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### Course: Precalculus (Eureka Math/EngageNY)>Unit 4
Lesson 1: Topic A: Lessons 1-2: Special values and properties of trigonometric functions
# Sine & cosine identities: symmetry
Sal finds several trigonometric identities for sine and cosine by considering horizontal and vertical symmetries of the unit circle. Created by Sal Khan.
## Want to join the conversation?
• Do I have to memorize all of these formulas? Is it important enough to know?
(37 votes)
• No, you can get through a lot of math without memorizing, but it just takes a lot longer to do the problems. Sometimes it is just plain easier to memorize a couple of formulas than to try to dig back to the basics and reconstruct the formulas.
In the case of the symmetry relationships, it is a great time-saver to know these. There are ways of reconstructing the information if you forget. One way is to memorize the signs for the different trig functions in the four quadrants. The way I remind myself of these formulas is to think of a point in the first quadrant (both x and y will be positive, so all sine and cosine values will be positive, as will tangent).
Then I think of a point in the second quadrant (x will be negative, since all the values for x will be less than zero, and y will be positive. As a result, sine will be positive, but cosine will be negative, and all tangent values will be negative.) In the third quadrant, all x and y values will be negative, so all sine and cosine values will be negative. Tangent will be positive because a negative divided by a negative is positive.) The final quadrant is the fourth quadrant, and there, all x values are positive, but all y values are negative, so sine will be negative, cosine will be positive and tangent values will be negative.
So, you CAN recreate the information by logic. In the meantime, others can use the symmetries and be done with the problem and maybe with the next problem as well. Also, there are some ways that the questions can be asked that make it difficult to use this method, and if you are not very conversant with unit circle and translating points to sine and cosine, then you may have some tough slogging ahead. Knowing this set of symmetries becomes handy in Physics and many other applications.
(166 votes)
• What is a radian? I think I have some vague description of a ratio and something with PI but I don't honestly remember.
(40 votes)
• A radian is the angle you get with its vertex at the center of a circle, and the two line segments that contain the angle each connect the vertex to an endpoint of an arc on the circle, and the arc length is equal to the radius of the circle. The arc is opposite the angle. Converting to degrees, a radian is approximately 57.3 degrees; pi radians equals 180 degrees.
(102 votes)
• so, this is basic trig right? still amaze me with its complications! i still don't get it :( any websites that helps for all basic trig ?? and college math ... thanks!
(1 vote)
• Honestly, you don't need any other websites. I have been looking at the Khan Academy for Trig, and it does a great job on explaining. Maybe just reviewing it over and over will help.
(86 votes)
• I must say that was a lot to take in for one video, usually there's the odd "There's this reason" or "There's that" , not much made alot of sense to me with without examples or references .
(22 votes)
• I thought it was mostly review, but just pulled together as a different way of looking at the symmetries around a unit circle. So, what exactly was confusing? There are a number of more basic videos on trigonometry that may help, but I am not sure where to recommend that you start. Then after you are familiar with the basics, it would be easier to see how the concepts fit together. The key was showing how to name the reflected angles, and then how to construct the cosine and sine statements for each angle, and finally how to relate the pairs. And it all starts with the unit circle, so if you are hazy on that, it would be a great place to start your review.
For example, let's say that we are looking at an angle of π/3 on the unit circle. The value of `sin (π/3) is ½√3` while `cos (π/3) has a value of ½`
The value of
`sin (-π/3) is -½√3` while `cos (-π/3) has a value of ½`
Already we can see that `cos theta = cos -theta` with this example.
And look at that: `sin -theta = -sin theta` just like Sal said.
If we go through all the other reflected angles, with this specific example of an angle, we will get the same relationships that Sal just walked us through.
One other example is π - theta
This looks mysterious until you realize that in our situation, theta equals π/3
and π - π/3 gives us an angle of ⅔π (In degrees, that is 120 degrees)
cos ⅔π is going to be a negative value because it is in the second quadrant, and from working with unit circles, we should remember that it is evaluated as cos π/3 except for the negative. So, cos (π - π/3) = - cos π/3 and `cos π/3 = - cos (π - π/3)`
Basically, if you have these symmetries, you have a multitude of sine and cosine values as long as you know what sine of theta is and cosine of theta is.
It may help you to continue around the circle with common angles like π/6 and π/4 (not to mention the rest of the π/3 gang). That may help you to see how the symmetrical relationships are great time savers.
(19 votes)
• why sin(-theta)=-sin(theta)
(11 votes)
• Sin(theta) produces a positive value on the y axis.
-Sin(theta) or, the negative of Sin(theta) takes that value, and multiplies it by -1, giving us a negative value on the y axis.
That same negative value on the y axis, can be produced with Sin(-theta)
(10 votes)
• What is SOH CAH TOA?
(0 votes)
• (S)ine
(O)pposite
(H)ypotenuse
(C)osine
(A)djacent
(H)ypotenuse
(T)angent
(O)pposite
(A)djacent (the non-hypotenuse adjacent)
The first letter in each of the letter triplets names the ratio the second letter of each triplet is the numerator, the third letter of each triplet is the denominator.
(33 votes)
• at , wouldnt cos of theta be positive because the x is positive? and the sin be negative because the y is negative? why are they both negative?
(10 votes)
• It's only the angle, or theta, that's negative. If you were to simplify the negative angles, then cos(-θ) = cos(θ) and sin(-θ) = -sin(θ).
(7 votes)
• what is the diffrence between -data and +data
(3 votes)
• - Theta is when you rotate clockwise to the right.
+Theta is when you rotate counterclockwise to the left.
(7 votes)
• I know this might be a dumb question but is Theta?
(3 votes)
• theta is just the variable used when referring to an angle a lot of the time
(11 votes)
• Why, why, why was this video placed AFTER solving sinusoidal equations?
You literally CANNOT solve them without these identities, but they don't teach you said identities until after you spend an hour failing to solve those equations and moving on. Staff members, please consider repositioning this video in the curriculum.
(6 votes)
• A few times when I took KA's courses I noticed weird orderings like that. You should try making a request at their help center at Zendesk for your concern to get noticed.
Happy learning,
- Convenient Colleague
(1 vote)
## Video transcript
Voiceover:Let's explore the unit circle a little bit more in depth. Let's just start with some angle theta, and for the sake of this video, we'll assume everything is in radians. This angle right over here, we would call this theta. Now let's flip this, I guess we could say, the terminal ray of this angle. Let's flip it over the X and Y-axis. Let's just make sure we have labeled our axes. Let's flip it over the positive X-axis. If you flip it over the positive X-axis, you just go straight down, and then you go the same distance on the other side. You get to that point right over there, and so you would get this ray. You would get this ray that I'm attempting to draw in blue. You would get that ray right over there. Now what is the angle between this ray and the positive X-axis if you start at the positive X-axis? Well, just using our conventions that counterclockwise from the X-axis is a positive angle, this is clockwise. Instead of going theta above the X-axis, we're going theta below, so we would call this, by our convention, an angle of negative theta. Now let's flip our original green ray. Let's flip it over the positive Y-axis. If you flip it over the positive Y-axis, we're going to go from there all the way to right over there then we can draw ourselves a ray. My best attempt at that is right over there. What would be the measure of this angle right over here? What was the measure of that angle in radians? We know if we were to go all the way from the positive X-axis to the negative X-axis, that would be pi radians because that's halfway around the circle. This angle, since we know that that's theta, this is theta right over here, the angle that we want to figure out, this is going to be all the way around. It's going to be pi minus, it's going to be pi minus theta. Notice, pi minus theta plus theta, these two are supplementary, and they add up to pi radians or 180 degrees. Now let's flip this one over the negative X-axis. If we flip this one over the negative X-axis, you're going to get right over there, and so you're going to get an angle that looks like this, that looks like this. Now what is going to be the measure of this angle? If we go all the way around like that, what is the measure of that angle? To go this far is pi, and then you're going another theta. This angle right over here is theta, so you're going pi plus another theta. This whole angle right over here, this whole thing, this whole thing is pi plus theta radians. Pi plus theta, let me just write that down. This is pi plus theta. Now that we've figured out these have different symmetries about them, let's think about how the sines and cosines of these different angles relate to each other. We already know that this coordinate right over here, that is sine of theta, sorry, the X-coordinate is cosine of theta. The X-coordinate is cosine of theta, and the Y-coordinate is sine of theta. Or another way of thinking about it is this value on the X-axis is cosine of theta, and this value right over here on the Y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the X-coordinate is cosine of pi minus theta. That's what this angle is when we go from the positive X-axis. This is cosine of pi minus theta. And the Y-coordinate is the sine of pi minus theta. Then we could go all the way around to this point. I think you see where this is going. This is cosine of, I guess we could say theta plus pi or pi plus theta. Let's write pi plus data and sine of pi plus theta. Now how do these all relate to each other? Notice, over here, out here on the right-hand side, our X-coordinates are the exact same value. It's this value right over here. So we know that cosine of theta must be equal to the cosine of negative theta. That's pretty interesting. Let's write that down. Cosine of theta is equal to ... let me do it in this blue color, is equal to the cosine of negative theta. That's a pretty interesting result. But what about their sines? Well, here, the sine of theta is this distance above the X-axis, and here, the sine of negative theta is the same distance below the X-axis, so they're going to be the negatives of each other. We could say that sine of negative theta, sine of negative theta is equal to, is equal to the negative sine of theta, equal to the negative sine of theta. It's the opposite. If you go the same amount above or below the X-axis, you're going to get the negative value for the sine. We could do the same thing over here. How does this one relate to that? These two are going to have the same sine values. The sine of this, the Y-coordinate, is the same as the sine of that. We see that this must be equal to that. Let's write that down. We get sine of theta is equal to sine of pi minus theta. Now let's think about how do the cosines relate. The same argument, they're going to be the opposites of each other, where the X-coordinates are the same distance but on opposite sides of the origin. We get cosine of theta is equal to the negative of the cosine of ... let me do that in same color. Actually, let me make sure my colors are right. We get cosine of theta is equal to the negative of the cosine of pi minus theta. Now finally, let's think about how this one relates. Here, our cosine value, our X-coordinate is the negative, and our sine value is also the negative. We've flipped over both axes. Let's write that down. Over here, we have sine of theta plus pi, which is the same thing as pi plus theta, is equal to the negative of the sine of theta, and we see that this is sine of theta, this is sine of pi plus theta, or sine of theta plus pi, and we get the cosine of theta plus pi. Cosine of theta plus pi is going to be the negative of cosine of theta, is equal to the negative of cosine of theta. Even here, and you could see, you could keep going. You could try to relate this one to that one or that one to that one. You can get all sorts of interesting results. I encourage you to really try to think this through on your own and think about how all of these are related to each other based on essentially symmetries or reflections around the X or Y-axis.
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# 10.15: Volume of Prisms Using Unit Cubes
Difficulty Level: At Grade Created by: CK-12
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Practice Volume of Prisms Using Unit Cubes
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Remember Jillian's box from an earlier Concept?
Jillian is working on her special box and is wondering how much she can fit inside of it. The box is a rectangular prism and has the following dimensions: 7" x 6" x 4".
Can you use unit cubes to figure out the volume of the box? How?
This Concept is about identifying volume. You will learn one strategy for accomplishing this task.
### Guidance
In this Concept, we will look at the volume of prisms.
Volume is the amount of space inside a solid figure.
In this Concept, we will look at the volume of prisms.
These cubes make up a rectangular prism. The cubes represent the volume of the prism.
This prism is five cubes by two cubes by one cube. In other words, it is five cubes long, by two cubes high by one cube wide.
We can multiply each of these values together to get the volume of the rectangular prism.
5 ×\begin{align*}\times\end{align*} 2 ×\begin{align*}\times\end{align*} 1 = 10 cubic units
If we count the cubes, we get the same result.
The volume of the rectangular prism is 10 cubic units or units3\begin{align*}^3\end{align*}.
Identify the volume of each prism.
#### Example A
Solution: 10 cubes
#### Example B
Solution: 16 cubes
#### Example C
Solution: 9
Now back to Jillian's box. Here is the original problem once again.
Jillian is working on her special box and is wondering how much she can fit inside of it. The box is a rectangular prism and has the following dimensions: 7" x 6" x 4".
Can you use unit cubes to figure out the volume of the box? How?
If you noticed in the three examples above, each measurement indicated the number of unit cubes that would be needed. We can apply this to Jillian's box.
There are seven 1 inch cubes for the length.
There are six 1 inch cubes for the width.
There are four 1 inch cubes for the height.
7×6×4=168\begin{align*}7 \times 6 \times 4 = 168\end{align*}
Jillian's box will hold 168 unit cubes.
### Vocabulary
Here are the vocabulary words in this Concept.
Surface area
the outer covering of a solid figure-calculated by adding up the sum of the areas of all of the faces and bases of a prism.
Net
diagram that shows a “flattened” version of a solid. Each face and base is shown with all of its dimensions in a net. A net can also serve as a pattern to build a three-dimensional solid.
Triangular Prism
a solid which has two congruent parallel triangular bases and faces that are rectangles.
Rectangular Prism
a solid which has rectangles for bases and faces.
Volume
the amount of space inside a solid figure
### Guided Practice
Here is one for you to try on your own.
What is the volume of this figure?
How many cubes are in this figure? We can see that if we count all the cubes, that we have 48 cubes.
The volume of this prism is 48 cubic units or units3\begin{align*}\text{units}^3\end{align*}.
### Video Review
Here is a video for review.
### Practice
Directions: Find the surface area and volume of each prism.
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Directions: Identify each type of prism.
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### Notes/Highlights Having trouble? Report an issue.
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Please to create your own Highlights / Notes
### Vocabulary Language: English
TermDefinition
Net A net is a diagram that shows a “flattened” view of a solid. In a net, each face and base is shown with all of its dimensions. A net can also serve as a pattern to build a three-dimensional solid.
Rectangular Prism A rectangular prism is a prism made up of two rectangular bases and four rectangular faces.
Surface Area Surface area is the total area of all of the surfaces of a three-dimensional object.
Triangular Prism A triangular prism is a prism made up of two triangular bases and three rectangular faces.
Volume Volume is the amount of space inside the bounds of a three-dimensional object.
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# Prove that, for any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms.
For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3.
-
Suppose that you add up the consecutive integers $n,n+1,\dots,n+m$; call the sum $S$. Now add up the numbers again, but in reverse numerical order, and arrange the sums like this:
$$\begin{array}{c} n&+&(n+1)&+&\ldots&+&(n+m-1)&+&(n+m)&=&S\\ (n+m)&+&(n+m-1)&+&\ldots&+&(n+1)&+&n&=&S\\ \hline (2n+m)&+&(2n+m)&+&\ldots&+&(2n+m)&+&(2n+m)&=&2S \end{array}$$
Each column must have the same sum: each time you move one column to the right, the top number increases by $1$ and the bottom number decreases by $1$, so the total remains the same. There are $m+1$ columns, so the bottom line can be summarized as
$$(m+1)(2n+m)=2S\;,$$
and $$S=\frac{(m+1)(2n+m)}2=\underbrace{(m+1)}_{\text{nr. of terms}}\cdot\frac{2n+m}2\;.$$
Since the number of terms is odd, $m+1$ is odd. But $S$ is an integer, so $(m+1)(2n+m)$ must be divisible by $2$. What does that tell you about $2n+m$?
-
Hint $\rm\,\ m = 2n\!+\!1\:$ consecutive integers form a complete set of remainders mod $\rm\:m,\:$ so they are congruent mod $\rm\:m\:$ to the complete set of remainders $\rm\:0,\pm1,\pm2,\ldots,\pm n,\:$ which have sum $\equiv 0.$
Remark $\$ This provides an arithmetical interpretation of the reflection around the middle term mentioned in some other answers. Here it becomes the negation reflection $\rm\ x \,\to\, -x \,\ (mod\ m),\:$ by which we can pair up each element with its negation, forcing them to cancel out of the sum. Such use of reflections (or involutions) to pair-up terms frequently proves handy, e.g. see prior posts here on Wilson's theorem (in groups), esp. this one to start.
-
Hint: Suppose there are $2t+1$ terms. Let $m$ be the "middle" term. Argue that the sum is $m$ times $2t+1$.
I suggest doing it by the "Gauss" method (which is older than Gauss by millenia) fpr summing an arithmetic progression. Go outwards from the middle. The two neighbours of the middle term add up to $2m$. So do the next two terms as you go outwards on both sides. And the next. And so on.
Remark: The same proof works for any arithmetic progression of integers with an odd number of terms.
-
Formula for sum of consecutive integers: $S = \frac{1}{2}n(f + l)$ where $S$ is the sum, $n$ is the number of terms, $f$ is the first term and $l$ is the last.
$S$ is an integer. So, if $n$ is odd, then $\frac{1}{2}(f + l)$ has to be an integer [Reason: $\frac{1}{2}(f + l)$ can either be an integer or half integer, but half integer is not an option]. So $n$ divides $S$.
-
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# math
posted by .
I don't understand the rules for adding and subtracting postive and negative ,can you please explain it to me? im having a test
• math -
1) When adding numbers of the same sign, we add their absolute values, and give the result the same sign.
Examples:
2 + 5.7 = 7.7
(-7.3) + (-2.1) = -(7.3 + 2.1) = -9.4
(-100) + (-0.05) = -(100 + 0.05) =
-100.05
2) When adding numbers of the opposite signs, we take their absolute values, subtract the smaller from the larger, and give the result the sign of the number with the larger absolute value.
Example:
7 + (-3.4) = ?
Subtracting the smaller from the larger gives 7 - 3.4 = 3.6, and since the larger value was 7, we give the result the same sign as 7, so 7 + (-3.4) = 3.6.
Example:
8.5 + (-17) = ?
Subtracting the smaller from the larger gives 17 - 8.5 = 8.5, and since the larger value was 17, we give the result the same sign as -17, so
8.5 + (-17) = -8.5.
Example:
-2.2 + 1.1 = ?
Subtracting the smaller from the larger gives 2.2 - 1.1 = 1.1, and since the larger value was 2.2, we give the result the same sign as -2.2, so
-2.2 + 1.1 = -1.1.
Example:
6.93 + (-6.93) = ?
Subtracting the smaller from the larger gives 6.93 - 6.93 = 0. The sign in this case does not matter, since 0 and -0 are the same. Note that 6.93 and -6.93 are opposite numbers. All opposite numbers have this property that their sum is equal to zero. Two numbers that add up to zero are also called additive inverses.
Subtracting Positive and Negative Numbers
Subtracting a number is the same as adding its opposite.
Examples:
In the following examples, we convert the subtracted number to its opposite, and add the two numbers.
7 - 4.4 = 7 + (-4.4) = 2.6
22.7 - (-5) = 22.7 + (5) = 27.7
-8.9 - 1.7 = -8.9 + (-1.7) = -10.6
-6 - (-100.6) = -6 + (100.6) = 94.6
• math -
thank you
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# What is 1 2 2 2 0
### What is a quadratic equation?
In a quadratic equation, the variable occurs to the power of two and not higher.
#### Examples
• \$\$ x ^ 2 = 3 \$\$
• \$\$ 2x ^ 2 + 1.5x = 0 \$\$
• \$\$ x ^ 2 + 2x - 3 = 0 \$\$
• \$\$ 0.5x ^ 2 - 3x = 1.5 \$\$
In addition to the quadratic term (\$\$ x ^ 2 \$\$), quadratic equations can contain a linear (\$\$ x \$\$) and an absolute term (a number).
#### example
\$\$ 0.5 x ^ 2 \$\$ (quadratic term) \$\$ - 3 x \$\$ (linear term) = \$\$ 1.5 \$\$ (absolute term)
Most of the time you should be doing quadratic equations to solve. You're looking for numbers for the variable that makes up the equation fulfill. These numbers are called solutions. All solutions make up the Solution set \$\$ L \$\$.
In a quadratic equation, the variable x occurs in the 2nd power, but not in a higher power.
• It's about equations with one variable (usually x).
• to the power of 2 means "square".
• "Fulfill" means: You insert a number for the variable in the equation and a true statement such as 2 = 2 emerges.
• The solutions quadratic equations are often infinite, non-periodic decimal fractions (irrational numbers).
The simplest quadratic equations take the form
\$\$ x ^ 2 = r, r in RR \$\$.
The \$\$ r \$\$ is any real number.
#### Example:
\$\$ x ^ 2 = 9 \$\$ with \$\$ r = 9 \$\$
You can do other quadratic equations equivalent transformations bring it into this shape.
#### Example:
\$\$ 3x ^ 2 - 4 = 8 | + 4 \$\$
\$\$ 3x ^ 2 = 12 |: 3 \$\$
\$\$ x ^ 2 = 4 \$\$
The simplest quadratic equations contain terms with \$\$ x ^ 2 \$\$ and real numbers. They can be transformed into the form \$\$ x ^ 2 = r \$\$ \$\$ (rinRR) \$\$.
With an equivalent transformation, the solution set of the equation does not change!
#### 1st example:
Solve the equation \$\$ x ^ 2 = 9 \$\$.
Solution:
\$\$ x_1 = 3 \$\$ and \$\$ x_2 = -3 \$\$, because \$\$ 3 ^ 2 = 9 \$\$ and \$\$ (- 3) ^ 2 = 9 \$\$.
Solution set: \$\$ L = {- 3; 3} \$\$
#### 2nd example:
Solve the equation \$\$ x ^ 2 = 1.69. \$\$
Solution:
\$\$ x_1 = 1.3 \$\$ and \$\$ x_2 = -1.3 \$\$,
because \$\$ 1.3 ^ 2 = 1.69 \$\$ and \$\$ (- 1.3) ^ 2 = 1.69. \$\$
Solution set: \$\$ L = {1.3; -1.3} \$\$
#### 3rd example:
Solve the equation \$\$ x ^ 2 = -4. \$\$
No solution, because \$\$ x ^ 2> 0 \$\$ for all real numbers x.
Solution set: \$\$ L = {} \$\$ (empty set)
If the quadratic equation is transformed into the form \$\$ x ^ 2 = r \$\$ and \$\$ r \$\$ is non-negative, the solutions of the equation can be determined by the square root of \$\$ r \$\$.
\$\$ x ^ 2 = 9 \$\$
\$\$ x_1 = + sqrt9 = 3 \$\$
\$\$ x_2 = - sqrt9 = - 3 \$\$
The square of a real number is always positive.
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### Form first
You can also solve more complicated equations if you can transform them into the form \$\$ x ^ 2 = r (r inRR) \$\$.
#### Example:
\$\$ 2x * (4-x) = 8 (x-1) \$\$
Forming:
Multiply the brackets on both sides.
\$\$ 2x * 4-2x * x = 8x-8 \$\$
\$\$ 8x-2x ^ 2 = 8x-8 \$\$ | \$\$ - 8x \$\$
\$\$ - 2x ^ 2 = -8 \$\$ | \$\$: (- 2) \$\$
\$\$ x ^ 2 = 4 \$\$ (purely square equation)
Solution:
\$\$ x_1 = 2 \$\$ and \$\$ x_2 = -2 \$\$
\$\$ L = {2; -2} \$\$
Sample:
\$\$ x_1 \$\$\$\$: \$\$ \$\$ 2 * 2 * (4-2) = 8 * (2-1) \$\$
\$\$4*2=8*1\$\$
\$\$8=8\$\$
Always try to simplify a given equation using equivalent transformation.
Multiply out: Each term in brackets is multiplied by the term in front of the brackets.
sample: Put the calculated solution in the variable.
### Solutions of the equation \$\$ x ^ 2 = r \$\$
What is the general solution?
An arbitrary equation of the form \$\$ x ^ 2 = r \$\$ is given.
Solutions: \$\$ x_1 = + sqrt (r) \$\$ and \$\$ x_2 = -sqrt (r) \$\$
The solvability of these equations only depends on the number \$\$ r \$\$.
There are 3 cases:
equation number
solutions
solution
\$\$ r> 0 \$\$\$\$: \$\$ \$\$ x ^ 2 = r \$\$2 solutions\$\$ x_1 = sqrt (r) \$\$
\$\$ x_2 = -sqrt (r) \$\$
\$\$ r = 0 \$\$\$\$: \$\$ \$\$ x ^ 2 = 0 \$\$1 solution\$\$ x = 0 \$\$
\$\$ r <0 \$\$\$\$: \$\$ \$\$ x ^ 2 = r \$\$no solution\$\$———\$\$
\$\$ (sqrt (r)) ^ 2 = r \$\$ and \$\$ (- sqrt (r)) ^ 2 = r \$\$
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# What is the limit of (1+2x)^(1/x) as x approaches infinity?
Feb 16, 2017
${\lim}_{x \to \infty} {\left(1 + 2 x\right)}^{\frac{1}{x}} = 1$
#### Explanation:
Write the function as:
${\left(1 + 2 x\right)}^{\frac{1}{x}} = {\left({e}^{\ln \left(1 + 2 x\right)}\right)}^{\frac{1}{x}} = {e}^{\ln \frac{1 + 2 x}{x}}$
Now evaluate:
${\lim}_{x \to \infty} \ln \frac{1 + 2 x}{x}$
This limit is in the indeterminate form $\frac{\infty}{\infty}$ so we can solve it using l'Hospital's rule:
${\lim}_{x \to \infty} \ln \frac{1 + 2 x}{x} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(1 + 2 x\right)}{\frac{d}{\mathrm{dx}} x} = {\lim}_{x \to \infty} \frac{2}{1 + 2 x} = 0$
As ${e}^{x}$ is a continuous function we have then:
${\lim}_{x \to \infty} {\left(1 + 2 x\right)}^{\frac{1}{x}} = {\lim}_{x \to \infty} {e}^{\ln \frac{1 + 2 x}{x}} = {e}^{\left({\lim}_{x \to \infty} \ln \frac{1 + 2 x}{x}\right)} = {e}^{0} = 1$
|
Subject:
Numbers and Operations
Material Type:
Lesson Plan
Level:
Middle School
6
Provider:
Pearson
Tags:
6th Grade Mathematics, Graphing, Word Problems
Language:
English
Media Formats:
# Gallery Overview
Allow students who have a clear understanding of the content thus far in the unit to work on Gallery problems of their choosing. You can then use this time to provide additional help to students who need review of the unit's concepts or to assist students who may have fallen behind on work.
# Gallery Description
## Diving
Chen stands on top of a cliff, and a woman scuba diver dives in the ocean below. Students will determine their positions on a vertical number line that represents distance above and below sea level.
## Negative Numbers?
Students will read about five students’ opinions about negative numbers and decide whose opinions they agree with, whose they disagree with, and why. Students will also share their own ideas about negative numbers.
## Temperatures in January
A map shows the lowest temperatures recorded in January since 2008 for five cities. Students will locate these temperatures on a number line and compare the temperatures.
## Greenwich Mean Time
Students will use positive and negative numbers and Greenwich Mean Time to find the times of different cities around the world.
## Numbers Timeline
Students will research the history of negative numbers and absolute value and create a timeline to show what they learned.
## Rational Numbers and Absolute Value Video
Students will create a video about rational numbers and absolute value.
# Diving
In this diagram, the vertical number line represents distances above and below sea level. Sea level is at 0 meters.
• Chen is standing on top of a cliff that is 20 meters above sea level. Write his position as a positive number on the number line.
• The diver is swimming 10 meters below sea level. Write her position as a negative number on the number line.
• The sea floor is 50 meters below the cliff top. Write this point as a negative number on the number line.
• Suppose the diver swims to a position that is 5 meters above the sea floor. Write her position as a positive or negative number on the number line.
# Negative Numbers?
4. Answers will vary. Possible answers: temperature, latitude, longitude, altitude, or negative cash flow
# Negative Numbers?
Jason, Emma, Denzel, Carlos, and Chen were discussing negative numbers.
Jason: “I think negative numbers are like ordinary numbers but with a negative sign.”
Emma: “Negative numbers are amounts less than 0.”
Chen: “I think of negative numbers as movements on a number line.”
Denzel: “Negative numbers are like debts, when you owe money, and positive numbers are like credits, when you have money.”
Carlos: “I do not think there is any such thing as a negative number. You can’t count negative numbers.”
1. Whose ideas do you agree with? Explain why you agree. Use drawings or diagrams if you wish.
2. Whose ideas do you disagree with? Explain why you disagree. Use drawings or diagrams if you wish.
3. Do you have another way of thinking about negative numbers? Write your idea.
4. Write at least two ways that negative numbers appear in the world around you.
5. How could you convince Carlos that negative numbers exist?
# Temperatures in January
1. Denver had the lowest temperature, –25°F.
2. Rosa is wrong. Possible answer: Since –3°F is farther away from 0°F than –2°F, –3°F is colder.
# Temperatures in January
This map shows five cities in the United States and their lowest recorded January temperatures (in degrees Fahrenheit) since 2008. Use the map to answer the following questions.
1. Draw a number line and place a point on the number line for the temperature of each city. Label each point with the temperature and the name of the city.
2. Which city had the lowest January temperature?
3. Rosa said that the lowest January temperature recorded in Portland is colder than the lowest January temperature recorded in New York. Do you agree? Explain how you know.
# Greenwich Mean Time
1. Answers will vary. Possible answers: Los Angeles is 1 hour behind Denver. Cairo is 7 hours ahead of New York City. Sydney is 10 hours ahead of London.
2. It will be 2 a.m. the next day in London and 11 a.m. the next day in Tokyo when it is 8 p.m. in Chicago.
# Greenwich Mean Time
Since 1884, people on Earth have used the time of the meridian at Greenwich, England—known as Greenwich Mean Time—as the basis for standard time around the globe. The time at different places varies from 12 hours ahead of Greenwich Mean Time to 12 hours behind it, as shown on the map. Greenwich is near London. Regions whose time zone is shown in two colors are halfway between the time zones represented by those colors. For example, New Delhi is 5$\frac{1}{2}$ hours ahead of Greenwich Mean Time.
1. Examine and interpret the map. Write three statements about the time in three different cities.
2. On the number line, label each city you discussed in problem 1 next to the number that represents its time zone. Use positive numbers for time zones ahead of Greenwich Mean Time and negative numbers for time zones behind it.
3. Use the number line from problem 2 to find what time it is in London and Tokyo when it is 8:00 p.m. in Chicago.
4. Complete the table. Find the times in the other cities when it is 8:00 p.m. in Chicago. Then find the number of hours each city is ahead of or behind London.
HANDOUT: Greenwich Mean Time
# Numbers Timeline
• Answers will vary. Have students present their timelines to the class.
# Numbers Timeline
• Research the history of negative numbers and absolute value.
• Create a timeline to show the information you found.
• Be prepared to present your timeline to the class.
|
# How to Calculate Tension in Physics
Co-authored by Bess Ruff, MA
Updated: January 2, 2020
In physics, tension is the force exerted by a rope, string, cable, or similar object on one or more objects. Anything pulled, hung, supported, or swung from a rope, string, cable, etc. is subject to the force of tension.[1] Like all forces, tension can accelerate objects or cause them to deform. Being able to calculate tension is an important skill not just for physics students but also for engineers and architects, who, to build safe buildings, must know whether the tension on a given rope or cable can withstand the strain caused by the weight of the object before yielding and breaking. See Step 1 to learn how to calculate tension in several physical systems.
### Method 1 of 2: Determining Tension On a Single Strand
1. 1
Define the forces on either end of the strand. The tension in a given strand of string or rope is a result of the forces pulling on the rope from either end. As a reminder, force = mass × acceleration. Assuming the rope is stretched tightly, any change in acceleration or mass in objects the rope is supporting will cause a change in tension in the rope. Don't forget the constant acceleration due to gravity - even if a system is at rest, its components are subject to this force. We can think of a tension in a given rope as T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any objects the rope is supporting and "a" is any other acceleration on any objects the rope is supporting.[2]
• For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can't be stretched or broken.
• As an example, let's consider a system where a weight hangs from a wooden beam via a single rope (see picture). Neither the weight nor the rope are moving - the entire system is at rest. Because of this, we know that, for the weight to be held in equilibrium, the tension force must equal the force of gravity on the weight. In other words, Tension (Ft) = Force of gravity (Fg) = m × g.
• Assuming a 10 kg weight, then, the tension force is 10 kg × 9.8 m/s2 = 98 Newtons.
2. 2
Account for acceleration after defining the forces. Gravity isn't the only force that can affect the tension in a rope - so can any force related to acceleration of an object the rope is attached to. If, for instance, a suspended object is being accelerated by a force on the rope or cable, the acceleration force (mass × acceleration) is added to the tension caused by the weight of the object.
• Let's say that, in our example of the 10 kg weight suspended by a rope, that, instead of being fixed to a wooden beam, the rope is actually being used to pull the weight upwards at an acceleration of 1 m/s2. In this case, we must account for the acceleration on the weight as well as the force of gravity by solving as follows:
• Ft = Fg + m × a
• Ft = 98 + 10 kg × 1 m/s2
• Ft = 108 Newtons.
3. 3
Account for rotational acceleration. An object being rotated around a central point via a rope (like a pendulum) exerts strain on the rope caused by centripetal force. Centripetal force is the added tension force the rope exerts by "pulling" inward to keep an object moving in its arc and not in a straight line. The faster the object is moving, the greater the centripetal force. Centripetal force (Fc) is equal to m × v2/r where "m" is mass, "v" is velocity, and "r" is the radius of the circle that contains the arc of the object's motion.[3]
• Since the direction and magnitude of centripetal force changes as the object on the rope moves and changes speeds, so does the total tension in the rope, which always pulls parallel to the rope towards the central point. Remember also that the force of gravity is constantly acting on the object in a downward direction. So, if an object is being spun or swung vertically, total tension is greatest at the bottom of the arc (for a pendulum, this is called the equilibrium point) when the object is moving fastest and least at the top of the arc when it is moving slowest.[4]
• Let's say in our example problem that our object is no longer accelerating upwards but instead is swinging like a pendulum. We'll say that our rope is 1.5 meters (4.9 ft) long and that our weight is moving at 2 m/s when it passes through the bottom of its swing. If we want to calculate tension at the bottom of the arc when it's highest, we would first recognize that the tension due to gravity at this point is the same as when the weight was held motionless - 98 Newtons.To find the additional centripetal force, we would solve as follows:
• Fc = m × v2/r
• Fc = 10 × 22/1.5
• Fc =10 × 2.67 = 26.7 Newtons.
• So, our the total tension would be 98 + 26.7 = 124.7 Newtons.
4. 4
Understand that tension due to gravity changes throughout a swinging object's arc. As noted above, both the direction and magnitude of centripetal force change as an object swings. However, though the force of gravity remains constant, the tension resulting from gravity also changes. When a swinging object isn't at the bottom of its arc (its equilibrium point), gravity is pulling directly downward, but tension is pulling up at an angle. Because of this, tension only has to counteract part of the force due to gravity, rather than its entirety.
• Breaking gravitational force up into two vectors can help you visualize this concept. At any given point in the arc of a vertically swinging object, the rope forms an angle "θ" with the line through the equilibrium point and the central point of rotation. As the pendulum swings, gravitational force (m × g) can be broken up into two vectors - mgsin(θ) acting tangent to the arc in the direction of the equilibrium point and mgcos(θ) acting parallel to the tension force in the opposite direction. Tension only has to counter mgcos(θ) - the force pulling against it - not the entire gravitational force (except at the equilibrium point, when these are equal).
• Let's say that when our pendulum forms an angle of 15 degrees with the vertical, it's moving 1.5 m/s. We would find tension by solving as follows:
• Tension due to gravity (Tg) = 98cos(15) = 98(0.96) = 94.08 Newtons
• Centripetal force (Fc) = 10 × 1.52/1.5 = 10 × 1.5 = 15 Newtons
• Total tension = Tg + Fc = 94.08 + 15 = 109.08 Newtons.
5. 5
Account for friction. Any object being pulled by a rope that experiences a "drag" force from friction against another object (or fluid) transfers this force to the tension in the rope. Force from friction between two objects is calculated as it would be in any other situation - via the following equation: Force due to friction (usually written Fr) = (mu)N, where mu is the friction coefficient between the two objects and N is the normal force between the two objects, or the force with which they are pressing into each other. Note that static friction - the friction that results when trying to put a stationary object into motion - is different than kinetic friction - the friction that results when trying to keep a moving object in motion.
• Let's say that our 10 kg weight is no longer being swung but is now being dragged horizontally along the ground by our rope. Let's say that the ground has a kinetic friction coefficient of 0.5 and that our weight is moving at a constant velocity but that we want to accelerate it at 1 m/s2. This new problem presents two important changes - first, we no longer have to calculate tension due to gravity because our rope isn't supporting the weight against its force. Second, we have to account for tension caused by friction, as well as that caused by accelerating the weight's mass. We would solve as follows:
• Normal force (N) = 10 kg × 9.8 (acceleration from gravity) = 98 N
• Force from kinetic friction (Fr) = 0.5 × 98 N = 49 Newtons
• Force from acceleration (Fa) = 10 kg × 1 m/s2 = 10 Newtons
• Total tension = Fr + Fa = 49 + 10 = 59 Newtons.
### Method 2 of 2: Calculating Tensions On Multiple Strands
1. 1
Lift parallel vertical loads using a pulley. Pulleys are simple machines consisting of a suspended disk that allows the tension force in a rope to change direction. In a simple pulley configuration, the rope or cable runs from a suspended weight up to the pulley, then down to another, creating 2 lengths of rope or cable strands. However, the tension in both sections of rope is equal, even if both ends of the rope are being pulled by forces of different magnitudes. For a system of two masses hanging from a vertical pulley, tension equals 2g(m1)(m2)/(m2+m1), where "g" is the acceleration of gravity, "m1" is the mass of object 1, and "m2" is the mass of object 2.[5]
• Note that, usually, physics problems assume ideal pulleys - massless, frictionless pulleys that can't break, deform, or become separated from the ceiling, rope, etc. that supports them.
• Let's say we have two weights hanging vertically from a pulley in parallel strands. Weight 1 has a mass of 10 kg, while weight 2 has a mass of 5 kg. In this case, we would find tension as follows:
• T = 2g(m1)(m2)/(m2+m1)
• T = 2(9.8)(10)(5)/(5 + 10)
• T = 19.6(50)/(15)
• T = 980/15
• T = 65.33 Newtons.
• Note that, because one weight is heavier than the other, all other things being equal, this system will begin to accelerate, with the 10 kg moving downward and the 5 kg weight moving upward.
2. 2
Lift loads using a pulley with non-parallel vertical strands. Pulleys are often used to direct tension in a direction other than up or down. If, for instance, a weight is suspended vertically from one end of the rope while the other end is attached to a second weight on a diagonal slope, the non-parallel pulley system takes the shape of a triangle with points at the first weight, the second weight, and the pulley. In this case, the tension in the rope is affected both by the force of gravity on the weight and by the component of the pulling force that's parallel to the diagonal section of rope.[6]
• Let's say we have a system with a 10 kg weight (m1) hanging vertically connected by a pulley to a 5 kg weight (m2) on a 60 degree ramp (assume the ramp is frictionless).To find the tension in the rope, it's easiest to find equations for the forces accelerating the weights first. Proceed as follows:
• The hanging weight is heavier and we're not dealing with friction, so we know it will accelerate downward. The tension in the rope is pulling up on it, though, so it's accelerating due to the net force F = m1(g) - T, or 10(9.8) - T = 98 - T.
• We know the weight on the ramp will accelerate up the ramp. Since the ramp is frictionless, we know that the tension is pulling it up the ramp and only its own weight is pulling it down. The component of the force pulling it down the ramp is given by sin(θ), so, in our case, we can say that it's accelerating up the ramp due to the net force F = T - m2(g)sin(60) = T - 5(9.8)(.87) = T - 42.63.[7]
• Acceleration of the two weights are the same, thus we have (98 - T)/m1 = (T - 42.63) /m2. After a little trivial work to solve this equation, finally we have T = 60.96 Newton.
3. 3
Use multiple strands to support a hanging object. Finally, let's consider an object hanging from a "Y-shaped" system of ropes - two ropes are attached to the ceiling, which meet at a central point from which a weight hangs by a third rope. The tension in the third rope is obvious - it's simply tension resulting from the gravitational force, or m(g). The tensions in the other two ropes are different and must add up to equal the gravitational force in the upward vertical direction and to equal zero in either horizontal direction, assuming the system is at rest. The tension in the ropes is affected both by the mass of the hanging weight and by the angle at which each rope meets the ceiling.[8]
• Let's say in our Y-shaped system that the bottom weight has a mass of 10 kg and that the two upper ropes meet the ceiling at 30 degrees and 60 degrees respectively. If we want to find the tension in each of the upper ropes, we'll need to consider each tension's vertical and horizontal components. Nonetheless, in this example, the two ropes happens to be perpendicular to each other, making it easy for us to calculate according to the definitions of trigonometric functions as follows:
• The ratio between T1 or T2 and T = m(g) is equal to the sine of the angle between each supporting rope and the ceiling. For T1, sin(30) = 0.5, while for T2, sin(60) = 0.87
• Multiply the tension in the lower rope (T = mg) by the sine of each angle to find T1 and T2.
• T1 = .5 × m(g) = .5 × 10(9.8) = 49 Newtons.
• T2 = .87 × m(g) = .87 × 10(9.8) = 85.26 Newtons.
## Community Q&A
Search
• Question
What will the dimension of tension be?
Bess Ruff, MA
Master's Degree, Environmental Science & Management, University of California Santa Barbara
Bess Ruff is a PhD student of Geography in Florida. She received her MA in Environmental Science and Management from Bren School of Environmental Science & Management, UC Santa Barbara in 2016.
Master's Degree, Environmental Science & Management, University of California Santa Barbara
Tension is measured in Newtons.
• Question
What is the main formula for tension?
Bess Ruff, MA
Master's Degree, Environmental Science & Management, University of California Santa Barbara
Bess Ruff is a PhD student of Geography in Florida. She received her MA in Environmental Science and Management from Bren School of Environmental Science & Management, UC Santa Barbara in 2016.
Master's Degree, Environmental Science & Management, University of California Santa Barbara
Tension (Ft) = Force of gravity (Fg) = m × g
• Question
Does tension always act in the opposite direction of an applied force?
This is one of Newton's laws! It doesn't just apply to tension, but to ANY force on an object, there is an equal force in the opposite direction. In the case of tension, it can only act in the direction parallel to the object it is in (like a rope or truss member).
• Question
What if I am not given the mass?
If you are not given the mass of an object, you most likely would be given the already calculated force. For example, 10kg x 9.8 = 98N, therefore you should have a force of 98 newtons shown in the diagram or in the question.
• Question
How would calculation be done if the multiple strands of ropes weren't perpendicular?
Mathwizurd29
You would solve the horizontal and vertical components separately. Gravity equals the sum of the vertical components of the strings, and the horizontal components equal each other.
• Question
If I am given mass and density, what formula will I use?
F = m x a still applies. Ignore density unless you have the volume, in which case you must first solve for mass using the density.
• Question
Why does the tension force have to be the same on both ends of a rope?
The tension must be even for accurate results.
• Question
How do I calculate tension on a pivot?
Force x distance= Distance x t1. Then solve for t1. The distances are from the pivot you are trying to work out. Then subtract this answer from the weight of the beam and it should give you the answer.
• Question
How do I find tension if I only know the weight and the angle?
James Wnek
Solve for the vertical component first. You then should use trig to calculate the true tension based on the angle that is given.
• Question
If a rope is bent over a pulley or hook, with a weight of 10 tons on each end of the rope, what is the tension in the rope over the pulley or hook? Is it 20 tons?
James Wnek
The tension would be 5 tons on each "side" of the rope. That way, the vertical components would cancel out, and the rope would not sway to one side or the other.
• What if the tension force has volume?
• How do I calculate tension of multiple objects at once?
• How do I find the tension on the top of a rope if the weight of the box is given and I know the newtons of my string?
• In tension physics, what is T=mg-ma for?
• Is the tension always equal or the same for two connected objects?
200 characters left
## Video.By using this service, some information may be shared with YouTube.
Master's Degree, Environmental Science & Management, University of California Santa Barbara
This article was co-authored by Bess Ruff, MA. Bess Ruff is a PhD student of Geography in Florida. She received her MA in Environmental Science and Management from Bren School of Environmental Science & Management, UC Santa Barbara in 2016.
Co-authors: 35
Updated: January 2, 2020
Views: 1,348,014
Categories: Classical Mechanics
Article SummaryX
To calculate the tension on a rope holding 1 object, multiply the mass and gravitational acceleration of the object. If the object is experiencing any other acceleration, multiply that acceleration by the mass and add it to your first total. To calculate the tension when a pulley is lifting 2 loads vertically, multiply gravity time 2, then multiply it by both masses. Divide that by the combined mass of both objects. When you’re done, remember to write your answer in Newtons! For examples and formulas for different situations, read on!
Thanks to all authors for creating a page that has been read 1,348,014 times.
• PM
Pratibha Mahlawat
Dec 26, 2019
"I was not capable of solving problems based on tension. But after going through this page, I am now capable of handling those problems very easily."..." more
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Oct 22, 2019
"I didn't know about tension and its formula, but this helped me to know it well."
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Oct 17, 2019
"I'm a 6th grader who is obsessed with learning new math, physics, calculus and quantum mechanics. I didn't know how to calculate, so this helped a ton!"..." more
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Lukasz Kedzior
Oct 23, 2016
"The article clearly, and step-by-step, explains all the forces involved in linear motion combined with rotational motion. Excellent tutorial for myself."..." more
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Clint Jones
Feb 12, 2017
"I used the pulley example to prove that the nut and saddles on guitar don't divide the string tension-wise."
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Bilqees Abiola
Mar 31, 2018
"The pictures are so explanatory. Now I can solve for tension better. A big thanks to wikiHow."
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Sep 5, 2017
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Jim Bestie
May 24, 2017
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Dec 23, 2017
"Clear diagrams and well-explained calculations."
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Samiran Das
Aug 2, 2016
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# Area Between Two Functions
If two functions $f(x)$ and $g(x)$ intersect at $x=1$ and $x=3$, and $f(x) \ge g(x)$ for all $1 \le x \le 3$, then the area of the shaded region between their points of intersection is given by:
\begin{align} \displaystyle A &= \int_{1}^{3}{f(x)}dx-\int_{1}^{3}{g(x)}dx \\ &= \int_{1}^{3}{\Big[f(x)-g(x)\Big]}dx \end{align}
### Example 1
Find the area bounded by the $x$-axis and $y=x^2-4x+3$.
\begin{align} \displaystyle x\text{-intercepts:} \\ x^2-4x+3 &= 0 \\ (x-1)(x-3) &=0 \\ x &=1 \text{ and } x=3 \\ \end{align}
\begin{align} \displaystyle A &= \int_{1}^{3}{\Big[0-(x^2-4x+3)\Big]}dx \\ &= \int_{1}^{3}{(-x^2+4x-3)}dx \\ &= \Big[\dfrac{-x^3}{3} + 2x^2- 3x\Big]_{1}^{3} \\ &= \Big(\dfrac{-3^3}{3} + 2 \times 3^2-3 \times 3 \Big)-\Big(\dfrac{-1^3}{3} + 2 \times 1^2- 3 \times 1\Big) \\ &= \dfrac{4}{3} \text{ units}^2 \end{align}
### Example 2
Find the area bounded by the $y=x+2$ and $y=x^2+x-2$.
\begin{align} \displaystyle \text{intersections:} \\ x^2+x-2 &= x+2 \\ x^2-4 &= 0 \\ (x+2)(x-2) &= 0 \\ x &=-2 \text{ and } x=2 \\ \end{align}
\begin{align} \displaystyle A &= \int_{-2}^{2}{\Big[(x+2)-(x^2+x-2)\Big]}dx \\ &= \int_{-2}^{2}{(x+2-x^2-x+2)}dx \\ &= \int_{-2}^{2}{(4-x^2)}dx \\ &= \Big[4x-\dfrac{x^3}{3}\Big]_{-2}^{2} \\ &= \Big(4 \times 2-\dfrac{2^3}{3}\Big)-\Big(4 \times (-2)-\dfrac{(-2)^3}{3}\Big) \\ &= \dfrac{32}{3} \text{ units}^2 \end{align}
### Example 3
Find the area bounded by the $x$-axis and $y=x^3-x^2-2x$.
\begin{align} \displaystyle x\text{-intercepts:} \\ x^3-x^2-2x &= 0 \\ x(x^2-x-2) &= 0 \\ x(x+1)(x-2) &= 0 \\ x &=-1, x=0 \text{ and } x=2 \\ \end{align}
\begin{align} \displaystyle A &= A_1 + A_2 \\ &= \int_{-1}^{0}{\big[(x^3-x^2-2x)-0\big]}dx + \int_{0}^{2}{\big[0-(x^3-x^2-2x)\big]}dx\\ &= \int_{-1}^{0}{(x^3-x^2-2x)}dx-\int_{0}^{2}{(x^3-x^2-2x)}dx\\ &= \Big[\dfrac{x^4}{4}-\dfrac{x^3}{3}-x^2\Big]_{-1}^{0}-\Big[\dfrac{x^4}{4}-\dfrac{x^3}{3}-x^2\Big]_{0}^{2} \\ &= 0-\Big[\dfrac{(-1)^4}{4}-\dfrac{(-1)^3}{3}-(-1)^2\Big]-\Big[\dfrac{2^4}{4}-\dfrac{2^3}{3}-2^2\Big]+0 \\ &= \dfrac{37}{12} \text{ units}^2 \\ \end{align}
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# Solve equations by taking square roots
It’s important to keep them in mind when trying to figure out how to Solve equations by taking square roots. We can solve math problems for you.
## Solving equations by taking square roots
When you try to Solve equations by taking square roots, there are often multiple ways to approach it. The three main branches of trigonometry are Plane Trigonometry, Spherical Trigonometry, and Hyperbolic Trigonometry. Plane Trigonometry is concerned with angles and sides in two dimensions, while Spherical Trigonometry deals with angles and sides on the surface of a sphere. Hyperbolic Trigonometry is concerned with angles and sides in three dimensions. The applications of trigonometry are endless, making it a vital tool for anyone who wants to pursue a career in mathematics or science.
This a website that enables you to get detailed solutions to your math word problems. Just enter the problem in the text box and click on the "Solve" button. will then show you step-by-step how to solve the problem. You can also use the site to check your answers to make sure you are on the right track. is a great resource for students of all levels who are struggling with math word problems.
Solving by square roots Solving by square roots Solving by square roots Solving by square Solving by square Solving Solving by Solving Solving Solving Solving Solvingsolving solving Equation Assume the given equation is of the form: ax^2 + bx + c = 0. Then, the solution to the equation can be found using the following steps: 1) Determine the value of a, b, and c. 2) Find the discriminant, which is equal to b^2 - 4ac. 3) If the discriminant is negative, then there are no real solutions to the equation. 4) If the discriminant is equal to zero, then there is one real solution to the equation. 5) If the discriminant is positive, then there are two real solutions to the equation. 6) Use the quadratic formula to find the value of x that solves the equation. The quadratic formula is as follows: x = (-b +/-sqrt(b^2-4ac))/2a.
In other words, if you know the lengths of two sides of a right triangle, you can use this theorem to find the length of the third side. For example, if you know that one leg is 3 inches long and the other leg is 4 inches long, you can use the Pythagorean theorem to find that the length of the hypotenuse is 5 inches. In general, solving for a side in a right triangle is a matter of applying simple algebra to the Pythagorean theorem. With a little practice, you will be able to solve for sides in right triangles with ease.
