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Saturday , June 10 2023
# 10 Examples of Distance and Displacement
Distance is generally used to refer separation between two things. It has a more precise definition in physics as, the physical length of separation between two points. We use the Google maps distance many times while travelling. In physics, we need to calculate distance between various points in order to find important information. There is another term displacement that generally refers to the shortest distance between two points. Here we have given 10 examples of distance and displacement for complete understanding.
## Difference between Distance and Displacement
The SI units of distance and displacement are same, meter. But distance is a scalar quantity that refers to the path length covered by an object. Displacement is a vector quantity that refers to the difference in the initial and final position of the object.
Through the following points it will get clear to us on how is displacement different from distance.
We can find distance formula, by considering the distance covered by an object in unit time. This gives us
### Distance= Speed x Time
The displacement formula is different from that of distance.
When we differentiate displacement, the result is velocity (which refers to both speed and direction of the object).
Here, we should note that distance is always positive whereas displacement can be negative or positive.
We will see some of the examples of distance and displacement in the following section.
# 10 Examples of Distance and Displacement
(1) A runner starts at the start line and runs 100 meters on a straight lane to complete the race. Let us find the distance travelled and displacement of the runner.
The runner runs on a straight path in a single direction. Thus his displacement and distance travelled will be same, and equal to 100 meters.
(2) A car starts from home and travels 40 kilometers to a hospital and returns back 20 km to a grocery. Let us consider the distance travelled and displacement of the car.
Distance travelled in this case will be 40+20=60 kms.
Displacement = 40-20=20 kms.
(3) In the above example, consider that the car starts from home to cover 40 kms till hospital and returns back to home.
Distance travelled will be 40 kilometers till hospital plus 40 kms back to home, equaling to 80 kms.
Since the car returns back to its original position, there is zero displacement.
(4) Starting from rest, a person moves 20km west, then 30kms north from there, and turn east and moves 20kms.
Displacement will be the difference between the final position and the initial position of the person.In this case, the distance will be the total path travelled which is, 20+30+20=70.
Finally, the person will be 30 km north from its starting position. Thus, its displacement is 30km towards north.
(5) An object starts from A and reaches C. Find its distance and displacement.
Distance= AB +BC = 20 kms +20 kms = 40 kilometers.
Displacement =AC
We can see that since AB is perpendicular to BC, AC is the hypotenuse.
AC =Displacement = 28.2 km from A to C
(6) In following figure, object starts from A to B to C to D to A.
Distance of the sum AB+BC+CD+DA=20m+30m+20m+30m=100 meters
Since the object starts from A and reaches back to A, its displacement is zero.
(7) If a satellite makes one revolution about the earth. What would be the total distance travelled by it, also find the displacement.
Radius of earth=6,371 kilometers ( Assume satellite revolves around this diameter)
The distance covered by a satellite in revolution would be equal to the circumference of earth.
Thus given the radius of earth, its circumference will be = 2π x radius of earth
Distance travelled= 2π x 6,371 kilometers = 40,07 kilometers
In one revolution the satellite comes back to the place from where it started, thus its displacement will be zero.
(8) In the above question, let us consider the satellite makes one and a half revolution. Let us find the distance travelled and displacement in this case.
The distance travelled will be = 2πr +2πr/2 = 40,07 km + 20,03 km= 6010 km
The displacement will be the shortest distance between the final and initial position of the satellite. The satellite is diametrically opposite to earth, therefore the displacement will be equals to the diameter of earth.
Thus, displacement = 2 x radius of earth = 2 x 6,371 km = 12742 km.
(9) Consider a car moving at a speed of 30km/h for duration of 2 hours in a straight road. What will be the distance travelled and displacement?
Since, Distance = Speed x Time
Therefore, distance travelled by the car will be 30km/h x 2h = 60km.
Since the particle is travelling in a linear direction, its displacement will be equal to its distance travelled, equals to 60km.
(10) Now consider the car moves in a direction for 2 hours at 30km/h and returns back and travels at 20km/h for 1 hour. Let us find the distance travelled and the displacement of the car.
The distance travelled by the car will be 30km/h x 2h + 20km/h x 1h =60km +20km = 80km.
Displacement = 60km – 20km = 40km from the position of start.
### Authored by Kiran Mandia
I am a computer science student at Indian Institute of Technology, Kharagpur; and a writer who loves to read about various topics.I also like making short films.
## 10 Interesting Facts To Know About Photosynthesis
All living things on this planet need energy to live and energy comes from food. ...
## Calculate Power Consumption and Average Monthly Electricity Bills of Your Home
Every household that uses electricity needs to pay monthly electricity bills, which is important part ... |
```Pythagorean Theorem and its
Converse
Cornell Notes due by:
Monday, January 12, 2015
Essential Question and
Objective
Essential Question
• How do I solve problems
using the Pythagorean
Theorem?
Objective
• Students will be able to
solve problems using the
Pythagorean Theorem.
Students will be able to solve problems using the
Pythagorean Theorem.
There are special names for the sides of a right triangle like
the one in the Solve It. The side opposite the right angle is
the hypotenuse. It is the longest side. Each of the sides
forming the right angle is a leg. The Pythagorean Theorem,
named after the Greek mathematician Pythagoras, relates
the lengths of the legs and the length of the hypotenuse.
Students will be able to solve problems using the
Pythagorean Theorem.
Focus Question
Why is the Pythagorean Theorem useful?
You can use the Pythagorean Theorem to find the length
of the third side of a right triangle if you know the length of
any two sides.
Students will be able to solve problems using the
Pythagorean Theorem.
You can use the Pythagorean
Theorem to find the length of a right
triangle’s hypotenuse given the
lengths of its legs. Using the
Pythagorean Theorem to solve for a
side length involves finding a
principal square root because side
lengths are always positive.
Students will be able to solve problems using the
Pythagorean Theorem.
Students will be able to solve problems using the
Pythagorean Theorem.
Students will be able to solve problems using the
Pythagorean Theorem.
An if-then statement such as “If an
animal is a horse, then it has four legs”
is called a conditional. Conditionals
have two parts. The part following if is
the hypothesis. The part following then
is the conclusion. The converse of a
conditional switches the hypothesis
and the conclusion.
Students will be able to solve problems using the
Pythagorean Theorem.
You can write the Pythagorean
Theorem as a conditional: “If a
triangle is a right triangle with legs of
lengths a and b and a hypotenuse of
length c, then a2 + b2 = c2.” The
converse of the Pythagorean
Theorem is always true.
Students will be able to solve problems using the
Pythagorean Theorem.
You can use the Pythagorean Theorem and its converse to
determine whether a triangle is a right triangle. If the side
lengths make the equation a2 + b2 = c2 true, then the
triangle is a right triangle. If they do not, then it is not a
right triangle.
Students will be able to solve problems using the
Pythagorean Theorem.
Students will be able to solve problems using the
Pythagorean Theorem.
``` |
# A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is a spade.
Given:
A card is drawn at random from a pack of 52 cards.
To do:
We have to find the probability that the card drawn is a spade.
Solution:
A pack of cards contains 52 cards of four suits and two colours red and black.
Four suits are named as spades, hearts, diamonds, and clubs.
Each suit consists of one ace, one king, one queen, one jack and 9 other cards numbered from 2 to 10.
This implies,
The total number of possible outcomes $n=52$.
Number of cards that are spades $=13$
Total number of favourable outcomes $=13$.
We know that,
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
Therefore,
Probability that the card drawn is a spade $=\frac{13}{52}$
$=\frac{1}{4}$
The probability that the card drawn is a spade is $\frac{1}{4}$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
40 Views |
# How to Square & Multiply Numbers Quick & Easily - Vedic Mathematics Method
[caption id="" align="aligncenter" width="210"] Addition, division, subtraction and multiplication symbols (Photo credit: Wikipedia)[/caption]
Part 1): A simple and quick method on How to square a number ending with 5.
In this tutorial we are going to learn one simple method of Vedic mathematics to make the square of a number which ends with 5.
Whenever a number ends with 5, definitely the last two digits will be 25. Now the remaining part of the answer can be calculated by multiplying a number with its next. For example if the given number is 35 then first part will be 25 and the remaining part will be 3×4=12. So the answer is 1225.
Example: Find square of 15.
Here the last two digits will be 25 and the first two digits are calculated by 1×2=02. So the answer is 225.
Let us take another example for clear understanding.
Example: Find square of 65.
Solution: first part will be 6×7=42. So the answer is 4225.
Part 2): Multiplication of two numbers having first digit same and last digits add up to 10.
Example: Multiply 46×44.
Solution:
• Bothe numbers having digit 4 same.
• Last digits that are 6 and 4 add up to 10.
• So we just multiply 4×5 =20 for getting first part of the answer.
• Also multiply last digits (6×4=24) for second part of the answer.
• So the answer is 2024.
Example: Multiply 71×79.
Solution:
• Bothe numbers having digit 7 same.
• Last digits that are 1 and 9 add up to 10.
• So we just multiply 7×8=56 for getting first part of the answer.
• Also multiply last digits (9×1=09) for second part of the answer.
• So the answer is 5609.
Related articles |
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# Vertical Asymptote
In Asymptote the distance between the curve and line approaches to zero but it never intersect and tends to infinity.
Let’s see different types of asymptote which are shown below:
Vertical asymptote,
Horizontal asymptote:
Let’s have a small introduction about both the asymptotes. First we will see the vertical asymptote. As we know that, equation of vertical line is given by:
⇒x = s;
The given equation is a vertical asymptotes of the graph which has a function y = f (s); this given function is applicable when one of the given condition is true.
The two conditions are shown below:
1. lim s → a- f (s) = + ∞;
2. lim s → a+ f (s) = + ∞;
At Point ‘a’ the given function f (s) may or may not be defined and at point x = s the value does not affect the asymptote.
Let’s see the example: let a function f (s) is given:
F (s) = 1 / s if the value of s > 0;
5 if the value of s ≤ 0
The given function has a limit of + ∞ as s → 0+, the function f (s) has the Vertical Asymptote s = 0, even the value of f (0) = 5. The graph of this function interests the vertical asymptote at point (0, 5).
Horizontal Asymptotes: Horizontal Asymptotes are the horizontal lines in which a graph of a function tends to s → + ∞.
The horizontal line is given by:
⇒s = c; the given equation is a Horizontal Asymptotes of a function s = f (p);
If it satisfy the given equation, and the equation is shown below.
⇒lim p → - ∞ f (p) = c or we can write it as:
⇒lim p → + ∞ f (p) = c.
## Vertical Asymptote Rules
In analytical Geometry, an Asymptote of a curve is a line in such a way that the distance between the curve and the line approaches to zero and the curve line tends to infinity.
In mathematics, there are two types of asymptote which are mention below:
Vertical asymptote,
Horizontal asymptote
Now we will see the Vertical Asymptote rules.
As we know that the...Read More |
# If 2 + sqrt(3) is a polynomial root, name another root of the polynomial, and explain how you know it must also be a root.
The aim of this question is to qualitatively evaluate the roots of a polynomial using prior knowledge of algebra.
As an example, let’s consider a standard quadratic equation:
$a x^{ 2 } \ + \ b x \ + \ c \ = \ 0$
The roots of such a quadric equation are given by:
$\lambda_{1,2} \ = \ \dfrac{ -b \ \pm \ \sqrt{ b^{ 2 } \ – \ 4 a c } }{ 2 a }$
Here, one may notice that the two roots are conjugates of each other.
A conjugate pair of roots is the one where two roots have the same non-square root term but their square root terms are equal and opposite in sign.
Given that:
$\lambda_1 \ = \ 2 \ + \ \sqrt{ 3 }$
If we assume that the polynomial has a degree of 2:
$a x^{ 2 } \ + \ b x \ + \ c \ = \ 0$
Then we know that the roots of such a quadric equation are given by:
$\lambda_{1,2} \ = \ \dfrac{ -b \ \pm \ \sqrt{ b^{ 2 } \ – \ 4 a c } }{ 2 a }$
This shows that the two roots $\lambda_1$ and $\lambda_2$ are conjugates of each other. So if $2 \ + \ \sqrt{ 3 }$ is one root then $2 \ – \ \sqrt{ 3 }$ must be the other root.
Here, we have assumed that the equation is quadratic. However, this fact is true for any polynomial of order higher than two.
## Numerical Result
If $2 \ + \ \sqrt{ 3 }$ is one root, then $2 \ – \ \sqrt{ 3 }$ must be the other root.
## Example
Given the equation $x^{ 2 } \ + \ 2 x \ + \ 4 \ = \ 0$, find its roots.
Comparing the given equation with the following standard quadratic equation:
$a x^{ 2 } \ + \ b x \ + \ c \ = \ 0$
We can see that:
$a \ = \ 1, \ b \ = \ 2 \text{ and } \ c \ = \ 4$
Roots of such a quadric equation are given by:
$\lambda_{1,2} \ = \ \dfrac{ -b \ \pm \ \sqrt{ b^{ 2 } \ – \ 4 a c } }{ 2 a }$
Substituting values:
$\lambda_{1,2} \ = \ \dfrac{ -2 \ \pm \ \sqrt{ 2^{ 2 } \ – \ 4 ( 1 ) ( 4 ) } }{ 2 ( 1 ) }$
$\lambda_{1,2} \ = \ \dfrac{ -2 \ \pm \ \sqrt{ 4 \ – \ 16 } }{ 2 }$
$\lambda_{1,2} \ = \ \dfrac{ -2 \ \pm \ \sqrt{ -12 } }{ 2 }$
$\lambda_{1,2} \ = \ -1 \ \pm \ \sqrt{ -3 }$
$\lambda_{1,2} \ = \ -1 \ \pm \ \sqrt{ 3 } i$
Which are the roots of the given equation. |
10.1 Non-right triangles: law of sines (Page 3/10)
Page 3 / 10
Given $\text{\hspace{0.17em}}\alpha =80°,a=120,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b=121,\text{\hspace{0.17em}}$ find the missing side and angles. If there is more than one possible solution, show both.
Solution 1
$\begin{array}{ll}\alpha =80°\hfill & a=120\hfill \\ \beta \approx 83.2°\hfill & b=121\hfill \\ \gamma \approx 16.8°\hfill & c\approx 35.2\hfill \end{array}$
Solution 2
$\begin{array}{l}{\alpha }^{\prime }=80°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}^{\prime }=120\hfill \\ {\beta }^{\prime }\approx 96.8°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{b}^{\prime }=121\hfill \\ {\gamma }^{\prime }\approx 3.2°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}^{\prime }\approx 6.8\hfill \end{array}$
Solving for the unknown sides and angles of a ssa triangle
In the triangle shown in [link] , solve for the unknown side and angles. Round your answers to the nearest tenth.
In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle $\text{\hspace{0.17em}}\gamma =85°,\text{\hspace{0.17em}}$ and its corresponding side $\text{\hspace{0.17em}}c=12,\text{\hspace{0.17em}}$ and we know side $\text{\hspace{0.17em}}b=9.\text{\hspace{0.17em}}$ We will use this proportion to solve for $\text{\hspace{0.17em}}\beta .$
To find $\text{\hspace{0.17em}}\beta ,\text{\hspace{0.17em}}$ apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for $\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$ It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.
$\begin{array}{l}\beta ={\mathrm{sin}}^{-1}\left(\frac{9\mathrm{sin}\left(85°\right)}{12}\right)\hfill \\ \beta \approx {\mathrm{sin}}^{-1}\left(0.7471\right)\hfill \\ \beta \approx 48.3°\hfill \end{array}$
In this case, if we subtract $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ from 180°, we find that there may be a second possible solution. Thus, $\text{\hspace{0.17em}}\beta =180°-48.3°\approx 131.7°.\text{\hspace{0.17em}}$ To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives
$\alpha =180°-85°-131.7°\approx -36.7°,$
which is impossible, and so $\text{\hspace{0.17em}}\beta \approx 48.3°.$
To find the remaining missing values, we calculate $\text{\hspace{0.17em}}\alpha =180°-85°-48.3°\approx 46.7°.\text{\hspace{0.17em}}$ Now, only side $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is needed. Use the Law of Sines to solve for $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ by one of the proportions.
The complete set of solutions for the given triangle is
Given $\text{\hspace{0.17em}}\alpha =80°,a=100,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=10,\text{\hspace{0.17em}}$ find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.
$\beta \approx 5.7°,\gamma \approx 94.3°,c\approx 101.3$
Finding the triangles that meet the given criteria
Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.
Using the given information, we can solve for the angle opposite the side of length 10. See [link] .
$\begin{array}{l}\text{\hspace{0.17em}}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{10}=\frac{\mathrm{sin}\left(50°\right)}{4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{10\mathrm{sin}\left(50°\right)}{4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha \approx 1.915\hfill \end{array}$
We can stop here without finding the value of $\text{\hspace{0.17em}}\alpha .\text{\hspace{0.17em}}$ Because the range of the sine function is $\text{\hspace{0.17em}}\left[-1,1\right],\text{\hspace{0.17em}}$ it is impossible for the sine value to be 1.915. In fact, inputting $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(1.915\right)\text{\hspace{0.17em}}$ in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.
Determine the number of triangles possible given $\text{\hspace{0.17em}}a=31,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=26,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta =48°.\text{\hspace{0.17em}}\text{\hspace{0.17em}}$
two
Finding the area of an oblique triangle using the sine function
Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as $\text{\hspace{0.17em}}\text{Area}=\frac{1}{2}bh,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is base and $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ is height. For oblique triangles, we must find $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ before we can use the area formula. Observing the two triangles in [link] , one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{\text{opposite}}{\text{hypotenuse}}\text{\hspace{0.17em}}$ to write an equation for area in oblique triangles. In the acute triangle, we have $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{h}{c}\text{\hspace{0.17em}}$ or $c\mathrm{sin}\text{\hspace{0.17em}}\alpha =h.\text{\hspace{0.17em}}$ However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ to form a right triangle. The angle used in calculation is $\text{\hspace{0.17em}}{\alpha }^{\prime },\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}180-\alpha .$
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function: 1--- g(f(x)) = x for all x in the domain of f 2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
False statement so you cannot prove it
Wilson
f(x)= 1 x f(x)=1x is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
24x^5
James
10x
Axmed
24X^5
Taieb
comment écrire les symboles de math par un clavier normal
SLIMANE |
# 10.10 Examples (Page 3/6)
Page 3 / 6
Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65.He samples ten statistics students and obtains the scores
• 65
• 65
• 70
• 67
• 66
• 63
• 63
• 68
• 72
• 71
. He performs a hypothesis test using a 5% level of significance. The data are from a normal distribution.
Set up the Hypothesis Test:
A 5% level of significance means that $\alpha =0.05$ . This is a test of a single population mean .
${H}_{o}$ : $\mu$ $=65\phantom{\rule{20pt}{0ex}}$ ${H}_{a}$ : $\mu$ $>65$
Since the instructor thinks the average score is higher, use a " $>$ ". The " $>$ " means the test is right-tailed.
Determine the distribution needed:
Random variable: $\overline{X}$ = average score on the first statistics test.
Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given $n=10$ sample data values. Notice also that the data come from a normal distribution. This means that thedistribution for the test is a student's-t.
Use ${t}_{\text{df}}$ . Therefore, the distribution for the test is ${t}_{9}$ where $n=10$ and $\text{df}=10-1=9$ .
Calculate the p-value using the Student's-t distribution:
$\text{p-value}=P\left($ $\overline{x}$ $>67$ ) $=0.0396$ where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.
Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 or more.
Compare $\alpha$ and the p-value:
Since $\alpha =.05$ and $\text{p-value}=0.0396$ . Therefore, $\alpha >\text{p-value}$ .
Make a decision: Since $\alpha >\text{p-value}$ , reject ${H}_{o}$ .
This means you reject $\mu =65$ . In other words, you believe the average test score is more than 65.
Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for ${\mu }_{0}$ , the name of the list where you put the data, and 1 for Freq: . Arrow down to $\mu :$ and arrow over to $>{\mu }_{0}$ . Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p-value ( $p=0.0396$ ) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standarddeviation. $\mu >65$ is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with $t=1.9781$ (test statistic) and $p=0.0396$ (p-value). Make sure when you use Draw that no other equations are highlighted in $Y=$ and the plots are turned off.
Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentageis the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level ofsignificance.
Set up the Hypothesis Test:
The 1% level of significance means that $\alpha =0.01$ . This is a test of a single population proportion .
${H}_{o}$ : $p$ $=0.50\phantom{\rule{20pt}{0ex}}$ ${H}_{a}$ : $p$ $\ne 0.50$
The words "is the same or different from" tell you this is a two-tailed test.
Calculate the distribution needed:
Random variable: $P\text{'}$ = the percent of of first-time brides who are younger than their grooms.
Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for $P\text{'}$ , the estimated proportion.
$P\text{'}$ ~ $N$ $\left(p,\sqrt{\frac{p\cdot q}{n}}\right)\phantom{\rule{20pt}{0ex}}$ Therefore, $P\text{'}$ ~ $N$ $\left(0.5,\sqrt{\frac{0.5\cdot 0.5}{100}}\right)$ where $p=0.50$ , $q=1-p=0.50$ , and $n=100$ .
Calculate the p-value using the normal distribution for proportions:
$\text{p-value}=P\left(\mathrm{p\text{'}}$ $()$ $0.47$ or $\mathrm{p\text{'}}>0.53$ ) $=0.5485$
where $x=53$ , $p\text{'}=\frac{x}{n}$ $=\frac{53}{100}=0.53$ .
Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion $p\text{'}$ is 0.53 or more OR 0.47 or less (see the graph below).
$\mu =p=0.50$ comes from ${H}_{o}$ , the null hypothesis.
$p\text{'}$ $=0.53$ . Since the curve is symmetrical andthe test is two-tailed, the $p\text{'}$ for the left tail is equal to $0.50-0.03=0.47$ where $\mu =p=0.50$ . (0.03 is the differencebetween 0.53 and 0.50.)
Compare $\alpha$ and the p-value:
Since $\alpha =0.01$ and $\text{p-value}=0.5485$ . Therefore, $\alpha$ $()$ $\text{p-value}$ .
Make a decision: Since $\alpha$ $()$ $\text{p-value}$ , you cannot reject ${H}_{o}$ .
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides that are younger than their grooms isdifferent from 50%.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for ${p}_{0}$ , 53 for $x$ and 100 for $n$ . Arrow down to Prop and arrow to not equals ${p}_{0}$ . Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the p-value ( $p=0.5485$ ) and the test statistic (z-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with $z=0.6$ (test statistic) and $p=0.5485$ (p-value). Make sure when you use Draw that no other equations are highlighted in $Y=$ and the plots are turned off.
The Type I and Type II errors are as follows:
The Type I error is to conclude that the proportion of first-time brides that are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%.(Reject the null hypothesis when the null hypothesis is true).
The Type II error is there is not enough evidence to conclude that the proportion of first time brides that are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
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
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
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
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types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
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many many of nanotubes
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what is the k.e before it land
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what is the function of carbon nanotubes?
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I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
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### number system
1. 1. NUMBER SYSTEM BY : IKA KOMALA SARI NILA PATMALA R. ULFAHTUL HASANAH VIERA VIRLIANI B.INGGRIS MATEMATIKA
2. 2. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 NUMBER SYSTEM Human beings have trying to have a count of their belonging, goods, ornaments, jewels, animals, trees, goats, etc. by using techniques. 1. putting scratches on the ground 2. by storing stones-one for each commodity kept taken out This was the way of having a count of their belongings without knowledge of counting
3. 3. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 NUMBER SYSTEM The questions of the type: HOW MUCH? HOW MANY? Need accounting knowledge
4. 4. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 The functions of learning number system Are 11 functions, that to: Illustrate the extension of system of number from natural number to real (rational and irrational) numbers Identify different types of numbers Express an integers as a rational number Express a rational number as a terminating or non-terminating repeating decimal and vice-versa
5. 5. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 The functions of learning number system Find rational numbers between any two rationals Represent a rational number on the number line Cites example of irrational numbers Represent 2, 3, 4 on the number line Find irrational numbers between any two given numbers 2 3 4
6. 6. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 The functions of learning number system Round off rational and irrational numbers to given number of decimal places Perform the four fundamental operation of addition, subtraction, multiplication, and division on real numbers
7. 7. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.1 EXPECTED BACKGROUND KNOWLEDGE It is about the accounting numbers in use on the day to day life Accounting numbers Day life
8. 8. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.2 Recall of Natural Numbers, Whole Numbers, and Integers Natural Numbers 1, 2, 3, … There is no greatest natural number, for if 1 added to any natural numbers. we get the next higher natural number, call its successor. Example : 4+2=6 12:2=6 22-6=16 12×3=36
9. 9. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.2 Recall of Natural Numbers, Whole Numbers, and Integers Addition and multiplication of natural numbers again yield a natural numbers But the subtraction and division of two natural number may or may not yield a natural numbers Example: Number line of natural numbers 1 2 3 4 5 6 7 8 9 … 2-6 = -4 6 : 4 = 3/2
10. 10. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.2 Recall of Natural Numbers, Whole Numbers, and Integers Whole Numbers The natural number were extended by zero (0) 0, 1, 2, 3, … There is no greatest whole numbers The number 0 has the following properties: a+0 = a = 0+a a-0 = a but 0-a is not defined in whole numbers a×0 = 0 = 0×a Division by 0 is not defined
11. 11. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.2 Recall of Natural Numbers, Whole Numbers, and Integers The whole number in four fundamental operation is same The line number of whole number 0 1 2 3 4 5 6 7 …
12. 12. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.2 Recall of Natural Numbers, Whole Numbers, and Integers Integers Another extension of numbers which allow such subtractions. It is begin from negative numbers until the whole number. The number line of integers … -3 -2 -1 0 1 2 3 4 …
13. 13. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.2 Recall of Natural Numbers, Whole Numbers, and Integers Representing Integers on number line A B C D … -4 -3 -2 -1 0 1 2 3 4 5 … Then A = -3 C = 2 B = -1 D = 3 A < B, D > C, B < C, C > A The rule: 1. A > B, if A is to the right of B 2. A < B, if A is to the left of B
14. 14. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.3 Rational Number Rational Numbers Consider the situation, when an integer a is divided by another non-zero integer b. The following case arise: 1. When A multiple of B A = MB, where M is natural number or integer. Then, A/B =M
15. 15. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.3 Rational Number 2. Rational number is when A is not A multiple B. A/B is not an integer. Thus, a number which can be put in the form p/q, where p and q are integers and q ≠ 0. Example: All Rational Numbers -2 5 6 11 3 -8 2 7
16. 16. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.3 Rational Number Positive and Negative Rational Number 1. p/q is said positive numbers if p and q are both positive or both negative integers 2. p/q is said negative if p and q are of different sign. Example: + - 3 4 -1 -5 -7 4 6 -5
17. 17. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.3 Rational Number Standard Form of a Rational Number We can see that -p/q = -(p/q) -p/-q = -(-p)/-(-q)= p/q p/-q= (-p)/q -p p -p p q -q -q q
18. 18. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.3 Rational Number Notes: A rational number is standard form is also referred to as “a rational lowest form” . There are two terms interchangeably Example: 18/27 can be written 2/3 in standard form (lowest form)
19. 19. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.3 Rational Number Some important result: 1. Every natural number is a rational number but vice-versa is not always true 2. Every whole number and integer is a rational number but vice-versa is not always true
20. 20. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS Addition of Rational Numbers 1. Consider the addition of rational numbers , + = for example : + = =
21. 21. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS 2. Consider the two rational numbers p/q and r/s p/q + r/s = ps/qs + rq/sq = for example : ¾ + 2/3 = = =
22. 22. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS from the above two cases, we generalise the following rule: (a)The addition of two rational numbers with common denominator is the rational number with common denominator and numerator as the sum of the numerators of the two rational numbers.
23. 23. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS b)The sum of two rational numbers with different denominator is a rational number with the denominator equal to the product of the denominators of two rational numbers and the numerator equal to sum of product of the numerator of first rational with the denominator of second and the product of numerator of second rational number and the denominator of the first rational number.
24. 24. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS Examples: Add the following rational numbers : (i) 2/7 and 6/7 (ii) 4/17 and -3/17 Solution: (i) 2/7 + 6/7 = 8/7 (ii) 4/17 + (-3)/17 = 1/17
25. 25. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS Add each of the following rational numbers, examples: (i) 3/4 and 1/7 Solution : (i) we have 3/4 + 1/7 = 3x7/4x7 + 1x4/7x4 = 21/28 + 4/28 = 25/28 3/4 + 1/7 = 25/28 or 3x7+4x1 / 4x7 = 21+4/728 = 25/28
26. 26. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS Subtraction of Rational Numbers (a) p/q – r/q = p-r/q Example : 7/4 – ¼ = … 7/4 –1/4 = 7 – 1 4 = 6/4 = 2x3 = 3/2 2x2 3/5 – 2/15 = … 3x12/5x12 – 2x5/12x5 = 36/60 – 10/60 = 26/60 = 13x2/30x2 = 13/30
27. 27. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS Multiplication and Division of Rational Numbers (i) Multiplication of two rational number (p/q) and (r/s) , q 0, s 0 is the rational number pr/ps where qs 0 = product of numerators/product of denominators (ii) Division of two rational numbers p/q and r/s , such that q 0, s 0, is the rational number ps/qr, where qr 0 In the other words (p/q) (r/s) = p/r x (s/r) Or (First rational number) x (Reciprocal of the second rational number) Let us consider some examples
28. 28. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS Examples : (i) 3/7 and 2/9 (ii) 5/6 and (-2/19) Solution : (i) 3/7 x 2/9 = 3x2/7x9 = 3x2/7x3x3 = 2/21 (3/7))x(2/9) = 2/21 (ii) 5/6 x (-2/19) = 5x(-2)/6x19 = - 2x5/2x3x19 = -5/57 5/6 x (-2/19) = -5/57
29. 29. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.7 FOUR FUNDAMENTAL OPERATIONS ON RATIONAL NUMBERS (i) (3/4) (7/12) Solution: (i) (3/4) (7/12) = (3/4) x (12/7) [Reciplocal of 7/12 is 12/7] = 3x12/4x7 = 3x3x4/7x4 = 9/7 (3/4) (7/12) = 9/7
30. 30. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.8 DECIMAL REPRESENTATION OF A RATIONAL NUMBER You are familiar with the division of an integer by another integer and expressing the result as a decimal number. The process of expressing rational number into decimal from is to carryout the process of long division using decimal notation. Example: Represent each one the following into a decimal number (i) (ii) : 5 12 25 27
31. 31. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.8 DECIMAL REPRESENTATION OF A RATIONAL NUMBER Solution: Using long division, we get (i) Hence , = 2,4 (ii) (-1, 08) hence, = -1, 08 4,2 0,2 0,2 10 ,12 5 x 5 12 x 200 200 25 27 25 25 27
32. 32. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.8 DECIMAL REPRESENTATION OF A RATIONAL NUMBER From the above example, it can be seen that the division process stops after a finite number of steps, when the remainder becomes zero and the resulting decimal number has a finite number of decimal places. Such decimals are known as terminating decimals. Note that in the above division, the denominators of the rational numbers had only 2 or 5 or both as the only prime factor
33. 33. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.8 DECIMAL REPRESENTATION OF A RATIONAL NUMBER Alternatively, we could have written as = = 2,4 Other examples: Here the remainder 1 repeats. The decimal is not a terminating decimal = 2,333… or 2,3 5 12 25 212 x x 10 24 33,2 00,1 9 0,1 6 00,7 3 3 7
34. 34. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.8 DECIMAL REPRESENTATION OF A RATIONAL NUMBER ` = 0,2 85714 Note: A bar over a digit or a group of digits implies that digit or group of digits starts repeating itself indefinitely. 28571428,0 4 56 60 14 20 28 30 7 10 49 50 35 40 56 60 14 000.2 7 7 2
35. 35. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.8 DECIMAL REPRESENTATION OF A RATIONAL NUMBER Expressing decimal expansion of rational number in p/q form Examples: Express in form p/q ! Express in in form p/q ! 0,48 =100 48 25 12 Let x = 0,666 (A) 10 x = 6,666 (B) (B)-(A) gives 9x = 6 or x = 2/3 0,666 The example above illustrates that: A terminating decimal or a non- terminating recurring decimal represents a rational number
36. 36. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.8 DECIMAL REPRESENTATION OF A RATIONAL NUMBER Note : The non-terminating recurring decimals like 0,374374374… are written as 0,374. The bar on the group of digits 374 indicate that group of digits repeats again and again.
37. 37. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.9 RATIONAL NUMBERS BETWEEN TWO RATIONAL NUMBERS Is it possible to find a rational number between two given rational numbers. To explore this, consider the following example. Example : Find rational number between and Let us try to find the number ( + ) ( ) = now, = = And = = 4 3 5 6 2 1 4 3 5 6 2 1 20 2415 40 39 4 3 104 103 x x 40 30 5 6 85 86 x x 40 48 abviously, < <40 30 40 39 40 48
38. 38. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.9 RATIONAL NUMBERS BETWEEN TWO RATIONAL NUMBERS is a rational number between the rational numbers and Note : = 0,75. = 0, 975 and = 1,2 Than: 0,75 < 0, 975 < 1,2 This can be done by either way : (i) reducing each of the given rational number with a common base and then taking their average (ii) by finding the decimal expansions of the two given rational numbers and then taking their average 4 340 39 5 6 4 3 40 39 5 6
39. 39. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.4 Equivalent Forms of a Rational Number A rational number can be written in an equivalent form by multiplying or dividing the numerator and denominator of the given rational number by the same number Example : 2/3 = 2x2 = 4/6 and 2/3 = 2x4 = 8/12 3x2 3x8 It’s mean 4/6 and 8/12 are equivalent form of the rational number 2/3
40. 40. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.5 Rational Numbers on the Number Line We know how to represent intergers on the number line. Let us try to represent ½ on the number line. The rational number ½ is positive and will be represented to the right of zero. As 0<½<1, ½ lies between 0 and 1. divide the distance OA in two equal parts. This can be done by bisecting OA at P
41. 41. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.5 Rational Numbers on the Number Line Let P represet ½. Similarly R, the mid- point of OA’, represents the rational number -½. A R 0 P A … -2 -1 0 1 2 3 … Similarly , can be represented on the number line as below: B’ A’ O A P B C … -2 -1 0 1 2 3 … As 1 < 4/3 < 2 therefore, 4/3 between 1 and 2 3 4 -1/2 1/2 4/3
42. 42. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.6 COMPARISON OF RATION NUMBER In order to compare to rational number, we follow any of the following methods: (i)If two rational numbers, to be compare have the same denominator compare their numerators. The number having the greater numerator is the greater rational number. Thus for the two rational numbers and , with the same positive denominator. as 9>5. so, 17 5 17 9 17 5 17 9 ,17 17 5 17 9
43. 43. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.6 COMPARISON OF RATION NUMBER (ii) If two rational number are having different denominator, make ther denominator equal by taking their equivalent form and then compare the numerator of the resulting rational numbers. The number having a greater numerator is greater rational number. For example, to compare two rational numbers and , we first make their denominator same in the following manner: 7 3 11 6
44. 44. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.6 COMPARISON OF RATION NUMBER (iii) By plotting two given rational numbers on the number line we see that rational number to the righ of the other rational number is greater. 77 33 117 113 x x 77 42 711 79 x x As 42>33, or 77 33 77 42 7 3 11 6
45. 45. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 For example, take and , we plot these number on the number line as below: -2 -1 0 1 2 3 3 2 4 3 1.6 COMPARISON OF RATION NUMBER
46. 46. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 1.6 COMPARISON OF RATION NUMBER 0<⅔<1 and 0< ¾<1. it means ⅔ and ¾ both lie between 0 and 1. by the method of diving a line Into equal number of parts, A represent ⅔ and B represent ¾ As B is to the right of A, ¾>⅔ or ⅔<¾ So, out of ⅔ and ¾, ¾ is greter number.
47. 47. Number System 1.1 1.2 1.5 1.6 1.3 1.7 1.8 1.9 Exit 1.4 Thank’s for your attention |
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This solution deals with quadratic equations.
Solution found
x=(4-sqrt(-8))/2=2-isqrt(2)=2.0000-1.4142i
x=(4+sqrt(-8))/2=2+isqrt(2)=2.0000+1.4142i
## Step by Step Solution
### Reformatting the input :
(1): "x2" was replaced by "x^2".
### Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
x^2-4*x-(-6)=0
## Step 1 :
#### Trying to factor by splitting the middle term
Factoring x2-4x+6
The first term is, x2 its coefficient is 1 .
The middle term is, -4x its coefficient is -4 .
The last term, "the constant", is +6
Step-1 : Multiply the coefficient of the first term by the constant 1 • 6 = 6
Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is -4 .
-6 + -1 = -7 -3 + -2 = -5 -2 + -3 = -5 -1 + -6 = -7 1 + 6 = 7 2 + 3 = 5 3 + 2 = 5 6 + 1 = 7
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
#### Equation at the end of step 1 :
x2 - 4x + 6 = 0
## Step 2 :
#### Parabola, Finding the Vertex :
Find the Vertex of y = x2-4x+6
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 2.0000
Plugging into the parabola formula 2.0000 for x we can calculate the y -coordinate :
y = 1.0 * 2.00 * 2.00 - 4.0 * 2.00 + 6.0
or y = 2.000
#### Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = x2-4x+6
Axis of Symmetry (dashed) {x}={ 2.00}
Vertex at {x,y} = { 2.00, 2.00}
Function has no real roots
#### Solve Quadratic Equation by Completing The Square
Solving x2-4x+6 = 0 by Completing The Square .
Subtract 6 from both side of the equation :
x2-4x = -6
Now the clever bit: Take the coefficient of x , which is 4 , divide by two, giving 2 , and finally square it giving 4
Add 4 to both sides of the equation :
On the right hand side we have :
-6 + 4 or, (-6/1)+(4/1)
The common denominator of the two fractions is 1 Adding (-6/1)+(4/1) gives -2/1
So adding to both sides we finally get :
x2-4x+4 = -2
Adding 4 has completed the left hand side into a perfect square :
x2-4x+4 =
(x-2) • (x-2) =
(x-2)2
Things which are equal to the same thing are also equal to one another. Since
x2-4x+4 = -2 and
x2-4x+4 = (x-2)2
then, according to the law of transitivity,
(x-2)2 = -2
We'll refer to this Equation as Eq. #2.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-2)2 is
(x-2)2/2 =
(x-2)1 =
x-2
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-2 = -2
Add 2 to both sides to obtain:
x = 2 + √ -2
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Since a square root has two values, one positive and the other negative
x2 - 4x + 6 = 0
has two solutions:
x = 2 + √ 2 i
or
x = 2 - √ 2 i
Solving x2-4x+6 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = 1
B = -4
C = 6
Accordingly, B2 - 4AC =
16 - 24 =
-8
4 ± √ -8
x = —————
2
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)
Both i and -i are the square roots of minus 1
Accordingly, -8 =
√ 8 • (-1) =
√ 8 • √ -1 =
± √ 8 • i
Can √ 8 be simplified ?
Yes! The prime factorization of 8 is
2•2•2
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
8 = √ 2•2•2 =
± 2 • √ 2
√ 2 , rounded to 4 decimal digits, is 1.4142
So now we are looking at:
x = ( 4 ± 2 • 1.414 i ) / 2
Two imaginary solutions :
x =(4+√-8)/2=2+i√ 2 = 2.0000+1.4142i
or: x =(4-√-8)/2=2-i√ 2 = 2.0000-1.4142i
## Two solutions were found :
1. x =(4-√-8)/2=2-i 2 = 2.0000-1.4142i
2. x =(4+√-8)/2=2+i 2 = 2.0000+1.4142i |
# Classical Probability: Example, Definition, and Uses in Life
Classical probability is the statistical concept that measures the likelihood (probability) of something happening. In a classic sense, it means that every statistical experiment will contain elements that are equally likely to happen (equal chances of occurrence of something). Therefore, the concept of classical probability is the simplest form of probability that has equal odds of something happening.
## Classical Probability Examples
Example 1: The typical example of classical probability would be rolling of a fair dice because it is equally probable that top face of die will be any of the 6 numbers on the die: 1, 2, 3, 4, 5, or 6.
Example 2: Another example of classical probability would be tossing an unbiased coin. There is an equal probability that your toss will yield either head or tail.
Example 3: In selecting bingo balls, each numbered ball has an equal chance of being chosen.
Example 4: Guessing a multiple choice quiz (MCQs) test with (say) four possible answers A, B, C or D. Each option (choice) has the same odds (equal chances) of being picked (assuming you pick randomly and do not follow any pattern).
## Formula for Classical Probability
The probability of a simple event happening is the number of times the event can happen, divided by the number of possible events (outcomes).
Mathematically $P(A) = \frac{f}{N}$,
where, $P(A)$ means “probability of event A” (event $A$ is whatever event you are looking for, like winning the lottery, that is event of interest), $f$ is the frequency, or number of possible times the event could happen and $N$ is the number of times the event could happen.
For example, the odds of rolling a 2 on a fair die are one out of 6, (1/6). In other words, one possible outcome (there is only one way to roll a 1 on a fair die) divided by the number of possible outcomes.
Classical probability can be used for very basic events, like rolling a dice and tossing a coin, it can also be used when occurrence of all events is equally likely. Choosing a card from a standard deck of cards gives you a 1/52 chance of getting a particular card, no matter what card you choose. On the other hand, figuring out will it rain tomorrow or not isn’t something you can figure out with this basic type of probability. There might be a 15% chance of rain (and therefore, an 85% chance of it not raining).
## Other Examples of classical Probability
There are many other examples of classical probability problems besides rolling dice. These examples include flipping coins, drawing cards from a deck, guessing on a multiple choice test, selecting jellybeans from a bag, and choosing people for a committee, etc.
## Classical Probability cannot be used:
Dividing the number of events by the number of possible events is very simplistic, and it isn’t suited to finding probabilities for a lot of situations. For example, natural events like weights, heights, and test scores need normal distribution probability charts to calculate probabilities. In fact, most “real life” things aren’t simple events like coins, cards, or dice. You’ll need something more complicated than classical probability theory to solve them.
## For further Detail see Introduction to Probability
Updated: Sep 18, 2017 — 8:41 am |
# MATH STUDY GUIDE
SLOPE
The slope of a line can be determined in two ways . 1) If you know the equation of the line , solve it for y. The slope is the coefficient of x. 2) If you know the coordinates of two points, and use the formula: Sec. 4.3, 4.4 Sec. 4.3
SIMPLIFY
Fractions: If there are no variables, see compute.
1 To add or subtract : a) Find the Lowest Common Denominator. b) Change each fraction to an equivalent fraction by multiplying numerator and denominator by the same value. c) Add the numerators and use the common denominator. If there is a “—“ in front of a fraction be sure to distribute it to EVERY TERM in the numerator. Sec. 8.3, 8.4 2 To multiply, factor numerators and denominators reducing where possible. Leave the answer in factored form unless it is part of a larger problem. (i.e. must be added to other terms.) Sec. 8.2 3 To divide, FIRST invert the divisor, and then proceed as in multiplication. Sec. 8.2 4 If there is a fraction within a numerator or denominator, either: a) Multiply numerator and denominator of the largest fraction by the LCD for all fractions Sec. 8.5 OR b) Treat numerator and denominator as a grouping and simplify, then divide as indicated by the larger fraction. Sec. 8.5 5 Remember, no denominator of any fraction may ever be zero. Sec. 8.1 6 Always reduce final answers where possible by dividing common factors from numerator and denominator. Sec. 8.1
Exponential expressions
You may apply any appropriate rule to the expression, but the following strategies may be useful: a) Are there powers of other expressions? Use the Power (of a Product) rule to remove parentheses . b) Are there powers of exponential expressions? Use Power (of a Power) rule where appropriate. c) Are there like bases in numerator or denominator? Use the product rule to simplify (add exponents.) d) Are there like bases in both numerator and denominator? Divide (by subtracting exponents.) e) Are there negative exponents ? Use the negative exponent rule to write the reciprocal. f) Write as a single fraction. g) Are you finished? Each exponent should apply to a single base. Each base should appear only once. There should be no negative exponents. Powers of numbers should be calculated . The fraction should be in lowest terms. Sec. 6.2, 6.5, 6.7
Scientific Notation
Numbers written in Scientific Notation are in the form (a number between 1 and 10) (a power of 10.) - To multiply or divide numbers written in scientific notation: 1) Multiply or divide the coefficients 2) Use the rules of exponents to multiply or divide the powers of 10. 3) Check to be sure the new coefficient is between 1 and 10. If not, rewrite it in Scientific Notation and simplify the powers of 10. Sec 6.7
FACTOR
To factor a number means to write it as a product of primes (numbers that cannot be factored further.) Begin with any product and then break each number down until none can be factored further.
To factor a polynomial: Sec. 7.5 1) Is there a factor common to all terms? Factor out the greatest common factor (term.) Sec. 7.1 2) Are there 4 terms? Try factoring by grouping Sec. 7.1 3) Is there a common pattern? a) Is this a difference of 2 squares? b) Is this a perfect square trinomial ? c) Is this the sum or difference of two cubes ? Sec. 7.4 When all else fails on a trinomial : 4) Try splitting the middle term into two and factoring by grouping. OR Sec. 7.3 5) Perform a structures search. (This is an organized version of the trial factors from the text.) a) List all the possible ways to factor the first (squared) term. These are the column headings. b) In each column, list all the possible arrangements of the factors for the last (constant) term. (These form the rows.) c) Test each entry in your table using FOIL to see if this makes the middle term possible. (If there are no candidates, report that it DOES NOT FACTOR.) d) If you have a candidate, insert signs to try to match original. i. If the last sign (constant) is negative, the signs are different. ii. If the last sign is positive , the two signs are alike, Use the sign of the middle term. iii. If none of the above works, go on searching for new candidates. iv. If you exhaust the list and none work, report that it DOES NOT FACTOR. e) Check your solution.Check to be sure that none of the factors can be factored further. Sec. 7.2
WORD PROBLEMS
“How to Solve Word Problems in Algebra” By Mildred Johnson is an excellent and inexpensive resource. It is available in the bookstore.1) Read through the problem to determine type. 2) Draw a picture, if possible. 3) Write “Let x be …” 4) Pick out the basic unknown and finish the above sentence. 5) Write as many other quantities as possible in terms of x and label them. Sec. 2.5 6) Is there STILL another unknown? If so, write, “Let y be …” and complete the sentence. Write all other quantities in terms of ‘x and y’. You may need one or more of the formulas below to complete this.Note: Tables are useful in many of these problems. Make one like the models in the text where appropriate. Sec. 5.4 7) Write any formula(s) that apply to this type of problem. a) d = rt (distance, time and speed) b) In wind or stream, when moving with the current, the speed is the sum of the speed of the craft and the current. Sec. 3.1 c) i = Pr (interest for 1 year) Sec. 3.1 d) Concentration of a solution (% target) (amount mixture) = amount target ingredient. e) (cost per item) (number of items) = value Sec. 3.1 f) (denomination of a bill) (# of bills) = value g) consecutive numbers x, x + 1, x + 2 , etc. h) consecutive ODD or EVEN numbers (The value of the first determines which) n, n + 2, n + 4 etc. i) In age problems, when they say “in 5 years,” write each age + 5 Sec. 2.5 j) Work rate problems convert the time to do a job into the work done per time period by taking the reciprocal. THESE quantities can be added or subtracted. Sec. 8.7 k) Geometric formulas for the perimeter of a triangle and the sum of its angle. Sec. 3.3 l) In a triangle with a right angle, you may use the Pythagorean Formula. Sec. 10.1 m) Basic %, A = PB 8) Use the formula or the words from the problem to write an equation. 9) Solve the equation for x (or x and y.) 10) REREAD the question. Write all the quantities from the original problem using the value for x as a key. 11) Answer the question asked. 12) Check the answer with the problem’s original words. Discard any answers that don’t fit. Sec. 2.4
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# Recurrence Relation
A recurrence relation for a sequence is a formula for the next term in the sequence as a function of the previous terms. A famous example that you may have heard of is the recurrence relation for the Fibonacci sequence where each term is the sum of the two previous terms. Here are the first few terms of the Fibonacci sequence: $$0,1,1,2,3,5,8,\dots$$
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## The Meaning of Recurrence Relations
Recurrence relations give a formula for the common rule that governs a given sequence of numbers. They are recursive, meaning that the recurrence relation formula for the next term in the sequence is given as a function of it's previous terms. They allow us to simplify sequences making it easier to analyse its characteristics and patterns.
A recurrence relation is a formula for the next term in a sequence as a function of its previous terms.
An example of a recurrence relation is $$u_{n+1}=4u_n+5$$. Where $$u_n$$ is the $$n^{th}$$ term in a sequence and the recurrence relation gives us a formula for working out the next term, $$u_{n+1}$$.
## Recurrence Relation Formula
Recurrence relation formulas can take many different forms. Commonly used notation uses $$u_{n}$$ to denote the $$n^{th}$$ term in a sequence and $$u_{n+1}$$ to denote the following (or next) term in a sequence. The formula of a recurrence relation depends on the order, also referred to as the degree, of the recurrence relation.
### Order/Degree of a Recurrence Relation
A recurrence relation of order $$k$$ is a formula for the $$n^{th}+1$$ term of the sequence as a function of the previous $$k$$ terms of the sequence. Another way to put it is that the order $$k$$ is the difference between the highest and lowest subscripts of the recurrence relation.
A recurrence relation of order/degree $$k$$ is an equation which is in the form:
$$u_{n}=a_{1}u_{n-1}+a_{2}u_{n-2}+a_{3}u_{n-3}+...+a_{k}u_{n-k} + f(n).$$
Where $$a_1,...,a_k$$ are constants and $$f(n)$$ is a function in terms of $$n$$.
A recurrence relation is non-homogenous if $$f(n)$$ is a polynomial or of the form $$a\times b^{n}$$.
If $$f(n)=0$$ then the recurrence relation is homogenous.
What is the order of the recurrence relation $$u_{n+1}=u_{n}+5n$$. Is it homogeneous or non-homogeneous?
The formula for recurrence relations is $$u_{n}=a_{1}u_{n-1}+a_{2}u_{n-2}+a_{3}u_{n-3}+...+a_{k}u_{n-k} + f(n)$$.
Comparing this with $$u_{n+1}=u_{n}+5n$$ gives:
• $$k=1$$ since the difference between $$n+1$$, the highest subscript, and $$n$$, the lowest subscript, is $$1$$,
• $$a_1=1$$ the coefficient of $$u_n$$,
• $$f(n)=5n$$.
Therefore, this is a non-homogenous first-order recurrence relation.
What is the order of the recurrence relation $$u_{n+1}=2u_{n}+3u_{n-1}$$? Is it homogeneous or non-homogeneous?
The formula for recurrence relations is $$u_{n}=a_{1}u_{n-1}+a_{2}u_{n-2}+a_{3}u_{n-3}+...+a_{k}u_{n-k} + f(n)$$.
Comparing this with $$u_{n+1}=2u_{n}+3u_{n-1}$$ gives:
• $$k=2$$ since the difference between $$n+1$$, the highest subscript, and $$n-1$$, the lowest subscript, is $$2$$,
• $$u_1=2$$ and $$u_2=3$$ the coefficients of $$u_n$$ and $$u_{n-1}$$ respectively,
• $$f(n)=0$$.
Therefore, this is a homogenous second-order recurrence relation.
When given a recurrence relation, to work out the terms in a sequence you will need initial conditions, sometimes called boundary conditions. You will need the same number of initial conditions as the order of a recurrence relation to be able to find the terms of the sequence, i.e. for a $$k^{th}$$ order recurrence relation you will need $$k$$ initial conditions to find the terms of the sequence. These initial conditions will normally be given as the first $$k$$ terms in the sequence.
The formula for the Fibonacci sequence is a second-order recurrence relation given by:
$$F_{n}=F_{n-1}+F_{n-2}$$
Suppose we are given two initial conditions $$F_0=0$$ and $$F_1=1$$.
Now we can work out the terms of the sequence.
Here are the first few terms written out with the formula:
\begin{align} F_0&=0\\F_1&=1\\F_2&=F_1+F_0=1+0=1\\F_3&=F_2+F_1=1+1=2\\F_4&=F_3+F_2=2+1=3\\F_5&=F_4+F_3=3+2=5\\F_6&=F_5+F_4=5+3=8. \end{align}
## Recurrence Relations: Examples
Let's look at some examples of sequences and find their recurrence relation formulas.
Can we find a recurrence relation for the following sequence of numbers?
$$3, 9, 21, 45, 93...$$
Solution:
The common notation used for sequences is given as:
\begin{align} u_1&=3\\u_2&=9\\u_3&=21\\u_4&=45\\u_5&=93. \end{align}
i.e. $$u_{n}$$ is the $$n^{th}$$ term of the sequence.
Note that some textbooks and questions start sequences with $$n=0$$ so make sure to check before attempting a question.
Now, we can find a formula for working out a particular term in the sequence. First look at the differences between consecutive terms:
\begin{align} u_2-u_1 &=9-3=6\\u_3-u_2&=21-9=12\\u_4-u_3&=45-21=24\\u_5-u_4&=93-45=48. \end{align}
This sequence is growing by a factor of 2 each time. With this in mind, look again at the original sequence:
\begin{align} u_1&=3\\u_2&=9=2\cdot3+3\\u_3&=21=2\cdot9+3\\u_4&=45=2\cdot21+3\\u_5&=93=2\cdot45+3. \end{align}
Therefore, with initial value $$u_{1}=3$$ the recurrence relation for this sequence is $$u_{n+1}=2u_{n}+3$$.
Suppose you have a sequence $$u_1,u_2,u_3, ...$$ defined by the recurrence relation $$u_{n+1}=u_n+2n+1$$ with initial value $$u_1=1$$.
Show that $$u_4=16$$ and hence find an expression for $$u_n$$ in terms $$n$$.
Solution:
The best way to go about this question is to write out the first four terms of the sequence.
\begin{align} u_1&=1\\u_2&=u_1+2\cdot1+1=1+2+1=4\\u_3&=u_2+2\cdot2+1=4+4+1=9\\u_4&=u_3+2\cdot3+1=9+6+1=16. \end{align}
Hence $$u_4=16$$.
From these values, we can also see that this is a sequence of square numbers. Therefore, an expression for $$u_n$$ in terms $$n$$ is $$u_n=n^{2}$$.
## Closed-Form of Recurrence Relations
Closed-form, or position-to-term, is the term we use to describe the formula for the $$n^{th}$$ term in terms of $$n$$. Either closed-form or position-to-term may be used in textbooks, either way is considered correct. These closed-form equations are useful if we want to find a particular term even when $$n$$ is large.
The closed-form is a formula for the $$n^{th}$$ term in the sequence in terms of $$n$$.
An example of closed-form for a sequence is $$u_{n}=4n-12$$.
Suppose you want to find the $$1000^{th}$$ term of this sequence, then you can simply substitute $$n=1000$$ into the equation and get:
$$u_{1000}=4(1000)-12=3988.$$
The closed form for the Fibonacci sequence is:
$$F_n=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2} \right)^{n}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n}.$$
Find the recurrence relation and closed-form for the sequence: $$27, 9, 3, 1, \frac{1}{3},\dots$$. Assume the initial value is denoted as $$u_0=27$$.
Solution:
Firstly, we can see that the terms in the sequence are decreasing by a factor of 3 each time. You can see this by dividing each term by its previous term which gives us 3 each time:
\begin{align} \frac{u_0}{u_1}&=\frac{27}{9}=3\\ \frac{u_1}{u_2}&=\frac{9}{3}=3 \\ \frac{u_2}{u_3}&=\frac{3}{1}=3\\ \frac{u_3}{u_4}&=\frac{1}{\frac{1}{3}}=3. \end{align}
Assuming this factor is constant throughout the sequence, the recurrence relation for the sequence is given by:
$$u_{n+1}=3u_n$$
with initial value $$u_{0}=27$$.
For the closed form of the sequence:
$$u_n=27\cdot \left(\frac{1}{3}\right)^{n}.$$
This sequence is also called a geometric sequence with a common ratio of 3.
### Proving Closed Forms of Recurrence Relations
The technique used for proving the closed-form of recurrence relations is proof by induction. You may have come across this technique before but the proof is structured as follows:
Let $$P(n)$$ be a mathematical statement such that $$n$$ is a natural number. Then $$P(n)$$ is true for values of $$n$$ if the following are satisfied:
Step 1. $$n=1$$ is true, i.e. $$P(1)$$ holds.
Step 2. Given/Assuming that $$n=k$$ is true, then $$n=k+1$$, i.e. if $$P(k)$$ holds then $$P(k+1)$$ holds.
Step 3: Conclusion: Since the statement was true for $$n=1$$ and assuming true for $$n=k$$, the statement is true for $$n=k+1$$. Therefore by the principal of mathematical induction, the statement holds for all natural numbers $$n=1,2,3,...$$.
For more information on using proof by induction "see Proof by Induction".
Prove by induction that $$u_{n}=5^{n-1}+1$$ is the closed form of the sequence defined by $$u_{n+1}=5u_{n}-4$$ with initial value $$u_{1}=2$$, for all natural numbers n.
Solution:
For this particular example $$P(n)=u_n=5^{n-1}+1$$.
