4}^{-3}}.$
${p}^{8}$ $64$
Simplify: $\frac{1}{{q}^{-7}}$ $\frac{1}{{2}^{-4}}.$
${q}^{7}$ $16$
Suppose now we have a fraction raised to a negative exponent. Let’s use our definition of negative exponents to lead us to a new property.
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\frac{3}{4}\right)}^{-2}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{−}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(\frac{3}{4}\right)}^{2}}\hfill \\ \text{Simplify the denominator.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{\frac{9}{16}}\hfill \\ \text{Simplify the complex fraction.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{16}{9}\hfill \\ \text{But we know that}\phantom{\rule{0.2em}{0ex}}\frac{16}{9}\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}{\left(\frac{4}{3}\right)}^{2}.\hfill & & & \\ \text{This tells us that:}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(\frac{3}{4}\right)}^{-2}={\left(\frac{4}{3}\right)}^{2}\hfill \end{array}$
Mario invested $475 in$45 and $25 stock shares. The number of$25 shares was five less than three times the number of $45 shares. How many of each type of share did he buy? Jawad Reply will every polynomial have finite number of multiples? cricket Reply a=# of 10's. b=# of 20's; a+b=54; 10a + 20b=$910; a=54 -b; 10(54-b) + 20b=$910; 540-10b+20b=$910; 540+10b=$910; 10b=910-540; 10b=370; b=37; so there are 37 20's and since a+b=54, a+37=54; a=54-37=17; a=17, so 17 10's. So lets check.$740+$170=$910.
. A cashier has 54 bills, all of which are $10 or$20 bills. The total value of the money is $910. How many of each type of bill does the cashier have? jojo Reply whats the coefficient of 17x Dwayne Reply the solution says it 14 but how i thought it would be 17 im i right or wrong is the exercise wrong Dwayne 17 Melissa wow the exercise told me 17x solution is 14x lmao Dwayne thank you Dwayne A private jet can fly 1,210 miles against a 25 mph headwind in the same amount of time it can fly 1,694 miles with a 25 mph tailwind. Find the speed of the jet Mikaela Reply Washing his dad’s car alone, eight-year-old Levi takes 2.5 hours. If his dad helps him, then it takes 1 hour. How long does it take the Levi’s dad to wash the car by himself? Sam Reply Ethan and Leo start riding their bikes at the opposite ends of a 65-mile bike path. After Ethan has ridden 1.5 hours and Leo has ridden 2 hours, they meet on the path. Ethan’s speed is 6 miles per hour faster than Leo’s speed. Find the speed of the two bikers. Mckenzie Reply Nathan walked on an asphalt pathway for 12 miles. He walked the 12 miles back to his car on a gravel road through the forest. On the asphalt he walked 2 miles per hour faster than on the gravel. The walk on the gravel took one hour longer than the walk on the asphalt. How fast did he walk on the gravel? Mckenzie Nancy took a 3 hour drive. She went 50 miles before she got caught in a storm. Then she drove 68 miles at 9 mph less than she had driven when the weather was good. What was her speed driving in the storm? Reiley Reply Mr Hernaez runs his car at a regular speed of 50 kph and Mr Ranola at 36 kph. They started at the same place at 5:30 am and took opposite directions. At what time were they 129 km apart? hamzzi Reply 90 minutes muhammad Melody wants to sell bags of mixed candy at her lemonade stand. She will mix chocolate pieces that cost$4.89 per bag with peanut butter pieces that cost $3.79 per bag to get a total of twenty-five bags of mixed candy. Melody wants the bags of mixed candy to cost her$4.23 a bag to make. How many bags of chocolate pieces and how many bags of peanut butter pieces should she use?
enrique borrowed $23,500 to buy a car he pays his uncle 2% interest on the$4,500 he borrowed from him and he pays the bank 11.5% interest on the rest. what average interest rate does he pay on the total $23,500 Nakiya Reply 13.5 Pervaiz Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tiles. She will use basic tiles that cost$8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be$10 per square foot?
The equation P=28+2.54w models the relation between the amount of Randy’s monthly water bill payment, P, in dollars, and the number of units of water, w, used. Find the payment for a month when Randy used 15 units of water.
Bridget
help me understand graphs
what kind of graphs?
bruce
function f(x) to find each value
Marlene
I am in algebra 1. Can anyone give me any ideas to help me learn this stuff. Teacher and tutor not helping much.
Marlene
Given f(x)=2x+2, find f(2) so you replace the x with the 2, f(2)=2(2)+2, which is f(2)=6
Melissa
if they say find f(5) then the answer would be f(5)=12
Melissa
I need you to help me Melissa. Wish I can show you my homework
Marlene
How is f(1) =0 I am really confused
Marlene
what's the formula given? f(x)=?
Melissa
It shows a graph that I wish I could send photo of to you on here
Marlene
Which problem specifically?
Melissa
which problem?
Melissa
I don't know any to be honest. But whatever you can help me with for I can practice will help
Marlene
I got it. sorry, was out and about. I'll look at it now.
Melissa
Thank you. I appreciate it because my teacher assumes I know this. My teacher before him never went over this and several other things.
Marlene
I just responded.
Melissa
Thank you
Marlene
-65r to the 4th power-50r cubed-15r squared+8r+23 ÷ 5r
Rich
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# CBSE Class 12 Maths Board Exam 2022: Important 4 marks questions
Important 4 marks questions for CBSE Class 12 Maths subject are provided here for students. The questions are prepared as per new latest exam pattern and syllabus (2021-2022). The Central Board of Secondary Education CBSE) provides students of class 12 with 4 marks question from the exam perspective. As Maths is the most logical and difficult subject amongst the students, it is important for students to practice these questions. The level of difficulty of 4 marks question varies from easy to difficult (sometimes). So solving this section can be cover a high percentage of marks in a shorter span of time.
Generally, the section contains 11 questions, containing 44% of the total marks. Thus, this section can be far more beneficial for the students to score more overall. We at BYJU’S provides students of CBSE class 12 with important 4 marks questions, which can be beneficial to excel in their examination.
## Class 12 Maths Important 4 Marks Questions
Important 4 Marks Questions for Class 12 Maths Board are as follows-
Question 1- The polynomials ax3 – 3x2 +4 and 2x3 – 5x +a when divided by (x – 2) leave the remainders p and q respectively. If p – 2q = 4, find the value of a.
Question 2- Construct a ΔABC in which BC = 3.8 cm, ∠B = 45oand AB + AC = 6.8cm.
Question 3- If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 +b3 +c3 − 3abc.
Question 4- Show that the function F in A = R – $$\left \{ \frac{2}{3} \right \}$$, defined as f(x) = $$\frac{4x + 3}{6x – 4}$$ is one-one and onto. Hence find $$f^{-1}$$.
Question 5- Find the value of the following-
$$\tan \frac{1}{2}\left [ \sin^{-1}\frac{2x}{1+x^{2}} + \cos^{-1}\frac{1 – y^{2}}{1 + y^{2}} \right ]$$,
where $$\left | x \right | < 1, y > 0\;\; and \;\;xy < 1$$
Question 6- Prove that $$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}$$
Question 7- Prove the following-
$$\begin{vmatrix} 1 & x & x^{2}\\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{vmatrix} = \left ( 1-x^{3} \right )^{2}$$
Question 8- Differentiate the following function with respect to x:
$$(\log x)^{x} + x^{\log x}$$
Question 9- If $$y = \log \left [ x + \sqrt{x^{2} + a^{2}} \right ]$$, show that $$(x^{2} + a^{2})\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} + x\frac{\mathrm{d} y}{\mathrm{d} x} = 0$$
Question 10- Evaluate: $$\int \frac{\sin (x-a)}{\sin (x+a)}dx$$
Question 11- Evaluate: $$\int_{0}^{4} (\left | x \right | + \left | x + 2 \right | + \left | x – 4 \right |) dx$$
Question 12- If $$\vec{a}$$ and $$\vec{b}$$ are two vectors such that $$\left |\vec{a} + \vec{b} \right | = \left |\vec{a} \right |$$, then prove that vector $$2\vec{a} + \vec{b}$$ is perpendicular to vector $$\vec{b}$$.
Question 13- A speaks 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact ? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A ?
Question 14- There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denotes the sum of the numbers on the two drawn cards. Find the mean & variance of X.
Question 15- Maximise Z = x + 2y,
Subject to the constraints
$$x + 2y \geq 100$$
$$2x – y \leq 0$$
$$2x + y \leq 200$$
$$x,y \geq 0$$
Solve the above LPP graphically
Question 16- A school wants to award its students for regularity and hard work with a total cash award of Rs. 6000. If three times the award money for hard work added to that given for regularity amounts to Rs. 11,000, represent the above situation algebraically and find the award money for each value, using matrix method. Suggest two more values, which the school must include for award.
Question 17- Find the intervals in which the function given by
f(x) = 2×3 -3×2 -36x+7 is
• Strictly increasing
• Strictly decreasing
Question 18- Bag A contains 3 red and 2 black balls, while bag B contains 2 red and 3 black balls. A ball drawn at random from bag A is transferred to bag B and then one ball is drawn at random from bag B. If this ball was found to be a red ball, find the probability that the ball drawn from bag A was red.
Question 19- If $$\tan^{-1}\frac{x-3}{x-4} + \tan^{-1}\frac{x+3}{x+4} = \frac{\pi}{4}$$, then find the value of x.
Question 20- If $$y = (\sec ^{-1}x)^{2}$$, then show that $$x^{2}(x^{2}-1)\frac{\mathrm{d} ^{2}y}{\mathrm{d} x} + (2x^{3}-x)\frac{\mathrm{d} y}{\mathrm{d} x} =2$$
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# If f(x)= cos(-2 x -1) and g(x) = 4x^2 -5 , how do you differentiate f(g(x)) using the chain rule?
Mar 18, 2016
$\frac{\mathrm{df}}{\mathrm{dx}} = - 16 x \sin \left(8 {x}^{2} - 9\right)$
#### Explanation:
Note that $\cos \left(- 2 x - 1\right) = \cos \left(2 x + 1\right)$ as $\cos \left(- \theta\right) = \cos \theta$.
If $f \left(x\right) = \cos \left(- 2 x - 1\right)$ and $g \left(x\right) = 4 {x}^{2} - 5$
$f \left(g \left(x\right)\right) = \cos \left(2 \left(4 {x}^{2} - 5\right) + 1\right)$
As according to chain rule, $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$
$\frac{\mathrm{df}}{\mathrm{dx}} = - 2 \sin \left(2 \left(4 {x}^{2} - 5\right) + 1\right) \times \frac{d}{\mathrm{dx}} \left(4 {x}^{2} - 5\right)$
= -2sin(2(4x^2-5)+1)xx8x)
= $- 16 x \sin \left(8 {x}^{2} - 9\right)$
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How do you evaluate \frac { 3x - 21} { 6x - 42}?
Sep 15, 2017
$\frac{1}{2}$
Explanation:
Notice that the denominator is two times the numerator.
Even if you don't notice that, you should notice that you can factor out a 2 from the denominator:
$\frac{3 x - 21}{6 x - 42} = \frac{3 x - 21}{2 \left(3 x - 21\right)}$
And now you've got an identical 3x-21 term in both numerator and denominator that cancels out, giving:
$\frac{1}{2}$
Sep 15, 2017
$\frac{1}{2}$
Explanation:
$\frac{3 x - 21}{6 x - 42}$
First we see that both terms in the denominator are divisible by ${2}^{1}$:
$\frac{3 x - 21}{2 \cdot 3 x - 2 \cdot 21}$
Since two is a common factor between the two terms, we can ${\text{factorize them}}^{2}$:
$\frac{3 x - 21}{2 \cdot \left(3 x - 21\right)}$
Simplifying:
$\frac{\cancel{3 x - 21}}{2 \cdot \cancel{3 x - 21}} = \frac{1}{2}$
$\Rightarrow \frac{3 x - 21}{6 x - 42} = \frac{1}{2}$
${.}^{1}$(To know when a number is divisible by 2, you look at the last digit, if it is divisible by 2 then the whole number is divisible by 2; 42 last digit 2 is divisible by 2 $\rightarrow$ 42 is divisible by 2).
${.}^{2}$ $\left(a x + a y\right) = a \left(x + y\right)$
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# The Meaning Of Fractions
Fractions can often be a source of frustration starting in elementary school. Many times this is because children have not previously been taught algorithms and procedures correctly. They may often confuse methods with others they have learned for whole numbers or have simply not practiced fractions enough to understand them fully. This is precisely why building and understanding the foundation of number sense has been shown to increase student achievement later on.
## Parent Involvement
Parents looking to help their children with a better understanding of mathematics also need to develop a more sound understanding of the foundations. This is why we involve parents in our programs at Dropkick Math. Our trained instructors will help build a parent’s mathematics capacity so they can adequately support their child’s journey in elementary math. We believe that success is achieved by learning together.
## Understanding Fractions
For students to really understand fractions, it is essential that they learn to view them as numbers. Specifically, numbers that represent different constructs based on the context. In the past, fractions education focused on the outcomes, memorizing procedures so that students could successfully operate with fractions. However, being a good mathematical thinker is no longer based on how quickly a child can produce an answer. It is more important that mathematical thinkers understand the process and have multiple pathways to a solution.
To become a good mathematical thinker, it is essential to understand the meaning of fractions. Fractions represent equal parts of a whole or of a collection.
Fraction of a whole: When a whole is divided into equal parts, each part is a fraction of the whole.
Fraction of a collection: Fractions can also represent parts of a set or a collection.
Fractions have two parts. The number on the top of the line is called the numerator and tells how many equal parts of the whole or collection are taken. The number below the line is called the denominator and shows the total divisible number of equal parts in a whole, or in a collection.
When explaining fractions to a child, some of the most common examples in real life are equal slices of pizza, fruit, cake, or a bar of chocolate. Children may also learn through these foods that when the parts of the whole are unevenly divided, they don’t form fractions.
Using examples of fractions in everyday life can help children understand and visualize the math concept. Some examples you can use for older children include:
• splitting a bill at a restaurant into halves, thirds, or quarters
• working out price comparisons in the grocery store when something is half price
• looking at a clock and teaching them about half an hour and a quarter past
When it comes to helping a child with their math homework, fractions are probably what you will struggle with the most. The best place to start when explaining fractions to a child is to offer a description such as, “a fraction is any part of a group, number, or whole.” Then, using real-life experiences, fractions can become a little less scary.
## Math Anxiety
For children, the world of math can be filled with despair and anxiety if they struggle to understand the concept of fractions truly. Fractions are known to be one of the main contributors to math anxiety and can be one of the most significant barriers to your child’s success in math. However, this can be avoided with the help of Dropkick Math Academy. Our programs are designed specifically to work with children to overcome any learning gaps they may have. Our programs include working with adding and subtracting fractions with like and unlike denominators. Children will also work with multiplying and dividing fractions with whole numbers and gain the ability to add and subtract fractions using mental math.
As children advance through our program, they will connect fractions, percents, and decimals and use each form flexibly. They will strengthen their proportional reasoning skills and develop proficiency with fractions. By the end of our program, children will have developed a solid foundation for secondary mathematics involving linear relationships, radian measures, and trigonometry.
## Get Back On Track
If your child is struggling with fractions or other math operations, Dropkick Math can help get them back on track! We focus on the critical gaps in learning where children often show difficulty and provide an exciting way for students to thrive in mathematics by applying newly discovered techniques. By focusing on the foundational concepts (number sense, operational sense, algebraic reasoning, and proportional reasoning), our engaging, innovative programs help students fully understand critical concepts that are the base fundamentals of mathematics.
We also address deficiencies through our innovative, research-based math learning techniques while correcting any underlying misconceptions about mathematics. All instructors are qualified Ontario Certified Teachers who can offer differentiated approaches, making it accessible for all learning needs.
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What is the coefficient matrix of the system of equations?
What is the coefficient matrix of the system of equations?
In linear algebra, a coefficient matrix is a matrix consisting of the coefficients of the variables in a set of linear equations. The matrix is used in solving systems of linear equations.
What is coefficient matrix and augmented matrix?
Solution: A coefficient matrix is a matrix made up of the coefficients from a system of linear equations. An augmented matrix is similar in that it, too, is a coefficient matrix, but in addition it is augmented with a column consisting of the values on the right-hand side of the equations of the linear system.
What is the determinant of the coefficient matrix of the system?
The denominator is the determinant of the coefficient matrix. And the numerator is the determinant of the matrix formed by replacing the column that represents the coefficients of y with the corresponding column of constants.
How do you find the coefficient of a linear equation?
If a, b, and r are real numbers (and if a and b are not both equal to 0) then ax+by = r is called a linear equation in two variables. (The “two variables” are the x and the y.) The numbers a and b are called the coefficients of the equation ax+by = r. The number r is called the constant of the equation ax + by = r.
What is an input coefficient matrix?
“Input coefficients” represent the scale of raw materials and fuels used can be obtained by dividing the input of raw materials and fuels utilized to generate one unit of production in each sector. They correspond to basic unit prices, and are obtained by dividing the amount of raw materials, fuel, etc.
How do you use the matrix method?
A system of equations can be solved using matrix multiplication. A is the coefficient matrix, X the variable matrix and B the constant matrix. The second method to find the solution for the system of equations is Row reduction or Gaussian Elimination. The augmented matrix for the linear equations is written.
How do you create a coefficient matrix?
To express this system in matrix form, you follow three simple steps:
1. Write all the coefficients in one matrix first. This is called a coefficient matrix.
2. Multiply this matrix with the variables of the system set up in another matrix.
What is coefficient in linear regression?
In linear regression, coefficients are the values that multiply the predictor values. Suppose you have the following regression equation: y = 3X + 5. In this equation, +3 is the coefficient, X is the predictor, and +5 is the constant.
What is a coefficient matrix?
In linear algebra, a coefficient matrix is a matrix consisting of the coefficients of the variables in a set of linear equations. The matrix is used in solving systems of linear equations .
What is the coefficient matrix of 2x + 3y?
Do It Faster, Learn It Better. A system of linear equations can be represented in matrix form using a coefficient matrix, a variable matrix, and a constant matrix. 2 x + 3 y = 8 5 x − y = − 2 . The coefficient matrix can be formed by aligning the coefficients of the variables of each equation in a row.
How do you find the matrix of a system of equations?
A system of linear equations can be represented in matrix form using a coefficient matrix, a variable matrix, and a constant matrix. 2 x + 3 y = 8 5 x − y = − 2 . The coefficient matrix can be formed by aligning the coefficients of the variables of each equation in a row.
What is the matrix of the equation 8-2?
On the right side of the equality we have the constant terms of the equations, 8 and − 2 . The two numbers in that order correspond to the first and second equations, and therefore take the places at the first and the second rows in the constant matrix. So, the matrix becomes [ 8 − 2] .
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The distinct factors of 162 are the numbers which when multiplied in pairs give the product together 162. There room in full 8 components of 162. These determinants of 162 can also be integers, together we have the right to have an adverse factors that 162 together well. The number 162 is an even composite number, that is because it has numerous factors. In this lesson, we will certainly calculate the determinants of 162, prime determinants of 162, and factors of 162 in pairs together with solved instances for a better understanding.
You are watching: What are the factors of 162
Factors the 162: 1, 2, 3, 9, 27, 54, 81 and also 162Prime Factorization of 162: 162 = 2 × 3 × 3 × 3 × 3
1 What space the determinants of 162? 2 How come Calculate factors of 162? 3 Factors the 162 by prime Factorization 4 Factors the 162 in Pairs 5 Important Notes 6 FAQs on factors of 162
The number 162 is an also composite number. As it is even, that will have actually 2 as its factor. To know why it is composite, let"s recall the meaning of a composite number. "If a number has an ext than two determinants then the is known as a composite number."Now, let"s take it the situation of 162. The factors of 162 are written together 1,2, 3, 6, 9, 18, 27, 54, 81, and also 162. There are more than two factors of 162, for this reason it"s composite. "The factors of 162 space all the integers that 162 can be split into."
To calculation the determinants of 162, we need to find all the numbers that divide 162, without leaving any type of remainder. We start with 1, and also then check the following numbers 2, 3, 4, 5, 6, 7, and so on, up to 81 (half of 162). We must note the 1 and also the number itself will constantly be a aspect of the provided number. Observe the adhering to table to inspect the determinants of 162 through division:
Division Factor 162÷1 Factor = 1 162÷2 Factor = 2 162÷3 Factor = 3 162÷6 Factor = 6 162÷9 Factor = 9 162÷18 Factor = 18 162÷27 Factor = 27 162÷54 Factor = 54 162÷81 Factor = 81 162÷162 Factor = 162
The numbers which us multiply to obtain 162 are the components of 162Factors of 162 are written together 1,2, 3, 6, 9, 18, 27, 54, 81 and also 162The number chin is a aspect of the number together it divides chin exactly.Factor pairs space the pairs of two numbers that, when multiplied, offer the number. Variable pairs of 162 space (1,162) (2,81) (3,54) (6,27) and also (9,18).
"Prime factorization method to refer a composite number together the product that its prime factors." To get the prime factors of 162, we division it by its the smallest prime element which is 2, 162÷2 = 81. Now, 81 is separated by its the smallest prime factor and also the quotient is obtained. This procedure goes top top till we get the quotient as 1. The factor tree that 162 or 162-factor tree is displayed below:
The over factorization is the tree diagram representation of primes factors of 162. Therefore, components of 162 are 2 × 3 × 3 × 3. Currently that we have done the prime factorization that 162, we deserve to multiply them and also get the various other factors. Deserve to you try and find out if all the components are covered or not? and as you could have already guessed, for prime numbers, there room no various other factors.
The pair of number which provides 162 as soon as multiplied is known as aspect pairs that 162. The complying with are the determinants of 162 in pairs.
The product type of 162 Pair factor 1 × 162 = 162 (1, 162) 2 × 81 = 162 (2, 81) 3 × 54 = 162 (3, 54) 6 × 27 = 162 (6, 27) 9 × 18 = 162 (9, 18) 18 × 9 = 162 (18, 9) 27 × 6 = 162 (27, 6) 54 × 3 = 162 (54, 3) 81 × 2 = 162 (81, 2) 162 × 1 = 162 (162, 1)
If we consider negative integers, then both the number in the pair components will it is in negative.72 is positive and also (- ve) × (- ve) = +ve.So, we can have aspect pairs that 162 as (-1,-162), (-2,-81), (-3,-54), (-6,-27), and (-9, -18).
Example 1 help Allan to perform the determinants of 162 and uncover the factor pairs.
Solution
Factors of 162 are the number that divide without any type of remainder.They room 1, 2, 3, 6, 9, 18, 27, 54, 81, 162Factors pairs are the bag of 2 numbers which, when multiplied, give 162
1×162=162
2×81=162
3×54=162
6×27=162
9×18=162
Example 2 deserve to you help Candy list the factors of 162?
Solution
Factors of 162 are the number that divides 162 exactly without any kind of remainder.
162÷1=162
162÷2=81
162÷3=54
162÷6=27
162÷9=18
∴ factors of 162 are 1, 2, 3, 6, 9, 18, 27, 54, 81 and 162.
See more: How Far Is Charleston Sc From Asheville Nc, How Far Is Charleston From Asheville
Example 3: Ross told James that -3 is one of the factors of 162. Have the right to you help him uncover the various other factor?
Solution:
162 = element 1 × factor 2∴162 = (−3) × element 2Factor 2 = 162 ÷ (−3) = (−54)The other aspect is -54.
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Introductory Statistics
# Practice
## 13.1One-Way ANOVA
Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled in order to perform a one-way ANOVA test. What are they?
1.
Write one assumption.
2.
Write another assumption.
3.
Write a third assumption.
4.
Write a fourth assumption.
5.
Write the final assumption.
6.
State the null hypothesis for a one-way ANOVA test if there are four groups.
7.
State the alternative hypothesis for a one-way ANOVA test if there are three groups.
8.
When do you use an ANOVA test?
## 13.2The F Distribution and the F-Ratio
Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in Table 13.13 are the weights for the different groups.
Group 1 Group 2 Group 3
216 202 170
198 213 165
240 284 182
187 228 197
176 210 201
Table 13.13
9.
What is the Sum of Squares Factor?
10.
What is the Sum of Squares Error?
11.
What is the df for the numerator?
12.
What is the df for the denominator?
13.
What is the Mean Square Factor?
14.
What is the Mean Square Error?
15.
What is the F statistic?
Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The one-way ANOVA results are shown in Table 13.14.
Team 1 Team 2 Team 3 Team 4
1 2 0 3
2 3 1 4
0 2 1 4
3 4 0 3
2 4 0 2
Table 13.14
16.
What is SSbetween?
17.
What is the df for the numerator?
18.
What is MSbetween?
19.
What is SSwithin?
20.
What is the df for the denominator?
21.
What is MSwithin?
22.
What is the F statistic?
23.
Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis?
## 13.3Facts About the F Distribution
24.
An F statistic can have what values?
25.
What happens to the curves as the degrees of freedom for the numerator and the denominator get larger?
Use the following information to answer the next seven exercise. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in Table 13.15.
Team 1 Team 2 Team 3 Team 4 Team 5
36 32 48 38 41
42 35 50 44 39
51 38 39 46 40
Table 13.15
26.
What is the df(num)?
27.
What is the df(denom)?
28.
What are the Sum of Squares and Mean Squares Factors?
29.
What are the Sum of Squares and Mean Squares Errors?
30.
What is the F statistic?
31.
What is the p-value?
32.
At the 5% significance level, is there a difference in the mean jump heights among the teams?
Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Table 13.16
Group A Group B Group C
101 151 101
108 149 109
98 160 198
107 112 186
111 126 160
Table 13.16
33.
What is the df(num)?
34.
What is the df(denom)?
35.
What are the SSbetween and MSbetween?
36.
What are the SSwithin and MSwithin?
37.
What is the F Statistic?
38.
What is the p-value?
39.
At the 10% significance level, are the scores among the different groups different?
Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
Northeast South West Central East
16.3 16.9 16.4 16.2 17.1
16.1 16.5 16.5 16.6 17.2
16.4 16.4 16.6 16.5 16.6
16.5 16.2 16.1 16.4 16.8
$x ¯ = x ¯ =$
$s 2 = s 2 =$
Table 13.17
Enter the data into your calculator or computer.
40.
p-value = ______
State the decisions and conclusions (in complete sentences) for the following preconceived levels of α.
41.
α = 0.05
a. Decision: ____________________________
b. Conclusion: ____________________________
42.
α = 0.01
a. Decision: ____________________________
b. Conclusion: ____________________________
## 13.4Test of Two Variances
Use the following information to answer the next two exercises. There are two assumptions that must be true in order to perform an F test of two variances.
43.
Name one assumption that must be true.
44.
What is the other assumption that must be true?
Use the following information to answer the next five exercises. Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker’s times have a variance of 12.1. The second worker’s times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times. Test the claim at the 10% level. Assume that commute times are normally distributed.
45.
State the null and alternative hypotheses.
46.
What is s1 in this problem?
47.
What is s2 in this problem?
48.
What is n?
49.
What is the F statistic?
50.
What is the p-value?
51.
Is the claim accurate?
Use the following information to answer the next four exercises. Two students are interested in whether or not there is variation in their test scores for math class. There are 15 total math tests they have taken so far. The first student’s grades have a standard deviation of 38.1. The second student’s grades have a standard deviation of 22.5. The second student thinks his scores are more consistent.
52.
State the null and alternative hypotheses.
53.
What is the F Statistic?
54.
What is the p-value?
55.
At the 5% significance level, do we reject the null hypothesis?
Use the following information to answer the next three exercises. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist records his or her speeds going up 35 hills. The first cyclist has a variance of 23.8 and the second cyclist has a variance of 32.1. The cyclists want to see if their variances are the same or different. Assume that commute times are normally distributed.
56.
State the null and alternative hypotheses.
57.
What is the F Statistic?
58.
At the 5% significance level, what can we say about the cyclists’ variances?
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# 6.2 Using the normal distribution (Page 4/25)
Page 4 / 25
## References
“Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013).
“403: NUMMI.” Chicago Public Media&Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013).
“Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013).
“Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).
## Chapter review
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ . A special normal distribution, called the standard normal distribution is the distribution of z -scores. Its mean is zero, and its standard deviation is one.
## Formula review
Normal Distribution: X ~ N ( µ , σ ) where µ is the mean and σ is the standard deviation.
Standard Normal Distribution: Z ~ N (0, 1).
Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation)
Calculator function for the k th percentile: k = invNorm (area to the left of k , mean, standard deviation)
How would you represent the area to the left of one in a probability statement?
P ( x <1)
What is the area to the right of one?
Is P ( x <1) equal to P ( x ≤ 1)? Why?
Yes, because they are the same in a continuous distribution: P ( x = 1) = 0
How would you represent the area to the left of three in a probability statement?
What is the area to the right of three?
1 – P ( x <3) or P ( x >3)
If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x ?
If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x ?
1 – 0.543 = 0.457
Use the following information to answer the next four exercises:
X ~ N (54, 8)
Find the probability that x >56.
Find the probability that x <30.
0.0013
Find the 80 th percentile.
Find the 60 th percentile.
56.03
X ~ N (6, 2)
Find the probability that x is between three and nine.
X ~ N (–3, 4)
Find the probability that x is between one and four.
0.1186
X ~ N (4, 5)
Find the maximum of x in the bottom quartile.
Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P (0< x <____________) = ___________ (Use zero for the minimum value of x .)
1. Check student’s solution.
2. 3, 0.1979
Find the probability that a CD player will last between 2.8 and six years.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P (__________< x <__________) = __________
Find the 70 th percentile of the distribution for the time a CD player lasts.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70%.
2. P ( x < k ) = __________ Therefore, k = _________
1. Check student’s solution.
2. 0.70, 4.78 years
does proportion alwaus give u a yes or no answer for data
frequency destribution
प्रायिकता सिध्दान्त पर आधारित प्रतिदर्श सिध्दान्त का विकास किसने किया?
7.The following data give thenumber of car thefts that occurred in a city in the past 12 days. 63711438726915 Calculate therange, variance, and standard deviation.
express the confidence interval 81.4% ~8.5% in interval form
a bad contain 3 red and 5 black balls another 4 red and 7 black balls, A ball is drawn from a bag selected at random, Find the probability that A is red?
The information is given as, 30% of customers shopping at SHOPNO will switch to DAILY SHOPPING every month on the other hand 40% of customers shopping at DAILY SHOPPING will switch to other every month. What is the probability that customers will switch from A to B for next two months?
Calculate correlation coefficient, where SP(xy) = 144; SS(x) = 739; SS(y) = 58. (2 Points)
The information are given from a randomly selected sample of age of COVID-19 patients who have already survived. These information are collected from 200 persons. The summarized information are as, n= 20; ∑x = 490; s^2 = 40. Calculate 95% confident interval of mean age.
Ashfat
The mode of the density of power of signal is 3.5. Find the probability that the density of a random signal will be more than 2.5.
Ashfat
The average time needed to repair a mobile phone set is 2 hours. If a customer is in queue for half an hour, what is the probability that his set will be repaired within 1.6 hours?
Ashfat
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 9 and variance 0.04. Do you think that the mean thickness of the spearmint gum it produces is 8.4
3. The following are the number of mails received in different days by different organizations: Days (x) : 23, 35, 38, 50, 34, 60, 41, 32, 53, 67. Number of mails (y) : 18, 40, 52, 45, 32, 55, 50, 48, 26, 25. i) Fit a regression line of y on x and test the significance of regression. ii) Estimate y
The number of problem creating computers of two laboratories are as follows: Number of computers: 48, 6, 10, 12, 30, 11, 49, 17, 10, 14, 38, 25, 15, 19, 40, 12. Number of computers: 12, 10, 26, 11, 42, 11, 13, 12, 18, 5, 14, 38. Are the two laboratories similar in respect of problem creating compute
Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?
null rejected
Pratik
a quality control specialist took a random sample of n=10 pieces of gum and measured their thickness and found the mean 7.6 and standered deviation 0.10. Do you think that the mean thickness of the spearmint gum it produces is 7.5?
99. A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct? a
A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct?
Niaz
what is null Hypothesis
Niaz
what is null Hypothesis
Niaz
when median is greater than mode?
hello
Amaano
is this app useful
Worthy
little bit 😭
G-
oh
Worthy
when tail is positive
Jungjoon
define hypothesis
Worthy
I'm struggling to type it's on my laptop...statistics
Yoliswa
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Indeterminations zero for infinity, infinite-infinite and 1 raised to infinity
# Indeterminations zero for infinity, infinity minus infinity and 1 raised to infinity
Now I will explain how to calculate limits with indeterminations zero for infinity, infinity minus infinity and 1 raised to infinity. We will see it in detail while with step-by-step exercises resolved.
The resolution of these indeterminations are chained with other types of indeterminations such as infinity among infinity, so you must also know how to resolve that indetermination (you have it explained in the Course of Limits, along with all indeterminations).
## Calculation of limits with zero by infinity indetermination
Any number multiplied by zero is zero, but in turn, any number multiplied by infinity is infinite. So What is zero for infinity? Is it zero? Is it infinity?
Well, it’s neither one thing nor the other. Therefore, zero for infinity is another indeterminacy.
The procedure for solving limits with zero indetermination by infinity is:
1. To reach zero indetermination by infinite by substituting the x for the number you shop for
2. Operate within the function to eliminate indeterminacy
3. Solve the infinite indeterminacy between the infinite that is left to us
Let’s solve an example of a limit with zero indetermination by infinity:
First of all, we replace the x with infinity and we come to the conclusion that we are facing a zero indeterminacy for infinity:
To solve the indeterminations of zero by infinity, what must be done first of all is to operate within the limit. In this case, we can multiply the fraction by the parenthesis and it remains:
We replace the x with infinity again and arrive at another indeterminacy, this time of infinity between infinity:
To solve the infinite indeterminations between infinity, we stay in the numerator and denominator with the term of the highest degree and then we operate:
We replace the x with infinity and arrive at the final solution of the limit, which is infinity:
We are going to solve another limit exercise with this kind of indetermination.
We begin by substituting the x for infinity and we reach the point where it has a zero indeterminacy for infinity:
We operate in the function by multiplying the fraction by the root and it remains:
We replace the x with infinity and arrive at the result of the infinite indeterminacy between infinity:
We keep the highest grade terms of the numerator and denominator and solve the root that is left in the numerator:
We can eliminate the x of both parts of the fraction and therefore, we reach the final result:
## Limit calculation with infinite indetermination minus infinity
Let me ask you a question: Infinity minus infinity is zero?
A lot of people would say yes, but not really. Although it is an abstract concept, there are many “sizes” of infinity, but we do not know. Since we do not know whether infinity is equal, infinity minus infinity is another indeterminacy:
The procedure for resolving boundaries with infinite indeterminacy minus infinite is as follows:
1. To reach infinite indeterminacy less infinite by substituting the x for the number you shop for
2. Multiply and divide by the function conjugate
3. Operate on the numerator of the resulting fraction to simplify term
4. Solve the infinite indeterminacy between the infinite that is left to us
Let’s solve an example of a limit with infinite indeterminacy less infinite less infinite slower:
First, we replace the x with infinity, which brings us to infinite indeterminacy less infinite:
To solve limits with infinite indeterminacy, less infinite, we must multiply and divide them by the conjugation of function:
We operate in the fraction numerator, where we have a difference of squares, since we had a multiplication of sum per difference:
We solve the squares and group similar terms in the numerator:
We replace the x with infinity. We arrive at another indeterminacy, which is now the one of infinity among infinity:
We keep the highest grade term in both the numerator and the denominator, solve the root in and group terms in the denominator:
Finally, we eliminate the x and y to reach the limit result:
## Calculation of limits with indeterminacy 1 raised to infinity
If I asked you how much 1 is elevated to infinity, would you say the result is 1? For you would not be right, for 1 elevated to infinity is another indeterminacy:
To calculate the limits with indeterminacy 1 raised to infinity, calculate using this formula:
Where the number to which the x tends can be any number or it can be more or less infinite:
We will follow the following procedure to resolve limits with indeterminacy 1 raised to infinity:
1. To reach indeterminacy 1 raised to infinity x by the number you are tending to
1. Solve, if necessary, the infinite indetermination between infinity of the limit of the function that forms the basis of power, to show that its result is 1
2. Apply the formula to solve indeterminations 1 raised to infinity
3. Perform operations on the function, within the limit
4. Solve the infinite indeterminacy between the infinite that is left to us
Let’s solve step by step an example of a limit with indeterminacy 1 raised to infinity:
We began to solve this limit by substituting the x for infinity and arrived at this result:
This indeterminacy is not 1 elevated to infinity, but in order to apply the above formula, we must demonstrate that the infinity in infinity that is in parentheses is equal to 1.
To do this we solve the limit of the function that forms the basis of the power, that is to say:
We replace the x with infinity and reach the infinite indeterminacy between infinity:
We keep the largest term of the numerator and denominator and operate, arriving at the result, which is equal to 1:
Knowing that the limit of the function, which forms the basis of the power, when x tends to infinity is 1, now we can say that this limit results in indeterminacy 1 raised to infinity:
Therefore, we apply the previous formula and we are left with it:
Once the formula is applied, we have to do operations. First, within the parenthesis, we subtract by reducing the common denominator and group terms in the numerator:
We now remove the parenthesis by multiplying it by the term before it:
When we can no longer operate, we replace the x with infinity and reach the infinite indeterminacy between infinity:
To resolve this indeterminacy, we leave the term of highest degree and operate:
Finally, we replace the x by infinite again, which is raised to less infinite by “e” than by properties of the powers, lower the denominator. “e” raised to infinity equals infinity and 1 split by infinity equals zero:
## Calculation of limits with infinity raised to infinity
Let’s see how to solve the limits in which once you replace the x with infinity, you have as a result infinity raised to infinity.
To begin with, infinity elevated to infinity is not an indeterminacy. Infinity elevated to infinity equals infinity:
Therefore, to solve this type of limit, we only have to replace the x by infinite and the result is obtained directly.
For example:
We replace the x by infinite, which remains infinite to infinite, whose result is infinite:
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# Objective Graph and solve systems of linear inequalities in two variables. A system of linear inequalities is a set of two or more linear inequalities.
## Presentation on theme: "Objective Graph and solve systems of linear inequalities in two variables. A system of linear inequalities is a set of two or more linear inequalities."— Presentation transcript:
Objective Graph and solve systems of linear inequalities in two variables. A system of linear inequalities is a set of two or more linear inequalities containing two or more variables. The solutions of a system of linear inequalities consists of all the ordered pairs that satisfy all the linear inequalities in the system.
Example 1A: Identifying Solutions of Systems of Linear Inequalities
Tell whether the ordered pair is a solution of the given system. y ≤ –3x + 1 (–1, –3); y < 2x + 2 (–1, –3) (–1, –3) y ≤ –3x + 1 y < 2x + 2 –3 –3(–1) + 1 –3 –2 + 2 < – (–1) + 2 (–1, –3) is a solution to the system because it satisfies both inequalities.
Example 1B: Identifying Solutions of Systems of Linear Inequalities
Tell whether the ordered pair is a solution of the given system. y < –2x – 1 (–1, 5); y ≥ x + 3 (–1, 5) (–1, 5) y < –2x – 1 5 –1 + 3 y ≥ x + 3 5 –2(–1) – 1 – 1 < (–1, 5) is not a solution to the system because it does not satisfy both inequalities.
An ordered pair must be a solution of all inequalities to be a solution of the system.
Remember!
To show all the solutions of a system of linear inequalities, graph the solutions of each inequality. The solutions of the system are represented by the overlapping shaded regions. Below are graphs of Examples 1A and 1B on p. 421.
Example 2B: Solving a System of Linear Inequalities by Graphing
Graph the system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions. –3x + 2y ≥ 2 y < 4x + 3 –3x + 2y ≥ 2 Write the first inequality in slope-intercept form. 2y ≥ 3x + 2
(0, 0) and (–4, 5) are not solutions.
Example 2B Continued Graph the system. (2, 6) (1, 3) y < 4x + 3 (0, 0) (–4, 5) (2, 6) and (1, 3) are solutions. (0, 0) and (–4, 5) are not solutions.
In Lesson 6-4, you saw that in systems of linear equations, if the lines are parallel, there are no solutions. With systems of linear inequalities, that is not always true.
Example 3A: Graphing Systems with Parallel Boundary Lines
Graph the system of linear inequalities. y ≤ –2x – 4 y > –2x + 5 This system has no solutions.
Example 3B: Graphing Systems with Parallel Boundary Lines
Graph the system of linear inequalities. y > 3x – 2 y < 3x + 6 The solutions are all points between the parallel lines but not on the dashed lines.
Example 3C: Graphing Systems with Parallel Boundary Lines
Graph the system of linear inequalities. y ≥ 4x + 6 y ≥ 4x – 5 The solutions are the same as the solutions of y ≥ 4x + 6.
Example 4: Application In one week, Ed can mow at most 9 times and rake at most 7 times. He charges \$20 for mowing and \$10 for raking. He needs to make more than \$125 in one week. Show and describe all the possible combinations of mowing and raking that Ed can do to meet his goal. List two possible combinations. Earnings per Job (\$) Mowing 20 Raking 10
Example 4 Continued Step 1 Write a system of inequalities. Let x represent the number of mowing jobs and y represent the number of raking jobs. x ≤ 9 He can do at most 9 mowing jobs. y ≤ 7 He can do at most 7 raking jobs. 20x + 10y > 125 He wants to earn more than \$125.
Example 4 Continued Step 2 Graph the system.
The graph should be in only the first quadrant because the number of jobs cannot be negative. Solutions
An ordered pair solution of the system need not have whole numbers, but answers to many application problems may be restricted to whole numbers. Helpful Hint
Check It Out! Example 4 At her party, Alice is serving pepper jack cheese and cheddar cheese. She wants to have at least 2 pounds of each. Alice wants to spend at most \$20 on cheese. Show and describe all possible combinations of the two cheeses Alice could buy. List two possible combinations. Price per Pound (\$) Pepper Jack 4 Cheddar 2
Check It Out! Example 4 Continued
Step 1 Write a system of inequalities. Let x represent the pounds of pepper jack and y represent the pounds of cheddar. x ≥ 2 She wants at least 2 pounds of pepper jack. y ≥ 2 She wants at least 2 pounds of cheddar. 4x + 2y ≤ 20 She wants to spend no more than \$20.
Check It Out! Example 4 Continued
Step 2 Graph the system. The graph should be in only the first quadrant because the amount of cheese cannot be negative. Solutions
Download ppt "Objective Graph and solve systems of linear inequalities in two variables. A system of linear inequalities is a set of two or more linear inequalities."
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# Equilateral triangle
Equilateral triangle
In geometry, an equilateral triangle is a triangle in which all three sides have equal lengths. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also equal to each other and are each 60°. They are regular polygons, and can therefore also be referred to as regular triangles.
Properties
* The area of an equilateral triangle with sides of length $a,!$ is $a^2frac\left\{sqrt\left\{3\left\{4\right\}$
*Perimeter is $P=3a,!$
*Circumscribed circle radius $r=afrac\left\{sqrt\left\{3\left\{3\right\}$
*Inscribed circle radius $r=afrac\left\{sqrt\left\{3\left\{6\right\}$
*Altitude is $afrac\left\{sqrt\left\{3\left\{2\right\}$.These formulas can be derived using the Pythagorean theorem.
An equilateral triangle is the most symmetrical triangle, having 3 lines of reflection and rotational symmetry of order 3 about its center.Its symmetry group is the dihedral group of order 6 "D"3.
Equilateral triangles are found in many other geometric constructs. The intersection of circles whose centers are a radius width apart is a pair of equilateral arches, each of which can be inscribed with an equilateral triangle. They form faces of regular and uniform polyhedra. Three of the five Platonic solids are composed of equilateral triangles. In particular, the regular tetrahedron has four equilateral triangles for faces and can be considered the three dimensional analogue of the shape. The plane can be tiled using equilateral triangles giving the triangular tiling.
A result finding an equilateral triangle associated to any triangle is Morley's trisector theorem.
Geometric construction
An equilateral triangle is easily constructed using a compass.Draw a straight line, and place the point of the compass on one end of the line, and swing an arc from that point past halfway of the line segment. Repeat with the other side of the line.Finally, connect the point where the two arcs intesect with each end of the line segment
Alternate method: Draw a circle with radius "r", place the point of the compass on the circle and draw another circle with the same radius. The two circles will intersect in two points. An equilateral triangle can be constructed by taking the two centres of the circles and either of the points of intersection.
Almost-equilateral Heronian triangles
A Heronian triangle is a triangle with rational sides and rational area. Since the area of an equilateral triangle with rational sides is an irrational number, no equilateral triangle is Heronian. However, there is a unique sequence of Heronian triangles that are "almost equilateral" because the three sides, expressed as integers, are of the form "n" − 1, "n", "n" + 1. The first few examples of these almost-equilateral triangles are set forth in the following table.
Subsequent values of "n" can be found by multiplying the last known value by 4, then subtracting the next to the last one (52 = 4 × 14 − 4, 194 = 4 × 52 − 14, etc), as expressed in:$q_n = 4q_\left\{n-1\right\} - q_\left\{n-2\right\}.,!$This sequence can also be generated from the solutions to the Pell equation "x"² − 3"y"² = 1, which can in turn be derived from the regular continued fraction expansion for √3. [Takeaki Murasaki (2004), [http://zmath.impa.br/cgi-bin/zmen/ZMATH/en/quick.html?first=1&maxdocs=3&bi_op=contains&type=pdf&an=02147316&format=complete "On the Heronian Triple (n+1, n, n−1)"] , Sci. Rep. Fac. Educ., Gunma Univ. 52, 9-15.]
In culture and society
Equilateral triangles have frequently appeared in man made constructions:
*Some archaeological sites have equilateral triangles as part of their construction, for example Lepenski Vir in Serbia.
*The shape also occurs in modern architecture such as Randhurst Mall and the Jefferson National Expansion Memorial.
*The Seal of the President of the Philippines and Flag of Junqueirópolis contain equilateral triangles.
*The shape has been given mystical significance, as a representation of the trinity in The Two Babylons and forming part of the tetractys figure used by the Pythagoreans.
ee also
* Trigonometry
* Viviani's theorem
References
* [http://mathworld.wolfram.com/GeometricConstruction.html MathWorld] - an overview of the Euclidean construction of an equilateral triangle
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Triangle - получить на Академике рабочий купон на скидку Geekbuying или выгодно triangle купить с бесплатной доставкой на распродаже в Geekbuying
• equilateral triangle — UK [ˌiːkwɪlæt(ə)rəl ˈtraɪæŋɡ(ə)l] / US [ˌɪkwɪlætərəl ˈtraɪæŋɡ(ə)l] / US [ˌekwɪlætərəl ˈtraɪæŋɡ(ə)l] noun [countable] Word forms equilateral triangle : singular equilateral triangle plural equilateral triangles maths a triangle whose sides are all … English dictionary
• equilateral triangle — triangle having all sides equal … English contemporary dictionary
• equilateral triangle — noun a three sided regular polygon • Syn: ↑equiangular triangle • Hypernyms: ↑triangle, ↑trigon, ↑trilateral, ↑regular polygon • Hyponyms: ↑delta … Useful english dictionary
• equilateral triangle — noun A triangle having all three sides equal … Wiktionary
• equilateral triangle — e|qui|lat|e|ral tri|an|gle [ˌi:kwılætərəl ˈtraıæŋgəl] n technical a ↑triangle whose three sides are all the same length … Dictionary of contemporary English
• equilateral triangle — noun (C) technical a triangle (1) whose three sides are all the same length … Longman dictionary of contemporary English
• Circle packing in an equilateral triangle — is a packing problem in discrete mathematics where the objective is to pack n unit circles into the smallest possible equilateral triangle. Optimal solutions are known for n < 13 and for any triangular number of circles, and… … Wikipedia
• Triangle isocèle — Triangle Pour les articles homonymes, voir Triangle (homonymie) … Wikipédia en Français
• Triangle scalène — Triangle Pour les articles homonymes, voir Triangle (homonymie) … Wikipédia en Français
• triangle — [ trijɑ̃gl ] n. m. • v. 1270; lat. triangulum 1 ♦ Figure géométrique, polygone plan à trois côtés. Les trois côtés, les trois sommets, les trois angles d un triangle. Triangle quelconque, scalène, isocèle, équilatéral. Triangle rectangle, qui a… … Encyclopédie Universelle
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1. ## Divisibility...
