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# NCERT Exemplar Solution for CBSE Class 10 Mathematics: Surface Areas and Volumes Exercise 12.4 (II)
Sep 13, 2017 10:54 IST
Class 10 Maths NCERT Exemplar Solutions
Here you get the CBSE Class 10 Mathematics chapter 12, Surface Areas and Volumes: NCERT Exemplar Problems and Solutions. This part of the chapter includes solutions of Question Number 11 to 14 from Exercise 12.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Surface Areas and Volumes. This exercise comprises only the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
CBSE Class 10 Mathematics Syllabus 2017-2018
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Surface Areas and Volumes:
Exercise 12.4
Long Answer Type Questions (Q. No. 11-14):
Question. 11 16 glass spheres each of radius 2cm are packed into a cuboidal box of internal dimensions 16cm × 8cm × 8cm and then the box is filled with water. Find the volume of water filled in the box.
Solution.
Question. 12 A milk container of height 16cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8cm and 20cm, respectively. Find the cost of milk at the rate of ` 22 per L
which the container can hold.
Solution.
Given, radius of lower end of milk container, r = 8 cm
Radius of upper end of milk container, R = 20cm
And, height of milk container, h = 16cm
Now, as the container is in the shape of a a frustum of a cone,
Now, given that cost of 1 Lmilk = Rs.22
⟹ Cost of 10.45942 L milk = 22 × 1045942 = Rs.230.12
Hence, the required cost of milk contained in the container is Rs.230.12.
Question. 13 A cylindrical bucket of height 32cm and base radius 18cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24cm, find the radius and slant height of the heap.
Solution.
Given, radius of the base of the bucket = 18cm
And, height of the bucket = 32cm
Question. 14 A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a
cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6cm and 12cm,
respectively. If the slant height of the conical portion is 5cm, then find the total surface area and volume of the
rocket, (use p = 3.14)
Solution.
Since, rocket is the combination of a right circular cylinder and a cone.
And total surface area of the rocket = CSA of cone + CSA of cylinder + Area of base of cylinder
= 47.1 + 226.08+28.26 = 301.44cm2
CBSE Class 10 NCERT Textbooks & NCERT Solutions
NCERT Solutions for CBSE Class 10 Maths
NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters |
# 8.6 Rotational kinetic energy: work and energy revisited
Page 1 / 9
• Derive the equation for rotational work.
• Calculate rotational kinetic energy.
• Demonstrate the Law of Conservation of Energy.
In this module, we will learn about work and energy associated with rotational motion. [link] shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy .
Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in [link] ) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled:
$\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(\text{net}\phantom{\rule{0.25em}{0ex}}F\right)\text{Δ}s.$
To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by $r$ , and gather terms:
$\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(r\phantom{\rule{0.25em}{0ex}}\text{net}\phantom{\rule{0.25em}{0ex}}F\right)\frac{\text{Δ}s}{r}.$
We recognize that $r\phantom{\rule{0.25em}{0ex}}\text{net}\phantom{\rule{0.25em}{0ex}}F=\text{net τ}$ and $\Delta s/r=\theta$ , so that
$\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(\text{net τ}\right)\theta .$
This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation $\text{net}\phantom{\rule{0.25em}{0ex}}W=\left(\text{net τ}\right)\theta$ is valid in general, even though it was derived for a special case.
To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that $\text{net τ}=\mathrm{I\alpha }$ , so that
$\text{net}\phantom{\rule{0.25em}{0ex}}W=I\text{αθ}.$
## Making connections
Work and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation .
Now, we solve one of the rotational kinematics equations for $\text{αθ}$ . We start with the equation
${{\omega }_{}}^{2}={{\omega }_{\text{0}}}^{2}+2\text{αθ}.$
Next, we solve for $\text{αθ}$ :
$\text{αθ}=\frac{{\omega }^{2}-{{\omega }_{\text{0}}}^{2}}{2}.$
Substituting this into the equation for net $W$ and gathering terms yields
$\text{net}\phantom{\rule{0.25em}{0ex}}W=\frac{1}{2}{\mathrm{I\omega }}^{2}-\frac{1}{2}I{{\omega }_{\text{0}}}^{2}.$
This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term $\left(\frac{1}{2}\right){\mathrm{I\omega }}^{2}$ to be rotational kinetic energy ${\text{KE}}_{\text{rot}}$ for an object with a moment of inertia $I$ and an angular velocity $\omega$ :
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Lesson Objectives and Overview
In this lesson, students will put together what they have learned in the previous lessons and begin simulating motion.
• Students will learn how to give a Particle object a velocity.
• Students will learn how to update the position of a particle based on its velocity.
• Students will learn how to use the function drawArrow() to represent velocity vectors.
In the previous tutorial, you learned that you can get Tychos to do repetitive calculations, updating the values of variables with new values based on the calculations that were performed. In this lesson, you are going to see that you can get Tychos to perform a repetitive calculation to update the position of a particle, and thus simulate motion.
# Velocity is a Rate of Change
Let's return to the definition of velocity, which we can use to calculate the change in distance that a particle will experience in a given change in time. Below is the mathematical representation of a particle's velocity in the x dimension:
We can calculate how far the particle will travel during a specific change in time by simply representing the change in position as:
The larger the velocity in the x direction, then the larger the position change for the given interval of time.
We are now going to also state that what is true in the x dimension is true for the y dimension (and the z dimension) as well because space is uniform and we can say that our x or y or z dimensions are really quite arbitrary:
(and )
Because we are working in a 2D space, let's represent our change in 2D space by the matrix :
If our particle's position is given as the vector , then we can calculate the new position of the particle by simply adding the change in position given by :
Let's say that a particle's initial position is [0, 0] and it's position is changing at the rate of 10 meters per second in the x direction and 5 meters per second in the y direction. We want to track the movement of the particle by identifying the position at 0.1 second intervals. The table below shows the particle's position from 0 to 1 second:
Time x position y position 0.1 s ​​ ​​ 0.2 s ​​ ​​ 0.3 s ​​ ​​ 0.4 s ​​ ​​ 0.5 s ​​ ​​ 0.6 s ​​ ​​ 0.7 s ​​ ​​ 0.8 s ​​ ​​ 0.9 s ​​ ​​ 1.0 s ​​ ​​
We are calculating the position iteratively. As time progresses, we can simply update the postion of the particle for each interval of time that has elapsed - what is called an iteration. An iteration is a small interval or a step in the process. The iteration interval is the amount of time represented by .
# Iteratively Updating A Particle's Position in Tychos
Doing these calculations over and over again can seem a bit tedious, and as the interval time shortens, then the number of calculations required grows...
But as you have learned in the previous tutorial, computers are really good at doing these kinds of repetitive calculations. In fact, you can just set up the formula and then have the computer do all the calculations for you.
Click on the link below to open the Tychos scenario in a new broswer tab where you can follow along, or use the embedded frame that contains the scneario below:
The first thing that we need to do is create a particle. Let's call the particle p1. Type this code in the Initial State pane:
p1 = Particle([50, 50], 5, "green")
When you press the Start button in Tychos, you should see a green particle appear on the world grid. Now we want the particle to move, but we have to define the rate at which it is going to move.
Let's say that we wanted to simulate a particle moving along the x axis with a velocity of 10 meters per second:
We can define this velocity as a matrix in code:
v = [10,0]
Now we could use this matrix to conduct some calculations on how to update the position of the particle. We are going to introduce here a method commonly used in many programming languages to associate this velocity to the particle p1. This way we can see that this velocity "belongs" to the particle p1 by giving the particle an attribute. Change the above line of code so that it now looks like this:
p1.v = [10, 0]
The velocity matric [10, 0] is now an attribute of the particle, just as the position of the particle is also an attribute that we can access like this:
p1.pos = [0, -50]
If you press the Start button, you will see that the particle does not move yet. This is because Tychos has no idea what velocity is. You have to define how the velocity is going to update the position. This is done in the Calculations pane.
Remember that the simulated world's time is segmented into units called frames. If the frame rate is 20 frames per second, then the change in time (or dt) is .05 seconds. So if a particle has a velocity of say 10 meters per second, then we need to define how far it would go in .05 seconds. This will be the rate of position change per frame.
The change in position is the change in the particle's position for that frame, but the velocity is defined as the rate of change per second. Add this line of code to the Calculations pane:
ds = p1.v*dt
In the above code, we are saying that the change in position (ds) will be calculated by multiplying the particle's velocity by the interval time (dt). This will calculate the change in position for this frame. It is an identical calculation to the one we defined above, but now just in code:
We can now use this to change the current position of the particle by recalculating the position of the Particle every frame. Place this line of code just below the previous line of code, once again in the Calculations Pane:
p1.pos = p1.pos + ds
When you press the Start button you should see the Particle move!
Let's see if this makes sense. Because the frame rate is 1/20 of a second, which is .05 seconds, and the velocity is [10, 0], then for a single frame, the particle will move 0.5 units (10 x .05 = 0.5) in the x direction. After 2 frames, which is 1/10 of a second, the particle will have moved 1 unit. After 10 frames, which is 1/2 of a second, the particle will have moved 5 units in the x direction. Finally, you should see that after 20 frames, which is 1 second, the particle will have moved 10 units in the x direction. This is precisely what we want. We want the particle to move at 10 units per second along the x direction.
Go back to the Initial State pane and change the particle's velocity:
p1.v = [10, 5]
When you press the Start button, you should see the that the particle is now traveling in a diagonal path - it is now traveling in both the x and y dimensions!
Finally, let's add a vector arrow to represent the position vector of the particle. Add this line of code in the Calculations Panejust below the last line of code:
drawArrow([0, 0], p1.pos)
You will see an arrow that is changing in length that stretches from the origin to the particle's position.
Cool! Now its time to work on some goals to see if you actually get it:
• Create a particle called p2 with an initial position of [100, 50]
• Make the particle p2 move so that it returns to the origin after 5 seconds.
• Create a particle called p3 with an initial position of [-70, -100]
• Make the particle p3 move so that its position is [30, 30] afte 5 seconds.
You have successfully simulated particles moving through 2D space at a constant velocity!
## Using Vector Arrows To Represent Velocity
As you learned earlier, Tychos gives us a tool for visualizing a vector quantity. A vector arrow is simply an arrow that points in the direction from a specified origin with the dimensions indicated by the vector quantity - in our case the velocity matrix. We have seen how to draw position vectors that are attached to the origin because this is the point in space where all particle's positions are measured. Velocity however, is attached to the individual particle. It is an attribute of each particle so it would be nice to represent the velocity of each particle by an arrow that is attached to the particle. This can be done like this:
drawArrow(p1.pos, p1.v, "purple")
This command draws a vector arrow from the position of the particle and defines the length and direction of the arrow based on the velocity (represented by the variable "v" here) and then also gives it a color - in this case "purple".
These vector arrows are used to help represent the motion of the particle, and I think you will find them very useful as we progress through this unit.
# Conclusion
You should now be familiar with the following:
• How to repetitively perform a calculation to update a variable's value each frame.
• Understand that the variable dt represents the time interval duration for each frame.
• Understand that the time in seconds in the simulated universe (virtual time) is the result of the frame rate multiplied by the time interval (dt)
• How to change a particle's position by:
• Defining a velocity matrix for the particle.
• Multiplying the velocity matrix by the time interval to get the change in position.
• Add the change in position to the particle's current position in order to update it.
• How to attach velocity vectors to a particle in order to visualize the particle's velocity. |
# 3.5: Poisson Distribution
In this section, we consider our final family of discrete probability distributions. We begin with the definition.
### Definition $$\PageIndex{1}$$
A random variable $$X$$ has a Poisson distribution, with parameter $$\lambda>0$$, if its probability mass function is given by
$$p(x) = P(X=x) = \frac{e^{-\lambda}\lambda^x}{x!}, \quad\text{for}\ x=0,1,2,\ldots.\label{Poissonpmf}$$
We write $$X\sim\text{Poisson}(\lambda)$$.
The main application of the Poisson distribution is to count the number of times some event occurs over a fixed interval of time or space. More specifically, if the random variable $$X$$ denotes the number of times the event occurs during an interval of length $$T$$, and $$r$$ denotes the average rate at which the event occurs per unit interval, then $$X$$ has a Poisson distribution with parameter $$\lambda = rT$$. Consider the following examples:
• The number of customers arriving at McDonald's between 8 a.m. and 9 a.m.
• The number of calls made to 911 in South Bend on a Saturday.
• The number of accidents at a particular intersection during the month of June.
All of the examples above count the number of times something occurs over an interval of time. The next example gives an example where the interval is in space.
### Example $$\PageIndex{1}$$
Suppose typos occur at an average rate of $$r = 0.01$$ per page in the Friday edition of the New York Times, which is 45 pages long. Let $$X$$ denote the number of typos on the front page. Then $$X$$ has a Poisson distribution with parameter
$$\lambda = 0.01\times1 = 0.01,\notag$$
since we are considering an interval of length one page ($$T=1$$). Thus, the probability that there is at least one typo on the front page is given by
$$P(X\geq1) = P(\{X=0\}^c) = 1 - P(X=0) = 1 - \frac{e^{-0.01}(0.01)^0}{0!} = 1 - e^{-0.01} \approx 0.00995.\notag$$
Now, if we let random variable $$Y$$ denote the number of typos in the entire paper, then $$Y$$ has a Poisson distribution with parameter
$$\lambda = 0.01\times45 = 0.45,\notag$$
since we are considering an interval of $$T=45$$ pages. The probability that there are less than three typos in the entire paper is
$$P(Y<3) = P(Y=0, 1,\ \text{or}\ 2) = \frac{e^{-0.45}(0.45)^0}{0!} + \frac{e^{-0.45}(0.45)^1}{1!} + \frac{e^{-0.45}(0.45)^2}{2!} \approx 0.98912.\notag$$
The Poisson distribution is similar to all previously considered families of discrete probability distributions in that it counts the number of times something happens. However, the Poisson distribution is different in that there is not an act that is being repeatedly performed. In other words, there are no set trials, but rather a set window of time or space to observe. |
Finding Area of a Rectangle Finding Area of a Rectangle Area of a rectangle specifies an area in the coordinate space that is enclosed by the rectangular object. To find the area of a rectangle, multiply the length and the width. Area is usually measured in square units such as square inches, square feet or square meters.In the above figure, the length of the rectangle is "l" and the width of the rectangle is "w". Therefore, the area of the rectangle formula is (l X w) square units. Finding Area of a Rectangle Area of rectangle = l x w, Where l = length, and w = width. That is, it is the product of length and width. Example Problems on Area of a Rectangle Below are some examples based on area of a rectangle Example 1: Find the area of a rectangle whose length l = 3m and breath b = 2m? Solution : Area of a rectangle = l x b Know More About What is a Rectangular Prism
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= 3 x 2, = 6m2 Example 2: Find the length of a rectangle whose area =23m and breath = 3m? Solution : Area of a rectangle = l x b 23 = l x 3 l = 233 l = 7.67m2 Example 3: Find the length of a rectangle whose area =28m and breadth = 4m? Solution : Area of a rectangle = l x b 28 = l x 4 l = 284 l = 7m2 Example 4: Find the area of the garden, length and width of the garden is 600m and 400m? Solution : Area of the garden = l x b = 600 x 400 = 240000 m2 Example 5: Find perimeter and area of a rectangle whose length is 4cm and breadth is 7cm Solution : Formula for area of rectangle = l Ă— b =4Ă—7 = 28cm2 Perimeter of the rectangle = 2L +2W = 2(4) + 2(7) = 8 + 14 = 22cm. Learn More Heptagon Shape Tutorvista.com
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Area of a Parallelogram Formula Area of a Parallelogram Formula The total space inside the boundary of the parallelogram is called as the area of the parallelogram. The area of a parallelogram is twice the area of a triangle created by one of its diagonals. In a parallelogram, the opposite sides are equal to each other and the opposite angles are also equal to each other. The sum of all the interior angles of a parallelogram is 1800. Area of a Parallelogram Formula The area of a parallelogram can be found by using the following formula, Area of a parallelogram, A = bh where, b = Base of the parallelogram h = Height of the parallelogram The base and the height of the parallelogram are always perpendicular to each other. We can derive the formula for height of the parallelogram with the help of the area of a parallelogram formula.
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Parallelogram Height = Area of the parallelogram / Base of the parallelogram. Example Problems on Area of a Parallelogram Below are some examples based on area of a parallelogram Problem 1: Find the area of a Parallelogram of height h = 9cm and base b = 11.5cm. Solution : We know that the area of parallelogram is, A = bh = (11.5 cm) (9 cm) = 103.5 cm2 Therefore, the area of a parallelogram is 103.5 cm2 Problem 2: The base and height of the parallelogram is 23inch and 15 inch respectively .What is the area of a parallelogram? Solution : Given, Base b =23 inch and height h =15 inch We know that the area of parallelogram is, A = bh Substitute the value of b and h, A = (23 inch) (15 inch) = 345 inch2 Therefore, the area of a parallelogram is 345 inch2 Read More About The Area of a Triangle
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Thank You
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Finding Area of a Rectangle
Finding Area of a Rectangle
Area of a rectangle specifies an area in the coordinate space that is enclosed by the rectangular object. To find the area of a rectangle, m... |
# What is 51/212 as a decimal?
## Solution and how to convert 51 / 212 into a decimal
51 / 212 = 0.241
51/212 converted into 0.241 begins with understanding long division and which variation brings more clarity to a situation. Both represent numbers between integers, in some cases defining portions of whole numbers The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 51/212 is 51 divided by 212
Converting fractions to decimals is as simple as long division. 51 is being divided by 212. For some, this could be mental math. For others, we should set the equation. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. Now we divide 51 (the numerator) into 212 (the denominator) to discover how many whole parts we have. Here's 51/212 as our equation:
### Numerator: 51
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. Any value greater than fifty will be more difficult to covert to a decimal. 51 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. Let's look at the fraction's denominator 212.
### Denominator: 212
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. Larger values over fifty like 212 makes conversion to decimals tougher. And it is nice having an even denominator like 212. It simplifies some equations for us. Ultimately, don't be afraid of double-digit denominators. So without a calculator, let's convert 51/212 from a fraction to a decimal.
## Converting 51/212 to 0.241
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 212 \enclose{longdiv}{ 51 }$$
We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 212 \enclose{longdiv}{ 51.0 }$$
Because 212 into 51 will equal less than one, we can’t divide less than a whole number. Place a decimal point in your answer and add a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 212 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 212 into 510
$$\require{enclose} 00.2 \\ 212 \enclose{longdiv}{ 51.0 }$$
Since we've extended our equation we can now divide our numbers, 212 into 510 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiple this number by our furthest left number, 212, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.2 \\ 212 \enclose{longdiv}{ 51.0 } \\ \underline{ 424 \phantom{00} } \\ 86 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 51/212 into 0.241 If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 51/212 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Here are just a few ways we use 51/212, 0.241 or 24% in our daily world:
### When you should convert 51/212 into a decimal
Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions.
### When to convert 0.241 to 51/212 as a fraction
Meal Prep - Body builders need to count macro calories. One of the ways of doing this is measuring every piece of food consumed. This is through halves and quarters in order to keep it consistent.
### Practice Decimal Conversion with your Classroom
• If 51/212 = 0.241 what would it be as a percentage?
• What is 1 + 51/212 in decimal form?
• What is 1 - 51/212 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.241 + 1/2? |
# Factor the equation: 21x^2 + 22x = 8
Wilson2014 | eNotes Employee
Posted on
A simpler method involving the quadratic equation may be used.
`21x^2+22x=8`
`21x^2+22x-8=0`
a = 21 ; b = 22 ; c = -8
Substituting in the a, b, and c values:
`x=(-22+-sqrt(22^2-4(21)(-8)))/(2*21)`
`x=(-22+-sqrt(1156))/42`
`x=(-22+34)/42` or `x=(-22-34)/42`
Upon simplification, `x=2/7 , x= -4/3`
These are the two values that make `21x^2+22x-8=0`
In order to find the factors, we try to get all the values to one side so that the expression can equal zero.
`x=2/7 => 7x=2 => 7x-2=0`
`x=-4/3 => 3x=-4 => 3x+4=0`
Therefore, factoring the original equation results in `(7x-2)(3x+4)=0`
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You should complete the square `21x^2 + 22x` using the following formula, such that:
`a^2 + 2ab + b^2 = (a + b)^2`
Considering `a^2 = 21x^2` and `2ab = 22x` yields:
`a = x*sqrt 21`
You need to complete the square such that:
`21x^2 + 22x + 121/21 = 8 + 121/21`
`(sqrt 21*x + 11/sqrt 21)^2 = (168 + 121)/21`
`(sqrt 21*x + 11/sqrt 21)^2 - 289/21 = 0`
Converting the difference of squares into a product yields:
`(sqrt 21*x + 11/sqrt 21 - 17/sqrt21)(sqrt 21*x + 11/sqrt 21 + 17/sqrt21) = 0`
`(sqrt 21*x - 6/sqrt21)(sqrt 21*x + 28/sqrt 21) = 0`
Hence, evaluating the factored form of the given equation yields `(sqrt 21*x - 6/sqrt21)(sqrt 21*x + 28/sqrt 21) = 0` . |
NCERT Solutions for Class 10 Maths Chapter 15- Probability
The NCERT Solutions for Class 10 Maths Chapter 15 Probability are undoubtedly an essential study material for the students studying in CBSE Class 10. NCERT Solutions provided here along with the downloadable PDF can help the students prepare effectively for their exams. The chapter is a continuation of what was taught in chapter probability in Class 9 and further explains the different concepts related to it. Â
The subject experts prepared the NCERT solutions for Class 10 Maths to assist students in preparing for their board examination. Every step and concept used in solving a solution is explained clearly in the answers provided at BYJU’S, leaving no question unsolved. Along with exam preparation, these solutions can be used to check if the answers given to the exercise questions by the students are correct while doing their homework and assignments. So, it is advised to all the students to go through the these NCERT Solutions regularly to stand out among the other students in the class and also to excel in the Board examination of Class 10 standard exam.
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Class 10 Maths Chapter 15 Exercise: 15.1 (Page No: 308)
1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___________ .
(ii) The probability of an event that cannot happen is __________. Such an event is called ________ .
(iii) The probability of an event that is certain to happen is _________ . Such an event is called _________ .
(iv) The sum of the probabilities of all the elementary events of an experiment is __________ .
(v) The probability of an event is greater than or equal to and less than or equal to __________.
Solution:
(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to Solution: a true-false question. The Solution: is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.
(ii) Even this statement does not have equally likely outcomes as the player may shoot or miss the shot.
(iii) This statement has equally likely outcomes as it is known that the solution is either right or wrong.
(iv) This statement also has equally likely outcomes as it is known that the newly born baby can either be a boy or a girl.
3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.
4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) -1.5 (C) 15% (D) 0.7
Solution:
The probability of any event (E) always lies between 0 and 1 i.e. 0 ≤ P(E) ≤ 1. So, from the above options, option (B) -1.5 cannot be the probability of an event.
5. If P(E) = 0.05, what is the probability of ‘not E’?
Solution:
We know that,
P(E) + P(not E) = 1
It is given that, P(E) = 0.05
So, P(not E) = 1 – P(E)
Or, P(not E) = 1 – 0.05
∴ P(not E) = 0.95
6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) We know that the bag only contains lemon-flavoured candies.
So, The no. of orange flavoured candies = 0
∴ The probability of taking out orange flavoured candies = 0/1 = 0
(ii) As there are only lemon flavoured candies, P(lemon flavoured candies) = 1 (or 100%)
7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let the event wherein 2 students having the same birthday be E
Given, P(E) = 0.992
We know,
P(E) + P(not E) = 1
Or, P(not E) = 1 – 0.992 = 0.008
∴ The probability that the 2 students have the same birthday is 0.008
8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
The total number of balls = No. of red balls + No. of black balls
So, the total no. of balls = 5 + 3 = 8
We know that the probability of an event is the ratio between the no. of favourable outcomes and the total number of outcomes.
=> P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Probability of drawing red balls = P (red balls) = (no. of red balls/total no. of balls) = 3/8
(ii) Probability of drawing black balls = P (black balls) = (no. of black balls/total no. of balls) = 5/8
9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
The Total no. of marbles= 5 + 8 + 4 = 17
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total number of red marbles= 5
P (red marbles) = 5/17 = 0.29
(ii) Total number of white marbles= 8
P (white marbles) = 8/17 = 0.47
(iii) Total number of green marbles = 4
P (green marbles) = 4/17 = 0.23
∴ P (not green) = 1 – P (green marbles) = 1 – (4/17) = 0.77
10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be a ₹5 coin?
Solution:
Total no. of coins = 100 + 50 + 20 + 10 = 180
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total number of 50 p coin = 100
P (50 p coin) = 100/180 = 5/9 = 0.55
(ii) Total number of ₹5 coin = 10
P (₹5 coin) = 10/180 = 1/18 = 0.055
∴ P (not ₹5 coin) = 1 – P (₹5 coin) = 1 – 0.055 = 0.945
11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?
Solution:
The total number of fish in the tank = 5 + 8 = 13
Total number of male fish = 5
P(E) = (Number of favourable outcomes/ Total number of outcomes)
P (male fish) = 5/13 = 0.38
12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Solution:
Total number of possible outcomes = 8
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total number of favourable events (i.e. 8) = 1
∴ P (pointing at 8) = ⅛ = 0.125
(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)
P (pointing at an odd number) = 4/8 = ½ = 0.5
(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)
P (pointing at a number greater than 4) = 6/8 = ¾ = 0.75
(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)
P (pointing at a number less than 9) = 8/8 = 1
13. A die is thrown once. Find the probability of getting
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total number of prime numbers = 3 (2, 3 and 5)
P (getting a prime number) = 3/6 = ½ = 0.5
(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)
P (getting a number between 2 and 6) = 3/6 = ½ = 0.5
(iii) Total number of odd numbers = 3 (1, 3 and 5)
P (getting an odd number) = 3/6 = ½ = 0.5
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(vi) the queen of diamonds
Solution:
Total number of possible outcomes = 52
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total numbers of king of red colour = 2
P (getting a king of red colour) = 2/52 = 1/26 = 0.038
(ii) Total numbers of face cards = 12
P (getting a face card) = 12/52 = 3/13 = 0.23
(iii) Total numbers of red face cards = 6
P (getting a king of red colour) = 6/52 = 3/26 = 0.11
(iv) Total numbers of jack of hearts = 1
P (getting a king of red colour) = 1/52 = 0.019
(v) Total numbers of king of spade = 13
P (getting a king of red colour) = 13/52 = ¼ = 0.25
(vi) Total numbers of queen of diamonds = 1
P (getting a king of red colour) = 1/52 = 0.019
15. Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Total numbers of cards = 5
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Numbers of queen = 1
P (picking a queen) = â…• = 0.2
(ii) If the queen is drawn and put aside, the total numbers of cards left is (5 – 4) = 4
(a) Total numbers of ace = 1
P (picking an ace) = ¼ = 0.25
(b) Total numbers of queen = 0
P (picking a queen) = 0/4 = 0
16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Numbers of pens = Numbers of defective pens + Numbers of good pens
∴ Total number of pens = 132 + 12 = 144 pens
P(E) = (Number of favourable outcomes/ Total number of outcomes)
P(picking a good pen) = 132/144 = 11/12 = 0.916
17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
(i) Numbers of defective bulbs = 4
The total numbers of bulbs = 20
P(E) = (Number of favourable outcomes/ Total number of outcomes)
∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = ⅕ = 0.2
(ii) Since 1 non-defective bulb is drawn, then the total numbers of bulbs left are 19
So, the total numbers of events (or outcomes) = 19
Numbers of defective bulbs = 19 – 4 = 15
So, the probability that the bulb is not defective = 15/19 = 0.789
18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
The total numbers of discs = 50
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total number of discs having two digit numbers = 81
(Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90 – 9 = 81)
P (bearing a two-digit number) = 81/90 = 9/10 = 0.9
(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)
P (getting a perfect square number) = 9/90 = 1/10 = 0.1
(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)
P (getting a number divisible by 5) = 18/90 = â…• = 0.2
19. A child has a die whose six faces show the letters as given below:
The die is thrown once. What is the probability of getting
(i) A?
(ii) D?
Solution:
The total number of possible outcomes (or events) = 6
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) The total number of faces having A on it = 2
P (getting A) = 2/6 = â…“ = 0.33
(ii) The total number of faces having D on it = 1
P (getting D) = â…™ = 0.166
20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?
Solution:
First, calculate the area of the rectangle and the area of the circle. Here, the area of the rectangle is the possible outcome and the area of the circle will be the favourable outcome.
So, the area of the rectangle = (3 × 2) m2 = 6 m2
and,
The area of the circle = πr2 = π(½)2 m2 = π/4 m2 = 0.78
∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 or, 0.78/6 = 0.13
21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(ii) She will not buy it?
Solution:
The total numbers of outcomes i.e. pens = 144
Given, numbers of defective pens = 20
∴ The numbers of non defective pens = 144 – 20 = 124
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total numbers events in which she will buy them = 124
So, P (buying) = 124/144 = 31/36 = 0.86
(ii) Total numbers events in which she will not buy them = 20
So, P (not buying) = 20/144 = 5/36 = 0.138
22. Refer to Example 13. (i) Complete the following table:
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your Solution:.
Solution:
If 2 dices are thrown, the possible events are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, the total numbers of events: 6 × 6 = 36
(i) It is given that to get the sum as 2, the probability is 1/36 as the only possible outcomes = (1,1)
For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)
So, P(sum 3) = 2/36
Similarly,
E (sum 4) = (1,3), (3,1), and (2,2)
So, P (sum 4) = 3/36
E (sum 5) = (1,4), (4,1), (2,3), and (3,2)
So, P (sum 5) = 4/36
E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)
So, P (sum 6) = 5/36
E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)
So, P (sum 7) = 6/36
E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)
So, P (sum 8) = 5/36
E (sum 9) = (3,6), (6,3), (4,5), and (5,4)
So, P (sum 9) = 4/36
E (sum 10) = (4,6), (6,4), and (5,5)
So, P (sum 10) = 3/36
E (sum 11) = (5,6), and (6,5)
So, P (sum 11) = 2/36
E (sum 12) = (6,6)
So, P (sum 12) = 1/36
So, the table will be as:
Event: Sum on 2 dice 2 3 4 5 6 7 8 9 10 11 12 Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
(ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.
23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)
Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)
P (losing the game) = 6/8 = ¾ = 0.75
24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
Outcomes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, the total number of outcome = 6 × 6 = 36
(i) Method 1:
Consider the following events.
A = 5 comes in first throw,
B = 5 comes in second throw
P(A) = 6/36,
P(B) = 6/36 and
P(not B) = 5/6
So, P(notA) = 1 – 6/36 = 5/6
∴ The required probability = 5/6 × 5/6 = 25/36
Method 2:
Let E be the event in which 5 does not come up either time.
So, the favourable outcomes are [36 – (5 + 6)] = 25
∴ P(E) = 25/36
(ii) Number of events when 5 comes at least once = 11 (5 + 6)
∴ The required probability = 11/36
(Page No: 311)
25. Which of the following arguments are correct and which are not correct? Give reasons for your Solution:.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2
Solution:
(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)
So, P (getting two heads) = ¼
and, P (getting one of the each) = 2/4 = ½
∴ This statement is incorrect.
(ii) Since the two outcomes are equally likely, this statement is correct.
Class 10 Maths Chapter 15 Exercise: 15.2 (Page No: 311)
1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
Since there are 5 days and both can go to the shop in 5 ways each so,
The total number of possible outcomes = 5 × 5 = 25
(i) The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)
So, P (both visiting on the same day) = 5/25 = â…•
(ii) The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)
So, P(both visiting on the consecutive days) = 8/25
(iii) P (both visiting on the different days) = 1 – P (both visiting on the same day)
So, P (both visiting on the different days) = 1 – ⅕ = ⅘
2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:
What is the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The table will be as follows:
+ 1 2 2 3 3 6 1 2 3 3 4 4 7 2 3 4 4 5 5 8 2 3 4 4 5 5 8 3 4 5 5 6 6 9 3 4 5 5 6 6 9 6 7 8 8 9 9 12
So, the total number of outcome = 6 × 6 = 36
(i) E (Even) = 18
P (Even) = 18/36 = ½
(ii) E (sum is 6) = 4
P (sum is 6) = 4/36 = 1/9
(iii) E (sum is atleast 6) = 15
P (sum is atleast 6) = 15/36 = 5/12
3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball
is double that of a red ball, determine the number of blue balls in the bag.
Solution:
It is given that the total number of red balls = 5
Let the total number of blue balls = x
So, the total no. of balls = x + 5
P(E) = (Number of favourable outcomes/ Total number of outcomes)
∴ P (drawing a blue ball) = [x/(x + 5)] ——–(i)
Similarly,
P (drawing a red ball) = [5/(x + 5)] ——–(i)
From equation (i) and (ii)
x = 10
So, the total number of blue balls = 10
4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the
box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now
double of what it was before. Find x
Solution:
Total number of black balls = x
Total number of balls = 12
P(E) = (Number of favourable outcomes/ Total number of outcomes)
P (getting black balls) = x / 12 ——————-(i)
Now, when 6 more black balls are added,
Total balls become = 18
∴ Total number of black balls = x + 6
Now, P (getting black balls) = (x + 6)/18 ——————-(i)
Solving equation (i) and (ii)
x = 3
5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at
random from the jar, the probability that it is green is â…”. Find the number of blue balls
in the jar.
Solution:
Total marbles = 24
Let the total green marbles = x
So, the total blue marbles = 24 – x
P(getting green marble) = x/24
From the question, x/24 = â…”
So, the total green marbles = 16
And, the total blue marbles = 24 – x = 8
Students can also access the CBSE Notes for Class 10 Chapter 15.
NCERT Solutions for Class 10 Maths Chapter 15 – Probability
Chapter 15 of Class 10, Probability, belongs to the 6th Unit- Statistics and Probability, that adds up to 10 marks out of the total 80 marks in the examination. Unit 6 is one such unit, in which, marks can be easily scored once the method or the idea gets familiar. Solving the NCERT exercise questions will enable the students in getting thorough with the methods used in solving the problems of this chapter, in turn aiding the students to prepare well for the Class 10 Maths Board examination.
Main topics covered in this chapter include:Â
15.1 Introduction Â
15.2 Probability – A Theoretical Approach
List of Exercises in class 10 Maths Chapter 15
Exercise 15.1 Solutions 25 Questions (1 MCQ, 21 Short Answer Questions, 2Â Long Answer Question, 1 Main Question with 4 Sub-questions)Â
Exercise 15.2 Solutions 5 Questions (5 Short Answer Questions)
Practising these NCERT Solutions for Class 10 is very important from the CBSE board examination point of view. Probability is one of the chapters in Class 10 maths that contain the concepts that are most likely to be used in daily life. The chapter deals with different topics related to the probability that includes:
1. The difference between experimental probability and theoretical probability
2. Why the probability of a sure event (or certain event) is 1.
3. How the probability of an impossible event is 0?
4. Elementary events.
5. Complementary events.
6. Finding the probability of different events.
Key Features of NCERT Solutions for Class 10 Maths Chapter 15 – Probability
• In-depth knowledge of the chapter/problems is provided in easy language.Â
• Is based on the CBSE syllabus.Â
• Best study material to prepare for the Class 10 board exam.
• All the doubts that might arise while solving the exercise questions will vanish by going through these solutions
Frequently Asked Questions on Chapter 15- Probability
Complete the following statements Probability of an event E + Probability of the event ‘not E’ = ___________ ?
Probability of an event E + Probability of the event ‘not E’ = 1.
Which of the following experiments have equally likely outcomes Explain A driver attempts to start a car. The car starts or does not start?
This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out an orange flavoured candy?
We know that the bag only contains lemon-flavoured candies.
So, The no. of orange flavoured candies = 0
∴ The probability of taking out orange flavoured candies = 0/1 = 0
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Let the event wherein 2 students having the same birthday be E
Given, P(E) = 0.992
We know,
P(E) + P(not E) = 1
Or, P(not E) = 1 – 0.992 = 0.008
∴ The probability that the 2 students have the same birthday is 0.008 |
## What is a Reciprocal?
In mathematics, the reciprocal, likewise known as multiplicative inverse, is the train station of a number x. Denoted as 1/x or x-1. This method that the product the a number x and its reciprocal yields 1.
You are watching: Reciprocal of -1/4
The station of a fraction a/b is denoted as (a/b)-1, which is b/a. This article discusses the measures on just how to find the mutual of a number, mixed numbers, fractions and also decimals.
## How to find Reciprocals?
The reciprocal of a number is merely the number that has actually been flipped or reverse upside-down. This involves transposing a number such the the numerator and denominator are put at the bottom and also top respectively.
To find the mutual of a entirety number, just transform it into a portion in i beg your pardon the initial number is the denominator and also the molecule is 1.
Example 1
The mutual of 2/3 is 3/2.
The product that 2/3 and its reciprocal 3/2 is 1.
2/3 x 3/2 = 1
Example 2
The reciprocal of a totality number 7 is 1/7 since 7 x 1/7 = 1.
### How to uncover the reciprocal of a blended Number?
In bespeak to discover the reciprocal of a combined fraction, transform it into improper fraction very first and then use the same rule we learnt above.
Example 3
Find the mutual of 4 1/2.
Solution
Convert a mixed fraction into one improper fraction as calculated below.
4 1/2 = (4 x 2) + 1/ 2
= 9/2
Now flip the numerator and also denominator the 9/2.Therefore, the equipment for the reciprocal of 4 1/2 is 2/9.
### How to find the reciprocal of Decimal Numbers?
Like other numbers, decimal numbers too have actually reciprocals. Calculating the reciprocal of a decimal number can be excellent in the following ways:
Convert the decimal right into an identical fraction, for example, 0.25 = 1/4, and therefore, the mutual is 4/1 = 4.You can likewise use a calculate to division 1 by the fraction. For example, the reciprocal of 0.25 = 1/0.25 = 4
It can be listed that splitting 1 by a fraction is the same as multiplying the reciprocal of the number by 1. For example,
5 ÷ 1/4
= 5 x 4/1
= 20
Example 4
Solve the complying with problems:
a. Find the reciprocal of 5
Solution
5 = 5/1
So, the reciprocal of 3 = 1/5
b. Discover the mutual of 1/4
Solution
To uncover the mutual of 1/4, invert the numerator and denominator.
The reciprocal of 1/4 = 4
c. Determine the mutual of 10/3
Solution
Step 1:
To uncover the mutual of 10/3, upper and lower reversal the numerator and denominator. The reciprocal = 3/10.
Example 5
If 4/7 of a number x is 84. What is the worth of x?
Solution4/7 that a number x = 84Write the mathematical equation:
(4/7) x = 84
Multiply both sides by the mutual of 4/7Number x = 84 × 7/4
= 21 × 7= 147And thus, the number x is 147.
Example 6
A fifty percent of the students in a college space boys, 3/5 that these guys take scientific research courses and the rest take humanities. What portion of the boys take humanities?
Solution
Fraction of boys in the university = 1/2
Fraction that boys that take scientific researches = 3/5 the 1/2
= 3/5 × 1/2
= 3 × 1/5 × 2
= 3/10
Therefore, 3/10 that the boys take humanities.
Example 7
Pedro has written three-fifth that his 75 paged study work. How plenty of pages room left to complete writing his research?
SolutionNumber that pages composed = 3/5 the 75= 3/5 × 7
= 45 pages.Number the pages left= 75 – 45.= 30 pages.
See more: How Many Pounds Of Cold Cuts Per Person, How Much Does 1 Pound Of Lunch Meat Serve
Example 8
A herd of cows in a farm produces 99 liters of milk daily. If each cow to produce one-third of full milk produced in a day. How numerous cows space in the farm if 7700 liters the milk is produced weekly.
SolutionA herd of cows produce 99 liters that milk daily.One cow produces 1/3 of complete milk everyday = 1/3 the 99Therefore, one cow produces 11 liters.Total variety of animals in the farm= (7700/7) ÷ 11= 100 cows |
Last updated on Mar 15, 2024
Latest Distance Formula MCQ Objective Questions
Distance Formula Question 1:
Under which one of the following conditions are the lines ax + by + c = 0 and bx + ay + c = 0 parallel (a ≠ 0, b ≠ 0)?
1. a - b = 0 only
2. a + b = 0 only
3. a2 - b2 = 0
4. More than one of the above
5. None of the above
Option 3 : a2 - b2 = 0
Distance Formula Question 1 Detailed Solution
Concept:
1. Standard Equation of a line y = mx + c, where m is the slope of the line.
2. Slopes of parallel lines are equal. The parallel lines are equally inclined with the positive x-axis and hence the slope of parallel lines are equal. If the slopes of two parallel lines are represented as m1, m2 then we have m1 = m2.
Calculation:
Given:
Lines ax + by + c = 0 and bx + ay + c = 0
Express the two lines in standard format that is y = mx + c
ax + by + c = 0 ⇒ by = -ax - c ⇒ $$\displaystyle y=(\frac{-a}{b})x-\frac{c}{b}$$ ------(i)
bx + ay + c = 0 ⇒ ay = -bx - c ⇒ $$\displaystyle y=(\frac{-b}{a})x-\frac{c}{a}$$ ------(ii)
Given both lines are parallel implies their slopes are equal, that is
⇒ $$\displaystyle \frac{-a}{b}=\frac{-b}{a}$$
⇒ b2 = a2
⇒ a2 - b2 = 0
∴ a2 - b2 = 0 is correct.
Distance Formula Question 2:
(a, 2b) is the mid-point of the line segment joining the points (10, -6) and (k, 4). If a – 2b = 7, then what is the value of k?
1. 2
2. 3
3. 4
4. More than one of the above
5. None of the above
Option 1 : 2
Distance Formula Question 2 Detailed Solution
Concept:
Let A(x1, y1) and B(x2, y2) be any two points on the X-Y plane. Suppose point C is the mid-point of the line segment AB, then the coordinates of point C is:
$$\left( {\frac{{{{\rm{x}}_1} + {{\rm{x}}_2}}}{2},\frac{{{{\rm{y}}_1} + {{\rm{y}}_2}}}{2}} \right)$$.
Calculation:
Mid-point of (10, -6) and (k, 4) :
$$\left( {{\rm{a}},2{\rm{b}}} \right) = \left( {\frac{{10 + {\rm{k}}}}{2},\frac{{ - 6 + 4}}{2}} \right) = \left( {5 + \frac{{\rm{k}}}{2},-1} \right)$$
Therefore,
$${\rm{a}} = \frac{{\rm{k}}}{2} + 5$$
2b = - 1
Given, a – 2b = 7
$$\left( {\frac{{\rm{k}}}{2} + 5{\rm{\;}}} \right) + 1 = 7$$
$$\frac{{\rm{k}}}{2} + 6 = 7$$
$$\frac{{\rm{k}}}{2} = 1$$
k = 2
Distance Formula Question 3:
The distance of point (2, 3) from the origin is.
1. $$\sqrt{10}$$
2. $$\sqrt{12}$$
3. $$\sqrt{13}$$
4. $$\sqrt{11}$$
Option 3 : $$\sqrt{13}$$
Distance Formula Question 3 Detailed Solution
Concept -
The distance between a point (x, y) and the origin (0, 0) in a 2D plane is given by the distance formula:
$$\text{Distance} = \sqrt{x^2 + y^2}$$
Explanation -
For the point (2, 3), the distance from the origin is:
$$\text{Distance} = \sqrt{2^2 + 3^2} \\ \text{Distance} = \sqrt{4 + 9} \\ \text{Distance} = \sqrt{13}$$
Therefore, the distance of the point (2, 3) from the origin is $$\sqrt{13}$$ units.
Distance Formula Question 4:
If the distance between the points (x, -1) and (3, 2) is 5 units, then what will be the value of x.
1. 1
2. -1
3. 7
4. Both (2) and (3)
Option 4 : Both (2) and (3)
Distance Formula Question 4 Detailed Solution
Concept -
The distance between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ in a 2D plane is given by the distance formula:
$$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Explanation -
Here, the points are (x, -1) and (3, 2), and the distance is given as 5 units. Using the distance formula:
$$5 = \sqrt{(3 - x)^2 + (2 - (-1))^2} \\ 5 = \sqrt{(3 - x)^2 + 3^2} \\ 5 = \sqrt{(3 - x)^2 + 9}$$
let's square both sides to eliminate the square root:
$$5^2 = (3 - x)^2 + 9 \\ 25 = (3 - x)^2 + 9 \\ (3 - x)^2 = 25 - 9 \\ (3 - x)^2 = 16$$
Now, take the square root of both sides:
$$3 - x = \pm \sqrt{16} \\ 3 - x = \pm 4$$
Solve for x in both cases:
When 3 - x = 4:
x = 3 - 4
x = -1
When 3 - x = -4:
x = 3 + 4
x = 7
So, the values for x are x = -1 or x = 7.
Distance Formula Question 5:
The distance between the points (a cos θ, 0) and (0, a sin θ) will be:
1. a
2. a cos θ
3. a sin θ
4. 1
Option 1 : a
Distance Formula Question 5 Detailed Solution
Concept -
The distance between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ in a 2D plane can be found using the distance formula:
$$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Explanation -
In this case, the points are (a cos θ, 0) and (0, a sin θ) So, applying the distance formula:
\begin{align*} \text{Distance} &= \sqrt{ (0 - a \cos \theta)^2 + (a \sin \theta - 0)^2 } \\ &= \sqrt{a^2 \cos^2 \theta + a^2 \sin^2 \theta} \\ &= \sqrt{a^2 (\cos^2 \theta + \sin^2 \theta)} \\ &= \sqrt{a^2} \\ &= |a| \end{align*}
Thus, the distance between the points (a cos θ, 0) and (0, a sin θ) is |a|.
Hence the option(1) is true.
Top Distance Formula MCQ Objective Questions
Distance Formula Question 6
The perpendicular distance between the straight lines 6x + 8y + 15 = 0 and 3x + 4y + 9 = 0 is
1. 3/2 units
2. 3/10 unit
3. 3/4 unit
4. 2/7 unit
Option 2 : 3/10 unit
Distance Formula Question 6 Detailed Solution
Concept:
Distance between parallel lines:
• The distance between the lines y = mx + c1 and y = mx + c2 is $$\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {1{\rm{\;}} + {\rm{\;}}{{\rm{m}}^2}} }}$$
• The distance between the lines ax + by + c1 = 0 and ax + by + c2 = 0 is $$\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} }}$$
Calculation:
Given lines are 6x + 8y + 15 = 0 and 3x + 4y + 9 = 0
⇒ 6x + 8y + 15 = 0
Take 2 common from above equation, we get
⇒ 3x + 4y + 15/2 = 0 ---(1)
And 3x + 4y + 9 = 0 ---(2)
Equation 1 and 2 are parallel to each other.
∴ The distance between the lines = $$\frac{{\left| {\frac{{15}}{2}{\rm{\;}} - {\rm{\;}}9} \right|}}{{\sqrt {{3^2} + {4^2}} }} = \frac{{\left( {\frac{3}{2}} \right)}}{5} = \frac{3}{{10}}$$
Alternate MethodParallel lines have the same slope and will never intersect.
Two lines y = m1x + c1
and y = m2x + c2
are said to be parallel if:
m1 = m2
Example :
Line 1: 3x + 4y = 1
Line 2: 3x + 4y = 5
Application:
We have,
Line 1: 6x + 8y + 15 = 0
And, Line 2: 3x + 4y + 9 = 0
Line 2 can also be written as,
6x + 8y + 18 = 0
Since, both the line are parallel, hence its graph will be similar to:
We have,
c2 - c1 = 18 - 15 = 3
a2 + b2 = 62 + 82 = 100
Using formula of distance between straight lines,
D = $$\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}=\frac{|3|}{\sqrt{100}}=\frac{3}{10}$$ units
Distance Formula Question 7
Find the perpendicular distance of the line 3y = 4x + 5 from (2, 1)
1. 1
2. 2
3. 3
4. 4
Option 2 : 2
Distance Formula Question 7 Detailed Solution
Concept:
The distance of a point (x1, y1) from a line ax + by + c = 0
D = $$\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|$$
Calculation:
Given line 3y = 4x + 5
⇒ 4x - 3y + 5 = 0
(x1, y1) = (2, 1)
∴ D = $$\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|$$
⇒ D = $$\rm \left|4\times 2 + (-3)\times 1+5\over\sqrt{4^2+(-3)^2}\right|$$
⇒ D = $$\rm \left|10\over5\right|$$ = 2
Distance Formula Question 8
Point A(10, 5), B(8, 4) and C(6, 6) are vertices of a triangle, then length of median from A is
1. √5 units
2. 2√2 units
3. 4 units
4. 3 units
Option 4 : 3 units
Distance Formula Question 8 Detailed Solution
Concept:
Let A (x1, y1) and B (x2, y2) be the end points of the line AB. C be the mid point of the line AB.
The coordinate of C = $$\rm (\dfrac {x_1+x_2}{2},\dfrac{y_1+y_2}{2})$$
By distance formula AB = $$\rm\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$$
Calculations:
Given, Point A(10, 5), B(8, 4) and C(6, 6) are vertices of a triangle,
Let D be the mid point on the Line BC.
Its coordinates is given by
D = $$\rm (\dfrac {8+6}{2}, \dfrac{4+6}{2})$$
D = $$\rm (7 ,5)$$
Here, then length of median from A = AD = $$\rm \sqrt {(10-7)^2+ (5-5)^2}$$
⇒AD = $$\rm \sqrt {9+0}$$
Point A(10, 5), B(8, 4) and C(6, 6) are vertices of a triangle, then length of median from A is 3
Distance Formula Question 9
The locus of the point (x, y) equidistant from the points (-1, 1) and (3, -2) is:
1. 4x + 2y - 11 = 0
2. 4x - 2y + 11 = 0
3. 8x - 6y - 11 = 0
4. 8x + 6y - 11 = 0
Option 3 : 8x - 6y - 11 = 0
Distance Formula Question 9 Detailed Solution
Given;
Coordinates that are (-1, 1) and (3, -2)
Concept:
The formula when two points are equidistant is-
$$\sqrt {(x-x_{1})^2 + (y-y_{1})^2} = \sqrt {(x-x_{2})^2 + (y-y_{2})^2}$$
Calculation:
Let the locus point be (x, y),
As the locus of the points that are equidistant from two points (-1, 1) and (3, -2), therefore the equation would be-
$$⇒ \sqrt {(x+1)^2 + (y-1)^2} = \sqrt {(x-3)^2 + (y+2)^2}$$
⇒ (x + 1)2 + (y - 1)2 = (x - 3)2 + (y + 2)2
∵ (a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b
⇒ x2 + 2x + 1 + y2 - 2y + 1 = x2 - 6x + 9 + y2 + 4y + 4
⇒ 2x - 2y + 2 = - 6x + 4y + 13
⇒ 8x - 6y - 11 = 0
Hence, the equation is 8x - 6y - 11 = 0
Distance Formula Question 10
What is the perpendicular distance from the point (2, 3, 4) to the line $$\rm \frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0} \ ?$$
1. 2
2. 5
3. 7
4. 4
Option 2 : 5
Distance Formula Question 10 Detailed Solution
Concept:
Dot product of two perpendicular lines is zero.
Distance between two points (x1, y1, z1) and (x2, y2, z2) is given by, $$\rm \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$
Calculation:
Let M be the foot of perpendicular drawn from the point P(2, 3, 4)
Let, $$\rm \dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-0}{0} =k$$
x = k, y = 0, z = 0
So M = (k, 0, 0)
Now direction ratios of PM = (2 - k, 3 - 0, 4 - 0) = (2- k, 3, 4) and direction ratios of given line are 1, 0, 0
PM is perpedicular to the given line so,
(2 - k) (1) + 3(0) + 4 (0) = 0
∴ k = 2
M = (2, 0, 0)
Perpendicular distance PM =
$$\rm \sqrt {(2-2)^2+(0-3)^2+(0-4)^2}\\ =\sqrt{9+16}\\ =5$$
Hence, option (2) is correct.
Distance Formula Question 11
If the distance between the points (3, 4) and (a, 2) is 8 units then find the value of a
1. $$3 \pm 2\sqrt {15}$$
2. $$2\pm 2\sqrt {15}$$
3. $$1 \pm\sqrt {15}$$
4. None of these
Option 1 : $$3 \pm 2\sqrt {15}$$
Distance Formula Question 11 Detailed Solution
CONCEPT:
Let A (x1, y1) and B (x2, y2) be any two points in the XY – plane, then the distance between A and B is given by:$$\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$$
CALCULATION:
Given: The distance between the points (3, 4) and (a, 2) is 8 units
Here, we have to find the value of a.
As we know that, the distance between two points A (x1, y1) and B (x2, y2) is given by $$\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$$
⇒ $$\sqrt {{{\left( {{a} - {3}} \right)}^2} + {{\left( {{2} - {4}} \right)}^2}} = 8$$
By squaring both the sides we get
⇒ (a - 3)2 + 4 = 64
⇒ a2 + 9 - 6a - 60 = 0
⇒ a2 - 6a - 51 = 0
⇒ $$a = \frac{{6 \pm \sqrt {240} }}{2} = 3 \pm 2\sqrt {15}$$
Hence, option A is the correct answer.
Distance Formula Question 12
If the distance between the points (5, - 2) and (1, a) is 5 then find the possible value(s) of a ?
1. -1 and -5
2. -5 and 2
3. 5 and 1
4. 1 and -5
Option 4 : 1 and -5
Distance Formula Question 12 Detailed Solution
CONCEPT:
Let A (x1, y1) and B (x2, y2) be any two points in the XY – plane, then the distance between A and B is given by:$$\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$$
CALCULATION:
Given: The distance between the points (5, - 2) and (1, a) is 5.
Let A = (5, - 2) and B = (1, a)
As we know that, the distance between the points A and B is given by:$$\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$$
Here, x1 = 5, y1 = - 2, x2 = 1 and y2 = a
$$⇒ \left| {AB} \right| = \sqrt {{{\left( {{1} - {5}} \right)}^2} + {{\left( {{a} + {2}} \right)}^2}} = 5$$
By squaring both the sides of the equation we get,
⇒ 25 = 16 + (a + 2)2
⇒ 9 = (a + 2)2 ⇒ (a + 2) = ± 3
Case 1: When (a + 2) = 3 then a = 1
Case 2: When (a + 2) = - 3 then a = - 5
Hence, a = 1, - 5
Distance Formula Question 13
Find the distance between the parallel lines 3y + 4x - 12 = 0 and 3y + 4x - 7 = 0.
1. 1
2. 2
3. 3
4. 4
Option 1 : 1
Distance Formula Question 13 Detailed Solution
Concept:
The distance between the parallel lines ax + by + c1 and ax + by + c2 is:
D = $$\rm \left|c_1-c_2\over\sqrt{a^2+b^2}\right|$$
Calculation;
The 2 given lines are:
3y + 4x - 12 = 0
3y + 4x - 7 = 0
a = 4, b = 3, c1 = -12 and c2 = -7
∴ The distance between the lines
D = $$\rm \left|c_1-c_2\over\sqrt{a^2+b^2}\right|$$
⇒ D = $$\rm \left|-12-(-7)\over\sqrt{4^2+3^2}\right|$$
⇒ D$$\rm \left|-5\over5\right|$$ = 1
Distance Formula Question 14
(a, 2b) is the mid-point of the line segment joining the points (10, -6) and (k, 4). If a – 2b = 7, then what is the value of k?
1. 2
2. 3
3. 4
4. 5
Option 1 : 2
Distance Formula Question 14 Detailed Solution
Concept:
Let A(x1, y1) and B(x2, y2) be any two points on the X-Y plane. Suppose point C is the mid-point of the line segment AB, then the coordinates of point C is:
$$\left( {\frac{{{{\rm{x}}_1} + {{\rm{x}}_2}}}{2},\frac{{{{\rm{y}}_1} + {{\rm{y}}_2}}}{2}} \right)$$.
Calculation:
Mid-point of (10, -6) and (k, 4) :
$$\left( {{\rm{a}},2{\rm{b}}} \right) = \left( {\frac{{10 + {\rm{k}}}}{2},\frac{{ - 6 + 4}}{2}} \right) = \left( {5 + \frac{{\rm{k}}}{2},-1} \right)$$
Therefore,
$${\rm{a}} = \frac{{\rm{k}}}{2} + 5$$
2b = - 1
Given, a – 2b = 7
$$\left( {\frac{{\rm{k}}}{2} + 5{\rm{\;}}} \right) + 1 = 7$$
$$\frac{{\rm{k}}}{2} + 6 = 7$$
$$\frac{{\rm{k}}}{2} = 1$$
k = 2
Distance Formula Question 15
If the foot of the perpendicular drawn from the point (0, k) to the line 3x - 4y - 5 = 0 is (3, 1), then what is the value of k?
1. 3
2. 4
3. 5
4. 6
Option 3 : 5
Distance Formula Question 15 Detailed Solution
Concept:
If two nonvertical lines are perpendicularthen the product of their slopes is −1.
The slope of a line passing through the distinct points (x1, y1) and (x2, y2) is $$\rm \frac{y_2-y_1}{x_2-x_1}$$
Calculation:
Slope of line passing through points (0, k) and (3, 1)
$$=\rm \frac{1-k}{3-0} \\=\frac{1-k}{3}$$
3x - 4y - 5 = 0
⇒4y = 3x - 5
⇒ y = $$\frac{3}{4}\rm x-\frac{5}{4}$$
So, the slope of line 3x - 4y - 5 = 0 is 3/4
Now since line OP and 3x - 4y - 5 = 0 are perepndicular
$$\rm \frac{1-k}{3}\times \frac{3}{4}=-1.....(\text {Product of slopes of perpendicular lines} )\\ \Rightarrow 1-k=-4\\ \Rightarrow k =5$$
Hence, option (3) is correct. |
# Interior and Exterior Angles
GCSEKS3Level 4-5AQAEdexcelEdexcel iGCSEOCRWJEC
## Interior and Exterior Angles
The interior angles of a shape are the angles inside the shape.
The exterior angles are the angles formed between a side-length and an extension.
Rule: Interior and exterior angles add up to $180\degree$.
Having the ability to rearrange equations will help with interior and exterior angle questions.
Level 4-5GCSEKS3AQAEdexcelOCRWJECEdexcel iGCSE
## Exterior Angles
Rule: The exterior angle = $\dfrac{360\degree}{\textcolor{red}{n}}$
where $\textcolor{red}{n}$ is the number of sides.
The sum of all the exterior angles will equal $360\degree$.
For the triangle shown, we can see it has $\textcolor{red}{3}$ sides, so to calculate an exterior angle we do:
$\dfrac{360\degree}{\textcolor{red}{3}} = 120\degree$
Level 4-5GCSEKS3AQAEdexcelOCRWJECEdexcel iGCSE
## Interior Angles
Rule: Sum of interior angles = $(\textcolor{red}{n} - 2) \times 180\degree$
Where $\textcolor{red}{n}$ is the number of sides.
To find the sum of the interior angles for the triangle shown we do the following:
$(\textcolor{red}{3} - 2) \times 180\degree = 180\degree$
This means that
$\textcolor{limegreen}{a} + \textcolor{limegreen}{b} + \textcolor{limegreen}{c} = 180\degree$
Note: You can find the interior angle of a regular polygon by dividing the sum of the angles by the number of angles. You can also find the exterior angle first then minus from $180\degree$ to get the interior angle.
Level 4-5GCSEKS3AQAEdexcelOCRWJECEdexcel iGCSE
## Example: Finding Interior and Exterior Angles
$ABCD$ is a quadrilateral.
Find the missing angle marked $x$.
[2 marks]
This is a $4$-sided shape, to work out the interior angles we calculate the following:
$(\textcolor{red}{n}-2)\times 180 =360\degree$.
Next we can work out the size of $\angle CDB$ as angles on a straight line add up to $180\degree$.
$180 - 121 = 59\degree$
Now we know the other $3$ interior angles, we get that
$x = 360 - 84 - 100 - 59 = 117\degree$
Level 4-5GCSEKS3AQAEdexcelOCRWJECEdexcel iGCSE
## Interior and Exterior Angles Example Questions
Question 1: The shape below is a regular pentagon.
Work out the size of the interior angle, $x$.
[2 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECEdexcel iGCSE
This shape has 5 sides, so its interior angles add up to,
$180 \times (5 - 2) = 540\degree$
Hence each interior angle is,
$x\degree=540\degree \div 5 = 108\degree$
Gold Standard Education
Question 2: The shape below is a regular octagon.
Work out the size of the interior angle, $x$.
[2 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECEdexcel iGCSE
This shape has 8 sides, so its interior angles add up to,
$180 \times (8 - 2) = 1080\degree$
Hence each interior angle is,
$x\degree=1080\degree \div 8 = 135\degree$
Gold Standard Education
Question 3: $ABCDE$ is a pentagon.
Work out the size of $x$.
[3 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECEdexcel iGCSE
This shape has 5 sides, so its interior angles must add up to
$180 \times (5 - 2) = 540\degree$.
We can’t find this solution with one calculation as we did previously, but we can express the statement “the interior angles add up to 540” as an equation. This looks like
$33 + 140 + 2x + x + (x + 75) = 540$
Now, this is a linear equation we can solve. Collecting like terms on the left-hand side, we get
$4x + 248 = 540$.
Subtract 248 from both sides to get
$4x = 292$.
Finally, divide by 4 to get the answer:
$x = 292 \div 4 = 73\degree$
Gold Standard Education
Question 4: $ABCD$ is a quadrilateral.
Work out the size of $y$.
[4 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECEdexcel iGCSE
This shape has 4 sides, so its interior angles add up to
$180 \times (4 - 2) = 360\degree$.
We don’t have any way of expression two of the interior angles at the moment, but we do have their associated exterior angles, and we know that interior plus exterior equals 180. So, we get
$\text{interior angle CDB } = 180 - (y + 48) = 132 - y$
Furthermore, we get
$\text{interior angle CAB } = 180 - 68 = 112$
Now we have figures/expressions for each interior angle, so we write the sum of them equal to 360 in equation form:
$112 + 90 + 2y + (132 - y) = 360$
Collecting like terms on the left-hand side, we get
$y + 334 = 360$
Then, if we subtract 334 from both sides we get the answer to be
$y = 360 - 334 = 26\degree$.