Step 1: Let $$n=1$$,
$$u_{1}=5^{0}+1=1.$$
Therefore the statement holds for $$n=1$$.
Step 2: Assume the statement is true for $$n=k$$, i.e. assume $$u_{k}=5^{k-1}+1$$.
Show that $$u_{n}=5^{n-1}+1$$ is true for $$n=k+1$$:
\begin{align} u_{k+1}&=5u_{k}-4\\ &=5(5^{k-1}+1)-4\\ &=5^{1+k-1}+5-4\\ &=5^{k}+1.\end{align}
Therefore the statement holds for $$n=k+1$$.
Step 3: Conclusion: Since the statement was true for $$n=1$$ and assuming true for $$n=k$$, the statement is true for $$n=k+1$$. Therefore by the principal of mathematical induction, the statement holds for all natural numbers $$n=1,2,3,...$$.
## Solving Recurrence Relations
Solving recurrence relations involves finding the closed-form of a recurrence relation given certain initial values or boundary conditions. There are different methods for solving recurrence relations. Sometimes they are simple enough to solve by inspection (as we have seen above) but for more complex first-order and second-order recurrence relations, the best methods to use are iteration or the Characteristic Root Technique. If you would like to go through these techniques in depth, have a look at the following articles on Solving First-Order Recurrence Relations and Solving Second-Order Recurrence Relations.
## Recurrence Relations - Key takeaways
• A recurrence relation gives a formula for the next term in a sequence as a function of its previous terms.
• The closed-form of a recurrence relation is a formula for the $$n^{th}$$ term in the sequence in terms of $$n$$.
• A recurrence relation of order $$k$$ is a formula for the $$n^{th}+1$$ term of the sequence as a function of the previous $$k$$ terms of the sequence.
• Mathematical induction may be used to prove results relating to recurrence relations.
#### Flashcards in Recurrence Relation 24
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##### Frequently Asked Questions about Recurrence Relation
What is meant by recurrence relation?
A recurrence relation for a sequence is a formula for the next term in a sequence as a function of it's previous terms.
What is the formula of recurrence relation?
Depending on the order of a recurrence relation, the formula for a recurrence relation of order k is a formula for the (n+1)th term of the sequence as a function of the previous k terms of the sequence.
How to solve recurrence relations?
There are many methods to solving recurrence relations. For simple recurrence relations we can use inspection and for more complex recurrence relations we use iteration or the Characteristic Root Technique.
Why do we use recurrence relations?
We use recurrence relations to simplify a sequence of numbers using a common rule that governs them.
What is an example of a recurrence relation?
The Fibonacci sequence has the recurrence relation
F= Fn-1 + Fn-2.
## Test your knowledge with multiple choice flashcards
What is the order of the recurrence relation un+1 = 5un + 6n. Is it homogeneous or non-homogeneous?
Which term best describes the following recurrence relation? $$u_{n+2}=2u_{n+1}+3u_n+4n+12$$.
Which term best describes the following recurrence relation? $$u_{n+2}=6u_{n+1}-9u_n$$.
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# Using the square root table, find the square root of 25725.
Last updated date: 15th Jun 2024
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384.6k+ views
Hint: Recall that ${{16}^{2}}=256$ , or ${{160}^{2}}=25600$ , which is quite close to 25725.
Factorize the given number into its prime factors and group them into even powers.
Look up the square root table for square roots of leftover single prime factors.
The given number 25725 is a multiple of 25 (last two digits are a multiple of 25). It can be factorized as follows:
$25725=1029\times 25$
Now, 1029 is divisible by 3 because the sum of its digits $(1+0+2+9=12=1+2=3)$ is a multiple of 3.
⇒ $25725=3\times 343\times 25$
Recall that $343={{7}^{3}}$ .
∴ $25725=3\times {{5}^{2}}\times {{7}^{3}}$
Separating the even powers, we get:
⇒ $25725=3\times 7\times {{5}^{2}}\times {{7}^{2}}$
Taking the square root, we get:
⇒ $\sqrt{25725}=\sqrt{3}\times \sqrt{7}\times 5\times 7$
From the table of square roots, we see that $\sqrt{3}=1.732$ and $\sqrt{7}=2.645$ .
⇒ $\sqrt{25725}=1.732\times 2.645\times 5\times 7$
On multiplying the numbers together, we get:
⇒ $\sqrt{25725}=160.3399$
Note: Square roots of prime numbers from 1 to 100: (total 25 primes). Remembering the given table will help us to solve it easily.
$\sqrt{2}$ 1.4142 $\sqrt{13}$ 3.6056 $\sqrt{31}$ 5.5678 $\sqrt{53}$ 7.2801 $\sqrt{73}$ 8.544 $\sqrt{3}$ 1.7321 $\sqrt{17}$ 4.1231 $\sqrt{37}$ 6.0828 $\sqrt{59}$ 7.6811 $\sqrt{79}$ 8.8882 $\sqrt{5}$ 2.2361 $\sqrt{19}$ 4.3589 $\sqrt{41}$ 6.4031 $\sqrt{61}$ 7.8102 $\sqrt{83}$ 9.1104 $\sqrt{7}$ 2.6458 $\sqrt{23}$ 4.7958 $\sqrt{43}$ 6.5574 $\sqrt{67}$ 8.1854 $\sqrt{89}$ 9.434 $\sqrt{11}$ 3.3166 $\sqrt{29}$ 5.3852 $\sqrt{47}$ 6.8557 $\sqrt{71}$ 8.4261 $\sqrt{97}$ 9.8489 |
# Prove $\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$
I want to prove this equation,
$$\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$$
I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I’m stuck. Help me please.
(Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.
#### Solutions Collecting From Web of "Prove $\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$"
We have a group of $n$ people. We want to select a committee of $b$ people, and choose $a$ of them to be a steering subcommittee.
The right-hand side counts the number of ways to pick the committee of $b$ people, and then choose the steering subcommittee from this committee.
The left-hand side picks the steering subcommittee first, then the rest of the committee.
Both sides count the same thing, so they are the same.
Or else we want to choose $b$ people from a class of $n$ to go on a trip. Of these $b$ people, $a$ will ride in the limousine, and the rest in an old bus. We can choose the $b$ people, and then choose the $a$ of them who will ride in the limousine.
Or else we can choose the $a$ people who will ride in the limousine , and then pick $b-a$ people from the remaining $n-a$ to ride in the bus.
Directly, we find that
\begin{align*}
{n \choose a}{n – a \choose b – a} &= \frac{n!}{(n – a)!a!} \frac{(n – a)!}{(n – a – (b – a))! (b – a)!} \\
&= \frac{n!}{a!} \frac{1}{(n – b)!(b – a)!} \\
&= \frac{n!}{(n – b)! b!} \frac{b!}{(b – a)!a!} \\
&= {n \choose b}{b \choose a}
\end{align*}
The third equality follows by multiplying and dividing by $b!$.
These are just two different ways to express the trinomial coefficient $\binom n{a,b-a,n-b}$ as a product of two binomial coefficients. The relevant formula (not obviously present on the wikipedia page) is
$$\binom n{k,l,m} = \binom nk\binom {n-k}l \qquad\text{whenever k+l+m=n,}$$
together with the symmetry with respect to $x,y,z$ of the trinomial coefficient, and similarly for binomial coefficients (so that $\binom nb=\binom n{n-b}$). The formula above is a consequence of $(X+Y+Z)^n=(X+(Y+Z))^n$, and similar formulas hold for higher multinomial coefficients.
Given $n$ people, we can form a committee of size $b$ in ${n\choose b}$ ways. Once the committee is formed we can form a sub-committee of size $a$ in ${b\choose a}$ ways. Thus we can form a committee of size $b$ with a sub-committee of size $a$ in ${n\choose b}{b\choose a}$ ways. We can count the same thing by forming the sub-committee first and then forming the committee that contains the sub-committee. Given $n$ people we can form a sub-committee of size $a$ in ${n\choose a}$ ways. Once the sub-committee is formed we must form the committee of size $b-a$ from the remaining $n-a$ people in ${n-a\choose b-a}$ ways. Thus we can form a sub-committee of size $a$ while forming the committee of size $b-a$ that contains the sub-committee in ${n\choose a}{n-a\choose b-a}$ ways. Hence ${n\choose b}{b\choose a}={n\choose a}{n-a\choose b-a}$.
This combinatorial identity is known as the Subset-of-a-Subset identity. |
# Tutorial on Slope of a Line
This is a tutorial on the slope of a line with examples and detailed solutions, and matched exercises also with solutions. If you wish to go first through a tutorial on the concept of the slope of a line Go here.
Example 1 : Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical. (2 , 1) , (4 , 5) (-1 , 0) , (3 , -5) (2 , 1) , (-3 , 1) (-1 , 2) , (-1 ,- 5) Solution to Example 1: The slope of the line is given by m = ( 5 - 1 ) / (4 - 2) = 4 / 2 = 2 Since the slope is positive, the line rises as x increases. The slope of the line is given by m = ( -5 - 0 ) / ( 3 - (-1) ) = -5 / 4 Since the slope is negative, the line falls as x increases. We first find the slope of the line m = ( 1 - 1 ) / ( -3 - 2 ) = 0 Since the slope is equal to zero, the line is horizontal (parallel to the x axis). The slope of the line is given by m = ( -5 - 2 ) / ( -1 - (-1) ) Since ( -1 - (-1) ) = 0 and the division by 0 is not defined, the slope of the line is undefined and the line is vertical. (parallel to the y axis). Matched Exercise to Example 1: Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.Solution (3 , -1) , (3 , 5) (-1 , 0) , (3 , 7) (2 , 1) , (6 , 0) (-5 , 2) , (9 , 2) Example 2: A line has a slope of -2 and passes through the point (2 , 5). Find another point A through which the line passes. (many possible answers)Solution to Example 2: Let x1 and y1 be the x and y coordinates of point A. According to the definition of the slope ( y1 - 5 ) / (x1 - 2) = -2 We need to solve this equation in order to find x1 and y1. This equation has two unknowns and therefore has an infinite number of pairs of solutions. We chose x1 and then find y1. if x1 = -1, for example, the above equation becomes ( y1 - 5 ) / (-1 - 2) = -2 We obtain an equation in y1 ( y1 - 5 ) / -3 = -2 Solve for y1 to obtain y1 = 11 One possible answer is point A at (-1 , 11) Check that the two points give a slope of -2 (11 - 5 ) / (-1 - 2) = 6/-3 = -2 Matched Exercise to Example 2: A line has a slope of 5 and passes through the point (1 , -4). Find another point A through which the line passes. (many possible answers) Example 3: Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular? L1: (1 , 2) , (3 , 1) L2: (0 , -1) , (2 , 0) L1: (0 , 3) , (3 , 1) L2: (-1 , 4) , (-7 , -5) L1: (2 , -1) , (5 , -7) L2: (0 , 0) , (-1 , 2) L1: (1 , 0) , (2 , 0) L2: (5 , -5) , (-10 , -5) L1: (-2 , 5) , (-2 , 7) L2: (5 , 1) , (5 , 13) Solution to Example 3:In what follows, m1 is the slope of line L1 and m2 is the slope of line L2. Find the slope m1 of line L1 and the slope m2 of line L1 m1 = ( 1 - 2 ) / ( 3 - 1 ) = -1 / 2 m2 = ( 0 - (-1) ) / ( 2 - 0 ) = 1/2 The two slopes m1 and m2 are not equal and their products is not equal to -1. Hence the two lines are neither parallel nor perpendicular. m1 = ( 1 - 3 ) / ( 3 - 0 ) = -2 / 3 m2 = ( -5 - 4 ) / ( -7 - (-1) ) = -9 / -6 = 3/2 The product of the two slopes m1*m2 = (-2 / 3)(3 / 2) = -1, the two lines are perpendicular. m1 = ( -7 - (-1) ) / ( 5 - 2 ) = -6 / 3 = -2 m2 = ( 2 - 0 ) / ( -1 - 0 ) = -2 The two slopes are equal, the two lines are parallel. m1 = ( 0 - 0 ) / ( 2 - 1 ) = 0 / 1 = 0 m2 = ( -5 - (-5) ) / ( -10 - 5 ) = 0 / -15 = 0 The two slopes are equal , the two lines are parallel. Also the two lines are horizontal m1 = ( 7 - 5 ) / ( -2 - (-2) ) m2 = ( 13 - 1 ) / ( 5 - 5 ) The two slopes are both undefined since the denominators in both m1 and m2 are equal to zero. The two lines are vertical lines and therefore parallel. Matched Exercise to Example 3: Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular? L1: (1 , 2) , (1 , 1) L2: (-4 , -1) , (-4 , 0) L1: (2 , 3) , (3 , 1) L2: (1 , -2) , (7 , -5) L1: (1 , -1) , (2 , -2) L2: (0 , 0) , (1 , 1) L1: (1 , 9) , (-2 , 9) L2: (18 , -1) , (0 , -1) Solution Example 4: Is it possible for two lines with negative slopes to be perpendicular?Solution to Example 4:No. If both slopes are negative, their product can never be equal to -1. Matched Exercise to Example 4: Is it possible for two lines with positive slopes to be perpendicular?Solution More References and links to topics on slopes. Slope of a Line - Interactive Java Applet. Easy to use calculator to find slope and equation of a line through two points.Two Points Calculator Another calculator to find slope, x and y intercepts given the equation of a line.Find Slope and Intercepts of a Line - Calculator |
What is 5/12 + 4/6?
5 12
+
4 6
Step 1
Now we need to make our denominators match.
12 goes into 6 evenly, so we only need to multiply one term to get a common denominator.
Multiply 4 by 2, and get 8, then we multiply 6 by 2 and get 12.
We now have a new problem, that looks like this:
5 12
+
8 12
Step 2
Since our denominators match, we can add the numerators.
5 + 8 = 13
13 12
Step 3
Now, do we need to simplify this fraction?
First, we attempt to divide it by 2...
Nope! So now we try the next greatest prime number, 3...
Nope! So now we try the next greatest prime number, 5...
Nope! So now we try the next greatest prime number, 7...
Nope! So now we try the next greatest prime number, 11...
Nope! So now we try the next greatest prime number, 13...
No good. 13 is larger than 12. So we're done reducing.
There you have it! Here's the final answer to 5/12 + 4/6
5 12
+
4 6
=
13 12 |
Home/All Courses/CBSE/NCERT/NCERT Solutions For Class 9/NCERT Solution For Class 9 Maths/NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes
## NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes
Ncert Solutions for Class 9 Maths Chapter 13 pdf covers the topic of surface areas and volumes class 9 and the formulas to calculate the areas and volumes of Square, Rectangle, Triangle, Circle. To get more in-depth knowledge on surface areas and volumes class 9 refer to ncert solutions for class 9 maths chapter 13 pdf download
## Area
The space occupied by a two-dimensional flat surface. It is measure in square units.
Generally, Area can be of two types
(i) Total Surface Area
(ii) Curved Surface Area
## Total Surface Area
Total surface area refers to area including the base(s) and the curved part.
Curved surface area (lateral surface area)
Refers to area of only the curved part excluding it’s base(s).
## Volume
The amount of space, measured in cubic units, that an object or substance occupies.
### Surface area and volume all formulas
The Surface area and volume all formulas are given below:
## Areas
square = a2
rectangle = ab
parallelogram = bh
trapezoid = h/2 (b1+ b2)
circle = pir 2
ellipse = pir1 r2
• triangle = (1/2) b h
• equilateral triangle = (1/4)(3) a2
• triangle given SAS = (1/2) a b sin C
• triangle given a,b,c = [s(s-a)(s-b)(s-c)] when s = (a+b+c)/2 (Heron’s formula)
• regular polygon = (1/2) n sin(360°/n) S2
when n = # of sides and S = length from center to a corner
## Volumes
cube = a3
rectangular prism = a b c
irregular prism = bh
cylinder = bh = r2 h
pyramid = (1/3) bh
Cone = (1/3) bh = 1/3 r2 h
sphere = (4/3) r3
ellipsoid = (4/3) pir1 r2 r3
## Surface Areas
cube = 6 a2
prism:
(lateral area) = perimeter(b) L
(total area) = perimeter(b) L + 2b
sphere = 4 r2
2018-10-31T11:15:53+00:00 Categories: NCERT Solution For Class 9 Maths||0 Comments
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Use adaptive quiz-based learning to study this topic faster and more effectively.
# Simplifying surds
We generally simplify surds as the product of positive integers times the surd of the product of prime numbers.
$\sqrt{9}$ simplifies to $3$ and $\sqrt{8}$ simplifies to $2\sqrt{2}$.
We illustrate the simplification process with $\sqrt{24}$.
• Find the prime factorisation of the number $$24 = 2^3\times 3$$
• Put aside the even powers of the prime numbers. $$24 = \Tblue{2^2}\times \Tgreen{2}\times \Tgreen{3}$$
• Take the square root of the even powers by dividing the power by 2. Multiply the numbers with multiplicity 1 and leave them as a surd. $$\sqrt{24} = \sqrt{\Tblue{2^2}}\times\sqrt{\Tgreen{2\times 3}} = \Tblue{2}\sqrt{\Tgreen{6}}$$
Here are other simplified surds.
\begin{align*} &\qquad\sqrt{4} = \Tblue{2},\quad \sqrt{75} = \sqrt{5^3} = \sqrt{\Tblue{5^2}\times \Tgreen{5}} = \Tblue{5}\sqrt{\Tgreen{5}},\\ &\sqrt{72} = \sqrt{2^3\times 3^2} = \sqrt{\Tblue{2^2}\times \Tgreen{2}\Tblue{\times 3^2}} = \Tblue{2\times 3}\sqrt{\Tgreen{2}} = \Tblue{6}\sqrt{\Tgreen{2}}. \end{align*} |
# 5.4: The Exponential Distribution
The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.
Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.
The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.
Example $$\PageIndex{1}$$
Let $$X$$ = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.
$$X$$ is a continuous random variable since time is measured. It is given that $$\mu = 4$$ minutes. To do any calculations, you must know $$m$$, the decay parameter.
$$m = \dfrac{1}{\mu}$$. Therefore, $$m = \dfrac{1}{4} = 0.25$$.
The standard deviation, $$\sigma$$, is the same as the mean. $$\mu = \sigma$$
The distribution notation is $$X \sim Exp(m)$$. Therefore, $$X \sim Exp(0.25)$$.
The probability density function is $$f(x) = me^{-mx}$$. The number $$e = 2.71828182846$$... It is a number that is used often in mathematics. Scientific calculators have the key "$$e^{x}$$." If you enter one for $$x$$, the calculator will display the value $$e$$.
The curve is:
$$f(x) = 0.25e^{-0.25x}$$ where $$x$$ is at least zero and $$m = 0.25$$.
For example, $$f(5) = 0.25e^{-(0.25)(5)} = 0.072$$. The value 0.072 is the height of the curve when x = 5. In Example $$\PageIndex{2}$$ below, you will learn how to find probabilities using the decay parameter.
The graph is as follows:
Notice the graph is a declining curve. When $$x = 0$$,
$$f(x) = 0.25e^{(-0.25)(0)} = (0.25)(1) = 0.25 = m$$. The maximum value on the y-axis is m.
Exercise $$\PageIndex{1}$$
The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.
$$X \sim Exp(0.125)$$;
$$f(x) = 0.125e^{-0.125x}$$;
Example $$\PageIndex{2}$$
1. Using the information in Exercise, find the probability that a clerk spends four to five minutes with a randomly selected customer.
2. Half of all customers are finished within how long? (Find the 50th percentile)
3. Which is larger, the mean or the median?
a. Find $$P(4 < x < 5)$$.
The cumulative distribution function (CDF) gives the area to the left.
$P(x < x) = 1 – e^{-mx}$
$P(x < 5) = 1 – e(–0.25)(5) = 0.7135$
and
$(P(x < 4) = 1 – e^{(-0.25)(4)} = 0.6321$
You can do these calculations easily on a calculator.
The probability that a postal clerk spends four to five minutes with a randomly selected customer is
$P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814.$
On the home screen, enter (1 – e^(–0.25*5)) – (1 – e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5).
b. Find the 50th percentile.
$$P(x < k) = 0.50$$, $$k = 2.8$$ minutes (calculator or computer)
Half of all customers are finished within 2.8 minutes.
You can also do the calculation as follows:
$P(x < k) = 0.50$
and
$P(x < k) = 1 – e^{-0.25k}$
Therefore,
$0.50 = 1 − e^{-0.25k}$
and
$e^{-0.25k} = 1 − 0.50 = 0.5$
Take natural logs:
$\ln(e^{-0.25k}) = \ln(0.50).$
So,
$-0.25k = ln(0.50).$
Solve for $$k: k = \dfrac{ln(0.50)}{-0.25} = 0.28$$ minutes. The calculator simplifies the calculation for percentile k. See the following two notes.
A formula for the percentile $$k$$ is $$k = ln(1 − \text{Area To The Left}) - mk = ln(1 - \text{Area To The Left}) - m$$ where $$ln$$ is the natural log.
c. From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.
Note
Collaborative Exercise
On the home screen, enter ln(1 – 0.50)/–0.25. Press the (-) for the negative.
Exercise $$\PageIndex{2}$$
The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?
$$P(x < 10) = 0.4866$$
50th percentile = 10.40
Collaborative Exercise
Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.
Let $$X =$$ the amount of money a student in your class has in his or her pocket or purse.
The distribution for $$X$$ is approximately exponential with mean, $$\mu =$$ _______ and $$m =$$ _______. The standard deviation, $$\sigma =$$ ________.
Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than \$.40 in his or her pocket or purse. (Shade $$P(x < 0.40)$$).
Example $$\PageIndex{3}$$
On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.
1. What is the probability that a computer part lasts more than 7 years?
2. On the average, how long would five computer parts last if they are used one after another?
3. Eighty percent of computer parts last at most how long?
4. What is the probability that a computer part lasts between nine and 11 years?
a. Let $$x =$$ the amount of time (in years) a computer part lasts.
$\mu = 10$
so
$m = \dfrac{1}{\mu} = \dfrac{1}{10} = 0.1$
Find $$P(x > 7)$$. Draw the graph.
$P(x > 7) = 1 – P(x < 7).$
Since $$P(X < x) = 1 – e^{-mx}$$ then
$P(X > x) = 1 –(1 –e^{-mx}) = e^{-mx}$
$P(x > 7) = e^{(–0.1)(7)} = 0.4966.$
The probability that a computer part lasts more than seven years is 0.4966.
On the home screen, enter e^(-.1*7).
b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.
c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile.
Solve for $$k: k = \dfrac{ln(1-0.80)}{-0.1} = 16.1$$ years
Eighty percent of the computer parts last at most 16.1 years.
On the home screen, enter $$\dfrac{ln(1-0.80)}{-0.1}$$
d. Find $$P(9 < x < 11)$$. Draw the graph.
$P(9 < x < 11) = P(x < 11) - P(x < 9) = (1 - e^{(–0.1)(11)}) - (1 - e^{(–0.1)(9)}) = 0.6671 - 0.5934 = 0.0737.$
The probability that a computer part lasts between nine and 11 years is 0.0737.
On the home screen, enter e^(–0.1*9) – e^(–0.1*11).
Exercise $$\PageIndex{3}$$
On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?
$$P(x > 15) = 0.4346$$
Six pairs of running shoes would last 108 months on average.
80th percentile = 28.97 months
Example $$\PageIndex{4}$$
Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = $$\dfrac{1}{12}$$. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let $$X$$ = the length of a phone call, in minutes.
What is $$m$$, $$\mu$$, and $$\sigma$$? The probability that you must wait more than five minutes is _______ .
• $$m = \dfrac{1}{12}$$
• $$\mu = 12$$
• $$\sigma = 12$$
$$P(x > 5) = 0.6592$$
Exercise $$\PageIndex{4}$$
Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter $$\dfrac{1}{20}$$. Let $$S =$$ the distance people are willing to commute in miles. What is $$m$$, $$\mu$$, and $$\sigma$$? What is the probability that a person is willing to commute more than 25 miles?
$$m = \dfrac{1}{20}$$; $$\mu = 20$$; $$\sigma = 20$$; $$P(x > 25) = 0.2865$$
Example $$\PageIndex{5}$$
The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.
1. On average, how many minutes elapse between two successive arrivals?
2. When the store first opens, how long on average does it take for three customers to arrive?
3. After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.
4. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.
5. Seventy percent of the customers arrive within how many minutes of the previous customer?
6. Is an exponential distribution reasonable for this situation?
1. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.
2. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.
3. Let $$X =$$ the time between arrivals, in minutes. By part a, $$\mu = 2$$, so $$m = \dfrac{1}{2} = 0.5$$.
Therefore, $$X \sim Exp(0.5)$$.
The cumulative distribution function is $$P(X < x) = 1 – e(–0.5x)^{e}$$.
Therefore $$P(X < 1) = 1 - e^{(–0.5)(1)} \approx 0.3935$$.
$$1 - e^(–0.5) \approx 0.3935$$
Figure $$\PageIndex{8}$$.
$$P(X > 5) = 1 - P(X < 5) = 1 - (1 - e^{(-5)(0.5)}) = e^{-2.5} \approx 0.0821$$.
Figure $$\PageIndex{9}$$.
$$1 - (1 - e^{( – 5*0.5)}$$ or $$e^{(-5*0.5)}$$
4. We want to solve $$0.70 = P(X < x)$$ for $$x$$.
Substituting in the cumulative distribution function gives $$0.70 = 1 – e^{–0.5x}$$, so that $$e^{–0.5x} = 0.30$$. Converting this to logarithmic form gives $$-0.5x = ln(0.30)$$, or $$x = \dfrac{ln(0.30)}{-0.5} \approx 2.41$$ minutes.
Thus, seventy percent of customers arrive within 2.41 minutes of the previous customer.
You are finding the 70th percentile $$k$$ so you can use the formula $$k = \dfrac{ln(1-\text{Area To The left Of k})}{-m}$$
$$k = \dfrac{ln(1-0.70)}{(-0.5)} \approx 2.41$$ minutes
Figure $$\PageIndex{10}$$.
This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.
Exercise $$\PageIndex{5}$$
Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution.
1. On average, how many seconds elapse between two successive cars?
2. After a car passes by, how long on average will it take for another seven cars to pass by?
3. Find the probability that after a car passes by, the next car will pass within the next 20 seconds.
4. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds.
1. At a rate of five cars per minute, we expect $$\dfrac{60}{5} = 12$$ seconds to pass between successive cars on average.
2. Using the answer from part a, we see that it takes $$(12)(7) = 84$$ seconds for the next seven cars to pass by.
3. Let $$T =$$ the time (in seconds) between successive cars.
The mean of $$T$$ is 12 seconds, so the decay parameter is $$\dfrac{1}{12}$$ and $$T \sim Exp\dfrac{1}{12}$$. The cumulative distribution function of $$T$$ is $$P(T < t) = 1 – e^{−\dfrac{t}{12}}$$. Then $$P(T < 20) = 1 –e^{−\dfrac{20}{12}} \approx 0.8111$$.
Figure $$\PageIndex{11}$$.
$$P(T > 15) = 1 – P(T < 15) = 1 – (1 – e^{−\dfrac{15}{12}}) = e^{−\dfrac{15}{12}} \approx 0.2865$$.
## Memorylessness of the Exponential Distribution
In Example recall that the amount of time between customers is exponentially distributed with a mean of two minutes ($$X \sim Exp (0.5)$$). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says that
$P(X > r + t | X > r) = P(X > t)$
for all $$r \geq 0$$ and $$t \geq 0$$.
For example, if five minutes has elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using $$r = 5$$ and $$t = 1$$ in the foregoing equation.
$$P(X > 5 + 1 | X > 5) = P(X > 1) = e(–0.5)(1) e(–0.5)(1) \approx 0.6065$$.
This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival.
The exponential distribution is often used to model the longevity of an electrical or mechanical device. In Example, the lifetime of a certain computer part has the exponential distribution with a mean of ten years ($$X \sim Exp(0.1)$$). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is $$P(X > 17 | X > 10) = P(X > 7) = 0.4966$$.
Example $$\PageIndex{6}$$
Refer to Example where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?
The decay parameter of $$X$$ is $$m = \dfrac{1}{4} = 0.25$$, so $$X \sim Exp(0.25)$$.
The cumulative distribution function is $$P(X < x) = 1 - e^{–0.25x}$$.
We want to find $$P(X > 7 | X > 4)$$. The memoryless property says that $$P(X > 7 | X > 4) = P(X > 3)$$, so we just need to find the probability that a customer spends more than three minutes with a postal clerk.
This is $$P(X > 3) = 1 - P(X < 3) = 1 - (1 - e^{-0.25 \cdot 3}) = e^{–0.75} \approx 0.4724$$.
$$1 - (1 - e^(-0.25*2)) = e^(-0.25*2)$$.
Exercise $$\PageIndex{6}$$
Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.
Let $$T =$$ the lifetime of the light bulb. Then $$T \sim Exp\left(\dfrac{1}{8}\right)$$.
The cumulative distribution function is $$P(T < t) = 1 − e^{-\dfrac{t}{8}}$$
We need to find $$P(T > 19 | T = 12)$$. By the memoryless property,
$$P(T > 19 | T = 12) = P(T > 7) = 1 - P(T < 7) = 1 - (1 - e^{-7/8}) = e^{-7/8} \approx 0.4169$$.
1 - (1 – e^(–7/8)) = e^(–7/8).
## Relationship between the Poisson and the Exponential Distribution
There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of $$\mu$$ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean $$\lambda = \dfrac{1}{\mu}$$. Recall from the chapter on Discrete Random Variables that if $$X$$ has the Poisson distribution with mean $$\lambda$$, then $$P(X = k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}$$. Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. $$(k! = k*(k-1*)(k - 2)*(k - 3) \dotsc 3*2*1)$$
Suppose $$X$$ has the Poisson distribution with mean $$\lambda$$. Compute $$P(X = k)$$ by entering 2nd, VARS(DISTR), C: poissonpdf$$(\lambda, k$$). To compute $$P(X \leq k$$), enter 2nd, VARS (DISTR), D:poissoncdf($$\lambda, k$$).
Example $$\PageIndex{7}$$
At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution.
1. Find the average time between two successive calls.
2. Find the probability that after a call is received, the next call occurs in less than ten seconds.
3. Find the probability that exactly five calls occur within a minute.
4. Find the probability that less than five calls occur within a minute.
5. Find the probability that more than 40 calls occur in an eight-minute period.
1. On average there are four calls occur per minute, so 15 seconds, or $$\dfrac{15}{60} = 0.25$$ minutes occur between successive calls on average.
2. Let $$T =$$ time elapsed between calls. From part a, $$\mu = 0.25$$, so $$m = \dfrac{1}{0.25} = 4$$. Thus, $$T \sim Exp(4)$$.
The cumulative distribution function is $$P(T < t) = 1 - e^{–4t}$$.
The probability that the next call occurs in less than ten seconds (ten seconds $$= \dfrac{1}{6}$$ minute) is $$P(T < \dfrac{1}{6}) = 1 - e^{-4 (\dfrac{1}{6})} \approx 0.4866)$$.
Figure $$\PageIndex{13}$$
3. Let $$X =$$ the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute.
Therefore, $$X \sim Poisson(4)$$, and so $$P(X = 5) = \dfrac{4^{5}e^{-4}}{5!} \approx 0.1563$$. ($$5! = (5)(4)(3)(2)(1)$$)
$$\text{poissonpdf}(4, 5) = 0.1563$$.
4. Keep in mind that $$X$$ must be a whole number, so $$P(X < 5) = P(X \leq 4)$$.
To compute this, we could take $$P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$$.
Using technology, we see that $$P(X \approx 4) = 0.6288$$.
$$\text{poisssoncdf}(4, 4) = 0.6288$$
5. Let $$Y =$$ the number of calls that occur during an eight minute period.
Since there is an average of four calls per minute, there is an average of $$(8)(4) = 32$$ calls during each eight minute period.
Hence, $$Y \sim Poisson(32)$$. Therefore, $$P(Y > 40) = 1 - P(Y \leq 40) = 1 - 0.9294 = 0.0707$$.
$$1 - \text{poissoncdf}(32, 40). = 0.0707$$
Exercise $$\PageIndex{7}$$
In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.
1. Calculate the probability that there are at most 2 accidents occur in any given week.
2. What is the probability that there is at least two weeks between any 2 accidents?
1. Let $$X =$$ the number of accidents per week, so that $$X \sim Poisson(3)$$. We need to find $$P(X \leq 2) \approx 0.4232$$
$$\text{poissoncdf}(3, 2)$$
2. Let $$T =$$ the time (in weeks) between successive accidents.
Since the number of accidents occurs with a Poisson distribution, the time between accidents follows the exponential distribution.
If there are an average of three per week, then on average there is $$\mu = \dfrac{1}{3}$$ of a week between accidents, and the decay parameter is $$m = \dfrac{1}{\left(\dfrac{1}{3}\right)} = 3$$.
To find the probability that there are at least two weeks between two accidents, $$P(T > 2) = 1 - P(T < 2) = 1 – (1 – e(–3)(2)) = e^{–6} \approx 0.0025$$.
e^(-3*2).
## Review
If $$X$$ has an exponential distribution with mean $$\mu$$, then the decay parameter is $$m = \dfrac{1}{\mu}$$, and we write $$X \sim Exp(m)$$ where $$x \geq 0$$ and $$m > 0$$ . The probability density function of $$X$$ is $$f(x) = me^{-mx}$$ (or equivalently $$f(x) = \dfrac{1}{\mu}e^{-\dfrac{x}{\mu}}$$). The cumulative distribution function of $$X$$ is $$P(X \leq X) = 1 - e^{-mx}$$.
The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that $$P(X > x + k | X > x) = P(X > k)$$.
If $$T$$ represents the waiting time between events, and if $$T \sim Exp(\lambda)$$, then the number of events $$X$$ per unit time follows the Poisson distribution with mean $$\lambda$$. The probability density function of $$PX$$ is $$(X = k) = \dfrac{\lambda^{k}e^{-k}}{k!}$$. This may be computed using a TI-83, 83+, 84, 84+ calculator with the command $$\text{poissonpdf}(\lambda, k)$$. The cumulative distribution function $$P(X \leq k)$$ may be computed using the TI-83, 83+,84, 84+ calculator with the command $$\text{poissoncdf}(\lambda, k)$$.
## Formula Review
Exponential: $$X \sim Exp(m)$$ where $$m =$$ the decay parameter
• pdf: $$f(x) = me^{(–mx)}$$ where $$x \geq 0$$ and $$m > 0$$
• cdf: $$P(X \leq x) = 1 - e^{(–mx)}$$
• mean $$\mu = \dfrac{1}{m}$$
• standard deviation $$\sigma = \mu$$
• percentile $$k: k = \dfrac{ln(1 - \text{Area To The Left Of k})}{-m}$$
• $$P(X > x) = e^{(–mx)}$$
• $$P(a < X < b) = e^{(–ma)} - e^{(–mb)}$$
• Memoryless Property: $$P(X > x + k | X > x) = P(X > k)$$
• Poisson probability: $$P(X = k) = \dfrac{\lambda^{k}e^{k}}{k!}$$ with mean $$\lambda$$
• $$k! = k*(k - 1)*(k - 2)*(k - 3) \dotsc 3*2*1$$
## References
1. Data from the United States Census Bureau.
2. Data from World Earthquakes, 2013. Available online at http://www.world-earthquakes.com/ (accessed June 11, 2013).
3. “No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/bullpen/No-hitter (accessed June 11, 2013).
4. Zhou, Rick. “Exponential Distribution lecture slides.” Available online at www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf (accessed June 11, 2013).
## Glossary
decay parameter
The decay parameter describes the rate at which probabilities decay to zero for increasing values of $$x$$ . It is the value $$m$$ in the probability density function $$f(x) = me^{(-mx)}$$ of an exponential random variable. It is also equal to $$m = \dfrac{1}{\mu}$$ , where $$\mu$$ is the mean of the random variable.
memoryless property
For an exponential random variable $$X$$, the memoryless property is the statement that knowledge of what has occurred in the past has no effect on future probabilities. This means that the probability that $$X$$ exceeds $$x + k$$, given that it has exceeded $$x$$, is the same as the probability that $$X$$ would exceed $$k$$ if we had no knowledge about it. In symbols we say that $$P(X > x + k | X > x) = P(X > k)$$
Poisson distribution
If there is a known average of $$\lambda$$ events occurring per unit time, and these events are independent of each other, then the number of events $$X$$ occurring in one unit of time has the Poisson distribution. The probability of k events occurring in one unit time is equal to $$P(X = k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}$$. |
## Matrices and Determinants Questions and Answers Part-2
1. The inverse of a symmetric matrix (if it exists) is
a) a symmetric matrix
b) a skew symmetric matrix
c) a diagonal matrix
d) none of these
Answer: a
Explanation: Let A be an invertible symmetric matrix
2. The inverse of a skew symmetric matrix (if it exists) is
a) a symmetric matrix
b) a skew symmetric matrix
c) a diagonal matrix
d) none of these
Answer: b
Explanation: We have A' = – A
3. The inverse of a skew symmetric matrix of odd order is
a) a symmetric matrix
b) a skew symmetric matrix
c) diagonal matrix
d) does not exist
Answer: d
Explanation: Let A be a skew symmetric, matrix of order n
4. If A is an orthogonal matrix, then |A| is
a) 1
b) -1
c) $\pm1$
d) 0
Answer: c
Explanation: As A is an orthogonal matrix
5. If A is a $3\times3$ non-singular matrix, then adj (adj A) is equal to
a) $\mid A\mid A$
b) $\mid A\mid^{2} A$
c) $\mid A\mid^{-1} A$
d) 0
Answer: a
Explanation:
6. If A and B are two square matrices such that $B=-A^{-1} BA$ , then $\left(A+B \right)^{2}$ is equal to
a) O
b) $A^{2}+B^{2}$
c) $A^{2}+2AB+B^{2}$
d) A+B
Answer: b
Explanation: As B = – A–1 BA, we get AB = – BA
7. If $A=\begin{bmatrix}\alpha & \beta \\\gamma & -\alpha \end{bmatrix}$ is such that $A^{2}=I$ , then
a) $1+\alpha^{2}+\beta\gamma=0$
b) $1-\alpha^{2}-\beta\gamma=0$
c) $1-\alpha^{2}+\beta\gamma=0$
d) $1+\alpha^{2}-\beta\gamma=0$
Answer: b
Explanation:
8. The value of x for which the matrix $A=\begin{bmatrix}2 & 0 & 7 \\0 & 1 & 0 \\1 & -2 & 1\end{bmatrix}$ is inverse of $B=\begin{bmatrix}-x & 14x & 7x \\0 & 1 & 0 \\x & -4x & -2x\end{bmatrix}$
is
a) $\frac{1}{2}$
b) $\frac{1}{3}$
c) $\frac{1}{4}$
d) $\frac{1}{5}$
Answer: d
Explanation:
9. If $A\left(\alpha,\beta\right)=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\-\sin\alpha & \cos\alpha & 0 \\0 & 0 & e^{\beta}\end{bmatrix}$ , then $A\left(\alpha,\beta\right)^{-1}$ is equal to
a) $A\left(-\alpha,\beta\right)$
b) $A\left(-\alpha,-\beta\right)$
c) $A\left(\alpha,-\beta\right)$
d) $A\left(\alpha,\beta\right)$
Answer: b
Explanation:
10. If A is a $3\times 3$ skew-symmetric matrix, then trace of A is equal to
a) 1
b) 3
c) -1
d) $\mid A\mid$
Answer: d
Explanation: As A is a skew symmetric matrix |
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Last updated date: 25th Nov 2023
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Tara is $66$ inches tall and her son, Tom is $59\dfrac{7}{{12}}$ inches tall. How much taller is tara?
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Hint: According to the question, we shall see the difference in height of a mother with her son tom. We can see the difference by subtracting the height of Tom from the height of Tara. By this we can find out who is taller.
Complete step-by-step solution:
By the given statistics the mother is $66$ inches tall and her son Tom is $59\dfrac{7}{{12}}$ inches tall.
As per the analytics the equation can be written as,
$66 - 59\dfrac{7}{{12}}$ inches tall
In the above equation, the mother looks taller by the analytics. So, that is the reason why we are putting her height first.
Now, by solving the equation written above, we get:
$= 66 - 59\dfrac{7}{{12}}$
Now, when we solve the mixed fraction, we get:
$= 66 - 59.5833$
$= 6.4166$
Therefore, this is the difference between the height of Tara and Tom. So, we can say that Tara is $6.4166$inches taller than Tom.
Note: To find out the difference between their height, we subtracted the height of Tom from the height of Tara, because Tara’s height is much more than Tom’s height. If we had done the opposite by subtracting the height of Tara from the height of Tom, then the answer would have come in a negative value, which would be incorrect. |
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Pre-Calc Homework Solutions 110
# Pre-Calc Homework Solutions 110 - 110 49 Section 3.7 xy...
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49. xy 1 2 x 2 y 5 0 } d d x } ( xy ) 1 } d d x } (2 x ) 2 } d d x } ( y ) 5 0 x } d d y x } 1 ( y )(1) 1 2 2 } d d y x } 5 0 ( x 2 1) } d d y x } 5 2 2 2 y } d d y x } 5 } 2 x 2 2 2 1 y } 5 } 2 1 1 2 y x } Since the slope of the line 2 x 1 y 5 0 is 2 2, we wish to find points where the normal has slope 2 2, that is, where the tangent has slope } 1 2 } . Thus, we have } 2 1 1 2 y x } 5 } 1 2 } 2(2 1 y ) 5 1 2 x 4 1 2 y 5 1 2 x x 5 2 2 y 2 3 Substituting 2 2 y 2 3 in the original equation, we have: xy 1 2 x 2 y 5 0 ( 2 2 y 2 3) y 1 2( 2 2 y 2 3) 2 y 5 0 2 2 y 2 2 8 y 2 6 5 0 2 2( y 1 1)( y 1 3) 5 0 y 5 2 1 or y 5 2 3 At y 5 2 1, x 5 2 2 y 2 3 5 2 2 3 5 2 1. At y 5 2 3: x 5 2 2 y 2 3 5 6 2 3 5 3. The desired points are ( 2 1, 2 1) and (3, 2 3). Finally, we find the desired normals to the curve, which are the lines of slope 2 2 passing through each of these points. At ( 2 1, 2 1), the normal line is y 5 2 2( x 1 1) 2 1 or y 5 2 2 x 2 3. At (3, 2 3), the normal line is y 5 2 2( x 2 3) 2 3 or y 5 2 2 x 1 3. 50. x 5 y 2 } d d x } ( x ) 5 } d d x } ( y 2 ) 1 5 2 y } d d y x } } d d y x } 5 } 2 1 y } The normal line at ( x , y ) has slope 2 2 y . Thus, the normal
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# Algebra: Week 5 Problems
I've been working on this for the last few days. Some of the problems I understand and some I do not.
1. Perform the indicated operation:
2 (see the attached file).
3 5
2. Use order of operations to simplify: 25 - 3(22 + 1) =
3. Simplify: 5(2a - 3b) + 3(4a - b) =
4. Evaluate 2(5x + 1) when x = 2.
For problems 5 and 6, solve for x:
5. 6x + 3 = 15
6. 4(x + 3) =.3(x - 2)
7. Solve the inequality 5(x + 2) > 25
8. Solve the proportion:
9. Determine if the ordered pair (1,3) is a solution of 5x + y = 8.
For the equation 6x + 3y = 12,
10. Solve it for y.
11. Identify the slope.
12. Identify the y intercept.
13. Solve the system by the substitution method.
2x + y = 4
4x . ; 3y = .2
Solve each system by the addition method. If there is no solution or an infinite number of solutions, so state.
14. x + 3y = 2
4x + 12y = 8
15. 3x . 2y = 14
2x + 4y = 4
Solve the system by any method. If there is no solution or an infinite number of solutions, so state.
.7x + y = 2
21x . 3y = 4.
#### Solution Preview
Please see the attachment.
1. Perform the indicated operation:
2 −(−3 )
3 5
2. Use order of operations to simplify: 25 - 3(22 + 1)
25 - 3(22 + 1) = 25-3(4+1)=25-3*5=25-15=10
3. Simplify: 5(2a - 3b) + 3(4a - b) =
5(2a - 3b) + 3(4a - b) =(10a-15b)+12a-3b
= 10a+12a-15b-3b
=22a-18b
4. Evaluate 2(5x + 1) when x = 2.
2(5*2 + 1)=2(10+1)=2*11=22
For problems 5 and 6, solve for x:
5. 6x + 3 = 15
6x+3=15
6x=15-3=12
x=12/6=2
6. 4(x + 3) = −3(x - 2)
4(x + 3) = −3(x - 2)
4x+4*3= -3x+(-3)*(-2)
4x+12= -3x+6
4x+3x= -12+6
7x=-6
X=-6/7
7. Solve the inequality 5(x + 2) > ...
#### Solution Summary
The solution provides step-by-step method for solving equations and inequalities using substitution method. Formula for the calculation is also included.
\$2.19 |
# How do you solve 8- 2x = - 7?
Jun 11, 2018
$x = 7.5 \mathmr{and} 7 \frac{1}{2}$
#### Explanation:
To solve for the variable $x$, we have to make it by itself. To do so, first subtract $\textcolor{b l u e}{8}$ from both sides of the equation:
$8 - 2 x \quad \textcolor{b l u e}{- \quad 8} = - 7 \quad \textcolor{b l u e}{- \quad 8}$
$- 2 x = - 15$
Now divide both sides by $\textcolor{b l u e}{- 2}$:
$\frac{- 2 x}{\textcolor{b l u e}{- 2}} = \frac{- 15}{\textcolor{b l u e}{- 2}}$
Therefore,
$x = 7.5 \mathmr{and} 7 \frac{1}{2}$
Hope this helps! |
# 3.1: Introduction to Integers (Part 1)
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##### Learning Objectives
• Locate positive and negative numbers on the number line
• Order positive and negative numbers
• Find opposites
• Simplify expressions with absolute value
• Translate word phrases to expressions with integers
##### be prepared!
Before you get started, take this readiness quiz.
1. Plot $$0$$, $$1$$, and $$3$$ on a number line. If you missed this problem, review Example 1.1.1.
2. Fill in the appropriate symbol: ($$=$$, $$<$$, or $$>$$ ): $$2$$___$$4$$ If you missed this problem, review Example 2.1.2.
## Locate Positive and Negative Numbers on the Number Line
Do you live in a place that has very cold winters? Have you ever experienced a temperature below zero? If so, you are already familiar with negative numbers. A negative number is a number that is less than $$0$$. Very cold temperatures are measured in degrees below zero and can be described by negative numbers. For example, $$-1^{\circ}$$ F (read as “negative one degree Fahrenheit”) is $$1$$ degree below $$0$$. A minus sign is shown before a number to indicate that it is negative. Figure $$\PageIndex{1}$$ shows $$-20^{\circ}$$ F, which is $$20$$ degrees below $$0$$.
Figure $$\PageIndex{1}$$: Temperatures below zero are described by negative numbers.
Temperatures are not the only negative numbers. A bank overdraft is another example of a negative number. If a person writes a check for more than he has in his account, his balance will be negative.
Elevations can also be represented by negative numbers. The elevation at sea level is $$0$$ feet. Elevations above sea level are positive and elevations below sea level are negative. The elevation of the Dead Sea, which borders Israel and Jordan, is about $$1,302$$ feet below sea level, so the elevation of the Dead Sea can be represented as $$−1,302$$ feet. See Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{2}$$: The surface of the Mediterranean Sea has an elevation of 0 ft. The diagram shows that nearby mountains have higher (positive) elevations whereas the Dead Sea has a lower (negative) elevation.
Depths below the ocean surface are also described by negative numbers. A submarine, for example, might descend to a depth of $$500$$ feet. Its position would then be $$−500$$ feet as labeled in Figure $$\PageIndex{3}$$.
Figure $$\PageIndex{3}$$: Depths below sea level are described by negative numbers. A submarine 500 ft below sea level is at −500 ft.
Both positive and negative numbers can be represented on a number line. Recall that the number line created in Add Whole Numbers started at $$0$$ and showed the counting numbers increasing to the right as shown in Figure $$\PageIndex{4}$$. The counting numbers ($$1, 2, 3, …$$) on the number line are all positive. We could write a plus sign, $$+$$, before a positive number such as $$+2$$ or $$+3$$, but it is customary to omit the plus sign and write only the number. If there is no sign, the number is assumed to be positive.
Figure $$\PageIndex{4}$$
Now we need to extend the number line to include negative numbers. We mark several units to the left of zero, keeping the intervals the same width as those on the positive side. We label the marks with negative numbers, starting with $$-1$$ at the first mark to the left of $$0, -2$$ at the next mark, and so on. See Figure $$\PageIndex{5}$$.
Figure $$\PageIndex{5}$$: On a number line, positive numbers are to the right of zero. Negative numbers are to the left of zero. What about zero? Zero is neither positive nor negative.
The arrows at either end of the line indicate that the number line extends forever in each direction. There is no greatest positive number and there is no smallest negative number.
##### Example $$\PageIndex{1}$$: plot on the number line
Plot the numbers on a number line:
1. $$3$$
2. $$-3$$
3. $$-2$$
Solution
Draw a number line. Mark $$0$$ in the center and label several units to the left and right.
1. To plot $$3$$, start at $$0$$ and count three units to the right. Place a point as shown in Figure $$\PageIndex{6}$$.
Figure $$\PageIndex{6}$$
1. To plot $$-3$$, start at $$0$$ and count three units to the left. Place a point as shown in Figure $$\PageIndex{7}$$.
Figure $$\PageIndex{7}$$
1. To plot $$-2$$, start at $$0$$ and count two units to the left. Place a point as shown in Figure $$\PageIndex{8}$$.
Figure $$\PageIndex{8}$$
##### Exercise $$\PageIndex{1}$$
Plot the numbers on a number line.
$$1$$, $$-1$$, $$-4$$
##### Exercise $$\PageIndex{2}$$
Plot the numbers on a number line.
$$-4$$, $$4$$, $$-1$$
## Order Positive and Negative Numbers
We can use the number line to compare and order positive and negative numbers. Going from left to right, numbers increase in value. Going from right to left, numbers decrease in value. See Figure $$\PageIndex{9}$$.
Figure $$\PageIndex{9}$$
Just as we did with positive numbers, we can use inequality symbols to show the ordering of positive and negative numbers. Remember that we use the notation $$a < b$$ (read $$a$$ is less than $$b$$) when $$a$$ is to the left of $$b$$ on the number line. We write $$a > b$$ (read $$a$$ is greater than $$b$$) when $$a$$ is to the right of $$b$$ on the number line. This is shown for the numbers $$3$$ and $$5$$ in Figure $$\PageIndex{10}$$.
Figure $$\PageIndex{10}$$: The number 3 is to the left of 5 on the number line. So 3 is less than 5, and 5 is greater than 3.
The numbers lines to follow show a few more examples.
$$4$$ is to the right of $$1$$ on the number line, so $$4 > 1$$. $$1$$ is to the left of $$4$$ on the number line, so $$1 < 4$$.
$$-2$$ is to the left of $$1$$ on the number line, so $$−2 < 1$$. $$1$$ is to the right of $$−2$$ on the number line, so $$1 > −2$$.
$$−1$$ is to the right of $$−3$$ on the number line, so $$−1 > −3$$. $$−3$$ is to the left of $$−1$$ on the number line, so $$−3 < − 1$$.
##### Example $$\PageIndex{2}$$: order the pairs
Order each of the following pairs of numbers using $$<$$ or $$>$$:
1. $$14$$___$$6$$
2. $$−1$$___$$9$$
3. $$−1$$___$$−4$$
4. $$2$$___$$−20$$
Solution
Begin by plotting the numbers on a number line as shown in Figure $$\PageIndex{11}$$.
Figure $$\PageIndex{11}$$
Compare 14 and 6. 14___6 14 is to the right of 6 on the number line. 14 > 6
Compare −1 and 9. −1___9 −1 is to the left of 9 on the number line. −1 < 9
Compare −1 and −4. −1___−4 −1 is to the right of −4 on the number line. −1 > −4
Compare 2 and −20. 2___−20 2 is to the right of −20 on the number line. 2 > −20
##### Exercise $$\PageIndex{3}$$
Order each of the following pairs of numbers using $$<$$ or $$>$$.
1. $$15$$___$$7$$
2. $$−2$$___$$5$$
3. $$−3$$___$$−7$$
4. $$5$$___$$−17$$
$$>$$
$$<$$
$$>$$
$$>$$
##### Exercise $$\PageIndex{4}$$
Order each of the following pairs of numbers using $$<$$ or $$>$$.
1. $$8$$___$$13$$
2. $$3$$___$$−4$$
3. $$−5$$___$$−2$$
4. $$9$$___$$−21$$
$$<$$
$$>$$
$$<$$
$$>$$
## Find Opposites
On the number line, the negative numbers are a mirror image of the positive numbers with zero in the middle. Because the numbers $$2$$ and $$−2$$ are the same distance from zero, they are called opposites. The opposite of $$2$$ is $$−2$$, and the opposite of $$−2$$ is $$2$$ as shown in Figure $$\PageIndex{12 a}$$. Similarly, $$3$$ and $$−3$$ are opposites as shown in Figure $$\PageIndex{12 b}$$.
Figure $$\PageIndex{12}$$
##### Definition: opposite
The opposite of a number is the number that is the same distance from zero on the number line, but on the opposite side of zero.
##### Example $$\PageIndex{3}$$:
Find the opposite of each number:
1. $$7$$
2. $$−10$$
Solution
1. The number $$−7$$ is the same distance from $$0$$ as $$7$$, but on the opposite side of $$0$$. So $$−7$$ is the opposite of $$7$$ as shown in Figure $$\PageIndex{13}$$.
Figure $$\PageIndex{13}$$
1. The number $$10$$ is the same distance from $$0$$ as $$−10$$, but on the opposite side of $$0$$. So $$10$$ is the opposite of $$−10$$ as shown in Figure $$\PageIndex{14}$$.
Figure $$\PageIndex{14}$$
##### Exercise $$\PageIndex{5}$$
Find the opposite of each number:
1. $$4$$
2. $$−3$$
$$-4$$
$$3$$
##### Exercise $$\PageIndex{6}$$
Find the opposite of each number:
1. $$8$$
2. $$−5$$
$$-8$$
$$5$$
### Opposite Notation
Just as the same word in English can have different meanings, the same symbol in algebra can have different meanings.
The specific meaning becomes clear by looking at how it is used. You have seen the symbol “$$−$$”, in three different ways.
10 - 4 Between two numbers, the symbol indicates the operation of subtraction. We read 10 − 4 as 10 minus 4. -8 In front of a number, the symbol indicates a negative number. We read −8 as negative eight. -x In front of a variable or a number, it indicates the opposite. We read −x as the opposite of x . − (−2) Here we have two signs. The sign in the parentheses indicates that the number is negative 2. The sign outside the parentheses indicates the opposite. We read − (−2) as the opposite of −2.
##### Definition: Opposite Notation
$$−a$$ means the opposite of the number $$a$$.
The notation $$−a$$ is read the opposite of $$a$$.
##### Example $$\PageIndex{4}$$: simplify
Simplify: $$−(−6)$$.
Solution
−(−6) The opposite of −6 is 6. 6
##### Exercise $$\PageIndex{7}$$
Simplify: $$−(−1)$$
$$1$$
##### Exercise $$\PageIndex{8}$$
Simplify: $$−(−5)$$
$$5$$
### Integers
The set of counting numbers, their opposites, and $$0$$ is the set of integers.
##### Definition: integers
Integers are counting numbers, their opposites, and zero.
$\ldots −3, −2, −1, 0, 1, 2, 3 \ldots \nonumber$
We must be very careful with the signs when evaluating the opposite of a variable.
##### Example $$\PageIndex{5}$$: evaluate
Evaluate $$−x$$:
1. when $$x = 8$$
2. when $$x = −8$$
Solution
To evaluate −x when x = 8, substitute 8 for x. $$-x$$ Substitute $$\textcolor{red}{8}$$ for x. $$-(\textcolor{red}{8})$$ Simplify. $$-8$$
To evaluate −x when x = −8, substitute 8 for x. $$-x$$ Substitute $$\textcolor{red}{-8}$$ for x. $$-(\textcolor{red}{-8})$$ Simplify. $$-8$$
##### Exercise $$\PageIndex{9}$$
Evaluate $$−n$$:
1. when $$n = 4$$
2. when $$n = −4$$
$$-4$$
$$4$$
##### Exercise $$\PageIndex{10}$$
Evaluate $$−m$$:
1. when $$m = 11$$
2. when $$m = −11$$
$$-11$$
$$11$$
## Simplify Expressions with Absolute Value
We saw that numbers such as $$5$$ and $$−5$$ are opposites because they are the same distance from $$0$$ on the number line. They are both five units from $$0$$. The distance between $$0$$ and any number on the number line is called the absolute value of that number. Because distance is never negative, the absolute value of any number is never negative. The symbol for absolute value is two vertical lines on either side of a number. So the absolute value of $$5$$ is written as $$|5|$$, and the absolute value of $$−5$$ is written as $$|−5|$$ as shown in Figure $$\PageIndex{15}$$.
Figure $$\PageIndex{15}$$
##### Definition: Absolute Value
The absolute value of a number is its distance from $$0$$ on the number line.
