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Home » 12 Helpful Tips For Doing explain in detail the triangle law of addition
# 12 Helpful Tips For Doing explain in detail the triangle law of addition
In addition to the two-dimensional triangle law of addition, I have two other important terms; the triangle law of the sum of squares and the sum of squares. These terms are often used with math. With the sum of squares, we use the triangle law of the sum of squares to describe how the sum of squares can change. The triangle law of addition is not a rule of art, but rather a mathematical formula that can be used to define your plan better.
The sum of squares is defined as the sum of the squares of the sides of a triangle. The sum of squares is the number of ways you can arrange three or more objects such that their sum is equal to the total number of objects. In the triangle law of addition, the sum of squares of two squares is one. The sum of squares of three squares is three.
If there are more sides than the sum of the squares of two squares, then the sum of squares of three is three. The sum of squares of two squares is three, but the sum of squares of three squares is less than the sum of the squares of two squares. The sum of squares of three squares is three.
The triangle law of addition can be applied to any number of objects to show that it is a perfect mathematical theorem. It’s the basis for the Pythagorean Theorem, which states that three and four times the same number equals the same number.
The triangle law of addition can be applied to any number of objects to show that it is a perfect mathematical theorem. Its the basis for the Pythagorean Theorem, which states that three and four times the same number equals the same number. |
# How to calculate the perimeter of an oval
Written by joan whetzel
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Ovals, or ellipses, look like horizontally elongated circles. Accurately finding the perimeter (circumference) requires some rather complicated calculus formulas. However, a much simpler formula provides a rough estimate that falls within 5 per cent of the value found by using the calculus equations. The rough estimate equation --- circumference≈2 π√1/2(a squared + b squared) -- begins with finding: (a) the semi-major axis (longer, horizontal radius) and (b) the semi-minor axis (shorter, vertical radius). The mathematical functions used in this equation include squaring and then adding the axes as well as division, square root and multiplication.
Skill level:
Moderate
• Ruler
• Oval diagram
## Instructions
1. 1
Find the oval's (a) semi-major axis by using the ruler to measure the longer, horizontal diameter, measuring from one side of the horizontal perimeter to the other, going through the centre point of the oval. For example: a = 10 feet. Make a note of the semi-major axis length on the oval diagram.
2. 2
Find the (b) semi-minor axis by using the ruler to measure the shorter, vertical diameter, measuring from one side of the vertical perimeter to the other, going through the centre point of the oval. Example: b = 6 feet. Jot down the semi-minor axis length on the oval diagram.
3. 3
Square both the semi-major and semi-minor axes and then add them together. Example: (a squared + b squared); (10 squared + 6 squared) = (100 + 36) = 136. Write down this number next to the oval diagram or on a separate piece of paper.
4. 4
Taking the value found, either multiply it by 1/2 or divide it by 2. Example 1/2(a squared + b squared); 1/2 x (100 + 36) = 1/2 x 136 or 136 / 2 = 68. Record this value.
5. 5
Using a calculator with the square root function, find the square root for the quotient, which will give a decimal value. Example: √1/2(a squared + b squared); √1/2(100+36) = √68 = 8.2462113. Write down the square root.
6. 6
For this final step, multiply 2π by the square root value. Note that this value will also contain decimals. Example: 2π√1/2(a squared + b squared); 2π√1/2 x (100 + 36) = 2π√68 = 2π x 8.2462113 = 2 x 3.14 x 8.2462113 = 51.786207. The circumference, or perimeter, of the oval is 51.786207 feet. Record the final answer inside or next to the diagram of the oval.
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# Math & Money
The girls are finally at the moment where they are very interested in money. We recently had a lesson on the different types on money in a general sense from coins to paper…etc. I compared the different bills in money to their decimal system tray and they seemed to have a good first understanding of the comparison. Today we took out the introduction to the decimal system tray again but we added the ‘fetch game’ to our lessons. I pulled the units (representing ones), Tens, Hundreds and thousand blocks out and lined them on the table as the ‘Bank’. I labeled each for visual reminders and also to help with the symbolic nature of learning the symbols themselves. I then gave the girls a tray and a simple number from each of the decimal values…100, 200, 10, 20, 1-9, 1000,2000 etc.
Today what we did with that was ‘build a number’. Then we broke that number down and saw it in it’s parts. Then rebuilt it to understand the place values we were reading. I am noticing that the girls have a harder time verbalizing the written symbol versus understand what the number actually represents. They seem to understand the quantity much more naturally. Future activities will include variations of this with the ‘fetch game’. Where will be given a number from each of the place values example 2345 and be expected to build that number in visuals. Or vice versa. This is also helping them with addition and writing formulas such as addition problems down as they are learning how to line up the numbers according to their place values by naturally catching on to the quantity of numbers (or place values) a number has.
Math materials pictures above are representing thousands, hundreds, tens, and units (ones). You’ll see an abacus there too which has coordinating place values. On the floor are the number rods that I like to do simple math with and building the formula next to them with the math box pieces that are sitting next to the rods. Or we use the chalkboard.
Lily is counting units to match the number she has been given.
The girls working on matching up their quantities.
After working with the ‘fetch game’ we moved onto the cash register where the girls got to put to use their decimal system skills they have been learning. We set up a grocery store and a bank (because using the wallets became too much at one time-and it was easier to see the bills laid out in a drawer rather than wadded up in a wallet). We started out as a dollar store where everything was a dollar and the girls took turns being at the cash register and as a customer. They gave exact change as it was one dollar bills. Then we started varying the prices and the amounts became more challenging. They began subtraction at this point. But instead of deciding on the register (calculator) or figuring the amount out in their heads or on their fingers we included the abacus to give them a visual of subtraction. And we discussed the steps of problem solving to understand the answer. That we use simple addition to add up the items and if the money we are given is more than the amount then we subtract. The girls have not used the addition snake game often (Montessori) so we are learning how to exchange numbers and quantities through our money game. So they are understanding that 5 and 5 is 10 and the same works with money. And they are learning that everything has a worth to it. And that that worth is a certain number. So this particular activity has reached a lot of areas. And there are many more areas to learn from it. |
# Table of 7: Master the Multiplication Table of Seven | 7 Times Table
Table for 7 is provided below so that students can do the fast calculation. This helps them to save a lot of time in the time-based competitive examination. We have provided here the multiplication table of 7 from 1 to 25 times in different formats to help students in memorising this table.
## Multiplication Table of 7
7 x 1 = 7 7 x 2 = 14 7 x 3 = 21 7 x 4 = 28 7 x 5 = 35 7 x 6 = 42 7 x 7 = 49 7 x 8 = 56 7 x 9 = 63 7 x 10 = 70 7 x 11 = 77 7 x 12 = 84 7 x 13 = 91 7 x 14 = 98 7 x 15 = 105 7 x 16 = 112 7 x 17 = 119 7 x 18 = 126 7 x 19 = 133 7 x 20 = 140 7 x 21 = 147 7 x 22 = 154 7 x 23 = 161 7 x 24 = 168 7 x 25 = 175
## How To Read Table Of 7
Master the multiplication table of 7 effortlessly with these structured tables. Below, you'll find how to read multiplication expressions using "times", "one is", "twos are", and "7 x 2 = " formats. Whether you're learning to say "Seven times two is 14" or "Seven twos are Fourteen," these tables provide a clear and systematic approach to understanding and memorizing the multiplication facts of 7.
### Learn the Multiplication Table of 7
Learn how to pronounce the table of 7. Perfect for students and educators alike, these tables make learning engaging and effective, helping build a solid mathematics foundation.
### Method 1: How to Read The Table Of 7 with "One is", "Twos are"
7 one is 7 Seven one is Seven 7
7 twos are 14 Seven twos are Fourteen 14
7 threes are 21 Seven threes are Twenty-One 21
7 fours are 28 Seven fours are Twenty-Eight 28
7 fives are 35 Seven fives are Thirty-Five 35
7 sixes are 42 Seven sixes are Fourty-Two 42
7 sevens are 49 Seven sevens are Fourty-Nine 49
7 eighths are 56 Seven eighths are Fifty-Six 56
7 nines are 63 Seven nines are Sixty-Three 63
7 tens are 70 Seven tens are Seventy 70
### Method 2: How to Read The Table Of 7 with "Times"
1 time 7 is 7 One time Seven is Seven 7
2 times 7 is 14 Two times Seven is Fourteen 14
3 times 7 is 21 Three times Seven is Twenty-One 21
4 times 7 is 28 Four times Seven is Twenty-Eight 28
5 times 7 is 35 Five times Seven is Thirty-Five 35
6 times 7 is 42 Six times Seven is Fourty-Two 42
7 times 7 is 49 Seven times Seven is Fourty-Nine 49
8 times 7 is 56 Eight times Seven is Fifty-Six 56
9 times 7 is 63 Nine times Seven is Sixty-Three 63
10 times 7 is 70 Ten times Seven is Seventy 70
These tables provide multiple ways to understand and memorize the multiplication facts of 7, enhancing learning through different expressions and values. Whether you're a student practicing or an educator teaching, these tables are invaluable tools for mastering basic arithmetic skills. |
Square root of 4 divided by square root of 4
What is the square root of 4 divided by the square root of 4? We start by displaying the problem we are solving in mathematical terms as follows:
4 ÷ √4 =
√4 √4
Looking at the problem above, you see that both the numerator and the denominator have a radical (√). It is tempting to conclude that the radicals even each other out, and that you can simply divide 7 by 8 to get the answer. However, that will not give you the correct answer unless the numbers are the same.
Here we will show you two different methods you can use to calculate the square root of 4 divided by the square root of 4.
Method 1
This is the most obvious method, where you simply first calculate the square root of each number, and then divide the result, like this:
√4 √4
=
2 2
2 2
= 1
Method 2
This is our favorite method, because it is easier and faster to enter into a handheld calculator. Since √4/√4 is the same as √(7/8), you can first divide the numbers, and then find the square root of the quotient, like this:
√4 ÷ √4 = √1
√1 = 1
As we said, we like this method because it is easy to enter into a calculator. This is what you enter into a calculator to get the answer to square root of 4 divided by square root of 4.
[4] [÷] [4] [=] [√x]
Square Root Divided by Square Root Calculator
Please enter a similar problem for us to solve for you.
√ ÷ √
Square root of 4 divided by square root of 5
Here is the next problem on our list that we have explained and calculated. |
# Area of a sector
The formula to find the area of a sector is A =
/ 360°
π × r2
Just use π = 3.14 and replace the value of r and n into the formula to get the area
However, what is a sector?
A sector is part of a circle bounded by two radii and their intercepted arc.
Examples of sectors are illustrated below. The portion of the circle shaded in blue is the sector
The best real life example of a sector I can think of is a slice of pizza.
In the example above, one-fourth of the pizza is removed and we can call it a sector. Next time you talk to a friend, you can tell them that you ate a sector of a pizza. Just kidding!
Therefore, to get the area of this slice of pizza, you will need to find the area of the circle and then divide the result by 4
Visualizing things this way may make it a little easier to see how they arrived to the formula
For a 360° circle, A =
360° / 360°
π × r2
Notice that
360° / 360°
= 1 and 1 × π × r2 = π × r2.
The formula is not changed by incorporating 360° / 360°
For a 180° circle or half a circle, A =
180° / 360°
π × r2
For a 90° circle or one-fourth of a circle, A =
90° / 360°
× π × r2
In general for a n° circle, A = n° / 360° π × r2
Caution when calculating the area of a sector: n must be positive is less than 360°
Example #1:
A circle has a radius of 5 cm. Calculate the area of a sector when the angle made by the radii is 60°
A =
60° / 360°
× 3.14 × 52
A =
60° / 360°
× 3.14 × 25
A =
60° / 360°
× 78.5
A = 0.166 × 78.5 = 13.06 cm2
Example #2:
A pizza has a diameter of 14 inches.
a. Calculate the area of a slice of pizza when the chef made all the slices with an angle of 45°
b. How many slices you think this pizza has?
c. Without using the formula of the area of a circle, what is the area of the whole pizza
A =
45° / 360°
× 3.14 × 142
A =
45° / 360°
× 3.14 × 196
A =
45° / 360°
× 615.44
A = 0.125 × 615.44 = 76.93 inches2
b. If all slices were cut with the same angle, then all slices must have the same size.
Just ask yourself "How many 45° will give 360°?"
45° + 45° + 45° + 45° + 45° + 45° + 45° + 45° = 360°
Or 8 × 45° = 360°
The pizza has 8 slices
c. Since the piza has 8 slices, the area of the whole pizza is 8 × 76.93 inches2 = 615.44
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## Projectiles Revision
In these questions, take g to be 10m/s2
### Question 1
A particle is projected with a velocity of 30 m/s at an angle of arctan(0.75) to the horizontal. It hits the ground at a point which is level to the point of projection. Find the time for which it is in air.
### Solution 1
As long as a projectile lands at a point level to the point of projection, we can use the general projectile equations.
This question requires the time of flight so we use the following equation:
### Question 2
A particle is projected with a velocity of 10m/s at an angle of 45° to the horizontal. It hits the ground at a point 3m below its point of projection. Find the time for which it is in air and the horizontal distance covered by the particle in this time.
### Solution 2
Since the projectile is landing at a point that is not level to its point of projection, we cannot use the general projectile equations, so we use the general equations of motion.
From what we can gather here: initial velocity (u) is 10m/s, the angle of projection is 45 ° and the final vertical displacement is -3m (i.e. 3m below the point of projection).
initial vertical velocity (uy) = u sin θ
initial horizontal velocity (ux) = u cos θ
With the data above, we can use the following equation to find time:
Since the particle is thrown upwards which is against the force of gravity, the value of the acceleration due to gravity is negative.
Now that it has reduced to a quadratic equation, we then solve the quadratic equation for the value of t.
Then we discard the negative value of t
Now that we found the time for which the particle is in the air, we can then use that time to find the horizontal distance covered as follows:
Since in projectile we take air resistance to be negligible, there is no acceleration or deceleration in the horizontal direction, which means a = 0. This leaves the equation as:
#### Sydney Chako
Mathematics, Chemistry and Physics teacher at Sytech Learning Academy. From Junior Secondary School to Tertiary Level Engineering Mathematics and Engineering Science.
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# Calculus
## Finding the Area between Two Curves using Vertical Elements
To find the area between two curves, we will follow the process of using the length of the rectangle multiplied with the width and adding them all up, which is similar to finding the area under a curve. However, there is one difference: the length of the rectangle is no longer the distance from the curve to the axis. Now, the length of the rectangle is the distance from the lower curve to the upper curve. In general, the area between 2 curves, $$f(x)$$ and $$g(x)$$, is given by evaluating the equation:
$$A=\int_a^b(f(x)-g(x)) \ dx$$
Example
Take these graphs for the curves $$y=\sqrt{x}$$ and $$y=x^2$$:
From the first graph, the length of the rectangles from the axis to curve, at any point, is $$y=\sqrt{x}$$. From the second graph, the length of the rectangles from the axis to curve, at any point, is $$y=x^2$$. However, to find the area in between the curves, we need to know the length of the rectangles in between the curves. To find this, we need to find the difference between them, by subtracting the lower curve from the upper curve:
So the length of the rectangles in between the curves, at any point, is $$length=\sqrt{x}-x^2$$. The width of the rectangles is $$dx$$, as in the example from
The calculations for this example would be as follows:
$$Area=\int_0^1(\sqrt{x}-x^2) \ dx$$
$$=(\frac{2}{3}x^{\frac{3}{2}}-\frac{x^3}{3})|_{x=0}^{x=1}$$
$$=\frac{2}{3}(1)^{\frac{3}{2}}-\frac{1^3}{3}$$
$$=\frac{2}{3}-\frac{1}{3}$$
$$=\frac{1}{3}$$
Try this interactive tool!
Adjust $$a$$ and $$b$$ to change $$g(x)$$. As you move the slider, you will see different definite integrals for $$f(x)$$, $$g(x)$$, and $$[f(x) - g(x)]$$.
## Finding the Area between Two Curves using Horizontal Elements
In the above example, the length of the rectangle was in the vertical direction. But we can also calculate the area between two curves by drawing these rectangles in the horizontal direction. We'll demonstrate this idea below:
Example
Let’s take the case of the 2 equations, $$x=4$$ and $$x=(y-3)^2$$.
If we try to use vertical rectangles again, we run into a problem. What is the length of the rectangles? The top of the rectangle and the bottom of the rectangle both touch the same curve, so we can’t subtract to find the difference.
The solution is that we have to use horizontal rectangles instead. This way, one end of the rectangle is touching one curve, and the other end of the rectangle is touching the other. This way, we can subtract the two graphs to find the difference between them, which is the length of the rectangles in between the curves.
When we were using vertical rectangles, we subtracted the lower curve from the upper curve. Now that we’re using horizontal rectangles, we will subtract the left curve from the right curve. So, in our example case, the length of the rectangles at any point between the curves is $$length=4-(y-3)^2$$.
When we were using vertical rectangles, the width of the rectangle was an infinitely small length along the x-axis, and so we named it $$dx$$. Now, with horizontal rectangles, the width of the rectangle is an infinitely small length along the y-axis, which is named $$dy$$.
So the area of one of these rectangles is $$length \times width = (4-(y-3)^2) \cdot dy$$. To add all of these areas together to find the total area between the curves, we integrate:
$$\int_b^a4-(y-3)^2 \ dy$$
Before we can evaluate this, we need to know the bounds of the integration, $$a$$ and $$b). Since we are using horizontal rectangles, instead of our bounds being on the x-axis, like they were in the previous examples, now our bounds are on the y-axis. Looking at our graph, we can see that the 2 bounds of our integration will be at the points where the lines intersect. To find these points of intersection, we need to substitute one of the variables, and solve for the remaining one: Equation 1: \(x=4$$
Equation 2: $$x=(y-3)^2$$
Since we are looking for the bounds on the y-axis, we know we need to solve for the y variable, and can eliminate x. Therefore:
$$4=(y-3)^2$$
$$y=3 \pm 2$$
$$y=1, y=5$$
Now that we’ve solved for the points of intersection, giving us the bounds of our integration, we can perform the following calculations to determine the area between these two curves:
$$Area=\int_1^5 4-(y-3)^2 \ dy$$
$$=\int_1^5 4-(y^2-6y+9) \ dy$$
$$=(-\frac{y^3}{3}+3y^2-5y)|_{x=1}^{x=5}$$
$$=8.33$$ |
Chapter 11 %28pages 153 -171%29
# Chapter 11 %28pages 153 -171%29 - MSIT 3000 Chapter 11...
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MSIT 3000 Chapter 11 Confidence Intervals and Hypothesis Tests for Means Section 11.1 Sampling Distributions for Means What other distributions have we talked about? How do we describe distributions? Previously, we’ve examined data and calculated statistics. The distributions were about data. Now we’ll consider the distribution, sampling distribution, that is associated with the statistic, y . This statistic has a distribution with a shape, center and variation. This distribution provides probabilities for the possible values of the statistic. 153
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Class Exercise: Working in pairs, toss a die 5 times. Record the number of dots on each of the 5 tosses, and calculate the average number of dots for the 5 tosses. Toss 1 2 3 4 5 Average # dots We will look at the THREE different distributions that are part of this exercise. Population Distribution : The distribution from which we took our sample. Data Distribution : The distribution of the data from ONE sample. Sampling Distribution of the Sample Mean: The distribution of the sample averages. Population Distribution 154
| | 1/6 | | | |_______________________________ Y 1 2 3 4 5 6 Shape: Center: Spread: Data Distribution - One Sample | | 1/6 | | | |_______________________________ y 1 2 3 4 5 6 Shape: Center: Spread: Sampling Distribution: Distribution of y values 155
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Shape: Look at histogram of sample means given in class Center : The mean of the sampling distribution of y equals the mean of the population. Spread : The standard deviation of y is smaller than the standard deviation of the population. Notation and Terminology: Term Pop. Sample Sampling Distn Mean µ y μ Std. Dev. σ s s n Note: The standard deviation of a statistic is called the standard error . Shape – two cases : If the population has a normal distribution, then the sampling distribution of y will have a normal distribution. Central Limit Theorem: For large n, the sampling distribution of y is approximately a normal distribution regardless of the distribution of the population . (How large is large? For this class, use n > 30) 156
In summary, it’s possible to use the normal model for quantitative and categorical situations. Ex: Consider a normal population with mean μ = 30 and standard deviation σ = 6. Suppose samples of size 4 are selected. 157 Mean (quantitative) Proportion (Categorical) Population Parameter µ p Sample Statistic sample mean, y sample proportion, ˆ p Mean of Sampling Distribution µ p Standard Error OR s n n σ (1 ) p p n - Can use bell-shaped Distribution IF Pop normal OR Data normal OR CLT: n large np >= 15 n(1-p) >= 15
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a. Find the mean and standard error of the distribution of the sample mean, y Mean Standard Error b. If a population is approximately N (30, 6), what is the probability that the average of a sample of size 4 is less than 32? c.
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## This note was uploaded on 01/10/2012 for the course MIST 3000 taught by Professor Kim during the Fall '11 term at University of Georgia Athens.
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# Pre-Algebra 6 – Commutative Property of Multiplication
Hello. I’m Professor Von Schmohawk
and welcome to Why U. In our last lecture, we saw how the people
on my primitive island of Cocoloco discovered addition and subtraction. Once we had invented addition and subtraction the Cocoloconians could calculate very complicated
coconut transactions with great precision. But we soon found out that with only
addition and subtraction some calculations could take a very long time. For instance, once a year, everyone on Cocoloco must donate three coconuts for the
annual feast of Mombozo. So if all 87 inhabitants of Cocoloco
each donate three coconuts then how many coconuts will we have
for the feast? Before we discovered multiplication,
can be found with a single calculation. Multiplication is just a tricky way to do
repeated addition. For instance, when the king of Cocoloco wanted
to tile the floor of his rectangular hut with very expensive imported
Bongoponganian tiles we needed to know exactly how many square
tiles to buy from the Bongoponganians. We knew how big each tile was so we could
have marked the floor into little squares and counted all the squares. But with multiplication, it was much easier. All we had to do was to figure out how many
rows of tiles we would need and how many tiles were in each row and then multiply the two numbers. Since we figured it would take
six rows of ten tiles we knew that we would need six times ten,
or sixty tiles. But then someone suggested that it would be
better to lay the tiles down in vertical rows instead of horizontal rows. We would then need ten rows of six tiles. At first we thought that this might require
a different number of tiles. Then we realized that ten times six
is also sixty so you will still have to buy sixty very expensive
imported Bongoponganian tiles. It doesn’t make any difference if you multiply
six times ten, or ten times six. You get the same number. We originally called this The Commutative Property of Multiplication of
Very Expensive Imported Bongoponganian Tiles. After a while we decided to shorten the name to
The Commutative Property of Multiplication. We can write this property as
A times B equals B times A. In Algebra, a dot is often used as a
multiplication symbol to avoid confusion with the letter X. Just like addition,
multiplication is a binary operation which, as you may recall
from our previous lecture is a mathematical calculation involving
two numbers. These numbers are called “operands” and in the case of multiplication these operands are multiplied together to
produce a result called the “product”. In multiplication operations, the operands
are sometimes referred to as “factors”. Even though multiplication is defined as a
binary operation you may often see multiplications involving
more than two operands. Just as in addition, this is possible because pairs of operands can be multiplied
one at a time with each product replacing the pair. In this way an unlimited number of operands
can be multiplied sequentially. On Cocoloco, we soon discovered that the
commutative property also applied to situations where more than two numbers were
multiplied together. For example, let’s say that you had
24 boxes. You can stack these boxes in several
different ways. For instance, you could arrange them in
three rows of four boxes and stack them two levels high. Or you could arrange them in
four rows of two boxes and stack them three levels high. Or you could arrange them in
two rows of three boxes and stack them four levels high. It doesn’t matter in which order you multiply
the dimensions of the stack. It will always add up to the same
number of boxes. We can apply the commutative property to multiplication
operations involving any number of operands. By swapping adjacent pairs of numbers, the
operands can be reordered in any way we please. For instance, in this multiplication
involving four operands the two at the end
could be moved up to the front. Or the five could be moved to the back. So two or more numbers which are multiplied
can be reordered in any way without affecting the result. As we saw in the previous lecture, the same
holds true for numbers which are added. Addition and multiplication are both commutative. Commutative properties are important
algebraic tools that allow us to rearrange groups of numbers
which are added or multiplied. In the next chapter, we will discover
several more properties which we will add to our tool chest of
mathematical tricks. |
Work and Energy: Chapter Outline || About the Tutorial || Tutorial Topics || Usage Policy || Feedback
### Lesson 1: Basic Terminology and Concepts
Work:
Definition and Mathematics of Work
Calculating the Work Done by Forces
### Lesson 2: The Work-Energy Relationship
The Work-Energy Connection:
Situations Involving External Forces
Situations in Which Energy is Conserved
Application and Practice
## Lesson 2: The Work-Energy Relationship
### Bar Chart Illustrations
One tool which can be utilized to express an understanding of the work-energy theorem is a bar chart. A work-energy bar chart represents the amount of energy possessed by an object by means of a vertical bar. The length of the bar is representative of the amount of energy present, with a longer bar representing a greater amount of energy. In a work-energy bar chart, a bar is constructed for each form of energy. Consistent with the work-energy relationship discussed in this lesson, the sum of all forms of initial energy plus the work done on the object by external forces equals the sum of all forms of final energy.
## KEi + PEi + Wext = KEf + PEf
In a work-energy bar chart, a bar is used to represent the amount of each term in the above equation. Consequently, the sum of the bar heights for the initial condition (initial energy + external work) must equal the sum of the bar heights for the final condition (final energy). Since the potential energy comes in two forms - the elastic potential energy stored in springs (PEspring) and the gravitational potential energy (PEgrav) - the above equation is rewritten as
## KEi + PEi-grav + PEi-spring + Wext = KEf + PEf-grav + PEf-spring
In this portion of Lesson 2, we will investigate the use and meaning of work-energy bar charts and make an effort to apply this understanding to a variety of motions involving energy changes or energy transformations.
Procedure for Constructing Bar Charts
The following procedure might be useful for constructing work-energy bar charts:
• analyze the initial and final states of the object in order to make decisions about the presence or absence of the different forms of energy
• analyze the forces acting upon the object during the motion to determine if external forces are doing work and whether the work (if present) is positive or negative
• construct bars on the chart to illustrate the presence and absence of the various forms of energy for the initial and final state of the object; the exact height of the individual bars is not important; what is important is that the sum of the heights on the left of the chart is balanced by the sum of the heights on the right of the chart
The above procedure is illustrated below.
Consider a ball falling from the top of a pillar to the ground below; ignore air resistance. The initial state is the ball at rest on top of the pillar and the final state is the ball just prior to striking the ground. Given this motion and the identification of the initial and final state of the ball, decisions can be made about the presence and absence of each form of energy. Since there is no motion at the top of the pillar, there is no initial kinetic energy. Since the ball is elevated above the ground while on top of the pillar, there is an initial gravitational potential energy (PEgrav). There are no springs involved; thus, there is neither initial nor final elastic potential energy (PEspring). In the final condition (just prior to striking the ground), the ball is moving. Thus, there is a final kinetic energy. And finally, the ball is no longer elevated above the ground so there is no final gravitational potential energy. The ball falls under the influence of gravity (an internal force) alone. Thus there are no external forces present nor doing work. The diagram at the right summarizes this analysis.
To complete the bar chart, an arbitrarily decided height for each bar is decided upon and a bar is constructed for each form of energy. As mentioned before, it is not important exactly how high each bar is. It is only important that the sum of the bar heights on the left balance the sum of the bar heights on the right.
Observe that this work-energy bar chart reveals that
• there is no kinetic and elastic potential energy in the initial state
• there is no gravitational and elastic potential energy in the final state
• there is no work done by external forces
• the sum of the heights on the right (5 units) equals the sum of the heights on the left (5 units)
It is not important as to how high the two bars are in the above bar chart. If the bars were 4 units high instead of 5 units high, then it would be an equally acceptable bar chart. The decision about bar height is entirely arbitrary.
Perhaps at this time you may want to review the lessons on work, potential energy and kinetic energy. Use the links below.
The Example of a Skidding Car
Now we will repeat the process for a car which skids from a high speed to a stop across level ground with its brakes applied. The initial state is the car traveling at a high speed and the final state is the car at rest. Initially, the car has kinetic energy (since it is moving) but does not have gravitational potential energy (since the height is zero) nor elastic potential energy (since there are no springs). In the final state of the car, there is neither kinetic energy (since the car is at rest) nor potential energy (since there is no height nor springs). The force of friction between the tires of the skidding car and the road does work on the car. Friction is an external force. Friction does negative work since its direction is opposite the direction of the car's motion.
Now that the analysis is complete, the bar chart can be constructed. The chart must be consistent with the above analysis.
Observe that the bar for work is a downward bar. This is consistent with the fact that the work done by friction is negative work. Whenever negative work is done by external forces, the Wext bar will be a downward bar. Note also that the sum of the bar height on the left side (+5 plus -5) is the same as the sum of the bar heights on the right side of the chart. One final comment is in order: even though the height of all bars on the left equals the height of all bars on the right, energy is not conserved. The bar chart includes both energy and work on the left side of the chart. If work is done by external forces, then the only reason that the sum of the bar heights are equal on both sides is that the Wext makes up for the difference between the initial and final amounts of total mechanical energy.
Perhaps at this time you may want to review the lessons on work, potential energy and kinetic energy. Use the links below.
The Example of a Skier
As a final example consider a skier that starts from rest on top of hill A and skis into the valley and back up onto hill B. The skier utilizes her poles to propel herself across the snow, thus doing work to change her total mechanical energy. The initial state is on top of hill A and the final state is on top of hill B. Suppose that friction and air resistance have a negligible affect on the motion. In the initial state, the skier has no kinetic energy (the skier is said to be at rest). There is no elastic potential energy in both the initial and the final states (since there are no springs). The skier has gravitational potential energy in both the initial and the final states (since the skier is at an elevated position). Finally, work is being done by external forces since the skier is said to be using "her poles to propel herself across the snow." This work is positive work since the force of the snow on her poles is in the same direction as her displacement.
Now that the analysis is complete, the bar charts can be constructed. The charts must be consistent with the above analysis.
Observe that the bar for work is an upward bar. This is consistent with the fact that the work done by the poles is positive work. Whenever positive work is done by external forces, the Wext bar will be an upward bar. Note also that the sum of the bar height on the left side (+5 plus +2) is the same as the sum of the bar heights on the right side (+4 plus +3) of the chart. As mentioned earlier, the exact heights of the individual bars is not important. It is only important that the bars exist, that they are in the correct direction (upward) and that their sum on the left is the same as the sum on the right.
Perhaps at this time you may want to review the lessons on work, potential energy and kinetic energy. Use the links below.
A Different Type of Bar Chart
On occasions it is customary to utilize a different type of work-energy bar chart that looks like the diagram below. If external forces are not doing work, then the total mechanical energy is conserved. The Wext term cancels from the work-energy equation leaving the equation
## KEi + PEi = KEf + PEf
This equation shows that the total mechanical energy (potential energy plus kinetic energy) is the same in the initial state as it is in the final state. In fact, if external forces are not doing work, then the total mechanical energy will be the same throughout the entire motion. For such situations it is customary to use the different style of bar chart to depict energy conservation. A series of bars are shown for a variety of positions throughout the motion; each set of bars reveals that the total mechanical energy (TME) is always the same while the potential energy (PE) and the kinetic energy (KE) are constantly changing.
As an example of the use of this bar chart, consider a roller coaster car in the ideal situation in which the force of air resistance is assumed negligible (indeed, an idealized situation). Since the normal force acts at right angles to the motion at all times, it does not do work. The only force doing work on the roller coaster car is gravity. And since the force of gravity is an internal or conservative force, the total mechanical energy is conserved (i.e., not changing). The energy may change forms - transforming from potential to kinetic and vice versa. Yet the total amount will never change. The diagram below depicts the conservation of total mechanical energy and the transformation of potential and kinetic energy for a roller coaster car at five positions along a track.
In conclusion, bar charts are a useful tool for depicting the influence of external forces (if present) upon the total mechanical energy. It is a conceptual tool for representing one's understanding of the work-energy relationship.
Construct analyses and work-energy bar charts for the following motions. Then check the answers by clicking the button and by clicking on the See Bar Chart link.
1. A ball is dropped from rest from a bridge. As the ball falls through the air, it encounters a small amount of air resistance. The final state of the ball is the instant before it strikes the water.
See Bar Chart
2. A volleyball player spikes the ball at just above net level and drives it over the net. The initial state is the ball just prior to the spike. The final state of the ball is the instant before it strikes the ground.
See Bar Chart
3. A spring gun is used to project a sponge dart into the air at an angle to the horizontal. The gun is held at a height of 1-meter before the trigger is pulled. The loaded spring gun is the initial state and the sponge dart at its peak is the final state.
See Bar Chart
4. A baseball is caught by a catcher after passing over home plate. The initial state is the baseball moving at high speed just prior to hitting the catcher's mitt. The final state is the baseball just after the catcher has applied the force to stop the ball. Assume that the ball does not change height as the catcher is catching it.
See Bar Chart
5. In a physics lab, a Hot Wheels car starts at an elevated position, moves down an incline to the level ground, strikes a box and skids to a stop. Consider three states for the car: state A is the top of the incline; state B is the bottom of the incline before striking the box; state C is after the car has been brought to a stop. Use the diagram at the right and your understanding of the work-energy theorem to construct bar charts for the motion from A to B and from B to C.
See Bar Charts
1. Here is the bar chart for question #1 above.
2. Here is the bar chart for question #2 above.
3. Here is the bar chart for question #3 above.
4. Here is the bar chart for question #4 above.
5. Here is the bar chart for question #5 above.
### Lesson 2: The Work-Energy Relationship
Work and Energy: Chapter Outline || About the Tutorial || Tutorial Topics || Usage Policy || Feedback |
# Difference between revisions of "2015 AMC 12A Problems"
## Problem 1
What is the value of $(2^0-1+5^2-0)^{-1}\times5?$
$\textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{5}{24}\qquad\textbf{(E)}\ 25$
## Problem 2
Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72$
## Problem 3
Mr. Patrick teaches math to 15 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 80. after he graded Payton's test, the class average became 81. What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$
## Problem 4
The sum of two positive numbers is 5 times their difference. What is the ratio of the larger number to the smaller?
$\textbf{(A)}\ \frac54 \qquad\textbf{(B)}\ \frac32 \qquad\textbf{(C)}\ \frac95 \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac52$
## Problem 5
Amelia needs to estimate the quantity $\frac{a}{b} - c$, where $a, b,$ and $c$ are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of $\frac{a}{b} - c$?
$\textbf{(A)}\ \text{She rounds all three numbers up.}\\ \qquad\textbf{(B)}\ \text{She rounds } a \text{ and } b \text{ up, and she rounds } c \text{ down.}\\ \qquad\textbf{(C)}\ \text{She rounds } a \text{ and } c \text{ up, and she rounds } b \text{ down.} \\ \qquad\textbf{(D)}\ \text{She rounds } a \text{ up, and she rounds } b \text{ and } c \text{ down.}\\ \qquad\textbf{(E)}\ \text{She rounds } c \text{ up, and she rounds } a \text{ and } b \text{ down.}$
## Problem 6
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2 : 1$?
## Problem 8
The ratio of the length to the width of a rectangle is $4$ : $3$. If the rectangle has diagonal of length $d$, then the area may be expressed as $kd^2$ for some constant $k$. What is $k$?
## Problem 10
Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$. What is $x$?
## Problem 12
The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$. What is $a+b$?
## Problem 14
What is the value of $a$ for which $\frac{1}{\text{log}_2a} + \frac{1}{\text{log}_3a} + \frac{1}{\text{log}_4a} = 1$?
## Problem 15
What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
## Problem 17
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
## Problem 19
For some positive integers $p$, there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C$, $AB=2$, and $CD=AD$. How many different values of $p<2015$ are possible?
## Problem 21
A circle of radius $r$ passes through both foci of, and exactly four points on, the ellipse with equation $x^2+16y^2=16$. The set of all possible values of $r$ is an interval $[a,b)$. What is $a+b$?
## Problem 23
Let $S$ be a square of side length 1. Two points are chosen independently at random on the sides of $S$. The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$, where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$. What is $a+b+c$?
## Problem 25
A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k\ge1$, the circles in $\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is $$\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?$$
$[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair cent1 = coord[k-skip], cent2 = coord[k+skip]; real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2); real shiftx = cent1.x + sqrt(4*r1*rn); coord[k] = (shiftx,rn); } // Draw the remaining 63 circles } for(int i=1;i<=63;i=i+1) { filldraw(circle(coord[i],coord[i].y),gray(0.75)); }[/asy]$
$\textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}$ |
# 3.4: Separable Equations - Mathematics
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Learning Objectives
• Use separation of variables to solve a differential equation.
• Solve applications using separation of variables.
We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.
## Separation of Variables
Definition: Separable Differential Equations
A separable differential equation is any equation that can be written in the form
[y'=f(x)g(y). label{sep}]
The term ‘separable’ refers to the fact that the right-hand side of Equation ef{sep} can be separated into a function of (x) times a function of (y). Examples of separable differential equations include
[ egin{align} y' =(x^2−4)(3y+2) label{eq1} [4pt] y' =6x^2+4x label{eq2}[4pt] y' =sec y+ an y label{eq3} [4pt] y' =xy+3x−2y−6. label{eq4} end{align}]
Equation ef{eq2} is separable with (f(x)=6x^2+4x) and (g(y)=1), Equation ef{eq3} is separable with (f(x)=1) and (g(y)=sec y+ an y,) and the right-hand side of Equation ef{eq4} can be factored as ((x+3)(y−2)), so it is separable as well. Equation ef{eq3} is also called an autonomous differential equation because the right-hand side of the equation is a function of (y) alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables.
Problem-Solving Strategy: Separation of Variables
1. Check for any values of (y) that make (g(y)=0.) These correspond to constant solutions.
2. Rewrite the differential equation in the form [ dfrac{dy}{g(y)}=f(x)dx.]
3. Integrate both sides of the equation.
4. Solve the resulting equation for (y) if possible.
5. If an initial condition exists, substitute the appropriate values for (x) and (y) into the equation and solve for the constant.
Note that Step 4 states “Solve the resulting equation for (y) if possible.” It is not always possible to obtain (y) as an explicit function of (x). Quite often we have to be satisfied with finding y as an implicit function of (x).
Example (PageIndex{1}): Using Separation of Variables
Find a general solution to the differential equation (y'=(x^2−4)(3y+2)) using the method of separation of variables.
Solution
Follow the five-step method of separation of variables.
1. In this example, (f(x)=x^2−4) and (g(y)=3y+2). Setting (g(y)=0) gives (y=−dfrac{2}{3}) as a constant solution.
2. Rewrite the differential equation in the form
[ dfrac{dy}{3y+2}=(x^2−4)dx.]
3. Integrate both sides of the equation:
[ ∫dfrac{dy}{3y+2}=∫(x^2−4)dx.]
Let (u=3y+2). Then (du=3dfrac{dy}{dx}dx), so the equation becomes
[ dfrac{1}{3}∫dfrac{1}{u}du=dfrac{1}{3}x^3−4x+C]
[ dfrac{1}{3}ln|u|=dfrac{1}{3}x^3−4x+C]
[ dfrac{1}{3}ln|3y+2|=dfrac{1}{3}x^3−4x+C.]
4. To solve this equation for (y), first multiply both sides of the equation by (3).
[ ln|3y+2|=x^3−12x+3C]
Now we use some logic in dealing with the constant (C). Since (C) represents an arbitrary constant, (3C) also represents an arbitrary constant. If we call the second arbitrary constant (C_1), the equation becomes
[ ln|3y+2|=x^3−12x+C_1.]
Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base (e)).
[ egin{align} e^{ln|3y+2|} =e^{x^3−12x+C_1} |3y+2| =e^{C_1}e^{x^3−12x} end{align}]
Again define a new constant (C_2=e^{c_1}) (note that (C_2>0)):
[ |3y+2|=C_2e^{x^3−12x}.]
This corresponds to two separate equations:
[3y+2=C_2e^{x^3−12x}]
and
[3y+2=−C_2e^{x^3−12x}.]
The solution to either equation can be written in the form
[y=dfrac{−2±C_2e^{x^3−12x}}{3}.]
Since (C_2>0), it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant (C) is entirely arbitrary, and can be dropped. Therefore the solution can be written as
[ y=dfrac{−2+Ce^{x^3−12x}}{3}.]
5. No initial condition is imposed, so we are finished.
Exercise (PageIndex{1})
Use the method of separation of variables to find a general solution to the differential equation
[ y'=2xy+3y−4x−6. onumber]
Hint
First factor the right-hand side of the equation by grouping, then use the five-step strategy of separation of variables.
[ y=2+Ce^{x^2+3x} onumber]
Example (PageIndex{2}): Solving an Initial-Value Problem
Using the method of separation of variables, solve the initial-value problem
[ y'=(2x+3)(y^2−4),y(0)=−1.]
Solution
Follow the five-step method of separation of variables.
1. In this example, (f(x)=2x+3) and (g(y)=y^2−4). Setting (g(y)=0) gives (y=±2) as constant solutions.
2. Divide both sides of the equation by (y^2−4) and multiply by (dx). This gives the equation
[dfrac{dy}{y^2−4}=(2x+3)dx.]
3. Next integrate both sides:
[∫dfrac{1}{y^2−4}dy=∫(2x+3)dx. label{Ex2.2}]
To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity
[dfrac{1}{y^2−4}=dfrac{1}{4}left(dfrac{1}{y−2}−dfrac{1}{y+2} ight).]
Then Equation ef{Ex2.2} becomes
[dfrac{1}{4}∫left(dfrac{1}{y−2}−dfrac{1}{y+2} ight)dy=∫(2x+3)dx]
[dfrac{1}{4}left (ln|y−2|−ln|y+2| ight)=x^2+3x+C.]
Multiplying both sides of this equation by (4) and replacing (4C) with (C_1) gives
[ln|y−2|−ln|y+2|=4x^2+12x+C_1]
[ln left|dfrac{y−2}{y+2} ight|=4x^2+12x+C_1.]
4. It is possible to solve this equation for y. First exponentiate both sides of the equation and define (C_2=e^{C_1}):
[left|dfrac{y−2}{y+2} ight|=C_2e^{4x^2+12x}.]
Next we can remove the absolute value and let (C_2) be either positive or negative. Then multiply both sides by (y+2).
[y−2=C_2(y+2)e^{4x^2+12x}]
[y−2=C_2ye^{4x^2+12x}+2C_2e^{4x^2+12x}.]
Now collect all terms involving y on one side of the equation, and solve for y:
[y−C_2ye^{4x^2+12x}=2+2C_2e^{4x^2+12x}]
[y(1−C_2e^{4x^2+12x})=2+2C_2e^{4x^2+12x}]
[y=dfrac{2+2C_2e^{4x^2+12x}}{1−C_2e^{4x^2+12x}}.]
5. To determine the value of (C_2), substitute (x=0) and (y=−1) into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation (dfrac{y−2}{y+2}=C_2e^{4x^2+12}). This is much easier to solve for (C_2):
[dfrac{y−2}{y+2}=C_2e^{4x^2+12x}]
[dfrac{−1−2}{−1+2}=C_2e^{4(0)^2+12(0)}]
[C_2=−3.]
Therefore the solution to the initial-value problem is
[y=dfrac{2−6e^{4x^2+12x}}{1+3e^{4x^2+12x}}.]
A graph of this solution appears in Figure (PageIndex{1}).
Exercise (PageIndex{2})
Find the solution to the initial-value problem
[ 6y'=(2x+1)(y^2−2y−8) onumber]
with (y(0)=−3) using the method of separation of variables.
Hint
Follow the steps for separation of variables to solve the initial-value problem.
[ y=dfrac{4+14e^{x^2+x}}{1−7e^{x^2+x}} onumber]
## Applications of Separation of Variables
Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.
#### Solution concentrations
Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations.
Example (PageIndex{3}): Determining Salt Concentration over Time
A tank containing (100,L) of a brine solution initially has (4,kg) of salt dissolved in the solution. At time (t=0), another brine solution flows into the tank at a rate of (2,L/min). This brine solution contains a concentration of (0.5,kg/L) of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of (2,L/min), so that the level of liquid in the tank remains constant (Figure (PageIndex{2})). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.
Solution
First we define a function (u(t)) that represents the amount of salt in kilograms in the tank as a function of time. Then (dfrac{du}{dt}) represents the rate at which the amount of salt in the tank changes as a function of time. Also, (u(0)) represents the amount of salt in the tank at time (t=0), which is (4) kilograms.
The general setup for the differential equation we will solve is of the form
[dfrac{du}{dt}=INFLOW RATE−OUTFLOW RATE.]
INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of (2)L/min, and each liter of solution contains (0.5) kilogram of salt, every minute (2(0.5)=1kilogram) of salt enters the tank. Therefore INFLOW RATE = (1).
To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time (t) is equal to (u(t)). Thus, the concentration of salt is (dfrac{u(t)}{100} kg/L), and the solution leaves the tank at a rate of (2) L/min. Therefore salt leaves the tank at a rate of (dfrac{u(t)}{100}⋅2=dfrac{u(t)}{50}) kg/min, and OUTFLOW RATE is equal to (dfrac{u(t)}{50}). Therefore the differential equation becomes (dfrac{du}{dt}=1−dfrac{u}{50}), and the initial condition is (u(0)=4.) The initial-value problem to be solved is
[dfrac{du}{dt}=1−dfrac{u}{50},u(0)=4.]
The differential equation is a separable equation, so we can apply the five-step strategy for solution.
Step 1. Setting (1−dfrac{u}{50}=0) gives (u=50) as a constant solution. Since the initial amount of salt in the tank is (4) kilograms, this solution does not apply.
Step 2. Rewrite the equation as
[dfrac{du}{dt}=dfrac{50−u}{50}.]
Then multiply both sides by (dt) and divide both sides by (50−u:)
[dfrac{du}{50−u}=dfrac{dt}{50}.]
Step 3. Integrate both sides:
[egin{align} ∫dfrac{du}{50−u} =∫dfrac{dt}{50} −ln|50−u| =dfrac{t}{50}+C. end{align}]
Step 4. Solve for (u(t)):
[ln|50−u|=−dfrac{t}{50}−C]
[e^{ln|50−u|}=e^{−(t/50)−C}]
[|50−u|=C_1e^{−t/50}.]
Eliminate the absolute value by allowing the constant to be either positive or negative:
[50−u=C_1e^{−t/50}.]
Finally, solve for (u(t)):
[u(t)=50−C_1e^{−t/50}.]
Step 5. Solve for (C_1):
[egin{align} u(0) =50−C_1e^{−0/50} 4 =50−C_1 C_1 =46. end{align}]
The solution to the initial value problem is (u(t)=50−46e^{−t/50}.) To find the limiting amount of salt in the tank, take the limit as (t) approaches infinity:
[egin{align} lim_{t→∞}u(t) =50−46e^{−t/50} =50−46(0)=50. end{align}]
Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is (50) kilograms, then it remains constant. If it starts at less than (50) kilograms, then it approaches 50 kilograms over time.
Exercise (PageIndex{3})
A tank contains (3) kilograms of salt dissolved in (75) liters of water. A salt solution of (0.4,kg salt/L) is pumped into the tank at a rate of (6,L/min) and is drained at the same rate. Solve for the salt concentration at time (t). Assume the tank is well mixed at all times.
Hint
Follow the steps in Example (PageIndex{3}) and determine an expression for INFLOW and OUTFLOW. Formulate an initial-value problem, and then solve it.
Initial value problem:
[ dfrac{du}{dt}=2.4−dfrac{2u}{25},, u(0)=3 onumber]
[u(t)=30−27e^{−t/50} onumber]
## Newton’s law of Cooling
Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let (T(t)) represent the temperature of an object as a function of time, then (dfrac{dT}{dt}) represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by (T_s). Then Newton’s law of cooling can be written in the form
[ dfrac{dT}{dt}=k(T(t)−T_s)]
or simply
[ dfrac{dT}{dt}=k(T−T_s).]
The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature (T_0). Therefore the initial-value problem that needs to be solved takes the form
[ dfrac{dT}{dt}=k(T−T_s) label{newton}]
with (T(0)=T_0), where (k) is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example (PageIndex{4}).
Example (PageIndex{4}): Waiting for a Pizza to Cool
A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is (350°F.) The temperature of the kitchen is (75°F), and after (5) minutes the temperature of the pizza is (340°F). We would like to wait until the temperature of the pizza reaches (300°F) before cutting and serving it (Figure (PageIndex{3})). How much longer will we have to wait?
Solution
The ambient temperature (surrounding temperature) is (75°F), so (T_s=75). The temperature of the pizza when it comes out of the oven is (350°F), which is the initial temperature (i.e., initial value), so (T_0=350). Therefore Equation ef{newton} becomes
[dfrac{dT}{dt}=k(T−75) ]
with (T(0)=350.)
To solve the differential equation, we use the five-step technique for solving separable equations.
1. Setting the right-hand side equal to zero gives (T=75) as a constant solution. Since the pizza starts at (350°F,) this is not the solution we are seeking.
2. Rewrite the differential equation by multiplying both sides by (dt) and dividing both sides by (T−75):
[dfrac{dT}{T−75}=kdt. onumber]
3. Integrate both sides:
[egin{align} ∫dfrac{dT}{T−75} =∫kdt onumber ln|T−75| =kt+C. onumber end{align} ]
4. Solve for (T) by first exponentiating both sides:
[egin{align}e^{ln|T−75|} =e^{kt+C} onumber |T−75| =C_1e^{kt} onumber T−75 =C_1e^{kt} onumber T(t) =75+C_1e^{kt}. onumber end{align} ]
5. Solve for (C_1) by using the initial condition (T(0)=350:)
[egin{align}T(t) =75+C_1e^{kt} onumber T(0) =75+C_1e^{k(0)} onumber 350 =75+C_1 onumber C_1 =275. onumber end{align} ]
Therefore the solution to the initial-value problem is
[T(t)=75+275e^{kt}. onumber]
To determine the value of (k), we need to use the fact that after (5) minutes the temperature of the pizza is (340°F). Therefore (T(5)=340.) Substituting this information into the solution to the initial-value problem, we have
[T(t)=75+275e^{kt} onumber]
[T(5)=340=75+275e^{5k} onumber]
[265=275e^{5k} onumber]
[e^{5k}=dfrac{53}{55} onumber]
[ln e^{5k}=ln(dfrac{53}{55}) onumber]
[5k=ln(dfrac{53}{55}) onumber]
[k=dfrac{1}{5}ln(dfrac{53}{55})≈−0.007408. onumber]
So now we have (T(t)=75+275e^{−0.007048t}.) When is the temperature (300°F)? Solving for t, we find
[T(t)=75+275e^{−0.007048t} onumber]
[300=75+275e^{−0.007048t} onumber]
[225=275e^{−0.007048t} onumber]
[e^{−0.007048t}=dfrac{9}{11} onumber]
[ln e^{−0.007048t}=lndfrac{9}{11} onumber]
[−0.007048t=lndfrac{9}{11} onumber]
[t=−dfrac{1}{0.007048}lndfrac{9}{11}≈28.5. onumber]
Therefore we need to wait an additional (23.5) minutes (after the temperature of the pizza reached (340°F)). That should be just enough time to finish this calculation.
Exercise (PageIndex{4})
A cake is removed from the oven after baking thoroughly, and the temperature of the oven is (450°F). The temperature of the kitchen is (70°F), and after (10) minutes the temperature of the cake is (430°F).
1. Write the appropriate initial-value problem to describe this situation.
2. Solve the initial-value problem for (T(t)).
3. How long will it take until the temperature of the cake is within (5°F) of room temperature?
Hint
Determine the values of (T_s) and (T_0) then use Equation ef{newton}.
Initial-value problem [dfrac{dT}{dt}=k(T−70),T(0)=450 onumber]
[T(t)=70+380e^{kt} onumber]
Approximately (114) minutes.
## Key Concepts
• A separable differential equation is any equation that can be written in the form (y'=f(x)g(y).)
• The method of separation of variables is used to find the general solution to a separable differential equation.
## Key Equations
• Separable differential equation
(y′=f(x)g(y))
• Solution concentration
(dfrac{du}{dt}=INFLOW RATE−OUTFLOW RATE)
• Newton’s law of cooling
(dfrac{dT}{dt}=k(T−T_s))
## Glossary
autonomous differential equation
an equation in which the right-hand side is a function of (y) alone
separable differential equation
any equation that can be written in the form (y'=f(x)g(y))
separation of variables
a method used to solve a separable differential equation
## Separable Space
A topological space $left( ight)$ is said to be a separable space if it has a countable dense subset in $X$ i.e., $A subseteq X$, $overline A = X$, or $A cup U e phi$, where $U$ is an open set.
In other words, a space $X$ is said to be a separable space if there is a subset $A$ of $X$ such that (1) $A$ is countable (2) $overline A = X$ ($A$ is dense in$X$).
Let $X = left < <1,2,3,4,5> ight>$ be a non-empty set and $au = left< ,left < <3,4> ight>,left < <2,3> ight>,left < <2,3,4> ight>> ight>$ be a topology defined on $X$. Suppose a subset $A = left < <1,3,5> ight> subseteq X$. The closed sets are $X,phi ,left < <1,2,4,5> ight>,left < <1,2,5> ight>,left < <1,4,5> ight>,left < <1,5> ight>$. Now we have $overline A = X$. Since $A$ is finite and dense in $X$, then $X$ is a separable space.
Consider that the set of rational numbers $mathbb$ a subset of $mathbb$ (with usual topology), then the only closed set containing $mathbb$ is $mathbb$, which shows that $overline = mathbb$. Since $mathbb$ is dense in $mathbb$, then $mathbb$ is also separable in $mathbb$. However, the set of irrational numbers is dense in $mathbb$ but not countable.
Theorems
• Every second countable space is a separable space.
• Every separable space is not a second countable space.
• Every separable metric space is a second countable space.
• The continuous image of a separable space is separable.
Many problems involving separable differential equations are word problems. These problems require the additional step of translating a statement into a differential equation.
When reading a sentence that relates a function to one of its derivatives, it's important to extract the correct meaning to give rise to a differential equation. The key is to search for phrases like "rate of change" because they indicate that there is a derivative involved. In fact, the term "derivative" and phrase "rate of change" are synonymous, so these should be kept in mind when constructing a differential equation that models a word problem.
• The rate of growth of a bacteria population is directly proportional to the current bacteria population.
• The rate of change of an object's temperature is directly proportional to the difference of the object's current temperature and the temperature of the surrounding environment.
• The force felt by an object in free-fall is the difference of its weight and drag force, where the drag force is proportional to the object's current velocity.
Once a word problem has been written as a differential equation, it can be solved using the techniques of the previous section.
The exact same approach can be used on other problems as well.
The rate of growth of a beanstalk is proportional to the square root of its current height. If the height is 100 feet initially and it grows to 400 feet after 5 days, how tall will it be after 20 more days?
Let the height of the beanstalk (in feet) be denoted by h h h , and the time (in days) by t t t . Then The rate of growth of a beanstalk ⏟ d h d t is ⏟ = proportional to ⏟ k the square root of its current height. ⏟ h underbrace < ext> _< huge< frac< dh > < dt >> > underbrace < ext< is >> _ < huge< = >> underbrace < ext< proportional to >> _ < huge< k >> underbrace < ext < the square root of its current height.>> _ > > d t d h
The rate of growth of a beanstalk =
is k
proportional to h
the square root of its current height.
Now, d h h = k d t ⟹ ∫ d h h = ∫ k d t ⟹ 2 h = k t + C . frac< dh > < sqrt< h >> = k dt implies int frac< dh > < sqrt< h >> = int k dt implies 2 sqrt < h >= kt + C. h
d h = k d t ⟹ ∫ h
d h = ∫ k d t ⟹ 2 h
= k t + C .
Since h = 100 h = 100 h = 1 0 0 at t = 0 t = 0 t = 0 and h = 400 h = 400 h = 4 0 0 at t = 5 t = 5 t = 5 , we have
< 2 100 = k × 0 + C 2 400 = k × 5 + C . egin 2 sqrt < 100 >& = k imes 0 + C 2 sqrt < 400 >& = k imes 5 + C. end < 2 1 0 0
2 4 0 0
= k × 0 + C = k × 5 + C .
From the first equation we have C = 20 C = 20 C = 2 0 . Substituting this into the second equation gives 40 = 5 k + 20 40 = 5k + 20 4 0 = 5 k + 2 0 , which implies k = 4 k = 4 k = 4 .
Thus, we now have the particular solution for the differential equation: 2 h = 4 t + 20 ⟹ h = ( 2 t + 10 ) 2 . 2 sqrt < h >= 4t + 20 implies h = (2t + 10) ^ < 2 >. 2 h
= 4 t + 2 0 ⟹ h = ( 2 t + 1 0 ) 2 .
Now finding the height of the beanstalk after 20 more days is very easy. We only need to find the height of the beanstalk at t = 5 + 20 = 25 days t = 5 + 20 = 25 ext < days>t = 5 + 2 0 = 2 5 days . Simply substitute t = 25 t = 25 t = 2 5 in the equation and obtain: h = ( 2 × 25 + 10 ) 2 = 6 0 2 = 3600 (feet) . h = ( 2 imes 25 + 10 ) ^ < 2 >= 60 ^ < 2 >= 3600 ext < (feet)>. h = ( 2 × 2 5 + 1 0 ) 2 = 6 0 2 = 3 6 0 0 (feet) .
The population of a country grows at a rate proportional to the size of the population. If the population doubles in 50 years, in how many years (from now) will it triple?
## Contents
Suppose a differential equation can be written in the form
which we can write more simply by letting y = f ( x ) :
As long as h(y) ≠ 0, we can rearrange terms to obtain:
so that the two variables x and y have been separated. dx (and dy) can be viewed, at a simple level, as just a convenient notation, which provides a handy mnemonic aid for assisting with manipulations. A formal definition of dx as a differential (infinitesimal) is somewhat advanced.
### Alternative notation Edit
Those who dislike Leibniz's notation may prefer to write this as
but that fails to make it quite as obvious why this is called "separation of variables". Integrating both sides of the equation with respect to x , we have
If one can evaluate the two integrals, one can find a solution to the differential equation. Observe that this process effectively allows us to treat the derivative d y d x >> as a fraction which can be separated. This allows us to solve separable differential equations more conveniently, as demonstrated in the example below.
(Note that we do not need to use two constants of integration, in equation (A1) as in
∫ 1 h ( y ) d y + C 1 = ∫ g ( x ) d x + C 2 , >,dy+C_<1>=int g(x),dx+C_<2>,>
### Example Edit
Population growth is often modeled by the differential equation
Separation of variables may be used to solve this differential equation.
To evaluate the integral on the left side, we simplify the fraction
and then, we decompose the fraction into partial fractions
Therefore, the solution to the logistic equation is
### Generalization of separable ODEs to the nth order Edit
Much like one can speak of a separable first-order ODE, one can speak of a separable second-order, third-order or nth-order ODE. Consider the separable first-order ODE:
The derivative can alternatively be written the following way to underscore that it is an operator working on the unknown function, y:
Thus, when one separates variables for first-order equations, one in fact moves the dx denominator of the operator to the side with the x variable, and the d(y) is left on the side with the y variable. The second-derivative operator, by analogy, breaks down as follows:
The third-, fourth- and nth-derivative operators break down in the same way. Thus, much like a first-order separable ODE is reducible to the form
a separable second-order ODE is reducible to the form
and an nth-order separable ODE is reducible to
### Example Edit
Consider the simple nonlinear second-order differential equation:
The method of separation of variables is also used to solve a wide range of linear partial differential equations with boundary and initial conditions, such as the heat equation, wave equation, Laplace equation, Helmholtz equation and biharmonic equation.
The analytical method of separation of variables for solving partial differential equations has also been generalized into a computational method of decomposition in invariant structures that can be used to solve systems of partial differential equations. [1]
### Example: homogeneous case Edit
Consider the one-dimensional heat equation. The equation is
The variable u denotes temperature. The boundary condition is homogeneous, that is
Let us attempt to find a solution which is not identically zero satisfying the boundary conditions but with the following property: u is a product in which the dependence of u on x, t is separated, that is:
Substituting u back into equation (1) and using the product rule,
Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value −λ. Thus:
λ here is the eigenvalue for both differential operators, and T(t) and X(x) are corresponding eigenfunctions.
We will now show that solutions for X(x) for values of λ ≤ 0 cannot occur:
Suppose that λ < 0. Then there exist real numbers B, C such that
and therefore B = 0 = C which implies u is identically 0.
Suppose that λ = 0. Then there exist real numbers B, C such that
From (7) we conclude in the same manner as in 1 that u is identically 0.
Therefore, it must be the case that λ > 0. Then there exist real numbers A, B, C such that
From (7) we get C = 0 and that for some positive integer n,
This solves the heat equation in the special case that the dependence of u has the special form of (3).
In general, the sum of solutions to (1) which satisfy the boundary conditions (2) also satisfies (1) and (3). Hence a complete solution can be given as
where Dn are coefficients determined by initial condition.
Given the initial condition
This is the sine series expansion of f(x). Multiplying both sides with sin n π x L < extstyle sin >> and integrating over [0, L] result in
### Example: nonhomogeneous case Edit
Suppose the equation is nonhomogeneous,
with the boundary condition the same as (2).
Expand h(x,t), u(x,t) and f(x) into
where hn(t) and bn can be calculated by integration, while un(t) is to be determined.
Substitute (9) and (10) back to (8) and considering the orthogonality of sine functions we get
which are a sequence of linear differential equations that can be readily solved with, for instance, Laplace transform, or Integrating factor. Finally, we can get
If the boundary condition is nonhomogeneous, then the expansion of (9) and (10) is no longer valid. One has to find a function v that satisfies the boundary condition only, and subtract it from u. The function u-v then satisfies homogeneous boundary condition, and can be solved with the above method.
### Example: mixed derivatives Edit
For some equations involving mixed derivatives, the equation does not separate as easily as the heat equation did in the first example above, but nonetheless separation of variables may still be applied. Consider the two-dimensional biharmonic equation
Proceeding in the usual manner, we look for solutions of the form
and we obtain the equation
Writing this equation in the form
E ( x ) + F ( x ) G ( y ) + H ( y ) = 0 ,
we see that the derivative with respect to x and y eliminates the first and last terms, so that
### Curvilinear coordinates Edit
In orthogonal curvilinear coordinates, separation of variables can still be used, but in some details different from that in Cartesian coordinates. For instance, regularity or periodic condition may determine the eigenvalues in place of boundary conditions. See spherical harmonics for example.
Partial differential equations
For many PDEs, such as the wave equation, Helmholtz equation and Schrodinger equation, the applicability of separation of variables is a result of the spectral theorem. In some cases, separation of variables may not be possible. Separation of variables may be possible in some coordinate systems but not others, [2] and which coordinate systems allow for separation depends on the symmetry properties of the equation. [3] Below is an outline of an argument demonstrating the applicability of the method to certain linear equations, although the precise method may differ in individual cases (for instance in the biharmonic equation above).
Hence, the spectral theorem ensures that the separation of variables will (when it is possible) find all the solutions.
The matrix form of the separation of variables is the Kronecker sum.
As an example we consider the 2D discrete Laplacian on a regular grid:
Some mathematical programs are able to do separation of variables: Xcas [5] among others.
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## 3.4: Separable Equations - Mathematics
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We solve separable differential equations and initial value problems.
### Differential Equations
A differential equation is an equation that involves one or more derivatives of an unknown function. Solving a differential equation entails determining the unknown function.
An initial value problem is a differential equation along with other information about the solution, usually the value of the function at a point. The purpose of the initial value is to determine one specific solution of the differential equation, in the event that there was more than one solution.
The general solution of the differential equation is
The solution to the initial value problem is
### Separable Differential Equations
A separable differential equation is a differential equation that can be put in the form . To solve such an equation, we separate the variables by moving the ’s to one side and the ’s to the other, then integrate both sides with respect to and solve for . In general, the process goes as follows: Let for convenience and we have
Integrating both sides with respect to using a the substitution , we have and so Now we solve for algebraically to get the final answer. To simplify the computations, we will use instead of and then make a slight abuse of notation to get to the same end result. From the beginning: again using for convenience. Now we write this last equation in differential form and put an integral sign on both sides(!): yielding as before.
This book presents a comprehensive introduction to the theory of separable algebras over commutative rings. After a thorough introduction to the general theory, the fundamental roles played by separable algebras are explored. For example, Azumaya algebras, the henselization of local rings, and Galois theory are rigorously introduced and treated. Interwoven throughout these applications is the important notion of étale algebras. Essential connections are drawn between the theory of separable algebras and Morita theory, the theory of faithfully flat descent, cohomology, derivations, differentials, reflexive lattices, maximal orders, and class groups.
The text is accessible to graduate students who have finished a first course in algebra, and it includes necessary foundational material, useful exercises, and many nontrivial examples.
Graduate students and researchers interested in algebra.
The book is neatly arranged. It can be used as a textbook for self-study and as a reference text for most of the topics related to separability. It will be a valuable resource for students and researchers and has the potential to be a standard reference on separable algebras for many years.
-- Wolfgang Rump, Mathematical Reviews
The thorough and comprehensive treatment of separable, Azumaya, and tale algebras, Hensel rings, the Galois theory of rings, and Galois cohomology of rings makes the book under review an indispensable reference for the graduate student interested in these topics. As an added bonus, the book comes with a rich, 155 item, bibliography, well-chosen examples, calculations, and sets of exercises in each chapter, which makes this book an excellent textbook for self-study or for a topics course on separable algebras.
-- Felipe Zaldivar, MAA Reviews
## Numerical solution of non-separable elliptic equations by the iterative application of FFT methods
A method for the numerical solution of non-separable (self-adjoint) elliptic equations is described in which the basic approach is the iterative application of direct methods. Such equations may be transformed into Helmholtz form and this Helmholtz problem is solved by the iterative application of FFT methods. An equation which is ‘near’ (in some sense) to the Helmholtz equation is appropriately chosen from the general class of equations soluble directly by FFT methods (see, for example, Le Bail, 1972) and the iteration (of block-Jacobi form) consists of corrections to the relevant Fourier harmonic amplitudes of the solution of this ‘nearby’ equation. It is also shown that the method is equivalent to a D' Yakonov-Gunn iteration [D' Yakonov (1961), Gunn(1964)] with a particular choice of iteration parameter and it is well known that, for self-adjoint problems with smooth coefficients, this form of iteration has a convergence rate which is essentially independent of grid-size.
In the Concus and Golub (1973) method for solving non-separable elliptic equations such equations are transformed to Helmholtz form and Poisson's equation is employed as the ‘nearby’ equation. At each iterative stage this equation is solved by the Bunemann (cyclic reduction) algorithm. Their method also uses shifted iterations to improve convergence rates and we adopt a similar approach in the present study. However, our choice of nearby equation allows the use of a form of variable shift (in addition to the constant shift used by Concus et al.,) and it will be demonstrated that the use of such a variable shift can considerably improve rates of convergence in some examples.
Numerical results are presented for illustrative examples in the unit square with Dirichlet boundary conditions and the general computational behaviour of the method was found to be in very good agreement with that predicted by a theoretical analysis. The general iterative approach may be extended to more general linear elliptic equations.
Separation of variables works on regions that are rectangular in some particular coordinate system. The underlying operator must be separable in that coordinate system. For the Laplacian, that generally means an orthogonal coordinate system such as spherical coordinates, cylindrical coordinates, elliptic coordinates, etc.. A rectangular region in spherical coordinates can be a sphere, a spherical thick shell, a wedge, etc.. There are a couple of dozen such coordinate systems. You can add various potentials, provided their dependence is also separable (usually meaning that the it depends on only one of the coordinate variables.)
So the number of configurations where you can separate variables is limited, but it includes an important class of problems. Once you are able to separate variables, the result is an ODE in each coordinate on an interval in the corresponding variable, which is why the region where you're solving needs to be a rectangular box in the chosen coordinate system. Sturm and Liouville did a nice job in the early 1800's of characterizing the ODEs coming out of separation of variables eigenvalue problems. These are the Sturm-Liouville eigenfunction equations with eigenvalue $lambda$ on an interval $[a,b]$: $frac<1>left[fracleft(p(x)frac ight)+q(x) ight]=lambda f, cosalpha f(a)+sinalpha f'(a) = 0, coseta f(b)+sineta f'(b) = 0.$ The function $w$ determines a weight for the space $L^2_w(a,b)$ where the problem is properly posed as a selfadjoint one. The inner product on this space involves this weight function, and as given by $(f,g)_w=int_^f(t)overlinew(t)dt.$ The function $p$ generally comes out of a scale factor associated with the coordinate change and is positive on $(a,b)$ and $q$ is a potential. Infinite and semi-infinite regions may occur (i.e., where $a=-infty$ and/or $b=infty$.) If $p$ vanishes at an endpoint of $(a,b)$, then the problem is singular, and there may or may not be some type of endpoint condition at the singular endpoint.