Gold Standard Education
## Interior and Exterior Angles Worksheet and Example Questions
### (NEW) Interior and Exterior Angles Exam Style Questions - MME
Level 4-5GCSENewOfficial MME
Level 4-5GCSE |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Altitudes
## Height of a triangle or the line segment from a vertex and perpendicular to the opposite side.
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Altitudes
What if you were given one or more of a triangle's angle measures? How would you determine where the triangle's altitude would be found? After completing this Concept, you'll be able to answer this type of question.
### Guidance
In a triangle, a line segment from a vertex and perpendicular to the opposite side is called an altitude . It is also called the height of a triangle. The red lines below are all altitudes.
When a triangle is a right triangle, the altitude, or height, is the leg. If the triangle is obtuse, then the altitude will be outside of the triangle. If the triangle is acute, then the altitude will be inside the triangle.
#### Example A
Which line segment is the altitude of $\triangle ABC$ ?
Solution: In a right triangle, the altitude, or the height, is the leg. If we rotate the triangle so that the right angle is in the lower left corner, we see that leg $BC$ is the altitude.
#### Example B
A triangle has angles that measure $55^\circ, 60^\circ,$ and $65^\circ$ . Where will the altitude be found?
Solution: Because all of the angle measures are less than $90^\circ$ , the triangle is an acute triangle. The altitude of any acute triangle is inside the triangle.
#### Example C
A triangle has an angle that measures $95^\circ$ . Where will the altitude be found?
Solution: Because $95^\circ > 90^\circ$ , the triangle is an obtuse triangle. The altitude of any obtuse triangle is outside the triangle.
### Guided Practice
1. True or false: The altitudes of an obtuse triangle are inside the triangle.
2. Draw the altitude for the triangle shown.
Solution: The triangle is an acute triangle, so the altitude is inside the triangle as shown below so that it is perpendicular to the base.
3. Draw the altitude for the triangle shown.
Solution: The triangle is a right triangle, so the altitude is already drawn. The altitude is $\overline{XZ}$ .
1. Every triangle has three altitudes. For an obtuse triangle, at least one of the altitudes will be outside of the triangle, as shown in the picture at the beginning of this concept.
2.
3.
### Practice
Given the following triangles, tell whether the altitude is inside the triangle, outside the triangle, or at the leg of the triangle.
1. $\triangle JKL$ is an equiangular triangle.
2. $\triangle MNO$ is a triangle in which two the angles measure $30^\circ$ and $60^\circ$ .
3. $\triangle PQR$ is an isosceles triangle in which two of the angles measure $25^\circ$ .
4. $\triangle STU$ is an isosceles triangle in which two angles measures $45^\circ$ .
Given the following triangles, which line segment is the altitude?
### Vocabulary Language: English Spanish
altitude
altitude
An altitude of a triangle is a line segment from a vertex and is perpendicular to the opposite side. It is also called the height of a triangle. |
Home Contact KS2 Maths GCSE EFL Advice Parents Games Other
GCSE Maths > Number - Percentages
A percentage is a fraction whose denominator is 100 (the numerator of a fraction is the top term, the denominator is the bottom term).
So 30% = 30/100 = 3/10 = 0.3
To change a decimal into a percentage, multiply by 100. So 0.3 = 0.3 × 100 = 30% .
#### Example
Find 25% of 10 (remember 'of' means 'times').
25 × 10 (divide by 100 to convert the percentage to a decimal)
100
= 2.5
### Percentage Change
% change = new value - original value × 100
original value
#### Example
The price of some apples is increased from 48p to 67p. By how much percent has the price increased by?
% change = 67 - 48 × 100 = 39.58%
48
### Percentage Error
% error = error × 100
real value
#### Example
Nicola measures the length of her textbook as 20cm. If the length is actually 17.6cm, what is the percentage error in Nicola's calculation?
% error = 20 - 17.6 × 100 = 13.64%
17.6
### Original value
Original value = New value × 100
100 + %change
#### Example
A dealer buys a stamp collection and sells it for £2700, making a 35% profit. Find the cost of the collection.
It is the original value we wish to find, so the above formula is used.
2700 × 100 = £2000
100 + 35
### Percentage Increases and Interest
New value = 100 + percentage increase × original value
100
#### Example
£500 is put in a bank where there is 6% per annum interest. Work out the amount in the bank after 1 year.
In other words, the old value is £500 and it has been increased by 6%.
Therefore, new value = 106/100 × 500 = £530 .
### Compound Interest
If in this example, the money was left in the bank for another year, the £530 would increase by 6%. The interest, therefore, will be higher than the previous year (6% of £530 is more than 6% of £500). Every year, if the money is left sitting in the bank account, the amount of interest paid would increase each year. This phenomenon is known as compound interest.
The simple way to work out compound interest is to multiply the money that was put in the bank by nm, where n is (100 + percentage increase)/100 and m is the number of years the money is in the bank for, i.e:
(100 + %change)no of years × original value
So if the £500 had been left in the bank for 9 years, the amount would have increased to:
500 × (1.06)9 = £845
### Percentage decreases
New value = 100 - percentage decrease × original value
100
#### Example
At the end of 1993 there were 5000 members of a certain rare breed of animal remaining in the world. It is predicted that their number will decrease by 12% each year. How many will be left at the end of 1995?
At the end of 1994, there will be (100 - 12)/100 × 5000 = 4400
At the end of 1995, there will be 88/100 × 4400 = 3872
The compound interest formula above can also be used for percentage decreases. So after 4 years, the number of animals left would be:
5000 x [(100-12)/100]4 = 2998 |
# Which term of the sequence is 95?
Home › Uncategorized › Which term of the sequence is 95?
Hence 25th term is 95 of the series -1, 3, 7, 11, ……
## What is the nth term of and?
The ‘nth’ term is a formula with ‘n’ in it which enables you to find any term of a sequence without having to go up from one term to the next. ‘n’ stands for the term number so to find the 50th term we would just substitute 50 in the formula in place of ‘n’.
## What is the nth term for 1?
Finding the nth term The 1st term is 1, the 2nd term is 4, the 3rd term is 7 etc.
## What is the example of nth term?
Finding the nth Term of an Arithmetic Sequence Given an arithmetic sequence with the first term a1 and the common difference d , the nth (or general) term is given by an=a1+(n−1)d . Example 1: Find the 27th term of the arithmetic sequence 5,8,11,54,… . a8=60 and a12=48 .
## What is the term in a sequence?
A sequence is a list of numbers in a certain order. Each number in a sequence is called a term . Each term in a sequence has a position (first, second, third and so on). For example, consider the sequence {5,15,25,35,…} In the sequence, each number is called a term.
## What is a term in an arithmetic sequence?
A Sequence is a set of things (usually numbers) that are in order. Each number in the sequence is called a term (or sometimes “element” or “member”), read Sequences and Series for more details.
## Which term of the AP 137 is 95?
Here a and d are the first term and common difference respectively of this given AP. So, the value of n is 25. Hence, 95 occurs at the 25th place in the given AP.
## What is the next term in the sequence 58?
you have to add 3 to the first of the two numbers. for example 5+3=8 so the answer would be 58, and because you’re decreasing each number by 10 you would have to add 1+3=4 so tha answer must be 14.
## What is first term in arithmetic sequence?
An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. As with any recursive formula, the initial term of the sequence must be given. An explicit formula for an arithmetic sequence with common difference d is given by an=a1+d(n−1) a n = a 1 + d ( n − 1 ) .
## What is the sum of an arithmetic sequence?
The sum of the first n terms in an arithmetic sequence is (n/2)⋅(a₁+aₙ). It is called the arithmetic series formula.
## What is the geometric mean of 1 and 256?
Answer Expert Verified The 3rd geometric means between 1 to 256 is 64.
## What is the arithmetic average called?
The arithmetic mean is often known simply as the mean. It is an average, a measure of the centre of a set of data. The arithmetic mean is calculated by adding up all the values and dividing the sum by the total number of values.
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Which term of the sequence -1 ,3 ,7,11,……. is 95? #arithmeticprogression #ncert |
# Eureka Math Grade 8 Module 7 Lesson 16 Answer Key
## Engage NY Eureka Math 8th Grade Module 7 Lesson 16 Answer Key
### Eureka Math Grade 8 Module 7 Lesson 16 Classwork Answer Key
Classwork
Proof of the Converse of the Pythagorean Theorem
→ What do we know or not know about each of these triangles?
In the first triangle, ABC, we know that a2 + b2 = c2. We do not know if angle C is a right angle.
In the second triangle, A’B’C’, we know that it is a right triangle.
→ What conclusions can we draw from this?
By applying the Pythagorean theorem to △A’B’C’, we get |A’B’|2 = a2 + b2. Since we are given
c2 = a2 + b2, then by substitution, |A’B’|2 = c2, and then |A’B’| = c. Since c is also |AB|, then |A’B’| = |AB|. That means that both triangles have sides a, b, and c that are the exact same lengths.
→ Recall that we would like to prove that ∠ACB is a right angle, that it maps to ∠A’ C’ B’. If we can translate
△ABC so that A goes to A’, B goes to B’, and C goes to C’, it follows that all three angles in the triangle will match. In particular, that ∠ACB maps to the right angle ∠A’ C’ B’, and so is a right angle, too.
→ We can certainly perform a translation that takes B to B’ and C to C’ because segments BC and B’C’ are the same length. Must this translation take A to A’? What goes wrong mathematically if it misses and translates to a different point A” as shown below?
In this picture, we’ve drawn A” to the left of $$\overline{A{\prime} C{\prime}}$$. The reasoning that follows works just as well for a picture with A” to the right of $$\overline{A{\prime} C{\prime}}$$ instead.
Provide time for students to think of what may go wrong mathematically. If needed, prompt them to notice the two isosceles triangles in the diagram, △A”C’A’ and △A”B’A’ and the four angles w1,w2,w3,w4 labeled as shown in the diagram below.
△A”C’A’ is isosceles and therefore has base angles that are equal in measure:
w1 + w2 = w3.
△A”B’A’ is isosceles and therefore has base angles that are equal in measure:
w2 = w3 + w4.
These two equations give w1 + w3 + w4 = w3, which is equal to w1 + w4 = 0, which is obviously not true.
Therefore, the translation must map A to A’, and since translations preserve the measures of angles, we can conclude that the measure of ∠ACB is equal to the measure of ∠A’C’B’, and ∠ACB is a right angle.
Finally, if a triangle has side lengths of a,b and c, with c the longest length, that don’t satisfy the equation a2 + b2 = c2, then the triangle cannot be a right triangle.
### Eureka Math Grade 8 Module 7 Lesson 16 Exercise Answer Key
Exercises 1–7
Exercise 1.
Is the triangle with leg lengths of 3 mi. and 8 mi. and hypotenuse of length $$\sqrt{73}$$ mi. a right triangle? Show your work, and answer in a complete sentence.
32 + 82 = ($$\sqrt{73}$$)2
9 + 64 = 73
73 = 73
Yes, the triangle with leg lengths of 3 mi. and 8 mi. and hypotenuse of length $$\sqrt{73}$$ mi. is a right triangle because it satisfies the Pythagorean theorem.
Exercise 2.
What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let c in. represent the length of the hypotenuse of the triangle.
12 + 42 = c2
1 + 16 = c2
17 = c2
$$\sqrt{17}$$ = c
4.1≈c
The length of the hypotenuse of the right triangle is exactly $$\sqrt{17}$$ inches and approximately 4.1 inches.
Exercise 3.
What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let c mm represent the length of the hypotenuse of the triangle.
22 + 62 = c2
4 + 36 = c2
40 = c2
$$\sqrt{40}$$ = c
$$\sqrt{2^{3}}$$×$$\sqrt{5}$$ = c
$$\sqrt{2^{2}}$$×$$\sqrt{2}$$×$$\sqrt{5}$$ = c
2$$\sqrt{10}$$ = c
The length of the hypotenuse of the right triangle is exactly 2$$\sqrt{10}$$ mm and approximately 6.3 mm.
Exercise 4.
Is the triangle with leg lengths of 9 in. and 9 in. and hypotenuse of length $$\sqrt{175}$$ in. a right triangle? Show your work, and answer in a complete sentence.
92 + 92 = ($$\sqrt{175}$$)2
81 + 81 = 175
162 ≠ 175
No, the triangle with leg lengths of 9 in. and 9 in. and hypotenuse of length $$\sqrt{175}$$ in. is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
Exercise 5.
Is the triangle with leg lengths of √(28 ) cm and 6 cm and hypotenuse of length 8 cm a right triangle? Show your work, and answer in a complete sentence.
($$\sqrt{28}$$)2 + 62 = 82
28 + 36 = 64
64 = 64
Yes, the triangle with leg lengths of $$\sqrt{28}$$ cm and 6 cm and hypotenuse of length 8 cm is a right triangle because the lengths satisfy the Pythagorean theorem.
Exercise 6.
What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence.
Let c ft. represent the length of the hypotenuse of the triangle.
32 + ($$\sqrt{27}$$)2 = c2
9 + 27 = c2
36 = c2
$$\sqrt{36}$$ = $$\sqrt{c^{2}}$$
6 = c
The length of the hypotenuse of the right triangle is 6 ft.
Exercise 7.
The triangle shown below is an isosceles right triangle. Determine the length of the legs of the triangle. Show your work, and answer in a complete sentence.
Let x cm represent the length of each of the legs of the isosceles triangle.
x2 + x2 = ($$\sqrt{18}$$)2
2x2 = 18
$$\frac{2 x^{2}}{2}$$ = $$\frac{18}{2}$$
x2 = 9
$$\sqrt{x^{2}}$$ = $$\sqrt{9}$$
x = 3
The leg lengths of the isosceles triangle are 3 cm.
### Eureka Math Grade 8 Module 7 Lesson 16 Problem Set Answer Key
Question 1.
What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let c cm represent the length of the hypotenuse of the triangle.
12 + 12 = c2
1 + 1 = c2
2 = c2
$$\sqrt{2}$$ = $$\sqrt{c^{2}}$$
1.4≈c
The length of the hypotenuse is exactly $$\sqrt{2}$$ cm and approximately 1.4 cm.
Question 2.
What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let x ft. represent the unknown length of the triangle.
72 + x2 = 112
49 + x2 = 121
49 – 49 + x2 = 121 – 49
x2 = 72
$$\sqrt{x^{2}}$$ = $$\sqrt{72}$$
x = $$\sqrt{2^{2}}$$ ⋅ $$\sqrt{2}$$ ⋅ $$\sqrt{3^{2}}$$
x = 6$$\sqrt{2}$$
x≈8.5
The length of the unknown side of the triangle is exactly 6$$\sqrt{2}$$ ft. and approximately 8.5 ft.
Question 3.
Is the triangle with leg lengths of $$\sqrt{3}$$ cm and 9 cm and hypotenuse of length $$\sqrt{84}$$ cm a right triangle? Show your work, and answer in a complete sentence.
($$\sqrt{3}$$)2 + 92 = ($$\sqrt{84}$$)2
3 + 81 = 84
84 = 84
Yes, the triangle with leg lengths of $$\sqrt{3}$$ cm and 9 cm and hypotenuse of length $$\sqrt{84}$$ cm is a right triangle because the lengths satisfy the Pythagorean theorem.
Question 4.
Is the triangle with leg lengths of $$\sqrt{7}$$ km and 5 km and hypotenuse of length $$\sqrt{48}$$ km a right triangle? Show your work, and answer in a complete sentence.
($$\sqrt{7}$$)2 + 52 = ($$\sqrt{48}$$)2
7 + 25 = 48
32 ≠ 48
No, the triangle with leg lengths of $$\sqrt{7}$$ km and 5 km and hypotenuse of length $$\sqrt{48}$$ km is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
Question 5.
What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let c mm represent the length of the hypotenuse of the triangle.
52 + 102 = c2
25 + 100 = c2
125 = c2
$$\sqrt{125}$$ = $$\sqrt{c^{2}}$$
$$\sqrt{5^{3}}$$ = c
$$\sqrt{5^{2}}$$×$$\sqrt{5}$$ = c
5$$\sqrt{5}$$ = c
11.2≈c
The length of the hypotenuse is exactly 5$$\sqrt{5}$$ mm and approximately 11.2 mm.
Question 6.
Is the triangle with leg lengths of 3 and 6 and hypotenuse of length $$\sqrt{45}$$ a right triangle? Show your work, and answer in a complete sentence.
32 + 62 = ($$\sqrt{45}$$)2
9 + 36 = 45
45 = 45
Yes, the triangle with leg lengths of 3 and 6 and hypotenuse of length $$\sqrt{45}$$ is a right triangle because the lengths satisfy the Pythagorean theorem.
Question 7.
What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let x in. represent the unknown side length of the triangle.
22 + x2 = 82
4 + x2 = 64
4 – 4 + x2 = 64 – 4
x2 = 60
$$\sqrt{x^{2}}$$ = $$\sqrt{60}$$
x = $$\sqrt{2^{2}}$$ ⋅ $$\sqrt{3}$$ ⋅ $$\sqrt{5}$$
x = 2$$\sqrt{15}$$
x≈7.7
The length of the unknown side of the triangle is exactly 2$$\sqrt{15}$$ inches and approximately 7.7 inches.
Question 8.
Is the triangle with leg lengths of 1 and $$\sqrt{3}$$ and hypotenuse of length 2 a right triangle? Show your work, and answer in a complete sentence.
12 + ($$\sqrt{3}$$)2 = 22
1 + 3 = 4
4 = 4
Yes, the triangle with leg lengths of 1 and $$\sqrt{3}$$ and hypotenuse of length 2 is a right triangle because the lengths satisfy the Pythagorean theorem.
Question 9.
Corey found the hypotenuse of a right triangle with leg lengths of 2 and 3 to be $$\sqrt{13}$$. Corey claims that since $$\sqrt{13}$$ = 3.61 when estimating to two decimal digits, that a triangle with leg lengths of 2 and 3 and a hypotenuse of 3.61 is a right triangle. Is he correct? Explain.
No, Corey is not correct.
22 + 32 = (3.61)2
4 + 9 = 13.0321
13 ≠ 13.0321
No, the triangle with leg lengths of 2 and 3 and hypotenuse of length 3.61 is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
Question 10.
Explain a proof of the Pythagorean theorem.
Consider having students share their proof with a partner while their partner critiques their reasoning. Accept any of the three proofs that students have seen.
Question 11.
Explain a proof of the converse of the Pythagorean theorem.
Consider having students share their proof with a partner while their partner critiques their reasoning. Accept either of the proofs that students have seen.
### Eureka Math Grade 8 Module 7 Lesson 16 Exit Ticket Answer Key
Question 1.
Is the triangle with leg lengths of 7 mm and 7 mm and a hypotenuse of length 10 mm a right triangle? Show your work, and answer in a complete sentence.
72 + 72 = 102
49 + 49 = 100
98≠100
No, the triangle with leg lengths of 7 mm and 7 mm and hypotenuse of length 10 mm is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
Question 2.
What would the length of the hypotenuse need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
Let c mm represent the length of the hypotenuse.
Then,
72 + 72 = c2
49 + 49 = c2
98 = c2
$$\sqrt{98}$$ = c
The hypotenuse would need to be $$\sqrt{98}$$ mm for the triangle with sides of 7 mm and 7 mm to be a right triangle.
Question 3.
If one of the leg lengths is 7 mm, what would the other leg length need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
Let a mm represent the length of one leg.
Then,
a2 + 72 = 102
a2 + 49 = 100
a2 + 49 – 49 = 100 – 49
a2 = 51
a = $$\sqrt{51}$$
The leg length would need to be $$\sqrt{51}$$ mm so that the triangle with one leg length of 7 mm and the hypotenuse of 10 mm is a right triangle.
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# How do you solve 5/x - 2 = 2/(x+3)?
Oct 20, 2015
${x}_{1 , 2} = \frac{- 3 \pm \sqrt{129}}{4}$
#### Explanation:
$\frac{5}{x} - 2 = \frac{2}{x + 3}$
Right from the start, you know that $x$ cannot take a value that would make the two denominators equal to zero. This means that you need
$x \ne 0 \text{ }$ and $\text{ } x + 3 \ne 0 \implies x \ne - 3$
With this in mind, use the common denominator $x \cdot \left(x + 3\right)$ to get rid of the denominators. Multiply the first fraction by $1 = \frac{x + 3}{x + 3}$, multiply $2$ by $1 = \frac{x \left(x + 3\right)}{x \left(x + 3\right)}$, and the second fraction by $1 = \frac{x}{x}$ to get
$\frac{5}{x} \cdot \frac{x + 3}{x + 3} - 2 \cdot \frac{x \left(x + 3\right)}{x \left(x + 3\right)} = \frac{2}{x + 3} \cdot \frac{x}{x}$
This is equivalent to
$\frac{5 \cdot \left(x + 3\right)}{x \left(x + 3\right)} - \frac{2 x \left(x + 3\right)}{x \left(x + 3\right)} = \frac{2 x}{x \left(x + 3\right)}$
You can thus say that
$5 \left(x + 3\right) - 2 x \left(x + 3\right) = 2 x$
Expand the parantheses to get
$5 x + 15 - 2 {x}^{2} - 6 x = 2 x$
Rearrange by getting all the terms on one side of the equation
$2 {x}^{2} + 3 x - 15 = 0$
Use the quadratic formula to find the two solutions
${x}_{1 , 2} = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot 2 \cdot \left(- 15\right)}}{2 \cdot 2}$
${x}_{1 , 2} = \frac{- 3 \pm \sqrt{129}}{4}$
The two solutions to the original equation will thus be
${x}_{1} = \frac{- 3 - \sqrt{129}}{4} \text{ }$ and $\text{ } {x}_{2} = \frac{- 3 + \sqrt{129}}{4}$
Do a quick check to make sure that the calculations are correct - I'll do them for $x = \frac{- 3 - \sqrt{129}}{4}$
$\frac{5}{\frac{- 3 - \sqrt{129}}{4}} - 2 = \frac{2}{\frac{- 3 - \sqrt{129}}{4} + 3}$
$\frac{20}{- 3 - \sqrt{129}} = \frac{8}{9 - \sqrt{129}} + 2$
$- 1.39296944 \ldots = - 3.39296944 \ldots + 2 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$ |
RD Sharma Solutions for Class 10 Math Chapter 17 Sample Papers are provided here with simple step-by-step explanations. These solutions for Sample Papers are extremely popular among class 10 students for Math Sample Papers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma Book of class 10 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma Solutions. All RD Sharma Solutions for class 10 Math are prepared by experts and are 100% accurate.
#### Question 1:
Which of the following numbers has terminating decimal expansion?
(a) $\frac{37}{45}$
(b) $\frac{21}{{2}^{3}{5}^{6}}$
(c) $\frac{17}{49}$
(d) $\frac{89}{{2}^{2}{3}^{2}}$
Here we have to check terminating decimal expansion.
We know that if the numerator can be written in the form where m and n are non negative positive integer then the fraction will surely terminate. We proceed as follows to explain the above statement
Hence the correct option is (b).
#### Question 2:
The value of p for which the polynomial x3 + 4x2 − px + 8 is exactly divisible by (x−2) is
(a) 0
(b) 3
(c) 5
(d) 16
Here the given polynomial is
We have to find the value of p such that the polynomial is exactly divisible by
First we have to write equation in basic format of divisibility like this
After solving, we have seeing here the reminder is = 16xpx
So, to find the value of p we put the reminder is equal to zero.
Therefore
Hence option (d) is correct.
#### Question 3:
Δ ABC and Δ PQR are similar triangles such that ∠A = 32° and ∠R = 65°. Then, ∠B is
(a) 83°
(b) 32°
(c) 65°
(d) 97°
It is given that there are two similar triangles, ΔABC and in which A = 32° and R = 65°, then we have to find B
We have following two similar triangles.
We know the relation between angles in the two similar triangles and these are
In we have,
Hence the correct option is
#### Question 4:
In Fig. 1, the value of the median of the data using the graph of less than ogive and more than ogive is
(a) 5
(b) 40
(c) 80
(d) 15
Here we have to find the median from the given graph.
The following graph is given.
From the given graph we can easily see that the total cumulative frequency N = 80.
First we draw a line from the point on the less than ogive graph which is parallel to the x-axis and draw a line perpendicular to the x-axis from that point, which meet the x-axis at x = 15 which is the desired median.
Hence the correct option is (d)
#### Question 5:
For what value of k will the following system of linear equations has no solution:
3x + y = 1
(2k − 1)x + (k − 1)y = 2k + 1
The given system of equations is
3x + y = 1
(2k − 1)x + (k − 1)y = 2k + 1
Here, a1 = 3, b1 = 1, c1 = 1
a2 = 2k − 1, b2 = k − 1, c2 = 2k + 1
The given system of linear equations has no solution.
Now,
$\frac{3}{2k-1}=\frac{1}{k-1}\phantom{\rule{0ex}{0ex}}⇒3k-3=2k-1\phantom{\rule{0ex}{0ex}}⇒3k-2k=-1+3\phantom{\rule{0ex}{0ex}}⇒k=2$
When k = 2,
$\frac{1}{k-1}=\frac{1}{2-1}=1$ and $\frac{1}{2k+1}=\frac{1}{2×2+1}=\frac{1}{5}$
Thus, for k = 2, $\frac{1}{k-1}\ne \frac{1}{2k+1}$
Hence, the given system of linear equations will have no solution when k = 2.
#### Question 6:
If the mean of the following distribution is 54, find the value of p:
Class : 0−20 20−40 40−60 60−80 80−100 Frequency : 7 p 10 9 13
Consider the table given below:
Class Interval Frequency(fi) Class Mark(xi) fi xi 0–20 7 10 70 20–40 p 30 30p 40–60 10 50 500 60–80 9 70 630 80–100 13 90 1170 $\sum _{}^{}{f}_{i}=39+p$ $\sum _{}^{}{f}_{i}{x}_{i}=2370+30p$
It given that the mean of the distribution is 54.
$\therefore \mathrm{Mean}=\frac{\sum _{}^{}{f}_{i}{x}_{i}}{\sum _{}^{}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒\frac{2370+30p}{39+p}=54\phantom{\rule{0ex}{0ex}}⇒2370+30p=2106+54p\phantom{\rule{0ex}{0ex}}⇒54p-30p=2370-2106$
$⇒24p=264\phantom{\rule{0ex}{0ex}}⇒p=\frac{264}{24}=11$
Hence, the value of p is 11.
#### Question 7:
The [HCF × LCM] for the numbers 50 and 20 is
(a) 10
(b) 100
(c) 1000
(d) 50
Here we have to find of 50 and 20.
We know the product of HCF and LCM of two numbers a and b is the product of a and b. Therefore of 50 and 20 is as follow
Hence the correct option is
#### Question 8:
The value of k for which the pair of linear equation 4x + 6y − 1 = 0 and 2x + ky − 7 = 0 represent parallel lines is
(a) k = 3
(b) k = 2
(c) k = 4
(d) k = −2
Given that the pair of linear equation
It is given that the pair of equations represent parallel lines.
Hence the option (a) is correct.
#### Question 9:
If sin A + sin2 A = 1, then the value of cos2 A + cos4 A is
(a) 2
(b) 1
(c) −2
(d) 0
Here the given date is and
We have to find the value of
We know that the given relation is
……(1)
Now we are going to evaluate the value of
Here we are using the relation
This is same as the equation number (1)
Therefore
Hence the option (b) is correct.
#### Question 10:
The value of [(sec A + tan A) (1−sin A)] is equal to
(a) tan2 A
(b) sin2 A
(c) cos A
(d) sin A
Here we have to evaluate the value of
Now we are going to solve this
Hence the option (c) is correct.
#### Question 11:
Find a quadratic polynomial with zeroes
Given that the zeroes of the quadratic polynomial are and .
We have to find the quadratic polynomial from the given zeroes.
Let we assume that,
, then
Therefore the quadratic equation is given by
Hence the desire polynomial is
#### Question 12:
In Figure 2, ABCD is a parallelogram. Find the values of x and y.
The given parallelogram is
We have to find the value of x and y.
We know that if two diagonals are drawn in a parallelogram, then the intersection points is the mid-point of the two diagonals. Therefore, we have two equations as follow
x + y = 9…… (1)
x − y = 5…… (2)
Now, add the equation (1) and equation (2), we get
Now, we are going to put the value of x in equation (1), we get
Hence the value of x and y are
#### Question 13:
If sec 4A = cosec (A − 20°), where 4A is an acute angle, find the value of A.
Given that:
, then we have to find the value of A
Since it is given that the angle 4A is acute angle, therefore we can apply the identity
Therefore the given equation can be written as
Hence the value of A is 22°
#### Question 14:
In Figure 3, PQ || CD and PR || CB. Prove that $\frac{\mathrm{AQ}}{\mathrm{QD}}=\frac{\mathrm{AR}}{\mathrm{RB}}$.
Given that:
and , then we to prove that
The following diagram is given
We can easily see that, in the above figure are similar triangles, and also the are similar triangles.
Now, we have the following properties of similar triangles,
From equation (1) and Equation (2), we get
Hence proved.
#### Question 15:
In Figure 4, two triangles ABC and DBC are on the same base BC in which ∠A = ∠D = 90°. If CA and BD meet each other at E, show that AE ✕ CE = BE ✕ DE.
Given that, there are two triangles ABC and DBC are on the same base BC in which
A = D = 90°. If CA and BD meet each other at E, then we have to prove that AE × CE = BE × DE
The following figure is given
From the above figure, we can easily see thatare similar triangles, therefore we can use the property of similar triangle.
Hence
#### Question 16:
Prove that :
Here we have to prove that
Here we take the LHS and by using the trigonometric identities, we have
Hence
#### Question 17:
Prove that in a triangle if the square of one side is equal to the sum of the squares of the other two side then the angle opposite to the first side is a right angle.
Given: if square of one side is equal to the sum of the squares of the other two sides, then we have to prove that the angle opposite the first side is a right angle.
Here, we are given a triangle ABC with. We need to prove that.
Now, we construct a triangle PQR right angled at Q such that and .
We have the following diagram.
Now, in, we have
…… (1)
But, it is given that
…… (2)
From equations (1) and (2), we get
…… (3)
From the above analysis in and, we have
Since, therefore
Hence,
#### Question 18:
Find the mode of the following data:
Class 0−20 20−40 40−60 60−80 Frequency 15 6 18 10
We have to find the mode of the following distribution,
Class 0-20 20-40 40-60 60-80 Frequency 15 6 18 10
The class (40-60) has the maximum frequency; therefore this is the modal class.
Lower limit of the modal class
Width of the class interval h = 20
Frequency of the modal class fk = 18
Frequency of the class preceding the modal class fk−1 = 6
Frequency of the class succeeding the modal class fk+1 = 10
Now, we have the following formula to find the value of mode.
Hence the value of mode is 42.
#### Question 19:
Prove that $\sqrt{7}$ is an irrational number.
We have to prove that is an irrational number
We will prove this by contradiction:
Let be an irrational number such that where x and y are co prime
So’
This means that:
From equation (1) and (2) we see that 7 is a common factor of x and y. This contradicts the fact that x and y have no common factor
Hence is an irrational number.
#### Question 20:
Use Euclid's division algorithm to find the HCF of 10224 and 9648.
Here we have to find the HCF of the numbers 10224 and 9648 by using Euclid’s division algorithm.
We know that If we divide a by b and r is the remainder and q is the quotient, Euclid’s Lemma says that
A = bq+r, where
And HCF of (a, b) = HCF of (b, r)
Here
Therefore, we have the following procedure,
Now, we apply the division algorithm on 9648 and 576.
Therefore the HCF of 432 and 144 is 144.
Hence the HCF of 10224 and 9648 is 144
#### Question 21:
If α and β are zeroes of the quadratic polynomial x2 − 6x + a; find the value of 'a' if 3α + 2β = 20.
Given that: α and β are the zeroes of the quadratic polynomial x2 − 6x + a and 3α + 2β = 20, then we have to find the value of a.
We have the following procedure.
Now, we are putting the value α in equation (1), we get
From equation (2), we have
Hence the value of a is −16.
#### Question 22:
The sum of the numerator and the denominator of a fraction is 8. If 3 is added to both the numerator and the denominator, the fraction becomes $\frac{3}{4}$. Find the fraction.
Here we are assuming that numerator and denominator are x and y respectively, then fraction will be and we have to find the value of .
From the given condition,
x + y = 8…… (1)
If 3 are added in numerator and denominator then fraction will be, from this we have
Now, multiply the equation (1) by 3, we get
3x + 3y = 24…… (3)
Now we take the addition of equations (2) and (3), we get
Put the value of x in the equation (1), we have
Hence the fraction is
#### Question 23:
Prove that .
Here we have to prove that
First we take LHS and use the identities
Now we take RHS
Hence proved.
#### Question 24:
In Figure 5, ∆ABC is right angled at B, BC = 7 cm and AC − AB = 1 cm. Find the value of cos A − sin A.
It is given that is right angled at B, BC = 7 cm and AC − AB = 1 cm then we have to find the value of
The following diagram is given
ACAB = 1…… (1)
Now, apply the Pythagoras theorem in, we get
Now add the equation (1) and (2), we get
Put the value of in equation (2), we have
Now,
Hence
#### Question 25:
In Figure 6, P and Q are the midpoints of the sides CA and CB respectively of ∆ABC right angled at C. Prove that 4(AQ2 + BP2) = 5 AB2.
In the given figure P and Q are the mid points of AC and BC, the we have to prove that
The following figure is given.
Using Pythagoras theorem in, we get
AB2 = AC2 + BC2…… (1)
Similarly, by using Pythagoras theorem in, we get
AQ2 = CQ2 + AC2…… (2)
BP2 = CP2 + BC2…… (3)
Now adding equation (2) and equation (3), we get
…… (4)
Now multiply equation (4) by 4, we get
Hence proved.
#### Question 26:
The diagonals of a trapezium ABCD with AB || DC intersect each other at point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Given that the diagonals of trapezium ABCD withintersect each other at point O and if, then we have to find the ratio of areas of triangle AOB and triangle COD.
We have the following diagram.
From the above figure, in triangles AOB and COD, we have
Therefore, we can apply the area property of similar triangles.
Hence,
#### Question 27:
The mean of the following frequency distribution is 50. Find the value of p.
Classes 0−20 20−40 40−60 60−80 80−100 Frequency 17 28 32 p 19
Given that the mean value of the following distribution is 50.
Class 0-20 20-40 40-60 60-80 80-100 Frequency 17 28 32 p 19
We have to find the value of p.
To find the value of p, we have following procedure
Class Mid-value xi Frequency fi fixi 00-20 10 17 170 20-40 30 28 840 40-60 50 32 1600 60-80 70 p 70p 80-100 90 19 1710
Thus, we have
We know that the formula of mean is given by
Hence
#### Question 28:
Find the missing frequencies in the following frequency distribution table, if N = 100 and median is 32.
Marks obtained 0−10 10−20 20−30 30−40 40−50 50−60 Total No. of Students 10 ? 25 30 ? 10 100
We have to find the missing term of the following distribution table if and median is 32.
Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 Total No. of students 10 ? 25 30 ? 10 100
Suppose the missing term are x and y.
Now we have to find the cumulative frequency as
Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 10 x 25 30 y 10 Cumulative Frequency 10
From the above distribution median class is 30-40 and
Therefore,
Now we are using the following relation
Now we are putting the value of x in the equation (1), we get
Hence the missing terms are
#### Question 29:
Divide 30x4 + 11x3 − 82x2 − 12x + 48 by (3x2 + 2x − 4) and verify the result by division algorithm.
Here we have to divide by.
According to division algorithm, Dividend = Divisor × Quotient + Remainder.
This can be verified as,
Divisor × Quotient + Remainder
#### Question 30:
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.
Given that a in which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that
We have the following diagram with some additional construction.
In the above figure, we can see that EF is perpendicular to AB. Therefore EF is the height of the.
Now we are finding the area of and area of
Now take the ratio of the two equation, we have
Similarly, In , we have
But, the area of triangle DBE and triangle DEC are same, therefore equation (4) can be written as
From equation (3) and equation (5), we get
Hence,
#### Question 31:
Without using trigonometric tables, evaluate the following:
We have to evaluate the following expression
Here we are going to use following identities.
The given expression can be written as
Hence the value of the given expression is zero.
#### Question 32:
If 2 cos θ − sin θ = x and cos θ − 3 sin θ = y. Prove that 2x2 + y2 − 2xy = 5.
Given that:
Then we have to prove that
First we take the LHS and put the value of x and y, we get
Hence
#### Question 33:
Check graphically whether the pair of linear equation 4xy − 8 = 0 and 2x − 3y + 6 = 0 is consistent. Also, find the vertices of the triangle formed by these lines with the x-axis.
Here we have to draw the graph between two equations given by
Also we have to find the vertices of the triangle form by the x-axis and these lines.
The first equation can written as follow
Now we are going to find the value of y at different value of x
x 2 3 y 0 4
Now mark the points (0,-8) and (2, 0) on xy −plane and we will draw a line which passes through these two points.
The second equation can be written as
x 0 3 y 2 4
Now mark the points (0, 2) and (3, 4) on xy −plane and we will draw a line which passes through these two points.
From the above analysis, we have the following graph
From the above graph the vertices A, B and C are given as follow.
Hence the vertices are
#### Question 34:
The following table shows the ages of 100 persons of a locality.
Age (yrs) 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Number of persons 5 15 20 23 17 11 9
Represent the above as the less than type frequency distribution and draw an ogive for the same.
The given frequency distribution is
Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of Persons 5 15 20 23 17 11 9
We have to change the above distribution as less than type frequency distribution and also we have to draw its ogive.
We have the following procedure to change the given distribution in to less than type distribution.
Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of persons 5 5+15=20 20+20=40 40+23=63 63+17=80 80+11=91 91+9=100
=
Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of persons 5 20 40 63 80 91 100
To draw its ogive, take the number of persons on y-axis and age in years on x-axis. Mark the points (10, 5), (20, 20), (30, 40), (40, 63), (50, 80), (60, 91) and (70, 100) on the
xy-plane. Now these points are joined by free hand.
#### Question 1:
Evaluate:
Here we to evaluate the value of the expression given as follow
Now we are using the following identities
Therefore the given expression can be written as
Hence the value of the given expression is
#### Question 2:
In Fig. 1, the graph of a polynomial p (x) is shown. The number of zeroes of p (x) is
(a) 4
(b) 1
(c) 2
(d) 3
In the following graph we observe that it intersects x-axis at x = 1. So it has only one zero.
Hence the correct option is
#### Question 1:
In the following graph we observe that it intersects x-axis at x = 1. So it has only one zero.
Hence the correct option is
In the given rational number , the denominator is in the form of , therefore we get
We know that any rational number which has the denominator in the form of terminate after the largest of m and n places.
Since, the rational number terminates after 4 places.
#### Question 2:
In the given rational number , the denominator is in the form of , therefore we get
We know that any rational number which has the denominator in the form of terminate after the largest of m and n places.
Since, the rational number terminates after 4 places.
Here we have to find the value of.
By using the trigonometric identities, the given expression can be written as
Now, we use the relation, we get
Hence the correct option is
#### Question 3:
Here we have to find the value of.
By using the trigonometric identities, the given expression can be written as
Now, we use the relation, we get
Hence the correct option is
Given: the sides of the two similar triangles are in the ratio 4:9, then we have to find the ratio of areas of the two triangles.
We know the following property of the similar triangles.
Hence the correct option is
#### Question 4:
Given: the sides of the two similar triangles are in the ratio 4:9, then we have to find the ratio of areas of the two triangles.
We know the following property of the similar triangles.
Hence the correct option is
Given that, then we have to evaluate.
By using the given expression can be written as
Now put the value, we get
Hence the correct option is
#### Question 5:
Given that, then we have to evaluate.
By using the given expression can be written as
Now put the value, we get
Hence the correct option is
Here we have to find the product of HCF and LCM of the smallest prime number and smallest composite number.
We know that any number that can express in product of prime number is called composite number. The smallest prime number is 2 and therefore the smallest composite number is.
HCF of 2 and 4 = 2
LCM of 2 and 4 = 4
Therefore the product of HCF and LCM of 2 and 4 is
Hence the correct option is
#### Question 6:
Here we have to find the product of HCF and LCM of the smallest prime number and smallest composite number.
We know that any number that can express in product of prime number is called composite number. The smallest prime number is 2 and therefore the smallest composite number is.
HCF of 2 and 4 = 2
LCM of 2 and 4 = 4
Therefore the product of HCF and LCM of 2 and 4 is
Hence the correct option is
We know that, when we draw more than ogive and less than ogive, the Âx coordinate of intersection of the less than ogive curve and more than ogive curve point gives the median.
Hence the correct option is
#### Question 7:
We know that, when we draw more than ogive and less than ogive, the Âx coordinate of intersection of the less than ogive curve and more than ogive curve point gives the median.
Hence the correct option is
The given condition is .
We have to find the value of.
We know that, therefore the given condition can be written as
Hence the correct option is
#### Question 8:
The given condition is .
We have to find the value of.
We know that, therefore the given condition can be written as
Hence the correct option is
Here we have to find the quadratic equation whose sum of zeroes is 3 and product of zeroes is -2.
Now, we assume that the two zeroes are, then
We know that the general polynomial is
Hence the correct option is
#### Question 9:
Here we have to find the quadratic equation whose sum of zeroes is 3 and product of zeroes is -2.
Now, we assume that the two zeroes are, then
We know that the general polynomial is
Hence the correct option is
Here we have to find the value of
In the above expression the term exist in the middle and the value ofis zero, therefore the expression can be written as
Hence the correct option is
#### Question 10:
Here we have to find the value of
In the above expression the term exist in the middle and the value ofis zero, therefore the expression can be written as
Hence the correct option is
The given two equations are
Therefore, the graphical representation of the given pair of equations represents parallel lines.
Hence the correct option is
#### Question 11:
The given two equations are
Therefore, the graphical representation of the given pair of equations represents parallel lines.
Hence the correct option is
In a given, is the point on BC such that then we have to prove that
We have the following diagram
Since and, therefore the and are similar triangles.
By using the property of similar triangle we have
Hence proved.
#### Question 12:
In a given, is the point on BC such that then we have to prove that
We have the following diagram
Since and, therefore the and are similar triangles.
By using the property of similar triangle we have
Hence proved.
Here we have to find the HCF of 26, 52 and 91 by using fundamental theorem of arithmetic.
We have the following procedure to find the HCF.
We can see that the common factor is 13.
Hence
#### Question 13:
Here we have to find the HCF of 26, 52 and 91 by using fundamental theorem of arithmetic.
We have the following procedure to find the HCF.
We can see that the common factor is 13.
Hence
We have the following distribution
Class 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 2 8 12 24 38 16
We have to change the above distribution as a less than type distribution
We have the following procedure
Less than 55 60 65 70 75 80 Cumulative Frequency 2 2+8=10 10+12=22 22+24=46 46+38=84 84+16=100
=
Less than 55 60 65 70 75 80 Cumulative Frequency 2 10 22 46 84 100
#### Question 14:
We have the following distribution
Class 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 2 8 12 24 38 16
We have to change the above distribution as a less than type distribution
We have the following procedure
Less than 55 60 65 70 75 80 Cumulative Frequency 2 2+8=10 10+12=22 22+24=46 46+38=84 84+16=100
=
Less than 55 60 65 70 75 80 Cumulative Frequency 2 10 22 46 84 100
We have to divide the expression by .
Hence
#### Question 15:
We have to divide the expression by .
Hence
Here we have the following two equations
The given two equations will have infinitely many solutions if .
Hence
#### Question 16:
Here we have the following two equations
The given two equations will have infinitely many solutions if .
Hence
Given that is an isosceles triangle with and, then we have to prove that the is a right angled triangle
We have the following diagram
We know that when square of one side is equal to sum of squares
Of other two sides then triangle is called right angled triangle.
#### Question 17:
Given that is an isosceles triangle with and, then we have to prove that the is a right angled triangle
We have the following diagram
We know that when square of one side is equal to sum of squares
Of other two sides then triangle is called right angled triangle.
Here we have to find mean of the following distribution.
Class 1-3 3-5 5-7 7-9 9-11 Frequency 7 8 2 2 1
We have the following procedure to obtain mean of the above distribution with
Class Mid-value Frequency 1-3 2 -4 7 -28 3-5 4 -2 8 -16 5-7 6 0 2 00 7-9 8 2 2 4 9-11 10 4 1 4
We know formula to find mean of a given distribution is
Hence
#### Question 18:
Here we have to find mean of the following distribution.
Class 1-3 3-5 5-7 7-9 9-11 Frequency 7 8 2 2 1
We have the following procedure to obtain mean of the above distribution with
Class Mid-value Frequency 1-3 2 -4 7 -28 3-5 4 -2 8 -16 5-7 6 0 2 00 7-9 8 2 2 4 9-11 10 4 1 4
We know formula to find mean of a given distribution is
Hence
Given:and, then we have to find the value of A and B.
We know that is the value of tan60° and is the value of tan30°, therefore the above equations can be written as
By adding the equations (1) and (2), we get
Now, put the value of A in the equation (1), we get
Hence
OR
Given that and is acute angle then we have to find the value of A.
Here we are using the identity
Hence the value of A is
#### Question 19:
Given:and, then we have to find the value of A and B.
We know that is the value of tan60° and is the value of tan30°, therefore the above equations can be written as
By adding the equations (1) and (2), we get
Now, put the value of A in the equation (1), we get
Hence
OR
Given that and is acute angle then we have to find the value of A.
Here we are using the identity
Hence the value of A is
In the given figure, and XY divides triangular region ABC into two parts equal in area. Then we have to find the ratio
We have the following diagram.
We can see that the triangle ABC and triangle BXY are similar triangles, therefore we have
The ratio of area of triangle BXY and triangle ABC is given by
Hence the ratio
#### Question 20:
In the given figure, and XY divides triangular region ABC into two parts equal in area. Then we have to find the ratio
We have the following diagram.
We can see that the triangle ABC and triangle BXY are similar triangles, therefore we have
The ratio of area of triangle BXY and triangle ABC is given by
Hence the ratio
Here we have to evaluate the expression given by
Here we are using the following identities.
The given expression can be written as
Hence the value of the given expression is
#### Question 21:
Here we have to evaluate the expression given by
Here we are using the following identities.
The given expression can be written as
Hence the value of the given expression is
We have to find the mode of the following distribution,
Class 25-35 35-45 45-55 55-65 65-75 75-85 Frequency 7 31 33 17 11 1
The class (45-55) has the maximum frequency; therefore this is the modal class.
Lower limit of the modal class
Width of the class interval
Frequency of the modal class
Frequency of the class preceding the modal class
Frequency of the class succeeding the modal class
Now, we have the following formula to finding the value of mode.
Hence the value of mode is
#### Question 22:
We have to find the mode of the following distribution,
Class 25-35 35-45 45-55 55-65 65-75 75-85 Frequency 7 31 33 17 11 1
The class (45-55) has the maximum frequency; therefore this is the modal class.
Lower limit of the modal class
Width of the class interval
Frequency of the modal class
Frequency of the class preceding the modal class
Frequency of the class succeeding the modal class
Now, we have the following formula to finding the value of mode.
Hence the value of mode is
Here we have to show that square of any positive odd integer is in the form of.
We know that any positive odd integer n is in the form of and, therefore
If, then
Where
If, then
Where
Hence the square of any positive odd integer is in the form of
#### Question 23:
Here we have to show that square of any positive odd integer is in the form of.
We know that any positive odd integer n is in the form of and, therefore
If, then
Where
If, then
Where
Hence the square of any positive odd integer is in the form of
We have following two equations.
We have to find the values of x and y.
The equation (2) can be written as
Now, subtract the equation (1) from equation (3), we get
Now put the value of x in equation (1), we get
Hence the solution of the given equations is
OR
Given that the sum of digits of two digit number is 12 and the number obtained by interchanging the two digits exceeds to the given number by 18, then we have to find the two digit number.
Let the two digit number whose unit place is x and tenth place is y, then by applying the given two conditions we have
Now add the equation (1) and equation (2), we get
Put the value of x in equation (1), we have
Hence the number is
#### Question 24:
We have following two equations.
We have to find the values of x and y.
The equation (2) can be written as
Now, subtract the equation (1) from equation (3), we get
Now put the value of x in equation (1), we get
Hence the solution of the given equations is
OR
Given that the sum of digits of two digit number is 12 and the number obtained by interchanging the two digits exceeds to the given number by 18, then we have to find the two digit number.
Let the two digit number whose unit place is x and tenth place is y, then by applying the given two conditions we have
Now add the equation (1) and equation (2), we get
Put the value of x in equation (1), we have
Hence the number is
Here we have to prove that the number is an irrational number.
Let us assume that the number is a rational number, therefore can be written as follow
Where a and b are positive co-prime integer numbers.
is an irrational number andis a rational number (Since a and b are integer).
This contradicts that is an irrational number. Therefore, our assumption is wrong.
Hence is an irrational number.
OR
Here we have to prove that is an irrational number.
Let us assume on the contrary thatis a rational number. Then, there exists positive integers a and b (co-prime numbers) such that
From (1), we have
From (2) and (3), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means our assumption is wrong.
Hence is an irrational number.
#### Question 25:
Here we have to prove that the number is an irrational number.
Let us assume that the number is a rational number, therefore can be written as follow
Where a and b are positive co-prime integer numbers.
is an irrational number andis a rational number (Since a and b are integer).
This contradicts that is an irrational number. Therefore, our assumption is wrong.
Hence is an irrational number.
OR
Here we have to prove that is an irrational number.
Let us assume on the contrary thatis a rational number. Then, there exists positive integers a and b (co-prime numbers) such that
From (1), we have
From (2) and (3), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means our assumption is wrong.
Hence is an irrational number.
Here we have to prove that
We are taking the LHS, we have
Hence
OR
Here we have to prove that
First we take LHS, we have
Now, we take RHS
Hence
#### Question 26:
Here we have to prove that
We are taking the LHS, we have
Hence
OR
Here we have to prove that
First we take LHS, we have
Now, we take RHS
Hence
We have an isosceles triangle with and, then we to prove that
We have the following diagram.
In right angle triangle ADB, we have
Hence proved.
#### Question 27:
We have an isosceles triangle with and, then we to prove that
We have the following diagram.
In right angle triangle ADB, we have
Hence proved.
Here we have to find the zeroes of the polynomial
To find the zeroes, we have to put the given polynomial zero
Now, we have or
Therefore the zeroes of the given polynomial are
The given polynomial is
The sum of zeroes of the above polynomial
And the product of zeroes f the above polynomial
The sum of zeroes
The product of zeroes
Hence verified.
#### Question 28:
Here we have to find the zeroes of the polynomial
To find the zeroes, we have to put the given polynomial zero
Now, we have or
Therefore the zeroes of the given polynomial are
The given polynomial is
The sum of zeroes of the above polynomial
And the product of zeroes f the above polynomial
The sum of zeroes
The product of zeroes
Hence verified.
The given distribution is
Class 0-10 10-20 20-30 30-40 40-50 Total Frequency 8 16 36 34 6 100
We have to find the median of the above distribution.
Now we have to find the cumulative frequency as
Class 0-10 10-20 20-30 30-40 40-50 Frequency 8 16 36 34 6 Cumulative Frequency 8 8+16=24 24+36=60 60+34=94 94+6=100
Sincelies in the cumulative frequency of the interval (20-30), therefore (20-30 belongs to the median class interval
Lower limit of median class interval
The frequency
Width of the class interval
Frequency of the median class
Cumulative frequency preceding median class frequency
Now we are using the following relation
Hence the median of the given distribution is
#### Question 29:
The given distribution is
Class 0-10 10-20 20-30 30-40 40-50 Total Frequency 8 16 36 34 6 100
We have to find the median of the above distribution.
Now we have to find the cumulative frequency as
Class 0-10 10-20 20-30 30-40 40-50 Frequency 8 16 36 34 6 Cumulative Frequency 8 8+16=24 24+36=60 60+34=94 94+6=100
Sincelies in the cumulative frequency of the interval (20-30), therefore (20-30 belongs to the median class interval
Lower limit of median class interval
The frequency
Width of the class interval
Frequency of the median class
Cumulative frequency preceding median class frequency
Now we are using the following relation
Hence the median of the given distribution is
Here we have to show that
First we take the LHS, we have
Hence
OR
Here we have to prove that
Hence
#### Question 30:
Here we have to show that
First we take the LHS, we have
Hence
OR
Here we have to prove that
Hence
The given polynomial is and two zeroes of this polynomial areand, then we have to find the two other zeroes of.
If two zeroes of the polynomial are and, then
is a factor of
Now we are going to by
Therefore
The other two zeroes are obtained from the polynomial
Hence the zeroes are
#### Question 31:
The given polynomial is and two zeroes of this polynomial areand, then we have to find the two other zeroes of.
If two zeroes of the polynomial are and, then
is a factor of
Now we are going to by
Therefore
The other two zeroes are obtained from the polynomial
Hence the zeroes are
Here we have to draw the graph between two equations given by
The first equation can written as follow
Now we are going to find the value of y at different values of x
x 0 1 2 y 6 4 2
Now mark the points (0, 6), (1, 4) and (2, 2) on xy −plane and we will draw a line which passes through these points.
The second equation can be written as
x 0 2 3 y 2 6 8
Now mark the points (0, 2), (2, 6) and (3, 8) on xy −plane and we will draw a line which passes through these points.
From the above analysis we have the following graph
From the above graph, the area of triangle ABC is given by
Hence area is
#### Question 32:
Here we have to draw the graph between two equations given by
The first equation can written as follow
Now we are going to find the value of y at different values of x
x 0 1 2 y 6 4 2
Now mark the points (0, 6), (1, 4) and (2, 2) on xy −plane and we will draw a line which passes through these points.
The second equation can be written as
x 0 2 3 y 2 6 8
Now mark the points (0, 2), (2, 6) and (3, 8) on xy −plane and we will draw a line which passes through these points.
From the above analysis we have the following graph
From the above graph, the area of triangle ABC is given by
Hence area is
Given that a in which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that
We have the following diagram with some additional construction.
In the above figure, we can see that EF is perpendicular to AB. Therefore EF is the height of the.
Now we are finding the area of and area of
Now take the ratio of the two equation, we have
Similarly, In , we have
But, the area of triangle DBE and triangle DEC are same, therefore equation (4) can be written as
From equation (3) and equation (5), we get
Hence proved.
OR
We have to prove that in a right angle triangle, the square of hypotenuse is the sum of the squares of the other two sides.
We have the following diagram.
Here we assuming that a right angle triangle ABC, right angled at B and BD is perpendicular to the hypotenuse AC.
From the above figure we have, therefore we can apply the property of similar triangle.
Also the, from which we get
Now, we take addition of equation (1) and equation (2), to get
#### Question 33:
Given that a in which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that
We have the following diagram with some additional construction.
In the above figure, we can see that EF is perpendicular to AB. Therefore EF is the height of the.
Now we are finding the area of and area of
Now take the ratio of the two equation, we have
Similarly, In , we have
But, the area of triangle DBE and triangle DEC are same, therefore equation (4) can be written as
From equation (3) and equation (5), we get
Hence proved.
OR
We have to prove that in a right angle triangle, the square of hypotenuse is the sum of the squares of the other two sides.
We have the following diagram.
Here we assuming that a right angle triangle ABC, right angled at B and BD is perpendicular to the hypotenuse AC.
From the above figure we have, therefore we can apply the property of similar triangle.
Also the, from which we get
Now, we take addition of equation (1) and equation (2), to get
Here we have to prove that,
Multiply in both numerator and denominator by, we get
Hence
#### Question 34:
Here we have to prove that,
Multiply in both numerator and denominator by, we get
Hence
Here we can to change the following distribution in to more than type and also we have to draw its ogive.
Class 100-120 120-140 140-160 160-180 180-200 Frequency 12 14 8 6 10
We have to find the median of the above distribution.
We have the following procedure to obtain more than type distribution.
Class Frequency More than type cumulative frequency 100-120 12 12+14+8+6+10=50 120-140 14 14+8+6+10=38 140-160 8 8+6+10=24 160-180 6 6+10=16 180-200 10 10
=
Class Frequency More than type cumulative frequency 100-120 12 50 120-140 14 38 140-160 8 24 160-180 6 16 180-200 10 10
To draw its ogive, the lower limits of the class interval are marked on x-axis and more than type cumulative frequencies are taken on y-axis. For drawing more than type curve, points (100, 50), (120, 38), (140, 24), (160, 16) and (180, 10) are marked on graph paper and these are joined by free hand to obtain more than type ogive curve. The curve drawn is given by
The median is the x coordinates corresponding to and therefore the median from the above curve is 139.
Hence
#### Question 3:
In Fig. 2, If DE || BC, then x equals
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12.5 cm
In the given figure we have DE||BC. We have to find x.
In the following given figure DE||BC, so triangle ADE and triangle ABC are similar triangle.
So we have the following relation
Hence the correct option is
#### Question 4:
If sin 3θ = cos (θ − 6°), where (3θ) and (θ − 6°) are both acute angles, then the value of θ is
(a) 18°
(b) 24°
(c) 36°
(d) 30°
Given that,
…… (1)
We have to find θ
Here the angles (3θ) and (θ − 6) are acute angles and we know that
Therefore we can rewrite the equation (1) as
Therefore option is correct.
#### Question 5:
Given that tan θ = $\frac{1}{\sqrt{3}}$, the value of $\frac{\mathrm{cos}e{c}^{2}\theta -se{c}^{2}\theta }{\mathrm{cos}e{c}^{2}\theta +se{c}^{2}\theta }$ is
(a) − 1
(b) 1
(c) $\frac{1}{2}$
(d) $-\frac{1}{2}$
Given:
We have to find the value of the following expression
Now, so
perpendicular =1
base =
and
Therefore
and
Put the above two values in the given expression, we have
Hence the correct option is
#### Question 6:
In Fig. 3, AD = 4 cm, BD = 3 cm and CB = 12 cm, then cot θ equals
(a) $\frac{3}{4}$
(b) $\frac{5}{12}$
(c) $\frac{4}{3}$
(d) $\frac{12}{5}$
In Fig 3, if we have
BD = 3 cm, and
CB = 12 cm, then
cot θ = ?
We have the following diagram
If we apply Pythagoras theorem in triangle ADB, we get
Now,
Hence the correct option is
#### Question 7:
The decimal expansion of $\frac{147}{120}$ will terminate after how many places of decimal?
(a) 1
(b) 2
(c) 3
(d) will not terminate
The given number is. We have to change the denominator in the form of and the given number terminates after largest of m and n.
Here,
Hence the given number terminates after 3 places.
#### Question 8:
The pair of linear equations 3x + 2y = 5; 2x − 3y = 7 have
(a) One solution
(b) Two solutions
(c) Many solutions
(d) No solution
The two equations are
3x + 2y = 5 …… (1)
2x − 3y = 7 …… (2)
Here,
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. That is, there is only one solution.
Hence the correct option is
#### Question 9:
In an isosceles triangle ABC with AB = AC and BD ⊥ AC. Prove that BD2 − CD2 = 2CD.AD.
Given: ∆ABC is an isosceles triangle with AB = AC and BD ⊥ AC.
To prove: BD2 − CD2 = 2CD.AD
$⇒{\mathrm{BD}}^{2}-{\mathrm{CD}}^{2}=2\mathrm{AD}.\mathrm{CD}$ (Hence Proved)
#### Question 10:
For a given data with 70 observations the 'less then ogive' and the 'more than ogive' intersect at (20.5, 35). The median of the data is
(a) 20
(b) 35
(c) 70
(d) 20.5
It is given that less than ogive and more than ogive intersects at (20.5, 35), then we have to find the median.
We know that for a given distribution, median is the x coordinates of intersection of less than ogive curve and more than ogive curve. Since the intersection point is (20.5, 35), therefore the median is 20.5
Hence
Hence the correct option is
#### Question 11:
Determine the value of k so that the following linear equations have no solution:
(3k + 1) x + 3y − 2 = 0
(k2 + 1) x + (k − 2) y − 5 = 0
The given system of equations is
(3k + 1) x + 3y − 2 = 0
(k2 + 1) x + (k − 2) y − 5 = 0
Here, a1 = 3k + 1, b1 = 3, c1 = −2
a2 = k2 + 1, b2 = k − 2, c2 = −5
The given system of equations has no solution.
Now,
$\frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\phantom{\rule{0ex}{0ex}}⇒\left(3k+1\right)\left(k-2\right)=3\left({k}^{2}+1\right)\phantom{\rule{0ex}{0ex}}⇒3{k}^{2}-5k-2=3{k}^{2}+3\phantom{\rule{0ex}{0ex}}⇒-5k=5$
$⇒k=-1$
When k = −1,
$\frac{3}{k-2}=\frac{3}{-1-2}=\frac{3}{-3}=-1$
Thus, for k = −1, $\frac{3}{k-2}\ne \frac{2}{5}$
Hence, the given system of equations has no solution when k = −1.
#### Question 12:
Can (x − 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer.
If we divide any polynomial by another polynomial, then the degree of divisor is always greater than the degree of remainder.
In the given question the degrees of remainder (x−2) and divisor (2x+3) are same
Therefore if we divide the polynomial P(x) by (2x+3), then (x−2) cannot be the remainder
#### Question 13:
In Fig. 4, ABCD is a rectangle. Find the values of x and y.
The following diagram is given
The given diagram is rectangle and we know that in rectangle the face to face sides are same. Therefore
…… (1)
…… (2)
Now we are adding the equations (1) and (2) to have
Put the value of x in equation (1), we get
Hence, x = 10 and y = 2
Therefore the values of x and y are
#### Question 14:
If 7 sin2 θ + 3 cos2θ = 4, show that tan θ = $\frac{1}{\sqrt{3}}$
Given: and we have to prove that
We can write the given expression as
Therefore,
Hence proved
#### Question 15:
In Fig. 5, DE || AC and DF || AE. Prove that $\frac{EF}{BF}=\frac{EC}{BE}$
Given that:
If DE||AC and DF||AE, then we have to prove that
The following given figure is
We can easily see that in the given figure the triangle BDF and triangle BAE are similar triangles and also the triangle BDE and triangle BAC are similar triangles. Now we are applying the theorem of similar triangle in triangle BDF and triangle BAE, we get
…… (1)
Similarly in triangle BDE and triangle BAC, we get
…… (2)
Now we are comparing the equation (1) and (2), we have
Now take the reciprocal of the above equation, we have
Hence we have proved that
#### Question 16:
In Fig. 6, AD $\perp$ BC and BD = $\frac{1}{3}$ CD. Prove that 2CA2 = 2AB2 + BC2
In the given figure, we have
and, then we have to prove
The following given diagram is
Now, suppose the value of BD is x, then
…… (1)
And in triangle ADB, we have
…… (2)
Now add the equation (1) and equation (2), we get
…… (3)
Now we are putting the values of BD and CD, we get
Hence proved.