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# How to Convert a Percent into a Decimal: A Step-by-Step Guide
## Introduction
Calculating percentages and decimals is a common part of our everyday lives. From calculating sales tax to interest rates, knowing how to convert percentages into decimals is an essential skill. In this article, we will explore the process of converting a percent into decimal form using simple step-by-step instructions.
## Understanding Percentages and Decimals
A percentage is a way of expressing a fraction as a portion of 100. For example, if we say that 50% of a class is female, it means that 50 out of 100 students are female. We can also express percentages as fractions or decimals. For instance, 50% is the same as 0.5 or 1/2.
A decimal is a way of representing a fraction or a whole number based on powers of 10. For example, the number 10 is written in decimal form as 10.0, since 10 is equal to 1 x 10.0. Similarly, the number 0.25 is written in decimal form as 0.25, since 0.25 is equal to 2.5 x 10^-1.
## How to Convert a Percentage into a Decimal
To convert a percentage into a decimal, simply follow these steps:
1. Take the percentage and divide it by 100. For example, to convert 25% into a decimal form, we divide 25 by 100. This gives us 0.25.
2. Remove the percent sign and write the decimal form. In our example, we remove the percent sign from 25%, and write 0.25 as the decimal equivalent of 25%.
Let’s consider another example:
Example: Convert 75% into decimal form.
1. 75 ÷ 100 = 0.75
2. The decimal equivalent of 75% is 0.75
You can also use a variety of methods to convert percentages into decimals, such as:
• Multiplying the percentage by 0.01, or
• Moving the decimal point two places to the left.
## Visual Aids
Charts and graphs can help illustrate the process of converting a percentage into a decimal. Here is an example of how to convert 50% to a decimal:
Percentage Divide by 100 Decimal Form
50% 50 ÷ 100 0.5
You can also use visual examples to help readers understand the math behind percentage to decimal conversion. For example:
## Interactive Calculator or Tool
Readers can also use an interactive calculator or tool to convert percentages into decimals. Here is a handy tool that allows you to input percentages and instantly convert them to decimals:
## Common Uses for Converting a Percentage into a Decimal
Converting percentages to decimals is useful in a variety of scenarios, such as:
• Calculating sales tax or interest rates
• Determining the discounted price of an item
• Calculating a tip at a restaurant
• Calculating profit or loss on an investment
Understanding this concept can help readers in their day-to-day financial life.
## Common Errors and Mistakes
When converting percentages into decimals, common mistakes include:
• Forgetting to divide the percentage by 100
• Dividing the percentage by 10 instead of 100
To avoid these mistakes, double-check your work and use the method that works best for you.
## Conclusion
Converting percentages into decimals is an important skill to have, especially in financial situations. By following the simple steps outlined in this article, readers can easily convert percentages into decimals. Remember to double-check your work to avoid common mistakes, and utilize the visual aids and interactive tools provided to aid in your understanding of this topic.
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Rs Aggarwal 2019 Solutions for Class 9 Math Chapter 1 Number System are provided here with simple step-by-step explanations. These solutions for Number System are extremely popular among Class 9 students for Math Number System Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.
#### Question 1:
What is the difference between a theorem and an axiom?
An axiom is a basic fact that is taken for granted without proof.
Examples:
i) Halves of equals are equal.
ii) The whole is greater than each of its parts.
Theorem: A statement that requires proof is called theorem.
Examples:
i) The sum of all the angles around a point is ${360}^{\circ }$.
ii) The sum of all the angles of triangle is ${180}^{\circ }$.
#### Question 2:
Define the following terms:
(i) Line segment
(ii) Ray
(iii) Intersecting lines
(iv) Parallel lines
(v) Half line
(vi) Concurrent lines
(vii) Collinear points
(viii) Plane
(i) Line segment :A line segment is a part of line that is bounded by two distinct end-points. A line segment has a fixed length.
(ii) Ray: A line with a start point but no end point and without a definite length is a ray.
(iii) Intersecting lines: Two lines with a common point are called intersecting lines.
(iv) Parallel lines: Two lines in a plane without a common point are parallel lines.
(v) Half line: A straight line extending from a point indefinitely in one direction only is a half line.
(vi) Concurrent lines: Three or more lines intersecting at the same point are said to be concurrent.
(vii) Collinear points: Three or more than three points are said to be collinear if there is a line, which contains all the points.
(viii) Plane: A plane is a surface such that every point of the line joining any two point on it, lies on it.
#### Question 3:
(i) six points
(ii) five lines segments
(iii) four rays
(iv) four lines
(v) four collinear points
(i) Points are A, B, C, D, P and R.
(ii)
(iii)
(iv)
(v) Collinear points are M, E, G and B.
#### Question 4:
(i) two pairs of intersecting lines and their corresponding points of intersection
(ii) three concurrent lines and their points of intersection
(iii) three rays
(iv) two line segments
(i) Two pairs of intersecting lines and their point of intersection are
(ii) Three concurrent lines are
(iii) Three rays are
(iv) Two line segments are
#### Question 5:
From the given figure, name the following:
(a) Three lines
(b) One rectilinear figure
(c) Four concurrent points
(a) $Line\stackrel{↔}{RS}$ and $Line\stackrel{↔}{AB}$
(b) $CEFG$
(c) No point is concurrent.
#### Question 6:
(i) How many lines can be drawn through a given point?
(ii) How many lines can be drawn through two given points?
(iii) At how many points can two lines at the most intersect?
(iv) If A, B and C are three collinear points, name all the line segments determined by them.
(i) Infinite lines can be drawn through a given point.
(ii) Only one line can be drawn through two given points.
(iii) At most two lines can intersect at one point.
(iv) The line segments determined by three collinear points A, B and C are
#### Question 7:
Which of the following statements are true?
(i) A line segment has no definite length.
(ii) A ray has no end-point.
(iii) A line has a definite length.
(iv) A line $\stackrel{↔}{AB}$ is same as line $\stackrel{↔}{BA}$.
(v) A ray $\underset{AB}{\to }$ is same as ray $\underset{BA}{\to }$.
(vi) Two distinct points always determine a unique line.
(vii) Three lines are concurrent if they have a common point.
(viii) Two distinct lines cannot have more than one point in common.
(ix) Two intersecting lines cannot be both parallel to the same line.
(x) Open half-line is the same thing as ray.
(xi) Two lines may intersect in two points.
(xii) Two lines are parallel only when they have no point in common.
(i) False. A line segment has a definite length.
(ii) False. A ray has one end-point.
(iii) False. A line has no definite length.
(iv) True
(v) False. $\stackrel{↔}{BA}$ and $\stackrel{↔}{AB}$ have different end-points.
(vi) True
(vii) True
(viii) True
(ix) True
(x) True
(xi) False. Two lines intersect at only one point.
(xii) True
#### Question 8:
In the given figure, L and M are the mid- points of AB and BC respectively.
(i) If AB = BC, prove that AL = MC.
(ii) If BL = BM, prove that AB = BC.
Hint
(i) $AB=BC⇒\frac{1}{2}AB=\frac{1}{2}BC⇒AL=MC$.
(ii) $BL=BM⇒2BL=2BM⇒AB=BC$.
(i) It is given that L is the mid-point of AB.
∴ AL = BL = $\frac{1}{2}$AB .....(1)
Also, M is the mid-point of BC.
∴ BM = MC = $\frac{1}{2}$BC .....(2)
AB = BC (Given)
⇒ $\frac{1}{2}$AB = $\frac{1}{2}$BC (Things which are halves of the same thing are equal to one another)
⇒ AL = MC [From (1) and (2)]
(ii) It is given that L is the mid-point of AB.
∴ AL = BL = $\frac{1}{2}$AB
⇒ 2AL = 2BL = AB .....(3)
Also, M is the mid-point of BC.
∴ BM = MC = $\frac{1}{2}$BC
⇒ 2BM = 2MC = BC .....(4)
BL = BM (Given)
⇒ 2BL = 2BM (Things which are double of the same thing are equal to one another)
⇒ AB = BC [From (3) and (4)]
#### Question 1:
In ancient India, the shapes of altars used for household rituals were
(a) squares and rectangles
(b) squares and circles
(c) triangles and rectangles
(d) trapeziums and pyramids
(b) squares and circles
#### Question 2:
In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for
(a) household rituals
(b) public rituals
(c) both (a) and (b)
(d) none of (a), (b) and (c)
The construction of altars (or vedis) and fireplaces for performining vedic rituals resulted in the origin of the geometry of vedic period. Square and circular altars were used for household rituals whereas the altars with combination of shapes like rectangles, triangles and trapezium were used for public rituals.
Hence, the correct answer is option (b).
#### Question 3:
The number of interwoven isosceles triangles in a Sriyantra is
(a) five
(b) seven
(c) nine
(d) eleven
(c) nine
#### Question 4:
In Indus Valley Civilisation (about BC 3000), the bricks used for construction work were having dimensions in the ratio of
(a) 5 : 3 : 2
(b) 4 : 2 : 1
(c) 4 : 3 : 2
(d) 6 : 4 : 2
(b) 4 : 2 : 1
#### Question 5:
Into how many chapters was the famous treatise, 'The Elements' divided by Euclid?
(a) 13
(b) 12
(c) 11
(d) 9
The famous treatise, 'The Elements' by Euclid is divided into 13 chapters.
Hence, the correct answer is option (a).
#### Question 6:
Euclid belongs to the country
(a) India
(b) Greece
(c) Japan
(d) Egypt
(b) Greece
#### Question 7:
Thales belongs to the country
(a) India
(b) Egypt
(c) Greece
(d) Babylonia
(c) Greece
#### Question 8:
Pythagoras was a student of
(i) Euclid
(ii) Thales
(iii) Archimedes
(ii) Thales
#### Question 9:
Which of the following needs a proof?
(a) axiom
(b) postulate
(c) definition
(d) theorem
(d) theorem
#### Question 10:
The statement that 'the lines are parallel if they do not intersect' is in the form of
(a) a definition
(b) an axiom
(c) a postulate
(d) a theorem
(a) a definition
#### Question 11:
Euclid stated that 'all right angles are equal to each other', in the form of
(a) a definition
(b) an axiom
(c) a postulate
(d) a proof
(b) an axiom
#### Question 12:
A pyramid is a solid figure, whose base is
(a) only a triangle
(b) only a square
(c) only a rectangle
(d) any polygon
(d) any polygon
#### Question 13:
The side faces of a pyramid are
(a) triangles
(b) squares
(c) trapeziums
(d) polygons
(a) triangles
​
#### Question 14:
The number of dimensions of a solid are
(a) 1
(b) 2
(c) 3
(d) 5
A solid shape has length, breadth and height. Thus, a solid has three dimensions.
Hence, the correct answer is option (c).
#### Question 15:
The number of dimensions of a surface are
(a) 1
(b) 2
(c) 3
(d) 0
A plane surface has length and breadth, but it has no height. Thus, a plane surface has two dimensions.
Hence, the correct answer is option (b).
#### Question 16:
How many dimensions does a point have
(a) 0
(b) 1
(c) 2
(d) 3
A point is a fine dot which represents an exact position. It has no length, no breadth and no height. Thus, a point has no dimension or a point has zero dimension.
Hence, the correct answer is option (a).
#### Question 17:
Boundaries of solids are
(a) lines
(b) curves
(c) surfaces
(d) none of these
(c) surfaces
#### Question 18:
Boundaries of surfaces are
(a) lines
(b) curves
(c) polygons
(d) none of these
(b) curves
#### Question 19:
The number of planes passing through 3 non-collinear points is
(a) 4
(b) 3
(c) 2
(d) 1
(d) 1
#### Question 20:
Axioms are assumed
(a) definitions
(b) theorems
(c) universal truths specific to geometry
(d) universal truths in all branches of mathematics
(d) universal truths in all branches of mathematics
#### Question 21:
Which of the following is a true statement?
(a) The floor and a wall of a room are parallel planes.
(b) The ceiling and a wall of a room are parallel planes.
(c) The floor and the ceiling of a room are parallel planes.
(d) Two adjacent walls of a room are parallel planes.
(c) The floor and the ceiling of a room are parallel planes.
#### Question 22:
Which of the following is a true statement?
(a) Only a unique line can be drawn through a given point.
(b) Infinitely many lines can be drawn through two given points.
(c) If two circles are equal, then their radii are equal.
(d) A line has a definite length.
(c) If two circles are equal, then their radii are equal.
#### Question 23:
Which of the following is a false statement?
(a) An infinite number of lines can be drawn through a given point.
(b) A unique line can be drawn through two given points.
(c) Ray .
(d) A ray has one end-point.
(c)
#### Question 24:
A point C is called the mid-point of a line segment $\overline{)AB}$ if
(a) C is an interior point of AB
(b) AC = CB
(c) C is an interior point of AB, such that $\overline{)AC}=\overline{)CB}$
(d) AC + CB = AB
(c) C is an interior point of AB, such that AC = CB
#### Question 25:
A point C is said to lie between the points A and B if
(a) AC = CB
(b) AC + CB = AB
(c) points A, C and B are collinear
(d) None of these
(c) points A, C and B are collinear
#### Question 26:
Euclid's which axiom illustrates the statement that when x + y = 15, then x + y + z = 15 + z?
(a) first
(b) second
(c) third
(d) fourth
Euclid's second axiom states that if equals be added to equals, the wholes are equal.
x + y = 15
Adding z to both sides, we get
x + y + z = 15 + z
Thus, Euclid's second axiom illustrates the statement that when x + y = 15, then x + y + z = 15 + z.
Hence, the correct answer is option (b).
#### Question 27:
A is of the same age as B and C is of the same age as B. Euclid's which axiom illustrates the relative ages of A and C?
(a) First axiom
(b) second axiom
(c) Third axiom
(d) Fourth axiom
Euclid's first axiom states that the things which are equal to the same thing are equal to one another.
It is given that, the age of A is equal to the age of B and the age of C is equal to the age of B.
Using Euclid's first axiom, we conclude that the age of A is equal to the age of C.
Thus, Euclid's first axiom illustrates the relative ages of A and C.
Hence, the correct answer is option (a).
View NCERT Solutions for all chapters of Class 9
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## Metric Ruler Lesson Plan: Centimeters and Millimeters
written by: Margo Dill • edited by: Wendy Finn • updated: 9/11/2012
Here's a lesson idea for elementary grades that practices estimation and measuring with a metric ruler. Kids will like the game aspect.
• slide 1 of 3
For this lesson plan on teaching students to read a metric ruler, each student will need a ruler with millimeters and centimeters (or you can have one ruler for every two students), several objects for students to measure, pencils, and notebook paper.
Start the lesson by asking students to look at their rulers and find the side that measures centimeters and millimeters. This side should be facing up, so they can use it. (This will be true if you have rulers that have inches on one side and centimeters on the other.) Explain to students that measuring with centimeters and millimeters is very similar to measuring with inches. Review how to measure an object, especially starting at the zero mark and measuring straight across. Also, you will want to remind them that things don't always measure perfectly--an object might be 2 and 3/4 inches long, not just 3 inches.
Finally, tell students that it takes 10 millimeters to equal one centimeter. Therefore, if something isn't an exact amount of centimeters, you can use decimal points to show how many centimeters and how many millimeters an object is such as 3.4 cm or 5.6 cm.
• slide 2 of 3
### Lesson and Practice
During this lesson you will ask students to measure the same objects and compare what measurements students got. Before students start measuring, you will need to show them a poster of a metric ruler and show them how to read the millimeters and centimeters. Post the drawing of the ruler on the chalkboard and draw a line to 3 cm and 4 mm. (This line will be much larger than 3 cm 4 mm, and the ruler will be enlarged also, so all students can see.)
Show students how to start at the centimeter mark that's closest to the end of the line and then count the small millimeter marks to get the measurement. Also demonstrate how to write this measurement as 3.4 cm or 34 mm or 3 cm and 4 mm. You can draw several lines and practice with the large class ruler until you think students understand the process.
Next ask students to measure the same objects with their own rulers such as the width and length of their desks or different textbooks. Anything that all students have access to that is the same will work for this practice. You will be able to get an accurate measurement and check if students are measuring correctly. When students get a measurement, ask them to write it down on a piece of notebook paper in one of the three ways you have showed them and large enough, so you can quickly walk around the classroom and see the measurement to gauge if students are measuring correctly or not.
• slide 3 of 3
### Practice with a Game!
To provide more practice for students in this lesson plan on teaching students to read a metric ruler, you can play a game with students. Divide students into two teams. Each team sends a member with his ruler to the front of the class. You show them something to measure such as a pencil. Before students measure, they estimate how many centimeters and millimeters the pencil is. They write this estimation down on a scrap of paper, but they don't tell what it is. Then each student measures the object and writes on the chalk board their answer. They get one point for the correct measurement and one point if their estimation is within 5 mm of the actual measurement. Play the game until class time is up or determine a score a team must reach to win.
At the end of the game, ask students to tell you some things they learned while playing the game and emphasize teaching points such as 10 mm equals 1 cm or starting on the zero mark when measuring.
More To Explore
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# Mathematical proof
rigorous demonstration that a mathematical statement follows from its premises
A mathematical proof is a way to show that a mathematical theorem is true. To prove a theorem is to show that theorem holds in all cases (where it claims to hold). To prove a statement, one can either use axioms, or theorems which have already been shown to be true. Many techniques for proving a statements exist, and these include proof by induction, proof by contraction and proof by cases.[1][2][3]
## Proof by induction
One type of proof is called proof by induction. This is usually used to prove that a theorem holds for all numbers (or all numbers from some point onwards). There are 4 steps in a proof by induction.
1. State that the proof will be by induction, and state which variable will be used in the induction step.
2. Prove that the statement is true for some beginning case.
3. Assume that for some value n = n0, the statement is true and has all of the properties listed in the statement. This is called the induction step.
4. Show that the statement is true for the next value, n0+1.
Once that is shown, it would mean that for any value of n that is picked, the next one is true. Since it is true for some beginning case (usually n=1), it's true for the next one (n=2). And since it is true for 2, it must be true for 3. And since it is true for 3, it must be true for 4, etc. Induction shows that it is always true, precisely because it is true for whatever comes after any given number.
An example of proof by induction is as follows:
Prove that for all natural numbers n, 2(1+2+3+....+n-1+n)=n(n+1).
Proof: First, the statement can be written as "For all natural numbers n, 2${\displaystyle \sum _{k=1}^{n}k}$ =n(n+1)."
By induction on n,
First, for n=1, 2${\displaystyle \sum _{k=1}^{1}k}$ =2(1)=1(1+1), so this is true.
Next, assume that for some n=n0 the statement is true. That is, 2${\displaystyle \sum _{k=1}^{n_{0}}k}$ = n0(n0+1).
Then for n=n0+1, 2${\displaystyle \sum _{k=1}^{{n_{0}}+1}k}$ can be rewritten 2(n0+1) + 2${\displaystyle \sum _{k=1}^{n_{0}}k}$ .
Since 2${\displaystyle \sum _{k=1}^{n_{0}}k}$ = n0(n0+1), 2n0+1 + 2${\displaystyle \sum _{k=1}^{n_{0}}k}$ = 2(n0+1) + 2n0(n0+1).
So 2(n0+1) + 2n0(n0+1)= 2(n0+1)(n0 + 2), which completes the proof.
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Into Math Grade 1 Module 4 Lesson 7 Answer Key Develop Fluency in Subtraction
We included HMH Into Math Grade 1 Answer Key PDF Module 4 Lesson 7 Develop Fluency in Subtraction to make students experts in learning maths.
HMH Into Math Grade 1 Module 4 Lesson 7 Answer Key Develop Fluency in Subtraction
I Can quickly solve subtraction facts within 10.
Step It Out
1. Annie has 8 flowers. She wants to give some to Grace and the rest to Marco. What are two ways Annie can give her flowers away?
A. Think of a way to subtract from 8. Draw and write to show your thinking.
8 – ___ = ___
__ flowers to Grace
__ flowers to Marco
Explanation:
One way
Annie has 8 flowers
She wants to give some to Grace and the rest to Marco
I subtract 3 from 8
8 – 3 = 5
So, 3 flowers to Grace
5 flowers to Marco.
B. Think of another way to subtract from 8. Draw and write to show your thinking.
8 – ___ = ___
__ flowers to Grace
__ flowers to Marco
Explanation:
Another way
Annie has 8 flowers
She wants to give some to Grace and the rest to Marco
I subtract 4 from 8
8 – 4 = 4
So, 4 flowers to Grace
4 flowers to Marco.
Step It Out
2. Write three ways to subtract from 7.
A. Think of strategies you can use.
Explanation:
One way:
Subtract 4 from 7
Start at 7 count back till 4
I counted 3 ones
So, 7 – 4 = 3
Second way:
Subtract 5 from 7
Start at 5
Think how many to add to get 7
5 + 2 = 7
So, 7 – 5 = 2
Third way:
Subtract 6 from 7
Use related facts
6 + 1 = 7
1 + 6 = 7
So, 7 – 6 = 1.
Turn and Talk Why do the subtraction problems above start with the total on top?
Check Understanding Math Board
Write a subtraction equation to solve.
Question 1.
There are 9 bees. Then 2 fly away. How many bees are there now?
___ -___ = ___
__ bees
Explanation:
There are 9 bees
Then 2 fly away
9 – 2 = 7
So, there are 7 bees now.
Subtract. Write the difference.
Question 2.
6 – 1 = ___
Explanation:
The difference of 6 and 1 is 5
So, 6 – 1 = 5.
Question 3.
___ = 10 – 2
Explanation:
The difference of 10 and 2 is 8
So, 10 – 2 = 8.
Question 4.
8 – 8 = ___
Explanation:
The difference of 8 and 8 is 0
So, 8 – 8 = 0.
Write a subtraction equation to solve.
Question 5.
Model with Mathematics 7 horses are in a field. 4 are brown. The rest are white. How many horses are white?
___ – ___ = ___
___ white horses
Explanation:
7 horses are in a field
4 are brown
The rest are white
7 – 4 = 3
So, 3 horses are white.
Question 6.
Open Ended Write two subtraction equations that use the same three numbers.
___ – ___ = ___
___ – ___ = ___
7 – 4 = 3
7 – 3 = 4
Explanation:
The above two equations are subtraction equations
Both the equations use the same three numbers.
Subtract. Write the difference.
Question 7.
3 – 3 = ___
Explanation:
The difference of 3 and 3 is 0
So, 3 – 3 = 0
Question 8.
8 – 7 = ___
Explanation:
The difference of 8 and 7 is 1
So, 8 – 7 = 1.
Question 9.
___ = 10 – 1
Explanation:
The difference of 10 and 1 is 9
So, 10 – 1 = 9.
Question 10.
___ = 10 – 0
Explanation:
The difference of 10 and 0 is 10
So, 10 – 0 = 10.
Question 11.
3 – 2 = ___
Explanation:
The difference of 3 and 2 is 1
So, 3 – 2 = 1.
Question 12.
7 – 5 = ___
Explanation:
The difference of 7 and 5 is 2
So, 7 – 5 = 2
Question 13.
___ = 5 – 3
Explanation:
The difference of 5 and 3 is 2
So, 5 – 3 = 2.
Question 14.
___ = 10 – 8
Explanation:
The difference of 10 and 8 is 2
So, 10 – 8 = 2.
Question 15.
9 – 3 = ___
Explanation:
The difference of 9 and 3 is 6
So, 9 – 3 = 6.
Question 16.
___ = 8 – 4
Explanation:
The difference of 8 and 4 is 4
So, 8 – 4 = 4.
Question 17.
6 – 2 = ___
Explanation:
The difference of 6 and 2 is 4
So, 6 – 2 = 4.
Question 18.
___ = 9 – 8
Explanation:
The difference of 9 and 8 is 1
So, 9 – 8 = 1.
Subtract to solve.
Question 19.
Model with Mathematics 9 fire trucks are at the station. Then 5 fire trucks leave. How many fire trucks are at the station now?
___ fire trucks
Explanation:
9 fire trucks are at the station
Then 5 fire trucks leave
9 – 5 = 4
So, 4 fire trucks are at the station now.
Subtract. Write the difference.
Question 20.
Explanation:
The difference of 5 and 3 is 2
So, 5 – 3 = 2.
Question 21.
Explanation:
The difference of 4 and 3 is 1
So, 4 – 3 = 1.
Question 22.
Explanation:
The difference of 9 and 4 is 5
So, 9 – 4 = 5.
Question 23.
Explanation:
The difference of 10 and 7 is 3
So, 10 – 7 = 3.
Question 24.
Explanation:
The difference of 7 and 0 is 7
So, 7 – 0 = 7.
Question 25.
Explanation:
The difference of 8 and 3 is 5
So, 8 – 3 = 5.
Question 26.
Explanation:
The difference of 7 and 6 is 1
So, 7 – 6 = 1.
Question 27.
Explanation:
The difference of 10 and 7 is 3
So, 10 – 7 = 3.
Question 28.
Explanation:
The difference of 9 and 1 is 8
So, 9 – 1 = 8.
Question 29.
Explanation:
The difference of 7 and 3 is 4
So, 7 – 3 = 4.
Question 30.
Explanation:
The difference of 8 and 2 is 6
So, 8 – 2 = 6.
Question 31.
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# Find if a Tree is Subtree of another Tree
## Introduction
This is a simple yet non trivial problem. Given two trees we need to evaluate if one tree is the sub tree of the other. If the condition is satisfied, it returns true, else it returns false.
The definition of sub tree must not be ambiguous. We say that two trees are equal if they are structurally equal and their corresponding nodes have equal keys.
## Equality of trees
Remember that subtree and contains relation are not the same. It may be possible that a tree T1 contains another tree T2 and still T2 is not a sub tree of T1.
This also means that all subtree satisfies the contains relation but not all contains relation satisfies the subtree relation.
##### Subtree Relation
In the above animation, the green tree is a subtree of the grey tree.
##### Contains Relation
In the above animation, the red tree is contained in the grey tree.
Why is the red tree not a sub tree?
It is fairly simple to see that if we lift the red tree and place it on the grey tree to align its nodes with the corresponding nodes of the grey tree, the node with value 7 is still left over as shown in the second animation above.
Where as if we lift the green tree and place it on the grey tree to align its nodes with the corresponding nodes of the grey tree. It perfectly covers the section and nothing spills over.
##### Equality relation
Two trees are equal if placing one tree over the other covers up the complete tree. Which simplifies our definition to the following:
Two trees T1 and T2 are said to be equal if,
• The root of T1 is equal to the root of T2.
• The subtrees for each of T1’s children is equal to the corresponding subtrees for each of T2’s children.
## Idea behind the solution
The definition of equality itself is recursive. Hence it would be easy if we think in a recursive manner. So, let us say we have some mechanism of comparing two trees given their root, and it returns a true or a false depending on the success or failure of the comparison respectively.
##### Algorithm to find the existence of a subtree
The algorithm now reduces to the following:
• Start from the root of T1 and T2
• Invoke the comparison method passing both the roots.
• If the comparison evaluates the tree to be equal, return true.
• If the above comparison returned false, compare T1’s root with T2’s each child individually and return the OR of the results
As the whole process is recursive, hence it will end up after evaluating all the possibilities.
The question now is to develop a comparison algorithm, here it is.
##### Algorithm to compare two trees
• Compare the two roots. If they are unequal, return false
• If they are equal compare each child of T1 with the corresponding child of T2 recursively
• return true
## Source Code to Find if a Tree is Subtree of another Tree
Here is the important methods:
## Analysis
The comparison algorithm takes O(N) time, where N is the number of nodes in the smaller tree. However, for simplicity, let us consider that both the trees have same number of nodes and that is N.
The subtree algorithm needs to evaluate each node of tree T2 , for the root of T1. For each evaluation it invokes the comparison routine. SO ideally it is a O(N2) solution in the worst case.
## Conclusion
This problem can also be extended to test if a tree is contained into another tree. This is an excessive use of recursion, a non-recursive code is also possible, which might need some extra space.
Also, this piece of solution contains a solution for another problem which is to test equality of two trees.
Advice : Please do not get tempted to find sub array sequences for in order, pre order and post order strings. It doesn’t work in many test cases.
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# Density: Definition, Formula
#### Video Lesson on Density: Definition, Formula
This lesson video Not available at this time available video coming soon
### Density: Definition, Formula & Practice Problems
In this lesson, you will learn about density, the way in which density can be calculated from mass and volume, and that density of substances are constant. You will get to calculate densities of various materials in different situations.
### Definition
Let's say we have three identical cubes (each side of the cube has a 1.0 centimeter length). One cube is made out of lead, the second cube is made out of aluminum, and the third cube is made out of wood. If I asked you which cube would be the most dense of the three cubes, you will most likely say the one made out of lead. Simple enough. But just what exactly is density? Density is defined as the ratio between mass and volume or mass per unit volume. It is a measure of how much stuff an object has in a unit volume (cubic meters or cubic centimeters). Mass is the measure of how much stuff an object contains and volume is the measure of how much space an object occupies in three-dimensional space. The figure below shows a dense object versus a less dense object.
Dense versus Less Dense
As you can see in the figure, the dense object (A) has more stuff than the less dense object (B). Both objects are shown to occupy the same space.
### Calculating Density
So, how do we find the density of an object? As you saw in the definition, you need two pieces of information in order to calculate density: mass (in grams or kilograms), and volume (in cubic centimeter or cubic meter). The formula for density is:
Going back to the three cubes in the beginning of the lesson, let us calculate the densities of the three objects. Let's say we have 11.3 grams of lead, 2.7 grams of aluminum, and 0.67 grams of wood (pinewood). The densities are calculated as mass per unit volume (1 cubic centimeter).
As we indicated earlier, lead had the highest density of the three objects with the same volume. We also point out that no matter what volume of lead, aluminum, or wood you have, the densities of all three objects will always be the same. Densities of substances do not change. Once you know the density of lead, that density will never change. So, let's say you have an object made up of an unknown substance. You measure the mass of the object and the volume of the object and get 11.3 grams per cubic centimeter. Then that unknown substance is lead!
### Practice Problems
What else is unknown at this point? How well you've understood the concept of density! It may seem easy but let's just make sure that's really the case with a bunch of good old practice problems.
### Question 1
Q1. The mass of object X is 10 grams (g) and its volume is 1 cubic centimeter (cc). What is its density?
A1. This one is super easy. We have density (d) = m/v, so d = 10/1 = 10 g/cc.
### Question 2
Q2. It's 2050 and Ethorium is a new alien substance recently discovered by the Mars One mission. It has a mass of 1500 g in a volume of 100 cc. What is its density?
A2. d = 1500/100 = 15 g/cc.
### Question 3
Q3. What is the name of the element that has 42.9 grams of matter per 2 cubic centimeters?
A3. This would require some calculation and just a tad of research using the internet. We set d = m/v to d = 42.9/2 = 21.45 g/cc. Looking up this density online we see that it refers to platinum.
### Question 4
Q4. All else equal, substance Z expands with increasing temperature. What can you say about its density as the temperature increases?
A4. The density will decrease. This is because mass stays the same but volume increases. In d = m/v, the denominator increases while the numerator stays the same. That means d gets smaller.
### Question 5
Q5. Albert Edison is a famous scientist who recently figured out that the density of a new substance he created is 76 g/cc. He remembers that the mass was 7.6 g but can't, for the life of him, recall what volume of space the substance occupied when he figured that out. Help him out!
A5. We know that d = 76 g/cc. So now we set that equal to our mass divided by our unknown volume (x). In other words:
76 = 7.6/x
Next, we need to re-arrange the equation so that x is left on one side:
7.6/76 = x
Solve to get x = 0.10. In other words, the volume was 0.10 cubic centimeters. Double check by writing d = m/v, then filling in the mass and volume to get d = 7.6/0.1 = 76.
### Question 6
Q6. What is the density of a substance with a mass of 27 g found in a cube that has a length of 3 cm?
A6. A cube is a 3D object where the length, width, and height are all the same and a volume of a cube is its length x width x height. This means the volume here is 3 x 3 x 3 = 27 cc.
Therefore d = 27/27 = 1 g/cc; which happens to be the density of water (at 4 degrees Celsius).
Question 7
Q7. You are a pirate. You've found a secret pirate map where X marks the spot. You know there is treasure to be found. You set sail, reach shore, dig, and find a treasure chest. You open it up and you see what looks like a large gold rectangular bar with its mass stamped right on top. It reads 5000 g. Your find that the bar measures 20 cm x 10 cm x 5 cm.
You can't believe your luck. Everyone is happy. But, old Sour Pegleg, the pirate no one likes, is a bit sour on this one. He thinks its fool's gold cubes, or iron pyrite cubes. The other pirates disagree. They can smell real gold from a mile away. Sour Pegleg, pessimist he is, is quite smart though. He knows that the density of fool's gold is about 5 g/cc while the density of real gold is about 19 g/cc. Given this, how can we tell who is right?
A7. First, we need to know what the volume is of this bar. A rectangular prism has a volume of length x width x height or 20 cm x 10 cm x 5 cm in this case. This is a volume of 1000 cc. So d = 5000/1000 = 5 cc. Despite this huge'gold' bar, old Sour Pegleg was right to be suspicious: it's just fool's gold.
### Lesson Summary
Density gives you information about how much stuff is in an object divided by how much space the object occupies. So, you can measure the mass of an object (in grams or kilograms), measure how much space the object occupies in three-dimensional space (in cubic centimeters or cubic meters), then divide the mass by the volume to get the density of the object. Once you know the density of a substance (like lead, aluminum, or pinewood), that density will not change even if you pick a larger or smaller volume than the volume you used to make your initial calculation. Densities of substances are constant.
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WHT
# Similarity Examples | Similarity Example Problems | Similarity | Geometry | Khan Academy
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## Similarity Example Problems | Similarity | Geometry | Khan Academy
Transcript:
In the first task we want to find segment length, segment CE. We have these two parallel lines. AB is parallel to DE. We have these two intersecting, of these two triangles. Let’s see what we can do. The first thing that may come to mind is that this angle and this angle are opposite (vertex) angles. So, they will be equal. Another thing you can think of is that the angle CDE is named after the angle CBA. We have this intersection here and these will be cross angles – that is, they will be equal. If we continue this intersection, we will get the corresponding angle with CDE here. And this one is just the opposite. Either way, this angle and this angle will be equal. So, we found that we have two triangles. And they have two corresponding angles that are equal. That, in itself, suffice it to say that triangles are similar. In fact – In fact, we can say that this and that angle are also equal as cross angles. But it is not necessary. We already know that they are similar. In fact, we can say that at the cross angles, they will also be equal. But we already know enough to say they’re similar, even before we do that. We have already met this triangle. I will try to color it so we have it identical corresponding vertices. It is very important to know which angles correspond to which countries so as not to confuse the proportions, and to know what corresponds to each other. Thus, we know that triangle ABC is similar to triangle .. And this peak A corresponds to peak E right here – is similar to peak E. A peak B here corresponds to vertex D, EDC. How does this help us? This tells us the proportions of the respective countries will be equal. They will be equal. They will be a constant. So, we have a corresponding country .. The proportion, for example, the corresponding side of the BC will be DC. We can see it. Just about the way we proved the similarity. If this is true, then BC is corresponding to the DC side We know that BC, the length of BC on DC, right here, will be equal to the length of .. First we have to find how much CE is, this is what interests us. I use BC and DC because I know their values. Thus, BC over DC will be equal to of the respective CE country. The relevant country here will be CA .. It will be equal to CA on CE corresponding countries. This is the last and the first, the last and the first ARE on CE. We know how many aircraft, The aircraft here is 5. We know how much DC is She is 3. We know how many CAs or ACs are here, CA is 4 And now we can calculate the CE. We can there are many ways you can calculate it. Multiply by the cross, which is actually a multiplication of the numerator by the denominator. We get 5 along the length of the CE, which is equal to 3 by 4, which in turn is equal to 12. that’s how we get CE. CE is equal to 12 over 5. This is the same as 2 and 2 fifths. or 2.4 This will be 2 and 2 fifths We’re ready. We used the similarity to get this country, as we simply knew that the proportions between the respective countries will be the same. Let’s solve this problem here now. Let’s look at this one here. I will draw a small line here. This is a different task. Here we want to find how much is DE. Again we have these two parallel lines. We know that the corresponding angles are equal. We know that this angle will be equal to this angle. Because this can be considered intersecting here. Also, we know this angle here will be equal to this angle here. Once again, we have corresponding angles for the intersections. In both triangles. I’m looking at CBD and CAE triangles. They share a common corner here. In fact, we showed once again that we can stop at two angles. We really showed that all three angles of these two triangles, all three corresponding angles, are equal to each other. Now we know .. The important thing we need to do is once again to make sure we get what you write, in the right order, when we prove the similarity. Now, we know that triangle CBD is similar, not equal, is similar to triangle CAE. This means that the relationship between the parties concerned will be a constant. We know, for example, that the relationship between CDs it will be the same such as the relationship between CB to CA. Let’s write it. We know that CB on CA will be equal to the ratio between CD on CE. We know that CB is, CB here is 5. We know how much is CA? We have to be very careful now. Not 3. SA, this whole country will be 5 plus 3 This is equal to 8. We also know that CD is, CD will be 4. Now we can multiply crosswise again. We have 5 CE, 5 CE equals 8 to 4 8 by 4 is 32. So CE is equal to 32 over 5. Or we can present it in another way. 6 and 2 fifths. We are not ready because we are not looking for how much CE is. Only this part is wanted here. Search with is DE. We know the whole length, CE here. She is 6th and 2nd fifth. So, DE is here what we need to find. It will be equal to this length, 6 and 2 fifth minus, minus CD. This will be equal to 2 and 2 fifths. 6 and 2 fifths minus 4 is 2 and 2 fifths. We’re ready! DE is 2 and 2 fifths.
## 0.3.0 | Wor Build 0.3.0 Installation Guide
Transcript: [MUSIC] Okay, so in this video? I want to take a look at the new windows on Raspberry Pi build 0.3.0 and this is the latest version. It's just been released today and this version you have to build by yourself. You have to get your own whim, and then you...
## Youtube Neural Network | But What Is A Neural Network? | Chapter 1, Deep Learning
Transcript: These are three, sloppily written and provided at a very low resolution of 28 x 28 pixels But your mind has no problem recognizing it as three and I want you to take a moment to appreciate How your brain can do this effortlessly and smoothly I mean this...
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# Multiply Polynomial Worksheets
There are three sets of polynomial worksheets:
Examples, solutions, videos, and worksheets to help Grade 7 and Grade 8 students learn how to multiply polynomials.
### How to multiply polynomials?
There are five sets of multiplying polynomials worksheets.
• Multiply Monomials & Binomials
• Multiply Monomials & Polynomials
• Multiply Binomials
• Squaring Binomials
• Multiply Polynomials
Here’s a step-by-step guide on how to multiply polynomials:
Multiplying polynomials involves applying the distributive property and combining like terms to simplify the expression. To multiply two polynomials, follow these steps:
1. Write the two polynomials side by side.
2. Apply the Distributive Property: Distribute the terms of the first polynomial across the terms of the second polynomial by multiplying each term in the first polynomial by each term in the second polynomial.
3. Combine Like Terms: After distributing, combine like terms by adding or subtracting coefficients with the same variables and exponents.
Let’s work through an example to illustrate these steps:
Example: Multiply (2x + 3) and (x - 4).
1. Write down the polynomials side by side:
(2x + 3)(x - 4)
2. Apply the distributive property:
(2x)(x) + (2x)(-4) + (3)(x) + (3)(-4)
3. Multiply the terms:
2x2 - 8x + 3x - 12
4. Combine like terms:
2x2 - 5x - 12
So, the product of (2x + 3) and (x - 4) is 2x2 - 5x - 12.
Remember that practice is key to mastering polynomial multiplication. It’s important to carefully distribute and combine terms to simplify the expression correctly.
Have a look at this video if you need to review how to multiply monomials & binomials.
Have a look at this video if you need to review how to multiply polynomials.
Click on the following worksheet to get a printable pdf document.
Scroll down the page for more Multiply Polynomial Worksheets.
### More Multiply Polynomial Worksheets
Remainder Theorem
More Printable Worksheets
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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ASSIGNMENT # 1
PROBLEM # 5
It is typical in a Trigonometry class to discuss the sine function and the various attributes of its graph. The graph of y = sin x is referred to as the "parent" graph and will be the graph to which all of our comparisons are made. One period of the function y = sin x looks like this:
Let's see what would happen if a negative were located in front of this function.
In the above picture, the red graph represents the function y = sin x and the green graph represents the function y = - sin x. Although the x-intercepts did not change, it appears as though each y-ccordinate is being multiplied by a negative 1. Hence, a negative value reflects the graph about the x-axis.
At this point, we want to examine the general form y = a sin (bx + c), where a, b, and c are real numbers. Let's begin by looking at y = a sin x, where a is varied.
Notice that the graph is red is y = sin x and this represents our "parent" graph, whereas the green graph is y = 2 sin x. Although the x-intercepts do not change, it appears as though each y-intercept from the parent graph is being multiplied by 2. It follows that if the amplitude (or height) of y = sin x is 1 then y = 2 sin x has an amplitude of 2. Hence, varying the the value for a affects the height of the graph. At this point, let's vary the value of b when y = a sin(bx + c).
The red graph represents the function y = sin x which is our "parent" graph; the blue graph represents the function
.
One obvious difference is that the period (or one complete cycle) of y = sin x is given by
or approximately 6.28. Notice that the blue graph has a period of , where we are simply taking the standard period and multiplying times 2. It follows that the graph is stretched and the coefficient of x (which is (1/2))is causing this stretch. Furthermore, the green graph represents the function y = sin 2x. Notice here that the period or cycle has been reduced. One may easily look at the graph and notice that the period has been reduced from to just . Hence, the coefficient of x affects the stretching or shrinking of the graph. Indeed, if x has an integral coefficient then the graph will undergo a horizontal shrink, whereas if x has a fractional coefficient then the graph will undergo a horizontal stretch. In addition, the "critical" values along the x-axis are altered as well. The "parent" graph y = sin x has critical values of 0, (pi)/2, pi, (3 pi)/2, and 2 pi. As for the function y = sin (1/2)x, the new critical values are 0, pi, 2 pi, 3 pi, 4pi. It would appear as though each critical value along the x-axis is being multiplied by a factor of 2. If we examine the critical values of y = sin 2x, we notice that each critical value along of the parent graph is being multiplied by a factor of (1/2), which is equivalent to dividing each value by 2. Those critical values are 0, (pi)/4), (pi)/2, (3 pi)/4, and pi. Notice that the coefficient of x directly affected the period, but had no bearing on the amplitude whatsoever.
Now, let's find out what effect adding or subtracting a fraction of pi has on the graph. We shall begin by comparing the parent graph to the graph
.
What changed here? Indeed, we are able to look at the graph and notice that the amplitude remained unchanged and the period is still 2 pi. However, we can readily see that the graph of the function y = sin (x + (pi)/4) (in green) and its critical values have been shifted to the left simply by comparing the green graph with the parent graph. Indeed, the graph shifted (pi/4) units to left just by adding (pi/4). It seems somewhat counter-intuivtive in that a student would typically think that adding a value would transform the graph in a positive direction. Let's see what happens if we subtract some value. Will the transformation be to the right in the positive direction? Again, let's compare the parent graph with y = sin (x - pi). If our hypothesis is correct, the graph should be transformed pi units to the right. Notice the graph below.
Fortunately, the conjecture was correct. Notice that our parent graph in red has an initial position of 0 and a terminal position of 2 pi if we only want to examine one period of the function; however, the green graph has has an initial position of pi and a terminal position of 3 pi. Note that all of the critical values from the parent to y = sin (x + pi) may be found by adding pi to each value--hence transforming the graph pi units to the right.
Based on the information above, describe the transformation(s) that will occur with each of the following sine functions and compare it to the graph.
1) y = -2 sin x.
*If we read the rule from left to right, we first notice that there is a negative in front of the function so there should be a reflection over the x-axis. Furthermore, the 2 in front of the function should alter the amplitude by multiplying each y-coordinate by 2. The period of 2 pi will not change and there will be no phase shift. Does the graph below fit this description?
2) .
*In this case, the (1/3) will affect the amplitude by dividing each y-coordinate by 3. So, instead of the range being [-1,1] as with the parent graph, the range should by [(-1/3), (1/3)]. Also, the normal period of 2 pi will be affected as well due to the fact that x has a coefficient other than 1. We noted earlier that if the coefficient were an integer, then the graph would undergo a horizontal shrink and in this case our new period would be pi as opposed to 2 pi. Finally, this graph will be shifted pi/4 units to the right. Indeed, it would not be terribly difficult to very this description by using Algebra Xpresser.
\
Obviously, this holds true.
CONCLUSIONS
The attributes of a sine function may be described in the following manner. If
y = a sin (bx + c), a is called the amplitude and is given by |a|. The period of the sine function is normally 2pi; however, the period of the function may be found by ((2pi)/b). The phase shift may be found by setting-up the following equation: bx + c = 0. In fact, solving this equation for x will yield the initial position for the function and solving the equation bx + c = 2pi will provide the ending position (for one period of the graph).
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# Mathematics. Mathematical Practices
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## Transcription
1 Mathematical Practices 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Many of the graphics by
2 Ratios and Proportional Relationships Understand ratio concepts and use ratio reasoning to solve problems - Part 1 6.RP.1. Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak. For every vote candidate A received, candidate C received nearly three votes. 6.RP.2. Understand the concept of a unit rate a/b associated with a ratio a:b with b 0, and use rate language in the context of a ratio relationship. For example, This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar. We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.
3 Ratios and Proportional Relationships Understand ratio concepts and use ratio reasoning to solve problems - Part 2 6.RP.3. Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed? Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent. Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
4 The Number System Apply and extend previous understandings of multiplication and division to divide fractions fractions by fractions. 6.NS.1. Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. For example, create a story context for (2/3) (3/4) and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that (2/3) (3/4) = 8/9 because 3/4 of 8/9 is 2/3. (In general, (a/b) (c/d) = ad/bc.) How much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 3/4-cup servings are in 2/3 of a cup of yogurt? How wide is a rectangular strip of land with length 3/4 mi and area 1/2 square mi? Compute fluently with multi-digit numbers and find common factors and multiples.
5 The Number System Compute fluently with multi-digit numbers and find common factors and multiples. 6.NS.2. Fluently divide multi-digit numbers using the standard algorithm. 6.NS.3. Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation. 6.NS.4. Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express as 4 (9 + 2). Apply and extend previous understandings of numbers to the system of rational numbers.
6 The Number System Apply and extend previous understandings of numbers to the system of rational numbers - Part 1 6.NS.5. Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation. 6.NS.6. Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., ( 3) = 3, and that 0 is its own opposite. Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize that when two ordered pairs differ only by signs, the locations of the points are related by reflections across one or both axes. Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane.
7 6th Grade Common Core State Standards The Number System Apply and extend previous understandings of numbers to the system of rational numbers - Part 2 6.NS.7. Understand ordering and absolute value of rational numbers. Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. For example, interpret 3 > 7 as a statement that 3 is located to the right of 7 on a number line oriented from left to right. Write, interpret, and explain statements of order for rational numbers in realworld contexts. For example, write 3 o C > 7 o C to express the fact that 3 o C is warmer than 7 o C. Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of 30 dollars, write 30 = 30 to describe the size of the debt in dollars.
8 The Number System Apply and extend previous understandings of numbers to the system of rational numbers - Part 3 6.NS.8. Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate.
9 Expressions and Equations Apply and extend previous understandings of arithmetic to algebraic expressions. - Part 1 6.EE.1. Write and evaluate numerical expressions involving whole-number exponents. 6.EE.2. Write, read, and evaluate expressions in which letters stand for numbers. Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation Subtract y from 5 as 5 y. Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity. For example, describe the expression 2 (8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms. Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving wholenumber exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas V = s 3 and A = 6 s 2 to find the volume and surface area of a cube with sides of length s = 1/2.
10 Expressions and Equations Apply and extend previous understandings of arithmetic to algebraic expressions. - Part 2 6.EE.3. Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. 6.EE.4. Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for. Reason about and solve one-variable equations and inequalities.
11 Expressions and Equations Reason about and solve one-variable equations and inequalities. 6.EE.5. Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. 6.EE.6. Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. 6.EE.7. Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers. 6.EE.8. Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
12 Expressions and Equations Represent and analyze quantitative relationships between dependent and independent variables.. 6.EE.9.Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation d = 65t to represent the relationship between distance and time.
13 Geometry Solve real-world and mathematical problems involving area, surface area, and volume - Part 1 6.G.1. Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems. 6.G.2. Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.
14 Geometry Solve real-world and mathematical problems involving area, surface area, and volume - Part 2 6.G.3. Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems. 6.G.4. Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.
15 Statistics and Probability Develop understanding of statistical variability. 6.SP.1. Recognize a statistical question as one that anticipates variability in the data related to the question and accounts for it in the answers. For example, How old am I? is not a statistical question, but How old are the students in my school? is a statistical question because one anticipates variability in students ages. 6.SP.2. Understand that a set of data collected to answer a statistical question has a distribution which can be described by its center, spread, and overall shape. 6.SP.3. Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary with a single number.
16 Statistics and Probability Summarize and describe distributions. 6.SP.4. Display numerical data in plots on a number line, including dot plots, histograms, and box plots. 6.SP.5. Summarize numerical data sets in relation to their context, such as by: Reporting the number of observations. Describing the nature of the attribute under investigation, including how it was measured and its units of measurement. Giving quantitative measures of center (median and/or mean) and variability (interquartile range and/or mean absolute deviation), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context in which thedata were gathered. Relating the choice of measures of center and variability to the shape of the data distribution and the context in which the data were gathered.
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## About "Area of a parallelogram"
Area of a parallelogram :
A parallelogram is a quadrilateral in which opposite sides are parallel and equal in length. In other words opposite sides of a quadrilateral are equal in length,then the quadrilateral is called a parallelogram.
Area of parallelogram = base x height
## Relationship between a rectangle and parallelogram
Draw a large parallelogram on grid paper. Cut out the parallelogram.
Cut the parallelogram along the dashed segment as shown. Then move the triangular piece to the other side of the parallelogram.
From here we can observer that
base of parallelogram = length of the rectangle
height of parallelogram = width of the rectangle
area of parallelogram = area of the rectangle
Example 1 :
A mirror is made of two congruent parallelograms as shown in the diagram. The parallelograms have a combined area of 9 ⅓ square yards. The height of each parallelogram is 1 ⅓ yards.
a. How long is the base of each parallelogram?
Solution :
(a) Since the given parallelograms are congruent area of two parallelogram will be equal.
Combined area of parallelogram = 9 ⅓ square yards
Area of one parallelogram = (28/3) / 2
= 14/3
Area of parallelogram = base x height
base x 1 ⅓ = 14/3
base x 4/3 = 14/3
base = (14/3) / (4/3) ==> 7/2 ==> 3½ yards
Example 2 :
A watercolor painting is 20 inches long by 9 inches wide. Ramon makes a border around the watercolor painting by making a mat that adds 1 inch to each side of the length and the width.What is the area of the mat?
Solution :
Length of the mat = 20 + 1 + 1 = 22 inches
breadth of the mat = 9 + 1 + 1 = 11 inches
The area of the mat
= total area added - the original area of the water color
= (22 x 11) - (20 x 9)
= 242 - 180
= 62 square inches
Example 3 :
Find the base of a parallelogram if its area is 40 cm² and its altitude is 15 cm.
Solution :
Area of a parallelogram = 40 cm²
b x h = 40 cm²
Here altitude (or) height = 15 cm
b x 15 = 40 ==> b = 2.67 cm
Example 4 :
Find the height of a parallelogram if its area is 612 cm² and its base is 18 cm.
