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In the given figure, ABC is a right-angled triangle with AB = 6
# In the given figure, ABC is a right-angled triangle with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r, the radius of the inscribed circle.
It is given that ABC is a right angle triangle with AB = 6 cm and AC = 8 cm and a circle with centre O has been inscribed.
Using Pythagoras theorem, we get
BC2 = AC2 + AB2
= (8)2 + (6)2
= 64 + 36 = 100
⇒ BC = 10 cm
Tangents at any point of a circle is perpendicular to the radius through the point of contact
= 3r + 4r + 5r
= 12r ...(ii)
Comparing (i) and (ii), we get
24 = 12r ⇒ r = 2 cm
Method – II:
∵ ∠OPA = ∠ORA = 90°
∠PAR = 90°
⇒ ∠OPA = ∠ORA = ∠PAR = ∠POR = 90° ....(i)
and AP = AR (ii) (length of tangents drawn from an external point are equal)
Using result (i) and (ii), we get
APOR is a square
Therefore,
OR = AR = r [Sides of square] and OR = AP = r [Sides of square]
Now, BP = AB – AP = 6 – r and, CR = AC – AR = 8 – r
Since tangents from an external point are equal
CR = CQ = 8 – r and BP = BQ = 6 – r
Now, In ΔABC,
BC2 = AC2 + AB2
⇒ (CQ + BQ)= (8)2 + (6)2
⇒ (8 – r + 6 – r)2 = 64 + 36
⇒ (14 – 2r)2 = 100
⇒ (14 – 2r)2 = (10)2
⇒ 14 – 2r = 10
⇒ –2r = –4
⇒ r = 2
Hence radius of circle (r) = 2 cm.
421 Views
From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm. Then what is the radius of the circle?
Since, the tangent of any point of a line is perpendicular to the radius through the point of contact.
Therefore, ∠OPO =90°
It is given that PQ =15 cm
and OQ = 17 cm
In right ΔOPQ
OQ2 = OP2 + PQ2
[Using Pythagoras theorem]
⇒ (17)2 = OP2 + (15)2
OP2 = (17)– (15)2
= 289 – 225 = 64 ⇒ OP = 8 cm
Hence, radius of the circle = 8 cm.
3033 Views
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.
Given : Two concentric circles C1 and Cof radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C1) and BD is a tangent to the smaller circle (C2).
∴ The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Now, in right triangle BOD, we have
OB2 = OD2 = BD2 [ Using Pythagoras theorem]
(13)= (8)+ BD
169 = 64 + BD2
BD2 = 169 - 64
BD2 = 105
Since, perpendicular drawn from the centre to the chord bisects the chord.
(∵ ∠BEA = 90°, angle in semicircle is right angle] and, ∠OBD = ∠ABE (common)
Therefore, using AAS similar condition ΔBOD ~ ΔBAE
[Proportional sides of two similar triangles]
Now, in right triangle ADE, we have
= 256 + 105
= 361
3545 Views
A line intersecting a circle in two points is called a ___________.
• Diameter
• Chord
• Secant
• Secant
C.
Secant
118 Views
ABCD is a quadrilateral such that ∠D = 90°. A cirlce C (O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, Find r.
So, BP = BQ
(Tangents from external point B)
But BP = 27 cm
⇒ BQ = 27 cm
It is given that BC = 38 cm
⇒ BQ + CQ = 38
⇒ 27 + CQ = 38
⇒ CQ = 11 cm
⇒ CQ = CR (Tagents from an external point C)
But CQ = 11 cm
⇒ CR = 11 cm
It is given that : CD = 25 cm
⇒ CR + DR = 25
⇒ 11 + DR = 25
⇒ DR = 14 cm
Since, tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠ORD = ∠OSD = 90°
It is given that
∠D = 90° |
Descartes' Rule of Signs allows us to determine the possible number of positive real zeros and the possible number of negative real zeros for a polynomial function with real coefficients and a nonzero constant term. This rule will help us to narrow down our choices when looking for zeros of a polynomial function.
Test Objectives
• Demonstrate the ability to find the possible number of real zeros
Descartes' Rule of Signs Practice Test:
#1:
Instructions: State the possible number of positive real zeros and negative real zeros.
a) f(x) = 4x5 + 2x4 + 30x3 + 15x2 - 16x - 8
b) f(x) = 2x5 + 4x4 + 5x3 + 10x2 - 12x - 24
#2:
Instructions: State the possible number of positive real zeros and negative real zeros.
a) f(x) = 4x6 + 16x4 - 25x2 - 100
b) f(x) = 9x6 + 45x4 - 4x2 - 20
#3:
Instructions: State the possible number of positive real zeros and negative real zeros.
a) f(x) = 5x6 - 4x4 - 20x2 + 16
b) f(x) = 6x5 - 9x4 - 34x3 + 51x2 + 20x - 30
#4:
Instructions: State the possible number of positive real zeros and negative real zeros.
a) f(x) = 5x5 + 25x4 + 14x3 + 70x2 - 3x - 15
b) f(x) = 16x6 + 64x4 - 25x2 - 100
#5:
Instructions: State the possible number of positive real zeros and negative real zeros.
a) f(x) = x6 - 64
b) f(x) = 27x7 + 37x4 - 64x
Written Solutions:
#1:
Solutions:
a) Positive Real Zeros: 1, Negative Real Zeros: 4, 2, or 0
b) Positive Real Zeros: 1, Negative Real Zeros: 4, 2, or 0
#2:
Solutions:
a) Positive Real Zeros: 1, Negative Real Zeros: 1
b) Positive Real Zeros: 1, Negative Real Zeros: 1
#3:
Solutions:
a) Positive Real Zeros: 2 or 0, Negative Real Zeros: 2 or 0
b) Positive Real Zeros: 3 or 1, Negative Real Zeros: 2 or 0
#4:
Solutions:
a) Positive Real Zeros: 1, Negative Real Zeros: 4, 2, or 0
b) Positive Real Zeros: 1, Negative Real Zeros: 1
#5:
Solutions:
a) Positive Real Zeros: 1, Negative Real Zeros: 1
b) Positive Real Zeros: 1, Negative Real Zeros: 1 |
Question Video: Finding the Measure of an Arc Given Its Equal Arcβs Measure Mathematics
Find π arc π΅πΈ, where π is the center of the circle.
01:22
Video Transcript
Find the measure of arc π΅πΈ, where π is the center of the circle.
First, letβs identify arc π΅πΈ. After we highlight arc π΅πΈ, itβs important to notice that the segment π΅π΄ and the segment πΈπΆ are two parallel chords in this circle. We know by the measures of arcs between parallel chords theorem that the measure of the arcs between parallel chords of a circle are equal. This means that the arc π΅πΈ will be equal in measure to the arc π΄πΆ. We can also see that line segment π·πΆ intersects line segment π΅π΄ at the center of the circle π.
We can therefore say that angle π΄ππΆ is the opposite or vertical angle from angle π΅ππ· and is, therefore, also equal to 39 degrees. Since π is the center of the circle and arc π΄πΆ is subtended by the angle π΄ππΆ, we can say that the arc measure is equal to the central angle measure, which is 39 degrees. Since arc π΄πΆ is 39 degrees, arc π΅πΈ must also be equal to 39 degrees. |
# Cubes and Cuboids: Estimate and Compare Volumes
In this worksheet, students must estimate and compare the volumes of the given cubes and cuboids.
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Volume and Capacity
Difficulty level:
### QUESTION 1 of 10
You should know the following formulae to work out the volumes of these shapes:
Cube
You must know:
• the side length
Remember that the length, the width and the height will all be equal to the side length, because this is a CUBE.
Volume of cube = side length × side length × side length
Cuboid
You must know:
• the length
• the width
• the height
Volume of cuboid = length × width × height
Example 1
How many 1 cm3 cubes will fit into this cuboid?
The length is 15 cm, the width is 4 cm and the height is 8 cm.
Volume of cuboid = length × width × height = 15 × 4 × 8 = 60 × 8 = 480 cm3.
So 480 cubes will fit.
Example 2
Calculate the volume of this cube in mm3.
The side length is 7 cm, so the width and the height are also 7 cm.
This will be 70 mm.
NB: We want the answer in mm3 which is why we should work in mm not cm.
Volume of cube = side length × side length × side length = 70 × 70 × 70 = 49 × 7 × 10 × 10 × 10 = 343 × 1000 = 343000 mm3.
NB: The length, the width and the height are all interchangeable with each other, depending on how you look at a cuboid.
How many 1 cm3 cubes will fit into this cuboid?
How many 1 cm3 cubes will fit into this cube?
Which is larger in volume, the cuboid or the cube?
cuboid
cube
both the same
Jack says that the volume of this cube is about 8000 m3.
Is he right or wrong?
right
wrong
Which is larger?
The cuboid shown below, or a cube with side length 200 cm?
cube
cuboid
both the same
How many 1 m3 cubes will fit into this cuboid?
The diagrams show a cuboid and a cube.
How many of the cubes will fit into the cuboid?
12
just over 2
just over 3
A large swimming pool is shown below.
Estimate the volume of water in the pool.
300 m3
3000 m3
1500 m3
This cuboid is made from 1 cm cubes.
What is the volume of the cuboid in cm3?
6 cm3
3 cm3
8 cm3
This cuboid is made from 2 cm cubes.
What is the volume of the cuboid in cm3?
6 cm3
12 cm3
48 cm3
• Question 1
How many 1 cm3 cubes will fit into this cuboid?
336
EDDIE SAYS
Volume = 12 × 4 × 7 = 336 cm3
• Question 2
How many 1 cm3 cubes will fit into this cube?
125
EDDIE SAYS
Volume = 5 × 5 × 5 = 125 cm3
• Question 3
Which is larger in volume, the cuboid or the cube?
cuboid
EDDIE SAYS
Volume of cuboid = 5 × 9 × 3 = 135 mm3
Volume of cube = 5 × 5 × 5 = 125 mm3
• Question 4
Jack says that the volume of this cube is about 8000 m3.
Is he right or wrong?
wrong
EDDIE SAYS
2 km = 2000 m
Volume = 2000 × 2000 × 2000 which is much bigger than just 8000.
• Question 5
Which is larger?
The cuboid shown below, or a cube with side length 200 cm?
cuboid
EDDIE SAYS
Volume of cuboid = 6 × 2 × 1 = 12 m3
Volume of cube = 2 × 2 × 2 = 8 m3
• Question 6
How many 1 m3 cubes will fit into this cuboid?
8
EDDIE SAYS
Work in metres.
Volume = 2 × 4 × 1 = 8 m3
• Question 7
The diagrams show a cuboid and a cube.
How many of the cubes will fit into the cuboid?
just over 2
EDDIE SAYS
The cube fits into the square end of the cuboid exactly.
2 × 9 = 18 which is nearly 20.
So just over 2 will fit exactly.
• Question 8
A large swimming pool is shown below.
Estimate the volume of water in the pool.
1500 m3
EDDIE SAYS
Volume is approximately 30 × 10 × 5 = 300 × 5 = 1500 m3.
• Question 9
This cuboid is made from 1 cm cubes.
What is the volume of the cuboid in cm3?
6 cm3
EDDIE SAYS
Each cube has volume 1 cm3.
• Question 10
This cuboid is made from 2 cm cubes.
What is the volume of the cuboid in cm3?
48 cm3
EDDIE SAYS
Each cube has volume 2 × 2 × 2 = 8 cm3.
---- OR ----
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## Precalculus (6th Edition) Blitzer
Step I: To check even or odd. In the provided equation, if $h\left( -x \right)=h\left( x \right)$ then it is an even function, and if $h\left( -x \right)=-h\left( x \right)$ , then it is odd. \begin{align} & h\left( x \right)={{x}^{2}}+2{{x}^{4}} \\ & h\left( -x \right)={{\left( -x \right)}^{2}}+2{{\left( -x \right)}^{4}} \\ & ={{x}^{2}}+2{{x}^{4}} \\ & =h\left( x \right) \end{align} It is an even function. Step II: To check symmetry about the y-axis: In the provided equation put $x=-x$ , if the equation remains the same, then it has symmetry about the y-axis. \begin{align} & h\left( x \right)={{x}^{2}}+2{{x}^{4}} \\ & h\left( -x \right)={{\left( -x \right)}^{2}}+2{{\left( -x \right)}^{4}} \\ & ={{x}^{2}}+2{{x}^{4}} \\ & =h\left( x \right) \end{align} It is the same as the given equation, hence it has symmetry about the y-axis. Step III: To check symmetry about the x-axis: In the provided equation put $y=-y$ , if the equation remains the same, then it has symmetry about the x-axis. \begin{align} & h\left( x \right)={{x}^{2}}+2{{x}^{4}} \\ & y={{x}^{2}}+2{{x}^{4}} \\ & -y={{x}^{2}}+2{{x}^{4}} \\ & y=-{{x}^{2}}-2{{x}^{4}} \end{align} It is not the same as the given equation, hence it is not symmetric about the x-axis. Step IV: To check symmetry about the origin: In the provided equation put $x=-x,\text{ and }y=-y$ , if the equation remains the same, then it has symmetry about the origin. \begin{align} & y={{x}^{2}}+2{{x}^{4}} \\ & \left( -y \right)={{\left( -x \right)}^{2}}+2{{\left( -x \right)}^{4}} \\ & -y={{x}^{2}}+2{{x}^{4}} \\ & y=-{{x}^{2}}-2{{x}^{4}} \end{align} It is not the same as provided equation, hence it is not symmetric about the origin. Therefore, the provided function is even, has symmetry about the y-axis only. |
# Lesson 12
Connections between Graphs and Equations
• Let’s examine some situations, equations, and graphs.
### 12.1: Math Talk: Evaluating a Function
Here is a function: $$g(x)=100-5x$$
Evaluate mentally:
$$g(0)$$
$$g(1)$$
$$g(4)$$
$$g(20)$$
### 12.2: Bank Accounts
Each function represents the amount in a bank account after $$t$$ weeks.
$$A(t) = 500$$
$$B(t) = 500 + 40t$$
$$C(t) = 500 - 40t$$
$$D(t) = 500 \boldcdot (1.5)^t$$
$$E(t) = 500 \boldcdot (0.75)^t$$
1. Make a table for each bank account showing the money in the account at 0, 1, 2, and 3 weeks.
2. Describe in words how the money in the account is changing week by week.
3. Use technology to create a graph of each function. How can you see your description in each graph?
### 12.3: Build a New Function
Consider the same five functions:
$$A(t) = 500$$
$$B(t) = 500 + 40t$$
$$C(t) = 500 - 40t$$
$$D(t) = 500 \boldcdot (1.5)^t$$
$$E(t) = 500 \boldcdot (0.75)^t$$
1. Starting with one of the functions, change it so that it represents an account that . . .
1. Starts with a balance of $300, and loses$40 each week.
2. Starts with a balance of $500, and gains$15 each week.
3. Starts with a balance of $500, and loses $$\frac{1}{10}$$ of its value each week. 4. Starts with a balance of$700, and gains $$\frac{3}{10}$$ of its value each week.
2. Here are four graphs. Which graph matches each of your new equations?
3. To check, use technology to graph your equations. Make sure to use the same graphing window. Check that the graph of your equation matches the graph you chose. |
Lesson 3 Problem Set 5.1 Answer Key
Posted on
Are you looking for Lesson 3 Problem Set 5.1 Answer Key for NYS Common Core Mathematics Curriculum Engage NY? If so, here you may be able to find the answer so that you are able to make sure whether your answer is correct or not. Let’s find out the answer key below.
Answer Key of Lesson 3 Problem Set 5.1
This key answer can be found in this link here. Cited from that link, here is the key answer for Lesson 3 Problem Set 5.1.