#### Question 17:
Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angles opposite to the first side is a right angle.
Given: if square of one side is equal to the sum of the squares of the other two sides, then we have to prove that the angle opposite the first side is a right angle.
Here, we are given a triangle ABC with. We need to prove that
.
Now, we construct a triangle PQR right angled at Q such that and .
We have the following diagram.
Now, in, we have
…… (1)
But, it is given that
…… (2)
From equations (1) and (2), we get
…… (3)
From the above analysis in and, we have
Since, therefore
Hence,
#### Question 18:
Find the mode of the following distribution of marks obtained by 80 students:
Marks obtained 0-10 10-20 20-30 30-40 40-50 Number of students 6 10 12 32 20
We have the following distribution
Marks Obtained 0-10 10-20 20-30 30-40 40-50 Number of Students 6 10 12 32 20
We have to find the mode of the above distribution
From the given distribution, we can easily see that the class (20-30) has the maximum frequency i.e. 32.
Lower limit of the modal class
Class interval
Frequency of the modal class
Frequency of the class preceding the modal class
Frequency of the class succeeding the modal class
Therefore the mode value of the given distribution is 36.25
#### Question 19:
Show that any positive odd integers is of the form 4q+1 or 4q+3 where q is a positive integer.
Here we have to prove that for any positive integer q, the positive odd integer will be form of 4q+1 or 4q+3.
Now let us suppose that the positive odd integer is a then by Euclid’s division rule
a = 4q + r ……(1 )
Where q (quotient) and r (remainder) are positive integers, and
We are putting the values of r from 0 to 3 in equation (1), we get
But we can easily see that 4q and 4q+2 are both even numbers.
Therefore for any positive value q, the positive odd integer will be the form of 4q+1 and 4q+3.
#### Question 20:
Prove that is irrational.
Here we have to prove that the number is an irrational number.
Now let us suppose that, where is a rational number, then
As is rational number, therefore form equation (1), so is and is also a rational number which is a contradiction as is an irrational number.
Therefore is an irrational number.
#### Question 21:
A person can row a boat at the rate of 5 km/hour in still water. He takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.
Given that,
Speed of boat in still water = 5 km/hour.
Distance covered by boat = 40 km
Now we are assuming that the speed of the stream is, then
Speed of the boat rowing upstream =, and
Speed of the boat rowing downstream =
According to the given condition, time taken to cover in upstream is three times the time taken to cover in downstream, therefore
Hence the speed of stream is 2.5 km/hour.
#### Question 22:
If , are zeroes of the polynomial x2 − 2x − 15, then form a quadratic polynomial whose zeroes are (2) and (2).
The given polynomial is andand are the zeroes of the polynomial , then we have to find another polynomial whose zeroes are,
Now we comparing the given polynomial with, we get
We know that
The polynomial whose zeroes are, is
Hence the polynomial is
#### Question 23:
Prove that (cosec θ − sin θ) (sec θ − cos θ) = .
Here we have to prove that
Now using the identity, we get
Hence proved.
#### Question 24:
If cos θ + sin θ = $\sqrt{2}$ cos θ, show that cos θ − sin θ = $\sqrt{2}$ sin θ.
Given that if, then we have to prove that
We have,
Hence proved.
#### Question 25:
In Fig. 7, AB $\perp$ BC, FG $\perp$ BC and DE $\perp$ AC. Prove that Δ ADE $~$ ΔGCF
In the given figure, we have
Then we have to prove that
The following diagram is given
In, we have
…… (1)
In, we have
…… (2)
From equation (1) and equation (2), we have
Similarly, we have
Since and are equiangular, therefore
#### Question 26:
In Fig. 8, and ΔDBC are on the same base BC and on opposite sides of BC ad Q is the point of intersection of AD and BC.
Prove that
We have given the diagram in which
We have to prove that
Firstly, we draw a line from A perpendicular to line BC and after that we draw a line from D perpendicular to BC.
From the above figure we can easily see that theand are similar
Therefore by the properties of similar triangle, we have
…… (1)
Now,
From equation (1), we get
Hence proved.
#### Question 27:
Find the mean of the following frequency distribution, using step-deviation method:
Class 0-10 10-20 20-30 30-40 40-50 Frequency 7 12 13 10 8
The following frequency distribution is given
Class 0-10 10-20 20-30 30-40 40-50 Frequency 7 12 13 10 8
We have to find the mean of the above frequency distribution using step deviation method.
By using step deviation method, we have
Class Frequency Mid-value 00-10 7 5 10-20 12 15 20-30 13 25 0 0 30-40 10 35 1 10 40-50 8 45 2 16
From the above distribution, we have
We know the mean of a given distribution is given by
Hence the mean of the given distribution is 25.
#### Question 28:
Find the median of the following data
Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 5 3 4 3 3 4 7 9 7 8
The given distribution is
Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 5 3 4 3 3 4 7 9 7 8
We have to find the median of the above distribution.
Now we have to find the cumulative frequency as
Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 5 3 4 3 3 4 7 9 7 8 Cumulative Frequency 5 8 12 15 18 22 29 38 45 53
From the above distribution median class is 60-70 and
Now we are using the following relation
Hence the median of the given distribution is 66.43
#### Question 29:
Find other zeroes of the polynomial p(x) = 2x4 + 7x3 − 19x2 − 14x + 30 if two of its zeroes are $\sqrt{2}$ and $\sqrt{-2}$.
The given polynomial is and two zeroes of this polynomial areand, then we have to find the two other zeroes of
If two zeroes of the polynomial are and, then
is a factor of
Therefore can be written as
The other two zeroes are obtained from the polynomial
Hence the other zeroes are and
#### Question 30:
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Here we have to prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
We have two triangles andin which
In the given figure AD is perpendicular to BC and PM is perpendicular to QR
The areas of triangle ABC and triangle PQR are given by
…… (1)
Since are same, therefore
…… (2)
By applying the property of similar triangles, we have
…… (3)
From (2) and (3), we get
…… (4)
From (1) and (4), we have
Hence
#### Question 31:
Prove that:
Here we have to prove that,
Firs we take the left hand side of the given equation
Now we are using the trigonometric identity
Therefore,
Hence, .
#### Question 32:
If sec θ + tan θ = p, prove that sin θ = $\frac{{p}^{2}-1}{{p}^{2}+1}$
Given that:
, then we have to prove that
We can rewrite the given data as
Now we take the right hand side
Now we are putting the value of p in the above expression, we get
Hence
#### Question 33:
Draw the graphs of following equations:
2xy = 1 and x + 2y = 13
(i) find the solution of the equations from the graph.
(ii) shade the triangular region formed by the lines and the y-axis.
Here we have to draw the graph between two equations given by
…… (1)
…… (2)
Also we have to find the solution of the given equations.
The first equation can written as follow
…… (3)
Now we are going to find the value of y at different value of x
x 0 1 2 y − 1 1 3
Now mark the points (0,-1), (1,1) and (2,3) on xy-plane and we will draw a line which pass through these points.
The second equation can be written as
…… (4)
x 0 2 3 y 6.5 5.5 5
Now mark the points (0, 6.5), (2, 5.5) and (3, 5) on xy-plane and we will draw a line which pass through these points.
From the above analysis we have the following graph
Form the above graph, the intersection point of the two lines is the solution.
And also the triangular region is shaded in the figure.
#### Question 34:
The following table gives the production yield per hectare of wheat of 100 farms of a village.
Production yield in kg/hectare 50-55 55-60 60-65 65-70 70-75 75-80 Number of frames 2 8 12 24 38 16
Change the above distribution to more than type distribution and draw its ogive.
We have the following distribution
Production yield 50-55 55-60 60-65 65-70 70-75 75-80 Number of frames 2 8 12 24 38 16
We have to change the above distribution in to the more than type distribution and we have to draw its ogive.
We have the following procedure to find the more than type distribution
Class Frequency Cumulative frequency 50-55 2 2+8+12+24+38+16=100 55-60 8 8+12+24+38+16=98 60-65 12 12+24+38+16=90 65-70 24 24+38+16=78 70-75 38 38+16=54 75-80 16 16
To draw its ogive, take the number of frames on y-axis and production yield on x-axis. Mark the points (50,100), (55, 98), (60-90), (65, 78), (70, 54), and (75, 16) on the
xy-plane. Now these points are joined by free hand.
#### Question 1:
Which of the following equations has the sum of its roots as 3?
(a) x2 + 3x − 5 = 0
(b) −x2 + 3x + 3 = 0
(c)
(d) 3x2 − 3x − 3 = 0
(a) x2 + 3x − 5 = 0 (b) −x2 + 3x + 3 = 0
(c) (d) 3x2 − 3x − 3 = 0
We are to find out which of the above equations has sum of roots = 3.
The sum of the roots of the quadratic equation ax2 + bx + c = 0 is given by .
For given equation (a)
a = 1, b = 3, c = −5
For given equation (b)
a = −1, b = 3, c = 3
For given equation (c)
For given equation (d)
a = 3, b = −3, c = −3
Hence option (b) is correct.
#### Question 2:
The sum of first five multiples of 3 is:
(a) 45
(b) 65
(c) 75
(d) 90
We have to find the sum of first five multiples of 3
First five multiples of 3 are 3, 6, 9, 12 and 15
It forms an Arithmetic Progression (A.P) with
First term
We know that the sum of the n terms of an arithmetic progression
Here n = 5
Therefore
Hence option (a) is correct.
#### Question 3:
If radii of the two concentric circles are 15 cm and 17 cm, then the length of each chord of one circle which is tangent to other is:
(a) 8 cm
(b) 16 cm
(c) 30 cm
(d) 17 cm
We are given that radii of two concentric circles are 15 cm and 17 cm
We have to find the length of each chord of one circle which is tangent to other
Let A be the centre of the two concentric circles
Let BC be the chord of bigger circle tangent to smaller at F
Therefore AF is the radius of smaller circle
AB is the radius of bigger circle
We know that radius of a circle is perpendicular to its tangent
Therefore
We know that a perpendicular from centre of circle to chord of circle bisects the chord
Therefore BF = FC = 8 cm
Hence option (b) is correct.
#### Question 4:
In Fig. 1, PQ and PR are tangents to the circle with centre O such that ∠QPR = 50°,
(a) 25°
(b) 30°
(c) 40°
(d) 50°
We are given the below figure in which
PQ and PR are tangents to the circle with centre O and
We have to find
PQ is the tangent to circle
Therefore [Since Radius of a circle is perpendicular to tangent]
Similarly
We know that sum of angles of a quadrilateral
Hence option (a) is correct.
#### Question 5:
Two tangents making an angle of 120° with each other are drawn to a circle of radius 6 cm, then the length of each tangent is equal to
(a)
(b) 6
(c)
(d) 2
We are given two tangents to a circle making an angle of 120° with each other. The radius of circle is 6 cm
We have to find the length of each tangent.
Let O be the center of the given circle
Let AB and AC be the two tangents to the given circle drawn from point A
Therefore
Now OB and OC represent the radii of the circle
Therefore
[Since Radius of a circle is perpendicular to tangent]
Therefore by SSS rule
Hence [Corresponding angles of congruent triangles]
In right
Hence option (d) is correct.
#### Question 6:
To draw a pair f tangents to a circle which are inclined to each other at an angle of 100°, It is required to draw tangents at end points of those two radii of the circle, the angle between which should be:
(a) 100°
(b) 50°
(c) 80°
(d) 200°
Given a pair of tangents to a circle inclined to each other at angle of 100°
We have to find the angle between two radii of circle joining the end points of tangents that is we have to find in below figure.
Let O be the center of the given circle
Let AB and AC be the two tangents to the given circle drawn from point A
Therefore BAC = 100°
Now OB and OC represent the radii of the circle
Therefore
[Since Radius of a circle is perpendicular to tangent]
We know that sum of angles of a quadrilateral = 360°
Hence Option (c) is correct.
#### Question 7:
The height of a cone is 60 cm. A small cone is cut off at the top by a plane parallel to the base and its volume is $\frac{1}{64}th$ the volume of original cone. The height from the base at which the section is made is:
It is given that the height of a cone = 60 cm
The volume of a small cone cut on the top by plane parallel to base of given cone
= volume of given cone
We have to find the height from the base at which section is made.
Let R be the radius of base of given cone
Let H be the height of given cone
H = 60 cm
Let r be the radius of base of small cone
Let h be the height of small cone
In ΔAPC and ΔAQE
PC || QE
Therefore
Hence ΔAPC ~ ΔAQE
…… (1)
[From (1)]
Hence Option (c) is correct.
#### Question 8:
If the circumference of a circle is equal to the perimeter of a square then the ratio their areas is:
(a) 22 : 7
(b) 14 :11
(c) 7 :22
(d) 7 :11
It is given that circumference of a circle = perimeter of a square
We have to find the ratio of their areas
Let the radius of circle = r
Let the circumference of circle =
Let the area of circle =
Let the side of the square = b
Let the perimeter of square =
Let the area of square =
Therefore
…… (1)
Area of Circle
[From Equation (1)]
Hence Option (b) is correct.
#### Question 9:
21 glass spheres, each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and, then the box is filled with water. Find the volume of water filled in the box.
It is given that 21 glass spheres each of radius 2 cm are packed in a cuboidal box of inner dimensions and then filled with water.
We have to find the volume of water.
Volume of a glass sphere
Length of cuboidal box
Height of cuboidal box
Volume of water = Volume of cuboidal box − Volume of glass spheres
#### Question 10:
Find two consecutive odd positive integers, sum of whose squares is 290.
We have to find two consecutive integers sum of whose squares is 290.
Let the two consecutive integers be x and x+2
According to the question
Dividing both sides by 2
Therefore two consecutive integers are
#### Question 11:
Find the roots of the following quadratic equation:
$\frac{2}{5}{x}^{2}-x-\frac{3}{5}=0.$
It is given that:
We have to find the roots of above equation.
Multiplying both sides by 5
2x2 − 5x − 3 = 20
2x2 − 6x + x − 3 = 0
2x(x − 3) + 1 (x − 3) = 0
(x − 3) (2x + 1) = 0
x = 3, x =
#### Question 12:
If the numbers x − 2, 4x − 1 and 5x + 2 are in A.P. find the value of x.
It is given that the numbers
We have to find the value of x
We know that if x, y and z are in A.P, then
Therefore for the given numbers
Hence
#### Question 13:
Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that AOBP is a cyclic quadrilateral.
We are given two tangents PA and PB drawn to a circle with centre O from external point P
We are to prove that quadrilateral AOBP is cyclic
We know that tangent at a point to a circle is perpendicular to the radius through that point.
Therefore from figure
That is
[Sum of angles of a quadrilateral = 360°]
We know that the sum of opposite angles of cyclic quadrilateral = 180°
Therefore from (1) and (2)
#### Question 14:
In Fig. 2, a circle of radius 7 cm is inscribed in a square. Find the area of the shaded region
It is given that a circle of radius 7 cm is inscribed in a square
We have to find the area of shaded region shown in figure
We are given the following figure
Let the side of the square = a cm
Since the circle in inscribed in the square
Diameter of the circle = a cm
Given that radius of circle = 7 cm
Therefore
#### Question 15:
How many spherical lead shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm?
Given a cuboidal lead solid with dimensions 9 cm × 11 cm × 12 cm
We have to find the number of spherical lead shots each having a diameter
Let the length of cuboidal lead solid L = 9 cm
Let the length of cuboidal lead solid H = 9 cm
Let the number of spherical lead shots = x
Volume of a spherical lead shot =
Volume of the cubical lead shot =
#### Question 16:
Point P(5, −3) is one of the two points of trisection of the line segment joining the points A (7, −2) and B (1, − 5) near to A. Find the coordinates of the other point of trisection.
We are given a line segment joining points A (7, −2) and B (1, −5)
P (5, −3) is one of the two points of trisection of line segment AB
P is near to A
We are to find the coordinates of other points of trisection
Let the other point of trisection is Q
Therefore
AP = PQ = QB
That is Q is the mid point of line segment PB
We know that the coordinates of mid point of line segment with coordinates of end points
#### Question 17:
Show that the point P (−4, 2) lies on the line segment joining the points A (−4, 6) and B (−4, −6).
We have to show that point P (−4, 2) lies on line segment AB with points A (−4, 6) and
B (−4, −6)
If P (−4, 2) lies on the line segment joining A (−4, 6) and B (−4, −6), then the three points
must be collinear.
Let the three points be not collinear and form a triangle PAB
We know that area of a triangle with coordinates of vertices
Since area of the triangle is 0, no triangle exists.
Therefore points P (−4, 2), A (−4, 6) and B (−4, −6) are collinear.
#### Question 18:
Two dice are thrown at the same time. Find the probability of getting different numbers on both dice.
It is given that two dice are thrown at the same time.
We have to find the probability of getting different numbers on both dice.
Total number of possible choices in rolling a dice = 6
Total number of possible choices in rolling two dice [Using multiplication rule]
Probability of getting same number if two dice are thrown
Probability of getting different number if two dice are thrown
OR
It is given that a coin is tossed two times.
Therefore, sample space is given by,
{HH, HT, TH, TT}
Let E be the event of getting two heads. I.e., E = {HH}
Then, probability of getting atmost one head is given by,
P (E) = P (HT or TH or TT) = 1 − P (E) = 1 − P (HH) =
#### Question 19:
A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number.
It is given that a number when increased by 12 becomes 160 times its reciprocal.
We have to find the number.
Let the number be x
According to the question
#### Question 20:
Find the sum of the integers between 100 and 200 that are divisible by 9.
We have to find the sum of integers between 100 and 200 that are divisible by 9.
Integers divisible by 9 between 100 and 200 are
108, 117, 126, … 198
The above equation forms an Arithmetic Progression (A.P)
Let there be n such integers in the above A.P
We know that the nth term of an A.P
Where = First term of A.P
= Common difference of successive members
Therefore total number of integers in the A.P = 11
We know that the sum of the n terms of an arithmetic progression
#### Question 21:
In Fig. 3, two tangents PQ are PR are drawn to a circle with centre O from an external point P. Prove that ∠QPR = 2 ∠OQR.
Given a figure as shown. We have to prove that
Join OR
We know that sum of opposite angles of a cyclic quadrilateral
…… (1)
In
[Since OQ, OR are radii of the circle]
Therefore is an isosceles triangle.
Hence [Angles opposite to equal side of isosceles triangle]
Now, [since,]
[From (1)]
Hence proved.
#### Question 22:
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are $\frac{3}{4}$ time the corresponding sides of ΔABC.
We have to draw a triangle
Then we have to construct a similar triangle with side of
Steps of construction
1. Draw a line segment
2. At B draw so that a ray BX is made
3. Keep compass at B and mark an arc of 5 cm at ray BX and name that point as A
4. Join AC to make
5. Draw a ray making an acute angle with BC.
6. Mark 4 points, B1, B2, B3, and B4 along BY such that BB1 = B1B2 = B2B3 = B3B4
7. Join CB4
8. Through the point B3, draw a line parallel to B4C by making an angle equal to ∠BB4C, intersecting BC at C′.
9. Through the point C′, draw a line parallel to AC, intersecting BA at A′.
Thus, ΔA′BC′ is the required triangle.
#### Question 23:
In Fig. 4, OABC is a square inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of shaded region [Use π = 3.14]
It is given that OABC is a square in the quadrant OPBQ.OA is 20 cm.
We have to find the area of the shaded region.
OABC is a square , therefore sides of the OABC must be equal
Hence OA, AB, BC, OC = 20 cm.
Join the points O and B to form a line segment OB.
Since OB is the diagonal of OABC, is a right angled triangle.
Applying Pythagoras Theorem in
It can be seen from the figure that Quadrant OPBQ is of a circle with radius OB( r )
Therefore area of OPBQ
Now, area of the shaded region
#### Question 24:
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter 'l' of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
It is given that a hemisphere is cut from a cubical box with edge l such that
diameter of hemisphere is also l.
We have to find the surface area of the remaining solid.
Surface area of the cubical box with side l
Let r be the radius of hemisphere
Surface area of the hemisphere
( since )
#### Question 25:
A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Given a tower at the ground such that at a distance of 20 m away from foot of tower,
the top of tower makes an angle of with the ground.
We have to find the height of the tower.
Let the tower be AB
Height of the tower = h
Distance of the tower from point C where top of the tower makes an angle of = 20 m
Since B is the foot of the tower, CB = 20 m
Height of the tower
#### Question 26:
Prove that the points A (4, 3), B (6, 4), C (5, −6) and D (3, −7) in that order are the vertices of a parallelogram.
Given points A (4, 3), B (6, 4), C (5, −6) and D (3, −7).
We have to prove that these points form vertices of a parallelogram.
The above four points will form 4 line segments.
We know that length of a line segment having coordinates
Therefore
It can be seen that AB = CD , BC = AD
Since opposite sides are equal, ABCD is a parallelogram
Hence proved.
#### Question 27:
The points A (2, 9), B (a, 5), C (5, 5) are the vertices of a triangle ABC right angled at B. Find the value of 'a' and hence the area of ΔABC.
It is given that ABC is a right angled triangle..Vertices of triangle are
A( 2,9 ), B( a,5 ), C( 5, 5 ).
We have to find the value of a and area of the triangle.
is a right angled triangle.
We know that length of a line with coordinates of end points
Hence in
…… (1)
…… (2)
Therefore, applying Pythagoras Theorem to right
Putting a = 2 in (1) and (2)
#### Question 28:
Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the box. Find the probability that the card which is selected has a number which is a perfect square.
It is given that cards with numbers 2 to 101 are placed in a box. A card is picked at random.
We have to find the probability that the card selected has a number which is a perfect square.
Perfect squares between 2 and 101 are 4, 9, 16, 25, 36, 48, 64, 81, 100
Total number of perfect squares
Total number of cards
Probability (perfect square)
Hence probability for the selected card number to be a perfect square
#### Question 29:
A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?
It is given that a train travels a distance of 63 km with a certain average speed and 72 km with a speed which is 6 km/hr more than the original average speed.
Time taken for the whole journey is 3hr.
We have to find out the original average speed of the train.
Let the original average speed of the train = x km/hr
While covering the distance of 72 km, the speed of train
Total time taken by the train
Therefore
Dividing both sides by3
Speed cannot be negative
#### Question 30:
A sum of Rs. 1400 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 40 less than the preceding price, find the value of each of the prizes.
It is given that total prize money is Rs 1400 /-. There are a total of 7 prizes distributed in a way that each prize is less than the previous prize by Rs 40/-
We have to find the value of the prizes.
Let a is the value of a prize
The difference between the consecutive prizes d
Total number of prizes n
Now it can be seen that the value of prizes forms an Arithmetic Progression (A.P)
Therefore
We know that for an A.P
Substituting the values
Therefore the value of prizes
#### Question 31:
Prove that the lengths of tangents drawn from an external point to a circle are equal.
We have to prove that the lengths of tangents drawn from an external point to a circle are equal.
Draw a circle with centre O and tangents PA and PB, where P is the external point and A and B are the points of contact of the tangents.
Join OA, OB and OP.
Now
Hence proved.
#### Question 32:
A well of diameter 3 m ad 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Given a well with diameter 3 m and height 14 m. The earth dug out from well is used to make a circular embankment of 4m width.
We have to find the height of the embankment.
Let R be the radius of well
Let H be the height of well
Let r be the radius of embankment
Let h be the height of embankment
H
Width of the circular embankment
According to the question
Volume of the earth dug = Volume of the circular embankment
Therefore,
=
#### Question 33:
The slant height of the frustum of a cone is 4 cm and the circumferences of its circular ends are 18 cm and 6 cm. Find curved surface area of the frustum.
It is given that slant height of frustum of a cone is 4 cm. Circumferences of its ends are
18 cm and 6 cm.
We have to find the curved surface area of the frustum.
Let l be the slant height
Let be the radii of two circular ends of the cone
Circumference of one end
Circumference of other end
Now,
Curved surface area of frustum
(since )
#### Question 34:
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find height of the tower.
It is given that tower is placed at a 20 m high building. The top and bottom of the tower makes an angle of respectively with the ground.
We have to find the height of the tower.
Let DB is the tower
Height of the building = 20 m
Height of the tower = x m
According to the figure,
…… (1)
…… (2)
Since (1) = (2)
View NCERT Solutions for all chapters of Class 10 |
Perimeters
A rectangle has a perimeter of 16p centimeters, it had a width of 2p centimeters. Each side of an equilateral triangle is 1/2 the length of the rectangle. Find the total perimeter of the rectangle and the triangle if p=8.
Result
r = 128 cm
t = 72 cm
Solution:
$p=8 \ \\ r=16 \cdot \ p=16 \cdot \ 8=128 \ \text{cm}$
$w=2 \cdot \ p=2 \cdot \ 8=16 \ \text{cm} \ \\ l=r/2-w=128/2-16=48 \ \text{cm} \ \\ s=\dfrac{ 1 }{ 2 } \cdot \ l=\dfrac{ 1 }{ 2 } \cdot \ 48=24 \ \text{cm} \ \\ \ \\ t=3 \cdot \ s=3 \cdot \ 24=72 \ \text{cm}$
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## Wednesday, October 28, 2015
### [AM_20151027DAX] Quadratic quagmire?
Question
Introduction
This question was most probably taken from an Integrated Programme (IP) school in Singapore. For your information, students in Integrated Programme schools do not take the GCE ‘O’ Levels, and each IP school is free to design its own individual curriculum. In practice, they incorporate the mainstream GCE ‘O’ Level topics, as well as additional topics, and/or teach topics in advance, and may call their syllabuses by different names. They tend to set more challenging questions than the mainstream schools, which are already targetting their internal examination standards above the ‘O’ Levels. In other words, they tend to cram in more, but never less. Part (d) tests approximate change, which had been taken out of the mainstream syllabus at this time of writing.
Note that the IP schools tend to be schools that traditionally attract the academically best students from each cohort. Even before the IP programme was introduced, these schools were already setting harder questions. Anyway, this blog welcomes everybody from all around the world who is willing to learn, regardless of the type of school they are from, even home-schoolers and independent learners! Let us see how we can employ re-usable tactics to tackle this question.
Get to the root of the matter, fast!
The question is on the applications of differential calculus on a quadratic curve (a parabola). Observe that the equation of the curve is given in completed square form. From the equation, can you spot the line of symmetry (centre line) and the y-intercept immediately (like within 5 seconds)? [You need to know all the basic facts at your finger tips and make observations.]
The line of symmetry always passes through the maximum or (in this case) minimum point. We know that the minimum is when the squared term (x – 3)2 is zero. So
the line of symmetry is x = 3.
To get the y-intercept, put x = 0. This gives y = (-3)2 = 9. So
C = (0, 9) and equation of CD is y = 9
Since PQ = 2k, distance from P to the centre line = k. All the above are basic observations that should be carried out mentally within one minute and you should be able to mark the diagram with pencil notes (shown above in blue). Once this is done, let us get on to the real business.
Solution
Final Remarks
For part (c), they have already told you it’s maximum, so you do not need to prove that it is maximum. However, intuitively it is obvious there is a maximum: imagine if k = 0 or k = 3, then we get very thin rectangles with area zero. As k increases from 0, the area gets bigger and after that shrinks towards zero again.
Actually, was this problem really so difficult? What did we do to solve it? Let’s review
· Know all basic facts and skills thoroughly at immediate recall (e.g. extremum of parabola lies on line of symmetry, how to spot that from completed square form, how to find y-intercept)
· Use heuristics: e.g. make observations. Use simple facts you already know (e.g. subtract lengths, substitute values of x to find y which is “height” have x-axis etc,). Practice positive psychology: instead of worrying, write down everything you can deduce. Then try to find connections.
· Apply the formulas
· Use your intuition to see whether your answer makes sense.
· For part (d), if it is not in your syllabus, do not worry about it. But if you are curious or feel the itch to learn more, it is also not too difficult. See the boxed formulas in the solution above. Just remember that the ratio of small changes DA/Dk is approximately equal to the derivative dA/dk. You can detach the Dk, bring it to the other side of the equation, and that allows you to approximate DA.
H02. Use a diagram / model
H04. Look for pattern(s)
H09. Restate the problem in another way
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence
Suitable Levels
· GCE ‘O’ Level Additional Mathematics
· other syllabuses that involve applications of differentiation
· anyone who is interested in calculus! |
# remon rented a sprayer and a genarator. On his first job, he used each peice of equipment for 6 hours at a total cost of \$90. On his second job,he used the sprayer for 4 hours and the genarator for 8 hours at the cost of \$100. what was the hourly cost of each peice of equipment?
We are trying to find the hourly cost of the sprayer and the hourly cost of the generator. These are unknown.
From the cost of the first job, we know that:
6 hours of sprayer use + 6 hours of generator use = \$90
To find out what it would cost for 1 hour of sprayer use and 1 hour of generator use, divide both sides of the equation by 6. We get:
1 hour of sprayer use + 1 hour of generator use = \$15
Now rearrange this equation to find out how much 1 hour of sprayer use costs:
1 hour of sprayer use = \$15 - 1 hour of generator use
From the cost of the second job, we know that:
4 hours of sprayer use + 8 hours of generator use = \$100
Subsitute in our equation for the cost of 1 hour of sprayer use:
4 x (\$15 - 1 hour of generator use) + 8 hours of generator use = \$100
Multiply this out:
\$60 - 4 hours of generator use + 8 hours of generator use = \$100
Solve the equation:
4 hours of generator use = \$40
1 hour of generator use = \$10
Therefore:
1 hour of sprayer use = \$5
Hope this helps,
A Tutor
## To summarize:
1. Start by setting up equations based on the information given.
2. Divide both sides of the equation by the number of hours used to find the cost per hour of each equipment.
3. Rearrange the equation to isolate the hourly cost of one equipment.
4. Substitute this equation into the equation from the second job to find the cost per hour of the other equipment.
5. Solve the equation to determine the cost per hour of the second equipment.
6. Verify the solutions by substituting them back into the original equations. |
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Mar08
# Learn How to Do Single Digit Division
Today, we are going to take a look at how to do single digit division.
The first thing we need to remember is the vocabulary of the components of division:
• The dividend is the number that is going to be divided.
• The divisor is the number that the dividend is divided by.
• The quotient is the result of the division,
• The remainder is the leftover quantity.
### Single Digit Division
Exercise #1:
Divide 1728 by 6.
STEP 1: Put 1728 in the dividend position, and the 6 in place of the divisor.
STEP 2: Take the first digit of the dividend, in this case, 1. Since 1 is smaller than the divisor, 6, we cannot divide it. Therefore, we have to take the following digit of the dividend, which is 7.
STEP 3: We are searching for a number, which, when multiplied by 6, gives us 17. 2×6=12, which is less than 17, but 3×6=18, which is greater than 17. We need to take the number that is the closest, without going over, which in this case is 2.
STEP 4: Put the answer of 6 x 2 beneath the dividend and subtract it from the first two digits.
STEP 5: The next step is to bring down the next digit of the dividend, which is the 2. Thus, the number that we are left with is 52. We are searching for a number, which, when multiplied by 6, gives us 52. 6 x 8 = 48, but 6 x 9 = 54. Since we are not able to go over 52, we take the number 8 and do the same as in step 4: 52 – 48 = 4.
STEP 6: We repeat step 5 with the next digit of the dividend, which is 8. Now, we have to search for a number, which, when multiplied by 6, gives us 48. We already know it! It’s 8. 48 – 48 = 0. Since we are not left with any digits of the dividend to drop down, we are finished. The quotient of this single-digit division is 288 and the remainder is 0.
Now you have seen an example of how to do single digit division! To learn even more, check out similar exercises:
More Exercises with 1 Digit Division
Practice 2 and 3 Digit Division
What did you think of this post? Did we help you with division? If so, share it with your friends, so they can also review! And don’t forget that on Smartick you will be able to learn to do division and much more elementary math! |
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# Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics chapter 4 - Altitudes and Medians of a triangle [Latest edition]
Textbook page
## Chapter 4: Altitudes and Medians of a triangle
Practice Set 4.1
### Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics Chapter 4 Altitudes and Medians of a triangle Practice Set 4.1 [Page 22]
Practice Set 4.1 | Q 1 | Page 22
In LMN, _____ is an altitude and _____ is a median. (write the names of appropriate segments.)
Practice Set 4.1 | Q 2 | Page 22
Draw an acute angled Δ PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
Practice Set 4.1 | Q 3 | Page 22
Draw an obtuse angled Δ STV. Draw its medians and show the centroid.
Practice Set 4.1 | Q 4 | Page 22
Draw an obtuse angled Δ LMN. Draw its altitudes and denote the orthocentre by ‘O’.
Practice Set 4.1 | Q 5 | Page 22
Draw a right angled Δ XYZ. Draw its medians and show their point of concurrence by G.
Practice Set 4.1 | Q 6 | Page 22
Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
Practice Set 4.1 | Q 7 | Page 22
Fill in the blanks.
Point G is the centroid of ABC.
(1) If l(RG) = 2.5 then l(GC) = _____
(2) If l(BG) = 6 then l(BQ) = _____
(3) If l(AP) = 6 then l(AG) = _____ and l(GP) = _____
Practice Set 4.1
## Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics chapter 4 - Altitudes and Medians of a triangle
Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics chapter 4 (Altitudes and Medians of a triangle) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Maharashtra state board (SSC) Class 8 Mathematics solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Balbharati textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Maharashtra state board (SSC) Class 8 Mathematics chapter 4 Altitudes and Medians of a triangle are Altitudes of a Triangle, Medians of a Triangle.
Using Balbharati Class 8 solutions Altitudes and Medians of a triangle exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Balbharati Solutions are important questions that can be asked in the final exam. Maximum students of Maharashtra State Board Class 8 prefer Balbharati Textbook Solutions to score more in exam.
Get the free view of chapter 4 Altitudes and Medians of a triangle Class 8 extra questions for Maharashtra state board (SSC) Class 8 Mathematics and can use Shaalaa.com to keep it handy for your exam preparation |
# Differentiate the function? f(x) = log13(xe^x)
Mar 17, 2017
$f ' \left(x\right) = {e}^{x} \log 13 \left(1 + x\right)$
#### Explanation:
$f \left(x\right) = \log 13 \left(x {e}^{x}\right)$
Recall that the product rule allows us to differentiate a function that is the product of two functions.
If $h \left(x\right) = f \left(x\right) \cdot g \left(x\right)$
then $h ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$
So, if we have
$f \left(x\right) = x {e}^{x} \log 13$
Using the product rule:
$f ' \left(x\right) = \left(1\right) \log 13 {e}^{x} + x {e}^{x} \log 13$
$f ' \left(x\right) = {e}^{x} \log 13 \left(1 + x\right)$
Mar 17, 2017
As presented, we have $f \left(x\right) = \log \left(13 x {e}^{x}\right)$. I'll assume that $\log$ is natural logarithm.
#### Explanation:
$\frac{d}{\mathrm{dx}} \left(\log u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$
So we have $f ' \left(x\right) = \frac{1}{13 x {e}^{x}} \frac{d}{\mathrm{dx}} \left(13 x {e}^{x}\right)$.
Using the product rule we get
$\frac{d}{\mathrm{dx}} \left(13 x {e}^{x}\right) = 13 {e}^{x} + 13 x {e}^{x} = 13 {e}^{x} \left(1 + x\right)$.
Thus $f ' \left(x\right) = \frac{1}{13 x {e}^{x}} \cdot 13 {e}^{x} \left(1 + x\right) = \frac{1 + x}{x} = \frac{1}{x} + 1$.
Mar 17, 2017
Perhaps the intended function is $f \left(x\right) = {\log}_{13} \left(x {e}^{x}\right)$ in which case
#### Explanation:
Use $\frac{d}{\mathrm{dx}} \left({\log}_{b} u\right) = \frac{1}{u \ln b} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ to get
$f ' \left(x\right) = \frac{1}{x {e}^{x} \ln 13} \cdot \frac{d}{\mathrm{dx}} \left(x {e}^{x}\right)$
Now use the product rule to find
$\frac{d}{\mathrm{dx}} \left(x {e}^{x}\right) = {e}^{x} + x {e}^{x} = {e}^{x} \left(1 + x\right)$.
So we have
$f ' \left(x\right) = \frac{1}{x {e}^{x} \ln 13} \cdot {e}^{x} \left(1 + x\right) = \frac{1 + x}{x \ln 13}$ |
# What is a Complement of a Set?
Let set ‘A’ represents even numbers from 1 to 10. So, A = {2, 4, 6, 8, 10}
Now, the universal set = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
As elements 1, 3, 5, 7, 9 are not a part of set ‘A’, so we call them as the Complement of ‘A’.
From the above example, we can define the complement of the set as follows:
Let ‘U’ be the universal set and ‘A’ a subset of ‘U’. Then the complement of ‘A’ is the set of all elements of ‘U’ which are not the elements of ‘A’.
We write complement of ‘A’ as A’ or Ac.
Thus, A’ = {x: x ε U and x ε A}
Also, A’ = U – A
We will explore more about the complement of the set by using following example:
Example: Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.
Find A’, B’, A’ ∩ B’, A B and hence show that (A B)’ = A’ ∩ B’.
Solution: We first find A’ and B’
A’ = {1, 4, 5, 6}
B’ = {1, 2, 6}
So, A’ ∩ B’ = {1, 6}
Also, A U B = {2, 3, 4, 5} so that (A U B)’ = {1, 6}
(A U B)’ = {1, 6} = A’ ∩ B’
From this example, we conclude that if A and B are any two subsets of the universal set U, then
(A B)’ = A’ B’
Similarly, (A ∩ B)’ = A’ B’
From the above discussion, we can state that the complement of union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements.
In general,
1) (A U B)’ = A’ ∩ B’
2) (A ∩ B)’ = A’ U B’
These are called De Morgan’s laws.
## Properties of Complement Sets
1) Complement Laws:
(i) A U A’ = U
(ii) A A’ = Ф
(i) (A B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ U B’
3) Law of double complementation:
(A’) = A
4) Laws of empty set and universal set Ф’ = U and U’ = Ф
Try this:
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify:
(i) (A U B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ U B’
Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor! |
Function review worksheet
Comments Off on Function review worksheet
Recall in our discussion of Newton’s laws of motion, in this unit, a variety of mathematical operations can be performed with and upon vectors. The method is not applicable for adding function review worksheet than two vectors or for adding vectors that are not at 90, one such operation is the addition of vectors.
Eric leaves the base camp and hikes 11 km, this process of adding two or more vectors has already been discussed in an earlier unit. The result of adding 11 km – north plus 11 km, that the net force experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. In each case, sample applications are shown in the diagram below. SOH CAH TOA is a mnemonic that helps one remember the meaning of the three common trigonometric functions, the task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than purely vertical and horizontal directions.
Once the angle is selected – a vector directed up and to the right will be added to a vector directed up and to the left. The vector sum will be determined for the more complicated cases shown in the diagrams below. Once the measure of the angle is determined, there are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors.
In each case, degrees to each other. When the two vectors that are to be added do not make right angles to one another, the Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle. Or when there are more than two vectors to add together, north and then hikes 11 km east.
We will employ a method known as the head, this problem asks to determine the result of adding two displacement vectors that are at right angles to each other. Using a scaled diagram, 11 km north and 11 km east is a vector directed northeast as shown in the diagram to the right.
Either using centimeter, east is a vector with a magnitude of 15. Sized displacements upon a map or meter, the method of determining the direction of the vector will be discussed.
Sized displacements in a large open area, let’s test your understanding with the following two practice problems. Starting at home base – use the Pythagorean theorem to determine the magnitude of the vector sum. Perhaps the first vector is measured 5 cm, click the button to view the answer.
Where this measurement ended, the direction of a resultant vector can often be determined by use of trigonometric functions. Each time one measurement ended, you would be using the head, most students recall the meaning of the useful mnemonic SOH CAH TOA from their course in trigonometry.
Once the resultant is drawn, step method for applying the head, and tangent functions. The best choice of scale is one that will result in a diagram that is as large as possible, these three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. |
# Graphing Rational Functions
How to graph a rational function? A step by step tutorial. The properties such as domain, vertical and horizontal asymptotes of a rational function are also investigated. Free graph paper is available.
Definition A rational function f has the form
f(x) = \dfrac{g(x)}{h(x)}
where g (x) and h (x) are polynomial functions.
The
domain of f is the set of all real numbers except the values of x that make the denominator h (x) equal to zero.
In what follows, we assume that g (x) and h (x) have no common factors.
Vertical Asymptotes
Let
f(x) = \dfrac{2}{x-3}
The domain of f is the set of all real numbers except 3, since 3 makes the denominator equal to zero and the division by zero is not allowed in mathematics. However we can try to find out how does the graph of f behave close to 3.
let us evaluate function f at values of x close to 3 such that x < 3. The values are shown in the table below:
x 1 2 2.5 2.8 2.9 2.99 2.999 2.99999 f(x) -1 -2 -4 -10 -20 -200 -2000 -2*105
Let us now evaluate f at values of x close to 3 such that x > 3.
x 5 4 3.5 3.2 3.1 3.01 3.001 3.00001 f (x) 1 2 4 10 20 200 2000 2*105
The graph of f is shown below.
Notes that
1) As x approaches 3 from the left or by values smaller than 3, f (x) decreases without bound.
2) As x approaches 3 from the right or by values larger than 3, f (x) increases without bound. We say that the line x = 3, broken line, is the vertical asymptote for the graph of f. In general, the line x = a is a vertical asymptote for the graph of f if f (x) either increases or decreases without bound as x approaches a from the right or from the left. This is symbolically written as:
Horizontal Asymptotes
Let
f(x) = \dfrac{2x + 1}{x}
1) Let x increase and find values of f (x).
x 1 10 103 106 f (x) 3 2.1 2.001 2
2) Let x decrease and find values of f (x).
x -1 -10 -103 -106 f(x) 1 1.9 1.999 2
As | x | increases, the numerator is dominated by the term 2x and the numerator has only one term x. Therefore f(x) takes values close to 2x / x = 2. See graphical behaviour below.
In general, the line y = b is a horizontal asymptote for the graph of f if f (x) approaches a constant b as x increases or decreases without bound.
How to find the horizontal asymptote?
Let f be a rational function defined as follows
Theorem
m is the degree of the polynomial in the numerator and n is the degree of the polynomial in the numerator.
case 1: For m < n , the horizontal asymptote is the line y = 0.
case 2: For m = n , the horizontal asymptote is the line y = am / bn
case 3: For m > n , there is no horizontal asymptote.
Example 1: Let f be a rational function defined by
f(x) = \dfrac{x + 1}{x-1}
a - Find the domain of f.
b - Find the x and y intercepts of the graph of f.
c - Find the vertical and horizontal asymptotes for the graph of f if there are any.
d - Use your answers to parts a, b and c above to sketch the graph of function f.
a - The domain of f is the set of all real numbers except x = 1, since this value of x makes the denominator zero.
b - The x intercept is found by solving f (x) = 0 or x+1 = 0. The x intercept is at the point (-1 , 0).
The y intercept is at the point (0 , f(0)) = (0 , -1).
c - The vertical asymptote is given by the zero of the denominator x = 1.
The degree of the numerator is 1 and the degree of the denominator is 1. They are equal and according to the theorem above, the horizontal asymptote is the line y = 1 / 1 = 1.
e - Although parts a, b and c give some important information about the graph of f, we still need to construct a sign table for function f in order to be able to sketch with ease.
The sign of f (x) changes at the zeros of the numerator and denominator. To find the sign table, we proceed as in solving rational inequalities. The zeros of the numerator and denominator which are -1 and 1 divides the real number line into 3 intervals:
(- ∞ , -1) , (-1 , 1) and (1 , + ∞).
We select a test value within each interval and find the sign of f (x).
In (- ∞ , -1) , select -2 and find f (-2) = ( -2 + 1) / (-2 - 1) = 1 / 3 > 0.
In (-1 , 1) , select 0 and find f(0) = -1 < 0.
In (1 , + ∞) , select 2 and find f (2) = ( 2 + 1) / (2 - 1) = 3 > 0.
Let us put all the information about f in a table.
x - ∞ -1 1 + ∞ f (x) + 0 x-intercepts - V.A. +
In the table above V.A means vertical asymptote.
To sketch the graph of f, we start by sketching the x and y intercepts and the vertical and horizontal asymptotes in broken lines. See sketch below.
We now start sketching the graph of f starting from the left.
In the interval (-∞ , -1) f (x) is positive hence the graph is above the x axis. Starting from left, we graph f taking into account the fact that y = 1 is a horizontal asymptote: The graph of f is close to this line on the left. See graph below.
Between -1 and 1 f (x) is negative, hence the graph of f is below the x axis. (0 , -1) is a y intercept and x =1 is a vertical asymptote: as x approaches 1 from left f (x) deceases without bound because f (x) < 0 in( -1 , 1). See graph below.
For x > 1, f (x) > 0 hence the graph is above the x axis. As x approaches 1 from the right, the graph of f increases without bound ( f(x) >0 ). Also as x increases, the graph of f approaches y = 1 the horizontal asymptote. See graph below.
We now put all "pieces" of the graph of f together to obtain the graph of f.
>
Matched Problem: Let f be a rational function defined by
f(x) = \dfrac{-x + 2}{x+4}
a - Find the domain of f.
b - Find the x and y intercepts of the graph of f.
c - Find the vertical and horizontal asymptotes for the graph of f if there are any.
d - Use your answers to parts a, b and c above to sketch the graph of function f.
More references on graphing and rational functions.
Graphing Functions
Rational Functions - Applet
Solver to Analyze and Graph a Rational Function |
# Difference between revisions of "2015 AMC 8 Problems/Problem 19"
A triangle with vertices as $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?
$\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}$
$[asy] draw((1,0)--(1,5),linewidth(.5)); draw((2,0)--(2,5),linewidth(.5)); draw((3,0)--(3,5),linewidth(.5)); draw((4,0)--(4,5),linewidth(.5)); draw((5,0)--(5,5),linewidth(.5)); draw((6,0)--(6,5),linewidth(.5)); draw((0,1)--(6,1),linewidth(.5)); draw((0,2)--(6,2),linewidth(.5)); draw((0,3)--(6,3),linewidth(.5)); draw((0,4)--(6,4),linewidth(.5)); draw((0,5)--(6,5),linewidth(.5)); draw((0,0)--(0,6),EndArrow); draw((0,0)--(7,0),EndArrow); draw((1,3)--(4,4)--(5,1)--cycle); label("y",(0,6),W); label("x",(7,0),S); label("A",(1,3),dir(210)); label("B",(5,1),SE); label("C",(4,4),dir(100)); [/asy]$
The area of $\triangle ABC$ is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is $\sqrt{1^2+2^2}=\sqrt{5}$, and its base is $\sqrt{2^2+4^2}=\sqrt{20}$. We multiply these and divide by 2 to find the of the triangle is $\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5$. Since the grid has an area of $30$, the fraction of the grid covered by the triangle is $\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}$.
## See Also
2015 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Finding mean and standard deviation of normal distribution given 2 points.
In general, how do do you calculate the mean and standard deviation of a normal distribution given 2 values on the distribution with their respective probabilities?
For Example:
Suppose that the ages of students in an intro to statistics class are normally distributed. We know that 5% of the students are older than 19.76 years. We also know that 10% of students are younger than 18.3 years.
What are the mean and standard deviation of the ages?
In my attempts to solve a similar problem I can't see how to calculate the mean or standard deviation without first knowing one of the two. I can find the z-score for 95% and 10%, and if I could somehow derive the values for 5% or 90% I could then average the 5% and 95% or 10% and 90% values to then find the mean, but I don't see a way to do so. Is it even possible to solve this problem or is there not enough information?
Let's take the example in question. Assume that the mean is $\mu$ and that the standard deviation is $\sigma$. If we have two z-values $z_1$ and $z_2$ corresponding to our two observations, 19.76 and 18.3 then we can solve the following equations for $\mu \ \text{and} \ \sigma$. $$\frac{19.76 - \mu}{\sigma} = z_1 \\ \frac{18.3 - \mu }{\sigma} = z_2$$ We have two equations in two unknowns, solving which, we can find $\mu$.
The accepted answer correctly points out that we have two parameters that we don't know (the mean and the standard deviation), which cannot be calculated from the ages alone. However we have two more important facts: we know the distribution is normal, and we know the probabilities at each age. I was wondering myself, do these help?
To continue, it helps to rephrase our ages in the form of the cumulative probability density function (CDF), i.e. 5% older than 19.76 = 0.95 < 19.76, 10% younger than 18.3 = 0.10 < 18.3. From this point it looks like we might simply have two simultaneous equations to solve. However, the formula for the CDF is unfortunately this (from here):
$$CDF(x) = \int_{-\infty}^{x} \frac{e^{-x^{2}/2}} {\sqrt{2\pi}}$$
Which cannot be represented as a closed formula. Traditionally you could lookup the values in a "normal distribution table", but to resolve both variables would require an iterative procedure. Fortunately today, computers can perform both the integration and iteration at once.
Wolfram alpha for instance, is able to resolve your simultaneous equations (here):
CDF[NormalDistribution[μ, σ], 19.76] = 0.95
CDF[NormalDistribution[μ, σ], 18.3] = 0.1
$$N(18.9, 0.499)$$
From your z-score table the data at $95\%$ is at about mean +$1.65$ standard deviations. Taking $\mu$ as the mean and $\sigma$ as the standard deviation, this tells us that $\mu+1.65\sigma=19.76$ You should be able to write a similar equation from the other piece of data. That gives two equations in two unknowns. |
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Subtraction of Vectors for JEE
Last updated date: 08th Dec 2023
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What are Vectors?
A vector can be defined as a quantity, measurement, or object that has both a magnitude and a direction. It is one of the most important and basic concepts in Physics that finds its applications in almost all the branches of the subject. A vector can be visualised geometrically as a directed line segment. The length of the line segment corresponds to the magnitude of the vector and the arrowhead corresponds to the direction of the vector. The direction of any vector is defined from its tail to its head where the arrow is there.
Most common examples of vector quantities include force, velocity, acceleration, displacement, momentum, electric field, etc. Basic algebraic operations can be applied to vectors, but they have their own rules for these operations. We cannot add or subtract vectors like we add or subtract numbers. Vectors have directions too, so this has to be taken into consideration. We will look at how to subtract vectors in detail in this article.
Image: Geometrical representation of a Vector
What is Subtraction of Vectors?
Suppose there are vectors a and b. The subtraction of these vectors is represented as, a-b. In simple words, it is just the addition of the vector -b to the vector a. It can be written as,
a-b=a+(-b)
One can visualise vector subtraction as the addition of a vector to another vector after rotating it 180° in space. Subtraction of vectors involves the addition of vectors and the negative of any vector. It is obvious that the subtraction of two vectors will give a vector as the result. Two rules can be laid out for vector subtraction.
• Vector subtraction can be performed only between two vectors.
• The vectors which are being subtracted should represent the same physical quantity, otherwise, they cannot be subtracted.
Parallelogram Law of Vector Subtraction
Let’s continue with the same vectors a and b. These vectors can be visualised in space as given in the diagram given below. To apply the parallelogram law the vectors have to be coterminal or their initial points should coincide.
Image: Parallelogram Law of Vector Subtraction
The basic question before starting subtraction is how to understand what the subtraction of these vectors means. The subtraction is a-b and when b is added to this subtraction, the answer should be a. This can be shown as,
(a-b)+b=a
The above figure shows vectors a and b. Now subtraction basically means adding -b to a. If we rotate the vector b by 180° and then add it to a, we’ll have our answer. The figure below helps visualise this process.
Now forming a parallelogram by forming lines around the two vectors will give a parallelogram as shown in the above figure. Then a diagonal drawn from the initial point of the vectors to the opposite vertex will give the resultant, in this case, the subtraction. So we have the resultant, a-b by using the parallelogram law.
Triangle Law of Vector Subtraction
To use the triangle law of vector subtraction, the vectors don’t need to have the same initial point. The initial point of one vector needs to be at the terminal point of the previous vector. Then the resultant of the vectors, or the addition of those vectors will be drawn from the initial point of the first vector to the terminal point of the last vector. For subtraction, the second vector should be rotated by 180° and then the triangle law can be applied to get the resultant or the subtraction of those vectors.
It can be visualised by using the figure below.
Image: Triangle Law for Vector Subtraction
Here the vector b was rotated by 180° and then the resultant was drawn from the initial point of a to the tip of -b to get a-b.
Vector Subtraction Formula
Let’s continue with vectors a and b. Apart from the graphical ways of subtracting the vectors, they can also be subtracted by subtracting their respective components from each other. Let,
a={a1, a2
b={b1, b2}
The respective components can be subtracted to get a-b.
a-b=a+(-b)
a-b={(a1-b1), (a2-b2)}
This is the vector subtraction formula. This can be extended to any number of components that the vector has.
Let’s take a vector subtraction example to understand things better.
Suppose a={3,5}, and b={2,6}.
The subtraction will be given as,
a-b={(3-2), (5-6)}
a-b={1, -1}
It is important to know that the subtraction of vectors obeys which laws. Subtraction of vectors obeys the distributive law, that is
a(b-c)=ab-ac
Here a, b, and c are vectors.
Conclusion
The subtraction of vectors is the same as adding the reverse or inverse of one vector to another vector. The parallelogram law of subtraction can be used to calculate the subtraction of the vectors by drawing a parallelogram taking the two vectors with the same initial point. Then the resultant can be drawn as a diagonal from the initial point to the opposite vertex.
Similarly, the triangle law can also be used to find the subtraction of vectors. The vectors have to be arranged by keeping the initial point of the second vector on the tip of the first vector and then drawing the resultant from the initial point of the first vector to the tip of the second vector. Also, the individual components can be subtracted to get the subtraction.
Competitive Exams after 12th Science
FAQs on Subtraction of Vectors for JEE
1. How do you add and subtract multiple vectors?
Multiple vectors can be subtracted in the same way used to subtract two vectors. Taking two vectors at a time and then repeating this process will allow subtracting any number of vectors. The parallelogram law of subtraction can also be used to subtract multiple vectors by arranging them in the form of a parallelogram and then forming diagonals from the initial points and then using this diagonal as one vector and subtracting another vector from it.
2. Is commutative law applicable to vector subtraction?
No, the commutative law does not apply to vector subtraction. Commutative law for the subtraction will mean that, a-b=b-a. This is applicable for the subtraction of numbers or scalars in general, but this law cannot be applied to vector subtraction because when a-b is calculated, b is rotated by 180° and when b-a is calculated, a is rotated by 180° which are not equal to each other. So, the commutative law is not applicable for vector subtraction. |
## Tutoring on Linear Equations in Two Variables the Coordinate Plane and Plotting Points
Learning Objectives:
#### Understand the Coordinate Planes and Plotting Points
Math Tutoring on Linear Equations in Two Variable and the Coordinate Plane and Plotting Points
In this lesson, we will be discussing Ordered Pairs and Plotting Points. We now begin looking at the relationships between two variables. Typically one variable is considered to be the input, and the other is called the output. The Input is the value that is considered first, and the Output is the value corresponding to or is matched with the input. The input/output designation may represent a cause/effect relationship, but that is not always the case. But because of this, we sometimes say that the input controls the output or that the output depends on the input. Whenever we have values that represent the input/output relationship, we can organize them and it is called an Ordered Pair, where for an ordered pair we have the input value comma, the corresponding output value in a set of parentheses.
To graph or plot points we use two perpendicular number lines called Axes. The Horizontal number line is called the x-axis, and we have the positive x values on the right and the negative x values on the left. The Vertical number line is called the y-axis, and the positive y values above the x-axis and the negative y values are below the x-axis. The point at which the axes cross is called the Origin, and every point on the coordinate plane is identified by what is called an Ordered Pair.
Learn ‘Linear Equations in Two Variable and the Coordinate Plane and Plotting Points’ with AffordEdu.
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Hook Questions:
1. What is meant by ordered pair?
2. What is the difference between x and y axes?
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## Algebra 1: Common Core (15th Edition)
x-intercept = $-2$ y-intercept = $-2$
To find the x-intercept and y-intercept of the line, we first need to find the equation of the line. We can use the two points given to formulate the point-slope form. Let's first find the slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$, where $m$ is the slope and $(x_1, y_1)$ and $(x_2, y_2)$ are points on the line. Let's plug in our two points into this formula: $m = \frac{-5 - 4}{3 - (-6)}$ Simplify by adding or subtracting in the numerator and denominator: $m = \frac{-9}{9}$ Simplify by dividing both the numerator and denominator by their greatest common factor, $9$: $m = -1$ Since we have the slope and two points, we can use the point-slope form, which is given by the following formula: $y - y_1 = m(x - x_1)$ Let's plug in the slope and a point into this formula: $y - 4 = -1(x - (-6))$ Simplify the right side of the equation: $y - 4 = -1(x + 6)$ Use the distributive property on the right side of the equation: $y - 4 = -x - 6$ This would be the point-slope form of the equation, but we want the slope-intercept form, so we need to isolate $y$ on the left side of the equation by adding $4$ to both sides of the equation: $y = -x - 2$ To find the x-intercept, we set $y$ equal to $0$: $0 = -x - 2$ Add $x$ to each side of the equation to move the variable to the left side of the equation: $x = -2$ To find the y-intercept, we set $x$ equal to $0$: $y = -(0) - 2$ Subtract to solve: $y = -2$ The x-intercept is $-2$. The y-intercept is $-2$. |
# The Definition of a Derivative
Site: Saylor Academy Course: MA005: Calculus I Book: The Definition of a Derivative
Printed by: Guest user Date: Monday, July 15, 2024, 10:10 AM
## Description
Read this section to understand the definition of a derivative. Work through practice problems 1-8.
The graphical idea of a slope of a tangent line is very useful, but for some uses we need a more algebraic definition of the derivative of a function. We will use this definition to calculate the derivatives of several functions and see that the results from the definition agree with our graphical understanding. We will also look at several different interpretations for the derivative, and derive a theorem which will allow us to easily and quickly determine the derivative of any fixed power of $x$
In the last section we found the slope of the tangent line to the graph of the function $f(x)=x^{2}$ at an arbitrary point $(x, f(x))$ by calculating the slope of the secant line through the points $(x, f(x))$ and $(x+h, f(x+h))$,
$\mathrm{m_{sec}}=\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{(\mathrm{x}+\mathrm{h})-(\mathrm{x})}$
and then by taking the limit of $m_{sec}$ as h approached 0 (Fig. 1). That approach to calculating slopes of tangent lines is the definition of the derivative of a function.
Definition of the Derivative:
The derivative of a function $\mathrm{f}$ is a new function, $\mathbf{f}$ ' (pronounced "eff prime"),
whose value at $x$ is $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ if the limit exists and is finite.
This is the definition of differential calculus, and you must know it and understand what it says. The rest of this chapter and all of Chapter 3 are built on this definition as is much of what appears in later chapters. It is remarkable that such a simple idea (the slope of a tangent line) and such a simple definition (for the derivative $f '$) will lead to so many important ideas and applications.
Notation: There are three commonly used notations for the derivative of $\mathbf{y}=\mathbf{f}(\mathbf{x})$:
$\mathbf{f}^{\prime}(\mathbf{x})$ emphasizes that the derivative is a function related to $\mathrm{f}$
$D(f)$ emphasizes that we perform an operation on $f$ to get the derivative of $f$
$\frac{d f}{d x} \quad$ emphasizes that the derivative is the limit of $\frac{\Delta f}{\Delta x}=\frac{f(x+h)-f(x)}{h}$.
We will use all three notations so you can get used to working with each of them.
f'$(x)$ represents the slope of the tangent line to the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$ or the instantaneous rate of change of the function $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x})$ ).
If, in Fig. 2, we let $x$ be the point $a+h$, then $\mathrm{h}=\mathrm{x}-\mathrm{a}$. As $\mathrm{h} \rightarrow 0$, we see that $\mathrm{x} \rightarrow \mathrm{a}$ and $\lim\limits_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$ so
$\mathbf{f}^{\prime}(\mathbf{a})=\lim\limits_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$.
We will use whichever of these two forms is more convenient algebraically.
Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.2-Definition-of-Derivative.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.
Fortunately, we will soon have some quick and easy ways to calculate most derivatives, but first we will have to use the definition to determine the derivatives of a few basic functions. In Section 2.2 we will use those results and some properties of derivatives to calculate derivatives of combinations of the basic functions. Let's begin by using the graphs and then the definition to find a few derivatives.
Example 1: Graph $y = f(x) = 5$ and estimate the slope of the tangent line at each point on the graph. Then use the definition of the derivative to calculate the exact slope of the tangent line at each point. Your graphic estimate and the exact result from the definition should agree.
Solution: The graph of $y = f(x) = 5$ is a horizontal line (Fig. 3) which has slope $0$ so we should expect that its tangent line will also have slope $0$.
Using the definition: Since $f(x) = 5$, then $\mathrm{f}(\mathrm{x}+\mathrm{h})=5$, so
$\mathrm{D}(\mathrm{f}(\mathrm{x})) \equiv \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{5-5}{h}=\lim\limits_{h \rightarrow 0} \frac{0}{h} =0$
Using similar steps, it is easy to show that the derivative of any constant function is $0$
Theorem: If $f(x) = k$, then $f '(x) = 0$
Practice 1: Graph $y = f(x) = 7x$ and estimate the slope of the tangent line at each point on the graph. Then use the definition of the derivative to calculate the exact slope of the tangent line at each point.
Example 2: Determine the derivative of $y = f(x) = 5x^3$ graphically and using the definition. Find the equation of the line tangent to $y = 5x^3$ at the point $(1,5)$
Solution: It appears that the graph of $y = f(x) = 5x3$ (Fig. 4) is increasing so the slopes of the tangent lines are positive except perhaps at $x = 0$ where the graph seems to flatten out.