The cases grow in complexity with singularities, but the basic conclusion is this: you'll probably never have to worry about things not working the way you expect in regard to completeness. :)
## Watch the video: Q153, Separable Differential Equation (July 2022).
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Home » How Many Times Does 2 Go Into 16? Update
# How Many Times Does 2 Go Into 16? Update
Let’s discuss the question: how many times does 2 go into 16. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## How many times can 3 go in 16?
The quotient (the number of times that 3 goes into 16) is 16/ 3 = 5 and the remainder is 16 % 3 = 1. 15.
## How many times does 2 fit 17?
This guess is important for what follows in polynomial division. We could base the guess on a rough estimate, how many times does 2 divide into 17, and the answer is 8.
### 16 TK: How Often . . . – Thường Bao Lâu Thì . . .
16 TK: How Often . . . – Thường Bao Lâu Thì . . .
16 TK: How Often . . . – Thường Bao Lâu Thì . . .
## How many times can 2 fit in 32?
32 divided by 2 is 16.
## How many times can 2 Go in 13?
Notice that 2 * 6 = 12, so we can fit 2 into 13 six times.
## What times what can equal 16?
1 x 16 = 16. 2 x 8 = 16. 4 x 4 = 16. 8 x 2 = 16.
## What can 4 go into?
Simple, 4 goes into 4, 8, 12, and 16.
## How do you solve 16 divided by 2?
Basic Math Examples
Place this digit in the quotient on top of the division symbol. Multiply the newest quotient digit (8) by the divisor 2 . Subtract 16 from 16 . The result of division of 16÷2 16 ÷ 2 is 8 .
## How many twos are there in 24?
Answer: 2 to the power of 24 is 16777216.
Let’s figure out the solution step by step.
## How many 2s are in 22?
There are eleven 2s in 22.
## How do you solve 16 divided by 4?
16 divided by 4 equals 4.
## How do you solve 15 divided by 2?
Using a calculator, if you typed in 15 divided by 2, you’d get 7.5. You could also express 15/2 as a mixed fraction: 7 1/2. If you look at the mixed fraction 7 1/2, you’ll see that the numerator is the same as the remainder (1), the denominator is our original divisor (2), and the whole number is our final answer (7).
## How do you write 12 divided by 2?
12 divided by 2 is 6. What is 12 divided by 2?
### pure imaginary numbers notes
pure imaginary numbers notes
pure imaginary numbers notes
## What can 16 be divided by?
For example, 16 can be divided evenly by 1, 2, 4, 8, and 16. So the numbers 1, 2, 4, 8, and 16 are called factors of 16.
## What can divide 17?
Explanation: Since 17 is a prime number, it has only 2 factors namely 1 and itself. These are the only 2 numbers which can divide perfectly into it without leaving a remainder.
## Can 13 be divided?
When we list them out like this it’s easy to see that the numbers which 13 is divisible by are 1 and 13. What is this? You might be interested to know that all of the divisor numbers listed above are also known as the Factors of 13. Not only that, but the numbers can also be called the divisors of 13.
## What two numbers make 16?
We know 2 and 8 are factors of 16 because 2 x 8 = 16. 4 is a factor of 16 because 4 x 4 = 16.
## What are the factors of 16?
The factors of 16 are 1, 2, 4, 8 and 16. Similarly, the negative factors of 16 are -1, -2, -4, -8 and -16.
## Which of the following are factor pairs for 16?
Therefore, the factor pairs of 16 are (1, 16), (2, 8), and (4, 4).
## What can go into 8?
What are the factors of 8? The factors of 8 are 1, 2, 4 and 8.
## How many fours is 60?
Answer. therefore 15 times 4 goes in 60.
## What can go into 41?
The number 41 is a prime number. Being a prime number, 41 has just two factors, 1 and 41.
## How do you solve 16 divided by 3?
16 divided by 3 is 5 with a remainder of 1 (16 / 3 = 5 R.
### MTH 9 Unit 8 Day 3 Notes
MTH 9 Unit 8 Day 3 Notes
MTH 9 Unit 8 Day 3 Notes
## How do you solve 16 divided by 12?
16 divided by 12 is equal to 1 with a remainder of 4, or when written as a decimal it is 1 1/3: 16 / 12 = 1 R. 4. 16 / 12 = 1 1/3.
## What is the answer for 2 divided by 2?
we are dividing the number 2 or 2 apples or 2 oranges into two equal parts . Each part consist of number 1 or 1 apple or 1 orange . so dividing 2 by 2 , or 2 apples by 2 , or 2 oranges by 2 is number 1 , or 1 apple , or 1 orange .
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# Appendix A Essay
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• on October 5, 2011
• Category: English
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Below is an essay on "Appendix A" from Anti Essays, your source for research papers, essays, and term paper examples.
A60
Appendix A
Review of Fundamental Concepts of Algebra
A.6
Linear Inequalities in One Variable
Introduction
Simple inequalities were discussed in Appendix A.1. There, you used the inequality symbols , and ≥ to compare two numbers and to denote subsets of real numbers. For instance, the simple inequality x ≥ 3 denotes all real numbers x that are greater than or equal to 3. Now, you will expand your work with inequalities to include more involved statements such as 5x and 3 ≤ 6x 1 < 3. 7 < 3x 9
What you should learn
• Represent solutions of linear inequalities in one variable. • Solve linear inequalities in one variable. • Solve inequalities involving absolute values. • Use inequalities to model and solve real-life problems.
Why you should learn it
Inequalities can be used to model and solve real-life problems. For instance, in Exercise 101 on page A68, you will use a linear inequality to analyze the average salary for elementary school teachers.
As with an equation, you solve an inequality in the variable x by finding all values of x for which the inequality is true. Such values are solutions and are said to satisfy the inequality. The set of all real numbers that are solutions of an inequality is the solution set of the inequality. For instance, the solution set of x 1 < 4
is all real numbers that are less than 3. The set of all points on the real number line that represent the solution set is the graph of the inequality. Graphs of many types of inequalities consist of intervals on the real number line. See Appendix A.1 to review the nine basic types of intervals on the real number line. Note that each type of interval can be classified as bounded or unbounded.
Example 1
Intervals and Inequalities
Write an inequality to represent each interval, and state whether the interval is bounded or unbounded. a. b. d. a. b. d. 3, 5 3, , 3, 5 corresponds to 3, , corresponds to corresponds to 3 < x ≤ 5. 3 < x.
< x <
c. 0, 2...
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• on October 5, 2011
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You are HERE >> Mathematics > Grade 5
Divisibility Rules Worksheet: Reducing Fractions
By Lisa Williams
March 26, 2001
Write true or false.
(1) 3/12 = 1/4 True
(2) 4/10 = 1/2 False
(3) 2/3 = 4/6 True
Reduce these fractions to lowest terms by dividing both terms by the greatest common factor. Greatest common factor means the largest number that is a factor for both numerator and denominator. In 'Mississippiese', it means the biggest number that will go into both top and bottom of the fraction)
(4) 18/30 18 = 2 x 9 = 2 x 3x 3 30 = 2 x 15 = 2 x 3 x 5 Therefore 2 x 3 = greatest common factor 6 = " " " so 18/6 = 3 30/6 = 5 when reduced to lowest terms = 3/5 (5) 8/12 8 = 2 x 4 = 2 x 2 x 2 12 = 2 x 6 = 2 x 2 x 3 Therefore 2 x 2 = greatest common factor 4 = " " " 8/4 = 2 12/4 = 3 reduced to lowest terms = 2/3 (6) 6/24 6 =2 x 3 24 = 2 x 12 = 2 x 2 x 6 = 2 x 2 x 2 x 3 Therefore 2 x 3 = greatest common factor 6 = " " " 6/6 = 1 24/6 = 4 reduced fraction = 1/4 (7) 5/100 5 = 5 100 = 5 x 20 Therefore 5 = greatest common factor so 5/5 = 1 100/5 = 20 reduced fraction = 1/20 (8) 3/9 3 = 3 9 = 3 x 3 Therefore 3 = greatest common factor so 3/3 = 1 9/3 = 3 reduced terms = 1/3 (9) 12/20 12 = 2 x 6 = 2 x 2 x 3 20 = 2 x 10 = 2 x 2 x 5 Therefore 2 x 2 = greatest common factor 4 = " " " so 12/4 = 3 20/4 = 5 lowest possible term = 3/5 (10) 12/14 12 = 2 x 6 14 = 2 x 7 Therefore 2 = greatest common factor so 12/2 = 6 14/2 = 7 reduced fraction = 6/7 (11) 30/55 30 = 2 x 15 = 2 x 3 x 5 55 = 5 x 11 Therefore 5 = greatest common factor 30/5 = 6 55/5 = 11 reduced fraction = 6/11 (12) 15/20 15 = 3 x 5 20 = 2 x 10 = 2 x 2 x 5 Therefore 5 = greatest common factor 15/5 = 3 20/5 = 4 reduced fraction term = 3/4 (13) 12/16 12 = 2 x 6 = 2 x 2 x 3 16 = 2 x 8 = 2 x 2 x 4 Therefore 2 x 2 = greatest common factor 4 = " " " 12/4 = 3 16/4 = 4 reduced fraction = 3/4 (14) 18/27 18 = 2 x 9 = 2 x 3 x 3 27 = 3 x 9 = 3 x 3 x 3 Therefore 3 x 3 = greatest common factor 9 = " " " 18/9 = 2 27/9 = 3 reduced fraction = 2/3 (15) 54/63 54 = 2 x 27 = 2 x 3 x 9 = 2 x 3 x 3 x 3 63 = 3 x 21 = 3 x 3 x 7 Therefore 3 x 3 = greatest common factor 9 = " " " 54/9 = 6 63/9 = 7 lowest terms = 6/7 (16) 15/21 15 = 3 x 5 21 = 3 x 7 Therefore 3 = greatest common factor 15/3 = 5 21/3 = 7 lowest terms = 5/7 (17) 6/30 6 = 2 x 3 30 = 2 x 15 = 2 x 3 x 5 Therefore 2 x 3 = greatest common factor 6 = " " " 6/6 = 1 30/6 = 5 reduced fraction = 1/5 (18) 2/18 2 = 2 18 = 2 x 9 Therefore 2 = greatest common factor so 2/2 = 1 18/2 = 9 lowest terms = 1/9 (19) 4/22 4 = 2 x 2 22 = 2 x 11 Therefore greatest common factor = 2 4/2 = 2 22/2 = 11 reduced fraction = 2/11 (20)12/132 12 = 2 x 6 = 2 x 2 x 3 132 = 2 x 66 = 2 x 2 x 33 = 2 x 2 x 3 x 11 Therefore 2 x 2 x 3 = greatest common factor 12 = " " " 12/12 = 1 132/12 = 11 therefore 12/132 reduced to its lowest possible terms is 1/11
Submitted by: Copyright© Lisa Williams I'm a homeschooling mom of 3 elementary school age children. I run a clowning business from my home (Jingles the clown), am pianist for a local church, teach a Sunday School class, teach piano, tutor basic math see Math tutoring and algebra online and I enjoy making people smile.
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# NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-9/
Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 9 Chapter Name Some Applications of Trigonometry Exercise Ex 9.1 Number of Questions Solved 16 Category NCERT Solutions
## NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1
Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure).
Solution:
Given: length of the rope (AC) = 20 m, ∠ACB = 30°
Let height of the pole (AB) = h metres
Hence, height of the pole = 10 m
Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Let DB is a tree and AD is the broken part of it which touches the ground at C.
Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Let l1 is the length of slide for children below the age of 5 years and l2 is the length of the slide for elder children
Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Solution:
Let h be the height of the tower
In ∆ABC,
Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution:
Let the height of the pedestal AB = h m
Given: height of the statue = 1.6 m, ∠ACB = 45° and ∠DCB = 60°
Question 9.
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
Given: height of the tower AB = 50 m,
∠ACB = 60°, ∠DBC = 30°
Let the height of the building CD = x m
In ∆ABC,
Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.
Solution:
Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.
Solution:
Let the height of the tower AB = h m and BC be the width of the canal.
Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let height of the tower AB = (h + 7) m
Given: CD = 7 m (height of the building),
Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:
Given: height of the lighthouse = 75 m
Let C and D are the positions of two ships.
Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
Solution:
Let the first position of the balloon is A and after sometime it will reach to the point D.
The vertical height ED = AB = (88.2 – 1.2) m = 87 m.
Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let the height of the tower AB = h m |
# Applying The Basics!
33 teachers like this lesson
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## Objective
Students will fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction.
#### Big Idea
Using base-ten materials and place-value skills, students gain a better understanding of how to add and subtract numbers.
## Opener
10 minutes
Material: Base Ten Cut OutsPlace Value Model
The eventual goal of this standard is fluency. I explain to my students that this will not happen all at once; I hope to gradually move them towards having their own procedures and strategies by which they can fluently add and subtract. I invite students to the carpet to discuss place value. I want to make sure my students have a firm understanding of the values of digits in two-digit numbers before I move them into adding and subtracting three-digit numbers.
To do this, I write 56, a two-digit number, on the board. I ask students to tell me what digits are in the ones and tens place. Then I ask a volunteer to come up and illustrate how we would represent 56 using base-tens. The representation should show 5 ten-rods and 6 ones. I ask if there is another way we can represent the value of each digit. I give students a hint; I tell them we have used this chart before when we needed to understand the value of two-digit numbers. Students are quick to say a place value chart. I enter the numbers into a blank place value chart on the board by placing the 5 under the tens place and the 6 under the ones place. Just to make sure students understand the importance of digit value, I ask them to tell me if I have placed the numbers correctly. Some students explain that I placed each number correctly because there were 5 tens and 6 ones in the base-ten model. I continue to work with students using a couple more two-digit numbers just to make sure they fully understand the value of digits before I move forward. MP6- Attending to precision.
## Informative
10 minutes
Material: Students Work
In this portion of the lesson I want my students use what they know about place value to correctly identify which digits are in the ones, tens, and hundreds place. This will help them correctly align numbers. First, I demonstrate what they should be thinking and saying. For example: 64 + 5 is written in sentence form. In order for you all to add or subtract the two numbers using the standard algorithms, you need to correctly align the numbers. To start, the largest number should always go on top! Which number is the largest? (64) I ask them to explain why 64 is larger than 5. Some students notice that 64 is a two-digit number, and 5 is only a one-digit number. You are correct, but based on the value of the digits, 64 is larger because it has 6 tens and 4 ones, and 5 only has 5 ones. To make sure students fully understand I use my magnetic base-ten blocks to represent 64 and 5. What is the difference between the two numbers? The visual representations confirm students thinking. MP3-Construct viable arguments & critiquing the reasoning of others.
Example
After I have fully demonstrated what they will be doing and how they should be thinking, I ask students to move to their assigned partner. You all are going to work on identifying digits to help practice addition and subtraction through 100. As students are working, I circle the room to see if students can add and subtract accurately, and to see if they can explain their strategies by asking questions. Why did you align your numbers that way? Explain. Can you tell me the correct position of each number? Can you explain the reasoning behind the strategy you used? Can you think of another method that can help you correctly identify the value of digits? MP1- Make sense of problems and persevering in solving them.
Student work samples show how students used place value in order to solve addition and subtraction problems.
## Putting it to the Test!
15 minutes
Material: Spinner Activity
In this portion of the lesson, I want students to have additional work time. I want them to know there are different ways of thinking about properties of operations to add and subtract. By exposing students to different methods and viewpoints, students can pursue their own purpose. Therefore, many of my students who were struggling are coming up with different approaches, discussing various strategies, and learning from one another.
To see just what students are capable of doing on their own, I ask them to move into their assigned groups and provide them with base-ten materials. I explain that they will take turns spinning two numbers to either add or subtract. However, I ask them to first represent each number with base-ten blocks and to identify identify the value of each digit before aligning the numbers. This assures that they are adding hundreds with hundreds, tens with tens, and ones with ones. As students are working, I circle the room to check for understanding. For instance, I may ask: what digits are in the ones, tens, and hundreds places? Can you explain? Can you represent the number using base-tens? Why is it important to align the numbers according to their place value? How does place value help you solve addition and subtraction problems? Some students discover that place value helps them group digits according to the value. Other students can explain the value of each digit fluently. For example, one girl in my class was able to classify digits by their value, and another used her notepad to enter the digits into the correct places. As students in her group modeled numbers and aligned the digits, she explained why they are wrong or right. I continue monitoring students until their given time is up.
## Wrapping it Up!
15 minutes
Discussion Circle:
I want to end this lesson with students giving their own account to what was learned. I ask students to join me on the carpet, and I take a seat in the middle of the circle. I ask student volunteers to begin sharing. As students are sharing, I make sure to ask questions to check for understanding. For instance: What number is in the hundreds, tens, and ones places. Can you represent the number using base tens? Can you explain why your illustrations represent your number correctly? Students seem to know what materials to use to represent hundreds, tens, and ones. They also know the value of each digit in isolation. They use this information to help them correctly add and subtract two- and three-digit numbers. I also ask them about moments in their learning where they struggled. This helps me to collect meaningful data for re-teaching strategies. Some students recall just knowing where the numbers should go, but not actually knowing how much each digit was worth. I ask what helps them understand the value. They all agreed that allowing them time to use base-ten materials and place value charts helps them make sense of the worth of each number.
To end I ask students to complete this activity by writing their experiences in their math journals. |
Write an inequality that represents the graphics
Even though the topic itself is beyond the scope of this text, one technique used in linear programming is well within your reach-the graphing of systems of linear inequalities-and we will discuss it here.
In general, if two lines have slopes and m2: If the point chosen is not in the solution set, then the other halfplane is the solution set. Solve this system by the substitution method and compare your solution with that obtained in this section.
Procedures To sketch the graph of a linear equation find ordered pairs of numbers that are solutions to the equation. We indicate this solution set with a screen to the left of the dashed line.
This will result in the same line. If we denote any other point on the line as P x, y see Figure 7. Locate these points on the Cartesian coordinate system and connect them with a line. Example 2 Solve by addition: Usually, equations are written so the first term is positive.
Solution We designate 3, 5 as x2, y2 and -4, 2 as x1, y1. Since we are dealing with equations that graph as straight lines, we can examine these possibilities by observing graphs.
Second we know that if we add the same or equal quantities to both sides of an equation, the results are still equal. Next check a point not on the line. Solve the system by substitution. We will now study methods of solving systems of equations consisting of two equations and two variables.
Equations in two unknowns that are of higher degree give graphs that are curves of different kinds. The symbols introduced in this chapter appear on the inside front covers. Solution Step 1 Our purpose is to add the two equations and eliminate one of the unknowns so that we can solve the resulting equation in one unknown.
The slope indicates that the changes in x is 4, so from the point 0,-2 we move four units in the positive direction parallel to the x-axis.
Please read the " Terms of Use ". Therefore, draw a solid line to show that it is part of the graph. That is, If you want to impress your friends, you can write where the Greek letter Note that the change in x is 3 and the change in y is 2.
Section dealt with solving literal equations. Why do we need to check only one point? Show the linear inequality and its graph. The change in x is 1 and the change in y is 3. Draw a straight line through those points that represent the graph of this equation.
Step 5 Check the solution in both equations. Study them closely and mentally answer the questions that follow. However, is undefined, so that a vertical line does not have a slope.Write and graph inequalities. Write and Graph Inequalities FAIR Jessica is trying to County State 2.
Which fair’s ride pass costs less?
You can write an inequality to represent a situation. Write Inequalities with Write an inequality for each sentence. You must be over 12 years old to ride the go-karts.
Words Variable Inequality. 2 Solving Linear Inequalities SEE the Big Idea Writing and Graphing Inequalities Write an inequality that represents the graph.
–8 –7 –6 –5 –4 –3 –2 –1 YOU MUST BE THIS TALL TO RIDE Representing Linear Inequalities Words Algebra Graph x is less than 2 x. Graph inequalities or systems of inequalities with our free step-by-step math inequality solver Represent the Cartesian coordinate system and identify the origin and axes.
Write the equation of a line in slope-intercept form. The inequality we solve can get as complex as the linear equations we solved. We will use all the same patterns to solve these inequalities as we did for solving.
Section Writing and Graphing Inequalities 57 Writing Linear Inequalities from Graphs Writing Inequalities from Graphs The graphs show the height restrictions h (in inches) for two rides at an amusement park.
Write an inequality that represents the height restriction of each ride. Solving Equations and Inequalities Graphically OBJECTIVES 1. represents the solution to the original equation.
Step by Step: Step 4 Mark the x values that make the inequality a true statement. Step 5 Write the solutions in set notation. Step by Step.
Write an inequality that represents the graphics
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## Lesson: Finding Probability Developing the Concept
Now that students have had an opportunity to find the probability of simple events by using a sample space or the fundamental counting principle, they can apply these strategies on their own.
Materials: 5 red, 3 green, and 2 yellow tiles; a paper bag; paper and pencil for each student
Prerequisite Skills and Background: Students should be able to apply the fundamental counting principle in a problem-solving situation.
On the board, write “5 red,” “3 green,” and “2 yellow” to indicate the color of the tiles in the bag.
• Say: Today we are going to look at finding the probability of events by drawing tiles from a bag. I have placed 5 red, 3 green, and 2 yellow tiles in the bag. If I reach into the bag and draw a tile without looking, what is the probability it will be green? Students will probably respond that the probability is .
• Ask: That's right. What is the probability of drawing a yellow tile? (, or )
What is the probability of drawing a red tile? (, or )
• Say: That's good. Now for a more difficult problem. What is the probability of reaching in and not drawing a yellow tile? Students may struggle with this, but will probably say that there are 8 tiles that are not yellow, so the probability will be , or . If they don't say this, you might ask them how many tiles there are and how many are not yellow.
• Ask: What is the probability of drawing a blue tile? (zero, since there are no blue tiles in the bag)
• Ask: What is the probability of reaching into the bag and not drawing a purple tile? (Since there are no purple tiles, the probability would be 1.)
• Say: Now we are going to investigate events that are a little more complex. Let's think about reaching into the bag and drawing out a tile, recording its color, replacing the tile in the bag, and then reaching in to get a second tile.
Do this so students can see exactly what you are saying.
• Say: The question I would like us to solve is, “What is the probability that both tiles drawn will be red?” How might we solve this problem? (Students may suggest using the fundamental counting principle.)
• Say: The first thing to do is to find out how many ways I could select two red tiles by reaching in and drawing a tile, recording it, replacing it, and then selecting a second tile.
Students will say that you can select any of 5 red tiles each time you draw, since you replace the tile.
• Say: Since I can draw any of 5 tiles on the first draw and any of 5 tiles on the second draw, these are independent events. So I can draw two red tiles 5 x 5, or 25, ways according to the fundamental counting principle.
• Ask: So how do we find the probability of these two independent events? If I reach into the bag and draw a tile, how many different outcomes are possible? (10) If I replace that tile and reach in and draw again, how many different outcomes are possible? (10)
• Say: Good, so there are 10 x 10, or 100, ways I could draw 2 tiles from the bag by replacing the tile after I draw the first time. Therefore, the probability of drawing two red tiles is , or .
• Say: Now we are going to change the problem slightly. Instead of replacing the tile, we keep it and draw a second tile. Now the two events are dependent events. How many different ways can these events occur?
Students should say that the first draw can occur 10 different ways and the second draw, 9 different ways; therefore, you can draw twice without replacing the tile in 9 x 10, or 90, ways.
• Ask: Keeping in mind that we are not replacing the tile, what is the probability of drawing 2 red tiles?
Give students time to think about it, and then ask for a volunteer to explain what he or she did. There would be 5 x 4, or 20, ways of selecting two red tiles out of 90 ways, if the first tile drawn is red, so the probability would be or .
• Ask: Again, if we don't replace the tile, what is the probability of drawing a green tile followed by a red tile?
Students will probably say that there are 3 ways to draw a green tile and 5 ways to draw a red tile, so there are 15 ways to draw a green tile followed by a red tile. Therefore, the probability would be , or .
• Say: You can see how important it is to know whether or not you replace the tile when drawing more than one tile.
Continue asking similar questions. If students seem to be catching on, you can go to selecting three tiles with or without replacing them.
Wrap-Up and Assessment Hints
“What if” questions are good ways to assess the depth of a student's understanding of probability. Most problems, like those you have done, have several factors that could affect the outcome. In those problems, the number of tiles of each color, the number of draws, and drawing with or without replacement are variables that you can alter to assess the depth of students' understanding. Drawing two tiles with replacement is an example of two independent events. The outcome of the first event does not affect the second event. Drawing tiles without replacing them is an example of dependent events. These are concepts your students should become familiar with. |
# What is the angle of an octagon?
1,080 degrees Fahrenheit
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### In light of this, how can one determine the angle of an octagon?
Because an octagon has eight sides, you can subtract two from eight to obtain the number six. To get the total number of degrees in an octagon, multiply six by 180 and divide the result by six. The result is 1,080 degrees. The measure of each inner angle in an octagon that is regular may be calculated by dividing 1,080 by eight. Each angle in a normal octagon is 135 degrees in length and width.
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135°
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### So, what is the cut angle for an octagon, just for reference?
Set the angle of your mitre saw to 22.5 degrees. This is the angle at which you must cut in order to create an octagon form. Using a 22.5-degree angle cutter, cut one end of each 12 scrap piece.
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### What is the area of each angle in an octagon?
The General Rule is as follows:
Shape Sides are the sum of the interior angles of a shape.
Octagon 8 1080°
Nonagon 9 1260°
..
Any polygon with a n (n-2) angle greater than 180°
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### When it comes to internal angles, what is the formula?
An internal angle is a polygonal angle that is placed inside the polygonal border. The formula S = (n – 2)*180 may be used to calculate the sum of all of the internal angles of a triangle. Additionally, assuming the polygon is regular, it is simple to compute the measure of each angle by dividing the total by the number of sides of the triangle.
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### Is it possible for an octagon to have six right angles?
As a result, an octagon may have up to six right angles. The sum of the angles is 1260′. If you have a polygon with n sides, the total of internal angles is 180n minus 360 degrees.
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### What is the formula for calculating the sum of internal angles?
It is (n – 2) * 180 that may be used to calculate the sum of the measures of the internal angles. This formula is used to obtain the measure of a single interior angle, which is (n – 2) * 180 divided by the number of sides of the triangle.
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### What is the sum of all the angles in a hexagonal shape?
The total of all internal angles is 720 degrees (as seen from above), and there are six interior angles, therefore we can use this information to calculate the measure of the interior angles. As a result, the inner angle of a regular hexagon is 120 degrees in length and width.
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### What is the number of degrees in a sphere?
In a spherical, there are effectively 360 degrees of 360 degrees. Edit: Maybe 180 degrees, now that I think about it! A circle would only need to turn in half to complete the turn. The term “degrees” refers to the measurement of two-dimensional objects.
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### What is the angle of rotation of this octagon that is regular? What is the length of an internal angle in degrees?
Coxeter diagram of an octagon (regular octagon) Internal angle (degrees) 135° Dual polygon Self-contained symmetry group Dihedral (D8), order 28 Internal angle (degrees) 135°
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### When you look at a Heptagon, what is the sum of the angle measures?
In order to calculate the measure of the inner angles, we must first remember that the sum of all the angles is 900 degrees (as seen from above), and that there are seven angles in all. As a result, the interior angle of a normal heptagon has a measure of about 128.57 degrees.
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### What is the total number of angles in a quadrilateral?
there are four angles
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### What is the best way to cut a 22.5 degree angle?
To use the handsaw on both sides of the mitre box, place it in the 22.5-degree slots on the fences on both sides of the box. Make long and steady handsaw strokes to cut the baseboard or moulding at an angle of 22.5 degrees using a circular saw. After you’ve measured the baseboard, mark the spot where you want to make the cut.
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### What is the best way to cut a 60-degree angle?
How to Cut a 60-Degree Angle with a Miter Saw (with Pictures) Placing the mitre saw on a solid workbench that is clear of dirt and unsecured items is recommended. Placing the wood stock on the mitre table so that it is flat against the back fence is the next step. Raise the blade by pulling it up. Remove the wood stock from the clamps and place it on the table such that your workpiece will abut against it.
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### What is the hexagonal angle?
A regular hexagon has sides that are all congruent with one another and angles that are all 120 degrees in length. 720 degrees is equal to the sum of the angles of a regular hexagon, which is 720 degrees in total.
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### What is the best way to cut landscaping timbers at an angle?
The actual angle of each cut, however, is the inverse of that of a straight cut — a straight cut has a 90-degree angle, therefore remove 60 degrees from that to get 30 degrees. Set your mitre saw or mitre box to 30 degrees and cut both ends of the timbers so that there is a long and a short side on both ends of the timbers, as shown.
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### What is the angle formed by each of the eight corners of an octagon?
Each of the octagon’s eight angles measures 135 degrees.
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### How do you calculate the angle at which a cut should be made?
The corner angle is calculated by dividing 360 by the number of sides, which is represented by the formula. Afterwards, divide the result by two to get the mitre angle. You would divide 360 degrees by five to obtain 72 degrees if you were building a five-sided item with all sides being equal. So each joint or corner makes a 72-degree angle with the next joint or corner. |
Short Blocks
# Maths Year 3/4 Spring Decimals and Money
Each unit has everything you need to teach a set of related skills and concepts.
First time using Hamilton Maths?
The PowerPoint incorporates step-by-step teaching, key questions, an in-depth mastery investigation, problem-solving and reasoning questions - in short, everything you need to get started.
All the other resources are there to support as-and-when required. Explore at your leisure - and remember that we are always here to answer your questions.
## Unit 1 x and ÷ with money and 1-place decimals (suggested as 3 days)
### Objectives
x and ÷ with Money and 1-Place Decimals
Unit 3: ID# 34444
Y3: Multiply and divide by 10 and 100 including with money.
Y4: Divide numbers by 10 to give tenths (one place decimal). Multiply 1-place decimals to give whole numbers; Multiply and divide by 10 and 100 (one place decimals).
National Curriculum
Y3: Num/PV (ii); Mult/Div (iii)
Y4: Dec/Fr (vii)
Y3 Hamilton Objectives
18. Multiply 2-digit numbers by 10 or 1-digit numbers by 100; divide multiples of 10 or 100 by 10 or 100. Understand the effect of x or ÷ by 10 or 100.
Y4 Hamilton Objectives
26. Know that one-place decimal numbers represent ones and tenths.
29. Find the effect of dividing a one- or two-digit number by 10 and 100, identifying the value of the digits in the answer as ones, tenths and hundredths.
### Planning and Activities
Day 1 Teaching
Model multiplying and dividing by 10 using a place value grid.
Further Teaching with Y4
Divide by 10 to give tenths using decimal place value grid.
Group Activities: T with Y4
Y3 -- Create 2-digit numbers and multiply and divide them by 10. Record these calculations. Some children challenge others to guess a starting number by giving the answer to × 10 questions.
Y4 -- Move digit cards on a place value grid to divide 2-digit numbers by 10. Write a generalisation.
Day 2 Teaching
Model multiplying and dividing by 10 and 100 using a place value grid.
Further Teaching with Y4
Multiply 1 place decimals by 10, moving digits across the decimal point. Use function machine to x10 and ÷10.
Group Activities: T with Y3
Use the in-depth problem-solving investigation ‘Testing 10s’ as today’s group activity.
Or, use these activities:
Y3 -- Play a divide and multiply card game.
Y4 -- Use a function machine or play bingo to practise multiplying and dividing numbers by 10.
Day 3 Teaching
Model multiplying and dividing by 10 and 100 (one-place decimals) using ITP ‘Moving digits’.
Further Teaching with Y3
Multiplying and dividing by 10 and 100 using money.
Group Activities: T with Y3
Y3 -- Multiply and divide shop prices by 10 and 100.
Y4 -- Find different routes through a maze (× 10, ÷ 10, × 100, ÷ 100) where the same number will come out exactly as it was entered.
### You Will Need
• 100s, 10s 1s place value grid (see resources)
• 10s, 1s, 0.1s place value grid (see resources)
• Cardboard box with two slots
• Sticky notes labelled ÷ 10 and × 10
• ÷ and × card game function cards and number cards (see resources)
• ‘Party bag toy shop’ sheet (see resources)
• Money place value grid (see resources)
• ‘Multiplying by 10 and 100’ sheets 1 and 2 (see resources)
• ITP: ‘Moving digits
• Real money
• ‘Multiplication and division maze’ (see resources)
### Short Mental Workouts
Day 1
Division facts for 10 times table
Day 2
Division facts for 5 times table
Day 3
£ and p notation
### Procedural Fluency
Day 1
Y3: Multiply and divide by 10.
Y4: Divide by 10 with decimals.
Day 2
Y3: Multiply and divide by 10 and 100.
Y4: Multiply and divide by 10 with decimals.
Day 3
Y3: Multiply and divide money by 10 and 100.
Y4: Multiply and divide by 10 and 100 with decimals.
### Mastery: Reasoning and Problem-Solving
Y3
• Describe in words what happens to a number when we multiply by 10. Now explain why it happens. You may draw a picture if it helps.
• Write the missing numbers:
(a) ☐ × 10 = 550
(b) 100 × 39 = ☐
(c) ☐ ÷ 10 = 60
(d) 17 × ☐ = 170
(e) 500 ÷ ☐ = 5
(f) ☐ × 10 = 990
• Write the result number in each chain:
5 × 100 ÷ 10 × 10 ÷ 100 = ?
300 ÷ 10 ÷ 10 × 10 × 10 = ?
40 × 10 ÷ 100 × 10 = ?
Invent your own chain where you finish with the starting number.
• Write the missing numbers
£3.40 × 10 = ☐ ☐ ÷ 10 = £5.50
100 × 9p = ☐ £7 ÷ ☐ = 7p
Y4
• Complete each sentence:
3.4 × 10 = ☐ 65 ÷ 10 = ☐
☐ ÷ 10 = 12.3 ☐ × 10 = 81
• Write the value of one tenth of each number:
57 84 6 13
• Use this fact 56 = 7 × 8 to find the answer to each calculation:
7 × 80 7 × 0.8 7 × 800 560 ÷ 8
In-depth Investigation: Testing 10s
Children multiply (Y3) or divide (Y4) numbers by 10, carry out a given procedure and look for patterns in their answer.
### Extra Support
Y3: Treasure or Trap
Multiplying and dividing by 10
Zeros Matter
Writing amounts of money in pounds and pence, including amounts with 0 in the 10ps or 1ps column
Y4: Moving Digits
Multiplying and dividing by 10 (whole number answers)
Digit Dance
Multiplying and dividing by 10 (including numbers and answers with one decimal place)
## Unit 2 Decimals and money on a line (suggested as 3 days)
### Objectives
Decimals and Money on a Line
Unit 4: ID# 34460
Y3: Placing and ordering 3-digit numbers and money on a number line; rounding them to the nearest 10.
Y4: Placing and ordering 1-place decimals on a line and rounding.
National Curriculum
Y3: PV (iii)
Y4: Fr (viii) (ix)
Y3 Hamilton Objectives
1. Read, write and locate any 3-digit number on a landmarked line from 0–1000 and use this to order and compare numbers.
6. Round to the nearest ten and hundred.
Y4 Hamilton Objectives
26. Know that one-place decimal numbers represent ones and tenths.
27. Round decimals with one decimal place to the nearest whole number.
### Planning and Activities
Day 1 Teaching
Model placing numbers on a number line and rounding.
Further Teaching with Y4
Relate one place decimals to cm and mm, and round decimals.
Group Activities: T with Y3
Y3 -- Locate 3-digit numbers on landmarked lines.
Y4 -- Position 1-place decimals accurately on landmarked 0–10 lines and round to the nearest whole. Some children will measure items less than 10cm to the nearest mm.
Day 2 Teaching
Order decimals on a number line.
Further Teaching with Y3
Order money on a number line and round to the nearest pound.
Group Activities: T with Y3
Y3 -- Create amounts of money using number cards and round to the nearest pound.
Y4 -- Position 1-place decimals accurately on number lines; some children will round to the nearest whole.
Day 3 Teaching
Compare decimal numbers using a number line.
Further Teaching with Y3
Compare and order amounts money using a number line.
Group Activities: T with Y4
Use the in-depth problem-solving investigation ‘Counter Place Value’ as today’s group activity.
Or, use these activities:
Y3 -- Investigation: How many amounts of money from £1 to £10 have the digit 9?
Y4 -- Roll a dice to make numbers with 1 decimal place. Play ‘Higher or lower’ using the rolled digits. Some children write 1-place decimal numbers in a given range then order the group of numbers.
### You Will Need
• Rulers marked in cm and mm
• Blank landmarked line (see resources)
• Small sticky notes
• ‘Placing 3-digit numbers on a landmarked line’ sheet (see resources)
• Items less than 10cm long
• 0–9 digit cards
• 0–10 lines (see resources)
• Blank cards
• £0 to £10 number line (see resources)
• 1–6 and 0–9 dice
• Squared paper
• Calculators
### Short Mental Workouts
Day 1
Place 2-digit numbers on a 0–100 line
Day 2
Count on and back in 1s and 10s
Day 3
Tell the time to the nearest 5 minutes
### Procedural Fluency
Day 1
Y3: Place 3-digit numbers between hundreds.
Y4: Place decimals on lines.
Day 2
Y3: Place money on a line.
Y4: More placing decimals on lines.
Day 3
Y3: Find ‘in between’ amounts.
Y4: Compare numbers with 1 decimal place.
### Mastery: Reasoning and Problem-Solving
Y3
• Sketch a 0–1000 line. Mark 500. Now mark 350, 700 and 990. How can you demonstrate that you have marked these accurately?
• True or false?
- Between any pair of next-door multiples of 100, there are always 98 whole numbers.
- The middle of a 500–1000 line is 800.
- There are ten numbers ending in 3 between 300 and 400.
- The digit 0 is used 18 times between 600 and 700.
• Write numbers to make the sentences true:
100 < ☐ < 110
304 > ☐ > 302
999 > ☐ > 888
0 < 101 < ☐
459 < ☐ < 461
• Write the value of the 5 digit in these numbers:
652
591
905
Y4
• Write each number as a decimal:
(i) One and four tenths
(ii) 6/10
(iii) 102/10
(iv) One half
(v) One fifth
• Billy measured his mobile phone. These were its dimensions:
Length: 12cm and 8mm
Width: 64mm
Thickness: 8mm.
Write these as numbers of centimetres, with a decimal place if necessary.
• Write <, > or = between each pair of numbers:
4.5 5.4
0.6 1/2
7.1 7.8
0.3 3/10
2/5 0.5
In-depth Investigation: Counter place value
Children place counters on a place value grid to make 3-digit numbers and then use logic to ensure they have made all combinations.
Y3 make whole numbers.
Y4 make numbers with 1 decimal place.
### Extra Support
Y3: In-betweenies
Placing numbers on a 0–100 landmarked line
Make the number
Knowing the value of each digit in 3-digit numbers
Y4: Decimals are a snip
Understanding place value in numbers with 1 decimal place and how we can write tenths as fractions or decimals |
# Divide Decimals Using Place Value
In these lessons, we will learn how to divide decimals by single-digit whole numbers involving easily identifiable multiples using place value understanding and relate to a written method.
New York State Common Core Math Grade 5, Module 1, Lesson 13
Problems 1 - 3
0.9 ÷ 3 =
0.24 ÷ 4 =
0.032 ÷ 8 =
Problems 4–5
1.5 ÷ 5 =
1.05 ÷ 5 =
### Lesson 13 Exit Ticket
1. Complete the sentences with the correct number of units and complete the equation.
a. 2 groups of ____ tenths is 1.8
1.8 ÷ 2 = ____
b. 4 groups of ____ hundredths is 0.32
0.32 ÷ 4 = ____
c. 7 groups of ____ thousandths is 0.021
0.021 ÷ 7 = ____
2. Complete the number sentence.
Express the quotient in units and then in standard form.
a. 4.5 ÷ 5 = ____ tenths ÷ 5 = ____ tenths = ____
b. 6.12 ÷ 6 =____ ones ÷ 6 + ____hundredths ÷ 6
= ____ ones + ____ hundredths
= ____
### Lesson 13 Homework
1. Complete the sentences with the correct number of units and complete the equation.
a. 3 groups of ___ tenths is 1.5, 1.5 ÷ 3 =
b. 6 groups of ___ hundredths is 0.24, 0.24 ÷ 6 =
c. 5 groups of ___ thousandths is 0.045, 0.045 ÷ 5 =
2. Complete the number sentence. Express the quotient in units and then in standard form.
a. 9.36 ÷ 3 = _____ ones ÷ 3 + ____ hundredths ÷ 3
= ____ ones + ____ hundredths
= ______
b. 36.012 ÷ 3 = ones ÷ 3 + thousandths ÷ 3
= ones + thousandths
= ______
c. 3.55 ÷ 5 = tenths ÷ 5 + hundredths ÷ 5
= ______
d. 3.545 ÷ 5 = ____
3. Find the quotients. Then use words, numbers, or pictures to describe any relationships you notice between each pair of problems and quotients.
a. 21 ÷ 7 = ___, 2.1 ÷ 7 = ___
b. 48 ÷ 8 = ___, 0.048 ÷ 8 = ___
a. 0.54 ÷ 6 = 9
b. 5.4 ÷ 6 = 0.9
c. 54 ÷ 6 = 0.09
5. A toy airplane costs \$4.84. It costs 4 times as much as a toy car. What is the cost of the toy car?
6. Julian bought 3.9 liters of cranberry juice and Jay bought 8.74 liters of apple juice. They mixed the two juices together then poured them equally into 2 bottles. How many liters of juice are in each bottle?
Looking at the unit values within a decimal in order to divide with greater speed and accuracy.
1. d. 3.545 ÷ 5 =
Examples:
0.9 ÷ 3 =
0.8 ÷ 4 =
1.5 ÷ 3 =
0.32 ÷ 8 =
30.5 ÷ 5 =
30.15 ÷ 5 =
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# 15.4 Pendulums (Page 4/7)
Page 4 / 7
## Measuring the torsion constant of a string
A rod has a length of $l=0.30\phantom{\rule{0.2em}{0ex}}\text{m}$ and a mass of 4.00 kg. A string is attached to the CM of the rod and the system is hung from the ceiling ( [link] ). The rod is displaced 10 degrees from the equilibrium position and released from rest. The rod oscillates with a period of 0.5 s. What is the torsion constant $\kappa$ ?
## Strategy
We are asked to find the torsion constant of the string. We first need to find the moment of inertia.
## Solution
1. Find the moment of inertia for the CM:
${I}_{\text{CM}}=\int {x}^{2}dm={\int }_{\text{−}L\text{/}2}^{+L\text{/}2}{x}^{2}\lambda dx=\lambda {\left[\frac{{x}^{3}}{3}\right]}_{\text{−}L\text{/}2}^{+L\text{/}2}=\lambda \frac{2{L}^{3}}{24}=\left(\frac{M}{L}\right)\frac{2{L}^{3}}{24}=\frac{1}{12}M{L}^{2}.$
2. Calculate the torsion constant using the equation for the period:
$\begin{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{I}{\kappa }};\hfill \\ \hfill \kappa & =\hfill & I{\left(\frac{2\pi }{T}\right)}^{2}=\left(\frac{1}{12}M{L}^{2}\right){\left(\frac{2\pi }{T}\right)}^{2};\hfill \\ & =\hfill & \left(\frac{1}{12}\left(4.00\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(0.30\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}\right){\left(\frac{2\pi }{0.50\phantom{\rule{0.2em}{0ex}}\text{s}}\right)}^{2}=4.73\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\text{.}\hfill \end{array}$
## Significance
Like the force constant of the system of a block and a spring, the larger the torsion constant, the shorter the period.
## Summary
• A mass m suspended by a wire of length L and negligible mass is a simple pendulum and undergoes SHM for amplitudes less than about $15\text{°}$ . The period of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$ , where L is the length of the string and g is the acceleration due to gravity.
• The period of a physical pendulum $T=2\pi \sqrt{\frac{I}{mgL}}$ can be found if the moment of inertia is known. The length between the point of rotation and the center of mass is L .
• The period of a torsional pendulum $T=2\pi \sqrt{\frac{I}{\kappa }}$ can be found if the moment of inertia and torsion constant are known.
## Conceptual questions
Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.
A pendulum clock works by measuring the period of a pendulum. In the springtime the clock runs with perfect time, but in the summer and winter the length of the pendulum changes. When most materials are heated, they expand. Does the clock run too fast or too slow in the summer? What about the winter?
The period of the pendulum is $T=2\pi \sqrt{L\text{/}g}.$ In summer, the length increases, and the period increases. If the period should be one second, but period is longer than one second in the summer, it will oscillate fewer than 60 times a minute and clock will run slow. In the winter it will run fast.
With the use of a phase shift, the position of an object may be modeled as a cosine or sine function. If given the option, which function would you choose? Assuming that the phase shift is zero, what are the initial conditions of function; that is, the initial position, velocity, and acceleration, when using a sine function? How about when a cosine function is used?
## Problems
What is the length of a pendulum that has a period of 0.500 s?
Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?
24.8 cm
What is the period of a 1.00-m-long pendulum?
How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?
4.01 s
The pendulum on a cuckoo clock is 5.00-cm long. What is its frequency?
Two parakeets sit on a swing with their combined CMs 10.0 cm below the pivot. At what frequency do they swing?
1.58 s
(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is $9.79\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ is moved to a location where the acceleration due to gravity is $9.82\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ . What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.
A pendulum with a period of 2.00000 s in one location ( $g=9.80{\text{m/s}}^{2}$ ) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?
$9.82002\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$
(a) What is the effect on the period of a pendulum if you double its length? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?
whats a two dimensional force
Where is Fourier Theorem?
what is Boyle's law
Boyle's law state that the volume of a given mass of gas is inversely proportion to its pressure provided that temperature remains constant
Abe
how do I turn off push notifications on this crap app?
Huntergirl
what is the meaning of in.
identity of vectors?
what is defined as triple temperature
Triple temperature is the temperature at which melting ice and boiling water are at equilibrium
njumo
a tire 0.5m in radius rotate at constant rate 200rev/min. find speed and acceleration of small lodged in tread of tire.
hmm
Ishaq
100
Noor
define the terms as used in gravitational mortion 1:earth' satellites and write two example 2:parking orbit 3:gravitation potential 4:gravitation potential energy 5:escping velocity 6:gravitation field and gravitation field strength
can gravitational force cause heat?
SANT
what larminar flow
smooth or regular flow
Roha
Hii
scalar field define with example
what is displacement
the change in the position of an object in a particular direction is called displacement
Noor
The physical quantity which have both magnitude and direction are known as vector.
Malik
good
Noor
Describe vector integral?
Malik
define line integral
Malik
Examples on how to solve terminal velocity
what is Force?
Lumai
the sideways pressure exerted by fluid is equal and canceled out.how and why?
Chaurasia
Particles emitted by black holes
Lord
what is oscillation |
# Evaluate the following integrals:
Question:
Evaluate the following integrals:
$\int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x$
Solution:
Given $I=\int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x$
We know that $\cos 2 x=1-2 \sin ^{2} x$
$\Rightarrow \int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x=\int \frac{1}{1-2 \sin ^{2} x+3 \sin ^{2} x} d x$
$=\int \frac{1}{1+\sin ^{2} x} d x$
Dividing numerator and denominator by $\cos ^{2} x$,
$\Rightarrow \int \frac{1}{1+\sin ^{2} x} d x=\int \frac{\sec ^{2} x}{\sec ^{2} x+\tan ^{2} x} d x$
Replacing $\sec ^{2} x$ in denominator by $1+\tan ^{2} x$
$\Rightarrow \int \frac{\sec ^{2} x}{\sec ^{2} x+\tan ^{2} x} d x=\int \frac{\sec ^{2} x}{1+2 \tan ^{2} x} d x$
Putting $\tan x=t$ so that $\sec ^{2} x d x=d t$,
$\Rightarrow \int \frac{\sec ^{2} x}{1+2 \tan ^{2} x} d x=\int \frac{d t}{1+2 t^{2}}$
$=\frac{1}{2} \int \frac{1}{\frac{1}{2}+t^{2}} d t$
We know that $\int \frac{1}{\mathrm{a}^{2}+\mathrm{x}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$
$\Rightarrow \frac{1}{2} \int \frac{1}{\frac{1}{2}+t^{2}} d t=\frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{2}}} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)+c$
$=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+c$
$\therefore \mathrm{I}=\int \frac{1}{\cos 2 \mathrm{x}+3 \sin ^{2} \mathrm{x}} \mathrm{dx}=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan \mathrm{x})+\mathrm{c}$ |
# Discrete Maths (MATH1081): Section 2 — Integers, Modular Arithmetic, and Relations
## Factors
If a is a factor of b, we also say:
• b is a multiple of a
• a is a divisor of b
• a divides b
• b is divisible by a
### Properties of divisibility
• if a b and a c then a b + c and a b – c
• if a b and a c then a sb + tc
• if a b and b c then a c
### Prime numbers
If p is prime, and p = ab, then either a = 1 or b = 1.
### Fundamental theorem of arithmetic
A composite positive integer can be factorised as a product primes. Further, a given integer has only one such factorisation.
e.g. 14175 = 34 × 52 × 7
### Testing for primes
Three ways of expressing the same idea
• If n is composite, then n has a factor c such that 1 < c ≤ sqrt(n)
• Any composite number n has a prime factor ≤ sqrt(n)
• Let n be a positive integer greater than 1. If n has no prime factor ≤ sqrt(n), then n is prime.
### Greatest common divisor
Let a, b ∈ ℤ. Any integer d such that d | a and d | b is called a common divisor or common factor of a and b. The largest d is the greatest common divisor, written as gcd(a, b)
### Least common multiple
Any m > 0 such that a | m and b | m is a common multiple of a and b. The smallest common multiple is written lcm(a, b)
### Coprime, relatively prime
If the gcd of a and b is 1 then we say that a and b are coprime or relatively prime.
### Applying fundamental theorem of arithmetic
To find the gcd of two numbers 14175 and 16758, we multiply all of the common prime factors of both.
14175 = 34 × 52 × 7
16758 = 2 × 32 × 72 × 19
gcd(14175, 16758) = 32 × 7
To find the lcm of two numbers 14175 and 16758 we need the smallest product of prime factors that include everything in both numbers.
e.g. lcm(14175, 16758) = 2 × 34 × 52 × 72 × 19 = 3770550
• if a, b > 0 then gcd(a, b) × lcm(a, b) = ab
### The Euclidean algorithm
Let a and b be positive integers. Suppose that
a = q1b + r1
b = q2r1 + r2
r1 = q3r2 + r3
rn – 2 = qnrn – 1 + rn
rn – 1 = qn + 1rn
Then gcd(a, b) = rn
• Note the down and to the left movement of each part.
• rn is the last non-zero remainder.
e.g. Find the gcd of 854 and 651.
854 = 1 × 651 + 203
651 = 3 × 203 + 42
203 = 4 × 42 + 35
42 = 1 × 35 + 7
35 = 5 × 7
So gcd(854, 651) = 7
### Solving equations like ax + by = c (Bezout property)
Consider ax + by = c
• if c = gcd(a, b) then it has a solution
• if c is a multiple of gcd(a, b) then it has multiple solutions
• if c is not a multiple of gcd(a, b) then there are no solutions
e.g. Find a solution for 30x + 73y = 1
Compute gcd(73, 30)
73 = 2 × 30 + 13 30 = 2 × 13 + 4 13 = 3 × 4 + 1
so 1 is the gcd.
Reversing this,
1 = 13 – 12
= 13 – (3 × 4)
= 13 – 3 × (4)
= 13 – 3 × (30 – 2 × 13)
= 13 - 3 × 30 + 6 × 13
= –3 × 30 + 7 × 13
= –3 × 30 + 7 × (73 – 2 × 30)
= –3 × 30 + 7 × 73 - 14 × 30
= –17 × 30 + 7 × 73
Therefore x = –17, y = 7
### Modulo
If m | ab, then a and b are congruent modulo m
a ≅ b mod(m)
which can also be written as
• m a – b
• a = b + km, k ∈ ℤ
• a and b have the same remainder when divided by m
#### Properties
If a ≅ b(mod m) and c ≅ d(mod m) (i.e. they have the same mods)
• a + c ≅ b + d(mod m)
• a – c ≅ b – d(mod m)
• ac ≅ bd(mod m)
• ca ≅ cb (mod m)
• an ≅ bn(mod m)
• if a ≅ b(mod m) and n m then a ≅ b(mod n)
### Simplifying modulo
e.g. Simplify 10123456789 modulo 41
1. Start by trying things out, we want to find a 1. 102 = 100 ≅ 18(mod 41)
103 ≅ 180 ≅ 16(mod 41)
104 ≅ 160 ≅ –4(mod 41)
105 ≅ -40 ≅ 1(mod 41)
2. Now that we have a 1, try to rewrite the equation to use that to simplify
10123456789 = 10123456785 × 104
≅ 105q + 4 ≅ 105q + 104 (mod 41) ≅ 1q × (–4) (mod 41) ≅ –4(mod 41) ≅ 37(mod 41)
### Solving congruence
We can use congruences to solve equations and vice versa.
79x ≅ 5347(mod 45) is the same as 79x – 45y = 5347
Then we apply the same steps as above.
1. Apply the Euclidean algorithm
2. Reverse the process to obtain a Bezout identity
3. Determine the solution(s)
Number of solutions
Consider ax ≅ b(mod m)
• if a and m are coprime then the congruence has a unique solution
• if gcd(a, m) isn’t a factor of b then there is no solution
• if gcd(a, m) is a factor of b then the congruence has a unique solution mod m
• if g = gcd(a, m) is a factor of b then the congruence has g different solutions mod m
## Relations
See course notes for introduction, definitions.
### Reflexive
A relation is reflexive if for every a ∈ A it is true that a R a
i.e. if every element is related to itself.
How to prove
“Let x ∈ S. Then … and thus x ~ x”
### Symmetric
A relation is symmetric when for all a, b in A, if a R b then b R a
i.e. if a R b then b R a.
How to prove
It’s an “if then” proof, so we start by assuming that it is true (a R b), and then prove that b R a.
“Let x, y ∈ ℝ and assume x ~ y … “ then prove y ~ x
### Transitive
R is transitive if a R b and b R a then a R c
i.e. if a is related to b, and b is related to c, then a is related to c.
How to prove
Also an “if then” proof, this time involving three variables.
“Let x, y, z ∈ ℝ and assume x ~ y and y ~ z then …”
### Antisymmetric
R is antisymmetric if a R b and b R a then a = b
i.e. two different elements of A may be related by R one way around, or the other, or neither, but not both.
How to prove
As above, an “if then” proof
### Equivalence relations
A relation that is reflexive, symmetric, and transitive.
roughly means “the same”.
#### Equivalence class
Let ~ be an equivalence relation on a set A. For any a ∈ A, the equivalence class of a with respect to ~ is the set [ a ] = { x ∈ A | x ~ a}
• every element is contained in its own equivalence class
• equivalence classes cannot be empty
• equivalence classes are sets: we prove they are equal by proving LHS ⊆ RHS and RHS ⊆ LHS
### Partial order
A relation that is reflexive, anti-symmetric and transitive is called a partial order.
• tells us which of two elements “comes first”
• “poset” = partially ordered set
#### Hasse Diagram
• shows a partial order on a finite set
• if two sets S and T are related then we place S further down the page than T and draw a line between them
• don’t draw loops
• don’t draw lines that can be attained by transitivity
### Definitions in relation to partial order
Maximal
If it is related to no element except itself, i.e. has nothing above it.
Greatest
If every element is related to it, i.e. it is ‘greater than’ or equal everything else.
• any greatest element is maximal
• there is only one greatest element (if any)
• if there is a unique maximal element, then it is greatest
Minimal
If no element except itself is related to it, i.e. has nothing below it.
Least
If it is related to every element, i.e. ‘less than’ or equal to everything
• any least element is minimal
• there is only one least element (if any)
• if there is a unique minimal element, then it is least
### Low and upper bound for partial order
• most often asked to determine after sketching a Hasse diagram (which makes it easier)
The greatest lower bound of a and b is glb(a, b)
The least upper bound of a and b is lub(a, b)
(I rarely see them referred to as glb and lub, however.) |
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# Single Variable Division Equations
## Solve one - step equations using multiplication.
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Practice Single Variable Division Equations
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Single Variable Division Equations
Have you ever been to a theater with limited seating? Well, Marc and Kara had this happen with their grandparents.
Marc and Kara went to see a play with their grandparents. When they arrived at the theater, the manager divided their group and several other people into six smaller groups. Each of these groups was lead to a section of the theater where there were empty seats. Each group had six people in it.
If this was the division, how many people did the manager divide up to start with?
Write a division equation and solve it to complete this dilemma.
This Concept is all about single variable division equations. You will know how to do this by the end of the Concept.
### Guidance
Sometimes, you will see equations that have division in them. We can use a fraction bar to show division.
To solve an equation in which a variable is divided by a number, we can use the inverse of division––multiplication. We can multiply both sides of the equation by that number to solve it.
We must multiply both sides of the equation by that number because of the Multiplication Property of Equality , which states:
if $a=b$ and $c \neq 0$ , then $a \times c=b \times c$ .
So, if you multiply one side of an equation by a nonzero number, $c$ , you must multiply the other side of the equation by that same number, $c$ , to keep the values on both sides equal.
Now let's apply this information.
$k \div (-4)=12$ .
In the equation, $k$ is divided by -4. So, we can multiply both sides of the equation by -4 to solve for $k$ . You will need to use what you know about multiplying integers to help you solve this problem. It may help to rewrite $k \div (-4)$ as $\frac{k}{-4}$ .
$k \div (-4) &= 12\\\frac{k}{-4} &= 12\\\frac{k}{-4} \times (-4) &= 12 \times (-4)\\\frac{k}{-4} \times \frac{-4}{1} &= -48\\\frac{k}{\cancel{-4}} \times \frac{\cancel{-4}}{1} &= -48\\\frac{k}{1} &= -48\\k &= -48$
The –4's will cancel each other out when they are divided. Then we multiply.
The value of $k$ is –48.
Remember the rules for multiplying integers will apply when working with these equations! Think back and use them as you work.
$\frac{n}{1.5}=10$
In the equation, $n$ is divided by 1.5. So, we can multiply both sides of the equation by 1.5 to solve for $n$ .
$\frac{n}{1.5} &= 10\\\frac{n}{1.5} \times 1.5 &= 10 \times 1.5\\\frac{n}{1.5} \times \frac{1.5}{1} &= 15\\\frac{n}{\cancel{1.5}} \times \frac{\cancel{1.5}}{1} &= 15\\\frac{n}{1} &= 15\\n &= 15$
The value of $n$ is 15.
When an equation has division in it, you can use the Multiplication Property of Equality to solve it, and then the property has been useful. Always remember to think about the inverse operations and associate them with the different properties. This will help you to keep it all straight and not get mixed up.
Now it is time for you to practice solving a few of these equations.
Solve each equation for the missing variable.
#### Example A
$\frac{x}{-2}=5$
Solution: $x = -10$
#### Example B
$\frac{y}{5}=6$
Solution: $y = 30$
#### Example C
$\frac{b}{-4}=-3$
Solution: $b = 12$
Here is the original problem once again.
Marc and Kara went to see a play with their grandparents. When they arrived at the theater, the manager divided their group and several other people into six smaller groups. Each of these groups was lead to a section of the theater where there were empty seats. Each group had six people in it too.
If this was the division, how many people did the manager divide up to start with?
Write a division equation and solve it to complete this dilemma.
First, we can write an equation.
Some number of people divided by six is six.
$\frac{x}{6} = 6$
Now we can solve it by multiplying.
$6 \times 6 = 36$
$x = 36$
### Guided Practice
Here is one for you to try on your own.
Three friends evenly split the total cost of the bill for their lunch. The amount each friend paid for his share was $4.25. a. Write an equation to represent $c$ , the total cost, in dollars, of the bill for lunch. b. Determine the total cost of the bill. Answer Consider part $a$ first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. Because the friends split the bill evenly, write a division equation to represent the problem. $& \underline{Three \ friends} \ \underline{evenly \ split} \ the \ \underline{total \ cost} \ldots amount \ each \ friend \ paid \ \underline{was} \ \underline{\4.25} \ldots\\& \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\& \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\& \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\& \qquad \ c \qquad \qquad \qquad \ \div \qquad \qquad \quad 3 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ = \quad 4.25$ This equation, $c \div 3=4.25$ , represents $c$ , the total cost of the lunch bill. Next, consider part $b$ . Solve the equation to find the total cost, in dollars, of the lunch bill. $c \div 3 &= 4.25\\\frac{c}{3} &= 4.25\\\frac{c}{3} \times 3 &= 4.25 \times 3\\\frac{c}{\cancel{3}} \times \frac{\cancel{3}}{1} &= 12.75\\c &= 12.75$ The total cost of the lunch bill was$12.75.
### Explore More
Directions: Solve each single-variable division equation for the missing value.
1. $\frac{x}{5}=2$
2. $\frac{y}{7}=3$
3. $\frac{b}{9}=-4$
4. $\frac{b}{8}=-10$
5. $\frac{b}{8}=20$
6. $\frac{x}{-3}=-10$
7. $\frac{y}{18}=-20$
8. $\frac{a}{-9}=-9$
9. $\frac{x}{11}=-12$
10. $\frac{x}{3}=-3$
11. $\frac{x}{5}=-8$
12. $\frac{x}{1.3}=3$
13. $\frac{x}{2.4}=4$
14. $\frac{x}{6}=1.2$
15. $\frac{y}{1.5}=3$
### Vocabulary Language: English
Division Property of Equality
Division Property of Equality
The division property of equality states that two equal values remain equal if both are divided by the same number. For example: If $2x = 8$, then $1x = 4$.
Inverse Operation
Inverse Operation
Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Isolate the variable
Isolate the variable
To isolate the variable means to manipulate an equation so that the indicated variable is by itself on one side of the equals sign.
Multiplication Property of Equality
Multiplication Property of Equality
The multiplication property of equality states that if the same constant is multiplied to both sides of the equation, the equality holds true. |
### #ezw_tco-2 .ez-toc-title{ font-size: 120%; ; ; } #ezw_tco-2 .ez-toc-widget-container ul.ez-toc-list li.active{ background-color: #ededed; } chapter outline
The quadratic equation, written in the general form as ax2 + bx + c = 0 is derived using the steps involved in completing the square.
## Steps to Derive the Quadratic Formula
To derive the quadratic formula, complete the square and solve for the variable x.
Step 1: In a quadratic function y = ax2 + bx + c, where a, b, and c are real numbers and a ≠0, let y = 0, such that:
ax2 + bx + c = 0
Step 2: Move ‘c’ to the right of the equation by subtracting both sides by ‘c’
ax2 + bx + c – c = 0 – c
=> ax2 + bx = – c
Step 3: Divide both sides of the equation by ‘a’
=> ${\dfrac{1}{a}( ax^{2}+bx= -c)}$
=> ${x^{2}+\dfrac{b}{a}x=-\dfrac{c}{a}}$
Step 4: Identify the coefficient of x, divide by 2, square the total term and simplify
=> ${\left( \dfrac{\dfrac{b}{a}}{2}\right) ^{2}=\left( \dfrac{b}{2a}\right) ^{2}}$
=> ${\dfrac{b^{2}}{4a^{2}}}$
Step 5: Add the result of step 5 to both sides of the equation
=> ${x^{2}+\dfrac{b}{a}x+\dfrac{b^{2}}{4a^{2}}=-\dfrac{c}{a}+\dfrac{b^{2}}{4a^{2}}}$
Step 6: Simplify the right side of the equation
=> ${x^{2}+\dfrac{b}{a}x+\dfrac{b^{2}}{4a^{2}}=\dfrac{-c\times 4a}{a\times 4a}+\dfrac{b^{2}}{4a^{2}}}$
=> ${x^{2}+\dfrac{b}{a}x+\dfrac{b^{2}}{4a^{2}}=${\dfrac{-4ac}{4a^{2}}+\dfrac{b^{2}}{4a^{2}}}$=>${x^{2}+\dfrac{b}{a}x+\dfrac{b^{2}}{4a^{2}}= ${\dfrac{b^{2}-4ac}{a^{2}}}$
Step 7: Express the trinomial as a perfect square
=> ${[ +\dfrac{b}{2a}) ^{2}=b^{2}\dfrac{-4ac}{4a^{2}}}$
Step 8: Square root both sides to remove the exponent 2
=> ${\sqrt{\left( x+\dfrac{b}{2a}\right) ^{2}}=\sqrt{\dfrac{b^{2}-4ac}{4a^{2}}}}$
Step 9: Simplify and add ≠on the right side of the equation
=> ${x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^{2}-4ac}}{4a^{2}}}$
Step 10: Subtract both sides by ${\dfrac{b}{2a}}$ to keep only ‘x’ on the left side and simplify
=> ${x+\dfrac{b}{2a}-\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^{2}-4ac}}{2a}-\dfrac{b}{2a}}$
=> ${x=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^{2}-4ac}}{2a}}$
=> ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
This gives us the quadratic formula we know and use for our calculations. |
# Factor Fun! Math Multiplication Games
Facts Flip: Each player flips over a card and multiplies it by the target number. Player with greatest product takes both cards!
Learning multiplication facts boils down to practice. (Lots of practice!) Mix in games to help students practice multiplication skills while having fun. Most of these games require easily found items such as dice, dominoes, or a deck of cards.
Multiplication Facts Flip:
In this multiplication card game, pairs of students use playing cards or homemade cards.
1. Prepare cards numbered 0 through 9 (two or three sets per student).
2. Choose a target number and write it on the board.
3. Give each pair of students a supply of cards.
4. Students divide cards and put them facedown between them.
5. Each player turns over a top card, multiplies it by the target number on the board, and says the product.
6. The player with the greater product takes both cards.
Variation: Multiplication War:
Students flip over cards and try to be the first to correctly multiply the numbers on the two cards. The first player to call out the answer keeps both cards. If both students call out the correct answer, they each get a card.
Download printables for “Multiplication Facts Flip” and “Generate a Product” games here: Math multiplication games from Ten-Minute Activities, Grades 1–3.
We’re on a Roll:
1. In this math game, pairs of students use dice to add and multiply.
1. Divide the class into pairs.
2. Give each pair two dice.
3. Write on the board the operations you want each group to perform. (For example: x1, x3, x5, X7, x9)
4. Each pair rolls their dice. They all mentally add the numbers shown on the dice together and then perform each operation listed on the board.
For example: Roll the dice. One shows a 6, the other a 4. Add 6 and 4 together to get 10. Then 10 is multiplied by 1, 3, 5, 7, and 9.
Variation: You can use dominoes in place of dice. Dominoes are placed facedown. Students flip over one domino at a time, add the numbers together, and perform the pre-set operations.
Download printables for this game as well as “Multiplication Magic” and “More Nifty Nines” here: Math multiplication games from Ten-Minute Activities, Grades 4–6.
More Math Resources
See these other Evan-Moor math books and e-books for supplemental math practice.
Games and centers:
Math Games: Centers for Up to 6 Players
Daily math practice activities (also available as interactive lessons)
Daily Word Problems: Math, Grades 1–6
Supplemental practice: |
# How do you solve 12/(x+4)<=4?
Jun 24, 2018
The solution is $x \in \left(- \infty , - 4\right) \cup \left[- 1 , + \infty\right)$
#### Explanation:
You cannot do crossing over.
The inequality is
$\frac{12}{x + 4} \le 4$
$\iff$, $\frac{12}{x + 4} - 4 \le 0$
$\iff$, $\frac{12 - 4 \left(x + 4\right)}{x + 4} \le 0$
$\iff$, $\frac{12 - 16 - 4 x}{x + 4} \le 0$
$\iff$, $\frac{- 4 - 4 x}{x + 4} \le 0$
$\iff$, $\frac{4 \left(1 + x\right)}{x + 4} \ge 0$
Let $f \left(x\right) = \frac{4 \left(1 + x\right)}{x + 4}$
Build a sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$color(white)(aaa)-$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$
$\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 a}$$+$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$
Therefore,
$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 4\right) \cup \left[- 1 , + \infty\right)$
Jun 24, 2018
$\frac{12}{x + 4} \le 4$ for $x < - 4$ and $x \ge - 1$
#### Explanation:
As the expression is undefined for $x = - 4$, we want to stay away from that value.
