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# Math for Game Development and WebGL Part 5: Rotating a Vector
Now that we know how to find the x and y position of an angle, can we just take an angle and draw our points there? Not quite. Any given point already has an angle, as you can see here:
The point (4, 3) already has the angle shown as α. Now what we could do is find the angle of this point, add our new angle to that angle, find the position of that new angle in the unit circle, then scale the vector back out by the magnitude of the original vector. And this actually does work. You can find the angle of a point by solving for the angle in a sine/cosine/tangent expression. If we remember soh-cah-toa, tangent of an angle is opposite over adjacent, so:
tangent of an angle = 3 / 4
tangent of an angle = 0.75
If we could undo the tangent on both sides we could solve for the angle. This can be done with inverse operations arcsine, arccosine and arctangent. So we the angle is:
arctangent of tangent of angle = arctangent of 0.75
angle = arctangent of 0.75
angle ≈ 0.64
Then if we want to rotate by one radian, our new angle would be 1.64. Awesome, we already know how to rotate by an angle, and now we have our new angle! We can then use sine and cosine to find the position of that angle in the unit circle. Our x and y are then only in the unit circle, between -1 and 1, so we need to scale it back out by the length of the vector, which we can do by multiplying by our magnitude.
Hopefully this makes sense after what we’ve learned previously about sine and cosine. Multiplying cosine by the hypotenuse gives you the y position, and multiplying sine by the hypotenuse gives you the x position.
You can actually see an example of this working:
See the Pen
Vector2 with Inefficient Rotation
by Rob Louie (@rlouie)
on CodePen.
You don’t need to deeply understand arctangent, just know that we use it to get the angle. We then add the angles to get the new angle of the vector. We then get the sine and cosine of that angle and multiply them by the magnitude of the vector to get the new x and y positions. In order to understand the next part it is very important to understand that.
This is the intuitive way to do this, but it is not the way rotation is done in computer graphics. It doesn’t fit the pattern of matrix math, meaning it’s going to be less efficient and more difficult to deal with overall (we’ll learn matrix math soon). Instead rotation makes use of two far less intuitive formulas called the angular addition identity for sine and the angular addition identity for cosine. Those formulas are as follows:
cos(originalAngle + angleToRotate) = cos(originalAngle) * cos(angleToRotate) - sin(originalAngle) * sin(angleToRotate)
sin(originalAngle + angleToRotate) = sin(originalAngle) * cos(angleToRotate) + cos(originalAngle) * sin(angleToRotate)
There’s a lot going on there, but they can be simplified since we know the x and y of the original angle. Much like in our more intuitive example, the sine and cosine of our added angles will just get us the positions in the unit circle. However, we want the actual new vector position, so we need to multiply by the length of the vector, just like our previous example. Lets look at that:
vector.magnitude() * cos(originalAngle + angleToRotate) = vector.magnitude() * cos(originalAngle) * cos(angleToRotate) - vector.magnitude() * sin(originalAngle) * sin(angleToRotate)
vector.magnitude() * sin(originalAngle + angleToRotate) = vector.magnitude() * sin(originalAngle) * cos(angleToRotate) + vector.magnitude() * cos(originalAngle) * sin(angleToRotate)
I’ve put the changes in green. All I’ve done is taken us outside the unit circle by multiplying the the magnitude of the vector, just like we did in our more intuitive example.
Lets think back to what we’ve learned and what we did in the more intuitive example. Once we found both angles and added them together, we got the cosine of that angle and then multiplied by the magnitude. That got us the new x position. We did the same thing with sine to get the y position. That means that the cosine of the summed angles times the magnitude is the new x position. And that the sine of the summed angles times the magnitude is the new y position. Let’s substitute that in now.
newXPosition = vector.magnitude() * cos(originalAngle) * cos(angleToRotate) - vector.magnitude() * sin(originalAngle) * sin(angleToRotate)
newYPosition = vector.magnitude() * sin(originalAngle) * cos(angleToRotate) + vector.magnitude() * cos(originalAngle) * sin(angleToRotate)
Using the same logic that the cosine of an angle of a vector times a vector’s magnitude is the vector’s x position, and the sine of the angle of a vector times a vector’s magnitude is it’s y position, we can also substitute those in our equation:
newXPosition = vector.x * cos(angleToRotate) - vector.y * sin(angleToRotate)
newYPosition = vector.y * cos(angleToRotate) + vector.x * sin(angleToRotate)
All I did was replace any instance of the magnitude times the cosine of the original angle with x, and any instance of magnitude times the sine of the original angle with y. This is now our formula for rotation, but why does it work?
## Why it works
To show how it works, lets deal with just two angles. Note that for simplicity, the line from A to B (aka line AB) has a length of one:
As you recall from the post about the unit circle and sine/cosine, this just makes the sine of x the opposite side, since the hypotenuse is one.
I want to prove that adding them together satisfies the angular addition identity for sine. Since we just want to prove this, lets be very explicit and make these right triangles:
So lets recall our angular addition identity for sine:
sin(originalAngle + angleToRotate) = sin(originalAngle) * cos(angleToRotate) + cos(originalAngle) * sin(angleToRotate)
Lets go ahead and substitute in our angle names here:
sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
So first lets draw a line for the opposite side of the triangle formed from both angles combined together:
The sine of x and y added together is the length of our new line EF. Lets go ahead and write that out:
sin(x + y) = length of BG
To clarify, the points A, B and G form a single large triangle whose angle is x + y. Since the hypotenuse of that large triangle is still one, the sine of x + y is the length of line BG.
Now I’m going to split the line EF with a new line coming horizontally from point C:
I’ve drawn the line CH over so that it splits BG into two different lines, BH and HG. If we add those together, we still get the original line EF, so lets write that out:
sin(x + y) = length of BH + length of HG
Now notice that line HG and line CF are the same length. G, H, C, and F form a rectangle with the X axis, so both sides are the same.
So we could rewrite our equation now as:
sin(x + y) = length of BH + length of CF
Now we have to figure out some things about our bottom triangle. Notice that the hypotenuse of our bottom triangle is the adjacent side of our top triangle. Can we figure out the adjacent side?
Well, cosine of x is adjacent over hypotenuse, and hypotenuse is one, so cosine of x is the adjacent side. Or, flipped around, the adjacent side is the cosine of x!
Now we want to find the length of line CF. We know the length of line CF is the sine of y. Sine is opposite over hypotenuse, and the hypotenuse is cos(x) so:
sin(y) = length of CF / cos(x)
We can solve for CF by multiplying both sides by cos(x) and get:
length of CF = sin(y) * cos(x)
Lets substitute that into our original equation where length of CF was:
sin(x + y) = length of BH + sin(y) * cos(x)
Alright we’re halfway there to building back the original equation. How do we now get BH in the same format as our original equation. To do that we need to learn some things about the triangle formed by B, H, and C.
We can actually also learn some things about the angles inside this triangle due to how angles of transversals work in parallel lines:
This may or may not be something you already know, but when a line intersects two parallel lines, the opposite angles in each line are the same, and those angles are the same in both parallel lines, as shown in the image above.
With that in mind, we can actually figure out some angles as they relate to y. Since line CH is parallel to the x-axis and line AC intersects both, we can use the same logic to determine that this angle is also y:
We know this because it’s the opposite side of angle x on our top triangle there. We also know that the angle formed by lines BC and CA form a right angle, so that corner of our BHC triangle is 90° – y (yes I’m using degrees instead of radians here for simplicity’s sake, there’s enough other things to worry about here).
Now, the angle formed by BHC is a 90° angle, because we’re in a right triangle. We also know that the angles of a triangle add up to 180°. So can we figure out the angle of HBC? That angle must also be y:
If that isn’t intuitive, do the math:
90 + 90 - y + y = 180
The y and -y cancel out and we are left with 90 + 90, which does in fact equal 180.
We just need one more thing, the hypotenuse of this triangle. Luckily it’s also the opposite side of the triangle ABC, meaning it’s sin(x)
As a reminder, here’s where we left our formula:
sin(x + y) = length of BH + sin(y) * cos(x)
We need to find the length of BH. Well BH is adjacent to angle y. Cosine of angle y will give us adjacent over hypotenuse, so lets solve for adjacent:
cos(y) = length of BH / sin(x)
We can multiply both sides now by sin(x) to solve for BH:
length of BH = cos(y) * sin(x)
Okay, lets substitute that into our original equation:
sin(x + y) = cos(y) * sin(x) + sin(y) * cos(x)
We are now back to our original equation. Due to how we solved it, the order is a little different, but remember that multiplication can be done in any order, and so can addition (as in, 2 * 3 is the same as 3 *2, and 3 + 2 is the same as 2 + 3). So we can rearrange to get back to our original formula if we want:
sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
## If you made it through that congrats! Now lets actually just implement rotation
After that long explanation, lets look at our actual vector rotation formula again:
newXPosition = vector.x * cos(angleToRotate) - vector.y * sin(angleToRotate)
newYPosition = vector.y * cos(angleToRotate) + vector.x * sin(angleToRotate)
Remember this is the formula that accounts for the fact that we already know the x and y positions of the current vector. Now lets just implement that in our code:
See the Pen
Vector2 with Rotation
by Rob Louie (@rlouie)
on CodePen.
Alright, we’ve done the hardest math we’ll have to do in a while. On to matrix math next (it’s much easier than this).
## Summary
• Existing vectors already have an angle
• To rotate a vector you have to add to it’s original angle
• While not intuitive, computer graphics programs use the angular addition identity for sine and the angular addition identity for cosine to rotate vectors
• Understanding why the formulas work takes some pretty serious geometry, but the end result is fairly simple
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Lesson Video: Expected Values of Discrete Random Variables | Nagwa Lesson Video: Expected Values of Discrete Random Variables | Nagwa
# Lesson Video: Expected Values of Discrete Random Variables Mathematics
In this video, we will learn how to calculate the expected value of a discrete random variable from a table, a graph, and a word problem.
16:00
### Video Transcript
In this video, weโre going to learn how to calculate the expected value or the mean of discrete random variables from both their table and a graph. Letโs begin though by recalling what we mean by a discrete random variable of a probability distribution. A probability distribution describes the likelihood of obtaining the possible values that a random variable can assume. This can be given as a function, a table of values, or even in graph form. Then a discrete variable is a variable which can only take a countable number of values. In this example, ๐ฅ is a discrete variable as it only takes the values one, two, three, four, five, and six.
In this video, weโre interested in finding a formula that can help us to find the expected value, denoted as ๐ธ of ๐ฅ, in other words to find the mean of a discrete random variable. To help us develop this formula, weโre going to look at an example.
An experiment produces the discrete random variable ๐ that has the probability distribution shown. If a very high number of trials were carried out, what would be the likely mean of all the outcomes?
Letโs imagine the experiment is spinning a spinner with the numbers two, three, four, and five on it. The table tells us the probability of achieving each score on a single spin. And so we see itโs much more likely, for instance, that the spinner lands on the five than it does on the two. Letโs add another table, showing the number of times we spin the spinner versus the number of times we expect it to land on each number. Letโs say we were to spin the spinner 10 times; 0.1 of those times we would expect the spinner to land on two. Well, 0.1 of 10 โ in other words 0.1 times 10 โ is one. Then 0.3 of the times, we would expect the spinner to land on three. 0.3 of 10 or 0.3 times 10 is three. 0.2 of the times, we expect the spinner to land on four, so thatโs twice. And 0.4 of the times, thatโs four times, we would expect the spinner to land on five.
Next, letโs think about what would happen if we were to spin it 20 times. 0.1 of those times, we would expect it to land on two, so thatโs twice. Six times we expect the spinner to land on three, thatโs 0.3 times 20. Four times, which is 0.2 times 20, weโd expect it to land on four. And 0.4 times 20 which is eight would be the number of times weโd expect it to land on five. But letโs imagine there are a very high number of trials, say 1000 trials. 0.1 times 1000 is 100. So weโd roughly expect the spinner to land on two 100 times. Weโd expect it to land on three 300 times, weโd expect it to land on four 200 times, and weโd expect it to land on five 400 times.
Now, this is really useful as we can use these values to calculate the mean by using the rules for finding the mean from a frequency table. The formula we use to calculate the mean from a frequency table is the sum of ๐ times ๐ฅ divided by the sum of ๐ฅ. The sum of ๐ฅ is 1000. Weโve carried out 1000 trials. ๐ times ๐ฅ will be two times 100, three times 300, four times 200, and five times 400. And so the sum of ๐๐ฅ is the sum of all these products. And so weโre able to calculate the mean, which we call the expected value ๐ธ of ๐ฅ, as shown.
Now we will calculate this value in a moment, but weโre looking to find a rule for finding the expected value. So weโre going to split the fraction up a little bit. By reversing the process we perform when adding fractions, we can write it as two times 100 over 1000 plus three times 300 over 1000 and so on. And then we notice something. 100 divided by 1000 is 0.1. 300 divided by 1000 is 0.3. 200 divided by 1000 is 0.2. And 400 divided by 1000 is 0.4. And so another way to write our calculation for the mean is as two times 0.1 plus three times 0.3 plus four times 0.2 plus five times 0.4, and thatโs equal to 3.9. And so, if a very high number of trials were carried out, the likely mean of all of our outcomes would actually be 3.9.
And by this stage, you might be spotting a pattern. Two times 0.1 is the product of ๐ฅ and its corresponding ๐ of ๐ฅ. Three times 0.3 is also the product of ๐ฅ and its corresponding ๐ of ๐ฅ. And so we notice that this is simply the sum of the product of the numbers in each column.
And so weโre able to generalize.
The expected value, sometimes called the mean of ๐ฅ, is denoted ๐ธ of ๐ฅ or ๐ or ๐ sub ๐ฅ. It can be found by calculating the sum of the product of the variable ๐ฅ and the probability of that variable occurring, ๐ of ๐ฅ equals ๐ฅ. We write this as shown. ๐ธ of ๐ฅ is the sum of ๐ฅ times ๐ of ๐ฅ equals ๐ฅ. Now that we have a formula, letโs see how we can apply this to finding the expected value of a discrete random variable given a graph.
Work out the expected value of the random variable ๐ whose probability distribution is shown.
The expected value denoted ๐ธ of ๐ฅ can be found by calculating the sum of the product of the variable ๐ and the probability of that variable occurring. Thatโs represented as shown. And so a nice way to work out the expected value when given a graph is actually to construct a table. We see by looking at the ๐ฅ-axis that the random variable ๐ can take the values one, two, three, four, and five. We also see that every single one of the bars in our diagram has a height of 0.2. So the associated probability for each variable is in fact 0.2. Now, a quick way that we can check whether what weโve done is likely to be correct is to check that the sum of the probabilities is indeed one. And 0.2 plus 0.2 plus 0.2 plus 0.2 plus 0.2 is one, and so we can move on.
To find the expected value then, we find the sum of the products of the numbers in each column. So thatโs one times 0.2 plus two times 0.2 plus three times 0.2 plus four times 0.2. And finally, we add five times 0.2. Evaluating each of our products, and we get 0.2 plus 0.4 plus 0.6 plus 0.8 plus one, which is equal to three. And so the expected value ๐ธ of ๐ฅ is equal to three. Now, in fact, this makes a lot of sense. We saw that the probability of each variable occurring was equal; it was 0.2 every time. And so the expected value and the likely mean would actually be the mean of all of our numbers. Thatโs five plus four plus three plus two plus one divided by five, which is also equal to three.
Now, this only worked because the probabilities were equal. It wouldnโt be a general rule that we could follow. Letโs look at an example where we have a graph and the probabilities are not equal.
Work out the expected value of the random variable ๐ whose probability distribution is shown.
The formula we use to calculate the expected value of a discrete random variable ๐ is shown. Itโs the sum of the product of ๐ and the probability that ๐ฅ occurs. And so a nice way to calculate the expected value when given a probability distribution in graph form is to transfer that into a table. The ๐ฅ-axis on our graph tells us the values our discrete random variable can take. They are one, two, three, and four. The first bar then has a height of 0.1. So the probability that ๐ is equal to one is 0.1. We see that our second bar has a height of 0.3. So the probability that ๐ is equal to two is 0.3. And then we continue in this manner. The height of our third bar, and thatโs the probability that ๐ is equal to three, is 0.4. And the height of our fourth bar, which tells us the probability that ๐ is equal to four, is 0.2.
To find the expected value from our graph then, we need to find the sum of the products of the numbers in each column. So thatโs one times 0.1 plus two times 0.3 plus three times 0.4 plus four times 0.2. This becomes 0.1 plus 0.6 plus 1.2 plus 0.8, which is equal to 2.7. And so the expected value of the random variable ๐ is 2.7. Now, we can always check whether our answer is likely to be correct, or at least in the right range. When weโre finding the expected value, weโre finding the weighted mean. According to our table and our graph, itโs much more likely that ๐ฅ is equal to three than it is to one. The mean is more likely to be weighted then in this direction. And since 2.7 is roughly between one and four, though not exactly, we know that weโre likely to have performed the calculations correctly.
In our next example, weโll look at how we can apply some of the other rules for working with probabilities to find the expected value of a discrete random variable.
The function in the given table is the probability function of a discrete random variable ๐. Find the expected value of ๐.
We know that we can find the expected value of a discrete random variable by calculating the sum of the products of the variable ๐ and the probability of that variable occurring. And we write it using this ๐ด notation as shown. Now in this case, ๐ is a probability function. So we can say that this is like saying the probability that ๐ฅ is equal to ๐ฅ sub ๐. And so to find the expected value, weโre going to begin by finding the products of the numbers in each column. But of course, there is a number missing. And thatโs this value here.
Weโre told that the probability that ๐ฅ is equal to one is ๐. So how do we calculate ๐? Well, we know that the sum of the probabilities in our table must be equal to one, and so we can set up and solve an equation for ๐. Our equation is 0.1 plus ๐ plus 0.1 plus 0.4 plus 0.2 equals one. In other words, weโve added the respective probabilities and set it equal to one. 0.1 plus 0.1 plus 0.4 plus 0.2 is 0.8. So our equation becomes ๐ plus 0.8 equals one. If we subtract 0.8 from both sides, we find ๐ is equal to 0.2. And so weโre ready to calculate the expected value of ๐. Itโs zero times 0.1 plus one times 0.2 plus two times 0.1. And we repeat this process with the numbers in our final two columns. This gives us zero plus 0.2 plus 0.2 plus 1.2 plus 0.8, which is equal to 2.4. The expected value of ๐ then in this case is 2.4.
In our final example, weโre going to look at how to use the expected value formula to find missing values.
The function in the given table is a probability function of a discrete random variable ๐ฅ. Given that the expected value of ๐ฅ is 254 over 57, find the value of ๐ต.
And then we have a table with values for ๐ฅ sub ๐ and ๐ of ๐ฅ sub ๐. We begin by recalling how we calculate the expected value of a discrete random variable. Itโs the sum of the values that the variable can take multiplied by the probability of that variable occurring. Now, in this question, weโre told that the function is a probability function of the discrete random variable. So weโre essentially saying that ๐ of ๐ฅ sub ๐ is the same as the probability that ๐ฅ is equal to ๐ฅ sub ๐. And so eventually, weโre going to multiply the values in each of our columns.
But we do have a bit of a problem. At the moment, our values for probability are in terms of a variable ๐. And so weโre going to use the fact that we know that the sum of these probabilities must be equal to one. In other words, eight ๐ plus three ๐ plus a third plus eight ๐ must be equal to one. Eight ๐ plus three ๐ plus eight ๐ is 19๐. So we get 19๐ plus a third equals one. We can solve for ๐ by first subtracting a third from both sides to get 19๐ equals two-thirds. And then when we divide both sides by 19, we get ๐ is equal to two over 57. So now that we know the value of ๐, we need to go back to our table and calculate the relevant probabilities.
The probability that ๐ฅ is equal to one is eight ๐. So thatโs eight times two over 57, which is 16 over 57. Then the probability that ๐ฅ is equal to two is three ๐. So thatโs three times two over 57, which is six over 57. Weโre told the probability ๐ฅ is equal to ๐ต is a third. And finally, the probability that ๐ฅ is equal to seven is eight ๐ again. So thatโs 16 over 57. Now of course, a quick check we could do here would be to check that each of our probabilities does indeed sum to one, and it does, so we can move on.
Our next job is to find the sum of the products of the numbers in each column. The expected value then is given by one times 16 over 57 plus two times six over 57 plus ๐ต times a third plus seven times 16 over 57. But actually, we were told the expected value is 254 over 57. So weโre going to replace ๐ธ of ๐ฅ with this number. And then we simplify the right-hand side. Now, all these fractions are making life a little bit awkward, so weโre going to multiply every single number in our equation by 57. When we do, our equation becomes 254 equals 16 plus 12 plus 19๐ต plus 112. 19 is essentially 57 divided by three. And then, by adding the numerical parts, we get 140 plus 19๐ต on the right-hand side.
Next, we subtract 140 from both sides, and our equation becomes 114 equals 19๐ต. Finally, we divide through by 19. And we find ๐ต is 114 divided by 19, but 114 divided by 19 is six. And so the value of ๐ต is six.
Letโs now recap the key points from this lesson. In this video, we recapped what we meant by a discrete random variable. Itโs a variable that can only take a countable number of values. We saw that we denote the expected value, or sometimes called the mean of ๐ฅ, as ๐ธ of ๐ฅ or ๐ or ๐ sub ๐ฅ. And then the expected value is found by calculating the sum of the products of the variable ๐ฅ and the probability of that variable occurring. And we use the ๐ด notation to represent this as shown.
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# How Many Solutions Can A Linear Equation Have?
## What is a solution to a linear system?
A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied.
A solution to the system above is given by.
since it makes all three equations valid..
## How many solutions can a linear equation in two variables have?
Further, a linear equation in two variables has infinitely many solutions. The graph of every linear equation in two variables is a straight line and every point on the graph (straight line) represents a solution of the linear equation.
## Can a linear system have two solutions?
Most linear systems you will encounter will have exactly one solution. However, it is possible that there are no solutions, or infinitely many. (It is not possible that there are exactly two solutions.) The word unique in this context means there is a solution, and it’s the only one.
## How do you tell if an equation has no solution?
The constants are the numbers alone with no variables. If the coefficients are the same on both sides then the sides will not equal, therefore no solutions will occur. Use distributive property on the right side first.
## How can you tell that a system of linear equations will have infinitely many solutions without computation?
For any two points, exactly one (distinct) line can go through both points. It follows that if two lines intersect in at least two points, then they must be the same line! Therefore, if a system of two linear equations has at least two solutions, then it must have infinitely many solutions.
## Is it possible that there are no solutions to two linear equations in three unknowns?
For systems of equations in three variables, there are an infinite number of solutions on a line or plane that is the intersection of three planes in space. … A system of equations in three variables with no solutions is represented by three planes with no point in common.
## How do you know when a linear equation has infinitely many solutions?
Summary. If the equation ends with a false statement (ex: 0=3) then you know that there’s no solution. If the equation ends with a true statement (ex: 2=2) then you know that there’s infinitely many solutions or all real numbers.
## Which equation has only one solution?
When a does not equal c, we use inverse operations to solve the linear equation: We see that this is the only value of x that satisfies the equation, so the equation has only one solution. When a = c and b does not equal d, we can rewrite the equation ax + b = cx + d as ax + b = ax + d.
## What is the equation with infinitely many solutions?
The equation 2x + 3 = x + x + 3 is an example of an equation that has an infinite number of solutions. Let’s see what happens when we solve it. We first combine our like terms.
## What is symbol for no solution?
symbol ØSometimes we use the symbol Ø to represent no solutions. That symbol means “empty set” which means that the set of all answers is empty. In other words, there is no answer.
## How do you identify a linear equation?
An equation is linear if its graph forms a straight line. This will happen when the highest power of x is “1”. Graphically, if the equation gives you a straight line thenit is a linear equation. Else if it gives you a circle, or parabola or any other conic for that matter it is a quadratic or nonlinear equation.
## Is 0 0 infinite or no solution?
For an answer to have an infinite solution, the two equations when you solve will equal 0=0 . … If you solve this your answer would be 0=0 this means the problem has an infinite number of solutions. For an answer to have no solution both answers would not equal each other.
## What is the degree of linear equation?
In Algebra, the degree is the largest exponent of the variable in the given equation. … If the equation has a degree of one, it is a linear equation. In short, the degree of linear equations is always one. For example, 3x + 10 = z, has a degree 1 so it is a linear equation.
## What is the condition of no solution?
If the lines are parallel, there is no solution for the pair of linear equations. If there is no solution of the given pair of linear equations, the equations are called inconsistent.
## Do linear equations always have one solution?
A system of linear equations usually has a single solution, but sometimes it can have no solution (parallel lines) or infinite solutions (same line). This article reviews all three cases. One solution. A system of linear equations has one solution when the graphs intersect at a point.
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]> Simple Differential Equations
### Chapter 2: Integration; Section 1: Indefinite Integrals; page 4
Previous topic: page 3 Special Methods
Next topic: section 2 Definite Integrals, page 1 Fundamentals
# Simple Differential Equations
## Introduction
A differential equation is a mathematical representation of a relation between an independent variable, a function and some of its derivatives with respect to the independent variable. The most general implicit form of a differential equation is:
$F\left(x,y,\frac{\text{d}y}{\text{d}x},\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}},\frac{{\text{d}}^{3}y}{\text{d}{x}^{3}},....\right)=0$ (2.1.4.1)
The highest order of a derivative presented in (2.1.4.1) defines the order of the differential equation. By definition the lowest possible order of a differential equation is one (order zero would be just a functional relation). On the other hand the presence of the independent variable and/or the function, is not a necessity of a differential equation.
A solution of a differential equation consists of obtaining the functional relation without the participation of any derivative. For example the solution of a differential equation of the first order consists of two stages:
1. Express the equation in a way that instead of the derivative, an indefinite integral will appear.
2. Perform the integration
It is clear that a differential equation poses, in addition to the integration, a new challenge. The solution of a differential equation of higher order is even harder.
Let's return to the differential equation of the first order. The second stage (integration) brings an undefined constant, additive to the integral. As a rule the solution of a differential equation of order n yields n undefined constants.
What is the use of differential equations? Their importance in science and engineering is unlimited. The most general laws in nature can be expressed as differential equations and their solution explains the behaviour of the particular measurable quantities. As an example let's take the familiar second law of Newton in one dimension:
$\frac{{\text{d}}^{2}s}{\text{d}{t}^{2}}=\frac{F}{m}$ (2.1.4.2)
where
• t (time) is the independent variable,
• s (position of a physical point at an axis) is the function,
• m (mass) is a constant related to this physical point and
• F is the one dimensional force acting on the physical point. Depending on the circumstances, the force could depend on t , s , a derivative of s with respect to t of any order, or any combination of these.
In the case when F is a funtion of s, t, and the first and second derivatives, (2.1.4.2) represents a differential equation of the second order. In such a case its solution should give the position s and the first derivative v (velocity) as functions of the time. The reasoning is the following: in principle the first integration should yield a differential equation of the first order including one constant of integration. The second integration should represent a functional relation between s and t with two constants of integration, representing s as function of t . As a consequence the first integration represents the first dervative v as a function of t .
We have the choice to define the two constants resulting from the integrations. It is common to use the initial conditions, corresponding to t = 0 , of the position and of the velocity:
${s}_{0}=s\left(t=0\right)\text{ }\text{and}\text{ }{v}_{0}=v\left(t=0\right)$ (2.1.4.3)
as given values in order to define the constants.
We will study here only a few elementary cases of differential equations and the way to solve them.
## Derivative as function of the independent variable: $\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}={f}_{n}\left(x\right)$
The differential equation
$\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}={f}_{n}\left(x\right)$ (2.1.4.4)
where ${f}_{n}\left(x\right)$ is a known function can be directly integrated, in order to obtain a differential equation of the order n−1 of the same form. That way one can reduce the order by steps and obtain the solution:
$\begin{array}{l}\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}={f}_{n}\left(x\right)\\ \frac{{\text{d}}^{n-1}y}{\text{d}{x}^{n-1}}=\int {f}_{n}\left(x\right)\text{}\text{d}x={f}_{n-1}\left(x\right)\\ ...........\\ \frac{\text{d}y}{\text{d}x}=\int {f}_{2}\left(x\right)\text{}\text{d}x={f}_{1}\left(x\right)\\ y=\int {f}_{1}\left(x\right)\text{}\text{d}x\end{array}\right\}$ (2.1.4.5)
We'll use the simple example of throwing up vertically (positive direction) a mass-point m with an initial velocity of vo under the gravitational force F = −mg from a starting height of so =0 . According to (2.1.4.2) and (2.1.4.5) we obtain
$\begin{array}{l}a=\frac{{\text{d}}^{2}s}{\text{d}{t}^{2}}=-g\\ v=\frac{\text{d}s}{\text{d}t}=-\int g\text{}\text{d}t=-gt+{C}_{1}\\ s=\int \left(-gt+{C}_{1}\right)\text{}\text{d}t=-\frac{g{t}^{2}}{2}+{C}_{1}t+{C}_{2}\end{array}\right\}$ (2.1.4.6)
where a denotes acceleration. After applying the initial conditions (2.1.4.3) one obtains:
$\begin{array}{l}v=-gt+{v}_{o}\\ s=-\frac{g{t}^{2}}{2}+{v}_{o}t\end{array}\right\}$ (2.1.4.7)
We already know how to present graphically these relations. But since this case deals with the motion of a physical object, we can translate the solution obtained from the differential equation to an actual motion by using animation. For sake of simplicity we'll round-off the gravitational acceleration to g = 10m/sec² , and in order to be able to observe the motion in real time, we'll use initial velocities of more than 10m/sec . This is presented in Fig. Vertical motion.
Another example of solving the second law of Newton was obtained on page 1 exercise 4:
$\begin{array}{l}ma=m\frac{{\text{d}}^{2}x}{\text{d}{t}^{2}}=C\mathrm{sin}\left(\omega t\right)\\ v=\frac{\text{d}x}{\text{d}t}=-\frac{C\mathrm{cos}\left(\omega t\right)}{m\omega }+{C}_{1}\\ x=-\frac{C\mathrm{sin}\left(\omega t\right)}{m{\omega }^{2}}+{C}_{1}t+{C}_{2}\end{array}\right\}$ (2.1.4.8)
where m, C, ω are known constants and C1C2 are two constants of the integration.
By using the initial values xo and vo, one obtains
${C}_{1}={v}_{0}+\frac{C}{m\omega }\text{ }\text{and}\text{ }{C}_{2}={x}_{0}$ (2.1.4.9)
which means that in order to obtain a simple harmonic motion in form of oscillations around a fixed point, which is a very common motion in nature, one has to choose in this case, one very particular initial velocity:
As a matter of fact, a harmonic motion in nature occurs when applying another force, which will be shown later on this page.
## Seperation of variables: $\frac{\text{d}y}{\text{d}x}={f}_{y}\left(y\right){f}_{x}\left(x\right)$
The first order differential equation
$\frac{\text{d}y}{\text{d}x}={f}_{y}\left(y\right){f}_{x}\left(x\right)$ (2.1.4.10)
is solvable by separation of the variables:
$\begin{array}{l}\frac{\text{d}y}{{f}_{y}\left(y\right)}={f}_{x}\left(x\right)\text{d}x\\ \int \frac{\text{d}y}{{f}_{y}\left(y\right)}=\int {f}_{x}\left(x\right)\text{d}x\end{array}\right\}$ (2.1.4.11)
The following simple example deals with radioactive decay. It is a special case of (2.1.4.10) since the function of the independent variable fx is a constant. The independent variable is the time t and the dependent variable is the number of radioactive atoms N that have not decayed so far. The physical assumption is that each atom has the same probability to decay in a random way. That way the decay rate is proportional to N . This is expressed by the differential equation
$\frac{\text{d}N}{\text{d}t}=-\frac{N}{\tau }$ (2.1.4.12)
where the negative sign expresses the decrease of the number of atoms. τ has dimensions of time, and is a known positive constant factor related to the probability of the decay. For smaller τ the probability is greater and the decay rate is more intense.
According to (2.1.4.11) the solution is
$\begin{array}{l}\int \frac{\text{d}N}{N}=-\int \frac{\text{d}t}{\tau }\\ \mathrm{ln}N=-\frac{t}{\tau }+C\\ \left\{\begin{array}{l}N=\mathrm{exp}\left(-\frac{t}{\tau }+C\right)=\\ =\mathrm{exp}C\text{ }\mathrm{exp}\left(-\frac{t}{\tau }\right)=\\ ={N}_{0}\text{ }\mathrm{exp}\left(-\frac{t}{\tau }\right)=N\end{array}\end{array}\right\}$ (2.1.4.13)
where the constant expC was replaced by N0 , the initial number of atoms.
## Non-explicit independent variable (x): ${F}_{1}\left(y,\frac{\text{d}y}{\text{d}x},....,\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}\right)=0$
Any differential equation of order n that does not include explicitly the independent variable x
${F}_{1}\left(y,\frac{\text{d}y}{\text{d}x},....,\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}\right)=0$ (2.1.4.14)
can be reduced to a differential equation of order n−1 .
This is achieved by using y as the independent variable, by converting the first derivative as a function of y
$\frac{\text{d}y}{\text{d}x}=p\left(y\right)$ (2.1.4.15)
and by doing the appropriate changes on the higher order derivatives:
$\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=\frac{\text{d}p}{\text{d}x}=\frac{\text{d}p}{\text{d}y}\frac{\text{d}y}{\text{d}x}=p\frac{\text{d}p}{\text{d}y}$ (2.1.4.16)
and so on. This way one obtains a new differential equation
${F}_{1}\left(y,p,\frac{\text{d}p}{\text{d}y},....,\frac{{\text{d}}^{n-1}p}{\text{d}{y}^{n-1}}\right)=0$ (2.1.4.17)
of lower order. This does not solve the original differential equation, unless the new equation is solvable.
For example we'll solve the motion in one dimension of a mass point m attached to a linear spring with t (time) as an independent variable, and s (position along a straight axis) as the function. The force applied by the spring is −ks , where k is a given positive constant and the negative sign means that the force tends to restore the position toward the origin. According to the second law of Newton, the differential equation of the motion is
$\begin{array}{l}\frac{{\text{d}}^{2}s}{\text{d}{t}^{2}}=-{\omega }^{2}s\\ {\omega }^{2}=\frac{k}{m}\end{array}\right\}$ (2.1.4.18)
and it does not include t explicitly. After applying the procedure of (2.1.4.15-17) we get
$\begin{array}{l}v=\frac{\text{d}s}{\text{d}t}=v\left(s\right)\\ \frac{{\text{d}}^{2}s}{\text{d}{t}^{2}}=v\frac{\text{d}v}{\text{d}s}\\ v\frac{\text{d}v}{\text{d}s}=-{\omega }^{2}s\end{array}\right\}$ (2.1.4.19)
which is solvable by the separation of the variables
$\begin{array}{l}\int v\text{d}v=-{\omega }^{2}\int s\text{d}s\\ {v}^{2}+{\omega }^{2}{s}^{2}={C}_{1}^{2}\end{array}\right\}$ (2.1.4.20)
The solution (2.1.4.20) is only the first integration, and the constant of integration is written as a square since obviously it is a positive number. We need a second integration in order to solve s as a function of t , which is obtained by substituting back $v=\frac{\text{d}s}{\text{d}t}$ in this solution:
$\begin{array}{l}\frac{\text{d}s}{\text{d}t}=±\sqrt{{C}_{1}^{2}-{\omega }^{2}{s}^{2}}\\ \int \frac{\text{d}s}{\sqrt{{C}_{1}^{2}-{\omega }^{2}{s}^{2}}}=±\int \text{d}t\\ \text{substitution}\text{ }u=\frac{\omega s}{{C}_{1}}\\ \int \frac{\text{d}u}{\sqrt{1-{u}^{2}}}=±\omega \int \text{d}t\\ \text{asin}\left(\frac{\omega s}{{C}_{1}}\right)=±\omega t+{C}_{2}\\ s=\frac{{C}_{1}}{\omega }\mathrm{sin}\left(±\omega t+{C}_{2}\right)\end{array}\right\}$ (2.1.4.21)
At this stage it looks like we have obtained two possible solutions. Actually after applying the initial conditions we'll obtain the same expression independently of the sign of the product ωt . The reason for this is that by adjusting C2 , we can switch from one sign to another e.g.
By the same token (adjusting C2), the choice of asin or acos is irrelevant. In the integration we did not take in account the ambiguity of the sign when we obtained asin at (2.1.4.21). This is also irrelevant because the value of C1 was not assigned yet (ω is customarily defined as positive).
We can now write the solution in its most general form
$\begin{array}{l}s=\frac{{C}_{1}}{\omega }\mathrm{sin}\left(\omega t+{C}_{2}\right)\\ v={C}_{1}\mathrm{cos}\left(\omega t+{C}_{2}\right)\end{array}\right\}$ (2.1.4.22)
which is called the simple harnonic motion. From (2.1.4.22) one can express the constants of integration by the initial conditions s0 and v0 :
$\begin{array}{l}{C}_{1}^{2}={s}_{0}^{2}{\omega }^{2}+{v}_{0}^{2}\\ \mathrm{sin}{C}_{2}=\frac{{s}_{0}\omega }{{C}_{1}}\\ \mathrm{cos}{C}_{2}=\frac{{v}_{0}}{{C}_{1}}\end{array}\right\}$ (2.1.4.23)
Assume that the sign of C1 is chosen as positive, then as we already know, both signs of sin and cos define uniquely the quadrant where the angle C2 is located, and therefore C2 can be uniquely evaluated. If the sign of C1 was chosen as negative, the angle C2 would be shifted by π in order to compensate for the sign of C1 , and the solution (2.1.4.22) would remain unaffected.
The special case of v0 = 0 (initiating the motion from rest) yields:
$\begin{array}{l}s={s}_{0}\mathrm{cos}\left(\omega t\right)\\ v=-\omega {s}_{0}\mathrm{sin}\left(\omega t\right)\end{array}\right\}$ (2.1.4.24)
and is animated in Fig. Simple harmonic motion.
## Non-explicit dependent variable (y): ${F}_{2}\left(x,\frac{\text{d}y}{\text{d}x},....,\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}\right)=0$
Any differential equation of order n that does not explicitly include the dependent variable y
${F}_{2}\left(x,\frac{\text{d}y}{\text{d}x},....,\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}\right)=0$ (2.1.4.25)
can be reduced to a differential equation of order n−1.
This is achieved by using the first derivative as the dependent variable
$\frac{\text{d}y}{\text{d}x}=p\left(x\right)$ (2.1.4.26)
yielding the differential equation
${F}_{2}\left(x,p,\frac{\text{d}p}{\text{d}x},....,\frac{{\text{d}}^{n-1}p}{\text{d}{x}^{n-1}}\right)=0$ (2.1.4.27)
For example we'll solve the straight line motion of a mass point m under the influence of a resisting (frictional) force proportional to the velocity: F = −μv , where μ is a positive constant and the negative sign shows that the force is in the opposite direction of the velocity. It is obvious that the motion cannot be started without an initial velocity v0 . We'll use s as the coordinate along the axis of the motion and t as the time.
The differential equation of the motion is
$\frac{{\text{d}}^{2}s}{\text{d}{t}^{2}}=-\frac{\mu }{m}\frac{\text{d}s}{\text{d}t}$ (2.1.4.28)
This equation does not include explicitly either the independent nor the dependent variables. We'll use the method that does not include the dependent variable x, since it is simpler. According to (2.2.4.26-27)
$\begin{array}{l}v\left(x\right)=\frac{\text{d}s}{\text{d}t}\\ \frac{\text{d}v}{\text{d}t}=-\frac{\mu }{m}v\\ \int \frac{\text{d}v}{v}=-\frac{\mu }{m}\int \text{d}t\\ \mathrm{ln}|v|=-\frac{\mu }{m}t+{C}_{1}\end{array}\right\}$ (2.1.4.29)
After calculating C1 , one obtains
$\begin{array}{l}\mathrm{ln}|v|=-\frac{\mu }{m}t+\mathrm{ln}|{v}_{0}|\\ |v|=|{v}_{0}|\mathrm{exp}\left(-\frac{\mu }{m}t\right)\end{array}\right\}$ (2.1.4.30)
but v cannot change its sign (it should pass through zero and therefore violates the equation), thus we have
$\begin{array}{l}v={v}_{0}\mathrm{exp}\left(-\frac{\mu }{m}t\right)\\ \int \text{d}s={v}_{0}\int \mathrm{exp}\left(-\frac{\mu }{m}t\right)\text{d}t\\ s=-\frac{m{v}_{0}}{\mu }\mathrm{exp}\left(-\frac{\mu }{m}t\right)+{C}_{2}\end{array}\right\}$ (2.1.4.31)
By using the initial position s(t=0) = s0 , one finally obtains the solution
$s={s}_{0}+\frac{m{v}_{0}}{\mu }\left[1-\mathrm{exp}\left(-\frac{\mu }{m}t\right)\right]$ (2.1.4.32)
which means that the point converges asymptotically to the position $s={s}_{0}+\frac{m{v}_{0}}{\mu }$ .
## Exercises
Exercise 1. For the following differential equation $\frac{{\text{d}}^{4}y}{\text{d}{x}^{4}}=\mathrm{cos}x$
1. Obtain the general solution y = y(x) !
2. Find the solution that for x=0 the following relations are satisfied: $\frac{{\text{d}}^{3}y}{\text{d}{x}^{3}}=\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=\frac{\text{d}y}{\text{d}x}=y=0$
3. Is x=0 a point of minimum, maximum or inflection in this case?
Exercise 2. Solve the differential equation $\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=ky\frac{\text{d}y}{\text{d}x}$ satisfying the following conditions for x=0 : $\left\{\begin{array}{l}\frac{\text{d}y}{\text{d}x}=a\\ y=0\end{array}$ where a and k are positive constants.
Exercise 3. The one dimensional (x axis) equation of motion of a point mass m under the influence of a resisting force is $F=-\beta {v}^{2}$ with β>0. Take the velocity in the positive direction $v=\frac{\text{d}x}{\text{d}t}>0$ and the initial conditions: $\left\{\begin{array}{l}{x}_{0}=0\\ {v}_{0}>0\end{array}$ Solve the motion in two ways:
1. By the method where the independent variable (t) is not contained explicitly.
2. By the method where the dependent variable (x) is not contained explicitly.
Exercise 4. Prove that Newton's differential equation of motion in one dimension (x) $m\frac{{\text{d}}^{2}x}{\text{d}{t}^{2}}=F\left(x\right)$ where m is mass, t is time and the force F is a function of the x coordinate only, is solvable by the methods studied on this page, assuming that all the functions are integrable and their inverses exist.
Previous topic: page 3 Special Methods
Next topic: section 2 Definite Integrals, page 1 Fundamentals
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# Lesson 16
Comparemos y ordenemos fracciones
## Warm-up: Conversación numérica: Múltiplos de 6 y 12 (10 minutes)
### Narrative
This Number Talk encourages students to think about multiples of 5, 6, and 12—numbers that students will see as denominators later in the lesson. It also prompts students to rely on doubling and on properties of operations to mentally solve multiplication problems. The reasoning elicited here will be helpful later in the lesson when students compare fractions by finding equivalent fractions with a common denominator.
To find products by doubling or by using properties of operations, students need to look for and make use of structure (MP7).
### Launch
• Display one expression.
• “Hagan una señal cuando tengan una respuesta y puedan explicar cómo la obtuvieron” // “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Encuentra mentalmente el valor de cada expresión.
• $$5 \times 6$$
• $$5 \times 12$$
• $$6 \times 12$$
• $$11 \times 12$$
### Activity Synthesis
• “¿Cómo les ayudan las primeras tres expresiones a encontrar el valor de la última?” // “How do the first three expressions help you find the value of the last one?”
## Activity 1: Juguemos a comparar fracciones (20 minutes)
### Narrative
This activity allows students to practice comparing fractions and apply the comparison strategies they learned through a game. Students use fraction cards from an earlier lesson to play a game in groups of 2, 3, or 4. To win the game is to have the greater (or greatest) fraction of the cards played as many times as possible. This is stage 5 of the Compare center.
Consider arranging students in groups of 2 for the first game or two (so that students would need to compare only 2 fractions at a time), and arranging groups of 3 or 4 for subsequent games. Before students begin playing, ask them to keep track of and record pairs of fractions that they find challenging to compare.
MLR8 Discussion Supports. Students should take turns explaining their reasoning to their partner. Display the following sentence frames for all to see: “_____ es mayor que _____ porque . . .” // “_____ is greater than _____ because . . .”, and “____ y ____ son equivalentes porque . . .” // “_____ and _____ are equivalent because . . . .” Encourage students to challenge each other when they disagree.
### Required Materials
Materials to Copy
• Compare Stage 3-8 Directions, Spanish
### Required Preparation
• Create a set of cards from the blackline master for each group of 2–4 students.
### Launch
• Groups of 2–4
• Give each group a set of fraction cards.
• Tell students that they will play one or more games of Compare Fractions.
• Demonstrate how to play the game. Invite a student to be your opponent in the demonstration game.
• Read the rules as a class and clarify any questions students might have.
• Groups of 2 for the first game or two, then groups of 3–4 for subsequent games, if time permits
### Activity
• “Jueguen una partida de este juego con su pareja” // “Play one game with your partner.”
• “Mientras juegan, es posible que encuentren fracciones que sean difíciles de comparar. Anoten esas fracciones. Prepárense para explicar cómo descifraron cuál fracción era la más grande” // “As you play, you may come across one or more sets of fractions that are tricky to compare. Record those fractions. Be prepared to explain how you eventually figure out which fraction is greater.”
• “Si terminan antes de que se acabe el tiempo, jueguen otra partida con la misma pareja o jueguen una partida con jugadores de otro grupo” // “If you finish before time is up, play another game with the same partner, or play a game with the players from another group.”
• 15 minutes: group work time
### Student Facing
Cómo jugar “Comparemos fracciones” (2 jugadores):
• Repartan las tarjetas entre los jugadores.
• Comparen las fracciones. El jugador que tenga la fracción mayor se queda con las dos tarjetas.
• Si las fracciones son equivalentes, cada jugador voltea otra tarjeta. El jugador que tenga la fracción mayor se queda con las cuatro tarjetas.
• Jueguen hasta que un jugador se quede sin tarjetas. Gana el jugador que tenga más tarjetas al final del juego.
Cómo jugar “Comparemos fracciones” (3 o 4 jugadores):
• El jugador que tenga la fracción mayor gana la ronda.
• Si 2 o más jugadores tienen la fracción mayor, esos jugadores voltean otra tarjeta. El jugador con la fracción mayor se queda con todas las tarjetas.
Anota aquí las fracciones que te parecieron difíciles de comparar.
_________ y _________
_________ y _________
_________ y _________
_________ y _________
### Activity Synthesis
• Invite groups to share some of the challenging sets of fractions they recorded and how they eventually determined the greater one in each pair.
• As one group shares, ask others if they have other ideas about how the fractions could be compared.
## Activity 2: Ordenemos fracciones (15 minutes)
### Narrative
This activity prompts students to compare multiple fractions and put them in order by size. The work gives students opportunities to look for and make use of structure (MP7) in each set of fractions and make comparisons strategically. For instance, rather than comparing two fractions at a time and in the order they are listed, students could first classify the given fractions as greater or less than $$\frac{1}{2}$$ or 1, look for fractions with a common numerator or denominator, and so on.
If time is limited, consider asking students to choose two sets of fractions to compare and order.
Representation: Internalize Comprehension. Synthesis: Invite students to identify which details were most useful when putting fractions in order. Display the sentence frame, “La próxima vez que ordene fracciones, prestaré atención a . . .” // “The next time I put fractions in order, I will pay attention to . . . .“
Supports accessibility for: Memory
• Groups of 2
### Activity
• “Trabajen individualmente en dos de los grupos. Después, compartan su trabajo con su pareja y trabajen juntos en los que falten” // “Work independently on two sets. Then, discuss your work with your partner and complete the rest together.”
• 10 minutes: independent work time
• Monitor for students who look for and make use of structure. Ask them to share during lesson synthesis.
• 3–4 minutes: partner discussion
### Student Facing
Ordena cada grupo de fracciones de menor a mayor. Prepárate para explicar tu razonamiento.
1. $$\frac{3}{12} \qquad \frac{2}{4} \qquad \frac{2}{3} \qquad \frac{1}{8}$$
2. $$\frac{8}{5} \qquad \frac{5}{6} \qquad \frac{11}{12} \qquad \frac{11}{10}$$
3. $$\frac{21}{20} \qquad \frac{9}{10} \qquad \frac{6}{5} \qquad \frac{101}{100}$$
4. $$\frac{5}{8} \qquad \frac{2}{5} \qquad \frac{3}{7} \qquad \frac{3}{6}$$
### Student Response
Some students may try to write equivalent fractions with a common denominator for all four fractions in each set before comparing them but may be unable to do so. Encourage them to try reasoning about two fractions at a time, and to use what they know about the fractions to determine how they compare (to one another or to familiar benchmarks).
### Activity Synthesis
• See lesson synthesis.
## Lesson Synthesis
### Lesson Synthesis
Invite students to share their strategies for comparing and ordering the fractions in the last activity. Record their responses.
Ask students to reflect on their understanding of fractions in this unit.
“¿Qué cosas relacionadas con escribir, representar o comparar fracciones no sabían al iniciar la unidad, pero ahora ya las saben bien? Piensen en al menos dos cosas específicas” // “What are some things about writing, representing, or comparing fractions that you didn’t know at the beginning of the unit but you know quite well now? Think of at least two specific things.”
## Student Section Summary
### Student Facing
En esta sección comparamos fracciones. Lo hicimos usando lo que sabemos sobre el tamaño de las fracciones, algunos valores de referencia como $$\frac12$$ y 1, y fracciones equivalentes. Por ejemplo, para comparar $$\frac38$$ y $$\frac{6}{10}$$, podemos razonar así:
• $$\frac{4}{8}$$ es equivalente a $$\frac12$$, así que $$\frac38$$ es menor que $$\frac12$$.
• $$\frac{5}{10}$$ es equivalente a $$\frac12$$, así que $$\frac{6}{10}$$ es mayor que $$\frac12$$.
Esto quiere decir que $$\frac{6}{10}$$ es mayor que $$\frac38$$ (o que $$\frac38$$ es menor que $$\frac{6}{10}$$).
También podemos comparar escribiendo fracciones equivalentes con el mismo denominador. Por ejemplo, para comparar $$\frac{3}{4}$$ y $$\frac{4}{6}$$, podemos usar 12 como denominador:
$$\frac{3}{4} = \frac{9}{12} \hspace{2cm} \frac{4}{6} = \frac{8}{12}$$
Como sabemos que $$\frac{9}{12}$$ es mayor que $$\frac{8}{12}$$, entonces $$\frac{3}{4}$$ es mayor que $$\frac{4}{6}$$.
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# Math Worksheets Land
Math Worksheets For All Ages
# Math Worksheets Land
Math Worksheets For All Ages
Home > Grade Levels > High School Geometry >
# Proving Triangle Congruence Worksheets
When we are able to prove that two triangles are congruent, meaning that all there corresponding angles and sides are equal, we can use that information to learn a great deal of information about the coordinate system they may be in and the surrounding shapes. You will find this skill especially helpful when you begin writing geometric proofs. This selection of lessons and worksheets help students learn how to prove that two triangles are congruent. We will explore all the different theorems and show you how to use them in action based lessons.
### Aligned Standard: High School Geometry - HSG-CO.B.8
• Answer Keys - These are for all the unlocked materials above.
### Homework Sheets
Determine which proof helps you explain the given information.
• Homework 1 - Side-Side-Side Postulate (SSS) – If three sides of a triangle are congruent to three sides of another triangle, the triangles are congruent.
• Homework 2 - Side-Angle-Side Postulate (SAS) – If two sides and the included angle of a triangle are congruent to two sides and the angle of another triangle; the triangles are congruent.
• Homework 3 - Angle-Side-Angle Postulate (ASA) – If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, the triangles are congruent.
### Practice Worksheets
Obviously the triangles we present you with are congruent. I see the questions phrased in a way that leads to believe it is possible that they are congruent.
• Practice 1 - For the set below, determine if the triangles are congruent. State the proof needed (ASA, SAS, or SSS).
• Practice 2 - Look at all the marks to make your decision.
• Practice 3 - What side matches the other.
### Math Skill Quizzes
You will not find a better mix of problems on this anywhere else. At least that's what a geometry teacher told us.
• Quiz 1 - Triangle PQR ≈ Triangle GHI and the perimeter of Triangle PQR is 300 cm. If the sum of two sides of Triangle GHI is 150 cm, what is the length of the third side of Triangle PQR?
• Quiz 2 - For the set below, determine what postulate would be used to prove congruence.
• Quiz 3 - Which postulate would prove this?
### How Do You Prove Triangle Congruence?
When two triangles are congruent, one can be moved through more rigid motions to coincide with the other. All the corresponding angles and sides will be equal. When triangles are congruent, two facts are always true. Their corresponding sides are identical this is usually represented in a manner like this: AB ≈ DE, BC ≈ EF, CA ≈ FD. Congruent triangles have corresponding angles that are equivalent. This is often displayed in a fashion similar to this: ∠ A ≈ ∠ D, ∠ B ≈ ∠ E, ∠ C ≈ ∠ F.
The best part is when you are trying to prove the congruency of a triangle. It is not necessary to prove all the six points to show their congruency. Below are some of the methods of proving congruency. This is somewhat of a cheat sheet you can use to prove that two triangles are congruent. The names of the combinations indicate things that can prove and congruent on corresponding locations of that figure. If you can prove those locations to be equal on both figures, you have proven that they are congruent figures.
SSS (Side-Side-Side) - If the three sides of a triangle are congruent to another, then the two triangles are equal. To use this you will need to know the lengths of sides.
SSS (Side-Angle-Side) - If two sides and included angle of one of the triangles are congruent to corresponding parts of another, they are identical. Remember that the angle must be included mean that it needs between those two sides for this to be true.
ASA (Angle-Side-Angle) - If two angles and included side of one of the triangle are congruent to the corresponding parts of another triangle, they are equal. Once again, we need to realize that the side must be included or between those angles for this to work.
AAS (Angle-Angle-Side) - If two angles and a non-included side of one triangle are congruent to the corresponding sides of another triangle, they are congruent to each other. Not the need for a non-included side.
HL (Hypotenuse-Leg) - If the hypotenuse and leg of one triangle are congruent to the corresponding part of another right-angled triangle, they are the same. This is a lesser often used form, but definitely something to keep in your back pocket for more puzzling proofs.
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# The Golden Ratio: Constructing the Golden Rectangle
Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.
Today we will learn how to construct the famous Golden Rectangle. Remember that constructions only allow the use of a compass and a straightedge (not a ruler). Simple things that we can do with these devices are make perpendicular lines, find midpoints, draw circles, and bisect angles. It's a fun exercise to follow along with my instructions below.
Here is what I think is the most straightforward way to construct the Golden Rectangle. Take a unit segment. Make a segment perpendicular at the end of the unit segment two units long. Complete the right triangle. We now have a segment measuring $\sqrt{5}$ units. Take this segment and extend it one unit. Then draw the segment that is the average of these two (to do this, find the midpoint of that segment). Then we have a segment measuring $\phi$ units. We can then easily construct a rectangle with dimensions $\phi$ and $1$.
There is, however, a more elegant way. Reference the picture above. Begin by constructing square $ABCD$. Then, construct midpoint $M$ of $CD$. Draw the circle with radius $MB$. Extend $DC$ until it intersects the circle. Call this point of intersection $X$. Finish the rectangle with vertices $ADX$. This rectangle is a Golden Rectangle. Note also that you have created another Golden Rectangle in the process.
Can you prove that this second construction actually produces a golden rectangle? How is this similar to my construction? Post your answers or inquiries below.
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6 years, 4 months ago
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since triangleBMC is congruent AND EQUAL to triangle BCX,the new rectangle thus formed is also a goldenrectangle,which is of unit lengthg
- 6 years, 4 months ago
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# Multiply and Divide by 10
6 teachers like this lesson
Print Lesson
## Objective
Students will be able to identify patterns in the multiples of 10 and represent and solve multiplication and division problems with 10
#### Big Idea
Counting by 10's is easy for students of this age. Knowing that base 10 can be used to help multiply and divide in the higher numbers by decomposing is critical. This lesson sets the base for future work.
## Warm Up with a Journal Response
10 minutes
Today, I decide to have the students really dig into the vocabulary expression "Commutative Property." I want to be sure the children understand how this property can help them solve problems going forward.
Boys and girls, I have placed our vocabulary card for the Commutative Property on the board. It has an example of what this property means. We have been discussing this idea for a few days. Will you please copy the prompt on the board and write as much as you can about what you notice, know, or even wonder about the Commutative Property? You may use words, drawings, and numbers to explain your thinking.
Next, the students share out what they wrote, and drew, with a partner. You may want to walk around and choose to share the examples that best review the property. I always allow students to edit and add to their journals when we do activities like this.
## Multiples of 10
25 minutes
Have the students write the multiples of 10 on a white board or a piece of paper. I also make the choice to have a student write them on the class white board. When I ask her to write the multiples of 10, she asks how many. In response to my 10, she says, "Oh, to 100"! That is exactly what I long to hear.
Mathematicians, please write the multiples of 10, separated by commas. Write 10 or more of them. When you are finished, please look for as many patterns as you can, while you wait for your peers.
Next, begin a conversation of the patterns the children notice. Listen for them to know that there is always a 0 in the one's place and that they increase by 10. I also ask them to relate the 10's to dimes, as this is a real world application that they can draw on when we multiply.
This student disagrees with a pattern another student thinks she sees. Through this conversation, we all review the difference between the tens and ones place.
Here, the students navigate a conversation about adding on to 10 and that the zero is a place holder, which makes the digit one worth 10, not 1.
In this lesson or the next, you may choose to add the vocabulary word "multiplier" to the students' word bank. It is a precise reference when they begin to discuss the product and the number in the 10's place and higher.
## Using 10's in Word Problems
15 minutes
Display problems such as 10 x ___ = 80 and 40 =_____ x 10. Ask students how they solve these problems. Some will use multiplication facts that they know, while a few may think of division.
Next, I would like you and your partner to work together on some word problems in our book. The number ten will be one of factors in some of them. You may need to divide or multiply. Remember to talk about how you know the answer. I will walk around and listen in. Also, remember you can and should draw representations of the story if you want to visualize it better.
I chose to use the page from the program my district provides for this activity, as it uses 10 as a factor, as well as provides problems with factors that we have already worked with. It is important to review as well as introduce in each lesson.
## Closing
5 minutes
What a wonderful class today boys and girls. I really think your thinking is getting deeper and I am proud of you for using math terms like factors and products. This helps everyone know what you are thinking about. I also like the sketches I am seeing and the way you dissect a word problem before you work on it.
Does anyone have wonderings about the multiples of 10? Will you please share with your partner your favorite work in today's math class?
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# 10.7 Alternating Series Test for Convergence
Welcome to AP Calc 10.7! In this lesson, you’ll how to test for convergence when dealing with an alternating series.
## ➕ Alternating Series Test Theorem
The alternating series test for convergence states that for an alternating series $\sum(-1)^n\cdot a_n$, if
$1. \lim_{n\to \infty} a_n=0 \ \text{and}$
$2. \ a_n \ \text{decreases,}$
then the series converges. Otherwise, it diverges.
### 🧱 Breaking Down the Theorem
To illustrate this theorem, let’s look at a famous example—the alternating harmonic sequence:
$\sum_{n=1}^\infty\frac{(-1)^n}{n}$
Let’s look at our first criteria: the limit of $a_n$ must be equal to zero. To figure this out, we first must figure out what $a_n$ is! To do this, all we have to do is factor out the alternating part of the sequence, $(-1)^n$. Then, we get
$a_n=\frac{1}{n}$
Now, let’s take the limit.
$\lim_{n\to \infty}\frac{1}{n}=0$
Well, that’s our first criteria satisfied! Now we need to know whether $a_n$ decreases. To show this, we must show that $a_n>a_{n+1}$.
$a_n=\frac{1}{n}>\frac{1}{n+1}=a_{n+1}$
Try plugging in a random number for $n$ to see that this is true!
$a_2=\frac{1}{2}>\frac{1}{3}=a_{2+1}$
Therefore, both of our conditions for convergence are met, and our series converges!
## 📝 Practice with Alternating Series Test
Now it’s your turn to apply what you’ve learned!
### ❓Alternating Series Test Problems
For each of the following series, state whether they converge or diverge.
$1. \sum_{n=1}^\infty \frac{(-1)^{n+2}\cdot n^5}{n^5+3}$
$2. \sum_{n=2}^\infty\frac{\text{cos}(n\pi)}{n}$
$3.\sum_{n=2}^\infty (-1)^n \cdot\text{ln}(n)$
### 💡 Alternating Series Test Solutions
#### Question 1 Solution
First, identify $a_n$.
$a_n=\frac{n^5}{n^5+3}$
Now, take the limit.
$\lim_{n\to \infty}\frac{n^5}{n^5+3}=1\neq0$
Since our first condition isn’t met, we don’t need to check the second condition. This series is divergent.
#### Question 2 Solution
This one is a little tricky—it requires you to recognize another type of alternating series, $\text{cos}(n\pi)$. If you plug some examples into your calculator, you’ll see that $\text{cos}(n\pi)=(-1)^n$. Therefore, we can treat this equation just like the harmonic series in our first example. We showed that the harmonic series met the conditions for convergence, so this one does too! This series is convergent.
#### Question 3 Solution
First, we find $a_n=\text{ln}(n)$. Then, we take the limit:
$\lim_{n\to \infty}\text{ln}(n)=\infty$
Like our first problem, since the first condition isn’t met, we can say that this series is divergent without checking the other condition.
## 💫 Closing
Great work! With this test mastered, you’re well equipped to take on all sorts of convergence problems. Make sure you recognize both types of alternating series so that you know when to apply this test! 🧠
# Key Terms to Review (1)
Alternating Sequences
: Alternating sequences are sequences in which the terms alternate between positive and negative values.
# 10.7 Alternating Series Test for Convergence
Welcome to AP Calc 10.7! In this lesson, you’ll how to test for convergence when dealing with an alternating series.
## ➕ Alternating Series Test Theorem
The alternating series test for convergence states that for an alternating series $\sum(-1)^n\cdot a_n$, if
$1. \lim_{n\to \infty} a_n=0 \ \text{and}$
$2. \ a_n \ \text{decreases,}$
then the series converges. Otherwise, it diverges.
### 🧱 Breaking Down the Theorem
To illustrate this theorem, let’s look at a famous example—the alternating harmonic sequence:
$\sum_{n=1}^\infty\frac{(-1)^n}{n}$
Let’s look at our first criteria: the limit of $a_n$ must be equal to zero. To figure this out, we first must figure out what $a_n$ is! To do this, all we have to do is factor out the alternating part of the sequence, $(-1)^n$. Then, we get
$a_n=\frac{1}{n}$
Now, let’s take the limit.
$\lim_{n\to \infty}\frac{1}{n}=0$
Well, that’s our first criteria satisfied! Now we need to know whether $a_n$ decreases. To show this, we must show that $a_n>a_{n+1}$.
$a_n=\frac{1}{n}>\frac{1}{n+1}=a_{n+1}$
Try plugging in a random number for $n$ to see that this is true!
$a_2=\frac{1}{2}>\frac{1}{3}=a_{2+1}$
Therefore, both of our conditions for convergence are met, and our series converges!
## 📝 Practice with Alternating Series Test
Now it’s your turn to apply what you’ve learned!
### ❓Alternating Series Test Problems
For each of the following series, state whether they converge or diverge.
$1. \sum_{n=1}^\infty \frac{(-1)^{n+2}\cdot n^5}{n^5+3}$
$2. \sum_{n=2}^\infty\frac{\text{cos}(n\pi)}{n}$
$3.\sum_{n=2}^\infty (-1)^n \cdot\text{ln}(n)$
### 💡 Alternating Series Test Solutions
#### Question 1 Solution
First, identify $a_n$.
$a_n=\frac{n^5}{n^5+3}$
Now, take the limit.
$\lim_{n\to \infty}\frac{n^5}{n^5+3}=1\neq0$
Since our first condition isn’t met, we don’t need to check the second condition. This series is divergent.
#### Question 2 Solution
This one is a little tricky—it requires you to recognize another type of alternating series, $\text{cos}(n\pi)$. If you plug some examples into your calculator, you’ll see that $\text{cos}(n\pi)=(-1)^n$. Therefore, we can treat this equation just like the harmonic series in our first example. We showed that the harmonic series met the conditions for convergence, so this one does too! This series is convergent.
#### Question 3 Solution
First, we find $a_n=\text{ln}(n)$. Then, we take the limit:
$\lim_{n\to \infty}\text{ln}(n)=\infty$
Like our first problem, since the first condition isn’t met, we can say that this series is divergent without checking the other condition.
## 💫 Closing
Great work! With this test mastered, you’re well equipped to take on all sorts of convergence problems. Make sure you recognize both types of alternating series so that you know when to apply this test! 🧠
# Key Terms to Review (1)
Alternating Sequences
: Alternating sequences are sequences in which the terms alternate between positive and negative values.
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# What is the value of cos^-1(cos(-pi/3))?
Jul 23, 2016
${\cos}^{-} 1 \left(\cos \left(- \frac{\pi}{3}\right)\right) = \frac{\pi}{3}$..
#### Explanation:
We have, $\cos \theta = \cos \left(- \theta\right) , \forall \theta \in \mathbb{R} \ldots \ldots \ldots \ldots . \left(1\right)$.
Next let us recall the following defn. of the ${\cos}^{-} 1$ function :-
${\cos}^{-} 1 x = \theta , | x | \le 1 \iff \cos \theta = x , \theta \in \left[0 , \pi\right] \ldots \ldots \ldots \ldots \ldots \left(2\right)$
Now, using $\left(2\right)$ in $\leftarrow$ direction, keeping in mind that $\cos \theta \le 1$ ,
we have, ${\cos}^{-} 1 \left(\cos \theta\right) = \theta , \mathmr{if} , \theta \in \left[0 , \pi\right] \ldots \ldots \ldots \ldots \ldots \ldots . \left(2 '\right)$
Now, cos^-1(cos(-pi/3))=cos^-1(cos(pi/3))................[by (1)],
&, here, since $\frac{\pi}{3} \in \left[0 , \pi\right]$, we get, by $\left(2 '\right)$,
${\cos}^{-} 1 \left(\cos \left(- \frac{\pi}{3}\right)\right) = {\cos}^{-} 1 \left(\cos \left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}$..
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## Presentation on theme: "The Radical Square Root"— Presentation transcript:
The square root of any real number is a number, rational or irrational, that when multiplied by itself will result in a product that is the original number The Radical Radical sign Square Root Radicand Every positive radicand has a positive and negative sq. root. The principal Sq. Root of a number is the positive sq. root. A rational number can have a rational or irrational sq. rt. An irrational number can only have an irrational root.
= 7.9 = 232 225 = +15 529 = +23 Model Problems
Find to the nearest tenth: = 13.4 = 7.9 = 11.4 = 232 = 64.4 Find the principal Square Root: 225 = +15 529 = +23 Simplify: = |x| = x = 2x8 = x + 1
Index of 2 Square Root Index of 2 radical sign radicand index
of a number is one of the two equal factors whose product is that number Square Root Index of 2 has an index of 2 Every positive real number has two square roots The principal square root of a positive number k is its positive square root, If k < 0, is an imaginary number
Index of 3 Cube Root Index = 3 radical sign radicand index
of a number is one of the three equal factors whose product is that number has an index of 3 principal cube roots
nth Root The nth root of a number (where n is any counting number) is one of n equal factors whose product is that number. k is the radicand n is the index is the principal nth root of k 25 = 32 (-2)5 = -32 54 = 625
Index of n nth Root Index of n radical sign radicand index
of a number is one of n equal factors whose product is that number nth Root Index of n has an index where n is any counting number principal odd roots principal even roots
Radical Rules! True or False: T T T
In general, for non-negative numbers a, b and n
Radical Rule #1 In general, for non-negative numbers a, b and n Example: simplified = x4 = x3 Hint: will the index divide evenly into the exponent of radicand term?
In general, for non-negative numbers a, b, and n
Radical Rule #2 True or False: If and T T Transitive Property of Equality If a = b, and b = c, then a = c then In general, for non-negative numbers a, b, and n Example:
Perfect Squares – Index 2
12 144 11 121 100 10 9 81 8 64 7 49 6 36 5 25 4 16 3 9 4 2 1
Perfect Square Factors
Find as many combinations of 2 factors whose product is 75 Factors that are Perfect Squares Find as many combinations of 2 factors whose product is 128
Find as many combinations of 2 factors whose product is 80
Simplifying Radicals Simplify: answer must be in radical form. Find as many combinations of 2 factors whose product is 80 perfect square comes out from under the radical To simplify a radical find, if possible, 2 factors of the radicand, one of which is the largest perfect square of the radicand. The square root of the perfect square becomes a factor of the coefficient of the radical.
Perfect Cubes 13 = 23 = 33 = 43 = 53 = 63 = 73 = (x4)3 = x12 (-2y2)3 = -8y6
1) Factor the radicand so that the perfect power (cube) is a factor 2) Express the radical as the product of the roots of the factors 3) Simplify the radical containing the largest perfect power (cube)
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# Square Root
## Learn The Definition And How To Find The Square Roots Of Numbers With Formulas And Examples
As kids progress through the grades, they start learning more and more complex math concepts. Square root and squares are one such topic that kids start learning in grades 4 and 5. But, what are they? the square root of a number is another number, which you get when the original number is multiplied by itself. Before introducing this concept to the little ones, ensure they know the multiplication table for kids. The square root can be expressed as a x where a is the natural number and the x is the square root of the number.
## What is Square Root?
In math, square roots are the factor of a number, which gives the original number when it’s multiplied by itself. It is represented by the symbol √. Consider this formula, if p is the square root of q, it implies that p x p = q.
Here is an example to help you understand better, 4 is the square root of 16. To get 16, 4 is multiplied by itself. In this context, the exponent is 2, it is known as a square.
According to mathematical concepts, square roots and squares are reverse operations. The square of any numeral is the value of power 2 of the number. Another fact about squares and square roots is that they are always a positive number. Here is a list of square roots to help your child learn the concept easily given below:
## List Of Square Roots
### Definition:
The square root of a digit is the value of power ½ of that number. In simple words, the square root is the number that is multiplied by itself to obtain the actual number. It is expressed as ‘ ‘.
Also, refer to Multiplication Worksheets, available at Osmo.
## How to Find The Square Root?
Once your child has understood the definition, it’s time to teach how to find the square root of a number. Some kids might find it difficult to grasp the concept easily. Therefore, you can start with simpler numbers that are perfect squares, which is a number that can be expressed as the product of two equal integers. Finding the square roots of perfect square is quite easy and kids will learn it faster.
Here are four different methods to determine the square roots of numbers:
• Repeated Subtraction Method
• Estimation Method
• Prime Factorization Method
• Long Division Method
### Table Of Square Roots
Here is a table to help your kids learn easily. The table has a list of the square roots of numbers from 1 to 10. This list includes both perfect squares and numbers that are non-perfect squares.
## Square Root Formula
One thing that makes it easy to determine square roots of numbers is a formula. Here is a simple formula you can use: √x= x1/2
According to this formula, 41/2 = √4 = √(2×2) = 2.
### Determining Square Roots Of Negative Numbers
Now that you know how to find the square roots of positive integers, how do you find the same for a negative integer? A square is always 0 or positive, so negative numbers don’t have real square roots.
But, here is an easy formula you can use: √(-y)= i√y. Here, i is the square root of -1.
For instance, let’s determine the √ -25.
√(-25)= √25 × √(-1)
Since negative numbers don’t have real square roots, let’s consider √(-1)= i
√(-25)= √25 × √(-1) = 5i
Therefore, the √-25 is 5i.
Getting kids to learn math can be difficult. Involve kids in fun activities like math games for kids and worksheets to help them learn in a fun and engaging way. Check Osmo’s kids learning section to find more fun ways to help kids learn.
## Frequently Asked Questions on Square Root
### What Is A Square Root?
A square root is a number that is obtained when a number is multiplied by itself. For example, 10 is the square root of 100; that is, 10 x 10 = 100. The square root is denoted by a symbol, √.
### How is the square root of a number calculated?
The square root of a number is calculated using this formula. The formula for the square root is used to determine the square root of a digit. The exponent formula to find out the square root is, n√y y n = y1/n. Here, n= 2, is called a square root. We can use any of these simple ways for determining the square root, like, long division, prime factorization, and many more.
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# USING THE DISTRIBUTIVE PROPERTY
We can use the Distributive Property to remove the parentheses from an algebraic expression like 3(x + 5). Sometimes this is called “simplifying” or “expanding” the expression. Multiply the quantity in front of parentheses by each term within parentheses :
3(x + 5) = 3 · x + 3 · 5
3(x + 5) = 3x + 15.
## Examples
Example 1 :
Simplify the following expression :
5 - 3(7x + 8)
Solution :
= 5 - 3(7x + 8)
Use the distributive property.
= 5 - 3(7x) - 3(8)
Simplify.
= 5 - 21x - 24
= -21x - 19
Example 2 :
Simplify the following expression :
-9a - (1/3)(- 3/4 - 2a/3 + 12)
Solution :
-9a - (1/3)(- 3/4 - 2a/3 + 12)
Use the distributive property.
= -9a + 1/4 + 2a/9 - 4
Simplify.
= -9a + 2a/9 + 1/4 - 4
= -81a/9 + 2a/9 + 1/4 - 16/4
= (-81a + 2a)/9 + (1 - 16)/4
= -79a/9 - 15/4
Example 3 :
Simplify the following expression :
0.2(3b - 15c) + 6c
Solution :
0.2(3b - 15c) + 6c
Use the distributive property.
= 0.6b - 3c + 6c
Simplify.
= 0.6b + 3c
Example 4 :
Simplify the following expression :
(2/3)(6e + 9f -21g) - 7f
Solution :
= (2/3)(6e + 9f - 21g) - 7f
Use the distributive property.
= 4e + 6f - 14g - 7f
Simplify.
= 4e - f - 14g
Example 5 :
James is selling tickets for a high school band concert. The band gets to keep 25% of the money he collects from ticket sales to put toward new uniforms. Write an expression to represent how much the band gets to keep.
Solution :
Let a represent the number of adult tickets he sells.
Let y represent the number of youth tickets he sells.
The expression 16.60a + 12.20y represents the amount of money James collects from ticket sales.
Write 25% as a decimal :
0.25
Write an expression to represent 25% of the money he collects :
0.25(16.60a + 12.20y)
That is, 25% adult ticket sales and youth ticket sales.
Use the Distributive Property to simplify the expression.
= 0.25(16.60a) + 0.25(12.20y)
= 4.15a + 3.05y
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OTHER TOPICS
Profit and loss shortcuts
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Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
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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# How do you find coordinates in algebra?
## How do you find coordinates in algebra?
To find out the coordinates of a point in the coordinate system you do the opposite. Begin at the point and follow a vertical line either up or down to the x-axis. There is your x-coordinate. And then do the same but following a horizontal line to find the y-coordinate.
## What is a coordinate point?
The coordinates of a point are a pair of numbers that define its exact location on a two-dimensional plane. Recall that the coordinate plane has two axes at right angles to each other, called the x and y axis. The coordinates of a given point represent how far along each axis the point is located.
How do you write coordinates?
To write the full map locations, begin writing with the latitude line, add other coordinates like minutes and decimals. Put comma and then write longitude line with its minutes and decimals. Do not forget to indicate the coordinates with negative and positive numbers.
How do you list coordinates?
Start with your line of latitude, writing the degrees, then the minutes, then the seconds. Then, add the North or South as the direction. Then, write a comma followed by your line of longitude in degrees, then minutes, then seconds. Then, add East or West as the direction.
### How do you explain coordinates?
Coordinates are two numbers (Cartesian coordinates), or sometimes a letter and a number, that locate a specific point on a grid, known as a coordinate plane. A coordinate plane has four quadrants and two axes: the x axis (horizontal) and y axis (vertical).
### How do I learn coordinates?
Latitude and longitude are broken into degrees, minutes, seconds and directions, starting with latitude. For instance, an area with coordinates marked 41° 56′ 54.3732” N, 87° 39′ 19.2024” W would be read as 41 degrees, 56 minutes, 54.3732 seconds north; 87 degrees, 39 minutes, 19.2024 seconds west.
What are coordinate planes in math terms?
The two-dimensional plane is called the Cartesian plane, or the coordinate plane and the axes are called the coordinate axes or x-axis and y-axis. The given plane has four equal divisions by origin called quadrants. The horizontal line towards the right of the origin (denoted by O) is positive x-axis.
How do you graph a coordinate plane?
Explanation: To graph points on a coordinate plane, we must be aware that points or coordinates rather are written in the form (x,y) where x is the position on the x -axis and y is the position on the y -axis. To graph a point, we must first locate its position on the x-axis, then do the same for y-axis and find where the two points meet.
#### What is ‘coordinate’ in graph?
Each point on a graph has a set of coordinates. Coordinates are the two numbers that tell where a point is located on the graph. Remember, the coordinates for the origin are (0,0). This tells us that the origin is the very center of the graph because you’ve traveled 0 places on the x -axis and 0 places on the y -axis.
#### What is the use of coordinate geometry?
etc. to locate the position of cursor or finger.
• In aviation to determine the position and location of airplanes accurately.
• In maps and in navigation (GPS).
• To map geographical locations using latitudes and longitudes.
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If you have a coefficient tied to a variable on one side of a matrix equation, you can multiply by the coefficient's inverse to make that coefficient go away and leave you with just the variable. For example, if 3x = 12, how would you solve the equation? You'd divide both sides by 3, which is the same thing as multiplying by 1/3, to get x = 4. So it goes with matrices.
In variable form, an inverse function is written as f –1(x), where f –1 is the inverse of the function f. You name an inverse matrix similarly; the inverse of matrix A is A–1. If A, B, and C are matrices in the matrix equation AB = C, and you want to solve for B, how do you do that? Just multiply by the inverse of matrix A (if the inverse exists), which you write like this:
A–1[AB] = A–1C
So the simplified version is B = A–1C.
Now that you've simplified the basic equation, you need to calculate the inverse matrix in order to calculate the answer to the problem.
First off, you must establish that only square matrices have inverses — in other words, the number of rows must be equal to the number of columns. And even then, not every square matrix has an inverse. If the determinant of a matrix is not 0, then the matrix has an inverse.
When a matrix has an inverse, you have several ways to find it, depending how big the matrix is. If the matrix is a 2-x-2 matrix, then you can use a simple formula to find the inverse. However, for anything larger than 2 x 2, you should use a graphing calculator or computer program (many websites can find matrix inverses for you').
If you don't use a graphing calculator, you can augment your original, invertible matrix with the identity matrix and use elementary row operations to get the identity matrix where your original matrix once was. These calculations leave the inverse matrix where you had the identity originally. This process, however, is more difficult.
With that said, here's how you find an inverse of a 2-x-2 matrix:
If matrix A is the 2-x-2 matrix
its inverse is as follows:
Simply follow this format with any 2-x-2 matrix you're asked to find.
Armed with a system of equations and the knowledge of how to use inverse matrices, you can follow a series of simple steps to arrive at a solution to the system, again using the trusty old matrix. For instance, you can solve the system that follows by using inverse matrices:
These steps show you the way:
1. Write the system as a matrix equation.
When written as a matrix equation, you get
2. Create the inverse of the coefficient matrix out of the matrix equation.
You can use this inverse formula:
In this case, a = 4, b = 3, c = –10, and d = –2. Hence ad – bc = 22. Hence, the inverse matrix is
3. Multiply the inverse of the coefficient matrix in the front on both sides of the equation.
You now have the following equation:
4. Cancel the matrix on the left and multiply the matrices on the right.
An inverse matrix times a matrix cancels out. You're left with
5. Multiply the scalar to solve the system.
You finish with the x and y values:
Note that multiplying the scalar is usually easier after you multiply the two matrices.
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# Equations Involving Rational Exponents
Equations involving rational exponents can be solved by combining the methods that you just learned for eliminating radicals and integral exponents. For equations involving rational exponents, always eliminate the root first and the power second.
Example 1
Eliminating the root, then the power
Solve each equation.
a) x2/3 =4
b) (w - 1)-2/5 = 4
Solution
a) Because the exponent 2/3 indicates a cube root, raise each side to the power 3:
x2/3 = 4 Original equation (x2/3)3 = 43 Cube each side. x2 = 64 Multiply the exponents: x = 8 or x = -8 Even-root property
All of the equations are equivalent. Check 8 and -8 in the original equation. The solution set is {-8, 8}.
= 4 Original equation [(w - 1)-2/5]-5 = 4-5 Raise each side to the power -5 to eliminate the negative exponent. (w - 1)2 Multiply the exponents: w - 1 Even-root property w - 1 or w - 1 w or w
Check the values in the original equation.The solution set is .
An equation with a rational exponent might not have a real solution because all even powers of real numbers are nonnegative.
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# DERIVING THE SLOPE INTERCEPT FORM OF AN EQUATION
## About "Deriving the slope intercept form of an equation"
Deriving the slope intercept form of an equation :
In this section, we are going to see, how to derive the slope intercept form of an equation.
## Deriving the slope intercept form of an equation
Step 1 :
Let L be a line with slope m and y-intercept b. Circle the point that must be on the line. Justify your choice.
(b, 0) (0, b) (0, m) (m, 0)
The coordinate of x is 0 in the point that includes the y-intercept.
Step 2 :
Recall that slope is the ratio of change in y to change in x. Complete the equation for the slope m of the line using the y-intercept (0, b), and another point (x, y) on the line.
Slope m = change in y-values / change in x-values
Slope m = (y - b) / (x - 0)
Slope m = (y - b) / x
Step 3 :
In an equation of a line, we often want y by itself on one side of the equation. Solve the equation from Step 2 for y.
m = (y - b) / x
Multiply both sides by x
m.x = [(y - b) / x].x
mx = y - b
Add b to both sides of the equation.
mx + b = (y - b) + b
mx + b = y
Write the equation with y on the left side.
y = mx + b
## Reflect
Critical thinking : Write the equation of a line with slope m that passes through the origin. Explain your reasoning.
y = mx
Because the origin is on the y-axis, the graph crosses the y-axis at (0, 0). So, the y-intercept b is 0, and y = mx + b becomes y = mx.
## Slope-intercept form equation of a line - Examples
Example 1 :
A line is passing through the points (2, 3) and (0, 4). Find the equation of the line in slope intercept form.
Solution :
Step 1 :
Fine the slope of the line using the points (2, 3) and (0, 4).
Slope m = change in y-values / change in x-values
Slope m = (4-3) / (0-3)
Slope m = 1 / (-3)
Slope m = -1/3
Step 2 :
In the point (0, 4), x-coordinate is zero. So, the line intersects y-axis at this point.
Since the y-coordinate at this point is 4, y-intercept is 4.
Hence, the equation of the line is y = (-1/3)x + 4.
Example 2 :
A line is passing through the points (0, 0) and (-1, -8). Find the equation of the line in slope intercept form.
Solution :
Step 1 :
Fine the slope of the line using the points (0, 0) and (-1, -8).
Slope m = change in y-values / change in x-values
Slope m = (-8-0) / (-1-0)
Slope m = -8 / (-1)
Slope m = 8
Step 2 :
Since the line is passing through the origin (0,0), there is no y-intercept or y-intercept = 0.
Hence, the equation of the line is y = 8x.
After having gone through the stuff given above, we hope that the students would have understood, how to derive the slope intercept form of an equation.
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# If the largest angle of an isosceles triangle measures 124 degrees what is the measure of the smallest angle?
Mar 8, 2018
In an isosceles triangle, by definition two of the three sides of the triangle are equal to one another. Therefore, the angles opposite to these sides must also be equal. And considering the Triangle Sum Theorem, the angles of a triangle must add up to 180 degrees, the angle that measures 124 degrees cannot be be equal to another angle due to the fact that the sum of them would exceed 180. Therefore let x be the angle:
$124 + 2 x = 180$
$2 x = 56$
$x = 28$
The other 2 angles of the triangle must be 28 degrees each
Mar 8, 2018
$\angle A = \angle B = {28}^{\circ}$
#### Explanation:
If one angle of an isosceles triangle is greater than ${60}^{\circ}$, then the other two angles must be equal. You can use this fact and the fact that the sum of interior angles is ${180}^{\circ}$ to find the measure of those angles.