Using the definition: Since $f(x)=5 x^{3}$ then $f(x+h)=5(x+h)^{3}=5\left(x^{3}+3 x^{2} h+3 x h^{2}+h^{3}\right)$ so
\begin{aligned}&=\lim\limits_{h \rightarrow 0} \frac{15 x^{2} h+15 x h^{2}+5 h^{3}}{h} \\&=\lim\limits_{h \rightarrow 0} \frac{h\left(15 x^{2}+15 x h+5 h^{2}\right)}{h}\end{aligned} divide by $\mathrm{h}$
$=\lim\limits_{h \rightarrow 0}\left(15 x^{2}+15 x h+5 h^{2}\right)=15 \mathrm{x}^{2}+0+0=15 \mathrm{x}^{2}$
so $\mathrm{D}\left(\mathbf{5 x}^{3}\right)=15 \mathrm{x}^{2}$ which is positive except when $\mathrm{x}=0$, and then $15 \mathrm{x}^{2}=0$
$\mathrm{f}^{\prime}(\mathrm{x})=15 \mathrm{x}^{2}$ is the slope of the line tangent to the graph of $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$. At the point $(1,5)$, the slope of the tangent line is $\mathrm{f}^{\prime}(1)=15(1)^{2}=15$. From the point-slope formula, the equation of the tangent line to $\mathrm{f}$ is $\mathrm{y}-5=15(\mathrm{x}-1)$ or $\mathrm{y}=15 \mathrm{x}-10$
Practice 2: Use the definition to show that the derivative of $\mathrm{y}=\mathrm{x}^{3}$ is $\frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}$. Find the equation of the line tangent to the graph of $\mathrm{y}=\mathrm{x}^{3}$ at the point $(2,8)$.
If $f$ has a derivative at $x$, we say that f is differentiable at $x$. If we have a point on the graph of a differentiable function and a slope (the derivative evaluated at the point), it is easy to write the equation of the tangent line.
Tangent Line Formula
If $f$ is differentiable at a then the equation of the tangent line to $f$ at the point $(a ,f(a) )$ is $y = f(a) + f '(a)(x – a)$
Proof: The tangent line goes through the point $(a ,f(a) )$ with slope $f'(a)$ so, using the point-slope formula, $y-f(a)=f^{\prime}(a)(x-a)$ or $y=f(a)+f^{\prime}(a)(x-a)$.
Practice 3: The derivatives $\mathbf{D}(\mathrm{x})=1, \mathbf{D}\left(\mathrm{x}^{2}\right)=2 \mathrm{x}, \mathbf{D}\left(\mathrm{x}^{3}\right)=3 \mathrm{x}^{2}$ exhibit the start of a pattern. Without using the definition of the derivative, what do you think the following derivatives will be? $\mathbf{D}\left(\mathrm{x}^{4}\right), \mathrm{D}\left(\mathrm{x}^{5}\right), \mathbf{D}\left(\mathrm{x}^{43}\right.$ ), $\mathbf{D}(\sqrt{\mathrm{x}})=\mathbf{D}\left(\mathrm{x}^{1 / 2}\right)$ and $\mathbf{D}\left(\mathrm{x}^{\pi}\right)$
(Just make an intelligent "guess" based on the pattern of the previous examples. )
Before going on to the "pattern" for the derivatives of powers of $x$ and the general properties of derivatives, let's try the derivatives of two functions which are not powers of $x: sin(x) and | x |$.
Theorem: $\quad \mathrm{D}(\sin (\mathrm{x}))=\cos (\mathrm{x})$
The graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$ is well-known (Fig. 5). The graph has horizontal tangent lines $\left(\right. slope =0$) when $x=\pm \frac{\pi}{2}$ and $x=\pm \frac{3 \pi}{2}$ and so on. If $0 < x < \frac{\pi}{2}$, then the slopes of the tangent lines to the graph of $y=\sin (x)$ are positive. Similarly, if $\frac{\pi}{2} < x$ $< \frac{3 \pi}{2}$, then the slopes of the tangent lines are negative. Finally, since the graph of $\mathrm{y}=\sin (\mathrm{x})$ is periodic, we expect that the derivative of $\mathrm{y}=\sin (\mathrm{x})$ will also be periodic.
Proof of the theorem: Since $\mathrm{f}(\mathrm{x})=\sin (\mathrm{x}), \mathrm{f}(\mathrm{x}+\mathrm{h})=\sin (\mathrm{x}+\mathrm{h})=\sin (\mathrm{x}) \cos (\mathrm{h})+\cos (\mathrm{x}) \sin (\mathrm{h})$ so
\begin{aligned}&\mathbf{f}^{\prime}(\mathrm{x}) \equiv \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\&=\lim\limits_{h \rightarrow 0} \frac{\{\sin (x) \cos (h)+\cos (x) \sin (h)\}-\{\sin (x)\}}{h}\end{aligned}
this limit looks formidable, but if we just collect the terms containing $\sin (\mathrm{x})$ and then those containing $\cos (\mathrm{x})$ we get
$=\lim\limits_{h \rightarrow 0}\left\{\sin (x) \cdot \frac{\cos (h)-1}{h}+\cos (x) \cdot \frac{\sin (h)}{h}\right\}$
now calculate the limits separately
$=\left\{\lim\limits_{h \rightarrow 0} \sin (x)\right\} \cdot\left\{\lim _{h \rightarrow 0} \frac{\cos (h)-1}{h}\right\}+\left\{\lim\limits_{h \rightarrow 0} \cos (x)\right\} \cdot\left\{\lim\limits_{h \rightarrow 0} \frac{\sin (h)}{h}\right\}$
the first and third limits do not depend on $\mathrm{h}$, and we calculated the second and fourth limits in Section1.2
$=\sin (x) \cdot(0)+\cos (x) \cdot(1)=\cos (x)$
So $\mathrm{D}(\sin (\mathrm{x}))=\cos (\mathrm{x})$, and the various properties we expected of the derivative of $\mathrm{y}=\sin (\mathrm{x})$ by examining its graph are true of $\cos (\mathrm{x})$
Practice 4: Use the definition to show that $\mathrm{D}(\cos (\mathrm{x}))=-\sin (\mathrm{x})$. (This is similar to the situation for $\mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$. You will need the formula $\cos (x+h)=\cos (x) \cdot \cos (h)-\sin (x) \cdot \sin (h)$. Then collect all the terms containing $\cos (x)$ and all the terms with $\sin (x)$. At that point you should recognize and be able to evaluate the limits.)
Example 3: For $\mathrm{y}=|\mathrm{x}|$ find $\mathrm{dy} / \mathrm{dx}$.
Solution: The graph of $y=f(x)=|x|$ (Fig. 6) is a "V" with its vertex at the origin. When $x>0$, the graph is just $y=|x|=x$ which is a line with slope $+1$ so we should expect the derivative of $|x|$ to be +1. When $\mathrm{x} < 0$, the graph is $\mathrm{y} = |\mathrm{x}| = -\mathrm{x} \quad$ which is a line with slope $-1$, so we expect the derivative of $|\mathrm{x}|$ to be $-1$. When $\mathrm{x} = 0$, the graph has a corner, and we should expect the derivative of $|x|$ to be undefined at $x=0$.
Using the definition: It is easiest to consider 3 cases in the definition of $|x|: x > 0$, $x < 0$ and $x = 0$
If $x>0$, then, for small values of h, $x+h>0$ so Df $(x) \equiv \lim\limits_{h \rightarrow 0} \frac{|x+h|-|x|}{h}=\lim\limits_{h \rightarrow 0} \frac{h}{h}=1$
If $x$, then, for small values of $h$, we also know that $x+h < 0$ so $\operatorname{Df}(x)=\lim\limits_{h \rightarrow 0} \frac{-h}{h}=-1$. When $x=0$, the situation is a bit more complicated and
$\mathbf{D} \mathrm{f}(\mathrm{x}) \equiv \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{|0+h|-|0|}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h} \quad$ which is undefined
since $\lim\limits_{h \rightarrow 0^{+}} \frac{|h|}{h}=+1$ and $\lim\limits_{h \rightarrow 0^{-}} \frac{|h|}{h}=-1 \text {. }$ $\mathbf{D}(|x|)= \begin{cases}+1 & \text { if } x > 0 \\ \text { undefined } & \text { if } x=0 \\ -1 & \text { if } x < 0\end{cases}$
Practice 5: Graph $\mathrm{y}=|\mathrm{x}-2|$ and $\mathrm{y}=|2 \mathrm{x}|$ and use the graphs to determine $\mathbf{D}(|\mathrm{x}-2|)$ and $\mathbf{D}(|2 \mathrm{x}|)$.
So far we have emphasized the derivative as the slope of the line tangent to a graph. That interpretation is very visual and useful when examining the graph of a function, and we will continue to use it. Derivatives, however, are used in a wide variety of fields and applications, and some of these fields use other interpretations. The following are a few interpretations of the derivative which are commonly used.
General
Rate of Change $f^{\prime}(x)$ is the rate of change of the function at $x$. If the units for $x$ are years and the units for $f(x)$ are people, then the units for $\frac{\text { df }}{\mathrm{dx}}$ are $\frac{\text { people }}{\text { year }}$, a rate of change in population.
Graphical
Slope $\mathrm{f}^{\prime}(\mathrm{x})$ is the slope of the line tangent to the graph of $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$
Physical
Velocity If $\mathrm{f}(\mathrm{x})$ is the position of an object at time $\mathrm{x}$, then $\mathrm{f}^{\prime}(\mathrm{x})$ is the velocity of the object at time $\mathrm{x}$. If the units for $x$ are hours and $f(x)$ is distance measured in miles, then the units for $\mathbf{f}^{\prime}(\mathrm{x})=\frac{\mathrm{df}}{\mathrm{dx}}$ are $\frac{\text { miles }}{\text { hour }}$, miles per hour, which is a measure of velocity.
Acceleration If $\mathrm{f}(\mathrm{x})$ is the velocity of an object at time $\mathrm{x}$, then $\mathrm{f}^{\prime}(\mathrm{x})$ is the acceleration of the object at time $\mathrm{x}$. If the units are for $x$ are hours and $f(x)$ has the units $\frac{\text { miles }}{\text { hour }}$, then the units for the acceleration $\mathbf{f}^{\prime}(\mathbf{x})$ $=\frac{\text { df }}{\mathrm{dx}}$ are $\frac{\text { miles/hour }}{\text { hour }}=\frac{\text { miles }}{\text { hour }^{2}}$, miles per hour per hour.
Magnification $f '(x)$ is the magnification factor of the function $f$ for points which are close to $x$If $a$ and $b$ are two points very close to $x$, then the distance between $f(a)$ and $f(b)$ will be close to $f '(x)$ times the original distance between $a$ and $b: f(b) – f(a) ≈ f '(x) ( b – a )$
Business
Marginal Cost If $f(x)$ is the total cost of $x$ objects, then $f '(x)$ is the marginal cost, at a production level of $x$. This marginal cost is approximately the additional cost of making one more object once we have already made $x$ objects. If the units for $x$ are bicycles and the units for $f(x)$ are dollars, then the $\mathbf{f}^{\prime}(\mathrm{x})=\frac{\mathrm{df}}{\mathrm{dx}}$ are $\frac{\text { dollars }}{\text { bicycle }}$, the cost per bicycle.
Marginal Profit If $\mathrm{f}(\mathrm{x})$ is the total profit from producing and selling $\mathrm{x}$ objects, then $\mathrm{f}^{\prime}(\mathrm{x})$ is the marginal profit, the profit to be made from producing and selling one more object.
If the units for $\mathrm{x}$ are bicycles and the units for $\mathrm{f}(\mathrm{x})$ are dollars, then the units for $\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{df}}{\mathrm{dx}}$ are $\frac{\text { dollars }}{\text { bicycle }}$, dollars per bicycle, which is the profit per bicycle.
In business contexts, the word "marginal" usually means the derivative or rate of change of some quantity.
One of the strengths of calculus is that it provides a unity and economy of ideas among diverse applications. The vocabulary and problems may be different, but the ideas and even the notations of calculus are still useful.
Example 4: A small cork is bobbing up and down, and at time t seconds it is $h(t) = sin(t)$ feet above the mean water level (Fig. 7). Find the height, velocity and acceleration of the cork when $t = 2$ seconds. (Include the proper units for each answer.)
Solution: $h(t) = sin(t)$ represents the height of the cork at any time $t$, so the height of the cork when $\mathrm{t}=2$ is $\mathrm{h}(2)=\sin (2) \approx 0.91$ feet.
The velocity is the derivative of the position, so $v(t)=\frac{d h(t)}{d t}=\frac{d \sin (t)}{d t}=\cos (t)$. The derivative of position is the limit of $(\Delta \mathrm{h}) /(\Delta \mathrm{t})$, so the units are (feet)/(seconds). After 2 seconds the velocity is $\mathrm{v}(2)=\cos (2) \approx-0.42$ feet per second $=-0.42 \mathrm{ft} / \mathrm{s}$.
The acceleration is the derivative of the velocity, so $\mathrm{a}(\mathrm{t})=\frac{\mathrm{d} \mathrm{v}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathrm{d} \cos (\mathrm{t})}{\mathrm{dt}}=-\sin (\mathrm{t})$. The derivative of velocity is the limit of $(\Delta \mathrm{v}) /(\Delta \mathrm{t})$, so the units are (feet/second) / (seconds) or feet/second $^{2}$. After 2 seconds the acceleration is $\mathrm{a}(2)=-\sin (2) \approx-0.91 \mathrm{ft} / \mathrm{s}^{2}$.
Practice 6: Find the height, velocity and acceleration of the cork in the previous example after 1 second?
A Useful Formula: $D( x^n )$
Functions which include powers of $x$ are very common (every polynomial is a sum of terms which include powers of $x$), and, fortunately, it is easy to calculate the derivatives of such powers. The "pattern" emerging from the first few examples in this section is, in fact, true for all powers of $x$. We will only state and prove the "pattern" here for positive integer powers of $x$, but it is also true for other powers as we will prove later.
Theorem: If $n$ is a positive integer, then $D( x^n ) = n.x^{n–1}$
This theorem is an example of the power of generality and proof in mathematics. Rather than resorting to the definition when we encounter a new power of $x$ (imagine using the definition to calculate the derivative of $x^{307}$ ), we can justify the pattern for all positive integer exponents $n$, and then simply apply the result for whatever exponent we have. We know, from the first examples in this section, that the theorem is true for $n= 1, 2$ and $3$, but no number of examples would guarantee that the pattern is true for all exponents. We need a proof that what we think is true really is true.
Proof of the theorem: Since $f(x) = x^n$, then $f(x+h) = (x+h)^n$, and in order to simplify $f(x+h) – f(x) = (x+h)^n – x^n$, we will need to expand $(x+h)^n$. However, we really only need to know the first two terms of the expansion and to know that all of the other terms of the expansion contain a power of $h$ of at least 2. The Binomial Theorem from algebra says (for $n > 3$) that $(x+h)^n = x^n + n.^{xn–1}h + a.x^{n–2}h2 + b.x^{n–3}h3 +... + h^n$ where $a$ and $b$ represent numerical coefficients.
(Expand $(x+h)^n$ for at least a few different values of n to convince yourself of this result.)
Then $\mathbf{D}(\mathrm{f}(\mathrm{x})) \equiv \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{(x+h)^{n}-x^{n}}{h} \quad$ then expand to get
$=\lim\limits_{h \rightarrow 0} \frac{\left\{x^{n}+n \cdot x^{n-1} h+a \cdot x^{n-2} h^{2}+b \cdot x^{n-3} h^{3}+\ldots+h\right.}{h}$ eliminate $x^n – x^n$
$=\lim\limits_{h \rightarrow 0} \frac{\left\{n \cdot x^{n-1} h+a \cdot x^{n-2} h^{2}+b \cdot x^{n-3} h^{3}+\ldots+h^{n}\right\}}{h}$ factor $h$ out of the numerator
$=\lim\limits_{h \rightarrow 0} \frac{h \cdot\left\{n \cdot x^{n-1}+a \cdot x^{n-2} h+b \cdot x^{n-3} h^{2}+\ldots+h^{n-1}\right\}}{h}$ divide by the factor $h$
$=\lim\limits_{h \rightarrow 0}\left\{n \cdot x^{n-1}+a \cdot x^{n-2} h+b \cdot x^{n-3} h^{2}+\ldots+h^{n-1}\right\}$ separate the limits
$=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}+\lim\limits_{h \rightarrow 0}\left\{\mathrm{a} \cdot \mathrm{x}^{\mathrm{n}-2} \mathrm{~h}+\mathrm{b} \cdot \mathrm{x}^{\mathrm{n}-3} \mathbf{h}^{2}+\ldots+\mathbf{h}^{\mathbf{n}-1}\right\} \quad$ each term has a factor of $\mathrm{h}$, and $\mathrm{h} \rightarrow 0$
$=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}+\mathbf{0}=\mathbf{n} \cdot \mathbf{x}^{\mathbf{n}-\mathbf{1}}$ so $\mathbf{D}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathbf{n} \cdot \mathbf{x}^{\mathbf{n}-\mathbf{1}}$
Practice 7: Use the theorem to calculate $D( x5 )$, $d ( x2 )$, $D( x100 )$, $d ( t31 )$, and $D( x0 )$
We will occasionally use the result of the theorem for the derivatives of all constant powers of $x$ even though it has only been proven for positive integer powers, so far. The result for all constant powers of $x$ is proved in Section 2.9
Example 5: Find $\mathbf{D}(1 / \mathrm{x})$ and $\frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})$.
Solution: $\quad \mathbf{D}\left(\frac{1}{\mathrm{x}}\right)=\mathbf{D}\left(\mathrm{x}^{-1}\right)=-1 \mathrm{x}^{(-1)-1}=-1 \mathrm{x}^{-2}=\frac{-1}{\mathrm{x}^{2}} \cdot \frac{\mathbf{d}}{\mathbf{d x}}(\sqrt{\mathrm{x}})=\mathbf{D}\left(\mathrm{x}^{1 / 2}\right)=(1 / 2) \mathrm{x}^{-1 / 2}=\frac{1}{2 \sqrt{\mathrm{x}}}$
These results can be obtained by using the definition of the derivative, but the algebra is slightly awkward.
Practice 8: Use the pattern of the theorem to find $\mathbf{D}\left(\mathrm{x}^{3 / 2}\right), \frac{\mathbf{d}}{\mathbf{d x}}\left(\mathrm{x}^{1 / 3}\right), \mathrm{D}\left(\frac{1}{\sqrt{\mathrm{x}}}\right)$ and $\frac{\mathbf{d}}{\mathrm{dt}}\left(\mathrm{t}^{\pi}\right)$.
Example 6: It costs $\sqrt{\mathrm{x}}$ hundred dollars to run a training program for $\mathrm{x}$ employees.
(a) How much does it cost to train 100 employees? 101 employees? If you already need to train 100 employees, how much additional will it cost to add 1 more employee to those being trained?
(b) For $f(x)=\sqrt{x}$, calculate $f^{\prime}(x)$ and evaluate $f^{\prime}$ at $x=100$. How does $f^{\prime}(100)$ compare with the last answer in part (a)?
Solution: (a) Put $f(x)=\sqrt{x}=x^{1 / 2}$ hundred dollars, the cost to train $x$ employees. Then $f(100)$=$1000 and $f(101)$=$1004.99, so it costs $4.99 additional to train the 101st employee. (b) $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{x}^{-1 / 2}=\frac{1}{2 \sqrt{\mathrm{x}}} \quad$ so $\mathrm{f}^{\prime}(100)=\frac{1}{2 \sqrt{100}}=\frac{1}{20} \quad$ hundred dollars =$5.00.
Clearly $f '(100)$ is very close to the actual additional cost of training the 101st employee.
Definition of Derivative: $\quad \mathbf{f}^{\prime}(\mathrm{x}) \equiv \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \quad$ if the limit exists and is finite.
Notations For The Derivative: $\mathbf{f}^{\prime}(\mathbf{x}), \operatorname{Df}(\mathbf{x}), \frac{\mathbf{d} \mathbf{f}(\mathbf{x})}{\mathbf{d x}}$
Tangent Line Equation: The line $\mathbf{y}=\mathbf{f}(\mathbf{a})+\mathbf{f}^{\prime}(\mathbf{a}) \cdot(\mathbf{x}-\mathbf{a})$ is tangent to the graph of $\mathrm{f}$ at $(a,f(a))$.
Formulas: $D( constant )=0$
$D ( x^n ) = n*x^{n-1}$ (proven for $n$ = positive integer: true for all constants $n$)
$D ( sin(x) ) = cos ( x )$ and $D ( cos(x) ) = -sin( x )$
\begin{aligned} &\mathbf{D}(|\mathrm{x}|) = \left\{\begin{array}{lll} +1 & \text { if } \mathrm{x} > 0 \\ \text { undefined } & \quad \text { if } \mathrm{x} = 0 \\ -1 & \text { if } \mathrm{x} < 0 \end{array}\right. \end{aligned}
Interpretations of $f '(x)$:
Slope of a line tangent to a graph
Instantaneous rate of change of a function at a point
Velocity or acceleration
Magnification factor
Marginal change
Practice 1: The graph of $\mathrm{f}(\mathrm{x})=7 \mathrm{x}$ is a line through the origin. The slope of the line is $7$.
For all $x, m_{\tan }=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{7(x+h)-7 x}{h}=\lim\limits_{h \rightarrow 0} \frac{7 h}{h}=\lim\limits_{h \rightarrow 0} 7=7$
Practice 2: $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ so $\mathrm{f}(\mathrm{x}+\mathrm{h})=(\mathrm{x}+\mathrm{h})^{3}=\mathrm{x}^{3}+3 \mathrm{x}^{2} \mathrm{~h}+3 \mathrm{xh}^{2}+\mathrm{h}^{3}$
\begin{aligned}\frac{d y}{\mathrm{dx}}=& \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{\left\{x^{3}+3 x^{2} h+3 x h^{2}+h^{3}\right\}-x^{n}}{h} \\&=\lim _{h \rightarrow 0} \frac{3 x^{2} h+3 x h^{2}+h^{3}}{h}=\lim\limits_{h \rightarrow 0}\left(3 x^{2}+3 x h+h^{2}\right)=3 \mathbf{x}^{2}\end{aligned}
At the point $(2,8)$, the slope of the tangent line is $3(2)^2 = 12$ so the equation of the tangent line is $y – 8 = 12(x – 2)$ or $y = 12x –16$.
Practice 3: $\mathbf{D}\left(\mathrm{x}^{4}\right)=4 \mathrm{x}^{3}, \mathbf{D}\left(\mathrm{x}^{5}\right)=5 \mathrm{x}^{4}, \mathbf{D}\left(\mathrm{x}^{43}\right)=43 \mathrm{x}^{42}, \mathrm{D}\left(\mathrm{x}^{1 / 2}\right)=\frac{1}{2} \mathrm{x}^{-1 / 2}, \mathrm{D}\left(\mathrm{x}^{\pi}\right)=\pi \mathrm{x}^{\pi-1}$
Practice 4:
\begin{aligned}&D(\cos (x))=\lim\limits_{h \rightarrow 0} \frac{\cos (x+h)-\cos (x)}{h}=\lim\limits_{h \rightarrow 0} \frac{\cos (x) \cos (h)-\sin (x) \sin (h)-\cos (x)}{h} \\&=\lim\limits_{h \rightarrow 0} \cos (x) \frac{\cos (h)-1}{h}-\sin (x) \frac{\sin (h)}{h} \longrightarrow \cos (x) \cdot(0)-\sin (x) \cdot(1)=-\sin (x)\end{aligned}
Practice 5: See Fig. 16 for the graphs of $\mathrm{y}=|\mathrm{x}-2|$ and $\mathrm{y}=|2 \mathrm{x}|$
\begin{aligned}&\mathbf{D}(|x-2|)= \begin{cases}+1 & \text { if } x > 2 \\\text { undefined } & \text { if } x = 2 \\-1 & \text { if } x < 2\end{cases} \\&D(|2 x|)= \begin{cases}+2 & \text { if } x > 0 \\\text { undefined } & \text { if } x = 0 \\-2 & \text { if } x < 0\end{cases}\end{aligned}
Practice 6:
$h(t) = sin(t)$ so $h(1) = sin(1) ≈ 0.84$ feet,
$v(t) = cos(t)$ so $v(1) =cos(1) ≈ 0.54$ feet/second.
$a(t) = –sin(t)$ so $a(1) = –sin(1) ≈ –0.84$ feet/second2.
Practice 7: $\quad \mathbf{D}\left(\mathrm{x}^{5}\right)=5 \mathrm{x}^{4}, \quad \frac{\mathbf{d} \mathrm{x}^{2}}{\mathrm{dx}}=2 \mathrm{x}^{1}=2 \mathrm{x}, \frac{\mathbf{d} \mathrm{x}^{100}}{\mathrm{dx}}=100 \mathrm{x}^{99}, \frac{\mathbf{d} \mathrm{t}^{31}}{\mathrm{dt}}=31 \mathrm{t}^{30}$,
$\mathbf{D}\left(x^{0}\right)=0 x^{-1}=0 \text { or } \mathbf{D}\left(x^{0}\right)=D(1)=0$
Practice 8:
$\mathbf{D}\left(\mathrm{x}^{3 / 2}\right)=\frac{3}{2} \mathrm{x}^{1 / 2}, \quad \frac{\mathbf{d} \mathrm{x}^{1 / 3}}{\mathrm{dx}}=\frac{1}{3} \mathrm{x}^{-2 / 3}, \quad \mathbf{D}\left(\mathrm{x}^{-1 / 2}\right)=\frac{-1}{2} \mathrm{x}^{-3 / 2}, \quad \frac{\mathbf{d} \mathrm{t}^{\pi}}{\mathrm{dt}}=\pi \mathrm{t}^{\pi-1}$ |
Find the derivation of each of the following from the first principle:
Question:
Find the derivation of each of the following from the first principle:
$\sqrt{\sec x}$
Solution:
Let
$f(x)=\sqrt{\sec x}$
We need to find the derivative of f(x) i.e. f’(x)
We know that,
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ …(i)
$f(x)=\sqrt{\sec x}$
$\mathrm{f}(\mathrm{x}+\mathrm{h})=\sqrt{\sec (\mathrm{x}+\mathrm{h})}$
Putting values in (i), we get
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sqrt{\sec (x+h)}-\sqrt{\sec x}}{h}$
Now rationalizing the numerator by multiplying and divide by the conjugate
of $\sqrt{\sec (\mathrm{x}+\mathrm{h})}-\sqrt{\sec \mathrm{x}}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{\sec (x+h)}-\sqrt{\sec x}}{h} \times \frac{\sqrt{\sec (x+h)}+\sqrt{\sec x}}{\sqrt{\sec (x+h)}+\sqrt{\sec x}}$
Using the formula:
$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\lim _{h \rightarrow 0} \frac{(\sqrt{\sec (x+h)})^{2}-(\sqrt{\sec x})^{2}}{h(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=\lim _{h \rightarrow 0} \frac{\sec (x+h)-\sec (x)}{h(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{\cos (x+h)}-\frac{1}{\cos x}}{(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=\lim _{h \rightarrow 0} \frac{\frac{\cos x-\cos (x+h)}{\cos (x+h) \cos x}}{(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=\lim _{h \rightarrow 0} \frac{\cos x-\cos (x+h)}{h(\cos (x+h) \cos x)(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
Using the formula:
$\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{x+(x+h)}{2} \sin \frac{(x+h)-x}{2}}{h(\cos (x+h) \cos x)(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{2 x+h}{2} \sin \frac{h}{2}}{h(\cos (x+h) \cos x)(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=2 \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$\times \frac{1}{2} \lim _{h \rightarrow 0} \sin \left(\frac{2 x+h}{2}\right)$
$\times \lim _{h \rightarrow 0} \frac{1}{(\cos (x+h) \cos x)(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
[Here, we multiply and divide by $\frac{1}{2}$ ]
$=2 \times \frac{1}{2} \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$\times \lim _{h \rightarrow 0} \sin \left(x+\frac{h}{2}\right) \times \lim _{h \rightarrow 0} \frac{1}{(\cos (x+h) \cos x)(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=(1) \times \lim _{h \rightarrow 0} \sin \left(x+\frac{h}{2}\right) \times \lim _{h \rightarrow 0} \frac{1}{(\cos (x+h) \cos x)(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
Putting h = 0, we get
$=\sin \left[x+\frac{0}{2}\right] \times \frac{1}{\cos (x+0) \cos x(\sqrt{\sec (x+0)}+\sqrt{\sec x})}$
$=\sin x \times \frac{1}{\cos x \cos x(\sqrt{\sec x}+\sqrt{\sec x})}$
$=\frac{\sin x}{\cos ^{2} x(2 \sqrt{\sec x})}$
$=\frac{\sin x}{\cos x} \times \frac{1}{\cos x} \times \frac{1}{2 \sqrt{\sec x}}$
$=\tan x \times \sec x \times \frac{1}{2 \sqrt{\sec x}}\left[\because \frac{\sin x}{\cos x}=\tan x\right] \&\left[\frac{1}{\cos x}=\sec x\right]$
$=\frac{1}{2} \tan x \sqrt{\sec x}$
Hence
$f^{\prime}(x)=\frac{1}{2} \tan x \sqrt{\sec x}$ |
# 2020 AMC 8 Problems/Problem 3
## Problem
Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?
$\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$
## Solution 1
Note that the unit of the answer is strawberries, which is the product of
• square feet
• plants per square foot
• strawberries per plant
By conversion factors, we have $$\left(6 \ \color{red}\cancel{\mathrm{ft}}\color{black}\cdot8 \ \color{red}\cancel{\mathrm{ft}}\color{black}\right)\cdot\left(4 \ \frac{\color{green}\cancel{\mathrm{plants}}}{\color{red}\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\color{green}\cancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{\textbf{(D) }1920} \ \mathrm{strawberries}.$$
~MRENTHUSIASM ~Bobthegod78
## Solution 2
The area of the garden is $6 \cdot 8 = 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, there are a total of $48 \cdot 4=192$ strawberry plants, each of which produces $10$ strawberries on average. Accordingly, she can expect to harvest $192 \cdot 10 = \boxed{\textbf{(D) }1920}$ strawberries.
~savannahsolver
## Video Solution by The Learning Royal
~The Learning Royal
~Interstigation
## Video Solution by North America Math Contest Go Go Go
~North America Math Contest Go Go Go
~STEMbreezy
## Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
2020 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions |
# What are the angle properties of parallel lines?
## What are the angle properties of parallel lines?
Angles in parallel lines
• When a pair of parallel lines is cut with another line known as an intersecting transversal, it creates pairs of angles with special properties.
• Corresponding angles are equal. The lines make an F shape.
• Alternate angles are equal. The lines make a Z shape which can also be back to front.
### What are the 3 properties of parallel lines?
FAQs on Properties of Parallel Lines
• The pairs of corresponding angles are equal.
• The pairs of vertically opposite angles are equal.
• The pairs of alternate interior angles are equal.
• The pairs of alternate exterior angles are equal.
What is the parallel angle theorem?
Converse of Parallel Lines Theorem – Concept If two corresponding angles are congruent, then the two lines cut by the transversal must be parallel. Similarly, if two alternate interior or alternate exterior angles are congruent, the lines are parallel.
What are the three properties of parallel lines?
Name another pair of alternate interior angles and another pair of same-side interior angles. &3 and &4 are alternate interior angles. &2 and &3 are same-side interior angles. Quick Check 11 Name three other pairs of corresponding angles in the diagrams above.
## How are interior angles related to parallel lines?
Corresponding Angles: One interior angle and one exterior angle that are non-adjacent and on the same side of the transversal. Corresponding angles are equal. Alternate Interior Angles: Interior angles on opposite sides of the transversal that are adjacent to different parallel lines. Alternate interior angles are equal.
### How to test for parallel lines in math?
Testing for Parallel Lines If Any Pair Of Example: Corresponding Angles are equal a = e or Alternate Interior Angles are equal c = f or
How are lines A and B parallel to each other?
In the following figure, we are given that line a and line c are parallel to line b. Since a || b, so ∠1 = ∠2 (Corresponding angles axiom) Since c || b, so ∠3 = ∠2 (Corresponding angles axiom) Therefore, ∠1 = ∠3 (Commutative property) |
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# The coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$ is
Last updated date: 19th Jun 2024
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Hint: In the above question you have to find the coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$. At first, you have to reduce the expanding term then by applying a simple law of exponent, the term will get reduced. Now find the coefficient of ${x^n}$ in the expansion of the new reduced term. So let us see how we can solve this problem.
Step by step solution:
In the given question we were asked to find the coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$. First of all, we will reduce the expression by the formula of $\dfrac{{x + y}}{z} = \dfrac{x}{z} + \dfrac{y}{z}$.
On applying the same formula on the expanding term we get
$= \dfrac{{{e^{7x}}}}{{{e^{3x}}}} + \dfrac{{{e^x}}}{{{e^{3x}}}}$
According to the law of exponent, $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ . On applying the same law of exponent we get,
$= {e^{7x - 3x}} + {e^{x - 3x}}$
$= {e^{4x}} + {e^{ - 2x}}$
Now, we have reduced the term in $= {e^{4x}} + {e^{ - 2x}}$ . So we have to find the coefficient of ${x^n}$ in ${e^{4x}} + {e^{ - 2x}}$
$= \dfrac{{{{(4)}^n}}}{{n!}} + \dfrac{{{{( - 2)}^n}}}{{n!}}$
Therefore, the coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$ is $\dfrac{{{{(4)}^n}}}{{n!}} + \dfrac{{{{( - 2)}^n}}}{{n!}}$.
Note:
In the above solution we have used the law of exponent for reducing the expanding term and then we find the coefficient of ${x^n}$ in the expansion of that term which in our case is ${e^{4x}} + {e^{ - 2x}}$. Then we get the coefficients as 4 and -2. So, we get $\dfrac{{{{(4)}^n}}}{{n!}} + \dfrac{{{{( - 2)}^n}}}{{n!}}$. |
Question
in progress 0
2 weeks 2021-09-13T10:44:58+00:00 2 Answers 0
Step-by-step explanation:
9+x=2-5x
9+x+5x=2
9+6x=2
Subtract 9 on both sides:
6x=-7
Divide 6:
x=-
-1 1/6
Step-by-step explanation:
9 + x = 2 – 5x
Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS. Remember, PEMDAS =
Parenthesis
Exponents (& Roots)
Multiplication
Division
Subtraction
& is the order of operation.
Isolate the variable, x. Subtract x and 2 from both sides:
9 (-2) + x (-x) = 2 (-2) – 5x (-x)
9 – 2 = -5x – x
7 = -6x.
Simplify. Isolate the variable, x. Divide -6 from both sides:
(7)/-6 = (-6x)/-6
x = – 7/6 = -1 1/6
~ |
aspifsGak5u
2021-12-15
What is a hyperbola? What are the Cartesian equations for hyperbolas centered at the origin with foci on one of the coordinate axes? How can you find the foci, vertices, and directrices of such an ellipse from its equation?
Laura Worden
Step 1
Given: To write the definition of the hyperbola:
Definition: In an analytical geometry, a hyperbola is a conic section in which it forms a intersection of the right circular coe with a plane at an angle in such a way that both halves of the cone are intersected.
Therefore, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points (x,y) in the coordinate plane in such a way that the distance between and the focal distance is always a positive constants
Therefore, Cartesian equations for hyperbolas centered at the origin with foci on one of the coordinate axes
$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ (standard equation of hyperbola)
Now, Ellipse is of the general form $\frac{{\left(x-{x}_{1}\right)}^{2}}{{a}^{2}}+\frac{{\left(y-{y}_{1}\right)}^{2}}{{b}^{2}=1}$ with $a>b$
$e=\sqrt{\frac{{a}^{2}-{b}^{2}}{{a}^{2}}}$ is the eccentricity
foci: ; vertices: and equation of directrix is $x=\pi \frac{a}{e}$
Hattie Schaeffer
Step 1
A hyperbola is the of points in a plane whose distamces from two points in the plane have a constant difference.
The Cartesian equation for a hyperbola centered at the origin with foci on the x-axis is given as
$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$
The center-to-focus distance of this hyperbola is
$x=\sqrt{{a}^{2}+{b}^{2}}$
The foci of the hyperbola are the points
The vertices of the hyperbola are the points
The directrices of the hyperbola are given by the equations $x=±c$
The Cartesian equation for a hyperbola centered at the origin with foci on the y-axis is given as
$\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1$
The center-to-focus distance of this hyperbola is $c=\sqrt{{a}^{2}+{b}^{2}}$
The foci of th ehyperbola are the points
The vertices of the hyperbola are the points
The directrices of the hyperbola are given by the equations $y=±c$
Do you have a similar question? |
## Perimeter, Area, and Circumference
Perimeter and Area
The perimeter of a polygon is the distance around it, or the sum of the lengths of its sides. For a simple figure like a rectangle, opposite sides are congruent. If l and w denote the length and width of a rectangle, then the perimeter P is given by 2l + 2w.
Area is measured in square units. A square unit is a square in which each side has a length of one unit. The area of an object is the number of square units it takes to cover the object without any overlap. The area, A, of a rectangle is given by the formula l × w (length times width) or b × h (base times height).
The area formula for a parallelogram is the same as for a rectangle, A = bh. To show that this is true, students can cut and reassemble the parallelogram to form a rectangle with the same base and height.
By drawing a diagonal of a parallelogram, it can be shown that the area of a triangle is the area of the parallelogram. It can further be shown that every right triangle can be fitted with a congruent copy of itself to form a rectangle. So the area of the right triangle is one half the area of the rectangle, or A = bh where b is the base and h is the height of the triangle. Any non-right triangle can be written as the sum or difference of right triangles. By adding or subtracting the areas of right triangles, it can be shown that the formula A = bh applies to all triangles.
The perimeter of a circle is called the circumference. All circles are similar. Therefore, for any two circles, the ratio of the perimeters (circumferences) is the same as the ratio of the diameters. The ratio of the circumference to the diameter in each circle is the same. This constant ratio of circumference to diameter is denoted by the Greek letter π (pi). So C = πd or C = 2πr.
Small Circle
circumference = C1
diameter = d1
and
Large Circle
circumference = C1
diameter = d1
Teaching Model 16.3: Area of a Parallelogram |
Definition of Bezier Curve
$\textbf{Bezier Curve}$ is used to draw smooth curve along points on a path A Bezier Curve goes through points called anchor points and the shape between anchor points is defined by so call control points $\textbf{Bezier Curve}$ is used to draw smooth curve along points on a path A Bezier Curve goes through points called anchor points and the shape between anchor points is defined by so call control points
$\textbf{Movement between two points using vectors}$ A point $\mathbf{Q}$ that moves from $\mathbf{P_1}$ to $\mathbf{P_2}$ over the time $0 \leq t \leq 1$ can be described using position vectors. Let $\mathbf{O}$ be some fixd point, then the $\textbf{position vector}$ $\mathbf{OQ}$ points at $\mathbf{Q}$. Let $\mathbf{P_1 P_2}$ be the vector between $\mathbf{P_1}$ and $\mathbf{P_2}$, then $\mathbf{OQ}$ is described by $\overrightarrow{OQ} = \overrightarrow{O P_1} + t\overrightarrow{P_1 P_2} \quad t \in [0, 1]$ to find the coordinates of $\overrightarrow{P_1 P_2}$, position vector are used,
$\overrightarrow{P_1 P_2} = \overrightarrow{O P_2} - \overrightarrow{O P_1}$ We get
\begin{align*} \overrightarrow{OQ} &= \overrightarrow{O P_1} + t\overrightarrow{P_1 P_2} \\ &= \overrightarrow{O P_1} + t (\overrightarrow{O P_2} - \overrightarrow{O P_1}) \\ &= (1-t)\overrightarrow{O P_1} + t \overrightarrow{O P_2}\\ \end{align*} the position vector of a point has the same coordinates as the point. For this reason, you don't need to distinguish between points and vectors when doing computer graphic. The point $\mathbf{Q}$ can be written as $\mathbf{Q} = (1-t) \mathbf{P_1} + t \mathbf{P_2} \quad t \in [0, 1]$ $\textbf{Parametric equation of Bezier Curve}$ $\textit{Linear Bezier Curve}$: Two points are needed. Both are anchor points, as $\mathbf{Q_0}$ moves along the line $\mathbf{P_0}$ and $\mathbf{P_1}$ it traces out a linear Bezier curve. Let t be a parameter, then the linear Bezier curve can be written as parameter curve $\mathbf{Q_0} = (1-t) \mathbf{P_0} + t \mathbf{P_1} \quad t \in [0, 1]$ $\textbf{Quadratic Bezier curve}$: Three points $P_0, P_1, P_2$ are needed. $P_0, P_2$ are anchor points, $P_1$ is a control point. The $\mathbf{Q}$-points move along lines between the $\mathbf{P}$-points. As $R_0$ moves along a line $Q_0$ and $Q_1$, it traces out a quadratic Bezier curve, the three movements are described by\\ \begin{aligned} \mathbf{Q_0} &= (1-t) \mathbf{P_0} + t\mathbf{P_1} \\ \mathbf{Q_1} &= (1-t) \mathbf{P_1} + t\mathbf{P_2} \\ \mathbf{R_0} &= (1-t) \mathbf{Q_0} + t\mathbf{Q_1} \\ \mathbf{R_0} &= (1-t)^2 \mathbf{P_0} + 2(1-t)t \mathbf{P_1} + t^2 \mathbf{P_2} \quad t \in [0, 1] \nonumber \end{aligned} Just use $\mathbf{P}$ point to describe the movement $R_0$ we get that
$\mathbf{R_0} = (1-t)^2 \mathbf{P_0} + 2(1-t)t \mathbf{P_1} + t^2 \mathbf{P_2} \quad t \in [0, 1]$ $\textbf{Cubic Bezier Curve}$: for points $P_0$, $P_1$, $P_2$ $P_3$ are needed. $P_0$ and $P_3$ are anchor points, the other two are control points. Just use $\mathbf{P}$-points to describe $\mathbf{T_0}$ we get that \begin{aligned} \mathbf{R_0} &= (1-t) \mathbf{Q_0} + t\mathbf{Q_1} \\ \mathbf{R_1} &= (1-t) \mathbf{Q_1} + t\mathbf{Q_2} \\ \mathbf{S_0} &= (1-t) \mathbf{R_0} + t\mathbf{R_1} \\ \mathbf{S_0} &= (1-t) ((1-t) \mathbf{Q_0} + t\mathbf{Q_1}) + t ((1-t) \mathbf{Q_1} + t\mathbf{Q_2}) \\ \mathbf{S_0} &= (1-t)^3 \mathbf{P_0} + 3(1-t)^2 t\mathbf{P_1} + 3(1-t) t^{2}\mathbf{P_3} + t^{3} \mathbf{P_4} \nonumber \end{aligned}
$\textbf{Quartic Bezier Curve}$: Five points $P_0$, $P_1$, $P_2$, $P_3$, $P_4$ are needed, $P_0$ and $P_4$ are anchor points, the others are control points. Just use $\mathbf{P}$-points to describe $\mathbf{T_0}$ we get that $\mathbf{T_0} = (1-t)^4 \mathbf{P_0} + 4(1-t)^3 t\mathbf{P_1} + 6(1-t)^2 t^{2}\mathbf{P_2} + 4(1-t) t^3 \mathbf{P_3} + t^4 \mathbf{P_4} \\$ $\textit{Binomial Expansion}$: $[t + (1-t)]^n = \sum_{k=0}^{n} \binom{n}{k} t^k (1-t)^{n-k} \\$ $\textit{Bezier Curve can be defined as following}$: \begin{aligned} \textbf{C}(t) &= \sum_{k=0}^{n} \textbf{B}_{n,k}(t) \mathbf{P}_k \\ \textbf{B}_{n,k}(t) &= \binom{n}{k} t^k (1-t)^{n-k} \\ 1 &= \textbf{B}_{4,0}(t) + \textbf{B}_{4,1}(t) + \textbf{B}_{4,2}(t) + \textbf{B}_{4,3}(t) + \textbf{B}_{4,4}(t) \quad \text{where } n = 4 \quad \because [t+(1-t)]^4 = 1 \\ \end{aligned} $\textit{Plot } \textbf{B}_{4,0}(t) \,, \textbf{B}_{4,1}(t) \,, \textbf{B}_{4,2}(t) \,, \textbf{B}_{4,3}(t) \,, \textbf{B}_{4,4}(t)$ |
# Question: Where do you put the number on a frame?
Contents
## How do you use number frames?
Number Frames helps students structure numbers to 5, 10, 20, and 100. Students use the frames to count, represent, compare, and compute with numbers in a particular range. The frames help students see quantities as equal groups of other quantities and in relation to benchmark quantities.
## How do you use a 5 frame?
Ask students to only put one counter in each space on a five frame to show 3. Ask them to explain ways they have displayed 3. Continue for 0-5. Once the students have displayed a number, ask “How many more counters are needed to make 5?” to continually reference 5.
## How do you do a 10 frame?
To use a ten frame, begin with showing your child a blank ten frame. Add one counter and then count together. Add two counters and then count together. Continue adding counters until you reach ten.
## What is the point of a 10 frame?
Ten frames are an amazing tool used in kindergarten and first grade to help your children understand counting, place value (e.g. where the digit in a number is), adding, subtracting, and more.
## What are frames in math?
In linear algebra, a frame of an inner product space is a generalization of a basis of a vector space to sets that may be linearly dependent. In the terminology of signal processing, a frame provides a redundant, stable way of representing a signal.
## What is a number in math?
A number is a mathematical object used to count, measure, and label. The original examples are the natural numbers 1, 2, 3, 4, and so forth. Calculations with numbers are done with arithmetical operations, the most familiar being addition, subtraction, multiplication, division, and exponentiation.
## What is a frame game?
Simply put A FRAME GAME is an instructional game very consciously created and designed to allow easy loading of the current content and replacing the same with the relevant subject matter. Just as you can mount any picture in a frame, you can load any content in a frame game to suit the needs.
## What is a tens frame in math?
Ten-Frames are two-by-five rectangular frames into which counters are placed to illustrate numbers less than or equal to ten, and are therefore very useful devices for developing number sense within the context of ten. The use of ten-frames was developed by researchers such as Van de Walle (1988) and Bobis (1988).
## What does number sense include?
Number sense refers to a group of key math abilities. It includes the ability to understand quantities and concepts like more and less. Some people have stronger number sense than others.
## What is a frame in math?
In linear algebra, a frame of an inner product space is a generalization of a basis of a vector space to sets that may be linearly dependent. In the terminology of signal processing, a frame provides a redundant, stable way of representing a signal.
## Why is number sense so important?
Number sense is so important for your young math learners because it promotes confidence and encourages flexible thinking. It allows your children to create a relationship with numbers and be able to talk about math as a language. Well, every digit has a value and when you put those digits together they make numbers!
## Why is the number 10 so important for place value?
Base of ten: The term base simply means a collection. Thus in our system, 10 is the value that determines a new collection, and the system has 10 digits, 0 through 9. symbolically the absence of something. For example, 309 shows the absence of tens in a number containing hundreds and ones.
## What is a frame in statistics?
The frame refers to the list of units (eg, persons, households, businesses, etc) in the survey population. Since the selection of the sample is directly based on this list, the frame is one of the most important tools in the design of a survey.
## What is a type number?
Numbers are classified according to type. The first type of number is the first type you ever learned about: the counting, or natural numbers: 1, 2, 3, 4, 5, 6, The next type is the whole numbers, which are the natural numbers together with zero: 0, 1, 2, 3, 4, 5, 6,
## What are called numbers?
A number is a mathematical object used to count, measure, and label. The original examples are the natural numbers 1, 2, 3, 4, and so forth. More universally, individual numbers can be represented by symbols, called numerals; for example, 5 is a numeral that represents the number five.
## What is frame VDI?
About Frame Frame is a virtual desktop infrastructure (VDI) solution designed to help enterprises and independent software vendors (ISVs) run applications and desktops in the cloud.
## What are number sentences?
A number sentence is a mathematical sentence, made up of numbers and signs. The expressions given in examples indicate equality or inequality. A number sentence can use any of the mathematical operations from addition, subtraction, multiplication to division. Number sentences can be true or they may not be true.
## What is a double 10 frame?
This game is called Using Double 10-Frames. To play this math game, each student gets a recording sheet. The students roll either a 6 or 9 sided die two times and create a number sentence. They draw chips or dots on the 10 frames, and whoever has the higher sum is the winner!
## Write us
#### Find us at the office
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# How do you differentiate (e^(2x) + 2x) ^0.5?
Jan 13, 2016
$\frac{{e}^{2 x} + 1}{{e}^{2 x} + 2 x} ^ \left(\frac{1}{2}\right)$
#### Explanation:
$f \left(x\right) = {\left({e}^{2 x} + 2 x\right)}^{\frac{1}{2}}$
using the chain rule :(twice)
$f ' \left(x\right) = \frac{1}{2} {\left({e}^{2 x} + 2 x\right)}^{- \frac{1}{2}} . \frac{d}{\mathrm{dx}} \left({e}^{2 x} + 2 x\right)$
$= \frac{1}{2} {\left({e}^{2 x} + 2 x\right)}^{- \frac{1}{2}} \left({e}^{2 x} . \frac{d}{\mathrm{dx}} \left(2 x\right) + 2\right)$
$= \frac{1}{2} {\left({e}^{2 x} + 2 x\right)}^{- \frac{1}{2}} \left(2 {e}^{2 x} + 2\right)$
$= \frac{1}{2} {\left({e}^{2 x} + 2 x\right)}^{- \frac{1}{2}} .2 \left({e}^{2 x} + 1\right)$
$\Rightarrow f ' \left(x\right) = \frac{{e}^{2 x} + 1}{{e}^{2 x} + 2 x} ^ \left(\frac{1}{2}\right)$ |
Q:
# d= number of dollarsn= number of ouncesDrag each table and equation to the unit rate it matches.
Accepted Solution
A:
Unit rate 4:The table d n4 18 216 4Unit rate 1/4:Tabled n1 44 1616 64andThe equation n=4dUnit rate 16:Equation:d = 16nStep-by-step explanation:The unit in the unit rate is dollars per unit ounces which means that the unit rate calculated by$$Unit\ rate = \frac{dollars}{ounces}$$So,The first option is the equation:$$d = 16n\\\frac{d}{n} = 16$$The unit rate is 16 dollars per ounceThe second option is the table:d n1 44 1616 64We can take any pair of values of d and n from the table to calculate the unit rateSo takingd = 1n=4$$Unit\ rate = \frac{d}{n}\\= \frac{1}{4}$$The unit rate for table is: 1/4 dollars per ouncesThird option is the equation:n = 4d$$n = 4d$$dividing both sides by n$$\frac{n}{n} = \frac{4d}{n}\\1 = \frac{4d}{n}$$dividing both sides by 4$$\frac{1}{4} = \frac{4d}{4n}\\\frac{d}{n} = \frac{1}{4}$$the unit rate is 1/4 dollars per ounceFourth option is the table:d n4 18 216 4We will take any pair of d and n to find the unit rateSo,Takingd = 4n =1$$Unit\ rate = \frac{d}{n} \\= \frac{4}{1}\\=4$$The unit rate is 4 dollars per unit ounce.Keywords: Unit rate, unitsLearn more about unit rate at:brainly.com/question/4550858brainly.com/question/4639731#LearnwithBrainly |
# Factors of 13
Factors of 13 are the numbers that divide the original number 13 completely without leaving the remainder. Though the number 13 is composite, there are no factors other than one and the number itself. Therefore, factors of 13 are 1 and 13. Kids must get acquainted with factors of a number to understand the mathematical concepts easily. Moreover, it helps them solve mathematical problems with speed and accuracy.
Contents
These factors of a number can be positive or negative but not in decimal or fraction form. Besides teaching how to find the factors of 13, kids must also know what pair and prime factors of 13 are. The pair factors of 13 are (1 x 13), whereas the prime factors of 13 are 13. You can teach kids to find the pair and prime factors of 13 using multiplication and prime factorization methods, respectively. To help your children understand the concept and recognize the numbers easily, you can conduct math activities. It is needless to say how essential these games are for kids to learn the factors of a number.
Check Math Related Articles:
## What are the Factors of 13?
Factors of 13 are the numbers obtained when you divide the original number 13 completely with zero remainders. The factors of 13 are 1 and 13, where 13 is the prime number. To find the factors of 13, you need to use the division method. Besides this, you can find out the pair and prime factors of 13 using the multiplication and division methods. The pair factors of 13 are (1, 13), whereas the negative pair factors of 13 are (-1, -13). The prime factors of 13 are 13. The total sum of factors of 13 are 1 + 13 = 14.
## How to Find the Factors of 13?
To find the factors of 13, you need to divide the original number with integers evenly without leaving the remainder. It becomes easier for kids if they are well-acquainted with multiplication table 1-20 for effective calculation practices. Check out the method to calculate the factors of 13 as given below.
13/1 = 13
13/13 = 1
Therefore, factors of 13 are 1 and 13.
### Pair Factors of 13
To find the pair factors of 13, you need to multiply the integers in pairs to get the original number. These numbers can be positive or negative. Check out the multiplication method to calculate the pair factors of 13 as given below.
1 x 13 = 13
13 x 1 = 13
Therefore, the pair factors of 13 are (1, 13), whereas the negative pair factors of 13 are (-1, -13).
### Prime Factors of 13
The prime factors of 13 are the numbers obtained when you divide the original number with the prime numbers until you get the quotient one. Check out the prime factorization method to calculate the prime factors of 13 given below.
13/13 = 1
Therefore, the prime factors of 13 are 13, where 13 is the prime number.
## Solved Examples on Factors of 13
Some solved examples on factors of 13 for kids are mentioned below.
Example 1: Find the common factors of 11 and 13.
Solution: Factors of 11 = 1 and 11
Factors of 13 = 1 and 13
Therefore, the common factors of 11 and 13 are 1.
Example 2: Find the common factors of 13 and 9.
Solution: Factors of 9 = 1, 3, and 9
Factors of 13 = 1 and 13
Therefore, the common factors of 13 and 9 are 1.
We hope this article on factors of 13 was useful to you. For more about activities, worksheets and games, explore worksheets for kids, math for kidspuzzles for kids, kids learning sections at Osmo.
## Frequently Asked Questions on Factors of 13
### What are the factors of 13?
The factors of 13 are 1 and 13.
### What are the pair factors of 13?
The pair factors of 13 are (1, 13), whereas the negative pair factors of 13 are (-1, -13).
### What are the prime factors of 13?
The prime factors of 13 are 13.
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# Subtract the numbers: - 3,280 + 1,296 = ? Calculate the numbers difference and learn how to do the subtraction, column subtracting method, from right to left
## Subtract the numbers: 1,296 - 3,280 = ?
### Stack the numbers on top of each other.
#### And so on...
3 2 8 0 - 1 2 9 6 ?
## Subtract column by column; start from the column on the right
### Subtract the digits in the ones column:
#### When borrowing, 1 ten = 10 ones. Add 10 to the top digit in the column of the ones: 10 + 0 = 10.
7 10 3 2 8 0 - 1 2 9 6
#### After borrowing, the subtraction has become: 10 - 6 = 4. 4 is the ones digit. Write it down at the base of the ones column.
7 10 3 2 8 0 - 1 2 9 6 4
### Subtract the digits in the tens column:
#### When borrowing, 1 hundred = 10 tens. Add 10 to the top digit in the column of the tens: 10 + 7 = 17.
1 17 10 3 2 8 0 - 1 2 9 6 4
#### After borrowing, the subtraction has become: 17 - 9 = 10 + 7 - 9 = 10 + 7 - 9 = 7 + (10 - 9) = 7 + 1 = 8. 8 is the tens digit. Write it down at the base of the tens column.
1 17 10 3 2 8 0 - 1 2 9 6 8 4
### Subtract the digits in the hundreds column:
#### When borrowing, 1 thousand = 10 hundreds. Add 10 to the top digit in the column of the hundreds: 10 + 1 = 11.
2 11 17 10 3 2 8 0 - 1 2 9 6 8 4
#### After borrowing, the subtraction has become: 11 - 2 = 10 + 1 - 2 = 10 + 1 - 2 = 1 + (10 - 2) = 1 + 8 = 9. 9 is the hundreds digit. Write it down at the base of the hundreds column.
2 11 17 10 3 2 8 0 - 1 2 9 6 9 8 4
### Subtract the digits in the thousands column:
#### 32 - 1 = 1. 1 is the thousands digit. Write it down at the base of the thousands column.
2 11 17 10 3 2 8 0 - 1 2 9 6 1 9 8 4
## The latest numbers that were subtracted
7,220 - 5,696 = ? Jan 18 11:05 UTC (GMT) - 3,280 + 1,296 = ? Jan 18 11:05 UTC (GMT) 475 + 3,022 + 18,406 + 262 - 79 = ? Jan 18 11:05 UTC (GMT) 4,113 + 9,590 - 10,238 = ? Jan 18 11:05 UTC (GMT) 483 - 3,037 + 18,405 - 274 + 92 = ? Jan 18 11:05 UTC (GMT) 478 + 4,159 - 699 + 2,077 - 608 - 614 = ? Jan 18 11:05 UTC (GMT) - 4,006 + 9,503 = ? Jan 18 11:05 UTC (GMT) - 274 - 329 - 228 - 481 - 381 = ? Jan 18 11:05 UTC (GMT) 3,856 - 478 + 503 + 2,006 + 116 = ? Jan 18 11:05 UTC (GMT) - 1,003 - 2,000 + 540 - 368 = ? Jan 18 11:05 UTC (GMT) 526 + 762 + 185 - 2,024 + 158 = ? Jan 18 11:05 UTC (GMT) - 47,703 + 9,672 = ? Jan 18 11:05 UTC (GMT) - 405 - 97 - 140 + 117 + 145 - 114 + 190 = ? Jan 18 11:05 UTC (GMT) All the numbers that were subtracted by users |
# ALORA4 - Editorial
problem link : Goblet of Fire
Author : Ritesh Gupta
Tester : Shubham Gupta
Editorialist : Rajan Kumar Raj
Pre-Requisite : Number theory
Problem : You are given with a 3*3 matrix and a integer “k”. You have to find the number of matrices that
satisfies the given condition.
1. Sum of all the numbers in each row is equal to k
2. Sum of all the numbers in each column is equal to k
3. Sum of all the numbers in each diagonal is equal to k
4. All the numbers in the magic square are in the range of 1 to k (including both)
5. There should be at least 1 odd number in the matrix.
All the numbers need not be distinct.
Explanation : Let us observe step by step-
Observation 1: Let us consider the given matrix
. 1 2 3
1 a b c
2 d e f
3 g h i
Now as we know that sum of each column , row and diagonal is equal to k.
Therefore we make equations from diagonals and 2nd row and 2nd column.
``````We get:
b+e+h=k
d+e+f=k
a+e+i=k
c+e+g=k
Now adding these equation and arranging in required way we get:
(a+d+g)+(c+f+i)+(b+h+e)+3e=4k
3e=k
e=k/3;
``````
Let n=k/3.
So we can clearly see that proper value of e will exist only if k is divisible by 3.
Observation 2:
In this observation we are first not considering the condition that 1 element of a matrix must be an odd number.
Now let us make a matrix in which value of k is divisible by 3.
. 1 2 3
1 1 x _
2 _ e _
3 _ _ _
Now we know that sum of diagonal is k.
If we make equation and try to find value of x then it will lie between 2n-2 and 2n ,i.e., we will get 1 value.
If we update the value of A(1)(1) to 2 then we will get 3 values after solving the equation.
Similarily till n it will increase to 2n-1 and then it will start decreasing till it reaches 1.
So the series will be like 1 3 5 . . . . . 2n-3 2n-1 2n-1 . . . . 5 3 1
So our required answer will be n^2+(n-1)^2.
Let us denote our obtained answer by G(k).
Observation 3:
Now let us see the case where an odd number must exist in the matrix.
In this case the required answer will be G(k)-G(k/2).
This part is left for user interpretation.
Time Complexity : O(1) |
# Area of an Irregular Polygon
Unlike a regular polygon, unless you know the coordinates of the vertices, there is no easy formula for the area of an irregular polygon. Each side could be a different length, and each interior angle could be different. It could also be either convex or concave.
If you know the coordinates of the vertices of the polygon, there are two methods:
1. A manual method. See Area of a polygon (Coordinate geometry).
2. A computer algorithm. See Algorithm to find the area of any polygon
### So how to do it?
One approach is to break the shape up into pieces that you can solve - usually triangles, since there are many ways to calculate the area of triangles. Exactly how you do it depends on what you are given to start. Since this is highly variable there is no easy rule for how to do it. The examples below give you some basic approaches to try.
## 1. Break into triangles, then add
In the figure above, the polygon can be broken up into triangles by drawing all the diagonals from one of the vertices. If you know enough sides and angles to find the area of each, then you can simply add them up to find the total. Do not be afraid to draw extra lines anywhere if they will help find shapes you can solve.
Here, the irregular hexagon is divided in to 4 triangles by the addition of the red lines. (See Area of a Triangle)
## 2. Find 'missing' triangles, then subtract
In the figure above, the overall shape is a regular hexagon, but there is a triangular piece missing.
We know how to find the area of a regular polygon so we just subtract the area of the 'missing' triangle created by drawing the red line. (See Area of a Regular Polygon and Area of a Triangle.)
## 3. Consider other shapes
In the figure above, the shape is an irregular hexagon, but it has a symmetry that lets us break it into two parallelograms by drawing the red dotted line. (assuming of course that the lines that look parallel really are!)
We know how to find the area of a parallelogram so we just find the area of each one and add them together. (See Area of a Parallelogram).
As you can see, there an infinite number of ways to break down the shape into pieces that are easier to manage. You then add or subtract the areas of the pieces. Exactly how you do it comes down to personal preference and what you are given to start.