1. In the number one, you are required to write the following in exponential form (e.g., 100 = 102)
a. 10,000 = 104
b. 1000 = 103
c. 10 x 10 = 102
d. 100 x 100 = 104
e. 1,000,000 = 106
f. 1000 x 1000 = 106
2. In the number two, you are required to write the following in standard form, for example 5 x 102= 500
a. 9 x 103 = 9,000
b. 39 x 104= 390,000
c. 7200 : 102 = 72
d. 7,200,000 : 103 = 7,200
e. 4.025 x 103 = 4,025
f. 40.25 x 104 = 402,500
g. 725 : 103 = 0.725
h. 7.2 : 102 = 0.72
3. In this number, you have to think about the answers to Problem 2 (a-d). You are required to explain the pattern used to find an answer when you multiply or divide a whole number by a power of 10.
Answer: When multiplying by a power of 10, I add zeros onto my whole number that equals my exponent. When dividing by a power of 10, I subtract zeros from my whole number that equals my exponent.
4. In this number, you are required to think about the answers to Problem 2(e-h). You have to be able to give your explanation about the pattern which is used to place the decimal in the answer when you multiply or divide a decimal by a power of 1.
Answer: When multiplying by a power of 10, the decimal point is moved to the right equal to my exponent to make my answer larger. When dividing by a power of 10, the decimal point is moved to the left equal to my exponent to make my answer smaller.
5. In this number, you are required to complete the patterns.
a. 0.03 0.3 3.0 300 3000
b. 6,500,000 65,000 650 6.5 0.065
c. 94,300 9,430 943 94.3 9.43 0.943
d. 999 9990 99,900 999,000 9,990,000 99,900,000
e. 0.075 7.5 750 75,000 7,500,000 750,000,000
f. Here, you are required to explain how you found the missing numbers in set b and you have to make sure that you include your reasoning about the number of zeros in your numbers and how you placed the decimal.
Answer: The next number in the pattern had 2 zeros less, so the numbers were being divided by 102 which makes it smaller by 2 places.
g. Here, you are required to explain how you found the missing numbers in set d and you have to make sure that you give your reasoning about the number of zeros in your numbers and how you put the decimal.
Answer: The whole numbers added one zero, so the numbers were being multiplied by 101 which makes it larger by one place or column each time in the pattern.
6. In this number, there is a story where Shauunie and Marlon missed the lesson on exponents and then Shaunnie incorrectly wrote 105 = 50 on her paper, and Marlon mistakenly wrote 2.5 x 102 = 2.500 on his paper.
a. Here, you are asked what mistake that has been made by Shaunnie and you have to explain it using words, numbers and pictures why her thinking is incorrect and what she needs to do to correct the answer that she made.
Answer: Shaunnie thinks 105 is 10 x 5 which is 50, but 105 means 10 x 10 x 10 x 10 x 10 which equals 100,000.
b. Here, you are asked about the mistake that Marlon has made and you have to explain it using words, numbers and pictures why his thinking is not correct and what he needs to do to correct the answer that he made.
Answer: Marlon made the mistake of adding two zeros at the end of 2.5. That would have worked for a whole number, but 2.5 is a decimal so he needs to move the decimal point to the right two shifts making 250. |
naming-angles-in-geometry
# Interactive video lesson plan for: Naming angles in geometry
#### Activity overview:
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http://www.moomoomath.com/Classifying-Angles.html
In Geometry there are four main angles you will be working with. They are: acute angle, obtuse angle,right angle, and straight angle. Video gives a quick summary of each angle.
Transcript
Today we are going to look at angles. The first angle we are going to look at is the ninety degree angle. That makes an L shape and the symbol is a little square box in the corner. Then we have what is called a straight angle which is straight across without an actual angle. I think of it as just a straight line. Then you have an angle that falls smaller than ninety degrees. That one is an acute angle, and then you have an angle that is larger than ninety, but less than straight and that is an obtuse angle so let's label those. This first one is ninety, and that is kind of your reference and you always want to be familiar with a ninety degree in order to see if an angle is smaller than it. That one would be an acute angle because it is less than ninety and this one is larger than ninety and you can sketch in your L if you want to in order to compare it, and that one is greater so it is obtuse and then a straight line is one hundred and eighty because it is kind of like two L's put together and ninety and ninety is one eighty so this is your straight angle. So you are always comparing to your ninety degree angle or your right angle. So that's how angles classify. So let's look at the rules. An acute angle is less than ninety degrees. A right angle is equal to ninety degrees and it is your reference angle. An obtuse angle is greater than ninety degrees and a straight angle which is equal to one hundred and eighty degrees. You can also use a protractor to measure an angle and when you use a protractor take the dot and line it up with the vertex. What is the vertex? It is the corner of the angle. Line it up, read it from the bottom up. This is how you classify angles.
-~-~~-~~~-~~-~-
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Tagged under: acute,moomoomath,geometry,math ,straight angles,obtuse,homework ,geometry
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Students can download Maths Chapter 7 Mensuration Ex 7.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.2
Question 1.
A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.
Radius of the well (r1) = 5 m
Depth of the well (h) = 14 m
Width of the embankment = 5 m
Outer radius (R) = 5 + 5 = 10 m
Let the height of the embankment be “H”
Volume of Earth in the embankment = Volume of the well
πH(R2 – r2) = $$\pi r_{1}^{2} h$$
H(102 – 52) = 5 × 5 × 14
H (100 – 25) = 5 × 5 × 14
H = $$\frac{5 \times 5 \times 14}{75}$$ = 4.67 m
Height of the embankment = 4.67 m
Question 2.
A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the rise of the water in the glass?
Radius of the cylindrical glass (r) = 10 cm
Height of the water (h) = 9 cm
Radius of the cylindrical metal (R) = 5 cm
Height of the metal (H) = 4 cm
Let the height of the water raised be “h”
Volume of the water raised in the cylinder = Volume of the cylindrical metal
πr2h = πr2H
10 × 10 × h = 5 × 5 × 4
h = $$\frac{5 \times 5 \times 4}{10 \times 10}$$ = 1 cm
Raise of water in the glass = 1 cm
Question 3.
If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.
Circumference of the wooden piece = 484 cm
2πr = 484
2 × $$\frac{22}{7}$$ × r = 484 cm
r = $$\frac{484 \times 7}{2 \times 22}$$
r = 77 cm
Height of the wooden piece (h) = 105 cm
Volume of the conical wooden piece = $$\frac{1}{3} \pi r^{2} h$$ cu.units
= $$\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 \mathrm{cm}^{3}$$
= 22 × 11 × 77 × 35 cm3
= 652190 cm3
Volume of the wooden piece = 652190 cm3
Question 4.
A conical container is fully filled with petrol. The radius is 10m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.
The radius of the conical container (r) = 10 m
Height of the container (h) = 15 m
Question 5.
A right-angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.
Three sides of a triangle are 6 cm, 8 cm and 10 cm.
Case (i): If the triangle is revolved about the side 6 cm, the cone will be formed with radius 6 cm and height 8 cm.
Volume of the cone = $$\frac{1}{3} \pi r^{2} h$$ cu. units
= $$\frac{1}{3}$$ × π × 6 × 6 × 8 = 96π cm3
Case (ii): If the triangle is revolved about the side 8 cm, the cone will be formed with radius 8 cm and height 6 cm.
Volume of the cone = $$\frac{1}{3}$$ × π × 8 × 8 × 6 = 128π cm3
Difference in volume of the two solids = (128π – 96π) cm3 = 32π cm3 = 32 × $$\frac{22}{7}$$ cm3 = 100.57 cm3
The difference in the volume of the two solids = 100.57 cm3
Question 6.
The volumes of two cones of same base radius are 3600 cm3 and 5040 cm3. Find the ratio of heights.
Let the radius of the two cones be ‘r’
Let the height of the two cones be h1 and h2
Ratio of their volumes = 3600 : 5040 (÷ 10)
$$\frac{1}{3} \pi r^{2} h_{1}: \frac{1}{3} \pi r^{2} h_{2}$$ = 360 : 504 (÷4)
h1 : h2 = 90 : 126 (÷3)
= 30 : 42 (÷3)
= 10 : 14 (÷2)
h1 : h2 = 5 : 7
Ratio of heights = 5 : 7
Question 7.
If the ratio of radii of two spheres is 4 : 7, find the ratio of their volumes.
Let the ratio of their radii is r1 : r2
r1 : r2 = 4 : 7
Ratio of their volumes
V1 : V2 = $$\frac{4}{3} \pi r_{1}^{3}: \frac{4}{3} \pi r_{2}^{3}$$
= $$r_{1}^{3}: r_{2}^{3}$$
= 43 : 73
Ratio of their volumes = 64 : 343
Question 8.
A solid sphere and a solid hemisphere have an equal total surface area. Prove that the ratio of their volume is 3√3 : 4.
Total surface area of a sphere = $$4 \pi r_{1}^{2}$$ sq. units
Total surface area of a hemisphere = $$3 \pi r_{2}^{2}$$ sq. units
Ratio of Total surface area = $$4 \pi r_{1}^{2}: 3 \pi r_{2}^{2}$$
1 = $$\frac{4 \pi r_{1}^{2}}{3 \pi r_{2}^{2}}$$ (Same Surface Area)
1 = $$\frac{4 r_{1}^{2}}{3 r_{2}^{2}}$$
Ratio of their volumes = 3√3 : 4
Hence it is proved.
Question 9.
The outer and the inner surface areas of a spherical copper shell are 576π cm2 and 324π cm2 respectively. Find the volume of the material required to make the shell.
Outer surface area of a spherical shell = 576π cm2
4πR2 = 576π
4 × R2 = 576
R2 = $$\frac{576}{4}$$ = 144
R = √144 = 12 cm
Inner surface area of a spherical shell = 324π cm2
4πr2 = 324π
4r2 = 324
r2 = 81
r = √81 = 9
Volume of the material required = Volume of the hollow hemisphere = $$\frac{4}{3}$$ π(R3 – r3) cm3
Volume of the material required = 4186.29 cm3
Question 10.
A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of ₹ 40 per litre. |
Question Video: Finding the Size of an Angle Using the Properties of Vertically Opposite Angles | Nagwa Question Video: Finding the Size of an Angle Using the Properties of Vertically Opposite Angles | Nagwa
# Question Video: Finding the Size of an Angle Using the Properties of Vertically Opposite Angles Mathematics
What is the measure of β ππΉπ?
03:19
### Video Transcript
What is the measure of angle ππΉπ?
The first thing we want to do is identify angle ππΉπ. Angles can be named by three points. This one is named ππΉπ and that means this is the angle weβre interested in. We want to measure this space. To do that, weβll have to remember some rules about intersecting lines.
One of the rules weβll need to remember is that when two lines intersect, opposite angles have the same measure. And we call those opposite congruent angles βverticals angles.β Hereβs an example: angle π΄πΆπ΅ is an opposite angle to angle πΉπΆπ. So angle πΉπΆπ is equal. It has the same measure as angle π΄πΆπ΅. These two angles are vertical angles. We can label angle ππΆπΉ as 32 degrees.
Another rule that can help us here is the fact that the sum of the interior angles in a triangle measure 180 degrees. Inside this figure, we have a triangle β πΆπΉπ. And if we add up all the angles inside this triangle, they will equal 180 degrees. We have 32 degrees. This symbol represents a right angle. And right angles measure 90 degrees plus our unknown measure of angle ππΉπΆ is equal to 180 degrees.
We can add our first two angles together: 32 plus 90 equals 122 degrees. 122 degrees plus our missing angle has to equal 180 degrees. If we take 180 degrees and subtract the 122 degrees, we come up with 58 degrees. Angle ππΉπΆ has a measure of 58 degrees.
But we are still looking for the measure of angle ππΉπ, here in pink. To find that, weβll go back to the first rule we started with. We need to go back to what we know about vertical angles. When two lines intersect, opposite angles have the same measure. Our unknown angle ππΉπ is a vertical angle of ππΉπΆ. Angle ππΉπ must be equal to a measure of 58 degrees.
Angle ππΉπ equals 58 degrees.
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# NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 6 Maths Chapter 11 Algebra Exercise 11.1
Q.1 Find the rule which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write the rule.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) Pattern of letter = 2n (as two matchstick used in each letter)
(b) Pattern of letter = 3n (as three matchstick used in each letter)
(c) Pattern of letter = 3n (as three matchstick used in each letter)
(d) Pattern of letter = 2n (as two matchstick used in each letter)
(e) Pattern of letter = 5n (as five matchstick used in each letter)
(f) Pattern of letter = 5n (as five matchstick used in each letter)
(g) Pattern of letter = 6n (as six matchstick used in each letter)
Q.2 We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
The letter ‘T’ and ‘V’ that has pattern 2 , n since 2 matchsticks are used in all these letters. 2. Number of rows = n
Q.3 Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Number of rows = n
Cadets in each row = 5
Therefore, total number of cadets = 5n
Q.4 The teacher distributes 5 pencils per students. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Number of students = s
Number of pencils to each student = 5
Therefore, total number of pencils needed are = 5s
Q.6 A bird flies 1 kilometer in one minute. Can you express the distance covered by the birds in terms of its flying time in minutes? (Use t for flying time in minutes.)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Let t minutes be the flying times
Distance covered in one minute = 1 km
Distance covered in t minutes = Distance covered in one minute × Flying time = 1 × t
= t km
Q.7 Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Number of dots in a row = 9
Number of rows = r
Total number of dots in r rows
= Number of dots in a row × number of rows = 9r
Number of dots in 8 rows = 8 × 9 = 72
Number of dots in 10 rows = 10 × 9 = 90
Q.8 Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Let Radha’s age be x years
Leela’s age = 4 years younger than Radha
= (x – 4) years
Q.9 Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Number of laddus mother gave = l
Total number of laddus = number of laddus given away by mother + number of laddus remaining
Q.10 Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Number of oranges in a small box = x
Number of oranges in two small boxes = 2x
Number of oranges remained = 10
Number of oranges in large box = number of oranges in two small boxes + number of oranges remained
= 2x + 10
Q. 11 (a) Look at the following matchstick pattern of squares (Fig 11.6). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks
in terms of the number of squares. (Hint: If you remove vertical stick at the end, you will get a pattern of Cs)
(b) Fig 11.7 gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) We may observe that in the given matchstick pattern, the number of matchsticks are 4, 7, 10 and 13, which is 1 more than the thrice of the number of squares in the pattern
Therefore the pattern is 3x + 1, where x is the number of squares
(b) We may observe that in the given matchstick pattern, the number of matchsticks are 3, 5, 7 and 9 which is 1 more than the twice of the number of triangles in the pattern.
Therefore the pattern is 2x + 1, where x is the number of triangles.
## NCERT solutions for class 6 Maths Chapter 11 Algebra Exercise 11.2
Q.1 The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Side of equilateral triangle = l
Perimeter = l + l + l = 3l
Q.2 The side of the regular hexagon (Fig 11.10) is denoted by l. Express the perimeter of the hexagon using l.
(Hint: A regular hexagon has all its six sides equal in length.)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Side of a regular hexagon = l
Perimeter = l + l + l + l + l + 1 = 6l
Q.3 A cube is three dimensional figure as shown in Fig 11.11. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Length of an edge of the cube = l
Number of edges = 12
Total length of the edges
= Number of edges × length of an edge
=12l
Q.4 The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure (Fig 11.2) AB is a diameter of a circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Diameter = AB
= AC + CB = r + r = 2r
Q.5 To find sum of three numbers 14, 27 and 13 we can have two ways:
(a) We may first add 14 and 27 to get 41and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus, (14 + 27) + 13 = 14 + (27 + 13)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
For any three whole numbers a, b and c
(a + b) + c = a + (b + c)
## NCERT solutions for class 6 Maths Chapter 11 Algebra Exercise 11.3
Q.1 Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(i) 8 + (5 + 7)
(ii) 5 + (8 – 7)
(iii) 8 + (5 x 7)
(iv) 7 – (8 – 5)
(v) 7 x (8 + 5)
(vi) 5 x (8 + 7)
(vii) 8 x (5 + 7)
(viii) 7 + (8 – 5)
(ix) (5 x 7) – 8
(x) 7 + (8 x 5)
Q.2 Which out of the following are expressions with numbers only?