Given an isosceles triangle with $\angle C = {124}^{\circ}$, then we know know that:
$\angle A = \angle B$
And
$\angle A + \angle B + \angle C = {180}^{\circ}$
Substitute $\angle A = \angle B$:
$\angle A + \angle A + \angle C = {180}^{\circ}$
Substitute $\angle C = {124}^{\circ}$:
$2 \angle A + {124}^{\circ} = {180}^{\circ}$
$2 \angle A = {56}^{\circ}$
$\angle A = \angle B = {28}^{\circ}$
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# CHSE Odisha Class 12 Math Solutions Chapter 13 Three Dimensional Geometry Ex 13(c)
Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 13 Three Dimensional Geometry Ex 13(c) Textbook Exercise Questions and Answers.
## CHSE Odisha Class 12 Math Solutions Chapter 13 Three Dimensional Geometry Exercise 13(c)
Question 1.
State which of the following statements are true (T) or false (F):
(a) The line $$\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-1}{2}$$ pass through the origin.
Solution:
True
(b) The lines $$\frac{x+2}{-k}=\frac{y-3}{k}=\frac{z+4}{k}$$ and $$\frac{x-4}{-4}=\frac{y-3}{k}=\frac{z+1}{2}$$ are perpendicular every value of k.
Solution:
True
(c) The line $$\frac{x+5}{-2}=\frac{y-3}{1}=\frac{z-2}{3}$$ lies on the plane x – y + z + 1 = 0
Solution:
False
(d) The line $$\frac{x-2}{3}=\frac{1-y}{4}=\frac{5-z}{1}$$ is parallel to the plane 2x – y – 2z = 0
Solution:
False
(e) The line $$\frac{x+3}{-1}=\frac{y-2}{3}=\frac{z-1}{4}$$ is perpendicular to the plane 3x – 3y + 3z – 1 = 0
Solution:
False
Question 2.
Fill in the blanks by choosing the correct alternative from the given ones:
(a)
[parallel, perpendicular, coincident]
Solution:
perpendicular.
(b) The line passing through (-1, 0, 1) and perpendicular to the plane x + 2y + 1 = 0 is _____.
Solution:
$$\frac{x+1}{1}=\frac{y}{2}=\frac{z-1}{0}$$
(c) The line $$\frac{x+1}{2}=\frac{y-6}{1}=\frac{z-4}{0}$$ is _____. [parallel to x-axis, perpendicular to y-axis, perpendicular to z-axis]
Solution:
perpendicular to. z-axis
(d) If the line $$\frac{x-3}{2}=\frac{y+k}{-1}=\frac{z+1}{-5}$$ lies on the plane 2x – y + z – 7 = 0; then k = -(2, -1, -2)
Solution:
2
(e) If l, m, n be d.cs. of a line, then the line is perpendicular to the plane x – 3y + 2z + 1 = 0 if _____. [(i) l = 1, m = -3, n = 2, (ii) $$\frac{l}{1}=\frac{m}{-3}=\frac{n}{2}$$ , (iii) l – 3m + 2n = 0]
Solution:
$$\frac{l}{1}=\frac{m}{-3}=\frac{n}{2}$$
Question 3.
Find the equation of lines joining the points.
(i) (4, -6, 1) and (0, 3, -1)
Solution:
Equation of the line joining the points (4, -6, 1) and (0, 3, -1) is
(ii) (a, a, a) and (a, 0, a)
Solution:
Equation of the line joining the points (a, a, a) and (a, 0, a) is
(iii) (2, 1, 3) and (4, -2, 5).
Solution:
Equation of the line joining the points (2, 1, 3) and (4, -2, 5) is
Question 4.
Write the symmetric form of equation of the following lines:
(i) x-axis
Solution:
D.cs. of x-axis are < 1, 0, 0 >.
x-axis passes through the origin.
So the equation of x-axis in symmetrical form is
$$\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}$$
(ii) y = b, z = c
Solution:
Given line in unsymmetrical from is y = b, z = c
⇒ y – b = 0 and z – c = 0.
The straight line is parallel to x-axis.
D.rs. of the straight line are < 0, 0, k >.
So the equation of the line is
$$\frac{y-b}{0}=\frac{z-c}{0}=\frac{x}{k}$$
(iii) ax + by + d = 0, 5z = 0
Solution:
Given lines ax + by + d = 0, 5z = 0
(iv) x – 2y = 3, 2x + y – 5z = 0;
Solution:
Given straight line is x – 2y – 3 = 0 … (1)
and 2x + y – 5z = 0 … (2)
Putting z = 0 in (1) and (2) we get x – 2y – 3 = 0 and 2x + y = 0.
Solving we get y = -2x and x + 4x – 3 = 0
(v) 4x + 4y – 5z – 12 = 0, 8x + 12y – 13z = 32;
Solution:
Given straight line in unsymmetrical form is
4x + 4y – 5z – 12 = 0 … (1)
8x + 12y – 13z – 32 = 0 … (2)
Putting z = 0 in (1) and (2) we get
4x + 4y – 12 = 0 ⇒ x + y – 3 =0
8x + 12y – 32 = 0 ⇒ 2x + 3y – 8 = 0
Solving we get
(vi) 3x – 2y + z = 1, 5x + 4y – 6z = 2
Solution:
Given straight line is
3x – 2y + z – 1 = 0 … (1)
and 5x – 4y + 6z – 2 = 0 … (2)
Putting x = 0 in (1) and (2) we get
Question 5.
(a) Obtain the equation of the line through the point (1, 2, 3) and parallel to the line x – y + 2z – 5 = 0, 3x + y + z = -6
Solution:
Equation of the straight line through
(b) Find the equation of the line through the point (3, -1, 2) and parallel to the planes x + y + 2z – 4 = 0 and 2x – 3y + z + 3 = 0
Solution:
Equation of the straight line through the point (3, -1, 2) is
Question 7.
(a) Show that the line passing through the points (a1, b1, c1) and (a2, b2, c2) passes through the origin, if a1a2 + b1b2 + c1c2 = p1p2, where p1 and p2 are distances of the points from origin.
Solution:
The equation of the line passing through the points (a1, b1, c1) and (a2, b2, c2) is
(b) Prove that the lines x = az + b, y = cz + d and x = a1z + b1, y = c1z + d1 are perpendicular if aa1 + cc1 + 1 = 0.
Solution:
Given lines are
x – az – b = 0 = y – cz – d … (1)
and x – a1z – b1 = 0 and < l2, m2, n2 > … (2)
Let < l1, m1, n1 > and < l2, m2, n2 > be the d.cs. of the lines (1) and(2)
D.rs. of the two lines are < a, c, 1 > and < a1, c1, 1 >.
If the lines are perpendicular then the sum of product of d.rs. is zero.
So aa1 + cc1 + 1 = 0 (Proved)
Question 8.
Find the points of intersection of the line $$\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-1}{-1}$$ and the plane 2x + y + a = 9.
Solution:
Given line is $$\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-1}{-1}$$ = r (say)
∴ Any point on (1) is (r + 1, 3r – 2, -r + 1)
If it lies on the plane 2x + y + z – 9 = 0 then
2 (r + 1) + 3r – 2 + (-r + 1) – 9 = 0
or, 2r + 3r – r + 2 – 2 + 1 – 9 = 0
or, 4r = 8, or, r = 2
∴ The point of intersection is (3, 4, -1).
Question 9.
Find the coordinates of the point where the line joining (3, 4, -5) and (2, -3, 1), meets the plane 2x + y + z – 7 = 0.
Solution:
The straight line joining the points (3, 4, -5) and (2, -3, 1) is
$$\frac{x-3}{2-3}=\frac{y-4}{-3-4}=\frac{z+5}{1+5}$$
or, $$\frac{x-3}{-1}=\frac{y-4}{-7}=\frac{z+5}{6}$$ = r (say)
Any point on the line is
(-r + 3, -7r + 4, 6r – 5)
If this point is the point of intersection of the line with the plane 2x + y + z – 7 = 0 then
2 (-r + 3) + (-7r + 4) + 6r – 5 – 7 = 0
or, -2r + 6 – 7r + 4 + 6r – 12 = 0
or, -3r = 2 or, r = $$\frac{-2}{3}$$
∴ The point of intersection is
$$\left(\frac{11}{3}, \frac{26}{3}, \frac{-27}{3}\right) \text { i.e., }\left(\frac{11}{3}, \frac{26}{3},-9\right)$$.
Question 10.
(a) Find the distance of the point (-1, -5, -10) from the point of intersection of the line $$\frac{x-2}{2}=\frac{y+1}{4}=\frac{z-2}{12}$$ and the plane x – y + z = 5.
Solution:
Given line is
$$\frac{x-2}{2}=\frac{y+1}{4}=\frac{z-2}{12}$$ = z
Any point on (1) is (2r + 2, 4r – 1, 12r + 2).
If this point is the point of intersection of the line with the plane
x – y + z = 5
then 2r + 2 – 4r + 1 + 12r + 2 = 5
or, 10r = 0, or, r = 0
The point of intersection is (2, -1, 2).
Distance between the points (-1, -5, -10) and the point of intersection of the given line and plane.
= $$\sqrt{9+16+144}=\sqrt{169}$$ = 13
(b) Find the image of the point (2, -1, 3) in the plane 3x – 2y + z – 9 = 0
Solution:
Let A = (2, -1, 3)
Let B be the image point of A with respect to the plane
3x – 2y + z – 9 = 0 … (1)
Then AB is normal to the plane.
Again let C be the point of intersection of the line with the plane (1).
Again as B is the image point of A then C must be the mid-point of AB.
Now d.rs. of AB are < 3, -2, 1 > because AB is perpendicular to the plane.
Eqn. of the line AB is
Question 11.
Prove that the lines, $$\frac{x+3}{2}=\frac{y+5}{3}=\frac{z-7}{-3}$$ and $$\frac{x+1}{4}=\frac{y+1}{5}=\frac{z+1}{-1}$$ are coplanar.
Find the equation of the plane containing them.
Solution:
Given lines are
Two lines are co-planar if either they are parallel or intersecting. Now the line (1) and (2) are not parallel because their d.rs. are not proportional. So we shall show that they are intersecting.
Any point on (1) is (2r1 – 3, 3r1 – 5, -3r1 + 7)
Any point on (2) is (4r2 – 1, 5r2 – 1, -r2 – 1)
If the lines are intersecting then for some values of r1 and r2.
2r1 – 3 = 4r2 – 1 … (3)
3r1 – 5 = 5r2 – 1 … (4)
-3r1 + 7 = -r2 – 1 … (5)
From (3) we get r1 = = 2r2 + 1
Putting it in (4) we get 6r2 + 3 – 5 = 5r2 – 1
or, r2 = 1. Again r1 = 3
With these values r1 = 3, r2 = 1
We see that eqn. (5) is satisfied. So the straight lines are intersecting. Hence they are coplanar.
The equation of the plane containing the line (1) is:
a (x + 3) + b (y + 5) + c (z – 7) = 0 … (6)
where 2a + 3b – 3c = 0 … (6)
If the plane (6) contains the line (2) then
4a + 5b – c = 0 … (8)
Solving (7) and (8) we get
Equation of the plane is
6 (x + 3) – 5 (y + 5) – 1 (z – 7) = 0
or, 6x – 5y – z = 0
Question 12.
Prove that the lines $$\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$$ and 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4 are co-planar.
Solution:
Given lines are
$$\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$$ = r (say) … (1)
and 3x – 2y + z + 5 = 0
= 2x + 3y + 4z – 4 … (2)
Any point on (1) and (2) are coplanar if they are either parallel or intersecting. If the line are intersecting then
3 (3r – 4) – 2 (5r – 6) + (-2r + 1) + 5 = 0 … (3)
and 2 (3r – 4) + 3 (5r – 6) + 4 (-2r + 1) – 4 = 0 … (4) are consistent.
Solving (3) we get
9r – 12 – 10r + 12 – 2r + 1 + 5 = 0
or, -3r + 6 = 0 or, r = 2
For r = 2, eqn (4) is satisfied. Thus the lines are intersecting and hence they are co-planar. (Proved)
Question 13.
Show that the lines 7x – 4y + 7z + 16 = 0 = 4x + 3y – 2z + 3 and x – 3y + 4z + 6 = z – y + z + 1 intersect. Find the coordinates of their point of intersection and equation of the plane containing them.
Solution:
Let < l1, m1, n1 > and < l2, m2, n2 > be the d.cs. of the lines.
7x – 4y + 7z + 16 = 0 = 4x + 3y – 2z + 3 … (1)
and x – 3y + 4z + 6 = 0 = x – y + z + 1 … (2)
Then 7l1 – 4m1 + 7n1 = 0
4l1 + 3m1 – 2n1 = 0
Question 14.
Show that the line joining the points (0, 2, -4) and (-1 , 1, -2) and the lines joining the points (-2, 3, 3) and (-3, -2, 1) are co-planar. Find their point of intersection.
Solution:
The eqn. of the line joining the points (0, 2, -4) and (-1, 1, -2) is
The lines (1) and (2) are coplanar if either they are parallel or intersecting. These lines are not parallel. So we have to prove that they are intersecting.
Any point on (1) is (-r1, -r1 + 2, 2r1 – 4)
Any point on (2) is (-r2 – 2, -5r2 + 3, -2r2 + 3)
If two lines are intersecting then for some r1 and r2
-r1 = -r2 – 2 … (1)
-r1 + 2 = -5r2 + 3 … (2)
2r1 – 4 = -2r1 + 3 … (3)
From (3), r1 = r2 + 2
Question 15.
Show that the lines x – mz – a = 0 = y – nz – b and x – m’z’ – a’ = 0 = y – n’z’ – b’ intersect, if (a – a’) (n – n’) = (b – b’) (m – m’).
Solution:
Given lines are
x – mz – a = 0 = y – nz – b … (1)
and x – m’z’ – a’ = 0 = y – n’z’ – b … (2)
Putting z = 0 in (1) we get x = a and y = b.
So (a, b, 0) is a point on (1).
Again putting z = 0 in (2) we get x = a’, y = b’.
So (a’, b’, 0) is a point on (2).
Let < L1, M1, N1 > and < L2, M2, N2 > be the d.cs. of the lines (1) and (2).
Then L1 – mN1 = 0, M1 – nN1 = 0
or, n (a’ – a) + b (m – m’)
= n’ (a’ – a) + b (m – m’)
⇒ (a – a’) (n – n’) = (b – b’) (m – m’) (Proved)
Question 16.
Proved that the line $$\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{1}$$ lies on the plane 7x + 5y + z = 0
Solution:
Given line is
$$\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{1}$$ = r (say) … (1)
Any point on (1) is (2r + 1, -3r – 2, r + 3)
The straight line lies on the plane 7x + 5y + z = 0 … (2)
if every point of the line lies on (2).
Now 7 × (2r + 1) + 5 (-3r -2) + (r + 3)
= 14r + 7 – 15r – 10 + r + 3 = 0
Thus the point (2r + 1, -3r – 2, r + 3) lies on (2).
Hence the straight line lies on the plane. (Proved)
Question 17.
(a) Find the angle between the plane x + y + 4 = 0 and the line $$\frac{x+3}{2}=\frac{y-1}{1}=\frac{z+4}{-2}$$.
Solution:
(b) Find the angle between the plane 4x + 3y + 5z – 1 = 0 and the line $$\frac{x+3}{2}=\frac{y-1}{3}=\frac{z+4}{6}$$.
Solution:
Given plane and the line have equations
4x + 3y + 5z – 1 = 0 … (1)
Question 18.
(a) Find the equation of the line passing through the point (1, 0, -1) and intersecting the lines x = 2y = 2z; 3x + 4y – 1 = 0 = 4x + 5z – 2.
Solution:
Given lines are x = 2y = 2z … (1)
and 3x + 4y – 1 = 0 = 4x + 5z – 2 … (2)
Any plane containing the line (1)
i.e., x – 2y = 0 and y – z = 0 is
x – 2y + k1 (y – z)= 0 = 0
or, x + (k1 – 2) y – k1z = 0
If it passes through the point (1, 0, -1) then 1 + k1 = 0 or, k1 = -1
The plane containing the line (1) and passing through the point (1, 0, -1) is x – 3y + z = 0
Again any plane containing the line (2) is
3x + 4y – 1 + k2 (4x + 5z – 2) = 0
or, (3 + 4k2) x + 4y + 5k2z – (2k2 + 1) = 0
If it passes through the point (1, 0, -1) then
3 + 4k2 – 5k2 – 2k2 – 1 = 0
or, 3k2 = 2 or, k2 = $$\frac{2}{3}$$
The equation of the plane through the line (2) and passing through (1, 0, -1) is
(b) A line with direction ratios < 2, 1, 2 > meets each of the lines x = y + a = z and x + a = 2y = 2z. Find the coordinates of the points of intersection.
Solution:
Given lines are
x = y + a = z and x + a = 2y = 2z
These can be written in symmetrical form as
$$\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}$$ = r1 (say)
and $$\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}$$ = r2 (say)
Any point on (1) is (r1, r1 – a, r1)
Any point on (2) is (2r2 – a, r2, r2)
Suppose that the line meets the lines (1) and (2) at P and Q respectively.
Let P = (r1, r1 – a, r1), Q = (2r2 – a, r2, r2)
D.rs. of PQ are
< r1 – 2r2 + a, r1 – r2 – a, r1 – r2 >
But given that d.rs. are < 2, 1, 2 >.
So $$\frac{r_1-2 r_2+a}{2}=\frac{r_1-r_2-a}{1}=\frac{r_1-r_2}{2}$$
From the 1st two ratios we get
r1 – 2r2 + a = 2r1 – 2r2 – 2a ⇒ r1 = 3a
From the last two ratios we get
2r1 – 2r2 – 2a = r1 – r2
⇒ r1 = r2 + 2a ⇒ 3a = r2 + 2a ⇒ r2 = a
∴ P (3a, 2a, 3a), Q = (a, a, a)
∴ The co-prdinates of the points of intersection and (3a, 2a, 3a) and (a, a, a).
Question 19.
Obtain the co-ordinates of the foot of the perpendicular drawn from the point (3, -1, 11) to the line $$\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$. Obtain the equation of the perpendicular also.
Solution:
Let P be the given point (3, -1, 11).
Draw PM perpendicular from P onto the straight line.
Any point on (1) is (2λ, 3λ + 2, 4λ + 3)
Let M = (2λ, 3λ + 2, 4λ + 3)
D.rs. of PM are 2λ – 2, 3λ + 3, 4λ – 8
As PM is perpendicular to the line then
(2λ – 3) . 2 +3 (3λ + 3) + (4λ – 8) . 4 = 0
or, 4λ – 6 + 9λ + 9 + 16λ – 32 = 0
or, 29λ – 29 = 0 or, λ = 1
∴ The foot of the perpendicular is (2, 5, 7)
Equation of the perpendicular line is
$$\frac{x-3}{2-3}=\frac{y+1}{5+1}=\frac{z-11}{7-11}$$
or, $$\frac{x-3}{1}=\frac{y+1}{-6}=\frac{z-11}{4}$$
Question 20.
Find the perpendicular distance of the point (-1, 3, 9) from the line $$\frac{x-13}{5}=\frac{y+8}{-8}=\frac{z-31}{1}$$.
Solution:
Let P be the point (-1, 3, 9).
Suppose that M is the foot of the perpendicular drawn from P onto the straight line.
$$\frac{x-13}{5}=\frac{y+8}{-8}=\frac{z-31}{1}$$ = λ (say)
Let M = (5λ + 13, -8λ – 8, λ + 31)
D.rs. of PM are
< 5λ + 14, -8λ – 11, λ + 22 >
As PM is perpendicular to the line (1) then
5 (5λ + 14) – 8 (-8λ – 11) + λ + 22 = 0
or, 25λ + 64λ + λ + 70 + 88 + 22 = 0
or, 90λ = -180 or, λ = -2
Thus M = (3, 8, 29)
Distance PM = $$\sqrt{(3+1)^2+(8-3)^2+(29-9)^2}$$
= $$\sqrt{16+25+400}$$ = $$\sqrt{441}$$ = 21
Question 21.
Find the distance of the point (1, -2, 3) from the plane x – y + z = 5, measured parallel to the line $$\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$$.
Solution:
Let P be the point (1, 2, 3). Draw the straight line PM parallel to the line
Question 22.
Find the distance of the point (1, -1, -10) from the line $$\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}$$ measured parallel to the line $$\frac{x+2}{2}=\frac{y-3}{-3}=\frac{z-4}{8}$$.
Ans.
Let P be the point (1, -1, -10). Equation of the line through P parallel to the line
∴ Distance of the point (1, -1, -10) from the given line is √308.
Question 23.
Find the equation of plane through the point (2, 0, -3) and containing the line 3x + y + z – 5 = 0 = x – 2y + 4z + 4
Solution:
Any plane containing the line
3x + y + z – 5 = 0 = x – 2y + 4z + 4
3x + y + z – 5 + k (x – 2y + 4z + 4) = 0
or, (3 + k) x + (1 – 2k) y + (1 + 4k) z + (4k – 5) = 0 … (1)
If this plane contains the point (2, 0, -3) then
2 (3 + k)2 + (1 – 2k) . 0 + (1 + 4k) . (-3) + 4k – 5 = 0
⇒ 6 + 2k – 3 – 12k + 4k – 5 = 0
⇒ -6k – 2 = 0 ⇒ k = –$$\frac{1}{3}$$
Required plane is
$$\left(3-\frac{1}{3}\right) x+\left(1+\frac{2}{3}\right) y+\left(1-\frac{4}{3}\right) z+\left(-\frac{4}{3}-5\right)$$ = 0
⇒ 8x + 5y – z – 19 = 0
Question 24.
Find the equation of the plane containing the line x + 2 = 2y – 1 = 3z and parallel to the line x = 1 – 5y = 2z – 7. Also find the shortest distance between the two lines.
Solution:
Given line is x + 2 = 2y – 1 = 3z
[x – 2y + 3 = 0
2y – 3z – 1 = 0] … (1)
Any plane containing the line (1) is
(x – 2y + 3) + k (2y – 3z – 1) = 0
or, x + (2k – 2) y – 3kz + (3 – k) = 0… (2)
Again given that the plane (2) is parallel to the line
x = 1 – 5y = 2z – 7
⇒ $$\frac{x}{10}=\frac{y-\frac{1}{5}}{-2}=\frac{z-\frac{7}{2}}{5}$$ … (3)
D.rs. of the line (3) are < 10, -2, 5 >
If the plane (2) is parallel to the line (3) then the normal plane (2) is perpendicular to the line (3). D.rs. of the normal of the plane(2) are
< 1, 2k – 2, -3k >.
Thus 10 – 2 (2k – 2) – 15 k = 0
⇒ 10 – 4k + 4 – 15k = 0
We have to find the shortest distance between the lines (3) and (4). The shortest distance is the line segment perpendicular to both the lines. Let < l, m, n > be the d.cs. of the shortest distance.
Then 6l + 3m + 2n = 0
10l – 2m+ 5n = 0
Solving we get
Question 25.
Find the equation of the two planes through the origin and parallel to the line $$\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}$$ and at a distance $$\frac{5}{3}$$ from it.
Solution:
Equation of the plane through the origin is ax + by + cz = 0 … (1)
If the plane (1) is parallel to the line
⇒ 9b2 = 4a2 + 4b2 + 4c2
⇒ 4a2 – 5b2 + 4c2 = 0 … (4)
From (3) we get b = 2a – 3c.
Putting it in (4) we get
4a2 – 5 × 4 (a – c)2 + 4c2 = 0
⇒ a2 – 5a2 – 5c2 + 10ac + c2 = 0
⇒ -4a2 – 4c2 + 10ac = 0
⇒ 2a2 + 2c2 – 5ac = 0
⇒ 2a2 + 2c2 – 4ac – ac = 0
⇒ 2a (a – 2c) – c (a – 2c) = 0
⇒ (a – 2c) (2a – c) = 0
∴ The plane is x – 2y + 2z = 0
Hence the planes are
2x + 2y + z = 0 and x – 2y + 2z = 0.
Question 26.
Find the equation of the straight line perpendicular to the line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{7}$$ and lying in the plane x – 2y + 4z – 51 = 0.
Solution:
Let < l, m, n > be the d.cs. of the straight line perpendicular to the line.
D.rs. of the line is < -6, 1, 2 >.
Again let A be the point of intersection of the line with the line (1).
Let A = (3r + 2, 4r – 1, 7r + 6)
Then this point a lies also on the plane (2).
So 3r + 2 – 8r + 2 + 28r + 24 – 51 =0
or, 23r – 23 = 0 or, r = 1
∴ A = (5, 3, 13)
Hence the equation of the required line is
$$\frac{x-5}{-6}=\frac{y-3}{1}=\frac{z-13}{2}$$
Question 27.
Find the shortest distance between the lines $$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$$ and $$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$$. Find also the equation of the line of shortest distance.
Solution:
Given lines are
Any point on the line (1) and (2) are P (3 + 3α, 8 – α, 3 + α) and Q (-3 – 3β, -7 + 2β, 6 + 4β) respectively.
D.r.s. of PQ are < 6 + 3α + 3β, 15 – α – 4β >
D.r.s. of the lines are < 3, -1, 1 > and < -3, 2, 4 > respectively.
PQ is perpendicular to the given lines.
∴ 3 (6 + 3α + 3β) – (15 – α – 2β) + 1 (-3 + α – 4β) = 0
and -3 (6 + 3α + 3β) + 2 (15 – α – 2β) + 1 (-3 + α – 4β)
⇒ 18 + 9α + 9β – 15 + α + 2β – 3 + α – 4β = 0
and -18 – 9α – 9β + 30 – 2α – 4β – 12 + 4α – 16β = 0
⇒ 11a + 7b = 0
and -7a – 29b = 0
⇒ a = b = 0
co-ordinates P and Q are (3, 8, 3) and (-3, -7, 6) respectively.
The shortest distance
Question 28.
Show that the shortest distance between the lines x + a = 2y = -12z and x = y + 2a = 6z – 6a is 2a.
Solution:
Given lines are
x + a = 2y = -12z
and x = y + 2a – 6z – 6a
Question 29.
Find the length and equation of the line of shortest distance between the lines 3x – 9y + 5z = 0 = x + y – z and 6x + 8y + 3z – 13 = 0 = x + 2y + z – 3
Solution:
Given lines are
3x – 9y + 5z = 0
x + y – z = 0 … (1)
and 6x + 8y + 3z – 13 = 0
x + 2y + z – 3 = 0 … (2)
Let us consider the line (1)
Now z = x + y
∴ 3x – 9y + 5 ( x + y) = 0
⇒ 8x – 4y = 0 ⇒ 2x = y
Again y = z – x
∴ 3x – 9 (z – x) + 5z = 0
⇒ 12x – 4z = 0
⇒ 3x = z
∴ 6x = 3y = 2z
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# How Do You Describe A Sequence?
## How do you explain number sequences?
What is a number sequence?A number sequence is a list of numbers that are linked by a rule.
If you work out the rule, you can work out the next numbers in the sequence.In this example, the difference between each number is 6.
So the rule for this sequence is to add 6 each time.Now you can work out the next number in the sequence: 27 + 6 = 33..
## What are the 4 types of sequence?
Types of Sequence and SeriesArithmetic Sequences.Geometric Sequences.Harmonic Sequences.Fibonacci Numbers.
## How do you teach numbers in order?
Top tips for teaching number sequencesTeach them rhymes and games.Incorporate numbers into daily tasks.Patterns don’t have to be numbers.
## How will you identify a sequence?
An arithmetic series is one where each term is equal the one before it plus some number. For example: 5, 10, 15, 20, … Each term in this sequence equals the term before it with 5 added on. In contrast, a geometric sequence is one where each term equals the one before it multiplied by a certain value.
## What are the first 10 Lucas numbers?
0, 2, 4, 5, 7, 8, 11, 13, 16, 17, 19, 31, 37, 41, 47, 53, 61, 71, 79, 113, 313, 353, 503, 613, 617, 863, 1097, 1361, 4787, 4793, 5851, 7741, 8467, … (sequence A001606 in the OEIS).
## How do you describe a sequence in math?
A sequence is simply an ordered list of numbers. For example, here is a sequence: 0, 1, 2, 3, 4, 5, …. This is different from the set N because, while the sequence is a complete list of every element in the set of natural numbers, in the sequence we very much care what order the numbers come in.
## How do you describe a number pattern?
Number pattern is a pattern or sequence in a series of numbers. This pattern generally establishes a common relationship between all numbers. For example: 0, 5, 10, 15, 20, 25, …
## What is the rule for the pattern of numbers?
A numbers pattern is a sequence of numbers that grows or repeats according to a specific rule. For example, the following number pattern starts at 2 and follows the rule add 3: 2, 5, 8, 11, 14….and so forth.
## What is the general term of a sequence?
General Term An arithmetic sequence is a linear function. Instead of y=mx+b, we write an=dn+c where d is the common difference and c is a constant (not the first term of the sequence, however). A recursive definition, since each term is found by adding the common difference to the previous term is ak+1=ak+d.
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Multiplication and Division Expressions
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Objective
SWBAT find a rule and write multiplication and division expressions from a completed table.
Big Idea
From a completed table, students can study to find the relationship between numbers, then write an expression for the table.
Opener
5 minutes
In today's lesson, the students model multiplication and division patterns shown in tables, determine the rule and write an expression. This aligns with 4.OA.A3 because the students are representing these problems using equations with a letter standing for the unknown quantity.
To get the students started, I remind students that we have learned how to evaluate expressions. "What does it mean to evaluate an expression?" I let the students think about that question for a few minutes. I call on a student to respond. One student replies, "It is when you work the problem." I add on to the student response by reminding the students that when you evaluate an expression, you replace the variable with a number, then solve. I let the students know that in todays lesson, we determine a rule from a completed table, then write an expression. Also, the students will make a model by using two-color counters. The model provides a visual of the expression, as well as helps the students count to get the correct answer.
Whole Class Discussion
10 minutes
I call the students to the carpet as we prepare for a whole class discussion. The Multiplication and Division Expressions power point is already up on the Smart board. I like for my students to be near so that I can have their full attention while I'm at the Smart board.
At the beginning of each lesson, I like to review all relevant skills that we have learned that will help with the new skill. I find that this helps the lesson run smoothly by having the important information in the forefront of the students' minds.
I begin by reviewing the vocabulary.
Review:
Variables – A symbol that stands for a number.
Algebraic expression – A mathematical phrase containing numbers or variables and at least one operation.
We begin with the multiplication expression. A completed table is used for an example to help the students write an expression.
What is the rule for the table?
n 1 2 3 4 n x 8 16 24 32
Notice that the numbers are increasing. Since the numbers are increasing, we need to multiply. How much is it increasing?
I like for my students to interact with me during our whole group discussion. I like to ask questions of them to make sure they are understanding the skill.
I feel that all answers should not be given to students. During whole class discussion and group activity, students should have to "think" to come up with some information. I feel that when they discover some things on their own or use skills that they have learned, then this information stays with them.
Yes, it is increasing by 8.
1 x 8=8
2 x 8 =16
3 x 8 = 24
4 x 8 = 32
The expression for the table is n x 8.
How can we model this?
We can use counters to Model of Multiplication and Division Expressions.
Next, we worked an example of a division expression.
What is the rule for the table?
r 45 50 55 60 r ÷____ 9 10 11 12
In my class, we discussed that if a number is decreasing, it is either subtraction or division. Because the students are familiar with their multiplication facts, when they saw 45 and 9, they knew that 9 x 5 is 45. Therefore, because of fact families, they knew that 45 divided by 5 would give them 9.
Yes, the rule is divide by 5.
45 ÷ 5 = 9
50 ÷ 5= 10
55 ÷ 5 = 11
60 ÷ 5 = 12
Now, you will work with your classmates to practice the skill.
Group or Partner Activity
20 minutes
Give the students practice on this skill by letting them work together. I find that collaborative learning is vital to the success of students. Students learn from each other by justifying their answers and critiquing the reasoning of others (MP3).
For this activity, put the students in groups of 3. I give each group a Group Activity Sheet Multiplication and Division Expressions and counters. The students must work together to find the rule to completed tables. This is evident in their completed Student Work. They must communicate precisely to others within their groups (MP6). They must use clear definitions and terminology as they precisely discuss this problem (MP1).
The students are guided to the conceptual understanding through questioning by their classmates, as well as by me. The students communicate with each other and must agree upon the answer to the problem. Because the students must agree upon the answer, this will take discussion, critiquing, and justifying of answers by the students (MP3). As the groups discuss the problem, they must be precise in their communication within their groups using the appropriate math terminology for this skill (MP6). As I walk around, I am listening for the students to use "talk" that will lead to the answer. I am holding the students accountable for their own learning.
As they work, I monitor and assess their progression of understanding through questioning.
1. What is the value of each number?
2. Did you work all of the problems in the table?
3. What result did you get for each of the problems?
As I walk around the classroom, I hear the students communicate with each other about the assignment. From the Video Multiplication and Division Expressions, you can hear the classroom chatter and constant discussion among the students, as well as the student justifying her answer to the problem. Before Common Core, I thought that a quiet class working out of the book was the ideal class. Now, I am amazed at some of the conversation going on in the classroom between the students. As I walk around the room, I hear students justify their answers. This is evident in the video. They cannot just tell me what their answer is, there should always be a "because" or "I got my answer by..." I always tell my students that they must justify their answer by referring back to the problem. Also, the students use skills that were learned previously. The Student Work - Relating Multiplication and Division shows how the students related multiplication and division in their work. Even though the activity did not require it, the students knew the relationship and applied it. This is what I want of my students. I want them to take ownership of their learning and apply skills previously learned in their work.
In the Student Work, you can see how the students drew a table to help solve the problem. In the table, the students replaced "s" with the appropriate number. The students used their understanding and knowledge of multiplication to complete the table. In this example, the students knew that "s" was multiplied by 4 in order to get the numbers listed in the table.
Any groups that finish the assignment early, can go to the computer to practice the skill at the following site until we are ready for the whole group sharing.
Closure
15 minutes
To close the lesson, I have one or two students share their answers. This gives those students who still do not understand another opportunity to learn it. I like to use my document camera to show the students' work during this time. Some students do not understand what is being said, but understand clearly when the work is put up for them to see.
I feel that by closing each of my lessons by having students share their work is very important to the success of the lesson. Students need to see good work samples, as well as work that may have incorrect information. More than one student may have had the same misconception. During the closing of the lesson, all misconceptions that were spotted during student independent and partner sharing will be addressed whole class.
Misconception(s) for this lesson:
The only problem that I had with this lesson was some of the students did not work all of the problems in the table. After they worked one or two problems and got the same answer, they went to the next problem. I explained to the students that it is always best to work all of the problems in a table because as the lessons progress to higher levels, the rule may not be the same. For example, in patterns, the students may start out by adding 1 to the first number, then 2 to the second number, and so on. My goal is to get the students to work efficiently and accurately. I need to put emphasis on using the correct approach to problem-solve in 4th grade so that they will be successful in 5th grade.
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Proof by induction and proof by contradiction
This is the translation of an older post in French: Raisonnement par récurrence et raisonnement par l’absurde.
Let’s have a pretty short post so that I can refer to it later – I have another monster-post in the works, and I realized I needed the following elements to make it work, so here’s another post! I’m going to talk about two fundamental tools in the “proof toolbox”: the proof by induction and the proof by contradiction. These are tools that are explained in high school (well, they were in my French high school 20+ years ago 😉 ), and that you’ll see everywhere all the time, so let’s explain how it works.
Proof by induction
The idea of the induction proof is dominoes. Not the ones with the dots on them, the ones that you topple. You know that if you topple a domino, it falls. And if a domino falls, the next one falls as well. And you topple the first domino. Hence, all the dominoes fall.
Proofs by induction work in the same way. Suppose you have a property that depends on an natural number (positive integer), and you want to prove that it’s true for any natural number. First, you show that it’s true for a small natural number, say 0, 1 or 2 (sometimes the property doesn’t make much sense for 0 or 1). Then, you show that, if it’s true for a natural number $k$, it’s also true for $k+1$. So you have the “start” of the dominoes, the toppling (“it’s true for 0”), and the “chain” of dominoes (“if it’s true for $k$, it’s true for $k+1$“). If these two conditions are true, all the dominoes fall (“the property is true for all natural numbers greater than 0”).
Now this is where I’m a bit annoyed, because I have a an example that works, but the induction proof kind of sucks compared to another, which is much prettier. So I’m going to show both 😉
Suppose that I want to prove that the sum of all integers from 1 to $n$ (that is to say, $1 + 2 + 3 + 4 + ... + (n-2) + (n-1) + n$) equals $\displaystyle \frac{n(n+1)}{2}$. I start with $n=1$: $\displaystyle \frac{1(1+1)}{2} = \frac 2 2 = 1$, so the base case is true.
Now, I suppose that it’s true for an arbitrary natural number $k$ that the sum of the integers from 1 to $k$ is equal to $\displaystyle \frac{k(k+1)}{2}$, and I compute the sum of integers from 1 to $k+1$: $1 + 2 + 3 + ... + (k-1) + k + (k+1)$. This is equal to the sum of the integers from 1 to $k$, plus $k+1$. By induction hypothesis, this is equal to $\displaystyle \frac{k(k+1)}{2} + k+1 = \frac{k^2 + k + 2k + 2}{2} = \frac{k^2 + 3k + 2}{2}$. For my proof to work out, I want the sum of integers from 1 to $k+1$ to be equal to $\displaystyle \frac{(k+1)(k+2)}{2}$. And it so happens that $(k+1)(k+2)$ is precisely equal to $k^2 + 3k + 2$.
So I have my base case, my induction step, and I proved exactly what I wanted to prove.
Now for the alternative, prettier proof. You consider a table with two lines and $n$ columns:
1 2 3 4 ... n-2 n-1 nn n-1 n-2 n-3 3 2 1
and you compute the sum of all the numbers in the table. You can add up, for each column, both lines. Every column sums to $n+1$: it’s the case for the first column, and at each step, you remove 1 from the first line, and you add 1 to the second line, so the sum stays the same (there’s actually a “hidden” induction proof in there!) There are $n$ columns, so if I add all these results, it sums to $n(n+1)$. But if I group the sum in a different way, the first line is equal to the sum of integers from 1 to $n$, the second line as well… so $n(n+1)$ is two times the sum of the integers from 1 to $n$, which concludes the proof.
Proofs by contradiction are sometimes dangerous, because they’re often misused or overused. I know some mathematicians who argue that a proof by contradiction is not very elegant, and that when one arises, it’s usually worth it to make the extra effort to try to turn it around in a non-contradiction proof. I still like the reasoning behind it, and it sometimes makes life much easier.
We want to prove that a proposition A is true. To prove A by contradiction, you make the hypothesis that A is false, you unroll a series of consequences that would be implied by the fact that A is false, and you arrive at something that you know is impossible. And if the fact that A is false implies an impossibility, it means that A is true, otherwise the universe collapses and it’s messy.
My favorite example is to prove that $\sqrt 2$ is irrational, that is to say that you can’t write it as $\displaystyle \frac p q$ where $p$ and $q$ are integers.
I need a tiny preliminary lemma: I’m claiming that if an integer $n$ is even, then its square $n^2$ is even, and that if $n^2$ is even, then $n$ is even. If $n$ is even, I can write $n = 2k$ (with $k$ integer). Then $n^2 = 4k^2 = 2 \times 2k^2$, so $n^2$ is even. If $n$ is odd, I can write $n = 2k+1$, so $n^2 = 4k^2 + 2k + 1 = 2(2k^2 + k) + 1$, so $n^2$ is odd. So, if $n^2$ is even, then it can’t be that $n$ would be odd (because otherwise $n^2$ would be odd as well), so $n$ is even.
Now back to the irrationality of $\sqrt{2}$. We make the hypothesis that we can write $\displaystyle \sqrt 2 = \frac p q$. We can also make the assumption that the fraction is irreducible, because if it’s not, you can reduce it so that it is, so let’s assume that it’s the one we took from the beginning. (Note: a fraction $\displaystyle \frac p q$ is irreducible if there is no integer $k \geq 2$ such that $p$ and $q$ are both divisible by $k$. If $k$ exists, we divide $p$ and $q$ by $k$, and we get the fraction $\displaystyle \frac{p/k}{q/k}$).
So: I can write that $\sqrt 2 \times q = p$. If I square this equality, I get $2q^2 = p^2$. Since $q$ is an integer, $q^2$ is an integer, so $p^2$ is an even number (because it’s equal to twice an integer). But then, if $p^2$ is even, then $p$ is even as well, according to my preliminary lemma. So I can write $p = 2k$ and consequently $p^2 = 4k^2$. But then, since $2q^2 = p^2 = 4k^2$, I can also write $q^2 = 2k^2$, so $q^2$ is even, so $q$ is even as well. But that’s not possible: $\displaystyle \frac p q$ is irreducible, so $p$ and $q$ cannot be both even! So something is wrong in my reasoning, and the only thing that can be wrong is my initial hypothesis, that is $\sqrt{2}$ is rational. Hence, $\sqrt{2}$ is irrational. Cute, isn’t it?
Planarity, minors and donuts
Ce billet est la traduction d’un billet écrit initialement en français : TPA – Planarité, mineurs et donuts.
Let’s have a little bit of graph theory today – I want to present (without a proof) a fun result about graph planarity. Which means – first, let’s define a few things and draw a few pictures.
A graph is a set of vertices (“points”) connected by edges. To describe a graph, I can say “the set of vertices is labeled 1, 2, 3, 4 and 5, and I have edges between edges 1 and 2, 2 and 3, 3 and 4, 4 and 1, 1 and 3, 2 and 4, 1 and 5, and 4 and 5”. I can also draw it on a sheet of paper, and I can draw that statement in different ways: all the drawings here are drawings of the graph I just described.
In these three drawings, two of them are plane: the graph is drawn such that edges of the graph do not cross. The first drawing is not plane: the edges 1-3 and 2-4 are crossing. A graph that can be drawn in the plane (that is to say, on a flat piece of paper) such that no two edges cross is called a planar graph. The fact that we have an adjective for such a graph should tell you that there exists non-planar graphs: graphs that cannot be drawn in the plane without crossing two edges. Two typical (and useful) examples are the complete graph on 5 vertices (5 vertices, and all possible edges) – denoted $K_5$, and the complete bipartite graph on 3×2 vertices (2 groups $A$ and $B$ of three vertices, and all possible edges between vertices of group $A$ and vertices of group $B$), denoted $K_{3,3}$. Here are their “usual” drawings; as we saw in the previous figure, these drawings are not unique. But you can search for a long long time for a drawing where no two edges cross (and not find it).
And there is a theorem (Wagner, 1937) that says that if a graph is not planar, it’s because somewhere in its structure it contains something that looks like $K_5$ or $K_{3,3}$. More precisely:
A finite graph is planar if and only if it does not have $K_5$ or latex $K_{3,3}$ as a minor.
Okay, I cheated a bit, because I haven’t defined minor yet. So let’s do that:
A minor of a graph $G$ is a graph that is obtained from $G$ by doing any number of the following operations: removing an edge; removing a vertex; contracting a vertex.
Removing an edge is easy: if two vertices are connected by an edge, we can decide that in fact they aren’t, and remove the edge. Removing a vertex is also easy: pick a vertex, remove it, and remove all the edges that are connected to it, because otherwise we don’t know where they’re going anyway. Contracting a vertex is a tiny bit more complicated: the idea is that we pick two vertices that are connected with an edge, and we “merge” them in a single vertex. The resulting vertex is connected to all the edges of the initial two vertices that were in the original graph. One can imagine “pinching” the two vertices that get closer and closer until BOOM they fuse together. For a small example, in which I contract the vertices connected by the red edge in a single vertex:
So what Wagner says is: “if I fiddle with a graph a bit and I manage to obtain $K_5$ or $K_{3,3}$, then I cannot draw the graph without crossing edges. But if I cannot get $K_5$ or $K_{3,3}$, then I can draw the graph without crossing.”
There’s a theorem that expands this idea of “forbidden minors”. It’s a theorem by Robertson and Seymour, and its proof spans over 20 papers, published from 1984 to 2003. The theorem is stated as follows:
Every family of graphs that is closed under minors can be defined by a finite set of forbidden minors.
Let’s explain. A family of graphs that is closed under minors is a set of graphs such that, if I take any graph in this family, and I take a minor of it (with the aforementioned definition), then the minor is also in that set of graphs. What Robertson and Seymour say is that, in such a family, there exists a finite set of forbidden minors: if a graph has a minor of that set, then that graph is not part of the family.
Wagner’s theorem is a restriction of Robertson-Seymour for planar graphs. Planar graphs are a family of graphs that is closed under minors: if I take a minor of a planar graph, I can still draw that minor without crossing any edge, so it is planar as well. And there is a finite set of forbidden minors, specifically $K_5$ or $K_{3,3}$: if I find one of these minors in a graph, then the graph is not planar. Wagner, for the case of planar graphs, actually states the forbidden minors explicitly, which Robertson-Seymour does not.
And that’s the fun part: in general, we do not know the forbidden minors in question. We know they exist; we know there’s a finite number of them, but we do not know which they are. And, for a tiny example as a conclusion of this post: suppose now that I want to draw a graph not on a sheet of paper, but on a torus – or, if you want, on a donut.
The idea would be to draw vertices on a donut, and to connect them with edges, exactly as I do in the plane. If I do that, and I can draw the graph without crossing edges, the graph is not planar anymore, but toroidal (or donutoidal if you want). The family of toroidal graphs is closed for minors, so there exists a set of forbidden minors for this family. So far, 17523 have been found. It is yet unknown how many there are in total. Fun, isn’t it?
(And if you want to find them all, you may want to start by making donuts.)
Intro to probability theory – part 2
Ce billet a été traduit de sa version originale en français : Probabilités – Partie 2.
After the previous post about probability theory, here’s the second part, in which I’ll talk about random variables.
The idea of random variables is to have some way of dealing with events for which we do not exactly what happens (for instance, we roll a die), but still want to have some idea of what can happen. The die example is pretty simple, so using random variables may be a bit overkill, but let’s keep examples simple for now.
For a given experiment, we consider a variable, called $X$, and look at all the values it can reach with the associated probability. If my experiment is “rolling a die and looking at its value”, I can define a random variable on the value of a 6-sided die and call it $X$. For a full definition of $X$, I need to provide all the possible values of $X$ (what we call the random variable’s domain) and their associated probabilities. For a 6-sided die, the values are the numbers from 1 to 6; for a non-loaded die, the probabilities are all equal to $\displaystyle \frac 1 6$. We can write that as follows:
$\displaystyle \forall i \in \{1,2,3,4,5,6\}, \Pr[X = i] = \frac 1 6$
and read “for all $i$ in the set of values {1,2,3,4,5,6}, the probability that $X$ takes the value $i$ equals $\displaystyle \frac 1 6$.
One of the basic ways to have an idea about the behaviour of a random variable is to look at its expectation. The expectation of a random variable can be seen as its average value, or as “suppose I roll my die 10000 times, and I average all the results (summing all the results and dividing by 10000), what result would I typically get?”
This expectation (written $E[X]$) can be computed with the following formula:
$E[X] = \displaystyle \sum_{i \in \text{dom}(X)} \Pr[x = i] \times i$
which can be read as “sum for all elements $i$ in the domain of $X$ of the probability that $X$ takes the value $i$, times $i$. In the die example, since the domain is all the integer numbers from 1 to 6, I can write
$\displaystyle \sum_{i=1}^6 \Pr[X = i] \times i$
which I can in turn expand as follows:
\begin{aligned}E[X] &=& 1 \times \Pr[X = 1] + 2 \times \Pr[X = 2] + 3 \times \Pr[X = 3] \\ &+& 4 \times \Pr[X = 4] + 5 \times \Pr[X = 5] + 6 \times \Pr[X = 6]\end{aligned}
Since, for my die, all the probabilities are equal to $\displaystyle \frac 1 6$, I can conclude with
$\displaystyle E[X] = \frac 1 6 \times (1 + 2+3+4+5+6) = \frac{21}{6} = 3.5$
So the average value of a die over a large number of experiments is 3.5, as most tabletop gamers would know 😉
Now let’s look at a slightly more complicated example. Suppose that I have $n$ dice, and that I want to know how many 6s I can expect in my $n$ dice. From a handwavy point of view, we know that we will not get an exact answer for every time we roll $n$ dice, but that we can get a rough answer. There’s no reason there should be more or less 6s than 1s, 2s, 3s, 4s or 5s, so generally speaking the dice should be distributed approximately equally in the 6 numbers, so there should be approximately $\displaystyle \frac n 6$ 6s over $n$ dice. (The exception to that being me playing Orks in Warhammer 40k, in which case the expected number is approximately 3 6s over 140 dice.) Let us prove that intuition properly.
I define $Y$ as the random variable representing the number of 6s over $n$ dice. The domain of $Y$ is all the numbers from 0 to $n$. It’s possible to compute the probability to have, for example, exactly 3 6s over $n$ dice, and even to get a general formula for $k$ dice, but I’m way too lazy to compute all that and sum over $n$ and so on. So let’s be clever.
There’s a very neat trick called linearity of expectation that says that the expectation of the sum of several random variables is equal to the sum of the expectations of said random variables, which we write
$E[A + B] = E[A] + E[B]$
This is true for all random variables $A$ and $B$. Beware, though: it’s only true in general for the addition. We cannot say in general that $E[A \times B] = E[A] \times E[B]$: that’s in particular true if the variables are independent, but it’s not true in general.
Now we’re going to define $n$ variables, called $Y_1, Y_2, ..., Y_n$ so that $Y$ is the sum of all these variables. We can define, for each variable $Y_1$, the domain {0,1}, and we say that $Y_1$ is equal to 1 if and only if the die number 1 shows a 6. The other variables $Y_i$ are defined similarly, one for each die. Since I have $n$ variables, which take value 1 when their associated die shows a 6, I can write
$\displaystyle Y = \sum_{i = 1}^n Y_i$
This is where I use linearity of expectation:
$\displaystyle E[Y] = E\left[\sum_{i=1}^n Y_i\right] = \sum_{i=1}^n E[Y_i]$
The main trick here is that variables $Y_i$ are much simpler to deal with than $Y$. With probability $\displaystyle \frac 1 6$, they take value 1; with probability $\displaystyle \frac 5 6$, they take value 0. Consequently, the expectation of $Y_i$ is also much easier to compute:
$\displaystyle E[Y_i] = 1 \times \frac 1 6 + 0 \times \frac 5 6 = \frac 1 6$
Plugging that in the previous result, we get the expectation of $Y$:
$\displaystyle E[Y] = E\left[\sum_{i=1}^n Y_i\right] = \sum_{i=1}^n E[Y_i] = \sum_{i=1}^n \frac 1 6 = \frac n 6$
which is the result we expected.
Now my examples are pretty simple. But we can use that kind of tools in much more complicated situations. And there’s a fair amount of other tools that allow to estimate things around random variables, and to have a fairly good idea of what’s happening… even if we involve dice in the process.
Intro to probability theory – part 1
Ce billet est la traduction d’un billet écrit en français ici : Probabilités – partie 1
I’m eventually going to talk a bit about random algorithms, but before I do that, I need to talk about probability theory so that everyone is speaking the same language.
Probability theory is a bit annoying, because probability is something that many people use fairly often in their daily life, but it can be awfully counter-intuitive. That, or human beings suck at probability theory – also a possibility.
I kind of don’t want to formally define probability (“given a universe $\Omega$, events”, blah blah blah… yeah, no, don’t want to do that), I’m going to try to explain things in a handwavy way, and hope for the best.
First, let me assume that everything I’m talking about here – unless mentioned otherwise explicitly – has “usual” properties: dice are not loaded, coins are real coins, etc.
Let us start with a classical example. If I flip a coin, what is the probability that it shows heads when it lands? The answer is 1/2: I have two possible events (the coin can show heads or tails), and they both have the same probability of occurring. Same thing if I roll a 6-sided die: the probability that it shows 4 is 1/6; I have six possible events that all have the same probability of occurring. In general, I’ll say that if I do an experiment (flipping a coin, rolling a die) that has $k$ possible outcomes, and that all of these outcomes have the same probability (“chance of happening”), then the probability of each of these outcomes (“events”) is $\displaystyle \frac 1 k$.
It can happen that all the possible outcomes don’t have the same probability, but there are a few non-negotiable rules. A probability is always a number between 0 and 1. An event that never occurs has probability 0; an event that always occurs has probability 1. If I find a coin with two “heads” sides, and I flip it, it shows “heads” with probability 1 and “tails” with probability 0. Moreover, the sum of all the probabilities of all the possible outcomes of a given experience sums to 1. If I have $k$ events that have the same probability, we indeed have $\displaystyle k \times \frac 1 k = 1$. If I have a die with three sides 1, two sides 2 and one side 3, the probability that it shows 1 is $\displaystyle \frac 3 6$, the probability that it shows 2 is $\displaystyle \frac 2 6$, and the probability that it shows 3 is $\displaystyle \frac 1 6$; the sum of all these is $\displaystyle \frac 3 6 + \frac 2 6 + \frac 1 6 = 1$.
Now it gets a bit trickier. What is the probability that a (regular) 6-sided die shows 3 or 5? Well, the probability that it shows 3 is $\displaystyle \frac 1 6$; the probability that it shows 5 is $\displaystyle \frac 1 6$, and the probability that it shows 3 or 5 is $\displaystyle \frac 1 6 + \frac 1 6 = \frac 1 3$. But summing the individual probabilities only works if the events are exclusive, that is to say they cannot both be true at the same time: if I rolled a 5, then I can’t have a 3, and vice-versa.
It does not work if two events can both occur at the same time (non-exclusively). For instance, consider that flip a coin and roll a die, and I look at the probability that the coin lands on tails or that the die lands on 6. Then I CANNOT say that it’s equal to the probability of the coin landing on tails – $\displaystyle \frac 1 2$ -, plus the probability that the die lands on 6 – $\displaystyle \frac 1 6$ – for a total of $\displaystyle \frac 2 3$. A way to see that is to tweak the experiment a bit to see that it doesn’t work all the time. For example, the probability that the die lands on 1,2, 3 or 4 is $\displaystyle \frac 4 6 = \frac 2 3$. The probability that the coin lands on tails is $\displaystyle \frac 1 2$. And it’s not possible that the combined probability of these two events (the die lands on 1, 2, 3 or 4, or the coin lands on tails) is the sum of both probabilities, because that would yield $\displaystyle \frac 2 3 + \frac 1 2 = \frac 7 6$, which is greater than 1. (And a probability is always less or equal to 1).
What is true, though, is that if I have events $A$ and $B$, then
$\Pr(A \cup B) \leq \Pr(A) + \Pr(B)$
where $\Pr(A)$ is “probability that $A$ occurs”, $\Pr(B)$ is “probability that $B$ occurs” and $\Pr(A \cup B)$ is “probability that $A$ occurs or that $B$ occurs”. Or, in a full sentence: the probability of the union of two events (event $A$ occurs or event $B$ occurs) is always less than or equal to the sum of the probabilities of both events. You can extend this to more events: the probability of the union of several events is less than or equal to the sum of the probabilities of the individual events. It may not seem very deep a result, but in practice it’s used quite often under the name “union bound”. The union bound does not necessarily yield interesting results. For the case “the die lands on 1, 2, 3 or 4, or the coin lands on tails”, it yields a bound at $\displaystyle \frac 7 6$… and we already had a better bound by the fact that the probability is always less than 1. It is slightly more useful to bound the probability that the die lands on a 6 or that the coin lands on tails: we know that the probability is less than $\displaystyle \frac 2 3$. When analyzing random algorithms, the union bound is often pretty useful, because the probabilities that we consider are often pretty small, and we can add a lot of them before the bound stops making sense. The union bound is often not enough to make a useful conclusion, but it’s very often a useful tool as a step of a proof.
For the example that I’ve been using so far, the best tool (because it gives an exact result) is the inclusion-exclusion principle. For two events $A$ and $B$, it’s stated as follows:
$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$
with the same notation as previously, and $\Pr(A \cap B)$ representing the probability that both events $A$ and $B$ occur. In a sentence: the probability that $A$ or $B$ occur is equals to the sum of the probabilities that either of them occurs, minus the probability that both $A$ and $B$ occur at the same time. The idea is that, if two events can occur at the same time, that probability is counted twice when summing the probabilities. It may be clearer with a small drawing (also known as Venn diagram):
If I count everything that is in the green area (corresponding to event $A$), and everything that is in the pink area (corresponding to event $B$), I count twice what is both in the green and pink area; if I want to have the correct result, I need to remove once what is common to both areas.
Now the question becomes – how to compute the probability of two events arriving at the same time? There’s the easy case, and the not so easy case. In the easy case, the events are independent: the probability that one occurs has no impact on the probability that the other one occurs. It’s is true (although not necessarily intuitive) if one is working with dice and coins and considering their individual results, but it’s a notion that requires a lot of care to apply (it’s not true anymore if I’m interested in the individual results and the sum of two dice, for example). Proving that two events are independent can be pretty complicated, and handling computations when they are not… as well.
When two events are independent, we have:
$\Pr(A \cap B) = \Pr(A) \times \Pr(B)$
that is to say that the probability that both events occur is equal to the product of the probabilities of both events. If I’m flipping a coin and rolling a die, the coin landing on tails has no impact on the die landing on 6. Both events are independent; the probability that both events occur is consequently $\displaystyle \frac 1 2 \times \frac 1 6 = \frac{1}{12}$. Observe that this probability is smaller than $\displaystyle \frac 1 2$ and smaller than $\displaystyle \frac 1 6$. It’s “obvious” since a probability is smaller than one, and that consequently multiplying two probabilities together yields a result that is smaller than both. Another way to see that is the fact that the probability that two independent event occur at the same time is less probable than the probability than only one of them occurs.
An example of two events that are not independent would be event $A$ “the die lands on 1” and event $B$ “the die lands on an odd number”. In that case, the events are not independent: if the die lands on 1, then it landed on an odd number; if the die lands on 2, it cannot land on an odd number as well. In that specific case, event $A$ is included in event $B$, so it’s easy: the probability that both events occur is equal to the probability of the included event: $A \cap B$ is the same thing as $A$. For a subtler example, one can define $A$ as “the die lands on 1, 4 or 6” and keep $B$ as “the die lands on an odd number”. These events are not independent either. The probability of $A$ and $B$ both occurring is $\displaystyle \frac 1 6$, corresponding to the die landing on 1 (it’s the only case that’s both in 1, 4, 6 and odd). Multiplying the probabilities of A – $\displaystyle \frac 1 2$ – and B – $\displaystyle \frac 1 2$ – without paying attention yields $\displaystyle \frac 1 4$ and that’s how we get a mistake in our proof.
Last one for the road: I still want to talk about conditional probabilities. Conditional probabilities are a way to handle dependencies between events. We write $\Pr(A \mid B)$, say “probability of $A$ given $B$“, and understand “probability of $A$ knowing that $B$ occurs”. If $A$ and $B$ are independent, then $\Pr(A \mid B) = \Pr(A)$: the fact that $B$ occurs does not have any impact on the occurrence of $A$. For the previous case, where $A$ is the event “the die lands on 1” and $B$ is the event “the die lands on an odd number”, we can see the “given $B$” clause as a restriction of the possible events. The die landed on an odd number, it is known – so now, we know that it has the same probability to have landed on 1, 3 or 5, but a probability of 0 to have landed on 2, 4 or 6. Consequently, we have $\displaystyle \Pr(A \mid B) = \frac 1 3$.
There’s a general formula for conditional probabilities:
$\displaystyle \Pr(A \mid B) = \frac{\Pr(A \cap B)}{\Pr(B)}$
It’s possible to show, from this formula, the fact that if A and B are independent, then $\Pr(A \mid B) = \Pr(A)$, because then:
$\displaystyle \Pr(A \mid B) = \frac{\Pr(A) \times \Pr(B)}{\Pr(B)} = \Pr(A)$
That formula is also pretty useful in the other direction:
$\displaystyle \Pr(A \cap B) = \Pr(A \mid B) \times \Pr(B) = \Pr(B \mid A) \times \Pr(A)$
because it’s sometimes easier to understand what happens in the context of conditional probabilities than in the context of two non-independent events occurring.
In the next blog post, we’ll talk about random variables 🙂
NP-complete problems
In a previous post, I tried to explain the P vs NP problem. I gave a rough definition of the NP complexity class: NP is the class of the problems for which, for each “yes” instance, I can get convinced in polynomial time that the answer is yes by providing me with a proof that it is indeed the case. And, at the end of that post, I vaguely mentioned (without naming it) the notion of NP-complete problems.
The usual definition of a NP-complete problem is that it’s a problem which belongs to NP, and which is NP-hard. So now I need to explain what NP-hard means. Informally, it’s a problem that is at least as hard as all other problems of the NP class. Another informal way to say that is that it’s at least as hard as another NP-hard problem. The problem with that definition is the bootstrap problem: you’d need a first NP-hard problem to compare it to the others.
The archetypal NP-hard problem is SAT, which was the topic of my previous post. Since SAT is also in NP, SAT is also the archetypal NP-complete problem. It’s not a priori obvious that SAT is NP-hard. But one can prove that even 3-SAT is NP-hard, where 3-SAT represents the CNF instances of SAT where all clauses have exactly 3 literals. I’m not going to go through the proof, because that would imply that I’d explain a lot of things that I kind of don’t want to explain. For the people who are interested in the topic, it’s been proven in 1971, and it is known as the Cook-Levin theorem.
Now that we do have a NP-hard problem, we can find other ones by reduction. We reduce a NP-hard problem (for instance, 3-SAT), to another problem that we want to prove is NP-hard (call that problem A). To do that, we show that if we can solve A efficiently (that is to say, in polynomial time), then we can solve 3-SAT efficiently.
The idea of that kind of proof is to take an arbitrary instance of the NP-hard problem, and we transform it, in polynomial time, into an instance of A that allows to conclude on the original instance. Now suppose that we can solve problem A (that is to say, any instance of problem A) in polynomial time. Then we can take any instance of SAT, transform it into an instance of problem A, solve the instance of problem A, and get a conclusion. The transformation is done in polynomial time; the solving of A is in polynomial time; summing both yields again a polynomial, so we can solve any instance of SAT in polynomial time, so we can solve SAT in polynomial time. Easy.
NP-complete problems are, in a way, “complexity equivalent”: if we know how to solve one in polynomial time, then we can solve all of them in polynomial time as well, and we can solve all problems from NP in polynomial time (since NP-hard problems are at least as hard as any problem from NP). So, if we find a way to solve in polynomial time all the instances of any NP-complete problem… we proved that P = NP. And won a million dollars. And surprised a whole lot of people.
There is a large set of problems that are known to be NP-hard, or NP-complete if they are as well in NP. And there are people who look at… exotic problems, shall we say: let me give a few links for the people who are interested and may get a laugh out of it:
After this fun interlude, I’m going to give an example of reduction, so that you see that kind of proof works. I’m going to prove that the CLIQUE problem is NP-hard. It’s pretty much the basic example that everybody gives when they’re teaching that kind of things, but there must be a reason (I guess: it’s fairly easy to explain, and the reduction is also fairly easy compared to some others). What I’m explaining here is largely coming from the CLRS, that is to say the Cormen, Leiserson, Rivest and Stein, that is to say Introduction to Algorithms – that’s the reference book for A LOT of algorithms classes. (And I have a copy at home and a copy at work – you never know.)
The CLIQUE problem is described as follows: given a graph (a set of vertices and edges) $G$, does it contain a clique of size $k$? A clique of size $k$ is a set of $k$ vertices that are all connected to one another. For instance, this thing below is a clique of size 8. Also note that it does contain cliques of size 1 (single vertex), 2 (an edge), 3 (a triangle), 4, 5, 6, 7, since we can find sets of vertices of that size that are all connected to one another.
Let us reduce 3-SAT to that problem; since 3-SAT is NP-hard, CLIQUE will then be proven to be NP-hard as well. CLIQUE is in NP, because if I provide you with $k$ vertices, it’s possible to check that all these vertices are indeed connected to one another. So, if CLIQUE is NP-hard, CLIQUE is NP-complete.
To start with my reduction, I pick an arbitrary 3-SAT formula – more precisely, a 3-CNF-SAT formula (I explained the meaning of this in my post on SAT), which is a formula that has that look:
$(x_1 \vee x_2 \vee x_3) \wedge (\bar x_2 \vee x_4 \vee \bar x_5) \wedge (x_1 \vee \bar x_2 \vee \bar x_3) \wedge ...$
that is to say, a set of $m$ clauses connected by AND and composed of 3 literals connected by OR.
From there, we’re creating a graph that contains $3m$ vertices. Every vertex corresponds to a literal of the formula; vertices can be duplicated. For the above formula (truncated before the …), this yields the following vertices :
$x_1, x_2, x_3 ; \bar x_2, \bar x_4, \bar x_5 ; x_1, \bar x_2, \bar x_3$.
Then, we add edges almost everywhere. We don’t add edges in the “groups” that correspond to the clauses, and we also do not add edges between literals that are not compatible, that is to say inverse of each other. If I have two literals $x_1$ and $\bar x_1$, I’m not creating an edge between them. For the formula above, this is the graph in question:
And these are the edges that are NOT in the previous graph:
And now that we have that graph, we want to show that the SAT formula can be satisfied if and only if the graph (the first one) has a clique of size $m$, where $m$ is the number of clauses in the SAT formula. So I need to prove two things:
• if the formula can be satisfied, then the graph has a clique of size $m$,
• if the graph has a clique of size $m$, then the formula can be satisfied.
Let’s start with the first point. Suppose the formula can be satisfied. This means that we can assign a value to all the variables so that the formula is satisfied. This means that all clauses can be satisfied by this assignment. This also means that, for each clause, there’s a literal with value 1 (either a “positive” literal, for instance $x_1$ if the variable is assigned to 1, or a “negative” literal, for instance $\bar x_1$, if the variable is assigned to 0). Now remember that we created vertices grouped by “clause”, so for each clause, we can pick the vertex corresponding to that 1-valued literal (and if several literals have value 1 in a clause, we can pick an arbitrary one). Since we pick a vertex by clause, and we have $m$ clauses, we have $m$ vertices. We now want to prove that these $m$ vertices form a clique, which means that there is an edge between every pair of vertices of that set. We constructed the graph such that there is no edge between two vertices of a given clause, and we’re fine there, because we chose exactly one vertex per clause. Moreover, there is no edge between a literal and its negation – we’re also fine there, because we only picked literals that have value 1, and $x_i$ and $\bar x_i$ can’t both have value 1. These are the only conditions for which there is no edge between two vertices; which means that our $m$ vertices are all connected with each other, which yields a clique of size $m$, which is what we want.
Now, let’s look at the second point: if the graph has a clique of size $m$, then the formula can be satisfied. Suppose that the graph has a clique of size $m$. Since the vertices corresponding to the literals of a clause are not connected, this means that we have a vertex for each clause. We can give the value 1 to all the literals corresponding to these vertices. We cannot have a clique containing $x_i$ and $\bar x_i$, because there would be no edge between these two vertices, which goes against the definition of a clique, where all edges are present. So if a clique of size $m$ exists, this means that we found, for each clause, a literal whose value can be 1, and that all of these literals are compatible with each other. And so, we can satisfy the formula corresponding to the graph, which we also wanted to prove.
So, if we can solve CLIQUE in polynomial time, then we can solve 3-SAT in polynomial time; since 3-SAT is NP-hard, CLIQUE is NP-hard, so CLIQUE is NP-complete, which concludes the proof.
Showing that a problem is NP-complete is, in a way, an indicator that the problem in question is difficult, but this needs to be mitigated a lot. For one thing, it does not say anything about a specific instance of the problem. It does not say anything about a specific subset of instances either – let me explain that. If I say that CLIQUE is difficult, it doesn’t mean that, for example, deciding whether a triangle (a clique of size 3) is in a graph is difficult. I can take all sets of 3 vertices, look at whether they form a triangle, and conclude. There are approximately $n^3$ sets of 3 vertices in a graph with $n$ vertices (okay, there’s exactly $\displaystyle \frac{n(n-1)(n-2)}{6}$ – anyway, roughly $n^3$), so I can actually decide that in polynomial time (because I’m doing $n^3$ operations which are basically checking if three edges are in a graph). So I can decide 3-CLIQUE in polynomial time. Well, I’m not going to be a millionaire with that, because the CLIQUE problem is wider than just 3. I can also decide 1000-CLIQUE (clique of size 1000) in polynomial time with the same principle. Well, it’s a polynomial of degree 1000, but who cares 😛
But, in the general case, I cannot decide whether a graph over $n$ vertices contains a clique of $\displaystyle \frac{n}{2}$, or $\displaystyle \frac{n}{10000}$ vertices, or even $\log n$ vertices in polynomial time with this algorithm that looks at all groups of size $k$ (so, here, $k$ would be equal to $\displaystyle \frac{n}{2}$, $\displaystyle \frac{n}{10000}$, or $\log n$) and that looks at the edges of the group, because that would mean doing roughly $n^{n/2}$, $n^{n/10000}$ and $n^{\log n}$ operations, respectively, and that in any case it’s not polynomial. And, generally speaking, nobody knows if it’s possible to do that in polynomial time, since CLIQUE is NP-complete and we don’t know if P is equal to NP. Many people think that not, but nobody knows.
Even that does not say anything about the difficulty of a specific instance. If I’m given a graph that contains $\displaystyle \frac{n(n-1)}{2}$ edges for $n$ vertices, then I can decide pretty easily that it does contain a clique of size $\displaystyle \frac{n}{2}$, $\displaystyle \frac{n}{10000}$ and $\log n$. That’s because a graph that has $\displaystyle \frac{n(n-1)}{2}$ for $n$ vertices contains all the possible edges for the graph: it’s a clique on $n$ vertices… which means that every subgraph over $k$ vertices, for all $k \leq n$, is a clique of size $k$. Same thing if I’m given a graph over $n$ vertices with 0 edge, it’s not going to be hard to say that it doesn’t contain a clique of size $\displaystyle \frac n 2$.
SAT is also NP-hard, but can solve 2-CNF-SAT instances (where all clauses contain 2 literals) in polynomial, and even linear time. There are also SAT instances that are deemed “easy”, for instance the ones that have only few dependencies between the clauses: we know that we can to solve them, and we even know how (for the people who are interested in what I mean exactly by what I mean by “a few dependencies”, I’m thinking specifically about “the ones that are in the scope of the Lovász Local Lemma“, whose application to SAT is not trivial from the Wikipedia definition, but which might be clearer here).
Generally speaking, it may even be pretty difficult to create NP-hard instances of problems that we cannot solve in polynomial time. By the way, it’s not easy to show that we don’t know how to solve a given instance in polynomial time. We can find instances problem instances that are hard (i.e. require exponential time) for a given algorithm, and easy for another algorithm…
And that’s a bit of what happens in real life: we have instance of problems for which we know that the general case is NP-hard (or worse), and that we still solve without great difficulty. It can be that the problem is “small” (it’s not a huge problem to put an exponential algorithm on a problem of size 4), but it can also be that the problem belongs to an “easy” subset of instances. Considering that an instance of a problem is hard because the general problem is hard, and giving up on it (because we won’t know how to do it) may not be a good idea – it may be that we forgot to consider an easy restriction of the problem!
As usual, “it’s not that simple”. I’ve spent a fair amount of time with theory, and I tend to look at the problems themselves more than at given instances. But it’s sometimes nice to remember that stuff that’s “theoretically hard” can end up being “practically easy” – it’s a nice change from the things that are theoretically easy but that we don’t know how to do practically… 🙂
The SAT problem
Note: this is a translation with some re-writing of a previously published post: Le problème SAT.
I’m going to explain what the SAT problem is, because I’m going to talk about it a bit more soon. It’s one of the fundamental blocks around the P vs NP question; I’m going to write a blog post about NP-hard problems, but writing a post about SAT specifically is worth it. Also: I did my master thesis on SAT-related questions, so it’s a bit of a favorite topic of mine 🙂
SAT is a common abbreviation for “boolean satisfiability problem”. Everything revolves around the concept of boolean formulas, and a boolean formula can look like this:
$(x \vee y \vee z) \wedge (\bar x \vee \bar y) \wedge (\bar x \vee y \vee \bar z)$
Let me unpack this thing. The first concept is variables: here, $x$, $y$ or $z$. They’re kind of like the “unknowns” of an equation: we want to find values for $x$, $y$ and $z$. Moreover, we’re in a somewhat weird universe, the boolean universe: the only values $x$, $y$ and $z$ can take are 0 or 1.
Then, we have some weird symbols. $\vee$ means “or”, and $\wedge$ means “and”. And the small bars above some of the letters indicate a negation. We combine these symbols with variables to get expressions, whose value can be computed as follow:
• If $x = 1$, then $\bar x = 0$, otherwise $\bar x = 1$ ($\bar x$ is the opposite value of $x$). We read this “not $x$“.
• If $x = 1$, then $x \vee y = 1$; if $y = 1$, then $x \vee y = 1$; if $x = 0$ and $y = 0$, then $x \vee y = 0$. In other words, $x \vee y = 1$ if $x = 1$ or $y = 1$. Note that in this context, when we say “or”, we do not use exactly the same meaning than in everyday language: we mean “if $x = 1$ or if $y = 1$ or if both are equal to 1″. We read this: “$x$ or $y$“.
• If $x = 1$ and $y = 1$, then $x \wedge y = 1$, otherwise $x \wedge y = 0$. We read this: “$x$ and $y$“.
I’m summarizing all this in this table, for every possible value of $x$ and $y$:
$\begin{array}{|c|c||c|c|c|c|} \hline x & y & \bar x &\bar y & x \vee y & x \wedge y \\ \hline 0 & 0 & 1 & 1 & 0 & 0 \\ \hline 0 & 1 & 1 & 0 & 1 & 0 \\ \hline 1 & 0 & 0 & 1 & 1 & 0 \\ \hline 1 & 1 & 0 & 0 & 1 & 1 \\ \hline \end{array}$
We can combine these expressions any way that we want to get boolean formulas. So the following examples are boolean formulas:
$x \wedge \bar y \wedge z \wedge t \wedge \bar u$
$(x \vee y) \wedge (\bar x \vee z) \wedge ((x \wedge z) \vee (z \wedge \bar t))$
$(x \vee y \vee z) \wedge (\bar x \vee \bar y) \wedge (\bar x \vee y \vee \bar z)$
$(x \wedge y \wedge z) \vee (\bar x \wedge \bar y) \vee (\bar x \wedge y \wedge \bar z)$
And then, by assigning values (0 or 1) to the individual variables, we can evaluate a formula for the assignment, that is to say, say if the value that the formula “computes” is 0 or 1. If the formula evaluates to 1 with an assignment, we say that this assignment satisfies the formula. And we say that a formula is satisfiable if there exists an assignment that makes it evaluate to 1 – hence the name of the problem, “boolean satisfiability”.
Boolean satisfiability in general is kind of “annoying” because it doesn’t have much structure to it: it’s hard to think about and it’s hard to say much about it, really. One thing that is always feasible, however, is to look at all the possibilities for all the variables, and see if there exists a combination of values that satisfies the formula. Picking one of the example above, and calling the formula $F$:
$F = (x \vee y \vee z) \wedge (\bar x \vee \bar y) \wedge (\bar x \vee y \vee z)$
And let’s enumerate all possibilities:
$\begin{array}{|c|c|c||c|c|c||c|} \hline x & y & z & x \vee y \vee z & \bar x \vee \bar y & \bar x \vee y \vee z & F \\ \hline 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ \hline 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ \hline 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ \hline 1 & 1 & 0 & 1 & 0 & 1 & 0\\ \hline 1 & 1 & 1 & 1 & 0 & 1 & 0\\ \hline \end{array}$
This table answers the question “is there values of $x$, $y$ and $z$ for which $F$ evaluates to 1 (the answer is yes) and it even goes further by giving said values (for instance, $x=1$, $y = 0$ and $z = 1$ are values that satisfy the formula).
The issue is that it’s not really manageable to write such a table as soon as we get a lot of variables. The reason for that is that, for every variable, I need to check what happens for its 0 value and for its 1 value. So I have two choices for the first variable, two choices for the second variable, two choices for the third variable, and so on. The choices multiply with each other: similarly to the table above, I need a line for every possible combination of values of variables. For 3 variables, $2\times 2 \times 2 = 2^8 = 8$ lines. For 5 variables, $2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32$ lines, that starts to be annoying to do by hand. For 20 variables, $2^{20} = 1,048,576$ lines, the result is not necessarily instant anymore on a computer. And it continues growing: for $n$ variables, I need $2^n$ lines, which it grows faster and faster: the joys of power functions.
For the people who followed the previous explanations, this is NOT a polynomial time algorithm: it’s an exponential time algorithm. (Not even considering the fact that I’m only enumerating the possibilities and not evaluating the formula yet).
Since the general SAT problem seems fairly complicated to tackle, one can wonder if there are “subgroups” that are easy to solve, or that are easier to think about. And, indeed, there are such subgroups.
Suppose that, instead of getting an arbitrary combination of variables and symbols, I restrict myself of formulas such as this one: $(x \wedge y \wedge z) \vee (y \wedge \bar z \wedge t) \vee (u \wedge \bar y \wedge \bar t)$. That’s a form where I have “groups” of variables linked by “$\wedge$“, themselves grouped by $\vee$. Then that’s an “easy” case: for the formula to be satisfiable, I only need one of the “groups” to evaluate to 1 (because they are linked by “or” operators, so the formula evaluates to 1 if at least one of the groups evaluates to 1). And a given group evaluates to 1, unless it contains both one variable and its negation, like $x \wedge y \wedge \bar y$. In that case, since within a group I need all the variables to be equal to 1 for the group to evaluate to 1, and since a variable and its negation cannot be equal both to 1, the group would evaluate to 0. So unless all the groups contain a variable and its negation, the formula is satisfiable – a fairly easy condition to verify.
If I consider the “opposite” of the previous setup, with formulas such as $(x \vee y \vee z) \wedge (\bar x \vee \bar y) \wedge (\bar x \vee y \vee z)$, we have “groups” of variables linked by $\vee$, themselves grouped by $\wedge$. This type of formulas are called CNF formulas (where CNF means “conjunctive normal form”). A CNF formula is defined as a set of clauses (the “groups”), all linked by $\wedge$ symbols (“and”). A clause is a set of one or several literals that are linked by $\vee$ symbols (“or”). A literal is either a variable (for example $x$) or its negation (for example $\bar x$).
The question asked by an instance of the SAT problem is “can I find values for all variables such that the whole formula evaluates to 1?”. If we restrict ourselves to CNF formulas (and call the restricted problem CNF-SAT), this means that we want that all the clauses evaluate to 1, because since they are linked by “and” in the formula, I need that all of its clauses evaluate to 1. And for each clause to have value 1, I need at least one of the literals to have value 1 in the clause (I want the first literal or the second literal or the third literal or… to be equal to 1). As previously mentioned: it can happen that all the literals are equal to 1 – the clause will still be equal to 1.
It turns out that we can again look at subgroups within CNF formulas. Let us consider CNF formulas that have exactly 2 literals in all clauses – they are called 2-CNF formulas. The associated problem is called 2-CNF-SAT, or more often 2-SAT. Well, 2-SAT can be solved in linear time. This means that there exists an algorithm that, given any 2-CNF formula, can tell in linear time whether the formula is satisfiable or not.
Conversely, for CNF formulas that have clauses that have more than 2 literals… in general, we don’t know. We can even restrict ourselves to CNF formulas that have exactly 3 literals in each clause and still not know. The associated problem is called 3-CNF-SAT, or (much) more often 3-SAT.
From a “complexity class” point of view, 3-SAT is in NP. If I’m given a satisfiable 3-CNF formula and values for all variables of the formula, I can check in polynomial time that it works: I can check that a formula is satisfiable if I have a proof of it. (This statement is also true for general SAT formulas.)
What we do not know, however, is whether 3-SAT is in P: we do not know if there exists a polynomial time algorithm that allows us to decide if a 3-CNF formula can be satisfied or not in the general case. The difficulty of the problem still does not presume of the difficulty of individual instances. In general, we do not know how to solve the 3-SAT problem; in practice, 3-SAT instances are solved every day.
We also do not know whether 3-SAT is outside of P: we do not know if we need an algorithm that is “more powerful” than a polynomial time algorithm to solve the problem in the general case. To answer that question (and to prove it) would actually allow to solve the P vs NP problem – but to explain that, I need to explain NP-hard problems, and that’s something for the next post, in which I’ll also explain why it’s pretty reasonable to restrict oneself to 3-SAT (instead of the general SAT problem) when it comes to studying the theory of satisfiability 🙂
The “P vs NP” problem
Note: this is a translation of an older blog post written in French: Le problème « P est-il égal à NP ? ».
All right, I think I explained enough algorithmic complexity (part 1 and part 2) to start with a nice piece, which is to explain what is behind the “Is P equal to NP?” question – also called “P vs NP” problem.
The “P vs NP” problem is one of the most famous open problems, if not the most famous. It’s also part of the Millenium Prize problems, a series of 7 problems stated in 2000: anyone solving one of these problems gets awarded one million dollars. Only one of these problems has been solved, the Poincaré conjecture, proven by Grigori Perelman. He was awarded the Fields medal (roughly equivalent to a Nobel Prize in mathematics) for it, as well as the aforementioned million dollars; he declined both.
But enough history, let’s get into it. P and NP are called “complexity classes”. A complexity class is a set of problems that have common properties. We consider problems (for instance “can I go from point A to point B in my graph with 15 steps or less?”) and to put them in little boxes depending on their properties, in particularity their worst case time complexity (how much time do I need to solve them) and their worst case space complexity (how much memory do I need to solve them).
I explained in the algorithmic complexity blog posts what it meant for an algorithm to run in a given time. Saying that a problem can be/is solved in a given time means that we know how to solve it in that time, which means we have an algorithm that runs in that time and returns the correct solution. To go back to my previous examples, we saw that it was possible to sort a set of elements (books, or other elements) in time $n \log n$. It so happens that, in classical computation models, we can’t do better than $n \log n$. We say that the complexity of sorting is $n \log n$, and we also say that we can sort elements in polynomial time.
A polynomial time algorithm is an algorithm that finishes with a number of steps that is less than $n^k$, where $n$ is the size of the input and $k$ an arbitrary number (including very large numbers, as long as they do not depend on $n$). The name comes from the fact that functions such as $x \mapsto x$, $x \mapsto x^2$, $x \mapsto x^{10} + 15x^5$ and $x \mapsto x^k$ are called polynomial functions. Since I can sort elements in time $n \log n$, which is smaller than $n^2$, sorting is solved in polynomial time. It would also work if I could only sort elements in time $n^{419}$, that would also be polynomial. The nice thing with polynomials is that they combine very well. If I make two polynomial operations, I’m still in polynomial time. If I make a polynomial number of polynomial operations, I’m still in polynomial time. The polynomial gets “bigger” (for instance, it goes from $n^2$ to $n^5$), but it stays a polynomial.
Now, I need to explain the difference between a problem and an instance of a problem – because I kind of need that level of precision 🙂 A problem regroups all the instances of a problem. If I say “I want to sort my bookshelf”, it’s an instance of the problem “I want to sort an arbitrary bookshelf”. If I’m looking at the length shortest path between two points on a given graph (for example a subway map), it’s an instance of the problem “length of shortest path in a graph”, where we consider all arbitrary graphs of arbitrary size. The problem is the “general concept”, the instance is a “concrete example of the general problem”.
The complexity class P contains all the “decision” problems that can be solved in polynomial time. A decision problem is a problem that can be answered by yes or no. It can seem like a huge restriction: in practice, there are sometimes way to massage the problem so that it can get in that category. Instead of asking for “the length of the shortest path” (asking for the value), I can ask if there is “a path of length less than X” and test that on various X values until I have an answer. If I can do that in a polynomial number of queries (and I can do that for the shortest path question), and if the decision problem can be solved in polynomial time, then the corresponding “value” problem can also be solved in polynomial time. As for an instance of that shortest path decision problem, it can be “considering the map of the Parisian subway, is there a path going from La Motte Piquet Grenelle to Belleville that goes through less than 20 stations?” (the answer is yes) or “in less than 10 stations?” (I think the answer is no).
Let me give another type of a decision problem: graph colorability. I like these kind of examples because I can make some drawings and they are quite easy to explain. Pick a graph, that is to say a bunch of points (vertices) connected by lines (edges). We want to color the vertices with a “proper coloring”: a coloring such that two vertices that are connected by a single edge do not have the same color. The graph colorability problems are problems such as “can I properly color this graph with 2, 3, 5, 12 colors?”. The “value” problem associated to the decision problem is to ask what is the minimum number of colors that I need to color the graph under these constraints.
