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### Library Home CBSE - IX (Change Class)
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Basic Constructions
In earlier classes you might have learnt to construct perpendicular bisector of a line segment, angles of 300,450 600, 900, 1200 and also to draw bisector of the given angle. An angle bisector is a ray, which divides an angle in to two equal parts. A line bisector is a line that cuts a line segment into two equal halves. A perpendicular bisector is a line, which divides a given line segment into two equal halves and is also perpendicular to the line segment.
To construct the bisector of a given angle.
Let’s consider angle DEF, we want to construct the bisector of angle DEF.
Steps of construction:
1. With E as centre and small radius draw arcs on the rays DE and EF.
2. Let the arcs intersect the rays DE and EF at G and H respectively.
3. With centres G and H draw two more arcs with the same radius such that they intersect at a point . Let the intersecting point be I
4. Now draw a ray with E as the starting point passing through I
5. EI is the bisector of the angle DEF
To construct a perpendicular bisector of a line segment
Lets consider the line segment as PQ. We have to construct the perpendicular bisector of PQ.
Steps of Construction:
1. Draw a line segment PQ.
2. With P as centre draw two arcs on either sides of PQ with radius more the half the length of the given line segment.
3. Similarly draw two more arcs with same radius from point Q such that they intersect the previous arcs at R and S respectively.
4. Join the Points R and S. RS is the required perpendicular bisector of the given line segment PQ.
To Construct an angle of 600 at the initial point of a given ray.
Let us take ray PQ with P as the initial point. We have to construct a ray PR such that it makes angle of 600 with PQ.
Steps of Constructions:
1. Draw a ray PQ
2. With P as centre draw an arc with small radius such that it intersects ray PQ at C.
3. With C as centre and same radius draw another arc to intersect the previous arc at D.
4. Draw a ray PR from point P through D. Hence the angle RPQ is equal to 60 degrees.
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# L.C.M concepts and formula
## L.C.M concepts and formula
One of the most prevalent quantitative aptitude topics asked in government tests is L.C.M. This is one of the most fundamental topics that students are already aware of before they begin studying for competitive exams.
Although the concepts of LCM remain the same, the kind of questions presented in exams may vary. L.C.M is used in all the topic of Quantitative Aptitude, this makes this topic very important.
2-3 word problems related to LCM are asked in exams, but candidates may also expect questions about data sufficiency and data interpretation related to LCM topic.
So, we are providing you with concepts and formulas for the LCM topic to help you prepare properly for the upcoming exams.
When we travel from one point to another, there will be three Quantity that defines this journey – LCM.
#### What is L.C.M (Least Common Multiple)
The least number which is exactly divisible by each one of the given numbers is called their L.C.M
Let’s see an example of L.C.M
Least Common Multiple of 3 and 5
Step-1 List the Multiples of each number,
The multiples of 3 are 3, 6, 9, 12, 15, 18, … etc
The multiples of 5 are 5, 10, 15, 20, 25, … etc
Step-2 Now find the first common value from these multiples this will be the L.C.M.
The first common multiple is 15. So the L.C.M of 3 and 5 is 15.
#### How to calculate the L.C.M.
Let’s find the LCM(10, 12, 15, 75)
Step-1 Write down your numbers in a row
10 12 15 75
Step-2 Divide the layer numbers by a prime number that is evenly divisible into two or more numbers in the layer and bring down the result into the next layer.
Step-3 If any number in the layer is not evenly divisible just bring down that number.
Step-4 Continue dividing layers by prime numbers. When there are no more primes that evenly divided into two or more numbers you are done.
Step-5 The LCM is the product of the numbers in the L shape, left column and bottom row. 1 is ignored.
LCM = 2 × 3 × 5 × 2 × 5
LCM = 300
Therefore, LCM(10, 12, 15, 75) = 300
Q1 Lets find the L.C.M of 12 and 15
Ans
L.C.M of 12 and 15 = 2 x 3 x 2 x 5 = 60
Q2 Find L.C.M of 5, 7, 14, and 20
Ans
L.C.M = 5 x 2 x 7 x 2 = 140
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# Lesson Explainer: Möbius Transformations Mathematics
In this explainer, we will learn how to interpret Möbius transformation in the complex plane.
Möbius transformations form an interesting class of the transformations of the complex plane. They have a number of deep properties and connections to other areas of mathematics, such as group theory, and even have a deep connection to relativity via Lorentz transformations. In this explainer, we will introduce Möbius transformations and consider some of their basic properties including the effect they have on lines and circles in the complex plane.
Before we consider Möbius transformations, we will recap some of the basic transformations of the complex plane.
### Basic Transformations of the Complex Plane
Let , where are constant.
1. The transformation represents a translation by the vector .
2. The transformation represents a dilation by scale factor and a counterclockwise rotation about the origin by .
These transformations have the property that they map straight lines to straight lines and circles to circles. In this explainer, we will consider a more general class of transformations that map circles and straight lines to circles and straight lines, possibly mapping a straight line to a circle and vice versa.
We will start by considering the effect of the reciprocal map .
### Example 1: Reciprocal Transformation
A transformation which maps the -plane to the -plane is defined by , where .
1. Find an equation for the image of under the transformation.
2. Find an equation for the image of .
3. Find a Cartesian equation for the image of .
4. Find a Cartesian equation for the image of .
To find an equation representing the image of a particular curve under the transformation , first express in terms of , and then substitute this into the equations defining the particular curve; then we can rearrange to get the expression.
Part 1
We begin by expressing in terms of as follows:
We can now substitute this into the equation to get
Using the properties of the modulus, we can rewrite this as
Rearranging the equation, we get
This is the equation of a circle of radius centered at the origin. We can represent the effect of this transformation visually as follows.
Part 2
Substituting into yields
Using the properties of the argument, we can rewrite this as
Since , we have
This is the half-line centered at the origin which makes an angle of with the positive real axis. We can represent the effect of this transformation visually as follows.
Part 3
Substituting into yields
To find a Cartesian equation, we begin by substituting into this equation as follows:
Multiplying the numerator and denominator by the complex conjugate of the denominator, we can rewrite the fraction as follows:
Hence,
Now we can manipulate this equation to put it into a standard form. To do this, we first multiply through by , which gives
Adding to both sides and dividing through by two gives
We can now complete the square in to get
Hence,
This represents a circle of radius centered at . We can visualize the effect of this transformation on the line as follows.
Part 4
Substituting into gives
Rewriting the subject of the modulus as a single fraction we have
Factoring out of the denominator, we have
We can now use the properties of the modulus to rewrite this as
Hence,
This is the equation of a circle. However, to find its Cartesian equation, we need to substitute into the equation as follows:
Squaring both sides yields
We can now use the definition of the modulus to rewrite this as
Expanding the parentheses, we get
Gathering together our like terms, we have
We now divide through by 3, which gives us
Finally, we can complete the square in to get
Hence,
This is the equation of a circle of radius centered at . We can visually represent the effect of this transformation on the circle as follows.
The previous example showed that the transformation maps some lines to circles and some circles to circles. By looking carefully at the previous example, we can piece together what the transformation is doing. The first part generalizes to tell us that the transformation maps a circle of radius centered at the origin to a circle of radius centered at the origin. The second part shows us that a ray from the origin gets mapped to another ray which has been reflected in the real axis. The third part shows us that lines which do not pass through the origin get mapped to circles which pass through the origin. Finally, the fourth part shows us that circles which do not pass through the origin get mapped to other circles. Putting these facts together and recalling some geometry, we can see that the transformation is a combination of inversion in the unit circle and reflection in the real axis.
We now turn our attention to the general definition of Möbius transformations.
Möbius transformations are the transformations we get by composing the following three basic types of transformation in the complex plane:
1. Translations: ,
2. Rotations and dilations: ,
3. Inversion with reflection in the real axis: .
Formally speaking, we define Möbius Transformations as follows.
### Möbius Transformations
A transformation of the form where and , is called a Möbius transformation.
In the definition of a Möbius transformation, we gave the restriction that . In the next example, we will explore why we impose this restriction.
### Example 2: Degenerate Case
Given that , simplify the expression where .
To use the property that , we can multiply the expression by to get
We can use the property and substitute for as follows:
Given that , we can cancel the expression from the numerator and denominator to get
As we saw in the previous example, we include the restriction that to avoid the trivial case where the function reduces to a constant function. We now consider what happens if we conform to Möbius transformations.
### Example 3: Composition of Möbius Transformations
Consider the two Möbius transformations and . Write an expression for the composition in terms of , , , , , , , and and determine whether the resulting transformation is a Möbius transformation.
To solve a problem like this, we simply need to substitute one equation into the other and work through the algebra. Hence, substituting into the formula defining , we have
We can simplify this expression by multiplying the numerator and denominator by as follows:
We now gather the terms and constant terms in the numerator and denominator to get
To confirm that this is a Möbius transformation, we need to check that . To do this, we expand the parentheses as follows:
Since and are Möbius transformations, we know that and ; therefore, , which implies that is a Möbius transformation.
The last example showed us the relationship between the four complex coefficients that define a Möbius transformation under composition. A careful examination of the relationship highlights an interesting connection between Möbius transformations and matrices. In particular, if we write the coefficients of in the matrix and the coefficients of in another matrix the coefficients of are the elements of the matrix . Furthermore, the condition that is simply the condition that the matrix is not singular. Although this relationship between matrices and Möbius transformations is interesting to explore in more depth, we will not explore this further here. Instead we will turn our attention to the effect of Möbius transformations on lines and circles in the complex plane.
### Example 4: Properties of Möbius Transformations
A transformation which maps the -plane to the -plane is given by , where .
1. Find the Cartesian equation of the image of under the transformation.
2. Find the Cartesian equation of the image of under the transformation.
Part 1
To find the equations of the image of , we first express in terms of . Starting from the definition of the transformation we multiply through by which gives
Gathering the terms of the left-hand side, we have
Dividing both sides of the equation by gives
We can now substitute this into the equation , to find an equation for its image in the -plane. Hence,
Factoring out from the numerator, we have
Using the properties of the modulus, we can rewrite this as
|3𝑖||𝑤−2𝑖||𝑤−𝑖|=3. (1)
Rearranging and using the fact that , we have
This is the equation of a line; in particular, it is the perpendicular bisector of the line segment between and . We can write the Cartesian equation for this line as . It is also possible to derive this equation by substituting into the equation as we will show. Starting from the equation we can square both sides to get
Using the definition of the modulus, we can rewrite this as
Canceling the terms and expanding the parentheses, we have
Canceling and rearranging, we get the equation for the line as . We can represent the effect of this Möbius transformation on this circle visually as follows.
Part 2
We can use the calculations we made in the previous part of the question to simplify finding the equation of the image of . In particular, we can use the left-hand side of equation (1), since this is equal to , and set it equal to 1 as follows:
Multiplying by and simplifying, we have
This is the equation of a circle. To find its Cartesian equation, we substitute into the equation to get
Squaring both sides of the equation yields
Using the definition of the modulus, we can rewrite this as
We now expand the parentheses to get
Rearranging, we have
Dividing through by 8 gives
We can now complete the square in to get
Hence, which represents a circle of radius centered at . We can represent the effect of this Möbius transformation on this circle visually as follows.
As we can see, much like the transformation , Möbius transformations map circles to both lines and circles. In the final example, we will consider what happens to lines that pass through the point where the denominator vanishes—we often refer to this point as a poll of the transformation.
### Example 5: Properties of Möbius Transformations
A transformation, , which maps the -plane to the -plane is given by , where .
1. Find a Cartesian equation for the image of the imaginary axis under the transformation .
2. Hence, find the image of the region under the transformation .
Part 1
We begin by expressing in terms of . Starting from the equation defining the transformation we multiply both sides by which give us
Gathering the terms on the left-hand side, we have
Hence, by dividing by , we have
Since we would like to consider the image of the imaginary axis (), we would need to find an expression for the real part of . To find this, we substitute in as follows: Gathering the real and imaginary parts in the numerator and denominator, we have
Multiplying the numerator and denominator by the complex conjugate of the denominator, we get
Expanding the parentheses in the numerator, we have
Hence,
We can now set the real part to be equal to zero to get
0=(4−𝑣)(𝑢−2)+𝑢(𝑣−1)(𝑢−2)+(𝑣−1). (2)
Multiplying by yields
Finally we can expand these parentheses to get
Hence, which is equivalent to the equation
Therefore, we can represent the effect of the transformation on the imaginary axis as follows.
Part 2
We would like to find the image of the region defined by . We have already found the equation of the boundary line in the previous part, we just need to consider the direction of the inequality. Looking back to equation (2), this is the point where we used the fact that the real part was equal to zero. Instead, we can use the fact that the real part is greater than or equal to 0. Hence, replacing the equality sign with an inequality sign, we get
Since the denominator is positive, we can multiply both sides of the equation by the denominator and it will not affect the direction of the inequality. Hence,
Working through the same algebra as in the last part, we get which we can rearrange to
We can represent this on the -plane as follows.
### Key Points
• Möbius transformations are a special type of transformations of the complex plane that may transform lines and circles to other circles and lines.
• A transformation of the form where and , is called a Möbius transformation.
• Composition of Möbius transformations results in another Möbius transformation.
• There are some interesting connections between Möbius transformations and other areas of mathematics.
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# Class 8 RD Sharma Solutions – Chapter 15 Understanding Shapes Polygons – Exercise 15.1
Last Updated : 18 May, 2021
### Question 1. Draw a Rough diagram to illustrate the Following.
(i) Open curve
(ii) Closed curve
Solution:
i) Open curve: A curve in which the beginning point and the ending point do not meet each other is known as an open curve.
ii) Closed curve: A curve in which the beginning point and endpoint meet at each other or cuts each other is known as a closed curve.
### Question 2. Classify the following as open or closed.
Solution:
• open curve: A curve in which starting point and ending point are different or do not cut each other .
• closed curve: A curve in which starting point and ending point are same and cut each other.
Using the above definition we can classify the given figures as follows :
i) open curve (as both stating and ending points are different)
ii) closed curve (as both the points are same )
iii) closed curve (as both the points cut each other)
iv) open curve (as both starting point and ending point are different)
v) open curve (as both starting point and ending point are different)
vi) closed curve (as both starting point and ending point meet at same point)
### Question 3. Draw a polygon and shade it’s interior. Also draw its diagonals, if any:
Solution:
In polygon ABCD, AC and BD are the diagonals of a polygon
### Question 4. Illustrate if possible, each one of the following with a rough diagram.
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
Solution:
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
A polygon with two sides is not possible because, a polygon should have minimum three sides.
### Question 5. Following are some figures: Classify each of these figures on the basis of the following:
(i) Simple curve
(ii) Simple closed curve
(iii) Polygon
(iv) Convex polygon
(v) Concave polygon
(vi) Not a curve
Solution:
(i) It is a Simple Closed curve and a concave polygon. This is a simple closed curve and as a concave polygon all the vertices are not pointing outwards.
(ii) It is a Simple closed curve and a convex polygon. This is a simple closed curve and as a convex polygon all the vertices are pointing outwards.
(iii) It is Not a curve and hence it is not a polygon.
(iv) It is Not a curve and hence it is not a polygon.
(v) It is a Simple closed curve but not a polygon.
(vi) It is a Simple closed curve but not a polygon.
(vii) It is a Simple closed curve but not a polygon.
(viii) It is a Simple closed curve but not a polygon.
### Question 6. How many diagonals does each of the following have?
(ii) A regular hexagon
(iii) A triangle
Solution:
For a convex quadrilateral we shall use the formula n(n-3)/2
So, number of diagonals = 4(4-3)/2 = 4/2 = 2
A convex quadrilateral has 2 diagonals.
(ii) A regular hexagon
For a regular hexagon we shall use the formula n(n-3)/2
So, number of diagonals = 6(6-3)/2 = 18/2 = 9
A regular hexagon has 9 diagonals.
(iii) A triangle
For a triangle we shall use the formula n(n – 3)/2
So, number of diagonals = 3(3 -3)/2 = 0/2 = 0
A triangle has no diagonals.
### Question 7. What is a regular polygon? State the name of a regular polygon of
(i) 3 sides
(ii) 4 sides
(iii) 6 sides
Solution:
Regular Polygon: A regular polygon is an enclosed figure. In a regular polygon minimum sides are three.
(i) 3 sides
A regular polygon with 3 sides is known as Equilateral triangle.
(ii) 4 sides
A regular polygon with 4 sides is known as Rhombus.
(iii) 6 sides
A regular polygon with 6 sides is known as Regular hexagon.
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?> Identity Property Worksheet | Problems & Solutions
# Identity Property Worksheet
Identity Property Worksheet
• Page 1
1.
Which symbol would you use to balance the number sentence?
a. I don′t know. b. × c. - d. +
#### Solution:
The product of any number and 1 is the number itself.
So, 8 × 1 = 8.
2.
Multiply:
a. 33 b. 31 c. 32 d. 1
#### Solution:
The product of any number and 1 is the number itself.
So the product of 32 and 1 is 32.
3.
Select the expression that is an example for the identity property of addition.
a. 4 + 0 = 4 b. 4 × 0 = 0 c. 4 × 1 = 4 d. 4 + 3 = 3 + 4
#### Solution:
According to identity property of addition, the sum of any number and zero is the same number.
So, among the number sentences listed, only '4 + 0 = 4' is an example of the identity property of addition.
4.
Pick out the number sentence that correctly shows the use of the identity property of multiplication.
a. 8 + 1 = 9 b. 8 × 0 = 0 c. 8 × 1 = 8 d. 8 + 0 = 8
#### Solution:
According to the identity property of multiplication, the product of any number and 1 is the same number.
So, among the number sentences listed, only '8 × 1 = 8' correctly shows the use of the identity property of multiplication.
5.
Choose the appropriate symbol to balance the number sentence.
a. > b. = c. < d. +
#### Solution:
The product of any number and zero is zero.
So, the product of 3800 and zero is zero.
The product of 5 and zero is also zero.
So, 3800 × 0 is equal to 5 × 0.
6.
Choose the operation that balances the equation.
a. - b. + c. × d. ÷
#### Solution:
The product of any number and zero is zero.
So, 3200 × 0 = 0.
7.
a. 0 and 1 cannot be added b. 10 c. 1
#### Solution:
The sum of any number and zero is the same number.
So, 1 + 0 is equal to 1.
8.
a. 55 b. 550 c. none of these
#### Solution:
The sum of any number and zero is the same number.
So, 55 + 0 is equal to 55.
9.
Divide:
14 ÷ 1
a. 13 b. 1 c. 14
#### Solution:
The quotient of any number divided by 1 is the same number, with no remainder.
So, 14 ÷ 1 = 14.
10.
Multiply:
9 × 1
a. 1 b. 10 c. 9
#### Solution:
The product of any number and 1 is the number itself.
So the product of 9 and 1 is 9.
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# RS Aggarwal Class 8 Math Ninth Chapter Percentage Exercise 9A Solution
## EXERCISE 9A
(1) Express each of the following as a fraction:
(2) Express each of the following as a decimal:
(3) Express each of the following s a percentage:
(4) Convert the ratio 4 : 5 to percentage.
(5) Express 125% as a ratio.
(7)(i) What per cent of 150 is 96?
(ii) What per cent of 5 kg is 200 g?
(iii) What per cent of 2 litres is 250 ml?
(9) If 16% of a number is 72, find the number.
Solution: Let the required number be x. Then,
16% of x = 72
Hence, the required number is 450.
(10) A man saves 18% of his monthly income. If he saves Rs 3780 per month, what is his monthly income?
Solution: Let his monthly income be Rs x.
Hence, the monthly income of the man is Rs 21000.
(11) A football team wins 7 games, which is 35% of the total games played. How many games were played in all?
Solution: Let the number of playing game be x.
Hence, the team 20 games were played in all.
(12) Amit was given an increment of 20% on his new salary is Rs 30600, what was his salary before the increment?
Solution: Let Amit’s salary before his increment be Rs x.
(13) Sonal attended her school on 204 days in full year. If her attendance is 85%, find the number of days on which the school was opened.
Solution: Let the numbers of days on which the school was opened be x days.
Then, (85/100)×x
⇒ 85x = 20400
⇒ x = 240
Hence, the number of days which Sonal’s school was opened is 240 days.
(14) A’s income is 20% less than that B. By what per cent is B’s income more than A’s?
Solution: Let B’s income be Rs 100.
Then, A’s income = Rs (100 – 20) = Rs 80.
If A’s income is Rs 80, then B’s income = Rs 100.
If A’s income is Rs 100, then B’s income = Rs [(100/80)×100]=Rs 125.
∴ B’s income more than A’s income by (125 – 100) = 25%.
(15) The price of petrol goes up by 10%. By how much per cent must a motorist reduce the consumption of petrol so that expenditure on it remains unchanged?
Solution: Let the consumption of petrol originally be 1 and let its cost be Rs 100.
New cost of 1 unit = Rs 110.
Now, Rs 110 yield 1 unit of petrol.
(16) The population of a town increases by 8% annually. If the present population is 54000, what was it a year ago?
Solution: Let the population of the town a year ago be x.
(17) The value of a machine depreciates every year by 20%. If the present value of the machine be Rs 160000, what was its value last year?
Solution: Let the value of the machine last year be Rs x.
(18) An alloy contains 40% copper, 32% nickel and rest zinc. Find the mass of zinc in one kg of the alloy.
Solution: mass of the alloy = 1 kg = 1000 g
Sum of percentage of the copper and nickel = (40 + 32) % = 72%
∴ Percentage of zinc = (100 – 72)% = 28%
∴ Mass of the zinc in 1 kg alloy = [(28/100)×1000]g=280 g.
(19) Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in his food daily, find in calories the amount of each of these in his daily food intake.
(20) Gunpowder contains 75% nitre and 10% sulphur. Find the amount of gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.5 kg sulphur?
Solution: Let the amount of the gunpowder be x.
Now, x kg be the amount of gunpowder containing 9 kg of nitre.
Hence, 12 kg of gunpowder contains 9 kg of nitre.
Now, x kg be the amount of gunpowder containing 2.5 kg of sulphur.
Hence, 25 kg of gunpowder contains 2.5 kg sulphur.
(21) Divide Rs 7000 among A, B and C such that A gets 50% of what B gets 50% of what C gets.
Solution: Let Rs x be the amount of money received by C.
⇒ 175x = 700000
⇒ x = 40000
∴ C get Rs 4000.
(22) Find the percentage of pure gold in 22 carat gold, if 24-carat gold is 100% pure.
(23) The salary of an officer is increased by 25%, By what per cent should the new salary be decreased to restore the original salary?
Solution: Let the salary be Rs 100.
⇒ – x = 80 – 100
⇒ – x = – 20
⇒ x = 20
Therefore, the new salary be decreased to restore the original salary by 20%.
For more exercise solution, Click below –
Updated: May 30, 2022 — 2:20 pm
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# Computing the area under a curve
Page 1 / 2
Basic Numerical Integration with MATLAB
This chapter essentially deals with the problem of computing the area under a curve. First, we will employ a basic approach and form trapezoids under a curve. From these trapezoids, we can calculate the total area under a given curve. This method can be tedious and is prone to errors, so in the second half of the chapter, we will utilize a built-in MATLAB function to carry out numerical integration.
## A basic approach
There are various methods to calculating the area under a curve, for example, Rectangle Method , Trapezoidal Rule and Simpson's Rule . The following procedure is a simplified method.
Consider the curve below:
Each segment under the curve can be calculated as follows:
$\frac{1}{2}({y}_{0}+{y}_{1})\mathrm{\Delta x}+\frac{1}{2}({y}_{1}+{y}_{2})\mathrm{\Delta x}+\frac{1}{2}({y}_{2}+{y}_{3})\mathrm{\Delta x}$
Therefore, if we take the sum of the area of each trapezoid, given the limits, we calculate the total area under a curve. Consider the following example.
Given the following data, plot an x-y graph and determine the area under a curve between x=3 and x=30
Index x [m] y [N]
1 3 27.00
2 10 14.50
3 15 9.40
4 20 6.70
5 25 5.30
6 30 4.50
First, let us enter the data set. For x, issue the following command x=[3,10,15,20,25,30]; . And for y, y=[27,14.5,9.4,6.7,5.3,4.5]; . If yu type in [x',y'] , you will see the following tabulated result. Here we transpose row vectors with ' and displaying them as columns:
ans = 3.0000 27.000010.0000 14.5000 15.0000 9.400020.0000 6.7000 25.0000 5.300030.0000 4.5000
Compare the data set above with the given information in the question .
To plot the data type the following:
plot(x,y),title('Distance-Force Graph'),xlabel('Distance[m]'),ylabel('Force[N]'),grid
The following figure is generated:
To compute dx for consecutive x values, we will use the index for each x value, see the given data in the question .:
dx=[x(2)-x(1),x(3)-x(2),x(4)-x(3),x(5)-x(4),x(6)-x(5)];
dy is computed by the following command:
dy=[0.5*(y(2)+y(1)),0.5*(y(3)+y(2)),0.5*(y(4)+y(3)),0.5*(y(5)+y(4)),0.5*(y(6)+y(5))];
dx and dy can be displayed with the following command: [dx',dy'] . The result will look like this:
[dx',dy'] ans =7.0000 20.7500 5.0000 11.95005.0000 8.0500 5.0000 6.00005.0000 4.9000
Our results so far are shown below
x [m] y [N] dx [m] dy [N]
3 27.00
10 14.50 7.00 20.75
15 9.40 5.00 11.95
20 6.70 5.00 8.05
25 5.30 5.00 6.00
30 4.50 5.00 4.90
If we multiply dx by dy, we find da for each element under the curve. The differential area da=dx*dy, can be computed using the 'term by term multiplication' technique in MATLAB as follows:
da=dx.*dy da =145.2500 59.7500 40.2500 30.0000 24.5000
Each value above represents an element under the curve or the area of trapezoid. By taking the sum of array elements, we find the total area under the curve.
sum(da) ans =299.7500
The following illustrates all the steps and results of our MATLAB computation.
x [m] y [N] dx [m] dy [N] dA [Nm]
3 27.00
10 14.50 7.00 20.75 145.25
15 9.40 5.00 11.95 59.75
20 6.70 5.00 8.05 40.25
25 5.30 5.00 6.00 30.00
30 4.50 5.00 4.90 24.50
299.75
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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A set of equations for which we get a common solution is known as system of equations. System of equations can be either linear or non-linear.
Two set of equations will have two variables, three set of equations will have three variables and so on. System of equations can have any number of variables which can be linear or non-linear.
## Definition of System of Equations
A system of equations is a collection of two or more unknowns where we try to find values for each of the unknowns which satisfies every equation in the system. Linear System of Equations
Linear equations use only linear functions and operations. Exponents will not be higher than one. When graphed, it will be a straight line.
Example: x + y + z = 6
They are mostly used in analysis.
Non-Linear System of Equations
It can consist of trignometric functions, multiplication or division by variables. One or more of the variables may contain an exponent larger than one. When graphed it represents some sort of curves.
Example: $x ^{2}$ + $y^{2}$ + z = 4
Non-linear equations dominate the realm of higher math and science.
A general system of m linear equations with n unknowns can be written as
$a_{11}x_{1}$ + $a_{12}x_{2}$ + .........+ $a_{1n}x_{n}$ = $b_{1}$
$a_{21}x_{1}$ + $a_{22}x_{2}$ + .........+ $a_{2n}x_{n}$ = $b_{2}$
...... ........
....... .......
$a_{m1}x_{1}$ + $a_{m2}x_{2}$ + .........+ $a_{mn}x_{n}$ = $b_{m}$
## Types of Systems of Equations
There are three types of system of linear equations. They are:
• Independence
• Consistency and
• Homogeneous
Independence: A system of linear equations are independent if there is no possibility of deriving equation from the others. When the equations are independent each equation will contain new information and removing any equation will lead to change in the size and solution set.
Consistency: A system of linear equations are consistent if they have a common solution else it will be inconsistent.
Example: -2x + 3y = 8
3x - y = -5
Given above is an example for consistent system of linear equations.
Homogeneous
: A system of linear equations are homogeneous if they can be derived algebraically with the same set of variables from the other else they are non-homogeneous.
Example: $x_{1}$ +$x_{2}$ - $2x_{3}$ = 0
$3x_{1}$ +$2x_{2}$ + $4x_{3}$ = 0
$4x_{1}$ +$3x_{2}$ + $3x_{3}$ = 0
## How To Solve Systems of Equations
System of equations can contain any number of variables and equations. There are three main methods to solve them, and you can use any one of them.
• Substitution method.
• Graphical method.
• Elimination method.
## Solving Systems of Equations by Substitution
There are two necessary conditions in solving system of equations by substitution
1. Number of equations should be equal to number of variables. If there are two variables, there must be two equations; 3 Variables = 3 Equations etc.,
2. One of the equations can easily be solved for one variable.
Given below are the steps to be considered for solving system of equations using substitution method.
1. Select one equation and isolate one variable and name it as first equation.
2. In second equation substitute the value of the first equation and solve for the variable.
3. Again by substituting in the given equations the value of the second variable is found. This process is continued until the values for the given variables are known.
### Solved Example
Question: x + y =3
y - 2x = 5
Solution:
The given equations are : x + y = 3 .................(1)
y - 2x = 5 ................(2)
Step -1 : From the first equation we isolate x (y can also be isolated).
$\Rightarrow$ x = 3 - y.
Step -2 : Substitute x = 3 - y in the second equation.
$\Rightarrow$ y - 2(3 - y) = 5
Step -3 : Solve for y
y - 6 + 2y = 5
$\Rightarrow$ 3y = 11
$\Rightarrow$ y = $\frac{11}{3}$
Step -4 : As the value of y is known. The value of x can be easily found now by substituting in any of the given equations.
Substituting in x = 3 - y we get y as
x = 3 - $\frac{11}{3}$
3x = 9 - 11
x = $\frac{-2}{3}$
Therefore, x = $\frac{-2}{3}$ and y = $\frac{11}{3}$.
Verification can be done by plugging the values of x and y in one of the given equations.
## Solving Systems of Equations by Graphing
When there are two equations graphical method is one of the easiest and most convenient method to solve system of equations. For complex numbers it is not reliable and is not the preferred method. In graphical method there can be one, none or infinitely many solutions. If for the given system of equations, we can graph a straight line then it possible to solve them graphically.
Graph the two lines and look for the point where they intersect(cross). The intersection point is termed as the solution.
### Solved Example
Question: Using graphical method solve the system of equations:
x + 2y = 3
4x + 5y = 6
Solution:
In order to graph them we solve each equation for y.
Consider the first equation, x + 2y = 3
Solve for y
$\Rightarrow$ 2y = 3 - x
y = $\frac{3}{2}$ -$\frac{x}{2}$
The second equation is
4x + 5y = 6
$\Rightarrow$ 5y = 6 - 4x
$\Rightarrow$ y = $\frac{6}{5}$ - $\frac{4}{5}$ x
Now plot y = $\frac{6}{5}$ -$\frac{4}{5}$ x and y = $\frac{3}{2}$ -$\frac{x}{2}$
We get
From the above graph we see that the two lines intersect at the point (-1, 2).
## Solving Systems of Equations by Elimination
The elimination method of solving systems of equations is also called the addition method. To solve a system of equations by elimination we transform the system such that one variable "cancels out".
### Solved Example
Question: Solve the system of equations using elimination method
6x - 5y = 10
x + 5y = 12
Solution:
For the given system of equations we can easily cancel out the y terms by adding two equations together.
6x - 5y = 10
x + 5y = 12
_________________
7x + 0 = 22
From the equation 7x = 22 we can now find x.
Therefore x= $\frac{22}{7}$
To find the value of y substitute x =$\frac{22}{7}$ in one of the given equations.
Substituting x = $\frac{22}{7}$ in x + 5y = 12, we get
$\frac{22}{7}$ + 5y = 12
Simplify and find the value of y
5y = 12 - $\frac{22}{7}$
$\Rightarrow$ 35y = 84 -22
$\Rightarrow$ y = $\frac{62}{35}$
Therefore, The solution is (x, y) = ( $\frac{22}{7}$ , $\frac{62}{35}$ )
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# How do you factor m^2-10my+25y^2?
##### 1 Answer
May 11, 2016
$\left(m - 5 y\right) \left(m - 5 y\right) = {m}^{2} - 10 m y + 25 {y}^{2}$
#### Explanation:
One of our first clues in factoring this expression is that the first and last terms are perfect squares:
$\sqrt{{m}^{2}} = m$
and
$\sqrt{25 {y}^{2}} = 5 y$
In the most general case, we are looking for a solution in the form of
$\left(a m + b y\right) \left(c m + \mathrm{dy}\right) = {m}^{2} - 10 m y + 25 {y}^{2}$
$a c {m}^{2} + \left(b c + a d\right) m y + b {\mathrm{dy}}^{2} = {m}^{2} - 10 m y + 25 {y}^{2}$
From the first term we can see that $a \cdot c = 1$. Assuming $a$ and $c$ are integers, they must be both either $+ 1$ or $- 1$. Let's make them $+ 1$ for now and continue (it actually doesn't matter which we choose at this point - can you see why?):
${m}^{2} + \left(b + d\right) m y + b {\mathrm{dy}}^{2} = {m}^{2} - 10 m y + 25 {y}^{2}$
From the last term we see that $b \cdot d = 25$. From the second term, we must also have $\left(b + d\right) = - 10$. The obvious solution for this is to have $b = d = - 5$. Therefore we have the solution
$\left(m - 5 y\right) \left(m - 5 y\right) = {m}^{2} - 10 m y + 25 {y}^{2}$
Once we have worked through this type of problem, we could have probably started by guessing this solution from the first two observations.
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Section 14.4 Trees
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# Section 14.4 Trees - PowerPoint PPT Presentation
## Section 14.4 Trees
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##### Presentation Transcript
1. Section 14.4Trees
2. What You Will Learn • Trees • Spanning Trees • Kruskal’s Algorithm
3. Tree • A tree is a connected graph in which each edge is a bridge.
4. Examples Not Trees • Trees
5. Spanning Tree • A spanning tree is a tree that is created from another graph by removing edges while still maintaining a path to each vertex.
6. Example 3: A Spanning Tree Problem • Schoolcraft College is considering adding awnings above its sidewalks to help shelter students from the snow and rain while they walk between some of the buildings on campus. A diagram of the buildings and the connecting sidewalks where the awnings are to be added is on the next slide.
7. Example 3: A Spanning Tree Problem • Originally, the president of the college wished to have awnings placed over all the sidewalks shown, but that was found to be too costly. Instead, the president has proposed to place just enough awnings over a select number of sidewalks so that, by moving from building to building, students would still
8. Example 3: A Spanning Tree Problem • be able to reach any location shown without being exposed to the elements. • a) Represent all the buildings and sidewalks shown with a graph. • b) Create three different spanning trees from this graph that would satisfy the president’s proposal.
9. Example 3: A Spanning Tree Problem • Solution • a) Using letters to represent the building names, vertices to represent the buildings, andedges to representthe sidewalksbetween buildings, wegenerate this graph.
10. Example 3: A Spanning Tree Problem • Solution • b) To create a spanning tree we remove nonbridge edges until a tree is created.
11. Example 3: A Spanning Tree Problem • Solution • b) Here are two more possibilities.
12. Minimum-cost spanning tree • A minimum cost spanning tree is the least expensive spanning tree of all spanning trees under consideration.
13. Kruskal’s Algorithm • To construct the minimum-cost spanning tree from a weighted graph: • 1. Select the lowest-cost edge on the graph. • 2. Select the next lowest-cost edge that does not form a circuit with the first edge. • 3. Select the next lowest-cost edge that does not form a circuit with the previously selected edges.
14. Kruskal’s Algorithm 4. Continue selecting the lowest-cost edges that do not form circuits with the previously selected edges. 5. When a spanning tree is complete, you have the minimum-cost spanning tree.
15. Example 7 • Schools in Budville, Fairplay, Happy Corners, Kieler, Louisburg, and Sinsinawa, Wisconsin, all wish to establish a fiber-optic computer network to share information and to obtain Internet access. The most efficient method of establishing such a network would be to install fiber-optic cable along roadsides.
16. Example 7 • The weighted graph shows the distance in miles between schools along existing roads.
17. Example 7 • a) Determine the shortest distance to link these six schools. • b) The cost to install fiber-optic cable is \$1257 per mile. What is the minimum cost to install the fiber-optic cable along the roadsides determined in part (a)?
18. Example 7 • Solution • a) We are seeking the minimum-cost spanning tree. • Use Kruskal’s algorithm. • Select edge HB, 1 mi; • edge BL, 1.5 mi; • edge FS, 2 mi
19. Example 7 Solution Selecting HL, 2.5 mi creates a circuit between H, B, L; so we must select LF, 3 mi. Finally, select edge KL, 3.5 mi.
20. Example 7 • Solution • According to Kruskal’s algorithm, this figure shows the minimum-cost spanning tree. • Place the fiber-optic cable along this path.
21. Example 7 • Solution • b) From the figure, there are • 1 + 1.5 + 3.5 + 3 + 2 = 11miles of fiber-opticcable needed. • At \$1257 per mile,the cost is • \$1257 × 11 = \$13,827.
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English
# Solving Cubic Equations With Worked Examples
Cubic equation is a third degree polynomial equation. To solve this equation, write down the formula for its roots, the formula should be an expression built with the coefficients a, b, c and fixed real numbers using only addition, subtraction, multiplication, division and the extraction of roots. Here given are worked examples for solving cubic equations.
#### Example 1:
Let us consider the problem with a cubic equation 5x3 + 4x2+ 2x + 2
###### Solution:
We can calculate the value using the given formula.
#### Cubic Equation Formula:
x1=(- term1 + r13 x cos(q3/3) ) x2=(- term1 + r13 x cos(q3+(2 x Π)/3) ) x3=(- term1 + r13 x cos(q3+(4 x Π)/3) ) Where,
discriminant(Δ) = q3 + r2 term1 = √(3.0) x ((-t + s)/2) r13= 2 x √(q) q = (3c- b2)/9 r = -27d + b(9c-2b2) s = r + √(discriminant) t = r - √(discriminant)
Substituting the values in the formula,
discriminant(Δ) = q3 + r2
= [(3x2 - 42) / 9]3 + [-27x2 + 3(9x2 - 2x32)]2
= [(6-16)/9]3 + [-54 + 3(18 - 62)]2
= [-10 / 9]3 + [-54 + 3(18 - 36)]2
= [-1.11111]3 + [-54 + 3(-18)]2
= -1.3717 + [-54 - 54]2
= -1.3717 + -11664
(Δ) = -11665.3717
Find the value of s,
s = r + √(discriminant)
= -108 + √-11665.3717
Hence, s = -108 + 108.0063502762684783i
Find the value of t,
t = r - √(discriminant)
= -108 - √-11665.3717
Hence, t = -108 - 108.0063502762684783i
Find the value of term1,
term1 = √(3.0) x ((-t + s)/2)
= √(3.0) x ((-(-108 - 108.0063502762684783i) + (-108 + 108.0063502762684783i)/2)
= 1.732 x ((108 + 108.0063502762684783i - 108 - 108.0063502762684783i) / 2)
Hence, term1 = 1.732 x 0 = 0
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# Question 4 & 5, Review Exercise 1
Solutions of Question 4 & 5 of Review Exercise 1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
If ${{z}_{1}}=2-i$,${{z}_{2}}=1+i,$ find $|\dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}|$.
\begin{align}{{z}_{1}}&=2-i,\\ {{z}_{2}}&=1+i,\\ \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}&=\dfrac{\left( 2-i \right)+\left( 1+i \right)+1}{\left( 2-i \right)-\left( 1+i \right)+1}\\ &=\dfrac{2-i+1+i+1}{2-i-1-i+1}\\ &=\dfrac{4}{2-2i}\\ &=\dfrac{2}{1-i}\\ &=\dfrac{2}{1-i}\times \dfrac{1+i}{1+i}\\ &=\dfrac{2\left( 1+i \right)}{1+1}\\ &=\dfrac{2\left( 1+i \right)}{2}\\ &=\left( 1+i \right)\end{align} Now
\begin{align}|\dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}|&=\sqrt{{{1}^{2}}+{{1}^{2}}}\\ &=\sqrt{2}\end{align}
Find the modulus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$.
\begin{align}\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}&=\dfrac{{{\left( 1+i \right)}^{2}}-{{\left( 1-i \right)}^{2}}}{1+1}\\ &=\dfrac{\left( 1+2i-1 \right)-\left( 1-2i-1 \right)}{2}\\ &=\dfrac{2i+2i}{2}\\ &=2i\end{align} Modulus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$ is
\begin{align}|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}|&=\sqrt{{{2}^{2}}}\\ |\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}|&=\sqrt{4}\\ |\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}|&=2\end{align}
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Exploring Box Plots - PowerPoint PPT Presentation
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Exploring Box Plots. Using the Key components of the Box plot to compare variability . Bell Ringer (review - calculate median). Ashley’s Dress shop posted sales for one week of \$874, \$1471, \$1275, \$1078, \$993, \$471, and \$1125. Find the median sales for the week.
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Exploring Box Plots
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Exploring Box Plots
Using the Key components of the Box plot to compare variability
Bell Ringer(review - calculate median)
Ashley’s Dress shop posted sales for one week of \$874, \$1471, \$1275, \$1078, \$993, \$471, and \$1125. Find the median sales for the week.
Ashley’s Dress shop posted sales for one week of \$874, \$1471, \$1275, \$1078, \$993, \$471, and \$1125. Find the median sales for the week.
Solution:
Step 1: Order the numbers in ascending order
\$471 \$874 \$993 \$1078 \$1230 \$1657
Step 2: Since there are 6 numbers, we must average the two middle numbers todeterminethe median value.
\$471 \$874 \$993\$1078 \$1230 \$1657
Step3: Add the 3rd and 4th numbers.
\$993 + \$1078 = \$2071
Step 4: Divide that answer by 2
\$2071/2 =
Step 5: Solution
Ashley’s Dress shop weekly median total is \$1035.50 dollars.
IntroductionHow to create Box Plots
Lesson Objectives
Learning Goals -
*Calculate and identify the values of the key components of a box plot
*Construct and compare multiple data sets using Box Plots
IntroductionHow to create Box Plots
Key Terms
• Minimum value
• Maximum Value
• Range
• Median value
• Lower Quartile (Q1)
• Upper Quartile (Q3)
• Interquartile Range (IQR)
The girls basketball team sold cookies for a fundraiser. The members of the team sold the following amount of cookies: 7, 13, 11, 5, 9, 19, 4, and 2. The coach wants to represent this data using a box plot.
Let’s create a box plot.
Step 1: Order the data
2, 4, 5, 7, 9, 11, 13, 19
Step 2: Separate the value in two equal groups and label them.
2, 4, 5, 7 9, 13, 11, 19
The girls basketball team sold cookies for a fundraiser. The members of the team sold the following amounts of cookies: 7, 13, 11, 5, 9, 19, 4, and 2. The coach wants to represent this data using a box plot.
Step 3: Label each group of values
Q1: 2, 4, 5, 7 M: 7, 9 Q3: 9, 11, 13, 19
Step 4: Determine the median, lower quartile, and upper quartile values. Solve for the Interquartile Range (IQR)
Q1 = 4.5 M = 8 Q3 = 12
IQR = Q3-Q1 = 10.5-4.5 = 6
Lets try it together using the steps that were outlined
The list below shows the ages of the children in the Alexander Family. Create a Box Plot.
1, 14, 5, 10, 3, 7, 12, 17
Step 1: Order the data in ascending order
1, 3, 5, 7, 10, 12, 14, 17
Step 2: Separate the data into equal groups
1, 3, 5, 7 10, 12, 14, 17
The list below shows the ages of the children in the Alexander Family. Find the Q1, Q3, median and IQR values of the different ages.
1, 14, 5, 10, 3, 7, 12, 17
Step 3: Label the groups
Q1: 1, 3, 5, 7 M: 7, 10 Q3: 10, 12, 14, 17
Step 4: Determine the median value for each of the three groups and the Interquartile Range (IQR)
Q1 = 4 M = 8.5 Q3 = 13 IQR = 9
Using the previous data from the Girls basketball cookie fundraiser, Create a Box Plot
2, 4, 5, 7, 9, 11, 13, 19
Step 1: Plot the points on a real number line
Using the previous data from the Girls basketball cookie fundraiser, Create a Box Plot
2, 4, 5, 7, 9, 11, 13, 19
Step 2:Q1 = 4.5 M = 8 Q3 = 12 IQR = 6
Step 3: Identify the Q1, M, and Q3 values on the number line
Step 4: Create the Box Plot
Lets try it together using the steps that were outlined
Using the ages of the children in the Alexander Family, create a Box Plot
1, 14, 5, 10, 3, 7, 12, 17
Step 1: Plot the points on a real number line
Using the ages of the children in the Alexander Family, create a Box Plot
1, 14, 5, 10, 3, 7, 12, 17
Step 2:
Q1 = 4 M = 8.5 Q3 = 13 IQR = 9
Step 3:
Identify the Q1, M, and Q3 values on the number
Step 4: Create the Box Plot
The precipitation data for the percentage of rainfall in the city of Miami during the month of May is listed below. Create a
Box Plot to display the range of the data
29, 17, 36, 10, 53, 20, 48, 42, 37, 60
Remember to use your steps when solving for the key components of the box plot.
Q1 = 20 M = 36.5 Q3 = 48 IQR = 28
Comparing two Box Plots
The Box plot displays the average grades for students in Mr. Gayle’s and Mrs. Smith’s Reading Class. Box A represents Mr. Gayle and Box B represents Mrs. Smith.
Compare the all of the components of the two box plots. What conclusions can you make?
Create two box plots and make comparisons.
Let’s try this one as a group
Box 1 will represent the number of letters in the first name.
Box 2 will represent the number of letters in the last name.
Use this data to create a box plot in your journals. (Remember to use your step as you begin to get your components.
Compare the components of the box plot, write your result in your journals.
Complete the Independent Practice worksheet.
The assignment must be completed on your own
Raise your hand if you need additional assistance while completing the activity
Closure
What did I learn today?
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# 2003 AMC 12B Problems/Problem 21
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
An object moves $8$ cm in a straight line from $A$ to $B$, turns at an angle $\alpha$, measured in radians and chosen at random from the interval $(0,\pi)$, and moves $5$ cm in a straight line to $C$. What is the probability that $AC < 7$?
$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$
## Solution 1 (Trigonometry)
By the Law of Cosines, \begin{align*} AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\ \cos \alpha &> \frac 12\\ \end{align*}
It follows that $0 < \alpha < \frac {\pi}3$, and the probability is $\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}$.
## Solution 2 (Analytic Geometry)
$WLOG$, let the object turn clockwise.
Let $B = (0, 0)$, $A = (0, -8)$.
Note that the possible points of $C$ create a semi-circle of radius $5$ and center $B$. The area where $AC < 7$ is enclosed by a circle of radius $7$ and center $A$. The probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ}$.
The function of $\odot B$ is $x^2 + y^2 = 25$, the function of $\odot A$ is $x^2 + (y+8)^2 = 49$.
$O$ is the point that satisfies the system of equations: $\begin{cases} x^2 + y^2 = 25 \\ x^2 + (y+8)^2 = 49 \end{cases}$
$x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25$, $64 + 16y =24$, $y = - \frac52$, $x = \frac{5 \sqrt{3}}{2}$, $O = (\frac{5 \sqrt{3}}{2}, - \frac52)$
Note that $\triangle BDO$ is a $30-60-90$ triangle, as $BO = 5$, $BD = \frac{5 \sqrt{3}}{2}$, $DO = \frac52$. As a result $\angle CBO = 30 ^\circ$, $\angle ABO = 60 ^\circ$.
Therefore the probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{\textbf{(D) } \frac13 }$
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# How do you find the sides of a triangle given the angle and hypotenuse?
## How do you find the sides of a triangle given the angle and hypotenuse?
If you have the hypotenuse, multiply it by sin(θ) to get the length of the side opposite to the angle. Alternatively, multiply the hypotenuse by cos(θ) to get the side adjacent to the angle.
Does 9 40 41 make a right triangle?
Yes, 9, 40, 41 is a Pythagorean Triple and sides of a right triangle.
### How do I find the length of the sides of a triangle?
The Pythagorean Theorem, a2+b2=c2, a 2 + b 2 = c 2 , is used to find the length of any side of a right triangle.
Is 234 a right triangle?
Any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples.
READ: Why do you put a ball of aluminum foil in your dryer?
#### Does 8 15 17 make a right triangle?
Yes, 8, 15, 17 is a Pythagorean Triple and sides of a right triangle.
How do you find the angle of a right triangle?
Their angles are also typically referred to using the capitalized letter corresponding to the side length: angle A for side a, angle B for side b, and angle C (for a right triangle this will be 90°) for side c, as shown below.
## How to find the length of the right triangle side lengths?
Given angle and hypotenuse; Apply the law of sines or trigonometry to find the right triangle side lengths: a = c * sin(α) or a = c * cos(β) b = c * sin(β) or b = c * cos(α) Given angle and one leg; Find the missing leg using trigonometric functions: a = b * tan(α) b = a * tan(β) Given area and one leg
What is the ratio of a 45 45 90 triangle?
45°-45°-90° triangle: The 45°-45°-90° triangle, also referred to as an isosceles right triangle, since it has two sides of equal lengths, is a right triangle in which the sides corresponding to the angles, 45°-45°-90°, follow a ratio of 1:1:√ 2. Like the 30°-60°-90° triangle, knowing one side length allows you to determine the lengths
READ: How do you deal with someone who cheats and lies?