(a) y + 3
(b) (7 × 20) – 8z
(c) 5 (21 – 7) + 7 × 2
(d) 5
(e) 3x
(f) 5 – 5n
(g) (7 × 20) – (5 × 10) – 45 + p
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(c) and (d)
Q.3 Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
(a) z + 1, z – 1, y + 17, y – 17
(b) 17y, $$\frac{y}{17}$$ , 5z
(c) 2y + 17, 2y – 17
(d) 7m, -7m + 3, -7m – 3
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) z + 1 $$\rightarrow$$ Addition, z ? 1 $$\rightarrow$$ Subtraction
y + 17 $$\rightarrow$$ Addition , y ? 17 $$\rightarrow$$ Subtraction
(b) 17 y $$\rightarrow$$ Multiplication
$$\frac{y}{17}$$ $$\rightarrow$$ Division
5z $$\rightarrow$$ Multiplication
(c) 2y +17 $$\rightarrow$$ Multiplication and Addition
2y ? 17 $$\rightarrow$$ Multiplication and Subtraction
(d) 7m $$\rightarrow$$ Multiplication
?7m + m $$\rightarrow$$ Multiplication and Addition
?7m? 3 $$\rightarrow$$ Multiplication and Subtraction
Q.4 Give expressions for the following cases.
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from –m
(f) –p multiplied by 5
(g) –p divided by 5
(h) p multiplied by -5
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) p + 7
(b) p – 7
(c) 7p
(d) $$\frac{p}{7}$$
(e) -m – 7
(f) -5p
(g) $$\frac{-p}{5}$$
(h) 5p
Q.5 Give expressions in the following cases.
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by -8
(f) y is multiplied by -8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by -5 and the result is added to 16.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) 2m + 11
(b) 2m – 11
(e) 5y + 3
(d) 5y – 3
(e) -8y
(f) -8y+5
(g) 16 – 5y
(h) -5y + 16
Q.6 (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) (t + 4), (t – 4), 4t, $$\frac{t}{4}$$ , $$\frac{4}{t}$$, (4 – t), (4 + t)
(b) 2y + 7, 2y – 7, 7y + 2 , 7y - 2 , $$\frac{7y}{2}$$ , $$\frac{2y}{7}$$ , $$\frac{2}{7y}$$ , $$\frac{7}{2y}$$
## NCERT solutions for class 6 Maths Chapter 11 Algebra Exercise 11.4
(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is two year younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
(b) The length of a rectangular hall is 4 meters less than three times the breadth of the hall. What is the length, if the breadth is b meters?
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d) Meena, Beena and Reena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) Sarita’s age is given y years.
(i) After 5 years from now, her age will be (y + 5) years.
(ii) 3 years back from now, she was (y – 3) years of age.
(iii) Age of her grandfather = 6y years.
(iv) Age of her grandmother = (6y – 2) years.
(v) Sarita’s father’s age = (3y + 5) years.
(b) Let T be the length of the rectangular hall
$$\therefore$$ length = (3b – 4) metre
(c) Height of the rectangular box is ‘h’
$$\therefore$$ Length = 5h cm
and Breadth = (5h – 10) cm.
(d) Meena is at step s.
$$\therefore$$Beena is at (s + 8) steps and Leena is at (s – 7) steps.
Total number of steps on to the hill top = (4s – 10)
(e) Distance travelled by Bus in 5 hours = 5v km.
$$Distance from Daspur to Beespur = (5v + 20) km Q.2 Change the following statements using expressions into statements in ordinary language. (For example, Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.) (a) A notebook costs Rs p. A book costs Rs 3p (b) Tony put q marbles on the table. He has 8 q marbles in his box. (c) Our class has n students. The school has 20 n students. (d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old. (e) In an arrangement of dots there are r rows. Each row contains 5 dots NCERT Solutions for Class 6 Maths Chapter 11 Algebra Answer : (a) A book costs 3 times the cost of a notebook. (b) Tony has 8 times the number of marbles put on the table by him. (c) The school has 20 times the number of students in a class. (d) Jaggu’s uncle’s age is 4 times his age and his aunt’s age is 3 years less than the age of his uncle. (e) Number of dots in a row is 5 times the number of rows. Q.3 (a) Given Mannu’s age to be x years, Can you guess what (x – 2) may show? (Hint: Think of Mannu’s younger brother) can you guess what (x + 4) may now? What (3x + 7) may show? (b) Given Sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate? y + 7, y – 3, y + \(4\frac{1}{2}$$ , y – $$2\frac{1}{2}$$ .
(c) Given n students in the class like football, what may 2n show? What may $$\frac{n}{2}$$ show?
(Think of games other than football).
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) Given that Mannu’s age = x years.
$$\therefore$$ (x -2) years may be the age of her younger brother or younger sister.
(x + 4) years show the age of her elder brother or elder sister.
(3x + 7) years may be the age of her father, mother or uncle.
(b) y represents the age of Sara in years.
$$\therefore$$ y + 7 shows her future age.
y – 3 shows her past age.
y + $$4\frac{1}{2}$$ show her future age i.e., the age after four and half years.
y – $$2\frac{1}{2}$$ shows her past age i.e., the age before two and half years.
(c) Number of students who like football = n
$$\therefore$$ 2n = twice the number of football players may like to play cricket.
and $$\frac{n}{2}$$ = half of the number of football 2 players may like to play basket ball.
## NCERT solutions for class 6 Maths Chapter 11 Algebra Exercise 11.5
Q.1 State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 17
(b) (t – 7) > 5
(c) $$\frac{4}{2}$$ = 2
(d) (7 × 3) – 19 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) – (12 × 4)
(j) 7 = (11 × 2) + p
(k) 20 = 5y
(l) $$\frac{2q}{2}$$< 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3 × 5
(o) 7 – x = 5
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) An equation with variable x
(b) An inequality equation
(c) No, it’s a numerical equation
(d) No, it’s a numerical equation
(e) An equation with variable x
(f) An equation with variable x
(g) An inequality equation
(h) An equation with variable n
(i) No, it’s a numerical equation
(j) An equation with variable p
(k) An equation with variable y
(l) An inequality equation
(m) An inequality equation
(n) No, it’s a numerical equation
(o) An equation with variable x
Q.2 Complete the entries in the third column of the table.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) 10y = 80
y = 10 is not a solution for this equation because if y = 10,
10y = 10 × 10
= 100 $$\neq$$80
(b) 10y = 80
y = 8 is a solution for this equation because if y = 8,
10y = 10 × 8
= 80
$$\therefore$$Equation satisfied
(c) 10y = 80
y = 5 is not a solution for this equation because if y = 5,
10y = 10 × 5
= 50 $$\neq$$ 80
(d) 4l = 20
l = 20 is not a solution for this equation because if l = 20,
4l = 4 × 20
= 80 $$\neq$$20
(e) 4l = 20
l = 80 is not a solution for this equation because if l = 80,
4l = 4 × 80
= 320 $$\neq$$ 20
(f) 4l = 20
l = 5 is a solution for this eqaution because if l = 5,
4l = 4 × 5
= 20
$$\therefore$$ Equation satisfied
(g) b + 5 = 9
b = 5 is not a solution for this equation because if b = 5,
b + 5 = 5 + 5
= 10 $$\neq$$ 9
(h) b + 5 = 9
b = 9 is not a solution for this equation because if b = 9,
b + 5 = 9 + 5
= 14 $$\neq$$ 9
(i) b + 5 = 9
b = 4 is a solution for this equation because if b = 4,
b + 5 = 4 + 5
= 9
$$\therefore$$ Equation satisfied
(j) h – 8 = 5
h = 13 is a solution for this equation because if h = 13,
h – 8 = 13 – 8
= 5
$$\therefore$$ Equation satisfied
(k) h – 8 = 5
h = 8 is not a solution for this equation because if h = 8,
h – 8 = 8 – 8
= 0 $$\neq$$ 5
(l) h – 8 = 5
h = 0 is not a solution for this equation because if h = 0,
h – 8 = 0 – 8
= – 8 $$\neq$$ 5
(m) p + 3 = 1
p = 3 is not a solution for this equation because if p = 3,
p + 3 = 3 + 3
= 6 $$\neq$$ 1
(n) p + 3 = 1
p = 1 is not a solution for this equation because if p = 1,
p + 3 = 1 + 3
= 4 $$\neq$$ 1
(o) p + 3 = 1
p = 0 is not a solution for this equation because if p = 0,
p + 3 = 0 + 3
= 3 $$\neq$$ 1
(p) p + 3 = 1
p = -1 is not a solution for this equation because if p = – 1,
p + 3 = -1 + 3
= 2 $$\neq$$ 1
(q) p + 3 = 1
p = -2 is a solution for this equation because if p = -2,
p + 3 = -2 + 3
= 1
$$\therefore$$ Equation satisfied
Q.3 Pick out the solution from the values given in the bracket next to each equation.
Show that the other values do not satisfy the equation.
(a) 5m = 60 (10, 5, 12, 15)
(b) n + 12 (12, 8, 20, 0)
(c) p – 5 = 5 (0, 10, 5 ,– 5)
(d) $$\frac{q}{2}$$ = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4, -4, 8, 0)
(f) x + 4 = 2 (-2, 0, 2, 4)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(a) 5m = 60 (10, 5 , 12, 15) For m = 10, LHS = 5 x 10 = 50, RHS = 60 Here, LHS $$\neq$$RHS
$$\therefore$$m = 10 is not the solution of the equation
For m = 5, LHS = 5×5 = 25, RHS = 60
Here, LHS $$\neq$$ RHS
$$\therefore$$ m = 5 is not the solution of the equation
For m = 12, LHS = 5 x 12 = 60, RHS = 60
Here, LHS = RHS
$$\therefore$$ m = 12 is the solution of the equation
For m = 15 LHS = 5 x 15 = 75, RHS = 60
Here, LHS $$\neq$$ RHS
$$\therefore$$ m = 15 is not the solution of the equation
(b) n + 12 = 20 (12, 8, 20, 0)
For n = 12, LHS = 12 + 12 = 24, RHS = 20
Here, LHS $$\neq$$ RHS
$$\therefore$$ n = 12 is not the solution of the equation
For n = 8, LHS = 8 + 12 = 20, RHS = 20
Here, LHS = RHS
$$\therefore$$ n = 8 is the solution of the equation
For n = 20, LHS = 20 + 12 = 32, RHS = 20
Here, LHS $$\neq$$ RHS
$$\therefore$$ n = 20 is not the solution of the equation
For n = 0, LHS = 0 + 12 – 12, RHS = 20
Here, LHS $$\neq$$ RHS
$$\therefore$$ n= 0 is not the solution of the equation
(c) p – 5 = 5 (0, 10, 5, -5)
For p = 0, LHS = 0 – 5 = -5, RHS = 5
Here, LHS $$\neq$$ RHS
$$\therefore$$ p = 0 is not the solution of the equation
For p = 10, LHS = 10 – 5 = 5, RHS = 5
Here, LHS = RHS
$$\therefore$$ p = 10 is the solution of the equation
For p = 5, LHS = 5-5-0, RHS = 5
Here LHS $$\neq$$ RHS
$$\therefore$$ p = 5 is not the solution of the equation
For p = 5, LHS = 5 – 5 = 0, RHS = 5
Here, LHS $$\neq$$ RHS
$$\therefore$$ p = -5 is not the solution of the equation
(d) q/2 = 7 (7, 2, 10, 14)
For q = 7, LHS = 7/2 , RHS = 7
Here LHS $$\neq$$ RHS
$$\therefore$$ q = 7 is not the solution of the equation
For q = 2, LHS = 2/2 = 1, RHS = 7
Here, LHS $$\neq$$ RHS
$$\therefore$$ q = 2 is not the solution of the equation
For q = 10, LHS = 10/2 = 5, RHS = 7
Here, LHS $$\neq$$ RHS
For q = 14, LHS = 14/2 = 7, RHS = 7
Here, LHS = RHS
$$\therefore$$ q = 14 is the solution of the equation
(e) r – 4 = 0 (4, -4, 8, 0)
For r = 4, LHS = 4 – 4 = 0, RHS = 0
Here, LHS = RHS
$$\therefore$$ r = 4 is the solution of the equation
For r = -4, LHS = -4 – 4 = -8, RHS = 0
Here, LHS $$\neq$$ RHS
$$\therefore$$ r = -4 is not the solution of the equation
For r = 8, LHS = 8 – 4 = 4, RHS = 0
Here, LHS $$\neq$$ RHS
For r = 8 is not the solution of the equation
For r = 0, LHS = 0 – 4 = – 4, RHS = 0
Here, LHS $$\neq$$ RHS
$$\therefore$$ r = 0 is not the solution of the equation
(f) x + 4 = 2 (-2, 0, 2, 4)
For x = -2, LHS = -2 + 4 = 2, RHS = 2
Here, LHS – RHS
$$\therefore$$ x = -2 is the solution of the equation
For x = 0, LHS = 0 + 4 – 4, RHS = 2
Here, LHS $$\neq$$ RHS
$$\therefore$$ x = 0 is not the solution of the equation
For x = – 2, LHS = -2 + 4 – 6, RHS = 2
Here, LHS $$\neq$$ RHS
$$\therefore$$ x = 2 is not the solution of the equation
For r = 4, LHS = 4 + 4 = 8, RHS = 2
Here, LHS $$\neq$$ RHS
$$\therefore$$x = 4 is not the solution of the equation
Q.4 (a)Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.
(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35
(c) Complete the table and find the solution of the equation z / 3 = 4 using the table.
(d) Complete the table and find the solution to the equation m – 7 = 3.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Q.5 Solve the following riddles, you may yourself construct such riddles.
Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
To get exactly thirty four!
(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!
(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!
(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!
NCERT Solutions for Class 6 Maths Chapter 11 Algebra
(i) According to the condition,
I + 12 = 34 or x + 12 = 34
$$\therefore$$ By inspection, we have
22 + 12 = 34
So, I am 22.
(ii) Let I am ‘x’.
We know that there are 7 days in a week.
$$\therefore$$ upcounting from x for 7, the sum = 23
By inspections, we have
16 + 7 = 23
$$\therefore$$ x = 16
Thus I am 16.
(iii) Let the special number be x and there are 11 players in cricket team.
$$\therefore$$Special Number -6 = 11
$$\therefore$$ x – 6 = 11
By inspection, we get
17 – 6 = 11
$$\therefore$$x = 17
Thus I am 17.
(iv) Suppose I am ‘x’.
$$\therefore$$ 22 – I = I
or 22 – x = x
By inspection, we have
22 – 11 = 11
$$\therefore$$x = 11
Thus I am 11.
##### FAQs Related to NCERT Solutions for Class 6 Maths Chapter 11 Algebra
There are total 29 questions present in ncert solutions for class 6 maths chapter 11 algebra
There are total 2 long question/answers in ncert solutions for class 6 maths chapter 11 algebra
There are total 5 exercise present in ncert solutions for class 6 maths chapter 11 algebra |
# Reviewing Math Concepts Covered In The Math Monks Activity
Reviewing Math Concepts Covered in the Math Monks Activity
## Introduction
Math is a subject that many people find difficult, but it is also one that is essential in our daily lives. The Math Monks activity is a great way to learn and review math concepts in a fun and engaging way. In this blog post, we will review some of the math concepts covered in the Math Monks activity and provide some tips on how to master them.
## Algebra
Algebra is a branch of mathematics that deals with variables and symbols. In the Math Monks activity, you will learn how to solve equations, simplify expressions, and graph linear equations. One tip for mastering algebra is to practice, practice, practice! The more you practice, the more familiar you will become with the concepts and the easier it will be to solve problems.
### Example:
If x + 3 = 7, what is the value of x?
To solve this equation, we need to isolate x. We can do this by subtracting 3 from both sides of the equation:
x + 3 - 3 = 7 - 3
x = 4
## Geometry
Geometry is a branch of mathematics that deals with shapes, sizes, and positions of objects. In the Math Monks activity, you will learn about angles, triangles, circles, and more. One tip for mastering geometry is to visualize the shapes and objects in your mind. This will help you understand the relationships between them and make it easier to solve problems.