Show that if integers a and b both are not divisible by 5, then the product ab is also not divisible by 5....where a=5k+r such as (1 <= r <= 4)
2. Originally Posted by Vedicmaths
Show that if integers a and b both are not divisible by 5, then the product ab is also not divisible by 5....where a=5k+r such as (1 <= r <= 4)
$a = 5k_1 + r_1$ and $b = 5k_2 + r_2$. Since 5 does not divide, $r_1 r_2 \neq 0$.
$ab= (5k_1 + r_1)(5k_2 + r_2) = 25k_1 k_2 + 5k_1 r_2 + 5k_2 r_1 + r_1 r_2$
Now we want to prove that 5 does not divide ab. Lets do what Jhevon has taught you before.... the contradiction method....
So let us assume 5 does divide ab, that means 5 divides $25k_1 k_2 + 5k_1 r_2 + 5k_2 r_1 + r_1 r_2 = 5(5k_1 k_2 + k_1 r_2 + k_2 r_1) + r_1 r_2$. But this means 5 divides $r_1 r_2$. But since $r_1$ and $r_2$ are both less than 5 and and nonzero, 5 cannot divide $r_1 r_2$. This contradicts our assumption that 5 divides ab. So 5 does not divide ab
3. Originally Posted by Vedicmaths
Show that if integers a and b both are not divisible by 5, then the product ab is also not divisible by 5....where a=5k+r such as (1 <= r <= 4)
Another way to do it.
Lets write this as an if then statement
if $5 \not | a$ and $5 \not | b$
then $5 \not | (ab)$
in this form we can find p and q
$p \implies q$ is logically equivelent to (the contraposition)
$\neg q \implies \neg p$
or in words
if $5 | (ab)$ then $5|a$ or $5|b$
this is easier to prove
since 5|ab $5=(ab)q,q \in \mathbb{Z}$
suppose that 5|a then we are done...
now suppose that $5 \not | a$ we need to show that 5|b
but 5=(ab)q 5=(aq)b so 5|b
QED
I hope this helps.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$-\dfrac{5}{4}\lt x \lt \dfrac{5}{2}$
$\bf{\text{Solution Outline:}}$ Use the properties of inequality to solve the given, $-15\lt -4x-5\lt0 .$ Then graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of equality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -15\lt -4x-5\lt0 \\\\ -15+5\lt -4x-5+5\lt0+5 \\\\ -10\lt -4x \lt 5 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol), the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-10}{-4}\lt \dfrac{-4x}{-4} \lt \dfrac{5}{-4} \\\\ \dfrac{5}{2}\gt x \gt -\dfrac{5}{4} \\\\ -\dfrac{5}{4}\lt x \lt \dfrac{5}{2} .\end{array}
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# Nine-point center
(Redirected from Nine-point centre)
A triangle showing its circumcircle and circumcenter (black), altitudes and orthocenter (red), and nine-point circle and nine-point center (blue)
In geometry, the nine-point center is a triangle center, a point defined from a given triangle in a way that does not depend on the placement or scale of the triangle. It is so-called because it is the center of the nine-point circle, a circle that passes through nine significant points of the triangle: the midpoints of the three edges, the feet of the three altitudes, and the points halfway between the orthocenter and each of the three vertices. The nine-point center is listed as point X(5) in Clark Kimberling's Encyclopedia of Triangle Centers.[1][2]
## Properties
The nine-point center N lies on the Euler line of its triangle, at the midpoint between that triangle's orthocenter H and circumcenter O. The centroid G also lies on the same line, 2/3 of the way from the orthocenter to the circumcenter,[2][3] so
${\displaystyle NO=NH=3NG.}$
Thus, if any two of these four triangle centers are known, the positions of the other two may be determined from them.
Andrew Guinand proved in 1984, as part of what is now known as Euler's triangle determination problem, that if the positions of these centers are given for an unknown triangle, then the incenter of the triangle lies within the orthocentroidal circle (the circle having the segment from the centroid to the orthocenter as its diameter). The only point inside this circle that cannot be the incenter is the nine-point center, and every other interior point of the circle is the incenter of a unique triangle.[4][5][6][7]
The distance from the nine-point center to the incenter I satisfies
${\displaystyle IN<{\tfrac {1}{2}}IO,}$
${\displaystyle IN={\tfrac {1}{2}}(R-2r)<{\frac {R}{2}},}$
${\displaystyle 2R\cdot IN=OI^{2},}$
The nine-point center is the circumcenter of the medial triangle of the given triangle, the circumcenter of the orthic triangle of the given triangle, and the circumcenter of the Euler triangle.[3] More generally it is the circumcenter of any triangle defined from three of the nine points defining the nine-point circle.
The nine-point center lies at the centroid of four points: the triangle's three vertices and its orthocenter.[8]
The Euler lines of the four triangles formed by an orthocentric system (a set of four points such that each is the orthocenter of the triangle with vertices at the other three points) are concurrent at the nine-point center common to all of the triangles.[9]:p.111
Of the nine points defining the nine-point circle, the three midpoints of line segments between the vertices and the orthocenter are reflections of the triangle's midpoints about its nine-point center. Thus, the nine-point center forms the center of a point reflection that maps the medial triangle to the Euler triangle, and vice versa.[3]
According to Lester's theorem, the nine-point center lies on a common circle with three other points: the two Fermat points and the circumcenter.[10]
The Kosnita point of a triangle, a triangle center associated with Kosnita's theorem, is the isogonal conjugate of the nine-point center.[11]
## Coordinates
Trilinear coordinates for the nine-point center are[1][2]
{\displaystyle {\begin{aligned}&\cos(B-C):\cos(C-A):\cos(A-B)\\={}&\cos A+2\cos B\cos C:\cos B+2\cos C\cos A:\cos C+2\cos A\cos B\\={}&\cos A-2\sin B\sin C:\cos B-2\sin C\sin A:\cos C-2\sin A\sin B\\={}&bc[a^{2}(b^{2}+c^{2})-(b^{2}-c^{2})^{2}]:ca[b^{2}(c^{2}+a^{2})-(c^{2}-a^{2})^{2}]:ab[c^{2}(a^{2}+b^{2})-(a^{2}-b^{2})^{2}].\end{aligned}}}
The barycentric coordinates of the nine-point center are[2]
{\displaystyle {\begin{aligned}&a\cos(B-C):b\cos(C-A):c\cos(A-B)\\={}&a^{2}(b^{2}+c^{2})-(b^{2}-c^{2})^{2}:b^{2}(c^{2}+a^{2})-(c^{2}-a^{2})^{2}:c^{2}(a^{2}+b^{2})-(a^{2}-b^{2})^{2}.\end{aligned}}}
Thus if and only if two of the vertex angles differ from each other by more than 90°, one of the barycentric coordinates is negative and so the nine-point center is outside the triangle.
## References
1. ^ a b Kimberling, Clark (1994), "Central Points and Central Lines in the Plane of a Triangle", Mathematics Magazine, 67 (3): 163–187, JSTOR 2690608, MR 1573021, doi:10.2307/2690608.
2. ^ a b c d Encyclopedia of Triangle Centers, accessed 2014-10-23.
3. ^ a b c Dekov, Deko (2007), "Nine-point center" (PDF), Journal of Computer-Generated Euclidean Geometry.
4. ^ Stern, Joseph (2007), "Euler’s triangle determination problem" (PDF), Forum Geometricorum, 7: 1–9.
5. ^ Euler, Leonhard (1767), "Solutio facilis problematum quorundam geometricorum difficillimorum", Novi Commentarii academiae scientiarum Petropolitanae (in Latin), 11: 103–123.
6. ^ Guinand, Andrew P. (1984), "Euler lines, tritangent centers, and their triangles", American Mathematical Monthly, 91 (5): 290–300, JSTOR 2322671, doi:10.2307/2322671.
7. ^ Franzsen, William N. "The distance from the incenter to the Euler line", Forum Geometricorum 11, 2011, 231-236. http://forumgeom.fau.edu/FG2011volume11/FG201126index.html
8. ^ The Encyclopedia of Triangle Centers credits this observation to Randy Hutson, 2011.
9. ^ Altshiller-Court, Nathan, College Geometry, Dover Publications, 2007 (orig. Barnes & Noble 1952).
10. ^ Yiu, Paul (2010), "The circles of Lester, Evans, Parry, and their generalizations", Forum Geometricorum, 10: 175–209, MR 2868943.
11. ^ Rigby, John (1997), "Brief notes on some forgotten geometrical theorems", Mathematics and Informatics Quarterly, 7: 156–158.
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• Related Books
0 answersPost by hani shuman on April 4 at 05:44:51 PM
Angles of a Triangle
• All triangles have three angles, and the three angles in a triangle add up to 180 degrees
Angles of a Triangle
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Angles of a Triangle 0:05
• All Triangles Have Three Angles
• Measure of Angles
• Extra Example 1: Find the Missing Angle Measure 5:39
• Extra Example 2: Angles of a Triangle 7:18
• Extra Example 3: Angles of a Triangle 9:24
Transcription: Angles of a Triangle
Welcome back to Educator.com.0000
For the next lesson, we are going to go over angles of a triangle.0002
Remember a triangle is a polygon with three sides; three straight sides.0009
Which means that there are three angles; those sides form three angles.0015
All triangles have three angles.0021
Here is one; here is another one; there is a third.0027
To name this angle here, we can say angle BAC.0031
That would be this angle right here; angle BAC.0040
But since the A is a vertex and there is only one angle0045
that this is a vertex for, we can just call this angle, angle A.0052
This one, I can just call angle B; this is angle C.0059
Again only if the point A is a vertex for just a single angle.0065
Let me give you an example of what it is not.0071
If I have an angle like that, I have two adjacent angles; this is A.0075
I can't call this angle, angle A, because there is three different angles formed here.0084
There is this small angle; there is this angle; there is this big angle.0089
This point, this vertex, is a vertex for three different angles.0095
In this case, you cannot call it angle A; you can't say angle A.0099
You would have to name the other three points like this one.0106
You would have to name, if this is B and this is C, then you have to say angle BAC or like that.0110
But again this one, because in a triangle, there is only three angles and three vertex.0119
You can just name this as angle A.0129
Within the three angles of a triangle, remember each angle has an angle measure, the number of degrees.0139
All three angle measures is going to add up to 180,0148
like the supplementary angles where we have two angles that form a straight line.0152
Here the three angles of a triangle also add up to 180.0159
If this is 60, this is 60, then what I can do is add these two up and subtract it from 180.0168
Here if I want to write an equation, I can say measure of angle A.0179
Remember this M is for measure; it is to show the number of degrees.0186
Measure of angle A plus the measure, the number of degrees, of angle B0190
plus the measure of angle C is going to equal 180 degrees.0199
We know what the measure of angle A is; how many degrees is angle A?0212
We know it is 60; this whole thing is just 60 degrees.0216
Measure of angle A is just 60; I can just replace this with 60.0221
Do I know measure of angle B?--no; I can just leave that there.0225
Plus the measure of angle C is also 60.0231
That is all going to add up to 180.0236
Again I can just add these two together which is this and this.0240
That is going to be 120; plus this unknown adds to 180.0244
I can subtract this from 180; 180 minus these two; whatever is left over.0256
From the 180 total, if I add these two together0263
and then figure out how many degrees are left over from the 180,0269
then all of that, all of those left over degrees have to go to angle B.0273
I am going to subtract; measure of angle B is going to be 60 degrees.0278
The leftover degrees from the 180 is 60; then this also has to be 60.0290
That is how you are going to solve for the missing angle measure.0300
Remember if we are going to be solving for the missing angle measure,0305
then we have to know two of the three angle measures.0309
I can't only have the measure of angle A and then find both B and C0317
because they are going to be different angles; they could be different angle measures.0323
I don't know how many are going to go here and how many are going to go here.0330
To find the missing angle measure, you have to have two out of the three like this one.0334
I have measure of angle A, 70 degrees.0343
I have the measure of angle B; that is 60 degrees.0348
I want to find the measure of angle C, meaning I want to find how many degrees is in angle C.0352
Again I can just take these two, add them together; how many from the 180?0359
I know that this plus this plus this all have to add up to 180.0364
This and this are used up.0371
However many are left over all have to go to angle C.0373
I can say 70 degrees plus this 60 plus the measure of angle C.0379
This is the proper way to write it.0387
I can't just write C because you are talking about the measure, meaning how many degrees.0389
It is all going to add up to 180.0394
Again I am going to add these two together.0398
This will be 130 plus the measure of angle C.0400
130 being used up plus the leftovers is going to equal 180.0410
Remember I subtract 180 with this number.0415
That way measure of angle C is going to be 50 degrees.0422
That means this has to be 50.0426
60 plus 70 plus 50 is going to add up to 180.0429
That is the missing angle measure.0434
Determine the angle measures if the angle measures could be the angle measures of a triangle.0441
Three angle measures for the three angles of a triangle.0448
If they add up to 180, then they can be the correct angle measures of a triangle.0455
But if not, if they don't add up to 180,0460
that means they can't be the three angle measures of a triangle.0462
The first one, I am going to take 50 plus the 90 plus the 40.0466
Just add them all up; I know that 0 plus 0 plus 0 is 0.0473
5 plus 9 is 14; plus 4 is 18; yes, they add up to 180.0479
That means these three angle measures can be the angle measures of a triangle.0489
This one is yes.0497
The next one, 45 plus 48 plus the 95.0504
5 plus... you can add this 5.0516
5 plus 5 is 10; plus 8 is 18; put up the 1; 8.0520
Already I know that it is not going to add up to 1800528
because the last digit has to be 0 and it is not.0533
This is 1 plus 4 is 5; plus 4 is 9; that plus 9 is 18.0537
This is 188; this is too much.0546
That means it can't be the angles of a triangle; this one is no.0550
Remember the angles of a triangle have to add up to 180.0557
The third example, find X.0565
We want to find the measure of this angle right here.0568
I have this triangle.0574
Remember all three angles of a triangle have to add up to 180.0578
But this one is what I am looking for; this is the missing angle measure.0583
I don't have this angle measure either.0586
If I need to find the third angle measure, I need to have the other two.0589
I have this one; I need to have this one also.0594
If I don't have this, then I don't know how many goes here.0598
I need to find this one first.0603
I have to use another method to find this angle measure.0606
I know that this right here, this straight line...0615
This is from the last lesson, the previous lesson on angles and lines.0621
If this is the line here, this one doesn't have an arrow.0629
Just do that; here is where it goes up.0635
Remember this, two angles right here, they are adjacent angles.0643
But they are also supplementary because it is a straight line.0652
It is straight; a straight line has an angle measure of 180.0656
This whole thing together is 180; that means this one plus this one is 180.0662
This is given that it is 135 degrees.0671
If this one together with this small one is 180, then I can just subtract it.0675
180 minus the 135 to see what this angle measure is going to be.0680
180 minus 135; this is going to be 45 degrees.0686
That means this has to be 45 because again this angle with this angle together forms a straight line.0698
That has to be 180; they are supplementary angles.0705
Now that I found this angle and I have this angle, I need to find the measure of this angle.0711
I can just say that X... this is just angle measure so I can just leave it as X.0719
I don't have to say measure of angle X because that is not a name.0728
That is the number of degrees.0732
X degrees plus 53 degrees plus 45 degrees all add up to 180 degrees.0734
See how they are all in degrees.0745
Again I am going to add these two together to see how many of the 180 I am using up.0749
Then see how many are left over to be X.0753
This is 53 plus 45 is 98 degrees.0761
That means X degrees, this many degrees, plus 90 degrees together is 180 degrees.0768
Again I am going to subtract this from 98; I get 82 degrees.0778
Right here, X is 82 degrees; this has to be 82.0796
That way this plus this plus this, the three angles of a triangle, are going to add up to 180.0804
That is it for this lesson; thank you for watching Educator.com.0812
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# Multiplying fractions – showing why
It was a long time ago, so I can’t be certain, but when I first learned to multiply fractions, it was a procedure that involved turning mixed numbers into improper fractions, multiplying numerators and denominators with maybe some cancelling down along the way. It was a procedure with no understanding.
You could just apply that procedure to these questions. But there is scope for a greater depth of understanding not to mention some creativity in showing why these work. Bar models are one way to demonstrate and calculate. Here are two examples:
A worthwhile exercise is to go through each of these questions attempting a drawing to show why (squared paper is a must).
Depending on your class, you will probably need to show some examples first. Or maybe you would prefer to give the completed statements so the focus is on drawing the representation rather doing the calculations.
Here is the lyx file for these questions and the pdf.
# Decimal Place Value Charts
I use these a lot at the early stages of understanding of place value. In my opinion, place value is the single most important mathematical concept that children need to master in upper KS2 to prepare them for secondary school maths. Many don’t, so it is incumbent on secondary maths teachers to ensure that any gaps are filled in Year 7.
As a pdf. Or if you want to change things, as a Google Sheets or MS Excel file.
If you print them out on card, you can put them inside transparent A4 plastic wallets and then use mini-whiteboard pens to write on them. There are different types of plastic wallet – you want the ones that have relatively thick plastic which is smooth, not textured. Otherwise you will have a hard time rubbing off the pen!
### Initial Activities
1. Show me 3 tens. Now show me Thirty. Why do we need to put the zero in there?
2. Show me 5 tenths. Show me 5 hundreths. Why do we need the zeros?
3. Show me 34. What is 34 made up of?
4. Show me 6.75. What is 6.75 made up of?
From here, of course, you might want to look at multiplying and dividing by powers of 10 and then eventually 4 operations involving decimals. But don’t rush into that until you are confident they have secured a depth of understanding. These questions are good for really testing that:
For me these are a crucial AfL tool to ensure the building blocks are in place before doing more complicated things with decimals.
# The Wisdom of the Crowd
Look, stats isn’t boring! I love doing these Wisdom of the Crowd exercises when teaching averages. Another way of doing it is to display a random scattering of dots on the screen (about 40 or so). Get everyone individually to provide an estimate then calculate the mean.
A really interesting addition to this is to do it once getting everybody to call out their estimate. Then, before calculating the mean, give everyone the chance to change their estimate and record them all a second time.
This can lead to regression to the mean, and standard deviation if you like.
See – stats isn’t boring!
Sir Francis Galton was a statistician in the 19th century. Thanks to him we have concepts such as correlation and standard deviation. Galton, it would seem, thought through the filter of statistics, a genius who produced hundreds of papers and books on fields as diverse as meteorology, historiometry and psychometrics and who pioneered the use of questionnaires to gather better information for his statistical analyses.
Last week, at my school’s Open Evening, we conducted a mathematical experiment based on one of Galton’s observations.
View original post 208 more words
This is an update of the original post. There were a couple of mistakes in the first one, which are now corrected. And I’ve learned some new maths in the process. I’ve credited those below who helped me. The power of Twitter – Thank you!
What happens when you take sequences of odd or even numbers and make fractions out of them?
These investigations provide some low threshold, high ceiling (in some cases too high for me!) rich investigations. I have been playing with these and encourage you to do so too!
These tasks have lots of benefits in the classroom if used well:
• Purposeful practise. In continuing the sequence and generating subsequent terms, students will repeatedly practice key skills. But rather than just working down a boring list of questions in a text book, they are practicing with a greater sense of purpose, i.e. to try to spot something else. There is a big range here from addition of consecutive numbers and cancelling down fractions to finding the nth term of a quadratic sequence (and harder).
• At key points stop everyone and get a whole classroom discussion going. Ask students to explain their “noticings”. By verbalising their reasoning pupils can grow their individual mathematical confidence. And it builds a classroom culture where other pupils’ noticings are highly valued. Children realise that we can learn by collaborating and listening to each others’ ideas.
• Always look for patterns and aim to generalise. Ask why this happens, does it happen every time and can we build a proof? This is hard. Sometimes what happens is that the students that aren’t ready to make this leap yet. Often they just continue generating more examples. This isn’t a problem as hopefully they are continuing to benefit from purposeful practice of an underlying mathematical skill.
I really encourage you to do some Mathematics and play with them before using them, but if time is short, here are some notes on each one. I have put them in order of difficulty.
### Sum of Odd Numbers
This is probably the most accessible of the four in terms of getting to a generalisation, although actually proving that algebraically is no mean feat!
Before we even get to the fractions, there is some good discussion to be had on mental methods for adding series of odd numbers and spotting that this generates square numbers:
```1+3=4
1+3+5=9
1+3+5+7=8+8=16
1+3+5+7+9=10+10+5=25```
To generalise this, we need to know that the nth odd number is (2n-1). Working from the last term backwards, we can write out the sequence as:
`1, 3, 5, ..., 2n-5, 2n-3, 2n-1`
By adopting the standard approach to find the sum of an arithmetic series, i.e. adding the first to last, second to second last, etc. we see that we get a whole bunch of “2n”s. How many “2n”s? Well there are n numbers so there must be n/2 pairs. So:
`2n × n/2 = n²`
Now you can start examining the fractions themselves. There is some good practice here of cancelling down fractions and students will realise quite quickly that they all cancel down to 1/3
At this point you might ask some students to generalise whilst some might prefer to continue generating examples.
The generalisation for the denominator builds on the generalisation for numerator. This time we with start with the odd number after the nth odd number and then add a series of odd numbers. Again think about what the last term would be and work backwards.
`2n+1, 2n+3, 2n+5, ..., 2n+(2n-5), 2n+(2n-3), 2n+(2n-1)`
By combining first and last, second and second last, etc. we can see we now have pairs of “6n”. How many “6n”s? Again, n/2. So the denominator becomes:
`6n × n/2 = 3n²`
and:
`3n²/n² = 3`
### Sum of Even numbers
With this one, cancelling down the fractions doesn’t help.
You end up with a pair of quadratic sequences (now corrected – thanks )
Which neatly cancels down to:
### Product of Even Numbers
This one provides lots of practice in “cancelling down” of fractions. Each time you end up with a unitary fraction (i.e. a numerator of one). But does this always happen? And why?
I made a mistake first time on this so I couldn’t find any pattern in the numbers that formed the denominators. Thanks to @mathforge and @wjhornby for pointing out by error.
So the sequence of denominators is 2, 6, 20, 70,… That’s beyond my knowledge of Integer sequences (I did Engineering, you know, not pure Maths!). But @mathsforge sent me this link to oeis.org. That’s another web-site I’ve learned about through this process!
If I had played around a bit longer with this and thought Factorials! then I might have eventually got here (thanks to @MrMattock for sending me this)
### Product of Odd Numbers
This one again provides some cancelling down practice although you are going to be reaching for the calculator pretty quickly.
I’m struggling to spot any pattern in here (no, the next term doesn’t have a 9 for its numerator…), but there must be something, right?
And this is no bad place to take a discussion with your class.
There must be something here to be found. We haven’t found it today. Your maths teacher is finding this very hard. Maybe nobody has ever found it. But if we start off with something so simple there must be a way of generalising it. Surely…?
Postscript: Again, I got a helpful response from @MrMattock. You can see it here, but don’t spoil it, have a go for yourself. The clue is to look for factorials again and don’t express all fractions in their simplest form. Good luck, but I warn you – it’s not pretty!
# The Power of Boxes and Circles
Here is a presentation that I gave at today’s #teachmeet held in Oakwood School, Surrey organised by Paul Collins, @mrprcollins.
I have been using this technique a lot recently so I collated some example questions from Year 1 to Year 12 (!) really just to show how versatile this technique can be. Here is a sample:
The full set are here as a pdf . For any Lyx.org fans out there, the original Lyx file.
# Reflections on Pythagoras
I was with a Year 10 class doing Pythagoras’ Theorem recently. This was a low attaining group who had encountered it before but were shaky. They had recently been doing rounding to d.p. and s.f. They had made progress with that but were getting a bit bored so the teacher decided to go over Pythagoras which I thought was a nice way of interleaving topics. Calculator answers needed to be rounded (for those triangles which were not pythagorean triples).
It got me thinking about the knowledge and understanding required to apply Pythagoras’ Theorem and also about planning topics. In this lesson, they stuck to finding the hypotenuse and didn’t do any problems where they had to find the shorter side. This decision was made by the teacher because she knew the class and the context. By the end all students had practised it a few times and had built self-belief that they could do it. In other contexts, another teacher might decide to introduce both cases side-by-side.
Is there a “best” approach? Has there been educational research looking into such a finer point of teaching this one topic? If there has, please point me to it because I wouldn’t have the time to look for it. Nor the belief that if I did eventually find something, reading it would actually improve the outcome of my learners. There is no single perfect way of teaching any topic. As teachers, we need to keep our eyes and minds open to approaches that we hadn’t thought of or used before. But we must not agonise over trying to find the holy grail, the golden nugget that will suddenly enlighten our students.
As I wasn’t actually teaching the lesson, just helping out, I had time to draw a quick mindmap in the lesson. I was trying to put myself in one of the students shoes.
`What are the things I need to know, understand and be able to do to be successful in this topic?`
This, fundamentally is what they care about.
I’ve included a picture of my mindmap here, not because it is a stunningly useful breakdown of this topic, but just to show the complexity of what our students need to learn. And this is without much page space taken up on the “And then…” topics of proofs, pythagorean triples, etc., some may say the “interesting bit”.
I am going to try to do this mindmapping exercise more often in my lesson planning. I found it quite easy to do when I was sitting in this lesson, but I find it harder when faced with a blank sheet and maybe some resources and ideas that I might have used before. The point is, I think, not to agonise over the perfect mindmap showing the perfect sequencing of “learning nuggets”. It needs to be done whilst thinking about the class and the context. And if there is ever time for such luxuries, doing it with a colleague surely makes the process more satisfying and enjoyable.
# The ones that didn’t get away
This week, most of Year 7 are away on a residential trip leaving 20 pupils who, for whatever reason, are in school. I taught them on Monday and much to my relief they didn’t seem particularly aggrieved by this state of affairs.
I don’t know what they are doing in other lessons, but in Maths, well, they’re going to do some mathematics!
I actually love the freedom of not having to follow a scheme of work for a bit (although I’m sure I would eventually feel lost without it!) I wanted to do something that was proper maths but felt a bit different from a normal maths lesson.
As luck would have it, I attended an fantastic session run by John Mason and Anne Watson on Friday. I’m not going to attempt to write about that session. All I will say is that if you ever get the chance to meet these two, you should. A great opportunity to reflect deeper on the nature of mathematics and mathematics teaching.
So, like all things you pick up at CPD, I decided I would use something soon before I forgot it. And here is what occupied us for nearly an hour last thing on Monday afternoon.
This is a classic low ceiling, high threshold task. Everyone can see some patterns in this, some will notice that the first term of the nth row is n², some will determine that the last term in every row is…
On Friday, however, I picked up something more valuable than this piece of mathematics. I got to play around with techniques of how to present it to students. To start with a blank board and write very deliberately and quite slowly stopping and asking questions like, “What am I going to write next?”, “What should we do now?”, “What are you thinking now?”
I really felt that we began to see the power of getting students to explain their reasoning, especially in a mixed group like this.
So, moving on. What else can I do in the remaining lessons I have with this group?
My first instinct was to look at my own blog under investigations. This is, after all the main reason I do this blog – as a reference for me!
Some good stuff, and it triggered memories of lessons I would otherwise have forgotten about. And then I thought, what about those CPD events I went on in the years before I started this blog? I looked back through my notes and found this from the 2012 ATM conference.
## Factor Game
Write out numbers 1-12.
1st person choses a number. 2nd person gets all the remaining factors of that number.
1st person choses another number (but it has to be one that still has factors remaining). 2nd person gets all the remaining factors of that number. Keep going until there are no numbers left to chose that still have factors – the second person gets all the remaining number. Sum the totals of each – highest score wins.
Nice way to practice factors of a number as well as introduce Gauss’s trick to sum an arithmetic progression of n terms: n(n+1)/2 as you only need to calculate one of the totals, you can then take it away from the sum to get the other total. So, in this case the total is always 78 because the sum of the first 12 numbers is 78 and that’s all we are doing.
e.g.
Person 1 Person 2 11 1 4 2 9 3 10 5 12 6 7,8 Total = 46 Total = 32
Can you always win? What is your strategy? What’s the perfect game, i.e. highest possible score. Try it for other totals.
I tried:
• 16 a bit more tricky, but the player choosing the numbers should still win
• 20 was interesting as it was a draw – 95 each!
• 19 was easy to win
# Mathematical Behaviour
An interesting Twitter chat tonight on behaviour in Maths lessons. And my first time hosting!
Lots of questions and even a few answers in the Storify of the chat here.
http://tinyurl.com/zztmfqr
# Rounding Errors
To investigate rounding more deeply, here is a structure for some questions that hopefully prompt the question, why does this happen?
Here are three examples as a pdf and as a Google doc. It doesn’t happen in the first example, but it does in the second two.
Next, see if students can create their own to investigate why this happens. I have created a blank sheet (3 on each page). Again, as a pdf document and as a Google doc.
# Formal methods – the devil’s work!
I’ve been thinking a lot recently about “formal methods”, e.g. Column Addition or Long Multiplication.
As secondary maths teachers we don’t pay much attention to these. I assess their ability to carry out the procedure and have sometimes attempted to teach long multiplication explicitly. But more often than not it is assumed that these methods have been learnt at primary school have been pretty well-practised and are therefore secure.
We have been doing a lot of number work with Year 7 so far this term including exploring in some detail the Laws of Arithmetic.
Here is an example of the type of question we have been looking at:
`127 + 54 + 73`
Most of my students’ first instinct with a question like this is to draw up a nice column addition and solve it. Would you agree that most students do this?
I want them to look at the structure of the numbers first. To realise that addition is commutative and that it’s much easier if we add the 127 to the 73 first because that’s a number bond to 200.
I am having 2 issues with this approach:
1. My students are not convinced that this is a valid exercise. Basically they think it is contrived (which, of course all exercises are). They think I have chosen numbers to make this work, it will only work for those certain numbers and basically they feel like I have tricked them.
2. Some students perceive the message that the formal methods they spent hours practising at primary school are now not the way we do things at secondary school. I have said that this is not the case, they are not the devil’s work, and that those methods still have a place. But we need to have smarter ways to work. The fact that we have spent many lessons not solving calculations using formal methods leads me to believe that they are confused about what I want them to do.
Fundamentally I want my students to rely less on the authority of the teacher and rely more on their own understanding. I want them to see the structure in everything they do, to be certain that what they have done is valid because the mathematics tells them it is so, not because I tell them that they are right.
How should we bridge the gap between informal methods which require (and develop) depth of understanding and formal methods which are efficient and accurate?
I have an idea using manipulatives. I am not sure if it will work, but if I get chance to try it with the class, I will write about it again later.
In the meantime, does this resonate with anyone else? How do you bridge the gap? I’d love to read your comments.
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# Proofs of the Product Rule of Derivatives
The Product Rule is one of the most helpful tools in Differential Calculus (or Calculus I) to derive two functions that are being multiplied. It can be used along with any existing types of functions as long as multiplication operations are present within the given derivation problem.
However, as easy as it seems to just use a standard formula in deriving functions with multiplication operations, it is essential to learn the concepts behind this standard formula that will satisfy the principles of the product rule. Hence, in this chapter, we will focus mainly on the proofs of the product rule formula by applying the concepts of derivation through limits and the chain rule.
##### CALCULUS
Relevant for
Learning about the different proofs of the product rule.
See proofs
##### CALCULUS
Relevant for
Learning about the different proofs of the product rule.
See proofs
## What is the Product Rule?
The product rule is defined as the derivative of the product of at least two functions. The product rule can be used to derive any given product of functions such as but not limited to:
$latex (fg)'(x) = \frac{d}{dx}(f(x) \cdot g(x))$
where f(x) and g(x) can be equivalent to any types of functions.
But how exactly do we derive that given function using the product rule?
The product rule states that the derivative of a product of two functions is equal to the first function f(x) in its original form multiplied by the derivative of the second function g(x) and then added to the original form of the second function g(x) multiplied by the derivative of the first function f(x).
To better illustrate, when you are given two functions f(x) and g(x) and then you are asked to get the derivative of fg(x) or the derivative of the product of f(x) and g(x), we have:
$$(fg)'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)$$
Easy, right? But we should not take this formula superficially if we aim to be able to derive any product of functions. In order to learn and understand the concepts behind the development of this product rule formula, we need to be familiarized with any proof which would satisfy the statement of the product rule.
## Proof of The Product Rule Using Limits
In this article, you are highly recommended to be familiarized with the topics, The Slope of a Tangent Line and Derivatives Using Limits, as a pre-requisite to better understand The Proof of The Product Rule Using Limits.
We can recall that
$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$
By applying limits, we can derive a function f(x). But how about a product of two functions f(x) and g(x)?
For instance, we are given two functions f(x) and g(x) and then, we are asked to get the derivative of fg(x); we have
$$(fg)'(x) = \frac{d}{dx}(f(x) \cdot g(x))$$
To derive the product of these two functions using limits, we have the following:
Let
$$\Upsilon(x) = f(x) \cdot g(x)$$
Then we have,
$$\Upsilon'(x) = \frac{d}{dx}(f(x) \cdot g(x))$$
and we can derive it by
$$\Upsilon'(x) = \lim \limits_{h \to 0} {\frac{\Upsilon(x+h)-\Upsilon(x)}{h}}$$
By substituting the equation $latex \Upsilon(x) = f(x) \cdot g(x)$, we have
$$\Upsilon'(x) = \lim \limits_{h \to 0} {\frac{f(x+h) \cdot g(x+h) – f(x) \cdot g(x)}{h}}$$
which can be clearly re-written as:
$$\frac{d}{dx}(f(x) \cdot g(x)) =\lim \limits_{h \to 0} {\frac{f(x+h) \cdot g(x+h) – f(x) \cdot g(x)}{h}}$$
This represents the derivative of a product in terms of limits. Now, this equation cannot be algebraically manipulated easily to arrive at the product rule formula that we are trying to prove.
However, we can add and subtract $latex f(x+h) \cdot g(x)$ to the numerator. Hence, we have
$$\frac{d}{dx}(f(x) \cdot g(x)) = \lim \limits_{h \to 0} {\frac{f(x+h) \cdot g(x+h) – f(x+h) \cdot g(x) + f(x+h) \cdot g(x) – f(x) \cdot g(x)}{h}}$$
Given that $$+ f(x+h) \cdot g(x) – f(x+h) \cdot g(x) = 0$$, we didn’t change the equation at all.
Now, we can factor $latex f(x+h)$ from the first two terms and $latex g(x)$ from the last two terms. Then, we can split it into two parts and by simplifying, we have,
$$\frac{d}{dx}(f(x) \cdot g(x)) = \lim \limits_{h \to 0} {\frac{f(x+h) \cdot (g(x+h)-g(x))}{h} +\lim \limits_{h \to 0} \frac{g(x) \cdot (f(x+h)-f(x))}{h}}$$
$$\frac{d}{dx}(f(x) \cdot g(x)) = \lim \limits_{h \to 0} {f(x+h) \cdot \frac{g(x+h)-g(x)}{h} +\lim \limits_{h \to 0} g(x) \cdot \frac{f(x+h)-f(x)}{h}}$$
By applying the properties of limits to solve the equation, we have
$$\frac{d}{dx}(f(x) \cdot g(x)) = \lim \limits_{h \to 0} {f(x+h)} \cdot \lim \limits_{h \to 0} {\frac{g(x+h)-g(x)}{h}} + \lim \limits_{h \to 0} {g(x)} \cdot \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$
The limits $latex \lim \limits_{h \to 0} {f(x+h)}$ and $latex \lim \limits_{h \to 0} {g(x)}$ can be easily solved. When h tends to zero, we will simply get $latex f(x)$ and $latex g(x)$ respectively.
The limits $latex \lim \limits_{h \to 0} {\frac{g(x+h)-g(x)}{h}}$ and $latex \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$ look more complicated, but they are simply the derivatives of $latex g(x)$ and $latex h(x)$ expressed in limits. Therefore, we have:
$$\frac{d}{dx}(f(x) \cdot g(x)) = f(x) \cdot \frac{d}{dx}(g(x)) + g(x) \cdot \frac{d}{dx}(f(x))$$
or it can be simply illustrated as
$$(fg)'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)$$
which is now The Product Rule Formula.
## Proof of The Product Rule Using The Chain Rule
Another way that might make the product rule easier to prove and formulate is by applying the chain rule formula. Hence, you are highly recommended to be familiarized with the topics, The Chain Rule FormulaThe Sum/Difference of Derivatives, as a prerequisite to better understand this proof.
We can recall that the chain rule formula is
$$\frac{d}{dx}[(f(x))^n] = n \cdot (f(x))^{n-1} \cdot \frac{d}{dx}(f(x))$$
To make the derivation of the product rule easier, we can consider the following expression:
$$\Upsilon(x) = {{(f+g)}^2}$$
where, f and g are two valid functions.
We are going to derive $latex \Upsilon(x)$ by applying the chain rule. Therefore, we have:
$latex \Upsilon'(x)= 2(f+g)(f’+g’)$
If we multiply and expand the parentheses, we have:
$latex \Upsilon'(x)= 2(ff’+fg’+gf’+gg’)$
$latex = 2ff’+2fg’+2gf’+2gg’$
Now, instead of using the chain rule, we can expand the expression $latex {{(f+g)}^2}$ to get:
$latex \Upsilon(x) = {{f}^2}+fg+{{g}^2}$
If we differentiate term by term, we get:
$latex \Upsilon'(x)= 2ff’+2(fg)’+2gg’)$
Since the two expressions we got for $latex \Upsilon'(x)$ are equivalent, we have:
$$2ff’+2(fg)’+2gg’)= 2ff’+2fg’+2gf’+2gg’$$
$latex (fg)’= fg’+gf’$
We have clearly arrived at the product rule.
## Proof of the Product Rule using logarithmic differentiation
A third method we can use to prove the product rule is by using logarithmic differentiation and implicit differentiation. This is a quicker method but requires you to be familiarized with both topics.
We can start by writing the following expression:
$latex y=f(x)g(x)$
Now, we can take the natural logarithm of both sides:
$latex \ln(y)=\ln(f(x)g(x))$
By using the laws of logarithms on the right side, we can write:
$latex \ln(y)=\ln f(x)+\ln g(x)$
Taking the derivative of both sides, we have:
$$\frac{y’}{y}=\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}$$
Solving for $latex y’$, we have:
$$y’=y\left(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\right)$$
Now, we can substitute $latex y=f(x)g(x)$ to get:
$$(fg)’=f(x)g(x)\left(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}\right)$$
If we multiply and expand the parentheses, we get:
$latex (fg)’=g(x)f'(x)+f(x)g'(x)$
We have arrived at the product rule.
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# A Jacked-Up House
Here's your nightly math! Just 5 quick minutes of number fun for kids and parents at home. Read a cool fun fact, followed by math riddles at different levels so everyone can jump in. Your kids will love you for it.
# A Jacked-Up House
September 18, 2014
Have you ever tried to build a house of cards using a deck of playing cards? Stacking cards is tricky business: you have to stand some of those skinny rectangles up on their edges and lean them against each other before they fall over. Then you rest other cards flat on top and build a new “story” above that, hoping the new cards don’t take the whole house down. Well, an architecture group decided to make a giant house of cards the size of a real building. They used regular building materials to make it strong, but painted the pieces to look like playing cards — and better yet, they built lights into the pieces so the cards flash at night. We’re not sure whether a person could make a house of playing cards this shape that wouldn’t fall over, but luckily this giant one is glued and bolted together.
Wee ones: Playing cards have different shapes to show what “suit” they are: diamonds, hearts, spades and clubs. How many different shapes is that?
Little kids: For any suit, like diamonds, there’s a card for each number from 2 to 10, plus an ace (which is like a 1), a king, a queen and a jack. How many cards does each suit have in total? Bonus: If the 1st floor of this house has 10 cards along one wall, including 2 cards with spades, 3 with diamonds and 1 with clubs, how many heart cards does that wall have?
Big kids: The 1st floor of the house appears to have 10 cards on each of the 2 long zigzag walls, and 6 cards on each of the short walls. How many cards do the 1st floor walls have? Bonus: If each floor has 4 fewer wall cards than the level below it, how many wall cards do the 4 floors have all together?
The sky’s the limit: Suppose that the first level uses all the diamond cards from 2 to 10, and 3 of those cards flash at night; that together those 3 cards have 16 diamonds painted on them; and only the middle card of the three (by value) is an even number. How many possible sets of 3 flashing cards can there be?
Wee ones: 4 shapes, or suits.
Little kids: 13 cards. Bonus: 4 hearts, since the other suits cover 6 of the 10 cards.
Big kids: 32 cards (10+6+10+6). Bonus: 104 cards all together: 32 cards on the 1st floor, 28 on the 2nd, 24 on the 3rd, and 20 on the 4th.
The sky’s the limit: There are only 2 possible combinations. The even card can be only a 4, 6 or 8, since 2 and 10 can’t be the middle card. Each of those then needs two odd cards that bring the total to 16. This gives us:
– The 4, 3 and 9 cards. 4, 5 and 7 won’t work because then 4 isn’t in the middle.
– The 6, 3 and 7 cards. 6 can’t be grouped with 9 since then we’re already at 15, nor with 5 because then we’d need another 5 and there’s only 1 of each card.
– Nothing works with 8 because 9 already brings the total past 16, and 8 has to be the middle card.
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# How To Find The Inverse Of A 2×2 Matrix?
## How do you find the inverse of a matrix?
Algebra – Finding the Inverse of a Matrix (1 of 2) A 3X3 Matrix
## Do all 2×2 matrices have an inverse?
A . Not all 2 × 2 matrices have an inverse matrix. If the determinant of the matrix is zero, then it will not have an inverse; the matrix is then said to be singular. Only non-singular matrices have inverses.
## How do you find the ADJ of a 2×2 matrix?
Inverse of a 2×2 Matrix using Adjoint –
## What is a 1 Matrix?
Matrix Inverse. Multiplicative Inverse of a Matrix. For a square matrix A, the inverse is written A-1. When A is multiplied by A-1 the result is the identity matrix I. Non-square matrices do not have inverses.
## What is Cramer’s rule matrices?
Cramer’s Rule for a 2×2 System (with Two Variables) Cramer’s Rule is another method that can solve systems of linear equations using determinants. In terms of notations, a matrix is an array of numbers enclosed by square brackets while determinant is an array of numbers enclosed by two vertical bars.
## How do you know if an inverse exists?
How to Determine if a 2 x 2 Matrix has an Inverse –
## How do you find the inverse?
Finding the Inverse of a Function
• First, replace f(x) with y .
• Replace every x with a y and replace every y with an x .
• Solve the equation from Step 2 for y .
• Replace y with f−1(x) f − 1 ( x ) .
• 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.
## How do you calculate the inverse?
We can calculate the Inverse of a Matrix by:
1. Step 1: calculating the Matrix of Minors,
2. Step 2: then turn that into the Matrix of Cofactors,
3. Step 3: then the Adjugate, and.
4. Step 4: multiply that by 1/Determinant.
We recommend reading: How To Find The Vertex Of A Parabola In Standard Form?
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The diagram is a paper of micrometer screw gauge. What is the reading shown? Assume the pitch is $0.5mm$ A. $4.01mm$ B. $4.51mm$ C. $5.00mm$ D. None of these
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Hint :In this question, we need to determine the reading of the micrometer screw gauge such that the pitch is 0.5 millimeters. First, we need to calculate the least count of the screw gauge. For this, we will recall the terms involved while measurements using a micrometer screw gauge.
A micrometer screw gauge is used for measuring very small distances. It is used to measure small thickness of papers, wires and other minute items. A micrometer screw gauge has two scales – the linear scale and the circular scale.
The pitch of the screw gauge is calculated by dividing the distance moved by the linear scale in one turn of the circular scale by the number of divisions of the circular scale. In this question, the pitch is directly given to use.
Given,
The pitch of the micrometer screw gauge is $0.5mm$
Thus the least count of the screw gauge is given by
$L.C = \dfrac{{pitch}}{\text{no.of circular divisions}} \\ = \dfrac{{0.5mm}}{{50}} \\ = 0.01mm \\$
Now we see that the linear scale reading is $4.5mm$ and the circular scale has turned $1$ divisions
Hence the final reading can be calculated by the formula
$M = L.S.D + C.S.D \times L.C$
We put all the values here and then get the reading
$M = 4.5mm + 1 \times 0.01mm \\ = 4.51mm \\$
Hence option (B) is correct.
Note :
Vernier calipers may also be used to measure small distances. However a micrometer screw gauge can measure even smaller distances compared to the Vernier Calipers. Students should also know how to calculate the pitch of the micrometer screw gauge. Also the students should make sure there is no zero error. If any zero error is present it should be subtracted from the final reading.
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## MATHEMATICS 10-3
[C] Communication
[CN] Connections
[ME] Mental Mathematics
and Estimation
[PS] Problem Solving
[R] Reasoning
[T] Technology
[V] Visualization
### Geometry
General Outcome: Develop algebraic and graphical reasoning through the study of relations.
Specific Outcomes: It is expected that students will:
1. Demonstrate an understanding of similarity of convex polygons, including regular and irregular polygons.
[C, CN, PS, V]
3.1 Determine, using angle measurements, if two or more regular or irregular polygons are similar.
## Similar and/or Congruent - congruent not part of the program of studies, but perhaps important enough to include - iFrame
3.2 Determine, using ratios of side lengths, if two or more regular or irregular polygons are similar.
3.3 Explain why two given polygons are not similar.
3.4 Explain the relationships between the corresponding sides of two polygons that have corresponding angles of equal measure.
3.5 Draw a polygon that is similar to a given polygon.
3.6 Explain why two or more right triangles with a shared acute angle are similar.
3.7 Solve a contextual problem that involves similarity of polygons.
1. Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent) by:
• applying similarity to right triangles
• generalizing patterns from similar right triangles
• applying the primary trigonometric ratios
• solving problems.
[CN, PS, R, T, V]
[ICT: C6–4.1]
4.1 Show, for a specified acute angle in a set of similar right triangles, that the ratios of the length of the side opposite to the length of the side adjacent are equal, and generalize a formula for the tangent ratio.
## Naming Sides of a Right Triangle
4.2 Show, for a specified acute angle in a set of similar right triangles, that the ratios of the length of the side opposite to the length of the hypotenuse are equal, and generalize a formula for the sine ratio.
## Sine - Math Glossary - Under Construction - Consider showing reference angle and side lengths. Display opp/hyp ratio.
4.3 Show, for a specified acute angle in a set of similar right triangles, that the ratios of the length of the side adjacent to the length of the hypotenuse are equal, and generalize a formula for the cosine ratio.