### What is the ratio of sides of a right triangle?
In this type of right triangle, the sides corresponding to the angles 30°-60°-90° follow a ratio of 1:√ 3 :2. Thus, in this type of triangle, if the length of one side and the side’s corresponding angle is known, the length of the other sides can be determined using the above ratio. For example, given that the side corresponding to
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# Evaluate
Question:
Evaluate
$\lim _{h \rightarrow 0}\left(\frac{\sqrt{x+h}-\sqrt{x}}{h}\right)$
Solution:
To evaluate:
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$
Formula used:
L'Hospital's rule
Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval $I$ except at a point a where
$\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$
then
$\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}$
As $\mathrm{X} \rightarrow 0$, we have
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{0}{0}$
This represents an indeterminate form. Thus applying L'Hospital's rule, we get
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim _{h \rightarrow 0} \frac{\frac{d}{d h}(\sqrt{x+h}-\sqrt{x})}{\frac{d}{d h}(h)}$
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim _{h \rightarrow 0} \frac{\frac{1}{2 \sqrt{x+h}}}{1}$
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{1}{2 \sqrt{x}}$
Thus, the value of $\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$ is $\frac{1}{2 \sqrt{x}}$
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Home >> Brackets >> Square Bracket >>
## Square Bracket
Parentheses Braces Square Bracket
Square Brackets are the Last one to be solved; when there are more than one type of brackets. Square brackets are denoted with the symbol "[ ]".
Following are some examples which shows the working of Square Brackets, Curly Braces & Parentheses
Example 1 = Solve 80 + [10{10 - (3 + 2)}]
Answer = Here we have three types of brackets, so we solve the Parentheses first and we get:
= 80 + [10{10-(5)}] = 80 + [10{10-5}]
Now, solve the Curly Braces and we get:
= 80 + [10 × 5]
Now, solve the Square Brackets and we get:
= 80 + 50 = 130
Example 2 = Solve 2 [10 + {5 - (2 × 3)}]
Answer = Here we have three types of brackets, so we solve the Parentheses first and we get:
= 2 [10 + {5 - (6)}] = 2 [10 + {5 - 6}]
Now, solve the Curly Braces and we get:
= 2 [10 + {-1}] = 2 [10 - 1]
Now, solve the Square Brackets and we get:
= 2 × 9 = 18
Example 3 = Solve 100 - [20 + {5(40 ÷ 4) - (2 × 2)}]
Answer = Here we have three types of brackets, so we solve the Parentheses first and we get:
= 100 - [20 + {5 × 10 - 4}]
Apply the BODMAS Rule and solve Curly Braces and we get:
= 100 - [20 + 46]
Now, solve the Square Brackets and we get:
= 100 - 66 = 34
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# Net force word problems
Find here in this lesson some easy and challenging net force word problems.
Problem #1:
What is the net force on the airplane in the figure below?
The airplane is moving with a force of 800 N. However, there are two forces moving in the opposite direction on the airplane.
Just add these two forces: 40 N + 60 N = 100 N
Subtract to get the net force: 800 N - 100 N = 700 N
The net force is 700 N.
The airplane will move with a force of 700 N as a result of air friction and wind.
Problem #2: You and your brother are pushing a car with a dead battery with forces of 20 N and 25 N in the same direction. What is the net force applied on the car?
Solution:
Since you are pushing the car in the same direction, the forces will be added together.
Net force = 20 N + 25 N
Net force = 45 N.
Problem #3: A brother is pulling a toy from his sister with a force of 6 N. The sister is pulling back with a force of 8 N.
Who gets the toy?
What is the net force?
Solution:
The sister gets the toy of course since she is pulling with a stronger force.
Net force = 8 N - 6 N
Net force = 2 N.
## More challenging net force word problems
Problem #4:
4 people are playing a tug of war. Two are pulling on the right side. Two are pulling on the left side. On the right side, one is pulling with a force of 60 N and the other with a force of 70 N. On the left side, one is pulling with a force of 30 N. How much force should the second person on the left apply to keep the rope in equilibrium?
The rope will be in equilibrium is the net force is 0.
The forces on the right is equal to 60 N + 70 N = 130 N
Let x be the force that must be applied by the second person on the left.
30 N + x = 130 N
Since 30 N + 100 N = 130 N, x = 100 N
The other person should pull with a force of 100 N to keep the rope in equilibrium.
Problem #5:
Your friend is pulling upward on an object with force of 3 N. You are pulling to the right with a force of 4 N.
Find the net force and the direction the object moves.
Just build the rectangle and find the resultant. The red arrow shows the direction the object will move.
This net force word problem is a little challenging. To find the net force, we need use the Pythagorean Theorem.
This net force word problem is a little challenging. To find the net force, we need use the Pythagorean Theorem.
Net force2 = 3 2 + 4 2
Net force2 = 9 + 16
Net force2 = 25
Since 52 = 25, net force = 5 N
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# Chapter1.7
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### Chapter1.7
1. 1. Commutative andCommutative and Associative PropertiesAssociative Properties
2. 2. Commutative and Associative Properties • Properties refer to rules that indicate a standard procedure or method to be followed. • A proof is a demonstration of the truth of a statement in mathematics. • Properties or rules in mathematics are the result from testing the truth or validity of something by experiment or trial to establish a proof. • Therefore, every mathematical problem from the easiest to the more complex can be solved by following step by step procedures that are identified as mathematical properties.
3. 3. Commutative and Associative Properties • Commutative Property means changing the order in which you add or subtract numbers does not change the sum or product. • Associative Property means changing the grouping of numbers when adding or multiplying does not change their sum or product. • Grouping symbols are typically parentheses (),but can include brackets [] or Braces {}.
4. 4. Commutative Property of addition - (Order) Commutative Property of addition - (Order) Commutative Property of multiplication - (order) Commutative Property of multiplication - (order) For any numbers a and b , a + b = b + a.For any numbers a and b , a + b = b + a. For any numbers a and b , a • b = b • a.For any numbers a and b , a • b = b • a. 45 + 5 = 5 + 4545 + 5 = 5 + 45 6 • 8 = 8 • 66 • 8 = 8 • 6 50 = 5050 = 50 48 = 4848 = 48 Commutative PropertiesCommutative Properties
5. 5. Associative Property of addition - (grouping symbols) Associative Property of addition - (grouping symbols) Associative Property of multiplication - (grouping symbols) Associative Property of multiplication - (grouping symbols) For any numbers a, b, and c, (a + b) + c = a + (b + c). For any numbers a, b, and c, (a + b) + c = a + (b + c). For any numbers a, b, and c, (ab) c = a (bc). For any numbers a, b, and c, (ab) c = a (bc). (2 + 4) + 5 = 2 + (4 + 5)(2 + 4) + 5 = 2 + (4 + 5) (2 • 3) • 5 = 2 • (3 • 5)(2 • 3) • 5 = 2 • (3 • 5) (6) + 5 = 2 + (9)(6) + 5 = 2 + (9) 11 = 1111 = 11 (6) • 5 = 2 • (15)(6) • 5 = 2 • (15) 30 = 3030 = 30 Associative PropertiesAssociative Properties
6. 6. Evaluate: 18 + 13 + 16 + 27 + 22 + 24Evaluate: 18 + 13 + 16 + 27 + 22 + 24 Rewrite the problem by grouping numbers that can be formed easily. (Associative property) This process may change the order in which the original problem was introduced. (Commutative property) Rewrite the problem by grouping numbers that can be formed easily. (Associative property) This process may change the order in which the original problem was introduced. (Commutative property) (18 + 22) + (16 + 24) + (13 + 27)(18 + 22) + (16 + 24) + (13 + 27) (40) + (40) + (40) = 120(40) + (40) + (40) = 120 Commutative and Associative PropertiesCommutative and Associative Properties • Commutative and Associative properties are very helpful to solve problems using mental math strategies.
7. 7. Evaluate: 4 • 7 • 25Evaluate: 4 • 7 • 25 Rewrite the problem by changing the order in which the original problem was introduced. (Commutative property) Group numbers that can be formed easily. (Associative property) Rewrite the problem by changing the order in which the original problem was introduced. (Commutative property) Group numbers that can be formed easily. (Associative property) 4 • 25 • 74 • 25 • 7 (4 • 25) • 7(4 • 25) • 7 (100) • 7 = 700(100) • 7 = 700 Commutative and Associative PropertiesCommutative and Associative Properties • Commutative and Associative properties are very helpful to solve problems using mental math strategies.
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Let's continue exploring integer addition, but this time we will progress from using colour alone to represent the opposite signs of positive and negative values and explicitly write out these expressions and equations mathematically!
In the previous posts, we have primarily used colour to differentiate between positive and negative values. In this second last video of the adding and subtracting integers series, we look to explicitly introduce the symbolic notation used to represent negative integers.
The video shown at the top of the post has the entire visual progression of integer addition using symbolic notation.
See the instructional guide below aimed to help teachers and/or parents use this activity with their own students/children.
## Visual Prompts: Question 1, Act 1
Watch the first 52 seconds of the video above before pausing.
The video shows the following integer situation which should be familiar if the previous integer posts have been explored already:
For most students, this first prompt will be a fairly easy warm-up with or without the number line with a result of 7.
Then, students are given a visual of 4 red squares added with 6 black squares.
Students will recall from the previous integer posts that combining a red and black square will make them disappear. This is the visual representation of the zero principle, which is very important for students to understand conceptually what is happening when we work with integers.
In the video, the red and black squares are laid out on the number line as we see below.
At this point, I would recommend that you pause the video and allow your students or your children attempt coming up with the solution.
After they have been given a chance to work through this problem, you can resume the video.
The Math Is Visual animation will show that 4 red squares combine with 4 black squares and disappear leaving only 2 black squares remaining.
This is where we finally formally introduce the symbolic notation used to represent the red squares. Instead of using the colour red, we are going to replace the colour red with a hyphen or a “negative” sign.
It is important for the teacher or parent to clarify to the participant that the hyphen can be used to represent the operation of subtraction, but also to represent a negative value algebraically. This is very heavy and complex for students to understand initially, so that is why I believe in doing a lot of work visually to ensure they build their fluency with this idea.
Next up in the video, we let students wrestle with some symbolic notation for the first time. Pause the video at the 1 minute, 45 second mark when you see the following:
By this point, I would be anticipating that students will be able to see that 5 red plus 3 red results in 8 red. While I’d likely continue speak in terms of colour at least initially, I would want to make the connection that this is the same as saying -5 + (-3) will result in -8 total.
While I would expect some students to struggle with the new symbolic notation we have just introduced, I would hope that they could come up with “8 red” as a result. If they are still struggling to identify how many squares in terms of colour will result, then you probably want to back up the bus and do more work there prior to introducing the symbolic notation.
After giving you students an opportunity to solve this problem, you can resume the video and see the 8 red squares or -8 on the number line.
Finally, we will give students one more problem to wrestle with.
Pause the video at the 2 minutes, 22 second mark when you see the following:
Giving your children an opportunity to use physical square tiles or counters would be beneficial here and even drawing out their own number line would be useful.
Have them line up the tiles like we see here:
Then, have them use the zero principle to make two black and two red tiles go away.
In other words, we had 4 and subtracted 4.
Or, 4 – 4.
Or, 4 + (-4).
Or, -4 + 4.
In any case, we end up with 4 of each “type” cancelling out because of the zero principle.
Watch in the next few days for the final video that will use symbolic notation for integer subtraction.
Let me know in the comments how you used this activity!
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# Video: Differentiating a Combination of Rational Functions Using the Quotient Rule
If π¦ = ((π₯ + 5)/(π₯ β 5)) β ((π₯ β 5)/(π₯ + 5)), find dπ¦/dπ₯.
01:48
### Video Transcript
If π¦ is equal to π₯ plus five over π₯ minus five minus π₯ minus five over π₯ plus five, find dπ¦ by dπ₯.
Our function, π¦, consists of two rational expressions, π₯ plus five over π₯ minus five and π₯ minus five over π₯ plus five. And we could find dπ¦ by dπ₯ by using the quotient rule on these two rational expressions. However, this would require using the quotient rule twice. We can make our work a little easier by combining the two rational expressions into one. We obtain that π¦ is equal to π₯ plus five squared minus π₯ minus five squared all over π₯ minus five times π₯ plus five. We can expand the brackets and then simplify to obtain that π¦ is equal to 20π₯ over π₯ squared minus 25.
And now, our function consists of only one rational expression. Weβre ready to use the quotient rule to differentiate this function. The quotient rule tells us that π’ over π£ prime is equal to π£π’ prime minus π’π£ prime over π£ squared. Setting π¦ equal to π’ over π£, we obtain that π’ is equal to 20π₯ and π£ is equal to π₯ squared minus 25. Next, we can find π’ prime and π£ prime, which gives us that π’ prime is equal to 20 and π£ prime is equal to two π₯.
Now, we can substitute them into the quotient rule in order to find that dπ¦ by dπ₯ is equal to π₯ squared minus 25 times 20 minus 20π₯ times two π₯ all over π₯ squared minus 25 squared. We simplify this to obtain that dπ¦ by dπ₯ is equal to negative 20π₯ minus 500 all over π₯ squared minus 25 squared.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Greatest Common Factor Using Factor Trees
## Multiplying only the same prime factors on each tree
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Practice Greatest Common Factor Using Factor Trees
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Greatest Common Factor Using Factor Trees
Richard is making gift bags. He has 36 pencils and 28 pens. How many gift bags can Richard make if there are the same number of pencils and pens in each bag? Use factor trees to solve this problem. How many pencils and pens will be in each bag?
In this concept, you will learn to find the greatest common factor using factor trees.
### Guidance
The greatest common factor (GCF) is the greatest factor that two or more numbers have in common. The GCF can be found by making a list and comparing all the factors. A factor tree can also be used to find the GCF. The GCF is the product of the common prime factors.
Let’s find the GCF of 20 and 30 using a factor tree.
First, make a factor tree for each number.
Then, identify the common factors. The numbers 20 and 30 have the factors 2 and 5 in common.
Next, multiply the common factors to find the GCF. If there is only one common factor, there is no need to multiply.
The GCF of 20 and 30 is 10.
Note that if the numbers being compared have no factors in common using a factor tree, they still have the factor 1 in common.
### Guided Practice
Find the GCF of 36 and 54 using factor trees.
First, make a factor tree for each number.
Then, identify the common factors. The numbers 36 and 54 have the factors 2 and two 3s in common.
Next, multiply the common factors to find the GCF.
The GCF of 36 and 54 is 18.
### Examples
Find the greatest common factor using factor trees.
#### Example 1
First, make a factor tree for each number.
Then, identify the common factors. The numbers 14 and 28 have the factors 2 and 7 in common.
Next, multiply the common factors to find the GCF.
The GCF of 14 and 28 is 14.
#### Example 2
First, make a factor tree for each number.
Then, identify the common factors. The numbers 24 and 34 have the factor 2 in common.
The GCF of 12 and 24 is 12.
#### Example 3
First, make a factor tree for each number.
Then, identify the common factors. The numbers 19 and 63 have the factor 1 in common.
The GCF of 19 and 63 is 1.
Richard needs to make gift bags with 36 pencils and 28 pens. Use factor trees to find the most number of bags he can make that have the same number of pencils and pens in each.
First, make a factor tree for each number.
Then, identify the common factors. The common factors are two 2s.
Next, multiply to common factors to find the GCF.
Finally, divide the number of pencils and pens by the GCF, 4.
Richard can make 4 gift bags that have 9 pencils and 7 pens in each bag.
### Explore More
Find greatest common factor for each pair of numbers.
1. 14 and 28
2. 14 and 30
3. 16 and 36
4. 24 and 60
5. 72 and 108
6. 18 and 81
7. 80 and 200
8. 99 and 33
9. 27 and 117
10. 63 and 126
11. 89 and 178
12. 90 and 300
13. 56 and 104
14. 63 and 105
15. 72 and 128
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 5.6.
### Vocabulary Language: English
factor
factor
Factors are the numbers being multiplied to equal a product. To factor means to rewrite a mathematical expression as a product of factors.
Greatest Common Factor
Greatest Common Factor
The greatest common factor of two numbers is the greatest number that both of the original numbers can be divided by evenly.
Product
Product
The product is the result after two amounts have been multiplied.
1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
5. [5]^ License: CC BY-NC 3.0
6. [6]^ License: CC BY-NC 3.0
### Explore More
Sign in to explore more, including practice questions and solutions for Greatest Common Factor Using Factor Trees.
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## Perimeter
It is the outside boundary of any closed shape. To find the perimeter we need to add all the sides of the given shape.
The perimeter of a rectangle is the sum of its all sides. Its unit is same as of its length.
Perimeter = 3 + 7 + 3 + 7 cm
Perimeter of rectangle = 20 cm
## Area
Area of any closed figure is the surface enclosed by the perimeter. Its unit is square of the unit of the length.
### Area of a triangle
The general formula to find the area of a triangle, if the height is given, is
### Area of a Right Angled Triangle
If we have to find the area of a right-angled triangle then we can use the above formula directly by taking the two sides having the right angle one as the base and one as height.
Here base = 3 cm and height = 4 cm
Area of triangle = 1/2 × 3 × 4
= 6 cm 2
Remark: If you take base as 4 cm and height as 3 cm then also the area of the triangle will remain the same.
### Area of Equilateral Triangle
If all the three sides are equal then it is said to be an equilateral triangle.
In the equilateral triangle, first, we need to find the height by making the median of the triangle.
Here the equilateral triangle has three equal sides i.e. 10 cm.
If we take the midpoint of BC then it will divide the triangle into two right angle triangle.
Now we can use the Pythagoras theorem to find the height of the triangle.
AB2 = AD2 + BD2
(10)2 = AD2 + (5)2
AD2 = (10)2 – (5)2
AD2 = 100 - 25 = 75
AD = 5√3
Now we can find the area of triangle by
Area of triangle = 1/2 × base × height
= 1/2 × 10 × 5√3
25√3 cm2
### Area of Isosceles Triangle
In the isosceles triangle also we need to find the height of the triangle then calculate the area of the triangle.
Here,
## Area of a Triangle — by Heron’s Formula
The formula of area of a triangle is given by heron and it is also called Hero’s Formula.
where a, b and c are the sides of the triangle and s is the semiperimeter
Generally, this formula is used when the height of the triangle is not possible to find or you can say if the triangle is a scalene triangle.
Here the sides of triangle are
AB = 12 cm
BC = 14 cm
AC = 6 cm
### Application of Heron’s Formula in Finding Areas of Quadrilaterals
If we know the sides and one diagonal of the quadrilateral then we can find its area by using the Heron's formula.
Find the area of the quadrilateral if its sides and the diagonal are given as follows.
Given, the sides of the quadrilateral
AB = 9 cm
BC = 40 cm
DC = 28 cm
AD = 15 cm
Diagonal is AC = 41 cm
Here, ∆ABC is a right angle triangle, so its area will be
Area of Quadrilateral ABCD = Area of ∆ABC + Area of ∆ADC
= 180 cm2 + 126 cm2
= 306 cm2
Categories: Class IX
### 1 Comment
#### Sakshi · October 1, 2021 at 7:33 pm
thank helped alot
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## Algebraic Equations
### Introduction
The fundamental idea on which algebraic equation are based is that an equation is exactly like a set of balance scales. If you do exactly the same thing to both sides of the equation then the equation will always balance. If you add the same amount to both sides or subtract the same amount to both sides the equation will balance. If you multiply the Left Hand Side (LHS) by the same amount as the Right Hand Side (RHS), then the equation will still balance, and so on. This is all analogous with a set of balance scales. With an algebraic equation we can go further and do exotic things and, for example, take the square root of both side and the equation will still balance.
Once you have this fundamental idea in your mind, we can go on to talk about how to use these allowed opersations systematically to solve an equation.
Another thing in passing - since we commonly use the letter 'x' in algebra, the use of the 'X' sign for multiplication is strongly discouraged.
### Basic Equations - No 1
Consider
x + 5 = 8
Although you might be able to guess what the answer is, I would to explain the algebraic method. The algebraic method would be (as always) to do exactly the same thing to both LHS and RHS.
Here I will subtract 5 from both sides
x = 3
As a second example, consider
x -4 = 6
The algebraic method would be to do exactly the same thing to both LHS and RHS.
Here I will add 4 to both sides
x = 10
Try these examples before proceeding
Quick Quiz a) x + 3 = 8 b) x + 4 = 7 c) x + 9 = 3 d) x - 11 = 15 e) x - 23 = -51
### Basic Equations - No 2
Consider
2x = 8
Again, although you might be able to guess what the answer is, I would to explain the algebraic method. The algebraic method would be (as always) to do exactly the same thing to both LHS and RHS.
Here I will divide both sides by 2
x = 4
As a second example, consider
$\frac{x}{6} = 6$
The algebraic method would be to do exactly the same thing to both LHS and RHS.
Here I will multiply both sides by 6
x = 36
Try these examples before proceeding
Quick Quiz a) $4x = 8$ b)$\frac{x}{3} = 5$ c) $3x = 12$ d)$\frac{x}{4} = 2$ e) $12x = 48$
### Basic Equations - No 3
If I can now make use of the methods used in the last two sections simultaneously in order to show how to tackle a more 'complicated' expression in a systematic way.
Consider
3x - 7 = x - 5
Firstly, I am going to use the methods of addition or subtraction to get every term containing an x on to the same side, and everything which is just a number on to the other side.
Secondly, I am going to use the operations of multiplication or division to obtain x
The first process involves two steps
Subtract x from both sides
2x - 7 = - 5
2x = 2
The second process involves just one step. We are told what 2x equals but I actually want to know what x equals, so
Divide both sides by 2
x = 1
As a second example, consider
6x + 9 = 25 - 2x
The first process involves two steps
8x + 9 = 25
Subtract 9 from both sides
8x = 16
The second process involves just one step. We are told what 8x equals but I actually want to know what x equals, so
Divide both sides by 8
x = 2
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# How do you integrate int dx/(x^2+25) using trig substitutions?
Dec 8, 2016
Let $x = 5 \tan \left(\theta\right)$, then dx = sec^2(theta)d"theta. To reverse the substitution, use $\theta = {\tan}^{-} 1 \left(\frac{x}{5}\right)$
#### Explanation:
$\int \frac{\mathrm{dx}}{{x}^{2} + 25}$
Let $x = 5 \tan \left(\theta\right)$, then dx = sec^2(theta)d"theta:
intsec^2(theta)/((5tan(theta))^2 + 25)d"theta =
intsec^2(theta)/(25tan^2(theta) + 25)d"theta =
1/25intsec^2(theta)/(tan^2(theta) + 1)d"theta =
Use the identity ${\sec}^{2} \left(\theta\right) = {\tan}^{2} \left(\theta\right) + 1$
1/25intsec^2(theta)/sec^2(theta)d"theta =
1/25intd"theta =
$\frac{1}{25} \theta + C =$
Reverse the substitution; substitute ${\tan}^{-} 1 \left(\frac{x}{5}\right) \text{ for } \theta$
$\frac{1}{25} {\tan}^{-} 1 \left(\frac{x}{5}\right) + C$
Dec 8, 2016
#### Explanation:
A good way to sort out trig substitutions is to recall the pythagorean identities in trig.
$1 - {\sin}^{2} x = {\cos}^{2} x$ is useful if we see $a - b {u}^{2}$. (We try to get $k \left(1 - {\sin}^{2} \theta\right) = k {\cos}^{2} \theta$)
In this case, we have a square plus a number. We think about how to use $t r i {g}^{2} + 1$
Eventually (perhaps by going through the list of pythagorean identities) we recall that
${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$. That gives us our substitution.
We want to have $25 {\tan}^{2} \theta + 25$, so we'll use $x = 5 {\tan}^{2} \theta$. The denominator becomes
${\left(5 \tan \theta\right)}^{2} + 25 = 25 \left({\tan}^{2} \theta + 1\right) = 25 \left({\sec}^{2} \theta\right)$
We also get $\mathrm{dx} = 5 {\sec}^{2} \theta d \theta$, so out integral becomes
$\int \frac{\mathrm{dx}}{{x}^{2} + 25} = \int \frac{5 {\sec}^{2} \theta d \theta}{25 {\sec}^{2} \theta}$
$= \frac{1}{5} \int d \theta = \frac{1}{5} \theta + C$
Since $x = 5 \tan \theta$, we have $\theta = {\tan}^{-} 1 \left(\frac{x}{5}\right) + C$
So
$\int \frac{\mathrm{dx}}{{x}^{2} + 25} = \frac{1}{5} {\tan}^{-} 1 \left(\frac{x}{5}\right) + C$
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# Lesson 13
Exponential Functions with Base $e$
## 13.1: $e$ on a Calculator (5 minutes)
### Warm-up
In this lesson, students are expected to use calculators with an $$e$$ button. This warm-up is meant to smooth out any technical difficulties with using their calculator to perform desired calculations.
Students may not have had much experience with entering an expression (rather than a single number) for the exponent and may not realize that parentheses may be needed to group the parts of the expression together. Expect calculation errors that stem from this issue.
Depending on the number of digits their calculator allows and the settings for displaying decimals, students may also notice that their calculation results have a different number of decimal places than those on the task statements.
### Launch
Arrange students in groups of 1. Provide access to calculators that have buttons for $$e$$ and exponent.
### Student Facing
The other day, you learned that $$e$$ is a mathematical constant whose value is approximately 2.718. When working on problems that involve $$e$$, we often rely on calculators to estimate values.
1. Find the $$e$$ button on your calculator. Experiment with it to understand how it works. (For example, see how the value of $$2e$$ or $$e^2$$ can be calculated.)
2. Evaluate each expression. Make sure your calculator gives the indicated value. If it doesn’t, check in with your partner to compare how you entered the expression.
1. $$10\boldcdot e^{(1.1)}$$ should give approximately 30.04166
2. $$5 \boldcdot e^{(1.1)(7)}$$ should give approximately 11,041.73996
3. $$e^{\frac{9}{23}}+7$$ should give approximately 8.47891
### Activity Synthesis
Invite students to share some issues they came across when evaluating the expressions with a calculator and how to resolve them. Make sure that students recognize that grouping symbols may be needed when entering an expression as an exponent, and that calculation results may vary because of the level of precision of the calculator.
## 13.2: Same Situation, Different Equations (20 minutes)
### Activity
This activity exposes students to growth factors written as $$e^r$$. An important takeaway for students is understanding that when we assume $$r$$ is a continuous growth rate, that is, the rate $$r$$ is happening at every moment and not just each unit of time $$t$$, then $$e^r$$ is how we can write the growth factor. If, however, we assume that the rate $$r$$ is happening for each unit of time $$t$$, then using a growth factor of $$1+r$$ is appropriate.
This work prepares students to work with equations of the form $$f(t)= P \boldcdot e^{rt}$$, which they will encounter in future activities throughout the rest of the unit. In the next activity, students will focus on interpreting the parameters of exponential equations written in this form. At this stage, students are not expected to understand $$e$$ in depth. That work is reserved for future, more-advanced courses. The goal here is to connect this new form of expressing an exponential function to students’ prior work by centering student thinking around the idea of how a growth rate is being applied.
### Launch
Read the introduction to the activity and display the equations for $$P$$ and $$Q$$. Ask students to think of at least one thing they notice and at least one thing they wonder about the functions. Give students 1 minute of quiet think time, and then invite students to share, recording and displaying their responses for all to see. Tell students that during this activity they will investigate pairs of functions that model the exponential growth of the same population, but make different assumptions about how the population is growing.
Arrange students in groups of 2–4 and provide access to graphing technology and a spreadsheet tool, in case requested. Ask groups to complete the first question and then pause for a brief whole-class discussion.
Invite students to share their tables and observations. Highlight the observations that the predictions produced by the two models are fairly close, but not identical, and that the difference seems to increase for larger values of $$t$$.
Next, tell students that they will now use graphs to compare the predictions of the two models for colonies that are growing at slower and faster rates. Suggest that they identify the two graphs for each colony with labels and different colors (if possible and simple to do). Consider asking students to split up the graphing work to optimize time.
Action and Expression: Develop Expression and Communication. Invite students to talk about their ideas with a partner before writing them down. Display sentence frames to support students when they explain their ideas. Begin by providing frames that encourage comparison of $$P(t)$$ and $$Q(t)$$, such as “How are _____ and _____ different?” and “What do _____ and _____ have in common?” Once students have agreed on some key comparisons, then display sentence frames that push their language toward more clearly articulating function behavior and understanding that $$P(t)$$ assumes the growth rate of 0.02 is applied for each interval of time $$t$$, while $$Q(t)$$ assumes 0.02 is a continuous growth rate applied at every moment. Provide sentence frames such as, “What does this part of _____ mean?” and “Another way to look at it is. . .” to help students articulate the meaning of the functions beyond surface comparisons.
Supports accessibility for: Language; Organization
### Student Facing
The population of a colony of insects is 9 thousand when it was first being studied. Here are two functions that could be used to model the growth of the colony $$t$$ months after the study began.
$$P(t) = 9 \boldcdot (1.02)^t$$
$$Q(t) = 9 \boldcdot e^{(0.02t)}$$
1. Use technology to find the population of the colony at different times after the beginning of the study and complete the table.
$$t$$ (time in months) $$P(t)$$ (population in thousands) $$Q(t)$$ (population in thousands)
6
12
24
48
100
2. What do you notice about the populations in the two models?
3. Here are pairs of equations representing the populations, in thousands, of four other insect colonies in a research lab. The initial population of each colony is 10 thousand and they are growing exponentially. $$t$$ is time, in months, since the study began.
$$\text{Colony 1}\\ f(t) = 10 \boldcdot (1.05)^t\\ g(t) = 10 \boldcdot e^{(0.05t)}$$
$$\text{Colony 2}\\k(t) = 10 \boldcdot (1.03)^t\\ l(t) = 10 \boldcdot e^{(0.03t)}$$
$$\text{Colony 3}\\p(t) = 10 \boldcdot (1.01)^t\\ q(t) = 10 \boldcdot e^{(0.01t)}$$
$$\text{Colony 4}\\v(t) = 10 \boldcdot (1.005)^t\\ w(t) = 10 \boldcdot e^{(0.005t)}$$
1. Graph each pair of functions on the same coordinate plane. Adjust the graphing window to the following boundaries to start: $$0 < x < 50$$ and $$0 < y < 80$$.
2. What do you notice about the graph of the equation written using $$e$$ and the counterpart written without $$e$$? Make a couple of observations.
### Activity Synthesis
Select groups of students to share their observations and graphs (or consider displaying the graphs in the Student Response for all to see). Here are some questions for discussion:
• “Which pair of graphs seemed most alike?” (Colony 4's graphs were almost the same until you got to large values of $$t$$. At the end of 2 year2, $$v(24)$$ and $$w(24)$$ were both about 11.3 thousand.)
• “Which pair of graphs seemed least alike?” (Colony 1's graphs looked different after just a few months. After 2 years, $$f(24) \approx32.3$$ while $$g(24)\approx33.2$$.)
• “For Colony 2, what is the growth factor for 1 month predicted by each model?”($$k$$ predicts a growth factor of 1.03 while $$l$$ predicts a growth factor of $$e^{0.03}$$, which is a bit more than 1.03.)
Tell students that exponential functions that involve small but ongoing growth (such as population growth or inflation) can be modeled in different ways. A factor written in the form of $$e^{0.03}$$ highlights that the $$r$$ percent growth rate (for example, 3%) is applied continuously. The $$(1+r)^t$$ model predicts a change by a factor of $$1 + r$$ for each unit of time. While these sound similar, they are not equivalent.
Students will learn more about $$e$$ in future courses. For now, it is sufficient simply to know that scientists often find it helpful to use $$e^r$$ to model exponential growth and decay in cases where the growth or decay is assumed to happen continuously.
Speaking: MLR8 Discussion Supports. Use this routine to support whole-class discussion. To help students share their observations about the graphs of the equations written using $$e$$ and their counterparts written without $$e$$, provide sentence frames such as: “I notice that . . .”, “_____ and _____ are alike because. . .”, “_____ and _____ are different because. . . .”
Design Principle(s): Optimize output (for comparison); Support sense-making
## 13.3: $e$ in Exponential Models (10 minutes)
### Activity
This activity further familiarizes students with the way continuously compounded models are typically written. It makes explicit the format of an equation used to represent such a model, which students will encounter more of in future lessons. Students also interpret the parameters of such equations in context and use the given structure to complete partially built equations. In this course, though, students are not assessed on generating a model to represent a situation characterized by continuous growth.
Only examples of growth situations are included in the activity. Decay situations are intentionally excluded so students don’t have to make sense of a negative factor in the exponent of an expression while also deciphering a new form for writing exponential functions. In synthesizing the activity, however, the teacher could choose to mention that in decay situations the $$r$$ takes on a negative value to represent a percent decrease.
### Launch
Arrange students in groups of 2. Give students a few minutes of quiet work time, and then ask them to discuss their responses with their partner. Follow with a whole-class discussion.
Representation: Internalize Comprehension. Use color coding and annotations to highlight connections between representations in a problem. For example, invite students to use the same colors in the diagram and color code corresponding elements of the questions in the activity.
Supports accessibility for: Visual-spatial processing
### Student Facing
Exponential models that use $$e$$ often use the format shown in this example:
Here are some situations in which a percent change is considered to be happening continuously. For each function, identify the missing information and the missing growth rate (expressed as a percentage).
1. At time $$t=0$$, measured in hours, a scientist puts 50 bacteria into a gel on a dish. The bacteria are growing and the population is expected to show exponential growth.
• function: $$b(t) = 50 \boldcdot e^{(0.25t)}$$
• continuous growth rate per hour:
2. In 1964, the population of the United States was growing at a rate of 1.4% annually. That year, the population was approximately 192 million. The model predicts the population, in millions, $$t$$ years after 1964.
• function: $$p(t) = \underline{\hspace{0.5in}} \boldcdot e^{\underline{\hspace{0.25in}}t}$$
• continuous growth rate per year: 1.4%
3. In 1955, the world population was about 2.5 billion and growing. The model predicts the population, in billions, $$t$$ years after 1955.
• function: $$q(t) = \underline{\hspace{0.5in}} \boldcdot e^{(0.0168t)}$$
• continuous growth rate per year:
### Anticipated Misconceptions
Some students may struggle relating the given exponential expression to the given situation. Draw their attention to what each part of the expression means. For example, 13 is the value of the expression when $$t = 0$$. For the second question, the value when $$t = 0$$ is the United States population, in millions, in 1964. What is this? (192). The coefficient of $$t$$ is the continuous growth rate. What is the continuous growth rate predicted by the model? (It's 1.4% which corresponds to 0.014.)
### Activity Synthesis
Select students to share how they completed and interpreted the missing information, where possible drawing attention to the structure of the exponential model given. Invite students to discuss how this form is like and unlike the equations students have seen prior to this point ($$f(x) = a \boldcdot b^x$$). In particular, highlight that the growth factor, $$e^r$$, is expressed in terms of both $$e$$ and $$r$$. This is different than growth at a rate of $$r$$ per unit time which is expressed only in terms of $$r$$ (as $$1+r$$).
Speaking, Representing: MLR8 Discussion Supports. Use this routine to support whole-class discussion. After each student shares, provide the class with the following sentence frames to help them respond: "I agree because . . .” or "I disagree because . . . .” If necessary, revoice student ideas to demonstrate mathematical language use by restating a statement as a question in order to clarify, apply appropriate language, and involve more students. For example, a statement such as, “The growth rate is 25%” can be restated as a question such as, “Is the continuous growth rate per hour 25%?” This will help students listen and respond to each other as they explain how they use the structure of the exponential model to fill in the missing information.
Design Principle(s): Support sense-making
## 13.4: Graphing Exponential Functions with Base $e$ (15 minutes)
### Optional activity
This activity is optional. Use this activity to allow students to practice analyzing graphs of exponential functions involving $$e$$ and setting an appropriate graphing window on their graphing technology in order to produce useful graphs.
Students have graphed exponential functions and analyzed the graphs to solve problems prior to this point. Though the functions here use base $$e$$, the analytical work is fundamentally the same as that when the base is another number. Students will also have opportunities to graph exponential functions with base $$e$$ later in the unit.
### Launch
Provide access to devices that can run Desmos or other graphing technology.
Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their written responses to the first question. Give students time to meet with 2–3 partners to share and receive feedback on their responses. Display feedback prompts that will help students strengthen their ideas and clarify their language. For example, “How did you get . . . ?”, “Why did you . . . ?”, and “How do you know . . . ?” Invite students to go back and revise or refine their written responses based on the feedback from peers. This will help students justify their reasoning for the population of bacteria 10 hours after the scientist put the initial bacteria in the dish.
Design Principle(s): Optimize output (for justification); Cultivate conversation
### Student Facing
1. Use graphing technology to graph the function defined by $$f(t) = 50 \boldcdot e^{(0.25t)}$$. Adjust the graphing window as needed to answer these questions:
1. The function $$f$$ models the population of bacteria in $$t$$ hours after it was initially measured. About how many bacteria were in the dish 10 hours after the scientist put the initial 50 bacteria in the dish?
2. About how many hours did it take for the number of bacteria in the dish to double? Explain or show your reasoning.
2. Use graphing technology to graph the function defined by $$p(t)=192 \boldcdot e^{(0.014t)}$$. Adjust the graphing window as needed to answer these questions:
1. The equation models the population, in millions, in the U.S. $$t$$ years after 1964. What does the model predict for the population of the U.S. in 1974?
2. In which year does the model predict the population will reach 300 million?
### Student Facing
#### Are you ready for more?
Research what the population of the U.S. was in the year the model predicted 300 million people. How far off was the model? What factors do you think account for the actual population in that year being different from the prediction of the model? In what year did the U.S. actually reach 300 million people?
### Activity Synthesis
Focus the discussion on how students set the graphing window and how they used the graphs to answer the questions.
Depending on the graphing technology used, students may use the following strategies:
• Visually estimate the value of the output given a certain input, or the input given an output.
• Use a tracing tool to trace the graph and identify the coordinates when the input or output has a certain value.
• To find an unknown input: graph a horizontal line at a given $$y$$-value (for example, $$y=100$$, a doubled bacteria population), find the intersection of that line and the graph of the exponential function, and identify the $$x$$-value at that intersection.
• To find an unknown output: use a table or type something like $$b(10)$$ after $$b$$ was defined.
Some students may also notice that they could calculate the output value given an exponential equation and an input value, but had to rely on the graph to find an unknown input value because they were unsure how to solve for $$x$$ when the equation has $$e$$ as a base. Tell students that in a future lesson they will learn how to use logarithms to solve this type of equation algebraically similar to how they have used logarithms previously.
## Lesson Synthesis
### Lesson Synthesis
In this lesson, we have seen that an exponential growth situation can be represented with models of different forms. Highlight the similarities and differences in the structure of the two forms students have seen so far. Consider using the equations representing Colony 4 from the first activity: $$v(t) = 100 \boldcdot (1.005)^t$$ and $$w(t) = 100 \boldcdot e^{(0.005t)}$$.
Explain to students that while the first form can be used to represent both discrete and exponential functions, the second form (using base $$e$$) is used only to represent situations where change happens "continuously" or at every moment. The constant $$e$$ is not used, for example, to describe how many layers of paper there are if we keep folding it in half $$n$$ times.
## 13.5: Cool-down - Two Population Predictions (5 minutes)
### Cool-Down
We learned earlier that the area, in square feet, can be modeled with a function such as $$a(d) = 24 \boldcdot (1+0.08)^d$$ or $$a(d) = 24 \boldcdot (1.08)^d$$, where $$d$$ is the number of days since the area was 24 square feet. This model assumes that the growth rate of 0.08 happens once each day.
In this lesson, we looked at a different type of exponential function, using the base $$e$$. For the algae growth, this might look like $$A(d)=24 \boldcdot e^{(0.08d)}$$. This model is different because the 8% growth is not just applied at the end of each day: it is successively divided up and applied at every moment. Because the growth is applied at every moment or "continuously," the functions $$a$$ and $$A$$ are not the same, but the smaller the growth rate the closer they are to each other.
Many functions that express real-life exponential growth or decay are expressed in the form that uses $$e$$. For the algae model $$A$$, 0.08 is called the continuous growth rate while $$e^{0.08}$$ is the growth factor for 1 day. In general, when we express an exponential function in the form $$P \boldcdot e^{rt}$$, we are assuming the growth rate (or decay rate) $$r$$ is being applied continuously and $$e^r$$ is the growth (or decay) factor. When $$r$$ is small, $$e^{rt}$$ is close to $$(1+r)^t$$.
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# Ordinal Numbers From Left T
It is possible to enumerate infinite sets using ordinal numbers. They also aid in generalize ordinal numbers.
## 1st
Ordinal numbers are among the most fundamental ideas in mathematics. It is a number that indicates the place of an object within an array of objects. Ordinal numbers are a number between 1 to 20. Ominal numbers serve many functions however, they are most often employed to show the sequence of items on an agenda.
Charts as well as words and numbers to represent ordinal numbers. They may also be used to indicate how a collection of pieces are arranged.
The majority of ordinal numbers fall into one of two categories. Transfinite ordinals can be represented using lowercase Greek letters, while finite ordinals can be represented by Arabic numbers.
As per the axioms, every well-ordered set should contain at least one ordinal. For instance, the top grade is awarded to the person who is enrolled in a class. The contest’s second-placed student who scored the highest.
## Combinational ordinal figures
Compounded ordinal numbers are those that have multiple digits. They are generated by multiplying ordinal number by its ultimate number. These numbers are typically used for dating and ranking. They don’t provide a unique ending for each number, as with cardinal numbers.
Ordinal numerals are utilized to identify the sequence of elements within the collection. The names of items in the collection are also indicated with these numbers. You can locate normal and suppletive numbers for ordinal numbers.
Prefixing a cardinal numeral with the suffix -u can create regular ordinals. Then, you type the number in an expression. Then , you can add an hyphen to it. There are also other suffixes. For example, “-nd” is for numbers ending with 2, and “-th” for numbers ending with 4 or 9.
Suppletive ordinals result from prefixing words by -u. The suffix, used to count is broader than the conventional one.
## Limit of the ordinal magnitude
Limits on ordinal numbers are ordinal numbers that don’t have zero. Limit ordinal numbers have a disadvantage: they do not have a limit on the number of elements. These numbers can be created by joining sets that are not empty and do not have any maximal elements.
Furthermore, the transcendent rules of recursion utilize limited ordinal numbers. Based on the von Neumann model, every infinite number of cardinal numbers is an ordinal limit.
A limit ordinal equals the sum of all other ordinals beneath. Limit ordinal figures are expressed using arithmetic, but they could also be expressed as an order of natural numbers.
Data is organized by ordinal number. They provide an explanation of the numerical location of objects. They are frequently utilized in set theory and arithmetic. Despite sharing a similar structure, they are not considered to be in the same class as natural numbers.
The von Neumann model uses a well-ordered set. It is assumed that fyyfy represents an element of g’, a subfunction of a function that is defined as a single operation. If fy’s subfunction is (i I, II) and g fulfills the requirements the g function is a limit ordinal.
The Church-Kleene ordinal is a limit ordinal in a similar manner. A Church-Kleene ordinal is a limit or limit ordinal that can be a properly ordered collection, that is comprised of smaller ordinals.
## Common numbers are often used in stories
Ordinal numbers are commonly utilized to establish the hierarchy between entities and objects. They are vital to organize, count, ranking, and other reasons. They can also be utilized to determine the order of things as well as the position of objects.
The ordinal number is often indicated by the letter “th”. Sometimes, however, the “nd” letter could be used instead of “th”. You will often find ordinal numbers on the title of books.
Although ordinal figures are generally employed in lists, you can write them down in the form of words. They may also be expressed as numbers and acronyms. In comparison, these numbers are much simpler to comprehend than the traditional ones.
Ordinary numbers are available in three distinct flavors. Through practice, games, and other exercises you can discover more about the different kinds of numbers. It is essential to know about them to increase your ability to arithmetic. Coloring exercises are an enjoyable and straightforward way to improve. Review your work using the handy marking sheet.
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# Find the sum of the series:
Question:
Find the sum of the series:
$(3 \times 8)+(6 \times 11)+(9 \times 14)+\ldots$ to $n$ terms
Solution:
In the given question we need to find the sum of the series.
For that, first, we need to find the nth term of the series so that we can use summation of the series with standard identities and get the required sum.
The series given is (3 × 8) + (6 × 11) + (9 × 14) + … to n terms.
The series can be written as, [(3 x 1) x (3 x 1 + 5)), (3 x 2) x (3 x 2 + 5))… (3n x (3n + 5))].
So, $\mathrm{n}^{\text {th }}$ term of the series,
$a_{n}=3 n(3 n+5)$
$a_{n}=9 n^{2}+15 n$
Now, we need to find the sum of this series, Sn.
$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{n}}$
$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(9 \mathrm{n}^{2}+15 \mathrm{n}\right)$
Note:
I. Sum of first $n$ natural numbers, $1+2+3+\ldots n$,
$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$
II. Sum of squares of first $n$ natural numbers, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,
$\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{2}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$
III. Sum of cubes of first $n$ natural numbers, $1^{3}+2^{3}+3^{3}+\ldots . n^{3}$,
$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$
IV. Sum of a constant k, N times,
$\sum_{k=1}^{N} k=N k$
So, for the given series, we need to find,
$\mathrm{S}_{\mathrm{n}}=9 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)+\sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})$
From, the above identities,
$S_{n}=\sum_{n=1}^{n}\left(9 n^{2}\right)+\sum_{n=1}^{n}(15 n)$
$\mathrm{S}_{\mathrm{n}}=9\left(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\right)+15\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)$
$=\left(\frac{n(n+1)}{2}\right)(6 n+18)$
$\mathrm{S}_{\mathrm{n}}=3 \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+3)$
So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=3 \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+3)$
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# A weight of 500 N is supported by two metallic ropes as shown in the figure. The values of tensions T1 and T2 are respectively:
1. 433 N and 250 N
2. 250 N and 433 N
3. 353.5 N and 250 N
4. 250 N and 353.5 N
## Answer (Detailed Solution Below)
Option 1 : 433 N and 250 N
## Detailed Solution
Concept:
Lami's Theorem: It is an equation that relates the magnitude of the three co-planner, concurrent and non-collinear forces that keeps a body in equilibrium. It states that each force is proportional to the sine of the angle between the other two forces.
Calculation:
$$\frac{{{{\rm{T}}_1}}}{{\sin 120^\circ }} = \frac{{{{\rm{T}}_2}}}{{\sin 150^\circ }} = \frac{{500}}{{\sin 90^\circ }}$$
T1 = 500 × sin 120° and T2 = 500 sin 150°
T1 = 433 N and T2 = 250 N
# If the sum of all the forces acting on a body is zero, it may be concluded that the body
1. Must be in equilibrium
2. May be in equilibrium
3. May be in equilibrium provided that forces are concurrent
4. May be in equilibrium provided that forces are parallel
## Answer (Detailed Solution Below)
Option 3 : May be in equilibrium provided that forces are concurrent
## Detailed Solution
Explanation:
If the sum of all forces acting on the body is zero, then it is not necessary that the body will be in equilibrium. For the explanation, let us take an example.
Let us assume that two equal and opposite forces having their line of action at a certain distance apart act on a body. Then the body in this case will not have translational motion but have rotation due to torque produced. Hence not in equilibrium. If two equal and opposite force act a point or concurrent, then the torque produced will be zero. The body will not have translational and rotational motion and will be in equilibrium.
# An attempt is made to pull a roller of weight W over a curb (step) by applying a horizontal force F as shown in the figure.The coefficient of static friction between the roller and the ground (including the edge of the step) is μ. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.
Option 2 :
## Detailed Solution
Explanation:
When the roller is just about to climb the step then,
• Static friction at point A is zero because when the cylinder is about to make out of the curb, it will lose its contact at A, the only contact will be at point B.
• At point B, the roller will be in a state of pure rolling so even the surfaces are rough there will be no friction at point B.
• Force F, contact force at point B, and weight of the roller W will be concurrent at point C.
Therefore the free body diagram will be as follows
# Condition of static equilibrium of a planar force system is written as
1. ΣF = 0
2. ΣM = 0
3. ΣF = 0 and ΣM = 0
4. None of these
## Answer (Detailed Solution Below)
Option 3 : ΣF = 0 and ΣM = 0
## Detailed Solution
Explanation:
Equilibrium of planar forces:
If the forces acting on the free-body are non-concurrent then the equivalent resultant will be a single force acting at a common point and a moment about the same point. The effect of such a force system will be to translate the body as well as to rotate it. Hence, for equilibrium to exist, both the force and moment must be null vectors.
i.e ΣF = 0, ΣM = 0
When the forces acting on a body lie in a plane (X - Y plane) but are non-concurrent, the body will have rotational motion perpendicular to the plane in addition to the translational motion along the plane, Hence necessary and sufficient condition for static equilibrium are,
ΣFx = 0, ΣFy = 0, ΣMz = 0
# When a body is subjected to two forces, the body will be in equilibrium if the two forces are:
1. collinear, equal and act in the same direction
2. collinear, unequal and opposite
3. non collinear, equal and opposite
4. collinear, equal and opposite
## Answer (Detailed Solution Below)
Option 4 : collinear, equal and opposite
## Detailed Solution
Explanation:
Principles of equilibrium
1. Two force principle: If only two forces act on a body that is in equilibrium, then they must be equal in magnitude, co-linear and opposite in sense.
2. Three force principle: If a body in equilibrium is acted upon by three forces, then the resultant of any two forces must be equal, opposite and collinear with the third force. If a three-force member is in equilibrium and the forces are not parallel, they must be concurrent. Therefore, the lines of action of all three forces acting on such a member must intersect at a common point; any single force is, therefore, the equilibrant of the other two forces.
If it does not pass through a common point, it will produce a couple.