Before we work on the expression algebraically, let's draw a graph:
graph{-(x+1)/(x+4) [-13.21, 6.79, -5.72, 4.28]}
Based on the graph we can see that the unequality is fulfilled for
$x < - 4$ and $x \ge - 1$
Let us clean up the expression to make it easier to work with:
An equivalent expression is
$\frac{3}{x + 4} \le 1$
$\frac{3}{x + 4} - 1 \le 0$
$\frac{3 - \left(x + 4\right)}{x + 4} \le 0$
$- \frac{x + 1}{x + 4} \le 0$
As this is undefined for $x = - 4$, we need to consider two situations: $x > - 4$ and $x < - 4$
1) $x > - 4$: As $x + 4 > 0$ we can multiply both sides with the denominator $x + 4$ and still keep the sign of inequality:
$- \frac{\left(x + 1\right) \left(x + 4\right)}{x + 4} \le 0$
$- \left(x + 1\right) \le 0$
$x + 1 \ge 0$
$x \ge - 1$
Therefore $\frac{12}{x + 4} \le 4$ when $x \ge - 1$
2) $x < - 4$: Now the denominator $\left(x + 4\right)$ is negative, so if we multiply the unequality with the value of denominator, we have to turn the unequal sign around:
$- \frac{\left(x + 1\right) \left(x + 4\right)}{x + 4} \ge 0$
$- \left(x + 1\right) \ge 0$
$x + 1 \le 0$
$x \le - 1$
As the starting point was that $x < - 4$, this means that $\frac{12}{x + 4} \le 4$ for all $x < - 4$ |
Sequences and Series
The sum of an infinite geometric sequence, infinite geometric series
The sum of an infinite geometric sequence, infinite geometric series
An infinite geometric series converges (has a finite sum even when n is infinitely large) only if the absolute ratio of successive terms is less than 1 that is, if -1 < r < 1
The sum of an infinite geometric series can be calculated as the value that the finite sum formula takes (approaches) as number of terms n tends to infinity,
first rewrite Sn, into
so that since | r | < 1, then rn ® 0 as n ® oo
thus, the sum of an infinite converging geometric series.
The sum of an infinite converging geometric series, examples
Example: Given a square with side a. Its side is the diagonal of the second square. The side of this square is then the diagonal of the third square and so on, as shows the figure below. Find the sum of areas of all these squares.
Solution: Using the given conditions,
Example: Given an equilateral triangle with the side a. Its height is the side of another equilateral triangle. The height of this triangle is then the side of the third equilateral triangle and so on, as shows the figure below. Find the sum of areas of all these triangles.
Solution: a1 = h, a2 = h1, a3 = h2, and so on. Thus,
Example: In a circle of the radius r inscribed is a square, in the square inscribed is another circle and in that circle another square and so on, as is shown in the figure below. Find the sum of areas of all the circles and the sum of areas of all the squares.
Solution: It follows that the radius r is the half of the diagonal d of the first inscribed square,
Example: In an isosceles triangle with the base 2a and the angle 2a, opposite to the base, inscribed are infinite sequence of circles such that first circle touches the base and opposite sides of the triangle while other circles touch opposite sides of the triangle and the preceding circle, as shows the figure below. Find the sum of all circumferences and areas of all inscribed circles.
Solution: In the shown right triangles we use following relations,
Therefore, the sum of all circumferences, SC = 2pr1 + 2pr2 + 2pr3 + . . .
The sum of areas of all circles, SA = r12p + r22p + r32p + . . .
Example: Into the regular square pyramid, with the side of the base a and the height h, inserted is the square with the base lying in the base of the pyramid, and the upper base of which is cross section of the pyramid. Into the space above this square inserted is, the same way, another square, and above of this third square and so on, to infinity, as is shown in the figure below. Find the volume of all inserted squares.
Solution: Following relations hold between similar right triangles shown in the figure,
The sum of volumes of all the squares, SV = V1 + V2 + V3 + . . . = a13 + a23 + a33 + . . .
Intermediate algebra contents |
# How do you differentiate f(x)=cot(3x) using the chain rule?
Feb 9, 2016
$\frac{\mathrm{du}}{\mathrm{dx}} = - 3 {\csc}^{2} \left(3 x\right)$
#### Explanation:
First, I assume you know that the derivative of $\cot x$ is $- {\csc}^{2} x$.
We substitute $u = 3 x$.
Therefore,
$\frac{\mathrm{du}}{\mathrm{dx}} = 3$.
Now we use the chain rule.
$\frac{d}{\mathrm{dx}} \left(\cot \left(3 x\right)\right) = \frac{d}{\mathrm{dx}} \left(\cot \left(u\right)\right)$
$= \frac{d}{\mathrm{du}} \left(\cot \left(u\right)\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$= - {\csc}^{2} \left(u\right) \cdot 3$
$= - 3 {\csc}^{2} \left(3 x\right)$ |
# 2007 AMC 12A Problems/Problem 16
## Problem
How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
$\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256$
## Solution 1
We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.
Common difference Sequences possible Number of sequences 1 $012, \ldots, 789$ 8 2 $024, \ldots, 579$ 6 3 $036, \ldots, 369$ 4 4 $048, \ldots, 159$ 2
This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$.
However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.
## Solution 2
Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives $3! = 6$ possible 3-digit numbers; otherwise, $4$ possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.
Case 1: 0 is not in the number. Then there are $\binom{5}{2} + \binom{4}{2} = 16$ ways to choose two nonzero digits of the same parity, and each choice generates $3! = 6$ 3-digit numbers, giving $16 \times 6 = 96$ numbers.
Case 2: 0 is in the number. Then there are $4$ ways to choose the largest digit (2, 4, 6, or 8), and each choice generates $4$ 3-digit numbers, giving $4 \times 4 = 16$ numbers.
Thus the total is $96 + 16 = 112 \Longrightarrow \mathrm{(C)}$. (by scrabbler94)
~ pi_is_3.14
## Closely-related question and solution (podcast)
https://www.buzzsprout.com/56982/episodes/415913 starts with a variation on this question (plus solution) |
# The area of a trapezium is 91 cm
Question:
The area of a trapezium is 91 cm2 and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
Solution:
Given:
Area of the trapezium $=91 \mathrm{~cm}^{2}$
Height $=7 \mathrm{~cm}$
Let the length of the smaller side be $x .$
Then, the length of longer side will be 8 more than smaller side, i.e. $8+x .$
Area of trapezium $=\frac{1}{2} \times($ Sum of the parallel sides $) \times($ Height $)$
$\Rightarrow 91=\frac{1}{2} \times[(8+x)+x] \times(7)$
$\Rightarrow 91=\frac{7}{2} \times[8+x+x]$
$\Rightarrow 91 \times 2=7 \times[8+2 x]$
We can rewrite it as follows:
$7 \times[8+2 \mathrm{x}]=182$
$\Rightarrow[8+2 x]=\frac{182}{7}=26$
$\Rightarrow 8+2 x=26$
$\Rightarrow 2 x=26-8=18$
$\Rightarrow x=\frac{18}{2}=9 \mathrm{~cm}$
$\therefore$ Length of the shorter side of the trapezium $=9 \mathrm{~cm}$
And, length of the longer side $=8+\mathrm{x}=8+9=17 \mathrm{~cm}$ |
## Elementary Statistics (12th Edition)
Published by Pearson
# Chapter 9 - Inferences from Two Samples - 9-3 Two Means: Independent Samples - Basic Skills and Concepts - Page 465: 14
#### Answer
a) Fail to reject the null hypothesis. b)$\mu_1-\mu_2$ is between -18.17 and 10.23.
#### Work Step by Step
a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1\ne\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(70.29-74.26)-(0)}{\sqrt{22.09^2/30+18.15^2/32}}=-0.77.$ The degree of freedom: $min(n_1-1,n_2-1)=min(30-1,32-1)=29.$ The corresponding P-value by using the table: p is more than 0.2. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is more than $\alpha=0.01$, because it is more than 0.2, hence we fail to reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.005}=2.756.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.756\sqrt{\frac{22.09^2}{30}+\frac{18.15^2}{32}}=14.2.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(70.29-74.26)-14.2=-18.17 and$\overline{x_1}-\overline{x_2}+E$=(70.29-74.26)+14.2=10.23.
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# Checking a Solution for a Linear System
## Substitute given values or graph to compare coordinates to intersections
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Checking a Solution for a Linear System
The Hiking Club is buying nuts to make trail mix for a fundraiser. Three pounds of almonds and two pounds of cashews cost a total of $36. Three pounds of cashews and two pounds of almonds cost a total of$39. Is (a, c) = ($6,$9) a solution to this system?
### Solution to a System of Linear Equations
A system of linear equations consists of the equations of two lines.The solution to a system of linear equations is the point which lies on both lines. In other words, the solution is the point where the two lines intersect. To verify whether a point is a solution to a system or not, we will either determine whether it is the point of intersection of two lines on a graph or we will determine whether or not the point lies on both lines algebraically.
Let's determine whether the given points are a solution to the systems of equations below.
1. Is the point (5, -2) the solution of the system of linear equations shown in the graph below?
Yes, the lines intersect at the point (5, -2) so it is the solution to the system.
1. Is the point (-3, 4) the solution to the system given below?
No, (-3, 4) is not the solution. If we replace the and in each equation with -3 and 4 respectively, only the first equation is true. The point is not on the second line; therefore it is not the solution to the system.
Now, let's find the solution to the system below.
Because the first line in the system is vertical, we already know the x-value of the solution, . Plugging this into the second equation, we can solve for y.
The solution is (5, -5). Check your solution to make sure it's correct.
You can also solve systems where one line is horizontal in this manner.
### Examples
#### Example 1
Earlier, you were asked if (a, c) = ($6,$9) is a solution to the system of equations.
The system of linear equations represented by this situation is:
If we plug in $6 for a and$9 for c, both equations are true. Therefore ($6,$9) is a solution to the system.
#### Example 2
Is the point (-2, 1) the solution to the system shown below?
No, (-2, 1) is not the solution. The solution is where the two lines intersect which is the point (-3, 1).
#### Example 3
Verify algebraically that (6, -1) is the solution to the system shown below.
By replacing and in both equations with 6 and -1 respectively (shown below), we can verify that the point (6, -1) satisfies both equations and thus lies on both lines.
#### Example 4
Explain why the point (3, 7) is the solution to the system:
The horizontal line is the line containing all points where the coordinate 7. The vertical line is the line containing all points where the coordinate 3. Thus, the point (3, 7) lies on both lines and is the solution to the system.
### Review
Match the solutions with their systems.
1. (1, 2)
1. (2, 1)
1. (-1, 2)
1. (-1, -2)
Determine whether each ordered pair represents the solution to the given system.
1. .
1. .
1. .
1. .
1. .
1. .
Find the solution to each system below.
1. .
1. .
1. .
1. .
1. Describe the solution to a system of linear equations.
2. Can you think of why a linear system of two equations would not have a unique solution?
To see the Review answers, open this PDF file and look for section 3.1.
### Notes/Highlights Having trouble? Report an issue.
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Games
Problems
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A student asked this question today: "Why is 0! (zero factorial) equal to one, instead of zero?"
Good question! Let's begin by making sure everyone knows what the "!" (factorial) notation means. n! means "the product of all the integers that lie between n and 1, inclusive."
Thus, 4! = 24, because 4! = 4(3)(2)(1) = 24.
6! = 6(5)(4)(3)(2)(1) = 720
The strange thing, though, is that 0! = 1, and that doesn't really seem to match our definition. After all, the integers between 0 and 1 inclusive are 0 and 1, and when you multiply them together, you get zero, not one!
Okay, so maybe our definition is flawed. We'll come back to that later.
The thing is, though, we don't want 0! to be equal to zero, because it's not useful. You see, we use factorials when we're calculating combinations of things, or when we're expanding a binomial to a power.
If you have something like (x + 1)5, the nth term in the expansion of that is (5Cn-1)x6-n. That's the binomial theorem.
The problem is, that theorem doesn't work if we say that 0! = 0. Why? Because 5C0 = 5!/(0!·5!), and if 0! is zero, then we have a division by zero problem! On the other hand, if we say that 0! = 1, then this works out perfectly to 5C0 = 1.
And really, if you think about it, that makes sense: If you have five objects, in how many different ways can you choose none of them? Uh, one!
We can see that 0! = 1 makes sense using patterns, too. Consider this:
7!/6! = 7
6!/5! = 6
5!/4! = 5
4!/3! = 4
3!/2! = 3
2!/1! = 2
Now to continue this pattern, what do we need next?
1!/0! = 1
Solve this equation for 0!, and you get: 0! = 1.
Okay, so it makes sense with the combination notation to say 0! = 1, and we can even see from patterns that it must equal 1. So the real problem is our definition. So maybe we should reword our definition a little bit.
I like this way of saying it: For all non-negative n, n! is the product of 1 with all the positive integers less than or equal to n.
Does that work? Sure! It keeps everything else the same, but since there no positive integers less than zero, we're left with 1.
That's one way of getting around it. Another way is to just say For all non-negative n, n! is the product of all the positive integers less than or equal to n. This way of defining it forces us to use the "empty product" definition, which says that multiplying together zero factors gives a result of one.
Or you can define it recursively by saying : 0! = 1, and for all integer n > 0, n! = n(n - 1)!.
Or, you can simply do this: For all positive n, n! is the product of 1 with all the positive integers less than or equal to n, and 0! = 1.
This last one defines n! when n>0, and then gives a special definition for 0!.
No matter how you choose to define it, the real point is that mathematicians chose to define it to be one instead of zero simply because it was of practical use to do so!
# Blogs on This Site
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# What is the range of the function f(x) = -3x^2 + 3x - 2?
May 14, 2018
$\left(- \infty , - \frac{5}{4}\right]$
#### Explanation:
$\text{we require to find the vertex and it's nature, that is}$
$\text{maximum or minimum}$
$\text{the equation of a parabola in "color(blue)"vertex form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$
$\text{to obtain this form use "color(blue)"completing the square}$
• " the coefficient of the "x^2" term must be 1"
$\text{factor out } - 3$
$y = - 3 \left({x}^{2} - x + \frac{2}{3}\right)$
• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - x$
$y = - 3 \left({x}^{2} + 2 \left(- \frac{1}{2}\right) x \textcolor{red}{+ \frac{1}{4}} \textcolor{red}{- \frac{1}{4}} + \frac{2}{3}\right)$
$\textcolor{w h i t e}{y} = - 3 {\left(x - \frac{1}{2}\right)}^{2} - 3 \left(- \frac{1}{4} + \frac{2}{3}\right)$
$\textcolor{w h i t e}{y} = - 3 {\left(x - \frac{1}{2}\right)}^{2} - \frac{5}{4} \leftarrow \textcolor{red}{\text{in vertex form}}$
$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{2} , - \frac{5}{4}\right)$
$\text{to determine if vertex is max/min}$
• " if "a>0" then minimum "uuu
• " if "a<0" then maximum "nnn
$\text{here "a=-3<0" hence maximum}$
$\text{range } y \in \left(- \infty , - \frac{5}{4}\right]$
graph{-3x^2+3x-2 [-8.89, 8.89, -4.444, 4.445]} |
# LESSON 10.1 PROBLEM SOLVING DEVELOPING FORMULAS FOR TRIANGLES AND QUADRILATERALS
Multiply the binomials FOIL. To find the area of a trapezoid, rhombus and kite. Each grid square has a side length of 1 in. Formula for area of a rhombus Substitute. Area of kite Substitute 48 for d1 and 42 for d2.
Over Lesson 11—1 A. You can use the Area Addition Postulate to see that a parallelogram has the same lewson as a rectangle with the same base and height.
Each grid square has a side length of 1 in. Areas of Trapezoids, Rhombuses and Kites Objectives: The area of a figure made with all the pieces is the sum of the areas of the formulax.
## 9-1 Developing Formulas for Triangles and Quadrilaterals Warm Up
Each grid square has a side length of 1 cm. Finding Measurements of Parallelograms Find the area of the parallelogram. The diagonals in a kite make a right angle.
IPS THESIS GUIDELINE USM
Formula for area of a rhombus Substitute.
Circle — Formulas Radius of the circle is generally denoted by letter R. My presentations Profile Feedback Log out.
FeatureLesson Geometry Lesson Main 1. Bell Work Find the area of each figure. Use the Pythagorean Theorem to find x and y. If you wish to download it, please recommend it to your friends in any social system. Area of a triangle Substitute 16 for b and 12 for h. Finding Measurements fornulas Rhombuses and Kites Find the area of the kite Step 1 The diagonals d1 and d2 form four right triangles. Step 1 Use the Pythagorean Theorem to find the height h. Area of a triangle Substitute 15×2 for A and 5x for h.
Use the grid to find the area of the shaded kite. Qudrilaterals with social network: Half of d2 is equal to 21, so d2 is equal to The pieces can be rearranged to form many different shapes. Two sides of the parallelogram are vertical and the other two sides are diagonals of a square of the grid.
THESIS FARNBOROUGH EQUITY FUND
Over Lesson 11—1 5-Minute Check 1 A. Finding Measurements of Parallelograms Find the area of the parallelogram.
To make this website work, we log user data and share it with processors. About project SlidePlayer Terms of Service. Share buttons are trianglss little bit lower. The base and height of the leftmost shaded parallelogram each measure 1 in.
Example 4 In the tangram, find the perimeter and area of the large green triangle. Round to the nearest tenth if necessary. |
# Derivatives
This unit covers the following ideas. In preparation for the quiz and exam, make sure you have a lesson plan containing examples that explain and illustrate the following concepts.
1. Find limits, and be able to explain when a function does not have a limit by considering different approaches.
2. Compute partial derivatives. Explain how to obtain the total derivative from the partial derivatives (using a matrix).
3. Find equations of tangent lines and tangent planes to surfaces. We'll do this three ways.
4. Find derivatives of composite functions, using the chain rule (matrix multiplication).
You'll have a chance to teach your examples to your peers prior to the exam.
# Limits
See Larson 13.2 for more about limits and continuity.
In the previous chapter, we learned how to describe lots of different functions. In first-semester calculus, after reviewing functions, you learned how to compute limits of functions, and then used those ideas to develop the derivative of a function. The exact same process is used to develop calculus in high dimensions. One glitch that will prevent us from developing calculus this way in high dimensions is the epsilon-delta definition of a limit. We'll review it briefly. Those of you who want to pursue further mathematical study will spend much more time on this topic in future courses. In first-semester calculus, you learned how to compute limits of functions. Here's the formal epsilon-delta definition of a limit.
Let $f:\RR\to\RR$ be a function. We write $\ds \lim_{x\to c} f(x)=L$ if and only if for every $\epsilon>0$, there exists a $\delta>0$ such that $0<|x-c|<\delta$ implies $|f(x)-L|<\epsilon$.
We're looking at this formal definition here because we can compare it with the formal definition of limits in higher dimensions. The only difference is that we just put vector symbols above the input $x$ and the output $f(x)$.
Let $\vec f:\RR^n\to\RR^m$ be a function. We write $\ds \lim_{\vec x\to \vec c} \vec f(\vec x)=\vec L$ if and only if for every $\epsilon>0$, there exists a $\delta>0$ such that $0<|\vec x-\vec c|<\delta$ implies $|\vec f(\vec x)-\vec L|<\epsilon$.
We'll find that throughout this course, the key difference between first-semester calculus and multivariate calculus is that we replace the input $x$ and output $y$ of functions with the vectors $\vec x$ and $\vec y$.
The point to this problem is to help you learn to recognize the dimensions of the domain and codomain of the function. If we write $\vec f:\RR^n\to \RR^m$, then $\vec x$ is a vector in $\RR^n$ with $n$ components, and $\vec y$ is a vector in $\RR^m$ with $m$ components.
For the function $f(x,y)=z$, we can write $f$ in the vector notation $\vec y=\vec f(\vec x)$ if we let $\vec x=(x,y)$ and $\vec y=(z)$. Notice that $\vec x$ is a vector of inputs, and $\vec y$ is a vector of outputs. For each of the functions below, state what $\vec x$ and $\vec y$ should be so that the function can be written in the form $\vec y = \vec f (\vec x)$.
1. $f(x,y,z)=w$
2. $\vec r(t)=(x,y,z)$
3. $\vec r(u,v)=(x,y,z)$
4. $\vec F(x,y)=(M,N)$
5. $\vec F(\rho,\phi,\theta)=(x,y,z)$
You learned to work with limits in first-semester calculus without needing the formal definitions above. Many of those techniques apply in higher dimensions. The following problem has you review some of these technique, and apply them in higher dimensions.
See 14.2: 1-30 for more practice.
Do these problems (without using L'Hopital's rule).
1. Compute $\ds \lim_{x\to 2} x^2-3x+5$ and then $\ds\lim_{(x,y)\to (2,1)} 9-x^2-y^2$.
2. Compute $\ds\lim_{x\to 3}\frac{x^2-9}{x-3}$ and then $\ds\lim_{(x,y)\to (4,4)} \frac{x-y}{x^2-y^2}$.
3. Explain why $\ds\lim_{x\to 0}\frac{x}{|x|}$ does not exist. [Hint: graph the function.]
In first semester calculus, we can show that a limit does or does not exist by considering what happens from the left, and comparing it to what happens on the right. You probably used the following theorem extensively.
If $y=f(x)$ is a function defined on some open interval containing $c$, then $\ds\lim_{x\to c}f(x)$ exists if and only if $\ds\lim_{x\to c^-}f(x) = \ds\lim_{x\to c^+}f(x)$.
A limit exists precisely when the limits from every direction exists, and all directional limits are equal. In first-semester calculus, this required that you check two directions (left and right). This theorem generalizes to higher dimensions, but it becomes much more difficult to apply.
Consider the function $\ds f(x,y)=\frac{x^2-y^2}{x^2+y^2}$. Our goal is to determine if the function has a limit at the origin $(0,0)$. We can approach the origin along many different lines. One line through the origin is the line $y=2x$. If we stay on this line, then we can replace each $y$ with $2x$ and then compute $$\ds\lim_{\substack{(x,y)\to(0,0)\\ y=2x }}\frac{x^2-y^2}{x^2+y^2} = \lim_{x\to 0} \frac{x^2-(2x)^2}{x^2+(2x)^2} = \lim_{x\to 0} \frac{-3x^2}{5x^2} = \lim_{x\to 0} \frac{-3}{5} =\frac{-3}{5}.$$ This means that if we approach the origin along the line $y=2x$, we will have a height of $-3/5$ when we arrive at the origin.
If the function $\ds f(x,y)=\frac{x^2-y^2}{x^2+y^2}$ has a limit at the origin, the previous problem suggests that limit will be $-3/5$.
Please read the previous example. Recall that we are looking for the limit of the function $\ds f(x,y)=\frac{x^2-y^2}{x^2+y^2}$ at the origin (0,0).
You may want to look at a graph in Sage or Wolfram Alpha (try using the “contour lines” option). As you compute each limit, make sure you understand what that limit means in the graph.
Our goal is to determine if the function has a limit at the origin $(0,0)$.
1. In the $xy$-plane, how many lines pass through the origin $(0,0)$? Give an equation a line other than $y=2x$ that passes through the origin. Then compute $$\ds\lim_{\substack{(x,y)\to(0,0)\\ \text{your line} }}\frac{x^2-y^2}{x^2+y^2} = \lim_{x\to 0} \frac{x^2-(?)^2}{x^2+(?)^2}=\ldots.$$
2. Give another equation a line that passes through the origin. Then compute $$\ds\lim_{\substack{(x,y)\to(0,0)\\ \text{your line}}}\frac{x^2-y^2}{x^2+y^2}.$$
3. Does this function have a limit at $(0,0)$? Explain.
See 14.2: 41-50 for more practice.See Larson 13.2:23–36 and example 4 for more practice.
The theorem from first-semester calculus generalizes as follows.
If $\vec y=\vec f(\vec x)$ is a function defined on some open region containing $\vec c$, then $\ds\lim_{\vec x\to \vec c}\vec f(\vec x)$ exists if and only if the limit exists along every possible approach to $\vec c$ and all these limits are equal.
There's a fundamental problem with using this theorem to check if a limit exists. Once the domain is 2-dimensional or higher, there are infinitely many ways to approach a point. There is no longer just a left and right side. To prove a limit exists, you must check infinitely many cases—that usually takes a really long time. The real power to this theorem is it allows us to show that a limit does not exist. All we have to do is find two approaches with different limits.
See Sage.
See 14.2: 41-50 for more practice.See Larson 13.2:9–36 for more practice.
Consider the function $\ds f(x,y) = \frac{xy}{x^2+y^2}$. Does this function have a limit at $(0,0)$?
1. Examine the function at $(0,0)$ by considering the limit as you approach the origin along several lines.
2. Convert $\frac{xy}{x^2+y^2}$ to polar coordinates (i.e., a function of $r$ and $\theta$). As $(x,y)$ approaches the origin, what does $r$ approach? Take the limit of your polar coordinate function as $r$ approaches that value and interpret your result.
In all the examples above, we considered approaching a point by traveling along a line. However, even if a function has a consistent limit along every line, that is not enough to always guarantee the function has a limit. The theorem requires every approach be consistent, which includes parabolic approaches, spiraling approaches, and more. Sometimes the straight-line paths happen to be consistent with each other, but a different path gives a different limit. Give some thought to this in the optional challenge problem below.
Give an example of a function $f(x,y)$ so that the limit at $(0,0)$ along every straight line $y=mx$ exists and equals 0. However, show that the function has no limit at $(0,0)$ by considering an approach that is not a straight line.
# The Derivative
Before we introduce derivatives, let's recall the definition of a differential. If $y=f(x)$ is a function, then we say the differential $dy$ is the expression $dy=f'(x) dx$ (we could also write this as $dy = \frac{dy}{dx}dx$). Think of differential notation $dy=f'(x)dx$ in the following way:
A small change in the output $y$ equals the derivative multiplied by a small change in the input $x$. In other words, if the input $x$ changes by a small amount $dx$, then we multiply that small change by the derivative to determine how much $y$ changes (i.e., $dy$).
To get the derivative in all dimensions, we just substitute in vectors to obtain the differential notation $d\vec y = f'(\vec x) d\vec x$. The derivative is precisely the thing that tells us how to get $d\vec y$ from $d\vec x$. We'll quickly see that the derivative $f'(\vec x)$ must be a matrix, and we'll start writing it as $Df$ instead of $f'$. We've actually already dealt with problems involving derivatives of multiple-variable functions in first-semester calculus. The next few problems are very similar to related rates or differential problems from first-semester calculus, and we'll see how the derivative $f'$ in $dy=f'(x) dx$ naturally generalizes to a matrix.
See 3.10 for more practice.See Larson 3.7 and 4.8.
The volume of a right circular cylinder is $V(r,h)= \pi r^2 h$. Imagine that each of $V$, $r$, and $h$ depends on $t$ (we might be collecting rain water in a can, or crushing a cylindrical concentrated juice can, etc.).
1. If the height remains constant, but the radius changes, what is $dV/dt$ in terms of $dr/dt$? Use this to find a formula for $dV$ in terms of $dr$ when $h$ is constant.
2. If the radius remains constant, but the height changes, what is $dV/dt$ in terms of $dh/dt$? What is $dV$ when $r$ is constant?
3. If both the radius and height change, what is $dV/dt$ in terms of $dh/dt$ and $dr/dt$? Solve for $dV$.
4. Make sure you ask me in class to show you physically exactly how you can see these differential formulas.
Show that we can write $dV$ as the matrix product of a 1-row by 2-column matrix with a 2-row by 1-column matrix: $$dV = \begin{bmatrix}2\pi rh& \pi r^2\end{bmatrix}\begin{bmatrix}dr\\dh\end{bmatrix}.$$ How do the columns of the first matrix relate to the calculations you did above?
The matrix $\begin{bmatrix}2\pi rh& \pi r^2\end{bmatrix}$ is the derivative of $V$. The columns of this matrix are the partial derivatives of $V$.
5. If we know that $r=3$ and $h=4$, and we know that $r$ increases by about $.1$ and $h$ increases by about $.2$, then approximate how much $V$ will increase. Use your formula for $dV$ to approximate this.
The volume of a box is $V(x,y,z)=xyz$. Imagine that each variable depends on $t$.
1. If both $y$ and $z$ remain constant, what is $dV/dt$? Use this to find a formula for $dV$ in terms of $dx$, assuming that $y$ and $z$ are constant.
2. Repeat the last step for when $y$ is the only variable that changes, and then for when $z$ is the only variable that changes.
3. What is $dV/dt$ in terms of $dx/dt$, $dy/dt$, and $dz/dt$ when all three variables are changing? Solve for $dV$.
4. Show that we can write $dV$ as the matrix product of a 1-row by 3-column matrix with a 3-row by 1-column matrix: $$dV = \begin{bmatrix}yz& ?&?\end{bmatrix}\begin{bmatrix}dx\\dy\\dz\end{bmatrix}.$$ How do the columns of the first matrix relate to the previous portions of the problem.
The matrix $\begin{bmatrix}yz& ?&?\end{bmatrix}$ is the derivative of $V$. The columns of this matrix are the partial derivatives of $V$.
5. If the current measurements of a box are $x=2$, $y=3$, and $z=5$, and we know that $x$ increases by .01, $y$ increases by .02, and $z$ decreases by .03, then by about how much will the volume change? Use your formula for $dV$ to approximate this.
Part 4 in each problem above, expressing relationships between changes in terms of differentials, is the KEY idea, let me repeat, THE KEY IDEA, to the rest of this course. The essential thing is that we can use differentials to understand and approximate how small changes in the inputs of a function will change the outputs of the function. For example, we can approximate the change in a function $f(x,y)$ if we know how much $x$ and $y$ will change.
Using differentials to analyze how outputs of a function change when the inputs change comes up in many applications, such as analyzing numerical roundoff error in calculations or analyzing manufacturing tolerances.
Consider the function $f(x,y) = x^2y +3x+4\sin(5y)$.
1. If both $x$ and $y$ depend on $t$, then use implicit differentiation to obtain a formula for $df/dt$ in terms of $dx/dt$ and $dy/dt$.
2. Solve for $df$, and write your answer as the matrix product (fill in the question mark) $$df = \begin{bmatrix}?& x^2+20\cos(5y)\end{bmatrix}\begin{bmatrix}dx\\dy\end{bmatrix}.$$
3. If you hold $y$ constant (so $dy=0$), then what is $df/dx$? If you are at $(x,y)=(1,2)$, and if $x$ changes by $.2$, then about how much does $z=f(x,y)$ change?
4. If you hold $x$ constant (so $dx=0$), then what is $df/dy$? If you are at $(x,y)=(1,2)$, and if $y$ changes by $-.3$, then about how much does $z=f(x,y)$ change?
5. If you are at the point $(x,y)=(1,2)$, and you move to $(1.2, 1.7)$, then what is $dx$ and $dy$? Approximate how much $z=f(x,y)$ changes.
6. (Challenge) In the last part, why was our calculation an approximation, rather than an exact calculation of how much $f(x,y)$ changed?
We need to add some vocabulary to make it easier to talk about what we just did. Let's introduce the vocabulary in terms of the problem above, and then make a formal definition.
• The derivative of $f$ in the previous problem is the matrix $$Df(x,y) = \begin{bmatrix} 2xy+3& x^2+20\cos(5y) \end{bmatrix}.$$ Some people call $Df$ the total derivative or the matrix derivative of $f$.
• The first column of this matrix is just part of the whole derivative—the part that deals with how changes in $x$ affect the output. We can get the first column by holding $y$ constant, and then differentiating with respect to $x$. We call this the partial derivative of $f$ with respect to $x$. We'll write this as $\frac{\partial f}{\partial x} = 2xy+3$ or $f_x = 2xy+3$.
• The second column of the derivative is the partial derivative of $f$ with respect to $y$—it tells us how changes in $y$ affect the output. We can get the second column by holding $x$ constant and differentiating with respect to $y$. We'll write this as $\frac{\partial f}{\partial y} = x^2+20\cos(5y)$ or $f_y = x^2+20\cos(5y)$.
• Remember, the derivative of $f$ is a matrix. The columns of the matrix are the partial derivatives with respect to the corresponding input variables—the first column is the partial derivative with respect to the first input variable, the second column with respect to the second input variable, etc.
Now we generalize the above example.
The textbook only talks about partial derivatives (see Larson 13.3). We emphasize the total derivative because it is more powerful, simpler, and helps us understand the concepts much better.
Let $f$ be a function.
• The partial derivative of $f$ with respect to $x$ is the normal first-semester calculus derivative of $f$, provided we hold every every input variable constant except $x$. We'll use the notations $$\frac{\partial f}{\partial x}, \quad \frac{\partial}{\partial x}[f], \quad f_x, \quad \text{ and }D_x f$$ to mean the partial of $f$ with respect to $x$.
• The partial of $f$ with respect to $y$, written $\ds \frac{\partial f}{\partial y}$, $\ds \frac{\partial }{\partial y}[f]$, $f_y$, or $D_yf$, is the normal first-semester calculus derivative of $f$ assuming that $y$ is the only variable changing (all other variables are constant). A similar definition holds for partial derivatives with respect to any variable.
• The derivative of $f$ is a matrix. The columns of the derivative are the partial derivatives. When there's more than one input variable, we'll use the notation $Df$ rather than $f'$ for the derivative. The order of the columns must match the order of the variables that are inputs to the function. If the function is $f(x,y)$, then the derivative is $Df(x,y) = \begin{bmatrix}\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\end{bmatrix}.$ If the function is $V(x,y,z)$, then the derivative is $DV(x,y,z) = \begin{bmatrix}\frac{\partial V}{\partial x}&\frac{\partial V}{\partial y}&\frac{\partial V}{\partial z}\end{bmatrix}.$
It's time to practice these new words in some problems. Remember, we're doing the exact same thing as we did at the start of this section. We're just using the vocabulary of differentiation.
Let's apply the problem and definition above. Fill in the blanks
1. If $f(x,y)=9-x^2-y^2$, then $df = f_xdx+f_ydy=\blank{1in}\,dx+\blank{1in}\,dy$. If we are on the surface at the point $(x,y,z)=(2,1,4)$, then the differential is $df=\blank{1cm}\,dx+\blank{1cm}\,dy$ (just plug $x=2$ and $y=1$ into the partial derivatives). If we move along the surface to $(x,y)=(2.1,1.1)$, then our change in $x$ is $\Delta x=\blank{1cm}$, our change in $y$ is $\Delta y=\blank{1cm}$, and the differential $df$ at $(x,y)=(2,1)$ estimates our change in height $\Delta f$ to be about $\Delta f\approx \blank{1cm}$ (just plug in $\Delta x$ for $dx$ and $\Delta y$ for $dy$ to get a single number).
2. If $f(x,y,z)=xy^2+yz^2$, then $$df = \blank{1in}\,dx+\blank{1in}\,dy+\blank{1in}\,dz.$$ If we are at the input vector $(x,y,z)=(1,-2,3)$, then the differential is $df = \blank{1cm}\,dx+\blank{1cm}\,dy+\blank{1cm}\,dz$. If we move to $(x,y,z)=(0.9, -2.2, 2.8)$, then the change in $x$ is $\Delta x=\blank{1cm}$, the change in $y$ is $\Delta y=\blank{1cm}$, and the change in $z$ is $\Delta z=\blank{1cm}$. The differential $df$ at $(x,y,z)=(1,-2,3)$ helps us estimate the change in $f$ to be about $\Delta f\approx \blank{1cm}$. [Hint: plug in our numeric $x$, $y$, $z$, $\Delta x$, $\Delta y$, and $\Delta z$.]
3. When a function has multiple outputs, we view the differential as a multiple-row and 1-column matrix, where each row has the partial derivative of one of the outputs. If $\vec r(t)=(3t^2, 2/t)$, then $$d\vec r = \begin{bmatrix}(3t^2)' \\ (2/t)'\end{bmatrix} dt=\begin{bmatrix}\blank{1in}\\\blank{1in}\end{bmatrix}dt.$$ If we are at $t=2$, then our differential becomes $d\vec r = \begin{bmatrix}\blank{1cm}\\\blank{1cm}\end{bmatrix}$. If we move to $t=2.1$, then the change in $t$ is $\Delta t=\blank{1cm}$ and the change in $\vec r$, as estimated by the differential $d\vec r$ at $t=2$, is approximately $\Delta \vec r\approx \begin{bmatrix}\blank{1cm}\\\blank{1cm}\end{bmatrix}$.
4. If $\vec r(u,\theta)=(u\cos(\theta), u\sin(\theta), u^2)$, then $$d\vec r = \begin{bmatrix}\blank{1in}\\\blank{1in}\\\blank{1in}\end{bmatrix}du + \begin{bmatrix}\blank{1in}\\\blank{1in}\\\blank{1in}\end{bmatrix}d\theta.$$ If we are at $(u,\theta)=(2,\pi/3)$, then the differential is $$d\vec r = \begin{bmatrix}\blank{1cm}\\\blank{1cm}\\\blank{1cm}\end{bmatrix}du + \begin{bmatrix}\blank{1cm}\\\blank{1cm}\\\blank{1cm}\end{bmatrix}d\theta.$$ If we move to $(u,\theta)=(1.9, \pi/2)$, then the change in $u$ is $\Delta u=\blank{1cm}$, the change in $\theta$ is $\Delta \theta=\blank{1cm}$, and the change in $\vec r$, as estimated by the differential $d\vec r$ at $(u,\theta)=(2,\pi/3)$, is approximately $\Delta \vec r\approx \begin{bmatrix}\blank{1cm}\\\blank{1cm}\\\blank{1cm}\end{bmatrix}$.
Use Sage to check your answers. See 14.3: 1-40 for more practice.See Larson 13.3:9–40 for more practice in doing partial derivatives. I strongly suggest you practice a lot of this type of problem until you can compute partial derivatives with ease.
Compute the requested partial and total derivatives.
1. For $f(x,y)=x^2+2xy+3y^2$, compute both $\ds\frac{\partial f}{\partial x}$ and $f_y$. Then write down $Df(x,y)$.
2. For $f(x,y,z)=x^2y^3z^4$, compute $f_x$, $\ds\frac{\partial f}{\partial y}$, and $D_z f$. Then write down $Df(x,y,z)$.
When a function has multiple outputs, its partial derivatives will have multiple components, which we write as a column vector. For example, if $f(x,y)=(3x^2, \sin(x)+xy)$, then $f_x=\left(\begin{matrix}6x\\\cos(x)+y\end{matrix}\right)$. This is similar to when we were computing derivatives of space curves, for example in this problem.
Let $\vec F(x,y)=(-y^3,2xy)$, a 2D vector field. Then $\vec F_x=\begin{bmatrix}0\\2y\end{bmatrix}$, $\vec F_y=\begin{bmatrix}-3y^2\\2x\end{bmatrix}$, and $D\vec F = \begin{bmatrix}0&-3y^2\\2y&2x\end{bmatrix}$. Also, $d\vec F=\begin{bmatrix}0&-3y^2\\2y&2x\end{bmatrix}\begin{bmatrix}dx\\dy\end{bmatrix}$. Remember that we can visualize a vector field by drawing arrows on the plane (for example, like wind velocities at every point on the plane). We can interpret this last equation as saying that if we are at a point $(x,y)$ on the plane, and we move a small distance away to $(x+dx, y+dy)$, then $d\vec F$ approximates how much the wind vector changes between the two points (i.e., how much the output of $\vec F$ changes). The derivative $D\vec F$ relates a small change in inputs (moving a small distance) to changes the outputs (the wind vector). We can then ask questions like: what direction should I move so that the velocity of the wind goes down? What direction should I move so that the wind blows more in a northern direction? If I walk in this direction, will the wind keep pushing me in the same direction? Will it get stronger or weaker?
Use Sage to check your work in each part.
Do the following for each of the functions below:
• Compute the partial derivatives and the total (matrix) derivative.
• Tell what the range and domain are (i.e., $\mathbb{R}^?\to\mathbb{R}^?$).
• Write down the relationship between small changes in inputs and outputs (in the form analagous to $d\vec y = D\vec f d\vec x$) and interpret this relationship graphically.
1. The parametric curve $\vec r(t)=(t,\cos t,\sin t)$.
2. The parametric surface $\vec r(u,v) = (u,v,u\cos(v))$. [Hint: compute partial derivatives $\vec r_u$, $\vec r_v$ and the total derivative $D\vec r(u,v)$, a 3-row by 2-column matrix].
3. The vector field $\vec F(x,y) = (-y,xe^{3y})$. [Hint: compute partial derivatives $\vec F_x$ and $\vec F_y$ and the total derivative $D\vec F(x,y)$, a 2-row by 2-column matrix.]
4. The parametric surface $\vec f(u,v)=(u^2,v^2,u-v)$.
5. The space transformation $\vec T(r,\theta,z)=(r\cos\theta, r\sin\theta, z)$.
As you completed the problems above, did you notice any connections between the size of the matrix and the size of the input and output vectors? Make sure you ask in class about this. We'll make a connection. We've now seen that the derivative of $z=f(x,y)$ is a matrix $Df(x,y) = \begin{bmatrix}f_x & f_y\end{bmatrix}$. This means that $Df$ is itself a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ that has inputs $x$ and $y$ and outputs $f_x$ and $f_y$. Therefore we can draw $Df$ as a 2d vector field.
Consider the function $f(x,y)=y-x^2$.
1. In the $xy$ plane, please draw several level curves of $f$ (maybe $z=0$, $z=2$, $z=-4$, etc.) Write the height on each curve (so you're making a contour plot).
2. Compute the derivative $Df$ (which we'll think of as a vector field).
3. We'll examine the connection between the derivative and level curves much more when we study optimization later.
Pick 8 points in the $xy$ plane that lie on the level curves you drew above. At these 8 points, add the vector given by the derivative evaluated at that point. For example, at $(0,0)$, draw the vector $Df(0,0)=(0,1)$, and at the point $(1,1)$, draw the vector $Df(1,1)=(-2,1)$. What do you observe about the relationship between the vectors and the contour lines?
# Tangent planes of $z=f(x,y)$
We promised earlier in this chapter that you can obtain most of the results in multivariate calculus by replacing the $x$ and $y$ in $dy=f'dx$ with $\vec x$ and $\vec y$. Let's review how to find the tangent line for functions of the form $y=f(x)$, and then generalize to finding tangent planes for functions of the form $z=f(x,y)$.
Consider the function $y=f(x)=x^2$.
1. The derivative is $f'(x) = ?$. At the point $x=3$ the derivative is $f'(3)=?$ and the output $y$ is $y=f(3)=?$.
2. If we move from the point $(3,f(3))$ to the point $(x,y)$ along the tangent line, then a small change in $x$ is $dx=x-3$. What is $dy$?
3. Differential notation states that a change in the output $dy$ equals the derivative times a change in the input $dx$, which gives us the equation $dy=f'(3)dx$. Replace $dx$, $dy$, and $f'(3)$ with what we know they equal, to obtain an equation $y-?=?(x-?)$. What line does this equation represent?
4. Draw both $f$ and the equation from the previous part on the same axes.
In first semester calculus, differential notation says $dy=f' dx$. A small change in the inputs times the derivative gives the change in the outputs. For the next problem, the output is $z$, and input is $(x,y)$, which means differential notation says $dz = Df \begin{bmatrix}dx\\dy\end{bmatrix}$.
See Sage for a picture.
See 14.6: 9-12 for more practice.See Larson 13.7:17–30 for more practice.
Let $z=f(x,y)=9-x^2-y^2$.
1. The derivative is $Df(x,y) = \begin{bmatrix}-2x&?\end{bmatrix}$. At the point $(x,y)=(2,1)$, the derivative is $Df(2,1) = \begin{bmatrix}-4&?\end{bmatrix}$ and the output $z$ is $z=f(2,1)=?$.
2. If we move from the point $(2,1,f(2,1))$ to the point $(x,y,z)$ along the tangent plane, then a small change in $x$ is $dx=x-2$. What are $dy$ and $dz$?
3. Explain why an equation of the tangent plane is
We'll construct a graph of $f$ and it's tangent plane in class.
$$z-4=\begin{bmatrix}-4 & -2 \end{bmatrix}\begin{bmatrix}x-2\\y-1\end{bmatrix} \quad \text{or}\quad z-4=-4(x-2)-2(y-1).$$ [Hint: What does differential notation tell us?]
Look back at the previous two problems. The first semester calculus tangent line equation, with differential notation, generalized immediately to the tangent plane equation for functions of the form $z=f(x,y)$. We just used the differential notation $dy=f'dx$ in 2D, and generalized to $dz = Df \begin{bmatrix}dx\\dy\end{bmatrix}$. Let's repeat this on another problem.
See Sage.
See 14.6: 9-12 for more practice.See Larson 13.7:17–30 for more practice.
Let $f(x,y)=x^2+4xy+y^2$. Give an equation of the tangent plane at $(3,-1)$. [Hint: Just as in the previous problem, find $Df(x,y)$, $dx$, $dy$, and $dz$. Then use differential notation.]
# Partial Derivatives of $z=f(x,y)$ functions
We can also understand tangent lines to surfaces using partial derivatives. The next problem will help you visualize what a partial derivative means in the graph of a surface.
See Sage.
See Larson 13.3:53–58 for more practice.
Consider the function $f(x,y)=9-x^2-y^2$. Construct a 3D surface plot of $f$. We'll focus on the point $(2,1)$.
1. Let $y=1$ and construct a graph in the $xz$ plane of the curve $z=f(x,1)=9-x^2-1^2$. Find an equation of the tangent line to this curve at $x=2$. Write the equation in the form $(z-z_0)=m(x-x_0)$ (find $z_0,m,x_0$). Also, find a direction vector $(1,0,?)$ for this line.
2. Let $x=2$ and construct a graph in the $yz$ plane of the curve $z=f(2,y)=9-2^2-y^2$. Find an equation of the tangent line to this curve at $y=1$. Write the equation in the form $(z-z_0)=m(y-y_0)$ (find $z_0,m,y_0$). Also, find a direction vector $(0,1,?)$ for this line.
3. Compute $f_x$ and $f_y$ and then evaluate each at $(2,1)$. What does this have to do with the previous two parts?
4. If the slope of a line $y=mx+b$ is $m$, then we know that an increase of $1$ unit in the $x$ direction will increase $y$ by $m$ units. Fill in the blanks by using the slopes of tangent lines calculated above for the function $z=f(x,y)=9-x^2-y^2$.
• Increasing $x$ by 1 unit when $y$ does not change will cause $z$ to increase by about ? units.
• Increasing $y$ by 1 unit when $x$ does not change will cause $z$ to increase by about ? units.
• Increasing $x$ by 1 unit and $y$ by 1 unit will cause $z$ to increase by about ? units.
5. In the previous part, we said that $z=9-x^2-y^2$ increased by about a certain amount each time. Why did we not say that $z=9-x^2-y^2$ increases by exactly that amount?
We'll conclude this section with a note about taking derivatives of higher orders. Since a partial derivative is a function, we can take partial derivatives of that function as well. If we want to first compute a partial with respect to $x$, and then with respect to $y$, we would use one of the following notations: $$f_{xy}=\ds\frac{\partial}{\partial y}\frac{\partial}{\partial x}f = \frac{\partial}{\partial y}\frac{\partial f}{\partial x} = \frac{\partial^2 f}{\partial y \partial x}.$$
See Larson 13.3:71–80 for more practice.
Complete the following:
1. Let $f(x,y)=3xy^3+e^{x}.$ Compute the four second partials $$\ds \frac{\partial^2 f}{ \partial x^2},\quad \ds\frac{\partial^2 f}{\partial y \partial x},\quad \ds\frac{\partial^2 f}{\partial y^2}, \quad \text{ and }\ds\frac{\partial^2 f}{\partial x \partial y}.$$
2. For $f(x,y)=x^2\sin(y)+y^3$, compute both $f_{xy}$ and $f_{yx}$.
3. Make a conjecture about a relationship between $f_{xy}$ and $f_{yx}$. Then use your conjecture to quickly compute $f_{xy}$ if $$f(x,y)=3xy^2+\tan^{2}(\cos(x)) (x^{49}+x)^{1000}.$$
Clairaut's theorem implies that if a function $f$ is “nice”—if $f$ and its partial derivatives and second partials are defined and continuous around a point $(a,b)$, then $f_{xy}=f_{yx}$ at that point. We will be dealing with nice functions of this sort in this class, so we will have this relationship between $f_{xy}$ and $f_{yx}$.
Let $z=f(x,y)=9-x^2-y^2$. We'll look at the point $(2,1)$, like above.
1. Compute $f_{xx}$, $f_{yy}$, $f_{xy}$, and $f_{yx}$ at the point $(2,1)$.
2. How do you interpret each of those second partials graphically on this function?
Let $f(x,y)=3xy^2+y\ln x$.
1. Compute $Df$, the matrix derivative.
2. We can think of $Df$ as being a function with some inputs and some outputs. How many inputs and how many outputs does $Df$ have?
3. Let $g(x,y)=Df(x,y)$. Compute $Dg$ (which is the second derivative of $f$, or $D^2f$). How does each entry in $D^2f$ relate to $f(x,y)$.
# Partial derivatives of parametric functions
Now let's examine computations similar to those in this problem in the light of parametric surfaces. With parametric functions, partial derivatives are vectors instead of just numbers. But they still represent how the outputs change relative to changes in the inputs.
See Sage for a picture.
See 16.5: 27-30 for more practice.See Larson 15.5:35–38 for more practice.
Let $z=f(x,y)=9-x^2-y^2$. We'll parameterize this function by writing $x=x, y=y, z=9-x^2-y^2$, or in vector notation we'd write $$\vec r(x,y) = (x,y,f(x,y)).$$
1. Compute $\ds \frac{\partial \vec r}{\partial x}$ and $\ds \frac{\partial \vec r}{\partial y}$. Then evaluate these partials at $(x,y)=(2,1)$. What do these vectors mean? [Hint: Draw the surface, and at the point $(2,1,4)$, draw these vectors. See the Sage plot. Think about how the vectors are telling you about how changes in the inputs are related to changes in the outputs.]
2. The vectors above are tangent to the surface. Use them to obtain a normal vector to the tangent plane, and then give an equation of the tangent plane. (You should compare it to your equation from this problem.)
If the vectors you found in the previous problem matched up with the direction vectors of the lines in this problem, you are doing things right. Partial derivatives of parametric functions tell us tangent directions. We can interpret this also in terms of motion.
Consider the change of coordinates $\vec T(r,\theta) = (r\cos \theta, r\sin \theta)$.
1. Use Sage to check your work.
Compute the partial derivatives $\ds\frac{\partial \vec T}{\partial r}$ and $\ds\frac{\partial \vec r}{\partial \theta}$, and then state the derivative $D\vec T(r,\theta)$. [Hint: $D\vec T$ is a 2 by 2 matrix, and each partial derivative is a column. Use Sage to check your answer (see the link to Sage in the margin of this problem for help with how to do this)]
2. Consider the polar point $(r,\theta) = (4,\pi/2)$:
1. Compute $T(4,\pi/2)$ (i.e., the $x,y$ coordinates for the polar point). Draw the point.
2. Compute $\ds \frac{\partial \vec T}{\partial r}(4,\pi/2)$, i.e., the partial with respect to $r$ evaluated at $r=4$, $\theta=\pi/2$. Plot this vector on the graph you drew in the previous part, starting at the point you drew.
3. Compute $\ds \frac{\partial \vec T}{\partial \theta}(4,\pi/2)$, i.e., the partial with respect to $\theta$ evaluated at $r=4$, $\theta=\pi/2$. Plot this vector on the graph you drew in the previous part, starting at the point you drew.
3. If you were standing at the polar point $(4,\pi/2)$ and someone said, “Hey you, keep your angle constant, but increase your radius,” then which direction would you move? What if someone said, “Hey you, keep your radius constant, but increase your angle”, which direction would you move?
4. Now change the polar point to $(r,\theta) = (2,3\pi/4)$. Try, without doing any computations, to repeat part 2 (at the point draw both partial derivatives vectors). Explain.
If your answers to the 2nd and 3rd part above were the same, then you're doing this correctly. The partial derivatives of parametric functions tell us about motion and tangents. The next problem reinforces this concept. But first, a short review about equations of lines.
If you know that a line passes through the point $(1,2,3)$ and is parallel to the vector $(4,5,6)$, give a vector equation, and parametric equations, of the line.
Answer: A vector equation is $\vec r(t) = (4,5,6)t+(1,2,3)$ or $\vec r(t) = (4t+1, 5t+2, 6t+3)$. Parametric equations for this line are $x=4t+1$, $y=5t+2$, and $z=6t+3$.
Consider the parametric surface $\vec r(a,t) = (a\cos t, a\sin t, t)$ for $2\leq a\leq 4$ and $0\leq t\leq 4\pi$. We encountered this parametric surface in chapter 5 when we considered a smoke screen left by multiple jets.
1. Compute the partial derivatives $\vec r_a$ and $\vec r_t$ (they are vectors), and state the total derivative. (How big is the matrix? What is the domain and range of $\vec r$?)
2. Please see Sage or Wolfram Alpha for a plot of the surface. Click on either link.
Look at a plot of the surface (use one of the links to the right). Now, suppose an object is on this surface at the point $\vec r(3,\pi) = (-3,0,\pi)$. At that point, please draw the partial derivatives $\vec r_a(3,\pi)$ and $\vec r_t(3,\pi)$.
3. If you were standing at $\vec r(3,\pi)$ and someone told you, “Hey you, hold $t$ constant and increase $a$,” then in which direction would you move? What if someone told you, “Hey you, hold $a$ constant and increase $t$”?
4. Give vector equations for two tangent lines to the surface at $\vec r(3,\pi)$. [Hint: You've got the point by plugging $(3,\pi)$ into $\vec r$, and you've got two different direction vectors from $D\vec r$. Once you have a point and a vector, we know (from chapter 2) how to get an equation of a line.]
In the previous problem, you should have noticed that the partial derivatives of $\vec r(a,t)$ are tangent vectors to the surface. Because we have two tangent vectors to the surface, we should be able to use them to construct a normal vector to the surface, and from that, a tangent plane. That's just cool.
If you know that a plane passes through the point $(1,2,3)$ and has normal vector $(4,5,6)$, then give an equation of the plane.
An equation of the plane is $4(x-1)+5(y-2)+6(y-3)=0$. If $(x,y,z)$ is any point in the plane, then the vector $(x-1,y-2,z-3)$ is a vector in the plane, and hence orthogonal to $(4,5,6)$. The dot product of these two vectors should be equal to zero, which is why the plane's equation is $(4,5,6)\cdot (x-1,y-2,z-3)=0$.
Consider again the parametric surface $\vec r(a,t) = (a\cos t, a\sin t, t)$ for $2\leq a\leq 4$ and $0\leq t\leq 4\pi$. We'd like to obtain an equation of the tangent plane to this surface at the point $\vec r(3,2\pi)$. Once you have a point on the plane, and a normal vector to the surface, we can use the concepts in chapter 2 to get an equation of the plane. Give an equation of the tangent plane. [Hint: To get the point, what is $\vec r(3,2\pi)$? The partial derivatives at $(3,2\pi)$ give us two tangent vectors. How do I obtain a vector orthogonal to both?]
See Sage.
See 16.5: 27-30 for more practice.See Larson 15.5:35–38 for more practice.
Consider the cone parametrized by $\vec r(u,v)=(u\cos v, u\sin v,u)$.
1. Give vector equations of two tangent lines to the surface at $\vec r(2,\pi/2)$ (so $u=2$ and $v=\pi/2$).
2. Give a normal vector to the surface at $\vec r(2,\pi/2)$.
3. Give an equation of the tangent plane at $\vec r(2,\pi/2)$.
We now have two different ways to compute tangent planes. One way generalizes differential notation $dy=f'dx$ to $dz = Df \begin{bmatrix}dx\\dy\end{bmatrix}$ and then uses matrix multiplication. This way will extend to tangent objects in EVERY dimension. It's the key idea needed to work on really large problems. The other way requires that we parametrize the surface $z=f(x,y)$ as $\vec r(x,y)=(x,y,f(x,y))$ and then use the cross product on the partial derivatives. Both give the same answer. The next problem has you give a general formula for a tangent plane. To tackle this problem, you'll need to make sure you can use symbolic notation. The review problem should help with this.
Joe wants to to find the tangent line to $y=x^3$ at $x=2$. He knows the derivative is $y=3x^2$, and when $x=2$ the curve passes through $8$. So he writes an equation of the tangent line as $y-8=3x^2(x-2)$. What's wrong? What part of the general formula $y-f(c) = f'(c) (x-c)$ did Joe forget?
Joe forgot to replace $x$ with $2$ in the derivative. The equation should be $y-8=12(x-2)$. The notation $f'(c)$ is the part he forgot. He used $f'(x)=3x^2$ instead of $f'(2)=8$.
Consider the function $z=f(x,y)$. Explain why an equation of the tangent plane to $f$ at $(x,y)=(a,b)$ is given by $$z-f(a,b) = \frac{\partial f}{\partial x}(a,b) (x-a) + \frac{\partial f}{\partial y}(a,b) (y-b).$$ Then give an equation of the tangent plane to $f(x,y) = x^2+3xy$ at $(3,-1)$. [Hint: Use either differential notation or a parametrization, or try both ways.]
# The Chain Rule
We'll now see how the chain rule generalizes to all dimensions. Just as before, we'll find that the first semester calculus rule will generalize to all dimensions if we replace $f'$ with the matrix $Df$. Let's recall the chain rule from first-semester calculus.
Let $x$ be a real number and $f$ and $g$ be functions of a single real variable. Suppose $f$ is differentiable at $g(x)$ and $g$ is differentiable at $x$. The derivative of $f\circ g$ at $x$ is $$(f\circ g)'(x) = \frac{d}{dx}(f\circ g)(x) = f'(g(x))\cdot g'(x).$$
Some people remember the theorem above as “the derivative of a composition is the derivative of the outside (evaluated at the inside) multiplied by the derivative of the inside.” If $u=g(x)$, we sometimes write $\ds \frac{df}{dx}=\frac{df}{du}\frac{du}{dx}$. The following problem should help us master this notation.
Suppose we know that $\ds f'(x) = \frac{\sin(x)}{2x^2+3}$ and $g(x)=\sqrt{x^2+1}$. Notice we don't know $f(x)$.
Not knowing a function $f$ is actually quite common in real life. We can often measure how something changes (a derivative) without knowing the function itself.
1. State $f'(x)$ and $g'(x)$.
2. State $f'(g(x))$, and explain the difference between $f'(x)$ and $f'(g(x))$.
3. Use the chain rule to compute $(f\circ g)'(x)$.
We now generalize to higher dimensions. If I want to write $\vec f(\vec g(\vec x))$, then $\vec x$ must be a vector in the domain of $g$. After computing $\vec g(\vec x)$, we must get a vector that is in the domain of $f$. Since the chain rule in first semester calculus states $(f(g(x))'=f'(g(x))g'(x)$, then in high dimension it should state $D(f(g(x)) = Df(g(x))Dg(x)$, the product of two matrices.
In this problem, we showed that for a circular cylinder with volume $V=\pi r^2 h$, the derivative is $$DV(r,h)=\begin{bmatrix} 2\pi rh & \pi r^2 \end{bmatrix}.$$ Suppose that the radius and height are both changing with respect to time, where $r=3t$ and $h=t^2$. We'll write this parametrically as $\vec g(t) =(3t, t^2)$ (i.e., $\vec g(t)=(r,h)$).
1. In $V=\pi r^2 h$, replace $r$ and $h$ with what they are in terms of $t$. Then compute $\dfrac{dV}{dt}$. This is a first-semester calculus derivative; we'll use it to check our work below.
2. We know $DV(r,h)=\begin{bmatrix} 2\pi rh & \pi r^2 \end{bmatrix}$ and $Dg(t)= \begin{bmatrix} 3\\ 2t \end{bmatrix}.$ In first semester calculus, the chain rule was the product of derivatives. Multiply these matrices together to get $$\dfrac{dV}{dt}=DV(g(t))\, D(r,h)(t).$$ Did you get the same answer as the first part?
3. To get the correct answer to the previous part, you had to replace $r$ and $h$ with what they equaled in terms of $t$. What part of the notation $\dfrac{dV}{dt}=DV(g(t))\, Dg(t)$ tells you to replace $r$ and $h$ with what they equal in terms of $t$?
Let's look at some physical examples involving motion and temperature, and try to connect what we know should happen to what the chain rule states.
Consider $f(x,y)=9-x^2-y^2$ and $\vec r(t)=(2\cos t, 3\sin t)$. Imagine the following scenario: a horse runs around outside in the cold. The horse's position at time $t$ is given parametrically by the elliptical path $\vec r(t)$. The function $T=f(x,y)$ gives the temperature of the air at any point $(x,y)$.
1. At time $t=0$, what is the horse's position $\vec r(0)$, and what is the temperature $f(\vec r(0))$ at that position? Find the temperatures at $t=\pi/2$, $t=\pi$, and $t=3\pi/2$ as well.
2. If you end up with an ellipse and several concentric circles, then you've done this right.
In the plane, draw the path of the horse for $t\in [0,2\pi]$. Then, on the same 2D graph, include a contour plot of the temperature function $f$. Make sure you include the level curves that pass through the points in this part, and write the temperature on each level curve you draw.
3. This idea leads to an optimization technique, Lagrange multipliers, later in the semester.
As the horse runs around, the temperature of the air around the horse is constantly changing. At which $t$ does the temperature around the horse reach a maximum? A minimum? Explain, using your graph.
4. As the horse moves past the point at $t=\pi/4$, is the temperature of the surrounding air increasing or decreasing? In other words, is $\dfrac{df}{dt}$ positive or negative? Use your graph to explain.
5. We'll complete this part in class, but you're welcome to give it a try yourself.
See Sage.
Draw the 3D surface plot of $f$. In the $xy$-plane of your 3D plot (so $z=0$) add the path of the horse. In class, we'll project the path of the horse up into the 3D surface.
Consider again $f(x,y)=9-x^2-y^2$ and $\vec r(t)=(2\cos t, 3\sin t)$, which means $x=2\cos t$ and $y=3\sin t$.
1. At the point $\vec r(t)$, we'd like a formula for the temperature $f(\vec r(t))$. What is the temperature of the horse at any time $t$? [In $f(x,y)$, replace $x$ and $y$ with what they are in terms of $t$.]
2. Compute $df/dt$ (the derivative as you did in first-semester calculus).
3. Construct a graph of $f(t)$ (use software to draw this if you like). From your graph, at what time values do the maxima and minima occur?
4. What is $df/dt$ at $t=\pi/4$?
5. Compare your work with the previous problem.
Consider again $f(x,y)=9-x^2-y^2$ and $\vec r(t)=(2\cos t, 3\sin t)$.
1. Compute both $Df(x,y)$ and $D\vec r(t)$ as matrices. One should have two columns. The other should have one column (but two rows).
2. The temperature at any time $t$ we can write symbolically as $f(r(t))$. First semester calculus suggests the derivative should be the produce $(f(\vec r(t))) ' = f'(\vec r(t))\vec r'(t)$. Write this using $D$ notation instead of prime notation.
3. Compute the matrix product $DfD\vec r$, and then substitute $x=2\cos t$ and $y=3\sin t$.
4. What is the change in temperature with respect to time at $t=\pi/4$? Is it positive or negative? Compare with the previous problem.
The previous three problems all focused on exactly the same concept. The first looked at the concept graphically, showing what it means to write $(f\circ \vec r)(t)=f(\vec r(t))$. The second reduced the problem to first-semester calculus. The third tackled the problem by considering matrix derivatives. In all three cases, we wanted to understand the following problem.
If $z=f(x,y)$ is a function of $x$ and $y$, and both $x$ and $y$ are functions of $t$ (i.e., $\vec r(t)=(x(t),y(t))$), then how do we discover how do changes in $t$ affect $f$? In other words, what is the derivative of $f$ with respect to $t$? Notationally, we seek $\ds \frac{df}{dt}$ which we formally write as $\ds \frac{d}{dt}[f(x(t),y(t))]$ or $\ds \frac{d}{dt} [f(\vec r(t))].$
To answer this problem, we use the chain rule, which is just matrix multiplication.
The Chain Rule Let $\vec x$ be a vector and $\vec f$ and $\vec g$ be functions so that the composition $\vec f(\vec g(\vec x))$ makes sense (we can use the output of $g$ as an input to $f$). Suppose $\vec f$ is differentiable at $\vec g(\vec x)$ and that $\vec g$ is differentiable at $\vec x$. Then the derivative of $\vec f\circ \vec g$ at $\vec x$ is $$D(\vec f\circ \vec g)(\vec x) = D\vec f(\vec g(\vec x))\cdot D\vec g(\vec x).$$ The derivative of a composition is equal to the derivative of the outside (evaluated at the inside), multiplied by the derivative of the inside.
This is exactly the same as the chain rule in first-semester calculus. The only difference is that now we have vectors above every variable and function, and we replaced the one-by-one matrices $f'$ and $g'$ with potentially larger matrices $Df$ and $Dg$. If we write everything in vector notation, the chain rule in all dimensions is the EXACT same as the chain rule in one dimension.
See 14.4: 1-6 for more practice. Don't use the formulas in the chapter, rather practice using matrix multiplication. The formulas are just a way of writing matrix multiplication without writing down the matrices, and only work for functions from $\RR^n\to\RR$. Our matrix multiplication method works for any function from $\RR^n\to\RR^m$.See Larson 13.5:1–6 for more practice (you can check answers in the back of the book). Don't use the formulas on pages 925–930. Instead, use matrix multiplication. The formulas are just a way of writing matrix multiplication without writing down the matrices, and only work for functions from $\RR^n\to\RR$. Our matrix multiplication method works for any function from $\RR^n\to\RR^m$.
Suppose that $f(x,y) = x^2+xy$ and that $x=2t+3$ and $y=3t^2+4$.
1. Rewrite the parametric equations $x=2t+3$ and $y=3t^2+4$ in vector form, so we can apply the chain rule. This means you need to create a function $\vec r(t) = (\blank{1in}, \blank{1in})$.
2. Compute the derivatives $Df(x,y)$ and $D\vec r(t)$, and then multiply the matrices together to obtain $\dfrac{df}{dt}$. How can you make your answer only depend on $t$ (not $x$ or $y$)?
3. The chain rule states that $D(f\circ \vec r)(t) = Df(\vec r(t))D\vec r(t)$. Explain why we write $Df(\vec r(t))$ instead of $Df(x,y)$.
If you'd like to make sure you are correct, try the following: replace $x$ and $y$ in $f=x^2+xy$ with what they are in terms of $t$, and then just use first-semester calculus to find $df/dt$. Is it the same?
See 14.4: 7-12 for more practice.See Larson 13.5:7–10 for more practice (remember to use matrix multiplication, not the formulas from the book).
Suppose $f(x,y,z) = x+2y+3z^2$ and $x=u+v$, $y=2u-3v$, and $z=uv$. Our goal is to find how much $f$ changes if we were to change $u$ (so $\partial f/\partial u$) or if we were to change $v$ (so $\partial f/\partial v$). Try doing this problem without looking at the steps below, but instead try to follow the patterns in the previous problem on your own.
1. Rewrite the equations for $x$, $y$, and $z$ in vector form $\vec r(u,v)=(x,y,z)$. If you were to graph $\vec r$, what kind of graph would you make?
2. Compute the derivatives $Df(x,y,z)$ and $D\vec r(u,v)$, and then multiply them together. Notice that since this composite function has 2 inputs, namely $u$ and $v$, we should expect to get two columns when we are done.
3. What are $\partial f/\partial u$ and $\partial f/\partial v$? [Hint: remember, each input variable gets a column.]
Let $\vec F(s,t) = (2s+t,3s-4t,t)$ and $s=3pq$ and $t=2p+q^2$. This means that changing $p$ and/or $q$ should cause $\vec F$ to change. Our goal is to find $\partial \vec F/\partial p$ and $\partial \vec F/\partial q$. Note that since $\vec F$ is a vector-valued function, the two partial derivatives should be vectors. Try doing this problem without looking at the steps below, but instead try to follow the patterns in the previous problems.
1. Rewrite the parametric equations for $s$ and $t$ in vector form.
2. Compute $D\vec F(s,t)$ and the derivative of your vector function from the previous part, and then multiply them together to find the derivative of $\vec F$ with respect to $p$ and $q$. How many columns should we expect to have when we are done multiplying matrices?
3. What are $\partial \vec F/\partial p$ and $\partial \vec F/\partial q$?
(Optional challenge) Suppose $\vec F(u,v) = (3u-v,u+2v,3v)$, $\vec G(x,y,z)=(x^2+z, 4y-x)$, and $\vec r(t) = (t^3, 2t+1, 1-t)$. We want to examine $\vec F(\vec G(\vec r(t))$. This means that $\vec F\circ \vec G\circ \vec r$ is a function from $\RR^n\to\RR^m$ for what $n$ and $m$? Similar to first-semester calculus, since we have two functions nested inside of each other, we'll just need to apply the chain rule twice. Our goal is to find $d\vec F/dt$. Try to do this problem without looking at the steps below.
1. Compute $D\vec F(u,v)$, $D\vec G(x,y,z)$, and $D\vec r(t)$.
2. Use the chain rule (matrix multiplication) to find the derivative of $\vec F$ with respect to $t$. What size of matrix should we expect for the derivative?
Suppose $f(x,y)=x^2+3xy$ and $(x,y) = \vec r(t) = (3t,t^2)$. Compute both $Df(x,y)$ and $D\vec r(t)$. Then explain how you got your answer by writing what you did in terms of partial derivatives and regular derivatives.