### Example:
What is the area of a circle with a radius of 5?
To find the area of a circle, we use the formula:
Area = πr²
Substituting the value of the radius, we get:
Area = π(5)² = 78.5
## Trigonometry
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. In the Math Monks activity, you will learn about sine, cosine, and tangent functions, as well as the Pythagorean theorem. One tip for mastering trigonometry is to use visual aids, such as diagrams or graphs, to help you understand the concepts.
### Example:
What is the length of the hypotenuse of a right triangle with legs of 3 and 4?
Using the Pythagorean theorem, we can find the length of the hypotenuse:
c² = a² + b²
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = 5
## Conclusion
The Math Monks activity is a great way to learn and review math concepts, whether you are a student or an adult looking to refresh your knowledge. By practicing regularly and using visual aids, you can master algebra, geometry, and trigonometry, and improve your math skills. |
# Chapter_2_Notes-Bittinger
Document Sample
``` CHAPTER 2
APPLICATIONS OF DIFFERENTIATION
In this chapter we will use derivatives to graph functions while finding maximums and minimums,
increasing and decreasing intervals, concavity intervals, and points of inflections. We will then look at
optimization applications where we can find the price to charge to maximize our profit. Also, we will
look at maximizing area given a certain perimeter.
(2.1) Using First Derivatives to find Max and Min Values
In this section we will use derivatives to find intervals of increasing and decreasing for a function which
will then lead us into finding relative extrema (maximums and minimums).
Let us begin by defining critical values:
Critical Values are values for x where f ( x) 0 or where f (x) is undefined.
The critical values are boundaries values that will define where a function is increasing and decreasing.
Critical values can also determine the relative extrema.
Relative Extrema: the relative maximum and minimum values of a function. Graphically:
To find extremas:
1) Find f (x)
2) Find the critical values, where f ( x) 0 or where f (x) is undefined
3) Using the critical values as boundaries, test each interval using f (x) to determine if the
function is increasing or decreasing and thus determine the relative extrema.
------------|-------------|------------|------------
c1 c2 c3
4) Find the ordered pair (x, y) using f(x)
Examples: Find the intervals where the function is increasing or decreasing and any relative extrema.
1) f ( x) 5 3x x²
x5
2) f ( x)
x3
3) f ( x) x 4 4 x ³ 20 x ² 12
(2.2) Using Second Derivatives to find Max and Min
In this section we will see that the second derivative, f ( x) ,also provides useful information about the
graph of f ( x) . We will look at the concavity of the graph.
CONCAVE UP: CONCAVE DOWN:
f is increasing as x increases f is decreasing as x increases
the slope of the tangent lines is increasing the slope of the tangent lines is decreasing
Test for Concavity:
1) If f ( x) 0 for all x in the interval, then f(x) is concave up
2) If f ( x) 0 for all x in the interval, then f(x) is concave down
Inflection Points: points where the graph changes from concave up to concave down. These will occur
where f ( x) 0 or is undefined. Graphically:
To Find Concavity Intervals and Points of Inflection:
1) Find where f ( x) 0 or is undefined
2) Using the values as boundaries, test each interval using f (x) to determine if the function is
concave up or concave down
* if f ( x) 0 then f(x) is concave up
* if f ( x) 0 then f(x) is concave down
3) Find any Points of Inflection (x, y) using f(x)
Examples: Find where the functions are concave up or down, and any points of inflection.
1) f ( x) x 3 3x 2 5 x 1
2) f ( x) x 4 20 x 3
Examples: Graph the following functions using the 1st and 2nd derivatives
1) f ( x) x³ 3x² 9 x 7
2) f ( x) x 4 8 x ³ 18 x ² 8
3) f ( x) (2 x 4) 5
1st DERIVATIVE 2nd DERIVATIVE
increasing and decreasing intervals concavity intervals
relative extremas relative extrema
critical points points of inflection
(2.3) Curve Sketching: Asymptotes and Rational Functions
We have looked at sketching continuous functions using the tools of calculus.. We now will look at
some discontinuous functions, most of which are rational functions.
P( x)
Rational Functions are in the form: f ( x) where P(x) and Q(x) are polynomials.
Q( x)
The graphs of Rational Functions typically involve Asymptotes. The following are definitions when
asymptotes occur.
VERTICAL ASYMPTOTE
The line x = a is a vertical asymptote if any of the following limit statements are true:
lim f ( x) lim f ( x) lim f ( x) lim f ( x)
x a x a x a x a
The graph will never cross a vertical asymptote. When the rational function is completely simplified,
vertical asymptotes will occur at values that make the denominator zero.
Examples:
5x
1) Determine the vertical asymptotes of the function: f ( x)
x 25
2
x2
2) Determine the vertical asymptotes of the function: f ( x)
x 6x 8
2
HORIZONTAL ASYMPTOTES
The line y = b is a horizontal asymptote if either or both the following limit statements is true:
lim f ( x) b lim f ( x) b
x x
The graph of a rational function may or may not cross a horizontal asymptote. Horizontal asymptotes
occur when the degree of the numerator is less than or equal to the degree of the denominator.
Examples:
2x
1) Determine the horizontal asymptote of the function: f ( x)
3x x 2
3
4 x3 3x 2
2) Determine the horizontal asymptote of the function: f ( x) 3
x 2x 4
NOTE:
1) When the degree of the numerator is the same as the degree of the denominator, the line y = a/b is a
horizontal asymptote. Where a is the leading coefficient of the numerator and b is the leading
coefficient of the denominator.
2) When the degree of the numerator is less than the degree of the denominator, the x-axis, or the line
y = 0, is the horizontal asymptote.
SKETCHING RATIONAL FUNCTIONS:
a) Find the x and y intercepts of the graph. Recall that the x-intercept is the points where
y f ( x) 0 and the y-intercepts is the points where x = 0.
b) Find all asymptotes (vertical and horizontal)
c) Use the first derivative to find relative extrema, increasing and decreasing intervals
d) Use the second derivative to find any points of inflections, concave up and concave down
intervals
e) Sketch the graph finding extra points if needed.
Examples:
2x 1
1) Sketch the graph of the function: f ( x)
x
x 1
2) Sketch the graph of the function: f ( x)
x2
2x2
3) Sketch the graph of the function: f ( x)
x 2 16
(2.4) Using Derivatives to find Absolute Extrema
ABSOLUTE EXTREMA:
Absolute Max: the largest value of a function in its domain
Absolute Min: the smallest value of a function in its domain
The Absolute Extrema exists on a closed interval [a, b] either at the endpoints of the domain or the
critical values. Graphically:
Finding Absolute Extrema:
1) Find all critical values, where f ( x) 0 or is undefined
2) Evaluate f at the critical values and the endpoints to find the functional value
3) Absolute Max exists at the largest f value
Absolute Min exists at the smallest f value
Examples: Find any Absolute Extrema
1) f ( x) x 2 4 x 5 on the interval [-1, 5]
2) f ( x) x 4 8 x 2 3 on the interval [-3, 0]
In this section we will look at optimization problems. Optimizing is finding the maximum or minimum
value where we can maximize profits, maximize revenue, or minimize costs.
Formulas to Use:
Profit = Revenue – Costs P( x ) R ( x ) C ( x )
Revenue = quantity x price R( x) x p( x)
R( x) C ( x) at maximum profit
Examples:
1. A company wants to build a parking lot along the side of one of its buildings using 800 feet of
fence. If the side along the building needs no fence, what are the dimensions of the largest
rectangle possible to maximize the area? What is the maximum area?
2. A farmer wants to make 2 identical adjoining rectangular enclosures along a straight river. If he
has 600 yards of fence, and if the sides along the river need no fence, what should be the
dimensions of each enclosure if the total area is to be maximized?
3. Find the maximum profit and the number of units that must be produced and sold in order to
yield the maximum profit given the following revenue, R(x) and cost, C(x) functions.
R( x) 5 x C ( x) 0.001 x 2 1.2 x 60
4. Riverside Appliance is marketing a new refrigerator. It determines that in order to sell x
refrigerators, the price per refrigerator must be p( x) 280 0.4 x . It also determines that the
total cost of producing x refrigerators is given by C ( x) 5000 0.6 x 2 .
a) Find the total revenue R(x).
b) Find the total profit P(x).
c) How many refrigerators must the company produce and sell in order to maximize profit?
d) What is the maximum profit?
e) What price per refrigerator must be charged in order to maximize profit?
5. A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at \$10, the
average attendance had been 27,000. When ticket prices were lowered to \$8, the average
attendance rose to 33,000. How should ticket prices be set to maximize revenue?
6. During the summer months Terry makes and sells necklaces on the beach. Last summer he sold
the necklaces for \$10 each and his sales averaged 20 per day. When he increased the price by \$1,
he found that he lost two sales per day. If the material for each necklace costs Terry \$6, what
should the selling price be to maximize his profit?
(2.6) Marginals and Differentials
R(x): Revenue Function – the income for selling x-units R(x) = price x quantity
C(x): Cost Function - the cost for producing x-units
P(x): Profit Function - the profit for selling x-units P(x) = R(x) – C(x)
When we take the derivative of the R(x), C(x), and the P(x) we get the rate of change in revenue, cost,
and profit. This is called the Marginal Revenue, Marginal Cost, and Marginal Profit.
MR(x): Marginal Revenue - the approximate revenue from the (x+1)st item.
MC(x): Marginal Cost – the approximate cost of the (x+1)st item.
MP(x): Marginal Profit – the approximate profit from the (x+1)st item.
Example:
1) Suppose that the daily cost, in dollars, of producing x radios is:
C ( x) 0.002 x3 0.1x 2 42 x 300
and currently 40 radios are produced daily.
a) What is the current daily cost?
b) What would be the additional daily cost of increasing production to 41 radios daily?
c) What is the marginal cost when x = 40?
d) Use marginal cost to estimate the daily cost of increasing production to 42 radios daily.
(2.7) Implicit Differentiation and Related Rates
y
Implicit differentiation is a techniques used to find when the function is not and can not be put in
x
y
the form y f (x) . That is, the y can not be isolated in the equation. If we need to find it is much
x
easier if the y is isolated.
y
Consider the function xy 1 to find
x
y
Consider the function x² y ² 1 to find
x
y
Consider the function x² y ³ y 0 to find
x
Examples:
y
1. Use implicit differentiation to find for: x² y ² 25
x
p
2) Use implicit differentiation to find for: p 2 p 2 x 40
x
y
3) Find and then find the slope of the curve at the given point: 2 x3 4 y 2 12 at (-2, -1)
x
y
4) Find and then find the slope of the curve at the given point: 2 x3 y 2 18 at (2, -1)
x
y
5) Evaluate at (6, 2) for the function xy y ² 12
x
6) Find the rate of change of profit with respect to time given the function:
R( x) 50 x 0.5 x 2 and C ( x) 10 x 3
when x = 10 and dx/dt = 20 units per day.
3
7. The number of traffic accidents per year in a city of population p is predicted to be A 0.002 p 2 . If
the population is growing by 500 people a year, find the rate at which traffic accidents will be rising
when the population is p=40,000.
8. A rocket fired straight up is being tracked by a radar station 3 miles from the launching pad. If the
rocket is traveling at 2 miles per second, how fast is the distance between the rocket and the tracking
station changing at the moment when the rocket is 4 miles up?
```
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## Three Popular Data Displays
This section elaborates on how to describe data. In particular, you will learn about the relative frequency histogram. Complete the exercises and check your answers.
### Relative Frequency Histograms
In our example of the exam scores in a statistics class, five students scored in the 80s. The number 5 is the frequency of the group labeled "80s". Since there are 30 students in the entire statistics class, the proportion who scored in the 80s is 5/30. The number 5/30, which could also be expressed as 0.16–≈.16670.16-≈.1667, or as 16.67%, is the relative frequency of the group labeled "80s". Every group (the 70s, the 80s, and so on) has a relative frequency. We can thus construct a diagram by drawing for each group, or class, a vertical bar whose length is the relative frequency of that group. For example, the bar for the 80s will have length 5/30 unit, not 5 units. The diagram is a relative frequency histogram for the data, and is shown in Figure 2.4 "Relative Frequency Histogram". It is exactly the same as the frequency histogram except that the vertical axis in the relative frequency histogram is not frequency but relative frequency.
Figure 2.4 Relative Frequency Histogram
The same procedure can be applied to any collection of numerical data. Classes are selected, the relative frequency of each class is noted, the classes are arranged and indicated in order on the horizontal axis, and for each class a vertical bar, whose length is the relative frequency of the class, is drawn. The resulting display is a relative frequency histogram for the data. A key point is that now if each vertical bar has width 1 unit, then the total area of all the bars is 1 or 100%.
Although the histograms in Figure 2.3 "Frequency Histogram" and Figure 2.4 "Relative Frequency Histogram" have the same appearance, the relative frequency histogram is more important for us, and it will be relative frequency histograms that will be used repeatedly to represent data in this text. To see why this is so, reflect on what it is that you are actually seeing in the diagrams that quickly and effectively communicates information to you about the data. It is the relative sizes of the bars. The bar labeled "70s" in either figure takes up 1/3 of the total area of all the bars, and although we may not think of this consciously, we perceive the proportion 1/3 in the figures, indicating that a third of the grades were in the 70s. The relative frequency histogram is important because the labeling on the vertical axis reflects what is important visually: the relative sizes of the bars.
When the size n of a sample is small only a few classes can be used in constructing a relative frequency histogram. Such a histogram might look something like the one in panel (a) of Figure 2.5 "Sample Size and Relative Frequency Histograms". If the sample size n were increased, then more classes could be used in constructing a relative frequency histogram and the vertical bars of the resulting histogram would be finer, as indicated in panel (b) of Figure 2.5 "Sample Size and Relative Frequency Histograms". For a very large sample the relative frequency histogram would look very fine, like the one in (c) of Figure 2.5 "Sample Size and Relative Frequency Histograms". If the sample size were to increase indefinitely then the corresponding relative frequency histogram would be so fine that it would look like a smooth curve, such as the one in panel (d) of Figure 2.5 "Sample Size and Relative Frequency Histograms".
Figure 2.5 Sample Size and Relative Frequency Histograms
It is common in statistics to represent a population or a very large data set by a smooth curve. It is good to keep in mind that such a curve is actually just a very fine relative frequency histogram in which the exceedingly narrow vertical bars have disappeared. Because the area of each such vertical bar is the proportion of the data that lies in the interval of numbers over which that bar stands, this means that for any two numbers a and b, the proportion of the data that lies between the two numbers a and b is the area under the curve that is above the interval (a,b) in the horizontal axis. This is the area shown in Figure 2.6 "A Very Fine Relative Frequency Histogram". In particular the total area under the curve is 1, or 100%.
Figure 2.6 A Very Fine Relative Frequency Histogram |
# Congruent Triangle Proofs (Part 1)
When two triangles are said to be congruent, there is a correspondence that matches each angle to a congruent angle and each side to a congruent side.
Here, ΔADC is congruent to ΔXZY. So we write ΔADC ≅ ΔXZY.
What if we are not told that one triangle is congruent to another? There are several ways to tell if two triangles are congruent. Let's take a look at two of the methods.
Method 1: SSS (Side, Side, Side)
To use this method, we need to show that each side of one triangle is congruent to a side in the second triangle.
In this example, side AB is congruent to side QR. Side AC is congruent to QP and side BC is congruent to side RP.
These two triangles are congruent because there are three pairs of congruent sides.
We use triangle congruence in mathematical proofs. Sometimes we will just need to show that two triangles are congruent. Other times, we will need to use the congruence to then show that some other fact about the triangles is also true.
Example #1:
Prove: < A ≅ < C
There are many triangles in this diagram. We are going to focus on only two of them. Here, we need to first show that ΔADE is congruent to ΔCED. We can then say that the corresponding parts of two congruent triangles are congruent in order to show that the angles are congruent.
Step 1: Set up two columns to show statements and reasons.
StatementsReasons
Step 2: Start to fill in the table with the given information.