A solid body applied to three forces whose lines of action are not parallel, is in equilibrium if the three following conditions satisfies:
1. The lines of action are coplanar (in the same plane).
2. The lines of action are meeting at a point.
3. The vector sum of these forces is equal to the zero vector.
3. Four force principle: If a body in equilibrium is acted upon by four forces, then the resultant of any two forces must be equal, opposite and collinear with the resultant of the other two forces.
# For any system of coplanar forces, the condition of equilibrium is that the
1. algebraic sum of the horizontal components of all the forces should be zero
2. algebraic sum of the vertical components of all the forces should be zero
3. algebraic sum of moments of all the forces about any point should be zero
4. all of the above
## Answer (Detailed Solution Below)
Option 4 : all of the above
## Detailed Solution
Explanation:
Coplanar force system:
A system in which all the forces acts in the same plane is termed as coplanar force system.
For any system of coplanar forces to be in equilibrium following condition should be satisfied:
• The algebraic sum of the horizontal components of all the forces should be zero.
• The algebraic sum of the vertical components of all the forces should be zero.
• The algebraic sum of moments of all the forces about any point should be zero.
ΣFx = 0, ΣFy = 0 and ΣM = 0
Concurrent force system:
A system in which all the forces acts in the same line is termed as concurrent force system.
For any system of Concurrent forces to be in equilibrium following condition should be satisfied:
• The algebraic sum of the horizontal components of all the forces should be zero.
• The algebraic sum of the vertical components of all the forces should be zero.
ΣFx = 0, ΣFy = 0
# A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure. Block R is tied to the wall by a massless and inextensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F (in kN) needed to move the block S is
1. 0.69
2. 0.88
3. 0.98
4. 1.37
Option 4 : 1.37
## Detailed Solution
Concept:
Making free body diagram of each block:
Balancing forces in positive x direction
∴ F = μRN1 + μRN2 = μ(RN1 + RN2)
where, RN1 = Normal reaction force on block R and RN2 = Normal reaction force on block S
Calculation:
Given:
μ = 0.4
RN1 = 100g = 100 × 9.81 = 981 N
RN2 = (100 + 150)g = 250 × 9.81 = 2452.5 N
Minimum force (F) is
F = μ(RN1 + RN2)
F = 0.4 × (981 + 2452.5) = 1373.4 N = 1.37 kN
# A steel ball of mass 2.4 kg is tied to a string and whirled it in a horizontal plane in a circle of diameter 2 m at a constant speed of 20 rpm. The tension in the string is (ignore gravity)
1. 5 N
2. 10.5 N
3. 100 N
4. 50.5 N
## Answer (Detailed Solution Below)
Option 2 : 10.5 N
## Detailed Solution
Concept:
Tension in the string = centrifugal force developed due to circular motion
T = mrω2
where T = tension in the string, m = mass of the ball, ω = angular velocity of rotation, r = radius of the circle of rotation
Angular velocity $$ω = \frac{{2π N}}{{60}}$$
where N is the speed in rpm.
Calculation:
Given:
m = 2.4 kg, Speed (N) = 20 rpm, diameter of circle of rotation (d) = 2 m, so radius of circle of rotation (r) = 1 m
angular speed of rotation $$ω = \frac{{2π N}}{{60}}$$
$$ω = \frac{{2π~ ×~20}}{{60}}$$
ω = 2.094 rad/sec
Now Tension T = 2.4 × 1 × 2.0942
T = 10.52 N
# A 1 m long uniform beam of 2 kg mass is being lifted vertically up by a force F at the 100 cm mark. What is the minimum force required to do so?
1. 1 N
2. 2 N
3. 10 N
4. 20 N
Option 3 : 10 N
## Detailed Solution
Concept:
Conditions for the system to be in equlibrium
ΣFx = 0, ΣFy = 0, ΣM = 0
Calculation:
Given:
m = 2 kg, Assume g = 10 m / s2
Lager will be the moment, smaller will be the force required to lift the rod. Hence, Applying Moment about 0 cm point we get.
w × 50 = F × 100
m × g × 50 = F × 100
2 × 10 × 50 = F × 100
F = 10 N
# An inextensible massless string goes over a frictionless pulley. Two weights of 100 N and 200 N are attached to the two ends of the string. The weights are released from rest, and start moving due to gravity. The tension in the string (in N) is __________
## Detailed Solution
Concept:
Using Newton’s second law. i.e. F = ma
Where F = Net force on the body, m = mass of body and a = acceleration
Calculation:
Free body diagram of above mass-pulley system
$${m_1}g = 100\;N \Rightarrow {m_1} = \frac{{100}}{g} = 10.19kg$$
$${m_2}g = 200\;N \Rightarrow {m_2} = \frac{{200}}{g} = 20.38kg$$
Consider block of 100 N:
From Newton’s 2nd law
T - 100 = m1a
∴ T = 10.19 a + 100 …i)
Now, Consider block of 200 N:
∴ 200 - T = m2a
∴ T = 200 - 20.38 a …ii)
From equation i) & ii)
10.19 a + 100 = 200 - 20.38 a
⇒ a = 3.27 m/s2
Now, substituting the value of a in equation (ii)
T = 200 - 20.38 × (3.27)
⇒ T = 133.32 N
# A force F is acting on a bent bar which is clamped at one end as shown in the figure.The CORRECT free body diagram is
Option 1 :
## Detailed Solution
Explanation:
When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.
Following is the conclusion:
Option (B) is wrong because the ground support is shown which we never show in FBD.
Option (C) is wrong because the X component is not shown.
Option (D) is wrong because the moment produced due to the eccentric force is not shown.
Hence only option (A) is correct.
# If point A is in equilibrium under the action of the applied forces, the values of tension. TAC = ?
1. 520 N
2. 150 N
3. 400 N
4. 450 N
Option 3 : 400 N
## Detailed Solution
Concept:
Lami's theorem:
It states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Consider three forces FA, FB, FC acting on a particle or rigid body making angles α, β and γ with each other.
Therefore, $$\frac{{{F_A}}}{{sin\ \alpha }} = \frac{{{F_B}}}{{sin\ \beta }} = \frac{{{F_C}}}{{sin\ \gamma }}$$
Calculation:
Given:
We have angle BAC = 180° - (30° + 60°)
angle BAC = 90°
By Lami’s Theorem,
$$\frac{{{800}}}{{sin~90°}} = \frac{{{T_{AC}}}}{{sin~150° }}$$
TAC = 400 N
# The force in member AB is ______. (take angle BAC as 60° and angle BCA as 30°)
1. 5√3 kN compressive
2. 2√3 kN tensile
3. 3√5 kN tensile
4. 2√5 kN compressive
## Answer (Detailed Solution Below)
Option 1 : 5√3 kN compressive
## Detailed Solution
Explanation:
Given:
∠BAC = 60°
∠BCA = 30°
Let FAB = compressive
FBC = compressive
FAB = ?
Considering joint B,
∑FH = 0
FAB cos 60° = FBC cos 30°
$$F_{AB} = \sqrt{3} \ F_{BC}$$
∑FV = 0
FAB sin 60° + FBC sin 30° = 10
$$\dfrac{\sqrt{3}}{2} F_{AB} + \dfrac{F_{AB}}{\sqrt{3}} \times \dfrac{1}{2} = 10$$
Multiplying both sides by $$\sqrt {3}$$
$$3 F_{AB} + F_{AB} = 20\sqrt{3}$$
$$4F_{AB} = 20\sqrt{3}$$
$$F_{AB} = 5 \sqrt{3}$$
Compressive (As direction assumed comp)
# Tension in the cable supporting a lift of weight ‘W’ and having an acceleration of ‘a’ while going upward is (g is acceleration due to gravity)
1. $$W\left( {1 + \frac{a}{g}} \right)$$
2. $$W\left( {1 - \frac{a}{g}} \right)$$
3. $$W\left( {2 + \frac{a}{g}} \right)$$
4. $$W\left( {1 + \frac{{2a}}{g}} \right)$$
## Answer (Detailed Solution Below)
Option 1 : $$W\left( {1 + \frac{a}{g}} \right)$$
## Detailed Solution
Concept:
For this type of problem, draw FBD and equate forces in the X and Y-axis respectively.
Calculation:
Given:
When the lift is moving upward:
Let R1 be the tension of cable and W be the weight of the lift
∴ R1 - W = ma
$$∴\;{R_1} = W + ma = W + \frac{W}{g}a = W\left( {1 + \frac{a}{g}} \right)$$
Additional Information
When the lift is moving downward:
Let R2 be the tension of cable and W be the weight of the lift
∴ W - R2 = ma
$$\therefore\;{R_2} = W - ma = W - \frac{W}{g}a = W\left( {1 - \frac{a}{g}} \right)$$
# Two forces P and P√2 act on a particle in directions inclined at an angle of 135° to each other. Find the magnitude of the resultant.
1. P
2. P√2
3. 5P
4. None of these
Option 1 : P
## Detailed Solution
CONCEPT:
Law of Parallelogram of forces: This law is used to determine the resultant of two coplanar forces acting at a point.
• It states that “If two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through that common point.”
Let two forces F1 and F2, acting at the point O be represented, in magnitude and direction, by the directed line OA and OB inclined at an angle θ with each other.
Then if the parallelogram OACB be completed, the resultant force R will be represented by the diagonal OC.
$$R = \sqrt{F_1^2 + F_2^2 + 2{F_1}{F_2}cosθ }$$
CALCULATION:
Given F1 = P, F2 = √2P, θ = 135
Then the resultant force is given by
$$F_{total} = \sqrt{P^{2}+(\sqrt2P)^{2}+2\times P\times \sqrt 2P\times cos135^∘}$$
$$F_{total} = \sqrt{P^2+2P^{2}-2P^{2}} = P$$
# A force f = (10î + 8ĵ - 5k̂) acts a point P(2, 5, 6). What will be the moment of the force about the point Q(3, 1, 4)?
1. – 36î + 15ĵ - 48k̂
2. 18î + 19ĵ - 37k̂
3. 32î - 31ĵ - 4k̂
4. Zero
## Answer (Detailed Solution Below)
Option 1 : – 36î + 15ĵ - 48k̂
## Detailed Solution
Solution
Concept:
Moment of a force about any point is given by
$$\overrightarrow {{m_0}} = \vec F \times \vec r$$
where $$\vec r = \overrightarrow {QP} = \;\vec P - \vec Q$$
Calculation:
Given:
$${\rm{\vec F}} = \left( {10\hat i + 8{\rm{\hat j}} - 5{\rm{\hat k}}} \right)$$P( 2, 5, 6), Q(3, 1, 4).
$$\vec r = \overrightarrow {QP} = \;\vec P - \vec Q$$
$$\vec r = \overrightarrow {PQ} = \;\vec Q - \vec P$$
$$\vec r=\left( {2\hat i + 5\hat j + 6\hat k} \right)- \left( {3\hat i + 1\hat j + 4\hat k} \right)$$
∴ $$\vec r = \left( {-1\hat i +4\hat j + 2\hat k} \right)$$
$$\therefore \;{\vec m_0} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ -1&{ 4}&{ 2}\\ {10}&8&{ - 5} \end{array}} \right|$$
$${\vec m_0} = \hat i\left( {-20 - 16} \right) - \hat j\left( { 5 - 20} \right) + \hat k\left( {-8 - 40} \right)$$
$${\vec m_0} = - 36\hat i + 15\hat j - 48\hat k$$
# A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-Y axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip it about point Q without slipping. What are the minimum values of the force (in Newton) and the static coefficient of friction μ between the floor and the wardrobe, respectively?
1. 490.5 and 0.5
2. 981 and 0.5
3. 1000.5 and 0.15
4. 1000.5 and 0.25
## Answer (Detailed Solution Below)
Option 1 : 490.5 and 0.5
## Detailed Solution
Given:
Mass of wardrobe (m) = 100 kg, height = 4 m, width = 2 m, depth = 1 m
The wardrobe will be on the verge of tip if the normal reaction passes through the point Q as shown in the free body diagram:
Taking moments about Q, ∑MQ = 0
⇒ W × 1 = P × 2
$$\Rightarrow {\rm{P}} = \frac{{100 \times 9.81}}{2} = 490.5{\rm~{N}}$$
∑H = 0
⇒ FF = P = 490.5 N
∑V = 0 ⇒ RN = mg = 981 N
Friction Force = μRN
$${\rm{\mu }} = \frac{{490.5}}{{981}} = 0.5$$
# If three coplanar concurrent forces acting at a point ‘O’ are in equilibrium then the ratio of T1 /T2 और T1 /T3 respectively will be?
1. $$\sqrt3$$ and $$\sqrt{\frac{3}{2}}$$
2. $$\sqrt {\frac{3}{2}} and \sqrt 3$$
3. 1 and $$\frac{1}{2}$$
4. $$\frac{1}{2}$$ and 1
## Answer (Detailed Solution Below)
Option 1 : $$\sqrt3$$ and $$\sqrt{\frac{3}{2}}$$
## Detailed Solution
Concept:
Lami's theorem:
Lami's theorem is an equation relating the magnitudes of three coplanar, concurrent and non-collinear forces, which keeps an object in static equilibrium, with the angles directly opposite to the corresponding forces. According to the theorem:
$$\frac{A}{{\sin \alpha }} = \frac{B}{{\sin\beta }} = \frac{C}{{\sin\gamma }}$$
Calculation:
Given:
From the given figure we have
$$\frac{T_1}{sin~(120)}~=~\frac{T_2}{sin~(150)}~=~\frac{T_3}{sin~(90)}$$
By solving the above equation we have,
$$\frac{T_1}{T_2}~=~\sqrt3$$ and $$\frac{T_1}{T_3}~=~\sqrt{\frac{3}{2}}$$
# The two forces of 9 Newtons and 12 Newtons which are acting at right angles to each other will have a resultant of
1. 8 N
2. 10 N
3. 15 N
4. 20 N
Option 3 : 15 N
## Detailed Solution
Concept:
Methods of finding resultant of forces
A) Law of a parallelogram of forces
Where P and Q are two forces acting on a body with angle θ between them
α = Angle made by resultant with respect to force P
α is the direction of resultant R
$${\rm{R}} = \sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQ}}\cos {\rm{\theta }}}$$
$${\rm{\alpha }} = {\tan ^{ - 1}}\left[ {\frac{{{\rm{Q}}\sin {\rm{\theta }}}}{{{\rm{P}} + {\rm{Q}}\cos {\rm{\theta }}}}} \right]{\rm{\;\;}}$$
B) Method of resolution of forces
ΣFX = summation of forces acting in X-direction
ΣFY = Summation of forces acting in Y-direction
$${\rm{R}} = \sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}} \right)}^2} + {{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}} \right)}^2}}$$
$${\rm{\theta }} = {\tan ^{ - 1}}\left| {\frac{{{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}}}{{{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}}}} \right|$$
Calculation:
Let the force P = 9 N and Q = 12 N and θ = 90°
By law of parallelogram of forces
$${\rm{R}} = \sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQ}}\cos {\rm{\theta }}}$$
$${\rm{R}} = \sqrt {{9^2} + {{12}^2} + 2 \times 9 \times 12 \times \cos 90^\circ }$$
$${\rm{R}} = \sqrt {{9^2} + {{12}^2}}$$
R = 15 N
Alternate method:
By the method of resolution
ΣFX = 9 N, ΣFY = 12 N
$${\rm{R}} = \sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}} \right)}^2} + {{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}} \right)}^2}}$$
$${\rm{R}} = \sqrt {{9^2} + {{12}^2}}$$
R = 15 N
# The resultant of two equal forces P making an angle θ, is given by:
1. $$2P\sin \left( {\frac{\theta }{2}} \right)$$
2. $$2P\tan \left( {\frac{\theta }{2}} \right)$$
3. $$2P\cot \left( {\frac{\theta }{2}} \right)$$
4. $$2P\cos \left( {\frac{\theta }{2}} \right)$$
## Answer (Detailed Solution Below)
Option 4 : $$2P\cos \left( {\frac{\theta }{2}} \right)$$
## Detailed Solution
Concept:
Law of Parallelogram of forces:
This law is used to determine the resultant of two coplanar forces acting at a point.
• It states that “If two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through that common point.”
• Let two forces F1 and F2, acting at the point O be represented, in magnitude and direction, by the directed line OA and OB inclined at an angle θ with each other.
Then if the parallelogram OACB be completed, the resultant force R will be represented by the diagonal OC.
$$R = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos\theta }$$
Calculation:
Given:
F1 = F2 = P
Then the resultant force is given by
$$F_{total} = \sqrt{P^{2}+P^{2}+2P^{2}\cosθ}$$
$$F_{total} = \sqrt{2P^{2}+2P^{2}\cosθ}$$
$$R=\sqrt{2P^2(1+\cos\theta)}$$
$$=\sqrt{2P^2(2\cos^2\frac{\theta}{2})}$$
$$=2P\cos\left(\frac{\theta}{2}\right)$$
Thus, option 4 is the correct Answer
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### PRODUCT DESCRIPTION
4th Grade MAFS Study Guide is a tool to review the Mathematics Florida Standards as a Test Prep. The study guide uses examples from the FSA Test Item Specification. This study guide covers MAFS.4.NF.2.3; MAFS.4.NF.2.4. It is a 5 days review that includes problems with multiple choice, multi-response and table response. It uses repetition as an instructional strategy. Students will solve similar problems every day with the goal of mastering these standards after the 5th day.
Common Core Standards:
MAFS.4.NF.2.3
Understand a fraction a/b with a > 1 as a sum of fractions 1/b.
a. Understand addition and subtraction of fractions as joining and
separating parts referring to the same whole.
b.Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model.
Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 + 1/8 = 8/8 + 8/8 + 1/8.
c. Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction.
d. Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem.
MAFS.4.NF.2.4 Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.
a. Understand a fraction a/b as a multiple of 1/b.
For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4).
b. Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number.
For example, use a visual fraction
model to express 3 × (2/5) as 6 × (1/5), recognizing this product as 6/5. (In general, n × (a/b) = (n × a)/b.)
c.Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie?
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Finding the Area of Irregular Figures
Media Type:
Video
Running Time: 3m 44s
Size: 5.0 MB
Collection Funded by:
Area is the measure of the surface inside a flat figure. Area is measured in square units. Partition irregular figures into regular shapes to find their area.
These resources are part of KET's Measurement and Geometry collection.
This video segment originally appeared in KET’s GED Connection series.
Background Essay
To tile a floor, you have to know how much area there is to be covered. Area is the measure of the surface inside a flat figure. Area is measured in square units. How do you find the surface area of a floor? You could cut a square foot out of cardboard and see how many times it will fit in the space. But instead, you can measure the length and width of the room and multiply. We express this as a formula:
• Area = length X width
A square is simply a rectangle with sides that are the same length. You can measure one side and multiply by four, but you can also use the formula for finding the area of a square:
• Area =side2
Rectangles and squares are “regular” shapes, but not all shapes are regular. “Irregular” shapes are a combination of several shapes. To find the area of an irregular shape, you will need to use logic and common sense to partition the shape into regular shapes that can be measured.
Discussion Questions
• How would you figure out the square footage of your home? It may be helpful to sketch a diagram.
• What irregular shapes do you see in the classroom? How would you partition and measure? Is there more than one way?
Teaching Tips
Have learners work in teams to draw irregular shapes. They must be able to explain how to partition the shape to determine area. But first, have them trade drawings with other teams to see if others can figure out how to partition. Encourage them to have fun and be creative with this exercise!
Have learners draw a regular or irregular figure and label it with measurements. Then have them describe the figure to another learner who must make as accurate a sketch as possible from the description.
Have learners sketch a diagram to help them solve for area of the following problem featured on p. 145 of the GED Connection “Mathematics” workbook:
The Park family is planning to carpet their family room, which measures 18 by 24 feet. They will not need to carpet the fireplace hearth, a 3-ft by 6–ft area centered on one of the 18-ft walls. How many square yards of carpet will they need?
Step 1. Sketch a diagram of the family room floor.
Step 2. Figure the total area of the room (18 X 24 = 432 sq ft) and subtract the floor area covered by the hearth (3 X 6 = 18 sq ft): 432 sq ft – 18 sq ft = 414 sq ft.
Step 3. Convert 414 square feet to square yards (9 sq ft = 1 sq yd).
Standards
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# HCF AND LCM WORKSHEET
HCF and LCM Worksheet :
Worksheet given in this section will be much useful to the students who would like to practice solving problems on HCF and LCM.
Before look at the worksheet, if you would like to learn the stuff HCF and LCM,
## HCF and LCM Worksheet - Problems
Problem 1 :
Find the HCF of :
(23 ⋅ 32 ⋅ ⋅ 74), (22 ⋅ 3⋅ 5⋅ 76) and (2⋅ 5⋅ 72)
Problem 2 :
Find the LCM of :
(23 ⋅ 33 ⋅ 5), (23 ⋅ 3⋅ 52) and (2 ⋅ 3 ⋅ 52)
Problem 3 :
Find the LCM of :
0.63, 1.05 and 2.1
Problem 4 :
Two number are in the ratio 15 : 11 and their HCF is 13. Find the numbers.
Problem 5 :
The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other number.
Problem 6 :
Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.
Problem 7 :
Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Problem 8 :
Find the largest number which divides 62, 132 and 237 leaves the same remainder in each case.
Problem 9 :
Find the greatest number of four digits which is divisible by 15, 25, 40 and 75.
Problem 10 :
Two numbers are in the ratio 6 : 7. If their LCM is 84, then find the two numbers.
## HCF and LCM Worksheet - Solutions
Problem 1 :
Find the HCF of :
(23 ⋅ 32 ⋅ ⋅ 74), (22 ⋅ 3⋅ 5⋅ 76) and (2⋅ 5⋅ 72)
Solution :
In the above given numbers, we find 2, 5 and 7 in common.
Take 2, 5 and 7 with minimum power.
That is 22, 5 and 72.
The HCF of the given numbers is
= 22 ⋅ ⋅ 72
= 4 ⋅ ⋅ 49
= 980
So, the HCF of the given numbers is 980.
Problem 2 :
Find the LCM of :
(23 ⋅ 33 ⋅ 5), (23 ⋅ 3⋅ 52) and (2 ⋅ 3 ⋅ 52)
Solution :
In the above given numbers, each base is a prime number.
Take all the different prime numbers with maximum powers.
They are 23, 33 and 52.
The LCM of the given numbers is
= 23 ⋅ 33 ⋅ 52
= 8 ⋅ 27 ⋅ 25
= 5400
So, the LCM of the given numbers is 5400.
Problem 3 :
Find the LCM of :
0.63, 1.05 and 2.1
Solution :
We we look at the given numbers 0.63, 1.05, and 2.1, maximum number of digits after the decimal is 2.
So, let us multiply each number by 100 to get rid of the decimal.
When the given numbers are multiplied by 100, we get 63, 105 and 210.
Now, decompose 63, 105 and 210 into prime factors.
63 = 3⋅
105 = 5 ⋅ 7
210 = 2 ⋅ ⋅ ⋅ 7
Take all the different prime factors with maximum powers.
They are 2, 32, 5 and 7.
The LCM of (63, 105 and 210 ) is
= 2 ⋅ 32 ⋅ 5 ⋅ 7
= 2 ⋅ 9 ⋅ 5 ⋅ 7
= 630
To get L.C.M of (0.63, 1.05, 2.1), divide 630 by 100.
630 / 100 = 6.3
So, the L.C.M of (0.63, 1.05, 2.1) is 6.3
Problem 4 :
Two number are in the ratio 15 : 11 and their HCF is 13. Find the numbers.
Solution :
From the given ratio, the two numbers can be assumed as 15x and 11x.
H.C.F of 15x and 11x is x.
Given : H.C.F of two numbers is 13.
Then, we have
x = 13
Therefore, we have
15x = 15 ⋅ 13 = 195
11x = 11 ⋅ 13 = 143
So, the two numbers are 195 and 143.
Problem 5 :
The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other number.
Solution :
Let x be the other number.
Product of two numbers = H.C.F ⋅ L.C.M
Substitute.
77 ⋅ x = 11 ⋅ 693
Divide each side 77.
x = 99
So, the other number is 99.
Problem 6 :
Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.
Solution :
Required greatest possible length is
= H.C.F of (495, 900, 1665)
To find HCF, decompose 495, 900 and 1665 into prime factors.
495 = 3⋅ ⋅ 11
900 = 2⋅ 3⋅ 52
1665 = 3⋅ ⋅ 37
In the prime factors of (495, 900, 1665), we find 3 and 5 in common.
Take 3 and 5 with minimum power.
They are 32 and 5.
Then the H.C.F is
= 3⋅ 5
= 9 ⋅ 5
= 45
So, the required greatest possible length is 45 cm.
Problem 7 :
Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Solution :
Required greatest number is
= H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of (1651, 2032)
= 127
So, the required number is 127.
Problem 8 :
Find the largest number which divides 62, 132 and 237 leaves the same remainder in each case.
Solution :
Required greatest number is
= H.C.F. of (132 - 62), (237 - 132) and (237 - 62)
= H.C.F. of (70, 105 and 175)
= 35
So, the required number is 35.
Problem 9 :
Find the greatest number of four digits which is divisible by 15, 25, 40 and 75.
Solution :
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
If a four digit number is divisible by 15, 25, 40 and 75, then the same four digit number should also be divisible by their L.C.M 600.
On dividing 9999 by 600, the remainder is 399.
To find the greatest four digit number exactly divisible by 600, we need to subtract the remainder 399 from 9999.
9999 - 399 = 9600
Therefore, 9600 is exactly divisible by 600.
Because, 9600 is exactly divisible by 600, the it should also be divisible by 15, 25, 40 and 75.
So, the greatest number of four digits which is divisible by 15, 25, 40 and 75 is 9600.
Problem 10 :
Two numbers are in the ratio 6 : 7. If their LCM is 84, then find the two numbers.
Solution :
From the given ratio, the numbers can be assumed as 6x and 7x.
We can find LCM of 6x and 7x using synthetic division as given below.
Therefore, LCM of (6x, 7x) is
= x ⋅ 6 ⋅ 7
= 42x
Given : LCM of the two numbers is 84.
Then, we have
42x = 84
Divide each side by 42.
x = 2
Substitute 2 for x in 6x and 7x to find the two numbers.
6x = 6 ⋅ 2 = 12
7x = 7 ⋅ 2 = 14
So, the two numbers are 12 and 14.
After having gone through the stuff given above, we hope that the students would have understood, how to solve HCF and LCM problems.
Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
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Algebra word problems
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Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
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Complementary and supplementary angles word problems
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Word problems on mixed fractrions
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Word problems on sets and venn diagrams
Word problems on ages
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OTHER TOPICS
Profit and loss shortcuts
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Times table shortcuts
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Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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# Helping solve mod problems
I am having trouble solving the below problems. My teacher taught us to write out the solutions by hand.. but I really think there is an easier way to do the higher numbers. Thanks!
• The last one you can do it using Chinese Remainder Theorem. – Alexei0709 Jun 3 '14 at 20:46
Hints: $$\text{(a) } 2 \cdot 5 \equiv -1 \text{ (mod 11)}$$ $$\text{(b) } 5x=10+25k \Leftrightarrow x \equiv 2 \text{ (mod 5)}$$ $$\text{(c) } 8x=6+16k \Leftrightarrow 4x \equiv 3 \text{ (mod 8) but what happens to numbers multiplied with 4 and then modded by 8?}$$ $$\text{(d) Have you heard of the Chinese Remainder Theorem?}$$
For (d) you can proceed as follows. From the second equation you get $x=50 + 121k$ $(*)$, where $k$ is an integer. Now mod this equation by $49$, this gives you $x\equiv1+23k \text{ (mod 49)}$, and using the first equation you find $10\equiv1+23k \text{ (mod 49)}$, which means $9\equiv23k \text{ (mod 49)}$. Now (and this requires a bit of calculation (see also here)), the inverse mod $49$ of $23$ is $32$. This yields $k\equiv32\cdot9\equiv43 \text{ (mod 49)}$, write $k=43+49l$, where again $l$ is an integer. So plugging this in $(*)$ we get $x=50+121(43+49l)=5253 +121\cdot49l$ , in other words $x \equiv 5253 \text{ (mod 5929)}$.
This method works always thanks to the fact that gcd$(49,121)=1$ and in general for any pair of positive integers being relatively prime. I leave it to you to do it the other way around, that is, to start with the first equation and proceed in a similar way. Of course it should give you the same solution.
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# An Inequality with Two Sets of Cevians
### Solution 1
Let be $x,y,z\in [0,1];$ $BA'=xa;$ $CA'=(1-x)a;$ $CB'=yb;$ $AB'=(1-y)b;$ $AC'=zc$; $BC'=(1-z)c.$
\displaystyle\begin{align} \frac{[A'B'C']}{[ABC]}&=\frac{[ABC]-[AB'C']-[BA'C']-[CA'B']}{[ABC]}\\ &=1-\frac{\frac{1}{2}z(1-y)bc\sin A }{\frac{1}{2}bc \sin A}-\frac{\frac{1}{2}x(1-z)ac\sin B}{\frac{1}{2}ac \sin B}\\ &\qquad\qquad-\frac{\frac{1}{2}y(1-x)ab\sin C}{\frac{1}{2}ab\sin C}\\ &=1-z(1-y)-x(1-z)-y(1-x)\\ &=1-(x+y+z)+xy+xz+yz \end{align}
To sum up:
(1)
$\displaystyle\frac{[A'B'C']}{[ABC]}=1-(x+y+z)+xy+xz+yz.$
From Ceva's theorem, $\displaystyle\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z}=1.\,$ Further:
\begin{align} xyz&=(1-x)(1-y)(1-z)\\ &=(1-x-y+xy)(1-z)\\ &=1-z-x+xz-y+yz+xy-xyz \end{align}
So that
(2)
$2xyz=1-(x+y+z)+(xz+xy+yz).$
From (1),(2) it follows that
$\displaystyle\frac{[A'B'C']}{[ABC]}=2xyz=\frac{2(xa)(yb)(zc)}{abc},$
i.e.,
(3)
$\displaystyle\frac{[A'B'C']}{[ABC]}=\frac{2BA'\cdot CB'\cdot AC'}{abc}.$
Similarly,
(4)
$\displaystyle\frac{[A''B''C'']}{[ABC] }=\frac{2BA''\cdot CB''\cdot AC''}{abc}$
From (3) and (4) and the AM-GM inequality, it follows that
\displaystyle\begin{align} \frac{[A'B'C']}{[A''B''C'']}&=\frac{BA'\cdot CB'\cdot AC'}{BA''\cdot CB'' \cdot AC''}\\ &=\frac{BA'}{BA''}\cdot \frac{CB'}{CB''}\cdot \frac{AC'}{AC''}\\ &\leq \frac{1}{27}\left(\frac{BA'}{BA''}+\frac{CB'}{CB''}+\frac{AC'}{AC''}\right)^3. \end{align}
Finally,
$\displaystyle \frac{27[A'B'C']}{[ABC]}\leq \left(\frac{BA'}{BA''}+\frac{CB'}{CB''}+\frac{AC'}{AC''}\right)^3.$
### Solution 2
We will work in areal (normalized barycentric) coordinates. The coordinates of the triangle vertices are
$A\rightarrow(1,0,0),~B\rightarrow(0,1,0),~C\rightarrow(0,0,1).$
If the point of concurrency of $AA'$, $BB'$, and $CC'$ has coordinates $(p,q,r)$, then the coordinates of the end points of the cevians are
$\displaystyle A'\rightarrow\left(0,\frac{q}{q+r},\frac{r}{q+r}\right),~ B'\rightarrow\left(\frac{p}{p+r},0,\frac{r}{p+r}\right),~ C'\rightarrow\left(\frac{p}{p+q},\frac{q}{p+q},0\right).$
The lengths in terms of the sides of the triangle $\{a,b,c\}$ are,
$\displaystyle BA'=\frac{ar}{q+r},~ CB'=\frac{bp}{p+r},~ AC'=\frac{cq}{p+q},$
and
\displaystyle\begin{align} [A'B'C']&=&(\text{Determinant of the coordinates})\cdot[ABC] \\ &=&\frac{2pqr}{(p+q)(q+r)(r+p)}[ABC]. \end{align}
Let the point of concurrency of $AA''$, $BB''$, and $CC''$ be $(u,v,w)$. Thus, the inequality can be written as
$\displaystyle 27~\frac{pqr(u+v)(v+w)(w+u)}{uvw(p+q)(q+r)(r+p)}\leq\left[\frac{r(v+w)}{w(q+r)}+ \frac{p(w+u)}{u(r+p)}+\frac{q(u+v)}{v(p+q)}\right]^3,$
and follows from the AM-GM inequality.
### Acknowledgment
The problem, with a solution of his solution 1, was kindly communicated to me by Dan Sitaru. The problem has been published in the MathProblems Journal (Vol 6, n 2, p 592, #62.)
Note that the formula for the area of the cevian triangle has several incarnations each of which leads to a slightly different derivation of the penultimate step. Ultimately, all proofs reduce to an application of the AM-GM inequality. Solution 2 (by Amit Itagi) is one of the examples.
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toseeifitworksforream
2022-04-04
Let a and b be coprime integers, and let m be an integer such that a | m and b | m. Prove that ab | m
alenahelenash
To prove that $ab\mid m$, we need to show that $m$ is a multiple of $ab$, which means that $m=abn$ for some integer $n$.
Since $a\mid m$, we can write $m=ak$ for some integer $k$. Similarly, since $b\mid m$, we can write $m=bl$ for some integer $l$.
Since $a$ and $b$ are coprime, we have $gcd\left(a,b\right)=1$. Therefore, there exist integers $x$ and $y$ such that $ax+by=1$ (this is the Bezout's identity).
Multiplying both sides by $m$, we get $\left(ax+by\right)m=m$. Since $m=ak=bl$, we can substitute these values to get $\left(ax+by\right)ak=\left(ax+by\right)bl$.
Expanding both sides, we get $axak+byak=axbl+bybl$, which simplifies to ${a}^{2}xk+abyk=abxl+{b}^{2}yl$.
Since $a\mid m$ and $b\mid m$, we know that $m$ is a common multiple of $a$ and $b$. Therefore, $ab$ is a common multiple of $a$ and $b$.
Using the definition of divisibility, we can say that $ab\mid {a}^{2}$ and $ab\mid {b}^{2}$. Hence, $ab$ divides the left-hand side of the above equation.
Since $ab\mid {a}^{2}xk$, $ab\mid abyk$, and $ab\mid abxl$, we can conclude that $ab\mid {b}^{2}yl$.
Since $a$ and $b$ are coprime, $b$ does not divide $a$. Therefore, $b$ must divide $yl$, which means that there exists an integer $n$ such that $yl=bn$.
Substituting this value, we get $ab\mid {b}^{2}bn$, which simplifies to $abn=m$.
Hence, we have shown that if $a$ and $b$ are coprime integers, and if $m$ is an integer such that $a\mid m$ and $b\mid m$, then $ab\mid m$.
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# SAT Math: Problem Solving and Data Analysis ๐
kritika gautam
kritika gautam
## SAT Math Overview
Welcome to another SAT Math guide! The Problem Solving & Data Analysis portion of the SAT Math exam makes up 17/58 questions, or about 30% of the math section. You should be able to change/convert to common units, understand percentages, read data from charts, graphs, or tables, and determine whether data is linear or exponential. Before we jump into the practice questions, let's break down the skills you need to be successful. You got this! ๐
## ๐บ๏ธ Main Problem Solving & Data Analysis Topic Areas
College Board has provided a list of skills that may be tested under "Problem Solving & Data Analysis." We've put this list of skills into three groups for you to focus on! Don't worry, we'll take you through all of these. Here is a checklist for you to work off of:
๐งพ Group 1: Ratio, Proportion, Units, and Percentage
• These skills are usually based on comparing two numbers and predicting unknown values.
โ๏ธ Group 2: Interpreting Relationships Presented in Scatterplots, Graphs, Tables, and Equations
• In these types of questions, you will need to look at how two variables interact and change. These relationships can be modeled as linear, quadratic, or exponential graphs and equations.
• This group includes table data, scatterplots, and linear and exponential growth.
๐ Group 3: More Data and Statistics
• These questions may ask you to interpret the provided data, sometimes by identifying key values or calculating probabilities.
• This group includes data inferences, statistical distributions, and data collection and conclusions.
## ๐งพ Ratio, Proportion, Units, and Percentage
Ratios and proportions are usually based on comparing two numbers. These questions can become tricky if you mix up numbers.ย
### ๐ง What You Need to Know: Ratio, Proportion, Units, and Percentage
For the first group of skills, you should be able to:
• Identify the difference between ratios, proportions, and percents.
• A ratio is a comparison of two different values, or parts. ๐๐๐๐:๐๐๐
• A proportion represents one part of a whole. โ
โ
โ
โโ
• A percent is a part of 100. ๐ฏ
• Apply the properties of proportion to word problems.
• Manipulate fractions and decimals.
• Convert to common units and between percentages, fractions, and decimals.
• Simply ratios. Ratios are always simplified.
• Calculate the change in percentage. Change in % = the difference in percentages / the total * 100
### ๐ค Applying Your Knowledge: Ratio, Proportion, Units, and Percentage
#### Unit Conversion Practice
Some common conversions are between measures of distance, such as from yards to inches. You can create a conversion table like the one below to work through the problem.
4 yards 3 feet 12 inches = 144 inches 1 yard 1 foot
๐ Let's practice converting 4 yards into inches. Here are the steps you should follow:
1. First, place the given, initial value on the very left! This value is 4 yards.
2. Next, we can use the identity property by multiplying that initial value by a fraction equal to 1. This allows the units to cancel out. For instance, since 3 feet equals 1 yard, we can write the fraction as 3 feet / 1 yard.
3. Once we multiply 4 yards x (3 feet / 1 yard), we get 12 feet. Notice how the unit, "yards," got canceled out. Make sure to place the value with the original unit (yards) in the denominator so that the unit cancels and we're left with the new unit (feet). ๐ถโโ๏ธ
4. You can then repeat this until you get the value in your desired unit. To convert 4 yards to inches, we did that process twiceโby multiplying by two fractions equivalent to oneโto get the final answer of 144 inches.
#### Percentage Practice
To calculate a percentage, we divide the value of the part by the total value and multiply by 100.
Letโs say there were 32 students in a class and 12 of them were on a field trip. We were then asked to calculate the percentage of students that were PRESENT.๐
1. We can begin by calculating the number of students present: 32-12 = 20 students in class. This tells us that a part of the whole class, 20 students out of 32, were present.
2. Then, we take that part and divide it by the total number of students: 20/32 = 0.625.
3. After multiplying by 100, we find that 62.5% of students were in class that day. ๐
## โ๏ธ Interpreting Relationships Presented in Scatterplots, Graphs, Tables, and Equations
In these types of questions, you will need to look at how two variables interact and change. These relationships can be modeled as linear, quadratic, or exponential graphs and equations.ย
### ๐ง What You Need to Know: Scatterplots, Graphs, Tables, and Equations
For the second group of skills, you should be able to:
• Interpret data from charts, graphs, or tables.
• Find and interpret a line of best fit to given data.
• Place data within given contexts of world problems
• Determine whether a function is linear or exponential:
• If there is a common difference in data values, the relationship is linear.ย Linear relationships can be modeled by y=mx + b. โ
• If there is a common product between data, the relationship is exponential.ย Exponential relationships can be modeled by y=a(b)x โคด
### ๐ค Applying Your Knowledge: Scatterplots, Graphs, Tables, and Equations
#### Linear vs. Exponential Functions
One tip for finding a line of best fit is to draw a line that has approximately an equal number of points above and below the line. This doesnโt have to be exact, but should serve as a good indicator of any patterns in the data. Letโs also look at finding differences in data values.
Table A:
If you look at the changes in outputs and inputs of Table A, you may notice that each Y value increases by 3 to make the next Y value. Since there is a common difference, where 3 is being added each time, we can conclude that this is a linear relationship.
Table B:
If you take a look at Table B, you may find that adding and subtracting a constant value does not provide the next output value. Instead, there is a common product between the values. You may have noticed that multiplying a Y value by 2 provides the next Y value. In this case, 2 is the common product, and Table B is modeling an exponential relationship.
## ๐ More Data and Statistics
These questions may ask you to interpret the provided data, sometimes by identifying key values or calculating probabilities.ย
### ๐ง What You Need to Know: More Data and Statistics
Congrats, you made it to the last group of skills! For the third and final group of skills, you should be able to:
• Answer questions that involve the measures of central tendency:
• The mean number of a data set is the mathematical average.ย
• The median number of a data set in order from least to greatest is the middle number.ย
• The mode of a data set is the most commonly occurring number.ย
• Use data to calculate probability:
• If event A and event B must occur, multiply the individual probabilities together.ย
• If event A or event B can occur, add the individual probabilities together.ย
• Understand sampling procedures in statistics:
• Samples must be randomly chosen and representative of the whole population in order to generalize the conclusions to the whole population.ย
• Samples must be randomly assigned to treatment groups for an experiment if determining cause and effect.ย
### ๐ค Applying Your Knowledge: More Data and Statistics
#### Measures of Central Tendency Practice
Take a look at this data set: 3, 6, 7, 9, 1, 22, 4, 6, 8.
• To find the mean, we can add up all the numbers and divide by the number of values in the set (9). Since 3 + 6 + 7 + 9 + 1 + 22 + 4 + 6 + 8 = 66, the mean equals 66 / 9 = 7.33.
• To find the median, we first have to place the data set in order: 1, 3, 4, 6, 6, 7, 8, 9, 22. Since there are 9 values in the set and the median is the middle value, the fifth value in this set is the median: 6.
• The mode is just the most occurring value. Since 6 occurs twice in this data set, and all of the other numbers occur only once, 6 is the mode.
#### Probability Practice
๐ช Letโs review some probability too. Imagine a jar of 20 cookies, where 12 are chocolate chip, 5 are peanut butter, and 3 are Snickerdoodles.
• If you were asked the probability of randomly choosing a peanut butter or Snickerdoodle cookie, you would have to add the probabilities of choosing either together.
• Since the probability of choosing peanut butter is 5/20 and choosing Snickerdoodle is 3/20, we calculate that the probability of choosing either is 5/20 + 3/20 = 8/20.
๐ฒ You may also be asked to calculate the probability of event A and event B happening. In this case, you would multiply the two probabilities. For instance, imagine you rolled two dies and wanted to calculate the probability that both would provide a number less than 3.
• The probability of getting a number less than 3 on dice A is 2/6.
• The probability of getting a number less than 3 on dice B is also 2/6.
• The probability of both events occurring can be found by multiplying the separate probabilities together. Therefore, the probability that both would provide a number less than 3 equals 2/6 * 2/6 = 4/36.
## ๐ SAT Math Sample Questions + Explanations
### SAT Math Questions 1-3: Line of Best Fit Practice
The following graph will be used for questions 1-3. These three questions fit under the second group of skills we outlined above! Let's take a look and go over some of its key points. You can open the image in a new tab to see it in a larger size! โ๏ธ
Image Courtesy of College Board's SAT Student Guide
One of the main things to note about this graph is that the line is a line of best fit and is only meant to represent a pattern. The dots are the real data points that represent individuals and their characteristics.
#### Question 1: Line of Best Fit
1. This question is considered "easy" in difficulty. Analyzing the line of best fit and comparing it to individual data points is important for solving this question.
Image Courtesy of College Board's SAT Student Guide
The individual's height, given by the y value of the data point, must be at least 3 centimeters away from the height predicted by the line of best fit. There are four individuals with a height that is at least 3 centimeters greater or less than their predicted height: the people with metacarpal bones of 4, 4.3, 4.8, and 4.9 cm.
#### Question 2: Analyzing Slope
2. This question is considered "easy" in difficulty. You will need to analyze the slope in context.
Image Courtesy of College Board's SAT Student Guide
Slope can be thought of as the change in the y value for each change in the x value. In this situation, the height is represented on the y-axis, and the length of the metacarpal bone is represented on the x-axis, so the slope must represent a change in height based on a change in the length of the metacarpal bone.
This eliminates answer choices B and D, which state the opposite. Answer choice C can also be eliminated because the slope never identifies a single value when the input is 0 - that is actually the y-intercept! Therefore, choice A is the only correct answer. ๐
#### Question 3 - Estimating with the Line of Best Fit
3. This question is considered "easy". You will need to use the line of best fit to make an estimate.
Image Courtesy of College Board's SAT Student Guide
For this question, you've already been given the input, or the x value, which is 4.45 cm. You can then find this value on the x-axis as being between the line that marks 4.4 cm and the line that marks 4.5 cm.ย
If you draw a straight vertical line where x=4.45 and find the point of intersection with the line of best fit,ย you will find that the line of best fit predicts that an individual with a metacarpal bone 4.45 cm long is predicted to be 170 cm in height. Don't be afraid to mark up your test paper! โ๏ธ
### SAT Math Questions 4: Unit Conversion Practice
4. The following example is considered โmediumโ in difficulty. Conversion between megabytes and gigabytes and converting between hours, minutes, and seconds is necessary for success in this problem. This fits under the first group of skills we outlined. Remember the conversion from yards to inches? ๐ค
Image Courtesy of College Board's SAT Practice Problems
Before we start our dimensional analysis, lets review some important facts about the scenario: 1 hour = 3600 seconds, 1 gigabit = 1024 megabits, and 1 image = 11.2 gigabits.
3 megabits 1 gigabit 1 image 3600 seconds 11 hours = 10.3585 images 1 second 1024 megabits 11.2 gigabit 1 hour
Your conversion table should be set up similarly to this! The key to solving these types of questions is to make sure your original values all cancel out as you multiply the conversion factors to finally reach the intended units.
Using the table, we were able to figure out that approximately 10 images per day (rounded down) can be sent daily. ๐ธ
## ๐ Closing
Congratulations! Youโve made it to the end of SAT Math - Problem Solving and Data Analysis prep ๐ You should have a better understanding of the Math sections for the SATยฉ, topic highlights, what you will have to be able to do in order to succeed, as well as have seen some practice questions that put the concepts in action. Good luck studying for the SAT Math section ๐
Need more SAT resources? Check out our SAT Math Section Tips and Tricks. Want to see more practice questions. Take a look at "What are the SAT Math Test Questions Like?"
Keep up the great work!! ๐ฅณ
# SAT Math: Problem Solving and Data Analysis ๐
kritika gautam
kritika gautam
## SAT Math Overview
Welcome to another SAT Math guide! The Problem Solving & Data Analysis portion of the SAT Math exam makes up 17/58 questions, or about 30% of the math section. You should be able to change/convert to common units, understand percentages, read data from charts, graphs, or tables, and determine whether data is linear or exponential. Before we jump into the practice questions, let's break down the skills you need to be successful. You got this! ๐
## ๐บ๏ธ Main Problem Solving & Data Analysis Topic Areas
College Board has provided a list of skills that may be tested under "Problem Solving & Data Analysis." We've put this list of skills into three groups for you to focus on! Don't worry, we'll take you through all of these. Here is a checklist for you to work off of:
๐งพ Group 1: Ratio, Proportion, Units, and Percentage
• These skills are usually based on comparing two numbers and predicting unknown values.
โ๏ธ Group 2: Interpreting Relationships Presented in Scatterplots, Graphs, Tables, and Equations
• In these types of questions, you will need to look at how two variables interact and change. These relationships can be modeled as linear, quadratic, or exponential graphs and equations.
• This group includes table data, scatterplots, and linear and exponential growth.
๐ Group 3: More Data and Statistics
• These questions may ask you to interpret the provided data, sometimes by identifying key values or calculating probabilities.
• This group includes data inferences, statistical distributions, and data collection and conclusions.
## ๐งพ Ratio, Proportion, Units, and Percentage
Ratios and proportions are usually based on comparing two numbers. These questions can become tricky if you mix up numbers.ย
### ๐ง What You Need to Know: Ratio, Proportion, Units, and Percentage
For the first group of skills, you should be able to:
• Identify the difference between ratios, proportions, and percents.
• A ratio is a comparison of two different values, or parts. ๐๐๐๐:๐๐๐
• A proportion represents one part of a whole. โ
โ
โ
โโ
• A percent is a part of 100. ๐ฏ
• Apply the properties of proportion to word problems.
• Manipulate fractions and decimals.
• Convert to common units and between percentages, fractions, and decimals.
• Simply ratios. Ratios are always simplified.
• Calculate the change in percentage. Change in % = the difference in percentages / the total * 100
### ๐ค Applying Your Knowledge: Ratio, Proportion, Units, and Percentage
#### Unit Conversion Practice
Some common conversions are between measures of distance, such as from yards to inches. You can create a conversion table like the one below to work through the problem.
4 yards 3 feet 12 inches = 144 inches 1 yard 1 foot
๐ Let's practice converting 4 yards into inches. Here are the steps you should follow:
1. First, place the given, initial value on the very left! This value is 4 yards.
2. Next, we can use the identity property by multiplying that initial value by a fraction equal to 1. This allows the units to cancel out. For instance, since 3 feet equals 1 yard, we can write the fraction as 3 feet / 1 yard.
3. Once we multiply 4 yards x (3 feet / 1 yard), we get 12 feet. Notice how the unit, "yards," got canceled out. Make sure to place the value with the original unit (yards) in the denominator so that the unit cancels and we're left with the new unit (feet). ๐ถโโ๏ธ
4. You can then repeat this until you get the value in your desired unit. To convert 4 yards to inches, we did that process twiceโby multiplying by two fractions equivalent to oneโto get the final answer of 144 inches.
#### Percentage Practice
To calculate a percentage, we divide the value of the part by the total value and multiply by 100.
Letโs say there were 32 students in a class and 12 of them were on a field trip. We were then asked to calculate the percentage of students that were PRESENT.๐
1. We can begin by calculating the number of students present: 32-12 = 20 students in class. This tells us that a part of the whole class, 20 students out of 32, were present.