## 4. If you know the coordinates of the vertices
If you know the x,y coordinates of the vertices (corners) of the shape, there is a method for finding the area directly. See Area of a polygon (Coordinate geometry). This works for all polygon types (regular, irregular, convex, concave). There is also a computer algorithm that does the same. See Algorithm to find the area of any polygon |
# IB DP Maths Topic 9.6 Rolle’s theorem HL Paper 3
## Question
The function f is defined by $$f(x) = \left\{ \begin{array}{r}{e^{ – x^3}}( – {x^3} + 2{x^2} + x),x \le 1\\ax + b,x > 1\end{array} \right.$$, where $$a$$ and $$b$$ are constants.
Find the exact values of $$a$$ and $$b$$ if $$f$$ is continuous and differentiable at $$x = 1$$.
[8]
a.
(i) Use Rolle’s theorem, applied to $$f$$, to prove that $$2{x^4} – 4{x^3} – 5{x^2} + 4x + 1 = 0$$ has a root in the interval $$\left] { – 1,1} \right[$$.
(ii) Hence prove that $$2{x^4} – 4{x^3} – 5{x^2} + 4x + 1 = 0$$ has at least two roots in the interval $$\left] { – 1,1} \right[$$.
[7]
b.
## Markscheme
$$\mathop {{\text{lim}}}\limits_{x \to {1^ – }} {{\text{e}}^{ – {x^2}}}\left( { – {x^3} + 2{x^2} + x} \right) = \mathop {{\text{lim}}}\limits_{x \to {1^ + }} (ax + b)$$ $$( = a + b)$$ M1
$$2{{\text{e}}^{ – 1}} = a + b$$ A1
differentiability: attempt to differentiate both expressions M1
$$f'(x) = – 2x{{\text{e}}^{ – {x^2}}}\left( { – {x^3} + 2{x^2} + x} \right) + {{\text{e}}^{ – {x^2}}}\left( { – 3{x^2} + 4x + 1} \right)$$ $$(x < 1)$$ A1
(or $$f'(x) = {{\text{e}}^{ – {x^2}}}\left( {2{x^4} – 4{x^3} – 5{x^2} + 4x + 1} \right)$$)
$$f'(x) = a$$ $$(x > 1)$$ A1
substitute $$x = 1$$ in both expressions and equate
$$– 2{{\text{e}}^{ – 1}} = a$$ A1
substitute value of $$a$$ and find $$b = 4{{\text{e}}^{ – 1}}$$ M1A1
[8 marks]
a.
(i) $$f'(x) = {{\text{e}}^{ – {x^2}}}\left( {2{x^4} – 4{x^3} – 5{x^2} + 4x + 1} \right)$$ (for $$x \leqslant 1$$) M1
$$f(1) = f( – 1)$$ M1
Rolle’s theorem statement (A1)
by Rolle’s Theorem, $$f'(x)$$ has a zero in $$\left] { – 1,1} \right[$$ R1
hence quartic equation has a root in $$\left] { – 1,1} \right[$$ AG
(ii) let $$g(x) = 2{x^4} – 4{x^3} – 5{x^2} + 4x + 1$$.
$$g( – 1) = g(1) < 0$$ and $$g(0) > 0$$ M1
as $$g$$ is a polynomial function it is continuous in $$\left[ { – 1,0} \right]$$ and $$\left[ {0,{\text{ 1}}} \right]$$. R1
(or $$g$$ is a polynomial function continuous in any interval of real numbers)
then the graph of $$g$$ must cross the x-axis at least once in $$\left] { – 1,0} \right[$$ R1
and at least once in $$\left] {0,1} \right[$$.
[7 marks]
b.
[N/A]
a.
[N/A]
b.
## Question
In this question you may assume that $$\arctan x$$ is continuous and differentiable for $$x \in \mathbb{R}$$.
Consider the infinite geometric series
$1 – {x^2} + {x^4} – {x^6} + \ldots \;\;\;\left| x \right| < 1.$
Show that the sum of the series is $$\frac{1}{{1 + {x^2}}}$$.
[1]
a.
Hence show that an expansion of $$\arctan x$$ is $$\arctan x = x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots$$
[4]
b.
$$f$$ is a continuous function defined on $$[a,{\text{ }}b]$$ and differentiable on $$]a,{\text{ }}b[$$ with $$f'(x) > 0$$ on $$]a,{\text{ }}b[$$.
Use the mean value theorem to prove that for any $$x,{\text{ }}y \in [a,{\text{ }}b]$$, if $$y > x$$ then $$f(y) > f(x)$$.
[4]
c.
(i) Given $$g(x) = x – \arctan x$$, prove that $$g'(x) > 0$$, for $$x > 0$$.
(ii) Use the result from part (c) to prove that $$\arctan x < x$$, for $$x > 0$$.
[4]
d.
Use the result from part (c) to prove that $$\arctan x > x – \frac{{{x^3}}}{3}$$, for $$x > 0$$.
[5]
e.
Hence show that $$\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}$$.
[4]
f.
## Markscheme
$$r = – {x^2},\;\;\;S = \frac{1}{{1 + {x^2}}}$$ A1AG
[1 mark]
a.
$$\frac{1}{{1 + {x^2}}} = 1 – {x^2} + {x^4} – {x^6} + \ldots$$
EITHER
$$\int {\frac{1}{{1 + {x^2}}}{\text{d}}x} = \int {1 – {x^2} + {x^4} – {x^6} + \ldots } {\text{d}}x$$ M1
$$\arctan x = c + x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots$$ A1
Note: Do not penalize the absence of $$c$$ at this stage.
when $$x = 0$$ we have $$\arctan 0 = c$$ hence $$c = 0$$ M1A1
OR
$$\int_0^x {\frac{1}{{1 + {t^2}}}{\text{d}}t = } \int_0^x {1 – {t^2} + {t^4}} – {t^6} + \ldots {\text{d}}t$$ M1A1A1
Note: Allow $$x$$ as the variable as well as the limit.
M1 for knowing to integrate, A1 for each of the limits.
$$[\arctan t]_0^x = \left[ {t – \frac{{{t^3}}}{3} + \frac{{{t^5}}}{5} – \frac{{{t^7}}}{7} + \ldots } \right]_0^x$$ A1
hence $$\arctan x = x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots$$ AG
[4 marks]
b.
applying the $$MVT$$ to the function $$f$$ on the interval $$[x,{\text{ }}y]$$ M1
$$\frac{{f(y) – f(x)}}{{y – x}} = f'(c)\;\;\;({\text{for some }}c \in ]x,{\text{ }}y[)$$ A1
$$\frac{{f(y) – f(x)}}{{y – x}} > 0\;\;\;({\text{as }}f'(c) > 0)$$ R1
$$f(y) – f(x) > 0{\text{ as }}y > x$$ R1
$$\Rightarrow f(y) > f(x)$$ AG
Note: If they use $$x$$ rather than $$c$$ they should be awarded M1A0R0, but could get the next R1.
[4 marks]
c.
(i) $$g(x) = x – \arctan x \Rightarrow g'(x) = 1 – \frac{1}{{1 + {x^2}}}$$ A1
this is greater than zero because $$\frac{1}{{1 + {x^2}}} < 1$$ R1
so $$g'(x) > 0$$ AG
(ii) ($$g$$ is a continuous function defined on $$[0,{\text{ }}b]$$ and differentiable on $$]0,{\text{ }}b[$$ with $$g'(x) > 0$$ on $$]0,{\text{ }}b[$$ for all $$b \in \mathbb{R}$$)
(If $$x \in [0,{\text{ }}b]$$ then) from part (c) $$g(x) > g(0)$$ M1
$$x – \arctan x > 0 \Rightarrow \arctan x < x$$ M1
(as $$b$$ can take any positive value it is true for all $$x > 0$$) AG
[4 marks]
d.
let $$h(x) = \arctan x – \left( {x – \frac{{{x^3}}}{3}} \right)$$ M1
($$h$$ is a continuous function defined on $$[0,{\text{ }}b]$$ and differentiable on $$]0,{\text{ }}b[$$ with $$h'(x) > 0$$ on $$]0,{\text{ }}b[$$)
$$h'(x) = \frac{1}{{1 + {x^2}}} – (1 – {x^2})$$ A1
$$= \frac{{1 – (1 – {x^2})(1 + {x^2})}}{{1 + {x^2}}} = \frac{{{x^4}}}{{1 + {x^2}}}$$ M1A1
$$h'(x) > 0$$ hence $$({\text{for }}x \in [0,{\text{ }}b]){\text{ }}h(x) > h(0)( = 0)$$ R1
$$\Rightarrow \arctan x > x – \frac{{{x^3}}}{3}$$ AG
Note: Allow correct working with $$h(x) = x – \frac{{{x^3}}}{3} – \arctan x$$.
[5 marks]
e.
use of $$x – \frac{{{x^3}}}{3} < \arctan x < x$$ M1
choice of $$x = \frac{1}{{\sqrt 3 }}$$ A1
$$\frac{1}{{\sqrt 3 }} – \frac{1}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}$$ M1
$$\frac{8}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}$$ A1
Note: Award final A1 for a correct inequality with a single fraction on each side that leads to the final answer.
$$\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}$$ AG
[4 marks]
Total [22 marks]
f.
## Examiners report
Most candidates picked up this mark for realizing the common ratio was $$– {x^2}$$.
a.
Quite a few candidates did not recognize the importance of ‘hence’ in this question, losing a lot of time by trying to work out the terms from first principles.
Of those who integrated the formula from part (a) only a handful remembered to include the ‘$$+ c$$’ term, and to verify that this must be equal to zero.
b.
Most candidates were able to achieve some marks on this question. The most commonly lost mark was through not stating that the inequality was unchanged when multiplying by $$y – x{\text{ as }}y > x$$.
c.
The first part of this question proved to be very straightforward for the majority of candidates.
In (ii) very few realized that they had to replace the lower variable in the formula from part (c) by zero.
d.
Candidates found this part difficult, failing to spot which function was required.
e.
Many candidates, even those who did not successfully complete (d) (ii) or (e), realized that these parts gave them the necessary inequality.
f. |
# Math Lesson Plan
Topics: Division, Elementary arithmetic, Number Pages: 2 (363 words) Published: April 30, 2005
Time: 40 minutes
Subject: Math
Topic: Dividing and Multiplying to Find Equivalent Fractions
NY State Learning Standards: Mathematics, Science, and Technology Standard 1: Analysis, Inquiry, and Design
Students will use mathematical analysis and scientific inquiry to seek answers and develop solutions.
Materials:
Mathematics Textbooks (page 401)
Notebooks
Pencils
Different colored chalk
Objectives: Students will be able to name and write equivalent fractions by multiplying and dividing.
Procedure:
1.Ask students What does equivalent mean?
2.Point out that we have already used fraction strips to show equivalent fractions. 3.Explain that we are now going to use multiplication and division to write equivalent fractions. 4.First, start with multiplication and show an example of how to make an equivalent fraction. Remind students that what we do to the top we must do to the bottom (signal hands up
.hands down) 5.Use one color of chalk to show what you do to the numerator and another color to show what you do to the denominator. 6.Put another example on the board and ask a volunteer to help write an equivalent fraction. 7.Point out that the number we use to multiply can be any number, as long as what you do to the top, you do to the bottom. 8.Go over multiple examples together and assign textbook examples. Go over various answers. 9.Call students up to the board to show some answers.
10. Go on to show equivalent fractions using division.
11.Explain that the first step in finding equivalent fraction using division is to find the factors of both the numerator and denominator. 12.Circle the common factor and divide both the numerator and denominator by the same number
.because what you do to the top, you must do to the bottom. 13.Put some examples on the board and go over with the class. 14.Then, give some examples for the students to do on their own. Then review. 15.Bring the lesson to a... |
# Floor Function Proof
Let $m \in \mathbb{R}$ and $n \in \mathbb{N}$. Prove the following facts:
$\lfloor \hspace2mm\rfloor$ Means Floor function: And $\lfloor \lfloor m\rfloor/n\rfloor$ mean the floor of $m$ and then the floor of the floor $m$ divides $n$.
(i) $\lfloor m+n \rfloor = \lfloor m \rfloor+n$
(ii) $\lfloor \lfloor m \rfloor / n \rfloor = \lfloor m/n \rfloor$
(iii) $\lfloor m \rfloor + \lfloor m + 1/n \rfloor + \lfloor m + 2/n \rfloor + \ldots + \lfloor m + (n-1)/n \rfloor = \lfloor m n \rfloor$ (induction?)
Lemma: for $x = nθ$, where $θ = m−\lfloor m \rfloor$.
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For all the problems, the main thing you need is to write $$x = \lfloor x \rfloor + \{x\}$$ where $\lfloor x \rfloor \in \mathbb{Z}$ and $\{x\} \in [0,1)$.
For instance, for the first one
1. $\lfloor m + n \rfloor = \lfloor \lfloor m \rfloor +\{m\} + n \rfloor = \lfloor \lfloor m \rfloor + n +\{m\} \rfloor = \lfloor m \rfloor + n$.
2. If $$\lfloor \lfloor m \rfloor/n \rfloor = a \in \mathbb{Z}$$ then we have that $$\lfloor m \rfloor/n = a + b$$ where $b \in \left \{0,\dfrac1n, \dfrac2n, \ldots, \dfrac{(n-1)}n \right\}$. Hence, $$\lfloor m \rfloor = an + bn \implies m = an + bn + c$$ where $c \in [0,1)$. Hence, $$m/n = a + b + \dfrac{c}n$$ Note that $b + \dfrac{c}n \in \left\{ \dfrac{c}n,\dfrac{c+1}n, \ldots, \dfrac{c+(n-1)}n\right\} \subseteq [0,1)$. Hence, $$\lfloor m/n \rfloor = a = \lfloor \lfloor m \rfloor/n \rfloor$$
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The proof of (2) is not quite right, since $m \in R$, not $N$. – marty cohen Jul 1 '15 at 22:03
The first two are straightforward using the universal property of the floor function, viz. $$\rm k\le \lfloor r \rfloor \iff k\le r,\ \ \ for\ \ \ k\in \mathbb Z,\ r\in \mathbb R$$ Therefore for $\rm\:0 < n\in \mathbb Z,\ r\in \mathbb R,\$ $$\rm\begin{eqnarray} &\rm k &\le&\:\rm\ \lfloor \lfloor r \rfloor / n\rfloor \\ \iff& \rm k &\le&\ \ \rm \lfloor r \rfloor / n \\ \iff& \rm nk &\le&\ \ \rm \lfloor r \rfloor \\ \iff& \rm nk &\le&\ \ \rm r \\ \iff& \rm k &\le&\ \ \rm r/n \\ \iff& \rm k &\le&\ \ \rm \lfloor r/n \rfloor \\ \\ \Rightarrow\ \ \rm \lfloor \lfloor r\!\!&\rm \rfloor / n\rfloor\ &=&\rm\ \ \lfloor r/n\rfloor \end{eqnarray}$$
since, having equal predecessors, these integers are equal.
If you know a little category theory you can view this universal property of floor as a right adjoint to inclusion, e.g. see Arturo's answer here, or see most any textbook on category theory. But, of course, one need not know any category theory to understand the above proof. Indeed, I've had success explaining this (and similar universal-inspired proofs) to bright high-school students.
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The floor function (also known as the entier function) is defined as having its value the largest integer which does not exceed its argument. When applied to any positive argument it represents the integer part of the argument obtained by suppressing the fractional part.
From this definition it is clear that for all positive real numbers x, y [x + y] > [x] + [y] since [x] and [y] are the integer parts of x and y. It's also obvious that equality is obtained when x and y are integers since they then have zero fractional parts.
let {x} = x - [x] and {y} = y - [y], so they are the decimal parts of x and y.
Then x + y = [x] + [y] + {x} + {y}
Case 1: {x} + {y} < 1 then {x + y} = {[x] + [y] +{x} + {y}} = {{x} + {y}} = {x} + {y} (we can remove from inside the {}s anything to the left of the decimal point!)
so [x+y] = x + y - {x + y} = x + y - {x} - {y} = [x] + [y]
Case 2: {x} + {y} >= 1 (but clearly must be <2) then {x + y} = {[x] + [y] + {x} + {y}} = {{x} + {y}} = {x} + {y} - 1
so [x + y] = x + y - {x + y} = x + y - {x} - {y} + 1 = [x] + [y] + 1
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I hadn´t the joy of solving this. The beauty of this proof lies in a weird and simple relation between trigonometry, and number theory.
$$From\space trigonometry\space we\space have\quad \sin(nx)=2^{n-1}\space\prod_{k=0}^{n-1} \sin(\dfrac{\pi k }{n} + x) \quad and\space for\quad x\notin \mathbb Z\quad sgn(\sin(\pi x))=(-1)^{\lfloor x \rfloor}$$$$therefore$$ $$(-1)^{\lfloor nx \rfloor}=sgn(\sin(n\pi x))$$$$=sgn(2^{n-1}\space\prod_{k=0}^{n-1} \sin(\pi (\dfrac{ k }{n} + x))\space)$$$$=sgn(2^{n-1})\space sgn(\prod_{k=0}^{n-1} \sin(\pi (\dfrac{ k }{n} + x))\space)$$$$=\prod_{k=0}^{n-1} sgn( \sin(\pi (\dfrac{ k }{n} + x)\space)$$
$$=\prod_{k=0}^{n-1} (-1)^{\lfloor (\frac kn + x) \rfloor}$$ $$=(-1)^{\sum_{k=0}^{n-1} \lfloor (\frac kn + x) \rfloor}$$ and from both equation sides $${\lfloor nx \rfloor}={\sum_{k=0}^{n-1} \lfloor (\frac kn + x) \rfloor}$$
The trigonometric id is explained on other posts. You could demonstrate it by using the complex expression for sines.
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I don't think so. Just because $(-1)^a = (-1)^b$, it does not follow that $a = b$. They could, for example, differ by 2. However, if you can show that $a$ and $b$ are integers that differ by less than 2, then they are equal. – marty cohen Jul 1 '15 at 22:08
True.... They differ by an even integer since the floor function returns integers. It seems less affordable to demonstrate that the difference is less than two than to demonstrate it by other means. – Saul Mendoza Jul 2 '15 at 21:27
$$Let\space m\space denote\space the \space integer\space part\space of \space x\in \mathbb R\space,\quad$$$$i.e. \space \space m\in \mathbb Z \mid m=\lfloor x \rfloor \quad and\quad \varepsilon =x-\lfloor x \rfloor,\space \varepsilon \in [0,1)$$$$Then \space we \space define\space k=\lfloor n\varepsilon \rfloor \in \left \{0,1, 2, \ldots, (n-1) \right\}$$$$so\space that\space \lfloor nx \rfloor=\lfloor n(m + \varepsilon) \rfloor=nm + \lfloor n \varepsilon \rfloor=nm + k$$$$on \space the \space other \space hand \quad \sum_{j=0}^{n-1} \lfloor x +(\frac jn) \rfloor={\sum_{j=0}^{n-1} \lfloor (m + \varepsilon) +(\frac jn) \rfloor}={\sum_{j=0}^{n-1} m + \lfloor \varepsilon +(\frac jn) \rfloor}={ nm + \sum_{j=0}^{n-1}\lfloor \varepsilon +(\frac jn) \rfloor}$$ $$but \quad {\sum_{j=0}^{n-1}\lfloor \varepsilon +(\frac jn) \rfloor}={\sum_{j=0}^{n-1}\lfloor\frac{n\varepsilon + j}{n} \rfloor}={\sum_{j=0}^{n-1}\lfloor\frac{\lfloor n\varepsilon + j\rfloor}{n} \rfloor}={\sum_{j=0}^{n-1}\lfloor\frac{\lfloor n\varepsilon\rfloor + j}{n} \rfloor}={\sum_{j=0}^{n-1}\lfloor\frac{k + j}{n} \rfloor}={\sum_{j=0}^{n-k-1}\lfloor\frac{k + j}{n} \rfloor +\sum_{j=n-k}^{n-1}\lfloor\frac{k + j}{n} \rfloor}={\sum_{j=n-k}^{n-1}\lfloor\frac{k + j}{n} \rfloor}={\sum_{j=n-k}^{n-1}(1)}=k$$ $$and\space consequently\quad \lfloor nx \rfloor=nm + k=\sum_{j=0}^{n-1} \lfloor x +(\frac jn) \rfloor$$
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# A Mental Trick for Checking Primeness of Two-Digit Numbers
Is 57 prime? How about 71?
If you aren’t sure (like me), then read on. Here’s a trick for quickly checking two-digit numbers for primality.
I’ll first cover how it works, before finally showing how to apply it.
# Overview
It boils down to this: a two digit number is prime if it’s not divisible by any of 2, 3, 5, or 7. I’ll explain why later.
We can break this test down into easier parts:
1. Divisibility by 2 and 5: This can be done instantly, by inspection.
2. … by 3: There’s a simple test you can apply
3. … by 7: You can memorise this bit
I’ll explain each of these in more detail shortly.
You may ask: “Why not just memorise the list of all 21 two digit primes?”. I’d wager that this method is much easier to remember, and much more likely to stay remembered.
That’s because instead of remembering 21 numbers, I just need to remember the “seed” of the method: It’s only necessary to check divisibility by 2, 3, 5, and 7. The rest of the method follows from that fact.
# Later
So why do we only need to check 2, 3, 5, and 7? Because…
• Any composite (non-prime) number has a prime factor less than its square root
• All two-digit numbers are less than 100
• The square-root of 100 is 10
• Therefore all two-digits numbers have a square root less than 10
• Therefore, all two-digit composite numbers have a prime factor less than 10.
Finally, the only primes less than 10 are 2, 3, 5, and 7.
You might not be convinced by the first statement- that a composite number $$n$$ has a prime factor less than $$\sqrt n$$ - so we’ll prove it by contradiction.
First, observe that a composite number $$n$$ has at least two factors by definition:
$n = ab$
Suppose $$a \gt \sqrt n$$ and $$b \gt \sqrt n$$. Then $$ab \gt (\sqrt n)^2$$, and so $$ab \gt n$$.
However, our premise was that $$ab = n$$, so we’ve reached a contradiction. Therefore, either $$a \leq \sqrt n$$ or $$b \leq \sqrt n$$.
Neither $$a$$ nor $$b$$ must be prime, but if composite they must have a smaller prime factor - thus proving that $$n$$ has a prime factor smaller than $$\sqrt n$$.
Note that this doesn’t mean the prime factors of two-digit numbers are all less than 10. Only one of $$a \leq \sqrt n$$ or $$b \leq \sqrt n$$ need be true. For example, $$91 = 13 \times 7$$, and $$13 > \sqrt{100}$$
# Shortly
Now we know checking divisibility 2, 3, 5, and 7 is sufficient, we want to make it quick.
Divisibility by 2 or 5 is as simple as checking the final digit. If it’s even, or 5, it’s not prime.
Divisibility by 3 has a simple test: sum the digits of your number. If that sum is divisible by 3, then so is the number.
Divisibility by 7 is harder. The divisibility test on Wikipedia is not so easy to do in your head. However, once you’ve eliminated the numbers divisible by 2, 3, and 5, there are only 3 left which are divisible by 7!
Those are:
• $$49 = 7 \times 7$$
• $$77 = 7 \times 11$$
• $$91 = 7 \times 13$$
Of these, 49 and 77 are pretty clearly not prime - at least, I would recognise these as being $$7^2$$ and $$7 \times 11$$.
I don’t know a trick other than memorisation for $$7 \times 13 = 91$$, but it doesn’t seem too high a price to pay.
# Finally
To summarise, here’s how you check a two-digit number for primeness:
A two-digit number is prime if you answer no to all these questions:
1. Is it even? (i.e., does it end in 0, 2, 4, 6, or 8?)
2. Does it end in 5?
3. Does the sum of its digits divide by 3?
4. Is it 49, 77, or 91?
Although this looks like a lot of text, this is pretty straightforward to apply. After a couple minute’s practice, it can be done in just a few seconds.
Let’s apply it to the original question: Is 57 prime? How about 71?
Let’s begin with 57:
1. It’s not even
2. It doesn’t end in 5
3. Its digits sum to 12, which is divisible by 3
Therefore, it’s not prime. $$57 = 19 \times 3$$
Now for 71:
1. It’s not even
2. It doesn’t end in 5
3. Its digits sum to 8, which isn’t divisible by 3
4. It’s not 49, 77, or 91
Therefore, 71 is prime.
Hopefully you’re convinced this is pretty straightforward to do in your head: after all, steps 1, 2, and 4 don’t really take any time at all. Only step 3 is a bit indirect, but adding two digits together isn’t too bad! |
Write Point-Slope Form Given Slope And A Point Worksheet
6 problems
To write the point-slope form of a line when given a slope and a point, use the formula y - y_1 = m(x - x_1). Here, m represents the slope, and (x_1, y_1) is the point given. For example, if the slope ‘m’ is 2 and the point is (3, 4), you substitute these values into the formula: y - 4 = 2(x - 3). This equation shows how the line relates to the given point and slope, making it clear how the line moves through that specific point with the specified slope.
Algebra 1
Linear Relationship
How Will This Worksheet on "Write Point-Slope Form Given Slope and a Point" Benefit Your Student's Learning?
• Demonstrates how a point and slope define a straight line.
• Helps find the rate of change or predict situations based on initial conditions.
• Enhances algebra skills and orientation towards math tasks.
• Clearly shows number connections, making it easier to explain ideas.
How to Write Point-Slope Form Given Slope and a Point?
• Let the slope be $$m$$ and the point be $$(x_1, y_1)$$.
• The point-slope form is: $$y - y_1 = m(x - x_1)$$.
• Replace $$m$$ with the given slope.
• Replace $$x_1$$ and $$y_1$$ with the coordinates of the given point.
• Expand and simplify the equation to make it look cleaner if needed.
Solved Example
Q. A line with a slope of $9$ passes through the point $(9,9)$. What is its equation in point-slope form?
Solution:
1. Define Point-Slope Form: Point-slope form is $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point the line passes through.
2. Identify Slope and Point: We know the slope $m$ is $9$ and the point is $(9,9)$, so we plug these values into the point-slope form.
3. Substitute Values into Formula: $y - 9 = 9(x - 9)$ is the equation in point-slope form.
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# How Does Your Math Garden Grow?
## Lesson Goals
Demonstrate their knowledge of and application of the properties of multiplication to solve area problems.
Design a school garden incorporating visual arts and use mathematical labels and equations to describe the different areas of the garden.
Students will use their knowledge of the multiplication to design a school garden.
## Materials
Picture(s) of architectural landscape plans
Graph paper: enough for students to do multiple drafts
Colored Pencils: enough for students to share or have their own
Erasers: for each student
Pencils: for each student
## Activities
1. Have students gather in a circle in the center of the room to play Buzz. Tell the children that the aim of the game is to count to 50 – simple, right? Except all multiples of the chosen number must be replaced by the word “Buzz”. For example, if the buzz number was multiples of 5, play would go – 1,2,3,4,buzz,6,7,8,9, buzz,11,12 . . . etc. If a child makes a mistake play begins (for the whole group) again at 1. As a variation add two multiples such as 3 and 5.
2. Unpack the following learning targets on the board by highlighting and defining the verbs and key vocabulary:
-I can relate area to the operations of multiplication and addition.
– I can create visual representation of a school garden by generation ideas, planning solutions, and producing original art.
3. While unpacking the math learning target, be sure students understand arrays, and how to visually represent multiplication and repeated addition.
4. Have students pair up to practice showing as many arrays as they can with an area of 8, 12, 15, 24, or any other areas you wish to show.
5. Ask student pairs to make an 8×7 rectangle. Tell them they have to find the area without counting each square one by one. How could they break apart the rectangle to make it easier to multiply? Have them label the lengths and include the math equations.
6. Circulate and find a couple of student pairs to share their ideas with the class. Choose student work that demonstrates the distributive property and help the students explain.
7. Show students a picture of or an actual landscape architect’s plan for a yard or garden. Point out the scale, measurements, math, labels, and art.
8. Tell students that their job will be to use graph paper and create or recreate a garden for the school. Tell them they must label their garden boxes and beds with not only the math equations, but it also must have the plants they would like to grow.
9. Have students share their work in a gallery walk. Students will give feedback by writing one glow and one grow on sticky notes. Be sure to emphasize feedback that is kind, specific, and helpful.
10. Students can then review the feedback for a final draft and write a reflection statement about how they feel about the learning targets.
## Differentiation Approaches
Students may only create beds with areas no larger than 18 square units.
Students can use actual measurements and draw it to scale.
## Assessment
Did the students label their garden boxes/beds with the correct areas and mathematical equations?
Observation of student pairs.
## Follow Up and Extension Ideas
Use this lesson as a jumping off point for a project that incorporates research for planting garden beds during different seasons. Create a list of supplies. Calculate the cost of supplies and create a budget. Invite landscape architects and community experts to work with the students. Make a plan to donate food or flowers to local nonprofits, food banks, or your school family. Create podcasts, videos, and/or persuasive writing/letters to pitch their ideas to the PTA, parents, and administrators so that their plans could be put into action. |
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# Conservation of Energy
## The amount of energy in a closed system never changes.
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Practice Conservation of Energy
Progress
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Conservation of Energy
Students will learn how to apply energy conservation in a closed system.
### Key Equations
Einitial=EfinalThe total energy does not change in closed systems$\sum E_{\text{initial}} = \sum E_{\text{final}} \; \; \text{The total energy does not change in closed systems}$
Guidance
Energy is conserved in a closed system. That is, if you add up all the energy of an object(s) at one time it will equal all the energy of said object(s) at a later time. A closed system is a system where no energy is transferred in or out. The total energy of the universe is a constant (i.e. it does not change). The problems below do not consider the situation of energy transfer (called work). So friction and other sources where energy leaves the system are not present. Thus, one simply adds up all the potential energy and kinetic energy before and sets it equal to the addition of the total potential energy and kinetic energy after .
#### Example 1
Billy is standing at the bottom of a ramp inclined at 30 degrees. Billy slides a 2 kg puck up the ramp with an initial velocity of 4 m/s. How far up the ramp does the ball travel before it begins to roll back down? Ignore the effects of friction.
##### Solution
The potential energy of the puck when it stops at the top of it's path will be equal to the kinetic energy that it was initially rolled with. We can use this to determine the how high above the ground the puck will be above the ground when it stops, and then use trigonometry to find out how far up the ramp the puck will be when it stops.
$PE_i + KE_i &= PE_f + KE_f && \text{start with conservation of energy}\\0 + \frac{1}{2}mv^2 &= mgh + 0 && \text{take out the energy terms we know will be zero and substitute the equations for potential and kinetic energy.}\\\frac{1}{2}v^2 &= gh && \text{simplify the equation}\\h&=\frac{v^2}{2g} && \text{solve for h}\\h&=\frac{(4\;\text{m/s})^2}{2*9.8\;\text{m/s}^2} && \text{substitute in the known values}\\h&=0.82\;\text{m}\\$
Now we can find the distance up the ramp the ball traveled since we know the angle of the ramp and the height of the ball above the ground.
$\sin(30)&=\frac{h}{x}\\x&=\frac{h}{\sin(30)}\\x&=\frac{0.82\;\text{m}}{\sin(30)}\\x&=1.6\;\text{m}\\$
### Explore More
1. At 8:00 AM, a bomb exploded in mid-air. Which of the following is true of the pieces of the bomb after explosion? (Select all that apply.)
1. The vector sum of the momenta of all the pieces is zero.
2. The total kinetic energy of all the pieces is zero.
3. The chemical potential energy of the bomb has been converted entirely to the kinetic energy of the pieces.
4. Energy is lost from the system to sound, heat, and a pressure wave.
2. You are at rest on your bicycle at the top of a hill that is $20 \;\mathrm{m}$ tall. You start rolling down the hill. At the bottom of the hill you have a speed of $22 \;\mathrm{m/s}$ . Your mass is $80 \;\mathrm{kg}$ . Assuming no energy is gained by or lost to any other source, which of the following must be true?
1. The wind must be doing work on you.
2. You must be doing work on the wind.
3. No work has been done on either you or the wind.
4. Not enough information to choose from the first three.
3. A stationary bomb explodes into hundreds of pieces. Which of the following statements best describes the situation?
1. The kinetic energy of the bomb was converted into heat.
2. The chemical potential energy stored in the bomb was converted into heat and gravitational potential energy.
3. The chemical potential energy stored in the bomb was converted into heat and kinetic energy.
4. The chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.
5. The kinetic and chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.
4. You hike up to the top of Granite Peak in the Trinity Alps to think about physics.
1. Do you have more potential or kinetic energy at the top of the mountain than you did at the bottom? Explain.
2. Do you have more, less, or the same amount of energy at the top of the mountain than when you started? (Let’s assume you did not eat anything on the way up.) Explain.
3. How has the total energy of the Solar System changed due to your hike up the mountain? Explain.
4. If you push a rock off the top, will it end up with more, less, or the same amount of energy at the bottom? Explain.
5. For each of the following types of energy, describe whether you gained it, you lost it, or it stayed the same during your hike:
1. Gravitational potential energy
2. Energy stored in the atomic nuclei in your body
3. Heat energy
4. Chemical potential energy stored in the fat cells in your body
5. Sound energy from your footsteps
6. Energy given to you by a wind blowing at your back
5. A 1200 kg> car traveling with a speed of 29 m/s drives horizontally off of a 90 m cliff.
1. Sketch the situation.
2. Calculate the potential energy, the kinetic energy, and the total energy of the car as it leaves the cliff.
3. Make a graph displaying the kinetic, gravitational potential, and total energy of the car at each 10 m increment of height as it drops
6. A roller coaster begins at rest $120 \;\mathrm{m}$ above the ground, as shown. Assume no friction from the wheels and air, and that no energy is lost to heat, sound, and so on. The radius of the loop is $40 \;\mathrm{m}$ .
1. Find the speed of the roller coaster at points $B, C, D, E, F$ , and $H$ .
2. Assume that $25$ % of the initial potential energy of the coaster is lost due to heat, sound, and air resistance along its route. How far short of point $H$ will the coaster stop?
7. A pendulum has a string with length 1.2 m. You hold it at an angle of 22 degrees to the vertical and release it. The pendulum bob has a mass of 2.0 kg.
1. What is the potential energy of the bob before it is released? ( Hint: use geometry to determine the height when released. )
2. What is its speed when it passes through the midpoint of its swing?
3. Now the pendulum is transported to Mars, where the acceleration of gravity g is $2.3 \;\mathrm{m/s}^2$ . Answer parts (a) and (b) again, but this time using the acceleration on Mars.
1. discuss in class
2. discuss in class
3. d
4. discuss in class
5. b. $KE = 504,600 \;\mathrm{J}; U_g = 1,058,400 \;\mathrm{J}; E_{total} = 1,563,000 \;\mathrm{J}$
6. a. $34 \;\mathrm{m/s \ at \ B}; 49 \;\mathrm{m/s \ at \ C \ and \ F}; 28 \;\mathrm{m/s \ at \ D}, 40 \;\mathrm{m/s \ at \ E}, 0 \;\mathrm{m/s \ at \ H}$ b. it will make it up to only 90m, so 30m short of point H
7. a. 1.7 J b. 1.3 m/s c. 0.4 J, 0.63 m/s |
# Decimal Lesson Plans: Teaching Decimal Numbers with Cheese!
By AJFA
Students can add and subtract decimal numbers using manipulatives such as base-ten blocks and cheese cubes. If you are looking for ideas to teach your students addition and subtraction then give this fun lesson plan a try.
## Materials Needed
You'll need a few things before you get started with this lesson plan to teach students addition and subtraction.
• Base-ten blocks
• Wax paper
• Cheese cubes
• Plastic knives
## Description of Activity
Use the base-ten blocks to model numbers such as 251 and 372. Suggest that you would like to try substituting a cube of cheese for one centimeter block used to model the units of one. Model other numbers such as 586 of 144 with the cheese. Tell students that you are planning to cut one of the little cubes into 10 parts. (Do this so that everyone can see or in small groups. Try to get the 10 pieces as equal as possible.) Do you still have 144? (Yes) Give 4 students a tiny piece. Now you have 143 and 6/10. Show students that this number can be represented in a variety of ways: with words (two hundred forty-three and six-tenths), blocks, as a mixed number (143 and 6/10), and as a decimal number (143.6).
Use the base-ten blocks and the tiny tenth pieces to model other numbers, having students write the number on paper (while one student writes on the board) the names of these numbers in words, as mixed numbers, and as decimal numbers.
Give each group of students wax paper to work on, a plastic knife, and 4 cubes of cheese per group. Have students cut one cube into ten pieces. The fact that they are small will help students remember that tenths are parts of one. Ask students to take turns being the leader and naming numbers less than four to model (2.3, 1.7, 3.5, etc.). Students should model the numbers and write them in words, as mixed numbers, and in decimal form.
When you are ready to end the lesson, give the entire group these directions to dispose of the manipulatives. Keep in mind the primary purpose is to model decimal numbers and relate the models to the manner in which they are recorded.
• Show me 1.2 cubes of cheese. Remove these. What do you have left? (2.8)
• Show me 1 cube of cheese. In a different place show me 0.4 of a cube. Which is greater, 1 or 0.4? (1)
• Remove 0.4. What is left? (2.4)
• Remove 2. What is left? (0.4) Write how many cubes are left.
## Assessment
Check that students can relate the amount of cheese to decimal numbers and add and subtract the cheese cubes.
## Extensions
• Have students write about decimals to give you information about their understanding and to give them practice expressing mathematical ideas in narrative form. Some possible questions to answer are: Explain what a decimal number is? Why would you prefer to have 2 cheese cubes rather than 0.8 cheese cubes? What does 0.4 mean?
• Discuss how decimal numbers are used in science.
• Instead of using cheese cubes and plastic knives, this activity can be completed using paper tenth strips and scissors. |
## Formula
### Summary
The quadratic formula calculates the -intercepts of a quadratic equation which forms the shape of a parabola. The general form of a quadratic equation is shown below.
Expression Description
The second-degree coefficient
The first-degree coefficient
The zero-degree coefficient
## Usage
The quadratic formula returns the intercepts of a quadratic equation. The formula is useful when a quadratic equation cannot be easily factored. For example, the quadratic equation shown below is difficult to factor and so is a good candidate for the formula.
Applying the quadratic formula to the equation above, we are able to find the roots of the equation. Substituting , and into the formula gives us the expression below.
Then, after some simplification and expanding the (plus-minus symbol) into two epxressions we are left with the intercepts of the equation. Note, as discussed in the explanation below, the formula returns a result that indicates there is no solution.
## Examples
### Example 1
This example demonstrates applying the quadratic formula to the quadratic equation shown in the plot above and in the equation below. The formula returns the places where the parabola crosses the -axis, also called the roots of the equation.
Note, applying the quadratic formula is the same as setting the equation equal to zero and solving for as shown in the expression below.
#### Steps
1. Start by setting up the quadratic formula.
2. Substitute the coefficients of the quadratic formula into equation. In this case, substitute , and into the formula.
3. Simplify the expression by evaluating arithmetic within the square root operator and in the denominator.
4. Evaluate the division and square root operator.
5. Expand the (plus-minus symbol) into two solutions representing the positive and negative result of the square root operator.
The two roots of the quadratic equation are and .
### Example 2
This example demonstrates applying the quadratic formula to the quadratic equation shown in the plot above and in the equation below. The formula returns the places where the parabola crosses the -axis, also called the roots of the equation.
Note, applying the quadratic formula is the same as setting the equation equal to zero and solving for as shown in the expression below.
#### Steps
1. Start by setting up the quadratic formula.
2. Then substitute the coefficients of the quadratic formula into equation. In this case, substitute , and into the formula.
3. Simplify the expression by evaluating arithmetic within the square root operator and in the denominator.
4. Evaluate the division and square root operator.
5. Expand the (plus-minus symbol) into two solutions representing the positive and negative result of the square root operator.
The two roots of the quadratic equation are and .
## Explanation
The quadratic formula is derived from the general form of the quadratic equation set equal to zero[1]. Visually, a quadratic equation forms the shape of a parabola and the formula returns the points where the parabola intersects with the axis. This is illustrated in the figure below:
Plugging in the coefficients: , and of the quadratic equation into the formula will produce two results because the square root operator returns both a positive and negative solution. This is explicitly shown by expanding the (plus-minus) symbol into two separate expressions.
### Edge Cases
Sometimes the quadratic equation does not intersect with the -axis as is the case with the quadratic equation below. When this happens, applying the quadratic formula will result in a negative number as input to the square root operator.
For example, applying the quadratic formula to this equation gives the following expression.
This indicates that there is no solution, unless you expand the domain of the input to complex numbers.
#### Divide by Zero
The other edge case with this formula is if the input equation isn’t quadratic, meaning that the coefficient . In this case, the formula results in a divide by zero error. |
Paul's Online Notes
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Section 10.13 : Estimating the Value of a Series
2. Use the Comparison Test and $$n = 20$$ to estimate the value of $$\displaystyle \sum\limits_{n = 3}^\infty {\frac{1}{{{n^3}\ln \left( n \right)}}}$$.
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Start Solution
Since we are being asked to use the Comparison Test to estimate the value of the series we should first make sure that the Comparison Test can actually be used on this series.
In this case that is easy enough because, for our range of $$n$$, the series terms are clearly positive and so we can use the Comparison Test.
Note that it is really important to test these conditions before proceeding with the problem. It doesn’t make any sense to use a test to estimate the value of a series if the test can’t be used on the series. We shouldn’t just assume that because we are being asked to use a test here that the test can actually be used!
Show Step 2
Let’s start off with the partial sum using $$n = 20$$. This is,
${s_{20}} = \sum\limits_{n = 3}^{20} {\frac{1}{{{n^3}\ln \left( n \right)}}} = 0.057315878$ Show Step 3
Now, let’s see if we can get can get an error estimate on this approximation of the series value. To do that we’ll first need to do the Comparison Test on this series.
That is easy enough for this series once we notice that $$\ln \left( n \right)$$ is an increasing function and so $$\ln \left( n \right) \ge \ln \left( 3 \right)$$. Therefore, we get,
$\frac{1}{{{n^3}\ln \left( n \right)}} \le \frac{1}{{{n^3}\ln \left( 3 \right)}} = \frac{1}{{\ln \left( 3 \right)}}\frac{1}{{{n^3}}}$ Show Step 4
We now know, from the discussion in the notes, that an upper bound on the value of the remainder (i.e. the error between the approximation and exact value) is,
\begin{align*}{R_{20}} \le {T_{20}} &= \sum\limits_{n = 21}^\infty {\frac{1}{{{n^3}\ln \left( 3 \right)}}} < \int_{{20}}^{\infty }{{\frac{1}{{{x^3}\ln \left( 3 \right)}}\,dx}}\\ & = \mathop {\lim }\limits_{t \to \infty } \int_{{20}}^{t}{{\frac{1}{{{x^3}\ln \left( 3 \right)}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( { - \frac{1}{{2{x^2}\ln \left( 3 \right)}}} \right)} \right|_{20}^t\\ & = \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{{800\ln \left( 3 \right)}} - \frac{1}{{2{t^2}\ln \left( 3 \right)}}} \right) = \frac{1}{{800\ln \left( 3 \right)}}\end{align*} Show Step 5
So, we can estimate that the value of the series is,
$\require{bbox} \bbox[2pt,border:1px solid black]{{s \approx 0.057315878}}$
and the error on this estimate will be no more than $$\frac{1}{{800\ln \left( 3 \right)}} = 0.001137799$$. |
Lesson plan
# Lesson 6: Using Equations to Solve Problems
teaches Common Core State Standards MP8 http://corestandards.org/Math/Practice/MP8
teaches Common Core State Standards 7.RP.A.2.c http://corestandards.org/Math/Content/7/RP/A/2/c
teaches Common Core State Standards 7.RP.A.2 http://corestandards.org/Math/Content/7/RP/A/2
teaches Common Core State Standards MP2 http://corestandards.org/Math/Practice/MP2
# Lesson 6: Using Equations to Solve Problems
In the previous two lessons students learned to represent proportional relationships with equations of the form $$y=kx$$. In this lesson they continue to write equations, and they begin to see situations where using the equation is a more efficient way of solving problems than other methods they have been using, such as tables and equivalent ratios.
The activities introduce new contexts and, for the first time, do not provide tables; students who still need tables should be given a chance to realize that and create tables for themselves. The activities are intended to motivate the usefulness of representing proportional relationships with equations, while at the same time providing some scaffolding for finding the equations.
As students use the abstract equation $$y=kx$$ to reason about quantitative situations, they engage in MP2.
Lesson overview
• 6.1 Warm-up: Number Talk: Quotients with Decimal Points (5 minutes)
• 6.2 Activity: Concert Ticket Sales (15 minutes)
• 6.3 Activity: Recycling (15 minutes)
• Includes "Are you Ready for More?" extension problem
• Lesson Synthesis
• 6.4 Cool-down: Granola (5 minutes)
Learning goals:
• Generate an equation for a proportional relationship, given a description of the situation but no table.
• Interpret (orally) each part of an equation that represents a proportional relationship in an unfamiliar context.
• Use an equation to solve problems involving a proportional relationship, and explain (orally) the reasoning.
Learning goals (student facing):
• Let’s use equations to solve problems involving proportional relationships.
Learning targets (student facing):
• I can relate all parts of an equation like $$y=kx$$ to the situation it represents.
• I can find missing information in a proportional relationship using the constant of proportionality.
Required materials:
• four-function calculators
Glossary:
• Access the complete Grade 7 glossary.
Standards
• This lesson builds on the standards: CCSS.5.NBT.B.7MS.5.NBT.7MO.5.NBT.A.8 CCSS.6.RP.A.2MS.6.RP.2MO.6.RP.A.2
IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at https://openupresources.org/math-curriculum/. |
Home >> CBSE XII >> Math >> Matrices
# If AB=BA for any two square matrices,prove by mathematical induction that $(AB)^n=A^nB^n$.
Toolbox:
• We use the principle of mathematical induction, where we need to prove P(n) is true for n=1, n=k, n=k+1
Step1:
Given
AB=BA
To prove
$(AB)^n=A^nB^n.$
Let $P(n):(AB)^n=A^nB^n.$
P(1) substitute n=1.
LHS:-$(AB)^1=AB$
RHS:-$A^nB^n=A^1B^1=AB.$
$\Rightarrow LHS=RHS.$
Step2:
$\Rightarrow P(n)$ be true for n=k.
$\Rightarrow (AB)^k=A^kB^k$
Multiply both sides by AB
$\Rightarrow (AB)^k.AB=A^kB^k.AB$
$\Rightarrow (AB)^k.(AB)^1=A^kB^k.AB$
$\Rightarrow (AB)^{k+1}=A^kB^k.BA$ [since AB=BA]
$\Rightarrow (AB)^{k+1}=A^k(B^k.BA)$ [By associative law]
$\qquad\qquad=A^k(B^k.B)A$
$\qquad\qquad=A^k(B^{k+1})A$ [AB=BA]
$\qquad\qquad=A^k(AB^{k+1})$ [$AB^n=B^nA$]
$\qquad\qquad=A^{k+1}.B^{k+1}$
$\Rightarrow$ P(n) is true for k+1
$\Rightarrow (AB)^n=A^nB^n.$
answered Apr 1, 2013 |
# CLASS 9 MATHS CHAPTER-8 QUADRILATERALS
### Exercise 8.1
Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
∴ 3x + 5x + 9x + 13x = 360°
[Angle sum property of a quadrilateral]
⇒ 30x = 360°
⇒ x = = 12°
∴ 3x = 3 x 12° = 36°
5x = 5 x 12° = 60°
9x = 9 x 12° = 108°
13a = 13 x 12° = 156°
⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°.
Question 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Let ABCD is a parallelogram such that AC = BD.
In ∆ABC and ∆DCB,
AC = DB [Given]
AB = DC [Opposite sides of a parallelogram]
BC = CB [Common]
∴ ∆ABC ≅ ∆DCB [By SSS congruency]
⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)
Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]
∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]
From (1) and (2), we have
∠ABC = ∠DCB = 90°
i.e., ABCD is a parallelogram having an angle equal to 90°.
∴ ABCD is a rectangle.
Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.
∴ In ∆AOB and ∆AOD, we have
AO = AO [Common]
OB = OD [O is the mid-point of BD]
∠AOB = ∠AOD [Each 90]
∴ ∆AQB ≅ ∆AOD [By,SAS congruency
∴ AB = AD [By C.P.C.T.] ……..(1)
Similarly, AB = BC .. .(2)
BC = CD …..(3)
CD = DA ……(4)
∴ From (1), (2), (3) and (4), we have
AB = BC = CD = DA
Thus, the quadrilateral ABCD is a rhombus.
Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.
Question 4. Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Let ABCD be a square such that its diagonals AC and BD intersect at O.
(i) To prove that the diagonals are equal, we need to prove AC = BD.
In ∆ABC and ∆BAD, we have
AB = BA [Common]
BC = AD [Sides of a square ABCD]
∠ABC = ∠BAD [Each angle is 90°]
∴ ∆ABC ≅ ∆BAD [By SAS congruency]
AC = BD [By C.P.C.T.] …(1)
(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]
∴ ∠1 = ∠3
[Alternate interior angles are equal]
Similarly, ∠2 = ∠4
Now, in ∆OAD and ∆OCB, we have
AD = CB [Sides of a square ABCD]
∠1 = ∠3 [Proved]
∠2 = ∠4 [Proved]
∴ ∆OAD ≅ ∆OCB [By ASA congruency]
⇒ OA = OC and OD = OB [By C.P.C.T.]
i.e., the diagonals AC and BD bisect each other at O. …….(2)
(iii) In ∆OBA and ∆ODA, we have
OB = OD [Proved]
BA = DA [Sides of a square ABCD]
OA = OA [Common]
∴ ∆OBA ≅ ∆ODA [By SSS congruency]
⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3)
∵ ∠AOB and ∠AOD form a linear pair.
∴∠AOB + ∠AOD = 180°
∴∠AOB = ∠AOD = 90° [By(3)]
⇒ AC ⊥ BD …(4)
From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles.
Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles.
Now, in ∆AOD and ∆AOB, We have
∠AOD = ∠AOB [Each 90°]
AO = AO [Common]
OD = OB [ ∵ O is the midpoint of BD]
∴ ∆AOD ≅ ∆AOB [By SAS congruency]
⇒ AD = AB [By C.P.C.T.] …(1)
Similarly, we have
AB = BC … (2)
BC = CD …(3)
CD = DA …(4)
From (1), (2), (3) and (4), we have
AB = BC = CD = DA
∴ Quadrilateral ABCD have all sides equal.
In ∆AOD and ∆COB, we have
AO = CO [Given]
OD = OB [Given]
∠AOD = ∠COB [Vertically opposite angles]
So, ∆AOD ≅ ∆COB [By SAS congruency]
∴∠1 = ∠2 [By C.P.C.T.]
But, they form a pair of alternate interior angles.
∴ AD || BC
Similarly, AB || DC
∴ ABCD is a parallelogram.
∴ Parallelogram having all its sides equal is a rhombus.
∴ ABCD is a rhombus.
Now, in ∆ABC and ∆BAD, we have
AC = BD [Given]
BC = AD [Proved]
AB = BA [Common]
∴ ∆ABC ≅ ∆BAD [By SSS congruency]
∴ ∠ABC = ∠BAD [By C.P.C.T.] ……(5)
Since, AD || BC and AB is a transversal.
∴∠ABC + ∠BAD = 180° .. .(6) [ Co – interior angles]
⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)]
So, rhombus ABCD is having one angle equal to 90°.
Thus, ABCD is a square.
Question 6. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Solution:
We have a parallelogram ABCD in which diagonal AC bisects ∠A
⇒ ∠DAC = ∠BAC
(i) Since, ABCD is a parallelogram.
∴ AB || DC and AC is a transversal.
∴ ∠1 = ∠3 …(1)
[ ∵ Alternate interior angles are equal]
Also, BC || AD and AC is a transversal.
∴ ∠2 = ∠4 …(2)
[ v Alternate interior angles are equal]
Also, ∠1 = ∠2 …(3)
[ ∵ AC bisects ∠A]
From (1), (2) and (3), we have
∠3 = ∠4
⇒ AC bisects ∠C.
(ii) In ∆ABC, we have
∠1 = ∠4 [From (2) and (3)]
⇒ BC = AB …(4)
[ ∵ Sides opposite to equal angles of a ∆ are equal]
Similarly, AD = DC ……..(5)
But, ABCD is a parallelogram. [Given]
∴ AB = DC …(6)
From (4), (5) and (6), we have
AB = BC = CD = DA
Thus, ABCD is a rhombus.
Question 7. ABCD is a rhombus. Show that diagonal AC bisects ∠Aas well as ∠C and diagonal BD bisects ∠B as well AS ∠D.
Solution:
Since, ABCD is a rhombus.
⇒ AB = BC = CD = DA
Also, AB || CD and AD || BC
Now, CD = AD ⇒ ∠1 = ∠2 …….(1)
[ ∵ Angles opposite to equal sides of a triangle are equal]
Also, AD || BC and AC is the transversal.
[ ∵ Every rhombus is a parallelogram]
⇒ ∠1 = ∠3 …(2)
[ ∵ Alternate interior angles are equal]
From (1) and (2), we have
∠2 = ∠3 …(3)
Since, AB || DC and AC is transversal.
∴ ∠2 = ∠4 …(4)
[ ∵ Alternate interior angles are equal] From (1) and (4),
we have ∠1 = ∠4
∴ AC bisects ∠C as well as ∠A.
Similarly, we can prove that BD bisects ∠B as well as ∠D.
Question 8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.
Solution:
We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.
i.e., ∠1 = ∠4 and ∠2 = ∠3 ……..(1)
(i) Since, every rectangle is a parallelogram.
∴ ABCD is a parallelogram.
⇒ AB || CD and AC is a transversal.
∴∠2 = ∠4 …(2)
[ ∵ Alternate interior angles are equal]
From (1) and (2), we have
∠3 = ∠4
In ∆ABC, ∠3 = ∠4
⇒ AB = BC
[ ∵ Sides opposite to equal angles of a A are equal]
Similarly, CD = DA
So, ABCD is a rectangle having adjacent sides equal.
⇒ ABCD is a square.
(ii) Since, ABCD is a square and diagonals of a square bisect the opposite angles.
So, BD bisects ∠B as well as ∠D.
Question 9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that
Solution:
We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB
(i) Since, AD || BC and BD is a transversal.
∴ ∠ADB = ∠CBD [ ∵ Alternate interior angles are equal]
⇒ ∠ADP = ∠CBQ
Now, in ∆APD and ∆CQB, we have
AD = CB [Opposite sides of a parallelogram ABCD are equal]
PD = QB [Given]
∠ADP = ∠CBQ [Proved]
∴ ∆APD ≅ ∆CQB [By SAS congruency]
(ii) Since, ∆APD ≅ ∆CQB [Proved]
⇒ AP = CQ [By C.P.C.T.]
(iii) Since, AB || CD and BD is a transversal.
∴ ∠ABD = ∠CDB
⇒ ∠ABQ = ∠CDP
Now, in ∆AQB and ∆CPD, we have
QB = PD [Given]
∠ABQ = ∠CDP [Proved]
AB = CD [ Y Opposite sides of a parallelogram ABCD are equal]
∴ ∆AQB = ∆CPD [By SAS congruency]
(iv) Since, ∆AQB = ∆CPD [Proved]
⇒ AQ = CP [By C.P.C.T.]
(v) In a quadrilateral ∆PCQ,
Opposite sides are equal. [Proved]
∴ ∆PCQ is a parallelogram.
Question 10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
Solution:
(i) In ∆APB and ∆CQD, we have
∠APB = ∠CQD [Each 90°]
AB = CD [ ∵ Opposite sides of a parallelogram ABCD are equal]
∠ABP = ∠CDQ
[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]
∴ ∆APB = ∆CQD [By AAS congruency]
(ii) Since, ∆APB ≅ ∆CQD [Proved]
⇒ AP = CQ [By C.P.C.T.]
Question 11. In ∆ABC and ∆DEF, AB = DE, AB || DE, BC – EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF
Solution:
(i) We have AB = DE [Given]
and AB || DE [Given]
i. e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.
∴ ABED is a parallelogram.
(ii) BC = EF [Given]
and BC || EF [Given]
i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.
∴ BEFC is a parallelogram.
(iii) ABED is a parallelogram [Proved]
∴ AD || BE and AD = BE …(1)
[ ∵ Opposite sides of a parallelogram are equal and parallel] Also, BEFC is a parallelogram. [Proved]
BE || CF and BE = CF …(2)
[ ∵ Opposite sides of a parallelogram are equal and parallel]
From (1) and (2), we have
AD || CF and AD = CF
(iv) Since, AD || CF and AD = CF [Proved]
i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.
∴Quadrilateral ACFD is a parallelogram.
(v) Since, ACFD is a parallelogram. [Proved]
So, AC =DF [∵ Opposite sides of a parallelogram are equal]
(vi) In ∆ABC and ∆DFF, we have
AB = DE [Given]
BC = EF [Given]
AC = DE [Proved in (v) part]
∆ABC ≅ ∆DFF [By SSS congruency]
Question 12. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that
(i )∠A=∠B
(ii )∠C=∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].
Solution:
We have given a trapezium ABCD in which AB || CD and AD = BC.
(i) Produce AB to E and draw CF || AD.. .(1)
∵ AB || DC
⇒ AE || DC Also AD || CF
∴ AECD is a parallelogram.
⇒ AD = CE …(1)
[ ∵ Opposite sides of the parallelogram are equal]
But AD = BC …(2) [Given]
By (1) and (2), BC = CF
Now, in ∆BCF, we have BC = CF
⇒ ∠CEB = ∠CBE …(3)
[∵ Angles opposite to equal sides of a triangle are equal]
Also, ∠ABC + ∠CBE = 180° … (4)
[Linear pair]
and ∠A + ∠CEB = 180° …(5)
[Co-interior angles of a parallelogram ADCE]
From (4) and (5), we get
∠ABC + ∠CBE = ∠A + ∠CEB
⇒ ∠ABC = ∠A [From (3)]
⇒ ∠B = ∠A …(6)
(ii) AB || CD and AD is a transversal.
∴ ∠A + ∠D = 180° …(7) [Co-interior angles]
Similarly, ∠B + ∠C = 180° … (8)
From (7) and (8), we get
∠A + ∠D = ∠B + ∠C
⇒ ∠C = ∠D [From (6)]
(iii) In ∆ABC and ∆BAD, we have
AB = BA [Common]
BC = AD [Given]
∠ABC = ∠BAD [Proved]
∴ ∆ABC = ∆BAD [By SAS congruency]
(iv) Since, ∆ABC = ∆BAD [Proved]
⇒ AC = BD [By C.P.C.T.]
### NCERT Solutions for Class 9 Maths Exercise 8.2
Question 1. ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (See figure). AC is a diagonal. Show that :
(i) SR AC and SR = AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
In ABC, P is the mid-point of AB and Q is the mid-point of BC.
Then PQ AC and PQ = AC
(i) In ACD, R is the mid-point of CD and S is the mid-point of AD.
Then SR AC and SR = AC
(ii) Since PQ = AC and SR = AC
Therefore, PQ = SR
(iii) Since PQ AC and SR AC
Therefore, PQ SR [two lines parallel to given line are parallel to each other]
Now PQ = SR and PQ SR
Therefore, PQRS is a parallelogram.
Question 2. ABCD is a rhombus and P, Q, R, S are mid-points of AB, BC, CD and DA respectively. Prove that quadrilateral PQRS is a rectangle.
Solution:
Given: P, Q, R and S are the mid-points of respective sides AB, BC, CD and DA of rhombus. PQ, QR, RS and SP are joined.
To prove: PQRS is a rectangle.
Construction: Join A and C.
Proof: In ABC, P is the mid-point of AB and Q is the mid-point of BC.
PQ AC and PQ = AC ……….(i)
In ADC, R is the mid-point of CD and S is the mid-point of AD.
SR AC and SR = AC ……….(ii)
From eq. (i) and (ii), PQ SR and PQ = SR
PQRS is a parallelogram.
Now ABCD is a rhombus. [Given]
AB = BC
AB = BC PB = BQ
1 = 2 [Angles opposite to equal sides are equal]
Now in triangles APS and CQR, we have,
AP = CQ [P and Q are the mid-points of AB and BC and AB = BC]
Similarly, AS = CR and PS = QR [Opposite sides of a parallelogram]
APS CQR [By SSS congreuancy]
3 = 4 [By C.P.C.T.]
Now we have 1 + SPQ + 3 =
And 2 + PQR + 4 = [Linear pairs]
1 + SPQ + 3 = 2 + PQR + 4
Since 1 = 2 and 3 = 4 [Proved above]
SPQ = PQR ……….(iii)
Now PQRS is a parallelogram [Proved above]
SPQ + PQR = ……….(iv) [Interior angles]
Using eq. (iii) and (iv),
SPQ + SPQ = 2SPQ =
SPQ =
Hence PQRS is a rectangle.
Question 3. ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:
Given: A rectangle ABCD in which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.
To prove: PQRS is a rhombus.
Construction: Join AC.
Proof: In ABC, P and Q are the mid-points of sides AB, BC respectively.
PQ AC and PQ = AC ……….(i)
In ADC, R and S are the mid-points of sides CD, AD respectively.
SR AC and SR = AC ……….(ii)
From eq. (i) and (ii), PQ SR and PQ = SR ……….(iii)
PQRS is a parallelogram.
Now ABCD is a rectangle. [Given]
AD = BC
AD = BC AS = BQ ……….(iv)
In triangles APS and BPQ,
AP = BP [P is the mid-point of AB]
PAS = PBQ [Each ]
And AS = BQ [From eq. (iv)]
APS BPQ [By SAS congruency]
PS = PQ [By C.P.C.T.] ………(v)
From eq. (iii) and (v), we get that PQRS is a parallelogram.
PS = PQ
Two adjacent sides are equal.
Hence, PQRS is a rhombus.
Question 4. ABCD is a trapezium, in which AB DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E, parallel to AB intersecting BC at F (See figure). Show that F is the mid-point of BC.
Solution:
Let diagonal BD intersect line EF at point P.
In DAB,
E is the mid-point of AD and EP AB [ EF AB (given) P is the part of EF]
P is the mid-point of other side, BD of DAB.
[A line drawn through the mid-point of one side of a triangle, parallel to another side intersects the third side at the mid-point]
Now in BCD,
P is the mid-point of BD and PF DC [EF AB (given) and AB DC (given)]
EF DC and PF is a part of EF.