## Cosine - Math Glossary - Under Construction - Consider showing reference angle and side lengths. Display adj/hyp ratio.
4.4 Identify situations where the trigonometric ratios are used for indirect measurement of angles and lengths.
4.5 Solve a contextual problem that involves right triangles, using the primary trigonometric ratios.
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# Finding Trigonometric Ratios04:29 minutes
Video Transcript
## TranscriptFinding Trigonometric Ratios
Let me tell you a tail of the Pharaoh SOH-CAH-TOA. To honor Pharaoh SOH-CAH-TOA, a huge PURR-amid...pyramid was constructed. The pharaoh, who is paw-sitively crazy for cats, ordered a pyramid from the pyramid-building company, Cleo-CAT-ra, for his cat to live in since pyramids are the PURR-fect shape. To determine the dimensions of the miniature pyramid, we gato use trigonometric ratios.
### Three Trig Ratios
Let’s review the ratios to help the Pharoah so we can help him avoid a cat-astrophy. For right triangles, the most common trig ratios are sine, cosine and tangent. Let’s take a look at the three ratios. You should remember that the sine of ∠A is the length of the opposite side divided by the length of the hypotenuse. The cosine of ∠A is the length of the adjacent side divided by the length of the hypotenuse. And finally, the tangent of ∠A is the length of the opposite side divided by the length of the adjacent side.
You won’t believe this, but the pharaoh’s name is a mnemonic device we can use to remember these three trig ratios! Let's paws and have a look: SOH, CAH, TOA. It’s easy to get confused about which side is which. The hypotenuse is always located opposite the right angle. The other two sides are named depending on the angle in question. The opposite side is across from the target angle and the adjacent side is between the target angle and the right angle.
### Calculating Trig Ratios
The Pharoah is not kitten around. Since he already knows the trig ratios, he can figure out the trig ratios for his pyramid by using the measurements he knows. Look at the triangle face of the pyramid. Dividing a side of the base by 2 and drawing in an altitude gives us two right triangles. Since each side of the base is 755 ft we can divide the base by 2, to calculate the length of the side adjacent to ∠A. Now we know all three lengths: The length adjacent to ∠A is 377.5 feet. The length of the hypotenuse is 610 feet, and the length opposite ∠A is 479.16 feet. Let's calculate the trig ratios!
To calculate the sine of an angle, simply divide the length of the opposite side, 479.16, by the length of the hypotenuse, 610. To get the cosine, divide the length of the adjacent side, 377.5, by the length of the hypotenuse, 610. And last, but not least, divide the length of the opposite side, 479.16, by the length of the adjacent side, 377.5, to get the tangent.
### Calculating the side lengths with given trig ratios
Now the pharaoh can use this information to calculate the measurements for the miniature pyramid. Because the kitty cats' pyramid will be a similar version of the pharaoh's, the trig ratios will be the same. If the miniature will have a height of 20 feet, what are the other lengths? Chose the trig ratio that will help you to calculate the unknown length with the fewest steps.
Let's use the tangent ratio, which is 1.269, to set up a proportion using 20 as our opposite side length. Now we have to *solve for the adjacent side. Using opposite operations and isolating our variable, we find that the adjacent side is equal to 15.76 feet. This is just half the base of the face of the pyramid, so we multiply by 2 to determine the full length of the base. Pharoah SOH-CAH-TOA's looks at the plans he's feline pretty good about the mini pyramid right about now!
The Pharoah is speechless! I guess the cat's got his tongue! Ugh...All these cat puns are freakin' meowt
1. lol
Posted by user 2., about 3 years ago
2. That cat pun counter really worried me with the amount of numbers available!
Posted by Plano Kellmeyer, over 3 years ago
## Videos in this Topic
Trigonometry (7 Videos)
## Finding Trigonometric Ratios Exercise
### Would you like to practice what you’ve just learned? Practice problems for this video Finding Trigonometric Ratios help you practice and recap your knowledge.
• #### Determine the trigonometric ratios of the pyramid.
##### Hints
The Pharoah's name, SOHCAHTOA, is a mnemonic device that helps us remember the trig ratios for $sin$, $cos$, and $tan$. It breaks into three parts, one for each trig ratio:
SOH, CAH, and TOA.
S represents sin, C represents cos, and T represents tan. O represents the side length Opposite to the angle, A represents the side length Adjacent to the angle, and H represents the hypotenuse.
The trigonometric ratios, or trig ratios, can be applied to right triangles.
The mnemonic device for each trig ratio starts with the first letter of the name of that ratio. For example, TOA starts with T, which stands for tan.
Next is the first letter of the side whose length is in the numerator of the trig ratio. Last is the first letter of the side whose length is in the denominator of the trig ratio. For example TOA means that the Opposite side is in the numerator of this trig ratio, and the Adjacent side is in the denominator.
##### Solution
In order to determine the trig ratios, we need a right triangle. The pyramid can be cut in half by drawing a line from the peak to the center of the base. This gives us two right triangles. We can use one of these right triangles to determine trig ratios.
The mnemonic device for each trig ratio starts with the first letter of the name of that ratio. For example, SOH starts with S, which stands for $sin$.
Next is the first letter of the side whose length is in the numerator of the trig ratio. Last is the first letter of the side whose length is in the denominator of the trig ratio. For example SOH means that the Opposite side is in the numerator of this trig ratio, and the Hypotenuse is in the denominator. Or in other words:
$~$
$sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$
$~$
Similarly, CAH means that:
$cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
$~$
And TOA means that:
$tan = \dfrac{\text{opposite}}{\text{adjacent}}$
$~$
Let's identify the three side lengths that we need to calculate these ratios.
The opposite side to angle $A$ is $479.16$ ft long. This is the side that does not have the angle $A$ at either of its end points.
The adjacent side is the side between angle $A$ and the right angle. So its length is $377.50$ ft.
The hypotenuse is always the side that is opposite to the right angle. In this case it is $610$ ft long.
Now we can use the equations for the trig ratios, and the required side lengths, to calculate each ratio:
$~$
$sin = \dfrac{\text{opposite}}{\text{hypotenuse}}= \dfrac{479.16}{610}$
So we have that $sin = 0.786$.
$~$
$cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}= \dfrac{377.50}{610}$
So we have that $cos = 0.619$.
$~$
$tan = \dfrac{\text{opposite}}{\text{adjacent}}= \dfrac{479.16}{377.50}$
So we have that $tan = 1.27$.
• #### Find the missing length.
##### Hints
Remember the mnemonic device "TOA".
The tangent ratio for an angle is equal to the length of the opposite side over the length of the adjacent side.
Once you calculate the length of the adjacent side using the tangent ratio, you will need to double it to get the full length of the base of the pyramid.
##### Solution
The Pharaoh knows that for angle $A$, he has
$tan = 1.269$.
The height of the pyramid is opposite side to the angle $A$, and measures $20$ ft.
The mnemonic $TOA$ reminds him that $tan$ is equal to the opposite side over the adjacent side length:
$tan = \dfrac{\text{opposite}}{\text{adjacent}}$.
He can put in the known values and solve for the missing value:
$\begin{array}{rcl} 1.269 & = & \dfrac{20}{\text{adjacent}}\\ \color{#669900}{\times \, \text{adjacent} \, \div \, 1.269} & &\color{#669900}{\times \, \text{adjacent} \, \div \, 1.269}\\ \text{adjacent} & = & \dfrac{20}{1.269}\\ \text{adjacent} & = & 15.76 \end{array}$
He knows that the adjacent side length to angle $A$ in the right triangle is only half the length of the base of the pyramid. So he multiplies this adjacent side length by two to get the length of the base of the pyramid, $31.52$ ft.
• #### Highlight the Pharaoh's mistakes.
##### Hints
Don't assume that anything is correct, even the first equation.
When the Pharaoh is distracted, he sometimes flips the numerator and denominator in his equations.
The base is equal to two times the length of the side adjacent to angle $A$.
The Pharaoh should use the tangent ratio to start his calculations.
##### Solution
To start with, we know the correct values of the three trig ratios for angle $A$, and we know the height of the desired pyramid. We want to find the length of one side of the base of the pyramid.
The Pharaoh's mistakes are noted below with an '✗' next to them. The rest is the correct solution to this problem.
We can find the base, $b$, as being equal to two times the adjacent side to angle $A$:
$b = 2 \times adj$.
The height of the pyramid is the opposite side to angle $A$. We have the length of the opposite side, and we need the length of the adjacent side. So we know that we need a trig ratio that includes both the opposite and adjacent sides. The tangent ratio meets this requirement:
$tan = \dfrac{\text{opp}}{\text{adj}}$.
✗ The Pharaoh wrote sin instead of $tan$.
We know that $tan A = 1.269$, and that the opposite side length is $12$. So:
$1.269 = \dfrac{12}{\text{adj}}$.
Using opposite operations:
$\begin{array}{rcl} 1.269 & = & \dfrac{12}{\text{adj}}\\ \color{#669900}{\div \, 1.269 \, \times \, \text{adj}} & &\color{#669900}{\div \, 1.269 \, \times \, \text{adj}}\\ \text{adj} & = & \dfrac{12}{1.269}\\ \text{adj} & = & 9.5 \end{array}$
✗ The Pharaoh switched the numerator and the denominator.
We know that $b = 2 \times adj$. Now we know the length of the adjacent side, so we can find the side length of the base $b$:
$b = 2 \times 9.5$, so
$b = 19$.
✗ The Pharaoh multiplied the adjacent side by 3, instead of two. As a result, he got the wrong side length for the base, even though he found the correct length of the side adjacent to angle $A$.
• #### Determine the appropriate trig ratio.
##### Hints
The hypotenuse is always the side that is opposite the right angle. The opposite and adjacent sides are always relative to the target angle. Here, the hypotenuse is $5$ units long. The opposite side is $4$ units long, and the adjacent side is not labeled with a length.
Each triangle has two known side lengths. The sine, cosine, or tangent ratio will include both of the known side lengths.
##### Solution
For each triangle we have a target angle and we know two side lengths. We must identify a trig ratio that can be found for the target angle using the known side lengths.
Let's start by identifying which side lengths we know, with respect to the target angle.
For the first pendant, the side with a length of $4$ units is opposite angle $b$. We can verify this by observing that angle $b$ is not at either end point of this side. Similarly, the side with a length of $5$ is the hypotenuse because it is opposite the right angle.
Sin is the trig ratio that includes the opposite side, and the hypotenuse. So we should use sin to characterize angle $b$ in the first pendant.
For the second pendant, we have an hourglass shape. The top half of the hourglass is a right triangle. Angle $b$ is the target angle. The side with a length of $2$ units is the adjacent side to angle $b$, because is goes between angle $b$ and the right angle. The side with a length of $3$ units is the hypotenuse because it is opposite the right angle. The cos trig ratio includes both the adjacent side and the hypotenuse, so we should use this ratio for angle $b$ in this triangle.
For the third pendant, the side with a length of $4$ units is adjacent side because it is between angle $b$ and the right angle. The side with a length of two is opposite to angle $b$; $tan$ is the trig ratio that includes both the opposite and the adjacent sides.
For the fourth pendant, we have a rectangle that is split to form two right triangles. Looking at angle a in the upper triangle, the side with a length of $3$ units is the adjacent side because it is between angle $b$ and the right angle. The side with a length of $4$ units is the hypotenuse because it is opposite the right angle.
• #### Identify the sides.
##### Hints
Remember that the Pharaoh's name, SOHCAHTOA, can help you remember the trig ratios.
The Pharaoh's name breaks into three syllables, each of which helps you remember one of the trig ratios. The three letters stand for the trig ratio, and the sides that belong in the numerator and denominator of that ratio.
##### Solution
The mnemonic device SOHCAHTOA breaks into three parts, one for each of the trig ratios. They are SOH, CAH, and TOA. Each starts with the first letter of the name of that ratio. For example, SOH starts with S, which stands for $sin$.
The next letter stands for the side whose length is in the numerator of the trig ratio. The last letter stands for the side whose length is in the denominator of the trig ratio. For example SOH means that the opposite side is in the numerator of this trig ratio, and that the hypotenuse is in the denominator. In other words:
$~$
$sin = \dfrac{\text{opposite}}{\text{adjacent}}$
$~$
Similarly, CAH means that:
$cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
$~$
And TOA means that:
$tan = \dfrac{\text{adjacent}}{\text{opppsite}}$
$~$
The opposite side to angle $A$ is the side that does not have the angle $A$ at either of its end points.
The adjacent side is the side between angle $A$ and the right angle.
The hypotenuse is always the side that is opposite to the right angle.
• #### Complete the trig ratios.
##### Hints
Remember that the opposite and adjacent sides are named with respect to the target angle. So the opposite side for angle $B$ is different than the opposite side for angle $A$.
It may help to write out the equations in numbers and symbols, in order to understand how to fill in the blanks in the sentences.
Remember, the Pharaoh's name, SOH-CAH-TOA, can help you remember what sides are in the numerator and the denominator of each trig ratio.
##### Solution
Recall the trig ratios:
$sin = \dfrac{\text{opp}}{\text{hyp}}$
$cos = \dfrac{\text{adj}}{\text{hyp}}$
$tan = \dfrac{\text{opp}}{\text{adj}}$
For the first triangle, the hypotenuse is 206 ft long, and does not change depending on the angle. For angle $A$, the opposite side is 100 ft long because this is the side that does not have angle $A$ at either of its end points. The adjacent side is 180 ft long because this is the side that goes between angle $A$ and the right angle. For angle $B$, the opposite side is 180 ft long, and the adjacent side is 100 ft long.
So we can find cosine ratio as:
$cos \, A = \dfrac{180}{206}$
$cos \, A = 0.874$
So the cosine of angle A is equal to 180 divided by 206. The ratio is 0.874.
Similarly:
$tan \, B = \dfrac{180}{100}$
$tan \, B = 1.8$
$sin \, A = \dfrac{100}{206}$
$sin \, A = 0.485$
And for the second triangle:
$sin \, A = \dfrac{40}{50}$
$sin \, A = 0.8$
$tan \, B = \dfrac{30}{40}$
$tan \, B = 0.75$
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Section1.2
# Section1.2 - Increasing A function is increasing on an...
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Unformatted text preview: Increasing A function is increasing on an interval I if F (x1 ) < f (x2 ) whenever x1 < x2 in I . Decreasing A function is decreasing on an interval I if F (x1 ) > f (x2 ) whenever x1 < x2 in I . Section 1.2 Mathematical Models Polynomial Functions A function is called a polynomial if it is of the form: P (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 where n is a nonnegative integer and the a’s are constants called the coefficients of the polynomial. The domain of the polynomial is the reals, and the degree of the polynomial is n. A polynomial of degree one is a linear function: P (x) = mx + b. A polynomial of degree 2 is called quadratic: P (x) = ax2 + bx + c. A polynomial of degree 3 is a cubic function: P (x) = ax3 + bx2 + cx + d. Power Functions A power function is of the form: f (x) = xa . And we have several cases: a = n, n is a positive integer This reduces in essence to the polynomials. a = 1/n, n is a positive integer This is a root function. f (x) = x1/n =n For n = 1 it is the square root. √ x. a = −1 This is the reciprocal function. It is a hyperbola with the coordinate axes as its asymptotes. yx = 1. Rational Functions are just the ratio of two polynomials. f (x) = and has the domain all values x such that Q(x) = 0. P (x) Q(x) Chapter 1 Review Algebraic Functions are functions that are constructed using algebraic operations on polyno functions can have extremely bizarre looking graphs! Trigonometric Functions Know the graphs! sin x, cos x, tan x (2π radians = 360 degrees) Exponential Functions Exponential functions are of the form f (x) = ax where a is called is a positive constant. Much more later. Logarithmic Functions The log functions are the inverse of the exponential functions. Muc Transcendental Functions Any function which is not algebraic is transcendental. The trig, and their inverses are examples of transcendental functions. A transcendental function is usua ized by an infinite series expansion, and there are a huge number of transcendental functions. Section 1.3: New Functions From Old Functions Translations Vertical and Horizontal Shifts: Suppose c > 0, to obtain the graph of y = f (x) + c, shift the graph of y = f (x) a distance c units upward y = f (x) − c, shift the graph of y = f (x) a distance c units downward y = f (x − c), shift the graph of y = f (x) a distance c units to the right y = f (x + c), shift the graph of y = f (x) a distance c units to the left Vertical and Horizontal Stretching and Reflecting: Stretching: If c > 1 then the graph is the graph of y = f (x) stretched by a factor of c in the vertical direction. Reflection: The graph of y = −f (x) is the graph of y = f (x) reflected about the x-axis, ordered pair has changed from (x, f (x)) to (x, −f (x)). ...
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# Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises: 55
$-9+40i$
#### Work Step by Step
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(4+5i)^2 ,$ use the special product on squaring binomials. Then use $i^2=-1$ and combine like terms. $\bf{\text{Solution Details:}}$ Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (4)^2+2(4)(5i)+(5i)^2 \\\\= 16+40i+25i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 16+40i+25(-1) \\\\= 16+40i-25 .\end{array} Combining like terms results to \begin{array}{l}\require{cancel} (16-25)+40i \\\\= -9+40i .\end{array}
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# What do the interior angles of a polygon have to add up to?
Nov 24, 2016
See below.
#### Explanation:
The sum of interior angles of a polygon is equal to
$A = \left(\frac{\pi}{2}\right) n$
where $n$ is the number of triangles needed to compose the polygon.
Occours that the number of triangles is equal to the number of sides minus $2$ or
$n = l - 2$ where $l$ is the number of polygon sides.
The final formula is
$A = \left(\frac{\pi}{2}\right) \left(l - 2\right)$
Examples.
A triangle has $l = 3$ so $A = \frac{\pi}{2}$
A quadrilateral has $l = 4$ so $A = \pi$
A heptagon has $l = 7$ so $A = 5 \left(\frac{\pi}{2}\right)$
Nov 24, 2016
To determine the sum of the internal angles of a polygon, take the number of sides on the polygon, subtract 2 and multiply by 180 degrees.
#### Explanation:
A triangle has three sides. $\left(3 - 2\right) \cdot 180 = 180$
The sum of the internal angles of a triangle is 180 degrees.
A quad lateral has four sides. $\left(4 - 2\right) \cdot 180 = 360$
The sum of the internal angles of a square is 360 degrees.
A pentagon has 5 sides. $\left(5 - 2\right) \cdot 180 = 540$
The sum of the internal angles of a pentagon is 540 degrees.
A hexagon has six sides. $\left(6 - 2\right) \cdot 180 = 720$
The sum of the internal angles of a hexagon is 720 degrees
A heptagon has seven sides. $\left(7 - 2\right) \cdot 180 = 900$
The sum of the internal angles of a heptagon is 900 degrees.
and so on!
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# Module 3: Mathematics Solution4: d / product of times = difference in speed /
difference in time
Let the initial time = T hour
Part A: TIME AND Then the new time = [ T+2(1/2)] hour
=> 600/T(T+5/2) = 12/(5/2)
DISTANCE
=> T(T+5/2) = 125 => T = 10
:. The required time = 10+2(1/2) = 12(1/2) hours
AVERAGE SPEED:
Example1: A car travels 600 km in 11 hours and another
DISTANCE COVERED IS DIFFERENT
800 km in 17 hours. Find the
Example5: A car travels for 11 hours. Out of this, it travels
average speed of the car
100 km at a certain speed and then it increases its speed
during the entire journey.
by 15 km/hr to cover the remaining 280 km. Find the time
Solution1: Here, distance
it takes to travel the span of 280 km.
and time are known.
Solution5: Given that
average speed =(d1 + d2) /(
Speed difference total time d2 d1
T1 + T2)
V1 - V2 T1 + T2 100 280
= (600 + 800) / (11+17) = 50
15 11
km/hour
Use (V1 - V2), because, here V1 and V2 are not to be found
out separately.
Example2: A bus moves 300 km at a speed of 45 km per
V1- V2 = d1/T1 - d2/T2
hour and then it increases its speed to 60 km per hour to
15= 280/x - 100/(11-x) (assume, x = time to travel 280 km)
travel another 500 km. Find the average speed of the bus.
3= 56/x - 20/(11-x) :. x = 7
Solution2: Here distance and speed are known. Time is not
:. the car travels the span of 280 km in 7 hours
known.
:. We use the relation,
Example6: A
average speed = [d1 + d2] /
car travels a
[d1/V1+ d2/V2]
distance of
average speed = (300+500) /
170 km in 2
(300/45+500/60)
hours partly at
= 800 / (20/3+ 25/3) = (800 *
a speed of 100
3) /45 = 53(1/3) km per hour
km/hr and
partly at 50
DISTANCE COVERED IS
km/hr. Find the distance travelled at speed of 100 km/hr.
SAME
Solution6: Given that:
Example3: An increase in the speed of a car by 10 km per
total distance total time
hour saves 30 minutes in a journey of 100 km. Find the
(d1 + d2) (T1 + T2) V1 V2
initial speed of the car.
170 2 100 50
Solution3: Let the initial speed of the car = V km/hr
Use (T1 + T2), because, here, T1 and T2 are not to be
The new speed of the car = (V + 10) km/hr.
found out separately.
Product of speeds / distance = difference in speed /
:. T1 + T2 = d1/ V1 + d2/ V2 (Time reference is used)
difference in time
2 = (x/100)+(170-x/50) (assume x = distance covered
V(V+10) / 100 = 10/ (30/60) [:. 30 minutes = 30/60 hour]
at 100 km/hr)
V (V + 10) = 2,000
x= 140
V (V + 10) = 40 * 50
:. The car travelled 140 km
Without further solving, it appears that V = 40 km/h.
at the speed of 100 km/hr.
Example7: If a truck travels
a distance of 240 km in 6
motorcycle is reduced by
hours, partly at a speed of
12 km/hr and it now takes
60 km/hr and partly at 30
2(1/2) hours more to
km/hr, then find the time
cover the same distance
for which it travels at 60
of 600km. Find the time it
km/hr.
now takes to cover the
Solution7: Given that
distance.
total distance total time
d1 + d2 t1+ t2 V1 V2
240 6 60 30
Use (d1 + d2), because here d1 and d2 are not to be found will it travel in 15 minutes?
out separately. 4. If a man walks to his office at 3/4 of his usual rate, he
So we find, reaches office 1/3 of an hour later than usual.
d1 + d2 = V1*T1 + V2*T2· What is his usual time to reach office?
240 = 60 * x + 30 * (6 - x), (assume x = time of travel at 5. If a man walks to his office at 5/4 of his usual rate, he
60 km/hr) reaches office 30 min. early than usual. What is his usual
x=2 :.The truck travels 2 hours at 60 km/hr time to reach office?
6. If I walk at 3km/h, I miss a train by 2 minutes. If
Example8: A man takes 4 however, I walk at 4km/h, I reach the station 2 minutes
hours 30 minutes in before the arrival of the train. How far do I walk to reach
walking to a certain place the station?
and riding back. He would 7. Excluding stoppages, the speed of a train is 45km/h, and
have gained 1 hour 45 including stoppages, it is 36km/h, For how many minutes
minutes by riding both does the train stop per hour?
ways. How long would he 8. A man covers a certain distance on auto rickshaw. Had it
take to walk both ways? moved 3 km/h faster, he would have taken 40 minutes less.
Solution8: Time for walking both ways - time by mixed (i.e. If it had moved 2 km/h slower, he would have taken 40
walking + riding) = time gained minutes more. Find the distance.
time for walking both ways -4(1/2) = +1(3/4) 9. A thief escapes
time for walking both ways = 4(1/2) +1(3/4) in a car at a speed
= 6(1/4) hours. of 40 km/h. After
:. The man will take 6(1/4) hours to cover the same distance half an hour a police
if he walks both ways. car sets off to
catch the thief at
TIME AND DISTANCE BETWEEN TWO MOVING 50 km/h . After
BODIES what time from the
Example9: Two men, P and Q, start walking from a hotel at theft will the police
2 km/hr and 2(1/2) km/hr respectively. By how many km car overtake the
will they be apart at the end of 3(1/2) hours, if thief?
(i) they walk in opposite directions Hint: Distance to be covered by the thief and by the
(ii) they walk in the same direction. police car is same.
Solution9: Let after time 't' police catches the thief, V1*T1 = V2*T2
(i) When they walk in opposite direction, then P and Q will :. T*40= (T-1/2) * 50 :. T = 2(1/2)hours.
be (2 + 2(1/2)) km or 4(1/2) km apart in 1 hour. 10. A man travels in a car
:.at the end of 3(1/2) hours, they will be 3(1/2) * 4(1/2) km from X to Y at a speed of 77
= 15(3/4) km apart. km/h and returns back at 33
(ii) When they walk in the same direction, then P and Q km/h from Y to X. Find the
will be [2(1/2) - 2] km or (1/2) km apart in 1 hour average speed of the journey.
:. at the end of 3(1/2) hours, they will be [(1/2) * 3(1/2) km 11. A man travels on a
= 1(3/4) km, apart. scooter from A to B at a speed
of 30 km/h and returns back
Practice Exercises. from B to A at 20 km/h. The total! journey was performed
Questions: by him in 10 hours. Find the distance from A to B.
1. Find the distance covered by a car moving at 20 meters Hint: Let the distance be 'd' km
per second for 3 hours. d/30+d/20 = T1+T2 = 10(given)
Hint: Remember 1 d = (10*30*20)/(30+20)=120km.
meter per second = 12. A long distance runner runs 9 laps of a 400 meters
18/5 km/h. track everyday. His timings (in min.) for four consecutive
:. d= Vt days are 88, 96, 89, and 87 respectively. On an average,
= 20 * 18/5 * 3 km, = how many meters/minute does the runner cover?
216 km Hint: Average speed (meter/minutes) = Total distance
2. Find the time taken /Total time =(9*400*4)/(88 + 96 + 89 + 87) = 40
to cover a distance of meter/minutes
0.9 km by a bullock cart 13. A man performs 2/25 fraction of his total journey by
moving at 0.25 meter/s. bus, 21/50 by car and the remaining 2 km on foot. Find the
3. A train travels at 90 distance of the total journey.
km/h. How many meters Hint: Let total journey = x km.
(2/25)x + (21/50)x +2 = x . days and then Maurice leaves. Mary finishes the remaining
x=4 work alone. In how many days is the total work finished?
14. Normally it takes 3 hours for a train to run from A to Solution3: Let the total work be finished in 'T' days.
B. One day, due to a minor trouble, the train had to reduce no. of days Mary worked/ alone time + no. of days Maurice
the speed by 12 km/h and so it took 3/4 of an hour more worked /alone time = 1
than usual. What is the distance from A to B. (T/10)+(3/15)=1 :.(T/10)=1-(1/15) :.T= 10*(4/5)= 8
Hint: Let d be the distance from A to B. :. Total work is finished in 8 days.
(d/T1) - (d2/T2) = V1 - V2
(d/3) -[d/3(3/4)] = 12 Example4: Singvi and Ravi can do a job alone in 10 days and
d = 180km. 12 days respectively. Singvi starts the work and after 6
Answer key: days Ravi also joins to finish the work together. For how
1. 216 km many days Ravi actually worked on the job?
2. 1h Solution4: Let the work be finished in T days,
3. 22,500m (no. of days Singvi worked /alone time) + (no. of days Ravi
4. 1h worked/alone time)=1
5. 2(1/2) (T/10)+(T-6)/12= 1
6. 4/5km T=(90/11) :. Ravi worked for T - 6 i.e. (90/11) - 6 = 2(2/11)
7. 12min days.
8. 40km
9. 2(1/2)hr after the theft Example5: A
10. 46.2km/h and B can do a
11. 120km. piece of work in
12. 40m 12 days, B and C
13. 4km in 15 days, C
14. 180km and A in 20
days. How long
Part B: TIME, WORK AND WAGES would each take
separately to do
BASIC CONCEPTS: Total amount of a complete job (or the same work?
assigned job) =1, always, unless otherwise specified. Solution5:
1 day's work by any person = ( 1/alone time)th part of total L.C.M. of 12, 15 and 20 = 60 = L, say A and B can do in T1 =
work 12 days, C absent; B and C can do in T2 = 15 days, A
absent; C and A can do in T3 = 20 days, B absent. Alone
Example1: Tuktuki time for A =(2L)/ [(L/T1)+(L/T3)-(L/(T2)] =
and Rasmani can do a ((2*60)/((60/12)+(60/20)-(60/15))=(120/(5+3-4)=30 days
job alone in 20 days Similarly, alone times for B and C can be found out.
and 30 days
respectively. In how Example6: A Husband and Wife can do a job in 15 hours
many days the job and 10 hours respectively. They began the work together
will be finished if but Husband leaves
they work together? after some hours and
Solution1: Here, we Wife finishes the
can use a direct remaining job in 5
formula, if a = 20, b hours. After how many
=30, then Combined hours did Husband
required time =ab/(a+b) =(20 * 30)/(20 + 30) = 12 days. leave?
Solution6: In this
Example2: Mohan and Sohan can do a job in 12 days. Sohan problem, the total time
alone can finish it in 28 days. In how many days can Mohan for completion is
alone finish the work? neither known nor to
Solution2: Short-Cut if T = 12, a = 28 & b=? be found out. Hence
1/12=1/28+1/b total time T is not be
Then b =(28*12)/(28-12)= 21 considered, but assume
Mohan can alone finish the work in 21 days. that P works for x
hours, then using the
Example3: Mary and Maurice can do a piece of work in 10
days and 15 days respectively. They work together for 3
concept, (no. of hours P worked /alone time for P) + (no. of If they are working for 1 hour atternately, Anil beginning,
hours Q worked /alone time for Q)=1 (x/15)+(x+5)/10=1 in how hours will the computer be assembled?
x = 3 :. P leaves after 3 hours. Solution10: After 10 hours of combined, but alternate
working, we get the part of computer assembled = AniI's 5
Example7: Ramesh is thrice as good a workman as Bipan, hours work + Sunil's 5 hours work. = 5*(1/10)+5*(1/12) =
and is therefore able to finish a piece of work in 40 days (11/12) th Remaining part = 1-(11/12)=(1/12) th Now, at the
less than Bipan. Find the time in which they can do it start of 11th hour, Anil will work. Time taken by Anil to
working together. do(1/12) th work = 10 *(1/12) hour =(5/6) hour .. Total time
Solution7: Since Ramesh is thrice as good a workman as = 10 hours + (5/6) hours = 10(5/6) hours. :. the computer
Bipan, then if Ramesh does a job in 1 day, Bipan will do the will be assembled in 10(5/6) hours.
same in 3 days and the difference is 3 - 1 = 2 days. :. For
40 days difference (20 * 2), Ramesh does in 20 days, Bipan Example11: Two friends
in (20 * 3) = 60 days. Now, (Ramesh + Bipan)'s 1 day work = take a piece of work for
(1/20)+(1/60)=(1/15) :. the required time is 15 days .. Rs 960. One alone could
do it in 12 days, the other
Example8: Two workers A in 16 days. With the
and B working together assistance of an expert
completed a job in 5 days. they finish it in 4 days.
If A worked twice as How much renumeration
efficiently as he actually the expert should get?
did and B worked (1/3) as Solution11: First
efficiently as he actually friend's 4 day's work =
did, the work would have (4/12)=(1/3) (Since, the
been completed in 3 days. work is finished in 4 days, when expert assists) second
Find the time taken by A to friend's 4 day's work =(4/16)=(1/4)
complete the job alone. The expert's 4 day's work =1-(1/3)+(1/4)=(5/12)
Solution8: Let 'A' alone Now, total wages of Rs 960 is to be distributed among two
complete the job in 'a' days friends and the expert in proportion to the amount of work
and B in 'b' days. done by each of them. So, 960 is to be divided in the
Then, (1/a)+(1/b)=(1/5) (eqn 1) proportion of (1/3):(1/4):(5/12) or 4:3:5 :. Share of expert
(i.e. 1 day's combined work) and in second case, A works
twice efficiently, so A's 1 day work = 2 * (1/a) and B works
(1/3) as efficiently, so B's 1 day work = (1/3)*(1/b) :.
combined 1 day's work = (1/3) =(2/a)+(1/3b) (6/a)+(1/b)=1
(eqn 2)
Solving (1) and (2), we get, a = 6(1/4) :. A completes the job
alone in 6(1/4)days.
## Example9: Kaberi takes twice as much time as Kanti and
thrice as much as Kalpana to finish a piece of work. They
together finish the work in one day. Find the time taken by
each of them to finish the work. Solution9 Here, the alone
time of Kaberi is related to the alone times of other two
persons, so assume the alone time of Kaberi = x, then,
alone time of Kanti = (x/2) and of Kalpana =(x/3) Kaberi's =(5/12)*960=Rs 400 Hence, the expert should get Rs 400.
1 day work + Kanti's 1 day work + Kalpana's 1 day work =
combined 1 day's work 1/x+1/(x/2)+1/(x/3)=1/1 x=6 :. Practice Questions:
Alone time for Kaberi = 6 days, for Kanti =(6/2)= 3 days, 1. Ram can polish the floor of a building in 16 days. Find
Kalpana =(6/3) = 2 the work done by Ram in one day.
days. 2. Mukesh can do (2/7) th of an work in 1 day. In how
many days can he complete the same work? Example: A
Example10: Anil and person 'P' can alone do a work in 10 days and 'Q' can do it
Sunil working in 15 days. What amount of work is done by P and Q
separately can together in one day?
assemble a computer 3. A person 'P' can alone do a work in 10 days and 'Q' can
in 10 hours and 12 do it in 15 days. What amount of work is done by P and Q
hours respectively. together in one day?
4. Three persons Ramesh, Suresh and Kana can do a job
alone in 10 days, 12 days and 15 days respectively. In how
many days they can finish the job working together?
5. A person 'M' can do a job in 15 days. How much of the
job is done by him in 7 days?
6. Two friends A and B can do a work alone in 12 days and
8 days respectively. Find the amount of work done by them
in 4 days.
7. Two persons P and Q can do a piece of work alone in 10
days and 15 days respectively. If P works for 2 days and Q
works for 5 days, then find the total amount of work done.
8. Dipa and Avik can do a piece of work in 20 days and 30
days respectively. They work together and Dipa leaves 5
days before the work is finished. Avik finishes the
remaining work alone. In how many days is the total work
finished?
|
# Theorem – There is one and only one circle passing through three given non-collinear points | Class 9 Maths
Last Updated : 06 May, 2021
### Theorem Statement: There is one and only one circle passing through three given non-collinear points.
Required Diagram:
Given: Three non-collinear points P, Q, and R.
To Prove: There is one and only one circle passing through P, Q, and R.
Construction: Join PQ and QR. Draw perpendicular bisectors AL and BM of PQ and RQ respectively. Since P, Q, R is not collinear. Therefore, the perpendicular bisectors AL and BM are not parallel. Let AL and BM intersect at O. Join OP, OQ, and OR.
### Proof
Since O lies on the perpendicular bisector of PQ. Therefore,
OP = OQ
Again, O lies on the perpendicular bisector of QR. Therefore,
OQ = OR
Thus, OP = OQ = OR = r (say).
Taking O as the centre draw a circle of radius s. Clearly, C (0, s) passes through P, Q, and R. This proves that there is a circle passing through the points P, Q, and R. We shall now prove that this is the only circle passing through P. Q and R. If possible, let there be another circle with centre O’ and radius r, passing through the points P, Q and R. Then, O’ will lie on the perpendicular bisectors AL of PQ and BM of QR. Since two lines cannot intersect at more than one point, so O’ must coincide with O. Since OP = r, O’P = s and O and O’ coincide, which means that,
r = s
Therefore,
C(O, r) ≅ (O’, s)
Hence, there is one and only one circle passing through three non-collinear points P, Q and R.
### Examples
Example 1: Write down the step-by-step construction procedures to find out the centre of the circle?
Solution:
Let the circle be C1. We need to find its centre.
Step 1: Take points P, Q, R on the circle
Step 2: Join PR and RQ.
We know that the perpendicular bisector of a chord passes through the centre. So, we construct perpendicular bisectors of PR and RQ.
Step 3: Take a compass. With point P as the pointy end and R as the pencil end of the compass, mark an arc above and below PR. Do the same with R as pointy end P as pencil end of the compass.
Step 4: Join points intersected by the arcs. The line formed is the perpendicular bisector of PR.
Step 5: Take a compass, with point R as the pointy end and Q as pencil end of the compass mark an arc above and below RQ. Do the same with Q as the pointy end and R as the pencil end of the compass.
Step 6: Join the points intersected by the arcs. The line formed is the perpendicular bisector of RQ.
Step 7: The point where two perpendicular bisectors intersect is the centre of the circle. Mark it as point O. Thus, O is the centre of the given circle.
Example 2: Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
(i) No point common:
(ii) One point common:
(iii) Two points common:
As we can analyze from above, two circles can cut each other maximum at two points.
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# Factors of 1085
Factors of 1085 are 1, 5, 7, 31, 35, 155, 217, and 1085
#### How to find factors of a number
1. Find factors of 1085 using Division Method 2. Find factors of 1085 using Prime Factorization 3. Find factors of 1085 in Pairs 4. How can factors be defined? 5. Frequently asked questions 6. Examples of factors
### Example: Find factors of 1085
• Divide 1085 by 1: 1085 ÷ 1 : Remainder = 0
• Divide 1085 by 5: 1085 ÷ 5 : Remainder = 0
• Divide 1085 by 7: 1085 ÷ 7 : Remainder = 0
• Divide 1085 by 31: 1085 ÷ 31 : Remainder = 0
• Divide 1085 by 35: 1085 ÷ 35 : Remainder = 0
• Divide 1085 by 155: 1085 ÷ 155 : Remainder = 0
• Divide 1085 by 217: 1085 ÷ 217 : Remainder = 0
• Divide 1085 by 1085: 1085 ÷ 1085 : Remainder = 0
Hence, Factors of 1085 are 1, 5, 7, 31, 35, 155, 217, and 1085
#### 2. Steps to find factors of 1085 using Prime Factorization
A prime number is a number that has exactly two factors, 1 and the number itself. Prime factorization of a number means breaking down of the number into the form of products of its prime factors.
There are two different methods that can be used for the prime factorization.
#### Method 1: Division Method
To find the primefactors of 1085 using the division method, follow these steps:
• Step 1. Start dividing 1085 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 1085, which is 5. Divide 1085 by 5 to obtain the quotient (217).
1085 ÷ 5 = 217
• Step 3. Repeat step 1 with the obtained quotient (217).
217 ÷ 7 = 31
31 ÷ 31 = 1
So, the prime factorization of 1085 is, 1085 = 5 x 7 x 31.
#### Method 2: Factor Tree Method
We can follow the same procedure using the factor tree of 1085 as shown below:
So, the prime factorization of 1085 is, 1085 = 5 x 7 x 31.
#### 3. Find factors of 1085 in Pairs
Pair factors of a number are any two numbers which, which on multiplying together, give that number as a result. The pair factors of 1085 would be the two numbers which, when multiplied, give 1085 as the result.
The following table represents the calculation of factors of 1085 in pairs:
Factor Pair Pair Factorization
1 and 1085 1 x 1085 = 1085
5 and 217 5 x 217 = 1085
7 and 155 7 x 155 = 1085
31 and 35 31 x 35 = 1085
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 1085. They are called negative pair factors.
Hence, the negative pairs of 1085 would be ( -1 , -1085 ) , ( -5 , -217 ) , ( -7 , -155 ) and ( -31 , -35 ) .
#### What are factors?
In mathematics, a factor is that number which divides into another number exactly, without leaving a remainder. A factor of a number can be positive or negative.
#### Properties of factors
• Every factor of a number is an exact divisor of that number, example 1, 5, 7, 31, 35, 155, 217, 1085 are exact divisors of 1085.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Each number is a factor of itself. Eg. 1085 is a factor of itself.
• 1 is a factor of every number. Eg. 1 is a factor of 1085.
• What are the pair factors of 1085?
Pair factors of 1085 are (1,1085), (5,217), (7,155), (31,35).
• What are five multiples of 1085?
First five multiples of 1085 are 2170, 3255, 4340, 5425, 6510.
• What two numbers make 1085?
Two numbers that make 1085 are 5 and 217.
• What is the sum of all factors of 1085?
The sum of all factors of 1085 is 1536.
• What are factors of -1085?
Factors of -1085 are -1, -5, -7, -31, -35, -155, -217, -1085.
• What are factors of 1085?
Factors of 1085 are 1, 5, 7, 31, 35, 155, 217, 1085.
• Is 1085 a perfect square?
No 1085 is not a perfect square.
• Which is the smallest prime factor of 1085?
Smallest prime factor of 1085 is 5.
• Which is greatest factor of 1085?
The greatest factor of 1085 is 217.
#### Examples of Factors
Can you help Sammy find out the product of the odd factors of 1085?
Factors of 1085 are 1, 5, 7, 31, 35, 155, 217, 1085.
Odd factors of 1085 are 1, 5, 7, 31, 35, 155, 217, 1085.
Hence product of odd factors of 1085 is; 1 x 5 x 7 x 31 x 35 x 155 x 217 x 1085 = 1385858700625.
Joey wants to write all the prime factors of 1085 in exponential form, but he doesn't know how to do so can you assist him in this task?
Prime factors of 1085 are 5, 7, 31.
So in exponential form it can be written as 5 x 7 x 31.
How many factors are there for 1085?
Factors of 1085 are 1, 5, 7, 31, 35, 155, 217, 1085.
So there are in total 8 factors.
Kevin has been asked to write 7 factor(s) of 1085. Can you predict the answer?
7 factor(s) of 1085 are 1, 5, 7, 31, 35, 155, 217.
Sammy is puzzled while calculating the prime factors of 1085. Can you help him find them?
Factors of 1085 are 1, 5, 7, 31, 35, 155, 217, 1085.
Prime factors of 1085 are 5, 7, 31
What is prime factorization of 1085?
Prime factorization of 1085 is 5 x 7 x 31 = 5 x 7 x 31.
Can you help Rubel to find out the product of the even factors of 1085?
Factors of 1085 are 1, 5, 7, 31, 35, 155, 217, 1085.
Even factors of 1085 are 0.
Hence, product of even factors of 1085 is; 0 = 0.
|
# Ratios
### Ratios
What are two-term and three term ratios? After watching this lesson, you will be able to answer this question by writing ratios in the lowest terms as well as in fraction and percent form. Practice your understanding by doing some real-life word questions too.
#### Lessons
In this lesson, we will learn:
• Ratios With Decimals
• Ratios With Fractions
• Ratios in Different Units
• Application of Ratios
• A ratio should always be in its simplest/most reduced form (no common factors).
• The values in a ratio should always be integers if possible.
• A ratio can be scaled up/down by multiplying/dividing the ratio numbers by the same value. This process is called scaling, and the value used for scaling is called the scale factor.
• 1.
Write the following ratios using ratio notation in lowest terms.
a)
5 km in 30 minutes.
b)
4 cases for \$500.
c)
5 cupcakes need 325 g of flour.
d)
130 mm of precipitation in 2 days.
• 2.
Write the following ratios in fraction form in lowest terms.
a)
3 hours in 1 week.
b)
35 g sugar in a 335 ml can of pop.
c)
4 servers to serve 6 tables with 4 guests each.
d)
22 sheep and 30 cows. What is the ratio of sheep to the total number of animals?
• 3.
The following table shows the sugar and flour needed for different kinds of bakery. Bakery Sugar (g) Flour (g) Cake 40 120 Cookie 20 60 bread 30 450
a)
Which 2 kinds of bakery have the same sugar-flour ratio? Show your work.
b)
What is the ratio of sugar needed for cookie to the total weight of sugar needed for all 3 kinds of bakery?
##### Do better in math today
5.
Introduction to Ratios, Rates and Percentages
5.1
Ratios
5.2
Rates
5.3
Proportions
|
# Modeling with Linear Functions by hcj
VIEWS: 36 PAGES: 4
• pg 1
``` Name: ____________________________________ Date: _________________
Modeling with Linear Functions
Algebra 1A
We have learned that slope is used to describe real life rates of change. We also know that the y-
intercept is where the line begins on the y-axis. The y-intercept always occurs where the independent
variable has a value of zero. Using these two quantities, slope and y-intercept, we can model and solve
many real life problems.
Exercise #1: The Arlington Freshmen class wants to have a fundraiser. The class wants to buy a
number of \$4 flip-flops and \$5 bracelets. The class has a total of \$100 to spend.
(a) If x represents the number of flip-flops and y represents the number of bracelets, complete the table
below.
# of flip-flops, x 0
# of bracelets, y 0
(b) Using the two points from part (a), write a linear equation in y mx b form that gives the
number of bracelets that can be bought as a function of the number of flip-flops bought.
(c) Using your equation from (b), determine the number of bracelets that can be bought if 10 flip-flops
were purchased.
Exercise #2: From 2000 to 2007 the number of coffee shops in a certain country increased by 100
shops per year. In 2002, there were 1100 coffee shops.
(a) Write a linear equation for the number of coffee shops, y, as a function of time, t, where t = 0
represents the year 2000.
(b) Based on your linear model from part (a), predict the number of coffee shops that will be in that
country in 2025.
Algebra 1A, Unit #2 – L11
Exercise #3: The cost to subscribe to an online internet service consists of a \$15 per month flat-fee
and a \$4.00 per hour additional charge.
Cost of Subscription, C
(a) Create a linear model to represent the total 100
cost per month, C, as a function of the
number of hours, h, that are used.
75
50
(b) Using your calculator to generate a table of
values, graph the model you formed in part
(a) on the grid provided.
25
(c) Lucy was charged \$75 after signing up and
using the service for one month. How
many hours did she use? Justify your
answer both algebraically and graphically. 5 10 15 20
Number of Hours Used, h
Exercise #4: Shirley’s Workout Club charges \$6 to sign up and \$3 each time a person works out.
Total Cost, C
(a) Write an equation representing the cost, C1 , to workout at
40
Shirley’s as a function of the number of workouts a person has
worked out, w.
30
(b) Tommy’s Exercise Center charges \$14 to sign up and \$2 for
each workout. Create another linear function, as in part (a), for
the cost, C 2 , of attending Tommy’s Center. 20
(c) Graph both equations on the grid to the right. What number of
10
workouts will result in the same cost for both gyms?
5 10
Algebra 1A, Unit #2 – L11
Number of Workouts, w
Name: ____________________________________ Date: _________________
Modeling with Linear Functions
Algebra 1A Homework
Applications (Use a scientific calculator at home to assist you with calculations.)
1. Hamal Rental Cars charges a flat-fee of \$25 per day to rent a new Chevy Impala, plus a mileage
charge of \$0.25 per mile.
Cost of Renting, C
(a) Write a linear equation to represent the cost, C1 , of
renting an Impala as a function of the number of 35
miles driven, m.
30
(b) On the grid to the right, graph and label the linear
function your created in part (a).
25
(c) Ike’s Rentals charges a flat fee of \$20 per day to rent
an Impala plus a mileage charge of \$0.50 per mile.
As in (a), write a linear equation to represent the
cost, C 2 , of renting an Impala from Ike’s and graph 20
this function on the grid at the right.
10 20 30 40
Number of Miles Driven, m
(d) For what number of miles, m, will the rental costs be
equal for the two places?
2. Kael wants to install a new toilet. Luigi the plumber charges \$100 for the cost of the toilet plus an
additional \$75 per hour.
(a) Write an linear equation that gives the cost, C1 , as a function of the hours, h, that Luigi works.