The absolute value of a number $$n$$ is written as $$|n|$$.
$|n| \geq 0\; for\; all\; numbers \nonumber$
##### Example 3.6:
Simplify:
1. $$|3|$$
2. $$|−44|$$
3. $$|0|$$
Solution
|3| 3 is 3 units from zero. 3
|-44| -44 is 44 units from zero. 44
|0| 0 is already at zero. 0
##### Exercise $$\PageIndex{11}$$
Simplify:
1. $$|12|$$
2. $$− |−28|$$
$$12$$
$$-28$$
##### Exercise 3.12:
Simplify:
1. $$|9|$$
2. $$− |37|$$
$$9$$
$$-37$$ |
Introduction to P.M.I
Question
# The identity ${1}^{3}+{2}^{3}+{3}^{3}+\dots +{\mathrm{n}}^{3}$ is equal to
Easy
Solution
## Let the given statement be P(n).$\mathrm{P}\left(\mathrm{n}\right):{1}^{3}+{2}^{3}+{3}^{3}+\dots +{\mathrm{n}}^{3}={\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]}^{2}$Step I : For n=1, $\mathrm{P}\left(1\right):{\left[\frac{1\left(1+1\right)}{2}\right]}^{2}={\left[\frac{1×2}{2}\right]}^{2}={1}^{2}=1={1}^{3}$ which is trueStep ll: Let it is true for n = k, ${1}^{3}+{2}^{3}+{3}^{3}+\dots +{\mathrm{k}}^{3}={\left[\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}\right]}^{2}------\left(\mathrm{i}\right)$Step lll: For n=k+1, $\left({1}^{3}+{2}^{3}+{3}^{3}+{4}^{3}+\dots +{\mathrm{k}}^{3}\right)+\left(\mathrm{k}+1{\right)}^{3}$ $={\left[\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}\right]}^{2}+\left(\mathrm{k}+1{\right)}^{3}$ [using Eq. (i)] $\begin{array}{r}=\frac{{\mathrm{k}}^{2}\left(\mathrm{k}+1{\right)}^{2}}{4}+\frac{\left(\mathrm{k}+1{\right)}^{3}}{1}\\ =\frac{{\mathrm{k}}^{2}\left(\mathrm{k}+1{\right)}^{2}+4\left(\mathrm{k}+1{\right)}^{3}}{4}\end{array}$ On taking (k + 1)2 common in numerator Part, $\begin{array}{l}=\frac{\left(\mathrm{k}+1{\right)}^{2}\left[{\mathrm{k}}^{2}+4\left(\mathrm{k}+1\right)\right]}{4}\\ =\frac{\left(\mathrm{k}+1{\right)}^{2}\left({\mathrm{k}}^{2}+4\mathrm{k}+4\right)}{4}\\ =\frac{\left(\mathrm{k}+1{\right)}^{2}\left(\mathrm{k}+2{\right)}^{2}}{4}\\ =\frac{\left(\mathrm{k}+1{\right)}^{2}\left[\left(\mathrm{k}+1\right)+1{\right]}^{2}}{4}\\ ={\left[\frac{\left(\mathrm{k}+1\right)\left\{\left(\mathrm{k}+1\right)+1\right\}}{2}\right]}^{2}\end{array}$Therefore, P(k + 1) is true when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.
Get Instant Solutions |
# A circle has a chord that goes from ( 3 pi)/2 to (7 pi) / 4 radians on the circle. If the area of the circle is 99 pi , what is the length of the chord?
Jun 21, 2016
7.62 units
#### Explanation:
First, use a unit circle to determine the end points of the chord on the circle.
If each endpoint on the chord is connected to the center of the circle, an isosceles triangle is formed whose congruent sides each have a length of $r$, the length of the radius.
The angle between the two equivalent sides of the triangle is equal to the difference between the angles given in the problem:
$\theta = \frac{7 \pi}{4} - \frac{3 \pi}{2} = \frac{\pi}{4} r a \mathrm{di} a n s$
Finally, the law of cosines can be used to determine an equation for the length of the chord:
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \theta$
Since both $a$ and $b$ are equal to $r$, the formula can be rewritten as:
${c}^{2} = {r}^{2} + {r}^{2} - 2 \cdot r \cdot r \cdot \cos \theta$
${c}^{2} = 2 {r}^{2} - 2 {r}^{2} \cos \theta$
${c}^{2} = 2 {r}^{2} \cdot \left(1 - \cos \theta\right)$
The problem states that the area of the circle is $99 \pi$. This allows us to solve for ${r}^{2}$:
$A = \pi {r}^{2}$
$\frac{A}{\pi} = {r}^{2}$
${r}^{2} = \frac{99 \pi}{\pi} = 99$
Plug this value into the equation for the chord:
${c}^{2} = 2 {r}^{2} \cdot \left(1 - \cos \theta\right)$
${c}^{2} = 2 \cdot 99 \cdot \left(1 - \cos \left(\frac{\pi}{4}\right)\right)$
${c}^{2} = 198 \cdot \left(1 - 0.707\right)$
${c}^{2} = 58.014$
$c = 7.62$
Note: Since the units of length are not provided, just use "units." |
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# How Many Millions in a Billion?
Sep 21, 2021
Mathematics is an interesting subject. But that is a very subjective statement someone can make since you may find something easy and interesting while others may find it difficult and boring. No matter who finds it easy or difficult, one thing is sure people often get confused between millions and billions. At times, we lose track of the number of zeros that will come after 1.
How many millions are in a billion? Can you answer that right now? Without using anything? If yes, congratulations! You are a genius. But some people would not be able to do that.
Let us say you won a lottery worth one billion, or your long-distance uncle asks you to convert a million into a billion to test your knowledge. What would you do? Refer to your mathematics book to check the number of zeros? Or you would start guessing, “How many millions in a billion? How many billions are in a trillion?”
Is it six, seven, eight, or ten zeros after “1” or more? No idea? Worry no more because we are here to break it down for you.
With this blog, we will talk about all the ins and outs of a million and a billion that will help you remember everything about them. So, get ready to master your ability to count zeroes in a million and a billion and get your basics clear. But before we talk about how many millions in a billion, we need to understand what a million and billion are. Let us start with the definition of number systems.
## Number Systems
• There are many number systems out there that you can use to represent a given number in numerals. For example, we have the Indian number system and the International number system.
• Both the systems are quite different from each other since they have different expressions and forms for representing numerals.
• Hence, to convert billions into trillions, you must know both the Indian number system and the International number system.
We can define a numeral system as a number system that represents a mathematical form for consistently expressing the numbers of a given range by using digits or other signs.
1. As per the Indian number system, we use terms like thousands, lakhs, and crores.
2. While in the International number system, on the other hand, we use terms like billion, million, and trillion.
## How Much is a Million?
The first step towards getting the answer to how many millions in a billion.
One million is written as 1,000,000, which is equal to one thousand thousand. It is the natural number that comes after 999,999 and before 1,000,001. The word million has been derived from the early Italian millione. It is commonly abbreviated as m. Remember, a million has six zeroes after the 1.
The word “million” has the same meaning in the short scale and long scale numbering systems. But the larger numbers have different names in the two systems. The mathematical notation of a million is 106. It means that we need to multiply 10 by 10, six times like:
10 × 10 × 10 × 10 × 10 × 10.
And the answer will be a million.
Sometimes, we use “million” as a metaphor for a very large number. For example, “Not in a million years” and “You’re one in a million.” In the examples, it doesn’t necessarily mean that a million years have passed. It could be less than that.
### Fun Facts About a Million
We have successfully understood everything about a million and have covered the first step of getting the answer to our question: how many millions in a billion? Here are some fun facts:
• 1,000,000 is also the square of 1000.
• 1, 000,000 is the cube of 100.
Don’t trust us? Go ahead and multiply 100 by 100 and see what’s the result. You can do the same to find its cube: 100 × 100 × 100.
## What is a Billion?
Now that we have covered the first step towards understanding how many millions in a billion, we are good to start with our next step which is the definition of a billion. But before we move ahead, tell us how many zeroes are there in a million? Kudos to you if you got that right? If not, scroll up a bit to get the answer.
As we keep counting in number systems, we go from 0 to 9, 9 to 99, 100 to 999, and so on. We start getting five-digit, six-digit, and ten-digit numbers, and it keeps going up as we keep counting. A billion is the first and smallest 10-digit number that is formed by a 1 followed by nine zeros.
• A billion is a very large number that has nine zeros after the 1.
• It is also known as a thousand million because if you multiply a thousand with a million, you will get a billion as your answer. Try it: 1,000 × 1000,000 = ?
Have you ever heard the word “gigameter?” Well, now you have. The prefix “Giga” in this word refers to billion. It means that 1 gigameter is equal to a billion meters.
### Fun Facts About a Billion
• Billion has two notations. One is 109 which is followed in the American system, including India.
• The other one is 1012, which means that 1 is followed by 12 zeros, which is followed in many non-American countries like France and Germany.
## Technique to Remember Zeroes in Billion
Reading about the definitions of a million and billion is not enough. To answer our question, “how many millions in a billion,” we need to keep in the mind the number of zeros in both figures. Hence, we have to remember the number of zeros in a million and a billion. Here are some points to help you remember this:
• We have six zeroes in a million.
• The mathematical notation of a million is 106.
• There are nine zeros in a billion: 1,000,000,000.
• The mathematical notation (American) of a million is 109.
We can only convert a million into a billion if we know the number of zeros in both figures. Hence, this step is very important for answering: how many millions in a billion?
## How Many Millions in a Billion?
And finally, the moment has arrived when we answer your question: how many millions in a billion. Since we have covered the definitions of a million and a billion, it should not be difficult to convert them into each other. But before we go ahead and break the answer down for you, step-by-step, we want you to try it on your own. Once done, you can match your answer with our solution to see if you were right.
Let us give you some hints to make it easier:
• Here, we need to convert 1 billion into a million so that we can see how many millions in a billion.
• To do this, we can either use the fact that 1 billion is equal to 1,000,000,000 and 1 million is equal to 1,000,000 and then do some simple calculations.
• Or, there is another way: we can simply multiply the given million amount or value by 1000. It will give us the value of 1 billion in million.
1. Here, we will convert 1 billion into millions.
2. As we discussed above in the definitions of a million and a billion, 1 billion is equal to 1,000,000,000 and 1 million is equal to 1,000,000.
1 million = 1,000,000 and
1 billion = 1,000,000,000
3.Since we need the 1 billion value, we need to multiply 1000 in the equation of 1 million that is: 1 million = 1000000.
4. It will give us the desired value: 1 million×1000 = 1000000×1000
5. Now we will simplify both sides to get the value of 1 billion as, 1000 million = 1000000000
6. Since 1 billion = 1000000000 and 1000 million = 1000000000, we can say that 1 billion = 1000 million.
Therefore, we can say that 1000 million will make 1 billion and 1 billion has 1000 million.
### Alternate Method
We can also solve this question by using the relation: 1 billion = 1000 million directly. In this relation,
1. 1 will be multiplied on both sides to get the value of 1 billion in million as a billion terms is 1.
2. There is one more direct relationship that we can use to get the required answer for this question: 1 million = 0.001 billion.
3. A billion terms are 0.001. To convert it into 1 and get the value of 1 billion, we need to multiply it by 1000.
4. Using both the relations, we will get the same answer that is 1000 million will make 1 billion.
## How Many Millions in a Billion: Quick Answer
To convert a million value into a billion value, i.e., to go from million to billion, all you need to do is multiply the million term by 1,000. In other words, there are 1,000 million in a billion. Look below for more clarity.
1,000,000 × 1,000 = 1,000,000,000
It means that if you multiply one million with one thousand, the answer will be one million. Let us say you need to convert four million into billions. In such a case, just multiply it with 1000 like:
4,000,000 × 1,000 = 4,000,000,000
It means that one billion has one thousand million.
### Remember
• There are six zeroes in a million (or two groups of three zeroes).
• There are nine zeroes in a billion (or three groups of three zeroes).
## First Million, Then Billion, What Is Next — A Trillion?
Well, we have already discussed enough to answer our question: how many millions in a billion. After carefully understanding the definitions of a million and a billion, we did some calculations and found our answer. But what if the same uncle asks you to convert a million into a trillion? Or a billion into a trillion? What would you do?
To save the day, we are going to discuss a trillion, too. It will boost your knowledge further give you an edge over others. Some of your friends may not know about a trillion, but you will. So get ready to learn about a trillion and show off a little bit.
## What is a Trillion?
One trillion is written as 1,000,000,000,000 in which the number “1” is followed by 12 zeroes Its mathematical notation is 1012. You can remember it as 4 groups of 3 zeros after the 1.
## How Many Billions in a Trillion?
1. As we discussed in the definition of a trillion, one trillion is written as 1,000,000,000,000 in the International number system. It has 12 zeros after the 1.
2. Also, we know that one billion is written as 1,000,000,000. It has 9 zeroes after the 1.
3. If we multiply 1000 with one billion, the answer will be one trillion. See below:
One thousand × One Billion = One Trillion
1000 × 1,000,000,000 = 1,000,000,000,000
It means that there are one thousand billion in one trillion.
One trillion = One thousand billion
## How Many Billions in a Trillion: Quick Answer
As we solve above and found that there are 1,000 million in a billion, there are 1,000 billion in a trillion.
1,000,000,000 × 1,000 = 1,000,000,000,000
Don’t get confused by so many zeroes. Remember the tricks that we have discussed:
• There are nine zeroes in a billion (or three groups of three zeroes).
• There are 12 zeroes in a trillion (or four groups of three zeroes).
## How Many Crores in a Trillion?
In the given question, we need to convert a number from one number system to another. Here is the complete step-by-step answer:
1. As we discussed in the definition of a trillion, one trillion is written as 1,000,000,000,000 in the International number system. It has 12 zeros after the 1.
2. Now, we have to convert the number one trillion into the Indian number system.
3. In the Indian number system, 1 crore is equal to 1,00,00,000. It has seven zeroes after the 1.
4. To find out the number of crores in one trillion, we need to divide one trillion by one crore.
1000000000000/10000000 = 100000
Hence, we can see that after dividing one trillion by one crore, the answer is 1,00,000. It means that there is 1,00,000 crore in one trillion.
One trillion = Ten thousand crore
## Million to Billion to Trillion: Comparison
When we deal with numbers, it becomes difficult to remember them all. And when we talk about big numbers like a million, billion, and trillion, it becomes even hard. To this point, we have already talked about the definitions of these three big numbers. We also converted them into each other through some calculations and found our answer to how many millions in a billion.
Now, it is time for some revision, and make some short but crisp notes to help you remember them all. We are going to compare all three — a million, billion, and trillion — with some fun facts that will make the process of remembering them easy for you.
### One Million
• One million is one thousand thousand.
• If you stacked one million pennies on top of each other, trying to make a big tower, it would go as high as a mile.
• If we decide to divide a million dollars between the total population of the United States, each person would get around ⅓ of one cent.
### One Billion
• One billion is one thousand million.
• If you stacked one billion pennies on top of each other, trying to make a big tower, it would go as high as 870 miles.
• If we decide to divide a billion dollars between the total population of the United States, each person would get around \$3.33.
### One Trillion
• One trillion is one thousand billion.
• If you stacked one trillion pennies on top of each other, trying to make a big tower, the tower would go as high as 870,000 miles. It means that your tower would reach the moon and back and back to the moon again.
• If we decide to divide a trillion dollars between the total population of the United States, each person would get around \$3,000.
The population of the United States is 32.82 crores. Now convert it into a million, billion, and a trillion. You know everything you need. Once done, verify the statements we made above. Do your answers match ours?
## Difference Between Billion, Million, and Trillion
In a number system, the counting always goes up in increasing order. So, the increasing order here is million, billion, and trillion. That means,
• We get a billion when we multiply 1000 by million
• We get a trillion when we multiply 1000 to a billion.
### Points to Remember About a Million
• A million is a thousand times a thousand.
• It can be written as 1,000 × 1,000 = 1,000,000.
• In scientific notation, it is represented as 1 × 106.
• In simple words, it has 6 zeroes.
### Points to Remember About a Billion
• A billion is a thousand times a million.
• It can be written as 1,000 × 1,000,000 =1,000,000,000.
• In scientific notation, it is represented as 1 × 109.
• In simple words, it has 9 zeroes.
### Points to Remember About a Trillion
• A trillion is a thousand times a billion.
• It can be written as 1,000 × 1,000,000,000 =1,000,000,000,000.
• In scientific notation, it is represented as 1 × 1012.
• In simple words, it has 12 zeroes.
Point One Million One Billion One Trillion Definition A million is a thousand times a thousand. A billion is a thousand times a million. A trillion is a million times a million. Written Form 1,000 × 1,000 = 1,000,000. 1,000 × 1,000,000 =1,000,000,000. 1,000,000 × 1,000,000 =1,000,000,000,000. Scientific Notation 1 × 106 1 × 109 1 × 1012 Number of Zeros 6 zeroes 9 zeroes 12 zeroes
## Solved Example on Billion
Question: How Much is 1 Billion Dollars in Rupees?
To solve this question, you need to use all the knowledge you have gathered about a million and a billion from this article. We will solve this step by step. You can solve it on your own and then can match the answer and procedure with ours. Let us get started!
### Some Key Points:
• In Indian rupees, one billion is equal to 10,000 lakhs.
One billion = Ten thousand lakhs × 1,00,000
1,000,000,000 = 10,000
• We need to convert the American number system into the Indian number system:
From dollars to Indian rupees
### Step 1: One Billion in Rupees
Currently, 1 dollar = 73.45 rupees
Value of 1 billion in rupees = 1,000,000,000 rupees
To get the value of 1 billion dollars in rupees, we need to multiply one billion with 73.45. Look below:
1,000,000,000 x 73.45 =7.34 x 1010 Indian Rupees
Similarly, if we ask you to convert 5 billion dollars into Indian rupees, you can follow the same process:
5 billion dollars in rupees =73.45 x 5,000,000,000 = 367 x 1011
### Step 2: Converting the American number system into the Indian number system
1 billion in rupees = 1,000,000,000 Rupees
We know that 1 lakh = 1,00,000
So, we can write one billion as:
1 Billion = 10,000 Lakhs (Just divide one billion by one lakh)
It means that one billion in lakhs is equal to 10,000 lakhs.
One billion = Ten thousand lakhs
## Practice Questions on Billion
Try these questions to test your knowledge:
• How many millions are there in a billion?
• How many billions are there in a million?
• How many lakhs are there in a trillion?
• How many crores are there in a billion?
#### Public vs Private Colleges – Which is better for you?
With approximately 4,300 colleges and universities in the United States, …
#### Allusion Definition and Examples
An allusion is a short reference to a person, object, …
#### PEMDAS: What Does It Mean and Why Does It Matter?
In mathematics, the order of operations is important to solve … |
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### Course: Algebra (all content)>Unit 3
Lesson 2: x-intercepts and y-intercepts
# Intercepts of lines review (x-intercepts and y-intercepts)
The x-intercept is where a line crosses the x-axis, and the y-intercept is the point where the line crosses the y-axis. Thinking about intercepts helps us graph linear equations.
## What are intercepts?
The $x$-intercept is the point where a line crosses the $x$-axis, and the $y$-intercept is the point where a line crosses the $y$-axis.
Want a deeper introduction to intercepts? Check out this video.
## Example: Intercepts from a graph
Looking at the graph, we can find the intercepts.
The line crosses the axes at two points:
The point on the $x$-axis is $\left(5,0\right)$. We call this the $x$-intercept.
The point on the $y$-axis is $\left(0,4\right)$. We call this the $y$-intercept.
Want to learn more about finding intercepts from graphs? Check out this video.
## Example: Intercepts from a table
We're given a table of values and told that the relationship between $x$ and $y$ is linear.
$x$$y$
$1$$-9$
$3$$-6$
$5$$-3$
Then we're asked to find the intercepts of the corresponding graph.
The key is realizing that the $x$-intercept is the point where $y=0$, and the $y$-intercept is where $x=0$.
The point $\left(7,0\right)$ is our $x$-intercept because when $y=0$, we're on the $x$-axis.
To find the $y$-intercept, we need to "zoom in" on the table to find where $x=0$.
The point $\left(0,-10.5\right)$ is our $y$-intercept.
Want to learn more about finding intercepts from tables? Check out this video.
## Example: Intercepts from an equation
We're asked to determine the intercepts of the graph described by the following linear equation:
$3x+2y=5$
To find the $y$-intercept, let's substitute $x=0$ into the equation and solve for $y$:
$\begin{array}{rl}3\cdot 0+2y& =5\\ 2y& =5\\ y& =\frac{5}{2}\end{array}$
So the $y$-intercept is $\left(0,\frac{5}{2}\right)$.
To find the $x$-intercept, let's substitute $y=0$ into the equation and solve for $x$:
$\begin{array}{rl}3x+2\cdot 0& =5\\ 3x& =5\\ x& =\frac{5}{3}\end{array}$
So the $x$-intercept is $\left(\frac{5}{3},0\right)$.
Want to learn more about finding intercepts from equations? Check out this video.
## Practice
Problem 1
Determine the intercepts of the line graphed below.
$x$-intercept:
$\left($
$,$
$\right)$
$y$-intercept:
$\left($
$,$
$\right)$
Want more practice? Check out these exercises:
## Want to join the conversation?
• im in 8th and its hard to keep all this stuff in your head
• I agree. I'm in eighth and confused.
• help me solve this problem step by step 1/3x-2 find the x,y intercept
• there is no interception points because that isn't a linear equation
• Math can be fun sometimes if you do it right lol
• it was sort of an obligation for me to be here but by seeing the progress I made in only 9 days ( i used to know almost nothing about math) I'm now addicted to learning it and i can't stop it's really fun
(my eyes are burning from the screen rn cuz i've been studying for hours straight)
• How do i find the y and x intercepts of an equation in standard form??
• You can always find the X-intercept by setting Y to 0 in the equation and solve for X.
Similarly, you can always find the Y-intercept by setting X to 0 in the equation and solve for Y.
Hope this helps.
• if the question is y=5x+random number how to find x intercept?
• In all equations, you find the x-intercept by using y=0 in the equation and solving for x.
• how do i put a fraction in
• this type of stuff is soooo confusing and too me it give off little explaination when it be like "well we r gon' to zoom in" like child what in da world how do we "zoom in" or "zoom out"? i am i 8th grade but sometime when oing this math it makes me feel like a 9th grader in the 8th grade!! does anyone else agree?
• i mean, teachers do say 8th grade is just a transition to 9th, or maybe thats just my school, who knows.
• what is the x- intercept in the equation y=8/-1x-22
• To find x-intercept, take y=0
0 = 8/-1x-22
-x-22 = 8
-x = -8 + 22
-x = 14
x = -14
Therefore, x-intercept = (-14,0) [Assuming I got your question right] |
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## Class 11 Math (India) - Hindi
### Course: Class 11 Math (India) - Hindi>Unit 7
Lesson 1: Limits intro
# Limits intro
Limits describe how a function behaves near a point, instead of at that point. This simple yet powerful idea is the basis of all of calculus.
To understand what limits are, let's look at an example. We start with the function $f\left(x\right)=x+2$.
The limit of $f$ at $x=3$ is the value $f$ approaches as we get closer and closer to $x=3$. Graphically, this is the $y$-value we approach when we look at the graph of $f$ and get closer and closer to the point on the graph where $x=3$.
For example, if we start at the point $\left(1,3\right)$ and move on the graph until we get really close to $x=3$, then our $y$-value (i.e. the function's value) gets really close to $5$.
Similarly, if we start at $\left(5,7\right)$ and move to the left until we get really close to $x=3$, the $y$-value again will be really close to $5$.
For these reasons we say that the limit of $f$ at $x=3$ is $5$.
You might be asking yourselves what's the difference between the limit of $f$ at $x=3$ and the value of $f$ at $x=3$, i.e. $f\left(3\right)$.
So yes, the limit of $f\left(x\right)=x+2$ at $x=3$ is equal to $f\left(3\right)$, but this isn't always the case. To understand this, let's look at function $g$. This function is the same as $f$ in every way except that it's undefined at $x=3$.
Just like $f$, the limit of $g$ at $x=3$ is $5$. That's because we can still get very very close to $x=3$ and the function's values will get very very close to $5$.
So the limit of $g$ at $x=3$ is equal to $5$, but the value of $g$ at $x=3$ is undefined! They are not the same!
That's the beauty of limits: they don't depend on the actual value of the function at the limit. They describe how the function behaves when it gets close to the limit.
Problem 1
This is the graph of $h$.
What is a reasonable estimate for the limit of $h$ at $x=3$?
We also have a special notation to talk about limits. This is how we would write the limit of $f$ as $x$ approaches $3$:
The symbol $lim$ means we're taking a limit of something.
The expression to the right of $lim$ is the expression we're taking the limit of. In our case, that's the function $f$.
The expression $x\to 3$ that comes below $lim$ means that we take the limit of $f$ as values of $x$ approach $3$.
Problem 2
This is the graph of $f$.
What is a reasonable estimate for $\underset{x\to 6}{lim}f\left(x\right)$ ?
Problem 3
Which expression represents the limit of ${x}^{2}$ as $x$ approaches $5$?
## In limits, we want to get infinitely close.
What do we mean when we say "infinitely close"? Let's take a look at the values of $f\left(x\right)=x+2$ as the $x$-values get very close to $3$. (Remember: since we're dealing with limits we don't care about $f\left(3\right)$ itself.)
$x$$f\left(x\right)$
$2.9$$4.9$
$2.99$$4.99$
We can see how, when the $x$-values are smaller than $3$ but become closer and closer to it, the values of $f$ become closer and closer to $5$.
$x$$f\left(x\right)$
$3.1$$5.1$
$3.01$$5.01$
We can also see how, when the $x$-values are larger than $3$ but become closer and closer to it, the values of $f$ become closer and closer to $5$.
Notice that the closest we got to $5$ was with $f\left(2.999\right)=4.999$ and $f\left(3.001\right)=5.001$, which are $0.001$ units away from $5$.
We can get closer than that if we want. For example, suppose we wanted to be $0.00001$ units from $5$, then we would pick $x=3.00001$ and then $f\left(3.00001\right)=5.00001$.
This is endless. We can always get closer to $5$. But that's exactly what "infinitely close" is all about! Since being "infinitely close" isn't possible in reality, what we mean by $\underset{x\to 3}{lim}f\left(x\right)=5$ is that no matter how close we want to get to $5$, there's an $x$-value very close to $3$ that will get us there.
If you find this hard to grasp, maybe this will help: how do we know there are infinite different integers? It's not like we've counted them all and got to infinity. We know they are infinite because for any integer there's another integer that's even larger than that. There's always another one, and another one.
In limits, we don't want to get infinitely big, but infinitely close. When we say $\underset{x\to 3}{lim}f\left(x\right)=5$, we mean we can always get closer and closer to $5$.
Problem 4
$x$$g\left(x\right)$
$-7.1$$6.32$
$-7.01$$6.1$
$-7.001$$6.03$
$-6.999$$6.03$
$-6.99$$6.1$
$-6.9$$6.32$
What is a reasonable estimate for $\underset{x\to -7}{lim}g\left(x\right)$?
## Another example: $\underset{x\to 2}{lim}{x}^{2}$
Let's analyze $\underset{x\to 2}{lim}{x}^{2}$, which is the limit of the expression ${x}^{2}$ when $x$ approaches $2$.
We can see how, when we approach the point where $x=2$ on the graph, the $y$-values are getting closer and closer to $4$.
We can also look at a table of values:
$x$${x}^{2}$
$1.9$$3.61$
$1.99$$3.9601$
$x$${x}^{2}$
$2.1$$4.41$
$2.01$$4.0401$
We can also see how we can get as close as we want to $4$. Suppose we want to be less than $0.001$ units from $4$. Which $x$-value close to $x=2$ can we choose?
Let's try $x=2.001$:
${2.001}^{2}=4.004001$
That's more than $0.001$ units away from $4$. Alright, so let's try $x=2.0001$:
${2.0001}^{2}=4.00040001$
That's close enough! By trying $x$-values that are closer and closer to $x=2$, we can get even closer to $4$.
In conclusion, $\underset{x\to 2}{lim}{x}^{2}=4$.
## A limit must be the same from both sides.
Coming back to $f\left(x\right)=x+2$ and $\underset{x\to 3}{lim}f\left(x\right)$, we can see how $5$ is approached whether the $x$-values increase towards $3$ (this is called "approaching from the left") or whether they decrease towards $3$ (this is called "approaching from the right").
Now take, for example, function $h$. The $y$-value we approach as the $x$-values approach $x=3$ depends on whether we do this from the left or from the right.
When we approach $x=3$ from the left, the function approaches $4$. When we approach $x=3$ from the right, the function approaches $6$.
When a limit doesn't approach the same value from both sides, we say that the limit doesn't exist.
Problem 5
This is the graph of function $g$.
Which of the limits exists?
## Want to join the conversation?
• In the last question, how does
x→7
lim
g(x)
exist? It has two locations right?
• The limit exists because the same y-value is approached from both sides. It does not have two locations because the open circle is a just gap in the graph. The closed circle is the actual y-value for when x=7.
• first day learning calculus :D
• Same
• In problem 5 why can one of the answers be x→6? but not x→3?
• I assume you are talking about the last example. As you approach x=6 from the left you move closer to 3; AND as you approach x=6 from the right, you also moce closer to 3.
As you move closer to x=3 form the left you move closer to 3, BUT when you move closer to x=3 from the right, you move closer to 6.
They must be moving to the same value of y from both sides if you are not going to specify the side in th limit notation.
• For the last question, how is there a limit for x-->6? There is a point there. I thought there could only be limits if there were open dots.
• As Sal explained both in the video Limits intro, and in the text, the beauty of limits, and one property of limits, is that they do not explain the actual point of the graph, but the behavior leading up to that point. Whether there is a point at f(x)= 6 or a hole, that would not change that there still is a limit, unless a jump occurs between the two leading lines.
• learning this calculus concept here was actually super fun. although i know this will take many hours, the hours will be exhilarating! i'm going to try and conquer calculus in 2 weeks. (today is june 24th, i have until july 10th)
• How is calc going so far for ya'll
• the plot has open dots and closed dots. what does that symbolize in limits?
• Open dots means it doesn't include that point and closed dots mean it does include that point. For example, a open dot at 6 means it can't be 6 but it can be 5.999999 or 6.000001, just not 6. A closed circle for the point 6 would mean it includes 6.
• over a year ago is was on algebra 1 videos now I'm on calculus finally. lets goo
• Bro what? Congratulations, but be sure you have all the fundamentals down from Algebra 1 and 2 along with Geometry and precal. You don't want to jump ahead into something that you don't understand. Be sure to have strong foundations before continuing.
(1 vote)
• what is the difference between APcalculus AB and APcalculus BC? |
# Ordered Trees
Wow, there sure are a lot of trees.
## Definition
An ordered tree is an oriented tree in which the children of a node are somehow "ordered."
If T1 and T2 are ordered trees then T1 ≠ T2 else T1 = T2.
## Types of Ordered Trees
There are several types of ordered trees:
• k-ary tree
• Binomial tree
• Fibonacci tree
## Fibonoacci Trees
A Fibonacci Tree (Fk) is defined by
• F0 is the empty tree
• F1 is a tree with only one node
• Fk+2 is a node whose left subtree is a Fk+1 tree and whose right subtree is a Fk tree.
Exercise: Draw the trees F0 through F5.
## Binomial Trees
The Binomial Tree (Bk) consists of a node with k children. The first child is the root of a Bk-1 tree, the second is the root of a Bk-2 tree, etc.
Exercise: Draw the trees B0 through B5.
## k-ary Trees
A k-ary tree is a tree in which the chilren of a node appear at distinct index positions in 0..k-1. This means the maximum number of children for a node is k.
Some k-ary trees have special names
• 2-ary trees are called binary trees.
• 3-ary trees are called trinary trees or ternary trees.
• 1-ary trees are called lists.
You have to draw k-ary trees carefully. In a binary tree, if a node has one child, you can't draw a vertical line from the parent to the child. Every child in a binary tree is either a left or a right child; that must be made clear.
### Representations of k-ary Trees
Since the subtrees of a node are both indexed and bounded, we can use an array for them. Here is an example for a 4-ary tree:
```class Node {
private Object data;
private Node[] children;
.
.
.
}
```
You can also use an array for the whole tree
1 A 2 B 3 4 C 5 F 6 7 8 9 10 11 E 12 D 13 G
In a binary tree
• parent(i) is i / 2
• leftChild(i) is 2i
• rightChild(i) is 2i + 1
In a k-ary tree
• parent(i) is (k + i – 2) / k
• jth child of i is ki – (k–2) + j
Exercise: Prove each of these statements.
### Complete k-ary Trees
The array representation may waste a lot of space. The maximally space efficient k-ary tree is called a complete tree. A complete tree is completely filled out on every level, except perhaps on the last one, on which all we require is that all its nodes are "as far to the left as possible." Examples:
Complete trees can be packed into an array with no wasted space at all.
### Perfect Trees
A perfect k-tree is a k-ary tree in which every node is either a 0-node or a k-node and all leaves are on the same level.
Perfect trees are sometimes called full trees, though some people say a full tree is just one in which every node is a 0-node or a k-node, without caring whether all leaves are on the same level or not.
### The Size of k-ary Trees
For k = 2:
Level Nodes on
Level
Nodes on levels
up to and
including this one
0 1 = 20 1 = 20+1–1
1 2 = 21 3 = 21+1–1
2 4 = 22 7 = 22+1–1
3 8 = 23 15 = 23+1–1
4 16 = 24 31 = 24+1–1
h 2h 2h+1–1
For arbitrary k:
Number of nodes on level h = kh, obviously
Let S(h) = the number of nodes in a perfect binary tree of height h. Let's find a closed form.
``` S(h) = 1 + k + k2 + k3 + ... + kh
k × S(h) = k + k2 + k3 + k4 + ... + kh+1
k × S(h) - S(h) = kh+1-1
(k-1) × S(h) = kh+1-1
kh+1 - 1
S(h) = ---------
k - 1
``` |
# calculus 08 Techniques of Integration
```8
Techniques of Integration
Over the next few sections we examine some techniques that are frequently successful when
seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be
apparent that the function you wish to integrate is a derivative in some straightforward
way. For example, faced with
Z
x10 dx
we realize immediately that the derivative of x11 will supply an x10 : (x11 )′ = 11x10 . We
don’t want the “11”, but constants are easy to alter, because differentiation “ignores” them
in certain circumstances, so
d 1 11
1
x =
11x10 = x10 .
dx 11
11
From our knowledge of derivatives, we can immediately write down a number of antiderivatives. Here is a list of those most often used:
Z
xn+1
+ C, if n 6= −1
xn dx =
n+1
Z
x−1 dx = ln |x| + C
Z
ex dx = ex + C
Z
sin x dx = − cos x + C
163
164
Chapter 8 Techniques of Integration
Z
Z
Z
cos x dx = sin x + C
sec2 x dx = tan x + C
sec x tan x dx = sec x + C
1
dx = arctan x + C
1 + x2
Z
1
√
dx = arcsin x + C
1 − x2
Z
8.1
Substitution
Needless to say, most problems we encounter will not be so simple. Here’s a slightly more
complicated example: find
Z
2x cos(x2 ) dx.
This is not a “simple” derivative, but a little thought reveals that it must have come from
an application of the chain rule. Multiplied on the “outside” is 2x, which is the derivative
of the “inside” function x2 . Checking:
d
d
sin(x2 ) = cos(x2 ) x2 = 2x cos(x2 ),
dx
dx
so
Z
2x cos(x2 ) dx = sin(x2 ) + C.
Even when the chain rule has “produced” a certain derivative, it is not always easy to
see. Consider this problem:
Z
p
x3 1 − x2 dx.
p
There are two factors in this expression, x3 and 1 − x2 , but it is not apparent that the
chain rule is involved. Some clever rearrangement reveals that it is:
Z
x
3
p
1−
x2
dx =
Z
p
1
(−2x) −
(1 − (1 − x2 )) 1 − x2 dx.
2
This looks messy, but we do now have something that looks like the result of the chain
√
rule: the function 1 − x2 has been substituted into −(1/2)(1 − x) x, and the derivative
8.1
Substitution
165
of 1 − x2 , −2x, multiplied on the outside. If we can find a function F (x) whose derivative
√
is −(1/2)(1 − x) x we’ll be done, since then
p
1
d
2
′
2
(1 − (1 − x2 )) 1 − x2
F (1 − x ) = −2xF (1 − x ) = (−2x) −
dx
2
p
= x3 1 − x2
But this isn’t hard:
√
1
− (1 − x) x dx =
2
1
− (x1/2 − x3/2 ) dx
2
1 2 3/2 2 5/2
+C
=−
x − x
2 3
5
1
1
x3/2 + C.
x−
=
5
3
Z
Z
(8.1.1)
So finally we have
Z
p
1
1
2
2
x 1 − x dx =
(1 − x2 )3/2 + C.
(1 − x ) −
5
3
3
So we succeeded, but it required a clever first step, rewriting the original function so
that it looked like the result of using the chain rule. Fortunately, there is a technique that
makes such problems simpler, without requiring cleverness to rewrite a function in just the
right way. It does sometimes not work, or may require more than one attempt, but the
idea is simple: guess at the most likely candidate for the “inside function”, then do some
algebra to see what this requires the rest of the function to look like.
One frequently good guess is any complicated expression inside a square root, so we
start by trying u = 1 − x2 , using a new variable, u, for convenience in the manipulations
that follow. Now we know that the chain rule will multiply by the derivative of this inner
function:
du
= −2x,
dx
so we need to rewrite the original function to include this:
Z
x
3
p
1 − x2 =
Z
√ −2x
x u
dx =
−2x
3
Z
x2 √ du
dx.
u
−2
dx
Recall that one benefit of the Leibniz notation is that it often turns out that what looks
like ordinary arithmetic gives the correct answer, even if something more complicated is
166
Chapter 8 Techniques of Integration
going on. For example, in Leibniz notation the chain rule is
dy dt
dy
=
.
dx
dt dx
The same is true of our current expression:
Z
x2 √ du
u
dx =
−2
dx
Z
x2 √
u du.
−2
Now we’re almost there: since u = 1 − x2 , x2 = 1 − u and the integral is
Z
√
1
− (1 − u) u du.
2
It’s no coincidence that this is exactly the integral we computed in (8.1.1), we have simply
renamed the variable u to make the calculations less confusing. Just as before:
Z
√
1
1
1
u3/2 + C.
u−
− (1 − u) u du =
2
5
3
Then since u = 1 − x2 :
Z
p
1
1
3
2
x 1 − x2 dx =
(1 − x2 )3/2 + C.
(1 − x ) −
5
3
To summarize: if we suspect that a given function is the derivative of another via the
chain rule, we let u denote a likely candidate for the inner function, then translate the
given function so that it is written entirely in terms of u, with no x remaining in the
expression. If we can integrate this new function of u, then the antiderivative of the
original function is obtained by replacing u by the equivalent expression in x.
Even in simple cases you may prefer to use this mechanical procedure, since it often
helps to avoid silly mistakes. For example, consider again this simple problem:
Z
2x cos(x2 ) dx.
Let u = x2 , then du/dx = 2x or du = 2x dx. Since we have exactly 2x dx in the original
integral, we can replace it by du:
Z
Z
2
2x cos(x ) dx = cos u du = sin u + C = sin(x2 ) + C.
This is not the only way to do the algebra, and typically there are many paths to the
correct answer. Another possibility, for example, is: Since du/dx = 2x, dx = du/2x, and
8.1
Substitution
167
then the integral becomes
Z
Z
Z
du
2
= cos u du.
2x cos(x ) dx = 2x cos u
2x
The important thing to remember is that you must eliminate all instances of the original
variable x.
Z
EXAMPLE 8.1.1 Evaluate (ax + b)n dx, assuming that a and b are constants, a 6= 0,
and n is a positive integer. We let u = ax + b so du = a dx or dx = du/a. Then
Z
Z
1 n
1
1
n
(ax + b) dx =
u du =
un+1 + C =
(ax + b)n+1 + C.
a
a(n + 1)
a(n + 1)
EXAMPLE 8.1.2
Evaluate
Z
sin(ax + b) dx, assuming that a and b are constants and
a 6= 0. Again we let u = ax + b so du = a dx or dx = du/a. Then
Z
Z
1
1
1
sin(ax + b) dx =
sin u du = (− cos u) + C = − cos(ax + b) + C.
a
a
a
EXAMPLE 8.1.3
Evaluate
Z
4
x sin(x2 ) dx. First we compute the antiderivative, then
2
evaluate the definite integral. Let u = x2 so du = 2x dx or x dx = du/2. Then
Z
Z
1
1
1
2
x sin(x ) dx =
sin u du = (− cos u) + C = − cos(x2 ) + C.
2
2
2
Now
Z
2
4
1
x sin(x ) dx = − cos(x2 )
2
4
2
2
1
1
= − cos(16) + cos(4).
2
2
A somewhat neater alternative to this method is to change the original limits to match
the variable u. Since u = x2 , when x = 2, u = 4, and when x = 4, u = 16. So we can do
this:
Z 4
Z 16
16
1
1
1
1
2
x sin(x ) dx =
sin u du = − (cos u) = − cos(16) + cos(4).
2
2
2
2
2
4
4
An incorrect, and dangerous, alternative is something like this:
Z 4
Z 4
4
4
1
1
1
1
1
2
2
x sin(x ) dx =
sin u du = − cos(u) = − cos(x ) = − cos(16) + cos(4).
2
2
2
2
2
2 2
2
2
Z 4
1
This is incorrect because
sin u du means that u takes on values between 2 and 4, which
2 2
1
is wrong. It is dangerous, because it is very easy to get to the point − cos(u)
2
4
and forget
2
168
Chapter 8 Techniques of Integration
1
1
to substitute x2 back in for u, thus getting the incorrect answer − cos(4) + cos(2). A
2
2
somewhat clumsy, but acceptable, alternative is something like this:
Z
4
2
x sin(x ) dx =
2
Z
x=4
x=2
x=4
1
1
sin u du = − cos(u)
2
2
x=2
4
1
= − cos(x2 )
2
2
=−
cos(16) cos(4)
+
.
2
2
1/2
cos(πt)
dt. Let u = sin(πt) so du = π cos(πt) dt or
2
1/4 sin (πt)
√
du/π = cos(πt) dt. We change the limits to sin(π/4) = 2/2 and sin(π/2) = 1. Then
EXAMPLE 8.1.4
Z
1/2
1/4
Evaluate
cos(πt)
dt =
sin2 (πt)
Z
1
√
2/2
Z
1 1
du =
π u2
Z
1
√
2/2
1 −2
1 u−1
u du =
π
π −1
1
√
2/2
√
2
1
.
=− +
π
π
Exercises 8.1.
Find the antiderivatives or evaluate the definite integral in each problem.
Z
Z
9
1.
(1 − t) dt ⇒
2.
(x2 + 1)2 dx ⇒
Z
Z
1
2
100
√
dt ⇒
3.
x(x + 1) dx ⇒
4.
3
1 − 5t
Z
Z p
3
5.
sin x cos x dx ⇒
6.
x 100 − x2 dx ⇒
Z
Z
x2
√
dx ⇒
8.
cos(πt) cos sin(πt) dt ⇒
7.
1 − x3
Z
Z
sin x
9.
dx
⇒
10.
tan x dx ⇒
cos3 x
Z π
Z
5
11.
sin (3x) cos(3x) dx ⇒
12.
sec2 x tan x dx ⇒
0
13.
Z
15.
Z
17.
Z
19.
Z
√
π/2
2
2
2
x sec (x ) tan(x ) dx ⇒
0
4
3
1
dx ⇒
(3x − 7)2
6x
dx ⇒
(x2 − 7)1/9
14.
Z
π/6
16.
Z
1
18.
Z
20.
Z
0
−1
1
−1
sin7 x dx ⇒
sin(tan x)
dx ⇒
cos2 x
(cos2 x − sin2 x) dx ⇒
(2x3 − 1)(x4 − 2x)6 dx ⇒
f (x)f ′ (x) dx ⇒
8.2
8.2
Powers of sine and
Powers of sine and cosine
169
osine
Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is
straightforward. Some examples will suffice to explain the approach.
Evaluate
EXAMPLE 8.2.1
Z
5
sin x dx =
Z
Z
sin5 x dx. Rewrite the function:
4
sin x sin x dx =
Z
2
2
sin x(sin x) dx =
Z
sin x(1 − cos2 x)2 dx.
Now use u = cos x, du = − sin x dx:
Z
2
2
sin x(1 − cos x) dx =
=
Z
Z
−(1 − u2 )2 du
−(1 − 2u2 + u4 ) du
2
1
= −u + u3 − u5 + C
3
5
1
2
= − cos x + cos3 x − cos5 x + C.
3
5
EXAMPLE 8.2.2
Evaluate
function:
Z
6
sin x dx =
Z
Z
2
3
sin6 x dx. Use sin2 x = (1 − cos(2x))/2 to rewrite the
(1 − cos 2x)3
dx
8
Z
1
=
1 − 3 cos 2x + 3 cos2 2x − cos3 2x dx.
8
Z
(sin x) dx =
Now we have four integrals to evaluate:
Z
and
Z
1 dx = x
3
−3 cos 2x dx = − sin 2x
2
170
Chapter 8 Techniques of Integration
are easy. The cos3 2x integral is like the previous example:
Z
Z
3
− cos 2x dx = − cos 2x cos2 2x dx
Z
= − cos 2x(1 − sin2 2x) dx
Z
1
= − (1 − u2 ) du
2
u3
1
u−
=−
2
3
sin3 2x
1
sin 2x −
.
=−
2
3
And finally we use another trigonometric identity, cos2 x = (1 + cos(2x))/2:
Z
Z
1 + cos 4x
sin 4x
3
2
3 cos 2x dx = 3
x+
.
dx =
2
2
4
So at long last we get
Z
x
sin3 2x
3
sin 4x
3
1
6
sin x dx = −
sin 2x −
+
x+
+ C.
sin 2x −
8 16
16
3
16
4
EXAMPLE 8.2.3 Evaluate
Z
sin2 x cos2 x dx. Use the formulas sin2 x = (1−cos(2x))/2
and cos2 x = (1 + cos(2x))/2 to get:
Z
Z
1 − cos(2x) 1 + cos(2x)
2
2
·
dx.
sin x cos x dx =
2
2
The remainder is left as an exercise.
Exercises 8.2.
Find Zthe antiderivatives.
1.
3.
Z
5.
Z
7.
Z
9.
Z
sin2 x dx ⇒
2.
Z
sin3 x dx ⇒
sin4 x dx ⇒
4.
Z
cos2 x sin3 x dx ⇒
6.
Z
sin2 x cos2 x dx ⇒
8.
Z
sin x(cos x)3/2 dx ⇒
10.
Z
tan3 x sec x dx ⇒
3
cos x dx ⇒
3
2
2
2
cos x sin x dx ⇒
sec x csc x dx ⇒
8.3
8.3
Trigonometri
Trigonometric Substitutions
171
Substitutions
So far we have seen that it sometimes helps to replace a subexpression of a function by
a single variable. Occasionally it can help to replace the original variable by something
more complicated. This seems like a “reverse” substitution, but it is really no different in
principle than ordinary substitution.
EXAMPLE 8.3.1
Evaluate
Z p
1 − x2 dx. Let x = sin u so dx = cos u du. Then
Z p
Z p
Z √
2
2
1 − x dx =
1 − sin u cos u du =
cos2 u cos u du.
√
We would like to replace cos2 u by cos u, but this is valid only if cos u is positive, since
√
cos2 u is positive. Consider again the substitution x = sin u. We could just as well think
of this as u = arcsin x. If we do, then by the definition of the arcsine, −π/2 ≤ u ≤ π/2, so
cos u ≥ 0. Then we continue:
Z √
cos2
u cos u du =
=
Z
2
cos u du =
Z
u sin 2u
1 + cos 2u
du = +
+C
2
2
4
arcsin x sin(2 arcsin x)
+
+ C.
2
4
This is a perfectly good answer, though the term sin(2 arcsin x) is a bit unpleasant. It is
possible to simplify this. Using the identity sin 2x = 2 sin x cos x, we can write sin 2u =
q
p
p
2
2 sin u cos u = 2 sin(arcsin x) 1 − sin u = 2x 1 − sin2 (arcsin x) = 2x 1 − x2 . Then the
full antiderivative is
√
√
arcsin x x 1 − x2
arcsin x 2x 1 − x2
+
=
+
+ C.
2
4
2
2
This type of substitution is usually indicated when the function you wish to integrate
contains a polynomial expression that might allow you to use the fundamental identity
sin2 x + cos2 x = 1 in one of three forms:
cos2 x = 1 − sin2 x
sec2 x = 1 + tan2 x
tan2 x = sec2 x − 1.
If your function contains 1 − x2 , as in the example above, try x = sin u; if it contains 1 + x2
try x = tan u; and if it contains x2 − 1, try x = sec u. Sometimes you will need to try
something a bit different to handle constants other than one.
172
Chapter 8 Techniques of Integration
EXAMPLE 8.3.2
Z p
Evaluate
4 − 9x2 dx. We start by rewriting this so that it looks
more like the previous example:
Z p
4−
9x2
Z p
Z p
2
dx =
4(1 − (3x/2) ) dx = 2 1 − (3x/2)2 dx.
Now let 3x/2 = sin u so (3/2) dx = cos u du or dx = (2/3) cos u du. Then
Z
2
p
1−
(3x/2)2
dx =
=
=
=
=
=
Z
4
2 1 − sin u (2/3) cos u du =
cos2 u du
3
4u 4 sin 2u
+
+C
6
12
2 arcsin(3x/2) 2 sin u cos u
+
+C
3
3
2 arcsin(3x/2) 2 sin(arcsin(3x/2)) cos(arcsin(3x/2))
+
+C
3
3
p
2 arcsin(3x/2) 2(3x/2) 1 − (3x/2)2
+
+C
3
3
√
2 arcsin(3x/2) x 4 − 9x2
+
+ C,
3
2
Z
p
2
using some of the work from example 8.3.1.
EXAMPLE 8.3.3
Z p
1+
Evaluate
x2
Z p
1 + x2 dx. Let x = tan u, dx = sec2 u du, so
Z p
Z √
2
2
dx =
1 + tan u sec u du =
sec2 u sec2 u du.
Since u = arctan(x), −π/2 ≤ u ≤ π/2 and sec u ≥ 0, so
Z √
sec2
2
u sec u du =
Z
√
sec2 u = sec u. Then
sec3 u du.
In problems of this type, two integrals come up frequently:
Z
sec3 u du and
Both have relatively nice expressions but they are a bit tricky to discover.
R
sec u du.
8.3
First we do
R
Trigonometric Substitutions
sec u du, which we will need to compute
Z
sec u du =
=
Z
Z
sec u
Z
173
sec3 u du:
sec u + tan u
du
sec u + tan u
sec2 u + sec u tan u
du.
sec u + tan u
Now let w = sec u + tan u, dw = sec u tan u + sec2 u du, exactly the numerator of the
function we are integrating. Thus
Z
Z
Z
1
sec2 u + sec u tan u
du =
dw = ln |w| + C
sec u du =
sec u + tan u
w
= ln | sec u + tan u| + C.
Now for
Z
sec3 u du:
sec3 u sec3 u
sec3 u (tan2 u + 1) sec u
sec u =
+
=
+
2
2
2
2
3
2
3
sec u sec u tan u sec u
sec u + sec u tan2 u sec u
=
+
+
=
+
.
2
2
2
2
2
3
We already know how to integrate sec u, so we just need the first quotient. This is “simply”
a matter of recognizing the product rule in action:
Z
sec3 u + sec u tan2 u du = sec u tan u.
So putting these together we get
Z
sec u tan u ln | sec u + tan u|
+
+ C,
sec3 u du =
2
2
and reverting to the original variable x:
Z p
sec u tan u ln | sec u + tan u|
+
+C
1 + x2 dx =
2
2
sec(arctan x) tan(arctan x) ln | sec(arctan x) + tan(arctan x)|
+
+C
2
2
√
√
ln | 1 + x2 + x|
x 1 + x2
+
+ C,
=
2
2
q
p
using tan(arctan x) = x and sec(arctan x) = 1 + tan2 (arctan x) = 1 + x2 .
=
174
Chapter 8 Techniques of Integration
Exercises 8.3.
Find Zthe antiderivatives.
1.
3.
5.
7.
9.
11.
8.4
csc x dx ⇒
2.
Z
csc3 x dx ⇒
Z p
4.
9 + 4x2 dx ⇒
Z
p
6.
x2 1 − x2 dx ⇒
Z p
x2 + 2x dx ⇒
8.
Z p
x2 − 1 dx ⇒
Z p
x 1 − x2 dx ⇒
Z
1
√
dx ⇒
1 + x2
Z
1
dx ⇒
2
x (1 + x2 )
√
Z
x
√
dx ⇒
1−x
x2
√
dx ⇒
4 − x2
Z
x3
√
dx ⇒
12.
4x2 − 1
10.
Z
Integration by Parts
We have already seen that recognizing the product rule can be useful, when we noticed
that
Z
sec3 u + sec u tan2 u du = sec u tan u.
As with substitution, we do not have to rely on insight or cleverness to discover such
antiderivatives; there is a technique that will often help to uncover the product rule.
d
f (x)g(x) = f ′ (x)g(x) + f (x)g ′ (x).
dx
We can rewrite this as
f (x)g(x) dx +
Z
f (x)g ′ (x) dx,
f (x)g (x) dx = f (x)g(x) −
Z
f ′ (x)g(x) dx.
f (x)g(x) =
and then
Z
Z
′
′
This may not seem particularly useful at first glance, but it turns out that in many cases
we have an integral of the form
Z
f (x)g ′ (x) dx
but that
Z
f ′ (x)g(x) dx
is easier. This technique for turning one integral into another is called integration by
parts, and is usually written in more compact form. If we let u = f (x) and v = g(x) then
8.4
Integration by Parts
175
du = f ′ (x) dx and dv = g ′ (x) dx and
Z
u dv = uv −
Z
v du.
To use this technique we need to identify likely candidates for u = f (x) and dv = g ′ (x) dx.
EXAMPLE 8.4.1
Z
Evaluate
let dv = x dx so v = x2 /2 and
Z
x2 ln x
x ln x dx =
−
2
EXAMPLE 8.4.2
Z
Evaluate
dv = sin x dx so v = − cos x and
Z
x sin x dx = −x cos x −
EXAMPLE 8.4.3
Z
Z
x ln x dx. Let u = ln x so du = 1/x dx. Then we must
x2 1
x2 ln x
dx =
−
2 x
2
Evaluate
x
x2 ln x x2
dx =
−
+ C.
2
2
4
x sin x dx. Let u = x so du = dx. Then we must let
− cos x dx = −x cos x +
Z
Z
Z
cos x dx = −x cos x + sin x + C.
sec3 x dx. Of course we already know the answer to this,
but we needed to be clever to discover it. Here we’ll use the new technique to discover the
antiderivative. Let u = sec x and dv = sec2 x dx. Then du = sec x tan x dx and v = tan x
and
Z
Z
3
sec x dx = sec x tan x − tan2 x sec x dx
Z
= sec x tan x − (sec2 x − 1) sec x dx
Z
Z
3
= sec x tan x − sec x dx + sec x dx.
176
Chapter 8 Techniques of Integration
At first this looks useless—we’re right back to
Z
Z
3
sec x dx +
2
Z
Z
Z
sec3 x dx. But looking more closely:
3
sec x dx = sec x tan x −
3
sec x dx = sec x tan x +
3
sec x dx = sec x tan x +
Z
Z
3
sec x dx +
Z
sec x dx
sec x dx
Z
sec x dx
Z
Z
sec x tan x 1
3
sec x dx =
sec x dx
+
2
2
=
sec x tan x ln | sec x + tan x|
+
+ C.
2
2
Z
Evaluate
x2 sin x dx. Let u = x2 , dv = sin x dx; then du = 2x dx
Z
Z
2
2
and v = − cos x. Now
x sin x dx = −x cos x + 2x cos x dx. This is better than the
EXAMPLE 8.4.4
original integral, but we need to do integration by parts again. Let u = 2x, dv = cos x dx;
then du = 2 and v = sin x, and
Z
2
2
x sin x dx = −x cos x +
Z
2x cos x dx
Z
2
= −x cos x + 2x sin x − 2 sin x dx
= −x2 cos x + 2x sin x + 2 cos x + C.
Such repeated use of integration by parts is fairly common, but it can be a bit tedious to
accomplish, and it is easy to make errors, especially sign errors involving the subtraction in
the formula. There is a nice tabular method to accomplish the calculation that minimizes
the chance for error and speeds up the whole process. We illustrate with the previous
example. Here is the table:
sign
u
dv
u
dv
x2
sin x
x2
sin x
−
2x
− cos x
−2x
− cos x
−
0
0
cos x
2
− sin x
cos x
or
2
− sin x
8.4
Integration by Parts
177
To form the first table, we start with u at the top of the second column and repeatedly
compute the derivative; starting with dv at the top of the third column, we repeatedly
compute the antiderivative. In the first column, we place a “−” in every second row. To
form the second table we combine the first and second columns by ignoring the boundary;
if you do this by hand, you may simply start with two columns and add a “−” to every
second row.