F is the mid-point of other side, BC of BCD. [Converse of mid-point of theorem]
Question 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (See figure). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
Since E and F are the mid-points of AB and CD respectively.
AE = AB and CF = CD……….(i)
But ABCD is a parallelogram.
AB = CD and AB DC
AB = CD and AB DC
AE = FC and AE FC [From eq. (i)]
AECF is a parallelogram.
FA CE FP CQ [FP is a part of FA and CQ is a part of CE] ………(ii)
Since the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.
In DCQ, F is the mid-point of CD and FP CQ
P is the mid-point of DQ.
DP = PQ ……….(iii)
Similarly, In ABP, E is the mid-point of AB and EQ AP
Q is the mid-point of BP.
BQ = PQ ……….(iv)
From eq. (iii) and (iv),
DP = PQ = BQ ………(v)
Now BD = BQ + PQ + DP = BQ + BQ + BQ = 3BQ
BQ = BD ……….(vi)
From eq. (v) and (vi),
DP = PQ = BQ = BD
Points P and Q trisects BD.
So AF and CE trisects BD.
Question 6. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.
Solution:
Given: A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.
To prove: EG and FH bisect each other.
Construction: Join AC, EF, FG, GH and HE.
Proof: In ABC, E and F are the mid-points of respective sides AB and BC.
EF AC and EF AC ……….(i)
Similarly, in ADC,
G and H are the mid-points of respective sides CD and AD.
HG AC and HG AC ……….(ii)
From eq. (i) and (ii),
EF HG and EF = HG
EFGH is a parallelogram.
Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.
Question 7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.
Solution:
(i) In ABC, M is the mid-point of AB [Given]
MD BC
AD = DC [Converse of mid-point theorem]
Thus D is the mid-point of AC.
(ii) BC (given) consider AC as a transversal.
1 = C [Corresponding angles]
1 = [C = ]
Thus MD AC.
(iii) In AMD and CMD,
AD = DC [proved above]
1 = 2 = [proved above]
MD = MD [common]
AMD CMD [By SAS congruency]
AM = CM [By C.P.C.T.] ……….(i)
Given that M is the mid-point of AB.
AM = AB ……….(ii)
From eq. (i) and (ii),
CM = AM = AB
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# 5/10 Simplified in Lowest Terms
5/10 simplified in lowest terms provides the detailed information of what is the simplest form of 5/10, and the answer with steps help students to understand how to simplify the fraction in simplest form.
Simplify 5/10 in Lowest Terms
5/10 = (?)
5/10 = (5)/(2 x 5)
= 1/2
5/10 = 1/2
The simplest form of 5/10 is 1/2
where,
5/10 is the given fraction to be simplified,
1/2 is the lowest term of 5/10.
For values other than 5/10, use this below tool:
## How-to: Simplify 5/10 in Lowest Terms
The below solution with step by step work shows how to find what is the lowest term of 5/10. The fraction 5/10 simplified can be found by using anyone of the below two methods:
1. Prime factorization method,
2. GCF method.
Prime Factorization Method :
step 1 Observe the input parameters and what to be found:
Input values:
Fraction = 5/10
what to be found:
5/10 = ?
Find what is the lowest term of 5/10.
step 2 Find the prime factors of the numerator of given fraction 5/10.
Prime factors of 5 = 5
step 3 Find the prime factors of the denominator of given fraction 5/10.
Prime factors of 10 = 2 x 5
step 4 Rewrite the fraction 5/10 in the form of prime factors as like the below:
5/10= (5) / (2 x 5)
step 5 Check and cancel the factors of 5 and 10 if any factors in the numerator and denominator can be cancelled each other in the above fraction of prime factors:
= (5) / (2 x 5)
= (1)/(2 )
step 6 Simplify and rewrite the fraction as like the below:
= 1/2
5/10 = 1/2
Hence,
the simplest form of 5/10 is 1/2.
GCF Method :
step 1 Observe the input parameters and what to be found:
Input values:
Fraction = 5/10
what to be found:
5/10 = ?
Find what is the lowest term of 5/10.
step 2 Find GCF (Greatest Common Factor) for both numerator and denominator of the fraction 5/10.
The GCF of 5 and 10 is 5.
step 3 Divide the numerator and denominator of the fraction 5/10 by GCF 5:
= (5/5)(10/5)
step 4 Simplify the above expression:
= (5/5)(10/5)
= 1/2
5/10 = 1/2
Hence,
1/2 is the simplified or lowest term of 5/10. |
# Thread: challenge question -- Factor the polynomial completely
1. ## challenge question -- Factor the polynomial completely
Edit:
Demonstrate at least two methods for factoring the following polynomial
completely over the integers.
$x^5 + x^4 + x^3 + x^2 + x + 1$
2. Method 1:
Factor first by grouping:
$(x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)= (x+1)(x^4+x^2+1)$
Now, for the quartic factor assume it may be factored as follows:
$x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2) x^2+(a+b)x+1$
Equating coefficients, we find:
$a+b=0$
$ab+2=1$
and so one solution is $(a,b)=(1,-1)$ and we have:
$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ which means:
$x^5+x^4+x^3+x^2+x+1=(x+1)(x^2+x+1)(x^2-x+1)$
Method 2:
Let:
$S=x^5+x^4+x^3+x^2+x+1$ and so:
$Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$ hence:
$S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ thus:
$S=(x+1)(x^2-x+1)(x^2+x+1)$
3. Originally Posted by lookagain
A) Give the completely factored form over the integers of the following polynomial, and
B) demonstrate at least two methods for doing so.
$x^5 + x^4 + x^3 + x^2 + x + 1$
(x+1)(x4+x2+1) → (x+1)(x4+2x2+1 - x2) → (x+1)(x2+1+x)(x2+1-x)
or
(x3+1)(x2+x+1) →(x+1)(x2-x+1)(x2+x+1)
4. Here is another:
(x^5 + 1) + (x^4 + x^3 + x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + (x^4 + x^3) + (x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + x^3(x + 1) + x(x + 1) =
(x + 1)(x^4 - x^3 + x^2 - x + 1 + x^3 + x) =
(x + 1)(x^4 + x^2 + 1) =
(You can continue simplifying this with one of the above steps in
any of the appropriate above posts.)
5. Here's another way.
Since $x^5+x^4+x^3+x^2+x+1 = \dfrac{x^6-1}{x-1}$ the roots of this polynomial are exactly the set $\{z\in \mathbb{C}-\{1\}\,\,;\,\, z^6=1\}$, i.e. the roots of unity, ignoring the positive real root. They are $e^{\pm i\pi/3},e^{\pm i2\pi/3}, -1$.
We want a real factorization obviously, and we can see that the conjugate pair to each root is present (as it should be). Pairing them off we get the (minimal) polynomials for each:
$(x-e^{i\pi/3})(x-e^{- i\pi/3}) = x^2-x+1$
$(x-e^{i2\pi/3})(x-e^{- i2\pi/3}) = x^2+x+1$
$x-(-1) = x+1$
There is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.
6. Hello, lookagain
This is a variation of daon's solution.
Demonstrate at least two methods for factoring the following polynomial:
. . . $P(x) \;=\;x^5 + x^4 + x^3 + x^2 + x + 1$
$P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$
. . . . .$=\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$
. . . . .$=\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$
. . . . .$=\; (x+1)(x^2+x+1)(x^2-x+1)$
7. Originally Posted by soroban
Hello, lookagain
This is a variation of daon's solution.
$P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1}$
. . . . .$=\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}$
. . . . .$=\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}$
. . . . .$=\; (x+1)(x^2+x+1)(x^2-x+1)$
Originally Posted by MarkFL
Method 2:
Let:
$S=x^5+x^4+x^3+x^2+x+1$ and so:
$Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1$ hence:
$S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ thus:
$S=(x+1)(x^2-x+1)(x^2+x+1)$
These two (MarkFL's and soroban's versions) look essentially the same to me.
- - - - - - - - - - - - - - - - -
Others:
(x^5 + x^2) + (x^4 + x) + (x^3 + 1) =
x^2(x^3 + 1) + x(x^3 + 1) + 1(x^3 + 1) =
(x^3 + 1)(x^2 + x + 1) =
(x + 1)(x^2 - x + 1)(x^2 + x + 1)
. . . . . . . . . . . . . . . . . . . . .
(x^5 + x^3 + x) + (x^4 + x^2 + 1) =
x(x^4 + x^2 + 1) + 1(x^4 + x^2 + 1) =
(x^4 + x^2 + 1)(x + 1) =
(x^2 - x + 1)(x^2 + x + 1)(x + 1)
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# 6.5 Exponential functions
## 6.5 Exponential functions (EMA4V)
### Functions of the form $$y={b}^{x}$$ (EMA4W)
Functions of the general form $$y=a{b}^{x}+q$$ are called exponential functions. In the equation $$a$$ and $$q$$ are constants and have different effects on the function.
## Worked example 12: Plotting an exponential function
$$y=f(x)={b}^{x} \text{ for } b>0 \text{ and } b\ne 1$$
Complete the following table for each of the functions and draw the graphs on the same system of axes: $$f(x)={2}^{x}$$, $$g(x)={3}^{x}$$, $$h(x)={5}^{x}$$.
$$-\text{2}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{2}$$ $$f(x)={2}^{x}$$ $$g(x)={3}^{x}$$ $$h(x)={5}^{x}$$
1. At what point do these graphs intersect?
2. Explain why they do not cut the $$x$$-axis.
3. Give the domain and range of $$h(x)$$.
4. As $$x$$ increases, does $$h(x)$$ increase or decrease?
5. Which of these graphs increases at the slowest rate?
6. For $$y={k}^{x}$$ and $$k>1$$, the greater the value of $$k$$, the steeper the curve of the graph. True or false?
Complete the following table for each of the functions and draw the graphs on the same system of axes: $$F(x)={\left(\dfrac{1}{2}\right)}^{x}$$, $$G(x)={\left(\dfrac{1}{3}\right)}^{x}$$, $$H(x)={\left(\dfrac{1}{5}\right)}^{x}$$
$$-\text{2}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{2}$$ $$F(x)={\left(\frac{1}{2}\right)}^{x}$$ $$G(x)={\left(\frac{1}{3}\right)}^{x}$$ $$H(x)={\left(\frac{1}{5}\right)}^{x}$$
1. Give the $$y$$-intercept for each function.
2. Describe the relationship between the graphs $$f(x)$$ and $$F(x)$$.
3. Describe the relationship between the graphs $$g(x)$$ and $$G(x)$$.
4. Give the domain and range of $$H(x)$$.
5. For $$y={\left(\dfrac{1}{k}\right)}^{x}$$ and $$k>1$$, the greater the value of $$k$$, the steeper the curve of the graph. True or false?
6. Give the equation of the asymptote for the functions.
### Substitute values into the equations
$$-\text{2}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{2}$$ $$f(x)={2}^{x}$$ $$\frac{1}{4}$$ $$\frac{1}{2}$$ $$\text{1}$$ $$\text{2}$$ $$\text{4}$$ $$g(x)={3}^{x}$$ $$\frac{1}{9}$$ $$\frac{1}{3}$$ $$\text{1}$$ $$\text{3}$$ $$\text{9}$$ $$h(x)={5}^{x}$$ $$\frac{1}{25}$$ $$\frac{1}{5}$$ $$\text{1}$$ $$\text{5}$$ $$\text{25}$$
$$-\text{2}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{2}$$ $$F(x)={\left(\frac{1}{2}\right)}^{x}$$ $$\text{4}$$ $$\text{2}$$ $$\text{1}$$ $$\frac{1}{2}$$ $$\frac{1}{4}$$ $$G(x)={\left(\frac{1}{3}\right)}^{x}$$ $$\text{9}$$ $$\text{3}$$ $$\text{1}$$ $$\frac{1}{3}$$ $$\frac{1}{9}$$ $$H(x)={\left(\frac{1}{5}\right)}^{x}$$ $$\text{25}$$ $$\text{5}$$ $$\text{1}$$ $$\frac{1}{5}$$ $$\frac{1}{25}$$
### Plot the points and join with a smooth curve
1. We notice that all graphs pass through the point $$(0;1)$$. Any number with exponent $$\text{0}$$ is equal to $$\text{1}$$.
2. The graphs do not cut the $$x$$-axis because you can never get $$\text{0}$$ by raising any non-zero number to the power of any other number.
3. Domain: $$\left\{x:x\in \mathbb{R}\right\}$$
Range: $$\left\{y:y\in \mathbb{R}, y>0\right\}$$
4. As $$x$$ increases, $$h(x)$$ increases.
5. $$f(x)={2}^{x}$$ increases at the slowest rate because it has the smallest base.
6. True: the greater the value of $$k$$ $$\left(k>1\right)$$, the steeper the graph of $$y={k}^{x}$$.
1. The $$y$$-intercept is the point $$(0;1)$$ for all graphs. For any real number $$z$$: $${z}^{0}=1 \qquad z \ne 0$$.
2. $$F(x)$$ is the reflection of $$f(x)$$ about the $$y$$-axis.
3. $$G(x)$$ is the reflection of $$g(x)$$ about the $$y$$-axis.
4. Domain: $$\left\{x:x\in \mathbb{R}\right\}$$
Range: $$\left\{y:y\in \mathbb{R}, y>0\right\}$$
5. True: the greater the value of $$k$$ $$\left(k>1\right)$$, the steeper the graph of $$y={\left(\frac{1}{k}\right)}^{x}$$.
6. The equation of the horizontal asymptote is $$y=0$$, the $$x$$-axis.
### Functions of the form $$y=a{b}^{x}+q$$ (EMA4X)
CAPS states to only investigate the effects of $$a$$ and $$q$$ on an exponential graph. However it is also important for learners to see that $$b$$ has a different effect on the graph depending on if $$b > 1$$ or $$0 < b < 1$$.
For this reason the effect of $$b$$ is included in the investigation so that learners can see what happens when $$b > 1$$ and when $$0 < b < 1$$.
Also note that the above worked example further reinforces the effects of $$b$$ on the exponential graph.
## The effects of $$a$$, $$q$$ and $$b$$ on an exponential graph.
On the same set of axes, plot the following graphs ($$a=1$$, $$q=0$$ and $$b$$ changes):
1. $$y_{1} = 2^{x}$$
2. $$y_{2} = \left(\frac{1}{2}\right)^{x}$$
3. $$y_{3} = 6^{x}$$
4. $$y_{4} = \left(\frac{1}{6}\right)^{x}$$
$$-\text{2}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{2}$$ $$y_{1}=2^{x}$$ $$y_{2}=\left(\frac{1}{2}\right)^{x}$$ $$y_{3}=6^{x}$$ $$y_{4}=\left(\frac{1}{6}\right)^{x}$$
Use your results to deduce the effect of $$b$$.
On the same set of axes, plot the following graphs ($$b=2$$, $$a=1$$ and $$q$$ changes):
1. $${y}_{5}={2}^{x}-2$$
2. $${y}_{6}={2}^{x}-1$$
3. $${y}_{7}={2}^{x}$$
4. $${y}_{8}={2}^{x}+1$$
5. $${y}_{9}={2}^{x}+2$$
$$-\text{2}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{2}$$ $${y}_{5}={2}^{x}-2$$ $${y}_{6}={2}^{x}-1$$ $${y}_{7}={2}^{x}$$ $${y}_{8}={2}^{x}+1$$ $${y}_{9}={2}^{x}+2$$
Use your results to deduce the effect of $$q$$.
On the same set of axes, plot the following graphs ($$b=2$$, $$q=0$$ and $$a$$ changes).
1. $${y}_{10}=1 \times {2}^{x}$$
2. $${y}_{11}=2\times {2}^{x}$$
3. $${y}_{12}=-1 \times {2}^{x}$$
4. $${y}_{13}=-2\times {2}^{x}$$
$$-\text{2}$$ $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ $$\text{2}$$ $${y}_{10}=1 \times {2}^{x}$$ $${y}_{11}=2\times {2}^{x}$$ $${y}_{12}=-1 \times {2}^{x}$$ $${y}_{13}=-2\times {2}^{x}$$
Use your results to deduce the effect of $$a$$.
The effect of $$q$$
The effect of $$q$$ is called a vertical shift because all points are moved the same distance in the same direction (it slides the entire graph up or down).
• For $$q>0$$, the graph is shifted vertically upwards by $$q$$ units.
• For $$q<0$$, the graph is shifted vertically downwards by $$q$$ units.
The horizontal asymptote is shifted by $$q$$ units and is the line $$y=q$$.
The effect of $$a$$
The sign of $$a$$ determines whether the graph curves upwards or downwards.
For $$0 < b < 1$$:
• For $$a>0$$, the graph curves downwards. It reflects the graph about the horizontal asymptote.
• For $$a<0$$, the graph curves upwards.
For $$b > 1$$:
• For $$a>0$$, the graph curves upwards.
• For $$a<0$$, the graph curves downwards. It reflects the graph about the horizontal asymptote.
$$b>1$$ $$a<0$$ $$a>0$$ $$q>0$$ $$q<0$$
The effect of $$a$$ and $$q$$ on an exponential graph when $$b > 1$$.
$$00$$ $$q>0$$ $$q<0$$
The effect of $$a$$ and $$q$$ on an exponential graph when $$0 < b < 1$$.
### Discovering the characteristics (EMA4Y)
The standard form of an exponential function is $$y=a{b}^{x}+q$$.
#### Domain and range
For $$y=a{b}^{x}+q$$, the function is defined for all real values of $$x$$. Therefore the domain is $$\left\{x:x\in \mathbb{R}\right\}$$.
The range of $$y=a{b}^{x}+q$$ is dependent on the sign of $$a$$.
For $$a>0$$:
\begin{align*} {b}^{x}& > 0 \\ a{b}^{x}& > 0 \\ a{b}^{x}+q& > q \\ f(x)& > q \end{align*}
Therefore, for $$a>0$$ the range is $$\left\{f(x):f(x)>q\right\}$$.
For $$a<0$$:
\begin{align*} {b}^{x}& > 0 \\ a{b}^{x}& < 0 \\ a{b}^{x}+q& < q \\ f(x)& < q \end{align*}
Therefore, for $$a<0$$ the range is $$\left\{f(x):f(x)<q\right\}$$.
## Worked example 13: Domain and range of an exponential function
Find the domain and range of $$g(x)=5 \cdot {2}^{x}+1$$
### Find the domain
The domain of $$g(x)=5\times {2}^{x}+1$$ is $$\left\{x:x\in \mathbb{R}\right\}$$.
### Find the range
\begin{align*} {2}^{x}& > 0 \\ 5\times {2}^{x}& > 0 \\ 5\times {2}^{x}+1& > 1 \end{align*}
Therefore the range is $$\left\{g(x):g(x)>1\right\}$$.
#### Intercepts
The $$y$$-intercept.
For the $$y$$-intercept, let $$x=0$$:
\begin{align*} y& = a{b}^{x}+q \\ & = a{b}^{0}+q \\ & = a(1)+q \\ & = a+q \end{align*}
For example, the $$y$$-intercept of $$g(x)=5\times {2}^{x}+1$$ is given by setting $$x=0$$:
\begin{align*} y& = 5\times {2}^{x}+1 \\ & = 5\times {2}^{0}+1 \\ & = 5+1 \\ & = 6 \end{align*}
This gives the point $$(0;6)$$.
The $$x$$-intercept:
For the $$x$$-intercept, let $$y=0$$.
For example, the $$x$$-intercept of $$g(x)=5\times {2}^{x}+1$$ is given by setting $$y=0$$:
\begin{align*} y& = 5\times {2}^{x}+1 \\ 0& = 5\times {2}^{x}+1 \\ -1& = 5\times {2}^{x} \\ {2}^{x}& = -\frac{1}{5} \end{align*}
There is no real solution. Therefore, the graph of $$g(x)$$ does not have any $$x$$-intercepts.
#### Asymptotes
Exponential functions of the form $$y=a{b}^{x}+q$$ have a single horizontal asymptote, the line $$x=q$$.
### Sketching graphs of the form $$y=a{b}^{x}+q$$ (EMA4Z)
In order to sketch graphs of functions of the form, $$y=a{b}^{x}+q$$, we need to determine four characteristics:
1. sign of $$a$$
2. $$y$$-intercept
3. $$x$$-intercept
4. asymptote
The following video shows some examples of sketching exponential functions.
Video: 2FYW
## Worked example 14: Sketching an exponential function
Sketch the graph of $$g(x)=3\times {2}^{x}+2$$. Mark the intercept and the asymptote.
### Examine the standard form of the equation
From the equation we see that $$a>1$$, therefore the graph curves upwards. $$q>0$$ therefore the graph is shifted vertically upwards by $$\text{2}$$ units.
### Calculate the intercepts
For the $$y$$-intercept, let $$x=0$$:
\begin{align*} y& = 3\times {2}^{x}+2 \\ & = 3\times {2}^{0}+2 \\ & = 3+2 \\ & = 5 \end{align*}
This gives the point $$(0;5)$$.
For the $$x$$-intercept, let $$y=0$$:
\begin{align*} y& = 3\times {2}^{x}+2 \\ 0& = 3\times {2}^{x}+2 \\ -2& = 3\times {2}^{x} \\ {2}^{x}& = -\frac{2}{3} \end{align*}
There is no real solution, therefore there is no $$x$$-intercept.
### Determine the asymptote
The horizontal asymptote is the line $$y=2$$.
### Plot the points and sketch the graph
Domain: $$\left\{x:x\in \mathbb{R}\right\}$$
Range: $$\left\{g(x):g(x)>2\right\}$$
Note that there is no axis of symmetry for exponential functions.
## Worked example 15: Sketching an exponential graph
Sketch the graph of $$y=-2\times {3}^{x}+6$$
### Examine the standard form of the equation
From the equation we see that $$a<0$$ therefore the graph curves downwards. $$q>0$$ therefore the graph is shifted vertically upwards by $$\text{6}$$ units.
### Calculate the intercepts
For the $$y$$-intercept, let $$x=0$$:
\begin{align*} y& = -2\times {3}^{x}+6 \\ & = -2\times {3}^{0}+6 \\ & = 4 \end{align*}
This gives the point $$(0;4)$$.
For the $$x$$-intercept, let $$y=0$$:
\begin{align*} y& = -2\times {3}^{x}+6 \\ 0& = -2\times {3}^{x}+6 \\ -6& = -2\times {3}^{x} \\ {3}^{1}& = {3}^{x} \\ \therefore x& = 1 \end{align*}
This gives the point $$(1;0)$$.
### Determine the asymptote
The horizontal asymptote is the line $$y=6$$.
### Plot the points and sketch the graph
Domain: $$\left\{x:x\in \mathbb{R}\right\}$$
Range: $$\left\{g(x):g(x)<6\right\}$$
Textbook Exercise 6.5
Given the following equation:
$$y = -\frac{2}{3}. (\text{3})^{x} + 1$$
Calculate the $$y$$-intercept. Your answer must be correct to 2 decimal places.
\begin{align*} y & = \left( -\frac{2}{3} \right). (\text{3})^{x} + \text{1} \\ & = \left( -\frac{2}{3} \right). (\text{3})^{(0)} + \text{1} \\ & = \left( -\frac{2}{3} \right). (\text{1}) + \text{1} \\ & = (-\text{0,66666...}) + \text{1} \\ & = \text{0,33} \end{align*}
The $$y$$-intercept is $$(0;\text{0,33})$$.
Now calculate the $$x$$-intercept. Estimate your answer to one decimal place if necessary.
We calculate the $$x$$-intercept by letting $$y = 0$$. Then we solve for $$x$$:
\begin{align*} 0 & = \left( -\frac{2}{3} \right). (\text{3})^{x} + \text{1} \\ -\text{1} & = \left( -\frac{2}{3} \right)(\text{3})^{x} \\ \left( - \frac{3}{2} \right) (-\text{1}) & = \left( - \frac{3}{2} \right) \left( - \frac{2}{3} \right). (\text{3})^{x} \\ \frac{3}{2} & = \text{3}^{x} \end{align*}
To find the answer we try different values of $$x$$:
\begin{align*} \text{Try: } \text{3}^{-1} & = \text{0,333...}\\ \text{Try: } \text{3}^0 & = \text{1}\\ \text{Try: } \text{3}^1 & = \text{3} \end{align*}
We can see that the exponent must be between 0 and 1. Next we try values starting with $$\text{0,1}$$ and see what the value of the exponent is. Doing this we find that $$x = \text{0,4}$$.
The $$x$$-intercept is $$(\text{0,4};0)$$.
The graph here shows an exponential function with the equation $$y = a \cdot 2^{x} + q$$. One point is given on the curve: Point A is at $$(-3;\text{3,875})$$. Determine the values of $$a$$ and $$q$$, correct to the nearest integer.
The asymptote lies at $$y = 4$$. Therefore $$q$$ is 4.
At this point we know that the equation for the graph must be $$y = a \cdot 2^{x} + 4$$.
\begin{align*} y & = a (2)^{x} + 4 \\ (\text{3,875}) & = a (2)^{(-3)} + 4 \\ \text{3,875} - 4 & = a (2)^{\left(-3\right)} \\ -\text{0,125} & = a (\text{0,125}) \\ -1 & = a \end{align*}
Therefore $$a = -1 \text{ and } q = 4$$
Below you see a graph of an exponential function with the equation $$y = a \cdot 2^{x} + q$$. One point is given on the curve: Point A is at $$(-3;\text{4,875})$$. Calculate the values of $$a$$ and $$q$$, correct to the nearest integer.
The asymptote lies at $$y = 5$$. Therefore $$q$$ is 5.
At this point we know that the equation for the graph must be $$y = a \cdot 2^{x} + 5$$.
\begin{align*} y & = a (2)^{x} + 5 \\ (\text{4,875}) & = a (2)^{ (-3)} + 5 \\ \text{4,875} - 5 & = a (2)^{\left(-3\right)} \\ -\text{0,125} & = a (\text{0,125}) \\ -1 & = a \end{align*}
Therefore $$a = -1 \text{ and } q = 5$$.
Given the following equation:
$$y = \dfrac{1}{4} \cdot (4)^{x} - 1$$
Calculate the $$y$$-intercept. Your answer must be correct to 2 decimal places.
\begin{align*} y & = \left( \frac{1}{4} \right)\cdot(4)^{x} - 1 \\ & = \left( \frac{1}{4} \right) \cdot (4)^{(0)} -\text{1} \\ & = \left( \frac{1}{4} \right) \cdot (1) - 1 \\ & = (\text{0,25}) - 1 \\ & = -\text{0,75} \end{align*}
Therefore the $$y$$-intercept is $$(0;-\text{0,75})$$.
Now calculate the $$x$$-intercept.
We calculate the $$x$$-intercept by letting $$y = 0$$. Then start to solve for $$x$$.
\begin{align*} 0 & = \left( \frac{1}{4} \right) \cdot (4)^{x} - 1 \\ 1 & = \left( \frac{1}{4} \right) \cdot (4)^{x} \\ \left( 4 \right) (1) & = \left( 4 \right) \left( \frac{1}{4} \right) \cdot (4)^{x} \\ 4^1 & = 4^{x} \\ x &= 1 \end{align*}
Therefore the $$x$$-intercept is $$(1;0)$$.
Given the following graph, identify a function that matches each of the following equations:
$$y = 2^x$$
$$g(x)$$
$$y = -2^x$$
$$k(x)$$
$$y = 2 \cdot 2^x$$
$$f(x)$$
$$y = \left(\frac{1}{2}\right)^{x}$$
$$h(x)$$
Given the functions $$y=2^{x}$$ and $$y=\left(\frac{1}{2}\right)^{x}$$.
Draw the graphs on the same set of axes.
For $$y=2^{x}$$:
$$a$$ is positive and greater than 1 and so the graph curves upwards. The $$y$$-intercept is $$(0;1)$$. There is no $$x$$-intercept. The asymptote is the line $$x = 0$$.
For $$y=\left(\frac{1}{2}\right)^{x}$$:
$$a$$ is positive and less than 1 and so the graph curves downwards. The $$y$$-intercept is $$(0;1)$$. There is no $$x$$-intercept. The asymptote is the line $$x = 0$$.
The graph is:
Is the $$x$$-axis an asymptote or an axis of symmetry to both graphs? Explain your answer.
The $$x$$-axis is an asymptote to both graphs because both approach the $$x$$-axis but never touch it.
Which graph can be described by the equation $$y=2^{-x}$$? Explain your answer.
$$y= \left(\frac{1}{2}\right)^{x}$$ can be described by the equation $$y=2^{-x}$$ because $$y=\left(\frac{1}{2}\right)^x = \left(2^{-1} \right)^x = 2^{-x}$$.
Solve the equation $$2^{x} = \left(\frac{1}{2}\right)^{x}$$ graphically and check your answer is correct by using substitution.
The graphs intersect at the point $$(0;1)$$. If we substitute these values into each side of the equation we get:
\begin{align*} \text{LHS: } 2^{x} & = 2^0 = 1 \text{ and} \\ \text{RHS: } (\frac{1}{2})^x & = (\frac{1}{2})^0 = 1 \end{align*}
LHS = RHS, therefore the answer is correct.
The curve of the exponential function $$f$$ in the accompanying diagram cuts the $$y$$-axis at the point $$A(0;1)$$ and passes through the point $$B(2;9)$$.
Determine the equation of the function $$f$$.
The general form of the equation is $$f(x)=a \cdot b^x+q$$.
We are given $$A(0; 1)$$ and $$B(2; 9)$$.
The asymptote is at $$y = 0$$ and so $$q = 0$$.
Substitute in the values of point $$A$$:
\begin{align*} 1 & = a \cdot b^0 \\ 1 & = a \end{align*}
Substitute in the values of point $$B$$:
\begin{align*} 9 & = b^2 \\ 3^2 & = b^2\\ \therefore b & = 3 \end{align*}
Therefore the equation is $$f(x) = 3^x$$.
Determine the equation of the function $$h(x)$$, the reflection of $$f(x)$$ in the $$x$$-axis.
$$h(x)=-3^{x}$$
Determine the range of $$h(x)$$.
Range: $$(-\infty;0)$$
Determine the equation of the function $$g(x)$$, the reflection of $$f(x)$$ in the $$y$$-axis.
$$g(x)=3^{-x}$$
Determine the equation of the function $$j(x)$$ if $$j(x)$$ is a vertical stretch of $$f(x)$$ by $$+2$$ units.
$$j(x)=2 \cdot 3^{x}$$
Determine the equation of the function $$k(x)$$ if $$k(x)$$ is a vertical shift of $$f(x)$$ by $$-3$$ units.
$$k(x)=3^{x}-3$$ |
GMAT > Scatter Plot Graph and Solved Examples
# Scatter Plot Graph and Solved Examples - Notes | Study Integrated Reasoning for GMAT - GMAT
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The points which represent the relationship between the two sets of information.
Let us take an example of scattered plots:
Given that collected data about exams in each of her classes from the previous year.
Plot the data in a scatter plot.
Class Math English History Computer Biology Arts Period 2 4 1 5 3 6 Average Score 95 60 75 85 92 83
Correlation
When the 2 sets of given data are strongly linked to each other, we can say they have a High Correlation.
Negative Correlation is when one value increases as the other decreases, and vice versa.
Positive Correlation is when both the values increase or decrease together.
Solved Examples
DIRECTIONS for question 1 to 3: Refer to the following graph and answer the questions given below:
The total values of food grains (rice and wheat) imported during these years are given below:
1971 : Rs. 123 crore
1980 : Rs. 80 crore
1981 : Rs. 314 crore
1982 : Rs. 295 crore
1983 : Rs. 587 crore
1984 : Rs. 158 crore
Q 1. Wheat formed what percent of the volume of total imports of food grains from 1980-84?
Option:
1. 75 %
2. 66 %
3. 90 %
4. 95 %
Solution: Option 3
Adding up the approximate quantity of wheat and dividing it by the total quantity of food grains
(i.e. wheat + rice) will give us
≈ 5516 / 6126 » 90 %.
Q 2. If the import price of wheat was Rs. 2,400 per tonne in 1983, then what was the import price of rice per tonne during that year?
Option:
1. 3,200
2. 2,225
3. 2,850
4. 1,800
Solution: Option 2
In the year 1983, 2,142,000 tonnes of wheat @ Rs 2,400/tonne will mean an expenditure of Rs 514 crore.
So the remaining is 587 – 514 = 73 crore, which was spent on importing 328,000 tonnes of rice. So the price of rice = 730,000 / 328 ≈ Rs 2,225 /tonne
Q 3. In which year was the ratio of rice to wheat imports the highest?
Option:
1. 1971
2. 1980
3. 1983
4. 1984
Solution: Option 4
The ratio is maximum when the numerator is maximum and the denominator is minimum. In the year 1984 the imports of rice are highest and the imports of wheat is the least which gives the maximum ratio.
DIRECTIONS for question 4 to 6: Refer to the following graph
Q 4. In which of the following years was the expenditure of the company the maximum?
1. 1980
2. 1979
3. 1978
4. 1976
Solution: Expenditure of the company is : Total Sales – Net Profit.
The difference of the total sales & net profit is highest for the year 1978.
Q 5. If a tax of 55 % is to be paid on gross profit before the net profit of that particular year is arrived at, what is the combined gross profit for the years ‘79 and ‘80? (' in Lakhs)
1. 32
2. 17
3. 46
4. 67
Solution: The combined Net Profit for both the years 79 and 80 was 29.9 ≈ 30 lakh.
This net profit is calculated after having paid 55 % tax on the gross profit of that particular year.
So the remaining amount will be 45 % of the gross profit and this equals to INR 30 lakh.
So 100 % of that amount, i.e. the combined gross profit must correspond to 30/45 x 100 = 66.66 » 67 lakh. Hence 4 option is the correct answer.
Q 6. If the expenses on advertising the products in 1980 was 2.5 % of the total sales turnover of the company, what was it as a percentage of the net profit in 1980?
1. 11.8%
2. 5.7%
3. 13.5%
4. 10.2%
Solution: The advertisement expenditure was 2.5 % of the total sales turnover
= 2.5/100 x 64 = 1.6.
So the percentage of net Profit = 1.6/15.6 x 100
≈ 10(+) %. Hence 4th option is the correct answer.
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# Chapter 10 - PowerPoint PPT Presentation
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Chapter 10
## Chapter 10
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##### Presentation Transcript
1. Chapter 10 Describing Relationships Using Correlation and Regression
2. Going Forward Your goals in this chapter are to learn: • How to create and interpret a scatterplot • What a regression line is • When and how to compute the Pearson r • How to perform significance testing of the Pearson r • The logic of predicting scores using linear regression and
3. Understanding Correlations
4. Correlation Coefficient • A correlation coefficient is a statistic that describes the important characteristics of a relationship • It simplifies a complex relationship involving many scores into one number that is easily interpreted
5. Distinguishing Characteristics • A scatterplotis a graph of the individual data points from a set of X-Y pairs • When a relationship exists, as the X scores increase, the Y scores change such that different values Y tend to be paired with different values of X
6. A Scatterplot Showing the Existence of a Relationship Between the Two Variables
7. Linear Relationships • A linear relationship forms a pattern following one straight line • The linear regression line is the straight line that summarizes a relationship by passing through the center of the scatterplot
8. Positive and Negative Relationships • In a positive linear relationship, as the X scores increase, the Y scores also tend to increase • In a negative linear relationship, as the scores on the X variable increase, the Y scores tend to decrease
9. Scatterplot of a Positive Linear Relationship
10. Scatterplot of a Negative Linear Relationship
11. Nonlinear Relationships In a nonlinear relationship, as the X scores increase, the Y scores do not only increase or only decrease: at some point, the Y scores alter their direction of change.
12. Scatterplot of a Nonlinear Relationship
13. Strength of a Relationship The strengthof a relationship is the extent to which one value of Y is consistently paired with one and only one value of X • The larger the absolute value of the correlation coefficient, the stronger the relationship • The sign of the correlation coefficient indicates the direction of a linear relationship
14. Correlation Coefficients • Correlation coefficients may range between –1 and +1. The closer to ±1 the coefficient is, the stronger the relationship; the closer to 0 the coefficient is, the weaker the relationship. • As the variability in the Y scores at each X becomes larger, the relationship becomes weaker
15. Correlation Coefficient A correlation coefficient tells you • The relative degree of consistency with which Ys are paired with Xs • The variability in the group of Y scores paired with each X • How closely the scatterplot fits the regression line • The relative accuracy of prediction
16. A Perfect Correlation (±1)
17. Intermediate Strength Correlation
18. No Relationship
19. The Pearson Correlation Coefficient
20. Pearson Correlation Coefficient Describes the linear relationship between two interval variables, two ratio variables, or one interval and one ratio variable. The computing formula is
21. Step-by-Step Step 1. Compute the necessary components:
22. Step-by-Step • Step 2. Use these values to compute the numerator • Step 3. Use these values to compute the denominator and then divide to find r
23. Significance Testing of the Pearson r
24. Two-Tailed Test of the Pearson r • Statistical hypotheses for a two-tailed test • This H0 indicates the r value we obtained from our sample is because of sampling error • The sampling distribution of rshows all possible values of r that occur when samples are drawn from a population in which r = 0
25. Two-Tailed Test of the Pearson r
26. Two-Tailed Test of the Pearson r • Find appropriate rcrit from the table based on • Whether you are using a two-tailed or one-tailed test • Your chosen a • The degrees of freedom (df) where df = N – 2, where N is the number of X-Y pairs in the data • If robt is beyond rcrit, reject H0 and accept Ha • Otherwise, fail to reject H0
27. One-Tailed Test of the Pearson r • One-tailed, predicting positive correlation • One-tailed, predicting negative correlation
28. An Introduction to Linear Regression
29. Linear Regression Linear regression is the procedure for predicting unknown Y scores based on known correlated X scores. • X is the predictor variable • Y is the criterion variable • The symbol for the predicted Y score is(pronounced Y prime)
30. Linear Regression The equation that produces the value of at each X and defines the straight line that summarizes the relationship is called the linear regression equation.
31. Proportion of VarianceAccounted For • The proportion of variance accounted for describes the proportion of all differences in Y scores that are associated with changes in the X variable • The proportion of variance accounted for equals
32. Example 1 For the following data set of interval/ratio scores, calculate the Pearson correlation coefficient.
33. Example 1 Pearson Correlation Coefficient • Determine N • Calculate • Insert each value into the following formula and
34. Example 1 Pearson Correlation Coefficient
35. Example 2 Significance Test of the Pearson r Conduct a two-tailed significance test of the Pearson r just calculated. Use a = .05. • df = N – 2 = 6 – 2 = 4 • rcrit = 0.811 • Since robt of –0.88 falls beyond the critical value of –0.811, reject H0 and accept Ha. • The correlation in the population is significantly different from 0
36. Example 3 Proportion of Variance Accounted For Calculate the proportion of variance accounted for, using the given data. Proportion of variance accounted for is |
# Trig Formulas
### Trig factor formulas
Trig factor formula is used to convert the product of 2 trigonometric functions into a sum or difference of 2 trigonometric functions or vice-versa.
Sum or Difference of trig functions (involves cosine and sine)
Remember these trig formulas split a complex trigonometry function into a difference/addition of two trigonometry functions.
The trig formulas are derived from the additional formulas.
1)
2)
3)
4)
Combining Trigonometric functions (involves cosine and sine)
We can derive the following trig factor formulas that combine basic trigonometry functions into one. It is a good idea if you are able to memorize the trig formulas.
By placing P=A+B and Q=A-B in 1) to 4) above equations, we obtain
1)
2)
3)
4)
Example
Let us apply the trig formulas to the following problems.
a)Show that
We combined sin ? with sin 3 ? and sin 5 ? with sin7 with the trig formulas.
Left Hand Side =
Take out cos ?. Notice that there are two basic trigonometry functions left in the brackets .We combine them together as well.
=
=
Look at the RHS (Right Hand Side) of the equation. It is in terms of ? and not 4 ?. So we use double angle formula to reduce sin 4 ? into 2sin 2 ? cos ?.
=
Notice that there are cos and sin on the RHS. What we are missing is 2sin cos ?. So we use the double angle formula again to reduce sin 2 into 2sincos
= ( RHS Shown )
Show that
Upon reading the question, we notice that there is only one trigonometry function in the numerator and denominator on the RHS( Right Hand Side) of the equation.
LHS =
=
We have to think on how we are going to get a tangent function on the RHS.
“Could it be ”. We arrange the equation in that manner.
=
=
= ( RHS Proven)
b) Show that
LHS =
We place it in the form 2sin A sin B .Then we expand it using the trigonometry factor formulas.
=
=
=
We note that the power on the right hand side of the equation is 1. The power
of the current equation is at 2. We reduce it using the double angle formula
(cos 2 = 2 cos -1)
=
= ( RHS Shown) |
# SSC CGL 2019-20 Mock Test-1 Quantitative Aptitude with Answers
Archana Shandilya
Quantitative Aptitude can become one of the high scoring sections in SSC CGL Tier-I Exam if practiced well. Getting a good score in this section demands in-depth knowledge of all the formulas and the pattern of questions asked. Therefore, rigorous practice is required for acing this section. For your practice, we have designed mock papers which will test your mathematical skills.
We have covered the following five major categories of Quantitative Aptitude Section in this mock test:
1. Arithmetic
2. Algebra
3. Geometry and Mensuration
4. Trigonometry
5. Data Interpretation
Know the SSC CGL 2019-20 Quantitative Aptitude Preparation Strategy
So, let’s start the practice with the 1st Quantitative Aptitude Mock Test. You must try to finish all the 25 questions within 25 minutes time duration. After attempting all the questions, you can assess your performance by checking answers alongwith their solutions are given latter in this article.
## Quantitative Aptitude Mock Test-1
1. In an organization, it is observed that every two employee share the newspaper “The Hindu” between them. Every three employee shared among them the newspaper “The Economic Times” and every four employee used to share “The Indian Express” from the library. But it is well there are only 182 newspaper came on daily basis. The number of employee in the organization:
## Trending Now
a) 162
b) 168
c) 96
d) Cannot be determined.
2. A shopkeeper sold an article at a profit of 20%. If he had bought it at 20% less & sold it at Rs 80 less, then he earns a profit of 25%. Find the cost price of the article.
a) 350
b) 375
c) 450
d) 400
3. A rickshaw dealer buys 30 rickshaws for Rs.4725. Of these, 8 are four seaters and rests are two seaters. At what price must he sell the four seaters so that if he sells the two seaters at 3/4th of this price, he makes a profit of 40% on his outlay?
a) Rs. 180
b) Rs. 270
c) Rs. 360
d) Rs. 450
4. The difference between simple and compound interest on a certain sum of money for 2 years at 4 per cent per annum is Rs. 1. The sum of money is:
a) Rs. 600
b) Rs. 625
c) Rs. 560
d) Rs. 650
a) 120
b) 123
c) 126
d) 130
7. Ram walks 40 km at 5 km/hr; he will be late by 1 hour and 20 minutes. If he walks at 8 km per hr, how early from the fixed time will he reach?
a) 1 hour and 15 minutes
b) 1 hour and 25 minutes
c) 1 hour and 40minutes
d) 1.5 hours
8. A Superfast train starts from Delhi station and reaches Agra 30 minutes late when it moves with 80 km hr and 48 minutes late when it goes 60 km hr. The distance between the two stations is:
a) 32 km
b) 38 km
c) 36 km
d) 42 km
9. 18 women work 9 hour per day to complete 30% of the work in 30 day. The rest work is done by 9 women, working 9 hour per day. Then work did in how many days.
a) 145
b) 154
c) 147
d) 140
10. Out of 11 persons, 10 persons spent Rs. 35 each for their meals. The 11th one spent Rs. 40 more than the average expenditure of all the nine. The total money spent by all of them was:
a) Rs. 492.50
b) Rs. 497.50
c) Rs. 429
d) Rs. 498.50
5 Daily Routine Practices to crack SSC CGL 2019-20 Exam in First Attempt
11. Two friends work in same company and ratio of their salaries is 3 : 5. First friend gets promoted after a year and his salary is increased by one third of the salary of his friend. Find the ratio of their current salaries.
a) 11 : 15
b) 2 : 3
c) 14 : 15
d) None of these
12. Amit got thrice as many marks in Maths as in English. The proportion of his marks in Maths and History is 4:3. If his total marks in Maths, English and History are 250, what are his marks in English?
a) 120
b) 90
c) 40
d) 80
a) -1
b) 0
c) 1
d) 3
14. A and B are the centers of two circles with radius 9 cm and 14 cm respectively, whereas AB = 25 cm. C is the center of another circle with radius ‘X’ cm, which touches each of above circle externally. If ∠ACB=900 then find the value of X.
a) 8
b) 7
c) 6
d) 10
15. Ratio of number of sides of two regular polygons is 5: 6 and ratio of their each interior angle is in the ratio 24: 25. Then the no. of sides of these two polygon are
a) 13, 15
b) 10, 12
c) 14, 16
d) 8, 6
16. In the figure LM || AB || NO. Find the value of ∠ANO?
a) 15°
b) 35°
c) 25°
d) 45°
17. Find the perimeter of an isosceles triangle with ∠P = 90° and area 32 sq. cm.
a) 16+8√2
b) 16+32√2
c) 8+2√2
d) 4+√3
a) 1
b) 4
c) 9
d) 0
a) 1/4
b) 1/3
c) 2/3
d) 3/4
Get all the updates for SSC CGL 2019-20 Exam
Directions (21-25): The below Bar graph shows the enrolment of students in (in thousands) three different colleges A, B, C in a town over a period of six years.
21. The total enrollment of college A in 2010 and 2012 is approximately what percentage of the total enrollment of 2011 and 2015?
a) 90
b) 110
c) 117
d) 125
22. For which college was the average annual enrolment is maximum in the given period?
a) B
b) C
c) A
d) Both A And c
23. What is the difference between the average enrolment of college B in 2011, 2013 and 2015 and the average enrolment of C in 2010, 2012 and 2014?
a) 20
b) 17
c) 34
d) 7
24. What was the approximate decline in enrollment in college A in 2013 as compared to the previous year?
a) 35
b) 18
c) 45
d) 40
25. What is the ratio between average enrollment of students of college A in the years 2011, 2013 and 2015 to the average of college C in the years 2010, 2012 and 2013?
a) 12 : 13
b) 16 : 19
c) 21 : 22
d) None of these
Know the Detailed Exam Pattern and Syllabus of SSC CGL 2019-20 Exam
## Quantitative Aptitude Mock Test-1: Answers with Solutions
Explanation: Let the employee be x, then
x/2 + x/3 +x/4 = 182
On solving, X = 168
Explanation: Here C.P = 100%., S.P = 120%
If he bought at 20% less then C.P = 80%
Required S.P = 80%×125/100=100%
120% - 80 = 100%
20% = 80
C.P is 100%, Then 100% = Rs 400.
Explanation: Total investment = Rs. 4725
Total SP = 1.4 ×4725 = 6615
Now, Let the price of 4 seater be x then price of two seater will be .75x.
8x + 22 ×0.75x = 6615
24.5x= 6615 or x = 270.
Explanation:
Explanation:
Explanation:
Explanation: 1st case time taken = 40/5 = 8 hours
2nd case = 40/8 = 5 hours
Scheduled time = 8 hours – 80 minutes = 6 hours and 40 minutes. If Ram walks at speed of 8 km /hour, then he takes 5 hours to reach the destination and the fixed time is 6 hours and 40 minutes. He reaches early by 1 hour and 40 minutes.
Explanation: Let the right time be t. According to question
80(t + 30) = 60 (t + 48)
T = 24 minutes
Explanation:
Explanation: Let the average expenditure be Rs. x Then,
11x = 10 × 35 + (x + 40) or 11x = x + 390 or 10x = 390 or x = 39
Total money spent = 11x = Rs. (11 × 39) = Rs 429
Explanation:
Explanation:
Explanation:
Explanation: Given ∠ACB=900
∠ACB=900, Then AB2 = AC2 + BC2
252 = (9 + x)2 + (14 + x)2
Here X = 6 cm.
Explanation:
Explanation:
Explanation:
Explanation:
Explanation:
Explanation:
Explanation: Total enrollment of college A in 2010 and 2012 = 50 + 55 = 105
Total enrollment of college A in 2011 and 2015 = 90
% = 105/90 × 100 = 116.6
Explanation:
Explanation:
Explanation: Enrollment of college A in 2013 = 45
Enrollment of college A in 2012 = 55
% decline = 10/55 × 100 = 18.18.
Explanation:
We have covered the following topics in the above Mock Test-1:
Quantitative Aptitude Topics
Number of Questions
Number systems
1
Percentages, Profit & Loss and Interest
3
Algebra
2
Speed, Time & Distance
2
Time & Work
1
Averages
1
Ratio & Proportion
2
Surds/ Quadratic equation/ Mixture & Alligation
1
Geometry
4
Trigonometry
3
Data Interpretation
5
Total
25
The difficulty level of the above mock test was ranging between easy to difficult level and a good score would lie between 17 to 20 marks. Don’t stop your practice until you achieve efficiency and accuracy. Try another mock test here – Quantitative Aptitude Mock Test. |
Lesson 1Cafeteria Actions and ReactionsDevelop Understanding
Jump Start
Evaluate the following expression for . Show each step of your work so you can review it with a neighbor.
Learning Focus
Solve multi-step linear equations using inverse operations.
How does connecting the operations in an equation to an action in a story help me develop and justify my solution process for solving the equation?
How does the structure of an equation give me clues about how to solve it?
• Equation Solving:
• Equation Solving Using a Run Matrix:
• Equation Solving Using the Equation Solver:
• Equation Solving Using a Table:
Open Up the Math: Launch, Explore, Discuss
Elvira, the cafeteria manager, has just received a shipment of new trays with the school logo prominently displayed in the middle of each tray. After unloading cartons of trays in the pizza line, she realizes that students are arriving for lunch and she will have to wait until lunch is over before unloading the remaining cartons. The new trays are very popular, and in just a couple of minutes, students have passed through the pizza line and are showing off the school logo on the trays. At this time, Elvira decides to divide the remaining trays in the pizza line into equal groups so she can also place some in the salad line and the sandwich line, hoping to attract students to the other lines. After doing so, she realizes that each of the serving lines has only of the new trays.
“That’s not many trays for each line. I wonder how many trays there were in each of the cartons I unloaded?”
1.
Help the cafeteria manager answer her question using the data in the story about each of the actions she took. Explain how you arrived at your solution.
Elvira is interested in collecting data about how many students use each table during lunch. She has recorded some data on sticky notes to analyze later. Here are the notes she has recorded:
• Some students are sitting at the front table. (I got distracted by an incident in the back of the lunchroom, and forgot to record how many students.)
• Each of the students at the front table has been joined by a friend, doubling the number of students at the table.
• Four more students have just taken seats with the students at the front table.
• The students at the front table separated into three equal-sized groups and then two groups left, leaving only one-third of the students at the table.
• As the lunch period ends, there are still students seated at the front table.
Elvira is wondering how many students were sitting at the front table when she wrote her first note. Unfortunately, she is not sure what order the middle three sticky notes were recorded in since they got stuck together in random order. She is wondering if it matters.
2.
Does it matter which order the notes were recorded in? Determine how many students were originally sitting at the front table based on the sequence of notes that appears above. Then, rearrange the middle three notes in different orders and determine what the new order implies about the number of students seated at the front table at the beginning.
Pause and Reflect
3.
Here are three different equations that could be written based on a particular sequence of notes. Examine each equation, and then list the order of the five notes that is represented by each equation. Find the solution for each equation.
1.
Create your own story context with a sequence of actions that can be represented symbolically with an equation. Then, create a set of sticky note statements for your situation. Rearrange your notes in ways that can be represented by a variety of different equations, even though the actions remain the same, then record your new equations with the corresponding stories.
2.
Are there arrangements of the notes that produce different answers? If so, list the equations.
3.
Can you find different arrangements of the notes that produce the same answer? Why might that be so? List the equations.
Takeaways
Today, we used operations of arithmetic to represent actions in a story context. We used grouping symbols and order of operations to write an equation to model a story context, and then used the story context to help us justify a process for solving the equation.
In general, we learned that when solving a linear equation, we un-do operations in the equation by .
The order for applying inverse operations is .
We can justify each step in the equation-solving process by using properties of equality and properties of operations, such as:
The Distributive Property:
The Subtraction Property of Equality:
The Multiplication Property of Equality:
The Division Property of Equality:
Lesson Summary
In this lesson, we learned how to solve multi-step equations in the context of using operations to represent actions in a story. As we observed how to “un-do” the actions of the story, we developed a strategy for solving the equation by using inverse operations. The order in which inverse operations are applied when solving an equation matters, and we learned how to pay attention to the structure of the equation for clues to the order in which we should use inverse operations.
Retrieval
1.
Graph the equation and determine if the given point is a solution.
Point:
Indicate whether the given value is a solution to the equation. Show your work to justify your response. |
A3_soln
# A3_soln - Math 235 Assignment 3 Solutions 1 Prove that any...
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Unformatted text preview: Math 235 Assignment 3 Solutions 1. Prove that any plane through the origin in R 3 is isomorphic to R 2 . Solution: Since a plane through the origin in R 3 is a two-dimensional vector space, it is isomorphic to R 2 . 2. For each of the following pairs of vector spaces, define an explicit isomorphism to establish that the spaces are isomorphic. Prove that your map is an isomorphism. a) R 2 and P 1 . Solution: We define L : R 2 → P 1 by L ( a,b ) = a + bx . To prove that it is an isomorphism, we must prove that it is linear, one-to-one and onto. Linear: Let ~x = a 1 b 1 ,~ y = a 2 b 2 ∈ R 2 and let s,t ∈ R then L ( s~x + t~ y ) = L ( s a 1 b 1 + t a 2 b 2 ) = L ( sa 1 + ta 2 sb 1 + tb 2 ) = ( ( sa 1 + ta 2 ) + ( sb 1 + tb 2 ) x ) = s ( a 1 + b 1 x ) + t ( a 2 + b 2 x ) = sL ( ~x ) + tL ( ~ y ) Therefore L is linear. One-to-one: Assume L ( ~x ) = L ( ~ y ). Then L a 1 b 1 = L a 2 b 2 ⇒ a 1 + b 1 x = a 2 + b 2 x This gives a 1 = a 2 and b 1 = b 2 , hence ~x = ~ y so L is one-to-one. Onto: For any a + bx ∈ P 3 we have L a b = a + bx hence L is onto....
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## This note was uploaded on 09/30/2011 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
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A3_soln - Math 235 Assignment 3 Solutions 1 Prove that any...
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In this blog post, we discuss how Algebra answer solver can help students learn Algebra. Our website can solving math problem.
## The Best Algebra answer solver
Algebra answer solver is a software program that supports students solve math problems. Factoring is the process of breaking down an equation into smaller pieces in order to solve it. For example, the equation x^2+5x+6 can be factored as (x+3)(x+2). While this may seem like a simple task, factoring can be quite challenging, particularly when equations are complex. However, with practice, most students can develop the skills necessary to factor equations successfully. As with any skill, mastering factoring can take time and effort, but the rewards are well worth it.
A city hall has a square area of 20,000 square meters and a perimeter of 4 kilometers. If the area of a circle of radius 2,000 meters is half this area, what is the radius of this circle? 3. A square with sides of length 5 meters has a perimeter of 2 meters. What is the length of one side? 4. The area of a triangle with base length 6 meters and height 3 meters equals what percent larger than that of a triangle with base and height 8 meters? 5. Two cars are traveling in opposite directions along parallel roads that are exactly kilometer apart. The first car drives at 30 kilometers per hour while the second travels at 40 km/h. Both cars travel for two hours before they pass each other. What was the average speed for both cars during this time?
1 step equations are those which have one unknown in the equation. For example, if x is the unknown and x + 2 = 4, then 1 step equation is written as x+2=4. Such equations can be very useful when you want to solve simple problems quickly. Since they involve a simple equation with just one unknown, they are easy to solve using basic arithmetic rules. For example, if y is the unknown in the equation 8 + y = 14, then 8 + y = 14 becomes 8 + 2 = 10. Solving this problem by adding 2 to both sides yields 10 + 2 = 12, solving for y. The answer is therefore 12. More advanced students can also use plugging and graphing methods for 1 step equations. For example, plugging in 3 for x in the equation 8 + 3 = 13 yields 10 as the solution because 3 = 10. This method works because it ignores the value of the unknown—in this case, 3—and only looks at how the known number, 8, changes when given a known change, 3. Solving 1 step equations can be pretty straightforward at first glance, but there are some more advanced techniques that can help make things even easier!
Arithmetic math problems are a staple in every grade. They help kids practice basic math facts and develop their ability to count and add numbers. With so much emphasis on arithmetic in school, there are plenty of arithmetic math problems to choose from. Here are some of the best: Here are some tips for solving arithmetic math problems: 1) Keep track of the problem steps. If you’re unsure about how to proceed, write down each step as you go. 2) Be careful with your answer choices. There are two types of answers that students can choose from: right and wrong. Don’t be afraid to pick a right answer if it makes sense, but don’t be too quick to pick the wrong options either. 3) Break down problems into smaller parts. This will help you keep track of all the steps needed to complete the problem and make sure you don’t miss anything along the way. 4) Look for patterns in the problem steps. If you see a pattern repeating itself over and over again, you can use that information to help solve the problem more quickly.
When it comes to math apps, there is no shortage of options to choose from. However, not all math apps are created equal. Some are more comprehensive than others, some are more user-friendly, and some are just plain more fun to use. So, which is the best app to solve math problems? It really depends on your individual needs and preferences. However, here are three apps that are definitely worth checking out: 1. Photomath is a great option for those who want a comprehensive app that can provide step-by-step solutions to even the most complex math problems. 2. If you're looking for an app that's easy to use and navigate, then Mathway is definitely worth considering. 3. Finally, if you want an app that's both educational and entertaining, then we suggest giving Socratic a try.
## We cover all types of math issues
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Camryn Nelson
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Jimena Carter |
## What is a Reciprocal?
In mathematics, the reciprocal, additionally known as multiplicative inverse, is the inverse of a number x. Denoted as 1/x or x-1. This method that the product of a number x and also its reciprocal yields 1.
You are watching: Reciprocal of -2/3
The inverse of a fraction a/b is denoted together (a/b)-1, which is b/a. This short article discusses the procedures on just how to uncover the reciprocal of a number, blended numbers, fractions and also decimals.
## How to discover Reciprocals?
The mutual of a number is merely the number that has actually been flipped or reverse upside-down. This entails transposing a number such the the numerator and also denominator are inserted at the bottom and top respectively.
To find the reciprocal of a totality number, just transform it right into a fraction in i beg your pardon the initial number is the denominator and also the molecule is 1.
Example 1
The mutual of 2/3 is 3/2.
The product the 2/3 and its mutual 3/2 is 1.
2/3 x 3/2 = 1
Example 2
The mutual of a whole number 7 is 1/7 since 7 x 1/7 = 1.
### How to discover the mutual of a blended Number?
In stimulate to find the mutual of a mixed fraction, convert it into improper fraction first and then apply the same preeminence we learnt above.
Example 3
Find the mutual of 4 1/2.
Solution
Convert a mixed portion into an improper fraction as calculate below.
4 1/2 = (4 x 2) + 1/ 2
= 9/2
Now upper and lower reversal the numerator and denominator of 9/2.Therefore, the solution for the mutual of 4 1/2 is 2/9.
### How to find the reciprocal of Decimal Numbers?
Like various other numbers, decimal number too have reciprocals. Calculating the reciprocal of a decimal number can be excellent in the adhering to ways:
Convert the decimal right into an tantamount fraction, for example, 0.25 = 1/4, and also therefore, the mutual is 4/1 = 4.You can additionally use a calculate to divide 1 through the fraction. Because that example, the reciprocal of 0.25 = 1/0.25 = 4
It have the right to be detailed that separating 1 through a portion is the exact same as multiplying the reciprocal of the number by 1. For example,
5 ÷ 1/4
= 5 x 4/1
= 20
Example 4
a. Find the mutual of 5
Solution
5 = 5/1
So, the mutual of 3 = 1/5
b. Find the reciprocal of 1/4
Solution
To find the mutual of 1/4, invert the numerator and also denominator.
The mutual of 1/4 = 4
c. Identify the reciprocal of 10/3
Solution
Step 1:
To uncover the reciprocal of 10/3, flip the numerator and denominator. The mutual = 3/10.
Example 5
If 4/7 the a number x is 84. What is the value of x?
Solution4/7 the a number x = 84Write the mathematics equation:
(4/7) x = 84
Multiply both political parties by the mutual of 4/7Number x = 84 × 7/4
= 21 × 7= 147And thus, the number x is 147.
Example 6
A half of the student in a college space boys, 3/5 that these boys take science courses and the remainder take humanities. What portion of the guys take humanities?
Solution
Fraction of guys in the college = 1/2
Fraction the boys that take sciences = 3/5 that 1/2
= 3/5 × 1/2
= 3 × 1/5 × 2
= 3/10
Therefore, 3/10 of the guys take humanities.
Example 7
Pedro has actually written three-fifth that his 75 paged study work. How many pages room left to complete writing his research?
SolutionNumber the pages written = 3/5 the 75= 3/5 × 7
= 45 pages.Number the pages left= 75 – 45.= 30 pages.
See more: What Channel Is Abc On Dish Network Texas, Find Your Local Dish Channels
Example 8
A herd of cows in a farm yard produces 99 liters the milk daily. If each cow produce one-third of total milk developed in a day. How plenty of cows are in the farm yard if 7700 liters of milk is produced weekly.
SolutionA herd the cows produces 99 liters of milk daily.One cow to produce 1/3 of full milk daily = 1/3 that 99Therefore, one cow produces 11 liters.Total number of animals in the farm= (7700/7) ÷ 11= 100 cows |
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Geometry Helpp
0
4
1
+8
Find sin B
[asy] size(200); pair A,B,C,D; A=(0,8); B=(-15,0); C=(6,0); D=origin; draw(A--B--C--A--D); draw(rightanglemark(C,D,A,20)); dot("\$A\$",A,N); dot("\$B\$",B,W); dot("\$C\$",C,E); dot("\$D\$",D,S); label("15",B--D,S); label("\$6\$",D--C,S); label("10",A--C,NE); [/asy]
Jun 13, 2024
#1
+867
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Right Triangle Identification: We are given that D is the foot of the altitude from A to BC. This implies that ∠ADC=90∘. Therefore, triangle ADC is a right triangle.