(b) Being very exact with his hours, Luigi charges Kael \$750. Determine, to the nearest tenth of an
hour, how long Luigi worked on this job. Justify your answer using algebra.
Algebra 1A, Unit #2 – L11
3. Javier is trying to find a linear equation for the cost of his cell-phone plan. The first month he talks
for only 32 minutes and is charged \$14.10. The second month he talks for 420 minutes and is
charged \$33.50.
(a) Write two ordered pairs, where the minutes are the independent variable (first coordinate) and the
charge is the dependent variable (second coordinate), that model the information given in the
problem.
(b) Using these two points, write a linear equation that gives Javier’s charge, C, as a function of the
number of minutes, m, that he talks.
(c) What does the slope of this linear function represent?
4. Miguel is driving towards New York City at a constant rate of speed. After 2 hours he notices that
he is 127 miles away and after 3 hours he notices that he is 69 miles away.
(a) Write the information above as two ordered pairs, with time being the independent variable (first
coordinate) and the distance from New York City being the dependent variable (second
coordinate).
(b) Using your ordered pairs from part (a), write a linear equation in which the distance Miguel is
away, D, as a function of the time he has been driving, t.
(c) Why is the slope of your linear equation from part (b) negative? Explain in terms of the real-life
scenario that the linear equation is modeling.
(d) How far from NYC was Miguel when he started his trip at t = 0 hours? Justify.
Algebra 1A, Unit #2 – L11
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We think you are located in South Africa. Is this correct?
# Chapter 7: Measurement
• Use paper or cardboard for the net of solids to help learners see the different heights, particularly perpendicular and slanted heights.
• Units are compulsory when working with real life contexts.
• Sketches are valuable and important tools.
• Rounding off should only be done in the last step and level of accuracy should be relevant to the context.
This chapter is a revision of perimeters and areas of two dimensional objects and volumes of three dimensional objects. We also examine different combinations of geometric objects and calculate areas and volumes in a variety of real-life contexts.
## 7.1 Area of a polygon (EMBHV)
Square $$\text{Area}={s}^{2}$$ Rectangle $$\text{Area}=b\times h$$ Triangle $$\text{Area}=\frac{1}{2}b\times h$$
Trapezium $$\text{Area}=\frac{1}{2}\left(a+b\right)\times h$$ Parallelogram $$\text{Area}=b\times h$$ Circle $$\text{Area}=\pi {r}^{2}$$ $$\left(\text{Circumference}=2\pi r\right)$$
## Worked example 1: Finding the area of a polygon
$$ABCD$$ is a parallelogram with $$DC = \text{15}\text{ cm}$$, $$h = \text{8}\text{ cm}$$ and $$BF = \text{9}\text{ cm}$$.
Calculate:
1. the area of $$ABCD$$
2. the perimeter of $$ABCD$$
### Determine the area
The area of a parallelogram $$ABCD =$$ base $$\times$$ height:
\begin{align*} \text{Area} &= \text{15} \times \text{8} \\ &= \text{120}\text{ cm$^{2}$} \end{align*}
### Determine the perimeter
The perimeter of a parallelogram $$ABCD = 2DC + 2BC$$.
To find the length of $$BC$$, we use $$AF \perp BC$$ and the theorem of Pythagoras.
\begin{align*} \text{In } \triangle \text{ ABF:} \quad AF^2 &= AB^2 - BF^2 \\ &= \text{15}^2 - \text{9}^2 \\ &= \text{144} \\ \therefore AF &= \text{12}\text{ cm} \end{align*}\begin{align*} \text{Area} ABCD &= BC \ \times AF \\ \text{120} &= BC \times \text{12} \\ \therefore BC &= \text{10}\text{ cm} \end{align*}\begin{align*} \therefore \text{Perimeter} ABCD &=2(\text{15})+2(\text{10}) \\ &= \text{50}\text{ cm} \end{align*}
# Success in Maths and Science unlocks opportunities
## Area of a polygon
Exercise 7.1
Vuyo and Banele are having a competition to see who can build the best kite using balsa wood (a lightweight wood) and paper. Vuyo decides to make his kite with one diagonal $$\text{1}$$ $$\text{m}$$ long and the other diagonal $$\text{60}$$ $$\text{cm}$$ long. The intersection of the two diagonals cuts the longer diagonal in the ratio $$\text{1}:\text{3}$$.
Banele also uses diagonals of length $$\text{60}$$ $$\text{cm}$$ and $$\text{1}$$ $$\text{m}$$, but he designs his kite to be rhombus-shaped.
Draw a sketch of Vuyo's kite and write down all the known measurements.
Determine how much balsa wood Vuyo will need to build the outside frame of the kite (give answer correct to the nearest cm).
\begin{align*} AD &= AB \\ &= \sqrt{30^2+25^2} \quad (\text {Pythagoras}) \\ AD&=\text{39}\text{ cm} \\ DC&=BC \\ &=\sqrt{30^2+75^2} \quad (\text {Pythagoras}) \\ &=\text{81}\text{ cm} \\ \therefore \text{Balsa wood: } &= 2(81+39)\\ &=\text{240}\text{ cm} \end{align*}
Calculate how much paper he will need to cover the frame of the kite.
\begin{align*} \text{Area } &= 60 \times 100 \\ &= \text{6 000}\text{ cm$^{2}$}\\ &=\text{0,6}\text{ m$^{2}$} \end{align*}
Draw a sketch of Banele's kite and write down all the known measurements.
Determine how much wood and paper Banele will need for his kite.
\begin{align*} \text{Side length} &= \sqrt{50^2+30^2} \\ &=\text{58,3}\text{ cm}\\ \therefore \text{Wood for frame } &= 4 \times \text{58,3}\text{ cm} \\ &= \text{233,2}\text{ cm}\\ \text{Area }&= 60 \times 100 \\ &= \text{6 000}\text{ cm$^{2}$}\\ &=\text{0,6}\text{ m$^{2}$} \end{align*}
Compare the two designs and comment on the similarities and differences. Which do you think is the better design? Motivate your answer.
Same amount of paper is required for both designs. Vuyo's designs uses more balsa wood.
$$O$$ is the centre of the bigger semi-circle with a radius of $$\text{10}$$ $$\text{units}$$. Two smaller semi-circles are inscribed into the bigger one, as shown on the diagram. Calculate the following (in terms of $$\pi$$):
The area of the shaded figure.
\begin{align*} \text{Area } &=\frac{1}{2}\pi(10)^2-2\left (\frac{1}{2}\pi(5)^2 \right )\\ &=50\pi-25\pi \\ &=25\pi \text{units$^{2}$} \end{align*}
The perimeter enclosing the shaded area.
\begin{align*} \text{Perimeter }&=\frac{1}{2}(2\pi(10))+2\pi(5) \\ &=10\pi+10\pi \\ &=20\pi \text{units$^{2}$} \end{align*}
Karen's engineering textbook is $$\text{30}$$ $$\text{cm}$$ long and $$\text{20}$$ $$\text{cm}$$ wide. She notices that the dimensions of her desk are in the same proportion as the dimensions of her textbook.
If the desk is $$\text{90}\text{ cm}$$ wide, calculate the area of the top of the desk.
\begin{align*} \text{Ratio } &= \frac{\text{table width}}{\text{book width}} \\ &=\frac{90}{20} \\ &=\text{4,5} \\ \therefore \text{Length of table } &= 30 \times \text{4,5} \text{cm} \\ &=\text{135}\text{ cm}\\ \text{Area table} &= 135 \times 90 \\ &=\text{12 150}\text{ cm$^{2}$}\\ &=\text{1,2}\text{ m$^{2}$} \end{align*}
Karen uses some cardboard to cover each corner of her desk with an isosceles triangle, as shown in the diagram:
Calculate the new perimeter and area of the visible part of the top of her desk.
\begin{align*} x^2&=15^2+15^2 \\ x&=\text{21,2}\text{ cm}\\ \text{New length}&=135-2(15) \\ &=\text{105}\text{ cm} \\ \text{New breadth}&=90-2(15) \\ &=\text{60}\text{ cm}\\ \text{New perimeter }&=2(105)+2(60)+4(\text{21,2})\\ &=\text{414,8}\text{ cm} \end{align*} \begin{align*} \text{Area cut off}&=2\times(15^2) \\ &=\text{450}\text{ m$^{2}$} \\ \text{New area } &= \text{12 150}\text{ cm$^{2}$} - \text{450}\text{ m$^{2}$} \\ &= \text{11 700}\text{ cm$^{2}$} \end{align*}
Use this new area to calculate the dimensions of a square desk with the same desk top area.
\begin{align*} s^2&= \text{11 700}\text{ cm$^{2}$} \\ \therefore s&= \text{108,2}\text{ cm}\\ \text{Square table of length} &\approx 108 \times 108 \text{cm$^{2}$} \end{align*}
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# Zone or Frustum of a Sphere
The portion of a sphere intercepted between two parallel planes is called a zone (i.e. a frustum).
(i) The volume of the zone (or frustum) of a sphere may be found by taking the difference between segment $EBD$ and the segment $ABC$ (see figure), that is:
Where $h$ is the altitude, and ${r_1}$ and ${r_2}$ are respectively the radii of the small circle (bases).
(ii) The surface area of the zone or frustum is equal to the circumference of the great circle of the sphere times the altitude of the same.
i.e. $S = 2\pi rh$
Where $S = {\text{area of a zone}}$
$h = {\text{altitude}}$
$r = {\text{radius of the sphere}}$
(iii) The total surface area of a zone $= 2\pi rh + \pi {r_1}^2 + \pi {r_2}^2$
Example:
The sphere of radius 8cm is cut by two parallel planes, one passing 2cm from the center and the other 6cm from the center. Find the area of the zone and the volume of the segment between the two planes if both planes are on the same side of the center.
Solution:
The surface area of the zone $S = 2\pi rh\,\,\,\,\, = 2 \times 3.1415 \times 8 \times 4$
$= 201.06\,{\text{sq}}{\text{.cm}}$
Now ${r_1} = \sqrt {{{\left( {OB} \right)}^2} - {{\left( {OE} \right)}^2}}$
$= \sqrt {{8^2} - {2^2}} \,\,\,\, = \sqrt {60}$
${r_2} = \sqrt {{{\left( {OD} \right)}^2} - {{\left( {OF} \right)}^2}}$
$= \sqrt {{8^2} - {6^2}} \,\,\,\, = \sqrt {28}$
$\therefore$ $V = \frac{{\pi h}}{6}\left( {{h^2} + 3{r_1}{h^2} + 3{r_2}{h^2}} \right)$
$= 586.43\,{\text{cu}}{\text{.cm}}$
Example:
A stone was rolled into a hemispherical basin 8cm in diameter having a $3\frac{1}{2}{\text{m}}$ depth of water in it, when the water immediately rose to the top of the basin. What was the cubic content of the stone?
Solution:
Let the figure represent a vertical mid-section of the basin and stone. Let $DE$ and $AB$ indicate the level of the water before and after the stone rolled into the basin. Then
$GC = 3\frac{1}{2}{\text{m}},\,\,\,FG = \frac{1}{2}{\text{m}},\,\,\,FB = 4{\text{m}}$
If $GE = x{\text{m}}$, then
${x^2} = 3\frac{1}{2} \times 4\frac{1}{2}\,\,\,\, = \sqrt {\frac{{63}}{4}} {\text{m}}$
$\Rightarrow$ $x = \frac{{\sqrt {63} }}{2}{\text{m}}$
Now, the cubical content of the stone:
$= \,{\text{the cubical content of the water displaced}}$
$= \,{\text{the cubical content of the same ABED}}$
$\frac{{\pi \left( {\frac{1}{2}} \right)}}{6}\left[ {3\left( {\frac{{63}}{4} + 16} \right) + {{\left( {\frac{1}{2}} \right)}^2}} \right]{\text{cu}}{\text{. m}}$
$= \frac{\pi }{{12}} \times \frac{{121}}{2} = 25\,{\text{cu}}{\text{.m}}$ nearly.
Segment of a Sphere
For a special segment of one base, the radius of the lower base ${r_1}$ is equal to zero. Therefore,
$V = \frac{{\pi h}}{6}\left( {{h^2} + 3{r_2}^2} \right)$
$\therefore$ In this case, the total surface area of the segment $= 2\pi rh + \pi {r^2}$
Example:
Find the volume of a segment of a sphere whose height is $4\frac{1}{2}{\text{cm}}$ and the diameter of whose base is 8cm.
Solution:
Given that:
$h = 4\frac{1}{2}{\text{cm}},\,\,\,\,\,{r_1} = 4{\text{cm}}$
$\therefore$ The volume of the segment $= \frac{{\pi h}}{6}\left( {{h^2} + 3{r_2}^2} \right)$
$= \frac{{\pi \times 9}}{{2 \times 6}}\left[ {{{\left( {\frac{9}{2}} \right)}^2} + 3 \times {4^2}} \right]$
$= \frac{{22 \times 9 \times 27 \times 3}}{{7 \times 2 \times 6 \times 4}}$
$= 160.8\,{\text{cu}}{\text{.cm}}$
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Multiplying polynomials
Multiplying polynomials is easy enough, but it can get a bit messy. Especially when dealing with a few variables. But if you know what you are doing, you will manage quite nicely. So let us get down to business.
There is a simple logic behind multiplying polynomials – just multiply every term in the first polynomial with every term of the second polynomial. After that, just tidy up the remaining terms by performing the necessary mathematical operations, such as addition and subtraction. Apart from that, all other rules for multiplication and the order of operations still apply and they should be observed.
Example 1.
So, if you have a polynomial like this:
(4m + 3) * (3m – 2m)
…the first thing you should do is to multiply the terms from the first polynomial with each term in the second one. The process looks like this:
4m*3m + 4m *(-2m) + 3*3m + 3*(-2m)
12m2 – 8m2 + 9m – 6m
After a bit of tidying up, it should look like this:
4m2 + 3m
And that is it. That is the process of multiplying polynomials. It is easy, right? Now we are going to solve a bit more complicated example to show you how to deal with the clutter that appears in these cases.
Example 2.
Let us assume that we have to simplify the product of these polynomials.
(-a + 3b) * (-a2 + ab + 3b2)
Again, we have to start by multiplying each term from the first polynomial with the terms in the second one.
(-a)*(-a2) + (-a)*ab + (-a)*3b2 + 3b2*(-a2) + 3b*ab + 3b*3b2
a3 – a2b – 3ab2 – 3a2b + 3ab2 + 9b3
Now it is time to perform the addition and subtraction to get this mathematical expression in order. Keep in mind that these operations can only be performed with terms whose variables are exactly the same. We will rewrite this expression in a way that these variables are next to each other. So, we get something like this:
a3 – a2b – 3a2b – 3ab2 + 3ab2 + 9b3
We can leave out the two terms that have the same value, but opposite signs since their sum is 0. That means the result of our simplification is:
a3 – 4a2b + 9b3
As you can see, even the most complicated examples are not that difficult to solve as viagra safe. However, a considerable amount of concentration is required because mistakes can happen pretty easily. When you deal with multiplying polynomials, be sure to check your calculations before going further with an assignment. It is worth the extra effort.
So, this is all there is to multiplying polynomials. They can get more complicated by adding more variables or extra polynomials, but if you follow these basic rules and focus on your calculations, you can solve them all. If you wish to practice multiplying polynomials, feel free to use the worksheets below.
Multiplying polynomials exams for teachers
Exam Name File Size Downloads Upload date Single variable – Integers Multiplication of polynomials – Integers with a single variable – very easy 557.8 kB 3987 September 3, 2019 Multiplication of polynomials – Integers with a single variable – easy 556.4 kB 2826 September 3, 2019 Multiplication of polynomials – Integers with a single variable – medium 580.7 kB 3937 September 3, 2019 Multiplication of polynomials – Integers with a single variable – hard 567.1 kB 3409 September 3, 2019 Multiplication of polynomials – Integers with a single variable – very hard 589.3 kB 3954 September 3, 2019 Single variable – Decimals Multiplication of polynomials – Decimals with a single variable – very easy 570.4 kB 1658 September 3, 2019 Multiplication of polynomials – Decimals with a single variable – easy 573.5 kB 1432 September 3, 2019 Multiplication of polynomials – Decimals with a single variable – medium 610.6 kB 1640 September 3, 2019 Multiplication of polynomials – Decimals with a single variable – hard 600.3 kB 1587 September 3, 2019 Multiplication of polynomials – Decimals with a single variable – very hard 642.1 kB 1590 September 3, 2019 Single variable – Fractions Multiplication of polynomials – Fractions with a single variable – very easy 579.5 kB 1748 September 3, 2019 Multiplication of polynomials – Fractions with a single variable – easy 587.8 kB 1619 September 3, 2019 Multiplication of polynomials – Fractions with a single variable – medium 626.1 kB 1664 September 3, 2019 Multiplication of polynomials – Fractions with a single variable – hard 633.1 kB 1669 September 3, 2019 Multiplication of polynomials – Fractions with a single variable – very hard 680.5 kB 1567 September 3, 2019 Two variables – Integers Multiplication of polynomials – Integers with two variables – very easy 560 kB 2481 September 3, 2019 Multiplication of polynomials – Integers with two variables – easy 559.8 kB 1980 September 3, 2019 Multiplication of polynomials – Integers with two variables – medium 593.7 kB 2779 September 3, 2019 Multiplication of polynomials – Integers with two variables – hard 583.5 kB 2385 September 3, 2019 Multiplication of polynomials – Integers with two variables – very hard 618.7 kB 2393 September 3, 2019 Two variables – Decimals Multiplication of polynomials – Decimals with two variables – very easy 582.8 kB 1407 September 3, 2019 Multiplication of polynomials – Decimals with two variables – easy 581 kB 1446 September 3, 2019 Multiplication of polynomials – Decimals with two variables – medium 625.3 kB 1289 September 3, 2019 Multiplication of polynomials – Decimals with two variables – hard 624.3 kB 1815 September 3, 2019 Multiplication of polynomials – Decimals with two variables – very hard 670.5 kB 1566 September 3, 2019 Two variables – Fractions Multiplication of polynomials – Fractions with two variables – very easy 591 kB 1503 September 3, 2019 Multiplication of polynomials – Fractions with two variables – easy 599.3 kB 1297 September 3, 2019 Multiplication of polynomials – Fractions with two variables – medium 643.3 kB 1437 September 3, 2019 Multiplication of polynomials – Fractions with two variables – hard 640.9 kB 1437 September 3, 2019 Multiplication of polynomials – Fractions with two variables – very hard 689.5 kB 1650 September 3, 2019
Multiplying polynomials worksheets for students
Worksheet Name File Size Downloads Upload date Single variable – Integers Integers – Simplify product of monomials and binomials 140.3 kB 3175 September 3, 2019 Integers – Simplify product of monomials and trinomials 199.7 kB 2429 September 3, 2019 Integers – Simplify product of binomials 176.1 kB 3612 September 3, 2019 Integers – Simplify product of binomials and trinomials 260.4 kB 3323 September 3, 2019 Single variable – Decimals Decimals – Simplify product of monomials and binomials 148.8 kB 1624 September 3, 2019 Decimals – Simplify product of monomials and trinomials 213.1 kB 1560 September 3, 2019 Decimals – Simplify product of binomials 188.2 kB 1583 September 3, 2019 Decimals – Simplify product of binomials and trinomials 275.5 kB 1435 September 3, 2019 Single variable – Fractions Fractions – Simplify product of monomials and binomials 274.9 kB 1544 September 3, 2019 Fractions – Simplify product of monomials and trinomials 397.9 kB 1368 September 3, 2019 Fractions – Simplify product of binomials 372.3 kB 1679 September 3, 2019 Fractions – Simplify product of binomials and trinomials 2.2 MB 1566 September 3, 2019 Two variables – Integers Integers – Simplify product of monomials and binomials 167.8 kB 3270 September 3, 2019 Integers – Simplify product of monomials and trinomials 257.7 kB 3591 September 3, 2019 Integers – Simplify product of binomials 241 kB 4916 September 3, 2019 Integers – Simplify product of binomials and trinomials 327.8 kB 5130 September 3, 2019 Single variable – Decimals Decimals – Simplify product of monomials and binomials 179.2 kB 1543 September 3, 2019 Decimals – Simplify product of monomials and trinomials 269.3 kB 1630 September 3, 2019 Decimals – Simplify product of binomials 252.8 kB 1540 September 3, 2019 Decimals – Simplify product of binomials and trinomials 346.3 kB 1306 September 3, 2019 Single variable – Fractions Fractions – Simplify product of monomials and binomials 337.2 kB 1642 September 3, 2019 Fractions – Simplify product of monomials and trinomials 483.9 kB 1631 September 3, 2019 Fractions – Simplify product of binomials 446 kB 1560 September 3, 2019 Fractions – Simplify product of binomials and trinomials 817.7 kB 1629 September 3, 2019
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# Difference of two
squares
## In mathematics, the difference of two
squares is a squared (multiplied by itself)
number subtracted from another squared
number. Every difference of squares may
be factored according to the identity
in elementary algebra.
Proof
Proof
The proof of the factorization identity is
straightforward. Starting from the left-
hand side, apply the distributive law to
get
terms cancel:
leaving
## The resulting identity is one of the most
commonly used in mathematics. Among
many uses, it gives a simple proof of the
AM–GM inequality in two variables.
The proof just given indicates the scope
of the identity in abstract algebra: it will
hold in any commutative ring R.
## Conversely, if this identity holds in a ring
R for all pairs of elements a and b of the
ring, then R is commutative. To see this,
apply the distributive law to the right-
hand side of the original equation and
get
## and for this to be equal to , we
must have
for all pairs a, b of elements of R, so the
ring R is commutative.
Geometrical demonstrations
The difference of two
squares can also be
illustrated geometrically as
the difference of two square
areas in a plane. In the diagram, the
between the areas of the two squares,
i.e. . The area of the shaded part
can be found by adding the areas of the
two rectangles; ,
which can be factorized to
. Therefore
## Another geometric proof proceeds as
the first diagram below, a large square
with a smaller square removed from it.
The side of the entire square is a, and the
side of the small removed square is b.
The area of the shaded region is
. A cut is made, splitting the region into
two rectangular pieces, as shown in the
second diagram. The larger piece, at the
top, has width a and height a-b. The
smaller piece, at the bottom, has width a-
b and height b. Now the smaller piece
can be detached, rotated, and placed to
the right of the larger piece. In this new
arrangement, shown in the last diagram
below, the two pieces together form a
rectangle, whose width is and
whose height is . This rectangle's
area is . Since this
rectangle came from rearranging the
original figure, it must have the same
area as the original figure. Therefore,
.
Uses
Factorization of polynomials
and simplification of
expressions
## The formula for the difference of two
squares can be used for factoring
polynomials that contain the square of a
first quantity minus the square of a
second quantity. For example, the
polynomial can be factored as
follows:
## As a second example, the first two terms
of can be factored as
, so we have:
Moreover, this formula can also be used
for simplifying expressions:
of two squares
## The difference of two squares is used to
find the linear factors of the sum of two
squares, using complex number
coefficients.
## For example, the complex roots of
can be found using difference of
two squares:
(since )
Therefore the linear factors are
and .
## Since the two factors found by this
method are complex conjugates, we can
use this in reverse as a method of
multiplying a complex number to get a
real number. This is used to get real
denominators in complex fractions.[1]
Rationalising denominators
## The difference of two squares can also
be used in the rationalising of irrational
denominators.[2] This is a method for
removing surds from expressions (or at
least moving them), applying to division
by some combinations involving square
roots.
## Here, the irrational denominator
has been rationalised to .
Mental arithmetic
## The difference of two squares can also
be used as an arithmetical short cut. If
you are multiplying two numbers whose
average is a number which is easily
squared the difference of two squares
can be used to give you the product of
the original two numbers.
For example:
## Which means using the difference of two
squares can be restated as
which is .
perfect squares
## The difference of two consecutive
perfect squares is the sum of the two
bases n and n+1. This can be seen as
follows:
Therefore the difference of two
consecutive perfect squares is an odd
number. Similarly, the difference of two
arbitrary perfect squares is calculated as
follows:
## Therefore the difference of two even
perfect squares is a multiple of 4 and the
difference of two odd perfect squares is
a multiple of 8.
Generalizations
Vectors a (purple), b (cyan) and a + b (blue) are
shown with arrows
## The identity also holds in inner product
spaces over the field of real numbers,
such as for dot product of Euclidean
vectors:
## The proof is identical. By the way,
assuming that a and b have equal norms
(which means that their dot squares are
equal), it demonstrates analytically the
fact that two diagonals of a rhombus are
perpendicular.
## If a and b are two elements of a
commutative ring R, then
.
## Note that binomial coefficients do not
appear in the second factor, and the
summation stops at n-1, not n.
Congruum, the shared difference of
three squares in arithmetic
progression
Conjugate (algebra)
Factorization
Notes
1. Complex or imaginary numbers
TheMathPage.com, retrieved 22
December 2011
TheMathPage.com, retrieved 22
December 2011
References
James Stuart Stanton: Encyclopedia of
Mathematics. Infobase Publishing,
2005, ISBN 9780816051243, p. 131
(online copy )
Alan S. Tussy, Roy David Gustafson:
Elementary Algebra, 5th ed.. Cengage
Learning, 2011, ISBN 9781111567668,
pp. 467 - 469 (online copy )
difference of two squares at
mathpages.com
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# Evaluating the sum of geometric series
Possible Duplicate:
Value of $\sum\limits_n x^n$
I’m trying to understand how to evaluate the following series:
$$\sum_{n=0}^\infty {\frac{18}{3^n}}.$$
I tried following this Wikipedia Article without much success. Mathematica outputs 27 for the sum.
If someone would be kind enough to show me some light or give me an explanation I would be grateful.
#### Solutions Collecting From Web of "Evaluating the sum of geometric series"
Let’s assume that the series converges, and let
$$S=\sum_{n=0}^\infty\frac{18}{3^n}=\sum_{n=0}^\infty\frac{18}{3^n}=\sum_{n=0}^\infty18\left(\frac13\right)^n=\color{blue}{18\left(\frac13\right)^0}+\color{red}{18\left(\frac13\right)^1+18\left(\frac13\right)^2+18\left(\frac13\right)^3+\ldots}\;.$$
Multiply by $\frac13$:
\begin{align*} \frac13S&=\frac13\left(18\left(\frac13\right)^0+18\left(\frac13\right)^1+18\left(\frac13\right)^2+18\left(\frac13\right)^3+\ldots\right)\\ &=\color{red}{18\left(\frac13\right)^1+18\left(\frac13\right)^2+18\left(\frac13\right)^3+18\left(\frac13\right)^4+\ldots}\\ &=S-\color{blue}{18\left(\frac13\right)^0}\\ &=S-18\;. \end{align*}
Now solve the equation $\frac13S=S-18$: $\frac23S=18$, and $S=\frac32\cdot18=27$. Similar reasoning works whenever the series converges. It’s cheating a bit, though, because justifying the assumption that $S$ exists requires being able to sum the finite series $\sum_{n=0}^m\frac{18}{3^n}$ for arbitrary $m\in\Bbb N$.
Of course once you know the general formula $$\sum_{n=0}^\infty ar^n=\frac{a}{1-r}$$ when $|r|<1$, you merely observe (as I did in the first calculation) that in the sum $\displaystyle\sum_{n=0}^\infty\frac{18}{3^n}$ the terms have the form $18\left(\dfrac13\right)^n$, so $a=18$ and $r=\dfrac13$, and the formula yields
$$S=\frac{18}{1-\frac13}=\frac{18}{2/3}=27\;.$$
It is indeed a geometric series:
$$\sum_{n=0}^\infty \frac{18}{3^n}=18\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n.$$
Can you take it from here?
$$\sum^{\infty}_{n=0} \frac{18}{3^n} = 18 \sum^{\infty}_{n=0} \frac{1}{3^n} = \frac{18}{1-1/3} = 27$$
Where the last equality follows from the fact that $\sum^{\infty}_{n=0} x^n =\frac{1}{1-x}$ if $|x| < 1$.
The sum of a geometric series is
$$\sum_{k=1} ^\infty ar^k = \frac{a}{1-r}$$
where $r<1$, as the Wikipedia article says.
now simply plug in the numbers in your case, $a=18, r=\frac{1}{3}$. and you’ll get
$$\frac{18}{1-\frac{1}{3}} = \frac{18}{\frac{2}{3}} = 27$$
$r=\text{common ratio}=1/3$.
$a=\text{first term} = 18$.
$$\text{sum} = \frac{a}{1-r} = \frac{18}{1-(1/3)} = 27.$$
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# The math tutor recommends a little light factoring on this beautiful Sunday morning….
Last post I discussed common factor, which we will be using in concert with difference of squares. Difference of squares factors x2 – 36 into (x + 6)(x – 6). By the foil method you can confirm:
F: x*x=x2
O: x*-6=-6x
I: 6*x=6x
L: 6*-6=-36
Displaying the terms in a row we get
x2 – 6x + 6x – 36 = x2 – 36
Example: Factor x2 – 49
Solution: We notice that the square root of 49 is 7. Therefore we write
x2 – 49 = (x + 7)(x – 7)
Difference of squares is easy to spot and factor if it’s plain. However, it may be “hidden” by a common factor:
Example: Factor 2x3 – 50x
Solution: We notice that 50x isn’t square rootable. However, we also notice that 2x can be taken out front as a common factor:
2x3 – 50x = 2x(x2 – 25)
Now, we’re getting somewhere: we follow with
2x(x2 – 25) = 2x(x + 5)(x – 5)
Removing the common factor of 2x allowed us to apply the difference of squares technique.
Have a nice day:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# As a math tutor, you notice the importance of this technique.
Factoring means breaking a number or expression into a product. For instance, we’ll factor 45:
45=9×5
In earlier posts I’ve mentioned prime factorization:
45=3x3x5
Now we’ll look at factorization of polynomials using common factor.
Example: factor -2x6 + 8x5-12x2
Solution: With the common factor method, we look for the expression that divides into all the terms, then write it out front. What remains in the brackets is each term divided by the common factor.
In this case we notice that 2x2 divides into all the terms. Therefore, we “take it out front”. Actually, we take out -2x2 because whenever the lead term is negative, you take out the negative with the common factor. Inside the brackets we write each term divided by -2x2:
-2x2(x4 -4x3+6)
Common factoring doesn’t have to be as complicated as the example above. Consider the following:
3x – 15 factors to 3(x – 5)
Working with polynomials, factoring is constantly used. There are at least five factoring techniques, of which common factor is the first. I’ll discuss the other techniques in future posts:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# As a math tutor, you’ll likely introduce this concept. It’s used even more in physics and chemistry.
So often in my university science courses I’d read “the mass is directly proportional to the volume” or “the distance is directly proportional to the time”, etc. Science people love direct proportionality because predicting the result of a given input is so easy.
Direct proportionality means that if you double the input, the output will also double. If y is directly proportional to x, it follows that
y=kx
In the above equation, is called the constant of proportionality. Once you know k, you can find the result of any input.
Example: The distance travelled by a long haul train is directly proportional to the time traveled. The train travels 600 km in 9 hours.
a) Find the equation to model this situation.
b) How far will the train travel in 12 hours?
Solution: y is always the “output”, while x is the “input”. Some people like to use different letters in order to reflect the actual wording of the question. In that case:
(1)
where, of course, d stands for distance, t for time.
To find k, we use the idea that the train travels 600 km in 9 hours:
(2)
Dividing both sides by 9, we get
(3)
(4)
We know now that the equation to model the train’s travel is
(5)
To predict the train’s distance over 12 hours, we simply put 12 in for t:
(6)
Therefore,
(7)
We conclude that over 12 hours, the train will travel 800 km.
More will be said about direct proportionality in future posts:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# Tutoring biology, you need to be aware of the connection between ATP and energy.
In a factory or a mill, there likely is a power plant where fuel is burnt en masse. The energy is captured there (often in the form of electricity), then channeled to the other locations as needed.
The human body, though, releases and consumes energy in a different way. Each cell receives fuel (glucose), burns it internally, then captures the released energy by using it to synthesize a high energy chemical bond. That bond can be broken at will when energy is needed.
ADP is adenosine diphosphate (adenosine bonded to two phosphate groups). When the cell burns glucose (in its mitochondria), the energy released is used to bond another phosphate to the ADP, so it becomes ATP (adenosine triphosphate).
When the cell needs energy, it breaks an ATP into an ADP and a phosphate. The breaking of that bond releases energy the cell can use to power any life process. Later, when the cell burns more glucose, it will use the energy to bond the ADP and the phosphate back into ATP.
The burning of one glucose molecule produces 36 or 38 ATP.
Hope this helps:)
Source: Inquiry into Life, Eleventh Edition, Sylvia S. Mader. McGraw-Hill: 2006.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# The math tutor continues to appreciate prime factorization for all it yields.
Let’s imagine you need to determine the square root of a number without a calculator. This challenge is part of the curriculum for local high school students.
Example: Determine if each number can be square rooted (to a whole number). If so, find its square root.
a) 540
b) 576
Before tackling the above problem, let’s dissect a number we know to be a perfect square.
Example: Confirm, by prime factorization, that the square root of 900 is 30.
Solution: We recall that a prime number is one that cannot be divided into smaller numbers, then break 900 down into primes:
900=10×90=(2×5)(9×10)=(2×5)(3x3x2x5)=2x5x3x3x2x5
Rearranging, we get
900=2x2x3x3x5x5=(2x3x5)(2x3x5)
We notice 900 can be broken into two identical groupings like so:
900 = (2x3x5)(2x3x5)
Therefore, the square root of 900 is 2x3x5=30
We now know what to seek: if a number is square rootable, its prime factorization can be organized into two equal groups. The square root is simply the product of one of the groups.
Back to our example:
Determine the whole number square root (if it exists) of the following:
a) 540
b) 576
Solution:
a) First we break 540 into primes:
540=10×54=(2×5)(6×9)=(2×5)(2x3x3x3)=2x2x3x3x3x5
With only one 5 in the prime factorization, we can’t separate it into two equal groups. 540 doesn’t have a whole number square root.
b) 2 and 4 both go into 576. Without a calculator, you either do it mentally or else use long division. To get started, just break it in half:
576=2×288=2(2×144)=2(2x12x12)=2(2x(3×4)(3×4))=2x2x3x4x3x4
Since we have only multiplication here, we can add and rearrange brackets at will. However, with mixed operations we wouldn’t be able to do so:)
Rearranging, we get
576=2x2x3x3x4x4=(2x3x4)(2x3x4)
Clearly, the prime factorization of 576 is separable into two equal groupings of 2x3x4. 2x3x4 = 24, so the square root of 576 is 24.
If its prime factorization can be separated into three equal groupings, the number is a perfect cube:
Example: Confirm that 9261 is a perfect cube.
Solution: We’ll break this one down using short division. Since 9+2+6+1=18, we know 9 divides into it:
Since 1+0+2+9=12, we know 3 divides into it (because 3 divides into 12).
Since 3+4+3=10 (which 3 doesn’t divide into), and 343 doesn’t end in 5 or 0, the next number to try is 7:
So we see that we can break down 9261 as follows:
9261=9(1029)=9(3×343)=9(3x7x49)=(3×3)(3x7x7x7)=3x3x3x7x7x7
Rearranging, we separate the prime factorization into three equal groupings:
9261=3x3x3x7x7x7=(3×7)(3×7)(3×7)
Therefore, 9261 is a perfect cube with cube root=3×7=21. The cube root of 9261 is 21.
Once again, short division was used to break down 9261. For more explanation about that very handy technique, please check future posts:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# As a math tutor, you deal with polynomials half the weeks of the year.
A polynomial is an expression in which the variables can have only positive, whole-number exponents. Examples of polynomials are 3x7-12x5+1 or -2x – 12.
In a polynomial, terms are separated by plus or minus signs. Therefore, 7x – 12 has two terms, whereas -13x7yz2 has only one.
Polynomials are often named by how many terms they have, as follows:
7x2 One term: monomial
3x – 12 Two terms: binomial
-4x4 + 3x -5 Three terms: trinomial
A polynomial with more than three terms is just called a polynomial.
Consider the following monomial:
-11x5
-11 is called the coefficient. x is called the variable. 5 is the exponent.
The degree of a polynomial is the highest exponent found in one of its terms. For example, 5x3 has degree 3. The trinomial
x7 + 3x3 – 12
has degree 7.
The constant term of a polynomial is the term with no variable attached. In 3x – 12, the constant term is -12.
There are two facts about polynomials that might be a little surprising:
Fact 1:
A constant term has degree zero. Reason: x0= 1 by definition. The result:
3x0 = 3(1) = 3.
We just write 3, but the degree of the term is still zero.
Fact 2:
If the coefficient is not written, it is 1. Reason: x5 = 1x5.
There are a couple of finer points, but the above is good for a start. Much more will be said about polynomials in future posts.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
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The Number of Elements in a Left (Right) Coset
# The Number of Elements in a Left (Right) Coset
Recall from the Left and Right Cosets of Subgroups page that if $(G, \cdot)$ is a group, $(H, \cdot)$ is a subgroup, and $g \in G$ then the left coset of $H$ with representative $g$ is defined as:
(1)
\begin{align} \quad gH = \{ gh : h \in H \} \end{align}
The right coset of $H$ with representative $g$ is defined as:
(2)
\begin{align} \quad Hg = \{ hg : h \in H \} \end{align}
We will now look at a rather simple theorem which will tell us that the number of elements in a left (or right coset) will equal to the number of elements in $H$. This seems rather obvious since $gH$ contains elements of the form $gh$ where we range through all of $h$. So $gH$ has at most the same number of elements in $H$. Of course, $gH$ may have less elements if $gh_1 = gh_2$ for distinct $h_1, h_2 \in H$. Of course this cannot be since by cancellation we would then have that $h_1 = h_2$. We make this argument more rigorous below.
Theorem 1: Let $(G, \cdot)$ be a group, $(H, \cdot)$ a subgroup, and let $g \in G$. Then the number of elements in $gH$ equals the number of elements in $H$, i.e., $\mid gH \mid = \mid H \mid$. Similarly, $\mid Hg \mid = \mid H \mid$.
We only prove the case of this theorem for left cosets. The case for right cosets is analogous.
• Proof: Define a function $f : H \to gH$ for all $h \in H$ by:
(3)
\begin{align} \quad f(h) = gh \end{align}
• We will show that $f$ is bijective. First we show that $f$ is injective. Let $h_1, h_2 \in H$ and assume that $f(h_1) = f(h_2)$. Then we have that:
(4)
\begin{align} \quad gh_1 = gh_2 \end{align}
• By left cancellation this implies that $h_1 = h_2$ and so $f$ is injective. We now show that $f$ is surjective. Let $gh \in gH$. Then $f(h) = gh$ (somewhat trivially) so $f$ is surjective.
• Since $f$ is bijective we have that $\mid H \mid = \mid gH \mid$ so the number of elements in the left coset $gH$ is equal to the number of elements in $H$. $\blacksquare$
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# Multiplication of Rational Numbers
Multiply fractions: multiply straight across
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Practice Multiplication of Rational Numbers
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Multiplication of Real Numbers
Jacob received tips of $4.00 each from three of his paper route customers. How much did he receive in total? ### Watch This ### Guidance Multiplication of two integers with the same signs produces a positive result and multiplication of two integers with unlike signs results in a negative answer. These rules can be applied to the multiplication of all real numbers. To multiply fractions, you multiply the numerators and then you multiply the denominators. The product of the numerators over the product of the denominators is the answer to the problem. Sometimes the answer can be expressed as an equivalent fraction. The rules for multiplying integers also apply to multiplying decimals. The sum of the number of digits after the decimal points determines the placement of the decimal point in the answer. #### Example A Sam spent$2.00 for a bottle of chocolate milk at the school cafeteria every school day. At the end of the week, how does this affect his net worth?
Solution: The result of $(+5) \times (-2)$ is –10. The product of a positive integer and a negative integer is always negative.
#### Example B
What is $(-2) \times (-3)$ ?
Solution: The result of $(-2) \times (-3)$ is +6. The product of two negative integers is always positive.
#### Example C
i) $\left(\frac{2}{3}\right) \times \left(\frac{5}{7}\right)$
ii) $\left(\frac{7}{8}\right) \times \left(3 \frac{3}{4}\right)$
iii) $\left(5 \frac{3}{4} \right) \times \left(2 \frac{3}{5}\right)$
Solution: Remember, there are three simple steps to follow to multiply fractions:
1. Multiply the numerators of the fractions
2. Multiply the denominators of the fractions.
3. Simplify the fraction if necessary.
i) $\left(\frac{2}{3}\right) \times \left(\frac{5}{7}\right)$
$&= \frac{2 \times 5}{3 \times 7}\\&= \frac{10}{21}$
ii) $\left(\frac{7}{8}\right) \times \left(3 \frac{3}{4}\right)$ Express the mixed number as an improper fraction.
$&= \left(\frac{7}{8}\right) \times \left(\frac{15}{4}\right) \rightarrow \frac{(4 \times 3)+3}{4}\\&= \frac{7 \times 15}{8 \times 4}\\&= \frac{105}{32}=3 \frac{9}{32}$
iii) $\left(5 \frac{3}{4}\right) \times \left(2 \frac{3}{5}\right)$ Express the mixed numbers as improper fractions.
$&= \left(\frac{23}{4}\right) \times \left(\frac{13}{5}\right) \rightarrow \frac{(4 \times 5)+3}{4} \ \text{and} \ \frac{(5 \times 2)+3}{5}\\&= \frac{23 \times 13}{4 \times 5}\\&= \frac{299}{20}=14 \frac{19}{20}$
#### Example D
$(14.65) \times (2.7)$
Solution: Multiply the numbers as you would whole numbers. To place the decimal point in the answer, count the number of digits after the decimal points in the problem. There are two digits after the decimal point in 14.65 and one digit after the decimal point in 2.7. This is a total of three digits after the decimal points. From the right of the answer, count three places to the left and insert the decimal point.
$& 14.65\\& \underline{\times \; 2.7 \;\;}\\& \ \ 10255\\& \underline{+29300}\\& \ \ \underset{\quad \ {\color{red}\longleftarrow}}{39 {\color{red}.} 555}$
Jacob received tips of $4.00 each from three of his paper route customers. How much did he receive in total? The result of $(+3) \times (+4)$ is +12. The product of two positive integers is always positive. ### Guided Practice Multiply the following fractions: 1. $\left(\frac{5}{9}\right) \times \left(\frac{-4}{7}\right)$ 2. $\left(3\frac{2}{3}\right) \times \left(4 \frac{1}{5}\right)$ 3. Determine the answer to $(-135.697) \times (-34.32)$ Answers: 1. Multiply the numerators. Multiply the denominators. Simplify the fraction. $& \left(\frac{5}{9}\right) \times \left(\frac{-4}{7}\right)=\frac{5 \times (-4)}{9 \times 7}=-\frac{20}{63}$ The answer can be written as $\frac{-20}{63}$ or $-\frac{20}{63}$ . 2. Write the two mixed numbers as improper fractions. Multiply the denominator and the whole number. Add the numerator to this product. Write the answer over the denominator. Follow the steps for multiplying fractions. Simplify the fraction if necessary. $& \left(3 \frac{2}{3}\right) \times \left(4 \frac{1}{5}\right)\\& \left(\frac{11}{3}\right) \times \left(\frac{21}{5}\right)\\& \left(\frac{11}{3}\right) \times \left(\frac{21}{5}\right)=\frac{231}{15}=15 \frac{2}{5}$ 3. Multiply the numbers as you would whole numbers. Remember the rule for multiplying integers. When you multiply two integers that have the same sign, the product will always be positive. $& \ \ -135.697\\& \underline{\times \; -34.32 \;\;\;\;}\\& \quad \quad \ \ 271394\\& \quad \quad \ 407091 {\color{blue}0}\\& \quad \ \ 542788 {\color{blue}00}\\& \underline{\;\;\;\; 407091 {\color{blue}000} \;}\\& \quad \underset{\quad \ {\color{red}\longleftarrow}}{4657{\color{red}.}12104}$ There are three digits after the decimal point in 135.697 and two digits after the decimal point in 34.32. Beginning at the right of the product, count five places to the left and insert the decimal point. ### Explore More Multiply. 1. $(-7) \times (-2)$ 2. $(+3) \times (+4)$ 3. $(-5) \times (+3)$ 4. $(+2) \times (-4)$ 5. $(+4) \times (-1)$ Match each given phrase with the correct multiplication statement. Then, determine each product. 1. take away six groups of 3 balls 2. net worth after losing seven$5 bills
3. take away nine sets of 8 forks
4. take away four sets of four plates
5. receive eight groups of 4 glasses
6. buy seven sets of 12 placemats
1. $(+8) \times (+4)$
2. $(+7) \times (-5)$
3. $(-4) \times (+4)$
4. $(-9) \times (+8)$
5. $(+7) \times (+12)$
6. $(-6) \times (+3)$
Use the rules that you have learned for multiplying real numbers to answer the following problems.
1. $(-13) \times (-9)$
2. $(-3.68) \times (82.4)$
3. $\left(\frac{4}{9}\right) \times \left(\frac{5}{7}\right)$
4. $\left(7 \frac{2}{3} \right) \times \left(6 \frac{1}{2}\right)$
5. $(15.734) \times (-8.1)$
### Vocabulary Language: English
Associative Property
Associative Property
The associative property states that you can change the groupings of numbers being added or multiplied without changing the sum. For example: (2+3) + 4 = 2 + (3+4), and (2 X 3) X 4 = 2 X (3 X 4).
Commutative Property
Commutative Property
The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example $a+b=b+a \text{ and\,} (a)(b)=(b)(a)$.
Denominator
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
distributive property
distributive property
The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
fraction
fraction
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
improper fraction
improper fraction
An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator.
Integer
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Mixed Number
Mixed Number
A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
multiplicative identity property
multiplicative identity property
The product of any number and one is the number itself.
Numerator
Numerator
The numerator is the number above the fraction bar in a fraction.
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# GMAT Tip: Strategies for Data Sufficiency Questions
This tip on improving your GMAT score was provided by Vivian Kerr at Veritas Prep.
Data Sufficiency is a unique question type and has the potential to look quite abstract, but as you’ll see, half the battle is making these questions concrete. These two strategies will help:
Count the Variables; Don’t Do the Math
Remember the “n equations with n variables” rule: If you have three unknowns, you’ll need three equations to solve for all three. If you only need the sum of two of those variables, however, you may need only two equations to solve. Let’s take a look at how this strategy can help us find a shortcut.
A total of 2,000 T-shirts was divided among a soccer team, two baseball teams, and a track team. How many shirts did the track team receive?
(1) The track team and one of the baseball teams together received 5/7 as many shirts as the other baseball team and the soccer team combined; and the two baseball teams each received the same number of shirts.
(2) Each baseball team received 400 fewer shirts than the soccer team and 400 more than the track team.
This is a value question for which we have four unknowns: the soccer shirts, baseball team A shirts, baseball team B shirts, and track team shirts. Statement (1) doesn’t allow us to solve for these four unknowns.
For (2), we’re given the relationship between one unknown (the soccer team) and the other three unknowns. That would allow us to choose a variable for the soccer team T-shirts and write the other unknowns with that same variable. We don’t even have to do any math to see that this is sufficient.