To compute with this second table we begin at the top. Multiply the first entry in
column u by the second entry in column dv to get −x2 cos x, and add this to the integral
of the product of the second entry in column u and second entry in column dv. This gives:
2
−x cos x +
Z
2x cos x dx,
or exactly the result of the first application of integration by parts. Since this integral is
not yet easy, we return to the table. Now we multiply twice on the diagonal, (x2 )(− cos x)
and (−2x)(− sin x) and then once straight across, (2)(− sin x), and combine these as
2
−x cos x + 2x sin x −
Z
2 sin x dx,
giving the same result as the second application of integration by parts. While this integral
is easy, we may return yet once more to the table. Now multiply three times on the diagonal
to get (x2 )(− cos x), (−2x)(− sin x), and (2)(cos x), and once straight across, (0)(cos x).
We combine these as before to get
2
−x cos x + 2x sin x + 2 cos x +
Z
0 dx = −x2 cos x + 2x sin x + 2 cos x + C.
Typically we would fill in the table one line at a time, until the “straight across” multiplication gives an easy integral. If we can see that the u column will eventually become zero,
we can instead fill in the whole table; computing the products as indicated will then give
the entire integral, including the “+C ”, as above.
Exercises 8.4.
Find Zthe antiderivatives.
1.
3.
Z
5.
Z
x cos x dx ⇒
2.
Z
x2 cos x dx ⇒
xex dx ⇒
4.
Z
xex dx ⇒
6.
Z
ln x dx ⇒
2
sin x dx ⇒
2
178
Chapter 8 Techniques of Integration
7.
Z
x arctan x dx ⇒
9.
Z
x3 cos x dx ⇒
11.
Z
x sin x cos x dx ⇒
13.
Z
√
sin( x) dx ⇒
8.5
8.
Z
x3 sin x dx ⇒
10.
Z
x sin2 x dx ⇒
12.
Z
√
arctan( x) dx ⇒
14.
Z
sec2 x csc2 x dx ⇒
Rational Fun tions
A rational function is a fraction with polynomials in the numerator and denominator.
For example,
x3
1
x2 + 1
,
,
,
x2 + x − 6
(x − 3)2
x2 − 1
are all rational functions of x. There is a general technique called “partial fractions”
that, in principle, allows us to integrate any rational function. The algebraic steps in the
technique are rather cumbersome if the polynomial in the denominator has degree more
than 2, and the technique requires that we factor the denominator, something that is not
always possible. However, in practice one does not often run across rational functions with
high degree polynomials in the denominator for which one has to find the antiderivative
function. So we shall explain how to find the antiderivative of a rational function only
when the denominator is a quadratic polynomial ax2 + bx + c.
We should mention a special type of rational function that we already know how to
integrate: If the denominator has the form (ax + b)n , the substitution u = ax + b will
always work. The denominator becomes un , and each x in the numerator is replaced by
(u − b)/a, and dx = du/a. While it may be tedious to complete the integration if the
numerator has high degree, it is merely a matter of algebra.
8.5
Find
EXAMPLE 8.5.1
Z
3
1
x
dx
=
(3 − 2x)5
−2
=
1
16
1
=
16
1
=
16
=−
Z
Z
Z
x3
dx. Using the substitution u = 3 − 2x we get
(3 − 2x)5
u−3
−2
u5
179
Rational Functions
3
1
du =
16
Z
u3 − 9u2 + 27u − 27
du
u5
u−2 − 9u−3 + 27u−4 − 27u−5 du
u−1
9u−2
27u−3
27u−4
−
+
−
−1
−2
−3
−4
+C
(3 − 2x)−1
9(3 − 2x)−2
27(3 − 2x)−3
27(3 − 2x)−4
−
+
−
−1
−2
−3
−4
1
9
9
27
+
−
+
+C
2
3
16(3 − 2x) 32(3 − 2x)
16(3 − 2x)
64(3 − 2x)4
+C
We now proceed to the case in which the denominator is a quadratic polynomial. We
can always factor out the coefficient of x2 and put it outside the integral, so we can assume
that the denominator has the form x2 + bx + c. There are three possible cases, depending
on how the quadratic factors: either x2 + bx + c = (x − r)(x − s), x2 + bx + c = (x − r)2 ,
or it doesn’t factor. We can use the quadratic formula to decide which of these we have,
and to factor the quadratic if it is possible.
EXAMPLE 8.5.2 Determine whether x2 + x + 1 factors, and factor it if possible. The
quadratic formula tells us that x2 + x + 1 = 0 when
x=
−1 ±
√
2
1−4
.
Since there is no square root of −3, this quadratic does not factor.
EXAMPLE 8.5.3 Determine whether x2 − x − 1 factors, and factor it if possible. The
quadratic formula tells us that x2 − x − 1 = 0 when
x=
Therefore
x2 − x − 1 =
1±
√
1+4
1± 5
=
.
2
2
√
√ !
√ !
1− 5
1+ 5
x−
.
x−
2
2
180
Chapter 8 Techniques of Integration
If x2 + bx + c = (x − r)2 then we have the special case we have already seen, that can
be handled with a substitution. The other two cases require different approaches.
If x2 + bx + c = (x − r)(x − s), we have an integral of the form
Z
p(x)
dx
(x − r)(x − s)
where p(x) is a polynomial. The first step is to make sure that p(x) has degree less than
2.
x3
dx in terms of an integral with a numerator
(x − 2)(x + 3)
that has degree less than 2. To do this we use long division of polynomials to discover that
EXAMPLE 8.5.4 Rewrite
so
Z
x3
x3
7x − 6
7x − 6
= 2
=x−1+ 2
=x−1+
,
(x − 2)(x + 3)
x +x−6
x +x−6
(x − 2)(x + 3)
Z
x3
dx =
(x − 2)(x + 3)
Z
x − 1 dx +
Z
7x − 6
dx.
(x − 2)(x + 3)
The first integral is easy, so only the second requires some work.
Now consider the following simple algebra of fractions:
B
A(x − s) + B(x − r)
(A + B)x − As − Br
A
+
=
=
.
x−r x−s
(x − r)(x − s)
(x − r)(x − s)
That is, adding two fractions with constant numerator and denominators (x−r) and (x−s)
produces a fraction with denominator (x − r)(x − s) and a polynomial of degree less than
2 for the numerator. We want to reverse this process: starting with a single fraction, we
want to write it as a sum of two simpler fractions. An example should make it clear how
to proceed.
x3
7x − 6
EXAMPLE 8.5.5 Evaluate
dx. We start by writing
(x − 2)(x + 3)
(x − 2)(x + 3)
as the sum of two fractions. We want to end up with
Z
7x − 6
A
B
=
+
.
(x − 2)(x + 3)
x−2 x+3
If we go ahead and add the fractions on the right hand side we get
(A + B)x + 3A − 2B
7x − 6
=
.
(x − 2)(x + 3)
(x − 2)(x + 3)
So all we need to do is find A and B so that 7x − 6 = (A + B)x + 3A − 2B, which is to
say, we need 7 = A + B and −6 = 3A − 2B. This is a problem you’ve seen before: solve a
8.5
Rational Functions
181
system of two equations in two unknowns. There are many ways to proceed; here’s one: If
7 = A+B then B = 7−A and so −6 = 3A−2B = 3A−2(7−A) = 3A−14+2A = 5A−14.
This is easy to solve for A: A = 8/5, and then B = 7 − A = 7 − 8/5 = 27/5. Thus
Z
Z
7x − 6
27 1
8
27
8 1
dx =
+
dx = ln |x − 2| +
ln |x + 3| + C.
(x − 2)(x + 3)
5x−2
5 x+3
5
5
The answer to the original problem is now
Z
Z
Z
x3
7x − 6
dx = x − 1 dx +
dx
(x − 2)(x + 3)
(x − 2)(x + 3)
=
8
27
x2
− x + ln |x − 2| +
ln |x + 3| + C.
2
5
5
Now suppose that x2 + bx + c doesn’t factor. Again we can use long division to ensure
that the numerator has degree less than 2, then we complete the square.
x+1
dx. The quadratic denominator does not
+ 4x + 8
factor. We could complete the square and use a trigonometric substitution, but it is simpler
to rearrange the integrand:
Z
Z
Z
x+2
1
x+1
dx =
dx −
dx.
2
2
2
x + 4x + 8
x + 4x + 8
x + 4x + 8
EXAMPLE 8.5.6
Evaluate
Z
x2
The first integral is an easy substitution problem, using u = x2 + 4x + 8:
Z
Z
x+2
1
du
1
dx =
= ln |x2 + 4x + 8|.
2
x + 4x + 8
2
u
2
For the second integral we complete the square:
!
2
x
+
2
+1 ,
x2 + 4x + 8 = (x + 2)2 + 4 = 4
2
making the integral
1
4
Using u =
Z
1
x+2 2
2
+1
dx.
x+2
we get
2
Z
Z
1
1
1
2
1
x+2
.
dx =
du = arctan
x+2 2
4
4
u2 + 1
2
2
+
1
2
Z
1
1
x+2
x+1
2
+ C.
dx = ln |x + 4x + 8| − arctan
x2 + 4x + 8
2
2
2
182
Chapter 8 Techniques of Integration
Exercises 8.5.
Find the antiderivatives.
Z
1
dx ⇒
1.
4 − x2
Z
1
3.
dx ⇒
x2 + 10x + 25
Z
x4
5.
dx ⇒
4 + x2
Z
x3
7.
dx ⇒
4 + x2
Z
1
dx ⇒
9.
2
2x − x − 3
8.6
x4
dx ⇒
4 − x2
Z
x2
4.
dx ⇒
4 − x2
Z
1
6.
dx ⇒
2
x + 10x + 29
Z
1
8.
dx ⇒
x2 + 10x + 21
Z
1
10.
dx ⇒
2
x + 3x
2.
Z
Numeri al Integration
We have now seen some of the most generally useful methods for discovering antiderivatives,
and there are others. Unfortunately, some functions have no simple antiderivatives; in such
cases if the value of a definite integral is needed it will have to be approximated. We will
see two methods that work reasonably well and yet are fairly simple; in some cases more
sophisticated techniques will be needed.
Of course, we already know one way to approximate an integral: if we think of the
integral as computing an area, we can add up the areas of some rectangles. While this
is quite simple, it is usually the case that a large number of rectangles is needed to get
acceptable accuracy. A similar approach is much better: we approximate the area under a
curve over a small interval as the area of a trapezoid. In figure 8.6.1 we see an area under
a curve approximated by rectangles and by trapezoids; it is apparent that the trapezoids
give a substantially better approximation on each subinterval.
....
........ ..............
.....
.....
....
.....
.
.
....
.
....
...
.
.
...
..
...
.
.
...
.
.
...
.
.
...
.
.
.
...
.
.
...
.
...
.
...
...
...
...
...
...
.
.
...
.
....
...
...
...
....
...
....
.
.
.....
...
.....
....
.....
.......
.....
................
Figure 8.6.1
.........
....... ...........
..... ... ..............
... ....
........
.
.
..
.. .....
..
... ....
... ...
... ...
... ......
.
......
.
.
.
.
.......
. ...
.
.
.......
. ....
.
.....
.
....
.
......
.
....
....
..
......
.....
.. ..
.......
......
.....
.
.
.......
... .
......
.. ...
... ...
... ...
.... ...
... ....
.
..... ...
... ...
..... ...
........ ...... ......
.
........ ..
......................
Approximating an area with rectangles and with trapezoids.
As with rectangles, we divide the interval into n equal subintervals of length ∆x. A
f (xi ) + f (xi+1 )
typical trapezoid is pictured in figure 8.6.2; it has area
2
8.6
Numerical Integration
183
the areas of all trapezoids we get
f (x1 ) + f (x2 )
f (xn−1 ) + f (xn )
f (x0 ) + f (x1 )
∆x +
∆x + · · · +
∆x =
2
2
2
f (x0 )
f (xn )
∆x.
+ f (x1 ) + f (x2 ) + · · · + f (xn−1 ) +
2
2
This is usually known as the Trapezoid Rule. For a modest number of subintervals this
is not too difficult to do with a calculator; a computer can easily do many subintervals.
(xi , f (xi )) ...................
... .....
... ....
... .....
... ....
... ...
.......
......
......
......
....
xi
Figure 8.6.2
(xi+1 , f (xi+1 ))
xi+1
A single trapezoid.
In practice, an approximation is useful only if we know how accurate it is; for example,
we might need a particular value accurate to three decimal places. When we compute a
particular approximation to an integral, the error is the difference between the approximation and the true value of the integral. For any approximation technique, we need an
error estimate, a value that is guaranteed to be larger than the actual error. If A is an
approximation and E is the associated error estimate, then we know that the true value
of the integral is between A − E and A + E. In the case of our approximation of the
integral, we want E = E(∆x) to be a function of ∆x that gets small rapidly as ∆x gets
small. Fortunately, for many functions, there is such an error estimate associated with the
trapezoid approximation.
THEOREM 8.6.1 Suppose f has a second derivative f ′′ everywhere on the interval
[a, b], and |f ′′ (x)| ≤ M for all x in the interval. With ∆x = (b − a)/n, an error estimate
for the trapezoid approximation is
E(∆x) =
Let’s see how we can use this.
b−a
(b − a)3
M (∆x)2 =
M.
12
12n2
184
Chapter 8 Techniques of Integration
EXAMPLE 8.6.2
Approximate
Z
1
2
e−x dx to two decimal places. The second deriva-
0
−x2
2
−x2
2
tive of f = e
is (4x −2)e
, and it is not hard to see that on [0, 1], |(4x2 −2)e−x | ≤ 2.
We begin by estimating the number of subintervals we are likely to need. To get two decimal places of accuracy, we will certainly need E(∆x) < 0.005 or
1
1
(2) 2 < 0.005
12
n
1
(200) < n2
6
r
100
<n
5.77 ≈
3
With n = 6, the error estimate is thus 1/63 < 0.0047. We compute the trapezoid approximation for six intervals:
f (0)
f (1)
+ f (1/6) + f (2/6) + · · · + f (5/6) +
2
2
1
≈ 0.74512.
6
So the true value of the integral is between 0.74512 − 0.0047 = 0.74042 and 0.74512 +
0.0047 = 0.74982. Unfortunately, the first rounds to 0.74 and the second rounds to 0.75,
so we can’t be sure of the correct value in the second decimal place; we need to pick a larger
n. As it turns out, we need to go to n = 12 to get two bounds that both round to the same
value, which turns out to be 0.75. For comparison, using 12 rectangles to approximate
the area gives 0.7727, which is considerably less accurate than the approximation using six
trapezoids.
In practice it generally pays to start by requiring better than the maximum possible
error; for example, we might have initially required E(∆x) < 0.001, or
1
1
(2) 2 < 0.001
12
n
1
(1000) < n2
6
r
500
<n
12.91 ≈
3
Had we immediately tried n = 13 this would have given us the desired answer.
The trapezoid approximation works well, especially compared to rectangles, because
the tops of the trapezoids form a reasonably good approximation to the curve when ∆x is
fairly small. We can extend this idea: what if we try to approximate the curve more closely,
8.6
Numerical Integration
185
by using something other than a straight line? The obvious candidate is a parabola: if we
can approximate a short piece of the curve with a parabola with equation y = ax2 + bx + c,
we can easily compute the area under the parabola.
There are an infinite number of parabolas through any two given points, but only
one through three given points. If we find a parabola through three consecutive points
(xi , f (xi)), (xi+1 , f (xi+1 )), (xi+2 , f (xi+2 )) on the curve, it should be quite close to the
curve over the whole interval [xi , xi+2 ], as in figure 8.6.3. If we divide the interval [a, b]
into an even number of subintervals, we can then approximate the curve by a sequence of
parabolas, each covering two of the subintervals. For this to be practical, we would like a
simple formula for the area under one parabola, namely, the parabola through (xi , f (xi )),
(xi+1 , f (xi+1 )), and (xi+2 , f (xi+2 )). That is, we should attempt to write down the parabola
y = ax2 + bx + c through these points and then integrate it, and hope that the result is
fairly simple. Although the algebra involved is messy, this turns out to be possible. The
algebra is well within the capability of a good computer algebra system like Sage, so we
will present the result without all of the algebra; you can see how to do it in this Sage
worksheet.
To find the parabola, we solve these three equations for a, b, and c:
f (xi ) = a(xi+1 − ∆x)2 + b(xi+1 − ∆x) + c
f (xi+1 ) = a(xi+1 )2 + b(xi+1 ) + c
f (xi+2 ) = a(xi+1 + ∆x)2 + b(xi+1 + ∆x) + c
Not surprisingly, the solutions turn out to be quite messy. Nevertheless, Sage can easily
compute and simplify the integral to get
Z
xi+1 +∆x
ax2 + bx + c dx =
xi+1 −∆x
∆x
(f (xi ) + 4f (xi+1 ) + f (xi+2 )).
3
Now the sum of the areas under all parabolas is
∆x
(f (x0 ) + 4f (x1 ) + f (x2 ) + f (x2 ) + 4f (x3 ) + f (x4 ) + · · · + f (xn−2 ) + 4f (xn−1 ) + f (xn )) =
3
∆x
(f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + 2f (x4 ) + · · · + 2f (xn−2 ) + 4f (xn−1 ) + f (xn )).
3
This is just slightly more complicated than the formula for trapezoids; we need to remember
the alternating 2 and 4 coefficients; note that n must be even for this to make sense. This
approximation technique is referred to as Simpson’s Rule.
As with the trapezoid method, this is useful only with an error estimate:
186
Chapter 8 Techniques of Integration
(xi , f (xi ))
..
....
.....
.....
.
.
.....
......
.....
.......
.
.
.
......
......
.......
.........
.
.
.
.
.
.
......
........
.....
.........
..... ... ... . . .................
....... .. .........
................
xi
Figure 8.6.3
xi+1
(xi+2 , f (xi+2 ))
xi+2
A parabola (dashed) approximating a curve (solid).
THEOREM 8.6.3 Suppose f has a fourth derivative f (4) everywhere on the interval
[a, b], and |f (4) (x)| ≤ M for all x in the interval. With ∆x = (b − a)/n, an error estimate
for Simpson’s approximation is
E(∆x) =
EXAMPLE 8.6.4
b−a
(b − a)5
M (∆x)4 =
M.
180
180n4
Let us again approximate
Z
1
2
e−x dx to two decimal places. The
0
−x2
2
2
2
fourth derivative of f = e
is (16x − 48x + 12)e−x ; on [0, 1] this is at most 12 in
absolute value. We begin by estimating the number of subintervals we are likely to need.
To get two decimal places of accuracy, we will certainly need E(∆x) < 0.005, but taking
a cue from our earlier example, let’s require E(∆x) < 0.001:
1
1
(12) 4 < 0.001
180
n
200
< n4
3
r
[4] 200
2.86 ≈
<n
3
So we try n = 4, since we need an even number of subintervals. Then the error estimate
is 12/180/44 < 0.0003 and the approximation is
(f (0) + 4f (1/4) + 2f (1/2) + 4f (3/4) + f (1))
1
≈ 0.746855.
3·4
So the true value of the integral is between 0.746855 − 0.0003 = 0.746555 and 0.746855 +
0.0003 = 0.7471555, both of which round to 0.75.
8.7
187
Exercises 8.6.
In the following problems, compute the trapezoid and Simpson approximations using 4 subintervals, and compute the error estimate for each. (Finding the maximum values of the second
and fourth derivatives can be challenging for some of these; you may use a graphing calculator
or computer software to estimate the maximum values.) If you have access to Sage or similar
software, approximate each integral to two decimal places. You can use this Sage worksheet to
get started.
Z 3
Z 3
1.
x dx ⇒
2.
x2 dx ⇒
1
0
4
3.
Z
2
5.
Z
Z
5
x3 dx ⇒
2
1
1
dx ⇒
1 + x2
x
7.
dx ⇒
1 1+x
Z 1p
x4 + 1 dx ⇒
9.
3
4.
Z
1
8.
Z
4
10.
Z
1
dx ⇒
x
1
Z 1
√
6.
x 1 + x dx ⇒
0
0
1
0
p
x3 + 1 dx ⇒
p
1 + 1/x dx ⇒
11. Using Simpson’s rule on a parabola f (x), even with just two subintervals, gives the exact value
of the integral, because the parabolas used to approximate f will be f itself. Remarkably,
Simpson’s rule also computes the integral of a cubic function f (x) = ax3 + bx2 + cx + d
exactly. Show this is true by showing that
Z x2
x2 − x0
(f (x0 ) + 4f ((x0 + x2 )/2) + f (x2 )).
f (x) dx =
3·2
x0
This does require a bit of messy algebra, so you may prefer to use Sage.
8.7
These problems require the techniques of this chapter, and are in no particular order. Some
problems may be done in more than one way.
1.
Z
(t + 4) dt ⇒
3.
Z
(et + 16)tet dt ⇒
5.
Z
tan t sec2 t dt ⇒
Z
3
2
2
1
dt ⇒
t(t2 − 4)
Z
cos 3t
√
dt ⇒
9.
sin 3t
Z
et
√
11.
dt ⇒
et + 1
7.
2.
Z
t(t2 − 9)3/2 dt ⇒
4.
Z
sin t cos 2t dt ⇒
6.
Z
12.
Z
2t + 1
dt ⇒
t2 + t + 3
Z
1
8.
dt ⇒
(25 − t2 )3/2
Z
10.
t sec2 t dt ⇒
cos4 t dt ⇒
188
Chapter 8 Techniques of Integration
13.
Z
1
dt ⇒
2
t + 3t
15.
Z
sec2 t
dt ⇒
(1 + tan t)3
17.
Z
et sin t dt ⇒
t3
dt ⇒
(2 − t2 )5/2
Z
arctan 2t
21.
dt ⇒
1 + 4t2
Z
23.
sin3 t cos4 t dt ⇒
Z
1
dt ⇒
25.
t(ln t)2
Z
27.
t3 et dt ⇒
19.
Z
Z
1
dt ⇒
t2 1 + t2
Z p
16.
t3 t2 + 1 dt ⇒
Z
√
18.
(t3/2 + 47)3 t dt ⇒
Z
1
dt ⇒
20.
t(9 + 4t2 )
Z
t
22.
dt ⇒
2
t + 2t − 3
Z
1
24.
dt ⇒
t2 − 6t + 9
Z
26.
t(ln t)2 dt ⇒
Z
t+1
28.
dt ⇒
2
t +t−1
14.
√
```
Arab people
15 Cards |
# Geometry Honors Section 9.1 Segments and Arcs of Circles
## Presentation on theme: "Geometry Honors Section 9.1 Segments and Arcs of Circles"— Presentation transcript:
Geometry Honors Section 9.1 Segments and Arcs of Circles
A *circle is a set of points, in a plane, that are equidistant from a given point. This given point is called the _______ of the circle. center
A circle can be named by using the symbol _____ and naming the center of the circle. The circle to the right is __________.
A *radius (plural: radii) is a segment from the center to a point on the circle.
A *chord is a segment whose endpoints are on the circle.
A *diameter is a chord which contains the center of the circle.
An arc is an unbroken part of a circle
An arc is an unbroken part of a circle. Any two distinct points on a circle divide the circle into two arcs. The two points are called the _________of the arc. endpoints
If the two points are the endpoints of a diameter, then each of the two arcs formed is called a __________ A semicircle is named by its two endpoints and another point that lies on the arc. Example: Name two semicircles _____ & _____ semicircle.
If the two points are not the endpoints of a diameter, then a minor arc and a major arc are formed.
A. minor arc is an arc which is shorter than a semicircle
A *minor arc is an arc which is shorter than a semicircle. A minor arc is named by its two endpoints. Example: Name two minor arcs. _____ & _____
A. major arc is an arc which is longer than a semicircle
A *major arc is an arc which is longer than a semicircle. A major arc is named by its two endpoints and another point that lies on the arc. . Example: Name two major arcs. _______ & _______
A *central angle of a circle is an angle whose vertex is at the center and whose sides are radii. The arc between the outer endpoints of the two radii is called the __________ arc of the central angle. intercepted
The degree measure of a minor arc is equal to the measure of its central angle. The degree measure of a major arc is equal to 360⁰ - the measure of the associated minor arc. The degree measure of a semicircle is ______. 180⁰
When referring to the measure of an arc, use the notation __________
100⁰ 38⁰
The following theorem mentions congruent circles
The following theorem mentions congruent circles. Two circles are congruent iff their radii are congruent. Chords and Arcs Theorem In a circle (or in congruent circles), two chords are congruent iff the minor arcs they determine are congruent.
Radius and Chord Theorem If a radius is perpendicular to a chord, then the radius bisects the chord and its arc. |
# Complex Numbers Consider the quadratic equation x2 + 1 = 0.
## Presentation on theme: "Complex Numbers Consider the quadratic equation x2 + 1 = 0."— Presentation transcript:
Complex Numbers Consider the quadratic equation x2 + 1 = 0.
Solving for x , gives x2 = – 1 We make the following definition:
Complex Numbers Note that squaring both sides yields: therefore and so
And so on…
Real numbers and imaginary numbers are subsets of the set of complex numbers.
Definition of a Complex Number
If a and b are real numbers, the number a + bi is a complex number, and it is said to be written in standard form. If b = 0, the number a + bi = a is a real number. If a = 0, the number a + bi is called an imaginary number. Write the complex number in standard form
Addition and Subtraction of Complex Numbers
If a + bi and c +di are two complex numbers written in standard form, their sum and difference are defined as follows. Sum: Difference:
Perform the subtraction and write the answer in standard form.
( 3 + 2i ) – ( i ) 3 + 2i – 6 – 13i –3 – 11i 4
Multiplying Complex Numbers
Multiplying complex numbers is similar to multiplying polynomials and combining like terms. Perform the operation and write the result in standard form. ( 6 – 2i )( 2 – 3i ) F O I L 12 – 18i – 4i + 6i2 12 – 22i + 6 ( -1 ) 6 – 22i
Consider ( 3 + 2i )( 3 – 2i ) 9 – 6i + 6i – 4i2 9 – 4( -1 ) 9 + 4 13 This is a real number. The product of two complex numbers can be a real number. This concept can be used to divide complex numbers.
Complex Conjugates and Division
Complex conjugates-a pair of complex numbers of the form a + bi and a – bi where a and b are real numbers. ( a + bi )( a – bi ) a 2 – abi + abi – b 2 i 2 a 2 – b 2( -1 ) a 2 + b 2 The product of a complex conjugate pair is a positive real number.
To find the quotient of two complex numbers multiply the numerator and denominator by the conjugate of the denominator.
Perform the operation and write the result in standard form.
Perform the operation and write the result in standard form.
Expressing Complex Numbers in Polar Form
Now, any Complex Number can be expressed as: X + Y i That number can be plotted as on ordered pair in rectangular form like so…
Expressing Complex Numbers in Polar Form
Remember these relationships between polar and rectangular form: So any complex number, X + Yi, can be written in polar form: Here is the shorthand way of writing polar form:
Expressing Complex Numbers in Polar Form
Rewrite the following complex number in polar form: 4 - 2i Rewrite the following complex number in rectangular form:
Expressing Complex Numbers in Polar Form
Express the following complex number in rectangular form:
Expressing Complex Numbers in Polar Form
Express the following complex number in polar form: 5i
Products and Quotients of Complex Numbers in Polar Form
The product of two complex numbers, and Can be obtained by using the following formula:
Products and Quotients of Complex Numbers in Polar Form
The quotient of two complex numbers, and Can be obtained by using the following formula:
Products and Quotients of Complex Numbers in Polar Form
Find the product of 5cis30 and –2cis120 Next, write that product in rectangular form
Products and Quotients of Complex Numbers in Polar Form
Find the quotient of 36cis300 divided by 4cis120 Next, write that quotient in rectangular form
Products and Quotients of Complex Numbers in Polar Form
Find the result of Leave your answer in polar form. Based on how you answered this problem, what generalization can we make about raising a complex number in polar form to a given power?
[r(cos F+isin F]n = rn(cos nF+isin nF)
De Moivre’s Theorem De Moivre's Theorem is the theorem which shows us how to take complex numbers to any power easily. De Moivre's Theorem – Let r(cos F+isin F) be a complex number and n be any real number. Then [r(cos F+isin F]n = rn(cos nF+isin nF) What is this saying? The resulting r value will be r to the nth power and the resulting angle will be n times the original angle.
Remember to save space you can write it in compact form.
De Moivre’s Theorem Try a sample problem: What is [3(cos 45°+isin45)]5? To do this take 3 to the 5th power, then multiply 45 times 5 and plug back into trigonometric form. 35 = 243 and 45 * 5 =225 so the result is 243(cos 225°+isin 225°) Remember to save space you can write it in compact form. 243(cos 225°+isin 225°)=243cis 225°
De Moivre’s Theorem Find the result of:
Because of the power involved, it would easier to change this complex number into polar form and then use De Moivre’s Theorem.
De Moivre’s Theorem De Moivre's Theorem also works not only for integer values of powers, but also rational values (so we can determine roots of complex numbers).
De Moivre’s Theorem Simplify the following:
De Moivre’s Theorem Every complex number has ‘p’ distinct ‘pth’ complex roots (2 square roots, 3 cube roots, etc.) To find the p distinct pth roots of a complex number, we use the following form of De Moivre’s Theorem …where ‘n’ is all integer values between 0 and p-1. Why the 360? Well, if we were to graph the complex roots on a polar graph, we would see that the p roots would be evenly spaced about 360 degrees (360/p would tell us how far apart the roots would be).
De Moivre’s Theorem Find the 4 distinct 4th roots of i
De Moivre’s Theorem Solve the following equation for all complex
number solutions (roots): |
# Applications Using Linear Models
## Solve story problems using linear equations
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Using Algebraic Models
Bailey is planting an outdoor garden. She wants the length of the planting area to be twice as large as the width. If she has 128 square feet of soil to work with, what dimensions must the garden be?
### Algebraic Models
Word problems are some of the hardest types of problems for students to grasp. There are a few steps to solving any word problem:
Step 1: Read the problem at least twice.
Step 2: Cross out any unnecessary words, circle any numbers or words that represent mathematical operators, or translate words into mathematical expressions.
Step 3: Write an equation and solve.
To help you with Steps 2 and 3, generate a list of words that represent: add, subtract, multiply, divide, equal, etc. Here are a few to get you started.
Operation words Operation words
add sum, plus, and increase, more (than) multiply times, of, product double (x2), triple (x3)
subtract difference, minus decrease, less (than) equal is, total, to made/make, spend/spent
divide quotient, half, ÷2 split, share variable
how many __
how much __
what amount (of) __
See if you can add anything to these lists.
Let's use this chart to help you with decoding the following word problems.
1. Two consecutive numbers add up to 55. What are the two numbers?
First, translate the statement. “Consecutive” means numbers that are one after the other. So if the first number is \begin{align*}x\end{align*}, then the second number will be \begin{align*}x + 1\end{align*}, and they add up to (equal) 55. The equation is: \begin{align*}x+(x+1)=55.\end{align*}
We put \begin{align*}x + 1\end{align*} in parenthesis to show that it is a separate value. Solve the equation.
\begin{align*}x+(x+1) &=55\\ 2x+1 &= 55\\ 2x &= 54\\ x &= 27\end{align*}
The smaller number is 27, and the larger number will be 28, since \begin{align*}27 + \color{red}28 \color{black}= 55 \end{align*}
Sometime you may encounter problems with “consecutive even numbers” or “consecutive odd numbers.” All even numbers are divisible by 2, so the smallest should be \begin{align*}2x\end{align*}, then the next even number would be \begin{align*}2x + 2\end{align*}. For consecutive odd numbers, they will always be an even number plus 1, 3, 5, etc. So, the smaller odd number would be \begin{align*}2x + 1\end{align*} and the larger odd number would be \begin{align*}2x + 3\end{align*}.
1. Over the Winter Break, you worked at a clothing store and made 9.00 an hour. During the two weeks you worked 65 hours at regular pay and 10 hours of overtime (1.5 times regular pay, or 'time and a half'). How much money did you make? First, we need to figure out how much you make for overtime. 'Time and a half' would be \begin{align*}\ 9.00 + \ 4.50 = \ 13.50\end{align*} per hour. So, you made: \begin{align*}& \ 9.00(65)+ \ 13.50(10) \\ & = \ 585.00+ \ 135.00 \\ & = \ 720.00\end{align*} 1. Elise is taking piano lessons. The first lesson is twice as expensive as each additional lesson. Her mom spends270 for 8 lessons. How much was the first lesson?
Translate each statement.
Call the regularly priced lessons \begin{align*}l\end{align*}. Then, the first lesson will be \begin{align*}2l\end{align*}.
You need to know how much money Javier needs to fill up his gas tank. Gas costs 3.79 per gallon and he needs 16 gallons of gas. It will cost \begin{align*}\ 3.79 \cdot 16= \ 60.64\end{align*} to fill up his tank. ### Review Answer each question to the best of your ability. 1. The average speed on highway 101 is 65 miles per hour (mph). Assuming you drive the speed limit, how long will it take you to drive 350 miles? Use the formula \begin{align*}distance = rate \cdot time\end{align*}. Round your answer to two decimal places. 2. Using the information in #1, how many miles did you drive on highway 101 if you drove for 2.5 hours? 3. The sum of two consecutive numbers is 79. Find the two numbers. 4. The sum of two consecutive odd numbers is 44. Find the two odd numbers. 5. You borrowed350 from your parents for a new Wii and games. They are not going to charge you interest, but you need to pay them back as quickly as possible. If you pay them $15 per week, how long will it take you to pay them back? 6. George is building a rectangular, fenced-in, dog run. He has 120 feet of fencing and wants the length to be 20 feet greater than the width. If you use all the fencing, find the length and width of the dog run. 7. Cynthia is selling chocolate bars for a fundraiser for school. Each bar costs$1.50. If she needs to raise $225, how many chocolate bars does she need to sell? 8. Harriet bakes and sells cookies to local stores. Her cost for one dozen cookies is$2.75 and she sells them to stores for $7.00 per dozen. How many dozen cookies does she need to make to earn$500? Round to the nearest dozen.
9. A football field is a rectangle where the length is 100 yards. If the total perimeter is 1040 feet, what is the width of a football field? Leave your answer in feet.
10. Challenge The sum of three consecutive even numbers is 138. What are the three numbers?
To see the Review answers, open this PDF file and look for section 1.16.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
# Ratio of sides
Calculate the area of a circle that has the same circumference as the circumference of the rectangle inscribed with a circle with a radius of r 9 cm so that its sides are in ratio 2 to 7.
Correct result:
S = 157.617 cm2
#### Solution:
$r=9 \ \text{cm} \ \\ r^2=(a/2)^2 + (b/2)^2 \ \\ a:b=2:7 \ \\ \ \\ r^2=(a/2)^2 + ((7/2 \cdot \ a)/2)^2 \ \\ r^2=a^2/4 + a^2 \cdot \ (7/4)^2 \ \\ \ \\ a=r / (\sqrt{ 1/4+(7/4)^2 })=9 / (\sqrt{ 1/4+(7/4)^2 }) \doteq 4.945 \ \text{cm} \ \\ \ \\ b=7/2 \cdot \ a=7/2 \cdot \ 4.945 \doteq 17.3074 \ \text{cm} \ \\ \ \\ o=2 \cdot \ (a+b)=2 \cdot \ (4.945+17.3074) \doteq 44.5048 \ \text{cm} \ \\ \ \\ o=2 \pi \cdot \ r_{1} \ \\ \ \\ r_{1}=o / (2 \pi)=44.5048 / (2 \cdot \ 3.1416) \doteq 7.0832 \ \text{cm} \ \\ \ \\ \ \\ S=\pi \cdot \ r_{1}^2=3.1416 \cdot \ 7.0832^2=157.617 \ \text{cm}^2$
Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!
Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors.
Tips to related online calculators
Looking for help with calculating roots of a quadratic equation?
Check out our ratio calculator.
Pythagorean theorem is the base for the right triangle calculator.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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# NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6
NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6 covers the answers to the questions present in this exercise. Angles and their construction, bisector of an angle and angles of special measures along with illustrative examples are covered under Exercise 14.6 of Chapter 14. Students are provided with exercise wise NCERT Solutions to get better acquainted with the concepts, which are important from the exam point of view.
## NCERT Solutions for Class 6 Chapter 14: Practical Geometry Exercise 14.6 Download PDF
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### Access NCERT Solutions for Class 6 Chapter 14: Practical Geometry Exercise 14.6
1. Draw ∠POQ of measure 75° and find its line of symmetry.
Solutions:
Following steps are followed to construct an angle of 750 and its line of symmetry
(i) Draw a line l and mark two points O and Q on it. Draw an arc of convenient radius, while taking centre as O. Let this intersect line l at R
(ii) Taking R as centre and with same radius as before, draw an arc such that it is intersecting the previously drawn arc at S
(iii) By taking same radius as before and S as centre, draw an arc intersecting the arc at point T as shown in figure
(iv) Take S and T as centre, draw an arc of same radius such that they intersect each other at U
(v) Join OU. Let it intersect the arc at V. Now, take S and V as centres draw arcs with radius more than 1 / 2 SV. Let these intersect each other at P. Join OP. Now OP is the ray making 750 with the line l.
(vi) Let this ray intersect our major arc at point W. By taking R and W as centres, draw arcs with radius more than 1 / 2 RW in the interior angle of 750. Let these intersect each other at point X. Join OX
OX is the line of symmetry for the ∠POQ = 750
2. Draw an angle of measure 147° and construct its bisector.
Solutions:
Following steps are followed to construct an angle of measure 1470 and its bisector
(i) Draw a line l and mark point O on it. Place the centre of protractor at point O and the zero edge along line l
(ii) Mark a point A at an angle of measure 1470. Join OA. Now OA is the required ray making 1470 with line l
(iii) By taking point O as centre, draw an arc of convenient radius. Let this intersect both rays of angle 1470 at points A and B.
(iv) By taking A and B as centres draw arcs of radius more than 1 / 2 AB in the interior angle of 1470. Let these intersect each other at point C. Join OC.
OC is the required bisector of 1470 angle
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3. Draw a right angle and construct its bisector.
Solutions:
Following steps are followed to construct a right angle and its bisector.
(i) Draw a line l and mark a point P on it. Draw an arc of convenient radius by taking point P as centre. Let this intersect line l at R
(ii) Draw an arc by taking R as centre and with the same radius as before such that it is intersecting the previously drawn arc at S
(iii) Take S as centre and with the same radius as before, draw an arc intersecting the arc at T as shown in figure
(iv) By taking S and T as centres draw arcs of same radius such that they are intersecting each other at U.
(v) Join PU. PU is the required ray making a right angle with the line l. Let this intersect major arc at point V.
(vi) Now take R and V as centres, draw arcs with radius more than 1 / 2 RV to intersect each other at point W. Join PW.
PW is the required bisector of this right angle.
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4. Draw an angle of measure 153° and divide it into four equal parts.
Solutions:
Following steps are followed to construct an angle of measure 1530 and its bisector
(i) Draw a line l and mark a point O on it. Place the centre of protractor at point O and the zero edge along line l
(ii) Mark a point A at the measure of angle 1530. Join OA. Now OA is the required ray making 1530 with line l
(iii) Draw an arc of convenient radius by taking point O as centre. Let this intersect both rays of angle 1530 at points A and B.
(iv) Take A and B as centres and draw arcs of radius more than 1 / 2 AB in the interior of angle of 1530. Let these intersect each other at C. Join OC
(v) Let OC intersect major arc at point D. Draw arcs of radius more than 1 / 2 AD with A and D as centres and also D and B as centres. Let these are intersecting each other at points E and F, respectively. Now join OE and OF
OF, OC, OE are the rays dividing 1530 angle into four equal parts.
5. Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°
Solutions:
(a) 600
Following steps are followed to construct an angle of 600
(i) Draw a line l and mark a point P on it. Take P as centre and with convenient radius, draw an arc of a circle such that it intersects the line l at Q.
(ii) Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
(iii) Join PR. PR is the required ray making 600 with the line l.
(b) 300
Following steps are followed to construct an angle of 300
(i) Draw a line l and mark a point P on it. By taking P as centre and with convenient radius, draw an arc of a circle such that it is intersecting the line l at Q.
(ii) Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
(iii) By taking Q and R as centres and with radius more than 1 / 2 RQ draw arcs such that they are intersecting each other at S. Join PS which is the required ray making 300 with the line l.
(c) 900
Following steps are followed to construct an angle of measure 900
(i) Draw a line l and mark a point P on it. Take P as centre and with convenient radius, draw an arc of a circle such that it is intersecting the line l at Q.
(ii) Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R
(iii) By taking R as centre and with the same radius as before, draw an arc intersecting the arc at S as shown in figure
(iv) Now take R and S as centre, draw arc of same radius to intersect each other at T.
(v) Join PT, which is the required ray making 900 with the line l.
(d) 1200
Following steps are followed to construct an angle of measure 1200
(i) Draw a line l and mark a point P on it. Taking P as centre and with convenient radius, draw an arc of circle such that it is intersecting the line l at Q.
(ii) By taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
(iii) Take R as centre and with the same radius as before, draw an arc such that it is intersecting the arc at S as shown in figure.
(iv) Join PS, which is the required ray making 1200 with the line l
(e) 450
Following steps are followed to construct an angle of measure 450
(i) Draw a line l and mark a point P on it. Take P as centre and with convenient radius, draw an arc of a circle such that it is intersecting the line l at Q.
(ii) Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R
(iii) By taking R as centre and with the same radius as before, draw an arc such that it is intersecting the arc at S as shown in figure.
(iv) Take R and S as centres and draw arcs of same radius such that they are intersecting each other at T
(v) Join PT. Let this intersect the major arc at point U.
(vi) Now take Q and U as centres and draw arcs with radius more than 1 / 2 QU to intersect each other at point V. Join PV.
PV is the required ray making 450 with the line l
(f) 1350
Following steps are followed to construct an angle of measure 1350
(i) Draw a line l and mark a point P on it. Taking P as centre and with convenient radius, draw a semicircle which intersects the line l at Q and R, respectively.
(ii) By taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S
(iii) Taking S as centre and with the same radius as before, draw an arc such that it is intersecting the arc at T as shown in figure
(iv) Take S and T as centres, draw arcs of same radius to intersect each other at U.
(v) Join PU. Let this intersect the arc at V. Now take Q and V as centres and with radius more than 1 / 2 QV, draw arcs to intersect each other at W.
(vi) Join PW which is the required ray making 1350 with the line l
6. Draw an angle of measure 45° and bisect it.
Solutions:
Following steps are followed to construct an angle of measure 450 and its bisector.
(i) Using the protractor ∠POQ of 450 measure may be formed on a line l
(ii) Draw an arc of convenient radius with centre as O. Let this intersect both rays of angle 450 at points A and B
(iii) Take A and B as centres, draw arcs of radius more than 1 / 2 AB in the interior of angle of 450. Let these intersect each other at C. Join OC
OC is the required bisector of 450 angle
7. Draw an angle of measure 135° and bisect it.
Solutions:
Following steps are followed to construct an angle of measure 1350 and its bisector.
(i) By using a protractor ∠POQ of 1350 measure may be formed on a line l
(ii) Draw an arc of convenient radius by taking O as centre. Let this intersect both rays of angle 1350 at points A and B, respectively.
(iii) Take A and B as centres, draw arcs of radius more than 1 / 2 AB in the interior of angle of 1350. Let these intersect each other at C. Join OC.
OC is the required bisector of 1350 angle
8. Draw an angle of 700. Make a copy of it using only a straight edge and compasses.
Solutions:
Following steps are followed to construct an angle of measure 700 and its copy.
(i) Draw a line l and mark a point O on it. Now place the centre of protractor at point O and the zero edge along line l.
(ii) Mark a point A at an angle of measure 700. Join OA. Now OA is the ray making 700 with line l. With point O as centre, draw an arc of convenient radius in the interior of 700 angle. Let this intersect both rays of angle 700 at points B and C, respectively
(iii) Draw a line m and mark a point P on it. Again draw an arc with same radius as before and P as centre. Let it cut the line m at point D
(iv) Adjust the compasses up to the length of BC. With this radius draw an arc taking D as centre which intersects the previously drawn arc at point E.
(v) Join PE. Here PE is the required ray which makes same angle of measure 700 with the line m
9. Draw an angle of 400. Copy its supplementary angle.
Solutions:
Following steps are followed to construct an angle of measure 450 and a copy of its supplementary angle
(i) Draw a line segment
and mark a point O on it. Place the centre of protractor at point O and the zero edge along line segment
.
(ii) Mark a point A at an angle of measure 400. Join OA. Here OA is the required ray making 400 with
. ∠POA is the supplementary angle of 400
(iii) With point O as centre, draw an arc of convenient radius in the interior of ∠POA. Let this intersect both rays of ∠POA at points B and C, respectively.
(iv) Draw a line m and mark a point S on it. Again draw an arc by taking S as centre with the same radius as used before. Let it cut the line m at point T.
(v) Now adjust the compasses up to the length of BC. Taking T as centre draw an arc with this radius which will intersect the previously drawn arc at point R.
(vi) Join RS. Here RS is the required ray which makes same angle with the line m, as the supplementary of 400 [i.e 1400] |
# What is the domain and range of y=4-x^2?
Jun 22, 2018
Domain: All Real Numbers
Range: [4, -$\infty$)
#### Explanation:
Lets begin by finding the domain of the function. As a parabola, and consequently a polynomial, the function 4-${x}^{2}$ is defined for all real numbers – there is no point on its domain where the function is undefined. To better understand why the function is defined for all real numbers, see its graph: graph{4-x^2 [-10, 10, -5, 5]}
The range is also relatively simple to find.
y = 4 - ${x}^{2}$ can be re-written as y = $- {x}^{2}$+ 4, which has the form $a {\left(x - h\right)}^{2}$+ k , meaning that it is the vertex form of the parabola. This form tells us two important properties of the parabola.
1. The parabola's vertex is at (0,4), as, in the parabola's equation,
h = 0 and k = 4.
2. The parabola is concave down (it opens downwards) as a is negative.
As a result, the range of the parabola is all real values of y such that y$\le$4.
Were you to encounter a quadratic equation of a different form (such as, say, $y = {x}^{2} + 5 x + 6$), you would need to either complete the square to derive the vertex form of that parabola or find the roots (x-intercepts) of the parabola, use those to find the midpoint between them (take the average of the two roots, and then find the y value of that point. Once you know the parabola's vertex and concavity, you can easily determine it's range, as shown above. |
# Parallel Line Calculator
Created by Bogna Szyk
Last updated: Jan 18, 2024
If you're scratching your head while trying to figure out some parallel line equations, stop worrying: this parallel line calculator is precisely the tool you need. In just a few seconds, it will determine the equation of a line that is parallel to a given line and passes through a given point. That's not all, though; our calculator can also determine the distance between the two lines.
Read on to discover how to find the slope of a parallel line or what a y-intercept is.
🙋 Are you interested in other calculators like this? Check out our perpendicular line calculator!
## How to find the slope of a parallel line?
Every straight line in a two-dimensional space can be described by a simple line equation:
$y = ax + b,$
where a and b are coefficients, x is the x-coordinate, and y is the y-coordinate. Every line is uniquely defined if the values of a and b are known.
Let's assume that you know the following information:
1. The equation of the given line is y = mx + r. You know the values of m and r and are looking for a line parallel to this one.
2. You also know the coordinates of the point your line is supposed to pass through. They are x₀ and y₀.
The slope of any line is equal to the value of a coefficient. If two lines are parallel, then they must have the same slope. From this, we can deduce that.
a = m
We also recommend checking our average rate of change calculator.
💡 To calculate the slope of any line, use our slope calculator.
## Parallel line equation
Once you know the a coefficient of the line, all that is left to do is determine the b coefficient (also known as the y-intercept).
The method is straightforward: you have to substitute the coordinates (x₀, y₀) and the value of a into the equation of your line.
$y = ax + b$
$y₀ = mx₀ + b$
$b = y₀ - mx₀$
## Finding the distance between two parallel lines
Now that you know the equation of your new line, you can easily use it to determine the distance between it and the first line. In this case, the distance is defined as the length of the shortest possible segment that would join the two lines together.
Our parallel line calculator finds this distance automatically. If, however, you would like to check whether the result is correct, you can use the distance formula:
$D = \frac{|b - r|}{\sqrt{m² + 1}}$
## Parallel line calculations: an example
If you're still not sure how to find the equation of a parallel line, take a look at the example below!
1. Write down the equation of the first line. Let's say it's:
y = 3x - 5
2. Write down the coordinates of the given point P that the second line will pass through. Let's assume it is (1,6). In other words, x₀ = 1 and y₀ = 6.
3. Write down the equation of your new line: y = ax + b. You will try to determine the values of coefficients a and b.
4. Coefficient a is equal to m. Hence,
a = m = 3.
5. Plug the coordinates of point P into the equation of your new line to determine b:
y₀ = ax₀ + b
6 = 3 × 1 + b
b = 6 - 3 × 1 = 3
6. Knowing the values of the slope and y-intercept, you can now write down the full equation of the new line: y = 3x + 3.
7. You can also calculate the distance between the two lines:
D = |b - r| / √(m² + 1)
D = |3 - 6| / √(3² + 1) = |-3| / √(10) = 2.53
The distance between the two lines is equal to 2.53.
## FAQ
### How do I calculate the distance between two parallel lines?
To find the distance between two parallel lines in the Cartesian plane, follow these easy steps:
1. Find the equation of the first line: y = m1 × x + c1.
2. Find the equation of the second line y = m2 × x + c2.
3. Calculate the difference between the intercepts: (c2 − c1).
4. Divide this result by the following quantity: sqrt(m² − 1):
d = (c2 − c1) / √(m² − 1)
This is the distance between the two parallel lines.
### How do I identify two parallel lines on the Cartesian plane?
Two parallel lines on the Cartesian plane have the same angular coefficient. Knowing this, you can exclude the trivial case where the intercept is the same (the two lines coincide).
It's harder to define parallelism in three dimensions: two lines can be nonparallel yet never intersect!
### How do I find the parallel line passing through a point?
The parallel to the line y = m × x + b passing through the point (p, q) can be found with the following steps:
1. Find the slope of the parallel line — it's m!
2. Find the intercept of the parallel line's equation substituting the coordinates of the point in the original line's equation:
q = m × p + b
Hence:
b = q / (m × p)
### What are some examples of parallel lines?
The sides of a road are an excellent example of parallel lines: before an intersection, the two sides never meet! Other examples are the parallels (but not the meridians!) on a globe: those are three-dimensional parallel lines. You can find parallel lines on many human creations, but rarely in nature: with a keen eye, you can see them in some geological formations and for brief lengths in trees and other plants.
Bogna Szyk
First line equation y = mx + r
m
r
Second line passes through point...
x
y
Parallel line equation y = ax + b
a
b
Distance between the lines
Distance
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Definitions of surds: A root of a positive real quantity is called a surd if its value cannot be exactly determined. It is a number that can’t be simplified to remove a square root (or cube root etc). For example, each of the quantities √3, ∛7, ∜19, (16)^2/5 etc. is a surd.
More Examples:
• √2 (square root of 2) can’t be simplified further so it is a surd
• √4 (square root of 4) CAN be simplified to 2, so it is NOT a surd
Multiplication of Surds
Follow the following steps to find the multiplication of two or more surds.
Step I: Express each surd in its simplest mixed form.
Step II: Observe whether the given surds are of the same order or not.
Step III: If they are of the same order then the required product is obtained by multiplying the product of the rational co-efficient by the product of surd-factors.
If they are of different orders then the product is obtained by the above method after reducing them to surds of the same order.
If different order surds have the same base then their product can easily be obtained using the laws of indices.
Examples of multiplication of surds:
Find the product of 74 and 53
Solution:
The product of 7∜4 and 5∜3
= (7∜4) × (5∜3)
= (7 × 5) × (∜4 × ∜3)
= 35 × 4√(4⋅3)
= 35 × ∜12
= 35∜12
Find the product of 2√12, 7√20 and √32
Solution:
The product of 2√12, 7√20 and √32
= (2√12) × (7√20) × (√32)
= (2√(2⋅2⋅3) × (7√(2⋅2⋅5) × √(2⋅2⋅2⋅2⋅2)
= (4√3) × (14√5) × (4√2)
= (4 × 14 × 4) × (√3 × √5 × √2)
= 224 × √(3⋅5⋅2)
= 224 × √30
= 224√30
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## 0.6 Inequalities
### QUICK REFERENCE
Inequality Signs
Symbol Words Example
$\neq$ not equal to ${2}\neq{8}$, 2 is not equal to 8.
$\gt$ greater than ${5}\gt{1}$, 5 is greater than 1
$\lt$ less than ${2}\lt{11}$, 2 is less than 11
$\geq$ greater than or equal to ${4}\geq{ 4}$, 4 is greater than or equal to 4
$\leq$ less than or equal to ${7}\leq{9}$, 7 is less than or equal to 9
Addition and Subtraction Properties of Inequalities
If $a>b$, then $a+c>b+c$.
If $a>b$, then $a−c>b−c$.
Multiplication and Division Properties of Inequality
Start With Multiply By Final Inequality $a>b$ $c$ $ac>bc$ $a>b$ $-c$ $ac Start With Divide By Final Inequality [latex]a>b$ $c$ $\displaystyle \frac{a}{c}>\frac{b}{c}$ $a>b$ $-c$ $\displaystyle \frac{a}{c}<\frac{b}{c}$
## Inequalities
First, let’s define some important terminology. An inequality is a mathematical statement that compares two expressions using the ideas of greater than or less than. Special symbols are used in these statements. When you read an inequality, read it from left to right—just like reading text on a page. In algebra, inequalities are used to describe large sets of solutions. Sometimes there are an infinite amount of numbers that will satisfy an inequality, so rather than try to list off an infinite amount of numbers, we have developed some ways to describe very large lists in succinct ways.
• ${x}\lt{9}$ indicates the list of numbers that are less than 9. Would you rather write ${x}\lt{9}$ or try to list all the possible numbers that are less than 9? (hopefully, your answer is no)
• $-5\le{t}$ indicates all the numbers that are greater than or equal to $-5$.
Note how placing the variable on the left or right of the inequality sign can change whether you are looking for greater than or less than.
For example:
• $x\lt5$ means all the real numbers that are less than 5, whereas;
• $5\lt{x}$ means that 5 is less than x, or we could rewrite this with the x on the left: $x\gt{5}$ note how the inequality is still pointing the same direction relative to x. This statement represents all the real numbers that are greater than 5, which is easier to interpret than 5 is less than x.
### Inequality Signs
The box below shows the symbol, meaning, and an example for each inequality sign. Sometimes it’s easy to get tangled up in inequalities, just remember to read them from left to right.
Symbol Words Example
$\neq$ not equal to ${2}\neq{8}$, 2 is not equal to 8.
$\gt$ greater than ${5}\gt{1}$, 5 is greater than 1
$\lt$ less than ${2}\lt{11}$, 2 is less than 11
$\geq$ greater than or equal to ${4}\geq{ 4}$, 4 is greater than or equal to 4
$\leq$ less than or equal to ${7}\leq{9}$, 7 is less than or equal to 9
The inequality $x>y$ can also be written as ${y}<{x}$. The sides of any inequality can be switched as long as the inequality symbol between them is also reversed.
## Solve Single-Step Inequalities
### Solve inequalities with addition and subtraction
You can solve most inequalities using inverse operations as you did for solving equations. This is because when you add or subtract the same value from both sides of an inequality, you have maintained the inequality. These properties are outlined in the box below.
### Addition and Subtraction Properties of Inequality
If $a>b$, then $a+c>b+c$.
If $a>b$, then $a−c>b−c$.
### Example
Solve for x.
${x}+3\lt{5}$
Just as you can check the solution to an equation, you can check a solution to an inequality. First, you check the endpoint by substituting it in the related equation. Then you check to see if the inequality is correct by substituting any other solution to see if it is one of the solutions. Because there are multiple solutions, it is a good practice to check more than one of the possible solutions.
The example below shows how you could check that $x<2$ is the solution to $x+3<5$.
### Example
Check that $x<2$ is the solution to $x+3<5$.
Substitute the end point 2 into the related equation, $x+3=5$.
$\begin{array}{r}x+3=5 \\ 2+3=5 \\ 5=5\end{array}$
Pick a value less than 2, such as 0, to check into the inequality. (This value will be on the shaded part of the graph.)
$\displaystyle \begin{array}{r}x+3<5 \\ 0+3<5 \\ 3<5\end{array}$
It checks!
$x<2$ is the solution to $x+3<5$.
The following examples show inequality problems that include operations with negative numbers.
### Example
Solve for x: $x-10\leq-12$
Check the solution to $x-10\leq -12$
### Example
Solve for a. $a-17>-17$
Check the solution to $a-17>-17$
What would you do if the variable were on the right side of the inequality? In the following example, you will see how to handle this scenario.
### Example
Solve for x: $4\geq{x}+5$
Check the solution to $4\geq{x}+5$
### Solve inequalities with multiplication and division
Solving an inequality with a variable that has a coefficient other than 1 usually involves multiplication or division. The steps are like solving one-step equations involving multiplication or division EXCEPT for the inequality sign. Let’s look at what happens to the inequality when you multiply or divide each side by the same number.