Solution :
Area of a parallelogram = 612 cm²
b x h = 612 cm²
base = 18 cm
18 x h = 612 ==> h = 34 cm
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How do you solve sqrt[x+5]=3x-7 and find any extraneous solutions?
Jun 30, 2017
$x = \setminus \frac{43 + \setminus \sqrt{265}}{18} \setminus \approx 3.293$
Explanation:
Simplifying the equation would involve first getting rid of the square root:
$\setminus \sqrt{x + 5} = 3 x - 7$
${\left(\setminus \sqrt{x + 5}\right)}^{2} = {\left(3 x - 7\right)}^{2}$
$x + 5 = {\left(3 x - 7\right)}^{2}$
Simplifying further, we can expand out the square on the right side of the equation:
$x + 5 = 9 {x}^{2} - 42 x + 49$
Since this is a quadratic equation, we want to put everything on one side and then find the zeros of the expression:
$0 = 9 {x}^{2} - 43 x + 44$
Using the quadratic formula (since we cannot factor the expression),
$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \setminus \frac{- \left(- 43\right) \setminus \pm \setminus \sqrt{{\left(- 43\right)}^{2} - 4 \left(9\right) \left(44\right)}}{2 \left(9\right)}$
Simplifying gives:
$x = \setminus \frac{43 \setminus \pm \setminus \sqrt{1849 - 1584}}{18}$
$x = \setminus \frac{43 \setminus \pm \setminus \sqrt{265}}{18}$
Using a calculator to find the value of this expression,
$x \setminus \approx 3.293$ or $1.485$
Plugging the first value back into the equation, we get a valid equation:
$\setminus \sqrt{3.293 + 5} = 3 \cdot 3.293 - 7$
$2.879 \setminus \approx 2.879$
However, when we plug the second value back into the equation, we have this:
$\setminus \sqrt{1.485 + 5} = 3 \cdot 1.485 - 7$
$\setminus \sqrt{6.485} = - 2.545$
Since the right side is negative, the second value that we got for $x$ is an extraneous solution. Thus, $x$ can only equal $\setminus \frac{43 + \setminus \sqrt{265}}{18} \setminus \approx 3.293$
Jun 30, 2017
all values of $x < - 5$ are extraneous.
$x = \frac{43}{18} \pm \frac{\sqrt{265}}{18} \text{ }$ as exact values
$x \approx + 3.293 \text{ }$to 3 decimal places
$x$ only has one answer if you use what is called the Principle Square Root. That is: only positive!
Explanation:
For the solution to remain within the set of numbers called Real the square root must be of a value that is not negative.
Thus all values of $x < - 5$ are extraneous.
Given:$\text{ "sqrt(x+5)=3x-7" } \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$
Square both sides
$x + 5 \text{ "=" "(3x-7)^2" "=" } 9 {x}^{2} - 42 x + 49$
Subtract $\left(x + 5\right)$ from both sides
$0 = 9 {x}^{2} - 43 x + 44$
To comply with convention write as:
$9 {x}^{2} - 43 x + 44 = 0 \text{ } , \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare to $a {x}^{2} + b x d + c = 0$ where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
In this case: $a = 9 \text{; "b=-43"; } c = 44$ giving:
$x = \frac{+ 43 \pm \sqrt{{\left(- 43\right)}^{2} - 4 \left(9\right) \left(44\right)}}{2 \left(9\right)}$
$x = \frac{+ 43 \pm \sqrt{265}}{18}$
$x = \frac{43}{18} \pm \frac{\sqrt{265}}{18} \text{ }$ as exact values
$x \approx + 3.293 \mathmr{and} 1.485$ to 3 decimal places
However, as per smartspot2, substituting 1.485 back into the original question proves this value to fail if you use the 'Principle Square Root'. Thus by his approach there is only 1 value that is correct:$\text{ } x = 3.293$ to 3 decimal places
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Mental Math
• Mental math is a skill that helps students to do math in their heads without using paper and pencil.
• Mental Math is useful in school and everyday life.
• It also helps to do calculations faster.
• Here, we will discuss the tricks and tips for addition.
Method 1:
• To add two or more numbers, split the second addend as per its place value.
Example :
458 + 243
= 458 + 200 + 40 + 3 = 658 + 40 + 3 = 698 + 3 = 701
Method 2:
• To add two or more numbers, split both addends as per their place value.
Example :
Method 3:
• To add two or more numbers, first round off the numbers to their nearest hundreds or tens, and then add.
• After that add/subtract the deficiency of numbers from the result.
Example :
a) 123 → 120
244 → 240
120 + 240 = 360
3 + 4 = 7
360 + 7 = 367
b) 286 → 290
479 → 480
290 + 480 = 770
Then subtract the deficiency.
4 + 1 = 5
770 – 5 = 765
Method 4:
• This method is called ‘Number Thievery’
• To add the numbers, redistribute the numbers such that the numbers are easier to add.
Example :
a) 141 + 426 =
Turn 141 to 140 and 426 to 427 by subtracting and adding 1 respectively.
Now, 140 + 427 = 567
Hence, 141 + 426 = 567
b) 589 + 356 =
Turn 589 to 590 and 356 to 355 by adding and subtracting 1 respectively.
Now, 590 + 355 = 945
Hence, 589 + 356 = 945.
Method 5:
• This method is called the benchmark method.
• Here, the numbers are splitted into two parts such that the addition of any part with the other addend results in any multiple of 10.
• This makes the addition easier.
Example :
a) 518 + 5 = 518 + 2 + 3
= 520 + 3 = 523
b) 729 + 6 = 729 + 1 + 5
= 730 + 5 = 735
• -
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# Factorize, More Boxes
Boxes and arrows, I know, it doesn’t look like a program yet. Still, it all is going somewhere, I promise. Let’s take a minute and talk about how factorization works. It’s pretty straightforward: either a number is evenly divisible by some other number (a factor) or it isn’t. Let’s look at an example: 48. What numbers go into 48? Well, I always think of it as 6 times 8, but it’s also 4 times 12. I’m the one with the keyboard, so let’s do 6 times 8. Now, let’s look at 6. Is it evenly divisible by anything? Sure, 2 and 3. How about them? Nope, they are not evenly divisible into smaller integer factors. That’s what it means to be prime, and two and three are the first two prime numbers. Eight, on the other side, divides down into twos. Here’s a tree, where each number gets split down into factors. When a factor can’t be split any more, that factor is prime:
The way the program is going to work (refer back to the previous drawing) is to start out with a number to factorize — let’s repeat this example and use 48 — and a list of primes. It’s going to start at the small end of the list of primes and keep trying until it runs out of primes or until it divides the number all the way down to primes. Here’s the tree the program would draw:
Here, the program would try the first prime, two. That works, so it would divide 48 by 2 and get 12. It would start over with 12 and see if two goes into 12. It does, and it would try again on 6. That works, too, and it would try again to divide 2 into 3. That doesn’t go evenly, so it would go to the next prime. The next prime is 3, which does go into 3 evenly. 3 divided by 3 is 1, so the program would know that it was done. This kind of drawing is called a tree (it’s upside down, don’t worry about it, but 48 is the root and all the primes are leaves). I’ve colored the primes green to highlight how they’re not being divided, and because they’re the leaves.
This process is pretty darned simple to explain. The only thing missing is, gosh, a list of prime numbers! It’s getting late and once again, I’m going to have to leave that for the next time. Here’s a bit more detail on the flowchart, though. This is the detail for the “let p be the next prime number” box. The first time we run through, we want to start at the circle labeled, “A,” but on all the following times, we want to come in at the circle marked, “B.”
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# Arithmetic Sequences Calculator
Instructions: This algebra calculator will allow you to compute elements of an arithmetic sequence. You need to provide the first term of the sequence ($$a_1$$), the difference between two consecutive values of the sequence ($$d$$), and the number of steps ($$n$$). Please provide the information required below:
First term ($$a_1$$)
Difference ($$d$$)
Num. of steps ($$n$$)
## What is an Arithmetic Sequence?
Learn more about this arithmetic sequences calculator so you can better interpret the results provided by this solver: An arithmetic sequence is a sequence of numbers $$a_1, a_2, a_3, ....$$ with the specific property that the difference between two consecutive terms of the sequence is ALWAYS constant, equal to a certain value $$d$$. The value of the $$n^{th}$$ term of the arithmetic sequence, $$a_n$$ is computed by using the following formula:
$a_n = a_1 + (n-1)d$
This means that in order to get the next element in the sequence we add $$d$$, to the previous one. So then, the first element is $$a_1$$, the next one is $$a_1 + d$$, the next one is $$a_1 + d + d$$, which can be rewritten as $$a_1 + 2d$$, etc.
### How to use this sequence solver?
This calculator will only need you to provide the initial value, the difference between values and the number of terms you want to add. With that information about this arithmetic progression you will be presented with the step-by-step procedure for its solution.
### Is the arithmetic sequence recursive?
The arithmetic sequence can be solved directly, without implementing any recursive procedure. There is a degree of recursiveness by the fact that $$a_{n+1} - a_{n} = d$$, for all successive terms in the progression.
### How about a geometric progression?
If instead of having that the difference between consecutive terms is constant, and you have the ratio of consecutive terms is constant, you will want to use instead a geometric sequence calculator.
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# RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes
In Chapter 19, we shall discuss the visualisation of 3-D shapes from their plane figures. We will also learn about the representation of three-dimensional shapes on the plane of the paper. For a better understanding of the concepts, students can solve the exercise wise problems using the solutions, which are formulated by our expert faculty team at BYJU’S. Students aspiring to secure high marks in their examination are advised to practice the solutions on a regular basis. RD Sharma Class 8 pdf can be downloaded from the links provided below.
Chapter 19- Visualising Shapes contains two exercises and the RD Sharma Class 8 Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.
• Polyhedra.
• Prisms and pyramids.
• Platonic solids.
• Visualisation of 3-D shapes through nets.
## Download the Pdf of RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes
### Access answers to Maths RD Sharma Solutions For Class 8 Chapter 19 Visualising Shapes
EXERCISE 19.1 PAGE NO: 19.9
1. What is the least number of planes that can enclose a solid? What is the name of the solid?
Solution:
The least number of planes that are required to enclose a solid is 4.
The name of solid is tetrahedron.
2. Can a polyhedron have for its faces?
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution:
(i) 3 triangles?
No, because a polyhedron is a solid shape bounded by polygons.
(ii) 4 triangles?
Yes, because a tetrahedron as 4 triangles as its faces.
(iii) a square and four triangles?
Yes, because a square pyramid has a square and four triangles as its faces.
3. Is it possible to have a polyhedron with any given number of faces?
Solution:
Yes, if number of faces is four or more.
4. Is a square prism same as a cube?
Solution:
Yes. We know that a square is a three dimensional shape with six rectangular shaped sides, out of which two are squares. Cubes are of rectangular prism length, width and height of same measurement.
5. Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Solution:
No.
Let us use Euler’s formula
V + F = E + 2
15 + 10 = 20 + 2
25 ≠22
Since the given polyhedron is not following Euler’s formula, therefore it is not possible to have 10 faces, 20 edges and 15 vertices.
6. Verify Euler’s formula for each of the following polyhedrons:
Solution:
(i) Vertices = 10
Faces = 7
Edges = 15
By using Euler’s formula
V + F = E + 2
10 + 7 = 15 + 2
17 = 17
Hence verified.
(ii) Vertices = 9
Faces = 9
Edges = 16
By using Euler’s formula
V + F = E + 2
9 + 9 = 16 + 2
18 = 18
Hence verified.
(iii) Vertices = 14
Faces = 8
Edges = 20
By using Euler’s formula
V + F = E + 2
14 + 8 = 20 + 2
22 = 22
Hence verified.
(iv) Vertices = 6
Faces = 8
Edges = 12
By using Euler’s formula
V + F = E + 2
6 + 8 = 12 + 2
14 = 14
Hence verified.
(v) Vertices = 9
Faces = 9
Edges = 16
By using Euler’s formula
V + F = E + 2
9 + 9 = 16 + 2
18 = 18
Hence verified.
7. Using Euler’s formula find the unknown:
Faces ? 5 20 Vertices 6 ? 12 Edges 12 9 ?
Solution:
(i)
By using Euler’s formula
V + F = E + 2
6 + F = 12 + 2
F = 14 – 6
F = 8
∴ Number of faces is 8
(ii)
By using Euler’s formula
V + F = E + 2
V + 5 = 9 + 2
V = 11 – 5
V = 6
∴ Number of vertices is 6
(iii)
By using Euler’s formula
V + F = E + 2
12 + 20 = E + 2
E = 32 – 2
E = 30
∴ Number of edges is 30
EXERCISE 19.2 PAGE NO: 19.12
1. Which among of the following are nets for a cube?
Solution:
Figure (iv), (v), (vi) are the nets for a cube.
2. Name the polyhedron that can be made by folding each net:
Solution:
(i) From figure (i), a Square pyramid can be made by folding each net.
(ii) From figure (ii), a Triangular prism can be made by folding each net.
(iii) From figure (iii), a Triangular prism can be made by folding each net.
(iv) From figure (iv), a Hexagonal prism can be made by folding each net.
(v) From figure (v), a Hexagonal pyramid can be made by folding each net.
(vi) From figure (vi), a Cuboid can be made by folding each net.
3. Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice?
Solution:
Figure (i), is a dice. Since the sum of numbers on opposite faces is 7 (3 + 4 = 7 and 6 + 1 = 7).
4. Draw nets for each of the following polyhedrons:
Solution:
(i) The net pattern for cube is
(ii) The pattern for triangular prism is
(iii) The net pattern for hexagonal prism is
(iv) The net pattern for pentagonal pyramid is
5. Match the following figures:
Solution:
(a)-(iv) Because multiplication of numbers on adjacent faces are equal, where 6×4 = 24 and 4×4 = 16
(b)-(i) Because multiplication of numbers on adjacent faces are equal, where 3×3 = 9 and 8×3 = 24
(c)-(ii) Because multiplication of numbers on adjacent faces are equal, where 6×4 = 24 and 6×3 = 18
(d)-(iii) Because multiplication of numbers on adjacent faces are equal, where 3×3 = 9 and 3×9 = 27
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# Express 140 as a product of its prime factors.
Last updated date: 03rd Mar 2024
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Hint:
In order to solve this question, we are required to express the given number as a product of its factors. Hence, we would simply use the prime factorization method with which, we would easily find the prime factors of 140. By multiplying all of them, we could express the given number as a product of its prime factors.
Complete step by step solution:
In this question, we are required to express the given number as a product of its prime factors.
First of all, prime factors are those factors which are greater than 1 and have only two factors, i.e. factor 1 and the prime number itself.
Now, in order to express the given number as a product of its prime factors, we are required to do the prime factorization of the given number.
Now, factorization is a method of writing an original number as the product of its various factors.
Hence, prime factorization is a method in which we write the original number as the product of various prime numbers.
Therefore, prime factorization of $140$ is:
$\begin{matrix} 2 \| \underline 140 \\ 2 \| \underline 70 \\ 5 \| \underline 35 \\ 7 \| \underline 7 \\ {} \| \underline 1 \\ \end{matrix}$
Hence, 140 can be written as:
$140 = 2 \times 2 \times 5 \times 7$
Hence, we have expressed the given number as a product of its prime factors.
$140 = 2 \times 2 \times 5 \times 7$
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# How do you find the equations of the tangent and normal of the curve at x=t^2, y=t+3, t=1?
Jun 24, 2017
The Tangent equation is:
$y = \frac{1}{2} x + \frac{7}{2}$
The Normal equation is:
$y = - 2 x + 6$
#### Explanation:
The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is $- 1$).
We have:
$x = {t}^{2}$
$y = t + 3$
Firstly, let us find the coordinates where $t = 1$:
$\implies x = 1$, $y = 4$ ie $\left(1 , 4\right)$
Then differentiating wrt $t$ we have:
$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}} = \frac{1}{2 t}$
So at the parametric coordinate $t = 1$, we have;
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$
So the tangent passes through $\left(1 , 4\right)$ and has gradient ${m}_{T} = \frac{1}{2}$, so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the tangent equation we seek is;
$y - 4 = \frac{1}{2} \left(x - 1\right)$
$\therefore y - 4 = \frac{1}{2} x - \frac{1}{2}$
$\therefore y = \frac{1}{2} x + \frac{7}{2}$
Similarly, the normal passes through $\left(1 , 4\right)$ and has gradient ${m}_{N} = - 2$, so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the normal equation we seek is;
$y - 4 = - 2 \left(x - 1\right)$
$\therefore y - 4 = - 2 x + 2$
$\therefore y = - 2 x + 6$
We can verify this graphically:
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# Surface area and volume of triangular prisms
## 17.5 Surface area and volume of triangular prisms
A triangular prism has two identical triangular bases and three rectangular sides joined at right angles to the bases.
We can calculate the surface area of the prism by finding the area of the triangular faces and the rectangular faces, and then adding them all together.
The net of a triangular prism consists of two triangular faces and three rectangular faces.
## Worked Example 17.7: Finding the surface area of a triangular prism
Find the surface area of the triangular prism.
### Find the areas of the middle rectangle in the net.
\begin{align} \text{Area rectangle} &= l \times b \\ &= 8 \times 12 \\ &= 96 \end{align}
### Use the theorem of Pythagoras to find the length of the hypotenuse.
The theorem of Pythagoras is covered in detail in Chapter 15, but it states that the square of the longest side in a right-angled triangle is equal to the sum of the squares of the other two sides. We use this theorem to work out the length of one side of a right-angled triangle if given the other two sides.
For the two smaller rectangular faces, we know that the breath is $$12 \text{ cm}$$ but we do not know the length. The length of a small rectangle is equal to the length of the hypotenuse of one of the right-angled triangles (orange line in diagram below) that is formed by the $$3 \text{ cm}$$ height of the triangle.
To use the theorem of Pythagoras for one half of the triangular faces:
Let the length of the longest side $$= x$$
\begin{align} x^2 &= 4^2 + 3^2 \\ x^2 &= 16 + 9 \\ x^2 &= 25 \\ x &= 5 \end{align}
The longest side is equal to $$5 \text{ cm}$$.
\begin{align} \text{Area of small rectangles} &= l \times b \\ &= 5 \times 12 \\ &= 60 \end{align}
### Find the area of the triangular faces.
\begin{align} \text{The area of a triangle} &= \frac{1}{2} b \times h \\ &= \frac{1}{2} (8 \times 3) \\ &= 12 \end{align}
### Find the sum of the areas of the faces.
\begin{align} \text{Surface area of triangular prism} &= 2(\text{area small rectangle}) + (\text{area middle rectangle}) + 2(\text{area triangle}) \\ &= 2(60) + 96 + 2(12) \\ &= 120 + 96+ 24 \\ &= 240 \end{align}
The surface area of the triangular prism is $$240 \text{ cm}^2$$.
temp text
### Calculating volume of a triangular prism
The volume of a triangular prism is calculated by multiplying the area of the triangular base by the height of the prism. To calculate the volume of a triangular prism, we must use the triangular face as the base of the prism so that the cross section of the prism is uniform (in other words, the same shape and area all the way up).
Base: Triangular face Cross section: Uniform Base: Rectangular face Cross section: Not uniform
## Worked Example 17.8: Calculating volume of a triangular prism
Calculate the volume of the prism.
### Use the formula for the area of the base.
\begin{align} \text{Area of triangle} &= \frac{1}{2}(b \times h) \\ &= \frac{1}{2}(8 \times 10) \\ &= 40 \end{align}
### Calculate the volume of the triangular prism.
\begin{align} \text{Volume of prism} &= \text{area of base} \times h \\ &= 40 \times 20 \\ &= 800 \end{align}
Volume of the triangular prism $$= 800 \text{ cm}^3$$.
## Worked Example 17.9: Using volume to find an unknown length of a triangular prism
The volume of the rectangular prism is $$1\ 650 \text{ cm}^3$$. Calculate the value of $$x$$. All measurements are given in centimetres.
### Write the formula for volume of a triangular prism.
\begin{align} \text{Volume of prism} &= \text{area of base} \times \text{height} \\ &= \frac{1}{2}(b \times h) \times h \end{align}
### Use the volume of the rectangular prism to find the value of $$x$$.
\begin{align} \text{Volume of prism} &= 1\ 650 \text{ cm}^3 \\ 1\ 650 &= \frac{1}{2} (44 \times 15) \times x \\ 1\ 650 &= 330x \\ 5 &= x \end{align}
Height of the triangular prism $$x = 5 \text{ cm}$$.
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# How do you find the slope and y intercept for: x +8y=0?
Nov 12, 2015
Slope: $\left(- \frac{1}{8}\right)$
y-intercept: $0$
#### Explanation:
Method 1
A linear equation in the form:
$\textcolor{w h i t e}{\text{XXX}} A x + B y = C$
the slope is $\left(- \frac{A}{B}\right)$
and
the y-intercept can be evaluated by setting $x$ to $0$ and solving for $y$
For the given equation
$\textcolor{w h i t e}{\text{XXX}} x + 8 y = 0$
the slope is $\left(- \frac{\left(1\right)}{8}\right) = - \frac{1}{8}$
and with $x = 0$
$\textcolor{w h i t e}{\text{XXX}} 0 + 8 y = 0 \Rightarrow y = 0$
Method 2
Convert the equation in its given form into slope-intercept form.
$x + 8 y = 0$
$\rightarrow 8 y = - x$
$\rightarrow y = - \frac{1}{8} x$
$\rightarrow y = \left(- \frac{1}{8}\right) x + 0$
$\textcolor{w h i t e}{\text{XXX}}$which is the slope-intercept form for a linear equation
with slope $\left(- \frac{1}{8}\right)$ and y-intercept $0$
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Chapter7(0211)
# Chapter7(0211) - 1 Ch7/MATH0211/YMC/2009-10 Chapter 7...
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Ch7/MATH0211/YMC/2009-10 1 Chapter 7. Integration 7.1. Indefinite Integrals We have studied how to differentiate a given function f ( x ) in the last two chapters. Starting with this section, we would like to know what function we diffferentiate to get f ( x ) , or equivalently, we would like to find a function F whose derivative is the given function f . If such a function F exists, it is called the antiderivative of f : Definition 7.1 An antiderivative of the function f is a function F such that F 0 ( x ) = f ( x ) For example, x 2 is the antiderivative of 2 x as d dx x 2 = 2 x . However, it is not the only antiderivative of 2 x : d dx ( x 2 + 1) = 2 x and d dx ( x 2 - 3) = 2 x. In fact, any function of the form x 2 + C , for any constant C , gives an antiderivative of 2 x . It follows that 2 x has infinitely many antiderivatives, and any two of them differ only by a constant. We can refer to x 2 + C the most general antiderivative of 2 x and denote it by R 2 x dx , that is, Z 2 x dx = x 2 + C. This is the called an indefinite integral of 2 x . The symbol R is called the integral sign , the function 2 x is called the integrand and C is the constant of integration . The dx is part of the integral notation and indicates the variable involved. If F is any antiderivative of f , then Z f ( x ) dx = F ( x ) + C where C is a constant Here are some important properties of indefinite integrals: Properties of the Indefinite Integration 1. Z k dx = kx + C where k is constant. 2. Z kf ( x ) dx = k Z f ( x ) dx where k is constant. 3. Z ( f ( x ) ± g ( x )) dx = Z f ( x ) dx ± Z g ( x )) dx .
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Ch7/MATH0211/YMC/2009-10 2 To actually compute indefinite integrals we begin with some of the basic integration formulas: Basic Integration Formulas 1. Z x n dx = x n +1 n + 1 + C for n 6 = - 1 . 2. Z 1 x dx = Z dx x = ln | x | + C . 3. Z e x dx = e x + C Example 7.1 Find the following indefinite integrals: ( a ) Z (3 + 5 x - x 4 ) dx ( b ) Z 1 t dt ( c ) Z x 3 - 1 x + 4 e x dx Solution . (a) Z (3 + 5 x - x 4 ) dx = Z 3 dx + Z 5 x dx - Z x 4 dx = (3 x + C 1 ) + 5 x 2 2 + C 2 - x 5 5 + C 3 = 3 x + 5 2 x 2 + 1 5 x 5 + ( C 1 + C 2 - C 3 ) For convenience, we will replace the constant C 1 + C 2 - C 3 by a single constant C . We then have Z (3 + 5 x - x 4 ) dx = 3 x + 5 2 x 2 + 1 5 x 5 + C. From now on when we integrate an expression involving more than one term, we will not write the constant of integration for each term. Only one constant is needed.
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# Rational Expressions
Rational Expression : A rational expression is the quotient of two polynomials.
Example:
• Domain of An Algebraic Expression : The domain of an algebraic expression is
the set of all numbers for which the expression is defined. (i.e. the set of all numbers
that can be used in the expression without having a zero in a denominator or a negative
Example: The domain of is all real numbers except -2. Since -2 would cause
the denominator of this expression to be zero, we exclude it from the domain. All other
real numbers may be substituted without any problems.
Example: The domain of is all real numbers greater than or equal to 1. Since
we cannot take the square root of a negative number, we need x-1 ≥ 0 which implies
that x ≥ 1.
Example: The domain of can be determined by finding the real numbers
that will cause the denominator to be zero. Therefore, if we factor the denominator it
will be clear what those numbers are: x^2 - x + 6 = (x - 3)(x + 2). We can see then,
that 3 and -2 will cause the denominator to be zero. Hence, we exclude these numbers
from the domain. The domain of the expression is all reals except 3,-2.
Simplifying Rational Expressions: To simplify a rational expression, follow these
steps:
1. Factor the numerator and denominator completely.
2. Cancel any factor that appears in both the numerator and the denominator.
Example:
Multiplying Rational Expressions: This is really just an extension of simplifying.
The steps are as follows :
1. Factor the numerator and denominator of each rational expression.
2. Cancel any factor that appears in a numerator and a denominator.
Example:
Dividing Rational Expressions: After the first step, follow the same strategy as in
multiplying rational expressions.
1. Invert (flip over) the rational expression you’re dividing by (i.e. the second expression
or the one in the denominator of the large fraction )
2. Factor
3. Cancel
Example:
Adding and Subtracting Rational Expressions: Since rational expressions are
basically fractions, in order to be able to add them or subtract them, we must first
have common denominators , and preferably the LEAST common denominator.
Strategy:
1. Factor the denominators.
2. The Least Common Denominator is the product of all the factors present raised
to the highest exponent that appears on the factor in the factorizations.
3. For each fraction, multiply the numerator and denominator by the factors that
are in the LCD
, but not in the fraction’s denominator.
4. Add or subtract the numerators as appropriate to the problem.
5. Simplify.
Example:
Try these: pp43-44: 4, 10, 16, 34
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# Simultaneous Equations: Solved Examples and Shortcut Tricks
Simultaneous equations are equations containing two unknown quantities whose values are to be determined in two equations at the same time. Solved Examples and Shortcut Tricks of simultaneous equations are well explained here.
I will try to bring this lesson down to a lay man’s understanding such that after reading this post, you will never find it difficult to solve simultaneous equations again.
## Terminologies Used in Simultaneous Equations
There are some important terms one needs to understand before moving on to solve simultaneous equations. The terms are listed here under:
1. Coefficient
2. Term
3. Variable
Let’s quickly explain the terms listed above.
Coefficient: This is a constant by which an algebraic term is multiplied. For instance, If we have 2x, 2 is the coefficient of x.
Variable: A variable has no fixed quantitative value. Example: From the example above, x is a variable. Any alphabet can be regarded as a variable. It changes. That is to say, its values are not constant.
Term: Any value (variable or constant) or expression separated from another term by a space or an appropriate character, in an overall expression or table. Put more aptly, a term is a combination of coefficient and variable. Example: 2x is a term.
Having explained the necessary terms used in solving simultaneous equations, we can progress to stating the difference between simultaneous equations and linear equation.
## Difference between Simultaneous equations and Linear equations
Linear equation has only one equation and only one unknown. Example: 2x+1 = 1
While Simultaneous equations have at least two equations and two unknown variables. Example:
2x+3y = 5 ————————————— equation 1
3x+3y = 2 —————————————- equation 2
Good! I believe you can proudly tell the difference between linear and simultaneous equations now.
## Methods of solving simultaneous equations
There are three most commonly used methods. Namely:
1. Substitution method
2. Elimination method
3. Matrix method
For the scope of this lesson, I will emphasize on substitution and elimination methods.
### Substitution Method Solving Simultaneous Equation
In this method, one of the two unknowns is made the subject of the formula. This is later substituted for in the second equation to have a simple equation with one unknown variable. Then the equation is solved linearly to obtain one value and the value obtained is therefore substituted for to get the other unknown.
### Steps to solve simultaneous equations using substitution method:
1. Call the first equation ‘equation 1 and second one ‘equation 2’
2. Make either x or y the subject of the formula in any of the two equations.
3. Substitute for the variable made the subject of the formula in the other equation.
4. Collect like terms and solve linearly for the one unknown in the simple equation.
5. Substitute for the solved unknown in either equation 1 or 2 to get the second unknown
Example 1
3x+2y = 10
x+2y = 2
Step 1
3x+2y = 10 ————————————— equation 1
x+2y = 2 —————————————- equation 2
Step 2: Let’s make x the subject of the formula in equation 2
x = 2 – 2y ————————————————– equation 3
Step 3: Substitute for x in equation 1 ( this means, put 2 – 2y in equation 1 wherever you see x)
This implies;
3(2 – 2y) + 2y = 10
open the bracket
6 – 6y + 2y = 10
Step 4
Collecting like terms and solving for the unknown
-6y + 2y = 10 – 6
-4y = 4
y = -4/4
y = -1
Step 5: To get x, substitute y in equation 2
x+2y = 2 —————————————- equation 2
x + 2(-1) = 2
x – 2 = 2
x = 4
Therefore, X = 4, y = -1
That’s how simple it is.
Read Also: Divide 1423 by 24 in base 5
### Elimination Method of Solving Simultaneous Equation
Here, one of the two unknowns is made to be removed either by addition or subtraction being performed on the two equations.
Note: Any unknown that is to be eliminated have the same coefficient so as to enable addition or subtraction that is to be performed.
### Steps to solve simultaneous equations using elimination method:
1. Call the first equation ‘equation 1 and second one ‘equation 2’
2. Check if any of the unknowns in the two equations have the value
3. If they have, subtract or add to eliminate one of the unknown variables
4. If they are not, use either the coefficient of x or y in equation 1 to multiply everything in equation 2 and call it equation 4. Also use either the coefficient of x or y in equation 2 to multiply everything in equation one to make one the unknowns equal and call it equation 3
5. Check if addition or subtraction will eliminate the variables that are equal and carry out the one that eliminates one of the unknowns.
6. Collect like terms and solve linearly for the one unknown in the simple equation.
7. Substitute for the solved unknown in either equation 1 or 2 to get the second unknown.
Let’s pick up the example 1 above and solve using elimination method
3x+2y = 10
x+2y = 2
Step 1
3x+2y = 10 ————————————— equation 1
x+2y = 2 —————————————- equation 2
Step 2: the coefficient of x in equation 1 is not the same with the coefficient of x in equation 2. The same thing applies to y. Skip step 3 since none is equal.
Step 4: multiply equation 1 by 1 (coefficient of x in equation 2) and multiply equation 2 by 3 (coefficient of x in equation 1)
This implies;
3x + 2y = 10 ——————————————— equation 3
3x + 6y = 6 ——————————————— equation 4
Step 5
subtracting equation 4 from equation 3
-4y = 4
y = -1
Step 5
To find x, substitute for y in equation
x+2y = 2 —————————————- equation 2
x + 2(-1) = 2
x – 2 = 2
x = 4
That’s it.
### Shortcut to solve simultaneous equations
Suppose you have set of simultaneous equations as shown below:
a1x+b1y = c1 ————————————- 1
a2x+b2y = c2 ————————————- 2
X = ( c1 b2 – c2 b1 ) / ( a1 b2 – a2 b1 )
Using this to solve for x from the example 1 above
3x+2y = 10 ————————————— equation 1
x+2y = 2 —————————————- equation 2
a1 = 3
a2 = 1
b1 = 2
b2 = 2
c1 = 10
c2 = 2
putting it in the above formula
x = (10*2 – 2*2) / (3*2 – 1*2)
x = 20 – 4 / (6 – 2)
x = 16 / 4
x = 4.
You can now substitute for x in either equation 1 or 2 to get y.
Please share this lesson with your friends by clicking the Facebook share button or other social media button at the left corner of this post. thanks for reading and keep coming back for more mathematics lessons.
See:
Subtract 265 from 317 in base eight
Evaluate 2115 base seven Divided by 12 base seven
Last Updated on July 6, 2019 by Admin
### 8 thoughts on “Simultaneous Equations: Solved Examples and Shortcut Tricks”
1. Interesting, am loving maths more.
2. Thank you
3. Thanks sir
4. Thank you sir
5. Wow am very glad that,
i learn through alot most especially how to know the differences
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# Class 8 RD Sharma Solutions – Chapter 21 Mensuration II (Volumes and Surface Areas of a Cuboid and a Cube)- Exercise 21.4 | Set 1
Last Updated : 21 Jul, 2021
### Question 1. Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.
Solution:
Given, the length of room = 12 m
The breadth of room = 9 m
The height of room = 8 m
We need to find the longest rod that can be placed in the room i.e we need to find the diagonal of the room
So, the diagonal of the room = √[l2 + b2 + h2]
= √[122 + 92 + 82] = 17 m
Hence, the longest rod that can be placed in the room is 17 m
### Question 2. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that 1/V = 2/S (1/a + 1/b + 1/c)
Solution:
Given, the dimensions of the cuboid are a, b, c
V is the volume of the cuboid and S is the surface area of the cuboid
We know, the surface area of cuboid = 2(l × b + b × h + h × l)
So, S = (a × b + b × c + c × a)
And the volume of cuboid = lbh
So, V = a × b × c
S/V = 2 [a×b + b×c + c×a] / a×b×c
= 2[(a×b/a×b×c) + (b×c/a×b×c) + (c×a/a×b×c)]
Solving further we get,
1/V = 2/S (1/a + 1/b + 1/c)
Hence, proved
### Question 3. The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V, prove that V2 = xyz.
Solution:
Given, x, y, and z are the adjacent faces of the cuboid
Let l be the Length, b be breadth, h be the Height and V be the volume of the cuboid
So, x = l × b
y = b × h
z = l × h
Multiplying x, y, and z we get,
x × y × z = l × b × b × h × h × l
So, xyz = (l × b × h)2
xyz = V2
Hence, proved
### Question 4. A rectangular water reservoir contains 105 m3 of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.
Solution:
Given, the volume of reservoir = 105 m3
The length of reservoir = 12 m
The breadth of reservoir = 3.5 m
Let h be the depth of the reservoir
So, Volume of reservoir = l × b × h
105 = 12 × 3.5 × h
So, h = 2.5 m
Hence, the depth of water in the reservoir is 2.5 m
### Question 5. Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D.
Solution:
Given, the edges of cube A, B, and C are 18 cm, 24 cm, and 30 cm
So, the Volume of cube A = (edge)3
= (18)3 = 5832 cm3
The Volume of cube B = (edge)3
= (24)3 = 13824 cm3
The Volume of cube C = (edge)2
= (30)3 = 27000 cm3
Let b be the edge of Cube D
The sum of volumes of cube A, B, and C will be equal to the volume of cube D
So, 5832 + 13824 + 27000 = a3
So, a = 36 cm
Hence, the edge of cube D is 36 cm
### Question 6. The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. Dm. Find its dimensions.
Solution:
Let l, b, and h be the Length, Breadth, and Height of the room
Given, the volume of the room is 512 dm3
Also, the breadth = 2 × h and b = l/2
So, l = 2 × b
And h = b/2
The volume of room = l × b × h
512 = 2b × b × (b/2)
So, b = 8 dm
Also, length = 2b = 16 dm
And, height = b/2 = 4 dm
Hence, the dimensions of the room are 16 dm, 8 dm, and 4 dm
### Question 7. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs. 5 per meter sheet, sheet being 2 m wide.
Solution:
Given, the length of tank = 12 m
The breadth of tank = 9 m
The height of tank = 4 m
Also, the area of iron sheet will be equal to surface area of cuboid
= 2(length × breadth + breadth × height + height × length)
= 2(12 × 9 + 9 × 4 + 4 × 12) = 384 m2
Now, let the length of iron sheet is a m
So, length of sheet × width of sheet = 384 m2
a × 2 = 384
a = 192 m
Cost of iron sheet will be 192 × 5 = Rs 960
Hence, the cost of iron sheet used is Rs 960
### Question 8. A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12m×8m×6m, find the cost of iron sheet at Rs. 17.50 per meter.
Solution:
Given, the length of tank = 12 m
The breadth of tank = 8 m
The height of tank = 6 m
So, the area of sheet required = The surface area of tank with only one top
= 12 × 8 + 2(12 × 6 + 8 × 6)
= 336 m2
Now, let the length of iron sheet is a m
So, length of sheet × width of sheet = 336 m2
a × 4 = 336
a = 84 m
Cost of iron sheet will be 84 × 17.50 = Rs 1470
Hence, the cost of iron sheet used is Rs 1470
### Question 9. Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let a be the edges of three cubes placed adjacently
So, the sum of areas of 3 cubes will be 3 × 6 (edge)2
= 3 × 6a2 = 18a2
Also, when these cubes are placed adjacently they form a cuboid
The length of cuboid so formed = a + a + a = 3a m
And, the breadth of cuboid so formed = a m
And, the height of cuboid so formed = a m
We know surface area of cuboid = 2(length × breadth + breadth × height + height × length)
= 2 (3a × a + a × a + a × 3a)
= 14a2
And finally, the ration of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes = 14a2/18a2
= 14/18 = 7 : 9
Hence, the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes is 7 : 9
### Question 10. The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs. 3.50 per square meter.
Solution:
Given, the length of room = 12.5 m
The breadth of room = 9 m
The height of room = 7 m
And, the dimensions of each door is 2.5 m × 1.2 m
And, the dimensions of each window is 1.5 m × 1 m
Now calculating area of four walls in which doors and windows are included,
= 2 (12.5×7 + 9×7) = 301 m2
Now calculating area of 2 doors and 4 windows,
= 2 [2.5 × 1.2] + 4 [1.5 × 1] = 12 m2
So, the area of four walls will be = 301 m2 – 12 m2
= 289 m2
Now, the cost of painting four walls = Rs 3.50 × 289 = Rs 1011.50
Hence, the cost of painting four walls is Rs 1011.50
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# What is the maximum number of 3-digit conscequetive integers that have atleast one odd digit?
Mar 8, 2016
997, 998 and 999.
#### Explanation:
If the numbers have at least one odd digit, in order to get the highest numbers let's chose 9 as the first digit. There is no restriction on the other digits, so the integers can be 997, 998 and 999.
Or you wanted to say at THE MOST one odd digit.
So let's chose 9 again. The other digits can not be odd. Since in three consecutive numbers, at least one must be odd, we cannot have three consecutive numbers in which 9 is the first digit.
So, we have to reduce the first digit to 8. If the second digit is 9, we can't have three consecutive numbers only with even numbers,unless the last of these numbers i 890, and the others are 889 and 888.
Mar 9, 2016
$111$
#### Explanation:
If I am interpreting the question correctly, it is asking for the length of the longest sequence of consecutive $3$-digit integers such that each integer contains at least one odd digit.
Any such sequence would necessarily include either $100 - 199$, $300 - 399$, $500 - 599$, $700 - 799$, or $900 - 999$.
We can discard $100 = 199$ as for any other sequence we gain additional values by subtracting from the lower end, whereas for $100$ we would go into $2$-digit integers, which are not allowed.
As adding $1$ to any of $399 , 599 , 799 , 999$ generates either an integer with no odd digits or with more than $3$ digits, one of those will be the greatest integer in the sequence. As there is no benefit to choosing one over another, we can choose one at random, say, $399$.
Counting down, as all of the $300$s have the first digit as odd, we only need to pay attention when we enter the $200$s. As we count down, all of the $290$s have the second digit as odd, and $289$ has the third digit as odd. Beyond that, we hit $288$ which would break the sequence. Similarly, if we tried with any other starting point, we would find that the longest sequence we could generate would be one of
$289 - 399$, $489 - 599$, $689 - 799$, or $889 - 999$.
each of which has a length of $111$.
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# Chapter R - Section R.5 - Factoring Polynomials - R.5 Exercises: 100
$2(x+1)(4x-1)$
#### Work Step by Step
Factor out $2$ to obtain: $=2(4x^2+3x-1)$ RECALL: A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$ If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping. The trinomial above has $a=4$, $b=3$, and $c=-1$. Thus, $ac = 4(-1) = -4$ Note that $-4=4(-1)$ and $4+(-1)=3$ This means that $d=4$ and $e=-1$. Rewrite the middle term of the trinomial as $4x$ +($-x$) to obtain: $4x^2+3x-1 =4x^2+4x+(-x)-1$ Group the first two terms together and the last two terms together. Then, factor out the GCF in each group to obtain: $=(4x^2+4x)+(-x-1) \\=4x(x+1) +(-1)(x+1)$ Factor out the GCF $x+1$ to obtain: $=(x+1)(4x-1)$ Therefore the completely factored form of the given expression is: $=2(x+1)(4x-1)$
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Language menu Subtract fractions: 7/8 - 1/2 = ? subtraction of ordinary (simple, common) fractions, result explained
You are watching: 7/8 - 1/2
### Reduce (simplify) fountain to their lowest state equivalents:
To alleviate a fraction: division the numerator and denominator by your greatest usual factor, GCF.
Fraction: 7/8 already reduced come the shortest terms. The numerator and denominator have no typical prime factors. Your prime factorization: 7 is a element number; 8 = 23; gcf (7; 23) = 1;
minimize (simplify) fractions to their simplest form, online calculator
## To operate fractions, build up your denominators the same.
### Calculate LCM, the least typical multiple of the platform of the fractions:
LCM will be the typical denominator that the fractions that we occupational with.
The prime factorization that the denominators: 8 = 23; 2 is a element number; Multiply every the unique prime factors, by the biggest exponents: LCM (8; 2) = 23 = 8
Divide LCM by the numerator of every fraction. Because that fraction: 7/8 is 8 ÷ 8 = 1; because that fraction: - 1/2 is 8 ÷ 2 = 23 ÷ 2 = 4;
Expand each portion - main point the numerator and denominator through the expanding number. Then job-related with the molecule of the fractions.
7/8 - 1/2 = (1 × 7)/(1 × 8) - (4 × 1)/(4 × 2) = 7/8 - 4/8 = (7 - 4)/8 = 3/8
### Reduce (simplify) the fraction to that lowest terms equivalent:
To alleviate a fraction: division the numerator and denominator by their greatest typical factor, GCF.
3/8 currently reduced to the lowest terms. The numerator and also denominator have no usual prime factors. Your prime factorization: 3 is a element number; 8 = 23; gcf (3; 23) = 1;
reduce (simplify) fountain to their easiest form, virtual calculator
## As a percentage: 7/8 - 1/2 = 37.5%
### how to subtract the simple fractions: - 13/17 + 3/11
Writing numbers: comma "," offered as a thousands separator; allude "." used as a decimal mark; Symbols: / fraction bar; ÷ divide; × multiply; + plus; - minus; = equal; ≈ approximation;
## Subtract plain fractions, digital calculator
Enter simple fractions come subtract, ie: 6/9 - 8/36 - 12/-90 + 5/20:
## The recent subtracted fractions
7/8 - 1/2 = ? Oct 03 22:25 UTC (GMT) - 1 + 1/4 - 7/18 = ? Oct 03 22:25 UTC (GMT) - 21/45 - 24/48 - 20/63 = ? Oct 03 22:25 UTC (GMT) - 16/41 - 17/32 = ? Oct 03 22:25 UTC (GMT) - 7,711/1,393 + 22,064/26 = ? Oct 03 22:25 UTC (GMT) 7/8 - 1/2 = ? Oct 03 22:25 UTC (GMT) - 41/3,565 - 59/19 = ? Oct 03 22:25 UTC (GMT) 14/10 + 7/25 = ? Oct 03 22:25 UTC (GMT) 11/15 - 7/6 = ? Oct 03 22:25 UTC (GMT) 1,068/336 - 68/23 = ? Oct 03 22:25 UTC (GMT) 31/241 + 6 = ? Oct 03 22:25 UTC (GMT) 25/23 + 48/75 = ? Oct 03 22:25 UTC (GMT) 50/63 - 46/50 = ? Oct 03 22:25 UTC (GMT) view more... Subtracted fractions
There space two cases concerning the denominators as soon as we subtract ordinary fractions:
A. The fractions have actually like denominators; B. The fractions have unlike denominators.
### A. How to subtract ordinary fractions that have like denominators?
Simply subtract the numerators of the fractions. The denominator of the resulting fraction will be the typical denominator that the fractions. Minimize the resulting fraction.
### An instance of subtracting plain fractions that have like denominators, v explanations
3/18 + 4/18 - 5/18 = (3 + 4 - 5)/18 = 2/18; We merely subtracted the numerators of the fractions: 3 + 4 - 5 = 2; The denominator of the resulting portion is: 18; The resulting portion is being lessened as: 2/18 = (2 ÷ 2)/(18 ÷ 2) = 1/9.
### B. To subtract fountain with different denominators (unlike denominators), construct up the fountain to the exact same denominator. Exactly how is it done?
1. Minimize the fountain to the lowest state (simplify them). Element the numerator and also the denominator of every fraction, break them under to prime factors (run their prime factorization). Calculation GCF, the greatest usual factor of the numerator and also of the denominator of every fraction. GCF is the product of all the unique usual prime factors of the numerator and also of the denominator, multiply by the lowest exponents. Division the numerator and the denominator that each portion by their GCF - after ~ this operation the portion is decreased to the lowest state equivalent. 2. Calculate the least typical multiple, LCM, of every the fractions" brand-new denominators: LCM is walk to it is in the common denominator that the added fractions, likewise called the lowest usual denominator (the least usual denominator). Element all the brand-new denominators of the lessened fractions (run the prime factorization). The least usual multiple, LCM, is the product of all the distinct prime components of the denominators, multiplied by the largest exponents. 3. Calculate each fraction"s broadening number: The broadening number is the non-zero number that will be offered to multiply both the numerator and the denominator of each fraction, in order to develop all the fractions as much as the same typical denominator. Divide the least typical multiple, LCM, calculation above, through the denominator of each fraction, in stimulate to calculate each fraction"s broadening number. 4. Increase each fraction: Multiply every fraction"s both numerator and also denominator by the expanding number. At this point, fountain are built up to the exact same denominator. 5. Subtract the fractions: In order come subtract all the fractions just subtract all the fractions" numerators. The end portion will have actually as a denominator the least usual multiple, LCM, calculated above. 6. Alleviate the end fraction to the lowest terms, if needed. ... Check out the remainder of this article, here: exactly how to subtract ordinary (common) fractions?
See more: Calories In A Bacon Egg And Cheese Biscuit, Regular, Bacon, Egg, & Cheese Biscuit Breakfast Sandwich
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# 5th Class Mathematics Geometrical Figures Angle
## Angle
Category : 5th Class
### Angle
Inclination between two rays having common end point is called angle.
In the above given picture, OA and OB are two rays which have a common end point 0. Point 0 is called vertex and rays OA and OB are called arms. The inclination between the rays OA and OB is called angle AOB, and it is denoted as $\angle \text{AOB}\text{.}$
Angle is measured in degree. Symbol of the degree is $~{{''}^{o}}''$ and written as ${{a}^{o}}.$
Types of Angle
There are different types of angles.
(a) Acute angle
(b) Right Angle
(c) Obtuse angle
(d) Straight angle
Acute Angle
An angle which measures between 0° and 90° is called acute angle.
Measure the given below angle and find is it an acute angle.
Explanation
Measure of the above given angle is ${{40}^{o}}.$
Therefore, the angle is an acute angle
Right Angle
An angle of ${{90}^{o}}$ is called right angle.
Obtuse Angle
An angle which measures between ${{90}^{o}}$ and ${{180}^{o}}$ is called obtuse angle.
Straight Angle
An angle which measures ${{180}^{o}}$ is called straight angle.
Triangle
The geometrical shapes having three sides are called triangle.
Properties of Triangle
Triangle has:
(i) Three sides,
(ii) Three angles
(iii) Three vertices
Three sides of the triangle $\text{XYZ}$are$\text{ }\!\!~\!\!\text{ XY, YZ,}$ and $\text{ZX}$
Three angles of the triangle are $\angle \text{X,}\angle \text{Y,}$and $\angle Z$
Three vertices of the triangle are point $\text{X,}$ point Y, and point Z.
Types of Triangle
Triangle has been classified:
(a) On the basis of sides
(b) On the basis of angles
Sides Based Classification
On the basis of sides, triangles are of three types
(i) Equilateral Triangle
(ii) Isosceles Triangle
(iii) Scalene Triangle
Equilateral Triangle
A triangle whose all sides are of equal length is called equilateral triangle.
$\Delta$ ABC is an equilateral triangle as AB = BC = AC = 4 cm.
Note: All the angles of an equilateral triangles are of ${{60}^{o}}$
Isosceles Triangle
A triangle whose any two sides are of equal length is called isosceles triangle.
$\Delta$ ABC is an isosceles triangle as AB = AC 5 cm.
Note: In an isosceles triangle, opposite angles of equal sides are equal
Scalene Triangle
A triangle whose all sides are of different length is called scalene triangle.
$\Delta$ PQR is a scalene triangle as $PQ\ne QR\ne PR.$
Note: In a scalene triangle, no angles are equal
Angle Based Classification
On the basis of angles, triangle are of three types
(i) Acute-angled Triangle
(ii) Right-angled Triangle
(iii) Obtuse-angled Triangle
Acute-Angled Triangle
A triangle having all angles between 90° and 0° is called acute-angled triangle.
ABC is an acute-angled triangle as its each angle ($\angle A,\angle B,\angle C$) measures between ${{0}^{o}}$ and ${{90}^{o}}.$
Right-Angled Triangle
A triangle having an angle of 90° is called a right-angled triangle.
$\Delta$ABC is a right-angled triangle as it contains a right angle($\Delta ABC$)
Obtuse-Angled Triangle
A triangle having one obtuse angle is called obtuse-angled triangle.
$\Delta$MNP is an obtuse-angled triangle as it contains an obtuse angle ($\angle MNP$)
#### Other Topics
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# How do you integrate x^2(sinx)dx?