If we wanted to try this out algebraically, we could pick “x” for the number of shirts given out to the baseball teams. Because the baseball team received 400 more than one team and 400 less than another, that makes the other two totals (x + 400) and (x – 400). Since a total of 2,000 shirts were given out, x + x + (x + 400) + (x – 400) = 2,000. We have a linear equation with a single variable, so we know this choice is sufficient, according to the “n equations, n variable” rule.
Data sufficiency questions come with five answer choices asking test-takers to determine if one or both of the statements, alone or in combination, are sufficient to answer the question. In this case the answer is (B): statement (2) alone is sufficient, but statement (1) alone is not.
Plug in Values
This strategy comes in handy especially for “yes or no” data sufficiency questions. Here’s an example:
Given that A > 0 and C > 0, is (A + B) / (C + B) > A/C?
(1) B > 0
(2) A
For (1), we are given only that B is a positive number. Picking values is a good strategy here. If we can choose values such that we get a “yes” and choose values so that we get a “no,” we can quickly eliminate choices. If A = 1 and C = 1 and B = 1, then the inequality is NOT correct. If, however, A = 1, C = 2, and B = 3, we get 4/5 > 1/2, which IS correct. Statement (2) can also be proved insufficient. Combined, if B is positive and AC, then it will continue to INCREASE the left-hand side of the inequality no matter what values we pick. The answer is (C): Both statements (1) and (2) together are sufficient to answer the question, but neither statement alone is sufficient.
Remember, even the most seemingly complex data sufficiency questions can be overcome with solid strategy.
Vivian Kerr has been teaching and tutoring in the Los Angeles area since 2005. She graduated from the University of Southern California, studied abroad in London, and has worked for several test-prep giants tutoring, writing content, and blogging about all things SAT, ACT, GRE, and GMAT.
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# Difference of Squares
The difference of squares formula is one of the primary algebraic formulas used to expand a term in the form of $\left({a}^{2}-{b}^{2}\right)$ . Basically, it is an algebraic form of an expression used to equate the differences between two square values. The formula helps make a complex equation into a simple one.
## Formula to calculate the difference of squares
Any polynomial that can be written as ${a}^{2}-{b}^{2}$ can be factored as the difference of a square.
$\left({a}^{2}-{b}^{2}\right)=\left(a+b\right)\left(a-b\right)$
The reason is that when you use FOIL to expand the right side, the ab terms cancel out as follows:
$\left(a+b\right)\left(a-b\right)={a}^{2}-ab+ab-{b}^{2}={a}^{2}-{b}^{2}$
Example 1
Take the equation ${12}^{2}-{8}^{2}$
Using the difference of squares formula,
${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
where
$a=12$
$b=8$
To calculate the left hand side (LHS),
${a}^{2}-{b}^{2}$
$={12}^{2}-{8}^{2}$
$=144-64$
$=80$
To calculate the right hand side (RHS),
$\left(a+b\right)\left(a-b\right)$
$=\left(12+8\right)\left(12-8\right)$
$=\left(20\right)\left(4\right)$
$=80$
$\mathrm{RHS}=\mathrm{LHS}$
This demonstrates that the formula works correctly.
Example 2
What is the value of $10{2}^{}-4{2}^{}$ ?
The formula for the difference of squares is:
${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
Therefore, from the given expression,
$a=10$
$b=4$
${a}^{2}-{b}^{2}\to {10}^{2}-{4}^{2}$
$=\left(10+4\right)\left(10-4\right)$
$=\left(14\right)\left(6\right)$
$=84$
We can double check that this is correct by performing the calculations on the original problem.
${10}^{2}-{4}^{2}$
$=100-16$
$=84$
And we've simplified the problem correctly again.
## Factoring the difference of squares with variables
Frequently in algebra problems, the point of the difference of squares identity is to factor the statement, not to find the value of an expression as in the previous example. This can be done with variables, which you will usually find in algebraic problems.
Example 3
Factor, if possible.
${x}^{2}-49$
This is a difference of squares where $a=x$ and $b=7$
${x}^{2}-{7}^{2}=\left(x-7\right)\left(x+7\right)$
Example 4
Factor, if possible.
$16{p}^{2}-64{q}^{2}$
This is a difference of squares where $a=4p$ and $b=8q$ .
$16{p}^{2}-64{q}^{2}=\left(4p-8q\right)\left(4p+8q\right)$
We can be confident that these examples are factored correctly because of the first example where we solved the entire problem, making sure that the formula works properly.
## Interesting facts about difference of squares
• The difference of squares has a geometric interpretation. When you draw a square with sides of "a" length, you can "cut out" a square from one of the corners of that square with sides of "b" length. Then you can take the rectangle that's left from the "b" cutout, turn it 90 degrees, and connect it to the "a" rectangle. You now have a rectangle with side lengths $\left(a+b\right)$ and $\left(a-b\right)$ .
• The difference of squares is helpful in solving certain algebraic equations. One of these is the quadratic equation in the form ${x}^{2}-{c}^{2}=0$ , where ${x}^{2}$ and ${c}^{2}$ are perfect squares. Using the difference of squares, we can rewrite the equation as $\left(x+c\right)\left(x-c\right)=0$ , which allows us to find the solutions $x=c$ and $x=-c$ easily.
• The difference of squares has applications in number theory, as well. For example, it can help to prove that every odd integer can be expressed as the difference between two squares. For example, the odd integer 7 can be written as ${4}^{2}-{3}^{2}=16-9=7$ . Similarly, the odd integer 9 can be written as ${5}^{2}-{4}^{2}=25-16=9$ .
## Flashcards covering the Difference of Squares
Algebra 1 Flashcards
## Get help learning about the difference of squares
Tutoring is an excellent way for your student to learn about the difference of squares. A tutor can supplement in-class teaching by providing continuous, in-the-moment feedback so your student doesn't allow bad habits to take root. Your student's tutor can also help them learn the ways in which they learn best and customize lessons based on that learning style. Not only can that help your student learn about the difference of squares, but it can help them in future studies. A tutor can break down the complex processes involved in solving difference of squares problems into simple steps that they can then follow each time they run across similar problems. Working side by side as your student completes their homework problems, their tutor is available to answer any questions that come up right away and help keep your student on task with their work. As they learn about the difference of squares, your student can also learn independent learning skills and time management from their tutor.
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# How do you factor 2d² +d -21?
Jun 8, 2015
$2 {d}^{2} + d - 21$
We can Split the Middle Term of this expression to factorise it.
In this technique, if we have to factorise an expression like $a {d}^{2} + b d + c$, we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 2 \cdot - 21 = - 42$
and
${N}_{1} + {N}_{2} = b = 1$
After trying out a few numbers we get ${N}_{1} = 7$ and ${N}_{2} = - 6$
$7 \cdot - 6 = - 42$, and $7 + \left(- 6\right) = 1$
$2 {d}^{2} + d - 21 = 2 {d}^{2} - 6 d + 7 d - 21$
$= 2 d \left(d - 3\right) + 7 \left(d - 3\right)$
$\left(d - 3\right)$ is common to both terms.
=color(green)((2d+7)(d-3)
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### Course: 6th grade>Unit 11
Lesson 5: Mean and median challenge problems
# Mean as the balancing point
Explore how we can think of the mean as the balancing point of a data distribution.
You know how to find the mean by adding up and dividing. In this article, we'll think about the mean as the balancing point. Let's get started!
## Part 1: Find the mean
Find the mean of $\left\{5,7\right\}$.
Find the mean of $\left\{5,6,7\right\}$.
Interesting! In the first two problems, the data was "balanced" around the number six. Try the next one without finding the total or dividing. Instead, think about how the numbers are balanced around the mean.
Find the mean of $\left\{1,3,5\right\}$.
Notice how the $1$ and $5$ were "balanced" on either side of the $3$:
Find the mean of $\left\{4,7,10\right\}$.
Can you see how the data points are always balanced around the mean? Let's try one more!
Find the mean of $\left\{2,3,5,6\right\}$.
## Part 2: A new way of thinking about the mean
You might have noticed in Part 1 that it's possible to find the mean without finding the total or dividing for some simple data sets.
Key idea: We can think of the mean as the balancing point , which is a fancy way of saying that the total distance from the mean to the data points below the mean is equal to the total distance from the mean to the data points above the mean.
### Example
In Part 1, you found the mean of $\left\{2,3,5,6\right\}$ to be $4$. We can see that the total distance from the mean to the data points below the mean is equal to the total distance from the mean to the data points above the mean because $1+2=1+2$:
#### Reflection questions
What is the total distance $\text{below}$ the mean in this example?
What is the total distance $\text{above}$ the mean in this example?
## Part 3: Is the mean always the balancing point?
Yes! It is always true that the total distance below the mean is equal to the total distance above the mean. It just happens to be easier to see in some data sets than others.
For example let's consider the data set $\left\{2,3,6,9\right\}$.
Here's how we can calculate the mean:
$\frac{2+3+6+9}{4}=5$
And we can see that the total distance below the mean is equal to the total distance above the mean because $2+3=1+4$:
## Part 4: Practice
### Problem 1
Which of the lines represents the mean of the data points shown below?
### Problem 2
Which of the lines represents the mean of the data points shown below?
## Challenge problem
The mean of four data points is $5$. Three of the four data points and the mean are shown in the diagram below.
Choose the fourth data point.
## Want to join the conversation?
• I do not get where the mean is located on a number line. Can someone help me i'm stuck
• the mean is a measure of central data so it will always be in the middle of the data plot. this is why your data plot should be balanced on both sides. but don't do the literal middle of the data plot because that won't always be correct. you have to consider all of the data you have.
• can someone please explain the last question to me
• You can take "mean" as a balance point between left and right, making the both side has exactly the same strength.
In this case, we can see that the energy above the mean number is (7-5)*2=4, so you need 4 energy below the mean to balance the strength. Then, below the mean, we already have (5-4) = 1 energy, therefore we still need 3 energy to make it balanced.
As a result, if you plot a dot at "2", you will have (5-2) = 3 energy.
• i get it but the bar is messing me up
• The bar is just another way of finding the mean
• the is fun
I love Khan Academy
• same
• I dont really understand this
• Probably too late but here's an example:
1 2. 3. 4. 5 6. 7. 8. 9. 10. 11.
the numbers with dots are points.
5 is the mean
the distance from 5 to 4 is one. the distance from 5 to 3 is two. the distance from 5 to 2 is three. 1+2+3=6 so the distance to the left of the mean is 6. the same on the other side.
Does that make sense?
• 42 is the answer to life, the universe, and everything.
• can you elaborate
(1 vote)
• Hi! How can I prove that in all cases the total distance below the mean is equal to the total distance above the mean?
• The statement that the total distance below the mean equals the total distance above the mean is equivalent to the statement that the sum of the directed distances from the mean is 0.
Let n be the number of data values, and let x_1, x_2, ... , x_n be the data values. Let m be the mean of these values.
We need to prove that sum j from 1 to n of (x_j - m) = 0.
By definition of mean, m = (sum j from 1 to n of x_j)/n.
So mn = sum j from 1 to n of x_j.
Therefore,
sum j from 1 to n of (x_j - m)
= (sum j from 1 to n of x_j) - (sum j from 1 to n of m)
= (sum j from 1 to n of x_j) - mn
= (sum j from 1 to n of x_j) - (sum j from 1 to n of x_j)
= 0.
Have a blessed, wonderful day!
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# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 53
$-(3x+5)(3x+1)$
#### Work Step by Step
Factoring the negative $GCF= -1 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -9x^2-18x-5 \\\\= -(9x^2+18x+5) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -(9x^2+18x+5) \end{array} has $ac= 9(5)=45$ and $b= 18 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 15,3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -(9x^2+15x+3x+5) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -[(9x^2+15x)+(3x+5)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -[3x(3x+5)+(3x+5)] .\end{array} Factoring the $GCF= (3x+5)$ of the entire expression above results to \begin{array}{l}\require{cancel} -[(3x+5)(3x+1)] \\\\= -(3x+5)(3x+1) .\end{array}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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## Fractal Basics
### Learning Outcomes
• Define and identify self-similarity in geometric shapes, plants, and geological formations
• Generate a fractal shape given an initiator and a generator
• Scale a geometric object by a specific scaling factor using the scaling dimension relation
• Determine the fractal dimension of a fractal object
Fractals are mathematical sets, usually obtained through recursion, that exhibit interesting dimensional properties. We’ll explore what that sentence means through the rest of the chapter. For now, we can begin with the idea of self-similarity, a characteristic of most fractals.
### Self-similarity
A shape is self-similar when it looks essentially the same from a distance as it does closer up.
Self-similarity can often be found in nature. In the Romanesco broccoli pictured below[1], if we zoom in on part of the image, the piece remaining looks similar to the whole.
Likewise, in the fern frond below[2], one piece of the frond looks similar to the whole.
Similarly, if we zoom in on the coastline of Portugal[3], each zoom reveals previously hidden detail, and the coastline, while not identical to the view from further way, does exhibit similar characteristics.
## Iterated Fractals
This self-similar behavior can be replicated through recursion: repeating a process over and over.
### Example
Suppose that we start with a filled-in triangle. We connect the midpoints of each side and remove the middle triangle. We then repeat this process.
If we repeat this process, the shape that emerges is called the Sierpinski gasket. Notice that it exhibits self-similarity—any piece of the gasket will look identical to the whole. In fact, we can say that the Sierpinski gasket contains three copies of itself, each half as tall and wide as the original. Of course, each of those copies also contains three copies of itself.
In the following video, we present another explanation of how to generate a Sierpinski gasket using the idea of self-similarity.
We can construct other fractals using a similar approach. To formalize this a bit, we’re going to introduce the idea of initiators and generators.
### Initiators and Generators
An initiator is a starting shape
A generator is an arranged collection of scaled copies of the initiator
To generate fractals from initiators and generators, we follow a simple rule:
### Fractal Generation Rule
At each step, replace every copy of the initiator with a scaled copy of the generator, rotating as necessary
This process is easiest to understand through example.
### Example
Use the initiator and generator shown to create the iterated fractal.
This tells us to, at each step, replace each line segment with the spiked shape shown in the generator. Notice that the generator itself is made up of 4 copies of the initiator. In step 1, the single line segment in the initiator is replaced with the generator. For step 2, each of the four line segments of step 1 is replaced with a scaled copy of the generator:
This process is repeated to form Step 3. Again, each line segment is replaced with a scaled copy of the generator.
Notice that since Step 0 only had 1 line segment, Step 1 only required one copy of Step 0.
Since Step 1 had 4 line segments, Step 2 required 4 copies of the generator.
Step 2 then had 16 line segments, so Step 3 required 16 copies of the generator.
Step 4, then, would require $16\cdot4=64$ copies of the generator.
The shape resulting from iterating this process is called the Koch curve, named for Helge von Koch who first explored it in 1904.
Notice that the Sierpinski gasket can also be described using the initiator-generator approach.
### Example
Use the initiator and generator below, however only iterate on the “branches.” Sketch several steps of the iteration.
We begin by replacing the initiator with the generator. We then replace each “branch” of Step 1 with a scaled copy of the generator to create Step 2.
Step 1, the generator. Step 2, one iteration of the generator.
We can repeat this process to create later steps. Repeating this process can create intricate tree shapes.[4]
### Try It
Use the initiator and generator shown to produce the next two stages.
Using iteration processes like those above can create a variety of beautiful images evocative of nature.[5][6]
More natural shapes can be created by adding in randomness to the steps.
### Example
Create a variation on the Sierpinski gasket by randomly skewing the corner points each time an iteration is made.
Suppose we start with the triangle below. We begin, as before, by removing the middle triangle. We then add in some randomness.
We then repeat this process.
Continuing this process can create mountain-like structures. This landscape[7] was created using fractals, then colored and textured.
The following video provides another view of branching fractals, and randomness.
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# Precalculus : Evaluating Trig Functions
## Example Questions
← Previous 1
### Example Question #1 : Evaluating Trig Functions
Find the value of to the nearest tenth if and .
Explanation:
Rewrite in terms of sine and cosine.
Substitute the known values and evaluate.
The answer to the nearest tenth is .
### Example Question #2 : Evaluating Trig Functions
Determine the value of in decimal form.
Explanation:
Ensure the calculator is in radian mode since the expression shows the angle in terms of . Also convert cotangent to tangent.
### Example Question #3 : Evaluating Trig Functions
Find the decimal value of
Explanation:
To determine the decimal value of the following trig function, , make sure that the calculator is in radian mode.
Compute the expression.
### Example Question #4 : Evaluating Trig Functions
Determine the correct value of
.
Explanation:
The question asks for the y-coordinate on the unit circle when the degree angle is .
Be careful not to confuse finding the value of the angle when the y-value of the coordinate of the unit circle is .
Ensure that the calculator is in degree mode.
### Example Question #5 : Evaluating Trig Functions
Find the value of
.
Explanation:
Before beginning this problem on a calculator, though this is not necessary since these are special angles, ensure that the mode of the calculator is in degrees.
Input the values of the expression and solve.
### Example Question #1 : Find The Degree Measure Of An Angle For Which The Value Of A Trigonometric Function Is Known
Solve for all x on the interval
,
,
,
,
,
Explanation:
Solve for all x on the interval
We can begin by recalling which two quadrants have a positive sine. Because sine corresponds to the y-value, we know that sine is positive in quadrants I and II.
Next, recall where we get .
always corresponds to our -increment angles. In this case, the angles we are looking for are and , because those are the two -increment angles in the first two quadrants.
Now, you might be saying, "what about ? That is an increment of 45."
While that is true, , not
,
### Example Question #2 : Find The Degree Measure Of An Angle For Which The Value Of A Trigonometric Function Is Known
Solve for all x on the interval
Explanation:
Solve for all x on the interval
Remember Soh, Cah, Toa?
For this problem it helps to recall that
Since our tangent is equal to 1 in this problem, we know that our opposite and adjacent sides must be the same (otherwise we wouldn't get "1" when we divided them)
Can you think of any angles in the first quadrant which yield equal x and y values?
If you guessed you guessed right! Remember that your angle in the unit circle will give you a triangle, which will have equal height and base.
### Example Question #8 : Evaluating Trig Functions
The above triangle is a right triangle. Find the value of (in degrees).
Explanation:
One can setup the relationship
After taking the arccosine,
the arccosine cancels out the cosine leaving just the value of .
### Example Question #9 : Evaluating Trig Functions
What is the value of (in degrees)?
Explanation:
One can setup the relationship
.
After taking the arctangent,
the arctangent cancels out the tangent and we are left with the value of .
Solve for :
or
or
or
or
or
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# Study Guide - Chapter 6
## Math 116
1. Sketch the graph of an exponential and logarithmic function on the same coordinate system. Look at problems 35 - 38 in section 6.2.
2. Rewrite an exponential equation in logarithmic form (look at 31-32 in the review*) and a logarithmic equation in exponential form. Two parts.
3. Evaluate the given expression without a calculator. Actually, you can use a calculator, I just want exact answers. Five parts. Look at problems 33 - 39 in the review* and problems 13 - 16 in section 6.4.
4. Evaluate each logarithm using the change of base formula. Round answers to four decimal places. Two parts. Look at problems 41 - 44 in the review* for one of the two parts.
5. Use properties of logarithms to write the expression as the sum, difference, and/or multiple of logarithms. Three parts. Look at problems 45 - 50 in the review*.
6. Use properties of logarithms to write the expressions as the logarithm of a single quantity. Three parts. Look at problems 51 - 56 in the review*.
7. Approximate the logarithm using properties of logarithms and the approximations of key logarithms given. Then find the base used. Three parts. Look at problems 61 - 64 in the review. The logarithms are the same as in 61 - 64, but the base is different than what the book used.
8. Solve each equation. Round the answers to three decimal places. Five parts. Look at problems 67 - 76 in the review*.
9. Find the exponential function which passes through the given points. Look at problems 77 - 80 in the review*.
10. Application problem involving exponentials. Look at problems 19 - 22 in the review*.
11. Application problem involving logarithms. Look at problems 49 - 50 in section 6.5*. Hint: The percent change (increase or decrease) is the change (New-Old) divided by the Old times 100%. A change from 50 to 60 is a (60-50)/50x100%=20% increase.
* means the problem is straight from the text.
Be sure to bring your calculator to the exam. Graphing calculators are allowed on this exam.
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# How do you find the derivative of sin(arccosx)?
Aug 1, 2015
Use the Chain Rule, resulting in $\cos \left({\cos}^{-} 1 \left(x\right)\right) \cdot \frac{- 1}{\sqrt{1 - {x}^{2}}}$
#### Explanation:
The Chain Rule involves breaking a function up into a function of a function, sometimes expressed as $y = f \left(u\right)$ and $u = g \left(x\right)$.
Step 1. Let $y = \sin \left(u\right)$ and $u = {\cos}^{-} 1 \left(x\right)$.
Step 2. Chain Rule says $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.
Step 3. Apply the Chain Rule to the equations in Step 1.
$\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left(u\right)$ and (du)/(dx)=(-1)/(sqrt(1-x^2).
Step 4. Get rid of the $u$'s in $\frac{\mathrm{dy}}{\mathrm{du}}$. Recall that $u = {\cos}^{-} 1 \left(x\right)$, so then
$\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left({\cos}^{-} 1 \left(x\right)\right)$
Step 5. Plug your derivatives back into Chain Rule of Step 2.
dy/dx=(dy)/(du)*(du)/(dx)=cos(cos^-1(x))*(-1)/(sqrt(1-x^2)
Step 6 is often optional, but you can then try to simplify as much as you can.
dy/dx=(dy)/(du)*(du)/(dx)=(-cos(cos^-1(x)))/(sqrt(1-x^2).
Aug 1, 2015
I would use the fact that $\sin \left(\arccos x\right) = \sqrt{1 - {x}^{2}}$
#### Explanation:
$\arccos x$ is an angle (or, a number) between $0$ and $\pi$ whose cosine is $x$.
The sine of an angle (or, a number) between $0$ and $\pi$ whose cosine is $x$ is $\sqrt{1 - {x}^{2}}$.
(Use ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$. So $\cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$
and with $\theta$ in $\left[0 , \pi\right]$, we have $\cos \theta > 0$.)
$\sin \left(\arccos x\right) = \sqrt{1 - {x}^{2}}$
Now i'll use $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \frac{\mathrm{du}}{\mathrm{dx}}$ to get:
$\frac{d}{\mathrm{dx}} \left(\sin \left(\arccos x\right)\right) = \frac{d}{\mathrm{dx}} \left(\sqrt{1 - {x}^{2}}\right)$
$= \frac{1}{2 \sqrt{1 - {x}^{2}}} \cdot \left(- 2 x\right)$
$= \frac{- x}{\sqrt{1 - {x}^{2}}}$
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### 62nd Putnam 2001
Problem B2
Find all real solutions (x, y) to the simultaneous equations:
1/x - 1/(2y) = 2 y4 - 2 x4
1/x + 1/(2y) = (3 x2 + y2)(x2 + 3 y2).
Solution
Answer: There is just one solution: x = (31/5 + 1)/2, y = (31/5 - 1)/2.
We need to find some way of eliminating one of the variables. The rhs of both equations is homogeneous, so this suggests using the variable t = y/x. Dividing through by x4, the equations become: 1/x5 - 1/(2 t x5) = 2t4 - 2, 1/x5 + 1/(2 t x5) = 3t4 + 10t2 + 3. It is now easy to eliminate x5 to get a quintic in t: 2t5 - 5t4 + 20t3 - 10t2 + 10t - 1 = 0.
However, it is not clear that this leads anywhere, if we differentiate we find that the derivative is the sum of two squares (10(t2 - t + 1)2 + 30t2), so the quintic is monotonic increasing and has just one real root. But a little trial makes clear that the root is not anything particularly nice (it lies between 0 and 1, but is not 1/2).
The trick is to play around until you recognize the binomial coefficients for n = 5. If you multiply out the second equation and then add and subtract the two equations, you get: 2/x = x4 + 10x2y2 + 5y4, 1/y = 5x4 + 10x2y2 + y4. The 1, 5, 10 pattern should look familiar at this point. Multiply across by the x, y to get: 2 = x5 + 10x3y2 + 5xy4, 1 = 5x4y + 10x2y3 + y5. Adding and subtracting gives: 3 = (x + y)5, 1 = (x - y)5.
We want real solutions, so x + y = k, x - y = 1, where k = 31/5.
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## Calculus with Applications (10th Edition)
$${\text{The limit does not exist}}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{8{x^6} + 3{x^4} - 9x}}{{9{x^7} - 2{x^4} + {x^3}}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {8{x^6} + 3{x^4} - 9x} \right) = 8{\left( 0 \right)^6} + 3{\left( 0 \right)^4} - 9\left( 0 \right) = 0 \cr & \mathop {\lim }\limits_{x \to 0} \left( {9{x^7} - 2{x^4} + {x^3}} \right) = 9{\left( 0 \right)^7} - 2{\left( 0 \right)^4} + {\left( 0 \right)^3} = 0 \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr & {\text{denominator}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{D_x}\left( {8{x^6} + 3{x^4} - 9x} \right)}}{{{D_x}\left( {9{x^7} - 2{x^4} + {x^3}} \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{48{x^5} + 12{x^3} - 9}}{{63{x^6} - 8{x^3} + 3{x^2}}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{48{{\left( 0 \right)}^5} + 12{{\left( 0 \right)}^3} - 9}}{{63{{\left( 0 \right)}^6} - 8{{\left( 0 \right)}^3} + 3{{\left( 0 \right)}^2}}} \cr & = \frac{{ - 9}}{0} \cr & {\text{Since the limits does not leads to the indeterminate form }}\frac{0}{0}{\text{ or }}\frac{{ \pm \infty }}{{ \pm \infty }} \cr & {\text{We cannot apply l'Hospital's rule}}. \cr & {\text{}} \cr & {\text{The limit does not exist}} \cr}
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Sunday, December 18, 2005
Continued Fractions and Matrices
Matrices are very useful when analyzing continued fractions. In previous blogs, I wrote about the continued fractions approximation function and the continued fractions basic properties. Today, I will show how matrices can be used to reason with continued fractions.
Today's blog is based on H. M. Stark's An Introduction to Number Theory. If you are not familiar with Continued Fractions, start here. If you are not familiar with the Continued Fraction approximation algorithm, start here.
A continued fraction breaks a real number into a series of integers such as a0, a1, ... an. It makes sense that we can use matrices to reason about them.
In today's blog, I will focus on the simplest properties of matrices. For example, I will only deal with two types of matrices:
1 x 2 matrices such as (5,6)
2 x 2 matrices
For those who would need a review of simple matrices, matrix products, determinants, inverses, or some basic lemmas, start here.
To represent, continued fractions as matrices, we define the following:
(a) An
NOTE: an is taken directly from the continued fraction. See here for details on how this is generated.
(b) Mn
NOTE: pn, qn are generated based on the continued fraction approximation function. See here for details on how these two values are generated.
(c) γn
γn = qn-2 + αnqn-1
NOTE: αn is generated based on the continued fraction for α (see here). qn is generated based on the continued fraction approximation algorithm (see here).
Using these definitions, I will go through some lemmas that will be useful in solving Pell's equation.
Lemma 1: Mn+1 = AnMn
(1) AnMn =
NOTE: By the definitions above.
NOTE: By carrying out the matrix product [see here for review if needed]
NOTE: By applying the definitions of pn and qn [see here for details]
= Mn+1 [By the definition above]
QED
Lemma 2: γn(1,α) = (1,αn)Mn
(1) (1,αn)Mn =
= (qn-2 + αnqn-1 , pn-2 + αnpn-1) [See here to review matrix products]
= (qn-2 + αnqn-1)(1,α) [Since α * (qn-2 + αnqn-1) = pn-2 + αnpn-1; see here]
= γn(1,α)
QED
Lemma 3: α is irrational → γn ≠ 0
(1) Since α is irrational, αn is irrational. (See here for proof; only rational numbers give way to rational αn)
(2) Assume γn = 0.
(3) Then, qn-2 + αnqn-1 = 0.
(4) Then qn-2=0 and qn-1 = 0. [See here for proof]
(5) q-1 = q1 - q0a1 = a1 - a1(1) = 0
q-2 = q0 - q-1a0 = 1 - 0*a0 = 1
NOTE: See here for review of qn
(6) We also know for all values n ≥ 1, that qn ≥ 1 (see here).
(7) So (#4) is impossible and we reject our assumption.
QED
Lemma 4: det(Mn) = (-1)n
(1) det(Mn) = pn-1qn-2 - pn-2qn-1 (See here for a review of determinants)
(2) So, det(Mn) = (-1)n by a previous result. (See here)
QED
Lemma 5: Mn+1 = AnAn-1...A1A0
(1) Mn+1 = AnMn (From Lemma 1 above)
(2) Applying Lemma 1 again gives us:
Mn+1 = AnAn-1Mn-1
(3) Since n is an integer greater than 0, we can keep on doing this until eventually we get:
Mn+1 = AnAn-1...A1A0
NOTE:
(a) p-1 = p1 - p0a1 = a0a1 + 1 - a0a1 = 1.
(b) q-1 = 0 (see Lemma 3 above)
(c)
QED
Lemma 6: Mn-1 is a 2 x 2 matrix of integers.
Mn-1 =
NOTE: det(Mn) = (-1)n from Lemma 4 above.
QED
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# Less than Zero
In the world of numbers, you will usually work with whole numbers. Those are the numbers that are zero and greater on a number line. There are no parts of numbers to worry about with whole or natural numbers. When you start to work with integers, you will start using numbers that are less than zero. For example, you might have -3, -25, or -147. When you divide positive and negative numbers, you don't worry about the signs until you are done with the problem. Just set them aside while you work on your problem. The rules are the same as the ones you learned for multiplication.
# It's All About Being Negative
When you are done with a division problem that has positive and negative values, you don't care how many positive values there are. You only need to pay attention to the number of negative values. Negative numbers have the power of negation. They can change signs of quotients. After you count them up, you need to figure out if you have an odd or an even number of negative signs. If it is even, you have a positive answer. If you have an odd number of negative values, your quotient will be negative.
Problem:
Positive or Negative Quotients?
56 ÷ 47÷ (-58) ÷ (-12) = ? (Positive: Even number of negative factors)
(-15) ÷ (-15) ÷ 98 ÷ (-3) = ? (Negative: Odd number of negative factors)
That's kind of it. We'll do an easier example to show you the process, but you really only need to do the division and then count the number of negative values.
Example:
50 ÷ 2 ÷ (-5) = ?
Steps to Solve:
(1) Write out the problem without the signs.
50 ÷ 2 ÷ 5 = ?
(2) Solve the division problems.
50 ÷ 2 = 25
25 ÷ 5 = 5
So...
50 ÷ 2 ÷ 5 = 5
(3) Count the number of negative factors in the original problem :1.
(4) Since the number of negative values is odd, the final answer is negative.
50 ÷ 2 ÷ (-5) = -5
Example (with decimals):
8.4 ÷ (-1.2) ÷ 3.5 = ?
Steps to Solve:
(1) Write out the problem without the signs.
8.4 ÷ 1.2 ÷ 3.5 = ?
(2) Solve the division problems. (You need to count the decimal places in this example).
8.4 ÷ 1.2 = 7
7 ÷ 3.5 = 2
So...
8.4 ÷ 1.2 ÷ 3.5 = 2
(3) Count the number of negative factors: 1
(4) Since the number of negative factors is odd, the final answer is negative.
8.4 ÷ (-1.2) ÷ 3.5 = -2
If you solved the last example on your own, you did a lot of division steps.
• Divided two-digit numbers.
• Came up with a quotient of two decimals.
• Figured out if the quotient was positive or negative.
Good work! If you didn't get the right answer, get a piece of scratch paper and try to work it out again. It will be worth the effort so that you understand all of the steps in the process.
## Related Activities
One and Two-Digit Division Quiz (With Remainders) - Play Activity One and Two-Digit Division Quiz (No Remainders) - Play Activity
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hw5solutions
# hw5solutions - Math 215 HW#4 Solutions 1 Problem 2.3.6...
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Unformatted text preview: Math 215 HW #4 Solutions 1. Problem 2.3.6. Choose three independent columns of U . Then make two other choice. Do the same for A . You have found bases for which spaces? U = 2 3 4 1 0 6 7 0 0 0 0 9 0 0 0 0 and A = 2 3 4 1 0 6 7 0 0 0 0 9 4 6 8 2 . Solution: The most obvious way to choose three independent columns of U is to pick the three pivot columns, namely 2 , 3 6 , and 1 9 . Another choice of three independent columns would be the first, third and fourth columns, which are independent because, if = c 1 2 + c 2 4 7 + c 3 1 9 = 2 c 1 + 4 c 2 + c 3 7 c 2 9 c 3 , then it must be the case that c 2 = c 3 = 0 (from the second and third components), so the first component reduces to 0 = 2 c 1 , so c 1 = 0 as well. We could also have chosen the second, third, and fourth columns. To see that these are independent, suppose = c 1 3 6 + c 2 4 7 + c 3 2 9 = 3 c 1 + 4 c 2 + c 3 6 c 1 + 7 c 2 9 c 3 . Then the third component implies that c 3 = 0 and the first two component give the system of equations 3 c 1 + 4 c 2 = 0 6 c 1 + 7 c 2 = 0 . Interpreting via the augmented matrix 3 4 6 7 and reducing by subtracting twice row 1 from row 2 yields 3 4- 1 . 1 Then c 2 = 0 and, since 3 c 1 + 4 c 2 = 0, this means that c 1 = 0 as well. So, as indicated, the second, third, and fourth columns of U are also independent. Since U is just the echelon form of A (in particular, you can get from A to U by subtracting twice row 1 from row 4), the relationship between the columns of A is the same as the relationship between the columns of U . Hence, we can translate the three choices made above into three different choices of independent columns of A . Namely, the first, second, and fourth columns of A are independent, as are the first, third, and fourth columns of A , and the second, third, and fourth columns of A are also independent. 2. Problem 2.3.10. Find two independent vectors on the plane x + 2 y- 3 z- t = 0 in R 4 . Then find three independent vectors. Why not four? This plane is the nullspace of what matrix? Solution: Notice that 2- 1 and 3 1 are vectors in the plane x + 2 y- 3 z- t = 0. They’re independent since = c 1 2- 1 + c 2 3 1 = 2 c 1 + 3 c 2- c 1 c 2 implies that c 1 = 0 = c 2 . The vector 1 1 is also in this plane and we can see that this collection of three vectors is linearly independent as follows: suppose...
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Notes On Solutions of a Differential Equation - CBSE Class 12 Maths
Consider the differential equation dy/dx = ex - 4 We claim that function y = ex - 4x + 3 is the solution of the differential equation. The solution of a differential equation is the function that satisfies it. There exist two types of solutions for a differential equation: 1. General solution 2. Particular solution dy/dx = ex - 4 y = ex - 4x + 3 Suppose function y is of the forms as shown. y = ex - 4x + 5 ⇒ dy/dx = ex - 4 y = ex - 4x - 5 ⇒ dy/dx = ex - 4 y =ex - 4x + 122 ⇒ dy/dx = ex - 4 Each of the three forms of the function y entitles to be the solution of the differential equation, since their derivatives are the same. y = ex - 4x + C We can represent the constant in the equations with a letter. We chose C here. So, this function represents all the solutions of the differential equation. Such a function is called the general solution. And, the function obtained by replacing the value of 'C' by a number is called the particular solution.
#### Summary
Consider the differential equation dy/dx = ex - 4 We claim that function y = ex - 4x + 3 is the solution of the differential equation. The solution of a differential equation is the function that satisfies it. There exist two types of solutions for a differential equation: 1. General solution 2. Particular solution dy/dx = ex - 4 y = ex - 4x + 3 Suppose function y is of the forms as shown. y = ex - 4x + 5 ⇒ dy/dx = ex - 4 y = ex - 4x - 5 ⇒ dy/dx = ex - 4 y =ex - 4x + 122 ⇒ dy/dx = ex - 4 Each of the three forms of the function y entitles to be the solution of the differential equation, since their derivatives are the same. y = ex - 4x + C We can represent the constant in the equations with a letter. We chose C here. So, this function represents all the solutions of the differential equation. Such a function is called the general solution. And, the function obtained by replacing the value of 'C' by a number is called the particular solution.
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## Basics on the topicAdding 2-Digit Numbers
Imagine you own 14 toys and your sibling owns 17 toys. Now you want to calculate how many toys you own together! In order to do this, you can use standard algorithm to solve 2 digit addition without regrouping and 2 digit addition with regrouping.
In this text, we will learn how to do 2 digit addition using the standard algorithm.
The standard algorithm is a method to use where the numbers are set up vertically, based on place values. This is an efficient way to find sums when there are several equations you need to solve.
In the explanation, there will be examples that have you adding 2 digit numbers without regrouping as well as adding 2 digit numbers with regrouping.
## Adding Two Digit Numbers – Example
To solve this 2-digit addition problem, we set the equation up vertically, with one addend on top of the other in order of their place values. We will find the sum of the numbers by adding the digits from right to left, beginning with the ones place.
What is six plus two more?
Place the eight below the line directly under the ones place.
Next, we move one place over to the left and now we add the tens place.
How many are three tens and two tens? We can write the digit five below the line under the tens place because the five represents five tens or fifty.
The sum is 58.
In this problem, we will need to do 2-digit addition with regrouping. We are going to find the sum of fourteen and forty-seven.
Starting with the ones place, four plus seven equals eleven. What do you notice about the sum of the ones place?
Eleven is a two-digit digit number made up of a one and one ten. So, one is written underneath the ones place and one is regrouped on top of the tens place.
Now we add the tens, including this new value added at the top. What is the sum of ten, forty, and ten? Sixty, which means we have six tens.
The sum is 61.
Do you now know how many toys your sibling and you own together? Correct, it is 31 toys in total! For this equation, you needed to regroup a number.
How do you add two digit numbers using the standard algorithm?
Remember: The standard algorithm is a method to use where the numbers are set up vertically, based on place values.
With the standard algorithm you can represent an addition of two numbers neatly to solve it with ease. Have another look at our example addition from earlier:
Tens Ones
1 4
+ 4 7
6 1
At the end of the video, you can find exercises for the continued practice of 2 digit addition as well as a 2 digit addition worksheets. Enjoy!
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Adding 2-Digit Numbers .
Hints
Add the digits in the ones column first.
Solution
Add the digits in the ones column first.
4 + 5 = 9
Then add the digits in the tens.
3 tens + 2 tens = 5 tens
34 + 29 = 59
• ### Place the digits onto the grid.
Hints
Make sure the tens number is in the T column.
Make sure the ones are in the O column.
The H column represents hundreds.
Solution
We need to make sure the numbers are in the correct column.
If it has 3 digits it will have a hundred, tens, and ones digit.
310 has 3 hundreds, 1 ten and 0 ones
27 has 2 tens and 7 ones
0 + 7 = 7
1 ten + 2 tens = 3 tens
3 hundreds + 0 hundreds = 3 hundreds
310 + 27 = 337
• ### Calculate the answers to the equations.
Hints
Add up the ones column first.
Make sure if your answer is 2* or more digits you carry it across.
After adding the ones column, move on to the tens column.
Don't forget to check if there are any numbers that you carried across from the ones column.
Solution
See image for the 1st problem
For the 2nd problem:
45 + 63 = 108
For the 3rd problem:
44 + 12 = 56
For the 4th problem:
69 + 72 = 141
• ### Match the question to the correct answer.
Hints
Layout the numbers in the format above.
Add the digits in the ones column first, followed by the digits in the tens column.
Add up all of the digits in the ones column first. If the answer is greater than 9, carry across the extra ten.
If the number in the tens column is greater than 9, make sure you carry across to the hundreds column.
Solution
Using the layout above, workout the answers to the equations.
4. If your answer from the tens column is more than 2 digits, carry across the hundreds digit (the first digit from the number).
38 + 64 = 102 see image above
27 + 35 = 62
39 + 54 = 93
74 + 57 = 131
• ### Identify the place value of the numbers below.
Hints
If the number has two digits and ends in a zero, it is a ten.
If it is a single digit it is a one.
The blue lines represent ten.
The yellow cubes represent one.
In this example, you can see how all the one numbers are in the O column. We have a mixture of words, numerals and yellow ones cubes.
In the tens column, we have a mixture of words, numerals and base 10 blocks. They all show values of ten
Solution
Two digit numbers have tens in them.
Single digit numbers are classed as ones.
• ### Calculate the answers to the equations.
Hints
Solution
First, add the digits in the ones column.
4 + 7 = 11. 1 ten would go into the tens column with the other tens digits, and the 1 representing the ones would go in the answer box under the ones
Then add all of the digits in the tens column.
1 + 8 + 3 = 12
As the answer to this is greater than 9, we need to carry across to the hundreds column.
33 + 42 = 75
18 + 71 = 89
84 + 37 = 121
69 + 33 = 102
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# The equation of the tangent line to the curve at the given point.
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 2.6, Problem 8E
To determine
## To find: The equation of the tangent line to the curve at the given point.
Expert Solution
The equation of the tangent line to the curve y=2x+1x+2 at the point (1,1) is y=13x+23.
### Explanation of Solution
Given:
The equation of the curve is y=2x+1x+2.
The curve passing through the point (1, 1).
Formula used:
The slope of the tangent curve y=f(x) at the point P(a,f(a)) is,
m=limxaf(x)f(a)xa (1)
The equation of the tangent line to the curve y=f(x) at the point (a,f(a)) is,
yf(a)=f(a)(xa) (2)
Calculation:
Obtain the slope of the tangent line to the parabola at the point (1, 1).
Substitute a=1 and f(a)=1 in equation (1),
m=limx1f(x)f(1)x1=limx1(2x+1x+2)1x1=limx1(2x+1)1(x+2)(x+2)x1=limx1((2x+1)(x+2)(x1)(x+2))
=limx12x+1x2(x1)(x+2)=limx1x1(x1)(x+2)
Since the limit x approaches 1 but not equal to 1, cancel the common term (x1)(0) from both the numerator and the denominator,
m=limx11(x+2)=11+2=13
Thus, the slope of the tangent line to the curve at the point (1, 1) is m=13_.
Obtain the equation of the tangent line.
Since the tangent line to the curve y=f(x) at (a,f(a)) is the line through the point (a,f(a)) whose slope is equal to the derivative of a, the value of f(a)=13.
Substitute a=1, f(a)=1 and f(a)=13 in equation (2),
(yf(a))=f(a)(xa)y(1)=13(x1)y1=13x13
Isolate y as shown below.
y=13x13+1=13x+23
Thus, the equation of the tangent line is y=13x+23.
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Optimization Calculus.. a box/shelter with sides missing..
I'm solving a problem involving calculus optimization. The problem is the following:
"We plan to build a boxshaped shelter with no floor and one side open. (Hence we need a roof and three sides). The base of the shelter is a square. The volume of the shelter is 144 cubic meters. Find the dimensions of the shelter that minimizes the material, i.e. the surface area. Use differentiation."
Everything I say from here on out are my conclusions:
I'm assuming that the shelter is a rectangular prism. Since the base of the shelter is a square, I assume the roof is as well.
So since the roof is a square, and a square's sides are all of equal length, that means that the width/length of the roof is really the same variable. So let x = length = width. Let y denote the height.
We want to minimize the surface area of the rectangle. The surface area of a rectangle is as follows:
$$SA = 2LW + 2HW + 2LH$$
However, since we are missing the bottom of the rectangle, $L*W$, and one of the sides, either $L*W$ or $H*W$, our SA formula becomes:
$$SA = x^2 + yx + 2xy$$ or $$SA = x^2 + 3xy$$
Remember that $x = length = width$ and $y = height$.
We have more than one variable in that equation, let us connect the two. We're given that the volume of the shelter is 144. So:
$$144 = x^2 * y$$ $$y = \frac{144}{x^2}$$
And this means that:
$$SA = x^2 + \frac{432}{x}$$
Now we differentiate the function and attempt to find the critical points (where $f'(x) = 0$ or $f'(x)$ is undefined. I.e. we want the local minimum..
Differentiating SA, it becomes:
$$SA' = 2x - 432x^{-2}$$
Is this really correct or did I go very wrong somewhere along the way? If by some miracle this is correct, how would I go about finding the local minimum?
• Thanks, this helped. One thing about the first derivative though, I find it very hard to see when it's zero. I tried to factor it as follows: $SA' = 2x(1 - 216x^{-3})$. Seeing this, I know if x is zero the function is undefined. So presumably $216x^{-3}$ must become one, but how can I find this out in an easy way? Oct 4, 2014 at 13:29
• Alright so we know $x \neq 0$, so when does $2x - 432x^(-2)$ become zero? (In order to find the local minimum). Sorry if I'm missing something.. Oct 4, 2014 at 13:37
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# Maths Year 5 Spring Decimals and Fractions
Each unit has everything you need to teach a set of related skills and concepts.
## Place value in decimals; rounding(suggested as 3 days)
### Planning and Activities
Day 1 Teaching
Revise place value, quizzing children about the digits in 4.03. Find 4.03 + 0.7. Rehearse writing 2-place decimal numbers. Use place value additions to emphasise the value of each digit, e.g. start with 4.56, add one hundredth, then add two tenths, subtract one whole, subtract four hundredths, subtract four tenths. Ask children to show you the new number each time.
Group Activities
Use the in-depth problem-solving investigation ‘Spiralling Decimals’ from NRICH as today’s group activity.
Or, use these activities:
-- Use place value to add and subtract numbers with two decimal places, one calculation at a time or in longer chains.
Day 2 Teaching
Use the place value grid on the ITP Moving digits to show the effect of multiplying and dividing by 10, 100 and 1000. The digits move to the left when multiplying and to the right when dividing. Emphasise that the decimal point does not move.
Group Activities
-- Explore decimal place value and number order when multiplying and dividing by 10, 100 and 1000.
Day 3 Teaching
Show children a counting stick. Say that one end represents 2; the other 3. Use this to demonstrate that we can round 1-place decimals to the nearest whole number, e.g. we round 2.4 to 2 and 2.7 to 3. Write some 2-place decimals between 2.3 and 2.4 and round these to the nearest tenth, then to the nearest whole number. Probe understanding of rounding: Write a number less than 6 with two decimal places that rounds up to 6, etc.
Group Activities
-- Explore rounding 2-place decimals marked on number lines, or while playing Bingo.
### You Will Need
• Place value grid 10–0.01 (see resources)
• Number cards 0–9
• Flipchart
• Whiteboards and pens
• A4 paper for each child
• ITP: Moving digits
• 3-digit number cards (see resources)
• Place value stars image (see resources)
• Place value dice
• × and ÷ starting numbers (see resources)
• Counting stick
### Mental/Oral Maths Starters
Day 1
1-place decimals (pre-requisite skills)
Suggested for Day 2
Adding to the next whole number from a 1-place decimal number (simmering skills)
Suggested for Day 3
Difference between negative numbers (simmering skills)
### Worksheets
Day 1
Place value additions and subtraction; add and subtract 0.1 and 0.01 and multiples of 0.1 and 0.01.
Day 2
Multiply and divide by 10 and 100.
Multiply and divide by 10, 100 and 1000.
Day 3
Round decimals to the nearest whole number.
Round decimals to the nearest tenth and whole number.
### Mastery: Reasoning and Problem-Solving
• Divide 47,310 by 10 repeatedly until you get a number that is less than 100. Write that number.
• Write the next two numbers in each sequence.
0.41 4.1 ____ ____
2.05 20.5 ____ ____
43020 4302 ____ ____
• True or false?