Answer: We have $Df(x,y) = \begin{bmatrix}2x+3y&3y\end{bmatrix}$ and $D\vec r(t) = \begin{bmatrix}3\\2t\end{bmatrix}$. We just computed $f_x$ and $f_y$, and $dx/dt$ and $dy/dt$, which gave us $Df(x,y) = \begin{bmatrix}\partial f/\partial x&\partial f/\partial y\end{bmatrix}$ and $D\vec r(t) = \begin{bmatrix}dx/dt\\dy/dt\end{bmatrix}$.
See 14.4: 13-24 for more practice. Practice these problems by using matrix multiplication. The examples problems in the text use a “branch diagram,” which is just a way to express matrix multiplication without having to introduce matrices.See Larson 13.5:7–10 for more practice.
Complete the following:
1. Suppose that $w=f(x,y,z)$ and that $x,y,z$ are all function of one variable $t$ (so $x=g(t), y=h(t), z=k(t)$). Use the chain rule with matrix multiplication to explain why $$\frac{dw}{dt} = \frac{\partial f}{\partial x}\frac{dg}{dt}+\frac{\partial f}{\partial y}\frac{dh}{dt}+\frac{\partial f}{\partial z}\frac{dk}{dt} .$$ which is equivalent to writing $$\frac{dw}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt} .$$ [Hint: Rewrite the parametric equations for $x$, $y$, and $z$ in vector form $\vec r(t) = (x,y,z)$ and compute $Dw(x,y,z)$ and $D\vec r(t)$.]
2. Suppose that $R=f(V,T,n,P)$, and that $V,T,n,P$ are all functions of $x$. Give a formula (similar to the above) for $\dfrac{dR}{dx}.$
See Larson 13.5:19–26 for more practice.
Make sure you practice problems 14.4: 13-24. Use matrix multiplication, rather than the “branch diagram” referenced in the text.See Larson 13.5:7–10 for more practice.
Suppose $z=f(s,t)$ and $s$ and $t$ are functions of $u$, $v$ and $w$. Use the chain rule to give a general formula for $\partial z/\partial u$, $\partial z/\partial v$, and $\partial z/\partial w$.
If $w=f(x,y,z)$ and $x,y,z$ are functions of $u$ and $v$, obtain formulas for $\dfrac{\partial f}{\partial u}$ and $\dfrac{\partial f}{\partial v}$.
We have $Df(x,y,z) =\begin{bmatrix}\dfrac{\partial f}{\partial x}&\dfrac{\partial f}{\partial y}&\dfrac{\partial f}{\partial z}\end{bmatrix}$. The parametrization $\vec r(u,v)=(x,y,z)$ has derivative $D\vec r =\begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\\ \dfrac{\partial z}{\partial u}&\dfrac{\partial z}{\partial v} \end{bmatrix}$. The product is $D(f(\vec r(u,v))) =\begin{bmatrix} \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial u}+ \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial u}+ \dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial u}& \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial v}+ \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial v}+ \dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial v} \end{bmatrix}$. The first column is $\dfrac{\partial f}{\partial u}$, and the second column is $\dfrac{\partial f}{\partial v}$.
You've now got the key ideas needed to use the chain rule in all dimensions. You'll find this shows up many places in upper-level math, physics, and engineering courses. The following problem will show you how you can use the general chain rule to get an extremely quick way to perform implicit differentiation from first-semester calculus.
See 14.4: 25-32 to practice using the formula you developed. To practice the idea developed in this problem, show that if $w=F(x,y,z)$ is held constant at $w=c$ and we assume that $z=f(x,y)$ depends on $x$ and $y$, then $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ and $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$. This is done on page 798 at the bottom.
See Larson 13.5:27–30 for more practice, and see pages 929–930 for how the book derives these formulas.
Suppose $z=f(x,y)$. If $z$ is held constant, this produces a level curve. As an example, if $f(x,y) = x^2+3xy-y^3$ then $5=x^2+3xy-y^3$ is a level curve. Our goal in this problem is to find $dy/dx$ in terms of partial derivatives of $f$.
1. Suppose $x=x$ and $y=y(x)$, so $y$ is a function of $x$. We can write this in parametric form as $\vec r(x) = (x,y(x))$. We now have $z=f(x,y)$ and $\vec r(x)=(x,y(x))$. Compute both $Df(x,y)$ and $D\vec r(x)$ symbolically. Don't use the function $f(x,y)=x^2+3xy-y^3$ until the last step.
2. Use the chain rule to compute $D(f(\vec r(x)))$. What is $dz/dx$ (i.e., $df/dx$)?
3. Since $z$ is held constant, we know that $dz/dx=0$. Use this fact, together with previous part, to explain why $\ds \frac{dy}{dx} = -\frac{f_x}{f_y} = -\frac{\partial f/ \partial x}{\partial f/ \partial y}$.
4. For the curve $5=x^2+3xy-y^3$, use this formula to compute $dy/dx$. |
# Math Expressions Grade 3 Student Activity Book Unit 3 Lesson 3 Answer Key Metric Units of Liquid Volume
This handy Math Expressions Grade 3 Student Activity Book Answer Key Unit 3 Lesson 3 Metric Units of Liquid Volume provides detailed solutions for the textbook questions.
## Math Expressions Grade 3 Student Activity Book Unit 3 Lesson 3 Metric Units of Liquid Volume Answer Key
Choose the Appropriate Unit
Choose the unit you would use to measure the liquid volume of each. Write mL or L.
liquid volume
milliliter (mL)
liter (L)
Question 1.
a kitchen sink ______
liquid volume
Explanation:
Liquid volume is the amount of space a liquid takes up in a container.
Liter is the basic metric unit for measuring liquid volume.
The S.I units of liter is L.
Question 2.
a soup spoon ______
milliliter mL
Explanation:
A metric unit used to measure capacity, which is equivalent to 1/1,000 of 1 liter.
Milliliters are used to measure the smaller units.
Question 3.
a teacup ______
milliliters mL
Explanation:
A metric unit used to measure capacity, which is equivalent to 1/1,000 of 1 liter.
Milliliters are used to measure the smaller units.
Question 4.
a washing machine ______
Liters L
Explanation:
Capacity of washing machine is measured in liters.
A liter is a metric unit of volume used to measure liquids.
Standard units of liter is L.
1 liter = 1000 mL
Liter is used to measure the capacity of big volumes.
Circle the better estimate.
Question 5.
a juice container 1L 1 mL
Explanation:
1 liter = 1000 mL
1 L is greater than 1 mL
So, a juice container is 1 mL
Question 6.
a bowl of soup 500L 500 mL
Explanation:
1 liter = 1000 mL
500 L is greater than 500mL.
So, a bowl of soup contains 500 mL
Use Drawings to Represent Problems
Use the drawing to represent and solve the problem.
Question 7.
There were 900 milliliters of water in a pitcher. Terri poured 500 milliliters of water into a bowl. How many milliliters of water are left in the pitcher?
400 ml
Explanation:
There were 900 milliliters of water in a pitcher.
Terri poured 500 milliliters of water into a bowl.
Total milliliters of water left in the pitcher,
900 – 500 = 400 milliliters.
Question 8.
Mr. Rojo put 6 liters of fuel into a gas can that can hold 10 liters. Then he added more liters to fill the can. How many liters of fuel did he add to the can?
4 L
Explanation:
Mr. Rojo put 6 liters of fuel into a gas can that can hold 10 liters.
Then he added more liters to fill the can.
Total liters of fuel he add to the can,
10 – 6 = 4 L
Question 9.
Shelby needs to water each of her 3 plants with 200 milliliters of water. How many milliliters of water does she need?
600 ml
Explanation:
Shelby needs to water each of her 3 plants with 200 milliliters of water.
Total milliliters of water she need for all,
3 x 200 = 600 ml
Make Sense of Problems Involving Liquid Volume
Use the drawing to represent and solve the problem.
Question 10.
The deli sold 24 liters of lemonade in 3 days. The same amount was sold each day. How many liters of lemonade did the deli sell each day?
8 liters of lemonade each day.
Explanation:
The deli sold 24 liters of lemonade in 3 days.
The same amount was sold each day.
Total liters of lemonade the deli sell each day,
24 ÷ 3 = 8 liters.
Question 11.
Tim has a bucket filled with 12 liters of water and a bucket filled with 20 liters of water. What is the total liquid volume of the buckets?
32 liters.
Explanation:
Tim has a bucket filled with 12 liters of water,
and a bucket filled with 20 liters of water.
The total liquid volume of the buckets are,
12 + 20 = 32 liters.
Question 12.
Bella made a smoothie and gave her friend 250 milliliters. There are 550 milliliters left. How many milliliters of smoothie did Sara make?
800 milliliters.
Explanation:
Bella made a smoothie and gave her friend 250 milliliters.
There are 550 milliliters left.
Total milliliters of smoothie Sara made,
250 + 550 = 800 milliliters.
Solve. Use a drawing if you need to.
Question 13.
Diane has 36 cups of lemonade to divide equally among 4 tables. How many cups should she put at each table?
9 cups.
Explanation:
Diane has 36 cups of lemonade to divide equally among 4 tables.
Total cups she put at each table,
36 ÷ 4 = 9
Question 14.
Mr. Valle filled 7 jars with his famous barbeque sauce. Each jar holds 2 pints. How many pints of sauce did he have? |
Chapter1Sets
Every mathematics, science, and engineering course uses sets, the basic building blocks of mathematics, so we start here. It might seem that we should start with numbers. However, set theory is required to do a mathematically rigorous development of the numbers as well, so sets are the best starting point.
We won't define the words “element,” “set” and “universe.” Rather, we will rely on our intuition. We will consider a set to be a collection of elements coming from some universe $\U$ of elements. For example, if we are talking about numbers, our universe might be the set of all real numbers or might be the set of all integers. If we are talking about animals, our universe might be all animals in the San Diego Zoo or it might be all animals on this planet.
Definition1.2.
If $\A$ and $\B$ are sets and every element of $\A$ is also in $\B\text{,}$ then we say that $\A$ is a subset of $\B$ and write $\A \subseteq \B$ .
Definition1.5.
The set $\R$ is the set of all real numbers.
Example1.6.
Are the sets $\dsp \A = \{ (x,y) \mid x\in \R \mbox{ and } x=\sqrt{y} \}$ and $\dsp \B = \{ (x,y) \mid x\in \R \mbox{ and } x^2 = y \}$ the same sets? What are some elements of $\A\text{?}$ Of $\B\text{?}$
Definition1.7.
The empty set is the set having no members,
\begin{equation*} \emptyset = \{x \in \U \mid x \ne x\}\text{.} \end{equation*}
We can construct new sets from old using the Set Specification Axiom.
Definition1.8.
Let $\A$ and $\B$ be sets. The intersection of $\A$ and $\B$ is the set
\begin{equation*} \A \cap \B = \{x \in \U \mid x \in \A \ \mbox{ and } \ x \in \B\}\text{.} \end{equation*}
Definition1.9.
Let $\A$ and $\B$ be sets. The union of $\A$ and $\B$ is the set
\begin{equation*} \A \cup \B = \{x \in \U \mid x \in \A \ \mbox{ or } \ x \in \B\}\text{.} \end{equation*}
Definition1.10.
Let $\B$ be a set. The complement of $\B$ is the set
\begin{equation*} \sim \B = \{x \in \U \mid x \not \in \B\}\text{.} \end{equation*}
Definition1.11.
Let $\A$ and $\B$ be sets. The difference between $\A$ and $\B$ is the set
\begin{equation*} \A \sim \B = \A \cap (\sim \B)\text{.} \end{equation*}
The following illustrations are Venn diagrams for the sets just defined.
Different expressions might represent the same set as illustrated by the next example, which together with Problem 1.12 will show that
\begin{equation*} \A \cap (\B \cup \C) = (\A \cap \B) \cup (\A \cap \C)\text{.} \end{equation*}
Example1.12.
For every choice of sets $\A, \B$ and $\C\text{,}$ $\dsp \A \cap (\B \cup \C) \subseteq (\A \cap \B) \cup (\A \cap \C)\text{.}$
For the remainder of this chapter we will assume that $\A, \B\text{,}$ and $\C$ are sets of elements from some universe $\U\text{.}$
Problem1.13.
Show that $(\A \cap \B) \cup (\A \cap \C) \subseteq \A \cap (\B \cup \C)\text{.}$
You are already familiar with operations on numbers such as addition, subtraction, multiplication, and division ($+\text{,}$ $-\text{,}$ $*\text{,}$ $\div$). We have introduced the operations on sets such as intersection, union, and difference ($\cap\text{,}$ $\cup\text{,}$ and $\sim$). Soon, we will introduce more set operations, including ×, $\oplus$ and $\circ\text{.}$ All of these number and set operations are referred to as binary operations because each operation takes two inputs.
For the next few problems, illustrate each of the following identities with Venn Diagrams and write down a proof using the Set Equality Axiom.
Problem1.14.
The Commutative Laws
1. $\displaystyle \A \cap \B = \B \cap \A$
2. $\displaystyle \A \cup \B = \B \cup \A$
Problem1.15.
The Associative Laws
1. $\displaystyle \A \cap (\B \cap \C) = (\A \cap \B) \cap \C$
2. $\displaystyle \A \cup (\B \cup \C) = (\A \cup \B) \cup \C$
Problem1.17.
The Distributive Laws
1. $\displaystyle \A \cup (\B \cap \C) = (\A \cup \B) \cap (\A \cup \C)$
2. $\displaystyle \A \cap (\B \cup \C) = (\A \cap \B) \cup (\A \cap \C)$
Problem1.18.
The Absorption Laws
1. $\displaystyle \A \cup (\A \cap \B) = \A$
2. $\displaystyle \A \cap (\A \cup \B) = \A$
For the next problem, it is helpful to make the observation about the empty set that any statement starting with “If $x \in \emptyset\text{,}$ then …” is a true statement. For example, the statement “If $x \in \emptyset\text{,}$ then $x$ is a seven-headed dog.” is a true statement. Why? Because the only way that statement could be false would be to have some value of $x$ such that $x \in \emptyset$ but $x$ was not a seven-headed dog. But there is no such $x$ since there is nothing in the empty set.
Problem1.19.
The Identity Laws
1. $\displaystyle \A \cup \emptyset = \A$
2. $\displaystyle \A \cap \U = \A$
Problem1.20.
The Inverse Laws
1. $\displaystyle \A \cup \sim \A = \U$
2. $\displaystyle \A \cap \sim \A = \emptyset$
Problem1.21.
DeMorgan's Laws
1. $\displaystyle \sim(\A \cap \B) = \sim\A \cup \sim \B$
2. $\displaystyle \sim(\A \cup \B) = \sim\A \cap \sim \B$
Definition1.22.
The symmetric difference between sets $\A$ and $\B$ is defined by
\begin{equation*} \A \oplus \B = (\A \cup \B) \sim (\A \cap \B)\text{.} \end{equation*}
We illustrate $\A \oplus \B$ via a Venn diagram.
Problem1.23.
$\A \oplus \B = (\A \sim \B) \cup (\B \sim \A)$
Example1.24.
Does
\begin{equation*} \A \oplus (\B \cup \C) = (\A \oplus \B) \cup (\A \oplus \C) ? \end{equation*}
Restated, does $\oplus$ distribute over $\cup\text{?}$
Problem1.25.
$\A \oplus (\B \oplus \C) = (\A \oplus \B) \oplus \C$
It is likely that you “know” what an ordered pair is from previous courses, but unlikely that you have ever seen a precise definition. The next two problems make precise the notion of ordered pairs.
Problem1.26.
Suppose that each of $a\text{,}$ $b\text{,}$ $c$ and $d$ is an element.
1. Suppose $\{a, b\} = \{c, d\}\text{.}$ Must it be true that $a = c$ and $b = d$ ?
2. Suppose $a = c$ and $b = d\text{.}$ Must it be true that $\{a, b\} = \{c, d\}$ ?
Definition1.27.
If each of $a$ and $b$ is an element, then by the ordered pair $(a, b)$ we mean the set $\{\{a\}, \{a, b\}\}\text{.}$ The elements $a$ and $b$ are known as the first and second coordinates of $(a,b)\text{,}$ respectively.
Problem 1.26 hinted at the property that we want from our new definition for ordered pairs. We want that $(a, b) = (c, d)$ if and only if $a = c$ and $b = d\text{.}$ The next problem allows you to show that our definition for ordered pairs, $(a, b) = \{ \{a\}, \{a, b\}\}\text{,}$ satisfies this property and therefore is a valid definition.
Problem1.28.
Suppose each of $a\text{,}$ $b\text{,}$ $c$ and $d$ is an element. Use Definition 1.27 to prove the following two statements:
1. If $(a, b) = (c, d)\text{,}$ then $a = c$ and $b = d\text{.}$
2. If $a = c$ and $b = d\text{,}$ then $(a, b) = (c, d)\text{.}$
Definition1.29.
If $\A$ and $\B$ are sets, the Cartesian product of $\A$ and $\B\text{,}$ denoted by $\A\times \B\text{,}$ consists of all ordered pairs $(x,y)$ where $x\in \A$ and $y\in \B\text{.}$ Restated,
\begin{equation*} \A\times \B = \{\,(x,y) \mid x\in \A \mbox{ and } y\in \B\,\}\text{.} \end{equation*}
Problem1.30.
Prove or give a counter-example:
\begin{equation*} \A\cap (\B\times \C) = (\A\cap \B) \times (\A\cap \C) \end{equation*} |
## 17Calculus Precalculus - Domain and Range of Logarithms
##### 17Calculus
Domain and Range of Logarithms
For this discussion, we assume that the base of the logarithm is positive. This is a common assumption but it is not always stated in textbooks and by instructors. We will use the natural logarithm here since you will use the natural log almost exclusively in calculus.
If you think about the relationship between exponents and logarithms, it looks like this.
$$x=e^y \to \ln(x) = y$$
Since $$e \approx 2.71828$$, $$e$$ is positive. So taking $$e$$ to any power $$y$$ always results in a positive number $$x$$. Therefore going in the other direction, $$x$$ must always be positive. Do you see that?
Now, what about zero? Can we take the logarithm of zero? Well, to answer that, let's look back at $$x=e^y$$. If we want $$x$$ to be zero, is there any number that we can choose for the exponent $$y$$ to get $$e^y = 0$$?
The answer is no. Even if we try very, very large negative numbers for $$y$$, $$x$$ will always be greater than zero. So in the logarithm equation, $$x$$ can never equal zero.
So, for our domain of the logarithm, we have
Domain of $$y=\ln(x)$$
$$\{ x ~|~ x ~ \in \mathbb{R} ~ and ~ x \gt 0 \}$$
Just a note on notation here. This set notation for the domain says, $$x$$ must be a real number, $$x ~ \in \mathbb{R}$$ and $$x$$ is greater than zero, $$x \gt 0$$. This notation is pretty formal. So you might see something like this $$D: x ~\in (0,\infty)$$ which implies that $$x$$ must be a real number without actually stating it. The capital 'D' stands for domain.
For the range, we have no restrictions on what we can use as the exponent $$y$$ in $$x=e^y$$, giving us $$\{ y ~|~ y ~ \in \mathbb{R} \}$$ as the range in $$y=\ln(x)$$. This can be written as $$R: y ~\in (-\infty, +\infty)$$ where the capital 'R' stands for range.
Okay, let's try some practice problems.
Practice
Unless otherwise instructed, determine the domain of these logarithmic functions.
$$f(x) = \ln(x-3)$$
Problem Statement
Find the domain of the logarithm function $$f(x) = \ln(x-3)$$
The domain of $$f(x) = \ln(x-3)$$ is $$\{ x ~|~ x ~ \in \mathbb{R} ~ and ~ x \gt 3 \}$$
Problem Statement
Find the domain of the logarithm function $$f(x) = \ln(x-3)$$
Solution
Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is
$$x-3 \gt 0$$. Solving for $$x$$, we get $$x \gt 3$$.
The domain of $$f(x) = \ln(x-3)$$ is $$\{ x ~|~ x ~ \in \mathbb{R} ~ and ~ x \gt 3 \}$$
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$$F(x) = \log_2(x^2)$$
Problem Statement
Find the domain of the logarithm function $$F(x) = \log_2(x^2)$$
The domain of $$F(x) = \log_2(x^2)$$ is $$x \neq 0$$, i.e. all real numbers except for $$x=0$$.
Problem Statement
Find the domain of the logarithm function $$F(x) = \log_2(x^2)$$
Solution
Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is $$x^2 \gt 0$$. Since $$x$$ is squared, it is always positive except when $$x=0$$. So $$x$$ can take on any value as long as it is not zero.
The domain of $$F(x) = \log_2(x^2)$$ is $$x \neq 0$$, i.e. all real numbers except for $$x=0$$.
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$$y = \log_2(5-x) + 3$$
Problem Statement
Find the domain of the logarithm function $$y = \log_2(5-x) + 3$$
Solution
### 3031 video solution
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$$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$
Problem Statement
Find the domain of the logarithm function $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$
The domain of $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$ is all real numbers that satisfy $$x \gt 10$$.
Problem Statement
Find the domain of the logarithm function $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$
Solution
Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is $$x/2 - 5 \gt 0$$. No other term in the function affects the domain. Solving for $$x$$, we have $$x/2 \gt 5 \to x \gt 10$$.
The domain of $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$ is all real numbers that satisfy $$x \gt 10$$.
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$$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$
Problem Statement
Find the domain of the logarithm function $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$
The domain of $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$ is $$x \gt -1$$.
Problem Statement
Find the domain of the logarithm function $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$
Solution
Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is $$\displaystyle{ \frac{1}{x+1} \gt 0 }$$. Since the numerator is always positive, we also need the denominator to be positive, i.e. $$x+1 \gt 0 \to x \gt -1$$. Looking back at the original function, we also have to make sure that the denominator of the fraction inside the logarithm is never zero. This is taken care of as long as $$x \neq -1$$. We are already including that in the inequality $$x \gt -1$$, so no more work needs to be done.
The domain of $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$ is $$x \gt -1$$.
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$$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$
Problem Statement
Find the domain of the logarithm function $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$
The domain of $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$ is $$x \gt 0$$ OR $$x \lt -1$$. This can also be stated as all real numbers except for $$-1 \leq x \leq 0$$.
Problem Statement
Find the domain of the logarithm function $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$
Solution
For this one, let's start with the rational expression inside the logarithm. We need to make sure we never have a zero in the denominator. So $$x \neq 0$$. Now need to make sure that the expression inside the logarithm is always positive. For rational expressions, the numerator and denominator need to be of the same sign, i.e. they are either both positive or both negative.
So for both to positive, we need $$x+1 \gt 0 \to x \gt -1$$ AND $$x \gt 0$$. The inequality $$x \gt 0$$ covers both cases.
Now for both to be negative, we need $$x+1 \lt 0 \to x \lt -1$$ AND $$x \lt 0$$. The first inequality, $$x \lt -1$$ covers both cases.
Now let's put all three cases together.
$$x \neq 0$$
$$x \gt 0$$
$$x \lt -1$$
Okay, so the first statement is covered in both of the other two, so we can ignore it. Combining the last two inequalities gives us the domain.
The domain of $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$ is $$x \gt 0$$ OR $$x \lt -1$$. This can also be stated as all real numbers except for $$-1 \leq x \leq 0$$.
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$$f(x) = \sqrt{\ln x}$$
Problem Statement
Find the domain of the logarithm function $$f(x) = \sqrt{\ln x}$$
The domain of $$f(x) = \sqrt{\ln x}$$ is $$x \geq 1$$.
Problem Statement
Find the domain of the logarithm function $$f(x) = \sqrt{\ln x}$$
Solution
Let's start with the logarithm. Since we can't take logarithms of negative numbers, we need $$x \gt 0$$. So that was pretty easy. Now we want to make sure that we don't take the square root of negative numbers. So we need $$\ln(x) > 0$$. Can this ever happen? Yes, it can since the range of a logarithm is all real numbers. So when will the logarithm be negative? By looking at a graph, we know that $$\ln 1 = 0$$ and values of $$x$$ less than $$1$$ produce negative values for the logarithm. So we need $$x \geq 1$$. Notice here that it is okay to take the square root of zero, so we include $$x = 1$$.
Okay, so putting the two inequalities together, we have $$x \gt 0$$ and $$x \geq 1$$. The first one is covered by the second one, so we just need the second one as our final answer.
The domain of $$f(x) = \sqrt{\ln x}$$ is $$x \geq 1$$.
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# How to write absolute value inequalities in interval notation
Minus two and plus four are six units apart.
### Absolute value equations and inequalities worksheet
And you really should visualize a number line when you do this, and you'll never get confused then. They're asking you for the x-values that will make the absolute-value expression greater than a negative number. Since the absolute value will always be greater than any negative number, the solution must be "all x" or "all real numbers". Or you divide both sides of this by 7, you get x is greater than or equal to 3. The something is the piece in the middle, where the variable is. It's going to be less than 12 away from 0. In this inequality, they're asking me to find all the x-values that are less than three units away from zero in either direction, so the solution is going to be the set of all the points that are less than three units away from zero. You subtract 3 from both sides. However, the number 2. Let's do a harder one. If you have something like f of x, the absolute value of f of x is less than, let's say, some number a.
And then we can draw the solution set. Example 1 Solve each of the following.
In interval notation, it would be everything between negative 12 and positive 12, and not including those numbers. Since the absolute value will always be greater than any negative number, the solution must be "all x" or "all real numbers".
## How to solve inequalities with absolute values on both sides
So in order to satisfy this thing in this absolute value sign, it has to be-- so the thing in the absolute value sign, which is 5x plus it has to be greater than negative 7 and it has to be less than 7, in order for its absolute value to be less than 7. And actually, we've solved it, because this is only a one-step equation there. This is an inequality. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero. This process can feel a bit weird, so I'll give a couple examples of how it works. So remember what I told you about the meaning of absolute value. We can draw a number line to represent the condition to be satisfied. The second method is to leave the compound inequality intact and perform solving procedures on the three parts at the same time. That translates to that, which translates to f of x greater than negative a and f of x less than a.
To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently. If you subtract 63 from both sides of this equation, you get 2x is less than negative This is an inequality.
Therefore, in this case there is no solution since it is impossible for an absolute value to be strictly less than zero i. Then, when you take its absolute value, it'll become a plus 5.
## Absolute value inequalities graph
If this thing, this 5x plus 3, evaluates anywhere over here, its absolute value, its distance from 0, will be less than 7. We solve them independently. Instead, we will mostly use the geometric definition of the absolute value: The absolute value of a number measures its distance to the origin on the real number line. Let's subtract 63 from both sides of this equation, and you get 2x-- let's see. Translate into English: we are looking for those real numbers x whose distance from the origin is less than 5 units. Oh, I just realized I made a mistake here. And then we can just solve these. This is an inequality. It's going to be less than 12 away from 0. It means how far away you are from 0.
Because the variable is contained within one interval. You have to be in this range. And I really want you to internalize this visualization here.
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# What Is The GCF Of 3 15 And 9?
## How do you find the GCF on a calculator?
Greatest Common Factor and Greatest Common Divisor The TI-84 Plus CE will find the GCF/GCD of two numbers.
Example 1: To find the GCF of 24 and 30, press math, arrow over to NUM, and select 9:gcd( —either by moving the cursor down to option 9 and pressing enter, or by simply pressing 9)..
## What’s the GCF of 3 and 2?
Multiply the circled numbers together. This is the greatest common factor! In the example, gcf(18,36,90)=2⋅3⋅3=18 gcf ( 18 , 36 , 90 ) = 2 ⋅ 3 ⋅ 3 = 18 .
## What is the GCF of 3 and 18?
We found the factors and prime factorization of 3 and 18. The biggest common factor number is the GCF number. So the greatest common factor 3 and 18 is 3.
## What is the HCF of 9 15 and 24?
The GCF is the highest number that is a factor of all three of our “test subject numbers”. Ok, now let’s find what is common. There are 2s in 24 but not in 9 or 15, so there are no 2s in our GCF. There is one 3 in 24 and in 15 and two in 9 – so one 3 is common to all three numbers.
## What is the GCF of 3 and 6?
The prime factors and multiplicities 3 and 6 have in common are: 3. 3 is the gcf of 3 and 6. gcf(3,6) = 3.
## What is the GCF of 9 12 and 15?
find gcf (9, 12, 15, . . . . , 21) = ? step 3The common prime factor or the product of all common prime factors between all integers is the highest common factor of this group of integers (9, 12, 15, 18 & 21).
## What is the GCF of 12 and 18?
In terms of numbers, the greatest common factor (gcf) is the largest natural number that exactly divides two or more given natural numbers. Example 1: 6 is the greatest common factor of 12 and 18.
## What is the GCF of 9 27?
We found the factors and prime factorization of 9 and 27. The biggest common factor number is the GCF number. So the greatest common factor 9 and 27 is 9.
## What is the GCF of 15 and 3?
Greatest common factor (GCF) of 3 and 15 is 3.
## What are the factors of 15 and 9?
The factors of 9 are 9, 3, 1. The factors of 15 are 15, 5, 3, 1. The common factors of 9 and 15 are 3, 1, intersecting the two sets above.
## What is the GCF of 3 and 9?
Greatest common factor (GCF) of 3 and 9 is 3.
## What is the GCF of 15 and 9?
Greatest common factor (GCF) of 9 and 15 is 3.
## What is the GCF of 15 and 20?
We found the factors and prime factorization of 15 and 20. The biggest common factor number is the GCF number. So the greatest common factor 15 and 20 is 5.
## What’s the GCF of 15 and 25?
5The Greatest Common Factor of 15and25 is 5. (the number itself and 1 do not count). The factor(s) of 25 is 5 . |
MAY 24, 2007
# Algebra Tiles Explained: What Do They Mean Exactly?
BY CAMILA TORRES-BERTRAND
Many people believe that algebra tiles are new in mathematics instruction, however, algebra tiles have been used in classrooms since the mid 1980s. Since then, algebra tiles have gained popularity in teaching circles and are regularly presented in conjunction with traditional methods in most textbooks. Unlike a calculator or an answer key, which does the thinking for you, algebra tiles are a tool that can guide the learner towards a greater understanding of the concepts involved with a particular skill.
Studies show that over 80% of students are visual learners rather than auditory learners. Therefore, it makes sense that the majority of students benefit from a visual, concrete representation of abstract algebraic concepts. Algebra tiles do just this ?they allow students to match a concept to a tile versus trying to imagine everything in their minds. (For more information visit our conceptual math techniques page.)
## Area: How can students use what they already know?
Algebra Tiles are created to allow students to view symbolic representations through concrete models while applying previously explored concepts of area.
By the time students are introduced to algebra, they should have already explored the concepts of area and perimeter. However, assuming as student has not solidified their understanding of area and perimeter, consider showing them this figure:
figure 1a.
When asked “What is the difference between area and perimeter?? students unfamiliar with the concepts will give responses such as “The perimeter is the outside of the rectangle, the area is the stuff inside the rectangle.? This response or similar responses provide teaching opportunities to review area/perimeter concepts and smoothly transition to algebraic concepts.
Consider this diagram which depicts figure 1a recreated with units inside. (Note: the units are square because area measure space in two dimensions: lw.) The measurement of this rectangle is now 5 units (l) by 3 units (w).
figure 1b.
Instead of simply giving them the formula of the perimeter of a rectangle P = 2l + 2w, I encouraged exploration so they could apply the definition to a variety of shapes in the future. To explore perimeter, students would simply count the sides that have a black border: 5 sides across the top, 3 sides down the right, 5 sides across the bottom and 3 sides up the left. After counting carefully, students will determine that perimeter of the figure is 16 units. Make sure that students know and understand why perimeter is a single dimensional unit ?length (l). The space it is measuring is only being measured in one direction. (For younger students, useful images and tools for measuring one-dimensional space is measuring distances using a string or toothpicks.)
As for area, it is a very different concept. This time, both length and width is being measured. (Perimeter only measured length.) Again, to encourage student thinking, I never gave them the formula for area (lw), rather I allowed the students to explore area for themselves. We ask them to count up the squares in figure 1b. and they quickly determine that there are 15 squares in the figure. If given different rectangles of varying dimensions and units, they will create the formula for themselves. Teachers may further coach the students to discover the concepts by asking them to write or journal about relationships or shortcuts they discovered within the lesson. When working with area, make sure students know and understand why area is a two-dimensional unit ?length (l) and width (w). The space area measures in being measured in two directions. Answers should be given in squared units (i.e. square feet, feet2 , square miles, miles2 , square meters, meters2 , etc.)
It has been a concern that these methodologies are time-consuming and tedious; however they show a tremendous opportunity for students to gain ownership of their own learning. Exploration can solidify this learning in a more meaningful way because they have discovered these relationships on their own. Furthermore, once students start to integrate these concepts volume (measurement in three dimensions) is innate. Teachers may further coach the students to discover the concepts by asking them to write or journal about relationships or shortcuts they discovered within the lesson. As for the formulas for perimeter and area, perhaps one of the most powerful discoveries for some students will ever discover is that multiplication is a shortcut for addition and addition is a shortcut for counting. It refutes the argument that math gets “harder?as you get progress. Rather, in this situation, multiplication (third/fourth grade) made a counting problem (kindergarten/first grade math) easier. Believe it or not, these shortcuts continue all the way into calculus and beyond!
The following is an introduction to the different types of algebra tiles in a typical algebra tile set.
## The Ones
The “ones?piece is a small square. Typically, one side is yellow and the reverse is red. The dimensions of the square are one (1) unit by one (1) unit. Therefore using the area model:
POSITIVE ONE: The positive one tile is represented by a small, yellow square. The dimensions of the square are 1 unit by 1 unit so therefore the area model would show that the small, yellow square has an area of 1 square unit. In addition to reinforcing the ideas behind the area model of a four sided figure, the notion of when a number is “squared?it actually can be represented as the shape of a square.
NEGATIVE ONE: Students need to understand the concept of negative being the opposite. When using algebra tiles, a negative one is represented by the same tile as a positive one. However, the negative is denoted by the red color of the same small square piece. When the yellow square is flipped over to its opposite side, the color is red which represents negative.
## The X s
The ?x ?piece is a rectangle. Typically one side is green and the reverse is red. The orientation of the piece in space, vertical or horizontal does not affect the significance of piece. The dimensions of the rectangle are one (1) unit by x unit(s).
Therefore using the area model:
POSITIVE X: The positive tile is represented by a green rectangle. The dimensions of the rectangle are 1 unit by an unknown length of the same unit. Since the sides do not have the same length, a rectangle is formed instead of a square.
If one were to line up the one tiles against the long side of x tile, if would not line up perfectly because the length is unknown. It is important for students to understand that there is no specific length associated with the unknown length side. That’s the point, it can be anything. However, if a one tile was lined up against the shorter side of the rectangle, it would line up perfectly because it shares the same dimensions ?one unit.
NEGATIVE X: The negative x is represented by the same tile, however just as for the ones tiles, the red color denotes the negative. When the green (positive) tile is flipped to its opposite side and displays the color red (negative) because the opposite of positive x is negative x.
## The X2
The x2 piece is a large square. Typically one side is blue and the reverse is red. The dimensions of the square are x unit(s) by unit(s).
POSITIVE X2: The positive x2 tile is represented by a large blue square. The dimensions of the blue square are one unknown length by the same unknown length with similar units
If one were to line up ones tiles against a side of the x2 tile, it would not line up perfectly because the length is supposed to be unknown. It is important for students to understand that there is no specific length associated with the unknown length side. However, if long side of the green tile was lined up against either side of the blue tile, it would line up perfectly because it shares the same unknown length. In other words, the unknown length of the blue square is the same length as the unknown length of the long side of the green rectangle.
NEGATIVE X2: The negative x2 tile is represented by the same blue square, but just as in all the previous examples, the red color denotes the negative. When the blue (positive) tile is flipped, the tile becomes red (negative) because the opposite of positive x2 is negative x2
The colors used in our explanations of algebra tiles are most commonly used in textbooks and manipulative sets sold commercially. It is important to note that previous and older versions of algebra manipulatives denoted the negative side with the color black. The positive side was usually represented using a variety of bright colors like red, yellow, green and blue. We at www.aplusalgebra.com have used the most widely-used color scheme for the A+ Algebra TabletTM used to create our lessons. |
Date: May 14, 2023
# Adding and Subtracting Fractions with Unlike Denominators Worksheet Test
Instructions: Solve the following problems and show all of your work.
## Part 1: Adding Fractions with Unlike Denominators
a. 1/3 + 2/5 + 3/4
b. 5/6 + 3/8 + 1/2
c. 2/7 + 4/9 + 3/5
d. 1/2 + 3/4 + 5/6
e. 7/8 + 2/3 + 4/5
2. Find the value of x:
a. 1/4 + x/6 = 1/3
b. 2/5 + x/10 = 3/4
c. 5/6 + x/4 = 4/5
d. 3/8 + x/4 = 5/6
e. 2/3 + x/9 = 7/12
3. Solve the following word problems:
a. John ate 1/4 of a pizza and his friend ate 2/5 of the same pizza. What fraction of the pizza did they eat together?
b. A recipe calls for 3/4 cup of flour and 1/2 cup of sugar. What is the total amount of flour and sugar needed for the recipe?
c. A field hockey team won 3/5 of their games this season and lost 2/7 of their games. What fraction of the total games did they win?
d. A pizza recipe calls for 2/3 cup of cheese and 1/4 cup of pepperoni. What is the total amount of cheese and pepperoni needed for the recipe?
e. A company spent 3/4 of their budget on marketing and 1/6 of their budget on research and development. What fraction of their budget is left?
## Part 2: Subtracting Fractions with Unlike Denominators
1. Subtract the following fractions:
a. 3/4 – 2/5
b. 5/6 – 1/3
c. 2/3 – 1/4
d. 3/5 – 1/8
e. 7/8 – 3/5
2. Find the value of x:
a. x/4 – 1/6 = 1/3
b. x/5 – 2/3 = 1/4
c. x/6 – 2/5 = 1/3
d. x/7 – 3/4 = 2/5
e. x/8 – 3/10 = 1/2
3. Solve the following word problems:
a. A pizza recipe calls for 3/4 cup of cheese and you only have 1/3 cup of cheese. How much more cheese do you need?
b. A recipe calls for 1/2 cup of milk and you only have 1/4 cup of milk. How much more milk do you need?
c. If you have 5/6 of a cake and you eat 1/4 of it, how much cake is left?
d. A recipe calls for 1/3 cup of sugar and you only have 1/8 cup of sugar. How much more sugar do you need?
e. If you have 2/3 of a bottle of soda and you pour out 1/4 of it, how much soda is left?
## Part 3: Adding and Subtracting Fractions with Unlike Denominators
1. Simplify the following fractions:
a. 4/9 + 1/6 – 2/3
b. 2/5 – 1/6 + 3/4
c. 5/6 – 2/3 + 1/4
d. 3/8 + 1/4 – 5/6
e. 1/2 – 1/3 + 1/4
2. Find the value of x:
a. 1/3 + x/4 – 1/6 = 5/12
b. x/5 – 2/3 + 1/4 = 1/2
c. 5/6 + x/4 – 2/5 = 1/3
d. 1/2 – x/6 + 1/4 = 3/8
e. 2/3 – x/5 + 1/8 = 1/4
3. Solve the following word problems:
a. Jane had 1/2 of a cake and she gave 1/4 of it to her friend. What fraction of the cake does Jane have left?
b. A recipe calls for 1/3 cup of flour, 1/4 cup of sugar, and 1/2 cup of milk. If you only have 1/6 cup of flour, 1/8 cup of sugar, and 1/4 cup of milk, how much more of each ingredient do you need?
c. A company spent 1/3 of their budget on salaries, 1/4 of their budget on rent, and 1/6 of their budget on supplies. What fraction of their budget is left?
d. A recipe calls for 2/3 cup of flour, 1/4 cup of sugar, and 1/2 cup of milk. If you have 3/4 cup of flour, 1/8 cup of sugar, and 1/3 cup of milk, how much of each ingredient do you have left?
e. John had 7/8 of a candy bar and he gave 1/4 of it to his friend. What fraction of the candy bar does John have left? |
SOLVING WORD PROBLEMS BASED ON TWO DIGIT NUMBER
Solving Word Problems Based on Two Digit Number :
We consider two digit number as xy. Here y is the number at the unit place and x is the number at the ten's place.
By writing xy in the expanded form, we get
xy = 10x + 1y
Example 1 :
A two digit number is four times the sum of its digits and twice the product of the digits. Find the number.
Solution :
Let xy be the required two digit number.
xy = 4(x + y)
10x + y = 4x + 4y
10x - 4x + y - 4y = 0
6x - 3y = 0
2x - y = 0
y = 2x-----------(1)
xy = 2 ⋅ x ⋅ y
10x + 1y = 2xy -----------(2)
Applying (1) in (2), we get
10x + 1(2x) = 2x(2x)
10x + 2x = 4x2
12x = 4x2
x = 12/4
x = 3
By applying the value of x in (1), we get
y = 2(3)
y = 6
Hence the required two digit number is 36.
Example 2 :
A two digit number such that the product of its digits is 21. When 36 is subtracted from the number the digits are interchanged. Find the number.
Solution :
Let xy be the required two digit number.
A two digit number such that the product of its digits is 21
⋅ y = 21
y = 21/x ---- (1)
When 36 is subtracted from the number the digits are interchanged
xy - 36 = yx
10x + y - 36 = 10y + x
10x - x + y - 10 y = 36
9x - 9y = 36
Dividing it by 9, we get
x - y = 4 --- (2)
By applying (1) in (2), we get
x - (21/x) = 4
x2 - 21 = 4x
x- 4x - 21 = 0
x2 - 7x + 3x - 21 = 0
x(x - 7) + 3(x - 7) = 0
(x - 7) (x + 3) = 0
x - 7 = 0x = 7 x + 3 = 0x = -3
By applying the value of x in (1), we get
y = 21/x
y = 21/7
y = 3
Therefore the required two digit number is 73.
Example 3 :
A two digit number is such that the product of its digits is 12. When 36 is added to this number the digits are interchanged. Find the numbers.
Solution :
Let xy be the required two digit number
A two digit number such that the product of its digits is 12
x ⋅ y = 12
y = 12/x ---- (1)
When 36 is added to the number the digits are interchanged
xy + 36 = yx
10x + y + 36 = 10y + x
10x - x + y - 10y = -36
9x - 9y = -36
Dividing it by 9, we get
x - y = -4 ---- (2)
By applying (1) in (2)
x - (12/x) = -4
x2 - 12 = -4 x
x+ 4x - 12 = 0
(x - 2) (x + 6) = 0
x - 2 = 0x = 2 x + 6 = 0x = -6
By applying the value of x in (1), we get
y = 12/x
y = 12/2
y = 6
Therefore the required two digit number is 26.
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Decimal representation of rational numbers
Finding square root using long division
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Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
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Resources tagged with Working systematically similar to Crossed Ends:
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There are 128 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
Sticky Numbers
Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
Maths Trails
Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
9 Weights
Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
Consecutive Numbers
Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Weights
Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
Consecutive Negative Numbers
Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
Special Numbers
Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
Number Daisy
Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
First Connect Three for Two
Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
More Magic Potting Sheds
Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
Summing Consecutive Numbers
Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
Making Maths: Double-sided Magic Square
Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
Two and Two
Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
First Connect Three
Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
Twinkle Twinkle
Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
Ben's Game
Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
Oranges and Lemons, Say the Bells of St Clement's
Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
Games Related to Nim
Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
Problem Solving, Using and Applying and Functional Mathematics
Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
Where Can We Visit?
Stage: 3 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
Tetrahedra Tester
Stage: 3 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
Pair Sums
Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
Squares in Rectangles
Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
Plum Tree
Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
Stage: 3 and 4 Challenge Level:
This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set.
Teddy Town
Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
Factor Lines
Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
Tea Cups
Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
LCM Sudoku II
Stage: 3, 4 and 5 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
Warmsnug Double Glazing
Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
Colour Islands Sudoku 2
Stage: 3, 4 and 5 Challenge Level:
In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9.
Counting on Letters
Stage: 3 Challenge Level:
The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern?
You Owe Me Five Farthings, Say the Bells of St Martin's
Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
Product Sudoku
Stage: 3 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
Crossing the Bridge
Stage: 3 Challenge Level:
Four friends must cross a bridge. How can they all cross it in just 17 minutes?
Bochap Sudoku
Stage: 3 and 4 Challenge Level:
This Sudoku combines all four arithmetic operations.
A First Product Sudoku
Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
Stage: 3 Challenge Level:
Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".
Intersection Sudoku 1
Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
Stage: 3 and 4 Challenge Level:
Four small numbers give the clue to the contents of the four surrounding cells.
Fence It
Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
Peaches Today, Peaches Tomorrow....
Stage: 3 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
The Great Weights Puzzle
Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
Difference Sudoku
Stage: 4 Challenge Level:
Use the differences to find the solution to this Sudoku.
More on Mazes
Stage: 2 and 3
There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.
Colour Islands Sudoku
Stage: 3 Challenge Level:
An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.
Stage: 3 Challenge Level:
You need to find the values of the stars before you can apply normal Sudoku rules. |
Home Explore MAPLE G05 INTEGRATED TEXTBOOK_Combine
MAPLE G05 INTEGRATED TEXTBOOK_Combine
Description: MAPLE G05 INTEGRATED TEXTBOOK_Combine
Let us see a few examples. Example 18: a) Find the sum of 173.809 and 23.617. b) Subtract 216.735 from 563.726. Solution: a) b) 12 16 11 5 2/ 6/ 12 17 3 . 8 0 9 5 6/ 3/ . 7/ 2/ 6 + 23 . 617 –2 1 6 . 7 3 5 197 . 426 346 . 991 Example 19: Solve: a) 294.631 + 306.524 b) 11.904 – 6.207 Solution: a) 1 1 1 b) 11 8 9 14 294 . 631 1/ 1/ . 9/ 0/ 4/ +3 0 6 . 5 2 4 – 6 . 20 7 601 . 155 5 . 69 7 15 1/7/2019 3:24:19 PM NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 56
Chapter Decimals - II 12 Let Us Learn About • multiplying and dividing decimals by 1-digit and 2-digit numbers. • m ultiplying decimals by 10, 100 and 1000. • multiplying and dividing a decimal number by another decimal number. • the relationship between percentages, decimals and fractions. Concept 12.1: Multiply and Divide Decimals Think Pooja bought six different types of toys for ` 236.95 each. She calculated the total cost and paid the amount to the shopkeeper. Pooja then went to a sweet shop where 410.750 kg of a sweet was prepared. She wanted to know the number of 250 g packs that can be made from it. Do you know how to find the total cost of the toys? Can you calculate how many packs of sweets can be made? Recall Multiplication and division of decimal numbers are similar to that of usual numbers. Let us recall multiplication and division of numbers by answering the following. Solve: a) 267 × 14 b) 3218 × 34 c) 7424 × 14 d) 576 ÷ 12 e) 265 ÷ 5 f) 384 ÷ 4 19 NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 60 1/7/2019 3:24:19 PM
& Remembering and Understanding Multiplication of decimals is similar to multiplication of numbers. When two decimal numbers are multiplied, a) count the total number of digits after decimal point in both the numbers. Say it is ‘n’. b) multiply the two decimal numbers as usual and place the decimal point in the product after ‘n’ digits from the right. Multiply decimals by 1-digit and 2-digit numbers Let us understand the multiplication of decimals through a few examples. Example 1: Solve: a) 25.146 × 23 b) 276.32 × 6 Solution: a) 25.146 × 23 T Th Th H TO 1 1 1 46 11 23 25 1 38 20 × 58 1 + 75 4 502 9 5 7 8.3 Therefore, 25.146 × 23 = 578.358 b) 276.32 × 6 Step 1: To multiply the given numbers, follow the steps outlined here. Multiply the numbers without considering the decimal point. T Th Th H T O 4 3 11 2 7 6 32 ×6 1 6 5 7 92 Decimals - II 20 NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 61 1/7/2019 3:24:19 PM
Step 2: Count the number of decimal places in the given number. The number of decimal places in 276.32 is two. Step 3: Count from the right, the number of digits in the product as the number of decimal places in the given number. Then place the decimal point. Therefore, 276.32 × 6 = 1657.92. Multiply decimals by 10,100 and 1000 Example 2: Solve: a) 3.4567 × 10 b) 3.4567 × 100 c) 3.4567 × 1000 Solution: To multiply a decimal number by 10, 100 and 1000, follow the steps. Step 1: Write the decimal number as it is. Step 2: Shift the decimal point to the right by as many digits as the number of zeros in the multiplier. Therefore, a) 3 .4567 × 10 = 34.567 (The decimal point is shifted to the right by one digit as the multiplier is 10.) b) 3.4567 × 100 = 345.67 (The decimal point is shifted to the right by two digits as the multiplier is 100.) c) 3 .4567 × 1000 = 3456.7 (The decimal point is shifted to the right by three digits as the multiplier is 1000.) Multiply a decimal number by another decimal number Multiplication of a decimal number by another decimal number is similar to multiplication of a decimal number by a number. Let us understand this through an example. Example 3: Solve: 7.12 × 3.7 Solution: Step1: Multiply the numbers without considering the decimal point. 1 7 12 × 37 11 4 9 84 +2 1 3 6 0 26 3 44 21 1/7/2019 3:24:19 PM NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 62
Step 2: Count the number of decimal places in both the multiplicand and the multiplier and add them. The number of decimal places in 7.12 is two The number of decimal places in 3.7 is three Total number of decimal places = 2 + 1 = 3 Step 3: Count as many digits in the product from the right as the total number of decimal places. Then place the decimal point. Therefore, 7.12 × 3.7 = 26.344. Divide decimal numbers by 1-digit and 2-digit numbers Division of decimal numbers is similar to the division of usual numbers. Let us understand this through a few examples. Example 4: Divide: a) 147.9 ÷ 3 b) 64.2 ÷ 6 Solution: Step 1: Follow the steps to divide. Divide the decimal number (dividend) by the 1-digit number (divisor) as usual. Step 2: Place the decimal point in the quotient exactly above the decimal point in the dividend. a) 49.3 b) 10.7 )3 147.9 )6 64.2 −12 −6 27 042 − 27 − 42 09 00 − 09 Example 5: 00 b) 56.96 ÷ 32 Divide: a) 20.475 ÷ 25 Solution : a) 0.819 b) 1.78 )25 20.475 )32 56.96 − 200 − 32 47 249 − 25 − 224 225 256 − 225 − 256 000 000 Decimals - II 22 NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 63 1/7/2019 3:24:19 PM
Divide decimals by 10,100 and 1000 Example 6: Solve: a) 3.4567 ÷ 10 b) 3.4567 ÷ 100 c) 3.4567 ÷ 1000 Solution: To divide a decimal number by 10, 100 and 1000, follow these steps: Step 1: Write the decimal number as it is. Step 2: Shift the decimal point to the left by as many digits as the number of zeros in the divisor. Therefore, a) 3.4567 ÷ 10 = 0.34567 (The decimal point is shifted to the left by one digit as the divisor is 10.) b) 3.4567 ÷ 100 = 0.034567 (The decimal point is shifted to the left by two digits as the divisor is 100.) c) 3.4567 ÷ 1000 = 0.0034567 (The decimal point is shifted to the left by three digits as the divisor is 1000.) Use decimals to continue division of numbers resulting in remainders Recall that sometimes we get remainders in the division of numbers. We can use the decimal point to divide the remainder up to the desired number of decimal places. Let us understand this through a few examples. Example 7: Solve: 54487 ÷ 46 Solution: To divide the given numbers, follow the steps given here. Step 1: Divide as usual, till you get a remainder. 1184 )46 54487 − 46 84 − 46 388 − 368 207 − 184 23 Step 2: Place a point to the right of the quotient. Add a zero to the right of the remainder and continue the division. 23 1/7/2019 3:24:19 PM NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 64
1184.5 )46 54487 − 46 84 − 46 388 − 368 207 − 184 230 − 230 000 In this case, the division is stopped after one decimal place as the remainder is zero. In some cases, the division continues for more than three decimal places. But usually, we divide up to three decimal places. We then round off the quotient to two decimal places. Example 8: Divide the following up to two decimal places. a) 91158 ÷ 28 b) 78323 ÷ 15 Solution: a) 3255.642 b) 5221.533 )28 91158 )15 78323 − 84 − 75 71 33 − 56 − 30 155 32 − 140 − 30 158 23 − 140 − 15 180 80 − 168 − 75 120 50 − 112 − 45 80 50 − 56 − 45 4 5 Therefore, 91158 ÷ 28 = 3255.64 and 78323 ÷ 15 = 5221.53 after rounding off to two decimal places. Decimals - II 24 NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 65 1/7/2019 3:24:19 PM
Divide a decimal number by another decimal number Let us understand the division of a decimal number by another through an example. Example 9: Solve: 3.0525 ÷ 5.5 Solution: To divide a decimal number by another, follow these steps. Step 1: Convert the decimals into fractions. Step 2: Step 3: 3.0525 = 30525 and 5.5 = 55 Step 4: 10000 10 Find the reciprocal of the divisor. 55 10 Reciprocal of is . 10 55 Multiply dividend by the reciprocal of divisor. 30525 10 30525 555 × = = 10000 55 55´1000 1000 Convert the fraction to a decimal number. 555 = 0.555 1000 Therefore, 3.0525 ÷ 5.5 = 0.555 25 1/7/2019 3:24:19 PM NR_BGM_9789386663511 MAPLE G05 INTEGRATED TEXTBOOK TERM 3_Text.pdf 66 |
# HSPT Math : How to simplify expressions
## Example Questions
← Previous 1
### Example Question #564 : Problem Solving
You are given that are whole numbers.
Which of the following is true of if and are both odd?
is always even.
None of the other statements are true.
is always odd if is odd, and always even if is even.
is always odd.
is always odd if is even, and always even if is odd.
is always odd if is even, and always even if is odd.
Explanation:
If is odd, then is odd, since the product of two odd whole numbers must be odd. When the odd number is added, the result, , is even, since the sum of two odd numbers must be even.
If is even, then is even, since the product of an odd number and an even number must be even. When the odd number is added, the result, , is odd, since the sum of an odd number and an even number must be odd.
### Example Question #1 : How To Simplify Expressions
Simplify the expression:
Explanation:
Combine all the like terms.
The terms can be combined together, which gives you .
When you combine the terms together, you get .
There is only one term so it doesn't get combined with anything. Put them all together and you get
.
### Example Question #11 : How To Simplify An Expression
Simplify the following expression:
Explanation:
First distribute the 2:
Combine the like terms:
### Example Question #2 : How To Simplify Expressions
Simplify the expression:
x
x + 1
2x + 1
2x
x+ 2x + 1
x + 1
Explanation:
Factor out a (2x) from the denominator, which cancels with (2x) from the numerator. Then factor the numerator, which becomes (+ 1)(+ 1), of which one of them cancels and you're left with (+ 1).
### Example Question #21 : Simplifying Expressions
Simplify the following expression: x3 - 4(x2 + 3) + 15
Explanation:
To simplify this expression, you must combine like terms. You should first use the distributive property and multiply -4 by x2 and -4 by 3.
x3 - 4x2 -12 + 15
You can then add -12 and 15, which equals 3.
You now have x3 - 4x2 + 3 and are finished. Just a reminder that x3 and 4x2 are not like terms as the x’s have different exponents.
### Example Question #22 : Simplifying Expressions
Simplify the following expression:
2x(x2 + 4ax – 3a2) – 4a2(4x + 3a)
12a– 22a2x + 8ax2 + 2x3
–12a– 14ax2 + 2x3
–12a3 – 14a2x + 2x3
–12a– 22a2x + 8ax2 + 2x3
12a– 16a2x + 8ax2 + 2x3
–12a– 22a2x + 8ax2 + 2x3
Explanation:
Begin by distributing each part:
2x(x2 + 4ax – 3a2) = 2x * x2 + 2x * 4ax – 2x * 3a2 = 2x3 + 8ax2 – 6a2x
The second:
–4a2(4x + 3a) = –16a2x – 12a3
Now, combine these:
2x3 + 8ax2 – 6a2x – 16a2x – 12a3
The only common terms are those with a2x; therefore, this reduces to
2x3 + 8ax2 – 22a2x – 12a3
This is the same as the given answer:
–12a– 22a2x + 8ax2 + 2x3
### Example Question #1 : How To Simplify Expressions
Which of the following does not simplify to ?
All of these simplify to
Explanation:
5x – (6x – 2x) = 5x – (4x) = x
(x – 1)(x + 2) - x2 + 2 = x2 + x – 2 – x2 + 2 = x
x(4x)/(4x) = x
(3 – 3)x = 0x = 0
### Example Question #1 : How To Simplify Expressions
Simplify:
Explanation:
In order to simplify this expression, distribute and multiply the outer term with the two inner terms.
### Example Question #2 : How To Simplify Expressions
Simplify:
Explanation:
When the same bases are multiplied, their exponents can be added. Similarly, when the bases are divided, their exponents can be subtracted. Apply this rule for the given problem.
Simplify: |
# How to algebraically solve $\frac{1}{x} < 5$ inequality to obtain two solutions? [duplicate]
Given the inequality:
$\frac{1}{x} < 5$
In order to find a solution, I would normally multiply both sides by $x$:
$1 < 5x$
Then I would divide by $5$
$\frac{1}{5} < x$
To obtain the solution: $x > \frac{1}{5}$.
Now, the thing is, the solutions are actually two: $x > \frac{1}{5}$ and $x < 0$
How am I supposed to reach this conclusion algebraically? It seems I'm not able to obtain the second solution ($x < 0$).
Thanks!
## marked as duplicate by Foobaz John, GNUSupporter 8964民主女神 地下教會, Rohan, JonMark Perry, kingW3Dec 27 '17 at 8:08
• When you're multiplying by $x$, the inequality sign changes depending on the sign of $x$! – Matija Sreckovic Dec 26 '17 at 19:19
Note that by going from $\dfrac{1}{x} < 5$ to $1 < 5x$, you are assuming that $x > 0$. You see that by assuming $x > 0$, you obtain $x > \dfrac{1}{5}$.
Now assume that $x < 0$. Then $\dfrac{1}{x} < 5 \implies 1 > 5x\implies x < \dfrac{1}{5}$ (flipping the inequality). But, remember, we assumed that $x < 0$. Thus, if $x < \dfrac{1}{5}$ and $x < 0$, we can succinctly write this as $x < 0$.
If you multiply with $x^2$ you get $x<5x^2$ so $x(5x-1)>0$ so $$x\in (-\infty, 0)\cup ({1\over 5},\infty)$$
$$\frac { 1 }{ x } <5\quad \Rightarrow \frac { 1 }{ x } -5<0\quad \Rightarrow \quad \frac { 1-5x }{ x } <0\quad \Rightarrow \frac { x\left( 1-5x \right) }{ { x }^{ 2 } } <0\\ x\left( 5x-1 \right) >0\quad \Rightarrow x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 5 } ,+\infty \right)$$
you must do case work, if we assume $x>0$ then we get $$\frac{1}{5}<x$$ in the other case $$x<0$$ we get $$\frac{1}{5}>x$$
For $x>0$
$$\frac{1}{x} < 5\implies x>\frac15$$
For $x<0$
$$\frac{1}{x} < 5\implies x\frac{1}{x} > 5x\implies x<\frac15$$ |
# When To Use Absolute Value (3 Things To Know)
Absolute value is used throughout mathematics and in other disciplines as well, including physics and finance. However, it still helps to know exactly when it might be used, along with how to solve the equations or inequalities that arise.
So, when do you use absolute value? Absolute value is used to express the distance between two numbers on a number line. It is also used to find the speed of an object and the distance it travels over time. Absolute value can express a range of values, and we can use it to compute percent error.
Of course, absolute value can never be negative, so we would not use it if we want to know whether an estimate is above or below the actual value.
In this article, we’ll talk about absolute value and when to use it. We’ll also look at some examples of how to solve absolute value equations and inequalities.
Let’s get started.
## When To Use Absolute Value
Absolute value has many uses in mathematics and in various disciplines, including:
• Distance from zero on a number line: |x| = |x – 0| is the distance from zero to x (in units).
• Distance between two points on a number line: |x – y| is the distance from x to y (in units).
• Speed of an object: graph the absolute value of velocity over time to get the graph of speed.
• Distance traveled by an object: graph the absolute value of velocity over time, then find the total area under curve and above the x-axis to get the distance.
• Range of Values: |x| <= 4 is compact way to express the range of value -4 <= x <= 4 or the interval [-4, 4].
• Error: comparing actual vs. expected values, or to see how far off we are, without regard to sign (absolute error or percent error).
Note: in Excel or Google sheets, the notation for the absolute value function is =ABS(x), where x is a number.
### Absolute Value For Distance On A Number Line
We can think of absolute value as a distance function with two inputs, given by:
• D(x, y) = |x – y| = distance between x and y on a number line
If x and y have the same value (x = y), then the distance between them is zero, since:
• D(x, y) = D(x, x) = |x – x| = 0
Also, if y = 0, then the distance function is simply the absolute value of x, since:
• D(x, y) = D(x, 0) = |x – 0| = |x|
### Absolute Value For Distance & Speed Of An Object
We can also use absolute value to find an object’s speed and the distance it has traveled in a time interval.
First, remember that speed is the absolute value of velocity (unsigned velocity). That is, speed tells us how fast an object is moving, without regard to the direction of the movement.
The graphs below show velocity and speed of an object.
To find displacement of an object, we find the area between the velocity curve and the x-axis. We count areas above the x-axis as positive, and areas below the x-axis as negative.
So, an object could have zero displacement if it moves 10 feet north from its starting position and then moves 10 feet south back to the starting position (10 + -10 = 0 feet of displacement).
However, the distance traveled would be |10| + |-10| = 10 + 10 = 20 feet. We can also see this by finding the area under the speed curve.
### Absolute Value For A Range Of Values
If we wish to express a range of values (interval), we can use absolute value instead of a compound inequality. For example, we can use |x| <= 4 instead of -4 <= x <= 4 or, as an interval, [-4, 4].
The general formula for the range of values C – R <= x <= C + R is:
• |x – C| <= R
where C is the center of the interval and R is the “radius” (half the width of the interval).
If the interval is [a, b], then we have:
• C = (a + b) / 2 [the center of an interval is the average of the endpoints]
• R = (b – a) / 2 [the radius of an interval is half of the width of the interval]
#### Example: Express A Range Of Values Using Absolute Value
Let’s say we want to express the interval [-6, 10] or -6 <= x <= 10 using absolute values. Then we have:
• C = (a + b) / 2 = (10 + -6) / 2 = 4 / 2 = 2
• R = (b – a) / 2 = (10 – (-6)) / 2 = 16 / 2 = 8
So, the center of the interval is 2, and the radius is 8 (the width is 16).
The inequality is:
• |x – C| <= R
• |x – 2| <= 8
We can check this by solving:
• |x – 2| <= 8
• -8 <= x – 2 <= 8
• -8 + 2 <= x <= 8 + 2
• -6 <= x <= 10
### Absolute Value For Error
We use absolute value when we want to find out how much an actual value differs from an expected value, without regard to sign.
For example, let’s say we are predicting the sale price of houses that go up for sale in a neighborhood. The following table shows the actual values (sales prices), our predicted values (expected), the differences (absolute value), and the percent error.
Note: in some cases, it will matter whether the actual value is above or below the expected value. In these cases, it is not appropriate to use absolute value.
For example, let’s say we want to find out if our estimates are high or low. If we often or always get a negative value for actual minus expected, then our home price predictions are too high.
In that case, we would want to revise our projections downward.
## How Do You Know When To Use Absolute Value?
The wording in a problem often gives us context and indicates when to use absolute value. The following vocabulary can tell us when to use absolute value:
• Distance
• Speed
• Range
• Error (Actual vs Expected)
• Percent Error
## How Do You Solve Problems With Absolute Value?
Solving problems with absolute value often involves two separates “halves”: the positive half and the negative half. For example, there are two solutions to |x| = 2: x = +2 and x = -2.
This means that an absolute value that is equal to a positive number will be an equation with two solutions.
Let’s take a look at how to solve absolute value problems, starting with equations.
### Solving Absolute Value Equations
Solving absolute value equations usually involves two halves: the positive and negative halves. However, there are some special cases where there is only one half (when the absolute value is equal to zero).
There are also cases when there is no solution (when the absolute value is equal to a negative number). Let’s see some examples.
#### Example 1: Solving An Absolute Value Equation With Two Solutions
Consider the absolute value equation
• |x – 2| = 5
Since |-5| = 5 and |5| = 5, there are two “halves”, or solutions:
x – 2 = -5 and x – 2 = 5
x = -3 and x = 7
#### Example 2: Solving An Absolute Value Equation With One Solution
Consider the absolute value equation
• |x + 4| = 0
Since only |0| = 0, there is only one solution:
• x + 4 = 0
• x = -4
#### Example 3: An Absolute Value Equation With No Solution
Consider the absolute value equation
• |x – 1| = -7
Since an absolute value is always nonnegative (either positive or zero), there is no way that an absolute value can be negative. So, there is no solution to this equation.
### Solving Absolute Value Inequalities
Solving absolute value inequalities usually involves two halves: the positive and negative halves. However, there are also cases when there is no solution. Let’s see some examples.
#### Example 1: Solving An Absolute Value Inequality With Two “Halves”
Consider the absolute value inequality
• |x – 3| <= 4
We have to consider two halves for the solution simultaneously:
• -4 <= x – 3 <= 4
• -4 + 3 <= x – 3 + 3 <= 4 + 3
• -1 <= x <= 7
So, the solution is the interval [-1, 7].
#### Example 2: Solving An Absolute Value Inequality With Two “Halves”
Consider the absolute value inequality
• |x + 1| >= 6
We have to consider two halves for the solution simultaneously:
• x + 1 >= 6 or x + 1 <= -6
• x + 1 – 1 >= 6 – 1 or x + 1 – 1 <= -6 – 1
• x >= 5 or x <= -7
So the solution is the compound inequality x >= 5 or x <= -7.
#### Example 3: Absolute Value Inequality With No Solution
Consider the absolute value inequality
• |x – 3| <= -2
Since an absolute value is always nonnegative, there is no way that the left side of this equation (an absolute value) can be less than -2.
So, there is no solution.
### Graphing Absolute Value Functions
Graphing absolute value functions is easier after converting to one or more piecewise functions and graphing from there.
#### Example 1: Graphing An Absolute Value Function
Consider the absolute value function f(x) = |x|.
As a piecewise function, we have:
• +x for x >= 0
• -x for x < 0
The graph is pictured below.
Note that this graph is always on or above the x-axis.
#### Example 2: Graphing An Absolute Value Function
Consider the absolute value function f(x) = |2x – 1|.
As a piecewise function, we have:
• 2x – 1 for x >= 1/2
• 1 – 2x for x < 1/2
The graph is pictured below.
Note that this graph is always on or above the x-axis.
### How Do You Solve Absolute Value Equations With Two Absolute Values?
To solve an absolute value equation with two (or more) absolute values, convert the absolute value functions to piecewise functions. Then, combine them, and graph the resulting “master” piecewise function to see where the solutions lie.
#### Example: Graphing An Absolute Value Equation With Two Absolute Values
Consider the absolute value equation
• |x – 2| + |2x – 6| = 4
For the first part, |x – 2|, the piecewise graph is:
• x – 2, x >= 2
• 2 – x, x < 2
Its graph is pictured below:
For the second part, |2x – 6|, the piecewise graph is:
• 2x – 6, x >= 3
• 6 – 2x, x < 3
Its graph is pictured below:
Taking the sum of these two, we get:
• 8 – 3x, x < 2
• 4 – x, 2 <= x < 3
• 3x – 8, x >= 3
The graph of this function is pictured below.
We can see where this graph intersects the line y = 4 by solving the first part:
• 8 – 3x = 4, x < 2
• -3x = -4, x < 2
• x = 4/3, x < 2
• x = 4/3
the second part:
• 4 – x = 4, 2 <= x < 3
• – x = 4, 2 <= x < 3
• x = -4, 2 <= x < 3
• No solution
and the third part:
• 3x – 8 = 4, x >= 3
• 3x = 12, x >= 3
• x = 4, x >= 3
• x = 4
So, the solutions are x = 4/3 and x = 4, as we can verify from the graph.
## Conclusion
Now you know when to use absolute value and how to solve the resulting equations and inequalities. You also know when these equations and inequalities have no solutions. |
# What is 2 minus 1/8 as a fraction?
## What is 2 minus 1/8 as a fraction?
2 – 1/8 = 158 = 1 78 = 1.875.
## How do I subtract fractions?
There are 3 simple steps to subtract fractions
1. Make sure the bottom numbers (the denominators) are the same.
2. Subtract the top numbers (the numerators). Put the answer over the same denominator.
3. Simplify the fraction (if needed).
## What is 1/3 minus 2/5 as a fraction?
1/3 – 2/5 = -115 ≅ -0.06666667.
## What is 3/4 divided by 2 in a fraction?
3/8
Answer: 3/4 divided by 2 in the fraction is equal to 3/8. Let us proceed step by step.