Then teach the underlying concepts
Don't copy without citing sources
preview
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#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
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mason m Share
Jul 26, 2016
$\int {x}^{2} \sin \left(x\right) \mathrm{dx} = - {x}^{2} \cos \left(x\right) + 2 x \sin \left(x\right) + 2 \cos \left(x\right) + C$
#### Explanation:
Use integration by parts, which takes the form:
$\int u \mathrm{dv} = u v - \int v \mathrm{du}$
For $\int u \mathrm{dv} = \int {x}^{2} \sin \left(x\right) \mathrm{dx}$, we let:
$u = {x}^{2} \text{ "=>" "(du)/dx=2x" "=>" } \mathrm{du} = 2 x \mathrm{dx}$
$\mathrm{dv} = \sin \left(x\right) \mathrm{dx} \text{ "=>" "intdv=intsin(x)dx" "=>" } v = - \cos \left(x\right)$
Thus, substituting these into the integration by parts formula, we see that:
$\int {x}^{2} \sin \left(x\right) \mathrm{dx} = - {x}^{2} \cos \left(x\right) - \int \left(- 2 x \cos \left(x\right)\right) \mathrm{dx}$
Simplifying the negative signs:
$\int {x}^{2} \sin \left(x\right) \mathrm{dx} = - {x}^{2} \cos \left(x\right) + 2 \int x \cos \left(x\right) \mathrm{dx}$
Now, do integration by parts once more on the remaining integral:
$u = x \text{ "=>" "(du)/dx=1" "=>" } \mathrm{du} = \mathrm{dx}$
$\mathrm{dv} = \cos \left(x\right) \mathrm{dx} \text{ "=>" "intdv=intcos(x)dx" "=>" } v = \sin \left(x\right)$
Thus:
$\int x \cos \left(x\right) \mathrm{dx} = x \sin \left(x\right) - \int \sin \left(x\right) \mathrm{dx}$
Since $\int \sin \left(x\right) \mathrm{dx} = - \cos \left(x\right)$, this becomes:
$\int x \cos \left(x\right) \mathrm{dx} = x \sin \left(x\right) + \cos \left(x\right)$
Now, returning to before:
$\int {x}^{2} \sin \left(x\right) \mathrm{dx} = - {x}^{2} \cos \left(x\right) + 2 \int x \cos \left(x\right) \mathrm{dx}$
Substitute in $\int x \cos \left(x\right) \mathrm{dx} = x \sin \left(x\right) + \cos \left(x\right)$:
$\int {x}^{2} \sin \left(x\right) \mathrm{dx} = - {x}^{2} \cos \left(x\right) + 2 \left(x \sin \left(x\right) + \cos \left(x\right)\right)$
$\int {x}^{2} \sin \left(x\right) \mathrm{dx} = - {x}^{2} \cos \left(x\right) + 2 x \sin \left(x\right) + 2 \cos \left(x\right) + C$
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# 1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation
## Karnataka 1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation
### 1st PUC Statistics Classification and Tabulation Two Marks Questions and Answers
Question 1.
What is classification of the data?
Solution:
Classification is the process of arranging the data in to groups or classes according to common
Question 2.
What are the objectives of classification?
Solution:
• To reduce the size of the data
• To bring the similarities together.
• To enable further statistical analysis
Question 3.
What are the basis/types of classification?
Solution:
1. Chronological classification
2. Geographical classification
3. Qualitative classification
4. Quantitative classification.
Question 4.
Give the formula of Sturge’s to find the number of classes. Or Give the formula used to determine the number of classes.
Solution:
The numbers of classes are obtained using Sturges’s Rule: k = 1 + 3.22 log N;
N-number of items.
Question 5.
For what purpose is correction factor used in frequency distribution?
Solution:
To get a better continuity between the class interval of a frequency distribution exclusive class intervals are used, so, if the frequency distribution is in inclusive class intervals isconverted into exclusive class intervals using correction factor.
Question 6.
What are the guidelines of classification?
Solution:
following are some guidelines following while classification:
• The number of classes should generally between 4 and 15.
• Exclusive classes should be formed for better continuity between the class intervals.
• The width of the classes should be usually kept constant throughout the distribution.
• Avoid open-end classes.
• The classes should be arranged in ascending or descending order.
• The lower limit of the first class should be either ‘o’ or multiple of 5.
Question 7.
Define the term tabulation
Solution:
Tabulation is a systematic arrangement of the classified data in to columns and rows of a table
Question 8.
Mention the parts of a Table
Solution:
Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source.
Question 9.
What are the objects of Tabulation?
Solution:
The object/Purpose of tabulation are:-
• It simplifies the complex data
• To facilitate for comparison
• To give an identity to the data
• To reveals trend and patterns of the data
Question 10.
How the number of classes using Prof. H.A.Sturge’s?
Solution:
The number of classes are obtained using Sturges’s Rule: k = 1 + 3.22 Log N
Question 11.
What are inclusive & exclusive class intervals?
Solution:
If in a class, both lower and upper limits are included in the same class are inclusive class intervals, eg. 0-9, 10-19, 20-29…
If in a class, the lower limit is included in the same class but the upper limit is included in the next class are exclusive class intervals, eg. 0-10, 10-20, 20-30…
Question 12.
Define frequency distribution
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called Frequency distribution.
Question 13.
What are Marginal and Conditional frequency distributions?
Solution:
If in a bivariate frequency distribution, if the distribution of only one variable is considered, the distribution is called marginal frequency distribution.
If in a bivariate frequency distribution, if the distribution of only one variable is formed subject to the condition of the other variable it is called conditional frequency distribution.
Question 14.
What is the tabulation of the data?
Solution:
Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.
Question 15.
What are the parts of a table?
Solution:
The parts of a table are: Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source
Question 16.
What is the purpose of ‘table number’ in tabulation?
Solution:
A number should be given to each and every table, in order to distinguish and also for easy reference.
Question 17.
What are captions and stubs of a table?
Solution:
Captions: Column headings are called captions. They explain what the column represents. Captions are always written in one or two words on the top of each column.
Stubs: Row headings are called Stubs. They explain what the row represents. Stubs are usually written in one or two words at the left extreme sid/of each row.
Question 18.
Solution:
It is a brief explanatory note or statement given just below the heading of the table put in a bracket. The statistical units of measurements, such as in ‘000s, Rs., Million tones, crores, Kgs., etc., are usually put in bracket.
Question 19.
What is indicated by source note of a table?
Solution:
Below foot note or below the table, source of the data may be mentioned for verification to the reader, regarding publications, organizations, pages, Journals etc,
### 1st PUC Statistics Classification and Tabulation Five Marks Questions and Answers
Question 1.
Explain chronological classification and geographical classification of data with examples.
Solution:
Temporal/Chronological classification: when the data enumerated over a different period of time, the type of classification is called chronological classification. The above type of classification is called as time series, e.g. Time series of the population, is listed in chronological order starting with the earliest period.
Table showing the population of India
Geographical classification :
In this type of classification the data are classified on the basis of geographical or locational or area wise differences between various items. Such as cities, districts, states etc. e.g., production of sugar in India may be presented state wise in the following manner:-
Table showing the production of sugar in India
Question 2.
Explain qualitative and quantitative classification with examples.
Solution:
Qualitative classification: Classification of the different units on the basis of qualitative characteristics (Called Attributes). Such as sex, literacy, employment etc.
e. g., the members of a club can be classified on the basis of sex wise distribution as follows:-
Table showing the sex distribution of members of a club
Quantitative classification: classification of the number of units on the basis of quantitative data, such as according to Height, weight, Wages, Age (years), Number of children, etc, Thus the groups of a student may be classified on their heights as follows:-
Table showing the heights of students
Online midpoint calculator and find the midpoint of aline segment joining two points using the midpoint calcultor in just one click.
Question 3.
Define the following terms :
i. Frequency, class frequency:
Solution:
Frequency refers to the number of times an observation repeated (f). The number of observations corresponding to a particular class is known as the class Frequency Class frequency is a positive integer including zero
ii. Class limits:
Solution:
Class limits- Lowest and the highest values that are taken to define the boundaries of the class are called class limits
iii. Range of the class: It is the difference between highest and lowest value in the data, i.e., Range = H.V. – L.V.
iv. Width of the class: The difference between the upper and lower limits of class called width of the class. It is denoted by c or i.
i/c = Upper limit(UL) – Lower limit(LL)
v. Class mid point
Solution:
The central value of a class called mid value/point; $$\mathrm{m} / \mathrm{x}=\frac{\mathrm{LL}+\mathrm{UL}}{2}$$
vi. Define Frequency Density:
Solution:
The frequency per unit of class interval is the frequency density.
i.e. frequency density = $$\frac{\text { Frequency the class }}{\text { width of the class }}=f / w$$
vii. Relative frequency:
Solution:
Relative frequency.. is the ratio of frequency of class to the total frequency of the distribution
Relative Frequency $$\frac{\mathrm{f}}{\mathrm{N}}$$
viii. Class interval-inclusive, exclusive and open-end classes:
Solution:
If in a class, lower as well as upper limits are included in the same class are called Inclusive class
e. g. 30-39,40-49….
If in a class, the lower limit is included in the same class and upper limit is included in the next class, such a class is called Exclusive class, eg. 30-40, 40-50…
If in a class, the lower and upper limits of the class is not specified are called open end classes” e.g. less than/below, or more than/ above 100
ix. Cumulative frequency- less than and more than cumulative frequency:
Solution:
The added up frequencies are called cumulative frequencies.
The number of observations with values less than upper limit is less than cumulative frequency. (l.c.f) i.e. Frequencies added from the top.
The number of observations with values more than lower class limit is more than cumulative frequency (m.c.f)
x. Correction factor:
Solution:
It is half of the difference between lower limit of a class and upper limit of the preceding class. Thus,
Solution:
Correction facctor (C.F) = $$\frac{\text { Lower limit of a class-Upper limit of the precending class }}{2}$$
xi. A Frequency distribution Discrete, Continuous, Bi-variate, Marginal
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called frequency is called Frequency distribution.
While framing a frequency distribution, if class intervals are not considered, is called discrete frequency distribution.
1. Example:
The number of families according to number of children .
While framing a frequency distribution, if class intervals are considered, is called continuous frequency distribution.
2. Example:
The following table showing the weight (kgs.) of persons
A frequency distribution formed on the basis of two related variables is called bi-variate frequency distribution.
For example, we want to classify data relating to the Height and Weights of a group of individuals, Income and Expenditure of a group of individuals, Ages of Husbands and Wives, Ages of mothers and Number of children, etc .
In a Bivariate frequency distribution, the frequency distribution of only one of the variables is considered, it is marginal frequency distribution.
Question 4.
Mention/what are the rules/principles of formation of Frequency of distribution?
Solution:
1. The lower limit of the first class should be either 0 or a multiple of 5
2. Exclusive classes should be formed for better continuity
3. The number of classes should be generally between 4 & 15
4. The width of the classes should be kept constant throughout the distribution
5. Avoid open-end classes
6. The classes should be arranged in ascending or descending order.
Question 5.
Mention the rules/principles of the tabulation
Solution:
1. The size of the table should be according to the size of the paper
2. The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
3. If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
4. The table should not be overloaded with number of characteristics, rather can be prepare another table.
5. The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ‘o’ but give (—) dash marks or write N.A
6. Miscellaneous column can be provided for the data which do not fit to the data, such as Ratio, percentages.
7. Ditto ( “ ) marks should not be used, as they may confuse with the no. 11
9. Sources if provided regarding publications, organizations, pages, Journals etc.
Question 6.
The employees of a college can be classified on the basis of sex wise distribution as follows:-
Solution:
Table showing the sex distribution of employess of a college.
Question 7.
The Employees of a college can be classified according to their occupations as :
Solution:
Table-1
The blank table given below represents the number of employees with different occupation in a college
Occupations Number of employees Teaching staff – clerks – Attenders – Security men – Total –
Question 8.
The employees of a commercial Bank can be classified according to their occupations and sex is :
Solution:
Table-2
The blank table given below represents the number of employees with different occupation and their sex in a commercial Bank
Question 9.
The employees of a college can be classified according to their occupations , sex and their marital status is :
Solution:
Table-3 : The blank table given below represents the number of employees with different occupation, sex and their marital status in a college
Question 10.
In a survey of 40 families in a certain locality, the number of children per family was recorded and the following data were obtained.
Represent the data in the form of a discrete frequency distribution.
Solution:
Frequency distribution of the number of children.
Question 11.
Prepare a frequency table from the following table regarding the number fatal accidents occurred in a day in Bangalore in June 2010.
Solution:
Frequency distribution of the number of children.
Question 12.
From the following paragraph prepare a discrete frequency table with the number of letters present in the words .
“Success in the examination confers no right to appointment unless government is satisfied, after such enquiry as may be deemed necessary that the candidate is suitable for appointment to the public service”
Solution:
The number of digits in the above statement: Highest digit = 11 and Lowest digit = 2
Frequency distribution of the number of letters in the words present in the statement
Question 13.
The following are the marks obtained by 50 college students in a certain test.
Take suitable width of the class interval marks using struge’s rule.
Solution:
Here, N= 50; Range = H.V.-L. V. = 49-12 = 37
The number classes as per Sturge’s rule are obtained as follows:
Number of class intervals (K) = 1 + 3.322 logN= 1 + 3.22 log 50 = 1 + (3.22 × 1.6990) = 6.47=7
classes (Approx.) Size/width of class intervals – e = $$\frac{\text { Range }}{\text { Number of class intervals }}=\frac{37}{7}$$
5.28 = 6 (Approx.)
The size/width of each class is 6 and there are 7 classes. Thus, the required continuous frequency distribution with exclusive class intervals width is prepared as :
Frequency distribution of marks of students
Question 14.
The following data gives ages of 32 individuals in a locality. Using Sturge’s rule form a frequency table with exclusive type of class intervals.
Solution:
Here N= 32; Range = H .V.- L.V. = 59- 01 = 58
The number classes as per Sturge’s rule are:
Number of class intervals (K) =1 + 3.22 logN = 1 + 3.22 log (32) = 1+(3.22 × 1.5051) = 5.85=6
classes (Approx.) Size/width of class intervals – e = $$\frac{\text { Range }}{\text { Number of class intervals }}=\frac{58}{6}$$ = 10
(approx). Teh size / width of each class is 10 and there are 6 classes.
Frequency distribution of marks of students
Question 15.
The following are the marks obtained by 50 students in statistics; prepare a frequency table with class intervals of 10 marks.
Solution:
Range: H.V-L.V = 93 – 23 = 70; take width as 10 marks, and then the number of classes will be: 70/10=7.
Frequency distribution of marks of students in statistics test
Question 16.
Prepare a bivariate frequency distribution of the marks in Accountancy & Statistics:
Solution:
Here the both variables are discrete in nature no need to prepare class interval.
Bi-variate frequency table showing Ages (years) of Mothers and Number of children
Question 17.
Below are the ages of husbands and wives prepare a bivariate frequency distribution with suitable width :
Here both are continuous variables form the class intervals as below :
Solution:
Ages of Husbands: Highest age = 47, Lowest age = 25, Difference = 22/(i)5width =5. classes. Ages of wife: Highest age = 47, Lowest age = 21, Difference = 26/(i)5width = 6 (Approx.) classes.
Let X and Y be the ages of Husbands and ages of Wives.
Question 18.
Below are given the marks obtained by a batch of 20 students in mathematics and statistics:
Solution:
Marks in Mathematics: Highest Marks = 72, Lowest marks = 25, Difference = 47/(i)l Owidth 5 = 5 (Approx.) classes.
Marks in statistics: Highest marks = 85, Lowest marks = 20, Difference = 65/(i)10width = 7 (Approx.) classes.
Let x and y be marks inmathematics and marks in statistics.
TABULATION
Question 19.
Give a general format of a table.
Solution:
General format of a table
Question 20.
What are the requisites of a good table? Or What are the General rules to the tabulation?
Solution:
• Size:- the size of the table should be according to the size of the paper with more rows than columns. Exchange of the data can be done by altering the column and rows. A sufficient space should be provided in a particular to enter any new or to alter any affected figures.
• Logical order:- The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
• Identity:- If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
• The table should not be overloaded with number of characteristics, rather can be prepare another table.
• The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ’o’ but give (—) dash marks or write N.A
• Miscellaneous column can be provided for the data which do not fit to the data, such as Radius, percentages.
• Ditto ( “ ) marks should not be used,as they may confuse with the number 11
• Footnote may contain about errors, omissions, remarks
• Sources if provided regarding publications, organizations pages Journals etc.
Question 21.
Elucidate the difference between classification and tabulation.
Solution:
Comparison between Classification and Tabulation:-
The following points may be given as comparison:
• Classification and Tabulation are not two distinct processes. Before tabulation, data are classified and then displayed under different columns and rows of a table.
• Classification is the process of arranging the data in to groups or classes according to common characteristics possessed by the items of the data;
Whereas Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.
• Table contains precise and accurate information, where as classification gives only a classified groups of data.
• Classification reduces the size of the data and brings the similarities together and tabulation facilitates comparison, reveal the trend and tendencies of the data.
Question 22.
In the college out of total of 1200 applications received for I puc admission, 450
were applied for science and 580 applied for commerce and remaining applied for
arts faculty. Tabulate the above information.
Solution:
Table 1
Table showing the distribution of applicants for I year puc to different faculties
Question 23.
In the college out of total of 1200 applications received for I puc admission 780 were boys. 450 were applied for science of which 300 were boys and 580 applied for commerce 250 were girls and remaining applied for arts faculty. Tabulate the above information.
Solution
Table 2
Table showing the sex-wise distribution of applicants for I year puc to different faculties
Question 24.
In a Sigma multinational accountant consultants there are 180 were accountants, 210 were article helpers, 300 were practitioner trainees. Of all the members 30% were women among accountants, 20% in Articles helpers and 15% among trainees. Tabulate the data.
Solution:
Table 3
Table represents the members of Sigma accountants according to cadre and sex
Footnote: * 180 × 30%= 54, 210 × 20% = 42
Question 25.
In a trip organized by a college there were 80 persons each of whom paid Rs.15.50 on an average? There were 60 students each of whom paid Rs.16. members of the teaching staff were charged at higher rate. The number of ser ants was 6 (all males) and they were not charged anything. The number of females was 20 percent of the total of which one was a lady staff member. Tabulate the information.
Solution:
Table 4
Table showing the contribution (in Rs.) details made by members of a college trip.
Question 26.
In a state, there are 30 Medical colleges, 10 Dental colleges and 50 Engineering colleges. Among the Medical colleges, 5 are government colleges, 10 are aided private colleges and remaining are the unaided private colleges. Of the unaided colleges, 5 colleges are run by minority institutions.
Among the Engineering colleges, 10 are government colleges, 20 are aided private colleges and the rest are unaided private colleges. Of the unaided colleges, 10 colleges are run by minority institutions.
Among the dental colleges, 2 are aided private colleges and the rest are the unaided private colleges of which one is run by a minority institutions.
Tabulate the above information.
Solution:
Table showing the Medical, Engineering and Dental colleges run by Government, Private Aided and Private Unaided colleges in a State
Question 27.
The number of cases filed, hearing made and disposed by different bench judges in a day at High court of Karnataka are as given:
(i) Criminal cases filed 12, hearings made in 8 cases and disposed 3,
(ii) Land dispute cases filed 18, hearings made in 12cases and disposed 5,
(iii) Government service cases filed 6, hearings made in 4 cases and disposed 3,
(iv) Cheating cases filed 15; hearing made in 12 cases and disposed 8.
Tabulate the above information.
Solution:
Table showing the different types of cases filed, heard and disposed in a day at High court of Karnataka
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# The Commutative and Associative Properties Homework
```Name: ______________________________ Block: _______
The Commutative and Associative Properties Homework
Commutative Property of Multiplication
Associative Property of Multiplication
Part I: Complete the equations below. Then, identify the property you used.
A. 6 + 3 = 3 + ____
_______________________________________
B. 4 • (7 • 5) = (______) • 5
_______________________________________
C. 4 • ____ = 2 • 4
_______________________________________
D. (2 + 4) + 7 = 2 + (______)
_______________________________________
3 • (6 • 4) = (______) • 4
_______________________________________
F. (3 + 4) + 8 = 3 + (______)
_______________________________________
G. -7 + 3 = 3 + ____
_______________________________________
H. 3 • ____ = 5 • 3
_______________________________________
E.
Part II: Determine if each statement below is true or false. If it is false, correct the statement.
1) The Commutative Property says that numbers being multiplied and divided can be reordered.
2) The Commutative Property of Multiplication states that numbers being multiplied can be
regrouped without affecting the product.
3) 25 – 14 = 14 – 25.
Name: ______________________________ Block: _______
Part III: Consider the expressions below.
Circle any examples of the Commutative Property.
Put a star by any examples of the Associative Property.
Put an “x” through any examples that are neither commutative nor associative.
Hint: Keep in mind that it is possible for a set of expressions to model two different properties.
A)
6+8
8+6
B)
(2 • -3) • 4
2 •(-3 • 4)
C)
5 + (3 + 2) + -1
5 + 3 + (2 + -1)
D)
5-4+2
4-5+2
E)
(9 • 3) + 5
(3 • 9) + 5
F)
(7 ÷ 4) • 9
(4 ÷ 7) • 9
G)
3 + -5 • 4
3 + 4 • -5
H)
(3 • 4) • (2 • 6)
(4 • 2) • (6 • 3)
I)
(7 + 2) + 9
(2 + 7) + 9
J)
7-8-3
8-7–3
K)
(10 + -32) • 16
10 + (-32 • 16)
L)
8 + (6 + 4) + 2
8 + 2 + (4 + 6)
```
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We've all seen those emails go around that go something like this:
Take the month and day of your birthday and
1. Multiply the number of the month by 5
3. Multiply by 4
5. Multiply by 5
6. Add the day of the month
7. Subtract 205
The last two digits are the day of the month, and the first digit(s) is the month number.
So, you have your doubts and go through the calculation to humor the sender. Let's see, my birthday is May 7th, so I should get 507.
1. 5 times 7 is 35
2. 35 plus 7 is 42
3. 42 times 4 is 168
4. 168 plus 13 is 181
5. 181 times 5 is 905
6. 905 plus 7 is 912
7. 912 minus 205 is 507
Amazing!!! How did they do that? It must be math magic.
Well, it's actually pretty straightforward and you can easily make up your own math magic tricks once you see how it's done.
Let's first start at the end and think about the resulting number. The last two digits are the day of the month and the first digit or digits are the month number.
• May 7th = 507
• September 20th = 920
• November 12th = 1112
• January 30 = 130
To get those numbers, we basically multiply the number of the month by 100 and then add the day of the month.
• May 7th = 5 * 100 + 7 = 507
• September 20th = 9 * 100 +20 = 920
• November 12th = 11 * 100 + 12 = 1112
• January 30 = 1 * 100 + 30 = 130
Whatever magic the seven steps above are doing, it must boil down to multiplying the number of the month by 100 and then adding the day of the month. The easiest way to prove that is what is happening is to write out the steps using variables. Lets use 'm' for month and 'd' for day.
Multiply the number of the month by 5 5m Add 7 5m + 7 Multiply by 4 (5m + 7) * 4 Add 13 (5m + 7) * 4 + 13 Multiply by 5 ((5m + 7) * 4 + 13) * 5 Add the day of the month ((5m + 7) * 4 + 13) * 5 + d Subtract 205 ((5m + 7) * 4 + 13) * 5 + d - 205
So, for this trick to work ((5m + 7) * 4 + 13) * 5 + d - 205 must be equal to 100m + d. Let's simplify our equation and see.
((5m + 7) * 4 + 13) * 5 + d - 205 = 100m + d (20m + 28 + 13) * 5 + d - 205 = 100m + d (20m + 41) * 5 + d - 205 = 100m + d 100m + 205 + d - 205 = 100m + d 100m + d = 100m + d
It works! All those 7 steps are really doing is multiplying the month by 100 and adding the day. You can see that it doesn't matter what the values of 'm' and 'd' are, the trick will always work for anyone's birthday.
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# Notes 31 Simple Interest 6-6.
## Presentation on theme: "Notes 31 Simple Interest 6-6."— Presentation transcript:
Notes 31 Simple Interest 6-6
Vocabulary Interest- the amount of money charged for borrowing or using money. Simple interest- is money paid only on the principal. Principal- is the percent charged or earned. Rate of interest- the percent charged or earned.
Additional Example 1: Finding Interest and Total Payment on a Loan
To buy a car, Jessica borrowed \$15,000 for 3 years at an annual simple interest rate of 9%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay? First, find the interest she will pay. I = P r t Use the formula. I = 15,000 0.09 3 Substitute. Use 0.09 for 9%. I = Solve for I.
Jessica will pay \$4050 in interest. You can find the total amount A to be repaid on a loan by adding the principal P to the interest I. P + I = A principal + interest = amount 15, = A Substitute. 19,050 = A Solve for A. Jessica will repay a total of \$19,050 on her loan.
Check It Out: Example 1 To buy a laptop computer, Elaine borrowed \$2,000 for 3 years at an annual simple interest rate of 5%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay?
Additional Example 2: Determining the Amount of Investment Time
Nancy invested \$6000 in a bond at a yearly rate of 3%. She earned \$450 in interest. How long was the money invested? I = P r t Use the formula. 450 = 6,000 0.03 t Substitute values into the equation. 450 = 180t 2.5 = t Solve for t. The money was invested for 2.5 years, or 2 years and 6 months.
Check It Out: Example 2 TJ invested \$4000 in a bond at a yearly rate of 2%. He earned \$200 in interest. How long was the money invested?
Additional Example 3: Computing Total Savings
John’s parents deposited \$1000 into a savings account as a college fund when he was born. How much will John have in this account after 18 years at a yearly simple interest rate of 3.25%? I = P r t Use the formula. I = 1000 Substitute. Use for 3.25%. I = Solve for I. Now you can find the total.
P + I = A Use the formula. = A Substitute. 1585 = A Solve for A. John will have \$1585 in the account after 18 years.
Check It Out: Example 3 Bertha deposited \$1000 into a retirement account when she was 18. How much will Bertha have in this account after 50 years at a yearly simple interest rate of 7.5%?
Additional Example 4: Finding the Rate of Interest
Mr. Johnson borrowed \$8000 for 4 years to make home improvements. If he repaid a total of \$10,320, at what interest rate did he borrow the money? P + I = A Use the formula. I = 10,320 Substitute. I = 10,320 – 8000 = 2320 Subtract from both sides. He paid \$2320 in interest. Use the amount of interest to find the interest rate.
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###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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# Finding an Inverse Algebraically - Concept
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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Once we learn the definition of a function's inverse we learn how to find the algebraic inverse, or how to find the inverse using algebraic methods. There are different methods for finding the inverse, the most common of which is to switch the dependent and independent variables and solve for the dependent variable. This is an important step in learning how to prove the inverse of a function.
Finding the inverse of a funtion Algebraically. So for this particular example, so what we want to do is find an equation for a inverse function. We're given a function here. In this case we know that our equation is a line. 3x-2 we know that's a line therefore we know it's 1 to 1 and it's going to have an inverse.
So how we find a inverse is by switching our x and our y, for this particular problem we don't have a y but what we need to remember is f of x is the same thing as y. So just by replacing f of x with y, 3x-2, I now have an equation using x and y. To find the inverse we switch x and y's, okay? So every time we see an x, we throw in a y, every time we see a y we throw in an x. Okay? And we're used to seeing equations of y equals. So really from here all we want to do is solve for y. Okay?
So do that as we would any other equqtion, add 2 to both sides x+2=3y divided by 3, x+2 divide by 3 is equal to y. And then we found y but we're asked to find f inverse of x. Once we switch our variables solve for y, this is now the inverse okay? So we can replace our y with our inverse notation. x+2 over 3. Okay?
So to find the inverse, it's quite simple. So make sure you if you have an f of x replace it with y. solve through sorry. Switch your x and y's. Solve through for y making sure to replace it with your inverse notation at the end.
Now some teachers do this a little different and I just want to touch on that for a second. You don't have to switch x and y right here, okay? What you could do is actually solve for this x and then switch it. In general, I tend to shy away from that just because I'm so used to solving for y where what I've done is I've done problems where I haven't switched it. Halfway down the problem I forget that I'm solving for x, go back solve for y and I end up with exactly what I started with. Okay?
But if you can keep track what you're solving for, that's perfectly acceptable. For me, I always switch it. Fairly one of my first steps solve for y and then just make sure you replace your inverse notation.
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0 0
The admission fee at an amusement park is \$3.75 for children and \$6.40 for adults.
The admission fee at an amusement park is \$3.75 for children and \$6.40 for adults. On a certain day, 295 people entered the park, and the admission fees collected totaled 1464 dollars. How many children and how many adults were admitted?
number of children equals
This is a repeat of a question that I just answered.
Hi Treena,
Let us assume that the number of children who entered the park is x.
Now, we know that 295 people entered the park on a certain day. Which means the number of adults who entered the park is (295 - x). You want to interpret this as, off the 295 people who entered the park on a given day, we know that x of them were children. Therefore, the balance must be adults.
Therefore, we have
Number of children who entered the park = x
Number of adults who entered the park = 295 - x
The fee for children is \$3.75 and for adults is \$6.40.
Therefore, total amount that x children paid to enter the park is \$3.75 * x
and, total amount that adults paid to enter the park is \$6.40 * (295 - x) = 1888 - 6.40 * x.
Therefore, combining the two amounts we will have the total admission fees collected. That is,
(Total amount paid by children) + (Total amount paid by adults) = Total admission fees collected.
Putting our numbers,
(3.75 * x) + (1888 - 6.40 * x) = 1464.
The expression on the left hand side will simplify to 1888 - (6.40 * x) + (3.75 * x). Combining the x term coefficients we have -6.40 + 3.75 = 2.65.
The expression on the left hand side simplifies to 1888 - 2.65 * x and is equal to 1464.
That is, 1888 - 2.65 * x = 1464. Let us move the 1464 to the left hand side and move the 2.65 * x term to the right. Since we are moving them to opposite sides of the equation, each term will reverse sign.
Therefore, now 1888 - 1464 = 2.65 * x. Or, 424 = 2.65*x.
Dividing both sides by 2.65, gives us (424 / 2.65) = (2.65 / 2.65)*x
Or, 160 = x. Therefore, 160 children entered the park. And (295 - x) adults entered the park - or, (295 - 160) adults - which is 135 adults.
Therefore, number of children = 160 and number of adults = 135
I hope this helps
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# Differentiating Trigonometric Functions
by Batool Akmal
My Notes
• Required.
Learning Material 2
• PDF
DLM Differentiation of Trigonometric Functions Calculus Akmal.pdf
• PDF
Report mistake
Transcript
00:01 Okay, here we're going to look at actually differentiating a function.
00:04 Now, this shouldn't be frightening because you already did this.
00:07 You've done the differential of sin, cos, and tan.
00:10 So anything after that is just going to be easy.
00:13 Let's have a look at the differential of cot of x and what we can do to differentiate it.
00:19 So, we are finding the derivative d by dx of cot of x.
00:25 Now, cot is a new identity for us. So, we don't really know what the answer to this is yet but you can break it down. So, we can write this as d/dx and instead of using cot, why don't we change it to 1 over tan x because we're more familiar with that.
00:43 So, using the identities at the start, we know that cot is 1 over tan and now we just have to differentiate 1 over tan x.
00:50 A couple of ways of doing this but observe that I have a function at the top and a function at the bottom. Hopefully, this is ringing some bells.
01:00 You have a function that is dividing a function, I have just called them u and v.
01:05 So, hopefully, we'll remember that we will have to use the quotient rule.
01:09 Quotient rule, dy/dx is vdudx minus udvdx over v squared.
01:17 In our case, our u is 1 and our v is tan x. When we differentiate u dashed, that's zero, that's always good news, and then when we differentiate tan we get sec squared x.
01:32 Let's apply the quotient rule now. So, we're doing dy/dx, vdudx, so we'll have tan of x multiplied by zero and then minus 1 times with sec squared of x over our v squared, so that's going to be tan x all squared. We're basically done but there are things we can do to simplify this to give us a nice answer.
02:01 So, we have minus sec squared of x at the top because the first term just becomes zero and then we have tan squared of x at the bottom.
02:10 Now instead of leaving it like this, let's just use some more identities.
02:14 So you know that sec from the start is cos, okay? So, if sec is cos, sec squared will be 1 over cos squared.
02:23 Apology, sec is a 1 over cos, so sec squared will be 1 over cos, all squared.
02:28 It's better if I write that down. So, we have 1 over cos x.
02:33 Remember, that it's squared because we're dealing with sec squared and then we also know that we were timesing this with 1 over tan squared x.
02:44 Tan is the same as sin over cos. So, if we write that here, we've got tan x is the same as sin x over cos x, one of the identities that we should know from previous use.
02:57 So, tan is sin over x. So, we can change this time to sin over x but remember that because it's 1 over tan squared, it would be cos over sin.
03:08 So if I did 1 over tan x that should be cos x over sin x. We just swap it.
03:14 So we're flipping the fraction. So if we do 1 over this, it becomes that.
03:19 Okay? So let me just rewrite that. So, I've got minus 1 over cos x, all squared and then if I multiply this with 1 over tan squared x, that should be cos x over sin x, all squared.
03:35 I hope you can see what's about to happen. Let me just do this in 2 steps.
03:40 So we have minus 1 when you square 1, you just get 1.
03:44 At the bottom, I'll have cos x squared, I have cos x squared when I square this and then I have sin x squared here. The cos x squared, cos x squared cancels out.
03:57 We now end up with minus 1 over sin squared x and this is done but you also can show your knowledge off a little bit more by using the fact that 1 over sin is cosec. So, you can write this as minus cosec squared x and if you go right to the beginning, you will see that we mentioned that the derivative of cot x is minus cosec squared x and I said that these are the kind of functions that you don't actually need to learn the derivative for because you can always just derive it. So our last example now.
04:33 We are differentiating 5 plus sin x divided by 2 plus cos x.
04:38 Quite obvious looking, it's a function being divided by a function.
04:43 So well talk about what rules we have to use.
04:45 You'll have to recall the methods that we've learned previously in order to differentiate this.
04:50 So we have a function y equals to 5 plus sin x over 2 plus cos x.
04:58 The easiest thing to do without having to rearrange or over complicate things is just to split it as u and v and obviously, from here you can see that in order to differentiate, we are going to have to use the quotient rule.
05:15 So, we know that the quotient rule is dy/dx is vdudx minus udvdx over v squared.
05:23 Let's put it all together here. So u equals to 5 plus sin x, v equals to 2 plus cos x.
05:32 Let's differentiate each one. 5 is a constant, so that goes away and sin x remember, differentiates to cos x. V dashed, 2 is a constant so that goes away, and cos x differentiates to minus sin x. So, nice and straightforward so far.
05:50 Let's put it all into dy/dx. Vdudx minus vdudx, so we can multiply cos x with 2 plus cos x, then we have a minus in the middle and then we have minus sin x multiplied by 5 plus sin x and it's all over v squared. So that's 2 plus cos x squared.
06:17 We are basically done with the differentiation, all we have to do is tidy this up.
06:22 So I can multiply cos through. So I've got 2 cos x plus cos squared x, leave the minus for now, this little minus here. Let's just expand this out first.
06:34 So if I leave the minus out, leave my brackets. So, I have minus 5 sin x and then I have minus sin squared x, all over 2 plus cos x, multiply this 2 now with the minus just to get rid of the brackets.
06:54 So we have 2 cos x, plus cos squared x, plus 5 sin x, and then plus sin squared x.
07:04 Okay I was hoping that this would happen, which is good news.
07:10 If you've observe here, you have a cos squared x plus sin squared x.
07:14 So if you remember previously, one of the identities says that cos squared x if I add this with this. Cos squared x plus sin squared x equals to 1.
07:23 So that now leaves us with 2 cos x plus 5 sin x. Again, this term here and this term here, adds together to give you 1. So we've got plus 1. Let me just write that here, the identity.
07:38 So sin squared x plus cos squared x equals to 1. So, when you write them next to each other, you'll be able to see that that goes to 1 and then it's all being divided by 2 cos x squared.
07:49 So the derivative of something this complicated is just this answer here and again, you can find your differential at different points by just substituting values in.
08:01 So now it's your turn to practice everything that we've learned.
08:05 It's now a chance for you to apply the differentials of sin, cos, and tan, and also all the other rules that we've learned previously.
08:12 So as you differentiate, don't forget the chain rule, don't forget the product rule, don't forget the quotient rule, and we're bringing it all together in this exercise. Good luck.
### About the Lecture
The lecture Differentiating Trigonometric Functions by Batool Akmal is from the course Differentiation of Trigonometric Functions.
### Included Quiz Questions
1. 5[sin²(x) + 2xsin(x)cos(x)]
2. [sin²(x) + 2xsin(x)cos(x)]
3. 5[sin²(x) - 2xsin(x)cos(x)]
4. -5[sin²(x) + 2xsin(x)cos(x)]
5. 5[cos²(x) + 2xsin(x)cos(x)]
1. 2tan(x)sec²(x)
2. 2tan(x)
3. tan(x)sec²(x)
4. 2sec²(x)
5. -2tan(x)sec²(x)
1. 0
2. cos²(x)/sin²(x) -1
3. -2
4. 1
5. sec²(x) - cot²(x) - 1
1. dy/dx = (cos(x) - sin(x) + 1)/(1-sin(x))²
2. dy/dx = (cos(x) + sin(x) - 1)/(1-sin(x))²
3. dy/dx = (cos(x) - sin(x) + sin²(x) - cos²(x))/(1-sin(x))²
4. dy/dx = (cos(x) - sin(x) - sin²(x) + cos²(x))/(1-sin(x))²
5. dy/dx = (cos(x) - sin(x) - 1)/(1-sin(x))²
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#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 10 Subquestion (iv)
Discontinuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} \frac{e^{x}-1}{\log (1+2 x)}, \text { if } x \neq 0 \\\\ 7, \text { if } x=0 \end{array}\right)$
We observe,
\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\log (1+2 x)} \\\\ &=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\frac{2 x \log (1+2 x)}{2 x}} \end{aligned} [Multiplying and dividing the denominator by $2x$]
\begin{aligned} &=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\left(\frac{e^{x}-1}{x}\right)}{\left(\frac{\log (1+2 x)}{2 x}\right)} \\\\ \end{aligned}
$=\frac{1}{2} \frac{\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x}\right)}=\frac{1}{2} \times \frac{1}{1}=\frac{1}{2}$
And $f\left ( 0 \right )=7$
$\lim _{x \rightarrow 0} f(x) \neq f(0)$
Thus, $f\left ( x \right )$ is discontinuous at $x=0$.
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# Problem of the Week Problem A and Solution Balancing Act
## Problem
Bailey is in charge of sending out boxes from a distribution centre. The contents of the boxes are identified by shapes stamped on them: a heart, a moon, or a sun. All boxes with the same stamp have the same mass, and the cost of sending a box depends on its mass. Bailey has a balance scale and a few standard weights to help with the job. The following diagrams show what Bailey observed when arranging some of the boxes and standard weights on the scales.
Find the mass of each box.
## Solution
From the diagrams we notice the following.
• One heart box has a mass of $$2$$ kg.
• One moon box and one sun box have a total mass of $$24$$ kg.
• One moon box and one sun box have the same total mass as one moon box and two heart boxes.
From this, we can conclude that one moon box and two heart boxes have a total mass of $$24$$ kg. Also, two heart boxes have the same mass as one sun box.
Since one heart box has a mass of $$2$$ kg, then two heart boxes have a mass of $$4$$ kg. Therefore, one sun box has a mass of $$4$$ kg.
This means $$4 \text{kg} + \text{(mass of a moon box)} = 24$$ kg. Since $$4 + 20 = 24$$, we can determine that one moon box must have a mass of $$20$$ kg.
Teacher’s Notes
The idea of a balance scale is a nice analogy for an algebraic equation. We can represent the information in the problem using equations with variables to represent the masses of the different types of boxes. Here is one way to solve the problem algebraically.
Let $$x$$ represent the mass of a heart box.
Let $$y$$ represent the mass of a sun box.
Let $$z$$ represent the mass of a moon box.
From the information in the diagrams, we can write the following equations: \begin{align} x &= 2 \tag{1}\\ y + z &= 24 \tag{2}\\ y + z &= 2x + z \tag{3} \end{align}
From equation $$(1)$$, we already know that a heart box has a mass of $$2$$ kg. From equations $$(2)$$ and $$(3)$$, we notice that the left sides are the same, so the right sides of the equations must be equal to each other. This means we know: $2x + z = 24 \tag{4}$
Now, substituting $$x = 2$$ into equation $$(4)$$, we get $2(2) + z = 24$
Subtracting $$4$$ from each side of this equation, we get $z = 20$
Finally, substituting $$z = 20$$ into equation $$(2)$$, we get $y + 20 = 24$
Subtracting $$20$$ from each side of this equation, we get $y = 4$
So, a heart box has a mass of $$2$$ kg, a sun box has a mass of $$4$$ kg, and a moon box has a mass of $$20$$ kg.
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# Congruence Statements
## Learn how to write congruence statements and use congruence statements to determine the corresponding parts of triangles.
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Congruence Statements
What if you were told that \begin{align*}&\triangle ABC \cong \triangle XYZ\end{align*}? How could you determine which side in \begin{align*}\triangle XYZ\end{align*} is congruent to \begin{align*}\overline{BA}\end{align*} and which angle is congruent to \begin{align*}\angle{C}\end{align*}? After completing this Concept, you'll be able to use congruence statements to state which sides and angles are congruent in congruent triangles.
### Watch This
Watch the first part of this video.
### Guidance
When stating that two triangles are congruent, use a congruence statement. The order of the letters is very important, as corresponding parts must be written in the same order. Notice that the congruent sides also line up within the congruence statement.
\begin{align*}\overline{AB} \cong \overline{LM}, \overline{BC} \cong \overline{MN}, \overline{AC} \cong \overline{LN}\end{align*}
We can also write this congruence statement several other ways, as long as the congruent angles match up. For example, we can also write \begin{align*}\triangle ABC \cong \triangle LMN\end{align*} as:
\begin{align*}& \triangle ACB \cong \triangle LNM \qquad \triangle BCA \cong \triangle MNL\\ & \triangle BAC \cong \triangle MLN \qquad \triangle CBA \cong \triangle NML\\ & \triangle CAB \cong \triangle NLM\end{align*}
One congruence statement can always be written six ways. Any of the six ways above would be correct.
#### Example A
Write a congruence statement for the two triangles below.
To write the congruence statement, you need to line up the corresponding parts in the triangles: \begin{align*}\angle R \cong \angle F, \angle S \cong \angle E,\end{align*} and \begin{align*}\angle T \cong \angle D\end{align*}. Therefore, the triangles are \begin{align*}\triangle RST \cong \triangle FED\end{align*}.
#### Example B
If \begin{align*}\triangle CAT \cong \triangle DOG\end{align*}, what else do you know?
From this congruence statement, we can conclude three pairs of angles and three pairs of sides are congruent.
\begin{align*}& \angle C \cong \angle D && \angle A \cong \angle O && \angle T \cong \angle G\\ & \overline{CA} \cong \overline{DO} && \overline{AT} \cong \overline{OG} && \overline{CT} \cong \overline{DG}\end{align*}
#### Example C
If \begin{align*}\triangle BUG \cong \triangle ANT\end{align*}, what angle is congruent to \begin{align*}\angle{N}\end{align*}?
Since the order of the letters in the congruence statement tells us which angles are congruent, \begin{align*}\angle{N} \cong \angle{U}\end{align*} because they are each the second of the three letters.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
If \begin{align*} \triangle ABC \cong \triangle XYZ\end{align*}, then \begin{align*} \overline{BA} \cong \overline{YX}\end{align*} and \begin{align*} \angle C \cong \angle Z\end{align*}.
### Vocabulary
To be congruent means to be the same size and shape. Two triangles are congruent if their corresponding angles and sides are congruent. The symbol \begin{align*}\cong\end{align*} means congruent.
### Guided Practice
1. If \begin{align*}\triangle ABC \cong \triangle DEF\end{align*}, what else do you know?
2. If \begin{align*}\triangle KBP \cong \triangle MRS\end{align*}, what else do you know?
3. If \begin{align*}\triangle EWN \cong \triangle MAP\end{align*}, what else do you know?
1. From this congruence statement, we know three pairs of angles and three pairs of sides are congruent. \begin{align*} \angle{A} \cong \angle{D}, \angle{B} \cong \angle{E}, \angle{C} \cong \angle{F}\end{align*}, \begin{align*}\overline{AB} \cong \overline{DE}, \ \overline{BC} \cong \overline{EF}, \ \overline{AC} \cong \overline{DF}\end{align*}.
2. From this congruence statement, we know three pairs of angles and three pairs of sides are congruent. \begin{align*} \angle{K} \cong \angle{M}, \angle{B} \cong \angle{R}, \angle{P} \cong \angle{S}\end{align*}, \begin{align*}\overline{KB} \cong \overline{MR}, \ \overline{BP} \cong \overline{RS}, \ \overline{KP} \cong \overline{MS}\end{align*}.
3. From this congruence statement, we know three pairs of angles and three pairs of sides are congruent. \begin{align*} \angle{E} \cong \angle{M}, \angle{W} \cong \angle{A}, \angle{N} \cong \angle{P}\end{align*}, \begin{align*}\overline{EW} \cong \overline{MA}, \ \overline{WN} \cong \overline{AP}, \ \overline{EN} \cong \overline{MP}\end{align*}.
### Practice
For questions 1-4, determine if the triangles are congruent using the definition of congruent triangles. If they are, write the congruence statement.
1. Suppose the two triangles to the right are congruent. Write a congruence statement for these triangles.
2. Explain how we know that if the two triangles are congruent, then \begin{align*}\angle{B} \cong \angle{Z}\end{align*}.
Suppose \begin{align*}\triangle TBS \cong \triangle FAM\end{align*}.
1. What angle is congruent to \begin{align*}\angle B\end{align*}?
2. What side is congruent to \begin{align*}\overline{FM}\end{align*}?
3. What side is congruent to \begin{align*}\overline{SB}\end{align*}?
Suppose \begin{align*}\triangle INT \cong \triangle WEB\end{align*}.
1. What side is congruent to \begin{align*}\overline{IT}\end{align*}?
2. What angle is congruent to \begin{align*}\angle W\end{align*}?
3. What angle is congruent to \begin{align*}\angle I\end{align*}?
Suppose \begin{align*}\triangle ADG \cong \triangle BCE\end{align*}.
1. What side is congruent to \begin{align*}\overline{CE}\end{align*}?
2. What side is congruent to \begin{align*}\overline{DA}\end{align*}?
3. What angle is congruent to \begin{align*}\angle G\end{align*}?
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# square root of 800
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We all have a standard way of calculating a number. A couple weeks ago I did a little experiment. I decided to see the effect of rounding a number. I rounded to the next whole number, and then I rounded it to the next whole number. I think I got a decent result and it’s something that I don’t think about. When it comes to numbers, I round to the next whole number and I do that in all of my calculations.
So if we round a number to the next whole number it would show that the answer is 800. Then if we round it to the next whole number it would show that the answer is 800. If we round it to the next whole number it would show that the answer is 800. How does this affect our calculations? When we round numbers to the next whole number, the whole number is reduced by the number. For example, if we round 3.
In this case, 3 is rounded to the next whole number (4), reducing it to the next whole number (4). This causes the answer to be 800.
The reason we rounded to the next whole number is to make our calculations easier. We do this by reducing the number to the next whole number. For example, we use the fact that 1.5 is 1.5. When we round 3 to the next whole number 4, we get 1.5. This means that 1.5 is the answer. In another example, we round 1.5 to the next whole number 2. This causes the answer to be 800.
This trick can be used to solve many problems. The question is, how many steps should we take to reach your house? We can use the fact that 2 divided by 3 is 2. The answer is 6. This is because 2 is the answer to the previous problem. This is because we were able to use the fact that 2 divided by 3 is 2.
This can be used in a lot of different ways. It’s used as a base for your speedometer or a base for the fact that the speedometer is on the right side of the car. We used it to solve the number 7 puzzle in the math class. I also made a YouTube video on the number 7 puzzle. But I’m not going to say too much about it.
The way we solved the number 7 problem was by using the fact that 2 divided by 3 is 2. This is because 2 is the answer to the previous problem. This is because we were able to use the fact that 2 divided by 3 is 2.
So how do we solve the number 7 puzzle? Well, we were able to do it using a calculator. Now, if you don’t remember the name of the calculator, the first thing you’ll find is that it’s actually a calculator. A calculator is a number system that uses a number to represent the number of digits in a string. But it’s also a way of saying that you’re supposed to be solving the number 7 puzzle.
To solve the number 7 puzzle, we’re going to need a number system that has 7 digits, and in our case, we’re going to use a number system that has a leading zero. So in our case, our number system will have seven numbers.
Since the calculator itself is a very simple number system, if we were to just use it to represent 800, we would get 800 as the number, but since we are supposed to solve the number 7 puzzle, we need to round it down to get to 800.
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# 4.16 Matrix nullspace basis
We are ready to prove that the fundamental solutions of $A\mathbf{x}=\mathbf{0}$ are a basis for $N(A)$. To set up the notation, suppose $A$ is a $m\times n$ matrix, $R$ is a RREF matrix obtained by doing row operations to $A$, the number of columns of $R$ with a leading entry is $r$ and the number of columns with no leading entry is $k$, so $r+k=n$. Let the numbers of the columns with no leading entry be
$j_{1}
and the numbers of the columns that do have a leading entry be
$i_{1}
so that the numbers $j_{1},\ldots,j_{k},i_{1},\ldots,i_{r}$ contain all of the numbers $1,\ldots,n$ exactly once. The variable corresponding to a column with no leading entry is called a free parameter because its value can be chosen freely. There are $k$ fundamental solutions to $A\mathbf{x}=\mathbf{0}$, with the $j$th fundamental solution defined to be the one where the $j$th free parameter is 1 and all the other free parameters are 0.
###### Theorem 4.16.1.
The fundamental solutions to $A\mathbf{x}=\mathbf{0}$ are a basis of the nullspace $N(A)$.
Before we do the proof, let’s work an example to illustrate how it will go. Free parameters and their coefficients will be coloured red. Take
$R=\begin{pmatrix}{\color[rgb]{1,0,0}0}&1&{\color[rgb]{1,0,0}2}&0&{\color[rgb]{% 1,0,0}3}\\ {\color[rgb]{1,0,0}0}&0&{\color[rgb]{1,0,0}0}&1&{\color[rgb]{1,0,0}4}\\ {\color[rgb]{1,0,0}0}&0&{\color[rgb]{1,0,0}0}&0&{\color[rgb]{1,0,0}0}\end{pmatrix}$
so that we have $m=3$ and $n=5$. The columns with no leading entry are 1, 3, and 5 so $k=3$, ${\color[rgb]{1,0,0}j_{1}}=1,{\color[rgb]{1,0,0}j_{2}}=3,{\color[rgb]{1,0,0}j_{% 3}}=5$, and variables ${\color[rgb]{1,0,0}x_{1}}$, ${\color[rgb]{1,0,0}x_{3}}$, and ${\color[rgb]{1,0,0}x_{5}}$ are the free parameters. The columns with leading entries are 2 and 4, so $r=2$ and $i_{1}=2,i_{2}=4$. The equations corresponding to $R\mathbf{x}=\mathbf{0}$ are
$\displaystyle x_{2}+{\color[rgb]{1,0,0}2x_{3}}+{\color[rgb]{1,0,0}3x_{5}}$ $\displaystyle=0$ $\displaystyle x_{4}+{\color[rgb]{1,0,0}4x_{5}}$ $\displaystyle=0$
(missing off the final one, since it just says $0=0$). These equations show that once values for the free parameters are known, the other variables $x_{2}$ and $x_{4}$ are completely determined by those values — we have $x_{2}={\color[rgb]{1,0,0}-2x_{3}-3x_{5}}$ and $x_{4}={\color[rgb]{1,0,0}-4x_{5}}$.