2. Then, we take that part and divide it by the total number of students: 20/32 = 0.625.
3. After multiplying by 100, we find that 62.5% of students were in class that day. ๐
## โ๏ธ Interpreting Relationships Presented in Scatterplots, Graphs, Tables, and Equations
In these types of questions, you will need to look at how two variables interact and change. These relationships can be modeled as linear, quadratic, or exponential graphs and equations.ย
### ๐ง What You Need to Know: Scatterplots, Graphs, Tables, and Equations
For the second group of skills, you should be able to:
• Interpret data from charts, graphs, or tables.
• Find and interpret a line of best fit to given data.
• Place data within given contexts of world problems
• Determine whether a function is linear or exponential:
• If there is a common difference in data values, the relationship is linear.ย Linear relationships can be modeled by y=mx + b. โ
• If there is a common product between data, the relationship is exponential.ย Exponential relationships can be modeled by y=a(b)x โคด
### ๐ค Applying Your Knowledge: Scatterplots, Graphs, Tables, and Equations
#### Linear vs. Exponential Functions
One tip for finding a line of best fit is to draw a line that has approximately an equal number of points above and below the line. This doesnโt have to be exact, but should serve as a good indicator of any patterns in the data. Letโs also look at finding differences in data values.
Table A:
If you look at the changes in outputs and inputs of Table A, you may notice that each Y value increases by 3 to make the next Y value. Since there is a common difference, where 3 is being added each time, we can conclude that this is a linear relationship.
Table B:
If you take a look at Table B, you may find that adding and subtracting a constant value does not provide the next output value. Instead, there is a common product between the values. You may have noticed that multiplying a Y value by 2 provides the next Y value. In this case, 2 is the common product, and Table B is modeling an exponential relationship.
## ๐ More Data and Statistics
These questions may ask you to interpret the provided data, sometimes by identifying key values or calculating probabilities.ย
### ๐ง What You Need to Know: More Data and Statistics
Congrats, you made it to the last group of skills! For the third and final group of skills, you should be able to:
• Answer questions that involve the measures of central tendency:
• The mean number of a data set is the mathematical average.ย
• The median number of a data set in order from least to greatest is the middle number.ย
• The mode of a data set is the most commonly occurring number.ย
• Use data to calculate probability:
• If event A and event B must occur, multiply the individual probabilities together.ย
• If event A or event B can occur, add the individual probabilities together.ย
• Understand sampling procedures in statistics:
• Samples must be randomly chosen and representative of the whole population in order to generalize the conclusions to the whole population.ย
• Samples must be randomly assigned to treatment groups for an experiment if determining cause and effect.ย
### ๐ค Applying Your Knowledge: More Data and Statistics
#### Measures of Central Tendency Practice
Take a look at this data set: 3, 6, 7, 9, 1, 22, 4, 6, 8.
• To find the mean, we can add up all the numbers and divide by the number of values in the set (9). Since 3 + 6 + 7 + 9 + 1 + 22 + 4 + 6 + 8 = 66, the mean equals 66 / 9 = 7.33.
• To find the median, we first have to place the data set in order: 1, 3, 4, 6, 6, 7, 8, 9, 22. Since there are 9 values in the set and the median is the middle value, the fifth value in this set is the median: 6.
• The mode is just the most occurring value. Since 6 occurs twice in this data set, and all of the other numbers occur only once, 6 is the mode.
#### Probability Practice
๐ช Letโs review some probability too. Imagine a jar of 20 cookies, where 12 are chocolate chip, 5 are peanut butter, and 3 are Snickerdoodles.
• If you were asked the probability of randomly choosing a peanut butter or Snickerdoodle cookie, you would have to add the probabilities of choosing either together.
• Since the probability of choosing peanut butter is 5/20 and choosing Snickerdoodle is 3/20, we calculate that the probability of choosing either is 5/20 + 3/20 = 8/20.
๐ฒ You may also be asked to calculate the probability of event A and event B happening. In this case, you would multiply the two probabilities. For instance, imagine you rolled two dies and wanted to calculate the probability that both would provide a number less than 3.
• The probability of getting a number less than 3 on dice A is 2/6.
• The probability of getting a number less than 3 on dice B is also 2/6.
• The probability of both events occurring can be found by multiplying the separate probabilities together. Therefore, the probability that both would provide a number less than 3 equals 2/6 * 2/6 = 4/36.
## ๐ SAT Math Sample Questions + Explanations
### SAT Math Questions 1-3: Line of Best Fit Practice
The following graph will be used for questions 1-3. These three questions fit under the second group of skills we outlined above! Let's take a look and go over some of its key points. You can open the image in a new tab to see it in a larger size! โ๏ธ
Image Courtesy of College Board's SAT Student Guide
One of the main things to note about this graph is that the line is a line of best fit and is only meant to represent a pattern. The dots are the real data points that represent individuals and their characteristics.
#### Question 1: Line of Best Fit
1. This question is considered "easy" in difficulty. Analyzing the line of best fit and comparing it to individual data points is important for solving this question.
Image Courtesy of College Board's SAT Student Guide
The individual's height, given by the y value of the data point, must be at least 3 centimeters away from the height predicted by the line of best fit. There are four individuals with a height that is at least 3 centimeters greater or less than their predicted height: the people with metacarpal bones of 4, 4.3, 4.8, and 4.9 cm.
#### Question 2: Analyzing Slope
2. This question is considered "easy" in difficulty. You will need to analyze the slope in context.
Image Courtesy of College Board's SAT Student Guide
Slope can be thought of as the change in the y value for each change in the x value. In this situation, the height is represented on the y-axis, and the length of the metacarpal bone is represented on the x-axis, so the slope must represent a change in height based on a change in the length of the metacarpal bone.
This eliminates answer choices B and D, which state the opposite. Answer choice C can also be eliminated because the slope never identifies a single value when the input is 0 - that is actually the y-intercept! Therefore, choice A is the only correct answer. ๐
#### Question 3 - Estimating with the Line of Best Fit
3. This question is considered "easy". You will need to use the line of best fit to make an estimate.
Image Courtesy of College Board's SAT Student Guide
For this question, you've already been given the input, or the x value, which is 4.45 cm. You can then find this value on the x-axis as being between the line that marks 4.4 cm and the line that marks 4.5 cm.ย
If you draw a straight vertical line where x=4.45 and find the point of intersection with the line of best fit,ย you will find that the line of best fit predicts that an individual with a metacarpal bone 4.45 cm long is predicted to be 170 cm in height. Don't be afraid to mark up your test paper! โ๏ธ
### SAT Math Questions 4: Unit Conversion Practice
4. The following example is considered โmediumโ in difficulty. Conversion between megabytes and gigabytes and converting between hours, minutes, and seconds is necessary for success in this problem. This fits under the first group of skills we outlined. Remember the conversion from yards to inches? ๐ค
Image Courtesy of College Board's SAT Practice Problems
Before we start our dimensional analysis, lets review some important facts about the scenario: 1 hour = 3600 seconds, 1 gigabit = 1024 megabits, and 1 image = 11.2 gigabits.
3 megabits 1 gigabit 1 image 3600 seconds 11 hours = 10.3585 images 1 second 1024 megabits 11.2 gigabit 1 hour
Your conversion table should be set up similarly to this! The key to solving these types of questions is to make sure your original values all cancel out as you multiply the conversion factors to finally reach the intended units.
Using the table, we were able to figure out that approximately 10 images per day (rounded down) can be sent daily. ๐ธ
## ๐ Closing
Congratulations! Youโve made it to the end of SAT Math - Problem Solving and Data Analysis prep ๐ You should have a better understanding of the Math sections for the SATยฉ, topic highlights, what you will have to be able to do in order to succeed, as well as have seen some practice questions that put the concepts in action. Good luck studying for the SAT Math section ๐
Need more SAT resources? Check out our SAT Math Section Tips and Tricks. Want to see more practice questions. Take a look at "What are the SAT Math Test Questions Like?"
Keep up the great work!! ๐ฅณ
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# Graphs of Functions based on Rules
## Graph functions using value tables
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Graphs of Functions based on Rules
What if you were given a function rule like . How could you graph that function? After completing this Concept, you'll be able to create a table of values to graph functions like this one in the coordinate plane.
### Try This
Once you’ve had some practice graphing functions by hand, you may want to use a graphing calculator to make graphing easier. If you don’t have one, you can also use the applet at http://rechneronline.de/function-graphs/. Just type a function in the blank and press Enter. You can use the options under Display Properties to zoom in or pan around to different parts of the graph.
### Guidance
Of course, we can always make a graph from a function rule, by substituting values in for the variable and getting the corresponding output value.
#### Example A
Graph the following function:
Solution
Make a table of values. Pick a variety of negative and positive values for . Use the function rule to find the value of for each value of . Then, graph each of the coordinate points.
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
It is wise to work with a lot of values when you begin graphing. As you learn about different types of functions, you will start to only need a few points in the table of values to create an accurate graph.
#### Example B
Graph the following function:
Solution
Make a table of values. We know can’t be negative because we can't take the square root of a negative number. The domain is all positive real numbers, so we pick a variety of positive integer values for . Use the function rule to find the value of for each value of .
0
1
2
3
4
5
6
7
8
9
Note that the range is all positive real numbers.
#### Example C
The post office charges 41 cents to send a letter that is one ounce or less and an extra 17 cents for each additional ounce or fraction of an ounce. This rate applies to letters up to 3.5 ounces.
Solution
Make a table of values. We can’t use negative numbers for because it doesn’t make sense to have negative weight. We pick a variety of positive values for , making sure to include some decimal values because prices can be decimals too. Then we use the function rule to find the value of for each value of .
Watch this video for help with the Examples above.
### Vocabulary
We represent functions graphically by plotting points on a coordinate plane (also sometimes called the Cartesian plane). The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at a point called the origin. The horizontal number line is called the axis and the vertical line is called the axis.
The and axes divide the coordinate plane into four quadrants. The quadrants are numbered counter-clockwise starting from the upper right, so the plotted point for (a) is in the first quadrant, (b) is in the second quadrant, (c) is in the fourth quadrant, and (d) is in the third quadrant.
### Guided Practice
Graph the following function:
Solution
Make a table of values. Even though can’t be negative inside the square root, because we are squaring first, the domain is all real numbers. So we integer values for which are on either side of zero. Use the function rule to find the value of for each value of .
-2
-1
0
1
2
Note that the range is all positive real numbers, and that this looks like an absolute value function.
### Practice
Graph the following functions.
1. Vanson spends $20 a month on his cat. 2. Brandon is a member of a movie club. He pays a$50 annual membership and \$8 per movie.
### Vocabulary Language: English
Coordinate Plane
Coordinate Plane
The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate plane is also called a Cartesian Plane.
coordinate point
coordinate point
A coordinate point is the description of a location on the coordinate plane. Coordinate points are written in the form (x, y) where x is the horizontal distance from the origin, and y is the vertical distance from the origin.
Function
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise.
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# How do you solve 3x - 3= 8- 2( x + 3)?
May 29, 2018
Work backwards PESADM to find the lost unknown.
#### Explanation:
To find something that is lost it usually best to work backwards.
The order of operation is PEMDAS backwards PESADM.
Think about it this way if you lose something in PE you are a SAD M(an, am)
First it necessary to remove parenthesis and exponents. (PE)
Then either add or subtract doing the opposite (SA)
Then either divide or multiply doing the opposite ( DM)
In this equation start by removing the parenthesis on the left.
$3 x - 3 = 8 - 2 \left(x + 3\right) = 3 x - 3 = 8 - 2 x - 6$
Now combine the constants by adding 3 to both sides and using the associative property
$3 x - 3 + 3 = 8 + 3 - 6 - 2 x$ This results in
$3 x = 5 - 2 x$ Now add 2x to both sides
$3 x + 2 x = 5 - 2 x + 2 x$ this leaves
$5 x = 5$ last divide both sides by 5
$5 \frac{x}{5} = \frac{5}{5}$ this gives the answer
$x = 1$
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# What is the Greatest Common Factor (GCF) of 2 and 399?
Are you on the hunt for the GCF of 2 and 399? Since you're on this page I'd guess so! In this quick guide, we'll walk you through how to calculate the greatest common factor for any numbers you need to check. Let's jump in!
Want to quickly learn or show students how to find the GCF of two or more numbers? Play this very quick and fun video now!
First off, if you're in a rush, here's the answer to the question "what is the GCF of 2 and 399?":
GCF of 2 and 399 = 1
## What is the Greatest Common Factor?
Put simply, the GCF of a set of whole numbers is the largest positive integer (i.e whole number and not a decimal) that divides evenly into all of the numbers in the set. It's also commonly known as:
• Greatest Common Denominator (GCD)
• Highest Common Factor (HCF)
• Greatest Common Divisor (GCD)
There are a number of different ways to calculate the GCF of a set of numbers depending how many numbers you have and how large they are.
For smaller numbers you can simply look at the factors or multiples for each number and find the greatest common multiple of them.
For 2 and 399 those factors look like this:
• Factors for 2: 1 and 2
• Factors for 399: 1, 3, 7, 19, 21, 57, 133, and 399
As you can see when you list out the factors of each number, 1 is the greatest number that 2 and 399 divides into.
## Prime Factors
As the numbers get larger, or you want to compare multiple numbers at the same time to find the GCF, you can see how listing out all of the factors would become too much. To fix this, you can use prime factors.
List out all of the prime factors for each number:
• Prime Factors for 2: 2
• Prime Factors for 399: 3, 7, and 19
Now that we have the list of prime factors, we need to find any which are common for each number.
Since there are no common prime factors between the numbers above, this means the greatest common factor is 1:
GCF = 1
## Find the GCF Using Euclid's Algorithm
The final method for calculating the GCF of 2 and 399 is to use Euclid's algorithm. This is a more complicated way of calculating the greatest common factor and is really only used by GCD calculators.
Hopefully you've learned a little math today and understand how to calculate the GCD of numbers. Grab a pencil and paper and give it a try for yourself. (or just use our GCD calculator - we won't tell anyone!)
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "Greatest Common Factor of 2 and 399". VisualFractions.com. Accessed on January 19, 2022. http://visualfractions.com/calculator/greatest-common-factor/gcf-of-2-and-399/.
• "Greatest Common Factor of 2 and 399". VisualFractions.com, http://visualfractions.com/calculator/greatest-common-factor/gcf-of-2-and-399/. Accessed 19 January, 2022.
• Greatest Common Factor of 2 and 399. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/greatest-common-factor/gcf-of-2-and-399/.
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# If two chords of a circle are equally inclined to the diameter through their point of intersection, prove the chords are equal.
Given:
Two chords of a circle are equally inclined to the diameter through their point of intersection.
To do:
We have to prove that the chords are equal.
Solution:
$AB$ and $AC$ are two chords which are equally inclined to the diameter $AD$ of the circle with centre $O$.
Draw $OL \perp AB$ and $OM \perp AC$.
In $\triangle OLA$ and $\triangle OMA$,
$\angle OLA=\angle OMA=90^o$
$OA=OA$ (Common side)
$\angle LAO=\angle MAO$ (Given)
Therefore,
$\triangle \mathrm{OLA} \cong \triangle \mathrm{OMA}$ (By AAS congruence)
$\Rightarrow \mathrm{OL}=\mathrm{OM}$ (CPCT)
This implies, the chords are equidstant from the centre $O$.
$\Rightarrow A B=A C$ (Equal chords of a circle are equidistant from the centre of the circle)
Hence proved.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
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## 2 + 3
means;
If we bring together 2 apples and 3 apples, we have 5 apples.
If we put them all in a bag, we have 5 apples.
## Before starting:
You should know the place values of the numbers.
In modeling, shapes and their meanings.
In modeling, shapes and their meanings can be used to represent mathematical concepts or quantities.We generally use cubes to represent 1000, squares for 100, rectangles for 10, and small squares for the units in modeling.
Modeling can help you understand what addition operations mean by providing a visual representation of the numbers and the process of combining them. It can make abstract mathematical concepts more tangible and easier to grasp, especially for those who are visual learners or new to the subject
# 24+52
If we bring these together;
# 1256+132
1256 >>>1256 consists of one thousand, two hundreds, five tens, and six ones.
132 >>>132 consists of one hundred, three tens, and two ones.
When we bring these together, the sum is 1388, which consists of one thousand, three hundreds, eight tens, and eight ones.
Now, let's add these numbers as you have learned before, by writing them vertically and aligning the same place values under each other.
I'm writing the numbers to be added, making sure that the units, tens, hundreds, and thousands are lined up correctly.
If you write the ones place values under each other, the other place values will automatically align vertically as well.
# Why should the same place values be aligned vertically?
Units should be added to units, tens to tens, hundreds to hundreds. Numbers take their value according to the place they are written, that is, their place value. If you write them in different places, their meaning changes.
Aligning the same place values ensures that you are adding like terms together, preserving the value and meaning of each digit according to its position in the number. It's a fundamental principle in arithmetic that helps maintain the integrity of the mathematical system.
## Incorrect vertical alignment example
In this way of writing, you are adding 1256 and 1320, it's because you wrote the 1 in the thousands place, making its value 1000. You wrote the 3 in the hundreds place, making its value 300, and you wrote the 2 in the tens place, making its value 20.
Place value is essential in understanding the value of a number, and each digit's value is determined by its position. Writing the numbers in the correct columns ensures that you are adding the corresponding units, tens, hundreds, and thousands together, preserving the true value of the numbers. If the digits are not aligned correctly, it can lead to a misunderstanding of the numbers being added.
6 units plus 2 units equals 8 units, and it is written in the ones place.
5 tens plus 3 tens equals 8 tens, and it is written in the tens place.
2 hundreds plus 1 hundred equals 3 hundreds, and it is written in the hundreds place.
I have 1 thousand and no other thousands to add to it, so the total is 1 thousand, and it is written in the thousands place.
In elementary school, you did carry-over addition. Let's see what this "carry-over" actually means.
# 3567 + 458
Paying attention to the place values, I wrote 3567 and 458 vertically and am starting to add them, beginning with the ones place.
If we add 7 units with 8 units, it makes 15 units. I can't write 15 in the ones place (since only one digit can be written in a place), so I write 5 in the ones place, and I say "carry over 1" for the remaining 10, to add to the other tens. "Carry over 1" means that I now have one ten in my hand And I will add this ten to the other tens.
If we add 6 tens with 5 tens, it makes 11 tens, and there was also one "carried over," so in total, I have 12 tens, which equals 120. Again, I can't write 12 tens in the tens place, so I take the 1 hundred from 120 and say "carry over 1." I write the remaining 20, or 2 tens, in the tens place. Now I have one hundred in my hand, and it will be added to the other hundreds.
If we add 5 hundreds with 4 hundreds, it makes 9 hundreds, and I also had one hundred carried over, so in total, I have 10 hundreds, which equals 1000. I can't write 10 hundreds in the hundreds place, so I say "carry over 1" to add the 1000 to the thousands. Right now, I have no other numbers except 1000, and that 1000 will be added to the other thousands, so I write 0 in the hundreds place.
I have no other thousands to add to the 3 thousands, but I had one thousand carried over from the hundreds. If I add that thousand to the 3 thousands, I have 4 thousands in total, and I write 4 in the thousands place.
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# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y=x^2 +2x -5?
Jul 11, 2015
The vertex is at ($- 1 , - 2$).
The axis of symmetry is $x = 1$.
The $y$-intercept is at ($0 , 5$).
The $x$-intercepts are at $x = - 1 - \sqrt{6}$ and $x = - 1 + \sqrt{6}$.
#### Explanation:
The standard form for the equation of a parabola is
$y = {a}^{2} + b x + c$
$y = {x}^{2} + 2 x - 5$
So
$a = 1$, $b = 2$, and $c = 5$.
Vertex
Since $a > 0$, the parabola opens upwards.
The $x$-coordinate of the vertex is at
x = –b/(2a) = -2/(2×1) = -2/2 = -1.
Insert this value of $x$ back into the equation.
$y = {x}^{2} + 2 x - 5 = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 5 = 1 - 2 - 5 = - 6$
The vertex is at ($- 1 , - 2$).
Axis of symmetry
The axis of symmetry must pass through the vertex, so
The axis of symmetry is $x = - 1$.
$y$-intercept
To find the $y$-intercept, we set $x = 0$ and solve for $y$.
y = x^2 + 2x – 5 = 0^2 + 2×0 -5 = 0 + 0 – 5 = -5
The $y$-intercept is at ($0 , 5$).
$x$-intercepts
To find the $x$-intercepts, we set $y = 0$ and solve for $x$.
y = x^2 + 2x – 5
0 = x^2 + 2x – 5
x = (-b ±sqrt(b^2-4ac))/(2a)
x = (-2 ± sqrt(2^2 - 4×1×(-5)))/(2×1)
x = (-2 ± sqrt(2^2 - 4×1×(-5)))/(2×1)
x = (-2 ± sqrt(4 + 20))/2
x = (-2 ± sqrt24)/2
x = -1 ± 1/2sqrt24
x = -1 ± 1/2(sqrt(4×6))
x = -1 ± 1/2(2sqrt6)
x = -1 ± sqrt6
The $x$-intercepts are at $x = - 1 - \sqrt{6}$ and $x = - 1 + \sqrt{6}$.
Graph
Now we prepare a chart.
The axis of symmetry passes through $x = - 1$.
Let's prepare a table with points that are 5 units on either side of the axis, that is, from $x = - 6$ to $x = 4$.
Plot these points.
And we have our graph.
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# An Introduction to Mathematical Induction: The Sum of the First n Natural Numbers, Squares and Cubes.
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Sigma Notation
In math, we frequently deal with large sums. For example, we can write
${\displaystyle 1+2+3+4+5+6+7+8+9+10+11+12+13,}$
which is a bit tedious. Alternatively, we may use ellipses to write this as
${\displaystyle 1+2+\cdots +13.}$
However, there is an even more powerful shorthand known as sigma notation. When we write
${\displaystyle {\displaystyle \sum _{i=1}^{13}\,i,}}$
this means the same thing as the previous two mathematical statements. Here, the index below the capital sigma, ${\displaystyle \Sigma }$, is the letter
[/itex]i [/itex], and the ${\displaystyle i}$ that follows the ${\displaystyle \Sigma }$ is our rule to
apply to each value for ${\displaystyle i}$. The values ${\displaystyle 1}$ and ${\displaystyle 13}$ tell us how many times to repeat the rule, i.e., to follow the rule for ${\displaystyle i=1,}$ then add the rule for ${\displaystyle i=2,}$ then for ${\displaystyle i=3,}$ and continue in this manner until you reach 13. In other words,
${\displaystyle {\displaystyle \sum _{i=1}^{13}}\,i\,=\,1+2+3+4+5+6+7+8+9+10+11+12+13.}$
Of course, we can change the rule and/or the index. For example,
${\displaystyle {\displaystyle \sum _{i=1}^{5}}\,i^{2}\,=\,1^{2}+2^{2}+3^{2}+4^{2}+5^{2}.}$
Most importantly, we frequently don't have the luxury of bounds that are actual values. We can also write something like
${\displaystyle {\displaystyle \sum _{i=n}^{2n}}\,i\,=\,n+(n+1)+\cdots +(2n-1)+2n,}$
or
${\displaystyle {\displaystyle \sum _{i=1}^{n}}\,i^{3}\,=\,1^{3}+2^{3}+3^{3}+\cdots +n^{3}.}$
These non-fixed indices allow us to find rules for evaluating some important sums.
## Proof by (Weak) Induction
When we count with natural or counting numbers (frequently denoted
[/itex]\mathbb{N} [/itex]), we begin with one, then keep adding one unit at a
time to get the next natural number. In other words,
The Natural Numbers [/itex]\,=\,\mathbb{N}\,=\,\{1,2,3,\ldots\}\,=\,\{1,1+1,1+1+1,1+1+1+1,\ldots\}. [/itex]
This is the basis for weak, or simple induction; we must first prove our conjecture is true for the lowest value (such as 1), and then show whenever it's true for an arbitrary ${\displaystyle n,}$ it's true for the ${\displaystyle n+1}$ as well. This mimics our expression of the natural numbers as just keep adding one.
It is also equivalent to prove that whenever the conjecture is true for ${\displaystyle n-1,}$ it's true for ${\displaystyle n.}$ Which approach you choose can depend on which is more convenient, or frequently which is more appealing to the teacher grading the work.
Although we won't show examples here, there are induction proofs that require strong induction. This occurs when proving it for the
[/itex]n^{\mathrm{th}} [/itex] case requires assuming more than just the ${\displaystyle (n-1)^{\textrm {th}}}$
case. In such situations, strong induction assumes that the conjecture is true for ALL cases of lower value than ${\displaystyle n}$ down to our base case.
## The Sum of the first ${\displaystyle n}$ Natural Numbers
Claim. The sum of the first ${\displaystyle n}$ natural numbers is
${\displaystyle {\displaystyle \sum _{i=1}^{n}i\,=\,1+2+\cdots +n\,=\,{\frac {n(n+1)}{2}}.}}$
Proof. We must follow the guidelines shown for induction arguments. Our base step is ${\displaystyle n=1,}$ and plugging in we find that
${\displaystyle {\displaystyle {\frac {n(n+1)}{2}}\,=\,{\frac {1(1+1)}{2}}\,=\,1.}}$
This gives us our starting point. For the induction step, let's assume the claim is true for ${\displaystyle n-1,}$ so
${\displaystyle {\displaystyle \sum _{i=1}^{n-1}i\,=\,{\frac {(n-1)\left(\left(n-1\right)+1\right)}{2}}\,=\,{\frac {(n-1)n}{2}}.}}$
Now, we have
${\displaystyle {\begin{array}{rcl}{\displaystyle \sum _{i=1}^{n}i}&=&{\displaystyle \sum _{i=1}^{n-1}i+n}\\\\&=&{\displaystyle {\frac {(n-1)n}{2}}+n\qquad \qquad {\mbox{(by the induction assumption)}}}\\\\&=&{\displaystyle {\frac {n^{2}-n}{2}}+{\frac {2n}{2}}}\\\\&=&{\displaystyle {\frac {n^{2}-n+2n}{2}}}\\\\&=&{\displaystyle {\frac {n^{2}+n}{2}}}\\\\&=&{\displaystyle {\frac {n(n+1)}{2}}},\end{array}}}$
as required.
${\displaystyle \square }$
## The Sum of the first ${\displaystyle n}$ Squares
Claim. The sum of the first ${\displaystyle n}$ squares is
${\displaystyle {\displaystyle \sum _{i=1}^{n}i^{2}\,=\,1^{2}+2^{2}+\cdots +n^{2}\,=\,{\frac {n(n+1)(2n+1)}{6}}.}}$
Proof. Again, our base step is ${\displaystyle n=1,}$ and plugging in we find that
${\displaystyle {\displaystyle {\frac {n(n+1)(2n+1)}{6}}\,=\,{\frac {1(1+1)(2+1)}{6}}\,=\,1.}}$
This gives us our starting point. For the induction step, let's assume the claim is true for ${\displaystyle n-1,}$ so
${\displaystyle {\displaystyle \sum _{i=1}^{n-1}i\,=\,{\frac {(n-1)\left(\left(n-1\right)+1\right)\left(2\left(n-1\right)+1\right)}{6}}\,=\,{\frac {(n-1)n(2n-1)}{6}}\,=\,{\frac {2n^{3}-3n^{2}+n}{6}}.}}$
Now, we have
${\displaystyle {\begin{array}{rcl}{\displaystyle \sum _{i=1}^{n}i^{2}}&=&{\displaystyle \sum _{i=1}^{n-1}i^{2}+n^{2}}\\\\&=&{\displaystyle {\frac {2n^{3}-3n^{2}+n}{6}}+n^{2}\qquad \qquad {\mbox{(by the induction assumption)}}}\\\\&=&{\displaystyle {\frac {2n^{3}-3n^{2}+n}{6}}+{\frac {6n^{2}}{6}}}\\\\&=&{\displaystyle {\frac {2n^{3}+3n^{2}+n}{6}}}\\\\&=&{\displaystyle {\frac {n(2n^{2}+3n+1)}{6}}}\\\\&=&{\displaystyle {\frac {n(n+1)(2n+1)}{6}}},\end{array}}}$
as required.
${\displaystyle \square }$
## The Sum of the first ${\displaystyle n}$ Cubes
Claim. The sum of the first ${\displaystyle n}$ cubes is
${\displaystyle {\displaystyle \sum _{i=1}^{n}i^{3}\,=\,1^{3}+2^{3}+\cdots +n^{3}\,=\,{\frac {n^{2}(n+1)^{2}}{4}}.}}$
Notice that the formula is really similar to that for the first ${\displaystyle n}$ natural numbers.
Proof.. Plugging in ${\displaystyle n=1,}$ we find that
${\displaystyle {\displaystyle {\frac {n^{2}(n+1)^{2}}{4}}\,=\,{\frac {1^{2}(1+1)^{2}}{4}}\,=\,1,}}$
completing our base step.
For the induction step, let's assume the claim is true for ${\displaystyle n-1,}$ so
${\displaystyle {\displaystyle \sum _{i=1}^{n-1}i^{3}\,=\,{\frac {(n-1)^{2}\left(\left(n-1\right)+1\right)^{2}}{4}}\,=\,{\frac {(n-1)^{2}n^{2}}{2}}.}}$
Now, we have
${\displaystyle {\begin{array}{rcl}{\displaystyle \sum _{i=1}^{n}i^{3}}&=&{\displaystyle \sum _{i=1}^{n-1}i^{3}+n^{3}}\\\\&=&{\displaystyle {\frac {(n-1)^{2}n^{2}}{4}}+n^{3}\qquad \qquad {\mbox{(by the induction assumption)}}}\\\\&=&{\displaystyle {\frac {n^{4}-2n^{3}+n^{2}}{4}}+{\frac {4n^{3}}{4}}}\\\\&=&{\displaystyle {\frac {n^{4}+2n^{3}+n^{2}}{4}}}\\\\&=&{\displaystyle {\frac {n^{2}(n^{2}+2n+1)}{4}}}\\\\&=&{\displaystyle {\frac {n^{2}(n+1)^{2}}{4}}},\end{array}}}$
as required.
${\displaystyle \square }$
Aside from being good examples of simple or weak induction, these formulas are frequently used to find an integral as a limit of a Riemann sum. \end{document}
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# Lines and Planes
Lines and planes are perhaps the simplest of curves and surfaces in three dimensional space. They also will prove important as we seek to understand more complicated curves and surfaces.
The equation of a line in two dimensions is $$ax+by=c$$; it is reasonable to expect that a line in three dimensions is given by $$ax + by +cz = d$$; reasonable, but wrong---it turns out that this is the equation of a plane.
A plane does not have an obvious "direction'' as does a line. It is possible to associate a plane with a direction in a very useful way, however: there are exactly two directions perpendicular to a plane. Any vector with one of these two directions is called normal to the plane. So while there are many normal vectors to a given plane, they are all parallel or anti-parallel to each other.
Suppose two points $$(v_1,v_2,v_3)$$ and $$(w_1,w_2,w_3)$$ are in a plane; then the vector $$\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$$ is parallel to the plane; in particular, if this vector is placed with its tail at $$(v_1,v_2,v_3)$$ then its head is at $$(w_1,w_2,w_3)$$ and it lies in the plane. As a result, any vector perpendicular to the plane is perpendicular to $$\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$$. In fact, it is easy to see that the plane consists of precisely those points $$(w_1,w_2,w_3)$$ for which $$\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$$ is perpendicular to a normal to the plane, as indicated in figure 12.5.1.
Turning this around, suppose we know that $$\langle a,b,c\rangle$$ is normal to a plane containing the point $$(v_1,v_2,v_3)$$. Then $$(x,y,z)$$ is in the plane if and only if $$\langle a,b,c\rangle$$ is perpendicular to $$\langle x-v_1,y-v_2,z-v_3\rangle$$. In turn, we know that this is true precisely when $$\langle a,b,c\rangle\cdot\langle x-v_1,y-v_2,z-v_3\rangle=0$$. That is, $$(x,y,z)$$ is in the plane if and only if
\eqalign{ \langle a,b,c\rangle\cdot\langle x-v_1,y-v_2,z-v_3\rangle&=0\cr a(x-v_1)+b(y-v_2)+c(z-v_3)&=0\cr ax+by+cz-av_1-bv_2-cv_3&=0\cr ax+by+cz&=av_1+bv_2+cv_3.\cr }
Working backwards, note that if $$(x,y,z)$$ is a point satisfying $$ax+by+cz=d$$ then
\eqalign{ ax+by+cz&=d\cr ax+by+cz-d&=0\cr a(x-d/a)+b(y-0)+c(z-0)&=0\cr \langle a,b,c\rangle\cdot\langle x-d/a,y,z\rangle&=0.\cr }
Namely, $$\langle a,b,c\rangle$$ is perpendicular to the vector with tail at $$(d/a,0,0)$$ and head at $$(x,y,z)$$. This means that the points $$(x,y,z)$$ that satisfy the equation $$ax+by+cz=d$$ form a plane perpendicular to $$\langle a,b,c\rangle$$. (This doesn't work if $$a=0$$, but in that case we can use $$b$$ or $$c$$ in the role of $$a$$. That is, either $$a(x-0)+b(y-d/b)+c(z-0)=0$$ or $$a(x-0)+b(y-0)+c(z-d/c)=0$$.)
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Geometric Progression, GP
Geometric progression (also known as geometric sequence) is a sequence of numbers where the ratio of any two adjacent terms is constant. The constant ratio is called the common ratio, r of geometric progression. Each term therefore in geometric progression is found by multiplying the previous one by r.
Eaxamples of GP:
• 3, 6, 12, 24, … is a geometric progression with r = 2
• 10, -5, 2.5, -1.25, … is a geometric progression with r = -1/2
The nth term of geometric progression
Given each term of GP as a1, a2, a3, a4, …, am, …, an, expressing all these terms according to the first term a1 will give us...
$a_1 = a_1$
$a_2 = a_1 r$
$a_3 = a_2 r = (a_1 r) \, r = a_1 r^2$
$a_4 = a_3 r = (a_1 r^2) \, r = a_1 r^3$
$\dots$
$a_m = a_1 r^{\, m - 1}$
$\dots$
$a_n = a_1 r^{\, n - 1}$
Where
a1 = the first term, a2 = the second term, and so on
an = the last term (or the nth term) and
am = any term before the last term
Sum of Finite Geometric Progression
The sum in geometric progression (also called geometric series) is given by
$S = a_1 + a_2 + a_3 + a_4 + \, \dots \, + a_n$
$S = a_1 + a_1 r + a_1 r^2 + a_1 r^3 + \, \dots \, + a_1 r^{\, n - 1}$ ← Equation (1)
Multiply both sides of Equation (1) by r will have
$Sr = a_1 r + a_1 r^2 + a_1 r^3 + a_1 r^4 + \, \dots \, + a_1 r^{\, n}$ ← Equation (2)
Subtract Equation (2) from Equation (1)
$S - Sr = a_1 - a_1 r^{\, n}$
$(1 - r)S = a_1 (1 - r^{\, n})$
$S = \dfrac{a_1 (1 - r^{\, n})}{1 - r}$
The above formula is appropriate for GP with r < 1.0
Subtracting Equation (1) from Equation (2) will give
$Sr - S = a_1 r^{\, n} - a_1$
$(r - 1)S = a_1 (r^{\, n} - 1)$
$S = \dfrac{a_1 (r^{\, n} - 1)}{r - 1}$
This formula is appropriate for GP with r > 1.0.
Sum of Infinite Geometric Progression, IGP
The number of terms in infinite geometric progression will approach to infinity (n = ∞). Sum of infinite geometric progression can only be defined at the range of -1.0 < (r ≠ 0) < +1.0 exclusive.
From
$S = \dfrac{a_1 (1 - r^{\, n})}{1 - r}$
$S = \dfrac{a_1 - a_1 r^{\, n}}{1 - r}$
$S = \dfrac{a_1}{1 - r} - \dfrac{a_1 r^{\, n}}{1 - r}$
For n → ∞, the quantity (a1rn) / (1 - r) → 0 for -1.0 < (r ≠ 0) < +1.0, thus,
$S = \dfrac{a_1}{1 - r}$
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# Indefinite integral
Let $f$ be any function and $F$ its antiderivative. The set of all antiderivatives of $f$ is called indefinite integral of the function $f$ denoted by
$$\int f(x) dx = F(x) + C.$$
Every two primitive functions differ by a constant $C$. This means, if we know the one antiderivative $F$ of the function $f$, then the another we can write in the form $F(x) + C$.
We do not have strictly rules for calculating the antiderivative (indefinite integral). The most antiderivatives we know is derived from the table of derivatives, which we read in the opposite direction.
Table of basic integrals
$$\int dx = x + C$$
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n\neq 1$$
$$\int \frac{1}{x} dx = \ln |x| + C$$
$$\int \frac{1}{x^n} dx = – \frac{1}{(n-1)x^{n-1}} + C, \quad n \neq 1$$
$$\int \frac{1}{\sqrt{x}} dx = 2 \sqrt{x} + C$$
$$\int e^x dx = e^x + C$$
$$\int a^x dx = \frac{a^x}{\ln a} + C, \quad a>0, a\neq 1$$
$$\int \sin x dx = – \cos x + C$$
$$\int \cos x dx = \sin x + C$$
Table of integrals of rational functions
Example 1. Determine
$$\int (2 – 3x + x^2 )dx.$$
Solution.
By the additivity and linearity property of the integral, we have
$$\int (2 – 3x + x^2 )dx = 2 \int dx – 3 \int x dx + \int x^2 dx.$$
From the table of integral, we read
$$\int (2 – 3x + x^2 )dx = 2x – \frac{3}{2} x^2 + \frac{1}{3}x^3 + C.$$
Example 2. Determine
$$\int \frac{x^3 -1}{2x^2} dx.$$
Solution.
$$\int \frac{x^3 -1}{2x^2} dx = \int \left(\frac{x^3}{2x^2} – \frac{1}{2x^2} \right) dx$$
$$= \int \left(\frac{x}{2} – \frac{1}{2x^2} \right) dx$$
$$= \frac{1}{2} \left( \int x – \frac{1}{x^2} \right) dx.$$
By the additivity property of the integral, we have
$$\frac{1}{2} \left( \int x – \frac{1}{x^2} \right) dx = \frac{1}{2} \left (\int x dx – \int \frac{1}{x^2} dx \right).$$
From the table of indefinite integrals, we read:
$$\frac{1}{2} \left (\int x dx – \int \frac{1}{x^2} dx \right) = \frac{1}{2} \left ( \frac{1}{2}x^2 + \frac{1}{x} \right) + C$$
$$= \frac{x^2}{4} + \frac{1}{2x} + C,$$
that is
$$\int \frac{x^3 -1}{2x^2} dx = \frac{x^2}{4} + \frac{1}{2x} + C.$$
Example 3. Determine
$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx .$$
Solution.
Since $\sin 2x = 2 \sin x \cos x$, we can write
$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx = \int \frac{2 \sin x \cos x + 3 \sin^2 x}{\sin x} dx$$
$$= \int \left( \frac{2 \sin x \cos x}{\sin x} + \frac{3 \sin^2 x}{\sin x} \right) dx$$
$$= \int \left( 2 \cos x + 3 \sin x \right) dx.$$
By the linearity and additivity property of the integral, we have
$$\int \left( 2 \cos x + 3 \sin x \right) = 2 \int \cos x dx + 3 \int \sin x dx.$$
From the table of indefinite integrals we read:
$$2 \int \cos x dx + 3 \int \sin x dx = 2 \sin x – 3 \cos x + C,$$
that is
$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx = 2 \sin x – 3 \cos x + C.$$
Integration by substitution
We will use the reverse chain rule of differentiation of composite functions and thus obtain the method which is called the method of substitution.
The chain rule:
$$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x).$$
The reverse of the chain rule:
$$\int f’ (g(x)) g'(x) dx = f(g(x)) + C.$$
The integration rule: if $g: I \to \mathbb{R}, I \subseteq \mathbb{R}$ is a differentiable and continuous function on $I$, then
$$\int f(g(x)) g'(x) dx = \int f(u) du,$$
where $g(x) = u$ and $g'(x) dx = du$.
Firstly, we need to choose a substitution to make. A substitution is not determined in advance, just by using the exercises we can discover the simplest way. Then we have to express the variable $x$ by $u$ and connect $du$ and $dx$. We change the integrand and $dx$ with expressions by the variable $u$ and calculate the integral. The result we write as the function of the variable $x$, that is, we return a substitution.
Example 4. Determine
$$\int \sqrt{3 + x} dx.$$
Solution.
We choose a substitution $u= \sqrt{3+x}, (u \ge 0).$ It follows
$$x= u^2 – 3$$
and
$$dx = 2u du.$$
Now, by changing $x$ with $u$, we have:
$$\int \sqrt{3 + x} dx = \int u 2u du = \int 2u^2 du.$$
From the linearity property of integrals we can write
$$\int 2u^2 du = 2 \int u^2 du.$$
From the table of integrals we read:
$$2 \int u^2 du = \frac{2}{3} u^3 + C.$$
Finally, the result we need to write as the function of the variable $x$:
$$\frac{2}{3} u^3 + C = \frac{2}{3} \sqrt{(3+x)^3} + C,$$
that is
$$\int \sqrt{3 +x } dx = \frac{2}{3} \sqrt{(3+x)^3} + C.$$
Example 5. Determine
$$\int \frac{2x + 1}{x^2 + x -3} dx.$$
Solution.
We choose a substitution $u = x^2 + x – 3$. Therefore, we have
$$\frac{du}{dx} = 2x +1 \Rightarrow dx = \frac{du}{2x+1}.$$
By replacing $x^2 + x – 3$ with $u$ and $dx$ with $\frac{du}{2x+1}$ we obtain:
$$\int \frac{2x + 1}{x^2 + x -3} dx = \int \frac{2x+1}{u} \cdot \frac{du}{2x + 1} = \int \frac{du}{u}.$$
From the table of integrals we read:
$$\int \frac{du}{u} = \ln |u|+ C.$$
By returning a substitution, we obtain the finally result
$$\int \frac{2x + 1}{x^2 + x -3} dx = \ln |x^2 + x – 3| + C.$$
Example 6. Determine
$$\int \frac{\ln x}{x} dx.$$
Solution.
We choose a substitution $u = \ln x$. It follows
$$\frac{du}{dx} = \frac{1}{x} \Rightarrow dx = x du.$$
By replacing $\ln x$ with $u$ and $dx$ with $x du$, we have
$$\int \frac{\ln x}{x} dx = \int \frac{u}{x} x du = \int u du.$$
From the table of integrals we read:
$$\int u du = \frac{1}{2} u^2 + C,$$
that is
$$\int \frac{\ln}{x} dx = \frac{1}{2} \ln^2 x + C.$$
Example 7. Determine
$$\int \cot x dx.$$
Solution.
Since $\cot x = \frac{\cos x}{\sin x}$, we can write
$$\int \cot x dx = \int \frac{\cos x}{\sin x} dx.$$
We choose a substitution $u = \sin x$. It follows
$$\frac{du}{dx} = \cos x \Rightarrow dx = \frac{du}{\cos x}.$$
Therefore, we have
$$\int \frac{\cos x}{\sin x} dx = \int \frac{\cos x}{u} \cdot \frac{du}{\cos x} = \int \frac{du}{u}.$$
From the table of integrals we read
$$\int \frac{du}{u} = \ln|u| + C.$$
By returning a substitution, we obtain
$$\int \cot x dx = \ln |\sin x| + C.$$
The same principle we use for calculating the definite integral. However, the result we do not need write as the function of the variable $x$, because the result is a number. Instead, together with the integrand we are changing the lower and upper limit of integration.
Example 8. Determine
$$\int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^3 x dx.$$
Solution.
We choose $u = \cos x$. It follows
$$\frac{du}{dx} = – \sin x \Longrightarrow dx = – \frac{du}{\sin x}.$$
The lower and upper limit are transformed in the following way:
$$x = 0 \Longrightarrow u = \cos 0 = 1,$$
$$x = \frac{\pi}{2} \Longrightarrow u = \cos \frac{\pi}{2} = 0.$$
Now we have:
$$\int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^3 x dx = \int_{0}^{1}u^4 \cdot \sin^3 x \cdot \left( – \frac{du}{\sin x} \right)$$
$$= \int_{0}^{1} u^4 \cdot (-\sin^2) du.$$
From the relation $\sin^2 x + \cos ^2 x = 1$, we can write $\sin^2 x = 1 – \cos^2 x$, that is
$$\int_{0}^{1} u^4 \cdot (-\sin^2) du = \int_{0}^{1} u^4 \cdot (-1)(1 – \cos^2 x) du$$
$$= \int_{0}^{1} u^4 \cdot (u^2 -1) du$$
$$= \int_{0}^{1} (u^6 – u^4) du$$
$$= \int_{0}^{1} u^6 du – \int_{0}^{1} u^4 du.$$
By using the first fundamental theorem of calculus, finally we have:
$$\int_{0}^{1} u^6 du – \int_{0}^{1} u^4 du = \frac{u^7}{7} \bigg|_{0}^{1} – \frac{u^5}{5}\bigg|_{0}^{1} = – \frac{2}{35}$$
Integration by parts
We cannot calculate all integrals by using the method of substitution. Therefore, for calculate the simple integral as $\int x \ln x$ we need to use different methods.
Integration by parts is the one useful method for calculating integrals. Let $u$ and $v$ be functions of the variable $x$. The formula for derivative of the product
$$[u(x)v(x)]’ = u'(x)v(x) + u(x)v'(x)$$
can be written in the equivalent form:
$$u(x)v(x) = \int [u'(x)v(x) + u(x)v'(x)] dx = \int u'(x) v(x) dx + \int u(x)v'(x) dx.$$
Thus we obtain:
$$\int u(x)v'(x) dx = u(x)v(x) – \int u'(x)v(x) dx,$$
or more simply
$$\int u v’ dx = uv – \int u’ v dx.$$
The formula above is called the formula for integration by parts.
The method of the integration by parts
• The integrand write as the product of functions $u$ and $v’$.
• Calculate the extra integral $v= \int v’ dx$. The function $v’$ must be chosen in the way that its integral is easy to determine.
• Calculate the derivative $u’$.
• Write the formula for integration by parts:
$$\int uv’ dx = uv – \int u’v dx.$$
• Calculate the integral $\int u’v dx$.
Example 9. Determine
$$\int x \sin x dx.$$
Solution.
We choose $v’ = \sin x dx$ and $u = x$. Now we calculate the integral $v= \int v’ dx$, that is
$$v’ = \sin x \Rightarrow v= \int \sin x dx = – \cos x + C.$$
The derivative of $u$ is equal to
$$u = x \Rightarrow u’ = 1.$$
By using the formula for integration by parts we have
$$\int x \sin x dx =x \cdot( – \cos x) + \int \cos x.$$
From the table of integrals we read: $\int \cos x = \sin x + C$. Finally, we have
$$\int x \sin x dx = – x \cdot \cos x + \sin x + C.$$
In some cases, integration by parts must be repeated.
Example 10. Determine
$$\int e^x \cos x dx.$$
Solution.
We choose $u = e^x$ and $v’ = \cos x dx$. It follows
$$v = \int \cos x dx \Longrightarrow v = \sin x ,$$
$$du = e^x dx.$$
By using the formula for integration by parts, we have:
$$\int e^x \cos x dx =e^x \cdot \sin x – \int e^x \sin x dx.$$
Now we choose $t = e^x$ and $w’ = \sin x dx$. It follows
$$w = \int \sin x dx \Longrightarrow w = – \cos x ,$$
$$dt = e^x dx,$$
that is
$$\int e^x \cos x dx =e^x \cdot \sin x – \int e^x \sin x dx$$
$$= e^x \cdot \sin x – \left[e^x \cdot ( – \cos x) – \int e^x \cos x dx \right]$$
$$= e^x \cdot \sin x + e^x \cdot \cos x – \int e^x \cos x dx.$$
Finally, we have
$$2 \int e^x \cos x dx = e^x ( \sin x + \cos x) + C,$$
that is
$$\int e^x \cos x dx = \frac{1}{2} e^x ( \sin x + \cos x) + C.$$
For the definite integral is valid the formula for integration by parts:
$$\int_{a}^b u(x) v'(x) dx = u(x) v(x) \bigg|_{a}^{b} – \int_{a}^{b} u'(x) v(x) dx.$$
Example 11. Determine
$$\int_{2}^{3} x e ^{x} dx.$$
Solution.
We choose $u= x$ and $v’= e^x dx$. It follows
$$v = \int e^x dx \Longrightarrow v = e^x,$$
$$du =dx.$$
By using the formula for integration by parts for definite integral and first fundamental theorem of calculus, we have:
$$\int_{2}^{3} x e ^{x} dx = x \cdot e^x \bigg|_{2}^3 – \int_{2}^{3} e^x dx$$
$$= x \cdot e^x \bigg|_{2}^3 – e^x \bigg|_2^3$$
$$=(3 \cdot e^3 – 2 \cdot e^2) – (e^3 – e^2)$$
$$=2 e^3 – e^2$$
$$=e^2(2e -1).$$
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# 5.10: Proportions Using Cross Products
Difficulty Level: At Grade Created by: CK-12
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Practice Proportions Using Cross Products
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Jamaya dropped bottles and wrappers whenever and wherever she wanted to. In order to break this bad habit, Jamaya’s mother made her help with a local clean-up effort. For every piece of litter that Jamaya had dropped in and around her home, she had to pick up 10 pieces with the group. If Jamaya dropped a total of 15 bottles, cans, and wrappers, how many pieces of litter does she have to clean up?