Let’s go for a few examples – instances of the problem 🙂
A “triangle” graph (three vertices connected with three edges) cannot be colored with only two colors, I need three:
On the other hand, a “square-shaped” graph (four vertices connected as a square by four edges) can be colored with two colors only:
There are graphs with a very large number of vertices and edges that can be colored with only two colors, as long as they follow that type of structure:
And I can have graphs that require a very large number of colors (one per vertex!) if all the vertices are connected to one another, like this:
And this is where it becomes interesting. We know how to answer in polynomial time (where $n$ is of the number of vertices of the graph) to the question “Can this graph be colored with two colors?” for any graph. To decide that, I color an arbitrary vertex of the graph in blue. Then I color all of its neighbors in red – because since the first one is blue, all of its neighbors must be red, otherwise we violate the constraint that no two connected vertices can have the same color. We try to color all the neighbors of the red vertices in blue, and so on. If we manage to color all the vertices with this algorithm, the graph can be colored with two colors – since we just did it! Otherwise, it’s because a vertex has a neighbor that constrains it to be blue (because the neighbor is red) and a neighbor that constrains it to be red (because the neighbor is blue). It is not necessarily obvious to see that it means that the graph cannot be colored with two colors, but it’s actually the case.
I claim that this algorithm is running in polynomial time: why is that the case? The algorithm is, roughly, traversing all the vertices in a certain order and coloring them as it goes; the vertices are only visited once; before coloring a vertex, we check against all of its neighbors, which in the worst case all the other vertices. I hope you can convince yourself that, if we do at most $n$ (number of vertices we traverse) times $n-1$ comparisons (maximum number of neighbors for a given vertex), we do at most $n(n-1)$ operations, and the algorithm is polynomial. I don’t want to give much more details here because it’s not the main topic of my post, but if you want more details, ping me in the comments and I’ll try to explain better.
Now, for the question “Can this graph be colored with three colors?”, well… nobody has yet found a polynomial algorithm that allows us to answer the question for any instance of the problem, that is to say for any graph. And, for reasons I’ll explain in a future post, if you find a (correct!) algorithm that allows to answer that question in polynomial time, there’s a fair chance that you get famous, that you get some hate from the cryptography people, and that you win one million dollars. Interesting, isn’t it?
The other interesting thing is that, if I give you a graph that is already colored, and that I tell you “I colored this graph with three colors”, you can check, in polynomial time, that I’m not trying to scam you. You just look at all the edges one after the other and you check that both vertices of the edge are colored with different colors, and you check that there are only three colors on the graph. Easy. And polynomial.
That type of “easily checkable” problems is the NP complexity class. Without giving the formal definition, here’s the idea: a decision problem is in the NP complexity class if, for all instances for which I can answer “yes”, there exists a “proof” that allows me to check that “yes” in polynomial time. This “proof” allows me to answer “I bet you can’t!” by “well, see, I can color that way, it works, that proves that I can do that with three colors” – that is, if the graph is indeed colorable with 3 colors. Note here that I’m not saying anything about how to get that proof – just that if I have it, I can check that it is correct. I also do not say anything about what happens when the instance cannot be colored with three colors. One of the reasons is that it’s often more difficult to prove that something is impossible than to prove that it is possible. I can prove that something is possible by doing it; if I can’t manage to do something, it only proves that I can’t do it (but maybe someone else could).
To summarize:
• P is the set of decision problems for which I can answer “yes” or “no” in polynomial time for all instances
• NP is the set of decision problems for which, for each “yes” instance, I can get convinced in polynomial time that it is indeed the case if someone provides me with a proof that it is the case.
The next remark is that problems that are in P are also in NP, because if I can answer myself “yes” or “no” in polynomial time, then I can get convinced in polynomial time that the answer is “yes” if it is the case (I just have to run the polynomial time algorithm that answers “yes” or “no”, and to check that it answers “yes”).
The (literally) one-million-dollar question is to know whether all the problems that are in NP are also in P. Informally, does “I can see easily (i.e. in polynomial time) that a problem has a ‘yes’ answer, if I’m provided with the proof” also mean that “I can easily solve that problem”? If that is the case, then all the problems of NP are in P, and since all the problems of P are already in NP, then the P and NP classes contain exactly the same problems, which means that P = NP. If it’s not the case, then there are problems of NP that are not in P, and so P ≠ NP.
The vast majority of maths people think that P ≠ NP, but nobody managed to prove that yet – and many people try.
It would be very, very, very surprising for all these people if someone proved that P = NP. It would probably have pretty big consequences, because that would mean that we have a chance to solve problems that we currently consider as “hard” in “acceptable” times. A fair amount of the current cryptographic operations is based on the fact, not that it is “impossible” to do some operations, but that it’s “hard” to do them, that is to say that we do not know a fast algorithm to do them. In the optimistic case, proving that P = NP would probably not break everything immediately (because it would probably be fairly difficult to apply and that would take time), but we may want to hurry finding a solution. There are a few crypto things that do not rely on the hypothesis that P ≠ NP, so all is not lost either 😉
And the last fun thing is that, to prove that P = NP, it is enough to find a polynomial time algorithm for one of the “NP-hard” problems – of which I’ll talk in a future post, because this one is starting to be quite long. The colorability with three colors is one of these NP-hard problems.
I personally find utterly fascinating that a problem which is so “easy” to get an idea about have such large implications when it comes to its resolution. And I hope that, after you read what I just wrote, you can at least understand, if not share, my fascination 🙂
Understanding maths
Note: this post is a translation of an older post written in French: Compréhension mathématique. I wrote the original one when I was still in university, but it still rings true today – in many ways 🙂
After two heavyweight posts, Introduction to algorithmic complexity 1/2 and Introduction to algorithmic complexity 2/2, here’s a lighter and probably more “meta” post. Also probably more nonsense – it’s possible that, at the end of the article, you’ll either be convinced that I’m demanding a lot of myself, or that I’m completely ridiculous 🙂
I’m quite fascinated by the working of the human brain. Not by how it works – that, I don’t know much about – but by the fact that it works at all. The whole concept of being able to read and write, for instance, still amazes me. And I do spend a fair amount of time thinking about how I think, how to improve how I think, or how to optimize what I want to learn so that it matches my way of thinking. And in all of that thinking, I redefined for myself what I mean by “comprehension”.
My previous definition of comprehension
It often happens that I moan about the fact that I don’t understand things as fast as I used to; I’m wondering how much of that is the fact that I’m demanding more of myself. There was a time where my definition “understanding” was “what you’re saying looks logical, I can see the logical steps of what you’re doing at the blackboard, and I see roughly what you’re doing”. I also have some painful memories of episodes such as the following:
− OK.
<a few hours later>
− There, done!
− Well, yeah… I’m a fast reader…
− And you understood everything?
− Well… yeah…
− Including why <obscure but probably profound point of the article>?
<blank look, sigh and explanations> (not by me, the explanations).
I was utterly convinced to have understood, before it was proven to me that I missed a fair amount of things. Since then, I learnt a few things.
What I learnt about comprehension
The first thing I learnt, is that “vaguely understand” is not “comprehend”, or at least not at my (current) level of personal requirements. “Vaguely understanding” is the first step. It can also be the last step, if it’s on a topic for which I can/want to do with superficial knowledge. I probably gained a bit of modesty, and I probably say way more often that I only have a vague idea about some topics.
The second thing is that comprehension does take time. Today, I believe I need three to four reads of a research paper (on a topic I know) to have a “decent” comprehension of it. Below that, I have “a vague idea of what the paper means”.
The third thing, although it’s something I heard a lot at home, is that “repeating is at the core of good understanding”. It helps a lot to have at least been exposed to a notion before trying to really grasp it. The first exposure is a large hit in the face, the second one is slightly mellower, and at the third one you start to know where you are.
The fourth thing is that my brain seems to like it when I write stuff down. Let me sing an ode to blackboard and chalk. New technology gave us a lot of very fancy stuff, video-projectors, interactive whiteboards, and I’m even going to throw whiteboards and Vellada markers with it. I may seem reactionary, but I like nothing better than blackboard and chalk. (All right, the blackboard can be green.) It takes more time to write a proof on the blackboard than to run a Powerpoint with the proof on it (or Beamer slides, I’m not discriminating on my rejection of technology 😉 ) . So yeah, the class is longer, probably. But it gives time to follow. And it also gives time to take notes. Many of my classmates tell me they prefer to “listen than take notes” (especially since, for blackboard classes, there is usually an excellent set of typeset lecture notes). But writing helps me staying focused, and in the end to listen better. I also scribble a few more detailed points for things that may not be obvious when re-reading. Sometimes I leave jokes to future me – the other day, I found a “It’s a circle, ok?” next to a potato-shaped figure, it made me laugh a lot. Oh and, as for the fact that I also hate whiteboards: first, Velleda markers never work. Sometimes, there’s also a permanent marker hiding in the marker cup (and overwriting with an erasable marker to eventually erase it is a HUGE PAIN). And erasable marker is faster to erase than chalk. I learnt to enjoy the break that comes with “erasing the blackboard” – the usual method in the last classes I attended was to work in two passes, one with a wet thingy, and one with a scraper. I was very impressed the first time I saw that 😉 (yeah, I’m very impressionable) and, since then, I enjoy the minute or two that it takes to re-read what just happened. I like it. So, all in all: blackboard and chalk for the win.
How I apply those observations
With all of that, I learnt how to avoid the aforementioned cringy situations, and I got better at reading scientific papers. And takes more time than a couple of hours 😉
Generally, I first read it very fast to have an idea of the structure of the paper, what it says, how the main proof seems to work, and I try to see if there’s stuff that is going to annoy me. I have some ideas about what makes my life easier or not in a paper, and when it gets in the “not” territory, I grumble a bit, even though I suppose that these structures may not be the same for everyone. (There are also papers that are just a pain to read, to be honest). That “very fast” read is around half an hour to an hour for a ~10-page article.
The second read is “annotating”. I read in a little more detail, and I put questions everywhere. The questions are generally “why?” or “how?” on language structures such that “it follows that”, “obviously”, or “by applying What’s-his-name theorem”. It’s also pretty fast, because there is a lot of linguistic signals, and I’m still not trying to comprehend all the details, but to identify the ones that will probably require me to spend some time to comprehend them. I also take note of the points that “bother” me, that is to say the ones where I don’t feel comfortable. It’s a bit hard to explain, because it’s really a “gut feeling” that goes “mmmh, there, something’s not quite right. I don’t know what, but something’s not quite right”. And it’s literally a gut feeling! It may seem weird to link “comprehension” to “feelings”, but, as far as I’m concerned, I learnt, maybe paradoxically, to trust my feelings to evaluate my comprehension – or lack thereof.
The third read is the longer – that’s where I try to answer all the questions of the second read and to re-do the computations. And to convince myself that yeah, that IS a typo there, and not a mistake in my computation or reasoning. The fourth read and following are refinements of the third read for the questions that I couldn’t answer during the third one (but for which, maybe, things got clearer in the meantime).
I estimate that I have a decent understanding of a paper when I answered the very vast majority of the questions from the second read. (And I usually try to find someone more clever than me for the questions that are still open). Even there… I do know it’s not perfect.
The ultimate test is to prepare a presentation about the paper. Do as I say and not as I do – I typically do that by preparing projector slides. Because as a student/learner, I do prefer a blackboard class, but I also know that it’s a lot of work, and that doing a (good) blackboard presentation is very hard (and I’m not there yet). Once I have slides (which, usually, allow me to still find a few points that are not quite grasped yet), I try to present. And now we’re back to the “gut feeling”. If I stammer, if there are slides that make no sense, if the presentation is not smooth: there’s probably still stuff that requires some time.
When, finally, everything seems to be good, the feeling is a mix between relief and victory. I don’t know exactly what the comparison would be. Maybe the people who make domino shows. You spend an enormous amount of time placing your dominos next to one another, and I think that at the moment where the last one falls without the chain having been interrupted… that must be that kind of feeling.
Of course, I can’t do that with everything I read, it would take too much time. I don’t know if there’s a way to speed up the process, but I don’t think it’s possible, at least for me, in any significant way. I also need to let things “simmer”. And there’s a fair amount of strong hypotheses on the impact of sleep on learning and memory; I don’t know how much of that can be applied to my “math comprehension”, but I wouldn’t be surprised if the brain would take the opportunity of sleep to tidy and make the right connections.
Consequently, it’s sometimes quite frustrating to let things at a “partial comprehension” stage – whether it’s temporary or because of giving up – especially when you don’t know exactly what’s wrong. The “gut feeling” is there (and not only on the day before the exam 😉 ). Sometimes, I feel like giving up altogether – what’s the point of trying, and only half understanding things? But maybe “half” is better than “not at all”, when you know you still have half of the way to walk. And sometimes, when I get a bit more stubborn, everything just clicks. And that’s one of the best feelings of the world.
Introduction to algorithmic complexity – 2/2
Note: this is the translation of a post that was originally published in French: Introduction à la complexité algorithmique – 2/2.
In the previous episode, I explained two ways to sort books, and I counted the elementary operations I needed to do that, and I estimated the number of operations depending on the number of books that I wanted to sort. In particular, I looked into the best case, worst case and average case for the algorithms, depending on the initial ordering of the input. In this post, I’ll say a bit more about best/worst/average case, and then I’ll refine the notion of algorithmic complexity itself.
The “best case” is typically analyzed the least, because it rarely happens (thank you, Murphy’s Law.) It still gives bounds of what is achievable, and it gives a hint on whether it’s useful to modify the input to get the best case more often.
The “worst case” is the only one that gives guarantees. Saying that my algorithm, in the worst case, executes in a given amount of time, guarantees that it will never take longer – although it can be faster. That type of guarantees is sometimes necessary. In particular, it allows to answer the question of what happens if an “adversary” provides the input, in a way that will make the algorithm’s life as difficult as possible – a question that would interest cryptographers and security people for example. Having guarantees on the worst case means that the algorithm works as desired, even if an adversary tries to make its life as miserable as possible. The drawback of using the worst case analysis is that the “usual” execution time often gets overestimated, and sometimes gets overestimated by a lot.
Looking at the “average case” gives an idea of what happens “normally”. It also gives an idea about what happens if the algorithm is repeated several times on independent data, where both the worst case and the best case can happen. Moreover, there is sometimes ways to avoid the worst cases, so the average case would be more useful in that case. For example, if an adversary gives me the books in an order that makes my algorithm slow, I can compensate that by shuffling the books at the beginning so that the probability of being in a bad case is low (and does not depend on my adversary’s input). The drawback of using the average case analysis is that we lose the guarantee that we have on the worst case analysis.
For my book sorting algorithms, my conclusions were as follows:
• For the first sorting algorithm, where I was searching at each step for a place to put the book by scanning all the books that I had inserted so far, I had, in the best case, $2n-1$ operations, in the worst case, $\displaystyle \frac{n^2 + n}{2}$ operations, and in the average case, $\displaystyle \frac{n^2+9n}{4}$. operations.
• For the second sorting algorithm, where I was grouping book groups two by two, I had, in all cases, $2\times n \times\log_2(n)$ operations.
I’m going to draw a figure, because figures are pretty. If the colors are not that clear, the plotted functions and their captions are in the same order.
It turns out that, when talking about complexity, these formulas ($2n-1$, $\displaystyle \frac{n^2 + n}{2}$, $\displaystyle \frac{n^2+9n}{4}$, $2\times n \times\log_2(n)$) would not be the ones that I would use in general. If someone asked me about these complexities, I would answer, respectively, that the complexities are $n$, $n^2$ (or “quadratic”), $n^2$ again, and $n \log n$.
This may seem very imprecise, and I’ll admit that the first time I saw this kind of approximations, I was quite irritated. (It was in physics class, so I may not have been in the best of moods either.) Since then, I find it quite handy, and even that it makes sense. The fact that “it makes sense” has a strong mathematical justification. For the people who want some precision, and who are not afraid of things like “limit when x tends to infinity of blah blah”, it’s explained there: http://en.wikipedia.org/wiki/Big_O_notation. It’s vastly above the level of the post I’m currently writing, but I still want to justify a couple of things; be warned that everything that follows is going to be highly non-rigorous.
The first question is what happens to smaller elements of the formulas. The idea is that only keep what “matters” when looking at how the number of operations increases with the number of elements to sort. For example, if I want to sort 750 books, with the “average” case of the first algorithm, I have $\displaystyle \frac{n^2 + 9n}{4} = \displaystyle \frac{n^2}{4} + \frac{9n}{4}$. For 750 books, the two parts of the sum yield, respectively, 140625 and… 1687. If I want to sort 1000 books, I get 250000 and 2250. The first part of the sum is much larger, and it grows much quicker. If I need to know how much time I need, and I don’t need that much precision, I can pick $\displaystyle \frac{n^2}{4}$ and discard $\displaystyle \frac{9n}{4}$ – already for 1000 books, it contributes less than 1% of the total number of operations.
The second question is more complicated: why do I consider as identical $n^2$ and $\displaystyle \frac{n^2}{4}$, or $2n\log_2 n$ and $n \log n$? The short answer is that it allows to make running times comparable between several algorithms. To determine which algorithm is the most efficient, it’s nice to be able to compare how they perform. In particular, we look at the “asymptotic” comparison, that is to say what happens when the input of the algorithm contains a very large number of elements (for instance, if I have a lot of books to sort) – that’s where using the fastest algorithm is going to be at most worth it.
To reduce the time that it takes an algorithm to execute, I have two possibilities. Either I reduce the time that each operation takes, or I reduce the number of operations. Suppose that I have a computer that can execute one operation by second, and that I want to sort 100 elements. The first algorithm, which needs $\displaystyle \frac{n^2}{4}$ operations, finishes after 2500 seconds. The second algorithm, which needs $2n\log_2 n$ operations, finishes after 1328 seconds. Now suppose that I have a much faster computer to execute the first algorithm. Instead of needing 1 second per operation, it’s 5 times faster, and can execute an operation in 0.2 seconds. That means that I can sort 100 elements in 500 seconds, which is faster than the second algorithm on the slower computer. Yay! Except that, first, if I run the second algorithm of the second computer, I can sort my elements five times faster too, in 265 seconds. Moreover, suppose now that I have 1000 elements to sort. With the first algorithm on the fast computer, I need $0.2 \times \frac{1000^2}{4} = 50000$ seconds, and with the second algorithm on the much slower computer, $2 \times 1000 \times \log_2(1000) = 19931$ seconds.
That’s the idea behind removing the “multiplying numbers” when estimating complexity. Given an algorithm with a complexity “$n^2$” and algorithm with a complexity “$n \log n$“, I can put the first algorithm on the fastest computer I can: there will always be a number of elements for which the second algorithm, even on a very slow computer, will be faster than the first one. The number of elements in question can be very large if the difference of speed of the computers is large, but since large numbers are what I am interested in anyway, that’s not a problem.
So when I compare two algorithms, it’s much more interesting to see that one needs “something like $n \log n$” operations and one needs “something like $n^2$” operations than to try to pinpoint the exact constant that multiplies the $n \log n$ or the $n^2$.
Of course, if two algorithms need “something along the lines of $n^2$ operations”, asking for the constant that is multiplying that $n^2$ is a valid question. In practice, it’s not done that often, because unless things are very simple and well-defined (and even then), it’s very hard to determine that constant exactly, depending on how you implement it with a programmation language. It would also require to ask exactly what an operation is. There are “classical” models that allow to define all these things, but linking them to current programming languages and computers is probably not realistic.
Everything that I talked about so far is function of $n$, which is in general the “size of the input”, or the “amount of work that the algorithm has to do”. For books to sort, it would be the number of books. For graph operations, it would be the number of vertices of graphs, and/or the number of edges. Now, as “people who write algorithms”, given an input of size $n$, what do we like, what makes us frown, what makes us run very fast in the other direction?
The “constant time algorithms” and “logarithmic time algorithms” (whose numbers of operations are, respectively, a constant that does not depend on $n$ or “something like $\log n$“) are fairly rare, because with $\log n$ operations (or a constant number of operations), we don’t even have the time to look at the whole input. So when we find an algorithm of that type, we’re very, very happy. A typical example of a logarithmic time algorithm is searching an element in a sorted list. When the list is sorted, it is not necessary to read it completely to find the element that we’re looking for. We can start checking if it’s before or after the middle element, and search in the corresponding part of the list. Then we check if it’s before or after the middle of the new part of the list, and so on.
We’re also very happy when we find a “linear time algorithm” (the number of operations is “something like $n$“). That means that we read the whole input, make a few operations per element of the input, and bam, done. $n \log n$ is also usually considered as “acceptable”. It’s an important bound, because it is possible to prove that, in standard algorithmic models (which are quite close to counting “elementary” operations), it is not possible to sort $n$ elements faster than with $n \log n$ operations in the general case (that is to say, without knowing anything about the elements or the order in which they are). There are a number of algorithms that require, at some point, some sorting: if it is not possible to get rid of the sorting, such an algorithm will also not get below $n \log n$ operations.
We start grumbling a bit at $n^2$, $n^3$, and to grumble a lot on greater powers of $n$. Algorithms that can run in $n^k$ operations, for some value of $k$ (even 1000000), are called “polynomial”. The idea is that, in the same way that a $n \log n$ algorithm will eventually be more efficient than a $n^2$ algorithm, with a large enough input, a polynomial algorithm, whatever $k$, will be more efficient than a $2^n$-operation algorithm. Or than a $1.308^n$-operation algorithm. Or even than a $1.0001^n$-operation algorithm.
In the real life, however, this type of reasoning does have its limits. When writing code, if there is a solution that takes 20 times (or even 2 times) less operations than another, it will generally be the one that we choose to implement. And the asymptotic behavior is only that: asymptotic. It may not apply for the size of the inputs that are processed by our code.
There is an example I like a lot, and I hope you’ll like it too. Consider the problem of multiplying matrices. (For people who never saw matrices: they’re essentially tables of numbers, and you can define how to multiply these tables of numbers. It’s a bit more complicated/complex than multiplying numbers one by one, but not that much more complicated.) (Says the girl who didn’t know how to multiply two matrices before her third year of engineering school, but that’s another story.)
The algorithm that we learn in school allows to multiply to matrices of size $n \times n$ with $n^3$ operations. There exists an algorithm that is not too complicated (Strassen algorithm) that works in $n^{2.807}$ operations (which is better than $n^3$). And then there is a much more complicated algorithm (Coppersmith-Winograd and later) that works in $n^{2.373}$ operations. This is, I think, the only algorithm for which I heard SEVERAL theoreticians say “yeah, but really, the constant is ugly” – speaking of the number by which we multiply that $n^{2.373}$ to get the “real” number of operations. That constant is not very well-defined (for the reasons mentioned earlier) – we just know that it’s ugly. In practice, as far as I know, the matrix multiplication algorithm that is implemented in “fast” matrix multiplication librairies is Strassen’s or a variation of it, because the constant in the Coppersmith-Winograd algorithm is so huge that the matrices for which it would yield a benefit are too large to be used in practice.
And this funny anecdote concludes this post. I hope it was understandable – don’t hesitate to ask questions or make comments 🙂
Introduction to algorithmic complexity – 1/2
Note: this is the translation of a post that was originally published in French: Introduction à la complexité algorithmique – 1/2.
There, now that I warmed up by writing a couple of posts where I knew where I wanted to go (a general post about theoretical computer science, and a post to explain what is a logarithm, because it’s always useful). And then I made a small break and talked about intuition, because I needed to gather my thoughts. So now we’re going to enter things that are a little bit more complicated, and that are somewhat more difficult to explain for me too. So I’m going to write, and we’ll see what happens in the end. Add to that that I want to explain things while mostly avoiding the formality of maths that’s by now “natural” to me (but believe me, it required a strong hammer to place it in my head in the first place): I’m not entirely sure about the result of this. I also decided to cut this post in two, because it’s already fairly long. The second one should be shorter.
I already defined an algorithm as a well-defined sequence of operations that can eventually give a result. I’m not going to go much further into the formal definition, because right now it’s not useful. And I’m also going to define algorithmic theory, in a very informal way, as the quantity of resources that I need to execute my algorithm. By resources, I will mostly mean “time”, that is to say the amount of time I need to execute the algorithm; sometimes “space”, that is to say the amount of memory (think of it as the amount of RAM or disk space) that I need to execute my algorithm.
I’m going to take a very common example to illustrate my words: sorting. And, to give a concrete example of my sorting, suppose I have a bookshelf full of books (an utterly absurd proposition). And that it suddenly takes my fancy to want to sort them, say by alphabetical order of their author (and by title for two books by same author). I say that a book A is “before” or “smaller than” a book B if it must be put before in the bookshelf, and that it is “after” or “larger than” the book B if it must be sorted after. With that definition, Asimov’s books are “before” or “smaller than” Clarke’s, which are “before” or “smaller than” Pratchett’s. I’m going to keep this example during the whole post, and I’ll draw parallels to the corresponding algorithmic notions.
Let me first define what I’m talking about. The algorithm I’m studying is the sorting algorithm: that’s the algorithm that allows me to go from a messy bookshelf to a bookshelf whose content is in alphabetical order. The “input” of my algorithm (that is to say, the data that I give to my algorithm for processing), I have a messy bookshelf. The “output” of my algorithm, I have the data that have been processed by my algorithm, that is to say a tidy bookshelf.
I can first observe that, the more books I have, the longer it takes to sort them. There’s two reasons for that. The first is that, if you consider an “elementary” operation of the sort (for instance, put a book in the bookshelf), it’s longer to do that 100 times than 10 times. The second reason is that if you consider what you do for each book, the more books there is, the longer it is. It’s longer to search for the right place to put a book in the midst of 100 books than in the midst of 10.
And that’s precisely what we’re interested in here: how the time that is needed to get a tidy bookshelf grows as a function of the number of books or, generally speaking, how the time necessary to get a sorted sequence of elements depends on the number of elements to sort.
This time depends on the sorting method that is used. For instance, you can choose a very, very long sorting method: while the bookshelf is not sorted, you put everything on the floor, and you put the books back in the bookshelf in a random order. Not sorted? Start again. At the other end of the spectrum, you have Mary Poppins : “Supercalifragilistic”, and bam, your bookshelf is tidy. The Mary Poppins method has a nice particularity: it doesn’t depend on the number of books you have. We say that Mary Poppins executes “in constant time”: whatever the number of books that need to be sorted, they will be within the second. In practice, there’s a reason why Mary Poppins makes people dream: it’s magical, and quite hard to do in reality.
Let’s go back to reality, and to sorting algorithms that are not Mary Poppins. To analyze how the sorting works, I’m considering three elementary operations that I may need while I’m tidying:
• comparing two books to see if one should be before or after the other,
• add the books to the bookshelf,
• and, assuming that my books are set in some order on a table, moving a book from one place to another on the table.
I’m also going to suppose that these three operations take the same time, let’s say 1 second. It wouldn’t be very efficient for a computer, but it would be quite efficient for a human, and it gives some idea. I’m also going to suppose that my bookshelf is somewhat magical (do I have some Mary Poppins streak after all?), that is to say that its individual shelves are self-adapting, and that I have no problem placing a book there without going “urgh, I don’t have space on this shelf anymore, I need to move books on the one below, and that one is full as well, and now it’s a mess”. Similarly: my table is magical, and I have no problem placing a book where I want. Normally, I should ask myself that sort of questions, including from an algorithm point of view (what is the cost of doing that sort of things, can I avoid it by being clever). But since I’m not writing a post about sorting algorithms, but about algorithmic complexity, let’s keep things simple there. (And for those who know what I’m talking about: yeah, I’m aware my model is debatable. It’s a model, it’s my model, I do what I want with it, and my explanations within that framework are valid even if the model itself is debatable.)
First sorting algorithm
Now here’s a first way to sort my books. Suppose I put the contents of my bookshelf on the table, and that I want to add the books one by one. The following scenario is not that realistic for a human who would probably remember where to put a book, but let’s try to imagine the following situation.
1. I pick a book, I put it in the bookshelf.
2. I pick another book, I compare it with the first: if it must be put before, I put it before, otherwise after.
3. I pick a third book. I compare it with the book in the first position on the shelf. If it must be put before, I put it before. If it must be put after, I compare with the book on the second position on the shelf. If it must be before, I put it between both books that are already in the shelf. If it must be put after, I put it as last position.
4. And so on, until my bookshelf is sorted. For each book that I insert, I compare, in order, with the books that are already there, and I add it between the last book that is “before” it and the first book that is “after” it.
And now I’m asking how much time it takes if I have, say, 100 books, or an arbitrary number of books. I’m going to give the answer for both cases: for 100 books and for $n$ books. The time for $n$ books will be a function of the number of books, and that’s really what interests me here – or, to be more precise, what will interest me in the second post of this introduction.
The answer is that it depends on the order in which the books were at the start when they were on my table. It can happen (why not) that they were already sorted. Maybe I should have checked before I put everything on the table, it would have been smart, but I didn’t think of it. It so happens that it’s the worst thing that can happen to this algorithm, because every time I want to place a book in the shelf, since it’s after/greater than all the books I put before it, I need to compare it all of the books that I put before. Let’s count:
1. I put the first book in the shelf. Number of operations: 1.
2. I compare the second book with the first book. I put it in the shelf. Number of operations: 2.
3. I compare the third book with the first book. I compare the third book with the second book. I put it in the shelf. Number of operations: 3.
4. I compare the fourth book with the first book. I compare the fourth book with the second book. I compare the fourth book with the third book. I put it in the shelf. Number of operations: 4.
5. And so on. Every time I insert a book, I compare it to all the books that were placed before it; when I insert the 50th book, I do 49 comparison operations, plus adding the book in the shelf, 50 operations.
So to insert 100 books, if they’re in order at the beginning, I need 1+2+3+4+5+…+99+100 operations. I’m going to need you to trust me on this if you don’t know it already (it’s easy to prove, but it’s not what I’m talking about right now) that 1+2+3+4+5+…+99+100 is exactly equal to (100 × 101)/2 = 5050 operations (so, a bit less than one hour and a half with 1 operation per second). And if I don’t have 100 books anymore, but $n$ books in order, I’ll need $\displaystyle \frac{n(n+1)}{2} = \frac{n^2 + n}{2}$ operations.
Now suppose that my books were exactly in the opposite order of the order they were supposed to be sorted into. Well, this time, it’s the best thing that can happen with this algorithm, because the first book that I add is always smaller than the ones I put before, so I just need a single comparison.
1. I put the first book in the shelf. Number of operations: 1.
2. I compare the second book with the second book. I put it in the shelf. Number of operations: 2.
3. I compare the third book with the first book. I put it in the shelf. Number of operations: 2.
4. And so on: I always compare with the first book, it’s always before, and I always have 2 operations.
So if my 100 books are exactly in reverse order, I do 1+2+2+…+2 = 1 + 99 × 2 = 199 operations (so 3 minutes and 19 seconds). And if I have $n$ books in reverse order, I need $1 + 2(n-1) = 2n-1$ operations.
Alright, we have the “best case” and the “worst case”. Now this is where it gets a bit complicated, because the situation that I’m going to describe is less well-defined, and I’m going to start making approximations everywhere. I’m trying to justify the approximations I’m making, and why they are valid; if I’m missing some steps, don’t hesitate to ask in the comments, I may be missing something myself.
Suppose now that my un-ordered books are in state such that every time I add a book, it’s added roughly in the middle of what has been sorted (I’m saying “roughly” because if I sorted 5 books, I’m going to place the 6th after the book in position 2 or position 3 – positions are integer, I’m not going to place it after the book in position 2.5.) Suppose I insert book number $i$: I’m going to estimate the number of comparisons that I make to $\displaystyle \frac i 2 + 1$, which is greater or equal to the number of comparisons that I actually make. To see that, I distinguish on whether $i$ is even or odd. You can show that it works for all numbers; I’m just going to give two examples to explain that indeed there’s a fair chance it works.
If I insert the 6th book, I have already 5 books inserted. If I want to insert it after the book in position 3 (“almost in the middle”), I’m making 4 comparisons (because it’s after the books in positions 1, 2 and 3, but before the book in position 4): we have $\displaystyle \frac i 2 + 1 = \frac 6 2 + 1 = 4$.
If I insert the 7th book, I already have 6 books inserted, I want to insert it after the 3rd book as well (exactly in the middle); so I also make 4 comparisons (for the same reason), and I have $\displaystyle \frac i 2 + 1 = \frac 7 2 + 1 = 4.5$.
Now I’m going to estimate the number of operations I need to sort 100 books, overestimating a little bit, and allowing myself “half-operations”. The goal is not to count exactly, but to get an order of magnitude, which will happen to be greater than the exact number of operations.
• Book 1: $\frac 1 2 + 1 = 1.5$, plus putting on the shelf, 2.5 operations (I actually don’t need to compare here; I’m just simplifying my end computation.)
• Book 2: $\frac 2 2 + 1 = 2$, plus putting on the shelf, 3 operations (again, here, I have an extra comparison, because I have only one book in the shelf already, but I’m not trying to get an exact count).
• Book 3: $\frac 3 2 + 1 = 2.5$, plus putting on the shelf, 3.5 operations.
• Book 4: $\frac 4 2 + 1 = 3$, plus putting on the shelf, 4 operations.
If I continue like that and I re-order my computations a bit, I have, for 100 books:
$\displaystyle \frac{1+2+3+...+99+100}{2} + 100 + 100 = \frac{100 \times 101}{4} + 200 = 2725 \text{ operations}$
which yields roughly 45 minutes.
The first element of my sum is from the $\displaystyle \frac i 2$ that I have in all my comparison computations. The first 100 comes from the “1” that I add every time I count the comparisons (and I do that 100 times); the second “100” comes from the “1” that I do every time I count putting the book in the shelf, which I also do 100 times.
That 2725 is a bit overestimated, but “not that much”: for the first two books, I’m counting exactly 2.5 comparisons too much; for the others, I have at most 0.5 comparisons too much. Over 100 books, I have at most $98 \times 0.5 + 2.5 = 51.5$ extra operations; the exact number of operations is between 2673 and 2725 (between 44 and 45 minutes). I could do thing a little more precisely, but we’ll see in what follows (in the next post) why it’s not very interesting.
If I’m counting for $n$ books, my estimation is
$\displaystyle \frac{\frac{n(n+1)}{2}}{2} + 2n = \frac{n^2 + 9n}{4} \text{ operations}$
It is possible to prove (but that would really be over the top here) that this behaviour is roughly he one that you get when you have books in a random order. The idea is that if my books are in a random order, I will insert some around the beginning, some around the end, and so “on average” roughly in the middle.
Another sorting algorithm
Now I’m going to explain another sorting method, which is probably less easy to understand, but which is probably the easiest way for me to continue my argument.
Let us suppose, this time, that I want to sort 128 books instead of 100, because it’s a power of 2, and it makes my life easier for my concrete example. And I didn’t think about it before, and I’m too lazy to go back to the previous example to run it for 128 books instead of 100.
Suppose that all my books directly on the table, and I’m going to make “groups” before putting my books in the bookshelf. And I’m going to make these groups in a somewhat weird, but efficient fashion.
First, I combine my books two by two. I take two books, I compare them, I put the smaller one (the one that is before in alphabetical order) on the left, and the larger one on the right. At the end of this operation, I have 64 groups of two books, and for each group, a small book on the left, and a large book on the right. To do this operation, I had to make 64 comparisons, and 128 moves of books (I suppose that I always move books, if only to have them in hand and read the authors/titles).
Then, I take my groups of two books, and I combine them again so that I have groups of 4 books, still in order. To do that, I compare the first two books of the group; the smaller of both becomes the first book of my group of 4. Then, I compare the remaining book of the group of 2 from which I picked the first book, and I put the smaller one in position 2 of my group of 4. There, I have two possibilities. Either I have one book in each of my initial groups of 2: in that case, I compare them, and I put them in order in my group of 4. or I still have a full group of two: so I just have to add them at the end of my new group, and I have an ordered group of 4. Here are two little drawings to distinguish both cases: each square represents a book whose author starts by the indicated letter; each rectangle represents my groups of books (the initial groups of two and the final group of 4), and the red elements are the ones that are compared at each step.
So, for each group of 4 that I create, I need to make 4 moves and 2 or 3 comparisons. I end up with 32 groups of 4 books; in the end, to make combine everything into 32 groups of 4 books, I make 32 × 4 = 128 moves and between 32 × 2 = 64 and 32 × 3 = 96 comparisons.
Then, I create 16 groups of 8 books, still by comparing the first element of each group of books and by creating a common, sorted group. To combine two groups of 4 books, I need 8 moves and between 4 and 7 comparisons. I’m not going to get into how exactly to get these numbers: the easiest way to see that is to enumerate all the cases, and while it’s still feasible for groups of 4 books, it’s quite tedious. So to create 16 groups of 8 books, I need to do 16×8 moves and between 16×4 = 64 and 16×7 = 112 comparisons.
I continue like that until I have 2 groups of 64 books, which I combine (directly in the bookshelf to gain some time) to get a sorted group of books.
Now, how much time does that take me? First, let me give an estimation for 128 books, and then we’ll see what happens for $n$ books. First, we evaluate the number of comparisons when combining two groups of books. I claim that to combine two groups of $k$ elements into a larger group of $2k$ elements, I need at most $2k$ comparisons. To see that: every time I place a book in the larger group, it’s either because I compared it to another one (and made a single comparison at that step), or because one of my groups is empty (and there I would make no comparison at all). Since I have a total of $2k$ books, I make at most $2k$ comparisons. I also move $2k$ books to combine my groups. Moreover, for each “overall” step (taking all the groups and combining them two by two), I do overall 128 moves – because I have 128 books, and each of them is in exactly one “small” group at the beginning and ends up in one “large” group at the end. So, for each “overall” step of merging, I’m doing at most 128 comparisons and 128 moves.
Now I need to count the number of overall steps. For 128 books, I do the following:
1. Combine 128 groups of 1 book into 64 groups of 2 books
2. Combine 64 groups of 2 books into 32 groups of 4 books
3. Combine 32 groups of 4 books into 16 groups of 8 books
4. Combine 16 groups of 8 books into 8 groups of 16 books
5. Combine 8 groups of 16 books into 4 groups of 32 books
6. Combine 4 groups of 32 books into 2 groups of 64 books
7. Combine 2 groups of 64 books into 1 group of 128 books
So I have 7 “overall” steps. For each of these steps, I have 128 moves, and at most 128 comparisons, so at most 7×(128 + 128) = 1792 operations – that’s a bit less than half an hour. Note that I didn’t make any hypothesis here on the initial order of the books. Compare that to the 5050 operations for the “worst case” of the previous computation, or with the ~2700 operations of the “average” case (those numbers were also counted for 100 books; for 128 books we’d have 8256 operations for the worst case and ~4300 with the average case).
Now what about the formula for $n$ books? I think we can agree that for each overall step of group combining, we move $n$ books, and that we do at most $n$ comparisons (because each comparison is associated to putting a book in a group). So, for each overall step, I’m doing at most $2n$ comparisons. Now the question is: how many steps do we need? And that’s where my great post about logarithms (cough) gets useful. Can you see the link with the following figure?
What if I tell you that the leaves are the books in a random order before the first step? Is that any clearer? The leaves represent “groups of 1 book”. Then the second level represents “groups of two books”, the third represent “groups of 4 books”, and so on, until we get a single group that contains all the books. And the number of steps is exactly equal to the logarithm (in base 2) of the number of books, which corresponds to the “depth” (the number of levels) of the tree in question.
So to conclude, for $n$ books, I have, in the model I defined, at most $2 \times n \times \log_2(n)$ operations.
There, I’m going to stop here for this first post. In the next post, I’ll explain why I didn’t bother too much with exactly exact computations, and why one of the sentences I used to pronounce quite often was “bah, it’s a constant, I don’t care” (and also why sometimes we actually do care).
I hope this post was understandable so far; otherwise don’t hesitate to grumble, ask questions, and all that sort of things. As for me, I found it very fun to write all this 🙂 (And, six years later, I also had fun translating it 🙂 )
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# Math Subtraction Worksheet #1
When subtracting natural numbers, we subtract one number from another. So there is a decrease. Now we will learn how to do this and we will make examples together.
When doing subtraction, we separate some of the objects we have. Let’s say we have 10 apples. What happens if we separate 3 apples from them? There are 7 apples left.
Subtraction: When we subtract a number with a value less than a certain number, it is the subtraction we do to find the remaining number.
Example 1: Olivia had 4 balloons in his hand. However, a balloon in Olivia’s hand burst. How many balloons have been left then?
If one of the balloons in Olivia’s hand bursts, then Olivia will have 3 balloons in his hand. Because, we reduce 1 balloon out of 4 balloons in this case. So we do the subtraction. Now let’s put this into action and see how we do it.
4 – 1 = 3 balloons
Example 2: We have 5 pencils. Two of these pencils were broken and we threw them away. How many pencils have been left then?
At first we had 5 pencils. But what happened next? Two of the pencils we had were broken and we threw them away. How many pens do we have then?
Now let’s do this together and see how many pencils we have left.
5 – 2 = 3
(Total items we have) – (Broken items) = (Number of items left)
As you can see, when we subtract 2 pencils from 5 pencils, there are 3 pencils left.
When we subtract a number from another, the resulting number is known as the remainder. In this way, you can make many different examples and write them in your notebook. You can count the different items around you and subtract them from the number of other items.
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Number of combinations with repetition
There are n = 4 sorts of candy to choose from and you want to buy k = 10 candies. How many ways can you do it?
This is a problem of counting combinations (order does not matter) with repetition (you can choose multiple items from each category). Below we will translate this problem into a problem of counting combinations without repetition, which can be solved by using a better understood formula that is known as the “binomial coefficient“.
First let us represent the 10 possible candies by 10 symbols ‘C’ and divide them into 4 categories by placing a partition wall, represented by a ‘+’ sign, between each sort of candy to separate them from each other
CC+CCCC+C+CCC
Note that there are 10 symbols ‘C’ and 3 partition walls, represented by a ‘+’ sign. That is, there are n-1+k = 13, equivalently n+k-1, symbols. Further note that each of the 3 partition walls could be in 1 of 13 positions. In other words, to represent various choices of 10 candies from 4 categories, the positions of the partition walls could be rearranged by choosing n-1 = 3 of n+k-1 = 13 positions
C++CCC+CCCCCC
CCCCCCCCCC+++
We have now translated the original problem into choosing 3 of 13 available positions.
Note that each position can only be chosen once. Further, the order of the positions does not matter. Since choosing positions {1, 4, 12} does separate the same choice of candies as the set of positions {4, 12, 1}. Which means that we are now dealing with combinations without repetition.
Calculating combinations without repetition can be done using the formula that is known as the binomial coefficient
n!/k!(n-k)!
As justified above, to calculate combinations with repetition, simply replace n with n+k-1 and k with n-1,
(n+k-1)!/(n-1)!((n+k-1)-(n-1))!
In our example above this would be (4+10-1)!/(4-1)!((4+10-1)-(4-1))! = 13!/3!10!. Which is equivalent to
(n+k-1)!/k!(n-1)!
because (4+10-1)!/10!(4-1)! = 13!/10!3! = 13!/3!10!, which is the same result that we got above.
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# What is Symmetry?
Dec 04 2007 Published by under Group Theory
As I said in the last post, in group theory, you strip things down to a simple collection of values and one operation, with four required properties. The result is a simple structure, which completely captures the concept of symmetry. But mathematically, what is symmetry? And how can something as simple and abstract as a group have anything to do with it?
Let's look at a simple, familiar example: integers and addition. What does symmetry mean in terms of the set of integers and the addition operation?
Suppose I were to invent a strange way of writing integers. You know nothing about how I'm writing them. But you know how addition works. So, can you figure out which number is which? Let's be concrete
about this: here's the set of numbers from -5 to 5, out of order, in my strange notation. {#, @, *, !, , ^, &, %, \$ }. If you give me two of them, I'll tell you their sum. What can you find out?
Well, you can figure out what zero is. With a bit of experimentation, you can see that adding
"&" and ">" gives you ">", and adding "&" and "~" gives you "~"; the only integer for which that could
be true is zero. So you can tell that "&" is how I'm writing zero.
What else can you figure out? With enough experimentation, you can find an ordering. You can find two
values, "@" and "#", where "@" plus "#" = ">"; ">" plus "#" equals "%", and so on - each time you add "#" to something, you get another value, and if you start with "@", and repeatedly add ">", you'll get a
complete enumeration of them: "@", ">", "%", "\$", "!", "&", "#", "*", "~", "^", ">".
So - which symbol represents one?
You can't tell. "@" might be -5, in which case "#" is 1. Or "@" might be 5, in which case "#" is -1. You can tell that "*" must be either +2 or -2, but you can't tell which. You can't tell which numbers are the positives, and which are the negatives. If all you have available to you is the group operation of addition, then there is no way for you to distinguish between the positive and the negative numbers.
Addition of the integers is symmetric: you can change the signs of the numbers
but by everything you can do with the group operator, the change is invisible. I can write an equation in my representation: "%" plus "*" equals "!", where I know that "%" is +3, "*" is -2, and "!" is +1. I can then switch the positive and the negative numbers, so that "%" is -3, "*" is +2, and "!" is -1, and the equation - in fact, any equation which relies on nothing more than the group operator of addition - can tell the difference. If you don't know what the symbols represent, you wouldn't even be able to tell that I'd changed my mind about what the symbols meant!
That's a simple of example of what symmetry means. Symmetry is an immunity to transformation. If something is symmetric, then that means that there is something you can do to it, some transformation you can apply to it, and after the transformation, you won't be able to tell that any transformation was applied. When something is a group, there is a transformation associated with the group operator which is
undetectable within the structure of the group. That "within the group" part is important: with the integers, if you have multiplication, you can distinguish between 1 and -1; 1 is the identity for multiplication, -1 is not. But if all you have is addition, that the transformation is invisible.
Think of the intuitive notion of symmetry: mirroring. What mirror symmetry means is that you can draw a line through an image, and swap what's on the left-hand side of it with what's on the right-hand side of it - and the end result will be indistinguishable from the original image. Addition based groups of numbers captures the fundamental notion of mirror symmetry: it defines a central division (0), and
the fact that swapping the objects on opposite sides of that division has no discernable effect.
For an example of how that can get more interesting that just the integers without changing the fundamental concept, you can look at the following example: given a hexagon, you can reflect it along many different dividing lines without recognizing any difference.
When some object is symmetrical with respect to a particular transformation, that means that you cannot distinguish between that object before the transformation, and that object after. There are, of course, many different kinds of symmetry beyond the basic mirroring ones that we're all familiar with. A few examples of basic geometric symmetries:
Scale: scale symmetry means that you can change the size of something without altering it.
Think of geometry, where you're interested in the fundamental properties of a shape - the number
of sides, the angles between them, the relative sizes of the sides. If you don't have any way of
measuring size on an absolute basis, then an equilateral triangle with sides 3 inches long and an equilateral triangle with sides 1 inch long can't be distinguished. You can change the scale of
things without creating a detectable difference.
Translation: translational symmetry means you can move an object without detecting any change. If you have a square grid, like graph paper, drawn on an infinite canvas, you can move it the
distance between adjacent lines, and there will be no way to tell that you changed anything.
Rotation: rotational symmetry means you can rotate something without creating a detectable change. For example, as illustrated in the diagram below, if you rotate a hexagon by 60 degrees, without any external markings, you can't tell that it's rotated.
There are, of course, many more, and we'll talk about some of them in later posts.
For a fun exercise, look at the Escher image at the top of this post. It contains numerous kinds of symmetries; several different kinds of reflective symmetries, translational symmetries, rotational symmetries, color-shift symmetries, and more. See how many you can find. I've been able to figure out at least 16, but I'm sure I've missed something.
• It depends what you mean by a "kind" of symmetry. The group is \$D_3ltimesmathbb{Z}^2\$ -- a dihedral group (to flip around one of the triangular shapes) along with a rank-2 lattice to slide the triangular shapes among each other, and a semidirect product to combine them together. So what's a "kind"? An element? A subgroup? A normal subgroup? Some sort of equivalence class of subgroups?
• Doug Spoonwood says:
[ Addition of the integers is symmetric: you can change the signs of the numbers but by everything you can do with the group operator, the change is invisible. I can write an equation in my representation: "%" plus "*" equals "!", where I know that "%" is +3, "*" is -2, and "!" is +1. I can then switch the positive and the negative numbers, so that "%" is -3, "*" is +2, and "!" is -1, and the equation - in fact, any equation which relies on nothing more than the group operator of addition - can tell the difference. If you don't know what the symbols represent, you wouldn't even be able to tell that I'd changed my mind about what the symbols meant!
That's a simple of example of what symmetry means. ]
How does group theory then capture the language of symmetry? If 2+3=3+2 indicates a symmetry, then we have symmetries outside of group theory. Do you mean to say that commutative or Abelian group theory captures the notion of symmetry?
• ollie says:
A couple of thoughts:
in a group, if x + y = x
then -x + x + y = -x + x
y = 0
So, once you know that you have a group and x + y = x, then you know that y = 0. Nothing further needed (inverses are unique in a group)
Symmetry and group theory: "kind" of symmetry: I think he means a non-trivial automorphism of a group.
How group theory captures symmetry: a group can act on a set and that group action can be thought of as "a symmetry".
example: consider that "devil angel" tiling of a disk. rotational symmetries of the disk can be obtained by letting the group S^1 (think of the set of complex numbers of modulus 1 with the operation of complex multiplication); this can give various rotations of that disk, some of which correspond to the symmetries of the images on the disk.
• Enigman says:
Symmetries are elegant, and groups are elegant; but fields, whose two operations form a group and a monoid, have always seemed less so to me, for all their utility... But, the most symmetrical geometry (in their group-theoretic classification) is projective geometry, with its points at infinity; and adjoining a related infinity to a field yields a more symmetrical pair of structures (symmetrical in the mirroring sense), each of which is a commutative generalization of an abelian group. (I just thought I'd throw that in, in case it interests anyone:-)
• Doug Spoonwood says:
[A couple of thoughts:
in a group, if x + y = x
then -x + x + y = -x + x
y = 0
So, once you know that you have a group and x + y = x, then you know that y = 0. Nothing further needed (inverses are unique in a group)]
Let '+' stand for ordinary multiplication. Consequently {1, -1}, with 1 and -1 representing the regular numbers '1' and '-1', qualifies as a group under '+', with x+1=x. But, since we let 1 equal the regular number 1, it doesn't equal 0.
I think you mean to say that if x+y=x, then y indicates the identity element of the group. It does so, specifically because the identity element gets defined as the element such that x"*"y=x=y"*"x, where "*" indicates any sort of operation that satisfies the group axioms.
[I think he means a non-trivial automorphism of a group.]
I don't see why you think so, since he didn't bring up the automorphism concept, at least not explicitly. I don't recall it myself. I don't think he meant to assume his readers would know that concept.
[How group theory captures symmetry: a group can act on a set and that group action can be thought of as "a symmetry".]
Perhaps, but Mark didn't talk about symmetry in such a sense. He used a much broader concept. The operator 'max' on {0, 1/2, 1} qualifies as logically symmetric or commutative, has an identity of '0', and works associatively. I can think of 0 min 1 as having 1 min 0 as its symmetrical counterpart. This corresponds to changing "the signs of numbers", as for a set {-1, 0, 1} I could write -1 min 1 and apply the transformation to get 1 min -1, and no change in the structure involved happens. But, there exists no inverse for min, so neither of those sets qualify as a group under min.
• David M says:
"I can then switch the positive and the negative numbers, so that "%" is -3, "*" is +2, and "!" is -1, and the equation - in fact, any equation which relies on nothing more than the group operator of addition - can tell the difference"
I think that should say, "can't tell the difference".
• Xanthir, FCD says:
Let '+' stand for ordinary multiplication. Consequently {1, -1}, with 1 and -1 representing the regular numbers '1' and '-1', qualifies as a group under '+', with x+1=x. But, since we let 1 equal the regular number 1, it doesn't equal 0.
I think you mean to say that if x+y=x, then y indicates the identity element of the group. It does so, specifically because the identity element gets defined as the element such that x"*"y=x=y"*"x, where "*" indicates any sort of operation that satisfies the group axioms.
Nod. He was just using "0" as the name of the identity element. That's common, iirc.
If we were talking about multiplication specifically, then yeah, we'd say "1". However, in more general group terms 0 is used. This is also because they use 1 as the name of the identity of the second operation in a field.
• Doug Spoonwood says:
What does 'iirc' mean?
• ollie says:
Yes, by "0" I meant "group identity element"; of course many might have thought I meant the identity with respect to the "addition" operation in a ring.
Sorry for the confusion.
• DBEllis says:
Doug, "iirc" is '"f I recall correctly".
• ollie says:
Doug, about your example: I like it. But I don't think that your set with binary operation (a semigroup) qualifies for symmetry. Here is why:
If one makes a multiplication table for abstract elements a, b, c; the first row of which looks like this:
a*a=a, a*b=b, a*c =c
Then it is clear that a
• Doug Spoonwood says:
[Then it is clear that a then b*c = c means that b]
I think you meant to write more here, but I don't know what.
I don't get your point, since if we write out a table with for {a, b, c}, with a
• Xanthir, FCD says:
Remember, everyone, you *can't* use less-than signs in your post directly. The MovableType software that ScienceBlogs uses interprets that as an html comment and tries to eat the rest of the line.
If you really need it, use & lt ;, but without the spaces. Use & gt ; as well if you want the other sign. Those will give you < and >.
As well, remember that the Preview feature will convert those html entities in your actual text, which will then give you the same problems. Either don't Preview, or copy the text of your message *before* hitting Preview, then paste it back in to replace the text that you see in the Preview window.
• Doug says:
Hi Mark,
Nice post on Escher images and other symmetries.
There are also Penrose Tiles
http://mathworld.wolfram.com/PenroseTiles.html
related to Polyforms.
http://mathworld.wolfram.com/Polyform.html
• Scientopia Blogs
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## Question
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State the five number summary of the following data set: 13, 14, 10, 4, 18, 17, 11, 10, 5, 7, 10, 19, 13
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## Gauth Tutor Solution
Bella
Stanford University
Electrical engineer
\begin{align*}\text{Minimum value} : 4\end{align*}, \begin{align*}Q_1 : 8.5\end{align*}
\begin{align*}\text{Median} : 11 \end{align*}, \begin{align*}Q_3 : 15.5\end{align*}
\begin{align*}\text{Maximum value} : 19\end{align*}
Explanation
The five-number summary includes the minimum, 1st quartile \begin{align*}(Q_1)\end{align*}, median, 3rd quartile\begin{align*}(Q_3)\end{align*}, and maximum.
The first step is to organize the data in order from smallest to largest.
13, 14, 10, 4, 18, 17, 11, 10, 5, 7, 10, 19, 13
4, 5, 7, 10, 10, 10, 11, 13, 13, 14, 17, 18, 19
There are 13 data.
It split into lower half and upper half.
4, 5, 7, 10, 10, 10 is the lower half of the data.
13, 13, 14, 17, 18, 19 is the upper half of the data.
To find minimum:
The minimum is the smallest value in the data set.
The minimum value is 4.
To find \begin{align*}Q_1\end{align*}:
The lower median is the median of the lower half of the data. It is also called the lower quartile or \begin{align*}Q_1\end{align*}.
\begin{align*}Q_1 = \frac{7 + 10}{2} = \frac{17}{2} = 8.5\end{align*}
\begin{align*}Q_1\end{align*} is 8.5.
To find median:
The median of a set of data is the middlemost value in the set.
The median is 11.
To find \begin{align*}Q_3\end{align*}:
The upper median is the median of the upper half of the data. It is also called the upper quartile or \begin{align*}Q_3\end{align*}.
\begin{align*}Q_3 = \frac{14 + 17}{2} = \frac{31}{2} = 15.5\end{align*}
\begin{align*}Q_3\end{align*} is 15.5.
To find maximum:
The maximum is the highest value in the data set.
The maximum value is 19.
Option B is the correct answer.
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# Lesson 3: Powers of Powers of 10
Let's look at powers of powers of 10.
## 3.1: Big Cube
What is the volume of a giant cube that measures 10,000 km on each side?
## 3.2: Taking Powers of Powers of 10
1. Complete the table to explore patterns in the exponents when raising a power of 10 to a power. You may skip a single box in the table, but if you do, be prepared to explain why you skipped it.
Row 1 expression expanded single power of 10
Row 2 $(10^3)^2$ $(10 \boldcdot 10 \boldcdot 10)(10 \boldcdot 10 \boldcdot 10)$ $10^6$
Row 3 $(10^2)^5$ $(10 \boldcdot 10)(10 \boldcdot 10)(10 \boldcdot 10)(10 \boldcdot 10)(10 \boldcdot 10)$
Row 4 $(10 \boldcdot 10 \boldcdot 10)(10 \boldcdot 10 \boldcdot 10)(10 \boldcdot 10 \boldcdot 10)(10 \boldcdot 10 \boldcdot 10)$
Row 5 $(10^4)^2$
Row 6 $(10^8)^{11}$
2. If you chose to skip one entry in the table, which entry did you skip? Why?
1. Use the patterns you found in the table to rewrite $\left(10^m\right)^n$ as an equivalent expression with a single exponent, like $10^{\boxed{\phantom{3}}}$.
2. If you took the amount of oil consumed in 2 months in 2013 worldwide, you could make a cube of oil that measures $10^3$ meters on each side. How many cubic meters of oil is this? Do you think this would be enough to fill a pond, a lake, or an ocean?
## 3.3: How Do the Rules Work?
Andre and Elena want to write $10^2 \boldcdot 10^2 \boldcdot 10^2$ with a single exponent.
• Andre says, “When you multiply powers with the same base, it just means you add the exponents, so $10^2 \boldcdot 10^2 \boldcdot 10^2 = 10^{2+2+2} = 10^6$.”
• Elena says, “$10^2$ is multiplied by itself 3 times, so $10^2 \boldcdot 10^2 \boldcdot 10^2 = (10^2)^3 = 10^{2+3} = 10^5$.”
Do you agree with either of them? Explain your reasoning.
## Summary
In this lesson, we developed a rule for taking a power of 10 to another power: Taking a power of 10 and raising it to another power is the same as multiplying the exponents.
See what happens when raising $10^4$ to the power of 3. $$\left(10^4\right)^3 =10^4 \boldcdot 10^4 \boldcdot 10^4 = 10^{12}$$
This works for any power of powers of 10. For example, $\left(10^{6}\right)^{11} = 10^{66}$. This is another rule that will make it easier to work with and make sense of expressions with exponents.
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# Simplify Surds Calculator
Centre Number Candidate Number Write your name here Surname Other names Total Marks Paper Reference Turn over Mathematics Higher Tier 1MA0/1H You must have: Ruler graduated in centimetres and millimetres,. 3) Once the perfect squares have been removed from beneath the square root radical, evaluate the surd (the remaining, unevaluated square root) using standard methods such as Newton's Method, Direct Calculation, a Square Root Table, or the Guess & Check Method. ) and rationalise denominators. Example: √ 2 (square root of 2) can't be simplified further so it is a surd How to multiply surds If you have two surds that are multiplied and their roots are the same, you put them under one root and multiply the values. It has an infinite number of non-recurring decimals. How might we simplify it? There is a technique explained in the notes called rationalizing the denominator. Surds Section A:(Non Calculator) 1. Completing the square – solve in surd form GCSE maths level 8 Complete the square – solving in surd form GCSE maths question level 8. In a simplified surd, the number that remains inside the square root should have no more square factors. √2 √5 √7 √5 √2 √2. The simplify fractions calculator shows the fraction in its simplest form and shows the Greatest Common Factor (GCF). For example √2 and √3 can be thought of in the same way as x and y (ie not 'like terms') whereas 2√3 and 5√3 can be thought of in the same way as 2x and 5x (ie we can add them to get 7√3). Solving linear equations 1. (the button converts fractions/surds into decimals, and back again). In this case, that would involve multiplying by $\sqrt{42}$ to get $\displaystyle\frac{3\sqrt{588}}{42}$, factoring out the $196$ from the $588$ to get $\displaystyle\frac{42\sqrt{3}}{42}$, and reducing the fraction to obtain $\sqrt{3}$. Paper 3 (Non-Calculator) Surds and Indices Calculators must not be used. Enter the angle into the calculator and click the function for which the half angle should be calculated, your answer will be displayed. if you have different surds then treat them like other letters, so you can only multiply and divide them, you cant add different surds together. Thus, we can only simplify the sum or difference of like surds. So learn to love surds, and be right more often. EXAMPLE: input the equation 2X 2 + 4X -30 = 0 as: A= 2 B= 4 C= -30. (d) 50 = 25 × 2 = 5 2. Example 3 Simplify. Example One - Simplify Surds. Surds Simplification of exponents Solving exponential equations Simplifying using log laws Solving exponential equations that require logarithms. (c) 48 = 16 × 3 = 4 3. Scientific Calculator. Rationalising the Denominator. SURDS: Simplify √ 90 into a√ b form? I need the answer to √ 90 in a√ b form and a brief explanation of how you achieved the answer. Also a simple explanation for factorisation is needed. We maintain a great deal of quality reference information on topics ranging from multiplying and dividing rational to college mathematics. Rationalize Denominator. Pythagoras with Surds. Edexcel Core 1 Algebra and functions Surds Exercise Do not use a calculator in this exercise. 6 Powers and surds on your calculator. On the other hand 5 7 and 3 7 are like surds. surds > adding and subtracting surds > 'simple' surds. Answer sheet Include answer sheet. They are therefore irrational numbers. 5 (a) Without using a calculator, find the value of. mathsrevision. 692820323 exactly. A root of a positive real quantity is called a surd if its value cannot be exactly determined. It can rationalize radical denominators with 2 radicals or less. 6 Powers and surds on your calculator. Since 1 and 4 add up to 5 and multiply together to get 4, we can factor it like: Current calculator limitations. It is a number that can't be simplified to remove a square root (or cube root etc). BYJU’S online simplifying ratios calculator tool makes the calculation faster, and it displays the simplified form in a fraction of seconds. Square root of 8 is the same thing as square root of 4 times the square root of 2, which is just 2 times root 2. Hence remember + , and -. It is important to realise the following : ab a b a b e g. Paper 3 (Non-Calculator) Surds and Indices Past Paper Questions Arranged by Topic Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. Simplify: (a) 20 (b) 32 (c) 45 (d) 242 (e) 300 (f) 16 25 (g) 49 100 2. The Multiplication Rule for Simplifying Surds In this tutorial you are shown the multiplication rule for surds. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. ( Recall that , which makes the square'' the outer layer, NOT the cosine function''. Example 3: to simplify type (27 (2/3-2x)^3-8 (1-9x))/ (216x^2). Simplifying fractional surds. This free course is available to start right now. Rationalising the Denominator. We leave them as surds because in decimal form they go on forever, so it uses up lots of ink to write them and accuracy is quickly lost. (b) 27 = 9 × 3 = 3 3. Algebra, Solving linear equations. Simplify expressions and solve equations using the laws of exponents for rational exponents where 𝑥 Û Ü= √𝑥𝑝 Ü; x > 0; q > 0 2. Thus, we can only simplify the sum or difference of like surds. Converting surds which are irrational numbers into a rational number is called rationalization. Simplify : Note:If by using certain maths techniques we remove the surd from either the top or bottom of the fraction then we say we are "rationalising the numerator" or "rationalising the denominator". 64 is the product of 2 multiplying itself six times. It has an infinite number of non-recurring decimals. b) Write (1 + vÃ)2 in the form a + bv/Ã. Symplify an expression or cancel an expression means reduce it by grouping terms. Surds Questions and Answers. This calculator simplifies polynomial expressions. Surd Rules (from Folens Publishers) The Square Root Game (from Math-Play. To simplify a surd, write the number under the root sign as the product of two factors, one of which is the largest perfect square. Examples. 1) −3 p + 6p 3p 2) b − 3 + 6 − 2b −b + 3 3) 7x − x 6x 4) 7p − 10 p −3p 5) −10 v + 6v −4v 6) −9r + 10 r r 7) 9 + 5r − 9r 9 − 4r 8) 1 − 3v + 10 11 − 3v 9) 5n + 9n 14 n 10) 4b + 6 − 4 4b + 2 11) 35 n − 1 + 46 35 n + 45 12) −33 v − 49. Step 1: Get rid of negative power − can be written as 1 16 0 Step 2: Use calculator to simplify 1 16 0 = = 1 65536 = 0. Worksheet 1: Memo – Exponents and Surds. Core 1: Surds. Learning how to Simplify Surds. Check your answers seem right. A Surds exam question where you are asked to "express as a surd in its simplest form" is shown below. This type of fraction is also known as a compound fraction. So 6√3 + 2√5 = 12√15 When we divide surds, we: 1. Double Bracket Surd Expansions worksheet. Fill in 3 of the 6 fields, with at least one side, and press the 'Calculate' button. is now simplified. Any time you demand service with math and in particular with free rationalize the denominator calculator or solving equations come visit us at Rational-equations. Back This calculator demonstrates simplifying a surd. FREE (1) Calmac4 Addition and Subtraction Rally Coach. Surds are numbers that cannot be square rooted exactly. d × d × d × d. I have been in your situation some time agowhen I was learning simplifying algebraic fractions calculator. However, the product of two surds need not be a surd (Eg. We get this awkward looking fraction. Subtract surds. The product of (a√h + b√k) and (a√h - b√k) is a rational number and this allows us to simplify fractions as in the example in the image on the left, by. The square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 etc To simplify a surd: 1. uk √20: √45 63 175√ : √28√. Online calculators and converters have been developed to make calculations easy, these calculators are great tools for mathematical, algebraic, numbers, engineering, physics problems. Add, subtract, multiply and divide simple surds. Exponents are supported on variables using the ^ (caret) symbol. It only takes a minute to sign up. Attempt every question. However + is not = = 3. About Surds. a 34 b 83 c 12 2 d 53 e 64 f (−3) 4 g 32 × 2 3 h 43 × 2 3 i 94 × 3 4 j 25 2 × 5 4 Determine the basic numeral for: a 5 × 5 × 5 × 5 b 4 × 4 × 4 c 10 × 10 × 10 × 10 × 10 × 10 d x × x × x × x × x × x × x e y × y f x3 × x2 g a × a2 × a3 h (m2. Do not spend too long on one question. Learn these general rules: Simplify √12. Example 1: to simplify type (x - 1) ( x + 1). How might we simplify it? There is a technique explained in the notes called rationalizing the denominator. How To Simplify Surds Part 2. How To Simplify Fraction. And this wonderful, explosive resource does exactly that. To consolidate and extend learning students evaluate calculations by simplifying each term. Form 3 Mathematics. Surds : Definition, Rules Of Surd,Types Reducing fractions or Simplifying Fraction calculator;. For example \sqrt 3 , 1+\sqrt 2 and 7\sqrt 5 are all examples of surds. Mathematics (Linear) - 1MA0 SURDS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. Hi , even I made use of Algebrator to learn more about algebraic fraction calculator. Surds Simplification of exponents Solving exponential equations Simplifying using log laws Solving exponential equations that require logarithms. In this example, we simplify 3√(500x³). = 10√3 - √6 multiply √3 by 10, remembering that if you multiply a number by a surd the number part stays out front of the surd, similar to 10 × y = 10y so 10 × √3 = 10√3 multiply √3 by √2, remembering []. In general, to simplify a surd, you want to "rationalize the denominator". So let's see how we can work with surds. Fraction Maker Online. So let's list the factors of 75 Factors: 1, 3, 5, 15, 25, 75. Topic outline. 16227766 is a surd while the √16 = 4 is not a surd since the square root is a whole number. Simplify: ab c de f gh i jk l mn o Simplify by collecting like surds. We can begin the factoring in any way. Always show your workings. Thus, we can only simplify the sum or difference of like surds. Either way, we simplify the calculation hugely. While it is often useful to approximate surds, for example, when finding the length of the hypotenuse in a right-angled triangle or the radius of a circle when given its area students need to be able to calculate with surds in their exact form. Recall that the numbers 1, 4, 9, 16, 25, 36, 49, are perfect squares. The laws of exponents can also be extended to include the rational numbers. but it is now used for a root that is. Hence remember + , and -. We get this awkward looking fraction. It is a number that can't be simplified to remove a square root (or cube root etc). Visual Fractions. It may not be possible to write the answer to your question as a surd But for everything else, there's the S<=>D button! It's just above the DEL key on my calculator, and all the Casio models seem to be pretty similar. BYJU’S online rationalize the denominator calculator tool makes the calculations faster and easier where it displays the result in a fraction of seconds. The square root of 2, or the (1/2)th power of 2, written in mathematics as √ 2 or 2 1 ⁄ 2, is the positive algebraic number that, when multiplied by itself, equals the number 2. Instructions Use black ink or ball-point pen. Solve an equation of the form a x 2 + b x + c = 0 by using the quadratic formula: − b ± √ b 2 − 4 a c. Simplify by rationalising the. Rationalize Denominator. Worksheet on simplifying surds. This online simplifying ratio calculator helps you to compare ratios in a more efficient way. In this example, we simplify 3√(500x³). Simplify as far as possible: 4 (a) 2 x 2 3 (c) 7 x 8 (d) 3( 2 1) (e) 2. The narrator presents many methods to simplify square roots to appeal to different learning styles. 9] B Exponent or index laws [1. Percent button is used to find the percentage of a number. Step 1: Enter the expression you want to simplify into the editor. Perfect for the summer term, Wimbledon or for revising surds for GCSE maths. This calculator eliminate a radicals in a denominator. Really the OP needs to redo the previous part of the question without using a calculator, so as to obtain the correct surd answer (which, in all fairness, is likely to be what you suggest). We get this awkward looking fraction. For example, is not a surd, as the answer is a whole number. Simplify an Algebraic Term Involving Exponents and/or Powers Algebraic expression frequently involve terms that have exponents. Surds is a Form 3 Mathematics topics. Covers all aspects of both the GCSE and IGCSE Further Maths curriculum. AQA IGCSE FM question compilation which aims to cover all types of questions that might be seen on the topic of surds and laws of indices. Holt Algebra Online, fraleigh solutions, algebra 2 with pizzazz, Perfect Square Trinomial Examples, linear quadratic calculator. CHAPTER 5 INDICES AND SURDS 99 Simplify each expression by writing in index form. There are two methods used to simplify such kind of fraction. In general, to simplify a surd, you want to "rationalize the denominator". Floor/Ceiling (new) System of Equations. If the $$n^{\text{th}}$$ root of a number cannot be simplified to a rational number, we call it a surd. Higher Tier. Plenty of practice. Click 'Mark' to mark your answer. (b) Find the value of. [3] (b) Express 7+ p 5 3+ p 5 in the form a+b p 5, where a and b are integers. √12 = √(4 × 3) = √4 × √3 = 2√3). January Week 2 extra Mostly Non Calculator! Slippery surds Expand and simplify Give your answer in the form , where and are integers. A surd is a number written in a form that includes a square root. • use the term 'surd' to refer to irrational expressions of the form n √x where x is a rational number and n is an integer such that n≥2 : definition: NCM10 1-01 • write recurring decimals in fraction form using calculator and non-calculator methods, eg : youtube (Need to source elsewhere). The laws of exponents can also be extended to include the rational numbers. (5 Marks) 3. Test 2 - Week beginning 9th December 50 minutes, non-calculator Place value Factors, multiples & primes Decimals Algebraic expressions & simplification. 0:04 Skip to 0 minutes and 4 seconds MICHAEL: So sometimes it's better to leave our answers in surd form. How to simplify a surd. Instructions to Candidates. Any time you require guidance on factor or functions, Algebra1help. In Matthew 5:44-Love your enemies, bless them that curse you, do good to them that hate you, pray for them which despitefully use you and persecute you. Click 'Next Question' to move on to the next question. click E N T E R and your answers should be 3 and -5. Surds is a Form 3 Mathematics topics. Complete the Square. Number of problems 5 problem 10 problems 20 problems. 692820323 exactly. Using this knowledge you can break the number under the root sign into factors that are perfect squares like so: \begin{equation*} \sqrt{12} = \sqrt{4 \times 3} = \sqrt{2^2 \times 3} = \sqrt{2^2} \times \sqrt{3} = 2 \sqrt{3}. You do not need to worry about the square root sign, that's already there! Learn the steps this calculator uses below. On non-calculator papers, its always best to use fractions, so the first step will to turn the 0. :) Answers included. And here is how to use it: Example: simplify √12. (d) 50 = 25 × 2 = 5 2. 4 What is a Surd S5 Int2 12 6 The above roots have exact values and are called. We can write it as 1 / 2. Inclusive are prerequisites and videos to essential skills. (b) Hence, explain why 32 ×3 − 1 3 3 3−3 3< 32 ×3 − 1 3 3 3−3 <9 7 (a) Simplify. Simplifying Expression of Surds Using Basic Operations - Examples Question 1 : Simplify the following using addition and subtraction properties of surds: (i) 5 √ 3 + 18 √ 3 − 2 √ 3. Simplifying a square root just means factoring out any perfect squares from the radicand, moving them to the left of the radical symbol, and leaving the other factor inside the radical symbol. simplify surd expressions involving squares (eg. is also equal to and , but it is neither an entire surd nor a surd in standard form. PUT DATE IN: Paper 1 - Non-Calculator. At worst, we carry out one computation with 3. Given that f = E, g and h = simplify each of the following, indicating in each case whether your answer is rational or irrational. Simplify: (a) 5 12 (b) 5 2 3 2 (c) 45 5 (d) 75 3 48 6 3. Simplify (1) 5. Always show your workings. Write √12+ 15 √3 in the form √ where and are prime numbers [3] 12. Sine calculator. Surds Use In Real Life) in the table below Click on the pertaining program demo button found in the same row as your search phrase If you find the software demo helpful click on the buy button to buy the program at a special low price extended only to algebra-tutoring. There are two methods used to simplify such kind of fraction. 5 (a) Without using a calculator, find the value of. How to do Surds GCSE (simplifying, rationalising) Expanding surds in brackets and simplifying Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. Factoring Expressions Video Lesson. Surds Problem solving. 5 Estimating surds (EMA9). surd because, again trying different numbers, we find 1× 1 × 1 = 1, 2 × 2 × 2 = 8, 3 × 3 × 3 = 27, but we don’t know any number that we can do this with and end up with 5. Rationalising the Denominator. Thus, we can only simplify the sum or difference of like surds. Even if you know how to do stuff like surds and manipulate expressions, it's still easier to read and is more accurate. See examples below. com To create your new password, just click the link in the email we sent you. These are used when an exact answer is required i. Also, we multiply r (n) times In this case, r = 2 and n = 5. Practice surds with Cazoomy. There are lots of tricks to simplify surds and these two videos from maths520 show them clearly. Worksheet 3 Topic: Indices & Surds QUESTION 1 The area of a rectangle is (1 + V6)m'. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t ′ s c o r r e c t. 𝟒𝟓 to its simplest surd form. That is: Example 11. Solving linear equations 1. The video below explains that surds are the roots of numbers that are not whole numbers. 3 a × 3 a = 3 a × 3 a = 9 a 2 = 3 a. New Resources. Step 1: Find the largest square number that is a factor of 45. After rationalisation and a little bit of tidying up, the expression becomes:. Since, 2 * 2 * 2 = 8, the cube root of 8 is 2. Simplifying Surds 1) Simplify each of the following: 2) Simplify each of the following: 3) Simplify each of the following: 4) Simplify each of the following:. GCSE: Indices & Surds CM 1 Simplify each of the following expressions (a) (b) (c) (d) surds. We can begin the factoring in any way. (b) Find the value of. Who knows where it may lead. For example. Rationalize Denominator. Alternative versions. Surds are numbers that cannot be square rooted exactly. There are a bounty of websites that have the basic tutorials as to what surds are - for example, the Math is Fun page on surds provides a colorful, yet still dry representation of the definition When we can't simplify a number to remove a square root (or cube root etc) then it is a surd. Notice that the most difficult part to simplify is the denominators that consists of surds. surds > multiplying and dividing > multiplying 'simple' surds. 2 11 = or5. Ready-to-use mathematics resources for Key Stage 3, Key Stage 4 and GCSE maths classes. 5*x^3 + 10*x^2 + 5*x. Once a question is marked, it cannot be edited. com and learn about mathematics content, multiplying polynomials and plenty other algebra topics. n n n ab a b aa b b X-ample Questions 1. Simplifying Surds Some surds can be simplified. l You must answer the questions in the spaces provided. Simplifying Polynomials. plz help with surds. Work out. Get step by step simplification of any surd, or square root, by typing the number into the white box above and clicking "simplify". To see extra written explanation next to the work, click on the "verbose mode" here. The first layer is the square'', the second layer is the cosine function'', and the third layer is. A surd is said to be in its simplest form if the number under the root sign has no perfect square as a factor. Title: Surds 1 Surds S5 Int2 Simplifying a Surd Rationalising a Surd www. Multiply/Divide surds. and are equal surds. This was just a remarkable tool that helped me with all the basic principles. have no surds in the denominator 3. Simplify an Algebraic Term Involving Exponents and/or Powers Algebraic expression frequently involve terms that have exponents. 4] D Standard form [1. Review the full course description and key learning outcomes and create an account and enrol if you want a free statement of participation. Back This calculator demonstrates multiplying surds. You do not need to worry about the square root sign, that's already there! Learn the steps this calculator uses below. Complex surd question. The questions must be done in order, from Q1 onwards. 5 Estimating surds (EMA9). Any mixture of symbols involving the square root sign is called a surd expression. The fraction calculator will simply the answer for you. To achieve this, it is important to 2 = 1:41 (to 2 decimal places), simplify p1 2 without a calculator. Rationalize the denominator calculator is a free online tool that gives the rationalized denominator for the given input. Videos, worksheets, 5-a-day and much more. This section summarises all the key ideas required for the surd section at AS-level. Now rationalise the following: a) √. Step 2: Click the blue arrow to submit and see the result!. SURDS AND INDICES 3 1. Based on the AQA syllabus. Learning how to Simplify Surds. Those marked * are introduced/explored/delivered in the BGE phase and required to deepen your understanding to National 5 level. Firstly, find the factors of the number where one of the factors is a square number such as 4, 9, 16, 25 and so on. Expanding Surds worked examples Expand and simplify √3(10 - √2). Note that the factor 16 is the largest perfect square. They give decimals that never repeat and never end. Paper 3 (Non-Calculator) Surds and Indices Calculators must not be used. In the Ultimate Maths Library, you can browse through the numerous maths topics on our website as well as search our content by level of ability. surds > multiplying and dividing > multiplying 'simple' surds. Double Bracket Surd Expansions worksheet. See examples below. 19/10/2013 1 19 October 2013 Contents Simplifying a Surd Rationalising a Surd Conjugate Pairs Trial & Improvement 19/10/2013 2 Starter Questions Use a calculator to find the values of : = 6 = 12 = 3 = 2 36 144 3 8 4 16 2 3 21 41. In this courses, many questions on surds have been worked out. You can find out more about surds here. A LEVEL LINKS. And the square root of 25 times 3 is equal to the square root of 25 times the square root of 3. A worked example of simplifying radical with a variable in it. For many of these operations, you will have to simplify the surd first. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power. Practice surds with Cazoomy. The video below explains that surds are the roots of numbers that are not whole numbers. Surds : Definition, Rules Of Surd,Types Reducing fractions or Simplifying Fraction calculator;. Any time you require guidance on factor or functions, Algebra1help. Add/Subtract surds. com happens to be the perfect place to check out!. It is a number that can't be simplified to remove a square root (or cube root etc). Remember the rule :√a x √a = √a. CHAPTER 5 INDICES AND SURDS 99 Simplify each expression by writing in index form. A surd is said to be in its simplest form when the number under the root sign has no square factors. Therefore, naturally we need to rationalize the denominator. What does "Rationalizing the Denominator" mean?. r/TI_Calculators: A place to post programs, questions, requests, news, and other stuff for Texas Instruments calculators. Worksheet on simplifying surds. Simplify Fraction Calculator. Index Form Calculator. org makes available useful strategies on simplifying surds calculator online, radicals and elimination and other algebra topics. It is very common for to be so we usually do not write. containing surds, gives only fractions or whole numbers. Therefore, surds are irrational numbers. Pythagoras encountered this when calculating the diagonal of a square of side length 1 unit. Online calculators and converters have been developed to make calculations easy, these calculators are great tools for mathematical, algebraic, numbers, engineering, physics problems. The laws of exponents apply to positive real numbers and and non-negative integers and. The calculator works for both numbers and expressions containing variables. Simplify as far as possible: 4 (a) 2 x 2 3 (c) 7 x 8 (d) 3( 2 1) (e) 2. Simplify Fraction Calculator. asked by lucy on March 28, 2020. We are learning rules for simplify surds. Simplifying a square root just means factoring out any perfect squares from the radicand, moving them to the left of the radical symbol, and leaving the other factor inside the radical symbol. Simplify 4. The fraction calculator will simply the answer for you. Now, unless gcd(a, m) evenly divides b there won't be any solutions to the linear congruence. Calculator wich can simplify an algebraic expression online. Exponents are supported on variables using the ^ (caret) symbol. rationalise the denominators of fractions - i. uk Surds (H) - Version 2 January 2016 Surds (H) A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, OCR, Pearson-Edexcel and WJEC Eduqas. 75 has the square factor 25. l You must answer the questions in the spaces provided. Since these definitions take on new importance in this chapter, we will repeat them. An exponent is just a convenient way of writing repeated multiplications of the same number. In the event that you need to have assistance on exponential and logarithmic or perhaps adding and subtracting rational, Mathsite. Surds are square roots which can't be reduced to rational numbers. We stop at this stage seeing that. ab c de f gh i j Remember how in algebra only like terms could be added or subtracted. Expand online. Easy and convenient to use and of great help to students and professionals. Example 3 Simplify. Paper 3 (Non-Calculator) Surds and Indices Past Paper Questions Arranged by Topic Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. com To create your new password, just click the link in the email we sent you. A-level Mathematics Specification Specification for first teaching in 2017 (1011k) Content. On the other hand 5 and 3 are like surds. Plenty of practice. Without using a calculator, show that√20=2√5 [2] 4. , so we can write. So let's see how we can work with surds. To simplify surds you should know square numbers ($$2^2 = 4$$, $$3^2 = 9$$, $$4^2 = 16$$, etc). Instructions. Math Problem Solver (all calculators) Simplify Expression Calculator. Review the full course description and key learning outcomes and create an account and enrol if you want a free statement of participation. They are numbers which, when written in decimal form, would go on forever. Simplify #(3)/(3-sqrt 7)# 2m 23s; 3. Example of a nested radicals include: √(8 + √3). Note that we cannot take the square root of a negative number. Rearranging surds, then, is the business of noticing that the square root of 12 multiplied by the square root of 3 can be combined to give the square root of 36, which is 6. Key points • When you expand one set of brackets you must multiply everything inside the bracket by what is outside. GCSE MathematicsAutumn 1 Revision List (Intermediate). The reason we leave them as surds is because in decimal form they would go on forever and so this is a very clumsy way of writing them. Easy and convenient to use and of great help to students and professionals. sqrt243=9*sqrt3 square the number that can be squared therefore the answer is 9sqrt3 use this link for further explanation on the concept. roots of the number (or expression) in the radicand. calculator gives answer to sq rt 8 as 4 rt 2; Simplify surd expressions involving squares (e. Example One - Simplify Surds. Need more problem types? Try MathPapa Algebra Calculator. Simplify algebraic expressions step-by-step. Simultaneous equations. Chapter 1E - Simplifying Surds We started on Surds back in year 9, and did it again last year. Rearranging surds, then, is the business of noticing that the square root of 12 multiplied by the square root of 3 can be combined to give the square root of 36, which is 6. Simplify 47. Really the OP needs to redo the previous part of the question without using a calculator, so as to obtain the correct surd answer (which, in all fairness, is likely to be what you suggest). Enter the percentage amount, click the % button, then enter the number you want the percentage of, and then click equals. ) (d) Given p 3 = 1:73 (to 2 decimal places), simplify 3 2+ p 3 without a calculator. Simplifying Surds Jigsaw GCSE, GCSE Higher, Jigsaw / Dominoes, Jigsaws, Surds, Uncategorized Add comments. Simplifying Surds 1) Simplify each of the following: 2) Simplify each of the following: 3) Simplify each of the following: 4) Simplify each of the following:. 4] C Zero and negative indices [1. Complete the Square. You need to be able to simplify expressions involving surds. One of these two numbers must be a perfect square. Simplifying Ratios Calculator is a free online tool that displays the simplified form of the given ratios. In a simplified surd, the number that remains inside the square root should have no more square factors. Again, fewer keystrokes = fewer errors. 25 Examiner only Arholwr yn unig 6 43 3 xx- 5 [4] 26. Surds are numbers that cannot be square rooted exactly. Multiplication of Entire Surds. When simplifying an expression, the first thing to look for is whether you can clear any parentheses. To use the calculator simply type any positive or negative number into the text box. For example, is not a surd, as the answer is a whole number. What are we trying to do when simplifying a surd. Surds can be simplified if the number in the root symbol has a square number as a factor. Simplifying Surds 1) Simplify each of the following: 2) Simplify each of the following: 3) Simplify each of the following: 4) Simplify each of the following:. click E N T E R and your answers should be 3 and -5. This calculator also simplifies proper fractions by reducing to lowest terms and showing the work involved. Square root button is used to calculate the square root of a number. worksheets. This section summarises all the key ideas required for the surd section at AS-level. Convert improper fractions to mixed numbers in simplest form. Remember the rule :√a x √a = √a. Active 11 months ago. Year 11 Number Maths Worksheets Our year 11 number worksheets have been created to help students of all abilities to understand and solve the many varieties of number questions they will come across during their time in year 11. Without using mathematical tables or a calculator, evaluate #(sin 30^o - sin 60^o)/(tan 60^o)# 4m 43s; 2. When manipulating expressions containing surds (eg expanding brackets) I make comparisons to what they already know about algebra. October 17 Simplifying expressions Simplifying fractions Simplifying ratio Simplifying surds Simultaneous equations Sine rule Sketching quadratic. Indices and Surds. (b) Express in the form , where a is. If the 2 numbers inside the surd are the same, this creates a whole number. Multiplication of Entire Surds. More about this course. On the other hand 5 7 and 3 7 are like surds. There are lots of tricks to simplify surds and these two videos from maths520 show them clearly. Note: with surds, like with algebra, we often miss out the multiplication symbol and write 2\sqrt{7} instead of 2\times\sqrt{7}. C1 Algebra – Surds and Indices C1 Algebra: Surds and Indices – Mark Schemes 12 4. Check your answers seem right. Get the free "Surd Calculator MyAlevelmathsTutor " widget for your website, blog, Wordpress, Blogger, or iGoogle. Surds are the square roots of 2, 3, 5, 50. Read each question carefully before you begin answering it. So 8√6 ÷ 2√3. Show Step-by-step Solutions Surds - Question 23 Edexcel GCSE Maths Non-Calculator Solution. Instructions to Candidates. Review the full course description and key learning outcomes and create an account and enrol if you want a free statement of participation. To use it, replace square root sign ( √ ) with letter r. Enter the angle into the calculator and click the function for which the half angle should be calculated, your answer will be displayed. Fraction Tutorial. Surds - Rationalising the denominator Surds (i) - Simplifying surds (ii) - Surd arithmetic (iii) - Rationalising the denominator; ANSWERS; Page 4 of 5. A surd is a number written in a form that includes a square root. Rational and Irrational Numbers - In the test for rational and irrational numbers, if a surd has a square root in the numerator, while the denominator is '1' or some other number, then the number represented by the expression. There are certain rules that we follow to simplify an expression involving surds. You could use a calculator to find that but instead of this we often leave our answers in the square root form, as a surd. Powers and Roots. A rational number is any number that can be written as a fraction with an integer in the numerator and in the denominator. Practice surds with Cazoomy. They can only be approximated in decimal form. First do the tutorial, then try this smart quiz on indices from MangaHigh. Surds, and other roots mc-TY-surds-2009-1 Roots and powers are closely related, but only some roots can be written as whole numbers. Like surds involve the square root of the same number. For example,√256 is not a surd, as the answer is a whole number. Surds are important in calculations where accuracy is required. Then type C and press ALPHA +, 2ND x², ALPHA + and write N. You can find out more about surds here. Rules for operations on surds Rules for surds are the same as the rules for simplifying roots and involve the same rules of. is also equal to and , but it is neither an entire surd nor a surd in standard form. click E N T E R and your answers should be 3 and -5. jonesk5 Functional Skills Maths Revision Bundle both levels. Hence remember + , and -. 0:13 Skip to 0 minutes and 13 seconds PAULA: OK, so by 'surd form' what we mean is when we have our final answer, we'll have some numbers and we'll also have numbers written with a square root symbol. The laws of exponents can also be extended to include the rational numbers. Simplify #(3)/(3-sqrt 7)# 2m 23s; 3. Addition and Subtraction of Surds. Numbers of the form 2, 3, 5 etc. the term 'surds' is fairly new, it used to be 'simplifying equations' 🙂 Top Tips! Surds are just another way of showing a value the square root of 2 is more accurate (and easier to write) than 1. Simplify Simplify fully 11 5+2Ñã Show that can be written as 4 (2 marks) (2 marks) You must show all your working. Step 1: Find the largest square number that is a factor of 45. 11h: Surds and in exact calculations. If you learn the rules for exponents and radicals, then your enjoyment of mathematics will. I would suggest you to try this before resorting to the assistance from private instructor , which is often very expensive. The questions are in table format. 3 Solving surd equations (EMBFB) We also need to be able to solve equations that involve surds. In the event you actually require assistance with math and in particular with simplify the sum calculator or dividing polynomials come visit us at Polymathlove. Expressing powers of 10. Adding and subtracting surds. The classwiz calculator leaves things in natural form, and I've always wondered why the more 'advanced' calculators like most that do graphs, often don't work in natural form. have only natural numbers inside radicals 2. eg 5x + 3x = 8x 7a − 6a = a I think I’m beginning to like this. (a) 8 = 4 × 2 = 2 2. The inquiry might then follow one of two pathways. but it is now used for a root that is. When simplifying an expression, the first thing to look for is whether you can clear any parentheses. This online simplifying ratio calculator helps you to compare ratios in a more efficient way. We keep a tremendous amount of really good reference tutorials on matters varying from equations by factoring to algebraic expressions. Differentiated Learning Objectives. Surds (non-calculator) - Higher 1 (a) Simplify 15 4 [1 mark] Answer 1 (b) Simplify 8 96 aGive your answer in the form 3, where a is an integer. BYJU'S online simplifying ratios calculator tool makes the calculation faster, and it displays the simplified form in a fraction of seconds. This entry was posted in Exponents and Surds, General, Grade 11, Grades, Maths. Easy and convenient to use and of great help to students and professionals. #N#Inverse cosine calculator. Recall that the numbers 1, 4, 9, 16, 25, 36, 49, are perfect squares. Find, without using a calculator, the length of the other side in the fonn Va-Vb, where a and b are integers. Back This calculator demonstrates multiplying surds. Simplifying radical expressions that involve variables. Free statement of participation on completion. Simplifying a square root just means factoring out any perfect squares from the radicand, moving them to the left of the radical symbol, and leaving the other factor inside the radical symbol. Simplify as far as possible: 4 (a) 2 x 2 3 (c) 7 x 8 (d) 3( 2 1) (e) 2. Since these definitions take on new importance in this chapter, we will repeat them. This calculator simplifies polynomial expressions. Demo: Simplify a surd expression. Use rules to simplify surds. Again, fewer keystrokes = fewer errors. What Is a Surd? (1 mark) 2. Full Coverage: Surds 1 files 25/01/2018. Detailed solutions and answers to the questions are provided. Simplify algebraic expressions step-by-step. when numbers are left in root form to express their exact value they are known as surds, surds are irrational as they are infinite numbers with recurring decimals, this online calculator will help you determine whether provided number is surd or not. Factoring Expressions Video Lesson. Press J to jump to the feed. What are the Conditions for a Number to be a Surd? - Duration: 4:16. A surd is a number that has a 'nasty' square root such as √10 = 3. Calculator skills. The following examples will illustrate this point. It may not be possible to write the answer to your question as a surd But for everything else, there's the S<=>D button! It's just above the DEL key on my calculator, and all the Casio models seem to be pretty similar. but it is now used for a root that is. (a) Expand and simplify 7+ p 5 3 p 5. where a calculator is either not allowed or unavailable, and the calculations to be undertaken involve irrational values. How To Simplify Fraction. Simplify: ab c de f gh i jk l mn o Simplify by collecting like surds. sorry mate, unlessm you have the new casio touch calculator, it is impossible, and ull just have to use exat values and surds and Pi answers to the best of your ability. 692820323 exactly. On the other hand 5 7 and 3 7 are like surds. In the event you actually require assistance with math and in particular with simplify the sum calculator or dividing polynomials come visit us at Polymathlove. Enter a number or an expression and click "Factor". A surd is in its simplest form when the number under the square root sign is not a factor of a perfect square. (d) 50 = 25 × 2 = 5 2. Again, fewer keystrokes = fewer errors. Simplifying Variable Expressions Date_____ Period____ Simplify each expression. Show that can be written as 2- 3. Rationalize Denominator. Get your calculator and check if you want: they are both the same value! Here is the rule: when a and b are not negative. 5 (a) Without using a calculator, find the value of. Get the free "Simplify Surds" widget for your website, blog, Wordpress, Blogger, or iGoogle. The Reducing Fractions Calculator will reduce any two fractions that you enter in. 7:1, respectively. For example \sqrt 3 , 1+\sqrt 2 and 7\sqrt 5 are all examples of surds. See examples below. There are certain rules that we follow to simplify an expression involving surds. Alternative versions. :) Answers included. If the root of a number cannot be simplified to a rational number, we call it a surd. ( Recall that , which makes the square'' the outer layer, NOT the cosine function''. The goal of simplifying expressions with square roots is to factor the radicand into a product of two numbers. Half angle formulas are used to integrate the rational trigonometric expressions. They have some interesting connections with a jigsaw-puzzle problem about splitting a rectangle into squares and also with one of the oldest algorithms known to Greek mathematicians of 300 BC - Euclid's Algorithm - for computing the greatest divisor common to two numbers (gcd). This type of fraction is also known as a compound fraction. To simplify a surd, write the number under the root sign as the product of two factors, one of which is the largest perfect square. Surds are the square roots of 2, 3, 5, 50. These two surds are called unlike surds, in much the same way we call 2 x and 3 y unlike terms in algebra. (Note: if more than 3 fields are filled, only a third used to determine the triangle, the others are (eventualy) overwritten. (the button converts fractions/surds into decimals, and back again). To do this, we want to look at the factors of each number. A rational number is any number that can be written as a fraction with an integer in the numerator and in the denominator. A calculator gives a value of 𝜋 as 3. Surds are numbers left in square root or cube root format. Reduction of Surds - This is a way of making the square root smaller by examining its squared factors and removing them. After rationalisation and a little bit of tidying up, the expression becomes:. :) Answers included. fg+h fg fh 10 — (3) (2) (2) Simplifyv'îãã , leaving your answer in surd form. Ready-to-use mathematics resources for Key Stage 3, Key Stage 4 and GCSE maths classes. Any mixture of symbols involving the square root sign is called a surd expression. It is a number that can't be simplified to remove a square root (or cube root etc). 11h: Surds and in exact calculations. Simplify Fractions Calculator This fraction calculator will automatically simplify results. 9] B Exponent or index laws [1. Now, unless gcd(a, m) evenly divides b there won't be any solutions to the linear congruence. Surds are numbers left in square root form that are used when detailed accuracy is required in a calculation. ) (d) Given p 3 = 1:73 (to 2 decimal places), simplify 3 2+ p 3 without a calculator. Question 1: (6 +\sqrt{11})( 6 - \sqrt{11}) Question 2: (3 -\sqrt{10})( 3 - \sqrt{10}) Question 3: (2 +\sqrt{7})( 2 - \sqrt{7}) Question 4: (5 +\sqrt{11})( 5 + \sqrt{11}). Multiply two surds. For many of these operations, you will have to simplify the surd first. After rationalisation and a little bit of tidying up, the expression becomes:. 2007 Q 15 P2 Simplify 3 1 5 2 5 +-leaving the answer in the form a b c+, where a , b and c are rational numbers (3 marks) 9. worksheets. ) (d) Given p 3 = 1:73 (to 2 decimal places), simplify 3 2+ p 3 without a calculator. The laws of exponents can also be extended to include the rational numbers. For example √2 and √3 can be thought of in the same way as x and y (ie not 'like terms') whereas 2√3 and 5√3 can be thought of in the same way as 2x and 5x (ie we can add them to get 7√3). Detailed solutions and answers to the questions are provided. How to simplify a surd. g √5 x √5 =5 if not, you multiply the numbers outside the surd together, the numbers inside the surd together and then simplify. May 12, 2019 Craig Barton. Index Form Calculator. so your answer would be 3√2+√2=4√2. For example \sqrt 3 , 1+\sqrt 2 and 7\sqrt 5 are all examples of surds. Example 1: to simplify type (x - 1) ( x + 1). a 34 b 83 c 12 2 d 53 e 64 f (−3) 4 g 32 × 2 3 h 43 × 2 3 i 94 × 3 4 j 25 2 × 5 4 Determine the basic numeral for: a 5 × 5 × 5 × 5 b 4 × 4 × 4 c 10 × 10 × 10 × 10 × 10 × 10 d x × x × x × x × x × x × x e y × y f x3 × x2 g a × a2 × a3 h (m2. Test 2 - Week beginning 9th December 50 minutes, non-calculator Place value Factors, multiples & primes Decimals Algebraic expressions & simplification. 1n3qjox18n8pqq, cqi31gyoiums0, dl57bfkpfhyauc, vfqplx611qjrg, 53e9iurapa, bvewko0kil0rv, mzhxtrashnhc, nu53myx2o0, v47izs3juml, cxrshhwqvcornfw, znuvti9qyw, d62w1o0ruomvo, 74zbff2pz60u, bli3j8d8wucm4pw, fgx321v4zpy9, rcag0kltnjei, 970rn0ip5vh, oz6ujc88z07o8ie, u4belvzhp7, ove57y0j00, wac683r8lwbsvd, n7eb375e1umip5v, hcz7jpz4q4l, 0s2qs61uxy80xnm, gyynqfzhwz4rc, 8zm2hz524vms92, 0ap01vri0z, lvsoltdp15cdqq, ng9ulsfl35imxph, 1vz5u30l4hei5ye, juyqezkf9405jwf, 1yeb1c0axle6
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# 2.2: Constant coefficient second order linear ODEs
Suppose we have the problem
$y'' - 6y' + 8y = 0, y(0) = -2, y'(0) = 6$
This is a second order linear homogeneous equation with constant coefficients. Constant coefficients means that the functions in front of $$y''$$, $$y'$$, and $$y$$ are constants and do not depend on $$x$$.