The three fundamental solutions are
$\mathbf{s}_{1}=\begin{pmatrix}{\color[rgb]{1,0,0}1}\\ 0\\ {\color[rgb]{1,0,0}0}\\ 0\\ {\color[rgb]{1,0,0}0}\end{pmatrix},\mathbf{s}_{2}=\begin{pmatrix}{\color[rgb]{% 1,0,0}0}\\ -2\\ {\color[rgb]{1,0,0}1}\\ 0\\ {\color[rgb]{1,0,0}0}\end{pmatrix},\mathbf{s}_{3}=\begin{pmatrix}{\color[rgb]{% 1,0,0}0}\\ -3\\ {\color[rgb]{1,0,0}0}\\ -4\\ {\color[rgb]{1,0,0}1}\end{pmatrix}.$
These are linearly independent, because if
$a_{1}\mathbf{s}_{1}+a_{2}\mathbf{s}_{2}+a_{3}\mathbf{s}_{3}=\mathbf{0}$
then row 1 of this equation shows that $a_{1}=0$, row 3 shows that $a_{2}=0$, and row 5 shows that $a_{3}=0$. To show that the fundamental solutions span, suppose $\mathbf{s}=\begin{pmatrix}s_{1}\\ \vdots\\ s_{5}\end{pmatrix}$ is any vector such that $R\mathbf{s}=\mathbf{0}$. We claim that $\mathbf{s}$ is equal to $s_{1}\mathbf{s}_{1}+s_{3}\mathbf{s}_{2}+s_{5}\mathbf{s}_{3}$. Certainly these two vectors have the same entries in rows 1, 3, and 5, since the entries in these rows are $s_{1},s_{3}$, and $s_{5}$. What about the entries in rows 2 and 4, that is, the values of $x_{2}$ and $x_{4}$? As above, these entries are completely determined by the values of the free parameters. The free parameters are the same for both vectors, so the values of $x_{2}$ and $x_{4}$ are the same as well. We have shown that any solution $\mathbf{s}$ is in the span of the fundamental solutions, so we are done.
###### Proof.
Theorem 3.9.1 shows that $N(A)=N(R)$, so we will show that the fundamental solutions are a basis of $N(R)$.
First we show that the fundamental solutions $\mathbf{s}_{1},\ldots,\mathbf{s}_{k}$ are linearly independent. Suppose that
$\sum_{j}a_{j}\mathbf{s}_{j}=\mathbf{0}.$ (4.6)
The first fundamental solution corresponds to the variable for column $j_{1}$, so $\mathbf{s}_{1}$ has a 1 in row $j_{1}$ and all the other fundamental solutions have a 0 there. Thus the entry in row $j_{1}$ on the left hand side of (4.6) is $a_{1}$, so $a_{1}=0$. Similarly all the other coefficients are zero.
Now let $\mathbf{s}$ be any solution of $A\mathbf{x}=\mathbf{0}$ and let the entry of $\mathbf{s}$ in row $i$ be $s_{i}$. We are going to show that
$\mathbf{s}=\sum^{k}_{i=1}s_{j_{i}}\mathbf{s}_{i}.$
The left hand side and right hand side of this claimed equation are solutions to $R\mathbf{x}=\mathbf{0}$ with the same free parameter values $s_{j_{1}},\ldots,s_{j_{k}}$. We just have to show that the values of the other variables are the same. But the values of the variables that aren’t free parameters are uniquely determined by the values of the free parameters, because by the RREF property, for $1\leqslant a\leqslant r$ the only equation containing $x_{i_{a}}$ is the one coming from row $a$ of $R$ which has the form
$x_{i_{1}}+\cdots=0$
where the only other variables occurring with nonzero coefficient are free parameters. Since $\mathbf{s}$ and $\sum^{k}_{i=1}s_{j_{i}}\mathbf{s}_{i}$ are solutions to $R\mathbf{x}=\mathbf{0}$ with the same free parameter values $s_{j_{1}},\ldots,s_{j_{k}}$, they are equal.
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# Difference between revisions of "2017 AMC 10A Problems/Problem 11"
## Problem
The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$. What is the length $\textit{AB}$?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$
## Solution 1
In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$. However, we need to find the region containing all points within $3$ units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$):
$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$, where $x$ is equal to the length of our line segment.
Solving, we find that $x = \boxed{\textbf{(D)}\ 20}$.
## Solution 2
Because this is just a cylinder and $2$ "half spheres", and the radius is $3$, the volume of the $2$ half spheres is $\frac{4(3^3)\pi}{3} = 36 \pi$. Since we also know that the volume of this whole thing is $216 \pi$, we do $216-36$ to get $180 \pi$ as the volume of the cylinder. Thus the height is $180 \pi$ divided by the area of the base, or $\frac{180 \pi}{9\pi}=20$, so our answer is $\boxed{\textbf{(D)}\ 20}.$
~Minor edit by virjoy2001
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# How do you solve 2j/7 - 1/7 = 3/14?
Jul 14, 2015
Rewrite with a common denominator or multiply through to remove the denominators then solve using standard arithmetic manipulations.
#### Explanation:
Note that there are two possible answers depending upon the interpretation of $2 \frac{j}{7}$
Interpretation 1 : $2 \frac{j}{7}$ is intended to be a multiplication: $2 \left(\frac{j}{7}\right)$
$2 \left(\frac{j}{7}\right) - \frac{1}{7} = \frac{3}{14}$
If we multiply all terms on both sides by $14$ we get
$\textcolor{w h i t e}{\text{XXXX}}$$4 j - 2 = 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow j = \frac{5}{4}$
Interpretation 2: $2 \frac{j}{7}$ is intended to be a mixed fraction$2 + \left(\frac{j}{7}\right)$
(It is unusual, but not impossible, that a variable is part of a mixed fraction).
$2 + \left(\frac{j}{7}\right) - \frac{1}{7} = \frac{3}{14}$
Again, if we multiply all terms on both sides by $14$ we get
$\textcolor{w h i t e}{\text{XXXX}}$$28 + 2 j - 2 = 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow 2 j + 26 = 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow 2 j = - 23$
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow j = - \frac{23}{2}$
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# Given matrix A and matrix B. Find (if possible) the matrices: (a) AB (b) BA. A=begin{bmatrix}1 & 2 &3&4end{bmatrix} , B=begin{bmatrix}1 2 3 4 end{bmatrix}
Given matrix A and matrix B. Find (if possible) the matrices: (a) AB (b) BA. $A=\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right],B=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$
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Step 1
the given matrices are:
$A=\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right],B=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$
we have to find (if possible) the matrices:
(a)AB
(b)BA
Step 2
as we know the product of the two matrices is possible if the former matrix in the product has no. of columns equal to the number of rows of the latter matrix.
for example if we have two matrices A and B having orders respectively then for product AB to be defined the number of columns of matrix A should be equal to the number of rows of matrix B that implies n=p and for product BA to be defined the number of columns of matrix B should be equal to the number of rows of matrix A that implies q=m therefore as the given matrices are:
matrix A has order $1×4$, that implies matrix A has 1 row and 4 columns.
matrix B has order $4×1$, that implies matrix B has 4 rows and 1 column.
Step 3
now as no. of columns of matrix A is equal to number of rows of matrix B.
therefore the product AB is defined.
therefore,$AB=\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$
$=\left[\begin{array}{cccc}1×1& 2×2& 3×3& 4×4\end{array}\right]$
$=\left[\begin{array}{cccc}1& 4& 9& 16\end{array}\right]$
$\left[\begin{array}{c}30\end{array}\right]$
therefore, $AB=\left[\begin{array}{c}30\end{array}\right]$
Step 4
now as number of columns of matrix B is equal to number of rows of matrix A. therefore the product BA is defined.
therefore, $BA=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]$
$=\left[\begin{array}{cccc}1×1& 1×2& 1×3& 1×4\\ 2×1& 2×2& 2×3& 2×4\\ 3×1& 3×2& 3×3& 3×4\\ 4×1& 4×2& 4×3& 4×4\end{array}\right]$
$=\left[\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 6& 8\\ 3& 6& 9& 12\\ 4& 8& 12& 16\end{array}\right]$
therefore, $BA=\left[\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 6& 8\\ 3& 6& 9& 12\\ 4& 8& 12& 16\end{array}\right]$
Jeffrey Jordon
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Parabolas: Definition, Reflectors/Collectors, Derivation of Equations
# Parabolas: Definition, Reflectors/Collectors, Derivation of Equations
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
• PRACTICE (online exercises and printable worksheets)
Parabolas were introduced in the Algebra II curriculum. Links to these earlier lessons are given below, together with concepts that you must know from each section.
Be sure to check your understanding by doing some of each problem type in these earlier exercises.
These concepts are then further explored in this current lesson.
• Parabolas
You must know the:
• definition of a parabola as the set of points in a plane that are the same distance from a fixed point and a fixed line
• terminology: focus (the fixed point); directrix (the fixed line); vertex
• reflecting property of parabolas
• collecting property of parabolas
• Equations of Simple Parabolas
• Know how to derive the equation ‘$\,x^2 = 4py\,$’ using the definition.
The equation $\,x^2 = 4py\,$ is one of the two standard forms for a parabola.
The other standard form, $\,y^2 = 4px\,,$ is derived on this page (below).
• The parabola described by $\,x^2 = 4py\,$ is a function of $\,x\,$;
it can be equivalently written as $\displaystyle\,y = \frac{1}{4p}x^2\,.$
• Know how to use graphical transformations to describe all parabolas with directrix parallel to the $x$-axis.
• Know the significance of the parameter $\,p\,$ in the equations for a parabola.
• Be able to graph the equation $\,x^2 = 4py\,$ (and shifted versions).
In particular, be able to:
• determine if the parabola is concave up or concave down
• find the coordinates of the focus
• find the equation of the directrix
• plot an additional point that gives a sense of the ‘width’ of the parabola
You can play with parabolas at right: Move the focus. How does the shape change as the focus moves closer to the directrix? How does the shape change as the focus moves farther away from the directrix? Move the directrix. When the directrix is parallel to the $y$-axis, which way does the parabola open? When the directrix is parallel to the $x$-axis, which way does the parabola open?
Below is some information that wasn't covered in the Algebra II curriculum.
However, it is assumed that you've mastered the exercises from the two earlier parabola lessons
## Axis of Symmetry for a Parabola
Recall that a parabola is determined by two pieces of information:
• a line (the directrix)
• a point not on the line (the focus)
The axis of symmetry for a parabola is the line through the focus that is perpendicular to the directrix.
If the parabola is ‘folded’ along this line, then ‘half’ the parabola perfectly coincides with the other half.
All conic sections have (at least one) axis of symmetry.
For the standard forms of conics, the axes of symmetry are always the $x$-axis and/or the $y$-axis.
For the standard form $\,x^2 = 4py\,$ of a parabola, the axis of symmetry is the $\,y\,$-axis.
For the standard form $\,y^2 = 4px\,$ of a parabola (derived next), the axis of symmetry is the $\,x\,$-axis.
## Deriving the standard form $\,y^2 = 4px\,$ for a parabola
Place a parabola with its vertex at the origin, as shown at right. If we put the focus on the $\,x$-axis, then the directrix will be parallel to the $\,y$-axis. Why? The axis of symmetry passes through the vertex (the origin) and the focus, hence is the $x$-axis. The directrix is perpendicular to the axis of symmetry. Or, if we put the directrix parallel to the $\,y$-axis, then the focus will be on the $\,x$-axis. Why? The axis of symmetry is perpendicular to the directrix and contains the vertex (the origin), hence is the $x$-axis. The focus always lies on the axis of symmetry. In either case, let $\,p \ne 0\,$ denote the $\,x$-value of the focus. Thus, the focus has coordinates $\,(p,0)\,.$ Although the sketch at right shows the situation where $\,p\gt 0\,,$ the following derivation also holds for $\,p \lt 0\,.$
Since the vertex is a point on the parabola, the definition of parabola dictates that it must be the same distance from the focus and the directrix.
Thus, the directrix must cross the $\,x\,$-axis at $\,-p\,$; indeed, every $\,x$-value on the directrix equals $\,-p\,.$
Let $\,(x,y)\,$ denote a typical point on the parabola.
The distance from $\,(x,y)\,$ to the focus $\,(p,0)\,$ is found using the distance formula: $$\cssId{s70}{\sqrt{(x-p)^2 + (y-0)^2 } = \sqrt{(x-p)^2 + y^2}} \tag{1}$$
To find the distance from $\,(x,y)\,$ to the directrix, first drop a perpendicular from $\,(x,y)\,$ to the directrix.
This perpendicular intersects the directrix at $\,(-p,y)\,.$
The distance from $\,(x,y)\,$ to the directrix is therefore the distance from $\,(x,y)\,$ to $\,(-p,y)\,$: $$\tag{2} \cssId{s75}{\sqrt{(x-(-p))^2 + (y-y)^2} = \sqrt{(x+p)^2}}$$
From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal: $$\cssId{s77}{\sqrt{(x-p)^2 + y^2} = \sqrt{(x+p)^2}}$$
This equation simplifies considerably, as follows:
Squaring both sides: $(x-p)^2 + y^2 = (x+p)^2$ Multiplying out: $x^2 - 2px + p^2 + y^2 = x^2 + 2px + p^2$ Subtracting $\,x^2 + p^2\,$ from both sides: $y^2 - 2px = 2px$ Adding $\,2px\,$ to both sides: $y^2 = 4px$
This is the second standard form for a parabola, where the axis of symmetry is the $x$-axis.
The most critical thing to notice is the coefficient of $\,x\,,$ since it holds the key to locating the focus of the parabola.
As an example, consider the equation $\,y^2 = 10x\,.$
Comparing with $\,y^2 = 4px\,,$ we see that $\,10 = 4p\,,$ or $\,p = \frac{10}4 = \frac 52\,.$
Thus, $\,y^2 = 10x\,$ graphs as a parabola with vertex at the origin and focus $\displaystyle\,(\frac 52,0)\,.$
## Standard Forms for Parabolas
In summary, we have:
STANDARD FORM OF PARABOLA AXIS OF SYMMETRY VERTEX FOCUS DIRECTRIX $x^2 = 4py$ $y$-axis $(0,0)$ $(0,p)$ $y = -p$ $y^2 = 4px$ $x$-axis $(0,0)$ $(p,0)$ $x = -p$
## Sketching Parabolas
You do not need to memorize lots of things to sketch parabolas in standard form!
The key thing to memorize is that the important coefficient is $\,4p\,,$
where the size (absolute value) of $\,p\,$ gives the distance from the focus to the origin.
You'll be letting the equation tell you the proper shape, as indicated next:
• recognize form:
Recognize that you have an equation of the form $\,x^2 = 4py\,$ or $\,y^2 = 4px\,.$
If needed, find the discriminant to jog your memory: $\,B^2 - 4AC = 0\,$ indicates a parabola.
There's only one interesting coefficient in these equations ($\,4p\,$), so it is referred to as ‘the coefficient’ below.
• isolate squared term:
Isolate the squared term.
That is, get the squared term (with a coefficient of $\,1\,$) all by itself on one side of the equation.
• determine shape (opening up/down/left/right):
Since a perfect square is always nonnegative, the other side of the equation must also be nonnegative.
Depending on the equation, this will force one of the following four situations:
• all the $\,x$-values positive (parabola opens to the right)
• all the $\,x$-values negative (parabola opens to the left)
• all the $\,y$-values positive (parabola is concave up)
• all the $\,y$-values negative (parabola is concave down)
The examples below illustrate how this works.
Use this information to (lightly) sketch the proper shape.
Remember that the vertex is the origin.
• plot the focus:
The focus is always inside the parabola.
To find its distance from the origin, throw away any minus signs (as needed) on the coefficient and divide by $\,4\,.$
Why? The coefficient is $\,4p\,,$ and $\,|p|\,$ gives the distance from the focus to the origin.
Or, you can set the coefficient to $\,4p\,$ and solve for $\,p\,$ (which will include the correct sign).
• draw in the directrix:
Sketch the directrix.
It is the same distance from the origin as the focus, on the opposite side.
• plot easy points to get the proper ‘width’:
There are always three easy points to get on a parabola, once you know the focus and the directrix. Refer to the sketch at right: The vertex is always halfway between the focus and directrix. In the standard forms, the vertex is the origin. Use two fingers to mark the distance from the focus to the directrix (shown in yellow at right). Keeping one finger on the focus, sweep around ($\,90^\circ\,$ in each direction) to get the two ‘easy points’. Why are they points on the parabola? Their distances to the focus and directrix both equal the radius of the circle shown (see the red segments at right). These two easy points give a sense of the ‘width’ of the parabola. three easy points on a parabola: the vertex; two points to give a sense of the ‘width’
## Example
Graph $\,y^2 = 3x\,.$
Solution:
The equation $\,y^2 = 3x\,$ is in the form $\,y^2 = 4px\,.$ The $\,y^2\,$ term is already isolated, with a coefficient of $\,1\,.$ $y^2\ge 0 \implies 3x\ge 0 \implies x\ge 0$ Thus, the $x$-value of every point on the parabola is nonnegative; the parabola opens to the right. $4p = 3 \implies p = \frac 34\implies \text{focus } = (\frac 34,0)$ equation of directrix: $\,x = -\frac 34\,$ two easy points: distance from focus to directrix: $\,2(\frac 34) = \frac 32\,$ coordinates of easy points: $\,(\frac 34,\frac 32)\,$ and $\,(\frac 34,-\frac 32)\,$ It's confidence-boosting to check that these points satisfy the equation $\,y^2 = 3x\,$: $$\begin{gather} \cssId{s175}{\bigl(\pm \frac 32\bigr)^2\ \ \overset{?}{=}\ \ 3\bigl(\frac 34\bigr)}\cr\cr \cssId{s176}{\frac 94 = \frac 94}\qquad \cssId{s177}{\text{Yep!}} \end{gather}$$
## Example
Graph $\,2x^2 + 5y = 0\,.$
Solution:
The equation $\,2x^2 + 5y = 0\,$ has only $\,x^2\,$ and $\,y\,$ terms. It can be put in the form $\,x^2 = 4py\,.$ Isolate the square term, and get a coefficient of $\,1\,$: $$\begin{gather} \cssId{s184}{2x^2 + 5y = 0}\cr \cssId{s185}{2x^2 = -5y}\cr \cssId{s186}{x^2 = -\frac 52y} \end{gather}$$ $x^2\ge 0 \implies -\frac 52y\ge 0 \implies y\le 0$ Thus, the $y$-value of every point on the parabola is negative (or zero); the parabola opens down. $4p = -\frac 52 \implies p = -\frac 58\implies \text{focus } = (0,-\frac 58)$ equation of directrix: $\,y = \frac 58\,$ two easy points: distance from focus to directrix: $\,2(\frac 58) = \frac 54\,$ coordinates of easy points: $\,(\frac 54,-\frac 58)\,$ and $\,(-\frac 54,-\frac 58)\,$ It's confidence-boosting to check that these points satisfy the equation $\,2x^2 + 5y = 0\,$: $$\begin{gather} \cssId{s199}{2\bigl(\pm \frac 54\bigr)^2 + 5(-\frac 58)\ \ \overset{?}{=}\ \ 0}\cr\cr \cssId{s200}{2\bigl(\frac {25}{16}\bigr) - \frac{25}{8}\ \ \ \overset{?}{=}\ \ 0}\cr\cr \cssId{s201}{0 = 0}\qquad \cssId{s202}{\text{Yep!}} \end{gather}$$
Master the ideas from this section
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# Video: AQA GCSE Mathematics Higher Tier Pack 2 β’ Paper 3 β’ Question 4
Make π the subject of the equation π = (3π‘)/(2π). Circle your answer. [A] π = (2π‘)/(3π ) [B] π = (3π‘)/(2π ) [C] π = (2π )/(3π‘) [D] π = (3π )/(2π‘)
02:05
### Video Transcript
Make π the subject of the equation π is equal to three π‘ over two π. Circle your answer. Is it π is equal to two π‘ over three π , π is equal to three π‘ over two π , π is equal to two π over three π‘, or π is equal to three π over two π‘?
It should be fairly obvious from the answers weβve been given that to make π the subject weβre looking for an equation which is π is equal to some other expression. Just like weβre solving equations, to achieve this, weβre going to need to perform a series of inverse, that means opposite, operations.
Letβs start with π is equal to three π‘ over two π. At the moment, this fraction is making things look a little bit nasty. So weβre going to multiply both sides by two π. Now this isnβt the only first step that we can take, and weβll look at slightly different method in a moment. Multiplying both sides by two π gives us two π multiplied by π on the left-hand side of this equation. Remember, we try to avoid using the multiplication symbol in algebra, so we just write two ππ . Three π‘ divided by two π and then multiply it by two π is simply three π‘. To get π to be the subject of this equation, we need to divide by two and π since at the moment π is being multiplied by both two and π .
Two ππ divided by two π is simply π, and three π‘ divided by two π is three π‘ over two π . And we can see that the correct answer is therefore π is equal to three π‘ over two π . Now you might have noticed that we started by multiplying by two π and then dividing by two π . Since multiplying by two and then dividing by two cancel each other out, we could have begun by multiplying both sides by π. Thatβs ππ is equal to three π‘ over two. Then we would divide both sides by π to once again give us π is equal to three π‘ over two π . Either method is absolutely fine. The answer is π is equal to three π‘ over two π .
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## Content description
Multiply decimals by whole numbers and perform divisions by non-zero whole numbers where the results are terminating decimals, with and without digital technologies (ACMNA129)
Source: Australian Curriculum, Assessment and Reporting Authority (ACARA)
### Multiply and divide decimals
A strong understanding of the multiplication and division algorithm used with whole numbers and a clear knowledge of place value are necessary when multiplying and dividing decimal numbers.
#### Multiply decimals by whole numbers
With the introduction of decimals into the multiplication process, students need lots of experience in using materials to demonstrate what is happening.
For a multiplication such as 0.5 × 6, modelling the situation with objects, such as lemons, can help with understanding.
From working with fractions, we know that 0.5 is the same as one half. We can use lemons to show six lots of one half:
Six lots of $$\dfrac{1}{2}$$ a lemon = three whole lemons.
6 × $$\dfrac{1}{2}$$ = 3
Six lots of 0.5 a lemon = three whole lemons.
6 × 0.5 = 3
This demonstrates that even though we are multiplying by 6, the result is smaller than 6.
For the multiplication 0.5 × 6 we might convert to fractions:
\begin{align*} 0.5 \times 6 &= \frac{1}{2} \times 6 \\ &= \frac{1}{2} \times \frac{6}{1} \\ &= \frac{6}{2} \\ &= 3 \end{align*}
\begin{align*} 0.5 \times 6 &= 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 \\ &= 3 \end{align*}
We can do a simple multiplication:
0.5 × 6 = 3.0
In our head we think 5 × 6 = 30, but the solution must have the same number of decimal places as the numbers we are multiplying. Thinking of 3.0 instead of 30 helps with this.
Or the equation could be set out using the multiplication algorithm we use for whole numbers:
0 . 5 × 6 3 . 0
Note that there is a total of one decimal place in the two numbers being multiplied and one decimal place in the solution.
Let’s look at another example: If we buy four 2.75 litre bottles of juice, how much juice is there?
0.75 of a litre is the same as $$\dfrac{3}{4}$$ of a litre. So if we convert to fractions:
\begin{align*}2.75 \times 4 &= 2 \frac{3}{4} \times 4 \\ &= \frac{11}{4} \times \frac{4}{1} \\ &= 11\end{align*}
$$^3 2$$ . $$^2 7$$ 5 × 4 1 1 . 0 0
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# Resolution of a Vector
Resolution of a vector is just opposite to the composition of vectors. It is the process of splitting a single vector into two or more vectors in different directions which together produce the same effect as produced by the single vector alone.
Let a vector $\overrightarrow{R}$ be represented by $\overrightarrow{OB}$. Let $\overrightarrow{P}$ and $\overrightarrow{Q}$ be represented by $\overrightarrow{OA}$ and $\overrightarrow{AB}$.
$\therefore\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}$ $\text{or,}\;\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$
Here, the single vector or vector sum $(\overrightarrow{R})$ is also known as the resultant of two vectors. And, the two vectors $(\overrightarrow{P}$ and $\overrightarrow{Q})$ whose sum is a given vector $(\overrightarrow{R})$ are called the components of the given vector.
This process of splitting the resultant vector $(\overrightarrow{R})$ into the components $(\overrightarrow{P}$ and $\overrightarrow{Q})$ is called resolving or resolution of the vector. We often find components parallel to the x-axis and y-axis. Such components are called rectangular components.
## Rectangular Components of a Vector in a Plane
When a vector is splitting into two component vectors at right angles to each other, the component vectors are called rectangular components of the vector.
Consider a vector $\overrightarrow{P}$ represented by $\overrightarrow{OC}$, has to be resolved into two rectangular component vectors along the direction of x-axis and y-axis. From the point $O$, draw $OX$ and $OY$ at right angles to represent x-axis and y-axis respectively. From $C$, draw $CA$ and $CB$ perpendiculars to $OX$ and $OY$ respectively.
Then, $\overrightarrow{OA}$ $(=\overrightarrow{P_x})$ and $\overrightarrow{OB}$ $(=\overrightarrow{P_y})$ are the rectangular components of $\overrightarrow{P}$. The vector $\overrightarrow{P_x}$ (parallel to x-axis) is called the horizontal component and the vector $\overrightarrow{P_y}$ (parallel to y-axis) is called the vertical component.
Here, $AC$ is equal and parallel to $OB$. Hence, $\overrightarrow{AC}$ also represents $\overrightarrow{P_y}$ in magnitude and direction. From the triangle law of vector addition, in $\Delta OAC$, we have,
$\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}$ $\therefore\overrightarrow{P}=\overrightarrow{P_x}\:\hat{i}+\overrightarrow{P_y}\:\hat{j}$
Let the angle between $\overrightarrow{P}$ and $\overrightarrow{P_x}$ be $\theta$, i.e. $\angle AOC=\theta$. In right angled $\Delta AOC$, $\frac{OA}{OC}=\cos\theta$ $\therefore OA=OC\cos\theta$ $\text{or,}\;P_x=P\cos\theta$
Similarly, $\frac{AC}{OC}=\sin\theta$ $\therefore AC=OC\sin\theta$ $\text{or,}\;P_y=P\sin\theta$ Also, $OC^2=OA^2+AC^2$ $P^2=P_x^2+P_y^2$ $\therefore P=\sqrt{P_x^2+P_y^2}$ and, $\tan\theta=\frac{AC}{OC}=\frac{P_y}{P_x}$ $\therefore\theta=\tan^{-1}\left(\frac{P_y}{P_x}\right)$
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# Trigonometry - modelling and problem solving
• May 12th 2013, 08:34 PM
elmidge
Trigonometry - modelling and problem solving
I am having trouble finding the length of a sloping edge in the following problem:
A right pyramid, ht 6 cm stands on a square base of side 5cm. Find the legth of a sloping edge and the area of a triangular face.
• May 17th 2013, 10:04 AM
x3bnm
Re: Trigonometry - modelling and problem solving
Quote:
Originally Posted by elmidge
I am having trouble finding the length of a sloping edge in the following problem:
A right pyramid, ht 6 cm stands on a square base of side 5cm. Find the length of a sloping edge and the area of a triangular face.
Suppose the right pyramid is like the one in the following image:
Attachment 28398
Right pyramid $ABCDE$ has a square base $BCDE$. This square base $BCDE$ has all equal sides where each side has length $5 \text{ cm}$. We also know that the height of the right pyramid is $6\text{ cm}$ which is $AO$
Now in sqare $BCDE$, $\triangle BED$ is a right triangle with $\angle BED$ a right angle.
So by Pythogorean theorem:
$BD^2 = BE^2 + ED^2$
But $BE = ED = 5 \text{ cm}$
So $BD = 5\sqrt{2}$
$BO = \frac{1}{2}BD = \frac{5}{\sqrt{2}} \text{ cm}$
We know that $\text{height} = AO = 6\text{ cm}$
Now $\triangle AOB$ is a right triangle with $\angle AOB$ a right angle.
So:
$AB^2 = AO^2 + BO^2$
$\therefore AB = \frac{\sqrt{97}}{\sqrt{2}}......\text{[because AO = } 6 \text{ cm and BO }= \frac{5}{\sqrt{2}}\text{ cm]}$
That's the length of the slopping edge which is $\frac{\sqrt{97}}{\sqrt{2}}\text{ cm}$
So the area of:
\begin{align*}\triangle ABC =& \frac{1}{2} \times BC \times AO\\=& 15\text{ cm}^2...\text{[because BC} = 5 \text{ cm and AO} = 6 \text{ cm]}\end{align*}
So the area of the triangular face is $15\text{ cm}^2$
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# Bell Work: Find the values of all the unknowns: R T = R T T + T = 60 R = 3 R = 6 1 1 2 2 1 2 1 2.
## Presentation on theme: "Bell Work: Find the values of all the unknowns: R T = R T T + T = 60 R = 3 R = 6 1 1 2 2 1 2 1 2."— Presentation transcript:
Bell Work: Find the values of all the unknowns: R T = R T T + T = 60 R = 3 R = 6 1 1 2 2 1 2 1 2
Answer: T = 20 T = 40 2121
Quadratic equations are second degree polynomial equations. Second degree in x means that the greatest exponent of x in any term is 2. both of these equations are quadratic equations in x because the greatest exponent of x is 2. 4 – 3x = 2x3x – 2x + 4 = 0 2
The first equation was in standard form because all nonzero terms are on the left of the equals sign and the terms are written in descending order of the variable. The coefficient of x cannot be zero, but either of the other two numbers can be zero. Thus, each of the following equations is also a quadratic equation in x: 4x = 0 4x + 2x = 0 4x – 3 = 0 2 2 2 2
To designate a general quadratic equation, we use the letter a to represent the coefficient of x, the letter b to represent the coefficient of x, and the letter c to represent the constant term. Using these letters to represent the constants in the equation, we can write a general quadratic equation in standard form as ax + bx + c = 0 2 2
If we let a = 1, b = -3, and c = -10, we have the equation x – 3x – 10 = 0 If we substitute either 5 or -2 for the variable x in the quadratic equation, the equation will be transformed into a true equation. If x = 5If x = -2 (5) – 3(5) – 10 = 0 (-2) – 2(-2) – 10 = 0 0 = 0 2 2
The numbers 5 and -2 are the only numbers that will satisfy the equation before. Every quadratic equation has at most two distinct numbers that will make the equation a true statement.
Some quadratic equations can be solved by using the zero factor theorem. Zero factor theorem: if p and q are any real numbers and if p x q = 0, then either p = 0 or q = 0, or both p and q equal 0.
For example, (x – 3)(x + 5) = 0 Here we have two quantities multiplies and the product is equal to zero. From the zero factor theorem, we know that at least one of the quantities must equal zero if the product is to equal zero.
So either x – 3 = 0x + 5 = 0 x = 3x = -5 Thus the two values of x that satisfy the condition stated are 3 and -5.
We can use the zero factor theorem to help us solve quadratic equations that can be factored. We do this by first writing the equation in standard form and factoring the polynomial; then we set each of the factors equal to zero and solve for the values of the variable.
Example: Use the factor method to find the roots of x – 18 = 3x. 2
Example: Find the roots of -25 = -4x 2
Example: Find the values of x that satisfy x – 56 = -x. 2
Example: Solve 3x – 6x = 9 2
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The sum of the legs of a right triangle is 36 cm. For what lengths of the sides will the square of the hypotenuse be a minimum?
Jan 8, 2016
We can do this in two ways: by lateral thinking or in the robust mathematical way,
Explanation:
Let's do the first way , assuming both legs are 18 cm. Then the square of the hypotenuse will be ${18}^{2} + {18}^{2} = 648$
If we change this to $17 \leftrightarrow 19$ it will be $650$
Even $10 \leftrightarrow 26$ will give a greater number: $686$
And $1 \leftrightarrow 35$ will lead to $1226$
The mathematical way:
If one leg is $a$ then the other one is $36 - a$
The square of the hypotenuse is then:
${a}^{2} + {\left(36 - a\right)}^{2} = {a}^{2} + 1296 - 72 a + {a}^{2}$
Now we have to find the minimum of:
$2 {a}^{2} - 72 a + 1296$ by setting the derivative to 0:
$4 a - 72 = 0 \to 4 a = 72 \to a = 18$
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# Algebra 1 : How to use FOIL in the distributive property
## Example Questions
Evaluate
Explanation:
### Example Question #91 : How To Use Foil In The Distributive Property
Expand the following expression.
Explanation:
In order to expand the equation given, we need to use the FOIL method for distribution. Let's look at the expression we need to simplify, and work through the steps of FOIL:
F stands for FIRST - we multiply together the first terms inside each set of parantheses, which in this case is , giving us
O stands for OUTER - we multiply together the first term in the first set of parentheses and the second term in the second pair of parentheses, which in this case is , giving us
I stands for INNER - we multiply together the second term in the first set of parentheses and the first term in the second pair of parentheses, which in this case is , giving us
L stands for LAST - we multiply together the second terms inside each set of parantheses, which in this case is , giving us .
Now, we add together all four values that we got from using FOIL to get:
We can combine like terms to reach our final answer of:
### Example Question #91 : How To Use Foil In The Distributive Property
Expand the following expression using FOIL
Explanation:
Expand the following expression using FOIL
Let's begin with a recap of what FOIL stands for:
First
Outer
Inner
Last
The point of foil is to be a helpful reminder to multiply all the terms and to not leave anything out.
So, up above we did the multiplying, to wrap up, let's order the final terms in standard (decreasing by exponent) order to get...
### Example Question #92 : How To Use Foil In The Distributive Property
Which answer demonstrates the use of FOIL correctly to expand the question?
Explanation:
FOIL stands for:
F: First (First set of terms in each equation) Eg. and
O: Outside (Outside sets of terms in each equation) Eg. and
I: Inside (Inside sets of terms in each equation) Eg. and
L: Last (Last set of terms in each equation) Eg. and
When using FOIL multiply each set of terms together.
After multiplying each set of terms together you should have
You must then simplify terms to get
When a variable has no number in front of it it means 1 of them.
So + is or just
### Example Question #92 : How To Use Foil In The Distributive Property
Distribute and simplify.
Explanation:
Let's use FOIL to distribute.
F: Multiply first terms in each binomial
O: Multiply the outer terms from both binomials
I: Multiply the inner terms from both binomials
L: Multiply the last terms in each binomial
Finally, we add them all up and we get . The middle terms are the same and can be added up and simplified to .
### Example Question #94 : How To Use Foil In The Distributive Property
Distribute and simplify.
Explanation:
Let's use FOIL to distribute.
F: Multiply first terms in each binomial
O: Multiply the outer terms from both binomials
I: Multiply the inner terms from both binomials
L: Multiply the last terms in each binomial
Finally, we add them all up and we get . The plus and minus signs all become minus signs. The middle terms are the same and can be added up and simplified to .
### Example Question #94 : How To Use Foil In The Distributive Property
Distribute and simplify.
Explanation:
Let's use FOIL to distribute.
F: Multiply first terms in each binomial
O: Multiply the outer terms from both binomials
I: Multiply the inner terms from both binomials
L: Multiply the last terms in each binomial
Finally, we add them all up and we get . The plus and minus signs all become minus signs. The middle terms are the same and are cancelled out and simplified to .
### Example Question #94 : How To Use Foil In The Distributive Property
Distribute and simplify.
Explanation:
Let's use FOIL to distribute.
F: Multiply first terms in each binomial
O: Multiply the outer terms from both binomials
I: Multiply the inner terms from both binomials
L: Multiply the last terms in each binomial
Finally, we add them all up and we get . The plus and minus signs all become minus signs. The middle terms are the same and can be added up and simplified to .
### Example Question #4911 : Algebra 1
Distribute and simplify.
Explanation:
Let's use FOIL to distribute.
F: Multiply first terms in each binomial
O: Multiply the outer terms from both binomials
I: Multiply the inner terms from both binomials
L: Multiply the last terms in each binomial
Finally, we add them all up and we get . The plus and minus signs all become minus signs. We now have: .
### Example Question #96 : How To Use Foil In The Distributive Property
Distribute and simplify.
Explanation:
Let's use FOIL to distribute.
F: Multiply first terms in each binomial
O: Multiply the outer terms from both binomials
I: Multiply the inner terms from both binomials
L: Multiply the last terms in each binomial
Finally, we add them all up and we get . The plus and minus signs all become minus signs. The middle terms are the same and simplified to .
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## Do you learn decimals in 4th grade?
Students in fourth grade are developing an understanding of both large numbers and numbers less than 1, like fractions and decimals. In fourth grade, students will compare two decimals to the hundredths place.
Figuring fractions Fourth graders gain a deeper understanding of fractions. They add and subtract fractions with the same denominator (the bottom number). They also add and subtract mixed numbers with the same denominator. Fourth graders learn how to find common denominators when those numbers are different.
How do you explain decimals?
In algebra, a decimal number can be defined as a number whose whole number part and the fractional part is separated by a decimal point. The dot in a decimal number is called a decimal point. The digits following the decimal point show a value smaller than one.
Fourth graders gain a deeper understanding of fractions. They add and subtract fractions with the same denominator (the bottom number). They also add and subtract mixed numbers with the same denominator. Fourth graders learn how to find common denominators when those numbers are different.
How to convert a fraction to a decimal?
Try another way of converting a fraction into a decimal.
• Understand power of 10 denominators. A “power of 10” denominator is a denominator consisting of any positive number that can be multiplied to make a multiple of 10.
• Learn to spot the easiest fractions that can be converted.
• Multiply your fraction by another fraction.
• How do you compare fractions and decimals?
Determine whether or not the fractions have the same denominator. This is the first step to comparing fractions.
• Find a common denominator. To be able to compare the fractions,you’ll need to find a common denominator so you can figure out which fraction is greater.
• Change the numerators of the fractions.
• Compare the numerators of the fractions.
• ## How do you convert fractions to decimals?
To convert fractions to decimals, look at the fraction as a division problem. Take the top number, or the numerator, of the fraction and divide it by the bottom number, or the denominator. You can do this in your head, by using a calculator, or by doing long division. For example, ΒΌ is just 1 divided by 4, or 0.25.
How do you change whole numbers into fractions?
Divide the numerator with the denominator
• Note down the whole number
• Later make a note of the remainder as a numerator above the denominator
|
### Set Square
A triangle PQR, right angled at P, slides on a horizontal floor with Q and R in contact with perpendicular walls. What is the locus of P?
### Pole Vaulting
Consider the mechanics of pole vaulting
# Five Circuits, Seven Spins
### Why do this problem?
This is a good problem for discussion and developing clear visualisation and mathematical communication. It relates the angle of rotation of a circle to a distance and is therefore of use in exploring radians and the formula $s=r\theta$.
### Possible approach
Students' abilities to visualise the meaning of this problem might vary considerably. As such, this problem can appear to be difficult until a clear approach to the solution is found. The behaviour of the disc at the corners is likely to cause the most difficulty in imagining the rotation. As a result, students might need to be given a variety of visual devices to allow them to get started. For example:
• Imagine looking down onto the tray and watching the disc rotate about its centre.
• Imagine breaking the journey into a series of straight line trips.
• Imagine that the disc is pinned down in the centre and the tray is a track moved around the disc.
• Imagine that the edge of the disc is coated in ink. Which parts of the tray would be coloured following a lap of the track?
• Roll a coin around a book and use the head on the coin as a reference. Does the head rotate as it moves through a corner (i.e. when moving from a horizontal to a vertical part of the the journey).
This is the sort of problem which becomes much clearer once a solution has been found. Once students have solved the problem they should try to rewrite their answer and method as clearly as possible, in a way which is both simple but complete.
It is possible to tackle this problem using degrees and the formula for the circumference of a circle, but it is much simpler to solve using radians and the formula $s=r\theta$.
### Key questions
How far does the centre of a disc move in one revolution when the plate is in contact with a straight edge?
What mathematics allows us to relate this distance to an angle?
What units should we measure the angle of rotation in? Why?
If the disc has rotated $7$ full times, how far must it have rolled?
As the disc makes a single lap of the tray, what parts of the tray will have made contact with the disc? How far is this?
### Possible support
Consider the distance a bicycle travels when the wheels rotate once.
Read the article A Rolling Disc - Periodic Motion.
### Possible extension
Try the problem Contact
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# Area word problems
You will encounter these area word problems often in math. Many of them will require familiarity with basic math, algebra skills, or a combination of both to solve the problems.
As I solve these area word problems, I will make an attempt to give you some problems solving skills
Word problem #1:
The area of a square is 4 inches. What is the length of a side?
Important concept: Square. It means 4 equal sides.
Area = s × s= 4 × 4 = 16 inches2
Word problem #2:
A small square is located inside a bigger square. The length of one side of the small square is 3 inches and the length of one side of the big square is 7 inches
What is the area of the region located outside the small square, but inside the big square?
Important concept: Draw a picture and see the problem with your eyes. This is done below:
The area that you are looking for is everything is red. So you need to remove the area of the small square from the area of the big square
Area of big square = s × s = 7 × 7 = 49 inches2
Area of small square = s × s = 3 × 3 = 9 inches2
Area of the region in red = 49 - 9 = 40 inches2
Word problem #3:
A classroom has a length of 20 feet and a width of 30 feet. The headmaster decided that tiles will look good in that class. If each tile has a length of 24 inches and a width of 36 inches, how many tiles are needed to fill the classroom?
Important concept:
Find area occupied by entire classroom
Find area for one tile
Decide which unit to use. In this case, we could use feet
Area of classroom = length × width = 20 × 30 = 600 ft2
Before we get the area of each tile, convert the dimensions to feet as already stated
Since 1 foot = 12 inches, 36 inches = 3 feet and 24 inches = 2 feet
Area of each tile = length × width = 2 × 3 = 6 ft2 If one tile takes 6 ft2, it takes 100 tiles to cover 600 ft2 (6 × 100 = 600)
Word problem #4:
Sometimes area word problems may require skills in algebra, such as factoring and solving quadratic equations
A room whose area is 24 ft2 has a length that is 2 feet longer than the width. What are the dimensions of the room?
Solution #1 Use of basic math skills and trial and error
Pretend width = 1, then length = 3 ( 1 + 2)
1 × 3 is not equal to 24 and 3 is not close at all to 24. Try bigger numbers
let width = 3, then length = 5 ( 3 + 2)
3 × 5 = 15. It is still not equal to 24
let width = 4, then length = 4 ( 4 + 2)
4 × 6 = 24. We are done!
Solution #2 Use of algebra
Let width = x, so length = x + 2 Area = length × width
24 = x × ( x + 2)
24 = x2 + 2x
x2 + 2x = 24 (swap the left side with the right side)
x2 + 2x - 24 = 0
( x + 6) × ( x - 4 ) = 0
x = -6 and x = 4
So width = 4 and length = 4 + 2 = 6
As you can see area word problems can become very complicated as shows in the last problem. And the use of basic math skills may not be the best way to go to solve area word problems. Sometimes algebra is better because trial and error can take a long time!
Have a question about these area word problems? send me a note.
## Recent Articles
1. ### Optical Illusions with Geometry
Oct 22, 18 04:14 PM
Take a good look at these optical illusions with geometry
New math lessons
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## Recent Lessons
1. ### Optical Illusions with Geometry
Oct 22, 18 04:14 PM
Take a good look at these optical illusions with geometry
Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Everything you need to prepare for an important exam!
K-12 tests, GED math test, basic math tests, geometry tests, algebra tests.
Real Life Math Skills
Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball.
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### Mathematics Class 10 Lab Manual | 21 Lab Activities
Mathematics Lab Manual Class X lab activities for class 10 with complete observation Tables strictly according to the CBSE syllabus also very useful & helpful for the students and teachers.
## Mathematics Assignment Class IX Chapter 3 Coordinate Geometry
Mathematics Assignment for Class IX Ch -3, Co-ordinate Geometry, strictly according to the CBSE syllabus. Math worksheet based on the topic Co-ordinate Geometry.
Note: First of all students should solve the NCERT question and then answer the following questions.
Question 1:
What is Cartesian coordinate system
Answer: A system which involves the x-axis and y-axis on the two dimensional surface is called a cartesian coordinate system.
Question 2:
What we call the horizontal line in Cartesian coordinate system.
Answer: Horizontal line is called the x - axis.
Question 3:
What we call the vertical line in Cartesian coordinate system.
Answer: Vertical Line is called the y - axis.
Question 4:
Which point is called the origin and what are its coordinates.
Answer: The point of intersection of x - axis and y - axis is called the Origin. Coordinates of origin are (0,0)
Question 5:
What is the point on x – axis.
Answer: Any point on the x - axis is (a, 0)
Question 6:
What is the point on y-axis.
Answer: Any point on the y - axis is (0, b)
Question 7:
What do you mean by ordinates.
Answer: Point on y - axis is called ordinate.
Question 8:
What do you mean by abscissa.
Answer: Point on the x-axis is called abscissa.
Question 9:
Write the coordinates of the points given in the following graph. Also write In which quadrant or on which axis the following points lie.
A(3, 4) lie in I quadrant
B(-4, 3) lie in II quadrant
C(-2. -3) lie in III quadrant
D(3, -2) lie in IV quadrant
E(2, 0) lie on the x-axis
F(-2, 0) lie on the x-axis
G(0, 2) lie on y-axis
H(0, -2) lie on y-axis
Question 10:
Write the abscissa and ordinate of the following points
(-3, 5), (0, 9), (-5, 0), (3,3), (7, -9), (12, 13)
Abscissa: -3, 0, -5, 3, 7, 12
Ordinate: 5, 9, 0, 3, -9, 13
Question 11:
Write the coordinates of the point :
a) Whose ordinate is -5 and which lies on y- axis.
b) Which lies on x-axis and y – axes both.
c) Whose abscissa is -3 and which lies on x – axis.
a) (0,-5)
b) Point on the x-axis is (a, 0) and Point on the y-axis is (0, b)
c) (-3, 0)
Question 12:
Write the coordinates of the points given in the following graph. Also write In which quadrant or on which axis the following points lie.
Question 13:
What is the perpendicular distance of the point (3, 4) from
a) x – axis b) Y – axis
a) 4 unit b) 3 unit
Question 14:
What is the perpendicular distance of the point (7, 26) from
a) x – axis b) Y – axis
a) 26 unit b) 7 unit
Question 15:
Name the quadrant in which abscissa and ordinate have different sign.
Ans: In II and IV quadrant abscissa and ordinate have different sign.
Question 16:
Plot the points P(1,0), Q(4,0), and S(1, 3). Find the coordinates of point R such that PQRS is a square. Also find the area of the square.
Coordinates of R(4, 3), Area of square PQRS = 9 square units)
Question 17:
Mark the points A(2, 2), B(2, -2), C(-2, -2) and D(-2, 2) on the graph paper and join these points in order. Identify the figure so obtained. Also find the area of the figure.
Square, Area = 16cm2]
Question 18:
Plot the points P(-1, 0), Q(0, 1) and R(2, 3) on the graph paper and check whether they are collinear or not.
Answer: If P, Q, R all points lie on a straight line then points are collinear.
Question 19:
If the points A(0, -6), B(0, 2 and C (a, 3) lie on the y-axis, then find the value of a.
Answer: For points on the y-axis the abscissa is "0" So the value of a = 0
Question 20:
From the given graph write :
a) The coordinates of the points B and F
b) The abscissas of points D and H
c) The ordinates of the points A and C
d) The perpendicular distance of the point G from x-axis
a) B(-5, -4), F(6, 1)
b) D(1, 1), H(0, 4)
c) A(-4, 1), C(-2, 0)
d) G(6, 4) , Perpendicular distance of poin G from x-axis is 4
1. It is very helpful for students.
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RS Aggarwal Solutions: Circles
# RS Aggarwal Solutions: Circles | Mathematics (Maths) Class 6 PDF Download
``` Page 1
Points to Remember :
Circle. It is the set of all those points in a plane whose distance from a fixed point remains constant.
The fixed point is called the centre of the circle and the constant distance is called the radius of the
circle.
All the radii of a circle are equal.
Diameter of a circle
A line segment passing through the centre of circle and having its end
points on the circle is diameter of the circle.
Chord of a circle. A line segment with its end points
lying on a circle is called the chord of the circle.
Secent of a Circle. A line passing through a circle and
intersecting the circle at two points is called a secent of
the circle.
Circumference of a circle. The perimeter of a circle is
called the circumference of the circle
Semi-circle. The end points of a diameter of a circle
divide the circle into two parts ; each part is called a
semi-circle.
Arc. Any part of a circle is called an arc of the circe.
Segments of a Circle
A chord AB of a circle divides the circular region into
two parts. Each part is called a segment of the circle.
The segment containing the centre of the circle is called
the major segment, while the segment not containing the
centre is called the minor segment of the circle.
Sector of a Circle. The area bounded by an arc and the
two radii joining the end points of the arc with the centre,
is called a sector. If the sector is formed by a major arc
then it is called a major sector. If the sector is formed by
a minor arc, it is called a minor sector.
Concentric circles. Two or more circles with the same
centre are called concentric circles.
Page 2
Points to Remember :
Circle. It is the set of all those points in a plane whose distance from a fixed point remains constant.
The fixed point is called the centre of the circle and the constant distance is called the radius of the
circle.
All the radii of a circle are equal.
Diameter of a circle
A line segment passing through the centre of circle and having its end
points on the circle is diameter of the circle.
Chord of a circle. A line segment with its end points
lying on a circle is called the chord of the circle.
Secent of a Circle. A line passing through a circle and
intersecting the circle at two points is called a secent of
the circle.
Circumference of a circle. The perimeter of a circle is
called the circumference of the circle
Semi-circle. The end points of a diameter of a circle
divide the circle into two parts ; each part is called a
semi-circle.
Arc. Any part of a circle is called an arc of the circe.
Segments of a Circle
A chord AB of a circle divides the circular region into
two parts. Each part is called a segment of the circle.
The segment containing the centre of the circle is called
the major segment, while the segment not containing the
centre is called the minor segment of the circle.
Sector of a Circle. The area bounded by an arc and the
two radii joining the end points of the arc with the centre,
is called a sector. If the sector is formed by a major arc
then it is called a major sector. If the sector is formed by
a minor arc, it is called a minor sector.
Concentric circles. Two or more circles with the same
centre are called concentric circles.
Q. 1. Take a point O on your note book and
draw circles of radii 4 cm., 5.3 cm. and
6.2 cm., each having the same centre.
Sol. Method :
Take a point O on the paper as shown in
the figure. With the help of the rular,
open out compasses in such a way that
the distance between the metal point and
pencil point is 4 cm. Take the compasses
in the same position and put its metal
point at O and draw the circle.
Remove the compasses and again open
out the compasses in such a way that
the distance between the metal point and
pencil point is 5.3 cm. Taking O as the
centre, draw another circle. Again
remove the compasses and similarly
draw the third circle with radius 6.2 cm.
Then the required circles are as shown
in the figure which have radius OA = 4
cm., OB = 5.3 cm. and OC = 6.2 cm.
Q. 2. Draw a circle with centre C and radius
4.5 cm. Mark points P, Q and R such
that P lies in the interior of the circle, Q
lies on the circle and R lies in the exterior
of the circle.
Sol. Method : Take a point C on the paper.
With the help of the rular, open out the
compasses in such a way that the
distance between its metal point and
pencil point is 4.5 cm. Take the
compasses in the same position and put
its metal point at C and draw the circle.
Mark points P, Q and R as shown in the
figure as required.
Q. 3. Draw a circle, with centre O and radius
4 cm. Draw a chord AB of the circle.
Indicate by marking point X and Y, the
minor are AX B and the major are AYB.
Sol. Method : Take a point O on the paper.
With the help of the rular, open out the
compasses in such a way that the
distance between the metal point and
pencil point is 4 cm. Take the compasses
in the same position and put the metal
point at O and draw the circle.