In this concept, you will learn how to solve proportions using cross products.
### Solving Proportions Using Cross Products
ratio represents a comparison between two quantities. Equivalent ratios are ratios that are equal. A proportion is made up of two equivalent ratios.
Proportional reasoning, or examining the relationship between two numbers, can be used to determine the value of x.
A proportion can be expressed as two equivalent fractions.
\begin{align*}\frac{a}{b}=\frac{c}{d}\end{align*}
A proportion can be expressed with colons.
\begin{align*}a:b = c:d\end{align*}
In a proportion, the means are the two terms that are closest together when the proportion is written with colons. So, in \begin{align*}a:b = c:d\end{align*}, the means are \begin{align*}b\end{align*} and \begin{align*}c\end{align*}.
The extremes are the terms in the proportion that are furthest apart when the proportion is written with colons. So, in \begin{align*}a:b = c:d\end{align*}, the extremes are \begin{align*}a\end{align*}and \begin{align*}d\end{align*}.
The diagram below shows how to identify the means and the extremes in a proportion.
The Cross Products Property of Proportions states that the product of the means is equal to the product of the extremes.
\begin{align*}\frac{a}{b}&=\frac{c}{d}\\ b\cdot c&=a\cdot d\end{align*}
Here is an example.
Solve for a.
\begin{align*}\frac{a}{4}=\frac{6}{8}\end{align*}
First, multiply the means and the extremes and set them equal to one another.
\begin{align*}a\cdot 8&=4\cdot6\\ 8a&=24\end{align*}
Next, solve the equation for the missing variable.
\begin{align*}\frac{8a}{8}&=\frac{24}{8}\\ a&=3\end{align*}
The answer is a = 3.
### Examples
#### Example 1
Earlier, you were given a problem about Jamaya and her littering habit.
She had dropped a total of 15 pieces of litter in and around her home, and her mother said she had to pick up 10 pieces for every piece that she dropped. How many pieces of litter does Jamaya have to clean up?
First, write a proportion.
\begin{align*}\frac{10 \ pieces \ to \ pick \ up }{1 \ piece \ dropped}= \frac{x \ pieces \ to \ pick \ up}{15 \ pieces \ dropped}\end{align*}
Next, cross multiply.
\begin{align*}x = 150\end{align*}
Jamaya must pick up 150 pieces of litter.
#### Example 2
Solve for x.
\begin{align*}\frac{x}{5}=\frac{6}{10}\end{align*}
First, cross multiply and set the products equal to one another.
\begin{align*}10x = 30\end{align*}
Next, solve for x.
\begin{align*}x = 3\end{align*}
#### Example 3
Solve for a.
\begin{align*}\frac{a}{9}=\frac{15}{27}\end{align*}
First, cross multiply and set the products equal to one another.
\begin{align*}27a = 135\end{align*}
Next, solve for a.
\begin{align*}a = 5\end{align*}
The answer is a = 5.
Example 4
Solve for b.
\begin{align*}\frac{b}{4}=\frac{12}{16}\end{align*}
First, cross multiply and set the products equal to one another.
\begin{align*}16b = 48\end{align*}
Next, solve for b.
\begin{align*}b = 3\end{align*}
The answer is b = 3.
Example 5
Rudy was a silly kitten who loved to play with ping pong balls. It took him just 15 minutes to swat 12 of them underneath the sofa. How many ping pong balls could Rudy hit under the sofa in 1 hour?
First, write a proportion.
\begin{align*}\frac{15 \ minutes}{12 \ balls} = \frac{60 \ minutes}{x \ balls}\end{align*}
Next, cross multiply.
\begin{align*}15x = 720\end{align*}
Then, solve for x.
\begin{align*}x = 48\end{align*}
The answer is 48. In one hour, 60 minutes, Rudy could swat 48 ping pong balls under the sofa.
### Review
Use cross products to find the value of the variable in each proportion.
1. \begin{align*}\frac{6}{10} = \frac{x} {5}\end{align*}
2. \begin{align*}\frac{2}{3} = \frac{x} {9}\end{align*}
3. \begin{align*}\frac{4}{9} = \frac{a} {45}\end{align*}
4. \begin{align*}\frac{7}{8} = \frac{a} {4}\end{align*}
5. \begin{align*}\frac{b}{8} = \frac{5} {16}\end{align*}
6. \begin{align*}\frac{6}{3} = \frac{x} {9}\end{align*}
7. \begin{align*}\frac{4}{x} = \frac{8} {10}\end{align*}
8. \begin{align*}\frac{1.5}{y} = \frac{3} {9}\end{align*}
9. \begin{align*}\frac{4}{11} = \frac{c} {33}\end{align*}
10. \begin{align*}\frac{2}{6} = \frac{5} {y}\end{align*}
11. \begin{align*}\frac{2}{10} = \frac{5} {x}\end{align*}
12. \begin{align*}\frac{4}{12} = \frac{6} {n}\end{align*}
13. \begin{align*}\frac{5}{r} = \frac{70} {126}\end{align*}
14. \begin{align*}\frac{4}{14} = \frac{14} {k}\end{align*}
15. \begin{align*}\frac{8}{w} = \frac{6} {3}\end{align*}
16. \begin{align*}\frac{2}{5} = \frac{17} {a}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Cross Product Property of Proportions The cross product property of proportions states that the cross products of two ratios will be equal if the two ratios form a proportion.
Cross Products To simplify a proportion using cross products, multiply the diagonals of each ratio.
Extremes In a proportion, the extremes are the values of the proportion that are furthest apart when written in ratio form using a colon. For example: In the proportion a : b = c : d, a and d are the extremes.
Means In a proportion, the means are the values of the proportion that are close to each other when written in ratio form using a colon.
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Instructor: Maria Blojay
Maria has taught College Algebra and has a master's degree in Education Administration.
This lesson will show how to solve radical inequalities using radical definitions and properties. Key terms and radical notes are included to share step by step the method to solve these special types of inequalities.
There are some mathematical definitions and properties for radicals that we need to review. These properties will help us to understand how to solve radical inequalities.
The definition of a radical inequality is an inequality that holds a variable expression within it. This means that the variable expression sits underneath the radical, and is called a radicand.
For example:
Let's take a look at this property:
The radicand cannot have a calculated final value that is negative WHEN the index of the radical is an even number.
Having a negative under the radical when the index is an even number, such as 2, 4, 6, etc. means that there is no solution. This must be checked for each radical inequality problem. We will see some problems using these extra checks.
We cannot mathematically solve the problem with the radical symbol in it. It must be removed. So don't forget that we have this property that tells us how to do that:
This means that we take the index value and use this same value as an exponent. In doing so, it cancels out the radical symbol and leaves the radicand variable expression by itself.
Let's take a look at an example of this:
If the index is not shown in the radical, it means that the index is equal to two.
Now that we have looked at our radical definitions and properties, we will take a look at three sample problems.
## Example 1
Let's take a look at our first problem:
In simplifying:
Squaring each side, we get,
Leaving us with,
To check, choose an x-value greater than 40. If we chose a number such as 68 and replaced this selected value for x:
which is greater than 8. So x >40 is the correct inequality answer.
## Example 2
Given our second example:
To get rid of the radical, we square each side of the inequality:
We then simplify the inequality and get:
Remember that our radicand can NOT be negative, or another way of saying this is that the radicand must be positive:
To check this ... we get:
Let's check our example with x-values of 3 and 5:
Here we have shown this is a true inequality, 0 is less than 2. Now let's try the x value 5:
Yes, we have a true inequality with an x value of 3 which is equal to 2.
The values of x that are 3 and 5 AND all values of x in between 3 and 5 will make the inequality true.
Here, 3 is our lower interval value, and 5 is our upper interval value.
To write all of the values that would solve this inequality concisely, we would write it as a compound inequality.
A compound inequality is written by using the lower and upper interval values with the same inequality symbols in between those values.
So, we would write:
## Example 3
In our last example, we have a radical expression on each side of the inequality.
Given:
Squaring each side:
Leaving us with:
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1) Arrange both the polynomial is same order of exponent . It would be good to have terms arrange from highest exponent to lowest exponent i.e
For example
S(x) = x2 + 5x3 + 1 +10x
P(x) = x3 + x2 + 5 +10x
Arranging them as per suggestion above
S(x) = 5x3 +x2+ 10x +1
P(x) = x3 + x2 + 10x +5
2) Add the like term . By like term ,we mean same exponent terms. There are two method to add like terms. We either can group them horizontally and add it. We can add vertically
Horizontally
S(x) + P(x)
=x2 + 5x3 + 1 +10x + (x3 + x2 + 5 +10x)
Arranging both the polynomial in same order
=5x3 +x2+ 10x +1 + (x3 + x2 + 10x +5 )
Opening the parentthesis and grouping the like terms. In addition,no sign are changed when parentthesis are opened
= 5x3 + x3 + x2+x2 + 10x +10x + 1+5
=6x3+2x2+20x+6
Vertically
S(x) = x2 + 5x3 + 1 +10x
P(x) = x3 + x2 + 5 +10x
Arranging both the polynomial in same order
S(x) = 5x3 +x2+ 10x +1
P(x) = x3 + x2 + 10x +5
Similarly we can add three or more polynomials
Example-1
(2x3 - x+1+x2) + (x3 + 6x - 7) + (-3x2 - 11 + 2x)
Arranging them in same order
=(2x3 + x2 - x+1) + (x3 + 6x - 7) + (-3x2 + 2x - 11)
Opening the parentthesis and grouping the like terms
=2x3+x3 +x2-3x2 -x +6x+2x +1-7-11
=3x3-2x2+7x-16
## Subtracting polynomials
We many times need to subtract the two polynomial .It is very similar to adding polynomails only.Subtracting polynomials just means Subtracting the like terms.We need to follow below steps for Subtraction of polynomial
1) Arrange both the polynomial is same order of exponent . It would be good to have terms arrange from highest exponent to lowest exponent i.e
For example
S(x) = x2 + 5x3 + 1 +10x
P(x) = x3 + x2 + 5 +10x
Arranging them as per suggestion above
S(x) = 5x3 +x2+ 10x +1
P(x) = x3 + x2 + 10x +5
2) Subtract the like term . By like term ,we mean same exponent terms. There are two method to subtract like terms. We either can group them horizontally and subtract it. We can add vertically
Horizontally
S(x) - P(x)
=x2 + 5x3 + 1 +10x - (x3 + x2 + 5 +10x)
Arranging both the polynomial in same order
=5x3 +x2+ 10x +1 -(x3 + x2 + 10x +5 )
Opening the parentthesis and grouping the like terms. In Subtract, sign are reversed when parentthesis are opened i.e + becomes - and - becomes +
= 5x3 - x3 + x2 - x2 + 10x -10x + 1-5
Subtact the like term
=4x3 - 4
Vertically
S(x) = x2 + 5x3 + 1 +10x
P(x) = x3 + x2 + 5 +10x
Arranging both the polynomial in same order
S(x) = 5x3 +x2+ 10x +1
P(x) = x3 + x2 + 10x +5
Similarly we can add three or more polynomials
Example-1
(2x3 - x+1+x2) - (x3 + 6x - 7) - (-3x2 - 11 + 2x)
Arranging them in same order
=(2x3 + x2 - x+1) - (x3 + 6x - 7) - (-3x2 + 2x - 11)
Opening the parentthesis and grouping the like terms
=2x3- x3 +x2+3x2 -x -6x-2x +1+7+11
=x3+4x2-9x+18
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In this post we are going to learn to analyze, think about, and solve addition problems. Usually, failure in mathematics comes from not understanding well the problems that we are given. This is why we are going to analyze 5 problems which require addition to solve.
There were 8 cars in the parking lot of a hotel. 5 more cars have entered the parking lot. How many cars are in the parking lot now?
This is the simplest type of problem that we can encounter: it is a short statement, with only 2 facts which have the same units: 8 cars and 5 cars. Now the only thing left for us to think about is what operation we have to do. To do this, we are going to think about if in the beginning there were 8 cars and later 5 more cars enter the parking lot, are there now more or fewer cars than in the beginning? The answer to this question is that there are now more cars than in the beginning. Therefore, the operation that we need to do is ADDITION.
## 8+5=13
Now there are 13 cars in the parking lot.
The walls of Catherine’s class are full of drawings! Today I went to see the class and on the wall to the right I counted 5 drawings and on the wall to the left 7 more. How many drawings in total are there on the walls of Catherine’s class?
Although this problem is simple, it is a little more complicated than the previous problem because it is longer, but it only has 2 facts which also have the same units: 5 drawings and 7 drawings. Before thinking about the operation that we have to do, we are going to reason through the statement: in order to count how many drawings there are in total we could collect them and put them all on the same wall and then count them. The number that we would obtain would be greater than 5 and greater than 7 too. Thus, the operation we have to use is ADDITION.
## 5+7=12
There are 12 drawings in total.
Mary has 4 candies that her grandmother gave to her when she was visiting. Richard has 2 more candies than Mary, because he bought some and later realized that he had more in his house. How many candies does Richard have?
In this problem we have to compare two quantities: Mary has 4 candies and Richard has 2 more candies than María. Let’s think. Does Richard have more or fewer candies than Mary? If Richard has 2 candies more than Mary, this means that he has more candies. So the operation that we need to carry out is ADDITION.
## 4+2=6
Richard has 6 candies.
Yesterday many kids went to eat breakfast in the school dining hall and drank all the juice. If they had served 5 liters less of juice yesterday, they would have served as many liters as they did today. Today they served 6 liters of juice. How many liters of juice did they serve yesterday?
In this problem we have to match two quantities: today they served 6 liters of juice and yesterday if they had served 5 liters less, they would have served the same amount of liters as today. We have to decide whether they served more or less juice yesterday than today. As we are told that to match today’s quantity they would have had to serve 5 liters less, we deduce that yesterday they served more liters than today. Thus, the operation that we have to use is ADDITION.
## 5+6=11
Yesterday they served 11 liters of juice.
During recess I’ve been playing marbles with 4 friends. Since I have been careless, I have lost 3 marbles and now I am left with 6 marbles. How many marbles did I have before I began to play?
In this problem we have 3 numerical facts: 4 friends, 3 marbles and 6 marbles. But be careful! Not all the facts help solve the problem. In this case, the number of friends that I have does not influence the quantity of marbles I had. So this fact is not relevant to the solution of the problem. We focus on the facts that interest us: I have lost 3 marbles and now I have 6 left. We have to reflect about the following question: before beginning to play did I have more or fewer marbles than I have now? If I lost marbles during the game, it means that I had more marbles before than I have now, so the operation we have to use is ADDITION.
## 3+6=9
Before playing, I had 9 marbles.
And that’s all for today’s post. We hope that you have enjoyed and learned from this post.
Next week we will continue with another post of problems.
If you want to learn much more math, try Smartick for free!
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# Dao's Six Point Circle
### What Might This Be About?
6 June 2014, Created with GeoGebra
### Problem
The centers of the six circles tangent to the medians of $\Delta ABC$ at the centroid and through the vertices of the triangle are concyclic:
### Solution
Let $A', B', C'$ be the midpoints of the sides of $\Delta ABC;$ $A'',$ $B'',$ $C''$ the midpoints of $AG, BG, CG$, respectively, and let the perpendicular bisectors of $AG, BG, CG$ meet at $K, L, M;$ finally, let $A_{B},$ $A_{C},$ $B_{C},$ $B_{A},$ $C_{A},$ $C_{B}$ be the centers of the mentioned six circles.
Since $GA''\perp KM$ and $GC''\perp LM,$ quadrilateral $A''GC''M$ is cyclic, implying
(1)
$\angle A_{C}C_{A}M + \angle C_{A}A_{C}M = \angle AGB' + \angle B'GC.$
As $B'$ is the midpoint of $AC$, we have
$\displaystyle \frac{\sin\angle AGB'}{\sin\angle B'GC} = \frac{GC}{GA} =\frac{GC''}{GA''}=\frac{GC_{A}}{GA_{C}}$
because $\Delta GA''A_{C}\sim \Delta GC''C_{A}$. And, to contninue,
$\displaystyle\frac{GC_{A}}{GA_{C}}=\frac{MA_{C}}{MC_{A}} = \frac{\sin\angle C_{A}A_{C}M}{\sin\angle A_{C}C_{A}M},$
so that
(2)
$\displaystyle\frac{\sin\angle AGB'}{\sin\angle B'GC} = \frac{\sin\angle A_{C}C_{A}M}{\sin\angle C_{A}A_{C}M}.$
(1) and (2) give $\angle AGB' = \angle A_{C}C_{A}M$ as well as $\angle B'GC = \angle C_{A}A_{C}M.$ Further, $\angle AGB' = \angle MKL$ because the sides are pairwise perpendicular. Hence, $A_{C}C_{A}$ is an antiparallel in $\Delta KLM.$ By analogy, so are $A_{B}B_{A}$ and $B_{C}C_{B},$ showing that the six centers form a Tucker hexagon and lie on a Tucker circle.
### Acknowledgment
This problem has been posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page. The solution is a modification of the one posted at a Vietnamese math site under the moniker Vslmat.
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# Algebra: Solving Rational Inequalities
## Solving Rational Inequalities
At the end of Quadratic Equations and Inequalities, I showed you how to solve and graph one-variable quadratic inequalities. Do you remember the process? Basically, you factored the quadratic, found critical numbers, split the number line into intervals based on those critical numbers, and then tested those intervals to see which were solutions of the inequality.
You'll use a very similar process to solve rational inequalities. Actually, to be perfectly honest, the process is exactly the same—I just didn't give you a completely accurate definition of what a critical number was back then. It's not that I didn't want to tell you; I've just found that algebra students work best on a "need to know" basis. You really didn't need to know the full definition in Quadratic Equations and Inequalities, but to solve rational inequalities, you do.
##### Talk the Talk
A critical number either causes a function to equal 0 or makes it undefined.
I already mentioned that a critical number is an x value that makes an expression equal 0. However, there's another way that values earn the classification of "critical number," when they cause the function to be undefined.
If you follow these steps, solving rational inequalities is a piece of cake:
1. Rearrange the inequality so that only 0 remains on the right side. This means you should add and subtract terms from both sides to get everything moved to the left.
2. Create one fraction on the left side, if necessary. Use common denominators to combine any terms into one single fraction.
3. Factor the numerator and denominator. This makes finding critical numbers extremely easy.
4. Set each factor in the numerator equal to 0 and solve. Mark these critical numbers on the number line using an open dot (if the inequality symbol is < or >) or a closed dot (if the inequality symbol is ≤ or ≥).
5. Set each factor in the denominator equal to 0 and solve. These, too, are critical numbers, but should always be marked with an open dot on the number line, since they represent the values that cause the rational function to be undefined. (Remember, a 0 in the denominator is bad news, since dividing by 0 is illegal.)
6. Choose test points to find solution intervals. Once you've found the critical numbers, the process is identical to the steps you followed in Quadratic Equations and Inequalities.
Be extra careful with the dots you place on the number line. If you use the wrong dot, you'll get the inequality signs in your final answer wrong, not to mention that the graph will be inaccurate as well.
Example 5: Solve the inequality and graph its solution.
Solution: Start by moving that -x to the left side of the inequality by adding x to both sides—the right side has to be completely clear of any terms except 0.
Your goal now is to create only one fraction on the left side of the inequality by adding everything together. The least common denominator of those two terms is x + 4, so multiply the newly relocated x1 term's numerator and denominator by that value and combine the fractions.
Factor the numerator.
Set each factor of the numerator equal to 0 and solve to get two critical numbers.
• x + 1 = 0 or x + 5 = 0
• x = -1 or x = -5
##### You've Got Problems
Problem 5: Solve and graph the inequality x + 7x - 2 < 3.
You should mark both x = -1 and x = -5 on the number line with solid dots, since the inequality sign ≥ allows for equality, as shown in Figure 18.4. The final critical number is generated by setting the denominator's factor equal to 0.
• x + 4 = 0
• x = -4
Remember to use an open dot for x = -4, since its value comes from the denominator.
Now choose test points from each interval (I suggest x = -6, x = -4.5, x = -2, and x = 0). Both x = -4.5 and x = 0 make the inequality true, so their intervals make up the solution: -5 ≤x < -4 or x ≥-1. Darken those intervals on the number line to make the graph, shown in Figure 18.5
Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.
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Abstract Algebra
# Abstract Algebra (3rd Edition)Solutions for Chapter 1.3
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 8
Let be the permutation as follows:
Let us find the cycle decomposition of the permutation.
To start a cycle decomposition pick the smallest element here the smallest element is 1. Now note down , which is 3. That is 1 is sent to 3, 3 is sent to 5 and 5 is sent to 1.
So, we get one cycle.
• Step 2 of 8
Next, 2 is sent to 4 and 4 is sent to 2. Therefore, the other cycle is .
Therefore, the cycle decomposition of the permutation is.
• Step 3 of 8
Let be the permutation as follows:
Now write the permutation as shown below:
1 is sent to 5, 5 is sent to 1. Therefore, the cycle is.
• Step 4 of 8
Now, 2 is sent to 3 and 3 is sent to 2 and 4 is sent to 4 itself. Therefore, the cycle is.
Therefore, the cycle decomposition of the permutation is .
• Step 5 of 8
Now let us find .
First write as shown below:-
Therefore,
Start from right as shown below:
1 is sent to 3, in the next group 3 is sent to 5 therefore, 1 maps to 5.
2 is sent to 4, in the next group 4 is sent to 2 therefore, 2 maps to 2.
3 is sent to 5, in the next group 5 is sent to 1 therefore, 3 maps to 1.
4 is sent to 2, in the next group 2 is sent to 4 therefore, 4 maps to 4.
5 is sent to 1, in the next group 1 is sent to 3 therefore, 5 maps to 3.
Here 2 and 4 are fixed.
Therefore, the cycle decomposition of the permutation is .
• Step 6 of 8
Similarly, we can find as shown below:
Therefore, the cycle decomposition of the permutation is.
• Step 7 of 8
Similarly, find .
Therefore, the cycle decomposition of the permutation is.
• Step 8 of 8
Now find.
Therefore, the cycle decomposition of the permutation is.
Corresponding Textbook
Abstract Algebra | 3rd Edition
9780471433347ISBN-13: 0471433349ISBN: Authors:
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# Thread: Optimization Problem
1. ## Optimization Problem
Find the maximum horizontal overhang for a 15 foot ladder over an 8 foot fence
The ladder base is always on the ground, the fence is vertical and ground it horizontal
2. Originally Posted by Exiab
Find the maximum horizontal overhang for a 15 foot ladder over an 8 foot fence
The ladder base is always on the ground, the fence is vertical and ground it horizontal
1. Draw a sketch. Let x denote the base of the ladder and y the overhang.
2. Use proportions:
$\displaystyle \frac8x = \frac z{x+y}~\implies~z=\frac{8(x+y)}{x}$
3. Use Pythagorean theorem:
$\displaystyle (x+y)^2 + z^2=15^2~\implies~(x+y)^2+\left(\frac{8(x+y)}{x}\r ight)^2=225$
4. Solve for y. That means: Calculate the overhang with respect to the base. You'll get a positive and a negative result. The negative result isn't very plausible here. Thus:
$\displaystyle y = \frac{15x}{\sqrt{x^2+64}}-x$
5. Calculate the first derivative (it's easier (for me!) to use product rule and chain rule):
$\displaystyle y' = \frac{960}{\left(\sqrt{x^2+64}\right)^3}-1$
6. Solve for x: y' = 0. I've got $\displaystyle x = 4 \cdot \sqrt{\sqrt[3]{225}-4} \approx 5.77193...$
3. Let F denote the foot of the ladder which is moved horizontally.
Then the top of the ladder describes a curve which I have "recorded". The vertical tangent to this curve indicates the maximum overhang.
According to my previous calculations the overhang is y = 3.004583...
4. Originally Posted by earboth
Let F denote the foot of the ladder which is moved horizontally.
Then the top of the ladder describes a curve which I have "recorded". The vertical tangent to this curve indicates the maximum overhang.
According to my previous calculations the overhang is y = 3.004583...
This answer is consistent with that found using the method suggested by Opalg at http://www.mathhelpforum.com/math-he...imization.html.
5. Originally Posted by mr fantastic
This answer is consistent with that found using the method suggested by Opalg at http://www.mathhelpforum.com/math-he...imization.html.
Opalg's method is much more elegant than mine. Unfortunately such solutions never occur to me ...
6. Originally Posted by earboth
Opalg's method is much more elegant than mine. Unfortunately such solutions never occur to me ...
Well, on the other hand:
1. You always have lovely graphics (it really is your trademark)
2. You reply is dated 11 March 11:20 am and Oplag's is dated 14 March 12:32 am. So it took Opalg 3 more days than you to get the elegant solution Lives might have been lost if not for your less elegant but timely solution
(ps Only kidding, opalg )
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# How do you factor 12x^2+7x-5?
Sep 20, 2015
color(blue)((12x-5)(x+1) is the factorised form.
#### Explanation:
$12 {x}^{2} + 7 x - 5$
We can Split the Middle Term of this expression to factorise it.
In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 12 \cdot - 5 = - 60$
and
${N}_{1} + {N}_{2} = b = 7$
After trying out a few numbers we get ${N}_{1} = 12$ and ${N}_{2} = - 5$
$12 \cdot - 5 = - 60$, and $12 + \left(- 5\right) = 7$
$12 {x}^{2} + 7 x - 5 = 12 {x}^{2} + 12 x - 5 x - 5$
$= 12 x \left(x + 1\right) - 5 \left(x + 1\right)$
color(blue)((12x-5)(x+1) is the factorised form.
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### 26-SNC2D-MirrorAndMagnification
```(10.3/10.4) Mirror and
Magnification Equations
(12.2) Thin Lens and
Magnification Equations
Mirror and Magnification Equations
The characteristics of an image can be
predicted by using two equations:
◦ Mirror Equation:
Allows us to determine the focal point, distance of
the image, or the distance of the object
Must know two of the three variables in order to solve
the third
◦ Magnification Equation:
Allows us to determine the height of the object or
the height of the image.
This equation is usually used following the mirror
equation
Mirror Equation
The mirror equation is seen below
◦ do represents the distance of the object
◦ di represents the distance of the image
◦ f represents focal length
Mirror Equation
If the image distance di is negative,
then the image is behind the mirror
(a virtual image)
•
Example 1: A concave mirror has a focal length of 12
cm. An object with a height of 2.5 cm is placed 40.0 cm
in front of the mirror. Calculate the image distance
using GRASP. Is the image in front of the mirror or
behind? How do you know?
•
Example 1: A concave mirror has a focal length of 12
cm. An object with a height of 2.5 cm is placed 40.0 cm
in front of the mirror. Calculate the image distance
using GRASP. Is the image in front of the mirror or
behind? How do you know?
•
Given:
f = 12 cm
ho = 2.5 cm
do = 40.0 cm
•
Required:
di = ?
•
Analysis:
1 + 1= 1
do
di f
1 = 1 - 1
di
f do
•
•
•
•
•
Example 1: A concave mirror has a focal length of 12
cm. An object with a height of 2.5 cm is placed 40.0 cm
in front of the mirror. Calculate the image distance. Is
the image in front of the mirror or behind? How do you
know?
Solution:
Use GRASP …
Given:
f = 12 cm
ho = 2.5 cm
do = 40.0 cm
Required:
di = ?
Analysis:
1 + 1= 1
do
di f
1 = 1 - 1
di
f do
1 = 1 - 1
di 12cm 40.0cm
1 = 10
3
di
120cm
120cm
1= 7
di 120 cm
di = 120 cm
7
di = 17.14 cm
di = 17 cm
Paraphrase
The image is 17 cm from the mirror. The sign is
positive so the image is in front of the mirror.
Magnification Equation
The magnification (m) tells you the size, or
height of the image relative to the object,
using object and image distances.
◦ Therefore, in order to use this equation the
distance of the object and image must be known.
Magnification Equation
If the image height hi is negative, the image
is inverted relative to the object.
Example 2: A concave mirror has a focal length of
12 cm. An object with a height of 2.5 cm is placed
40.0 cm in front of the mirror. The image distance
has been calculated to be 17.14 cm. What is the
height of the image? Is the image inverted? Explain.
Example 2: A concave mirror has a focal length of
12 cm. An object with a height of 2.5 cm is placed
40.0 cm in front of the mirror. The image distance
has been calculated to be 17.14 cm. What is the
height of the image? Is the image inverted? Explain.
•
•
Use GRASP …
Given:
f = 12 cm
ho = 2.5 cm
do = 40.0 cm
di = 17.14 cm
•
Required:
hi = ?
•
Analysis:
hi = - di
ho
do
Example 2: A concave mirror has a focal length of
12 cm. An object with a height of 2.5 cm is placed
40.0 cm in front of the mirror. The image distance
has been calculated to be 17.14 cm. What is the
height of the image? Is the image inverted? Explain.
•
•
Use GRASP …
Given:
f = 12 cm
ho = 2.5 cm
do = 40.0 cm
di = 17.14 cm
•
Required:
hi = ?
•
Analysis:
hi = - di
ho
do
Solution:
hi
= - 17.14 cm
2.5 cm
40.0cm
hi
= (- 17.14 cm) (2.5 cm)
40.0cm
hi
= -1.07 cm
hi
= -1.1 cm
Paraphrase
The height of the image is 1.1 cm. The sign
is negative , so the image is inverted.
Example 3: A convex surveillance mirror in a
convenience store has a focal length of -0.40 m.
A customer, who is 1.7 m tall, is standing 4.5 m in
front of the mirror.
◦ a) Calculate the image distance.
◦ b) Calculate the image height.
di = -0.37 m
hi = 0.14 m
Thin Lens and Magnification Equations
There are two ways to determine
characteristics of images formed by
lenses:
◦ Using ray diagrams
◦ Using algebra!
The Thin Lens and Magnification
Equations are the same as the mirror
equations.
Thin Lens and Magnification Equations
Lens Terminology
d0 = distance from object to the optical centre
di = distance from the image to the optical centre
h0 = height of the object
hi = height of the image
f = principal focal length of the lens
Thin Lens Equation
Sign conventions:
d0 is always positive
di is positive for real images and negative for
virtual images.
f is positive for converging lenses and negative
for diverging lenses.
Magnification Equation
Sign conventions:
- Object and image heights are positive when
measured upward and negative when measured
downward.
- Magnification is positive for an upright image and
negative for an inverted image.
- Magnification has no units.
Example 1: A diverging lens has a focal length of
10.0 cm. A candle is located 36 cm from the lens.
What type of image will be formed? Where will it
be located? Use GRASP.
|
# Search by Topic
#### Resources tagged with Generalising similar to Flip Flop - Matching Cards:
Filter by: Content type:
Stage:
Challenge level:
### There are 81 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Generalising
### Nim-7
##### Stage: 1, 2 and 3 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Taking Steps
##### Stage: 2 Challenge Level:
In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
### Maths Trails
##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### Pentanim
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Triangle Pin-down
##### Stage: 2 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Stop the Clock for Two
##### Stage: 1 Challenge Level:
Stop the Clock game for an adult and child. How can you make sure you always win this game?
### Walking the Squares
##### Stage: 2 Challenge Level:
Find a route from the outside to the inside of this square, stepping on as many tiles as possible.
### Nim-7 for Two
##### Stage: 1 and 2 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### More Numbers in the Ring
##### Stage: 1 Challenge Level:
If there are 3 squares in the ring, can you place three different numbers in them so that their differences are odd? Try with different numbers of squares around the ring. What do you notice?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Fault-free Rectangles
##### Stage: 2 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Dotty Circle
##### Stage: 2 Challenge Level:
Watch this film carefully. Can you find a general rule for explaining when the dot will be this same distance from the horizontal axis?
### Stop the Clock
##### Stage: 1 Challenge Level:
This is a game for two players. Can you find out how to be the first to get to 12 o'clock?
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Spirals, Spirals
##### Stage: 2 Challenge Level:
Here are two kinds of spirals for you to explore. What do you notice?
### Centred Squares
##### Stage: 2 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### Circles, Circles
##### Stage: 1 and 2 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Cuisenaire Rods
##### Stage: 2 Challenge Level:
These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like?
### Move a Match
##### Stage: 2 Challenge Level:
How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
### Polygonals
##### Stage: 2 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### Steps to the Podium
##### Stage: 2 and 3 Challenge Level:
It starts quite simple but great opportunities for number discoveries and patterns!
### Snake Coils
##### Stage: 2 Challenge Level:
This challenge asks you to imagine a snake coiling on itself.
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Round the Three Dice
##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Unit Differences
##### Stage: 1 Challenge Level:
This challenge is about finding the difference between numbers which have the same tens digit.
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### The Add and Take-away Path
##### Stage: 1 Challenge Level:
Two children made up a game as they walked along the garden paths. Can you find out their scores? Can you find some paths of your own?
### Round the Four Dice
##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
### Roll over the Dice
##### Stage: 2 Challenge Level:
Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded?
### Odd Squares
##### Stage: 2 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Growing Garlic
##### Stage: 1 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### Counting Counters
##### Stage: 2 Challenge Level:
Take a counter and surround it by a ring of other counters that MUST touch two others. How many are needed?
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Magic Circles
##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
### Build it Up
##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Cut it Out
##### Stage: 2 Challenge Level:
Can you dissect an equilateral triangle into 6 smaller ones? What number of smaller equilateral triangles is it NOT possible to dissect a larger equilateral triangle into?
### Simple Train Journeys
##### Stage: 1 and 2 Challenge Level:
How many different journeys could you make if you were going to visit four stations in this network? How about if there were five stations? Can you predict the number of journeys for seven stations?
### Calendar Calculations
##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
|
How do you draw a 200 degree angle with a protractor?
1. Given : angle 200 degree.
2. To Find : draw angle 200 degree with protractor
3. Solution: Step 1 : Draw a line segment AB. Step 2 : Draw an angle of 160° at B on AB and mark C. ∠ABC = 160° Step 3 : Mark angle out side of angle that will be ∠ABC = 360° – 160° = 200° Learn More:
Is 270 degrees a reflex angle?
Angles such as 270 degrees which are more than 180 but less than 360 degrees are called reflex angles.
∠T
Which side of FGH is the shortest?
In FGH, mWhich side of FGH is the shortest? Note about triangles: The small sides are opposite the small angles, and the large sides are opposite the large angles. Since mside opposite, segment FG, would be the shortest.
Does the longer side has the smaller angle opposite to it in a triangle?
Theorem: If one side of a triangle is longer than another side, then the angle opposite the longer side will be larger than the angle opposite the shorter side.
Is it possible to construct a triangle with side lengths of 6/7 and 11 units if not explain why not?
ANSWER: No; 11. SOLUTION: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
Is it possible to draw a triangle the length of whose sides are 7 cm 8 cm and 15 cm?
Hence, it is not possible to draw a triangle having sides 7 cm, 8 cm and 15 cm. (iv) Consider the numbers 3.4, 2.1 and 5.3. Thus, the sum of any two sides is greater than the third side.
Is it possible to draw a triangle the lengths of whose sides are 7 cm 8 cm 15 cm explain?
Step-by-step explanation: One side of the triangle has to be less than the sum of the other two sides and greater than the magnitude of their difference. Lets check the fact by adding / subtracting two sides and comparing with the third one.. 15–8 = 7 = length of third side. 15–7=8 = length of third side.
Is it possible to construct a triangle with length of its side is 4 cm 3 cm and 7 cm?
It’s not possible to construct a triangle with the given lengths (4,3,7).
2020-08-17
|
# Algorithm and Flowchart to check whether a number is twisted prime or not
[166 views]
### Prime Numbers:
A number is said to be a prime number when it has only two factors, that is, when the factors of the number are 1 and the number itself. Example: 2, 3, 11, 17, etc.
### What are Twisted Prime Numbers?:
A prime number is said to be ‘twisted prime’ when on reversing the number, the new number is also a prime number. Twisted prime numbers are also known as Emirp numbers.
For example: 97 and 79.
97 is a prime number. On reversing 97, we get 79, which is also a prime number. Therefore, both 97 and 79 are twisted prime numbers.
Other examples include: 2, 3, 5, 7, 13, 17, 37, etc.
Now let us have a look at the algorithm and flowchart to check whether a given number is twisted prime or not.
To avoid redundancy, we will use the concept of functions. Here, we will use a function to check whether a number is prime or not. While writing the program, we can call this function as many times we want.
#### CheckPrime(num):
Assumption: For this algorithm, we assume that the function takes a number ‘num’, to be checked and returns true if the number is prime, otherwise, returns false.
Step 1: Start Step 2: Initialize number of factors of num, f = 0 Step 3. Initialize i = 1 Step 4. Repeat until i<=num: 4.1: If num % i == 0: 4.2: Increment f by 1 4.3: Increment i by 1 Step 5: If f = 2, then: 5.1: Return True Step 6: Else: 6.1 Return False Step 7. Stop
### Algorithm to check whether a number is twisted prime or not:
Step 1: Start Step 2: Read the number to be checked from the user, say n Step 3: original = CheckPrime(n) Step 4: Initialize rev = 0 Step 5. Repeat WHILE n ? 0: 5.1: Extract the last digit by: d = n % 10 5.2: Calculate reverse by: rev = (rev*10) + d 5.3: Remove the last digit from the number: n = n / 10 Step 6: reverse = CheckPrime(rev) Step 7: If original = True AND reverse = True, then: 7.1: Display “Given number is twisted prime” Step 8: Else: 8.1: Display “Given number is not twisted prime” Step 9: Stop
#### Explanation:
We start this algorithm by taking the number to be checked as input from the user. To check whether the number is twisted prime, we first need to check whether the number is prime, and then reverse the number.
To check whether the given number is prime, we count the number of factors of that number. We start a loop which runs from 1 to num. If the number is divisible by the loop variable, then the loop variable is a factor of that number. Therefore, we increment the number of factors by one. Once this check is done, we now have to reverse the number.
To reverse the number, we initialize a variable, say rev as 0. This variable will store the reverse. We now start a loop which will run until the number is not equal to zero. We extract the last digit of the number, by doing: d = n % 10. This last digit is added to (rev * 10); to maintain the place values. Now, the last digit is removed from the number by doing: n = n /10. Once this loop completes its execution, we get the reverse in rev.
Now, we will check whether the reversed number is prime or not, using the same method as before. If both the original and reversed numbers are prime, the given number is twisted prime, else, it is not twisted prime.
Let us consider an example:
n = 71
71 is a prime number.
Reverse of the number, rev = 17
17 is also a prime number.
Therefore, 71 is twisted prime.
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# Puzzle – The Hands of a Clock First Overlap
• Difficulty Level : Basic
• Last Updated : 18 Jan, 2023
Question:
An hour hand and a minute hand are standard on a clock. At 12 midnight the hands are exactly aligned. When will they next precisely align or overlap? How frequently will they cross paths each day?
Solution:
a) The hour hand revolves more slowly than the minute hand.
In one minute, the minute hand made an angle of 360/60 = 6°.
In one minute, an hour hand made an angle of 360/720 = 0.5°.
The difference between the angle made by an hour hand and a minute hand is 6° – 0.5° = 5.5°,
The difference increases by 5.5° each minute.
The difference will be 11° after 2 minutes and 16.5° after 3 minutes.
The angle between the hour hand and minute hand at 12 midnight is zero degrees.
The difference between both hands must be 360° for them to overlap once more.
As both hands rotate continuously, we can use a fundamental ratio to get the difference of 360°.
1: 5.5° =? : 360°
i.e., 360/5.5 = 65.4545 minutes.
Therefore, they will overlap each other after 65.4545 minutes. It will happen when you convert it in the time given.
12am + 65.4545 minutes = 01: 05: 27 am (approximately) {0.4545 × 60=27.27}
b) Let’s start at 12:00 a.m. Hands were therefore overlapping there. And then, it overlaps once more between 1-2 AM. So, it goes on until 10 or 11 AM. i.e., It overlaps once per hour.
However, it overlaps at 12 PM for 11 AM to 12 PM. They overlap 11 times in 12 hours.
1) 12:00 AM
2) 1AM – 2 AM
3) 2AM – 3AM
4) 3AM – 4AM
5) 4AM – 5AM
6) 5AM – 6AM
7) 6AM – 7AM
8) 7AM – 8AM
9) 8AM – 9AM
10) 9AM – 10ÀM
11) 10AM – 11AM
They won’t overlap after 11 AM until after 12 PM.
The cycle then repeats from noon until 11:59 at night. Since our first unit began at 12 a.m. (midnight), the work should have been finished by 12 AM. the following day.
11 + 11 = 22 in total.
That means an hour hand and a minute hand will overlap or meet 22 times in a day.
My Personal Notes arrow_drop_up
|
# 750 in words
750 in words is written as Seven hundred and Fifty. 750 represents the count or value. The article on Counting Numbers can give you an idea about count or counting. The number 750 is used in expressions that relate to money, days, distance, length, weight and so on. Let us consider a few examples for 750. “A competition on English Essay writing has a rule to have exactly Seven Hundred and Fifty words”. “I purchased a New Refrigerator with a capacity of 750L”.
750 in words Seven Hundred and Fifty Seven Hundred and Fifty in Numbers 750
## How to Write 750 in Words?
We can convert 750 to words using a place value chart. The number 750 has 3 digits, so let’s make a chart that shows the place value up to 3 digits.
Hundreds Tens Ones 7 5 0
Thus, we can write the expanded form as:
7 × Hundred + 5 × Ten + 0 × One
= 7 × 100 + 5 × 10 + 0 × 1
= 750
= Seven Hundred and Fifty.
750 is the natural number that is succeeded by 749 and preceded by 751.
750 in words – Seven Hundred and Fifty.
Is 750 an odd number? – No.
Is 750 an even number? – Yes.
Is 750 a perfect square number? – No.
Is 750 a perfect cube number? – No.
Is 750 a prime number? – No.
Is 750 a composite number? – Yes.
## Solved Example
1. Write the number 750 in expanded form
Solution: 7 × 100 + 5 × 10 + 0 × 1
We can write 750 = 750 + 0 + 0
= 7 × 100 + 5 × 10 + 0 × 1
## Frequently Asked Questions on 750 in words
Q1
### How to write the number 750 in words?
750 in words is written as Seven Hundred and Fifty.
Q2
### State True or False. 750 is divisible by 3?
True. 750 is divisible by 3.
Q3
### Is 750 a perfect square number?
No. 750 is not a perfect square number.
|
# Divisibility
Many people know about testing numbers for divisibility by 3 or 9: you add the digits and if (and only if) the sum is divisible by 3, so is the original number; if (and only if) the sum is divisible by 9, so is the original number. Not as many people know how to prove that, though. There are a number of ways, but one comes out of an easy procedure to create divisibility tests for other numbers, as follows.
Is 38752 a multiple of 7? We can write that number as 10 x 3875 + 2. More generally, if we have some number a we want to test, we can write it as a = 10b + c, where b is a with its last digit truncated, and c is the last digit of a. Now, it won’t change the divisibility by 7 if we add a multiple of 7, so let’s add 49c. We get a + 49c = 10b + 50c = 10(b + 5c). Nor will it change the divisibility by 7 if we divide by something that’s not a multiple of 7, namely 10. So a is divisible by 7 if and only if b + 5c is. In words, take the last digit of a, multiply by 5, and add it to the remaining digits; if and only if that’s a multiple of 7, then so is a. For example: 38752 becomes 3875+5×2 = 3885. We can do it again: 3885 becomes 388+5×5 = 413; and again: 413 becomes 41+5×3 = 56. We can stop here, we know 56 is a multiple of 7, so 38752 is too.
We could also subtract a multiple of 7 from a: a–21c = 10b–20c = 10(b–2c), from which we see a is a multiple of 7 if and only if twice the last digit subtracted from the rest of the number is. So 38752 becomes 3871, then 385, then 28, a multiple of 7.
You can come up with similar divisibility tests for any prime except 2 or 5, or in fact any number which is relatively prime to 10 (that is, any number whose last digit is 1, 3, 7, or 9). 11, for instance: adding 99c to a gives us an “addition coefficient” of 10, meaning you can test for divisibility by 11 by adding 10 times the last digit to the rest of the number; or subtract 11c to derive a “subtraction coefficient” of 1 — test by subtracting the last digit from the rest. 82115 becomes 8206 becomes 814 becomes 77, so 82115 is a multiple of 11.
The general recipe is, to derive addition and subtraction tests for divisibility by a number n ending in 1, 3, 7, or 9, multiply n by 9, 3, 7, or 1, respectively, and add 1 and divide by 10 to get the addition coefficient; multiply by 1, 7, 3, or 9, respectively, and subtract 1 and divide by 10 to get the subtraction coefficient. (Or use the fact that the addition and subtraction coefficients sum to n.)
You can do the same thing with more digits, although in most cases you probably wouldn’t want to. But, for instance, for divisibility by 11 you can write a = 100b + 10c + d, add 99(10cd) to get 100(b + (10cd)), and derive the test: Add the last two digits to the rest of the number, and if the result is divisible by 11, then so is the original. (82115 becomes 836 becomes 44, with fewer steps than before.)
For both 3 and 9 the 1-digit addition coefficient is 1, meaning that if you add the last digit to the rest of the number, the result has the same divisibility by 3 or 9. Clearly this implies if you add all the digits together the result has the same divisibility by 3 or 9 — proving the well known rules.
|
# How do you find the sum of the first 25 terms of the sequence: 7,19,31,43...?
$3775$
#### Explanation:
The given series:
$7 , 19 , 31 , 43 , \setminus \ldots$
Above series is an arithmetic progression with a common difference
$d = 19 - 7 = 31 - 19 = 43 - 31 = \setminus \ldots = 12$
First term: $a = 7$
The sum of first $n$ terms of an AP with term $a$ & a common difference $d$ is given as
${S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$
Hence, the sum of first $n = 25$ terms of an AP with term $a = 7$ & a common difference $d = 12$ is given as
${S}_{25} = \frac{25}{2} \left(2 \setminus \cdot 7 + \left(25 - 1\right) 12\right)$
$= 3775$
|
# How do you find all relative extrema and points of inflection for the equation: y=x^2 log_3x?
Jul 5, 2015
Find extrema using the derivative and points of inflection using the second derivative.
#### Explanation:
We will make use of: ${\log}_{3} x = \ln \frac{x}{\ln} 3$ and
$\frac{d}{\mathrm{dx}} {\log}_{3} x = \frac{d}{\mathrm{dx}} \left(\ln \frac{x}{\ln} 3\right) = \frac{1}{x \ln 3}$
Finding relative extrema
$f \left(x\right) = y = {x}^{2} {\log}_{3} x$
Note that the domain of this function is $\left(0 , \infty\right)$
Use the product rule to get:
$y ' = 2 x {\log}_{3} x + {x}^{2} \cdot \frac{1}{x \ln 3}$
$= 2 x {\log}_{3} x + \frac{x}{\ln} 3$
To find critical numbers, it is convenient to use all logarithms with the same base.:
$f ' \left(x\right) = y ' = \frac{2 x \ln x}{\ln} 3 + \frac{x}{\ln} 3 = \frac{x \left(2 \ln x + 1\right)}{\ln} 3$
$y '$ in never undefined in the domain, and is equal to $0$ at
$x = 0$ or $2 \ln x + 1 = 0$. $0$ is not in the domain, so we need only solve:
$2 \ln x + 1 = 0$.
$\ln x = - \frac{1}{2}$
$x = {e}^{- \frac{1}{2}} = \frac{1}{\sqrt{e}}$
The partitions on the domain are: $\left(0 , {e}^{- \frac{1}{2}}\right) , \left({e}^{- \frac{1}{2}} , \infty\right)$
Testing $\left(0 , {e}^{- \frac{1}{2}}\right)$.
Note that $\frac{1}{e} < \frac{1}{\sqrt{e}}$, so ${e}^{-} 1$ is in the first interval and
$f ' \left({e}^{- 1}\right) = \frac{{e}^{- 1} \left(2 \ln {e}^{- 1} + 1\right)}{\ln} 3 = \frac{\frac{1}{e} \left(- 2 + 1\right)}{\ln} 3$ which is negative.
So $f '$ is negative on $\left(0 , {e}^{- \frac{1}{2}}\right)$.
Testing $\left({e}^{- \frac{1}{2}} , \infty\right)$
Use $1$ as the test number.
$f ' \left(e\right) = \frac{1 \left(2 \ln 1 + 1\right)}{\ln} 3 = \frac{1 \left(0 + 1\right)}{\ln} 3$ which is positive, so
So $f '$ is positive on $\left({e}^{- \frac{1}{2}} , \infty\right)$.
This tells us that f(e^(-1/2)) is a relative minimum.
Finding points of inflection
Recall:
$f ' \left(x\right) = y ' =$ = 2xlog_3x + x/ln3
So
$f ' ' \left(x\right) = y ' ' = 2 {\log}_{3} x + 2 x \cdot \frac{1}{x \ln 3} + \frac{1}{\ln} 3$
$= \frac{2 \ln x}{\ln} 3 + \frac{2}{\ln} 3 + \frac{1}{\ln} 3 = \frac{2 \ln x + 3}{\ln} 3$
The only partition number for $f ' '$ is $\ln x = - \frac{3}{2}$ so $x = {e}^{- \frac{3}{2}}$
For $f ' '$ the parition intervals are $\left(0 , {e}^{- \frac{3}{2}}\right) , \left({e}^{- \frac{3}{2}} , \infty\right)$
Using ${e}^{-} 2$ and $1$ as test numbers we will find that:
$f ' '$ is negative on $\left(0 , {e}^{- \frac{3}{2}}\right)$.
$f '$ is positive on $\left({e}^{- \frac{3}{2}} , \infty\right)$.