Let’s start with the true statement: $10>5$ Let’s try again by starting with the same true statement: $10>5$ Next, multiply both sides by the same positive number: $10\cdot 2>5\cdot 2$ This time, multiply both sides by the same negative number: $10\cdot-2>5 \\ \,\,\,\,\,\cdot -2\,\cdot-2$ 20 is greater than 10, so you still have a true inequality: $20>10$ Wait a minute! $−20$ is not greater than $−10$, so you have an untrue statement. $−20>−10$ When you multiply by a positive number, leave the inequality sign as it is! You must “reverse” the inequality sign to make the statement true: $−20<−10$
Caution! When you multiply or divide by a negative number, “reverse” the inequality sign. Whenever you multiply or divide both sides of an inequality by a negative number, the inequality sign must be reversed in order to keep a true statement. These rules are summarized in the box below.
### Multiplication and Division Properties of Inequality
Start With Multiply By Final Inequality $a>b$ $c$ $ac>bc$ $a>b$ $-c$ $ac Start With Divide By Final Inequality [latex]a>b$ $c$ $\displaystyle \frac{a}{c}>\frac{b}{c}$ $a>b$ $-c$ $\displaystyle \frac{a}{c}<\frac{b}{c}$
Keep in mind that you only change the sign when you are multiplying and dividing by a negative number. If you add or subtract by a negative number, the inequality stays the same.
### Example
Solve for x. $3x>12$
There was no need to make any changes to the inequality sign because both sides of the inequality were divided by positive 3. In the next example, there is division by a negative number, so there is an additional step in the solution.
### Example
Solve for x. $−2x>6$
## Combine properties of inequality to solve algebraic inequalities
A popular strategy for solving equations, isolating the variable, also applies to solving inequalities. By adding, subtracting, multiplying and/or dividing, you can rewrite the inequality so that the variable is on one side and everything else is on the other.
### Example
Solve for p. $4p+5<29$
Check the solution.
### Example
Solve for x: $3x–7\ge 41$
Check the solution.
When solving multi-step equations, pay attention to situations in which you multiply or divide by a negative number. In these cases, you must reverse the inequality sign.
### Example
Solve for p. $−58>14−6p$
Check the solution.
## Simplify and solve algebraic inequalities using the distributive property
As with equations, the distributive property can be applied to simplify expressions that are part of an inequality. Once the parentheses have been cleared, solving the inequality will be straightforward.
### Example
Solve for x. $2\left(3x–5\right)\leq 4x+6$
Check the solution. |
# Distributive Property
11 teachers like this lesson
Print Lesson
## Objective
Students will be able to use the distributive property to re-write numerical expressions as a different yet equivalent numerical expression.
#### Big Idea
A Distribution Collusion: Obtaining the secrets to successfully using the distributive property.
## Curriculum Reinforcer
5 minutes
To begin today's lesson, my students will complete the following three problems to reinforce earlier learning:
1. A gardener has 27 pansies and 36 daisies. He plants an equal number of each type of flower in each row. What is the greatest possible number pansies in each row?
2. Fourteen boys and 21 girls will be equally divided into groups. Find the greatest number of groups that can be created if no one is left out.
3. Macy is painting a design that contains two repeating patterns. One pattern repeats every 8 inches. The other pattern repeats every 12 inches. If both patterns begin at the same place, in how many inches will they begin together again?
These three problems reinforce the concepts of Greatest Common Factor and Least Common Multiple learned in yesterday's lesson
Teaching Note: Greatest Common Factor is a prerequisite skill for today's lesson.
## Engagement
5 minutes
My classroom is normally set up in six groups of 4-5 students. In today's engagement activity, I will provide each of the six groups with 27 cubes presented as a set of 12 and a set of 15. I will then ask the students to create a rectangle with each set of cubes using the following criteria:
The rectangle created with each set of cubes needs to use all of the cubes. In addition, both rectangles need to be created with the same number of rows.
Once the student have figured out how to create the rectangles, I will ask the following questions:
• How did you create the first rectangle with the set of 12 cubes?
• How did you create the second rectangle with the set of 15 cubes?
• How many cubes do you have altogether?
• Look at the rectangles you have created. What are some ways that you can represent what you see using math?
• Can you represent what you see using a mathematical expression?
• How many different ways can you come up with to represent what you see?
As I listen to to my students answer these questions, I am looking for students to tell me that they can represent what they see in the following ways:
• 12 + 15 = 27
• (3 x 4) + (3 x 5) = 27
If I do not receive this response from my students, I will use more probing and strategic questions to help them arrive to this conclusion. For example, I might offer a leading questions like:
• Hey, do these rectangles remind you of an array?
• Can we use what we know about arrays to help us to come up with another way to represent what we see?
## Instruction & Teacher Modeling
10 minutes
In today's instructional piece, I want to truly help my students understand the concept of distributive property. To do this, I will first present the students with the actual explanation/definition of the distributive property. I share some of the teaching moves that I will use with my students in my Distributive Property video.
The distributive property involves the operations of multiplication and addition or multiplication and subtraction. When we use the distributive property, we are multiplying each term inside the parentheses with the term outside of the parentheses. The distributive property, which is displayed below, holds true for all real numbers ab, and c. Also notice that, if you view the formula in the opposite direction, we are just taking out the common factor of a.
## Examples
5(3 + 5)
Using the distributive property, we simplify as follows:
5(3) + 5(5) = 15 + 25 = 40
During the presentation of this explanation, I want to drive home the meaning of the word distribute and how it applies to this situation. I want to make sure that my students see that the factor on the outside of the parenthesis is being distributed evenly among the terms on the inside of the parenthesis.
After providing my students with this definition/explanation, I will ask them how does the activity that we completed today apply to the distributive property. It is my hope that they will see that the number of rows are representative of the a in the example above and that the number of columns are representative of the b and the c in the above example.
Next, I will complete one example to ensure their understanding of the presented material.
The example I will use is 16 + 24
Using the above expression, I will demonstrate to students how to factor using the distributive property. I want to make sure that the students understand the following:
• You must find the greatest common factor. Students need to understand that there is a possibility that there are more than one common factor between two quantities presented and that in order for the expression to be factored properly that the greatest amount should be used.
For example, 16 and 24 have a common factor of 2, 4, and 8... all of these factors can be used but, we should be using the greatest common factor at all times therefore,
• 2(8 + 12) is incorrect because more can be factored out.
• 4(4 + 6) is incorrect because more can be factored out.
• 8(2 + 3) is CORRECT because we have factored out the greatest amount possible.
## Try It Out
10 minutes
To try out this concept of distributive property, I will have my students complete the following 4 problems from Try It Out:
Factor the following expressions using distributive property.
1.) 21 + 12
2.) 28 + 16
As my students work on problems 1 and 2, I am on the lookout for whether or not my students identify the greatest common factor before factoring.
Use the distributive property to simplify the following expressions.
3.) 5(4 + 7)
4.) 9(7 + 3)
I hope that my students are careful when following the algorithm for distributing. There may be some students who distribute the number that is on the outside of the parenthesis to the first term on the inside of the parenthesis only. I like to start with arithmetic expressions, even though we are breaking order of operations, because my students are able to check their own work.
While my students are completing these problems, I will be traveling the room checking their work for accuracy. Should I find that several students are making similar mistakes then, I will take the time to reteach and/or address any misconceptions.
## Independent Exploration
20 minutes
To explore the Distributive Property a little more deeply further, my student will complete this Exploration Worksheet. I will ask them to complete this work in pairs.
The problems in this worksheet are similar to the work completed for the Try It Out section of the lesson. They will need to factor using the distributive property and they will also have to use the distributive property to simplify an expression. In this activity, I also ask my students to show what is happening in each problem using an area model.
## Closing Summary
20 minutes
To close out this lesson, I will have selected student pairs to come to place their work underneath the document camera. They will need to explain what they did to solve the problem that they are presenting. They will be expected to do the following:
• Present the problem.
• Explain what the problem was asking them to do.
• Describe step by step how they chose to solve the problem.
• Connect their visual representation to the mathematics.
The observing students will need to ask questions and critique the work of their peers. I will also be asking questions and critiquing the work. The type of questions that I will be asking will be strategic in nature, designed to provoke thought and and aide in students in being thorough and precise in their explanations.
Questions that I might ask to promote thoroughness and precision:
• Why did you do what you did at this point?
• So, what does that mean?
• So how does your area model represent the situation?
• Tell me what each part of your area model represents? How do you know?
• How did you know that the quantity you have there is the greatest common factor? |
# Ex.1.3 Q3 Real Numbers Solution - NCERT Maths Class 10
## Question
Prove that the following are irrationals:
(i) \begin{align}\frac{1}{{\sqrt 2 }}\end{align}
(ii) \begin{align}7\sqrt 5 \end{align}
(iii) \begin{align}6 +\sqrt 2 \end{align}
Video Solution
Real Numbers
Ex 1.3 | Question 3
## Text Solution
What is unknown?
(i) \begin{align}\frac{1}{{\sqrt 2 }}\end{align}
(ii) \begin{align}7\sqrt 5 \end{align}
(iii) \begin{align}6 +\sqrt 2 \end{align} are Irrationals
Steps:
(i) \begin{align} \frac{1}{{\sqrt 2 }}\end{align}
Let us assume, to the contrary \begin{align} \frac{1}{{\sqrt 2 }}\end{align} that is a rational number.
\begin{align} \frac{1}{{\sqrt 2 }} = \frac{p}{q}\end{align}
Let $$p$$ and $$q$$ have common factors, so by cancelling them we will get \begin{align} \frac{a}{b},\end{align} where $$a$$ and $$b$$ are co-primes.
\begin{align}\frac{1}{{\sqrt 2 }}& = \frac{a}{b} \\\sqrt 2 {a} &= {b}\\\sqrt 2 &= \frac{b}{a}\end{align}
(where $$a$$ and $$b$$ are co - primes and have no common factor otherthan $$1$$)
Since, $$b$$ and $$a$$ are integers, \begin{align} \frac{b}{a}\end{align} is rational number and so, \begin{align} \sqrt 2 \end{align} is rational.
But we know that \begin{align} \sqrt 2 \end{align} is irrational this contradicts the fact that \begin{align} \sqrt 2 \end{align} is rational.
So, our assumption was wrong. Therefore, \begin{align} \frac{1}{{\sqrt 2 }}\end{align} is a rational number.
(ii) \begin{align} 7\sqrt 5 \end{align}
Let us assume, to the contrary that \begin{align} 7\sqrt 5 \end{align} is a rational number.
\begin{align} 7\sqrt 5 = \frac{p}{q}\end{align}
Let $$p$$ and $$q$$ have common factors, so by cancelling them we will get \begin{align} \frac{a}{b},\end{align} where $$a$$ and $$b$$ are co-primes.
\begin{align} 7\sqrt 5 &= \frac{a}{b} \\7\sqrt 5 {\text{b}} &= {\text{a}}\\\sqrt 5 &= \frac{a}{{7b}}\end{align}
(where $$a$$ and $$b$$ are co - primes and have no common factor other than $$1$$)
Since, $$a,\; 7$$ and $$b$$ are integers. So, \begin{align} \frac{a}{{7b}}\end{align} is rational number and so, \begin{align} \sqrt 5 \end{align} is rational. But this contradict the fact that \begin{align} \sqrt 5 \end{align}
So, our assumption was wrong. Therefore, \begin{align} 7\sqrt 5 \end{align} is a rational number.
(iìi) \begin{align} 6{\text{ }} + \sqrt 2 \end{align}
Let us assume, to the contrary that \begin{align} 6{\text{ }} + \sqrt 2 \end{align} is a rational number.
\begin{align} 6{\text{ }} + \sqrt 2 = \frac{p}{q}\end{align}
Let $$p$$ and $$q$$ have common factors, so by cancelling them we will get \begin{align} \frac{a}{b},\end{align} where $$a$$ and $$b$$ are co-primes.
\begin{align} 6 + \sqrt 2 &= \frac{a}{b}\\\sqrt 2 &= \frac{a}{b} - 6 \end{align}
(where $$a$$ and $$b$$ are co - primes and have no common factor other than $$1$$)
Since, $$a, \,b$$ and $$6$$ are integers. So, \begin{align} \frac{a}{b} - 6\end{align} is rational number and so, \begin{align} \sqrt 2 \end{align} is also a rational number.
But this contradicts the fact that \begin{align} \sqrt 2 \end{align} is irrational. So, our assumption was wrong.
Therefore, \begin{align} 6{\text{ }} + \sqrt 2 \end{align} is a rational number.
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
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2021-12-30
Is the algebraic expression a polynomial? If it is, write the polynomial in standard form : ${x}^{2}-{x}^{3}+{x}^{4}-5$.
Jenny Bolton
Step 1
Definition:
The formula for the norm of a polynomial function of degree n is ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\dots +{a}_{1}x+{a}_{0}$ where ${a}_{n}\ne 0$
Step 2
${x}^{2}-{x}^{3}+{x}^{4}-5$ is the algebraic expression provided.
Yes. The algebraic expression ${x}^{2}-{x}^{3}+{x}^{4}-5$ is a polynomial of degree 4.
Its standard form is ${x}^{4}-{x}^{3}+{x}^{2}-5$.
Anzante2m
This fits the definition of a polynomial. It has multiple terms as well as non-negative integers for exponents. To put it in standard form, we put the exponents in descending order.
Result:
${x}^{4}-{x}^{3}+{x}^{2}-5$
Vasquez
A polynomial is a single term or the sum of two or more terms containing variables with whole-number exponents.
Each term has a variable with whole-number exponent so the algebraic expression is a polynomial. Note that the constant term has a variable with exponent 0.
A polynomial In standard form is written in the order of descending powers of the variable.
So, we rewrite the given as:
${x}^{4}-{x}^{3}+{x}^{2}-5$
Result:
yes; ${x}^{4}-{x}^{3}+{x}^{2}-5$
Do you have a similar question? |
The perimeter of a parallelogram is the total distance enclosed by its boundary. Since the parallelogram is a type of quadrilateral, thus it has four sides. The perimeter of the parallelogram will be equal to the sum of all these four sides. Geometry is a branch of mathematics that deals with the study of different geometrical shapes. It is of two types:
• Two-dimensional Geometry
• Three-dimensional Geometry
A parallelogram is a two-dimensional geometric shape bounded by four sides. The properties of a parallelogram are:
• The pair of opposite sides of a parallelogram are parallel and equal to each other.
• Opposite angles are equal in measure
• Diagonals of parallelogram bisect each other
• Pair of adjacent angles are supplementary
• Sum of all interior angles is equal to 360 degrees
Now, let us find here the perimeter of parallelogram along with formulas and also see some solved examples to understand better.
## What is the Perimeter of a Parallelogram?
The perimeter of a parallelogram is defined as the total length of the boundaries, surrounding it. The sum of all the sides of a parallelogram is known as the perimeter of a parallelogram.
The parallelogram perimeter is similar to the perimeter of the rectangle. Since, both the shapes having similar properties, the area and the perimeter of the parallelogram have more or less the same formulae. By adding all the boundaries or sides of the parallelogram, we can easily find the parallelogram perimeter.
## Perimeter of a Parallelogram Formula
Let “a” and “b” be the sides of a parallelogram. Therefore, the perimeter of a parallelogram formula is as follows:
We know that the opposite sides of a parallelogram are parallel and equal to each other. Thus, the formula for finding the perimeter of a parallelogram is given by:
So, the perimeter of Parallelogram, P = a + b + a + b units
P = 2a +2b
P = 2(a+b)
Therefore, the perimeter of a parallelogram, P = 2(a+b) units
## How to Find Perimeter of Parallelogram?
We have already discussed the formula to calculate the perimeter of a parallelogram, given the length of its parallel sides. Thus, combining it there are three cases to find the perimeter of a parallelogram, when:
• Length of adjacent sides are given
• Length of base and height along with one interior angle is given
Let us find the perimeter of a parallelogram based on the above three cases.
## Perimeter of a Parallelogram Given Two Adjacent Sides
This is the common scenario, when we have the length of two adjacent sides of a parallelogram given to us. Thus, we can simply the formula:
Perimeter of a parallelogram = 2 (a + b)
where a and b are the length of two adjacent sides of parallelogram
Example: If a = 10 cm and b = 5 cm, then the perimeter of a parallelogram;
P = 2(a + b) = 2(10 + 5) = 2 (15) = 30 cm
## Perimeter of a parallelogram with Base, Height and an Angle
The perimeter of the parallelogram with base and height is given using the property of the parallelogram. If “b” is the base of the parallelogram and “h” is the height of the parallelogram, then the formula is given as follows:
According to the property of the parallelogram, the opposite sides are parallel to each other, and the parallelogram perimeter is defined as two times of the base and height.
Thus, the formula for the perimeter of a parallelogram is:
P = 2 (b +h/cos θ)
where θ is the angle BAE, formed between the height and side of the parallelogram, i.e. AE and AB.
## Area and Perimeter of a Parallelogram
We know that the area of a parallelogram is equal to the product of base and height.
A = b x h square units ……(1)
The relationship between the area and perimeter of a parallelogram is:
P = 2 (a+b) units
Therefore, the value of b in terms of P is
P/2 = a + b
b=(P/2) – a
Now, substitute the value of b in (1)
A = ((P/2) – a)h Square units
## Related Articles
• Parallelogram
• Perimeter of a Parallelogram
• Angles of a Parallelogram
• Area of Parallelogram
• Perimeter of Rectangle
• Perimeter of Square
• Perimeter of Triangle
## Solved Examples on Perimeter of a Parallelogram
Example 1:
Find the perimeter of a parallelogram whose base and side lengths are 10cm and 5cm, respectively.
Solution:
Given:
Base length of a parallelogram = 10 cm
Side length of a parallelogram = 5 cm
We know that the perimeter of a parallelogram, P = 2(a+b) units.
Substitute the values
P = 2(10+5)
P = 2(15)
P = 30 cm
Therefore, the perimeter of a parallelogram is 30 cm.
Example 2:
Find the length of another side of the parallelogram whose base is 5 cm and the perimeter is 40 cm.
Solution:
Given:
Base, h = 5cm
Perimeter, p = 40cm
We know, that the perimeter of a parallelogram is,
p=2 (a+b) units
Now substitute the given values in the formula,
40 = 2 (a +5)
40 = 2a + 10
2a = 40-10
2a = 30
a = 30/2
a = 15 cm
Thus, the length of other side of the parallelogram is 15 cm.
## Practice Questions
1. Find the perimeter of a parallelogram whose two adjacent sides are:
1. 3cm , 5cm
2. 6cm, 10cm
3. 10cm, 15cm
4. 50mm, 100mm
2. What is the perimeter of a parallelogram whose one of its sides is 15 m, height is 20 m, and one of its interior angles is 30 degrees.
## Frequently Asked Questions on Perimeter of a Parallelogram
### What is the formula for perimeter of a parallelogram?
P = 2(Sum of adjacent sides)
### What is the perimeter of parallelogram using base and height?
P = 2 (b +h/cos θ)
where b = base, h = height of parallelogram with respect to base, θ is the interior angle.
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Deviation Demystified: Unraveling Averages, Means, and Standard Deviations
Comprehensive Definition, Description, Examples & RulesÂ
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Understanding Average Deviation and Mean Deviation
Both average deviation and mean deviation are an essential part of mathematical calculations. The average deviation is calculated while computing the mean and then the distance between every score and the mean without regard to whether the score is below or above the mean. The average deviation is also known as the average absolute deviation.Â
Mean deviation is a statistical measure that you use to calculate the average deviation from the mean value from a given data set. You use the mean deviation to calculate the average absolute difference between the data provided to you.Â
Significance in Statistics and Data Analysis
The significance of both average and main deviation in statistics and data analysis, include:
• It helps to calculate the central value in the distribution.Â
• You use the deviation values to signify that the data points are far from the mean.
• The difference in the range is higher than both the deviations.
• It simplifies the calculation of measurements when there is distributed data.
How to Calculate Average Deviation?
The step-by-step procedure you can follow while collecting average deviation includes:Â
• First, you must calculate the mean of all data points.Â
• Then, find the difference between each of the data points and the mean.
• Finally, calculate the average of all the absolute values of the differences.
Examples:
How to find Average Deviation for the given data: 2,4,6,8,10?
Solution:
n = 5
x= (2+4+6+8+10)/5
x ,= 6
Average Deviation is
(|2 – 6| + |4 – 6| + |6 – 6| + |8 – 6| + |10 – 6|) / 5
(4 + 2 + 0 + 2 + 4)/ 5
2.4
So,
The average Deviation is 2.40
Calculating Mean Deviation: Step-by-Step
The step-by-step guide that you can go through which will help you to find out the mean deviation is:
• Calculate the mean value for a particular data.
• Subtract the mean from every data value, which will help you find the absolute deviation.Â
• You need to find the mean of the absolute deviation to get the mean deviation value.
Examples:
How to find Mean Deviation using Mean and Standard Deviation?
Classes Frequencies 20-40 3 40-80 6 80-100 20
  Solution:
c
f
m
X−A(60)
|d|
f|d|
d = (x-a)/t
fd
d2
fd2
20-40
3
30
-30
30
90
-3
-9
9
27
40-80
6
60
0
0
0
0
0
0
0
80-100
20
90
30
60
1200
3
60
9
180
∑f=29
∑f|d|= 1290
∑fd = 51
∑fd2 =
207
x= A + ∑fd /∑f *i
60+ 51/29*10
7.76
MDX = ∑f|d|/∑f
1290/29
44.48
Deviation is:
20.12
So, Mean deviation is 20.12.
The Average Deviation Formula
The mathematical average deviation formula that you should use will help you to calculate is:
• 1/n Σ | xi – x |
If you want to describe the formula, then it will be:
xi is the data values that are given in a particular set
x is the mean of the set
n here is the total number of data values that are present in the sum.Â
The Mean Deviation Formula
The mathematical mean deviation formula you use for calculating is:
Grouped Data: (Σ1n | xi – x | )/(Σ1nfi)
Defining the formula for mean deviation:
fi here is the repetition frequency of xi.
xi denotes the middle value of the class interval.Â
Practical Examples: Finding Average and Mean Deviation
The real-world examples of applications where you use both average and main deviation are:
• Both these deviations have a significant role to play while calculating business and commerce performance, and companies also use them to understand the performance of their marketing strategies.Â
• One of the primary areas where the deviation is used is to calculate the percentage of marks, and the student’s performance is compared with the other students by making an average.Â
• The company wants to do a graphical analysis of data values and data sets and use this raw data to understand their performance.
Showing how these measures provide proper insights into data:
• These measures provide proper insights into data and represent the typical value, which is essential for observation.
• One of the primary insights that the data provide is showing how the company’s performance is then how well the employees perform.Â
Comparing Average Deviation and Mean Deviation
Differences
Average deviation and mean deviation have a basic difference where the mean deviation is a statistical measure that helps you to compute the average deviation from an average value of a data collection. On the other hand, the average deviation is the calculation of the mean from a particular distance.Â
Similarities
The primary similarity between an average and a mean deviation is that both measure the numerical data and help you find out the data sets’ lowest and highest values and the data group’s middle term. Â
The average deviation is considered a better major, which comes with absolute variability, but when you need to determine the size of the data and the data group has a lot of values, then using the mean variation might be helpful.Â
Interpreting Average and Mean Deviation
The significance of calculating the deviations is:Â
• It shows how much variation comes from the main group.
• The higher the deviation values, the more significant the data point will be from the mean.Â
• You can identify the central value of the distribution.Â
• You can do an indicated research among the data values that are provided to you among the data set.Â
To indicate the data in the form of a data analysis you need to keep a few things in mind:
• A division close to zero indicates that data points are much closer to the mean.Â
• If you have a large deviation, it will indicate that the data sets are spread far away from the mean.
 If you keep this data in mind, you can format it in the form of data analysis and the following data sets to represent the values.Â
Average and Mean Deviation in Descriptive Statistics
The average and mean deviation are important. In descriptive statistics, these divisions are known to be individual data points from which the mean is calculated, and the data sets are highly crusted around the mean. Â
Roles in Summarizing and Analyzing
The role of average and mean deviation in summarizing and analyzing includes:
• It has a primary role in summarizing, analyzing, and representing the data sets’ data.Â
• You can derive the data in tables and graphs, which involves measuring Central tendency.Â
• You can use the measure of dispersion or variance, which will help you represent the data sets using average and mean deviation.Â
Standard Deviation Formula for Ungrouped Data
The standard deviation formula for ungrouped data Ungrouped Data includes:
  (Σ1n | xi – x | )/n
Defining the formula:
xi is the ith observation of the group.
x is representing the central, whether it is the mean mode or median.
n is the total number of observations present in the group.Â
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Key Takeaways
1. Calculation with the average and the mean deviation is challenging enough compared to the other dispersion.Â
2. Giving more way to the extreme values and less to the main ones is important while calculating the deviation.Â
3. Calculating the mean and average deviation in open intervals is impossible.Â
4. Using average and mean deviation when two or more data sets are provided is impossible.
Quiz
Check your score in the end
Quiz
Check your score in the end
Question of
Question comes here
Practical examples where you can use standard deviation for calculating the ungrouped data include:
• Calculating weather forecasting.Â
• Calculating the performance of stock in real estate.
• It also helps calculate medicines’ performance in the healthcare department.
• In human resource management, using standard deviation is also very important to calculate the performance of employees.Â
The calculation for the standard deviation for ungrouped data is done with the formula:
 (Σ1n | xix | )/n
Here,
xi is the ith observation of the group.
x is representing the central, whether it is the mean mode or median.
n is the total number of observations present in the group.Â
Find the Average Deviation for the given data: 5, 6,8,10,12,14
Solution:
n = 6
x= (5+6+8+10,11,14)/6
x ,= 9
Average Deviation is
(|5 – 9| + |6 – 9| + |8 – 9| + |10 – 9| + |11 – 9| + |12 – 9| ) / 6
(4 + 3 + 1+1 + 2 + 3)/ 6
2.33
So,
The average Deviation is 2.37
The limitations you should be aware of are:
• If you compare it with the other measures of dispersion, then calculation using the standard deviation is more challenging.
• You need to give more weight to the extreme values and less to the near means.
• You cannot calculate the average and the mean deviation in open intervals.
• If two or more data sets are provided, then the calculation will not be possible for comparison.Â
Average Deviation is calculated when you compute the mean and a particular distance between the group data without regard for whether the mean is above or below the data.Â
The primary difference between average and mean deviation is that average deviation calculates the mean from a particular distance. In contrast, mean deviation is a statistical measure to compute the average deviation of the average value.Â
Standard deviation is a broader concept, and average deviation is a part of standard deviation.Â
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## College Algebra (10th Edition)
$b-a$
$\bf{\text{Solution Outline:}}$ To write the given expression, $\ln1.5 ,$ in terms of $a$ and $b,$ where $a=\ln2$ and $b=\ln3,$ use the laws of logarithms and substitution. $\bf{\text{Solution Details:}}$ The given expression is equivalent to \begin{array}{l}\require{cancel} \ln\dfrac{3}{2} .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent to: \begin{array}{l}\require{cancel} \ln3-\ln2 .\end{array} By substitution, since $a=\ln2$ and $b=\ln3,$ the expression above is equivalent to: \begin{array}{l}\require{cancel} b-a .\end{array} |
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All Lesson Plans
# Fraction Inequalities
## Lesson Outline and Objective
Students will compare fractions with different denominators.
Fraction bars offer a great concrete tool for working with equivalent fractions. This task encourages students to move from the concrete to the abstract by creating equations and inequalities with fraction bars and then explaining numerically what is happening. Polypad provides a safe place to explore and for students to generate their own examples at a level they are comfortable with, before gradually pushing themselves to create more complicated examples as they become more confident.
## Warm Up
Show students this fraction wall built in Polypad:
Invite them to write down as many sets of equivalent fractions as they can see in the picture. For example. they could write $4/6=2/3$ because the line for four of the groups of $1/6$ coincides with the line for two of the groups of $1/3$.
Students who complete this quickly could also write down other equivalent fraction relationships that do not appear on the wall by using denominators bigger than 10, or fractions bigger than 1. Encourage struggling students to use the bars to create equivalent fractions.
Once students have created their sets of equivalent fractions, take some time to talk about how they can check their answers numerically. If your students would benefit from a more concrete review of this idea, you could create something like this Polypad with them.
## Main Activity
Show students the picture below and pose the question: Which is bigger: $2/3$ or $5/7$? By how much?
Students are likely to agree from the picture that it is clear that $5/7>2/3$, but it’s worth pointing out that it might not always be so obvious, and that students won’t always be able to use Polypad to help them! Invite students to suggest ways of working out which is bigger, and by how much. Use the pen tool to demonstrate dividing each third into seven pieces, and each seventh into three pieces:
Ask students if they can now make a better estimate of how much bigger $5/7$ is than $2/3$. It may be somewhat clear that the bars differ by 1/21. Confirm this by showing the following: $5/7=15/21$ and $2/3=14/21$, so not only can we work out that $5/7>2/3$, we can also work out that $5/7-2/3=1/21$. As you work this out, you could label the Polypad as shown:
Share with students that the visual of $5/7$ and $2/3$ made it clear which fraction was bigger and that drawing in the lines to represent equivalent fractions also helped show the difference of $1/21$.
Next, invite students to create some fraction inequalities of their own. Consider any of the following options:
• Some students might just want to explore fraction combinations without a specific goal or challenge in mind. Invite them to label their diagram as shown above and share their Polypad link with you when finished. Maybe they’ll be ready now to move onto some of the challenges below.
• Challenge students to create examples in which the fractions differ by a certain amount. Make a list of unit fractions on the board ($1/2, 1/3, 1/4$ and so on) and challenge the class to find as many as they can. Check them off as students find solutions. Students can show their work on Polypad as shown above and share their link with you.
• Challenge students to create examples of equivalent fractions in which each row uses more than 1 type of unit fractions (The top and bottom fraction bar must have at least two different colors).
• Challenge students to create examples in which the two fraction representations are as close to each as possible, but not equivalent. Share with the class that based on the example above with 3rd and 7ths, the smallest difference the class has found so far is $1/21$. Challenge them to beat this.
Below are some examples of smaller ones. The 1st one has a difference of $1/180$. The 2nd one has a difference of $1/168$. The 3rd one has a difference of $1/840$. Use these to provide hints to students as needed.
## Support and Extension
To get students used to the idea of making inequalities with fractions, invite them to build a fraction wall of their own, and write down all the different inequalities they can make with it. Encourage reasoning that will help them to develop number sense and estimation skills – for example, when comparing $3/7$ with $5/9$, invite students to consider how each fraction would compare with $1/2$. Students might also benefit from thinking about what happens when we compare fractions with the same numerator, but different denominator. |
# Fractions as Division (Grade 5)
Videos and lessons to help Grade 5 students learn to interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
Common Core: 5.NF.3
### Suggested Learning Targets
• I can explain that fractions represent division.
• I can solve word problems that involve division of whole numbers and interpret the quotient in the context of the problem.
• I can explain or illustrate my solution using visual fraction models or equations.
Related Topics:
Common Core for Grade 5
More Lessons for Grade 5
Interpreting fractions as division and solving word problems involving division (5.NF.3)
In this lesson you will learn to solve division problems by partitioning the whole.
In this lesson you will learn to solve division story problems by representing the answer as an equal share of fractional pieces.
In this lesson you will learn to solve word problems with whole numbers that lead to fractional answers by partitioning the whole.
In this lesson you will learn to solve word problems that divide whole numbers with fractional answers by partitioning the whole.
In this lesson you will learn how to divide a whole number by a whole number to produce a fraction by partitioning a whole.
In this lesson you will learn how to solve a division problem by writing a multiplication equation and partitioning the remainder.
In this lesson you will learn how to divide whole numbers by unit fractions by using a number line.
Division Written as a Fraction; Improper Fractions to Mixed Number
5.NF.3 [interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b); solve for answers in mixed numbers]
Fractions as Division
In this video, we use visual models to represent improper fractions and then we use division to change from improper to mixed. We also look at a proper fraction and why if you divide, you still get the proper fraction.
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# Common Core: High School - Functions : Average Rate of Change: CCSS.Math.Content.HSF-IF.B.6
## Example Questions
### Example Question #1 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the fastest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Diane had the fastest pace for the relay.
### Example Question #2 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the slowest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Hector had the slowest pace for the relay.
### Example Question #3 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the fastest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Diane had the fastest pace for the relay.
### Example Question #4 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the slowest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Diane had the slowest pace for the relay.
### Example Question #5 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the fastest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Jane had the fastest pace for the relay.
### Example Question #6 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the slowest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Diane had the slowest pace for the relay.
### Example Question #7 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the fastest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Hector had the fastest pace for the relay.
### Example Question #8 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the fastest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Hector and Jane had the fastest pace for the relay.
### Example Question #9 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the fastest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Hector had the fastest pace for the relay.
### Example Question #10 : Average Rate Of Change: Ccss.Math.Content.Hsf If.B.6
Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.
Who had the slowest pace?
Possible Answers:
Correct answer:
Explanation:
This question is testing one's ability interpret a table and identify the rate of change.
For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Identify the average rate of change formula for calculations.
In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.
Step 2: Calculate the mile pace for Diane, Jane, and Hector.
Using the information provided by the table, calculate the pace of each individual.
Step 3: Compare the paces to arrive at a solution.
Since,
therefore Diane and Jane had the slowest pace for the relay. |
# RS Aggarwal Solutions for Class 8 Maths Chapter 22 - Introduction to Coordinate Geometry
RS Aggarwal Solutions For Class 8 Maths Chapter 22 Introduction to Coordinate Geometry are provided here. You can download the pdf of RS Aggarwal Solutions for Class 8 Maths Chapter 22 Introduction to Coordinate Geometry from the given links. Class 8 is an important phase of a student’s life. It is critical to thoroughly understand the concepts taught in Class 8 as they are continued in Class 9 and 10, and also build a foundation. Here we will learn about coordinate geometry, Cartesian plane.
RS Aggarwal Solutions helps students to get a good score in the examinations, while also providing extensive knowledge about the subject, as Class 8 is a critical stage in their academic career. Therefore, we at BYJU’S provide answers to all questions uniquely and briefly.
## Download pdf of RS Aggarwal Solutions For Class 8 Maths Chapter 22 Introduction To Coordinate Geometry
### Access Answers to Maths RS Aggarwal Solutions for Class 8 Chapter 22 – Introduction To Coordinate Geometry
Exercise 22A Page No: 247
1. Write the abscissa of the each of the following points:
(i) (0,5)
(ii) (3,7)
(iii) (-2,4)
(iv) (6, -3)
Solution:
(i) Abscissa of (0,5) is 0
(ii) Abscissa of (3,7) is 3
(iii) Abscissa of (-2, 4) is -2
(iv) Abscissa of (6, -3) is 6
2. Write the ordinate of the each of the following points:
(i) (4,0)
(ii) (5,2)
(iii) (1,-4)
(iv) (-10,-7)
Solution:
(i) Ordinate of (4,0) is 0
(ii) Ordinate of (5,2) is 2
(iii) Ordinate of (1,-4) is -4
(iv) Ordinate of (-10,-7) is -7
3. On a graph paper, plot the each of the following points:
(i) A(4,3)
(ii) B(-2,5)
(iii) C(0,4)
(iv) D(7,0)
(v) E(-3,-5)
(vi) F(5,-3)
(vii) G(-5,-5)
(viii) H(0,0)
Solution:
4. Plot each of the following points on the graph sheet. Verify that they lie on a line.
(i) A(4,0), B(4,2), C(4,2.5) and D(4,6)
(ii) P(1,1), Q(2,2), R(3,3) and S(4,4)
(iii) L(6,2), M(5,3), N(3,5) and O(2,6)
Solution:
(i) After joining the points A (4, 0), B (4, 2), C (4, 2.5) and D (4, 6) we can see the given points forms a straight line.
(ii) After joining the points P (1, 1), Q (2, 2), R (3, 3) and S (4, 4) we can see the given points forms a straight line.
(iii) After joining the points L (6, 2), M (5, 3), N (3, 5) and O (2, 6) we can see the given points forms a straight line.
5. Plot the given points on a graph sheet and check if they lie on a straight line. If not, name the shape when they joined in the order.
(i) A(0,2), B(0,3.5), C(0,5), D(0,6)
(ii) P(1,3), Q(1,5), R(3,3), S(3,5)
(iii) E(4,5), F(5,5), G(5,7),H(6,5)
(iv) L(2,0), M(2,3), N(0,3),O(0,0)
(v) J(4,3), K(6,1), L(6,5),M(4,7)
Solution:
(i) After joining the points A (0, 2), B (0, 3.5), C (0, 5), D (0, 6) we can see they form a straight line.
(ii) After joining the points P (1, 3), Q (1, 5), R (3, 3), S (3, 5) we can see they form a square.
(iii) After joining the points E (4, 5), F (5, 5), G (5, 7), H (6, 5) we can see they form a triangle.
(iv) After joining the points L (2, 0), M (2, 3), N (0, 3), O (0, 0) we can see they form a rectangle.
(v) After joining the points J (4, 3), K (6, 1), L (6, 5), M (4, 7) we can see they form a parallelogram.
6. Locate the points A (1, 2), B (4, 2) and C (1, 4) on graph sheet taking the suitable axes. Write the coordinates of the fourth point D in order to complete the rectangle ABCD.
Solution:
Coordinates of D = (4, 4)
Exercise 22B Page No: 249
Select the correct answer in each of the following:
1. In which of the following quadrants does the point P (3, 6) lie?
(a)I (b) II (c) III (d) IV
Solution:
(a)I
Explanation:
We know that abscissa and coordinates are positive in I quadrant.
2. In which of the following quadrants does the point P (-7, -1) lie?
(a)I (b) II (c) III (d) IV
Solution:
(c) III
Explanation:
We know that abscissa and coordinates are negative in III quadrant.
3. In which of the following quadrants does the point P (2,-3) lie?
(a)I (b) II (c) III (d) IV
Solution:
(d)IV
Explanation:
We know that abscissa is positive and coordinate is negative in IV quadrant.
## RS Aggarwal Solutions for Class 8 Maths Chapter 22 – Introduction To Coordinate Geometry
Chapter 22, Introduction to Coordinate Geometry, contains 2 Exercises. RS Aggarwal Solutions given here contains the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this Chapter.
Coordinate axis
Cartesian plane
Definition of order pair
Meaning of coordinates of a point |
Engineering Math - Calculus Vector Field Let's assume that you are asked o create a mathematical model for the water flow shown below. How easy/difficult you think it would be ? It would be almost impossible to make a accurate model which can explain very detailed movement (flow) of the water. But you can come out with big picture of the overall flow without much difficulty. Here comes one way. Let's suppose you are on a small boat on the water(Don't try riding a boat or even fishing in this kind of dangerous place -:). The boat will float along the water flowing under the boat. Now you can measure the direction and the speed of the boat at a point and you can draw an arrow (vector) representing the direction and the speed. (The direction of arrow head would represent the direction that the boat is moving and the size of arrow would represent the speed of the boat). You can repeat this kind of measurement on many different point (the best way would be to make a fine grid lines and perform the measurement and draw a vector at each grid point), you would get a lot of arrows (vectors). And then you would see the overall pattern of the water flow as shown below. This is the concept of vector field. Vector field is a method to represent 'degree of changes' in arrows (vectors) at many points along a specified line, surface or volumn. Now let's get into more mathematical context. In the above example, you 'felt' the flow with your body, now you are given a set of mathematical function as shown below. You see two functions P(x,y) and Q(x,y). Since this example is about the vectors on two dimensional plane, you have two variables x and y. P(x,y) is the function that gevern the vector component in x direction and Q(x,y) is the function that governs the vector component in y direction. The way you draw a vector field is as follows. Pick any points in the coordinate. Plug the coordinate value into the function that represents the x component of a vector (P(x,y) in this example) and get the value, and draw a line on the x direction (green line in the following illustration). Next, Plug the coordinate value into the function that represents the y component of a vector (Q(x,y) in this example) and get the value, and draw a line on the y direction (red line in the following illustration). Following illustration shows this process for three different points. Make it sure that you understand this process very clearly and make some practice on your own. Otherwise, you would have difficulties interpreting the vector field. If you repeat the process explained above for all the points on the grid, you will get a lot of arrows as shown below. This is called a vector field. Examples of Vector Field In this section, I will try to give you as many examples as possible so that you can make practise of plotting vector field. (I will keep adding the examples) For every example, try to draw a couple of arrows by yourself and imagine the final outcome in your mind first and then check if it is the same as the plot given here. Following is the matlab code to draw all the plots in this section. Of course, you have to change the equation for dx, dy in the code. I put various examples of the vector field plot in a slideshow format in my other notes here : www.freetechslide.com. v = -1:0.2:1; % you can change the range to best fit the plot [x,y] = meshgrid(v); dx=y; % you can define P(x,y) part according to following examples. dy=-x - 0.25 .* y; % you can define Q(x,y) part according to following examples. quiver(x,y,dx,dy); axis([-1 1 -1 1]); % you can change the range to best fit the plot Applications of Vector Field There is very important application of 'Vector Field'. You can get a solution to a differential equation without solving it if you use the vector field. Let's see an example as shown below. You can draw vector field from this set of differential equation by converting this equation into vector field format as shown below. As you see, just take the right hand side of differential equation and assign them to P(x,y) and Q(x,y) as shown below. From the vector field plot for this equation, take any one point you want (This first point is the initial condition for the set of differential equation). Go to the end of the arrow (vector) and find another arrow which starts closest to it and then go to the end of the next arrow. If you repeat the process and draw a continous line you followed through, that becomes the solution of the different equation. (The red curve on the plot in the right hand is the path we can draw by this way). I put another note showing the step by step procedure of how this graph is drawn in slideshow and animation at my other note : www.freetechslide.com Following is the matlab/octave code that produce the plots shown above. You can try with various other differential equations with this code. Just modify the parts in blue. v = -1:0.2:1; [y1,y2] = meshgrid(v); u1=y2; u2=-y1 - 0.25 .* y2; dy_dt = @(t,y) [y(2);... -y(1) - 0.25 .* y(2)]; odeopt = odeset ('RelTol', 0.00001, 'AbsTol', 0.00001,'InitialStep',0.5,'MaxStep',0.5); [t,y] = ode45(dy_dt,[0 25], [0.0 1.0],odeopt); subplot(1,2,1);plot(t,y(:,1),'r-',t,y(:,2),'b-'); xlabel('time'); legend('y2(t)','y1(t)'); subplot(1,2,2); hold on; quiver(y1,y2,u1,u2); plot(y(:,1),y(:,2),'r-'); xlabel('y(2)');ylabel('y(1)'); axis([-1 1 -1 1]); hold off; |
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# Lesson 20 Homework 5.4 Answer Key
Greetings, math enthusiasts! Today, we embark on a journey through Lesson 2: Solving Systems of Linear Equations. Systems of equations might appear complex, but fear not – we’re here to guide you through this mathematical terrain. In this blog post, we present to you the invaluable Lesson 2 Systems of Linear Equations Answer Key. With this key, you’ll unlock the secrets to solving these equations and gain a deeper understanding of their applications.
Blog Body:
## Cracking the Code of Systems of Linear Equations
Understanding Systems of Equations
A system of linear equations involves multiple equations with the same variables. The goal is to find values that satisfy all the equations simultaneously. These systems often represent real-world scenarios where multiple variables interact.
Methods of Solving
Several methods can be employed to solve systems of linear equations: graphing, substitution, and elimination. Each method has its strengths and suits different scenarios.
## Guiding You Through the Answer Key
Example 1: Substitution Method
Consider the system: 2x + y = 10 x – y = 2
The Lesson 2 Systems of Linear Equations Answer Key walks you through the substitution method. Replace one variable with its equivalent from another equation, and then solve for the remaining variable. The key illustrates each step, helping you grasp the concept effortlessly.
Example 2: Elimination Method
Now, let’s tackle the same system using the elimination method. Multiply one or both equations to make the coefficients of one variable equal, then subtract the equations to eliminate that variable. The answer key demonstrates this process, empowering you to solve even more complex systems.
## Practical Applications and Mastery
Real-World Relevance
Systems of linear equations find applications in diverse fields, such as economics, physics, and engineering. They model scenarios like balancing budgets, determining optimal solutions, and analyzing interactions between variables.
Mastery Through Practice
Practice is key to mastering systems of linear equations. The Lesson 2 Systems of Linear Equations Answer Key doesn’t just provide answers – it equips you with problem-solving strategies. With consistent practice, you’ll build confidence and improve your skills.
In Conclusion
Navigating systems of linear equations might initially seem like navigating a maze. However, armed with the Lesson 2 Systems of Linear Equations Answer Key, you have a trusted map to guide you through. Embrace the challenge, explore different methods, and hone your problem-solving abilities. Mathematics is a journey of discovery, and each system you solve adds another milestone to your path of proficiency. So, go forth, unravel equations, and enjoy the empowering adventure of mathematical exploration!
## Vocabulary Workshop Level B Unit 10 Answers
Vocabulary Workshop Level B Unit 10 Answers: A Helpful Guide for Language Learners
## Degrees Of Lewdity Apk
Introduction Dive into a world of adult-themed text-based gaming with “Degrees of Lewdity APK.” In this article, we…
## Chapter 21 Sentence Check 2 Answer Key
Blog Introduction: Are you struggling to prepare for your Chapter 2 test in Form 2A? Do you need… |
Measuring Capacity: How to compare units with Examples | Class 3
Measurements
Measuring Capacity for Class 3 Math
Here the student will recall their previous learning from measuring capacity and learn more about measuring cups and spoons. Also, it is explained with relevant word problems on the capacity for class 3, which will help the students relate the topic to real life.
The student will learn to
• Define the definition of capacity in maths
• Identify the capacity measurement units
• Convert L to mL
• Apply word problems on the capacity for class 3 in real-life situations
The learning concept is explained to class 3 students with examples, illustrations, and a concept map. At the end of the page, two printable measuring capacity worksheets for class 3 with solutions are attached for the students.
Download the worksheets and solutions to assess our knowledge of the concept.
Capacity Definition
Capacity: The capacity of a vessel/container is the quantity of liquid or matter it can hold. We use measuring spoons or measuring jugs to measure capacity.
• We often measure capacity in litre or millilitre.
• We measure a small quantity of liquid like perfume, medicine etc. in millilitres (mL) and a large quantity of liquid like milk, water, oil etc. in litres (L).
• The standard unit of measuring liquid is ‘Litre’.
Liquid Measuring Cup
• First place the measuring cup in a flat surface
• Then pour the liquid in the measuring cup. Measuring cups are marked with standard measuring units on their body.
• Now bend down eye level to the marks on the body of the measuring cup and watch carefully where the surface of the liquid touches the line.
• That line signifies the measurement of the liquid.
The picture given below show the process for measurement of liquid.
The cup above has 4.5 mL of liquid. In the similar way we can measure the liquid in a cylinder.
L to mL
We know that 1 L = 1000 mL.
So, to convert litre to millilitre, we need to multiply litre with 1000.
Examples:
Convert from L to ml: (i) 45 L (ii)10 L
1. 45 L = 45 × 1000 ml = 45000 ml
2. 10 L = 10 × 1000 ml = 10000 ml
Word Problems on Capacity for Class 3
Examples 1:
There are 6 bottles of oil in a retail shop and each bottle contains 375 mL oil. The shop keeper has sold 750 mL oil. How much oil is left in the shop?
In 6 bottles, there is total 6 × 375 mL = 2250 mL oil.
If 750 mL oil is sold, then the total amount of oil left is:
2250 ml – 750 ml = 1500 ml
∴ So, 1500 ml oil is left in the shop.
Examples 2:
Soma and Ram bought a packet of 500 ml milk each. Tarru and Jumbo bought a packet of 1L milk each. They poured the milk in a container and distributed the total milk between 6 children. How much milk did each of the children get?
Soma and Ram bought total 2 × 500 ml = 1000 ml
Tarru and Jumbo bought total 2 × 1 L = 2 L
Now, 2 L = 2000 ml
They poured the total milk in the container.
So, the total amount of milk in the container is:
1000 ml + 2000 ml = 3000 ml
Total milk is distributed between 6 children.
Now, each of the children will get 3000 ml ÷ 6 = 500 ml
Hence, each of the children got 500 ml.
• - |
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# LINEAR EQUATIONS PART I
Assignments. LINEAR EQUATIONS PART I. Basic Coordinate Plane Info Review on Plotting Points Finding Slopes x and y intercepts Slope-Intercept Form of a Line Graphing Lines Determine the equation of a line given two points, slope and one point, or a graph. COORDINATE PLANE. Y-axis.
## LINEAR EQUATIONS PART I
E N D
### Presentation Transcript
1. Assignments LINEAR EQUATIONS PART I Basic Coordinate Plane Info Review on Plotting Points Finding Slopes x and y intercepts Slope-Intercept Form of a Line Graphing Lines Determine the equation of a line given two points, slope and one point, or a graph.
2. COORDINATE PLANE Y-axis Parts of a plane X-axis Y-axis Origin Quadrants I-IV QUAD II QUAD I Origin ( 0 , 0 ) X-axis QUAD III QUAD IV
3. PLOTTING POINTS Remember when plotting points you always start at the origin. Next you go left (if x-coordinate is negative) or right (if x-coordinate is positive. Then you go up (if y-coordinate is positive) or down (if y-coordinate is negative) B C A D Plot these 4 points A (3, -4), B (5, 6), C (-4, 5) and D (-7, -5)
4. SLOPE Slope is the ratio of the vertical rise to the horizontal run between any two points on a line. Usually referred to as the rise over run. Run is 6 because we went to the right Slope triangle between two points. Notice that the slope triangle can be drawn two different ways. Rise is 10 because we went up Rise is -10 because we went down Run is -6 because we went to the left Another way to find slope
5. FORMULA FOR FINDING SLOPE The formula is used when you know two points of a line. EXAMPLE
6. Find the slope of the line between the two points (-4, 8) and (10, -4) If it helps label the points. Then use the formula
7. X AND Y INTERCEPTS The x-intercept is the x-coordinate of a point where the graph crosses the x-axis. The y-intercept is the y-coordinate of a point where the graph crosses the y-axis. The x-intercept would be 4 and is located at the point (4, 0). The y-intercept is 3 and is located at the point (0, 3).
8. SLOPE-INTERCEPT FORM OF A LINE The slope intercept form of a line is y = mx + b, where “m” represents the slope of the line and “b” represents the y-intercept. When an equation is in slope-intercept form the “y” is always on one side by itself. It can not be more than one y either. If a line is not in slope-intercept form, then we must solve for “y” to get it there. Examples
9. Put y – x = 10 into slope-intercept form Add x to both sides and would get y = x + 10 Put 2y – 8 = 6x into slope-intercept form. Add 8 to both sides then divide by 2 and would get y = 3x + 4 Put y + 4 = 2x into slope-intercept form. Subtract 4 from both sides and would get y = 2x – 4.
10. GRAPHING LINES I could refer to the table method by input-output table or x-y table. For now I want you to include three values in your table. A negative number, zero, and a positive number. Graph y = 3x + 2 BY MAKING A TABLE OR USING THE SLOPE-INTERCEPT FORM By making a table it gives me three points, in this case (-2, -4) (0, 2) and (1, 5) to plot and draw the line. See the graph.
11. Plot (-2, -4), (0, 2) and (1, 5) Then draw the line. Make sure your line covers the graph and has arrows on both ends. Be sure to use a ruler. Slope-intercept graphing
12. Slope-intercept graphing Steps Make sure the equation is in slope-intercept form. Identify the slope and y-intercept. Plot the y-intercept. From the y-intercept use the slope to get another point to draw the line. y = 3x + 2 Slope = 3 (note that this means the fraction or rise over run could be (3/1) or (-3/-1). The y-intercept is 2. Plot (0, 2) From the y-intercept, we are going rise 3 and run 1 since the slope was 3/1.
13. FIND EQUATION OF A LINE GIVEN 2 POINTS Find the equation of the line between (2, 5) and (-2, -3). Find the slope between the two points. Plug in the slope in the slope-intercept form. Pick one of the given points and plug in numbers for x and y. Solve and find b. Rewrite final form. Slope is 2. y = 2x + b Picked (2, 5) so (5) = 2(2) + b b = 1 y = 2x + 1 Two other ways
14. stop
15. Steps if given the slope and a point on the line. Substitute the slope into the slope-intercept form. Use the point to plug in for x and y. Find b. Rewrite equation. If given a graph there are three ways. One way is to find two points on the line and use the first method we talked about. Another would be to find the slope and pick a point and use the second method. The third method would be to find the slope and y-intercept and plug it directly into y = mx + b.
16. The End
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How do you use pascal’s triangle to expand ${(2x - 3y)^3}?$
Last updated date: 03rd Aug 2024
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Hint: As we know that we have to apply the pascal’s triangle to expand the binomial. We know that it is an infinite equilateral triangle which consists of sequence of numbers. It starts with $1$ . The second row consists of the sum of two numbers above it, Similarly we can find out the values of the next rows. It creates a pattern and it is shown below:
$1 \\ 1 - 1 \\ 1 - 2 - 1 \\ 1 - 3 - 3 - 1 \\ \\$
As we know that the main application of this triangle is to solve a binomial function. If the binomial equation is ${(a + b)^n}$ , then the expansion is
${C_1}{a^n}{b^0} + {C_2}{a^{n - 1}}{b^1} + ... + {C_n}{a^0}{b^n}$ .
Since we have $n = 3$ , and from the above triangle we can see that the third order term is $1,3,3,1$ .
For this we have the formula,
${(a - b)^3} = 1{a^3}{b^0} - 3{a^2}{b^1} + 3{a^1}{b^2} - 1{a^0}{b^3}$ .
By applying the above formula we can write
${(2x - 3y)^3} = 1{(2x)^3}{(3y)^0} - 3{(2x)^2}{(3y)^1} + 3{(2x)^1}{(3y)^2} - 1{(2x)^0}{(3y)^3}$ .
On further simplifying we have,
${(2x)^3} - 3{(2x)^2}{(3y)^{}} + 3{(2x)^1}{(3y)^2} - {(3y)^3}$ .
It gives us the expression:
$8{x^3} - 3 \times 4{x^2}{(3y)^{}} + 6x \times 9{y^2} - 27{y^3}$ .
Hence the required value is $8{x^3} - 36{x^2}y + 54x{y^2} - 27{y^3}$ .
So, the correct answer is “ $8{x^3} - 36{x^2}y + 54x{y^2} - 27{y^3}$ ”.
Note: We should note that pascal’s triangle is helpful only when the value of $n$ is small in the equation ${(a + b)^n}$ . If the value is large then it is very tedious to draw the triangle until we reach $n$ . The formula that we used above because the question has a cube of the difference. It there is cube of the sums then the formula that we use is ${(a + b)^3} = 1{a^3}{b^0} + 3{a^2}{b^1} + 3{a^1}{b^2} + 1{a^0}{b^3}$ . |
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# In the given figure, the square ABCD is divided into five equal; parts, all having the same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22cm. Find the perimeter of the part ABEF.
Last updated date: 25th Feb 2024
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Hint: Find the area of the given and find the area of the circle which is ${{\dfrac{1}{5}}^{th}}$of the total area. Take O as the central point. Find the value of OA, AE and find the value of arc EF. Add all the values to get the perimeter of part ABEF.
Complete step-by-step Solution:
Given a square ABCD, we know that in a square all sides are equal and all the angles are ${{90}^{\circ }}$.
Given that AB = 22cm, where AB is the side of a square.
We are told that all sides are equal.
So, AB = BC = CD = DA = 22cm.
Area of the square is given as the square of the side.
i.e. Area of square $={{\left( side \right)}^{2}}$.
Here, the sides of the square are AB, BC, CD and DA which is equal to 22cm.
$\therefore$Side of square = 22cm
$\therefore$Area of square $={{\left( 22 \right)}^{2}}=22\times 22=484c{{m}^{2}}$
It is said that the square ABCD is divided into 5 equal parts in terms of area.
We know the total area of the square as $484c{{m}^{2}}$. Out of these 5 equal parts, one part is the circle inscribed inside the square. To find the area of the circle we need to divide the total area by 5.
$\therefore$Area of circle$=\dfrac{1}{5}\times$area of square
$=\dfrac{1}{5}\times 484=96.8c{{m}^{2}}$
Area of the circle is given by the formula, $\pi {{r}^{2}}$. We need to find the radius, r of the circle.
\begin{align} & \therefore \pi {{r}^{2}}=96.8 \\ & \Rightarrow {{r}^{2}}=\dfrac{96.8}{\pi }=30.81 \\ & \therefore r=\sqrt{30.81}=5.55cm \\ \end{align}
$\therefore$Radius of the central part = 5.55cm.
Let us consider O as the center of the circle thus O also becomes the center of the square.
We know the diagonal of a square is given by $\sqrt{2}a$, where ‘a’ is the side of the square.
Let the diagonal of square AC = $\sqrt{2}a$.
\begin{align} & AC=OA+OC \\ & \therefore OA=OC=\dfrac{AC}{2} \\ \end{align}
i.e. Diagonals of the square are equal in length and bisect each other.