4030 ÷ 100 = 43
1.09 × 100 = 190
0.09 × 10 = 0.9
7000 ÷ 1000 = 0.7
• Rounding to nearest tenth
Tick (a) or (b)
(a) 20.07 → 20
(b) 20.07 → 20.1
Rounding to nearest whole number
(a) 20.09 → 21
(b) 20.09 → 20
In-depth investigation: Spiralling Decimals
Take turns to place a decimal number on a spiral number line. Spiralling Decimals from nrich.maths.org.
### Extra Support
Left or Right?
Multiplying numbers with one decimal place by 10 and dividing 2-digit whole numbers by 10
Left, Left or Right, Right?
Understanding place value in numbers with two decimal places;
Beginning to multiply numbers with two decimal places by 100 and divide 3-digit numbers by 100
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How to display points and lines on the coordinate axes
How to display points and lines on the coordinate axes
Here I will explain everything you need to know to represent points on Cartesian axes and how to represent lines on coordinate axes.
First of all, you need to be very clear about the Cartesian axes or coordinate axes, which is what I will explain to you in the following point.
The Cartesian axes
To represent points and lines we need the Cartesian axes or coordinate axes, which are two perpendicular lines that meet the following characteristics:
• The horizontal axis is called x-axis or x-axis
• The vertical axis is called y-axis or ordinate axis
• The coordinates of any point are represented by (x,y), where x is measured on the x and y axis and y is measured on the y axis.
• The point where the two lines are cut is called coordinate origin and their coordinates are (0,0).
How to get the coordinates of a point
We will now learn how to obtain the coordinates of a point that is already represented in the coordinate axes.
For example: What are the coordinates of point A?
To obtain the x-coordinate, draw a vertical line (perpendicular to the x-axis) from the point to the x-axis. The cut-off point will be the x-coordinate:
It cuts with the x-axis in the value 4, so the x-coordinate is 4.
To obtain the y-coordinate, draw a horizontal line (perpendicular to the y-axis) from the point to the y-axis. The cut-off point will be the y-coordinate:
It cuts with the y axis in the value 3, so the y-coordinate is 3.
The coordinates of point A are (4,3).
Positive and negative values in the coordinate axes
The coordinates don’t always have to be positive. They can also be negative.
Where are the coordinates positive or negative?
On the x-axis, the values are positive to the right of the coordinate center and negative to the left of it.
On the y-axis, the values are positive when they are above the coordinate center and negative when they are below it.
Note that to obtain the coordinates of a point, the line always starts from the point to the axes.
Depending on the quadrant in which the point is located, the coordinates may be positive or negative. In the example above, the two coordinates were positive.
Let’s see what happens in the rest of the quadrants.
What would be the coordinates of points B, C and D?
For point B, we draw a vertical and a horizontal line from the point to the axes and see where they cut, just like we did for point A:
Its coordinates are (-2.1). In this case, x takes a negative value, since it is to the left of the coordinate origin.
We do the same thing with point C:
The coordinates of C are (-3,-2)
And now for point D:
The coordinates of D are (1,-3).
Proposed Exercises
What are the coordinates of the following points?
How to represent a point on the coordinate axes
Now I am going to explain the opposite case to the previous one: we have the coordinates of the point and we have to represent it in the coordinate axes.
Let’s look at it with an example: Represent point A (2,3) on the coordinate axes.
In this case we have the x and y coordinates.
From the x-coordinate, i.e. from 2, we draw a vertical line and from the y-coordinate, from 3, a horizontal line. Where the two lines are cut, point A:
In this case the lines go from the axes to the point.
Another example: Represent point B (-4.2)
Proposed Exercises
Represent the following points on the coordinate axes:
How to represent a straight line in the coordinate axes
Now that you know how to represent points, I will explain how to represent a line.
What do you need to draw a line on the Cartesian axes?
Imagine you want to draw a line with a pencil and ruler.
You start by pressing the pencil down on the paper and then slide it along the ruler. But the ruler can turn and take infinite directions.
What is the correct direction for your period?
You need another point to place the ruler to define the line you want to draw.
In other words, you need to join two points to get a line.
But the straight line doesn’t start and end at those two points. Those two points just mark the direction of the line. A straight line has an infinite length, so you can lengthen it as much as you want.
Now, how to draw the equation of a line on the coordinated axes?
Let’s see how to represent a line on the coordinate axes with an example. They ask you to represent this line:
As I indicated in the previous section, you need two points. Therefore, first of all, we must find those two points that belong to the line.
To do this, we choose two random x-values, which for example can be 0 and 1 and from these values we will calculate two more y-values. Each pair of values (x,y) will be a point on the line.
To calculate the values of y, each value of x, we substitute it in the equation and operate:
From where we get the first point (0,2)
From where we get the second point (1.0)
And you put this on a chart for clarity:
Now all you have to do is draw those points on the coordinate axes:
And once you have the points, you can now join them together and lengthen the line at both ends:
Could you calculate more points to represent the line?
You can calculate all the points you want, but 2 is enough because you don’t need more points to draw it.
If you want it once you have it represented, give x the value 3 and get a third point and you will see how it falls into the line.
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Find solution of equations- Equations given
Chapter 4 Class 12 Determinants
Concept wise
### Transcript
Misc 7 Solve the system of the following equations 2/x + 3/y + 10/z = 4 4/x + 6/y + 5/z = 1 6/x + 9/y + 20/z = 2 The system of equations are 2/x + 3/y + 10/z = 4 4/x + 6/y + 5/z = 1 6/x + 9/y + 20/z = 2 Now let 𝟏/𝒙 = u , 𝟏/𝒚 = v , & 𝟏/𝒛 = w The system of equations become 2u + 3v + 10w = 4 4u – 6v + 5w = 1 6u + 9v – 20w = 2 Writing equation as AX = B [■8(2&3&10@4&−6&5@6&9&−20)] [■8(𝑢@𝑣@𝑤)] = [■8(4@1@2)] Hence A = [■8(2&3&10@4&−6&5@6&9&−20)] , X = [■8(𝑢@𝑣@𝑤)] & B = [■8(4@1@2)] Calculating |A| |A| = |■8(2&3&10@4&−6&5@6&9&−20)| = 2 |■8(−6&5@9&−20)| – 3 |■8(4&5@6&−20)| + 10 |■8(4&−6@6&9)| = 2 (120 – 45) –3 (–80 – 30) + 10 ( 36 + 36) = 2 (75) –3 (–110) + 10 (72) = 150 + 330 + 720 = 1200 ∴ |A|≠ 0 So, the system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj (A) = [■8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^′ = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(2&3&10@4&−6&5@6&9&−20)] M11 = |■8(−6&5@9&−20)| = 120 – 45 = 75 M12 = |■8(4&5@6&−20)| = (–80 – 30) = –110 M13 = |■8(4&−6@6&9)| = 36 –36 = 72 M21 = |■8(3&10@9&−20)| = −60 – 90 = –150 M22 = |■8(2&10@6&−20)| = –40 – 60 = –100 M23 = |■8(2&3@6&9)| = 18 – 18 = 0 M31 = |■8(3&10@−6&5)| = 15 + 60 = 75 M32 = |■8(2&10@4&5)| = 10 – 40 = –30 M33 = |■8(2&3@4&−6)| = –12 – 12 = –24 Now, A11 = 〖"(–1)" 〗^(1+1) M11 = (–1)2 . 75 = 75 A12 = 〖"(–1)" 〗^"1+2" M12 = 〖"(–1)" 〗^3 . (–110) = 110 A13 = 〖(−1)〗^(1+3) M13 = 〖(−1)〗^4 . (72) = 72 A21 = 〖(−1)〗^(2+1) M21 = 〖(−1)〗^3 . (–150) = 150 A22 = 〖(−1)〗^(2+2) M22 = (–1)4 . (–100) = –100 A23 = 〖(−1)〗^(2+3). M23 = 〖(−1)〗^5. 0 = 0 A31 = 〖(−1)〗^(3+1). M31 = 〖(−1)〗^4 . 75 = 75 A32 = 〖(−1)〗^(3+2) . M32 = 〖(−1)〗^5. (–30) = 30 A33 = 〖(−1)〗^(3+3) . M33 = (–1)6 . –24 = –24 Thus, adj A = [■8(75&150&75@110&−110&30@72&0&−24)] Now, A-1 = 1/(|A|) adj A A-1 = 𝟏/𝟏𝟐𝟎𝟎 [■8(𝟕𝟓&𝟏𝟓𝟎&𝟕𝟓@𝟏𝟏𝟎&−𝟏𝟏𝟎&𝟑𝟎@𝟕𝟐&𝟎&−𝟐𝟒)] Also, X = A−1 B Putting Values [■8(𝑢@𝑣@𝑤)]= 1/1200 [■8(75&150&75@110&−110&30@72&0&−24)] [■8(4@1@2)] [■8(𝑢@𝑣@𝑤)]= 1/1200 [■8(75(4)+150(1)+75(4)@110(4)+(−110)(1)+30(1)@72(4)+0(1)+(−24)2)] [■8(𝑢@𝑣@𝑤)] = 1/1200 [■8(300+150+150@440−100+60@288+0−48)] = 1/1200 [■8(600@400@140)] [■8(𝒖@𝒗@𝒘)] = [■8(𝟏/𝟐@𝟏/𝟑@𝟏/𝟓)] Hence u = 1/2 , v = 1/3 , & w = 1/5 Thus, x = 2, y = 3 & z = 5 Putting u = 𝟏/𝒙 1/2 = 1/𝑥 x = 2 Putting v = 𝟏/𝒚 1/3 = 1/𝑦 y = 3 Putting w = 𝟏/𝒛 1/5 = 1/𝑧 z = 5
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# Expected Area of a Triangle Paradox
There have been several recent problems about expected values and unit circles. I came up with another problem like this, but I noticed something interesting. Here's the problem.
It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius $1,$ find the expected area of such a triangle.
Seems pretty simple, right? Actually, it has two different answers, depending on how you interpret the problem. Consider the following.
It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius $1,$ a diameter and a random point not on the diameter are randomly picked and connected to form a triangle. Find the expected area of this triangle.
And
It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius $1,$ an $x\text{-value}$ is randomly selected from the diameter and its corresponding $y\text{-value}$ is plotted on the circle. Find the expected area of this triangle.
Assume WLOG that the diameters of these circles are on the $x\text{-axis}$
These seem like the exact same problem. But there's two different ways to solve them, given the wording of the problem.
$\textbf{Solving the first problem}$
The points on the diameter are $(-1,0)$ and $(1,0).$ These are the base of the triangle. We just have to find the average height of the circle. Any point on the unit circle can be represented by $(\cos\theta,\sin\theta),$ where $\theta$ is the angle formed by drawing the ray starting at the origin and passing through $(1,0)$ and the ray starting at the origin and passing through the point. The height of the triangle is going to be $\sin\theta.$ The angle will be anywhere in the range $[0,2\pi).$ We can just find the average of $|\sin\theta|$ over the range $[0,2\pi)$ to find the expected height. This will be the same as the average value of $\sin\theta$ over $[0,\pi).$ \begin{aligned} \dfrac{1}{\pi}\int_0^\pi\sin\theta\text{ }d\theta&=\dfrac{-1}{\pi}\left.\cos\theta\right|^\pi_0\\ &=\dfrac{-1}{\pi}(\cos\pi-\cos0)\\ &=\dfrac{-1}{\pi}(-1-1)\\ &=\dfrac{2}{\pi} \end{aligned} The height of this triangle has an average value of $\frac{2}{\pi}$ and the base is a constant $2,$ so the expected area of the triangle is $\frac{1}{2}\times2\times\frac{2}{\pi}=\boxed{\frac{2}{\pi}}$
$\textbf{Solving the second problem}$
In this problem, you are randomly picking an $x\text{-value}$ in the range $(-1,1)$ and plotting it on the circle, which we can again remove the parts of the circle in the third and fourth quadrants because of symmetry. We can model the semicircle by $y=\sqrt{1-x^2}.$ The average value of this function is the area of the semicircle divided by $2,$ which is $\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}.$ The base of the triangle is $2$ and the average height is $\frac{\pi}{4},$ so the expected area of the triangle is $\frac{1}{2}\times2\times\frac{\pi}{4}=\boxed{\frac{\pi}{4}}.$
Wait, we've gotten two different answers to the same problem. Why is that? Consider the length of the arc we are picking a point from. In the first problem, we are picking any point from an arc of length $\pi.$ In the second problem, the arc has length $2.$ But these points are still going to result in the same triangles. The variable arc in the first problem $(A_1)$ and the variable arc in the second problem $(A_2)$ are different; $\frac{A_1}{A_2}=\frac{\pi}{2}.$ So we can then expect to see a difference in area.
See if you can try to explain the difference mathematically!
Note by Trevor B.
5 years, 8 months ago
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In the first case $\theta$ is uniformly distributed over [0,$\pi$] so by transform of variable distribution function of $x= \cos \theta$ will be= $\frac {1}{\pi \sqrt{1-x^2}}$ over [-1,1]
In the second case x is distributed uniformly over [-1,1] .
- 5 years, 8 months ago
The difference in the two cases is that the probability distributions are different. Hence, the expected value is different in each case.
In the first case, you are picking the third point with uniform distribution on the circular arc, whereas in second case, the uniform distribution in on the x-axis (diameter). In the first case, the distribution over x-axis would be a bowl shaped distribution, i.e. it would produce higher probability for smaller-area triangles, whereas in second case, as stated earlier, the distribution is uniform over x-axis. Hence, you get lower expected area in first case.
- 5 years, 8 months ago
Can i ask you one thing....how much time did it take to write such a long note with latex
- 5 years, 8 months ago
This took maybe about $15$ minutes. LaTeX isn't that hard if you've used it a lot.
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# 3.2 Vector addition and subtraction: graphical methods
Page 1 / 16
## Learning objectives
By the end of this section, you will be able to:
• Understand the rules of vector addition, subtraction, and multiplication.
• Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.
The information presented in this section supports the following AP® learning objectives and science practices:
• 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
• 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1)
## Vectors in two dimensions
A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector's magnitude and pointing in the direction of the vector.
[link] shows such a graphical representation of a vector , using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction . We shall use the notation that a boldface symbol, such as $\text{D}$ , stands for a vector. Its magnitude is represented by the symbol in italics, $D$ , and its direction by $\theta$ .
## Vectors in this text
In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector $\text{F}$ , which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as $F$ , and the direction of the variable will be given by an angle $\theta$ .
The head-to-tail method is a graphical way to add vectors, described in [link] below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.
#### Questions & Answers
how can I read physics...am finding it difficult to understand...pls help
try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
Ju
hope that helps
Ju
I have a exam on 12 february
what is velocity
Jiti
the speed of something in a given direction.
Ju
what is a magnitude in physics
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
What is meant by dielectric charge?
what happens to the size of charge if the dielectric is changed?
omega= omega not +alpha t derivation
u have to derivate it respected to time ...and as w is the angular velocity uu will relace it with "thita × time""
Abrar
do to be peaceful with any body
the angle subtended at the center of sphere of radius r in steradian is equal to 4 pi how?
if for diatonic gas Cv =5R/2 then gamma is equal to 7/5 how?
Saeed
define variable velocity
displacement in easy way.
binding energy per nucleon
why God created humanity
Because HE needs someone to dominate the earth (Gen. 1:26)
Olorunfemi
why god made humenity
Ali
and he to multiply
Owofemi
stuff happens
Ju
God plays dice
Ju
Is the object in a conductor or an insulator? Justify your answer. whats the answer to this question? pls need help figure is given above
ok we can say body is electrically neutral ...conductor this quality is given to most metalls who have free electron in orbital d ...but human doesn't have ...so we re made from insulator or dielectric material ... furthermore, the menirals in our body like k, Fe , cu , zn
Abrar
when we face electric shock these elements work as a conductor that's why we got this shock
Abrar
how do i calculate the pressure on the base of a deposit if the deposit is moving with a linear aceleration
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I recently came across a pretty interesting puzzle that used a different kind of mathematics – called Ramsey theory – in order to tackle the problem.
Here’s the problem:
Suppose you have a group of six people. Show that you will necessarily have at least three mutual friends or three mutual strangers.
This is a sort of pigeonhole problem, which sort of means that we have to make a binary choice for each person. But what I love about this proof so much is that it’s entirely visual and fairly easy to grasp.
First, let’s imagine we have six people, who we can each represent as dots.
From there, we want to show the relationships between people. To do this, we will use lines to mark the type of relationship two people have. We will use orange to represent friends, and blue to represent strangers.
To begin, we need to start drawing lines. The first thing that you need to know is that there will always be at least three lines of any one colour that goes out from one person. The reason that this is true is because if we look at one person, the five others have no choice but to be friends or strangers. As such, every person will have at least three orange or blue lines connecting to other people, so we can simply choose one scenario of having three friends, which means three orange lines. We’ll connect them to other people at random, and so we get the following scenario.
Now that we’ve got three lines down, we know that each line represents a friendship between the people. But notice that, to show what we’ve set out to do, we only need to have one of those three friends also be friends. Put differently, if we can find a friendship such that we get a triangle of a single colour, we will prove our statement. However, it’s trivial to simply draw a triangle of one colour and say that we’re done. Instead, we need to show that we’ll be forced to make a triangle of one colour, no matter how much we don’t want to.
Suppose we consider one of those friends who were touched by our original rays. If we look at Person 4, they are friends with Person 1. Additionally, we see that Person 1 is friends with Person 3 and Person 5. Therefore, if we draw a friendship line between either Person 4 and Person 3, or Person 4 and Person 5, we will create a triangle of one colour, which means three people are friends. Since we’re trying to avoid that at all costs, we will draw two blue lines between these people, making sure that no orange triangles are created. This give us the following:
But what, we now have a problem. Look at the line we will need to eventually draw between Person 3 and Person 5. If we draw an orange line between them, then we will create an orange triangle between Person 1, Person 3, and Person 5. But on the other hand, if we draw a blue line between Person 3 and Person 5, we’ll be creating a blue triangle between Person 3, Person 4, and Person 5. Therefore, we’re stuck!
This means that no matter what, a group of six people will have at least three mutual friends or three mutual strangers. Using this technique, we’ve pigeonholed, or forced, ourselves into this situation.
What I like the most about this proof is that it’s visual and easy to understand. It’s what happens when you apply logic at each step, until you arrive at an undeniable conclusion. Furthermore, it’s a glimpse into a type of mathematics that wasn’t shown to me in school, so I thought it would be interesting to share. As I mentioned, I first saw this from a video by Raj Hansen who explains this problem in the introductory video to his Ramsey theory series. You should definitely check it out if you find this proof interesting. Personally, I’m excited to see what other kinds of applications we can find for Ramsey theory.
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# ISEE Lower Level Math : How to find the whole from the part
## Example Questions
← Previous 1 3
### Example Question #1 : How To Find The Whole From The Part
Find the whole from the part. Round to the nearest tenths place.
16 is 55% of what number?
Explanation:
To find the whole from a part, divide the "part" by the percent. But first, change the percent into a decimal (55% = 0.55)
Round to the nearest tenths place = 29.1.
### Example Question #2 : How To Find The Whole From The Part
Roman is ordering uniforms for the tennis team. He knows how many people are on the team and how many uniforms come in each box. Which equation can be used to solve for how many boxes Roman should order?
b = number of students - number of uniforms per box
b = number of students ÷ number of uniforms per box
b = number of boxes x number of uniforms per box
b = number of boxes ÷ number of uniforms per box
b = number of students x number of uniforms per box
b = number of students ÷ number of uniforms per box
Explanation:
The total number of uniforms needed equals the number of students divided by the number of uniforms per box.
### Example Question #2 : How To Find The Whole From The Part
52 is 25% of what number?
1,300
240
13
208
208
Explanation:
To find the whole from the part, divide the “part” by the percent. In order to do this, first change the percent into a decimal (25% becomes 0.25).
Therefore, 52 is 25% of 208.
### Example Question #3 : How To Find The Whole From The Part
At a party, of the cake was eaten. If each of the 12 attendees had one piece of cake, how many pieces were in the whole cake?
36
4
15
48
36
Explanation:
12 pieces of cake is equivalent to of the whole. Proportions can be used to determine how many pieces were in the entire cake. Let equal the number of pieces in the entire cake.
Using this set up, both fractions are equal to . Cross multiply.
### Example Question #4 : How To Find The Whole From The Part
is of . Solve for .
Explanation:
### Example Question #3 : How To Find The Whole From The Part
is of what number?
Explanation:
Solve by setting up a proportion.
Cross multiply.
is of .
### Example Question #5 : How To Find The Whole From The Part
Which of the following numbers is a multiple of ten?
Explanation:
A multiple is a number that can be divided by another number without a remainder. 30 can be divided by 10 without a remainder.
It is the only number from the list of answer choices for which this is true. Therefore, it is the correct answer.
### Example Question #6 : How To Find The Whole From The Part
Andrea and her three friends all pitched in to buy a puppy. Each person paid equal parts. If Andrea paid 20 dollars, what was the total cost of the puppy?
Explanation:
We can solve this question with multiplication.
First, we know that Andrea paid 20 dollars and that each person paid equal parts. That means that each person paid 20 dollars.
Second, we know that there are four people total: Andrea and three additional friends.
Multiply the cost per person by the number of people to get the total cost.
### Example Question #8 : How To Find The Whole From The Part
Together, Patricia, Alex and Steve compose 25 percent of their math club. How big is the math club?
4 people
9 people
10 people
12 people
12 people
Explanation:
If 3 people compose 25% of the math club, then 100% of the math club would consist of 4 times the number 3. This gives us a product of 12. Therefore, the correct answer is 12 people.
### Example Question #7 : How To Find The Whole From The Part
Bessy is making shirts for her daughters. If 2 square feet of fabric is enough for her to make 1 shirt, how many shirts can she make with 9 square feet of fabric?
9 shirts
5 shirts
4 shirts
7 shirts
6 shirts
4 shirts
Explanation:
Given that 2 square feet of fabric is enough for Bessy to make 1 shirt, 9 square feet of fabric is 4.5 times the amount needed to make a shirt. This means that Bessy could make 4.5 shirts. Because making 0.5 shirts does not count as a full shirt, the correct answer is therefore 4 shirts.
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Courses
# Notes - Force & Pressure Class 8 Notes | EduRev
## Class 8 : Notes - Force & Pressure Class 8 Notes | EduRev
``` Page 1
BAL BHARATI PUBLIC SCHOOL, PITAMPURA, DELHI – 110034
Class-VIII { U.T. - 1}
NOTES : FORCE & PRESSURE
Q.1. What is force? State the various effects of force.
Ans. A push or pull on an object is called force.
Effects of force:
(i) Force can change the speed of a moving object.
(ii) Force can change the direction of a moving object.
(iii) Force can change the shape of an object.
Q.2. Give one example where force can changes the
(a) Speed of a moving object
(b) Direction of a moving object.
© the shape of an object.
Ans. (a) The force applied by a hockey player on a moving ball changes the speed of ball.
(b) The force applied by a batsman changes the direction of moving cricket ball.
© Force applied on a spring can change its shape.
Q.3. Identify the actions involved in the following situations as push or pull, or both:
(a) Opening a drawer.
© A cricket ball hit by a batsman.
(d) Drawing a bucket of water from a well.
Ans. (a) Pull.
(b) Push; Pull.
(d) Pull.
Note: a) At least two objects must interact for a force to come into play.
b) Force applied on an object in the same direction add to one another.
c) If the forces act in the opposite directions on object, the net force acting on it is the
difference between the two forces.
d) The state of motion of an object is described by its speed & the direction of motion. So we
can say that a force can change the state of motion of an object.
e) S.I unit of force is Newton ( N).
Q.4. In a tug of war, when the two teams are pulling the rope, a stage comes when the rope
does not move to either side at all. What can you say about the magnitudes & directions
of the forces being applied to the rope by the two teams at this stage?
Ans. At this point of time, the two teams are applying equal but opposite forces to the rope due to
which the net force acting on the rope is zero & hence the rope does not move.
Page 2
BAL BHARATI PUBLIC SCHOOL, PITAMPURA, DELHI – 110034
Class-VIII { U.T. - 1}
NOTES : FORCE & PRESSURE
Q.1. What is force? State the various effects of force.
Ans. A push or pull on an object is called force.
Effects of force:
(i) Force can change the speed of a moving object.
(ii) Force can change the direction of a moving object.
(iii) Force can change the shape of an object.
Q.2. Give one example where force can changes the
(a) Speed of a moving object
(b) Direction of a moving object.
© the shape of an object.
Ans. (a) The force applied by a hockey player on a moving ball changes the speed of ball.
(b) The force applied by a batsman changes the direction of moving cricket ball.
© Force applied on a spring can change its shape.
Q.3. Identify the actions involved in the following situations as push or pull, or both:
(a) Opening a drawer.
© A cricket ball hit by a batsman.
(d) Drawing a bucket of water from a well.
Ans. (a) Pull.
(b) Push; Pull.
(d) Pull.
Note: a) At least two objects must interact for a force to come into play.
b) Force applied on an object in the same direction add to one another.
c) If the forces act in the opposite directions on object, the net force acting on it is the
difference between the two forces.
d) The state of motion of an object is described by its speed & the direction of motion. So we
can say that a force can change the state of motion of an object.
e) S.I unit of force is Newton ( N).
Q.4. In a tug of war, when the two teams are pulling the rope, a stage comes when the rope
does not move to either side at all. What can you say about the magnitudes & directions
of the forces being applied to the rope by the two teams at this stage?
Ans. At this point of time, the two teams are applying equal but opposite forces to the rope due to
which the net force acting on the rope is zero & hence the rope does not move.
Q.5. What is meant by a contact force?
Ans. A force that can be applied only, when an object comes in direct contact with an another object
is called contact force. Eg. Muscular force & Frictional force.
Q.6. What is muscular force?
Ans. The force resulting due to the action of muscles is called muscular force. Eg pushing an object
or lifting a bucket of water.
Note: Since muscular force can be applied to an object only when our body is in contact with
the object, therefore, muscular force is a contact force.
Q.7. What is meant by non-contact force?
Ans. A force that can be applied by an object on another object even from a distance (without
touching) is called a non-contact force. Eg. Magnetic force, Electrostatic force & Gravitational
force.
Q.8. A small device pulls iron nails from a distance. Which type of force is involved in this
process?
Ans. Magnetic force
Q.9. What is electrostatic force?
Ans. The force exerted by a charged body on another charged or uncharged body is known as
electrostatic force.
Q.10. When a plastic pen is rubbed in dry hair, it attracts tiny pieces of paper. Which force is
involved in this process?
Ans. Electrostatic force.
Q.11. What is meant by gravitational force?
Ans. Every object in this universe attracts every other object with a force, called gravitational force.
Q.12. What is meant by Force of gravity?
Ans. The force, with which the earth pulls the objects towards it, is called the force of gravity or
simply gravity.
Q.13. A coin falling to the ground on slipping from hand. Name the type of force involved in
this process?
Ans. Gravitational force.
Q.14. Define pressure.
Ans. The force acting on unit area of surface is called pressure.
Pressure = Force / Area.
Page 3
BAL BHARATI PUBLIC SCHOOL, PITAMPURA, DELHI – 110034
Class-VIII { U.T. - 1}
NOTES : FORCE & PRESSURE
Q.1. What is force? State the various effects of force.
Ans. A push or pull on an object is called force.
Effects of force:
(i) Force can change the speed of a moving object.
(ii) Force can change the direction of a moving object.
(iii) Force can change the shape of an object.
Q.2. Give one example where force can changes the
(a) Speed of a moving object
(b) Direction of a moving object.
© the shape of an object.
Ans. (a) The force applied by a hockey player on a moving ball changes the speed of ball.
(b) The force applied by a batsman changes the direction of moving cricket ball.
© Force applied on a spring can change its shape.
Q.3. Identify the actions involved in the following situations as push or pull, or both:
(a) Opening a drawer.
© A cricket ball hit by a batsman.
(d) Drawing a bucket of water from a well.
Ans. (a) Pull.
(b) Push; Pull.
(d) Pull.
Note: a) At least two objects must interact for a force to come into play.
b) Force applied on an object in the same direction add to one another.
c) If the forces act in the opposite directions on object, the net force acting on it is the
difference between the two forces.
d) The state of motion of an object is described by its speed & the direction of motion. So we
can say that a force can change the state of motion of an object.
e) S.I unit of force is Newton ( N).
Q.4. In a tug of war, when the two teams are pulling the rope, a stage comes when the rope
does not move to either side at all. What can you say about the magnitudes & directions
of the forces being applied to the rope by the two teams at this stage?
Ans. At this point of time, the two teams are applying equal but opposite forces to the rope due to
which the net force acting on the rope is zero & hence the rope does not move.
Q.5. What is meant by a contact force?
Ans. A force that can be applied only, when an object comes in direct contact with an another object
is called contact force. Eg. Muscular force & Frictional force.
Q.6. What is muscular force?
Ans. The force resulting due to the action of muscles is called muscular force. Eg pushing an object
or lifting a bucket of water.
Note: Since muscular force can be applied to an object only when our body is in contact with
the object, therefore, muscular force is a contact force.
Q.7. What is meant by non-contact force?
Ans. A force that can be applied by an object on another object even from a distance (without
touching) is called a non-contact force. Eg. Magnetic force, Electrostatic force & Gravitational
force.
Q.8. A small device pulls iron nails from a distance. Which type of force is involved in this
process?
Ans. Magnetic force
Q.9. What is electrostatic force?
Ans. The force exerted by a charged body on another charged or uncharged body is known as
electrostatic force.
Q.10. When a plastic pen is rubbed in dry hair, it attracts tiny pieces of paper. Which force is
involved in this process?
Ans. Electrostatic force.
Q.11. What is meant by gravitational force?
Ans. Every object in this universe attracts every other object with a force, called gravitational force.
Q.12. What is meant by Force of gravity?
Ans. The force, with which the earth pulls the objects towards it, is called the force of gravity or
simply gravity.
Q.13. A coin falling to the ground on slipping from hand. Name the type of force involved in
this process?
Ans. Gravitational force.
Q.14. Define pressure.
Ans. The force acting on unit area of surface is called pressure.
Pressure = Force / Area.
Unit of pressure is N/m
2
(Newton per square metre) which is also called Pascal (Pa).
Note: Pressure is inversely proportional to area. Therefore smaller the area, larger the
pressure on a surface for the same force.
Q.15. Explain why, school bags are provided with wide straps to carry them?
Ans. As a result the area of contact between the shoulders & the straps increases, thereby
decreasing the pressure on the shoulders, as pressure is inversely proportional to area.
Q.16. Explain why, wooden sleepers are kept below the railway line?
Ans. So as to increase the area of contact between the rails & the ground, thereby decreasing the
pressure on the ground, as pressure is inversely proportional to area. This prevents the sinking
of rails into the ground.
Q.17. Why is it easier for a camel to run in a desert?
Ans. Camel has board feet causes an increase in the area of contact between the sand & its feet. As a
result pressure exerted by camel on the sand decreases because pressure is inversely
proportional to area. This prevents the sinking of its feet’s into the sand.
Q.18. How does the pressure of a liquid depends on its depth?
Ans The pressure exerted by a liquid increases with increasing depth inside the liquid..
Q.19. Where is the pressure greater, 10 m below the surface of the sea or 20 m below the
surface of sea?
Ans. 20 m below the surface of sea.
Q.20. One student says that water exerts pressure on the bottom of the bucket but another
student says that water exerts pressure on the sides of the bucket. What would you like
to say?
Ans. Both the students are correct as water exerts pressure on the bottom as well as on the side
walls of the container.
Q21. Two tiny holes are made in a plastic bucket, one near the middle part & the other just
above the bottom. When this bucket is filled with water, the water rushes out from the
bottom hole much faster than from the upper hole. What conclusion do you get from
these observations?
Ans The pressure exerted by a liquid increases with increasing depth inside the liquid..
Q22. Explain why, the walls of dam are thicker near the bottom than at the top?
Ans The pressure exerted by a liquid increases with increasing depth inside the liquid. That’s why
the wall of a dam is made thicker at the bottom (than at the top) so as to tolerate very high
sideways pressure exerted by deep water.
Page 4
BAL BHARATI PUBLIC SCHOOL, PITAMPURA, DELHI – 110034
Class-VIII { U.T. - 1}
NOTES : FORCE & PRESSURE
Q.1. What is force? State the various effects of force.
Ans. A push or pull on an object is called force.
Effects of force:
(i) Force can change the speed of a moving object.
(ii) Force can change the direction of a moving object.
(iii) Force can change the shape of an object.
Q.2. Give one example where force can changes the
(a) Speed of a moving object
(b) Direction of a moving object.
© the shape of an object.
Ans. (a) The force applied by a hockey player on a moving ball changes the speed of ball.
(b) The force applied by a batsman changes the direction of moving cricket ball.
© Force applied on a spring can change its shape.
Q.3. Identify the actions involved in the following situations as push or pull, or both:
(a) Opening a drawer.
© A cricket ball hit by a batsman.
(d) Drawing a bucket of water from a well.
Ans. (a) Pull.
(b) Push; Pull.
(d) Pull.
Note: a) At least two objects must interact for a force to come into play.
b) Force applied on an object in the same direction add to one another.
c) If the forces act in the opposite directions on object, the net force acting on it is the
difference between the two forces.
d) The state of motion of an object is described by its speed & the direction of motion. So we
can say that a force can change the state of motion of an object.
e) S.I unit of force is Newton ( N).
Q.4. In a tug of war, when the two teams are pulling the rope, a stage comes when the rope
does not move to either side at all. What can you say about the magnitudes & directions
of the forces being applied to the rope by the two teams at this stage?
Ans. At this point of time, the two teams are applying equal but opposite forces to the rope due to
which the net force acting on the rope is zero & hence the rope does not move.
Q.5. What is meant by a contact force?
Ans. A force that can be applied only, when an object comes in direct contact with an another object
is called contact force. Eg. Muscular force & Frictional force.
Q.6. What is muscular force?
Ans. The force resulting due to the action of muscles is called muscular force. Eg pushing an object
or lifting a bucket of water.
Note: Since muscular force can be applied to an object only when our body is in contact with
the object, therefore, muscular force is a contact force.
Q.7. What is meant by non-contact force?
Ans. A force that can be applied by an object on another object even from a distance (without
touching) is called a non-contact force. Eg. Magnetic force, Electrostatic force & Gravitational
force.
Q.8. A small device pulls iron nails from a distance. Which type of force is involved in this
process?
Ans. Magnetic force
Q.9. What is electrostatic force?
Ans. The force exerted by a charged body on another charged or uncharged body is known as
electrostatic force.
Q.10. When a plastic pen is rubbed in dry hair, it attracts tiny pieces of paper. Which force is
involved in this process?
Ans. Electrostatic force.
Q.11. What is meant by gravitational force?
Ans. Every object in this universe attracts every other object with a force, called gravitational force.
Q.12. What is meant by Force of gravity?
Ans. The force, with which the earth pulls the objects towards it, is called the force of gravity or
simply gravity.
Q.13. A coin falling to the ground on slipping from hand. Name the type of force involved in
this process?
Ans. Gravitational force.
Q.14. Define pressure.
Ans. The force acting on unit area of surface is called pressure.
Pressure = Force / Area.
Unit of pressure is N/m
2
(Newton per square metre) which is also called Pascal (Pa).
Note: Pressure is inversely proportional to area. Therefore smaller the area, larger the
pressure on a surface for the same force.
Q.15. Explain why, school bags are provided with wide straps to carry them?
Ans. As a result the area of contact between the shoulders & the straps increases, thereby
decreasing the pressure on the shoulders, as pressure is inversely proportional to area.
Q.16. Explain why, wooden sleepers are kept below the railway line?
Ans. So as to increase the area of contact between the rails & the ground, thereby decreasing the
pressure on the ground, as pressure is inversely proportional to area. This prevents the sinking
of rails into the ground.
Q.17. Why is it easier for a camel to run in a desert?
Ans. Camel has board feet causes an increase in the area of contact between the sand & its feet. As a
result pressure exerted by camel on the sand decreases because pressure is inversely
proportional to area. This prevents the sinking of its feet’s into the sand.
Q.18. How does the pressure of a liquid depends on its depth?
Ans The pressure exerted by a liquid increases with increasing depth inside the liquid..
Q.19. Where is the pressure greater, 10 m below the surface of the sea or 20 m below the
surface of sea?
Ans. 20 m below the surface of sea.
Q.20. One student says that water exerts pressure on the bottom of the bucket but another
student says that water exerts pressure on the sides of the bucket. What would you like
to say?
Ans. Both the students are correct as water exerts pressure on the bottom as well as on the side
walls of the container.
Q21. Two tiny holes are made in a plastic bucket, one near the middle part & the other just
above the bottom. When this bucket is filled with water, the water rushes out from the
bottom hole much faster than from the upper hole. What conclusion do you get from
these observations?
Ans The pressure exerted by a liquid increases with increasing depth inside the liquid..
Q22. Explain why, the walls of dam are thicker near the bottom than at the top?
Ans The pressure exerted by a liquid increases with increasing depth inside the liquid. That’s why
the wall of a dam is made thicker at the bottom (than at the top) so as to tolerate very high
sideways pressure exerted by deep water.
Q23. What is meant by atmospheric pressure?
Ans. The envelop of air around us is known as atmosphere. Now the pressure exerted by this air is
known as atmospheric pressure.
Note:
(a) The atmospheric pressure is maximum at the sea level.
(b) As we go up in the atmosphere from the surface of earth, the atmospheric pressure goes on
decreasing.
Q24. What is the cause of atmospheric pressure?
Ans. The atmospheric pressure is due to the weight of air present in the atmosphere above us.
Q25. Why are our bodies not crushed by the large pressure exerted by the atmosphere?
Ans. The reason that we are not crushed is that the pressures inside our bodies is also equal to the
atmospheric pressure & cancel the pressure from outside.
Note: (a) At high altitudes, the atmospheric pressure becomes much less than our blood
pressure. Therefore some of the blood vessels in our body burst & nose bleeding
take place at high altitudes.
(b) Rubber suckers are usually used to hold objects with the help of suctions. When
we press the sucker, most of the air between its cup & the surface escapes out.
The sucker sticks to the surface because the pressure of atmosphere acts on it. If
there is no air between the cup & the surface, than it would be impossible to pull
the sucker off the surface.
NUMERICALS:
1. Calculate the pressure when a force of 200 N is exerted on an area of:
(a) 10 m
2
(b) 5 m
2
2. What force acting on area of 0.5 m
2
will produce a pressure of 500 Pa?
3. The tip of drawing pin has an area 1.0 x 10
-8
m
2
. Find the pressure exerted if force applied is
10 N.
```
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# The Pythagorean Theorem04:35 minutes
Video Transcript
## TranscriptThe Pythagorean Theorem
Damo, the son of Pythagoras, wants to build a vacation cottage for his mother and father. In keeping with the family business of triangles, Damo decides to incorporate the shape into the design of the cottage. What a perfect time to use his dad's theorem:The Pythagorean Theorem!
### Right Triangles and Pythagorean Triples
For the roof of the cottage, Damo will use a triangle design, but what size and shape should the triangle be? Damo's dad spoke of many combinations of integers to create the three sides of right triangles, called Pythagorean triples, but which one is best? Damo thinks back to his dad's class on triangles from school. His dad taught him well! He remembers that right triangles have one and only one right angle, and a right angle is equal to 90°. Even though one angle has to be 90°, there are several different combinations of side lengths that Damo can use.
These special relationships are called Pythagorean Triples. If the integer side lengths of a triangle create a Pythagorean triple, the formula a² + b² = c² will work. How about this triangle, is it a Pythagorean triple? Let's see. The triangle has a right angle, and if you sum the area of the two smaller squares, the sum is equal to the area of the larger square.
### Examples for Pythagorean Triples
Damo has a choice of several different roof shapes, but he wants one in the shape of a right triangle, and it must be just right for the cottage. The first Pythagorean triple Damo remembers is 3-4-5. Plugging this into dear ol' dad's theorem gives him, 3² + 4² = 5² ...which is 9 + 16 = 25. This is awesome! No wonder dad gets so excited about right triangles. He knows this one is in the shape of a right triangle, but it's too small.
Next, he tries side lengths 8, 15, and 17. Let’s see, is it a right triangle? Shall we try out the theorem? 8² + 15² = 17². So...64 + 225 = 289. Yup, this one works too, but darn – it looks just like the roof on top of Yorgos’ Yogurt Shop, and that just won’t do for his parents.
There are only a few more roof styles that the roof cutters offer: Damo checks the larger roof, with side measures 7-24-25, first. Does it satisfy the theorem? 7² + 24² = 25². This is the same as 49 + 576 = 625. Yes it works, but it’s just too big.
There's one last option, a triangle with side lengths of 5, 12 and 13. I sure hope this one works 25 + 144 = 169. Yeah! This shape is a right triangle, and it's just the right size and style. So Damo picked a roof that's just right. He also practiced the Pythagorean Theorem and learned some Pythagorean triples. Seems like a win, win to me!
The roof of the cottage looks great, and his dad looks very pleased with the shape. Oh no! Damos forgot that different materials have different densities. Oh Geez. Where was Damo when his dad was teaching the unit on density?? I guess he can still learn something from his dear ol' dad. Join us next time for more triangular adventures with Damo!
## Videos in this Topic
Trigonometry (7 Videos)
## The Pythagorean Theorem Exercise
### Would you like to practice what you’ve just learned? Practice problems for this video The Pythagorean Theorem help you practice and recap your knowledge.
• #### Define the Pythagorean Theorem.
##### Hints
The Pythagorean theorem focuses on the side lengths of right triangles.
The Pythagorean Theorem focuses on the relationships between the side lengths of right triangles.
Some options in the answer set are true statements about triangles, but are not part of the Pythagorean Theorem.
##### Solution
The Pythagorean Theorem relates the side lengths of right triangles. The theorem states that the side lengths of right triangles have a special relationship with each other that is described by the equation $a^2 + b^2 = c^2$. Further, there are sets of integers that satisfy this equation, which are known as "Pythagorean triples."
Knowing this, let's evaluate each of the statements and determine if they describe the Pythagorean theorem.
• The three side lengths of right triangles form sets called "Pythagorean triples." ✓ This is correct, and is part of the Pythagorean Theorem.
• A right triangle must have two acute angles. ✗ Although this statement is true, it is not part of the Pythagorean Theorem. This theorem focuses on the side lengths of a right triangle, not the other angles in a right triangle.
• If a triangle is a right triangle, the side lengths $a$, $b$, and $c$, will satisfy the equation $a^2 + b^2 = c^2$. ✓ This is correct, and is part of the Pythagorean Theorem.
• The hypotenuse will always be the longest side of a right triangle. ✗ Although this statement is true, it is not part of the Pythagorean Theorem. It can be proved using the Pythagorean Theorem, though!
• #### Determine if the side lengths are a Pythagorean triple.
##### Hints
Remember that the Pythagorean Theorem can be represented as an equation that uses the squares of the side lengths of a triangle.
The Pythagorean triples must not include fractions or decimals.
The set of side lengths is a Pythagorean triple, if they satisfy the equation $a^2 + b^2 = c^2$. This means that the left hand side of the equation must equal the right hand side.
##### Solution
The equation given by the Pythagorean Theorem is $a^2 + b^2 = c^2$. Integers that satisfy these equations are known as "Pythagorean triples."
We are given the side lengths of the triangle: 8, 15, and 17. All being whole numbers, they are definitely also integers. So if the set of numbers satisfies the equation $a^2 + b^2 = c^2$, then it must be a Pythagorean triple.
The value $c$ is the length of the hypotenuse, which is the longest side. It is also the side across from the right angle. Therefore this side is 17 feet long. Either of the other two sides can act as $a$ or $b$, since they are interchangeable.
Let's assign the values $a = 8$, $b = 15$, and $c = 17$.
Substituting these values in, we get:
$a^2 + b^2 = c^2$
$8^2 + 15^2 = 17^2$
$64 + 225 = 289$
$289 = 289$
The left and right hand sides of the equation are equal. This means that the values satisfy the equation. We can see that this triangle does have side lengths that are a Pythagorean triple.
• #### Find the areas.
##### Hints
How can you use the Pythagorean Theorem to help you with this question?
The area of each square that is connected to a side represents the literal "square" of that side length. This is equal to the side length, squared.
What is the relationship between the square of each side length?
You can use the equation $a^2 + b^2 = c^2$ to find the area of the square that is attached to the side that has an unknown length.
##### Solution
In each of the two cases, we have a right triangle, and we know the lengths of two sides. We must find the area of the square that is connected to the hypotenuse. The area of each square that is connected to a side represents the literal "square" of that side length. This is equal to the side length, squared. We do not know the length of the hypotenuse, but there is another way to find this area.
From the Pythagorean Theorem, we know that:
$a^2 + b^2 = c^2$
We know that $c^2$ represents the area that we must find. We know the side lengths $a$ and $b$.
$a^2 = 3^2$
$a^2 = 9$
$b^2 = 4^2$
$b^2 = 16$
Now we can find $c^2$:
$a^2 + b^2 = c^2$
$9 + 16 = c^2$
$25 = c^2$
Similarly, for the second triangle...
$a^2 = 36$
$b^2 = 64$
$100 = c^2$
• #### Decide if the ladder fits.
##### Hints
Remember: the hypotenuse is always the longest side of the a right triangle.
If the equation is satisfied by the given values, then the ladder is the perfect length to fit between the ground and the roof, with the base of the ladder sitting 10 feet from the cottage.
In this question we move straight from the equation $a^2 + b^2 = c^2$, to the square of the values we substitute in. For example, if we had $a = 3$, $b = 4$, and $c = 5$, we would write:
$a^2 + b^2 = c^2$
$9 + 16 = 25$
By doing this, we are skipping the middle step:
$3^2 + 4^2 = 5^2$
##### Solution
The ladder, the cottage, and the ground form a right triangle. We are certain about the length of two of the sides of this triangle. We know that the cottage is 24 feet tall. And we know that the bottom of the ladder must sit at least 10 feet from the base of the cottage. The farther away from the cottage the base of the ladder is placed, the longer the ladder will need to be. Let's assume we place the base of the ladder exactly 10 feet from the cottage, so that Damo is able to use the shortest possible ladder.
If the ladder is the correct length, then this set of side lengths should satisfy the Pythagorean theorem, $a^2 + b^2 = c^2$. We know that $c$ should be the hypotenuse. We can see that the ladder, which is 26 feet long, is opposite the right angle. Additionally, the ladder forms the longest side of this right triangle. Therefore it must be the hypotenuse. The other two sides can be labelled $a$ and $b$ interchangably.
$a = 10$
$b = 24$
$c = 26$
By substituting the values in, and simplifying the equation, we can check if the equation is satisfied by these values:
$\begin{array}{rclcl} a^2 + b^2 & = & c^2 &|&\text{subsituting values} \\ 10^2 + 24^2 & = & 26^2 \\ 100 + 576 & = & 676 \\ 676 & = & 676 &|&\text{✓ the two sides are equal} \\ \end{array}$
The two sides of this equation are equal. Therefore the values do satisfy the equation, and the ladder is the correct length.
• #### Identify the right triangles.
##### Hints
How can you determine if a triangle is a right triangle using the Pythagorean Theorem?
What equation can you use to check if a triangle is right, using it's side lengths, $a$, $b$, and $c$?
If a triangle's side lengths, $a$, $b$, and $c$, satisfy the equation $a^2 + b^2 = c^2$, then it is a right triangle.