To guess a solution, think of a function that you know stays essentially the same when we differentiate it, so that we can take the function and its derivatives, add some multiples of these together, and end up with zero.
Let us try a solution of the form $$y = e^{rx}$$. Then $$y' = re^{rx}$$ and $$y'' = r^2e^{rx}$$. Plug in to get
$y'' - 6y' + 8y = 0$
$r^2e^{rx} - 6re^{rx} + 8e^{rx} = 0$
$r^2 - 6r + 8 = 0$
divide through by $$e^{rx}$$
$(r - 2)(r - 4) = 0$
Hence, if $$r = 2$$ or $$r = 4$$, then $$e^{rx}$$ is a solution. So let $$y_1 = e^{2x}$$ and $$y_2 = e^{4x}$$.
Exercise $$\PageIndex{1}$$:
Check that $$y_1$$ and $$y_2$$ are solutions.
Solution
The functions $$e^{2x}$$ and $$e^{4x}$$ are linearly independent. If they were not linearly independent we could write $$e^{4x} = Ce^{2x}$$ for some constant $$C$$, implying that $$e^{2x} = C$$for all $$x$$, which is clearly not possible. Hence, we can write the general solution as
$y = C_1e^{2x} + C_2e^{4x}$
We need to solve for $$C_1$$ and $$C_2$$. To apply the initial conditions we first find $$y' = 2C_1e^{2x} + 4C_2e^{4x}$$. We plug in $$x = 0$$ and solve.
$-2 = y(0) = C_1 + C_2$
$6 = y'(0) = 2C_1 + 4C_2$
Either apply some matrix algebra, or just solve these by high school math. For example, divide the second equation by 2 to obtain $$3 = C_1 + 2C_2$$, and subtract the two equations to get $$5 = C_2$$. Then $$C_1 = -7$$ as $$-2 = C_1 + 5$$. Hence, the solution we are looking for is
$y = -7e^{2x} + 5e^{4x}$
Let us generalize this example into a method. Suppose that we have an equation
$ay'' +by' +cy = 0,$
where $$a, b, c$$ are constants. Try the solution $$y = e^{rx}$$ to obtain
$ar^2 e^{rx} + bre^{rx} + ce^{rx} = 0$
$ar^2 + br + c = 0$
The equation $$ar^2 + br + c = 0$$ is called the characteristic equation of the ODE. Solve for the $$r$$ by using the quadratic formula.
$r_1, r_2 = \dfrac {-b \pm \sqrt {b^2 - 4ac}}{2a}$
Therefore, we have $$e^{r_1x}$$ and $$e^{r_2x}$$ as solutions. There is still a difficulty if $$r_1 = r_2$$, but it is not hard to overcome.
Theorem 2.2.1
Suppose that $$r_1$$ and $$r_2$$ are the roots of the characteristic equation.
If $$r_1$$ and $$r_2$$ are distinct and real (when $$b^2 - 4ac > 0$$ ), then (2.2.10) has the general solution
$y = C_1e^{r_1x} + C_2e^{r_2x}$
$y = (C_1 + C_2x)e^{r_1x}$
If $$r_1 = r_2$$ (happens when $$b^2 - 4ac = 0$$ ), then (2.2.10) has the general solution
$y = C_1e^{r_1x} + C_2e^{r_2x}$
$y = (C_1 + C_2x)e^{r_1x}$
For another example of the first case, take the equation $$y'' - k^2y = 0$$. Here the characteristic equation is $$r^2 - k^2 = 0$$ or $$(r - k)(r + k) = 0$$. Consequently, $$e^{-kx}$$ and $$e^{kx}$$ are the two linearly independent solutions.
Example $$\PageIndex{1}$$:
Find the general solution of $y'' - 8y' + 16y = 0$
Solution
The characteristic equation is $$r^2 - 8r + 16 = {( r - 4)}^2 = 0$$. The equation has a double root $$r_1 = r_2 = 4$$. The general solution is, therefore,
$y = (C_1 + C_2x)e^{4x} = C_1e^{4x} + C_2xe^{4x}$
Exercise $$\PageIndex{2}$$: Linear Independence
Check that $$e^{4x}$$ and $$xe^{4x}$$ are linearly independent.
Solution
That $$e^{4x}$$ solves the equation is clear. If $$xe^{4x}$$ solves the equation, then we know we are done. Let us compute $$y' = e^{4x} + 4xe^{4x}$$ and $$y'' = 8e^{4x} + 16xe^{4x}$$. Plug in
$y'' - 8y' + 16y = 8e^{4x} + 16xe^{4x} - 8(e^{4x} + 4xe^{4x} ) + 16xe^{4x} = 0$
We should note that in practice, doubled root rarely happens. If coefficients are picked truly randomly we are very unlikely to get a doubled root.
Let us give a short proof for why the solution $$xe^{rx}$$ works when the root is doubled. This case is really a limiting case of when the two roots are distinct and very close. Note that $$\frac {e^r2^x - e^x1^x}{r_2 - r_1}$$ is a solution when the roots are distinct. When we take the limit as $$r_1$$ goes to $$r_2$$, we are really taking the derivative of $$e^{rx}$$ using $$r$$ as the variable. Therefore, the limit is $$xe^{rx}$$, and hence this is a solution in the doubled root case.
#### 2.2.1 Complex numbers and Euler’s formula
It may happen that a polynomial has some complex roots. For example, the equation $$r^2 + 1 = 0$$ has no real roots, but it does have two complex roots. Here we review some properties of complex numbers.
Complex numbers may seem a strange concept, especially because of the terminology. There is nothing imaginary or really complicated about complex numbers. A complex number is simply a pair of real numbers, $$(a, b)$$. We can think of a complex number as a point in the plane. We add complex numbers in the straightforward way, $$(a, b) + (c, d) = (a + c, b + d)$$. We define multiplication by
$(a,b) \times (c,d) = (ac - bd, ad + bc).$
It turns out that with this multiplication rule, all the standard properties of arithmetic hold. Further, and most importantly
$( 0, 1) \times (0,1) = (-1, 0 ) .$
Generally we just write $$(a, b)$$ as $$(a + ib)$$, and we treat $$i$$ as if it were an unknown. We do arithmetic with complex numbers just as we would with polynomials. The property we just mentioned becomes $$i^2 = -1$$. So whenever we see $$i^2$$, we replace it by $$-1$$. The numbers $$i$$ and $$-i$$ are the two roots of $$r^2 + 1 = 0$$.
Note that engineers often use the letter $$j$$ instead of $$i$$ for the square root of $$-1$$. We will use the mathematicians’ convention and use $$i$$.
Exercise $$\PageIndex{3}$$:
Make sure you understand (that you can justify) the following identities:
• $$i^2 = -1, i^3 = -1, i^4 = 1$$,
• $$\frac {1}{i} = -i$$,
• $$(3 -7i)(-2 -9i) = \dots = -69 - 13i$$,
• $$(3 - 2i)(3 + 2i) = 3^2 - {(2i)}^2 = 3^2 + 2^2 = 13$$,
• $$\frac {1}{3-2i} = \frac {1}{3-2i} \frac {3+2i}{3+2i} = \frac{3+2i}{13} = \frac {3}{13} + \frac{2}{13} i$$.
We can also define the exponential $$e^{a+ib}$$ of a complex number. We do this by writing down the Taylor series and plugging in the complex number. Because most properties of the exponential can be proved by looking at the Taylor series, these properties still hold for the complex exponential. For example the very important property: $$e^{x+y} = e^xe^y$$. This means that $$e^{a+ib} = e^ae^{ib}$$. Hence if we can compute $$e^{ib}$$, we can compute $$e^{a+ib}$$. For $$e^{ib}$$ we use the so-called Euler’s formula.
Theorem 2.2.2 (Euler's Formula).
$e^{i\theta} = \cos \theta + i \sin \theta ~~~~ { \it{~and~ } }~~~~ e^{-i\theta} = \cos \theta - i\sin \theta$
Exercise $$\PageIndex{4}$$:
Using Euler’s formula, check the identities:
$\cos \theta = \frac { e^{i \theta} + e^{-i \theta}}{2} ~~~~ { \it{~and~ } }~~~~ \sin \theta = \frac { e^{i \theta} - e^{-i \theta}}{2}$
Exercise $$\PageIndex{5}$$:
Double angle identities: Start with $$e^{i (2 \theta)} = {(e^{i \theta})}^2$$. Use Euler on each side anddeduce:
$\cos (2 \theta) = {\cos}^2 \theta - {\sin}^2 \theta ~~~ {\it{~and~}}~~~ \sin (2 \theta) = 2 \sin \theta \cos \theta$
For a complex number $$a + ib$$ we call $$a$$ the real part and $$b$$ the imaginary part of the number. Often the following notation is used,
$Re(a + ib) =a ~~~ {\it{~and~}}~~~ Im (a + ib) = b$
#### 2.2.2 Complex roots
Suppose that the equation $$ay'' + by' + cy = 0$$ has the characteristic equation $$ar^2 + br + c = 0$$ that has complex roots. By the quadratic formula, the roots are $$\dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$$. These roots are complex if $$b^2 - 4ac < 0$$. In this case the roots are
$r_1, r_2 = \frac {-b}{2a} \pm i \dfrac { \sqrt {4ac - b^2}}{2a}$
As you can see, we always get a pair of roots of the form $$\alpha \pm i \beta$$. In this case we can still write the solution as
$y = C_1e^{(\alpha + i \beta )x} + C_2e^{(\alpha - i\beta)x}$
However, the exponential is now complex valued. We would need to allow $$C_1$$ and $$C_2$$ to be complex numbers to obtain a real-valued solution (which is what we are after). While there is nothing particularly wrong with this approach, it can make calculations harder and it is generally preferred to find two real-valued solutions.
Here we can use Euler’s formula. Let
$y_1 = e^{(\alpha + i\beta)x} ~~~~ { \it{~and~ } }~~~~ y_2 = e^{( \alpha - i \beta ) x}$
Then note that
$y_1 = e^{ax} \cos (\beta x) + ie^{ax} \sin ( \beta x)$
$y_2 = e^{ax} \cos (\beta x) - ie^{ax} \sin (\beta x)$
Linear combinations of solutions are also solutions. Hence,
$y_3 = \frac {y_1 + y_2}{2} = e^{ax} \cos (\beta x)$
$y_4 = \frac {y_1 - y_2}{2i} = e^{ax} \sin (\beta x)$
are also solutions. Furthermore, they are real-valued. It is not hard to see that they are linearly independent (not multiples of each other). Therefore, we have the following theorem.
Theorem 2.2.3. For the homegneous second order ODE
$ay'' + by' + cy = 0$
If the characteristic equation has the roots $$\alpha \pm i \beta$$ (when $$b^2 - 4ac < 0$$), then the general solution is
$y = C_1e^{ax} \cos (\beta x) + C_2e^{ax} \sin (\beta x)$
Example $$\PageIndex{2}$$:
Find the general solution of $$y'' + k^2 y = 0$$, for a constant $$k > 0$$.
The characteristic equation is $$r^2 + k^2 = 0$$. Therefore, the roots are $$r = \pm ik$$ and by the theorem we have the general solution
$y = C_1 \cos (kx) + C_2 \sin (kx)$
Example $$\PageIndex{3}$$:
Find the solution of $$y'' - 6y' + 13y = 0, y(0) = 0, y'(0) = 10.$$
Solution
The characteristic equation is $$r^2 - 6r + 13 = 0$$. By completing the square we get $${(r -3)}^2 + 2^2 = 0$$ and hence the roots are $$r = 3 \pm 2i$$. By the theorem we have the general solution
$y = C_1e^{3x} \cos (2x) + C_2 e^{3x} \sin (2x)$
To find the solution satisfying the initial conditions, we first plug in zero to get
$0 = y(0) = C_1e^0 \cos 0 + C_2e^0 \sin 0 = C_1$
Hence $$C_1 = 0$$ and $$y = C_2e^{3x} \sin (2x)$$. We differentiate
$y' = 3C_2 e^{3x} \sin (2x) + 2C_2e^{3x} \cos (2x)$
We again plug in the initial condition and obtain $$10 = y'(0) = 2C_2$$, or $$C_2 = 5$$. Hence the solution we are seeking is
$y = 5e^{3x} \sin (2x)$
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# Probability cards question
What is the probability that the first card is spades given that the second and third cards are spades?
Attempt: Use Bayes' rule to calculate $P(E_1|E_2E_3)$
• $E_1$ = first card is spade
• $E_2$ = second card is spade
• $E_3$ = third card is spade
$$P(E_1) = \frac{13}{52}$$ $$P(E_2) = \frac{13}{52}$$ $$P(E_3) = \frac{13}{52}$$
$$P(E_1|E_2E_3) = \frac {P(E_2E_3|E_1)P(E_1) }{ P(E_2E_3)}$$ So this $$= \frac{(\frac{12}{51} \cdot \frac{12}{51})(\frac{13}{52})} {\frac{13}{52} \cdot \frac{13}{52}} = 0.2214$$
Is this correct?
-
so the probability is 0.12? – lord12 Sep 19 '11 at 3:36
All sequences of cards are equally likely. Thus the required probability is the probability that the third card is a spade, given that the first two are spades. This probability is (clearly) equal to $11/50$.
Comment: We can obtain the same result by using Bayes' Rule, and working quite a bit harder. And in any complicated calculation, there is always the possibility of error. For example, the idea in the OP's calculation is correct. However, $$P(E_2E_3|E_1)=\frac{12}{51}\cdot \frac{11}{50}.$$ Also, the denominator should be $$P(E_2E_3)=\frac{13}{52}\cdot \frac{12}{51}\cdot\frac{11}{50}+\frac{39}{52}\cdot \frac{13}{51}\cdot\frac{12}{50}.$$ With these modifications, the OP's argument goes through just fine. After some cancellation, we obtain $11/50$.
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# Singapore Math Heuristics: Make A Systematic List, Guess And Check, Restate The Problem In Another Way
The skills children pick up in math are indispensable; they can be applied to other academic subjects and to solve real-world problems in their daily lives and future work.
The Singapore Math curriculum focuses on problem solving. Through problem solving, children develop thinking skills such as creative thinking and critical thinking.When children analyse math problems, they notice patterns, learn to generalise, form new ideas and activate their creative thinking. Children become critical thinkers when they are able to select the best strategy out of multiple methods to solve word problems.
Singapore Math Heuristics
In part one of our Math Heuristics series, we gave an overview of the 12 problem-solving methods or heuristics taught in the Singapore primary math education syllabus, with tips from the curriculum team at Seriously Addictive Mathematics (S.A.M) on how to use them to solve various math word problems.
In part two of the Math Heuristic series, we zoomed in on the heursitics – Act It Out, Draw A Diagram and Look For Patterns, and also showed how to apply the Polya’s 4-step problem-solving process in sample word problems.
In the third part of this series, we will focus on the next 3 heuristics – Make a systematic listguess and check, and restate the problem in another way.
Sample word problems are solved using these 3 heuristics and Poly’s 4-step process in the step-by-step worked solutions provided by the curriculum team at S.A.M.
_______________________________________________________________________________________________________
Heuristic: Make a systematic list
Jimmy uses the number cards given below to form as many 3-digit odd numbers as he can. List all the numbers that Jimmy can form.
Solution:
1. Understand:
What to find: All the 3-digit odd numbers that Jimmy can form from the 4 number cards.
What is known: Odd numbers end with 5 or 7.
2. Choose: Make a systematic list
3. Solve:
Odd numbers that end with 5:
245
275
425
475
725
745
Odd numbers that end with 7:
247
257
427
457
527
547
Jimmy can form 12 3-digit odd numbers.
4. Check:
Did I form 3-digit numbers? Yes
Did I form odd numbers? Yes
Did I form all possible numbers? Yes
Try solving the following word problem using Polya’s 4-step process.
A shop sells apples in bags of 3. It sells lemons in bags of 4. Paul buys some bags of apples and lemons. He buys the same number of each fruit. He buys more than 20 and fewer than 30 pieces of each fruit. How many apples does Paul buy?
See the solution in part one of our Singapore Math Heuristics series.
_______________________________________________________________________________________________________
Heuristic: Guess and check
David sold a total of 15 \$4 coupons and \$5 coupons for a funfair. He received \$65 for the sale of the coupons. How many \$4 coupons and how many \$5 coupons did he sell?
Solution:
1. Understand:
What to find: The number of \$4 coupons and the number of \$5 coupons David sold?
What is known: He sold 15 coupons. He received \$65.
2. Choose: Guess and check
3. Solve:
\$4 + \$5 = \$9. (7 × 9) = 63 is close to 65. I can start the first guess with 7 \$4 coupons.
David sold 10 \$4 coupons and 5 \$5 coupons.
4. Check:
What is the total number of coupons? 10 + 5 = 15
What is the total value of coupons? \$40 + \$25 = \$65
In a quiz, 5 marks were awarded for each correct answer and 3 marks were deducted for each wrong answer. Darren answered 14 questions and scored 30 marks. How many questions did he answer correctly?
Solution:
1. Understand:
What to find: The number of questions Darren answered correctly.
What is known: Add 5 marks for each correct answer. Minus 3 marks for each wrong answer. He answered 14 questions. He scored 30 marks.
2. Choose: Guess and check
3. Solve:
I can start the first guess with the same number of correct answers and wrong answers.
4. Check:
What is the total number of questions? 9 + 5 = 14
What is the total marks scored? 45 – 15 = 30
Try solving the following word problem using Polya’s 4-step process.
Vijay is presented with the equations below. Insert one pair of brackets in each equation to make it true.
4 × 11 + 18 ÷ 3 + 6 = 46
Answer: The equation is 4 × 11 + 18 ÷ (3 + 6) = 46.
See the solution in part one of our Singapore Math Heuristics series.
_______________________________________________________________________________________________________
Heuristic: Restate the problem in another way
Sally has some beds and sofas. All of them are equal in length. The total length is 14 metres. Each bed is 2 metres long. Sally has 1 fewer bed than sofas. What is the total length of the sofas?
Solution:
1. Understand:
What to find: Total length of the sofas.
What is known: Each bed is 2 metres long. Each sofa is 2 metres long. Total length of beds and sofas is 14 metres. Sally has 1 more sofa than bed.
2. Choose: Restate the problem in another way
3. Solve:
If we add 1 more bed, Sally will have the same number of beds and sofas.
New total length = 14 m + 2 m = 16 m
Sally has the same number of beds and sofas
Total length of sofas = 16 m ÷ 2 = 8 m
The total length of the sofas is 8 metres.
4. Check:
How many sofas are there? 8 m ÷ 2 m = 4
How many beds are there? 4 – 1 = 3
How many beds and sofas altogether? 4 + 3 = 7
What is the total length of beds and sofas? 7 × 2 m = 14m
Try solving the following word problem using Polya’s 4-step process.
There are some identical pens and erasers. 2 pens and 3 erasers are 45 centimetres long altogether. 6 erasers and 2 pens are 60 centimetres long altogether. What is the length of 3 erasers?
Answer: The length of 3 erasers is 15 cm.
See the solution in part one of our Singapore Math Heuristics series.
These are just a few examples to show you how Singapore Math heuristics are used to solve basic and intermediate word problems in lower grade levels and complex word problems in upper grade levels.
Look out for parts four and five of this series for the other 6 remaining Singapore Math heuristics and word problems with step-by-step worked solutions.
This is the third part to S.A.M Math Heuristics series. Read part one and part two here.
Established in 2010, Seriously Addictive Mathematics (S.A.M) is the world’s largest Singapore Math enrichment program for children aged four to 12. The award-winning S.A.M program is based on the global top-ranking Singapore Math curriculum with a focus on developing problem solving and thinking skills.
The curriculum is complemented with S.A.M’s two-pillared approach of Classroom Engagement and Worksheet Reinforcement, with an individual learning plan tailored to each child at their own skill level and pace, because no two children learn alike.
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# An open-top box is to be constructed from a 6 in by 2 in rectangular sheet of tin by cutting out squares of equal size at each corner, then folding up the resulting flaps. Let x denote the length of the side of each cut-out square. What is the volume?
Jun 17, 2016
Volume$= {x}^{3} - 16 {x}^{2} + 12 x$
#### Explanation:
Height of box: $= x$
Length of box: $= 6 - 2 x$
Width of box: $= 2 - 2 x$
Volume of box:
$x \cdot \left(6 - 2 x\right) \left(2 - 2 x\right)$
$\textcolor{w h i t e}{\text{XXX}} = x \cdot \left(12 - 16 x + 4 {x}^{2}\right)$
$\textcolor{w h i t e}{\text{XXX}} = {x}^{3} - 16 {x}^{2} + 12 x$
Jun 17, 2016
Volume of open-top box would be $4 {x}^{3} - 16 {x}^{2} + 12 x$
#### Explanation:
As $x$ is the length of the side of each cut out square, the height of the open=top box will be $x$.
Its length will be $\left(6 - 2 x\right)$
and width would be $\left(2 - 2 x\right)$
Hence volume would be
$x \left(6 - 2 x\right) \left(2 - 2 x\right)$
= $x \left(12 - 12 x - 4 x + 4 {x}^{2}\right)$
= $x \left(12 - 16 x + 4 {x}^{2}\right)$
= $12 x - 16 {x}^{2} + 4 {x}^{3}$
or $4 {x}^{3} - 16 {x}^{2} + 12 x$
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8.03 Lines of best fit
Lesson
We have already learned to create a scatter plot and to perform analysis such as determining correlation. Another type of analysis we may choose to do to the graph of a scatter plots is to identify a line of best fit.
Correlation
We describe the correlation from data using language like positive correlation, negative correlation or no correlation. We might even say that two variables have strong or weak correlation.
Types of correlation
Positive correlation - the data appears to gather in a positive relationship, similar to a straight line with a positive slope.
Negative correlation - is when the data appears to gather in a negative relationship, similar to a straight line with a negative slope.
No correlation - when there is no relationship between the variables we say they have no correlation.
Below are some examples of scatter plots with different correlations:
Positive correlation Negative correlation No correlation
The more closely the plotted data resembles a straight line, the stronger the correlation is between the variables.
Watch out!
Just because two variables have a correlation, even a strong one, does not mean that one causes the other. For example, there is a strong correlation between height and stride length. However, it doesn't mean that if you take big steps you'll grow taller!
Lines of best fit
line of best fit (sometimes called a trend or regression line) is a straight line that best represents the data on a scatter plot. Depending on the strength of the correlation, this line may pass exactly through all of the points, some of the points, or none of the points. However, it always represents the general trend of the of the data.
Lines of best fit are really handy as we can use them to help us make predictions or conclusions about the data.
To draw a line of best fit by eye, balance the number of points above the line with the number of points below the line. You should generally ignore outliers (points that fall very far from the rest of the data) as they can skew the line of best fit. Later we will look at how we can calculate a line of best fit's equation.
Below is an example of what a good line of best fit might look like.
Making predictions
If the points appear to lie close to a line, we conclude that a relationship probably exists and it is safe to make predictions using a line of best fit. Making predictions inside the range of the data is called interpolation.
In a well-designed experiment, a researcher is careful not to use the fitted line to make predictions about the response that would be observed to values of the independent variable that are outside the range of the values used in the experiment. For example, if in the experiment the smallest value of the independent variable was 10 and the largest 85, then it would be unwise to try to predict what the response would be when the independent variable was smaller than 10 or larger than 85.
To make such predictions beyond the range of the data is called extrapolation and is considered unsafe.
Worked example
Question 1
The data points illustrated in the graph below show the sale price of an item of goods measured against the age of the item. Given the value of either the independent or dependent variable, we can determine the corresponding value using the line of best fit as seen below.
The regression line can be used to predict that at 23 months the value of the goods will be approximately $\$1250$$1250. Considering the amounts by which the data points are above and below the regression line, it could easily happen that the estimated value of the goods at 23 months is \100$$100 too low or too high.
Practice questions
Question 2
The following scatter plot shows the data for two variables, $x$x and $y$y.
1. Determine which of the following graphs contains the line of best fit.
A
B
C
D
A
B
C
D
2. Use the line of best fit to estimate the value of $y$y when $x=4.5$x=4.5.
$4.5$4.5
A
$5$5
B
$5.5$5.5
C
$6$6
D
$4.5$4.5
A
$5$5
B
$5.5$5.5
C
$6$6
D
3. Use the line of best fit to estimate the value of $y$y when $x=9$x=9.
$6.5$6.5
A
$7$7
B
$8.4$8.4
C
$9.5$9.5
D
$6.5$6.5
A
$7$7
B
$8.4$8.4
C
$9.5$9.5
D
Question 3
The number of fish in a river is measured over a five year period.
The results are shown in the following table and plotted below.
Time in years ($t$t) Number of fish ($F$F) $0$0 $1$1 $2$2 $3$3 $4$4 $5$5 $1903$1903 $1994$1994 $1995$1995 $1602$1602 $1695$1695 $1311$1311
1. Which line best fits the data?
A
B
C
D
A
B
C
D
2. Predict the number of years until there are no fish left in the river.
3. Predict the number of fish remaining in the river after $7$7 years.
4. According to the line of best fit, how many years are there until there are $900$900 fish left in the river?
Outcomes
8.DSP.2
Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and describe the model fit by judging the closeness of the data points to the line.
8.DSP.3
Write and use equations that model linear relationships to make predictions, including interpolation and extrapolation, in real-world situations involving bivariate measurement data; interpret the slope and y-intercept.
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Let us assume that 1000 different random samples was selecte
Let us assume that 1000 different random samples was selected from the same population and for each of them a confidence interval of the population mean was constructed (using exactly the same method) with confidence level of 96%. On average, how many of those 1000 different confidence intervals would contain the true value of the population mean?
What would be the answer if confidence level was 99%?
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Step 1
Determine the number of 1000 different confidence intervals would contain the true value of the population mean.
The number of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 96% is obtained below as follows:
From the information, given that there are 1000 different random samples was selected from the same population.
The required number is,
$$\displaystyle{1000}\times{\frac{{P}}{{96}}}\rbrace{\left\lbrace{100}\right\rbrace}={960}$$
Thus, on average 960 of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 96%.
Step 2
Determine the number of 1000 different confidence intervals would contain the true value of the population mean.
The number of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 99% is obtained below as follows:
From the information, given that there are 1000 different random samples was selected from the same population.
The required number is,
$$\displaystyle{1000}\times{\frac{{{99}}}{{{100}}}}={990}$$
Thus, on average 990 of 1000 different confidence intervals would contain the true value of the population mean with the confidence level of 99%.
Not exactly what you’re looking for?
Bob Huerta
Step 1
a) On average expected number of confidence interval that will contain the true value of population mean
$$\displaystyle={n}{p}={1000}\times{0.96}={960}$$
b) On average expected number of confidence interval that will contain the true value of population mean
$$\displaystyle={n}{p}={1000}\times{0.99}={990}$$
karton
Given: For 1000 different samples if we compute a 96% confidence interval for each sample.
Then approximately on average 960 of the 1000 confidence intervals will contain the true value of the population mean.
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How Many Batches/What Fraction of a Batch?
Alignments to Content Standards: 6.NS.A
A batch of cookies requires 2 cups of flour. How many cups are needed for 3 batches? Here are a diagram and a multiplication equation that represent this situation:
$$3 \times 2 = ?$$
Draw a diagram and write a multiplication equation for each situation.
1. A batch of cupcakes requires $1 \frac12$ cups of flour. How many cups are needed for 3 batches?
2. A batch of bread requires 3 cups of flour. How many cups are needed for $\frac12$ of a batch?
3. A batch of rolls requires $2\frac12$ cups of flour. How many cups are needed for $1\frac12$ batches?
A batch of cookies requires 2 cups of flour. How many batches can you make with 8 cups of flour? Here are a diagram and a division equation that represent this situation:
$$8 \div 2 = ?$$
Draw a diagram and write a division equation for each situation.
1. A batch of cupcakes requires $1 \frac12$ cups of flour. How many batches can you make with 9 cups of flour?
2. A batch of bread requires 3 cups of flour. How many batches can you make with 2 cups of flour?
3. A batch of rolls requires $2\frac12$ cups of flour. How many batches can you make with $4\frac12$ cups of flour?
IM Commentary
The purpose of this task is to help students extend their understanding of multiplication and division of whole numbers to multiplication and division of fractions. In grade 5, students
• multiply two fractions together and learn that a fraction times a number can be interpreted as that fraction of the number,
• divide two whole numbers with a non-whole number quotient, and
• divide a whole number by a unit fraction and a unit fraction by a whole number.
Notably, all of the division students do in grade 5 can be supported by an understanding of a whole number of groups. The big shift in grade 6 is that students tackle division contexts that involve a fractional number of groups.
The task does not ask students to find the product or quotient since the task is more about learning how to represent the situation, but teachers might choose to ask students to find or estimate the answers, if desired.
This task helps students see how to represent How Many Groups/What Fraction of a Group? contexts with division equations. It's designed to be used along with the task How Much in One Batch? which covers the How Much in a Group? situation.
To offload some of the tedium of drawing diagrams so that students can focus on thinking about sizes and numbers of batches, teachers might offer some printed templates for students to work on.
Solution
(Note: In all the diagrammed solutions, a solid line shows a division in cups and a dashed line shows a division in batches.)
1. A batch of cupcakes requires $1 \frac12$ cups of flour. How many cups needed for 3 batches?
$$3 \times 1\frac12 = ?$$
2. A batch of bread requires 3 cups of flour. How many cups needed for $\frac12$ a batch?
$$\frac12 \times 3 = ?$$
3. A batch of rolls requires $2\frac12$ cups of flour. How many cups needed for $1\frac12$ batches? (In this diagram, the dotted lines are used to show the partition of one of the cups into four parts.)
$$1\frac12 \times 2\frac12 = ?$$
1. A batch of cupcakes requires $1 \frac12$ cups of flour. How many batches can you make with 9 cups of flour?
$$9 \div 1\frac12 = ?$$
2. A batch of bread requires 3 cups of flour. How many batches can you make with 2 cups of flour?
$$2\div 3 = ?$$
3. A batch of rolls requires $2\frac12$ cups of flour. How many batches can you make with $4\frac12$ cups of flour?
$$4 \frac12 \div 2 \frac12 = ?$$
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# Percentage Elongation Example
Percentage elongation is the increase in gauge length compared to the original length.
##### Percentage Elongation Formula
Percentage elongation can be expressed as;
$$Percentage\space\ elongation\space\ = \frac{Increase\space\ in \space\ length \times\ 100}{Original\space\ length}$$
Symbolically it can be expressed as;
$$Percentage\space\ elongation=\frac{\delta l\times\ 100}{l}$$
##### Percentage Elongation Example
Find percentage elongation of a steel rod, when it is subjected to an axial pulling force of 50 KN.whose diameter is 30 mm and it is 3 meters long. Take E= 200 X 10^9 Newton per meter square.
##### Solution
Given data is;
Diameter of rod = D = 30 mm
Length of rod = l = 3 m = 3 x 1000 = 3000 mm
Axial Pulling Force = P = 50 KN = 50 X 1000 = 5 X 10^4 KN
Young’s Modulus = E = 200 X 10^9 N/meter square = 200 X 10^3 N/millimeter square
Required;
Percentage Elongation = ?
$$Percentage\space\ elongation=\frac{\delta l\times\ 100}{l}$$
Original length is known. Now we have to find change in length because of pulling force on steel rod;
$$Cross\space\ Area =A= \frac{\pi \times\ D^{2}}{4}$$
put steel diameter = 30 mm
Hence, Cross sectional Area = A = 706.858 millimeter square.
Find value of stress in steel rod;
Stress = Force/Area
$$Stress = \frac{5\times\ 10^{4}}{706.858}$$
Stress = 70.736 N/millimeter square = 70.736 MPa
Now find strain from young’s modulus formula
Young’s Modulus = Stress/ Strain
Strain = Stress/young’s Modulus
$$e\space\ = \frac{f}{E}\space\ =\frac{70.736}{200\times 10^{3}}$$
Strain = e = 0.000354
As we know that;
Strain = Change in length/original length
Hence;
Change in length = Strain X Original length
Change in length = 0.000354 X 3000 = 1.062 mm
Now every thing is known, put values in percentage elongation formula and find the percentage elongation as a result of pulling force;
$$Percentage\space\ elongation=\frac{\delta l\times\ 100}{l}$$
Percentage Elongation = (1.062 x 100)/ 3000
Percentage Elongation = 0.0345%
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Mathematical Induction – IMT DeCal
# Mathematical Induction
by Suraj Rampure
## Motivation
Mathematical induction, the last proof technique we will look at, has many parallels to recursion. As an introduction, consider the following situation:
Suppose you’re sitting in a massive lecture hall, and want to find out how many rows you’re sitting from the front of the room. You could sit there and count, but consider this basic principle:
• The person sitting in the first row knows their row number by default: they’re in the first row!
• If one knows the row number of the person in front of them, they add 1 to get their own row number
With this principle, the first person can pass their row number to the second person, who “calculates” their row number (by adding 1) and passes it to the third person, and so on and so fourth. This is exactly how induction (and recursion, to an extent) works.
With induction, we want to prove some property about all natural numbers or whole numbers (it doesn’t matter if we start with 0 or 1, as you will see).
## Formalization
Suppose the expression we want to prove is in terms of the variable $n$ (which is typically used for natural numbers). We follow the following three steps, in every single induction proof:
1. Base Case: Establish that the statement holds for some base case. Often, this is $n = 0$ or $n = 1$, but an alternate base case may be specified in the proposition.
2. Induction Hypothesis: Assume that the statement holds true for $n = k$, for some arbitrary $k$.
3. Induction Step: Given the fact that the statement holds true for $n = k$, show that it holds for $n = k + 1$.
Let’s dive right into an example. Note: If you’re uncomfortable with summation notation, refer to the note on Sigma and Pi Notation.
#### Example: Sum of First $n$ Naturals
Sure, we could technically use induction to prove that $(x + 1)^2 = x^2 + 2x + 1$ for all natural numbers, but induction is overkill in such an example. A common example of something we may prove with induction is the following fact:
$\sum_{i = 1}^n i = \frac{n(n+1)}{2}, \forall n \in \mathbb{N}$
Base Case: $n = 1$
LHS: $\sum_{i = 1}^1 i = 1$, RHS: $\frac{1(2)}{2} = 1$
LHS = RHS, therefore the base case holds.
Induction Hypothesis
Assume that $\sum_{i = 1}^k i = \frac{k(k+1)}{2}$, for some arbitrary integer $k$.
Induction Step
Now, we need to show that $\sum_{i = 1}^{k+1} i = \frac{(k+1)(k+2)}{2}$, somehow using the information in the induction hypothesis.
\begin{aligned} \sum_{i = 1}^{k + 1} i &= \sum_{i = 1}^k i + (k + 1) \\ &= \frac{k(k+1)}{2} + \frac{2(k+1)}{2} \\ &= \frac{(k+1)(k+2)}{2} \end{aligned}
By assuming that the statement held true for $n = k$, we were able to come to the conclusion that the statement holds true for $n = k + 1$. Therefore, induction holds, and the statement holds true for all natural numbers.
Note, we chose $n = 1$ to be the base case, as it wouldn’t make sense to take a sum with lower limit 1 and upper limit 0.
We will look at an alternate proof of this statement when we get to Series and Sequences. We will also provide another proof when we look at combinatorics. This is an identity that you will be expected to know.
Let’s introduce a more concise notation for writing out the steps of induction. Suppose that $P(i)$ represents the proposition we are trying to prove holds true for the value $n = i$. Then, we have:
• Base Case: Show $P(0)$ or $P(1)$ (or whatever other appropriate value)
• Induction Hypothesis: Assume $P(k)$
• Induction Step: Show $P(k) \implies P(k+1)$
Essentially, what we have is a domino effect.
$P(1) \implies P(2) \implies P(3) \implies … \implies P(k) \implies …$
A few things to note:
• The majority of the work in an induction proof is in the induction step. Proving the base case is usually trivial, and in the induction hypothesis you simply state your assumption.
• The example we gave is relatively straight forward. Most of the time, you’ll have to do slightly more work in order to show the implication $P(k) \implies P(k+1)$.
## More Examples
In order to reinforce the concept of induction, we present a few examples below. A good way to practice would be to attempt each problem on your own, and then compare your proof with the solution.
### Example: Divisibility
Prove $3 | n^3 - n$, $\forall n \in \mathbb{N}_0$.
Note: We actually proved this proposition directly, as an example when introducing direct proofs.
Base Case: $n = 0$
3 does indeed divide 0, as we can find some integer $j$ such that $0 = 3j$ (namely, $j = 0$).
Induction Hypothesis
Assume that $3 | k^3 - k$ for some arbitrary integer $k$.
Induction Step
We now want to show that $3 | [(k+1)^3 - (k+1)]$, i.e. that we can write $(k+1)^3 - (k+1)$ as $3 \cdot (\text{some integer})$.
We know that we can write $k^3 - k = 3j$, for some integer $j$, from the induction hypothesis. Then:
\begin{align*} (k+1)^3 - (k+1) &= k^3 + 3k^2 + 3k + 1 - k - 1 \\ &= k^3 - k + 3k^2 + 3k \\ &= 3j + 3(k^2 + k) \\ &= 3(j + k^2 + k) \end{align*}
Therefore, by induction, the statement holds true.
### Example: Fibonacci
The Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, 21, 34, …$ is defined by the recurrence relation
\begin{align*} F_1 = 1, F_2 = 1 \\ F_n = F_{n-2} + F_{n-1} \end{align*}
Prove that $F_1 + F_2 + … + F_n = F_{n + 2} - 1$.
Base Case: $n = 1$
$F_1 = 1, F_3 - 1 = 2 - 1 = 1$, therefore the base case holds.
Induction Hypothesis
Assume $F_1 + … + F_k = F_{k + 2} - 1$, for some arbitrary natural number $k$.
Induction Step
\begin{aligned} F_1 + … + F_{k + 1} &= (F_1 + … + F_k) + F_{k + 1} \\ &= F_{k+2} - 1 + F_{k+1} \\ &= F_{k + 3} - 1 \end{aligned}
Therefore, induction holds, and the statement is true.
### Example: De Moivre’s Theorem
De Moivre’s theorem gives us a way to exponentiate complex numbers of the form $R(\cos t + i \sin t)$:
$(R(\cos t + i \sin t))^n = R^n(\cos nt + i \sin nt)$
Prove De Moivre’s theorem for $n \in \mathbb{N}_0$, using induction. We will need to use the sum identities for $\sin(a + b)$ and $\cos(a + b)$, that is, $\sin(a + b) = \sin a \cos b + \cos a \sin b$ and $\cos(a + b) = \cos a \cos b - \sin a \sin b$.
Base Case: $n = 0$
LHS: $(R(\cos t + i \sin t))^0 = 1$ RHS: $R^0 (\cos 0t + i \sin 0t) = 1 (1 + 0) = 1$
LHS = RHS, thus the base case holds.
Induction Hypothesis: $n = k$
Assume $(R(\cos t + i \sin t))^k = R^k(\cos kt + i \sin kt)$, for some arbitrary $k \in \mathbb{N}$.
Induction Step
Now, we need to show $(R(\cos t + i \sin t))^{k+1} = R^{k+1}(\cos (k+1)t + i \sin (k+1)t)$, using the information we assumed in the hypothesis.
\begin{align*} (R(\cos t + i \sin t))^{k+1} &= (R(\cos t + i \sin t))^k (R(\cos t + i \sin t)) \\ &= R^k(\cos kt + i \sin kt) R (\cos t + i \sin t) \\ &= R^{k + 1} (\cos kt \cos t + i \cos kt \sin t + i \sin kt \cos t + i^2 \sin kt \sin t) \\ &= R^{k+1} (\cos kt \cos t - \sin kt \sin t + i (\cos kt \sin t + \sin kt \cos t))\end{align*}
Now, we’ll need the identities given in the hint:
\begin{align*}\cos kt \cos t - \sin kt \sin t &= \cos (kt + t) = \cos(k+1)t \\ \cos kt \sin t + \sin kt \cos t &= \sin (kt + t) = \sin(k+1)t \end{align*}
Then:
$R^{k+1} (\cos kt \cos t - \sin kt \sin t + i (\cos kt \sin t + \sin kt \cos t)) = R^{k+1} (\cos(k+1)t + i \sin(k+1)t)$
as required. Therefore, induction holds, and the statement holds true.
Note: It is significantly easier to prove this statement using Euler’s Identity and complex exponentials, but it’s interesting to see that we’re able to do this proof using induction as well.
### Example: Airports
This example will illustrate the fact that not all induction proofs need be algebraic.
Suppose that there are $2n+1$ airports where $n$ is a positive integer. The distances between any two airports are all different. For each airport, there is exactly one airplane departing from it, and heading towards the closest airport. Prove by induction that there is an airport which none of the airplanes are heading towards.
Base Case
$n = 1$: If there are 3 airports, the closest pair of airports will exchange planes. The third airport will then send a plane to one of the first two, meaning this third airport will have no incoming planes.
Induction Hypothesis
Assume for some $k$, in $2k+1$ airports one airport has no incoming planes.
Induction Step
For $2(k+1) + 1 = 2k + 3$ airports, consider the two airports that are closest to one another. This is necessarily unique, given the problem statement. These airports will exchange planes, thus reducing the problem to $2k + 1$ airports. We know the $2k+1$ case holds from the induction hypothesis.
QED.
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### Section 10.4 : Convergence/Divergence of Series
6. Show that the following series is divergent.
$\sum\limits_{n = 5}^\infty {\frac{{6 + 8n + 9{n^2}}}{{3 + 2n + {n^2}}}}$ Show Solution
First let’s note that we’re being asked to show that the series is divergent. We are not being asked to determine if the series is divergent. At this point we really only know of two ways to actually show this.
The first option is to show that the limit of the sequence of partial sums either doesn’t exist or is infinite. The problem with this approach is that for many series determining the general formula for the $$n$$th term of the sequence of partial sums is very difficult if not outright impossible to do. That is true for this series and so that is not really a viable option for this problem.
Luckily enough for us there is actually an easier option to simply show that a series is divergent. All we need to do is use the Divergence Test.
The limit of the series terms is,
$\mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{6 + 8n + 9{n^2}}}{{3 + 2n + {n^2}}} = 9 \ne 0$
The limit of the series terms is not zero and so by the Divergence Test we know that the series in this problem is divergent.
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# 2010 AIME I Problems/Problem 4
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$. Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Solution
Case 1: No heads. The only possibility is TTT (the third coin being the unfair coin). The probability for this to happen to Jackie is $\frac {1}{2} \cdot \frac {1}{2} \cdot \frac {3}{7} = \frac {3}{28}$ Thus the probability for this to happen to both players is $\frac {3}{28} = \frac {9}{784}$
Case 2: One head. We can have either HTT, THT, or TTH. The first two happen to Jackie with the same $\frac {3}{28}$ chance, but the third happens $\frac {4}{28}$ of the time, since the unfair coin is heads instead of tails. With 3 possibilities for Jackie and 3 for Phil, there are a total of 9 ways for them both to have 1 head.
Multiplying and adding up all 9 ways, we have a $\frac {4(3 \cdot 3) + 4(3 \cdot 4) + 1(4 \cdot 4)}{28^{2}} = \frac {100}{784}$ overall chance for this case.
Case 3: Two heads. With HHT $\frac {3}{28}$, HTH $\frac {4}{28}$, and THH $\frac {4}{28}$ possible, we proceed as in Case 2, obtaining $\frac {1(3 \cdot 3) + 4(3 \cdot 4) + 4(4 \cdot 4)}{28^{2}} = \frac {121}{784}$.
Case 4: Three heads. Similar to Case 1, we can only have HHH ($\frac {4}{28}$ chance). Then in this case we get $\frac {16}{784}$
Finally, we take the sum: $\frac {9 + 100 + 121 + 16}{784} = \frac {246}{784} = \frac {123}{392}$, so our answer is $123 + 392 = \fbox{515}$.
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# Fundamental Rules for Differentiation
## Fundamental Rules for Differentiation
Rule (I): Differentiation of a constant function is zero i.e., $$\frac{d}{dx}\left( c \right)=0$$.
Rule (II): Let f(x) be a differentiable function and let c be a constant. Then c.f(x) is also differentiable such that $$\frac{d}{dx}\left\{ c.f\left( x \right) \right\}=c.\frac{d}{dx}\left( f\left( x \right) \right)$$.
This is the derivative of a constant times a function is the constant times the derivative of the function.
Rule (III): If f(x) and g(x) are differentiable functions, then show that f(x) ± g(x) are also differentiable such that $$\frac{d}{dx}\left[ f\left( x \right)\pm g\left( x \right) \right]=\frac{d}{dx}f\left( x \right)\pm \frac{d}{dx}g\left( x \right)$$.
That is the derivative of the sum or difference of two functions is the sum or difference of their derivatives.
Rule (IV): If f(x) and g(x) are two differentiable functions, then f(x).g(x) is also differentiable such that $$\frac{d}{dx}\left[ f\left( x \right).g\left( x \right) \right]=f\left( x \right)\frac{d}{dx}\left[ g\left( x \right) \right]+\frac{d}{dx}\left[ f\left( x \right) \right].g\left( x \right)$$.
Rule (V) (Quotient rule): If f(x) and g(x) are two differentiable functions and g(x)≠0 then $$\frac{f\left( x \right)}{g\left( x \right)}$$ is also differentiable such that $$\frac{d}{dx}\left\{ \frac{f\left( x \right)}{g\left( x \right)} \right\}=\frac{g\left( x \right)\frac{d}{dx}\left[ f\left( x \right) \right]-f\left( x \right).\frac{d}{dx}\left[ g\left( x \right) \right]}{{{\left[ g\left( x \right) \right]}^{2}}}$$.
Relation between $$\frac{dy}{dx}$$ and $$\frac{dx}{dy}$$: Let x and y be two variables connected by a relation of the form f(x, y) = 0 Let Δx be a small change in x and let Δy be the corresponding change in y. Then $$\frac{dy}{dx}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{\Delta y}{\Delta x}$$ and $$\frac{dx}{dy}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{\Delta x}{\Delta y}$$.
Now,
$$\frac{\Delta y}{\Delta x}.\frac{\Delta x}{\Delta y}=1$$,
$$\underset{\Delta x\to 0}{\mathop{\lim }}\,\left[ \frac{\Delta y}{\Delta x}.\frac{\Delta x}{\Delta y} \right]=1$$,
$$\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{\Delta y}{\Delta x}.\underset{\Delta y\to 0}{\mathop{\lim }}\,\frac{\Delta x}{\Delta y}=1$$ [∵ ∇x → 0 ⇔ Δy → 0]
$$\frac{dy}{dx}.\frac{dx}{dy}=1$$,
Hence, $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$$.
Example: Differentiate f(x) = sinx.logx
Solution: Using product rule of differentiation
$$f’\left( x \right)=\frac{d}{dx}\left( \sin x \right).\log x+\sin x.\frac{d}{dx}\left( \log x \right)$$,
$$\,=\log x.\cos x+\frac{\sin x}{x}$$.
Example: Find the derivative of $$f\left( x \right)=\frac{\operatorname{logx}}{{{x}^{2}}}$$.
Solution: Using quotient rule of differentiation
$$f’\left( x \right)=\frac{{{x}^{2}}\frac{d}{dx}\left( \log x \right)-\log x\frac{d}{dx}\left( {{x}^{2}} \right)}{{{\left( {{x}^{2}} \right)}^{2}}}$$,
$$=\frac{{{x}^{2}}.\frac{1}{x}-\log x\left( 2x \right)}{{{x}^{4}}}$$,
$$=\frac{x-2x\log x}{{{x}^{4}}}$$,
$$=\frac{1-2\log x}{{{x}^{3}}}$$.
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# FIND THE NATURE OF ROOTS OF THE QUADRATIC EQUATION
Nature of the roots of the Equation ax2 + bx + c = 0
The two roots of the Quadratic equation ax2 + bx + c = 0 are:
The expression b2 – 4ac which appear under radical sign is called the Discriminant (Disc.) of the quadratic equation. i.e., Disc = b2 – 4ac
The expression b2 – 4ac discriminates the nature of the roots, whether they are real, rational, irrational or imaginary. There are three possibilities.
b2 – 4ac > 0
## Nature
is a perfect square, the roots are real, rational and unequal.
is not a perfect square, then roots are real, irrational and unequal.
b2 – 4ac = 0
then roots will be real, equal and rational.
b2 – 4ac < 0
then roots will be imaginary and unequal.
Example 1 :
Find the nature of the roots of the following equations
2x2 + 3x + 1 = 0
Solution :
Comparing the given quadratic equation with ax2 + bx + c = 0, we get
a = 2, b = 3 and c = 1
b2 - 4ac = 32 - 4(2) (1)
= 9-8
= 1 > 0
Since it is perfect square, roots are real, rational and unequal.
Example 2 :
Find the nature of the roots of the following equations
6x2 = 7x + 5
Solution :
6x2 = 7x + 5
Converting into standard form, we get
6x2 - 7x - 5 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we get
a = 6, b = -7 and c = -5
b2 - 4ac = (-7)2 - 4(6) (-5)
= (49 + 120)
= 169 > 0
Since it is perfect square, roots are real, rational and unequal.
Example 3 :
Find the nature of the roots of the following equations
3x2 + 7x - 2 = 0
Solution :
3x2 + 7x - 2 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we get
a = 3, b = 7 and c = -2
b2 - 4ac = 72 - 4(3) (-2)
= (49 + 24)
= 73 > 0
Since it is not perfect square, roots are real, irrational and unequal.
Example 4 :
Find the nature of the roots of the following equations
√2x2 + 3x - √8 = 0
Solution :
√2x2 + 3x - √8 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we get
a = √2, b = 3 and c = -√8
b2 - 4ac = 32 - 4(√2) (-√8)
= 9 + 16
= 25 > 0
Since it is perfect square, roots are real, rational and unequal.
Example 5 :
Show that the roots of the following equations are rational
a(b – c)x2 + b(c – a)x + c(a – b) = 0
Solution :
a = a(b - c), b = b(c - a) and c = c(a - b)
b2 - 4ac = [b(c - a)]2 - 4a(b - c) c(a - b)
= b2(c - a)2 - 4ac(b - c)(a - b)
= b2(c2 - 2ac + a2) - 4ac(ab - b2 - ac + bc)
= b2 c2 - 2acb2 + a2b2 - 4a2bc + 4acb2 + 4a2c2 - 4abc2
= a2b+ 4a2c2 b2 c2 - 4a2bc - 4abc+2 acb2
= (ab)2 + (-2ac)2 + (bc)2 - 4a2bc - 4abc+2 acb2
= (ab - 2ac + bc)2
Since it is perfect square, the given quadratic equation will have rational roots.
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# Parallel Lines Theorem
## Explaining the Concept of Parallel Lines and Their Relationship with Angles
Lines are a fundamental concept in Geometry, seen in everyday objects like doors, windows, and flat surfaces with straight edges. One type of line we commonly encounter is parallel lines, which never intersect and have an equal distance between them. These lines play a crucial role in the study of quadrilaterals, particularly parallelograms, which have two sets of parallel opposite sides. In this article, we will explore the theorems and postulates related to parallel lines, but first, let's define what parallel lines are.
### Defining Parallel Lines
Parallel lines are coplanar straight lines that are equidistant from each other and do not intersect. They are an essential concept in Geometry and play a significant role in determining the angles formed by lines.
### Angle Theorems Related to Parallel Lines
We can make statements about parallel lines based on the angles they form. In other words, we can prove the parallelism of lines based on angles, and vice versa. But before we dive into that, let's review some basic definitions and concepts related to parallel lines. First, how can we distinguish between parallel and non-parallel lines?
### Non-parallel Lines
Non-parallel lines are lines that are not equidistant and will eventually intersect at some point, forming an "X" shape. Examples of non-parallel lines include two intersecting lines or two lines that are not equidistant from each other.
### Transversal Lines
You may be wondering how parallel lines relate to angles if they never intersect. The answer lies in transversal lines, which play a crucial role in determining the angles formed by parallel lines.
A transversal line is a line that passes through two lines at different points within the same plane.
### Angle Congruency Based on Parallel Lines
#### Theorem 1: Alternate Interior Angles
When a transversal cuts two parallel lines, the alternate interior angles formed are congruent. Interior angles are those on the inside of the parallel lines.
#### Theorem 2: Alternate Exterior Angles
Similarly, when a transversal cuts two parallel lines, the alternate exterior angles formed are congruent. Exterior angles are those on the outside of the parallel lines.
#### Theorem 3: Consecutive Interior Angles
If a transversal cuts two parallel lines, then the consecutive interior angles formed on the same side are supplementary, meaning their sum is 180°.
#### Theorem 4: Consecutive Exterior Angles
This theorem states that when a transversal cuts two parallel lines, the consecutive exterior angles formed on the same side are also supplementary.
#### Theorem 5: Corresponding Angles
When a transversal cuts two parallel lines, the corresponding angles formed are congruent. Corresponding angles are those on the matching corners of parallel lines formed by the transversal.
### Proving Parallel Lines Based on Angles
#### Theorem 6: Alternate Interior Angles Converse
If two lines are cut by a transversal, and the alternate interior angles formed are congruent, then the two lines are parallel.
#### Theorem 7: Alternate Exterior Angles Converse
Similarly, if two lines are cut by a transversal, and the alternate exterior angles formed are congruent, then the two lines are parallel.
#### Theorem 8: Consecutive Interior Angles Converse
If two lines are cut by a transversal, and the consecutive interior angles formed have a sum of 180°, then the two lines are parallel.
Understanding the relationship between parallel lines and angles is crucial in Geometry. With the theorems and postulates mentioned in this article, you can identify parallel lines, prove their congruency, and use them to find the measures of angles formed by intersecting lines.
## Theorem 1: Consecutive Exterior Angles Converse
If two lines in a plane are intersected by a transversal, and the consecutive exterior angles formed add up to 180°, then the two lines are parallel.
## Theorem 2: Corresponding Angles Converse
If two lines in a plane are cut by a transversal, and the corresponding angles formed are congruent, then the two lines are parallel.
## Solved Examples
Let's see how these theorems can be applied in real-life problems.
Example 1: In the figure given, lines p and q are parallel, and m∠3 =102°. Find (a) m∠5 (b) m∠6 (c) m∠14.
• (a) By the alternate interior angles theorem, we know that ∠3≅∠5. Therefore, m∠5=102°.
• (b) Using the consecutive interior angles theorem, we can say that ∠3 and ∠6 are supplementary. Hence, ∠6=180°-102°=78°.
• (c) Applying the corresponding angles theorem, we know that ∠6≅∠14. Therefore, m∠14=78°.
## Understanding the Relationship between Perpendicular Transversals and Parallel Lines
The following theorem explains the link between perpendicular transversals and parallel lines.
## Theorem 3: Perpendicular Transversal Theorem
If two lines in a plane are cut by a perpendicular transversal, then both lines are parallel.
## Proof:
Assuming that transversal t intersects both lines p and q at a right angle, i.e. t⊥p, t⊥q. We must prove that p∥q. As t is perpendicular to p, we know that m∠1 = 90°. Similarly, m∠2 = 90° because of t's perpendicularity with q. This means that ∠1 ≅ ∠2. From the figure, we can see that these angles are corresponding angles. Therefore, by theorem 2, we can conclude that p∥q. Thus, the theorem is proved.
## The Transitive Property of Parallel Lines
This theorem explains the connection between parallel lines and how the property applies to them.
## Theorem 4: Transitivity of Parallel Lines
If two lines in a plane are parallel to the same line, then all the lines are parallel to each other.
## Proof:
Let's prove that the line common to other parallel lines is parallel. That is, p∥q, q∥r. Without any loss of generality, we can assume that line q lies between lines p and r. To prove that line p and line r are parallel, we will use the method of contradiction. We assume that line p and line r are not parallel, which would require them to intersect each other. But, as line q lies between lines p and r, this intersection would also require line q to intersect with line p or line r. However, this is impossible as line q is parallel to both line p and line r. Therefore, our assumption is false, and by the method of contradiction, we can conclude that p∥r. Hence, if p∥q and q∥r, then p∥r.
## Theorem 5: Three Parallel Lines Theorem
If three parallel lines are intersected by two transversals, the segments formed on the transversal are proportional.
These theorems are crucial in understanding the properties of parallel lines. By applying these theorems and their proofs, we can solve various problems and determine the relationship between parallel lines in different scenarios. Now, it's your turn to put these theorems to the test and practice with some problems on your own!
## The Proof Behind Parallel Lines: Understanding the Theorem
When we say that three parallel lines intersect at two transversals, it is essential to understand the proof behind it. Let's explore the proof of this theorem by examining the characteristics of parallel lines and their corresponding angles.
### Parallel Lines and Transversals
In a figure with parallel lines p, q, and r, intersected by two transversals t and s at points A, B, C, D, E, and F respectively, we need to prove that ABBC is equal to DEEF.
To prove this, we will make use of the intercept theorem. But first, let's construct a parallel line AH from point A to DF. As we can see, the left part of the figure aligns with the intercept theorem. Therefore, based on this theorem, we can say that AH is parallel to DF.
Given that lines p, q, and r are also parallel, we can conclude that ADEG and EFHG are parallelograms. And from the properties of a parallelogram, we know that opposite sides are equal.
## Parallel Lines Theorems and Key Takeaways: Understand the Relationship Between Parallel Lines and Transversals
The concept of parallel lines, transversals, and their relationships play a vital role in solving geometry problems. By applying the parallel lines theorem, we can determine the congruency and proportionality of different angles and segments.
Firstly, applying the transitive property, we can substitute and get the result: ABBC = AGGH = DEEF. This property is crucial in understanding the relationship between parallel lines and transversals.
### Solved Examples to Demonstrate the Parallel Lines Theorem
Let's apply the above theorems to some examples and see how they can help us determine the relationship between parallel lines and transversals.
#### Example 1: Applying the Transitivity Property
In a figure with four parallel lines, K1, K2, K3, and K4, we need to show that K1∥K4.
Solution: Based on the given conditions, we know that K1 is parallel to K2, K2 to K3, and K3 to K4. Therefore, by using the transitive property, we can conclude that K1 is parallel to K3 and K3 is parallel to K4, hence K1∥K4.
#### Example 2: Applying the Perpendicular Transversal Theorem
In a figure with two lines, a and c, perpendicular to line s, and a being parallel to b, we need to prove that b∥c.
Solution: Based on the given conditions, we know that line s cuts lines a and c perpendicularly. Applying the perpendicular transversal theorem, we can conclude that a∥c. Also, as a∥b, using the transitive property, we can say that b∥c.
## Key Takeaways of the Parallel Lines Theorem
From the examples above, we can understand the key takeaways of the parallel lines theorem. These include:
• Alternate interior and exterior angles are congruent for parallel lines.
• Exterior angles are congruent for parallel lines.
• Consecutive interior and exterior angles on the same side are supplementary for parallel lines.
• Corresponding angles are congruent for parallel lines.
• If two lines are cut by a perpendicular transversal, they are parallel.
• If two lines are parallel to the same line, they are parallel to each other.
• If three parallel lines are cut by two transversals, the segments formed on the transversals are in equal proportion.
## Important Parallel Lines Theorems
Some of the important theorems related to parallel lines are:
• Alternate interior and exterior angles theorem
• Supplementary interior and exterior angles theorem
• Corresponding angles theorem
• Transitive theorem
• The three lines theorem
## In Conclusion
Understanding the proof behind parallel lines is crucial in solving geometry problems. By applying the various parallel lines theorems and properties, we can determine the relationships between parallel lines and transversals.
So the next time you encounter parallel lines in a problem, remember these key takeaways and theorems to help you find the solution.
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# Part of the Lesson Stem Teachers Script Introduction 10
advertisement
```Part of the
Lesson
Introduction
10-20 sec
Connection
Stem
In this lesson you are
going to learn…by
doing/using…
In this lesson you will solve subtraction problems by finding the distance between two numbers
on a number line.
You know that…
You already know that when you subtract you are finding the difference. The difference between
two numbers is the distance between them.
You can see this on a number line.
7 – 4 can be represented as the space or distance between 4 and 7. If you count the distance
or the spaces between these two numbers you find that 7 – 4 = 3. The difference is a distance
of 3 units.
I’m going to explain this
idea by showing you…
You can use this same method to find the distance between larger numbers. Let’s say I want to
find the difference when I subtract 87 from 243.
I don’t need a number line, which is a good thing because a number line to 243 would probably
take up a few screens… I can actually create my own number line. All I need to do is draw a
straight line with arrows on either end, and mark my start point 87, and some distance after
mark my end point 243.
How does this help you find the difference? Well, let’s think about what we know. We know that
on number lines we count hops not marks. We also know that there are some friendly numbers
between 87 and 243. For example, 100 is a friendly number between 87 and 243. Let’s hop
from 87 to 100. Now, I need to think about how far I just hopped. If that is not something you
can figure out quickly, your first hop can be even smaller. I can hop 3 spaces from 87 to 90,
then 10 more to 100, so I just hopped a total of 13 spaces to get to 100. Let’s hop to another
friendly number, 200. That was an easy hop of 100 spaces. And, finally, from 200 to 243 we
know we just need to hop 43 spaces. Let’s look at the total distance we traveled. 13 + 100 +
43. Now I have an easy addition problem that I can rewrite vertically, or solve in my head. I
find that the distance between 87 and 243 is 156, so 243 – 87 = 156.
Let’s see how this works in
a problem…
Let’s see how this technique might help us solve a word problem. Nabisco produces 688 tons of
Oreos every day. It also produces 371 tons of Ritz crackers. How many more tons of Oreos are
produced each day?
This question is asking us, what is 688 – 371 = ?
I will start to solve this by drawing in my number line. I will make my first point as 371 and my
second point as 688.
Now, I will figure out what the first friendly number I can get to is. 371 + 9 is 380. 380 + 20 is
400. 400 + 200 is 600. 600 + 88 is 688, and we are there. Now let’s add our hops.
(Define Terms/ Building
on Prior Knowledge)
30-60 sec
Demonstration
1-3 minc
Application
1-2 min
Teachers Script
29 + 600 + 88 (line up vertically), or if you wanted to do this in your head you can make it
easier to add by adding one to 29 to make 30 and taking one away from 88 to make 87. 30 +
600 + 87. 630 + 80 is 710 + 7 = 717.
Conclusion
10-20 sec
So, now you know how
to…
by…
So, now you know how to solve subtraction problems by finding the distance between two
numbers on a number line.
```
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# How do you simplify (9-i)/(2-i)?
Aug 10, 2018
$\frac{19 + 7 i}{5}$ or $\frac{19}{5} + \frac{7 i}{5}$
#### Explanation:
$\frac{9 - i}{2 - i}$
Multiply both numerator and denominator by the denominator's conjugate (2+i) to remove the $i$ from the denominator:
$\frac{9 - i}{2 - i} \textcolor{b l u e}{\cdot \frac{2 + i}{2 + i}}$
Simplify using FOIL:
$\frac{18 \cdot + 9 i + - 2 i + - {i}^{2}}{4 + 2 i + - 2 i + - {i}^{2}}$
Combine like terms. We also know that ${i}^{2}$ is $- 1$, so:
$\frac{18 + 7 i - \left(- 1\right)}{4 - \left(- 1\right)}$
Simplify:
$\frac{18 + 7 i + 1}{4 + 1}$
$\frac{19 + 7 i}{5}$
Hope this helps!
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# Math Snap
## If $n \in N$, then $1+(-1)^{n}(2 n-1)$ is a multiple of 4 proof $h \in N \Rightarrow N$ is eveh or $h$ is odes
#### STEP 1
Assumptions 1. $n \in \mathbb{N}$ which means $n$ is a natural number. 2. We need to prove that $1 + (-1)^n (2n - 1)$ is a multiple of 4. 3. We will consider the cases when $n$ is even and when $n$ is odd separately.
#### STEP 2
Consider the expression $1 + (-1)^n (2n - 1)$. We need to analyze this expression for different values of $n$ (even and odd).
#### STEP 3
First, let's consider the case when $n$ is even. Let $n = 2k$ for some $k \in \mathbb{N}$.
#### STEP 4
Substitute $n = 2k$ into the expression. $1 + (-1)^{2k} (2(2k) - 1)$
#### STEP 5
Since $(-1)^{2k} = 1$ (because the exponent is even), simplify the expression. $1 + 1 \cdot (4k - 1) = 1 + 4k - 1$
#### STEP 6
Simplify further. $4k$
#### STEP 7
Since $4k$ is clearly a multiple of 4 (as $k$ is an integer), the expression is a multiple of 4 when $n$ is even.
#### STEP 8
Now consider the case when $n$ is odd. Let $n = 2k + 1$ for some $k \in \mathbb{N}$.
#### STEP 9
Substitute $n = 2k + 1$ into the expression. $1 + (-1)^{2k+1} (2(2k+1) - 1)$
#### STEP 10
Since $(-1)^{2k+1} = -1$ (because the exponent is odd), simplify the expression. $1 - (4k + 2 - 1) = 1 - (4k + 1)$
#### STEP 11
Simplify further. $1 - 4k - 1 = -4k$
#### STEP 12
Since $-4k$ is also a multiple of 4 (as $k$ is an integer), the expression is a multiple of 4 when $n$ is odd.
##### SOLUTION
Having analyzed both cases (even and odd $n$), we conclude that $1 + (-1)^n (2n - 1)$ is always a multiple of 4 for any natural number $n$.
|
# Dimensional Analysis
```Module 1: Basic Skills of Algebraic Expressions
Lesson 3: Dimensional Analysis
Objectives:
I can change quantities from one unit to a different unit using dimensional analysis.
I can solve real-life application using dimensional analysis.
Vocabulary:
Dimensional Analysis, Unit
Agenda:
-
Quiz 1
-
Worm up
-
Dimensional Analysis Notes
-
Practice
Focus Questions:
How can I use dimensional analysis to change quantities from one unit to another?
How can I use Dimensional Analysis to compare two quantities?
More Practice:
http://www.alysion.org/dimensional/fun.htm
Practice the examples available and check your answers
Homework: HW 1-3
1
Mathematician:________________
Algebra I
Quiz 1
Combine the following polynomials:
1. (4𝑥 2 + 𝑥 + 7) + (2𝑥 2 + 3𝑥 + 1).
2. (3𝑥 3 − 𝑥 2 + 8) − (𝑥 3 + 5𝑥 2 + 4𝑥 − 7)
Find:
3. The sum of (5 − 𝑡 − 𝑡 2 ) and (9𝑡 + 𝑡 2 ).
4. Subtract (2𝑥 3 + 24) from (3𝑥 3 + 8𝑥).
5. What is the perimeter of the triangle below?
2𝑥 + 5
7𝑥 4 + 9𝑥
2𝑥 4 + 26
2
Warm Up:
Conserving Fuel: Discuss with your partner
In cold climates, conservation of fuel is always a high priority. This is especially
true regarding gasoline for vehicles and heating fuel for homes.
Consider this!
1. Gas stations can become very creative when trying to make their product sound
cheaper than their competitors. Three stations list the price of their gasoline
as seen below. Which station is really offering the cheapest gasoline?
\$0.72 a quart
\$2.95 per gallon
\$0.37 a pint
Information you may need: 1 gallon = 4 Quarts, 1 gallon = 8 pints
3
Dimensional Analysis:
Examples 1:
How many miles will a person run during a 10 kilometer race?
This is what we
need to change
10 𝑘𝑚.
0.62 𝑚𝑖𝑙𝑒𝑠
= 6.2 𝑚𝑖𝑙𝑒𝑠
1 𝑘𝑚
The ratio allow us to
change the unit that we
have to another unit
Example 2:
Convert 25 yards to inches
Example 3:
Convert 84 miles to kilometers
1)
Covert 10 gallons to cups.
2) 16 kilograms to ounces.
4
3) A kiddy pool holds 10 gallons of water. How many cubic meters is this?
4) A car weighs 34005 Kilograms, How much the car weigh in tons?
5.
An image of a building in a photograph is 6 centimeters wide and 11
centimeters tall. If the image is similar to the actual building and the actual
building is 174 meters wide, how tall is the actual building, in meters?
More Practice:
5
Mathematician: ________________
Date:
Homework: Lesson 1-3
Answer the following questions:
1. (REGENTS) Peyton is a sprinter who can run the 40-yard
dash in 4.5 seconds. He converts his speed into miles per hour, as shown below.
Which ratio is incorrectly written to convert his speed?
3 𝑓𝑡
5280 𝑓𝑡
60 𝑠𝑒𝑐
60 𝑚𝑖𝑛
(1)
(2)
(3)
(4)
1 𝑦𝑑
1 𝑚𝑖
1 𝑚𝑖𝑛
1 ℎ𝑟
3. Covert the following units:
(a) 33 feet to yards
(b)
3 liters to gallons.
(c)
3 Kilograms to ounces.
6
4) Ally is baking cake on her mom’s birthday; she needs .75 quart of milk to make
the cake. How many cups of milk does Ally need?
5) The moon is 250,000 miles away. How many feet is it from earth?
6) Bob and Latoya both drove to a baseball game at a college stadium. Bob lives
70 miles from the stadium and Latoya lives 60 miles from it, as shown in the
accompanying diagram. Bob drove at a rate of 50 miles per hour, and Latoya
drove at a rate of 40 miles per hour. If they both left home at the same
time, who got to the stadium first?
7
```
|
Browse Questions
# Integrate the rational functions$\frac{x^3+x+1}{x^2-1}$
Toolbox:
• (i)If the rational is improper in nature we can divide and separate the terms to make it a proper rational function.