Take A and B any points on the circle
and join AB. Then AB is the chord of the
Page 3
Points to Remember :
Circle. It is the set of all those points in a plane whose distance from a fixed point remains constant.
The fixed point is called the centre of the circle and the constant distance is called the radius of the
circle.
All the radii of a circle are equal.
Diameter of a circle
A line segment passing through the centre of circle and having its end
points on the circle is diameter of the circle.
Chord of a circle. A line segment with its end points
lying on a circle is called the chord of the circle.
Secent of a Circle. A line passing through a circle and
intersecting the circle at two points is called a secent of
the circle.
Circumference of a circle. The perimeter of a circle is
called the circumference of the circle
Semi-circle. The end points of a diameter of a circle
divide the circle into two parts ; each part is called a
semi-circle.
Arc. Any part of a circle is called an arc of the circe.
Segments of a Circle
A chord AB of a circle divides the circular region into
two parts. Each part is called a segment of the circle.
The segment containing the centre of the circle is called
the major segment, while the segment not containing the
centre is called the minor segment of the circle.
Sector of a Circle. The area bounded by an arc and the
two radii joining the end points of the arc with the centre,
is called a sector. If the sector is formed by a major arc
then it is called a major sector. If the sector is formed by
a minor arc, it is called a minor sector.
Concentric circles. Two or more circles with the same
centre are called concentric circles.
Q. 1. Take a point O on your note book and
draw circles of radii 4 cm., 5.3 cm. and
6.2 cm., each having the same centre.
Sol. Method :
Take a point O on the paper as shown in
the figure. With the help of the rular,
open out compasses in such a way that
the distance between the metal point and
pencil point is 4 cm. Take the compasses
in the same position and put its metal
point at O and draw the circle.
Remove the compasses and again open
out the compasses in such a way that
the distance between the metal point and
pencil point is 5.3 cm. Taking O as the
centre, draw another circle. Again
remove the compasses and similarly
draw the third circle with radius 6.2 cm.
Then the required circles are as shown
in the figure which have radius OA = 4
cm., OB = 5.3 cm. and OC = 6.2 cm.
Q. 2. Draw a circle with centre C and radius
4.5 cm. Mark points P, Q and R such
that P lies in the interior of the circle, Q
lies on the circle and R lies in the exterior
of the circle.
Sol. Method : Take a point C on the paper.
With the help of the rular, open out the
compasses in such a way that the
distance between its metal point and
pencil point is 4.5 cm. Take the
compasses in the same position and put
its metal point at C and draw the circle.
Mark points P, Q and R as shown in the
figure as required.
Q. 3. Draw a circle, with centre O and radius
4 cm. Draw a chord AB of the circle.
Indicate by marking point X and Y, the
minor are AX B and the major are AYB.
Sol. Method : Take a point O on the paper.
With the help of the rular, open out the
compasses in such a way that the
distance between the metal point and
pencil point is 4 cm. Take the compasses
in the same position and put the metal
point at O and draw the circle.
Take A and B any points on the circle
and join AB. Then AB is the chord of the
circle. Mark points X and Y on the circle
as shown. Then arc AXB and arc AYB
are the required minor and major arcs
respectvely.
Q. 4. Which of the following statements are
true and which are false ?
(i) Each radius of a circle is also a chord of
the circle.
(ii) Each diameter of a circle is also a chord
of the circle.
(iii) The centre of a circle bisects each chord
of the circle.
(iv) Secent of a circle is a segment having
its end points on the circle.
(v) Chord of a circle is a segment having its
end points on the circle.
Sol. (i) False (ii) True
(iii) False (iv) False
(v) True.
Q. 5. Draw a circle with centre O and radius
3·7 cm. Draw a sector having the angle
72º.
Sol. Steps of construction :
(i) With centre O and radius 3·7 cm, draw a
circle.
(ii) Take a point A on the circumference of
the circle.
(iii) Join OA.
(iv) At O, draw another radius OB such that
AOB = 72º with the help of protractor.
Then sector AOB is the required one.
Q. 6. Fill in the blanks by using <, >, = or
(i) OP........OQ, where O is the centre of
the circle, P lies on the circle and Q is in
the interior of the circle.
(ii) OP........OR, where O is the centre of the
circle, P lies on the circle and R lies in
the exterior of the circle.
(iii) Major arc..........minor arc of the circle.
(iv) Major arc........semicircumference of the
circle.
Sol. (i) > (ii) < (iii) > (iv) >. Ans.
Q. 7. Fill in the blanks :
(i) A diameter of a circle is a chord
that..........the centre.
(ii) A radius of a circle is a line segment with
one end point.........and the other end
point......... .
(iii) If we join any two points of a circle by a
line segment, we obtain a..........of the
circle.
(iv) Any part of a circle is called an...........of
the circle.
(v) The figure bounded by an arc and the
two radii joining the end points of the arc
with the centre is called a...........of the
circle.
Sol. (i) Passes through (ii) at the centre, on
the circle (iii) chord (iv) arc (v) sector.
Ans.
```
## Mathematics (Maths) Class 6
94 videos|347 docs|54 tests
## Mathematics (Maths) Class 6
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# Math Help: Mixed Fraction Problems and Solutions
Now that you've mastered basic fractions, you'll learn to work with mixed fractions. You'll need to understand mixed fractions when you're doing advanced long division and some story problems. Keep reading for help with mixed fraction problems!
## Mixed Fraction Help
Fractions represent parts of a whole, like 1/2 of a pizza or 3/4 of an orange. But what if you have two and 1/2 pizzas, or three and 3/4 oranges? When you have an integer amount, like two or three, and a fractional amount left over, you represent this using a mixed fraction, like 2 1/2 or 3 3/4.
### Division
One important use of mixed fractions is writing the answers to division problems that have remainders. For instance, if you divide 25 by 12, you'll get an answer of two with a remainder of one. This means that 12 went into 25 two times, but you have 1/12 left over, so you could write this answer as the mixed fraction 2 1/12. The numerator of the fraction is the remainder, and the denominator is the divisor.
Tip: Sometimes, you'll need to simplify the fractional part of the mixed fraction. For example, 5 2/6 would be simplified to 5 1/3.
### Mixed Numbers to Improper Fractions
If you want to add, subtract, multiply or divide mixed fractions, you have to turn them into improper fractions first. An improper fraction is a fraction which has a numerator that's larger than the denominator. For instance, 11/5 and 21/8 are improper fractions. Just like mixed fractions, they represent amounts that are greater than one. To turn a mixed fraction into an improper fraction, there are three steps to follow:
1. Multiply the denominator of the fraction by the integer.
3. The answer becomes the numerator of the improper fraction, and the denominator remains the same.
For example, the first step toward converting 5 1/3 into an improper fraction is to multiply three and five to get 15 (3 x 5 = 15). Next, add 15 to one to get 16 (15 + 1 = 16). Last, 16 becomes the numerator and three remains the denominator: 16/3.
### Improper Fractions to Mixed Numbers
If you have an improper fraction and you need to change it to a mixed number, you reverse the steps described earlier for changing a mixed number to a fraction. Here's the process:
1. Divide the numerator of the improper fraction by the denominator. The result becomes the integer part of the mixed number.
2. The remainder, if any, becomes the numerator of the new mixed fraction.
3. The denominator of the mixed fraction is the denominator from the improper fraction. Reduce the fractional part of the mixed number if necessary.
As an example, let's change 15/6 into a mixed number. First, divide 15 by six to get two with a remainder of three. Two will become the integer in the mixed number, and three will become the numerator of the fraction in the mixed number. The denominator will still be six. The entire mixed number will look like this: 2 3/6. Since 3/6 can be simplified to 1/2, the final answer will be 2 1/2.
### Practice Problems
1. Nina has four whole watermelons and one quarter of a watermelon. Express this as a mixed fraction.
2. Change 4 1/3 to an improper fraction.
3. Write the improper fraction 18/7 as a mixed number.
#### Solutions
1. Four and one quarter is written as a mixed fraction like this: 4 1/4.
2. First, multiply three by four to get 12 (3 x 4 = 12). Add one to 12 to get 13 (1 + 12 = 13), which becomes the numerator of the improper fraction. The denominator will still be three, so the improper fraction will be 13/3.
3. To write 18/7 as a mixed number, first divide 18 by seven to get two with a remainder of four. Two will become the integer in the mixed number. The numerator will be the remainder, four, and the denominator will still be seven. The mixed number will be 2 4/7.
Did you find this useful? If so, please let others know!
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# Table of 99
Table of 99 is the multiplication table of the number ninety-nine (99). Here, we have provided the 99 times table from 1 to 20 times along with the chart and PDF. Students can refer to all Tables from 1 to 100 to work with numerical simplifications quickly.
What is the table of 99?
The table of 99 or 99 times table is called the multiplication table of 99. This can be defined using two basic operations such as addition and multiplication. For example, the value 3 times 99 can be expressed as:
99 × 3 = 297
We can also get this value by repeatedly adding 99 for 3 times such as,
99 + 99 + 99 = 297
In the same way, we can get the other multiples of 99. The below table shows the 99 times table from 1 to 20.
## Multiplication Table of 99
99 × 1 = 99 99 × 2 = 198 99 × 3 = 297 99 × 4 = 396 99 × 5 = 495 99 × 6 = 594 99 × 7 = 693 99 × 8 = 792 99 × 9 = 891 99 × 10 = 990 99 × 11 = 1089 99 × 12 = 1188 99 × 13 = 1287 99 × 14 = 1386 99 × 15 = 1485 99 × 16 = 1584 99 × 17 = 1683 99 × 18 = 1782 99 × 19 = 1881 99 × 20 = 1980
### How to Read the Table of 99
Read the below ninety-nine (99) times table from 1 to 10.
• One time ninety-nine is ninety nine
• Two times ninety-nine is one hundred and ninety eight
• Three times ninety-nine is two hundred and ninety seven
• Four times ninety-nine is three hundred and ninety six
• Five times ninety-nine is four hundred and ninety five
• Six times ninety-nine is five hundred and ninety four
• Seven times ninety-nine is six hundred and ninety three
• Eight times ninety-nine is seven hundred and ninety two
• Nine times ninety-nine is eight hundred and ninety one
• Ten times ninety-nine is nine hundred and ninety
From the above, we can observe that there is a pattern for memorizing the table of 99. That means, the unit digit is decreasing from 9 to 0, the tenth digits remain the same, i.e. 9 and the hundredth digits are increasing from 0 to 9. This trick is used to write the 99 times table quickly and accurately.
### More Multiplication Tables
Visit www.byjus.com for more maths tables and download BYJU’S – The Learning App for engaging videos of maths concepts.
|
# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/8, the angle between sides B and C is (3pi)/4, and the length of side B is 7, what is the area of the triangle?
Feb 9, 2018
$17.324$
#### Explanation:
From the diagram:
$\boldsymbol{\theta} = \pi - \left(\frac{3 \pi}{4} + \frac{\pi}{8}\right) = \frac{\pi}{8}$
$\boldsymbol{\alpha} = \pi - \frac{3 \pi}{4} = \frac{\pi}{4}$
Using The Sine Rule
bb(SinA/a=SinB/b=SinC/c
We only need to find side $\boldsymbol{c}$
We know angle B and side b, so:
$\sin \frac{\frac{\pi}{8}}{7} = \sin \frac{\frac{\pi}{8}}{c} \implies c = \frac{7 \sin \left(\frac{\pi}{8}\right)}{\sin} \left(\frac{\pi}{8}\right) = 7$
From diagram:
$\boldsymbol{h} = 7 \sin \left(\frac{\pi}{4}\right)$
Area of triangle is:
$\boldsymbol{\frac{1}{2}}$base x height
$\frac{1}{2} c \times h$
$\frac{1}{2} \left(7\right) \cdot 7 \sin \left(\frac{\pi}{4}\right) = \frac{49 \sqrt{2}}{4} = 17.324$
|
Practice with the help of Spectrum Math Grade 1 Answer Key Chapter 3 Lesson 3.15 Using Addition and Subtraction regularly and improve your accuracy in solving questions.
Think addition for subtraction. Solve each problem.
When seven is subtracted from twenty it gives the value 13 as shown below.
i.e. 20 -7 = 13
Let the unknown number be x
7 + x = 20
x = 20 -7
x = 13
Therefore 7 + 13 is equal to 20.
When seven is subtracted from Eighteen it gives the value 13 as shown below.
i.e. 18 -5 = 13
Let the unknown number be x
5 + x = 20
x = 18 – 5
x = 13
Therefore 5 + 13 is equal to 18.
When seven is subtracted from Nineteen it gives the value 12 as shown below.
i.e. 19 -7 = 12
Let the unknown number be x
7 + x = 19
x = 19 – 7
x = 12
Therefore 7 + 12 is equal to 19.
When six is subtracted from seventeen it gives the value 11 as shown below.
i.e. 17 -6 = 11
Let the unknown number be x
6 + x = 17
x = 17 – 6
x = 11
When four is subtracted from sixteen it gives the value 12 as shown below.
i.e. 16 – 4 = 12
Let the unknown number be x
4 + x = 16
x = 16 – 4
x = 12
When one is subtracted from twelve it gives the value 11 as shown below.
i.e. 12 – 1 = 11
Let the unknown number be x
1 + x = 12
x = 12 – 1
x = 11
When three is subtracted from fifteen it gives the value 12 as shown below.
i.e. 15 – 3 = 12
Let the unknown number be x
3 + x = 15
x = 15 – 3
x = 12
When three is subtracted from fifteen it gives the value 12 as shown below.
i.e. 14 – 3 = 11
Let the unknown number be x
3 + x = 14
x = 14 – 3
x = 11
|
# Find an equation of the tangent line to the curve at the given point. y = sin(3x) sin2 (3x) given the point (0,0)
Find an equation of the tangent line to the curve $y = sin(3x) + sin^2 (3x)$ given the point (0,0). Answer is $y = 3x$, but please explain solution steps.
Hint:
Do you know that the slope of the tangent line at a point of the graph of a function is the derivative of the function at this point?
So, for $y = \sin(3x) + \sin^2 (3x)$ find the derivative $y'$ ( can you do?), then evaluate this derivative for $x=0$
Now the line has equation $y=mx$ with $m=y'(0)$
Using the chain rule the derivative is: $$y'=\cos(3x)\cdot(3x)'+2\sin(3x)(\sin(3x))'=3\cos(3x)+2\sin(3x)\cos(3x)(3x)'$$$$=3\cos(3x)+6\sin(3x)\cos(3x)$$ so $y'(0)=3$.
• I know that I have to find the derivative, but I don't know how to. I also know that once you find m (by taking the derivative) you can plug it into the $y-y=m(x-x)$ or $y = mx + b$ slope formulas. – Chaniqua Ranson Oct 5 '16 at 2:04
• Added to my answer. – Emilio Novati Oct 5 '16 at 7:40
Goal
Find the slope $m$, and intercept $b$, for the line $$y = mx + b,$$ tangent at the origin to the curve $$f(x) = \sin ^2(3 x)+\sin (3 x).$$
Intercept $b$
Because the function goes through the origin, the tangent line will also go through the origin. Therefore the $y-$intercept $b=0$.
Slope $m$
The slope of the tangent line $m$ is, by definition, the same as the slope of the target function at the point of contact. The slope of the function is $$f'(x) = 6 \sin (3 x) \cos (3 x) + 3 \cos (3 x).$$ The problem specifies $x_{*} = 0.$ The point of contact is $$\left( x_{*}, f(x_{*}) \right) = \left( 0, 3 \right).$$ The slope of the function at the point of contact is $$f'( x_{*} ) = 6 \sin (3 x_{*}) \cos (3 x_{*}) + 3 \cos (3 x_{*}) = 6\cdot 0 \cdot 0 + 3 \cdot 1 = 3.$$ The slope of the tangent line is $$m = f'( x_{*} ) = 3.$$
Solution The equation of the line tangent to $f(x)$ at $x_{*} = 0$ is $$y = mx + b = 3x.$$
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# 1
Mashallah
Tuition Centre & Pre-Entry Test Centre.
MATHEMATICS -1 Chapter - 1
SETS
Definition of set
A collection well defined and distinct objects such as number, points, shapes ideas etc is
called a set which is denoted by Capital letters of Alphabets i.e. A, B, C… X, Y, Z. Its
symbol is { }.
Notations of Sets
1. Tabular Method
In this method, we define a set merely by listing its elements in closed within braces
{ }.
Example: N = {1, 2, 3…}
## 2. The Descriptive Method
In this method the element of the set are described by stating their common
characteristic which an object must posses in order to be an element of the set.
Example: Set of Odd Number (O)
A = {x | x is an odd integer}
## 3. Set builder Notation
In this method the elements of the set have to satisfy some condition
i.e. {x | x satisfy some condition}
Example: N = {x | x ∈ N}
If A be the set of rational numbers, then in Set Builder Notation we will write it as
A = {x | x is a rational number}
4. Venn Diagram
In this method the set is written in figure such as rectangle, square, circle or any
other.
Example: Draw a Venn Diagram to represent.
U = {1, 2, 3, … ,10} and A = {1, 3, 5, 7, 9}
Figure
U
A 2 4
1 3 5 6 8
7 9 10
2
Kinds of Sets
Universal sets (U)
Universal set is the set, which contains all the available elements.
Equal Sets
Equal set have exactly the same number of same elements.
For example set A = {a,b,c,d} and B = {b,c,a,d} are equal since they have the same
elements. We write A = B, if A and B are equal.
Equivalent Sets
Two sets A and B are said to be equivalent, if they have same number of elements, it is
denoted by A ∼ B.
Example: A = {a,b,c} , B = {1,2,3}
Then A ∼ B, because order of A is equal to order of B
Disjoint Sets
If two sets do not have an element in common, they are said to be disjoint as A∩B =φ
Finite Sets
A set is finite if it contains a limited number of different elements.
Example: A = {1, 2, 3, 4}, B = {a, b, c, d, e, f} etc.
Infinite Sets
A set which is not finite is called infinite set.
Example: A = {1, 3, 5 …}, B = {1, 2, 3 …}, C = {… -3,-2,-1,0,1,2,3 …}
## Null Or Empty Set
A Null set { } or empty set is the set which contains no element.
Example: A = {x | x > 5 and x < 2} = ∅
Power sets
The all of possible subset of a set A is called the power set of A and it is denoted by the
symbol P(A).
Example: If A = {1,2,3} then,
P(A) = {∅ {1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
Note: The number of subsets of power set can be obtained by using 2m where m is no. of
elements in set A.
Subset
If every element of a set A is also an element of a set B, then A is a subset of B and we
write A ⊆ B.
Example: If A = {1,2,3} and B = {1,2,3,4,5} then A ⊂ B
3
Superset
If B is a subset of a set A, then A is called a superset of B, denoted by A ⊂ B or B ⊃ A
Example: If A = {1,2,3} and B = {1,2,3,4,5} then B ⊃ A
Proper Subset
If A is subset of B and A ≠ B then we write A ⊂ B and say that A is Proper subset of B.
Example: If A = {1,2,3} and B = {1,2,3,4,5} then A ⊂ B
Improper Subset
If A ⊆ B and B ⊆ A, then sets A and B are said to be improper subsets of one another if
A=B
Example: If A = {1,2,3} and B = {1,2,3} then A ⊆ B or B ⊆ A
Set Operation
Complement
The complement of a set A relative to a universal set U is the set if all elements in U except
those in A denoted by A’ and A’ = U – A.
A’ = {x | x ∈ U and x ∉ A}
Exhaustive Sets
If A and B be subsets of a set U such that, A U B = U
Symmetric Difference
The symmetric difference of sets A and B, denoted by A ∆ B, is the set containing those
elements which are either in A or in B but not in both A and B.
Example: If A = {1,2,3,4,6} and B = {1,3,5,7} then A ∆ B = {2,4,5,6,7}
Note: A ∆ B = A U B – A ∩ B
## Difference of Two Sets
The difference of sets A and B, denoted by A-B or A/B, is the set containing those elements
that are in A but not in B.
A - B = {x | x ∈ A and x ∉ B}
Intersection of Sets
The intersection of two sets A and B Is the set of element which are common to both A and
B it is denoted by A ∩ B.
A ∩ B = {x | x ∈ A and x ∈ B}
Union of Sets
The union of two sets A and B is set of elements which are in A or B or both it is denoted by
A ∪ B.
A U B = {x | x ∈ A or x ∈ B}
4
Number of Elements n (A)
n (Ф) = 0
n (A U B) = n (A) + n(B) – n(A ∩ B)
## n (A ∩ B) = n (A) + n(B) – n(A U B)
n (A’) = n(U) – n(A)
## The Cartesian product of two Sets
The Cartesian production of any set A with other set B is the set all ordered pairs (a, b)
where a ∈ A and b ∈ B it is denoted by A x B = {(a, b) | a ∈ A, b ∈ B}.
## Some Important Conditions
1. A ⊆ B and B ⊆ A ⇒ A = B
2. if A ⊂ B then B ⊃ A
3. A = B and B = C ⇒ A = C
4. A ∼ B and B ∼ C ⇒ A ∼ C
5. if A = { } = φ then P(A) = {A}
6. the number of subset of A i.e. n{P(A)} is 2m where m is number of elements in A i.e.
n(A) = m.
## Some Important Laws
1. Idempotent Law
AUA=A A∩A=A
2. Associative Laws
If A, B and C are any three sets then
(A U B) U C = A U (B U C), (A ∩ B) ∩ C = A ∩ (B ∩ C)
3. Distributive Laws
If A, B and C are any three sets then
A ∩ (B U C) = (A ∩ B) U (A ∩ C)
(B U C) ∩ A = (B ∩ A) U (C ∩ A)
A U (B ∩ C) = (A U B) ∩ (A U C)
4. Identity Law
AUφ=A A∩U=A
AUU=U A∩φ =φ
5. Complement Law
A U A’ = U A ∩ A’ = φ
(A’)’ = A U’ = φ , φ’ = U
6. Commutative Law
A U B = B U A, A∩B=B∩A
7. De-Morgan’s Law
If A, B and C are any three sets then
(A U B)’ = A’ ∩ B’ (A ∩ B)’ = A’ U B’
5
## Notation for Sets of Numbers
N = {1, 2, 3, …}
i.e. the set of all natural numbers.
W = {0, 1, 2, 3…}
i.e., the set of all non-negative integers.
Or set of whole numbers.
## Z = {…., -3 , -2, -1, 0, 1, 2, 3,…}
i.e., the set of all integers.
P = {2, 3, 5, 7, 11 …}
i.e. the set of all Positive Prime numbers.
O = {±1, ±3, ±5 …}
i.e. the set of all Odd numbers.
## E = {0, ±2, ±4, ±6 …}
i.e. the set of all Even numbers.
## Q = {x|x = p/q, p and q ∈ Z, q ≠ 0}
i.e., the set of all rational numbers.
## Q’ = I = { x|x ≠ p/q, p and q ∈ Z, q ≠ 0}
i.e., the set of all irrational numbers.
## R = { x|x = p/q, x ≠ p/q, p and q ∈ Z, q ≠ 0}
i.e., the set of all real numbers.
Or
R = Q U Q’
Rational Number
Rational Number is a number which can be expressed as a terminating decimal fraction or a
recurring decimal fraction.
Irrational Number
Irrational Number is a number which can be expressed only as a non-recurring, non-
terminating decimal fraction.
6
## Set of Real Numbers
Union of the sets of rational numbers (Q) and irrational numbers (Q’) is called the set of
real numbers, denoted by R.
i.e. R = Q U Q’
or R = {x | x ∈ Q or x ∈ Q’}
The sets Q and Q’ are disjoint sets.
## Properties of Real Numbers
1. Closer property
Sum of any two real numbers is also a real number.
i.e. x, y ∈ R ⇒ x+y∈R
2. Commutative Property
x + y = y + x , ∀ x, y ∈ R (∀ is read as “for all”)
3. Associative property
x + (y + z) = (x + y) + z , ∀ x, y, z ∈ R
There exists a number 0 ∈ R such that
x + 0 = 0 + x = x, ∀x∈R
The element 0 is called the additive identity.
For every x ∈ R, there exists an element x’ ∈ R such that
x + x’ = 0 = x’ + x
x’ is called Additive Inverse of x and is denoted by “-x”
## Property w.r.t. Multiplication
1. Closer property
Products of any two real numbers is also a real number.
i.e. x, y ∈ R ⇒ x.y ∈ R
2. Commutative Property
x.y = y.x , ∀ x, y ∈ R
3. Associative property
x (y.z) = (x.y) z , ∀ x y ∈ R
4. Multiplicative Identity
There exists a number 1 ∈ R such that
7
x . 1 = 1 . x = x,
The number 1 is called the Multiplicative identity.
5. Multiplicative Inverse
For each x ∈ R, x ≠ 0, there exists an element x* ∈ R such that
x . x* = x* . x = 1
x* is called Multiplicative Inverse of x. It is also written as 1/x or x-1 Thus,
x. 1/x = x/x = 1/x .x = 1 or x . x-1 = x-1.x = 1
## Distributive property of Multiplication w.r.t. Addition
x (y + z) = xy + xz and (y + z) x = yx +zx ∀ x, y, z ∈ R
Trichotomy Property
For any two real numbers x and y, either x < y or x = y or x > y.
## Property of Equality of real Numbers
In the set R of real numbers, a relation of equality to be denoted by “=” is defined. This
relation satisfies the following properties:
1. Reflexive Property
x = x, for all x ∈ R
2. Symmetric property
x = y ⇒ y = x, for all x, y ∈ R
3. Transitive property
x = y and y = z ⇒ x = z for all x, y, z ∈ R
x = y ⇒ x + z = y + z and z + x = z + y, for all x, y, z ∈ R
i.e. addition of same no. to each side of an equality does not change the relation.
5. Multiplicative Property
∀ x, y, z ∈ R
x = y ⇒ xz = yz (Right Multiplication)
and zx = zy (Left Multiplication)
i.e. Multiplication by the same no. to each side of an equality does not change the relation.
## 6. Cancellation Property w.r.t. Addition
∀ x, y, z ∈ R
(a) x + z = y +z ⇒ x = y (Right Cancellation )
(b) z + x = z + y ⇒x=y (Left Cancellation)
## 7. Cancellation Property w.r.t. Multiplication
∀ x, y, z ∈ R, z ≠ 0
(a) xz = yz ⇒ x = y (Right Cancellation )
(b) zx = zy ⇒ x = y (Left Cancellation)
8
Note: Additive property and Cancellation property w.r.t. Addition are converse of each
other. Similarly, Multiplicative property and Cancellation property w.r.t. Multiplication are
converse of each other.
|
We will discuss here about the advantages of tabular data. When a larger number of data is given in a tabular form, it is easier to get information from it.
1. The heights of the women of a school are given below.
Height (in cm) 150 151 152 153 154 155 156 157 158 159 No. Of Women 5 4 10 3 3 2 2 3 6 4
This table tells us that the number of women whose height is 150 cm, is 5; 4 women are 151 cm tall & so on. We get a number of other information also. Some are given below.
(i) The total number of women in the clubs = 5 + 4 + 10 + 3 + 3 + 2 + 2 + 3 + 6 + 4 = 42.
(ii) The number of women who are 153 cm tall is 3.
(iii) The largest number of women of the same height is 10 and they are 152 cm tall.
(iv) The number of women who are shorten than 154 cm = 5 + 4 + 10 + 3 = 22
(v) If a women taller than 154 cm is considered tall, the number of tall women in the clubs = 2 + 2 + 3 + 6 + 4 = 17
So, we see that as data is arranged systematically in a table, it becomes more useful.
2. In order to find out the amount of money collected for charity the 16 students of a class, the monitor notes down the following:
Roll No1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Amount (in $) 62 78 80 62 85 58 90 62 70 15 20 40 50 20 10 25 First arrange the collected data in ascending order and then in a tabular form. Then answer the following: (i) How many students collected$ 30?
(ii) How many students collected less than $25? (iii) What percentage of students collected more than$ 30?
The data arranged in ascending order:
10, 10, 15, 15, 15, 20, 20, 20, 20, 25, 30, 30, 40, 40, 50, 50
Now putting the data in tabular form, we get the following:
Amount (in $) 10 15 20 25 30 35 40 45 50 No. Of Students 2 3 4 1 2 0 2 0 2 From the above table we can see that. (i) The number of students who collected$ 30 is 2.
(ii) The number of students who collected less than $30 is 2 + 3 + 4 + 1 = 10 (iii) The number of students who collected more than$ 35 is 2 + 0 + 2 = 4
Total number of students = 16
Then percentage of students who collected more than \$ 35 = 4/ 16 X 100 = 25%
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
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# Matrices Mathematics: Matrix
Algorithms
## Matrix System
In mathematics, a matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. Matrices are widely used in various branches of mathematics, physics, computer science, and engineering. Here is a basic overview of the matrix system:
### Matrix Notation:
A matrix is usually denoted by a capital letter. For example, let's use the letter A to represent a matrix.
$A=\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ \mathrm{⋮}\phantom{\rule{0em}{1.5em}}& \mathrm{⋮}\phantom{\rule{0em}{1.5em}}& \ddots & \mathrm{⋮}\phantom{\rule{0em}{1.5em}}\\ {a}_{m1}& {a}_{m2}& \cdots & {a}_{mn}\end{array}\right]$
In this matrix A, the element aij is located in the i-th row and j-th column.
### Matrix Size:
A matrix with m rows and n columns is called an m × n matrix. The size is often written as m × n.
### Special Matrices:
1. Square Matrix: A matrix where the number of rows is equal to the number of columns (i.e., m = n).
$B=\left[\begin{array}{ccc}{b}_{11}& {b}_{12}& {b}_{13}\\ {b}_{21}& {b}_{22}& {b}_{23}\\ {b}_{31}& {b}_{32}& {b}_{33}\end{array}\right]$
2. Identity Matrix (I): A square matrix with ones on the main diagonal and zeros elsewhere.
$I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
3. Zero Matrix (or Null Matrix): A matrix where all elements are zeros.
$O=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
### Matrix Operations:
1. Matrix Addition: If A and B are matrices of the same size, then their sum A + B is obtained by adding corresponding elements.
2. Scalar Multiplication: If A is a matrix and c is a scalar, then the product cA is obtained by multiplying each element of A by c.
3. Matrix Multiplication: If A is an m × p matrix and B is a p × n matrix, then their product AB is an m × n matrix.
### Transpose of a Matrix:
The transpose of a matrix A is denoted by AT and is obtained by swapping its rows and columns.
${A}^{T}=\left[\begin{array}{cccc}{a}_{11}& {a}_{21}& \cdots & {a}_{m1}\\ {a}_{12}& {a}_{22}& \cdots & {a}_{m2}\\ \mathrm{⋮}\phantom{\rule{0em}{1.5em}}& \mathrm{⋮}\phantom{\rule{0em}{1.5em}}& \ddots & \mathrm{⋮}\phantom{\rule{0em}{1.5em}}\\ {a}_{1n}& {a}_{2n}& \cdots & {a}_{mn}\end{array}\right]$
These are some fundamental concepts of matrices in the mathematical sense. They are extensively used in solving systems of linear equations, transformations, and various other mathematical operations.
#### Programmatic Expression:
Here's an example of a C program that calculates the transpose of a matrix. The program assumes a square matrix, but you can modify it to handle rectangular matrices as needed.
c Copy Code
#include ＜stdio.h>
#define MAX_SIZE 10
void transposeMatrix(int matrix[MAX_SIZE][MAX_SIZE], int rows, int cols, int result[MAX_SIZE][MAX_SIZE]) {
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
result[j][i] = matrix[i][j];
}
}
}
void displayMatrix(int matrix[MAX_SIZE][MAX_SIZE], int rows, int cols) {
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
printf("%d\t", matrix[i][j]);
}
printf("\n");
}
}
int main() {
int matrix[MAX_SIZE][MAX_SIZE], transpose[MAX_SIZE][MAX_SIZE];
int rows, cols;
printf("Enter the number of rows and columns of the matrix: ");
scanf("%d %d", &rows, &cols);
// Input matrix elements
printf("Enter the elements of the matrix:\n");
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
printf("Enter element matrix[%d][%d]: ", i, j);
scanf("%d", &matrix[i][j]);
}
}
// Calculate transpose of the matrix
transposeMatrix(matrix, rows, cols, transpose);
// Display original matrix
printf("\nOriginal Matrix:\n");
displayMatrix(matrix, rows, cols);
// Display transpose matrix
printf("\nTranspose Matrix:\n");
displayMatrix(transpose, cols, rows);
return 0;
}
Explanation:
This program first takes the size of the matrix as input, then reads the matrix elements, calculates the transpose, and finally displays both the original and transpose matrices. Adjust the 'MAX_SIZE' macro according to the maximum size of the matrix you want to handle in your program.
Output:
Enter the number of rows and columns of the matrix: 3 3
Enter the elements of the matrix:
Enter element matrix[0][0]: 1
Enter element matrix[0][1]: 2
Enter element matrix[0][2]: 3
Enter element matrix[1][0]: 4
Enter element matrix[1][1]: 5
Enter element matrix[1][2]: 6
Enter element matrix[2][0]: 7
Enter element matrix[2][1]: 8
Enter element matrix[2][2]: 9
Original Matrix:
1 2 3
4 5 6
7 8 9
Transpose Matrix:
1 4 7
2 5 8
3 6 9
This example demonstrates the program with a 3x3 matrix. You can input different matrices by adjusting the size and elements as needed.
#### What's Next?
We've now entered the finance section on this platform, where you can enhance your financial literacy.
|
Question
# Solution of $\dfrac{{dx}}{{dy}} + mx = 0,{\text{ where }}m < 0$ is:(A) $x = C{e^{my}}$ (B) $x = C{e^{ - my}}$ (C) $x = my + C$ (D) $x = C$
Hint: Separate the terms containing $x$ and $y$ on either side and then integrate both sides.
According to the question, the given differential equation is $\dfrac{{dx}}{{dy}} + mx = 0$.
If we separate $x$ and $y$ terms, we’ll get:
$\Rightarrow \dfrac{{dx}}{{dy}} + mx = 0, \\ \Rightarrow \dfrac{{dx}}{{dy}} = - mx, \\ \Rightarrow \dfrac{{dx}}{x} = - mdy \\$
Integrating both sides, we’ll get:
$\Rightarrow \int {\dfrac{{dx}}{x}} = - m\int {dy} ,$
We know that, $\int {\dfrac{{dx}}{x}} = \ln x$. Using this we’ll get:
$\Rightarrow \ln x = - my + c,{\text{ where }}c{\text{ is the constant of integration}} \\ \Rightarrow x = {e^{ - my + c}}, \\ \Rightarrow x = {e^c} \times {e^{ - my}}, \\ \Rightarrow x = C{e^{ - my}}{\text{ [}}\therefore {e^c}{\text{ = C (another constant) ]}} \\$
Thus, the solution of the given differential equation is $x = C{e^{ - my}}$. (B) is the correct option.
Note: The above differential equation is a first order differential equation. We can also solve it by calculating integrating factors. Suppose we have a first order differential equation:
$\Rightarrow \dfrac{{dx}}{{dy}} + Px = Q,$
We calculate the integrating factor as:
$\Rightarrow I = {e^{\int {Pdy} }}$.
The solution of the differential equation is:
$\Rightarrow Ix = \int {IQdy}$
If we solve by this method, we’ll get the same result.
|
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Adding two negative numbers together? Just add the absolute value of each number together, put a negative sign in front, and you have your answer! See how it's done in this tutorial.
#### How Do You Solve a Word Problem Where You Multiply and Subtract Whole Numbers and Fractions?
Word problems are a great way to see math in the real world. In this tutorial, you'll see how to translate a word problem to a mathematical equation. Then, see how to use the order of operations to get the answer!
#### How Do You Solve a Word Problem Where You Divide Fractions?
Working with word problems AND fractions? This tutorial shows you how to take a word problem and translate it into a mathematical equation involving a complex fraction. Then, you'll see how to simplify the complex fraction to get the answer. Check it out!
#### How Do You Subtract a Whole Number from a Fraction?
Subtracting a whole number from a fraction can be tricky. Luckily, watching this tutorial can make this subtraction no big deal!
#### How Do You Subtract Decimals?
Doing math with paper and pencil can come in real handy, so make sure you're comfortable subtracting decimals by hand. After all, you don't want the calculator to be a crutch!
#### How Do You Add Decimals?
Doing math with paper and pencil can come in real handy, so make sure you're comfortable adding decimals by hand. After all, you don't want the calculator to be a crutch!
#### How Do You Multiply Decimals?
Doing math with paper and pencil can come in real handy, so make sure you're comfortable multiplying decimals by hand. After all, you don't want the calculator to be a crutch!
#### How Do You Distribute With Whole Numbers and Fractions?
This tutorial shows you how to distribute a whole number into the sum of fractions. It's an important skill to have when you're solving equations, and you never know when it can come up. Be sure to check out this tutorial!
#### How Do You Solve a Word Problem with an Equation Using Addition?
Word problems are a great way to see math in action! See how to translate a word problem into an equation, solve to find the answer, and check your found answer all in this tutorial.
#### How Do You Solve a Word Problem with an Equation Using Subtraction?
Word problems are a great way to see math in action! See how to translate a word problem into an equation, solve to find the answer, and check your found answer all in this tutorial.
#### How Do You Solve a Word Problem Using an Equation Where You're Multiplying Fractions?
Working with word problems AND fractions? This tutorial shows you how to take a word problem and translate it into a mathematical equation involving fractions. Then, you'll see how to solve and get the answer. Check it out!
#### How Do You Solve a Word Problem with an Equation Using Multiplication?
Working with word problems AND fractions? This tutorial shows you how to take a word problem and translate it into a mathematical equation involving fractions. Then, you'll see how to solve and get the answer. Check it out!
#### How Do You Solve a Word Problem with an Equation Using Division?
Word problems are a great way to see math in action! See how to translate a word problem into an equation, solve to find the answer, and check your found answer all in this tutorial.
#### How Do You Find the Length of a Rectangle if You Know its Width and Area?
How do you find the length of a rectangle if you're given the width and the area? This tutorial shows you how!
#### How Do You Use an Equation with Consecutive Numbers to Solve a Word Problem?
Word problems are a great way to see math in the real world. In this tutorial, you'll see how to translate a word problem into a mathematical equation involving consecutive numbers. Then you'll see how to solve that equation and check your answer!
#### How Do You Solve a Word Problem Where You Multiply Fractions and Work Backwards?
Word problems are a great way to see math in the real world. In this tutorial, you'll see how to solve a word problem by working backwards using a table. Check it out!
#### How Do You Solve a Word Problem Using a Two-Step Equation with Decimals?
Word problems are a great way to see math in the real world. In this tutorial, you'll see how to translate a word problem into a mathematical equation. Then you'll see how to solve that equation and check your answer!
#### How Do You Solve a Word Problem Using a Percent Proportion?
Word problems allow you to see the real world uses of math! This tutorial shows you how to take a words problem and turn it into a percent proportion. Then see how to solve for the answer using the mean extremes property of proportions. Take a look!
#### How Do You Set Up a Proportion from a Word Problem?
Sometimes the hardest part of a word problem is figuring out how to turn the words into an equation you can solve. This tutorial let's you see the steps to take in order to turn a word problem involving a blueprint into a proportion. Take a look!
#### How Do You Use a Proportion to Find a Part of a Whole?
Taking a percent of a number? Trying to figure out the result? Use a percent proportion to solve! This tutorial will show you how!
#### How Do You Use the Formula for Simple Interest?
If you already have a bank account or if you plan to have one in the future, then this tutorial is a must see! Follow along as this tutorial goes through a word problem involving simple interest.
#### What is the Formula for Simple Interest?
Interest is found in a bunch of places: savings accounts, mortgages, loans, investments, credit cards, and more! Watch this tutorial and learn how to calculate simple interest!
#### How Do You Figure Out a Percent of Change?
Word problems allow you to see the real world uses of math! In this tutorial, learn how to calculate the percent of increase using the percent of change formula.
#### What's a Percent of Change?
Lots of things in this world change their value such as cars, video games, and computers. When something either increases or decreases in value, it can be useful to know the percent of that change in value. To figure out that percent, you'll need the percent of change formula. Learn it with this tutorial!
#### How Do You Figure Out How Much Something is Marked Down?
Going shopping? Is something you want on sale? Trying to figure out the sale price of that item? Follow along with this word problem and you'll see how to calculate that price!
#### How Do You Set Up a Percent Proportion from a Word Problem?
Sometimes the hardest part of a word problem is figuring out how to turn the words into an equation you can solve. This tutorial let's you see the steps to take in order to do just that! Take a look! You'll be glad you did!
#### How Do You Figure Out Sales Tax?
Going shopping can be tons of fun, but things can go sour when you get to the register and realize that the sales tax puts you over your budget. Always stay under budget by figuring out your total cost BEFORE you hit the check out. Watch this tutorial and learn how to calculate sales tax!
#### How Do You Use an Equation to Find a Part of a Whole?
Taking a percent of a number? Trying to figure out the result? Convert the percent to a decimal and multiply it by the number! This tutorial will show you how!
#### How Do You Solve a Word Problem Using the Direct Variation Formula?
Word problems allow you to see math in action! Take a look at this word problem involving an object's weight on Earth compared to its weight on the Moon. See how the formula for direct variation plays an important role in finding the solution. Then use that formula to see how much you would weigh on the Moon!
#### What's the Direct Variation or Direct Proportionality Formula?
Ever heard of two things being directly proportional? Well, a good example is speed and distance. The bigger your speed, the farther you'll go over a given time period. So as one variable goes up, the other goes up too, and that's the idea of direct proportionality. But you can express direct proportionality using equations, and that's an important thing to do in algebra. See how to do that in the tutorial!
#### How Do You Use the Formula for Direct Variation?
If two things are directly proportional, you can bet that you'll need to use the formula for direct variation to solve! In this tutorial, you'll see how to use the formula for direct variation to find the constant of variation and then solve for your answer.
#### How Do You Solve a Proportion Using Cross Products?
Want to solve a percent proportion? Just use the means extremes property of proportions to cross multiply! Solve for the variable, and you have your answer! Learn how with this tutorial.
#### How Do You Use a Proportion to Find What Percent a Part is of a Whole?
A part is some percent of a whole. Trying to calculate the percent? Use a percent proportion to solve! This tutorial will show you how!
#### What's a Proportion?
The idea of proportions is that a ratio can be written in many ways and still be equal to the same value. That's why proportions are actually equations with equal ratios. This is a bit of a tricky definition, so make sure to watch the tutorial!
#### What's the Means-Extremes Property of Proportions?
The means-extremes property of proportions allows you to cross multiply, taking the product of the means and setting them equal to the product of the extremes. This property comes in handy when you're trying to solve a proportion. Watch this tutorial to learn more!
#### What are the Means and Extremes of Proportions?
A proportion is just an equation where two ratios are equal, and each piece of the proportion has a special name. This tutorial will teach you those names, and this will help you understand cross multiplication when you learn it later!
#### How Do You Solve a Word Problem Using a Proportion?
This tutorial provides a great real world application of math. You'll see how to use the scale from a blueprint of a house to help find the actual height of the house. This tutorial shows you how to use a proportion to solve!
#### How Do You Use Subtraction to Solve an Inequality Word Problem?
Word problems allow you to see the real world uses of math! In this tutorial, learn how to translate a word problem into an inequality. Then see how to solve the inequality and understand the meaning of the answer.
#### How Do You Use Addition to Solve an Inequality Word Problem?
Word problems allow you to see math in action! This tutorial deals with inequalities and money in a bank account. See how to translate a word problem into an inequality, solve the problem, and understand the answer. Take a look!
#### How Do You Use Division with Positive Numbers to Solve an Inequality Word Problem?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Use Division with Negative Numbers to Solve an Inequality Word Problem?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Use Multiplication with Positive Numbers to Solve an Inequality Word Problem?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Use Multiplication with Negative Numbers to Solve an Inequality Word Problem?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve a Word Problem Using an Inequality Where You're Multiplying Positive Fractions?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve a Word Problem Using an Inequality Where You're Multiplying Negative Fractions?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve a Word Problem Using a Multi-Step Inequality?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve a Word Problem Using an Inequality With Variables on Both Sides?
This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### Why Can't You Divide By 0?
Why can't you divide by 0? This may be one of the most asked math questions. Get this question answered once and for all by watching this tutorial!
#### How Do You Determine Whether a Triangle Can Be Formed Given Three Side Lengths?
If you're given 3 side measurements, there's a quick way to determine if those three sides can form a triangle. Follow along with this tutorial and learn what relationship these sides need in order to form a triangle.
#### How Do You Solve a Problem by Making an Organized List?
Organization is a big part of math. In this tutorial, you'll see how organizing information given in a word problem can help you solve the problem and find the answer!
#### How Do You Figure Out a Tip?
If you need to leave a tip at a restaurant, you can quickly estimate the amount in your head! This tutorial shows you how to use estimation and mental math to estimate a tip!
#### How Do You Find Equivalent Ratios by Making a Table?
To master equivalent ratios, you need to practice. Follow along with this tutorial to practice filling in a table with equivalent ratios.
#### How Do You Estimate a Sale Price?
Sales are great, but how much are you really saving? This tutorial shows you how to estimate the sales price of an item.
#### How Do You Figure Out a Percent From a Part to Part Ratio?
Word problems and percents can be a fun combination! This tutorial shows you how to find the percent of something in a basket using ratios!
#### How Do You Figure Out a Percent From a Part to Whole Ratio?
If you want to find a percent in a word problem, you may be able to use a ratio to help you! This tutorial shows you how to do exactly that!
#### How Do You Add Integers Using a Number Line?
Combining integers? You could use a number line to help find the answer! In this tutorial, see how to use a number line to add together integers with the same sign and ones with opposite signs. Take a look!
#### How Do You Subtract Integers Using a Number Line?
Subtracting integers? You could use a number line to help find the answer! In this tutorial, see how to use a number line to subtract integers with the same sign and ones with opposite signs. Take a look!
#### What are the Rules for Using Absolute Values to Add Integers?
When you're combining numbers, there are some helpful rules to make that process a little easier. This tutorial shows you the rules for using absolute values to combine integers with the same sign or with opposite signs. Take a look!
#### What is the Opposite, or Additive Inverse, of a Number?
Have you ever combined two numbers together and found their sum to be zero? When that happens, those numbers are called additive inverses of each other! In this tutorial, you'll learn the definition for additive inverse and see examples of how to find the additive inverse of a given value.
#### How Do You Solve a Two-Step Equation?
Becoming a pro at solving equations takes practice! Follow along with this tutorial to see an example of solving an equation for a variable.
#### How Do You Factor a Greatest Common Factor Out of an Expression?
The greatest common factor (GCF) is the largest factor two or more numbers have in common. Finding the GCF can be very useful in simplifying an expression or solving an equation. In this tutorial, see how to identify the GCF of an expression and factor it out. Check it out!
#### How Do You Divide a Decimal by a Decimal?
Dividing decimals? Then this tutorial is a must see! Follow along and learn how you can divide decimals by rewriting the problem as a fraction and then using long division to solve. Check it out!
#### How Do You Divide Fractions?
Dividing fractions? Change that division to a multiplication by multiplying the dividend by the reciprocal of the divisor. Learn all about it by watching this tutorial!
#### How Do You Turn a Fraction Into a Terminating Decimal?
Did you know that a fraction just represents a division? To turn a fraction into a decimal, divide the numerator by the denominator. In this tutorial, see how to convert a fraction into the terminating decimal it represents.
#### How Do You Turn a Fraction Into a Repeating Decimal?
Did you know that a fraction just represents a division? To turn a fraction into a decimal, divide the numerator by the denominator. In this tutorial, see how to convert a fraction into the repeating decimal it represents.
#### What is a Terminating Decimal?
A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits. Did you know that all terminating decimals can be rewritten as fractions? Watch this tutorial to learn about terminating decimals and see some examples!
#### What is a Repeating Decimal?
A repeating decimal is a decimal that has a digit, or a block of digits, that repeat over and over and over again without ever ending. Did you know that all repeating decimals can be rewritten as fractions? To make these kinds of decimals easier to write, there's a special notation you can use! Learn about repeating decimals in this tutorial.
#### How Do You Find Equivalent Ratios?
Ratios are used to compare numbers. When you're working with ratios, it's sometimes easier to work with an equivalent ratio. Equivalent ratios have different numbers but represent the same relationship. In this tutorial, you'll see how to find equivalent ratios by first writing the given ratio as a fraction. Take a look!
#### How Do You Know If Two Ratios are Proportional?
Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional. To see this process in action, check out this tutorial!
#### How Do You Use Unit Rates to Compare Rates?
Word problems allow you to see the real world uses of math! This tutorial shows you how to use ratios to figure out which store has a better deal on cupcakes. Take a look!
#### How Do You Determine if Two Ratios are Proportional Using Cross Products?
Trying to figure out if two ratios are proportional? If they're in fraction form, set them equal to each other to test if they are proportional. Cross multiply and simplify. If you get a true statement, then the ratios are proportional! This tutorial gives you a great example!
#### How Do You Solve a Proportion by Finding an Equivalent Ratio?
Trying to find a missing value in order to create a proportion with two ratios? Take the ratios in fraction form and identify their relationship. Use that relationship to find your missing value. This tutorial will show you how!
#### How Do You Determine Whether Values in a Table are Proportional?
To see if multiple ratios are proportional, you could write them as fractions, reduce them, and compare them. If the reduced fractions are all the same, then you have proportional ratios. To see this process step-by-step, check out this tutorial!
#### How Do You Solve a Proportion Using the Multiplication Property of Equality?
Trying to find a missing value in a ratio to create proportional ratios? You could use the multiplication property of equality! In this tutorial, see how to use this property to find a missing value in a ratio. Take a look!
#### How Do You Solve a Scale Model Problem Using a Scale Factor?
This tutorial provides a great real world application of math! You'll see how to use the scale on a house blueprint to find the scale factor. Then, see how to use the scale factor and a measurement from the blueprint to find the measurement on the actual house! Check out this tutorial and see the usefulness blueprints and scale factor!
#### How Do You Find the Scale of a Model?
Want some practice with scale? Then check out this tutorial and you'll see how to find the scale of a model given the lengths of the model and the actual object. Take a look!
#### How Do You Use the Scale on a Map to Find an Actual Distance?
Maps help us get from one place to another. In this tutorial, you'll learn how to use a map to find an actual distance.
#### How Do You Figure Out What Size a Model Should Be if You Have a Scale?
Before tall sky scrapers are build, a scale model of the building is made, but how does the architect know what size the model should be? Follow along with this tutorial to find out!
#### How Do You Solve a Word Problem Using Ratios?
This tutorial shows you how to use a ratio to create equivalent ratios. Then, use a multiplier to find a missing value and solve the word problem. Take a look!
#### What are Equivalent Ratios?
Equivalent ratios are just like equivalent fractions. If two ratios have the same value, then they are equivalent, even though they may look very different! In this tutorial, take a look at equivalent ratios and learn how to tell if you have equivalent ratios.
#### What is a Scale?
In math, the term scale is used to represent the relationship between a measurement on a model and the corresponding measurement on the actual object. Without scales, maps and blueprints would be pretty useless. Check out this tutorial and learn about scale factor!
#### How Do You Figure Out the Price of a Marked Up Item?
The price of items is always changing. You've probably went to the store to buy an item and found that its price has been marked up. In this tutorial, learn how to figure out the new price of an item that was marked up. Take a look!