The concavity does change, so
$\left({e}^{- \frac{3}{2}} , f \left({e}^{- \frac{3}{2}}\right)\right)$ is a point of inflection.
|
## Wednesday, May 20, 2015
### Another method to prove $7 ≥ \sqrt 2+\sqrt 5 + \sqrt {11}$
Show with proof which of these two values is smaller:
$7$, or $\sqrt 2+\sqrt 5 + \sqrt {11}$
In my (few) previous blog post (Which is greater), I mentioned of how I proved for $\sqrt 2+\sqrt 5 + \sqrt {11}$ is smaller than $7$.
But that doesn't mean that solution is the only way out to prove for that kind of problem.
If you have not seen what I have observed, you could still be able to prove it, provided you know how to square any given math expression to get rid of the square root terms.
Without knowing which is greater in the beginning, you are safe to assume either way.
Let's say we assume that $\sqrt 2+\sqrt 5 + \sqrt {11}> 7$ is true.
We then keep squaring both sides of the inequality until we obtain numerical value on both sides of the inequality sign:
$\sqrt 2+\sqrt {11}> 7-\sqrt 5$
$(\sqrt 2+\sqrt {11})^2> (7-\sqrt 5)^2$
$2+11+2\sqrt {2(11)}> 49+5-2(7)\sqrt 5$
$2\sqrt {2(11)}+14\sqrt 5> 41$
$(2\sqrt {2(11)}+14\sqrt 5)^2> (41)^2$
$4(2(11))+14^2(5)+2(2)(14)\sqrt {2(11)(5)}> (41)^2$
$2(2)(14)\sqrt {2(11)(5)}> 613$
$(2(2)(14)\sqrt {2(11)(5)})^2> (613)^2$
$344,960> 375,769$
This is pure wrong therefore the converse of our previous assumption must be true, i.e. we can conclude by now that
$\sqrt 2+\sqrt 5 + \sqrt {11}< 7$
As you can probably tell, this method is not "grant" nor "superior" but you must know that in solving any given mathematical problem, solving it is what matters and you must not look down on a seemingly "weak" solution.
Instead, all schools around the world are continually seeking ways to improve their teaching methodology because we are eager and want to produce critical thinkers to push our country and economy moving forward. This is good news, as through the improved version of the teaching methods, we can always learn more, so we will never be satisfied with where we are today.
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# Ch 1: Real Numbers: Types and Properties
Watch video lessons and learn about the different types of numbers and their properties. Complete the quizzes that follow each lesson to measure your learning.
## Real Numbers: Types and Properties - Chapter Summary and Learning Objectives
The lessons in this chapter explore many types of numbers, including real, whole and natural numbers. Several lessons look at rational numbers and how to graph them and convert them from fractions to decimals. You'll get information on how to utilize the commutative and associative properties and work out the absolute value of a real number. By the end of this chapter, you should be familiar with:
• Showing inequalities between numbers
• Find the absolute value of a real number
• The multiplication property of zero
• Pythagorean triples
Video Objective
What Are the Different Types of Numbers? In this lesson, you'll learn about the various types of numbers, such as real, whole, natural, rational and irrational.
Graphing Rational Numbers on a Number Line This lesson explains how to use a number line to graph rational numbers.
Notation for Rational Numbers, Fractions and Decimals Here, we'll examine how to change rational numbers from fractional to decimal notation. You'll be given some examples to help you understand.
The Order of Real Numbers: Inequalities In this lesson, we'll look at inequalities, including how to use symbols to indicate which number is greater. You'll learn to determine which inequalities have the same meaning.
Finding the Absolute Value of a Real Number You'll learn how to determine a real number's absolute value.
The Commutative Property: Definition and Examples This lesson shows you what the commutative property is and how it can be useful.
The Associative Property: Definition and Examples Here, you'll get information on associating expressions through the use of this property.
The Multiplication Property of Zero: Definition and Examples We'll explore the rules regarding multiplying by zero.
Rationalize the Denominator You'll learn to simplify an expression by removing a radical symbol from the denominator.
Algebraic Numbers and Transcendental Numbers This lesson explains what these terms mean and gives well-known examples.
Pythagorean Triples Here, we'll learn to identify Pythagorean triples.
10 Lessons in Chapter 1: Real Numbers: Types and Properties
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
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# Section 1.1: Integer Operations and the Division Algorithm - PowerPoint PPT Presentation
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Section 1.1: Integer Operations and the Division Algorithm. MAT 320 Spring 2008 Dr. Hamblin. Addition. “You have 4 marbles and then you get 7 more. How many marbles do you have now?”. 4. 11. 7. Subtraction. “If you have 9 toys and you give 4 of them away, how many do you have left?”. 5.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Section 1.1: Integer Operations and the Division Algorithm
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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## Section 1.1: Integer Operations and the Division Algorithm
MAT 320 Spring 2008
Dr. Hamblin
• “You have 4 marbles and then you get 7 more. How many marbles do you have now?”
4
11
7
### Subtraction
• “If you have 9 toys and you give 4 of them away, how many do you have left?”
5
4
9
### Multiplication
• “You have 4 packages of muffins, and each package has 3 muffins. How many total muffins do you have?”
4
12
3
### Division
• “You have 12 cookies, and you want to distribute them equally to your 4 friends. How many cookies does each friend get?”
3
12
### Examining Division
• As you can see, division is the most complex of the four operations
• Just as multiplication is repeated addition, division can be thought of as repeated subtraction
### 28 divided by 4
• 28 – 4 = 24
• 24 – 4 = 20
• 20 – 4 = 16
• 16 – 4 = 12
• 12 – 4 = 8
• 8 – 4 = 4
• 4 – 4 = 0
• Once we reach 0, we stop. We subtracted seven 4’s, so 28 divided by 4 is 7.
### 92 divided by 12
• 92 – 12 = 80
• 80 – 12 = 68
• 68 – 12 = 56
• 56 – 12 = 44
• 44 – 12 = 32
• 32 – 12 = 20
• 20 – 12 = 8
• We don’t have enough to subtract another 12, so we stop and say that 92 divided by 12 is 7, remainder 8.
### Expressing the Answer As an Equation
• Since 28 divided by 4 “comes out evenly,” we say that 28 is divisible by 4, and we write 28 = 4 · 7.
• However, 92 divided by 12 did not “come out evenly,” since 92 12 · 7. In fact, 12 · 7 is exactly 8 less than 92, so we can say that 92 = 12 · 7 + 8.
remainder
dividend
quotient
divisor
### 3409 divided by 13
• Subtracting 13 one at a time would take a while
• 3409 – 100 · 13 = 2109
• 2109 – 100 · 13 = 809
• 809 – 50 · 13 = 159
• 159 – 10 · 13 = 29
• 29 – 13 = 19
• 19 – 13 = 3
• So 3409 divided by 13 is 262 remainder 3.
• All in all, we subtracted 262 13’s, so we could write 3409 – 262 · 13 = 3, or 3409 = 13 · 262 + 3.
### How Division Works
• Start with dividend a and divisor b (“a divided by b”)
• Repeatedly subtract b from a until the result is less than a (but not less than 0)
• The number of times you need to subtract b is called the quotient q, and the remaining number is called the remainder r
• Once this is done, a = bq + r will be true
### Theorem 1.1: The Division Algorithm (aka The Remainder Theorem)
• Let a and b be integers with b > 0. Then there exist unique integers q and r, with 0 r < b and a = bq + r.
• This just says what we’ve already talked about, in formal language
### Ways to Find the Quotient and Remainder
• Method 2: Guess and CheckFill in whatever number you want for q, and solve for r. If r is between 0 and b, you’re done. If r is too big, increase q. If r is negative, decrease q.
• Method 3: CalculatorType in a/b on your calculator. The number before the decimal point is q. Solve for r in the equation a = bq + r
### Negative Numbers
• Notice that in the Division Algorithm, b must be positive, but a can be negative
• How do we handle that?
### -30 divided by 8
• “You owe me 30 dollars. How many 8 dollar payments do you need to make to pay off this debt?”
• Instead of subtracting 8 from -30 (which would just increase our debt), we add 8 repeatedly
### -30 divided by 8, continued
• -30 + 8 = -22
• -22 + 8 = -14
• -14 + 8 = -6 (debt not paid off yet!)
• -6 + 8 = 2
• So we made 4 payments and had 2 dollars left over
• -30 divided by 8 is -4, remainder 2
• Check: -30 = 8 · (-4) + 2
### Caution!
• Negative numbers are tricky, be sure to always check your answer
• Be careful when using the calculator method
• Example: -41 divided by 7The calculator gives -5.857…, but if we plug in q = -5, we get r = -6, which is not a valid remainder
• The correct answer is q = -6, r = 1
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# Diagonal Count
The purpose of this page is to help you count a number of non-intersecting diagonals in a convex polygon. I won't give you the formula right away. You are to come up with the right one by experimenting with the applet below. Once you surmise the right result, try to prove it before looking into the explanation at the bottom of the page.
To draw diagonals just drag the mouse from one vertex to another. Remember that you are counting non-intersecting diagonals.
### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
What if applet does not run?
Explanation
## Explanation
It's obvious that from a given vertex one may only draw exactly (n-3) diagonals. For one can't draw a diagonal to the selected vertex itself and its two immediate neighbors. Such diagonals will not intersect one another. What's interesting is that, in a convex n-gon, the number of non-intersecting diagonals always equals (n-3) regardless of which diagonals are drawn and in what order.
## Proof #1
After all eligible diagonals have been drawn, the polygon will be split into a number T of triangles. Between them these triangles have 3T sides. Except that those that coincide with diagonals are counted twice. Let D be the number of diagonals. We thus have
3T = n + 2D.
Adding up interior angles of the triangles gives, on the one hand, 180°T = 180°(n + 2D)/3. On the other hand, since diagonals do not intersect, their angles fill up the angles of the polygon. Thus (by the distributive law),
180°(n - 2) = 180°(n + 2D)/3.
Simplifying we get D = n - 3.
As a byproduct of the proof, we obtain a formula for the number of triangles thus formed: T = (n + 2(n-3))/3 = n - 2.
## Proof #2 (By induction)
For n = 3 D = 0 since a triangle has no diagonals. Assume the formula has been established for all m<n. Consider a convex n-gon and draw its diagonals as expected. We may cut the polygon along one of its diagonals which will create two smaller polygons with their diagonals drawn according the conditions of our theorem. For, if we could draw an additional diagonal in one of the polygons, this line would serve as a diagonal in the original one as well. "Non-intersection" condition will also be inherited which would contradict our construction. Let the two polygons have n1 and n2 sides, respectively. Then n1 + n2 = n + 2; for the selected diagonal will be counted as a side of both small polygons. In obvious notations, D1 = n1 - 3 and D2 = n2 - 3 by the inductive assumption. On the other hand, D1 + D2 = D - 1; for the selected diagonal drops out of the count. Combining these two facts leads to
D1 + D2 = (n1 - 3) + (n2 - 3) = n1 + n2 - 6 = (n + 2) - 6 = D - 1
and finally to the desired result. Note that since both n1,n2≥3, n1 + n2 = n + 2 implies n1,n2<n as needed.
## Remark 1
If we are looking for the totality of all diagonals in an n-gon allowing for intersections then, of course, it does not matter in what order we count them - there is a unique set of all diagonals of a given polygon. From any of the n vertices emanate (n-3) diagonals which gives n(n-3) except that each diagonal, having two ends, is counted twice. Therefore the total number of diagonals in an n-gon is n(n-3)/2.
## Remark 2
The question of a number of ways in which a convex (n+2)-gon can be split into n triangles (Euler's problem) leads to the famous Catalan numbers
cn = (2n)!/[n!(n+1)!].
Catalan numbers pop up in a multitude of seemingly unrelated places. M. Gardner reports that in 1976 Henry. W. Gould counted 450 references. Popular acounts can be found in
1. J.H.Conway and R.K.Guy, The Book of Numbers, Springer-Verlag, NY, 1996.
2. H.Dorrie, 100 Great Problems Of Elementary Mathematics, Dover Publications, NY, 1965.
3. M.Gardner, Time Travel and Other Mathematical Bewilderments, W.H.Freeman and Co., NY, 1988.
• Chessboard
• Changing Colors, an Interactive Activity
• Plus or Minus Game
• Solitaire on a Circle
• Squares, Circles, and Triangles
• Calendar Magic
• Counting Diagonals in a Convex Polygon
• The Game of Fif
• Nim
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# Geometric Series Formula
## Geometric Series Formula
Number sequences follow certain patterns and laws. Any routine pattern that is transferable from one term to the next may be this pattern. The Geometric Series Formula is one in which more multiplication is required to obtain the following term. The sum of the sequence’s terms can be obtained by adding or subtracting symbols. The geometric series will result from this. One such geometric series is 1 + 2 + 4 + 8 + 16 for instance.
The geometric progression is another name for a geometric series. It is a series where the second term is obtained by multiplying the first term by a predetermined amount. This operation is repeated until the series has the necessary number of terms. Such a rise occurs in a particular way, resulting in geometric development.
The geometric series formula will determine the general term and the total of all its terms. For instance, if students add 2 to the first number in the series above, we will receive the second number, and so on.
Such Geometric Series Formula follow the straightforward rule that you must multiply one term by a certain integer to reach the next. As a result, students can produce any number of terms for this series. Thus, students can learn about the usual ratios or fixed values for multiplication by looking at such a series. It is represented by the letter “r.” One can study the Geometric Series Formula on the Extramarks website or mobile application.
## What is a Geometric Series?
The total of a geometric sequence’s finite or infinite terms is known as a geometric series. The analogous geometric series is a + ar + ar2 +…, arn-1 + for the geometric sequence a, ar, ar2,…, arn-1,… The word “series” is known to mean “sum.”
The Geometric Series Formula specifically refers to the total of phrases with a common ratio between every pair of neighbouring terms. Geometric series can be of two different types: finite and infinite. Below are a few illustrations of the Geometric Series Formula.
In a finite Geometric Series Formula, 1/2 + 1/4 +…. + 1/8192, the first term, a, is equal to 1/2 and the common ratio, r, is equal to 1/2. In an infinite geometric series, -4 + 2 – 1 + 1/2 – 1/4 +…, the first term, a, is equal to -4 and the common ratio, r, is equal to -1/2.
## Geometric Series Formula
Let’s review what a Geometric Series Formula is first before learning the formula. Every two successive terms in the series (sum of terms) have the same ratio. The formula for the geometric series includes A formula to determine a Geometric Series Formula nth term, calculate a finite geometric series’ sum, and calculate the total of an endless geometric sequence.
## Geometric Series Formulas
Calculating the Geometric Series Formula is necessary to understand the compounding that takes place over time. By using the geometric progression’s mean, it illustrates the fundamental behaviour of the progression. For instance, it is simple to use the geometric mean to examine the growth of bacteria. In other words, the compounding becomes more critical the longer the time horizon or the more different the values in the series, and as a result, the geometric mean is a better choice.
The formulas to determine the nth term, the sum of n terms, and the sum of infinite terms are all included in the formulas for geometric series. Consider a Geometric Series Formula where the common ratio is r, and the first term is a.
a + ar + ar2 + ar3 + …
### Convergence of Geometric Series
Any finite geometric series converges eventually. However, the common ratio’s value determines whether or not an infinite geometric series will eventually converge. A, Ar, Ar2,… is a part of an infinite geometric series.
converges when |r| 1. Thus, one can use the formula a / to find its sum (1 – r).
diverges when |r| > 1. Therefore, it is not possible to determine its total in this situation.
### Examples Using Formula for a Geometric Series
Example 1: Find the 10th term of the geometric series 1 + 4 + 16 + 64 + …
Solution:
To find The 10th term of the given geometric series.
In the given series,
The first term, a = 1.
The common ratio r = 4 / 1 (or) 16 / 4 (or) 64 / 16 = 4.
Using the Geometric Series Formula, the nth term is found using:
nth term = an rn-1
Substitute n = 10, a = 1, and r = 4 in the above formula:
10th term = 1 × 410-1 = 49 = 262,144
Answer: The 10th term of the given geometric sequence = 262,144.
Example 2: Find the sum of the following geometric series: i) 1 + (1/3) + (1/9) + … + (1/2187)
ii) 1 + (1/3) + (1/9) + … using the Geometric Series Formula:
Solution:
To find The sum of the given two geometric series.
In both of the given series,
the first term, a = 1.
The common ratio r = 1 / 3.
1. i) In the given series,
nth term = 1 / 2187
a rn-1 = 1 / 37
1 × (1 / 3) n-1 = 3-7
3-n + 1 = 3-7
-n + 1 = -7
-n = -8
n = 8
So we need to find the sum of the first 8 terms of the given series.
Using the sum of the finite geometric series formula:
Sum of n terms = a (1 – rn) / (1 – r)
Sum of 8 terms = 1 (1 – (1/3)8) / (1 – 1/3)
= (1 – (1 / 6561)) / (2 / 3)
= (6560 / 6561) × (3 / 2)
= 3280 / 2187
ii) The given series is an infinite Geometric Series Formula.
Using the sum of the infinite Geometric Series Formula:
Sum of infinite geometric series = a / (1 – r)
Sum of the given infinite geometric series
= 1 / (1 – (1/3))
= 1 / (2 / 3)
= 3 / 2
Answer: i) Sum = 3280 / 2187 and ii) Sum = 3 / 2
Example 3: Calculate the sum of the finite geometric series if a = 5, r = 1.5 and n = 10.
Solution:
To find: the sum of the Geometric Series Formula
Given: a = 5, r = 1.5, n = 10
sn = a(1−rn)/(1−r)
The sum of ten terms is given by S10 = 5(1−(1.5)10)/(1−1.5)
= 566.65
Answer: The sum of the geometric series is 566.65.
### 1. What Are the Mathematical Geometric Series Formulas?
The formulas for geometric series can be used to determine the sum of a finite geometric sequence, the sum of an infinite geometric series, and the nth term of a geometric sequence. The initial term “a” and the common ratio “r” in these Geometric Series Formula are given as
A rn-1 = nth term
The Sum of terms is equal to a (1 – rn) (1 – r)
The Sum of all geometric series is equal to a. (1 – r)
### 2. Is it always possible to determine the sum of a geometric series?
Students can always find the sum of a geometric series if it is finite. However, only when its common ratio’s absolute value is less than one can its total be calculated.
### 3. What is the point at which a geometric sequence converges?
The letter “r” represents the common ratio between each pair of consecutive entries in a geometric series. Only when |r| one does an infinite geometric series converge, and only then can its sum be determined.
### 4. What Purposes Do Geometric Series Serve?
Formulas for geometric series are often used in mathematics. These have significant applications in finance, queueing theory, computer science, biology, economics, and engineering.
### 5. How does the"r" work in Geometric Series Formula?
The term “r” in the formula for geometric series stands for the common ratio. The following are the formulas for geometric series with “n” terms with the term “a” as the first term: The equation for the nth term: A rn-1 = nth term, the Sum of terms is equal to a (1 – rn) (1 – r) and the Sum of all geometric series is equal to a. (1 – r)
### 6. What distinguishes geometric sequence from geometric series?
A geometric sequence is a collection of terms in which the ratio of every two successive terms is the same, whereas a geometric series is the geometric sequence’s “sum.”
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# Understanding Ordering of Numbers and Applying the Rules of Ascending & Descending Order Value
One of the most important topics for early Math learners is to identify the order of numbers and correctly arrange them from small to big and big to small. Ordering is basically arranging numbers in a sequence where the successive number either gets larger or smaller, depending upon the equation.
### Ascending Order
When a set of numbers are arranged from smallest to largest, those numbers are said to be in ascending order.
Explain to children that ascending means ‘moving up’. When you climb a mountain or stairs, you move in upward direction, this is termed as an ascend. Thus, ascending order is denoted by an upward arrow sign.
### Descending Order
When numbers are arranged from largest to smallest, it is said to be in descending order.
When you move from top to bottom or when you walk down the mountain, you are essentially descending. Thus, descending order is denoted by a downward arrow sign.
### Examples of Equations for Ascending Order
#1 Arrange the numbers in ascending order – when the numbers follow a particular sequence.
3, 4, 2, 1, 5
Answer: 1, 2, 3, 4, 5,
#2 Arrange numbers in ascending order where the starting digit is not 1.
7,9,10,8
#3 Arrange the following random numbers in ascending order from 7 to 13 – when the numbers do not follow a continuous sequence. Numbers that are not next to each other on a number line.
10, 13, 9, 7
#4 Arrange the following 5 times table in ascending order.
15, 10, 30, 45
### Examples of Equations for Descending Order
#1 Arrange the numbers in descending order – when the numbers follow a particular sequence.
2, 3, 5, 4, 1
Answer: 1, 2, 3, 4, 5
#2 Arrange numbers in descending order where the starting digit is not 1.
3, 2, 5, 4
#3 Arrange the following random numbers in descending order from 25 to 10 – when the numbers do not follow a continuous sequence. Numbers that are not next to each other on a number line.
23, 17, 10, 25
#4 Arrange the following 8 times table in descending order.
16, 32, 64, 40
### Pre-learning strategies before Introducing the Concept of Ascending & Descending
Directly exposing children to Mathematical terms such as ‘ascending’ and ‘descending’ can get a little intimidating. So, to warm them up and informally share the idea with your kids, here are some activities you can practice;
• Keep 4 to 5 balls of different sizes and ask your child to arrange them in order i.e.; from smallest to largest and then largest to smallest.
• Take 4 to 5 bottles of different heights and then encourage the learner to place the bottles in height-wise order starting with the shortest and proceeding to the tallest, and vice versa.
• Introduce objects of different weights. Then let your child pick up the object and estimate which one is the heaviest and which is lighter, and the lightest.
• Assign each child with a number from 1 to 5 or 15 to 20, and let them figure how to stand orderly in a row from lowest to the highest number and highest to lowest.
• Play a relay race, where children have to pass on the baton to the person wearing the jersey of the next number. Example, number 1 jersey passes on the baton to the participant with number 2 jersey.
Mathseeds is an excellent Math based platform available in the UAE, across Dubai and other Middle East countries. Mathseeds targets children between ages 3 years to 9 years and contributes in establishing a sound fundamental base for kids to learn important math concepts. The lessons are structured systematically and the level of difficulty advances gradually, depending on the learning capabilities of the school goers.
If you stay in GCC or other Middle Eastern country, and wish to introduce ascending and descending order concepts to your learners in a fun and simplistic manner, you can subscribe to the Mathseeds app, by learning more about it here: https://knowledge-hub.com/mathseeds/
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# Linear Thinking Solving First Degree Equations - Ship
Linear Thinking Solving First Degree Equations - Ship
0011 0010 1010 1101 0001 0100 1011
Linear Thinking
Solving First Degree Equations
9/21/09
MAT 400
Chessa Horomanski
Jessica DiPaul
4 2 1
0011 0010 1010 1101 0001 0100 1011
The Rhind Papyrus
• Named after A. Henry Rhind
• Collection of problems probably used for training
young scribes in Ancient Egypt
Source: HistoryofScience.com
4 2 1
0011 0010 1010 1101 0001 0100 1011
Sample Problem
A quantity; its half and its third are added to
it. It becomes ten.
• For us:
• A scribe is instructed to solve it as we
would:
Divide 10 by
4 2 1
Example:
0011 0010 1010 1101 0001 0100 1011
A Different Method
A quantity; its fourth is added to it. It becomes fifteen.
4 2 1
• Assume (posits) the quantity is 4
• Take 4 and add its fourth to it, you get 4+1=5, but we
wanted 15
• We need to multiply what we got (5) by 3 to get what
we wanted to get (15)
• Take our guess (4) and multiply by 3 to get an answer
of 12
0011 0010 1010 1101 0001 0100 1011
False Position
• We posit an answer that we don’t really
expect to be the right one
– Makes the computations easy
4 2 1
• Use the incorrect result find number we
need to multiply our guess to get the correct
0011 0010 1010 1101 0001 0100 1011
You try it!
Use the method of false position to solve this problem.
A quantity and its ½ added together become 16. What is
the quantity?
4 2 1
Assume 2. Take . We wanted 16 but we got 3.
We need to multiply what we got (3) by to get 16.
Take our guess 2 and multiply it by
The quantity
0011 0010 1010 1101 0001 0100 1011
• Using symbols
Symbols
• Only works on equations in the form of
• Does not work on equations of the form
4 2 1
Double False Position
0011 0010 1010 1101 0001 0100 1011
• Effective method for solving linear
equations that it continued to be used long
after the invention of algebraic notations
• Requires no algebra
• Taught in arithmetic
textbooks
– Daboll’s Schoolmaster’s
Assistant
4 2 1
Source: HarvestofHistory.org
Example of Double False Position
0011 0010 1010 1101 0001 0100 1011
A purse of 100 dollars is to be divided among
four men A, B, C and D, so that B may have
four dollars more than A, and C eight
dollars more than B, and D twice as many
as C. What is each one’s share of the
money?
4 2 1
0011 0010 1010 1101 0001 0100 1011
Modern Approach:
Let x be the amount given to A. Then B gets
x+4, C gets (x+4)+8=x+12 and D gets
2(x+12). Total is \$100.
4 2 1
0011 0010 1010 1101 0001 0100 1011
Daboll’s Method
• Make a guess – assume A gets 6 dollars
– Then B gets 10, C gets 18, and D gets 36
4 2 1
– Adding them together we get \$70, only \$30 off
• Let’s try again
– Assume A gets 8 dollars, B gets 12, C gets 20,
and D gets 40
– Total now is \$80, still off by \$20
MAGIC!!
6 30
0011 0010 1010 1101 0001 0100 1011
8 20
4 2 1
• Cross multiply and subtract the differences to
get 120
• Divide the difference of errors to get 10
• Divide 120 by 10 so the right choice for man A
is \$12
• Only works when both errors are same type
Isn’t this puzzling?? Why does it work?
0011 0010 1010 1101 0001 0100 1011
Let’s use some graphical thinking.
By simplifying the left side, the equation will
be something of the form mx + b =100
4 2 1
We need 2 points.
We will use our guesses: (6, 70) and (8, 80)
0011 0010 1010 1101 0001 0100 1011
We want to find x so that (x, 100) is on the same
line.
100
80
70
6 8
x
4 2 1
0011 0010 1010 1101 0001 0100 1011
• Our way of understanding equations as lines
is quite recent (17 th century)
• Double false position is very old
4 2 1
• Change in the output is proportional to the
change in the input “linearity”
Linear” and “Nonlinear”
Linear
Simple relation – a
constant ratio –
between changes in
the input and output
0011 0010 1010 1101 0001 0100 1011
Nonlinear
No such simple relation
– very small changes in
the input may produce
huge changes in the
output
•We use linear problems to find approx. solutions to
nonlinear ones.
•The methods we use are based on the fundamental
insight which served as the basis for the method of
false position.
4 2 1
1650 B.C.
0011 0010 1010 1101 0001 0100 1011
• Rhind Papyrus
-False position
-Double false position
Early 17 th century
Timeline
• Modern way of understanding equations as lines
Early 1800s
• “Daboll’s Schoolmaster’s Assistant” –most popular
arithmetic book in America prior to 1850
4 2 1
0011 0010 1010 1101 0001 0100 1011
References
• Berlinghoff, William P. and Fernando Q. Gouvea. Math Through the
Ages. Oxton House Publishers, Maine, 2002.
• “Daboll’s Schoolmaster’s Assistant.”
http://www.harvestofhistory.org/primary_source_detail.html?ps_id_69
(September 16, 2009).
4 2 1
• John Fauvel and Jeremy Gray, eds. History of Mathematics: A Reader.
Macmillan Press Ltd., Basingstoke, 1988.
• Lucas N. H. Bunt, Phillip S. Jones, and Jack D. Bediant. The
Historical Roots of Elementary Mathematics. Dover Publications, New
York, 1976.
• “Mathematics/Logic Timeline: From Cave Paintings to the Internet.”
http://historyofscience.com/G21/timeline/index.php?category=Mathem
atics+%2F+Logic (September 16, 2009).
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# Given: Q = 7m + 3n, R = 11 - 2m, S = n + 5, and T = -m - 3n + 8, how do you simplify R - S + T?
$14 - 3 m - 4 n$
#### Explanation:
Given that
$R = 11 - 2 m , \setminus S = n + 5$ & $T = - m - 3 n + 8$
$\setminus \therefore R - S + T$
$= 11 - 2 m - \left(n + 5\right) + \left(- m - 3 n + 8\right)$
$= 11 - 2 m - n - 5 - m - 3 n + 8$
$= 14 - 3 m - 4 n$
Jul 26, 2018
$14 - 3 m - 4 n$
#### Explanation:
$\text{Substitute the given values into the expression}$
$R - S + T$
$= 11 - 2 m - \left(n + 5\right) + \left(- m - 3 n + 8\right)$
$= 11 - 2 m - n - 5 - m - 3 n + 8$
$= \left(11 - 5 + 8\right) + \left(- 2 m - m\right) + \left(- n - 3 n\right)$
$= 14 - 3 m - 4 n$
|
Wiki
# Vở bài tập Toán lớp 3 Tập 2 trang 32, 33 Bài 53 Tiết 1 | Kết nối tri thức
Have you been looking for effective and detailed solutions for your 3rd grade math homework from Exercise Book 2, specifically on Pages 32-33, Lesson 1? Look no further! At Kienthucykhoa.com, we provide top-notch assistance to help students easily complete their math exercises at home.
## Solving Math Homework: Exercise Book 2, Pages 32-33, Lesson 1 – Kienthucykhoa.com
### Exercise 1: Filling in the Blanks
In the first exercise on Page 32, you are required to write the appropriate numbers in the given blanks according to the pattern. Here’s how you can do it:
• Side length of a square:
• 15 cm
• 9 cm
• …(missing value)… cm
• 10 dm
• Perimeter of the square:
• 60 cm
• …(missing value)… cm
• 36 cm
• …(missing value)… dm
### Solution:
The formula for calculating the perimeter of a square is: Perimeter = (Length of one side) × 4
To find the length of one side of the square, we can use the formula: Length of one side = Perimeter of the square ÷ 4
Let’s fill in the missing values:
• Side length of a square:
• 15 cm
• 9 cm
• 9 cm
• 10 dm
• Perimeter of the square:
• 60 cm
• 36 cm
• 36 cm
• 40 dm
### Exercise 2: Calculating the Perimeter of a Rectangle
In Exercise 2 on Page 32, you need to calculate the perimeter of a rectangle with given length and width. Here are the two parts of this exercise:
a) Calculate the perimeter of a rectangle with a length of 3 dm and a width of 5 cm.
b) Calculate the perimeter of a rectangle with a length of 3 dm and a width of 20 cm.
### Solution:
a) First, let’s convert 3 dm to cm: 3 dm = 30 cm
The formula for calculating the perimeter of a rectangle is: Perimeter = (Length + Width) × 2
For the given values, the perimeter of the rectangle is: (30 + 5) × 2 = 70 cm
b) Again, let’s convert 3 dm to cm: 3 dm = 30 cm
The perimeter of the rectangle is: (30 + 20) × 2 = 120 cm
### Exercise 3: Perimeter of a Composite Square
In Exercise 3 on Page 32, you are asked to calculate the perimeter of a square made up of four smaller squares. Each side of the smaller squares is 50 cm. Here’s how you can do it:
### Solution:
The length of a square used to cover the entire area is obtained by multiplying the side length of each smaller square by 4. So, the length of the composite square is: 50 × 4 = 200 cm
The perimeter of the composite square is simply 4 times the length of one side: 200 × 4 = 800 cm
### Exercise 4: Fencing the Flower Gardens
In Exercise 4 on Page 33, Ms. Hoa is fencing different pieces of land to plant roses, daisies, and orchids. It is known that the adjacent pieces of land are 1m apart (as shown in the diagram).
a) Let’s fill in the blanks:
Piece of land A has a fence that is …m long, piece of land B has a fence that is …m long, and piece of land C has a fence that is …m long.
b) Circle the correct letter:
Which piece of land is dedicated to growing orchids?
A. Piece of land A
B. Piece of land B
C. Piece of land C
### Solution:
Let’s calculate the lengths of the fences for each piece of land:
+) Garden A:
• Length of the fence = 1 × 4 = 4 m
• Width of the fence = 1 × 3 = 3 m
• Perimeter of the fence = (4 + 3) × 2 = 14 m
+) Garden B:
• Length of the fence = 1 × 5 = 5 m
• Width of the fence = 1 × 4 = 4 m
• Perimeter of the fence = (4 + 5) × 2 = 18 m
+) Garden C:
• Length of one side of the fence = 1 × 4 = 4 m
• Perimeter of the fence = 4 × 4 = 16 m
a) Based on the calculations, the lengths of the fences are:
• Piece of land A: 4 m
• Piece of land B: 5 m
• Piece of land C: 4 m
b) The piece of land dedicated to growing orchids is Piece of land C.
Circle “C”.
For more detailed and helpful solutions for Exercise Book 2, Pages 34-35, Lesson 2 and Lesson 3, check out the following links:
### Kiến Thức Y Khoa
Xin chào các bạn, tôi là người sở hữu website Kiến Thức Y Khoa. Tôi sử dụng content AI và đã chỉnh sửa đề phù hợp với người đọc nhằm cung cấp thông tin lên website https://kienthucykhoa.edu.vn/.
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Find a Nonsingular Matrix Satisfying Some Relation
Problem 280
Determine whether there exists a nonsingular matrix $A$ if
$A^2=AB+2A,$ where $B$ is the following matrix.
If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$.
(a) $B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 1 & 2 & -2 \end{bmatrix}$
(b) $B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.$
Solution.
Suppose that a nonsingular matrix $A$ satisfying $A^2=AB+2A$ exists.
Then $A$ is invertible since $A$ is nonsingular, and thus the inverse $A^{-1}$ exists.
Multiplying by $A^{-1}$ on the left, we have
\begin{align*}
A=&A^{-1}A^2=A^{-1}(AB+2A)\\
&=A^{-1}AB+2A^{-1}A\\
&=B+2I,
\end{align*}
where $I$ is the $3\times 3$ identity matrix.
Therefore, if such a nonsingular matrix exists, it must be
$A=B+2I. \tag{*}$
(a) The first case
Let us consider the case
$B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 1 & 2 & -2 \end{bmatrix}.$ In this case, we have from (*)
$A=B+2I=\begin{bmatrix} 1 & 1 & -1 \\ 0 &1 &0 \\ 1 & 2 & 0 \end{bmatrix}.$
We still need to check that this matrix is in fact a nonsingular matrix.
To check the non-singularity and to find the inverse matrix at once, we consider the augmented matrix $[A\mid I]$ and apply elementary row operations.
We have
\begin{align*}
&[A\mid I] =
\left[\begin{array}{rrr|rrr}
1 & 1 & -1 & 1 &0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
1 & 2 & 0 & 0 & 0 & 1 \\
\end{array} \right]\10pt] & \xrightarrow{R_3-R_1} \left[\begin{array}{rrr|rrr} 1 & 1 & -1 & 1 &0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & -1 & 0 & 1 \\ \end{array} \right] \xrightarrow{\substack{R_1-R_2 \\ R_3-R_2}} \left[\begin{array}{rrr|rrr} 1 & 0 & -1 & 1 & -1& 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 1 \\ \end{array} \right] \\[10pt] &\xrightarrow{R_1+R_3} \left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 0 & -2& 1 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 1 \\ \end{array} \right]. \end{align*} The left part of the last matrix is the identity matrix, and thus the matrix A is invertible and the inverse matrix is the right half: \[A^{-1}=\begin{bmatrix} 0 & -2 & 1 \\ 0 &1 &0 \\ -1 & -1 & 1 \end{bmatrix}.
(b) The second case
Next let us consider the case
$B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.$ By (*), if a nonsingular matrix $A$ exists, it must be
$A=B+2I=\begin{bmatrix} 1 & 1 & -1 \\ 0 &1 &0 \\ 2 & 1 & -2 \end{bmatrix}.$
We need to determine whether this matrix is actually nonsingular.
In fact, we prove that this matrix is singular.
That is, we show that $A\mathbf{x}=\mathbf{0}$ has a nonzero solution.
Consider the augmented matrix $[A\mid \mathbf{0}]$. By Gauss-Jordan elimination, we have
\begin{align*}
&[A\mid \mathbf{0}] = \left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 &1 & 0 & 0 \\
2 & 1 & -2 & 0
\end{array} \right] \xrightarrow{R_3-2R_1}\\[10pt] & \left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 &1 & 0 & 0 \\
0 & -1 & 0 & 0
\end{array} \right] \xrightarrow{\substack{R_1-R_2\\ R_3+R_2}}
\left[\begin{array}{rrr|r}
1 & 0& -1 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}
The last matrix is in reduced row echelon form and it has a zero row. From this, we see that $x_3$ must be a free variable, and the matrix $A$ is singular.
(The general solution is $x_1=x_3, x_2=0$. Thus for example, $x_1=1, x_2=0, x_3=1$ is a nonzero solution of $A\mathbf{x}=\mathbf{0}$.)
Thus, we conclude that there is no nonsingular matrix $A$ satisfying $A^2=AB+2A$ in this case.
Determine conditions on the scalars $a, b$ so that the following set $S$ of vectors is linearly dependent. \begin{align*} S=\{\mathbf{v}_1,...
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# Math Clip Art--Base Ten Blocks--50
Number Models
## Description
This image is a crucial component of the base ten blocks series, demonstrating the modeling of the number 100. It employs a template in the form of a placemat where base ten blocks representing thousands, hundreds, tens, and ones are strategically placed. This visual representation provides a concrete model of the number 100 using the base ten system, marking the beginning of our exploration into three-digit numbers.
The significance of math clip art, particularly base ten block models, in mathematics education is profound. These visual tools transform abstract numerical concepts into tangible, easy-to-grasp representations. They play a vital role in helping students understand place value, number composition, and the foundations of our decimal system, especially as we move into larger numbers. By integrating these images into Number lessons, educators can create a more engaging and accessible learning environment, fostering a deeper comprehension of fundamental numerical concepts and relationships.
Teacher's Script: "Now, let's explore how we represent the number 100 using our base ten blocks. We have one large flat square representing a hundred. This is our starting point for exploring three-digit numbers. How does this compare to the numbers we've seen before? What patterns do you think we'll notice as we build larger numbers in the hundreds? Can you predict what 150 might look like?"
For a complete collection of math clip art related to Number Models click on this link: Number Models: Base Ten Blocks Collection.
## The Base 10 Number System
When you first learned to count, you started at zero or one and counted to nine.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
After nine, the numbers started repeating.
10, 11, 12, 13, 14, 15, 16, 17, 18, 19
After 19, the pattern of numbers continues. In fact, counting numbers involves just using the digits from 0 to 9 in different combinations.
Why is that?
Our numbering system is called a base 10 system. This means there are ten digits, which you know as 0 to 9. All numbers in a base 10 system just use these digits.
## Counting in a Base 10 System
When you look at a number, each digit represents a certain place value. Take a look at this three-digit number
The digit 3 is in the ones place. The digit 2 is in the tens place. The digit 1 is in the hundreds place.
The digit 3 is in the ones place. The digit 2 is in the tens place. The digit 1 is in the hundreds place.
For whole numbers, place value increases 10-fold.
## Place Value in the Base 10 System
Place value determines the size of a number. Take a look at this number:
### 23
In the ones place is the digit 3, but int he tens place is the digit 2. Don't let the single digit fool you. The 2 in the tens place has a value of 20.
Now look at this number:
### 542
In the ones place is the digit 2. In the tens place is the digit 4. In the hundreds place is 5. If we were to separate these numbers, we would see this:
### 2
So, in going from left to right starting at the decimal point, each place value is ten times the value of the place value to the right.
If you're given a description of a number using just place value information, you can write a number. Here's an example.
The digit in the ones place is 5. The digit in the tens place is 3. The digit in the hundreds place is 9. What is the number?
Using this description we write the following number: 935.
## Number Systems with Other Bases
You've seen how the base 10 number system works, but why do we use a base 10 system? Are there other number systems?
Let's address the first question. The reason we use a base 10 system is probably because we have ten fingers to count with. Do you think it's a coincidence that our base 10 numbering system aligns with our 10 fingers to count?
But we have 10 fingers and 10 toes. Are there number systems that use base 20. Yes!
The Mayan culture used a base 20 system for counting. These are twenty symbols used to count.
This numbering system uses just three symbols in different combinations: dots, dashes, and the shell. The shell represents zero. The dash represents an increment of five. The dot represents a value of one. Different combinations of these symbols can model any number.
Another, even more popular system is the binary system, which is used extensively with computers. With the binary system, there are only two digits: 0 and 1. This combination of digits can be used any number equivalent to our base 10 numbers. Here are the first 10 numbers in binary and base 10.
Base 10 Binary 0 0 1 1 2 10 3 11 4 100 5 101 6 110 7 111 8 1000 9 1001
Why do computers use a binary, or base 2, system? What is the advantage of a binary system?
Computer circuitry and connections involve a lot of on-off switching. Many computer commands are basically combinations of these on-off switches. A binary system is ideal for modeling an on-off system. If you let 0 mean off and 1 mean on, then a binary code can be used to model not only numbers but any computer state.
Each number system has its own rules for writing numbers. Once you learn the rules, then you can write any number in any number system.
### Number Models: Base Ten Blocks
One way to model base ten numbers is to use number models, and one of the best ones to use is to use base ten blocks. Base ten blocks are a visual model to represent numbers. Here is what these models look like.
For each of these models, place the correct number of blocks to model the digit in that place value.
This is a model for the number 24. There are two tens blocks in the tens place and four ones blocks in the ones place.
Here is a number to the hundreds place.
To model 250, put two hundreds blocks in the hundreds place, five tens blocks in the tens place, and no blocks in the ones place.
Common Core Standards CCSS.MATH.CONTENT.K.NBT.A.1, CCSS.MATH.CONTENT.1.NBT.B.2, CCSS.MATH.CONTENT.2.NBT.A.1 K - 3 Arithmetic • Numbers and Patterns • Place Value 2021 place value, base ten blocks
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# The Second Derivative
The second derivative of a quadratic function is constant.
In calculus, the second derivative, or the second order derivative, of a function f is the derivative of the derivative of f. Roughly speaking, the second derivative measures how the rate of change of a quantity is itself changing; for example, the second derivative of the position of an object with respect to time is the instantaneous acceleration of the object, or the rate at which the velocity of the object is changing with respect to time. In Leibniz notation:
${\displaystyle \mathbf {a} ={\frac {d\mathbf {v} }{dt}}={\frac {d^{2}{\boldsymbol {x}}}{dt^{2}}},}$
where a is acceleration, v is velocity, t is time, x is position, and d is the instantaneous "delta" or change. The last expression ${\displaystyle {\tfrac {d^{2}{\boldsymbol {x}}}{dt^{2}}}}$ is the second derivative of position (x) with respect to time.
On the graph of a function, the second derivative corresponds to the curvature or concavity of the graph. The graph of a function with a positive second derivative is upwardly concave, while the graph of a function with a negative second derivative curves in the opposite way.
## Second derivative power rule
The power rule for the first derivative, if applied twice, will produce the second derivative power rule as follows:
${\displaystyle {\frac {d^{2}}{dx^{2}}}\left[x^{n}\right]={\frac {d}{dx}}{\frac {d}{dx}}\left[x^{n}\right]={\frac {d}{dx}}\left[nx^{n-1}\right]=n{\frac {d}{dx}}\left[x^{n-1}\right]=n(n-1)x^{n-2}.}$
## Notation
The second derivative of a function ${\displaystyle f(x)}$ is usually denoted ${\displaystyle f''(x)}$. That is:
${\displaystyle f''=\left(f'\right)'}$
When using Leibniz's notation for derivatives, the second derivative of a dependent variable y with respect to an independent variable x is written
${\displaystyle {\frac {d^{2}y}{dx^{2}}}.}$
This notation is derived from the following formula:
${\displaystyle {\frac {d^{2}y}{dx^{2}}}\,=\,{\frac {d}{dx}}\left({\frac {dy}{dx}}\right).}$
## Alternative notation
As the previous section notes, the standard Leibniz notation for the second derivative is ${\textstyle {\frac {d^{2}y}{dx^{2}}}}$. However, this form is not algebraically manipulable. That is, although it is formed looking like a fraction of differentials, the fraction cannot be split apart into pieces, the terms cannot be cancelled, etc. However, this limitation can be remedied by using an alternative formula for the second derivative. This one is derived from applying the quotient rule to the first derivative. Doing this yields the formula:
${\displaystyle y''(x)={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)={\frac {d\left({\frac {dy}{dx}}\right)}{dx}}={\frac {d^{2}y}{dx^{2}}}-{\frac {dy}{dx}}{\frac {d^{2}x}{dx^{2}}}}$
In this formula, ${\displaystyle du}$ represents the differential operator applied to ${\displaystyle u}$, i.e., ${\displaystyle d(u)}$, ${\displaystyle d^{2}u}$ represents applying the differential operator twice, i.e., ${\displaystyle d(d(u))}$, and ${\displaystyle du^{2}}$ refers to the square of the differential operator applied to ${\displaystyle u}$, i.e., ${\displaystyle (d(u))^{2}}$.
When written this way (and taking into account the meaning of the notation given above), the terms of the second derivative can be freely manipulated as any other algebraic term. For instance, the inverse function formula for the second derivative can be deduced from algebraic manipulations of the above formula, as well as the chain rule for the second derivative. Whether making such a change to the notation is sufficiently helpful to be worth the trouble is still under debate.
## Example
Given the function
${\displaystyle f(x)=x^{3},}$
the derivative of f is the function
${\displaystyle f^{\prime }(x)=3x^{2}.}$
The second derivative of f is the derivative of ${\displaystyle f^{\prime }}$, namely
${\displaystyle f^{\prime \prime }(x)=6x.}$
## Relation to the graph
A plot of ${\displaystyle f(x)=\sin(2x)}$ from ${\displaystyle -\pi /4}$ to ${\displaystyle 5\pi /4}$. The tangent line is blue where the curve is concave up, green where the curve is concave down, and red at the inflection points (0, ${\displaystyle \pi }$/2, and ${\displaystyle \pi }$).
### Concavity
The second derivative of a function f can be used to determine the concavity of the graph of f. A function whose second derivative is positive will be concave up (also referred to as convex), meaning that the tangent line will lie below the graph of the function. Similarly, a function whose second derivative is negative will be concave down (also simply called concave), and its tangent lines will lie above the graph of the function.
### Inflection points
If the second derivative of a function changes sign, the graph of the function will switch from concave down to concave up, or vice versa. A point where this occurs is called an inflection point. Assuming the second derivative is continuous, it must take a value of zero at any inflection point, although not every point where the second derivative is zero is necessarily a point of inflection.
### Second derivative test
The relation between the second derivative and the graph can be used to test whether a stationary point for a function (i.e., a point where ${\displaystyle f'(x)=0}$) is a local maximum or a local minimum. Specifically,
• If ${\displaystyle f^{\prime \prime }(x)<0}$, then ${\displaystyle f}$ has a local maximum at ${\displaystyle x}$.
• If ${\displaystyle f^{\prime \prime }(x)>0}$, then ${\displaystyle f}$ has a local minimum at ${\displaystyle x}$.
• If ${\displaystyle f^{\prime \prime }(x)=0}$, the second derivative test says nothing about the point ${\displaystyle x}$, a possible inflection point.
The reason the second derivative produces these results can be seen by way of a real-world analogy. Consider a vehicle that at first is moving forward at a great velocity, but with a negative acceleration. Clearly, the position of the vehicle at the point where the velocity reaches zero will be the maximum distance from the starting position – after this time, the velocity will become negative and the vehicle will reverse. The same is true for the minimum, with a vehicle that at first has a very negative velocity but positive acceleration.
## Limit
It is possible to write a single limit for the second derivative:
${\displaystyle f''(x)=\lim _{h\to 0}{\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}.}$
The limit is called the second symmetric derivative. Note that the second symmetric derivative may exist even when the (usual) second derivative does not.
The expression on the right can be written as a difference quotient of difference quotients:
${\displaystyle {\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}={\frac {{\frac {f(x+h)-f(x)}{h}}-{\frac {f(x)-f(x-h)}{h}}}{h}}.}$
This limit can be viewed as a continuous version of the second difference for sequences.
However, the existence of the above limit does not mean that the function ${\displaystyle f}$ has a second derivative. The limit above just gives a possibility for calculating the second derivative—but does not provide a definition. A counterexample is the sign function ${\displaystyle \operatorname {sgn}(x)}$, which is defined as:
${\displaystyle \operatorname {sgn}(x)={\begin{cases}-1&{\text{if }}x<0,\\0&{\text{if }}x=0,\\1&{\text{if }}x>0.\end{cases}}}$
The sign function is not continuous at zero, and therefore the second derivative for ${\displaystyle x=0}$ does not exist. But the above limit exists for ${\displaystyle x=0}$:
{\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {\operatorname {sgn}(0+h)-2\operatorname {sgn}(0)+\operatorname {sgn}(0-h)}{h^{2}}}&=\lim _{h\to 0}{\frac {\operatorname {sgn}(h)-2\cdot 0+\operatorname {sgn}(-h)}{h^{2}}}\\&=\lim _{h\to 0}{\frac {\operatorname {sgn}(h)+(-\operatorname {sgn}(h))}{h^{2}}}=\lim _{h\to 0}{\frac {0}{h^{2}}}=0.\end{aligned}}}
${\displaystyle f(x)\approx f(a)+f'(a)(x-a)+{\tfrac {1}{2}}f''(a)(x-a)^{2}.}$
|
## How do you find the roots of an equation?
The roots of any quadratic equation is given by: x = [-b +/- sqrt(-b^2 – 4ac)]/2a. Write down the quadratic in the form of ax^2 + bx + c = 0. If the equation is in the form y = ax^2 + bx +c, simply replace the y with 0. This is done because the roots of the equation are the values where the y axis is equal to 0.