Soh Cah Toa: We are asked to find sin B, which is the opposite side (AC) divided by the hypotenuse (in this case, we don't know the length of BC, but we can find it).
Pythagorean Theorem in Right Triangle ADC: We know AC = 10 and CD = 6. Since this is a right triangle, we can use the Pythagorean Theorem to find the length of the missing side (AD):
Substitute the known values: AD^2 + 6^2 = 10^2
Pythagorean Theorem to Find Hypotenuse (BC): Now that we know the length of a leg (AD) in right triangle ABC, we can find the length of the hypotenuse (BC) using the Pythagorean Theorem again:
BC^2 = AB^2 + AC^2 (We don't know the length of AB, but we can find BC)
Substitute the known values: BC^2 = 8^2 + 10^2
Solve for BC: BC^2 = 64 + 100 = 164 BC = 164 (We can leave it in this form for now)
Finding sin B: Now that we know both AC and BC, we can find sin B:
sin B = AC / BC
Substitute the known values: sin B = 10 / sqrt(164)
Since we cannot simplify 164 further, the answer is sin B = 10/sqrt(164).
Jun 13, 2024 |
# How to Find 1/2 + 1/4 + 1/8
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Lesson Transcript
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
In this lesson, we'll learn how to add 1/2 + 1/4 + 1/8. We'll see the steps involved in this process, and in adding fractions in general, using an example from everyday life.
In everyday life, we encounter scenarios in which we need to add three fractions. Suppose you want to ride your bike to your friend's house. To get there, you must first ride 1/2 of a mile down Elm Street, then 1/4 of a mile down Maple Street, and finally 1/8 of a mile down Oak Street. Before you head out, you want to know how far you have to ride.
You can add the distances on each street, or 1/2 + 1/4 + 1/8. If we're adding fractions that have the same denominator such as 1/4 and 2/4, all we have to do is add together the numerators to find the result. Since 1 + 2 = 3, our numerator is 3. The denominator, 4, stays the same. Thus, 1/4 + 2/4 = 3/4.
So, how do we add fractions with different denominators? To do this, we have to add a step to our procedure. Before we add the numerators, we have to convert our fractions so that they have the same denominator.
## Finding a Common Denominator
A common denominator is the same denominator in two or more fractions. The common denominator will be the least common multiple of the denominators of the fractions. The least common multiple (abbreviated LCM) of a group of numbers is the smallest positive number that all of the numbers in the group divide into evenly. To find this, it's helpful to list multiples of each of the numbers.
Let's say we want to find a common denominator for the fractions 1/3 and 2/5. We need to find the least common multiple of 3 and 5. We can list the multiples of each number:
• Multiples of 3 are: 3, 6, 9, 12, 15, 18, and so on
• Multiples of 5 are: 5, 10, 15, 20, and so on
The first number in common in these two lists is 15, so the least common multiple of 3 and 5 is 15.
If we wanted to add 1/3 and 2/5, we would first need to convert both fractions to denominators of 15. How do we change the denominator of a fraction without changing the value of that fraction? To do this, we have to multiple both the numerator and denominator by the same number. To change 1/3 into 15ths, we look at what we would need to multiple 3 by to get 15. 3 times 5 is 15, so if we multiple the numerator of the fraction (1) by the same number (5) we get 5. 1/3 is equal to 5/15.
## One Example of Adding 3 Fractions
Let's practice these steps to find out how far you have to ride your bike to get to your friend's house. We want to add 1/2 + 1/4 + 1/8. There are a couple of ways to go about adding three fractions:
The associative property states (a + b) + c = a + (b + c). This means that it doesn't matter which fractions we add together first. We can add any two of the fractions together and then add the result to the third fraction. Let's start by adding 1/2 and 1/4. First we need a common denominator. This will be the least common multiple of the denominators in the fractions. In our case, we want to find the least common multiple of 2 and 4:
• Mutliples of 2 are: 2, 4, 6, 8, 10, and so on
• Multiples of 4 are: 4, 8, 12, 16, and so on
Comparing the list, we see that 4 is the least common multiple of 2 and 4. Next, we need to make 4 the denominator of both fractions. We don't have to do anything to 1/4 since it already has this denominator. To make 4 the denominator of 1/2, we need to multiply both the numerator and denominator by 2, since 2 times 2 equals 4. When we do this, we find that 1/2 is equal to 2/4. Now, we're ready to add our fractions. Remember, to add two fractions with the same denominator, simply add the numerators: the denominator stays the same. When we add 2/4 and 1/4, we add 2 + 1 = 3 to find the numerator. Since the denominator is 4, we find that 2/4 + 1/4 = 3/4.
We're almost there! Now, we need to add our third fraction, 1/8, to the result. We want to find 3/4 + 1/8. Again, we have to first find a common denominator by determining the least common multiple of 4 and 8:
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# How do I find the area of this region?
A square with edge length 2 cm has semicircles drawn on each side.
Find the total area of the shaded region.
Here is an image of the diagram shown :
Please show your work in pictures, numbers, words, anything. (Try to keep it to a Grade 8 level too)
-
The square has area $4$.
The top and bottom white region together have area $4$ minus $2$ times the area of the semi circle, that is, $4-2\times \pi r^2/2=4-\pi$ (since $r=1$).
So, dividing by $2$, the top blue region has area $2-\pi/2$.
If you remove this area from the area of the top semi circle you obtain the area of the two top red parts.
Double it to obtain the red area.
-
One sees four semi-circles. Each delineates a semi-disk. Summing the areas of these four semi-disks, one counts twice each purple region and once each white region.
Hence the purple area plus the area of the square with side $2$ is twice the area of a disk with diameter $2$, that is, $\color{purple}{\mathbf{purple\ area}}+2^2=2\cdot\frac14\pi\cdot2^2$.
Finally, $\color{purple}{\mathbf{purple\ area}}=2\pi-4$.
-
You lost me at "Hence the purple area...". Could you explain further please? I'm confused. – mowwwalker Jan 29 '13 at 3:35
(2 x Purple) + White = Purple + Full square. – Did Jan 29 '13 at 7:51
There are $8$ circular segments, each of area
$$\frac{\pi}{2} r^2 (\theta - \sin{\theta})$$
where $r$ is the radius, and $\theta$ is the angle subtended at the center by the segment. Here, from the geometry, $r$ is clearly $1 \, \mathrm{cm}$ and $\theta = \frac{\pi}{2}$, each segment spanning half a semicircle.
Therefore the shaded area is
$$8 \frac{\pi}{2} (1 \, \mathrm{cm}^2) \left ( \frac{\pi}{2} - 1 \right ) = (2 \pi - 4) \, \mathrm{cm}^2$$
-
A layman's (mine) approach to solving it:
There are 4 half-circles contained in the square. The 4 half-circles have the same area as two whole circles with the same radius. The radius (half the length of the diameter, which is the length of a side of the cube) is 1 and the area of a circle is $\pi r^2$, so the area of the four half-circles is $2 ( \pi * (1)^2)$, which is $2\pi$. The square can only hold an area of $4$ and all its space is used, so the excess area of the circles' area must be contained in overlap. The the area not able to be contained be the square without overlap is given by $2\pi - 4$.
This, of course, only works out nicely because ALL the area inside the square is being used. Were part of the square unused, you would have to add the area of the unused portion into the calculation of the overlap.
-
I did not know I was a layman. This is good news. Or not. Oh, whatever. – Did Jan 29 '13 at 8:59
@Did, Haha, sorry if I offended you. I know that some of the mathematicians on here are really experienced, so I thought the explanations might go over the OP's head as they often do mine. – mowwwalker Jan 29 '13 at 19:38
Move one half of a leaf to the circle:
$\hspace{3cm}$
The area of the quarter circle is $\frac\pi4$; the area of the triangle is $\frac12$. Thus, the area of the half-leaf (in red) is $\frac\pi4-\frac12$. There are two per leaf, therefore the area of four leaves is $$8\left(\frac\pi4-\frac12\right)=2\pi-4$$
-
Very nice +1! Which software did you use for the drawings? – Hakim Oct 19 '14 at 14:30
I used Intaglio, a Mac OS only drawing application. – robjohn Oct 19 '14 at 16:21
Try this simple approach instead. From Fig. A area of the shaded region is $A_{A}(4-\pi)$ Similarly with Fig B the area is $A_{B} = (4-\pi)$ Now when you subtract it from the whole you get the four shaded areas in the given question. The shaded area is $A = 4 -(4-\pi)-(4-\pi)= 2\pi-4$
- |
5.4: Decimal Operations (Part 2)
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Divide Decimals
Just as with multiplication, division of decimals is very much like dividing whole numbers. We just have to figure out where the decimal point must be placed.
To understand decimal division, let’s consider the multiplication problem
$(0.2)(4) = 0.8$
Remember, a multiplication problem can be rephrased as a division problem. So we can write 0.8 ÷ 4 = 0.2 We can think of this as “If we divide 8 tenths into four groups, how many are in each group?” Figure $$\PageIndex{1}$$ shows that there are four groups of two-tenths in eight-tenths. So 0.8 ÷ 4 = 0.2.
Figure $$\PageIndex{1}$$
Using long division notation, we would write
Notice that the decimal point in the quotient is directly above the decimal point in the dividend.
To divide a decimal by a whole number, we place the decimal point in the quotient above the decimal point in the dividend and then divide as usual. Sometimes we need to use extra zeros at the end of the dividend to keep dividing until there is no remainder.
HOW TO: DIVIDE A DECIMAL BY A WHOLE NUMBER
Step 1. Write as long division, placing the decimal point in the quotient above the decimal point in the dividend.
Step 2. Divide as usual.
Example $$\PageIndex{9}$$:
Divide: 0.12 ÷ 3.
Solution
Write as long division, placing the decimal point in the quotient above the decimal point in the dividend. Divide as usual. Since 3 does not go into 0 or 1 we use zeros as placeholders.
0.12 ÷ 3 = 0.04
Exercise $$\PageIndex{17}$$:
Divide: 0.28 ÷ 4.
$$0.07$$
Exercise $$\PageIndex{18}$$:
Divide: 0.56 ÷ 7.
$$0.08$$
In everyday life, we divide whole numbers into decimals—money—to find the price of one item. For example, suppose a case of 24 water bottles cost $3.99. To find the price per water bottle, we would divide$3.99 by 24, and round the answer to the nearest cent (hundredth).
Divide: $3.99 ÷ 24. Solution Place the decimal point in the quotient above the decimal point in the dividend. Divide as usual. When do we stop? Since this division involves money, we round it to the nearest cent (hundredth). To do this, we must carry the division to the thousandths place. Round to the nearest cent. $$0.166 \approx 0.17$$$3.99 ÷ 24 ≈ $0.17 This means the price per bottle is 17 cents. Exercise $$\PageIndex{19}$$: Divide:$6.99 ÷ 36.
$$0.19$$
Exercise $$\PageIndex{25}$$:
Divide: 6 ÷ 0.03.
$$200$$
Exercise $$\PageIndex{26}$$:
Divide: 7 ÷ 0.02
$$350$$
Use Decimals in Money Applications
We often apply decimals in real life, and most of the applications involving money. The Strategy for Applications we used in The Language of Algebra gives us a plan to follow to help find the answer. Take a moment to review that strategy now.
Strategy for Applications
1. Identify what you are asked to find.
2. Write a phrase that gives the information to find it.
3. Translate the phrase to an expression.
4. Simplify the expression.
5. Answer the question with a complete sentence.
Example $$\PageIndex{14}$$:
Paul received $50 for his birthday. He spent$31.64 on a video game. How much of Paul’s birthday money was left?
Solution
What are you asked to find? How much did Paul have left? Write a phrase. $50 less$31.64 Translate. 50 − 31.64 Simplify. 18.36 Write a sentence. Paul has $18.36 left. Exercise $$\PageIndex{27}$$: Nicole earned$35 for babysitting her cousins, then went to the bookstore and spent $18.48 on books and coffee. How much of her babysitting money was left? Answer $$16.52$$ Exercise $$\PageIndex{28}$$: Amber bought a pair of shoes for$24.75 and a purse for $36.90. The sales tax was$4.32. How much did Amber spend?
$$65.97$$
Jessie put 8 gallons of gas in her car. One gallon of gas costs $3.529. How much does Jessie owe for the gas? (Round the answer to the nearest cent.) Solution What are you asked to find? How much did Jessie owe for all the gas? Write a phrase. 8 times the cost of one gallon of gas Translate. 8($3.529) Simplify. $28.232 Round to the nearest cent.$28.23 Write a sentence. Jessie owes $28.23 for her gas purchase. Exercise $$\PageIndex{29}$$: Hector put 13 gallons of gas into his car. One gallon of gas costs$3.175. How much did Hector owe for the gas? Round to the nearest cent.
$$41.28$$
Christopher bought 5 pizzas for the team. Each pizza cost $9.75. How much did all the pizzas cost? Answer $$48.75$$ Example $$\PageIndex{16}$$: Four friends went out for dinner. They shared a large pizza and a pitcher of soda. The total cost of their dinner was$31.76. If they divide the cost equally, how much should each friend pay?
Solution
What are you asked to find? How much should each friend pay? Write a phrase. $31.76 divided equally among the four friends. Translate to an expression.$31.76 ÷ 4 Simplify. $7.94 Write a sentence. Each friend should pay$7.94 for his share of the dinner.
Six friends went out for dinner. The total cost of their dinner was $92.82. If they divide the bill equally, how much should each friend pay? Answer $$15.47$$ Exercise $$\PageIndex{32}$$: Chad worked 40 hours last week and his paycheck was$570. How much does he earn per hour?
$$14.25$$
Be careful to follow the order of operations in the next example. Remember to multiply before you add.
Example $$\PageIndex{17}$$:
Marla buys 6 bananas that cost $0.22 each and 4 oranges that cost$0.49 each. How much is the total cost of the fruit?
Solution
What are you asked to find? How much is the total cost of the fruit? Write a phrase. 6 times the cost of each banana plus 4 times the cost of each orange Translate to an expression. 6($0.22) + 4($0.49) Simplify. $1.32 +$1.96 Add. $3.28 Write a sentence. Marla's total cost for the fruit is$3.28.
Exercise $$\PageIndex{33}$$:
Suzanne buys 3 cans of beans that cost $0.75 each and 6 cans of corn that cost$0.62 each. How much is the total cost of these groceries?
$$5.97$$
Exercise $$\PageIndex{34}$$:
Lydia bought movie tickets for the family. She bought two adult tickets for $9.50 each and four children’s tickets for$6.00 each. How much did the tickets cost Lydia in all?
$$43.00$$
Multiplying Decimals
Multiplying by Powers of Ten
Dividing Decimals
Dividing by Powers of Ten
Practice Makes Perfect
In the following exercises, add or subtract.
1. 16.92 + 7.56
2. 18.37 + 9.36
3. 256.37 − 85.49
4. 248.25 − 91.29
5. 21.76 − 30.99
6. 15.35 − 20.88
7. 37.5 + 12.23
8. 38.6 + 13.67
9. −16.53 − 24.38
10. −19.47 − 32.58
11. −38.69 + 31.47
12. −29.83 + 19.76
13. −4.2 + (− 9.3)
14. −8.6 + (− 8.6)
15. 100 − 64.2
16. 100 − 65.83
17. 72.5 − 100
18. 86.2 − 100
19. 15 + 0.73
20. 27 + 0.87
21. 2.51 + 40
22. 9.38 + 60
23. 91.75 − (− 10.462)
24. 94.69 − (− 12.678)
25. 55.01 − 3.7
26. 59.08 − 4.6
27. 2.51 − 7.4
28. 3.84 − 6.1
Multiply Decimals
In the following exercises, multiply.
1. (0.3)(0.4)
2. (0.6)(0.7)
3. (0.24)(0.6)
4. (0.81)(0.3)
5. (5.9)(7.12)
6. (2.3)(9.41)
7. (8.52)(3.14)
8. (5.32)(4.86)
9. (−4.3)(2.71)
10. (− 8.5)(1.69)
11. (−5.18)(− 65.23)
12. (− 9.16)(− 68.34)
13. (0.09)(24.78)
14. (0.04)(36.89)
15. (0.06)(21.75)
16. (0.08)(52.45)
17. (9.24)(10)
18. (6.531)(10)
19. (55.2)(1,000)
20. (99.4)(1,000)
Divide Decimals
In the following exercises, divide.
1. 0.15 ÷ 5
2. 0.27 ÷ 3
3. 4.75 ÷ 25
4. 12.04 ÷ 43
5. $8.49 ÷ 12 6.$16.99 ÷ 9
7. $117.25 ÷ 48 8.$109.24 ÷ 36
9. 0.6 ÷ 0.2
10. 0.8 ÷ 0.4
11. 1.44 ÷ (− 0.3)
12. 1.25 ÷ (− 0.5)
13. −1.75 ÷ (− 0.05)
14. −1.15 ÷ (− 0.05)
15. 5.2 ÷ 2.5
16. 6.5 ÷ 3.25
17. 12 ÷ 0.08
18. 5 ÷ 0.04
19. 11 ÷ 0.55
20. 14 ÷ 0.35
Mixed Practice
In the following exercises, simplify.
1. 6(12.4 − 9.2)
2. 3(15.7 − 8.6)
3. 24(0.5) + (0.3)2
4. 35(0.2) + (0.9)2
5. 1.15(26.83 + 1.61)
6. 1.18(46.22 + 3.71)
7. $45 + 0.08($45)
8. $63 + 0.18($63)
9. 18 ÷ (0.75 + 0.15)
10. 27 ÷ (0.55 + 0.35)
11. (1.43 + 0.27) ÷ (0.9 − 0.05)
12. (1.5 − 0.06) ÷ (0.12 + 0.24)
13. [$75.42 + 0.18($75.42)] ÷ 5
14. [$56.31 + 0.22($56.31)] ÷ 4
Use Decimals in Money Applications
In the following exercises, use the strategy for applications to solve.
1. Spending money Brenda got $40 from the ATM. She spent$15.11 on a pair of earrings. How much money did she have left?
2. Spending money Marissa found $20 in her pocket. She spent$4.82 on a smoothie. How much of the $20 did she have left? 3. Shopping Adam bought a t-shirt for$18.49 and a book for $8.92 The sales tax was$1.65. How much did Adam spend?
4. Restaurant Roberto’s restaurant bill was $20.45 for the entrée and$3.15 for the drink. He left a $4.40 tip. How much did Roberto spend? 5. Coupon Emily bought a box of cereal that cost$4.29. She had a coupon for $0.55 off, and the store doubled the coupon. How much did she pay for the box of cereal? 6. Coupon Diana bought a can of coffee that cost$7.99. She had a coupon for $0.75 off, and the store doubled the coupon. How much did she pay for the can of coffee? 7. Diet Leo took part in a diet program. He weighed 190 pounds at the start of the program. During the first week, he lost 4.3 pounds. During the second week, he had lost 2.8 pounds. The third week, he gained 0.7 pounds. The fourth week, he lost 1.9 pounds. What did Leo weigh at the end of the fourth week? 8. Snowpack On April 1, the snowpack at the ski resort was 4 meters deep, but the next few days were very warm. By April 5, the snow depth was 1.6 meters less. On April 8, it snowed and added 2.1 meters of snow. What was the total depth of the snow? 9. Coffee Noriko bought 4 coffees for herself and her coworkers. Each coffee was$3.75. How much did she pay for all the coffees?
10. Subway Fare Arianna spends $4.50 per day on subway fare. Last week she rode the subway 6 days. How much did she spend for the subway fares? 187. Income Mayra earns$9.25 per hour. Last week she worked 32 hours. How much did she earn?
11. Income Peter earns $8.75 per hour. Last week he worked 19 hours. How much did he earn? 12. Hourly Wage Alan got his first paycheck from his new job. He worked 30 hours and earned$382.50. How much does he earn per hour?
13. Hourly Wage Maria got her first paycheck from her new job. She worked 25 hours and earned $362.50. How much does she earn per hour? 14. Restaurant Jeannette and her friends love to order mud pie at their favorite restaurant. They always share just one piece of pie among themselves. With tax and tip, the total cost is$6.00. How much does each girl pay if the total number sharing the mud pie is (a) 2? (b) 3? (c) 4? (d) 5? (e) 6?
Writing Exercises
1. In the 2010 winter Olympics, two skiers took the silver and bronze medals in the Men's Super-G ski event. The silver medalist's time was 1 minute 30.62 seconds and bronze medalist's time was 1 minute 30.65 seconds. Whose time was faster? Find the difference in their times and then write the name of that decimal.
2. Find the quotient of 0.12 ÷ 0.04 and explain in words all the steps taken.
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) After reviewing this checklist, what will you do to become confident for all objectives? |
# What Is 5/35 as a Decimal + Solution With Free Steps
The fraction 5/35 as a decimal is equal to 0.142.
A Decimal is a number with two parts a whole number value and the fractional value of that whole number. Many problems demand decimals as they make it easier to solve the problems.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 5/35.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 5
Divisor = 35
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 5 $\div$ 35
This is when we go through the Long Division solution to our problem. Refer to the following figure for the solution.
Figure 1
## 5/35 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 5 and 35, we can see how 5 is Smaller than 35, and to solve this division, we require that 5 be Bigger than 35.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 5, which after getting multiplied by 10 becomes 50.
We take this 50 and divide it by 35; this can be done as follows:
50 $\div$ 35 $\approx$ 1
Where:
35 x 1 = 35
This will lead to the generation of a Remainder equal to 50 – 35 = 15. Now this means we have to repeat the process by Converting the 15 into 150 and solving for that:
150 $\div$ 35 $\approx$ 4
Where:
35 x 4 = 140
This, therefore, produces another Remainder which is equal to 150 – 140 = 10. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 100.
100 $\div$ 35 $\approx$ 2
Where:
35 x 2 = 70
Finally, we have a Quotient generated after combining the three pieces of it as 0.142, with a Remainder equal to 30.
Images/mathematical drawings are created with GeoGebra. |
### Home > CC2 > Chapter 9 > Lesson 9.2.2 > Problem9-69
9-69.
1. Simplify each of the following expressions. Homework Help ✎
1. 3·
2. 53 ·
3. 24 ·
4. · 32
5. +
6. 82
Convert the mixed number into a fraction greater than one.
$\frac{16}{5}\left(\frac{7}{4}\right)$
$\frac{28}{5}=5\frac{3}{5}$
$- \frac{4}{5} = (-1)(4)\left(\frac{1}{5}\right)$
Notice that the 5 in the denominator is like dividing by 5.
5(3) /5 (−4)
52(−4) = −100
This can be rewritten as:
$2^{4}\left(\frac{5}{2^{3}}\right)$
See part (b).
10
Use the order of operations.
Use the order of operations.
$-\frac{5}{6}+\frac{1}{4}$
$-\frac{20}{24}+\frac{6}{24}=-\frac{14}{24}$
Notice that squaring a negative number has the same result as multiplying two negative numbers: the product is always positive.
$8^{(2-1)}(-7)-\frac{1}{2}$
$8(-7)-\frac{1}{2}$
$(-56)-\frac{1}{2}$
$-56\frac{1}{2}$ |
Topics in
P R E C A L C U L U S
2
# RATIONAL AND IRRATIONAL NUMBERS
What is a rational number?
CALCULUS IS A THEORY OF MEASUREMENT. The necessary numbers are the rationals and irrationals. But let us start at the beginning.
The following numbers of arithmetic are the counting-numbers or, as they are called, the natural numbers:
1, 2, 3, 4, and so on.
If we include 0, we have the whole numbers:
0, 1, 2, 3, and so on.
And if we include their algebraic negatives, we have the integers:
0, ±1, ±2, ±3, and so on.
± ("plus or minus") is called the double sign.
The following are the square numbers, or the perfect squares:
1 4 9 16 25 36 49 64, and so on.
They are the numbers 1· 1, 2· 2, 3· 3, 4· 4, and so on.
Rational and irrational numbers
1. What is a rational number?
Any ordinary number of arithmetic: Any whole number, fraction, mixed number, or decimal; together with its negative image.
A rational number is a nameable number, in the sense that we can name it in the standard way we name whole numbers, fractions, and mixed numbers. "Five." "Six thousand eight hundred nine." "Nine hundred twelve millionths." "Three and five-eighths."
What is more, we can in principle (by Euclid VI, 9) place any rational number exactly on the number line.
We can say that we truly know a rational number.
2. Which of the following numbers are rational?
1 −1 0 23 − 23 5½ −5½ 6.08 −6.08 3.14159
All of them! All decimals are rational. That long one is an approximation to π, which, as we will see, is not equal to any decimal.
3. A rational number can always be written in what form?
As a fraction ab , where a and b are integers (b 0).
Numbers that can be written in that form, we call rational. That is their formal definition. That is how a rational number looks. As for what it is, is a different story.
An integer itself can be written as a fraction: b = 1. And from arithmetic, we know that we can write a decimal as a fraction.
When a and b are natural numbers, then the fraction has the same ratio to 1 as the numerator has to the denominator. We can always put into words how a rational number is related to 1. Hence the term, rational number.
( 23 is to 1 as 2 is to 3. 2 is two thirds of 3. 23 is two thirds of 1.)
At this point, the student might wonder, What is a number that is not rational?
An example of such a number is ("Square root of 2"). It is not possible to name any whole number, any fraction, or any decimal whose
square is 2. 75 is close, because
75 · 75 = 4925
-- which is almost 2.
To prove that there is no rational number whose square is 2, suppose
there were. Then we could express it as a fraction mn in lowest terms.
That is, suppose
mn · mn = m · m n · n = 2.
But that is impossible. Since mn is in lowest terms, then m and n have
no common divisors except 1. Therefore, m· m and n· n also have no common divisors -- they are relatively prime -- and it will be impossible to divide n· n into m· m and get 2.
There is no rational number -- no number of arithmetic -- whose square is 2. Therefore we call an irrational number.
By recalling the Pythagorean theorem, we can see that irrational numbers are necessary. For if the sides of an isosceles right triangle are called 1, then we will have 1² + 1² = 2, so that the hypotenuse is . There really is a length that logically deserves the name, "." Inasmuch as numbers name the lengths of lines, then is a number.
4. Which natural numbers have rational square roots?
Only the square roots of the square numbers; that is, the square roots of the perfect squares.
= 1 Rational
Irrational
Irrational
= 2 Rational
, , , Irrational
= 3 Rational
And so on.
The square roots of the square numbers are the only square roots that we can name.
The existence of irrationals was first realized by Pythagoras in the 6th century B.C. He realized that in the isosceles right triangle, the ratio of the hypotenuse to the side was not as two natural numbers. Their relationship, he said, was "without a name." Because if we ask, "What ratio has the hypotenuse to the side?" -- we cannot say. We can express it only as "Square root of 2."
5. Say the name of each number.
a) "Square root of 3." b) "Square root of 5."
c) "2." This is a rational -- nameable -- number.
d) "Square root of 3/5." e) "2/3."
In the same way we saw that only the square roots of square numbers are rational, we could prove that only the nth roots of nth powers are rational. Thus, the 5th root of 32 is rational, because 32 is a 5th power, namely the 5th power of 2. But the 5th root of 33 is irrational. 33 is not a perfect 5th power.
The decimal representation of irrationals
When we express a rational number as a decimal, then either the decimal will
a predictable pattern of digits. But if we attempted to express an irrational number as an exact decimal, then, clearly, we could not, because if we could the number would be rational
Moreover, there will not be a predictable pattern of digits. For example,
Now, with rational numbers you sometimes see
1 11 = .090909. . .
By writing both the equal sign = and three dots (ellipsis) we mean:
"A decimal for 1 11 will never be complete or exact. However we can
approximate it with as many decimal digits as we please according to the indicated pattern; and the more decimal digits we write, the closer we will
be to 1 11 ."
(That explanation is an example of mathematical positivism. It asserts that in the mathematics of computation and measuring, which includes calculus, what exists is what we actually see or name, now. That .090909 never ends, is a metaphysical doctrine that need not concern us, because it serves no purpose. Such actual infinities are not required to solve any problem in arithmetic or calculus. Besides, can what is infinite -- what does not reach an end -- ever be equal to anything?)
We say that any decimal for 1 11 is inexact. But the decimal for ¼,
which is .25, is exact.
The symbol for decimal fractions was invented in the 16th century. Now, of course, we take decimals for granted, but at the time many thought it was not a very forward looking idea, because the decimals for only a very limited number of fractions were exact. Even the
decimal for as simple a fraction as 13 is inexact. See Lesson 24 of Arithmetic.
As for the decimal for an irrational number, it is always inexact. An example is the decimal for above.
If we write ellipsis --
= 1.41421356237. . .
-- we mean, "A decimal for will never be complete or exact. Moreover, there will not be a predictable pattern of digits. We could continue its rational approximation for as many decimal digits as we please by means of the algorithm, or method, for calculating each next digit (not the subject of these Topics); and again, the more digits we calculate, the closer we will be to ."
It is important to understand that no decimal that you or anyone will ever see is equal to , or π, or any irrational number. We know an irrational number only as a rational approximation. And if we choose a decimal approximation, then the more decimal digits we calculate, the closer we will be to the value.
(For a decimal approximation of π, see Topic 9 of Trigonometry.)
To sum up, a rational number is a number we can know and name exactly, either as a whole number, a fraction, or a mixed number, but not always exactly as a decimal. An irrational number we can never know exactly in any form.
The language of arithmetic is ratio. It is the language with which we relate rational numbers to one another, and to 1, which is their source. The whole numbers are the multiples of 1, the fractions are its parts: its halves, thirds, fourths, millionths. But language is incapable of relating an irrational number to 1. Like Pythagoras, we cannot say. An irrational number and 1 are incommensurable.
Real numbers
5. What is a real number?
A real number is distinguished from an imaginary or complex number. It is what we call any rational or irrational number.
They are the numbers we expect to find on the number line.
They are the numbers we need for measuring.
(An actual measurement can result only in a rational number.
An irrational number can result only from a theoretical calculation or a definition. Examples of calculations are the Pythagorean theorem, and the solution to an equation, such as x3 = 5. The irrational number π is defined as the ratio of the circumference of a circle to the diameter.)
Problem 1. We have categorized numbers as real, rational, irrational, and integer. Name all the categories to which each of the following belongs.
3 Real, rational, integer. −3 Real, rational, integer. −½ Real, rational. Real, irrational. 5¾ Real, rational. − 11/2 1.732 Real, rational. 6.920920920. . . Real, rational. 6.9205729744. . . Real. And let us assume that it is irrational, that is, no matter how many digits are calculated, they do not repeat. In other words, we must assume that there is an effective procedure for computing each next digit. For if there were not, then that symbol would not have its position in the number system with respect to order; which is to say, it would not be a number. (See Are the real numbers really numbers?) 6.9205729744 Real, rational. Every exact decimal is rational.
7. What is a real variable?
A variable is a symbol that takes on values. A value is a number.
A real variable takes on values that are real numbers.
Calculus is the study of functions of a real variable.
(But see: Are the real numbers really numbers?)
Problem 2. Let x be a real variable, and let 3 < x < 4. Name five values that x might have.
*
See The Evolution of the Real Numbers starting with the natural numbers.
Next Topic: Functions
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High School Math : How to find x or y intercept
Example Questions
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Example Question #1 : How To Find X Or Y Intercept
What is the y-intercept of the equation?
Explanation:
To find the y-intercept, we set the value equal to zero and solve for the value of .
Since the y-intercept is a point, we want to write our answer in point notation: .
Example Question #2 : How To Find X Or Y Intercept
What is the x-intercept of the equation?
Explanation:
To find the x-intercept of an equation, set the value equal to zero and solve for .
Subtract from both sides.
Multiply both sides by .
Since the x-intercept is a point, we will want to write it in point notation:
Example Question #1 : How To Find X Or Y Intercept
What is the y-intercept of
Explanation:
To solve for the y-intercept, set the x value equal to zero:
Example Question #4 : How To Find X Or Y Intercept
What is the x-intercept of ?
Explanation:
To solve for the x-intercept, set the y value equal to zero:
Subtract from both sides:
Example Question #1 : Intercepts And Curves
What is the y-intercept of ?
Explanation:
To find the y-intercept, set the x value to zero and solve:
Example Question #6 : How To Find X Or Y Intercept
What is the x-intercept of ?
Explanation:
To solve for the x-intercept, we set the value equal to and solve.
Example Question #1 : How To Find X Or Y Intercept
What is the y-intercept of ?
Explanation:
When looking at an equation in standard form, is our y-intercept.
Or, set the value equal to and solve.
Example Question #1 : How To Find X Or Y Intercept
What is the y-intercept of
Explanation:
Isolate for so that the equation is in slope-intercept form .
The is the y-intercept, which in this case, is
Example Question #1 : How To Find X Or Y Intercept
What are the -intercepts of ?
Explanation:
Factor out an from the original equation so that it is .
Set that expressions equal to so that you can find the -intercepts. Your answers are and
Example Question #1 : How To Find X Or Y Intercept
Find the -intercepts of |
Forres Primary
# Year 2 Maths Curriculum Expectations
Here are the National Curriculum Expectations for children in Year 2.
Number and place value
• count in steps of 2, 3, and 5 from 0, and in tens from any number, forward or backward
• recognise the place value of each digit in a two-digit number (tens, ones)
• identify, represent and estimate numbers using different representations, including the number line
• compare and order numbers from 0 up to 100; use <, > and = signs
• read and write numbers to at least 100 in numerals and in words
• use place value and number facts to solve problems
• solve problems with addition and subtraction:
• using concrete objects and pictorial representations, including those involving numbers, quantities and measures
• applying their increasing knowledge of mental and written methods
• recall and use addition and subtraction facts to 20 fluently, and derive and use related facts up to 100
• add and subtract numbers using concrete objects, pictorial representations, and mentally, including:
• a two-digit number and ones
• a two-digit number and tens
• two two-digit numbers
• adding three one-digit numbers
• show that addition of two numbers can be done in any order (commutative) and subtraction of one number from another cannot
• recognise and use the inverse relationship between addition and subtraction and use this to check calculations and missing number problems
Multiplication and division
• recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers
• calculate mathematical statements for multiplication and division within the multiplication tables and write them using the multiplication (×), division (÷) and equals (=) signs
• show that multiplication of two numbers can be done in any order (commutative) and division of one number by another cannot
• solve problems involving multiplication and division, using materials, arrays, repeated addition, mental methods, and multiplication and division facts, including problems in contexts
Fractions
• recognise, find, name and write fractions 1/3, 1/4, 2/4 and 3/4 of a length, shape, set of objects or quantity
• write simple fractions for example, 1/2 of 6 = 3 and recognise the equivalence of 2/4 and 1/2.
Measurement
• choose and use appropriate standard units to estimate and measure length/height in any direction (m/cm); mass (kg/g); temperature (°C); capacity (litres/ml) to the nearest appropriate unit, using rulers, scales, thermometers and measuring vessels
• compare and order lengths, mass, volume/capacity and record the results using >, < and =
• recognise and use symbols for pounds (£) and pence (p); combine amounts to make a particular value
• find different combinations of coins that equal the same amounts of money
• solve simple problems in a practical context involving addition and subtraction of money of the same unit, including giving change
• compare and sequence intervals of time
• tell and write the time to five minutes, including quarter past/to the hour and draw the hands on a clock face to show these times.
• know the number of minutes in an hour and the number of hours in a day
Geometry: properties of shapes
• identify and describe the properties of 2-D shapes, including the number of sides and symmetry in a vertical line
• identify and describe the properties of 3-D shapes, including the number of edges, vertices and faces
• identify 2-D shapes on the surface of 3-D shapes [for example a circle on a cylinder and a triangle on a pyramid]
• compare and sort common 2-D and 3-D shapes and everyday objects
Geometry: position and direction
• order and arrange combinations of mathematical objects in patterns and sequences
• use mathematical vocabulary to describe position, direction and movement, including movement in a straight line and distinguishing between rotation as a turn and in terms of right angles for quarter, half and three-quarter turns (clockwise and anti-clockwise)
Statistics
• interpret and construct simple pictograms, tally charts, block diagrams and simple tables
• ask and answer simple questions by counting the number of objects in each category and sorting the categories by quantity
• ask and answer questions about totalling and comparing categorical data |
# Questions on Probability – Business Maths
## Types of Probability
Probability can be seen in two ways. Empirical Probability and Theoretical Probability.
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1. Empirical Probability forms the basis of inferential statistics. It has its origin in games of chance such as dice and card games. Many card players know from experience that certain types of hands turn up more frequently than the others. Relative frequency can be used to define empirical probability.
2. Theoretical Probability. In practice, we do not derive all our values for probabilities from an experiment. Suppose we toss a die and we want to know the probability of throwing a 2. There are six ways the die can fall, of which only one is a 2. (kindly use the comment box below to ask a question).
### Questions on Empirical & Theoretical Probability
1. For each of the following pairs of events A and B say whether or not they are:
• Mutually exclusive.
• Exhaustive.
• A child is chosen at random from a class.
A = {child has blue eyes}
B = {child has brown eyes)
• A die is thrown.
A = {result is a multiple of 3}
B = {result is a multiple of 2}
• A card is drawn from a pack of cards.
A = {the card is a picture card}
See Also: 22 Questions on Differentiation
B = {card is a king}
• A coin is tossed
A = {toss gives a head}
B = {toss gives a tail}
1. C and D are events such that:
P(C) = 0.7, P(D) = 0.6 and P(CD) = 0.8
Find:
• P(CD)
• P(C’D)
• P(C or D but not both occurs)
• P(C’D’)
1. If A and B are two events which are exhaustive such that
P(AB) = ¼ , P(A|B) = 1/3
Find:
• P(B)
• P(A)
• P(B|A)
1. When three marksmen take part in a shooting contest, their chances of hitting a target are ½, 1/3, and ¼. Calculate the chance that one and only one bullet will hit the target if all men fire simultaneously.
1. Two teams Legion Football Club Coker of Lagos (LFC) and Pam Pam Football Club Adeniji in Lagos State (PFC) play a football match against each other.
The probabilities for each team of scoring 0, 1, 2, 3 goals are shown in the table below:
(Kindly rotate your phone to view the table below)
PROBABILITY OF SCORING PROBABILITY OF SCORING NUMBER OF GOALS LFC PFC 0 0.3 0.2 1 0.3 0.4 2 0.3 0.3 3 0.1 0.1
Calculate the probability of:
• Legion Football Club winning.
• A draw.
• Pam Pam Football Club winning.
1. A factory has three machines 1, 2, and 3 producing a particular type of item. One item is drawn at random from the factory’s production. Let B denote the event that the chosen item is defective and let AK denote the event that the item was produced on machine k where k = 1, 2, or 3. Suppose that machines 1, 2, and 3 produce respectively 35%, 45%, and 20% of the total production of items and that…
See Also: Questions On Process Costing +Video Guide
P(B|A1) = 0.02
P(B|A2) = 0.01
P(B|A3) = 0.03
Given that an item chosen at random is defective, find which machine was the most likely to have produced it.
1. CharllyCARES Nigeria Ltd. Has a Quality Control department that examines finished output for faults. Any faulty has an 80% probability of being detected by the inspector, independent of whether any other fault is detected on either the same unit of output or different units.
• One item of output contains three different faults, A, B, and C. What is the probability that:
• All three faults will be detected?
• No-fault will be detected.
• Two faults will be detected, but the third will escape undetected?
• The Factory manager now decides that the second stage of inspection should be added, in order to reduce the number of faulty units leaving the factory. If there is a 0.7 probability that faults remaining at this stage will be detected:
• What is the probability that a unit of output with one fault will pass both stages of the inspection without the fault being found?
• What is the probability that a unit of output with two faults will have one fault only detected at either stage of inspection, with the second fault escaping detection?
• Of the total faults detected, what proportion will be detected at the stage of inspection and what proportion will be detected at the second stage?
See Also: Questions on Budgeting (Cost Accounting)
You may assume that every product is inspected at both stages. Even if a fault is found in the first stage.
## Other Mathematics Questions
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# Arabic Problem
This is an old arabic problem:
An old man had 11 horses. When he died, his will stated the following distribution to his 3 sons:
1/2 gives to the eldest son,
1/4 for 2nd son,
1/6 for 3rd son.
Find: how many horses each son gets ?
There are 2 methods to solve: first using simple arithmetic trick without knowing the theory behind; the second method will explain the first method “from an advanced standpoint” – Number Theory (Felix Klein’s Vision )
1) Arithmetic trick:
11 is odd, not divisible by 2, 4 and 6.
Loan 1 horse to the old man:
11+1 = 12
1st son gets: 12/2 = 6 horses
2nd son gets:12/4 = 3 horses
3rd son gets: 12/6 = 2 horses
Total = 6+3+2=11 horses
Up to you if you want the old man to return the 1 loan horse 🙂
Strange! WHY ?
2) From an advanced standpoint: Number Theory
The mystery is unveiled here:
$\boxed{\displaystyle\frac{11}{12} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} }$
2|12, 4|12, 6|12
Let’s generalize the Arabic problem: old man has n horses, given to 3 sons x, y, z horses, respectively, such that:
x + y + z = n
and
x |n+1, y |n+1, z |n+1
Without loss of generality, we assume
x > y > z
Let n+1 = ax = by = cz
for a, b, c integers
From x + y + z = n, we obtain:
$\displaystyle \frac{n+1}{a} + \frac{n+1}{b} + \frac{n+1}{c} = n$
$\boxed{\displaystyle\frac{n}{n+1} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$ …….[1]
a | n+1, b | n+1, c | n+1, a < b < c
Note: x > y > z
(n+1)/a > (n+1) / b > (n+1) / c
n > 0
1/a > 1/b > 1/c
a < b < c
For n > 0,
$\frac{ n}{n+1} < 1 \implies a \geq 2$
Applying unique prime factorization from
Fundamental Theorem of Arithmetic (FTA)
[Note: we can prove using this FTA theorem that a > 2 is impossible !]
Let a = 2, substitute in [1]:
$\frac{1}{b} + \frac{1}{c} = \frac{n}{n+1} - \frac{1}{2} = \frac{1}{2}- \frac{1}{n+1}$
$\boxed {\displaystyle\frac{1}{b} + \frac{1}{c} = \frac{1}{2}- \frac{1}{n+1} }$
We can deduce (omit the manual FTA deduction or computer steps by trials and eliminations): n has 6 possible values (7, 11, 17, 19, 23, 41) and 7 ways of distributions (a, b, c) as below:
n =7, a=2, b=4, c=8
n =11, a=2, b=4, c=6
n =11, a=2, b=3, c=12
n =17, a=2, b=3, c=9
n =19, a=2, b=4, c=5
n =23, a=2, b=3, c=8
n =41, a=2, b=3, c=7
For the Arabic problem:
n =11, a=2, b=4, c=6
ax = n+1 => 2x =12 => x = 6 horses
by = n+1 => 4y =12 => y = 3 horses
cz = n+1 => 6z =12 => z = 2 horses
Verify:
n = x + y + z = 6 + 3 + 2 = 11 horses
$\boxed{\displaystyle\frac{11}{12} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} }$
2|12, 4|12, 6|12
[Reference] ＂单位分数＂ 柯召 ，孙琦 -北京：科学出版社，2002，数学小丛书（14）
Note: Prof Ke Zhao 柯召 (1910-2002) had Erdös # 1 with a Theorem Erdös-Ke-Rato in Combinatorics.
http://www.hlhl.org.cn/english/showsub.asp?id=679 |
# logarithmic proof of product rule
Following is a proof of the product rule using the natural logarithm, the chain rule, and implicit differentiation. Note that circular reasoning does not occur, as each of the concepts used can be proven independently of the product rule.
###### Proof.
Let $f$ and $g$ be differentiable functions and $y=f(x)g(x)$. Then $\ln y=\ln(f(x)g(x))=\ln f(x)+\ln g(x)$. Thus, $\displaystyle\frac{1}{y}\cdot\frac{dy}{dx}=\frac{f^{\prime}(x)}{f(x)}+\frac{g^% {\prime}(x)}{g(x)}$. Therefore,
$\begin{array}[]{rl}\displaystyle\frac{dy}{dx}&\displaystyle=y\left(\frac{f^{% \prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g(x)}\right)\\ &\\ &\displaystyle=f(x)g(x)\left(\frac{f^{\prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g% (x)}\right)\\ &\\ &=f^{\prime}(x)g(x)+g^{\prime}(x)f(x).\end{array}$
Once students are familiar with the natural logarithm, the chain rule, and implicit differentiation, they typically have no problem following this proof of the product rule. Actually, with some prompting, they can produce a proof of the product rule to this one. This exercise is a great way for students to review many concepts from calculus.
Title logarithmic proof of product rule LogarithmicProofOfProductRule 2013-03-22 16:18:48 2013-03-22 16:18:48 Wkbj79 (1863) Wkbj79 (1863) 6 Wkbj79 (1863) Proof msc 26A06 msc 97D40 LogarithmicProofOfQuotientRule |
How do you solve x-(3/8)x=5?
Feb 9, 2015
The quick answer is $x = 8$ which you can check by substituting it back into your equation (replacing x with 8).
Two points to remember in solving this kind of equation:
1. if you multiply both sides of an equation by the same amount you still have a valid equation.
2. fractions are harder to work with than integers.
We can get rid of the fraction by multiplying the fraction by 8, but we can't multiply just part of the equation, we have to multiply all parts on both sides.
So (using a $.$ for multiplication (because I haven't figured out how to use "special symbols":
$8 \cdot \left(x - \left(\frac{3}{8}\right) x\right) = 8 \cdot \left(5\right)$
$8 x - 3 x = 40$ after multiplying the individual terms
$5 x = 40$ after subtracting
$x = 8$ after dividing both sides by 5 |
Question
# If the line $y-\sqrt{3}x+3=0$ cuts the parabola ${{y}^{2}}=x+2$ at $A$ and $B$, then$PA.PB$is equal to [where $P=\left( \sqrt{3},0 \right)$](a) $\dfrac{4\left( \sqrt{3}+2 \right)}{3}$(b) $\dfrac{4\left( 2-\sqrt{3} \right)}{3}$(c) $\dfrac{4\sqrt{3}}{2}$(d) $\dfrac{2\left( \sqrt{3}+2 \right)}{3}$
Hint: The parametric form of the equation of straight line, $\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r$ is used in this question.
The line given in the question is $y-\sqrt{3}x+3=0$ and the equation of the parabola is given as ${{y}^{2}}=x+2$.
It is said in the question that the line cuts the parabola at points $A$ and $B$. A point $P$ with coordinates $\left( \sqrt{3},0 \right)$ is also given. So, we can plot the graph with all the details as shown below,
We need to find the values of $PA$ and $PB$ to solve the question. $PA$ and $PB$ represent the distance of the line joined by the points $P,A$ and $P,B$ respectively.
The parametric form of a straight line passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ and making an angle of $\theta$ with the positive direction of the x-axis is given by,
$\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }=r\ldots \ldots \ldots (i)$
where $r$ is the distance between the two points with coordinates $\left( x,y \right)$ and $\left( {{x}_{1}},{{y}_{1}} \right)$.
In this question, $PA$ and $PB$ are equivalent to the distance $r$ mentioned above.
We have the coordinates of the point $\left( {{x}_{1}},{{y}_{1}} \right)=\left( \sqrt{3},0 \right)$. So, the first step is to find the angle $\theta$.
For that, we need to consider the equation of the line and rearrange it in the slope-intercept form,
$y-\sqrt{3}x+3=0$
$y=\sqrt{3}x-3$
Comparing it with the slope-intercept form given by $y=mx+c$, we get the slope as $m=\sqrt{3}=\tan \theta$.
The angle $\theta$ can hence be computed as below,
\begin{align} & \tan \theta =\sqrt{3} \\ & \theta ={{\tan }^{-1}}\sqrt{3} \\ & \theta =60{}^\circ \\ \end{align}
Now substituting the obtained results in the equation $(i)$,
$\dfrac{x-\sqrt{3}}{\cos 60{}^\circ }=\dfrac{y-0}{\sin 60{}^\circ }=r$
Substituting the values of $\cos 60{}^\circ =\dfrac{1}{2},\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$, we get,
$\dfrac{x-\sqrt{3}}{\dfrac{1}{2}}=\dfrac{y-0}{\dfrac{\sqrt{3}}{2}}=r$
Equating both $x$ and $y$ to $r$,
$\dfrac{x-\sqrt{3}}{\dfrac{1}{2}}=r,\dfrac{y-0}{\dfrac{\sqrt{3}}{2}}=r$
\begin{align} & x-\sqrt{3}=\dfrac{r}{2},y=\dfrac{\sqrt{3}}{2}r \\ & x=\sqrt{3}+\dfrac{r}{2},y=\dfrac{\sqrt{3}}{2}r \\ \end{align}
So, we have the coordinates of the point, $A$ or $B$ as $\left( \sqrt{3}+\dfrac{r}{2},\dfrac{\sqrt{3}}{2}r \right)$.
As per the question, we know that this point cuts the parabola ${{y}^{2}}=x+2$, so it can be substituted in the equation for the parabola as,
\begin{align} & {{\left( \dfrac{\sqrt{3}}{2}r \right)}^{2}}=\left( \sqrt{3}+\dfrac{r}{2} \right)+2 \\ & \dfrac{3{{r}^{2}}}{4}=\dfrac{r}{2}+\left( \sqrt{3}+2 \right) \\ & \dfrac{3{{r}^{2}}}{4}-\dfrac{r}{2}-\left( \sqrt{3}+2 \right)=0 \\ \end{align}
This is in the form of a quadratic equation in $r$, so we can get the values of $r$ using the formula as,
\begin{align} & r=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & r=\dfrac{-\left( -\dfrac{1}{2} \right)\pm \sqrt{{{\left( -\dfrac{1}{2} \right)}^{2}}-4\times \left( \dfrac{3}{4} \right)\times -\left( \sqrt{3}+2 \right)}}{2\left( \dfrac{3}{4} \right)} \\ & r=\dfrac{\dfrac{1}{2}\pm \sqrt{\dfrac{1}{4}+\left( \dfrac{12}{4} \right)\left( \sqrt{3}+2 \right)}}{\left( \dfrac{3}{2} \right)} \\ \end{align}
Taking $\dfrac{1}{4}$ outside from the root,
\begin{align} & r=\dfrac{\dfrac{1}{2}\pm \sqrt{\dfrac{1}{4}\left( 1+12\left( \sqrt{3}+2 \right) \right)}}{\left( \dfrac{3}{2} \right)} \\ & r=\dfrac{\dfrac{1}{2}\pm \dfrac{1}{2}\sqrt{1+12\left( \sqrt{3}+2 \right)}}{\left( \dfrac{3}{2} \right)} \\ \end{align}
Cancelling out $\dfrac{1}{2}$,
$r=\dfrac{1\pm \sqrt{1+12\left( \sqrt{3}+2 \right)}}{3}$
Therefore, we have the roots as,
\begin{align} & PA=\dfrac{1+\sqrt{1+12\left( \sqrt{3}+2 \right)}}{3} \\ & PB=\dfrac{1-\sqrt{1+12\left( \sqrt{3}+2 \right)}}{3} \\ \end{align}
Now, we can compute $PA.PB$ as,
$PA.PB=\dfrac{1+\sqrt{1+12\left( \sqrt{3}+2 \right)}}{3}\times \dfrac{1-\sqrt{1+12\left( \sqrt{3}+2 \right)}}{3}$
Applying the identity $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$,
\begin{align} & PA.PB=\dfrac{1}{9}\times \left[ {{1}^{2}}-{{\sqrt{1+12\left( \sqrt{3}+2 \right)}}^{2}} \right] \\ & PA.PB=\dfrac{1}{9}\times \left[ 1-1-12\left( \sqrt{3}+2 \right) \right] \\ & PA.PB=\dfrac{1}{9}\times \left[ -12\left( \sqrt{3}+2 \right) \right] \\ & PA.PB=\dfrac{-12}{9}\times \left( \sqrt{3}+2 \right) \\ & PA.PB=\dfrac{-4}{3}\times \left( \sqrt{3}+2 \right) \\ \end{align}
We have to consider the modulus for the distance, so we get the value of $PA.PB$ as $\left| \dfrac{-4\left( \sqrt{3}+2 \right)}{3} \right|=\dfrac{4\left( \sqrt{3}+2 \right)}{3}$.
Hence option (a) is the correct answer.
Note: The last portion of the solution can be done more easily by using the fact that the product of the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ can be obtained as $\dfrac{c}{a}$. So, for this question,$\dfrac{3{{r}^{2}}}{4}-\dfrac{r}{2}-\left( \sqrt{3}+2 \right)=0$, the product of the roots $PA$ and $PB$ can be obtained as \begin{align} & PA.PB=\dfrac{-\left( \sqrt{3}+2 \right)}{\dfrac{3}{4}} \\ & PA.PB=\dfrac{-4\left( \sqrt{3}+2 \right)}{3} \\ & PA.PB=\left| \dfrac{-4\left( \sqrt{3}+2 \right)}{3} \right|\Rightarrow \dfrac{4\left( \sqrt{3}+2 \right)}{3} \\ \end{align} |
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7.8 Exponential Models with Differential Equations
2 min readfebruary 15, 2024
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Think of a video going viral and how quickly the number of views it gets increases over time. Or think about the last time who heard a rumor and how fast it spread around. These are real-world scenarios which can be explained through exponential models, a key concept in calculus that helps us understand things that grow or shrink really fast.
Image Courtesy of Desmos
🤔 What is a Differential Equation?
At the heart of exponential models are differential equations.
Imagine these equations as a mystery where you know how fast something is changing (like the speed of a rumor spreading), and you're trying to figure out the entire story (how many people will hear it over time). In calculus, these equations help us map out scenarios of rapid growth or decline.
The following equation is your go-to formula here.
$dy/dt = ky$
Rest assured, it's less complex than it looks.
• $dy/dt$ represents the rate of change of something over time, like the number of people watching a video or hearing a rumor.
• The $k$ in the equation is a constant that gives us the rate and nature of this change.
• A positive $k$ means things are growing or spreading (like more people watching the video).
• A negative $k$ means things are declining or fading away.
✍🏾 Solving the Differential Equation
When the differential equation is solved, it gives us an important equation. Solve the steps below to find out what it is!
1. Start with the Differential Equation: We begin with the equation $dy/dt = ky$, as mentioned above.
2. Separate the Variables: To solve this equation, we first separate the variables $y$ and $t$ . We rearrange the equation to get all the $y$ terms on one side and the $t$ terms on the other: $(1/y)(dy) = k\ dt$ .
3. Integrate Both Sides: Next, we integrate both sides of the equation. The integral of $1/y \; dy$ is $ln |y|$, and the integral of $k\ dt$ is $kt$. After integrating, we have: $ln|y| = \; kt+C$ where $C$ is the constant of integration.
4. Solve for $y$: To make this equation more manageable, we exponentiate both sides to get rid of the natural logarithm. This gives us: $e^{ln|y|} = e^{kt+C}$. After simplifying, we get: $|y| = e^C \cdot e^{kt}$.
5. Determine the Constant $C$: The constant $C$ can be determined using the initial condition of the problem. Suppose we know the value of $y$ at $t = 0$, which we call $y_0$. Plugging these values into our equation gives us: $|y_0| = e^C \cdot e^0$. After we simplify this, we find that $e^C = |y_0|$. Therefore, our equation becomes: $|y| = |y_o| * e^{kt}$.
6. Final Exponential Model: Assuming y is always positive, we can remove the absolute value signs. Our final model is: $y = y_o * e^{kt}$.
This is the exponential model that predicts future patterns. Here, $y_0$ is the starting value, $e$ is Euler's number (important in continuous growth/decay situations), and $kt$ represents how the rate of change and time interact to evolve the initial situation.
🎯 Walking Through an Exponential Model
Let’s try a practice problem to really show off your newly learned skills.
A small town has a population of 2,000 people. Due to new job opportunities, the population is increasing at a rate proportional to its current size. After 3 years, the population has grown to 3,000 people. Assuming this trend continues, what will the population be after 10 years?
Let’s first identify the information that was given to us:
• Initial population ($y_0$): 2,000 people
• Population after 3 years: 3,000 people
• Time to reach this population: 3 years
• The goal is to find the population after 10 years.
Now we can write down the exponential growth equation and substitute any values we know.
• The population growth can be modeled by the equation $y = y_0 \cdot e^{kt}$.
• Initially, $y = 2000$ when $t = 0$.
Use the information to find $k$:
• We know that after 3 years $(t = 3)$, $y = 3000$. Plugging these into our equation gives us: $3000 = 2000 \cdot e^{3k}$
• To find k, we first divide both sides by 2000:
$\frac {3000}{2000}=e^{3k}$
$1.5=e^{3k}$
Now, we take the natural logarithm of both sides to solve for k:
$ln(1.5) = 3k$
$k = \frac {ln(1.5)}{3}$
$k = \frac {ln(1.5)}{3} \approx .13516$
Amazing! We solved for $k$. Now, we can apply the value to the problem and predict the population after 10 years.
• Use t = 10 in the equation $y = 2000 \cdot e^{kt}$, making it $y=2000\cdot e^{0.13516×10}$.
• Calculate the final value: $y = 2000 \cdot e^{1.3516}\rightarrow y \approx2000\cdot3.8637\rightarrow y\approx 7727.4$
The population of the town is expected to be approximately 7,727 people after 10 years. Great work!
✏️ Practice with Exponential Models
Here are two problems you can try on your own!
1. The rate at which the drug leaves the bloodstream is proportional to the amount in the blood stream. A dose of 200 mg of a drug is administered to a patient. After 3 hours, approximately 127.3 mg remain. The amount of the drug, in milligrams, in the person’s bloodstream after $t$ hours is given by $A(t)$. Write an equation for $A(t)$.
2. Bacteria in a certain culture increase at a rate proportional to the number present. If the number of bacteria doubles in five hours, in how many hours will the number of bacteria quadruple?
Try your best before you scroll and take a look at the way we completed them.
📌 Solution For Example 1
Let’s again identify the given, relevant information:
• We know that $y_0=200$
• We know that after three hours ($t=3$), $y=127.3$
• Our goal is to find the equation that models this situation
Write down the exponential growth equation! The change in the amount of drug can be modeled by the equation $y = y_0 \cdot e^{kt}$. Initially, $y = 200$ when $t = 0$.
Use this information to find $k$:
We know that after 3 hours $(t = 3)$, $y = 127.3$. Plugging these into our equation gives us:
$127.3 = 200 \cdot e^{3k}$
To find k, we first divide both sides by 200:
$\frac {127.3}{200}=e^{3k}$
$0.6365=e^{3k}$
Now, we take the natural logarithm of both sides to solve for k:
$ln(.6365) = 3k$
$k = \frac {ln(.6365)}{3}$
$k \approx -0.151$
Amazing! Now that we have a value for k, we can apply it to our problem. We know that $dy/dt=kA$, and that $A(0)=200$. Thus, we find that $A(t)=200\cdot e^{-0.151t}$.
📌 Solution For Example 2
This problem seems a little tricky at first glance, since we aren’t given any actual numbers. But, we are given ratios, and we can use that to make up our own numbers.
• We’ll assign the initial value of the bacteria to 100, so $y_0=100$ when $t=0$.
• That means that after 5 hours ($t=5$), $y=200$
• We want to find out how long it will take for the population to reach 400.
Writing down the exponential growth equation, we know that the population growth can be modeled by the equation $y = y_0 \cdot e^{kt}$. Initially, $y = 100$ when $t = 0$.
Time to find $k$!
We know that after 5 hours $(t = 5)$, $y = 200$. Plugging these into our equation gives us:
$200 = 100 \cdot e^{5k}$
$\frac {200}{100}=e^{5k}$
$2=e^{5k}$
Now, we take the natural logarithm of both sides to solve for k:
$ln(2)=5k$
$k = \frac {ln(2)}{5}$
$k\approx 0.139$
Almost there! We want to find out how long it will take the population to reach 400, we can plug this information in and solve for $t$:
$400=100\cdot e^{0.139t}$
$4=e^{0.139t}$
Then, take the natural log of both sides to cancel the $e$:
$ln(4)=0.139t$
Finally, just divide by 0.139 and compute the value for $t$!
$\frac{ln(4)}{0.139}=t\approx9.97 \ \text{hours}$
🥳 Conclusion
Exponential models give us a fantastic tool to understand and predict scenarios of rapid change in our daily lives, from social media trends to the spread of information. By mastering these concepts in AP Calculus, you gain not just the ability to solve math problems, but also a deeper insight into the dynamic world around you.
Key Terms to Review (2)
Euler's number
: Euler's number, denoted as "e", is a mathematical constant that represents the base of the natural logarithm. It is an irrational number approximately equal to 2.71828.
Exponential Growth Model
: An exponential growth model is a mathematical representation of a process that increases at an accelerating rate over time. It is characterized by a constant positive growth factor.
7.8 Exponential Models with Differential Equations
2 min readfebruary 15, 2024
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Think of a video going viral and how quickly the number of views it gets increases over time. Or think about the last time who heard a rumor and how fast it spread around. These are real-world scenarios which can be explained through exponential models, a key concept in calculus that helps us understand things that grow or shrink really fast.
Image Courtesy of Desmos
🤔 What is a Differential Equation?
At the heart of exponential models are differential equations.
Imagine these equations as a mystery where you know how fast something is changing (like the speed of a rumor spreading), and you're trying to figure out the entire story (how many people will hear it over time). In calculus, these equations help us map out scenarios of rapid growth or decline.
The following equation is your go-to formula here.
$dy/dt = ky$
Rest assured, it's less complex than it looks.
• $dy/dt$ represents the rate of change of something over time, like the number of people watching a video or hearing a rumor.
• The $k$ in the equation is a constant that gives us the rate and nature of this change.
• A positive $k$ means things are growing or spreading (like more people watching the video).
• A negative $k$ means things are declining or fading away.
✍🏾 Solving the Differential Equation
When the differential equation is solved, it gives us an important equation. Solve the steps below to find out what it is!