## What fraction is three quarters?
The fraction 3/4 or three quarters means 3 parts out of 4. The upper number, 3, is called the numerator and the lower number, 4, is the denominator.
## 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 does ¾ mean?
Definition of three-quarters : an amount equal to three of the four equal parts which make up something : seventy-five percent Three-quarters of the class will be going on the trip.
## What is three eighths plus two-eighths?
Three-eighths plus two-eighths equals five-eighths There are two popular methods to make the denominators the same: Least Common Denominator, or
## What is half the half of one and a half?
“half the half” is one quarter. One quarter = 1/4 One and a half = 11/2 = 3/2 One quarter of 1 and a half = 1/4 x 3/2 = 3/8 So, half the half of one and a half = three eighths What is 1 quarter added to 1 eighth?
## How many slices of pizza are in a quarter?
Slice a pizza, and we get fractions: 1 / 2: 1 / 4: 3 / 8 (One-Half) (One-Quarter) (Three-Eighths) The top number says how many slices we have. The bottom number says how many equal slices the whole pizza was cut into. Have a try yourself: Equivalent Fractions. Some fractions may look different, but are really the same, for example:
## What does Oone-quarter mean?
One-quarter plus one-quarter equals two-quarters, equals one-half |
# Percentage
PERCENTAGE
The Percentage problems are the basic methodology in the Numerical Ability section. These percentage problems are frequently asked in any levels of the difficulty in every competitive exam. Especially these questions are focused mainly on every bank examinations. The percentage problems are asked as separate questions are in the combination with the DI. Percentage problems are the easily understood with few concepts. We provide you with those concepts and shortcut methodologies to crack percentage questions in a few seconds. We also provide you with problems based on percentage quiz on a daily basis to improve your performance in the exam. The candidates preparing for banking and other competitive exams can validate your ability here.
Fraction Percentage 1/100 1% 1/20 5% 1/10 10% 1/8 121/2Â % 1/5 20% 1/4 25% 1/3 331/3% 1/2 50% 3/4 75% 4/5 80% 9/10 90% 99/100 99%
#### Check Below the Practice Questions
What is the Percentage: A fraction with its denominator as ‘100’ is called a percentage. Percentage means per hundred. So it is a fraction of the form 6/100, 37/100, 151/100 and these fractions can be expressed as 6%, 37% and 151% respectively. By a certain percent, we mean that many hundredths.
Thus x percent means x hundredths, written as x%.
To express x% as a fraction: We have, x% = x/100.
Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.
To express a/b as a percent: We have, a/b = ((a/b)*100)%
Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%
Why Percentage: Percentage is a concept evolved so that there can be a uniform platform for comparison of various things. (Since each value is taken to a common platform of 100)
Example: To compare three different students depending on the marks they scored we cannot directly compare their marks until we know the maximum marks for which they took the test. But by calculating percentages they can directly be compared with one another.
Important Points to Remember:
a)Â If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is
[R / (100+R))*100] %
b)Â If the price of the commodity decreases by R%, then the increase in consumption so as to decrease the expenditure is
[(R / (100-R)*100] %
c) If A is R% more than B, then B is less than A by
[(R/ (100+R))*100]%
d) If A is R% less than B, then B is more than A by
[(R/ (100-R))*100]%
Results on Population: Let the population of the town be P now and suppose it increases at the rate of R% per annum, then:
1.Population after n years = P [1+(R/100)]n
2.Population n years ago = P / [1+(R/100)]n
Results on Depreciation: Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,
1.Value of the machine after n years = P [1-(R/100)]n
2.Value of the machine n years ago = P / [1-(R/100)]n |
# Chain Rule
A LevelAQAEdexcelOCR
## Chain Rule
The chain rule is best used when we have a function too complex to differentiate in one fell swoop. We look to turn one function into a function of a function – it sounds worse, but I guarantee it will make your job much easier!
A LevelAQAEdexcelOCR
## Adapting to $\dfrac{dx}{dy}$
For equations of the form $\textcolor{blue}{x} = f(\textcolor{limegreen}{y})$, we can calculate $\dfrac{d\textcolor{blue}{x}}{d\textcolor{limegreen}{y}}$ in the same way as usual, and find its reciprocal to find an expression for $\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}$. We can then couple this with Integration to find an expression for $\textcolor{limegreen}{y} = f^{-1}(\textcolor{blue}{x})$, but we’ll not worry about this for the moment.
The most important thing to remember is that $\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} \times \dfrac{d\textcolor{blue}{x}}{d\textcolor{limegreen}{y}} = 1$, so we have
$\dfrac{d\textcolor{blue}{x}}{d\textcolor{limegreen}{y}} = \dfrac{1}{\left( \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}\right) }$
A LevelAQAEdexcelOCR
## How to Use the Chain Rule
Let’s say our complex function $f(\textcolor{blue}{x})$ can be split into a function of a function, i.e. $f(\textcolor{blue}{x}) = g(h(\textcolor{blue}{x}))$.
Then we have the Chain Rule:
$\dfrac{df(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = \dfrac{dg(h(\textcolor{blue}{x}))}{d\textcolor{blue}{x}} = \dfrac{dg(h(\textcolor{blue}{x}))}{dh(\textcolor{blue}{x})} \times \dfrac{dh(\textcolor{blue}{x})}{d\textcolor{blue}{x}}$
That might still seem a little confusing, so you may see it given where $y=g(u)$ and $u = h(x)$:
$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$
A LevelAQAEdexcelOCR
@mmerevise
## Connecting Rates of Change
You may see examples where there are several variables linked, for example distance, speed and acceleration or length, surface area and volume. If you know the rate of change of one variable, and the equations linking the variables, then you can use the chain rule to find the rates of change of the other variables.
Watch out for hidden derivatives given as words, such as “rate” or “per“.
Once you understand what the question is asking the calculations are usually pretty straightforward.
A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR
## Example 1: Using the Chain Rule
Let $f(\textcolor{blue}{x}) = \sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}$. Find the derivative of $f(\textcolor{blue}{x})$ with respect to $\textcolor{blue}{x}$.
[4 marks]
Take $g(\textcolor{red}{u}) = \sqrt{\textcolor{red}{u}}$ and $\textcolor{red}{u} = h(\textcolor{blue}{x}) = 2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}$, so $f(\textcolor{blue}{x}) = g(h(\textcolor{blue}{x}))$.
Then,
$\dfrac{dg(\textcolor{red}{u})}{d\textcolor{red}{u}} = \dfrac{1}{2\sqrt{\textcolor{red}{u}}} = \dfrac{1}{2\sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}}$ and $\dfrac{d\textcolor{red}{u}}{d\textcolor{blue}{x}} =\dfrac{dh(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = 2 + \dfrac{2}{\textcolor{blue}{x}^3} = \dfrac{2\textcolor{blue}{x}^3 + 2}{\textcolor{blue}{x}^3}$.
Multiplying the two gives
$\dfrac{df(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = \left( \dfrac{2\textcolor{blue}{x}^3 + 2}{\textcolor{blue}{x}^3}\right) \left( \dfrac{1}{2\sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}}\right) = \dfrac{\textcolor{blue}{x}^3 + 1}{\textcolor{blue}{x}^3 \sqrt{2\textcolor{blue}{x} - \dfrac{1}{\textcolor{blue}{x}^2}}}$
A LevelAQAEdexcelOCR
## Example 2: Connecting Rates of Change
The surface area of a cube of width $x \text{ cm}$, length $2x \text{ cm}$ and height $4x \text{ cm}$ decreases at a constant rate of $0.07 \text{ cm}^2 \text{s}^{-1}$. Find $\textcolor{purple}{\dfrac{dx}{dt}}$ at the point where $\textcolor{orange}{x = 5 \text{ cm}}$.
[3 marks]
The surface area of the cuboid is:
$A = 2 \times (x \times 2x) + 2 \times (x \times 4x) + 2 \times (2x \times 4x) = 28x^2$
So,
$\dfrac{dA}{dx} = 56x$
$A$ decreases at a constant rate of $0.07 \text{ cm}^2\text{s}^{-1}$, so we can write this as
$\dfrac{dA}{dt} = - 0.07$
Now, use the chain rule to find $\textcolor{purple}{\dfrac{dx}{dt}}$:
\begin{aligned} \textcolor{purple}{\dfrac{dx}{dt}} &= \dfrac{dx}{dA} \times \dfrac{dA}{dt} \\[1.2em] &= \dfrac{1}{\left( \dfrac{dA}{dx} \right)} \times \dfrac{dA}{dt} \\[1.2em] &= \dfrac{1}{56x} \times (-0.07) \\[1.2em] &= - \dfrac{0.00125}{x} \end{aligned}
When $\textcolor{orange}{x = 5}$,
$\textcolor{purple}{\dfrac{dx}{dt}} = - \dfrac{0.00125}{x} = - \dfrac{0.00125}{\textcolor{orange}{5}} = -0.00025 \text{ cms}^{-1}$
A LevelAQAEdexcelOCR
## Chain Rule Example Questions
Let $g(u) = \sin u$ and $u = h(x) = \dfrac{1}{x}$. Then we have $y = g(h(x))$.
$\dfrac{dy}{dx} = \dfrac{dg(u)}{du} \times \dfrac{du}{dx}$
$\dfrac{dg(u)}{du} = \cos u = \cos \dfrac{1}{x}$ and $\dfrac{du}{dx} = \dfrac{-1}{x^2}$
so, $\dfrac{dy}{dx} = \dfrac{-\cos \dfrac{1}{x}}{x^2}$.
Gold Standard Education
Set $x = 3y^2 - 5y + 2 = 0$ to get $y = \dfrac{2}{3}$ and $y = 1$.
$\dfrac{dx}{dy} = 6y - 5$
Therefore,
$\dfrac{dy}{dx} = \dfrac{1}{6y - 5}$
When $y = \dfrac{2}{3}$, $\dfrac{dy}{dx} = -1$.
Gold Standard Education
Since the metal is cubic, we have $V = x^3$.
Therefore, $\dfrac{dV}{dx} = 3x^2$.
We also have
$\dfrac{dx}{dt} = 0.1$
so
$\dfrac{dV}{dt} = \dfrac{dV}{dx} \times \dfrac{dx}{dt} = 3x^2 \times 0.1 = 0.3x^2$
Gold Standard Education
A Level
A Level
Product |
# Important Questions Class 9 Maths Chapter 4
## Important Questions Class 9 Mathematics Chapter 4 – Linear Equations in Two Variables
Mathematics Chapter 4 of Class 9 introduces students to Linear Equations In Two Variables. A required linear equation in two variables has two numbers that can satisfy the given equation. These two numbers are called the solution of the required linear equation in two variables.
Following are the key topics covered in Chapter 4 of CBSE Class 9 Mathematics syllabus:
Graphical representation of a linear equation in 2 variables
• Any given linear equation in the required standard form ax+by+c=0 has a couple of solutions in the required form (x,y) that can be illustrated in the coordinate plane.
• When a needed equation is represented graphically, it is a straight line that may or may not cut the coordinate axes.
Solutions of Linear equation in 2 variables on a graph
• A given linear equation ax+by+c=0 is illustrated graphically as a straight line.
• Each point on the given line is a solution for the linear equation.
• Each solution of the given linear equation is a given point on the required line.
Lines passing through the origin
• Particular linear equations exist such that their required solution is (0, 0). Such equations, when illustrated graphically, pass through the origin.
• The coordinate axes, that is, the x-axis and y-axis, can be defined as y=0 and x=0, respectively.
Lines parallel to coordinate axes
• Linear equations of the given form y=a, when represented graphically, are lines parallel to the x-axis, and a is the needed y-coordinate of the required points in the same line.
• Linear equations of the given form x=a, when represented graphically, are lines parallel to the y-axis, and a is the needed x-coordinate of the required points in the same line.
Extramarks’ credibility lies in providing reliable and trusted study material related to NCERT is the most useful study companion for students and enables students with their comprehensive online study solutions from Class 1 to Class 12. Our qualified Mathematics subject experts have prepared various NCERT explanations to help students in their studies and exam preparation. Students can direct to our Important Questions Class 9 Mathematics Chapter 4 to practise exam-oriented questions. We have collected questions from diverse sources such as NCERT textbooks and exemplars, CBSE sample papers, CBSE past year question papers, etc. Students can prepare excellently for their exams and quizzes by solving various chapter questions from our Important Questions Class 9 Mathematics Chapter 4.
To maximise their potential during exam preparations, students can register on our Extramarks website and get full access to Important Questions Class 9 Mathematics Chapter 4 and other study materials, including NCERT solutions, CBSE revision notes, etc.
## Important Questions Class 9 Mathematics Chapter 4 – With Solutions
Our in-house Mathematics faculty experts have collected an entire list of Important Questions Class 9 Mathematics Chapter 4 by referring to various sources. For every question, the team has prepared a step-by-step explanation that will aid students in understanding the concepts used in each question. Also, the given questions are prepared in a way that will cover the entire chapter.. By practising from our question bank, students should be able to revise the chapter and self-assess their strong and weak points. And improvise by further focusing on weak areas of the chapter and practising harder to maximise their potential.
Following are some of the questions and explanations from our question bank of Mathematics Class 9 Chapter 4 Important Questions:
Question 1: Define the following linear equations in the form ax + by + c = 0 and show the values of a, b and c in every individual case:
(i) x – y/5 – 10 = 0
(ii) -2x+3y = 6
(iii) y – 2 = 0
(i) The equation x-y/5-10 = 0
(1)x + (-1/5) y + (-10) = 0
Directly compare the above equation with ax + by + c = 0
Therefore, we get;
a = 1
b = -⅕
c = -10
(ii) –2x + 3y = 6
Re-arranging the provided equation, we obtain,
–2x + 3y – 6 = 0
The required equation –2x + 3y – 6 = 0 can be written as,
(–2)x + 3y +(– 6) = 0
Directly comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0
We obtain a = –2
b = 3
c = -6
(iii) y – 2 = 0
y – 2 = 0
The required equation y – 2 = 0 can be written as,
0x + 1y + (–2) = 0
Directly comparing 0x + 1y + (–2) = 0 with ax + by + c = 0
We obtain a = 0
b = 1
c = –2
Question 2: The price of a notebook is twice the cost of a pen. Note a linear equation in two variables to illustrate this statement.
(Taking the price of a notebook to be ₹ x and that of a pen to be ₹ y)
Answer 2: Let the price of one notebook be = ₹ x
Let the price of one pen be = ₹ y
As per the question,
The price of one notebook is twice the cost of one pen.
i.e., the price of one notebook = 2×price of a pen
x = 2×y
x = 2y
x-2y = 0
x-2y = 0 is the required linear equation in two variables to illustrate the statement, ‘The price of one given notebook is twice the cost of a pen.
Question 3: Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
(i) 2x + 9 = 0
We have, 2x + 9 = 0
2x = – 9
x = -9/2
which is the required linear equation in one variable, that is, x only.
Therefore, x= -9/2 is a unique solution on the number line as shown below:
(ii) 2x +9=0
We can write 2x + 9 = 0 in the two variables as 2x + 0, y + 9 = 0
or x = −9−0.y/2
∴ When y = 1, x = −9−0.(1)/2 = -9/2
y=2 , x = −9−0.(2)/2 = -9/2
y = 3, x = −9−0.(3)/2= -9/2
Therefore, we obtain the following table:
X -9/2 -9/2 -9/2 Y 1 2 3
Now, plotting the ordered pairs (−9/2,3), (−9/2,3) and (−9/2,3) on graph paper and connecting them, we get a line PQ as the solution of 2x + 9 = 0.
Question 4: Note four solutions individually for the following equations:
(i) 2x + y = 7
Answer 4: For the four answers of 2x + y = 7, we replace different values for x and y
Let x = 0
Then,
2x + y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Now,
2x + y = 7
(2×1)+y = 7
2+y = 7
y = 7 – 2
y = 5
(1,5)
Let y = 1
Now,
2x + y = 7
2x+ 1 = 7
2x = 7 – 1
2x = 6
x = 3
(3,1)
Let x = 2
Now,
2x + y = 7
2(2)+y = 7
4+y = 7
y = 7 – 4
y = 3
(2,3)
The answers are (0, 7), (1,5), (3,1), (2,3)
Question 5: The linear equation 2x -5y = 7 has
(A) A unique solution
(B) Two solutions
(C) Infinitely many solutions
Answer 5: (C) Infinitely many solutions
Solution:
Linear equation: The equation of two variables which gives a straight line graph is called a linear equation.
Here the linear equation is 2x – 5y = 7
Let y = 0, then the value of x is:
2x – 5(0)=7
2x =7
x = 7/2
Now, let y = 1, then the value of x is:
2x – 5 (1) =7
2x -5 =7
2x = 7 + 5
2x =12
x = 12/2
x = 6
Here for different values of y, we are getting different values of x
Therefore the equation has infinitely many solutions
Question 6: Represent the following linear equations in the form ax + by + c = 0 and show the required values of a, b and c in every case:
Answer 6: (i) x –(y/5)–10 = 0
The required equation x –(y/5)-10 = 0 can be written as,
1x+(-1/5)y +(–10) = 0
Comparing the given equation x+(-1/5)y+(–10) = 0 with ax+by+c = 0
We obtain,
a = 1
b = -(1/5)
c = -10
(ii) –2x+3y = 6
–2x+3y = 6
Rearranging the equation, we obtain,
–2x+3y–6 = 0
The required equation –2x+3y–6 = 0 can be written as,
(–2)x+3y+(– 6) = 0
Comparing the given equation (–2)x+3y+(–6) = 0 with ax+by+c = 0
We obtain a = –2
b = 3
c =-6
(iii) x = 3y
x = 3y
Rearranging the equation, we obtain,
x-3y = 0
The required equation x-3y=0 can be written as,
1x+(-3)y+(0)c = 0
Comparing the given equation 1x+(-3)y+(0)c = 0 with ax+by+c = 0
We obtain a = 1
b = -3
c =0
(iv) 2x = –5y
2x = –5y
Rearranging the equation, we obtain,
2x+5y = 0
The required equation 2x+5y = 0 can be written as,
2x+5y+0 = 0
Comparing the given equation 2x+5y+0= 0 with ax+by+c = 0
We obtain a = 2
b = 5
c = 0
(v) 3x+2 = 0
3x+2 = 0
The required equation 3x+2 = 0 can be written as,
3x+0y+2 = 0
Comparing the given equation 3x+0+2= 0 with ax+by+c = 0
We obtain a = 3
b = 0
c = 2
(vi) y–2 = 0
y–2 = 0
The required equation y–2 = 0 can be written as,
0x+1y+(–2) = 0
Comparing the given equation 0x+1y+(–2) = 0 with ax+by+c = 0
We obtain a = 0
b = 1
c = –2
(vii) 5 = 2x
5 = 2x
Rearranging the equation, we obtain,
2x = 5
i.e., 2x–5 = 0
The required equation 2x–5 = 0 can be written as,
2x+0y–5 = 0
Comparing the given equation 2x+0y–5 = 0 with ax+by+c = 0
We obtain a = 2
b = 0
c = -5
Question 7: Note four solutions individually for the following equations:
πx + y = 9
Answer 7: For the four answers of πx + y = 9, we replace other values for x and y
Let x = 0
Now,
πx + y = 9
(π × 0)+y = 9
y = 9
(0,9)
Let x = 1
Now,
πx + y = 9
(π×1)+y = 9
π+y = 9
y = 9-π
(1,9-π)
Let y = 0
Now,
πx + y = 9
πx +0 = 9
πx = 9
x =9/π
(9/π,0)
Let x = -1
Now,
Put x=2, we have
πx + y = 9
π(2) + y = 9
y = 9 – 2π
The answers are (0,9), (1,9-π),(9/π,0),(2,9 – 2π)
Question 8: Find out the value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k.
Answer 8: The provided equation is
2x + 3y = k
As per the given question, x = 2 and y = 1.
Then, Replacing the values of x and y in the equation 2x + 3y = k,
We get,
⇒(2 x 2)+ (3 × 1) = k
⇒4+3 = k
⇒7 = k
⇒k = 7
The required value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k, is 7.
Question 9: Establish that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the required linear equation y = 9x – 7.
Answer 9: We include the equation,
y = 9x – 7
For A (1, 2),
Replacing (x,y) = (1, 2),
We obtain,
2 = 9(1) – 7
2 = 9 – 7
2 = 2
For B (–1, –16),
Replacing (x,y) = (–1, –16),
We get,
–16 = 9(–1) – 7
-16 = – 9 – 7
-16 = – 16
For C (0, –7),
Replacing (x,y) = (0, –7),
We obtain,
– 7 = 9(0) – 7
-7 = 0 – 7
-7 = – 7
Therefore, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Therefore, A (1, 2), B (–1, –16) and C (0, –7) are answers to the linear equation y = 9x – 7
Thus, points A (1, 2), B (–1, –16), and C (0, –7) lie on the graph of the linear equation y = 9x – 7.
Question 10: Note the linear equation such that every point on its graph has a coordinate 3 times its abscissa.
As per the question,
A given linear equation such that every point on its graph has a coordinate(y) which is 3 times its
abscissa(x).
So we obtain
⇒ y = 3x.
Therefore, y = 3x is the required linear equation.
Question 11: Illustrate the graph of the given linear equation 3x + 4y = 6. At what points does the graph cut the X and Y-axis?
3x + 4y = 6.
We need at least 2 points on the graph to illustrate the graph of this equation,
Therefore, the points the graph cuts
(i) x-axis
The given point is on the x-axis. We have y = 0.
Replacing y = 0 in the equation, 3x + 4y = 6,
We get,
3x + 4×0 = 6
⇒ 3x = 6
⇒ x = 2
Therefore, the point at which the graph cuts the x-axis = (2, 0).
(ii) y-axis
Since the point is on the y-axis, we have x = 0.
Replacing x = 0 in the equation, 3x + 4y = 6,
We obtain,
3×0 + 4y = 6
⇒ 4y = 6
⇒ y = 6/4
⇒ y = 3/2
⇒ y = 1.5
Thus, the point at which the graph cuts the x-axis = (0, 1.5).
By plotting the points (0, 1.5) and (2, 0) on the graph.
we obtain,
image source: Online
Question 12: Show that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the given graph of the linear equation y = 9x – 7.
Answer 12: We have the given equation,
y = 9x – 7
For A (1, 2),
Substitute the values of (x,y) = (1, 2),
We obtain,
2 = 9 (1) – 7 = 9 – 7 = 2
For B (–1, –16),
Substitute the values of (x,y) = (–1, –16),
We obtain,
–16 = 9(–1) – 7 = – 9 – 7 = – 16
For C (0, –7),
Substitute the values of (x,y) = (0, –7),
We obtain,
– 7 = 9(0) – 7 = 0 – 7 = – 7
Thus, we locate that points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Thus, A (1, 2), B (–1, –16), and C (0, –7) are required solutions of the linear equation y = 9x – 7
Hence, the given points A (1, 2), B (–1, –16) and C (0, –7) lie on the graph of the required linear equation y = 9x – 7.
Question 13: Show the required geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Answer 13: (i) In one variable
y = 3
∵ y = 3 is the required equation in one variable, that is, y only.
∴ y = 3 is the required unique solution on the number line as displayed below:
(ii) In two variables
When an equation in two variables, it can be expressed as,
0.x + y – 3 = 0
which is a linear equation in the variables x and y.
When x = 0, y = 3
When x = 1, y = 3
Question 14: In countries like the USA and Canada, the temperature is measured In the required Fahrenheit, whereas in countries like India, it is calculated in Celsius. Given here is a linear equation that converts Fahrenheit to Celsius:
F = (95 )C + 32
(i) Draw the linear equation graph above using Celsius for the x-axis and Fahrenheit for the y-axis.
(ii) If the required temperature Is 30°C, what is the temperature in Fahrenheit?
(iii) If the required temperature is 95°F, what is the temperature in Celsius?
(iv) If the required temperature is 0°C, what is Fahrenheit? If the required temperature is 0°F, what Is the temperature In Celsius?
(v) Is a required temperature numerically the same in Fahrenheit and Celsius? If yes, find It.
When C=0 then F=32
also, when C=5 then F=41
C 0 5 F 32 41
(ii) Placing the value of C=30 in F=9/5C+32, we obtain
F=9/5×30+32
F=54+32
F=86
(iii) Putting the value of F=95 in F=9/5C+32, we get
95=9/5C+32
9/5C=95−32
C=63×5/9
C=35
Question 15: If the work accomplished by a body on application of a steady force is directly proportional to the required distance traversed by the body, convey this in the required form of an equation in two variables and sketch the exact graph by taking the steady force as 5 units. Likewise, read from the graph the work done when the distance traversed by the body is
(i) 2 units
(ii) 0 unit
Let the distance traversed = x units and work done = y units.
Work done = Force x Distance
y = 5 x x
y = 5x
For sketching the graph, we are having y = 5x
When x = 0,
y = 5(0) = 0
x = 1, then y = 5(1) = 5
x = -1, then y = 5(-1) = -5
∴ We obtain the following given table:
x 0 1 -1 y 0 5 -5
Plotting the required ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we acquire a straight line AB as shown.
From the required graph, we obtain
(i) The required distance traversed =2 units, that is, x = 2
∴ If x = 2,
then y = 5(2) = 10
The required work done = 10 units when the distance travelled by it is 2 units.
(ii) The required distance traversed = 0 unit ,that is , x = 0
∴ If x = 0
y = 5(0) – 0
The required work done = 0 unit when the distance travelled by it is 0 units.
.
Benefits Of Solving Important Questions Class 9 Mathematics Chapter 4
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Q.1 If the point (4, 3) lies on the graph of the equation 3x ay = 6, find whether (2, 6) also lies on the same graph.
Marks:3
Ans
$\begin{array}{l}\text{Since}\left(\text{4, 3}\right)\text{liesonline3x}\text{ay=6}\\ \text{So,puttingx=4andy=3,weget}\\ \text{3}\left(\text{4}\right)\text{a}\left(\text{3}\right)\text{=6}\\ \text{12}\text{3a=6}\\ \text{12}\text{6=3a}\\ \text{a=}\frac{\text{6}}{\text{3}}\\ \text{=2}\\ \text{So,equationbecomes}\\ \text{3x-2y=6}\\ \text{Puttingx=}\text{2andy=}\text{6inLHS,}\\ \text{LHS=3x}\text{y}\\ \text{=3}\left(\text{2}\right)\text{2}\left(\text{6}\right)\\ \text{=}\text{6+12}\\ \text{=6=RHS}\\ \text{Thus,}\left(\text{2,}\text{6}\right)\text{alsolieson the graph of the sameline.}\end{array}$
Q.2 For what value of k, the point (k, 5) lies on the line 4x 5y = 10
Marks:2
Ans
Point (k, 5) lies on the line 4x 5y = 10,
Since the point satisfies the line, so, we have
4(k) 5(5) =10
4k 25 = 10
4k = 35
k =
$\frac{35}{4}$
Q.3 Find four different solutions of the equation x+2y = 6.
Marks:3
Ans
Let us consider x = 0
Then the given equation reduces to 2y = 6
y = 3
So, (0,3) is a solution of x+2y = 6.
At x = 2, the given equation reduces to
2+2y = 6
y = 2
So, (0,2) is a solution of given equation.
At y = 0, the given eqution reduces to x = 6.
So (6,0) is a solution of given equation.
At y = 1 given equation reduces to x+2 = 6
x = 4.
So (4,1) is a solution of given equation.’
Thus (2,2), (0,3), (6,0) and (4,1) are the solutions of given equation.
Q.4 Draw the graph of x+y = 7.
Marks:5
Ans
$\begin{array}{l}\text{Todrawthegraph,weneedatleasttwosolutions.}\\ \text{x+y=7}\text{y=7}\text{x}\\ \text{x=0}\text{y=7}\text{0}\text{y=7}\\ \text{x=7}\text{y=7}\text{7}\text{y=0}\\ \left(\text{0,7}\right)\text{and}\left(\text{7,0}\right)\text{aresolutionsofx+y=7}\\ \text{We​canusethefollowingtabletodrawthegraph:}\\ \begin{array}{|ccc|}\hline \text{x}& \text{0}& \text{7}\\ \text{y}& \text{7}& \text{0}\\ \hline\end{array}\end{array}$
## Please register to view this section
### 1. What is the standard equation in Linear Equations In Two Variables?
The standard equation for Equations in two variables is ax + by +c =0
### 2. How can I get good grades in Class 9 Mathematics examinations?
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### 3. What can I get from the Extramarks website?
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### 4. How many total chapters will students study in Class 9 Mathematics?
There are 15 chapters in Class 9 Mathematics. The list is given below:
• Chapter 1- Number System
• Chapter 2 –Polynomials
• Chapter 3 – Coordinate Geometry
• Chapter 4 –Linear Equations In Two Variables
• Chapter 5 – Introduction To Euclid’s Geometry
• Chapter 6 – Lines And Angles
• Chapter 7 –Triangles
• Chapter 9 –Areas Of Parallelograms And Triangles
• Chapter 10 – Circles
• Chapter 11- Constructions
• Chapter 12- Heron’s Formula
• Chapter 13-Surface Area And Volumes
• Chapter 14- Statistics
• Chapter 15- Probability
### 5. What are the important chapters in Class 9 Mathematics Chapter 4?
The NCERT Mathematics book has 15 chapters. When it comes to grasping the fundamentals and taking the test, each and every chapter is equally important. Additionally, because CBSE does not specify marks distribution. Students need to study all the chapters in order to receive a good grade.
### 6. Where can I get important questions for Class 9 Mathematics Chapter 4 online?
On the Extramarks website, you can find all the important questions for Class 9 Mathematics Chapter 4, along with their answers. You can even find many other NCERT-based study solutions on the website for Classes 1 to 12. |
# 2010 AMC 12A Problems/Problem 14
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The following problem is from both the 2010 AMC 12A #14 and 2010 AMC 10A #16, so both problems redirect to this page.
## Problem
Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$, and $DC=8$. What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$
## Solution
By the Angle Bisector Theorem, we know that $\frac{AB}{BC} = \frac{3}{8}$. If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$, respectively), then $AB + BC = AD + DC = AC$, contradicting the Triangle Inequality. If we use the next lowest values ($AB = 6$ and $BC = 16$), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$, or choice $\textbf{(B)}$.
## Solution 2(Trick)
We find that $\frac{AB}{BC}=\frac{3}{8}$ by the Angle Bisector Theorem so we let the lengths be $3n$ and $8n$, respectively where $n$ is a positive integer. Also since $AD=3$ and $BC=8$, we notice that the perimeter of the triangle is the sum of these, namely $3n+8n+3+8=11n+11.$ This can be factored into $11(n+1)$ and so the sum must be a multiple of $11$. The only answer choice which is a multiple of $11$ is $\boxed{\textbf{(B)} 33}$. ~mathboy282 |
1. Chapter 2 Class 10 Polynomials (Term 1)
2. Serial order wise
3. Ex 2.2
Transcript
Ex2.2, 1 Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (iii) 6x2 – 3 – 7x Let p(x) = 6x2 – 7x – 3 Zero of the polynomial is the value of x where p(x) = 0 Putting p(x) = 0 6x2 – 7x – 3 = 0 We find roots using splitting the middle term method 6x2 – 9x + 2x – 3 = 0 3x(2x – 3) + 1(2x – 3) = 0 (3x + 1)(2x – 3) = 0 So 3x + 1 = 0 & 2x – 3 = 0 Therefore, α = – 1/3 & β = 3/2 are the zeroes of the polynomial p(x) = 6x2 – 7x – 3 Comparing with ax2 + bx + c So a = 6 , We verify Sum of zeroes = − (𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥)/(𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥2) i.e. α + β = - 𝑏/𝑎 |
## 305 -Homework set 1
January 26, 2009
This set is due February 6 at the beginning of lecture. Consult the syllabus for details on the homework policy.
1. a. Complete the proof by induction that if $a,b$ are integers and $(a,b)=1$, then $(a^n,b)=1$ for all integers $n\ge1$.
b. This result allows us to give a nice proof of the following fact: Let $n$ be a natural number and let $m$ be a positive integer. If the $m$-th root of $n$, $\root m\of n$, is rational, then it is in fact an integer. (The book gives a proof of a weaker fact.) Prove this result as follows: First verify that if $a/b=c/d$ and $b+d\ne0$, then $\displaystyle \frac ab=\frac{a+c}{b+d}$. Show that any fraction $a/b$ with $a,b$ integers, can be reduced so $(a,b)=1$. Assume that $\root m\of n$ is rational, say $a/b$. Then also $\root m\of n=n/(\root m\of n)^{m-1}$. Express this last fraction as a rational number in terms of $n,b,a$. Use that $(a^k,b)=1$ for all $k\ge1$ and the general remarks mentioned above, to show that $\root m\of n$ is in fact an integer.
2. Show by induction that for all integers $k\ge 1$ there is a polynomial $q(x)$ with rational coefficients, of degree $k+1$ and leading coefficient $1/(k+1)$, such that for all integers $n\ge1$, we have $\displaystyle \sum_{i=1}^n i^k =q(n).$ There are many ways to prove this result. Here is one possible suggestion: Consider $\displaystyle \sum_{i=1}^n [(i+1)^{k+1}-i^{k+1}]$.
3. Euclidean algorithm. We can compute the gcd of two integers $m,n,$ not both zero, as follows; this method comes from Euclid and is probably the earliest recorded algorithm. Fist, we may assume that $m, n$ are positive, since $(m,n)=(|m|,|n|)$, and also we may assume that $m\ge n$, so $m>0$. Now define a sequence $r_0,r_1,r_2,\dots$ of natural numbers as follows:
• $r_0=m$, $r_1=n$.
• Given $r_i,r_{i+1}$, if $r_{i+1}=0$, then ${\tt STOP}$.
• Otherwise, $r_{i+1}>0$, and we can use the division algorithm to find unique integers $q,r$ with $0\le r such that $r_i=r_{i+1}q+r$. Set $r_{i+2}=r$.
• Let $A$ be the set of those $r_k$ that are strictly positive. This set has a least element, say $r_k$. By the way the algorithm is designed, this means that $r_{k+1}=0$.
• Show that $r_k=(m,n)$, and that we can find from the algorithm, integers $x,y$ such that $r_k=mx+ny$.
(If the description above confuses you, it may be useful to see the example in the book.)
4. Assume that the application of the algorithm, starting with positive integers $m>n$, takes $k$ steps. [For example, if $m=35$ and $n=25$, the algorithm gives:
Step 1: $35=25.1+10$, so $r_0=35,r_1=25,r_2=10$.
Step 2: $25=10.2+5$, so $r_3=5$.
Step 3: $10=5.2$, so $r_4=0$, and $(35,25)=5$. Here, $k=3$]
Show that $n\ge F_{k+1}$, where the numbers $F_1,F_2,\dots$ are the Fibonacci numbers, see Exercises 15-22 in Chapter 1 of the book.
5. Extra credit problem. With $m,n,k$ as in the previous exercise, let $t$ be the number of digits of $n$ (written in base 10). Show that $k\le 5t.$
## 305 -Algebra and induction (3)
January 26, 2009
Definition. $p>1$ is prime iff the only positive divisors of $p$ are $1$ and $p$.
Proposition. A number $p>1$ is prime iff for all $m$, either $p\mid m$ or else $(p,m)=1$. $\Box$
Proposition. Let $p$ be prime, and let $m,n$ be integers. If $p\mid mn$ then either $p\mid m$ or $p\mid n$. $\Box$
In the more general setting of rings, of which the integers are an example, it is customary to call a number $p$ prime when it satisfies the second proposition, and to call it irreducible when it satisfies the definition above. Both notions coincide for the integers, but there are examples of rings where there are irreducible elements that are not prime. We will introduce rings later on in the course.
Mathematical induction. Let $N\in{\mathbb Z}$. Let $P(\cdot)$ be a statement about integers, so for each integer $m$, $P(m)$ may or may not hold. Assume
1. $P(N)$ holds, and
2. $\forall k\ge N,($ if $P(k)$ holds then $P(k+1)$ holds$).$
Then $\forall k\ge N,(P(k)$ holds$)$.
Strong induction. Let $N\in{\mathbb Z}$. Let $P(\cdot)$ be a statement about integers. Assume
• $\forall k\ge N,($ if $\forall m$ with $N\le m it is the case that $P(m)$ holds, then also $P(k)$ holds$).$
Then $\forall k\ge N,(P(k)$ holds$)$.
Both induction and strong induction are consequences of the well-ordering principle. In fact, all three statements are equivalent. Most properties about the natural numbers are established by inductive arguments. Here are two examples:
Example 1. For all $n\ge1$, $\displaystyle \sum_{ i=1}^n i=\frac {n(n+1)}2$.
We also have $\displaystyle \sum_{ i=1}^n i^2=\frac {n(n+1)(2n+1)}6$ and $\displaystyle \sum_{ i=1}^n i^3=\left(\frac {n(n+1)}2\right)^2$. In fact, for all $k\ge 1$, there is a polynomial $q(x)$ with rational coefficients, of degree $k+1$ and leading coefficient $\displaystyle \frac1{k+1}$ such that for all $n\ge1$$\displaystyle \sum_{ i=1}^n i^k=q(n)$.
Example 2. For all integers $a,b$, if $(a,b)=1$ then for all $n\ge1$, $(a^n,b)=1$. |
## Exercises - Confidence Intervals for Proportions
1. You play a game and win 136 out of 270 times.
• Estimate the probability of winning the game.
$\widehat{p} = 136\,/\,270 \doteq 0.5037$
• Find a 95% confidence interval for the probability of winning the game.
We first check the requirement that $np \ge 5$ and $nq \ge 5$ by estimating $p$ with the sample proportion $\widehat{p}$ and ensuring that we see at least $5$ successess and $5$ failures. Of course, here, we see $136$ successes (wins) and $134$ failures (losses), so the requirements are met.
Now note, for 95% confidence we have $z_{\alpha/2} = 1.96$, so the confidence interval is given by:
$$\frac{136}{270} \pm 1.96 \sqrt{\frac{\left(\frac{136}{270}\right) \left(\frac{134}{270}\right)}{270}}$$ or approximately $$(0.4441, 0.5633)$$
2. Find $z_{\alpha/2}$ for a 90% confidence interval for a proportion.
$$z_{\alpha/2} = 1.645$$ Note, on a TI-83 calculator, invNorm((1 - 0.90)/2) = -1.645
3. A CBS News/New York Times poll found that 329 out of 763 adults said they would travel to outer space in their lifetime, given the chance. Estimate the true proportion of adults who would like to travel to outer space with 92% confidence.
Check requirements: Substituting $\widehat{p}$ for $p$, we have $np \approx 329 \ge 5$ and $nq \approx 434 \ge 5$, so the requirements are met.
$\widehat{p} = 329\,/\,763$
$n = 763$
$92\% \rightarrow z_{\alpha/2} = 1.751$,
Thus, the confidence interval is given by:
$$\frac{329}{763} \pm 1.751 \sqrt{ \frac{ \left(\frac{329}{763}\right) \left(\frac{434}{763}\right)}{763}}$$
$$= (0.3998,0.4626)$$
4. The Gallup Poll found that 27% of adults surveyed nationwide said they had personally been in a tornado. How many adults should be surveyed to estimate the true proportion of adults who have been in a tornado with a 95% confidence interval 5% wide?
Recall $ME = z_{\alpha/2} \sqrt{\frac{\widehat{p}\widehat{q}}{n}}$, so solving for $n$ we have $$n = \widehat{p}\widehat{q} \left( \frac{z_{\alpha/2}}{ME} \right)^2$$
We don't know $\widehat{p}$ or $\widehat{q}$, as the survey hasn't happened yet -- but we can approximate these with the Gallup results. Thus,
$$n = (0.27)(0.73) \left(\frac{1.960}{0.05}\right)^2 \doteq 302.87$$
So one should survey 303 adults. (Actually, one should probably survey more than this to take into account response bias.)
5. A recent study indicated that 29% of the 100 women over age 55 in the study were widows.
• Find a 90% confidence interval for the true proportion of women over age 55 who are widows.
Checking requirements, note that $np = (100)(0.29) = 29 \ge 5$ and $nq = (100)(0.71) = 71 \ge 5$, so the requirements are met.
$90\%$ confidence $\rightarrow z_{\alpha/2} = 1.645$, so the confidence interval is given by
$$0.29 \pm 1.645 \sqrt{\frac{(0.29)(0.71)}{100}}$$ $$= (0.2154, 0.3646)$$
• How large a sample must one take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows?
$$n = pq\left(\frac{z_{\alpha/2}}{ME}\right)^2$$ Using the recent study percentages as approximations for $p$ and $q$, we have $$n = (0.29)(0.71)\left(\frac{1.645}{0.05}\right)^2 \doteq 222.86$$ Thus, one should take a sample of size $223$ women over $55$.
• If no estimate of the sample proportion is available, how large should the sample be to be 90% confident that the estimate is within 0.05 of the true proportion?
In the absence of an estimate of the sample proportion, we error on the side of being overly conservative. The maximum product of $pq$ is $1/4$, so take
$$n = \frac{1}{4} \left(\frac{z_{\alpha/2}}{ME}\right)^2 = 270.6$$
Thus, one should sample 271 women over 55.
6. An organization advertises that on a given poll, 43% answered "yes" to the question "Would you rather have a boring job than no job?", with a margin of error of $\pm1\%$. What did the organization fail to reveal?
The confidence level for the margin of error reported.
7. Upon taking a sample to estimate a population proportion, why is it better to report a confidence interval than $\widehat{p}$, the best point-estimate for this proportion.?
Just reporting $\widehat{p}$ does not indicate the accuracy of the estimate.
8. Find the indicated $z$-scores:
1. Find the $z$-scores associated with $92\%$ confidence
2. Find the $z$-scores associated with $99.5\%$ confidence
1. $\pm1.7507$
2. $\pm2.8070$
9. Express the confidence interval $0.200 \lt p \lt 0.500$ in the form of $\widehat{p} \pm E$
$0.350 \pm 0.150$
10. Find the confidence interval for a proportion if $\widehat{p} = 0.222$ and the margin of error is $0.044$.
$(0.178,0.266)$
11. If a sample is used to estimate a population proportion $p$, find the margin of error $E$ that corresponds to $n = 1000, x = 400$, and $95\%$ confidence.
$0.0304$
12. Construct a $95\%$ confidence interval to estimate a population proportion $p$ if $n=200, x=40$.
$0.145 \lt p \lt 0.255$
13. Construct a $99\%$ confidence interval for a population proportion $p$ if $n=1236, x = 109$
$(0.0674,0.109)$
14. What sample size should be used to estimate a population proportion within $0.045$ with $95\%$ confidence?
$475$
15. What sample size should be used to estimate a population proportion within $2\%$, with $99\%$ confidence, when a prior study estimated $\widehat{p} = 0.14$?
$1998$
16. A medication is suspected of increasing the likelihood of conceiving a girl. Of $574$ pregnancies where the mother was taking this medication during her pregnancy, $525$ of them gave birth to a girl. Construct a $95\%$ confidence interval for the proportion of births that result in a girl when the mother is taking this medication.
$0.892 \lt p \lt 0.937$
17. In one of Mendel's famous genetics experiments with peas, he predicted that $25\%$ of offspring peas would be yellow. He instead saw 152 yellow peas and 428 green peas. Find a $95\%$ confidence interval estimate for the percentage of yellow peas. Do the results contradict his hypothesis?
$0.226 \lt p \lt 0.298$; These results do not contradict his hypothesis, as $0.25$ is included in the confidence interval.
18. In a study of $420,095$ cell phone users, $135$ developed brain cancer or cancer of the nervous system. Prior to this study, it was found that the rate of such cancers was $0.034\%$ for people not using cell phones. Construct a $95\%$ confidence interval for the proportion of cell phone users that develop such cancers. Is there a significant difference between cell phone users and people that don't use cell phones?
$0.000267 \lt p \lt 0.000376$; There does not appear to be a significant difference as $0.00034$ is included in the confidence interval.
19. Assuming we wish to estimate the proportion of American adults who use the internet within $2\%$ and be $99\%$ confident in our results, how many randomly selected adults should we survey?
$4147$ |
# Common Core: 7th Grade Math : Understand Fraction of Outcomes: CCSS.Math.Content.7.SP.C.8a
## Example Questions
← Previous 1
### Example Question #1 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Joe has a bag of marbles: red marbles, , yellow marbles, and blue marbles. If the first marble he draws is a red marble, then what is the probability that he will draw a blue marble on his second try?
Explanation:
Joe starts out with marbles, and of the marbles are blue. This means that the probability of Joe drawing a blue marble from the bag on his first attempt is
Now that Joe has taken a red marble from the bag, we still have blue marbles left, but only a total of marbles left in the bag; thus, the probability of Joe drawing a blue marble on his second attempt is
### Example Question #2 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Dan has a bag of marbles: red marbles, , yellow marbles, and blue marbles, purple marble, and orange. If the first marble he draws is a purple marble, then what is the probability that he will draw a red marble on his second try?
Explanation:
Dan starts out with marbles, and of the marbles are red. This means that the probability of Dan drawing a blue marble from the bag on his first attempt is
Now that Dan has taken a purple marble from the bag, we still have red marbles left, but only a total of marbles left in the bag; thus, the probability of Dan drawing a red marble on his second attempt is
### Example Question #3 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Joe has a bag of marbles: red marbles, , yellow marbles, and blue marbles. If the first marble he draws is a red marble, then what is the probability that he will draw another red marble on his second try?
Explanation:
Joe starts out with marbles, and of the marbles are red. This means that the probability of Joe drawing a red marble from the bag on his first attempt is
Now that Joe has taken a red marble from the bag, we have only red marbles left, and a total of marbles left in the bag; thus, the probability of Joe drawing a red marble on his second attempt is
### Example Question #4 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Joe has a bag of marbles: red marbles, , yellow marbles, and blue marbles. If the first marble he draws is a yellow marble, then what is the probability that he will draw another yellow marble on his second try?
Explanation:
Joe starts out with marbles, and of the marbles are yellow. This means that the probability of Joe drawing a yellow marble from the bag on his first attempt is
Now that Joe has taken a yellow marble from the bag, we have only yellow marbles left, and a total of marbles left in the bag; thus, the probability of Joe drawing a yellow marble on his second attempt is
### Example Question #5 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Joe has a bag of marbles: red marbles, , yellow marbles, and blue marbles. If the first marble he draws is a blue marble, then what is the probability that he will draw another blue marble on his second try?
Explanation:
Joe starts out with marbles, and of the marbles are blue. This means that the probability of Joe drawing a blue marble from the bag on his first attempt is
Now that Joe has taken a blue marble from the bag, we have only blue marbles left, and a total of marbles left in the bag; thus, the probability of Joe drawing a blue marble on his second attempt is
### Example Question #6 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Joe has a bag of marbles: red marbles, , yellow marbles, and blue marbles. If the first marble he draws is a red marble, then what is the probability that he will draw a yellow marble on his second try?
Explanation:
Joe starts out with marbles, and of the marbles are yellow. This means that the probability of Joe drawing a yellow marble from the bag on his first attempt is
Now that Joe has taken a red marble from the bag, we still have yellow marbles left, but only a total of marbles left in the bag; thus, the probability of Joe drawing a yellow marble on his second attempt is
### Example Question #7 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Dan has a bag of marbles: red marbles, , yellow marbles, and blue marbles, purple marble, and orange. If the first marble he draws is a red marble, then what is the probability that he will draw a blue marble on his second try?
Explanation:
Dan starts out with marbles, and of the marbles are blue. This means that the probability of Dan drawing a blue marble from the bag on his first attempt is
Now that Dan has taken a red marble from the bag, we still have blue marbles left, but only a total of marbles left in the bag; thus, the probability of Dan drawing a blue marble on his second attempt is
### Example Question #8 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Dan has a bag of marbles: red marbles, , yellow marbles, and blue marbles, purple marble, and orange. If the first marble he draws is a yellow marble, then what is the probability that he will draw a purple marble on his second try?
Explanation:
Dan starts out with marbles, and of the marbles is purple. This means that the probability of Dan drawing a purple marble from the bag on his first attempt is
Now that Dan has taken a yellow marble from the bag, we still have purple marble left, but only a total of marbles left in the bag; thus, the probability of Dan drawing a purple marble on his second attempt is
### Example Question #9 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Dan has a bag of marbles: red marbles, , yellow marbles, and blue marbles, purple marble, and orange. If the first marble he draws is as orange marble, then what is the probability that he will draw a red marble on his second try?
Explanation:
Dan starts out with marbles, and of the marbles are red. This means that the probability of Dan drawing a red marble from the bag on his first attempt is
Now that Dan has taken an orange marble from the bag, we still have red marbles left, but only a total of marbles left in the bag; thus, the probability of Dan drawing a red marble on his second attempt is
### Example Question #10 : Understand Fraction Of Outcomes: Ccss.Math.Content.7.Sp.C.8a
Dan has a bag of marbles: red marbles, , yellow marbles, and blue marbles, purple marble, and orange. If the first marble he draws is an orange marble, then what is the probability that he will draw another orange marble on his second try? |
# Hasil perhitungan rumus
Rumus
Jawaban
$$2 x ^ { 2 } - 3 x - 1 = 0$$
$\begin{array} {l} x = \dfrac { 3 + \sqrt{ 17 } } { 4 } \\ x = \dfrac { 3 - \sqrt{ 17 } } { 4 } \end{array}$
Hitunglah menggunakan rumus akar
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 3 \right ) \pm \sqrt{ \left ( - 3 \right ) ^ { 2 } - 4 \times 2 \times \left ( - 1 \right ) } } { 2 \times 2 }$
Sederhanakan minus
$x = \dfrac { 3 \pm \sqrt{ \left ( - 3 \right ) ^ { 2 } - 4 \times 2 \times \left ( - 1 \right ) } } { 2 \times 2 }$
$x = \dfrac { 3 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 1 \right ) } } { 2 \times 2 }$
Karena kuadrat genap bilangan negatif adalah bilangan positif, maka hapuslah tanda (-)
$x = \dfrac { 3 \pm \sqrt{ 3 ^ { 2 } - 4 \times 2 \times \left ( - 1 \right ) } } { 2 \times 2 }$
$x = \dfrac { 3 \pm \sqrt{ \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } - 4 \times 2 \times \left ( - 1 \right ) } } { 2 \times 2 }$
Hitung kuadratnya
$x = \dfrac { 3 \pm \sqrt{ \color{#FF6800}{ 9 } - 4 \times 2 \times \left ( - 1 \right ) } } { 2 \times 2 }$
$x = \dfrac { 3 \pm \sqrt{ 9 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) } } { 2 \times 2 }$
Kalikan angkanya
$x = \dfrac { 3 \pm \sqrt{ 9 + \color{#FF6800}{ 8 } } } { 2 \times 2 }$
$x = \dfrac { 3 \pm \sqrt{ \color{#FF6800}{ 9 } \color{#FF6800}{ + } \color{#FF6800}{ 8 } } } { 2 \times 2 }$
Tambahkan $9$ dan $8$
$x = \dfrac { 3 \pm \sqrt{ \color{#FF6800}{ 17 } } } { 2 \times 2 }$
$x = \dfrac { 3 \pm \sqrt{ 17 } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
Kalikan $2$ dan $2$
$x = \dfrac { 3 \pm \sqrt{ 17 } } { \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 \pm \sqrt{ 17 } } { 4 } }$
Pisahkanlah jawabannya
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 + \sqrt{ 17 } } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 - \sqrt{ 17 } } { 4 } } \end{array}$
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### Teacher Guide to Addition and Subtraction
Learning addition and subtraction is easier than you may think. Whether you know it or not, you already know how. In fact, you knew how to add and subtract before you knew the words for numbers. For example, you know that, if you have three cookies in your lunch and your friend Jill has only two cookies, you have one more cookie than she does. Likewise, you know that, if you have three cookies, your friend Ron has none, and you give him one, you now only have two cookies. Addition and subtraction is the same thing. The only difference is that the rest of the story is missing, so, instead of you having cookies in front of you, all you have is the numbers themselves. There are several ways you can improve your math skills:
Counting
The easiest way to learn how to add and subtract is to count it out. Use your fingers and toes, a bowl of dry beans, toy blocks, or anything that you have around. To practice adding, count the number of blocks, for instance, of the first number and set those aside. Next, count out the second number and add those to your first pile. To get your sum, count all the blocks you have set aside.
Learning to subtract is just as easy. Begin by counting out the larger number of blocks. Next, take away the second number. Lastly, count the remaining blocks to get your answer.
Using a Reference Number
The hardest part about counting is knowing what to do if you do not have enough items to count. When this happens, you need to "lock" the biggest number in your head, then start counting on your fingers, or whatever else you have available, from there. For example, if you are adding 9 + 7, you would start by "locking" the number 9 in your head and then using your fingers to count out the rest.
You can also break your problem down into parts by drawing a ladder. For example, if you are subtracting 12 - 7, draw a ladder with 7 and 12 marked and then draw a step between them - say 10. You can then figure out the difference between steps, in this case
10 - 7 = 3 and
12 - 10 = 2.
Next, add 3 + 2 (it equals 5) - so, the answer to 12 - 7 is 5.
Play Games
For practice with addition and subtraction, try playing number-base games, such as Yahtzee or Uno. In each of these games you keep score by adding and subtracting. In addition to games you play with friends, you can always ask your teacher or parent to help you find these types of games online.
Tell a Story
Sometimes, addition and subtraction are easiest when you are able to tell a story with the numbers. For example, if your teacher assigns you a problem, like 3 + 2, you can find the answer by asking yourself a question like, "If I have three video games and I get two for my birthday, how many video games would I have?" |
# How do we find the term n^(th) and sum to n^(th) of the following series?
## 1)1*2^2+2*3^2+3*4^2+...n^(th) term 2)1^2+(1^2+2^2)+(1^2+2^2+3^2)+....n^(th))term
May 29, 2018
${s}_{n} = \frac{1}{12} n \left(n + 1\right) \left(n + 2\right) \left(3 n + 5\right)$
#### Explanation:
${s}_{n} = {\sum}_{k = 1}^{n} k {\left(k + 1\right)}^{2} = {\sum}_{k = 2}^{n + 1} \left(k - 1\right) {k}^{2}$
${s}_{n} = {\sum}_{k = 2}^{n + 1} {k}^{3} - {\sum}_{k = 2}^{n + 1} {k}^{2}$
${s}_{n} = {\sum}_{k = 1}^{n + 1} {k}^{3} - {\sum}_{k = 1}^{n + 1} {k}^{2}$
I won't derive these common sums:
The sum of the squares is
${\sum}_{k = 1}^{N} {k}^{2} = \frac{1}{6} N \left(N + 1\right) \left(2 N + 1\right)$
The sum of cubes is remarkably the square of the triangular numbers, the sums of the natural numbers.
${\sum}_{k = 1}^{N} {k}^{3} = {\left(\frac{1}{2} N \left(N + 1\right)\right)}^{2}$
Now,
${s}_{n} = {\sum}_{k = 1}^{n + 1} {k}^{3} - {\sum}_{k = 1}^{n + 1} {k}^{2}$
$= {\left(\frac{1}{2} \left(n + 1\right) \left(n + 2\right)\right)}^{2} - \frac{1}{6} \left(n + 1\right) \left(n + 2\right) \left(2 \left(n + 1\right) + 1\right)$
$= \frac{1}{4} {\left(n + 1\right)}^{2} {\left(n + 2\right)}^{2} - \frac{1}{6} \left(n + 1\right) \left(n + 2\right) \left(2 n + 3\right)$
$= \frac{1}{12} \left(3 {\left(n + 1\right)}^{2} {\left(n + 2\right)}^{2} - 2 \left(n + 1\right) \left(n + 2\right) \left(2 n + 3\right)\right)$
$= \frac{1}{12} \left(n + 1\right) \left(n + 2\right) \left(3 \left(n + 1\right) \left(n + 2\right) - 2 \left(2 n + 3\right)\right)$
$= \frac{1}{12} \left(n + 1\right) \left(n + 2\right) \left(3 {n}^{2} + 9 n + 6 - 4 n - 6\right)$
${s}_{n} = \frac{1}{12} n \left(n + 1\right) \left(n + 2\right) \left(3 n + 5\right)$
Is that a second sum? I'll leave that to others.
Jun 1, 2018
Second sum: sum_{i=1}^n sum_{j=1}^i j^2 = 1/12 n (n + 1)^2 (n + 2)
#### Explanation:
Let's do the second sum.
The $i$th term is ${\sum}_{j = 1}^{i} {j}^{2}$
There are $n$ of those terms
${S}_{n} = {\sum}_{i = 1}^{n} {\sum}_{j = 1}^{i} {j}^{2}$
The inner sum is known:
sum_{j=1}^i j^2 = 1/6 i (i+1)(2i+1) = i^3/3 + i^2/2 + i/6
${S}_{n} = \frac{1}{3} {\sum}_{i = 1}^{n} {i}^{3} + \frac{1}{2} {\sum}_{i = 1}^{n} {i}^{2} + \frac{1}{6} {\sum}_{i = 1}^{n} i$
The sum of cubes is the square of the triangular number,
${\sum}_{i = 1}^{n} {i}^{3} = {\left(\frac{1}{2} n \left(n + 1\right)\right)}^{2}$
The triangular numbers are the sums of the naturals,
${\sum}_{i = 1}^{n} i = \frac{1}{2} n \left(n + 1\right)$
${S}_{n} = \frac{1}{3} {\left(\frac{1}{2} n \left(n + 1\right)\right)}^{2} + \frac{1}{2} \left(\frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)\right) + \frac{1}{6} \left(\frac{1}{2} n \left(n + 1\right)\right)$
${S}_{n} = \frac{1}{12} n {\left(n + 1\right)}^{2} \left(n + 2\right)$ |
# ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 5 Sets Check Your Progress
## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 5 Sets Check Your Progress
Question 1.
State which of the given collections are set:
(i) Collection of all poor people of Dhanbad.
(ii) Collection of all difficult problems in your maths book.
(iii) Collection of all fools.
(iv) Collection of all countries of Asia.
(v) Collection of four countries of Asia.
(vi) Collection of three cities of India whose name start with the letter ‘J’.
(vii) Collection of all people in this world over 50 year of age.
Solution:
Question 2.
If A = (3, 5, 7, 9, 11}, then write which of the following statements are true. If a statement is not true, mention why.
(i) 3 ϵ A
(ii) 5, 9 ϵ A
(iii) 8 ∉ A
(iv) 7 ∉ A
(v) {3} ϵ A
(vi) {5, 9} ϵ A
Solution:
Question 3.
Write the following sets in the roster farm :
(i) A = (x | x is a month of a year having 30 days}
(ii) B = (x | x = 2n, n ϵ W and n < 5}
(iii) C = (x | x ϵ N and x2 < 40}
(iv) D = (all letters in the word PERMISSION}
(v) E = (x : x ϵ I and x2 < 10}
(vi) F = (x : x ϵ N, 15 < x < 50 and x is divisible by 6}
(vii) the set of whole numbers which are greater than 14 and divisible by 7.
(viii) the set of signs of four fundamental operation of arithmetic.
Solution:
Question 4.
Write the following sets in set builder form :
(i) A = (2, 3, 5, 7, 11, 13, 17, 19}
(ii) B = (all months of a year}
(iii) C = (Monday, Tuesday, Wednesday}
Solution:
Question 5.
Write the following sets in roster form and also in set builder form :
(i) A = {even whole numbers which are less than 50}
(ii) B = {two digit numbers which are perfect square}
(iii) The set of letters in the word MUSSOORIE
Solution:
Question 6.
The sets on the left are in tabular form while the sets on the right are in set builder form. Match them.
(i) {2, 3} – (a) {x/x ϵ N and x < 6}
(ii) {P, A, Y} – (b) {x: x is a prime factor of 6}
(iii) {1, 3, 5} – (c) {x | x is an odd natural number less than 6}
(iv) {1, 2, 3, 4, 5} – (d) {x/x is a letter in word PAPAYA}
Solution:
Question 7.
Classify the following sets as empty set, finite set or infinite set:
(i) The set of all even prime number > 2.
(ii) The set of even prime numbers.
(iii) The set of prime numbers less than one crore.
(iv) {All points on a line segment of length 3cm}.
Solution:
Question 8.
Find the cardinal number of the following sets.
(i) A = {x | x is a consonant in the word HUNDRED}
(ii) B = {x/x is a vowel in the word DEHRADOON}
(iii) C = {x | x ϵ W and x2 < 50} |
# Finite Geometric Sequence
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A finite geometric sequence is a list of numbers (terms) with an ending; each term is multiplied by the same amount (called a common ratio) to get the next term in the sequence.
For example: the sequence 5, 10, 20, 40, 80, … 320 ends at 320. Each term is multiplied by 2 to get the next term.
Note: A slightly different form is the geometric series, where terms are added instead of listed: a + ar + ar2 + ar3 + …. These behave differently, and their sums are different. This article is about the geometric sequence; If you want to learn about the series, see: What is a Geometric Series?
## Nth Term of a Geometric Sequence
The general form of the sequence is a1, a1r2, a1r3, a1r4,… a1r(n – 1)
.We don’t always want to write out the entire sequence all the time, so instead of writing everything out (5, 10, 20, 40, 80, 160…) we can use a much shorter formula. The general formula for the nth term of a geometric sequence is:
an = a1r(n – 1)
Where:
• a1 = the first term in the sequence,
• r = the common ratio.
• n = the nth term.
For the example sequence above, the common ratio is 2 and the first term is 5. We can find out the nth terms by plugging those into the formula:
an = 5 · 2(n – 1).
• First term: 5 · 2(1 – 1) = 5 · 2(0) = 5 · 1 = 5
• Second term: 5 · 2(2 – 1) = 5 · 2(1) = 5 · 2 = 10
• Third term: 5 · 2(3 – 1) = 5 · 2(2) = 5 · 4 = 20
• Fourth term: 5 · 2(4 – 1) = 5 · 2(3) = 5 · 8 = 40
Another example: let’s say you are given 6(3)n – 1 and you’re asked to find the first five terms. Note: The first term in the formula is always in the first position, you know the first term here is 6.
• First term: 6
• Second term: 6 · 32 – 1 = 6 · 31 = 18
• Third term: 6 · 33 – 1 = 6 · 3 2 = 54
• Fourth term: 6 · 34 – 1 = 6 · 3 3 = 162
• Fifth term: 6 · 35 – 1 = 6 · 3 4 = 486
## Sum of a Finite Geometric Sequence
The sum of a finite geometric sequence is given by the formula (Larson, 2014):
Where:
Example question: What is the sum of the first 7 terms of a finite geometric series if the first term (a1 = 1 and the common ratio (r) = 2?
## References
Harrison, B. (2020). Take Your Medicine. Adapted from Section 9.1 in Hughes-Hawlett, Deborah, et.al; Single Variable Calculus; John Wiley & Sons, Inc.; New York; 2002
Larson, R. (2014). College Algebra. Cengage Learning.
Seward, K. (2011). College Algebra: Tutorial 54D: Geometric Sequences and Series. Retrieved August 24, 2020 from: https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut54d_geom.htm
CITE THIS AS:
Stephanie Glen. "Finite Geometric Sequence" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/finite-geometric-sequence/
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## What is 3 percent of 7040?
3% of 7040 is 211.2
# Percentage calculators
Use these percentage calculators to work out percentages quickly. If you want to know how to work them out yourself, follow the examples below to learn how.
## Calculate what percent one value is of another
out of is how many percent?
## Calculate the percentage of a value
What is percent of ?
# The calculations explained
## How do I calculate what percent 75 is of 400?
To calculate 75 of 400 as a percentage, first calculate what one percent of 400 is. To do so, divide 400 by 100.
400 ÷ 100 = 4
Now divide 75 by 4 to get the answer.
75 ÷ 4 = 18%
## How do I calculate what 50% of 200 is?
First calculate what one percent of 200 is by dividing it by 100.
200 ÷ 100 = 2
Now multiply 50 by 2 to get the answer.
50 × 2 = 100
## Exercises
### Question
Find 39% of 316.
To figure out 39% of 316, first divide 316 by 100 to work out one percent: 316 ÷ 100 = 3.16. Now times that by 100 to get the answer. 39 × 3.16 = 123.24
### Question
Out of 600 pencils, if Cuthbert takes 132, what percent of the pencils does Cuthbert have?
Cuthbert has 132 pencils. There are 600 pencils altogether. One percent of 600 is 600 ÷ 100 = 6. Now, divide the amount Cuthbert has by one percent. 132 ÷ 6 = 22% . Cuthbert has 22% of the pencils.
### Question
Out of 125 fidget-spinners, if Faith takes 100, what percent of the fidget-spinners does Faith have?
Faith has 100 fidget-spinners. There are 125 fidget-spinners in total. One % of 125 is 125 ÷ 100 = 1.25. Next, divide the amount Faith has by one percent. 100 ÷ 1.25 = 80% . Faith has 80% of the fidget-spinners.
### Question
A drawer contains 220 fidget-spinners. Pietr is given 25. How many percent of the 220 fidget-spinners does Pietr have?
Pietr has 25 fidget-spinners. There are 220 fidget-spinners in total. One % of 220 is 220 ÷ 100 = 2.2. So, divide the amount Pietr has by one percent. 25 ÷ 2.2 = 11.36% . Pietr has 11.36% of the fidget-spinners.
### Question
From a pile of 300 toys, Jo takes 13 toys. What percent of the total toys does Jo have?
Jo has 13 toys. There are 300 toys in total. 1% of 300 is 300 ÷ 100 = 3. So, divide the amount Jo has by one percent. 13 ÷ 3 = 4.33% . Jo has 4.33% of the toys.
### Question
Out of 250 sweets, if Irene takes 244, what percent of the sweets does Irene have?
Irene has 244 sweets. There are 250 sweets combined. One % of 250 is 250 ÷ 100 = 2.5. Next, divide the amount Irene has by one percent. 244 ÷ 2.5 = 97.6% . Irene has 97.6% of the sweets.
### Question
Find 51 percent of 422.
To calculate 51 percent of 422, first divide 422 by 100 to get 1 percent: 422 ÷ 100 = 4.22. Now times that by 100 to get the answer. 51 × 4.22 = 215.22
### Question
Out of 18 toys, if Candida takes 8, what percent of the toys does Candida have?
Candida has 8 toys. There are 18 toys combined. 1 percent of 18 is 18 ÷ 100 = 0.18. Now, divide the amount Candida has by one percent. 8 ÷ 0.18 = 44.44% . Candida has 44.44% of the toys.
### Question
A bag contains 500 sweets. Bill is given 260. How many percent of the 500 sweets does Bill have?
Bill has 260 sweets. There are 500 sweets altogether. One percent of 500 is 500 ÷ 100 = 5. Next, divide the amount Bill has by one percent. 260 ÷ 5 = 52% . Bill has 52% of the sweets.
### Question
A bag contains 400 coins. Dave is given 137. How many percent of the 400 coins does Dave have?
Dave has 137 coins. There are 400 coins in total. One percent of 400 is 400 ÷ 100 = 4. Next, divide the amount Dave has by one percent. 137 ÷ 4 = 34.25% . Dave has 34.25% of the coins.
### Question
What is 59 percent of 346?
To figure out 59% of 346, first divide 346 by 100 to work out 1%: 346 ÷ 100 = 3.46. Now times that by 100 to get the answer. 59 × 3.46 = 204.14
### Question
80% of 465 is what?
To figure out 80 percent of 465, first divide 465 by 100 to work out 1%: 465 ÷ 100 = 4.65. Now times that by 100 to get the answer. 80 × 4.65 = 372
### Question
A bucket contains 88 sticks. Pietr is given 6. How many percent of the 88 sticks does Pietr have?
Pietr has 6 sticks. There are 88 sticks as a whole. One percent of 88 is 88 ÷ 100 = 0.88. So, divide the amount Pietr has by one percent. 6 ÷ 0.88 = 6.82% . Pietr has 6.82% of the sticks.
### Question
Find 53 percent of 249.
To figure out 53% of 249, first divide 249 by 100 to determine one%: 249 ÷ 100 = 2.49. Now multiply that by 100 to get the answer. 53 × 2.49 = 131.97
### Question
What is 6 percent of 184?
To calculate 6 percent of 184, first divide 184 by 100 to work out one%: 184 ÷ 100 = 1.84. Now multiply that by 100 to get the answer. 6 × 1.84 = 11.04
### Question
If Charlie takes 25 apples from a bag of 50 apples, what percent of the apples does Charlie have?
Charlie has 25 apples. There are 50 apples as a whole. 1% of 50 is 50 ÷ 100 = 0.5. So, divide the amount Charlie has by one percent. 25 ÷ 0.5 = 50% . Charlie has 50% of the apples.
### Question
What is 11% of 415? |
# Solving Equations and Problems Solving
Section 3.2
Solving Equations : The
Properties
Equation vs. Expression
Statements like 5 + 2 = 7 are called
equations.