## What is meant by root of equation?
A real number x will be called a solution or a root if it satisfies the equation, meaning . It is easy to see that the roots are exactly the x-intercepts of the quadratic function. , that is the intersection between the graph of the quadratic function with the x-axis. a<0. a>0.
## How do you find the roots of an equation with one root?
Hint: Here, one root is given for the quadratic equation x2−5x+6=0. Now, we can find the other root by the formula for sum and product of the roots. If α and β are the two roots of the quadratic equation ax2+bx+c=0 then the sum and product of the roots are given by the formula: α+β=−ba and αβ=ca.
## Why do we find roots of equations?
Finding roots are a means to an end in solving sets of equalities (and are useful for understanding inequalities as well). For example if you need to find where two lines meet, then you set up equalities and solve for the unknowns.
## How do you find the roots of a linear equation?
Solution of linear equation or Root of linear equation: The value of the variable which makes left hand side equal to right hand side in the given equation is called the solution or the root of the equation. If we put x = 3, then L.H.S. is 3 + 1 which is equal to R.H.S. 2. 5x – 2 = 3x – 4 is a linear equation.
## What is root math example?
A square root of a number is a value that, when multiplied by itself, gives the number. Example: 4 × 4 = 16, so a square root of 16 is 4. Note that (−4) × (−4) = 16 too, so −4 is also a square root of 16. The symbol is √ which always means the positive square root. Example: √36 = 6 (because 6 x 6 = 36)
## What does equation mean?
An equation is a mathematical statement that two things are equal. It consists of two expressions, one on each side of an ‘equals’ sign. For example: 12.
## What is the zero of an equation?
A root or a zero of a polynomial are the value(s) of X that cause the polynomial to = 0 (or make Y=0). It is an X-intercept. The root is the X-value, and zero is the Y-value. It is not saying that imaginary roots = 0.
## What are the roots of a quadratic equation?
The roots of a function are the x-intercepts. By definition, the y-coordinate of points lying on the x-axis is zero. Therefore, to find the roots of a quadratic function, we set f (x) = 0, and solve the equation, ax2 + bx + c = 0.
## How do you find the sum of the roots?
Example: What is an equation whose roots are 5 + √2 and 5 − √2. When a=1 we can work out that: Sum of the roots = −b/a = -b. Product of the roots = c/a = c.
## What is discriminant in math?
Discriminant, in mathematics, a parameter of an object or system calculated as an aid to its classification or solution. In the case of a quadratic equation ax2 + bx + c = 0, the discriminant is b2 − 4ac; for a cubic equation x3 + ax2 + bx + c = 0, the discriminant is a2b2 + 18abc − 4b3 − 4a3c − 27c2.
## Why are roots important in math?
Finding the roots of a function means you are finding solutions to an equation. Those solutions can be really important. For example, they can tell you what price you should charge customers to maximize your expected profits.
## Are roots and zeros the same?
A zero is of a function. A root is of an equation. But, when the equation only has numbers and one variable, the ONLY appropriate term is roots. However, when looking at just a polynomial (no equation) then either term is appropriate, because they both imply making the polynomial equal to zero first.
### Releated
#### Factoring an equation
How do you factor algebraic equations? So, if, in your equation, your b value is twice the square root of your c value, your equation can be factored to (x + (sqrt(c)))2. For example, the equation x2 + 6x + 9 fits this form. 32 is 9 and 3 × 2 is 6. So, we […]
#### The equation of exchange can be stated as
What is the equation of exchange equal to? So the equation of exchange says that the total amount of money that changes hands in the economy will always equal the total money value of the goods and services that change hands in the economy. So now the equation of exchange says that total nominal expenditures […]
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# How do you write a rule for the nth term of the geometric sequence given the two terms a_2=-153, a_4=-17?
Jan 19, 2018
${u}_{n} = \pm 459 \cdot {\left(\pm \frac{1}{3}\right)}^{n - 1}$
#### Explanation:
geometric sequence: ${u}_{n} = a {r}^{-} 1$
where $a$ is the starting term
and $r$ is the number by which one number is multiplied to make the next number in the sequence. (common ratio)
${u}_{2} = - 153$
${u}_{4} = - 17$
${u}_{2} : n = 2$
${u}_{2} = a {r}^{2 - 1} = a {r}^{1}$
${u}_{2} = a r$
${u}_{4} : n = 4$
${u}_{4} = a {r}^{4 - 1} = a {r}^{3}$
$a r = - 153$
$a {r}^{3} = - 17$
${r}^{2} = \frac{a {r}^{3}}{a r} = \frac{- 17}{-} 153$
${r}^{2} = \frac{1}{9}$
$r = \pm \frac{1}{3}$
$a r = - 153$
$r = \pm \frac{1}{3}$
$a = - \frac{153}{\pm \frac{1}{3}} = - 153 \cdot \pm 3$
$a = \pm 459$
the $n$th term of the geometric sequence is ${u}_{n} = \pm 459 \cdot {\left(\pm \frac{1}{3}\right)}^{n - 1}$
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# Multiplication Facts 0-10 Memory Tips
## MULTIPLICATION FACTS 0-12
When students learn multiplication, it starts out easy.
Too easy, as 0 times anything equals 0 (e.g. 0 x 8 = 0, while anything times 1 equals itself (e.g. 4 x 1 = 4).
But it doesn’t stay easy. Most students struggle to remember multiplication facts where one of the numbers is 6, 7, 8, or 9.
However, there are some tips to aid in memorization.
For one, mirror images don’t matter. The word for this is commutative.
This means 2 x 8 = 8 x 2, for example. Both equal 16.
So if you know 6 x 8 = 48, you also know that 8 x 6 = 48. You don’t need to memorize both.
One of the tricky multiplication facts is 7 x 8. But it’s easy if you remember the trick.
Remember 5678. These are 5 thru 8 in order. 56 = 7 x 8. Piece of cake, huh?
The 9’s are easy. The answer is one decade less than multiplying by 10, then make the two digits add up to 9.
For example, 9 x 7 is in the 60’s (because 10 x 7 = 70, one decade less is 60). It’s 63 since 6 + 3 = 9.
Another example is 5 x 9. It’s in the 40’s (one decade less than 5 x 10 = 50). It’s 45 since 4 + 5 = 9.
To do the trick, after you figure out the decade (by multiplying by 10 and then subtracting 10), subtract the tens digit from 9 to get the units digit.
For example, consider 9 x 6. Multiply 10 x 6 to get 60, and subtract 10 to make 50 (one decade less). Now subtract 5 (the tens digit of 50) from 9 to get 4. Therefore, 9 x 6 = 54.
Once you get the hang of it, this makes remembering the 9’s easy. Try out all the 9’s to get some practice.
If you’re good at doubling numbers quickly, try writing 6 as 2 x 3.
Then 6 x 7 = 2 x 3 x 7. If you know 3 x 7 = 21, double 21 to get 42.
Similarly, for 6 x 4 = 2 x 3 x 4, start with 3 x 4 = 12 and double 12 to make 24.
You can use the doubling trick for the 8’s, too. Just double the number 3 times.
For example, consider 5 x 8. Double 5 three times: 10, 20, 40. So 5 x 8 = 40.
Try 8 x 6. Double 6 three times: 12, 24, 48. Therefore, 8 x 6 = 48.
It works with the 4’s, also. Just double twice.
With 4 x 9, double 9 twice: 18, 36. So 4 x 9 = 36.
That leaves 7 x 7 = 49. You should know 7 x 9 from the 9’s trick.
You can make 7 x 6 and 7 x 8 from the doubling tricks. (The latter you can also know from the 5678 trick.)
5 and under are easier. So to complete the 7’s, you really just need to memorize 7 x 7 = 49.
This covers the 6 thru 9’s, which tend to be the trickier multiplication facts.
The 10’s and 11’s are easy. For the 10’s, just add a zero, as in 8 x 10 = 80 or 6 x 10 = 60.
For the 11’s, with 1 thru 9 just double the digits, like 3 x 11 =33 or 11 x 8 = 88. Get 11 x 10 = 110 from the 10’s trick (add a zero). Then you just need to memorize that 11 x 11 = 121 to complete the 11’s.
You can get the 12’s by doubling the 6’s. For example, knowing that 6 x 5 = 30, you can find that 12 x 5 = 60 by doubling 6 x 5.
Do you know any other tips for remembering multiplication facts 0-12? If so, please share them in the comments.
## CHRIS MCMULLEN, PH.D.
Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks
• arithmetic facts
• multi-digit arithmetic
• long division with remainders
• fractions, decimals, and percents
• algebra and trigonometry
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Prove that, for any polygon, taking all pair of adjacent angles, subtracting 180 from their sum, and adding all the results together equals $180(n-4)$
Take any simple polygon. Extend all sides in both directions. Note the angles where these sides meet, if at all. What is the sum of these angles?
For example, consider the figure formed from a regular hexagon, the Star of David:
Given that the angles of a regular hexagon are each $$120$$ degrees, it’s easy to calculate that the angles in question are $$60$$ degrees each, since the base angles of the triangle are supplementary to the angles of the hexagon, and the angles of a triangle must add to $$180$$.
Therefore, since there are six of these, they add up to $$360$$ degrees in total.
Exchanging the $$6$$ for $$n$$ (for general case), this can be written as
$$n\left(180-2\left(180-\frac{180(n-2)}{n}\right)\right)$$
Distributing and simplifying:
$$180n-2n\left(180-\frac{180(n-2)}{n}\right)$$ $$180n-360n+360n-720$$ $$180n-720$$ $$180(n-4)$$
Another way of wording this is that this gives a measure of how far any two adjacent angles in a polygon are from forming parallel lines; that is, how far their angles are from $$90$$ degrees, with the result being positive if they’re slanted inward, and negative if they’re slanted outward. You can calculate this by taking each angle in the pair, subtracting $$90$$ from them, and adding the results together.
Mathematically, it’s the same as above, but it’s conceptually very different. The benefit of thinking of it this way is that this justifies a square giving an output of $$0$$ and an equilateral triangle giving an output of $$-180$$. And...well, when I discuss crazier cases later, thinking of the angles this way will make more sense, especially when they don’t exist.
This proof hinges on the original polygon being regular, however. The formula is derived from multiplying the angle by the number of triangles, and that’s only true if there are that many triangles.
Consider the following pentagon, with its lines extended:
This pentagon has three right angles and two $$135$$-degree angles. By the same method as above, the four triangles formed can be shown to have points of $$45$$ degrees each. The bottom of the pentagon doesn’t form a triangle at all, but, more specifically, it forms two parallel lines; as this doesn’t stray from $$90$$ degrees at all, this gets us a $$0$$ for the bottom, and $$180$$ overall.
This is exactly what you’d get from the above formula: $$180(5-4)$$.
Consider the figure formed from a trapezoid:
Two pairs of angles yield parallel lines and therefore output $$0$$ each. One pair actually yields a triangle, with a third angle measuring $$90$$; that is to say, each of those angles on the trapezoid leans inward from a perfectly horizontal line by $$45$$ degrees. The fourth pair doesn’t form a triangle, but more importantly, the lines lean outward by $$45$$ degrees. These two effects cancel out, and overall, the figure nets a $$0$$ - no different than a square.
It’s not just irregular polygons - concave ones follow this pattern as well. It would be too hard to draw how the “triangles” work here, but consider a quadrilateral with angles $$90, 30, 30, 210$$. By the same logic as above - by subtracting $$90$$ from each angle in each adjacent pair - you still end up with a result of $$0$$!
I can justify $$180(n-4)$$ holding true by regular polygons. How do I prove that this is the case for all polygons?
For an n-gon, considering a line rotating around it rotates 360. This implies the standard result that the sum of the exterior angles in the direction of the rotation is 360 so the sum of the interior angles is n*180-360 = 180(n-2).
Therefore the sum of the exterior angles on both sides of each vertex is 720.
From your first diagram, each angle of the extended polygon is 180 minus the sum of the adjacent exterior angles, so their sum is 180*n-720 =180(n-4).
• I don’t follow how the first paragraph implies the second. – DonielF Dec 20 '18 at 4:35
• There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720. – marty cohen Dec 20 '18 at 14:10
• I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram? – DonielF Dec 20 '18 at 14:16
• I'm not sure whether my answer is the same as yours, just written differently. Can you confirm? – DonielF Dec 24 '18 at 18:07
I'm disappointed that I didn't see this earlier.
Another way to word my conclusion is that the formula I calculated is another way of saying that if you take every angle of the polygon and subtract $$90$$ from each, and you do this twice for each (one for each time you extend the line - one in each direction), you get $$180(n-4)$$. That is:
$$2\sum_{i=1}^n(a_i-90)=180(n-4)$$ where $$n$$ is the number of angles in the polygon.
Since addition is commutative, the summation can be restated as
$$2\left(\sum_{i=1}^na_i-\sum_{i=1}^n90\right)$$
The first sigma is just adding up the angles of the polygon, which we know is $$180(n-2)$$, and the second one is the same as adding $$90$$ to itself $$n$$ times, so this can be rewritten as
$$2(180(n-2)-90n)$$ $$360n-720-180n$$ $$180n-720$$ $$180(n-4)$$
Q.E.D.
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# Unlocking the Magic of 5.25 as a Fraction: Exploring Its Meaning, Uses, and Applications
## Understanding 5.25 as a Fraction: A Comprehensive Guide
### What is a Fraction?
A fraction is a way of representing a part of a whole. It consists of two numbers separated by a line, with the top number called the numerator and the bottom number called the denominator. The numerator represents the number of parts we have, while the denominator represents the total number of equal parts that make up the whole. In the case of 5.25, we need to understand how to express this decimal value as a fraction.
### Converting 5.25 to a Fraction
To convert a decimal to a fraction, we need to understand the place value of each digit in the decimal number. In 5.25, the decimal point separates the whole number part (5) from the decimal part (0.25). Since the decimal part is 0.25, we can write it as a fraction by putting the decimal part over a multiple of 10. In this case, the decimal part is 25, and the multiple of 10 is 100. So, 0.25 can be written as 25/100.
### Simplifying the Fraction
To simplify the fraction 25/100, we need to find the greatest common divisor (GCD) between the numerator and the denominator. In this case, both numbers can be divided by 25, which gives us 1/4. Therefore, 5.25 as a fraction is 1/4.
It’s important to understand the concept of fractions and how to convert decimals to fractions, as it allows us to work with various mathematical operations, comparisons, and solve real-world problems. Knowing how to express 5.25 as a fraction can be handy in situations where fractions are more commonly used and understood than decimals.
## Simplify and Convert 5.25 to a Fraction Step-by-Step
### Step 1: Understanding the Concept
To simplify and convert 5.25 to a fraction, it’s important to have a clear understanding of the concept. A fraction is a way of representing a part of a whole number or a ratio between two numbers. In this case, we have a decimal number, 5.25, that we want to convert to a fraction.
### Step 2: Converting the Decimal to a Fraction
To convert a decimal to a fraction, we need to determine the place value of the decimal. In the number 5.25, the 5 is in the ones place, the 2 is in the tenths place, and the 5 is in the hundredths place.
To convert 5.25 to a fraction, we can write it as a fraction with the decimal part as the numerator and the denominator as a power of 10 based on the number of decimal places. Since there are two decimal places, we use 100 as the denominator.
The fraction form of 5.25 is therefore 525/100.
### Step 3: Simplifying the Fraction
To simplify the fraction 525/100, we can divide both the numerator and denominator by their greatest common divisor (GCD). In this case, the GCD of 525 and 100 is 25.
Dividing both the numerator and denominator by 25, we get 21/4.
Therefore, 5.25 can be simplified and converted to the fraction 21/4.
## Discover the Common Equivalents of 5.25 as a Fraction
When working with fractions, it’s important to understand how to express mixed numbers as improper fractions or decimal numbers. In this case, we’ll focus on the common equivalents of 5.25 as a fraction. Let’s dive into it!
5.25 as an Improper Fraction: To convert a mixed number like 5.25 into an improper fraction, we multiply the whole number (5) by the denominator of the fraction (4 in this case) and add the numerator (25). This gives us a result of 5 multiplied by 4, which is 20, plus 25. So, the improper fraction equivalent of 5.25 is 125/4.
5.25 as a Decimal: Another way to express 5.25 is as a decimal. To do this, we divide the numerator of the fraction (25) by the denominator (4). This gives us a decimal value of 6.25. Therefore, 5.25 can also be written as 6.25 as a decimal.
Converting 5.25 to Other Common Fractions: Apart from the improper fraction and the decimal form, there are other common fraction equivalents of 5.25. Some of them include 5 1/4, 21/4, and 105/20. These fractions represent different ways to express the same numerical value of 5.25.
## Mastering the Decimal to Fraction Conversion: 5.25 Unveiled
You may also be interested in: Converting 8000 Meters to Miles: The Ultimate Guide for Accurate Measurements
### What is Decimal to Fraction Conversion?
Decimal to fraction conversion is the process of converting a decimal number into a fraction. It is a valuable skill to have, especially for mathematicians, engineers, and anyone working with measurements. By mastering this conversion, you can easily compare and manipulate fractional values, opening up a whole new world of possibilities in your calculations.
The Basics of Decimal to Fraction Conversion: To convert a decimal to a fraction, you need to understand the place values of decimals. In the given example of “5.25,” the digit 5 is in the whole number place, the digit 2 is in the tenths place, and the digit 5 is in the hundredths place. Understanding this notation is crucial for effective conversion.
### Converting 5.25 to Fraction:
To convert a decimal like 5.25 into a fraction, follow these steps:
1. Identify the decimal place value: In this case, we have tenths (2) and hundredths (5).
2. Rewrite the decimal as a fraction with a denominator equal to the place value. In this case, the denominator will be 10 for the tenths place and 100 for the hundredths place.
3. Simplify the fraction if needed: Since 25/100 can be reduced, we can simplify it to 1/4.
Finally, 5.25 can be expressed as the fraction 5 1/4.
Why Mastering Decimal to Fraction Conversion is Important: Converting decimals to fractions is useful for various reasons. It allows for easier comparison of values, particularly when solving mathematical problems or working with measurements. Moreover, understanding this skill is vital for converting decimals to percentages and vice versa, which is often required when dealing with statistics or financial calculations.
In conclusion, mastering decimal to fraction conversion is crucial for any individual working with numbers. Understanding the basics of decimal place values and following a step-by-step process can help you convert decimals into fractions accurately. This skill will not only enhance your mathematical abilities but also enable you to solve problems more efficiently and work with measurements more effectively.
## Pro Tips for Calculating and Manipulating 5.25 as a Fraction
### Understanding Decimal Fractions
You may also be interested in: Discover How to Convert 66.5 inches to Feet Easily: The Ultimate Conversion Guide
Decimals represent numbers that are not whole, but instead include fractions and decimals. When it comes to converting decimals into fractions, it’s important to have a solid understanding of basic fraction concepts. A decimal like 5.25 can be expressed as a fraction, where 25 is the numerator and 100 is the denominator. Simplifying the fraction to its lowest terms gives us 5/4.
### Converting Decimals to Fractions
To convert a decimal like 5.25 into a fraction, follow these steps:
1. Identify the decimal value (in this case, 5.25).
2. Count the number of decimal places (there are two decimal places in 5.25).
3. Write the decimal as the numerator, and place the denominator as 1 followed by the same number of zeros as there are decimal places (in this example, the denominator would be 100).
4. Simplify the fraction, if possible.
### Manipulating the Fraction
Once you have converted 5.25 into the fraction 5/4, you can perform various mathematical operations with it. For example, you can add or subtract this fraction from other fractions or whole numbers. When adding or subtracting fractions, ensure that they have a common denominator. If not, you’ll need to convert them to fractions with a common denominator before proceeding.
You may also be interested in: Converting 75°C to Fahrenheit Made Simple: Your Essential Temperature Conversion Guide
It’s also possible to multiply 5/4 by other fractions or whole numbers. Simply multiply the numerators and denominators together to get the new values. Similarly, dividing 5/4 by another fraction or whole number involves multiplying by its reciprocal.
Understanding how to calculate and manipulate a decimal like 5.25 as a fraction can be handy in various applications, such as solving mathematical problems or working with measurements. By following these pro tips, you can confidently convert and work with decimal fractions like 5.25 in a fraction form.
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# Show the identity can be derived from a sum or a difference identity and the pythagorean identity?
## $\cos \left(2 a\right) = 1 - 2 {\sin}^{2} a$
Jun 3, 2018
See explanation
#### Explanation:
Remember the angle sum identity
color(blue)(cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
Now let color(red)(x=y=a
$\cos \left(a + a\right) = \cos \left(a\right) \cos \left(a\right) - \sin \left(a\right) \sin \left(a\right)$
$\implies \cos \left(2 a\right) = {\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right)$
By the pythagorean trig identity color(blue)(cos^2(a)=1-sin^2(a)
$\cos \left(2 a\right) = \left(1 - {\sin}^{2} \left(a\right)\right) - {\sin}^{2} \left(a\right)$
$\implies \cos \left(2 a\right) = 1 - 2 {\sin}^{2} \left(a\right) \leftarrow \textcolor{red}{\text{What we wanted to show}}$
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# Search by Topic
#### Resources tagged with Addition & subtraction similar to Primary Advent Calendar 2010:
Filter by: Content type:
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##### Other tags that relate to Primary Advent Calendar 2010
Interactivities. Games. Generalising. Mental addition & subtraction. Working systematically. Addition & subtraction. Roadshow. Visualising.
### There are 254 results
Broad Topics > Calculations and Numerical Methods > Addition & subtraction
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Cubes Within Cubes
##### Stage: 2 and 3 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Pass the Peas, Please
##### Stage: 1 Challenge Level:
A game for 2 or more players. Practise your addition and subtraction with the aid of a game board and some dried peas!
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
##### Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Robot Monsters
##### Stage: 1 Challenge Level:
Use these head, body and leg pieces to make Robot Monsters which are different heights.
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Find the Difference
##### Stage: 1 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
### Making Maths: Double-sided Magic Square
##### Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Weighted Numbers
##### Stage: 1 Challenge Level:
Use the number weights to find different ways of balancing the equaliser.
### Double or Halve?
##### Stage: 1 Challenge Level:
Throw the dice and decide whether to double or halve the number. Will you be the first to reach the target?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Making Trains
##### Stage: 1 Challenge Level:
Can you make a train the same length as Laura's but using three differently coloured rods? Is there only one way of doing it?
### Carroll Diagrams
##### Stage: 1 Challenge Level:
Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers?
### Ring a Ring of Numbers
##### Stage: 1 Challenge Level:
Choose four of the numbers from 1 to 9 to put in the squares so that the differences between joined squares are odd.
### Fair Exchange
##### Stage: 1 Challenge Level:
In your bank, you have three types of coins. The number of spots shows how much they are worth. Can you choose coins to exchange with the groups given to make the same total?
### Domino Join Up
##### Stage: 1 Challenge Level:
Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Reach 100
##### Stage: 2 and 3 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### 2,4,6,8
##### Stage: 1 Challenge Level:
Using the cards 2, 4, 6, 8, +, - and =, what number statements can you make?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Mrs Beeswax
##### Stage: 1 Challenge Level:
In how many ways could Mrs Beeswax put ten coins into her three puddings so that each pudding ended up with at least two coins?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Number Balance
##### Stage: 1 Challenge Level:
Can you hang weights in the right place to make the equaliser balance?
### Cycling Squares
##### Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### The Brown Family
##### Stage: 1 Challenge Level:
Use the information about Sally and her brother to find out how many children there are in the Brown family.
### Here to There 1 2 3
##### Stage: 1 Challenge Level:
Move from the START to the FINISH by moving across or down to the next square. Can you find a route to make these totals?
### Totality
##### Stage: 1 and 2 Challenge Level:
This is an adding game for two players.
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Wild Jack
##### Stage: 2 Challenge Level:
A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.
### Criss Cross Quiz
##### Stage: 2 Challenge Level:
A game for 2 players. Practises subtraction or other maths operations knowledge.
### Two Dice
##### Stage: 1 Challenge Level:
Find all the numbers that can be made by adding the dots on two dice.
|
# Factorising
[mathjax]
Expanding Brackets
Removing the brackets is known as distributive law. To remove the brackets, multiply each term outside the bracket.
Example:
$$3(x + 2)$$ Multiply each term inside the bracket by 3.
$$3 (x) + 3 (2) = 3x + 6$$
Example 2: $$2x(x + 2)$$
$$2x (x) + 2x (2) = 2x^2 + 4x$$
Pairs of Bracket
Removing pairs of brackets using FOIL method.
FOIL Method
First – Multiply the first term in the first bracket with the first term in the second bracket.
Outside – Multiply the first term in the first bracket with the second term in the second bracket.
Inside – Multiply the second term in the first bracket with first term in the second bracket.
Last – Multiply the second term in the first bracket with the second term in the second bracket.
Example: (x + 2) (x + 3)
FOIL method will be:
$$x(x) + x(3) +2(x) + 2 (3)$$
$$x^2 + 5x + 6$$
Example 2: $$(x + 5) (x – 2)$$
$$x (x)+ x (-2) + 5 (x) – 2 (5)$$
$$x^2 + 3x – 10$$
Factorising
Factorising is simplifying a quadratic expression. It is the reverse of expanding the bracket. To factorise, look for two numbers that have the sum of the second term and product of the third term.
Example: $$x^2 + 9x + 12$$
Look for two numbers whose sum is 9 and if multiplied the answer is 12.
The factor pair of 12 are the following:
a. 12 and 1
b. 4 and 3
c. 6 and 2
In the pair of a factor, the product of 12 and 1, and 4 and 3 is 12 but the sum is not 9. The pair of the factor that suits the condition is 6 and 2. Factor of $$x^2 + 9x + 12$$ is $$(x + 6) and (x + 2)$$ .
$$x^2 + 9x + 12 = (x + 6) (x + 2)$$
Example 2: $$x^2 + x – 30$$
To factor the expression, think of two numbers that the product is -30 and added up is 1. Since the product is a negative, the two numbers must have a positive and a negative number.
These are all the factor of -30:
a. 15 and -5
b. -15 and 5
c. 10 and -3
d. -10 and 3
e. 6 and -5
f. -6 and 5
Only 6 and -5 have the sum of 1.
$$x^2 + x – 30 = (x + 6) (x – 5)$$
Factorising the Difference of Two Squares
Example: $$x^2 – 36$$
Notice that the only term is $$x^2$$ and a number. Some quadratics do not have the second term. Therefore we will look for a factor whose sum is 0.
The pair of factor of 36.
a. 36 and – 1
b. -36 and 1
c. 18 and -2
d. -18 and 2
e. 12 and -3
f. -12 and 3
g. 9 and -4
h. -9 and 4
i. 6 and -6
$$x^2 – 36$$ can be factor by 6 and -6. These examples of factoring are called the sum and difference of square or difference of two squares.
However, not all without a second term can be factorised.
Example 2: $$x^2 – 30$$
Here are all the factor of -30.
a. 15 and -5
b. -15 and 5
c. 10 and -3
d. -10 and 3
e. 6 and -5
f. -6 and 5
From the pair of factor, it does not have the sum equal to zero. So, $$x^2 – 30$$ does not have a pair of factor. Therefore, it cannot be factorized.
|
Paul's Online Notes
Home / Algebra / Graphing and Functions / Inverse Functions
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### Section 3-7 : Inverse Functions
5. Given $$\displaystyle f\left( x \right) = \frac{{4x}}{{5 - x}}$$ find $${f^{ - 1}}\left( x \right)$$ .
Show All Steps Hide All Steps
Hint : Just follow the process outlines in the notes and you’ll be set to do this problem!
Start Solution
For the first step we simply replace the function with a $$y$$.
$y = \frac{{4x}}{{5 - x}}$ Show Step 2
Next, replace all the $$x$$’s with $$y$$’s and all the original $$y$$’s with $$x$$’s.
$x = \frac{{4y}}{{5 - y}}$ Show Step 3
Solve the equation from Step 2 for $$y$$.
\begin{align*}x & = \frac{{4y}}{{5 - y}}\\ x\left( {5 - y} \right) & = 4y\\ 5x - xy & = 4y\\ 5x & = 4y + xy\\ 5x & = \left( {4 + x} \right)y\\ \frac{{5x}}{{4 + x}} & = y\end{align*} Show Step 4
Replace $$y$$ with $${f^{ - 1}}\left( x \right)$$.
${f^{ - 1}}\left( x \right) = \frac{{5x}}{{4 + x}}$ Show Step 5
Finally, do a quick check by computing one or both of $$\left( {f \circ {f^{ - 1}}} \right)\left( x \right)$$ and $$\left( {{f^{ - 1}} \circ f} \right)\left( x \right)$$ and verify that each is $$x$$. In general, we usually just check one of these and well do that here.
\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f\left[ {{f^{ - 1}}\left( x \right)} \right] & = f\left[ {\frac{{5x}}{{4 + x}}} \right]\\ & = \frac{{4\left( {\frac{{5x}}{{4 + x}}} \right)}}{{5 - \frac{{5x}}{{4 + x}}}}\frac{{4 + x}}{{4 + x}} = \frac{{20x}}{{5\left( {4 + x} \right) - 5x}} = \frac{{20x}}{{20}} = x\end{align*}
The check works out so we know we did the work correctly and have inverse.
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A B C
Solutions for Session 10, Part B
See solutions for Problems: B1 | B2 | B3 | B4 | B5
Problem B1 The largest possible perimeter is 22 units; the smallest is 14 units. This is because all but one pentomino have a perimeter of 12 units. One pentomino has a perimeter of 10 units. A method for determining perimeter when combining two pentominoes is to add the perimeters of the two pentominoes and then subtract the number of sides touching when the pentominoes are combined. To get the largest possible perimeter, combine two 12-unit pentominoes so that the least number of sides touch, which is two. So 12 + 12 - 2 = 22. To get the smallest perimeter, combine one 12-unit pentomino with the 10-unit pentomino so that as many sides as possible are touching. The maximum number of sides touching is eight, making the smallest possible combined perimeter 12 + 10 - 8, or 14 units.
Problem B2 One advantage of recording on grid paper is that it allows students to trace the perimeters of the shapes they've formed. Recording on grid paper also enables students to keep track of the different shapes. Recording all of their findings also helps them notice patterns. When tracing their combined shapes, however, if students do not also include the detail of all the squares in the pentomino, they will not be able to see the number of sides that are touching to make their new combined shape. This will lessen the power of recording their findings. Some of the problem-solving strategies students used included the guess-and-check strategy, as well as analyzing shapes and drawings for patterns (e.g., all perimeters must be even, and the combined perimeter will be the sum of the perimeters of both pentominoes less the total number of sides touching).
Problem B3 To determine the perimeter without counting, the students first added the perimeters of the two pentominoes being combined for the new shape and then subtracted the number of sides touching between the two pentominoes. To help students see this analytic method, Mr. Belber asked them questions that focused on the number of sides that touched in the new shape and how that number decreased the sum of the two pentomino perimeters. Additionally, you could ask students the following questions to confirm that they understand the concept: What patterns do you notice in the perimeters of the combined figure when the two pentominoes have two sides touching? Four sides touching? etc. Does your answer depend on which two pentominoes you choose? Can you explain why this pattern makes sense, or why increasing the number of sides touching makes the perimeter smaller?
Problem B4 This lesson has a clear mathematical purpose, which is the understanding that if area remains constant, the perimeter of shapes constructed having that area can vary. Using manipulatives helps students see that area stays constant. The tiles also allow students to see and feel the number of sides touching, and how that number affects the perimeter.
Problem B5 Mr. Belber extended learning by asking students to explain why they knew they had found the smallest or largest perimeter, how they knew they had found all possible perimeters, and why they got only even perimeters. Some generalizations he might expect students to make include the fact that only even perimeters can be made and that a rule for determining perimeter is that the perimeters of two pentominoes, minus the number of sides touching in the combined new shape, equals the perimeter of the new shape. Mr. Belber's homework assignment of repeating the task with three pentominoes provides additional generalizing opportunities and could ultimately lead to a rule for determining the maximum and minimum perimeters for any number of pentominoes.
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# 13.7 Quadratic Equations and Problem Solving
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1 13.7 Quadratic Equations and Problem Solving Learning Objectives: A. Solve problems that can be modeled by quadratic equations. Key Vocabulary: Pythagorean Theorem, right triangle, hypotenuse, leg, sum, difference, product, times, twice, more than, less than, square of, consecutive odd, consecutive even Consecutive Numbers 1. Consecutive Number. Consecutive Odd Number 3. Consecutive Even Number 1 st number 1 st number 1 st number nd number + 1 nd number + nd number + 3 rd number + 3 rd number rd number th number th number th number + 6 Pythagorean Theorem In any right triangle Leg a Hypotenuse c a + b = c Leg b Class notes: Eample 1. Find two consecutive odd integers whose product is 3 more than their sum.
2 Eample. The length of a rectangular garden is 5 feet more than its width. The area of the garden is 176 square feet. Find the length and the width. Eample 3. The hypotenuse of a right triangle is 6 inches more than the shorter leg. The longer leg is 3 inches more than the shorter leg. Find the lengths of all three sides.
3 13.7 Eercises Solve. 1. A rectangle has an area of 4 square inches. The width is represented by 3 and the length is +. Find the dimensions.. The length of a rectangle is 3 cm more than the width. The area is 70 cm. Find the dimensions of the rectangle. 3. The length of a proposed rectangular flower garden is 6 m more that its width. The area of the garden is 7 m. Find the dimensions of the proposed flower garden. 4. A square field had 5 m added to its length and m added to its width. The field then had an area of 130 m. Find the length of a side of the original field. 5. A rock is dropped from a 784 foot cliff. The height h of the rock after t seconds is given by the equation h = 16t How long will it take the ball to hit the ground? 6. One leg of a right triangle measures 6 m while the length of the other leg measures meters. The hypotenuse measures ( 6) m. Find the length of all three sides. 7. The longer leg of a right triangle measures two feet more than twice the length of the shorter leg. The hypotenuse measures 3 feet more than twice the shorter leg. Find the length of all three sides. 8. Find the length of a ladder leaning against a building if the top of the ladder touches the building at a height of 1 feet. Also, the length of the ladder is 4 feet more than its distance from the base of the building. 9. One leg of a right triangle is 14 inches longer than the other leg. The hypotenuse is 6 inches long. Find the length of each leg. 10. Eight more than the square of a number is the same as si times the number. Find the number. 11. Fifteen less than the square of a number is the same as twice the number. Find the numbers. 1. Seven less than 4 times the square of a number is 18. Find the number. 13. Find two consecutive positive odd integers whose product is The sum of the squares of two consecutive integers is 41. Find the integers. 15. Find two consecutive odd integers such that the square of the first added to 3 times the second is 4.
4 16. The square of a number minus twice the number is 63. Find the number. 17. The length of a rectangular garden is 5 feet more than its width. The area of the garden is 176 square feet. Find the length and the width. 18 A student dropped a ball from the top of a 64 foot building. The height of the ball after t seconds is given by the quadratic equation h = 16t How long will it take the ball to hit the ground? 19. The length of one leg of a right triangle is 7 meters less than the length of the other leg. The length of the hypotenuse is 13 m. Find the lengths of the legs. 0. The numerical difference between the area and the circumference of a circle is 8π. Find the radius of the circle. (Hint: first factor out π in your equation.)
5 Some challenging problems: Factor: y 7 y 17. Factor: 3. A bo is made from a rectangular piece of metal with length 0 inches and width 10 inches by cutting out square corners of length and folding up the sides. a. What are the limitations of the size of? Eplain. b. Write a polynomial that gives the volume of the bo. c. Factor out the greatest common factor for this epression you found in part (b). d. Find the volume of the bo when = 3 inches. e. Write a polynomial that gives the outside surface area of the bo. (Hint: consider the size of the metal sheet and how much was cut out.) f. Factor out the greatest common factor for the epression you found in part (e). g. If the outside surface area of the bo is 184 square inches, find.
6 4. Cut the squares apart. Match equivalent epressions. You should get a new 4 4 square.
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# Working With Money Yorubah Banks. Content Area: Mathematics Grade Level: Grade 2 Summary: The purpose of this power point is to give the students.
## Presentation on theme: "Working With Money Yorubah Banks. Content Area: Mathematics Grade Level: Grade 2 Summary: The purpose of this power point is to give the students."— Presentation transcript:
Working With Money Yorubah Banks
Content Area: Mathematics Grade Level: Grade 2 Summary: The purpose of this power point is to give the students the ability to learn about using money interacticely. Objective: Given a visual aid the student will be able to solve word problems using money, such as dimes, quarters, dollar bills, with 100% accuracy. Standard: 8. Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have?
Let’s Learn! How much is two dollar bills, three dimes, four nickels, and two pennies? This equals up to two dollars. One dime equals ten cent so you have to add ten three times to get a total of 30¢. One nickel equals five sent. In order to figure how much four nickels equal, you have to add five four times to get a total of 20¢. One penny equal one cent. Add the number one two time for a total of 2¢. Now that you know the amount of money that each description of bills and coin equals add up the amount…the amount you should have come up with equals \$2.52
How much money does one dollar, two dimes, three pennies equal? \$1.22 \$1.22 \$2.65 \$2.65 \$1.00 \$1.00
Correct
How much does five dimes equal? 60 ¢ 60 ¢ 35 ¢ 35 ¢ 50 ¢ 50 ¢
Correct
Six dollars, three nickels, two, pennies equal? \$6.17 \$6.17 \$4.50 \$4.50 \$3.46 \$3.46
Correct
Summary You now know how to solve simple word problems using some coins and dollar bills. Next week we will move into more difficult word problems adding more bills and coins.
Incorrect
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# Lesson 7
Graphs of Proportional Relationships
## 7.1: Notice These Points (5 minutes)
### Warm-up
This warm-up prepares students for graphing proportional relationships in the coordinate plane. They practice graphing coordinate points and notice that all points lie on a straight line.
### Launch
Give students 3 minutes quiet work time followed by a whole-class discussion.
### Student Facing
1. Plot the points $$(0,10), (1,8), (2,6), (3,4), (4,2)$$.
2. What do you notice about the graph?
### Launch
Give students 3 minutes quiet work time followed by a whole-class discussion.
### Student Facing
1. Plot the points $$(0,10), (1,8), (2,6), (3,4), (4,2)$$.
2. What do you notice about the graph?
### Activity Synthesis
Invite students to share their observations. Ask if other students agree. If some students do not agree that the points lie on a straight line, ask which points break the pattern and give students a chance to self-correct their work.
## 7.2: T-shirts for Sale (10 minutes)
### Activity
This introductory activity asks students to plot points using tables of values that represent scenarios familiar from previous lessons. This activity is intended as a review of the coordinate plane, its axes, and plotting ordered pairs.
Teacher Notes for IM 6–8 Accelerated
Adjust this activity to 15 minutes.
During the synthesis of this activity, display these graphs of proportional and nonproportional relationships for all to see.
Graphs of Proportional Relationships
Graphs of Nonproportional Relationships
Ask students, "What properties do the graphs representing proportional relationships have?" (The points are all in a line. The line goes through the point $$(0,0)$$.)
If necessary, use a straightedge to show that the points are all in a line.
Introduce the word origin to refer to the point $$(0,0)$$ if students are unfamiliar with the term.
### Launch
Arrange students in groups of 2. Provide access to rulers. Give students 5 minutes of quiet work time followed by students discussing responses with a partner, followed by whole-class discussion.
The second question reviews work from grades 5 and 6, but it is important that students understand how to plot points in the coordinate plane. Display the graph for all to see. Show students how to plot the pair from the first row in the table in the coordinate plane. Ask students to plot the remaining pairs and check with nearby students as they work. Be on the lookout for students plotting coordinates in the wrong order.
Some T-shirts cost $8 each. $$x$$ $$y$$ 1 8 2 16 3 24 4 32 5 40 6 48 1. Use the table to answer these questions. 1. What does $$x$$ represent? 2. What does $$y$$ represent? 3. Is there a proportional relationship between $$x$$ and $$y$$? 2. Plot the pairs in the table on the coordinate plane. 3. What do you notice about the graph? ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Launch Arrange students in groups of 2. Provide access to rulers. Give students 5 minutes of quiet work time followed by students discussing responses with a partner, followed by whole-class discussion. The second question reviews work from grades 5 and 6, but it is important that students understand how to plot points in the coordinate plane. Display the graph for all to see. Show students how to plot the pair from the first row in the table in the coordinate plane. Ask students to plot the remaining pairs and check with nearby students as they work. Be on the lookout for students plotting coordinates in the wrong order. ### Student Facing Some T-shirts cost$8 each.
$$x$$ $$y$$
1 8
2 16
3 24
4 32
5 40
6 48
1. Use the table to answer these questions.
1. What does $$x$$ represent?
2. What does $$y$$ represent?
3. Is there a proportional relationship between $$x$$ and $$y$$?
2. Plot the pairs in the table on the coordinate plane.
3. What do you notice about the graph?
### Activity Synthesis
Discuss the first question if students had trouble with it.
Ask students to share their observations about the plotted points. Ask, “Could we buy 0 shirts? 7 T-shirts? 10 T-shirts? Can we buy half of a T-shirt?” Note that the graph consists of discrete points because only whole numbers of T-shirts make sense in this context; however, people often connect discrete points with a line to make the relationship more clear, even when the in-between values don’t make sense.
Ask the students, “Suppose instead of price per shirt, this graph displayed the cost of cherries that are \$8 per pound. Given that context, how should we change the graph?” Weights need not have integer values, so the graph is not restricted to discrete points. If you haven’t done so already, draw the ray starting at $$(0,0)$$ that passes through the points. Speaking: MLR8 Discussion Supports. As students describe their observations about the proportional relationship represented in the graph, revoice student ideas to demonstrate mathematical language use. Press for details in students’ explanations by requesting that students challenge an idea, elaborate on an idea, or give an example. Show central concepts multi-modally by using different types of sensory inputs: acting out scenarios or inviting students to do so, using gestures, and talking about the context of selling T-shirts or cherries. This will help students to produce and make sense of the language needed to communicate their own ideas. Design Principle(s): Support sense-making, Optimize output (for explanation) ## 7.3: Tyler's Walk (15 minutes) ### Activity This activity is intended to further students’ understanding of the graphs of proportional relationships in the following respects: • points on the graph of a proportional relationship can be interpreted in the context represented (MP2) • for these points, the quotient of the coordinates is—excepting $$(0,0)$$—the constant of proportionality • if the first coordinate is 1, then the corresponding coordinate is $$k$$, the constant of proportionality. Students explain correspondences between parts of the table and parts of the graph. The graph is simple so that students can focus on what a point means in the situation represented. Students need to realize, however, that the axes are marked in 10-unit intervals. The discussion questions are opportunities for students to construct viable arguments and critique the reasoning of others (MP3). ### Launch Arrange students in groups of 2. Give students 5 minutes of quiet work time followed by students discussing responses with a partner, followed by whole-class discussion. Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organizational skills in problem solving. For example, present one question at a time and monitor students to ensure they are making progress throughout the activity. Supports accessibility for: Organization; Attention Speaking, Reading: MLR5 Co-Craft Questions. To help students make sense of graphs of proportional relationships, start by displaying only the first line of this task (“Tyler was at the amusement park. He walked at a steady pace from the ticket booth to the bumper cars.”) and the graph. Ask students to write down possible mathematical questions that could be asked about the situation. Invite students to compare their questions before revealing the remainder of the questions. Listen for and amplify any questions involving correspondences between parts of the table and parts of the graph. This helps students produce the language of mathematical questions and talk about the relationship between distance and time. Design Principle(s): Maximize meta-awareness; Support sense-making ### Student Facing Tyler was at the amusement park. He walked at a steady pace from the ticket booth to the bumper cars. 1. The point on the graph shows his arrival at the bumper cars. What do the coordinates of the point tell us about the situation? 2. The table representing Tyler's walk shows other values of time and distance. Complete the table. Next, plot the pairs of values on the grid. 3. What does the point $$(0, 0)$$ mean in this situation? 4. How far away from the ticket booth was Tyler after 1 second? Label the point on the graph that shows this information with its coordinates. 5. What is the constant of proportionality for the relationship between time and distance? What does it tell you about Tyler's walk? Where do you see it in the graph? time (seconds) distance (meters) 0 0 20 25 30 37.5 40 50 1 ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Student Facing #### Are you ready for more? If Tyler wanted to get to the bumper cars in half the time, how would the graph representing his walk change? How would the table change? What about the constant of proportionality? ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Extension Student Response. ### Anticipated Misconceptions These questions can be used for discussion or for students who need scaffolding. • "What quantities are shown in the graph?" (Distance in meters that Tyler is from the ticket booth and time elapsed in seconds since he started walking.) • "How far is the ticket booth from the bumper cars?" (50 meters, assuming that Tyler walked in a straight line.) This is an opportunity for attention to precision (MP6) and making explicit assumptions about a situation (MP4). • "Do the values in your table show a proportional relationship? How do you know?" (Based on prior lessons in this unit, students should identify the relationship as proportional because for every point the unit rate is the same.) • "What do the coordinates of the points on the graph show?" (The first coordinate gives amount of time in seconds that elapsed since Tyler started walking. Its corresponding second coordinate shows how many meters away from the ticket booth Tyler was at the corresponding time, assuming that Tyler walked in a straight line.) ### Activity Synthesis After students work on the task, it is important to discuss how the axis labels and the description in the task statement help us interpret points on the graph. Consider asking these questions: • "What quantities are shown in the graph?" (Distance in meters that Tyler is from the ticket booth and time elapsed in seconds since he started walking.) • "How far is the ticket booth from the bumper cars?" (50 meters, assuming that Tyler walked in a straight line.) This is an opportunity for attention to precision (MP6) and making explicit assumptions about a situation (MP4). • "Do the values in your table show a proportional relationship? How do you know?" (Based on prior lessons in this unit, students should identify the relationship as proportional because for every point the unit rate is the same.) • "What do the coordinates of the points on the graph show?" (The first coordinate gives amount of time in seconds that elapsed since Tyler started walking. Its corresponding second coordinate shows how many meters away from the ticket booth Tyler was at the corresponding time, assuming that Tyler walked in a straight line.) Ask students for the equation of this proportional relationship. Finally ask where students see $$k$$, the constant of proportionality, in each representation, the equation, the graph, the table, and the verbal description of the situation. ## Lesson Synthesis ### Lesson Synthesis At the end of the lesson, make sure that students know that the graph of a proportional relationship lies on a line through the origin. (They will be able to explain why this is true in grade 8.) Display one of the graphs of a proportional relationship from the activities. Choose a point on the graph and ask students to interpret its coordinates in the situation. Then, choose the point with $$x$$-coordinate 1 and ask about the significance of its $$y$$-coordinate. ## 7.4: Cool-down - Filling a Bucket (5 minutes) ### Cool-Down Teachers with a valid work email address can click here to register or sign in for free access to Cool-Downs. ## Student Lesson Summary ### Student Facing One way to represent a proportional relationship is with a graph. Here is a graph that represents different amounts that fit the situation, “Blueberries cost \$6 per pound.”
Different points on the graph tell us, for example, that 2 pounds of blueberries cost \$12, and 4.5 pounds of blueberries cost \$27.
Sometimes it makes sense to connect the points with a line, and sometimes it doesn’t. We could buy, for example, 4.5 pounds of blueberries or 1.875 pounds of blueberries, so all the points in between the whole numbers make sense in the situation, so any point on the line is meaningful.
If the graph represented the cost for different numbers of sandwiches (instead of pounds of blueberries), it might not make sense to connect the points with a line, because it is often not possible to buy 4.5 sandwiches or 1.875 sandwiches. Even if only points make sense in the situation, though, sometimes we connect them with a line anyway to make the relationship easier to see.
Graphs that represent proportional relationships all have a few things in common:
• There are points that satisfy the relationship lie on a straight line.
• The line that they lie on passes through the origin, $$(0,0)$$.
Here are some graphs that do not represent proportional relationships:
These points do not lie on a line.
This is a line, but it doesn’t go through the origin.
Here is a different example of a relationship represented by this table where $$y$$ is proportional to $$x$$. We can see in the table that $$\frac54$$ is the constant of proportionality because it’s the $$y$$ value when $$x$$ is 1.
The equation $$y = \frac54 x$$ also represents this relationship.
$$x$$ $$y$$
4 5
5 $$\frac{25}{4}$$
8 10
1 $$\frac{5}{4}$$
Here is the graph of this relationship.
If $$y$$ represents the distance in feet that a snail crawls in $$x$$ minutes, then the point $$(4, 5)$$ tells us that the snail can crawl 5 feet in 4 minutes.
If $$y$$ represents the cups of yogurt and $$x$$ represents the teaspoons of cinnamon in a recipe for fruit dip, then the point $$(4, 5)$$ tells us that you can mix 4 teaspoons of cinnamon with 5 cups of yogurt to make this fruit dip.
We can find the constant of proportionality by looking at the graph, because $$\frac54$$ is the $$y$$-coordinate of the point on the graph where the $$x$$-coordinate is 1. This could mean the snail is traveling $$\frac54$$ feet per minute or that the recipe calls for $$1\frac14$$ cups of yogurt for every teaspoon of cinnamon.
In general, when $$y$$ is proportional to $$x$$, the corresponding constant of proportionality is the $$y$$-value when $$x=1$$
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# In a college, 70 % students pass in physics, 75 % pass in mathematics and 10 % students fail in both. One student is chosen at random. What is the probability that the student passes in mathematics given that he passes in physics?
Last updated date: 12th Jul 2024
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Hint: Let us denote P be the event that the student passes in physics and let us denote the M be the event that the student passes in mathematics. The probability that the student passes in mathematics given that he passes in physics is given by the formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)}$. Using this formula, we can solve this question.