$\therefore OA=\dfrac{\sqrt{2}a}{2}=\dfrac{\sqrt{2}\times 22}{2}=11\sqrt{2}cm$
From the figure we can say that length of AE is equal to length of BF.
$AE=OA-OE$[from the figure]
$\therefore AE=BF=OA-OE$
OE is equal to the radius of the circle = 5.55cm.
\begin{align} & \therefore AE=BF=11\sqrt{2}-5.55\left[ \because \sqrt{2}=1.414 \right] \\ & AE=\left( 11\times 1.414 \right)-5.55 \\ & AE=BF=9.96cm \\ \end{align}
The arc of circle EF$=\dfrac{1}{4}\times$circumference of the circle
We know the circumference of the circle $=2\pi r$.
$\therefore arcEF=\dfrac{1}{4}\times 2\pi r=\dfrac{1}{4}\times 2\pi \times 5.55=8.72cm$
Hence, perimeter of point ABEF = AB + AE + EF + BF = 22 + 9.96 + 8.72 + 9.96 = 50.64cm
$\therefore$Perimeter of part ABEF = 50.64cm.
Note:
Since it is a square, the area of part ABEF is equal to the area of part BFGC, CGHD and DHEF. And their areas are the same as mentioned in the question.
The area of the circle is$96.8c{{m}^{2}}$, which was one part of the circle. Similarly, the area of ABEF, BFGC, CGHD and DHEF is equal to $96.8c{{m}^{2}}$. By adding all this we get the total area as $484c{{m}^{2}}$. |
# What is 3/261 Simplified?
Are you looking to calculate how to simplify the fraction 3/261? In this really simple guide, we'll teach you exactly how to simplify 3/261 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms).
To start with, the number above the line (3) in a fraction is called a numerator and the number below the line (261) is called the denominator.
So what we want to do here is to simplify the numerator and denominator in 3/261 to their lowest possible values, while keeping the actual fraction the same.
To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers.
In our case with 3/261, the greatest common factor is 3. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified:
3/3 = 1
261/3 = 87
1 / 87
What this means is that the following fractions are the same:
3 / 261 = 1 / 87
So there you have it! You now know exactly how to simplify 3/261 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below:
1/87
## Convert 3/261 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
3 / 261 = 0.0115
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• "What is 3/261 Simplified?". VisualFractions.com. Accessed on January 21, 2022. http://visualfractions.com/calculator/simplify-fractions/what-is-3-261-simplified/.
• "What is 3/261 Simplified?". VisualFractions.com, http://visualfractions.com/calculator/simplify-fractions/what-is-3-261-simplified/. Accessed 21 January, 2022. |
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# A man completes a journey in 10 hours. He travels the first half of the journey at the rate of 21 km/hr and the second half at the rate of 24 km/hr. Find the total journey in km.(a) 220 km(b) 224 km(c) 230 km(d) 234 km
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Hint: Let us assume that the total distance of the journey is x. Now, find the time taken to cover half of the journey with a speed of 21 km/hr and also the time taken to cover the other half of the journey with a speed of 24 km/hr. The time is being calculated using the formula between speed, distance and time $Speed=\dfrac{\text{Distance}}{\text{Time}}$. Now, add the time that we have got in each half of the journey and equate them to 10. Hence, we will find the value of x.
Let us assume that the total distance a man travelled in a journey is x km.
Now, it is given that the man travelled half of the distance i.e. $\dfrac{x}{2}$ with a speed of 21 km/hr. Then the time taken by the man to travel $\dfrac{x}{2}$ is going to be calculated by using the following formula:
$Speed=\dfrac{\text{Distance}}{\text{Time}}$
Substituting distance as $\dfrac{x}{2}$ and speed as 21 in the above formula we get,
\begin{align} & 21=\dfrac{\dfrac{x}{2}}{\text{Time}} \\ & \Rightarrow Time=\dfrac{x}{2\left( 21 \right)}=\dfrac{x}{42} \\ \end{align}
This is the time that a man took in travelling the first half of the journey.
Now, we are going to calculate the time taken by a man to travel the other half of the journey with a speed of 24 km/hr.
$Speed=\dfrac{\text{Distance}}{\text{Time}}$
Substituting distance as $\dfrac{x}{2}$ and speed as 24 in the above formula we get,
\begin{align} & 24=\dfrac{\dfrac{x}{2}}{\text{Time}} \\ & \Rightarrow Time=\dfrac{x}{2\left( 24 \right)}=\dfrac{x}{48} \\ \end{align}
It is given that the total time taken by a man to cover the total journey is 10 hours so adding the time of the two halves of the journey and adding them to 10.
\begin{align} & \dfrac{x}{42}+\dfrac{x}{48}=10 \\ & \Rightarrow x\left( \dfrac{8+7}{336} \right)=10 \\ & \Rightarrow x\left( \dfrac{15}{336} \right)=10 \\ & \Rightarrow x=\dfrac{10\times 336}{15}=224 \\ \end{align}
Hence, the distance of the journey is equal to 224 km.
So, the correct answer is “Option B”.
Note: In the above problem, the possibility of making calculation mistakes is pretty high. The other thing to keep in mind is that the units of speed, distance and time are in sync with each other.
For instance in the below formula,
$Speed=\dfrac{\text{Distance}}{\text{Time}}$
If speed is in km/hr, then units of distance should be in km and time should be in hour.
Luckily in this problem, time, distance and speed are already given in such a way that they are in sync with the above formula. But in other questions, you won’t be that lucky and might be time is given in minutes and speed is in km/hr and distance is in km then you have to convert time in hr to make the speed, distance and time sync with each other. |
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hw9-19 - (d How high is it when its velocity is-48 ft/s 1 6...
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Math 170-010 Homework 9/19 Fall 2011 Goal: Use your understanding of derivatives to work more complicated problems. Learn to recognize whether a problem requires that you compute slope, given location, or that you find location, given slope. Note: You can use any of the easy derivative rules from the list we have developed so far. If you encounter a function NOT on the list, you have to resort to secant slope methods. Note: Before you start each problem, try to identify each problem as Type 1: “Given location, find slope”, or Type 2: “Given slope, find location”. Exercises: 1. From Schaum’s , Chapter 9, Problem 19. 2. For each function in Problem 19, determine all locations where there is a slope of exactly - 2. 3. From Schaum’s , Chapter 9, Problem 21. 4. From Schaum’s , Chapter 9, Problem 22. 5. An object is dropped from a tower so that after t seconds its height above ground is h ( t ) = 100 - 16 t 2 feet. (a) What is its velocity after 2 seconds? (b) When is its velocity - 60 ft/s? (c) What is its velocity at the time when it hits the ground?
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Unformatted text preview: (d) How high is it when its velocity is-48 ft/s? 1 6. The graph of f ( x ) = 1 /x has a tangent line that passes through (0 , 1 . 5) as shown at right. (a) Where is the point of tangency? (b) Find the equation of the tangent line at the point (2 , . 5). Sketch this line in the graph at right. –1 –0.5 0.5 1 1.5 2 –1 1 2 3 4 x Challenge Problems: 1. In the ±gure the below the curve is y = sin x and the line is tangent to the curve at x = 3 π/ 4. Find the x-intercept of the line. 2 2. In the fgure the below the curve is y = sin x and the tangent line has x-intercept at 5 π/ 4. Find the point o± tangency. Hints and Answers Exercises: 1. All Type 1. 2. All Type 2. (a) x = 1 / 5; (b) x =-1 ± √ 2; (c) x =-2 ,-4. 3. Type 2. 4. Type 1. 5. (a) Type 1,-64 ±t/s; (b) Type 2, 1.875 s; (c) Type 1,-80 ±t/s; (d) Type 2, 64 ±t. 6. (a) Type 2, (4 / 3 , 3 / 4); (b) Type 1, y =-x/ 4 + 1. Challenge: 1. Type 1, x = 3 π/ 4 + 1 ≈ 3 . 356. 2. Type 2, x ≈ 2 . 063. 3...
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# 83. Determinants Solutions Part 4
1. $$\Delta = \begin{vmatrix}1 & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & 1 & \frac{\log y}{\log z} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & 1\end{vmatrix}$$
$$= \frac{1}{\log x\log y\log z}\begin{vmatrix}\log x & \log y & \log z \\ \log x & \log y & \log z \\ \log x & \log y & \log x\end{vmatrix}$$
$$= 0$$ because all three rows are identical.
2. $$\Delta = \begin{vmatrix}a^{2x} + a^{-2x} + 2 & a^{2x} + a^{-2x} - 2 & 1 \\ b^{2x} + b^{-2x} + 2 & b^{2x} + b^{-2x} - 2 & 1 \\ c^{2x} + c^{-2x} + 2 & c^{2x} + c^{-2x} - 2 & 1\end{vmatrix}$$
$$= \begin{vmatrix}a^{2x} + a^{-2x} & a^{2x} + a^{-2x} & 1 \\ b^{2x} + b^{-2x} & b^{2x} + b^{-2x} & 1 \\ c^{2x} + c^{-2x} & c^{2x} + c^{-2x} & 1\end{vmatrix} [C_1\rightarrow C_1 - 2C_3; C_2\rightarrow C_2 + 2C_3]$$
$$= 0$$ because first two columns are identical.
3. Considering first determinant only:
$$\Delta = \begin{vmatrix}115 & 114 & 103 \\ 108 & 106 & 111 \\ 113 & 116 & 104\end{vmatrix}[C_1\leftrightarrow C_2; C_2\leftrightarrow C_3]$$
Performing $$R_1\leftrightarrow R_3$$
$$\Delta = -\begin{vmatrix}113 & 116 & 104 \\ 108 & 106 & 111 \\ 115 & 114 & 103\end{vmatrix}$$
Thus, given condition is satisfied.
4. $$\sum{n = 1}^N U_n = \begin{vmatrix}\sum{n = 1}^N n & 1 & 5 \\ \sum{n = 1}^N n^2 & 2N + 1 & 2N + 1 \\ \sum{n = 1}^N n^3 & 3N^2 & 3N\end{vmatrix}$$
$$= \begin{vmatrix}\frac{n(n + 1)}{2} & 1 & 5 \\ \frac{n(n + 1)(2n + 1)}{6} & 2N + 1 & 2N + 1 \\ \left\{\frac{n(n + 1)}{2}\right\}^2 & 3N^2 & 3N\end{vmatrix}$$
Taking $$\frac{N(N + 1)}{2}$$ common from first column and then performing $$C_1\rightarrow C_1 - \frac{1}{6}(C_2 + C_3)$$
$$= \frac{N(N + 1)}{2}\begin{vmatrix}0 & 1 & 5 \\ 0 & 2N + 1 & 2N + 1 \\ 0 & 3N^2 & 3N\end{vmatrix}$$
Since first column has only $$0$$ as element the sum of determinants is zero.
5. $$\because A, B, C$$ are angles of a triangle, therefore $$A + B + C == \pi; \sin(A + B + C) = 0; \cos(A + B) = -\cos C$$
$$\therefore \Delta = \begin{vmatrix}0 & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ -\cos C & -\tan A & 0 \end{vmatrix}$$
Changing rows into corresponding columns
$$= \begin{vmatrix}0 & -\sin B & -\cos C \\ \sin B & 0 & -\tan A \\ \cos C & \tan A & 0\end{vmatrix}$$
Taking $$-1$$ common from second and third columns, we have
$$= \begin{vmatrix}0 & \sin B & \cos C \\ \sin B & 0 & \tan A \\ \cos C & -\tan A & 0\end{vmatrix} = -\Delta$$
$$\Rightarrow 2\Delta = 0 \Rightarrow \Delta = 0$$
6. Taking $$b - a$$ common from first and third columns
$$\Delta = (b - a)^2 \begin{vmatrix}b & b - c & c \\ a & a - b & b \\ c & c - a & a\end{vmatrix}$$
$$= (b - a)^2 \begin{vmatrix}b - c & b - c & c \\ a - b & a - b & b \\ c - a & c - a & a\end{vmatrix}[C_1 \rightarrow C_1 - C_3]$$
Since the first two columns are same the determinant is zero.
7. We can rewrite it as $$\sum_{j = 0}^{n - 1}\Delta_j = \begin{vmatrix}\sum_{j = 0}^{n - 1} j & n & 6 \\ \sum_{j = 0}^{n - 1} j^2 & 2n^2 & 4n - 2 \\ \sum_{j = 0}^{n - 1}j^3 & 3n^3 & 3n^2 - 3n\end{vmatrix}$$
$$= \begin{vmatrix}\frac{n(n - 1)}{2} & n & 6 \\ \frac{n(n - 1)(2n - 1)}{6} & 2n^2 & 4n - 2 \\ \left\{\frac{n(n - 1)}{2}\right\}^2 & 3n^3 & 3n^2 - 3n\end{vmatrix}$$
$$= \frac{n(n - 1)}{2}\begin{vmatrix}1 & n & 6 \\ \frac{2n - 1}{3} & 2n^2 & 4n - 2 \\ \frac{n(n - 1)}{2} & 3n^3 & 3n^2 - 3n\end{vmatrix}$$
$$= \frac{n(n - 1)}{2}\begin{vmatrix}0 & n & 6 \\ 0 & 2n^2 & 4n - 2 \\ 0 & 3n^3 & 3n^2 -3n\end{vmatrix}[C_1\rightarrow C_1 - \frac{C_3}{6}]$$
Since first column is entirely made up of zeros the value of determinant is zero, which is a constant as desired.
8. $$\sum_{r = 0}^m(2r - 1) = \frac{1}{2}(m + 1)(2m - 1 - 1) = m^2 - 1$$
$$\sum_{r = 0}^m({}^nC_r) = 2^m$$
$$\sum_{r = 0}^m1 = m + 1$$
Thus, first two rows of determinant become zero leading the desired sum to be $$0.$$
9. $$= \begin{vmatrix}{}^xC_r & {}^{x + 1}C_{r + 1} & {}^{x + 1}C_{r + 2} \\ {}^yC_r & {}^{y + 1}C_{r + 1} & {}^{y + 1}C_{r + 2} \\ {}^zC_r & {}^{z + 1}C_{r + 1} & {}^{z + 1}C_{r + 2}\end{vmatrix} [C_3 \rightarrow C_3 + C_2; C_2\rightarrow C_2 + C_1]$$
Performing $$C_3\rightarrow C_3 + C_2$$ we get the determinant on R.H.S.
10. $$\sum_{r = 1}^n \Delta_r = \begin{vmatrix}\sum_{r = 1}^nr & n + 1 & 1 \\ \sum_{r = 1}^nr^2 & 2n - 1 & \frac{2n + 1}{3} \\ \sum_{r = 1}^nr^3 & 3n + 2 \\ \frac{n(n + 1)}{2}\end{vmatrix}$$
$$= \begin{vmatrix}\frac{n(n + 1)}{2} & n + 1 & 1 \\ \frac{n(n + 1)(2n + 1)}{6} & 2n - 1 & \frac{2n + 1}{3} \\ \left\{\frac{n(n + 1)}{2}\right\}^2 & 3n + 2 & \frac{n(n + 1)}{2}\end{vmatrix}$$
If we take $$\frac{n(n + 1)}{2}$$ common from first column then first and third column becomes same. Thus, $$\sum_{r = 1}^n \Delta_r = 0$$
11. $$\sum_{r = 1}^n 2^{r - 1} = 1 + 2 + \ldots + 2^{n - 1} = \frac{2^n - 1}{2 - 1} = 2^n - 1$$
$$\sum_{r = 1}^n 2.3^{r - 1} = 2.\frac{3^n - 1}{3 - 1} = 3^n - 1$$
$$\sum_{r = 1}^n 4.5^{r - 1} = 4.\frac{35^n - 1}{5 - 1} = 5^n - 1$$
Thus, we see that first row and third rows are equal leading the sum of the determinants to zero.
12. $$\Delta = \begin{vmatrix}2x - 1 & 2x - 3 & x^2 - 4x + 4 \\ 2x - 3 & 2x - 5 & x^2 - 6x + 9 \\ 2x -5 & 2x -7 & x^2 - 8x + 16\end{vmatrix} [C_1 \rightarrow C_1 - C_1; C_2\rightarrow C_2 - C_3]$$
$$= \begin{vmatrix}2x - 1 & 2x - 3 & x^2 \\ 2x - 3 & 2x - 5 & x^2 \\ 2x - 5 & 2x - 7 & x^2\end{vmatrix} + \begin{vmatrix}2x - 1 & 2x - 3 & -4x \\ 2x - 3 & 2x - 5 & -6x \\ 2x - 5 & 2x - 7 & -8x\end{vmatrix} + \begin{vmatrix}2x - 1 & 2x - 3 & 4 \\ 2x - 3 & 2x - 5 & 9 \\ 2x - 5 & 2x - 7 & 16\end{vmatrix}$$
Clearly, if we perform $$R_1\rightarrow R_1- R_2; R_2\rightarrow R_2 - R_3$$ will make $$R_1$$ and $$R_3$$ same in the first determinant.
This is also true for second determinant.
$$= \begin{vmatrix}2 & 2 & -5 \\ 2 & 2 & -7 \\ 2x - 5 & 2x - 7 & 16\end{vmatrix}$$
Clearly, the determinant is independent of $$x$$
13. $$\Delta = \begin{vmatrix}2 & 1 + i & 3 \\ 1 - i & 0 & 2 + i \\ 3 & 2 - i & 1\end{vmatrix}$$
Taking complex conjugate and exchanging rows into corresponding columns
$$\overline{\Delta} = \begin{vmatrix}2 & 1 + i & 3 \\ 1 - i & 0 & 2 + i \\ 3 & 2 - i & 1\end{vmatrix} = \Delta$$
Since $$\overline{\Delta} = \Delta,$$ the determinant is purely real.
14. $$\Delta = \begin{vmatrix}x - 3 & 2x & 2 \\ 3x + 2 & x & 1 \\ 5x + 1 & 5x & 5\end{vmatrix} + \begin{vmatrix}x - 3 & 1 & 2 \\ 3x + 2 & 2 & 1 \\ 5x + 1 & 4 & 5\end{vmatrix}$$
If we take out $$x$$ common from second column of first determinant then second and third columns are same, making it zero. Now expandng second determinant
$$= \begin{vmatrix}x & 1 & 2 \\ 3x & 2 & 1 \\ 5x & 4 & 5 \end{vmatrix} +$$ a determinant of constants(say $$k$$)
$$= x \begin{vmatrix}0 & 1 & 2 \\ 1 & 2 & 1 \\ 1 & 4 & 5\end{vmatrix} [C_1\rightarrow C_1 - C_2] + k$$
$$= x \begin{vmatrix}0 & 1 & 2 \\ 1 & 2 & 1 \\ 0 & 2 & 4\end{vmatrix} [R_3\rightarrow R_3 - R_2] + k$$
$$= x\begin{vmatrix}0 & 1 & 2 \\ 1 & 2 & 1 \\ 0 & 0 & 0\end{vmatrix} [C_3\rightarrow C_3 - 2C_1] + k$$
$$= k$$
15. $$\Delta = \begin{vmatrix}a^n - x & a^n(a - 1) & a^{n + 1}(a - 1) \\ a^{n + 3} - x & a^{n + 3}(a - 1) & a^{n + 4}(a - 1) \\ a^{n + 6} - x & a^{n + 6}(a - 1) & a^{n + 7}(a - 1)\end{vmatrix}[R_2\rightarrow R_2 - R_1; R_3\rightarrow R_3 - R_2]$$
$$=a^{n(n + 1)}(a - 1)^2\begin{vmatrix}a^n - x & 1 & 1 \\ a^{n + 3} - x & a^3 & a^3 \\ a^{n + 6} - x & a^6 & a^6\end{vmatrix} = 0$$
Since second and third columns are same, the edterminant is zero.
16. $$\Delta = \sum_{r = 2}^n (-2)^r \begin{vmatrix}{}^nC_r & {}^{n - 2}C_{r - 1} & {}^{n - 2}C_r \\ 0 & 1 & 1 \\ 0 & -1 & 9\end{vmatrix} [C_1\rightarrow C_1 + 2C_2 + C_3]$$
$$= \sum_{r = 2}^n (-2)^r{}^nC_r$$
$$= \sum_{r = 0}^n (-2)^r{}^nC_r - ({}^nC_0 -2{}^nC_1)$$
$$= 2n - 1 + (-1)^n$$
17. Performing $$R_1\rightarrow aR-1, R_2\rightarrow bR_2, R_3\rightarrow cR_3$$ and then taking out $$abc$$ out from first two columns,
$$\Delta = abc\begin{vmatrix}bc & 1 & a(b + c) \\ ca & 1 & b(c + a) \\ ab & 1 & c(a + b)\end{vmatrix}$$
Performing $$C_3\rightarrow C_3 + C_1$$ and then taking $$ab + bc + ca$$ out
$$= abc(ab + bc + ca)\begin{vmatrix}bc & 1 & 1 \\ ca & 1 & 1 \\ ab & 1 & 1\end{vmatrix}$$
Since last two columns are same, the determinant is zero.
Rest of the problems are left as exercises. |
HomeTren&dHow Many Faces Does a Cone Have?
# How Many Faces Does a Cone Have?
A cone is a three-dimensional geometric shape that is commonly encountered in various fields, including mathematics, engineering, and everyday life. It is a fascinating object with unique properties and characteristics. One question that often arises when studying cones is: how many faces does a cone have? In this article, we will explore the answer to this question in detail, providing valuable insights and examples along the way.
## The Definition of a Cone
Before delving into the number of faces a cone possesses, let’s first establish a clear understanding of what a cone is. In geometry, a cone is a solid object that has a circular base and a pointed top, known as the apex. The base can be any size, and the apex is directly above the center of the base. The sides of the cone connect the apex to the points on the circumference of the base.
Cones are classified as a type of pyramid, as they share the characteristic of having a polygonal base and triangular sides that converge at a single point. However, unlike other pyramids, cones have a circular base instead of a polygonal one.
## The Faces of a Cone
Now that we have a clear understanding of what a cone is, let’s explore the number of faces it possesses. A face is a flat surface that forms part of the boundary of a solid object. In the case of a cone, the number of faces depends on how we define them.
### Definition 1: Faces as Flat Surfaces
If we define faces as flat surfaces, a cone has two faces: the curved surface and the base. The curved surface is the lateral surface of the cone, which connects the apex to the points on the circumference of the base. It is a single continuous surface that wraps around the cone. The base, on the other hand, is a flat circular surface that forms the bottom of the cone.
Therefore, when considering faces as flat surfaces, a cone has two faces: the curved surface and the base.
### Definition 2: Faces as Individual Triangles
Another way to define faces is by considering each individual triangle that makes up the cone. In this case, a cone has an infinite number of faces. Each face is a triangle formed by connecting a point on the circumference of the base to the apex of the cone.
As the number of triangles that can be formed is infinite, we can say that a cone has an infinite number of faces when considering faces as individual triangles.
## Real-Life Examples
Cones are not just abstract mathematical objects; they can be found in various real-life examples. Let’s explore a few examples to better understand the concept of faces in cones.
### 1. Ice Cream Cone
One of the most common examples of a cone is an ice cream cone. The ice cream cone consists of a conical-shaped wafer or sugar cone that serves as the curved surface, and the ice cream itself forms the base. In this case, the cone has two faces: the curved surface of the wafer cone and the flat circular base formed by the ice cream.
### 2. Traffic Cone
Traffic cones, also known as pylons or road cones, are used to redirect traffic or indicate hazards on the road. They are typically made of bright orange plastic and have a conical shape. In this example, the traffic cone has two faces: the curved surface and the flat circular base.
### 3. Volcanic Cone
A volcanic cone is a landform that is created by the eruption of a volcano. It is formed by layers of volcanic ash, lava, and other materials. The shape of a volcanic cone is typically conical, with a pointed summit and a circular base. In this case, the volcanic cone has two faces: the curved surface and the flat circular base.
## Q&A
### Q1: Can a cone have more than one base?
A1: No, a cone can only have one base. The base is a defining characteristic of a cone, and it is always a flat circular surface.
### Q2: Are all cones the same shape?
A2: No, not all cones are the same shape. While all cones have a circular base and a pointed apex, the size and proportions of the cone can vary. Some cones may be tall and slender, while others may be short and wide.
### Q3: Can a cone have a square base?
A3: No, a cone cannot have a square base. The base of a cone is always circular. If the base were square, it would be classified as a different geometric shape known as a square pyramid.
### Q4: Are all cones symmetrical?
A4: No, not all cones are symmetrical. A cone is considered symmetrical if its apex is directly above the center of the base, and the sides of the cone are evenly distributed. However, cones can also be asymmetrical, where the apex is not directly above the center of the base.
### Q5: Can a cone have a curved base?
A5: No, a cone cannot have a curved base. The base of a cone is always a flat circular surface. If the base were curved, it would be classified as a different geometric shape, such as a sphere or a cylinder.
## Summary
In conclusion, the number of faces a cone has depends on how we define them. If we consider faces as flat surfaces, a cone has two faces: the curved surface and the base. However, if we consider faces as individual triangles, a cone has an infinite number of faces. Real-life examples, such as ice cream cones, traffic cones, and volcanic cones, help illustrate the concept of faces in cones. Understanding the number of faces in a cone is essential for various applications in mathematics, engineering, and everyday life.
Remember, a cone is not just a delicious treat or a traffic safety device; it is a fascinating geometric shape with its own unique properties and characteristics.
Riya Sharma
Riya Sharma is a tеch bloggеr and UX/UI dеsignеr spеcializing in usеr еxpеriеncе dеsign and usability tеsting. With еxpеrtisе in usеr-cеntric dеsign principlеs, Riya has contributеd to crafting intuitivе and visually appеaling intеrfacеs.
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5
Q:
# Two boys and a girl can do a work in 5 days, while a boy and 2 girls can do it in 6 days. If the boy is paid at the rate of 28$a week, what should be the wages of the girl a week ? A) 24$ B) 22 $C) 16$ D) 14 $Answer: C) 16$
Explanation:
Let the 1 day work of a boy=b and a girl=g, then
2b + g = 1/5 ---(i) and
b + 2g = 1/6 ---(ii)
On solving (i) & (ii), b=7/90, g=2/45
As payment of work will be in proportion to capacity of work and a boy is paid $28/week, so a girl will be paid $28×245790$ = 16$.
Q:
Raghu can do a job in 12 days alone and Sam can do the same job in 15 days alone. A third person Aru whose efficiency is two-third of efficiency of both Ram and Shyam together, can do the same job in how many days alone?
A) 10 B) 12 C) 13 D) 15
Explanation:
One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20
Aru efficiency = 2/3 of (Raghu + Sam)
Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days.
3 332
Q:
12 men complete a work in 14 days. 5 days after they had started working, 3 men join them. How many more days will all of them take to complete the remaining work?
A) 5 B) 4 C) 3 D) 2
Explanation:
Let p be the required number of days.
From the given data,
12 x 14 = 12 x 5 + (12+3) x p
12 x 14 = 12[5 + 3p]
14 = 5 + 3p
3p = 9
p = 3 days.
Hence, more 3 days all of them take to complete the remaining work.
0 497
Q:
Lasya alone can do a work in 16 days. Srimukhi’s efficiency is 20 % lesser than that of Laya. If Rashmi and Srimukhi together can do the same work in 12 days, then find the efficiency ratio of Rashmi to that of Lasya?
A) 19 : 7 B) 30 : 19 C) 8 : 15 D) 31 : 17
Explanation:
Given Lasya can do a work in 16 days.
Now, time taken by Srimukhi alone to complete the work = 16 x 100/80
Time taken by Rashmi = n days
=> (12 x 20)/(20 - 12) = (12 x 20)/n
=> n= 30 days.
Required ratio of efficiencies of Rashmi and Lasya = 1/30 :: 1/16 = 8 : 15.
5 692
Q:
5 boys and 3 girls can together cultivate a 23 acre field in 4 days and 3 boys and 2 girls together can cultivate a 7 acre field in 2 days. How many girls will be needed together with 7 boys, if they cultivate 45 acres of field in 6 days?
A) 4 B) 3 C) 2 D) 1
Explanation:
Let workdone 1 boy in 1 day be b
and that of 1 girl be g
From the given data,
4(5b + 3g) = 23
20b + 12g = 23 .......(a)
2(3b + 2g) = 7
6b + 4g = 7 ........(b)
Solving (a) & (b), we get
b = 1, g = 1/4
Let number og girls required be 'p'
6(7 x 1 + p x 1/4) = 45
=> p = 2.
Hence, number of girls required = 2
5 607
Q:
70000 a year is how much an hour?
A) 80 B) 8 C) 0.8 D) 0.08
Explanation:
Given for year = 70000
=> 365 days = 70000
=> 365 x 24 hours = 70000
=> 1 hour = ?
70000/365x24 = 7.990 = 8
2 646
Q:
A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?
A) 1/5 B) 1/6 C) 1/7 D) 1/8
Explanation:
Total work is given by L.C.M of 72, 48, 36
Total work = 144 units
Efficieny of A = 144/72 = 2 units/day
Efficieny of B = 144/48 = 3 units/day
Efficieny of C = 144/36 = 4 units/day
According to the given data,
2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100
3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54
p = 198/16.5
p = 12 days.
Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day
(C+D) in p days = (4 + 6) x 12 = 120 unit
Remained part of work = (144-120)/144 = 1/6.
6 1820
Q:
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
A) 215 days B) 225 days C) 235 days D) 240 days
Explanation:
Given that
(10M + 15W) x 6 days = 1M x 100 days
=> 60M + 90W = 100M
=> 40M = 90W
=> 4M = 9W.
From the given data,
1M can do the work in 100 days
=> 4M can do the same work in 100/4= 25 days.
=> 9W can do the same work in 25 days.
=> 1W can do the same work in 25 x 9 = 225 days.
Hence, 1 woman can do the same work in 225 days.
9 2320
Q:
A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work?
Given A,B,C can complete a work in 15,20 and 30 respectively.
The total work is given by the LCM of 15, 20, 30 i.e, 60.
A's 1 day work = 60/15 = 4 units
B's 1 day work = 60/20 = 3 units
C's 1 day work = 60/30 = 2 units
(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units
Let B + C worked for x days = (3 + 2) x = 5x units
C worked for 2 days = 2 x 2 = 4 units
Then, 18 + 5x + 4 = 60
22 + 5x = 60
5x = 38
x = 7.6
Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days. |
# PHYS 170
Lecture 7 - Potential Energy
## Introduction
There are two types of mechanical energy, kinetic energy and potential energy. For now we will focus on gravitational potential energy: $$PE_g = mgh$$ The height $$h$$ is measure from an arbitrary reference level (say the ground). The higher an object is, the more gravitational potential energy it has. Together with kinetic energy $$KE=\frac{1}{2}mv^2$$ we can apply the law of conservation of energy to solve many problems. As with all other types of energy, its SI unit is $$J$$.
### Try It Yourself (click to show)
$$5m$$ above ground: $$PE = mgh = (10)(9.8)(5) = 490J$$ $$3m$$ below ground: $$PE = mgh = (10)(9.8)(-3) = -294J$$
## Conservation of Energy
When friction is absent, the total mechanical energy of an object is conserved, i.e. $$KE + PE = constant$$. A more useful way to restate this is to say:
$$KE_1 + PE_1 = KE_2 + PE_2$$
where point 1 and point 2 are any two points along the motion of an object.
A ball was initially at $$h_1$$ above ground and traveling downward at the speed of $$v_1$$. Find the speed of the ball when it is at height $$h_2$$ above ground.
We first calculate the initial energy. Note that the mass is not give, so we will call it $$m$$ for now. It will be seen below that the mass will drop out of the answer in the end. We begin by calculating the initial and final $$KE$$ and $$PE$$: $$\begin{eqnarray} KE_1 &=& \frac{1}{2} m v_1^2 \hspace{1cm} PE_1 &=& mgh_1 \\ KE_2 &=& \frac{1}{2} m v_2^2 \hspace{1cm} PE_2 &=& mgh_2 \end{eqnarray}$$ By conservation of energy: $$\begin{eqnarray} KE_1 + PE_1 &=& KE_2 + PE_2 \\ \Rightarrow \frac{1}{2} m v_1^2 + mgh_1 &=& \frac{1}{2} m v_2^2 + mgh_2\\ \Rightarrow \frac{1}{2} m v_2^2 &=& \frac{1}{2} m v_1^2 + mg(h_1-h_2)\\ \Rightarrow v_2 &=& \sqrt{v_1^2 + 2 g(h_1-h_2)} \end{eqnarray}$$
### Simulation - Sliding Down an Incline (click to hide)
Canvas not supported
Drag on the surface of the incline to change the angle.
The force diagram can also be dragged to a different location.
The length of the track is $$20m$$. The mass of the block is $$1kg$$.
### Try It Yourself (click to show)
$$\begin{eqnarray} KE_1 &=& \frac{1}{2} m v_1^2 \\ &=& \frac{1}{2} (2) (5^2) = 25J\\ PE_1 &=& mgh \\ &=& (2) (9.8) (10) = 196J \\ E_1 &=& KE_1 + PE_1 = 221 \end{eqnarray}$$
By conservation of energy: $$\begin{eqnarray} E_1 &=& E_2 \\ \Rightarrow KE_1 + PE_1 &=& KE_2 + PE_2 \\ \Rightarrow KE_2 + PE_2 &=& E_1\\ \Rightarrow PE_2 &=& E_1 - KE_2 = E_1 - \frac{1}{2} m v_2^2 \\ &=& 221 - \frac{1}{2} (2) (8)^2 \\ &=& 221 - 64 = 157J \\ \Rightarrow mgh_2 &=& 157 \\ h_2 &=& \frac{157}{(2)(9.8)} = 8.01m \end{eqnarray}$$
A ball of unknown mass was thrown upward from an intial height of $$3m$$ above ground with an initial speed of $$15m/s$$. Find its speed when it was at $$12m$$ above ground.
$$\begin{eqnarray} KE_1 + PE_1 &=& KE_2 + PE_2 \\ \Rightarrow \frac{1}{2} m v_1^2 + mgh_1 &=& \frac{1}{2} m v_2^2 + mgh_2\\ \Rightarrow \frac{1}{2} m v_2^2 &=& \frac{1}{2} m v_1^2 + mg(h_1-h_2)\\ \Rightarrow v_2 &=& \sqrt{v_1^2 + 2 g(h_1-h_2)} \\ &=& \sqrt{15^2 + 2 (9.8)(3 - 12)} = 6.97m/s \end{eqnarray}$$ Note that the second term in the square root gives a negative contribution, generating a lower final velocity.
From the figure, we have $$v_1 = 2m/s, h_1=20m, h_2=0m$$, and we are solving for $$v_2$$, so we apply conservation of energy: $$\begin{eqnarray} KE_1 + PE_1 &=& KE_2 + PE_2 \\ \Rightarrow \frac{1}{2} m v_1^2 + mgh_1 &=& \frac{1}{2} m v_2^2 + mgh_2\\ \Rightarrow \frac{1}{2} m v_2^2 &=& \frac{1}{2} m v_1^2 + mg(h_1-h_2)\\ \Rightarrow v_2 &=& \sqrt{v_1^2 + 2 g(h_1-h_2)} \\ &=& \sqrt{2^2 + 2 (9.8)(20 - 0)} = 19.90m/s \end{eqnarray}$$
We have $$v_1 = 4m/s, h_1=25, h_2=2R=20m$$, and we are solving for $$v_2$$, so we apply conservation of energy: $$\begin{eqnarray} KE_1 + PE_1 &=& KE_2 + PE_2 \\ \Rightarrow \frac{1}{2} m v_1^2 + mgh_1 &=& \frac{1}{2} m v_2^2 + mgh_2\\ \Rightarrow \frac{1}{2} m v_2^2 &=& \frac{1}{2} m v_1^2 + mg(h_1-h_2)\\ \Rightarrow v_2 &=& \sqrt{v_1^2 + 2 g(h_1-h_2)} \\ &=& \sqrt{4^2 + 2 (9.8)(25 - 20)} = 10.68m/s \end{eqnarray}$$
The potential energy of a pendulum is gravitational in nature, so it is still given by $$PE=mgh$$. The only extra step that is neccessary here is a mathematical one: how to write $$h$$ in terms of $$\theta$$?
The figure on the left shows the geometrical meaning of $$h$$ for the pendulum. From the triangle, one could work out the side adjacent to $$\theta$$ as $$L\cos\theta$$. Since the length of the pendulum does not change, we have: $$\begin{eqnarray} L&=& h + L\cos\theta \\ \Rightarrow h &=& L - L\cos\theta \\ &=& L(1 - \cos\theta) \\ \Rightarrow PE &=& mgh = mgL(1 - \cos\theta) \end{eqnarray}$$
### Simulation - Simple Pendulum (click to hide)
Canvas not supported
Drag on the ball to change the length.
Drag on the bar to change the angle.
Click on the clock to reset the timer.
The mass of the object is fixed to be $$m = 1kg$$.
The grey horizontal line represents the lowest level of the pendulum trajectory, used as a reference level for height measurement.
#### Activity
Use the clock to time 10 oscillations and deduce the period. Repeat for a different length and see how the period changes.
## Potential Energy in a Spring
There are other types of potential energy. For exmample, when a spring is stretched (or compressed), the energy you spent pulling it is stored inside the spring as potential energy. When the spring is released, the energy could be turned into other forms of energy, such as kinetic energy.
The potential energy in a spring in given by the equation: $$PE_{spring} = \frac{1}{2} k x^2$$ where $$k$$ [unit: $$N/m$$] is the spring constant, and $$x$$ [unit: $$m$$] is the extension of the spring. Note that the energy is not dependent on the sign of $$x$$.
A related equation, called Hooke's Law, which we will only use in later chapters, is the force from a spring: $$F_{spring} = - kx$$ The equation shows the force from a spring is proportional to the extension $$x$$. The negative sign in the fron means the force from the spring is always opposite in direction to the direction of $$x$$: when the spring is pulled to the right of the equilibrium position, the spring will pulled left; when $$x$$ is to the left of the equilibrium positionl, the spring will push to the right. The spring constant $$k$$ is the "stiffness" of the spring, the stiffer it is, the more force it can generate.
When a spring is in equilibrium (completely relaxed), its length is called the natural length. The extension $$x$$ is always measured from the equilibrium position as seen on the figure on the left.
### Simulation - Hooke's Law (click to hide)
Canvas not supported
Drag on the mass to change its position. Click "Play" to start the oscillation.
### Try It Yourself (click to show)
Because the object was at rest in the beginning, there was no kinetic energy. All potential energy is in the spring, no gravitational potential energy need to be considered because the object's height never changes. Note that $$x$$ was given in $$cm$$ so you will have to convert it into $$m$$ before putting it in the equation. $$\begin{eqnarray} KE_1 &=& 0J \\ PE_1 &=& \frac{1}{2}kx^2 \\ &=& \frac{1}{2}(8)(0.1^2) = 0.04J \\ E_1 &=& KE_1 + PE_1 = 0.04J \end{eqnarray}$$
Because the spring became completely relaxed in the end, there is no more potential energy in the end, so $$PE_2=0J$$. You can now use the total energy you calculated previously to find the velocity. $$\begin{eqnarray} KE_2 + PE_2 &=& KE_1 + PE_1 = E_1 \\ \Rightarrow \frac{1}{2}mv^2 + 0 &=& E_1 \\ \Rightarrow v &=& \sqrt{\frac{2E_1}{m}} \\ &=& \sqrt{\frac{2 (0.04)}{2}} = 0.2m/s \end{eqnarray}$$
## Notations
Name Symbol Unit Meaning
Energy $$E$$ $$J$$ a conserved quantity in nature
Kinetic energy $$KE$$ $$J$$ energy of motion
Potential energy $$PE$$ $$J$$ an invisible form of energy |
# What does it mean to be a parent function?
In mathematics, a mother or father operate is the most effective operate of a household of features that preserves the definition (or shape) of the whole family. For example, for the family of quadratic features having the general form. the most effective function is .
A parent function is the simplest function that also satisfies the definition of a undeniable form of function. For example, once we give some thought to the linear functions which make up a family of functions, the parent function would be y = x.
Additionally, what’s a mother or father graph? A parent graph is the graph of a quite simple function. Via remodeling the operate in several ways, the graph may well be translated, reflected, or or else changed.
Maintaining this in view, how do you write a parent function?
For example, the parent function for “y=x^+x+1” is simply “y=x^2,” often known as the quadratic function. Other parent functions include the simple varieties of the trigonometric, cubic, linear, absolute value, rectangular root, logarithmic and reciprocal functions.
What is the mother or father operate of a line?
A linear guardian operate is the equation y = x or f(x) = x. A parent operate is the most effective equation of a function. Thus, f(x) = x is the most effective of all linear functions and that is the reason why it is referred to as linear guardian function. You may make a table of values to graph this function.
### What are the qualities of a function?
A operate is a relation in which every possible input significance results in precisely one output value. We say “the output is a operate of the input.” The enter values make up the domain, and the output values make up the range.
### What are the 7 mother or father functions?
The following figures exhibit the graphs of guardian functions: linear, quadratic, cubic, absolute, reciprocal, exponential, logarithmic, square root, sine, cosine, tangent.
### How do features work?
A function is an equation that has only 1 solution for y for every x. A operate assigns exactly one output to every input of a precise type. It’s common to call a operate either f(x) or g(x) instead of y. f(2) means that we ought to discover the value of our operate whilst x equals 2.
### What are the eight guardian functions?
Graphs of eight ordinary guardian capabilities are proven below. Classify every function as constant, linear, absolute importance , quadratic , square root , cubic , rational, or exponential.
### What is the guardian operate of a constant?
Functions are often grouped into households according to the form of their defining formulas, or other commom characteristics. The Constant function. The graph of the fixed function f(x) =k is the graph of the equation y = k, which is the horizontal line. If we range okay then we achieve a household of horizontal lines.
### What are the qualities of some of the easy mother or father functions?
What are some features of the fundamental guardian functions? (Linear… Odd. End habit pass in several directions. If a operate is positive, the left part of the graph will point down and the right aspect will factor up (increasing from left to right). Immediately line. Constant. Has a slope.
### Can a parent operate be negative?
The absolute-value parent graph of the operate y = |x| turns all inputs non-negative (0 or positive). To graph absolute-value functions, you start at the beginning and then every positive quantity receives mapped to itself, while each adverse variety gets mapped to its effective counterpart.
### What are the different types of parent functions?
Types of Functions Linear. Quadratic. Absolute value. Exponential growth. Exponential decay. Trigonometric (sine, cosine, tangent) Rational. Exponential.
### How do you find Asymptotes?
The vertical asymptotes will occur at those values of x for which the denominator is the same as zero: x − 1=0 x = 1 Thus, the graph can have a vertical asymptote at x = 1. To find the horizontal asymptote, we word that the degree of the numerator is two and the measure of the denominator is one.
### What is a function family?
A family of capabilities is a collection of capabilities whose equations have a similar form. The “parent” of the household is the equation in the household with the best form. For example, y = x2 is a parent to different functions, such as y = 2×2 – 5x + 3.
### How do you graph a function?
Consider the operate f(x) = 2 x + 1. We realize the equation y = 2 x + 1 because the Slope-Intercept sort of the equation of a line with slope 2 and y-intercept (0,1). Think about a point moving at the graph of f. As the point moves in the direction of the right it rises. |
# Material Taken From: Mathematics for the international student Mathematical Studies SL Mal Coad, Glen Whiffen, John Owen, Robert Haese, Sandra Haese and.
## Presentation on theme: "Material Taken From: Mathematics for the international student Mathematical Studies SL Mal Coad, Glen Whiffen, John Owen, Robert Haese, Sandra Haese and."— Presentation transcript:
Material Taken From: Mathematics for the international student Mathematical Studies SL Mal Coad, Glen Whiffen, John Owen, Robert Haese, Sandra Haese and Mark Bruce Haese and Haese Publications, 2004
Objectives Describe what the first derivative yields. Find the first derivative of functions. Write equations of the line tangent to a curve.
Introductory Videos Brainpop.com o Calculus Khan Academy o Calculus (Derivatives 1) Section 19A-F – Introduction to Calculus
Consider point A (a, f(a)) and point B (x, f(x)) slope of the tangent line at A
Main Ideas: The derivative of a curve at a point is the slope of the tangent line at that point. The derivative function is the function that represents the slopes of all the tangents throughout the entire curve.
Graphing Example 1
Graphing Example 2 What is the function or equation for this graph? What is the derived function of this graph? What would that function look like? What would that function represent?
name of original functionname of derivative function f(x)f(x)f’(x) yy' y Notation and Terminology Differentiation is the process of finding the derivative or the derivative function (the slope function). Notation
FunctionDerivative Function f(x) = x f’(x) = f(x) = x 2 f’(x) = f(x) = x 3 f’(x) = f(x) = x 4 f’(x) = f(x) = x 5 f’(x) = f(x) = x -1 f’(x) = f(x) = x -2 f’(x) = f(x) = x -3 f’(x) = f(x) = f’(x) = f(x) = f’(x) =
Rules (Page 615) f(x)f’(x)in words: a0 The derivative of a constant is zero. xnxn nx n-1 Bring down the power, subtract one from the power. ax n anx n-1 Multiply by the coefficient (same as above). u(x) + v(x)u’(x) + v’(x) The derivative of the sum is the sum of the derivatives.
Find the first and second derivative of f(x) = 3x 4 + 2x 3 – 5x 2 + 7x + 6 Example 1
Find the first and second derivative of f(x) = 5x 3 + 6x 2 – 3x+ 2 Example 2
Find the derivative of Hint: First rewrite the function, then take the derivative. Example 3
Find the slope function of Hint: First rewrite the function, then take the derivative. Example 4
a) Find the gradient function of and then find: b) gradient of the tangent to the function where x = 2. c) equation of the tangent when x = 2. Example 5
a) Find the gradient function of and then find: The gradient function is the first derivative. Now find the gradient when x = 2. Therefore, m = 5 Finally, find the equation of the line. You need a point. We already have the slope. So, the point is (2, 2) and the slope is 5. Therefore, the equation of the tangent at x = 2 is y = 5x – 8. b) gradient of the tangent to the function where x = 2. c) equation of the tangent when x = 2. Example 5 Answer
Find the equation of the tangent to f(x) = x 2 + 1 at the point where x = 1 Example 6
Find the equations of any horizontal tangents to y = x 3 – 12x + 2 Example 7
Summary The rate of change at a point is the slope of the tangent line at that point. The slope of the tangent line at that point is known as the derivative. Therefore: – rate of change = slope of tangent line = derivative
Homework 19E, #1a-l, 2a-f 19F #1a-d, 2abc, 3abc Kahn Academy Video – Calculus: Derivatives 3
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Browse Questions
# The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation If it is replaced by 12.
$\begin{array}{1 1}(A)\;10.2,1.98\\(B)\;8.4,5.6\\(C)\;3,8\\(D)\;5,7\end{array}$
If it is replaced by 12,
Given mean $\bar{x}=10$
Standard deviation $(\sigma) =2$
Number of observations =20
Step 1:
$\therefore \large\frac{\sum x_i}{20} $$=10 \sum x_i=200 If observation 8 is replayed by 12 \sum x = 200-8+12 \qquad= 192+12=204 \therefore \sum x =204 Correct mean =\large\frac{204}{20} \qquad= 10.2 Step 2: Given standard deviation \sigma=2 Variance \sigma ^2=4 \large\frac{\sum x_i^2}{n} -\bigg(\large\frac{\sum x_i}{n}\bigg)^2$$=4$
$\therefore \large\frac{\sum x_i^2}{20} $$=4+100 => \sum x_i^2 = 104 \times 20 \qquad= 2080 If the observation 8 is replaced by 12,then \sum x_i^2 =2080 -8^2+12^2 \qquad = 2080 -64+144 \qquad= 2224-64 \qquad= 2160 Step 3: Correct standard deviation \sigma= \sqrt {\large\frac{\sum x_i^2}{n} -\bigg(\large\frac{\sum x_i}{n}\bigg)^2} \qquad= \sqrt { \large\frac{2160}{20} - \bigg( \large\frac{204}{20} \bigg)^2} \qquad= \sqrt {\large\frac{2160 \times 20 -(204)^2}{20 \times 20}} \qquad= \large\frac{1}{20}$$\sqrt{43200-41616}$
$\qquad= \large\frac{1}{20}$$\sqrt { 1584}$
$\qquad= 1.98$
Hence A is the correct answer. |
## Integration by Parts
Integration by parts is another technique for simplifying integrands. As we saw in previous posts, each differentiation rule has a corresponding integration rule. In the case of integration by parts, the corresponding differentiation rule is the Product Rule. The technique of integration by parts allows us to simplify integrands of the form: $$\int f(x) g(x) dx$$
Examples of this form include: $$\int x \cos{x} \space dx, \qquad \int e^x \cos{x} \space dx, \qquad \int x^2 e^x \space dx$$
As integration by parts is the product rule applied to integrals, it helps to state the Product Rule again. The Product Rule is defined as: $$\frac{d}{dx} \big[ f(x)g(x) \big] = f^{\prime}(x) g(x) + f(x) g^{\prime}(x)$$
When we apply the product rule to indefinite integrals, we can restate the rule as:
$$\int \frac{d}{dx} \big[f(x)g(x)\big] \space dx = \int \big[f^{\prime} g(x) + f(x) g^{\prime}(x) \big] \space dx$$
Then, rearranging so we get $f(x)g^{\prime}(x) \space dx$ on the left side of the equation:
$$\int f(x)g^{\prime}(x) \space dx = \int \frac{d}{dx} \big[f(x)g(x)\big] \space dx - \int f^{\prime}(x)g(x) \space dx$$
Which gives us the integration by parts formula! The formula is typically written in differential form:
$$\int u \space dv = uv - \int v \space du$$
## Examples¶
The following examples walkthrough several problems that can be solved using integration by parts. We also employ the wonderful SymPy package for symbolic computation to confirm our answers. To use SymPy later to verify our answers, we load the modules we will require and initialize several variables for use with the SymPy library.
In [1]:
from sympy import symbols, limit, diff, sin, cos, log, tan, sqrt, init_printing, plot, integrate
from mpmath import ln, e, pi, cosh, sinh
init_printing()
x = symbols('x')
y = symbols('y')
Example 1: Evaluate the integrand $\int x \sin{\frac{x}{2}} \space dx$
Recalling the differential form of the integration by parts formula, $\int u \space dv = uv - \int v \space du$, we set $u = x$ and $dv = \sin{\frac{x}{2}}$
Solving for the derivative of $u$, we arrive at $du = 1 \space dx = dx$. Next, we find the antiderivative of $dv$. To find this antiderivative, we employ the Substitution Rule.
$$u = \frac{1}{2}x, \qquad du = {1}{2} \space dx, \qquad \frac{du}{dx} = 2$$$$y = \sin{u}, \qquad dy = -\cos{u} \space du, \qquad \frac{dy}{du} = -\cos{u}$$
Therefore, $v = -2 \cos{\frac{x}{2}}$
Entering these into the integration by parts formula:
$$-2x\cos{\frac{x}{2}} - (-2)\int \cos{\frac{x}{2}}$$
Then, solving for the integrand $\int \cos{\frac{x}{2}}$, we employ the Substitution Rule again as before to arrive at $2\sin{\frac{x}{2}}$ (the steps in solving this integrand are the same as before when we solved for $\int \sin{\frac{x}{2}}$). Thus, the integral is evaluated as:
$$-2x\cos{\frac{x}{2}} + 4\sin{\frac{x}{2}} + C$$
Using SymPy's integrate, we can verify our answer is correct (SymPy does not include the constant of integration $C$).
In [2]:
integrate(x * sin(x / 2), x)
Out[2]:
$$- 2 x \cos{\left (\frac{x}{2} \right )} + 4 \sin{\left (\frac{x}{2} \right )}$$
Example 2: Evaluate $\int t^2 \cos{t} \space dt$
We start by setting $u = t^2$ and $dv = \cos{t}$. The derivative of $t^2$ is $2t$, thus $du = 2t \space dt$, or $\frac{du}{dt} = 2t$. Integrating $dv = \cos{t}$ gives us $v = \sin{t} \space du$. Entering these into the integration by parts formula:
$$t^2 \sin{t} - 2\int t \sin{t}$$
Therefore, we must do another round of integration by parts to solve $\int t \sin{t}$.
$$u = t, \qquad du = dt$$
$$dv = \sin{t}, \qquad v = -\cos{t} \space du$$
Putting these together into the integration by parts formula with the above:
$$t^2 \sin{t} - 2 \big(-t \cos{t} + \int \cos{t} \space dt \big)$$
Which gives us the solution:
$$t^2 \sin{t} + 2t \cos{t} - 2 \sin{t} + C$$
As before, we can verify that our answer is correct by leveraging SymPy.
In [6]:
t = symbols('t')
integrate(t ** 2 * cos(t), t)
Out[6]:
$$t^{2} \sin{\left (t \right )} + 2 t \cos{\left (t \right )} - 2 \sin{\left (t \right )}$$
Example 3: $\int x e^x \space dx$
Here, we set $u = x$ and $dv = e^x$. Therefore, $du = dx$ and $v = e^x \space dx$. Putting these together in the integration by parts formula:
$$xe^x - \int e^x$$
As the integral of $e^x$ is just $e^x$, our answer is:
$$xe^x - e^x + C$$
We can again verify our answer is accurate using SymPy.
In [7]:
integrate(x * e ** x, x)
Out[7]:
$$2.71828182845905^{x} \left(1.0 x - 1.0\right)$$ |
## Hexagon
Content Objective:
Students will learn the method of finding the perimeter of regular hexagons.
Students will discover two different methods for finding the area of regular hexagons as related to regular triangles, and the apothem.
Materials:
Rulers for pairs, hexagon perimeter worksheet, hexagon area worksheet
Procedure:
### Perimeter
1. Again, collectively recall the definition of perimeter given on review day and for regular triangles and squares. Have students get into pairs and distribute the hexagon perimeter worksheet. Once worksheet is complete and students have compared their answers, ask the class as a whole the following questions:
What do you notice about the side lengths of our hexagon? Is this what you expected?
What happens to the perimeter when the side lengths are changed?
Is there a way to find the perimeter of a regular hexagon without measuring all six sides?
Recalling the general expressions for the perimeter of regular triangles and squares. Do you notice a pattern? What can we expect when we discover the perimeter of a hexagon?
2. Demonstrate using Geometer's Sketchpad the properties of the side lengths of regular hexagon. Click here to use the GSP file and watch the animation for the changing perimeter.
### Area
We will show two ways of finding the area of a regular hexagon. The first method uses the students' previous knowledge of equilateral triangles.
METHOD 1:
Begin by showing the figures below and ask the following questions:
First, how can we break this figure into familiar shapes?
What special property do these six triangles have?
Given the side of our hexagon is length s, what is the side length of each of the triangles?
Recall the method for finding the area of a regular triangle. What is the area of ONE of the triangles in our hexagon?
What is a quick way to find the area for the entire hexagon?
So we have discovered a general formula for the area, using the smaller triangles inside the hexagon!
Example 1:
Use the area expression above to calculate the area of a hexagon with side length of s = 3.00cm and a height of h = 2.60cm for comparison with method 2 later.
METHOD 2:
Recall the formula for perimeter of our regular hexagon. How can we simplify the expression we found for area?
We will call the perimeter p. So now we have,
Now, we can look at the general method for finding the area of a regular polygon. Define apothem for students to begin the second method for finding the area of a regular hexagon.
Apothem - the distance of the line segment from the center of a regular polygon perpendicular to a side (i.e. when a regular polygon is broken into triangles, the apothem is the height of one of the triangles whose base is a side of the regular polygon).
Discuss the concept of apothem with students and ask the following questions:
What does the apothem of a regular hexagon look like?
How is the apothem related to the height we found in the regular hexagons above?
How many apothems does a regular hexagon have? Are the apothems all the same length?
How can we rewrite our formula for area, using the apothem a for the regular hexagon instead of the height h of the regular triangle?
Example 2:
Now we can fill in the values for p and a to find the area of this regular hexagon. Again, let s = 3.00cm and let a = 2.60cm. What is the area?
COMPARING METHODS:
Discuss with students the two different methods for finding the area of a regular hexagon.
As a class, discuss the advantages and disadvantages for each of the methods and ask the following questions:
Which method is easier to formulate?
Will these methods ever produce different answers?
Finally, distribute the hexagon area worksheet and have students complete in groups, using either method. Then, as a class, compare answers and discuss the methods for finding the solutions.
Demonstrate using Geometer's Sketchpad the properties of the side lengths and apothem lengths of regular hexagons. Click here to use the GSP file and watch the animation for the changing area. |
# Lesson video
In progress...
Hi, everyone.
It's me again, Ms. Jones.
And today we are looking at Inequalities.
Inequalities, that sounds similar to what we've looked at so far, maybe it's linked with equations, equality.
What do you think inequality means? And where do you think it comes up? That's what we'll be looking at today.
But before we do, please make sure you have a pen and some paper.
You have a nice quiet space to work, if you can find one and you remove all distractions away from you.
Pause now to make sure all of that is ready so that we can make start.
Okay, the first thing I would like you to do is to tell me, which of the cards below could you place in the inequality? So inequality is essentially just not equal, isn't it? So here it says zero is.
And this means less than something subtract something.
And that's what you need to fit in using these cards here.