StatementsReasons
1. AECD 1. Given
Step 3: Look for any other given information that could help show that the two triangles are congruent. We have been given two pairs of congruent sides, so we can look for a third pair to show that these triangles are congruent. In this case, side DE is the same as side ED in the triangles. We call this the reflexive property
StatementsReasons
1. AECD 1. Given
3. EDDE 3. Reflexive Property
Step 4: Show that the two triangles are congruent. We just showed that there are three pairs of congruent sides. Therefore, we used the SSS method.
StatementsReasons
1. AECD 1. Given
3. EDDE 3. Reflexive Property
4. ΔADE ≅ ΔCED 4. SSS
Step 5: Now that the two triangles are congruent, we can say that the corresponding side and corresponding angles are congruent. For the reason, we simplify this by just writing CPCTC which stands for "Corresponding parts of congruent triangles are congruent."
StatementsReasons
1. AECD 1. Given
3. EDDE 3. Reflexive Property
4. ΔADE ≅ ΔCED 4. SSS
5. < A ≅ < C 6. CPCTC
So by first showing that two triangles were congruent because they had three sets of congruent corresponding sides, we can then show that the corresponding angles are also congruent.
Related Links: Math Geometry Triangles Congruent Triangles
To link to this Congruent Triangle Proofs (Part 1) page, copy the following code to your site: |
×
# Degrees of Freedom
Top
FAQ
## What is Degrees of Freedom?
View Notes
Degrees of freedom definition is a mathematical equation used principally in statistics, but also in physics, mechanics, and chemistry. In a statistical calculation, the degrees of freedom illustrates the number of values involved in a calculation that has the freedom to vary. The degrees of freedom can be computed to ensure the statistical validity of t-tests, chi-square tests and even the more elaborated f-tests.
In this lesson, we will explore how degrees of freedom can be used in statistics to identify if outcomes are significant.
### Use of Degrees of Freedom
Tests like t-tests, chi-square tests are frequently used to compare observed data with data that would be anticipated to be obtained as per a particular hypothesis.
### Degrees of Freedom Example
Examples of how degrees of freedom can enter statistical calculations are the t-tests and chi-squared tests. There are a number of t-tests and chi-square tests which can be differentiated with the help of degrees of freedom.
Let’s consider a degree of freedom example. Suppose a medicinal trial is carried out on a group of patients and it is postulated that the patients receiving the medication would display increased heartbeat in comparison to those that did not receive the medication. The outcome of the test could then be evaluated to identify whether the difference in heart rates is considered crucial, and degrees of freedom are part of the computations.
### Degrees of Freedom Formula
The statistical formula to find out how many degrees of freedom are there is quite simple. It implies that degrees of freedom is equivalent to the number of values in a data set minus 1, and appears like this:
df = N-1
Where, N represents the number of values in the data set (sample size).
That being said, let’s have a look at the sample calculation.
If there is a data set of 6, (N=6).
Call the data set X and make a list with the values for each data.
For this example data, set X of the sample size includes: 10, 30, 15, 25, 45, and 55
This data set has a mean, or average of 30. Find out the mean by adding the values and dividing by N:
(10 + 30 + 15 + 25 + 45 + 55)/6= 30
Using the formula, the degrees of freedom will be computed as df = N-1:
In this example, it appears, df = 6-1 = 5
This further implies that, in this data set (sample size), five numbers contain the freedom to vary as long as the mean remains 30.
### Critical Values
Having the awareness of the degrees of freedom for a sample or the population size does not provide us a whole lot of substantial information by itself. This is because after we perform computations of the degrees of freedom, which are actually the number of values in a calculation that we can vary, it is essential to look up the critical values for our equation with the help of a critical value table. Note that these tables can be found online or in textbooks. When using a critical value table, the values found in the table identify the statistical significance of the outcomes.
### Solved Examples on Finding How Many Degrees of Freedom
Now that we know the degree of freedom meaning, let's get to learn how to find the degrees of freedom.
Example:
Evaluate the Degree of Freedom For a Given Sample or Sequence: x = 3, 6, 2, 8, 4, 2, 9, 5, 7, 12
Solution:
Given n= 10
Thus,
DF = n-1
DF = 10-1
DF = 9
Example:
Determine the Degree of Freedom For the Sequence Given Below:
x = 12, 15, 17, 25, 19, 26, 35, 46
y = 18, 32, 21, 43, 22, 11
Solution:
Given: n1 = 8 n2 = 6
Here, there are 2 sequences, so we require to apply DF = n1 + n2 – 2
DF = 8 + 6 -2
DF = 12
### Fun Facts
• Since degrees of freedom calculations determine the number of values in the final calculation, they are allowed to vary, and to even contribute to the validity of a result.
• Degree of freedom calculations are typically dependent upon the sample size, or observations, and the criterions to be estimated, but usually, degree of freedom mathematics and statistics equals the number of observations minus the number of criterion/parameter.
• There will be more degrees of freedom with a larger size of sample.
FAQ (Frequently Asked Questions)
1. What are the Formulas of Calculating the Degrees of Freedom?
Answer: There are several formulas to calculate degrees of freedom with respect to sample size. Following are the formulas to calculate degrees of freedom based on sample:
• One Sample T Test Formula : DF = n−1
• Two Sample T Test Formula: DF = n1 + n2 − 2
• Simple Linear Regression Formula: DF = n−2
• Chi Square Goodness of Fit Test Formula: DF = k−1
• Chi Square Test for Homogeneity Formula: DF = (r−1)(c−1)
2. How are Degrees of Freedom Used in Standard Deviation?
Answer: Another place where degrees of freedom occur is in the standard deviation formula. This appearance is not as clear and apparent, but we can notice it if we know where to look. To determine a standard deviation we are looking at the "average" deviation from the mean. However, after we do subtraction of the mean from each data value and squaring the differences, we end up dividing by n-1 instead of n as we might anticipate.
The occurrence of the n-1 takes place from the number of degrees of freedom. Because the sample mean and the n data values are being used in the formula, there are n-1 degrees of freedom.
3. How are Degrees of Freedom Used in Advanced Statistical Techniques?
Answer: More advanced statistical techniques apply more complex ways of counting the degrees of freedom. When computing the test statistic for two means having independent samples of n1 and n2 elements, the number of degrees of freedom consists of a little complicated formula. It can be calculated using the smaller of n1-1 and n2-1.
Another example of counting the degrees of freedom shows up with an F test. In carrying out an F test we have k samples each of size n—the degrees of freedom in the numerator will be k-1 and in the denominator will be k(n-1). |
☰ अधिक माहितीसाठी येथे क्लीक करा
#### Question 1:
Convert into improper fractions.
(i)
$7\frac{2}{5}$
(ii)
$5\frac{1}{6}$
(iii)
$4\frac{3}{4}$
(iv)
$2\frac{5}{9}$
(v)
$1\frac{5}{7}$
#### Answer 1:
(i)
$7\frac{2}{5}=\frac{5×7+2}{5}\phantom{\rule{0ex}{0ex}}=\frac{35+2}{5}\phantom{\rule{0ex}{0ex}}=\frac{37}{5}$
(ii)
$5\frac{1}{6}=\frac{6×5+1}{6}\phantom{\rule{0ex}{0ex}}=\frac{30+1}{6}\phantom{\rule{0ex}{0ex}}=\frac{31}{6}$
(iii)
$4\frac{3}{4}=\frac{4×4+2}{4}\phantom{\rule{0ex}{0ex}}=\frac{16+2}{4}\phantom{\rule{0ex}{0ex}}=\frac{18}{4}$
(iv)
$2\frac{5}{9}=\frac{9×2+5}{9}\phantom{\rule{0ex}{0ex}}=\frac{18+5}{9}\phantom{\rule{0ex}{0ex}}=\frac{23}{9}$
(v)
$1\frac{5}{7}=\frac{7×1+5}{7}\phantom{\rule{0ex}{0ex}}=\frac{7+5}{7}\phantom{\rule{0ex}{0ex}}=\frac{12}{7}$
#### Question 2:
Convert into mixed numbers.
(i)
$\frac{30}{7}$
(ii)
$\frac{7}{4}$
(iii)
$\frac{15}{12}$
(iv)
$\frac{11}{8}$
(v)
$\frac{21}{4}$
(vi)
$\frac{20}{7}$
#### Answer 2:
(i)
$\frac{30}{7}=\frac{28+2}{7}\phantom{\rule{0ex}{0ex}}=\frac{28}{7}+\frac{2}{7}\phantom{\rule{0ex}{0ex}}=4+\frac{2}{7}\phantom{\rule{0ex}{0ex}}=4\frac{2}{7}$
(ii)
$\frac{7}{4}=\frac{4+3}{4}\phantom{\rule{0ex}{0ex}}=\frac{4}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=1\frac{3}{4}$
(iii)
$\frac{15}{12}=\frac{12+3}{12}\phantom{\rule{0ex}{0ex}}=\frac{12}{12}+\frac{3}{12}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{12}\phantom{\rule{0ex}{0ex}}=1\frac{3}{12}$
(iv)
$\frac{11}{8}=\frac{8+3}{8}\phantom{\rule{0ex}{0ex}}=\frac{8}{8}+\frac{3}{8}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{8}\phantom{\rule{0ex}{0ex}}=1\frac{3}{8}$
(v)
$\frac{21}{4}=\frac{20+1}{4}\phantom{\rule{0ex}{0ex}}=\frac{20}{4}+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=5+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=5\frac{1}{4}$
(vi)
$\frac{20}{7}=\frac{14+6}{7}\phantom{\rule{0ex}{0ex}}=\frac{14}{7}+\frac{6}{7}\phantom{\rule{0ex}{0ex}}=2+\frac{6}{7}\phantom{\rule{0ex}{0ex}}=2\frac{6}{7}\phantom{\rule{0ex}{0ex}}$
#### Question 3:
Write the following examples using fractions.
(i) If 9 kg rice is shared amongst 5 people, how many kilograms of rice does each person get?
( ii) To make 5 shirts of the same size, 11 metres of cloth is needed. How much cloth is needed for one shirt?
#### Answer 3:
(i) If 9 kg rice is shared amongst 5 people, then each person will get
$\frac{9}{5}$
kilograms of rice.
( ii) If 11 metres of cloth is needed to make 5 shirts of the same size, then one shirt will need
$\frac{11}{5}$
metres of cloth.
#### Question 1:
Add.
(i)
$6\frac{1}{3}+2\frac{1}{3}$
(ii)
$1\frac{1}{4}+3\frac{1}{2}$
(iii)
$5\frac{1}{5}+2\frac{1}{7}$
(iv)
$3\frac{1}{5}+2\frac{1}{3}$
#### Answer 1:
(i)
$6\frac{1}{3}+2\frac{1}{3}=\frac{6×3+1}{3}+\frac{2×3+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{18+1}{3}+\frac{6+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{19}{3}+\frac{7}{3}$
$=\frac{19+7}{3}\phantom{\rule{0ex}{0ex}}=\frac{26}{3}\phantom{\rule{0ex}{0ex}}=\frac{24+2}{3}$
$=\frac{24}{3}+\frac{2}{3}\phantom{\rule{0ex}{0ex}}=8+\frac{2}{3}\phantom{\rule{0ex}{0ex}}=8\frac{2}{3}$
(ii)
$1\frac{1}{4}+3\frac{1}{2}=\frac{1×4+1}{4}+\frac{3×2+1}{2}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\frac{7}{2}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\frac{7×2}{2×2}$
$=\frac{5}{4}+\frac{14}{4}\phantom{\rule{0ex}{0ex}}=\frac{5+14}{4}\phantom{\rule{0ex}{0ex}}=\frac{19}{4}\phantom{\rule{0ex}{0ex}}=\frac{16+3}{4}$
$=\frac{16}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=4+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=4\frac{3}{4}$
(iii)
$5\frac{1}{5}+2\frac{1}{7}=\frac{5×5+1}{5}+\frac{2×7+1}{7}\phantom{\rule{0ex}{0ex}}=\frac{26}{5}+\frac{15}{7}\phantom{\rule{0ex}{0ex}}=\frac{26×7}{5×7}+\frac{15×5}{7×5}\phantom{\rule{0ex}{0ex}}$
$=\frac{182}{35}+\frac{75}{35}\phantom{\rule{0ex}{0ex}}=\frac{182+75}{35}\phantom{\rule{0ex}{0ex}}=\frac{257}{35}$
$=\frac{245+12}{35}\phantom{\rule{0ex}{0ex}}=\frac{245}{35}+\frac{12}{35}\phantom{\rule{0ex}{0ex}}=7+\frac{12}{35}\phantom{\rule{0ex}{0ex}}=7\frac{12}{35}$
(iv)
$3\frac{1}{5}+2\frac{1}{3}$
$3\frac{1}{5}+2\frac{1}{3}=\frac{3×5+1}{5}+\frac{2×3+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{16}{5}+\frac{7}{3}\phantom{\rule{0ex}{0ex}}=\frac{16×3}{5×3}+\frac{7×5}{3×5}\phantom{\rule{0ex}{0ex}}$
$=\frac{48}{15}+\frac{35}{15}\phantom{\rule{0ex}{0ex}}=\frac{48+35}{15}\phantom{\rule{0ex}{0ex}}=\frac{83}{15}$
$=\frac{75+8}{15}\phantom{\rule{0ex}{0ex}}=\frac{75}{15}+\frac{8}{15}\phantom{\rule{0ex}{0ex}}=5+\frac{8}{15}\phantom{\rule{0ex}{0ex}}=5\frac{8}{15}$
#### Question 2:
Subtract.
(i)
$3\frac{1}{3}-1\frac{1}{4}$
(ii)
(iii)
(iv)
#### Answer 2:
(i)
$3\frac{1}{3}-1\frac{1}{4}=\frac{10}{3}-\frac{5}{4}\phantom{\rule{0ex}{0ex}}=\frac{10×4}{3×4}-\frac{5×3}{4×3}\phantom{\rule{0ex}{0ex}}=\frac{40}{12}-\frac{15}{12}\phantom{\rule{0ex}{0ex}}=\frac{40-15}{12}$
$=\frac{25}{12}\phantom{\rule{0ex}{0ex}}=\frac{24+1}{12}\phantom{\rule{0ex}{0ex}}=\frac{24}{12}+\frac{1}{12}\phantom{\rule{0ex}{0ex}}=2+\frac{1}{12}\phantom{\rule{0ex}{0ex}}=2\frac{1}{12}$
(ii)
$=\frac{33-20}{6}\phantom{\rule{0ex}{0ex}}=\frac{13}{6}\phantom{\rule{0ex}{0ex}}=2\frac{1}{6}$
(iii)
$=\frac{285-244}{40}\phantom{\rule{0ex}{0ex}}=\frac{41}{40}\phantom{\rule{0ex}{0ex}}=1\frac{1}{40}$
(iv)
$=\frac{75-32}{10}\phantom{\rule{0ex}{0ex}}=\frac{43}{10}\phantom{\rule{0ex}{0ex}}=4\frac{3}{10}$
#### Question 3:
Solve.
(1) Suyash bought
$2\frac{1}{2}$
kg of sugar and Ashish bought
$3\frac{1}{2}$
kg. How much sugar did they buy altogether? If sugar costs 32 rupees per kg, how much did they spend on the sugar they bought?
(2) Aradhana grows potatoes in
$\frac{2}{5}$
part of her garden, greens in
$\frac{1}{3}$
part and brinjals in the remaining part. On how much of her plot did she plant brinjals?
(3) Sandeep filled water in
$\frac{4}{7}$
of an empty tank. After that, Ramakant filled
$\frac{1}{4}$
part more of the same tank. Then Umesh used
$\frac{3}{14}$
part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?
#### Answer 3:
(1) The amount of sugar they bought altogether =
$2\frac{1}{2}+3\frac{1}{2}$
$=\frac{5}{2}+\frac{7}{2}\phantom{\rule{0ex}{0ex}}=\frac{5+7}{2}\phantom{\rule{0ex}{0ex}}=\frac{12}{2}$
= 6 kg
Now, cost of 1 kg of sugar = Rs 32
Therefore, the cost of 6 kg of sugar is = 6 × 32
= Rs 192
Hence, they spend Rs 192 on the sugar they bought.