##### Solution
We know that if a triangle is right, it's side lengths, $a$, $b$, and $c$, must satisfy the equation:
$a^2 + b^2 = c^2$
We are given the side lengths for each triangle. Let's use the equation to check if the triangles are right angle triangles.
The side lengths of the first triangle are 6, 8, and 10. We know that the value for $c$ must be the longest side of the triangle.
So let's assign the values $a = 6$, $b = 8$, and $c = 10$.
Putting these values into the equation gives us:
$6^2 + 8^2 = 10^2$
$36 + 64 = 100$
$100 = 100$
Since the right hand side is equal to the left hand side, the values of $a$, $b$, and $c$ for this triangle satisfy the equation. That means that this is a right triangle!
Similarly, for the other triangles, we arrive at:
$a = 9$, $b = 40$, $c = 41$:
$81 + 1600 = 1681$ ✓ The left and right sides of this equation are equal, so the triangle is right.
$a = 25$, $b = 60$, $c = 65$:
$625 + 3600 = 4225$ ✓ The left and right sides of this equation are equal, so the triangle is right.
$a = 11$, $b = 60$, $c = 61$:
$121 + 3600 = 3721$ ✓ The left and right sides of this equation are equal, so the triangle is right.
$a = 26$, $b = 36$, $c = 45$:
$676 + 1296 = 2025$ ✗ The left and right sides of this equation are not equal, so the triangle is not right.
$a = 40$, $b = 75$, $c = 85$:
$1600 + 5625 = 7225$ ✓ The left and right sides of this equation are equal, so the triangle is right.
• #### Identify the Pythagorean triples.
##### Hints
Sometimes, the two sides of the simplified equation will not be equal. When that happens, the set is not a Pythagorean triple.
Be sure to carefully check that you have squared the terms correctly.
##### Solution
Damo wants to know if these sets of numbers are Pythagorean triples. We can check this using the equation $a^2 + b^2 = c^2$, where $c$ is the largest number in the set. $a$ and $b$ are the other two numbers, and are interchangeable.
Looking at the first set, $5, 4$ and $6$, let's set:
$a = 5$
$b = 4$
$c = 6$ (because $6$ is the largest number)
We can substitute these values into the equation:
\begin{array}{rclcl} a^2 + b^2 & = & c^2 &|&\text{substituting} \\ 5^2 + 4^2 & = & 6^2 &|&\text{simplifying} \\ 25 + 16 & = & 36 &|&\text{simplifying} \\ 41 & = & 36 &|&\text{✗ the two sides of the equation are not equal} \\ \end{array}
Therefore this set of numbers does not satisfy the equation, and these numbers are not a Pythagorean triple. So the statement in the question is false.
Similarly, for the second set, $12, 16$ and $20$:
$12^2 + 16^2 = 20^2$
$144 + 256 = 400$
Both sides of this equation are equal. Therefore the set is a Pythagorean triple.
And for the third set, $15, 20$ and $25$:
$15^2 + 20^2 = 25^2$
$225 + 400 = 625$
Both sides of this equation are equal. Therefore the set is a Pythagorean triple.
|
# What is 16 over 18 as a decimal?
## What is 16 over 18 as a decimal?
16/18 as a decimal is 0.88888888888889.
What is 16% as a fraction?
4/25
Answer: 0.16 as a fraction in its simplest form is 4/25. Let’s evaluate.
### What is 18% as a fraction?
9/50
Answer: 18% can be represented as 9/50 in fractional form.
What’s 18 percent as a decimal?
Why do we move the decimal point 2 places to the left?
Solution
Percent Decimal
18% .18
07% .07
82% .82
## What percent is a 16 out of 18?
What is this? What is this? Now we can see that our fraction is 88.888888888889/100, which means that 16/18 as a percentage is 88.8889%.
What is 0.125 as a fraction?
1/8
0.125 = 125/1000. We can reduce this to lowest terms by dividing the numerator and denominator by 125 to get the equivalent fraction 1/8.
### What is .17 as a fraction?
Similarly, if you divide 17 by 100 , you get 0.17 . Thus, the fraction form for 0.17 is 17100 . Since there is no common factor which can divide the numerator, 17 , and the denominator, 100 , besides 1 , 17100 is the final answer.
What is the decimal for 1 16?
0.0625
Answer: 1/16 as a decimal is written as 0.0625.
## What is 15 percent as a fraction?
3/20
Answer: 15% is expressed as 3/20 as a fraction.
How do you convert percent to a fraction?
To convert a percent to a fraction, we have to remove the percent sign and divide the given number by 100. And, then we express the fractional form of the percentage in the simplest form. For example, 1% is 1/100, 2% is 2/100 which can be reduced to 1/50.
### What is 15 out of 18 as a percentage?
Now we can see that our fraction is 83.333333333333/100, which means that 15/18 as a percentage is 83.3333%.
What grade is a 16 out of 20?
Now we can see that our fraction is 80/100, which means that 16/20 as a percentage is 80%. And there you have it!
0.16 or 16% as a Fraction Decimal to Fraction Number calculator – online basic math function tool to convert decimal point number 0.16 to fraction equivalent. Simplest fraction: 0.16 = 16/100 = 4/25; percentage: 0.16 = 16/100 or 16% Get it! Home Math Functions Decimal to Fraction Calculator for 0.16 How to Write 0.16 or 16% as a Fraction?
## What is 16/100 as a percentage?
0.16 or 16% as a Fraction. Decimal to Fraction Number calculator – online basic math function tool to convert decimal point number 0.16 to fraction equivalent. Simplest fraction: 0.16 = 16/100 = 4/25; percentage: 0.16 = 16/100 or 16%. Get it!
What is 16/100 as a fraction in simplest form?
It can be written as 16% = 16/100 or 16/100 step 4 To simplify 16/100 to its lowest terms, find LCM (Least Common Multiple) for 16 & 100. 4 is the LCM for 16 & 100 step 5 divide numerator & denominator by 4 16/100 = (16 / 4) / (100 / 4) = 4/25 4/25 is a simplest fraction for the decimal point number 0.16
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# Measures of Dispersion for Numerical Data Study Guide (page 2)
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#### Example
Find the interquartile range of the population of orchestra members.
#### Solution
The median, 66.85 inches, divided the population into two halves, those members who are less than 66.85 inches tall and those who are greater than 66.85 inches tall. There are 31 members in each half. Because an odd number of values exists in the half, the median of the lower half is the = 16th value in the ordered set of the 31 lower values. Thus, the first quartile is Q1 = 62.5 inches. Similarly, the third quartile is the median of the 31 upper half values. The 16th value in the ordered set of upper values is Q3 = 69.5 inches. The interquartile range is now found to be
Q3Q1 = 69.5 – 62.5 = 7 inches
which is a value much smaller than the range. This is the spread in the middle half of the population of heights.
## Mean Absolution Deviation
Before describing the next measure of dispersion, we need to define what is meant by a deviation. The quantity, (Xiμ) , is the deviation of the ith population value from the population. By taking the absolute value, the deviation is a measure of how far the value is from the mean. For the orchestra members, the shortest person has a deviation of 53.5 – 66.4 = –12.9 inches. The negative sign indicates that this person's height is below the mean; that is, the shortest member's height is 12.9 inches below the mean height of all orchestra members. The member who is 74 inches tall has a deviation of 74 – 66.4 = 7.6 inches. He is 7.6 taller than the average height of all orchestra members. If we add all of the deviations together, we get zero.
Mean absolute deviation is the mean (or average) distance of the population values from the population mean; that is,
Now, (Xiμ) is the deviation of the ith population value from the mean. The absolute deviation of the ith population value from the population mean is the distance of that value from the population mean, |Xi μ |. For the populations of X, Y, and Z, we have μ = 15. For a population value of 13, the distance of that value from the mean is |13 –15| = | –2 | =2. The mean absolute deviation is the population mean of these distances, E(|Xμ|).
The mean absolute deviation of the population distribution of Y is 0.75, and that of Z is 1.14.Notice that the greater dispersion in the distribution of Z as compared to Y is captured in the mean absolute deviation.
#### Example
Find the mean absolute deviation of the heights of the 62 orchestra members. Interpret the mean absolute deviation in the context of the problem.
#### Solution
We begin by finding the deviation and the absolute deviation from the mean for each band member. The mean height was 66.4, so the deviation for a particular orchestra member is that member's height minus 66.4. The absolute deviation is the absolute value of the deviation, or how far the member's height is from the mean. These are given in Table 6.1.
The mean absolute deviation is the average of the absolute deviations. For the orchestra members, the mean absolute deviation is 4.32 inches. This means that, on average, an orchestra member's height is 4.32 inches from the population mean height of 66.4 inches.
## Variance and Standard Deviation
Two more measures of dispersion are the variance and the standard deviation. The variance is the mean squared distance of the population values from the mean; that is,
Notice that we have expressed the variance as an expected value.Here, it represents the average squared deviation of a population value from the mean. For the population distribution of Y, displayed in Figure 6.1, the variance is
.
For the population distribution of Z, displayed in Figure 6.2, the variance is 2.2. The units associated with the variance are the square of the measurement units of the population values.As an illustration, if the population values are recorded in inches, as they are with the orchestra members' heights, the variance is in inches2. To obtain a measure of dispersion in the same units as the population values, the standard deviation is found to be the square root of the variance; that is, σ = √σ2.
Although not technically correct, the standard deviation is often described as being the average distance of a population value from the mean. For the population distribution of Y, the standard deviation is √1 = 1. For Z, displayed in Figure 6.2, the population standard deviation is √2.2 = 1.5. The mean absolute deviation and the standard deviation are the same for Y. This is very unusual.More commonly, the mean absolute deviation is close, but not equal to the standard deviation. For Z, the mean absolute deviation is 2, and the standard deviation is 1.5. Although these parameters are different, we will often interpret the standard deviation as we would the mean absolute deviation.We simply need to realize that this interpretation of the standard deviation is only an approximate one.
Which measure of dispersion should we use? It is always good to compute more than one measure of dispersion as each gives you slightly different information. The range is often reported. The mean absolute deviation is an intuitive measure of dispersion, but it is not used much in practice, primarily because it is difficult to answer more advanced statistical questions using mean absolute deviation. Most of the statistical methods are based on the standard deviation. However, the range, mean absolute deviation, and standard deviation are all inflated by unusually large or unusually small values in the population. In this case, the IQR is often the best measure of dispersion.
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1 Induction proofs To prove that some proposition P (n) is true for all n n 0 by induction on n n 0 . Basis : Check that P (n 0 ) is true. 2. Inductive step . Assume that P (n ) is true for n = k , where k is some integer, k n 0 Prove that P (n) is true for n = k +1 , in inductive step we prove that for any n n 0 P (n) P (n+1) lly this is simpler than prove directly that for an ).
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Induction proofs. To prove that some proposition P ( n ) is true for all n n 0 by induction on n n 0. 1. Basis : Check that P ( n 0 ) is true. 2. Inductive step . Assume that P ( n ) is true for n = k , where k is some integer, k n 0 - PowerPoint PPT Presentation
### Transcript of Induction proofs
• Induction proofs To prove that some proposition P (n) is true for all n n0 by induction on n n0 1. Basis : Check that P (n0) is true.2. Inductive step. Assume that P (n ) is true for n = k , where k is some integer, k n0 Prove that P (n) is true for n = k +1So, in inductive step we prove that for any n n0 P (n) P (n+1)Usually this is simpler than prove directly that for any n n0 P (n).
• Example. Let P (n) : n3n is divisible by 6 To prove that P (n) is true for all n 0:1. Basis. P (0): 03 0=0 is divisible by 6. P (0) is true 2. Inductive step P (n =k ) P (n =k +1), k 0. Assume P (n =k ) is true, where k is some integer, k 0. So, we assume that k 3 k is divisible by 6 i.e. k 3 k = 6 s (s is any integer)Goal: to prove that P (n =k +1) is true, i. e. that(k +1)3 (k +1) is divisible by 6.(k +1)3 (k +1) = k 3 +3 k2 + 3 k +1 k 1 = = (k 3 k ) + 3 k2 + 3 k = = 6 s + 3 k (k +1)
• Suppose we want to prove the formula for the sum of geometric progressionHow can we prove that the formula is correct for any n0 ?
• Show it for some n0 (prove P (n) for any n directly)
• 1) Base: n=0 2) Induction Hypothesis: Assume the formula is correct for n=k, where k is some integer, k0 Induction Step: prove the formula for n=k+1, i. e.
• Induction Principle . If A is a subset of N that satisfies twoproperties:1) n0 A 2) n n0 [n A(n+1)A], then A =N We just can think about A as a set of integers, that satisfy P:A ={nN | P (n)}Then,1) n0 A P (n0 ) 2) n n0 [nA(n+1)A] n n0 (P(n) P (n +1))
• Although it sounds very obvious, the induction principle hinges on the property of integers, so called Well-Ordering principle (accept it as an axiom). Well-Ordering Principle. Any nonempty subset of N contains a smallest element. Neither rational, no real numbers has this property!Induction principle is equivalent to the Well-ordering principle.
• Theorem. Well-ordering principle (WOP) implies Induction Principle(IP)Proof. Suppose WOP is true, i.e. any nonempty subset of N contains a smallest element. We want to prove that IP is true, i. e. if any set A satisfies propertiesi) 0Aii) n 0 [nA(n+1)A], then A =N.
• We are going to prove it by contradiction. 1) Assume that A satisfies both properties, but A N . 2) Then N A = B , i. e. B is nonempty subset of N .3) By WOP B contains the smallest element. Let denote it s, s B, so sA. 4) Since s is the smallest element of B, s1B. 5) Since 0A by i), 0B, so s >0 and s1 0, i. e. s1N.6) s1N and s1B imply s1A .7) By ii) s1A sA , contradiction with sA.
• n012...i) 0A A = {n N | P(n)} ii) n 0 [nA(n+1)A]Proof by contradiction:Assume that A N. s 1 A, since s is the smallest in contradiction with sB
• Example. Prove that the number of different binary strings of length n is 2n.Proof by induction on n1.We want to prove: n1 [P (n)], where P (n): the number of binary strings of length n is 2n. 1) Basis. n=1. We have two strings of the length one: 0 and 1.2) Assume P (n) for n=k, where k is some integer, k 1 (IH).So, we assume that for some k 1 there are 2k binary strings.In the IS we need to prove P (n) for n=k+1I. e. we need to show that there are 2(k+1) binary stringsof the length (k +1).
• Here is the proof that there are 2(k+1) binary strings of length (k+1).Any binary string of length (k+1) can be represented as a binarystring of length k with one extra bit (let it be the last one). There are two choices for the last bit: 0 or 1. By IH there are 2k strings of length k. Then, we have 2k strings of length (k+1) that end with 0 and 2k strings of length (k+1) that end with 1, so the total number is 2k + 2k =2 2k =2(k+1) QED. By IP we conclude, that for any n 1 the number of binary strings is 2n.
• Lets use the induction to prove the following known fact. Example. Prove that for any set A cardinality of a power set is |Power(A)|=2|A|.To apply induction we need to chose the integer variable.Here we can prove by induction on n = |A|. So, we want to prove that the propositionP : |Power(A)|=2|A| is true for any n=|A| 0.1) Basis: n=|A|=0, A= , There is only one subset of the empty set, so |Power()|=1=202) IH: Assume that for n=k, k is any integer k 0, we have that any set A with |A|=k has 2k subsets. IS: we need to imply that any set B with cardinality |B|=k+1 has the property P, i. e. has 2(k+1) subsets.
• Here we need to do the jobTake any set B of k+1 elements. Pick any element x. Then all subsets of B can be divided into two parts: i) subsets that do not contain x. How many such subsets exist? We can argue, that subsets of B that do not contain x constitute subsets of the set B {x}, that has cardinality k. | B {x}| = k and by IH there exist 2k subsets.ii) subsets that include x. How many such subsets exist?Again, the number of such subsets equals the number of Subsets of the set B {x}, this number is 2k. Now, the total number of subsets is the sum of two groups, 2k + 2k =2 2k= 2k+1. So, from assumption that the proposition is true for n=k ( k is any integer k 0) we can imply that the proposition is true for n =k+1.By Induction principle we conclude that the proposition is true for all integer n 0, i. e. for sets of any size |A|=n.
• What is wrong with the following reasoning? All horses are the same color P (n): n horses are the same color Basis. P (1)=true.IS . Assume that for some k 1 P (k) is true. We need to prove that P (k +1) is true.
• You are given a board divided into mm equal squares, where m is a power of 2. One arbitrary square is distinguished as special. Prove that with exception of one special square that may be arbitrarily located on the board, the board can be always covered with L-shaped tiles.Tiling problem.A special square
• Proof by induction on k1, where m=2k is the side of the board.
• Induction Hypothesis. Assume that for a board of 2k2k squares the tiling problem can be solved, i.e. for arbitrary location of a special square the board can be covered by L-shaped tiles. In induction step we need to prove that the tiling can be solved for the board of 2k+12k+1 squares.
• In accordance with IH, tiling problem can be solved for each of these four boards for arbitrary positions of special squares. In particular, it means that we can choose three of specials around the center, so that to use additional tile for them. The remaining special can be placed anywhere and in this way we proved that tiling problem can be solved for a 2k+12k+1 - board.
• Induction and Recursion. Explicit definitionRecursive definition Factorial Geometric progression Power of relation on a set
• The recursive definition of a function makes reference to earlier versions of itself. The main connection between recursion and induction is that objects are defined by means of a natural sequence. Induction is usually the best (possibly the only) way to prove results about recursively defined objects.
• The sequence named for Fibonacci (1202) is introduced in terms of rabbits. Suppose that every month a breeding pair of rabbits produce a pair of offspring. The offspring in turn start breeding two months later, and so on. Denote Fn the number of pairs in month n. Suppose you buy a pair of rabbits in month 1 (F0=0, F1=1) then you still have one pair in month 2 (F2=1), but in month 3 they start breeding, F3=2. The sequence begins 0, 1, 1, 2, 3, 5, 8, 13, 21, You can observe, that in a month n you have all pairs from the previous month (n1), plus offspring pairs for all pairs from month (n2), i. e. :Fibonacci Numbers
• Proof by induction on n 0.So, Fibonacci numbers can be defined recursively: These numbers have numerous applications in CS, mathematics, theory of games.
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Question Video: Stating the System of Inequalities That Describes a Given Situation | Nagwa Question Video: Stating the System of Inequalities That Describes a Given Situation | Nagwa
# Question Video: Stating the System of Inequalities That Describes a Given Situation Mathematics • First Year of Secondary School
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Anthony and Daniel went on a driving tour to Luxor and Aswan and took turns driving. Each day, Anthony drove for at least 4 hours and no more than 8 hours, while Daniel drove at least 2 hours and less than 7 hours. The total time they drove daily was no more than 12 hours. State the system of inequalities that describes the situation using π₯ to represent the number of hours that Anthony drove and π¦ to represent the number of hours that Daniel drove.
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### Video Transcript
Anthony and Daniel went on a driving tour to Luxor and Aswan and took turns driving. Each day, Anthony drove for at least four hours and no more than eight hours, while Daniel drove at least two hours and less than seven hours. The total time they drove daily was no more than 12 hours. State the system of inequalities that describes the situation using π₯ to represent the number of hours that Anthony drove and π¦ to represent the number of hours that Daniel drove.
In this question, we are told to let π₯ represent the number of hours that Anthony drove and π¦ represent the numbers that Daniel drove. We are given information about both drivers together with the total time they drove. Letβs begin by considering Anthony, we are told that he drove for at least four hours and no more than eight hours each day. This means that π₯ must be greater than or equal to four and less than or equal to eight.
We are told that Daniel drove at least two hours and less than seven hours. This means that π¦ must be greater than or equal to two and less than seven. It is important to note that we have a strict inequality at the upper end, as Daniel drove less than seven hours. Finally, we are told that the total time they drove daily was no more than 12 hours. This total time will be the sum of π₯ and π¦. And we therefore have the inequality π₯ plus π¦ is less than or equal to 12.
The system of inequalities that describes the situation is π₯ is greater than or equal to four and less than or equal to eight, π¦ is greater than or equal to two and less than seven, and π₯ plus π¦ is less than or equal to 12. Whilst it is not required in this question, we could also model the situation graphically.
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# Pure Mathematics 2: Solution of equation by Numerical Methods
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Introduction
Pure Mathematics 2: Solution of equation by Numerical Methods Introduction: In this coursework, I am going to solve equations by using the Numerical Methods. Numerical methods are used to solve equation that cannot be solved algebraically e.g. quadratic equations ax�+bx+c=0 can be solved using this formula: x= -b� V b� - 4ac 2a Therefore numerical methods would not be used for quadratic equations. I will be working with cubic equation because there is no formula to solve it. There are three methods, which I will be using: * Change of sign method * Newton-Raphson method * Rearranging f(x) = 0 in the form x = g(x) Change of sign method: This method is concerned with when a function crosses the x-axis, and by definition changes sign (+ and -). If we are looking the root of equation f(x) = 0. The point at which the curve crosses x-axis is the root. Once an interval where f(x) changes sign then the root must be in the interval. f(a) > 0 f(b) < 0 Therefore root must be between [a,b] f(a) < 0 f(b) > 0 Root is between the interval [a,b] To find the interval of each root for the equation, I'll be doing a decimal search first. Lets take the equation y = x� - 12x + 5 x -4 -3 -2 -1 0 1 2 3 4 f(x) -11 14 21 16 5 -6 -11 -4 21 There are 3 roots in this equation and they are in these intervals: [-4, -3] [ 0 , 1 ] [ 3 , 4 ] Now, I am going to use Interval Bisection to find out one of the root. This method is similar to decimal search but instead of dividing each interval into 10 parts, using interval bisection, only need to divide into 2, which is more simpler. f(x) = x� - 12x + 5 I am going to use the root in the interval [0,1] f(0) ...read more.
Middle
f(x(n)) f'(x(n)) x(n+1) 1 2 3 9 1.666667 2 1.666667 0.62963 5.333333 1.548611 3 1.548611 0.06804 4.194589 1.53239 4 1.53239 0.001218 4.044659 1.532089 5 1.532089 4.17E-07 4.04189 1.532089 6 1.532089 4.88E-14 4.041889 1.532089 7 1.532089 0 4.041889 1.532089 This shows the root is close to 1.532089 (6 d.p.) Failure of Newton-Raphson method: To test the failure of the Newton-Raphson method, I am going to choose another equation; y = x� -4x + 2 f(x) = x� - 4x + 2 xn+1 = xn - xn� - 4xn + 2 3xn� - 4 Let x1 = 1.1 This is close to the root in interval [0,1] and [1, 2], therefore using the Newton-Raphson method, it should converge to either one of these roots. x1 = 1.1 x2 = 1.1 - (1.1� - 4(1.1) + 2) 3(1.1�) - 4 = 1.1 - (1.331 - 4.4 + 2) 3.63 - 4 = 1.1 - (-1.069) -0.37 = 1.1 - 2.889189189 = -1.789189 x3 = -1.789189 - (-1.789189� - 4(-1.789189) + 2) 3(-1.789189�) - 4 = -1.789189 - (-5.7275456956 - (-7.156756) + 2) 9.603591833 - 4 = -1.789189 - 3.429210305 5.603591833 = -1.789189 - 0.611966468 = -2.401155 x4 = -2.401155 - (-2.401155� - 4(-2.401155) + 2) 3(-2.401155�) - 4 = -2.401155 - (-13.84396801 - (-9.60462) + 2) 17.296636 - 4 = -2.401155 - (-2.23934801) 13.296636 = -2.401155 - (-0.168414628) = -2.232740 n x(n) f(x(n)) f'(x(and)) x(n+1) 1 1.1 -1.069 -0.37 -1.789189 2 -1.789189 3.429208 5.603594 -2.401155 3 -2.401155 -2.239348 13.29664 -2.23274 4 -2.23274 -0.199539 10.95539 -2.214527 5 -2.214527 -0.002216 10.71238 -2.21432 6 -2.21432 -2.84E-07 10.70964 -2.21432 7 -2.21432 -5.33E-15 10.70964 -2.21432 8 -2.21432 0 10.70964 -2.21432 As you can see, the root is diverging towards the interval [-2, -3] instead of converging towards [0, 1]. This diagram shows clearly that when x1 is 1.1, the tangent diverts to -2, the reason that this happens is because the value that I chosen to draw the tangent is too close to the turning point of the curve, therefore when the tangent is drawn, it diverts to -2. ...read more.
Conclusion
However, the root that is found using this method is just as accurate as the other two methods, but using this method, the number of steps to work the root out is 20, and is the most steps taken in all three methods. * Newton-Raphson The Newton-Raphson method gave the fastest speed of convergence, although when calculating the root using the Newton-Raphson iteration formula takes a lot of time because xn, follows one after another. When the iterative formula is used, it can fail to find the root if the calculations show divergence (when x1 is too close to the turning point). Using the iterative formula, only 5 steps are needed to work out the root. * Rearranging method Rearranging the function of f(x) = 0 to g(x), using the iteration formula, and the staircase diagram, this converges to the root in [0, 1] but have to rearrange the formula again in order to work out the other two. Therefore, using this method may take longer to work out all three roots in the equation. Only 6 steps are needed to work out the root. All three methods are at 6 d.p. Uses of hardware and software: In this investigation, I had used quite a range of software packaging, which enables me to do complicated equations or graph instantly. A graph-drawing package (Autograph) benefited all the methods. I was able to create the graphs for the functions and equations. This package also enables me to zoom into a root to draw tangents. A spreadsheet software package (Excel) allowed me to customize the tables and implement formulae that helped to calculate the approximations at each value of x. On the graphics application, I could easily revise and correct the formulae to make changes where applicable. I was then able to import the formulae into the word processor and present them alongside the graphs and explanations. For hardware, I used a calculator to help me with complicated calculations to check for answers for Newton-Raphson and Rearranging formular. . ?? ?? ?? ?? Celene Leong 13.7 1 ...read more.
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1. ## Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...
0.025129 0.007813 9 1.523438 0.025129 1.53125 -0.00827 1.527344 0.008436 0.003906 10 1.527344 0.008436 1.53125 -0.00827 1.529297 8.43E-05 0.001953 11 1.529297 8.43E-05 1.53125 -0.00827 1.530273 -0.00409 0.000977 12 1.529297 8.43E-05 1.530273 -0.00409 1.529785 -0.002 0.000488 13 1.529297 8.43E-05 1.529785 -0.002 1.529541 -0.00096 0.000244 14 1.529297 8.43E-05 1.529541 -0.00096 1.529419 -0.00044 0.000122
2. ## Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ...
=(B8+C8)/2 =(D8-1)*(D8+2)*(D8-3)-1 =ABS(C8-B8)/2 =A8+1 =IF(E8>0,D8,B8) =IF(E8<0,D8,C8) =(B9+C9)/2 =(D9-1)*(D9+2)*(D9-3)-1 =ABS(C9-B9)/2 =A9+1 =IF(E9>0,D9,B9) =IF(E9<0,D9,C9) =(B10+C10)/2 =(D10-1)*(D10+2)*(D10-3)-1 =ABS(C10-B10)/2 =A10+1 =IF(E10>0,D10,B10) =IF(E10<0,D10,C10) =(B11+C11)/2 =(D11-1)*(D11+2)*(D11-3)-1 =ABS(C11-B11)/2 =A11+1 =IF(E11>0,D11,B11) =IF(E11<0,D11,C11) =(B12+C12)/2 =(D12-1)*(D12+2)*(D12-3)-1 =ABS(C12-B12)/2 =A12+1 =IF(E12>0,D12,B12) =IF(E12<0,D12,C12) =(B13+C13)/2 =(D13-1)*(D13+2)*(D13-3)-1 =ABS(C13-B13)/2 =A13+1 =IF(E13>0,D13,B13) =IF(E13<0,D13,C13) =(B14+C14)/2 =(D14-1)*(D14+2)*(D14-3)-1 =ABS(C14-B14)/2 =A14+1 =IF(E14>0,D14,B14)
1. ## I am going to solve equations by using three different numerical methods in this ...
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So, although we start with a value that is quite close to it, we fail to find the particular root we want. That is why the failure occurs. Method 3: Rearrangement This method aims to rearrange f(x)=0 in the form x=g(x)
1. ## Numerical integration can be described as set of algorithms for calculating the numerical value ...
+ f2n-1) + 2(f2 + f4 + f6 + .... + f2n-2)] However, seeing as I am carrying out the mid-point rule then the trapezium rule, I will be using a different version of Simpson's rule (2Mn + Tn)/3. This is simply due to the fact that it involves less arithmetic than using than using the formula stated in the page above.
2. ## C3 COURSEWORK - comparing methods of solving functions
Therefore, the root is 0.87939 to 5 decimal places. Lastly, we will use x=g(x) method to find the root in the interval [0, 1]: I need to rearrange the equation x³+3x²–3=0 in to the form of x=g(x), Let X1= 1 x³+3x²–3=0 n x 1 1 0.816496581 2 0.816496581 0.904741021 3
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AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Questions and Answers.
## AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.5
Question 1.
Solve the following equations.
i) $$\frac{n}{5}-\frac{5}{7}=\frac{2}{3}$$
Solution:
ii) $$\frac{x}{3}-\frac{x}{4}=14$$
⇒ $$\frac{4 x-3 x}{12}$$ = 14
⇒ $$\frac{x}{12}$$ = 14
⇒ x = 12 × 14 = 168
∴ x = 168
iii) $$\frac{z}{2}+\frac{z}{3}-\frac{z}{6}=8$$
iv) $$\frac{2 p}{3}-\frac{p}{5}=11 \frac{2}{3}$$
v) $$9 \frac{1}{4}=y-1 \frac{1}{3}$$
vi) $$\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}$$
vii) $$\frac{x}{2}-\frac{1}{4}=\frac{x}{3}+\frac{1}{2}$$
viii) $$\frac{2 x-3}{3 x+2}=\frac{-2}{3}$$
⇒ 3(2x – 3) = – 2(3x + 2)
⇒ 6x – 9 = -6x – 4
⇒ 6x + 6x = -4 + 9
⇒ 12x = 5
∴ x = $$\frac{5}{12}$$
ix) $$\frac{8 p-5}{7 p+1}=\frac{-2}{4}$$
Solution:
⇒ $$\frac{8 p-5}{7 p+1}=\frac{-2}{4}$$
⇒ 2(8p – 5) = – (7p + 1)
⇒ 16p – 10 = – 7p – 1
⇒ 16p + 7p = – 1 + 10
⇒ 23p = 9
∴ x = $$\frac{9}{23}$$
x) $$\frac{7 y+2}{5}=\frac{6 y-5}{11}$$
⇒ 11 (7y + 2) = 5 (6y-5)
⇒ 77y + 22 = 30y – 25
⇒ 77y – 30y = – 25 – 22
⇒ 47y = – 47
∴ y = $$\frac{-47}{47}$$
∴ y = -1
xi) $$\frac{x+5}{6}-\frac{x+1}{9}=\frac{x+3}{4}$$
⇒ 4(x + 13) = 18 (x + 3)
⇒ 4x + 52 = 18x + 54
⇒ 4x – 18x = 54-52
⇒ – 14x = 2
⇒ x = $$\frac{2}{-1}$$ = $$\frac{-1}{7}$$
∴ x = $$\frac{-1}{7}$$
xii) $$\frac{3 t+1}{16}-\frac{2 t-3}{7}=\frac{t+3}{8}+\frac{3 t-1}{14}$$
Solution:
⇒ -11t + 55 = 2(19t + 17) = 38t + 34
⇒ -11t – 38t = 34 – 55
⇒ -49t = – 21
⇒ $$\frac{-21}{-49}$$ = $$\frac{3}{7}$$
∴ t = $$\frac{3}{7}$$
Question 2.
What number is that of which the third part exceeds the fifth part by 4?
Solution:
Let the number be ‘x’ say.
$$\frac{1}{3}$$ rd of a number = $$\frac{1}{3}$$ x x = $$\frac{x}{3}$$
$$\frac{3}{7}$$ th of a number = $$\frac{1}{5}$$ x x = $$\frac{x}{5}$$
According to the sum
∴ The required number is 30.
Question 3.
The difference between two positive integers is 36. The quotient when one integer is
divided by other is 4. Find the integers.
(Hint: If one number is ‘X’, then the other number is ‘x – 36’)
Solution:
Let the two positive numbers be x, (x – 36) say.
If one number is divided by second tten the quotient is 4.
∴ $$\frac{x}{x-36}=4$$
⇒ x = 4(x – 36) = 4x – 144
⇒ 4x – x = 144
3x = 144
x = 48
∴ x – 36 = 48 – 36 = 12
∴ The required two positive intgers are 48, 12.
Question 4.
The numerator of a fraction is 4 less than the denominator. If 1 is added to both its
numerator and denominator, it becomes 1/2 . Find the fraction.
Solution:
Let the denominator of a fractin be x.
The numerator of a fraction is 4 less than the denominator.
∴ The numerator = x – 4
∴ Fraction $$\frac{x-4}{x}$$
If ‘1’ is added to both, its numerator and denominator, it becomes $$\frac{1}{2}$$
∴ $$\frac{1+x-4}{1+x}=\frac{1}{2}$$
2 + 2x – 8 = 1 + x
2x – x = 1 + 6 = 7
x = 7
∴ The denominator = 7
The numerator = 7 – 4 = 3
∴ Fraction = $$\frac{3}{7}$$
Question 5.
Find three consecutive numbers such that if they are divided by 10, 17, and 26 respectively,
the sum of their quotients will be 10.
(Hint: Let the consecutive numbers = x, x+ 1, x+ 2, then $$\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10$$)
Solution:
Let the three consecutive numbers be assume that x, (x + 1), (x + 2) respectively.
Given that x, (x + 1), (x + 2) are divided by 10, 17, 26 respectively, the sum of the quotients is 10. Then
⇒ $$\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10$$
⇒ $$\frac{x \times 221+130(x+1)+85(x+2)}{2210}=10$$
⇒ 221x + 130x + 85x + 130 + 170 = 22,100
⇒ 436x + 300 = 22,100
⇒ 436x = 22,100 – 300
⇒ 436x = 21,800
⇒ $$\frac{21800}{436}$$
∴ x = 50
∴ The required three consecutive num-bers are x = 50
x + 1 =50+ 1 = 51
x + 2 = 50 + 2 = 52
Question 6.
In class of 40 pupils the number of girls is three-fifths of the number of boys. Find the
number of boys in the class.
Solution:
Let the number of boys = x say.
Total number of students = 40
Number of girls = $$\frac{3}{5}$$ × x = $$\frac{3x}{5}$$
According to the sum 3x
∴ x = 25
∴ Number of boys in the class room = 25
Question 7.
After 15 years , Mary’s age will be four times of her present age. Find her present age.
Solution:
Let the present age of Mary = x years say.
After 15 years Mary’s age = (x + 15) years
According to the sum
(x + 15) = 4 x x
⇒ x + 15 = 4x
⇒ 4x – x =15
⇒ 3x = 15
⇒ x = 5
∴ The present age of Mary = 5 years.
Question 8.
Aravind has a kiddy bank. It is full of one-rupee and fifty paise coins. It contains 3 times
as many fifty paise coins as one rupee coins. The total amount of the money in the bank is
₹ 35. How many coins of each kind are there in the bank?
Solution:
Number of 1 rupee coins = x say.
Number of 50 – paise coins = 3 x x = 3x
The value of total coins = $$\frac{3x}{2}$$ + x
[∵50 paisa coins of 3x = ₹$$\frac{3x}{2}$$
According to the sum
⇒ $$\frac{3x}{2}$$ + x = 35
⇒ $$\frac{3 x+2 x}{2}$$ = 35
⇒ 5x = 2 × 35
⇒ x = 2 × $$\frac{35}{5}$$
∴ x = 14
∴ Number of 1 rupee coins = 14
Number of 50 paisa coins = 3 × x = 3 × 14 = 42
Question 9.
A and B together can finish a piece of work in 12 days. If ‘A’ alone can finish the same work in 20days , in how many days B alone can finish it?
Solution:
A, B can do a piece of work in 12 days.
(A + B)’s 1 day work = $$\frac{1}{12}$$ th part.
A can complete the same work in 20 days.
Then his one day work = $$\frac{1}{20}$$
B’s one day work = (A+B)’s 1 day work – A’s 1 day work
$$=\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{2}{60}=\frac{1}{30}$$
∴ Number of days to take B to com¬plete the whole work = 30.
Question 10.
If a train runs at 40 kmph it reaches its destination late by 11 minutes . But if it runs at 50 kmph it is late by 5 minutes only. Find the distance to be covered by the train.
Solution:
Let the distance to be reached = x km say. Time taken to travel ‘x’ km with speed x
40 km/hr = $$\frac{x}{40}$$ hr.
Time taken to travel ‘x’ km with speed 50 km/hr = $$\frac{x}{50}$$ hr.
According to the sum the difference between the times
∴ The required distance to be trav¬elled by a train = 20 kms‘.
Question 11.
One fourth of a herd of deer has gone to the forest. One third of the total number is
grazing in a field and remaining 15 are drinking water on the bank of a river. Find the total
number of deer.
Solution:
Number of deer = x say.
Number of deer has gone to the forest
= $$\frac{1}{4}$$ × x = $$\frac{x}{4}$$
Number of deer grazing in the field
= $$\frac{1}{3}$$ × x =$$\frac{x}{3}$$
Number of remaining deer =15
According to the sum
∴ x = 36
∴ The total number of deer = 36
Question 12.
By selling a radio for ₹903, a shop keeper gains 5%. Find the cost price of the radio.
Solution:
The selling price of a radio = ₹ 903
Profit % = 5%
C.P = ?
C.P = $$\frac{\mathrm{S.P} \times 100}{(100+\mathrm{g})}$$
= $$\frac{903 \times 100}{(100+5)}$$
= $$\frac{903\times 100}{105}$$
C.P. = 8.6 × 100 = 860
∴ The cost price of the radio = ₹ 860
Question 13.
Sekhar gives a quarter of his sweets to Renu and then gives 5 sweets to Raji. He has 7 sweets left. How many did he have to start with?
Solution:
Number of sweets with Sekhar = x say.
Number of sweets given to Renu
= $$\frac{1}{4}$$ × x = $$\frac{x}{4}$$
Number of sweets given to Raji = 5
Till he has 7 sweets left.
x – ( $$\frac{x}{4}$$ + 5) = 7
⇒ x – $$\frac{x}{4}$$ – 5 = 7
⇒ x – $$\frac{x}{4}$$ = 7 + 5 = 12
⇒ $$\frac{4 x-x}{4}$$ = 12
⇒ $$\frac{3x}{4}$$ = 12
⇒ x = 12 × $$\frac{4}{3}$$ = 16
∴ x = 4 × 4 = 16
∴ Number of sweets with Sekhar at the beginning = 16
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## 2. The goals lessons:
Educational goals: Study quadrilateral concepts of its
elements:
Developing goals: activation of cognitive activity of students
through the solution of practical problems, the ability to
choose the right decision, succinctly express their thoughs,
analuze, and draw conclusions.
Communication goals: mutual aid, reviewing the responses,
the organization of mutual contral.
Presentain
Blackboard
## 4. Type of lesson:
Summarizing the lesson
## 5. Plan of lesson:
1. Topics include. (1-2min)
2. Сhecking homework. (4-5min)
3. Theoretical number material. (13-14min)
4. solving the problem. (20-23min)
5. HOMEWORK. (2-3min)
1
3
2
4
9
5
6
7
8
10
11
12
13
14
## 7. What is a tetragon (Quadrilateral)?
The Tetragon (Quadrilaterals) is a flat figure, which
consists their four lengths (segments),
consecutively connecting four points. No from
these point not to rest upon one direct, but
connecting their length (segments) are not crossed
(intersect).
## 8. That is a parallelogram?
The Parallelogram is a quadrilateral, the opposite sides being parallel two and
two (two by two parallel).
Name the characteristic of the sides and the angles of a parallelogram:
The opposite sides and angles of a parallelogram are equal.
Name the characteristic of the diagonals of a parallelogram.
The diagonals of a parallelogram are crossed (intersect) and divided fifty-fifty
cross point (into two equal parts by the point of their intersection).
## 9. types of parallelograms
The Rectangle - a parallelogram, beside which all corners
direct. Diagonals rectangle are.
The Rhombus - a parallelogram, beside which all sides are.
Diagonals of a rhombus are crossed under right angle and
are a bisections of its angles.
The Square - a rectangle, beside which all sides are.
Diagonals square are, are crossed under direct angle and
are a bisections of its angles.
В
А
А
В
О
О
Е
К
В
С
D
А
О
С
М
## 11. Solving problems.
- a parallelogram М + В = 144 о
Find the angles of the parallelogram.
1. Data: ABCМ
Solution. Since angle М and angle В opposite, that is 144 degrees
divided by 2 equals 72 degrees.
Similarly, angle A is equal to angle С.
But М + С = 180 о about angles С and D - two internal
unilateral corners, formed when crossing two parallel direct lines.
Then С = 180 о - М = 180 о – 72 о = 108 о.
The answer is: angle М is equal to angle В and makes 72 о, angle А
is equal to angle С and makes 108 o.
HOMEWORK
Page: 92-93
Exercise №5,7,14
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Found in: Page 73
### Pre-algebra
Book edition Common Core Edition
Author(s) Ron Larson, Laurie Boswell, Timothy D. Kanold, Lee Stiff
Pages 183 pages
ISBN 9780547587776
# Are $2\left(x+1\right) \text{and} 2x+1$ equivalent variable expressions? Explain.
The two expressions are not same & they are not equivalent
See the step by step solution
## Step-1 – What are real numbers?
Real numbers are simply the combination of Whole numbers, Integers, Rational and Irrational numbers in the number system.
## Step-2 – Distributive Property of Multiplication
Suppose a, b, and c is three real numbers.
Distributive Property of Multiplication
Property: $\text{a} \text{(b} \text{+} \text{c} \text{)} = \text{a} · \text{b} + \text{a} · \text{c}$
Example: $12 \text{(3} \text{+} \text{4} \text{)} 12= \text{12} · \text{3} + \text{12} · \text{4} = 36 + 48 = 84$
## Step-3: Are 2(x+1) and 2x+1 equivalent variable expressions? Explain.
Using distributive property of multiplication
Property: $\text{a} \text{(b} \text{+} \text{c} \text{)} = \text{a} · \text{b} + \text{a} · \text{c}$
We can write by using distributive method of multiplication: $2\left(x+1\right)$ = $2x+2$
The second expression is $2\left(x+1\right)$
Since this two expressions are not same & they are not equivalent.
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# If the points (2, 1) and (1, −2) are equidistant
Question:
If the points (2, 1) and (1, −2) are equidistant from the point (xy), show that x + 3y = 0.
Solution:
The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula
$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{-}-y_{2}\right)^{2}}$
Here it is said that the points (21) and (12) are equidistant from (x, y).
Let be the distance between (21) and (x, y).
Let be the distance between (1, −2) and (x, y).
So, using the distance formula for both these pairs of points we have
$d_{1}=\sqrt{(2-x)^{2}+(1-y)^{2}}$
$d_{2}=\sqrt{(1-x)^{2}+(-2-y)^{2}}$
Now since both these distances are given to be the same, let us equate both and
$d_{1}=d_{2}$
$\sqrt{(2-x)^{2}+(1-y)^{2}}=\sqrt{(1-x)^{2}+(-2-y)^{2}}$
Squaring on both sides we have,
$(2-x)^{2}+(1-y)^{2}=(1-x)^{2}+(-2-y)^{2}$
$4+x^{2}-4 x+1+y^{2}-2 y=1+x^{2}-2 x+4+y^{2}+4 y$
$-4 x-2 y=-2 x+4 y$
$-2 x-6 y=0$
$x+3 y=0$
Hence we have proved that when the points $(2,1)$ and $(1,-2)$ are equidistant from $(x, y)$ we have $x+3 y=0$.
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Åk 6–9
English/Polski
4.1 Calculating with Formulas and Functions
It is important to be able to figure out the unknown variable in a function if you know the others. This you do by taking out the unknown variable.
We will look at two examples of formulas where you can do a little different calculation depending on which variables you know:
This is a well known formula which you use to calculate speed.
v: speed (v comes from the word velocity.)
d: distance
t: time
The formula looks like this:
v = d t
If you know the distance and the time, then it is not a problem to calculate the speed.
Example:
d = 240 km
t = 3 h
v = 240 = 80 km/h 3
If you know the speed and the time, then it is not a problem to calculate the distance.
v = 70 km/h
t = 2.5 h
The variable that you don’t know is d (the distance). How do you calculate the distance?
We begin by taking out the distance from the formula:
d · t v · t = We multiply both sides with t. t
d
= v · t
Then we get:
d =
70 · 2.5
d = 175 km
A formula we often use in geometry is the formula for the rectangle’s perimeter.
h: height b: base P: perimeter P = 2h + 2b
If you know the base and the height of the rectangle, it is no problem to calculate the perimeter. But how do you do it if you only know the perimeter and the base and want to calculate the rectangle’s height? We take an example:
P = 38 cm
b = 9 cm
We put these in the formula:
38 = 2h + 2 · 9
38 = 2h + 18
38 - 18 = 2h + 18 - 18
20 = 2h
h = 10 cm
Answer: The rectangle’s height is 10 cm.
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# Find The Domain of Functions with Square Root - Questions With Detailed Solutions
How to find the domain of functions with square root? Several examples are presented along with detailed solutions and explanations with graphical explanations and interpretation of the domain.