So zero is less than something subtract something.
Which ones, which of those cards, and which of those integers are going to work in this inequality here? Pause the video to have a go at that.
If for example, there were loads you could have had, you could have had just starting with 12, we could have had all of these 12 subtract one, 12 subtract negative three.
All of those would be greater than zero, and there are loads of others you could have got.
So really well done for getting lots of those.
We can use greater than, and less than, to express an inequality.
You can draw bar models to demonstrate an inequality involving positive numbers.
Here, this model is showing us that five is less n, 'cause we can see is smaller than, and we don't know how much more than, but we know it's smaller than some where somehow that end, which inequalities can you see in this image here? So we see in this image, the equations, which inequalities can you see? However, Xavier can see that P, which remember is that purple one is greater than three w's.
And if you almost draw that little line up there, you can see that those three w's are less than p.
Have a go at creating some other inequalities, pause the video to do that.
But then there were loads that you could have got.
So for example, we've got p is greater than g.
Which we've demonstrated with this bar model here.
You've got g is greater than w and you've got g, green add 2w is less than 3r.
If you want something a little bit more complex, and there's loads you could have got, you could have written an endless string of algebraic terms. That's brilliant, well done for those.
For your independent task, the first question asks you to just get used to the using these different symbols.
And you can see, we have got an equal sign as well as are greater than, or less than.
For question number two, you needed to write an algebraic inequality from the description.
So n is greater than negative three, again, we're just practising using our inequality symbols.
Then we were asked to take the inequalities that are true when a equals four.
So we need to substitute in, a equals four here.
Remember 2a would not be 24.
It would be two lots of four, which is eight.
So actually this symbol should have been an equal sign to improve it.
So that one was not in accord.
That was not true, when an a equals four.
Really well done if you've got all of those correct, or even some of those correct, great job.
Finally for our explore task, we have three facts up here, a equals four, b equals two and c equals nine.
Given that those are the values of a, b and c, which of the following inequalities are true? Is it true? That 2a is greater than c.
The ac is less than 17b and I'll let you read the rest of them.
So I don't give away what the symbols mean.
Create your own examples of true inequalities.
Pause the video here to have a go at that.
.
So, actually none of those were true, really well done if you've got that.
two lots of a be eight, which has in fact less than c, which was nine.
Ac means, remember a multiplied by c, so four lots of night, it was greater than 17 lots of b or two.
b subtract c got us negative two, which is not greater than three lots of four.
a squared got us a 16 and b to the power four also got us to 16, so those two were equal.
Remember b to the power of four does not mean be multiplied by four, it means b multiply by b, multiply by b, multiply by b.
Absolutely loads of examples of true inequalities, here are just some I created, a squared is less than c squared, negative b is greater than negative c.
So extra points and extra well done done, if you use some negative or even some decimals and fractions, that really good job.
Hopefully you now have a really good understanding of what inequalities are and how we can use them, especially with algebraic terms. Remember to complete your quiz at the end of this lesson and an amazing job, well done. |
# 2020 AMC 10A Problems/Problem 10
The following problem is from both the 2020 AMC 12A #7 and 2020 AMC 10A #10, so both problems redirect to this page.
## Problem
Seven cubes, whose volumes are $1$, $8$, $27$, $64$, $125$, $216$, and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$
## Solution 1
The volume of each cube follows the pattern of $n^3$ ascending, for $n$ is between $1$ and $7$.
We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\times 7\times 7$ cube (which is just $7 \times 7 = 49$). The sides areas can be measured as the sum $4\sum_{n=0}^{7} n^2$, giving us $560$. Structurally, if we examine the tower from the top, we see that it really just forms a $7\times 7$ square of area $49$. Therefore, we can say that the total surface area is $560 + 49 + 49 = \boxed{\textbf{(B) }658}$. Alternatively, for the area of the tops, we could have found the sum $\sum_{n=0}^{6}((n+1)^{2}-n^{2})$, giving us $49$ as well.
~ciceronii
## Solution 2
It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.
First, we will calculate the total surface area of the cubes, ignoring overlap. This value is $6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840$. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $2\sum_{n=1}^{6} n^2 = 182$. Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = \boxed{\textbf{(B) }658}$. ~emerald_block
## Solution 3 (a bit more tedious than other solutions)
It can be seen that the side lengths of the cubes using cube roots are all integers from $1$ to $7$, inclusive.
Only the cubes with side length $1$ and $7$ have $5$ faces in the surface area and the rest have $3$. Also, since the
cubes are stacked, we have to find the difference between each $n^2$ and $(n-1)^2$ side length as $n$ ranges from $7$ to
$2$.
We then come up with this: $5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)$.
We then add all of this and get $\boxed{\textbf{(B) }658}$.
~aryam
~IceMatrix
## Solution 5
Notice that the surface area of the top cube is $6s^2$ and the others are $4s^2$. Then we can directly compute. The edge length for the first cube is $7$ and has a surface area of $294$. The surface area of the next cube is $144$. The surface area of the next cube $100$. The surface area of the next cube is $64$. The surface area of the next cube is $36$. The surface area of the next cube is $16$. The surface area of the next cube is $4$. The surface area of the next cube is $4$. We then sum up $294+144+100+64+36+16+4$ to get $658$. -smartatmath
## See Also
2020 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 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 AMC 10 Problems and Solutions
2020 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 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 AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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The Pentium division Flaw - Chapter 1
Introduction Chapter 2
CHAPTER 1: REVIEW
Here we briefly review some simple concepts you probably already know.
1.1 INTERPRETING THE VALUE OF A DECIMAL NUMBER:
(100) ones - - + + - - tenths (10-1) (101) tens - + | | + - hundredths (10-2) (102) hundreds + | | | | + thousandths (10-3) | | | | | | 4 1 9 . 5 8 3 ^ -decimal point
1.2 INTERPRETING THE VALUE OF A BINARY NUMBER:
(20) ones - - - + + - - - halves (1/2) (21) twos - - + | | + - - fourths (1/4) (22) fours - + | | | | + - eighths (1/8) (24) eights + | | | | | | + sixteenths (1/16) | | | | | | | | 1 0 1 1 . 1 0 1 1 ^ -radix point
1.3 MOVING THE DECIMAL POINT OF A BASE 10 NUMBER:
Given an ordinary decimal number such as 14.195835 we can move the decimal point to the right or left by multiplying or dividing by powers of ten. For example, to move the decimal point right two places, we multiply by 102 = 100 14.195835 * 100 = 1419.5835
1.4 MOVING THE RADIX POINT OF A BINARY NUMBER:
Given an ordinary binary number such as 1011.1011 we can move the radix point to the right or left by multiplying or dividing by powers of two. For example, to move the decimal point right two places, we multiply by 22 = 4 1011.1011 * 4 = 101110.11
1.5 NORMALIZED NOTATION: BASE 10
Given an ordinary decimal number such as 14.195835 we can move the decimal point so it lies directly after the first digit by multiplying or dividing by an appropriate power of 10. In our example:
14.195835 = 1.4195835 * 101 mantissa exponent
In this representation 1.4195835 is called the mantissa, and 101 is called the exponent.
1.6 NORMALIZED NOTATION: BASE 2
Given an ordinary binary number such as 1011.1011 we can move the radix point so it lies directly after the first digit by multiplying or dividing by an appropriate power of 2. In our example:
1011.1011 = 1.0111011 * 23 mantissa exponent
In this representation 1.0111011 is called the mantissa, and 23 is called the exponent.
1.7 DIVISION NOMENCLATURE
``` dividend numerator 12
quotient = -------- = ----------- Example: 3 = --
divisor denominator 4
quotient 3
+--------- Example: +---
divisor | dividend 4 | 12
```
1.8 DIVISION OF NORMALIZED NUMBERS: BASE 10
Given any two decimal numbers that we wish to divide, such as:
``` 14.195835
----------
119.716320
```
we can normalize the values using exponential notation to obtain:
1.4195835 * 101 1.4195835 101 1.4195835
----------------- = ---------- * ---- = ---------- * 10(1-2)
1.19716320 * 102 1.19716320 102 1.19716320
Thus the problem has been reduced to the division of two normalized numbers followed by a shifting of the decimal point.
1.9 DIVISION OF NORMALIZED NUMBERS: BASE 2
Given any two binary numbers that we wish to divide, such as:
``` 1011.1011
---------
11.001000
```
we can normalize the values using exponential notation to obtain:
``` 1.0111011 * 2^3 1.0111011 2^3 1.0111011
--------------- = --------- * --- = --------- * 2(3-1)
1.1001000 * 2^1 1.1001000 2^1 1.1001000
```
Thus the problem has been reduced to the division of two normalized numbers followed by a shifting of the radix point. The important point to note is that any division operation can be reduced to dividing normalized mantissas. The exponents and signs can be handled separately.
1.10 PENTIUM REPRESENTATION OF FLOATING POINT NUMBERS
The Pentium uses IEEE standard 594 for representing floating point numbers. In this standard a single precision floating point variable is represented using a 24 bit mantissa and an exponent which can take on values between +127 and -126.
The Pentium also uses two's compliment notation to represent negative numbers. A positive number is negated by complementing all the bits (one's complement, i.e. "flip the bits") and then adding 1 to the result.
Carry-Save addition is a method of quickly reducing the sum of three variables A, B, and C, down to the sum of two variables PARTIAL_SUM and CARRY, using only bitwise logical operations (AND, OR, Exclusive-OR). With Carry-Save addition we operate on all of the columns at once, placing the partial sum of that column at the bottom, and saving any overflow as a carry digit at the top. The bottom is our PARTIAL_SUM, and the carry digits at the top become our CARRY. The true sum of (A + B + C) is thus reduced to the sum (PARTIAL_SUM + CARRY).
EXAMPLE #1:
``` CARRY= 0010.1100010
------------
A=101.11010011
B=001.01100110 CARRY= 0010.1100010
C=000.01100000 PARTIAL_SUM= 100.11010101
------------ -------------
PARTIAL_SUM= 100.11010101 TOTAL= 0111.10011001
```
Note that:
``` PARTIAL_SUM = (A ^ B ^ C)
CARRY = (A & B) | (A & C) | (B & C)
^ represents Exclusive-OR
& represents AND
| represents OR
```
Thus both PARTIAL_SUM and CARRY are calculated quickly using nothing but bitwise logical operations.
The one drawback to Carry-Save addition is if we want to know the true sum of (A + B + C) we still have to add (PARTIAL_SUM + CARRY), and this can take time. The classic method of adding two numbers is to start with the right most column and add it up, placing the partial sum for that column below and any overflow as carry digits above the next column to the left. Then move left one column and repeat. Continue in this manner, adding one column at a time, until all the columns have been added and the calculation is complete. This method of addition is slow since each number may have 64 bits, and we must loop through all of the columns, working on only one column at a time, propagating the overflow on to the next column as carry bits.
However, it may be that we don't need an exact answer but that an approximate answer will do fine. We can get an approximate answer by adding just the first few columns and ignoring the rest. For Example, we can calculate an approximate answer to example #1 above by considering just the first 7 columns:
``` CARRY= 0010.110
PARTIAL_SUM= 100.110
-------------
approx total= 0111.100
```
Here our total is approximate, but it is very close to the actual answer. In this case the total dropped down from the true answer to the first integral multiple of 0.001 below the true answer.
Consider now example #2 below which shows a slightly different PARTIAL_SUM and CARRY. Notice the actual sum is identical to the sum we had before, but the approximate total obtained by adding the first seven bits only is lower this time:
EXAMPLE #2:
``` CARRY= 0010.1011101 CARRY= 0010.101
PARTIAL_SUM= 100.11011111 PARTIAL_SUM= 100.110
------------- -------------
actual total= 0111.10011001 approx total= 0111.011
```
This time our approximate total dropped down to the second integral multiple of 0.001 rather than the first integral multiple as it did before. The difference is in example #1 the sum of the truncated parts is less than 0.001, whereas in example #2 the sum of the truncated parts is greater than 0.001:
``` Example #1: Example #2:
|truncated |truncated
|part |part
| |
CARRY= . 0010 CARRY= . 1101
PARTIAL_SUM= . 10101 PARTIAL_SUM= . 01111
------------- -------------
TOTAL= .___11001 TOTAL= .__101001
```
Mathematically our approximate total is equal to the true total minus the sum of the truncated parts:
``` approx total = total - (sum of truncated parts)
```
Consider now a worst case scenario in which all the truncated bits are 1's. Convince yourself that even in this case the sum of the truncated parts will always be less than (2 x 0.001). Consider now the other extreme scenario where all the truncated bits are 0's. Convince yourself that in this trivial case the sum of the truncated parts is 0.
We have thus determined the bounds on our approximate total:
``` total >= approx total > total - (2 x 0.001)
```
Our approximate total will always be equal to or lower than the true answer, and the error will always be less than (2 x 0.001). This is a crucial result so make sure you understand it.
Introduction Chapter 2 |
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Applications with Inequalities
Story problems, set notation, interval notation
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Practice Applications with Inequalities
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Write and Solve Multi-Step Inequalities Given Problem Situations
The marching band at Floyd Middle School always performs during half-time of the football games. The band usually performs different routines that are an average of 6 minutes for each arrangement. Mrs. Kline likes the students to have a variety of different pieces to perform.
During the game, they always perform for at least 42 minutes but not more than 60 minutes. If the average time for each piece is 8 minutes, what is the least number of pieces that the band can perform? What is the greatest number of pieces that the band will perform?
This problem has a compound inequality in it. You will notice that there are two different inequalities being mentioned. This Concept will teach you how to translate verbal language into a compound inequality so that you can solve the inequality. Pay close attention, because you will see this problem again at the end of the Concept.
Guidance
Sometimes, however, to truly understand the values represented by a variable, we need consider two inequalities together .
Two inequalities considered together form a compound inequality.
Think about it this way, suppose we know that $n>0$ and we also know that $n<5$ . In that case, we need to consider those two inequalities together. That means, we need to consider a compound inequality.
There are two types of compound inequalities you should know about.
• A conjunction is a compound inequality that contains the word and. A conjunction is true only if both inequalities are true.
$n>0$ and $n < 5$
• A disjunction is a compound inequality that contains the word or. A disjunction is true if either of the inequalities is true.
We can write compound inequalities when we have words that describe more than one inequality. Let’s look at one.
The sum of a number, $n$ , and 4 is at least 12 and at most 20.
a. Translate the sentence above into a compound inequality.
b. Solve the compound inequality.
Consider part a first.
Break the sentence into parts and translate each part into an inequality.
$& The \ \underline{sum \ of \ a \ number, \ n, \ and \ 4} \ is \ \underline{at \ least} \ \underline{12} \ldots\\& \qquad \qquad \qquad \quad \ \downarrow \qquad \qquad \qquad \qquad \ \downarrow \quad \ \ \downarrow\\& \qquad \qquad \quad \ \quad \ n+4 \qquad \qquad \qquad \quad \ \ge \ \ \ 12$
$& The \ \underline{sum \ of \ a \ number, \ n, \ and \ 4} \ is \ldots \underline{at \ most} \ \underline{20}.\\& \qquad \qquad \qquad \quad \ \downarrow \qquad \qquad \qquad \qquad \quad \ \downarrow \qquad \downarrow\\& \qquad \qquad \qquad \ \ n+4 \qquad \qquad \qquad \quad \quad \ \le \quad \ 20$
Look at the original sentence.
The sum of a number, $n$ , and 4 is at least 12 and at most 20.
The word and which is underlined above shows that this sentence represents a conjunction.
So, the compound inequality could be written as: $n+4 \ge 12$ and $n+4 \le 20$ .
Or, we could rewrite $n+4 \ge 12$ as $12 \le n+4$ and put the two inequalities together like this: $12 \le n+4 \le 20$ .
Next, consider part b .
To solve the compound inequality, treat each inequality separately.
Subtract 4 from each side.
$n+4 & \ge 12 \qquad \qquad \quad \ \ n+4 \le 20\\n+4-4 & \ge 12-4 \qquad \ n+4-4 \le 20-4\\n+0 & \ge 8 \qquad \quad \quad \qquad n+0 \le 16\\n & \ge 8 \qquad \quad \ \quad \qquad \quad \ n \le 16$
So, the solution could be written as: $n \ge 8$ and $n \le 16$ .
It could also be written as: $8 \le n \le 16$ .
Here is another real-world situation.
Brandon earns $7 per hour at his job. The number of hours he works varies from week to week. However, each week he always earns no less than$70 and no more than $140. Let $h$ represent the number of hours he works each week. a. Write a compound inequality to represent this problem situation. b. Solve the inequality to determine the range in the number of hours Brandon works each week. c. According to the problem, does Brandon ever work 25 hours in any given week? Explain. Consider part a first. Since Brandon earns$7 per hour, you can represent the number of dollars he earns each week by multiplying 7 by the number of hours he works. In other words, you can represent his total earnings as $7 \times h$ or $7h$ .
Now, break the problem into parts and translate each part into an inequality.
$& \ldots \underline{each \ week \ldots earns} \ \underline{no \ less \ than} \ \underline{\ 70} \ldots\\& \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \downarrow \qquad \quad \downarrow\\& \qquad \qquad \quad 7h \qquad \qquad \quad \ \ \ge \qquad \ \ 70$
$& \ldots \underline{each \ week \ldots earns} \ \underline{no \ more \ than} \ \underline{\ 140}.\\& \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \qquad \quad \ \ \downarrow\\& \qquad \qquad \quad 7h \qquad \qquad \qquad \le \qquad \quad 140$
Look at the original sentence.
...each week he always earns no less than $70 and no more than$140.
Because of the word and , this compound inequality is an example of a conjunction.
We could represent this conjunction as: $7h \ge 70$ and $7h \le 140$ . We could also write this conjunction as: $70 \le 7h \le 140$ .
Next, consider part b .
To solve the compound inequality, treat each inequality separately.
Divide each side by 7.
$7h & \ge 70 \qquad \quad \ 7h \le 140\\\frac{7h}{7} & \ge \frac{70}{7} \qquad \quad \frac{7h}{7} \le \frac{140}{7}\\1h & \ge 10 \qquad \quad \ 1h \le 20\\h & \ge 10 \qquad \quad \ \ h \le 20$
So, the solution could be written as: $h \ge 10$ and $h \le 20$ . It could also be written as: $10 \le h \le 20$ .
This solution shows that the number of hours Brandon works each week is greater than or equal to 10 and less than or equal to 20. In other words, Brandon works 10 to 20 hours in any given week.
Lastly, consider part c .
In part b , you determined that the number of hours Brandon works in any given week is always less than or equal to 20. This means he never works more than 20 hours in any given week. So, he would not work 25 hours during a particular week.
Example A
The sum of a number and 3 is at least 10 but not more than 25.
Solution: $10\le x+3 \le25$
Example B
The sum of a number and six is greater than four and less than 12.
Solution: $4
Example C
The product of a number times two is greater than fifteen and less than twenty.
Solution: $15<2x<20$
Now let's go back to the dilemma from the beginning of the Concept.
First, let’s write down the given information.
The average length of each piece is 6 minutes.
The band performs for no less than 42 minutes.
The band performs for no more than 60 minutes.
We need to figure out the range of the number of pieces performed. This is our variable $p$ .
Here is the inequality.
$42 \le 6p \ge 60$
Now we can solve each inequality for the range. We will have a low range and a high range for our number of pieces.
$42 & \le 6p\\7 &\le p\\6p &\ge 60\\p &\ge 10$
The band will perform between 7 and 10 pieces given the range of times that they are allowed to perform.
Vocabulary
Compound Inequality
an inequality that combines two inequalities when considering a problem.
Guided Practice
Here is one for you to try on your own.
Six less than half of a number, $x$ , is either less than -1 or greater than 10.
a. Translate the sentence above into a compound inequality.
b. Solve the compound inequality.
Solution
Consider part a first.
Break the sentence into parts and translate each part into an inequality.
$& \underline{Six} \ \underline{less \ than} \ \underline{half \ of \ a \ number, \ x,} \ is \ldots \underline{less \ than} \ \underline{-1} \ldots \\& \searrow \qquad \downarrow \qquad \qquad \qquad \swarrow \qquad \qquad \qquad \qquad \ \ \downarrow \qquad \downarrow\\& \searrow \qquad \downarrow \qquad \qquad \qquad \swarrow \qquad \qquad \qquad \qquad \ \ \downarrow \qquad \downarrow\\& \searrow \qquad \downarrow \qquad \qquad \qquad \ \swarrow \qquad \qquad \qquad \qquad \ \downarrow \qquad \downarrow\\& \ \frac{x}{2} \qquad - \qquad \qquad \qquad 6 \qquad \qquad \qquad \quad \quad \ < \quad \ -1$
$& \underline{Six} \ \underline{less \ than} \ \underline{half \ of \ a \ number, \ x}, \ is \ldots\underline{greater \ than} \ \underline{10}.\\& \searrow \qquad \downarrow \qquad \qquad \qquad \swarrow \qquad \qquad \qquad \qquad \quad \ \downarrow \qquad \ \ \downarrow\\& \searrow \qquad \downarrow \qquad \qquad \qquad \swarrow \qquad \qquad \qquad \qquad \quad \ \downarrow \qquad \ \ \downarrow\\& \searrow \qquad \downarrow \qquad \qquad \qquad \ \swarrow \qquad \qquad \qquad \qquad \quad \downarrow \qquad \ \ \downarrow\\& \ \frac{x}{2} \qquad - \qquad \qquad \qquad 6 \qquad \qquad \qquad \qquad \quad \ > \quad \ \ 10$
Look at the original sentence.
Six less than half of a number, $x$ , is either greater than 10 or less than -1.
The word or which is underlined above shows that this sentence represents a disjunction.
So, the compound inequality could be written as: $\frac{x}{2}-6<-1$ or $\frac{x}{2}-6>10$
Next, consider part b .
To solve the compound inequality, treat each inequality separately.
First, add 6 to each side.
$\frac{x}{2}-6 &< -1 && \frac{x}{2}-6>10\\\frac{x}{2}-6+6 &< -1+6 && \frac{x}{2}-6+6>10+6\\\frac{x}{2}+(-6+6) &< 5 && \frac{x}{2}+(-6+6) > 16\\\frac{x}{2}+0 &< 5 && \frac{x}{2}+0>16\\\frac{x}{2} &< 5 && \frac{x}{2}>16$
Next, multiply each side by 2.
$\frac{x}{2} &< 5 && \quad \ \ \frac{x}{2}>16\\\frac{x}{2} \times 2 &< 5 \times 2 && \frac{x}{2} \times 2 > 16 \times 2\\\frac{x}{\bcancel{2}} \times \frac{\bcancel{2}}{1} &< 10 && \frac{x}{\bcancel{2}} \times \frac{\bcancel{2}}{1} > 32\\\frac{x}{1} &< 10 && \quad \ \ \frac{x}{1}>32\\x &< 10 && \qquad x > 32$
So, the solution could be written as: $x<10$ or $x>32$ .
Practice
Directions: Use each example to work with compound inequalities.
Two less than a number, $x$ , is at least 16 and at most 25
1. Translate the sentence into an inequality.
2. Solve the inequality.
3. Write a mathematical statement to show the range of possibilities for the answer.
One-third of a number, $n$ , is either less than -5 or greater than 3.
1. Translate the sentence into an inequality.
2. Solve the inequality.
3. Write a mathematical statement to show the range of possibilities for the answer.
Seven more than twice a number, $n$ , is either less than -5 or at least 9.
1. Translate the sentence into an inequality.
2. Solve the inequality.
3. Write a mathematical statement to show the range of possibilities for the answer.
Directions: Solve each problem.
When Harriet goes to the diner for lunch, she buys exactly two items: a wrap sandwich and a $3 milkshake. The total cost of her lunch is always more than$8 and less than $12. Let $w$ represent the cost, in dollars, of any of the wrap sandwiches Harriet buys. 1. Write a compound inequality to represent this problem situation. 2. Solve the inequality you wrote in part a. 3. According to the problem, is it possible that Harriet sometimes buys a wrap sandwich that costs$10?
4. Why? Explain.
Mr. Jameson pays $3 per gallon for gasoline. Each week, the amount he spends on gas for his car is always at least$30 on gas and at most \$105. Let $g$ represent the number of gallons of gasoline he buys in any given week.
1. Write a compound inequality to represent this problem situation.
2. Solve the inequality you wrote in part a.
Vocabulary Language: English
Compound Inequality
Compound Inequality
A compound inequality is an inequality that combines two other inequalities. For example: $5 < x < 10$, meaning $x > 5$ and $x < 10$.
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 $\ne$. |
# Question Video: Studying the Collision of Two Bodies Moving on the Same Line in Opposite Directions Forming One Body Mathematics
Two spheres, A and B, are moving in a straight line on a smooth horizontal plane in opposite directions at 7.17 m/s. If their masses are 4 m and 8 m, respectively, find the velocity 𝑣_(AB) of sphere A relative to sphere B. Given that the two bodies coalesce on impact into one body, find the speed 𝑣 of this new body just after the collision.
02:46
### Video Transcript
Two spheres, A and B, are moving in a straight line on a smooth horizontal plane in opposite directions at 7.17 metres per second. If their masses are four m and eight m, respectively, find the velocity 𝑣 AB of sphere A relative to sphere B. Given that the two bodies coalesce on impact into one body, find the speed 𝑣 of this new body just after the collision.
So the first thing we’re asked to do here is find the velocity 𝑣 AB of sphere A relative to sphere B. Let’s draw a little sketch and see if we can work out what’s happening. We have sphere A with the mass of four m moving in one direction at 7.17 metres per second and a mass B moving in the opposite direction at the same speed. 𝑣 AB is the relative velocity of sphere A to sphere B. Now, sphere B is moving at 7.17 metres per second in one direction. We’re going to subtract the velocity of sphere A. Now, that’s negative 7.17 because that’s moving in the other direction. 7.17 minus negative 7.17 is 7.17 plus 7.17. That’s 14.34 and the units remain the same. So 𝑣 AB is 14.34 metres per second.
Now, the next part of this question asks us to find the speed of the new body, that’s once the bodies coalesce, just after the collision. We might assume, since the mass of B is greater than the mass of A, that the velocity is acting in the same direction as the velocity for B after the collision. The total mass of the new body is four m plus eight m. So it’s 12 m. And now, we apply the law of conservation of momentum. This says that the total momentum before the collision must be equal to the total momentum after. If we let the new body be C, we can say that the momentum of A plus the momentum of B must be equal to the momentum of C. Where 𝑃, which is momentum, is equal to mass times velocity.
We begin by calculating the momentum of sphere A. It’s four m times negative 7.17. And the momentum of B is eight m times 7.17. Now, of course, we could have assumed that the other direction was positive. It really doesn’t matter as long as we’re consistent throughout the question. The momentum of sphere C is 12 m times 𝑣. We divide through by m. And this simplifies to 28.68 equals 12𝑣. We solved this equation for 𝑣 by dividing through by 12. And we get 2.39. Again, the units remain unchanged. So the speed is 2.39 metres per second.
Now, had we, in fact, chosen the other direction to be positive, we would’ve ended up with a velocity of negative 2.39 metres per second. Of course, though, we’re interested in the speed which is the magnitude of velocity. So the sign doesn’t matter. The answer is 2.39 metres per second. |
# 22.4 Magnetic field strength: force on a moving charge in a magnetic (Page 2/7)
Page 2 / 7
## Calculating magnetic force: earth’s magnetic field on a charged glass rod
With the exception of compasses, you seldom see or personally experience forces due to the Earth’s small magnetic field. To illustrate this, suppose that in a physics lab you rub a glass rod with silk, placing a 20-nC positive charge on it. Calculate the force on the rod due to the Earth’s magnetic field, if you throw it with a horizontal velocity of 10 m/s due west in a place where the Earth’s field is due north parallel to the ground. (The direction of the force is determined with right hand rule 1 as shown in [link] .)
Strategy
We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation $F=\text{qvB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ to find the force.
Solution
The magnetic force is
$F=\text{qvb}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$
We see that $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$ , since the angle between the velocity and the direction of the field is $\text{90º}$ . Entering the other given quantities yields
$\begin{array}{lll}F& =& \left(\text{20}×{\text{10}}^{–9}\phantom{\rule{0.25em}{0ex}}C\right)\left(\text{10 m/s}\right)\left(5×{\text{10}}^{–5}\phantom{\rule{0.25em}{0ex}}T\right)\\ & =& 1×{\text{10}}^{\text{–11}}\phantom{\rule{0.25em}{0ex}}\left(C\cdot \text{m/s}\right)\left(\frac{N}{C\cdot \text{m/s}}\right)=1×{\text{10}}^{\text{–11}}\phantom{\rule{0.25em}{0ex}}N.\end{array}$
Discussion
This force is completely negligible on any macroscopic object, consistent with experience. (It is calculated to only one digit, since the Earth’s field varies with location and is given to only one digit.) The Earth’s magnetic field, however, does produce very important effects, particularly on submicroscopic particles. Some of these are explored in Force on a Moving Charge in a Magnetic Field: Examples and Applications .
## Section summary
• Magnetic fields exert a force on a moving charge q , the magnitude of which is
$F=\text{qvB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta ,$
where $\theta$ is the angle between the directions of $v$ and $B$ .
• The SI unit for magnetic field strength $B$ is the tesla (T), which is related to other units by
$1 T=\frac{\text{1 N}}{C\cdot \text{m/s}}=\frac{\text{1 N}}{A\cdot m}.$
• The direction of the force on a moving charge is given by right hand rule 1 (RHR-1): Point the thumb of the right hand in the direction of $v$ , the fingers in the direction of $B$ , and a perpendicular to the palm points in the direction of $F$ .
• The force is perpendicular to the plane formed by $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$ . Since the force is zero if $\mathbf{\text{v}}$ is parallel to $\mathbf{\text{B}}$ , charged particles often follow magnetic field lines rather than cross them.
## Conceptual questions
If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in that region is necessarily zero?
## Problems&Exercises
What is the direction of the magnetic force on a positive charge that moves as shown in each of the six cases shown in [link] ?
(a) Left (West)
(b) Into the page
(c) Up (North)
(d) No force
(e) Right (East)
(f) Down (South)
Repeat [link] for a negative charge.
What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases in [link] , assuming it moves perpendicular to $\mathbf{\text{B}}?$
(a) East (right)
(b) Into page
(c) South (down)
Repeat [link] for a positive charge.
What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases in the figure below, assuming $\mathbf{\text{B}}$ is perpendicular to $\mathbf{\text{v}}$ ?
(a) Into page
(b) West (left)
(c) Out of page
Repeat [link] for a negative charge.
What is the maximum force on an aluminum rod with a $0\text{.}\text{100}\text{-μC}$ charge that you pass between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s? In what direction is the force?
$7\text{.}\text{50}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{N}$ perpendicular to both the magnetic field lines and the velocity
(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a $0\text{.}\text{500}\text{-μC}$ charge and flies due west at a speed of 660 m/s over the Earth’s south magnetic pole, where the $8\text{.}\text{00}×{\text{10}}^{-5}\text{-T}$ magnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.
(a) A cosmic ray proton moving toward the Earth at $\text{5.00}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ experiences a magnetic force of $1\text{.}\text{70}×{\text{10}}^{-\text{16}}\phantom{\rule{0.25em}{0ex}}\text{N}$ . What is the strength of the magnetic field if there is a $\text{45º}$ angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss.
(a) $3\text{.}\text{01}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{T}$
(b) This is slightly less then the magnetic field strength of $5×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{T}$ at the surface of the Earth, so it is consistent.
An electron moving at $4\text{.}\text{00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ in a 1.25-T magnetic field experiences a magnetic force of $1\text{.}\text{40}×{\text{10}}^{-\text{16}}\phantom{\rule{0.25em}{0ex}}\text{N}$ . What angle does the velocity of the electron make with the magnetic field? There are two answers.
(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than $1\text{.}\text{00}×{\text{10}}^{-\text{12}}\phantom{\rule{0.25em}{0ex}}N$ . What is the greatest the charge can be if it moves at a maximum speed of 30.0 m/s in the Earth’s field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.
(a) $6\text{.}\text{67}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{C}$ (taking the Earth’s field to be $5\text{.}\text{00}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{T}$ )
(b) Less than typical static, therefore difficult
#### Questions & Answers
Who is the father of physics
Newton.Geliliyo and Einstein is called father of physics
Neha
what is wave
a wave is a distirbance that transmits energy from one place ro another within or without a medium
Vincent
wave is the transfer of energy from one medium to another without the transfer of particles
ZIFAC
wave is a disturbance which transfer energy from one medium to another without causing any permanent displacement by itself
Joyfulsounds
wave is a disturbance or oscillation that travel through space and matter,accompanied by a transfer of energy
Ridwan
A wave is any disturbances in an elastic medium which carries energy from one point to another through a medium
abdul
what is harmonic motion
Nozyani
is a restoring force
Joyfulsounds
a wave is a disturbance and there are trasfer energy from one medium to another and travel through any space without the trasfer of particles.
Neha
a wave is a disturbance that travels or carries energy from one point to another through a medium
Wisdom
A moving disturbance in the level of a body water, undulation
Jizel
what is thermodynamics
what is thermodynamics
Charity
thermodynamics is a heat and energy significant physics
Neha
Relating to the conversation of heat into other forms of energy
Jizel
Are the antimatters of Hadrons also Hadrons?!Does the same rule apply to Leptons?
yes. Hadrons are the elementary particles that take part in stong, electromagnetic and weak interactions. Infact only Hadrons are involved in Strong interactions and when an anti-particle of any hadron is produced, it would be a hadron-conservations laws. Leptons are involved in weak int and follow
Lalita
what is physics
physic is a pure science that deal with behavior of matter,energy & how it related to other physical properties
Ridwan
Owk. But am are Art student.
Hussaini
What happens when an aeroplanes window is opened at cruise altitude?
what is the minimum speed for any object to travel in time?
as per theory of relativity, minimum speed will be the speed of light
Mr.
what is physics
it is just a branch of science which deals with the reasons behind the daily activities taking place everyday in our lives. it clearly states the reason in the form of laws.
sandhya
?
lkpostpost2000@yahoo
like Newton's laws , Kepler's laws etc....
sandhya
physics is the study of motion or moving things. Usually the moving things are normal items like vars or planets but sometimes it's electricity or heat that moves.
Jake
physics is one of the most significant diciplines of natural science which describe the nature and its matter
Neha
I would describe it as the science that is interested in the fundamental laws of nature. For example, what is light, what is sound, what is electricity/magentism, what forces are at work on a specific body. The knowledge of the world around us makes it possible to fly, have cell phones, GPS, etc.
Robyn
what happens when an aeroplane takes off?
it flies
Mr.
the lift generated by the wing overcome the weight of the plane(in Newton)and a net force of upward is created
Phebilia
it is a direct application of Magnus effect (which helps in throwing curve balls) the wings of plane are made in such a way that the net flow of air is more below them rather than on their upper side. So when the plane accelerates, the flaps produce the upward lift when enough velocity is obtained
Mr.
then due to lower pressure on upper part of wings helps producing an additional lift because air flows from areaof lower to the area of higher pressure
Mr.
The engines located under the wings generate thrust .. in relation thrust is a force ... which ovwrcomes or becomes greater than the weight of the plane.. remember weight is a force Weight = m x g-2 So therefore F(thrust) becomes greater than F(weight) Even if by 1Newton the plane starts lifting o
Theophilus
what happens when a ship moves
Williams
What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?
If it reduces the magnitude of the velocity, the acceleration sign is the opposite compared to the velocity.
Nicolas
yes
Williams
what is accerelation
an objects tendency to speed up over time
RayRay
acceleration is the change in velocity over the change in time it would be written delta-v over delta-t.
Shii
the change in velocity V over a period of time T.
Matthew
Delta means "change in"...not period of
Shii
just kidding. it all works mathematically
Shii
except doesn't time really only change if the instantaneous speeds vary...?
Shii
and I assume we are all talking average acceleration
Shii
Hey shiii 😀
the rate of change of velocity is callaed acceleration
Amna
a=delta v/delta t
Amna
the rate of change in velocity with respect to time is acceleration
Nana
nana you r right
Indrajit
good
oguji
what is meant by lost volt
Lost volt. Lol. It is the electrical energy lost due to the nature or the envirommental conditions (temperature and pressure) that affect the cable across which the potential difference is measured.
Theophilus
What is physics?
physics is brance science concerned with nature and properties of matter and energy
George
sure
Okpara
yah....
kashif
physics is study of the natural phenomenon on the basis of certain laws and principles. it's like watching a game of chess and trying to understand its rules how it's played.
Ajit
awesome
Okpara
physics is study of nature and it's law
AMRITA
physics is a branch of science that deals with the study of matter ,properties of matter and energy
Lote
Branch of science (study) of matter, motion and energy
Theophilus
what is a double-slit experiment?Explain.
when you pass a wave of any kind ie sound water light ect you get an interface pattern forming on a screen behind it, where the peaks and troughs add and cancel out due to the diffraction caused by a wave traveling through the slits
Luke
double slit experiment was done by YOUNG. And it's to give out monochromatic coherent, if an incoherent wave is passing through it. And then the waves form interference fringes. The screen placed in front of the double slit is preferably a film and then in the middle where "p=0" a brighter color
navid
is formed and then the constructive interferences occur at 0 (which is the brightest band)... then a sequence of bright band (constructive interference) and dark band (destructive interference) happens and the further from the central band the lower the intensity of bright band(constructive interfe
navid
what is photoelectric effect
the emission of electrons in some materials when light of suitable frequency falls on them
Hardeyyemih
The phenomenon that involves the emission of electrons (photoelectrons) when light of appropriate wavelength and frequency is incident on the surface of a metal.
ibrahim
what is regelation
is the process of melting under pressure and freezing when pressure is reduce
bawire
poisons ratio is which chapter
STREET_
Regelation is the phenomenon of melting under pressure and freezing again when the pressure is reduced
Theophilus |
# On Area percentage -3 (with bonus challenge!)
Not go on the title, in this discussion I will not deal with percentage but instead measure of area
In the below diagram $$O$$ is the center of the circle with tangents $$\overline{BD}$$ and $$\overline{AE}$$
The area of $\triangle AOB=\frac{1}{2}\overline{OD}^2\tan(\angle AOB)\sec(\angle AOB)$
Proof:
Let $\overline{OD}=x;\angle AOB=\theta$
In $\triangle AEO$ $\tan(\theta)=\dfrac{\overline{AE}}{x}$ $\overline{AE}=x\tan(\theta)........[1]$ As $\angle AEO=90^\degree$, applying Pythagorus theorem $x^2+\overline{AE}^2=\overline{AO}^2$ $\overline{AO}^2=x^2+x^2\tan^2(\theta)=x^2(1+\tan^2(\theta))=x^2\sec^2(\theta)$ $\overline{AO}=x\sec(\theta)........[2]$
Now in $\triangle AEO$ and $\triangle BDO$ $\angle AEO=90^\degree=\angle BDO$ $\angle AOE=\theta=\angle BOD$ $\overline{OE}=x=\overline{OD}$ From RHS criterion of congruence $\triangle AEO\cong\triangle BDO$ $\therefore \overline{AO}=\overline{BO}........[3]$
Area of $\triangle ABO=\frac{1}{2}\overline{BO}\times\overline{AE}$
From $[3]$ and this
Area of $\triangle ABO=\frac{1}{2}\overline{AO}\times\overline{AE}$
From $[1],[2]$ and this
Area of $\triangle ABO=\frac{1}{2}x\sec(\theta)\times x\tan(\theta)=\frac{1}{2}x^2\tan(\theta)\sec(\theta)$
Bonus
In the above diagram if you know the value of $R_1,R_2$ and $R_3$, then find a formula to find the area of the $\red{red}$ region.
Note:
• I myself don't know the answer to the bonus problem.
Note by Zakir Husain
6 months, 2 weeks ago
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Centres of circles are joined,then we can find the area of the formed triangle by heron's formula and then we find each angle by cosine rule and area of each sector.Subtracting area of sectors from area of triangle gives our result.
@Zakir Husain,I will try to find shorter method.
- 6 months, 2 weeks ago
Or try writing it out as one formula and try to factorise & simplify :)
- 6 months, 2 weeks ago
If someone gets the answer with solution of bonus, please share! Thanks!
- 6 months, 2 weeks ago
I got it Where,s=r1+r2+r3
- 6 months, 2 weeks ago
Incredibly! long formula
- 6 months, 2 weeks ago
Ya,the earlier solution is longer than it.I'll try to short the formula
- 6 months, 2 weeks ago
Can you describe the parts of this formula? Did you "use Heron's formula and then try to subtract the sectors"... If so I think there is a small mistake in Heron's formula, it's: $\sqrt{s (s-r_{1})(s-r_{2})(s-r_{3})}$. Anyways please describe how you got till this formula. Thanks!
- 6 months, 2 weeks ago
@Mahdi Raza,check my solution
- 6 months, 2 weeks ago
Wait…but the triangle’s sides aren’t $r_1,r_2,r_3$, and it is equilateral (tangent of circle from same point)
- 6 months, 2 weeks ago
It might be the smallest among the radii of the circles, but we need more proof. :/
- 6 months, 2 weeks ago
@Jeff Giff, @Mahdi Raza, @Kriti Kamal - the red area is not a triangle so don't confuse, it is a shape made of three arcs not three sides.
- 6 months, 2 weeks ago
Yes, but I meant the smallest triangle containing the red area :)
- 6 months, 2 weeks ago
I have considered them as arcs not as linesegmets :-)
- 6 months, 2 weeks ago
Oic. My solution is different. I thought about the smallest triangle containing the red area :)
- 6 months, 2 weeks ago
So my solution is $sizeof\triangle$containing red area $-sizeofARCH1,2,3$
- 6 months, 2 weeks ago
Maybe I am wrong :D Sorry for being unhelpful :|
- 6 months, 2 weeks ago
See my solution..
- 6 months, 2 weeks ago
Here,is my solution.Please comment if you find any mistake.Sorry for delay,it happens because tommorow my hands were burn by fire.
- 6 months, 2 weeks ago
The angles can be calculated in sin inverse and cos inverse function
- 6 months, 2 weeks ago
@Zakir Husain,can i send another version of solution
- 6 months, 2 weeks ago
Sure!
- 6 months, 2 weeks ago
The area of the triangle can be found using Zakir’s formula above:
$\frac12r_1^2\tan\angle A\sec\angle A$
Or any $x$ and $\theta$ pair with the relation above.
- 6 months, 2 weeks ago
You are assuming that the perpendicular of $\overline{AB}$ at point $D$ will touch the point $C$, the same assumption you have to make to the perpendicular of $\overline{AC}$ and point $B$. Then only you can apply my formula to get the area of $\triangle ABC$. You first have to prove that these assumptions are true for all cases. Or you can build a new formula for this special case separately.
- 6 months, 2 weeks ago
Aha! @Zakir Husain, your formula can be simplified as $\frac12x^2\sin \theta$, since $\tan \theta \sec \theta=\tan \theta \frac{1}{\cos \theta}=\frac{\frac{\sin \theta}{\cos \theta}}{\cos \theta}=\sin \theta$ :)
- 6 months, 2 weeks ago
$\tan\theta\sec\theta=\dfrac{\sin\theta}{\cos\theta}\times\dfrac{1}{\cos\theta}=\red{\dfrac{\sin\theta}{\cos^2\theta}}$
- 6 months, 2 weeks ago
Oops!
- 6 months, 2 weeks ago
Slight improvement: express angles in radians. Then, the formula becomes $\sqrt{sr_1r_2r_3}-\sum _{cyc}r_1^2\tan^{-1}\sqrt{\frac{r_2r_3}{sr_1}}$
- 6 months, 2 weeks ago
I know it.but generally i use my most of work in degrees
- 6 months, 2 weeks ago |
## Precalculus (6th Edition) Blitzer
The limit of a polynomial function $f\left( x \right)={{b}_{n}}{{x}^{n}}+{{b}_{n-1}}{{x}^{n-1}}+\cdots +{{b}_{1}}x+{{b}_{0}}$ is $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$, which is the value of the polynomial evaluated at the $a$.
A polynomial function is a sum of monomials. $f\left( x \right)={{b}_{n}}{{x}^{n}}+{{b}_{n-1}}{{x}^{n-1}}+\cdots +{{b}_{1}}x+{{b}_{0}}$ Thus, to find the limit of a polynomial, find the limit of each of the monomials and add them up. So, \begin{align} & \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,\left( {{b}_{n}}{{x}^{n}}+{{b}_{n-1}}{{x}^{n-1}}+\cdots +{{b}_{1}}x+{{b}_{0}} \right) \\ & =\underset{x\to a}{\mathop{\lim }}\,{{b}_{n}}{{x}^{n}}+\underset{x\to a}{\mathop{\lim }}\,{{b}_{n-1}}{{x}^{n-1}}+\cdots +\underset{x\to a}{\mathop{\lim }}\,{{b}_{1}}x+\underset{x\to a}{\mathop{\lim }}\,{{b}_{0}} \end{align} Use $\underset{x\to a}{\mathop{\lim }}\,c=c\ \text{with }c={{b}_{0}}$, \begin{align} & \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)={{b}_{n}}{{a}^{n}}+{{b}_{n-1}}{{a}^{n-1}}+\cdots +{{b}_{1}}a+{{b}_{0}} \\ & =f\left( a \right) \end{align} which is equal to value of the polynomial evaluated at $a$. Thus, if $f$ is a polynomial function, then $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ for any number $a$. For example, $f\left( x \right)={{x}^{3}}-2{{x}^{2}}+x+3$ \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( {{x}^{3}}-2{{x}^{2}}+x+3 \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,{{x}^{3}}-\underset{x\to 0}{\mathop{\lim }}\,2{{x}^{2}}+\underset{x\to 0}{\mathop{\lim }}\,x+\underset{x\to 0}{\mathop{\lim }}\,3 \\ & ={{0}^{3}}-2\times {{0}^{2}}+0+3 \\ & =0-0+0+3 \\ & =3 \end{align} |
# Sine and cosine graphs
In this lesson, we will look at the sine and cosine graphs and compare them.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 6
Q1.Fill in the gap: You can use the inverse sine function to work out the unknown angle, when given the opposite and ................. in a right- angled triangle.
1/6
Q2.Fill in the gap: You can use the inverse cosine function to work out the unknown angle, when given the hypotenuse and ................. in a right- angled triangle.
2/6
Q3.Use the sine ratio to find out the missing angle in the triangle shown below.
3/6
Q4.Which is the correct method to work out the missing angle?
4/6
Q5.Which is the correct method to work out the missing angle?
5/6
Q6.Which is the correct method to work out the length of missing side?
6/6
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 6
Q1.Fill in the gap: You can use the inverse sine function to work out the unknown angle, when given the opposite and ................. in a right- angled triangle.
1/6
Q2.Fill in the gap: You can use the inverse cosine function to work out the unknown angle, when given the hypotenuse and ................. in a right- angled triangle.
2/6
Q3.Use the sine ratio to find out the missing angle in the triangle shown below.
3/6
Q4.Which is the correct method to work out the missing angle?
4/6
Q5.Which is the correct method to work out the missing angle?
5/6
Q6.Which is the correct method to work out the length of missing side?
6/6
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Sine and cosine graphs
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
## Question 7
Q1.The image below shows the graph of ......
1/7
Q2.The image below shows the graph of ......
2/7
Q3.Use the graph to give an approximate value for sin (40).
3/7
Q4.Use the graph to give an approximate value for cos (80).
4/7
Q5.Which statement is correct?
5/7
Q6.Is this true or false? sin 90 = 1
6/7
Q7.Is this true or false? Cos 90 = 1
7/7
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Sine and cosine graphs
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
## Question 7
Q1.The image below shows the graph of ......
1/7
Q2.The image below shows the graph of ......
2/7
Q3.Use the graph to give an approximate value for sin (40).
3/7
Q4.Use the graph to give an approximate value for cos (80).
4/7
Q5.Which statement is correct?
5/7
Q6.Is this true or false? sin 90 = 1
6/7
Q7.Is this true or false? Cos 90 = 1
7/7
# Lesson summary: Sine and cosine graphs
### It looks like you have not completed one of the quizzes.
To share your results with your teacher please complete one of the quizzes.
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# Can we predict the cost of subway fare from the price of a slice of pizza? In the recent NY Times article Will Subway Fares Rise? Check at Your Pizza Place,
## Presentation on theme: "Can we predict the cost of subway fare from the price of a slice of pizza? In the recent NY Times article Will Subway Fares Rise? Check at Your Pizza Place,"— Presentation transcript:
Can we predict the cost of subway fare from the price of a slice of pizza? In the recent NY Times article Will Subway Fares Rise? Check at Your Pizza Place, reporter Clyde Haberman wrote that in NY City, the subway fare and the cost of a slice of pizza have run remarkably parallel for decades. A random sample of costs (in dollar) of pizza and subway fares are listed in the table below. Year196019731986199520022003 Cost of Pizza 0.150.351.001.251.752.00 Subway Fare 0.150.351.001.351.502.00
To see the relationship between the price of a pizza slice and the subway fare, we can make a scatter plot in Excel:
What is correlation? Correlation measures the strength of a linear relationship between 2 variables (like the price of pizza and the subway fare) Variables can be positively or negatively correlated: – Positive correlation: As the value of one variable increases, so does the value of the other variable. – Negative correlation: As the value of one variable increases, the value of the other variable decreases. r = correlation coefficient – r is between -1 and 1 – Indicates the strength of the correlation, ignoring the sign
Examples of different r values r =1.00 r =.42 r =.85 r =.17 r =-0.98
In case of a non-linear relationship the value of r will be close to 0.
Back to pizza price & subway fare… Year196019731986199520022003 Cost of Pizza 0.150.351.001.251.752.00 Subway Fare 0.150.351.001.351.502.00 To find the correlation coefficient r in Excel, type in: =CORREL(A2:A7, B2:B7) In this case, r = 0.9878. This indicates that there is a strong positive linear relationship between the two variables. Column AColumn B Row 2 Row 7
In fact, we can go one step farther! Question: What proportion of the variation in the subway fare can be explained by the variation in the costs of a slice of pizza? Answer: Find r 2 With r = 0.988, we get r 2 = 0.976. This means that about 97.6% of the variation in the cost of subway fares can be explained by its linear relationship with the cost of pizza. This implies that about 2.4% of the variation in costs of subway fares cannot be explained by the costs of pizza. CAUTION: This does not mean that increases in pizza sales cause increases in subway fares. Both costs might be affected by some other variable lurking in the background!
How to see into the future In Excel scatter plot, go to Chart tools Layout Trendline Linear Trendline Excel can figure out the equation of this line for you! Just go to More Trendline Options…
How to see into the future, contd. Now you can use the trendline to make predictions! y = 0.945x + 0.0346 Example: When the cost of a pizza slice was \$1.25, what was the cost of subway fare? y = 0.945(1.25) + 0.0346 = \$1.22 Example: When the cost of a pizza slice is \$2.25, what will the cost of subway fare be? y = 0.945(2.25) + 0.0346 = \$2.16 Example: When the cost of a pizza slice is \$20, what will the cost of subway fare be? y = 0.945(20) + 0.0346 = \$18.94 Not an appropriate prediction!
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# What are Binary, Decimal and Hexadecimal Number Systems?
Humans have been seeking a means to keep track of things and count them since the beginning of time. Earlier civilizations utilized various forms and techniques to keep track of things, but as human history progressed, man developed a number system to make counting things quicker and simpler.
What is a Number System in Maths?
A number system is a way of expressing numbers through writing. It represents the arithmetic and algebraic structure of the figures and offers a unique representation for each number. We can also do mathematical operations like addition, subtraction, and division using it.
The value of any digit in a number may be calculated using the following formula:
• The number
• Its numerical position
• The number system’s foundation
Binary System
One form of the Number Representation method is the Binary Number System. The binary system is used to represent binary quantities that can be represented by any device with only two possible operating states. A switch, for example, has only two states: open and closed.
There are just two symbols or potential digit values in the Binary System, namely 0 and 1. The addition of a 0b prefix or a 2 suffix to a binary number indicates it is binary. By clicking here you can convert binary systems to other number systems.
Every digit’s position has a weight that is a power of two. Each place in the Binary system is worth twice as much as the preceding one, therefore the numeric value of a Binary number is calculated by multiplying each digit by the value of the position in which the digit appears, then adding the results.
Example-1 − The number 125 is interpreted as
125 = 1×26+1×25+1×24+1×23+1×22+0x21+1×20=1111101
The least significant bit (LSB) is the first bit on the right, while the most important bit is the first bit on the left (MSB).
Decimal Number System
When a number system’s base value is 10, it’s known as a decimal number system, and it plays a crucial part in the advancement of science and technology. The value of each digit is determined by its location (or weight) in a number in the weighted (or positional) number representation. This is also known as the base-10 number system, which contains ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and so on. Every digit’s position has a weight that is a power of ten. Each place in the decimal system has ten times the significance of the preceding position, which implies that the numeric value of a decimal number is calculated by multiplying each digit by the value of the position in which the digit appears, then adding the products.
Example-1 − The number 2025 may be written as
2×103+0x10x2+2x10x1+5×100 = 2000+0+20+5 = 2005.
The least significant bit (LSB) is the fifth bit from the right, while the most important bit is the second bit from the left (MSB)
A number represented in the base system has coefficients multiplied by the power of r in general. An aj coefficient is a number that varies from 0 to 1. (r-1).
The following is an example of how to represent a real number in base-r:
anxrn+a(n-1) xr(n-1) +… … +a1xr1+a0+a-1xr-1+a-2xr-2+… … +a-mxr-m
Where a0, a1, …, an (n-1) are integer component digits and an is the total number of integer digits, and n is the total number of integer digits. Fractional part digits a-1, a-2, …, and a-m, where m is the total number of fractional digits.
The Hexadecimal Number System is a kind of numerical representation in which the base number is 16. This indicates that there are only 16 potential digit values: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. Where A, B, C, D, E, and F represent the decimal values 10, 11, 12, 13, 14, and 15 in single bits. Any digit’s value can be represented using only four bits. The addition of a 0x prefix or an h suffix to a decimal number indicates it is hexadecimal.
Every digit’s position has a weight that is a power of 16. The numeric value of a hexadecimal number is calculated by multiplying each digit by the value of the place in which the digit appears and then adding the products in the Hexadecimal system. As a result, it’s also a weighted (or positional) number system.
Each Hexadecimal number may be represented with only four bits, with distich values ranging from 0000 (for 0) to 1111 (for F = 15 = 8+4+2+1). The binary equivalents of Hexadecimal numbers are listed below.
Because the Hexadecimal number system’s base value is 16, the maximum value of a digit is 15, and it cannot be greater than 15. The weights of 160, 161, 162, 163, and so on are assigned to the places to the left of the hexadecimal point in this number system.
Similarly, the weights of 16-1, 16-2, 16-3, and so on are assigned to the places to the right of the hexadecimal point. This is referred to as the base power of 16. Any hexadecimal number’s decimal value may be calculated by adding the sum of each digit’s positional value.
Example 1: The number 512 may be deduced as
512=2×162+0x161+0x160=200
The least significant bit (LSB) is 0 on the right, while the most significant bit is 2 on the left (MSB).
Conclusion
In Boolean algebra, the binary number system can also be utilized. The advantages of the Binary Number System are simplicity of coding, fewer computations, and fewer computational mistakes.
Because digital systems (e.g., computers) and hardware are based on the binary system (either 0 or 1), we need 4-bit space to store each bit of a decimal number, whereas a hexadecimal number requires only 4-bit and has more digits than a decimal number, which is a benefit of the Hexadecimal Number System.
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# Class Width: Simple Definition
A frequency distribution table showing a class width of 7 for IQ scores (e.g. 125-118 = 7)
Class width refers to the difference between the upper and lower boundaries of any class (category). Depending on the author, it’s also sometimes used more specifically to mean:
• The difference between the upper limits of two consecutive (neighboring) classes, or
• The difference between the lower limits of two consecutive classes.
Note that these are different than the difference between the upper and lower limits of a class.
## Calculating Class Width in a Frequency Distribution Table
In a frequency distribution table, classes must all be the same width. This makes it relatively easy to calculate the class width, as you’re only dealing with a single width (as opposed to varying widths). To find the width:
1. Calculate the range of the entire data set by subtracting the lowest point from the highest,
2. Divide it by the number of classes.
3. Round this number up (usually, to the nearest whole number).
## Example of Calculating Class Width
Suppose you are analyzing data from a final exam given at the end of a statistics course. The number of classes you divide them into is somewhat arbitrary, but there are a few things to keep in mind:
• Make few enough categories so that you have more than one item in each category.
• Choose a number that is easy to manipulate; usually, something between five and twenty is a good idea. For example, if you are analyzing a relatively small class of 25 students, you might decide to create a frequency table with five classes.
Example: Find a reasonable class with for the following set of student scores:
52, 82, 86, 83, 56, 98, 71, 91, 75, 88, 69, 78, 64, 74, 81, 83, 77, 90, 85, 64, 79, 71, 64, and 83.
1. Calculate the range by subtracting the lowest point from the highest: the difference between the highest and lowest score: 92 – 52 = 46.
2. Divide it by the number of classes: 46/5, = 9.2.
3. Round this number up: 9.2≅ 10.
## References
Gleaton, James U. Lecture Handout: Organizing and Summarizing Data. Retrieved from http://www.unf.edu/~jgleaton/LectureTransCh2.doc on August 27, 2018.