(2) The part of the garden in which Aradhana grew brinjals is given by
$1-\frac{2}{5}-\frac{1}{3}$
$=\frac{1×15}{1×15}-\frac{2×3}{5×3}-\frac{1×5}{3×5}\phantom{\rule{0ex}{0ex}}=\frac{15}{15}-\frac{6}{15}-\frac{5}{15}\phantom{\rule{0ex}{0ex}}=\frac{15-6-5}{15}\phantom{\rule{0ex}{0ex}}=\frac{4}{15}$
Hence, Aradhana grew brinjals in
$\frac{4}{15}$
part of her garden.
(3) The amount of water will be left in the tank is given by
$\frac{4}{7}\left(560\right)+\frac{1}{4}\left(560\right)-\frac{3}{14}\left(560\right)$
Hence, 340 l of water will be left in the tank.
Std 6 th Mathematics- 4. Operations on Fractions page 1 Reviewed by Amol Uge on March 02, 2019 Rating: 5
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# DIMENSIONS AND UNITS Dimensions and Units: A dimension is a property that can be measured, such as length, time, mass
, temperature, velocity (length/time). A unit is a mean of expressing a dimension, such as meters or feet for length, seconds or hours for time. Two of the most commonly systems of units are: 1. SI – Le Systeme Internationale d’Unites. 2. AE – American Engineering system of units. Fundamental (or basic) dimensions/units are those that can be measured independently such as mass, length, time and temperature. Derived dimensions/units are those that can be developed from the fundamental dimensions/units, such as area (m2 or ft2) or velocity (m/s or ft/s). A summary of fundamental and derived units in both the SI and the AE systems are listed in Tables 1 and 2.
1
Table 1: SI Units (Himmelblau & Riggs).
2
Table 2: AE Units (Himmelblau & Riggs).
3
Conversion of Units: An engineer has to be able to convert units from one system to another. The procedure is done by multiplying or dividing by conversion factors. The procedure described here is done by multiplying by 1. For example:
1in = 2.54 cm
(1)
Therefore:
1in =1 2.54 cm So, that 10 cm are: 1in 10 cm = 3.94 in 2.54 cm (3) (2)
Example 1.2 (H&R): Convert 400 in3/day to cm3/min.
400 in 3 ⎛ 2.54 cm ⎞ 1day 1 hr = 4.55 cm3/min (4) ⎜ ⎟ 1day ⎝ 1in ⎠ 24 hr 60 min
4
3
The gc Conversion Factor: In the past you used to write Newton’s second law of motion in the following wrong way:
F = ma
(5)
Where F is force, m is mass and a is the acceleration. However: N ≠ kg⋅m/s2. In order to make the equation dimensionally consistent, we introduce a conversion factor:
F= 1 ma gc
(6)
The force of 1 N is required to accelerate a mass of 1 kg at a rate of 1 m/s2. Therefore, in SI units the conversion factor is gc = 1 kg⋅m/N⋅s2. In the AE system gc = 32.2 lbm⋅ft/ lbf⋅s2. Most of the books do not place the gc in the equations. However, you will find out that this is essential, especially when working with the AE system.
5
Significant Figures: A measurement should include the following: (a) the magnitude of the variable being measured, (b) its units and (c) an estimate of its uncertainty.
When we have no idea of the accuracy of measurement we can do the following. For example, 1.43 indicates 1.43 ± 0.005, so the value can be between 1.425 and 1.435. Another interpretation is 1.43 ± 0.01. The numbers 81, 81.0 and 81.00 are different. The first one has 2 significant figures, the second one has 3 significant figures and the third one has 4 significant figures. The number 2300 has 2 significant figures and the number 23,040 has 4 significant figures. When we multiply or divide numbers, we should retain in the final answer the lowest number of significant figures among all the numbers involved. For example:
1.47 ⋅ 3.0926 = 4.54612 = 4.55
(7)
6
Consider the following example:
98 = 1.05365 = 1.1 93.01
(8)
If we follow the rule, the answer is a distortion of the true error. Since 98 ± 1 has an error of about 1% while the result 1.1 ± 0.1 has an error of about 10%. This time 1.05 may reflect a better answer. Avoid increasing the precision of your answer very much over the precision in your measurements. Also, use some common sense! When adding or subtracting, the number with the lowest decimal places will determine the number of digits in the solution. For example:
110.3 + 0.038 = 110.338 = 110.3
(9)
Finally, a rule of thumb for rounding off numbers in which the digit to be dropped is 5 is always to make the last digit an even number. For example:
1.35 → 1.4
(10)
7 |
Find Inverse Function - Tutorial
Examples on how to find inverse functions analytically are presented. Detailed solutions and matched exercises with answers at the end of this page are also included.
Examples with Detailed Solutions
Example 1
Find the inverse function of the linear function f given by
f(x) = 2x + 3
Solution to example 1
write the function as an equation.
y = 2x + 3
solve for x.
x = (y - 3)/2
now write f-1(y) as follows .
f
-1(y) = (y - 3)/2
or
f
-1(x) = (x - 3)/2
Check
f(f
-1(x)) = 2(f -1(x)) + 3
= 2 ((x - 3)/2) + 3
= (x - 3) + 3
= x
f
-1(f(x)) = f -1(2x+3)
= ((2x + 3) - 3)/2
= 2 x / 2
= x
conclusion: The inverse of function f given above is
f
-1(x) = (x - 3)/2
Matched Exercise 1
Find the inverse function of f given by
f(x) = - x - 4
Example 2
Find the inverse function of the quadratic function f given by
f(x) = (x - 3)2, if x ≥ 3
Solution to example 2
write the function as an equation.
y = (x - 3)
2
solve for x, two solutions .
x = 3 + √y
x = 3 - √y
the first solution is selected since x ≥ 3, write f-1(y) as follows.
f
-1(y) = 3 + √y
or
f
-1(x) = 3 + √x
Check
f(f
-1(x))=((3 + √x) - 3)2
= (√x)
2
= x
f
-1(f(x)) = 3 + √((x - 3)2)
simplifies to
= 3 + |x - 3| ( because x ≥ 3 implies x - 3 ≥ 0 which implies |x - 3| = x - 3)
= 3 + (x - 3)
= x
conclusion
The inverse of function f given above is f
-1(x) = 3 + √x
Matched Exercise 2
Find the inverse function of f given by
f(x) = (x + 1)2, if x ≥ -1
Example 3
Find the inverse function of the rational function f given by
f(x) = (x + 1)/(x - 2)
Solution to example 3
Write the function as an equation.
y = (x + 1) / (x - 2)
Multiply both sides of the above equation by x - 2 and simplify.
y (x - 2) = x + 1
Multiply and group.
y x - 2 y = x + 1
y x - x = 2 y + 1
Factor x on the left side and solve
x (y - 1) = 1 + 2 y
x = (1 + 2 y) / (y - 1)
Change x to y and y to x
y = (1 + 2 x) / (x - 1)
The inverse of function f given above is
f
-1(x) = (1 + 2 x) / (x - 1)
Matched Exercise 3
Find the inverse function of f given by
f(x) = (x + 1)/(x - 1
Example 4
Find the inverse function of the logarithmic function f given by
f(x) = ln(x + 2) - 3
Solution to example 4
Write the function as an equation.
y = ln(x + 2) - 3
Rewrite the equation so that it is easily solved for x.
ln(x + 2) = y + 3
Rewrite the above in exponential form.
x + 2 = e
y + 3
Solve for x.
x = e
y + 3 - 2
now write f-1(y) as follows .
f
-1(y) = ey + 3 - 2
or change x into y and y into x in the above to have
f
-1(x) = ex + 3 - 2
Check
f(f
-1(x)) = f(x) = ln(ex + 3 - 2 + 2) - 3
= ln(e
x + 3) - 3
= (x + 3) - 3
= x
f
-1(f(x))=f -1(2x + 3)
= e
(ln(x + 2) - 3) + 3 - 2
= e
ln(x + 2) - 2
= x + 2 - 2
= x
conclusion: The inverse of function f given above is f
-1(x) = ex + 3 - 2
Matched Exercise 4
Find the inverse function of f given by
f(x) = 2 ln(x + 4) - 4
Answer to Matched Exercises
Answer to Matched Exercise 1
f
-1(x) = - x - 4
Answer to Matched Exercise 2
f
-1(x) = - 1 + √x
Answer to Matched Exercise 2
f
-1(x) = (x + 1)/(x - 1)
Answer to Matched Exercise 4
f
-1(x) = ex / 2 + 2 - 4
More References and Links to Inverse Functions
Find the Inverse Functions - Calculator
Applications and Use of the Inverse Functions
Find the Inverse Function - Questions
Inverse of Quadratic Functions.
Definition of the Inverse Function - Interactive Tutorial
Find Inverse Of Cube Root Functions.
Find Inverse Of Square Root Functions.
Find Inverse Of Logarithmic Functions.
Find Inverse Of Exponential Functions.
Step by Step Solver Calculator to Find the Inverse of a Linear Function.
Find the Inverse of a Square Root Function.
Find the Inverse of a Cubic Function.
Find the Inverse of a Rational Function. |
# CBSE Class 10th Math 15 – Probability MCQs
#### The possible outcomes, when a coin is tossed are H and T.
Correct! Wrong!
When a coin is tossed, either a head or a tail comes up. Hence, the possible outcomes are H and T.
#### If a pair of dice is thrown simultaneously, then the possible outcomes are ____.
Correct! Wrong!
When a pair of dice is thrown simultaneously, the possible outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) and (6, 6). Total number of possible outcomes = 36
#### If two dice are thrown simultaneously, then the probability that the sum of the two numbers appearing on them is 6, is ____.
Correct! Wrong!
Total number of outcomes, when two dice are thrown simultaneously = 36 ⇒ Number of possible outcomes = 36 Let E be the event of getting the sum 6. The outcomes favorable to E are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3). ⇒ Number of outcomes favorable to E = 5 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 5/36 Hence, the required probability is 5/36.
#### If two dice are thrown simultaneously, then the probability that the sum of the two numbers appearing on them is 12, is ____.
Correct! Wrong!
Total number of outcomes, when two dice are thrown simultaneously = 36 ⇒ Number of possible outcomes = 36 Let E be the event of getting the sum 12. The outcome favorable to E is (6, 6). ⇒ Number of outcomes favorable to E = 1 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 1/36 Hence, the required probability is 1/36.
#### A box contains 100 toys. Out of which 84 are good, 12 have minor defects and 4 have major defects. Pavel, a shop keeper accepts only good toys and Valentina, another shop keeper rejects only the toys with major defects. If one toy is drawn at random from the box, then the probability that it is acceptable to Pavel is ____.
Correct! Wrong!
Total number of toys = 100 ⇒ Number of possible outcomes = 100 Let E be the event that a toy is acceptable to Pavel. Number of good toys = 84 ⇒ Number of outcomes favorable to E = 84 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 84/100 = 0.84 Hence, the required probability is 0.84.
#### A box contains 12 balls, out of which 5 are black. If one ball is drawn at random, then the probability of getting a black ball is ____.
Correct! Wrong!
Total number of balls = 12 ⇒ Number of possible outcomes = 12 Let E be the event of getting a black ball. Number of black balls = 5 ⇒ Number of outcomes favorable to E = 5 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 5/12 Hence, the required probability is 5/12.
#### John and Ray are friends. The probability that both will have the birthday on same day in a non-leap year is ____.
Correct! Wrong!
Number of days in a non-leap year = 365 ⇒ Total number of possible outcomes = 365 Let E be the event that John and Ray will have the birthday on same day. The number of favorable outcomes that John and Ray will have the same birthday is 1. P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 1/365 Hence, the probability that both will have different birthdays in a non-leap year is 1/365.
#### Two coins are tossed simultaneously for 1000 times. If the frequencies of two tails, one tail and no tail are 210, 550 and 240 respectively, then the probability of obtaining one tail is ____.
Correct! Wrong!
Number of possible outcomes = 1000 Let E be the event of getting one tail. Frequency of one tail in the experiment = 550 ⇒ Number of outcomes favorable to E = 550 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 550/1000 = 0.55 Hence, the required probability is 0.55.
#### If one card is drawn from a well shuffled deck of 52 cards, then the probability of getting a red card is ____.
Correct! Wrong!
Total number of cards = 52 ⇒ Number of possible outcomes = 52 Let E be the event of getting a red card. Number of red cards = 26 ⇒ Number of outcomes favorable to E = 26 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment =26/52 = 1/2 Hence, the required probability is 1/2.
#### If two dice are thrown simultaneously, then the probability of getting a doublet is ____.
Correct! Wrong!
Total number of outcomes, when two dice are thrown simultaneously = 36 ⇒ Number of possible outcomes = 36 Let E be the event of getting a doublet. The outcomes favorable to E are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6). ⇒ Number of outcomes favorable to E = 6 P(E) = Number of outcomes favorable to E/Number of all possible outcomes of the experiment = 6/36 = 1/6 Hence, the required probability is 1/6.
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CBSE Class 10th Math 15 - Probability MCQs
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Subtracting Fractions on a Number Line — Let's Practice!
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Basics on the topicSubtracting Fractions on a Number Line — Let's Practice!
Today we are practicing subtracting fractions with Razzi! This video contains examples to help you further practice and grow confident in this topic.
TranscriptSubtracting Fractions on a Number Line — Let's Practice!
Razzi says get these items ready, because today we're going to practice subtracting fractions on a number line. Using a number line, solve six eighths minus three eighths. Pause the video to work on the problem, and press play when you are ready to see the solution! Both denominators are eight, so divide the number line into eight equal parts. Now locate six eighths, which is here. Count back the numerator of the second fraction, which is three. Did you also get three eighths? Let's tackle the next problem! Using a number line, solve four fourths minus three fourths. Pause the video to work on the problem, and press play when you are ready to see the solution! Both denominators are four, so divide the number line into four equal parts. One whole is the same as four fourths, so start here. Count back the numerator of the second fraction, which is three. Did you also get one fourth? Let's tackle the final problem! Using a number line, solve two thirds minus two thirds. Pause the video to work on the problem, and press play when you are ready to see the solution! Both denominators are three, so divide the number line into three equal parts. Now locate two thirds, which is here. Count back the numerator of the second fraction, which is two. We land on zero, so we say the answer is zero. Did you also get zero? Razzi had so much fun practicing with you today! See you next time!
Subtracting Fractions on a Number Line — Let's Practice! exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Subtracting Fractions on a Number Line — Let's Practice!.
• Which number is the numerator?
Hints
The denominator is the bottom number in a fraction.
Solution
The numerator in the fraction is the 1! The top number is the numerator and it will be very important in subtracting fractions!
• Subtract these fractions and find the solution.
Hints
Here is how the number line should look to solve the problem!
Since the problem is $\frac{7}{10}$ - $\frac{3}{10}$ you should start counting back from the fraction $\frac{7}{10}$!
Solution
The correct answer is $\frac{4}{10}$! If you count back 3 points on the number line from $\frac{7}{10}$ you will see the answer is $\frac{4}{10}$!
• Subtract these fractions and find the solution.
Hints
Here is the number line you should use to solve the problem!
Since the problem is $\frac{4}{5}$ - $\frac{2}{5}$ you should start counting back from the fraction $\frac{4}{5}$!
Count back 2 from $\frac{4}{5}$!
Solution
The correct answer is $\frac{2}{5}$! If you count 2 points back on the number line starting at $\frac{4}{5}$ you will see the answer is $\frac{2}{5}$!
• Match the equations with the correct solutions.
Hints
Remember, when subtracting fractions if the denominators are the same in the problem it will also be the same in the solution.
When using a number line to subtract, use the numerator (top number) in the second fraction to count backwards!
Solution
• What is the missing numerator?
Hints
Start counting down from $\frac{9}{10}$!
Count back 5 from $\frac{9}{10}$!