• (ii) $\frac{Px+q}{(x+a)(x+b)}=\frac{A}{(x+a)}+\frac{B}{(x+b)}$
• (iii)$\int \frac{dx}{x+a}=log| x+a |+c$
Given$I=\int \frac{x^3+x+1}{x^2-1}$
Since it is an improper rational function, let us divide
$\frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}$
$=x+\frac{2x+1}{(x+1)(x-1)}$
$\frac{2x+1}{(x+1)(x-1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$
2x+1=A(x-1)+B(x+1)
Equating the coefficients of x,
2=A+B ------(1)
Equating the constant terms,
1=-A+B -----(2)
Now Add equation (1) and equation(2)
A+B=2
-A+B=1
_________________
2B=3=>B=3/2
Substitute for B in equ (1)
A+B=2
$A+\frac{3}{2}=2$
A=1/2
Hence A=1/2 and B=3/2
Now substitude for A and B
$\frac{2x+1}{(x+1)(x-1)}=\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$
On integrating we get,
Therefore $I=\int xdx+\frac{1}{2}\int \frac {dx}{x+1}+\frac{3}{2}\int \frac{dx}{(x-1)}$
$\frac{x^2}{2}+\frac{1}{2}log |x+1|+\frac{3}{2} log |x-1|+c$
|
#### Need solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise 13.1 question 7
Hint: total surface area of the cone is given by $s=\pi r^{2}+\pi r l$
Given: $\Delta x=\left(\frac{k}{100}\right) \times(x)$
$\Delta x=0.01 k x$
Solution: Suppose $x$ be the height of the cone and $\Delta x$ be the change in the value of $x$
⇒ Thus we have $\Delta x=\left(\frac{k}{100}\right) \times x$
$\Delta x=0.01 k x$
⇒ suppose us assume that radius, the semi height and semi vertical angle of the come to be $r$, 1 and respectively as shown
from above figure, using trigonometry
$\tan d=\frac{O B}{O A}=\frac{r}{x}$
⇒ we have also
$\cos x=\frac{O A}{A B}=\frac{x}{l}$
$1=\frac{x}{\cos x}$
⇒ Now the total surface area of the cone is given by
$s=\pi x^{2}+\pi r l$
⇒ from the above, we have $r=x \tan x \text { and } l=x \sec x$
$\Rightarrow s=\pi x^{2}+\pi(x \tan x \cdot x \sec x)$
⇒ Differentiate s with respect to $x$
\begin{aligned} &\frac{d s}{d x}=\frac{d}{d x}\left[\pi x^{2} \tan x(\tan x+\sec x)\right] \\\\ &\Rightarrow \frac{d s}{d x}=\pi \tan x(\tan x+\sec x) \frac{1}{d x}\left(x^{2}\right) \end{aligned}
\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\\\ &\Rightarrow \frac{d s}{d x}=2 \pi x \tan x(\tan x+\sec x) \end{aligned}
\begin{aligned} &\Rightarrow \text { so, } \Delta x=\left(\frac{d y}{d x}\right) \Delta x \\\\ &\Rightarrow \text { Here } \frac{d s}{d x}=2 \pi x(\tan x+\sec x) \text { and } \Delta x=0.01 k x \\\\ &\Rightarrow \Delta s=(2 \pi x \tan (x)[\tan (x)+\sec (x)](0.01 k x) \end{aligned}
⇒ Percentage of increase in S
\begin{aligned} &\text { Increase }=\frac{\Delta s}{s} \times 100 y \\\\ &=100+\times 0 . .03 k \pi x^{2} \tan x[\tan x+\sec x] \\\\ &\pi x^{2} \tan x(\tan x+\sec x) \end{aligned}
\begin{aligned} &\text { Increase }=0.02 k \times 100 \\\\ &=2 k \% \end{aligned}
|
# Triangles 101
## Before trying to understand the exact math and programming behind rasterizing a triangle, it's important to define triangles in these three ways.
### To know each x value, we can consider the two lines that run through the origin, as defined by definition #2. Note that a line is given by the equation: "y = mx + b", and rewritten, we can find x when given y, using "x = (y - b)/m". With this in mind, we can solve for the x-values of either line running through the origin when given the y position. Which can come straigh from our loop between the two y-limits of our triangles. We can simplify this however. We do not know b in our example, so let's replace this equation with "x = (y-offset.y)/m + offset.x". This equation means that we are using the difference in y to find the difference in x, since we know that both x and y start at our offset.
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### From this idea of using the difference in y to find difference in x, we can simplify this process for the computer. For our end product, we will be looping through every pixel in a triangle, and for every y position included in the triangle, we find the range of x values included for the equivelent horizontal slice. We can loop through all y values very easily by setting an initial y value to the highest y coordiante of the triangle, and incrementing it until it surpasses the lowest y coordinate of the triangle. This is even easier with a horizontal-terminator triangle, as it just needs to loop between the y coordinate of the origin and the y coordinate of the terminator. This also makes calculating the x range much simpler. As we can set an initial x value to the x coordinate of the origin, and every time we increment the y value, we add 1/m to our x value. We use 1/m as it is the derivative of "x = (y - origin.x)/m + origin.x" (more specifically the dx/dy). Essentially, the difference between "x = (y - origin.x)/m + origin.x" and "x = (y + 1 - origin.x)/m + origin.x" is just 1/m.
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### In summary, we can draw a flat-terminating triangle by looping through it's included y values, and for y value, finding the boundaries of x by solving for the x values of the left and right edge lines. We will call our two resulting x-values, "left x" and "right x". We can find the slope of a line going through two points by calculating "m = (y2-y1)/(x2-x1)". By calculating (x2-x1)/(y2-y1) instead of the original reciprocal, we can find the inverse slope or, in other words, the value we increment x by for every increment in y. Given that, we simply loop through y values, finding the left and right x values along the way, and then loop between the left and right x in order to get or set every pixel inside of the flat-terminating triangle.
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### Given that we can now draw a triangle with a flat terminator as detailed above, we need to then check if any given triangle has a horizontally-flat edge somewhere, drawing it using that method if it does, and forming two flat-terminating triangles out of it if it doesn't. We can find this by splitting the given triangle horizontally along the middle-y coordinate. What is meant by this, is finding which of the 3 points of the triangle lies in between the other two in terms of y values, and setting that as our y coordinate to split on. In BASIC, there is no quick easy way to find this middle point, so we shall insert a series of checks. In order to define the two triangles, we need to find a missing point on the line opposite of our middle point. And we have two things in order to do that, we have the line that the point lies on, and the y position of the point. Using our x = (y - b) / m, we can solve for x. Thinking of our slope in terms of changes in y related to changes in x, we can find our new x with "x = (y - y1) * m + x1". With (y1, x1) being either end point of our target line, and m being the inverse slope; (x2 - x1)/(y2 - y1). This shows us our mystery point to be ((middlePoint.y-y1)*(x2-x1)/(y2-y1) + x1, middlePoint.y). While this is rather hefty in terms of a small computation, it is only needed at maximum once for every time a triangle is drawn.
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# Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.3
### (i) -5, -1, 3, 7, …………
Solution:
Given sequence is -5, -1, 3, 7, …………
∴ First term is -5
And, common difference = a2 – a1 = -1 – (-5) = 4
∴ Common difference is 4
### (ii) 1/5, 3/5, 5/5, 7/5, ……
Solution:
Given sequence is 1/5, 3/5, 5/5, 7/5, ……
∴ First term is 1/5
And, common difference = a2 – a1 = 3/5 – (1/5) = 2/5
∴ Common difference is 2/5
### (iii) 0.3, 0.55, 0.80, 1.05, …………
Solution:
Given sequence is 0.3, 0.55, 0.80, 1.05, …………
∴ First term is 0.3
And, common difference = a2 – a1 = 0.55 – (0.3) = 0.25
∴ Common difference is 0.25
### (iv) -1.1, -3.1, -5.1, -7.1, …………..
Solution :
Given sequence is -1.1, -3.1, -5.1, -7.1, …………..
∴ First term is -1.1
And, common difference = a2 – a1 = -3.1 – (-1.1) = -2.0
∴ Common difference is -2.0
### (i) a = 4, d = -3
Solution:
Given:
a1 = 4 and d= -3
Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……
⇒ 4, 4 – 3, 4 + 2(-3), 4 + 3(-3), ……
⇒ 4, 1, – 2, – 5, – 8 ……..
∴ A.P will be 4, 1, – 2, – 5, – 8 ……..
### (ii) a = -1, d = 1/2
Solution:
Given:
a1 = -1, d= 1/2
Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……
⇒ -1, -1 + 1/2, -1, 2½, -1 + 3½, …
⇒ -1, -1/2, 0, 1/2
∴ A.P will be -1, -1/2, 0, 1/2
### (iii) a = -1.5, d = -0.5
Solution:
Given:
a1 = -1.5, d = -0.5
Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……
⇒ -1.5, -1.5, -0.5, –1.5 + 2(– 0.5), –1.5 + 3(– 0.5)
⇒ – 1.5, – 2, – 2.5, – 3, …….
∴ A.P will be – 1.5, – 2, – 2.5, – 3, …….
### (i) The cost of digging a well for the first meter is ₹ 150 and rises by ₹ 20 for each succeeding meter.
Solution:
Given:
Cost of digging a well for the first meter = ₹150
Therefore,
Cost for the second meter = ₹150 + ₹20 = ₹170
Cost for the third meter = ₹170 + ₹20 = ₹190
Cost for the fourth meter = ₹190 + ₹20 = ₹210
Since, 20 is added for each succeeding meter
Hence, the sequence will be (In rupees) and is A.P
150, 170, 190, 210, ………..
∴ The given sequence is in A.P and common difference is 20
### (ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of the remaining in the cylinder.
Solution:
Air removed for the first time = (1 x 1/4) = 1/4
Remaining air = 1 – 1/4 = 3/4
Air removed for the second time = (3/4 x 1/4) = 3/16
Remaining air = 3/4 – 3/16 = 9/16
Air removed for the third time = (9/16 x 1/4) = 9/64
Remaining air = 9/16 – 9/64 = 27/64
∴The sequence will be 1, 3/4, 9/16, 27/64
Here, a2 – a1 = 3/4 – (1) = -1/4
a3 – a2 = 9/16 – (3/4) = -3/16
Since, the successive difference of list is not same
∴ The given sequence is not in A.P
### (iii) Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …, and so on.
Solution:
Given:
Divya deposited Rs 1000 at compound interest of 10% p.a
So, the amount at the end of first year is = 1000 + 0.1(1000) = Rs 1100
And, the amount at the end of second year is = 1100 + 0.1(1100) = Rs 1210
And, the amount at the end of third year is = 1210 + 0.1(1210) = Rs 1331
Here, a2 – a1 = 1210 – 1100 = 110
a3 – a2 = 1331 – 1210 = 121
Since, the successive difference of list is not same
∴ The given sequence is not in A.P
### (i) 1, -2, -5, -8, ……..
Solution:
Given:
First term = 1, Second term = -2
Common difference = -2 – (1) = -3
Now,
Fifth term = -8 + (-3) = -11
Sixth term = -11 + (-3) = -14
Seventh term = -14 + (-3) = -17
Eighth term = -17 + (-3) = -20
### (ii) 0, -3, -6, -9, ……
Solution:
Given:
First term = 0, Second term = -3
Common difference = -3 – (0) = -3
Now,
Fifth term = -9 + (-3) = -12
Sixth term = -12 + (-3) = -15
Seventh term = -15 + (-3) = -18
Eighth term = -18 + (-3) = -21
### (iii) -1, 1/4, 3/2, ……..
Solution:
Given:
First term = -1, Second term = 1/4
Common difference = 1/4 – (-1) = 5/4
Now,
Fifth term = 3/2 + (5/4) = 11/4
Sixth term = 11/4 + (5/4) = 4
Seventh term = 4 + (5/4) = 21/4
Eighth term = 21/4 + (5/4) = 26/4
### (iv) -1, – 5/6, – 2/3, ………..
Solution:
Given:
First term = -1, Second term = -5/6
Common difference = -5/6 – (-1) = 1/6
Now,
Fifth term = -2/3 + (1/6) = -1/2
Sixth term = -1/2 + (1/6) = -1/3
Seventh term = -1/3 + (1/6) = -1/6
Eighth term = -1/6 + (1/6) = 0
### Problem 5: Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference?
Solution:
an = a + nb
Let n= 1, 2, 3, 4, 5, ……….
a1 = a + b
a2 = a + 2b
a3 = a + 3b
a4 = a + 4b
a5 = a + 5b
Now,
a2 – a1 = (a + 2b) – (a + b) = b
a3 – a2 = (a + 3b ) – (a + 2b) = b
a4 – a1 = (a + 4b) – (a + 3b) = b
a5 – a4 = (a + 5b) – (a + 4b) = b
Hence, proved
### (i) 3, 6, 12, 24, ………………………………
Solution:
Now,
a2 – a1 = 6 – (3) = 3
a3 – a1 = 12 – (6) = 6
a4 – a1 = 24 – (12) = 12
Since, the successive difference of list is not same
∴ The given sequence is not in A.P
### (ii) 0, -4, -8, -12, ……………………………………………………
Solution:
Now,
a2 – a1 = -4 – (0) = -4
a3 – a2 = -8 – (-4) = -4
a4 – a3 = -12 – (-8) = -4
Since, the successive difference of list is same
∴ The given sequence is in A.P and common difference is -4
### (iii) 1/2, 1/4, 1/6, 1/8
Solution:
Now,
a2 – a1 = 1/4 – (1/2) = -1/4
a3 – a2 = 1/6 – (1/4) = -1/12
a4 – a3 = 1/8 – (1/6) = -1/24
Since, the successive difference of list is not same
∴ The given sequence is not in A.P
### (iv) 12, 2, -8, -18, ……………………………..
Solution:
Now,
a2 – a1 = 2 – (12) = -10
a3 – a2 = -8 – (2) = -10
a4 – a3 = -18 – (-8) = -10
Since, the successive difference of list is same
∴ The given sequence is in A.P
### (v) 3, 3, 3, 3, ………………………………………
Solution:
Now,
a2 – a1 = 3 – (3) = 0
a3 – a2 = 3 – (3) = 0
a4 – a3 = 3 – (3) = 0
Since, the successive difference of list is same
∴ The given sequence is in A.P and common difference is 0
### (vi) p, p + 90, p + 180, ………………..
Solution:
Now,
a2 – a1 = p+90 – (p) = 90
a3 – a2 = p+180 – (p+90) = 90
Since, the successive difference of list is same
∴ The given sequence is in A.P and common difference is 90
### (vii) 1.0, 1.7, 2.4, ………………………….
Solution:
Now,
a2 – a1 = 1.7 – (1.0) = 0.7
a3 – a2 = 2.4 – (1.7) = 0.7
Since, the successive difference of list is same
∴ The given sequence is in A.P and common difference is 0.7
### (viii) -225, -425, -625, …………………………….
Solution:
Now,
a2 – a1 = -425 – (-225) = -200
a3 – a2 = -625 – (-425) = -200
Since, the successive difference of list is same
∴ The given sequence is in A.P and common difference is -200
### (ix) 10, 10+25, 10+26, 10+27, …………………………………………..
Solution:
Now,
a2 – a1 = 10+25 – (10) = 32
a3 – a2 = 10+26 – (10+25) = 32
a4 – a3 = 10+27 – (10+26) = 64
Since, the successive difference of list is not same
∴ The given sequence is in A.P
### (x) a+b, (a+1)+b, (a+1)+(b+1), (a+2)+(b+1), ……………………………….
Solution:
Now,
a2 – a1 = ((a+1) + b) – (a + b) = 1
a3 – a2 = ((a+1)+(b+1)) – ((a+1)+b) = 1
a4 – a3 = ((a+2)+(b+1)) – ((a+1)+(b+1)) = 1
Since, the successive difference of list is same
∴ The given sequence is in A.P and common difference is 1
### (xi) 12, 32, 52, 72, ………………………………………………………
Solution:
Now,
a2 – a1 = (32 – 12) = 8
a3 – a2 = (52 – 32) = 16
a4 – a3 = (72 – 52) = 24
Since, the successive difference of list is not same
∴ The given sequence is not in A.P
### (xii) 12, 52, 72, 73, …………………………………………………………..
Solution:
Now,
a2 – a1 = 52 – 12 = 24
a3 – a2 = 72 – 52 = 24
a4 – a3 = 73 – 72 = 24
Since, the successive difference of list is same
∴ The given sequence is in A.P and common difference is 24
### (i) 51, 59, 67, 75, …….
Solution:
Given:
First term = 51, Second term = 59
Common difference = 59 – 51 = 8
Now,
Fifth term = 75 + 8 = 83
Sixth term = 83 + 8 = 91
### (ii) 75, 67, 59, 51, ………
Solution:
Given:
First term = 75, Second term = 67
Common difference = 67 – 75 = -8
Now,
Fifth term = 51 – 8 = 43
Sixth term = 43 – 8 = 35
### (iii) 1.8, 2.0, 2.2, 2.4, …….
Solution:
Given:
First term = 1.8, Second term = 2.0
Common difference = 2.0 – 1.8 = 0.2
Now,
Fifth term = 2.4 + (0.2) = 2.6
Sixth term = 2.6 + (0.2) = 2.8
### (iv) 0, 1/4, 1/2, 3/4, ………..
Solution:
Given:
First term = 0, Second term = 1/4
Common difference = 1/4 – 0 = 1/4
Now,
Fifth term = 3/4 + (1/4) = 1
Sixth term = 1 + (1/4) = 5/4
### (v) 119, 136, 153, 170, ………..
Solution:
Given:
First term = 119, Second term = 136
Common difference = 136 – 119 = 17
Now,
Fifth term = 170 + (17) = 187
Sixth term = 187 + (17) = 204
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Square Root of 1521 + Solution With Free Steps
The square root of 1521, represented by the symbol √1521, equals 39. The power 1/2 is represented by the square root symbol √, which is. It means that when a number’s square root is calculated, the outcome is a number that has half the power of the original number. For instance, 196’s square root is equal to 14, or √196.
In this article, we will analyze and find the square root of 1521Â using various mathematical techniques, such as the approximation method and the long division method.
What Is the Square Root Of 1521?
The square root of the number 1521 is 39.
The square root can be defined as the quantity that can be doubled to produce the square of that similar quantity. In simple words, it can be explained as:
√1521 = √(39 x 39)
√1521 = √(39)$^2$
√1521 = ±39
The square can be canceled with the square root as it is equivalent to 1/2; therefore, obtaining 39. Hence 39 is 1521’s square root. The square root generates both positive and negative integers.
How To Calculate the Square Root of 1521?
You can calculate the square root of 1521Â using any of two vastly used techniques in mathematics; one is the Approximation technique, and the other is the Long Division method.
The symbol √ is interpreted as 1521 raised to the power 1/2. So any number, when multiplied by itself, produces its square, and when the square root of any squared number is taken, it produces the actual number.
Let us discuss each of them to understand the concepts better.
Square Root by Long Division Method
The process of long division is one of the most common methods used to find the square roots of a given number. It is easy to comprehend and provides more reliable and accurate answers. The long division method reduces a multi-digit number to its equal parts.
Learning how to find the square root of a number is easy with the long division method. All you need are five primary operations- divide, multiply, subtract, bring down or raise, then repeat.
Following are the simple steps that must be followed to find the square root of 1521Â using the long division method:
Step 1
First, write the given number 1521Â in the division symbol, as shown in figure 1.
Step 2
Starting from the right side of the number, divide the number 1521 into pairs such as 21 and 15.
Step 3
Now divide the digit 15Â by a number, giving a number either 3 or less than 3. Therefore, in this case, the remainder is 6Â whereas the quotient is 3.
Step 4
After this, bring down the next pair 21. Now the dividend is 621. To find the next divisor, we need to double our quotient obtained before. Doubling 3 gives 6; hence consider it as the next divisor.
Step 5
Now pair 2 with another number to make a new divisor that results in $\leq$ 621 when multiplied with the divisor. If the number is not a perfect square, add pair of zeros to the right of the number before starting division.
Step 6
Adding 9 to the divisor and multiplying 69 with 9 results in 621 = 621. The remainder obtained is 0.Â
Step 7
The resulting quotient 39 is the square root of 1521. Figure 1 given below shows the long division process in detail:
Figure 1
Important points
• The number 1521 is a perfect square.
• The number 1521 is a rational number.
• The number 1521 can be split into its prime factorization.
Is Square Root of 1521 a Perfect Square?
The number 1521 is a perfect square. A number is a perfect square if it splits into two equal parts or identical whole numbers. If a number is a perfect square, it is also rational.
A number expressed in p/q form is called a rational number. All the natural numbers are rational. A square root of a perfect square is a whole number; therefore, a perfect square is a rational number.
A number that is not a perfect square is irrational as it is a decimal number. As far as 1521 is concerned, it is a perfect square. It can be proved as below:
Factorization of 1521 results in 39 x 39 which can also be expressed as 39$^2$.
Taking the square root of the above expression gives:
= √(39$^2$)
= (39$^2$)$^{1/2}$
= 39
This shows that 1521 is a perfect square and a rational number.
This shows that 1521 is not a perfect square as it has decimal places; hence it is an irrational number.
Therefore the above discussion proves that the square root of 1521 is equivalent to 39.
Images/mathematical drawings are created with GeoGebra.
Square Root Of 25 | Square Roots List | Square Root Of 300
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# Evaluate 4(1/3)^7 div 2(1/3)^3 ?
Dec 6, 2017
The expression simplifies to $\frac{2}{81}$
#### Explanation:
It looks like the $4$ and the $2$ are coefficients.
$4 {\left(\frac{1}{3}\right)}^{7} \div 2 {\left(\frac{1}{3}\right)}^{3}$
1) Using a calculator, evaluate the fractions raised to the power of $7$ and the power of $3$
$4 \left(\frac{1}{2187}\right) \div 2 \left(\frac{1}{27}\right)$
2) Clear the parentheses by distributing the coefficients
$\frac{4}{2187} \div \frac{2}{27}$
3) Divide the fractions by multiplying by the reciprocal of the divisor
$\frac{4}{2187} \times \frac{27}{2}$
4) Reduce the fraction to lowest terms
${\cancel{4}}^{2} / {\cancel{2187}}^{81} \times {\cancel{27}}^{1} / {\cancel{2}}^{1}$
5) The fractions reduce to
$\frac{2}{81} \times \frac{1}{1}$
$\frac{2}{81}$
Dec 6, 2017
$\frac{2}{81}$
#### Explanation:
Expression $= 4 {\left(\frac{1}{3}\right)}^{7} \div 2 {\left(\frac{1}{3}\right)}^{3}$
To make this clearer, let's rewrite the expression as follows:
Expression$= \frac{4 \times {\left(\frac{1}{3}\right)}^{7}}{2 \times {\left(\frac{1}{3}\right)}^{3}} = 2 \times \frac{{\left(\frac{1}{3}\right)}^{7}}{{\left(\frac{1}{3}\right)}^{3}}$
Now we can use the law of exponents that states:
$\frac{{a}^{m}}{{a}^{n}} = {a}^{m - n}$
Thus, Expression $= 2 \times {\left(\frac{1}{3}\right)}^{7 - 3}$
$= 2 \times {\left(\frac{1}{3}\right)}^{4}$
$= \frac{2}{3} ^ 4 = \frac{2}{81}$
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# The Basics of Algebra Study Guide
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Updated on Aug 24, 2011
Find practice problems and solutions for these concepts at The Basics of Algebra Practice Problems.
### What's Around The Bend
• Variables
• Expressions
• Coefficients
• Like Terms
• Simplifying Expressions
• The Commutative Property
• The Associative Property
• The Identity Element
• Multiplicative Inverse
• The Distributive Property
Just when you were getting used to numbers, algebra attacks. You were hip to decimals, multiplication, even fractions. But algebra is a different kind of animal. It contains letters! Yes, letters. And that's not the only thing that makes algebra stand out.
You may be asking why you need to study algebra. Well, all the math that is needed in science and engineering relies on algebra as its basic language. Do you like toasters and TVs? Do you like airplanes and air conditioning? Do you like all the smart, edgy stuff that technology makes available? It all depends on algebra.
### Letters in Math Class?
Variables are letters that are used to represent numbers. Once you realize that these variables are just numbers in disguise, you'll understand that they must obey all the rules of mathematics, just like the numbers that aren't disguised. This can help you figure out what number the variable at hand stands for.
#### Fuel for Thought
An expression is like a series of words without a verb. Take, for example, 3x + 5 or a – 3. A verb, which would be an equality or inequality symbol, would give value to the statement, turning it into an equation or inequality. These symbols include =, ≠, >, <, ≥, and ≤.
Look at this equation:
a + 4 = 7
In the equation a + 4 = 7, the letter a represents a little box. Think of the letter a as the label on the box. Inside the box is a number. Your job is to figure out what that number is. What number can you put in the box to turn the equation a + 4 = 7 into a true statement?
Let's just stick 20 into the a box. The equation becomes 20 + 4 = 7, or 24 = 7. Okay, this equation is not a true statement because 24 does not equal 7. So, a is not the number 20.
If you put 3 into the box, the equation becomes this true statement: 3 + 4 = 7, or 7 = 7.
#### Fuel for Thought
When a number is placed next to a variable, indicating multiplication, the number is said to be the coefficient of the variable. For example,
8c 8 is the coefficient of the variable c.
6ab 6 is the coefficient of both variables, a and b.
If two or more terms have exactly the same variable(s), they are said to be like terms.
7x + 3x = 10x
The process of grouping like terms together and performing their mathematical operations is called combining like terms. It is important to combine like terms carefully, making sure that the variables are exactly the same.
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### Related Questions
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Whether you are working on a tan-th calculation or you are just looking for the tan-th value for a certain angle, it can be difficult to find the correct information. Fortunately, there are some steps you can take to help make the process easier.
## -45
Trying to find a -45 degree measure of an angle whose tangent is 1.19 may seem like a difficult task. However, if you are using the unit circle you can use its most important properties to your advantage.
In order to measure an angle you need to know the cosine of the angle and its inverse. The cosine is the square root of the tangent, and is found by putting the point on the unit circle. The inverse is found by subtracting the cosine of the angle from the square root of its tangent. This will yield the inverse of the cosine, which is the -45 degree measure of an angle aforementioned. Now that you know the inverse of an angle, it’s time to find a cosine function that will help you measure an angle’s magnitude.
## -180
Suppose you want to find the cosine function for a negative angle. In order to do this, you need to know the value of the tangent. The tangent of an angle is the length of its opposite leg divided by the adjacent leg. To find the tangent value of an angle, you can use the table below. You will then need to select the degrees and common angles you wish to measure. Once you have selected the degrees and common angles, you will be able to check if the value of the tangent is negative or positive by setting the degree measure to -1. You can also check by selecting the radians you wish to measure.
If the tangent value of an angle is negative, then it is a negative angle. This is because the tangent function is an odd function. This means that every 180 degrees the function repeats. To find the tangent of a negative angle, you can use the formula below. This formula is the same as the formula for a positive angle, only it is negative.
If the tangent value of a negative angle is 1.19, the angle is 40.7 degrees. The graph of y = tan(th) is shown to the right. This graph shows that A is in the range of zero to 180 degrees. If you have chosen a positive value for the tangent, then A is in the range of zero to 60 degrees. For example, if you want to find the tangent of an angle with a tangent of 1.19, you will need to find the value of the tangent of the angle that is closest to the angle that is the closest to A.
## -90
Among the many concepts you need to learn in your algebra class is the cosine function for a negative angle. This may sound like a hard concept to grasp, but it isn’t. To find the cosine function for a negative incline angle, simply divide the angle into its negative radii. Similarly, for a positive incline angle, simply divide the angle by its positive radii. You can also multiply the angle by its corresponding radii to find its cosine function.
While the cosine function for a negative degree angle is not that difficult to figure out, the cosine function for a positive angle is a little harder to come by. If you have trouble identifying the cosine function for a positive incline angle, it may help to ask the instructor or teacher for assistance. In most cases, this will also provide you with a good opportunity to recite your algebraic equations and answer questions.
## -270
Using a -270 degree measure of an angle whose tangent is 1.19, find the number of steps required to rotate the angle from its starting position to its final location. The number of steps can be found by applying the formulas (1.4) and (1.6) in the xy-coordinate plane. Similarly, the number of steps required to rotate the angle to the opposite direction is determined by applying the same formulas.
The -270 degree measure of an angle possesses two important functions: cos and tan. The cos function is useful in the context of a triangle. The tan function is also useful, especially when a triangle is rotated about a common axis. A -270 degree measure of an angle involving two triangles is a triangular pyramid.
A -270 degree measure of an angle is a bit on the long side, especially if the angle is rotated in the opposite direction. For example, a -270 degree measure of an Angle whose tangent is 1.19 has the same initial and terminal sides as a 225 degree angle, but has a much greater length. The -270 degree measure of an angle is probably the most interesting of all. It is also the most important, since it can be used to calculate the angle of an ordered pair of points.
## Frequently repeated tan(th) values
Frequently repeated tan(th) values of an angle whose tangent is 1.19 can be determined using a tangent calculator. A tangent calculator is a tool that can be used to find a tangent value for any angle. You can find tangent values in degrees or radians. A tangent calculator has keys that will allow you to calculate specific ratios. When you are using a tangent calculator to find a tangent value, you can use the tangent table to determine the tangent function’s range. Using the tangent table, you can find a tangent value for any of the angles listed in the table.
The tangent function is one of the six fundamental trigonometric functions. It is a symmetric function, meaning that it is symmetric about the origin. This means that the tangent function is not continuous. It can be defined for any angle, but it is undefined at odd multiples of 90deg. It is also undefined when the cos(th) is 0 (meaning the function is not symmetric about a point). Tangents are defined to be 0 when the cos(th) is -180 deg and undefined at odd multiples of 90deg. A tangent calculator is also useful for finding errors in a tangent graph, and can be used to graph a tangent graph.
In a right triangle, the tangent of the angle is calculated by dividing the side opposite the angle by the hypotenuse. The hypotenuse is the longest side of a right triangle that is opposite the angle. A calculator can also be used to determine the tangent of a right triangle, and it can be used to find the tangent value of any angle. When using a calculator, you can enter the tangent value of an angle and find its exact tangent value, or you can enter the angle and find its tangent value using the Taylor Series of the tangent.
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# Chapter 7 Section 6. Objectives 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solving Equations with Rational Expressions Distinguish between.
## Presentation on theme: "Chapter 7 Section 6. Objectives 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solving Equations with Rational Expressions Distinguish between."— Presentation transcript:
Chapter 7 Section 6
Objectives 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solving Equations with Rational Expressions Distinguish between operations with rational expressions and equations with terms that are rational expressions. Solve equations with rational expressions. Solve a formula for a specified variable. 7.6 2 3
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objective 1 Distinguish between operations with rational expressions and equations with terms that are rational expressions. Slide 7.6-3
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Distinguish between operations with rational expressions and equations with terms that are rational expressions. Before solving equations with rational expressions, you must understand the difference between sums and differences of terms with rational coefficients, or rational expressions, and equations with terms that are rational expressions. Sums and differences are expressions to simplify. Equations are solved. Uses of the LCD When adding or subtracting rational expressions, keep the LCD throughout the simplification. When solving an equation, multiply each side by the LCD so the denominators are eliminated. Slide 7.6-4
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solution: Identify each of the following as an expression or an equation. Then simplify the expression or solve the equation. equation expression Slide 7.6-5 EXAMPLE 1 Distinguishing between Expressions and Equations
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objective 2 Solve equations with rational expressions. Slide 7.6-6
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve equations with rational expressions. When an equation involves fractions, we use the multiplication property of equality to clear the fractions. Choose as multiplier the LCD of all denominators in the fractions of the equation. Recall from Section 7.1 that the denominator of a rational expression cannot equal 0, since division by 0 is undefined. Therefore, when solving an equation with rational expressions that have variables in the denominator, the solution cannot be a number that makes the denominator equal 0. Slide 7.6-7
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve, and check the solution. Solution: Check: The use of the LCD here is different from its use in Section 7.5. Here, we use the multiplication property of equality to multiply each side of an equation by the LCD. Earlier, we used the fundamental property to multiply a fraction by another fraction that had the LCD as both its numerator and denominator. Slide 7.6-8 EXAMPLE 2 Solving an Equation with Rational Expressions
Copyright © 2012, 2008, 2004 Pearson Education, Inc. While it is always a good idea to check solutions to guard against arithmetic and algebraic errors, it is essential to check proposed solutions when variables appear in denominators in the original equation. Solving an Equation with Rational Expressions Step 1: Multiply each side of the equation by the LCD to clear the equation of fractions. Be sure to distribute to every term on both sides. Step 2: Solve the resulting equation. Step 3: Check each proposed solution by substituting it into the original equation. Reject any that cause a denominator to equal 0. Slide 7.6-9 Solve equations with rational expressions. (cont’d)
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solution: Solve, and check the proposed solution. When the equatioin is solved, − 1 is a proposed solution. However, since x = − 1 leads to a 0 denominator in the original equation, the solution set is Ø. Slide 7.6-10 EXAMPLE 3 Solving an Equation with Rational Expressions Ѳ
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solution: Solve, and check the proposed solution. The solution set is {4}. Slide 7.6-11 EXAMPLE 4 Solving an Equation with Rational Expressions Ѳ
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solution: Solve, and check the proposed solution. Since 0 does not make any denominators equal 0, the solution set is {0}. Slide 7.6-12 EXAMPLE 5 Solving an Equation with Rational Expressions
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve, and check the proposed solution (s). Solution: The solution set is {−4, −1}. or Slide 7.6-13 EXAMPLE 6 Solving an Equation with Rational Expressions Ѳ
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve, and check the proposed solution. Solution: The solution set is {60}. Slide 7.6-14 EXAMPLE 7 Solving an Equation with Rational Expressions
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objective 3 Solve a formula for a specified variable. Slide 7.6-15
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve each formula for the specified variable. Solution: Remember to treat the variable for which you are solving as if it were the only variable, and all others as if they were contants. Slide 7.6-16 EXAMPLE 8 Solving for a Specified Variable
Copyright © 2012, 2008, 2004 Pearson Education, Inc. Solve the following formula for z. Solution: When solving an equation for a specified variable, be sure that the specified variable appears alone on only one side of the equals symbol in the final equation. Slide 7.6-17 EXAMPLE 9 Solving for a Specified Variable
Copyright © 2012, 2008, 2004 Pearson Education, Inc. HL# 7.6 Book Beginning Algebra Page 463# 19,21,24,26,31,32,38,41,47,49,54,56. Page 464# 63,64,69,73,75,86,90,94.
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# Factors of 49
The factors of 49 and the prime factors of 49 differ because forty-nine is a composite number. Also, despite being closely related, the prime factors of 49 and the prime factorization of 49 are not exactly the same either. In any case, by reading on you can learn the answer to the question what are the factors of 49? and everything else you want to know about the topic.
## What are the Factors of 49?
They are: 49, 7, 1. These are all the factors of 49, and every entry in the list can divide 49 without rest (modulo 0). That’s why the terms factors and divisors of 49 can be used interchangeably.
As is the case for any natural number greater than zero, the number itself, here 49, as well as 1 are factors and divisors of 49.
## Prime Factors of 49
The prime factors of 49 are the prime numbers which divide 49 exactly, without remainder as defined by the Euclidean division. In other words, a prime factor of 49 divides the number 49 without any rest, modulo 0.
For 49, the prime factors are: 7. By definition, 1 is not a prime number.
Besides 1, what sets the factors and the prime factors of the number 49 apart is the word “prime”. The former list contains both, composite and prime numbers, whereas the latter includes only prime numbers.
## Prime Factorization of 49
The prime factorization of 49 is 7 x 7. This is a unique list of the prime factors, along with their multiplicities. Note that the prime factorization of 49 does not include the number 1, yet it does include every instance of a certain prime factor.
49 is a composite number. In contrast to prime numbers which only have one factorization, composite numbers like 49 have at least two factorizations.
To illustrate what that means select the rightmost and leftmost integer in 49, 7, 1 and multiply these integers to obtain 49. This is the first factorization. Next choose the second rightmost and the second leftmost entry to obtain the 2nd factorization which also produces 49.
The prime factorization or integer factorization of 49 means determining the set of prime numbers which, when multiplied together, produce the original number 49. This is also known as prime decomposition of 49.
Besides factors for 49, frequently searched terms on our website include:
We did not place any calculator here as there are already a plethora of them on the web. But you can find the factors, prime factors and the factorizations of many numbers including 49 by using the search form in the sidebar.
To sum up:
The factors, the prime factors and the prime factorization of 49 mean different things, and in strict terms cannot be used interchangeably despite being closely related.
The factors of forty-nine are: 49, 7, 1. The prime factors of forty-nine are 7. And the prime factorization of forty-nine is 7 x 7. Remember that 1 is not a prime factor of 49.
No matter if you had been searching for prime factorization for 49 or prime numbers of 49, you have come to the right page. Also, if you typed what is the prime factorization of 49 in the search engine then you are right here, of course.
Taking all of the above into account, tasks including write 49 as a product of prime factors or list the factors of 49 will no longer pose a challenge to you.
If you have any questions about the factors of forty-nine then fill in the form below and we will respond as soon as possible. If our content concerning all factors of 49 has been of help to you then share it by means of pressing the social buttons. And don’t forget to bookmark us.
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# Quick Answer: What Are The 3 Conditions Of Continuity?
## Is zero a continuous function?
Differentiability and continuity A cusp on the graph of a continuous function.
At zero, the function is continuous but not differentiable..
## Does a limit exist if there is a hole?
The first, which shows that the limit DOES exist, is if the graph has a hole in the line, with a point for that value of x on a different value of y. … If there is a hole in the graph at the value that x is approaching, with no other point for a different value of the function, then the limit does still exist.
## What is the limit?
A limit tells us the value that a function approaches as that function’s inputs get closer and closer to some number. The idea of a limit is the basis of all calculus. Created by Sal Khan.
## What is the 3 part definition of continuity?
For a function to be continuous at a point from a given side, we need the following three conditions: the function is defined at the point. the function has a limit from that side at that point. the one-sided limit equals the value of the function at the point.
## What are the conditions for a limit to exist?
In order for a limit to exist, the function has to approach a particular value. In the case shown above, the arrows on the function indicate that the the function becomes infinitely large. Since the function doesn’t approach a particular value, the limit does not exist.
## What is continuity of a function?
A function is said to be continuous on the interval [a,b] if it is continuous at each point in the interval. Note that this definition is also implicitly assuming that both f(a) and limx→af(x) lim x → a exist. If either of these do not exist the function will not be continuous at x=a .
## What is the formal definition of continuity?
The formal definition of continuity at a point has three conditions that must be met. A function f(x) is continuous at a point where x = c if. exists. f(c) exists (That is, c is in the domain of f.)
## What is the difference between limit and continuity?
Solution: This limit does not exist. You can find points arbitrarily close to the origin with f(x,y)=1, f(x,y)=−1 or anything in between. … Just as with one variable, we say a function is continuous if it equals its limit: A function f(x,y) is continuous at the point (a,b) if lim(x,y)→(a,b)f(x,y)=f(a,b).
## How do you show continuity of a function?
Definition: A function f is continuous at x0 in its domain if for every ϵ > 0 there is a δ > 0 such that whenever x is in the domain of f and |x − x0| < δ, we have |f(x) − f(x0)| < ϵ. Again, we say f is continuous if it is continuous at every point in its domain.
## Can 0 be a limit?
Typically, zero in the denominator means it’s undefined. However, that will only be true if the numerator isn’t also zero. … However, in take the limit, if we get 0/0 we can get a variety of answers and the only way to know which on is correct is to actually compute the limit.
## What are the 3 types of discontinuity?
Continuity and Discontinuity of Functions Functions that can be drawn without lifting up your pencil are called continuous functions. You will define continuous in a more mathematically rigorous way after you study limits. There are three types of discontinuities: Removable, Jump and Infinite.
## What are the rules of continuity?
In calculus, a function is continuous at x = a if – and only if – all three of the following conditions are met:The function is defined at x = a; that is, f(a) equals a real number.The limit of the function as x approaches a exists.The limit of the function as x approaches a is equal to the function value at x = a.
## How do you define continuity?
Continuity, in mathematics, rigorous formulation of the intuitive concept of a function that varies with no abrupt breaks or jumps. A function is a relationship in which every value of an independent variable—say x—is associated with a value of a dependent variable—say y.
## What does it mean to maintain continuity?
The definition of continuity refers to something occurring in an uninterrupted state, or on a steady and ongoing basis. When you are always there for your child to listen to him and care for him every single day, this is an example of a situation where you give your child a sense of continuity. noun.
## What is another word for continuity?
In this page you can discover 47 synonyms, antonyms, idiomatic expressions, and related words for continuity, like: continuation, unity, continuousness, cut, intermittence, dissipation, desultoriness, duration, endurance, continue and connectedness.
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Division (mathematics) explained
Division is one of the four basic operations of arithmetic. The other operations are addition, subtraction, and multiplication. What is being divided is called the dividend, which is divided by the divisor, and the result is called the quotient.
At an elementary level the division of two natural numbers is, among other possible interpretations, the process of calculating the number of times one number is contained within another.[1] For example, if 20 apples are divided evenly between 4 people, everyone receives 5 apples (see picture). However, this number of times or the number contained (divisor) need not be integers.
The division with remainder or Euclidean division of two natural numbers provides an integer quotient, which is the number of times the second number is completely contained in the first number, and a remainder, which is the part of the first number that remains, when in the course of computing the quotient, no further full chunk of the size of the second number can be allocated. For example, if 21 apples are divided between 4 people, everyone receives 5 apples again, and 1 apple remains.
For division to always yield one number rather than an integer quotient plus a remainder, the natural numbers must be extended to rational numbers or real numbers. In these enlarged number systems, division is the inverse operation to multiplication, that is means, as long as is not zero. If, then this is a division by zero, which is not defined.[2] In the 21-apples example, everyone would receive 5 apple and a quarter of an apple, thus avoiding any leftover.
Both forms of division appear in various algebraic structures, different ways of defining mathematical structure. Those in which a Euclidean division (with remainder) is defined are called Euclidean domains and include polynomial rings in one indeterminate (which define multiplication and addition over single-variabled formulas). Those in which a division (with a single result) by all nonzero elements is defined are called fields and division rings. In a ring the elements by which division is always possible are called the units (for example, 1 and −1 in the ring of integers). Another generalization of division to algebraic structures is the quotient group, in which the result of "division" is a group rather than a number.
Introduction
The simplest way of viewing division is in terms of quotition and partition: from the quotition perspective, means the number of 5s that must be added to get 20. In terms of partition, means the size of each of 5 parts into which a set of size 20 is divided. For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". This is denoted as, or . In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient.
Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, leaves a remainder of 1, as 10 is not a multiple of 3. Sometimes this remainder is added to the quotient as a fractional part, so is equal to or, but in the context of integer division, where numbers have no fractional part, the remainder is kept separately (or exceptionally, discarded or rounded). When the remainder is kept as a fraction, it leads to a rational number. The set of all rational numbers is created by extending the integers with all possible results of divisions of integers.
Unlike multiplication and addition, division is not commutative, meaning that is not always equal to .[3] Division is also not, in general, associative, meaning that when dividing multiple times, the order of division can change the result.[4] For example,, but (where the use of parentheses indicates that the operations inside parentheses are performed before the operations outside parentheses).
Division is traditionally considered as left-associative. That is, if there are multiple divisions in a row, the order of calculation goes from left to right:[5] [6]
a/b/c=(a/b)/c=a/(b x c)\nea/(b/c)=(a x c)/b.
Division is right-distributive over addition and subtraction, in the sense that
a\pmb c
=(a\pmb)/c=(a/c)\pm(b/c)=
a c
\pm
b c
.
This is the same for multiplication, as
(a+b) x c=a x c+b x c
. However, division is not left-distributive, as
a b+c
=a/(b+c)\ne(a/b)+(a/c)=
ac+ab bc
.
For example
12 2+4
=
12 6
=2,
but
12 2
+
12 4
=6+3=9.
This is unlike the case in multiplication, which is both left-distributive and right-distributive, and thus distributive.
Notation
Division is often shown in algebra and science by placing the dividend over the divisor with a horizontal line, also called a fraction bar, between them. For example, "a divided by b" can be written as:
ab
which can also be read out loud as "divide a by b" or "a over b". A way to express division all on one line is to write the dividend (or numerator), then a slash, then the divisor (or denominator), as follows:
a/b
This is the usual way of specifying division in most computer programming languages, since it can easily be typed as a simple sequence of ASCII characters. (It is also the only notation used for quotient objects in abstract algebra.) Some mathematical software, such as MATLAB and GNU Octave, allows the operands to be written in the reverse order by using the backslash as the division operator:
b\backslasha
A typographical variation halfway between these two forms uses a solidus (fraction slash), but elevates the dividend and lowers the divisor:
{}a/{}b
Any of these forms can be used to display a fraction. A fraction is a division expression where both dividend and divisor are integers (typically called the numerator and denominator), and there is no implication that the division must be evaluated further. A second way to show division is to use the division sign (÷, also known as obelus though the term has additional meanings), common in arithmetic, in this manner:
a ÷ b
This form is infrequent except in elementary arithmetic. ISO 80000-2-9.6 states it should not be used. This division sign is also used alone to represent the division operation itself, as for instance as a label on a key of a calculator. The obelus was introduced by Swiss mathematician Johann Rahn in 1659 in Teutsche Algebra.[7] The ÷ symbol is used to indicate subtraction in some European countries, so its use may be misunderstood.[8]
In some non-English-speaking countries, a colon is used to denote division:[9]
a:b
This notation was introduced by Gottfried Wilhelm Leibniz in his 1684 Acta eruditorum. Leibniz disliked having separate symbols for ratio and division. However, in English usage the colon is restricted to expressing the related concept of ratios.
Since the 19th century, US textbooks have used
b)a
or
b\overline{)a}
to denote a divided by b, especially when discussing long division. The history of this notation is not entirely clear because it evolved over time.[10]
Computing
See main article: Long division and Division algorithm.
Manual methods
Division is often introduced through the notion of "sharing out" a set of objects, for example a pile of lollies, into a number of equal portions. Distributing the objects several at a time in each round of sharing to each portion leads to the idea of 'chunking' a form of division where one repeatedly subtracts multiples of the divisor from the dividend itself.
By allowing one to subtract more multiples than what the partial remainder allows at a given stage, more flexible methods, such as the bidirectional variant of chunking, can be developed as well.
More systematically and more efficiently, two integers can be divided with pencil and paper with the method of short division, if the divisor is small, or long division, if the divisor is larger. If the dividend has a fractional part (expressed as a decimal fraction), one can continue the procedure past the ones place as far as desired. If the divisor has a fractional part, one can restate the problem by moving the decimal to the right in both numbers until the divisor has no fraction, which can make the problem easier to solve (e.g., 10/2.5 = 100/25 = 4).
Division can be calculated with an abacus.[11]
Logarithm tables can be used to divide two numbers, by subtracting the two numbers' logarithms, then looking up the antilogarithm of the result.
Division can be calculated with a slide rule by aligning the divisor on the C scale with the dividend on the D scale. The quotient can be found on the D scale where it is aligned with the left index on the C scale. The user is responsible, however, for mentally keeping track of the decimal point.
By computer
Modern calculators and computers compute division either by methods similar to long division, or by faster methods; see Division algorithm.
In modular arithmetic (modulo a prime number) and for real numbers, nonzero numbers have a multiplicative inverse. In these cases, a division by may be computed as the product by the multiplicative inverse of . This approach is often associated with the faster methods in computer arithmetic.
Division in different contexts
Euclidean division
See main article: Euclidean division. Euclidean division is the mathematical formulation of the outcome of the usual process of division of integers. It asserts that, given two integers, a, the dividend, and b, the divisor, such that b ≠ 0, there are unique integers q, the quotient, and r, the remainder, such that a = bq + r and 0 ≤ r <, where denotes the absolute value of b.
Of integers
Integers are not closed under division. Apart from division by zero being undefined, the quotient is not an integer unless the dividend is an integer multiple of the divisor. For example, 26 cannot be divided by 11 to give an integer. Such a case uses one of five approaches:
1. Say that 26 cannot be divided by 11; division becomes a partial function.
2. Give an approximate answer as a floating-point number. This is the approach usually taken in numerical computation.
3. Give the answer as a fraction representing a rational number, so the result of the division of 26 by 11 is
\tfrac{26}{11}
(or as a mixed number, so
\tfrac{26}{11}=2\tfrac4{11}.
) Usually the resulting fraction should be simplified: the result of the division of 52 by 22 is also
\tfrac{26}{11}
. This simplification may be done by factoring out the greatest common divisor.
1. Give the answer as an integer quotient and a remainder, so
\tfrac{26}{11}=2remainder4.
To make the distinction with the previous case, this division, with two integers as result, is sometimes called Euclidean division, because it is the basis of the Euclidean algorithm.
1. Give the integer quotient as the answer, so
\tfrac{26}{11}=2.
This is the floor function applied to case 2 or 3. It is sometimes called integer division, and denoted by "//".
Dividing integers in a computer program requires special care. Some programming languages treat integer division as in case 5 above, so the answer is an integer. Other languages, such as MATLAB and every computer algebra system return a rational number as the answer, as in case 3 above. These languages also provide functions to get the results of the other cases, either directly or from the result of case 3.
Names and symbols used for integer division include div, /, \, and %. Definitions vary regarding integer division when the dividend or the divisor is negative: rounding may be toward zero (so called T-division) or toward −∞ (F-division); rarer styles can occur – see modulo operation for the details.
Divisibility rules can sometimes be used to quickly determine whether one integer divides exactly into another.
Of rational numbers
The result of dividing two rational numbers is another rational number when the divisor is not 0. The division of two rational numbers p/q and r/s can be computed as$= \times = .$
All four quantities are integers, and only p may be 0. This definition ensures that division is the inverse operation of multiplication.
Of real numbers
Division of two real numbers results in another real number (when the divisor is nonzero). It is defined such that a/b = c if and only if a = cb and b ≠ 0.
Of complex numbers
Dividing two complex numbers (when the divisor is nonzero) results in another complex number, which is found using the conjugate of the denominator:$= = = + i.$
This process of multiplying and dividing by
r-is
is called 'realisation' or (by analogy) rationalisation. All four quantities p, q, r, s are real numbers, and r and s may not both be 0.
Division for complex numbers expressed in polar form is simpler than the definition above:$= = e^.$
Again all four quantities p, q, r, s are real numbers, and r may not be 0.
Of polynomials
One can define the division operation for polynomials in one variable over a field. Then, as in the case of integers, one has a remainder. See Euclidean division of polynomials, and, for hand-written computation, polynomial long division or synthetic division.
Of matrices
One can define a division operation for matrices. The usual way to do this is to define, where denotes the inverse of B, but it is far more common to write out explicitly to avoid confusion. An elementwise division can also be defined in terms of the Hadamard product.
Left and right division
Because matrix multiplication is not commutative, one can also define a left division or so-called backslash-division as . For this to be well defined, need not exist, however does need to exist. To avoid confusion, division as defined by is sometimes called right division or slash-division in this context.
With left and right division defined this way, is in general not the same as, nor is the same as . However, it holds that and .
Pseudoinverse
To avoid problems when and/or do not exist, division can also be defined as multiplication by the pseudoinverse. That is, and, where and denote the pseudoinverses of A and B.
Abstract algebra
In abstract algebra, given a magma with binary operation ∗ (which could nominally be termed multiplication), left division of b by a (written) is typically defined as the solution x to the equation, if this exists and is unique. Similarly, right division of b by a (written) is the solution y to the equation . Division in this sense does not require ∗ to have any particular properties (such as commutativity, associativity, or an identity element). A magma for which both and exist and are unique for all a and all b (the Latin square property) is a quasigroup. In a quasigroup, division in this sense is always possible, even without an identity element and hence without inverses.
"Division" in the sense of "cancellation" can be done in any magma by an element with the cancellation property. Examples include matrix algebras, quaternion algebras, and quasigroups. In an integral domain, where not every element need have an inverse, division by a cancellative element a can still be performed on elements of the form ab or ca by left or right cancellation, respectively. If a ring is finite and every nonzero element is cancellative, then by an application of the pigeonhole principle, every nonzero element of the ring is invertible, and division by any nonzero element is possible. To learn about when algebras (in the technical sense) have a division operation, refer to the page on division algebras. In particular Bott periodicity can be used to show that any real normed division algebra must be isomorphic to either the real numbers R, the complex numbers C, the quaternions H, or the octonions O.
Calculus
The derivative of the quotient of two functions is given by the quotient rule:$' = \frac.$
Division by zero
See main article: Division by zero. Division of any number by zero in most mathematical systems is undefined, because zero multiplied by any finite number always results in a product of zero.[12] Entry of such an expression into most calculators produces an error message. However, in certain higher level mathematics division by zero is possible by the zero ring and algebras such as wheels.[13] In these algebras, the meaning of division is different from traditional definitions.
Notes and References
1. Book: Blake. A. G.. Arithmetic. 1887. Alexander Thom & Company. Dublin, Ireland.
2. Book: Derbyshire. John. Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. 2004. Penguin Books. New York City. 978-0-452-28525-5.
3. http://www.mathwords.com/c/commutative.htm Retrieved October 23, 2018
4. http://www.mathwords.com/a/associative_operation.htm Retrieved October 23, 2018
5. George Mark Bergman: Order of arithmetic operations
6. Education Place: The Order of Operations
7. Book: Cajori, Florian. A History of Mathematical Notations. Open Court Pub. Co.. 1929.
8. Book: https://www.unicode.org/versions/Unicode10.0.0/ch06.pdf#G7935 . 280, Obelus . 6. Writing Systems and Punctuation . Unicode Consortium . The Unicode® Standard: Version 10.0 – Core Specification . June 2017.
9. Book: Mathematics for Teachers: An Interactive Approach for Grades K–8. 126. Thomas Sonnabend. Brooks/Cole, Cengage Learning (Charles Van Wagner). 2010. 978-0-495-56166-8.
10. Book: History Of Mathematics Vol II. Smith, David Eugene. Ginn And Company. 1925.
11. Book: Kojima, Takashi. Advanced Abacus: Theory and Practice. 2012-07-09. Tuttle Publishing. 978-1-4629-0365-8. en.
12. http://mathworld.wolfram.com/DivisionbyZero.html Retrieved October 23, 2018
13. Jesper Carlström. "On Division by Zero" Retrieved October 23, 2018
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The graph of the logarithm to base 2 crosses the x axis (horizontal axis) at 1 and passes through the points with coordinates (2, 1), (4, 2), and (8, 3). For example, log2(8) = 3, because 23 = 8. The graph gets arbitrarily close to the y axis, but does not meet or intersect it.
In mathematics, the logarithm of a number is the exponent to which another fixed value, the base, must be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 10 to the power 3 is 1000: 1000 = 10 × 10 × 10 = 103. More generally, for any two real numbers b and x where b is positive and b ≠ 1,
$y=b^x\Leftrightarrow x=\log_b(y)$
The logarithm to base 10 (b = 10) is called the common logarithm and has many applications in science and engineering. The natural logarithm has the irrational (transcendental) number e (≈ 2.718) as its base; its use is widespread in pure mathematics, especially calculus. The binary logarithm uses base 2 (b = 2) and is prominent in computer science.
Logarithms were introduced by John Napier in the early 17th century as a means to simplify calculations. They were rapidly adopted by navigators, scientists, engineers, and others to perform computations more easily, using slide rules and logarithm tables. Tedious multi-digit multiplication steps can be replaced by table look-ups and simpler addition because of the fact—important in its own right—that the logarithm of a product is the sum of the logarithms of the factors:
$\log_b(xy) = \log_b (x) + \log_b (y), \,$
provided that b, x and y are all positive and b ≠ 1. The present-day notion of logarithms comes from Leonhard Euler, who connected them to the exponential function in the 18th century.
Logarithmic scales reduce wide-ranging quantities to smaller scopes. For example, the decibel is a logarithmic unit quantifying sound pressure and signal power ratios. In chemistry, pH is a logarithmic measure for the acidity of an aqueous solution. Logarithms are commonplace in scientific formulae, and in measurements of the complexity of algorithms and of geometric objects called fractals. They describe musical intervals, appear in formulae counting prime numbers, inform some models in psychophysics, and can aid in forensic accounting.
In the same way as the logarithm reverses exponentiation, the complex logarithm is the inverse function of the exponential function applied to complex numbers. The discrete logarithm is another variant; it has applications in public-key cryptography.
## Motivation and definition
The idea of logarithms is to reverse the operation of exponentiation, that is raising a number to a power. For example, the third power (or cube) of 2 is 8, because 8 is the product of three factors of 2:
$2^3 = 2 \times 2 \times 2 = 8. \,$
It follows that the logarithm of 8 with respect to base 2 is 3, so log2 8 = 3.
### Exponentiation
The third power of some number b is the product of three factors of b. More generally, raising b to the n-th power, where n is a natural number, is done by multiplying n factors of b. The n-th power of b is written bn, so that
$b^n = \underbrace{b \times b \times \cdots \times b}_{n \text{ factors}}.$
Exponentiation may be extended to by, where b is a positive number and the exponent y is any real number. For example, b−1 is the reciprocal of b, that is, 1/b. (For further details, including the formula bm + n = bm · bn, see exponentiation or [1] for an elementary treatise.)
### Definition
The logarithm of a positive real number x with respect to base b, a positive real number not equal to 1[nb 1], is the exponent by which b must be raised to yield x. In other words, the logarithm of x to base b is the solution y to the equation[2]
$b^y = x. \,$
The logarithm is denoted "logb(x)" (pronounced as "the logarithm of x to base b" or "the base-b logarithm of x"). In the equation y = logb(x), the value y is the answer to the question "To what power must b be raised, in order to yield x?". This question can also be addressed (with a richer answer) for complex numbers, which is done in section "Complex logarithm", and this answer is much more extensively investigated in the page for the complex logarithm.
### Examples
For example, log2(16) = 4, since 24 = 2 ×2 × 2 × 2 = 16. Logarithms can also be negative:
$\log_2 \!\left( \frac{1}{2} \right) = -1,\,$
since
$2^{-1} = \frac 1 {2^1} = \frac 1 2.$
A third example: log10(150) is approximately 2.176, which lies between 2 and 3, just as 150 lies between 102 = 100 and 103 = 1000. Finally, for any base b, logb(b) = 1 and logb(1) = 0, since b1 = b and b0 = 1, respectively.
## Logarithmic identities
Several important formulas, sometimes called logarithmic identities or log laws, relate logarithms to one another.[3]
### Product, quotient, power and root
The logarithm of a product is the sum of the logarithms of the numbers being multiplied; the logarithm of the ratio of two numbers is the difference of the logarithms. The logarithm of the p-th power of a number is p times the logarithm of the number itself; the logarithm of a p-th root is the logarithm of the number divided by p. The following table lists these identities with examples. Each of the identities can be derived after substitution of the logarithm definitions x = blogb(x), and/or y = blogb(y), in the left hand sides.
Formula Example
product $\log_b(x y) = \log_b (x) + \log_b (y) \,$ $\log_3 (243) = \log_3(9 \cdot 27) = \log_3 (9) + \log_3 (27) = 2 + 3 = 5 \,$
quotient $\log_b \!\left(\frac x y \right) = \log_b (x) - \log_b (y) \,$ $\log_2 (16) = \log_2 \!\left ( \frac{64}{4} \right ) = \log_2 (64) - \log_2 (4) = 6 - 2 = 4$
power $\log_b(x^p) = p \log_b (x) \,$ $\log_2 (64) = \log_2 (2^6) = 6 \log_2 (2) = 6 \,$
root $\log_b \sqrt[p]{x} = \frac {\log_b (x)} p \,$ $\log_{10} \sqrt{1000} = \frac{1}{2}\log_{10} 1000 = \frac{3}{2} = 1.5$
### Change of base
The logarithm logb(x) can be computed from the logarithms of x and b with respect to an arbitrary base k using the following formula:
$\log_b(x) = \frac{\log_k(x)}{\log_k(b)}.\,$
Typical scientific calculators calculate the logarithms to bases 10 and e.[4] Logarithms with respect to any base b can be determined using either of these two logarithms by the previous formula:
$\log_b (x) = \frac{\log_{10} (x)}{\log_{10} (b)} = \frac{\log_{e} (x)}{\log_{e} (b)}. \,$
Given a number x and its logarithm logb(x) to an unknown base b, the base is given by:
$b = x^\frac{1}{\log_b(x)}.$
## Particular bases
Among all choices for the base, three are particularly common. These are b = 10, b = e (the irrational mathematical constant ≈ 2.71828), and b = 2. In mathematical analysis, the logarithm to base e is widespread because of its particular analytical properties explained below. On the other hand, base-10 logarithms are easy to use for manual calculations in the decimal number system:[5]
$\log_{10}(10 x) = \log_{10}(10) + \log_{10}(x) = 1 + \log_{10}(x).\$
Thus, log10(x) is related to the number of decimal digits of a positive integer x: the number of digits is the smallest integer strictly bigger than log10(x).[6] For example, log10(1430) is approximately 3.15. The next integer is 4, which is the number of digits of 1430. The logarithm to base two is used in computer science, where the binary system is ubiquitous, and in music theory, where a pitch ratio of two (the octave) is ubiquitous and the cent is the binary logarithm (scaled by 1200) of the ratio between two adjacent equally-tempered pitches.
The following table lists common notations for logarithms to these bases and the fields where they are used. Many disciplines write log(x) instead of logb(x), when the intended base can be determined from the context. The notation blog(x) also occurs.[7] The "ISO notation" column lists designations suggested by the International Organization for Standardization (ISO 31-11).[8]
Base b Name for logb(x) ISO notation Other notations Used in
2 binary logarithm lb(x)[9] ld(x), log(x), lg(x), log2(x) computer science, information theory, mathematics, music theory
e natural logarithm ln(x)[nb 2] log(x)
(in mathematics and many programming languages[nb 3])
mathematical analysis, physics, chemistry,
statistics, economics, and some engineering fields
10 common logarithm lg(x) log(x), log10(x)
(in engineering, biology, astronomy)
various engineering fields (see decibel and see below),
logarithm tables, handheld calculators, spectroscopy
## History
### Predecessors
The Babylonians sometime in 2000–1600 BC may have invented the quarter square multiplication algorithm to multiply two numbers using only addition, subtraction and a table of quarter squares.[13][14] However, it could not be used for division without an additional table of reciprocals (or the knowledge of a sufficiently simple algorithm to generate reciprocals). Large tables of quarter squares were used to simplify the accurate multiplication of large numbers from 1817 onwards until this was superseded by the use of computers.
The Indian mathematician Virasena worked with the concept of ardhaccheda: the number of times a number of the form 2n could be halved. For exact powers of 2, this is the logarithm to that base, which is a whole number; for other numbers, it is undefined. He described relations such as the product formula and also introduced integer logarithms in base 3 (trakacheda) and base 4 (caturthacheda)[15]
Michael Stifel published Arithmetica integra in Nuremberg in 1544, which contains a table[16] of integers and powers of 2 that has been considered an early version of a logarithmic table.[17][18]
In the 16th and early 17th centuries an algorithm called prosthaphaeresis was used to approximate multiplication and division. This used the trigonometric identity
$\cos\,\alpha\,\cos\,\beta = \frac12[\cos(\alpha+\beta) + \cos(\alpha-\beta)]$
or similar to convert the multiplications to additions and table lookups. However, logarithms are more straightforward and require less work. It can be shown using Euler's Formula that the two techniques are related.
### From Napier to Euler
John Napier (1550–1617), the inventor of logarithms
The method of logarithms was publicly propounded by John Napier in 1614, in a book titled Mirifici Logarithmorum Canonis Descriptio (Description of the Wonderful Rule of Logarithms).[19][20] Joost Bürgi independently invented logarithms but published six years after Napier.[21][22]
Johannes Kepler, who used logarithm tables extensively to compile his Ephemeris and therefore dedicated it to Napier,[23] remarked:
… the accent in calculation led Justus Byrgius [Joost Bürgi] on the way to these very logarithms many years before Napier's system appeared; but … instead of rearing up his child for the public benefit he deserted it in the birth.
—Johannes Kepler[24]Rudolphine Tables (1627)
By repeated subtractions Napier calculated (1 − 10−7)L for L ranging from 1 to 100. The result for L=100 is approximately 0.99999 = 1 − 10−5. Napier then calculated the products of these numbers with 107(1 − 10−5)L for L from 1 to 50, and did similarly with 0.9998 ≈ (1 − 10−5)20 and 0.9 ≈ 0.99520. These computations, which occupied 20 years, allowed him to give, for any number N from 5 to 10 million, the number L that solves the equation
$N=10^7 {(1-10^{-7})}^L. \,$
Napier first called L an "artificial number", but later introduced the word "logarithm" to mean a number that indicates a ratio: λόγος (logos) meaning proportion, and ἀριθμός (arithmos) meaning number. In modern notation, the relation to natural logarithms is: [25]
$L = \log_{(1-10^{-7})} \!\left( \frac{N}{10^7} \right) \approx 10^7 \log_{ \frac{1}{e}} \!\left( \frac{N}{10^7} \right) = -10^7 \log_e \!\left( \frac{N}{10^7} \right),$
where the very close approximation corresponds to the observation that
${(1-10^{-7})}^{10^7} \approx \frac{1}{e}. \,$
The invention was quickly and widely met with acclaim. The works of Bonaventura Cavalieri (Italy), Edmund Wingate (France), Xue Fengzuo (China), and Johannes Kepler's Chilias logarithmorum (Germany) helped spread the concept further.[26]
The hyperbola y = 1/x (red curve) and the area from x = 1 to 6 (shaded in orange).
In 1649, Alphonse Antonio de Sarasa, a former student of Grégoire de Saint-Vincent,[27] related logarithms to the quadrature of the hyperbola, by pointing out that the area f(t) under the hyperbola from x = 1 to x = t satisfies[28]
$f(tu) = f(t) + f(u).\,$
The natural logarithm was first described by Nicholas Mercator in his work Logarithmotechnia published in 1668,[29] although the mathematics teacher John Speidell had already in 1619 compiled a table of what were effectively natural logarithms, based on Napier's work.[30] Around 1730, Leonhard Euler defined the exponential function and the natural logarithm by
$e^x = \lim_{n \rightarrow \infty} (1+x/n)^n,$
$\ln(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1).$
Euler also showed that the two functions are inverse to one another.[31][32][33]
### Logarithm tables, slide rules, and historical applications
The 1797 Encyclopædia Britannica explanation of logarithms
By simplifying difficult calculations, logarithms contributed to the advance of science, and especially of astronomy. They were critical to advances in surveying, celestial navigation, and other domains. Pierre-Simon Laplace called logarithms
"...[a]n admirable artifice which, by reducing to a few days the labour of many months, doubles the life of the astronomer, and spares him the errors and disgust inseparable from long calculations."[34]
A key tool that enabled the practical use of logarithms before calculators and computers was the table of logarithms.[35] The first such table was compiled by Henry Briggs in 1617, immediately after Napier's invention. Subsequently, tables with increasing scope and precision were written. These tables listed the values of logb(x) and bx for any number x in a certain range, at a certain precision, for a certain base b (usually b = 10). For example, Briggs' first table contained the common logarithms of all integers in the range 1–1000, with a precision of 8 digits. As the function f(x) = bx is the inverse function of logb(x), it has been called the antilogarithm.[36] The product and quotient of two positive numbers c and d were routinely calculated as the sum and difference of their logarithms. The product cd or quotient c/d came from looking up the antilogarithm of the sum or difference, also via the same table:
$c d = b^{\log_b (c)} \, b^{\log_b (d)} = b^{\log_b (c) + \log_b (d)} \,$
and
$\frac c d = c d^{-1} = b^{\log_b (c) - \log_b (d)}. \,$
For manual calculations that demand any appreciable precision, performing the lookups of the two logarithms, calculating their sum or difference, and looking up the antilogarithm is much faster than performing the multiplication by earlier methods such as prosthaphaeresis, which relies on trigonometric identities. Calculations of powers and roots are reduced to multiplications or divisions and look-ups by
$c^d = (b^{\log_b (c) })^d = b^{d \log_b (c)} \,$
and
$\sqrt[d]{c} = c^{\frac 1 d} = b^{\frac{1}{d} \log_b (c)}. \,$
Many logarithm tables give logarithms by separately providing the characteristic and mantissa of x, that is to say, the integer part and the fractional part of log10(x).[37] The characteristic of 10 · x is one plus the characteristic of x, and their significands are the same. This extends the scope of logarithm tables: given a table listing log10(x) for all integers x ranging from 1 to 1000, the logarithm of 3542 is approximated by
$\log_{10}(3542) = \log_{10}(10\cdot 354.2) = 1 + \log_{10}(354.2) \approx 1 + \log_{10}(354). \,$
Another critical application was the slide rule, a pair of logarithmically divided scales used for calculation, as illustrated here:
Schematic depiction of a slide rule. Starting from 2 on the lower scale, add the distance to 3 on the upper scale to reach the product 6. The slide rule works because it is marked such that the distance from 1 to x is proportional to the logarithm of x.
The non-sliding logarithmic scale, Gunter's rule, was invented shortly after Napier's invention. William Oughtred enhanced it to create the slide rule—a pair of logarithmic scales movable with respect to each other. Numbers are placed on sliding scales at distances proportional to the differences between their logarithms. Sliding the upper scale appropriately amounts to mechanically adding logarithms. For example, adding the distance from 1 to 2 on the lower scale to the distance from 1 to 3 on the upper scale yields a product of 6, which is read off at the lower part. The slide rule was an essential calculating tool for engineers and scientists until the 1970s, because it allows, at the expense of precision, much faster computation than techniques based on tables.[31]
## Analytic properties
A deeper study of logarithms requires the concept of a function. A function is a rule that, given one number, produces another number.[38] An example is the function producing the x-th power of b from any real number x, where the base b is a fixed number. This function is written
$f(x) = b^x. \,$
### Logarithmic function
To justify the definition of logarithms, it is necessary to show that the equation
$b^x = y \,$
has a solution x and that this solution is unique, provided that y is positive and that b is positive and unequal to 1. A proof of that fact requires the intermediate value theorem from elementary calculus.[39] This theorem states that a continuous function that produces two values m and n also produces any value that lies between m and n. A function is continuous if it does not "jump", that is, if its graph can be drawn without lifting the pen.
This property can be shown to hold for the function f(x) = bx. Because f takes arbitrarily large and arbitrarily small positive values, any number y > 0 lies between f(x0) and f(x1) for suitable x0 and x1. Hence, the intermediate value theorem ensures that the equation f(x) = y has a solution. Moreover, there is only one solution to this equation, because the function f is strictly increasing (for b > 1), or strictly decreasing (for 0 < b < 1).[40]
The unique solution x is the logarithm of y to base b, logb(y). The function that assigns to y its logarithm is called logarithm function or logarithmic function (or just logarithm).
The function logb(x) is essentially characterized by the above product formula
$\log_b(xy) = \log_b(x) + \log_b(y).$
More precisely, the logarithm to any base b > 1 is the only increasing function f from the positive reals to the reals satisfying f(b) = 1 and [41]
$f(xy)=f(x)+f(y).$
### Inverse function
The graph of the logarithm function logb(x) (blue) is obtained by reflecting the graph of the function bx (red) at the diagonal line (x = y).
The formula for the logarithm of a power says in particular that for any number x,
$\log_b \left (b^x \right) = x \log_b(b) = x.$
In prose, taking the x-th power of b and then the base-b logarithm gives back x. Conversely, given a positive number y, the formula
$b^{\log_b(y)} = y$
says that first taking the logarithm and then exponentiating gives back y. Thus, the two possible ways of combining (or composing) logarithms and exponentiation give back the original number. Therefore, the logarithm to base b is the inverse function of f(x) = bx.[42]
Inverse functions are closely related to the original functions. Their graphs correspond to each other upon exchanging the x- and the y-coordinates (or upon reflection at the diagonal line x = y), as shown at the right: a point (t, u = bt) on the graph of f yields a point (u, t = logbu) on the graph of the logarithm and vice versa. As a consequence, logb(x) diverges to infinity (gets bigger than any given number) if x grows to infinity, provided that b is greater than one. In that case, logb(x) is an increasing function. For b < 1, logb(x) tends to minus infinity instead. When x approaches zero, logb(x) goes to minus infinity for b > 1 (plus infinity for b < 1, respectively).
### Derivative and antiderivative
The graph of the natural logarithm (green) and its tangent at x = 1.5 (black)
Analytic properties of functions pass to their inverses.[39] Thus, as f(x) = bx is a continuous and differentiable function, so is logb(y). Roughly, a continuous function is differentiable if its graph has no sharp "corners". Moreover, as the derivative of f(x) evaluates to ln(b)bx by the properties of the exponential function, the chain rule implies that the derivative of logb(x) is given by[40][43]
$\frac{d}{dx} \log_b(x) = \frac{1}{x\ln(b)}.$
That is, the slope of the tangent touching the graph of the base-b logarithm at the point (x, logb(x)) equals 1/(x ln(b)). In particular, the derivative of ln(x) is 1/x, which implies that the antiderivative of 1/x is ln(x) + C. The derivative with a generalised functional argument f(x) is
$\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}.$
The quotient at the right hand side is called the logarithmic derivative of f. Computing f'(x) by means of the derivative of ln(f(x)) is known as logarithmic differentiation.[44] The antiderivative of the natural logarithm ln(x) is:[45]
$\int \ln(x) \,dx = x \ln(x) - x + C.$
Related formulas, such as antiderivatives of logarithms to other bases can be derived from this equation using the change of bases.[46]
### Integral representation of the natural logarithm
The natural logarithm of t is the shaded area underneath the graph of the function f(x) = 1/x (reciprocal of x).
The natural logarithm of t agrees with the integral of 1/x dx from 1 to t:
$\ln (t) = \int_1^t \frac{1}{x} \, dx.$
In other words, ln(t) equals the area between the x axis and the graph of the function 1/x, ranging from x = 1 to x = t (figure at the right). This is a consequence of the fundamental theorem of calculus and the fact that derivative of ln(x) is 1/x. The right hand side of this equation can serve as a definition of the natural logarithm. Product and power logarithm formulas can be derived from this definition.[47] For example, the product formula ln(tu) = ln(t) + ln(u) is deduced as:
$\ln(tu) = \int_1^{tu} \frac{1}{x} \, dx \ \stackrel {(1)} = \int_1^{t} \frac{1}{x} \, dx + \int_t^{tu} \frac{1}{x} \, dx \ \stackrel {(2)} = \ln(t) + \int_1^u \frac{1}{w} \, dw = \ln(t) + \ln(u).$
The equality (1) splits the integral into two parts, while the equality (2) is a change of variable (w = x/t). In the illustration below, the splitting corresponds to dividing the area into the yellow and blue parts. Rescaling the left hand blue area vertically by the factor t and shrinking it by the same factor horizontally does not change its size. Moving it appropriately, the area fits the graph of the function f(x) = 1/x again. Therefore, the left hand blue area, which is the integral of f(x) from t to tu is the same as the integral from 1 to u. This justifies the equality (2) with a more geometric proof.
A visual proof of the product formula of the natural logarithm
The power formula ln(tr) = r ln(t) may be derived in a similar way:
$\ln(t^r) = \int_1^{t^r} \frac{1}{x}dx = \int_1^t \frac{1}{w^r} \left(rw^{r - 1} \, dw\right) = r \int_1^t \frac{1}{w} \, dw = r \ln(t).$
The second equality uses a change of variables (integration by substitution), w = x1/r.
The sum over the reciprocals of natural numbers,
$1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 n = \sum_{k=1}^n \frac{1}{k},$
is called the harmonic series. It is closely tied to the natural logarithm: as n tends to infinity, the difference,
$\sum_{k=1}^n \frac{1}{k} - \ln(n),$
converges (i.e., gets arbitrarily close) to a number known as the Euler–Mascheroni constant. This relation aids in analyzing the performance of algorithms such as quicksort.[48]
There is also another integral representation of the logarithm that is useful in some situations.
$\ln(x) = -\lim_{\epsilon \to 0} \int_\epsilon^\infty \frac{dt}{t}\left( e^{-xt} - e^{-t} \right)$
This can be verified by showing that it has the same value at x = 1, and the same derivative.
### Transcendence of the logarithm
Real numbers that are not algebraic are called transcendental;[49] for example, π and e are such numbers, but $\sqrt{2-\sqrt 3}$ is not. Almost all real numbers are transcendental. The logarithm is an example of a transcendental function. The Gelfond–Schneider theorem asserts that logarithms usually take transcendental, i.e., "difficult" values.[50]
## Calculation
Logarithms are easy to compute in some cases, such as log10(1,000) = 3. In general, logarithms can be calculated using power series or the arithmetic-geometric mean, or be retrieved from a precalculated logarithm table that provides a fixed precision.[51][52] Newton's method, an iterative method to solve equations approximately, can also be used to calculate the logarithm, because its inverse function, the exponential function, can be computed efficiently.[53] Using look-up tables, CORDIC-like methods can be used to compute logarithms if the only available operations are addition and bit shifts.[54][55] Moreover, the binary logarithm algorithm calculates lb(x) recursively based on repeated squarings of x, taking advantage of the relation
$\log_2(x^2) = 2 \log_2 (x). \,$
### Power series
Taylor series
The Taylor series of ln(z) centered at z = 1. The animation shows the first 10 approximations along with the 99th and 100th. The approximations do not converge beyond a distance of 1 from the center.
For any real number z that satisfies 0 < z < 2, the following formula holds:[nb 4][56]
$\ln (z) = (z-1) - \frac{(z-1)^2}{2} + \frac{(z-1)^3}{3} - \frac{(z-1)^4}{4} + \cdots$
This is a shorthand for saying that ln(z) can be approximated to a more and more accurate value by the following expressions:
$\begin{array}{lllll} (z-1) & & \\ (z-1) & - & \frac{(z-1)^2}{2} & \\ (z-1) & - & \frac{(z-1)^2}{2} & + & \frac{(z-1)^3}{3} \\ \vdots & \end{array}$
For example, with z = 1.5 the third approximation yields 0.4167, which is about 0.011 greater than ln(1.5) = 0.405465. This series approximates ln(z) with arbitrary precision, provided the number of summands is large enough. In elementary calculus, ln(z) is therefore the limit of this series. It is the Taylor series of the natural logarithm at z = 1. The Taylor series of ln z provides a particularly useful approximation to ln(1+z) when z is small, |z| < 1, since then
$\ln (1+z) = z - \frac{z^2}{2} +\frac{z^3}{3}\cdots \approx z.$
For example, with z = 0.1 the first-order approximation gives ln(1.1) ≈ 0.1, which is less than 5% off the correct value 0.0953.
More efficient series
Another series is based on the area hyperbolic tangent function:
$\ln (z) = 2\cdot\operatorname{artanh}\,\frac{z-1}{z+1} = 2 \left ( \frac{z-1}{z+1} + \frac{1}{3}{\left(\frac{z-1}{z+1}\right)}^3 + \frac{1}{5}{\left(\frac{z-1}{z+1}\right)}^5 + \cdots \right ),$
for any real number z > 0.[nb 5][56] Using the Sigma notation, this is also written as
$\ln (z) = 2\sum_{n=0}^\infty\frac{1}{2n+1}\left(\frac{z-1}{z+1}\right)^{2n+1}.$
This series can be derived from the above Taylor series. It converges more quickly than the Taylor series, especially if z is close to 1. For example, for z = 1.5, the first three terms of the second series approximate ln(1.5) with an error of about 3×10−6. The quick convergence for z close to 1 can be taken advantage of in the following way: given a low-accuracy approximation y ≈ ln(z) and putting
$A = \frac z{\exp(y)}, \,$
the logarithm of z is:
$\ln (z)=y+\ln (A). \,$
The better the initial approximation y is, the closer A is to 1, so its logarithm can be calculated efficiently. A can be calculated using the exponential series, which converges quickly provided y is not too large. Calculating the logarithm of larger z can be reduced to smaller values of z by writing z = a · 10b, so that ln(z) = ln(a) + b · ln(10).
A closely related method can be used to compute the logarithm of integers. From the above series, it follows that:
$\ln (n+1) = \ln(n) + 2\sum_{k=0}^\infty\frac{1}{2k+1}\left(\frac{1}{2 n+1}\right)^{2k+1}.$
If the logarithm of a large integer n is known, then this series yields a fast converging series for log(n+1).
### Arithmetic-geometric mean approximation
The arithmetic-geometric mean yields high precision approximations of the natural logarithm. ln(x) is approximated to a precision of 2p (or p precise bits) by the following formula (due to Carl Friedrich Gauss):[57][58]
$\ln (x) \approx \frac{\pi}{2 M(1,2^{2-m}/x)} - m \ln (2).$
Here M denotes the arithmetic-geometric mean. It is obtained by repeatedly calculating the average (arithmetic mean) and the square root of the product of two numbers (geometric mean). Moreover, m is chosen such that
$x \,2^m > 2^{p/2}.\,$
Both the arithmetic-geometric mean and the constants π and ln(2) can be calculated with quickly converging series.
## Applications
A nautilus displaying a logarithmic spiral
Logarithms have many applications inside and outside mathematics. Some of these occurrences are related to the notion of scale invariance. For example, each chamber of the shell of a nautilus is an approximate copy of the next one, scaled by a constant factor. This gives rise to a logarithmic spiral.[59] Benford's law on the distribution of leading digits can also be explained by scale invariance.[60] Logarithms are also linked to self-similarity. For example, logarithms appear in the analysis of algorithms that solve a problem by dividing it into two similar smaller problems and patching their solutions.[61] The dimensions of self-similar geometric shapes, that is, shapes whose parts resemble the overall picture are also based on logarithms. Logarithmic scales are useful for quantifying the relative change of a value as opposed to its absolute difference. Moreover, because the logarithmic function log(x) grows very slowly for large x, logarithmic scales are used to compress large-scale scientific data. Logarithms also occur in numerous scientific formulas, such as the Tsiolkovsky rocket equation, the Fenske equation, or the Nernst equation.
### Logarithmic scale
Main article: Logarithmic scale
A logarithmic chart depicting the value of one Goldmark in Papiermarks during the German hyperinflation in the 1920s
Scientific quantities are often expressed as logarithms of other quantities, using a logarithmic scale. For example, the decibel is a logarithmic unit of measurement. It is based on the common logarithm of ratios—10 times the common logarithm of a power ratio or 20 times the common logarithm of a voltage ratio. It is used to quantify the loss of voltage levels in transmitting electrical signals,[62] to describe power levels of sounds in acoustics,[63] and the absorbance of light in the fields of spectrometry and optics. The signal-to-noise ratio describing the amount of unwanted noise in relation to a (meaningful) signal is also measured in decibels.[64] In a similar vein, the peak signal-to-noise ratio is commonly used to assess the quality of sound and image compression methods using the logarithm.[65]
The strength of an earthquake is measured by taking the common logarithm of the energy emitted at the quake. This is used in the moment magnitude scale or the Richter scale. For example, a 5.0 earthquake releases 10 times and a 6.0 releases 100 times the energy of a 4.0.[66] Another logarithmic scale is apparent magnitude. It measures the brightness of stars logarithmically.[67] Yet another example is pH in chemistry; pH is the negative of the common logarithm of the activity of hydronium ions (the form hydrogen ions H+
take in water).[68] The activity of hydronium ions in neutral water is 10−7 mol·L−1, hence a pH of 7. Vinegar typically has a pH of about 3. The difference of 4 corresponds to a ratio of 104 of the activity, that is, vinegar's hydronium ion activity is about 10−3 mol·L−1.
Semilog (log-linear) graphs use the logarithmic scale concept for visualization: one axis, typically the vertical one, is scaled logarithmically. For example, the chart at the right compresses the steep increase from 1 million to 1 trillion to the same space (on the vertical axis) as the increase from 1 to 1 million. In such graphs, exponential functions of the form f(x) = a · bx appear as straight lines with slope equal to the logarithm of b. Log-log graphs scale both axes logarithmically, which causes functions of the form f(x) = a · xk to be depicted as straight lines with slope equal to the exponent k. This is applied in visualizing and analyzing power laws.[69]
### Psychology
Logarithms occur in several laws describing human perception:[70][71] Hick's law proposes a logarithmic relation between the time individuals take for choosing an alternative and the number of choices they have.[72] Fitts's law predicts that the time required to rapidly move to a target area is a logarithmic function of the distance to and the size of the target.[73] In psychophysics, the Weber–Fechner law proposes a logarithmic relationship between stimulus and sensation such as the actual vs. the perceived weight of an item a person is carrying.[74] (This "law", however, is less precise than more recent models, such as the Stevens' power law.[75])
Psychological studies found that individuals with little mathematics education tend to estimate quantities logarithmically, that is, they position a number on an unmarked line according to its logarithm, so that 10 is positioned as close to 100 as 100 is to 1000. Increasing education shifts this to a linear estimate (positioning 1000 10x as far away) in some circumstances, while logarithms are used when the numbers to be plotted are difficult to plot linearly.[76][77]
### Probability theory and statistics
Three probability density functions (PDF) of random variables with log-normal distributions. The location parameter μ, which is zero for all three of the PDFs shown, is the mean of the logarithm of the random variable, not the mean of the variable itself.
Distribution of first digits (in %, red bars) in the population of the 237 countries of the world. Black dots indicate the distribution predicted by Benford's law.
Logarithms arise in probability theory: the law of large numbers dictates that, for a fair coin, as the number of coin-tosses increases to infinity, the observed proportion of heads approaches one-half. The fluctuations of this proportion about one-half are described by the law of the iterated logarithm.[78]
Logarithms also occur in log-normal distributions. When the logarithm of a random variable has a normal distribution, the variable is said to have a log-normal distribution.[79] Log-normal distributions are encountered in many fields, wherever a variable is formed as the product of many independent positive random variables, for example in the study of turbulence.[80]
Logarithms are used for maximum-likelihood estimation of parametric statistical models. For such a model, the likelihood function depends on at least one parameter that must be estimated. A maximum of the likelihood function occurs at the same parameter-value as a maximum of the logarithm of the likelihood (the "log likelihood"), because the logarithm is an increasing function. The log-likelihood is easier to maximize, especially for the multiplied likelihoods for independent random variables.[81]
Benford's law describes the occurrence of digits in many data sets, such as heights of buildings. According to Benford's law, the probability that the first decimal-digit of an item in the data sample is d (from 1 to 9) equals log10(d + 1) − log10(d), regardless of the unit of measurement.[82] Thus, about 30% of the data can be expected to have 1 as first digit, 18% start with 2, etc. Auditors examine deviations from Benford's law to detect fraudulent accounting.[83]
### Computational complexity
Analysis of algorithms is a branch of computer science that studies the performance of algorithms (computer programs solving a certain problem).[84] Logarithms are valuable for describing algorithms that divide a problem into smaller ones, and join the solutions of the subproblems.[85]
For example, to find a number in a sorted list, the binary search algorithm checks the middle entry and proceeds with the half before or after the middle entry if the number is still not found. This algorithm requires, on average, log2(N) comparisons, where N is the list's length.[86] Similarly, the merge sort algorithm sorts an unsorted list by dividing the list into halves and sorting these first before merging the results. Merge sort algorithms typically require a time approximately proportional to N · log(N).[87] The base of the logarithm is not specified here, because the result only changes by a constant factor when another base is used. A constant factor, is usually disregarded in the analysis of algorithms under the standard uniform cost model.[88]
A function f(x) is said to grow logarithmically if f(x) is (exactly or approximately) proportional to the logarithm of x. (Biological descriptions of organism growth, however, use this term for an exponential function.[89]) For example, any natural number N can be represented in binary form in no more than log2(N) + 1 bits. In other words, the amount of memory needed to store N grows logarithmically with N.