An equation is of the form
expression = expression
An equation can be labeled as
Addition Property of Equality
Let a, b, and c represent numbers.
If a = b, then
a + c = b + c
and
a – c = b − c
In other words, the same number may be
added to or subtracted from both sides
of an equation without changing the
solution of the equation .
Multiplication Property of Equality
Let a, b, and c represent numbers and
let c ≠ 0. If a = b, then
a · c = b · c and
In other words, both sides of an
equation may be multiplied or divided
by the same nonzero number without
changing the solution of the equation.
Solve for x.
x − 4 = 3
To solve the equation for x, we need to rewrite
the equation in the form
x = number.
To do so, we add 4 to both sides of the
equation.
x − 4 = 3
x − 4 + 4 = 3 + 4 Add 4 to both sides.
x = 7 Simplify.
Check
To check, replace x with 7 in the original
equation.
Original equation Replace x with 7. True.
Since 3 = 3 is a true statement, 7 is the
solution of the equation.
Solve for x
4x = 8
To solve the equation for x, notice that 4
is multiplied by x.
To get x alone, we divide both sides of
the equation by 4 and then simplify .
Check
To check, replace x with 2 in the
original equation.
Original equation Let x = 2. True.
The solution is 2.
Using Both Properties to
Solve Equations
2(2x – 3) = 10
Use the distributive property to simplify
the left side.
4x – 6 = 10
Add 6 to both sides of the equation
4x – 6 + 6 = 10 + 6
4x = 16
Divide both sides by 4.
x = 4
Check
To check, replace x with 4 in the original
equation.
Original equation Let x = 4. True.
The solution is 4.
Section 3.3
Solving Linear Equations in
One Variable
Linear Equations in One
Variable
3x - 2 = 7 is called a linear equation or
first degree equation in one variable.
The exponent on each x is 1 and there
is no variable below a fraction bar.
It is an equation in one variable
because it contains one variable, x.
Make sure you understand which property
to use to solve an equation.
To undo addition of 5, we subtract 5 from both sides. x + 5 - 5 = 8 - 5 Use Addition Property of Equality x = 3 To undo multiplication of 3, we divide both sides by 3. Use Multiplication Property of Equality x = 4
Steps for Solving an Equation
Step 1. If parentheses are present, use the
distributive property.
Step 2. Combine any like terms on each side of the
equation.
Step 3. Use the addition property to rewrite the
equation so that the variable terms are on
one side of the equation and constant terms
are on the other side.
Step 4. Use the multiplication property of equality to
divide both sides by the numerical
coefficient of x to solve .
Step 5. Check the solution in the original equation.
Key Words or Phrases that translate to an equal
sign when writing sentences as equations.
Key Words or Phrases Sentences Equations equals 5 equals 2 plus 3. 5 = 2 + 3 gives The quotient of 8 and 4 gives 2. is/was/will be x is 5. x = 5 yields y plus 6 yields 15. y + 6 = 15 amounts to Twice x amounts to - 8. 2x = - 8 is equal to 36 is equal to 4 times 9. 36 = 4(9)
Section 3.4
Linear Equations in One
Variable and Problem Solving
Problem-Solving Steps
1. UNDERSTAND the problem. During this step,
become comfortable with the problem.
Some ways of doing this are:
Choose a variable to represent the
unknown.
Construct a drawing .
Propose a solution and check it. Pay
careful attention to how you check your
proposed solution. This will help when
writing an equation to model the
problem.
Problem-Solving Steps . . .
2. TRANSLATE the problem into an equation.
3. SOLVE the equation.
4 INTERPRET the results Check the proposed
solution in the stated problem and state |
# Probability
## Probability
#### Introduction
With probability we are dealing with the probability of an event happening (or not happening). An event could be anything from 'obtaining a head when flipping a coin' to 'it raining next Thursday'.
The probability that an event, A, will happen is written as P(A).
The probability that the event A, does not happen is called the complement of A and is written as A'.
As either A must or must not happen then:
P(A') = 1 − P(A)
...as probability of a certainty is equal to 1
Set notation
If A and B are two events then:
A ∩ B represents the event 'both A and B occur'.
A ∪ B represents the event 'either A or B occur'.
#### Mutually Exclusive Events
Two events are mutually exclusive if the event of one happening excludes the other from happening. In other words, they both cannot happen simultaneously.
For exclusive events A and B then:
P(A or B) = P(A) + P(B)
This can be written in set notation as:
P(A ∪ B) = P(A) + P(B)
This can be extended for three or more exclusive events:
P(A or B or C) = P(A) + P(B) + P(C)
Example:
When a fair die is rolled find the probability of rolling a 4 or a 1.
P(4 or 1) = P(4) + P(1) = 1/6 + 1/6 = 2/6 = 1/3
Handy hint:
Exclusive events will involve the words 'or', 'either' or something which implies 'or'.
#### Independent Events
Two events are independent if the occurrence of one happening does not affect the occurrence of the other.
For independent events A and B then:
P(A and B) = P(A) × P(B)
This can be written in set notation as:
P(A ∩ B) = P(A) × P(B)
Again, this result can be extended for three or more events:
P(A and B and C) = P(A) × P(B) × P(C)
Example:
A coin is flipped at the same time as a dice is rolled. Find the probability of obtaining a head and a 5.
P(H and 5) = P(H) × P(5) = 1/2 × 1/6 = 1/12
Handy hint:
Independent events will involve the words 'and', 'both' or something which implies either of these.
Remember 'and' means 'multiply'.
#### Tree diagrams
Most problems will involve a combination of exclusive and independent events. One of the best ways to answer these questions is to draw a tree diagram to cover all the arrangements.
Example:
A bag contains 8 apples. 5 are eating apples and 3 are cooking apples. If 2 apples are drawn, without replacement, find the probability that at least 1 is an eating apple.
To answer this question we will draw a tree diagram. As there are to be two picks we will draw our tree diagram in two stages. The first pick and then the second pick.
We let E, stand for eating apple and C, for cooking apple.
1. In first pick, P(E) = 5/8 and P(C) = 3/8. We write the possibilities on each branch. These branches are exclusive.
2. If an eating apple is picked first, then for the second pick we can either get another eating apple or a cooking apple. Now only 7 apples are left in the bag.
3. However, if a cooking apple is picked first we can then pick either an eating apple or a cooking apple.
P(at least 1 Eating apple) = P(E and E) + P(E and C) + P(C and E) = 20/56 + 15/36 + 15/36 = 50/56
Note: We could have answered this question using the complement...
P(at least 1 Eating apple) = 1 - P(no eating apples) = 1 - P (C and C) = 1 - 6/56 = 50/56 as before
Note: Each pair of branches sum to 1 (they are exclusive). Events between the first and second set of branches are independent events.
If two events, A and B, are not mutually exclusive then the probability that A or B will occur is given by the addition formula:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Don't panic, this just means: The probability of A or B occurring is the probability of A add the probability of B minus the probability that they both occur. This is best seen by an example...
Example:
Pick a card at random from a pack of 52 cards. Find the probability that you pick an ace or a spade. |
Lesson Objectives
• Demonstrate an understanding of Linear Equations in Two Variables
• Demonstrate an understanding of an Ordered Pair (x,y)
• Learn how to create a Cartesian Coordinate Plane
• Learn how to identify the Quadrants in the Cartesian Coordinate Plane
• Learn how to plot an ordered pair (x,y) on the Cartesian Coordinate Plane
## How to Plot an Ordered Pair
In the last lesson, we introduced a new type of equation known as a linear equation in two variables. As an example, suppose we came across the equation:
5x - 2y = 20
For this type of equation, we see that we have two variables, x, and y. When we work with a linear equation in two variables, we have an infinite number of solutions. These solutions are most commonly written as (x,y) values known as an ordered pair. For our above equation, we can show that:
(0,-10), (4,0), and (2,-5) are solutions. Just like with a linear equation in one variable, we can plug in for the variables and simplify. We know our solution is correct when each side simplifies to the same value. Let's take a look at our three ordered pairs as solutions to our equation.
(0,-10)
Plug in a 0 for x and a (-10) for y:
5(0) - 2(-10) = 20
20 = 20
(0,-10) is a solution to our equation.
(4,0)
Plug in a 4 for x and a 0 for y:
5(4) - 2(0) = 20
20 = 20
(4,0) is a solution to our equation.
(2,-5)
Plug in a 2 for x and a (-5) for y:
5(2) - 2(-5) = 20
10 + 10 = 20
20 = 20
(2,-5) is a solution to our equation.
### Cartesian Coordinate Plane
In the next section, we will start to graph linear equations in two variables. In order to perform this action, we must first understand how the coordinate plane works. The Cartesian Coordinate Plane, also known as the Rectangular Coordinate Plane is made up of two number lines, one vertical and one horizontal: The horizontal axis is known as the "x" axis. We can see the blue "x" label at the far right of the horizontal number line. We should be fairly familiar with this horizontal number line. We have used this type of number line throughout pre-algebra and algebra to show integers, add integers, graph inequalities,...etc. We know that numbers increase as we move to the right and numbers decrease as we move to the left. New to us is the vertical number line, which is known as the "y" axis. We can see the purple "y" label at the top of the vertical number line. Numbers on the vertical number line increase as we move up and decrease as we move down. The origin is the point where both number lines intersect. This spot has an ordered pair of (0,0), which means x = 0, and y = 0. The origin is the exact center of our coordinate plane: The coordinate plane is split up into four quadrants. These quadrants are labeled with Roman Numerals as: I, II, III, and IV. Quadrant I starts at the top right and the rest are arranged moving counter-clockwise. It's important to note the values in each quadrant:
I: (+,+) » x and y values are positive
II: (-,+) » x values are negative and y values are positive
III: (-,-) » x and y values are negative
IV: (+,-) » x values are positive and y values are negative
### How to Plot an Ordered Pair
An ordered pair is often referred to as a point. When we plot a point (ordered pair), we are just finding the meeting point of the x location and the y location. We will draw a closed circle or dot at that location. In many cases, we will label the point with its ordered pair (x,y). Let's look at a few examples.
Example 1: Plot each ordered pair and determine its quadrant
(2,5)
To plot the point (2,5), we want an x location of 2. This means we move 2 units to the right on the x-axis (horizontal axis). We want a y location of 5, this means we move 5 units up on the y-axis (vertical axis). We will place and label a point at that location. Additionally, both x and y values are positive. This means our point lies in quadrant I.
Example 2: Plot each ordered pair and determine its quadrant
(-4,-7)
To plot the point (-4,-7), we want an x location of -4. This means we move 4 units left on the x-axis (horizontal axis). We want a y location of -7, this means we move 7 units down on the y-axis (vertical axis). We will place and label a point at that location. Additionally, both x and y values are negative. This means our point lies in quadrant III.
Example 3: Plot each ordered pair and determine its quadrant
(7,-3)
To plot the point (7,-3), we want an x location of 7. This means we move 7 units right on the x-axis (horizontal axis). We want a y location of -3, this means we move 3 units down on the y-axis (vertical axis). We will place and label a point at that location. Additionally, the x value is positive, while the y value is negative. This means our point lies in quadrant IV. |
Using the formula. The altitude of a triangle is the perpendicular distance from the base to the opposite vertex. \begin{align} h=\dfrac{2\sqrt{s(s-a)(s-b)(s-c)}}{b} \end{align}, \begin{align} h=\dfrac{2}{a} \sqrt{\dfrac{3a}{2}(\dfrac{3a}{2}-a)(\dfrac{3a}{2}-a)(\dfrac{3a}{2}-a)} \end{align}, \begin{align} h=\dfrac{2}{a}\sqrt{\dfrac{3a}{2}\times \dfrac{a}{2}\times \dfrac{a}{2}\times \dfrac{a}{2}} \end{align}, \begin{align} h=\dfrac{2}{a} \times \dfrac{a^2\sqrt{3}}{4} \end{align}, \begin{align} \therefore h=\dfrac{a\sqrt{3}}{2} \end{align}. Once you have the triangle's height and base, plug them into the formula: area = 1/2(bh), where "b" is the base and "h" is the height. Then, measure the height of the triangle by measuring from the center of the base to the point directly across from it. 2. In case of an equilateral triangle, all the three sides of the triangle are equal. $This follows from combining Heron's formula for the area of a triangle in terms of the sides with the area formula (1/2)×base×height, where the base is taken as side … The altitude of a triangle to side c can be found as: where S - an area of a triangle, which can be found from three known sides using, for example, Hero's formula, see Calculator of area of a triangle using Hero's formula. Speci cally, from the side to the orthocenter. [25] The sides of the orthic triangle are parallel to the tangents to the circumcircle at the original triangle's vertices. If one angle is a right angle, the orthocenter coincides with the vertex at the right angle. , Solution To solve the problem, use the formula … The altitude or the height from the acute angles of an obtuse triangle lie outside the triangle. This is illustrated in the adjacent diagram: in this obtuse triangle, an altitude dropped perpendicularly from the top vertex, which has an acute angle, intersects the extended horizontal side outside the triangle. Since, the altitude of an isosceles triangle drawn from its vertical angle bisects its base at point D. So, We can determine the length of altitude AD by using Pythagoras theorem. When we construct an altitude of a triangle from a vertex to the hypotenuse of a right-angled triangle, it forms two similar triangles. Also the altitude having the incongruent side as its base will be the angle bisector of the vertex angle. The altitude is the mean proportional between the … A triangle therefore has three possible altitudes. b-Base of the isosceles triangle. In the complex plane, let the points A, B and C represent the numbers Dover Publications, Inc., New York, 1965. Click here to see the proof of derivation and it will open as you click. Click here to see the proof of derivation. The side to which the perpendicular is drawn is then called the base of the triangle. The math journey around altitude of a triangle starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. sin A triangle with one interior angle measuring more than 90° is an obtuse triangle or obtuse-angled triangle. How to Find the Height of a Triangle. {\displaystyle z_{A}} Example 4: Finding the Altitude of an Isosceles Right Triangle Using the 30-60-90 Triangle Theorem. Solution: altitude of c (h) = NOT CALCULATED. Let's visualize the altitude of construction in different types of triangles. The process of drawing the altitude from the vertex to the foot is known as dropping the altitude at that vertex. To calculate the area of a triangle, simply use the formula: Area = 1/2 ah "a" represents the length of the base of the triangle. How to Find the Equation of Altitude of a Triangle - Questions. Let D, E, and F denote the feet of the altitudes from A, B, and C respectively. Here are a few activities for you to practice. A In geometry, the altitude is a line that passes through two very specific points on a triangle: a vertex, or corner of a triangle, and its opposite side at a right, or 90-degree, angle. 1 Here we are going to see, how to find the equation of altitude of a triangle. The side to which the perpendicular is drawn is then called the base of the triangle. In terms of our triangle, this … It is common to mark the altitude with the letter h (as in height), often subscripted with the name of the side the altitude is drawn to. Solving for altitude of side c: Inputs: length of side (a) length of side (b) length of side (c) Conversions: length of side (a) = 0 = 0. length of side (b) = 0 = 0. length of side (c) = 0 = 0. a-Measure of the equal sides of an isosceles triangle. The three (possibly extended) altitudes intersect in a single point, called the orthocenter of the triangle, usually denoted by H.[1][2] The orthocenter lies inside the triangle if and only if the triangle is acute (i.e. Edge a. ... Triangle Formula: The area of a triangle ∆ABC is equal to ½ × BD × AC = ½ × 5 × 8 = 20. Bell, Amy, "Hansen's right triangle theorem, its converse and a generalization", http://mathworld.wolfram.com/KiepertParabola.html, http://mathworld.wolfram.com/JerabekHyperbola.html, http://forumgeom.fau.edu/FG2014volume14/FG201405index.html, http://forumgeom.fau.edu/FG2017volume17/FG201719.pdf, "A Possibly First Proof of the Concurrence of Altitudes", Animated demonstration of orthocenter construction, https://en.wikipedia.org/w/index.php?title=Altitude_(triangle)&oldid=1002628538, Creative Commons Attribution-ShareAlike License. Edge b. a. {\displaystyle \sec A:\sec B:\sec C=\cos A-\sin B\sin C:\cos B-\sin C\sin A:\cos C-\sin A\sin B,}. 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# The Percentile – Explanation & Examples
The definition of percentile is:
“The percentile is the value below which a certain percent of numerical data falls.”
In this topic, we will discuss the percentile from the following aspects:
• What does percentile mean in statistics?
• How to find the percentile?
• Percentile formula.
• Practical questions.
## What does percentile mean in statistics?
The percentile is the value below which a certain percent of numerical data falls.
For example, if you score 90 out of 100 on a certain test. That score has no meaning unless you know what percentile you fall into.
If your score (90 out of 100) is the 90th percentile. This means that you score better than 90% of the test takers.
If your score (90 out of 100) is the 60th percentile. This means that you score better than only 60% of the test takers.
The 25th percentile is the first quartile or Q1.
The 50th percentile is the second quartile or Q2.
The 75th percentile is the third quartile or Q3.
## How to find the percentile?
We will go through several examples.
### – Example 1
For the 10 numbers,10,20,30,40,50,60,70,80,90,100. Find the 30th, 40th, 50th and 100th percentiles.
1. Order the numbers from smallest to largest number.
The data is already ordered, 10,20,30,40,50,60,70,80,90,100.
2. Assign a rank to each value of your data.
values rank 10 1 20 2 30 3 40 4 50 5 60 6 70 7 80 8 90 9 100 10
3. Calculate the ordinal rank for each required percentile. Round the obtained number to the next integer.
Ordinal rank = (percentile/100) X total number of data points.
4. The value with the next rank to the ordinal rank is the required percentile.
The ordinal rank for the 30th percentile = (30/100) X 10 = 3. The next rank is 4 with 40 data value, so 40 is the 30th percentile.
We note that 40 is higher than 10,20,30 or 3 data values/10 data values = 0.3 or 30% of the data.
The ordinal rank for the 40th percentile = (40/100) X 10 = 4. The next rank is 5 with 50 data value, so 50 is the 40th percentile.
We note that 50 is higher than 10,20,30,40 or 4/10 = 0.4 or 40% of the data.
The ordinal rank for the 50th percentile = (50/100) X 10 = 5. The next rank is 6 with 60 data value, so 60 is the 50th percentile.
We note that 60 is higher than 10,20,30,40,50 or 5/10 = 0.5 or 50% of the data.
The ordinal rank for the 100th percentile = (100/100) X 10 = 10. The next rank is 11 with no data value.
In that case, we assume that 100 is the 100th percentile, although it is also the 90th percentile.
It is always that the 100th percentile is the maximum value and the 0th percentile is the minimum value.
### – Example 2
The following is the age in years for 20 participants from a certain survey.
26 48 67 39 25 25 36 44 44 47 53 52 52 51 52 40 77 44 40 45.
Find the 10th, 30th, 60th, 80th percentiles.
1. Order the numbers from smallest to largest number.
25 25 26 36 39 40 40 44 44 44 45 47 48 51 52 52 52 53 67 77.
2. Assign a rank to each value of your data.
values rank 25 1 25 2 26 3 36 4 39 5 40 6 40 7 44 8 44 9 44 10 45 11 47 12 48 13 51 14 52 15 52 16 52 17 53 18 67 19 77 20
Note that repeated values or ties are ranked sequentially as usual.
3. Calculate the ordinal rank for each required percentile. Round the obtained number to the next integer.
Ordinal rank = (percentile/100) X total number of data points.
4. The value with the next rank to the ordinal rank is the required percentile.
The ordinal rank for the 10th percentile = (10/100) X 20 = 2. The next rank is 3 with 26 data value, so 26 is the 10th percentile.
We note that 26 is higher than 25,25 or 2 data values/20 data values = 0.1 or 10% of the data.
The ordinal rank for the 30th percentile = (30/100) X 20 = 6. The next rank is 7 with 40 data value, so 40 is the 30th percentile.
We note that 40 is higher than 25,25,26,36,39,40 or 6 data values/20 data values = 0.3 or 30% of the data.
The ordinal rank for the 60th percentile = (60/100) X 20 = 12. The next rank is 13 with 48 data value, so 48 is the 60th percentile.
We note that 48 is higher than 25,25,26,36,39,40,40,44,44,44,45,47 or 12 data values/20 data values = 0.6 or 60% of the data.
The ordinal rank for the 80th percentile = (80/100) X 20 = 16. The next rank is 17 with 52 data value, so 52 is the 80th percentile.
We note that 52 is higher (in rank) than 25,25,26,36,39,40,40,44,44,44,45,47,48,51,52,52 or 16 data values/20 data values = 0.8 or 80% of the data.
### – Example 2
The following is the daily temperature measurements for 50 days in New York, May to September 1973.
67 72 74 62 56 66 65 59 61 69 74 69 66 68 58 64 66 57 68 62 59 73 61 61 57 58 57 67 81 79 76 78 74 67 84 85 79 82 87 90 87 93 92 82 80 79 77 72 65 73.
Find the 10th, 20th, 30th, 40th, 50th, 60th, 70th, 80th, 90th percentiles.
1. Order the numbers from smallest to largest number.
56 57 57 57 58 58 59 59 61 61 61 62 62 64 65 65 66 66 66 67 67 67 68 68 69 69 72 72 73 73 74 74 74 76 77 78 79 79 79 80 81 82 82 84 85 87 87 90 92 93.
2. Assign a rank to each value of your data.
values rank 56 1 57 2 57 3 57 4 58 5 58 6 59 7 59 8 61 9 61 10 61 11 62 12 62 13 64 14 65 15 65 16 66 17 66 18 66 19 67 20 67 21 67 22 68 23 68 24 69 25 69 26 72 27 72 28 73 29 73 30 74 31 74 32 74 33 76 34 77 35 78 36 79 37 79 38 79 39 80 40 81 41 82 42 82 43 84 44 85 45 87 46 87 47 90 48 92 49 93 50
3. Calculate the ordinal rank for each required percentile. Round the obtained number to the next integer.
Ordinal rank = (percentile/100) X total number of data points.
4. The value with the next rank to the ordinal rank is the required percentile.
The ordinal rank for the 10th percentile = (10/100) X 50 = 5. The next rank is 6 with 58 data value, so 58 is the 10th percentile.
The ordinal rank for the 20th percentile = (20/100) X 50 = 10. The next rank is 11 with 61 data value, so 61 is the 20th percentile.
The ordinal rank for the 30th percentile = (30/100) X 50 = 15. The next rank is 16 with 65 data value, so 65 is the 30th percentile.
The ordinal rank for the 40th percentile = (40/100) X 50 = 40. The next rank is 21 with 67 data value, so 67 is the 40th percentile.
The ordinal rank for the 50th percentile = (50/100) X 50 = 25. The next rank is 26 with 69 data value, so 69 is the 50th percentile.
The ordinal rank for the 60th percentile = (60/100) X 50 = 30. The next rank is 31 with 74 data value, so 74 is the 60th percentile.
The ordinal rank for the 70th percentile = (70/100) X 50 = 35. The next rank is 36 with 78 data value, so 78 is the 70th percentile.
The ordinal rank for the 80th percentile = (80/100) X 50 = 40. The next rank is 41 with 81 data value, so 81 is the 80th percentile.
The ordinal rank for the 90th percentile = (90/100) X 50 = 45. The next rank is 46 with 87 data value, so 87 is the 90th percentile.
We can add this to the above table.
values rank percentile 56 1 57 2 57 3 57 4 58 5 58 6 10th 59 7 59 8 61 9 61 10 61 11 20th 62 12 62 13 64 14 65 15 65 16 30th 66 17 66 18 66 19 67 20 67 21 40th 67 22 68 23 68 24 69 25 69 26 50th 72 27 72 28 73 29 73 30 74 31 60th 74 32 74 33 76 34 77 35 78 36 70th 79 37 79 38 79 39 80 40 81 41 80th 82 42 82 43 84 44 85 45 87 46 90th 87 47 90 48 92 49 93 50
We can plot this data as a box plot with lines for different percentiles.
## Percentile formula
To calculate the percentile for a certain number (x) in your data, use the formula:
percentile = (number of ranks below x/total number of ranks) X 100.
For example, in the table above, the number 58 with a rank = 6.
Number of ranks below 58 = 5, total number of ranks = 50.
The percentile for 58 = (5/50)X 100 = 10th.
Using that formula, we can calculate the percentiles for all numbers in our data.
Generally speaking, the 0th percentile is the minimum value and the 100th percentile is the maximum value.
values rank percentile 56 1 0th 57 2 2th 57 3 4th 57 4 6th 58 5 8th 58 6 10th 59 7 12th 59 8 14th 61 9 16th 61 10 18th 61 11 20th 62 12 22th 62 13 24th 64 14 26th 65 15 28th 65 16 30th 66 17 32th 66 18 34th 66 19 36th 67 20 38th 67 21 40th 67 22 42th 68 23 44th 68 24 46th 69 25 48th 69 26 50th 72 27 52th 72 28 54th 73 29 56th 73 30 58th 74 31 60th 74 32 62th 74 33 64th 76 34 66th 77 35 68th 78 36 70th 79 37 72th 79 38 74th 79 39 76th 80 40 78th 81 41 80th 82 42 82th 82 43 84th 84 44 86th 85 45 88th 87 46 90th 87 47 92th 90 48 94th 92 49 96th 93 50 98th
Although 93 is the 98th percentile, it is also considered the 100th percentile as there is no value in our data that is larger than all our data values.
### Practical questions
1. The following are some percentiles for some daily ozone measurements in New York, May to September 1973.
percentile value 10% 11.00 30% 20.00 70% 49.50 75% 63.25
What percentage of data is less than 20?
What is the third quartile of this data or Q3?
2. The following are daily solar radiation measurements for 20 days in New York, May to September 1973.
236 259 238 24 112 237 224 27 238 201 238 14 139 49 20 193 145 191 131 223.
Construct a table with the rank and percentile for each value.
3. The following are murder rates per 100,000 population for 50 states of the United States of America in 1976.
state value Alabama 15.1 Alaska 11.3 Arizona 7.8 Arkansas 10.1 California 10.3 Colorado 6.8 Connecticut 3.1 Delaware 6.2 Florida 10.7 Georgia 13.9 Hawaii 6.2 Idaho 5.3 Illinois 10.3 Indiana 7.1 Iowa 2.3 Kansas 4.5 Kentucky 10.6 Louisiana 13.2 Maine 2.7 Maryland 8.5 Massachusetts 3.3 Michigan 11.1 Minnesota 2.3 Mississippi 12.5 Missouri 9.3 Montana 5.0 Nebraska 2.9 Nevada 11.5 New Hampshire 3.3 New Jersey 5.2 New Mexico 9.7 New York 10.9 North Carolina 11.1 North Dakota 1.4 Ohio 7.4 Oklahoma 6.4 Oregon 4.2 Pennsylvania 6.1 Rhode Island 2.4 South Carolina 11.6 South Dakota 1.7 Tennessee 11.0 Texas 12.2 Utah 4.5 Vermont 5.5 Virginia 9.5 Washington 4.3 West Virginia 6.7 Wisconsin 3.0 Wyoming 6.9
Construct a table with the rank and percentile for each value.
4. The following are some percentiles of temperature in certain months.
Month 10th 90th 5 57.0 74.0 6 72.9 87.3 7 81.0 89.0 8 77.0 94.0 9 67.9 91.1
For August or Month 8, what percent of temperatures are less than 94?
Which month has the highest spread in its temperatures?
5. The following are some percentiles of per capita income in 1974 for the 4 regions of the US.
region 10th 90th Northeast 3864.4 5259.2 South 3461.5 4812.0 North Central 4274.4 5053.4 West 4041.4 5142.0
Which region has the highest 90th percentile?
Which region has the highest 10th percentile?
1. The percentage of data that is less than 20 is 30% because 20 is 30% percentile.
The third quartile of this data or Q3 is 75% percentile or 63.25.
2. Following the above steps, we can construct the following table:
values rank percentile 14 1 0th 20 2 5th 24 3 10th 27 4 15th 49 5 20th 112 6 25th 131 7 30th 139 8 35th 145 9 40th 191 10 45th 193 11 50th 201 12 55th 223 13 60th 224 14 65th 236 15 70th 237 16 75th 238 17 80th 238 18 85th 238 19 90th 259 20 95th
3. Following the above steps, we can construct the following table:
state value rank percentile North Dakota 1.4 1 0th South Dakota 1.7 2 2th Iowa 2.3 3 4th Minnesota 2.3 4 6th Rhode Island 2.4 5 8th Maine 2.7 6 10th Nebraska 2.9 7 12th Wisconsin 3.0 8 14th Connecticut 3.1 9 16th Massachusetts 3.3 10 18th New Hampshire 3.3 11 20th Oregon 4.2 12 22th Washington 4.3 13 24th Kansas 4.5 14 26th Utah 4.5 15 28th Montana 5.0 16 30th New Jersey 5.2 17 32th Idaho 5.3 18 34th Vermont 5.5 19 36th Pennsylvania 6.1 20 38th Delaware 6.2 21 40th Hawaii 6.2 22 42th Oklahoma 6.4 23 44th West Virginia 6.7 24 46th Colorado 6.8 25 48th Wyoming 6.9 26 50th Indiana 7.1 27 52th Ohio 7.4 28 54th Arizona 7.8 29 56th Maryland 8.5 30 58th Missouri 9.3 31 60th Virginia 9.5 32 62th New Mexico 9.7 33 64th Arkansas 10.1 34 66th California 10.3 35 68th Illinois 10.3 36 70th Kentucky 10.6 37 72th Florida 10.7 38 74th New York 10.9 39 76th Tennessee 11.0 40 78th Michigan 11.1 41 80th North Carolina 11.1 42 82th Alaska 11.3 43 84th Nevada 11.5 44 86th South Carolina 11.6 45 88th Texas 12.2 46 90th Mississippi 12.5 47 92th Louisiana 13.2 48 94th Georgia 13.9 49 96th Alabama 15.1 50 98th
4. For August or Month 8, the percent of temperatures that are less than 94 is 90% since 94 is the 90th percentile.
To see the spread of temperatures for each month, we can see the difference between 90th and 10th percentiles.
Month 10th 90th difference 5 57.0 74.0 17.0 6 72.9 87.3 14.4 7 81.0 89.0 8.0 8 77.0 94.0 17.0 9 67.9 91.1 23.2
The highest difference is for Month 9 or September, so September has the highest spread in its temperatures.
5. Northeast has the highest 90th percentile of 5259.2.
North Central has the highest 10th percentile of 4274.4. |
# Division Property Of Equality - Definition with Examples
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## What is Division Property of Equality?
At times we refer to algebra as generalized arithmetic. Algebraic thinking plants seeds for many higher-order concepts as well helps in understanding the other domains of science.
The equation is a mathematical sentence with an equal sign, and it is one of the essential elements Algebra.
Example
The operations of addition, subtraction, multiplication and division do not change the truth value of any equation.
The division property of equality states that when we divide both sides of an equation by the same non-zero number, the two sides remain equal.
That is, if a, b, and c are real numbers such that a = b and c ≠0, then = .
Example: Consider the equation 12 = 12.
Divide both sides by 4.
12 =12
3 = 3
That is, the equation still remains true.
Note, that the divisor cannot be zero as the division by zero is not defined.
This property is used in solving equations.
Example: 6x = 24
Divide both sides by 6.
6x =24x
6x6=246
x=4
To check we can substitute the value of x in the original equation.
6×4=24
24 = 24
Example: Rhea bought 7 notebooks for \$21. What is the cost of each notebook?
Let a be the cost of each notebook. Then, 7 times a is the total cost, \$21.
That is, 7a = 21.
By the division property of equality, if you divide both sides by the same non-zero number, the equality still holds true. So, divide both sides by 7.
17a =21
a=3
Therefore, the cost of each notebook is \$3.
Fun Fact: Another form of the property is if a, b, c, and d are real numbers such that a = b, c = d and c, d ≠0, then a c = b d .
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Video: The Chain Rule for Multivariate Functions
Let π§ = π(π₯, π¦) over the curve π₯ = π‘Β² + 1, π¦ = π‘Β² β 1. Write an expression for dπ§/dπ‘, giving your answer in terms of the partial derivatives of π.
03:17
Video Transcript
Let π§ be equal to π of π₯, π¦ over the curve π₯ equals π‘ squared plus one, π¦ equals π‘ squared minus one. Write an expression for dπ§ by dπ‘, giving your answer in terms of the partial derivatives of π.
We see that π§ is a multivariate function. Itβs a function in both π₯ and π¦. And weβre asked to find an expression for dπ§ by dπ‘ giving our answers in terms of the partial derivatives of π. Those are ππ ππ₯ β β thatβs the partial derivative of π with respect to π₯ β β and ππ ππ¦ β the partial derivative of π with respect to π¦. Now, our functions for π₯ and π¦ are themselves differentiable functions of π‘. And so we can quote the chain rule for one independent variable. Here, thatβs π‘.
This says that suppose π₯ and π¦ are differentiable functions of π‘ and π§ is a function in π₯ and π¦ as a differentiable function. Then this means π§ is π of π₯ of π‘, π¦ of π‘ is a differentiable function of π‘. Such that dπ§ by dπ‘, the derivative of π§ with respect to π‘ is equal to the partial derivative of π§ with respect to π₯ times the derivative of π₯ with respect to π‘ plus the partial derivative of π§ with respect to π¦ times the ordinary derivative of π¦ with respect to π‘. Well, in this case, π₯ is the function π‘ squared plus one and π¦ is the function π‘ squared minus one. These are both polynomials in π‘, and we know polynomials are differentiable over their entire domain. So π₯ and π¦ are both differentiable functions.
Weβre going to evaluate dπ₯ by dπ‘, the ordinary derivative of π₯ with respect to π‘, and dπ¦ by dπ‘, the ordinary derivative of π¦ with respect to π‘. Now, we know that, to differentiate a power term, we multiply the entire term by the exponent, then reduce that exponent by one. So in both cases, the derivative of π‘ squared with respect to π‘ is two π‘. But we also know that the derivative of a constant is zero. So both one and negative one differentiate to zero. And we see that dπ₯ by dπ‘ and dπ¦ by dπ‘ are both equal to two π‘.
The question tells us to give our answer in terms of the partial derivatives of π. Those are ππ§ ππ₯ and ππ§ ππ¦. And so we can say that the derivative of π§ with respect to π‘ is equal to ππ§ ππ₯ times two π‘ plus ππ§ ππ¦ times two π‘. We simplify a little, and we can write this as two π‘ ππ§ ππ₯ plus two π‘ ππ§ ππ¦, where we said ππ§ ππ₯ is the first partial derivative of π§ with respect to π₯ and ππ§ ππ¦ is the first partial derivative of π§ with respect to π¦.
But remember, π§ is a function in π₯ and π¦, where π₯ is π‘ squared plus one and π¦ is π‘ squared minus one. So we can rewrite ππ§ ππ₯ as shown and ππ§ ππ¦ as shown. And so we found an expression for dπ§ by dπ‘ in terms of the partial derivatives of π. Itβs two π‘ times the first partial derivative of the function in π‘ squared plus one and π‘ squared minus one with respect to π₯ plus two π‘ times the first partial derivative of the function in π‘ squared plus one and π‘ squared minus one with respect to π¦.
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339 Developing Fraction Concepts LEARNER OUTCOMES After reading this chapter and engaging in the embedded activities and reflections, you should be able to: 15.1 Describe and give examples for fractions constructs. 15.2 Name the types of fractions models and describe activities for each. 15.3 Explain foundational concepts of fractional parts, including iteration and partitioning, and connect these ideas to CCSS-M expectations. 15.4 Illustrate examples across fraction models for developing the concept of equivalence. 15.5 Compare fractions in a variety of ways and describe ways to teach this topic conceptually. 15.6 Synthesize how to effectively teach fraction concepts. F ractions are one of the most important topics students need to understand in order to be successful in algebra and beyond, yet it is an area in which U.S. students struggle. NAEP test results have consistently shown that students have a weak understanding of fraction con- cepts (Sowder & Wearne, 2006; Wearne & Kouba, 2000). This lack of understanding is then translated into difficulties with fraction computation, decimal and percent concepts, and the use of fractions in other content areas, particularly algebra (Bailey, Hoard, Nugent, & Geary, 2012; Brown & Quinn, 2007; National Mathematics Advisory Panel, 2008). Therefore, it is absolutely critical that you teach fractions well, present fractions as interesting and important, and commit to helping students understand the big ideas. BIG IDEAS For students to really understand fractions, they must experience fractions across many constructs, including part of a whole, ratios, and division. Three categories of models exist for working with fractions—area (e.g., 1 3 of a garden), length (e.g., 3 4 of an inch), and set or quantity (e.g., 1 2 of the class). Partitioning and iterating are ways for students to understand the meaning of fractions, especially numerators and denominators. Equal sharing is a way to build on whole-number knowledge to introduce fractional amounts. Equivalent fractions are ways of describing the same amount by using different-sized fractional parts. Fractions can be compared by reasoning about the relative size of the fractions. Estimation and reasoning are important in teaching understanding of fractions. 15 CHAPTER
# Developing Fraction Concepts
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Developing Fraction ConceptsLearner OutCOmesAfter reading this chapter and engaging in the embedded activities and reflections, you should be able to:
15.1 Describe and give examples for fractions constructs.
15.2 Name the types of fractions models and describe activities for each.
15.3 Explain foundational concepts of fractional parts, including iteration and partitioning, and connect these ideas to CCSS- M expectations.
15.4 Illustrate examples across fraction models for developing the concept of equivalence.
15.5 Compare fractions in a variety of ways and describe ways to teach this topic conceptually.
15.6 Synthesize how to effectively teach fraction concepts.
Fractions are one of the most important topics students need to understand in order to be successful in algebra and beyond, yet it is an area in which U.S. students struggle. NAEP
test results have consistently shown that students have a weak understanding of fraction con-cepts (Sowder & Wearne, 2006; Wearne & Kouba, 2000). This lack of understanding is then translated into difficulties with fraction computation, decimal and percent concepts, and the use of fractions in other content areas, particularly algebra (Bailey, Hoard, Nugent, & Geary, 2012; Brown & Quinn, 2007; National Mathematics Advisory Panel, 2008). Therefore, it is absolutely critical that you teach fractions well, present fractions as interesting and important, and commit to helping students understand the big ideas.
Big IDEAS◆◆ For students to really understand fractions, they must experience fractions across many
constructs, including part of a whole, ratios, and division.
◆◆ Three categories of models exist for working with fractions— area (e.g., 13 of a garden), length (e.g., 34 of an inch), and set or quantity (e.g., 12 of the class).
◆◆ Partitioning and iterating are ways for students to understand the meaning of fractions, especially numerators and denominators.
◆◆ Equal sharing is a way to build on whole- number knowledge to introduce fractional amounts.
◆◆ Equivalent fractions are ways of describing the same amount by using different-sized fractional parts.
◆◆ Fractions can be compared by reasoning about the relative size of the fractions. Estimation and reasoning are important in teaching understanding of fractions.
15C h a p t e r
M15_VAND8930_09_SE_C15.indd 339 08/12/14 11:54 AM
340 Chapter 15 Developing Fraction Concepts
meanings of Fractions
Fraction understanding is developmental in nature. Fraction experiences should begin as early as first grade. In the Common Core State Standards students partition shapes and refer to the fractional amounts in grades 1 and 2 as “equal shares.” In grade 3, fractions are a major emphasis, with attention to using fraction symbols, exploring unit fractions (fractions with numerator 1), and comparing fractions. Grade 4 focuses on fraction equivalence and begins work on fraction operations (Chapter 16). This emphasis over years of time is an indication of both the complexity and the importance of fraction concepts. Students need significant time and experiences to develop a deep conceptual understanding of this important topic.
Understanding a fraction is much more than recognizing that 35 is three shaded parts of
a shape partitioned into five sections. Fractions have numerous constructs and can be repre-sented as areas, quantities, or on a number line. This section describes these big ideas. The next sections describe how to teach the concepts of fractions.
Fraction ConstructsUnderstanding fractions means understanding all the possible concepts that fractions can rep-resent. One of the commonly used meanings of fraction is part‐whole. But many who research fraction understanding believe students would understand fractions better with more emphasis across other meanings of fractions (Clarke, Roche, & Mitchell, 2008; Lamon, 2012; Siebert & Gaskin, 2006).
Pause & ReflectBeyond shading a region of a shape, how else are fractions represented? Try to name three ideas. ●
Part- Whole. Using the part‐whole construct is an effective starting point for building mean-ing of fractions (Cramer & Whitney, 2010). Part‐whole can be shading a region, part of a group of people (3
5 of the class went on the field trip), or part of a length (we walked 312 miles).
measurement. Measurement involves identifying a length and then using that length as a measurement piece to determine the length of an object. For example, in the fraction 58, you can use the unit fraction 18 as the selected length and then count or measure to show that it takes five of those to reach 58. This concept focuses on how much rather than how many parts, which is the case in part‐whole situations (Behr, Lesh, Post, & Silver, 1983; Martinie, 2007).
Division. Consider the idea of sharing \$10 with 4 people. This is not a part‐whole scenario, but it still means that each person will receive one‐fourth (1
4) of the money, or 212 dollars. Divi-
sion is often not connected to fractions, which is unfortunate. Students should understand and feel comfortable with the example here written as 10
4 , 4)10, 10 , 4, 224, and 21
2 (Flores, Samson, & Yanik, 2006).
Operator. Fractions can be used to indicate an operation, as in 45 of 20 square feet or 2
3 of the audience was holding banners. These situations indicate a fraction of a whole number, and students may be able to use mental math to determine the answer. This construct is not emphasized enough in school curricula (Usiskin, 2007). Just knowing how to represent frac-tions doesn’t mean students will know how to operate with fractions, which occurs in various other areas in mathematics ( Johanning, 2008).
ratio. Discussed at length in Chapter 18, the concept of ratio is yet another context in which fractions are used. For example, the fraction 1
4 can mean that the probability of an event is one in four. Ratios can be part‐part or part‐whole. For example, the ratio 34 could be the ratio of those wearing jackets (part) to those not wearing jackets (part), or it could be part‐whole, meaning those wearing jackets (part) to those in the class (whole).
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Meanings of Fractions 341
Why Fractions are DifficultStudents build on their prior knowledge, meaning that when they encounter situations with fractions, they naturally use what they know about whole numbers to solve the problems. Based on the research, there are a number of reasons students struggle with fractions. They include:
• There are many meanings of fractions (see later section “Fraction Constructs”).• Fractions are written in a unique way.• Students overgeneralize their whole- number knowledge (McNamara & Shaughnessy,
2010).
It is important for a teacher to help students see how fractions are alike and different from whole numbers. An explanation of common misapplications of whole- number knowledge to fractions follows, along with ways you can help.
misconception 1. Students think that the numerator and denominator are separate values and have difficulty seeing them as a single value (Cramer & Whitney, 2010). It is hard for them to see that 34 is one number.
How to Help: Find fraction values on a number line. This can be a fun warm- up activity each day, with students placing particular values on a classroom number line or in their math journals. Measure with various levels of precision (e.g., to the nearest eighth- inch). Avoid the phrase “three out of four” (unless talking about ratios or probability) or “three over four.” Instead, say “three fourths” (Siebert & Gaskin, 2006).
misconception 2. Students do not understand that 23 means two equal‐sized parts (although
not necessarily equal- shaped objects). For example, students may think that the following shape shows 34 green, rather than 12 green:
How to Help: Ask students to create their own representations of fractions across various ma-nipulatives and on paper. Provide problems like the one illustrated here, in which all the par-titions are not already drawn.
misconception 3. Students think that a fraction such as 15 is smaller than a fraction such as 110
because 5 is less than 10. Conversely, students may be told the reverse— the bigger the denom-inator, the smaller the fraction. Teaching such rules without providing the reason may lead students to overgeneralize that 15 is more than 7
10.
How to Help: Use many visuals and contexts that show parts of the whole. For example, ask students if they would rather go outside for 12 of an hour, 14 of an hour, or 1
10 of an hour. Use the idea of fair shares: Is it fair if Mary gets one- fourth of the pizza and Laura gets one- eighth? Ask students to explain why this is not fair and who gets the larger share.
misconception 4. Students mistakenly use the operation “rules” for whole numbers to com-pute with fractions— for example, 12 + 1
2 = 24.
How to Help: Use many visuals and contexts. Emphasize estimation (see later section in this chapter) and focus on whether answers are reasonable or not.
Students who make these errors do not understand fractions. Until they understand frac-tions meaningfully, they will continue to make errors by overapplying whole‐number con-cepts (Cramer & Whitney, 2010; Siegler et al., 2010). The most effective way to help students
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342 Chapter 15 Developing Fraction Concepts
reach higher levels of understanding is to use multiple representations, multiple approaches, and explanation and justification (Harvey, 2012; Pantziara & Philippou, 2012). This chapter is designed to help you help students deeply understand fractions.
Complete Self- Check 15.1: Meanings of Fractions
models for Fractions
There is substantial evidence to suggest that the effective use of visuals in fraction tasks is important (Cramer & Henry, 2002; Siebert & Gaskin, 2006). Unfortunately, textbooks rarely incorporate manipulatives, and when they do, they tend to be only area models (Hodges, Cady, & Collins, 2008). This means that students often do not get to explore fractions with a variety of models and/or do not have sufficient time to connect the visuals to the related con-cepts. In fact, what appears to be critical in learning is that the use of physical tools leads to the use of mental models, which builds students’ understanding of fractions (Cramer & Whitney, 2010; Petit, Laird, & Marsden, 2010).
Properly used, tools can help students clarify ideas that are often confused in a purely symbolic model. Sometimes it is useful to do the same activity with two different representa-tions and ask students to make connections between them. Different representations offer dif-ferent opportunities to learn. For example, an area model helps students visualize parts of the whole. A linear model shows that there is always another fraction to be found between any two numbers— an important concept that is underemphasized in the teaching of fractions. Some students are able to make sense of one representation, but not another. Importantly, students need to experience fractions in real- world contexts that are meaningful to them (Cramer & Whitney, 2010). These contexts may align well with one representation and not as well with another. For example, if students are being asked who walked the farthest, a linear model is more likely to support their thinking than an area model.
Table 15.1 provides an at‐a‐glance explanation of three types of models— area, length, and set— defining the wholes and their related parts for each model. Using appropriate rep-resentations and different categories of models broaden and deepen students’ (and teachers’) understanding of fractions.
An increasing number of Web resources are available to help represent fractions. One excellent source, though subscription based, is Conceptua Fractions (https://www.youtube .com/watch?v=7OJTjYxWCIU), developed by Conceptua Math. This site offers free tools that help students explore various fraction concepts using area, set, and length models (includ-ing the number line). The activities can be prescribed by the teacher and contain formative assessment resources.
standards for mathematical Practice
mP5. Use appropriate tools strategically.
Table 15.1 mODeLs FOr FraCtiOn COnCePts anD HOW tHey COmPare
model What Defines the Whole
What Defines the Parts What the Fraction means
Area The area of the defined region
Equal area The part of the area covered as it relates to the whole unit
Length or number line The unit of distance or length
Equal distance/length The location of a point in relation to 0 and other values on the number line
Set Whatever value is determined as one set
Equal number of objects
The count of objects in the subset as it relates to the defined whole
Source: Based on Petit, Laird, & Marsden (2010).
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Models for Fractions 343
area modelsWith these visuals, fractions are based on parts of an area. See Figure 15.1 for examples. Area is a good place to begin fraction explorations because it lends itself to equal sharing and partitioning.
Circular Fraction pieces are the most commonly used area model. One advantage of the circular model is that it emphasizes the part‐whole concept of fractions and the meaning of the relative size of a part to the whole (Cramer, Wyberg, & Leavitt, 2008). Other area models in Figure 15.1 demonstrate how different shapes can be the whole. Grid or Dot paper provides flexibility in selecting the size of the whole and the size of the parts (see Blackline Masters 5–11 for a selection). Many commercial versions of area manipulatives are available, including cir-cular and rectangular pieces, pattern blocks, geoboards, and tangrams. Activity 15.1 (adapted from Roddick & Silvas- Centeno, 2007) uses pattern blocks to help students develop concepts of partitioning and iterating.
Circular “pie” pieces Rectangular regionsAny piece can be selected as the whole.
Fourths on a geoboard
Paper foldingDrawings on grids or dot paper Pattern blocks
One-fifth or two-tenthsOne-third,5—15
Figure 15.1 Area models for fractions.
CCSSM: 1.G.a.3; 2.G.a.3; 3.NF.a.1
Playground FractionsCreate this “playground” with your pattern blocks (see pattern Block playground activity page). It is the whole. For each fraction below, find the pieces of the playground and draw it on your paper. For grades 1 and 2 use words, not fraction symbols (e.g., half of, one- half, or four- thirds).
12 playground 1
3 playground
112 playgrounds 2
3 playground
2 playgrounds 43 playground
Activity 15.1
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344 Chapter 15 Developing Fraction Concepts
Length modelsWith length models, lengths or measurements are compared instead of areas. Either physical materials are compared on the basis of length or number lines are subdivided, as shown in Figure 15.2. Length models are very important in developing student understanding of fractions, yet they are not widely used in classrooms. Recent reviews of research on fractions (Petit et al., 2010; Siegler et al., 2010) report that the number line helps students understand a fraction as a number (rather than one number over another number) and helps develop other fraction concepts. In a report completed by the Institute of Educational Sciences (IES), the researchers prepared recom-mendations for supporting the learning of fractions (Siegler et al., 2010), advising teachers to:
Help students recognize that fractions are numbers and that they expand the number system beyond whole numbers. Use number lines as a central representational tool in teaching this and other fraction concepts from the early grades on. (p. 1)
Linear models are closely connected to the real‐world contexts in which fractions are commonly used, such as measuring. Music, for example, is an excellent opportunity to explore 12, 14, 18, and 1
16 in the context of notes (Goral & Wiest, 2007). In fact, Courey, Balogh, Siker, and Paik (2012) found that connecting fractions to measures in music significantly improved stu-dent understanding of fractions.
One length manipulative, Cuisenaire rods, has pieces in lengths of 1 to 10 measured in terms of the smallest strip or rod. Each length is a different color for ease of identification. Strips of paper or adding‐machine tape, also a length model, can be folded to produce student‐made fraction strips.
Technology Note. Virtual Cuisenaire rods and accom-panying activities can be found at online at various web-
sites such as the University of Cambridge’s NRICH Project. ■
Rods or strips provide flexibility because any length can represent the whole. For example, if you wanted students to work with 14 and 18, select the brown Cuisenaire rod, which is 8 units long. Therefore, the four rod (purple) becomes 12, the two rod (red) becomes 1
4, and the one rod (white) becomes 18. For
exploring twelfths, put the orange rod and red rod together to make a whole that is 12 units long.
Cuisenaire rods consist of the following colors and lengths:
White
Red
Light green
Purple
Yellow
Dark green
Black
Brown
Blue
Orange
1
2
3
4
5
6
7
8
9
10
The number line is a significantly more sophisticated length model than the physical tools described previously (Bright, Behr, Post, & Wachsmuth, 1988), but it is an essen-tial model that needs be emphasized more in the teaching of fractions (Clarke et al., 2008; Flores et al., 2006; Siegler et al., 2010; Usiskin, 2007; Watanabe, 2006).
Fraction strips or Cuisenaire rods
0 1 214
12
34
54
64
74
Measurement tools
Folded paper strips
0 1 2 3 4 5 6 7 8 9 10
Figure 15.2 Length or measurement models for fractions.
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Models for Fractions 345
Like with whole numbers, the number line is used to compare the relative size of num-bers. Importantly, the number line reinforces that fact that there is always one more fraction to be found between two fractions. The following activity (based on Bay‐Williams & Martinie, 2003) is a fun way to use a real‐world context to engage students in thinking about fractions through a linear model.
CCSSM: 3.NF.a.2a, b; 3.NF.a.3a, b, d
Who Is Winning?Use Who Is Winning? activity page and give students paper strips or ask them to draw a number line. this activity can be done two ways (depending on your lesson goals). First, ask students to use reasoning to answer the question “Who is winning?” Students can use reasoning strategies to compare and decide. Second, students can locate each person’s position on a number line. explain that the friends below are playing “red Light, Green Light.” the fractions tell how much of the distance they have already moved. Can you place these friends on a line to show where they are between the start and finish? Second, rather than place them, ask students to use reasoning to answer the question “Who is winning?”
Mary: 34 harry: 12 Larry: 56
han: 58 Miguel: 59 angela: 23
this game can be differentiated by changing the value of the fractions or the number of friends (fractions). the game of “red Light, Green Light” may not be familiar to eLLs. Modeling the game with people in the class and using estimation are good ways to build background and support students with disabilities.
Activity 15.2
engLisH Language Learners
stuDents with
sPeCiaL neeDs
set modelsIn set models, the whole is understood to be a set of objects, and subsets of the whole make up fractional parts. For example, 3 objects are one‐fourth of a set of 12 objects. The set of 12 in this example represents the unit, the whole or 1. The idea of referring to a collection of counters as a single entity makes set models difficult for some students. Putting a piece of yarn in a loop around the objects in the set to help students “see” the whole. Figure 15.3 illustrates several set models for fractions.
A common misconception with set models is to focus on the size of a subset rather than the number of equal sets in the whole. For example, if 12 counters make a whole, then a sub-set of 4 counters is one‐third, not one‐fourth, because 3 equal sets make the whole. However, the set model helps establish important connections with many real‐world uses of fractions and with ratio concepts.
Two color counters are an effective set manipulative. Counters can be flipped to change their color to model vari-ous fractional parts of a whole set. Any countable objects (e.g., a box of crayons) can be a set model (with one box being the unit or whole). The following activity uses your students as the whole set. It can be done as an energizer, warm- up, or full lesson.
35
915
Two-color counters in arrays. Rows and columns help show parts. Each array makes a whole. Here or are yellow.
Two-color counters insets showing red. The whole must be clearly indicated.
131
12 makes 1 whole
23
69Objects. Shows are cars.or
Figure 15.3 Set models for fractions.
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346 Chapter 15 Developing Fraction Concepts
Students must be able to explore fractions across the three area, length, and set models. As a teacher, you will not know whether they really understand the meaning of a fraction such as 14 unless you have seen a student represent one- fourth using all three models.
FormaTive assessmenT Notes. A straightforward way to assess students’ knowledge of a fractional amount is to give them a piece of paper folded into thirds. Write area, length, and
set at the top of each section and give them a fractional value (e.g., 34). Observe as they (1) draw a
picture and (2) write a sentence describing a context or example for the selected fraction. This can be done exactly for commonly used fractions or can be an estimation activity with fractions like 31
58. ■
Technology Note. Virtual manipulatives are available for all three models. Virtual manipulatives have been found to positively affect student achievement, especially when
they are paired with using the actual manipulatives ( Moyer- Packenham, Ulmer, & Anderson, 2012). Recommended sites include:
Cyberchase (PBS): Cyberchase is a popular television series. Their website offers videos that model fractions with real‐world connections and activities such as “Thirteen Ways of Looking at a Half” (fractions of geometric shapes) and “Make a Match” (concept of equivalent fractions).Illuminations (NCTM) Fractions Model: Explore length, area, region, and set models of fractions, including fractions greater than one, mixed numbers, decimals, and percentages.Math Playground Fraction Bars: On this site you can explore fractional parts, the concepts of numerator and denominator, and equivalence.National Library of Virtual Manipulatives: This site offers numerous models for exploring fractions, including fraction bars and fraction pieces. There is also an applet for comparing and visualizing fractions. ■
Complete Self- Check 15.2: Models for Fractions
Fractional Parts
The first goal in the development of fractions should be to help students construct the idea of fractional parts of the whole— the parts that result when the whole or unit has been partitioned into equal‐sized portions or fair shares. (Recall that Table 15.1 describes the meanings of parts and wholes across each type of model.)
Students understand the idea of separating a quantity into two or more parts to be shared fairly among friends. This is the beginning of understanding fractions and in the CCSS- M
CCSSM: 3.NF.a.1; 3.NF.a.3b
Class FractionsUse a group of students as the whole— for example, six students if you want to work on halves, thirds, and sixths. Invite them to come to the front of the room. Say to the group, “If you [are wearing tennis shoes, have brown hair, etc.], move to the right. If not, move to the left.” ask everyone, “What fraction of our friends [are wearing tennis shoes]?” Invite them to whisper to a partner, then share with the class.
Change the number of selected students in the whole. You can also ask, “how many friends in one- half of this group? One- fourth of this group?” Connect to the symbols for fractions (e.g., 12, 13, and 16), asking students to write the fraction that tells how many students [are wearing tennis shoes]. also, if you have six students in your group, and you have three in your subgroup, then students are likely to say “ three- sixths” and “ one- half.” Discuss that these values are equivalent.
Activity 15.3
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Fractional Parts 347
occurs in grades 1 and 2. In grade 3 and beyond, students make connections between the idea of fair (equal) shares and fractional parts. The next three sections describe ideas foundational to finding equal shares.
Fraction size is relativeA fraction by itself does not describe the size of the whole or the size of the parts. A fraction tells us only about the relationship between the part and the whole. Consider the following situation:
pizza Fallacy: Mark is offered the choice of a third of a pizza or a half of a pizza. Because he is hungry and likes pizza, he chooses the half. his friend Jane gets a third of a pizza but ends up with more than Mark. how can that be?
One-half ofa pizza One-third of
a pizza
Do you wanthalf of a pizzaor a third of
a pizza?
The visual illustrates how Mark got misdirected in his choice. The point of the “pizza fallacy” is that whenever two or more fractions are discussed in the same context, one cannot assume (as Mark did) that the fractions are all parts of the same size whole. Teachers can help students understand fractional parts if they regularly ask, “What is the whole?” or “What is the unit?”
Comparing two fractions with any representation can be made only if both fractions are parts of the same size whole. For example, when using Cuisenaire rods, 2
3 of a light green strip cannot be compared to 25 of an orange strip.
PartitioningSectioning a shape into equal‐sized parts is called partitioning. When a brownie (or other area) has been partitioned into four equal shares, the parts are called fourths. Explain to students, “We call these fourths. The whole is cut into four parts. All of the parts are the same size— fourths.” The words for fractional parts (e.g., halves, thirds, fourths, eighths, and so on) are introduced before the symbols. In the CCSS- M the words for fractional parts (as they relate to equal shares) are introduced in grades 1 and 2; the sym-bols for fractions are introduced in grade 3. Figure 15.4 illus-trates sixths across area, length, and set models.
Partitioning with area models. When partitioning an area into fractional parts, students need to be aware that (1) the fractional parts must be the same size, though not nec-essarily the same shape; and (2) the number of equal‐sized parts that can be partitioned within the unit determines the fractional amount (e.g., partitioning into 4 parts means each
Sixths
Sixths
Sixths
One
One
One
Cuisenaire rods
Pattern blocks
Area Model
Length Model
Set Model
Counters
Figure 15.4 Which of these shapes are partitioned into sixths? Explain why or why not for each.
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part is one- fourth of the unit). It is important for students to understand, however, that some-times visuals do not show all the partitions. For example, consider the following picture:
Referring back to the two criteria, a student might think, “If I partitioned this so that all pieces were the same size, then there will be four parts; therefore, the smaller partitioned region represents one‐fourth”—not one‐third, as many students without a conceptual under-standing of fractional parts might suggest.
Some manipulatives, like fraction bars or fraction circles, can mislead students to believe that fractional parts must be the same shape as well as the same size. Color tiles can be used to create rectangles that address this misconception. Ask students to describe the fractional parts in a rectangle, such as the one illustrated here:
Students who recognize that each color represents thirds understand that fractional parts must be the same size, and that the shape of the thirds may be different.
Area models are the first types of models to use in teaching fractional parts. Young stu-dents, in particular, tend to focus on shape, when the focus should be on equal‐sized parts. Activity 15.1 is an example of how you can use pattern blocks to focus on partitioning into equal‐sized parts. You can build on this activity by building other shapes that use different pattern block pieces and then have students figure out how much each piece is of the whole. View how a classroom teacher helps his students understand parts of the whole using pattern blocks.
Activity 15.4 uses partitioned drawings to develop the concept of fractional parts.
CCSSM: 1.G.a.3; 2.G.a.3; 3.NF.a.1
Partitioning: Fourths or Not Fourths?Use Fourths or Not Fourths activity page showing examples and nonexamples (which are very important to use with students with disabilities) of fourths (see Figure 15.5).
ask students to identify the wholes that are correctly divided into fourths (equal shares) and those that are not. For each response, have students explain their reasoning. repeat with other fractional parts, such as thirds or eighths. to challenge students, ask them to draw shapes that fit each of the four categories listed on the next page for other fractional parts, such as sixths. (See Sixths or Not Sixths activity page.)
Activity 15.4
stuDents with
sPeCiaL neeDs
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Fractional Parts 349
In the preceding activity, the shapes fall in each of the following categories:
1. Same shape, same size: (a) and (f ) [equivalent]2. Different shape, same size: (e) and (g) [equivalent]3. Different shape, different size: (b) and (c) [not equivalent]4. Same shape, different size: (d) [not equivalent]
FormaTive assessmenT Notes. Activity 15.4 is a good diagnostic interview to assess whether students un-
derstand that it is the size that matters, not the shape. If students get all correct except (e) and (g), they hold the misconception that parts should be the same shape. Future tasks are needed that focus on equivalence. For example, you can ask students to take a square and subdivide a picture themselves, as in Activity 15.5. ■
(a) (b) (c)
(e)
(d)
(f) (g)
Figure 15.5 Given a whole, find fractional parts.
CCSSM: 1.G.a.3; 2.G.a.3; 3.NF.a.1
Finding (All the) Fair SharesGive students dot paper and ask them to enclose a region that lends itself to partitioning with a particular fractional part. For example, they might enclose a 3- by- 6 rectangle if they are going to partition into thirds. ask students to find a way to partition the rectangle into thirds. then redraw another rectangle that is the same size whole and partition it a different way to show thirds. ask students to find a way to show thirds where the thirds are different shapes. See how many ways they can find. For eLLs, fraction parts sound like whole numbers (e.g., fourths and fours). Be sure to emphasize the th on the end and explicitly discuss the difference between four areas and a fourth of an area.
Activity 15.5
engLisH Language Learners
Partitioning with Length models. The explanation of partitioning in the CCSS- M may be difficult to interpret. For example:
3.NF.A.2b: Represent a fraction a/b on a number line diagram by marking off a lengths 1/b from 0. Recognize that the resulting interval has size a/b and that its endpoint locates the number a/b on the number line. (CCSSO, 2010, p. 24)
Put more simply, students need to be able to partition a number line into fourths and realize that each section is one- fourth:
0 1 2
14
04 2
434
Size of interval: 34
44 5
464
74
84 9
4104
114
Number lines are difficult for students. Students may ignore the size of the interval (McNamara & Shaughnessy, 2010; Petit et al., 2010). Students can develop an understanding of the number line by folding paper strips. Provide examples where the shaded sections are in
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different positions and where partitioning isn’t already shown to strengthen students’ under-standing of equal parts. Activity 15.6 and Activity 15.7 provide such opportunities, one with paper strips and one with number lines.
CCSSM: 3.NF.a.1; 3.NF.a.2a, b
What Fraction Is Colored?prepare a set of paper strips prior to doing this activity (you can cut 1-inch wide pieces of 8.5" by 11" paper and shade, or cut pieces of adding machine tape). Color the strips so that they have a fractional amount shaded in various positions (not just left justified!) (Sarazen, 2012). here are a few examples:
explain that the strip represents one whole. Give each student a paper strip. ask students to explain what fraction is colored and explain how they know. a common misconception is for students to count parts and call each of these one- third. If a stu-dent makes this error, ask if the parts are the same- sized and if not to partition to make same-sized parts. Use toothpicks or uncooked spaghetti to illustrate the partitions:
Students can also justify their reasoning by measuring the length of each partition.
Activity 15.6
Using paper strips can help students better understand the number line, the focus of the next activity.
CCSSM: 3.NF.a.1; 3.NF.a.2a, b
How Far Did Nicole Go?Give students number lines partitioned such that only some of the partitions are showing. Use a context such as walking to school. For each number line, ask, “how far has Nicole gone? how do you know?”
10
10
Nicole’s distance
Nicole’s distance
10
Nicole’s distance
Students can justify their reasoning by measuring the length of each partition.
Activity 15.7
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Locating a fractional value on a number line is particularly challenging but very import-ant for students to be able to do. Shaughnessy (2011) found four common errors students make in working with the number line: They use incorrect notation, change the unit (whole), count the tick marks rather than the space between the marks, and count the ticks marks that appear without noticing any missing ones. This is evidence that we must use number lines more extensively in exploring fractions (most real- life contexts for fractions are measurement related).
Partitioning is a strategy commonly used in Singapore (a high- performing country on international mathematics assessments) as a way to solve story problems. Consider the follow-ing story problem (Englard, 2010):
a nurse has 54 bandages. Of those, 29 are white and the rest are brown. how many of them are brown?
A bar diagram can be used as a tool for solving the problem. A student first partitions a strip into nine parts and then figures out the equal shares of bandages for each partition:
6 6 6 6 6 6 6 6 6
Did you notice that this is an example of fraction as operator? These types of partitioning tasks are good building blocks for multiplying with fractions.
Partitioning with set models. Students can partition sets of objects such as coins, counters, or baseball cards. When partitioning sets, students may confuse the number of counters in a share with the name of the share. In the example in Figure 15.4, the 12 counters are partitioned into 6 sets— sixths. Each share or part has two counters, but it is the number of shares that makes the partition show sixths. As with the other models, when the equal parts are not already figured out, then students may not see how to partition. Students seeing a picture of two cats and four dogs might think 2
4 are cats (Bamberger, Oberdorf, & Schultz‐Ferrell, 2010). Consider the following problem:
eloise has 6 trading cards, andre has 4 trading cards, and Lu has 2 trading cards. What fraction of the trading cards does Lu have?
Understanding that parts of a whole must be partitioned into equal‐sized parts across dif-ferent models is an important step in conceptualizing fractions and provides a foundation for exploring sharing and equivalence tasks, all of which are prerequisites to performing fraction operations (Cramer & Whitney, 2010).
sharing tasksAn important recommendation by the IES research team on ways to help students learn frac-tions states, “Build on students’ informal understanding of sharing and proportionality to develop initial fraction concepts” (Siegler et al., 2010, p. 1). In particular, they suggest using equal- sharing activities to develop the concepts of fraction, equivalence, and ordering of frac-tions. See equal Sharing Stories expanded Lesson for a lesson designed for grades 1 or 2.
Students in the early grades partition by thinking about fair shares (division). Sharing tasks are generally posed in the form of a simple story problem. Four friends are sharing two cookies. How many cookies will each friend get? Then problems become slightly more difficult: Suppose there are four cookies to be shared fairly among three children. How much will each child get? See how eduardo reasons about this sharing situation. Students initially perform sharing tasks by
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distributing items one at a time. When this process leaves leftover pieces, students must figure out how to subdivide so that every group (or person) gets a fair share. Contexts that lend to subdividing an area include cookies, brownies, sandwiches, pizzas, and so on.
Pattern blocks are a good tool to focus on equal shares because each piece is not an equal share, so creating shapes with pattern blocks and asking about equal shares helps students focus on the important idea of fair (equal) shares. Ask students to create a “cookie” using the six different pattern block shapes and ask, “Can this cookie be shared fairly with 6 people?” (Ellington & Whit-enack, 2010). The answer is “no.” Then, ask students to build a cookie that can be shared fairly.
Sharing brownies is a classic activity that focuses on partitioning to make equal shares (see, for example, Empson, 2002). Using concrete tools such as dough can make sharing accessible even to kindergartners (Cwikla, 2014).
Figure 15.6 Ten brownies shared with four children.
CCSSM: 1.G.a.3; 2.G.a.3
Cookie Dough: Cut Me a Fair Share!Give students a ball of dough and a plastic knife. explain that they are going to be finding a way to share each group of cookies fairly with a group of students. Start with an example that is not too difficult. For example:
Four friends want to share ten brownies so that each friend gets the same amount of brownies. how much will each friend get?
Invite students to shape their dough into squares for brownies and then show how to share them fairly with four friends, using a paper knife if necessary. encourage students to share their ways of thinking about this problem. a strategy many students will use for this problem is to deal out two brownies to each child and then halve each of the remaining brownies (see Figure 15.6).
then, offer a selection of other sharing tasks with different numbers of brownies and different number of sharers (see addi-tional examples below).
Activity 15.8
“Kids and Cookies” is an excellent online tool for sharing cookies (both round and rectan-gular). Display the situations on an interactive whiteboard and ask for different ways to share fairly (you can begin with whole numbers and increase in difficulty) (Center for Technology and Teacher Education, n.d.).
The relationship between the number of things to be shared and the number of sharers determines problem difficulty. Students’ initial strategies for sharing involve halving, so a good place to begin is with two, four, or even eight sharers. Here are some examples:
5 brownies shared with 2 children5 brownies shared with 4 children7 brownies shared with 4 children2 brownies shared with 4 children4 brownies shared with 8 children3 brownies shared with 4 children
These can be prepared as Brownie Sharing Cards. The last example, three brownies shared with four children, is more challenging because there are more sharers than items, and it involves more than just finding halves. One strategy is to partition each brownie into four parts and give each child one- fourth from each brownie— a total of three- fourths. Stu-dents (even adults) are surprised at the relationship between the problem and the answer. Felisha explains the fractional amount each of 5 children get when sharing 2 cookies, but loses track of what the whole is in determining each person’s share.