Solution
The correct answer is $\frac{4}{10}$! If you count 5 points back on the number line from $\frac{9}{10}$ you will see the answer is $\frac{4}{10}$! The 5 comes from the fraction $\frac{5}{10}$ in the problem.
• What is the missing numerator?
Hints
We can use a number line to help us solve for the missing number too! Use this number line to help!
To solve for the missing number, we just need to find the difference between the numbers we do know. Subtract the smaller fraction from the larger fraction.
Solution
The missing number is 4! We found this by subtracting $\frac{7}{8}$ - $\frac{3}{8}$ leaving us with the solution $\frac{4}{8}$. |
How to Reduce Mixed Numbers & Improper Fractions to the Lowest Terms
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When you see the term "improper fraction," it doesn't have anything to do with etiquette. Instead, it means that the numerator, or top number, of the fraction is larger than the denominator, or bottom number. Depending on the instructions for the problem you're working on, you can keep an improper fraction in that form, or you can convert it to a mixed number: A whole number paired with a proper fraction. Either way, your math life will be a lot easier if you get into the habit of reducing all of those fractions to lowest terms.
Converting Improper Fractions to Mixed Numbers
Should you keep improper fractions the way they are, or convert them into a mixed number? That depends on the instructions you get and your ultimate goal. As a general rule, if you're still doing arithmetic with the fraction, it's easier to leave it in improper form. But if you're done with the arithmetic and ready to interpret your answer, it's easier to convert the improper fraction to a mixed number by working the division it represents.
Recall that you can also write a fraction as division. For example, 33/12 is the same as 33 ÷ 12. Work the division the fraction represents, leaving your answer in remainder form. To continue with the example given:
33 ÷ 12 = 2 \text{, remainder } 9
Write the remainder as a fraction, using the same denominator as your original fraction:
\text{remainder } 9 = \frac{9}{12}
because 12 was the original denominator
Finish writing the mixed number as a combination of the whole number result from Step 1, and the fraction from Step 2:
2 \enspace \frac{9}{12}
Simplifying Fractions to Lowest Terms
Whether you're dealing with improper fractions or the fraction part of a mixed number, simplifying the fraction to lowest terms makes them easier to read and easier to work arithmetic with. Consider the fraction part of the mixed number you just calculated
\frac{9}{12}
Look for factors that are present in both the numerator and the denominator of the fraction. You can either do this by examination (looking at the numbers and listing their factors in your head) or by writing out the factors for each number. Here's how you'd write the factors out:
Factors of 9: 1, 3, 9
Factors of 12: 1, 3, 4, 12
Whether you're using examination or a list, find the greatest factor that both numbers share. In this case, the greatest factor present in both numbers is 3.
Divide both numerator and denominator by the greatest common factor or, to think of it another way, factor that number out of both numerator and denominator and then cancel it. Either way, you end up with:
\frac{9 ÷ 3}{12 ÷ 3} = \frac{3}{4}
Because the numerator and denominator no longer have any common factors greater than 1, your fraction is now in lowest terms.
Simplifying Improper Fractions
The process works exactly the same for simplifying an improper fraction to lowest terms. Consider the improper fraction
\frac{25}{10}
Examine both numbers, or make a list, to find their factors:
Factors of 25: 1, 5, 25
Factors of 10: 1, 2, 5, 10
In this case, the greatest factor that's in both numbers is 5.
Divide both numerator and denominator by 5. This gives you:
\frac{5}{2}
Because 5 and 2 share no common factors greater than 1, the fraction is now in lowest terms.
Tips
• Note that your result is still an improper fraction. |
# How to integrate ? int_0^oo 1/(1+e^x)*dx
May 6, 2017
$\ln \left(2\right)$
#### Explanation:
First working with the indefinite integral, divide the numerator and denominator by ${e}^{x}$:
$\int \frac{1}{1 + {e}^{x}} \mathrm{dx} = \int {e}^{-} \frac{x}{{e}^{-} x + 1} \mathrm{dx}$
Let $u = {e}^{-} x + 1$, implying that $\mathrm{du} = - {e}^{-} x \mathrm{dx}$:
$= - \int \frac{- {e}^{-} x}{{e}^{-} x + 1} \mathrm{dx} = - \int \frac{1}{u} \mathrm{du} = - \ln \left\mid u \right\mid$
$= - \ln \left({e}^{-} x + 1\right) + C$
So:
${\int}_{0}^{\infty} \frac{1}{1 + {e}^{x}} \mathrm{dx}$
$= \left({\lim}_{x \rightarrow \infty} \left(- \ln \left({e}^{-} x + 1\right)\right)\right) - \left(- \ln \left({e}^{0} + 1\right)\right)$
Note that ${\lim}_{x \rightarrow \infty} {e}^{-} x = 0$:
$= - \ln \left(1\right) + \ln \left(1 + 1\right)$
$= \ln \left(2\right)$ |
## Multiplication of Three Digit Numbers (3 x 2 Digits)
Hi Kids,
Once we learn the basics of multiplication, it is easy to multiply even bigger numbers. Let us now see how to multiply a 3 (three) and a 2 (two) digit number.
Let us consider the numbers 463 and 53.
Write the numbers one below the other, as given in the below image.
As we did in the previous multiplication, we need to first multiply 3 of 53 with 463, in this problem. Let us do that one at a time.
Let us first concentrate on the “ones” column. Let us multiply 3 of 53 with 3 of the three digit number 463. We get 9 as the answer. Write 9 in the “ones” column.
Then multiply 3 with 6 of 463 in this elementary math problem. We get 18. Write 8 in the “tens” column and carry over 1 to the “hundreds” column.
Now multiply 3 with 4 of 463. We get 12. Along with this, add the 1 which we carried over, as we did in other problems. We get 13. Write 13 in the answer column. So we get the answer as 1389.
The next step in this multiplication is to multiply the number 5 of 53 with 463. As mentioned in the previous chapter, write 0 below 9 in the answer column. Now, let us multiply 5 and 3. The answer is 15. Write 5 below 8 of 1389 and carry over 1 to “tens” column.
Then multiply 5 of 53 with 6 of the three digit number 463. The answer is 30. Now add 30 with 1, which gives 31. Write 1 below 3 of 1389 and carry over 3 to “hundreds” column.
Now multiply 5 of 53 with 4 of 463. We get 20. Now add 20 with the carry over number 3. We get 23. Write 23 in the answer column, as given below.
Finally, add the two numbers 1389 and 23150 in the answer column.
We get 24539 as the answer. Practice with few more problems to understand this better. |
# Derivative of e^-3x
The function e to the power -3x is written as e-3x and its derivative is -3e-3x. Here, we will find the derivative of e-3x by the following methods:
• Logarithmic differentiation
• Chain rule of differentiation
• First principle of derivatives.
## Derivative of e-3x Formula
The derivative of e-3x is -3e-3x. Mathematically, we can express it as follows:
d/dx(e-3x) = -3e-3x or (e-3x)’ = -3e-3x.
## What is the derivative of e-3x?
Answer: The derivative of e to the power -3x is -3e-3x.
Proof: Let us use the logarithmic differentiation to find the derivative of e-3x. We put
y = e-3x
Taking logarithms with base e, we obtain that
loge y = loge e-3x
⇒ loge y = -3x by the logarithm rule loge ea = a.
Differentiating both sides with respect to x, we get that
$\dfrac{1}{y} \dfrac{dy}{dx}=-3$
⇒ $\dfrac{dy}{dx}=-3y$
⇒ $\dfrac{dy}{dx}=-3e^{-3x}$ as y=e-3x.
Thus, the derivative of e to the power -3x is -3e-3x which is obtained by the logarithmic differentiation method.
## Derivative of e-3x by Chain Rule
To find the derivative of a composite function, we use the chain rule. Let us now find the derivative of e to the power -3x by the chain rule.
Let u=-3x.
Thus, the derivative of e-3x by the chain rule is -3e-3x.
## Derivative of e-3x by First Principle
Recall the first principle of derivatives: Let f(x) be a function of the variable x. The derivative of f(x) from the first principle is equal to the following limit:
$\dfrac{d}{dx}(f(x))=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
We put f(x)=e-3x in the above formula. Then the derivative of e to the power -3x by the chain rule is given by
$\dfrac{d}{dx}(e^{-3x})= \lim\limits_{h \to 0} \dfrac{e^{-3(x+h)}-e^{-3x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{-3x-3h}-e^{-3x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{-3x} \cdot e^{-3h}-e^{-3x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{-3x}(e^{-3h}-1)}{h}$
=e-3x $\lim\limits_{h \to 0} \Big(\dfrac{e^{-3h}-1}{-3h} \times (-3) \Big)$
= -3e-3x $\lim\limits_{h \to 0} \dfrac{e^{-3h}-1}{-3h}$
[Let t = -3h. Then t→0 as x →0]
= -3e-3x $\lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$
= -3e-3x ⋅ 1 as the limit of (ex-1)/x is 1 when x→0.
= -3e-3x
∴ The differentiation of e-3x by the first principle is -3e-3x.
## FAQs on Derivative of e-3x
Q1: What is the derivative of e to the power -3x?
Answer: The derivative of e-3x is -3e-3x.
Q2: What is the integration of e-3x?
Answer: The integration of e-3x is -e-3x/3+c, where c is an integration constant.
Q3: What is the derivative of e3x+e-3x?
Answer: The derivative of e3x+e-3x is 3(e3x-e-3x).
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# 7.5-Using Proportional Relationships
## Presentation on theme: "7.5-Using Proportional Relationships"— Presentation transcript:
7.5-Using Proportional Relationships
2/5/13
Bell Work Convert each measurement. 1. 6 ft 3 in. to inches 75 in.
2. 5 m 38 cm to centimeters Find the perimeter and area of each polygon. 3. square with side length 13 cm 4. rectangle with length 5.8 m and width 2.5 m 75 in. 538 cm P = 52 cm, A =169 cm2 P =16.6 m, A = 14.5 m2
Definition 1 Indirect measurement is any method that uses formulas, similar figures, and/or proportions to measure an object. The following example shows one indirect measurement technique. Whenever dimensions are given in both feet and inches, you must convert them to either feet or inches before doing any calculations. Helpful Hint
Example 1 Tyler wants to find the height of a telephone pole. He measured the pole’s shadow and his own shadow and then made a diagram. What is the height h of the pole? Step 1 Convert the measurements to inches. AB = 7 ft 8 in. = (7 12) in. + 8 in. = 92 in. BC = 5 ft 9 in. = (5 12) in. + 9 in. = 69 in. FG = 38 ft 4 in. = (38 12) in. + 4 in. = 460 in.
Ex. 1 continued Step 2 Find similar triangles.
Because the sun’s rays are parallel, A F. Therefore ∆ABC ~ ∆FGH by AA ~. Step 3 Find h. Corr. sides are proportional. Substitute 69 for BC, h for GH, 92 for AB, and 460 for FG. Cross Products Prop. 92h = 69 460 Divide both sides by 92. h = 345 The height h of the pole is 345 inches, or 28 feet 9 inches.
Example 2 A student who is 5 ft 6 in. tall measured shadows to find the height LM of a flagpole. What is LM? Step 1 Convert the measurements to inches. GH = 5 ft 6 in. = (5 12) in. + 6 in. = 66 in. JH = 5 ft = (5 12) in. = 60 in. NM = 14 ft 2 in. = (14 12) in. + 2 in. = 170 in.
Ex. 2 continued Step 2 Find similar triangles.
Because the sun’s rays are parallel, L G. Therefore ∆JGH ~ ∆NLM by AA ~. Step 3 Find h. Corr. sides are proportional. Substitute 66 for BC, h for LM, 60 for JH, and 170 for MN. Cross Products Prop. 60(h) = 66 170 Divide both sides by 60. h = 187 The height of the flagpole is 187 in., or 15 ft. 7 in.
Definition 2 A scale drawing represents an object as smaller than or larger than its actual size. The drawing’s scale is the ratio of any length in the drawing to the corresponding actual length. For example, on a map with a scale of 1 cm : 1500 m, one centimeter on the map represents 1500 m in actual distance. A proportion may compare measurements that have different units. Remember!
Example 3 Cross Products Prop. Simplify. x 145
On a Wisconsin road map, Kristin measured a distance of 11 in. from Madison to Wausau. The scale of this map is 1inch:13 miles. What is the actual distance between Madison and Wausau to the nearest mile? To find the actual distance x write a proportion comparing the map distance to the actual distance. Cross Products Prop. Simplify. x 145 The actual distance is 145 miles, to the nearest mile.
Example 4 Find the actual distance between City Hall and El Centro College. To find the actual distance x write a proportion comparing the map distance to the actual distance. Cross Products Prop. 1x = 3(300) Simplify. x 900
Example 5 Lady Liberty holds a tablet in her left hand. The tablet is 7.19 m long and 4.14 m wide. If you made a scale drawing using the scale 1 cm:0.75 m, what would be the dimensions to the nearest tenth? Set up proportions to find the length l and width w of the scale drawing. 0.75w = 4.14 9.6 cm 5.5 cm w 5.5 cm
Example 6 The rectangular central chamber of the Lincoln Memorial is 74 ft long and 60 ft wide. Make a scale drawing of the floor of the chamber using a scale of 1 in.:20 ft. Set up proportions to find the length l and width w of the scale drawing. 20w = 60 w = 3 in 3.7 in. 3 in.
Rule for Similar Triangles & Theorem 1
Example 7: using ratios to find perimeters and areas
Given that ∆LMN:∆QRS, find the perimeter P and area A of ∆QRS. The similarity ratio of ∆LMN to ∆QRS is By the Proportional Perimeters and Areas Theorem, the ratio of the triangles’ perimeters is also , and the ratio of the triangles’ areas is
Example 7 continued Perimeter Area 13P = 36(9.1) 132A = (9.1)2(60) P = 25.2 A = 29.4 cm2 The perimeter of ∆QRS is 25.2 cm, and the area is 29.4 cm2.
Example 8 Perimeter Area 12P = 42(4) 122A = (4)2(96) P = 14 mm
∆ABC ~ ∆DEF, BC = 4 mm, and EF = 12 mm. If P = 42 mm and A = 96 mm2 for ∆DEF, find the perimeter and area of ∆ABC. Perimeter Area 12P = 42(4) 122A = (4)2(96) P = 14 mm The perimeter of ∆ABC is 14 mm, and the area is 10.7 mm2. |
# Lesson 3
Sumemos 1 o 2
## Warm-up: Cuántos ves: Dados de puntos (10 minutes)
### Narrative
The purpose of this How Many Do You See is for students to determine the number of dots in an arrangement without counting each dot. Dots are arranged in the formation they appear on a dot cube to build on the previous lessons. When students use the dot images to relate addition to counting on, they look for and make use of the structure of whole numbers (MP7).
### Launch
• Groups of 2
• “¿Cuántos ven? ¿Cómo lo saben?, ¿qué ven?” // “How many do you see? How do you see them?”
• Flash the image.
• 1 minute: quiet think time
### Activity
• Display the image.
• “Discutan con su pareja cómo pensaron” // “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Record responses.
### Student Facing
¿Cuántos ves?
¿Cómo lo sabes?, ¿qué ves?
### Activity Synthesis
• “¿Cómo supieron cuántos puntos había en total?” // “How did you know how many dots there are in all?”
• “¿Alguien puede expresar con otras palabras la forma en la que _____ vio los puntos?” // “Who can restate the way _____ saw the dots in different words?”
• “¿Alguien vio los puntos de la misma forma, pero lo explicaría de otra manera?” // “Did anyone see the dots the same way but would explain it differently?”
• “¿Alguien quiere compartir otra observación sobre la manera en la que _____ vio los puntos?” // “Does anyone want to add an observation to the way _____ saw the dots?”
## Activity 1: Conozcamos “Cinco en línea (suma y resta): Suma 1 o 2” (20 minutes)
### Narrative
The purpose of this activity is for students to learn the first stage in the center, Five in a Row. In this stage, students pick a card and choose to add 1 or 2 to the number on their card. They place a counter on the sum on their game board. The first person to get five counters in a row wins. Students begin to notice that when they add 1 to any number, the sum is the next number in the counting sequence, and when they add 2, the sum is two numbers more in the counting sequence (MP7, MP8).