#### How Do You Figure Out Whether a Percent of Change is an Increase or a Decrease?
Word problems are a great way to see the real world applications of math! In this tutorial, you'll see how the percent of change can be found from the information given in a word problem. Check it out!
#### How Do You Use Complementary Angles to Find a Missing Angle?
If two angles are complementary, that means that they add up to 90 degrees. This is very useful knowledge if you have a figure with complementary angles and you know the measurement of one of those angles. In this tutorial, see how to use what you know about complementary angles to find a missing angle measurement!
#### How Do You Use Supplementary Angles to Find a Missing Angle?
If angles combine to form a straight angle, then those angles are called supplementary. In this tutorial, you'll see how to use your knowledge of supplementary angles to set up an equation and solve for a missing angle measurement. Take a look!
#### How Do You Find Angle Measures for Adjacent Angles?
Learning how to find missing angle measurements is a very useful skill. In this tutorial, get some practice finding missing angle measurements by first creating an equation. Take a look!
#### What are Complementary Angles?
Do complementary angles always have something nice to say? Maybe. One thing complementary angles always do is add up to 90 degrees. In this tutorial, learn about complementary angles and see how to use this knowledge to solve a problem involving these special types of angles!
#### What are Supplementary Angles?
Knowing about supplementary angles can be very useful in solving for missing angle measurements. This tutorial introduces you to supplementary angles and shows you how to use them to solve for a missing angle measurement. Take a look!
#### What are Vertical Angles?
Vertical angles have a very special quality. They are always congruent to one another! Check out this tutorial to learn about and see how to identify vertical angles!
#### How Do You Find the Width of a Rectangle if You Know its Length and Perimeter?
Trying to figure out a missing side length of a rectangle? Got the perimeter and the other side length? Then you can use that information and the formula for the perimeter of a rectangle to find that missing length! This tutorial will show you how!
#### How Do You Find the Area of a Rectangle?
To find the area of a rectangle, multiply the length times the width! This tutorial will show you how to find the area of a rectangle. Check it out!
#### How Do You Find the Area of a Triangle?
Finding the area of a triangle? Know the length of the base and the height? Then just take those values and plug them into the formula for the area of a triangle and solve! This tutorial shows you how.
#### How Do You Find the Area of a Parallelogram?
Looking for the area of a parallelogram? Got the length of the base and the height? Then plug those values into the formula for the area of a parallelogram and solve. This tutorial takes you through the process!
#### How Do You Find the Area of a Trapezoid?
Want to find the area of a trapezoid? If you have the length of each base and the height, you can use them to find the area. In this tutorial, you'll see how to identify those values and plug them into the formula for the area of a trapezoid. Then see how to simplify to get your answer!
#### How Do You Find the Area of a Composite Figure?
Composite figures are just a combination of simpler figures in disguise! In this tutorial, you'll see how to break down a composite figure into simpler figures. Then, see how to find the area of each of those individual figures to find the area of the entire composite figure. Watch the whole process in this tutorial!
#### How Do You Find the Circumference of a Circle if You Know the Diameter?
Trying to find the circumference of a circle? Know the diameter? Then you can use the formula for the circumference of a circle to get the answer! Just plug the value for the diameter into the formula and solve. This tutorial shows you how!
#### How Do You Find the Circumference of a Circle if You Know the Radius?
Trying to find the circumference of a circle? Know the radius? Then you can use the formula for the circumference of a circle to get the answer! Just plug the value for the radius into the formula and solve. This tutorial shows you how!
#### How Do You Find the Radius of a Circle if You Know the Circumference?
Want to find the radius of a circle? Already have the circumference? Then you can use the formula for the circumference of a circle to solve! This tutorial shows you how to use that formula and the given value for the circumference to find the radius. Take a look!
#### How Do You Find the Area of a Circle if You Know the Radius?
If you know the radius of a circle, you can use it to find the area of that circle. Just plug that value into the formula for the area of a circle and solve. Watch this tutorial to see how it's done!
#### How Do You Find the Radius of a Circle if You Know the Area?
Want to find the radius of a circle? Already have the area? Then you can use the formula for the area of a circle to solve! This tutorial shows you how to use that formula and the given value for the area to find the radius. Take a look!
#### How Do You Find the Area of a Circle if You Know the Diameter?
If you have the diameter of a circle, you can use it to find the area of that circle. Just plug that value into the formula for the area of a circle and solve. Watch this tutorial to see how it's done!
#### How Do You Find the Volume of a Rectangular Prism?
Finding the volume of a rectangular prism isn't so bad, especially if you already know the length, width, and height. In this tutorial, you'll see how to use that information and the formula for the volume of a rectangular prism to get the answer. Check it out!
#### How Do You Find the Volume of a Triangular Prism?
Finding the volume of a triangular prism isn't so bad, especially if you already know the length and height of the base and the height of the prism. In this tutorial, you'll see how to use that information and the formula for the volume of a triangular prism to get the answer. Take a look!
#### How Do You Find the Volume of a Composite Figure?
Composite figures are just a combination of simpler figures in disguise! In this tutorial, you'll see how to break down a composite figure into simpler figures. Then, see how to find the volume of each of those individual figures to find the volume of the entire composite figure. Watch the whole process in this tutorial!
#### How Do You Find the Lateral and Surface Areas of a Rectangular Prism?
The lateral area of a three-dimensional solid is the area of all the lateral faces. In this tutorial, you'll see how to use the dimensions of a rectangular prism to find the lateral area. Take a look!
#### How Do You Find the Lateral and Surface Areas of a Triangular Prism?
Want to know how the find the lateral and surface areas of a triangular prism? Then check out this tutorial! You'll see how to apply each formula to the given information to find the lateral area and surface area. Take a look!
#### What is a Circle?
Circles are a fundamental part of math! In this tutorial, you'll be introduced to circles and see the different parts of a circle such as the diameter, radius, and chord. Check out this tutorial to learn about circles!
#### What is a Composite Figure?
Ever notice that some figures look like a combination of multiple other figures? These types of figures are called composite figures. This tutorial introduces you to composite figures and shows you how to break up a composite figure into multiple shapes. Take a look!
#### What is the Formula for the Area of a Rectangle?
Trying to find the area of a rectangle? There's a formula that can help! Check out this tutorial to learn about the formula for the area of a rectangle.
#### What is the Formula for the Area of a Triangle?
Did you know that the formula for the area of a triangle can be found by using the formula for the area of a parallelogram? In this tutorial, you'll see how it's done! Take a look!
#### What is the Formula for the Area of a Parallelogram?
Parallelograms and rectangles are pretty similar. In fact, you can turn a parallelogram into a rectangle to find the formula for the area of a parallelogram! Check out this tutorial to see how it's done!
#### What is the Formula for the Area of a Trapezoid?
Trying to figure out the formula for the area of a trapezoid? You could start by creating a parallelogram out of two trapezoids. Then, use the formula for the area of a parallelogram to figure out the formula for the area of one trapezoid. This tutorial shows you how!
#### What is Circumference?
The circumference of a circle is the distance around that circle. But what is the formula to find the circumference? In this tutorial, you'll learn the formulas for the circumference of a circle. Take a look!
#### What is the Formula for the Area of a Circle?
Did you know that you can figure out the formula for the area of a circle by first turning the circle into a parallelogram? It seems a little weird, but it really works! Watch this tutorial to see how it's done!
#### What is the Formula for the Volume of a Prism?
Trying to find the volume of a prism? Did you know that there's a formula to find that volume? In this tutorial, you'll learn about the formula for the volume of a prism. Check it out!
#### What is the Formula for the Surface Area of a Prism?
To find the lateral and surface areas of a prism, it’s important to know their formulas. In this tutorial, you’ll learn about each of these formulas and see them used in an example. Check it out!
#### How Do You Determine All the Possible Outcomes of an Experiment?
When you perform an experiment, how do you figure out all the possible outcomes? Follow along with this tutorial to see!
#### What is an Unbiased Sample?
When you're trying to learn about a population, it can be helpful to look at an unbiased sample. An unbiased sample can be an accurate representation of the entire population and can help you draw conclusions about the population. This tutorial introduces you to unbiased sampling!
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# Equilateral Triangles
## Properties of triangles with three equal sides.
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Equilateral Triangles
What if your parents want to redo the bathroom? Below is the tile they would like to place in the shower. The blue and green triangles are all equilateral. What type of polygon is dark blue outlined figure? Can you determine how many degrees are in each of these figures? Can you determine how many degrees are around a point?
### Equilateral Triangles
By definition, all sides in an equilateral triangle have exactly the same length.
#### Investigation: Constructing an Equilateral Triangle
Tools Needed: pencil, paper, compass, ruler, protractor
1. Because all the sides of an equilateral triangle are equal, pick a length to be all the sides of the triangle. Measure this length and draw it horizontally on your paper.
2. Put the pointer of your compass on the left endpoint of the line you drew in Step 1. Open the compass to be the same width as this line. Make an arc above the line.
3. Repeat Step 2 on the right endpoint.
4. Connect each endpoint with the arc intersections to make the equilateral triangle.
Use the protractor to measure each angle of your constructed equilateral triangle. What do you notice?
From the Base Angles Theorem, the angles opposite congruent sides in an isosceles triangle are congruent. So, if all three sides of the triangle are congruent, then all of the angles are congruent or each.
Equilateral Triangles Theorem: All equilateral triangles are also equiangular. Also, all equiangular triangles are also equilateral.
#### Solving for Variables
1. Find the value of .
Because this is an equilateral triangle . Now, we have an equation, solve for .
2. Find the values of and .
Let’s start with . Both sides are equal, so set the two expressions equal to each other and solve for .
For , we need to use two expressions because this is an isosceles triangle and that is the base angle measurement. Set all the angles equal to and solve.
#### Measuring the Sides of an Equilateral Triangle
Two sides of an equilateral triangle are units and units. How long is each side of this triangle?
The two given sides must be equal because this is an equilateral triangle. Write and solve the equation for .
To figure out how long each side is, plug in for in either of the original expressions. . Each side is units.
Bathroom Tile Problem Revisited
Let’s focus on one tile. First, these triangles are all equilateral, so this is an equilateral hexagon (6 sided polygon). Second, we now know that every equilateral triangle is also equiangular, so every triangle within this tile has angles. This makes our equilateral hexagon also equiangular, with each angle measuring . Because there are 6 angles, the sum of the angles in a hexagon are or . Finally, the point in the center of this tile, has angles around it. That means there are around a point.
### Examples
#### Example 1
Find the measure of .
The marking show that all angles are congruent. Since all three angles must add up to this means that each angle must equal . Write and solve an equation:
#### Example 2
Fill in the proof:
Given: Equilateral with
Prove: is equiangular
Statement Reason
1. 1. Given
2. 2. Base Angles Theorem
3. 3. Base Angles Theorem
4. 4. Transitive PoC
5. is equiangular 5.
Statement Reason
1. 1. Given
2. 2. Base Angles Theorem
3. 3. Base Angles Theorem
4. 4. Transitive PoC
5. is equiangular 5. Definition of equiangular.
#### Example 3
True or false: All equilateral triangles are isosceles triangles.
This statement is true. The definition of an isosceles triangle is a triangle with at least two congruent sides. Since all equilateral triangles have three congruent sides, they fit the definition of an isosceles triangle.
### Review
The following triangles are equilateral triangles. Solve for the unknown variables.
1. Find the measures of and .
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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## Math Expressions Common Core Grade 3 Unit 4 Lesson 14 Answer Key Relate Addition and Subtraction
Math Expressions Grade 3 Unit 4 Lesson 14 Homework
At the school play, 194 people attended the show on the first night and 185 people attended the show on the second night. How many people attended the show the two nights?
Relate Addition And Subtraction Grade 3 Unit 4 Math Expressions Question 1.
Solve the problem. _____
Given,
At the school play, 194 people attended the show on the first night and 185 people attended the show on the second night.
194+185 =379
it’s 379 people because you have to add 194 and 185 and it equals 379
Unit 4 Lesson 14 Math Expressions Grade 3 Answer Key Question 2.
Write a subtraction word problem related to the addition word problem. Then find the answer without doing any calculations.
379 – 185 = 194
Here is a subtraction word problem:
Rosi had 200 sheets of construction paper. She used 67 sheets to make invitations to a party and the rest to make decorations. How many sheets did she use to make decorations?
Lesson 14 Answer Key Math Expressions Grade 3 Unit 4 Question 3.
Solve the problem. _____
200-67 = 133
If rose had 200 sheets of construction paper and she used 67 sheets to make invitations to a party and rest to make decorations then she have used 133 sheets to make decorations
Write an addition word problem related to the subtraction word problem. Then find the answer without doing any calculations.
133 + 67 = 200
Math Expressions Grade 3 Unit 4 Lesson 14 Remembering
Write the times as minutes after an hour and minutes before an hour.
Lesson 14 Math Expressions Grade 3 Unit 4 Answer Key Question 1.
Answer: Twenty-seven minutes after six and thirty-three minutes before seven
Answer: Thirteen minutes after ten and forty-seven minutes before eleven
Question 3.
Answer: Fifty-six minutes after three and four minutes before four
Question 4.
Answer: forty-five minutes after eight and fifteen minutes before nine
Question 5.
Answer: Thirty – nine minutes after two and twenty one minutes before three
Question 6.
Answer: five minutes after five and fifty five minutes before six
Subtract.
Question 7.
Question 8.
Question 9.
Question 10.
Question 11.
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• RD Sharma Class 10 Solutions
# Class 10 RD Sharma Solutions – Chapter 16 Surface Areas and Volumes – Exercise 16.2 | Set 2
### Question 13. A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is 14/3 and the diameter of the hemisphere is 3.5 m. Calculate the volume and the internal surface area of the solid.
Solution:
According to the question
Diameter of the hemisphere(d) = 3.5 m,
So, the radius of the hemisphere(r) = 1.75 m,
Height of the cylinder(h) = 14/3 m,
Now we find the volume of the cylinder
V1 = πr2 h1
= π(1.75)2 x 14/3 m3
Now we find the volume of the hemisphere
V2= 2/3 × 22/7 × r3
V2 = 2/3 × 22/7 × 1.753 m3
So, the total volume of the vessel is
V = V1 + V2
V = π(1.75)2 x 14/3 + 2/3 × 22/7 × 1.753
V = 56 m3
Now we find the internal surface area of solid
S = 2πrh1 + 2πr2
= 2 π(1.75)(143) + 2 π(1.75)2 = 70.51 m3
Hence, the internal surface area of the solid is 70.51 m3 and the volume is 56 m3
### Question 14. Consider a solid which is composed of a cylinder with hemispherical ends. If the complete length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm. Find the cost of polishing its surface at the rate of Rs. 10 per dm2.
Solution:
According to the question
Radius of the hemispherical end (r) = 7 cm,
Height of the solid = h + 2r = 104 cm, h = 90 cm
Now we find the curved surface area of the cylinder
A1 = 2 πr h
= 2 π(7) h
= 2 π(7)(90)
= 3948.40 cm2
Now we find the curved surface area of the two Hemisphere
A2 = 2 (2πr2)
= 22π(7)2 = 615.75 cm2
Hence, the total curved surface area of the solid is
A = A1 + A2
= 3948.40 + 615.75 = 4571.8 cm2 = 45.718 dm2
So, he cost of polishing the 1 dm2 surface of the solid is Rs. 15
Therefore, the cost of polishing the 45.718 dm2 surface of the solid = 10 45.718 = Rs. 457.18
Hence, the cost of polishing is Rs. 457.18.
### Question 15. A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16cm and height of 42 cm. The total space between the two vessels is filled with Cork dust for heat insulation purposes. Find how many cubic cms of the Cork dust will be required?
Solution:
According to the question
Depth of the vessel = Height of the vessel = h = 42 cm,
Inner diameter of the vessel = 14 cm,
So, inner radius of the vessel = r1 = 14/2 = 7 cm
Outer diameter of the vessel = 16 cm,
So, the outer radius of the vessel = r2 = 16/2 = 8 cm
Now we find the volume of the vessel
V = π(r22 – r12)h
= π(82 – 72) x 42 = 1980 cm3
Hence, the volume of the vessel is 1980 cm3 which is equal to the amount of cork dust required.
### Question 16. A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making roller is 9 cm. Find the mass of the road roller if 1 cm3 of the iron has 7.8 gm mass.
Solution:
According to the question
Height of the road roller (h) = 1 m = 100 cm,
Internal Diameter of the road roller = 54 cm,
Thickness of the road roller (T) = 9 cm,
Let us considered R be the outer radii of the road roller. So,
T = R – r
9 = R – 27
R = 27 + 9 = 36 cm
Now we find the volume of the Iron Sheet
V = π × (R2 − r2) × h
= π × (362 − 272) × 100 = 1780.38 cm3
Hence the mass of 1 cm3 of the iron sheet = 7.8 gm [Given]
Therefore, the mass of 1780.38 cm3 of the iron sheet = 1388696.4 gm = 1388.7 kg
Hence, the mass of the road roller is 1388.7 kg
### Question 17. A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13cm. Find the inner surface area of the vessel.
Solution:
According to the question
Diameter of the hemisphere = 14 cm,
So, the radius of the hemisphere = 7 cm
And the total height of the vessel = = h + r = 13 cm
Now we find the inner surface area of the vessel
A = 2πr (h + r)
= 2 x 22/7 x (13) x (7)
= 572 cm2
Hence, the inner surface area of the vessel is 572 cm2
### Question 18. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
According to the question
Radius of the conical part of the toy(r) = 3.5 cm
Total height of the toy (h) = 15.5 cm
Slant height of the cone (L) = 15.5 – 3.5 = 12 cm
Now we find the curved surface area of the cone
A1 = πrL
= π(3.5)(12) = 131.94 cm2
Now we find the curved surface area of the hemisphere
A2 = 2πr2
= 2π(3.5)2 = 76.96 cm2
Hence, the total surface area of the toy
A = A1 + A2
= 131.94 + 76.96 = 208.90 cm2
Hence, the total surface area of the toy is 209 cm2
### Question 19. The difference between outside and inside surface areas of the cylindrical metallic pipe 14 cm long is 44 dm2. If pipe is made of 99 cm2 of metal. Find outer and inner radii of the pipe.
Solution:
According to the question
Length of the cylinder (h) = 14 cm
Difference between the outer and the inner surface area = 44 dm2
Volume of the metal used = 99 cm2,
Let’s assume that the inner radius of the pipe be r1 and r2 be the outer radius of the pipe
Now, the surface area of the cylinder is = 2π x 14 x (r2 – r1) = 44
(r2 − r1) = 1/2 —————-(i)
Volume of the cylinder is
πh(r22 – r12) = 99
πh(r2 – r1) (r2 + r1) = 99
= 22/7 x 14 x 1/2 x (r2 + r1) = 99
(r2 + r1) = 9/2 ————–(ii)
On Solving eq(i) and (ii), we get,
r2 = 5/2 cm
r1 = 2 cm
Hence, the inner radius is 2cm and outer radius of the pipe is 5/2 cm.
### Question 20. A right circular cylinder having diameter 12 cm and height 15 cm is full ice-cream. The ice-cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Solution:
According to the question
Radius of cylinder (r1) = 6 cm,
Radius of hemisphere (r2) = 3 cm,
Height of cylinder (h) = 15 cm,
slant height of the cones (l) = 12 cm,
Now we find the volume of cylinder
V = πr12h
= π × 62 ×15 ————–(i)
The volume of each ice cream cone = Volume of cone + Volume of hemisphere
= 1/3πr22h + 2/3πr23
= 1/3π x 62 x 12 + 2/3π x 33 —————-(ii)
Let’s assume that number of cones be ‘n’
n(Volume of each ice cream cone) = Volume of cylinder
n(1/3 π x 32 x 12 + 2/3 π x 33) = π(6)2 x15
n = 50/5 = 10
Hence, the number of cones being filled with ice-cream is 10
### Question 21. Consider a solid iron pole having cylindrical portion 110 cm high and the base diameter of 12 cm is surmounted by a cone of 9 cm height. Find the mass of the pole. Assume that the mass of 1 cm3 of iron pole is 8 gm.
Solution:
According to the question
Base diameter of the cylinder = 12 cm,
So, the radius of the cylinder (r) = 6 cm
Height of the cylinder (h) = 110 cm,
Slant height of the cone (L) = 9 cm,
Now we find the volume of the cylinder
V1 = π × r2 × h
= π × 62 × 110 cm3
Now we find the volume of the cone
V2 = 1/3 × πr2L
= 1/3 × π x 62 x 12 = 108π cm3
Hence, the volume of the pole (V)
V = V1 + V2
= 108π + π(6)2 110 = 12785.14 cm3
So, the mass of 1 cm3 of the iron pole = 8 gm [Given]
Then, the mass of 12785.14 cm3 of the iron pole = 8
12785.14 = 102281.12 gm = 102.2 kg
Hence, the mass of the iron pole is 102.2 kg
### Question 22. A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover?
Solution:
According to the question
Radius of the cone, cylinder, and hemisphere (r) = 2 cm
Height of the cone (l) = 2 cm
Height of the cylinder (h) = 4 cm
Now we find the volume of the cylinder
V1 = π × r2 × h
= π × 22 × 4 cm3
Now we find the volume of the cone
V2 = 1/3 π r2l
= 1/3 × π × 22 × 2
= 1/3 × π x 4 × 2 cm3
Now we find the volume of the hemisphere
V3 = 2/3 π r3
= 2/3 × π × 23 cm3
= 2/3 π × 8 cm3
Therefore, the remaining volume of the cylinder when the toy is inserted to it is
V = V1 – (V2 + V3)
= 16π – 8π = 8π cm3
Hence, remaining volume of the cylinder when toy is inserted into it is 8π cm3
### Question 23. Consider a solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm, is placed upright in the right circular cylinder full of water such that it touches bottoms. Find the volume of the water left in the cylinder, if radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
According to the question
Radius of the circular cone (r) = 60 cm,
Height of the circular cone (L) = 120 cm,
Radius of the hemisphere (r) = 60 cm,
Radius of the cylinder (R) = 60 cm,
Height of the cylinder (H) = 180 cm,
Now we find the volume of the circular cone
V1 = 1/3 × πr2l
= 1/3 × π × 602 × 120 = 452571.429 cm3
Now we find the volume of the hemisphere
V2 = 2/3 × πr3
= 2/3 × π × 603 = 452571.429 cm3
Now we find the volume of the cylinder
V3 = π × R2 × H
= π × 602 × 180 = 2036571.43 cm3
Hence, the volume of water left in the cylinder
V = V3 – (V1 + V2)
= 2036571.43 – (452571.429 + 452571.429)
= 2036571.43 – 905142.858 = 1131428.57 cm3 = 1.1314 m3
Hence, the volume of the water left in the cylinder is 1.1314 m3
### (ii) Left in the cylinder
Solution:
According to the question
Internal diameter of the cylindrical vessel (D)= 10 cm,
Radius of the cylindrical vessel (r) = 5 cm
Height of the cylindrical vessel (h) = 10.5 cm,
Base diameter of the solid cone = 7 cm,
Radius of the solid cone (R) = 3.5 cm
Height of the cone (L) = 6 cm,
(i) We find the volume of water displaced out from the cylinder which is equal to the volume of the cone
So, V1 = 1/3 × πR2L
V1 = 1/3 × π3.52 × 6 = 77 cm3
Hence, the volume of the water displaced out of the cylinder is 77 cm3
(ii) First we find the volume of the cylindrical vessel is
V2 = π × r2 × h
= π × 52 × 10.5
= 824.6 cm3 = 825 cm3
Now we find the volume of the water left in the cylinder
V = V2 – V1
V = 825 – 77 = 748 cm3
Hence, the volume of the water left in cylinder is 748 cm3
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# Area of a rectangle involving fractions
Area − The amount of surface a figure covers is its area.
For example, the perimeter measures the length of a fence going around a garden. Area measures the entire floor space that is going to be covered with a carpet.
A fraction is a number that is greater than zero but less than 1.
When two fractions that are both less than 1 are multiplied together, their product is smaller than either fraction.
In this lesson, we find areas of rectagular figures that have fractional lengths and widths.
Formula for the area of a rectangle involving fractions
If a rectangular figure has length and width of $\frac{a}{b}$ and $\frac{c}{d}$ where a, b, c and d are whole numbers, then the area of the rectangular figure is given by
Area = l × w = $\mathbf{\frac{a}{b}}$ × $\mathbf{\frac{c}{d}}$ = $\mathbf{\frac{ab}{cd}}$ square units
A lake is $\frac{2}{5}$ mile in length and $\frac{3}{7}$ mile in width. What is the area of the lake?
### Solution
Step 1:
Area of rectangle = l × w square units; l = length; w = width
Step 2:
Area of the lake = l × w = $\frac{2}{5}$ × $\frac{3}{7}$ = $\frac{6}{35}$ square mile
An island in the Pacific Ocean was $\frac{8}{13}$ miles wide and $\frac{9}{11}$ miles long. What is the area of the island?
### Solution
Step 1:
Area of rectangle = l × w square units; l = length; w = width
Step 2:
Area of the island = l × w = $\frac{8}{13}$ × $\frac{9}{11}$ = $\frac{72}{143}$ square miles
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NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2.
Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 6 Chapter Name Squares and Square Roots Exercise Ex 6.2 Number of Questions Solved 2 Category NCERT Solutions
## NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2
Question 1.
Find the square of the following numbers:
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Solution.
(i) 32
32 = 30 + 2
Therefore, $${ 32 }^{ 2 }$$ = $${ \left( 30+2 \right) }^{ 2 }$$
= 30 (30 + 2) + 2 (30 + 2)
= 900 + 60 + 60 + 4
= 1024
(ii) 35
35 = 30 + 5
Therefore, $${ 35 }^{ 2 }$$ = $${ \left( 30+5 \right) }^{ 2 }$$
= 30 (30 + 5) + 5 (30 + 5)
= 900 + 150 + 150 + 25
= 1225
(iii) 86
86 = 80 + 6
Therefore, $${ 86 }^{ 2 }$$= $${ \left( 80+6 \right) }^{ 2 }$$
= 80 (80 + 6) + 6 (80 + 6)
= 6400 + 480 + 480 + 36
= 7396
(iv) 93
93 = 90 + 3
Therefore, $${ 90 }^{ 2 }$$= $${ \left( 90+3\right) }^{ 2 }$$
= 90 (90 + 3) + 3 (90 + 3)
= 8100 + 270 + 270 + 9
= 8649
(v) 71
71 = 70 + 1
Therefore, $${ 70 }^{ 2 }$$= $${ \left( 70+1\right) }^{ 2 }$$
= 70 (70 + 1) + 1 (70 + 1)
= 4900 + 70 + 70 + 1
= 5041
(vi) 46
46 = 40 + 6
Therefore, $${ 40 }^{ 2 }$$= $${ \left( 40+6\right) }^{ 2 }$$
= 40 (40 + 6) + 6 (40 + 6)
= 1600 + 240 + 240 + 36
= 2116
Question 2.
Write a Pythagorean triplet whose one number is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution.
(i) 6
Let 2m = 6
⇒ $$m=\frac { 6 }{ 2 } =3$$
(ii) 14
Let 2m = 14
(iii) 16
Let 2m = 16
(iv) 18
Let 2m = 18
We hope the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2 are part help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2 are part, drop a comment below and we will get back to you at the earliest.
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# Angles formed by Parallel lines
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Category: Education
## Presentation Description
Three main types of angles are formed by parallel lines - 1. Corresponding angles 2. Alternate angles and 3. Interior angles Here it is explained how to Identify these three types of angles
## Presentation Transcript
### Slide 1:
PARALLEL LINES 6/11/2010 MATH GIRISH RAO If two lines are drawn in a plane such that they do not intersect, even when extended indefinitely in both directions, then such lines are called as parallel lines. A line which intersects two distinct lines in a plane in exactly two points is called a transversal.
### Slide 2:
When a transversal intersect two parallel lines, we get following types of angles Corresponding Angles Alternate Angles Co – interior angles Linear pair of Angles (Adjacent Angles) Vertically opposite Angles Out of the five types, let us see angles of first three types 6/11/2010 MATH GIRISH RAO
### Slide 3:
6/11/2010 MATH GIRISH RAO Corresponding angles They are “F” shaped angles.
### Slide 4:
6/11/2010 MATH GIRISH RAO
### Slide 5:
6/11/2010 MATH GIRISH RAO
### Slide 6:
6/11/2010 MATH GIRISH RAO
### Slide 7:
When a pair of parallel lines are intersected by the transversal we get four pairs of Corresponding angles as shown in the figure Corresponding angles have same measure 6/11/2010 MATH GIRISH RAO
### Slide 8:
6/11/2010 MATH GIRISH RAO Alternate angles. They are “Z” shaped angles
### Slide 9:
6/11/2010 MATH GIRISH RAO
### Slide 10:
6/11/2010 MATH GIRISH RAO When a pair of parallel lines are intersected by the transversal we get two pairs of interior Alternate angles as shown in the figure Alternate angles have same measure
### Slide 11:
6/11/2010 MATH GIRISH RAO Co – interior angles. This are “ [ “ shaped angles
### Slide 12:
6/11/2010 MATH GIRISH RAO
### Slide 13:
6/11/2010 MATH GIRISH RAO When a pair of parallel lines are intersected by the transversal we get two pairs of interior co – Interior angles as shown in the figure. The sum of co-interior angles is always 1800
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# Math Snap
## The Maclaurin series of the function $f(z)=\int_{0}^{z} \cos \left(2 t^{2}\right) d t$ is $\sum_{n=0}^{\infty} \frac{(-1)^{n} 4^{n}}{(2 n)!(4 n+1)} z^{4 n+1}$ b. $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n(2 n+1)!} z^{2 n+1}$ c. $\sum_{n=0}^{\infty} \frac{(-1)^{n} 4^{n}}{n!(4 n+1)} z^{4 n+1}$ d. $\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{n}}{(2 n)!(4 n+1)} z^{4 n+1}$
#### STEP 1
Assumptions 1. We are given the function $f(z) = \int_{0}^{z} \cos(2t^2) \, dt$. 2. We need to find the Maclaurin series for this function. 3. The Maclaurin series is a Taylor series expansion of a function about $z = 0$. 4. The general form of the Maclaurin series for a function $f(z)$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n$. 5. The options provided are different series expansions, and we need to determine which one correctly represents the Maclaurin series for $f(z)$.
#### STEP 2
First, we need to find the Maclaurin series for the integrand $\cos(2t^2)$. The Maclaurin series for $\cos(x)$ is given by: $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$
#### STEP 3
Substitute $x = 2t^2$ into the Maclaurin series for $\cos(x)$: $\cos(2t^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (2t^2)^{2n}}{(2n)!}$
#### STEP 4
Simplify the expression inside the summation: $\cos(2t^2) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n t^{4n}}{(2n)!}$
#### STEP 5
Now, we need to integrate this series term-by-term from 0 to $z$: $f(z) = \int_{0}^{z} \cos(2t^2) \, dt = \int_{0}^{z} \sum_{n=0}^{\infty} \frac{(-1)^n 4^n t^{4n}}{(2n)!} \, dt$
#### STEP 6
Interchange the summation and the integral (justified if the series converges uniformly): $f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{(2n)!} \int_{0}^{z} t^{4n} \, dt$
#### STEP 7
Integrate $t^{4n}$ with respect to $t$: $\int_{0}^{z} t^{4n} \, dt = \left[ \frac{t^{4n+1}}{4n+1} \right]_{0}^{z} = \frac{z^{4n+1}}{4n+1}$
#### STEP 8
Substitute the result of the integral back into the series: $f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{(2n)!} \cdot \frac{z^{4n+1}}{4n+1}$
#### STEP 9
Combine the constants and simplify the expression: $f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{(2n)!(4n+1)} z^{4n+1}$
##### SOLUTION
Compare the obtained series with the given options. The series we derived matches option (a): $\sum_{n=0}^{\infty} \frac{(-1)^{n} 4^{n}}{(2 n)!(4 n+1)} z^{4 n+1}$ Therefore, the correct answer is option (a).
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Chap 13 Interpolation and Approximation
# Chap 13 Interpolation and Approximation - FIRST YEAR...
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Unformatted text preview: FIRST YEAR CALCULUS W W L CHEN c W W L Chen, 1987, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 13 INTERPOLATION AND APPROXIMATION 13.1. Exact Fitting Example 13.1.1. We wish to find a polynomial through the points (0, 6), (1, 2) and (5, 6). To do this, consider a polynomial p(x) = ax2 + bx + c, where we shall determine suitable values for the three coefficients. Since the polynomial p(x) passes through (0, 6), (1, 2) and (5, 6), we must have p(0) = 6, p(1) = 2 and p(5) = 6. It follows that we must have 6 = c, 2 = a + b + c, 6 = 25a + 5b + c. This is a system of 3 linear equations in 3 unknowns. Solving this system, we get a = 1, b = −5 and c = 6. Hence p(x) = x2 − 5x + 6. Example 13.1.2. We wish to find a polynomial through the points (−1, 5), (0, 1), (1, −1) and (3, 49). To do this, consider a polynomial p(x) = ax3 + bx2 + cx + d, where we shall determine suitable values for the four coefficients. Since the polynomial p(x) passes through (−1, 5), (0, 1), (1, −1) and (3, 49), we must have p(−1) = 5, p(0) = 1, p(1) = −1 and p(3) = 49. It follows that we must have 5 = −a + b − c + d, 1 = d, −1 = a + b + c + d, 49 = 27a + 9b + 3c + d. This is a system of 4 linear equations in 4 unknowns. Solving this system, we get a = 2, b = 1, c = −5 and d = 1. Hence p(x) = 2x3 + x2 − 5x + 1. Chapter 13 : Interpolation and Approximation page 1 of 6 First Year Calculus c W W L Chen, 1987, 2005 The two examples above illustrate a very crude technique. When we attempt to fit a polynomial through k points, we use a polynomial of degree (k − 1). We then have to determine the k coefficients of this polynomial. This amounts to solving a system of k linear equations in the k unknowns. Clearly it is rather tedious, particularly when k is large. Let us therefore use a different approach on the same problems. Example 13.1.3. As in Example 13.1.1, let us find a polynomial through the points (0, 6), (1, 2) and (5, 6). Try p(x) = a(x − 1)(x − 5) + b(x − 0)(x − 5) + c(x − 0)(x − 1). Since p(0) = 6, we must have 6 = a(0 − 1)(0 − 5), so that a = 6/5. Since p(1) = 2, we must have 2 = b(1 − 0)(1 − 5), so that b = −1/2. Since p(5) = 6, we must have 6 = c(5 − 0)(5 − 1), so that c = 3/10. Hence p(x) = 6(x − 1)(x − 5) (x − 0)(x − 5) 3(x − 0)(x − 1) − + = x2 − 5x + 6. 5 2 10 Example 13.1.4. As in Example 13.1.2, let us find a polynomial through the points (−1, 5), (0, 1), (1, −1) and (3, 49). Try p(x) = a(x − 0)(x − 1)(x − 3) + b(x + 1)(x − 1)(x − 3) + c(x + 1)(x − 0)(x − 3) + d(x + 1)(x − 0)(x − 1). Since p(−1) = 5, we must have 5 = a(−1 − 0)(−1 − 1)(−1 − 3), so that a = −5/8. Since p(0) = 1, we must have 1 = b(0+1)(0 − 1)(0 − 3), so that b = 1/3. Since p(1) = −1, we must have −1 = c(1+1)(1 − 0)(1 − 3), so that c = 1/4. Since p(3) = 49, we must have 49 = d(3 + 1)(3 − 0)(3 − 1), so that d = 49/24. Hence p(x) = − 5x(x − 1)(x − 3) (x + 1)(x − 1)(x − 3) x(x + 1)(x − 3) 49x(x + 1)(x − 1) + + + 8 3 4 24 3 2 = 2x + x − 5x + 1. Let us look at one more example. However, we shall be a little more systematic. Example 13.1.5. We wish to find a polynomial through the points (1, −3), (3, 3) and (4, 9). Try p(x) = a(x − 3)(x − 4) + b(x − 1)(x − 4) + c(x − 1)(x − 3). Substituting x = 1, x = 3, x = 4, we obtain respectively a= Hence p(x) = p(1) (x − 3)(x − 4) (x − 1)(x − 4) (x − 1)(x − 3) + p(3) + p(4) . (1 − 3)(1 − 4) (3 − 1)(3 − 4) (4 − 1)(4 − 3) p(1) , (1 − 3)(1 − 4) b= p(3) , (3 − 1)(3 − 4) c= p(4) . (4 − 1)(4 − 3) Since p(1) = −3, p(3) = 3 and p(4) = 9, a little calculation gives p(x) = x2 − x − 3. Consider now the general situation. Suppose that we wish to find a polynomial through the points (x1 , y1 ), . . . , (xk , yk ). Then it is not too difficult to see that the polynomial k k (1) p(x) = i=1 yi j =1 j =i x − xj xi − xj Chapter 13 : Interpolation and Approximation page 2 of 6 First Year Calculus c W W L Chen, 1987, 2005 satisfies the requirements. To see that, note that for every i = 1, . . . , k , we have k j =1 j =i x − xj xi − xj = 1 if x = xi , 0 if x = x1 , . . . , xi−1 , xi+1 , . . . , xk . The polynomial (1) is called the Lagrange interpolation polynomial. 13.2. Approximate Fitting Fitting the points exactly is unsatisfactory from the numerical point of view, particularly so when the number of points is large. We therefore sometimes attempt to fit all points closely but not exactly. After all, experimental data are subject to errors anyway! Consider a given set of n points (x1 , y1 ), . . . , (xn , yn ). We now attempt to fit these points with a polynomial p(x) = ak−1 xk−1 + ak−2 xk−2 + . . . + a0 . Recall that when k ≥ n, this can always be done; for example, simply take p(x) to be the Lagrange interpolation polynomial. However, if k < n, an exact fit may not be possible; for example, it is not possible to fit a straight line (k = 2) to go through three non-collinear points (n = 3) exactly. We therefore consider the errors i = |p(xi ) − yi |, where i = 1, . . . , n. The problem now is to choose a0 , . . . , ak−1 in such a way in order to make the errors small. There are many ways to make errors small, and the following are examples: n (A) Choose a0 , . . . , ak−1 to minimize i=1 i. i (B) Choose a0 , . . . , ak−1 to minimize max (C) Choose a0 , . . . , ak−1 to minimize 1≤i≤n n 2 i i=1 – minimax approximation. – least squares approximation. Remark. It is generally considered that (A) is the best criterion but most awkward, and that (C) is the least satisfactory criterion but easiest to use. An analogous problem to that discussed in the previous section is the question of approximating a function f (x) by a polynomial p(x) = ak−1 xk−1 + ak−2 xk−2 + . . . + a0 in an interval a ≤ x ≤ b. Here we consider the errors (x) = |p(x) − f (x)|, where a ≤ x ≤ b. The problem now is to choose a0 , . . . , ak−1 in such a way in order to make the errors small. There are many ways to make errors small, and the following are examples: b (A) Choose a0 , . . . , ak−1 to minimize a (x) dx. a≤x≤b b 2 a (B) Choose a0 , . . . , ak−1 to minimize max (x) – minimax approximation. (C) Choose a0 , . . . , ak−1 to minimize (x) dx – least squares approximation. Remark. As before, it is generally considered that (A) is the best criterion but most awkward, and that (C) is the least satisfactory criterion but easiest to use. Chapter 13 : Interpolation and Approximation page 3 of 6 First Year Calculus c W W L Chen, 1987, 2005 13.3. Minimax Approximation We shall illustrate the technique by two of the simplest examples. Example 13.3.1. Consider the points (1, −3), (3, 3) and (4, 9). It was shown in Example 13.1.5 that we can fit the polynomial x2 − x − 3 precisely. Suppose now that we wish to find a minimax approximation by a polynomial of degree 1 (linear minimax approximation). Suppose that p(x) = ax + b. We then consider the errors 1 2 3 = |p(x1 ) − y1 | = |a + b + 3|, = |p(x2 ) − y2 | = |3a + b − 3|, = |p(x3 ) − y3 | = |4a + b − 9|, and minimize max{|a + b + 3|, |3a + b − 3|, |4a + b − 9|}. If we take a = 4 and b = −8, then max{|a+b+3|, |3a+b−3|, |4a+b−9|} = 1. Hence p(x) = 4x−8 is a linear approximation with maximum error 1. It can be shown that this is the best minimax approximation by a polynomial of degree 1, but demonstrating this point is not so straightforward. Example 13.3.2. Consider the function f (x) = x2 in the interval 0 ≤ x ≤ 2. Suppose that we wish to find a minimax approximation by a polynomial of degree 1 (linear minimax approximation). Suppose that p(x) = ax + b. We then consider the errors (x) = |ax + b − x2 |, where 0 ≤ x ≤ 2. Consider first of all the function h(x) = ax + b − x2 . Then h(x) has a maximum value when dh/dx = 0. This occurs when x = a/2, and this is in the interval 0 ≤ x ≤ 2 provided that 0 ≤ a ≤ 4. Note that h(a/2) = a2 /4 + b. Also h(0) = b and h(2) = 2a + b − 4. We now choose a and b such that h(0) = h(2) < 0 < h a = −h(0) 2 (the reader should draw a picture of h(x) in the interval 0 ≤ x ≤ 2 to illustrate these special requirements). Then we must have b = 2a + b − 4 < 0 < a2 + b = −b, 4 so that a = 2 and b = −1/2. Hence the linear polynomial p(x) = 2x − 1/2 gives max (x) = 1 . 2 0≤x≤2 It should by now be clear that minimax approximations are rather awkward to use, even in the simplest cases. 13.4. Least Squares Approximation We shall illustrate the technique by two simple examples. Chapter 13 : Interpolation and Approximation page 4 of 6 First Year Calculus c W W L Chen, 1987, 2005 Example 13.4.1. Consider the points (1, 1), (2, 3), (3, 4), (4, 3), (5, 4) and (6, 2). Suppose that we wish to find a least squares approximation by a polynomial of degree 1 (linear least squares approximation). Suppose that p(x) = ax + b. We then consider the errors i = |p(xi ) − yi | = |axi + b − yi |, where i = 1, 2, 3, 4, 5, 6, and choose a and b so as to minimize 6 6 2 i i=1 S (a, b) = = i=1 (axi + b − yi )2 . Let us now think of S (a, b) as a function of the two variables a and b. We then must have 0= ∂S xi (axi + b − yi ), =2 ∂a i=1 ∂S =2 (axi + b − yi ), ∂b i=1 6 6 0= so that 6 6 6 x2 i i=1 6 a+ i=1 6 xi b= i=1 6 xi yi , yi . i=1 xi i=1 a+ i=1 1 b= Substituting for (xi , yi ) for i = 1, 2, 3, 4, 5, 6, we have 91a + 21b = 63, 21a + 6b = 17, so that a = 1/5 and b = 32/15. Hence p(x) = is the best linear least squares approximation. Example 13.4.2. As in Example 13.3.2, consider the function f (x) = x2 in the interval 0 ≤ x ≤ 2. Suppose that we wish to find a least squares approximation by a polynomial of degree 1 (linear least squares approximation). Suppose that p(x) = ax + b. We then consider the errors (x) = |ax + b − x2 |, and choose a and b so as to minimize 2 2 2 0 1 32 x+ 5 15 where 0 ≤ x ≤ 2, T (a, b) = (x) dx = 0 (ax + b − x2 )2 dx. Let us now think of T (a, b) as a function of the two variables a and b. We then must have 0= ∂T =2 ∂a 2 x(ax + b − x2 ) dx = 2 0 2 8 a + 2b − 4 , 3 8 3 , page 5 of 6 ∂T 0= =2 ∂b (ax + b − x2 ) dx = 2 2a + 2b − 0 Chapter 13 : Interpolation and Approximation First Year Calculus c W W L Chen, 1987, 2005 so that a = 2 and b = −2/3. Hence p(x) = 2x − is the best linear least squares approximation. In general, if we try to fit a polynomial p(x) = ak xk + . . . + a0 to n points (x1 , y1 ), . . . , (xn , yn ), then we choose a0 , . . . , ak to minimize n 2 3 S (a0 , . . . , ak ) = i=1 (p(xi ) − yi )2 . The requirement that ∂S = 0 for every j = 0, . . . , k ∂aj gives rise to a system of (k + 1) linear equations in the (k + 1) unknowns a0 , . . . , ak . If we try to fit a polynomial p(x) = ak xk + . . . + a0 to a given function f (x) in an interval a ≤ x ≤ b, then we choose a0 , . . . , ak to minimize b T (a0 , . . . , ak ) = a (p(x) − f (x))2 dx. The requirement that ∂T = 0 for every j = 0, . . . , k ∂aj gives rise to a system of (k + 1) linear equations in the (k + 1) unknowns a0 , . . . , ak . Hence the determination of the best least squares approximations amounts to nothing more than solving a system of linear equations. Squaring the errors removes any ambiguity on the sign of the errors. Problems for Chapter 13 1. Find a polynomial to pass through the points (−2, 99), (−1, 11), (0, 1), (1, 3) and (2, 47). 2. Find the best linear least squares approximation to the points (−2, 99), (−1, 11), (0, 1), (1, 3) and (2, 47). 3. Find the best linear least squares approximation to the function ex in the interval [0, 1]. Chapter 13 : Interpolation and Approximation page 6 of 6 ...
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# Prime Factorization
Go back to 'Arithmetic-Integers'
### What is Prime Factorisation?
Prime factorisation allows us to write any number as a product of prime factors. For example, consider the number 360. Let us write this number as follows:
$\begin{array}{l}360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5\\\;\;\;\;\;\, = {2^3} \times {3^2} \times {5^1}\end{array}$
Basically, we have expressed 360 as a product of prime numbers. Let us take some other (composite) numbers and express them as products of prime numbers:
$\begin{array}{l}\;\,80 = 2 \times 2 \times 2 \times 2 \times 5\\\;\;\;\;\;\,\,\, = {2^4} \times {5^1}\\144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3\\\;\;\;\;\;\,\,\, = {2^4} \times {3^2}\\600 = 2 \times 2 \times 2 \times 3 \times 3 \times 5\\\;\;\;\;\;\,\,\,= {2^3} \times {3^2} \times {5^1}\end{array}$
Expressing a number this way (as a product of primes) is called the prime factorization of that number. Note that the prime factorization of a prime number is trivial, since the only non-unity divisor of any prime number is that number itself.
### Can every number be prime-factorised in a unique way?
Two questions now arise:
1. Can every composite number be expressed as a product of primes, that is, can every composite number be prime factorized?
2. If we prime factorize a composite number, is that factorization unique? That is, if we express a composite number as a product of primes, can it be expressed as a product of some other set of primes?
These questions are answered by one of the most important results in Mathematics, which we discuss next.
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This lesson will explore the regions bounded by two radii of a circle and their intercepted arc. Additionally, the formula for the area of such regions will be derived considering the part-whole relationships.
Catch-Up and Review
Here is some recommended reading before getting started with this lesson.
Investigating the Area of a Slice of Pizza
Three friends are sharing a inch pizza equally. Emily is becoming a nutritionist and is curious about how many calories are in a slice. To find out, she will calculate the area of one slice.
Help Emily to find the area of one slice. How can the measure of a central angle be used to find the area?
Sector of a Circle
A sector of a circle is a portion of the circle enclosed by two radii and their intercepted arc.
In the diagram, sector is created by and
Area of a Sector of a Circle
The area of a sector of a circle is calculated by multiplying the circle's area by the ratio of the measure of the central angle to
From the fact that equals an equivalent formula can be written if the central angle is given in radians.
Since the measure of an arc is equal to the measure of its central angle, the arc measures Therefore, by substituting for another version of the formula is obtained which can also be written in degrees or radians.
or
Proof
Consider sector bounded by and
Since a circle measures this sector represents of Therefore, the ratio of the area of a sector to the area of the whole circle is proportional to
Recall that the area of a circle is By substituting it into the equation and solving for the area of a sector, the desired formula can be obtained.
Therefore, the area of a sector of a circle can be found by using the following formula.
Using Sectors of Circles to Solve Problems
Like the pizza problem, numerous real-life problems can be modeled by sectors of a circle.
Tiffaniqua has a trapezoid-shaped yard whose side lengths are shown on the diagram. To water the lawn, she sets up a water sprinkler that can water the grass within a meter radius. as shown.
Tiffaniqua knows that measures
a Help Tiffaniqua calculate the area of the lawn covered by the sprinkler. If necessary, round the answer to the nearest square meter.
b What is the area of the lawn that is not watered? If necessary, round the answer to the nearest square meters.
Hint
a The sprinkler covers a degree sector of a circle with a radius of meters.
b The area of a trapezoid is half the product of the height and the sum of the two bases.
Solution
a The sprinkler covers a degree sector of a circle with radius meters. The area of a sector of a circle can be calculated using the following formula.
Substituting for and for into the formula will give the result.
Evaluate right-hand side
The area of the lawn covered is about square meters.
b To find the area of the lawn that is not watered, the area found in Part A should be subtracted from the area of the trapezoid.
On the diagram, the trapezoid's bases are and meters, and its height is meters. Substitute these values into the formula for the area of a trapezoid.
Evaluate right-hand side
This means that the area that Tiffaniqua should water, initially, is square meters. By subtracting the area of the sector from the total area, the lawn area that is not watered can be found.
Tiffaniqua needs to water an area of square meters.
Calculating the Central Angle of Pac-Man's Mouth
When the area of a sector is given, the measure of the corresponding central angle can be calculated
Consider a two-dimensional image of Pac-Man. The area covered by Pac-Man is about square millimeters.
If the radius of the circle used to draw Pac-Man is millimeters, find the measure of the central angle formed in the colored region. If necessary, round the answer to the nearest degree.
Hint
Pac-Man is essentially a sector of a circle. Use the formula for the area of a sector of a circle to find the measure of the angle.
Solution
To find the measure of the corresponding central angle, the formula for the area of a sector will be used.
The area of the sector and the radius are given. Substitute these values into the formula.
Solve for
The corresponding central angle is about
Comparing the Radius of Two Circles
The diagram below models the motion of two gears and Gear has a radius of inches.
a Find the radius of the larger gear.
b Find the area of the sector of Gear formed when Gear completes two revolutions. Write the answer in terms of
Hint
a How can the measure of the central angle formed in Gear be used?
b Use the formula for the area of a sector of a circle.
Solution
a When the smaller gear completes a single revolution, the central angle measured in the larger gear becomes Since the measure of an arc is the same as its corresponding central angle, the measure of the colored arc of is also
Furthermore, the length of the arc is equal to the circumference of Recall that the circumference is given by the formula Substituting into the formula will give the circumference of
The length of the arc of is, therefore, inches. Now that the measure and length of the arc is known, the radius of can be found. To do so, substitute the values into the formula for the arc length.
Solve for
The radius of the larger gear is inches.
b As can be seen on the diagram, when Gear makes two complete revolutions, the central angle measures
Note that the arc created due to revolutions also measures In the previous part, the radius of is found as inches. Using the formula for the area of a sector, the sector of can be calculated.
Evaluate right-hand side
The area of the sector of is square inches.
Using Areas Of Sectors to Solve Problems
In his free time, Dylan enjoys making decorative figures by hand. He has identical sectors and brings these sectors together as shown.
Dylan knows that the area of each sector is square millimeters.
a Find the perimeter of the star-shaped figure. If necessary, round the answer to one decimal place.
b Find the perimeter of the figure. If necessary, round the answer to one decimal place.