How to find the domain of a function with square root? We first need to understand that $\sqrt x$ is real only if the argument $x$ which is the quantity under the radical $\sqrt{}$ satisfies the condition: $x \ge 0$. This can easily be verified by examining the graph of $y = \sqrt x$ shown below: The graph exists only for $x \ge 0$. Example 1: Find the domain of the function $f(x) = \sqrt{x - 2}$. Solution: $f(x) = \sqrt{x - 2}$ takes real values if the argument $x - 2$, which is the quantity under the radical $\sqrt{}$, satifies the condition: $x - 2 \ge 0$. The solution of the inequality is $x \ge 2$ which is the domain of the function and this can be checked graphically as shown below where the graph of $f$ "exists" for $x \ge 2$. Example 2: Find the domain of the function $f(x) = \sqrt{|x - 1|}$. Solution: $f(x) = \sqrt{|x - 1|}$ takes real values if the argument $|x - 1|$, which is the quantity under the radical $\sqrt{}$, satifies the condition: $|x - 1| \ge 0$. The solution of the inequality is all real number because the absolute value expression $|x - 1|$ is always positive or zero for $x = 1$. The domain of the function is the set of real numbers $\mathbb{R}$ and this can be checked graphically as shown below where the graph of $f$ "exists" for all $x$ values.. Example 3: Find the domain of the function $f(x) = \dfrac{1}{ \sqrt{x + 3}}$. Solution: Taking into account that the function is the ratio of two functions and that division by zero is not allowed, the given function takes real values if the argument $x + 3$, which is the quantity under the radical $\sqrt{}$, satisfies the condition: $x + 3 \gt 0$. Note the symbol of the inequality used is $\gt$ and not $\ge$ because we do not want to have zero in the denominator. Solution of the inequality is $x \gt - 3$. The domain of the function is the set of real numbers greater than -3 and this can be checked graphically as shown below where the graph of $f$ "exists" for all $x > - 3$ values.. Example 4: Find the domain of the function $f(x) = \dfrac{\sqrt{x + 4}}{ \sqrt{x - 2}}$. Solution: The given function takes real values if two conditions are satisfied. 1) $x + 4 \ge 0$ , zero is allowed in the numerator, hence the use of the inequality symbol $\ge$. and 2) $x - 2 \gt 0$ , zero is not allowed in the denominator, hence the use of the inequality symbol $\gt$. The domain of the function is the intersection of the two solution sets of the two inequalities above.. $x \ge - 4$ and $x \gt 2$ The intersection of the two solution sets above is.. $x \gt 2$ Which is the domain of the given function as shown below in the graph of $f$.. Example 5: Find the domain of the function $f(x) = \sqrt{\dfrac{x + 4}{ {x - 2}}}$. Solution: The given function takes real values if $\dfrac{x + 4}{ {x - 2}} \ge 0$ We need to solve the above inequality. The zeros of the numerator and denominator are x = - 4 and x = 2 The zeros split the number line into 3 intervals over which the sign of the inequality is the same. Hence the intervals. $(-\infty , -4)$ , $(-4 , 2 )$ , $(2 , \infty)$ We select a value within each interval and use it to find the sign of the expression $\dfrac{x + 4}{ {x - 2}}$. 1) On the interval $(-\infty , -4)$ , select x = -6 and substitute x in $\dfrac{x + 4}{ {x - 2}}$ by -6 to determine its sign $\dfrac{ -6 + 4}{ {-6 - 2}} \gt 0$ 2) On the interval $(-4 , 2)$ , select x = 0 and substitute x in $\dfrac{x + 4}{ {x - 2}}$ by 0 to determine its sign $\dfrac{ 0 + 4}{ {0 - 2}} \lt 0$ 3) On the interval $(2 , \infty)$ , select x = 3 and substitute x in $\dfrac{x + 4}{ {x - 2}}$ by 3 to determine its sign $\dfrac{ 3 + 4}{ {3 - 2}} \gt 0$ Hence the domain is the union of all intervals over which $\dfrac{x + 4}{ {x - 2}} \gt 0$ and is given by. $(-\infty , -4] \cup (2 , \infty)$ The graph of $f$ is as shown below and we may easily check the domain found above. Note x = 2 is not the domain because the division by zero is not allowed.. Example 6: Find the domain of the function $f(x) = \sqrt{-x^2-4}$. Solution: The given function takes real values if $-x^2 - 4 \ge 0$ The expression $x^2 + 4$ is the sum of a square and a positive number. Hence $x^2 + 4 \ge 0$ Multiply all terms of the above inequality by -1 and change the symbol of inequality to obtain $- x^2 - 4 \le 0$ Hence the domain of the given function is an empty set and the given function has no graph. Try to graph it using a graphing calculator. Example 7: Find the domain of the function $f(x) = \dfrac{\sqrt{6 - x}}{ \sqrt{x - 2}}$. The given function takes real values if two conditions are satisfied. 1) $6 - x \ge 0$ , zero is allowed in the numerator, hence the use of the inequality symbol $\ge$. and 2) $x - 2 \gt 0$ , zero is not allowed in the denominator, hence the use of the inequality symbol $\gt$. The domain of the function is the intersection of the two solution sets of the two inequalities above.. $x \le 6$ and $x \gt 2$ The intersection of the two solution sets above is given by the interval. $(2 , 6]$ The graph of $f$ is as shown below and we may easily check the domain found above. Note x = 2 is not the domain because the division by zero is not allowed.. Example 8: Find the domain of the function $f(x) = \sqrt{x^2 - 4}$. Solution: The given function takes real values if $x^2 - 4 \ge 0$ which can also be written $(x - 2)(x + 2) \ge 0$ The solution set to the above quadratic inequality is given by the interval $(-\infty , -2] \cup [2 , \infty)$ The graph of $f$ is as shown below and we may easily check the domain found above.. Example 9: Find the domain of the function $f(x) = \sqrt{4 - x^2}$. Solution: The given function takes real values if $4 - x^2 \ge 0$ which can also be written as $(2 - x)(2 + x) \ge 0$ The solution set to the above quadratic inequality is given by the interval $[2 , 2]$ The graph of $f$ is as shown below and we may easily check the domain found above..
Updated: 20 January 2017 (A Dendane)
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# Thread: simultaneous equation using elimination method
1. ## simultaneous equation using elimination method
can you please solve this equatiion using elmination method.
$3x+2y=27$
$x-2y=-27$
2. Originally Posted by johnsy123
can you please solve this equatiion using elmination method.
$3x+2y=27$
$x-2y=-27$
Okay, so by the elimination method we just add the equations together. This always works because if A=B and C=D, then A+C=B+C because we are allowed to add the same thing to both sides...
But now we can say A+C=B+D because C=D, so looking back to our original equations we can see that we just added the left side of the first equation to the left of the second and the right of the first to the right of the second....
Onto this question, $3x+2y+x-2y=27-27$ the left side is 4x and the right is 0 so
$4x=0$ divide by 4 to get x=0
Now plug x=0 into one of the given equations, I'll use the first one: $3(0)+2y=27$ so 2y=27 so y= $\frac{27}{2}$
3. Originally Posted by johnsy123
can you please solve this equatiion using elmination method.
$3x+2y=27$
$x-2y=-27$
Just a couple of points to remember as well:
You can only eliminate terms if they have coefficients that are the same size.
In this case, you can recognise that the y-values both have coefficients of "size" 2.
Then you check their signs.
If they're the same sign, then you would have to subtract one equation from the other. If they are different sign, then you add them together.
In this case, you can see that one is positive, one is negative, so you would have to add the equations together, as artvandalay said.
Chookas for the exam
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# How do you solve 3/(x-2)<5/(x+2) using a sign chart?
Mar 6, 2017
The solution is x in ]-2,2 [ uu ]8, +oo [
#### Explanation:
We cannot do crossing over
Let's simplify the inequality
$\frac{3}{x - 2} < \frac{5}{x + 2}$
$\frac{3}{x - 2} - \frac{5}{x + 2} < 0$
$\frac{3 \left(x + 2\right) - 5 \left(x - 2\right)}{\left(x - 2\right) \left(x + 2\right)} < 0$
$\frac{3 x + 6 - 5 x + 10}{\left(x - 2\right) \left(x + 2\right)} < 0$
$\frac{16 - 2 x}{\left(x - 2\right) \left(x + 2\right)} < 0$
$\frac{2 \left(8 - x\right)}{\left(x - 2\right) \left(x + 2\right)} < 0$
Let $f \left(x\right) = \frac{2 \left(8 - x\right)}{\left(x - 2\right) \left(x + 2\right)}$
We, now, build the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a a a a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$8$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$8 - x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$
Therefore,
$f \left(x\right) < 0$ when x in ]-2,2 [ uu ]8, +oo [
|
# How do you find the area of the region bounded by the polar curve r=2-sin(theta) ?
Sep 17, 2014
The polar curve $r = 2 - \sin \theta$, $0 \le \theta < 2 \pi$ looks like this.
we can find the area $A$ of the enclosed region can be found by
$A = {\int}_{0}^{2 \pi} \setminus {\int}_{0}^{2 - \sin \theta} r \mathrm{dr} d \theta = \frac{9 \pi}{2}$
Let us evaluate the double integral above.
$A = {\int}_{0}^{2 \pi} \setminus {\int}_{0}^{2 - \sin \theta} r \mathrm{dr} d \theta$
$= {\int}_{0}^{2 \pi} {\left[{r}^{2} / 2\right]}_{0}^{2 - \sin \theta} d \theta$
$= \frac{1}{2} {\int}_{0}^{2 \pi} {\left(2 - \sin \theta\right)}^{2} d \theta$
$= \frac{1}{2} {\int}_{0}^{2 \pi} \left(4 - 4 \sin \theta + {\sin}^{2} \theta\right) d \theta$
by ${\sin}^{2} \theta = \frac{1}{2} \left(1 - \cos 2 \theta\right)$,
$= \frac{1}{2} {\int}_{0}^{2 \pi} \left(\frac{9}{2} - 4 \sin \theta - \frac{1}{2} \cos 2 \theta\right) d \theta$
$= \frac{1}{2} {\left[\frac{9}{2} \theta + 4 \cos \theta - \frac{1}{4} \sin 2 \theta\right]}_{0}^{2 \pi}$
$= \frac{1}{2} \left[9 \pi + 4 - 0 - \left(0 + 4 - 0\right)\right] = \frac{9 \pi}{2}$
|
# Lesson: The Complement
## Comment on The Complement
### Hello Brent,
Hello Brent,
If we only need 1 number to be even to have an even product, why is 1/2 an icorrect answer, since a pick of the "2" from the first set is sufficient to make the product even ?
### You are correct in saying
You are referring to the question that starts at 2:30.
You're correct in saying that P(selecting the "2" from the first set) = 1/2.
However, that is only one way to get an even product.
We can also get an even product by selecting the 3 from the first set and selecting the 8, 12, 18 or 20 from the second set.
A better way to illustrate the issue with your solution is to consider a very similar question:
{2, 3}
{2, 4, 6, 8, 10}
{1, 3, 5}
If one number is randomly selected from each set above, what is the probability that the product of the three selected numbers is even?
Since all five numbers in the second set are even, the product of the three selected numbers is guaranteed to be even, regardless of what happens with the selections from the other two sets.
### thank you ! that's clear now
thank you ! that's clear now
### In the last question why are
In the last question why are we not determining the probability of an odd number being selected in the first set, second set and the third set.
### We can use counting methods
We can use counting methods or probability rules to solve this question. In the video, we use counting methods, because we haven't yet covered probability rules in the module.
Using probability rules, we can say the following:
P(product is odd) = P(# from 1st set is odd AND # from 2nd set is odd AND # from 3rd set is odd)
= P(# from 1st set is odd) x P(# from 2nd set is odd) x P(# from 3rd set is odd)
= 1/2 x 1/5 x 1/4
= 1/40
So, P(product is even) = 1 - 1/40 = 39/40
Thanks.
### Sir how to solve this
Sir how to solve this question
A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected?
### The easiest solution is to
The easiest solution is to use the complement.
P(at least one green hat) = 1 - P(ZERO green hats)
Let's calculate P(ZERO green hats)
P(ZERO green hats) = P(1st hat is red AND 2nd hat is red)
= P(1st hat is red) x P(2nd hat is red)
= 6/11 x 5/10
= 30/110
= 3/11
So, P(at least one green hat) = 1 - 3/11
= 8/11
Cheers,
Brent
### hi Brent,
hi Brent,
How would you solve the product of 3 numbers question if you have 2 odd numbers in the 2nd group?
You're referring to the question that starts at 2:30.
Let's change the numbers in the second set to include TWO odd numbers to get:
{2, 3}
{5, 9, 12, 18, 20}
{4, 6, 9, 10}
So, P(product is EVEN) = 1 - P(product is ODD)
P(product is ODD) = P(1st number is odd AND 2nd number is odd AND 3rd number is odd)
= P(1st number is odd) x P(2nd number is odd) x P(3rd number is odd)
= (1/2) x (2/5) x (1/4)
= 1/20
So, P(product is EVEN) = 1 - 1/20
= 19/20
Cheers,
Brent
### For the question below, I was
For the question below, I was thinking a group of siblings of 4 and another group 3 (I.e 4 +3) as opposed to 2+2+ 3
What is the hint to crack the 2+2+3. Based on solutions it seems it is obviously but I Could not see it. Is it answer choices?
https://www.beatthegmat.com/siblings-t280457.html
Good idea, but "a group of siblings of 4 and another group 3" does meet the given condition that 4 people have exactly 1 sibling in the room.
If we have a family of 4 siblings, then each of those siblings will have 3 siblings in the room (not 1)
Cheers,
Brent
### Hi Brent,
Hi Brent,
Firstly apologies if I have been asking too many questions lately (as mentioned before, Counting/Probability are my weakest points).
Given this question: What is the probability of rolling three fair dice and having at least one of the three dice show an even number?
A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216
Could you please explain why we need to find the complement in order to obtain the correction solution? I can't seem to understand the logic behind this given that each dice roll/probability set contains an equal number of odd and even numbers.
For instance, why can't we go straight to the conclusion that each roll has a 3/6 chance of giving us an even number? Why is that we must first calculate the probability of an odd roll (which is also 3/6) in order to find the complement then subtract from 1.
AKA if one die has 3 even and 3 odd numbers, this implies that we have a 1/2 probability to roll an even or odd number. If we are to roll this one die 3 times, we get:
P(EVEN) --> (1/2) x (1/2) x (1/2) = 1/8
To me, this logically answers the question being asked. But I can't seem to understand why it is wrong. Appreciate any clarification!
Thanks as always :)
NEVER worry about asking a lot of questions. That's exactly what you're supposed to do when you're experiencing difficulties.
We want to determine P(at least one even)
If we don't use the complement here, we must consider all of the possible ways to get at least one even number.
So, we must consider:
- getting exactly one even number
- getting exactly two even numbers
- getting exactly three even numbers
In other words: P(get at least one even) = P(get 1 even number OR get 2 even numbers OR get 3 even numbers)
= P(get 1 even number) + P(get 2 even numbers) + P(get 3 even numbers)
As you can see, each of these calculations will take a lot of work.
However, if we use the complement (as I have done here https://gmatclub.com/forum/what-is-the-probability-of-rolling-three-fair...), we will have fewer calculations to make.
Important: When solving probability questions, it helps to rewrite the probability in terms that are clear to you (more here https://www.gmatprepnow.com/module/gmat-probability/video/754)
For example, above I rewrote the probability as follows:
P(get at least one even) = P(get 1 even number) + P(get 2 even numbers) + P(get 3 even numbers)
Then we should do the same for each of the above probabilities.
For example: P(get 3 even numbers) = P(1st roll is even AND 2nd roll is even AND 3rd roll is even)
= P(1st roll is even) x P(2nd roll is even) x P(3rd roll is even)
= (1/2) x (1/2) x (1/2)
= 1/8
I believe your solution, P(EVEN) = (1/2) x (1/2) x (1/2) = 1/8, is calculating the probability that all three rolls are even.
Cheers,
Brent
### Thank you so much! I'm
Thank you so much! I'm starting to piece this section together much better than counting. Also, thanks for always responding so diligently!
### Hi Brent,
Hi Brent,
Sorry my last qs was very stupid. I moved away from my desk and when I came back I re-read the qs. I guess doing so many qs at a time and not taking a break has made me tired and I overlook such simple details.
My apologies and sorry to have bothered you.
### No need to apologize. All
No need to apologize. All questions are welcome! :-)
Cheers,
Brent
### Hi Brent,
Hi Brent,
I solved the below question using complement but ain't getting the answer right. My method:
1 - 8C6/10C6 (8C6: Removed Alan and Becky from the group of 10 people). Where am I going wrong in this?
https://gmatclub.com/forum/alan-and-becky-are-among-the-ten-students-a-teacher-can-choose-from-to-285483.html
To better understand the ramifications of this question, let's say that I'm the leader of a very elite club.
In order to be in this club, you must meet BOTH of the following requirements:
1) Your birth month must be April
2) You must be right-handed
If you don't meet both requirements, then you cannot be in the club.
Now, let's say Joe applies, but does not get to be in the club.
What can we conclude about Joe?
Can we conclude that he wasn't born in April AND he isn't right-handed?
No, there are 3 possible scenarios that disqualify Joe from being in the club:
- Joe was born in April, and Joe is left-handed
- Joe was not born in April, and Joe is right-handed
- Joe was not born in April, and Joe is left-handed
-------------------------------
The same logic applies to the original question.
If Alan and Becky are both selected to be on the team, then Alan and Becky are BOTH on the team.
What if Alan and Becky are NOT both on the team?
There are 3 possible scenarios:
1) Alan was selected, and Becky was NOT selected
2) Alan was NOT selected, and Becky was selected
3) Alan was NOT selected, and Becky was NOT selected
You have accounted for scenario #3 only.
If we factor in scenarios 1 and 2, you'll arrive at the correct answer.
Cheers,
Brent
### https://gmatclub.com/forum/in
https://gmatclub.com/forum/in-a-room-filled-with-7-people-4-people-have-exactly-1-sibling-in-the-87550.html
There are two groups of siblings- (1234)
and (567). Total number of combinations of 2 students is 7C2=21.
Why can't I take total number of ways of selecting one sibling from each group as 4*3= 12
The various combinations are-
15
16
17
25
26
27
35
36
37
45
46
47
So 12/21
That's a good strategy, but you're missing four possible outcomes.
If 1&2 are siblings, and 3&4 are siblings, then there are 4 additional outcomes in which the two selected people are NOT siblings:
1) 13
2) 14
3) 23
4) 24
So, there are 16 allowable outcomes (not 12)
Cheers,
Brent
### Ohh yes! I totally missed
Ohh yes! I totally missed that part. Thank you!
### https://gmatclub.com/forum/8
https://gmatclub.com/forum/8-cities-including-memphis-are-finalists-to-be-chosen-to-host-a-poli-187957.html
In this question, statement 1 will not be helpful at all imo.
because it saying that one out of the 8 are selected at a 7/8 probability.
I think you may be reading statement 1 incorrectly.
Target question: What P(Memphis is NOT chosen)?
Statement 1: The probability that any one of the 8 cities does NOT win the competition is 7/8.
In other words, for EACH of the 8 cities, P(that city is NOT chosen) = 7/8
Since Memphis is one of those 8 cities, we can conclude that P(Memphis is NOT chosen) = 7/8
So statement 1 is sufficient.
Does that help?
### Yes, got it. Thank you so
Yes, got it. Thank you so much!
|
# What is the implicit derivative of 5=x-1/(xy^2)?
Apr 18, 2018
See below.
#### Explanation:
When finding the implicit derivative of an equation, we are finding the derivative in respect to $x$.
This means that when we find the derivative of any variable other than $x$, we need to tag on a $\frac{\mathrm{ds} q u a r e}{\mathrm{dx}}$ $\left(\square \text{is the variable}\right)$ $\text{to the end} .$
Knowing this, we will start with implicit differentiation.
Here is what we are given:
$5 = x - \frac{1}{x {y}^{2}}$
We can go ahead and avoid the Quotient Rule by multiply each term by $x {y}^{2}$.
$5 = x - \frac{1}{x {y}^{2}}$
$\implies 5 x {y}^{2} = {x}^{2} {y}^{2} - 1$
Here we can use the Product Rule to solve:
$5 x {y}^{2} = {x}^{2} {y}^{2} - 1$
$\implies 5 \left({y}^{2}\right) + 5 x \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x \left({y}^{2}\right) + {x}^{2} \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
Simplify:
$\implies 5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$
Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\implies 10 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} - 5 {y}^{2}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} \left(10 x y - 2 {x}^{2} y\right) = 2 x {y}^{2} - 5 {y}^{2}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} - 5 {y}^{2}}{10 x y - 2 {x}^{2} y}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y - 5 y}{10 x - 2 {x}^{2}}$
Apr 18, 2018
Please see below.
#### Explanation:
I assume that we want $\frac{\mathrm{dy}}{\mathrm{dx}}$ given that
5 = x-1/(xy^2
Method 1
To avoid the quotient rule, I would start by multiplying both sides be $x {y}^{2}$ to get
$5 x {y}^{2} = {x}^{2} {y}^{2} - 1$
Differentiating with respect to $x$ (and using the product rule) gets us
$\frac{d}{\mathrm{dx}} \left(5 x {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{2} - 1\right)$
$\left(5\right) {y}^{2} + 5 x \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = \left(2 x\right) {y}^{2} + {x}^{2} \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) - 0$
Now we'll do some algebra.
$5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$
$10 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} - 5 {y}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} - 5 {y}^{2}}{10 x y - 2 {x}^{2} y}$
$= \frac{2 x y - 5 y}{10 x - 2 {x}^{2}}$
Method 2
$\frac{d}{\mathrm{dx}} \left(5\right) = \frac{d}{\mathrm{dx}} \left(x - \frac{1}{x {y}^{2}}\right)$
$0 = 1 + \frac{1 {y}^{2} + 2 x y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{x}^{2} {y}^{4}}$
$0 = 1 + \frac{y + 2 x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{x}^{2} {y}^{3}}$
$0 = {x}^{2} {y}^{3} + y + 2 x \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{x}^{2} {y}^{3} + y}{2 x}$
Resolution
To see that these are equivalent, start with
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y - 5 y}{10 x - 2 {x}^{2}}$
$= \frac{2 x y - \left(5\right) y}{2 x \left(5\right) - 2 {x}^{2}}$
Now replace $5$ by $x - \frac{1}{x {y}^{2}}$
$= \frac{2 x y - \left(x - \frac{1}{x {y}^{2}}\right) y}{2 x \left(x - \frac{1}{x {y}^{2}}\right) - 2 {x}^{2}}$
Simplify algebraically to get
$= - \frac{{x}^{2} {y}^{3} + y}{2 x}$
|
# Alternate Solution to Example Problem on Series Circuits
Here is an alternate way of solving the example problem. Instead of first finding all the resistances, we can begin the problem by finding the emf voltage first.
By Ohm's Law, we know that the emf is equal to the product of the total current and the total resistance.
$\varepsilon = I R$
$\varepsilon = (1.0)(30) = 30V$
Now that we know the emf voltage, we also know the total voltage. Since $\varepsilon =V$, where V is the total voltage, then $V=30 V$.
Also, the total voltage, $V$, is equal to the sum of the voltages across the resistors in this circuit (because this is a series circuit).
$V = V_1 + V_2 + V_3 + V_4$
We also know all the voltages across the resistors except for $V_4$. So,
$V_4 = V - (V_1 + V_2 + V_3)$
$V_4 = 30 - (5.0 + 8.0 + 7.0)$
$V_4 = 10V$
The only unknowns now are the resistances. We know the voltages across each resistor. Also, because this is a series circuit, the current through each resistor is the same. Then, by Ohm's Law, we can find the resistances of each resistor:
$R_1 = \frac {V_1}{I}, R_2 = \frac {V_2}{I}, R_3 = \frac{V_3}{I}$
$R_1 = \frac {5.0}{1.0} =5.0 \quad \Omega \quad R _2 = \frac {8.0}{1.0} = 8.0 \quad \Omega \quad R_3 = \frac{7.0}{1.0} = 7.0 \Omega$
|
Properties of Complex Numbers MathBitsNotebook.com Terms of Use Contact Person: Donna Roberts
Properties of Complex Numbers: (Specifically those complex numbers involving "i ".)
1. Conjugates: When we worked with radicals and with binomials in general, we discovered expressions referred to as "conjugates". A binomial expression, when multiplied by its conjugate, results in the difference of the squares of each term (with the resulting "middle terms" dropping away). This "dropping away" proved very useful when working with radicals, and became our strategy for rationalizing denominators. In a similar fashion, when a complex number is multiplied by its conjugate, the "middle terms" will drop away giving us a purely real number as the product. This is yet another strategy which will prove useful.
(notice the notation of "z with a bar" for the conjugate)
(a + bi) • (a - bi) = a2 + b2 (a + bi)•(a - bi) = a2 - abi + abi - b2i2 = a2 - b2i2 = a2 - b2(-1) = a2 + b2
The squares of real numbers are positive.
The sum of the squares of two real numbers creates another real number.
Therefore, a2 + b2 is a positive real number, and we have the following rule:
Rule: The product of a complex number and its conjugate is a positive Real Number. Example: (2 + 3i)(2 - 3i) = 4 - 6i + 6i- 9i2 = 4 - 9(-1) = 13
2. Equality: Determining if two complex numbers are equal is exactly as you would think:
a + bi = c + di when a = c and b = d
Rule: Two complex numbers are equal if and only if their real parts and their imaginary parts are respectively equal. Example: 2 + 3i = 2 + 3i
3. Closure: The complex numbers are closed under addition, subtraction. multiplication and division - when not considering division by zero. Remember that closure means that when you perform an operation on two numbers in a set, you will get another number in that set. In this case, the set is complex numbers.
Rule: If two complex numbers are added (subtracted), the sum (difference) is a complex number.
Rule: If two complex numbers are multiplied (divided - not by 0), the product (quotient) is a complex number.
4. Commutative, Associative, Distributive Properties:
All complex numbers are commutative and associative under addition and multiplication, and multiplication distributes over addition.
Commutative: For all complex numbers z1 and z2, z1 + z2 = z2 + z1 (under addition) Ex. (3 + 2i) + (5 - 4i) = (5 - 4i) + (3 + 2i) z1• z2 = z2 • z1 (under multiplication) Ex. (3 + 2i) • (5 - 4i) = (5 - 4i) • (3 + 2i)
Associative: For all complex numbers z1, z2, z3, z1 + (z2 + z3) = (z1 + z2) + z3 (under addition) Ex. (7 + 5i) + [(3 + 2i) + (5 - 4i)] = [(7 + 5i) + (3 + 2i)] + (5 - 4i) z1• (z2• z3) = (z1• z2) • z3 (under multiplication) Ex. (7 + 5i) • [(3 + 2i) • (5 - 4i)] = [(7 + 5i) • (3 + 2i)] • (5 - 4i)
Complex multiplication distributes over addition: For all z1, z2, z3, z1• (z2 + z3) = z1• z2 + z1• z3 Ex. (7 + 5i) • [(3 + 2i) + (5 - 4i)] = (7 + 5i) • (3 + 2i) + (7 + 5i) • (5 - 4i)
5. Arithmetic Operations: For detailed information regarding the arithmetic of complex numbers see the following pages:
6. Tidbit of Info:
Rule: The square of any complex number and the square of its conjugate are also conjugates.
How to use your TI-83+/84+ calculator with complex numbers. Click here.
NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use".
Contact Person: Donna Roberts
|
# Factors of 22
Factors of 22 are those numbers that, when multiplied together, give the result of 22. We can also say that a factor is a number that divides a number ultimately, leaving zero as a remainder. We will use the multiplication method to find the factors of 22.
## Factors Calculator
Enter Number
Factors of 22: 1, 2, 11 and 22
Factors of -22: -1, -2, -11 and -22
Prime Factors of 22: 2, 11
Prime Factorization of 22: 2 × 11
Factors of 22 in Pairs: (1, 22) and (2, 11)
## What are the factors of 22?
The method of calculating the factors of 22 is as follows. First, each number can be divided by one and by itself.
Consequently, 1 and 22 are the factors of 22.
By dividing a number by 1, 2, 3, 4… we can discover all its factors.
(i) 22 ÷ 1 = 22
This division gives the remainder 0 and so is divisible by 22. So please put them 1 and 22 in your factor list.
1, …….. 22
(ii) 22 ÷ 2 = 11
This division gives the remainder 0 and so is divisible by 11. So please put them 2 and 11 in your factor list.
1, 2 …….. 11, 22
(iii) Since we don’t have any more numbers to calculate, we are putting the numbers so far.
So 1, 2, 11 and 22 are factors of 21.
## Factors of -22
To get the factors of -22 by the multiplication method, we will first write the number -22 multiplicatively in all possible ways. We can write -22 as follows and we have:
22 = -1 × -22 Therefore, -1, -22 are factors of -22
22 = -2 × -11 Therefore, -2, -11 are factors of -22
As there are no other ways we can express -22 multiplicative, we will stop now, and we have obtained all the factors of -22.
So, the factors of -22 are -1, -2, -11 and -22.
## All factors of 22
Here is a list of all the positive and negative factors of 22 in numerical order.
1, 2, 11, 22, -1, -2, -11 and -22
Therefore, 1, 2, 11, 22, -1, -2, -11 and -22 are All factors of 22.
## Factor pairs of 22
22 x 1 = 22
2 x 11 = 22
So, (1, 22) and (2, 11) are factor pairs of 22
## Prime Factorization of 22
22 ÷ 2 = 11
11 ÷ 11 = 1
Therefore, 2 × 11 are the Prime factorization of 22.
## Factor tree of 22
`````` 22
/ \
2 11``````
|
### Estimation methods
A typical estimation problem:
guess how many paper clips are in a jar?
Photo by Dean Terry
My daughter's homework was to estimate the number of 953 divided by 18, using front-end estimation, using rounding to estimate, and using compatible numbers to estimate. What are the differences between these three methods? Do we get the same result? Thanks!
You definitely don't get the same results as these three methods are quite different!
1. Front-end estimation means you keep the "front" or first digit of each number, and make the other digits to be zeros.
So, 953 ÷ 18 is estimated to 900 ÷ 10 = 90.
Another example: 56 × 295 would be estimated as 50 × 200 = 10,000.
2. Rounding means you round the numbers, usually to their biggest place values, but sometimes you can round "creatively". In any case, the numbers you round to should be easy to work with mentally.
So, in 953 divided by 18 we round 953 to nearest hundred, and 18 to nearest ten. 953 ÷ 18 becomes 1,000 ÷ 20 = 50.
Another example: 56 × 295 would become 60 × 300 = 18,000.
An example of rounding "creatively": with 24 × 32 you can round 24 to 25 (to the "middle five"), and 32 to 30. The estimated result is 25 × 30 = 750. The exact result in this case would be 768 so the estimation was fairly close.
Another principle to keep in mind with when using rounding to estimate is that if you have an addition or multiplication problem, it's best to round one number down, the other up, in order to minimize the rounding error. If you have a division or subtraction, it's best to round both numbers "the same direction", either up or down.
3. Compatible numbers means finding numbers that are close to the numbers in the problems but such as are easily to work with mentally.
So, 953 ÷ 18 could be estimated as either 960 ÷ 20 or 1000 ÷ 20, depending on your mental division skills.
960 ÷ 20 = 96 ÷ 2 = 48.
Another example: estimating 56 × 295 depends, again, on your mental multiplication skills. You could try to leave 56 as it is, and make 295 to be 300, to get 56 × 300 = 16,800. Or, you might make the numbers to be 60 and 300 (the same as in the rounding method) and get 18,000.
To compare all three methods, we check the exact result of our problem, which is 953 ÷ 18 = 52.944444444... From this we can see that the rounding method was most accurate in this case (it gave us 50), and front-end estimation did really bad (it gave us 90).
In the other example, the exact result is 56 × 295 = 16,520. Again, rounding (18,000) or compatible numbers (16,800 or 18,000) method did best, and front-end estimation the worst (10,000).
Anonymous said…
I feel frustrated now for not knowing how many paperclips are in the jar.
You explain this SOOOO well! I am so glad I found your blog. It just so happens I am working on front end estimations with my 7th grader, and I think this will help. Thank you!
Anonymous said…
Great entry, thanks! ~K
Maria said…
Thank you so much for the informations. I've used your website for helping my students before I go to textbook or worksheets. Thanks again
Ms. Guerrero
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Intuitive Trigonometry/Appendix 1: Solutions to Problems
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Appendix 1: Solutions to Problems
Radians and Arc Length
1. ${\displaystyle \ell _{\text{arc}}=2\pi r\left({\frac {\theta }{360^{\circ }}}\right)}$ is our formula for arc length; we can plug in our given values and solve. ${\displaystyle \ell =2\pi \times 2\left({\frac {10^{\circ }}{360^{\circ }}}\right)}$ which simplifies to ${\displaystyle \ell ={\frac {4\pi }{36}}}$ (notice the degree signs cancel) and further simplifies to ${\displaystyle \ell ={\frac {\pi }{9}}}$ which is our solution.
2. ${\displaystyle 2\pi {\text{ rad}}=360^{\circ }}$, so we can create a proportion
3. ${\displaystyle 2\pi {\text{ rad}}=360^{\circ }}$, so we can create a proportion
4. ${\displaystyle \ell _{\text{arc}}=2\pi r\left({\frac {\theta }{360^{\circ }}}\right)}$ is our formula for arc length; we can plug in our given values and solve. ${\displaystyle 5=2\pi \times 4\left({\frac {\theta }{360^{\circ }}}\right)}$ which simplifies to ${\displaystyle 5={\frac {8\pi \theta }{360^{\circ }}}}$ or ${\displaystyle 1800^{\circ }=8\pi \theta }$ or ${\displaystyle {\frac {1800}{8\pi }}^{\circ }={\frac {225}{\pi }}^{\circ }=\theta }$, which is our solution.
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Middle Years
# 1.08 HCF and LCM
Lesson
When we compare two numbers, we often want to know what their highest common factor and lowest common multiple are. These numbers help us understand how they are related, and are often the answer to problems involving them.
### Highest common factor
A number is a common factor of two other numbers if it divides both numbers without remainder. Below is a table with all the factors of $24$24 and $54$54, with their common factors listed in the last row:
Factors of $24$24 and $54$54
Factors of $24$24 $1,2,3,4,6,8,12,24$1,2,3,4,6,8,12,24
Factors of $54$54 $1,2,3,6,9,18,27,54$1,2,3,6,9,18,27,54
Common factors $1,2,3,6$1,2,3,6
The common factors tell us how we can break up both numbers equally.
For example, $24$24 students and $54$54 books can be divided into:
• $2$2 equal groups with $12$12 students and $27$27 books each
• $3$3 equal groups with $8$8 students and $18$18 books each
• $6$6 equal groups with $4$4 students and $9$9 books each
The largest number in the list of common factors, $6$6, is called the highest common factor (HCF) of $24$24 and $54$54.
To find the HCF of two numbers we can always create a list of factors, like we did above. But there is a faster way using factor trees. To find the highest common factor of $126$126 and $294$294, we start by drawing their factor trees:
This gives us their prime factorisations in expanded form:
$126=2\times3\times3\times7$126=2×3×3×7
$294=2\times3\times7\times7$294=2×3×7×7
We then find what factors appear in both factorisations - they have one $2$2, one $3$3, and one $7$7 in common. The highest common factor is the product of the common prime factors, $2\times3\times7=42$2×3×7=42.
### Lowest common multiple
When we multiply two numbers together, these two numbers are always factors of the new number. Another way to look at it is to say the new number is a common multiple of both of the original ones.
Below is a table with the first few multiples of $12$12 and $18$18, with their common multiples listed in the last row:
Multiples of $12$12 and $18$18
Multiples of $12$12 $12,24,36,48,60,72,84,96,108,\ldots$12,24,36,48,60,72,84,96,108,
Multiples of $18$18 $18,36,54,72,90,108,126,144,\ldots$18,36,54,72,90,108,126,144,
Common multiples $36,72,108,\ldots$36,72,108,
The common multiples are the numbers that can be broken up into both the original numbers equally.
• $36$36 can be broken up into $3$3 groups of $12$12, and $2$2 groups of $18$18
• $72$72 can be broken up into $6$6 groups of $12$12, and $4$4 groups of $18$18
• $108$108 can be broken up into $9$9 groups of $12$12, and $6$6 groups of $18$18
The smallest number in the list of common multiples, $36$36, is called the lowest common multiple (LCM) of $12$12 and $18$18.
To find the LCM of two numbers we can always create a list of multiples, like we did above. But just like for HCF, there is a faster way using factor trees. To find the lowest common multiple of $126$126 and $294$294, we look again at their factor trees:
We can instead jump straight to the prime factorisations in expanded form, since we found that already:
$126=2\times3\times3\times7$126=2×3×3×7
$294=2\times3\times7\times7$294=2×3×7×7
Now we create the number by multiplying one of the numbers by any prime factors it is missing from the other one.
There is only one $2$2 in each, there are two $3$3s in $126$126 (and only one in $294$294), but there are two $7$7s in $294$294.
If we multiply $126$126 by $7$7, the missing factor, we will find the lowest common multiple: $126\times7=882$126×7=882.
We would get the same result if we started with $294$294 - it has more than enough $7$7s, but it is missing the second $3$3 that $126$126 has. The multiplication $294\times3=882$294×3=882 gives us the same answer.
HCF and LCM
The highest common factor (HCF) of two numbers is the largest factor of both numbers. It is the product of the common prime factors between them.
The lowest common multiple (LCM) of two numbers is the smallest multiple of both numbers. Multiply one number by the prime factors it is missing from the other.
#### Practice questions
##### Question 1
Consider the following prime factorisations:
$180=2\times2\times3\times3\times5$180=2×2×3×3×5
$600=2\times2\times2\times3\times5\times5$600=2×2×2×3×5×5
1. Find the highest common factor of $180$180 and $600$600.
##### Question 2
Find the highest common factor of $150$150 and $560$560.
##### Question 3
Consider the following prime factorisations:
$2200=2\times2\times2\times5\times5\times11$2200=2×2×2×5×5×11
$2750=2\times5\times5\times5\times11$2750=2×5×5×5×11
1. Find the lowest common multiple of $2200$2200 and $2750$2750.
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# An equilateral triangle of side $a$ carries a current $I$. What is the magnitude and direction of the magnetic field at the point $P$?
Last updated date: 23rd Feb 2024
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Hint: First, we have to know about the magnetic field and how it can be generated from electricity. We can determine the magnetic field's magnitude thus produced if we know about the correlation between the electric field and the magnetic field. After that, we can find out the direction of the generated magnetic field with the right-hand rule.
Formula used:
$B = \dfrac{{{\mu _0}I}}{{2\pi r}}$
Complete step by step solution:
The vector field that describes the magnetic influence on moving electric charges or electric current or any other magnetic material is known as the magnetic field. A current-carrying wire can produce a magnetic field in its surroundings. The right-hand rule can determine the direction of the magnetic field. The correlation of the current and the produced magnetic field can be given as-
$B = \dfrac{{{\mu _0}I}}{{2\pi r}}$
where,
$B$ is the magnitude of the magnetic field
$I$ is the current flowing through the conductor
${\mu _0}$ is the magnetic permittivity of vacuum
$r$ is the distance from the conductor to the point of the magnetic field
However, here, we will think more deeply and use symmetry to solve our problem.
So the sides $PQ$ and $PR$ do not contribute to the generation of the magnetic field at $P$.
Here,
$\Rightarrow r = \sqrt {{a^2} - {{\left( {\dfrac{a}{2}} \right)}^2}} = \sqrt {{a^2} - \dfrac{{{a^2}}}{4}}$
Upon solving further we get,
$\Rightarrow r = \sqrt {\dfrac{{3{a^2}}}{4}}$
$\Rightarrow r = \dfrac{{\sqrt 3 a}}{2}$
Therefore,
$\Rightarrow B = \dfrac{{{\mu _0}}}{{4\pi }}.\dfrac{I}{r}\left[ {\sin {\alpha _1} + \sin {\alpha _2}} \right]$
We put ${\alpha _1} = {\alpha _2} = {30^ \circ }$ , and we get-
$\Rightarrow B = \dfrac{{{\mu _0}}}{{4\pi }}.\dfrac{I}{{\left( {\dfrac{{\sqrt 3 a}}{2}} \right)}}\left[ {\sin \left( {{{30}^ \circ }} \right) + \sin \left( {{{30}^ \circ }} \right)} \right]$
We put $\sin \left( {{{30}^ \circ }} \right) = \dfrac{1}{2}$-
$\Rightarrow B = \dfrac{{{\mu _0}I}}{{2\sqrt 3 \pi a}}\left[ {\dfrac{1}{2} + \dfrac{1}{2}} \right]$
$\Rightarrow B = \dfrac{{{\mu _0}I}}{{2\sqrt 3 \pi a}}$
Hence, the magnitude of the generated magnetic field is $\dfrac{{{\mu _0}I}}{{2\sqrt 3 \pi a}}$ , and with the right-hand rule, we can determine that the magnetic field's direction is upward, i.e., pointing outward of the page.
Note: In this problem, we have used the Biot-Savart Law to determine the magnetic field. We will always use the Biot-Savart Law to solve problems where we can use symmetry to find out the magnetic field generated due to the action of a current-carrying conductor.
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# Three numbers form an arithmetic sequence having a common difference of 4. If the first number is increased by 2, the second number by 3, and the 3rd number by 5, the resulting numbers form a geometric sequence. How do you find the original numbers?
May 1, 2016
$23 , 27 , 31$
#### Explanation:
The original sequence is:
${a}_{1} = a$
${a}_{2} = a + 4$
${a}_{3} = a + 8$
The modified sequence is:
${b}_{1} = a + 2$
${b}_{2} = a + 7$
${b}_{3} = a + 13$
Since ${b}_{1} , {b}_{2} , {b}_{3}$ is a geometric sequence, the middle term ${b}_{2}$ must be a geometric mean of ${b}_{1}$ and ${b}_{3}$. Hence:
${b}_{2}^{2} = {b}_{1} {b}_{3}$
${a}^{2} + 14 a + 49 = {\left(a + 7\right)}^{2} = \left(a + 2\right) \left(a + 13\right) = {a}^{2} + 15 a + 26$
Subtracting ${a}^{2} + 14 a + 26$ from both ends, we find:
$a = 23$
So ${a}_{1} , {a}_{2} , {a}_{3} = 23 , 27 , 31$
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# hw7 - 5.2.1 Consider y 00 y = 0 x = 0 We seek a series...
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Unformatted text preview: 5.2.1. Consider y 00- y = 0, x = 0. We seek a series solution about x = 0. i.e. We are looking for a solution of the form y ( x ) = ∞ X n =0 a n x n . (1) It follows then that y 00 ( x ) = ∞ X n =0 n ( n- 1) a n x n- 2 = ∞ X n =2 n ( n- 1) a n x n- 2 . (The last equality follows from the fact that the first two terms of the series were 0.) Shifting indices to start from n = 0, we have y 00 ( x ) = ∞ X n =0 ( n + 2)( n + 1) a n +2 x n . (2) Plugging (1) and (2) into our differential equation, we see ∞ X n =0 (( n + 2)( n + 1) a n +2- a n ) x n = 0 . Thus as the equality above must hold for all x , ( n + 2)( n + 1) a n +2- a n = 0 for all n . That is, we have the recurrence relation a n +2 = a n ( n + 1)( n + 2) . Note for n even, a n = a n- 2 ( n- 1) n = a n- 4 ( n- 3)( n- 2)( n- 1) n = ··· = a 1 · 2 ··· n = a n ! . Similarly for n odd, a n = a n- 2 ( n- 1) n = a n- 4 ( n- 3)( n- 2)( n- 1) n = ··· = a 1 2 · 3 ··· n = a 1 n ! . We use theorem 5.3.1 to find a lower bound for the radius of convergence. In the language of the theorem, p = 0 = ∑ x n and q =- 1 =- 1 + ∑ x n . These trivial series clearly have an infinite radius of convergence, and so our series has a radius of convergence of at least the minimum of ∞ and ∞ . i.e. We have an infinite radius of convergence. We are free to vary a and a 1 as we please. A reasonable way to get two linearly independent solutions is to first take a = 1 and a 1 = 0 and then to take a = 0 and a 1 = 1....
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# How do you solve the following linear system: -9 y + 2x = 2 , 5x - y = 3?
May 30, 2017
See a solution process below:
#### Explanation:
Step 1) Solve the second equation for $y$:
$5 x - y = 3$
$- \textcolor{red}{3} + 5 x - y + \textcolor{b l u e}{y} = - \textcolor{red}{3} + 3 + \textcolor{b l u e}{y}$
$- 3 + 5 x - 0 = 0 + \textcolor{b l u e}{y}$
$- 3 + 5 x = y$
$y = - 3 + 5 x$
Step 2) Substitute $\left(- 3 + 5 x\right)$ for $y$ in the first equation and solve for $x$:
$- 9 y + 2 x = 2$ becomes:
$- 9 \left(- 3 + 5 x\right) + 2 x = 2$
$\left(- 9 \times - 3\right) + \left(- 9 \times 5 x\right) + 2 x = 2$
$27 - 45 x + 2 x = 2$
$27 + \left(- 45 + 2\right) x = 2$
$27 - 43 x = 2$
$- \textcolor{red}{27} + 27 - 43 x = - \textcolor{red}{27} + 2$
$0 - 43 x = - 25$
$- 43 x = - 25$
$\frac{- 43 x}{\textcolor{red}{- 43}} = \frac{- 25}{\textcolor{red}{- 43}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 43}}} x}{\cancel{\textcolor{red}{- 43}}} = \frac{25}{43}$
$x = \frac{25}{43}$
Step 3) Substitute $\frac{25}{43}$ for $x$ in the solution to the second equation at the end of Step 1 and calculate $y$:
$y = - 3 + 5 x$ becomes:
$y = - 3 + \left(5 \times \frac{25}{43}\right)$
$y = - 3 + \frac{125}{43}$
$y = \left(- 3 \times \frac{43}{43}\right) + \frac{125}{43}$
$y = - \frac{129}{43} + \frac{125}{43}$
$y = - \frac{4}{43}$
The solution is: $x = \frac{25}{43}$ and $y = - \frac{4}{43}$ or $\left(\frac{25}{43} , - \frac{4}{43}\right)$
May 30, 2017
$x = \frac{25}{43} , y = - \frac{4}{43.}$
#### Explanation:
From the second eqn., $5 x - 3 = y .$
Subst.ing this value of $y$ in the first eqn., we get,
$- 9 \left(5 x - 3\right) + 2 x = 2.$
$\therefore - 45 x + 27 + 2 x = 2.$
$\therefore - 43 x = 2 - 27 = - 25.$
$\therefore x = \frac{25}{43.}$
$\therefore y = 5 x - 3 = 5 \left(\frac{25}{43}\right) - 3 = \frac{125}{43} - 3 = \frac{125 - 129}{43.}$
;. y=-4/43.
Hence, the Soln. $x = \frac{25}{43} , y = - \frac{4}{43.}$
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# Sum of digits | AMC-10A, 2020 | Problem 8
Try this beautiful problem from Algebra, based on Sum of digits from AMC-10A, 2020. You may use sequential hints to solve the problem
Try this beautiful problem from Algebra based on sum of digits
## Sum of digits – AMC-10A, 2020- Problem 8
What is the value nof
$1+2+3-4+5+6+7-8+……+197+198+199-200$?
• $9800$
• $9900$
• $10000$
• $10100$
• $10200$
### Key Concepts
Algebra
Arithmetic Progression
Series
Answer: $9900$
AMC-10A (2020) Problem 8
Pre College Mathematics
## Try with Hints
The given sequence is $1+2+3-4+5+6+7-8+……+197+198+199-200$. if we look very carefully then notice that $1+2+3-4$=$2$, $5+6+7-8=10$,$9+10+11-12=18$…..so on.so $2,10,18……$ which is in A.P with common difference $8$. can you find out the total sum which is given….
can you finish the problem……..
we take four numbers in a group i.e $(1+2+3-4)$,$(5+6+7-8)$,$(9+10+11-12)$……,$(197+198+199-200)$. so there are $\frac{200}{4}=50$ groups. Therefore first term is$(a)$= $2$ ,common difference$(d)$=$8$ and numbers(n)=$50$. the sum formula of AP is $\frac{n}{2}\{2a+(n-1)d\}$
can you finish the problem……..
$\frac{n}{2}\{2a+(n-1)d\}$=$\frac{50}{2}\{2.8+(50-1)8\}$=$9900$
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# Missing Numbers up to 10
Printable worksheets on missing numbers up to 10 help the kids to practice counting of the numbers. Number math activities for kids are very important and necessary in early childhood education, preschool, kindergarten, nursery school and also for homeschooled kids.
Here the children are asked to look at the grid of numbers and identify the missing numbers and then fill in the missing numbers to count up to 10. In these missing number worksheets some of the numbers are already filled in to help kids to recognize the missing number up to 10 in the box.
# Fill in the missing numbers up to 10:
Parents and teachers can help the children to figure out the missing numbers and write in the box in the worksheets on missing numbers up to 10.
Math Only Math is based on the premise that children do not make a distinction between play and work and learn best when learning becomes play and play becomes learning.
However, suggestions for further improvement, from all quarters would be greatly appreciated.
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