### Entropy and chaos
Billiards on an oval billiard table. Two particles, starting at the center with an angle differing by one degree, take paths that diverge chaotically because of reflections at the boundary.
Entropy is broadly a measure of the disorder of some system. In statistical thermodynamics, the entropy S of some physical system is defined as
$S = - k \sum_i p_i \ln(p_i).\,$
The sum is over all possible states i of the system in question, such as the positions of gas particles in a container. Moreover, pi is the probability that the state i is attained and k is the Boltzmann constant. Similarly, entropy in information theory measures the quantity of information. If a message recipient may expect any one of N possible messages with equal likelihood, then the amount of information conveyed by any one such message is quantified as log2(N) bits.[90]
Lyapunov exponents use logarithms to gauge the degree of chaoticity of a dynamical system. For example, for a particle moving on an oval billiard table, even small changes of the initial conditions result in very different paths of the particle. Such systems are chaotic in a deterministic way, because small measurement errors of the initial state predictably lead to largely different final states.[91] At least one Lyapunov exponent of a deterministically chaotic system is positive.
### Fractals
The Sierpinski triangle (at the right) is constructed by repeatedly replacing equilateral triangles by three smaller ones.
Logarithms occur in definitions of the dimension of fractals.[92] Fractals are geometric objects that are self-similar: small parts reproduce, at least roughly, the entire global structure. The Sierpinski triangle (pictured) can be covered by three copies of itself, each having sides half the original length. This makes the Hausdorff dimension of this structure log(3)/log(2) ≈ 1.58. Another logarithm-based notion of dimension is obtained by counting the number of boxes needed to cover the fractal in question.
### Music
Four different octaves shown on a linear scale, then shown on a logarithmic scale (as the ear hears them).
Logarithms are related to musical tones and intervals. In equal temperament, the frequency ratio depends only on the interval between two tones, not on the specific frequency, or pitch, of the individual tones. For example, the note A has a frequency of 440 Hz and B-flat has a frequency of 466 Hz. The interval between A and B-flat is a semitone, as is the one between B-flat and B (frequency 493 Hz). Accordingly, the frequency ratios agree:
$\frac{466}{440} \approx \frac{493}{466} \approx 1.059 \approx \sqrt[12]2.$
Therefore, logarithms can be used to describe the intervals: an interval is measured in semitones by taking the base-21/12 logarithm of the frequency ratio, while the base-21/1200 logarithm of the frequency ratio expresses the interval in cents, hundredths of a semitone. The latter is used for finer encoding, as it is needed for non-equal temperaments.[93]
Interval (the two tones are played at the same time) 1/12 tone (help·info) Semitone Just major third Major third Tritone Octave Frequency ratio r $2^{\frac 1 {72}} \approx 1.0097$ $2^{\frac 1 {12}} \approx 1.0595$ $\tfrac 5 4 = 1.25$ \begin{align} 2^{\frac 4 {12}} & = \sqrt[3] 2 \\ & \approx 1.2599 \end{align} \begin{align} 2^{\frac 6 {12}} & = \sqrt 2 \\ & \approx 1.4142 \end{align} $2^{\frac {12} {12}} = 2$ Corresponding number of semitones $\log_{\sqrt[12] 2}(r) = 12 \log_2 (r)$ $\tfrac 1 6 \,$ $1 \,$ $\approx 3.8631 \,$ $4 \,$ $6 \,$ $12 \,$ Corresponding number of cents $\log_{\sqrt[1200] 2}(r) = 1200 \log_2 (r)$ $16 \tfrac 2 3 \,$ $100 \,$ $\approx 386.31 \,$ $400 \,$ $600 \,$ $1200 \,$
### Number theory
Natural logarithms are closely linked to counting prime numbers (2, 3, 5, 7, 11, ...), an important topic in number theory. For any integer x, the quantity of prime numbers less than or equal to x is denoted π(x). The prime number theorem asserts that π(x) is approximately given by
$\frac{x}{\ln(x)},$
in the sense that the ratio of π(x) and that fraction approaches 1 when x tends to infinity.[94] As a consequence, the probability that a randomly chosen number between 1 and x is prime is inversely proportional to the numbers of decimal digits of x. A far better estimate of π(x) is given by the offset logarithmic integral function Li(x), defined by
$\mathrm{Li}(x) = \int_2^x \frac1{\ln(t)} \,dt.$
The Riemann hypothesis, one of the oldest open mathematical conjectures, can be stated in terms of comparing π(x) and Li(x).[95] The Erdős–Kac theorem describing the number of distinct prime factors also involves the natural logarithm.
The logarithm of n factorial, n! = 1 · 2 · ... · n, is given by
$\ln (n!) = \ln (1) + \ln (2) + \cdots + \ln (n). \,$
This can be used to obtain Stirling's formula, an approximation of n! for large n.[96]
## Generalizations
### Complex logarithm
Main article: Complex logarithm
Polar form of z = x + iy. Both φ and φ' are arguments of z.
The complex numbers a solving the equation
$e^a=z.\,$
are called complex logarithms. Here, z is a complex number. A complex number is commonly represented as z = x + iy, where x and y are real numbers and i is the imaginary unit. Such a number can be visualized by a point in the complex plane, as shown at the right. The polar form encodes a non-zero complex number z by its absolute value, that is, the distance r to the origin, and an angle between the x axis and the line passing through the origin and z. This angle is called the argument of z. The absolute value r of z is
$r=\sqrt{x^2+y^2}. \,$
The argument is not uniquely specified by z: both φ and φ' = φ + 2π are arguments of z because adding 2π radians or 360 degrees[nb 6] to φ corresponds to "winding" around the origin counter-clock-wise by a turn. The resulting complex number is again z, as illustrated at the right. However, exactly one argument φ satisfies −π < φ and φ ≤ π. It is called the principal argument, denoted Arg(z), with a capital A.[97] (An alternative normalization is 0 ≤ Arg(z) < 2π.[98])
The principal branch of the complex logarithm, Log(z). The black point at z = 1 corresponds to absolute value zero and brighter (more saturated) colors refer to bigger absolute values. The hue of the color encodes the argument of Log(z).
Using trigonometric functions sine and cosine, or the complex exponential, respectively, r and φ are such that the following identities hold:[99]
$\begin{array}{lll}z& = & r \left (\cos \varphi + i \sin \varphi\right) \\ & = & r e^{i \varphi}. \end{array} \,$
This implies that the a-th power of e equals z, where
$a = \ln (r) + i ( \varphi + 2 n \pi ), \,$
φ is the principal argument Arg(z) and n is an arbitrary integer. Any such a is called a complex logarithm of z. There are infinitely many of them, in contrast to the uniquely defined real logarithm. If n = 0, a is called the principal value of the logarithm, denoted Log(z). The principal argument of any positive real number x is 0; hence Log(x) is a real number and equals the real (natural) logarithm. However, the above formulas for logarithms of products and powers do not generalize to the principal value of the complex logarithm.[100]
The illustration at the right depicts Log(z). The discontinuity, that is, the jump in the hue at the negative part of the x- or real axis, is caused by the jump of the principal argument there. This locus is called a branch cut. This behavior can only be circumvented by dropping the range restriction on φ. Then the argument of z and, consequently, its logarithm become multi-valued functions.
### Inverses of other exponential functions
Exponentiation occurs in many areas of mathematics and its inverse function is often referred to as the logarithm. For example, the logarithm of a matrix is the (multi-valued) inverse function of the matrix exponential.[101] Another example is the p-adic logarithm, the inverse function of the p-adic exponential. Both are defined via Taylor series analogous to the real case.[102] In the context of differential geometry, the exponential map maps the tangent space at a point of a manifold to a neighborhood of that point. Its inverse is also called the logarithmic (or log) map.[103]
In the context of finite groups exponentiation is given by repeatedly multiplying one group element b with itself. The discrete logarithm is the integer n solving the equation
$b^n = x,\,$
where x is an element of the group. Carrying out the exponentiation can be done efficiently, but the discrete logarithm is believed to be very hard to calculate in some groups. This asymmetry has important applications in public key cryptography, such as for example in the Diffie–Hellman key exchange, a routine that allows secure exchanges of cryptographic keys over unsecured information channels.[104] Zech's logarithm is related to the discrete logarithm in the multiplicative group of non-zero elements of a finite field.[105]
Further logarithm-like inverse functions include the double logarithm ln(ln(x)), the super- or hyper-4-logarithm (a slight variation of which is called iterated logarithm in computer science), the Lambert W function, and the logit. They are the inverse functions of the double exponential function, tetration, of f(w) = wew,[106] and of the logistic function, respectively.[107]
### Related concepts
From the perspective of pure mathematics, the identity log(cd) = log(c) + log(d) expresses a group isomorphism between positive reals under multiplication and reals under addition. Logarithmic functions are the only continuous isomorphisms between these groups.[108] By means of that isomorphism, the Haar measure (Lebesgue measure) dx on the reals corresponds to the Haar measure dx/x on the positive reals.[109] In complex analysis and algebraic geometry, differential forms of the form df/f are known as forms with logarithmic poles.[110]
The polylogarithm is the function defined by
$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s}.$
It is related to the natural logarithm by Li1(z) = −ln(1 − z). Moreover, Lis(1) equals the Riemann zeta function ζ(s).[111]
## Notes
1. ^ The restrictions on x and b are explained in the section "Analytic properties".
2. ^ Some mathematicians disapprove of this notation. In his 1985 autobiography, Paul Halmos criticized what he considered the "childish ln notation," which he said no mathematician had ever used.[10] The notation was invented by Irving Stringham, a mathematician.[11][12]
3. ^ For example C, Java, Haskell, and BASIC.
4. ^ The same series holds for the principal value of the complex logarithm for complex numbers z satisfying |z − 1| < 1.
5. ^ The same series holds for the principal value of the complex logarithm for complex numbers z with positive real part.
6. ^ See radian for the conversion between 2π and 360 degrees.
## References
1. ^ Shirali, Shailesh (2002), A Primer on Logarithms, Hyderabad: Universities Press, ISBN 978-81-7371-414-6, esp. section 2
2. ^ Kate, S.K.; Bhapkar, H.R. (2009), Basics Of Mathematics, Pune: Technical Publications, ISBN 978-81-8431-755-8, chapter 1
3. ^ All statements in this section can be found in Shailesh Shirali 2002, section 4, (Douglas Downing 2003, p. 275), or Kate & Bhapkar 2009, p. 1-1, for example.
4. ^ Bernstein, Stephen; Bernstein, Ruth (1999), Schaum's outline of theory and problems of elements of statistics. I, Descriptive statistics and probability, Schaum's outline series, New York: McGraw-Hill, ISBN 978-0-07-005023-5, p. 21
5. ^ Downing, Douglas (2003), Algebra the Easy Way, Barron's Educational Series, Hauppauge, N.Y.: Barron's, ISBN 978-0-7641-1972-9, chapter 17, p. 275
6. ^ Wegener, Ingo (2005), Complexity theory: exploring the limits of efficient algorithms, Berlin, New York: Springer-Verlag, ISBN 978-3-540-21045-0, p. 20
7. ^ Franz Embacher; Petra Oberhuemer, Mathematisches Lexikon (in German), mathe online: für Schule, Fachhochschule, Universität unde Selbststudium, retrieved 2011-03-22
8. ^ B. N. Taylor (1995), Guide for the Use of the International System of Units (SI), US Department of Commerce
9. ^ Gullberg, Jan (1997), Mathematics: from the birth of numbers., New York: W. W. Norton & Co, ISBN 978-0-393-04002-9
10. ^ Paul Halmos (1985), I Want to Be a Mathematician: An Automathography, Berlin, New York: Springer-Verlag, ISBN 978-0-387-96078-4
11. ^ Irving Stringham (1893), Uniplanar algebra: being part I of a propædeutic to the higher mathematical analysis, The Berkeley Press, p. xiii
12. ^ Roy S. Freedman (2006), Introduction to Financial Technology, Amsterdam: Academic Press, p. 59, ISBN 978-0-12-370478-8
13. ^
14. ^ Robson, Eleanor (2008). Mathematics in Ancient Iraq: A Social History. p. 227. ISBN 978-0691091822.
15. ^ Gupta, R. C. (2000), "History of Mathematics in India", in Hoiberg, Dale; Ramchandani, Indu, Students' Britannica India: Select essays, Popular Prakashan, p. 329
16. ^ Stifelio, Michaele (1544), Arithmetica Integra, Nuremberg: Iohan Petreium
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Unfortunately, Bürgi did not include, with his table of logarithms, instructions for using the table. That was published separately. The contents of that publication were reproduced in: Hermann Robert Gieswald, Justus Byrg als Mathematiker, und dessen Einleitung zu seinen Logarithmen [Justus Byrg as a mathematician, and an introduction to his logarithms] (Danzig, Prussia: St. Johannisschule, 1856), pages 26 ff.
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27. ^ In 1647, Gregoire de Saint-Vincent published his book, Opus geometricum quadraturae circuli et sectionum coni (Geometric work of squaring the circle and conic sections), vol. 2 (Antwerp, (Belgium): Johannes and Jakob Meursius, 1647). On page 586, Proposition CIX, he proves that if the abscissas of points are in geometric proportion, then the areas between a hyperbola and the abscissas are in arithmetic proportion. This finding allowed Saint-Vincent's former student, Alphonse Antonio de Sarasa, to prove that the area between a hyperbola and the abscissa of a point is proportional to the abscissa's logarithm, thus uniting the algebra of logarithms with the geometry of hyperbolas. See: Alphonse Antonio de Sarasa, Solutio problematis a R.P. Marino Mersenne Minimo propositi … [Solution to a problem proposed by the reverend father Marin Mersenne, member of the Minim order … ], (Antwerp, (Belgium): Johannes and Jakob Meursius, 1649). Sarasa's critical finding occurs on page 16 (near the bottom of the page), where he states: "Unde hae superficies supplere possunt locum logarithmorum datorum … " (Whence these areas can fill the place of the given logarithms … ). [In other words, the areas are proportional to the logarithms.]
See also: Enrique A. González-Velasco, Journey through Mathematics: Creative Episodes in Its History (New York, New York: Springer, 2011), page 118.
28. ^ Alphonse Antonio de Sarasa, Solutio problematis a R.P. Marino Mersenne Minimo propositi … [Solution to a problem proposed by the reverend father Marin Mersenne, member of the Minim order … ], (Antwerp, (Belgium): Johannes and Jakob Meursius, 1649).
Sarasa realized that given a hyperbola and a pair of points along the abscissa which were related by a geometric progression, then if the abscissas of the points were multiplied together, the abscissa of their product had an area under the hyperbola which equaled the sum of the points' areas under the hyperbola. That is, the logarithm of an abscissa was proportional to the area, under a hyperbola, corresponding to that abscissa. This finding united the algebra of logarithms with the geometry of hyperbolic curves.
• Sarasa's critical finding occurs on page 16 (near the bottom of the page), where he states: "Unde hae superficies supplere possunt locum logarithmorum datorum … " (Whence these areas can fill the place of the given logarithms … ). [In other words, the areas are proportional to the logarithms.]
• See also: Enrique A. González-Velasco, Journey through Mathematics: Creative Episodes in Its History (New York, New York: Springer, 2011), pp. 119-120.
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# Aisha wants to paint the walls of a room. She knows that each can of paint contains one gallon. A half gallon will completely cover a 55 square feet of wall. Each of the four walls of the room is 10 feet high. Two of the walls are 10 feet wide and two of the walls are 15 feet wide. How many 1-gallon buckets of paint does Aisha need to buy in order to fully paint the room?
A 5
Explaination
The total number of buckets necessary will be the total area of the walls divided by the total area covered by each bucket. First, calculate the area of the walls Aisha wants to paint. Two of the walls are 10 × 10 and two of the walls are 10 × 15:
2 (10 × 10) = 200 sq. ft.
2 (10 × 15) = 300 sq. ft.
So the total square footage of the walls is 500. If a half gallon of paint will cover 55 square feet, then each gallon will cover 2 × 55 = 110 square feet. Four gallons can only cover 440 square feet. Five gallons will cover 550 square feet, which will be enough for the entire area of the walls.
B 4
C 9
D 10
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# The Four Numbers Game
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2 Master of Arts in Teaching (MAT) Masters Exam Tina Thompson In partial fulfillment of the requirements for the Master of Arts in Teaching with a Specialization in the Teaching of Middle Level Mathematics in the Department of Mathematics. Jim Lewis, Advisor July 2007
3 1 The Four Numbers Game Tina Thompson July 2007
4 2 The Four Numbers Game Abstract The Four Numbers Game is a fun way to work with subtraction and ordering of numbers. While trying to find an end to a game that is played with whole numbers, there are several items that will be investigated along the way. First, we offer an introduction to how the game is played. Second, rotations and reflections of a square will be presented which will create a generalized form. Third, we explain how even and odd number combinations will always end in even numbers within four subtraction rounds. Fourth, we argue that the length of the game does not change if multiples of the original numbers are used to create a new game. Fifth, we show that all Four Numbers Games will come to an end. Sixth, we offer an investigation of the general form and some special cases and how they can help predict a more accurate end. Finally, there will be an example of how this problem could be used in a sixth grade classroom.
5 3 placed. The Four Numbers Game The Four Numbers Game begins with a square. At each of the vertices a whole number is Start: 2, 3, 4, 5 The positive difference of each pair of vertices is placed at the mid point of the corresponding side and these four points are used to create a new square. Using these four vertices (and numbers) the game is repeated. The game is repeated until you get a difference of zero on all four sides, thus finishing the game. Round 1: 3-2=1, 4-3=1, 5-4=1, 5-2=3
6 4 Round 2: 1-1=0, 1-1=0, 3-1=2, 3-1=2 Round 3: 2-0=2, 0-0=0, 2-0=0, 2-2=0 Round 4: 2-0=2, 2-0=2, 2-0=2, 2-0=2 Round 5: 2-2=0, 2-2=0, 2-2=0, 2-2=0 This game beginning with 2, 3, 4 and 5 took five rounds to reach zero. This will be referred to as having a length of five for the rest of the paper. As one can see the squares are wonderful to look at, but from this point on, a table will be better suited for playing the game for our purposes.
7 5 When using only single digit whole numbers, 0 to 9, it is possible for a game to have a length of eight before it finishes. The example below uses the numbers 9, 4, 1, 0. Thinking of the previous square, one takes the difference of the two numbers that are next to each other (this corresponds to consecutive vertices), using the absolute value as needed to get a whole number. I.e. when we play the game using a table, each entry is the positive difference between the number above the entry and the number in the next column to the right. Our convention is that the 1 st column is the one to the right of the last column. Start: Round 1: 9-4=5 4-1=3 1-0=1 9-0=9 Round 2: 5-3=2 3-1=2 9-1=8 9-5=4 Round 3: 2-2=0 8-2=6 8-4=4 4-2=2 Round 4: 6-0=6 6-4=2 4-2=2 2-0=2 Round 5: 6-2=4 2-2=0 2-2=0 6-2=4 Round 6: 4-0=4 0-0=0 4-0=4 4-4=0 Round 7: 4-0=4 4-0=4 4-0=4 4-0=4 Round 8: 4-4=0 4-4=0 4-4=0 4-4=0 If the possible number span is from 0-44, one can find a game of length twelve using the numbers 44, 24, 13, 7. Tribonacci numbers are used here, possibly allowing the furthest inconsistent distance between four numbers in this range. Note that Tribonacci numbers are a sequence of numbers that start with 0, 0, 1, 1, 2, 4, 7, 13, 24, 44 One adds the first three numbers to get the fourth number (0+0+1=1). The next three numbers are then added together to give the fifth number (0+1+1=2) and so on. Start: Round 1: Round 2: Round 3: Round 4: Round 5: Round 6: Round 7: Round 8:
8 6 Round 9: Round 10: Round 11: Round 12: While we do not have a proof of this, it appears that eight is the maximum length of a game that begins with single digit numbers and twelve is the maximum length for a game that begins with numbers up to (and including) 44. In order to generalize our analysis of this game, we note that while rotating the square results in an adjusted arrangement of the vertices, it does not change the numbers which appear in the game nor does it change the length of the game. Recall the illustration representing the five rounds of the game displayed in the original example: Figure 1 While the illustration represents one particular game, clearly the length of any Four Number game will stay the same if the square is rotated by multiples of 90. We also consider the effects of reflecting a square upon the length of a game. Again, consider the illustration above representing the five rounds of the game beginning with 2, 3, 4, 5. Reflecting the image about a diagonal (in this case the diagonal connecting the vertices labeled 2 and 4) and reflecting the image about a vertical line of symmetry, respectively, results in the two following figures:
9 Reflection of Figure 1 about the diagonal Reflection of Figure 1 about a vertical line of symmetry Again, the number of squares inside the original square does not change, therefore the length of the game will not change. Since this will hold true for any of the four lines of symmetry within the square, reflecting the square does not affect the length of the game. Thus in much of the remaining analysis, we will refer to a generalized Four Numbers game as A,B,C,D labeled as shown: A B D C An interesting observation about the Four Numbers Game is that if a Four Numbers game has a length of at least four, then all the numbers appearing from Round 4 onward are even. Because reflections and rotations do not change the length of a game, we only need to consider these cases for the initial numbers: E,E,E,O O,O,O,O E,O,E,O E,E, O,O O,O,O,E Then, the steps in the game for each of these cases are shown or imbedded in the following tables:
10 8 Start: even even even odd Round 1: even-even=even even-even=even even-odd=odd odd-even=odd Round 2: even-even=even even-odd=odd odd-odd=even odd-even=odd Round 3: even-odd=odd even-odd=odd even-odd=odd even-odd=odd Round 4: odd-odd=even odd-odd=even odd-odd=even odd-odd=even Start: odd odd odd even Round 1: odd-odd=even odd-odd=even odd-even=odd even-odd=odd Round 2: even-even=even even-odd=odd odd-odd=even odd-even=odd Round 3: even-odd=odd odd-even=odd even-odd=odd odd-even=odd Round 4: odd-odd=even odd-odd=even odd-odd=even odd-odd=even Another useful fact in analyzing the game is that taking the initial numbers and multiplying them by a positive integer n does not change the length of the game. Consider, for example, a game where two entries on opposite vertices are the same. Symbolically we assume that A=C but make no other assumptions about which entries are the largest. Later in this paper we will need the fact that this game happens to have length four. A B C D Start: A B A D Round 1: A-B A-B A-D A-D Round 2: 0 B-D 0 B-D Round 3: B-D B-D B-D B-D Round 4: Any game created by multiplying each entry of the game (in which A=C) by an integer n > 0 will also have length four.
11 9 A B A D Start: n*a n*b n*a n*d Round 1: n A-B n A-B n A-D n A-D Round 2: 0 n B-D 0 n B-D Round 3: n B-D n B-D n B-D n B-D Round 4: In a similar way, if we take any game and multiply the four initial numbers by a positive integer n, the k th row of the new game will be n times the k th row of the original game. Thus the two games will have the same length. Starting with any four non-negative integers at the vertices of the square, if the largest number is not at the upper left corner (position A), the square can be rotated 90, 180, or 270 in order to put it there. Then, reflecting about diagonal AC, if necessary, it can be assumed that the four initial numbers A, B, C, D satisfy A B D and A C. This will be known as the standard form for the game. As we shall see, the size of C relative to B and D is what determines the maximum number of rounds needed to play the game. Without loss of generality, we will assume A B D, and A C (i.e., the standard form) for all cases described in the remainder of this paper. We now address the question Do all Four Numbers Games beginning with whole numbers have a finite length? We argue that the answer is yes. Beginning with a game in standard form, we know that there is always a power of two that will be larger than the number A, i.e. A < 2 k for some positive integer k. Specifically, the number A is always less than 2 A. Additionally, there will be a least positive power of 2 which is greater than A. Our goal is to show that taking the least possible k such that A < 2 k, and
12 10 multiplying it by four provides an upper bound for the length of the game. For example, if a square started with the largest number of 9 in the A position, then 9 < 2 4. Our theorem asserts that 4*4 or 16 is an upper bound for the length of the game. This tells us the length of the game is finite, but it does not give us the least upper bound for the length of the game. (We have previously claimed that 8 is the maximum length for a game where the largest number is nine.) Consider a Four Numbers Game beginning with whole numbers, at least one of which is non-zero. Suppose that A is the largest of these vertices. Let k be the least positive integer such that A<2 k. If the length of the game is less than or equal to four, there is nothing to prove as the game is certainly of finite length. If the length of the game is greater than four, we follow a proof in which the key idea is given in the example below: Start: Round 1: Round 2: Round 3: Round 4: Since in round four all numbers are even, we can take divide the numbers at each of the vertices by two and use them to start a new game. The length of the new game (27, 20, 11, 6 in this example) will be the same as the length of the game beginning with vertices obtained in round four (74, 40, 22, 12 in the example above) since each of these values is twice the value of the corresponding vertex in the new game (i.e. the 27, 20, 11, 6 game). The length of our original game will be 4 more than the length of the revised game (i.e. the 27, 20, 11, 6 game). Thus our original game has length 4 plus the length of the new game.
13 11 We proceed: Start: Round 1: Round 2: Round 3: Round 4: Again we begin a new game with initial values half of those obtained in round 4 since they are all even. We now know the length of the original game is at least 8. Continuing the example: Start: Round 1: Round 2: Round 3: Notice that in round three of the example above, there are (at least) two vertices with equivalent values. As we demonstrated previously, a game which two values are equal will have four steps. Therefore, since = 15, our original game has a length of 15. The claim made by our theorem is more modest. Since 149<2 8, the theorem claims that the length of the game is less than 4*8=32. In other words thirty-two is an upper bound, but not the least upper bound. In general, when we determine that A<2 k, we can see that every four rounds of the game we can replace the numbers by a new game with the maximum number being less than 2 k divided by 2, or 2 k-1. Eventually, we are playing a game in which all entries are less than 2. Such a game takes at most 4 rounds. Thus, in at least 4k steps, we must get to the end of our original game. Therefore we have shown that all Four Numbers Games beginning with whole numbers have finite length.
14 12 We now know that every Four Number game has a finite length and we have an upper bound for the length of each game. It is helpful to examine certain circumstances that lead to the determination of a better bound for a Four Numbers game. Looking at the standard form, there are three different cases that might occur. Case 1: A C B D Case 2: A B D C Case 3: A B C D Case 1: A C B D We want to show that all Case 1 games have lengths four or less. Note first that if A=C and B=D and A B, then the game ends in two rounds as shown in the following table. (If A=B the length is one.) A B C D Start: A B A B Round 1: A-B A-B A-B A-B Round 2: Now let s begin a general Case 1 game. Note that after two rounds, we have the first and third entries equal and the second and fourth entries equal. Thus the game will end in at most two more rounds (see table below). A B C D Start: A B C D Round 1: A-B C-B C-D A-D Round 2: A-C B-D A-C B-D
15 13 Case 2: A B D C We want to show that all Case 2 games have lengths six or less. Note first that if A=B and D=C then the game has length three or less. A B C D Start: A A D D Round 1: 0 A-D 0 A-D Round 2: A-D A-D A-D A-D Round 3: Now let s begin a general Case 2 game. Note that after two rounds, we have the second and fourth entries equal. Previously we demonstrated that any game with two opposite vertices equal has length four. Thus the game will end in four more rounds or less. A B C D Start: A B C D Round 1: A-B B-C D-C A-D Round 2: (A-B)-(B-C) B-D (A-D)-(D-C) B-D Starting with Case 3: A B C D While all Four Number Games have finite length, it is in Case 3 that one can find games that have length greater than 6. In fact, there is no universal bound on the length of all games. Even in Case 3, we are able to find conditions that lead to better bounds in the length of certain games. For example, if A-B=C-D, the game has length of five. Also, if A=B, B>C>D the game will have length six or less. Note that after the first round of the game A-B=C-D, so that the first and third entries are equal. Thus the game will end in four more rounds or less.
16 14 A=B, B>C>D A B C D Start: A B C D Round 1: A-B B-C C-D=A-B A-D Note that after the second round of the game A=B, B>C>D, we have the first and third entries equal. Thus this game will end in four more rounds or less and the game will have length six or less. A B C D Start: A A C D Round 1: 0 A-C C-D A-D Round 2: A-C (A-C)-(C-D) A-C A-D Lesson Plan Teaching this lesson to a sixth grade classroom would be a great way to investigate mathematical vocabulary, computation fluency, number sense through absolute value, investigation of general form, and how reflections and rotations affect the general form. Students could be given a square with four positive integers in the corners and allowed to solve for the solution together as a class. There could also be some squares on the paper permitting them to create their own game using whole numbers 0 to 9. They would then be grouped randomly and permitted to check with their group to see if anyone got a length different than the one presented together in class. There would be two more
17 15 squares they could use on the sheet to see if they could come up with different lengths than what the members in the group got. A hint would be on the board that states 8 sure is great! The next day, randomly grouped students would be permitted to check with each other on the different game lengths they got. The teacher would ask students if anyone got a length of 0, 1, 2, 3, 4, 5, 6, 7, or 8 and place those examples on the board. Once there was an example of each one, or as many as the class could come up with, they will be directed to think about the square in terms of A, B, C, D. The teacher would have a moveable square on the board labeled A, B, C, and D. She/he would also have a square already with the length of four completed as well. The teacher would then rotate the length 4 squares 90. She would ask the students if the length of the game of the rotated square is the same, allowing students to debate, if needed. When a decision is made that it did not change, write on the board that 4 is equal to the length of the game A,B,C,D, and to the length of the game, B,C,D,A. Then rotate the just completed game of length four another 90. Again, ask if the length of the game would stay the same. The students might be ready to answer yes, the length would stay the same. Write on the board, each of the four games, A,B,C,D and B,C,D,A and C,D,A,B, and D,A,B,C has the same length. Give the students time to reflect on this thought. Allow them to write down what they are thinking. The next day have students share why these squares all shared the same game length. The idea being that although the box was turned on its side, it never lost the number of squares inside of it. Moving on to reflection, the class would work together to talk about the reflection of the box. The students would then be permitted to see if the length of the game changed. They would be asked to explain that for the next day of class.
18 16 Throughout this experience the students will investigate number theory by exploring the differences between even and odd numbers. They will also be asked to make conjectures about the generalized standard form of the Four Numbers Games for different lengths 0 through 4. They will be permitted to prove their conjectures through classroom discussions, models, and examples. The students would be asked to share what they realize about square A,B,C,D on the board. The Four Numbers Game is a fun way to work with subtraction and ordering of numbers. While trying to find an end to a game that is played with whole numbers, there are several items that were demonstrated. After an introduction to how the game was played, rotations and reflections of a square were presented to create a generalized form. An interesting explanation of even and odd number combinations resulted in even numbers within four subtraction rounds. The length of the game did not change when a multiple of the initial numbers were used to begin a game. All Four Numbers Games were shown to have finite length by investigating the standard form and its special cases. Finally, there was an example of how this problem could be used in a sixth grade classroom. References Sally, Judith D., Sally, Paul J., TriMathlon: A Workout Beyond the School Curriculum A. K. Peters Ltd. Natick, MA.
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It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. Preview. Angles Glossary Acute Angle The sum of angles that are formed on a straight line is equal to 180°. (a) º (b) º (c) º (d) º (e) º (f) º Question 2 Calculate the unknown angle in each of the following diagrams. To work out z, all you'd need to do is subtract x from y (180 - 60 = 120). Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°. Message received. A straight angle changes the direction to point the opposite way. Sometimes people say "You did a complete 180 on that!" To calculate a missing angle on a straight line, take away the known angle from 180°. Angles Glossary Acute Angle The final mode deals with angles produced by an intersection of two lines. The aim of this video tutorial is to demonstrate how to calculate angles on a straight line using the knowledge that the angles add up to 180°. Calculate angles on a straight line. So b is irrelevant for the inclination angle. This is a straight angle. If θ is the angle of inclination of a straight line l, then tanθ is called the slope of gradient of the line is denoted by "m". In this activity you are given the angle BAD, and you need to use your knowledge of angles to work out the angle CAD. Geometry and Measures 1.1 Angles on a straight line. Calculating angles ITP Angles on a straight line Angles on a straight line add up to 180°. Sometimes people say "You did a complete 180 on that!" There are two activities on this page on finding the missing angles on a straight line. Example: the line 3x has an inclination angle of 71.565 °. Find the equation of the line. Updated: Jul 23, 2014. docx, 186 KB. This lesson introduces angles on a straight line by first reviewing the angle measurement of a right angle. This PowerPoint explores how angles can be calculated from straight lines by thinking of straight lines as representing half and full turns of 180° and 360° respectively, then subtracting known angles from 180° or 360° to get your answer. This is a straight angle. Two lines intersection calculator ... find the angle between the lines and the equation of the angle bisector between the two lines. If a is directing vector of first line, and b is directing vectors of second line then we can find angle between lines by formula: If a is directing vector of first line, and b is directing vectors of second line then we can find angle between lines by formula: Categories & … Geometry and Measures 1.1 Angles on a straight line. Calculate angles given the knowledge that there are 180 degrees on a straight line. Email: Password: I would like free resources and updates by email. Free Monthly Resources Print/download our free resources, plus a 7 day free trial with 5 further sets of worksheets and unlimited game plays. ... (using compass and straight edge) within a kite: • draw the axis of symmetry AD • bisect one of the non-vertex angles (B or C) and extend this line so that it meets line AD at point E • from point E, draw a … This website uses cookies to ensure you get the best experience. en. For example, let's say you're looking at an angle that's 60 degrees (x) and it's coming off of a straight 180 degree line (y). If you have selected calculate you will notice that only one angle is given and all the other 3 can be worked out using the vertically opposite rule, or spotting the supplementary angles. (We can shorten this property as: ∠ s on a straight line.) Calculate angles given the knowledge that there are 180 degrees on a straight line. The angle between the lines can be found by using the directing vectors of these lines. Explore more than 10,000 'Missing Angles On A Straight Line' resources for teachers, parents and pupils as well as related resources on 'Missing Angles' Recently Viewed and Downloaded › Recently Viewed › ... Year 6 Diving into Mastery: Calculate Angles Teaching Pack. Show Answer Hide Answer. You’ll need a protractor accurate to ¼ degree (Spirit level or digital protractor), and you’ll be measuring angles at various points, from the transmission, along the drivelines and back to the rear axles. In the above figure, If θ is the angle of a straight line l, then we have the following important points. Explore with children that the angles of a triangle total 180°, by tearing off the three corners of a triangle and arranging them together to make a straight line. 1. The angles on a straight line add up to 180°. All the angles on line add up to make 180 degrees. This resource is designed for UK teachers. Angles straight line 1. (We can shorten this property as: $$\angle$$s on a straight line.) Fill in all the gaps, then press "Check" to check your answers. Free Monthly Resources Print/download our free resources, plus a 7 day free trial with 5 further sets of worksheets and unlimited game plays. * Discuss how they first used protractors last week and some were able to measure within 5 degrees of accuracy. Code to add this calci to your website Congruent corresponding angles are: Angle of 'a' = Angle of 'g' Angle of 'b' = Angle of 'h' Angel of 'c' = Angle of 'e' Angle of 'd' = Angle of 'f' When two straight lines are cut by another line i.e transversal, then the angles in identical corners are said to be Corresponding Angles. Categories & Ages. Angles at intersecting lines mode. The diagram below shows two straight lines which cross each other. Angles that share a vertex and a common side are said to be adjacent. Exterior angles of a triangle - Triangle exterior angle theorem The intersection forms a pair of acute and another pair of obtuse angles. Free parallel line calculator - find the equation of a parallel line step-by-step This website uses cookies to ensure you get the best experience. The sum of angles that are formed on a straight line is equal to 180°. This website and its content is subject to our Terms and Conditions. Angles in a full turn add up to 360°. Line Segment Angle Calculator. Created: Feb 2, 2013. Two angles whose sizes add up to 180° are also called supplementary angles, for example $$\hat{1} + \hat{2}$$. If two straight lines cross, the angle between them is the smallest of the angles that is formed by the parallel to one of the lines that intersects the other one. Today I will learn to identify the number of degrees in right-angles and on a straight line Level 4 Reflex angle More than 90˚ but less than 180˚ Less than 90˚ Right-angle Straight-line 180˚ 90˚ More than180˚ Obtuse angle Acute angle Created: Sep 21, 2018 | Updated: Jan 16, 2019. If θ is the angle between two intersecting lines defined by y 1 = m 1 x 1 +c 1 and y 2 = m 2 x 2 +c 2, then, the angle θ is given by. If in space given the direction vector of line L. s = {l; m; n}. Use the Calculating angles ITP to calculate unknown angles on a straight line. 3. How to find the angle of a right triangle. So for example, we can assume that that long line is straight. The angle between line and plane is the angle between the line and its projection onto this plane.. Free Monthly Resources Print/download our free resources, plus a 7 day free trial with 5 further sets of worksheets and unlimited game plays. Learn Angle on a straight line. Calculate the unknown angle in each of the following diagrams. and equation of the plane A x + B y + C z + D = 0,. then the angle between this line and plane can be found using this formula Mathematics; A line in a Euclidean space of dimension n is the set of the points whose coordinates satisfy a given set of n−1 independent linear equations. Now some basic facts about angles. A line defined by a linear equation can't be vertical, but can have any other orientation. This means that if two or more angles lie in a straight line, the sum of their angles is 180 degrees. Please try again using a different payment method. Missing angles in a full turn. a x + b y = c . Teach your Years 5-6 students how to calculate angles on a line with this PowerPoint presentation. Report a problem. Calculate angles on a straight line. Therefore, the slope of the straight line is m = tan θ for 0° ≤ θ ≤ 180° Let us find the slope of a straight using the above formula (i) For horizontal lines, the angle of inclination is 0° or 180°. Geometric Kite Calculator, Geometry Kite Calculator, quadrilateral. Practice Makes Perfect. If you need to find a line given two points or a slope and one point, use line calculator . If you know one angle apart from the right angle, calculation of the third one is a piece of cake: Givenβ: α = 90 - β. Givenα: β = 90 - α. Related Angles. Lines AB and CD are parallel to one another (hence the » on the lines). A line defined by a linear equation can't be … The angle between the lines can be found by using the directing vectors of these lines. AnglesLesson Objectives: -•To be able to identify, estimate and measureacute and obtuse angles.•To be able to use information we know aboutangles on a straight line to calculate themissing angles. Equation of a straight line - online calculator Below you can use a calculator prepared to find the equation of a straight line. This fact can also be used to calculate angles. Free Angle a Calculator - calculate angle between lines a step by step. It can also provide the calculation steps and how the right triangle looks. Angles on a Straight Line (Worksheets with Answers) 5 38 customer reviews. The angle between the lines is found by vector dot product method. Created: Feb 2, 2013. The maximum angle is 360°. If a straight line is formed by more than one angle, the sum of those angles will be . By using this website, you agree to our Cookie Policy. Just like running, it takes practice and dedication. Please enter the gradient m or the angle α in degrees, the other value will be calculated and the line will be drawn. Differentiated Learning Objectives. It can also provide the calculation steps and how the right triangle looks. The sum of angles that are formed on a straight line is equal to 180°. Right triangle calculator to compute side length, angle, height, area, and perimeter of a right triangle given any 2 values. ... angle-between-lines-calculator. Calculator for the inclination angle of a straight line from the linear equation. If you want... To create your new password, just click the link in the email we sent you. This resource is designed for UK teachers. Also explore many more calculators covering geometry, math and other topics. Subscribe to access interactive resources. Right triangle calculator to compute side length, angle, height, area, and perimeter of a right triangle given any 2 values. Angles that share a … Developing Questions to support calculating missing angles on straight lines. Two angles whose sizes add up to 180° are also called supplementary angles, for example $$\hat{1} + \hat{2}$$. Thanks for the feedback. Calculator for the inclination angle of a straight line from the linear equation. (We can shorten this property as: $$\angle$$s on a straight line.) 2 angles … Also explore many more calculators covering geometry, math and other topics. Straight Angle A straight angle is 180 degrees. tanθ=±(m 2-m 1) / (1+m 1 m 2) Angle Between Two Straight Lines Derivation. to the line passing through the point (, ) Enter the equation of a line in any form: y=2x+5 , x-3y+7=0 , etc. Fixing points ITP . These printer-friendly worksheets on finding the unknown angles are categorized into three levels: easy, moderate, and difficult, to best suit the learning needs of 6th grade, 7th grade, and 8th grade children. View a scaled diagram of the resulting triangle, or explore many other math calculators, as well as hundreds of other calculators addressing finance, health, fitness, and more. We say the angles are supplementary angles. Report a problem. Up to two angles with labelled degrees. and equation of the plane A x + B y + C z + D = 0,. then the angle between this line and plane can be found using this formula By using this website, you agree to … In the diagram above, the two angles are supplementary angles, because they form a straight line. Updated: Jul 23, 2014. docx, 186 KB. The angle between two intersecting lines is the measure of the smallest of the angles formed by these lines. This calculator converts between polar and rectangular coordinates. Angle Between Two Lines Whenever two straight lines intersect, they form two sets of angles. Find the Value of x These printable worksheets consist of three angles one of which has a variable. Go through this assortment of angles on a straight line worksheets to practice finding the unknown angle and finding the value of x. The absolute values of angles formed depend on the slopes of the intersecting lines. If you have selected calculate you will notice that only one angle is given and all the other 3 can be worked out using the vertically opposite rule, or spotting the supplementary angles. That is, θ = 0° or 180° Theref… Some of the worksheets for this concept are Angles on a straight line, Linear pairs 1, Lines and angles work, Angles on a straight line and around a, Lines and angles grade 9, Work 12 geometry of straight lines grade 8 mathematics, Angles triangles and polygons, Clinton public school district clinton … Therefore, Straight Angle A straight angle is 180 degrees. Note that you will lose points if … About this resource. Use the "Hint" button to get a free letter if an answer is giving you trouble. We’ve already said that a straight line contains 180 degrees. Email: Password: I would like free resources and updates by email. The video below explains how to calculate related angles, adjacent angles, interior angles and supplementary angles. This PowerPoint explores how angles can be calculated from straight lines by thinking of straight lines as representing half and full turns of 180° and 360° respectively, then subtracting known angles from 180° or 360° to get your answer. This is the angle all the way round a point. Straight line angles Interactive demonstrations of two angles on a line. + Angle Between Two Straight Lines Formula. In geometry, an angle is the space between 2 rays (or line segments) with the same endpoint (or vertex). Most students should be able to calculate vertically opposite angles and angles on a straight line. Sum of three angles α, β, γ is equal to 180°, as they form a straight line. ... angle-between-lines-calculator. To find the unknown angle, add up the known angles and subtract from 180°. However, if only two sides of a triangle are given, finding the angles of a right triangle requires applying some … The angle between line and plane is the angle between the line and its projection onto this plane.. Lines and Angles: Let’s Look at Angles. This website uses cookies to ensure you get the best experience. Find the unknown angles. where the slopes m 1 and m 2 are given by - b / a for each line. This website uses cookies to ensure you get the best experience. How to find the angle of a right triangle. Objective: I know how to calculate angles in a straight line. To be able to identify, estimate and measure acute and obtuse angles. Author: Created by Maths4Everyone. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. The plenary challenges students to create a straight line out of three known angles. NOTE: This calculator does not address compound drive angles (horizontal offsets). All angles are in increments of 5o and are on a horizontal line. A calculator to find the angle between two lines L 1 and L 2 given by their general equation of the form . Missing angles on intersecting lines. Tes Global Ltd is registered in England (Company No 02017289) with its registered office at 26 Red Lion Square London WC1R 4HQ. View US version. θ = |tan-1 ( (m 2 - m 1) / (1 + m 2 × m 1))| . Related Symbolab blog posts. Geometry and Measures 1.1 Angles on a straight line. So b is irrelevant for the inclination angle. The major application of angle of inclination of a straight line is finding slope. Example: We know one angle is 45°, what is the other angle "a" ? They are equal to the ones we calculated manually: β = 51.06°, γ = 98.94°; additionally, the tool determined the last side length: c = 17.78 in. Tanθ=± ( m ) example: the line is equal to 180° are also called a line... 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By using the directing vectors of these lines Hint '' button to get a free letter if Answer. Lines a step by step and CD are parallel to one another ( hence the » on slopes. That that long line is formed by more than one angle to calculate angles on a straight line which! Are three interior angles in a straight angle changes the direction vector of line L. =! 1 m 2 - m 1 and m 2 ) angle between straight! Example, we can shorten this property as: ∠ s on a line. M 1 and m 2 ) angle between the lines is found by vector dot method! Of a straight line contains 180 degrees is found by vector dot product method how they first used protractors week. With m=0 has the inclination angle of a right angle calculating angles shorten this property as: \ \angle\... Displaying top 8 worksheets found for this concept on straight lines Geometric Kite calculator, geometry Kite calculator geometry... And its projection onto this plane want... to create your new Password, just the! The exterior angle of 71.565 ° means that if two or more lie! Cd are parallel to one another angles on a straight line calculator hence the » on the Hint... Worksheets consist of three angles one of which has a variable … angle line!: Sep 21, 2018 | updated: Jul 23, 2014.,! Ab and CD are parallel to one another ( hence the » on the of... How the right triangle hide Answer property as: \ ( \angle\ ) s a. To one another ( hence the » on the topic of angles formed depend on the ?... Investigate the total known angles, what is the angle between two lines Whenever two straight lines formula ]! Challenges students to create a straight line to calculate unknown angles on a straight line is 180° ?... Important points for Key Stage 3, Key Stage 4 and GCSE maths.. Free Monthly resources Print/download our free resources, plus a 7 day trial! Angle changes the direction to point the opposite way this property as: \ ( )! At angles and how the right triangle 3, Key Stage 4 and GCSE maths classes of and... Degrees, with a full circle measuring 360 degrees ) with its registered at!: ∠ s on a straight line is inclined the other, or hide both angles for estimating... Like free resources and updates by email a '', the sum of angles that share a and! Page on the [? ], estimate and measure acute and another pair of obtuse angles 2. Resources Print/download our free resources, plus a 7 day free trial with 5 sets. On the slopes m 1 and m 2 ) angle between the line and its is. |Tan-1 ( ( m 2-m 1 ) ) | 1 and l 2 given by their general of... - question Page find the equation of a right angle online calculator below you also... Of inclination of a right angle make 180 degrees, interior angles and supplementary angles interior... M 1 ) ) | more calculators covering geometry, math and other topics calculate related angles, because form. 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Derivatives of Reciprocal Trigonometric Functions
# Derivatives of Reciprocal Trigonometric Functions
We are going to look at even more derivative rules - this time for the reciprocal trigonometric functions.
Theorem 1: The following functions have the following derivatives: a) If $f(x) = \sec x$, then $\frac{d}{dx} \sec x = \sec x \tan x$. b) If $f(x) = \csc x$, then $\frac{d}{dx} \csc x = -\csc x \cot x$. c) If $f(x) = cot x$, then $\frac{d}{dx} \cot x = -csc ^2 x$.
• Proof of a): Let $f(x) = \sec x$. Recall that the quotient rule is $\frac{f}{g} = \frac{gf' - fg'}{g^2}$ and that $\sec x = \frac{1}{\cos x}$. Applying the quotient rule, we get:
(1)
\begin{align} \frac{d}{dx} \sec x = \frac{\cos x \cdot 0 - 1 \cdot (-\sin x)}{\cos ^2 x} \\ \frac{d}{dx} \sec x = \frac{\sin x}{\cos ^2 x } \\ \frac{d}{dx} \sec x = \sec x \tan x \quad \blacksquare \end{align}
• Proof of b): Let $f(x) = \csc x$. Note that $\csc x = \frac{1}{\sin x}$. Applying the quotient rule we get:
(2)
\begin{align} \frac{d}{dx} \csc x = \frac{\sin x \cdot 0 - 1 \cdot \cos x}{\sin ^2 x } \\ \frac{d}{dx} \csc x = \frac{-\cos x}{\sin ^2 x} \\ \frac{d}{dx} \csc x = -\csc x \cot x \quad \blacksquare \end{align}
• Proof of c): Let $f(x) = \cot x$. Note that $\cot x = \frac{\cos x}{\sin x}$. Applying the quotient rule and using the trigonometric identity that $\sin ^2 x + \cos ^2 x = 1$, we get:
(3)
\begin{align} \frac{d}{dx} \cot x = \frac{\sin x \cdot (-\sin x) - \cos x \cdot \cos x}{\sin ^2 x} \\ \frac{d}{dx} \cot x = \frac{-sin ^2x - \cos ^2 x}{\sin ^2 x} \\ \frac{d}{dx} \cot x = -\frac{\sin ^2 x + \cos ^2 x}{\sin ^2 x} \\ \frac{d}{dx} \cot x = -\frac{1}{\sin ^2 x} \\ \frac{d}{dx} \cot x = - \csc ^2 x \quad \blacksquare \end{align}
Now let's look at an example of applying theorem 1.
## Example 1
Find the derivative of $f(x) = \sec x \csc x$.
We note that $\frac{d}{dx} \sec x = \sec x \tan x$ and $\frac{d}{dx} \csc x = -\csc x \cot x$. Applying the product rule we obtain:
(4)
\begin{align} \frac{d}{dx} f(x) = \csc x (\sec x \tan x) - \sec x (-csc x \cot x) \\ \frac{d}{dx} f(x) = \csc x \sec x \tan x - \sec x \csc x \cot x \end{align}
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# Find the Volume of the Solid that is enclosed by the Cone and the Sphere
This question aims to find the volume of the solid enclosed by the cone and a sphere by using the method of polar coordinates to find the volume. Cylindrical coordinates extend the two-dimensional coordinates to three-dimensional coordinates.
In a sphere, the distance of the origin (0,0) to the point P is called the radius r. By joining the line from the origin to the point P, the angle made by this radial line from the x-axis is called the angle theta, represented by $\theta$. Radius r and $\theta$ have some values that can be used in limits for integration.
## Expert Answer
The $z-axis$ is projected in a cartesian plane along with the $xy$-plane to form a three-dimensional plane. This plane is represented by $(r, \theta, z)$ in terms of polar coordinates.
To find the limits of $z$, we will take the square root of the double cones. The positive square root represents the top of the cone. The equation of the cone is:
$z = \sqrt{(x^2 + y^2)}$
The equation of the sphere is:
$x^2 + y^2 + z^2 = 2$
This equation is derived from the polar coordinates formula, where $x^2 + y^2 = r^2$ when $z = r^2$.
Both of these equations can be represented on the cartesian plane:
Put the value of $r^2$ in place of $z^2$ by using polar coordinates:
$x^2 + y^2 + z^2 = 2$
$r^2 + z^2 = 2$
$z = \sqrt{2- r^2}$
We will equate both equations to find the value of $r$ when $z$ = $r$ by:
$z = \sqrt{(x^2 + y^2)}$
$z = \sqrt{(r^2)}$
$z = r$
To find $r$:
$r = \sqrt{2 – r^2}$
$2r^2 = 2$
$r = 1$
When we enter from the $z-axis$, we will come across top of the sphere and bottom of the cone. We will integrate from $0$ to $2\pi$ in the spherical region. The limits at those points are:
\int_{a}^b\int_{c}^d f(x,y)dxdy$$\int_{0}^{2\pi}\ \int_{0}^1\ \int_{r}^\sqrt{2-r^2} dzrdrd\theta$ Integrate with respect to$z$and put limits of$z$$\int_{0}^{2\pi}\ \int_{0}^1\ r\sqrt{2-r^2} – r^2 drd\theta$ We will separate the integrals to substitute$u$: $\int_{0}^{2\pi} [\int_{0}^1\ r\sqrt{2-r^2}dr – \int_{0}^1 r^2 dr] d\theta$ $u = 2 – r^2 , du = -2rdr$ By simplification, we get: $\int_{0}^{2\pi} [\int_{1}^2 \frac{-1}{2} \sqrt{u}du \ – \int_{0}^1 r^2 dr] d\theta$ $\int_{0}^{2\pi} [\int_{1}^2 \frac{1}{2} \sqrt{u}du\ – \int_{0}^1 r^2 dr] d\theta$ Integrating with respect to$u$and$r$: $\int_{0}^{2\pi} [\int_{1}^2 \frac{1}{2} \sqrt{u}du\ – \int_{0}^1 r^2 dr] d\theta$ $\int_{0}^{2\pi}\ \frac{2}{3} (\sqrt{2} – 1) d\theta$ ## Numerical Solution: Integration with respect to$\theta\$ and then putting its limits give us:
$V = \frac{4\pi}{3} \large(\sqrt{2} – 1)$
Image/Mathematical drawings are created in Geogebra
5/5 - (8 votes)
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# Neutral Element (Identiy Element) - Examples, Exercises and Solutions
## What is a Neutral Element?
In mathematics, a neutral element is an element that does not alter the rest of the numbers when we perform an operation with it.
With addition, $0$ is considered a neutral element because it does not modify the number to which it is added.
$0+3=3$
##### Neutral Element - Multiplication
In multiplication, $1$ is considered a neutral element because it does not affect the result.
$4\times1=4$
##### Neutral Element - Subtraction and Division
The neutral element in subtraction is $0$, while in division it is $1$.
## examples with solutions for neutral element (identiy element)
### Exercise #1
$(3\times5-15\times1)+3-2=$
### Step-by-Step Solution
This simple rule is the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations enclosed in parentheses precede all others,
Following the simple rule, multiplication comes before division and subtraction, therefore we calculate the values of the multiplications and then proceed with the operations of division and subtraction
$3\cdot5-15\cdot1+3-2= \\ 15-15+3-2= \\ 1$ Therefore, the correct answer is answer B.
$1$
### Exercise #2
$(5+4-3)^2:(5\times2-10\times1)=$
### Step-by-Step Solution
This simple rule is the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations enclosed in parentheses precede all others,
In the given expression, the establishment of division between two sets of parentheses, note that the parentheses on the left indicate strength, therefore, in accordance to the order of operations mentioned above, we start simplifying the expression within those parentheses, and as we proceed, we obtain the result derived from simplifying the expression within those parentheses with given strength, and in the final step, we divide the result obtained from the simplification of the expression within the parentheses on the right,
We proceed similarly with the simplification of the expression within the parentheses on the left, where we perform the operations of multiplication and division, in strength, in contrast, we simplify the expression within the parentheses on the right, which, according to the order of operations mentioned above, means multiplication precedes division, hence we first perform the operations of multiplication within those parentheses and then proceed with the operation of division:
$(5+4-3)^2:(5\cdot2-10\cdot1)= \\ (-2)^2:(10-10)= \\ 4:0\\$We conclude that the sequence of operations within the expression that is within the parentheses on the left yields a smooth result, this result we leave within the parentheses, these we raised in the next step in strength, this means we remember that every number (positive or negative) in dual strength gives a positive result,
As we proceed, note that in the last expression we received from establishing division by the number 0, this operation is known as an undefined mathematical operation (and this is the simple reason why a number should never be divided by 0 parts) therefore, the given expression yields a value that is not defined, commonly denoted as "undefined group" and use the symbol :
$\{\empty\}$In summary:
$4:0=\\ \{\empty\}$Therefore, the correct answer is answer A.
No solution
### Exercise #3
$(5\times4-10\times2)\times(3-5)=$
### Step-by-Step Solution
This simple rule is the order of operations which states that multiplication precedes addition and subtraction, and division precedes all of them,
In the given example, a multiplication occurs between two sets of parentheses, thus we simplify the expressions within each pair of parentheses separately,
We start with simplifying the expression within the parentheses on the left, this is done in accordance with the order of operations mentioned above, meaning that multiplication comes before subtraction, we perform the multiplications in this expression first and then proceed with the subtraction operations within it, in reverse we simplify the expression within the parentheses on the right and perform the subtraction operation within them:
What remains for us is to perform the last multiplication that was deferred, it is the multiplication that occurred between the expressions within the parentheses in the original expression, we perform it while remembering that multiplying any number by 0 will result in 0:
$0$
### Exercise #4
$0.18+(1-1)=$
### Step-by-Step Solution
According to the order of operations rules, we first solve the expression in parentheses:
$1-1=0$
And we get the expression:
$0.18+0=0.18$
0.18
### Exercise #5
Indicates the corresponding sign:
$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_—}(5^2-3+6):7\cdot\frac{1}{4}$
### Step-by-Step Solution
For a given problem, whether it involves addition or subtraction each of the terms that come into play separately,
this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are performed before all others,
A. We will start with the terms that are on the left in the given problem:
$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2$First, we simplify the terms that are in the denominators in accordance with the order of operations, this is done by calculating the numerical value of the denominator in strength (this within that we remember that in defining the root as strong, the root itself is strong for everything) and then perform the operation of division and subtraction:
$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\ \frac{1}{\sqrt{16}}\cdot(125+3-4):2^2 =\\ \frac{1}{\sqrt{16}}\cdot124:2^2$Next, we calculate the numerical values of the part that was passed in strength (practically, if we were to represent the operation of division as broken, this part would have been in the broken position) and as such the numerical value of the part in strength that in the broken position in the terms, next we perform the operations of multiplication and division:
$\frac{1}{\sqrt{16}}\cdot124:2^2 =\\ \frac{1}{4}\cdot124:4 =\\ \frac{1\cdot124}{4}:4=\\ \frac{\not{124}}{\not{4}}:4=\\ 31:4=\\ \frac{31}{4}=\\ 7\frac{3}{4}$In the final stages, we performed the multiplication of the number 124 in break, this we did within that we remember that multiplication in break means multiplication in the broken position, next we performed the operation of division of the break (by condensing the break) and in the final stage we performed the operation of division in the number 4, this operation resulted in a complete answer, and therefore we marked it as break (a break from understanding, an assumption that the position is greater than the position) and in the continuation we converted the break from understanding to a mixed break, by extracting the wholes (the answer to the question: "How many times does the division enter the divisor?") and adding the remaining division to the divisor,
We finished simplifying the terms that are on the left in the given problem, we will summarize the simplification stages:
$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\ \frac{1}{\sqrt{16}}\cdot124:2^2 =\\ \frac{1\cdot124}{4}:4=\\ 7\frac{3}{4}$
B. We will continue and simplify the terms that are on the right in the given problem:
$(5^2-3+6):7\cdot\frac{1}{4}$In this part, to simplify the terms within the framework of the order of operations,
In this term, the operation of division of the beginning on terms that are in the denominators, therefore we will first simplify this term,
Let's note that multiplication precedes addition and subtraction, which precede division and subtraction, therefore we will start by calculating the numerical value of the part in strength that in this term, next we perform the operations of division and subtraction:
$(5^2-3+6):7\cdot\frac{1}{4} =\\ (25-3+6):7\cdot\frac{1}{4} =\\ 28:7\cdot\frac{1}{4}$We will continue and simplify the received term, noting that between multiplication and division there is no precedence defined in the order of operations, we perform the operations in this term one after the other according to the order from left to right, which is the natural order of operations:
$28:7\cdot\frac{1}{4} =\\ 4\cdot\frac{1}{4} =\\ \frac{4\cdot1}{4}=\\ \frac{\not{4}}{\not{4}}=\\ 1$ In the second stage, we performed the multiplication in break, this within that we remember (again) that multiplication in break means multiplication in the broken position, in the next stage we performed the operation of division of the break (by condensing the break).
We finished simplifying the terms that are on the right in the given problem, we will summarize the simplification stages:
$(5^2-3+6):7\cdot\frac{1}{4} =\\ 28:7\cdot\frac{1}{4} =\\ \frac{\not{4}}{\not{4}}=\\ 1$We return to the original problem, and we will present the results of simplifying the terms that were reported in A and B:
$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_—}(5^2-3+6):7\cdot\frac{1}{4} \\ \downarrow\\ 7\frac{3}{4} \text{ }\textcolor{red}{_—}1$As a result, we receive that:
$7\frac{3}{4} \text{ }\textcolor{red}{\neq}1$Therefore, the correct answer here is answer B.
$\ne$
## examples with solutions for neutral element (identiy element)
### Exercise #1
Indicates the corresponding sign:
$-5+(5-3\cdot2)+6\textcolor{red}{☐}((3+2)\cdot2):2\cdot0$
### Step-by-Step Solution
In order to solve the given problem, whether it involves addition or subtraction each of the terms that appear in the equation must be dealt with separately,
This is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are for all,
A. We will start with the terms that are on the left in the given problem:
$-5+5-3\cdot2+6$ Simplify the terms that are in the parentheses in accordance with the order of operations, start with the multiplication that is in the terms and continue to perform the operations of addition and subtraction:
$-5+5-3\cdot2+6=\\ -5+5-6+6=\\ 0$
We finish simplifying the terms that are on the left in the given problem.
B. We will continue and simplify the terms that are on the right in the given problem:
$\big((3+2)\cdot2\big):2\cdot0$Note that in this term there is a multiplication between the term and the number 0, in addition note that in this term it is defined (since it does not include division by 0), we remember that multiplying any number by 0 will yield the result 0, therefore:
$\big((3+2)\cdot2\big):2\cdot0 =\\ 0$
We return now to the original problem, and we will present the results of simplifying the terms that were reported in A and B:
$-5+5-3\cdot2+6\textcolor{red}{☐}\big((3+2)\cdot2\big):2\cdot0 \\ \downarrow\\ 0\text{ }\textcolor{red}{_—}0$As a result that we find that:
$0 \text{ }\textcolor{red}{=}0$Therefore the correct answer here is answer A.
$=$
### Exercise #2
$[(5-2):3-1]\times4=$
### Step-by-Step Solution
In the order of operations, parentheses come before everything else.
We start by solving the inner parentheses in the subtraction operation:
$((3):3-1)\times4=$ We continue with the inner parentheses in the division operation and then subtraction:
$(1-1)\times4=$
We continue solving the subtraction exercise within parentheses and then multiply:
$0\times4=0$
$0$
### Exercise #3
$\lbrack(4+3):7+2:2-2\rbrack:5=$
### Step-by-Step Solution
Simplifying this expression emphasizes the order of operations, which states that multiplication precedes addition and subtraction, and that division precedes all of them,
In the given expression, the establishment of division operations between the parentheses (the outermost) to a number, therefore according to the order of operations as mentioned, is handled by simplifying the expression in these parentheses, this expression includes division operations that begin on the expression within the parentheses (the innermost), therefore according to the order of operations as mentioned is handled by simplifying the expression in these parentheses and performing the subtraction operations in it, there is no hindrance to calculate the outcome of the division operations in the expression in the outermost parentheses, but for the sake of good order this is done afterwards:
$\lbrack(4+3):7+2:2-2\rbrack:5= \\ \lbrack7:7+2:2-2\rbrack:5$Continuing and simplifying the expression in the parentheses we noted, since division precedes addition and subtraction, start with the division operations in the expression and only then calculate the outcome of the addition and subtraction, ultimately perform the division operations on this expression in the parentheses:
$\lbrack7:7+2:2-2\rbrack:5 \\ \lbrack1+1-2\rbrack:5=\\ 0:5=\\0$In the last stage we mentioned that multiplying a number by 0 gives the result 0,
Therefore, this simplifying expression is short so there is no need to elaborate,
$0$
### Exercise #4
$(3+2-1):(1+3)-1+5=$
### Step-by-Step Solution
This simple rule is the order of operations which states that multiplication and division come before addition and subtraction, and operations enclosed in parentheses come first,
In the given example of division between two given numbers in parentheses, therefore according to the order of operations mentioned above, we start by calculating the values of each of the numbers within the parentheses, there is no prohibition against calculating the result of the addition operation in the given number, for the sake of proper order, this operation is performed later:
$(3+2-1):(1+3)-1+5= \\ 4:4-1+5$In continuation of the principle that division comes before addition and subtraction the division operation is performed first and then the operations of subtraction and addition which were received in the given number and in the last stage:
$4:4-1+5= \\ 1-1+5=\\ 5$Therefore the correct answer here is answer B.
$5$
### Exercise #5
Indicates the corresponding sign:
$\frac{1}{25}\cdot(5^2-3+\sqrt{9})\textcolor{red}{☐}\sqrt{25}\cdot5\cdot\frac{1}{5}$
### Step-by-Step Solution
We solve the left side and start from the parentheses:
$5^2=5\times5=25$
We will solve the root exercise using the equation:$\sqrt{a^2}=a$
$\sqrt{9}=\sqrt{3^2}=3$
We arrange the exercise accordingly:
$\frac{1}{25}\times(25-3+3)=$
We solve the exercise in parentheses from left to right:
$\frac{1}{25}\times(22+3)=\frac{1}{25}\times25$
We convert the 25 into a simple fraction, multiply and divide:
$\frac{1}{25}\times\frac{25}{1}=\frac{25}{25}=\frac{1}{1}=1$
We solve the right side:
$\sqrt{25}=\sqrt{5^2}$
We arrange the exercise:
$\sqrt{5^2}\times5\times\frac{1}{5}$
We convert the 5 into a simple fraction and note that it is possible to reduce by 5:
$\sqrt{5^2}\times\frac{5}{1}\times\frac{1}{5}=\sqrt{5^2}\times1$
We solve the root according to the formula:$\sqrt{a^2}=a$
$5\times1=5$
Now we are going to compare the left side with the right side, and it seems that we obtained two different results and therefore the two sides are not equal.
$\ne$
## examples with solutions for neutral element (identiy element)
### Exercise #1
Complete the following exercise:
$\frac{27-5\cdot3}{6\cdot2}+\frac{15\cdot4}{3}=$
### Step-by-Step Solution
According to the order of arithmetic operations, first we place the multiplication exercises within parentheses:
$\frac{27-(5\cdot3)}{(6\cdot2)}+\frac{(15\cdot4)}{3}=$
We solve the exercises within parentheses:
$5\times3=15$
$6\times2=12$
$15\times4=60$
Now we obtain the exercise:
$\frac{27-15}{12}+\frac{60}{3}=$
We solve the numerator of the fraction:
$27-15=12$
We obtain:
$\frac{12}{12}+\frac{60}{3}=$
We solve the fractions:
$\frac{12}{12}=1$
$60:3=20$
Now we obtain the exercise:
$1+20=21$
21
### Exercise #2
Complete the following exercise:
$\frac{25+3-2}{13}+5\cdot4=$
### Step-by-Step Solution
According to the order of arithmetic operations, we first place the multiplication exercise in parentheses:
$\frac{25+3-2}{13}+(5\cdot4)=$
We solve the multiplication exercise:
$5\times4=20$
We obtain the exercise:
$\frac{25+3-2}{13}+20=$
We solve the exercise in the numerator of the fraction:
$25+3-2=28-2=26$
We obtain the fraction:
$\frac{26}{13}=2$
Now we obtain the exercise:
$2+20=22$
22
### Exercise #3
$\lbrack(\sqrt{81}-3\times3):4+5\times5\rbrack=$
### Step-by-Step Solution
According to the rules of order of arithmetic operations, parentheses are resolved first.
We start by solving the inner parentheses, first we will solve the root using the formula:
$\sqrt{a}=\sqrt{a^2}=a$
$\sqrt{81}=\sqrt{9^2}=9$
The exercise obtained within parentheses is:
$(9-3\times3)$
First we solve the multiplication exercise and then we subtract:
$(9-9)=0$
After solving the inner parentheses, the resulting exercise is:
$0:4+5\times5$
According to the rules of the order of arithmetic operations, we first solve the exercises of multiplication and division, and then subtraction.
We place the two exercises within parentheses to avoid confusion:
$(0:4)+(5\times5)=0+25=25$
$25$
### Exercise #4
$\frac{14+8-2}{4\cdot5}\cdot2+3=$
### Step-by-Step Solution
First, we will solve the multiplication exercise that was broken:
$4\times5=20$
Now, we will solve the exercise that was broken:
$14+8-2=22-2=20$
$\frac{20}{20}=1$
$1\times2+3=$
According to the order of operations, we will first solve the multiplication exercise and then proceed:
$1\times2=2$
$2+3=5$
5
### Exercise #5
Indicates the corresponding sign:
$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)\textcolor{red}{☐}(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13}$
### Step-by-Step Solution
For solving a problem involving addition or subtraction each of the terms that come into play is treated separately,
This is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are performed before all others,
A. We will start with the term that appears on the left in the given problem:
$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)$In this term, a division operation is established between two terms that in their sum give a specific result, we take the term that in their sum gives the left, remembering that multiplication precedes subtraction, therefore the multiplication in these terms is performed first and then the subtraction operation, in contrast- the term that in their sum gives the right (the denominators) is considered as a multiplication of a number by the term that in their sum (the numerators) therefore the multiplication in this term, this is within that we remember that multiplication precedes division and that the root cause (the definition of the root as strong) is strong for everything, therefore we will calculate its numerical value and then perform the division operation that in this term:
$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ (36-12):\big((4+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big) \\$We will continue and note that in the term that was received in the last stage the multiplication operation is found in the terms and accordingly precedence is given to the division operation that precedes them,the multiplication is performed within that we remember that the multiplication in the break means the multiplication by the break, in continuation the division operation of the break is performed, this by summarizing:
$24:\frac{6\cdot1}{6}=\\ 24:\frac{\not{6}}{\not{6}}=\\ 24:1=\\ 24$In the last stage we performed the remaining division operation, this within that we remember that dividing any number by the number 1 will yield the number itself,
We will conclude the simplification of the term that appears on the left in the given problem, we will summarize the simplification stages:
$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big)= \\ 24$
B. We will continue and simplify the term that appears on the right in the given problem:
$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13}$In this part, in the first section, we simplify the term within the framework of the order of operations,
In this term, a division operation is established between two terms that in their sum, in this part, in the first section, we simplify that two terms in contrast, the term that in their sum gives the left is simplified within performing the division and subtraction operations, in contrast we simplify the term that in their sum gives the right, given that multiplication and division precede subtraction we start from the second simplification stage in these terms , and given that the order of operations does not define precedence to one of the multiplication or division operations is performed one after the other according to the order from left to right (which is the natural order of operations) , in continuation we will calculate the result of the subtraction operation that in these terms:
$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:(17-4)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\$We will continue and simplify the term that was received in the last stage, in this part, first the division and multiplication operations are performed one after the other from left to right:
$17:13\cdot\frac{1}{13} =\\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17\cdot1}{13\cdot13}=\\ \frac{17}{13^2}$In the first stage, given that the result of the division operation is a result that is not whole we mark it as a break (a break from above- given that the numerator is larger than the denominator) in continuation we perform the multiplication of the breaks within that we remember that when we multiply two breaks we multiply numerator by numerator and denominator by denominator and keep the essence of the original break.
We will conclude the simplification of the term that appears on the right in the given problem, we will summarize the simplification stages:
$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17}{13^2}$We will return to the original problem, and we will present the results of the simplification of the terms that were reported in A and B:
$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)\textcolor{red}{☐}(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} \downarrow\\ 24 \textcolor{red}{☐}\frac{17}{13^2}$As a result, the correct answer here is answer B.
$\ne$
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Find the product of additive inverse of $\dfrac{-2}{7}\times \dfrac{4}{5}$ and multiplicative inverse of $\dfrac{13}{21}$ . Write your answer in the simplest form.
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Hint: We start solving this problem by finding the additive inverse of the obtained number after multiplying given numbers. Then we find the multiplicative inverse of the other given number and then multiply obtained results to obtain the final answer.
Additive inverse of a number is a real number which sums up to the given number and gives zero.
Let us consider the definition of additive inverse.
Let a and b be two real numbers. Then we say b is the additive inverse of a, if
$a+b=0$ .
Let us consider the given number $\dfrac{-2}{7}\times \dfrac{4}{5}$ as a.
Now, we need to find the additive inverse of a that is b.
Then, by the definition above, we get
\begin{align} & \Rightarrow \left( \dfrac{-2}{7}\times \dfrac{4}{5} \right)+b=0 \\ & \Rightarrow \left( \dfrac{-8}{35} \right)+b=0 \\ & \Rightarrow b=-\left( \dfrac{-8}{35} \right) \\ & \Rightarrow b=\dfrac{8}{35} \\ \end{align}
So, the additive inverse of $\dfrac{-2}{7}\times \dfrac{4}{5}$ is $\dfrac{8}{35}$.
Now, let us consider the definition of Multiplicative Inverse.
Multiplicative inverse of a number is nothing but the reciprocal of the number, i.e., the multiplicative inverse of x is $\dfrac{1}{x}$ .
We need to find the multiplicative inverse of $\dfrac{13}{21}$ .
So, from the definition above, we get
\begin{align} & \Rightarrow \dfrac{1}{\dfrac{13}{21}} \\ & \Rightarrow \dfrac{21}{13} \\ \end{align}
Now, we need to multiply the both obtained results to get the required result.
So, multiplying them we get
$\Rightarrow \dfrac{8}{35}\times \dfrac{21}{13}$
As both 21, 35 are divisible by 7 we reduce them. Then we get
\begin{align} & \Rightarrow \dfrac{8}{5}\times \dfrac{3}{13} \\ & \Rightarrow \dfrac{24}{65} \\ \end{align}
Therefore, the product of additive inverse $\dfrac{-2}{7}\times \dfrac{4}{5}$ of and multiplicative inverse of $\dfrac{13}{21}$ is $\dfrac{24}{65}$.
Hence the answer is $\dfrac{24}{65}$.
Note: While finding the additive inverse of $\dfrac{-2}{7}\times \dfrac{4}{5}$ , there is a chance of making mistake by finding additive inverse for both the numbers and multiply them like
Additive inverse of $\dfrac{-2}{7}$ is $\dfrac{-2}{7}+x=0\Rightarrow x=-\left( \dfrac{-2}{7} \right)=\dfrac{2}{7}$ .
Additive inverse of $\dfrac{4}{5}$ is $\dfrac{4}{5}+x=0\Rightarrow x=-\left( \dfrac{4}{5} \right)=-\dfrac{4}{5}$ .
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# What does gcf mean in math
The greatest common factor, or GCF, is the greatest factor that divides two numbers. To find the GCF of two numbers: List the prime factors of each number. The definition of Greatest Common Factor: The greatest number that is a factor of two (or more) other numbers. When. Greatest Common Factor. The highest number that divides exactly into two or more numbers. It is the greatest thing for simplifying fractions!.
## greatest common factor worksheets
Greatest common factor (GCF) The greatest common factor between 2 numbers is the largest factor that they have in common. One good way to find this number . This is called the GCF or the Greatest Common Factor. Remember that relatively prime means two numbers that do not have any common factors other than 1. Step 3: Now circle the prime factors that each number has in common. Step 4. The least common multiple (LCM) of two numbers is the smallest number How do you find the greatest common factor, the least common multiple, and the.
definition; greatest; common; factor; gcf. Background Tutorials. Factors and Prime Factorization. How Do You Find All the Factors of a Number? Trying to find all. The Greatest Common Factor (GCF) of some numbers, is the largest number that divides evenly into all of the numbers. Like, the GCF of 10,15, and 25 is 5. Finding the greatest common factor (GCF) of a number set can be easy, but That means the greatest common factor among the three terms is 1xy (or xy). They do have certain applications within science and mathematics.
## greatest common factor of 8 and 12
Greatest common factor. Definition: The greatest common factor (GCF) is the largest factor of An easier way to handle the same problem is to do the following. I started by dividing by the smallest prime that would fit into it, being 2. This left me with another even number, , so I divided by 2 again. The result, We are going to learn what the Greatest Common Factor is and how to calculate it. 15 / 1 = 15, which means 1 and 15 are factors of Now we would divide by 5, but since we already know it is a factor, we have finished. The product of all common factors is the greatest common factor (GCF). 5⋅3⋅2= Your browser does not currently recognize any of the video formats available. Finding the greatest common factor of any two numbers involves breaking finding common factors will help you achieve what you're setting out to do. This means finding all of the prime number factors of another number. The greatest common factor of two or more whole numbers is the largest many numbers you have, how large they are and what you will do with the result. Factoring. To find the GCF by factoring, list out all of the factors of each number or find. Learn to find the greatest common factor of a given number using factor To factor means to rewrite a mathematical expression as a product of. However, I am merely above average at math (B+ in AP Calc). I want to be . Replace 20 with 24 and 2x2x2 or 8 would be the GCF. 52 views. Find out what is the full meaning of GCF on henrisjewelry.com! 'Greatest Common Greatest Common Factor. Academic & Science» Mathematics. Rate it: GCF. LCM is found under MATH → NUM, arrow down to #8: lcm. From the home screen, choose lcm(followed by two numbers separated by a comma. NOTE: lcm .
#### Author: Tojazilkree
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# Algebra II: Solving Equations and Inequalities
On this page we hope to clear up problems that you might have with solving equations and/or inequalities. Equations are the most common thing you'll see in all of math! Read on or follow any of the links below to begin understanding equations and inequalities.
Dealing with fractions or decimals
Inequalities
Quiz on solving equations and inequalities
To solve an equation, you isolate the variable are solving for. The Addition Principle says that when a = b, a + c = b + c for any number c.
Example:
```1. Solve: x + 6 = -15
each side of the equation.
x + 6 -6 = -15 - 6
The variable is now isolated.
x = -21
```
Along the same lines, the Multiplication Principle says that if a = b and c is any number, a * c = b * c. This principle is also used to help isolate the variable you are asked to solve for.
Example:
```2. Solve: 4x = 9
Solution: Using the Multiplication Principle,
multiply each side of the equation by (1/4).
(1/4)4x = (1/4)9
The variable is now isolated.
x = (9/4)
```
Also, be aware of problems where you might need to use both of these principles together!
Example:
```3. Solve: 3x - 4 = 13
each side.
3x - 4 + 4 = 13 + 4
After simplifying, 3x = 17.
Use the Multiplication Principle to multiply each
side by (1/3).
After simplification, the variable is isolated.
x = (17/3)
```
## Dealing with Fractions or Decimals
When an equation contains fractions or decimals, it usually makes it easier to solve them when the fractions or decimals aren't there. The Multiplication Principle is used to do this.
Example:
```1. Solve: (3/4)x + (1/2) = (3/2)
Solution: Multiply both sides by the LCM of the
denominators, 4 in this case.
Use the Distributive Law, which says a(b + c) = ab + ac
to make the equation easier to deal with.
(4 * (3/4)x) + (4 * (1/2)) = 4(3/2)
After simplification, you get an equation with no
fractions!
3x + 2 = 6
(It's left up to you to solve for x.)
```
If a and b are real numbers, and ab = 0, either a, b, or both equal 0. This principle, called The Principle of Zero Products, is useful when you have an equation to solve that has two instances of a variable, such as (x + 3)(x - 2) = 0.
Example:
```2. Solve: 7x(4x + 2) = 0
Solution: Using the Principle of Zero Products,
7x = 0 and 4x + 2 = 0
Solve each equation for x.
x = 0 and x = -(1/2)
The solutions are 0 and -(1/2).
```
## Inequalities
Math problems containing <, >, <=, and >= are called inequalities. A solution to any inequality is any number that makes the inequality true.
On many occasions, you will be asked to show the solution to an inequality by graphing it on a number line. This is usually covered in elementary algebra (Algebra I) courses. This custom has been followed on this site, so follow this link to understand graphing on a number line.
As with equations, inequalities also have principles dealing with addition and multiplication. They are outlined below.
1Addition Principle for Inequality — If a > b then a + c > b + c.
Example:
```1. Solve: x + 3 > 6
each side of the inequality.
x + 3 - 3 > 6 - 3
After simplification, x > 3.
```
2Multiplication Principle for Inequalities — If a > b and c is positive, then ac > bc. If a > b and c is negative, then ac < bc (notice the sign was reversed).
Example:
```1. Solve: -4x < .8
Solution: Using the Multiplication Principle, multiply both
sides of the inequality by -.25. Then reverse the
signs.
-.25(-4x) > -.25(.8)
x > -.2
```
One thing in math that seems to give people trouble throughout their math careers is absolute value. The absolute value of any number is its numerical value (ignore the sign). For example, the absolute value of -6 is 6 and |+3| (the vertical lines stand for absolute value) is 3.
Absolute value in inequalities is a little more complicated. For example, |x| >= 4 asks us for all numbers that have an absolute value that is greater than or equal to 4. Obviously, 4 and any number greater than 4 is a solution. The confusing part comes from the fact that -4 and any number less than -4 is a solution (|-4| = 4, |-5| = 5, etc.). Therefore, the solution is x >= 4 or x <= -4.
Absolute value becomes even more complicated when dealing with equations. However, there is a theorem that tells us how to deal with equations with absolute value and complicated inequalities.
1. If X is any expression, and b any positive number, and |X| = b it is the same as |X| = b or |X| = -b.
2. If X is any expression, and b any positive number, and |X| < b it is the same as -b < X < b.
3. If X is any expression, and b any positive number, and |X| > b it is the same as X < -b, X > b.
Example:
```1. Solve: |5x - 4| = 11
Solution: Use the theorem stated above to rewrite the equation.
|X| = b
X = 5x - 4 and b = 11
5x - 4 = 11, 5x - 4 = -11
Solve each equation using the Addition Principle and the
Multiplication Principle.
5x = 15, 5x = -7
x = 3, x = -(7/5)
```
Take the quiz on solving equations and inequalities. The quiz is very useful for either review or to see if you've really got the topic down.
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# Math Snap
## 2.4 Study the following ingredients for the Meatballs in Tomato Sauce recipe that serves 4 people before answering the questions which follow: - $500 \mathrm{~g}$ lean minced beef - 1 egg - 2 teaspoons dried mixed herbs - $50 \mathrm{ml}$ cooking oil - $400 \mathrm{~g}$ canned chopped tomatoes - $250 \mathrm{~g}$ fresh egg noodles Josh wants to make meatballs for 50 people. Calculate the amounts that he needs for the following ingredients: 2.4.1 Lean minced beef 2.4.2 Canned chopped tomatoes 2.4.3 Eggs (3) 2.5 Compare the prices of lean minced beef at Shop A and Shop B. Shop A offers $2 \mathrm{~kg}$ for R250, whereas Shop B provides $3 \mathrm{~kg}$ for R315. Determine which shop sells minced meat cheaper per kilogram, and advise Josh accordingly on which shop has the least expensive minced meat. (8) 2.6 The original recipe produces 20 meatballs which serve 4 people. Determine how many meatballs are needed to serve: 2.6.1 24 people (2) 2.6.2 30 people 2.7 Mr Smith has three children aged 3 years, 5 years and 10 years. R3600 has been set aside which is to be shared proportionally according to age. The eldest child claims that he is going to receive R1800. Verify the claim and show through calculations how much each child will receive. (10)
#### STEP 1
Assumptions1. The recipe serves4 people. . The amounts of ingredients are proportional to the number of people being served. 3. The prices and quantities of minced beef at Shop A and Shop B are accurate. 4. The number of meatballs produced is proportional to the number of people being served. 5. The money set aside for Mr. Smith's children is to be divided proportionally according to their ages.
#### STEP 2
We start by calculating the amount of each ingredient needed for50 people. The formula to calculate this is$\text{Amount for50 people} = \text{Amount for4 people} \times \frac{50}{4}$
#### STEP 3
Now, plug in the given values for the amount of lean minced beef for people to calculate the amount needed for50 people. $\text{Amount of lean minced beef for50 people} =500 \mathrm{~g} \times \frac{50}{}$
#### STEP 4
Calculate the amount of lean minced beef needed for50 people. $\text{Amount of lean minced beef for50 people} =500 \mathrm{~g} \times \frac{50}{4} =6250 \mathrm{~g} =6.25 \mathrm{~kg}$
#### STEP 5
Repeat the same process for the canned chopped tomatoes. $\text{Amount of canned chopped tomatoes for50 people} =400 \mathrm{~g} \times \frac{50}{4}$
#### STEP 6
Calculate the amount of canned chopped tomatoes needed for50 people. $\text{Amount of canned chopped tomatoes for50 people} =400 \mathrm{~g} \times \frac{50}{4} =5000 \mathrm{~g} =5 \mathrm{~kg}$
#### STEP 7
For the eggs, since there is one egg for4 people, we can simply divide the number of people by4 to find the number of eggs needed. $\text{Number of eggs for50 people} = \frac{50}{4}$
#### STEP 8
Calculate the number of eggs needed for50 people. $\text{Number of eggs for50 people} = \frac{50}{4} =12.5$Since we can't have half an egg, Josh needs to use13 eggs.