Jones, James. Statistics: Frequency Distributions & Graphs. Retrieved from https://people.richland.edu/james/lecture/m170/ch02-def.html on August 27, 2018.
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Statistical concepts explained visually - Includes many concepts such as sample size, hypothesis tests, or logistic regression, explained by Stephanie Glen, founder of StatisticsHowTo. |
# Find the absolute maximum and absolute minimun, if they exist, for `y=f(x)=2x^3 -3x^2 -12x+24` A). `[-3,4]` B). `[-2,3]`
txmedteach | High School Teacher | (Level 3) Associate Educator
Posted on
To find the absolute minima and maxima for a function over a given interval, you need to check two categories of points:
1) At the boundaries of the interval
2) At points where `(df)/(dx) = 0`
For both, let's check the values at the boundaries. For A, this means checking `f(-3)` and `f(4)`, and for B, this means checking `f(-2)` and `f(3)`.
`f(-3) = 2(-3)^3-3(-3)^2-12(-3) + 24 = -21`
`f(4) = 56`
`f(-2) = 20`
`f(3) = 15`
Now that we have established the function values at the boundaries, we must now solve for where the function has derivatives equal to zero. Start by taking the derivative:
`(df)/(dx) = 6x^2-6x-12`
Now, solve for the x-values at which the derivative is zero:
`0 = 6x^2-6x-12`
`0 = x^2-x-2`
We can solve this equation for `x` by easily factoring:
`0 = (x-2)(x+1)`
This result gives us the two `x` values at which we might find other minima or maxima: `x = 2` and `x = -1`.
Now, we find the function values at `2` and `-1`:
`f(-1) = 31`
`f(2) = 4`
Both of these values are present on both intervals, so we need to take both into account when finding extrema.
Solving for the absolute minima and maxima for A, we simply find which `x` value gives the smallest and largest values for `f(x)` from the `x`-values `-3`, `-1`, `2`, and `4`. Clearly, the maxima and minima are found at the boundaries. `(-3, -21)` is the absolute minimum, and `(4, 56)` is the absolute maximum.
For B, we must do the same, but now we consider the function values at `x =` `-2`, `-1`, `2`, and `3`. Now, the absolute minimum and maximum are not at the boundaries. Instead, we see the absolute minimum at `(2, 4)` and the absolute maximum at `(-1, 31)`.
Recapping:
A) Minimum: (-3, -21); Maximum: (4, 56)
B) Minimum: (2, 4); Maximum: (-1, 31) |
# Length of a Line Segment
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The length of a line segment between two points is calculated with the distance formula:
This formula only works for linear (straight line) functions. If you have a curved function, use the arc length formula instead.
## Length of a Line Segment: Example
Example question: What is the length of the line segment for f(x) = 5x + 2 on the interval [0,1]?
Step 1: (Optional) Draw a Graph of the function. This step can help you see where the points you need for the formula are (x1, x2, y1, y2). I used Desmos.com to create this graph:
A quick look at the line segment between these two points and I’m going to guess that the length should be around 5. That gives me a way to check that my final answer is reasonable.
Step 2: Plug your two (x, y) coordinates into the distance formula.
If you’re having trouble deciding which point goes where, the following graph has the x and y points labeled.
Step 3: Simplify and solve:
• (1 – 0)2 = 1
• (7 – 2)2 = 25
• 1 + 25 = 26
• √(26) ≈ 5.1
So my guess at 5 was fairly close.
## Length of a Line Segment & Pythagorean Theorem
You might notice that the distance formula is really an application of the Pythagorean theorem. And in fact, you could get to the same answer by using the Pythagorean formula a2 + b2 = c2.
The following image shows that the hypotenuse of the triangle lies on the line segment you’re trying to find the length of:
You may be wondering why we use the distance formula to find the length of a line segment instead of the more simple Pythagorean formula. The answer is that while a, b, and c work perfectly well in geometry, it doesn’t translate well to the Euclidean plane of x’s and y’s, especially once you get to variations on the distance formula for curves. You can’t fit a triangle to a curve, but you can fit a series of small lines. This process of fitting small lines to approximate a curve uses integration, and (along with derivatives) it’s one of the two fundamental areas of calculus.
CITE THIS AS:
Stephanie Glen. "Length of a Line Segment" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/length-of-a-line-segment/
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Albert Teen
YOU ARE LEARNING:
Reverse Percentages
# Reverse Percentages
### Reverse Percentages
Reverse percentages allow us to work out the original quantity once we know the amount that it has changed by.
Sometimes we are told the percentage change, but need to work out the original quantity. This is a reverse percentage.
Consider a situation like this: you receive a pay rise of $5\%$. Your current salary is now $\pounds30600$. What was your salary before the raise?
1
Think about what this means mathematically
You are earning $105\%$ of your previous salary. Therefore, multiplying the original amount by $1.05$ equals $\pounds30600$. Let's call the original amount $x$
2
Rearrange to find the original amount
$x \times1.05=30600$. Therefore, $x=\dfrac{30600}{1.05}$
3
Find $x$
$x=\dfrac{30600}{1.05}=30000$
4
Awesome! 🙌
Your original salary was $\pounds30000$
A price increases by $30\%$. By what should you divide to find the original?
If the new quantity is less than the original, the decimal multiplying the original will be less than 1.
You see that a t-shirt is reduced by 15% to £17. How much was it originally?
1
We have $85\%$ of the original amount, which we will call $x$.
2
Rearrange to find the original amount
$x \times 0.85=17 \rightarrow x=\dfrac{17}{0.85}$
3
Use the equation to find the original amount
$\dfrac{17}{0.85}=20$
4
Great work! 💪
The t-shirt was originally worth $\pounds20$
A price decreases by $20\%$. By what should you divide to get to the original price?
If a price decreases by $33\%$ By what should you divide to get to the original price? |
# Class 6 Maths Exercise 1.1 Chapter 1 Knowing Your Numbers Solutions
NCERT Solutions Class 6 Maths Chapter 1 Knowing Your Numbers Exercise 1.1 are provided here. These solutions are prepared by expert teachers to guide you in your exam preparation. The solutions are always prepared by following NCERT guidelines, so that it covers the whole syllabus accordingly. Our CBSE Class 6 Maths Solutions Chapter 1 Exercise 1.1 Solutions are very helpful in scoring well in examinations.
## NCERT Solutions for Class 6 Maths Chapter 1 Exercise 1.1
EXERCISE: 1.1
Question 1: Fill in the blanks:
(a) 1 lakh = ………….. ten thousand.
(b) 1 million = ………… hundred thousand.
(c) 1 crore = ………… ten lakh.
(d) 1 crore = ………… million.
(e) 1 million = ………… lakh.
(a) 10
(b) 10
(c) 10
(d) 10
(e) 10
Question 2: Place commas correctly and write the numerals:
(a) Seventy-three lakh seventy-five thousand three hundred seven.
(b) Nine crore five lakh forty-one.
(c) Seven crore fifty-two lakh twenty-one thousand three hundred two.
(d) Fifty-eight million four hundred twenty-three thousand two hundred two.
(e) Twenty-three lakh thirty thousand ten.
(a) 73,75,307
(b) 9,05,00,041
(c) 7,52,21,302
(d) 5,84,23,202
(e) 23,30,010.
Question 3: Insert commas suitable and write the names according to Indian system of numeration:
(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701
(a) 8,75,95,762
Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.
(b) 85,46,283
Eighty-five lakh forty-six thousand two hundred eighty-three.
(c) 9,99,00,046
Nine crore ninety-nine lakh forty-six.
(d) 9,84,32,701
Nine crore eighty-four lakh thirty-two thousand seven hundred one.
Question 4: Insert commas suitable and write the names according to International system of numeration:
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831
(a) 78,921,092
Seventy-eight million nine hundred twenty-one thousand ninety-two
(b) 7,452,483
Seven million four hundred fifty-two thousand two hundred eighty-three
(c) 99,985,102
Ninety-nine million nine hundred eighty-five thousand one hundred two
(d) 48,049,831
Forty-eight million forty-nine thousand eight hundred thirty-one
### Access NCERT Solutions for Class 6 Maths Chapter 1 Knowing Your Numbers – All Exercises
Below we have provided the links to check solutions of the other exercises of the chapter. |
# Work and Electric Potential Energy
In this lecture we will learn about work and electric potential energy. You can watch the following video or read the written tutorial below.
## Work
In the previous lectures we’ve talked about electric fields and forces, and how they affect the things in their surroundings. If we understand forces, we can easily understand work!
Let’s say, we observe an object, and a force is applied on that object. The force is doing work if the object moves.
But what is work?
Work is a product of the magnitude of the displacement of an object and the component of the force parallel to the displacement.
Work is a scalar quantity defined only by its magnitude, and it can be positive or negative.
## Example for Work
Now, let’s take a look at a simple example that will help you easily understand and calculate work.
### Work When Force and Displacement Are Parallel
In the first scenario you’re pulling a box with a rope across the floor. The box is the object, and the force you’re using to pull the box is an external force. Let’s say the rope is parallel to the floor, and you’re pulling the box straight behind with a force of 60 Newtons. You’re moving the box 6 meters across the floor, which means you’re doing work on the box. But how much work did you do?
The equation shows that work is equal to the magnitude of the applied force, times the distance we moved the object, times the angle between the force and the displacement θ (theta).
In the example, the force and the displacement are parallel to each other, and in the same direction, the angle between them θ is 0°, and cosθ is 1.
Whenever the force and the displacement are in the same direction, θ=0°.
If you moved the box 6 meters with a constant force of 60 Newtons, you actually did a 360 Newton/meters of work on the box, which is equivalent to 360 Joules. Instead Newton/meters, the work is expressed in units called Joules.
### Work When Force and Displacement Are Not Parallel
In the second scenario, you’re pulling the box at 6 meters distance, with a constant force of 60 Newtons, but now you’re pulling at an upward angle of 30°, and the box moves parallel to the surface. So, the force you apply to the box is not in the same direction as the direction in which the box is moving. In that case, only the components of the applied force that are parallel to the displacement are part of the work done on the box.
The angle θ here is 30°, and cosθ is 0.866.
Now we have everything we need to calculate the work. We plug everything in and we get 311 Joules, which is less work in this case.
### Work When Force is Applied without Displacement of the Object
In the third scenario, you push a wall with a force of 60 Newtons. You may feel that you’re doing a lot of work, but the wall won’t move, so you’ve done zero work. This means that work can only be done when displacement occurs.
### Work When the Magnitude of the Force Varies
What if the magnitude of the force varies? For example you started pulling harder, but you got tired and the force got smaller the farther you dragged the box. In this case, to find the work, you need to calculate the force you applied by using integration. And this is the equation that will help you calculate the work:
The same principle that we explained so far in this video applies to charges as well!
## Energy
We already said that work is measured in Joules. But, Joules are also used as the units for energy, because work is just a change in energy.
In other words, energy is the ability, or the capacity to do work.
There are different kinds of energy, but in this lecture we will pay attention to potential energy.
Potential energy is energy that could be used to do work.
It is energy due to position. And the object needs to be positioned where we can get that energy back out again. Potential energy is denoted by capital U.
### Gravitational Potential Energy
There’s a gravitational potential energy, stored in an object as the result of its vertical position or height.
If you hold a book 1.5 meters above the ground, we can say that the book has gravitational potential energy. You have added energy to the book by lifting it up, or in other words, you have done a work on that book. And if you let it go, gravity will exert a force, and the book will fall down to the ground.
When the book is on the ground it has zero gravitational potential energy, because gravity can’t do work on it anymore.
We can calculate gravitational potential energy by multiplying the mass of the book times the constant for gravity times the height of the book.
For example, the mass of the book is 0.5 kilograms, and you’re holding it 1.5 meters above the ground, the gravitational potential energy will be 7.35 Joules.
This can only work if we’re using conservative forces.
A conservative force is a force where the work that we’re putting into the object is independent of the path which that object takes. The object needs to be positioned where we can get that energy back out again.
But what if I just take the book, and drag it across the floor? In that case, the book doesn’t have a potential energy, because it is not going to move back to its initial position.
This means that if you just add a force over a given distance, if you do work on an object, it doesn’t mean that the object will have a potential energy.
### Electric Potential Energy
We also have conservative forces when we’re looking at charges. In a gravitational field we have a mass that has a gravitational potential energy, and on the other side, in an electric field, we have charges that have an electric potential energy.
The electric potential energy describes how much stored energy a charge has, when moved by an electrostatic force.
We can calculate the potential energy only if we determine a reference point. The reference point is always arbitrary.
## Example for Electric Potential Energy
Let’s say we have a system of two positive charges: a point charge Q equal to 3 μC, and a test charge, q equal to 1 μC. The test charge is 5 cm away from the point charge. Q generates an electric field with the electric field lines pointing outward. This means that if we bring the test charge closer to the point charge it would be repelled.
Now, let’s say we want to bring the test charge closer, at 3 cm distance from the point charge. That would mean moving the test charge 2 cm against the direction of the electric field, which would require work by an external force.
The angle θ in this case is 0, and cosθ is 1.
Instead F, we will use the equation for electric force, and instead d, we will use r for the distance. R cancels out, and we’re left with this equation:
We use the same equation to calculate the electric potential energy U!
The electric potential energy U is equal to the Coulomb’s constant k, multiplied by the charge that creates the electric field Q, times the charge that would be placed at a reference point some distance away from the main charge q, and divided by the distance from the reference point r.
Now let’s go back to the example, and calculate the electric potential energy:
The potential energy at 3 cm distance from the point charge is 1.35 Joules.
## Summary
From the equation for electric potential energy, we can notice that the greater the charge on the test charge, the greater the repulsive force, and the more work would have to be done to move it closer to the positive point charge.
Opposite, the movement of a positive test charge in the direction of an electric field would occur without the need of work by an external force. This will further result in losing potential energy.
That would be all for this lecture. I hope it was helpful and you learned something new. In the next Basic Electronics tutorial we will talk about Electric Potential and Electric Potential Difference.
### 2 thoughts on “Work and Electric Potential Energy”
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# What does eight seven central mean theorem
In this chapter, you will study means and the Central Limit Theorem. example in class: Suppose 8 of you roll 1 fair die 10 times, 7 of you roll 2. Before illustrating the use of the Central Limit Theorem (CLT) we will first in which the population mean is 75 with a standard deviation of 8. Limit Theorem (i.e., min(np, n(1-p)) = min(10(), 10()) = min(3, 7) = 3). Sampling Distribution of the Sample Mean; Sampling Distribution of the Mean When the Population is Normal; Central Limit Theorem; Application of Sample.
Module 7: The Central Limit Theorem Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). The central limit theorem for sample means says that if you keep drawing larger and larger . An unknown distribution has a mean of 45 and a standard deviation of eight. Central Limit Theorem Examples: . 8: Convert the decimal in Step 7 to a. "8/7c" or "8 - 7 Central" means that this show is occurring at 8 o'clock Eastern Time, or 7 o'clock Central time. The reason Eastern Time is.
What are the mean, the standard deviation, and the sample size? Exercise 7. . 8. . Exercise The length of time a particular smartphone's battery lasts follows . According to the Central Limit Theorem, the larger the sample, the closer. So how do you figure out what the best estimate of the population mean is? scores sample mean sample first second () 1 2 2 2 2 2 4 3 3 2 6 4 4 2 8 5 5 4 2 3 6 4 4 4 7 4 6 (shape, mean, variability) are covered in the Central Limit Theorem. The central limit theorem states that the sample mean ¯X follows approximately the normal nσ), where µ and σ are the mean and standard deviation of the .. 7 . 8. 9. N(5,. 2. 4.) x f((x.)) 3. 4. 5. 6. 7. |
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# Subtracting Fractions
In mathematics, a fraction is a portion of a quantity out of the whole. The whole quantity can be any number, special value, or item. We can perform different arithmetic operations on fractions such as addition, subtraction, multiplication, and division.
A fraction consists of two parts-the numerator and the denominator. The upper part of the fraction is called the numerator and the lower part of the fraction is called the denominator. For example, $\frac{4}{9}$ is a fraction. Here, 4 is the numerator and 9 is the denominator. Based on the numerator and denominator, there are different types of fractions including proper fractions, improper fractions, mixed fractions, and more.
## What is meant by subtracting fractions?
Subtracting fractions means the process of subtraction of two fractional values. We have learned to subtract whole numbers. For example, the subtraction of 2 from 7 results in 5. We can also perform subtraction operations on fractions. We can:
• Subtract fractions with like denominators
• Subtract fractions with unlike denominators
• Subtract mixed fractions
• Subtract fractions with whole numbers
## Subtracting fractions with like denominators
To subtract fractions with like denominators, keep the denominators as they are, subtract the numerators, and write the difference over the denominator. If necessary, simplify the resulting fraction.
$\frac{5}{7}-\frac{4}{7}=\frac{1}{7}$
Example 1
Find $\frac{4}{5}-\frac{2}{5}$
Since the denominators are the same, subtract the numerators.
$=\frac{4-2}{5}$
$=\frac{2}{5}$
You may not get an answer that is in lowest terms, even if the fractions you were subtracting were. In that case, you must reduce the fraction.
Example 2
$\frac{8}{9}-\frac{2}{9}=\frac{6}{9}$
$=\frac{6÷3}{9÷3}$
$=\frac{2}{3}$
## Subtracting fractions with unlike denominators
If the denominators are not the same in the fractions you are subtracting, then you must use equivalent fractions that do have a common denominator. To do this, you have to find the least common multiple (LCM) of the two denominators.
To subtract fractions with unlike denominators, rename the fractions with a common denominator. Then subtract and simplify.
Example 3
For example, suppose you want to subtract
$\frac{6}{7}-\frac{2}{3}$
The LCM of 3 and 7 is 21. So we need to find fractions equivalent to $\frac{6}{7}$ and $\frac{2}{3}$ that have 21 in the denominator. To do this, you multiply the numerator and denominator of $\frac{6}{7}$ by 3 and multiply the numerator and denominator of $\frac{2}{3}$ by 7.
$\frac{6×3}{7×3}-\frac{2×7}{3×7}=\frac{18}{21}-\frac{14}{21}$
Now that we have like denominators, we can subtract as described above.
$\frac{18}{21}-\frac{14}{21}=\frac{4}{21}$
## Subtracting mixed fractions
While subtracting mixed fractions, use the following steps.
• Convert mixed fractions into improper fractions.
• Check the denominator values. If the fractions have like denominators, follow the instructions for subtracting fractions with like denominators. If the fractions have unlike denominators, follow the instructions for subtracting fractions with unlike denominators.
• Simplify, converting back into a mixed number if applicable.
Example 4
Subtract $8\frac{5}{6}$ from $15\frac{3}{4}$ .
Convert mixed fractions into improper fractions.
$\frac{63}{4}-\frac{53}{6}$
Next, find the LCM of 4 and 6, which is 12, and make the denominators equal.
$\frac{189}{12}-\frac{106}{12}=\frac{83}{12}$
Convert back to a mixed number.
$6\frac{11}{12}$
So $15\frac{3}{4}-8\frac{5}{6}=6\frac{11}{12}$
## Subtracting fractions with whole numbers
Follow these steps while subtracting fractions with whole numbers.
1. Convert the whole number into the fractional form. For example, 3 is a whole number, so you would convert it to 3/1.
2. Follow the procedure of subtracting fractions with unlike denominators.
3. Simplify the fraction if required.
Example 5
Subtract $2-\frac{1}{2}$ .
Convert the number 2 into the fractional form.
$\frac{2}{1}-\frac{1}{2}$
Find the LCM of 1 and 2, which is 2. Do the necessary multiplication of numerators to get
$\frac{4}{2}-\frac{1}{2}=\frac{3}{2}$
Simplify.
$1\frac{1}{2}=\frac{3}{2}$
So $2-\frac{1}{2}=1\frac{1}{2}$ .
## Get help learning about subtracting fractions
Subtracting fractions can get confusing because of the different behaviors of the numerators and denominators. If your student is having a hard time subtracting fractions, have them meet with a math tutor who can help them understand how and why the steps work. With the 1-on-1 attention a tutor provides, your student can quickly gain a thorough understanding of how to subtract fractions without distractions. Contact the Educational Directors at Varsity Tutors today to learn more about how working with a tutor can benefit your student.
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# A collection of limits
Páginas: 55 (13685 palabras) Publicado: 15 de junio de 2011
A Collection of Limits
March 28, 2011
Contents
1 Short theoretical introduction 2 Problems 3 Solutions 1 12 23
2
Chapter 1
Short theoretical introduction
Consider a sequence of real numbers (an )n≥1 , and l ∈ R. We’ll say that l represents the limit of (an )n≥1 if any neighborhood of l contains all the terms of the sequence, starting from a certain index. We write this fact aslim an = l, n→∞ or an → l. We can rewrite the above definition into the following equivalence:
n→∞
lim an = l ⇔ (∀)V ∈ V(l), (∃)nV ∈ N∗ such that (∀)n ≥ nV ⇒ an ∈ V .
One can easily observe from this definition that if a sequence is constant then it’s limit is equal with the constant term. We’ll say that a sequence of real numbers (an )n≥1 is convergent if it has limit and lim an ∈ R, ordivergent if it doesn’t have a limit or if it has the limit n→∞ equal to ±∞. Theorem: If a sequence has limit, then this limit is unique. Proof: Consider a sequence (an )n≥1 ⊆ R which has two different limits l , l ∈ R. It follows that there exist two neighborhoods V ∈ V(l ) and V ∈ V(l ) such that V ∩ V = ∅. As an → l ⇒ (∃)n ∈ N∗ such that (∀)n ≥ n ⇒ an ∈ V . Also, since an → l ⇒ (∃)n ∈ N∗ such that(∀)n ≥ n ⇒ an ∈ V . Hence (∀)n ≥ max{n , n } we have an ∈ V ∩ V = ∅. Theorem: Consider a sequence of real numbers (an )n≥1 . Then we have: (i) lim an = l ∈ R ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒ |an − l| < ε.
n→∞
1
2
A Collection of Limits
(ii) lim an = ∞ ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒ an > ε.
n→∞
(iii) lim an = −∞ ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒an < −ε
n→∞
Theorem: Let (an )n≥1 a sequence of real numbers. 1. If lim an = l, then any subsequence of (an )n≥1 has the limit equal to l.
n→∞
2. If there exist two subsequences of (an )n≥1 with different limits, then the sequence (an )n≥1 is divergent. 3. If there exist two subsequences of (an )n≥1 which cover it and have a common limit, then lim an = l.
n→∞
Definition: A sequence (xn)n≥1 is a Cauchy sequence if (∀)ε > 0, (∃)nε ∈ N such that |xn+p − xn | < ε, (∀)n ≥ nε , (∀)p ∈ N. Theorem: A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Theorem: Any increasing and unbounded sequence has the limit ∞. Theorem: Any increasing and bounded sequence converge to the upper bound of the sequence. Theorem: Any convergent sequence is bounded. Theorem(Cesarolemma): Any bounded sequence of real numbers contains at least one convergent subsequence. Theorem(Weierstrass theorem): Any monotonic and bounded sequence is convergent. Theorem: Any monotonic sequence of real numbers has limit. Theorem: Consider two convergent sequences (an )n≥1 and (bn )n≥1 such that an ≤ bn , (∀)n ∈ N∗ . Then we have lim an ≤ lim bn .
n→∞ n→∞
Theorem: Consider a convergentsequence (an )n≥1 and a real number a such that an ≤ a, (∀)n ∈ N∗ . Then lim an ≤ a.
n→∞
Theorem: Consider a convergent sequence (an )n≥1 such that lim an = a.
n→∞
Them lim |an | = |a|.
n→∞
Short teoretical introduction
3
Theorem: Consider two sequences of real numbers (an )n≥1 and (bn )n≥1 such that an ≤ bn , (∀)n ∈ N∗ . Then: 1. If lim an = ∞ it follows that lim bn = ∞.
n→∞n→∞
2. If lim bn = −∞ it follows that lim an = −∞.
n→∞ n→∞
Limit operations: Consider two sequences an and bn which have limit. Then we have: 1. lim (an + bn ) = lim an + lim bn (except the case (∞, −∞)).
n→∞ n→∞ n→∞
2. lim (an · bn ) = lim an · lim bn (except the cases (0, ±∞)).
n→∞ n→∞ n→∞
3. lim
lim an an = n→∞ (except the cases (0, 0), (±∞, ±∞)). n→∞ bn lim bn
n→∞ lim bn
4.lim abn = ( lim an )n→∞ n
n→∞ n→∞
(except the cases (1, ±∞), (∞, 0), (0, 0)).
5. lim (logan bn ) = log lim a ( lim bn ). n n→∞ n→∞
n→∞
Trivial consequences: 1. lim (an − bn ) = lim an − lim bn ;
n→∞ n→∞ n→∞
2. lim (λan ) = λ lim an (λ ∈ R);
n→∞ n→∞
3. lim
√ k
n→∞
an =
k
n→∞
lim an (k ∈ N);
Theorem (Squeeze theorem): Let (an )n≥1 , (bn )n≥1 , (cn )n≥1...
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Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
4.1 Order of Operations 4.2 Numbers 4.3 Factors 4.4 Multiples 4.5 Fractions 4.6 Decimals 4.7 Percents
4.8 Exponents 4.9 Roots and Radicals 4.10 Scientific Notation 4.11 Logarithms 4.12 Review Questions 4.13 Explanations
Percents
A percent is another way to describe a part of a whole (which means that percents are also another way to talk about fractions or decimals). Percent literally means “of 100” in Latin, so when you attend school 25 percent of the time, that means you only go to school 25/ 100 of the time (or .25).
You would probably fail all your classes if your attendance percentage was that low, so don’t get any ideas from our example. Instead, take a look at this question: 3 is what percent of 15?
This question presents you with a whole, 15, and then asks you to determine how much of that whole 3 represents in percentage form. Since a percent is “of 100,” to solve the question you have to set the fraction 3/ 15 equal to x100:
You then cross-multiply and solve for x:
Converting Percents into Fractions or Decimals
You should be skilled at converting percents into fractions and decimals, because these problems will definitely come up on the Math IC test.
Percents relate to decimal numbers very simply and directly. A percent is a decimal number with the decimal point moved two decimal places to the left.
For example:
To convert from a decimal number to a percent, move the decimal point two places to the right:
On an even more simplistic level, we can just say that 50% = .5 or 22.346% = .22346. Percentages greater than 100 exist, too. 235% = 2.35, for example.
To convert from a percent to a fraction, take the percentage number and place it as the numerator over the denominator 100. 58 percent is the same as 58/ 100.
To convert from a fraction back to a percent, the easiest method is to convert the fraction into a decimal first and then change the resultant decimal into a percent.
Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
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# Parallel Axis Theorem
Moment of Inertia Using the Parallel Axis Theorem – Irregular Shapes
Moment of Inertia Using the Parallel Axis Theorem – Irregular Shapes
## Introduction to Parallel Axis Theorem
Physics frequently deals with situations where objects rotate about an axis that passes through their centre of mass. It is usually simple to seek up the moment of inertia for the majority of objects of various forms when considering rotational motion about an object’s centre of mass.
The moment of inertia is the lowest when the rotational axis is across the centre of mass. As the rotational axis’s separation from the centre of mass widens, the moment of inertia grows.
The theorem makes it easy to connect any two parallel axes’ moments of inertia by going through the centre of mass.
## History of the Scientist/Theorem
The polymath Christiaan Huygens(14 April 1629 – 8 July 1695) first introduced the idea of the moment of inertia while researching a compound pendulum. Jakob Steiner(18 March 1796 – 1 April 1863) afterwards added his significant addition to this theorem. In honour of Christiaan Huygens and Jakob Steiner, the parallel axis theorem is frequently referred to as the Huygens-Steiner theorem.
Christiaan Huygens(14 April 1629 – 8 July 1695)
## Parallel Axis Theorem Definition
According to the “Parallel Axis Theorem,” a body’s moment of inertia about any axis is equal to the product of that body’s mass and the square root of the distance between the axes, as well as its moment of inertia about a parallel axis that passes through its centre of mass.
Parallel Axis Theorem
## What is the Formula of the Theorem of Parallel Axis?
Mathematically the parallel axis theorem can be expressed as,
$I = {I_0} + M{s^2}$
Where,
• I denote the body’s moment of inertia concerning any axis.
• Io denotes the body’s moment of inertia concerning the parallel axis through its centre of mass.
• M denotes the mass of the body.
• S denotes the distance between the two parallel axes.
## Parallel Axis Theorem Proof
Consider two parallel axes, one of which, designated $OY$, passes through the rigid body’s centre of mass, and the other, designated ${O_1}{Y_1}$, which is situated $s$ distance away from $OY$. Let’s take into account a small mass $dm$ that is located $R$from the axis $OY$ and ${R_1}$from the axis ${O_1}{Y_1}$.
Parallel Axis Theorem Proof
The rigid body’s total mass is $M = \int {dm}$. Therefore, the rigid body’s moment of inertia about the axes $OY$ and ${O_1}{Y_1}$ is,
$\begin{array}{l}I{}_0 = \int {{R^2}} dm{\rm{ and}}\\I = \int {{R_1}^2} dm\end{array}$
The above figure allows us to write,
$R_1^2 = {R^2} + {s^2} = 2sR\cos \theta$
Therefore,
$\begin{array}{l}I = {\int {{R_1}} ^2}dm\\I = \int {{R^2}} dm + \int {{s^2}} dm + \int {2sR} \cos \theta dm\end{array}$
$\begin{array}{l}I = {I_0} + M{s^2} + \int {2sR} \cos \theta dm\\{\rm{[}}{I_0} = \int {{R^2}} dm{\rm{ and }}\int {{s^2}} dm = M{s^2}{\rm{] }}\end{array}$
In this case, $R\cos \theta$ is the mass element’s x-coordinate about the mass centre.$\int {R\cos \theta } dm$, then, is the body’s moment of mass concerning the axis of its centre of mass, $OY$. Since the body’s centre of mass serves as the foundation for equilibrium,
$\int {R\cos \theta } dm = 0$
Therefore, the final equation will be as follows,
$I = {I_0} + M{s^2}$
Hence Proved.
## Limitations of Parallel Axis Theorem
• The usefulness of the parallel axis theorem is constrained.
• The object’s mass should be evenly distributed along the reference axis.
## Application of Parallel AxisTheorem
• The moment of inertia of a rigid body is determined using the parallel axis theorem in conjunction with any axis.
• Any object in rotation can have its moment of inertia determined using the parallel axis theorem.
### Solved Examples
1. If a body has a moment of inertia of 30 kgm2 along an axis perpendicular to its centre of gravity and a mass of 50kg. What is that body’s moment of inertia along a different axis that is parallel to and 50cm distant from the current axis?
Ans: Given data:
$\begin{array}{l}{I_0} = 30kg{m^2}\\M = 50kg\\s = 20cm = 0.2m\end{array}$
We have the formula,
$I = {I_0} + M{s^2}$
Replace every known value in the formula above.
$\begin{array}{l}I = 30 + 50 \times {(0.2)^2}\\ \Rightarrow I = 30 + 2\\ \Rightarrow I = 32kg{m^2}\end{array}$
2. How much inertia does a stick with a mass of 100 g, and a length of 10 cm have?
Ans. Given data:
$\begin{array}{l}M = 50g = 0.05kg\\s = 20cm = 0.2m\end{array}$
The general formula for the moment of inertia of a rod with mass (M) and length (L) and an axis that passes through the rod’s centre is;
$I = (\frac{1}{{12}})M{s^2}$
Therefore,
$\begin{array}{l}I = \frac{1}{{12}} \times 0.05 \times 0.2\\I = 0.000833Kg{m^2}\end{array}$
3. Find the moment of inertia of a solid sphere with a mass of 30kg and a radius of 2m around an axis that is 5m distant from the surface using the parallel axis theorem.
Ans. Given data:
$\begin{array}{l}M = 30Kg\\R = 2m\\s = 5m\end{array}$
The moment of inertia of a solid sphere with mass m and radius R about an axis that goes through its centre is ${I_0} = \frac{2}{5}m{R^2}$.
Solid Sphere
Therefore by the parallel axis theorem,
$I = {I_0} + M{s^2}$
Replace every known value in the formula above.
$\begin{array}{l}I = \frac{2}{5}(30){(2)^2} + 30 \times {(5)^2}\\ \Rightarrow I = 48 + 750\\ \Rightarrow I = 798kg{m^2}\end{array}$
## Important Points to Remember
To apply the parallel axis theorem, the following considerations must be made.
• Considered axis A and B must be parallel.
• The axis B must go through the body’s centre of mass, and
• The distance between them must be the shortest possible.
## Important Formulas to Remember From Theorem
• The general formula for the Parallel axis theorem with mass (M) and distance(s),
$I = {I_0} + M{s^2}$
• The general formula for the moment of inertia of a rod with mass (M) and length (L) and an axis that passes through the rod’s centre is;
$I = (\frac{1}{{12}})M{s^2}$
## Conclusion
The parallel axis theorem implies the sum of the moment of inertia through the mass centre and the product of the mass and square of the angle perpendicular to the rotational axis. The moments of inertia about various axes are connected by this theorem. But the usefulness of this theory is limited, though. For instance, the reference axis in the parallel axis theorem should pass through the object’s centre of mass.
## FAQs on Parallel Axis Theorem
1. Define the gyrating radius or radius of gyration when the centre of mass is located where the axis of rotation passes.
The distance from the axis of rotation to the location where the body’s total mass is thought to be concentrated, allowing the moment of inertia about the axis to remain constant, is known as the radius of gyration. The dispersion of an object’s parts is known as gyration, which is represented by the symbol K. And the value of the radius of gyration or gyrating radius when the centre of mass is located where the axis of rotation posses is minimal, but it is not exactly zero.
2. What is the moment of inertia, and what is the parallel axis theorem’s formula for it?
Each particle’s mass multiplied by the square of its separation from the axis of rotation yields the body’s moment of inertia. It follows that the moment of inertia is a physical characteristic that combines the mass and dispersion of the particles around the rotational axis. Observe that different moment of inertia result from rotation about various axes of the same body. And we can calculate the moment of inertia using the parallel axis theorem. Here is the equation for that:
$I = {I_0} + M{s^2}$
3. Differentiate between parallel axis theorem and perpendicular axis theorem.
Both theorems deal with the moment of inertia, but how they are applied differs. According to the perpendicular axis theorem, the sum of any two perpendicular axes of the body that meet the first axis determines the moment of inertia for any axis that is perpendicular to the plane. While the parallel axis theorem is used to determine the moment of inertia of a rigid body whose axis is parallel to the axis of the known moment and which is located via the body’s centre of gravity.
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# CLASS_11_MATHS_SOLUTIONS_NCERT
## Class XI Chapter 10 –
Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ 3 cos = - 2 , sin = 1 , and p =1 2 7 Since the value of sin and cos are negative, 6 6 Thus, the respective values of and p are 7 and 1. 6 Question 3: Find the equation of the line, which cut-off intercepts on the axes whose sum and product are 1 and -6, respectively. Solution 3: Let the intercepts cut by the given lines on the axes be a and b. It is given that a + b = 1 …(1) ab = -6 …(2) On solving equations (1) and (2) , we obtain a = 3 and b = -2 or a = -2 and b = 3 It is known that the equation of the line whose intercepts on the axes are a and b is x y 1 or bx + ay – ab = 0 a b Case I: a = 3 and b = -2 In case, the equation of the line is -2x + 3y + 6 = 0, i.e., 2x – 3 y = 6. Case II: a = - 2 and b = 3 In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., -3x + 2y = 6. Thus, the required equation of the lines are 2x – 3 y = 6 and -3x + 2y = 6. Question 4: x y What are the points on the y-axis whose distance from line 1is 4 units. 3 4 Solution 4: x y Let (0, b) be the point on y-axis whose distance from line 1 is 4 units. 3 4 The given line can be written as 4x + 3y – 12 = 0 …(1) On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, C = -12. Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point x y is given by Ax1By 1C d 2 2 A B x y Therefore, if (0, b) is the point on the y-axis whose distance from line 1 is 4 units, then: 3 4 403b12 4 2 2 4 3 3b 12 4 5 20 3b 12 b b or b 20 3 12 20 3 12 20 3 12 3b 20 12or3b 20 12 32 8 b orb 3 3 32 Thus, the required points are 0, 3 and 8 0, 3 1, 1 Question 5: Find the perpendicular distance from the origin to the line joining the points cos ,sin and cos ,sin Solution 5: The equation of the line joining the points cos ,sin and cos ,sin cos ,sin and cos ,sin sin sin ysin xcos cos cos y is given by cos cos sin cos cos xsin sin cos sin sin y sin sin ycos cos sin( ) 0 x sin sin cos cos cos sin cos sin sin cos sin cos 0 x Ax By C 0, whereA sin sin , B cos cos , andC sin( ) It is known that the perpendicular distance (d) of a line Ax +By +C =0 from a point x y is Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor. 1, 1
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# Negative Integers
Negative integers are numbers having values less than 0 and thus have a negative sign before them. Being an integer, they do not include a fraction or a decimal.
It starts from -1, the largest negative integer, and goes on endlessly. Thus {….- 5, -4, -3, -2, -1} is a set of negative numbers.
Thus, when represented on a number line, they are usually drawn on the left of zero.
Negative integers are widely used in different fields such as banking and finance for calculating profit and loss, science and medicine for showing temperature, calibrating instruments, and measuring blood pressure, body weight, and medical tests.
They are also used to calculate goal differences in sports such as football, hockey, and basketball.
## Rules
There are some rules followed while performing basic operations involving negative integers.
When adding a negative integer with a negative integer, we add the numbers and give the sign of the original values. Thus, we move to the left of the number line.
For example, (-4) + (-2) = -6
When represented on a number line, we get:
When adding a positive and a negative integer, we subtract one number from the other number and provide the sign of the larger absolute value.
For example,
(+4) + (-8) = -4
When represented on a number line, we move to its left:
Again, (-4) + (+8) = +4
When represented on a number line, we move to its right:
### Subtracting Integers
Subtracting a Positive Number from a Negative Number
When subtracting a positive number, it is the same as adding the negative value of that number.
For example, (-6) – (+4) is equivalent to (-6) + (-4) = -10
When represented on a number line, we get
Subtracting a Negative Number from a Positive Number
When subtracting a negative number, it is the same as adding the positive of that number.
For example, (+6) – (-4) is equivalent to (+6) + (+4) = +10
When represented on a number line, we get
### Subtracting a Negative Number from a Negative Number
When subtracting a negative number from a negative number is same as adding a positive number to a negative. It is nothing but subtraction between the two numbers and giving the sign of the greater number.
For example, (-8) – (-2) is equivalent to (-8) + (+2) = -6
### Multiplying and Dividing Integers
Multiplication or division of negative numbers with a like sign gives a result with a positive sign before it.
For example,
(-6) × (-2) = 12
(-6) ÷ (-2) = 3
The multiplication or division of numbers with unlike signs (positive with a negative number or vice versa) gives a result with a negative sign before it.
(-6) × (+2) = -12
(+6) × (-2) = -12
(-6) ÷ (+2) = -3
(+6) ÷ (-2) = -3
## Solved Examples
a) (-20) + (-7)
b) (+11) + (-5) Â
c) (-16) – (+4)
d) (-8) – (+10)Â Â
e) (+6) – (-4)
Solution:
a) (-20) + (-7) = -27
b) (+11) + (-5) = +6
c) (-16) – (+4)= -20
d) (-8) – (+10) = (-8) + (-10) = -18
e) (+10) – (-9) =(+10) + (+9) = +19
Simplify:
a) (-9) × (-7)
b) (-24) ÷ (-3)
Solution:
a) (-9) × (-7) = +63
b) (-24) ÷ (-3) = +8 |
# How do you differentiate f(x)=x(3x-9)^3 using the chain rule?
Feb 25, 2016
$f ' \left(x\right) = 3 {\left(3 x - 9\right)}^{2} \left(4 x - 3\right)$
#### Explanation:
The $\textcolor{b l u e}{\text{ Product rule " "will have to be used as well as the }}$
$\textcolor{b l u e}{\text{ chain rule }}$
There is a product of 2 functions here , x and ${\left(3 x - 9\right)}^{3}$
using the $\textcolor{b l u e}{\text{ Product rule }}$
If f(x) = g(x).h(x) then f'(x) = g(x).h'(x) + h(x).g'(x)
using the$\textcolor{b l u e}{\text{ chain rule }}$
$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$
hence: $f ' \left(x\right) = x \frac{d}{\mathrm{dx}} {\left(3 x - 9\right)}^{3} + {\left(3 x - 9\right)}^{3} \frac{d}{\mathrm{dx}} \left(x\right)$
$= x \left[3 {\left(3 x - 9\right)}^{2} \frac{d}{\mathrm{dx}} \left(3 x - 9\right)\right] + {\left(3 x - 9\right)}^{3} .1$
$= x \left[3 {\left(3 x - 9\right)}^{2} .3\right] + {\left(3 x - 9\right)}^{3}$
$= 9 x {\left(3 x - 9\right)}^{2} + {\left(3 x - 9\right)}^{3}$
( can be simplified by 'taking out' common factor )
$= {\left(3 x - 9\right)}^{2} \left[9 x + 3 x - 9\right] = {\left(3 x - 9\right)}^{2} . 3 \left(4 x - 3\right)$ |
### Bending the Rules to Square the Circle
Squaring the circle was one of the famous Ancient Greek mathematical problems. Although studied intensively for millennia by many brilliant scholars, no solution was ever found. The problem requires the construction of a square having area equal to that of a given circle. This must be done in a finite number of steps, using only ruler and compass.
Taking unit radius for the circle, the area is π, so the square must have a side length of √π. If we could construct a line segment of length π, we could also draw one of length √π. However, the only constructable numbers are those arising from a unit length by addition, subtraction, multiplication and division, together with the extraction of square roots.
In 1882, Ferdinand Lindemann proved that π is transcendental, that is, not a solution of any simple polynomial equation with integer coefficients. Therefore, π is not constructable with ruler and compass: the circle cannot be squared within the rules of classical geometry.
Bending the Rules
However, there are many simple ways to construct a line of length π, given one of unit length. Perhaps the simplest is to take the unit circle and roll it along a line for precisely half a revolution. Since the circumference is 2π, the point of contact traces a line segment of length π.
The next challenge is to construct a segment of length √π. This is a standard problem in Euclidean geometry. We can use a result known as the intersecting chords theorem. If two chords of a circle intersect, dividing each into two segments, then the product of the two segments of one chord equals the corresponding product for the other chord.
The figure below (left) shows the idea, and the result is easily demonstrated by using the properties of similar triangles. If the chord ab is a diameter, and chord cc is perpendicular to it, then cc is bisected (figure, right). If b = 1, then c = √a.
We use the properties of intersecting chords to find the square root of a given length. The line segment covered by the rolling unit circle has length π. We extend this by one unit and draw the circle of which it is a diameter. The diameter AB has two segments AC and CB, of length π and 1, so their product is π. We draw a chord through C perpendicular to AB. This has equal segments, whose product must be π. Therefore, the segment CD has length √π (Figure below).
Using CD as a side of a square, the area must be π, as required. Thus, by stretching the rules of the game, we have squared the circle.
Quad Erat Demonstrandum — but not by the methods permitted within Euclidean geometry.
Last week’s post: Bloom’s attempt to Square the Circle. |
What Is 19/38 as a Decimal + Solution With Free Steps
The fraction 19/38 as a decimal is equal to 0.5.
Repetitive decimal places is a unique property of a decimal number, depending on the two dividing numbers. When the decimal numbers repeat after a specific interval, they are considered repetitive, else, non-repetitive. All repetitive decimal numbers are non-terminating.
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 19/38.
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 = 19
Divisor = 38
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 = 19 $\div$ 38
This is when we go through the Long Division solution to our problem. Given is the Long division process in Figure 1:
Figure 1
19/38 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 19 and 38, we can see how 19 is Smaller than 38, and to solve this division, we require that 19 be Bigger than 38.
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 19, which after getting multiplied by 10 becomes 190.
We take this 190 and divide it by 38; this can be done as follows:
190 $\div$ 38 $\approx$ 5
Where:
38 x 5 = 190
This will lead to the generation of a Remainder equal to 190 – 190 = 0.
Finally, we have a Quotient generated as 0.5, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra. |
# A Prime Search
Purpose
In this unit we use rectangular models or arrays to explore numbers from one to fifty. We practice expressing numbers as the product of two smaller numbers and in doing so identify all the factors of numbers systematically. We are also introduced to prime numbers.
Achievement Objectives
NA4-1: Use a range of multiplicative strategies when operating on whole numbers.
NA4-8: Generalise properties of multiplication and division with whole numbers.
Specific Learning Outcomes
• Model the numbers from 1 to 100 as rectangular arrays.
• Connect the possible arrays for a given number to the factors of that number.
• Identify the factors of the numbers 1 to 100 using divisibility.
• Identify whether natural numbers from 1 to 100 are prime, composite, or a special case (i.e. 1).
Description of Mathematics
This unit looks at the number concepts of factors, multiples, and prime numbers. These fundamental ideas have a surprisingly wide range of applications. Searching for certain types of prime number has become a test for the speed of new computers and methods of protection of computers for unwanted access. Prime numbers are an integral part of modern coding theory. This allows the easy encryption of words and numbers but means that decoding is quite difficult. Codes are based around the fact that the prime factors of large numbers are hard to find. Such codes are used by banks and the military because they are very difficult to break, even in the age of computers.
Finding factors of a given number can always be done by a systematic search. A search for prime numbers firstly assumes that we are looking only at natural numbers {1, 2, 3, 4, 5, …} That means zero cannot be prime. Starting at 1, each consecutive number is tested to see if it is a factor of the number in question. The search for factors from above and below continues until they converge on the square root of the number. For example, to find the factors of 18, first check 1, then 2, then 3, then 4, and so on until 5, since five is just greater than the square root of 18 (√18 = 4.24, 2dp.). That way all the factors of 18, or any other number for that matter can be found. Systematic searches are important throughout mathematics, especially to verify that all possible answers have been found.
The convergence of factors from above and below to the square root works in this way. We need not test any numbers above 4.24. This is because any number above 4 will be paired with a factor less than 4 and all the smaller potential factors were tested. For 18 this means that we get 1 and 18 by testing for 1; 2 and 9 by testing for 2; 3 and 6 by testing for 3. Four is not a factor of 18 so the job is done.
The learning opportunities in this unit can be differentiated by providing or removing support to students and by varying the task requirements. Ways to differentiate include:
• Altering the degree of abstraction from working with cubes, to making diagrams of arrays, to working with only numbers.
• Altering the size of the numbers that students find factors for. Generally, larger natural numbers are more difficult to work with.
• Providing calculators so students can test for divisibility if their basic facts knowledge, and multiplicative thinking is not strong.
The context for this unit is about using cubes to create arrays. The context can be connected to everyday life through situations where arrays are useful. For example:
• Creating an array of kumara pits.
• Looking at arrays on gameboards, like Chess and Twister.
• Designing parking lots for cars.
• Setting out a group formation in dance or kapa haka.
Required Resource Materials
Activity
#### Getting Started
Today students investigate the possible rectangular arrays for given whole numbers. They record the appropriate factors for each array.
1. Give each student 12 cubes and ask them to form a rectangle. All 12 cubes must be used to form each array.
What size is the rectangle you have made? (Discuss the description of rectangles using rows and columns.)
Have we found all the rectangles? How do you know? (Expect the students to check each of the numbers to 12 although some may realise that you only need to check as so far as the factors start repeating, e.g. 3 x 4 and 4 x 3. Highlight that the factor pairs repeat after 3 (Square root of 12 = 3.46).
2. As a class record each of the rectangles using squared paper.
3. Attach these rectangles to an A3 page headed with a 12. Record the expression with each rectangle. Organise the rectangles from 1x12 to 12x1. (This will allow for more systematic comparison with the factors of other numbers.)
Are each of the rectangles unique? (‘Unique’ means unlike any others such as one array cannot be rotated or reflected to make another)
If we remove any rectangles that are copies, how many unique rectangles are there? (1 x 12, 2 x 6, 3 x 4).
What name do we give to numbers that multiply to give a number, like 12? (Factors of 12)
You might record the findings using formal notation:
Factors of 12 = {1, 2, 3, 4, 6, 12}
It is also interesting to consider why whole number are not factors of 12. For example, 5 is not a factor because all factors of 5 end in either 0 or 5 (in the ones place).
4. Give each pair of students a number in the range 10 to 24) and ask them to form as many unique rectangles as they can for their number. As they form the rectangles, first with cubes and then on squared paper, ask questions to focus on the factors of the number in a systematic way, such as arranging the first factor from lowest to highest.
How many rectangles have you found for your number?
How do you know you have found them all?
Why do some numbers have more rectangles than others?
Are there any numbers that form only one rectangle?
Is it possible to predict these ‘one rectangle’ numbers?
5. Ask each pair of students to glue their rectangles to the ‘page’ for their number. Alongside each rectangle, a corresponding multiplication should be written.
For example, 4 x 3 or 3 x 4 for this array:
6. Share the number pages as a class.
Which number has the most rectangles? (24) Why? Students may think that 24 has the most factors because it is the largest number. You might investigate 25 that has only 3 factors.
Which numbers have only one rectangle (two factors)? (The primes 11, 13, 17, 19, 23)
Which numbers have only 2 rectangles (four factors)? (4, 6, 8, 9, 10, 14)
Can a number have 3 rectangles (six factors)? (Yes, 12 does)
Can a number have 4 rectangles (eight factors)? (Yes, 24 does)
Note that students may not accept squares as a special class of rectangles, squares. For example, they may not include 4 x 4 in their search for arrays with 16 cubes. It is important to discuss this point as it has implications for geometry as well.
7. Formally list sets of factors, e.g. Factors of 10 = {1, 2, 5, 10} and Factors of 11 = {1,11}.
What happens when one of the rectangles is a square?
These numbers are known as square numbers and they have an odd number of factors, e.g. Factors of 9 = {1, 3, 9}.
Numbers that have more than two factors are called composite numbers. What numbers in the range 10-24 are composite?
(10, 12, 14, 15, 16, 18, 20, 21, 22, 24)
Numbers that have only two factors are called prime numbers. What numbers in the range 10-24 are prime? (11, 13, 17, 19, 23)
Can a number be neither (not) composite or prime? (Usually this is an ‘either or’ choice except for the special case of one)
How many rectangles can be made with one square? (only a 1 x 1 rectangle is possible, so 1 is a special case, neither composite or prime but it is a square number.)
#### Sessions Two, Three and Four
Over the next 2-3 days ask your students to create rectangle charts for each of the numbers from 1 to 100. Use the charts to develop their understanding of factors, multiples, and primes.
1. Look at the charts from the previous day.
I could say that the factors of 12 are 1, 2, 3, 4, 6 and 12. What does the word factor mean?
You might look up the internet to create a formal definition of the word “factor.”
I could say that 12 is a multiple of 1, 2, 3, 4, 6, and 12. What does the word multiple mean? Create a formal definition of the word “multiple.”
2. Choose other numbers in the range 10-24 and invite your students to make statements that use the words factor/s and multiple/s. The characteristics of multiples of a given number are used to test for divisibility by that number. For example, multiples of 2 are {2, 4, 6, 8, 10, 12,…} so to test to see if a number has 2 as a factor we only need to check that it is even, so has 0, 2, 4, 6 or 8 in the ones place. The divisibility rule for 5 is also easy but for other factors like 3, 6, and 9 the rules involve digital sums. For a lesson on divisibility by 3 or 9 go to Pages 72-74 of this book in the Numeracy Project series.
3. Put the numbers from 24 to 100 in a "hat". One student in the pair picks a number from the hat and then the two students work together to construct all rectangular arrays for the number. They record the rectangles on squared paper and then attach these rectangles to an A3 piece of paper. As the students work ask questions that focus on their identification of the factors of a number.
How many rectangles have you found for your number?
How do you know you have found them all?
What are the factors of your number?
Is your number prime or composite? How do you know?
Is your number a square number?
4. An interesting investigation is the relationship between square numbers and the square root function. For example, 64 is a square number since 8 x 8 is one possible array. The factors of 64 are {1, 2, 4, 8, 16, 32, 64}. An odd number of factors is one property of natural numbers that are squares. Using a calculator find that the square root of 64 is 8.
What does 8 as a factor have to do with an array of 64 cubes? (8 is the side length of a square with an area of 64 square units.)
5. When the students have completed a number, they select another from the hat.
6. At the end of each session look at the developing display of rectangles charts. Invite pairs of students to share their findings with the class.
7. You might use a hundred chart to colour code the composite and prime numbers. Provide students with their own chart and look for a virtual hundreds board online to use with the class. This may lead to a discussion about which kind of number is most common. Students may notice that primes become less frequent as the range is extended.
8. Possibly note the fact that primes tend to be located next to multiples of six. For example, both 5 and 7 are primes. However, being one more or less than a multiple of six does not guarantee that a number is prime. For example, 48 is a multiple of six. 47 is prime but 49 is a square composite.
#### Session Five
Today we look at our completed display of rectangle charts and create a newsletter for our families telling them about our findings.
1. Display the factor charts, in order, for the class to examine.
2. Encourage the students to look at the charts and write statements (in pairs) about their observations. The following questions may be used as prompts for the students.
Which number has the most factors?
How many prime numbers are there less than 100?
Is there a way to predict if a number, like 51 or 57, is prime?
What number do you think is the most interesting? Why?
Which decade has the most prime numbers? Why do you think it is the tens decade?
3. Share statements.
4. Investigate the method used by a mathematician from ancient Greece. His name was Eratosthenes, and he is most famous for estimating the circumference of the Earth. That was quite a feat for a person who lived 276 - 194BC. Eratosthenes used a sieve method to leave behind only the primes. You can find videos online explaining the method. Your students might be interested.
5. Use students’ statements to form the basis for the newsletter home.
6. In addition to the class statements you may like to include the following brainteaser that is very challenging.
Challenge: The Census Problem
A census taker approaches a house and asks the person who answers the door.
"How many children do you have, and what are their ages?"
Person: I have three children. The product of their ages is 36, the sum of their ages is equal to the address of the house next door."
The census taker walks next door, comes back and says to the woman.
Person: "I have to go. My oldest child is sleeping upstairs."
Census taker: "Thank you, I have everything I need."
Question: What are the ages of the each of the three children?
This is a very famous problem and the first reaction of people is often that insufficient information is given. However, it can be solved.
First make a systematic is of the sets of three factors that have a product of 36, since the ages of the three children have a product of 36. Begin with one as a factor and increase the size of factors from there:
1 x 1 x 36 (unlikely a child is 36 years old)
1 x 2 x 18, 1 x 3 x 12, 1 x 4 x 9, 1 x 6 x 6
2 x 2 x 9, 2 x 3 x 6,
3 x 3 x 4
Since the only combination possible with four is covered that is the end of the list. Next find the sums by adding the factors.
1 + 1 + 36 = 38,
1 + 2 + 18 = 21, 1 + 3 + 12 = 16, 1 + 4 + 9 = 14, 1 + 6 + 6 = 13,
2 + 2 + 9 = 13, 2 + 3 + 6 = 11,
3 + 3 + 4 = 10.
The only way the census worker needs more information is the possibility that two or more sets of factors might have the same sum. In this case 1, 6, and 6 have the same sum and 2, 2, and 9.
Knowing that there is an oldest child means that 2, 2, and 9 are the ages of the children. |
# 2016 AMC 12B Problems/Problem 20
## Problem
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$
## Solution 1
We use complementary counting. Firstly, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$, $B$ beat $C$, and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.
There are $21$ ways to choose the team that beat the two other teams, and $\binom{10}{2} = 45$ to choose two teams that the first team both beat. This is $21 * 45 = 945$ sets. There are $\binom{21}{3} = 1330$ sets of three teams total. Subtracting, we obtain $1330 - 945 = \boxed{385}$, thus $(\text{A})$ is our answer.
## Solution 2
As above, note that there are 21 teams, and call them A, B, C, ... T, U. WLOG, assume that A beat teams B-L and lost to teams M-U. We will count the number of sets satisfying the conditions of the problem that include A, then multiply by 21 (for each other team) and divide by 3 (since every set will be counted by each of the 3 teams that are apart of that set). To do this, let X$=\{B, ..., L\}$ and Y$=\{M, ..., U\}$. Since a total of $10*10=100$ losses total were suffered by teams in Y and $\binom{10}{2}=45$ losses were suffered by teams in Y from teams in Y, we have $100-45=55$ losses suffered by teams in Y from teams in X. Hence, for each of these $55$ losses, there is exactly one set of three teams that includes A that satisfies the problem conditions. Thus, the answer is $\frac{55\cdot 21}{3}=\boxed{385}$.
## Solution 3 (very risky if you're running out of time)
Note that there are $21$ teams total and $\binom{21}{3}=1330$ ways to pick ${A,B,C}.$ The possible arrangements are one team beats the other two or they each win/lose equally (we want the second case). Approximately $\frac{1}{4}$ of all the arrangements satisfy the second case, and $\frac{1330}{4}=332.5,$ which is by far the closest to $\boxed{(A)}.$
2016 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 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 AMC 12 Problems and Solutions |
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