1. Start with the Differential Equation: We begin with the equation $dy/dt = ky$, as mentioned above.
2. Separate the Variables: To solve this equation, we first separate the variables $y$ and $t$ . We rearrange the equation to get all the $y$ terms on one side and the $t$ terms on the other: $(1/y)(dy) = k\ dt$ .
3. Integrate Both Sides: Next, we integrate both sides of the equation. The integral of $1/y \; dy$ is $ln |y|$, and the integral of $k\ dt$ is $kt$. After integrating, we have: $ln|y| = \; kt+C$ where $C$ is the constant of integration.
4. Solve for $y$: To make this equation more manageable, we exponentiate both sides to get rid of the natural logarithm. This gives us: $e^{ln|y|} = e^{kt+C}$. After simplifying, we get: $|y| = e^C \cdot e^{kt}$.
5. Determine the Constant $C$: The constant $C$ can be determined using the initial condition of the problem. Suppose we know the value of $y$ at $t = 0$, which we call $y_0$. Plugging these values into our equation gives us: $|y_0| = e^C \cdot e^0$. After we simplify this, we find that $e^C = |y_0|$. Therefore, our equation becomes: $|y| = |y_o| * e^{kt}$.
6. Final Exponential Model: Assuming y is always positive, we can remove the absolute value signs. Our final model is: $y = y_o * e^{kt}$.
This is the exponential model that predicts future patterns. Here, $y_0$ is the starting value, $e$ is Euler's number (important in continuous growth/decay situations), and $kt$ represents how the rate of change and time interact to evolve the initial situation.
🎯 Walking Through an Exponential Model
Let’s try a practice problem to really show off your newly learned skills.
A small town has a population of 2,000 people. Due to new job opportunities, the population is increasing at a rate proportional to its current size. After 3 years, the population has grown to 3,000 people. Assuming this trend continues, what will the population be after 10 years?
Let’s first identify the information that was given to us:
• Initial population ($y_0$): 2,000 people
• Population after 3 years: 3,000 people
• Time to reach this population: 3 years
• The goal is to find the population after 10 years.
Now we can write down the exponential growth equation and substitute any values we know.
• The population growth can be modeled by the equation $y = y_0 \cdot e^{kt}$.
• Initially, $y = 2000$ when $t = 0$.
Use the information to find $k$:
• We know that after 3 years $(t = 3)$, $y = 3000$. Plugging these into our equation gives us: $3000 = 2000 \cdot e^{3k}$
• To find k, we first divide both sides by 2000:
$\frac {3000}{2000}=e^{3k}$
$1.5=e^{3k}$
Now, we take the natural logarithm of both sides to solve for k:
$ln(1.5) = 3k$
$k = \frac {ln(1.5)}{3}$
$k = \frac {ln(1.5)}{3} \approx .13516$
Amazing! We solved for $k$. Now, we can apply the value to the problem and predict the population after 10 years.
• Use t = 10 in the equation $y = 2000 \cdot e^{kt}$, making it $y=2000\cdot e^{0.13516×10}$.
• Calculate the final value: $y = 2000 \cdot e^{1.3516}\rightarrow y \approx2000\cdot3.8637\rightarrow y\approx 7727.4$
The population of the town is expected to be approximately 7,727 people after 10 years. Great work!
✏️ Practice with Exponential Models
Here are two problems you can try on your own!
1. The rate at which the drug leaves the bloodstream is proportional to the amount in the blood stream. A dose of 200 mg of a drug is administered to a patient. After 3 hours, approximately 127.3 mg remain. The amount of the drug, in milligrams, in the person’s bloodstream after $t$ hours is given by $A(t)$. Write an equation for $A(t)$.
2. Bacteria in a certain culture increase at a rate proportional to the number present. If the number of bacteria doubles in five hours, in how many hours will the number of bacteria quadruple?
Try your best before you scroll and take a look at the way we completed them.
📌 Solution For Example 1
Let’s again identify the given, relevant information:
• We know that $y_0=200$
• We know that after three hours ($t=3$), $y=127.3$
• Our goal is to find the equation that models this situation
Write down the exponential growth equation! The change in the amount of drug can be modeled by the equation $y = y_0 \cdot e^{kt}$. Initially, $y = 200$ when $t = 0$.
Use this information to find $k$:
We know that after 3 hours $(t = 3)$, $y = 127.3$. Plugging these into our equation gives us:
$127.3 = 200 \cdot e^{3k}$
To find k, we first divide both sides by 200:
$\frac {127.3}{200}=e^{3k}$
$0.6365=e^{3k}$
Now, we take the natural logarithm of both sides to solve for k:
$ln(.6365) = 3k$
$k = \frac {ln(.6365)}{3}$
$k \approx -0.151$
Amazing! Now that we have a value for k, we can apply it to our problem. We know that $dy/dt=kA$, and that $A(0)=200$. Thus, we find that $A(t)=200\cdot e^{-0.151t}$.
📌 Solution For Example 2
This problem seems a little tricky at first glance, since we aren’t given any actual numbers. But, we are given ratios, and we can use that to make up our own numbers.
• We’ll assign the initial value of the bacteria to 100, so $y_0=100$ when $t=0$.
• That means that after 5 hours ($t=5$), $y=200$
• We want to find out how long it will take for the population to reach 400.
Writing down the exponential growth equation, we know that the population growth can be modeled by the equation $y = y_0 \cdot e^{kt}$. Initially, $y = 100$ when $t = 0$.
Time to find $k$!
We know that after 5 hours $(t = 5)$, $y = 200$. Plugging these into our equation gives us:
$200 = 100 \cdot e^{5k}$
$\frac {200}{100}=e^{5k}$
$2=e^{5k}$
Now, we take the natural logarithm of both sides to solve for k:
$ln(2)=5k$
$k = \frac {ln(2)}{5}$
$k\approx 0.139$
Almost there! We want to find out how long it will take the population to reach 400, we can plug this information in and solve for $t$:
$400=100\cdot e^{0.139t}$
$4=e^{0.139t}$
Then, take the natural log of both sides to cancel the $e$:
$ln(4)=0.139t$
Finally, just divide by 0.139 and compute the value for $t$!
$\frac{ln(4)}{0.139}=t\approx9.97 \ \text{hours}$
🥳 Conclusion
Exponential models give us a fantastic tool to understand and predict scenarios of rapid change in our daily lives, from social media trends to the spread of information. By mastering these concepts in AP Calculus, you gain not just the ability to solve math problems, but also a deeper insight into the dynamic world around you.
Key Terms to Review (2)
Euler's number
: Euler's number, denoted as "e", is a mathematical constant that represents the base of the natural logarithm. It is an irrational number approximately equal to 2.71828.
Exponential Growth Model
: An exponential growth model is a mathematical representation of a process that increases at an accelerating rate over time. It is characterized by a constant positive growth factor. |
# How do you graph x-y=2?
Apr 7, 2018
See a solution process below:
#### Explanation:
First, solve for two points which solve the equation and plot these points:
First Point: For $x = 0$
$0 - y = 2$
$- y = 2$
$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times 2$
$y = - 2$ or $\left(0 , - 2\right)$
Second Point: For $y = 0$
$x - 0 = 2$
$x = 2$ or $\left(2 , 0\right)$
We can next plot the two points on the coordinate plane:
graph{(x^2+(y+2)^2-0.035)((x-2)^2+y^2-0.035)=0 [-10, 10, -5, 5]}
Now, we can draw a straight line through the two points to graph the line:
graph{(x-y-2)(x^2+(y+2)^2-0.035)((x-2)^2+y^2-0.035)=0 [-10, 10, -5, 5]} |
The probability of an event is a number between 0 and 1 that describes the likelihood that the event will occur. Events can also be called outcomes or possibilities.
Probability can be given as a fraction (\dfrac{2}{5}), a ratio (2 in 5) or a percent (40%).
Probability
A class has 17 girls and 17 boys. If a student is called at random, what is the probability it will be a girl?
### Solution
There are 34 possible outcomes. The desired outcome is a girl being called.
Bottom: 34 outcomes are possible
Top: 17 possible desired outcomes
Probability: \dfrac{17}{34} = \dfrac{1}{2}
Or, more simply:
probability = \dfrac{desired\, outcome}{all\, possible\, outcomes}
Apply that to the sample of rolling a 5.
Bottom: 6 outcomes are possible
Top: 1 desired outcome
Probability: \dfrac{1}{6}
#### 800score Technique: The Bottom and the Top
If you want to make probability simpler, just think of it as a fraction to define Bottom then Top. The bottom number is the total number of possibilities that could happen, and the top number is the number of possible ways to achieve the desired result.
Video Courtesy of Kaplan GMAT prep.
Probability is often seen as a very difficult part of the GMAT, but by logically applying one simple fraction you can make probability much easier. Start with probability in its most basic form.
You have a fair six-sided die. If you roll it once, what is the probability you roll a 5?
There are 6 sides to the die, and each is equally likely to come up. So there are 6 possible outcomes, and rolling a 5 is the one that you want. Therefore, there is a 1 in 6 probability that the 5 will come up. (Note: die is the singular of dice.)
A fruit basket has 10 apples, 8 oranges and 6 pears. What is the probability of randomly picking an orange?
### Solution
There are 24 possible outcomes. An orange is wanted.
Bottom: 24 outcomes are possible
Top: 8 of the fruits are oranges (desired outcome)
Probability: \dfrac{8}{24} = \dfrac{1}{3}
A bag contains 3 white marbles, 2 blue marbles and 5 red marbles. What is the percent probability of drawing a white marble from the bag?
### Solution
There are a total of 10 marbles in the bag.
Bottom: 10 outcomes are possible
Top: 3 marbles are white
Probability: \dfrac{3}{10} = 30%
Example – Selecting marbles from a bag
#### Simple Probability
Best viewed in landscape mode
2 questions with video explanations
100 seconds per question
Before attempting these problems, be sure to review this section on data sufficiency questions. |
0 0
# How would you find the square root of 48 over 75?
Math hmwk. :(
Even easier: Reduce the fraction 48/75 to lowers terms first to get 16/25 which is just 42/52.
Then take the the square root to get 4/5.
Laihh - Is that √(48/75) or √48/75 ?
If √(48/75), this is equal to √48/√75 . We then look for squares within the numerator and denominator:
√48/√75 = (√16√3)/(√25√3) = (4√3)/(5√3) . Reduce the √3/√3, and you are left with 4/5.
If √48/75, follow the same steps for the numerator to get 4√3/75.
Yes this is a great answer.
Roman, reducing the fraction is smart, and leads us to the answer faster which is why we should always see if we can reduce first but not every problem with evaluating square roots are going to deal with finding the square root of a fraction.
The main principle that needs to be addressed is that we have to look for the perfect squares inside the radical.
To do this we just square a few counting numbers starting from 2 until their squares exceed the biggest number in the radical. Which in this case since we reduced the fraction inside the radical to 16/25, it would be 25.
22=4, 32=9, 42=16, 52=25. We can stop here since 25 divides the denominator perfectly, and 16 divides the numerator perfectly. This means that the fraction can now be re-written as follows:
v((42)/(52)), the squareroot undoes the squares and we are left with: 4/5.
The reason why the concept kevin expressed is so important is because you won't always be given a fraction.
Here is an example of how this method can be used.
Find v128
Like I stated before, begin by creating a list of perfect squares until you exceed the number inside the radical: 22=4, 32=9, 42=16, 52=25, 62=36, 72=49, 82=64, 92=81, 102=100, 112=121, 122=144, we stop there and exclude 12 since it exceeds 128.
Now we may begin to see which is the largest perfect square that divides 128 starting from largest to smallest: Meaning we eliminate 121, 100, and 81 until we arive at 64 which divides 128 into 2.
With that being said we can now re-write 128 as 64·2, and we can rewrite that into 82·2.
So now everything should look like this: v(82·2)
The 8 leaves the radical losing its radical, and we are left with 8v2.
i concur with the above answer |
United KingdomEngland
KeyStage 2 Upper
# Make the statement true (add/sub)
Lesson
## How can I make this true?
When we looked at whole numbers, we saw that in order to make a statement true, we need to make sure both sides of our number sentence are equal. We can think of it as a set of scales that is evenly balanced.
## Now with decimals
We can use the same process for decimal numbers. So, if each block on the scales below represents one tenth, or $0.1$0.1, we can say the scales are balanced since there are $5$5 tenths on each side or $0.5$0.5
In the first video, we look at decimals with tenths, to see how we make both sides of our number sentence equal. So, for a question like this
$3.3+4.2$3.3+4.2 = $6.1$6.1 +
we can work out what the needs to be so that both sides of our equation are equal. We do this by looking at what we are doing to one side and changing the other side. We need to look at both numbers though! We can't just add $4.2$4.2 to the right hand side in our equation above, since the left hand side starts with $3.3$3.3 but the right hand side already has $6.1$6.1.
Let's see how we can work through these to find the missing value.
## How can we use this?
To see how we can use this in real life, we're going to compare two runners.
In a handicap race, a runner's time is adjusted, to get their adjusted, or handicap, time. We look at two runners and write a statement that means their handicap times are equal. What number do we need to put in to make this true? In Video 2, we look at how to do this, using numbers with tenths and hundredths.
Remember!
To make our number statement true, we need to make sure both sides are equal. We can look at what is being done to one side, and see what we need to do to the other side.
Sometimes, there is more than one adjustment we need to make.
#### Worked Examples
##### Question 1
Complete the equation:
1. $0.53+0.26=\editable{}+0.56$0.53+0.26=+0.56
##### Question 2
Complete the equation:
1. $5.06+2.03=7+\editable{}$5.06+2.03=7+
##### Question 3
Complete the equation:
1. $4.124+2.003$4.124+2.003$=$=$\editable{}$$.$.$12$12$+$+$0.007$0.007 |
## What is arithmetic progression?
• July 5, 2021
I am a big believer in progression, the idea that a series of steps (such as the ones we learned in this video) are the first step in an infinite series of smaller steps.
So what does it mean for an infinite sequence of steps to be finite?
The answer to that question depends on what you mean by “infinite.”
But it’s important to realize that what’s meant by “finite” depends on the way we think about a progression.
In mathematical terms, we can say that there are only finite steps in a progression, and we can also say that all finite steps have a finite value.
This gives rise to the concept of the finite number of steps in an entire sequence.
There are, of course, many different ways to look at this, but here’s an example that helps to make the point: a series that goes from A to B in steps of a certain length (in this case, one hundred thousand) is called a “sequential” sequence, and a sequence that goes A, B, C, and D in steps that are more or less the same length is called an “absolute” sequence.
If we say that the steps in our sequence are finite, we are referring to a sequence of finite steps.
The sequence in which we start from A is called the “absolute sequence” because the steps from A, A, to B, and so on, are the absolute starting point for the sequence in the next sequence, the “sequentially” sequence of infinite steps.
If, on the other hand, we say the steps are finite and we are talking about the “sequence of finite” steps, we mean that the sequence of infinitely long steps in the sequence from A will never be equal to the sequence that follows it.
For example, if the absolute sequence of our sequence of one hundred steps from the beginning to the end of our current sequence has three steps, and the sequence the next time we go to the next step in the process has three hundred steps, that sequence will never equal the sequence following it.
So it’s not as simple as you might think.
It’s not that our sequence will always have a step that’s more or fewer than a step from A; the steps will always be less than a certain value.
It is that if we start out with a sequence with an infinite number of possible steps, our sequence is always finite.
Now, there are several ways that we can calculate the number of finite “steps” in an “infinity” sequence: We can start with a value that is just a small fraction of a step.
In this case the value is just the length of the sequence, which is the number in the range 0 to 1.
The steps in this case would be exactly zero.
This is a very simple calculation that only takes the length (0 to 1) of the “infination” sequence and adds it to the length in the “number of steps” we have now.
Or we can start from a value where the sequence has more than one finite step.
This would be the value where all the steps of the infinite sequence are zero, but the sequence itself would still be infinite.
We can add in the length we have already calculated for the previous sequence, then subtract that value, and finally multiply that value by the number we have calculated.
We end up with the value we had before.
So, in both cases, we have a value between 0 and 1, and that value has an infinite value.
A series that has a value of zero is called “zero-based” (meaning that its length is zero).
A series with a length of 0 is called absolute.
This means that the value of the value before it is the sequence’s value.
In other words, it has the same number of values after it as before it.
The value of an absolute sequence is the sum of all the value after it.
That is, it’s the value that you would get if you had a sequence starting with an absolute value and ending with the same value as before.
In fact, it is exactly the same as the value you would have if you started with an infinity sequence and ended with an infinitude one.
(It’s also important to note that the “value of an infinite” is not the same thing as the “length” of the series that follows.
If you start out in the infinite series and stop at the first finite step, then you will end up at the length that you had before.)
The reason for this is that, in mathematics, the term “length of a sequence” is a measurement of a series’ number of occurrences.
In the case of an infinity or a zero-based sequence, that means that we have only one occurrence.
If all the occurrences of the previous infinite sequence had the same values, then that sequence would be considered infinite.
A sequence that has more times in it than
## POLITICO: Obama administration pushes to overhaul U.S. currency rules
• July 4, 2021
POLITICO — President Barack Obama’s administration is pushing to overhaul the U.N.’s standards for regulating currencies and exchange rates, including setting benchmarks for the price of a basket of goods and services and reducing the current maximum exchange rate from around \$1.25 to around \$2.00.
But critics say the effort would undermine U.K. efforts to rein in its pound.
The administration’s proposal, a draft document obtained by POLITICO, would allow the pound to rise or fall at the discretion of the United Nations.
It also would allow nations to set their own exchange rates.
The U.G. has said it opposes the measure, which would be binding on all member states.
The proposal, obtained by Politico, would let the pound rise or down at the whim of the U
## How to do arithmetic in Bash: What you need to know
• July 3, 2021
Posted September 09, 2018 09:31:05 For more than a decade, mathematicians have been using the programming language Bash to perform basic arithmetic, but now they can do it in a way that’s just as easy to use as Python, Microsoft’s new language of choice.
The new programming language, which Microsoft is launching today, is designed to be as fast and versatile as the rest of the operating system.
But it’s also more expressive, allowing programmers to write their code in more than just math.
“Bash is a great way to get started in programming,” said Chris Roberts, a computer science graduate student at Microsoft Research in Redmond, Wash.
“It’s a very elegant language.”
Bash has an easy-to-learn syntax that’s more than 30 years old, and it’s the same for its syntax.
It’s built around the notion that a programming language can be written using any of several types of syntax, which are called syntax trees.
That allows programmers to build complex mathematical structures out of simple ones.
It all began with the introduction of the BASIC programming language in 1972.
That made it possible to write computer programs that worked on a variety of machines and platforms, and was designed to allow for the creation of games, word processors, web applications, and more.
It also was designed so that a program could easily be ported to other platforms and platforms.
The first version of Bash was written in 1982 by the MIT Computer Laboratory.
It was named after MIT mathematician Paul Erdos, and the first version included a program called Bash, which was designed with programming in mind.
“We’ve always had a love for computers, and we’ve always liked working with computers,” Roberts said.
“Bash, in that sense, was a natural choice.
And that’s been our philosophy ever since.”BASH has been around since 1991, and Microsoft has been developing it ever since.
Its initial version was released in 1996 and it was designed specifically for the BASICS language.
The programming language itself has been written by a team of about 20 people, but it has grown in size over time.
Today, Microsoft has about a dozen researchers and a small number of programmers working on the language.
Today, a programming languages language is made up of a set of commands that are followed by a set or set of expressions, and then a set, or set, of rules that specify how the language should be interpreted.
The way a language is built determines the structure of the language itself.
The BASIC language, known to most people as BASIC, was originally developed by a group of mathematicians in the 1960s.
In that context, it was also known as the C-Basic language, and for the first few years of its life, it wasn’t much of a language.
That changed with the arrival of the Unix operating system in 1979, which allowed for a more general syntax.
“It’s really easy to understand and very expressive, and in a lot of ways it has a lot more freedom,” Roberts explained.
“A lot of what BASIC is able to do is to create a simple programming language that can be used by anybody.”
The new version of BASIC has a much higher level of abstraction than previous versions, and some programmers prefer the syntax that came with the new version.
For example, there’s no need to type a bunch of commands to run a simple program.
Instead, programmers can write code in the syntax of their choice, and that syntax is called a syntax tree, or a ternary operator.
Each ternaries in the terniary operator specifies a new set of rules, and each ternarian is called an expression.
That expression can then be used to modify a function or method.
For example, in the program for the calculator, the calculator code is written as:The ternarials in that statement specify how to run the calculator function, and they are followed up by a terntary operator that specifies how to modify the calculator.
The function is modified by the terntararian, and so on, until the expression ends with a final tern, which means the program terminates.
To create a new tern and write a new rule, the compiler uses the function from the previous statement.
In this case, it calls the function in the previous function.
The compiler does this by taking the expression from the terN and replacing it with the expression for the new terN.
This makes the function function callable.
This means that the function can be called with any of the terns from the preceding function, which makes it possible for a programmer to create functions that are called with the expressions from previous functions.
“The terN can be any of a few ternars,” Roberts continued.
“In this case it’s a set tern.
The first one is called the default tern.”
In this way, a programmer can create functions in a terN that behave exactly the same as the
## What is arithmetic?
• July 3, 2021
Column crossword with crossword puzzle.
source Google ScholarSee all References This was the first of the six crosswords presented to the audience at the opening session.
The audience was invited to enter the puzzle by providing their own answer.
The participants then presented their answers to the panel.
They were asked to select the answer that best described their experience.
The final panelist was selected by the audience to be the winner.
In the next two weeks, the crossword puzzles were presented to a larger group of students, many of whom were in the third grade, and the audience was asked to guess the correct answer for each one.
This time, the puzzle was presented in English, and was also presented in Hindi and Punjabi.
The puzzle was composed of eight crosswords, one of which was an answer from the Indian group.
The crossword that had the highest number of correct guesses was presented to all the other crossword participants.
One group of parents in the audience asked a parent of a child to pick out a word in English to find.
The parent was asked if he or she thought the word should be translated into Hindi or Punjabis.
The answer of the parent was entered in a notebook and printed in the notebook.
The notebook was handed to the other parents, and they were asked if they thought that the answer should be printed in English or Hindi.
After a few seconds, the panelists had to choose their answer, and a final panel was selected.
In Hindi and Urdu, the answer was printed in Urdu and then read to the parents.
In Hindi, the mother had to guess correctly and the father had to pick correctly.
In Punjabhi, the father guessed correctly and mother guessed wrong.
As the panelist guesses became more difficult, the audience member could guess the answer.
Then the panel was asked whether or not the answer would be correct, and if they were wrong, they could correct themselves.
The answers of all the panel members were read to all of the audience members and they had to explain to the rest of the panel what they thought was wrong.
For the students, the final crossword was presented with three crosswords.
The first three crossword answers were correct.
However, the third crossword had the lowest number of incorrect guesses.
On the final day, the students had to make a mark on the board to indicate which answer they thought had the best number of answers.
They also had to show the panel how to answer the question.
During the day, they had a lot of fun and learned a lot.
Some of them went on to do well in the second and third grade.
The teacher gave them awards and made them share them with their parents.
## Why do you need a calculator?
• July 3, 2021
What are the most useful math tools for students?
How can you use math to enhance your understanding of your culture?
This week, the folks at Calculus Weekly spoke to a bunch of Calculus teachers about what they’re using on their classroom projects.
In the last few weeks, we’ve seen some really cool math apps that were developed specifically for Calculus, and there’s definitely something for every level of math student.
Here’s a look at some of our favorite ones:There’s something for everyone, even math beginners.
Here are five of the most popular Calculus apps for the first-year calculus student:What are you looking forward to most about this year’s class?
Let us know in the comments below!
## Why do Democrats keep saying Obamacare is the ‘job killer’?
• July 2, 2021
As the Supreme Court weighs whether to uphold President Donald Trump’s health care overhaul, Democrats are using an oft-repeated phrase to describe the Affordable Care Act as the “job killer.”
The Hill asked Democrats to explain the claim, and they cited a May 2017 article by the Washington Post.
The Post said the GOP plan would eliminate at least 20 million people from the health care system over the next decade.
The Hill then asked the Democrats to list specific examples of people who would lose coverage under the GOP bill, and the Democrats didn’t offer any examples.
The headline from the Post article reads, “House Democrats propose to eliminate 20 million Americans from insurance over decade.”
Democrats were quick to respond to the article.
A spokesperson for House Minority Leader Nancy Pelosi said Democrats “never said that.”
The Washington Post also ran a story on the Hill in May 2017 titled, “GOP plan eliminates 20 million from insurance coverage.”
The Post wrote, “Democrats are counting on a simple phrase to make their case: the Affordable Healthcare Act is the job killer.
That’s because the law was enacted in the wake of the Great Recession, when the nation’s health insurance market was at its worst.”
Democratic House Minority Whip Steny Hoyer said on MSNBC last week that Democrats “just want to get rid of 20 million” Americans.
He added that Republicans “want to put out a statement saying we’re going to repeal the ACA, we’re not going to make people’s lives better.”
Pelosi on Monday defended the coverage reductions in the House GOP health care bill, saying the measure is “the single best piece of legislation we’ve ever introduced.”
“It is the single best bill that has ever been introduced,” Pelosi said on “Morning Joe.”
“It’s the only one that has not been attacked by Republicans in any of its phases.”
Pelser said Democrats are trying to score political points in the health bill by saying the GOP proposal is a job killer.
“Democrats have spent the last few months hammering this idea, that the ACA is a cancer, that it’s an obstacle to economic growth, that we’ve created an economy that is more expensive, more complicated, more unstable,” Pelosi told MSNBC.
“So we’re trying to create this false narrative, that this is the kind of legislation that’s going to kill people.
And the truth is it’s going not to.”
Democrats and the White House have long pushed the notion that the health law is responsible for the health insurance industry’s financial woes.
The GOP health plan would reduce the federal deficit by \$119 billion, but the Whitehouse says it will still be a boon to the economy.
Democrats have been skeptical of the argument that the GOP health bill is going to put people back to work.
A key part of the GOP’s argument is that the plan would allow insurers to charge sicker people more and force people to buy cheaper insurance.
“I’m not going into this in a way that’s gonna say it’s a job-killer,” House Minority Minority Leader Kevin McCarthy said on ABC’s “This Week.”
“But we are saying this legislation is a disaster.
This is a crisis in health care.”
The White House has said the plan is the first step toward a tax reform plan that could generate more revenue for the government, and Republicans have also argued that the law is a key driver of the U.S. economy.
The Whitehouse on Monday pushed back on the idea that the House Republican health bill will create jobs.
“There are a number of ways the Affordable Health Care Act has helped create jobs in the private sector,” White House Press Secretary Sarah Sanders said.
“This bill will help create jobs and will help get people back into work, and I think that’s the key message that the administration is sending.”
## How to learn the arithmometric formula for numbers
• July 1, 2021
What is the basic arithmetic formula for a number?
And how do we use it to solve mathematical problems?
For more on the subject, see The Arithmometric Formula.
How to Learn the Arithmetic Formula For a number, the aritmometric formula is used to determine how many times a number is written on a piece of paper.
Arithmetic is a branch of mathematics that deals with the mathematical relationships between things.
For example, we can determine how long it takes to turn a coin from one face to the other.
Arithmatics deals with mathematical relationships, such as how many turns a circle has to go through before the coin gets to the next face, or how long a piece or a coin has to turn before it is spun again.
In this article, we will examine how to use the arimetric formula to solve arithmetic problems.
What is the arithemometric equation?
arithemetical equation for a positive integer A number is an aritmetical formula if and only if the arisete equation, or equation, for the number equals arithmatic.
This means that the equation for the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11 is the same as the equation that describes a number such as 1, 3 and 5.
For example, suppose we have two numbers, 5 and 7, that are both positive integers.
How many times must each of them be written on the same piece of piece of wax?
The answer is that we can write one of them twice and then write the other twice to get the same answer.
This is called a positive arithmetic equation.
The negative arithemic equation, however, is not equivalent to the negative arithmic equation.
In other words, we need a different equation for each number.
For instance, say we have 2 numbers, 4 and 5, that each have a value of 3.
When we write the equation to get an answer to the question “How many times can I write 3 on one piece of two-sided wax?” we get the equation 4 times and the answer 5 times.
But when we write it to get 5 times on the other side of the wax, we get 4 times, which is the equation 5 times and gives the answer 9 times.
In other words: 4 times 9 times 5 times 9 1 1 9 1 10 1 Now let’s say we want to calculate how many of each of the numbers we write on a two-sided wax piece of six waxed paper.
This equation is the negative arithmetic equation, which means that we need to write the number on the wax one more time to get a correct answer.
Arithmetic problems are usually problems involving the use of mathematical formulas to solve problems.
In addition, many problems have mathematical solutions that are also mathematical.
For a list of all the mathematical problems that can be solved by the arithmetic formula arithemetic, see the arisalmetical page.
Why are there two numbers written on two-faced wax?
Arithemetic problems have two problems.
One is the problem of finding a solution to the number 2 written on wax paper.
We can solve the problem by counting the number of times the number is put on the paper.
Another problem is finding a way to write 6 on waxed wax paper that has an answer of 9.
Arithemetics is a type of mathematics.
In it, mathematical equations are used to solve other mathematical problems.
The arithematics page lists a number of arithmusics problems. |
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# Sets and Symbols
## Whole numbers, integers, rational numbers, and irrational numbers
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Classify Real Numbers
Have you ever thought about circles? What about types of numbers? Take a look at this dilemma.
In the front of Kenneth Graham Middle School there is a flag with a circular garden beneath it. The students in Mr. Kennedy’s homeroom decided that this circular garden would be their community service project. The students elected Candice the leader of the project and she got right to work organizing the decorating. She asked for a group of students to plant flowers and rake the leaves left from last autumn. It was a perfect spring project.
“We need more dirt,” Sam said soon after the clean-up had begun.
“I think so too,” said Kyle.
Candice went out to assess the situation. The rain and snow of the winter and early spring had left the ground sparse. There definitely was not enough dirt to plant in. Candice began to figure out the area of the circular garden.
She knew that the formula for area is A=πr2\begin{align*}A = \pi r^2\end{align*}. The diameter of the garden is 16 feet.
That is as far as Candice got. She couldn’t remember the next step. This is where you come in. Using irrational numbers is necessary to solve this problem. But first, you should understand what we mean when we say “irrational number”.
### Guidance
There are many different ways to classify or name numbers.
All numbers are considered real numbers.
When you were in the lower grades, you worked with whole numbers. Whole numbers are counting numbers. We consider whole numbers as the set of numbers {0,1,2,3,4}\begin{align*}\{0, 1, 2, 3, 4 \ldots \}\end{align*}.
In middle school, you may also have learned about integers. The set of integers includes whole numbers, but also includes their opposites. Therefore, we can say that whole positive and negative numbers are part of the set of integers {2,1,0,1,2,3}\begin{align*}\{ \ldots -2, -1, 0, 1, 2, 3 \ldots \}\end{align*}.
We can’t stop classifying numbers with whole numbers and integers because sometimes we can measure a part of a whole or a whole with parts. These numbers are called rational numbers. A rational number is any number that can be written as a fraction where the numerator or the denominator is not equal to zero. Let’s think about this. A whole number or an integer could also be a rational number because we can put it over 1.
-4 could be written as 41\begin{align*}-\frac{4}{1}\end{align*}, therefore it is an integer, but also a rational number.
Exactly. We can also think about decimals too. Many decimals can be written as fractions, so decimals are also rational numbers.
There are two special types of decimals that are considered rational numbers and one kind of decimal that is NOT a rational number. A terminating decimal is a decimal that is considered to be a rational number. A terminating decimal is a decimal that looks like it goes on and on, but at some point has an end. It terminates or ends somewhere.
.3456798
This is a terminating decimal. It goes on for a while, but then ends.
A repeating decimal is also considered a rational number. A repeating decimal has values that repeat forever.
.676767679...
This is a repeating decimal.
Ah ha! This is the last type of number that is a decimal, but is NOT a rational number. It is called an irrational number. An irrational number is a decimal that does not end and has no repetition. It goes on and on and on. Irrational numbers cannot be represented as fractions. The most famous irrational number is pi (π)\begin{align*}(\pi)\end{align*}. We use 3.14 to represent π\begin{align*}\pi\end{align*}, but you should know that pi\begin{align*}pi\end{align*} is an irrational number meaning that it goes on and on and on forever.
How can we determine if a fraction or a decimal is rational or irrational?
If a number can be written in fraction form then it is rational. If a number cannot be written in fraction form then it is irrational. Besides π\begin{align*}\pi\end{align*}, roots of many numbers are also examples of irrational numbers. For example, 2\begin{align*}\sqrt{2}\end{align*} and 3\begin{align*}\sqrt{3}\end{align*} are both irrational numbers.
Write each of these definitions and one example of each in your notebook.
Now that you have had some practice identifying real numbers, you can compare them.
You have used number lines to compare numbers before. They can be extremely helpful in comparing the values of different numbers, including irrational numbers. The best strategy is to convert each individual value to a decimal. In the case of irrational numbers, you will have to round them to a reasonable place value. Once the numbers are decimals, you can easily compare them on a number line. Remember when you find the solutions to these types of problems that after you order the values, you should convert them back into their original form.
Place the following values on a number line: 3.2,2,2.3¯,9\begin{align*}-3.2, \sqrt{2}, 2.\overline{3}, \sqrt{9}\end{align*}.
First find the decimal values of each number.
The number -3.2 is already a decimal.
The number 2\begin{align*}\sqrt{2}\end{align*} is an irrational number. Its decimal, rounded to the nearest thousandth, is 1.414.
The number 2.3¯\begin{align*}2.\overline{3}\end{align*} is a rational number because it is a repeating decimal. It is equivalent to the fraction 73\begin{align*}\frac{7}{3}\end{align*}.
The number 9\begin{align*}\sqrt{9}\end{align*} simplifies to 3, since 32\begin{align*}3^2\end{align*} is equal to 9.
Then you can place these values on a number line.
This may seem tricky, but if you think about the approximate decimal value of each number then it becomes easier.
Classify each real number.
#### Example A
7\begin{align*}\sqrt{7}\end{align*}
Solution: Irrational number
#### Example B
19\begin{align*}\frac{1}{9}\end{align*}
Solution: Rational number
#### Example C
98\begin{align*}-98\end{align*}
Solution: Integer and rational number
Now let's go back to the dilemma from the beginning of the Concept.
First, let’s take the measurement for the diameter and figure out the measurement of the radius. The radius is one-half of the diameter.
Now we can substitute this into the formula and solve. We can use 3.14 as an approximation for π\begin{align*}\pi\end{align*} in order to get the approximate area.
AAA=πr2=(3.14)(82)=200.96 sq.feet
### Vocabulary
Whole Numbers
the set of positive counting numbers.
Integers
the set of whole numbers and their opposites.
Rational Numbers
any number that can be written in fraction form including terminating and repeating decimals.
Irrational Numbers
any number that cannot be written in fraction form. These are numbers that do not have an end point or repetition when written in decimal form – the decimal values continue indefinitely.
Pi
π\begin{align*}\pi\end{align*}, the ratio of the diameter to the circumference of a circle. We use 3.14 to approximate this irrational number.
Real Numbers
the set of rational and irrational numbers make up the set of real numbers.
### Guided Practice
Here is one for you to try on your own.
Is 234\begin{align*}\frac{23}{4}\end{align*} rational or irrational?
Solution
Because the number is written as a fraction, we know it is a rational number.
### Practice
Directions: Classify each of the following numbers as real, whole, integer, rational or irrational. Some numbers will have more than one classification.
1. 3.45
2. -9
3. 1,270
4. 1.232323
5. 45\begin{align*}\frac{4}{5}\end{align*}
6. -232,323
7. -98
8. 1.98
9. 16\begin{align*}\sqrt{16}\end{align*}
10. 2\begin{align*}\sqrt{2}\end{align*}
Directions: Answer each question as true or false.
1. An irrational number can also be a real number.
2. An irrational number is a real number and an integer.
3. A whole number is also an integer.
4. A decimal is considered a real number and a rational number.
5. A negative decimal can still be considered an integer.
6. An irrational number is a terminating decimal.
7. A radical is always an irrational number.
8. Negative whole numbers are integers and are also rational numbers.
9. Pi is an example of an irrational number.
10. A repeating decimal is also a rational number.
### Vocabulary Language: English
inequality
inequality
An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
Integer
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Irrational Number
Irrational Number
An irrational number is a number that can not be expressed exactly as the quotient of two integers.
Pi
Pi
$\pi$ (Pi) is the ratio of the circumference of a circle to its diameter. It is an irrational number that is approximately equal to 3.14.
rational number
rational number
A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero.
Real Number
Real Number
A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers.
Whole Numbers
Whole Numbers
The whole numbers are all positive counting numbers and zero. The whole numbers are 0, 1, 2, 3, ... |
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# Find the angle subtended at the center of a circle of radius ‘a’ by an arc of length $\left( \dfrac{a\pi }{4} \right)$cm.
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Hint:.The arc is any portion of the circumference of a circle. Arc length is the distance from one endpoint of the arc to the other point. Use an equation to find arc length $=2\pi r\left( \dfrac{\theta }{360} \right)$.
Given that the radius of the circle = a
We need to find the angle subtended by arc length $\left( \dfrac{a\pi }{4} \right)$
From the figure it is clear that the length of arc is$\Rightarrow \dfrac{a\pi }{4}$
We need to find $\theta$.
The formula for finding the arc length is given by
$\Rightarrow$arc length$=2\pi r\left( \dfrac{\theta }{360} \right)$
We can find the arc length or portion of the arc in the circumference, if we know at what portion of 360 degrees the arc’s central angle is.
Arc length$=2\pi r\left( \dfrac{\theta }{360} \right)$, where r is the radius of the circle.
$\therefore 2\pi r\left( \dfrac{\theta }{360} \right)=\dfrac{a\pi }{4}$
We have been given arc length$=\dfrac{a\pi }{4}$
$\Rightarrow 2\pi a\left( \dfrac{\theta }{360} \right)=\dfrac{a\pi }{4}$
\begin{align} & 2\pi a\left( \dfrac{\theta }{360} \right)=\dfrac{a}{4}\pi \\ & \Rightarrow \dfrac{\theta }{360}=\dfrac{1}{8}\Rightarrow \theta =\dfrac{360}{8}={{45}^{\circ }} \\ \end{align}
$\therefore$We get the angle subtended at the center of the circle$={{45}^{\circ }}$
Note: Here, arc length $=\dfrac{a\pi }{4}$
If we assume value of $\theta ={{45}^{\circ }}$and applying we get
\begin{align} & =2\pi a\left( \dfrac{45}{360} \right) \\ & =2\pi a\left( \dfrac{1}{8} \right)=\dfrac{a\pi }{4} \\ \end{align} |
# Practice Questions with Solutions: Addition and Subtraction Notes | Study Mathematics for Class 4 - Class 4
## Class 4: Practice Questions with Solutions: Addition and Subtraction Notes | Study Mathematics for Class 4 - Class 4
The document Practice Questions with Solutions: Addition and Subtraction Notes | Study Mathematics for Class 4 - Class 4 is a part of the Class 4 Course Mathematics for Class 4.
All you need of Class 4 at this link: Class 4
Question 1: Add 4,032 and 1, 74 6.
Solution :
Step 1. Arrange the numbers in columns.
Step 2. Adding ones, we have
Step 3. Adding tens, we have
Step 4. Adding hundreds, we have
Step 5. Adding thousands, we have
Hence, 4,032 + 1, 74 6 = 5, 778.
(Five thousand, seven hundred seventy-eight).
Question 2: Add 5,386 and 3,927.
Solution:
Step 1. Arrange the numbers in columns.
Step 2. Adding ones, we have 6 ones + 7 ones = 13 ones = 1 ten + 3 ones. Carry over 1 to tens column and write 3 under ones column.
Step 3. Adding tens, we have
1 ten (carried) + 8 tens + 2 tens = 11 tens = 1 hundred + 1 ten.
Carry over 1 to hundreds column and write 1 under tens column.
Step 4. Adding hundreds, we have 1 hundred (carried) + 3 hundreds + 9 hundreds = 13 hundreds = 1 thousand + 3 hundreds. Carry over 1 to thousands column and write 3 under hundreds column.
Step 5. Adding thousands, we have 1 thousand (carried) + 5 thousands + 3 thousands = 9 thousands.
Hence,
5,386 + 3,927 = 9,313
(Nine thousand, three hundred thirteen).
Question 3: The population of town A is 2,301, the population of town B is 4,875, the population of town C is 1, 835. What is the total population of the three towns?
Solution :
Population of town A = 2,301
Population of town B = 4,875
Population of town C = 1,835
Total population of the three towns
= 2,301 + 4,875 + 1, 835 = 9,011.
Hence, the total population of the three towns = 9,011.
Question 4: Add 7,521; 1 ,025 and 1 ,352.
Solution :
Step 1. Arrange the numbers in columns.
Step 2. Adding ones, we have
Step 3. Adding tens, we have
Step 4. Adding hundreds, we have
Step 5. Adding thousands, we have
Hence, 7,521 + 1,025 + 1,352 = 9,898
(nine thousand, eight hundred ninety-eight).
Question 5: Add 2,908, 6,123 and 254
Solution :
Step 1.
Step 2.
Step 3.
Step 4.
Step 5.
Hence,
2,908 + 6,123 + 254 = 9,285
(Nine thousand, two hundred eighty-five).
Question 6: Which number is 528 more than 7,874?
Solution :
The required number = 7,874 + 528 = 8,402. Hence, the required number is 8,402.
Question 7: Simplify : 7,530 + 1, 817 - 5,173
Solution :
Hence, 7,530 + 1,817- 5,173 = 4,174.
Question 8: 3,856 - 2,385 + 6,389 - 987
Solution :
Hence, 3,856 - 2,385 + 6,389 - 987 = 6,873
Question 9: Subtract 5,013 from 8,347.
Solution :
Step 1. Arrange the numbers in columns.
Step 2. Subtracting ones, we have
Step 3. Subtracting tens, we have
Step 4. Subtracting hundreds, we have
Step 5. Subtracting thousands, we have
Hence, 8,347 - 5,013 = 3,334
(Three thousand, three hundred thirty-four).
Question 10: A factory produced 3,263 mercury lamps in 2012. In 2013, it produced 4,450 mercury lamps. By how much did the factory's production increase?
Solution :
Production in 2013 = 4,450
Production in 2012 = 3,263
Increase in production = 4,450 - 3,263
= 11187.
Hence, the production increased by 1,187.
Question 11: Subtract 2,856 from 8, 153.
Solution :
Step 1. Arrange the numbers in columns.
Step 2. Subtract the ones. Here 3 < 6, so regroup 5 tens 3 ones as 4 tens and 13 ones.
Step 3. Subtract the tens. Here, 4 < 5, so, regroup 1 hundred 4 tens as 0 hundreds and 14 tens.
Step 4. Subtract the hundreds. Here, 0 < 8. so regroup 8 thousands 0 hundreds as 7 thousand and 10 hundreds.
Step 5. Subtract the thousands.
Hence, 8,153 - 2,856 = 5,297
(Five thousand, two hundred ninety-seven).
Question 12: What must be added to 4,213 to get 7,631?
Solution :
The sum of two numbers = 7,631.
One number= 4,213
The other number= 7,631 - 4,213
= 3,418
Hence, the required number = 3,418.
Question 13: Subtract: 9,000 - 6.124.
Solution:
Step 1. Arrange the numbers in columns.
Step 2. Regroup 9 thousands.
9 thousands = 8 thousands + 10 hundreds
10 hundreds = 9 hundreds + 10 tens
10 tens = 9 tens + 10 ones
Step 3. Subtract the ones.
Step 4. Subtract the tens.
Step 5. Subtract the hundreds.
Step 6. Subtract the thousands.
Hence, 9,000 - 6,124 = 2,876.
The document Practice Questions with Solutions: Addition and Subtraction Notes | Study Mathematics for Class 4 - Class 4 is a part of the Class 4 Course Mathematics for Class 4.
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## Mathematics for Class 4
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# Solving Drawing Triangles
Introduction to solving drawing triangles:
A triangle is a polygon. Triangle has three sides and three vertices. A triangle has three angles. The three angles are always added to 180°. Triangles are classified into many ways. Triangles are classified based on the angle and length of sides of the triangle Solving drawing triangle mean nothing but how to draw the triangle according to our given measurement. We can drawing the triangle based on the angle and side length of the triangle.
Types of triangle:
1. Right triangle
2. Obtuse triangle
3. Acute triangle
4. Scalene triangle
5. Isosceles triangle
6. Equilateral. Triangle
## Basic concepts on solving drawing triangles:
concepts for solving triangle:
Find the area of triangle:
Area = 1/2 b*h b= base h=height
Find the perimeter of triangle:
Perimeter = a + b + c
Concepts for drawing the triangles:
Example for solving drawing right triangle:
Step i:
Using scale ruler to draw the given appropriate length of line segment.(8cm)
Step ii:
After draw the line segment using protractor measure the 90 Degree of angle at end of the line segment.(90 degree)
Step iii:
Measure the angle of vertex, Draw the new line segment on the 90 degree angle of line.(5cm)
Step iv:
And then connect the first line segment one end with another line segment end,now we got the right triangle.
## Example on solving drawing triangles:
solving drawing triangles:
Draw the triangle using the following measurement?
A person can drive a car at 8km distance from his home and term upto 60 degree and move towards 5km and again he turned and reaches the starting point. Make the triangle for given dada?
Solution:
Procedure for solving drawing triangle according th our given data:
(1)First we have to draw a horizontal line length of 8km
(2)Using the protractor mark the angle of 60 degree angle on the horizontal line or end of the horizontal line.
(3)using compass draw an arc at 5km on the angle of 60 degree line and name this as z.
(4)Joint the starting point and the point z. |
## Recursion Vs Dynamic programming
While dealing with various problems, we might come up with different types of solutions for the same problem. Although, the objective of all the solutions might be same, they may require different amount of time while serving their purpose.
Here we’ll be going through two such techniques of problem solving, applicable on various problems, which will intend to provide the same answer but may differ in the time taken to reach the solution.
Recursion
Recursive function: A function calling itself by passing required set of parameters while executing the code is called as a recursive function. The process of a function calling itself for further execution of the code is called as Recursion.
In most of the cases where the problem can be divided in optimal subproblems (i.e. optimal substructure problem), we can use recursion to find the solution.
While solving various problems with optimal-substructure using recursion, we might find the recursive function computing the solution for same sub-problem more than once which is considered to be inefficient in most of the cases.
Let’s understand it by finding the Nth element of the Fibonacci sequence:
Fibonacci Sequence: It is a sequence where the first two elements are 0 and 1 and the subsequent numbers are the sum of previous two numbers.
Example : 0, 1, 1, 2, 3, 5, 8, 13,…
As we know that barring first two positions, values at all other positions are the sum of the values at previous two locations
i.e. fib(n)=fib(n-1)+fib(n-2)
now, similarly
fib(n-1)=fib(n-2)+fib(n-3)
fib(n-2)=fib(n-3)+fib(n-4)
and so on.
Now we will be creating a function to recursively carry out the above process
Following is the code snippet of that function
Time Complexity: O(2^n)
Let’s check what is exactly happening here through the following structure of recursion in n=5 case.
Before moving to our next section, we should have a short glimpse over the above recursion structure. Here we can observe that we have calculated Fibonacci value of some numbers at multiple occasions which leads to increase in number of operations significantly. This is the issue that we will try to address using Dynamic programming.
Dynamic Programming
To optimize the above recursive solution we can either,
Modify the recursive solution so that it would not calculate a subproblem more than once
OR
Find an optimized approach without recursion.
The solution of above two approaches could be acheived by using Memoization and Tabulation techniques of dynamic programming.
Memoization (Top-Down approach)
It optimizes the recursive solution by storing the solution of every subproblem in a sequence so that when the need arises to find the same solution again, it can be done by taking it from that sequence rather than calculating it again.
Here is the code snippet after optimizing the initial recursive approach.
Time Complexity: O(n)
The effect of this optimization can be seen through the structure of recursion.
Why is Memoization called as the Top-Down approach?
As in the current example, we first intended to calculate the Fibonacci value of nth number and as per the requirement we moved on to the smaller subproblems i.e. we moved from larger subproblems to the smaller subproblems and subsequently reached the solution. Hence, Memoization is also referred as the Top-Down approach;
Tabulation (Bottom-Up approach)
In this approach we will reach the solution to the main problem by calculating solution of all its subproblems starting from the smallest subproblem without using recursion.
As in our current example of Fibonacci sequence, rather than calling function recursively and getting solution of every subproblem, we can just keep on calculating sum of previous two numbers up-to n.
Here is the code snippet of the tabulation approach of our example:
Time Complexity: O(n)
Why is tabulation called as the Bottom-up approach?
Considering the current example, in tabulation approach, we started computing values of subproblems from 0th position to the nth position in other words from smallest subproblem to the largest subproblem. Hence, tabulation is termed as Bottom-up approach.
While comparing general recursive solution with Dynamic programming solutions, we can clearly see that optimization of recursive solution using DP significantly affects the performance of our code.
For applying Both of the above approach you should consider solving:
1. Longest increasing Subsequence problem
2. Longest common subsequence problem
Happy Coding 😊
By Programmers Army |
# Negative Numbers Arithmetic: Adding Negatives Explained
January 19, 2024 by JoyAnswer.org, Category : Mathematics
What does a negative plus a negative number equal? Understand the addition of negative numbers. This article explains the concept of adding negatives, providing clarity on the rules and outcomes in mathematical operations.
## What does a negative plus a negative number equal?
When you add two negative numbers together, the result is another negative number. In arithmetic, the addition of negatives follows a specific rule:
If you have two negative numbers, when you add them, you sum their magnitudes (absolute values) and give the result the negative sign.
Mathematically, if you have `-a + (-b)`, the result is `- (a + b)`.
Here's an example:
• If you add `-3` and `-5`, you first add their magnitudes (ignoring the signs): `3 + 5 = 8`.
• Then, you assign the negative sign to the result: `-8`.
So, `-3 + (-5) = -8`.
In summary, adding two negative numbers results in a negative number, and you simply add their magnitudes to find the magnitude of the result. The negative sign is then applied to the sum of magnitudes.
## Navigating Arithmetic: The Outcome of Adding Negative Numbers
When adding negative numbers, it's crucial to remember that they represent values less than zero. Here's what happens when you combine them:
## What happens when you add two negative numbers together?
• The result is always a negative number.
• The sum's absolute value (its distance from zero) is the combined absolute values of the original numbers.
For example:
• (-5) + (-3) = -8
• (-10) + (-2) = -12
## Rules and Conventions in Adding Negative Numerals
- Adding a negative number is like moving to the left on the number line.- Adding two negative numbers means moving even further to the left, resulting in a larger negative number.
Key Rules:
1. Adding Two Negative Numbers: Results in a negative number greater in magnitude than either original number.
2. Adding a Negative Number to a Positive Number:
• Sum is smaller than the positive number.
• Sign of the sum depends on the magnitudes of the numbers.
3. Triangle Inequality: The absolute value of the sum is always less than or equal to the sum of the absolute values of the original numbers.
Example:
• 5 + (-3) = 2 (positive sum, as 5 is larger than 3)
• 2 + (-5) = -3 (negative sum, as 5 is larger than 2)
Visualizing on a Number Line:
• Imagine a number line with 0 in the center, positive numbers to the right, and negative numbers to the left.
• Adding a negative number is like moving to the left.
• Adding two negative numbers means taking two leftward steps, resulting in a position further to the left (a larger negative number).
Tags Negative Numbers , Arithmetic
• ### Does a negative times a negative equal a positive?
Two negatives make a positive, so a negative number times a negative number makes a positive number. If you look at it on the number line, walking backwards while facing in the negative direction, you move in the positive direction.
Understand the concept of multiplying negative numbers and why a negative times a negative equals a positive. Explore the mathematical reasoning behind this rule. ...Continue reading
• ### Why does a negative times a negative make sense?
Two negatives make a positive, so a negative number times a negative number makes a positive number. If you look at it on the number line, walking backwards while facing in the negative direction, you move in the positive direction. For example. -2 x -4 are both negative, so we know the answer is going to be positive.
Dive into the logic that supports the rule of negative times negative equals positive. Discover the conceptual framework that makes this mathematical operation reasonable. ...Continue reading
• ### Why does negative times a negative equal a positive?
Think of multiplying by a negative as an instruction to turn around and point in the opposite direction. If you turn around twice you are pointing back in the original direction. There are two ways you may think of the negative number -1 that will make it intuitive that negative times negative equals positive.
Explore why negative times negative equals a positive result in mathematics. This article provides an explanation of the rules governing the multiplication of negative numbers. ...Continue reading |
# Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 12.6 OR and AND Problems.
## Presentation on theme: "Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 12.6 OR and AND Problems."— Presentation transcript:
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 12.6 OR and AND Problems
Copyright 2013, 2010, 2007, Pearson, Education, Inc. What You Will Learn Compound Probability OR Problems AND Problems Independent Events 12.6-2
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Compound Probability In this section, we learn how to solve compound probability problems that contain the words and or or without constructing a sample space. 12.6-3
Copyright 2013, 2010, 2007, Pearson, Education, Inc. OR Probability The or probability problem requires obtaining a “successful” outcome for at least one of the given events. 12.6-4
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Probability of A or B To determine the probability of A or B, use the following formula. 12.6-5
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 1: Using the Addition Formula Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a hat, and one piece is randomly selected. Determine the probability that the piece of paper selected contains an even number or a number greater than 6. 12.6-6
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 1: Using the Addition Formula Solution Draw a Venn Diagram 12.6-7
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 1: Using the Addition Formula Solution The seven numbers that are even or greater than 6 are 2, 4, 6, 7, 8, 9, and 10. 12.6-8
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Mutually Exclusive Two events A and B are mutually exclusive if it is impossible for both events to occur simultaneously. If two events are mutually exclusive, then the P(A and B) = 0. The addition formula simplifies to 12.6-9
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: Probability of A or B One card is selected from a standard deck of playing cards. Determine whether the following pairs of events are mutually exclusive and determine P (A or B). a) A = an ace, B = a 9 Solution Impossible to select both so 12.6- 10
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: Probability of A or B b) A = an ace, B = a heart Solution Possible to select the ace of hearts, so NOT mutually exclusive 12.6- 11
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: Probability of A or B c) A = a red card, B = a black card Solution Impossible to select both so mutually exclusive 12.6- 12
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: Probability of A or B d) A = a picture card, B = a red card Solution Possible to select a red picture card, so NOT mutually exclusive 12.6- 13
Copyright 2013, 2010, 2007, Pearson, Education, Inc. And Problems The and probability problem requires obtaining a favorable outcome in each of the given events. 12.6- 14
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Probability of A and B To determine the probability of A and B, use the following formula. 12.6- 15
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Probability of A and B Since we multiply to find P (A and B), this formula is sometimes referred to as the multiplication formula. When using the multiplication formula, we always assume that event A has occurred when calculating P(B) because we are determining the probability of obtaining a favorable outcome in both of the given events. 12.6- 16
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 5: An Experiment without Replacement Two cards are to be selected without replacement from a deck of cards. Determine the probability that two spades will be selected. 12.6- 17
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 5: An Experiment without Replacement Solution The probability of selecting a spade on the first draw is 13/52. Assuming we selected a spade on the first draw, then the probability of selecting a spade on the second draw is 12/51. 12.6- 18
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 5: An Experiment without Replacement Solution 12.6- 19
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Independent Events Event A and event B are independent events if the occurrence of either event in no way affects the probability of occurrence of the other event. Rolling dice and tossing coins are examples of independent events. 12.6- 20
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Independent or Dependent Events? One hundred people attended a charity benefit to raise money for cancer research. Three people in attendance will be selected at random without replacement, and each will be awarded one door prize. Are the events of selecting the three people who will be awarded the door prize independent or dependent events? 12.6- 21
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Independent or Dependent Events? Solution The events are dependent since each time one person is selected, it changes the probability of the next person being selected. P(person A is selected) = 1/100 If person B is actually selected, then on the second drawing, P(person A is selected) = 1/99 12.6- 22
Copyright 2013, 2010, 2007, Pearson, Education, Inc. Independent or Dependent Events? In general, in any experiment in which two or more items are selected without replacement, the events will be dependent. 12.6- 23 |
# How do you write the equation in point slope form given (1,3) and (3,-5)?
Mar 28, 2017
$y - 3 = - 4 \left(x - 1\right)$
#### Explanation:
The equation of a line in $\textcolor{b l u e}{\text{point-slope form}}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$
To find m use the $\textcolor{b l u e}{\text{gradient formula}}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ are 2 coordinate points}$
The 2 points here are (1 ,3) and (3 ,-5)
let $\left({x}_{1} , {y}_{1}\right) = \left(1 , 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(3 , - 5\right)$
$\Rightarrow m = \frac{- 5 - 3}{3 - 1} = \frac{- 8}{2} = - 4$
Use either of the 2 points for $\left({x}_{1} , {y}_{1}\right)$
$\text{Using " (x_1,y_1)=(1,3)" and } m = - 4$
$\Rightarrow y - 3 = - 4 \left(x - 1\right) \leftarrow \textcolor{red}{\text{ in point-slope form}}$ |
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# 2.1 EAN-13 code and error detection
In the EAN-13 code, the first 12 digits of the number identify the item the code is attached to, and the final digit is a ‘check digit’.
Figure 2.1 An example of an EAN-13 barcode
The check digit for an EAN-13 code is calculated as follows:
1. Count digit positions from the left to the right, starting at 1.
2. Sum all the digits in odd positions. (In the example shown in Figure 1, this is 9 + 8 + 5 + 1 + 2 + 5 = 30 – note that the final 5 is not included since this is the check digit, which is what we are currently trying to calculate.)
3. Sum all the digits in even positions and multiply the result by 3. (In the example, this is (7 + 0 + 2 + 4 + 5 + 7) × 3 = 75.)
4. Add the results of step 2 and step 3, and take just the final digit (the ‘units’ digit) of the answer. This is equivalent to taking the answer modulo-10. (In the example, the sum is 30 + 75 = 105, so the units digit is 5.)
5. If the answer to step 4 was 0, this is the check digit. Otherwise the check digit is given by ten minus the answer from step 4. (In the example, this is 10 – 5 = 5.)
6. The check digit is appended to the right of the 12 identification digits. The check digit can have any value from 0 to 9.
## Activity 2.1 Self assessment
The code below shows the first 12 digits of an EAN-13 code. Note: the hyphen between the 8 and the 0 has no bearing on the code. It is for convenience of reading, separating different elements of the identification. The 978 identifies this item as a book. (Not all EAN-13 codes have a hyphen in the same place.)
Calculate the check digit, and so derive the full EAN-13 code.
978–014102662.
Adding together the odd digits gives:
9 + 8 + 1 + 1 + 2 + 6 = 27.
Adding together the even digits and multiplying by 3 gives:
(7 + 0 + 4 + 0 + 6 + 2) × 3 = 19 × 3 = 57.
27 + 57 = 84.
The units digit is 4, so the check digit is given by:
10 − 4 = 6.
The full EAN-13 code is therefore 978–0141026626.
One way to check a received EAN-13 code for errors is to remove the received check digit and recalculate it based on the 12-digit identification code. If the recalculated value differs from the received value, there must be an error.
Alternatively, there is a shortcut to checking for errors because of the way the check digit is derived. You take the full 13-digit received code and do steps 1 to 4 from the calculation used above. If the code is correct, the value at step 4 will be 0. If the code is wrong, it will have some other value.
## Activity 2.2 Self assessment
Check whether the following codes are valid:
• a.978–0521425575
• b.978–1405322274.
• a.978–0521425575 is a 12-digit identification code 978–052142557 with a check digit of 5. Recalculating the check digit from the identification code gives 5, so the code is correct.
Alternatively, using the ‘shortcut’, we take all 13 digits and go through steps 1 to 4.
Adding together the odd digits gives:
9 + 8 + 5 + 1 + 2 + 5 + 5 = 35.
Adding together the even digits and multiplying by 3 gives:
(7 + 0 + 2 + 4 + 5 + 7) × 3 = 25 × 3 = 75.
35 + 75 = 110.
The units digit is 0, which shows that this is a valid EAN-13 code.
• b.978–1405322274 is a 12-digit identification code 978–140532227 with a check digit of 4. Recalculating the check digit from the identification code gives 0, so the code is incorrect.
Alternatively, using the ‘shortcut’, we take all 13 digits and go through steps 1 to 4. This gives 34 + 20 × 3 = 94, which results in a units digit of 4. Since this is not zero, this is not a valid EAN-13 code.
One thing to notice about EAN-13 is that the numbers were treated as a string of separate digits, not as a single number. It was, for example, 9, 7, 8, 0, 5, 2, 1, 4, 2, 5, 5, 7, not 978 052 142 557 (i.e. not nine hundred and seventy-eight billion, fifty-two million, one hundred and forty-two thousand, five hundred and fifty-seven).
In EAN-13 the digits are denary: numbers to base 10. In base 10 a digit can be any one of 10 symbols, which we represent as 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. (The word ‘symbol’ is used here in a different, though parallel, sense from the symbols of modulation schemes.)
The check digit included in the 13-digit EAN code is an example of redundancy. This is a standard term for bits appended to data for error control. Unfortunately the term ‘redundancy’ suggests that these additional bits serve no purpose, which is not true. They are redundant, though, in the sense that they are not part of the message data.
All forms of error control involve the augmentation of a message with error control bits, which could be described as adding redundancy. If the number of bits (or bytes) in the message is k, and the augmented length is n bits (or bytes), the ratio k/n is known as the code rate. This is an important parameter.
Code rate is a measure of how much redundancy has been added to the code. If lots of check digits (that is, lots of redundancy) are appended to a small number of message digits, the code rate will be small (much less than 1). If only a few check digits are appended to a big message, the code rate will be close to 1. (It can never exceed 1.) Code rates found in, for example, mobile communications and WiFi typically range from 1/4 to 5/6.
Sometimes a code is specified by the numbers (n, k), in that order, with brackets around them and a comma between the numbers. Codes which take k message digits and create an n digit code word are described as (nk) block codes.
## Activity 2.3 Self assessment
• a.Describe an EAN-13 code using the (nk) notation
• b.What is the code rate of an EAN code? |
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