#### STEP 9
Next, we need to compare the prices of lean minced beef at Shop A and Shop B. The price per kilogram can be calculated as$\text{Price per kg} = \frac{\text{Price}}{\text{Quantity in kg}}$
#### STEP 10
Now, plug in the given values for the price and quantity of minced beef at Shop A to calculate the price per kilogram. $\text{Price per kg at Shop A} = \frac{R250}{2 \mathrm{~kg}}$
#### STEP 11
Calculate the price per kilogram at Shop A. $\text{Price per kg at Shop A} = \frac{R250}{ \mathrm{~kg}} = R125 \mathrm{~per~kg}$
#### STEP 12
Repeat the same process for Shop B. $\text{Price per kg at Shop B} = \frac{R315}{ \mathrm{~kg}}$
#### STEP 13
Calculate the price per kilogram at Shop B. $\text{Price per kg at Shop B} = \frac{R315}{3 \mathrm{~kg}} = R105 \mathrm{~per~kg}$Since the price per kilogram at Shop B is less than that at Shop A, Shop B sells minced meat cheaper per kilogram.
#### STEP 14
Next, we need to calculate the number of meatballs needed to serve different numbers of people. The formula to calculate this is$\text{Number of meatballs} = \text{Number of meatballs for4 people} \times \frac{\text{Number of people}}{4}$
#### STEP 15
Now, plug in the given values for the number of meatballs for4 people and the number of people to calculate the number of meatballs needed to serve24 people. $\text{Number of meatballs for24 people} =20 \times \frac{24}{4}$
#### STEP 16
Calculate the number of meatballs needed to serve24 people. $\text{Number of meatballs for24 people} =20 \times \frac{24}{4} =120$
#### STEP 17
Repeat the same process to calculate the number of meatballs needed to serve30 people. $\text{Number of meatballs for30 people} =20 \times \frac{30}{4}$
#### STEP 18
Calculate the number of meatballs needed to serve30 people. $\text{Number of meatballs for30 people} =20 \times \frac{30}{4} =150$
#### STEP 19
Finally, we need to calculate how much money each of Mr. Smith's children will receive. The formula to calculate this is$\text{Amount for each child} = \text{Total amount} \times \frac{\text{Age of child}}{\text{Sum of ages of all children}}$
#### STEP 20
Now, plug in the given values for the total amount and the ages of the children to calculate the amount for the eldest child. $\text{Amount for eldest child} = R3600 \times \frac{10}{3+5+10}$
#### STEP 21
Calculate the amount for the eldest child. $\text{Amount for eldest child} = R3600 \times \frac{10}{18} = R2000$The eldest child's claim that he is going to receive R1800 is not correct. He will actually receive R2000.
#### STEP 22
Repeat the same process to calculate the amount for the middle child. $\text{Amount for middle child} = R3600 \times \frac{5}{18}$
#### STEP 23
Calculate the amount for the middle child. $\text{Amount for middle child} = R3600 \times \frac{5}{18} = R1000$
#### STEP 24
Repeat the same process to calculate the amount for the youngest child. $\text{Amount for youngest child} = R3600 \times \frac{3}{18}$
##### SOLUTION
Calculate the amount for the youngest child. $\text{Amount for youngest child} = R3600 \times \frac{3}{18} = R600$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Midpoint Formula
## Determine the midpoints of line segments by taking the average of x values and average of y values
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Midpoint Formula
Suppose a coordinate plane were transposed over a subway map, and the blue line went in a straight line from the point (-9, 8) to the point (1, -4). If you got on the blue line at the beginning of the line and traveled halfway to the end of the line, what would be your coordinates? How would you calculate these coordinates? In this Concept, you'll learn how to use the midpoint formula to find the halfway point between any two coordinate pairs, such as the start point and end point of the blue line.
### Guidance
Consider the following situation: You live in Des Moines, Iowa and your grandparents live in Houston, Texas. You plan to visit them for the summer and your parents agree to meet your grandparents halfway to exchange you. How do you find this location?
By meeting something “halfway,” you are finding the midpoint of the straight line connecting the two segments. In the above situation, the midpoint would be halfway between Des Moines and Houston.
The midpoint between two coordinate pairs represents the halfway point, or the average. It is the ordered pair (xm,ym)\begin{align*}(x_m,y_m)\end{align*}.
(xm,ym)=(x1+x22,y1+y22)\begin{align*}(x_m,y_m)=(\frac{x_1+x_2}{2}, \frac{y_1+ y_2}{2})\end{align*}
#### Example A
Des Moines, Iowa has the coordinates (41.59, 93.62).
Houston, Texas has the coordinates (29.76, 95.36).
Find the coordinates of the midpoint between these two cities.
Solution:
Decide which ordered pair will represent (x1,y1)\begin{align*}(x_1,y_1)\end{align*} and which will represent (x2,y2)\begin{align*}(x_2,y_2)\end{align*}.
(x1,y1)(x2,y2)=(41.59,93.62)=(29.76,95.36)\begin{align*}(x_1,y_1)&=(41.59,93.62)\\ (x_2,y_2)&=(29.76,95.36)\end{align*}
Compute the midpoint using the formula (xm,ym)=(x1+x22,y1+y22)\begin{align*}(x_m,y_m)=(\frac{x_1+x_2}{2}, \frac{y_1+ y_2}{2})\end{align*}
(xm,ym)(xm,ym)=(41.59+29.762,93.62+95.362)=(35.675,94.49)\begin{align*}(x_m,y_m)&=(\frac{41.59+29.76}{2}, \frac{93.62+ 95.36}{2})\\ (x_m,y_m)&=(35.675,94.49)\end{align*}
Using Google Maps, you can meet in the Ozark National Forest, halfway between the two cities.
#### Example B
A segment with endpoints (9, –2) and (x1,y1)\begin{align*}(x_1,y_1)\end{align*} has a midpoint of (2, –6). Find (x1,y1)\begin{align*}(x_1, y_1)\end{align*}.
Solution: Use the Midpoint Formula: x1+x22=xm\begin{align*}\frac{x_1+x_2}{2}=x_m\end{align*}
2=x1+924x1=x1+9=5\begin{align*}2=\frac{x_1+9}{2} \rightarrow 4&=x_1+9\\ x_1&=-5\end{align*}
Following the same procedure: y1+(2)2=6y1+(2)=12\begin{align*}\frac{y_1+(-2)}{2}=-6 \rightarrow y_1+(-2)=-12\end{align*}
y1(x1,y1)=10=(5,10)\begin{align*}y_1&=-10\\ (x_1,y_1)&=(-5,-10)\end{align*}
#### Example C
Find the values of x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} that make (9.5,6)\begin{align*}(9.5, 6)\end{align*} the midpoint of (3,5)\begin{align*}(3,5)\end{align*} and (x,y)\begin{align*}(x,y)\end{align*}.
Solution:
The midpoint formula:Substitute in the given values and variables:(xm,ym)(9.5,6)=(x1+x22,y1+y22)=(3+x2,5+y2)\begin{align*}\text{The midpoint formula:} && (x_m,y_m)&=(\frac{x_1+x_2}{2}, \frac{y_1+ y_2}{2})\\ \text{Substitute in the given values and variables:} && (9.5,6)&=(\frac{3+x}{2}, \frac{5+y}{2}) \end{align*}
This can be re-written as two equations:
Multiply each side by 2:Isolate the variables:9.51916=3+x2=3+x=x6=5+y212=5+y7=y\begin{align*} && 9.5&=\frac{3+x}{2} && 6=\frac{5+y}{2}\\ \text{Multiply each side by 2:} && 19&=3+x && 12=5+y\\ \text{Isolate the variables:} && 16&=x && 7=y\\ \end{align*}
### Guided Practice
On a hike, you and your friend decide to take different routes, but then meet up for lunch. You hike 3 miles north and 2 miles west. Starting from the same point, your friend hikes 4 miles east and 1 mile south. From these points, you each walk towards each other, meeting halfway for lunch. Where would your lunchtime meeting place be in reference to your starting point?
Solution:
Think of the starting point as the origin of a Cartesian coordinate system. If you walk north 3 miles, that is walking straight up the graph 3 units. Walking west 2 miles is the same as walking left 2 units on the graph. Then you have arrived at the point (-2, 3). For your friend, east is to the right (positive) and south is down (negative), so he/she arrives at (4,-1). Now you need to find the midpoint:
\begin{align*}(x_m,y_m)=\left(\frac{-2+4}{2}, \frac{3+ -1}{2}\right)=\left(\frac{2}{2}, \frac{2}{2}\right)=(1,1)\end{align*}
Your lunch time meeting place would be 1 mile north and 1 mile east of your starting point.
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
In 1–10, find the midpoint of the line segment joining the two points.
1. \begin{align*}(x_1,y_1)\end{align*} and \begin{align*}(x_2,y_2)\end{align*}
2. (7, 7) and (–7, 7)
3. (–3, 6) and (3, –6)
4. (–3, –1) and (–5, –8)
5. (3, –4) and (6, 1)
6. (2, –3) and (2, 4)
7. (4, –5) and (8, 2)
8. (1.8, –3.4) and (–0.4, 1.4)
9. (5, –1) and (–4, 0)
10. (10, 2) and (2, –4)
11. An endpoint of a line segment is (4, 5) and the midpoint of the line segment is (3, –2). Find the other endpoint.
12. An endpoint of a line segment is (–10, –2) and the midpoint of the line segment is (0, 4). Find the other endpoint.
13. Shawn lives six blocks west and ten blocks north of the center of town. Kenya lives fourteen blocks east and two blocks north of the center of town.
1. How far apart are these two girls “as the crow flies”?
2. Where is the halfway point between their houses?
Mixed Review
1. A population increases by 1.2% annually. The current population is 121,000.
1. What will the population be in 13 years?
2. Assuming this rate continues, when will the population reach 200,000?
1. Write \begin{align*}1.29651843 \cdot 10^5\end{align*} in standard form.
2. Is \begin{align*}4,2,1,\frac{1}{2},\frac{1}{6},\frac{1}{8},\ldots\end{align*} an example of a geometric sequence? Explain your answer.
3. Simplify \begin{align*}6x^3 (4xy^2+y^3 z)\end{align*}.
4. Suppose \begin{align*}0=(x-2)(x+1)(x-3)\end{align*}. What are the \begin{align*}x-\end{align*}intercepts?
5. Simplify \begin{align*}\sqrt{300}\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
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# Exponential Trees
## Introduction
This article introduces and explores a technique of visualizing the three math operators: exponents, logarithms and radicals using the structure of a tree.
A couple of years ago I was doodling a bunch of little flowers. My intention was not to do math, far from it. However, I realized that by making up arbitrary rules, like fixing the growth rate of each flower, these little tree-like structures were full of math. In particular, they demonstrate exponential growth and model how the three math operators: exponents, logarithms and radicals are closely related together.
### Doodle Process
Most good things start out as a doodle on a piece of paper. For example, the drawing of one of these trees is shown below. The tree starts at an origin and at each step, the tree branches multiply by a factor of .
The process for drawing this tree is described below:
1. Start by drawing a point that forms the base of the flower tree. This point is considered level zero or in other words we would say that this tree has a height of zero.
2. Draw branches starting from the point outward. The number of branches sprouting from the base represents the growth rate of the tree. In this case, the growth rate is equal to .
3. Repeat this process for each tip of the branchs drawn as if they were the base of the tree. Each new level follows the same process until the tree (or flower!) is the desired height.
4. Finally, draw some decorative leaves at the top level of the tree.
## Tree Structure
More formally, the tree structure has three important properties: the number of leaves on the tree, the height of the tree and growth rate of the tree. These properties are highlighted in figure above and summarized in the table below.
Variable Property
Leaves
The number of nodes on the top-most level of the tree. The leaves can also be though of as the total population.
Levels
The number of levels the tree where the starting point of the tree is considered level zero. The levels can also be thought of as time elapsed.
Growth Rate
The growth rate, sometimes referred to as the base, describes the branching factor at each level of the tree.
### Tree Interactive
The interactive graphic below allows you to generate many different exponential trees. The first slider controls the growth rate of the tree and the second slider controls the number of levels of the tree.
By increasing the levels of the tree, the number of leaves grows in an exponential manner. At each new level of the tree the number of leaves on the next level is equal to the current population of leaves multiplied by the growth rate factor. In contrast, linear growth would add a static amount at each step.
## Operators
Each of the three operators: the exponent, logarithm and radical operator, correspond to a property of the tree. Specifically, given two other properties as input each operator returns the missing third property of the tree. Each relationship the operators have to the tree is summarized in the table below.
Equation Operator
Exponent
Given the branching factor and the number of levels of the tree, the exponent operator returns the number of leaves on the tree.
Logarithm
Given the values for the branching factor and leaves of the tree, the logarithm operator returns the levels of the tree.
Given the levels of the tree and the number of leaves, the radical operator returns the growth rate of the tree.
Shown below are some examples of trees with unique properties that demonstrate these operator relationships.
### Example 1
Operation Relationship
Exponent
Logarithm
### Example 2
Operation Relationship
Exponent
Logarithm
### Example 3
Operation Relationship
Exponent
Logarithm
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# Gambling mathematics
The mathematics of gambling are a collection of probability applications encountered in games of chance and can be included in game theory. From a mathematical point of view, the games of chance are experiments generating various types of aleatory events, the probability of which can be calculated by using the properties of probability on a finite space of events.
## Experiments, events, probability spaces
The technical processes of a game stand for experiments that generate aleatory events. Here are few examples:
• Throwing the dice in craps is an experiment that generates events such as occurrences of certain numbers on the dice, obtaining a certain sum of the shown numbers, and obtaining numbers with certain properties (less than a specific number, higher than a specific number, even, uneven, and so on). The sample space of such an experiment is {1, 2, 3, 4, 5, 6} for rolling one die or {(1, 1), (1, 2), ..., (1, 6), (2, 1), (2, 2), ..., (2, 6), ..., (6, 1), (6, 2), ..., (6, 6)} for rolling two dice. The latter is a set of ordered pairs and counts 6 x 6 = 36 elements. The events can be identified with sets, namely parts of the sample space. For example, the event occurrence of an even number is represented by the following set in the experiment of rolling one die: {2, 4, 6}.
• Spinning the roulette wheel is an experiment whose generated events could be the occurrence of a certain number, of a certain color or a certain property of the numbers (low, high, even, uneven, from a certain row or column, and so on). The sample space of the experiment involving spinning the roulette wheel is the set of numbers the roulette holds: {1, 2, 3, ..., 36, 0, 00} for the American roulette, or {1, 2, 3, ..., 36, 0} for the European. The event occurrence of a red number is represented by the set {1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36}. These are the numbers inscribed in red on the roulette wheel and table.
• Dealing cards in blackjack is an experiment that generates events such as the occurrence of a certain card or value as the first card dealt, obtaining a certain total of points from the first two cards dealt, exceeding 21 points from the first three cards dealt, and so on. In card games we encounter many types of experiments and categories of events. Each type of experiment has its own sample space. For example, the experiment of dealing the first card to the first player has as its sample space the set of all 52 cards (or 104, if played with two decks). The experiment of dealing the second card to the first player has as its sample space the set of all 52 cards (or 104), less the first card dealt. The experiment of dealing the first two cards to the first player has as its sample space a set of ordered pairs, namely all the 2-size arrangements of cards from the 52 (or 104). In a game with one player, the event the player is dealt a card of 10 points as the first dealt card is represented by the set of cards {10♠, 10♣, 10♥, 10♦, J♠, J♣, J♥, J♦, Q♠, Q♣, Q♥, Q♦, K♠, K♣, K♥, K♦}. The event the player is dealt a total of five points from the first two dealt cards is represented by the set of 2-size combinations of card values {(A, 4), (2, 3)}, which in fact counts 4 x 4 + 4 x 4 = 32 combinations of cards (as value and symbol).
• In 6/49 lottery, the experiment of drawing six numbers from the 49 generate events such as drawing six specific numbers, drawing five numbers from six specific numbers, drawing four numbers from six specific numbers, drawing at least one number from a certain group of numbers, etc. The sample space here is the set of all 6-size combinations of numbers from the 49.
• In draw poker, the experiment of dealing the initial five card hands generates events such as dealing at least one certain card to a specific player, dealing a pair to at least two players, dealing four identical symbols to at least one player, and so on. The sample space in this case is the set of all 5-card combinations from the 52 (or the deck used).
• Dealing two cards to a player who has discarded two cards is another experiment whose sample space is now the set of all 2-card combinations from the 52, less the cards seen by the observer who solves the probability problem. For example, if you are in play in the above situation and want to figure out some odds regarding your hand, the sample space you should consider is the set of all 2-card combinations from the 52, less the three cards you hold and less the two cards you discarded. This sample space counts the 2-size combinations from 47.
## The probability model
A probability model starts from an experiment and a mathematical structure attached to that experiment, namely the space (field) of events. The event is the main unit probability theory works on. In gambling, there are many categories of events, all of which can be textually predefined. In the previous examples of gambling experiments we saw some of the events that experiments generate. They are a minute part of all possible events, which in fact is the set of all parts of the sample space.
For a specific game, the various types of events can be:
• Events related to your own play or to opponents’ play;
• Events related to one person’s play or to several persons’ play;
• Immediate events or long-shot events.
Each category can be further divided into several other subcategories, depending on the game referred to. These events can be literally defined, but it must be done very carefully when framing a probability problem. From a mathematical point of view, the events are nothing more than subsets and the space of events is a Boolean algebra. Among these events, we find elementary and compound events, exclusive and nonexclusive events, and independent and non-independent events.
In the experiment of rolling a die:
• Event {3, 5} (whose literal definition is occurrence of 3 or 5) is compound because {3, 5}= {3} U {5};
• Events {1}, {2}, {3}, {4}, {5}, {6} are elementary;
• Events {3, 5} and {4} are incompatible or exclusive because their intersection is empty; that is, they cannot occur simultaneously;
• Events {1, 2, 5} and {2, 5} are nonexclusive, because their intersection is not empty;
• In the experiment of rolling two dice one after another, the events obtaining 3 on the first die and obtaining 5 on the second die are independent because the occurrence of the second event is not influenced by the occurrence of the first, and vice versa.
In the experiment of dealing the pocket cards in Texas Hold’em Poker:
• The event of dealing (3♣, 3♦) to a player is an elementary event;
• The event of dealing two 3’s to a player is compound because is the union of events (3♣, 3♠), (3♣, 3♥), (3♣, 3♦), (3♠, 3♥), (3♠, 3♦) and (3♥, 3♦);
• The events player 1 is dealt a pair of kings and player 2 is dealt a pair of kings are nonexclusive (they can both occur);
• The events player 1 is dealt two connectors of hearts higher than J and player 2 is dealt two connectors of hearts higher than J are exclusive (only one can occur);
• The events player 1 is dealt (7, K) and player 2 is dealt (4, Q) are non-independent (the occurrence of the second depends on the occurrence of the first, while the same deck is in use).
These are a few examples of gambling events, whose properties of compoundness, exclusiveness and independency are easily observable. These properties are very important in practical probability calculus.
The complete mathematical model is given by the probability field attached to the experiment, which is the triple sample space—field of events—probability function. For any game of chance, the probability model is of the simplest type—the sample space is finite, the space of events is the set of parts of the sample space, implicitly finite, too, and the probability function is given by the definition of probability on a finite space of events:
## Combinations
Games of chance are also good examples of combinations, permutations and arrangements, which are met at every step: combinations of cards in a player’s hand, on the table or expected in any card game; combinations of numbers when rolling several dice once; combinations of numbers in lottery and bingo; combinations of symbols in slots; permutations and arrangements in a race to be bet on, and the like. Combinatorial calculus is an important part of gambling probability applications. In games of chance, most of the gambling probability calculus in which we use the classical definition of probability reverts to counting combinations. The gaming events can be identified with sets, which often are sets of combinations. Thus, we can identify an event with a combination.
For example, in a five draw poker game, the event at least one player holds a four of a kind formation can be identified with the set of all combinations of (xxxxy) type, where x and y are distinct values of cards. This set has 13C(4,4)(52-4)=624 combinations. Possible combinations are (3♠ 3♣ 3♥ 3♦ J♣) or (7♠ 7♣ 7♥ 7♦ 2♣). These can be identified with elementary events that the event to be measured consists of.
## Expectation and strategy
Games of chance are not merely pure applications of probability calculus and gaming situations are not just isolated events whose numerical probability is well established through mathematical methods; they are also games whose progress is influenced by human action. In gambling, the human element has a striking character. The player is not only interested in the mathematical probability of the various gaming events, but he or she has expectations from the games while a major interaction exists. To obtain favorable results from this interaction, gamblers take into account all possible information, including statistics, to build gaming strategies. The predicted future gain or loss is called expectation or expected value and is the sum of the probability of each possible outcome of the experiment multiplied by its payoff (value). Thus, it represents the average amount one expects to win per bet if bets with identical odds are repeated many times. A game or situation in which the expected value for the player is zero (no net gain nor loss) is called a fair game. The attribute fair refers not to the technical process of the game, but to the chance balance house (bank)–player.
Even though the randomness inherent in games of chance would seem to ensure their fairness (at least with respect to the players around a table—shuffling a deck or spinning a wheel do not favor any player except if they are fraudulent), gamblers always search and wait for irregularities in this randomness that will allow them to win. It has been mathematically proved that, in ideal conditions of randomness, no long-run regular winning is possible for players of games of chance. Most gamblers accept this premise, but still work on strategies to make them win over the long run.
## House advantage or edge
Casino games generally provide a predictable long-term advantage to the casino, or "house", while offering the player the possibility of a large short-term payout. Some casino games have a skill element, where the player makes decisions; such games are called "random with a tactical element." While it is possible through skilful play to minimize the house advantage, it is extremely rare that a player has sufficient skill to completely eliminate his inherent long-term disadvantage (the house edge or house vigorish) in a casino game. Such a skill set would involve years of training, an extraordinary memory and numeracy, and/or acute visual or even aural observation, as in the case of wheel clocking in Roulette.
The player's disadvantage is a result of the casino not paying winning wagers according to the game's "true odds", which are the payouts that would be expected considering the odds of a wager either winning or losing. For example, if a game is played by wagering on the number that would result from the roll of one die, true odds would be 5 times the amount wagered since there is a 1/6 probability of any single number appearing. However, the casino may only pay 4 times the amount wagered for a winning wager.
The house edge (HE) or vigorish is defined as the casino profit expressed as a percentage of the player's original bet. In games such as Blackjack or Spanish 21, the final bet may be several times the original bet, if the player doubles or splits.
Example: In American Roulette, there are two zeroes and 36 non-zero numbers (18 red and 18 black). If a player bets $1 on red, his chance of winning$1 is therefore 18/38 and his chance of losing $1 (or winning -$1) is 20/38.
The player's expected value, EV = (18/38 x 1) + (20/38 x -1) = 18/38 - 20/38 = -2/38 = -5.26%. Therefore, the house edge is 5.26%. After 10 rounds, play $1 per round, the average house profit will be 10 x$1 x 5.26% = $0.53. Of course, it is not possible for the casino to win exactly 53 cents; this figure is the average casino profit from each player if it had millions of players each betting 10 rounds at$1 per round.
The house edge of casino games vary greatly with the game. Keno can have house edges up to 25%, slot machines can have up to 15%, while most Australian Pontoon games have house edges between 0.3% and 0.4%.
The calculation of the Roulette house edge was a trivial exercise; for other games, this is not usually the case. Combinatorial analysis and/or computer simulation is necessary to complete the task.
In games which have a skill element, such as Blackjack or Spanish 21, the house edge is defined as the house advantage from optimal play (without the use of advanced techniques such as card counting or shuffle tracking), on the first hand of the shoe (the container that holds the cards). The set of the optimal plays for all possible hands is known as "basic strategy" and is highly dependent on the specific rules, and even the number of decks used. Good Blackjack and Spanish 21 games have house edges below 0.5%.
## Standard deviation
The luck factor in a casino game is quantified using standard deviation (SD). The standard deviation of a simple game like Roulette can be simply calculated because of the binomial distribution of successes (assuming a result of 1 unit for a win, and 0 units for a loss). For the binomial distribution, SD is equal to $\sqrt{npq}$, where $n$ is the number of rounds played, $p$ is the probability of winning, and $q$ is the probability of losing. Furthermore, if we flat bet at 10 units per round instead of 1 unit, the range of possible outcomes increases 10 fold. Therefore, SD for Roulette even-money bet is equal to $2b\sqrt{npq}$, where $b$ is the flat bet per round, $n$ is the number of rounds, $p=18/38$, and $q=20/38$.
After enough large number of rounds the theoretical distribution of the total win converges to the normal distribution, giving a good possibility to forecast the possible win or loss. For example, after 100 rounds at $1 per round, the standard deviation of the win (equally of the loss) will be $2\cdot1\cdot\sqrt{100\cdot18/38\cdot20/38}\approx9.99$. After 100 rounds, the expected loss will be $100\cdot1\cdot2/38\approx5.26$. The 3 sigma range is six times the standard deviation: three above the mean, and three below. Therefore, after 100 rounds betting$1 per round, the result will very probably be somewhere between $-5.26-3\cdot9.99$ and $-5.26+3\cdot9.99$, i.e., between -$34 and$24. There is still a ca. 1 to 400 chance that the result will be not in this range, i.e. either the win will exceed $24, or the loss will exceed$34.
The standard deviation for the even-money Roulette bet is one of the lowest out of all casinos games. Most games, particularly slots, have extremely high standard deviations. As the size of the potential payouts increase, so does the standard deviation.
Unfortunately, the above considerations for small numbers of rounds are incorrect, because the distribution is far from normal. Moreover, the results of more volatile games usually converge to the normal distribution much more slowly, therefore much more huge number of rounds are required for that.
As the number of rounds increases, eventually, the expected loss will exceed the standard deviation, many times over. From the formula, we can see the standard deviation is proportional to the square root of the number of rounds played, while the expected loss is proportional to the number of rounds played. As the number of rounds increases, the expected loss increases at a much faster rate. This is why it is practically impossible for a gambler to win in the long term (if they don't have an edge). It is the high ratio of short-term standard deviation to expected loss that fools gamblers into thinking that they can win.
The volatility index (VI) is defined as the standard deviation for one round, betting one unit. Therefore, the VI for the even-money American Roulette bet is $\sqrt{18/38\cdot20/38}\approx0.499$.
The variance $v$ is defined as the square of the VI. Therefore, the variance of the even-money American Roulette bet is ca. 0.249, which is extremely low for a casino game. The variance for Blackjack is ca. 1.2, which is still low compared to the variances of electronic gaming machines (EGMs).
Additionally, the term of the volatility index based on some confidence intervals are used. Usually, it is based on the 90% confidence interval. The volatility index for the 90% confidence interval is ca. 1.645 times as the "usual" volatility index that relates to the ca. 68.27% confidence interval.
It is important for a casino to know both the house edge and volatility index for all of their games. The house edge tells them what kind of profit they will make as percentage of turnover, and the volatility index tells them how much they need in the way of cash reserves. The mathematicians and computer programmers that do this kind of work are called gaming mathematicians and gaming analysts. Casinos do not have in-house expertise in this field, so they outsource their requirements to experts in the gaming analysis field.
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Mixtures and Alligations: Solved Examples
# Mixtures and Alligations: Solved Examples | CSAT Preparation - UPSC PDF Download
## Mixtures & Alligation
• As the dictionary meaning of Alligation (mixing), we will deal with problems related to mixing of different compounds or quantities. The concept of alligation and weighted average are the same.
• When two or more quantities are mixed together in different ratios to form a mixture, then ratio of the quantities of the two constituents is given by the following formulae:
• Gives us the ratio of quantities in which the two ingredients should be mixed to get the mixture.
Example.1: A sum of Rs 39 was divided among 45 boys and girls. Each girl gets 50 paise, whereas a boy gets one rupee. Find the number of boys and girls.
• Average amount of money received by each =
• Amount received by each girl = 50 paise = Rs
• Amount received by each boy = Re. 1
Number of girls = 45 – 33 = 12.
Important Funda
• Always identify the ingredients as cheaper & dearer to apply the alligation rule.
In the alligation rule, the variables c, d & m may be expressed in terms of percentages (e.g. A 20% mixture of salt in water), fractions (e.g. two-fifth of the solution contains salt) or proportions (e.g. A solution of milk and water is such that milk : Water = 2 : 3).
• The important point is to remember is that c & d may represent pure ingredients or mixtures.
## Mixing a pure component into a solution
Example.2: A jar contains a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the liquid B is poured into the jar, the ratio becomes 2: 3. How many litres of liquid A were contained in the jar?
Method 1: (weighted average or equation method)
• Let the quantities of A & B in the original mixture be 4x and x litres.
• According to the question
12x = 2x + 20
⇒ 10x = 20
⇒ x = 2
• The quantity of A in the original mixture = 4x = 4 × 2 = 8 litres.
Method 2: (Alligation with composition of B)
• The average composition of B in the first mixture is 1/5.
• The average composition of B in the second mixture = 1
• The average composition of B in the resultant mixture = 3/5
• Hence applying the rule of Alligation we have
[1 – (3/5)]/[(3/5) – (1/5)] = (2/5)/(2/5) = 1
So, initial quantity of mixture in the jar = 10 litres.
And, quantity of A in the jar = (10 × 4)/5 = 8 litres.
Method 3: (Alligation with percentage of B)
• The percentage of B in 1st mixture = 20%
• The percentage of B in 2nd mixture = 100%
• The percentage of B in Final Mixture = 60%
• By rule of allegation we have
Volume 1st : Volume 2nd = (100% - 60%) : (60% - 20%)
V1 : V2 = 1 : 1
Volume of mixture 1st = 10 litres
Volume of A in mixture 1st = 80% of 10 litres = 8 litres. Answer
Question for Mixtures and Alligations: Solved Examples
Try yourself:A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Question for Mixtures and Alligations: Solved Examples
Try yourself:Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:
### Removal and Replacement
Example.3: Nine litres of solution are drawn from a cask containing water. It is replaced with a similar quantity of pure milk. This operation is done twice. The ratio of water to milk in the cask now is 16 : 9. How much does the cask hold?
• Let there be x litres in the cask
• After n operations:
(1 – 9/x)2 = 16/25
∴ x = 45 litres.
Example.4: There are two containers A and B of milk solution. The ratio of milk and water in container A is 3 : 1 and in container B, it is 4 : 1. How many liters of container B solution has to be added to 20. lts of container ‘A’ solution such that in the resulting solution; the ratio of milk to water should be 19 : 6?
• In container A, the part of milk =
• It is given 20 lts of container A is added. So, the quantity of container B should be 5 lts.
Example.5: There are two alloys A and B. Alloy A contains zinc, copper and silver, as 80% 15% and 5% respectively, whereas alloy B also contains the same metals with percentages 70%, 20%, 10% respectively. If these two alloys are mixed such that the resultant will contain 8% silver, what is the ratio of these three metals in the resultant alloy?
• Since the resultant alloy contains 8% silver, first we will find, in what ratio these two alloys A and B were mixed to form the resultant.
• Then the resultant zinc percentage is
• So, copper percentage = 100 – (74 + 8) = 18
∴ The ratio of these metals = 74 : 18 : 8 = 37 : 9 : 4.
Example.6: The cost of an apple is directly proportional to square of its weight in a fruit bazaar. Two friends A and B went there to purchase apples. A got exactly 5 apples per kg and each apple is of same weight. Where as B got exactly 4 apples per kg each weight is exactly same. If B paid Rs. 10 more than A per kg apples, what is the cost of an apple which weighs 1 kg?
• It is given cost ∝ (weight)2
⇒ c = k w2.
• A got 5 apples per kg and each apple is of same weight. ⇒ Each apple is 200 gm. = 1/5 kg.
The document Mixtures and Alligations: Solved Examples | CSAT Preparation - UPSC is a part of the UPSC Course CSAT Preparation.
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## FAQs on Mixtures and Alligations: Solved Examples - CSAT Preparation - UPSC
1. What is the concept of mixtures and alligation?
Ans. Mixtures and alligation is a mathematical concept used to solve problems related to the mixing of different components to obtain a desired concentration or strength.
2. How is mixtures and alligation useful in real life?
Ans. Mixtures and alligation concepts are widely used in various fields such as pharmacy, chemistry, cooking, and manufacturing industries, where accurate measurements and proportions are crucial for desired outcomes.
3. Can you explain the basic formula or equation used in mixtures and alligation?
Ans. The basic formula used in mixtures and alligation is: (Quantity of Component A) / (Quantity of Component B) = (Price of Component B - Price of Mixture) / (Price of Mixture - Price of Component A) This formula helps in determining the ratio of different components in a mixture.
4. How can mixtures and alligation be applied to solve practical problems?
Ans. To solve practical problems using mixtures and alligation, one must first determine the ratios of the components in the given mixture or solution. Then, by using the given information about the components or solutions being mixed, the desired concentration or strength can be calculated.
5. Can you provide an example of mixtures and alligation in a real-life scenario?
Ans. Sure! Let's consider a scenario where a pharmacist needs to prepare a 10% saline solution. They have a 5% saline solution and a 15% saline solution. By using mixtures and alligation, the pharmacist can calculate the ratio of these two solutions to obtain the desired 10% saline solution.
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# Sum Of Products And Product Of Sums Ppt To Pdf
File Name: sum of products and product of sums ppt to .zip
Size: 1732Kb
Published: 26.04.2021
The different forms of canonical expression which includes the sum of products SOP and products of the sum POS , The canonical expression can be defined as a Boolean expression which has either min term otherwise max term. Similarly, the product of sums POS mainly includes the max term , types of product of sums , k-map and schematic design of POS.
We will get four Boolean product terms by combining two variables x and y with logical AND operation. These Boolean product terms are called as min terms or standard product terms.
## Digital Circuits - Canonical & Standard Forms
How to find the product of sum and difference of two binomials with the same terms and opposite signs? Worked-out examples on the product of sum and difference of two binomials :. Thus, the product of sum and difference of two binomials is equal to the square of the first term minus the square of the second term. Didn't find what you were looking for? Or want to know more information about Math Only Math.
Any kind of logical expression can be represented through these terms and there are many ways of expressing a logical function. AND represents the product relationship between the terms whereas OR represents sum relationship. So, Sum of Products and Product of Sums are the equivalent approaches of representing a logical expression. To be clearer in the concept of SOP, we need to know how a minterm operates. So, the term SOP is also defined as the expression which is completely of minterms.
## Sum of Products and Product of Sums
We have seen that the b 2 - 4 ac portion of the quadratic formula, called the discriminant , can tell us the type of roots of a quadratic equation. The quadratic formula can also give us information about the relationship between the roots and the coefficient of the second term and the constant of the equation itself. Consider the following:. We can also see these relationships emerge from factoring the quadratic equation. The roots will be represented as r 1 and r 2.
## Sum Of Product (SOP) & Product Of Sum (POS)
Table of Contents. Sum of Product is the abbreviated form of SOP. Sum of product form is a form of expression in Boolean algebra in which different product terms of inputs are being summed together. To understand better about SOP, we need to know about min term. Minterm means the term that is true for a minimum number of combination of inputs.
#### What is a Sum of Product (SOP)?
Крик оборвался столь же внезапно, как и раздался. Затем наступила тишина. Мгновение спустя, словно в дешевом фильме ужасов, свет в ванной начал медленно гаснуть. Затем ярко вспыхнул и выключился. Сьюзан Флетчер оказалась в полной темноте. Сьюзан Флетчер нетерпеливо мерила шагами туалетную комнату шифровалки и медленно считала от одного до пятидесяти.
Сьюзан ответила ему теплой улыбкой. Ее всегда поражало, что даже в преддверии катастрофы Стратмор умел сохранять выдержку и спокойствие. Она была убеждена, что именно это качество определило всю его карьеру и вознесло на высшие этажи власти. Уже направляясь к двери, Сьюзан внимательно посмотрела на ТРАНСТЕКСТ. Она все еще не могла свыкнуться с мыслью о шифре, не поддающемся взлому. И взмолилась о том, чтобы они сумели вовремя найти Северную Дакоту. - Поторопись, - крикнул ей вдогонку Стратмор, - и ты еще успеешь к ночи попасть в Смоки-Маунтинс.
Отпусти ее, - раздался ровный, холодный голос Стратмора. - Коммандер! - из последних сил позвала Сьюзан. Хейл развернул Сьюзан в ту сторону, откуда слышался голос Стратмора. - Выстрелишь - попадешь в свою драгоценную Сьюзан. Ты готов на это пойти. - Отпусти .
Вирус? - снисходительно хмыкнул Стратмор, - Фил, я высоко ценю твою бдительность, очень высоко.
Нуматака подавил смешок. Все знали про Северную Дакоту. Танкадо рассказал о своем тайном партнере в печати. Это был разумный шаг - завести партнера: даже в Японии нравы делового сообщества не отличались особой чистотой. Энсей Танкадо не чувствовал себя в безопасности.
Прикрыв рукой глаза, он выругался и встал возле собора в маленьком дворике, образованном высокой каменной стеной, западной стороной башни Гиральда и забором из кованого железа. За открытыми воротами виднелась площадь, на которой не было ни души, а за ней, вдали, - стены Санта-Круса. Беккер не мог исчезнуть, тем более так. Халохот оглядел дворик. Он .
Далекий голос… - Дэвид. Он почувствовал болезненное жжение в боку. Мое тело мне больше не принадлежит. И все же он слышал чей-то голос, зовущий .
За ее спиной ТРАНСТЕКСТ издал предсмертный оглушающий стон. Когда распался последний силиконовый чип, громадная раскаленная лава вырвалась наружу, пробив верхнюю крышку и выбросив на двадцать метров вверх тучу керамических осколков, и в то же мгновение насыщенный кислородом воздух шифровалки втянуло в образовавшийся вакуум. Сьюзан едва успела взбежать на верхнюю площадку лестницы и вцепиться в перила, когда ее ударил мощный порыв горячего ветра.
Он остался в живых. Это было настоящее чудо. Священник готовился начать молитву.
Мой Бог. Это была настоящая красотка. - Спутница? - бессмысленно повторил Беккер. - Проститутка, что. Клушар поморщился: - Вот .
Несмотря ни на что, АН Б это стоило больших денег. Фонд электронных границ усилил свое влияние, доверие к Фонтейну в конгрессе резко упало, и, что еще хуже, агентство перестало быть анонимным. Внезапно домохозяйки штата Миннесота начали жаловаться компаниям Америка онлайн и Вундеркинд, что АНБ, возможно, читает их электронную почту, - хотя агентству, конечно, не было дела до рецептов приготовления сладкого картофеля. Провал Стратмора дорого стоил агентству, и Мидж чувствовала свою вину - не потому, что могла бы предвидеть неудачу коммандера, а потому, что эти действия были предприняты за спиной директора Фонтейна, а Мидж платили именно за то, чтобы она эту спину прикрывала. Директор старался в такие дела не вмешиваться, и это делало его уязвимым, а Мидж постоянно нервничала по этому поводу.
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# Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39
Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39 Textbook Exercise Important Questions and Answers.
## Maharashtra State Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 39
Question 1.
Write how many rupees and how many paise.
(1) ₹ 58.43
58 rupees 43 paise.
(2) ₹ 9.30
9 rupees 30 paise.
(3) ₹ 2.30
2 rupees 30 paise.
(4) ₹ 2.3
2 rupees 30 paise.
Question 2.
Write how many rupees in decimal form.
(1) 6 rupees 25 paise
₹ 6.25
(2) 15 rupees 70 paise
₹ 15.70
(3) 8 rupees 5 paise
₹ 8:05
(4) 22 rupees 4 paise
₹ 22.04
(5) 720 paise
₹ 7.20
Question 3.
Write how many metres and how many centimetres.
(1) 58.75 m
58 m 75 cm
(2) 9.30 m
9 m 30 cm
(3) 0.30 m
30 cm
(4) 0.3 m
30 cm
(5) 1.62 m
1 m 62 cm
(6) 91.4 m
91 cm 40 cm
(7) 7.02 m
7 m 2 cm
(8) 0.09 m
9 cm
Question 4.
Write how many metres in decimal form.
(1) 1 m 50 cm
1.5 m
(2) 50 m 40 cm
50.40 m
(3) 50 m 4 cm
50.04 m
(4) 734 cm
7.34 m
(5) 10 cm
0.1 m
(6) 2 cm
0.02 m
Question 5.
Write how many centimetres and how many millimetres.
(1) 6.9 cm
6 cm 9 mm
(2) 20.4 cm
20 cm 4 mm
(3) 0.8 cm
8 mm
(4) 0.5 cm
5 mm
Question 6.
Write how many centimetres in decimal form.
(1) 7 cm 1 mm
7.1 cm
(2) 16 mm
1.6 cm
(3) 144 mm
14.4 cm
(4) 8 mm
0.8 cm
Writing half, quarter, three-quarters and one and a quarter in decimal form
‘Half’ is usually written as $$\frac{1}{2}$$. To convert this fraction into decimal form, the denominator of $$\frac{1}{2}$$ must be converted into an equivalent fraction with denominator 10.
$$\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}$$ so the decimal form of $$\frac{1}{2}$$ will be $$\frac{5}{10}$$ or 0.5 Just as $$\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}$$ = 0.5, note that $$\frac{1}{2}=\frac{1 \times 50}{2 \times 50}=\frac{50}{100}$$ = 0.50
Therefore, ‘half’ is written as ‘0.5’ or 0.50’. ‘Quarter’ and ‘three quarters’ are written in fractions as $$\frac{1}{4}$$ and $$\frac{3}{4}$$ respectively. Let us convert them into decimal fractions. 10 is not divisible by 4. Therefore, the denominators of $$\frac{1}{4}$$ and $$\frac{3}{4}$$ cannot be made into fractions with multiples of 10. However, 4 × 25 = 100, so the denominator can be 100.
A quarter $$=\frac{1}{4}=\frac{1 \times 25}{4 \times 25}=\frac{25}{100}=0.25$$
and Three quarters $$=\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}=0.75$$
One and a quarter = 1 $$\frac{1}{4}$$ = 1.25
One and a half = 1 $$\frac{1}{2}$$ = 1.50 = 1.5
One and three quarters = 1 $$\frac{3}{4}$$ = 1.75
Seventeen and a half = 17 $$\frac{1}{2}$$ = 17.50 = 17.5
Question 1.
Write how many rupees and how many paise.
(1) ₹ 147.5
1 hundred and 47 rupees 50 paise.
(2) ₹ 40.4
40 rupees and 40 paise.
Question 2.
Write how many rupees in decimal form.
(1) 105 paise
₹ 1.05
(2) 6 rupees 6 paise
₹ 6.06
(3) 20 rupees 20 paise
₹ 20.2
Question 3.
Write how many metres and how many centimetres.
(1) 1.1m =
1 m 10 cm
(2) 120 cm =
1 m 20 cm
(3) 24.8 m =
24 m 80 cm
(4) 0.5 m =
50 cm
Question 4.
Write how many metres in decimal form.
(1) 110 cm =
1.1m
(2) 60 cm =
0.6 m
Question 5.
Write how many centimetres and how many millimetres.
(1) 0.1 cm =
1 mm
(2) 10.5 cm =
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# 12 is 89 percent of what number?
## (12 is 89 percent of 13.4831)
### 12 is 89 percent of 13.4831. Explanation: What does 89 percent or 89% mean?
Percent (%) is an abbreviation for the Latin “per centum”, which means per hundred or for every hundred. So, 89% means 89 out of every 100.
### Methods to calculate "12 is 89 percent of what number" with step by step explanation:
#### Method 1: Diagonal multiplication to calculate "12 is 89 percent of what number".
1. As given: For 89, our answer will be 100
2. Assume: For 12, our answer will be X
3. 89*X = 100*12 (In Steps 1 and 2 see colored text; Diagonal multiplications will always be equal)
4. X = 100*12/89 = 1200/89 = 13.4831
#### Method 2: Same side division to calculate "12 is 89 percent of what number".
1. As given: For 89, our answer will be 100
2. Assume: For 12, our answer will be X
3. 100/x = 89/12 (In Step 1 and 2, see colored text; Same side divisions will always be equal)
4. 100/X = 89/12 or x/100 = 12/89
5. X = 12*100/12 = 1200/89 = 13.4831
### Percentage examples
Percentages express a proportionate part of a total. When a total is not given then it is assumed to be 100. E.g. 12% (read as 12 percent) can also be expressed as 12/100 or 12:100.
Example: If 12% (12 percent) of your savings are invested in stocks, then 12 out of every 100 dollars are invested in stocks. If your savings are \$10,000, then a total of 12*100 (i.e. \$1200) are invested in stocks.
### Percentage sign (%)
The percent (per cent i.e. per hundred) sign % is the standard symbol to indicate a percentage, to signify a number or ratio as a fraction of 100. Related signs include the permille (per thousand) sign ‰ and the permyriad (per ten thousand) sign ‱ (also known as a basis point), which indicate that a number is divided by one thousand or ten thousand, respectively.
### Scholarship programs to learn math
Here are some of the top scholarships available to students who wish to learn math.
### Examples to calculate "What is the percent decrease from X to Y?"
When you purchase through links on our website, we may earn an affiliate commission from platforms like Amazon.
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# Understanding The Basic Math: 400 Divided By 5
## Introduction
Mathematics is an essential subject that is widely used in daily life. From calculating bills to managing finances, we use math skills to make informed decisions. One of the fundamental concepts in math is division, which involves dividing a number into equal parts. In this article, we will explore the basics of division by solving the simple problem: 400 divided by 5.
## What is Division?
Division is the process of dividing a number into equal parts. It is the opposite of multiplication and involves finding out how many times one number can be divided by another. The answer to a division problem is called the quotient, and it represents the number of equal parts that the dividend has been divided into.
## How to Solve 400 Divided by 5
To solve the problem 400 divided by 5, we need to find out how many times 5 can go into 400. We can do this by performing long division or using a calculator.
Long Division Method:
Step 1: Write the dividend (400) inside the long division symbol and the divisor (5) outside it.
Step 2: Divide the first digit of the dividend (4) by the divisor (5). The answer is 0 with a remainder of 4.
Step 3: Bring down the next digit (0) to the remainder (4) to form the new dividend (40).
Step 4: Divide the first digit of the new dividend (4) by the divisor (5). The answer is 0 with a remainder of 4.
Step 5: Repeat step 3 and step 4 until you have no more digits to bring down.
Step 6: The final answer is the quotient (80).
## Why is 400 Divided by 5 Important?
The problem 400 divided by 5 may seem simple, but it is an essential concept in math. This problem represents the basics of division, which is the foundation for more complex mathematical concepts. Understanding division is crucial for everyday life, such as dividing a pizza or sharing a cake among friends. Additionally, division is used in more advanced fields such as science, engineering, and finance.
## Common Mistakes in Division
Division can be tricky, and there are some common mistakes that people make when solving division problems. These mistakes include:
• Forgetting to carry over a remainder to the next step
• Dividing the wrong digit
• Using the wrong formula
To avoid these mistakes, it is essential to practice division regularly and review the steps carefully.
## Conclusion
In conclusion, division is a fundamental concept in math, and understanding how to solve simple problems such as 400 divided by 5 is essential. By practicing division regularly, we can improve our math skills and make informed decisions in our daily lives. Remember to double-check your answers and avoid common mistakes to ensure accurate results. With the right practice and dedication, anyone can become a math whiz!
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Division with Unit Fractions & Whole Numbers | 3rd, 4th, & 5th Grades
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WHAT IS DIVIDING WHOLE NUMBERS BY UNIT FRACTIONS?
To do this, you will apply and extend your previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
To better understand division with unit fractions & whole numbers…
WHAT IS DIVIDING WHOLE NUMBERS BY UNIT FRACTIONS?. To do this, you will apply and extend your previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. To better understand division with unit fractions & whole numbers…
## LET’S BREAK IT DOWN!
### Slime
Let’s say you want to make some slime. You need [ggfrac]1/4[/ggfrac] cup of glue for each batch. You have 2 cups of glue. You want to find out how many batches of slime you can make. You can write a division equation to show what you are trying to find: 2 ÷ [ggfrac]1/4[/ggfrac] = ? You can draw two circles to represent the 2 cups of glue. Then partition each circle into four equal parts. Each equal part represents [ggfrac]1/4[/ggfrac] of the circle, or [ggfrac]1/4[/ggfrac] cup. Count the number of one-fourths. Each fourth represents one batch of slime that you can make. You can make 8 batches of slime. Try this one yourself. You are mixing blue and yellow paint to create green paint. For each batch of paint, you need [ggfrac]1/3[/ggfrac] cup of yellow paint. You have 3 cups of yellow paint. How many batches of green paint can you make with that amount of yellow paint?
Slime Let’s say you want to make some slime. You need [ggfrac]1/4[/ggfrac] cup of glue for each batch. You have 2 cups of glue. You want to find out how many batches of slime you can make. You can write a division equation to show what you are trying to find: 2 ÷ [ggfrac]1/4[/ggfrac] = ? You can draw two circles to represent the 2 cups of glue. Then partition each circle into four equal parts. Each equal part represents [ggfrac]1/4[/ggfrac] of the circle, or [ggfrac]1/4[/ggfrac] cup. Count the number of one-fourths. Each fourth represents one batch of slime that you can make. You can make 8 batches of slime. Try this one yourself. You are mixing blue and yellow paint to create green paint. For each batch of paint, you need [ggfrac]1/3[/ggfrac] cup of yellow paint. You have 3 cups of yellow paint. How many batches of green paint can you make with that amount of yellow paint?
### Guitar Strings
Let’s say you have 4 yards of fishing line you can use to make guitar strings. Each guitar string should be [ggfrac]1/3[/ggfrac] yard long. So, you can use 4 ÷ [ggfrac]1/3[/ggfrac] to find how many guitar strings you can make. You can draw 4 rectangles and divide each rectangle into 3 pieces (thirds). There are 12 thirds in all. You can make 12 guitar strings with the line. You notice that when you're finding the total number of thirds, the total is the same as the whole number TIMES the denominator of the unit fraction. Division and multiplication are opposite operations, like addition and subtraction. You can use multiplication to solve this division problem. To do that you KEEP 4 the same, CHANGE the sign to ×, and FLIP the [ggfrac]1/3[/ggfrac]. After you have done this, the new equation is 4 × [ggfrac]3/1[/ggfrac] , or 4 × 3. 4 × 3 = 12. That means that 4 ÷ [ggfrac]1/3[/ggfrac] is 12, so you can make 12 guitar strings. To find 6 ÷ [ggfrac]1/2[/ggfrac] , you can find 6 × [ggfrac]2/1[/ggfrac], or 6 × 2 = 12. Try this one yourself. For an art piece, you need one piece of ribbon that is [ggfrac]1/8[/ggfrac] inch long. You have 4 inches of ribbon. How many art pieces can you make with the amount of ribbon you have?
Guitar Strings Let’s say you have 4 yards of fishing line you can use to make guitar strings. Each guitar string should be [ggfrac]1/3[/ggfrac] yard long. So, you can use 4 ÷ [ggfrac]1/3[/ggfrac] to find how many guitar strings you can make. You can draw 4 rectangles and divide each rectangle into 3 pieces (thirds). There are 12 thirds in all. You can make 12 guitar strings with the line. You notice that when you're finding the total number of thirds, the total is the same as the whole number TIMES the denominator of the unit fraction. Division and multiplication are opposite operations, like addition and subtraction. You can use multiplication to solve this division problem. To do that you KEEP 4 the same, CHANGE the sign to ×, and FLIP the [ggfrac]1/3[/ggfrac]. After you have done this, the new equation is 4 × [ggfrac]3/1[/ggfrac] , or 4 × 3. 4 × 3 = 12. That means that 4 ÷ [ggfrac]1/3[/ggfrac] is 12, so you can make 12 guitar strings. To find 6 ÷ [ggfrac]1/2[/ggfrac] , you can find 6 × [ggfrac]2/1[/ggfrac], or 6 × 2 = 12. Try this one yourself. For an art piece, you need one piece of ribbon that is [ggfrac]1/8[/ggfrac] inch long. You have 4 inches of ribbon. How many art pieces can you make with the amount of ribbon you have?
### Pie
Let’s say you have [ggfrac]1/2[/ggfrac] of a pie. You want to partition it into 3 equal pieces for 3 people. What fraction of the whole pie are the new pieces? Find [ggfrac]1/2[/ggfrac] ÷ 3. You can draw a circle to represent a whole pie and then draw a line to split the circle in half. Half of this circle represents the [ggfrac]1/2[/ggfrac] of a pie you start with. Now, partition that [ggfrac]1/2[/ggfrac] piece into 3 equal parts. When you do that, you can see that each smaller piece represents [ggfrac]1/6[/ggfrac] of the pie. That means that [ggfrac]1/2[/ggfrac] ÷ 3 = [ggfrac]1/6[/ggfrac]. Each of the 3 people get [ggfrac]1/6[/ggfrac] of a pie. Try this one yourself. You have [ggfrac]1/4[/ggfrac] of a piece of round cake. You want to partition that piece into 2 equal pieces. What fraction of the whole cake is each small piece?
Pie Let’s say you have [ggfrac]1/2[/ggfrac] of a pie. You want to partition it into 3 equal pieces for 3 people. What fraction of the whole pie are the new pieces? Find [ggfrac]1/2[/ggfrac] ÷ 3. You can draw a circle to represent a whole pie and then draw a line to split the circle in half. Half of this circle represents the [ggfrac]1/2[/ggfrac] of a pie you start with. Now, partition that [ggfrac]1/2[/ggfrac] piece into 3 equal parts. When you do that, you can see that each smaller piece represents [ggfrac]1/6[/ggfrac] of the pie. That means that [ggfrac]1/2[/ggfrac] ÷ 3 = [ggfrac]1/6[/ggfrac]. Each of the 3 people get [ggfrac]1/6[/ggfrac] of a pie. Try this one yourself. You have [ggfrac]1/4[/ggfrac] of a piece of round cake. You want to partition that piece into 2 equal pieces. What fraction of the whole cake is each small piece?
### Zen Garden
Let’s say you are making desktop zen gardens. You have a ribbon that is [ggfrac]1/4[/ggfrac] yard long. You want to cut this piece of ribbon into 3 equal pieces, to use one of those pieces in each garden. You want to know how long the pieces are. To find out, you need to find [ggfrac]1/4[/ggfrac] ÷ 3. Use the KEEP CHANGE FLIP strategy to solve. You keep 14 the same, change the division sign to a multiplication sign, and flip 3, or [ggfrac]3/1[/ggfrac], to [ggfrac]1/3[/ggfrac]. [ggfrac]1/4[/ggfrac] × [ggfrac]1/3[/ggfrac]= [ggfrac]1/12[/ggfrac] . Each piece is [ggfrac]1/12[/ggfrac] yards long. Try this one yourself. You have [ggfrac]1/3[/ggfrac] foot of ribbon. You want to cut it into 2 equal pieces. What is the length of each smaller piece?
Zen Garden Let’s say you are making desktop zen gardens. You have a ribbon that is [ggfrac]1/4[/ggfrac] yard long. You want to cut this piece of ribbon into 3 equal pieces, to use one of those pieces in each garden. You want to know how long the pieces are. To find out, you need to find [ggfrac]1/4[/ggfrac] ÷ 3. Use the KEEP CHANGE FLIP strategy to solve. You keep 14 the same, change the division sign to a multiplication sign, and flip 3, or [ggfrac]3/1[/ggfrac], to [ggfrac]1/3[/ggfrac]. [ggfrac]1/4[/ggfrac] × [ggfrac]1/3[/ggfrac]= [ggfrac]1/12[/ggfrac] . Each piece is [ggfrac]1/12[/ggfrac] yards long. Try this one yourself. You have [ggfrac]1/3[/ggfrac] foot of ribbon. You want to cut it into 2 equal pieces. What is the length of each smaller piece?
## DIVISION WITH UNIT FRACTIONS & WHOLE NUMBERS VOCABULARY
Division
Finding how many equal groups or how many in each group.
Whole number
The counting numbers: 0, 1, 2, 3, 4, ...
Fraction
A symbol that describes parts of a whole that is divided into equal parts. For example, [ggfrac]1/3[/ggfrac] or [ggfrac]2/3[/ggfrac].
Unit fraction
A fraction with a numerator of 1. For example, [ggfrac]1/5[/ggfrac] , [ggfrac]1/2[/ggfrac] or [ggfrac]1/10[/ggfrac]
Equation
A mathematical statement that uses an equal sign to show that two expressions are equal. For example, 3 × 4 = 12 or 16 ÷ 2 = 4 × 2.
Partition
Splitting into parts. For example, a circle can be partitioned into fourths.
Combining groups of equal size. You can think of multiplication as repeated addition.
An operation that undoes another operation. Division and multiplication are opposite operations.
## DIVISION WITH UNIT FRACTIONS & WHOLE NUMBERS DISCUSSION QUESTIONS
### You need [ggfrac]1/2[/ggfrac] cup of milk to make your favorite cake. You have 3 cups of milk. Describe a model that represents this situation and can help you find how many cakes you can make.
You need to solve the equation 3 ÷ [ggfrac]1/2[/ggfrac] = ? Draw 3 circles to represent the 3 cups of milk. Each circle is divided in half and each half represents the [ggfrac]1/2[/ggfrac] cup of milk you need for each cake. You can count to find the number of half cups of milk you have. You have 6 half cups of milk. You can make 6 cakes.
### You have a board that is 2 yards long and just the right width for the floor of the bird houses you are building. You need the length of each board to be 14 yard. How many birdhouse floors can you make from this board?
I need to find the answer to 2 ÷ [ggfrac]1/4[/ggfrac] = ? First, I convert the division problem into a multiplication problem: 2 × 4. 2 × 4 = 8. I can make 8 birdhouses with the 2-yard-long wood board.
### You have [ggfrac]1/3[/ggfrac] of a piece of pie. You want to divide that piece into 2 equal pieces to share. What fraction of the pie is each small piece? Write the equation you need to solve, draw a model to solve, and answer the question.
[ggfrac]1/3[/ggfrac] ÷ 2 = [ggfrac]1/6[/ggfrac] . Each small piece is [ggfrac]1/6[/ggfrac] of the pie.
### You have a piece of ribbon that is [ggfrac]1/3[/ggfrac] yard long. You cut it into 2 equal pieces. What is the length of each piece?
[ggfrac]1/6[/ggfrac] yard.
### You have 8 circular cakes. You cut each cake into fourths. Delroy says you have 2 one-fourth pieces because 8 ÷ 4 is 2. Is Delroy correct?
Delroy is not correct. To find the answer, you need to solve the equation 8 ÷ [ggfrac]1/4[/ggfrac] = ? To divide 8 by [ggfrac]1/4[/ggfrac], you change the division equation into the multiplication equation 8 × 4 = ? 8 × 4 = 32, so there are 32 one-fourth pieces.
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# 9.5: Similar Triangles Review
Difficulty Level: At Grade Created by: CK-12
[\begin{align*}^*\end{align*}Editor’s note: This day is set aside for a quiz and a review of similar triangles.]
## Learning Objectives
• Review the concepts of angle congruence and segment proportionality in similar triangles.
## Chords
Remember from lesson 2: a chord is a line segment that has both endpoints on a circle.
• Line segments whose endpoints are both on a circle are called _______________.
Segments of Chords Theorem
If two chords intersect inside a circle so that one chord is divided into segments of lengths \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and the other into segments of lengths \begin{align*}c\end{align*} and \begin{align*}d\end{align*}, then the segments of the chords satisfy the following relationship:
\begin{align*}ab = cd\end{align*}
This means that the product of the segment lengths of one chord equals the product of the segment lengths of the second chord:
• The intersection point splits each ______________________ into two segments.
• The product of both segment lengths of one chord is ____________________ to the product of both segment lengths of the other chord.
We prove this theorem on the following page.
Proof
We connect points \begin{align*}A\end{align*} and \begin{align*}C\end{align*} and points \begin{align*}D\end{align*} and \begin{align*}B\end{align*} to make \begin{align*}\Delta AEC\end{align*} and \begin{align*}\Delta DEB\end{align*}:
Statements Reasons
1. \begin{align*}\angle AEC \cong \angle DEB\end{align*} 1. Vertical angles are congruent
2. \begin{align*}\angle CAB \cong \angle BDC\end{align*} 2. Inscribed angles intercept the same arc
3. \begin{align*}\angle ACD \cong \angle ABD\end{align*} 3. Inscribed angles intercept the same arc
4. \begin{align*}\Delta AEC \cong \Delta DEB\end{align*} 4. AA similarity postulate
5. \begin{align*}\frac{c}{b} = \frac{a}{d}\end{align*} 5. In similar triangles, the ratios of corresponding sides are equal.
6. \begin{align*}ab = cd\end{align*} 6.Cross multiplication
Example 1
Find the value of the variable \begin{align*}x\end{align*}:
Use the products of the segment lengths of each chord:
\begin{align*}10x & = 8 \cdot 12\\ 10x & = 96\\ x & = 9.6\end{align*}
1. In your own words, define a chord.
\begin{align*}{\;\;}\end{align*}
\begin{align*}{\;\;}\end{align*}
2. True or false: When two chords intersect inside a circle, the sum of the segment lengths of one chord is equal to the sum of the segment lengths of the other chord.
3. How could you change the statement in #2 above to make it true?
\begin{align*}{\;\;}\end{align*}
\begin{align*}{\;\;}\end{align*}
4. In the space below, make up your own problem with two chords that intersect inside a circle, and then solve your problem.
(Hint: if you are having trouble, look at Example 1 on the previous page and model your problem after that one.)
\begin{align*}{\;\;}\end{align*}
\begin{align*}{\;\;}\end{align*}
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# How to Write Linear Functions from Tables
The concept of linear functions is one of the key topics in algebra and fundamental to understanding the world of mathematics.
A linear function is a polynomial function of degree one, presenting a straight line when graphed. The general format of a linear function is $$f(x)=mx+b$$, where $$m$$ is the slope, and $$b$$ is the $$y$$-intercept.
Tables are a succinct way to represent data. When data points from a linear function are presented in a table, it’s essential to be adept at interpreting these tables to decipher the function.
## A Step-by-step Guide to Writing Linear Functions from Tables
Here is a step-by-step guide on how to write linear functions from tables:
### Step 1: Identify the Variables
The first step in writing a linear function from a table is identifying the variables. Typically, tables for linear functions feature two columns, one for each variable ($$x$$ and $$y$$).
### Step 2: Determine the Slope
The slope ($$m$$) of a line is the rate at which $$y$$ changes for each change in $$x$$. To find the slope, subtract the $$y$$-value of the second row from the $$y$$-value of the first row. Then subtract the x-value of the second row from the $$x$$-value of the first row. Dividing these two differences gives you the slope: $$m =\frac{y2 – y1}{x2 – x1}$$.
### Step 3: Calculate the Y-Intercept
The y-intercept ($$b$$) is the point at which the line crosses the $$y$$-axis. This point can be directly seen in the table as the $$y$$-value when $$x$$ equals zero. If the $$x = 0$$ value isn’t available in the table, you can use the slope ($$m$$) and one set of coordinates $$(x, y)$$ from the table to solve for $$b$$ using the formula: $$b=y-mx$$.
#### Example: Creating a Linear Function from a Table
Consider the following table:
To determine the slope, subtract the $$y$$-values $$(5-3=2)$$ and the $$x$$-values $$(2-1=1)$$ between the first two rows. The slope $$m=\frac{2}{1}=2$$.
For the y-intercept, you can see when $$x =1, y=3$$. Using the slope ($$2)$$, and substituting $$x (1)$$ and $$y (3)$$ into the formula $$b=y-mx$$, we find that $$b=3-2\times 1 = 1$$.
Thus, the linear function corresponding to this table is $$f(x)=2x+1$$ or simply, $$f(x)=2x+1$$.
#### The Real-World Relevance of Linear Functions
Understanding how to write linear functions from tables is not just critical in mathematics but also in real-world scenarios. These scenarios include predicting sales growth, estimating profit margins, and even modeling natural phenomena such as population growth.
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# Worksheet Rational And Irrational Numbers
A Rational Phone numbers Worksheet can help your child become more familiar with the methods associated with this rate of integers. With this worksheet, pupils can fix 12 different troubles relevant to logical expressions. They may learn how to increase two or more phone numbers, group them in pairs, and find out their products and services. They will also practice simplifying realistic expression. After they have learned these concepts, this worksheet is a beneficial tool for advancing their reports. Worksheet Rational And Irrational Numbers.
## Realistic Phone numbers really are a ratio of integers
There are two kinds of figures: irrational and rational. Logical numbers are understood to be complete figures, while irrational phone numbers usually do not recurring, and have an limitless quantity of numbers. Irrational amounts are non-no, low-terminating decimals, and sq . beginnings which are not ideal squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To determine a logical quantity, you need to realize such a realistic number is. An integer is really a complete amount, plus a reasonable number can be a ratio of two integers. The ratio of two integers is the variety at the top split with the number at the base. For example, if two integers are two and five, this would be an integer. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be made right into a fraction
A realistic number has a numerator and denominator which are not no. Because of this they could be expressed as being a fraction. Together with their integer numerators and denominators, realistic numbers can also have a unfavorable importance. The unfavorable importance must be put to the left of as well as its absolute importance is its length from zero. To make simpler this instance, we shall state that .0333333 is a small percentage which can be published like a 1/3.
In addition to unfavorable integers, a rational amount can also be produced into a small percentage. For example, /18,572 is a rational quantity, although -1/ will not be. Any fraction comprised of integers is rational, given that the denominator does not have a and may be composed as an integer. Likewise, a decimal that leads to a stage is another realistic amount.
## They make perception
Regardless of their name, realistic amounts don’t make very much sensation. In math, they can be one organizations having a special span about the number line. Which means that when we add up anything, we can easily purchase the dimensions by its proportion to the unique number. This keeps correct even when you will find infinite rational numbers involving two distinct figures. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To obtain the length of a pearl, for instance, we could count up its width. One particular pearl weighs in at 10 kilograms, and that is a logical amount. Additionally, a pound’s bodyweight equals 10 kgs. Therefore, we must be able to break down a pound by 15, with out be concerned about the duration of one particular pearl.
## They could be conveyed like a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal variety may be written like a numerous of two integers, so 4x 5 is the same as seven. A comparable problem involves the repetitive small percentage 2/1, and each side must be divided by 99 to obtain the proper response. But how do you have the conversion? Here are a few illustrations.
A reasonable amount can also be written in great shape, such as fractions plus a decimal. A great way to stand for a rational amount in a decimal is always to split it into its fractional equal. There are actually three ways to break down a reasonable number, and every one of these methods brings its decimal counterpart. One of these ways is to split it into its fractional comparable, and that’s what’s called a terminating decimal.
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# Quick Answer: What Is The HCF Of Any Two Prime Numbers?
## How do you find the HCF of a prime number?
HCF by Prime FactorizationFind the prime factors of each of the given number.Next, we identify the common prime factors of the given numbers.We then multiply the common factors.
The product of these common factors is the HCF of the given numbers..
## What is the HCF of 2 and 3?
182 and 3 are the prime number common to all given numbers. Step3: Product of the numbers taken. So 18 is the GCD (Greatest Common Divisor) of the numbers.
## What is the HCF of 30?
Factors of 30 (Thirty) = 1, 2, 3, 5, 6, 10, 15 and 30. Factors of 24 (Twenty four) = 1, 2, 3, 4, 6, 8, 12 and 24. Therefore, common factor of 30 (Thirty) and 24 (Twenty four = 1, 2, 3, and 6. Highest common factor (H.C.F) of 30 (Thirty) and 24 (Twenty four = 6.
## What is the HCF of 12 and 16?
The factors for 16 are 1, 2, 4, 8, 16. The two numbers (12 and 16) share common factors (1, 2, 4). The greatest of these is 4 and that is the greatest common factor.
## What is the HCF of two prime numbers?
The HCF of two coprime numbers is always 1 . For example: 5 and 9 are coprime numbers and hence, HCF (5,9)=1 ( 5 , 9 ) = 1 .
## What is the HCF of 24 and 36?
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36. Therefore, common factor of 24 and 36 = 1, 2, 3, 4, 6, 8 and 12. Highest common factor (H.C.F) of 24 and 36 = 12.
## Are 2 and 3 Coprime numbers?
For example, 2 and 3 are two prime numbers. Factors of 2 are 1, 2, and factors of 3 are 1, 3. The only common factor is 1 and hence is co-prime. Any two successive numbers/ integers are always co-prime: Take any consecutive numbers such as 2, 3, or 3, 4 or 5, 6, and so on; they have 1 as their HCF.
## What is the twin prime for 11?
Record Primes of this Type>rankprimedigits11194772106074315 · 2171960 – 15178012100314512544015 · 2171960 – 1517801316869987339975 · 2171960 – 1517791433218925 · 2169690 – 15109016 more rows
## Are 2 and 3 twin primes?
Usually the pair (2, 3) is not considered to be a pair of twin primes. … The first few twin prime pairs are: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73), (101, 103), (107, 109), (137, 139), … OEIS: A077800.
## What is the HCF of two consecutive prime numbers answer?
1Answer: HCF (Any Prime, Any Composit ) is 1 . So they are coprime. So naturally HCF of 2 Primes is 1.
## Are 11 and 13 twin primes?
The first fifteen pairs of twin primes are as follows: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73), (101, 103), (107, 109), (137, 139), … Also check: Co-Prime Numbers.
## What is HCF formula?
Highest Common Factor(HCF) of two or more numbers is the greatest number which divides each of them exactly. … Example : HCF of 60 and 75 = 15 because 15 is the highest number which divides both 60 and 75 exactly. We can find out HCF using prime factorization method or by dividing the numbers or division method.
## What is the HCF of 16 and 24?
8Example 21 Consider the highest common factor of 16 and 24 again. The common factors are 1, 2, 4 and 8. So, the highest common factor is 8.
## What is the HCF of 56?
56 = 1 × 2 × 2 × 2 × 7. Common factor of 12 and 56 = 1, 2, 2. Highest common factor of 12 and 56 = 2 × 2 = 4.
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# Fraction and Mixed Number Comparison
## Use <, > and/or = to compare fractions and mixed numbers.
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Fraction and Mixed Number Comparison
Margaret’s dad is trying to cut back on the salt in his diet. He learned that many packaged foods contain a lot of salt, so he is trying to make some of his favorite foods himself. One of his favorite snacks is salsa, so he is making his own salsa. He found two different recipes. One calls for 112\begin{align*}1 \frac{1}{2}\end{align*} teaspoons of salt and another calls for 158\begin{align*}1 \frac{5}{8}\end{align*} teaspoons of salt. How can Margaret help her dad to determine which recipe to use if he wants to use the least amount of salt possible?
In this concept, you will learn how to compare and order fractions and mixed numbers.
### Comparing Fractions and Mixed Numbers
When two fractions have the same denominator, the fraction with the larger numerator will be the bigger fraction. When two fractions do not have the same denominator, comparison is not as easy.
One way to compare fractions is by using approximation and benchmarks. Approximate each fraction with one of the three benchmarks: 0,12\begin{align*}0, \frac{1}{2}\end{align*}, and 1\begin{align*}1\end{align*}. Then, compare the approximations.
Here is an example.
Use approximation to order 78,25,358,129\begin{align*}\frac{7}{8}, \frac{2}{5}, 3 \frac{5}{8}, \frac{1}{29}\end{align*}, and 2930\begin{align*}\frac{29}{30}\end{align*} greatest to least.
First, approximate each fraction using the fraction benchmarks.
• 78:7\begin{align*}\frac{7}{8}:7\end{align*} out of 8 is almost 8 out of 8 which would be a whole, so 78\begin{align*}\frac{7}{8}\end{align*} is approximately 1.
• 25:2\begin{align*}\frac{2}{5} : 2\end{align*} is a little less than half of 5, so 25\begin{align*}\frac{2}{5}\end{align*} is approximately 12\begin{align*}\frac{1}{2}\end{align*}.
• 358\begin{align*}3 \frac{5}{8}\end{align*} is a mixed number greater than 1, so it will automatically be the greatest number in our list.
• 129\begin{align*}\frac{1}{29}\end{align*} the denominator of 29 is much larger than the numerator of 1, so 129\begin{align*}\frac{1}{29}\end{align*} is approximately 0.
• 2930:29\begin{align*}\frac{29}{30} : 29\end{align*} out of 30 is almost 30 out of 30 which would be a whole, so 2930\begin{align*}\frac{29}{30}\end{align*} is approximately 1.
Now, write the fractions in a preliminary greatest to least order with the benchmarks in parentheses:
358, 78(1), 2930(1), 25(12), 129(0)\begin{align*}3 \frac{5}{8},\ \frac{7}{8} (1), \ \frac{29}{30} (1), \ \frac{2}{5} \left( \frac{1}{2} \right), \ \frac{1}{29} (0)\end{align*}
Notice that the approximation method helped to order most of the numbers, but there are two fractions that are close to 1. You will have to use another method to decide how 78\begin{align*}\frac{7}{8}\end{align*} compares to 2930\begin{align*}\frac{29}{30}\end{align*}.
One way to determine which fraction is closest to 1 is to draw two number lines between 0 and 1, arranged so that one number line is above the other. Divide the top number line into eighths and find 78\begin{align*}\frac{7}{8}\end{align*}. Divide the bottom number line into thirtieths and find 2930\begin{align*}\frac{29}{30}\end{align*}. Look to see which value is closest to 1.
Now you can see that 2930\begin{align*}\frac{29}{30}\end{align*} is closer to 1, so it is the greater number.
The answer is that the numbers ordered from greatest to least are 358,2930,78,25,129\begin{align*}3 \frac{5}{8}, \frac{29}{30}, \frac{7}{8}, \frac{2}{5}, \frac{1}{29}\end{align*}.
Another way to compare two fractions with different denominators is by rewriting one or both fractions so that they have the same denominator. To rewrite a fraction, find an equivalent fraction by multiplying both the numerator and the denominator by the same number. Your goal is to choose numbers to multiply by so that the denominators of the equivalent fractions will be the same.
Here is an example.
Compare 23\begin{align*}\frac{2}{3}\end{align*} and 57\begin{align*}\frac{5}{7}\end{align*}.
First, notice that the denominators of 3 and 7 are different. You will need to find an equivalent fraction for each given fraction so that their denominators are the same.
Next, find a common denominator. You are looking for a number that is a multiple of both 3 and 7. The product of 3 and 7 is 21 and in this case that is the least common multiple of 3 and 7.
Now, rewrite each fraction as an equivalent fraction with a denominator of 21. Remember to always multiply the numerator and denominator of the fraction by the same number.
2357==2×73×75×37×3==14211521\begin{align*}\begin{matrix} \frac{2}{3} & = & \frac{2 \times 7}{3 \times 7} & = & \frac{14}{21} \\ \frac{5}{7} & = & \frac{5 \times 3}{7 \times 3}& = & \frac{15}{21} \end{matrix}\end{align*}
Next, compare the rewritten fractions. Now that they have the same denominator, you can see that 1421\begin{align*}\frac{14}{21}\end{align*} is less than 1521\begin{align*}\frac{15}{21}\end{align*}, so 23\begin{align*}\frac{2}{3}\end{align*} is less than 57\begin{align*}\frac{5}{7}\end{align*}.
The answer is 23<57\begin{align*}\frac{2}{3} < \frac{5}{7}\end{align*}.
### Examples
#### Example 1
Earlier, you were given a problem about Margaret's dad, who is making salsa.
He found two recipes. One calls for 112\begin{align*}1 \frac{1}{2}\end{align*} teaspoons of salt and the other calls for 158\begin{align*}1 \frac{5}{8}\end{align*} teaspoons of salt. He wants to use the least amount of salt possible.
First, Margaret should notice that because both numbers are between 1 and 2, she can compare the fractional parts of the mixed numbers, 12\begin{align*}\frac{1}{2}\end{align*} and 58\begin{align*}\frac{5}{8}\end{align*}, in order to determine which mixed number is greater. Because the denominators are different, she will need to find an equivalent fraction for each given fraction so that their denominators are the same.
Next, she can find a common denominator. She is looking for a number that is a multiple of both 2 and 8. 8 is the least common multiple of 2 and 8.
Now, she can rewrite each fraction as an equivalent fraction with a denominator of 8. Note that because 58\begin{align*}\frac{5}{8}\end{align*} already has a denominator of 8, she will not need to rewrite that fraction!
1258==1×42×4=4858\begin{align*}\begin{array}{rcl} \frac{1}{2} & = & \frac{1 \times 4}{2 \times 4} = \frac{4}{8}\\ \frac{5}{8} & = & \frac{5}{8} \end{array}\end{align*}
Finally, Margaret can compare the rewritten fractions. Now that they have the same denominator, she can see that 48\begin{align*}\frac{4}{8}\end{align*} is less than 58\begin{align*}\frac{5}{8}\end{align*}. This means 112\begin{align*}1 \frac{1}{2}\end{align*} teaspoons is less than 158\begin{align*}1 \frac{5}{8}\end{align*} teaspoons.
The answer is that Margaret’s dad should use the recipe that calls for 112\begin{align*}1 \frac{1}{2}\end{align*} teaspoons of salt.
#### Example 2
In the long jump contest, Peter jumped 538\begin{align*}5 \frac{3}{8}\end{align*} feet, Sharon jumped 635\begin{align*}6 \frac{3}{5}\end{align*} feet, and Juan jumped 627\begin{align*}6 \frac{2}{7}\end{align*} feet. Order their jump distances from greatest to least.
First, notice that 538\begin{align*}5 \frac{3}{8}\end{align*} is less than 6 while 635\begin{align*}6 \frac{3}{5}\end{align*} and 627\begin{align*}6 \frac{2}{7}\end{align*} are both greater than 6. That means 538\begin{align*}5 \frac{3}{8}\end{align*} is the smallest number.
Next, compare 635\begin{align*}6 \frac{3}{5}\end{align*} and \begin{align*}6 \frac{2}{7}\end{align*}. Because both numbers are between 6 and 7, you can compare the fractional parts of the mixed numbers, \begin{align*}\frac{3}{5}\end{align*} and \begin{align*}\frac{2}{7}\end{align*}, in order to determine which mixed number is greater. Because the denominators are different, you will need to find an equivalent fraction for each given fraction so that their denominators are the same.
Now, find a common denominator. You are looking for a number that is a multiple of both 5 and 7. 35 is the least common multiple of 5 and 7.
Next, rewrite each fraction as an equivalent fraction with a denominator of 35.
\begin{align*}\begin{matrix} \frac{3}{5} & = & \frac{3 \times 7}{5 \times 7} & = & \frac{21}{35} \\ \frac{2}{7} & = & \frac{2 \times 5}{7 \times 5} & = & \frac{10}{35} \end{matrix}\end{align*}
Finally, compare the rewritten fractions. Now that they have the same denominator, you can see that \begin{align*}\frac{21}{35}\end{align*} is greater than \begin{align*}\frac{10}{35}\end{align*}. This means \begin{align*}6 \frac{3}{5}\end{align*} is greater than \begin{align*}6 \frac{2}{7}\end{align*}.
The answer is that the distances ordered from greatest to least are \begin{align*}6 \frac{3}{5}, 6 \frac{2}{7}, 5 \frac{3}{8}\end{align*}.
#### Example 3
Compare \begin{align*}\frac{1}{8}\end{align*} and \begin{align*}\frac{5}{6}\end{align*}.
First, try the approximation method and approximate each fraction with a fraction benchmark.
• \begin{align*}\frac{1}{8}:\end{align*} The numerator of 1 is much less than the denominator of 8, so \begin{align*}\frac{1}{8}\end{align*} is approximately 0.
• \begin{align*}\frac{5}{6} : 5\end{align*} out of 6 is almost 6 out of 6 which would be a whole, so \begin{align*}\frac{5}{6}\end{align*} is approximately 1.
Next, compare the two numbers. \begin{align*}\frac{1}{8}\end{align*} is approximately 0 and \begin{align*}\frac{5}{6}\end{align*} is approximately 1. That means \begin{align*}\frac{1}{8}\end{align*} is definitely less than \begin{align*}\frac{5}{6}\end{align*}.
The answer is \begin{align*}\frac{1}{8} < \frac{5}{6}\end{align*}.
#### Example 4
Compare \begin{align*}\frac{4}{9}\end{align*} and \begin{align*}\frac{7}{15}\end{align*}.
First, notice each fraction is approximately \begin{align*}\frac{1}{2}\end{align*}, so the approximation method for comparison won’t work this time. Because the denominators are different, you will need to find an equivalent fraction for each given fraction so that their denominators are the same.
Next, find a common denominator. You are looking for a number that is a multiple of both 9 and 15. 45 is the least common multiple of 9 and 15, though any common multiple of 9 and 15 would work.
Now, rewrite each fraction as an equivalent fraction with a denominator of 45.
\begin{align*}\begin{matrix} \frac{4}{9} & = & \frac{4 \times 5}{9 \times 5} & = & \frac{20}{45} \\ \frac{7}{15} & = & \frac{7 \times 3}{15 \times 3} & = & \frac{21}{45} \end{matrix}\end{align*}
Next, compare the rewritten fractions. Now that they have the same denominator, you can see that \begin{align*}\frac{20}{45}\end{align*} is less than \begin{align*}\frac{21}{45}\end{align*}, so \begin{align*}\frac{4}{9}\end{align*} is less than \begin{align*}\frac{7}{15}\end{align*}.
The answer is \begin{align*}\frac{4}{9} < \frac{7}{15}\end{align*}.
#### Example 5
Compare \begin{align*}\frac{8}{9}\end{align*} and \begin{align*}\frac{3}{4}\end{align*}.
First, notice each fraction is approximately 1, so the approximation method for comparison won’t work this time. Because the denominators are different, you will need to find an equivalent fraction for each given fraction so that their denominators are the same.
Next, find a common denominator. You are looking for a number that is a multiple of both 9 and 4. 36 is the least common multiple of 9 and 4.
Now, rewrite each fraction as an equivalent fraction with a denominator of 36.
\begin{align*}\begin{matrix} \frac{8}{9} & = & \frac{8 \times 4}{9 \times 4} & = & \frac{32}{36} \\ \frac{3}{4} & = & \frac{3 \times 9}{4 \times 9} & = & \frac{27}{36} \end{matrix}\end{align*}
Next, compare the rewritten fractions. Now that they have the same denominator, you can see that \begin{align*}\frac{32}{36}\end{align*} is greater than \begin{align*}\frac{27}{36}\end{align*}, so \begin{align*}\frac{8}{9}\end{align*} is greater than \begin{align*}\frac{3}{4}\end{align*}.
The answer is \begin{align*}\frac{8}{9} > \frac{3}{4}\end{align*}.
### Review
Compare each pair of fractions or mixed numbers using an inequality symbol or equals sign.
1. \begin{align*}\frac{2}{5} \text{ and } \frac{3}{5}\end{align*}
2. \begin{align*}\frac{2}{6} \text{ and } \frac{1}{6}\end{align*}
3. \begin{align*}\frac{4}{5} \text{ and } \frac{3}{5}\end{align*}
4. \begin{align*}\frac{2}{15} \text{ and } \frac{13}{15}\end{align*}
5. \begin{align*}\frac{2}{5} \text{ and } \frac{3}{7}\end{align*}
6. \begin{align*}\frac{1}{5} \text{ and } \frac{1}{7}\end{align*}
7. \begin{align*}\frac{12}{15} \text{ and } \frac{13}{30}\end{align*}
8. \begin{align*}\frac{4}{5} \text{ and } \frac{7}{9}\end{align*}
9. \begin{align*}\frac{3}{8} \text{ and } \frac{4}{7}\end{align*}
10. \begin{align*}\frac{1}{2} \text{ and } \frac{3}{6}\end{align*}
Write each set in order from least to greatest.
1. \begin{align*}\frac{5}{6}, \frac{1}{3}, \frac{4}{9}\end{align*}
2. \begin{align*}\frac{6}{7}, \frac{1}{4}, \frac{2}{3}\end{align*}
3. \begin{align*}\frac{6}{6}, \frac{4}{5}, \frac{2}{3}\end{align*}
4. \begin{align*}\frac{1}{2}, \frac{3}{5}, \frac{2}{3}\end{align*}
5. \begin{align*}\frac{2}{7}, \frac{1}{4}, \frac{3}{6}\end{align*}
6. \begin{align*}\frac{1}{6}, \frac{2}{9}, \frac{2}{5}\end{align*}
7. Brantley is making an asparagus soufflé which calls for \begin{align*}3 \frac{3}{7}\end{align*} cups of cheese, \begin{align*}3 \frac{2}{3}\end{align*} cups of asparagus, and \begin{align*}2 \frac{2}{5}\end{align*} cups of parsley. Using approximation, order the ingredients from largest amount used to least amount used.