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Fractional Parts 353
Pause & ReflectIs this pattern (that three brownies shared among four children means each gets three- fourths) true for any sharing tasks? Why is this true? ●
Because the level of difficulty of these sharing tasks varies, it is useful for creating a tiered lesson. In a tiered lesson, the goal (sharing) is the same, but the specific tasks vary in their challenge. Figure 15.7 shows how one teacher offers these three tiers for her lesson on sharing brownies (Williams, 2008).
Sharing into thirds or sixths is more challenging because students cannot rely on halving to get to the answer. Here are some examples:
4 pizzas shared with 3 children7 pizzas shared with 6 children5 pizzas shared with 3 children4 pizzas shared with 6 children
Figure 15.8 shows how a student partitioned to solve “5 pizzas shared with 3 children.” This took much guess and check, at which point the teacher asked, “Can you see a pattern in how you have divided the pizza and how many people are sharing?” At this point, the student noticed a pattern: If there are three people, the remaining pizzas need to be partitioned into thirds.
As students report their answers, it is important to empha-size the equivalence of different representations (Flores & Klein, 2005). For example, in the case of three people sharing four pizzas, the answer might be noted on the board this way:
43= 1
13= 1 +
13
iteratingIn whole- number learning, counting precedes and helps stu-dents to add and later subtract. This is also true with fractions. Counting fractional parts, or iterating, helps students understand the relationship between the parts (the numerator) and the whole (the denominator). The iterative concept is most clear when focusing on these two ideas about fraction symbols:
• The top number (numerator) counts.• The bottom number (denominator) tells what is being
counted.Figure 15.8 Elizabeth explains a pattern for finding equal shares of a pizza.
Figure 15.7 Example of a tiered lesson for the brownie- sharing problem.
Tier 1 task: For students who still need experience with halving
Tier 2 task: For students comfortable with halving and ready to try other strategies
Tier 3 task: For students ready to solve tasks in which students combine halving with new strategies
How can 2 people share 3 brownies? How can 4 people share 3 brownies? How can 3 people share 5 brownies?
How can 2 people share 5 brownies? How can 3 people share 4 brownies? How can 3 people share 2 brownies?
How can 4 people share 3 brownies? How can 3 people share 5 brownies? How can 6 people share 4 brownies?
How can 3 people share 4 brownies? How can 6 people share 4 brownies? How can 5 people share 4 brownies?
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354 Chapter 15 Developing Fraction Concepts
Students need to understand that 34, for example, can be thought of as a count of three parts
called fourths (Post, Wachsmuth, Lesh, & Behr, 1985; Siebert & Gaskin, 2006; Tzur, 1999).If you know the kind of part you are counting, you can tell when you get to one, when
you get to two, and so on. Students should be able to answer the question, “How many fifths are in one whole?” just as they know how many ones are in ten. However, the 2008 National Assessment of Education Progress (NAEP) results indicated that only 44 percent of students answered this question correctly (Rampey, Dion, & Donahue, 2009). This is the focus of Activities 15.9 through 15.11.
CCSSM: 3.NF.a.1; 3.NF.a.2a, b
More, Less, or Equal to One WholeGive students a collection of fractional parts (all the same-size pieces) and indicate the kind of fractional part they have. For example, if done with Cuisenaire rods, the collection might have seven light green rods/strips with a caption or note indicating “each piece is 18.” the task is to decide if the collection is less than one whole, equal to one whole, or more than one whole. ask students to draw pictures or use symbols to explain their answer. as students count each collection of parts, discuss the relationship to one whole. ask questions that help students focus on the meaning of the numerator and denominator, such as “Why did we get almost two wholes with seven-fourths, and yet we don’t even have one whole with ten-twelfths?”
Activity 15.9
After exploring Activity 15.9 with same- sized pieces, try Activity 15.10, which returns to using the pattern blocks to help students focus on the size of the parts, not the number of pieces or partitions (Champion & Wheeler, 2014; Ellington & Whitenack, 2010).
CCSSM: 3.NF.a.1; 3.NF.a.2a, b
Pattern Block Creaturesask students to build a pattern Block Creature that fits with a set of rules (a creature represents one- whole). these rules can begin with just stating a fractional quantity for a color, such as “the red trapezoid is one- fourth of the creature.” But, more constraints can be added to the rules. For example:
the blue parallelogram is one- sixth of the creature. Use at least two colors to build your creature.the yellow hexagon is one- half of the creature. Use three colors to build your creature.Green triangles are one- third of your creature. Use four different colors to build your creature.after a student creates their creature, they can sketch the creature on paper and write the rule below it. Other stu-
dents can critique the creature to see if it follows the rules it was given. alternatively, the student can write their rule as “the red trapezoid is ______ of my creature” and trade it with another student to see if they can figure out the fractional amount.
Activity 15.10
CCSSM: 3.NF.a.1; 3.NF.a.3a, c
Calculator Fraction CountingMany calculators, like the tI-15, display fractions in correct fraction format and offer a choice of showing results as mixed numbers or simple fractions. ask students to type in a fraction (e.g., 14) and then + and the fraction again. to count, press 0 , , , repeating to get the number of fourths wanted. the display will show the counts by fourths and also the number of times that the key has been pressed. ask students questions such as the following: “how many fourths to get to 3?” “how many fifths to get to 2?” these can get increasingly more challenging: “how many fourths to get to 41
2?” “how many two- thirds to get to 6? estimate and then count by two- thirds on the calculator.” Students, particularly students with disabilities, should coordinate their counts with fraction models, adding a new fourths piece to the pile with each count.
Activity 15.11
stuDents with
sPeCiaL neeDs
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Fractional Parts 355
The TI- 15 display can be switched back and forth from mixed number to fractions, rein-forcing the equivalence of values such as 12
3 and 53. A variation on Activity 15.11 is to show
students a mixed number such as 318 and ask how many counts of 18 on the calculator it will take
to count that high. The students should try to stop at the correct number, 258 , before pressing
the mixed‐number key.Iterating applies to all models but is particularly connected with length models because
iteration is much like measuring. Consider that you have 212 yards of ribbon and are trying to
figure out how many fourths you can cut. You can draw a strip and start counting (iterating) the fourths:
1 ft 1 ft ft1–2
Using a ribbon that is 14 of a yard long as a measuring tool, a student marks off ten fourths:
Students can participate in many tasks that involve iterating lengths, including ones where they are asked to find what the whole or unit is.
CCSSM: 3.NF.a.1; 3.NF.a.2a, b
A Whole Lot of FunUse a Whole Lot of Fun activity page and a strip of paper like the one here:
tell students that this strip is three- fourths of one whole (unit). ask students to sketch strips of the other lengths on their paper (e.g., 52). You can repeat this activity by selecting other values for the starting amount and selecting different fractional values to sketch. a context, such as walking, is effective in helping students make sense of the situation. Be sure to use fractions less than and greater than 1 and mixed numbers.
Activity 15.12
Notice that to solve the task in Activity 15.12, students first partition the piece into three sec-tions to find 14 and then iterate the 14 to find the other lengths.
Iterating can be done with area models. Display some circular fractional pieces in groups as shown in Figure 15.9. For each collection, tell students what type of piece is being shown and simply count them together: “One‐fourth, two‐fourths, three‐fourths, four‐fourths, five‐fourths.” Ask, “If we have five‐fourths, is that more than one whole, less than one whole, or the same as one whole?” To reinforce the piece size even more, you can slightly alter your language to say, “One one- fourth, two one- fourths, three one- fourths,” and so on.
Iteration can also be done with set models. For example, show a collection of two‐color counters and ask questions such as, “If 5 counters is one‐fourth of the whole, how much of the whole is 15 counters?” These problems can be framed as engaging puzzles for students.
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356 Chapter 15 Developing Fraction Concepts
For example: “Three counters represent 18 of my set; how big is my set?” If the fraction is not a unit fraction, then students first partition and then iterate. For example, “Twenty count-ers represents 2
3 of my set; how big is my set?” first requires finding find 13 (10 counters), then iterating that three times to get 30 counters in three- thirds (one whole). Counting Count-ers: Find the part Activity Page and Counting Counters: Find the Whole Activity Page provides such problems for students to solve.
Pause & ReflectWork through the exercises in Figures 15.10 and 15.11. If you do not have access to Cuisenaire rods or counters, just draw lines. What can you learn about student understanding of partitioning and iterating if they are able to solve problems in Figure 15.10 but not 15.11? If students are stuck, what con-texts for each model can be used to support their thinking? ●
The partitioning and iterating questions are challenging yet very effective at helping students reflect on the meanings of the numerator and denominator and understand the mean-ing of fractions.
FormaTive assessmenT Notes. The tasks in Figures 15.10 and 15.11 can be used as performance assessments.
If students are able to solve these types of tasks, they can parti-tion and iterate. That means they are ready to do equivalence and comparison tasks. If they are not able to solve problems such as these, provide a range of similar tasks, using real- life contexts and involving area, length, and set models. ■
Fraction notationThe way that we write fractions with a top and a bottom num-ber and a bar between is a convention— an arbitrary agree-ment for how to represent fractions. However, understanding of the convention can be clarified by giving explicit attention to the meaning of the numerator and the denominator as part of iterating activities. Students can understand the idea of halves and fourths, yet not understand the meaning of the symbols 12 and 14. In the CCSS- M, understanding the symbols for fractions is an emphasis in grade 3. Fractions should include examples that are less than 1, equal to one (e.g., 44), and greater than 1 (e.g., 48 and 43). Engage students in iterating tasks and using symbols, then pose questions to make sense of the symbols, such as:
What does the numerator in a fraction tell us?What does the denominator in a fraction tell us?What might a fraction equal to 1 look like?How do you know if a fraction is greater than or less than 1? Greater than or less than 2?
Here are some likely explanations for the top and bottom numbers from third graders:
5 fourths
3 fourths
10 twelfths
10 fourths
How much moreto get to a secondwhole quesadilla?
Wow! Ten-fourths:How many
Is ten-twelfths as much as ten-fourths? Is it as much as five-fourths?
Figure 15.9 Iterating fractional parts in an area model.
—find one-fourth.
—find two-thirds.
—find five-thirds.
If this rectangle is one whole,
If 8 counters are a whole set, how many are in one-fourth of a set?
If 15 counters are a whole, how many counters make three-fifths?
If 9 counters are a whole, how many are in five-thirds of a set?
If dark green is one whole, what rod is two-thirds?
If brown is the whole, find one-fourth.
If dark green is one whole, what rod is three-halves?
Figure 15.10 Given the whole and the fraction, find the part.
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Fractional Parts 357
• The numerator is the counting number. It tells how many shares or parts we have. It tells how many have been counted.
• The denominator tells what size piece is being counted. For example, if there are four parts in a whole, then we are counting fourths.
Making sense of symbols requires connections to visuals. Illustrating what 5
4 looks like in terms of pizzas (area), on a number line (length), or connected to filling bags with objects (set) will help students make sense of this value.
One of the best things we can do for students is to empha-size equivalence and different ways to write fractional amounts.
Pause & ReflectWhat fraction notation might you use for the visual here (the large square represents one unit)? ●
There are (at least) three ways to notate this quantity:
54 1
14
14
+14
+14
+14
+14
Do you think that students would be able to describe this quantity in all three ways? In the fourth National Assessment of Educational Progress (NAEP), fewer than half of the seventh graders assessed knew that 51
4 was the same as 5 + 14 (Kouba et al., 1988).
Throughout this chapter we have been including fractions less than 1 and fractions greater than 1. This helps students develop understanding of fractions as values that come between whole numbers (or can be equivalent to whole numbers). Too often, students aren’t exposed to numbers equal to or greater than 1 (e.g., 66, 52 or 41
4), so when these values are added into the mix (no pun intended!), students find them confusing.
The term improper fraction is used to describe fractions that are greater than one, such as 52. This term can be a source of confusion as the word improper implies that this representation is not acceptable, which is not the case at all— in fact, it is often the preferred representation in algebra. Instead, try not to use this phrase and instead use “fraction” or “fraction greater than 1.” Note that the word improper is not used in the CCSS- M content standards.
If you have counted fractional parts beyond a whole, as discussed in the previous section, your students already know how to write 13
6 or 135 . Ask students to use a model to illustrate these
values and find equivalent representations using wholes and fractions (mixed numbers). Using connecting cubes was the most effective way to help students see both forms for recording fractions greater than 1 (Neumer, 2007) (see Figure 15.12). Students identify one cube as the unit fraction (1
5) for the problem (125 ). They count out 12 fifths and build wholes. Conversely,
they could start with the mixed number, build it, and find out how many total cubes (or fifths) were used. Repeated experiences in building and solving these tasks will help students to notice a pattern that actually explains the algorithm for moving between mixed numbers and fractions greater than 1.
Help students move from physical models to mental images. Challenge students to figure out the two equivalent forms by just picturing the stacks in their heads. A good explanation for
If this rectangle is one-third, what could the whole look like?
If this rectangle is three-fourths, draw a shape that could be the whole.
If this rectangle is four-thirds, what rectangle could be the whole?
If purple is one-third, what rods are the whole?
If dark green is two-thirds, what rod is the whole?
If yellow is five-fourths, what rod is one whole?
If 4 counters are one-half of a set, how big is the set?
If 12 counters are three-fourths of a set, how many counters are in the full set?
If 10 counters are five-halves of a set, how many counters are in one set?
Figure 15.11 Given the part and the fraction, find the whole.
standards for mathematical Practice
mP1. Make sense of problems and persevere in solving them.
standards for mathematical Practice
mP8. Look for and express regularity in repeated reasoning.
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358 Chapter 15 Developing Fraction Concepts
314 might be that there are 4 fourths in one whole, so there are
8 fourths in two wholes and 12 fourths in three wholes. The extra fourth makes 13 fourths in all, or 13
4 . (Note the iteration concept playing a role.)
Do not push the standard algorithm (multiply the bot-tom by the whole number and add the top), as it can interfere with students making sense of the relationship between the two and their equivalence.
Complete Self- Check 15.3: Fractional parts
equivalent Fractions
As discussed in Chapter 14, equivalence is a critical but often poorly understood concept. This is particularly true with fraction equivalence. In the CCSS- M, fraction equivalence and com-parisons are emphasized in grade 3 and applied in grade 4 (and beyond) as students engage in computation with fractions. Students cannot be successful in fraction computation without a strong understanding of fraction equivalence.
Conceptual Focus on equivalence
Pause & ReflectHow do you know that 46 = 2
3? Before reading further, think of at least two different explanations. ●
Here are some possible answers to the preceding question:
1. They are the same because you can simplify 46 and get 23.2. If you have a set of 6 items and you take 4 of them, that would be 46. But you can make the
6 into 3 groups, and the 4 would be 2 groups out of the 3 groups. That means it’s 23.
3. If you start with 23, you can multiply the top and the bottom numbers by 2, and that will
give you 46, so they are equal.4. If you had a square cut into 3 parts and you shaded 2, that would be 2
3 shaded. If you cut all 3 of these parts in half, that would be 4 parts shaded and 6 parts in all. That’s 46, and it would be the same amount.
All of these answers are correct. But let’s think about what they tell us. Responses 2 and 4 are conceptual, although not as efficient. The procedural responses, 1 and 3, are efficient but do not indicate conceptual understanding. All students should eventually be able to write an equivalent fraction for a given fraction. At the same time, the procedures should never be taught or used until the students understand what the result means. Consider how different the procedure and the concept appear to be:
Whole (5 cubes) or 2 wholes and
5–5
5–5
2–5
12—5
2–5
Figure 15.12 Connecting cubes are used to represent the equivalence of 12
5 and 225.
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Equivalent Fractions 359
Concept: Two fractions are equivalent if they are representations for the same amount or quantity— if they are the same number.Procedure: To get an equivalent fraction, multiply (or divide) the top and bottom numbers by the same nonzero number.
Rushing too quickly to the algorithm can impede students’ conceptual understanding of fractions and fraction equivalence. Be patient!
equivalent Fraction modelsThe general approach to helping students create an understanding of equivalent fractions is to have them use contexts and models to find different names for a fraction (see Figure 15.13 for examples). This is the first time in students’ experience that they are seeing that a fixed quantity can have more than one name (actually an infinite number of names). Area models are a good place to begin understanding equivalence.
= 112
412
26
= 16
13
== 16
26
13
== 118
618
13
== 172
2472
13
== 19
39
13
13
14
18
=48
12
=24
Grid paper
Filling in regions with fraction pieces
Dot paper
Paper folding
23
46
812
or
Rectangle only partially subdivided
1326
first fold
second fold
third fold412
Figure 15.13 Area models for equivalent fractions.
CCSSM: 3.NF.a.1; 3.NF.3a, b, c
Making StacksSelect a manipulative that is designed for exploring fractions (e.g., pattern blocks, tangrams, fraction strips, or fraction circles). prepare fraction cards with different fractional amounts, such as 23, 12, 34, 1, 32, 43, and 2. (Note: you may want to begin with fractions less than 1, then move to fractions equal to and greater than 1.) Working individually or with a partner, students first identify the whole. then they see how many stacks they can make on top of the whole. a stack must use the same- sized piece. ask students to record all the possibilities they find (they can color the shapes and write the fraction). after completing several examples, have students look at the fractions they wrote for a stack and describe or write about the patterns they notice.
Activity 15.13
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360 Chapter 15 Developing Fraction Concepts
In a classroom discussion following this activity, you can help students reason about equiv-alent fractions by asking them to consider what other equivalencies are possible (and justify their thinking). For example, ask, “What equivalent fractions could you find if we had six-teenths in our fraction kit? If you could have a piece of any size at all, what other fraction names are possible?”
The following activity moves from using manipulatives to sketches on paper.
standards for mathematical Practice
mP3. Construct viable arguments and critique the reasoning of others.
CCSSM: 3.NF.a.1; 3.NF.3a, b, c
Dot Paper EquivalencesUse Fraction Names activity page, which includes three different grids with a fraction shaded (each enclosed area represents one whole). ask students how many fraction names they think the first problem has. then ask them to see how many they can find (working individually or in partners). Invite students to share and explain the fraction names they found for problem 1. repeat for the next two problems. alternatively, cut this page into three task cards, laminate the cards, and place each at a station along with an overhead pen. have students rotate in partners to a station and see how many fraction names they can find for that shape (using the pen as needed to show their ways). rotate to the next station. You may also want to refer to the Dot paper equivalences expanded Lesson.
to make additional pictures, create your own using your choice of Grid or Dot paper (see Blackline Masters 5–11). (Figure 15.13 includes an example drawn on an isometric grid). the larger the size of the whole, the more names the activity will generate.
Activity 15.14
The Dot Paper Equivalences activity involves what Lamon (2012) calls “unitizing”—that is, given a quantity, finding different ways to chunk the quantity into parts in order to name it. She points out that this is a key ability related not only to equivalent fractions but also to proportional reasoning, especially in the comparison of ratios.
Length models should be used in activities similar to the Making Stacks task. Asking stu-dents to locate 25 and 4
10 on a number line, for example, can help them see that the two fractions are equivalent (Siegler et al., 2010). Rods or paper strips can be used to designate both a whole and a part, as illustrated in Figure 15.14. Students use smaller rods to find fraction names for the given part. To have larger wholes or values greater than one whole, use a train of two or three rods. Folding paper strips is another method of creating fraction names. In the example shown in Figure 15.14, one‐half is subdivided by successive folding in half. Other folds would produce other names, and these possibilities should be discussed if no one tries to fold the strip in an odd number of parts.
CCSSM: 3.NF.a.3a, b, d; 4.NF.a.2
Stretching Number LinesUsing elastic strips has been effective in helping students understand equivalence of fractions and compare fractions (harvey, 2012).
Cut strips of elastic (about 1 meter or yard in length). hold the elastic taut and mark off ten partitions on each. hand one out to each pair of students. ask students to use their stretching number line to find the place on a table that represents the fraction of the distance across the table. For each pair, ask: ask “Which distance is greater, or are they equal?”
a. 510 of the distance across.
b. 12 of the distance across.
a. 310 of the distance across.
b. 38 of the distance across. (Note: they have to rethink the whole as 8 sections.)
a. 34 of the distance across. (Note: they have to rethink the whole as 8 sections.)
b. 68 of the distance across. (Note: they have to rethink the whole as 8 sections.)
For early finishers, invite them to find their own equivalencies using their elastic to test their ideas.
Activity 15.15
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Equivalent Fractions 361
Set models can also be used to develop the concept of equiva-lence. Legos, a highly motivating manipulative, can help stu-dents learn to write fraction equivalencies (for an excellent elaboration on this idea, see Gould, 2011). Lego bricks and can be viewed as an area (array) or as a set (students can count the studs).
Two color counters are an effective tool for fraction equivalencies. Click here to listen to John Van de Walle dis-cuss fractions equivalencies with two color counters. Activ-ity 15.17 uses the context of apples, which can be modeled using two color counters.
In the activities so far, there has only been a hint of a rule for finding equivalent fractions. Activity 15.18 moves a bit closer, but should still be done before developing an algorithm.
Students who have learning disabilities and other stu-dents who struggle with mathematics may benefit from using clocks to do equivalence; for example, to find equivalent frac-tions for 10
12, 34, 46, and so on (Chick, Tierney, & Storeygard, 2007).
NCTM’s Illuminations website offers an excellent set of three units called “Fun with Fractions.” Each unit uses one of the model types (area, length, or set) and focuses on comparing and ordering fractions and equivalences. The five to six lessons in each unit incorporate a range of manipulatives and engaging activities to support student learning.
CCSSM: 2.G.a.3; 3.NF.a.1; 3.NF.3a, b, c; 4.NF.B.3a, b
Lego Land: Building Optionshand out one 2- by- 6 Lego to each student. ask them to describe it (there are 12 studs, two rows of 6). For second grade or as a warm- up, ask students what same- colored pieces could cover their land (e.g., 6 of the 2- by- 1 pieces). ask students to imagine that 12 pieces of Legos represents one plot of land. It can be covered with various smaller pieces as shown here:
ask students to build the plot of land using different Lego pieces (1- by- 2, 1- by- 3, 2- by- 2, 1- by- 1, 2- by- 6, etc.). after they have completed their plot of land, ask students to tell the fraction of their land that is represented by a particular piece (e.g., the 2- by- 6 is 6
12 as well as 12 and 24).to focus on iteration and to build connections to addition (grade 4), students can write equations to describe their Lego
Land. In the one pictured here, that would be
612
+3
12+
212
+1
12 Or
12
+14
+16
+1
12.
Note that students have misconceptions about how to name fractions parts, naming the blue part as 13 rather than 14 because they see three pieces (Wilkerson, Bryan, & Curry, 2012). this becomes a good topic for a classroom discussion: What is the fractional value of the blue pieces (simplified)?
Activity 15.16
Cuisenaire rods
Folding paper strips
C B C A
One whole
Blue = = 34
912
First fold (A)
Second fold (B)24
816
12
Third fold (C)
Last fold
48
Figure 15.14 Length models for equivalent fractions.
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362 Chapter 15 Developing Fraction Concepts
Developing an equivalent-Fraction algorithmWhen students understand that fractions can have different (but equivalent) names, they are ready to develop a method for finding equivalent names for a particular value. An area model is a good visual for connecting the concept of equivalence to the standard algorithm for finding equivalent fractions (multiply both the top and bottom numbers by the same number to get an equivalent fraction). The approach suggested here is to look for a pattern in the way that the fractional parts in both the part and the whole are counted. Activity 15.21 is a beginning, but a good class discus-sion following the activity will also be required.
CCSSM: 3.NF.3a, b, c; 4.NF.a.1
Missing‐Number EquivalencesUse Missing- Number equivalences activity page or give students an equation expressing an equivalence between two fractions, but with an unknown value. ask students to use counters or rectangles to illustrate and find the equivalent fraction. example equations:
53
=6
23
=6
8
12=
3
912
=3
the missing value can be in a numerator or a denominator; the missing number can be either larger or smaller than the corresponding part of the equivalent fraction. (all four possibilities are represented in the examples.) Figure 15.17 illustrates how Zachary represented the equivalences with equations and partitioning rectangles. the examples shown involve simple whole-number multiples between equivalent fractions. Next, consider pairs such as 68 =
12 or 912 = 6 . In these equivalences,
one denominator or numerator is not a whole- number multiple of the other. In addition, include equivalencies for whole num-bers and fractions greater than one: 8x = 6
6 103 = x
9
Activity 15.18
CCSSM: 3.NF.3b; 4.NF.a.1
Garden Plotshave students draw a square “garden” on blank paper, or give each student a square of paper (like origami paper). Begin by explaining that the garden is divided into rows of various vegetables. In the first example, you might illustrate four rows (fourths) and designate 34 as corn. ask students to partition their square into four rows and shade three-fourths as in Figure 15.18. then explain that the garden is going to be shared with family and friends in a way that each person gets a harvest that is 34 corn. Show how the garden can be partitioned horizontally to represent two people sharing the corn (i.e., 68). ask what fraction of the newly divided garden is corn. Next, tell students to come up with other ways that friends can share the garden (they can choose how many friends, or you can). For each newly divided garden, ask students to record an equation showing the equivalent fractions.
Activity 15.19
standards for mathematical Practice
mP1. Make sense of problems and persevere in solving them.
CCSSM: 3.NF.a.1; 3.NF.3a, b, c
Apples and BananasUse apples and Bananas activity page or just have students set out a specific number of counters in two colors— for example, 24 counters, with 16 of them red (apples) and 8 of them yellow (bananas). the 24 counters make up the whole. the task is to group the counters into different fractional parts of the whole and use the parts to create fraction names for the fractions that are apples and fractions that are bananas. ask questions such as, “If we make groups of four, what part of the set is red?” to encourage students to think of different ways to form equal- sized groups. In Figure 15.15, 24 counters are arranged in different groups. You might also suggest arrays (see Figure 15.16). eLLs may not know what the term group means because when used in classrooms, the word usually refers to arranging students. Spend time before the activity modeling what it means to group objects.
Activity 15.17
engLisH Language Learners
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Equivalent Fractions 363
Apples and Bananas
16 and 824 counters = 1 whole
and 2 more sets of 4 makes 24.
groups are apples46
groups are bananas26
groups are apples23
groups are bananas13
groups are apples812
groups are bananas412
1624
812
46
824
1624
Make the 16 into 4 groups of 4
16 is 2 groups of 8.
8 groups of 4 groups of
apples
bananas
Apples are = 23
= =
824
412
26
= 13
= =
of the fruit
Bananas are of the fruit
Figure 15.15 Set models for illustrating equivalent fractions.
23
23
1624
1624
are apples
of the rows are apples
=
46
46
1624
1624
are apples
of the columns are apples
=
Figure 15.16 Arrays for illustrating equivalent fractions.
Figure 15.17 A student illustrates equivalence fractions by partitioning rectangles.
Garden plots activity
Figure 15.18 A third grader partitions a garden to model fraction equivalences.
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364 Chapter 15 Developing Fraction Concepts
After students have prepared their own examples, provide time for them to look at their fractions and gardens and notice pat-terns about the fractions and the diagrams. Once they have time to do this individually, ask students to share. Figure 15.19 pro-vides student explanations that illustrate the range of “noticing.”
As you can see, for some of these students more experi-ences are needed. You can also assist in helping students make the connection from the partitioned square to the procedure by displaying a square (for example, partitioned to show 4
5) (see Figure 15.20). Then, partition the square vertically into six parts, covering most of the square as shown in the figure. Ask, “What is the new name for my 45?”
The reason for this exercise is that it helps students see the connection to multiplication. With the covered square, students can see that there are four columns and six rows to the shaded part, so there must be 4 * 6 parts shaded. Simi-larly, there must be 5 * 6 parts in the whole. Therefore, the new name for 45 is 4 * 6
5 * 6, or 2430.
Examine examples of equivalent fractions that have been generated with other visuals (e.g., a number line) and see if the rule of multiplying top and bottom numbers by the same number holds. Ask students to explain why. Ask students, “If the rule is correct, how can 68 and 9
12 be equivalent?”
Writing Fractions in simplest terms. The multi-plication scheme for equivalent fractions produces fractions with larger denominators. But creating equivalent forms for 68 might involve multiplication to get 12
16 or division to get 34.
To write a fraction in simplest terms means to write it so that numerator and denominator have no common whole‐ number factors. One meaningful approach to this task of finding sim-plest terms is to reverse the earlier process, as illustrated in Figure 15.21. The search for a common factor or a simplified fraction should be connected to grouping. Texas Instruments offers a comparing fractions activity using the number line on their Classroom Activities Exchange.
Two additional things should be noted regarding fraction simplification:
1. Notice that the phrase reducing fractions was not used. Because this would imply that the fraction is being made smaller, this terminology should be avoided. Fractions are simplified, not reduced.
2. Teachers sometimes tell students that fraction answers are incorrect if not in simplest or lowest terms. This also mis-informs students about the equivalence of fractions. When students add 1
6 + 12, both 2
3 and 46 are correct. It is best to
reinforce that they are both correct and are equivalent.
multiplying by One. Mathematically, equivalence is based on the multiplicative identity (any number multiplied by 1 remains unchanged). Any fraction of the form n
n can be used as the iden-tity element. Therefore, 3
4 = 34 * 1 = 3
4 * 22 = 6
8. Furthermore, the numerator and denominator of the identity element can also be fractions. In this way, 6
12 = 612 * (1/6
1/6) = 12. Understanding this
idea is an expectation in the CCSS- M in grade 4.
Simpler terms
= =812
23
Higher terms
2 × ×
4
4 3
Figure 15.21 Using the equivalent-fraction algorithm to write fractions in simplest terms.
(a)
(b)
(c)
Figure 15.19 Students explain what they notice about fraction equivalences based on partitioning “gardens” in different ways.
45
45 = ?
Figure 15.20 How can you count the fractional parts if you cannot see them all?
M15_VAND8930_09_SE_C15.indd 364 08/12/14 11:55 AM
Comparing Fractions 365
Technology Note. Developing the concept of equivalence can be supported with the use of technology. In the NCTM e‐Examples, there is a fraction game (Fraction Track)
for two players (Applet 5.1, Communicating about Mathematics Using Games). The game uses a number‐line model, and knowledge of equivalent fractions plays a significant role. The Equiv-alent Fractions tool from NCTM’s Illuminations website is designed to help students create equivalent fractions by dividing and shading square or circular regions and then matching each fraction to its location on a number line. Students can use the computer‐generated fraction or build their own. Once the rectangular or circular shape is divided, the student fills in the parts or fractional region and then builds two models equivalent to the original fraction. The three equivalent fractions are displayed in a table and in the same location on a number line. ■
Complete Self- Check 15.4: equivalent Fractions
Comparing Fractions
When students are looking to see whether two or more fractions are equivalent, they are com-paring them. If they are not equivalent, then students can determine which ones are smaller and which ones are larger. As illustrated in ally’s interview about comparing fractions, students often have misconceptions about fractions and therefore are not able to compare— for example, think-ing that bigger numbers in the denominator mean the fraction is bigger. The ideas described previously for equivalence across area, length, and set models are appropriate for comparing fractions. The use of contexts, models, and mental imagery can help students build a strong understanding of the relative size of fractions (Bray & Abreu‐Sanchez, 2010; Petit et al., 2010). The next section offers ways to support students’ understanding of the relative size of fractions.
Comparing Fractions using number senseIn the National Assessment of Educational Progress (NAEP) test, only 21 percent of fourth‐grade students could explain why one unit fraction was larger or smaller than another— for example, 1
5 and 14 (Kloosterman et al., 2004). For eighth graders, only 41 percent were able to
correctly put in order three fractions given in simplified form (Sowder, Wearne, Martin, & Strutchens, 2004).
Comparing unit Fractions. As noted earlier, whole- number knowledge can interfere with comparing fractions. Students think, “Seven is more than four, so sevenths should be bigger than fourths” (Mack, 1995). The inverse relationship between number of parts and size of parts cannot be “told” but must be developed in each student through many experiences, including ones that were described earlier related to partitioning, iterating, estimation, and equivalence.
CCSSM: 3.NF.a.3d; 4.NF.a.2
Ordering Unit FractionsList a set of unit fractions such as 13, 18, 15, and 1
10 (assume same size whole for each fraction). ask students to use reasoning to put the fractions in order from least to greatest. Challenge students to explain their reasoning with an area model (e.g., circles) and on a number line. ask students to connect the two representations. (“What do you notice about 13 of the circle and 13 on the number line?”) Students with disabilities may need to use clothespins with the fractions written on them and place them on the line first.
repeat with all numerators equal to some number other than 1.repeat with fractions that have different numerators and different demonimators. You can vary how many
fractions are being compared to differentiate the task.
Activity 15.20
stuDents with
sPeCiaL neeDs
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366 Chapter 15 Developing Fraction Concepts
Students may notice that larger bottom numbers mean smaller fractions (this is an import-ant pattern to notice), but it only holds true when the numerators are the same. Still, it is con-jecture that can be posed to the class for testing. Eventually they will find cases where it is not true (e.g., 18 6 1
4 but 38 7 14).
Comparing any Fractions. You have probably learned rules or algorithms for compar-ing two fractions. The usual approaches are finding common denominators and using cross‐multiplication. These rules can be effective in getting correct answers but require no thought about the size of the fractions. If students are taught these rules before they have had the opportunity to think about the relative sizes of various fractions, they are less likely to develop number sense about fraction size. The goal is to select an efficient strategy for determining the larger fraction, and these two methods are not always the most efficient (CCSSO, 2010).
Pause & ReflectAssume for a moment that you do not know the common-denominator or cross-multiplication techniques. Now examine the pairs of fractions in Figure 15.22 and select the largest of each pair using a reasoning approach that a fourth grader might use. ●
The Which Is Greater? Activity Page can be used to do this activity with students. Figure 15.23 provides explanations from two students on the first column ( A– F). Both students are able to rea-son to determine which is larger, though one student is better able to articulate those ideas.
The following list summarizes ways that the fractions in Figure 15.22 might have been compared:
1. Same‐size whole (same denominators). To compare 38 and 5
8, think about having 3 parts of something and also 5 parts of the same thing. (This method can be used for problems B and G.)
2. Same number of parts (same numerators) but different‐sized wholes. Consider the case of 34
and 37. If a whole is divided into 7 parts, the parts will certainly be smaller than if divided
into only 4 parts. (This strategy can be used with problems A, D, and H.)3. More than/less than one‐half or one. The fraction pairs 37 versus 58 and 54 versus 78 do not lend
themselves to either of the previous thought processes. In the first pair, 37 is less than half of the number of sevenths needed to make a whole, and so 3
7 is less than a half. Similarly, 58 is
more than a half. Therefore, 58 is the larger fraction. The second pair is determined by not-ing that one fraction is greater than 1 and the other is less than 1. (This method could be used on problems A, D, F, G, and H.)4. Closeness to one‐half or one. Why is 9
10 greater than 34? Each is one fractional part away from one whole, and tenths are smaller than fourths. Similarly, notice that 58 is smaller than 46 because it is only one‐eighth more than a half, wheras 4
6 is a sixth more than a half. Can you use this basic idea to compare 3
5 and 59? (Hint: Each is half of a fractional part
more than 12.) Also try 5
7 and 79. (This is a good strategy for
problems C, E, I, J, K, and L.)
How did your reasons for choosing fractions in Fig-ure 15.22 compare to these ideas? It is important that you are comfortable with these informal comparison strategies as a major component of your own number sense as well as for helping students develop theirs. Notice that some of the com-parisons, such as problems D and H, could have been solved using more than one of the strategies listed.
Tasks you design for your students should assist them in developing these methods of comparing two fractions. The ideas should emerge from your students’ reasoning. To teach “the four ways to compare fractions” defeats the purpose of encouraging
standards for mathematical Practice
mP2. Reason abstractly and quantitatively.
A.
Which fraction in each pair is greater?Give one or more reasons. Try not to use drawings or models.Do not use common denominators or cross-multiplication.
or45
B.
C.
D.
E.
F.
G.
H.
I.
J.
K.
L.
49
or47
57
or38
410
or53
58
or34
910
or38
47
or712
512
or35
37
or58
610
or98
43
or46
712
or89
78
Figure 15.22 Comparing fractions using reasoning strategies.
M15_VAND8930_09_SE_C15.indd 366 08/12/14 11:55 AM
Comparing Fractions 367
students to apply their number sense. Instead, select pairs of fractions that will likely elicit desired comparison strategies. On one day, for example, you might have pairs of fractions with the same numerators. Ask students to tell which is greater and why they think so. Ask them to give an example to convince you. On another day, you might pick fraction pairs in which each fraction is exactly one part away from a whole. Try to build strategies over several days by the strategic choice of fraction pairs.
The use of an area or number‐line model may help students who are struggling to reason mentally. Place greater emphasis on students’ reasoning and connect it to the visual models.
using equivalent Fractions to CompareEquivalent‐fraction concepts can be used in making compar-isons. Smith (2002) suggests that the comparison question to ask is, “Which of the following two (or more) fractions is greater, or are they equal?” (p. 9). He points out that this ques-tion leaves open the possibility that two fractions that may look different can, in fact, be equal.
In addition to this point, with equivalent- fraction con-cepts, students can adjust how a fraction looks so that they can use ideas that make sense to them. Burns (1999) describes how fifth graders compared 6
8 to 45. (You might want to stop for a
moment and think how you would compare these fractions.) One child changed the 45 to 8
10 so that both fractions would be two parts away from the whole and he reasoned from there. Another changed both fractions to a common numerator of 12.
Be absolutely certain to revisit the comparison activities and include pairs such as 8
12 and 23 in which the fractions are
equal but do not appear to be.
estimating with FractionsNumber sense with fractions means that students have some intuitive feel about the relative size of fractions (knowing “about” how big a particular fraction is). As students are deciding about how big something is, they are comparing that fraction to benchmark numbers, such as 0, 1
2, or 1. As with whole numbers, students are less confident and less capable of estimating than they are at computing exact answers and a focus on estimation can strengthen their understanding of fractions (Clarke & Roche, 2009). Therefore, you need to provide many opportunities for students to estimate with fractions. In daily classroom discussions, ask questions like “About what fraction of your classmates are wearing sweaters?” Or after tallying survey data about a topic like favorite dinner, ask, “About what fraction of our class picked spa-ghetti?” Activity 15.13 offers examples of visual estimating activities with area and number lines.
(a)
(b)
Figure 15.23 Two students explain how they compared the fractions in problems A through F from Figure 15.22.
CCSSM: 3.NF.a.1; 3.NF.a.2a, b
About How Much?Draw a picture like one of those in Figure 15.24 (or prepare some ahead of time for the overhead). have each student write down a fraction that he or she thinks is a good estimate of the amount shown (or the indicated mark on the number line). Listen to the ideas of several students, and ask them whether a particular estimate is a good one. there is no single correct answer, but estimates should be in the “ballpark.” If students have difficulty coming up with an estimate, ask whether they think the amount is closer to 0, 12, or 1. For students with disabilities, you may want to give them a set of cards showing possible options for estimates. then they can match the card to one of the pictures.
Activity 15.21
stuDents with
sPeCiaL neeDs
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368 Chapter 15 Developing Fraction Concepts
The number line is a good model for helping students develop a better understanding for the relative size of a frac-tion (Petit, Laird, & Marsden, 2010). For example, if students think about where 3
20 might be by partitioning a number line between 0 and 1, they will see that 3
20 is close to 0, whereas 910 is quite close to 1. Number lines should also go beyond 1, asking students to tell a nearby benchmark fraction— for example, explaining that 33
7 is almost 312.
After students have experience with visuals, they should continue to reason about the relative size of fractions using mental strategies or creating their own visuals to reason about the fractions.
Finally, comparing fractions can include finding fractions that fall between two given fractions. An important idea in fractions that there is always one more fraction between any two given numbers. Fraction Find is an activity that focuses on this idea. The fractions selected can be varied to meet individ-ual students’ needs.
0 ? 1
0 ?1 2
Figure 15.24 About how much? Name a fraction for each drawing and explain why you chose that fraction.
CCSSM: 3.NF.a.3d; 4.NF.a.2
Zero, One‐Half, or OneOn a set of cards, write a collection of 10 to 15 fractions, one per card. a few should be greater than 19
8 or 1110, with the others ranging from 0 to 1. Let students sort the fractions into three
groups: those close to 0, close to 12, and close to 1. For those close to 12, have them decide whether the fraction is more or less than half. the difficulty of this task largely depends on the fractions you select. the first time you try this, use fractions that are very close to the three benchmarks, such as 1
20, 53100, or 9
10. On subsequent days, mostly use fractions with denominators less than 20. You might include a few fractions that are exactly in between the benchmarks, such as 28 or 34. ask students to explain how they are using the numerator and denominator to decide. For eLLs, be sure the term benchmark is understood and encourage illustrations as well as explanations.
as an extension or alternative to differentiate this activity, ask students to create their own fractions close to each benchmark fraction.
Activity 15.22
Complete Self- Check 15.5: Comparing Fractions
teaching Considerations for Fraction Concepts
Because the teaching of fractions is so important, and because fractions are often not well under-stood even by adults, a recap of the big ideas is needed. Hopefully you have recognized that one reason fractions are not well understood is that there is a lot to know about them, from part- whole relationships to division constructs, and understanding includes representing across area, length, and set models and includes contexts that fit these models. Many of these strategies may not have been part of your own learning experience, but they must be part of your teaching experience so that your students can fully understand fractions and be successful in algebra and beyond.
M15_VAND8930_09_SE_C15.indd 368 08/12/14 11:55 AM
Iterating and partitioning must be a significant aspect of fraction instruction. Equivalence, including comparisons, is a central idea for which students must have sound understanding and skill. Connecting visuals with the procedure and not rushing the algorithm too soon are important aspects of the process.
Clarke and colleagues (2008) and Cramer and Whitney (2010), researchers of fraction teaching and learning, offer research‐based recommendations that provide an effective sum-mary of this chapter:
1. Give a greater emphasis to number sense and the meaning of fractions, rather than rote procedures for manipulating them.
2. Provide a variety of models and contexts to represent fractions.3. Emphasize that fractions are numbers, making extensive use of number lines in represent-
ing fractions.4. Spend whatever time is needed for students to understand equivalences (concretely and
symbolically), including flexible naming of fractions.5. Link fractions to key benchmarks and encourage estimation.
Complete Self- Check 15.6: teaching Considerations for Fraction Concepts
Reflections on Chapter 15 369
reFLeCtiOns o N CHAPTER 15
Writing tO Learn
1. What is the goal of activities involving the concept of sharing? When would you implement sharing activities?
2. Give examples of manipulatives and contexts that fall into each of the three categories of fraction models (area, length, and set).
3. What does partitioning mean? Explain and illustrate.
4. What does iteration mean? Explain and illustrate.
5. What are two ways to build the conceptual relationship between 11
4 and 234?
6. Describe two ways to compare 512 and 58 (not using com-
mon denominator or cross‐product methods).
FOr DisCussiOn anD exPLOratiOn
◆◆ A common error that students make is to write 35 for the fraction represented here:
◆◆ Why do you think they do this? What activity or strat-egy would you use to try to address this misconception?
◆◆ Fractions are often named by adults (and depicted in cartoons) as a dreaded math topic. Why do you think this is true? How might your fraction instruction alter this perception for your students?
M15_VAND8930_09_SE_C15.indd 369 08/12/14 11:55 AM
370 Chapter 15 Developing Fraction Concepts
resOurCes Fo R CHAPTER 15
Literature COnneCtiOnsContext takes students away from rules and encourages them to explore ideas in a more open and meaningful manner. The way that students approach fraction concepts in these con-texts may surprise you.
How many snails? a Counting Book Giganti (1988)
Each page of this book has a similar pattern of questions. For example, the narrator wonders how many clouds there are, how many of them are big and fluffy, and how many of them are big and fluffy and gray. Students can look at the pictures and find the fraction of the objects (e.g., clouds) that have the particular characteristic (big and fluffy). Whitin and Whitin (2006) describe how a class used this book to write their own stories in this pattern and record the fractions for each subset of the objects.
the Doorbell rang Hutchins (1986)
Often used to investigate whole‐number operations of mul-tiplication and division, this book is also an excellent early introduction to fractions. The story is a simple tale of two children preparing to share a plate of 12 cookies. Just as they have figured out how to share the cookies, the doorbell rings and more children arrive. You can change the number of chil-dren to create a sharing situation that requires fractions (e.g., 5 children).
the man Who Counted: a Collection of mathematical adventures Tahan (1993)
This book contains a story, “Beasts of Burden,” about a wise mathematician, Beremiz, and the narrator, who are traveling together on one camel. They are asked by three brothers to solve an argument. Their father has left them 35 camels to divide among them: half to one brother, one‐third to another, and one‐ninth to the third brother. The story provides an excellent context for discussing fractional parts of sets and how fractional parts change as the whole changes. However, if the whole is changed from 35 to, say, 36 or 34, the problem of the indicated shares remains unresolved. The sum of 1
2, 13,
and 19 will never be one whole, no matter how many camels
are involved. Bresser (1995) describes three days of activities with his fifth graders.
apple Fractions Pallotta (2002)
This book offers interesting facts about apples while introduc-ing fractions as fair shares (of apples, a healthier option than books that focus on chocolate and cookies!). In addition, the
words for fractions are used and connected to fraction sym-bols, making it a good connection for fractions in grades 1–3.
reCOmmenDeD reaDingsarticlesClarke, D. M., Roche, A., & Mitchell, A. (2008). Ten practical
tips for making fractions come alive and make sense. Math-ematics Teaching in the Middle School, 13(7), 373–380.Ten excellent tips for teaching fractions are discussed and favorite activities are shared. An excellent overview of teaching fractions.
Flores, A., & Klein, E. (2005). From students’ problem‐ solving strategies to connections in fractions. Teaching Children Mathematics, 11(9), 452–457.This article offers a very realistic view (complete with photos of student work) of how children develop initial fraction concepts and an understanding of notation as they engage in sharing tasks like those described in this chapter.
BooksBurns, M. (2001). Teaching arithmetic: Lessons for introducing frac-
tions, grades 4–5. Sausalito, CA: Math Solutions Publications.This book offers well‐designed lessons with lots of details, sample student dialogue, and blackline masters. These are introductory ideas for fraction concepts. Five lessons cover one‐half as a bench-mark. Assessments are also included.
Lamon, S. (2012). Teaching fractions and ratios for understand-ing: Essential content knowledge and instructional strategies. New York, NY: Taylor & Francis Group.As the title implies, this book has a wealth of information to help with better understanding fractions and teaching fractions well. Many rich tasks and student work are provided throughout.
McNamara, J., & Shaughnessy, M. M. (2010). Beyond pizzas and pies: 10 essential strategies for supporting fraction sense (grades 3–5). Sausalito, CA: Math Solutions Publications.This book has it all— classroom vignettes, discussion of research on teaching fractions, and many activities, including student work.
WebsiteRational Number Project (website housed at the University of
Minnesota).This project offers numerous readings, activities for students, and other materials— a great collection of high quality resources for teaching fractions for understanding.
M15_VAND8930_09_SE_C15.indd 370 08/12/14 11:55 AM |
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Type Subject Issued-By Level Year
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### Credentials
Type Subject Issued-By Level Year
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## Herbert H
### Teaching Style
The teaching style I use is completely dependent on the particular student being taught. I'm a big fan of the adage 'Give a man a fish, and he will eat for a day. Teach a man to fish, and he will eat for the rest of his life'. My objective is to teach a child not only the material, but how they best learn so that they can teach themselves. I enjoy using applicable references that are relevant to the child so that they will more easily grasp the concepts. I believe a personalized approach works best.
### Experience Summary
I have earned a bachelor's degree in Computer Engineering with a software focus along with a Sales Engineering minor. I was 3 credit hours away from both my Math and Business minors. In the last 5 years, I have tutored students in Basic Math, Algebra 1 and 2, Geometry, Calculus, Visual Basic, Physics, and Differential Equations. I have taught math for the past 4 years at the Professional Academies Magnet at Loften High School in the subjects of Algebra I, Algebra I Honors, Geometry, Geometry Honors, Algebra II, Algebra II Honors, Statistics, Mathematics for College Readiness, and Liberal Arts Math.
### Credentials
Type Subject Issued-By Level Year
Other Mathematics Professional Academies Magnet 9-12 2007-current
Other Tutor Starke Church of God by Faith 9-12 2005-2007
Degree Computer Engineering University of Florida BA 2004
## Vinod V
### Teaching Style
The cornerstone of my teaching philosophy and personal teaching goals is to help students develop their own thinking skills. I believe all students should leave the school armed with the ability to think for them selves, to think critically and to think creatively. Understanding how people learn is one of the significant aspects of teaching. This is linked to their “knowledge” background and maturity. The key to teaching is to relate to the audience by starting from what they know and building upon it. As a teacher I am totally involved with the class, dedicated to my students and 100% prepared to devote time and energy for their intellectual growth. Love for teaching evokes passion and dedication within me. I believe that the enthusiasm of a motivated teacher rubs off on his/her students, who derive the inspiration and encouragement which actuates their desire to learn. A good teacher should have sound fundamentals and command over the concepts. Fundamentals are the foundation intrinsic for mastering the subject; only teachers who are strong in fundamentals will be able to pass it on to their students. I believe that my strong command over the fundamentals will rub off on my students. I believe that the role of a teacher is that of a leader where you have to show the path, motivate, encourage, and lead by example. In short, my success lies in seeing my students succeed.
### Experience Summary
My enthusiasm and love for education can be gauged from the fact that I pursued three Masters degrees in three distinct but related fields. One cannot pursue engineering as a profession without having an affinity for Math and Analysis. Math was a passion for me from my young days and still very much remains so. I have a thorough knowledge and understanding of math. Right from my school days I was involved and loved to teach math. I invariably obtained A+ scores in whatever math test I took in my lifetime. For instance my GRE math score was above 95% of test takers' scores. I have taught Middle school, High school and under-graduate students in Algebra, Geometry, Trigonometry, Quadratic Equations, Applied Probability and Calculus.
### Credentials
Type Subject Issued-By Level Year
Degree City Planning Kansas State University MRCP 2002
Degree Engineering Anna University ME 2000
Degree Civil Engineering Institution of Engineers BE 1994
## Jennifer J
### Teaching Style
I enjoy working with a student one-on-one. We know that every child develops on a different learning curve, which is why the tutoring process is so crucial. Working individually with a child can help them overcome some of the frustration and confusion that comes along with learning in a crowded classroom. I have a great amount of patience and understanding to help the child develop his own learning style and guide him to work through the problems at his own pace to ensure that he truly understands the material. Learning is a step by step process and I know that I can guide each child through the steps they need to take to complete any and all challenges they may face, not just in school, but in life.
### Experience Summary
I began tutoring algebra when I was in high school for my boss' kids in the back of the restaurant where I worked. I then went on to get my BA in mathematics at La Salle University, with a minor in Education. While in college, I continued tutoring my fellow students in math and chemistry for a peer tutoring program set up by the school. In addition I observed/student taught at various schools in the Philadelphia area. I have knowledge of a wide range of mathematics, my specialty being algebra and calculus. Due to my experience, I am also proficient in English and can assist in research paper writing.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics La Salle University Bachelor's 2007
## Farrah F
### Teaching Style
I enjoy teaching and working with students. I now understand that all children learn on their own level and at their own pace. I am very easy going, but I insist that students do their homework and stay after school for extra help if they need it. I grade work off of understanding. It is not always finding the correct answer, but the process to get to the correct answer. I believe that a good teacher makes a good student and I strive to be that good teacher and student.
### Experience Summary
I enjoyed tutoring while I was in college working in the math lab for three years. However, now that I have been actually teaching my very own students I have a better understanding of the basics of mathematics and I am a much better student myself.
### Credentials
Type Subject Issued-By Level Year
Certification Mathematics 6-12 Florida Department of Education Leon County 2005
Degree Mathematics Florida A&M University BS 2004
## Maria M
### Teaching Style
I love tutoring and consider myself to be effective at it. I approach it with enthusiasm. I apply theory to practical applications in my career and life. My approach is one of ease. I consider mathematics to be easy, as long as you accept what I call "the rules of the game"; i.e., there are certain principles that have to be accepted, and not questioned, then everything else falls in place. I believe that all people have the capability of learning, and I love the opportunity to provide a positive experience to students.
### Experience Summary
I have been tutoring for more than 5 years. My main focus is mathematics, but I also tutor students in Spanish I through IV. I have tutored Spanish, Algebra I, Algebra II, Geometry, Trigonometry, and Pre-Calculus. I have recently helped several students prepare for the SAT math test. I enjoy making a difference in the student’s life.
### Credentials
Type Subject Issued-By Level Year
Other Spanish Native speaker Fluent Current
Certification Project Management Studies Project Management Institute PMP Certification 2006
Degree Civil Engineering California State University at Long Beach MSCE 1987
Other Mathematics El Camino College AA 1981
## Robert R
### Teaching Style
I’ve always been interested in the application of math and science to the solution of real world problems. This led me to a very satisfying career in engineering. Therefore my approach to teaching is very application oriented. I like to relate the subject to problems that the students will encounter in real life situations. I've generally only worked with older students; high school or college age or older mature adults who have returned to school to get advance training or learn a new trade.
### Experience Summary
I’ve always been interested in math and science; especially in their application to solving real world problems. This led me to a very satisfying career in engineering. I have a BS in electrical engineering from General Motors Institute (now Kettering University) and an MS in electrical engineering from Marquette University. I am a registered professional engineer in Illinois. I have over 30 years of experience in the application, development, and sales/field support of electrical/electronic controls for industrial, aerospace, and automotive applications. I’m currently doing consulting work at Hamilton-Sundstrand, Delta Power Company, and MTE Hydraulics in Rockford. I also have college teaching and industrial training experience. I have taught several courses at Rock Valley College in Electronic Technology, mathematics, and in the Continuing Education area. I’ve done industrial technical training for Sundstrand, Barber Colman, and others. I’ve also taught math courses at Rasmussen College and Ellis College (online course). I’ve also been certified as an adjunct instructor for Embry-Riddle Aeronautical University for math and physics courses. I've tutored my own sons in home study programs. I'm currently tutoring a home schooled student in math using Saxon Math. I hope to do more teaching/tutoring in the future as I transition into retirement.
### Credentials
Type Subject Issued-By Level Year
Degree Electrical Engineering Marquette University MS 1971
Degree Electrical Engineering GMI (Kettereing University) BS 1971
## Marilyn P.
Lighthouse Point, FL
I just wanted to tell you how great Danny’s tutor is. He loves her and is responding extremely well to her teaching. He told his mother it only seems like he’s with her 10 minutes! That surely tells it all. He’s reading at home and his mom notice...
## Megan G.
Tampa, FL
I was extremely impressed with Patti on Tuesday, September 9. She was competely prepared for me and had looked over the assignments I sent her and made study cards as well as sheets that I could use. I would highly recommend her based on our first se... |
Inbrunstlr
2022-10-08
Determine whether the lines for each pair of equations 3x+2y=-5 y=-2/3x+6 are parallel, perpendicular, or neither
Mckenna Friedman
First, we get the two linear equations into y=mx+b form:
${L}_{1}:y=-\frac{2}{3}x+6\to m=-\frac{2}{3}$
${L}_{2}:3x+2y=-5$
${L}_{2}:2y=-3x-5$
${L}_{2}:y=-\frac{3}{2}x-5\to m=-\frac{3}{2}$
If the lines were parallell, they would have the same m-value, which they don't, so they cannot be parallell.
If the two lines are perpendicular, their m-values would be negative reciprocals of each other. In the case of ${L}_{1}$, the negative reciprocal would be:
$-\frac{1}{-\frac{2}{3}}=-\left(-\frac{3}{2}\right)=\frac{3}{2}$
This is almost the negative reciprocal, but we're off by a minus sign, so the lines are not perpendicular.
Denisse Fitzpatrick
Rearranging the 1 st equation as y=mx+c,we get,
$y=-\frac{3}{2}x-\left(\frac{5}{2}\right)$ hence, slope =$-\frac{3}{2}$
the other equation is, $y=-\frac{2}{3}x+6$ ,slope is $-\frac{2}{3}$
Now,slope of both the equations are not equal,so they are not parallel lines.
Again,product of their slope is $-\frac{3}{2}\cdot \left(-\frac{2}{3}\right)=1$
But,for two lines to be perpendicular, product of their slope has to be −1
So,they are not perpendicular as well.
Do you have a similar question? |
# Difference between revisions of "2008 AMC 12B Problems/Problem 9"
## Problem 9
Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?
$\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4$
## Solutions
### Solution 1
Let $\alpha$ be the angle that subtends the arc $AB$. By the law of cosines, $6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$.
The half-angle formula says that $\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The law of cosines tells us $AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$, which is answer choice $\boxed{\text{A}}$.
### Solution 2
$[asy] defaultpen(fontsize(8)); pair A=(-3,4), B=(3,4), C=(0,5), D=(0,4), O=(0,0); D(Circle(O,5)); D(O--B--A--O--C);D(A--C--B); label("A",A,(-1,1));label("O",O,(0,-1));label("B",B,(1,1));label("C",C,(0,1));label("D",D,(-1,-1)); [/asy]$ Figure 1
Define $D$ as the midpoint of line segment $\overline{AB}$, and $O$ the center of the circle. Then $O$, $C$, and $D$ are collinear, and since $D$ is the midpoint of $AB$, $m\angle ODA=90\deg$ and so $OD=\sqrt{5^2-3^2}=4$. Since $OD=4$, $CD=5-4=1$, and so $AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}$. |
# What Is The Slope Of Y 5x?
## What is the slope of Y 5x − 1y?
Step-by-step explanation: In this question the slope would be 5..
## What is the Y intercept for y 5x?
You could simply write it as y = 5x. Thus if we compare to the standard form equation y = ax + b, a=5 and b=0. Thus the y-inercept is zero.
## What is the slope of Y 4x 1?
Using the slope-intercept form, the slope is 4 .
## What is the slope of the equation y 5x 7?
Using the slope-intercept form, the slope is 5 .
## What is the slope and y intercept of y 3x 5?
Answer: The slope is 3 and the y-intercept is −5 .
## Is Y 5x a linear equation?
Ax + By = C, where A and B are not both 0. This form is called the standard form of a linear equation. Example 1: Determine whether the equation y = 5x – 3 is linear or not….xy = 5x – 3(x, y)1y = 5(1) – 3 = 2(1, 2)2 more rows•Jul 29, 2011
## What is the Y intercept of y =- 5x 4?
Explanation: The y-intercept is -4 because it is not in front of an x in the problem.
## What is the slope of y =- 5x 4?
Using the slope-intercept form, the slope is 5 .
## What is the slope of the equation 3x 4y 32?
When you isolate the y on one side, the slope is 3/4.
## What is the Y intercept of y =- 4x 1?
Explanation: y=4x+1 is a linear equation in the slope-intercept form y=mx+b , where m is the slope, and b is the y-intercept. For the equation y=4x+1 , m=4andb=1 .
## Is Y 4x 1 a linear function?
Explanation: Since the equation is linear/straight in the form y=mx+c , you can determine the slope/gradient ( m ) and y-intercept ( c ). To graph this equation, you’ll need to find 2 points on the line and join them together. We know one of the points is (0, 1) as the y-intercept is 1.
## What is the slope of Y 5x 9?
Using the slope-intercept form, the slope is −5 .
## Is 5x y 4 a linear equation?
Linear graphs are straight. Your equation, 5x=y−4 or y=5x+4 will be a straight line with a gradient (slope) of 5 and a y-intercept of 4 (where the line cuts through the y axis).
## What is the slope of Y 4x 7?
Algebra Examples Using the slope-intercept form, the slope is −4 .
## What is the slope of y =- 1 3x?
The slope-intercept form is y=mx+b y = m x + b , where m m is the slope and b b is the y-intercept. Combine 13 1 3 and x x . Rewrite in slope-intercept form. Using the slope-intercept form, the slope is 13 .
## What is the Y intercept of y =- 5x 1?
So for the linear equation y = 5x – 1, it has a y-intercept equal to -1.
## What is the slope of y =- 5x?
The slope-intercept form is y=mx+b y = m x + b , where m is the slope and b is the y-intercept. Using the slope-intercept form, the slope is 5 .
## What is the slope of the equation y 5 =- 3x?
Using the slope-intercept form, the slope is −3 .
## What does Y 5x mean?
In the equation y=5x the slope (m) is 5, and the y-intercept is the point (0,0) since there is nothing added or subtracted to the 5x, and thus b = 0.
## What is the slope of Y 4?
Using the slope-intercept form, the slope is 0 . |
# September 11, 2012 Properties and Integers Warm-up: Order of Operations 1.Simplify 10 + 2(3 – 5) – 4 3 2 2.Challenge: Your objective is to use the digits.
## Presentation on theme: "September 11, 2012 Properties and Integers Warm-up: Order of Operations 1.Simplify 10 + 2(3 – 5) – 4 3 2 2.Challenge: Your objective is to use the digits."— Presentation transcript:
September 11, 2012 Properties and Integers Warm-up: Order of Operations 1.Simplify 10 + 2(3 – 5) – 4 3 2 2.Challenge: Your objective is to use the digits 1, 2, 3, 4 ONLY ONCE and arrange them using addition, subtraction, multiplication, exponents, and parentheses (NOT division), so that they equal a specific number. You may also make a two digits to make one number, such as 12 or 34. EXAMPLE: An expression equal to 43. 43 = 42 + 1 3 HW: 1.9 WS Quiz next class!
Check HW 1.4-1.6 Today’s homework: Worksheet 1.9 Opposites and Absolute Value #3, Mixed Review #1-6 Quiz Friday,
The Commutative Property says the order in which two numbers are added or multiplied does not affect the answer. a b = b a x + (y + z) = (y + z) + x
The Associative Property says the sum or product of any three numbers is the same, no matter how they are grouped using parentheses and the order of the numbers always stays the same. (a + b) + c = a + (b + c) x (y z) = (x y) z
The Inverse Property of Addition says the sum of a number and its opposite equals 0. a + (-a) = 0 -x + x = 0
The Inverse Property of Multiplication says any number multiplied by its reciprocal equals 1.
The Additive Identity: 0 added to any number will always equal the same number. a + 0 = a 0 + x = x
The Multiplicative Identity: any number multiplied by 1 will always equal the same number. a 1 = a 1 x = x
The Distributive Property multiplies the expression outside parentheses to the expression inside. a(x + y) = ax + ay (a + b)(x + y) = ax + ay + bx + by
Identify the property that goes with the following expressions. 1. 3(x + y) = 3x + 3y2. a + (b + c) = (a + b) + c 3. -10 + 10 = 0 4. 5. 4 + 0 = 4
The Real Number Line Graph the following numbers on the number line: -8.56 positivenegative origin
Comparing Integers (positive and negative numbers) Use to compare the numbers. 1) 5.5 ____ -6 2) -3.2 _____ -3.8 3) -100 ____ 0 4) -4-2024-315-53 5) Rewrite in increasing order 3.6, 3.0, -3.2, 0, -3.5, 3.2, 2.0
Opposite − ( ) The negative sign outside parentheses means “opposite”. - ( girl ) =- ( hot ) = -( positive ) = Examples: Evaluate. 6. -(100)7. -(-6.2)
Absolute Value │x│ The Absolute Value of a number is the distance from the given number to zero. │ -4 │ “How many steps from -4 to zero?” 8. │ 100 │ 9. │-45.8│ 10. │0│+ │-6│ 11. 12. Determine │ -6 │____ |5|
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# Find the area of a parallelogram with vertices A(1,5,0), B(6,10,−3), C(−4,5,−2), and D(1,10,−5)? Show steps.
Jan 6, 2016
${S}_{\text{parallelogram}} = 15 \cdot \sqrt{6} \cong 36.742$
#### Explanation:
Note that it doesn't matter whether the parallelogram is formed by the vertices A, B, C, D in this order or by the vertices A, B, D. C in this order, any diagonal of the parallelogram divides it in two triangles with equal areas. Then S_("parallelogram" =2*S_(triangle). In this explanation the $\triangle A B C$ is selected (but any other possible triangle formed with the points A, B. C or D, would do).
Repeating the coordinates of the points of $\triangle A B C$:
A(1,5,0)
B(6,10,-3)
C(-4,5,-2)
Obtaining sides of the triangle ABC:
$A B = \sqrt{{\left(6 - 1\right)}^{2} + {\left(10 - 5\right)}^{2} + {\left(- 3 - 0\right)}^{2}} = \sqrt{25 + 25 + 9} = \sqrt{59}$
$A C = \sqrt{{\left(- 4 - 1\right)}^{2} + {\left(5 - 5\right)}^{2} + {\left(- 2 - 0\right)}^{2}} = \sqrt{25 + 0 + 4} = \sqrt{29}$
$B C = \sqrt{{\left(- 4 - 6\right)}^{2} + {\left(5 - 10\right)}^{2} + {\left(- 2 + 3\right)}^{2}} = \sqrt{100 + 25 + 1} = \sqrt{126}$
Using Heron's Formula
Triangle ABC ($a = A B , b = A C \mathmr{and} c = B C$)
$s = \frac{a + b + c}{2} = \frac{\sqrt{59} + \sqrt{29} + \sqrt{126}}{2} \cong 12.14564$
$\left(s - a\right) = \frac{- \sqrt{59} + \sqrt{29} + \sqrt{126}}{2} \cong 4.46450$
$\left(s - b\right) = \frac{\sqrt{59} - \sqrt{29} + \sqrt{126}}{2} \cong 6.76048$
$\left(s - c\right) = \frac{\sqrt{59} + \sqrt{29} - \sqrt{126}}{2} \cong 0.92067$
${S}_{\triangle A B C} = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)} = \sqrt{12.141564 \times 4.46450 \times 6.76048 \times 0.92067} = 18.371$
${S}_{\text{parallelogram}} = 2 \cdot {S}_{\triangle A B C} = 2 \cdot 18.371 = 36.712$
Other way to find the area of the kind of triangle involved in this question is described in:
when the triangle is embedded in three-dimensional space |
# 2020 AMC 10B Problems/Problem 3
## Problem 3
The ratio of $w$ to $x$ is $4:3$, the ratio of $y$ to $z$ is $3:2$, and the ratio of $z$ to $x$ is $1:6$. What is the ratio of $w$ to $y?$
$\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3$
## Solution 1
WLOG, let $w=4$ and $x=3$.
Since the ratio of $z$ to $x$ is $1:6$, we can substitute in the value of $x$ to get $\frac{z}{3}=\frac{1}{6} \implies z=\frac{1}{2}$.
The ratio of $y$ to $z$ is $3:2$, so $\frac{y}{\frac{1}{2}}=\frac{3}{2} \implies y=\frac{3}{4}$.
The ratio of $w$ to $y$ is then $\frac{4}{\frac{3}{4}}=\frac{16}{3}$ so our answer is $\boxed{\textbf{(E)}\ 16:3}$ ~quacker88
## Solution 2
We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two.
$z:x=1:6=2:12$, and since $y:z=3:2$, we can link them together to get $y:z:x=3:2:12$.
Finally, since $x:w=3:4=12:16$, we can link this again to get: $y:z:x:w=3:2:12:16$, so $w:y = \boxed{\textbf{(E)}\ 16:3}$ ~quacker88 |
Square Root Functions
Definition and Graph of the Square Root Function
The square root of a nonnegative real number $x$ is a number $y$ such $x = y^2$. For example, $3$ and $- 3$ are the square roots of $9$ because $3^2 = 9$ and $(-3)^2 = 9$
Of the two roots, the nonnegative root of a nonnegative number $x$ is written as function with the following notation $f(x) = \sqrt x$ where the symbol $\sqrt { \; }$ is called the radical and $x$ is called the radicand and must be nonnegative so that $f(x)$ is real.
The square root function defined above is evaluated for some nonnegative values of $x$ in the table below.
$x$ $f(x) = \sqrt x$
$0$ $f(x) = \sqrt 0 = 0$
$1$ $f(x) = \sqrt 1 = 1$
$2$ $f(x) = \sqrt 2 \approx 1.414$
$4$ $f(x) = \sqrt 4 = 2$
$7$ $f(x) = \sqrt 7 \approx 2.645$
$9$ $f(x) = \sqrt 9 = 3$
$16$ $f(x) = \sqrt 16 = 4$
The graph of the square root function is shown below with some points from the above table.
Properties of the Square Root Function
Some of the properties of the square root function may be deduced from its graph
1. The domain of the square root function $f(x) = \sqrt x$ is given in interval form by: $[0, + \infty)$
2. The range of the square root function $f(x) = \sqrt x$ is given in interval form by: $[0, + \infty)$
3. The x and y intercepts are both at $(0,0)$
4. The square root function is an increasing function
5. The square root function is a one-to-one function and has an inverse.
Common Mistakes to Avoid when Working with Square Root Functions
1. It is wrong to write $\sqrt{25} = \pm 5$. The radicand is the symbol of the square root function and a function has only one output which as defined above is equal to the positive root.