Complete step by step solution:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
If we are given two events A and B, then the probability of event A given that event B will also occur is given by the formula,
$P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ . . . . . . . . . . . . . (1)
Also, in probability, we have a formula $P\left( A\cap B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cup B \right)$ . . . . . . . . . (2)
For this question, let us denote P be the event that the student passes in physics and let us denote the M be the event that the student passes in mathematics.
It is given that 70 % students pass in physics, so, $P\left( P \right)=\dfrac{70}{100}$.
$\Rightarrow P\left( M \right)=\dfrac{7}{10}$
It is given that 75 % students pass in mathematics, so, $P\left( M \right)=\dfrac{75}{100}$.
$\Rightarrow P\left( P \right)=\dfrac{3}{4}$
Also, it is given that 10 % of the students fail in both subjects. So, we can say 90% of the students passed either in physics, or in mathematics, or in both.
\begin{align} & \Rightarrow P\left( M\cup P \right)=\dfrac{90}{100} \\ & \Rightarrow P\left( M\cup P \right)=\dfrac{9}{10} \\ \end{align}
Using formula (2), we get,
\begin{align} & P\left( M\cap P \right)=\dfrac{7}{10}+\dfrac{3}{4}-\dfrac{9}{10} \\ & \Rightarrow P\left( M\cap P \right)=\dfrac{28+30-36}{40} \\ & \Rightarrow P\left( M\cap P \right)=\dfrac{22}{40} \\ & \Rightarrow P\left( M\cap P \right)=\dfrac{11}{20} \\ \end{align}
Using formula (1), the probability that the student passes in mathematics given that he passes in physics is equal to,
\begin{align} & P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)} \\ & \Rightarrow P\left( M|P \right)=\dfrac{\dfrac{11}{20}}{\dfrac{7}{10}} \\ & \Rightarrow P\left( M|P \right)=\dfrac{11}{14} \\ \end{align}
Hence, the answer is $\dfrac{11}{14}$.
Note: There is a possibility that one may commit a mistake while using the formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)}$. Instead of the correct formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( P \right)}$, it is possible that one may use the formula $P\left( M|P \right)=\dfrac{P\left( M\cap P \right)}{P\left( M \right)}$ which will lead us to an incorrect answer.
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# 6.1 Point Estimation and Sampling Distributions
Learning Objectives
By the end of this chapter, the student should be able to:
• Understand point estimation
• Apply and interpret the Central Limit Theorem
• Construct and interpret confidence intervals for means when the population standard deviation is known
• Understand the behavior of confidence intervals
• Carry out hypothesis tests for means when the population standard deviation is known
• Understand the probabilities of error in hypothesis tests
# Statistical Inference
It is often necessary to “guess”, infer, or generalize about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs.
Statistical inference uses what we know about probability to make our best “guesses” or estimates from samples about the population they came from. The main forms of Inference are:
# Point Estimation
Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion.
The most natural way to estimate features of the population (parameters) is to use the corresponding summary statistic calculated from the sample. Some common point estimates and their corresponding parameters are found i n the following table:
Parameter Measure Statistic μ Mean of a single population x̅ p Proportion of a single population ˆp μD Mean difference of two dependent populations (MP) x̅ D μ1-μ2 Difference in means of two independent populations x̅ 1-x̅ 2 p1–p2 Difference in proportions of two populations ˆp1-ˆp2 σ2 Variance of a single population S2 σ Standard deviation of a single population S
Suppose the mean weight of a sample of 60 adults is 173.3 lbs; this sample mean is a point estimate of the population mean weight, µ. Remember this is one of many samples that we could have taken from the population. If a different random sample of 60 individuals were taken from the same population, the new sample mean would likely be different as a result of sampling variability. While estimates generally vary from one sample to another, the population mean is a fixed value.
Suppose a poll suggested the US President’s approval rating is 45%. We would consider 45% to be a point estimate of the approval rating we might see if we collected responses from the entire population. This entire-population response proportion is generally referred to as the parameter of interest. When the parameter is a proportion, it is often denoted by p, and we often refer to the sample proportion as ˆp (pronounced “p-hat”). Unless we collect responses from every individual in the population, p remains unknown, and we use ˆp as our estimate of p.
How would one estimate the difference in average weight between men and women? Suppose a sample of men yields a mean of 185.1 lbs and a sample of women men yields a mean of 162.3 lbs. What is a good point estimate for the difference in these two population means? We will expand on this in following chapters.
# Sampling Distributions
We have established that different samples yield different statistics due to sampling variability . These statistics have their own distributions, called sampling distributions, that reflect this as a random variable. The sampling distribution of a sample statistic is the distribution of the point estimates based on samples of a fixed size, n, from a certain population. It is useful to think of a particular point estimate as being drawn from a sampling distribution.
Recall the sample mean weight calculated from a previous sample of 173.3 lbs. Suppose another random sample of 60 participants might produce a different value of x, such as 169.5 lbs. Repeated random sampling could result in additional different values, perhaps 172.1 lbs, 168.5 lbs, and so on. Each sample mean can be thought of as a single observation from a random variable X. The distribution of X is called the sampling distribution of the sample mean, and has its own mean and standard deviation like the random variables discussed previously. We will simulate the concept of a sampling distribution using technology to repeatedly sample, calculate statistics, and graph them. However, the actual sampling distribution would only be attainable if we could theoretically take an infinite amount of samples.
Each of the point estimates in the table above have their own unique sampling distributions which we will look at in the future
# Unbiased Estimation
Although variability in samples is present, there remains a fixed value for any population parameter. What makes a statistical estimate of this parameter of interest a “Good” one? It must be both accurate and precise.
The accuracy of an estimate refers to how well it estimates the actual value of that parameter. Mathematically, this is true when that the expected value your statistic is equal to the value of that parameter. This can be visualized as the center of the sampling distribution appearing to be situated at the value of that parameter.
According to the law of large numbers, probabilities converge to what we expect over time. Point estimates follow this rule, becoming more accurate with increasing sample size. The figure below shows the sample mean weight calculated for random samples drawn, where sample size increases by 1 for each draw until sample size equals 500. The maroon dashed horizontal line is drawn at the average weight of all adults 169.7 lbs, which represents the population mean weight according to the CDC.
Note how a sample size around 50 may produce a sample mean that is as much as 10 lbs higher or lower than the population mean. As sample size increases, the fluctuations around the population mean decrease; in other words, as sample size increases, the sample mean becomes less variable and provides a more reliable estimate of the population mean.
In addition to accuracy, a precise estimate is also more useful. This means when repeatedly sampling, the values of the statistics seem pretty close together. The precision of an estimate can be visualized as the spread of the sampling distribution, usually quantified by the standard deviation. The phrase “the standard deviation of a sampling distribution” is often shortened to the standard errorA smaller standard error means a more precise estimate and is also effected by sample size.
## Image Credits
Figure 6.1: Michael Longmire (2019). “Coins spilling out of a jar.” Public domain. Retrieved from https://unsplash.com/photos/lhltMGdohc8
Figure 6.3: Kindred Grey (2020). “Figure 6.3.” CC BY-SA 4.0. Retrieved from https://commons.wikimedia.org/wiki/File:Figure_6.3.png
definition
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# Understanding the Ohm’s Law Wheel
If you’re looking for a comprehensive guide to the Ohm’s Law Wheel, look no further! In this blog post, we will discuss everything you need to know about electricity. We’ll start by talking about the different parts of Ohm’s Law Wheel and what they mean. Then, we’ll go over some common electrical terms and concepts. Finally, we’ll talk about how to use Ohm’s Law Wheel to solve problems. Are you ready to learn about electricity? Let’s get started!
## What is Ohm’s Law?
Ohm’s Law is the most important equation in all of the electricity, and it’s vital to understanding how electrical circuits work [1].
If you double the voltage, you will also double the current.
It is also worth noting that Ohm’s Law equation only applies to DC circuits. AC circuits are a bit more complicated, but they can be simplified using the power factor.
## Who Developed Ohm’s Law?
Physicist Georg Simon Ohm was a German who discovered the relationship between resistance, current, and voltage. This discovery became known as Ohm’s Law. In 1827, he published his findings.
Ohm’s law is represented by the equation: V = IR
This equation is known as Ohm’s law formula.
## How to Use Ohm’s Law
Now that we’ve gone over what each symbol in the Ohm’s Law equation represents, it’s time to put this knowledge to use and learn how to calculate electrical values [2].
The first thing you need to do is identify which two of the three values you know. Once you have done that, you can solve for the unknown value using the following equation:
For example, you want to calculate V across a resistor with a resistance of 12 Ω. In this case, you would use the equation V = IR. Plugging in the values, you get:
V = 12Ω * I
Now, all you have to do is solve for I to get the value of current. In this case, I = V/R, so:
I = 12/12
From this, we can see that the voltage across the resistor is equal to the amount of current flowing through it.
Now you can plug in different values to solve for the unknown quantity.
Just remember, this equation is only valid when resistance is constant. If the resistance changes, so will the current and voltage.
## What is Ohm’s Law Wheel?
The Ohm’s Law Wheel is a very useful tool for understanding electricity and how it works. It is a visual guide that illustrates the relationship between resistance, current, and voltage.
The wheel is divided into four quadrants. Each of them represents a different value of either V, I, or R. The values of the quadrants are:
• Voltage: The potential difference between two points. Measured in volts.
• Current: The rate at which charge flows through a conductor. Measured in amperes.
• Resistance: The opposition to the flow of current. Measured in ohms.
• Power (P): The rate at which work is done. Measured in watts.
To use the Ohm’s Law Wheel, simply find the quadrant that corresponds to the value you are trying to find.
Also, the Ohm’s Law Wheel can be used to find the value of any two variables if the third variable is known. Let’s say you know the values of resistance and voltage, you can find the value of current by finding the quadrant that contains those values.
If you are ever unsure about how electricity works, be sure to consult the Ohm’s Law Wheel.
It is important to note that the Ohm’s Law Wheel is not a substitute for actual knowledge of electricity. It is merely a tool to help you understand the concepts. If you want to learn more about electricity, be sure to consult other sources as well.
## When to Use Ohm’s Law Wheel
Now that we’ve gone over the basics of Ohm’s law wheel, you might be wondering when it would come in handy. The truth is, this tool can be used in a variety of situations!
Here are just a few examples:
• You’re working with a new electrical device and want to understand its capabilities – Ohm’s law will tell you how much current the device can handle.
• You want to know how to properly wire an electrical circuit – by using the wheel, you can determine the correct gauge of wire to use.
• You’re troubleshooting an electrical issue – Ohm’s law can help you pinpoint the problem area in a circuit.
• You need to calculate the power dissipation of a resistor – the power dissipation formula is included on Ohm’s law wheel!
In general, Ohm’s law wheel is a great tool to have on hand whenever you’re working with electricity. It’s easy to use and can be a real lifesaver in a pinch.
As you can see, Ohm’s law wheel is a versatile tool that can be used in a variety of situations. There are countless other uses for Ohm’s law wheel – these are just a few examples.
So whether you’re a beginner just learning the basics of electricity or a seasoned professional, this guide will come in handy. So, the next time you’re working with electricity, be sure to pull out your trusty Ohm’s law wheel!
## Ohm’s Law Example Problems
Now that you know how to use Ohm’s Law Wheel, let’s go over some example problems. These will help you see how the wheel can be used in practice to calculate different values related to electric current, voltage, and resistance.
Example Problem #01:
You have a light bulb with a resistance of 50 ohms. You know that the voltage in your home’s electrical system is 120 volts. What is the current flowing through the light bulb?
• To solve this problem, simply find the point on the wheel where “V” and “R” intersect. So that point is at “I” (current). And you have the current flowing through the light bulb is 0.24 amps.
Example Problem #02:
You have a toaster with a resistance of 30 ohms. You know that the current flowing through the toaster is 0.50 amps. What is the voltage across the toaster?
• To find the solution, look at the point on the wheel where “I” (current) and “R” (resistance) intersect. In this way, that point is at “V” (voltage). Now you know that the voltage across the toaster is 15 volts.
Example Problem #03:
You have a car battery with a voltage of 12 volts. You know that the current flowing through the car battery is 25 amps. What is the resistance of the car battery?
• For solving the problem, take the point on the wheel where “I” and “V” intersect. In this case, that point is at resistance. So the resistance of the car battery is 0.48 ohms.
Example Problem #04:
You want to build a circuit with a voltage of 12 volts and a current of 0.50 amps. What should the resistance of the circuit be?
• To solve this problem, find the point on the wheel where current and voltage intersect. Now that point is at “R”. Here we get that the resistance of the circuit should be 24 ohms.
As you can see, using Ohm’s Law Wheel is a quick and easy way to solve problems involving electric current, voltage, and resistance. In just a few seconds, you can figure out values that would otherwise take much longer to calculate.
## Principles of Ohm’s Law
There is two primary principles of Ohm’s Law: proportional and inverse proportionality.
• Proportional: This is when the voltage and current are equal. In this instance, the resistance remains the same.
• Inverse Proportionality: This is when the voltage and current are inversely proportional to each other. In this principle, the resistance changes.
These principles are what Ohm’s Law Wheel is based on.
## Ohm’s Law Wheel: A Comparison of Key Indicators
To fully grasp its importance, it is essential to understand the various indicators involved in its use. This table provides a comparison of the key indicators related to the Ohm’s Law Wheel, including their definitions, formulas, and units of measurement. By understanding these indicators, individuals can gain a deeper understanding of how to apply the Ohm’s Law Wheel in solving complex problems in electrical circuits.
Indicator Definition Formula Unit
Voltage The electrical potential difference between two points in a circuit V = IR Volts (V)
Current The flow of electric charge through a circuit I = V/R Amps (A)
Resistance The opposition to the flow of electric charge R = V/I Ohms (Ω)
Power The rate at which energy is transferred in a circuit P = IV Watts (W)
Ohm’s Law Wheel A tool used to solve for voltage, current, resistance, and power in a circuit N/A N/A
Note that the Ohm’s Law Wheel does not have a specific formula or unit, as it is a visual aid to help solve for voltage, current, resistance, and power using the three formulas listed above.
## FAQ
### Why do we use Ohm’s law?
Ohm’s law is used to describe the relationship between V, I, and R in an electrical circuit. By using this law, we can determine the amount of current flowing through a circuit given the voltage and resistance. Additionally, we can calculate the V drop across a resistor given the current and resistance. Finally, we can determine the resistance of a material given the voltage and current.
### How is Ohm’s law used in everyday life?
Ohm’s law is used in a variety of ways in our everyday lives. For example, when we charge our phones overnight, we are using Ohm’s law to ensure that the correct amount of current flows through the charging cord so that our phones do not overcharge. Additionally, when we adjust the volume on our stereo systems, we are using Ohm’s law to calculate the resistance in the circuit so that we can control the amount of current flowing through the speakers.
### Why is I current in electronics?
The letter “I” is used to represent current in electronics because it is the symbol for intensity. In an electrical circuit, a current is the measure of the flow of charge through a material. The SI unit for measuring current is the ampere (A).
### What is the difference between AC and DC current?
AC and DC current are two types of current that flow through electrical circuits.
• AC flows in one direction, then reverses direction periodically. The frequency of this reversal is measured in hertz (Hz).
• DC flows in one direction only.
AC is the type of current that is used in our homes, while DC is the type of current that is used in batteries.
### Why is the Ohm’s law wheel used in electrical engineering?
The Ohm’s law wheel is a graphical representation of the relationships between voltage, current, and resistance. It is a useful tool for understanding how these quantities interact with each other in electrical circuits.
The Ohm’s law wheel can be used to determine the values of unknown quantities in a circuit, as well as to analyze the behavior of circuits under various conditions. It is also a valuable teaching tool, as it can help students visualize the relationships between these important concepts.
### How can the Ohm’s law wheel be used to solve problems?
The Ohm’s law wheel can be used to solve problems by understanding the relationships between voltage, current, and resistance. By using the wheel, you can determine the values of any two variables if the third is known. For example, if you know the voltage and resistance in a circuit, you can use the wheel to find the value of the current.
Another way to use the Ohm’s law wheel is to find out what happens when one of the variables in a circuit is changed. For example, if you increase the resistance in a circuit, how will that affect the current? You can use the Ohm’s law wheel to find out.
Finally, you can use the Ohm’s law wheel to troubleshoot problems in a circuit. For example, if you know the voltage and current in a circuit, but the resistance is unknown, you can use the wheel to find out what the resistance might be.
### What are some of the limitations of the Ohm’s law wheel?
The Ohm’s law wheel is a great tool to help you visualize the relationships between voltage, current, and resistance. However, there are some limitations to using this tool. One of the biggest limitations is that it only applies to linear circuits. This means that if you have a circuit with multiple resistors or other non-linear components, the Ohm’s law wheel will not be accurate.
### What are some of the benefits of using the Ohm’s law wheel?
The Ohm’s law wheel is a great tool for understanding electrical circuits. It can help you determine the resistance of a circuit, the voltage drop across a resistor, and the current through a circuit. Additionally, the Ohm’s law wheel can be used to find the power dissipated in a circuit.
Another benefit of using the Ohm’s law wheel is that it can help you solve complex problems. For example, if you know the voltage across a resistor and the current flowing through it, you can use the Ohm’s law wheel to find the resistance of the resistor.
Finally, the Ohm’s law wheel is a handy tool to have around when working with electrical circuits. It can help you troubleshoot problems and make calculations quickly and easily.
### Are there any drawbacks to using the Ohm’s law wheel?
The main drawback of using the Ohm’s law wheel is that it can be confusing to look at and understand. The wheel is divided into different sections, each with its own symbol and value. If you don’t know what you’re looking at, it can be difficult to make sense of it all.
Ohm’s law also cannot be applied to non-linear systems. This includes systems where the resistance changes with voltage or current, such as in a transistor. In these cases, other laws must be used to determine the relationship between voltage, current, and resistance.
### How do you memorize the Ohm’s law wheel?
One common mistake that students make when trying to memorize the Ohm’s law wheel is trying to memorize all of the information at once. This is often overwhelming and can lead to frustration. Instead, it is important to focus on one small section at a time. For instance, start by memorizing the formula for Ohm’s law (V = IR). Then, once you have mastered that, move on to memorizing the definitions of voltage, current, and resistance. Once you have mastered those basics, you can then move on to more complicated concepts such as parallel and series circuits. Whichever method you choose, it is important to practice regularly so that you can recall the information when you need it.
### What does P stand for in Ohm’s law?
P stands for power and is measured in watts. Power is the rate at which work is done or energy is used. In Ohm’s law, P=IV, so power is equal to current times voltage.
The standard unit of measurement for electrical power is the watt (W). One watt is equal to one joule per second (J/s). Electrical power can be calculated using either the voltage and current values, or the resistance value with either voltage or current.
### Is Ohm’s law always true?
No, Ohm’s law is not always true. It is only true when the resistance in a circuit remains constant. If the resistance changes, then Ohm’s law no longer applies.
However, there are certain conditions where Ohm’s law can be approximated to be true. For example, if the voltage and current in a circuit are small and the temperature is constant, then Ohm’s law will hold true.
In such cases, Ohm’s law is a 100% accurate model of how current flows through a resistor.
## Final Thoughts
Whether you’re a student, a professional, or just someone who’s interested in electricity, understanding the Ohm’s Law Wheel is essential. With this guide, you should now have a clear understanding of how the wheel works and what it means.
Remember, the Ohm’s Law Wheel is a powerful tool that can be used to solve a variety of problems. By understanding the relationships between V, I, and R, you can troubleshoot electrical issues, design circuits, and much more.
If you’re ever stuck on a problem or don’t know where to start, refer back to this guide. With practice, you’ll be able to use Ohm’s Law Wheel with ease. Thanks for reading!
I hope this guide was helpful in understanding Ohm’s Law Wheel. If you have any questions, feel free to leave a comment below and I’ll do my best to answer them. Thanks for reading!
References:
1. https://phyxter.ai/blog/understanding-ohms-law-wheel
2. https://electricianapprenticehq.com/basic-ohms-law-formulas/
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### One-to-One
##### 1. One-to-one and uniqueness
After the discussion about the relation between onto and existence, we turn to uniqueness from the transformation viewpoint. Again the relation can be seen through everyday life questions.
Example For the transformation
Capital City: country → city
we already know the city of Beijing is in the range. Now we ask
How many cities has Beijing as the capital city?
The answer is that Beijing is the capital city of a unique contry - China. In other words,
For b =Beijing, the solution to the equation Capital City(x) = b is unique.
In fact, we know that the uniqueness for any capital city b in the world. This means that for any city b, we have
Capital City(country1) = Capital City(country2) ⇒ country1 = country2.
In other words, countries with the same capital must be the same country.
The discussion leads to the following definition.
A transformation T: XY is one-to-one (injective) if T(x) = T(x') implies x = x'.
We note that the condition is also equivalent to
xx'T(x) ≠ T(x')
(different elements are transformed to different elements)
If a transformation is not one-to-one, then it is several-to-one: Different elements of X may have the same image.
More discussion on one-to-one transformations can be found here.
Example For the transformation
Instructor: course → professor
the following are equivalent statements
• Instructor is one-to-one
• Instructor(course1) = Instructor(course2) ⇒ course1 = course2 (courses taught by the same professor are the same course)
• course1 ≠ course2Instructor(course1) ≠ Instructor(course2) (different courses are taught by different professors)
• No professor teaches more than one courses
The answer is (almost) definitely no.
Because there are millions of people at age 20, the transformation Age: people → number is not one-to-one.
Example The transformation Square: RR is two-to-one at nonzero numbers (and is one-to-one at 0). Overall, the transformation is not one-to-one.
For Reflection in x-axis, note that two points with the same reflections must be the same point. Therefore the transformation is one-to-one.
Similarly, the Rotation transformation is also one-to-one.
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BY DR. JULIA ARNOLD The Algebra of Functions What does it mean to add two functions? If f(x) = 2x + 3 and g(x) = -4x - 2 What would (f+g)(x) be? (f+g)(x)
Presentation on theme: "BY DR. JULIA ARNOLD The Algebra of Functions What does it mean to add two functions? If f(x) = 2x + 3 and g(x) = -4x - 2 What would (f+g)(x) be? (f+g)(x)"— Presentation transcript:
BY DR. JULIA ARNOLD The Algebra of Functions
What does it mean to add two functions? If f(x) = 2x + 3 and g(x) = -4x - 2 What would (f+g)(x) be? (f+g)(x) = f(x) + g(x) It means to add the two functions (f+g)(x) = -2x + 1 Likewise (f - g)(x) = f(x) - g(x) or (f - g)(x) = 6x +5 Multiplication of two functions is expressed like this: fg(x) = f(x)g(x) In our example, fg(x) = -8x 2 -16x -6
The Algebra of Functions (f+g)(x) = f(x) + g(x) (f - g)(x) = f(x) - g(x) fg(x) = f(x)g(x) Division also follows a logical path: In our example: f(x) = 2x + 3 and g(x) = -4x - 2
Application: In business you would have fixed costs, such as rent, and variable costs from producing your commodity. We will call the cost C(x) Revenue is the money you make in your business. We will call revenue R(x). Profit is what you hope to make from your business and is denoted as P(x) = R(x) - C(x).
Suppose your company manufactures water filters and has fixed costs of \$10,000 per month. The cost of producing the water filters is represented by -.0001x 2 + 10x where How do you represent the cost function? C(x) = -.0001x 2 + 10x +10000 Suppose the total revenue you make from the sale of x water filters is given by R(x) = -.0005x 2 + 20x What would the profit function be? How much would you make if you sold 10,000 water filters?
Suppose your company manufactures water filters and has fixed costs of \$10,000 per month. The cost of producing the water filters is represented by -.0001x 2 + 10x where How do you represent the cost function? C(x) = -.0001x 2 + 10x +10000 Suppose the total revenue you make from the sale of x water filters is given by R(x) = -.0005x 2 + 20x What would the profit function be? P(x) = R(x) - C(x) = -.0005x 2 + 20x - (-.0001x 2 + 10x +10000) P(x)= (-.0005 +.0001) x 2 +20x -10x - 10000 P(x) = -.0004x 2 +10x -10000 How much would you make if you sold 10,000 water filters? P(10000)= -.0004(10000) 2 +10(10000) - 10000 = 50,000 per month
Composition of Functions (one more operation) The easiest way to describe composition is to say it is like substitution. In fact Read f of g of x which means substitute g(x) for x in the f(x) expression. Suppose f(x)= 2x + 3 and g(x) = 8 - x f(g(x) )= 2 g(x) + 3 f(8 - x)= 2 (8 - x) + 3 f(g(x)) = 16 -2x + 3 or 19 - 2x
An interesting fact is that most of the time. Let’s see if this is the case for the previous example.
f(x) = 2x + 3, andg(x) = 8 - x Thus we will substitute f into g. g(x) = 8 - x g(f(x) ) = 8 - f(x) Now substitute what f(x) is: g(2x + 3) = 8 - (2x + 3) = 8 - 2x - 3 = 5 - 2x f(g(x)) = 19 - 2x while g(f(x)= 5 - 2x
Write the f function Substitute g(x) for x Replace g(x) with Simplify Step 1 Step 2 Step 3 Step 4
We can also evaluate the composition of functions at a number. Let: and Find
= f(g(3)) This says to insert the value for g(3) into f, so… Step 1 is to find g(3)
Find = f(g(3)) Now substitute the answer into f(x) for x. 12 34 5 1: Take the square root of top and bottom. 2: Find a number that rationalizes the denominator 3: multiply top and bottom4: Take the square root of 36 5: add the 1 as 6/6
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# Perimeter of Rectangle
A rectangle can be explained as a 4-sided quadrilateral which contains equal opposite sides. In a rectangle, the opposite sides are always parallel to each other and commonly classified as equiangular quadrilateral. As the rectangle has no equal adjacent sides, we always get 2 different measurements. In the below image, the corners are A, B, C and D. The sides of this rectangle are AB, BC, CD and AD. Finally, the diagonals are AC and BD.
M
## How to Calculate the Perimeter of a Rectangle?
Perimeter of a rectangle is nothing but the overall distance outside the rectangle as shown in the below image:
## Steps to Calculate the Perimeter of a Rectangle
1. Check the length and the width for the given rectangle
2. By short, we'll call the length as 'l' and width as 'w'.
3. Apply the formula, 2(l+w) or just add l + w + l + w and you've got the perimeter for the rectangle.
## Perimeter of a Rectangle Problems
Problem 1: A rectangle is 6m long and 5 m wide. What is its perimeter?
Solution: Length, l = 6m.
Width, w = 5m.
Perimeter of the rectangle = 2(l+w) = 2(6+5) = 2 x 11 = 22 m.
Problem 2: A rectangle is 5cm long and its perimeter is 18cm. What is its width?
Solution: Length, l = 5cm.
Width, w say.
Perimeter of the rectangle = 18 cm.
Perimeter of the rectangle = 2(l+w)= 2l + 2w = 2(5) + 2w = 10 + 2w
Thus, 10 + 2w = 18
2w = 18 - 10 = 8
w = 4
Thus the width of the rectangle is 4 cm.
Problem 3: In a rectangle's length is 5x + 2 and its width is 5x - 2. If its Perimeter is 40 cm, what is the value of x?
Solution: Length, l = 5x + 2.
Width, w = 5x - 2.
The Perimeter of the rectangle = 40 cm. ......(1)
Also by the definition of Perimeter of rectangle,
Perimeter of rectangle = 2(l*w) = 2[(5x + 2)+(5x - 2)] .....(2)
Equating (1) and (2),
2[(5x + 2)+(5x - 2)] = 40
(5x + 2)+(5x - 2) = 40/2 = 20
10x = 20
x = 20/10
x = 2
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# Anshul moves towards East a distance of 5 meters, then he turns to his left and walks 20 meters, then again he turns left and walks 15 meters. Now, he turns ${45^0}$ towards his right and goes straight to cover $20\sqrt 2$ meters. How far is he from his starting point?$(1){\text{ 40 m}} \\ {\text{(2) 30 m}} \\ {\text{(3) 50 m}} \\ {\text{(4) 55 m}} \\$
Last updated date: 11th Jun 2024
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Hint: Before starting with the solution, candidates should be aware of the directions where man is traveling through like east, west, north, and south. Below is the detailed diagram included for the direction analysis. In this question, all the directions are given in 90 degrees except one which is making an angle of 45 degrees with the horizontal. To start the solution, first of all, we need to draw the traveling trajectory of the man and use the Pythagoras theorem for the calculation of the distance.
Complete step by step solution:
Applying the Pythagoras theorem in the triangle ABC to determine the value of the distance between the points A and X.
Consider the distance between the points A and C be x.
$A{C^2} = A{B^2} + B{C^2} \\ {x^2} = {10^2} + {20^2} \\ {x^2} = 100 + 400 \\ x = \sqrt {500} \\ = 10\sqrt 5 \\ = 10 \times 2.236 \\ = 22.36{\text{ meters}} \\$
Now, to calculate the distance between the starting point and the endpoint $20\sqrt 2 = 20 \times 1.414 = 28.28{\text{ meters}}$ will be added to the value 22.36 meters.
$D = 28.28 + 22.36 \\ = 50.64{\text{ meters}} \\$
As we are doing with the approximations in the calculation, the nearest value to the number 50.64 meters given in the option is 50 meters. Hence, the distance between the starting point and the endpoint is 50 meters.
Note: It is very important to note here that, whenever the calculation part includes an approximation then, the value closest to the calculated result will be the desired result. Candidates can remember the direction convention by locating the sun and it is well known that “SUN rises in the EAST” and then follows the direction according to the question.
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# Chapter 4 First and second order ODEs
Not all ODEs are analytically solvable. In this section we discuss some types of first and second order ODEs that are analytically solvable and see some examples.
## 4.1 First order ODEs
A first order ODE has only the first derivative represented. The general implicit form for a first order ODE for the function $$x(t)$$ is: $G(t, x, \frac{dx}{dt}) = 0,$ and its explicit form is: $\frac{dx}{dt} = F(x, t).$
In the following we discuss some classes of first order ODEs and describe methods of obtaining a solution for them. These inculde separable and linear first order ODEs. The other types of First Order ODEs that can be solved are based on transformations or change of variables. We see two examples of this in the following. Another important class of first order ODEs that can be solved are Exact ODEs that will be discussed in part III of the course after introducing partial and total differentiation.
### 4.1.1 Separable First Order ODEs
A separable first order ODE can be written in the following form: $\frac{dx}{dt} = F_1(x) F_2(t).$ Solution Rearranging and integrating both sides we get: $\int \frac{dx}{F_1(x)} = \int F_2(t) dt + c_1.$
### 4.1.2 Linear First Order ODEs
First order linear ODEs have the following general form: $\frac{dy}{dx} + p(x)y = q(x).$ Solution This is solved by finding an integrating factor (IF). We look for $$I(x)$$ such that: $I(x)\left[\frac{dy}{dx} + p(x)y\right] = \frac{d[I(x)y]}{dx},$ Then, we have \begin{align*} \frac{d[I(x)y]}{dx} & = I(x)q(x),\\ \int d[I(x) y] & = \int q(x)I(x) \,dx + c_1,\\ y(x) & = \frac{1}{I(x)} \left[\int q(x)I(x) \, dx + c_1 \right]. \end{align*}
Integrating factors must fulfil: \begin{align*} \frac{d(Iy)}{dx} & = I \frac{dy}{dx} + Ipy, \\ I \frac{dy}{dx} + y \frac{dI}{dx} & = I\frac{dy}{dx} + Ipy,\\ \int \frac{dI}{I} & = \int p(x) \,dx + c'.\\ \end{align*} So we have: $I(x) = A e^{\int p(x)\, dx},$ where $$A$$ is a new arbitrary constant (of integration).
So, we have the following for the general solution: $y(x) = e^{-\int p(x)\, dx} \left[\int e^{\int p(x)\, dx} q(x) \, dx + c \right],$ where $$c = c_1/A$$ is a new arbitrary constant of integration.
### 4.1.3 Dimensionally Homogeneous
The dimensionally homogeneous have the following general form: $\frac{dy}{dx} = F\left( \frac{y}{x}\right).$ Solution Let $$u = y/x$$ we obtain: $\frac{dy}{dx} = u + x \frac{du}{dx}$ The ODE in terms of $$u(x)$$, which is separabale is $u + x \frac{du}{dx} = F(u),$ Finding general solution $$u_{GS}(x)$$ for this ODE then we find the general solution for the original ODE as $$y_{GS}(x) = u_{GS}(x) x$$.
### 4.1.4 Bernoulli ODEs
There are other examples of transformations can turn specific ODEs into separable or linear. Some such as Bernoulli are classic: $\frac{dy}{dx} + p(x) y = q(x) y^n,$ where $$n \in \mathbb{R}$$.
Solution We use the change of variable $$u = y^{1-n}$$. We obtain: $\frac{du}{dx} = (1-n) y^{-n} \frac{dy}{dx}.$ Writing the original ODE in terms of $$u$$ we have: $\frac{du}{dx} + (1-n) p(x) u = (1-n) q(x),$ which is a linear ODE for $$u(x)$$, so we obtain $$u_{GS}(x)$$ and then we have $y_{GS} = u_{GS}^{\frac{1}{1-n}}.$
## 4.2 Second Order ODEs
The general implicit form is: $G(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}) = 0,$ and the general explicit form is: $\frac{d^2y}{dx^2} = F(x, y, \frac{dy}{dx}).$
The second order ODEs are common in Mechanics as Newton’s second law is such ODE with independent variable as time $$t$$. They are difficult to solve for general $$F$$ but there are some special cases that can be solved as described in the following. Also, the linear case is discussed in the next chapter in detail.
### 4.2.1$$F$$ only depends on $$x$$
$\frac{d^2y}{dx^2} = F(x)$ Solution Let $$u = \frac{dy}{dx}$$ then we have $$\frac{du}{dx} = F(x)$$. A first integration gives us: $u = \int F(x) dx + c_1$ A second integration then gives us $$y_{GS}$$: $y_{GS} = \int \left[\int F(x) dx\right] dx + c_1x + c_2.$
### 4.2.2$$F$$ only depends on $$x$$ and $$\frac{dy}{dx}$$
$\frac{d^2y}{dx^2} = F(x, \frac{dy}{dx})$ Solution Let $$u = \frac{dy}{dx}$$ then we have $$\frac{du}{dx} = F(x, u)$$, which is a first order ODE. If we could obtain the general solution $$u_{GS}(x; c_1)$$ then we have: $y_{GS}(x) = \int u_{GS}(x; c_1) \, dx + c_2.$
### 4.2.3$$F$$ only depends on $$y$$
$\frac{d^2y}{dx^2} = F(y)$
Solution We let $$u = \frac{dy}{dx}$$ then $$\frac{du}{dx} = F(y)$$. Then we have: $\frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx} = u \frac{du}{dy} = \frac{d}{dy}\left( \frac{1}{2}u^2 \right) = F(y),$ which is a first order separable ODE for $$u(y)$$. We have: $\frac{1}{2}u^2 = \int F(y)dy + c_1 = G(y) + c_1.$ So we have: $\frac{dy}{dx} = u = \pm \sqrt{2 G(y) + 2c_1},$ which is a first order separable ODE for $$y(x)$$ and can be integrated to obtain $$y_{GS}(x; c_1, c_2)$$ as seen in the following example.
Example 4.1 (Mechanics Harmonic Oscillator) Hooke’s law states if $$x(t)$$ is displacement relative to an ideal spring relaxed position, the spring force is: $$F = -kx$$ Using second Newton Law we have: $$ma = F \quad \Longrightarrow \quad m\frac{d^2x}{dt^2} = - kx$$
Let velocity to be $$u = \frac{dx}{dt}$$, then we have: $a = \frac{du}{dt} = \frac{d}{dx}\left[\frac{1}{2}u^2\right] = -\frac{kx}{m}.$ Integrating both sides we obtain: $\frac{u^2}{2} = -\frac{k}{2m} x^2 + c_1.$ This equation gives us a constant of motion $$(E = c_1 m)$$, which is known as total energy, the sum of kinetic energy ($$1/2mu^2$$) and potential energy ($$1/2 k x^2$$).
$$u = \frac{dx}{dt} = \pm \sqrt{\frac{2E - kx^2}{m}} \quad \Longrightarrow \quad \displaystyle\int \frac{dx}{\pm \sqrt{\frac{2E - kx^2}{m}} } = \displaystyle\int dt$$
Sticking with the postive sign on the LHS we have: $\frac{1}{\sqrt{2E/m}}\displaystyle\int \frac{dx}{\sqrt{1 - \frac{k}{2E}x^2} } = \sqrt{\frac{m}{k}}\sin^{-1}\left(\sqrt{\frac{k}{2E}}x\right) = t + c_2$ Rearranging the solution we obtain: $x_{GS} = A \sin(\omega t + \phi),$ where, $$\omega = \sqrt{k/m}$$ is the frequency of oscillations and $$A = \sqrt{2E/k}$$ and $$\phi = \sqrt{k/m} c_2$$ are new constants of integration. We note that, if we had chosen to use the minus sign above, we would have obtained the same family of solutions but the constants of integrations would be differently defined.
### 4.2.4$$F$$ only depends on $$y$$ and $$\frac{dy}{dx}$$
$\frac{d^2y}{dx^2} = F(y, \frac{dy}{dx})$
let $$u = \frac{dy}{dx} \quad \Longrightarrow \quad \frac{du}{dx} = F(y, u).$$ So we have $\frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx} = u \frac{du}{dy} = \frac{d}{dy}\left( \frac{1}{2}u^2 \right).$ Therefore we have the following first order ODE for $$u(y)$$ to solve $\frac{d}{dy}\left( \frac{1}{2}u^2 \right) = F(y, u).$ Given $$u_{GS}(y; c_1)$$ being a general solution for the above ODE, we have the following first order ODE for $$y(x)$$: $\frac{dy}{dx} = u_{GS}(y; c_1).$
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# If y varies inversely with x and y = 8 when x = 5, then what is y when x = 2?
Apr 29, 2017
$y = 20$
#### Explanation:
"varies inversely means " 1/("variable")
$\text{the initial statement is } y \propto \frac{1}{x}$
To convert to an equation, multiply by k, the constant of variation.
$\Rightarrow y = k \times \frac{1}{x} = \frac{k}{x}$
$\text{to find k, use the condition given}$
$\text{that is " y=8" when } x = 5$
$y = \frac{k}{x} \Rightarrow k = x y = 5 \times 8 = 40$
$\text{equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = \frac{40}{x}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{when } x = 2 \to y = \frac{40}{2} = 20$
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# Class 7: Systems of linear equations and SAGEmath
Overview: This part introduces you to an essential skill in any mathematician's or cryptographer's tool chest. Using linear algebra (matrices) to find answers to linear systems. This talent will unlock the ability to break the LCGs and LFSRs from the last lesson.
### What is a linear system?
You're probably comfortable with old-school algebra problems like: on the line $$y = x + 8$$, what is the value of $$y$$ when $$x$$ is $$12$$?
No big deal right? Now if we have two lines $$y = x + 8$$ and $$y = 2x - 4$$ we might ask, where do they intersect?
\begin{align} y&= 2x - 4 \\ y&= x + 8 \end{align}
You probably have several ways to solve it in mind already. Substitute: $$x+8 = 2x - 4$$ leads to $$x = 12$$. Or subtract one equation from the other:
$$(y-y) = (2x - x) + (-4 -8)$$
The best solution for this course is to look at the system of linear equations as a matrix problem ( $$A \vec{x} = \vec{b}$$ ). It's much more awkward for the small cases but the only realistic way to code it when we're dealing with tens or hundreds of equations. Here is the basic idea, make the equations match $$A \cdot \begin{bmatrix} y \\ x \end{bmatrix} = \vec{b}$$ then find $$A^{-1}$$ and we'll know that $$\begin{bmatrix} y \\ x \end{bmatrix} = A^{-1} \cdot \vec{b}$$. Here are my steps:
\begin{align} y& -2x& = -4 \\ y& -x& = 8 \end{align}
$$\begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} y \\ x \end{bmatrix} = \begin{bmatrix} -4 \\ 8 \end{bmatrix}$$
$$A^{-1} = \begin{bmatrix} -1 & 2 \\ -1 & 1 \end{bmatrix}$$
$$\begin{bmatrix} y \\ x \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ -1 & 1 \end{bmatrix} \cdot \begin{bmatrix} -4 \\ 8 \end{bmatrix} = \begin{bmatrix} 20 \\ 12 \end{bmatrix}$$
Try it out: Solve the system $$3y = 2x + 4$$ and $$2y = 2x - 4$$. Now write it down as a matrix problem.
Generalize: Now try to solve the system $$ay + bx = e$$ and $$cy + dx = f$$ in full generality. When is there no solution?
Increase the size: Now take a stab at solving a three equation system in $$x,y,z$$. You'll have to research inverting a $$3 \times 3$$ matrix.
Overview: In this section I introduce you to a freely available and powerful tool called SAGE. It is basically Python mixed with high-powered math. I'll show you the basics of getting setup and how to use it to solve some of the number theory problems we face. After you're in I'll give you some problems to work through.
## Solving Number Theory Problems Online
Here is a screen-shot by screen-shot intro to signing up for SAGE math using your UDel email. This is a fast browser-based way to get high-powered number theory working for you and it will help when you need to solve tricky problems using exact math algorithms.
Here is a video demonstrating how to solve a problem in SAGE (matrix inversion).
## Now try these algebraic problems:
Now here are a set of number theory exercises that should push you a little bit, but get your mind right for using this new tool in your arsenal. These refer to the end of class 5 which has a bullet-point jump through some algebra and they help us set up an attack we will pull off later in the course.
Euler's Phi: Calculate $$\Phi(10)$$ using SAGE. What are the elements in $$\mathbb{Z}_{11}^{*}$$? How many generators of $$\mathbb{Z}_{11}^{*}$$ should there be? Find those elements that generate $$\mathbb{Z}_{11}^{*}$$ .4, the generators are 2, 6, 7, and 8
Cyclic Subgroups: Pick one of those generators which generates all ten elements and find the unique subgroup of size 5. Now find the unique subgroup of size 2.
Cosets by hand: Find all cosets of the subgroup of size 2 (multiply the elements of that subgroup by something not in that group).
Safe Primes: Generate a "safe prime" which is a prime number that happens to be 1 more than double another prime number, that is, $$p = 2\cdot q + 1$$. Find it by trial and error and using x.is_prime().
Generalize: Modulo your safe prime what are the orders of the elements in the multiplicative subgroup?
Generalize: How many generators are there? Find one.
Overview: In this part we use the skills from the last two parts to break the Linear Congruential Generator (a harder way) by figuring out the secret parameters and the current "state". Then we can predict correctly what the rest of the sequence will be (far from 50/50). This will be the simplest break we do but the others build on this example.
### Continuing with Systems of Linear Equations
Now one of the things about math (and math education) is that solving a problem that has been setup for you is much easier than finding the setup in the wild. In order for the skills of this lesson to be of any real value you'll want to develop some instincts for finding systems of linear equations in problems you face.
In this case I'm outright telling you that we can find linear algebra inside of the PRNGs we've seen so far. But let's make that more clear now. (If you like a challenge then pause and think about where the linear systems are so far.)
### Using Linear Algebra to tackle LCGs
Let's start with the Linear Congruential Generator. You know that each term is a function of the term before, $$t_i = a \cdot t_{i-1} + c \pmod{m}$$. If you see several terms in a row then to predict the next term you need to know $$a, c$$ and $$m$$. You often know these values and might just want to know the seed, or you know the modulus and only care to predict all future values. OK. So let's make it simple for now. Suppose $$m = 101$$ and you see the sequence $$\{40, 42, 49 \}$$. Can you figure out $$a$$ and $$c$$?
\begin{align} 42 &= a \cdot 40 &+ &c \pmod{101} \\ 49 &= a \cdot 42 &+ &c \pmod{101} \end{align}
Solve it: solve this simple LCG.
Target: the following four terms were generated by an LCG (modulo 2**32). Find the next term:
• 3904654068
• 897137925
• 4281244274
• 3573336907
### Fast and Dirty Analysis of Systems
When you're out in the wild thinking about solving a linear system you have to know if it will work. Here is the rule:
Count the number of equations and count the number of unknowns. If there are more equations than unknowns then there is either 1 exact answer or no answer. If you've got more unknowns than equations then there are an infinite number of possible answers.
In this previous example we could generate as many equations as we want. If there is an answer then any two will do.
#### LCG with less knowledge:
If you're a CTF trainer you might need to crack an LCG without knowing the modulus, the multiplier, or the increment. You still can you just need more data points (because if the number of equations is more than the number of unknowns you'll be alright).
Here is a script I use all of the time to do this:
Overview: In this part we extend the ideas of the last PRNG exploit to the larger system for Linear Feedback Shift Registers. In breaking these systems we'll see that designers of LFSRs could add just a bit of extra noise to thwart us and many stream ciphers do just that.
### World 1: We know the coefficients
So if someone is generating a stream of random bits with a Linear Feedback Shift Register and you know which register-bits are XORed to generate the next input bit then it's really easy to correctly predict all future bits. Let's do that now.
Recall that a general LFSR looks like:
Let's say that the length of the register, $$L$$, is $$5$$ and that every coefficient is $$1$$. That means that the next input bit is a "parity-check" of the current 5 (add 'em all up mod 2).
Anticipate: How many bits do you need to see before you can correctly predict all future bits? 5
What's the process? Wait until you know enough bits to reverse engineer the state of the register once, then you'll know all of the future states of the register.
Actually do it: Suppose that you got the opening bits 1, 0, 0, 1, 1. Can you tell me the next 10 bits that would follow? 1, 1, 0, 0, 1, 1, 1, 1, 0, 0
How good is that? When you see that output do you see an issue with the setup of that LFSR? What is the length of it's cycle? What is the maximum length of an LFSR cycle with $$L=5$$?
### World 2: We don't know the feedback coefficients
Now what do we do? Before, the moment we knew the entire register once we could correctly know it in all future cases. But what if you don't know how the next bit is made?
I claim that it takes $$2L$$ bits before you can break it, but that you need linear algebra.
First an observation about XOR. It is the same as addition modulo 2. $$1 \oplus 1 = 1 + 1 \pmod{2}$$.
Confirm: Confirm that XOR on the 0 and 1 has the exact same output as addition modulo 2.
So what does that get us? Well we can now look at register bits as linear combinations of products.
If $$c_1, \ldots, c_L$$ are unknown zeroes or ones then $$s_{t+L} = s_{t+L-1} \cdot c_1 + \cdots + s_{t} \cdot c_L \pmod{2}$$.
We can also reverse engineer as many of the $$s_i$$ as we need by just waiting and watching, just like in World 1.
So we have $$L$$ unknowns and we need $$L$$ equations. After waiting for $$L+1$$ bits we have 1 equation, but after $$L+2$$ bits we have 2 equations. So after $$L+L$$ bits we should have $$L$$ equations with $$L$$ unknowns and that's enough to solve the linear algebra.
Let's just do it. I've secretly selected 5 coefficients for an LFSR. The first 10 bits of output are: 0, 1, 1, 0, 0, 1, 1, 0, 0, 1
The first 6 give us one equation:
$$1=0\cdot c_5+ 0 \cdot c_4+ 1 \cdot c_3+ 1 \cdot c_2+ 0 \cdot c_1$$
Continue that one more time and you get:
$$1= 1\cdot c_5+ 0 \cdot c_4+ 0 \cdot c_3+ 1 \cdot c_2+ 1 \cdot c_1$$
Write down the equations: Go ahead and generate 3 more equations from the sequence of bits.
Now find the coeffs Use SAGE to invert a $$5 \times 5$$ matrix over GF(2) to solve for the coefficients.
Predict without linear algebra: Of course a pattern emerged and you can probably solve this one without using SAGE. What's the pattern?
Prove it: At what point can you guarantee that the pattern you found repeats forever?
Do it at full size: The first 38 bits of a length 19 LFSR's output are: 00100001000001000111010001001110010100. Find the coefficients. [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1]
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# Difference between revisions of "2011 AMC 10B Problems/Problem 23"
## Problem
What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$
## Solution 1
Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$
To compute this, we use a clever application of the binomial theorem.
\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}
In all of the other terms, the power of $10$ is greater than $3$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:
\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}
Therefore, the hundreds digit is $\boxed{\textbf{(D) } 6}.$
Sidenote: By Euler's Totient Theorem, $a^{\phi (1000)} \equiv a \pmod{1000}$ for any $a$, so $a^{400} \equiv a \pmod{1000}$ and $11^{2011} \equiv 11^{11} \pmod{1000}$. We can then proceed using the clever application of the Binomial Theorem.
## Solution 2
We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem, it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$
In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have $$2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.$$
In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have \begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}
After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{\textbf{(D) } 6}.$
## Solution 3
Notice that the hundreds digit of 2011^2011 won't be affected by 2000. Essentially we could solve the problem by finding the hundreds digit of 11^2011. Powers of 11 are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of 11 can be found by reading rows of the triangle and adding extra numbers up. (add source) For example, the sixth row of the triangle is 1, 5, 10, 10, 5, and 1. Adding all numbers from right to left, we get 161051, which is also 11^5. In other words, each number is 10^n steps from the right side of the row. The hundreds digit is 0. We can do the same for 11^2011, but we only need to find the 3 digits from the right. Observing, every 3 number from the right is 1 + 2 + 3... + n. So to find the third number from the right on the row of 11^2011, f(11^n) = 1 + 2 + 3... + (n-1), or (2010 * 2011)/2, or 2021055. The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously 2011. We must carry the 1 in 2011's tens digit to the 5 in 2021055's unit digit to get 6. The one at the very end of the row doesn't affect anything, so we can leave it alone.
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