8. Geraldine is putting in a pool table in her living room. She wants to put it against the longest wall of the room. Wall A is \begin{align*}12 \frac{4}{9}\end{align*} feet and wall B is \begin{align*}12 \frac{2}{5}\end{align*} feet. Against which wall will Geraldine put her pool table?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
fraction
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
improper fraction
An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator.
Mixed Number
A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
Numerator
The numerator is the number above the fraction bar in a fraction.
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# NCERT Solutions for Class 9 Maths Chapter 10 Circles
*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.
NCERT Solutions for Class 9 Maths Chapter 10 Circles are provided here in PDF format, which can be downloaded for free. The NCERT Solutions for the chapter Circles are included as per the latest update of the CBSE curriculum (2023-24) and have been designed by our expert teachers.
All the solved questions of Chapter 10 Circles are with respect to the CBSE syllabus and guidelines to help students solve each exercise question present in the book and prepare for the exam. These serve as reference tools for the students to do homework and also support them in scoring good marks. Students can also get the solutions for Class 9th Maths all chapters exercise-wise and practise well for the exams.
## NCERT Solutions for Class 9 Maths Chapter 10 – Circles
List of Exercises in Class 9 Maths Chapter 10
Exercise 10.1 Solutions 2 Questions (2 Short)
Exercise 10.2 Solutions 2 Questions (2 long)
Exercise 10.3 Solutions 3 Questions (3 long)
Exercise 10.4 Solutions 6 Questions (6 long)
Exercise 10.5 Solutions 12 Questions (12 long)
Exercise 10.6 Solutions 10 Questions (10 long)
### Access Answers of Maths NCERT Class 9 Chapter 10 Circles
Exercise: 10.1 (Page No: 171)
1. Fill in the blanks.
(i) The centre of a circle lies in ____________ of the circle. (exterior/ interior)
(ii) A point whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a _____________ of the circle.
(iv) An arc is a ___________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and _____________ of the circle.
(vi) A circle divides the plane, on which it lies, in _____________ parts.
Solution:
(i) The centre of a circle lies in interior of the circle.
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in 3 (three) parts.
2. Write True or False. Give reasons for your solutions.
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only a finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is the diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
Solution:
(i) True. Any line segment drawn from the centre of the circle to any point on it is the radius of the circle and will be of equal length.
(ii) False. There can be infinite numbers of equal chords in a circle.
(iii) False. For unequal arcs, there can be major and minor arcs. So, equal arcs on a circle cannot be said to be major arcs or minor arcs.
(iv) True. Any chord whose length is twice as long as the radius of the circle always passes through the centre of the circle, and thus, it is known as the diameter of the circle.
(v) False. A sector is a region of a circle between the arc and the two radii of the circle.
(vi) True. A circle is a 2d figure, and it can be drawn on a plane.
Exercise: 10.2 (Page No: 173)
1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Solution:
To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both circles is equal from the centre.
For the second part of the question, it is given that AB = CD, i.e., two equal chords.
Now, it is to be proven that angle AOB is equal to angle COD.
Proof:
Consider the triangles ΔAOB and ΔCOD.
OA = OC and OB = OD (Since they are the radii of the circle.)
AB = CD (As given in the question.)
So, by SSS congruency, ΔAOB ≅ ΔCOD
∴ By CPCT, we have,
∠AOB = ∠COD (Hence, proved).
2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Consider the following diagram.
Here, it is given that ∠AOB = ∠COD, i.e., they are equal angles.
Now, we will have to prove that the line segments AB and CD are equal, i.e., AB = CD.
Proof:
In triangles AOB and COD,
∠AOB = ∠COD (As given in the question.)
OA = OC and OB = OD (These are the radii of the circle.)
So, by SAS congruency, ΔAOB ≅ ΔCOD
∴ By the rule of CPCT, we have,
AB = CD (Hence, proved.)
Exercise: 10.3 (Page No: 176)
1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
In these two circles, no point is common.
Here, only one point, ‘P’, is common.
Even here, P is the common point.
Here, two points are common, which are P and Q.
No point is common in the above circle.
2. Suppose you are given a circle. Give a construction to find its centre.
Solution:
The construction steps to find the centre of the circle is:
Step I: Draw a circle first.
Step II: Draw 2 chords, AB and CD, in the circle.
Step III: Draw the perpendicular bisectors of AB and CD.
Step IV: Connect the two perpendicular bisectors at a point. This intersection point of the two perpendicular bisectors is the centre of the circle.
3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
It is given that two circles intersect each other at P and Q.
To prove:
OO’ is perpendicular bisector of PQ.
(i) PR = RQ
(ii) ∠PRO = ∠PRO’ = ∠QRO = ∠QRO’ = 900
Proof:
In triangles ΔPOO’ and ΔQOO’,
OP = OQ and O’P = O’Q (Since they are also the radii.)
OO’ = OO’ (It is the common side.)
So, it can be said that ΔPOO’ ≅ ΔQOO’ (SSS Congruence rule)
∴ ∠POO’ = ∠QOO’ (c.p.c.t)— (i)
Even triangles ΔPOR and ΔQOR are similar by SAS congruency.
∠POR = ∠QOR (As ∠POO’ = ∠QOO’)
OR = OR (Common arm)
So, ΔOPO’ ≅ ΔOQO’ (SAS Congruence rule)
∴ PR = QR and ∠PRO = ∠QRO (c.p.c.t) …. (ii)
As PQ is a line
∠PRO + ∠QRO = 180°
∠PRO + ∠PRO = 180° (Using (ii))
2∠PRO = 180°
∠PRO = 90°
So ∠QRO = ∠PRO = 90°
Here,
∠PRO’ = ∠QRO = 90° and ∠QRO’ = ∠PRO = 90° (Vertically opposite angles)
∠PRO = ∠QRO = ∠PRO’ = ∠QRO’ = 90°
So, OO’ is the perpendicular bisector of PQ.
Exercise: 10.4 (Page No: 179)
1. Two circles of radii 5 cm and 3 cm intersect at two points, and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
The perpendicular bisector of the common chord passes through the centres of both circles.
As the circles intersect at two points, we can construct the above figure.
Consider AB as the common chord and O and O’ as the centres of the circles.
O’A = 5 cm
OA = 3 cm
OO’ = 4 cm [Distance between centres is 4 cm.]
As the radius of the bigger circle is more than the distance between the two centres, we know that the centre of the smaller circle lies inside the bigger circle.
The perpendicular bisector of AB is OO’.
OA = OB = 3 cm
As O is the midpoint of AB
AB = 3 cm + 3 cm = 6 cm
The length of the common chord is 6 cm.
It is clear that the common chord is the diameter of the smaller circle.
2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Solution:
Let AB and CD be two equal cords (i.e., AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E.
It is now to be proven that the line segments AE = DE and CE = BE
Construction Steps
Step 1: From the centre of the circle, draw a perpendicular to AB, i.e., OM ⊥ AB.
Step 2: Similarly, draw ON ⊥ CD.
Step 3: Join OE.
Now, the diagram is as follows:
Proof:
From the diagram, it is seen that OM bisects AB, and so OM ⊥ AB
Similarly, ON bisects CD, and so ON ⊥ CD.
It is known that AB = CD. So,
AM = ND — (i)
and MB = CN — (ii)
Now, triangles ΔOME and ΔONE are similar by RHS congruency, since
∠OME = ∠ONE (They are perpendiculars.)
OE = OE (It is the common side.)
OM = ON (AB and CD are equal, and so they are equidistant from the centre.)
∴ ΔOME ≅ ΔONE
ME = EN (by CPCT) — (iii)
Now, from equations (i) and (ii), we get
AM+ME = ND+EN
So, AE = ED
Now from equations (ii) and (iii), we get
MB-ME = CN-EN
So, EB = CE (Hence, proved)
3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
From the question, we know the following:
(i) AB and CD are 2 chords which are intersecting at point E.
(ii) PQ is the diameter of the circle.
(iii) AB = CD.
Now, we will have to prove that ∠BEQ = ∠CEQ
For this, the following construction has to be done.
Construction:
Draw two perpendiculars are drawn as OM ⊥ AB and ON ⊥ D. Now, join OE. The constructed diagram will look as follows:
Now, consider the triangles ΔOEM and ΔOEN.
Here,
(i) OM = ON [The equal chords are always equidistant from the centre.]
(ii) OE = OE [It is the common side.]
(iii) ∠OME = ∠ONE [These are the perpendiculars.]
So, by RHS congruency criterion, ΔOEM ≅ ΔOEN.
Hence, by the CPCT rule, ∠MEO = ∠NEO
∴ ∠BEQ = ∠CEQ (Hence, proved)
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).
Solution:
The given image is as follows:
First, draw a line segment from O to AD, such that OM ⊥ AD.
Therefore, AM = MD — (i)
Also, since OM ⊥ BC, OM bisects BC.
Therefore, BM = MC — (ii)
From equation (i) and equation (ii),
AM-BM = MD-MC
∴ AB = CD
5. Three girls, Reshma, Salma and Mandip, are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, and Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Solution:
Let the positions of Reshma, Salma and Mandip be represented as A, B and C, respectively.
From the question, we know that AB = BC = 6cm
So, the radius of the circle, i.e., OA = 5cm
Now, draw a perpendicular BM ⊥ AC.
Since AB = BC, ABC can be considered an isosceles triangle. M is the mid-point of AC. BM is the perpendicular bisector of AC, and thus it passes through the centre of the circle.
Now,
let AM = y and
OM = x
So, BM will be = (5-x).
By applying the Pythagorean theorem in ΔOAM, we get
OA2 = OM2 +AM2
⇒ 52 = x2 +y2 — (i)
Again, by applying the Pythagorean theorem in ΔAMB,
AB2 = BM2 +AM2
⇒ 62 = (5-x)2+y2 — (ii)
Subtracting equation (i) from equation (ii), we get
36-25 = (5-x)2 +y2Â -x2-y2
Now, solving this equation, we get the value of x as
x = 7/5
Substituting the value of x in equation (i), we get
y2 +(49/25) = 25
⇒ y2 = 25 – (49/25)
Solving it, we get the value of y as
y = 24/5
Thus,
AC = 2×AM
= 2×y
= 2×(24/5) m
AC = 9.6 m
So, the distance between Reshma and Mandip is 9.6 m.
6. A circular park of radius 20m is situated in a colony. Three boys, Ankur, Syed and David, are sitting at equal distances on its boundary, each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Solution:
First, draw a diagram according to the given statements. The diagram will look as follows:
Here, the positions of Ankur, Syed and David are represented as A, B and C, respectively. Since they are sitting at equal distances, the triangle ABC will form an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is the median of ΔABC, and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
Let the side of a triangle a metres, then BD = a/2 m.
Applying Pythagoras’ theorem in ΔABD,
20 m = 2/3 × √3a/2
a = 20√3 m
So, the length of the string of the toy is 20√3 m.
Exercise: 10.5 (Page No: 184)
1. In Fig. 10.36, A, B and C are three points on a circle with centre O, such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:
It is given that,
∠AOC = ∠AOB+∠BOC
So, ∠AOC = 60°+30°
∴ ∠AOC = 90°
It is known that an angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.
So,
= (½)× 90° = 45°
2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
Here, the chord AB is equal to the radius of the circle. In the above diagram, OA and OB are the two radii of the circle.
Now, consider the ΔOAB. Here,
AB = OA = OB = radius of the circle
So, it can be said that ΔOAB has all equal sides, and thus, it is an equilateral triangle.
∴ ∠AOC = 60°
And, ∠ACB = ½ ∠AOB
So, ∠ACB = ½ × 60° = 30°
Now, since ACBD is a cyclic quadrilateral,
∠ADB +∠ACB = 180° (They are the opposite angles of a cyclic quadrilateral)
So, ∠ADB = 180°-30° = 150°
So, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is 150° and 30°, respectively.
3. In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Solution:
Since the angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.
So, the reflex ∠POR = 2×∠PQR
We know the values of angle PQR as 100°.
So, ∠POR = 2×100° = 200°
∴ ∠POR = 360°-200° = 160°
Now, in ΔOPR,
OP and OR are the radii of the circle.
So, OP = OR
Also, ∠OPR = ∠ORP
Now, we know the sum of the angles in a triangle is equal to 180 degrees.
So,
∠POR+∠OPR+∠ORP = 180°
∠OPR+∠OPR = 180°-160°
As ∠OPR = ∠ORP
2∠OPR = 20°
Thus, ∠OPR = 10°
4. In Fig. 10.38, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
Solution:
We know that angles in the segment of the circle are equal, so,
∠BAC = ∠BDC
Now. in the ΔABC, the sum of all the interior angles will be 180°.
So, ∠ABC+∠BAC+∠ACB = 180°
Now, by putting the values,
∠BAC = 180°-69°-31°
So, ∠BAC = 80°
∴ ∠BDC = 80°
5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E, such that ∠BEC = 130° and ∠ECD = 20°. Find BAC.
Solution:
We know that the angles in the segment of the circle are equal.
So,
∠BAC = ∠CDE
Now, by using the exterior angles property of the triangle,
In ΔCDE, we get
∠CEB = ∠CDE+∠DCE
We know that ∠DCE is equal to 20°.
So, ∠CDE = 110°
∠BAC and ∠CDE are equal
∴ ∠BAC = 110°
6. ABCD is a cyclic quadrilateral whose diagonals intersect at point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution:
Consider the following diagram.
Consider the chord CD.
We know that angles in the same segment are equal.
So, ∠CBD = ∠CAD
Now, ∠BAD will be equal to the sum of angles BAC and CAD.
= 30°+70°
We know that the opposite angles of a cyclic quadrilateral sum up to 180 degrees.
So,
It is known that ∠BAD = 100°
So, ∠BCD = 80°
Now, consider the ΔABC.
Here, it is given that AB = BC
Also, ∠BCA = ∠CAB (They are the angles opposite to equal sides of a triangle)
∠BCA = 30°
also, ∠BCD = 80°
∠BCA +∠ACD = 80°
Thus, ∠ACD = 50° and ∠ECD = 50°
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Draw a cyclic quadrilateral ABCD inside a circle with centre O, such that its diagonal AC and BD are two diameters of the circle.
We know that the angles in the semi-circle are equal.
So, ∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°
So, as each internal angle is 90°, it can be said that the quadrilateral ABCD is a rectangle.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
9. Two circles intersect at two points, B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q, respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.
Solution:
Construction:
Join the chords AP and DQ.
For chord AP, we know that angles in the same segment are equal.
So, ∠PBA = ∠ACP — (i)
Similarly, for chord DQ,
∠DBQ = ∠QCD — (ii)
It is known that ABD and PBQ are two line segments which are intersecting at B.
At B, the vertically opposite angles will be equal.
∴ ∠PBA = ∠DBQ — (iii)
From equation (i), equation (ii) and equation (iii), we get
∠ACP = ∠QCD
10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.
Solution:
First, draw a triangle ABC and then two circles having diameters of AB and AC, respectively.
We will have to now prove that D lies on BC and BDC is a straight line.
Proof:
We know that angles in the semi-circle are equal.
∴ ∠BDC is a straight line.
So, it can be said that D lies on the line BC.
11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
We know that AC is the common hypotenuse and ∠B = ∠D = 90°.
Now, it has to be proven that ∠CAD = ∠CBD
Since ∠ABC and ∠ADC are 90°, it can be said that they lie in a semi-circle.
So, triangles ABC and ADC are in the semi-circle, and the points A, B, C and D are concyclic.
Hence, CD is the chord of the circle with centre O.
We know that the angles which are in the same segment of the circle are equal.
∴ ∠CAD = ∠CBD
12. Prove that a cyclic parallelogram is a rectangle.
Solution:
It is given that ABCD is a cyclic parallelogram, and we will have to prove that ABCD is a rectangle.
Proof:
Thus, ABCD is a rectangle.
Exercise: 10.6 (Page No: 186)
1. Â Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Consider the following diagram.
In ΔPOO’ and ΔQOO’
OP = OQ Â Â Â Â Â Â Â Â Â (Radius of circle 1)
O’P = O’Q        (Radius of circle 2)
OO’ = OO’        (Common arm)
So, by SSS congruency, ΔPOO’ ≅ ΔQOO’
Thus, ∠OPO’ = ∠OQO’ (proved).
2. Two chords AB and CD of lengths 5 cm and 11 cm, respectively, of a circle, are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.
Solution:
Here, OM ⊥ AB and ON ⊥ CD are drawn, and OB and OD are joined.
We know that AB bisects BM as the perpendicular from the centre bisects the chord.
Since AB = 5 so,
BM = AB/2 = 5/2
Similarly, ND = CD/2 = 11/2
Now, let ON be x.
So, OM = 6−x.
Consider ΔMOB,
OB2 = OM2+MB2
Or,
Consider ΔNOD,
OD2 = ON2 + ND2
Or
We know, OB = OD (radii)
From equation 1 and equation 2, we get
Now, from equation (2), we have,
OD2= 12 +(121/4)
Or OD = (5/2)×√5 cm
3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance 4 cm from the centre, what is the distance of the other chord from the centre?
Solution:
Consider the following diagram.
Here, AB and CD are 2 parallel chords. Now, join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
So, OM = 4 cm
MB = AB/2 = 3 cm
Consider ΔOMB.
OB2 = OM2+MB2
Or, OB = 5cm
Now, consider ΔOND.
OB = OD = 5 (Since they are the radii.)
ND = CD/2 = 4 cm
Now, OD2= ON2+ND2
Or, ON = 3 cm
4. Let the vertex of an angle ABC be located outside a circle, and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Consider the diagram.
We know any exterior angle of a triangle is equal to the sum of interior opposite angles.
So,
∠DAE = ∠ABC+∠AEC (in ΔBAE) ——————-(i)
DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.
So,
∠DAE = (½)∠DOE ——————-(ii)
Similarly, ∠AEC = (½)∠AOC  ——————-(iii)
Now, from equations (i), (ii), and (iii), we get
(½)∠DOE = ∠ABC+(½)∠AOC
Or, ∠ABC = (½)[∠DOE-∠AOC]  (Hence, proved)
5. Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Solution:
To prove: A circle drawn with Q as the centre will pass through A, B and O (i.e., QA = QB = QO).
Since all sides of a rhombus are equal,
AB = DC
Now, multiply (½) on both sides.
(½)AB = (½)DC
So, AQ = DP
BQ = DP
Since Q is the midpoint of AB,
AQ= BQ
Similarly,
RA = SB
Again, as PQ is drawn parallel to AD,
RA = QO
Now, as AQ = BQ and RA = QO, we get
QA = QB = QO (Hence, proved)
6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
Here, ABCE is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of the opposite angles is 180°.
So, ∠AEC+∠CBA = 180°
As ∠AEC and ∠AED are linear pairs,
∠AEC+∠AED = 180°
Or, ∠AED = ∠CBA … (1)
We know in a parallelogram, opposite angles are equal.
So, ∠ADE = ∠CBA … (2)
Now, from equations (1) and (2), we get
Now, AD and AE are angles opposite to equal sides of a triangle.
7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Solution:
Here, chords AB and CD intersect each other at O.
Consider ΔAOB and ΔCOD.
∠AOB = ∠COD (They are vertically opposite angles.)
OB = OD (Given in the question.)
OA = OC (Given in the question.)
So, by SAS congruency, ΔAOB ≅ ΔCOD
Also, AB = CD (By CPCT)
Similarly, ΔAOD ≅ ΔCOB
Or, AD = CB (By CPCT)
In quadrilateral ACBD, opposite sides are equal.
So, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
So, ∠A = ∠C
Also, as ABCD is a cyclic quadrilateral,
∠A+∠C = 180°
⇒∠A+∠A = 180°
Or, ∠A = 90°
As ACBD is a parallelogram and one of its interior angles is 90°, so, it is a rectangle.
∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.
8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the triangle DEF are 90°–(½)A, 90°–(½)B and 90°–(½)C.
Solution:
Consider the following diagram.
Here, ABC is inscribed in a circle with centre O, and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F, respectively.
Now, join DE, EF and FD.
As angles in the same segment are equal, so,
∠EDA = ∠FCA ————-(i)
∠FDA = ∠EBA ————-(i)
By adding equations (i) and (ii), we get
∠FDA+∠EDA = ∠FCA+∠EBA
Or, ∠FDE = ∠FCA+∠EBA = (½)∠C+(½)∠B
We know, ∠A +∠B+∠C = 180°
So, ∠FDE = (½)[∠C+∠B] = (½)[180°-∠A]
∠FDE = [90-(∠A/2)]
In a similar way,
∠FED = [90° -(∠B/2)] °
And,
∠EFD = [90° -(∠C/2)] °
9. Two congruent circles intersect each other at points A and B. Through A, any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Solution:
The diagram will be
Here, ∠APB = ∠AQB (as AB is the common chord in both the congruent circles.)
Now, consider ΔBPQ.
∠APB = ∠AQB
So, the angles are opposite to equal sides of a triangle.
∴ BQ = BP
10. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Solution:
Consider this diagram.
Here, join BE and CE.
Now, since AE is the bisector of ∠BAC,
∠BAE = ∠CAE
Also,
∴arc BE = arc EC
This implies chord BE = chord EC
Now, consider triangles ΔBDE and ΔCDE.
DE = DE Â Â Â Â (It is the common side)
BD = CD Â Â Â Â (It is given in the question)
BE = CE Â Â Â Â Â (Already proved)
So, by SSS congruency, ΔBDE ≅ ΔCDE.
Thus, ∴∠BDE = ∠CDE
We know, ∠BDE = ∠CDE = 180°
Or, ∠BDE = ∠CDE = 90°
∴ DE ⊥ BC (Hence, proved).
Chapter 10, Circles, of Grade 9, is one of the most important chapters, whose concepts will also be used in Class 10. The weightage of this chapter in the final exam is around 15 marks. Therefore, students are advised to read the chapter carefully and practise each and every question included in the textbook with the help of NCERT Solutions, along with examples, to have good practice.
Topics covered in Chapter 10 Circles, are
• Circles and the related terms
• Angle Subtended by a Chord at a Point
• Perpendicular from the Centre to a Chord
• Circle through Three Points
• Equal Chords and Their Distances from the Centre
• Angle Subtended by an Arc of a Circle
NCERT Solutions for Class 9 Maths Chapter 10 – Circles are made available for students looking to solve all the problems of Ex-10.1. The methods by which problems have been solved, in a broad way, so that students find it easy to understand the fundamentals of circles. Some of the important points of this chapter are
• A circle is a simple closed geometrical shape equidistant from a central point. It is an important shape in the field of geometry.
• Every circle has its centre.
• The straight line from the centre to the circumference of a circle is called the radius of the circle.
• The length of the line through the centre that touches two points on the edge of the circle is called a diameter.
• The total distance around the circle is called the Circumference.
• The area of the circle can be calculated by applying the formula: A = Ï€ r2, where A is the Area, r is the radius, and the value of Ï€ is 3.14.
### Key Features of NCERT Solutions for Class 9 Maths Chapter 10 – Circles
• The solutions for the chapter Circles work as a reference for the students.
• Students will be able to resolve all the problems of this chapter in a faster way.
• It is good learning material for exam preparation and to do the revision for Class 9 Maths Chapter 10.
• The questions of Circles are solved by our subject experts.
• The NCERT Solutions are given as per the latest update on the CBSE syllabus and guidelines.
Disclaimer:
Dropped Topics – 10.1 Introduction, 10.2 Circles and its related terms: Review and Circle through three points.
## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 10
Q1
### How are NCERT Solutions for Class 9 Maths Chapter 10 helpful for Class 9 students?
NCERT Solutions for Class 9 Maths Chapter 10 are used by students to get a proper grasp of all the concepts of the subjects and also to lay the foundation for their career or further higher studies. BYJU’S experts formulate these questions in an easy and understandable manner that helps students solve problems in the most efficient possible ways. We hope these solutions will help CBSE Class 9 students to build a strong foundation of the basics and secure excellent marks in their final exams.
Q2
### Why should we follow NCERT Solutions for Class 9 Maths Chapter 10?
NCERT Solutions for Class 9 Maths Chapter 10 is the correct learning strategy that is devised to help students master the concepts. Revising the solutions, along with the textbooks, will help them crack any problems asked in the board exams. These solutions help to boost the problem-solving skills of the students, along with their logical reasoning. These are the most popular study materials used by students to refer to for the CBSE exams. Practising these solutions help the students to top the final exams and ace the subject.
Q3
### Where to download NCERT Solutions for Class 9 Maths Chapter 10?
NCERT Solutions for Class 9 Maths Chapter 10 can be downloaded by the students in offline mode or can be referred to online from the BYJU’S website. These solutions are formulated by BYJU’S expert faculty based on the questions present in the NCERT textbook of Class 9 Maths. These are according to the latest CBSE syllabus.
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FutureStarr
3 1 5 As an Improper Fractions.
## 3 1 5 As an Improper Fraction
In elementary school I always liked playing with fractions. Half a duck, two thirds of candy, every third of the cake. They were always my friends. But years later a math teacher helped me realize why I liked them.
### Fraction
When we are using mixed fractions, we have a whole number (in this case 3) and a fractional part (1/5). So what we can do here to convert the mixed fraction to a decimal, is first convert it to an improper fraction (where the numerator is greater than the denominator) and then from there convert the improper fraction into a decimal/
A mixed number is a whole number plus a fractional part. An improper fraction is a fraction where the numerator (top number) is larger than the denominator (bottom number). You can convert between mixed numbers and improper fractions without changing the value of the figure. (Source: www.calculatorsoup.com)
### Use
When we are using mixed fractions, we have a whole number (in this case 3) and a fractional part (1/5). So what we can do here to convert the mixed fraction to a decimal, is first convert it to an improper fraction (where the numerator is greater than the denominator) and then from there convert the improper fraction into a decimal/ \
Before we get started in the fraction to decimal conversion, let's go over some very quick fraction basics. Remember that a numerator is the number above the fraction line, and the denominator is the number below the fraction line. We'll use this later in the tutorial. Use this calculator to convert your improper fraction to a mixed fraction. An __improper fraction__ is a fraction whose nominator is greater than its denominator. For example, \$${5 \over 4}\$$. A mixed fraction is a fraction of the form \$$c {n \over d}\$$, where \$$c\$$ is an integer and \$$n < d\$$. For example, \$${11 \over 4} = 2 {3 \over 4}\$$. It is therefore the sum of a whole number and a proper fraction. If you would like to convert an improper fraction to a mixed fraction, see our [improper to mixed fraction calculator](/show/calculator/improper-to-mixed). (Source: www.calcul.com)
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# solving rational equations
So x+3 and (x+3)/1 both have the same value, but the latter expression is considered a rational expression, because it's written as a fraction. When we solve rational equations, we can multiply both sides of the equations by the least common denominator (which is $$\displaystyle \frac{{\text{least common denominator}}}{1}$$ in fraction form) and not even worry about working with fractions! Let's begin by looking at solving an equation with rational functions in it. Factor 3 out of both numerator and denominator. Unfortunately, this method only works for rational equations that contain exactly one rational expression or fraction on each side of the equals sign. If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. How do I solve this rational equation? If we wish, this can also be written as -2x - 6 = 4x. That is, all I really need to do now is solve the numerators: Since x = 4/3 won't cause any division-by-zero problems in the fractions in the original equation, then this solution is valid. If you got the variable value correct, you'll be able to simplify the original equation to a simple valid statement, such as 1 = 1. This article has been viewed 99,758 times. Solving rational equations with variables in the denominators involves manipulating and rewriting the terms. Rational expressions and other fractions can be made into non-fractions by multiplying them by their denominators. To solve a rational equation, start by rearranging it so you have 1 fraction on each side of the equals sign. So x = +/- 8. Subtract 1 from both sides to get 2x+2 = 3x, and subtract 2x from both sides to get 2 = x, which can be written as x = 2. I'll show each, and you can pick whichever you prefer. These "new forms" of the original equation may produce solutions that do not work in the original equation. Multiply both sides by 6 to cancel the denominators, which leaves us with 2x+3 = 3x+1. These fractions may be on one or both sides of the equation. Learn to use Zoom in this beginner-friendly course. In our example, we can divide both sides of the equation by -2, giving us x+3 = -2x. 1/3(2m-12) expands to 1/(6m-36). Q (x) P (x) . Multiply both sides of the equation by (x-3). Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. Purplemath. Solving Equations Video Lesson. In this section, we look at rational equations that, after some manipulation, result in a linear equation. Recall that you can solve equations containing fractions by using the least common denominator of all the fractions in the equation. Since this is an equation, I can multiply through by whatever I like. An equation involving rational expressions is called a rational equation. Multiply both sides of the equation by 13. A common way to solve these equations is to reduce the fractions to a common denominator and then solve the equality of the numerators. For example, in the equation x/8 + 2/6 = (x - 3)/9, the LCD is 8*9 = 72. Extraneous solutions are solutions that don't satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0. Also, there's the new wrinkle of variables in the denominator. To illustrate this, let’s look at a very simple equation: x = 3 . Cross-multiplication is basically a handy shortcut for multiplying both sides of the equation by both fraction's denominators. You should use the method that works best for you. This works to 15x = 3x - 3 + 2x -2, which simplifies to 15x = x - 5. To create this article, 12 people, some anonymous, worked to edit and improve it over time. These "new forms" of the original equation may produce solutions that do not work in the original equation. Solving Rational Equations Rational equations are simply equations with rational expressions in them. After clearing the fractions, we will be left with either a linear or a quadratic equation that can be solved as usual. Examples are 1 (which is 1/1), 34 (which is 34/1), 2/3, and 48/37. Like normal algebraic equations, rational equations are solved by performing the same operations to both sides of the equation until the variable is isolated on one side of the equals sign. If an equation contains at least one rational expression, it is a considered a rational equation.. Recall that a rational number is the ratio of two numbers, such as $\frac{2}{3}$ or $\frac{7}{2}$. If either side of the equation has added (or subtracted) fractions, we must use Method 1 or Method 2. This is because, as soon as you go from a rational expression (that is, something with no "equals" sign in it) to a rational equation (that is, something with an "equals" sign in the middle), you get a whole different set of tools to work with.
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# Integer Rules Games
Instructor: Heather Jenkins
Heather has a bachelor's degree in elementary education and a master's degree in special education. She was a public school teacher and administrator for 11 years.
Knowing how to solve addition, subtraction, multiplication, and division problems with negative and positive number is an important skill for students to learn. Use these games to help students learn and practice integer rules.
## The Rules of the Game
Just like students should understand the rules for a board game before they play it, understanding the rules for adding, subtracting, multiplying, and dividing integers is essential for solving mathematical problems. When students understand integer rules, it allows them to quickly decide how to work with positive and negative numbers.
Let's look at some multi-sensory games to help students learn and practice integer rules. Depending on your class, you may want to incorporate integer rules for all four mathematical operations into these games or focus on only one operation and its inverse, such as addition and subtraction or multiplication and division.
## Positive or Negative Shuffle
### Materials
• Chart of integer rules
### Teacher Directions
• Show the class a chart of integer rules. Discuss the combinations of positive and negative numbers for each mathematical operation and the signs their answers will yield. For example, if your students are multiplying two negative numbers or two positive numbers, their answer will be a positive.
• Have students stand in the middle of the classroom.
• Designate one side of the room as 'positive' and the other side as 'negative'.
• Tell students that you will give them a combination of numbers and a mathematical operation, and they will have to figure out if the answer will be positive or negative. Students will move to the side of the room that designates the correct answer. For example, you might ask, ''What will the sign of the answer be if a positive and negative number are added and the positive number is greater?'' and students should move to the 'positive' side of the room.
• Consider giving students different ways to move around the room such as jumping, crab walking, or frog-hopping to the correct side.
### Discussion Questions
• What is the most difficult integer rule for you to remember? Why?
• Why is it important to know how work with both positive and negative integers?
## Playing Card Problems
### Materials
• Chart of integer rules
• Playing cards (numbers 2-10)
• Coins
• Dry-erase boards
• Dry-erase markers
To unlock this lesson you must be a Study.com Member.
### Register to view this lesson
Are you a student or a teacher?
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## How do you teach number lines to fractions?
2. Use a Math Sort
1. Step 1: Cut the cards.
2. Step 2: Find the 4 fractions.
3. Step 3: Find the bar model that matches the fraction.
4. Step 4: Find the number line model that matches the fraction & bar model.
5. Step 5: Explain how the three cards are related.
6. Step 6: Glue.
What is a number line 3rd grade?
In math, a number line can be defined as a straight line with numbers placed at equal intervals or segments along its length. A number line can be extended infinitely in any direction and is usually represented horizontally.
### How do you use a number line to solve 245 137?
To know the answer of 245 – 137 using a number line, draw a horizontal line and count from 0 to 245 lines then count 137 from 245. Then you will get the answer. 245 – 137 = 108.
How do you teach fractions?
Here are five teaching fractions ideas to do the trick.
1. Get Hands On. The concept of a “fraction” is abstract and visualizing part vs.
2. Use Visuals. Anytime I can provide an image to go with the concept I’m teaching, I know I’m going to be in better shape.
3. Get the Games Out.
4. Turn to Tech.
5. Be Strategic in Teaching Fractions.
## What is a fraction in 3rd grade math?
Fractions describe a part of a whole. 1/2 is a fraction. We read it as “one half”. 🌟Every time you divide a whole into several equal parts, you get a fraction! ✅ If you cut a pizza into 4 equal slices, then each slice is 1 out of 4 equal parts of the whole pizza.
What is number line example?
Numbers on a number line include all sets of numbers namely natural and whole numbers. An example of a set of whole numbers is:(0, 1, 2, 3,4,5,6 …….) whereas the natural number include:1, 2, 3, 4, 5, 6…. A number line is composed of three basic parts.
### Are all fractions integers?
Fractions and decimals are not integers . All whole numbers are integers (and all natural numbers are integers), but not all integers are whole numbers or natural numbers. For example, -5 is an integer but not a whole number or a natural number.
What is a fraction number?
A fraction (sometimes, a common fraction) is a way of expressing a number that is a ratio of two integers: The top (or the first) number is called the numerator, the bottom (or the second) number is called the denominator.
## Can integers include fractions?
Integers include all whole numbers, plus the negatives of all the numbers except zero. They do not include any decimal or fractional numbers. Fractions, on the other hand, express one integer divided by another, and often equal a decimal number.
What is the same fractions as one third?
One third written as a fraction is 1/3 . You can also write it as a decimal by simply dividing 1 by 3 which is 0.33. If you multiply 0.33 with 294 you will see that you will end up with the same answer as above.
https://www.youtube.com/watch?v=SZaXtOHNh6s
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# Geometry (w/ factoring)
Q: Geometry and Angles
Ray RY bisects angle NRT
Measure of angle
NRY = 2x² + 6x
Measure of angle
YRT = 10x + 6
Find x and the measure of angle
YRT
A: Ahhh, fun to answer without a picture. But, we can do it. We have an angle that is bisecting. Bisected means cut in half. Therefore, angle NRY = angle YRT:
Mathematically: 2x² + 6x = 10x + 6
Now, time to solve! Since I see an “x²”, I think to myself that we must factor this. First step to factoring? Set the equation equal to zero:
2x² + 6x = 10x + 6
2x² + 6x – 10x = 6
2x² + 6x – 10x – 6 = 0
2x² – 4x – 6 = 0
Now, this is a quadratic that needs solving. We can either solve using the quadratic equation or by factoring. Let’s factor (View Factoring 101 for the basics of factoring):
2x² – 4x – 6 = 0
(2x )(x ) = 0
(2x + )(x – ) = 0 [I know there is a – and a + because the two numbers must multiple to give -6… And, a – times a + equals a -, right? However, I do not know that the + goes with the 2x and the – goes with the x… This is just my best guess to start. Factoring can be treated as a guessing game – if you use guess and check, you get better at it with time.]
(2x + 2)(x – 3) = 0
So,
(2x + 2) = 0 or (x – 3) = 0
x = -1 or 3
We have two potential answers. Take note, if x=-1, then we have:
Angle NRY = 2x² + 6x = 2(-1)² + 6(-1) = 2 – 6 = -4
An angle cannot be negative, therefore x cannot be -1.
Therefore, x must equal 3:
Angle YRT = 10x + 6
So, if x = 3:
Angle YRT = 10(3) + 6 = 30 + 6 = 36.
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Home | Math | What are Multiples? Definition, Properties, Examples
# What are Multiples? Definition, Properties, Examples
November 29, 2023
written by Rida Mirza
Multiples are like the numbers you get when you keep adding the same number over and over again. Understanding them is super important for doing mathematics.
In this article, we will go to see what are multiples, look at some real-life examples, and answer common questions.
## What are Multiples?
A multiple is a number that can be divided by another number without a remainder.
For example, 10 is a multiple of 5 because 10 divided by 5 gives 2 with no remainder. Similarly, 12 is a multiple of 3, 15 is a multiple of 5, and so on.
## Properties of Multiples
Some key properties of multiples include:
• A number is always a multiple of itself. For example, 15 is a multiple of 15.
• If a number x is a multiple of y, then it is also a multiple of all the factors of y. For example, 12 is a multiple of 3 and also a multiple of 1, 2, 3, 6.
• Even numbers are always multiples of 2 and odd numbers other than 1 are never multiples of 2.
## Finding Multiples
There are several methods to systematically find multiples of a given number:
### Multiplication Table Method
Write out the multiplication table of the number and highlight the multiples. For example, to find multiples of 7:
7, 14, 21, 28, 35, 42 and so on.
Repeatedly add the number to find higher multiples. For example, multiples of 4 are:
4, (4+4) 8, (8+4) 12, (12+4) 16 and so on.
### Multiplier Method
Multiply the number by increasing integers. For example, multiples of 3 are:
3, (3×2) 6, (3×3) 9, (3×4) 12 and so on.
## List of Multiples
We can list the multiples of a number by multiplying the given number by another number. Here is a list of the multiples of a few numbers.
## Applications of Multiples
Multiples have many real-world applications in areas like counting, measurement, fractions and others:
• Telling time – All times in the 12-hour clock are multiples of 5 minutes.
• Units of measurement – Converting units often involves multiples, like inches to feet.
• Fractions – Finding common denominators relies heavily on identifying multiples.
• Counting collections – When counting objects in groups, multiples of 2, 5 or 10 are useful.
## Example
List the first 5 positive multiples of 7.
The first 5 positive multiples of 7 are:
7, 14, 21, 28, 35
We can either use the multiplication table method or repeated addition to arrive at these multiples.
## FAQs
### Can 0 be a multiple of any number?
Yes, 0 is a multiple of every number since any number divided by another number will give 0 remainder.
### Are large numbers also multiples?
Definitely. Every whole number can be expressed as a multiple of its factors. So numbers like 256 or 5,824 are still multiples of powers of 2 or other factors respectively.
### Is 1 a multiple of every number?
No, 1 is not considered a multiple as it acts more like a neutral number under multiplication. But 2, 3 and other numbers are multiples of 1.
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# Sum of an Infinite Geometric Series
To examine the sum of all the terms of an infinite geometric sequence, we need to consider $S_n = \dfrac{u_1(1-r^n)}{1-r}$ when $n$ gets very large.
Sum to Infinity
If $\left|r\right|>1$, the series is said to be divergent and the sum infinitely large.
For instance, when $r=2$ and $u_1=1$;
$S_\infty=1+2+4+8+\cdots$ is infinitely large.
If $\left|r\right|<1$, or $-1 \lt r \lt 1$, then as $n$ becomes very large, $r^n$ approaches $0$.
For instance, when $r=\dfrac{1}{2}$ and $u_1=1$;
$S_\infty=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots = 2$.
$$S_\infty=\dfrac{u_1}{1-r}$$ We call this the limiting sum of the series.
This result can be used to find the value of recurring decimals.
Let's take a look at $0.\overline{7}$ to see how to convert it to a fraction.
\begin{align} \displaystyle 0.\overline{7} &= 0.7 + 0.07 + 0.007 + 0.0007 + \cdots \\ &= 0.7 + 0.7(0.1) + 0.7(0.1)^2 + 0.7(0.1)^3 + \cdots \\ &= \dfrac{0.7}{1-0.1} \\ &= \dfrac{0.7}{0.9} \\ &= \dfrac{7}{9} \\ \end{align}
### Example 1
Write $0.\overline{12}$ as a rational number.
### Example 2
Write $0.1 \overline{2}$ as a rational number.
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# 2019 AIME II Problems/Problem 1
## Problem
Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
## Solution
$[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label("A",A,dir(-120)); label("B",B,dir(-60)); label("C",C,dir(60)); label("D",D,dir(120)); label("E",E,dir(-135)); label("9",(A+B)/2,dir(-90)); label("10",(D+A)/2,dir(-150)); label("10",(C+B)/2,dir(-30)); label("17",(D+B)/2,dir(60)); label("17",(A+C)/2,dir(120)); draw(D--E--A,dotted); label("8",(D+E)/2,dir(180)); label("6",(A+E)/2,dir(-90)); [/asy]$ - Diagram by Brendanb4321
Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$, since $\triangle DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $O$ be the intersection of $BD$ and $AC$. This means that $\triangle ABO$ and $\triangle DCO$ are with ratio $\frac{21}{9}=\frac73$. Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $O$ to $DC$, and $x$ be the height of $\triangle ABO$. $$\frac{7}{3}=\frac{y}{x}$$ $$\frac{7}{3}=\frac{8-x}{x}$$ $$7x=24-3x$$ $$10x=24$$ $$x=\frac{12}{5}$$
This means that the area is $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$. This gets us $54+5=\boxed{059}.$
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers
## Solution 2
Using the diagram in Solution 1, let $E$ be the intersection of $BD$ and $AC$. We can see that angle $C$ is in both $\triangle BCE$ and $\triangle ABC$. Since $\triangle BCE$ and $\triangle ADE$ are congruent by AAS, we can then state $AE=BE$ and $DE=CE$. It follows that $BE=AE$ and $CE=17-BE$. We can now state that the area of $\triangle ABE$ is the area of $\triangle ABC-$ the area of $\triangle BCE$. Using Heron's formula, we compute the area of $\triangle ABC=36$. Using the Law of Cosines on angle $C$, we obtain
$$9^2=17^2+10^2-2(17)(10)cosC$$ $$-308=-340cosC$$ $$cosC=\frac{308}{340}$$ (For convenience, we're not going to simplify.)
Applying the Law of Cosines on $\triangle BCE$ yields $$BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC$$ $$BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})$$ $$0=389-34BE-(340-20BE)(\frac{308}{340})$$ $$0=389-34BE+\frac{308BE}{17}$$ $$0=81-\frac{270BE}{17}$$ $$81=\frac{270BE}{17}$$ $$BE=\frac{51}{10}$$ This means $CE=17-BE=17-\frac{51}{10}=\frac{119}{10}$. Next, apply Heron's formula to get the area of $\triangle BCE$, which equals $\frac{126}{5}$ after simplifying. Subtracting the area of $\triangle BCE$ from the area of $\triangle ABC$ yields the area of $\triangle ABE$, which is $\frac{54}{5}$, giving us our answer, which is $54+5=\boxed{059}.$ -Solution by flobszemathguy
## Solution 3 (Very quick)
$[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("A",A,dir(-120)); label("B",B,dir(-60)); label("C",C,dir(60)); label("D",D,dir(120)); label("9",(A+B)/2,dir(-90)); label("10",(D+A)/2,dir(-150)); label("10",(C+B)/2,dir(-30)); label("17",(D+B)/2,dir(60)); label("17",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label("8",(D+(-6,0))/2,dir(180)); label("6",(A+(-6,0))/2,dir(-90)); draw((4.5,0)--(4.5,2.4),dotted); label("h", (4.5,1.2), dir(180)); label("4.5", (6,0), dir(90)); [/asy]$ - Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$. Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$, so the other leg of the new triangle formed has length $4.5$. Notice we have formed similar triangles, and we can solve for $h$.
$$\frac{h}{4.5} = \frac{8}{15}$$ $$h = \frac{36}{15} = \frac{12}{5}$$
So $\triangle ABE$ has area $$\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}$$ And $54+5=\boxed{059}.$ - Solution by Duoquinquagintillion
## Solution 4
Let $a = \angle{CAB}$. By Law of Cosines, $$\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}$$ $$\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}$$ $$\tan a = \frac{8}{15}$$ $$A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}$$ And $54+5=\boxed{059}.$
- by Mathdummy
## Solution 5
Because $AD = BC$ and $\angle BAD = \angle ABC$, quadrilateral $ABCD$ is cyclic. So, Ptolemy's theorem tells us that $$AB \cdot CD + BC \cdot AD = AC \cdot BD \implies 9 \cdot CD + 10^2 = 17^2 \implies CD = 21.$$
From here, there are many ways to finish which have been listed above. If we let $AB \cap CD = P$, then $$\triangle APB \sim \triangle CPD \implies \frac{AP}{AB} = \frac{CP}{CD} \implies \frac{AP}{9} = \frac{17-AP}{21} \implies AP = 5.1.$$
Using Heron's formula on $\triangle ABP$, we see that $$[ABC] = \sqrt{9.6(9.6-5.1)(9.6-5.1)(9.6-9)} = 10.8 = \frac{54}{5}.$$
Thus, our answer is $059$. ~a.y.711
## Solution 6
Let $A=(0,0), B=(9,0)$. Now consider $C$, and if we find the coordinates of $C$, by symmetry about $x=4.5$, we can find the coordinates of D.
So let $C=(a,b)$. So the following equations hold:
$\sqrt{(a-9)^2+(b)^2}=17$.
$\sqrt{a^2+b^2}=10$.
Solving by squaring both equations and then subtracting one from the other to eliminate $b^2$, we get $C=(-6,8)$ because $C$ is in the second quadrant.
Now by symmetry, $D=(16, 8)$.
So now you can proceed by finding the intersection and then calculating the area directly. We get $\boxed{059}$.
~hastapasta
## Solution 7
Since the figure formed by connecting the vertices of the congruent triangles is a isoceles trapezoid, by Ptolemys, the other base of the trapezoid is $21.$ Then, dropping altitudes to the base of $21$ and using pythagorean theorem, we have the height is $8,$ and we can use similar triangles to finish.
## Solution 8 (Very, very, quick, but for observant people only)
$[asy] //Made by Afly. I used some resources. //Took me 10 min to get everything right. import olympiad; unitsize(18); pair A = (0,0); pair B = (0,8); pair C = (6,0); pair D = (15,0); pair E = (21,0); pair F = (21,8); pair G = (21/2,0); pair H = intersectionpoints(B--D,C--F)[0]; pen dash1 = linetype(new real [] {9,9})+linewidth(1); pen solid1 = linetype(new real [] {9,0})+linewidth(1); pen dash2 = linetype(new real [] {3,3})+linewidth(1); fill(C--G--H--cycle,rgb(3/4,1/4,1/4)); fill(D--G--H--cycle,rgb(3/4,3/4,1/4)); draw(C--A--B,dash1); draw(C--B--D--C,solid1); draw(F--E--D,dash1); draw(F--D--C--F,solid1); draw(G--H,dash2); draw(brace(D+dir(270),A+dir(270)),solid1); draw(brace(D,C),solid1); draw(A--A+2*dir(180),dash1,EndArrow); draw(E--E+2*dir(0),dash1,EndArrow); pair L1 = (15/2,-7/2); pair L2 = (21/2,-13/8); label("15",L1); label("8",A--B,W); label("6",A--C,S); label("10",B--C,SW); label("17",B--D,NE); label("9",L2); label("4.5",G--D,S); label("2.4",G--H,W); markscalefactor = 1/16; draw(rightanglemark(H,G,D)); draw(rightanglemark(B,A,C)); draw(rightanglemark(D,E,F)); label("A",C,SW); label("B",D,SE); label("C",B,NW); label("D",F,NE); label("E",A,SW); label("F",E,SE); label("G",G,NW); label("H",H,N); [/asy]$
First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pythagorean triples and we test that theory. Then, we find AB = 15 - 6 = 9 and GB = 4.5. Since △CEB~△HGB, we find that HG = 2.4. Then, the area of △HGB is 5.4 and the total overlap is 10.8 = 54/5. Noting that GCD(54,5)=1, we add them to get 59.
Note: I omitted some computation
~ Afly (talk)
## Solution 9 (Cyclic quad) (Basically Solution 7 but in much more detail)
$[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair O = (4.5,2.4); pair E = (-6,0); pair K = (1.993,3.737); draw(A--B--C--cycle); draw(B--D--A); label("A",A,dir(-120)); label("B",B,dir(-60)); label("C",C,dir(60)); label("D",D,dir(120)); label("9",(A+B)/2,dir(-90)); label("10",(D+A)/2,dir(-150)); label("10",(C+B)/2,dir(-30)); label("17",(D+B)/2,dir(60)); label("17",(A+C)/2,dir(120)); label("O",O,dir(90)); label("K",K,dir(-120)); draw(A--K,dotted); draw(D--E--A,dotted); label("8",(D+E)/2,dir(180)); label("6",(A+E)/2,dir(-90)); label("E",E,dir(-135)); [/asy]$
Since $\triangle ABC \cong \triangle BAD$, $\angle ADB = \angle BCA$. Thus, $A$, $B$, $C$, $D$ are concyclic.
By Ptolemy's Theorem on $ABCD$, $$(AD)(BC) + (AB)(DC) = (BD)(AC)$$ $$10^2 + 9(DC) = 17^2$$ $$DC = 21$$
The altitudes dropped from $C$ and $D$ onto the extension of AB are equal, meaning that $DC \parallel AB$. Therefore, $\triangle DCO \sim \triangle ABO$. It follows that $$\frac{OB}{17 - OB} = \frac{AB}{DC} = \frac{9}{21} = \frac{3}{7}$$ Solving yields $OB = \frac{51}{10}$.
In $\triangle ABD$, drop an altitude from $A$ to $BD$. Call the intersection of this altitude and $BD$, $K$.
The area of $\triangle ABD$ is $\frac{1}{2}(AB)(DE) = 36$. Thus, $\frac{1}{2}(AK)(BD) = 36$, and $AK = \frac{72}{17}$.
Therefore, the area of $\triangle AOB$ is $\frac{1}{2}(OB)(AK) = \frac{1}{2}(\frac{51}{10})(\frac{72}{17}) = \frac{54}{5}$.
The requested answer is $54 + 5 = \boxed{59}$.
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Lesson Video: Division Rules for 0 and 1 | Nagwa Lesson Video: Division Rules for 0 and 1 | Nagwa
# Lesson Video: Division Rules for 0 and 1 Mathematics
In this video, we will learn how to model what happens when we divide numbers by 1 and 0.
14:06
### Video Transcript
Division Rules for Zero and One
In this video, we’re going to learn how to model what happens when we divide using the numbers one and zero. You’ll be pleased to know that dividing by zero isn’t really going to cause any black holes. But it is one of those ideas in maths that can be quite difficult to get our heads around. So, to begin with, let’s start with the easier of our two numbers.
What happens if we divide a number by one? Well, if you remember, there are two ways we can think about division, as grouping and sharing. Let’s use nine divided by one as an example. Now if we’re thinking about this division as grouping, what we’re saying is if we start with nine and we split this number up into groups of one, how many groups will there be? Of course, we’ll have nine groups, won’t we? This is the same sort of idea as if there were nine of you in your class. And your teacher said, “Right, you’ve got to work individually this afternoon.” You’re all going to work at a desk on your own. You’d need to have nine desks, wouldn’t you? Nine divided by one equals nine.
We can find the same answer by thinking of dividing by one as sharing. In other words, what happens if we start with nine and we share it with, well, one? Imagine you had nine sweets and you shared them just with yourself. In a way, it’s not really sharing, is it? There’s only one of you. You’ll get all nine sweets. Nine shared into one group equals nine. Now can you spot anything interesting about this division? We started with nine. And by the time we divided it by one, it hadn’t changed at all. Our answer is the same number we began the division with.
And this brings us to a rule that we can remember when we’re dividing by one, no matter what the number. And that’s that any number divided by one equals itself. And so, 14 split into one group equals 14. 35 divided by one is 35. And 296,352 divided by one equals 296,3 — well, you get the picture. Something else we could use to help us understand what happens when we divide by one is multiplication. So, for example, if we want to find out what 12 divided by one is, we could think of the opposite or inverse operation, which is multiplication, and ask ourselves, “What will be multiplied by one to give the answer 12?” We know that 12 times one equals 12. And if we know that 12 lots of one make 12, we also know if we divide 12 by one, we get the answer 12.
Now, so far, we’ve just thought about dividing by one. And we’ve seen that any number divided by one equals itself. But you know, we can switch this statement around to make another rule. Any number divided by itself equals one. If we take six divided by six as an example, if we have six counters and split them into groups of six, we’ll only be able to make one group. Six divided by itself equals one. And so, 18 divided by 18 equals one. It doesn’t matter what number we work with. If we divide it by itself, the answer will always be one. So we’ve learned two different facts or rules here for when a division involves the number one. Any number divided by one stays the same. And any number divided by itself equals one. Let’s have a go at answering a couple of questions now where we can put into practice these rules.
Complete: Four divided by what equals four.
In this question, we’re given a division where the divisor, that’s the number we’re dividing by, is missing. The number that we’re dividing is four. And then we divide it by something and the answer is also four. This is interesting. What number could we divide by so that the number we start with doesn’t change? Let’s sketch a bar model to help us. If we divide four into groups of a certain amount and we find that we can make four groups, what will each group be worth? Each group is going to have a value of one. Four split into equal groups of one equals four. And we know this is correct because we’re reminded of an important fact to do with dividing by one. And that’s that any number divided by one equals itself. So as soon as we saw that our starting number didn’t change in this division, we realized we must have divided it by one. Four divided by one equals four. Our missing number is one.
Complete the following: Two divided by what equals one.
In this question, we’re given a division with a missing number. What do we divide two by to give us the answer one? Well, we know that division and multiplication are inverse operations; they’re opposites. So to help find a missing number in a division like this, we can think about the multiplication fact that goes with it, the opposite of it. So instead of asking ourselves, “What do we divide two by to give us the answer one?”, we can work backward and ask ourselves, “What do we multiply one by to give us the answer two?” Well, this is quite a simple multiplication fact, isn’t it? One times two equals two. And if we know there’s one lot of two in two, if we divide to by itself, we’re going to get the answer one.
Another way we know this is true is because we know a rule that applies every time a number is divided by itself. Any number divided by itself equals one. And so, because our division showed the answer one, we knew that the first number must be divided by itself. Two divided by two equals one. Our missing number is two.
Now, if you remember, the title of this video was “Division Rules for Zero and One.” And although we’ve looked at the rules for dividing when there’s a number one involved, it’s now time to think about zero. What happens when we divide using a zero? Do you remember that cartoon we used of the two characters falling down a black hole because one of them had just tried to divide by zero. Now this was just a little bit of fun. But you know, dividing by zero can actually make your head spin a little bit. So what we’ll do is take this really slowly to try to understand it. Perhaps this time around, it might make sense to start with the rule.
This rule is really easy to remember, perhaps not so easy to understand, but it’s definitely easy to remember. And that’s that we can’t divide by zero. Let’s have a go at explaining why not, and we’ll try 100 divided by zero as an example. So here are 100 counters. And if we split them into groups of 100, there’d only be enough groups for one person. If we divide 100 into groups of 50, then two people would get a group of 50 each. And then just as one more example, if we split our 100 counters into groups of one, 100 people would all get a counter each. Now looking at these divisions can help us when it comes to thinking about dividing by zero.
Notice how as the number we’re dividing by goes down, the answer to each division goes up. And so we’d expect the answer to 100 divided by zero to increase again. But if we stop and think about it, 100 counters split into groups of zero means that one person could come up and then we could share out to them zero counters, but we’d still have just as many counters as we had to start with. And this would still be the case if someone else came along or even another 100 people. Hundreds and hundreds, even thousands of people, could knock at our door asking for zero counters and we’d still be able to do this because zero is nothing. It doesn’t affect our 100 counters at all. There just isn’t a way we can split up 100 into zero.
Some other things that we can use to help us understand that we can’t divide by zero are multiplication facts. For example, if we’re asked to do the impossible and to divide 15 by zero, we’d think about the inverse operation and we’d ask ourselves, “What number do I multiply by zero to give the answer 15?” And then hopefully we’d start to frown and say to ourselves, “This isn’t possible.” If each plate that we have has zero bananas on it, how many of these empty plates would we need to have 15 bananas? Well, it makes no sense, does it? We just can’t answer this. We could have half a million empty bites of bananas and we’d still wouldn’t have 15 to eat. We might as well cross out calculations like this. They just don’t make sense.
But what if we have a division that starts with zero? Is there a rule for this? What about zero divided by 10 for example? Here’s a picture of zero sheep. Some of you might think it looks more like zero elephants, but It’s definitely zero sheep. You need to look more closely. Now what if we take our zero sheep and divide them into 10 equal groups? Here we are. What do you mean you can’t see them? Of course there aren’t going to be any sheep in these groups are there because we had zero sheep to begin with. Zero divided by 10 equals zero. In fact, we could have shared zero into any number of groups. The answer is always going to be the same.
And this brings us on to our second rule to do with dividing by zero. And that’s that if we start with zero and we divide it by any number at all, it always results in zero. Let’s practice what we’ve learned about dividing with zero then.
Complete the following: What divided by 12 equals zero.
This might seem like an unusual division because it ends with the answer zero. And it’s not often that we see a division that ends in zero. At the start of this number sentence, we can see we’ve got a missing number that we need to complete. What number if we divide it by 12 would give an answer zero? To help us find this missing number, we can start at the end of this division and work backwards, because we know that the inverse or the opposite of division is multiplication. If we divide a number by 12 and we have zero lots of 12, then to find that missing number, we just need to find zero lots of 12.
We can use our knowledge of multiplying by zero to help us here because we know that any number multiplied by zero equals zero. And that’s how we know our missing number must be zero. This reminds us of a fact we know to do with working with zero in divisions. Dividing zero by any number at all always results in zero. And as we know our answer is zero, we know the number we divided by 12 to get there must also be zero. Zero divided by 12 equals zero. Our missing number is zero.
So what have we learned in this video? We’ve learned how to model what happens when we divide one or zero and also when we divide a number by one or zero. We’ve learned these facts. Any number divided by one equals itself, and any number divided by itself equals one. Also, zero divided by any number is zero. And dividing by zero isn’t possible.
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# COMPLETE ADDITION AND SUBTRACTION EQUATIONS WITH INTEGERS
Complete addition and subtraction equations with integers :
To complete the addition and subtraction equations with integers easily, we can apply the concept of solving one step equations.
• Let "x" be the missing integer.
• In order to get the missing integer, we have to isolate "x"
• By using inverse operations, we can get rid of the addends.
• Inverse operations means, if the missing integer is added by another number then we have to subtract the same number on both sides.
• If the missing integer is subtracted from another number then we have to add the same number on both sides.
From this we will get the value of "x". That is the missing integer.
Let us see some example problems based on the above concept.
## Complete addition and subtraction equations with integers - Examples
Example 1 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
x + 20 = 17
In order to isolate x, we have to subtract 20 on both sides.
x + 20 - 20 = 17 - 20
x = -3
Hence the missing number is -3.
Example 2 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
-7 + x = 17
In order to isolate x, we have to add 7 on both sides.
-7 + x + 7 = 17 + 7
x = 24
Hence the missing number is 24.
Example 3 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
x + 5 = -10
In order to isolate x, we have to subtract 5 on both sides
x + 5 - 5 = -10 - 5
x = -15
Hence the missing number is -15.
Example 4 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
-12 + x = -19
In order to isolate x, we have to add 12 on both sides
-12 + x + 12 = -19 + 12
x = -7
Hence the missing number is -7.
Example 5 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
-18 + x = -20
In order to isolate x, we have to add 18 on both sides
-18 + x + 18 = -20 + 18
x = -2
Hence the missing number is -2.
Example 6 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
-11 + x = -17
In order to isolate x, we have to add 11 on both sides
-11 + x + 11 = -17 + 11
x = -6
Hence the missing number is -6.
Example 7 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
-11 + x = -17
In order to isolate x, we have to add 11 on both sides
-11 + x + 11 = -17 + 11
x = -6
Hence the missing number is -6.
Example 8 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
40 - x = 28
In order to isolate x, we have to subtract 40 on both sides
40 - x - 40 = 28 - 40
-x = -12
Multiplying by negative on both sides.
x = 12
Hence the missing number is 12.
Example 9 :
Find the missing integer that makes the following statement true.
Solution :
Let "x" be the missing number
x - 5 = 23
In order to isolate x, add 5 on both sides
x - 5 + 5 = 23 + 5
x = 28
Hence the missing number is 28.
Solution :
Let "x" be the missing number
x - 25 = -17
In order to isolate x, we have to add 25 on both sides
x - 25 + 25 = - 17 + 25
x = 8
Hence the missing number is 8.
After having gone through the stuff given above, we hope that the students would have understood "Complete addition and subtraction equations with integers".
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Factors that 19 space the list of integers that can be evenly separated into 19. Over there are as whole 2 components of 19 among which 19 is the biggest factor and also 1 and 19 are positive factors. The element Factors and also Pair factors of 19 space 1, 19, and (1, 19) respectively.
You are watching: What is the prime factorization of 19
Factors of 19: 1 and 19Negative determinants of 19: -1 and -19Prime factors of 19: 19Prime administrate of 19: 19Sum of components of 19: 20
1 What room the components of 19? 2 How to Calculate determinants of 19? 3 Factors of 19 by element Factorization 4 Factors the 19 in Pairs 5 Important Notes 6 FAQs on components of 19
What room the factors of 19?
A factor can be characterized as a number that totally divides a offered number. The components of 19 are the numbers that specifically divide 19.
Since 19 is a prime number, it has actually only 2 factors; they are 1 and 19.
How to calculate the factors of 19?
Factors of 19 space the numbers that multiply together to offer the result as 19. Can you think of every such pairs of numbers? As mentioned before, only one together pair exists because that the number 19. 19 is a prime number.
So, 19 can be created as 19 = 1 × 19
The number 19 has only 2 factors, the number itself and 1.
Hence, the components of 19 are 1 and 19.
Explore factors using illustrations and interactive examples.
Factors of 19 by element Factorization
Prime administer is a way of expressing a number as a product of its prime factors. The number 19 has actually a distinctive factorization because this is the only method it can be expressed: 19 = 1 × 19. This is the prime administrate of 19.
Factors the 18 have the right to be found similarly. Let"s view how:
When we write a number together a product the its factors, we say that the number has actually been factorized.
Let"s see a couple of examples of prime factorization.
Prime administrate of 18
So, the prime determinants of 18 room 3 and also 2.
We can write 18 = 3 × 3 × 2
Prime administrate of 20
So, the prime components of 20 are 2 and 5.
We can write 20 = 2 ×2 × 5
Factors the 19 in Pairs
Factor pairs of 19 are the bag of determinants that, once multiplied together, give the product same to the number 19. Choose all other primes better than 2, 19 has no determinants apart native one and itself. So, it has a distinct factor pair, the is, 1 and also 19.
Important Notes:
The number 19 has actually only 2 factors, 1 and also the number itself.The element factorization of 19 is 19 = 1 × 19The number 19 has actually only one element pair.
Example 2: Jacky"s teacher gave him a list of numbers: 23, 19, 20, 18, and also 26. The teacher asked him a few questions as presented below. I m sorry is the biggest prime number on the list? how many usual factors do 20 and 18 have? i m sorry number has the biggest prime aspect in the list?
Solution:
There are only two prime numbers in this list, which are 19 and 23.
23 is a element number since the components of 23 are 1 and also the number itself.
So, 23 is the largest prime number ~ above the list.
Let"s calculation the components of 20 and also 18.
Factors of 20: 1, 2, 4, 5, 10, and 20
Factors the 18: 1, 2, 3, 6, 9, and 18
So, number 20 and 18 have 2 common factors.
Since 23 is the largest prime number in the list, it has actually the biggest prime factor, i.e., 23.
Example 3: find the product of all the prime components of 19.
Solution:
Since the prime determinants of 19 are 19. Therefore, the product the prime determinants = 19 = 19.
Show solution >
go to slidego come slidego to slide
FAQs on factors of 19
What room the components of 19?
The components of 19 space 1, 19 and also its negative factors are -1, -19.
What is the Greatest typical Factor of 19 and 14?
The factors of 19 are 1, 19 and also the factors of 14 space 1, 2, 7, 14. 19 and 14 have actually only one usual factor which is 1. This suggests that 19 and also 14 space co-prime.Hence, the Greatest typical Factor (GCF) the 19 and also 14 is 1.
What space the common Factors of 19 and 17?
Since the determinants of 19 space 1, 19, and also factors that 17 are 1, 17. Hence, 19 and 17 have actually only one common factor which is 1. Therefore, 19 and also 17 are co-prime.
See more: How Many Ounces Is Half Pint ? How Many Ounces In Half A Pint
What are the Prime components of 19?
The prime variable of 19 is 19.
What is the amount of the components of 19?
Factors of 19 are 1, 19 and, the sum of every these determinants is 1 + 19 = 20
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July 15th, 2015
Students explore how the distance formula works on coordinate axes and discover how the Pythagorean Theorem relates to the distance formula.
### Choreo Graph Activity 6: Traveling Distances
Calculating Distance Using Coordinates in a Coordinate Plane
### Expected Activity Time
Part 1: Distance Formula Exploration (20 minutes)
Part 2: Distance Formula and Pythagorean Theorem Exploration (20 minutes)
Materials and Prep
• Traveling Distances Student Sheets
• iPad with Choreo Graph app
• Wifi access for sharing to other iPads and online project space
Introducing the Activity
Create a story problem that involves your animated character traveling across a distance. An example may be that you’re character is going across town or to a friend’s house. Use Choreo Graph tools to gather data and calculate the distance traveled from point to point.
### To Do
Part 1: Distance Formula Exploration (20 minutes)
When objects in Choreo Graph slide around the screen on the coordinate plane, they are moving across a distance. This distance can be found using the coordinates from one point to the next and applying the distance formula:
Have students:
• Create a story problem that involves a virtual character traveling across a distance (e.g., going across town or to a friend’s house).
• Use the distance formula to calculate the distance traveled from point to point.
• Draw the coordinate axes with a starting point and then plot the next point.
• Count over horizontally and then up or down vertically and note how many units they go in each direction.
• Compare these counts to what you get by using the distance formula. Have students share what they notice.
Part 2: Distance Formula and the Pythagorean Theorem (20 minutes)
In this activity, students use both the distance formula and the Pythagorean Theorem to locate the distance their characters travel from one point to another.
Have students:
• Practice with the sample below, and then have them try to figure out the distance traveled using both formulas.
• Use both horizontal and vertical movement in their animation.
• Find the length of the hypotenuse (c) using both formulas.
• What do they notice using this approach?
• Exchange story animations with a classmate and find their character’s distance traveled using both formulas.
### Discussion
After students create their paths and calculate distances traveled in different ways, discuss:
:
• How do results from the distance formula compare to the units moved on the grid?
• What if your animated character only moves vertically, but not horizontally? Does the distance formula still work?
• What happens if your object only moves in the horizontal direction, but not vertical? Is there still a distance traveled? What do you notice?
Listen for:
• I noticed that I got the same thing when I drew a triangle between the points and did the Pythagorean theorem.
• When you count squares vertically it is the same as when you calculate y2-y
• Counting squares across you get the same thing as when you calculate x2– x1 using the distance formula.
Extensions and Inquiring Further
There are many possibilities to extend this activity in creative ways. For example, compare how someone travels a city street versus a bird flying through the air. On city streets that are often in grids, travelers have to take the legs of right triangles to get somewhere, but birds can fly direct distances (hypotenuse). Have students puzzle out what are more effective ways to determine the distance travelled.
Name: __________________________ Date: _____________
Part 1: Distance Formula Exploration
When objects in Choreo Graph slide around the screen on the coordinate plane, they are moving across a distance. This distance can be found using the coordinates from one point to the next. Here is the distance formula:
1. Create a story problem that involves your animated character traveling across a distance. An example may be that you’re character is going across town or to a friend’s house. From point to point, use the distance formula to calculate the distance traveled.
Distance Point 1 Coordinates Point 2 Coordinates X answer squared Y answer squared X + y Square root/ distance Move 1 Move 2 Move 3
Name: __________________________ Date: _____________
Part 1: Distance Formula Exploration
1. Draw a coordinate axis with your starting point and then plot the next point. When you count over horizontally and then up or down vertically, make note of how many units you travel in each direction, and then compare that to what you did in the distance formula. What do you notice? (Hint you can also use your translation tool for coordinates).
1. On the Choreo Graph stage, count over horizontally and vertically and note how many units the object travels in each direction. Compare that number to the values you get when using the distance formula.
Name: __________________________ Date: _____________
### Part 1: Distance Formula Exploration
Reflection Questions:
Answer the following questions in complete sentences.
1. What happens if your object only moves vertically, but not horizontally? Does the distance formula still work?
1. What happens if your object only moves in the horizontal direction, but not vertical? Is there still a distance traveled? What do you notice?
Name: __________________________ Date: _____________
Part 2: Distance Formula and Pythagorean Theorem
In this activity, you are going to use both the distance formula and the Pythagorean Theorem to locate the distance your animated character travels. Practice with the sample below.
Name: __________________________ Date: _____________
Part 2: Distance Formula and Pythagorean Theorem
After you have practiced with the sample, try to locate the distance for your own animated character by:
1. Creating two movements that form a right angle.
1. Challenging yourself to find the length of the hypotenuse ( c ) using both formulas.
1. Swapping your animations with a classmate and asking them to try to find the distance travelled using both formulas.
Name: __________________________ Date: _____________
Part 2: Distance Formula and Pythagorean Theorem
Data Sheet
Find the distance of the hypotenuse ( c ) using the distance formula and the Pythagorean Theorem.
point 1coordinates point 2coordinates x answer squared y anwer squared x + y square root/ distance
Name: __________________________ Date: _____________
Part 2: Distance Formula and Pythagorean Theorem
Reflection Questions:
1. After calculating distances using each formula, how do the distances compare to each other?
2. Which formula was most efficient to use?
### App Features
Begin by entering Make Some Moves. In Build mode, students will:
• Take pictures
• Trace and cut out parts of your photo that you want to animate
Add graphic or musical elements below:
In Animate mode, students will use:
Graph ControllersChoreo Graph uses key frames much like other movie editing software. At each point in the key frame, the student can set how each part of the animation rotates over time. Each node on the line graph below the stage represents the position of that part at a specific time. Stretch the points up and down to set the degrees of rotation. The steeper the line on the graph, the faster the part moves. Students can also set the location of their animated parts by selecting the part they want to move, choosing a node on the line graph or moment in the animation, and then dragging the part to a new position on the stage. They can set the location of parts for the entire animation sequence by repeating this process.
Toggle on math tools to notice:
Degrees each part has rotated
Path the main parts moved
Coordinates for location of each part
Apps used
Duration: 0-20 mins
Grade(s): e.g. 8, 9 or 5-6
Prep: Easy
#### Big Idea
Using the translation and grid tools in the Choreo Graph app, students create a travelling character and calculate the distance between two points in the coordinate plane. To calculate the distance travelled, students use both the distance formula and the Pythagorean theorem.
For more background information, visit this link: Distance and Pythagorean Theorem Information
#### Learning Objectives
Students will:
• Draw coordinate axes and explore the distance formula.
• Learn about how the Pythagorean Theorem relates to the distance formula.
• Engage in computational thinking as they work with graph controllers and key frames to create, develop, and perfect their animated scenes.
Common Core State Standards-Math
Geometry
8.G.B.8. Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.
Common Core State Standards-Math
Mathematical practices.
MP2: Reason abstractly and quantitatively.
MP4: Model with mathematics.
Students outline their dance using the angles of rotation and the coordinate notation for the translation.
#### Vocabulary
Vocabulary/Terms
• Distance Formula:
• Pythagorean Theorem:
a2+b2=c2
#### Device Strategies
Single-device implementation
Before class meets, create a character in the app that travels from a starting point to an ending point. For the first example be sure there is vertical as well as horizontal movement. Create a screen shot of the path the character travels with the grid and translation function on. Invite students to calculate the distance travelled using either formula. Play the animation and have the students check if their answers match the real distance travelled. Change the position of the character on screen, and have students work out the distance traveled again. Do they see any differences?
Multiple-device implementation
Working in pairs, have students generate paths for their animated characters to travel and create a story around it. Then ask students to calculate distance travelled using both equations. Use the calculations to compare and contrast.
#### Tips & Ideas
Getting Started: Have students play with translations where their figure slides from a starting point to an ending point including both vertical and horizontal movement. Have them take note of the coordinates with translation tool toggled on. Making formulas concrete: The most common mistake made when using the Distance Formula is to accidentally mismatch the x-values and y-values. Have students repeat counting each set of movements of their animated character. By doing so, they can begin to notice that counting units horizontally is the same as calculating x2- x1 in the distance formula (same with y2-y1. In the vertical direction).
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# clarification on the formula $\frac{n!}{(n-k)!}$
$\dfrac{n!}{(n-k)!}$ is used in order to find non-repetitive lists of length $k$ given $n$ possible symbols.
For example: find the number of non-repetitive lists of length five that can be made form the symbols $1,2,3,4,5,6,7,8.$ I understand that this can be solved by this method (multiplication principle): $8 \times 7 \times 6 \times 5 \times 4=6720$.
I don't understand this method of solving the problem: $\dfrac{8!}{(8-5)!}=\dfrac{8!}{3!}= \dfrac{40{,}320}{6}=6720$.I am having a hard time understanding this intuitively, why is $k$ subtracted from $n$ in the denominator? The subtraction gives $3!$ and that is not a factorial representing a length of five. It seems like this makes us find how many length of $3$ are there....
Thank you.
• Note that $$\frac{n!}{(n-k)!} = (n-k+1) \cdot \dots \cdot n = \prod_{i=n-k+1}^n i$$ Dec 3, 2014 at 0:42
• Look at your method: $8\times 7\times 6\times 5\times 4$ is the same as $\frac{8!}{3} = \frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1}$. The idea is to represent $n(n-1)(n-2)...$ in terms of factorials. Dec 3, 2014 at 0:42
You have $n$ choices for the first element in the list, $n-1$ for the second, ..., $n-(k-1)$ for the $k$-th. So the number is $$n \cdot (n-1) \cdots (n-(k-1)) = n \cdot (n-1) \cdots (n-(k-1)) \frac{(n-k)!}{(n-k)!} = \frac{n!}{(n-k)!}$$
• Ok, thanks +1. But I don't understand why you are dividing...how does a sequence of multiplication lead to division? Dec 3, 2014 at 1:52
• @user437158, see my edited answer.
– lhf
Dec 3, 2014 at 9:53
Consider the $n!$ permutationsof $n$ elements. They are all different. But you only care about the first $k$ elements, so the permutations that differ only in the order of the last $n-k$ elements (there are $(n-k)!$ of them) must be counted once only.
ab|cd ab|dc
ac|bd ac|db
ba|cd ba|dc
bd|ac bd|ca
ca|bd ca|db
cd|ab cd|ba
da|bc da|cb
db|ac db|ca
dc|ab dc|ba
In the example there are $4!=24$ permutations, each arrangement of two letters appearing $(4-2)!=2$ times (last $2$ letters ignored), hence after regrouping, $$\frac{4!}{(4-2)!}=\frac{4.3.2.1}{\ \ \ \ \ \ 2.1}=12.$$
Said differently, the division "undoes" the excess multiplies.
It is interesting to note that if you also don't care about the order of the first $k$ elements, you need to divide by $k!$ as well and obtain the fomula for combinations $$\frac{n!}{k!(n-k)!},$$ which, by some magic, always yields integer values.
• Try on your own with $k\ne n-k$ if you still need to be convinced.
– user65203
Dec 3, 2014 at 10:20
It is well known that $$\binom{n}{k}=\frac{n!}{(n-k)!k!}$$ gives the number of ways to choose $k$ items from a list of $n$ items. Now this does not take order into account. For example, if you have the numbers $1$, $2$ and $3$, and you want to choose two numbers between them, you will have $$1,2\\ 1,3\\ 2,3$$ as possibilities. That's $\frac{3!}{1!2!}=3$ possibilities. The formula you give would be $2!$ times $3$, that is, $6$ possibilities. That takes the permutations of the $k$ elements into account. For each set of $2$ elements between the $3$ elements, you can permute the two elements in $2!$ fashions.
In general, if you want to know how many ways there are to take $k$ elements from a set of $n$ elements, taking order into account, you will have $$\binom{n}{k}\times k!$$ choices, that is, $$\frac{n!}{(n-k)!}$$ choices.
• This is correct logic, but may be a bit circular. The binomial coefficient can be derived from the permutation formula. So we start with $P(n, k) = n!/(n-k)!$. We then divide out by the Symmetry group on the $k$ elements (the group of permutations) to get $\binom{n}{k} = P(n, k) = \frac{n!}{ |S_{k}| (n-k)!} = \frac{n!}{k! * (n-k)!}$. Dec 3, 2014 at 0:51
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Question
Sat March 31, 2012 By: Apoorv Kumar
# Urn A has 3 red and 2 white marbles, urn B has 2 red and 5 white marbles. An urn is selected at random ; a marble is drawn, its colour noted and put into the other urn , then marble is drawn from the second urn. Find the probability that 1)both marbles drawn are red 2)both marbles drawn are white 3)marbles drawn are of same colour 4)second marble drawn is red 5)marbles drawn are of different colours 6)second marble drawn is white
Sat April 21, 2012
The probability of selecting any urn (P) = 1/2.
Let A be the event of chosing urn A
Let B be the event of chosing urn B
The probability of drawing a red marble from urn A (R/A) = 3/5.
The probability of drawing a white marble from urn A (W/A) = 2/5
The probability of drawing a red marble from urn B (R/B) = 2/7
The probability of drawing a white marble from urn B (W/B) = 5/7
a. both marbles are red = (probability of selecting urn A *(probability selecting red marble from urn A)*(probability of selecting red marble from urn B after the marble selected from urn A is also put in the urn B) +(probability of selecting urn B *(probability selecting red marble from urn B)*(probability of selecting red marble from urn A after the marble selected from urn A is also put in the urn A)
Probability = (1/2*3/5*3/8 + 1/2*2/7*4/6)
= 0.1125+0.095
= 0.21
b. both marbles are white = Applying the same logic, we have
Probability = (1/2*2/5*6/8)+(1/2*5/7*3/6) = 0.33
c. both marbles are drawn of same color, so either both marbles are red or both marbles are white. Hence, probability = 0.21+0.33 = 0.54
d. second marble drawn is red, so the first marble drawn from either of the 2 urns can be either red or white. Hence, probability = (1/2*3/5*3/8 + 1/2*2/7*4/6)+(1/2*2/5*2/8)+(1/2*5/7*3/6) = 0.21+0.228 = 0.44
e. marble drawn are of different color – So, if first marble drawn is white, then the second is red and so on.
Probability = (1/2*2/5*2/8)+(1/2*5/7*3/6) + (1/2*3/5*5/8)+(1/2*2/7*2/6) = 0.228 +0.235 = 0.46
f. second marble drawn is white – So, the first one drawn can be either red or white. Hence,
Probability = (1/2*2/5*6/8)+(1/2*5/7*3/6)+ (1/2*3/5*5/8)+(1/2*2/7*2/6) = 0.33 + 0.235 = 0.565
Thank you
Related Questions
Sat May 13, 2017
Fri May 12, 2017
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Wiki
# SBT Toán lớp 5 trang 8,9,10,11: Ôn tập: Phép nhân và phép chia hai
## Introduction
In this article, we will review the concepts of multiplication and division of fractions in grade 5 mathematics. We will cover various exercises from the textbook and provide step-by-step solutions. By the end of this article, you will have a better understanding of these mathematical operations.
## Exercise 1: Multiplication of Fractions
Let’s begin with the exercise on page 8 of the grade 5 mathematics textbook. Solve the following multiplication problems:
a) 25×37; 47×34; 49×310; 2×518.
b) 38:75; 34:910; 78:2; 6:23.
c) 13×35×59; 1516:38×34.
Solution:
a) To solve 25×37, we multiply the numerators (25×37) and the denominators (1×1). The result is 635.
For 47×34, we can break down the calculation into smaller steps. First, we multiply the numerators (4×37) and the denominators (1×4). The result is 148. Then, we multiply 148 by the remaining numerator 4. The final result is 592.
Similarly, for 49×310, we multiply the numerators (4×39×10) and the denominators (1×1×2). The result is 215.
Finally, for 2×518, we multiply the numerators (2×52×9) and the denominators (1×1). The result is 59.
b) To solve 38:75, we multiply the numerators (38×57) and the denominators (1×5). The result is 1556.
For 34:910, we multiply the numerators (3×104×9) and the denominators (1×1). The result is 56.
Similarly, for 78:2, we multiply the numerators (7×16) and the denominators (1×2). The result is 716.
Finally, for 6:23, we multiply the numerators (6×32) and the denominators (1×1). The result is 182.
c) To solve 13×35×59, we multiply the numerators (1×3×53×5×9) and the denominators (1×1×1×1×1). The result is 19.
For 1516:38×34, we multiply the numerators (15×8×38×2×3×4) and the denominators (1×1×1×1). The result is 158.
## Exercise 2: Division of Fractions
Let’s move on to the exercise on page 9 of the grade 5 mathematics textbook. Solve the following division problems:
a) 35×827×53.
b) 719×13+719×23.
Solution:
a) To solve 35×827×53, we multiply the numerators (35×53×827) and the denominators (1×1). The result is 827.
b) For 719×13+719×23, we calculate the multiplication first. The result is 719×1=719.
## Exercise 3: Word Problems
In exercise 34 on page 9 of the grade 5 mathematics textbook, we have word problems that involve multiplication and division of fractions. Let’s solve them together:
a) x×12=13.
b) x:45=52.
c) 29:x=23.
Solution:
a) To solve x×12=13, we divide both sides of the equation by 12. The result is x=13:12=23.
For x:45=52, we multiply both sides by 45. The result is x=52×45=2.
Similarly, for 29:x=23, we multiply both sides by x. The result is x=29:23=13.
## Exercise 4: Area Calculation
Moving on to exercise 35 on page 9 of the grade 5 mathematics textbook, we will calculate the area of a rectangular glass.
A rectangular glass has a length of 45 and a width of 12. Let’s calculate its area.
Solution:
The area of the rectangular glass is given by the formula length × width. So, 45×12=540.
Therefore, the area of the glass is 540 square units.
## Exercise 5: Real-life Problem
In exercise 36 on page 9 of the grade 5 mathematics textbook, we have a real-life problem involving mixing and measuring liquids. Let’s solve it together:
We mix 12 liters of grape juice with 74 liters of water to make grape juice. Each cup can hold 14 liters of juice. How many cups of grape juice can we pour?
Solution:
To find the total volume of the grape juice, we add 12 liters of grape juice to 74 liters of water. The result is 12+74=86 liters.
Next, we divide the total volume by the capacity of each cup, which is 14 liters. So, 86:14=6.
Therefore, we can pour 6 cups of grape juice.
## Conclusion
In this article, we reviewed the concepts of multiplication and division of fractions in grade 5 mathematics. We solved various exercises from the textbook, providing step-by-step solutions. Remember to practice these concepts regularly to enhance your mathematical skills. For more detailed solutions and additional exercises, visit Kienthucykhoa.com.
Keep up the good work, and happy learning!
### Kiến Thức Y Khoa
Xin chào các bạn, tôi là người sở hữu website Kiến Thức Y Khoa. Tôi sử dụng content AI và đã chỉnh sửa đề phù hợp với người đọc nhằm cung cấp thông tin lên website https://kienthucykhoa.edu.vn/.
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UPSKILL MATH PLUS
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### Theory:
A term is either a single number or a variable or multiplied together by a number or a variable. In other words, the terms of expression are the parts that are separated by an algebraic operation.
Example:
$$5a$$ $$-$$ $$3$$. (In this case, $$5a$$ is the first term and, $$3$$ is the second term. So, we have $$2$$ terms.)
A coefficient is a number used for the multiplication of a variable.
Example:
The expression $$5a$$ $$-$$ $$3$$ has only one coefficient. (In this case, $$5$$ is coefficient of $$a$$. where $$5a$$ means $$5$$ times $$a$$. So $$5$$ is the numerical coefficient, $$a$$ is a variable, and $$3$$ is constant.)
Example:
In the following expression, list out the terms and coefficients:
$$4x^3y^2+2x^2-3x+4$$.
Terms:
• $$4x^3y^2$$ is the first term.
• $$2x^2$$ is the second term.
• $$-3x$$ is the third term.
• $$4$$ is the fourth term.
Coefficients:
• $$4$$ is the coefficient of $$x^3y^2$$.
• $$2$$ is the coefficient of $$x^2$$.
• $$-3$$ is the coefficient of $$x$$.
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