Correct answer: $\sqrt{25} = 5$
2. It is wrong to write $\sqrt{x^2} = x$. The output of the square root is nonnenegative and $x$ in the given expression may be negative, zero or postive.
Correct answer: $\sqrt{x^2} = |x|$
Exploring Interactively the Square Root Function
Square root functions of the general form
$f(x) = a \sqrt{x - c} + d$ and the characteristics of their graphs such as domain, range, x intercept, y intercept are explored interactively. Square root equations are also explored graphically. There is also another tutorial on graphing square root functions in this site.
The exploration is carried out by changing the parameters $a, c$ and $d$ included in the expression of the square root function defined above. The Answers to the questions in the tutorial are included in this page.
click on the button above "draw" and start exploring.
a = 1 -10+10 c = 0 -10+10 d = 0 1+10
>
1. Use the sliders to set parameters $a$ and $c$ to some constant values and change $d$. What happens to the the graph when the value of parameter $d$ changes? Give an analytical explanation.
2. Use the sliders to set parameters $a$ and $d$ to some constant values and change $c$. What happens to the the graph when the value of parameter $c$ changes? Give an analytical explanation.
3. Use the sliders to set parameters $c$ and $d$ to some constant values and change parameters $a$. What happens to the graph when the value of parameter $a$ changes? Give an analytical explanation.
4. Use the sliders to set parameters $a$, $c$ and $d$ to different values and determine which parameters affect the domain of the square root function $f$ defined above? Find the domain analytically and compare it to the domain obtained graphically.
5. Use the sliders to set parameters $a$, $c$ and $d$ to different values and determine which parameters affect the range of the square root function $f$ defined above? Find the range analytically and compare it to the range obtained graphically.
6. How many x intercept the graph of $f$ has?
7. How many solutions an equation of the form
$a \sqrt{x - c} + d = 0$
has? (parameter $a$ not equal to zero).
Find the solution to this equation in terms of $a$, $c$ and $d$ and compare it to the x intercept given graphically.
8. Find the y intercept analytically and compare it to the one given by the app.
1. Changes in the parameter $d$ affect the y coordinates of all points on the graph hence the vertical translation or shifting. When $d$ increases, the graph is translated upward and when d decreases the graph is translated downward.
2. When $c$ increases, the graph is translated to the right and when $c$ decreases, the graph is translated to the left. This is also called horizontal shifting.
3. Parameter $a$ is a multiplicative factor for the y coordinates of all points on the graph of function $f$. Let $a$ be greater than zero. If $a$ gets larger than 1, the graph stretches (or expands) vertically. If $a$ gets smaller than 1, the graph shrinks vertically. If $a$ changes sign, a reflection of the graph on the x axis occurs.
4. Only parameter $c$ affects the domain. The domain of
$f(x) = a \sqrt{x - c} + d = 0$
may found by solving the inequality $x - c \ge 0$ hence the domain is the set of all values in the interval $[c , + \infty)$
5. Only parameters $a$ and $d$ affect the range. The range of function $f$ given above may found as follows: With x in the domain defined by interval $[c , + \infty)$ , the $\sqrt{x - c}$ is always positive or equal to zero hence
$\sqrt{x - c} \ge 0$
If parameter $a$ is positive then
$a \sqrt{x - c} \ge 0$
Add $d$ to both sides to obtain
$a \sqrt{x - c} + d \ge d$
Hence the range of the square root function defined above is the set of all values in the interval $[d , + \infty)$
If parameter $a$ is negative then
$a \sqrt{x - c} \le 0$
Add $d$ to both sides to obtain
$a \sqrt{x - c} + d \le d$
Hence the range of the square root function defined above is the set of all values in the interval $(- \infty , d]$
6. one solution or no solution.
7. Solve the equation
$a \sqrt{x - c} + d = 0$
add $-d$ to both sides of the equation
$a \sqrt{x - c} = - d$
If $d$ is positive, $- d$ is negative and the above equation has no solution. If $d$ is negative, we square both sides and solve to obtain
$x = (-d/a)^2 + c$
The above equation may have one solution or no solution.
8. If $c$ is positive, $\sqrt{-c}$ is not a real number and therefore the graph has no y intercept. If $c$ is negative or equal to zero then the y intercept is given by
$y = a \sqrt{-c} + d$
Square Root
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Found in: Page 437
### Linear Algebra and its Applications
Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384
# In Exercises 1-4, write y as an affine combination of the other point listed, if possible.{{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{\bf{0}}\end{aligned}} \right), {{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{6}}}\\{\bf{7}}\end{aligned}} \right), {{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{\bf{3}}\\{\bf{1}}\end{aligned}} \right), {\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{ - {\bf{4}}}\end{aligned}} \right)
The affine combination is $${\bf{y}} = \frac{1}{5}{{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{6}{5}{{\bf{v}}_3}$$.
See the step by step solution
## Step 1: Find the translated point
Write the translated points as shown below:
{{\bf{v}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}1\\8\\{ - 7}\end{aligned}} \right)
{{\bf{v}}_3} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}3\\1\\1\end{aligned}} \right)
{\bf{y}} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\2\\4\end{aligned}} \right)
Write the equation by using the translated matrix as shown below:
\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} = {c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\\\left( {\begin{aligned}{*{20}{c}}{ - 4}\\2\\4\end{aligned}} \right) = {c_2}\left( {\begin{aligned}{*{20}{c}}1\\{ - 8}\\7\end{aligned}} \right) + {c_3}\left( {\begin{aligned}{*{20}{c}}3\\1\\1\end{aligned}} \right)\end{aligned}
## Step 2: Write the augmented matrix
The augmented matrix can be written as shown below:
M = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\{ - 8}&1&2\\7&1&{ - 4}\end{aligned}} \right)
Row reduce the augmented matrix as shown below:
\begin{aligned}{c}M = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\{ - 8}&1&2\\7&1&{ - 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\0&{25}&{ - 30}\\0&{ - 20}&{24}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_2} \to {R_2} + 8{R_1}\\{R_3} \to {R_3} - 7{R_1}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\0&5&{ - 6}\\0&{ - 10}&{12}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_2} \to \frac{1}{5}{R_2}\\{R_3} \to \frac{1}{2}{R_3}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\0&1&{ - \frac{6}{5}}\\0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_3} \to {R_3} + 2{R_2}\\{R_2} \to \frac{1}{5}{R_2}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&{ - \frac{2}{5}}\\0&1&{\frac{6}{5}}\\0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} - 3{R_2}} \right)\end{aligned}
## Step 3: Write the system of equations
From the augmented matrix, the system of the equation is shown below:
$${c_2} = - \frac{2}{5}$$
And,
$${c_3} = \frac{6}{5}$$
Substitute the values in the equation of translated points as shown below:
\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} = - \frac{2}{5}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + \frac{6}{5}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\\{\bf{y}} = {{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{2}{5}{{\bf{v}}_1} + \frac{6}{5}{{\bf{v}}_3} - \frac{6}{5}{{\bf{v}}_1}\\{\bf{y}} = \frac{1}{5}{{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{6}{5}{{\bf{v}}_3}\end{aligned}
So, the vector $${\bf{y}}$$ is $$\frac{1}{5}{{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{6}{5}{{\bf{v}}_3}$$. |
# Mastering Place Value – Number Sense
Children of all ages need to develop a quality known as number sense. Number sense involves understanding how numbers relate to each other. One important aspect of number sense is mastering place value.
## Place Value Defined
Place value is breaking down each digit in a number into its true value. For example, the number 273 has a 2 in the hundreds place, a 7 in the tens place, and a 3 in the ones place. Therefore the 2 is worth 200, the 7 is worth 70, and the 3 is worth 3. 200 + 70 + 3 equals 273, our original number. Understanding place value will be a great help to your child as they go on to master other math concepts like addition, subtraction, multiplication, and division.
## Place Value Chart
In order to understand place value, a child must be familiar with the place value chart. While a child may not be required to memorize the chart, memorizing it will make place value much simpler. The link above has a color coded place value chart that you can download and print for your child to use.
## Place Value Basics
There are three ways to write numbers using place value. You can write numbers in standard form, expanded form, and word form. We will use the number 46,279 as an example for each type.
• Standard form – This is the normal way we write numbers every day. While it is what children are most familiar with, you may need to reinforce calling it standard form until they connect the term with way of writing numbers. (46,279 is already written in standard form.)
• Expanded form – This breaks down the number into the value of each digit and then adds the values together. The value of each digit is simply what the number is worth. For example the value of the 2 above is 200. (Expanded form for 46,279 would be 40,000 + 6,000 + 200 + 70 + 9.)
• Word form – Writing the number out in words is called word form. While this sounds difficult, a child just has to write the number exactly how they would read it. (46,279 would be read and written as forty-six thousand, two hundred seventy-nine.)
## Place Value Practice
Practicing place value is easy to do at home. You can write out any number and have your child tell you what place each digit is in. Then have your child write that number in expanded form and word form. Start with simple numbers and move up to larger numbers as your child achieves mastery.
Practicing place value increases your child’s number sense. It will allow them to do well in many other areas of math also. The next post on number sense will deal specifically with learning number words. |
# Why can't you divide by zero?
## Understanding the Division by Zero Conundrum
In the world of mathematics, few rules are as universally accepted and rigorously enforced as the prohibition against dividing by zero. This simple operation, which combines a basic arithmetic operation with a seemingly innocuous number, has profound implications that extend far beyond elementary mathematics into complex theoretical realms. Why exactly is dividing by zero such a forbidden territory in mathematics, and what happens when we attempt to break this rule?
To understand the intricacies of division by zero, let's first revisit the fundamental concept of division itself. Division is the mathematical operation that seeks to determine how many times one number (the divisor) is contained within another number (the dividend). For example, when we divide 10 by 2, we are essentially asking how many 2s fit into 10, which yields the result 5.
Now, let's consider what happens when we attempt to divide by zero. Take the expression 10 divided by 0. On the surface, one might speculate that the result should be infinity, since dividing a number by progressively smaller numbers tends to yield larger results. However, this intuition quickly runs into problems when we examine it more closely.
Mathematically, division by zero is undefined. This means there is no meaningful numerical value that can be assigned to the expression 10 / 0 within the framework of conventional arithmetic. To understand why, we need to delve into the properties of multiplication and inverse operations.
Every nonzero number has a multiplicative inverse, which is a number that, when multiplied by the original number, yields the identity element (usually 1). For instance, the multiplicative inverse of 2 is 1/2 because 2 * (1/2) = 1. However, zero does not have a multiplicative inverse. This is because any number multiplied by zero equals zero, not one. Therefore, there is no number that can be multiplied by zero to yield 1, which means zero lacks a well-defined inverse.
In mathematical terms, if we denote the multiplicative inverse of a number x as 1/x, then for any nonzero x, 1/x * x = 1. However, for zero, there is no such number that satisfies this condition. Consequently, division by zero results in a mathematical contradiction because it implies seeking a solution to an equation like 0 * y = 1, which has no valid solution in the realm of real numbers.
To illustrate further, consider the implications of allowing division by zero. If we were to define division by zero as yielding infinity (i.e., 10 / 0 = ∞), we would encounter inconsistencies when applying basic arithmetic operations. For example, the expression 0 * ∞ = 1 would conflict with established mathematical principles, leading to contradictions and paradoxes within the numerical system.
Moreover, the consequences of defining division by zero as infinity extend beyond elementary arithmetic into advanced mathematical fields such as calculus, number theory, and physics. In calculus, for instance, division by zero leads to indeterminate forms that require more sophisticated techniques, such as limits and derivatives, to resolve. These techniques involve approaching the division by zero scenario indirectly, rather than attempting to define it directly.
In physics, where mathematical models describe natural phenomena, division by zero can imply undefined physical quantities or breakdowns in theoretical frameworks. For instance, in equations describing the behavior of particles or waves, division by zero can lead to nonsensical predictions or invalid mathematical operations, highlighting the critical importance of maintaining mathematical rigor.
Interestingly, mathematicians and theorists have explored alternative mathematical structures where division by zero is defined or handled differently. For example, in some advanced algebraic structures or abstract mathematical spaces, division by zero may be interpreted within specific rules or frameworks that differ from traditional arithmetic. These explorations often involve creating new mathematical systems or extending existing ones to accommodate scenarios where division by zero is meaningful or permissible.
In conclusion, while the prohibition against dividing by zero may seem like a simple rule, its implications are profound and far-reaching within the realm of mathematics and its applications. By respecting this rule, mathematicians uphold the integrity and consistency of mathematical reasoning, ensuring that arithmetic operations yield meaningful results and that mathematical models accurately describe the world around us. Thus, division by zero stands as a testament to the foundational principles of mathematics, emphasizing precision, clarity, and logical coherence in mathematical discourse and exploration.
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# Diagram depicts which expression?.
Sure, here's a brief introduction for your blog post:
Welcome to Warren Institute! Have you ever wondered how diagrams can represent mathematical expressions? In this article, we'll explore the connection between visual representations and algebraic expressions. By the end, you'll be able to confidently identify which expression is represented by the diagram and understand the underlying concepts of mathematical symbolism. Let's dive into the world where math meets visualization!
## The Basics of Expressions in Mathematics
Understanding the fundamental concepts of expressions in mathematics is crucial for students to grasp more complex topics later on. Expressions are mathematical phrases that can contain numbers, variables, and operations. It's important for students to be able to identify and interpret expressions in various forms, such as diagrams, equations, and word problems.
## Analyzing the Given Diagram for Mathematical Expression
When presented with a diagram representing a mathematical expression, students need to develop the skills to analyze and translate the visual representation into a mathematical expression. This involves identifying the components of the diagram, such as shapes, symbols, and patterns, and understanding how they correspond to mathematical operations and variables.
## Connecting Visual Representations to Algebraic Expressions
Visual aids, like diagrams, play a significant role in helping students bridge the gap between concrete visual representations and abstract algebraic expressions. By making connections between the visual elements and the corresponding mathematical symbols, students can enhance their comprehension of expressions and their applications.
## The Importance of Contextualizing Mathematical Expressions
Contextualizing mathematical expressions within real-world scenarios or practical problems helps students understand the relevance and application of algebraic expressions. By relating expressions to everyday situations, students can see the utility of mathematics and develop a deeper appreciation for its significance in problem-solving and decision-making.
### Which expression is represented by the diagram in this geometry problem?
The expression represented by the diagram in this geometry problem is (x + 3)(x - 2).
### Can you explain which expression is represented by the diagram in the algebraic equation?
The expression represented by the diagram in the algebraic equation is a + b.
### How can students determine which expression is represented by the diagram in a word problem?
Students can determine which expression is represented by the diagram in a word problem by identifying the variables and operations used in the problem, and then translating the visual information into a mathematical expression.
### What strategies can be used to identify which expression is represented by the diagram in a mathematical model?
One strategy is to analyze the components of the diagram and look for patterns that match known mathematical expressions.
### Why is it important for students to understand which expression is represented by the diagram in a geometric proof?
It is important for students to understand which expression is represented by the diagram in a geometric proof because it helps strengthen their spatial reasoning skills and visualization abilities, which are crucial for solving complex mathematical problems.
In conclusion, understanding the representation of mathematical expressions through diagrams is crucial for developing students' visual and conceptual understanding of mathematical concepts. By visualizing and interpreting the diagrams, students can grasp the relationships between different mathematical expressions, fostering a deeper comprehension of the subject matter. This visual approach to learning also cultivates students' critical thinking skills and enhances their overall mathematical fluency. Incorporating diagrams into mathematics education not only reinforces traditional learning methods but also promotes a more holistic and multifaceted understanding of mathematical concepts.
See also Finding Length and Width of Rectangle Given Perimeter: Step-by-Step Guide
If you want to know other articles similar to Diagram depicts which expression?. you can visit the category General Education.
Michaell Miller
Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.
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# Two-Step Equations and Properties of Equality
## Maintain balance of an equation while solving using addition, subtraction, multiplication, or division.
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Taking Temperatures
Credit: Liz West
Source: http://www.flickr.com/photos/53133240@N00/3201760894/
License: CC BY-NC 3.0
If you’re watching the news and the meteorologist says that it’s going to be 30 degrees outside, how would you dress? Does the answer change if you’re visiting Italy? It should! 30 degrees in the U.S. is winter coat weather, but 30 degrees in Italy means shorts and a T-shirt. Why? Because the countries use two different scales of temperature.
#### Freezing, Boiling, and Everything In Between
The United States uses the Fahrenheit scale to measure temperature. On the Fahrenheit scale, water freezes at 32 degrees and boils at 212 degrees. 98.6 degrees is the human body temperature, and 30 degrees is a chilly winter day with a chance of snow. The Celsius scale sets the freezing point of water at 0 and the boiling point of water at 100. You can convert between the scales by solving a two-step equation: C=5(F32)9\begin{align*}C = \frac{5(F-32)}{9}\end{align*}. For example, you can figure out what 30°C is in degrees Fahrenheit by solving the following equation:
C2705486=5(F32)9=5(F32)=F32=F\begin{align*}C & = \frac{5(F-32)}{9}\\ 270 & = 5(F-32)\\ 54 & = F-32\\ 86 & = F\end{align*}
So, 30°C is equivalent to 86°F, or a hot summer day.
Credit: Laura Guerin
Source: CK-12 Foundation
License: CC BY-NC 3.0
Most scientists use a third temperature scale: the Kelvin scale. 0 degrees Celsius is 273 degrees on the Kelvin scale. 0 on the Kelvin scale is absolute zero, the coldest temperature possible. The temperature in space is about 2.73 K. Scientists working in labs have tried to reach temperatures very close to absolute zero, and they’ve made some fascinating discoveries.
See for yourself: http://www.youtube.com/watch?v=TNUDBdv3jWI
#### Explore More
Watch the following videos to learn more about the different temperature scales and absolute zero.
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### Image Attributions
1. [1]^ Credit: Liz West; Source: http://www.flickr.com/photos/53133240@N00/3201760894/; License: CC BY-NC 3.0
2. [2]^ Credit: Laura Guerin; Source: CK-12 Foundation; License: CC BY-NC 3.0
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3.6 – Inverse Functions Exchanging input and output values of a function or relation is called an inverse relation. If the inverse is a function, then we have an inverse function. The key idea of inverses is that for a point x , y on a graph, the point y , x is on its inverse graph. EXAMPLE 1: Sketch the graph of the quadratic function f which passes through the following points: (6,5) (0,5)
(5,0)
(1,0)
(4, 3)
(2, 3)
(3, 4) Graph the inverse on the same set of axes and describe the relationship between the function and its inverse.
Let f be a function. If a, b is a point on f, then b, a is a point on the graph of its inverse. The graph of the inverse of f is a reflection of the graph of f across the line y x . To find an inverse from an equation, write the function f (x) in terms of x and y, switch the x and y, and then solve for y. Call the result g(x) . EXAMPLE 2: Find g(x) , the inverse of f (x) 2x 1 .
EXAMPLE 3: Find g(x) , the inverse of f (x) 4 x 2 2x .
To determine whether an inverse of a function f is also a function, every output of f must correspond to exactly one input, making f a one-to-one function. A function f is one-to-one if f (a) f (b) implies that a b . If a function is one-to-one, then its inverse is also a function. A function is one-to-one if it passes the horizontal line test. (NOTE: a relation is a function if it passes the vertical line test and if it passes the horizontal line test, its inverse is also a function!!)
EXAMPLE 4: Graph each function below and determine whether the function is one-to-one. If so, graph its inverse function: a) f (x) 3x 5 2x 4 3x 3 x 2
Note that f is always ____________________, so it passes the ___________________________and is _____________________; its inverse is a ________________ that is also always ______________________. b) g(x) 0.5x 3 x 1
Note that g is ____________________ AND ____________________, so it does NOT pass the ________________________ and it is NOT _____________________ and its inverse is NOT a function!! c) h(x) 2 0.8 x 3
Note that h is always ____________________, so it passes the ________________________ and is _____________________ and its inverse is also always ______________________. Let f be a function. TFAE (the following are equivalent): The inverse of f is a function f is one-to-one the graph of f passes the Horizontal Line Test The inverse function, if it exists, is written as f 1 where: if y f (x) , then x f 1 (y) the notation f 1 does not mean
1 f
When a function is not one-to-one, it is possible to create an inverse function by restricting the domain of the function to the part that that is one-to-one. EXAMPLE 5: Find an interval on which the function f (x) x 2 3 is one-to-one, and find f 1 on that interval.
The inverse of a function f sends each output of f back to the input it came from, that is: f a b exactly when f 1 b a (a is sent to b under f exactly when b gets sent back to a under f 1 )
A one-to-one function f and its inverse function f 1 have these properties:
f
f
1
f
x
f 1 f x x for every x in the domain of f
f 1 x f f
1
x
x for every x in the domain of f 1
Also, any two functions having both properties are one-to-one and are inverses of each other. IN OTHER WORDS: If you compose a function with its inverse, you will get x.
EXAMPLE 6: Assume f is a one-to-one function. a) If f 3 10, find f 1 10
b) If f 1 7 3, find f 3
2x 3x . Use composition to verify that f and g are inverses of each and g(x) 3 x x 2 other. (In other words, compute f g and g f and show that the result is x.)
EXAMPLE 7: Let f (x)
/31.6__Inverse_Functions
http://www.houstonchristian.org/data/files/news/ClassHomework/31.6__Inverse_Functions.pdf |
# 3 Modular arithmetic
## 3.1 Division
In this section, instead of enlarging the number system , we do arithmetic with finite sets of integers, namely the sets of possible remainders when we divide by particular positive integers. This type of arithmetic is important in number theory (the study of the integers) and in cryptography. It is used frequently in group theory.
If we divide one positive integer by another we obtain a quotient and a remainder. For example, 29 divided by 4 gives quotient 7 and remainder 1 because 29 = 7 × 4 + 1. If we divide any positive integer by 4, the remainder will be one of the numbers 0, 1, 2, 3.
This idea can be extended to the division of a negative integer by a positive integer. For example, −19 divided by 4 gives quotient −5 and remainder 1 because −19 = (−5) × 4 + 1. If we divide any negative integer by 4, the remainder is again one of the numbers 0, 1, 2, 3.
This result can be generalised to the following theorem.
### Theorem 4 Division Algorithm
Let a and n be integers, with n > 0. Then there are unique integers q and r such that
Strictly speaking, this theorem is not an algorithm, but ‘division algorithm’ is the traditional name for it.
We say that dividing a by the divisor n gives quotient q and remainder r.
We do not give a formal proof of Theorem 4, but it can be illustrated as follows. We mark integer multiples of n along the number line as shown in the diagram below, and then observe in which of the resulting intervals of length n the integer a lies. Suppose that a lies in the interval [qn, (q +1)n)
so that qna < (q + 1)n.
Then, if we let r = aqn, we have a = qn + r and 0 ≤ r < n, which is the required result.
### Exercise 37
For each of the following integers a and n, find the quotient and remainder on division of a by n.
(a) a = 65, n =7
(b) a = −256, n =13
### Solution
(a) 65 = 9 ×7 + 2, so the quotient is 9 and the remainder is 2.
(b) −256 = −20 × 13 + 4, so the quotient is −20 and the remainder is 4.
### Exercise 38
(a) What are the possible remainders on division of an integer by 7?
(b) Find two positive and two negative integers all of which have remainder 3 on division by 7.
### Solution
(a) The possible remainders are 0, 1, 2, 3, 4, 5 and 6.
(b) There are many possible answers here; for example, 3, 10, −4 and −11. |
LEARNATHON
III
Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!
### Theory:
A copper wire is connected between the points $$X$$ and $$Y$$ in the given circuit. Imagine a nichrome wire used instead of copper wire to conduct the experiment, maintaining the same potential difference across the circuit.
Do you think any change will occur in the circuit? If so, what might have changed?
An electric circuit
Yes, there will be a change in the value of the current flowing through the circuit.
Now, change the nichrome wire to an aluminium wire and observe the same. Again, you will notice a different set of current values in the circuit.
The change in the value of current for the same amount of potential difference shows that the resistance varies depending on the material.
A material's resistance is its ability to resist the flow of charges and thus the current flow through it. The value of resistance is different for different materials.
From Ohm's law $V=\mathit{IR}$, the formula of resistance ($$R$$) of any material or a conductor is written as,
$R=\frac{V}{I}$
The resistance of a conductor can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it.
Unit:
The SI unit of resistance is ohm, and it is denoted by $\mathrm{\Omega }$.
Resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its ends.
$\mathit{1}\phantom{\rule{0.147em}{0ex}}\mathit{ohm}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{1}\phantom{\rule{0.147em}{0ex}}\mathit{volt}}{\mathit{1}\phantom{\rule{0.147em}{0ex}}\mathit{ampere}}$
Some other units of resistance are:
There are some factors that affect the resistance of a conductor. Let us study these factors in detail in the upcoming section. |
9.3 Systems of nonlinear equations and inequalities: two variables (Page 5/9)
Page 5 / 9
Verbal
Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers.
A nonlinear system could be representative of two circles that overlap and intersect in two locations, hence two solutions. A nonlinear system could be representative of a parabola and a circle, where the vertex of the parabola meets the circle and the branches also intersect the circle, hence three solutions.
When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution?
When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region?
No. There does not need to be a feasible region. Consider a system that is bounded by two parallel lines. One inequality represents the region above the upper line; the other represents the region below the lower line. In this case, no points in the plane are located in both regions; hence there is no feasible region.
If you graph a revenue and cost function, explain how to determine in what regions there is profit.
If you perform your break-even analysis and there is more than one solution, explain how you would determine which x -values are profit and which are not.
Choose any number between each solution and plug into $\text{\hspace{0.17em}}C\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}R\left(x\right).\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}C\left(x\right) then there is profit.
Algebraic
For the following exercises, solve the system of nonlinear equations using substitution.
$\left(0,-3\right),\left(3,0\right)$
$\left(-\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2}\right),\left(\frac{3\sqrt{2}}{2},-\frac{3\sqrt{2}}{2}\right)$
For the following exercises, solve the system of nonlinear equations using elimination.
$\begin{array}{l}\hfill \\ 4{x}^{2}-9{y}^{2}=36\hfill \\ 4{x}^{2}+9{y}^{2}=36\hfill \end{array}$
$\left(-3,0\right),\left(3,0\right)$
$\begin{array}{l}{x}^{2}+{y}^{2}=25\\ {x}^{2}-{y}^{2}=1\end{array}$
$\begin{array}{l}\hfill \\ 2{x}^{2}+4{y}^{2}=4\hfill \\ 2{x}^{2}-4{y}^{2}=25x-10\hfill \end{array}$
$\left(\frac{1}{4},-\frac{\sqrt{62}}{8}\right),\left(\frac{1}{4},\frac{\sqrt{62}}{8}\right)$
$\begin{array}{l}{y}^{2}-{x}^{2}=9\\ 3{x}^{2}+2{y}^{2}=8\end{array}$
$\begin{array}{l}{x}^{2}+{y}^{2}+\frac{1}{16}=2500\\ y=2{x}^{2}\end{array}$
$\left(-\frac{\sqrt{398}}{4},\frac{199}{4}\right),\left(\frac{\sqrt{398}}{4},\frac{199}{4}\right)$
For the following exercises, use any method to solve the system of nonlinear equations.
$\left(0,2\right),\left(1,3\right)$
$\left(-\sqrt{\frac{1}{2}\left(\sqrt{5}-1\right)},\frac{1}{2}\left(1-\sqrt{5}\right)\right),\left(\sqrt{\frac{1}{2}\left(\sqrt{5}-1\right)},\frac{1}{2}\left(1-\sqrt{5}\right)\right)$
$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9{x}^{2}+25{y}^{2}=225\hfill \\ {\left(x-6\right)}^{2}+{y}^{2}=1\hfill \end{array}$
$\left(5,0\right)$
$\left(0,0\right)$
For the following exercises, use any method to solve the nonlinear system.
$\left(3,0\right)$
No Solutions Exist
$\begin{array}{l}\hfill \\ -{x}^{2}+y=2\hfill \\ -4x+y=-1\hfill \end{array}$
No Solutions Exist
$\begin{array}{l}{x}^{2}+{y}^{2}=25\\ {x}^{2}-{y}^{2}=36\end{array}$
$\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right),\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right),\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right),\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$
$\left(2,0\right)$
$\left(-\sqrt{7},-3\right),\left(-\sqrt{7},3\right),\left(\sqrt{7},-3\right),\left(\sqrt{7},3\right)$
$\left(-\sqrt{\frac{1}{2}\left(\sqrt{73}-5\right)},\frac{1}{2}\left(7-\sqrt{73}\right)\right),\left(\sqrt{\frac{1}{2}\left(\sqrt{73}-5\right)},\frac{1}{2}\left(7-\sqrt{73}\right)\right)$
Graphical
For the following exercises, graph the inequality.
${x}^{2}+y<9$
${x}^{2}+{y}^{2}<4$
For the following exercises, graph the system of inequalities. Label all points of intersection.
$\begin{array}{l}{x}^{2}+y<1\\ y>2x\end{array}$
$\begin{array}{l}{x}^{2}+y<-5\\ y>5x+10\end{array}$
$\begin{array}{l}{x}^{2}+{y}^{2}<25\\ 3{x}^{2}-{y}^{2}>12\end{array}$
$\begin{array}{l}{x}^{2}-{y}^{2}>-4\\ {x}^{2}+{y}^{2}<12\end{array}$
$\begin{array}{l}{x}^{2}+3{y}^{2}>16\\ 3{x}^{2}-{y}^{2}<1\end{array}$
Extensions
For the following exercises, graph the inequality.
$\begin{array}{l}\hfill \\ y\ge {e}^{x}\hfill \\ y\le \mathrm{ln}\left(x\right)+5\hfill \end{array}$
$\begin{array}{l}y\le -\mathrm{log}\left(x\right)\\ y\le {e}^{x}\end{array}$
For the following exercises, find the solutions to the nonlinear equations with two variables.
$\begin{array}{l}\frac{4}{{x}^{2}}+\frac{1}{{y}^{2}}=24\\ \frac{5}{{x}^{2}}-\frac{2}{{y}^{2}}+4=0\end{array}$
$\begin{array}{c}\frac{6}{{x}^{2}}-\frac{1}{{y}^{2}}=8\\ \frac{1}{{x}^{2}}-\frac{6}{{y}^{2}}=\frac{1}{8}\end{array}$
$\left(-2\sqrt{\frac{70}{383}},-2\sqrt{\frac{35}{29}}\right),\left(-2\sqrt{\frac{70}{383}},2\sqrt{\frac{35}{29}}\right),\left(2\sqrt{\frac{70}{383}},-2\sqrt{\frac{35}{29}}\right),\left(2\sqrt{\frac{70}{383}},2\sqrt{\frac{35}{29}}\right)$
No Solution Exists
Technology
For the following exercises, solve the system of inequalities. Use a calculator to graph the system to confirm the answer.
$\begin{array}{l}xy<1\\ y>\sqrt{x}\end{array}$
$x=0,y>0\text{\hspace{0.17em}}$ and $0
$\begin{array}{l}{x}^{2}+y<3\\ y>2x\end{array}$
Real-world applications
For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions.
Two numbers add up to 300. One number is twice the square of the other number. What are the numbers?
12, 288
The squares of two numbers add to 360. The second number is half the value of the first number squared. What are the numbers?
A laptop company has discovered their cost and revenue functions for each day: $\text{\hspace{0.17em}}C\left(x\right)=3{x}^{2}-10x+200\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}R\left(x\right)=-2{x}^{2}+100x+50.\text{\hspace{0.17em}}$ If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit.
2–20 computers
A cell phone company has the following cost and revenue functions: $\text{\hspace{0.17em}}C\left(x\right)=8{x}^{2}-600x+21,500\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}R\left(x\right)=-3{x}^{2}+480x.\text{\hspace{0.17em}}$ What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit.
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert |
# How do you find the vertex and intercepts for x^2-10x-8y+33=0?
##### 1 Answer
Mar 11, 2017
${y}_{\text{intercept}} = \frac{33}{8}$
Vertex$\to \left(x , y\right) = \left(5 , 1\right)$
NO X-INTERCEPT
#### Explanation:
Moving $8 y$ to the other side of the = and change its sign
${x}^{2} - 10 x + 33 = + 8 y$
To get $y$ on its own divide both sides by 8
$\frac{{x}^{2}}{8} - \frac{10}{8} x + \frac{33}{8} = y$
Write as:
$y = \frac{{x}^{2}}{8} - \frac{10}{8} x + \frac{33}{8}$
Write as:
$y = \frac{1}{8} \left({x}^{2} \textcolor{red}{- 10} x\right) + \frac{33}{8}$
color(green)(y_("intercept")=+33/8 larr" read directly off the equation")
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- 10}\right) = + 5}$
The above line is part of the process of completing the square.
$\textcolor{g r e e n}{\text{The "x^2/8"is positive so the graph is of general shape } \cup}$
Substituting $x = + 5$ gives:
$\textcolor{g r e e n}{{y}_{\text{vertex}} = \frac{1}{8} \left[\textcolor{w h i t e}{\frac{.}{.}} {5}^{2} - 10 \left(5\right) \textcolor{w h i t e}{.}\right] + \frac{33}{8} = 1}$
$\textcolor{g r e e n}{\text{Vertex} \to \left(x , y\right) = \left(5 , 1\right)}$
As the graph is of general shape $\cup$ and ${y}_{\text{vertex}}$ is above the x-axis there is NO X-INTERCEPT |
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# Special Product Special Product
## 1. Square of a Binomial 1. Square of a Binomial
(𝑥 + 𝑦)2 = 𝑥 2 + 2𝑥𝑦 + 𝑦 2 (𝑥 + 𝑦)2 = 𝑥 2 + 2𝑥𝑦 + 𝑦 2
(𝑥 − 𝑦)2 = 𝑥 2 − 2𝑥𝑦 + 𝑦 2 (𝑥 − 𝑦)2 = 𝑥 2 − 2𝑥𝑦 + 𝑦 2
Steps in Squaring a Binomial: Steps in Squaring a Binomial:
1. Square the first term. 1. Square the first term.
2. Twice the product of the first and the second term. 2. Twice the product of the first and the second term.
3. Squaring the last term. 3. Squaring the last term.
2. Product of the Sum and Difference of the Same Two 2. Product of the Sum and Difference of the Same Two
Terms Terms
(𝑥 + 𝑦)(𝑥 − 𝑦) = 𝑥 2 − 𝑦 2 (𝑥 + 𝑦)(𝑥 − 𝑦) = 𝑥 2 − 𝑦 2
To get the product of the sum and difference of the same To get the product of the sum and difference of the same
two terms, just get the difference of the square of the first two terms, just get the difference of the square of the first
term and the square of the second term. term and the square of the second term.
3. Cube of a Binomial 3. Cube of a Binomial
(𝑥 + 𝑦)3 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 (𝑥 + 𝑦)3 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3
(𝑥 − 𝑦)3 = 𝑥 3 − 3𝑥 2 𝑦 + 3𝑥𝑦 2 − 𝑦 3 (𝑥 − 𝑦)3 = 𝑥 3 − 3𝑥 2 𝑦 + 3𝑥𝑦 2 − 𝑦 3
Steps in Cubing a Binomial: Steps in Cubing a Binomial:
1. Cube the first term. 1. Cube the first term.
2. Thrice the square of the first term times the last term. 2. Thrice the square of the first term times the last term.
3. Thrice the square if the second term times the first 3. Thrice the square if the second term times the first
term. term.
4. Cube the last term. 4. Cube the last term.
4. Special case of the Product of a Binomial and a Trinomial 4. Special case of the Product of a Binomial and a Trinomial
(𝒙 + 𝒚)(𝒙𝟐 − 𝒙𝒚 + 𝒚𝟐 ) = 𝒙𝟑 + 𝒚𝟑 (𝒙 + 𝒚)(𝒙𝟐 − 𝒙𝒚 + 𝒚𝟐 ) = 𝒙𝟑 + 𝒚𝟑
(𝒙 − 𝒚)(𝒙𝟐 + 𝒙𝒚 + 𝒚𝟐 ) = 𝒙𝟑 − 𝒚𝟑 (𝒙 − 𝒚)(𝒙𝟐 + 𝒙𝒚 + 𝒚𝟐 ) = 𝒙𝟑 − 𝒚𝟑
The product of a binomial and a trinomial will NOT always The product of a binomial and a trinomial will NOT always
lead to Sum of Two Cubes lead to Sum of Two Cubes
(x3 + y3) or Difference of Two cubes (x 3 - y3). We need to (x3 + y3) or Difference of Two cubes (x 3 - y3). We need to
inspect if the given trinomial factor is “fit” with our inspect if the given trinomial factor is fit with our binomial
binomial factor. factor.
1. The first term of the trinomial factor is the square of 1. The first term of the trinomial factor is the square of
the first term of the binomial. the first term of the binomial.
2. The second term of the trinomial is the negative 2. The second term of the trinomial is the negative
product of the first and the second term of the product of the first and the second term of the
binomial. binomial.
3. The last term of the trinomial factor is the square of 3. The last term of the trinomial factor is the square of
the last term of the binomial factor. the last term of the binomial factor.
(If these three conditions are satisfied then we can (If these three conditions are satisfied then we can
apply the special case of the product of a binomial apply the special case of the product of a binomial
and a trinomial) and a trinomial)
5. Square of a Trinomial 5. Square of a Trinomial
(𝑎 + 𝑏 + 𝑐)2 = 𝑎2 + 𝑏2 + 𝑐 2 + 2𝑎𝑏 + 2𝑎𝑐 + 2𝑏𝑐 (𝑎 + 𝑏 + 𝑐)2 = 𝑎2 + 𝑏2 + 𝑐 2 + 2𝑎𝑏 + 2𝑎𝑐 + 2𝑏𝑐
## Steps in squaring a trinomial: Steps in squaring a trinomial:
1. Square the first term. 1. Square the first term.
2. Square the second term. 2. Square the second term.
3. Square the last term. 3. Square the last term.
4. Twice the product of the first and second 4. Twice the product of the first and second
term. term.
5. Twice the product of the second and last 5. Twice the product of the second and last
term. term.
6. Twice the product of the first and last term. 6. Twice the product of the first and last term. |
# Matrices: An introduction
This tutorial discusses some of the ins and outs of matrices. Matrices can be fun, but more importantly, they can really save you time. Author: Lionel Brits
Introduction
A matrix is, by definition, a rectangular array of numeric or algebraic quantities which are subject to mathematical operations. Matrices can be defined in terms of their dimensions (number of rows and columns). Let us take a look at a matrix with 4 rows and 3 columns (we denote it as a 4×3 matrix and call it A):
Each individual item in a matrix is called a cell, and can be denoted by the particular row and column it resides in. For instance, in matrix A, element a32 can be found where the 3rd row and the 2nd column intersect.
What are they used for?
Matrices are used to represent complicated or time-consuming mathematical operations. A single matrix can hold an infinite number of calculations, which can then be applied to a number, vector, or another matrix. There are several operations that can be done on matrices, including addition, multiplication and inverse calculation; some of which will be discussed shortly. Operations done on one matrix can be transferred to another matrix simply by concatenating the two (by matrix multiplication). Matrices often find their use in 3 dimensional applications, were numerous identical operations are performed on thousands of vectors 30 or 40 times a second. Combining all these operations in one single matrix significantly improves the speed and functionality of a 3D rendering pipeline. Matrices are also used in financial processes (again, where a large number of data has to be processed in a similar fashion).
Operations on Matrices
Addition and subtraction operations can easily be performed on matrices, provided the matrices have the same dimensions. All that is required is to add or subtract the corresponding cells of each matrix involved in the operation. Let us take a look at the addition of two 2×3 matrices, A and B:
2.Scalar Multiplication
Multiplying a matrix by a scalar value involves multiplying every element of the matrix by that value. Here we multiply our 2×3 matrix A by a scalar value β:
3.Matrix Multiplication
The multiplication operation on matrices differs significantly from its real counterpart. One major difference is that multiplication can be performed on matrices with different dimensions. The first restriction is that the first matrix has to have the same amount of columns as the second has rows. The reason for this will become clear shortly. Another thing to note is that matrix multiplication is not commutative i.e., (CD) does not equal (DC).
The procedure for matrix multiplication is rather simple. First, we determine the dimensions of the resultant matrix. All we require is that there are as many columns in the first matrix as there are rows in the second. A simple way of determining is to look at the nearest and farthest dimensions of two matrix symbols written next to each other, for instance: C[2x3] D[3x2]. The nearest dimensions are both equal to 3, and so we know that the operation is possible. The farthest dimensions will give us the dimensions of the product matrix, so our result will be a 2×2 matrix. The general rule says that in order to perform the multiplication AB, where A is a (m x n) matrix and B a (k x l) matrix, we must have n=k. The result will be a (m x l) matrix.
Performing the operation product involves multiplying the cells of a particular rows in the first matrix by the cells of a particular column in the second matrix, adding the products, and storing the result in the cell of the resultant matrix whose coordinates correspond to the row of the first matrix and the column of the second matrix. For instance, in AB = C, if we want to find the value of c12, we must multiply the cells of row 1 in the first matrix by the cells of column 2 in the second matrix and sum the results.
4.Matrix/Vector Multiplication
We can also multiply matrices and vectors together, since a vector is nothing more than a 1-column matrix. Consider a 4×4 matrix M which comprises an arbitrary number of transformations (rotation, scaling, translation, etc.) which are to be applied to a 4×1 vector, S. The resultant 4×1 vector R obtained by multiplying M and S together is then a transformation of S.
5.Coding with matrices
How do we define a matrix in C++ you ask? Simple. If a matrix is a two dimensional array of numbers, by definition, then we should be able to define it in C++ that way too. Here’s how:
`typedef float MATRIX[4][4]; // A 4×4 matrix`
The simplest thing you would want to do is concatenate (multiply) two matrices. We perform the operation, as mentioned earlier, like so:
```// matmult: Multiplies matrices A and B, storing the result in A void matmult(MATRIX &A, MATRIX B) {```
```MATRIX conc; // Multiplies by rows and columns for (int i = 0; i <4; i++) for (int j = 0; j < 4; j++) {```
```conc[i][j] = (A[0][j] * B[i][0])+ (A[1][j] * B[i][1])+ (A[2][j] * B[i][2])+ (A[3][j] * B[i][3]); }```
```// Copy result matrix into matrix A for (int p = 0; p < 4; p++) for (int q = 0; q < 4; q++) a[p][q] = conc[p][q]; }```
Note: The code references cell yx as cell xy. This may be changed in the near future for the sake of clarity.
You also need to initialize a matrix before you begin to work with it (unless you are simply copying from one matrix to another). Here’s the code to do so:
```// matinit: Initializes a matrix, and sets to 'identity' void matinit(MATRIX &a) {```
```for (int i = 0; i < 4; i++) for (int j = 0; j < 4; j++) a[i][j] = 0.0f; a[0][0] = 1.0f;```
```a[1][1] = 1.0f; a[2][2] = 1.0f; a[3][3] = 1.0f; }```
Note: The code references cell yx as cell xy. This may be changed in the near future for the sake of clarity.
Matrices are essentially useless unless you apply them to something. In 3D programming, you would want to apply them to vectors. By applying say, a rotation matrix, to a vector, we can rotate that vector around the origin in 3D space. 3D vectors are simply 3×1 matrices. Here is the code to do so:
```// vecmatmult: Multiplies vector V and matrix B void vecmatmult(VECTOR &v, MATRIX b) {```
```float x, y, z, w = 1; x = (v.x * b[0][0]) + (v.y * b[0][1]) + (v.z * b[0][2]) + (w * b[0][3]); y = (v.x * b[1][0]) + (v.y * b[1][1]) + (v.z * b[1][2]) + (w * b[1][3]); z = (v.x * b[2][0]) + (v.y * b[2][1]) + (v.z * b[2][2]) + (w * b[2][3]);```
```v.x = x; v.y = y; v.z = z; }```
Note: The code references cell yx as cell xy. This may be changed in the near future for the sake of clarity.
What is w you ask? The w is simply to accommodate the translation cells, since the vector is only 3×1 and the matrix is 4×4.
Useful matrices
Here are a couple of useful matrices that you can use. Most of them are for 3D geometry and are 4×4 matrices. I will add some more in the near future and also show how they are derived. Here they are:
The identity matrix is a matrix that does not affect the contents stored in another matrix when multiplied. It can be used as a basis for for constructing other matrices. An identity matrix can be any size, as long as it’s diagonal consists of 1’s and all other cells are filled with zeros, as shown:
The translation matrix is a matrix that can be used to translate vectors, such points, in n-dimensional space. Here, translations are for 3-dimensional space, where tx, ty and tz relate to displacements in the x, y and z dimensions respectively. Each component of the translation is added to it’s respective vector component. Notice how this matrix is based on the identity matrix:
The scaling matrix is a matrix that can be used to scale vectors, such points, in n-dimensional space. Here, scaling values are for 3-dimensional space, where sx, sy and sz relate to scaling factors in the x, y and z dimensions respectively. Notice, once again, how this matrix is based on the identity matrix:
The x-rotation matrix is a matrix that can be used to rotate vectors around the x-axis in 3-dimensional space. Theta (q) is the angle at which the vector is rotated. Lets look at how this matrix is derived:
The process of rotating a vector around the x-axis in 3D space is as follows:
x’ = x
y’ = (y*cosq) + (z*sinq)
z’ = (z*-sinq) - (y*cosq)
When applying this matrix to a vector (by means of a dot product), each of those transformations is only applied once, to the appropriate vector component. Again, these matrices are based on the identity matrix:
The y-rotation matrix is a matrix that can be used to rotate vectors around the y-axis in 3-dimensional space. Theta (q) is the angle at which the vector is rotated. Lets look at how this matrix is derived:
Here, the process of rotation is similar to that of the x-rotation matrix:
x’ = (x*cosq) - (z*sinq)
y’ = y
z’ = (x*sinq) + (z*cosq)
Here is the matrix:
The z-rotation matrix is a matrix that can be used to rotate vectors around the z-axis in 3-dimensional space. Again, theta (q) is the angle at which the vector is rotated. Lets look at a simple derivation:
Here, the process of rotation is similar to that of the x-rotation matrix:
x’ = (x*cosq) + (y*sinq)
y’ = -(x*sinq) + (y*cosq)
z’ = z
Here it is:
### 3 Responses to “Matrices: An introduction”
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2. zerodtkjoe says:
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3. bet365 says:
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The median problem
The median of a set of numbers is widely used in various industries, and as such it's an everyday problem. Calculating the middle value is tightly related to sorting. In this post I will show a solution to the median problem.
Problem statement
You decide to buy a house in a fancy suburb: Big homes, white fence and large trees. While preparing to home opens you want to know how much homes cost in the suburb. You did some research and learned that the last nine homes were sold at the following prices (in thousand dollars):
435, 850, 394, 465, 443, 490, 450, 425, 524
Calculate the median of the given home prices.
Background
If you are familiar with the concept of various averages including the median, feel free to skip the following section.
The mean
One can use different types of averages to describe the centre of a set of numbers. The most well-known of these methods is the mean, which is equal to the sum of the scores divided by the number of scores. Most people associate the average with the mean.
However, mean has a disadvantage that cannot be neglected: Outliers can skew the result. If the set of numbers contains an extremely high (or extremely low) value, the mean won’t represent the set properly anymore because the extreme value will increase the average. This is the reason why the largest and the smallest scores given by the judges are ignored in certain sports.
The median
Some industries (like real estate) use the median instead, which can better represent the overall characteristics of the values.
The median is the middle value in an ordered set of numbers, i.e. the same number of values should occur both on the left and the right of the median in the set.
Finding the median involves two steps: 1) Order the numbers. 2) Find the element in the middle.
If we have five numbers in ascending order, we will need the third one from the left and this value will be the median. Out of nine values the median will be the 5th one.
But what if we have to find the median of a set of even numbers? There won’t be any middle value!
This time we will take middle two numbers. In a set of six numbers we will need to take the 3rd and the 4th values and calculate their mean (i.e. add them together and divide the result by 2).
This way we can calculate the median of a set of any numbers.
Breaking it down
As always, it’s a good idea to break down the problem to building blocks.
Calculating the median involves two important steps. First, we need to order the values from smallest to greatest. Then we need to pick the middle element.
The first part seems to be easy but step back and think for a moment. Do we really need to sort every number? No, we don’t. If we could somehow divide the set into two partitions, we could see which partition contains the required element (in our example it’s the 5th smallest value). If we have this piece of information, we can decide which partition can be abandoned and which should be kept and worked with.
Of course, it’s not a big deal for nine numbers but it can be so if millions of numbers are sorted, so we will follow the abovementioned technique and will only sort the relevant part of the set.
Sorting
It turned out that finding the median is a sorting problem. I will apply the principles of the quicksort algorithm to create a solution.
Again, we could simply `sort` the numbers and then get the index of the middle value but it involves quite a bit of unnecessary operations.
JavaScript code
Let’s see how each step can be assigned some JavaScript code.
We can use an array to store home prices.
We will need the k-th smallest number in the array where `k` is the middle value. We will use quicksort, which was covered in one of the past articles.
The solution will be very similar to the one described in the linked post. We will need to find a pivot element and partition the array into three parts (left, equal and right). But then we will decide which partition to work with based on the length of those.
If the length of the left partition is greater than or equal to `k` (the position or index of the median), the median must be in the left partition. Similarly, if the length of the right partition is greater than or equal to `k`, the median value must be in the right block. If we are lucky enough to get both left and right partitions to be of equal length, the median will be the pivot value.
We need to repeat this process until we get one value (the median), which means that some recursion will be involved in the algorithm, similar to quicksort.
Solution
Let’s start with the `prices` array and the declaration of our function:
``````const prices = [435, 850, 394, 465, 443, 490, 450, 425, 524];
const medianPosition = Math.floor(prices.length / 2) + 1;
function getMedian(arr, med) {
}
``````
We also define the position (index) of the median.
First we need to find a pivot element. We will use the first element in the array as it’s easy to work with. We also declare the partitions and start grouping the elements of the array into the partitions based on their value compared to the pivot.
This is the exact same process as described in the quicksort algorithm:
``````if (arr.length === 1) return arr[0];
let k = med;
const pivot = arr[0];
const left = [];
const right = [];
const equal = [pivot];
for (let i = 1; i < arr.length; i++) {
if (arr[i] < pivot) {
left.push(arr[i]);
} else if (arr[i] > pivot) {
right.push(arr[i]);
} else {
equal.push(arr[i]);
}
}
``````
The first line is important as this will end the recursion. When we have only one element left in any of the partitions, we will stop calling `getMedian` and return the element.
We also create the `k` variable, which is initially equal to the median position calculated above.
From this point on the solution will slightly differ from the quicksort algorithm. We need to find the k-th smallest element, so `k` needs to be compared to the length of the partitions.
Let’s see the code first:
``````const leftLength = left.length;
const equalLength = equal.length;
if (leftLength + equalLength === k) {
return pivot;
} else if (leftLength + equalLength > k) {
if (leftLength < k) return pivot;
return getMedian(left, k);
} else {
k = k - (leftLength + equalLength);
return getMedian(right, k);
}
``````
Because we need to find the k-th smallest element, we are only interested in the `left` and `equal` partitions as numbers increase from left to right. Based on their combined length and the value of `k`, we can observe three cases.
If `leftLength + equalLength === k`, i.e. the sum of the lengths of the `left` and `equal` partitions is exactly `k`, we have found the median and it will be the `pivot` element. Why? Because each element in `left` is less than the pivot and we will have `k` - 1 element that are less than the pivot, so the k-th smallest value will be `pivot` itself.
We get lucky here. We don’t even need to sort either `left` or `right` as we have found the median immediately. This case won’t occur very often though, so don’t bet the house on it.
What happens when the combined length of `left` and `equal` is greater than `k`? It means that the median will be in either in `left` or in `equal`. `equal` contains the `pivot` element, so if the length of `left` is less than `k` (the position of the median), the median must be in `equal` that is we need to return the `pivot` element again (same values can occur multiple times in the set). Otherwise the median must be in `left` and all we need to do is repeat the partitioning process on `left`.
The third scenario is when the median is in the `right` partition. In this case we need to redefine `k` as we are not interested in the k-th smallest number anymore on the new `right` array. In `right` the k-th smallest value has no sense, so the line
``````k = k - (leftLength + equalLength);
``````
is supposed to make this correction. We subtract the combined length of `left` and `equal` from `k` and this will be the element we will be looking for in the `right` partition when we call `getMedian` again. It’s like starting over the partitioning process with `right` being the array of values and the new `k` being the index whose value we are interested in.
The full code looks like this:
``````function getMedian(arr, med) {
if (arr.length === 1) return arr[0];
let k = med;
const pivot = arr[0];
const left = [];
const right = [];
const equal = [pivot];
for (let i = 1; i < arr.length; i++) {
if (arr[i] < pivot) {
left.push(arr[i]);
} else if (arr[i] > pivot) {
right.push(arr[i]);
} else {
equal.push(arr[i]);
}
}
const leftLength = left.length;
const equalLength = equal.length;
if (leftLength + equalLength === k) {
return pivot;
} else if (leftLength + equalLength > k) {
if (leftLength < k) return pivot;
return getMedian(left, k);
} else {
k = k - (leftLength + equalLength);
return getMedian(right, k);
}
}
``````
Let’s call `getMedian` with `prices` and `medianPosition`:
``````console.log(getMedian(prices, medianPosition)); // 450
``````
We get that the median price is \$450.000.
Considerations
What if we have even number of values?
One solution is to use the function we have just created and apply it on the two middle elements one by one.
We can check first if the length of the array is odd. If so, `getMedian` can be applied as is.
If the length is an even number, we will take the middle two values and call `getMedian` on each and calculate the mean of the results.
Thanks for reading and I hope that you liked this post. If so, see you next time. |
4.1: Fundamental Counting Rules – Counting
# 4.1: Fundamental Counting Rules – Counting
To begin, we will formalize two of the most basic rules in counting – the product rule and sum rule. We will then look at several examples that apply both of these rules.
## Product Rule
### Example: Number of Factors
How many factors does $1200$ have?
We could sit down and manually enumerate through all factors of 1200, one by one - 1, 2, 3, 4, 5, 6, … but that would take quite a bit of time. Instead, we will look at the prime factorization of 1200: $1200 = 2^4 \cdot 3 \cdot 5^2$.
We know each factor of 1200 will have some number of $2$s, some number of $3$s, and some number of $5$s. There are 5 options for the number of $2$s a factor could have: 0, 1, 2, 3 or 4 (meaning a factor of 1200 could either have a factor of $2^0$, or $2^1$, or $2^2$, or $2^3$, or $2^4$). Similarly, there are 2 options for the number of $3$s a factor could have (either 0 or 1) and 3 options for the number of $5$s (0, 1, or 2).
Since we’re making three successive choices, we take the product of the number of choices at each step, yielding
$\text{number of factors of } 1200 = (4 + 1) (1 + 1) (2 + 1) = 30$
In general, if we have $n = p_1^{e_1} \cdot p_2^{e_2} \cdot … \cdot p_k^{e_k}$, the following holds:
$\text{number of factors of } n = \prod_{i = 1}^k (e_i + 1) = (e_1 + 1) \cdot (e_2 + 1) \cdot ... \cdot (e_{k - 1} + 1) \cdot (e_k + 1)$
### Example: Cardinality of the Power Set
The power set $P(S)$ of a set $S$ is a set of all possible subsets of $S$. For example, if $S = {1, 2, 3}$, we have $P(S) = \{ \emptyset , \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\} \}$.
If $S$ is a set such that $|S| = n$, what is $|P(S)|$?
When creating a subset of $S$, for each of the $n$ items in $S$, we have two options: either we include it in the subset, or do not include it in the subset. Since we have two options for each of the $n$ items, the total number of ways we can create a subset of $S$ is $2 \cdot 2 \cdot … \cdot 2 = 2^n$. |
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# Right Triangle Formula
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Last updated date: 02nd Aug 2024
Total views: 376.8k
Views today: 8.76k
## Introduction to Right Triangle
When you are studying geometry, there is a chance that you might come across a wide range of figures that vary in shapes and sizes. Each and every single one of the figures has different properties that make them unique. In this section, we are going to discuss a particular figure known as the right triangle. You will get to know the right triangle formula, types and much more from here.
### Right Angle
An angle that is exactly equal to 90 degrees is called a right angle and there are various real-life examples of the right angles in our daily life. For example, the corner angle of a book, edges of the cardboard, etc. Any object with square and rectangular shapes will have its corners equal to 90 degrees or right angle.
Before you get into the formulas for the area and volume of the right angle triangle, you need to know about the properties that are exhibited in the triangle. The triangle can be defined as a figure that is closed. It is basically a polygon that has about 3 sides in total. There are 3 different vertices of a triangle. With the 3 enclosed sides of the triangle, there is a formation of the 3 interior angles. One of the main things to notice about the right angle triangle is that the sum of all the angles in the interior is always 180 degrees in total. Here we are going to discuss some of the important formulas that you need to know about.
### Right angled Triangle
A triangle that has one of its angles equal to 90 degrees is called a right-angled triangle. The right-angle side of the triangle is perpendicular and the base of the triangle. The third side of the triangle is called the hypotenuse, which is the longest side of all three sides.
### Formula for Area of Right-Triangle
The formula area of a right triangle,
Area of a triangle = $\frac{1}{2}$ bh
Where,
b is the base or adjacent side of the right triangle.
h is the height of the right triangle.
### Formula for Perimeter of Right-Triangle
The perimeter of a right triangle is the total length covered by the outer boundary or the sum of all three sides of the triangle.
The formula perimeter of a right triangle,
Perimeter of a right-angled triangle = a + b + c (sum of all the three sides.)
Where a, b, and c are the length of three sides of the right triangle.
### Pythagoras Theorem for a Right Angle Triangle
Pythagoras Theorem is one of the most important theorems in geometry. This is a theorem that is used in order to find the diagonal of the triangle formula. The three sides of the right triangle are interrelated to each other. This relationship between the sides of the triangle is expressed through the Pythagoras theorem.
Hypotenuse² = Perpendicular² + Base²
According to the theorem, if we are familiar with the two different sides of the right triangle, then it will be very easy to find the right side. The Pythagoras theorem states that.
## FAQs on Right Triangle Formula
1. What is pythagoras theorem in a right-angled triangle?
The Pythagoras theorem, also known as the Pythagorean theorem, states the relationship between the three sides of a right-angled triangle. According to the Pythagoras theorem, the square of the hypotenuse of the triangle is equal to the sum of the squares of the other perpendicular and base sides of a right-angled triangle. If ABC is a triangle, where AB is the perpendicular side, BC is the base and AC is the hypotenuse of the right-angled triangle.
The Pythagoras theorem is written as AC² = BC² + AB².
2. What is a right triangle?
An angle that is exactly equal to 90 degrees is called a right angle. The hypotenuse is the longest side of the right triangle, and the other two sides are the height and the base.
3. Is it possible to have two right angles?
No, It is impossible for a triangle to have 2 right angles. We know that a triangle consists of 3 sides and the sum of interior angles of the triangle, sums up to 180°. Consider, if a triangle has two right angles (90 degrees), then the third angle will have to be 0 degrees which means the third side will overlap with another side of the triangle. Hence, a triangle with 2 right angles is not possible. |
Equivalent Ratios – Definition with Examples
Home » Math Vocabulary » Equivalent Ratios – Definition with Examples
What Are Equivalent Ratios?
Two ratios that turn out to be the same in comparison are known as equivalent ratios. In order to check whether the given ratios are equivalent or not, we will have to simplify them or reduce them to their simplest form.
Example: Consider the ratios as 1:2, 2:4. 3:6.
If you reduce 3:6 to its simplest form, you get 1:2.
If you reduce 2:4 to its simplest form, you get 1:2.
Thus, these are equivalent ratios.
What Is a Ratio?
Ratio is defined as the quantitative relation or comparison between two different quantities of the same kind and same unit. In mathematics, the symbol “:” is used to express ratio, where a ratio is anything that compares two quantities of the same kind.
Ratio of a to $b = a : b = \frac{a}{b}$
The first quantity of the ratio is called antecedent, whereas the second quantity of the ratio is called consequent. In the above example, 1 is the antecedent and 3 is the consequent.
So, what are equivalent ratios? Let’s see an example.
Example: The ratio of the number of oranges to the number of apples in the fruit basket is 2:3 or $\frac{2}{3}$. Can you guess how many apples and oranges are there in the basket?
Here’s the interesting part! Note that this does not mean that there are 2 oranges and 3 apples in the basket. It means that the number of oranges will be some multiple of 2 and the number of apples will be the multiple of 3 with respect to the same number.
What are the possibilities then?
And so on, of course.
Here, the ratios 2:3, 4:6, 6:9 are equivalent ratios since they ultimately get reduced to the same value, 2:3.
Equivalent Ratios: Definition
Two ratios are said to be equivalent if they represent the same value when reduced to the simplest form.
Examples of equivalent ratios:
• 1:2 , 2:4, 3:6
• 3:7, 6:14, 9:21
• 4:3, 8:6, 20:15
What Is the Standard Form of Ratio?
The standard form of the ratio can be given as a:b, where a is the antecedent and b is the consequent.
When a ratio is expressed in the form of a fraction it can be written as a/b or $\frac{a}{b}$. Here, a is the numerator and b is the denominator.
How to Identify Equivalent Ratios
We can use two methods. The first method is the cross multiplication method and the second method is the Highest Common Factor (HCF) method. Let us understand both of these methods with the help of examples.
Cross Multiplication Method
This method is convenient to use when the numbers involved are small.
Check whether 12:18 and 10:15 are equivalent ratios or not using the cross multiplication method.
Step 1: Write the given ratios in the fractional form that is numerator by denominator form.
$12:18 = \frac{12}{18}$ and $10:15 = \frac{10}{15}$
Step 2: Cross multiply.
$12 \times 15 = 180$
$18 \times 10 = 180$
Step 3: If both products turn out to be equal, it would mean that the given ratios are equivalent ratios.
Here, $12 \times 15 = 18 \times 10 = 180$.
Therefore, the given ratios (12:18 and 10:15) are equivalent ratios.
HCF Method
Let’s use the same example.
Step 1: We will find the HCF of the antecedent and consequent of both the given ratios.
HCF $(12,18) = 6$
HCF $(10,15) = 5$
Step 2: Next, divide both the antecedent and consequent terms of both ratios by their respective HCF. So, we will get
$(12 \div 6) :(18 \div 6) = 2:3$
$(10 \div 5) :(15 \div 5)=2:3$
Step 3: If the reduced forms of both the given ratios are equal, it means that the given ratios are equivalent.
Here, $12:18 = 10:15$.
How to Find Equivalent Ratios?
If one of the ratios can be expressed as a multiple of the other given ratio, then they are said to be equivalent ratios. Thus, creating equivalent ratios is simple.
As is the case for equivalent fractions, we can easily find an equivalent ratio by multiplying the given ratio (both antecedent and consequent) with the same natural number.
Example: Find equivalent ratios of 1:4.
Equivalent ratios of 1:4 are
2:8 … multiply 1:4 by 2
3:12 … multiply 1:4 by 3
4:16 … multiply 1:4 by 4
and so on.
Table of Equivalent: Ratios
As we discussed earlier, we can easily find an equivalent ratio by multiplying the given ratio (both antecedent and consequent) with the same number. This number could be any natural number. We can find an infinite number of equivalent ratios for a given ratio.
These equivalent ratios for a given ratio, when combined together and presented in a tabular format, gives us the required “Equivalent Ratios Table.”
Let us make our own Equivalent Ratio Table when the given ratio is 5:3. All we have to do is to think of any natural number and then multiply both the terms of the given ratio with that number to obtain a unique equivalent ratio.
$5:3 = (5 \times 2) :(3 \times 2) = 10:6$
$5:3 = (5 \times 3) : (3 \times 3) = 15:9$
$5:3 = (5 \times 4) : (3 \times 4) = 20:12$
$5:3 = (5 \times 5) : (3 \times 5) = 25:15$
Equivalent Ratio Table for the ratio 5:3 can thus be represented as,
Visual Representation of Equivalent Ratios
When we represent the equivalent ratios visually, the shaded area (and thus the unshaded area) is the same for each ratio.
Example: 1: 3
2 : 6
4 : 12
Fun Facts!
• The quantities that are to be compared using ratios should be of the same kind.
• The quantities that are to be compared using ratios should have the same unit.
• Not only can we multiply the terms of the ratio to get an equivalent ratio, but we can also divide both the terms with the same natural number.
Conclusion
In this article, we learned about equivalent ratios definition and meaning in mathematics, what ratios are, what the standard form of ratio is, and how to find equivalent ratios. We also understood how to make the equivalent ratios table for a given ratio.
Solved Examples On Equivalent Ratios
1. Find one equivalent ratio of 3:22.
Solution: We will first write the given ratio in the form of a fraction.
$3 :22 \Rightarrow \frac{3}{22}$
Now we will multiply both the numerator and denominator by 2
$\frac{3 \times 2}{22 \times 2} = \frac{6}{44} = 6 : 44$
So, 8 :44 is an equivalent ratio of 3 :22.
2. Find any two equivalent ratios of 14 :21.
Solution:
We will first write the given ratio in the form of a fraction.
$14:21 \Rightarrow \frac{14}{21}$
Now we will multiply both the numerator and denominator by 3, to get the first equivalent fraction.
$= \frac{14}{21}$
$= \frac{14 \times 3}{21 \times 3}$
$= \frac{42}{63}= \frac{14}{21}$
Again, multiply and divide 1421 by another natural number, such as 5, as given below:
$= \frac{14}{21}$
$= \frac{14 \times 5}{21 \times 5}$
$= \frac{70}{105} = \frac{14}{21}$
Hence, the two equivalent ratios of 14 :21 are 42 :63 and 70 :105.
3. Find any four equivalent ratios of 2:9. Present them with the help of the Equivalent Ratios Table.
Solution:
$2:9 = (2 \times 2) : (9 \times 2) = 4:18$
$2:9 = (2 \times 3) : (9 \times 3) = 6:27$
$2:9 = (2 \times 4) : (9 \times 4) = 8:36$
$2:9 = (2 \times 5) : (9 \times 5) = 10:45$
Equivalent Ratio Table for ratio 2:9 can thus be represented as:
4. Are the ratios 18:10 and 63:35 equivalent?
Solution:
Let’s use the HCF method.
HCF $(18,10) = 2$
HCF $(63:35) = 7$
$\frac{18 \div 2}{10 \div 2} = \frac{9}{5}$ and $\frac{63 \div 7}{35 \div 7} = \frac{9}{5}$
Both ratios in their reduced form are equal.
Thus, the ratios 18:10 and 63:35 are equivalent.
5. What will be the value of x if 2:5 is equivalent to 12:x?
Solution:
It is given that $2 :5 = 12:x$.
In fraction form, we write $\frac{2}{5} = \frac{12}{x}$
By cross multiplying, we get
$2x = 12 \times 5$
$2x = 60$
$x = 30$
Practice Problems On Equivalent Ratios
1
The ratios 5:3 and 30:x are equivalent. Find x.
6
9
18
28
CorrectIncorrect
We can write the given ratios as
$\frac{5}{3} = \frac{30}{x}$ .
By cross multiplying, we get
$5x = 90$
Thus, $x = 18$
2
Equivalent ratios have the same ______________.
Numerator
Denominator
Antecedent and consequent
Reduced form
CorrectIncorrect
Equivalent ratios may have different numerator (antecedent) and denominator (consequent) but they have the same reduced form.
3
Which of the following ratios are not equivalent with 5:10?
14:28
2:6
6:12
10:20
CorrectIncorrect
$\frac{14}{28} = \frac{6}{12} = \frac{10}{20} = \frac{1}{2}$
$\frac{2}{6} = \frac{1}{3} (\text{HCF} = 2)$
4
Find the odd one out.
★:☆☆☆
⬚⬚ :◾◾◾◾◾◾
♣♣♣ : ♧♧♧♧♧♧♧♧♧
♠♠♠♠ : ♤♤♤♤♤♤♤
CorrectIncorrect
Correct answer is: ♠♠♠♠ : ♤♤♤♤♤♤♤
The first three options represent equivalent ratios 1:3, 2:6, and 3: 9.
Frequently Asked Questions On Equivalent Ratios
Fraction means a part of the whole.
Proportion is defined as the equality between two ratios.
The full form of HCF is the Highest Common Factor. It is the greatest factor that divides the given two or more numbers. For example, 4 is the HCF of 4 and 16.
The full form of LCM is the least common multiple. For example, LCM of 16 and 24 will be $2 \times 2 \times 2 \times 2 \times 3 = 48$, where 48 is the smallest common multiple for numbers 16 and 24.
Ratio is a number, so it has no unit. |
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