Hint
a Notice that each side of the star-shaped figure is a radius.
b Start by finding the corresponding arc length of a sector.
Solution
a Notice that the sides of the star-shaped figure are the radii of the sectors.
The value of the radius can be found using the formula for the area of a sector. Substitute for the area of the sector and for into the formula.
Solve for
Since identical radii form the figure, the perimeter is
The perimeter of the star-shaped figure is millimeters.
b The perimeter of the figure created by Dylan is the sum of identical arc lengths.
In Part A, the radii of the sectors were found to be about millimeters. The measure of the arc is because the corresponding central angle measures Now, the Arc Length Formula can be used.
Evaluate right-hand side
There are of these arc.
Therefore, the figure has a perimeter of approximately millimeters.
Solving Real Life Problems Using Sectors of Circles
Mark set up a lamp in his courtyard. He uses a light bulb that illuminates a circular area with a radius of meters. The diagram shows a bird's eye view of Mark's house.
If the measure of arc is what is the area of the region that is illuminated outside of the courtyard area? If necessary, round the answer to two decimal places.
Hint
The area of a triangle is half the product of the lengths of any two sides and the sine of the included angle.
Solution
From the diagram, it can be seen that the region bounded by and is a sector of
The region bounded by and is called segment of the circle To find the area of the segment, the area of the triangle should be subtracted from the area of the sector
The area of is half the product of the lengths of any two sides and the sine of the included angle.
Since and are radii of and are meters. Moreover, the included angle measures because it intercepts a arc. Substitute these values.
Evaluate right-hand side
Next, the area of the sector will be calculated.
Evaluate right-hand side
Finally, the area of the region is the difference of and
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#### Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 11
$y=\frac{c-x}{(1+c x)}$ is a solution of differential equation
Hint:
Differentiate the given solution and substitute in the differential equation.
Given:
$y=\frac{c-x}{(1+c x)}$ is a solution of the equation
Solution:
Differentiating on both sides with respect to $x$
\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{c-x}{(1+c x)}\right) \\\\ &\frac{d y}{d x}=\frac{\left[(1+c x) \frac{d}{d x}(c-x)\right]-\left[(c-x) \frac{d}{d x}(1+c x)\right]}{(1+c x)^{2}} \end{aligned}
\begin{aligned} &\frac{d y}{d x}=\frac{[(1+c x)(0-1)]-[(c-x)(0+c)]}{(1+c x)^{2}} \\\\ &\frac{d y}{d x}=\frac{[(1+c x)(-1)]-[(c-x)(c)]}{(1+c x)^{2}} \end{aligned}
\begin{aligned} &\frac{d y}{d x}=\frac{-1-c x-c^{2}+c x}{(1+c x)^{2}} \\\\ &\frac{d y}{d x}=\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}} \end{aligned} .............(i)
Put equation (i) in differential equation as follows
\begin{aligned} &\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right)=0 \\\\ &L H S=\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right) \end{aligned}
\begin{aligned} &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\left(1+y^{2}\right) \\\\ &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\left[1+\left(\frac{(c-x)}{(1+c x)}\right)^{2}\right] \end{aligned}
\begin{aligned} &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+1+\frac{(c-x)^{2}}{(1+c x)^{2}} \\\\ &=\left(1+x^{2}\right)\left[\frac{-\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\frac{(1+c x)^{2}+(c-x)^{2}}{(1+c x)^{2}} \end{aligned}
$=\left[\frac{-\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\frac{1+2 c x+c^{2} x^{2}+c^{2}-2 c x+x^{2}}{(1+c x)^{2}}$
\begin{aligned} &=\left[\frac{-\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\frac{1+c^{2} x^{2}+c^{2}+x^{2}}{(1+c x)^{2}} \\\\ &=\left[\frac{-\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right]+\left[\frac{\left(1+x^{2}\right)\left(1+c^{2}\right)}{(1+c x)^{2}}\right] \\\\ &=0 \\\\ &=R H S \end{aligned}
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## What is the meaning of rectangular coordinates?
A rectangular coordinate system is defined, originating at the center point between the plates with z in the direction of plate separation, x in the width direction, and y in the length direction.
### What is a rectangular coordinate system examples?
The x- and y-axes break the plane into four regions called quadrantsThe four regions of a rectangular coordinate plane partly bounded by the x- and y-axes and numbered using the roman numerals I, II, III, and IV., named using roman numerals I, II, III, and IV, as pictured. In quadrant I, both coordinates are positive.
Why is it called rectangular coordinate system?
The system is called rectangular because the angle formed by the axes at the origin is 90 degrees and the angle formed by the measurements at point p is also 90 degrees.
What is rectangular coordinate system composed of?
The rectangular coordinate system is formed by two intersecting perpendicular number lines. The horizontal number line is called the x-axis and the vertical number line is called the y-axis.
## How are Cartesian coordinates written?
The Cartesian coordinates of a point are usually written in parentheses and separated by commas, as in (10, 5) or (3, 5, 7). The origin is often labelled with the capital letter O. Each axis is usually named after the coordinate which is measured along it; so one says the x-axis, the y-axis, the t-axis, etc.
### Which of the following is the format for relative rectangular coordinates?
Relative Rectangular Coordinate system: Using relative coordinate, points entered by typing @x,y [Enter].
How do you find rectangular coordinates?
Converting from Polar Coordinates to Rectangular Coordinates
1. Given the polar coordinate (r,θ), write x=rcosθ and y=rsinθ.
2. Evaluate cosθ and sinθ.
3. Multiply cosθ by r to find the x-coordinate of the rectangular form.
4. Multiply sinθ by r to find the y-coordinate of the rectangular form.
How do you do coordinates?
Coordinates are written as (x, y) meaning the point on the x axis is written first, followed by the point on the y axis. Some children may be taught to remember this with the phrase ‘along the corridor, up the stairs’, meaning that they should follow the x axis first and then the y.
## What are the main features of rectangular coordinate system?
Notice that the rectangular coordinate system consists of 4 quadrants, a horizontal axis, a vertical axis, and the origin. The horizontal axis is usually called the x–axis, and the vertical axis is usually called the y–axis. The origin is the point where the two axes cross.
### How do you read a rectangular coordinate system?
In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the x-coordinate of the point, and the second number is the y-coordinate of the point. , gives the coordinates of a point in a rectangular coordinate system. The first number is the x-coordinate.
How to plot a point in a rectangular coordinate system?
To plot the point ( − 2, 4) ( − 2, 4), begin at the origin. The x -coordinate is –2, so move two units to the left. The y -coordinate is 4, so then move four units up in the positive y direction.
Is the Cartesian coordinate system a two dimensional system?
The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x -axis and the y -axis. Perpendicular to each other, the axes divide the plane into four sections.
## Is the y coordinate the same as the x coordinate?
The y -coordinate is also 3, so move three units up in the positive y direction. To plot the point ( 0, − 3) ( 0, − 3), begin again at the origin. The x -coordinate is 0. This tells us not to move in either direction along the x -axis.
### Why is an ordered pair called a coordinate pair?
An ordered pair is also known as a coordinate pair because it consists of x and y -coordinates. For example, we can represent the point (3, − 1) in the plane by moving three units to the right of the origin in the horizontal direction and one unit down in the vertical direction.
What is the meaning of rectangular coordinates? A rectangular coordinate system is defined, originating at the center point between the plates with z in the direction of plate separation, x in the width direction, and y in the length direction. What is a rectangular coordinate system examples? The x- and y-axes break the plane into…
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# Apps that do your homework
Apps that do your homework can be a helpful tool for these students. So let's get started!
## The Best Apps that do your homework
This Apps that do your homework helps to fast and easily solve any math problems. solving equations is a process that involves isolating the variable on one side of the equation. This can be done using inverse operations, which are operations that undo each other. For example, addition and subtraction are inverse operations, as are multiplication and division. When solving an equation, you will use these inverse operations to move everything except for the variable to one side of the equal sign. Once the variable is isolated, you can then solve for its value by performing the inverse operation on both sides of the equation. For example, if you are solving for x in the equation 3x + 5 = 28, you would first subtract 5 from both sides of the equation to isolate x: 3x + 5 - 5 = 28 - 5. This results in 3x = 23. Then, you would divide both sides of the equation by 3 to solve for x: 3x/3 = 23/3. This gives you x = 23/3, or x = 7 1/3. Solving equations is a matter of isolating the variable using inverse operations and then using those same operations to solve for its value. By following these steps, you can solve any multi-step equation.
Basic mathematics is the study of mathematics that is necessary for everyday life. It includes topics such as addition, subtraction, multiplication, and division. Basic mathematics also covers fractions, decimals, and percents. Basic mathematics is an important subject because it helps us to understand the world around us. It is used in everyday life, such as when we cook or do laundry. Basic mathematics is also used in more complicated situations, such as budgeting or investing. By understanding basic mathematics, we can make better decisions in all areas of our lives.
Calculus can be a difficult subject for many students. In addition to mastering a new set of concepts, students must also learn how to apply those concepts to solve complex problems. While some students may be able to do this on their own, others may find it helpful to use a calculus solver with steps. A calculus solver with steps can show students how to work through a problem from start to finish, allowing them to see the thought process behind each step. This can be a valuable tool for students who are struggling to understand the material or for those who simply want to check their work. Calculus solvers with steps are available online and in many textbooks. With a little bit of searching, students should be able to find a calculator that meets their needs.
Think Through Math is an app that helps students learn math by thinking through the problems. The app provides step-by-step instructions on how to solve each problem, and it also includes a variety of math games to help students practice their skills. Think Through Math is available for both iOS and Android devices, and it is a great way for students to improve their math skills.
Solving for x with fractions can be tricky, but there are a few steps that can make the process simpler. First, it is important to understand that when solving for x, the goal is to find the value of x that will make the equation true. In other words, whatever value is plugged into the equation in place of x should result in a correct answer. With this in mind, the next step is to create an equation using only fractions that has the same answer no matter what value is plugged in for x. This can be done by cross-multiplying the fractions and setting the two sides of the equation equal to each other. Once this is done, the final step is to solve for x by isolating it on one side of the equation. By following these steps, solving for x with fractions can be much less daunting.
## We cover all types of math problems
great app! really the steps in solving problems are clearly explained and it helps you understand more what you are doing. I really recommend this app to anyone who loves math but find it difficult to understand some concepts. Thanks a lot for the app
Giuliana Griffin
Lifesaver for sure! Loved it so much that I got the plus version and I don't regret it at all! It's fantastic! They not only give you an answer, they show you how they got that answer and walk you through it step-by-step. Again, this has been a lifesaver!
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Math helper app Step by step equations Trig solver with steps Partial fractions solver Pre calculus help
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There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Geometry. In the right column below are links to related online activities, videos and teacher resources.
A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Geometry Starters:
Angle Estimates: Estimate the sizes of each of the angles then add your estimates together.
Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines?
Christmas Tables: Which of the two shapes has the largest area? You will be surprised!
Cross Perimeter: Calculate the distance around the given shape
Dice Reflections: A dice is reflected in two mirrors. What numbers can be seen?
Hexagon: On a full page in the back of your exercise book draw a perfectly regular hexagon.
Mathterpiece: Memorise a picture made up of geometrical shapes
Oblongs: Find the dimensions of a rectangle given the perimeter and area.
Pentagon: On a full page in the back of your exercise book draw a perfectly regular pentagon.
Polygon Riddle 1: Solve the riddle to find the name of the polygon then sum the interior angles.
Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid.
Reflective Cat: On squared paper copy the drawing of the face then reflect it in three different lines.
Rotational Symmetry: Draw a pattern with rotational symmetry of order 6 but no line symmetry.
Sectors: Work out which sectors fit together to make complete circles. Knowledge of the sum of the angles at a point will help find more than one correct solution to this puzzle.
Square Angles: Find a trapezium, a triangle and a quadrilateral where all of the angles are square numbers.
Square Circle Kite: Write down the names of all the mathematical shapes you know.
Stair Perimeter: Use the information implied in the diagram to calculate the perimeter of this shape.
Step Perimeter: Is it possible to work out the perimeter of this shape if not all the side lengths are given?
Find The Radius: Find the radius of the circle from the small amount of information provided.
Geometry Snack: Find the value of the marked angle in this diagram from the book Geometry Snacks
Quad Midpoints: What shape is created when the midpoints of the sides of a quadrilateral are joined together?
Three Right Triangles: Calculate the lengths of the unlabelled sides of these right-angled triangles.
#### Volume
Use formulae to solve problems involving the volumes of cuboids, cones, pyramids, prisms and composite solids.
Transum.org/go/?to=volume
### Curriculum for Geometry:
#### Year 5
Pupils should be taught to use the properties of rectangles to deduce related facts and find missing lengths and angles more...
Pupils should be taught to distinguish between regular and irregular polygons based on reasoning about equal sides and angles more...
#### Year 6
Pupils should be taught to draw 2-D shapes using given dimensions and angles more...
Pupils should be taught to compare and classify geometric shapes based on their properties and sizes and find unknown angles in any triangles, quadrilaterals, and regular polygons more...
Pupils should be taught to recognise angles where they meet at a point, are on a straight line, or are vertically opposite, and find missing angles. more...
#### Years 7 to 9
Pupils should be taught to draw and measure line segments and angles in geometric figures, including interpreting scale drawings more...
Pupils should be taught to describe, sketch and draw using conventional terms and notations: points, lines, parallel lines, perpendicular lines, right angles, regular polygons, and other polygons that are reflectively and rotationally symmetric more...
Pupils should be taught to use the standard conventions for labelling the sides and angles of triangle ABC, and know and use the criteria for congruence of triangles more...
Pupils should be taught to derive and illustrate properties of triangles, quadrilaterals, circles, and other plane figures [for example, equal lengths and angles] using appropriate language and technologies more...
Pupils should be taught to identify and construct congruent triangles, and construct similar shapes by enlargement, with and without coordinate grids more...
Pupils should be taught to apply the properties of angles at a point, angles at a point on a straight line, vertically opposite angles more...
Pupils should be taught to derive and use the sum of angles in a triangle and use it to deduce the angle sum in any polygon, and to derive properties of regular polygons more...
Pupils should be taught to apply angle facts, triangle congruence, similarity and properties of quadrilaterals to derive results about angles and sides, including Pythagoras’ Theorem, and use known results to obtain simple proofs more...
Pupils should be taught to interpret mathematical relationships both algebraically and geometrically. more...
#### Years 10 and 11
Pupils should be taught to identify and apply circle definitions and properties, including: centre, radius, chord, diameter, circumference, tangent, arc, sector and segment more...
Pupils should be taught to {apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} more...
Pupils should be taught to construct and interpret plans and elevations of 3D shapes more...
Pupils should be taught to calculate arc lengths, angles and areas of sectors of circles more...
Pupils should be taught to apply the concepts of congruence and similarity, including the relationships between lengths, {areas and volumes} in similar figures more...
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### Notes:
Geometry is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. Geometry arose independently in a number of early cultures as a body of practical knowledge concerning lengths, areas, and volumes, with elements of a formal mathematical science emerging in the West as early 6th Century BC.
See also the topics of Angles, Area, Bearings, Circles, Enlargements, Mensuration, Pythagoras, Shape, Shape (3D), Symmetry, Transformations and Trigonometry.
### Geometry Teacher Resources:
Angle Theorems: Diagrams of the angle theorems with interactive examples.
Circle Parts Kim's Game: A memory game to be projected to help the whole class revise the names for the parts of a circle.
Circle Theorems: Diagrams of the circle theorems to be projected onto a white board as an effective visual aid.
Geometry Toolbox: Create your own dynamic geometrical diagrams using this truly amazing tool from GeoGebra.
Kite Maths: Can you make a kite shape from a single A4 size sheet of paper using only three folds?
Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes.
Polygons: Name the polygons and show the number of lines and order of rotational symmetry.
### Geometry Activities:
Angle Chase: Find all of the angles on the geometrical diagrams.
Angle Parallels: Understand and use the relationship between parallel lines and alternate and corresponding angles.
Angle Points: Apply the properties of angles at a point, angles on a straight line and vertically opposite angles.
Angles in a Triangle: A self marking exercise involving calculating the unknown angle in a triangle.
Area Maze: Use your knowledge of rectangle areas to calculate the missing measurement of these composite diagrams.
Area of a Triangle: Calculate the areas of the given triangles in this self marking quiz.
Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines?
Areas of Composite Shapes: Find the areas of combined (composite) shapes made up of one or more simple polygons and circles.
Circle Theorems Exercise: Show that you understand and can apply the circle theorems with this self marking exercise.
Congruent Triangles: Test your understanding of the criteria for congruence of triangles with this self-marking quiz.
Estimating Angles: Estimate the size of the given acute angles in degrees.
Formulae Pairs: Find the matching pairs of diagrams and formulae for basic geometrical shapes.
Geometry Toolbox: Create your own dynamic geometrical diagrams using this truly amazing tool from GeoGebra.
Measuring Units: Check your knowledge of the units used for measuring with this multiple choice quiz about metric and imperial units.
Polybragging: The Transum version of the Top Trumps game played online with the properties of polygons.
Polygon Angles: A mixture of problems related to calculating the interior and exterior angles of polygons.
Polygon Pieces: Arrange the nine pieces of the puzzle on the grid to make different polygons.
Polygon Properties: Connect the names of the polygons with the descriptions of their properties.
Polygons: Name the polygons and show the number of lines and order of rotational symmetry.
Proof of Circle Theorems: Arrange the stages of the proofs for the standard circle theorems in the correct order.
Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid.
Similar Shapes: Questions about the scale factors of lengths, areas and volumes of similar shapes.
Surface Area: Work out the surface areas of the given solid shapes.
Tantrum: A game, a puzzle and a challenge involving counters being placed at the corners of a square on a grid.
Transformations: Draw transformations online and have them instantly checked. Includes reflections, translations, rotations and enlargements.
Vectors: An online exercise on addition and subtraction of vectors, multiplication of vectors by a scalar, and diagrammatic representations of vectors.
Volume: Use formulae to solve problems involving the volumes of cuboids, cones, pyramids, prisms and composite solids.
Finally there is Topic Test, a set of 10 randomly chosen, multiple choice questions suggested by people from around the world.
### Geometry Investigations:
Area shapes: Investigate polygons with an area of 4 square units. This is your starting point, you can decide how to proceed.
How Many Rectangles?: Investigate the number of rectangles on a grid of squares. What strategies will be useful in coming up with the answer?
Maxvoltray: Find the maximum volume of a tray made from an A4 sheet of paper. A practical mathematical investigation.
Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes.
Polygon Areas: Investigate polygons with an area of 4 sq. units. Investigate polygons with other areas.
Rectangle Perimeters: The perimeter of a rectangle is 28cm. What could its area be?
Tantrum: A game, a puzzle and a challenge involving counters being placed at the corners of a square on a grid.
Tessellations: Drag the shapes onto the canvas to create tessellating patterns and investigate the laws of tessellations.
### Geometry Videos:
Construct a congruent triangle: Construction (with compass and straight edge) of a triangle congruent to a given triangle.
Different types of Triangle: Euclid and his friends explain how many different kinds of triangle there are.
Parallelogram: Instructional video showing how the area of a parallelogram can be determined.
### Geometry Worksheets/Printables:
Angle Chase Worksheets: A set of printable Angle Chase sheets on which pupils fill in the missing angles.
Polybragging Cards: Printable cards for the Polybragging game. Use the properties of polygons to win.
Links to other websites containing resources for Geometry are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below:
### Search
The activity you are looking for may have been classified in a different way from the way you were expecting. You can search the whole of Transum Maths by using the box below.
### Other
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#### Angle Chase
Find all of the angles on the geometrical diagrams.
Transum.org/go/?to=chase
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# Differentiation of Exponential Functions
Formulas and examples of the derivatives of exponential functions, in calculus, are presented. Several examples, with detailed solutions, involving products, sums and quotients of exponential functions are examined.
## First Derivative of Exponential Functions to any Base
The derivative of f(x) = b x is given by
f '(x) = b x ln b
Note: if f(x) = e
x , then f '(x) = e x
### Example 1
Find the derivative of f(x) = 2 x
### Solution to Example 1
• Apply the formula above to obtain
f '(x) = 2 x ln 2
### Example 2
Find the derivative of f(x) = 3 x + 3x 2
### Solution to Example 2
• Let g(x) = 3 x and h(x) = 3x 2, function f is the sum of functions g and h: f(x) = g(x) + h(x). Use the sum rule, f '(x) = g '(x) + h '(x), to find the derivative of function f
f '(x) = 3 x ln 3 + 6x
### Example 3
Find the derivative of f(x) = e x / ( 1 + x )
### Solution to Example 3
• Let g(x) = e x and h(x) = 1 + x, function f is the quotient of functions g and h: f(x) = g(x) / h(x). Hence we use the quotient rule, f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x) 2, to find the derivative of function f.
g '(x) = e x
h '(x) = 1
f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x) 2
= [ (1 + x)(e x) - (e x)(1) ] / (1 + x) 2
• Multiply factors in the numerator and simplify
f '(x) = x e x / (1 + x) 2
### Example 4
Find the derivative of f(x) = e 2x + 1
### Solution to Example 4
• Let u = 2x + 1 and y = e u, Use the chain rule to find the derivative of function f as follows.
f '(x) = (dy / du) (du / dx)
• dy / du = e u and du / dx = 2
f '(x) = (e u)(2) = 2 e u
• Substitute u = 2x + 1 in f '(x) above
f '(x) = 2 e 2x + 1
### Exercises
Find the derivative of each function.
1 - f(x) = e
x 2 x
2 - g(x) = 3
x - 3x 3
3 - h(x) = e
x / (2x - 3)
4 - j(x) = e
(x2 + 2)
### Solutions to the Above Exercises
1 - f '(x) = e
x 2 x ( ln 2 + 1)
2 - g '(x) = 3
x ln 3 - 9x 2
3 - h '(x) = e
x(2x -5) / (2x - 3) 2
4 - j '(x) = 2x e
(x2 + 2)
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# Area of a Semicircle
Home » Math Vocabulary » Area of a Semicircle
## Semicircle: Introduction
In geometry, a semicircle is defined as a half circle formed by cutting the circle into two halves. Every diameter of a circle divides it into two semicircles.
We get a semicircle when we fold a circular piece along its diameter. So, a circle, when divided in two equal halves, gives us a semicircle.
## What Is a Semicircle?
A semicircle is half of a circle. It is a plane figure formed when we divide a circle into two identical halves.
Take any two points on the circle such that the line joining these two points passes through the center of the circle. The two halves that we get are called semicircles. These two semicircles, when taken together, give us a complete circle.
In the following image, O is the center. The diameter BC divides the circle into two semicircles.
## Semicircle Shape
To obtain a semicircular shape, we can simply cut the circle from the center.
So we could also say that the area of the semicircle is just half of that of the circle.
This can be easily understood with the help of the figure given below.
Consider the diameter AB of the given circle. This diameter AB divides the given circle into two identical halves. These halves are semicircles and the area of these two semicircles, when combined, gives the area of the entire circle.
## Area of Semicircle
We know that the area of a circle is given by
Area of circle $= \pi r^{2}$
where,
$\pi = 3.1415$ or $\frac{22}{7}$ , and r is the radius of the circle.
The area of the semicircle is half of the area of a circle. This gives us the desired formula, that is,
Area of semicircle $= \frac{\pi r^{2}}{2} = \frac{1}{2} \pi r^{2}$
## How to Find the Area of a Semicircle
Step 1: Note the radius of the circle.
If the diameter is given, find the radius by dividing the diameter by 2.
For example, the diameter of the circle is 14 inches, then the radius of the circle is 7 inches.
Step 2: Find the area of the circle using the formula and by substituting the values of r and $\pi$.
$\text{A} = \pi r^{2}$
You can use the value of π as either 3.14 or $\frac{22}{7}$, when not specified in the question.
In our example, $r = 7$ inches and use $\pi = \frac{22}{7}$
So, area of circle $= \pi r^{2}$
area of circle $= \frac{22}{7} \times (7)^{2}$
area of circle $= 22 \times 7$
area of circle $= 154$ square inches.
Step 3: Divide the area of the circle by 2 to obtain the area of the semicircle.
Area of semicircle $= \frac{\pi r^{2}}{2}$
In the above example, Area of semicircle $= \frac{154}{2}$
Area of semicircle $= 77\; \text{inches}^{2}$
Area of a semicircle is measured in “square units.”
## Area of a Semicircle Using Diameter
We know that
Area of semicircle $= \frac{\pi r^{2}}{2}$
Here, $r = \frac{d}{2}$
Area of semicircle $=\frac{\pi}{2} \left[\frac{d}{2}\right]^{2}= \frac{πd^{2}}{8}$
## Derivation
We know that the area of any shape can also be determined with the help of the number of squares in that particular shape. So, we can say that the area of the circle can be measured by the number of square units fitting in that circle.
Consider a circle shown below, containing isosceles triangles.
All the radii of the circle are equal in length.
So, the polygon in the above figure can be easily divided into various (n) isosceles triangles (radius being two equal sides).
One such isosceles triangle can be represented as given below.
Area of a triangle can be given as half times the product of its height and base.
In this given isosceles triangle, base $= s$ and height $= h$.
The area of this isosceles triangle $= \frac{1}{2} (h \times s)$.
Similarly, for n such isosceles triangles the area would be n times the area of one such triangle.
Area of n such isosceles triangles $= \frac{1}{2} (n \times h \times s)$.
Here, the term $n \times s$ gives the perimeter of the polygon. As the polygon happens to look more and more like a circle, the perimeter term approaches the circumference of the circle, which is given as $2\pi r$.
When we substitute $2 \times \pi \times r$ in place of $n \times s$ we get,
Area of polygon $= 12 (2\pi r)$
Also, as s approaches zero, which happens as the number of sides increase making the triangle narrower, h approaches to r (the value of h becomes very close to r). So we substitute, r in place of h to get:
Area of polygon $= \frac{r}{2} (2 \times \pi \times r)$
Area of polygon $= \frac{r}{2} (2 \times \pi \times r)$
Area of polygon $= \frac{1}{2} (r \times 2 \times \pi \times r)$
Rearrange it to get,
Area of circle $= \pi r^{2}$
We know that the area of a semicircle equation is equal to half the area of the complete circle.
Therefore,
Area of a semicircle $= \frac{\pi r^{2}}{2}$
## Area of Semicircle Formula
So we derive the formula for the area of a semicircle as $(\pi r^{2})/2$, which is half the area of a circle.
Also, the surface area of the semicircle formula is the same as the area of the semicircle formula.
Area of semicircle $=\frac{\pi r^{2}}{2}$
## Fun Facts!
• Semicircle is obtained by cutting the circle into two exactly equal halves.
• Area of a semicircle is half the area of a circle with the same radius.
• Area of a semicircle with radius r can be given as $(\pi r^{2})/2$.
## Conclusion
In this article, we learned about semicircles. We derived the formula for calculating the area of a given semicircle with the help of the area of a circle with a given radius. We also learned a few amazing facts related to semicircles. The concept was complemented by a few solved and a few unsolved questions related to the topic.
## Solved Examples
1. If the radius of a semicircle is 49 inches, find its area.
Solution: Area of semicircle $= \frac{\pi r^{2}}{2}$
Area of semicircle $= \frac{22}{7} \times \frac{49^{2}}{2}$
Area of semicircle $= 11 \times 7 \times 49$
Area of semicircle $= 3773\; \text{inches}^{2}$
2. If the diameter of a semicircle is 70 inches, find its area of the semicircle with diameter.
Solution: Area of Semicircle $= \frac{\pi r^{2}}{2}$
Radius $= \frac{70}{2} = 35$ inches.
Now, Area of semicircle $= \frac{22}{7} \times \frac{35^{2}}{2}$
Area of circle $= 11 \times 5 \times 35$
Area of circle $= 1925\; \text{inches}^{2}$
3. Find the area of the figure in which ABCD is a square of side 42 inches and CPD is a semicircle. $(\text{Use}\; \pi = \frac{22}{7})$
Solution:
The required area $=$ Area of the square $\text{ABCD} +$ Area of the semicircle $\text{CPD}$
$= a^{2} + \frac{1}{2} \pi r^{2}$
$= 42^{2} + \frac{1}{2} \times \pi \times (\frac{1}{2} \times 42)^{2}$
$= (42^{2} +\frac{1}{2} \times \frac{22}{7} \times 21^{2})$
$= 1764 + 693$
$= 2457$ square inches
4. If the area of a circle is 100 square yards, what is the area of a semicircle?
Solution:
Area of circle $= 100$ square yards
Area of semicircle $= \frac{100}{2}$ square yards
5. Hazel has a circular garden outside her house with a diameter of 12 yards. Hazel wants to mow exactly half of the garden. Find the area of the part she wants to mow.
Solution:
Diameter $= 12$ yards
Area $= ?$
Area of a semicircle $= \frac{\pi r^{2}}{2}$
Radius $= \frac{12}{2} = 6$ yards
$\pi = 3.142$
Therefore, area $= \frac{3.142 \times 6 \times 6}{2} = 56.55$ square yards
## Practice Problems
1
### The area of a semicircle is ____ the area of the circle.
Twice
Half
Thrice
Equal to
CorrectIncorrect
Correct answer is: Half
Radius of the semicircle is the same as the radius of the circle.
2
### A circle has a diameter of 14 inches. What is the area of the semicircle?
7 inches
77 square inches
343 square inches
49 square inches
CorrectIncorrect
Correct answer is: 77 square inches
$d = 14$ inches. So, $r = \frac{14}{2} = 7$ inches
Area of semicircle $= \frac{\pi r^{2}}{2}$
Area of semicircle $= \frac{22}{7} \times \frac{7^{2}}{2}$
Area of semicircle $= 77\; \text{inches}^{2}$.
3
### The ____ of circle divides into two semicircles.
Radius
Chord
Tangent
Diameter
CorrectIncorrect
Correct answer is: Diameter
The diameter of a circle divides it into two equal parts, which are called semicircles.
4
### Area of the semicircle shown below is ____.
$\frac{9\pi}{2}$
$\frac{3\pi}{2}$
$\frac{\pi}{2}$
$\frac{12\pi}{2}$
CorrectIncorrect
Correct answer is: $\frac{9\pi}{2}$
Area of complete circle $= \pi r^2 = 9\pi$
Area of semicircle $= \frac{\pi r^2}{2} = \frac{9\pi}{2}$
## Frequently Asked Questions
If a circle is divided into four equal parts, then each part is called a “quarter circle.” Four quarter circles make a complete circle.
Circumference is the length of the boundary of the circle.
Perimeter of the semicircle $= \pi r + d$, where d is the diameter of the circle.
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# All About calculate volume of concrete for cylinder
Have you ever wondered how much concrete you would need for a cylindrical structure? Calculating the volume of concrete for a cylinder is a crucial step in construction projects, as it determines the amount of material needed and the cost of the project. Whether you are a construction professional or someone interested in learning about the process, this article will provide you with all the necessary information about calculating the volume of concrete for a cylinder. With a clear understanding of the concept and the right calculations, you can ensure that your construction project is completed efficiently and within budget.
## How to calculate volume of concrete for cylinder
Calculating the volume of a cylinder is an essential skill for any civil engineer working with concrete. The volume of a cylinder will determine the amount of concrete needed for a project, making it a critical calculation for both budget and construction planning. Here are the steps to calculate the volume of concrete for a cylinder:
Step 1: Measure the height and diameter of the cylinder
The first step is to accurately measure the height and diameter of the cylinder using a tape measure or ruler. Make sure to take the measurements in the same units, either feet or meters, for consistency.
To calculate the volume of a cylinder, you will need to know its radius, which is half of the diameter. The formula for calculating radius is: Radius = Diameter/2. For example, if the cylinder’s diameter is 10 feet, the radius will be 10/2 = 5 feet.
Step 3: Calculate the area of the base
The base of a cylinder is circular, and the area of a circle can be calculated using the formula A = πr2, where π is the mathematical constant, and r is the radius. Therefore, the area of the base of a cylinder with a radius of 5 feet is 3.14 x 52 = 78.5 square feet.
Step 4: Calculate the volume
The final step is to multiply the area of the base by the height of the cylinder to get the volume. The formula for calculating the volume of a cylinder is V = πr2h, where h is the height. For example, if the height of the cylinder is 15 feet, the volume will be 78.5 x 15 = 1177.5 cubic feet.
Step 5: Convert to the desired unit
The final step is to convert the volume to the desired unit, whether it is cubic inches, cubic yards, or any other unit, depending on the project requirements. You can use conversion factors or online converters to make this calculation.
In conclusion, calculating the volume of a cylinder for concrete is crucial for civil engineers to estimate the amount of material needed for construction projects accurately. By following these simple steps, civil engineers can determine the volume of concrete for any given cylinder, making the planning and budgeting process more efficient and effective.
## Conclusion
In conclusion, calculating the volume of concrete for a cylinder is a crucial step in construction projects. It ensures that the exact amount of concrete is used, reducing waste and saving costs. By using the formula V=πr^2h or utilizing online calculators, it is a simple and efficient process that can be easily mastered. Maintaining accuracy in volume calculation is essential for the durability and stability of any structure. With proper knowledge and understanding of the process, builders and engineers can ensure the success of their projects. So, before starting any construction involving cylinders, make sure to calculate the volume of concrete accurately, and the results will speak for themselves.
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# What is 8/9 in percentage form?
## Fraction to Percent Conversion Summary
We encourage you to check out our introduction to percentage page for a little recap of what percentage is. You can also learn about fractions in our fractions section of the website. Sometimes, you may want to express a fraction in the form of a percentage, or vice-versa. This page will cover the former case. Luckily for us, this problem only requires a bit of multiplication and division. We recommend that you use a calculator, but solving these problems by hand or in your head is possible too! Here’s how we discovered that 8 / 9 = 88.89% :
• Step 1: Divide 8 by 9 to get the number as a decimal. 8 / 9 = 0.89
• Step 2: Multiply 0.89 by 100. 0.89 times 100 = 88.89. That’s all there is to it!
## When are fractions useful?
Fractions are commonly used in everyday life. If you are splitting a bill or trying to score a test, you will often describe the problem using fractions. Sometimes, you may want to express the fraction as a percentage.
### Fraction Conversion Table
Percentage Fraction Decimal
88.89% 8 / 9 0.89
## Find the Denominator
A percentage is a number out of 100, so we need to make our denominator 100! If the original denominator is 9, we need to solve for how we can make the denominator 100. To convert this fraction, we would divide 100 by 9, which gives us 11.11.
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## Find the Numerator
Now, we multiply 11.11 by 8, our original numerator, which is equal to 88.89.
## Convert your Fraction to a Percent
Remember, a percentage is any number out of 100. If we can balance 8 / 9 with a new denominator of 100, we can find the percentage of that fraction!
## Understanding Fractions and Percentages
Fractions and percentages are both ways to represent a part of a whole. A fraction consists of a numerator (the top part) and a denominator (the bottom part), with the numerator representing the part and the denominator representing the whole. In contrast, a percentage represents a part out of 100. Converting between fractions and percentages can help us better understand and compare different quantities or proportions.
In this example, we have converted the fraction 8 / 9 into a percentage: 88.89%. This means that the fraction 8 / 9 is equivalent to 88.89 parts out of 100. It can be useful to know both the fraction and percentage forms when solving problems or interpreting data.
## Real-World Applications of Fractions and Percentages
Fractions and percentages are used in many real-world situations. For example, when calculating discounts on products, understanding both the fraction and percentage can help you quickly determine how much you will save. If a store offers a discount of 1/4 (1 / 4) off the original price, you can convert this fraction to a percentage (25%) to better understand the savings.
Another example could be in measuring ingredients for a recipe. If you know that you need 1/2 (1 / 2) of a cup of an ingredient, converting this fraction to a percentage (50%) can help you visualize how much you need relative to a full cup.
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## Tips for Converting Fractions to Percentages
When converting fractions to percentages, it’s essential to remember that percentages always have a denominator of 100. To convert a fraction like 8 / 9 into a percentage, you can follow these steps:
• Divide the numerator (8) by the denominator (9) to get the decimal equivalent.
• Multiply the decimal by 100 to get the percentage.
Keep in mind that you can also reverse these steps and still arrive at the correct solution. If you multiply the numerator (8) by 100 first and then divide the result by the denominator (9), you will still get the correct percentage (88.89%).
• What is the numerator of 8 / 9? The numerator is 8.
• What is the denominator of 8 / 9? The denominator is 9.
• When would you use 8 / 9 as a fraction? Give examples. You would use 8 / 9 as a fraction when dividing a whole into 9 equal parts and taking 8 of those parts.
• When would you use 8 / 9 as a decimal? Give examples. You would use 8 / 9 as a decimal when you need to express the fraction as a decimal number, such as in calculations or measurements.
• When would you use 8 / 9 as a percentage? Give examples. You would use 8 / 9 as a percentage when you want to express the fraction as a part out of 100, such as in comparisons or analyzing data.
• What are three other fractions that convert to 88.89%? Three other fractions that convert to 88.89% are 16/18, 24/27, and 40/45.
• Ask your students to think of three real-life examples of when to use fractions vs percentages. Students may consider examples like calculating grades (using percentages) or dividing a pizza into equal slices (using fractions).
• Which fraction is larger: 8 / 9 or 131 / 140? To compare fractions, we can either find a common denominator or convert them to decimals or percentages. In this case, it is easier to convert them to decimals or percentages. Converting 8 / 9 to a decimal, we get approximately 0.8889. Converting 131 / 140 to a decimal, we get approximately 0.9357. Therefore, 8 / 9 is smaller than 131 / 140.
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## The Circumference of a Cylinder: Understanding the Formula and its Applications
A cylinder is a three-dimensional geometric shape that consists of two parallel circular bases connected by a curved surface. It is a fundamental shape found in various fields, including engineering, architecture, and mathematics. Understanding the circumference of a cylinder is crucial for calculating its surface area, volume, and other important measurements. In this article, we will explore the formula for finding the circumference of a cylinder, its applications, and provide valuable insights into this topic.
## What is the Circumference of a Cylinder?
The circumference of a cylinder refers to the distance around the curved surface of the cylinder. It is similar to the concept of the circumference of a circle, but in the case of a cylinder, it is the measurement around the curved surface rather than just the base. The formula for finding the circumference of a cylinder is:
C = 2πr + 2πh
Where:
• C represents the circumference of the cylinder
• π (pi) is a mathematical constant approximately equal to 3.14159
• r is the radius of the circular base of the cylinder
• h is the height of the cylinder
## Calculating the Circumference of a Cylinder
Let’s consider an example to understand how to calculate the circumference of a cylinder. Suppose we have a cylinder with a radius of 5 units and a height of 10 units. Using the formula mentioned earlier, we can calculate the circumference as follows:
C = 2πr + 2πh
C = 2π(5) + 2π(10)
C = 10π + 20π
C = 30π
Therefore, the circumference of the given cylinder is 30π units. If we want to find the approximate numerical value, we can substitute the value of π as 3.14159:
C ≈ 30 × 3.14159
C ≈ 94.2478
So, the approximate circumference of the cylinder is 94.2478 units.
## Applications of the Circumference of a Cylinder
The concept of the circumference of a cylinder finds applications in various fields. Let’s explore a few examples:
### 1. Engineering and Architecture
In engineering and architecture, the circumference of a cylinder is essential for designing and constructing cylindrical structures such as pipes, columns, and storage tanks. By accurately calculating the circumference, engineers and architects can determine the amount of material required, estimate costs, and ensure structural integrity.
### 2. Manufacturing and Production
In manufacturing and production industries, the circumference of a cylinder is crucial for designing and producing cylindrical objects like cans, bottles, and tubes. Understanding the circumference helps in determining the dimensions, material requirements, and production efficiency.
### 3. Mathematics and Geometry
The concept of the circumference of a cylinder is an integral part of mathematics and geometry. It helps in understanding the relationship between the circumference, radius, and height of a cylinder. Additionally, it serves as a foundation for more complex calculations involving surface area, volume, and other geometric properties.
### Q1: Can the circumference of a cylinder be greater than its height?
A1: No, the circumference of a cylinder cannot be greater than its height. The circumference is the measurement around the curved surface, while the height is the distance between the two circular bases. Since the circumference is a curved path, it cannot be longer than a straight line distance, which is the height of the cylinder.
### Q2: How is the circumference of a cylinder different from the circumference of a circle?
A2: The circumference of a cylinder refers to the measurement around the curved surface of the cylinder, including both circular bases. On the other hand, the circumference of a circle is the measurement around the boundary of a single circular base. While the formulas for calculating both circumferences involve the mathematical constant π, the cylinder’s circumference formula includes an additional term for the height of the cylinder.
### Q3: Can the circumference of a cylinder be negative?
A3: No, the circumference of a cylinder cannot be negative. Circumference is a physical measurement that represents a distance, and distances cannot be negative. The circumference of a cylinder will always be a positive value or zero if the cylinder has no curved surface (i.e., it is a flat disk).
### Q4: How does the circumference of a cylinder relate to its surface area?
A4: The circumference of a cylinder is directly related to its surface area. The formula for calculating the surface area of a cylinder involves the circumference of its circular base. By multiplying the circumference by the height of the cylinder, we can find the total surface area. Therefore, understanding the circumference is crucial for accurately calculating the surface area of a cylinder.
### Q5: Can the circumference of a cylinder be used to find its volume?
A5: No, the circumference of a cylinder alone cannot be used to find its volume. The volume of a cylinder depends on its circular base’s area and the height of the cylinder. While the circumference is related to the surface area, it does not provide sufficient information to calculate the volume. To find the volume, one needs to use the formula V = πr^2h, where r is the radius of the circular base and h is the height of the cylinder.
## Summary
The circumference of a cylinder is a fundamental measurement that helps in understanding the curved surface of this three-dimensional shape. By using the formula C = 2πr + 2πh, we can calculate the circumference by considering the radius and height of the cylinder. This concept finds applications in various fields, including engineering, architecture, manufacturing, and mathematics. Understanding the circumference is crucial for accurately calculating the surface area, estimating material requirements, and ensuring structural integrity. By grasping the concept of the circumference of a cylinder, we can unlock a deeper understanding of this geometric shape and its practical applications.
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Contents
Cartesian coordinate system
The term 'Cartesian' is named with respect to the French mathematician Descartes in honor of his efforts to merge algebra and Euclidean geometry. This work was influential to the development of analytic geometry, calculus, and cartography.
The idea of this system was developed in 1637 in two writings by Descartes:
The modern Cartesian coordinate system in two dimensions (also called a rectangular coordinate system) is commonly defined by two axes, at right angles to each other, forming a plane (an xy-plane). The horizontal axis is labeled x, and the vertical axis is labeled y. In a three dimensional coordinate system, another axis, normally labeled z, is added, providing a sense of a third dimension of space measurement. The axes are commonly defined as mutually orthogonal to each other (each at a right angle to the other). (Early systems allowed "oblique" axes, that is, axes that did not meet at right angles.) All the points in a Cartesian coordinate system taken together form a so-called Cartesian plane.
The point of intersection, where the axes meet, is called the origin normally labeled O. With the origin labeled O, we can name the x axis Ox and the y axis Oy. The x and y axes define a plane that can be referred to as the xy plane. Given each axis, choose a unit length, and mark off each unit along the axis, forming a grid. To specify a particular point on a two dimensional coordinate system, you indicate the x unit first (abscissa), followed by the y unit (ordinate) in the form (x,y), an ordered pair. In three dimensions, a third z unit is added, (x,y,z).
The choices of letters come from the original convention, which is to use the latter part of the alphabet to indicate unknown values. The first part of the alphabet was used to designate known values.
An example of a point P on the system is indicated in the picture below using the coordinate (5,2).
The arrows on the axes indicate that they extend forever in the same direction (i.e. infinitely). The intersection of the two x-y axes creates four quadrants indicated by the roman numerals I, II, III, and IV. Conventionally, the quadrants are labeled counter-clockwise starting from the northeast quadrant. In Quadrant I the values are (x,y), and II:(-x,y), III:(-x,-y) and IV:(x,-y). (see table below.)
I > 0 > 0
II < 0 > 0
III < 0 < 0
IV > 0 < 0
Three dimensional coordinate system
Sometime in the early 19th century the third dimension of measurement was added, using the z axis.
A three dimensional coordinate system is usually depicted using what is called the right-hand rule, and the system is called a right-handed coordinate system (see handedness). By holding up the middle finger, index finger, and thumb of the right hand, you will see the orientation of the X, Y, and Z axes, respectively (the thumb being the Z axis). The fingers each point toward the positive direction of their representative axes. In the picture above, we see a right-handed coordinate system. Less common, but still in use (normally outside of the physical sciences) is the left-handed coordinate system.
When the z axis is depicted as pointing upward, this is sometimes called a world coordinates orientation. However, the important thing is which direction the axes point in the positive direction with respect to each other. If we drew an image in the right-handed system and then plotted the image, point for point in a left-handed system, you would have a mirror image.
The coordinates in a three dimensional system are of the form (x,y,z). An example of two points plotted in this system are in this picture, point P(5,0,2) and Q(-5,-5,10): . Notice that the axes are depicted in a world-coordinates orientation with the Z axis pointing up. With your right hand, point your thumb, index and middle finger out, tilting your hand back. Notice that your middle finger is pointing up, thumb to the right, and the index finger is point outward just like the y axis in the picture. Your thumb is pointing in the same direction as the x axes does when it moves in a positive direction. This is a right-hand coordinate system.
The three dimensional coordinate system is popular because it provides the physical dimensions of space, of height, width, and length, and this is often referred to as "the three dimensions". It is important to note that a dimension is simply a measure of something, and that, for each class of features to be measured, another dimension can be added. Attachment to visualizing the dimensions precludes understanding the many different dimensions that can be measured (time, mass, color, cost, etc.). It is the powerful insight of Descartes that allows us to manipulate multi-dimensional object algebraically, avoiding compass and protractor for analyzing in more than three dimensions.
### Further Notes
In analytic geometry the Cartesian Coordinate System is the foundation for the algebraic manipulation of geometrical shapes. Many other coordinate systems have been developed since Descartes. One common set of systems use polar coordinates; astronomers often use spherical coordinates, a type of polar coordinate system. In different branches of mathematics coordinate systems can be transformed, translated, rotated, and re-defined altogether to simplify calculation and for specialized ends.
It may be interesting to note that some have indicated that the master artists of the Renaissance used a grid, in the form of a wire mesh, as a tool for breaking up the component parts of their subjects they painted--a trade secret. That this may have influenced Descartes is merely speculative.
### References
Descartes, René. Oscamp, Paul J. (trans). Discourse on Method, Optics, Geometry, and Meteorology. 2001.
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# Quadratic Formula
Let's kick things off with a...
### Sample Problem!
What, you were expecting a party?
Solve the quadratic equation ax2 + bx + c = 0 by completing the square.
Here we know a, b, and c are numbers, but we don't know what any of them are. We do know that a can't be 0, or we wouldn't have a quadratic equation. We have a sneaking suspicion that b is 17, but that's only based on a dream we had last night, so we should probably do the math to be on the safe side.
First, we divide both sides by a, since we don't want a coefficient on x2 gumming up the works.
Then we subtract from both sides to get it out of the way.
Next we take (the coefficient on x), divide by 2, and square to find . It's not as nice-looking as what we've had in the past, but we'll go with it. This is the number we add to both sides:
The left-hand side of the equation can now be written as a square:
Since , we can add the numbers on the right-hand side of the equation:
It's a good feeling to look at such an ugly conglomeration of numbers, variables, fractions, and parentheses and know that you can make some sense out of it, eh? Not that you want to print it out and hang it on your bedroom wall, but still. We think it's nice.
Now, writing the left-hand side of the equation as a square and the right-hand side of the equation in its new form, we need to solve the equation:
Since we have the square of a first-degree polynomial on the left and an admittedly messy number on the right, we know where to go from here. We summon our radical powers and take the square root of both sides.
We can simplify the right-hand side:
Now, we need to solve these two equations:
Here's our first solution:
Ready for the second solution? Comin' atcha:
These solutions are usually written together as:
We call this gnarly equation the quadratic formula.
You may have already heard of this one. It's famous. It used to have its own talk show, and it was voted one of People's Sexiest Formulas Alive in 300 BCE.
Okay, maybe it hasn't achieved that kind of fame, but it is well-known. The quadratic formula is another way we can find solutions to quadratic equations. When given a quadratic equation, we figure out which numbers correspond to a, b, and c, then plug them into the quadratic formula to find our answers. It's like one of those factory machines where you throw in various ingredients, and then out comes a gloriously wonderful Oreo, or Twinkie, or wax lips. Yes, we went with "wax lips."
As complicated as it is to arrive at and understand the quadratic formula, it's one of those things that's best to memorize and use. It's unlikely that you'll be asked in school to explain where the quadratic formula comes from, but it's reassuring to know that, like babies, it does come from somewhere.
Try to make it feel at home, won't you?
### Sample Problem
Solve: x2 + 5x – 7 = 0.
The coefficient on x2 is 1, the coefficient on x is 5, and the constant term is -7, so we have:
All we do is plug these numbers into the quadratic formula, , and we'll have our solution in no time:
Since 53 is a prime number, we can't simplify any more. The two solutions to the equation are:
Be Careful: In order to use the quadratic formula, you need an equation of the form ax2 + bx + c = 0. Because we must have zero on one side of the equation, you can't go quadratic equation-happy and start plugging everything under the sun into it. By the way, if you ever go overseas, make sure you pack your European quadratic plug adapter.
### Sample Problem
A student, the same hopeless daydreamer who aspires to one day be a member of the National Audubon Society, was asked to use the quadratic formula to solve the equation x2 + 3x – 6 = 7.
She wrote down the following work. What did she do wrong?
The original equation didn't have zero on one side. D'oh! The student should have first subtracted 7 from each side, then solved the equation x2 + 3x – 13 = 0.
This is starting to become a problem. Her teacher might want to consider drawing the blinds.
### Sample Problem
A different student, one who's not interested in birds so much as he is captivated by the incoming texts on the cell phone he's discreetly concealing in one of his sleeves, was asked to use the quadratic formula to solve the equation 4x2 + 15x – 9 = 0.
He wrote down the following work. What did he do wrong?
In the original equation, c was -9, not +9. He might have noticed that if he hadn't been so busy lol'ing and jk'ing.
The student's first two lines should have been:
Sometimes a quadratic equation has no solutions. The quadratic equation says . But since we can't take square roots of negative numbers when we're looking for real number solutions, this only works if b2 – 4ac is positive (or zero). The expression b2 – 4ac is called the discriminant of the equation ax2 + bx + c = 0.
The quantity b2 – 4ac discriminates between those equations that have real number solutions and those that don't. That's a whole lot of power for one tiny little formula.
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# How do you factor 3bc-2b-10+15c?
Jan 20, 2017
$3 b c - 2 b - 10 + 15 c = \left(b + 5\right) \left(3 c - 2\right)$
#### Explanation:
This quadrinomial factors by grouping (at least once we swap a couple of the terms around)...
$3 b c - 2 b - 10 + 15 c = 3 b c - 2 b + 15 c - 10$
$\textcolor{w h i t e}{3 b c - 2 b - 10 + 15 c} = \left(3 b c - 2 b\right) + \left(15 c - 10\right)$
$\textcolor{w h i t e}{3 b c - 2 b - 10 + 15 c} = b \left(3 c - 2\right) + 5 \left(3 c - 2\right)$
$\textcolor{w h i t e}{3 b c - 2 b - 10 + 15 c} = \left(b + 5\right) \left(3 c - 2\right)$
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