The game board will be used again in upcoming lessons. Consider copying on cardstock or laminating for future use.
MLR7 Compare and Connect. Synthesis: After students share how they found sums, lead a discussion about how students decided to add 1 or 2. Ask, “¿Cuándo decidieron sumar 1? ¿Cuándo decidieron sumar 2? ¿Por qué?” // “When did you decide to add 1? When did you decide to add 2? Why?” To amplify student language and illustrate connections, record student strategies on a visible display.
### Required Materials
Materials to Gather
Materials to Copy
• Five in a Row Addition and Subtraction Stages 1 and 2 Gameboard, Spanish
### Required Preparation
• Each group of 2 needs a set of Number Cards 0–10.
### Launch
• Groups of 2
• Give each group a set of number cards, a game board, two-color counters, and access to 10-frames.
• “Vamos a conocer un juego llamado 'Cinco en línea: Sumemos 1 o 2'. Juguemos una ronda juntos” // “We are going to learn a game called Five in a Row, Add 1 or 2. Let’s play a round together.”
• Display the game board.
• “Primero debemos eliminar cualquier tarjeta con el número 10. No usaremos estas tarjetas en este juego. Ahora voy a voltear una tarjeta y a decidir si quiero sumarle 1 o 2 al número” // “First we need to remove any card with the number 10 on it. We will not use these cards in this game. Now I’m going to flip over a card and decide if I want to add 1 or 2 to the number.”
• Flip over a card.
• “Me salió un (5) y voy a elegir sumarle 2 a mi número. ¿Cuál es la suma?” // “I got a (5) and I’m going to choose to add 2 to my number. What is the sum?”
• “Ahora pongo una ficha sobre la suma en el tablero de juego. Después, es el turno de mi compañero” // “Now I put a counter on the sum on the game board. Then it is my partner’s turn.”
### Activity
• “Antes de empezar, ustedes y su compañero deben decidir quién usará el lado rojo de las fichas y quién usará el lado amarillo. Después, jueguen por turnos a voltear una tarjeta y sumarle 1 o 2 al número. La primera persona que logre cinco fichas en línea en el tablero gana. Las fichas pueden estar en una línea de lado a lado, de arriba a abajo o en diagonal” // “Before you begin, you and your partner need to decide who will use the red side of the counters, and who will use the yellow side. Then take turns flipping over a card and adding 1 or 2 to the number. The first person to get five counters in a row on the game board wins. The counters can be in a row across, up and down, or diagonal.”
• 10 minutes: partner work time
• As students work, consider asking:
• “¿Cómo encontraron la suma?” // “How did you find the sum?”
• “¿Cómo decidieron si sumaban 1 o 2?” // “How did you decide whether to add 1 or 2?”
### Activity Synthesis
• Display a game board with the center column covered with red counters except for the 8 at the bottom.
• Display the number card 6.
• “Este es mi tablero. Acabo de sacar el número 6. ¿Debería sumarle 1 o 2 al número? ¿Por qué?” // “This is my game board. I just picked the number 6. Should I add 1 or 2 to the number 6? Why?” (You should add 2 because $$6 + 2$$ is 8 and then you could cover the last number in the column and have five in a row.)
## Activity 2: Centros: Momento de escoger (20 minutes)
### Narrative
The purpose of this activity is for students to experience center choice time for the first time. Students choose from activities that offer practice counting up to 20 objects or adding within 10. Students choose from previously introduced centers and are encouraged to choose the center that will be most helpful for them at this time.
• Counting Collections
• Number Race
• Check it Off
• Five in a Row: Addition and Subtraction
Students will choose from these centers throughout the section. Keep materials from these centers organized to use each day.
Engagement: Provide Access by Recruiting Interest. Use visible timers or audible alerts to help students anticipate and prepare to transition between center activities.
Supports accessibility for: Social-Emotional Functioning, Organization
### Required Materials
Materials to Gather
### Required Preparation
• Gather materials from previous centers:
• Counting Collections, Stage 1
• Number Race, Stage 3
• Check it Off, Stage 1
• Five in a Row: Addition and Subtraction, Stage 1
### Launch
• Groups of 2
• “Ahora van a escoger centros de los que ya conocemos” // “Now you are going to choose from centers we have already learned.”
• Display the center choices in the student book.
• “Piensen qué les gustaría hacer primero” // “Think about what you would like to do first.”
• 30 seconds: quiet think time
### Activity
• Invite students to work at the center of their choice.
• 8 minutes: center work time
• “Escojan qué les gustaría hacer ahora” // “Choose what you would like to do next.”
• 8 minutes: center work time
### Student Facing
Escoge un centro.
Contar colecciones
Carrera con números
Márcalo
Cinco en línea
(suma y resta)
### Activity Synthesis
• “¿Cómo hicieron para trabajar bien hoy con su pareja en los centros?” // “How did you and your partner work well together during centers today?”
## Lesson Synthesis
### Lesson Synthesis
Math Community
Display the math community poster and read the student actions listed under “Hacer matemáticas” // “Doing Math.”
“¿Cuáles de estas hicieron hoy? ¿Cómo les ayudaron en clase?” // “Which of these did you do today? How did they help you in class?”
“¿Hay algo más que deberíamos agregar al póster?” // “Is there anything else we should add to the poster?” |
# JKBOSE Class 12th Statistics Notes | Study Materials
## JKBOSE Class 12th Statistics Notes
JKBOSE Class 12th Statistics Notes PDF Download. If you are the students of Jammu and Kashmir and are looking for important questions and Notes of Statistics Subject then you are at right place. Get JKBOSE important Study Materials Notes of all the subjects for Class 12th in this site but in this article we will provide you Statistics Notes for Class 12th. So keep visiting and get the free and best notes.
## JKBOSE Class 12th Statistics Notes Unitwise
Unit- 1 Probability – I
Introduction and Objectives
Probability: Probability is a fundamental concept in mathematics and statistics that quantifies the likelihood of an event occurring. It provides a framework for reasoning and making predictions in uncertain situations.
Probability is defined as a number between 0 and 1 where 0 represents impossibility and 1 represents certainty. The probability of an event A denoted as P(A) measures the relative likelihood of A occurring. It is determined by considering all possible outcomes and assigning numerical values to them based on their likelihood.
Axion of Peobability: The axioms of probability are a set of fundamental principles that govern the behavior of probabilities. There are three axioms:
Non-Negativity: The probability of any event is a non-negative number i.e. P(A) ≥ 0 for any event A.
Normalization: The probability of the entire sample space denoted as S is equal to 1 i.e. P(S) = 1.
Additivity: For any collection of mutually exclusive events (events that cannot occur simultaneously) the probability of their union is equal to the sum of their individual probabilities. If A and B are mutually exclusive events then P(A ∪ B) = P(A) + P(B).
Concept of Conditional Probability
Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B) where A and B are two events. The conditional probability of A given B is calculated by dividing the probability of the intersection of A and B by the probability of event B assuming that P(B) > 0. It is defined as P(A|B) = P(A ∩ B) / P(B).
Conditional probability allows us to update our knowledge or beliefs about an event based on new information or evidence. It plays a crucial role in various fields including statistics machine learning and decision theory.
Unit- 2 Probability – II
Introduction and Objectives
Random variable: A random variable in probability refers to a numerical quantity whose value is determined by the outcome of a random event or experiment. It is a way of quantifying uncertainty and capturing the variability in a particular situation. Random variables are typically denoted by capital letters such as X or Y and can represent a wide range of quantities such as the number of heads obtained when flipping a coin or the temperature measured in a specific location.
Discrete variable: A discrete variable is a type of random variable that can only take on a countable number of distinct values. In other words its values are typically integers or whole numbers. For example the number of students in a classroom the outcomes of rolling a die or the number of cars passing through a toll booth in a given time period are all examples of discrete variables. The probability distribution of a discrete random variable can be represented by a probability mass function (PMF) which assigns probabilities to each possible value of the variable.
Continuous Random Variable: A continuous random variable can take on any value within a certain interval or range. It is not restricted to specific individual values. Examples of continuous random variables include measurements like height weight time or temperature. The probability distribution of a continuous random variable is described by a probability density function (PDF) which specifies the relative likelihood of the variable taking on different values. Unlike the PMF of a discrete variable the PDF does not assign probabilities to specific values but instead provides the likelihood of the variable falling within certain intervals.
Unit- 3 Regression Analysis
Introduction and Objectives
Concept of Regression: Regression is a statistical concept used to analyze the relationship between variables. It aims to predict the value of a dependent variable based on one or more independent variables. In other words it helps us understand how changes in independent variables affect the dependent variable. Regression is widely used in various fields such as economics social sciences finance and machine learning.
In regression analysis we often use a technique called linear regression. It involves fitting a line (regression line) to a scatterplot of data points to approximate the relationship between variables. The regression line represents the best fit to the data minimizing the distance between the line and the actual data points.
Regression lines: The regression line is defined by an equation of the form y = a + bx where y is the dependent variable x is the independent variable a is the y-intercept and b is the slope of the line. The y-intercept (a) represents the predicted value of the dependent variable when the independent variable is zero while the slope (b) represents the change in the dependent variable for a unit change in the independent variable.
Regression Coefficients: Regression coefficients refer to the values of a and b in the regression equation. The coefficient a is also known as the intercept coefficient while the coefficient b is the slope coefficient. These coefficients provide important information about the relationship between variables. A positive slope coefficient indicates a positive relationship between the variables while a negative slope coefficient indicates a negative relationship. The coefficients can be used to make predictions by plugging in values for the independent variable into the regression equation.
Unit- 4 Theory of Attributes
Introduction and Objectives
Manifolds Classifications: Manifolds are fundamental objects in mathematics that play a crucial role in various fields including differential geometry topology and physics. A manifold can be loosely described as a space that locally looks like Euclidean space. One of the key aspects of studying manifolds is their classification.
Manifolds can be classified in different ways based on various criteria. One common classification is based on the dimension of the manifold. For example a 1-dimensional manifold is a curve a 2-dimensional manifold is a surface and a 3-dimensional manifold is our familiar three-dimensional space. Beyond three dimensions manifolds are more difficult to visualize but can still be mathematically defined.
Another classification is based on the smoothness of the manifold. A smooth manifold is one where there are no abrupt changes or singularities and the transition between local Euclidean spaces is smooth. These smooth manifolds are further classified into different categories such as compact manifolds (those that are closed and bounded) and non-compact manifolds (those that are either open or unbounded).
There are also specific types of manifolds that are studied extensively such as Riemannian manifolds which have a metric tensor that defines a notion of distance and angle. These play a crucial role in differential geometry and the formulation of physical theories like general relativity.
Ultimate Class Frequency: The term "ultimate class frequency" is not a standard term or concept in the context of manifold classification. It does not have a defined meaning and thus it is not possible to provide a precise explanation for it.
Unit- 5 Index Numbers
Introduction and Objectives
Index Number: Index numbers are statistical tools that measure changes in a variable over time. They provide a way to compare different observations or sets of data relative to a base period or reference point. Index numbers are commonly used in economics finance and other fields to track changes in prices quantities or other measurable factors.
Characteristics of Index numbers: The characteristics of index numbers include:
Base period: Index numbers require a base period which serves as a reference point for comparison. All subsequent observations are measured relative to this base period.
• Relative measurement: Index numbers provide a relative measurement by expressing the change in a variable as a percentage or ratio compared to the base period. This allows for meaningful comparisons between different periods or groups.
• Weighting: Depending on the application index numbers may incorporate weighting to reflect the importance of different components within a dataset. Weighting assigns greater significance to certain variables or groups resulting in a more accurate representation of the overall change.
• Uniqueness: Index numbers are unique to the variable they measure and the purpose they serve. Different variables may require different formulas and methodologies to construct appropriate index numbers.
Uses of Index numbers: Some common uses of index numbers include:
• Inflation measurement: Consumer price indices (CPI) are widely used to measure changes in the average price level of a basket of goods and services over time. These indices help track inflation rates and are essential for economic policy-making.
• Stock Market Analysis: Stock indices such as the S&P 500 or Dow Jones Industrial Average provide a snapshot of the overall performance of a selected group of stocks. Investors and analysts use these indices to assess market trends and make investment decisions.
• Economic Indicators: Index numbers are used to track changes in various economic indicators like industrial production employment levels and business activity. These indicators provide insights into the overall health and performance of an economy.
• Cost-of-living Adjustments: Index numbers are used to calculate cost-of-living adjustments (COLAs) for wages pensions and social security benefits. They ensure that income levels keep pace with changes in the cost of living maintaining the purchasing power of individuals over time.
Unit- 6 Vital Statistics
Introduction and Objectives
Vital Statistics: Vital statistics refer to numerical data and information related to events that are vital or essential to individuals and populations. They primarily focus on three main aspects: births deaths and marriages. These statistics provide valuable insights into population dynamics health and demographic characteristics which are crucial for planning and policy-making.
Nature of Vital Statistics: The nature of vital statistics is primarily quantitative as they involve numerical measurements and analysis. They are collected through the registration of vital events by governmental authorities such as birth and death certificates marriage licenses and related documents. These records contain important details such as names dates locations and other demographic information.
Uses of Vital Statistics: The uses of vital statistics are multifaceted. First and foremost they are essential for demographic analysis and studying population trends. They provide information on birth rates death rates and marriage rates allowing researchers and policymakers to understand population growth fertility patterns mortality rates and changes in marital status.
Vital statistics also play a crucial role in public health. They provide data on causes of death which helps in the identification and monitoring of diseases identifying risk factors and developing appropriate public health interventions. They also aid in assessing the effectiveness of healthcare systems and interventions.
Furthermore vital statistics are vital for administrative purposes. They serve as legal documents for individuals allowing them to establish their identity citizenship and other legal rights. They also facilitate the functioning of government agencies and enable the allocation of resources and services based on population needs.
Unit- 7 Sampling Theory
Introduction and Objectives
Meaning of Sampling: Sampling is a statistical technique used to gather data from a subset of a larger group known as a population in order to make inferences about the whole population. It is impractical and time-consuming to collect data from an entire population so sampling allows researchers to study a representative sample and draw conclusions about the population as a whole.
Objectives of Sampling: The primary objective of sampling is to obtain accurate and reliable information about a population while minimizing costs and resources. By selecting a sample that is representative of the population researchers aim to generalize the findings from the sample to the larger population. This is possible when the sample is chosen using randomization techniques and when the sample size is sufficiently large to minimize sampling errors.
Concept of Statistical Population: The concept of a statistical population refers to the entire group of individuals objects or events that researchers are interested in studying. It represents the larger target group from which a sample is drawn. The population can be finite such as the number of students in a school or infinite such as all the possible outcomes when rolling a dice.
It is crucial to define the population accurately to ensure the findings from the sample can be applied to the intended population. The characteristics of the population such as size variability and homogeneity influence the sampling process. Researchers use various sampling methods including random sampling stratified sampling cluster sampling and convenience sampling depending on the research objectives and constraints.
Sampling involves selecting a subset of a larger population to gather data and make inferences about the whole population. The objective is to obtain reliable information while minimizing costs and the statistical population refers to the entire group of interest from which the sample is drawn.
Unit- 8 Time Series and Computers
Introduction and Objectives
Time series analysis is a statistical technique that deals with data collected over time at regular intervals. It involves studying the pattern behavior and trends exhibited by the data to make predictions and forecasts about future values. Time series analysis is widely used in various fields such as finance economics weather forecasting stock market analysis sales forecasting and many others.
The importance of time series analysis lies in its ability to uncover valuable insights and patterns hidden within the temporal data. By analyzing historical data time series models can capture seasonality trends and cyclic patterns allowing us to understand the underlying dynamics of a system and make informed decisions. Here are some key reasons why time series analysis is significant:
• Forecasting: Time series analysis helps in predicting future values based on historical patterns and trends. By understanding the past behavior of a time series we can develop accurate forecasts for future time points. This is particularly valuable in areas such as sales forecasting demand planning and resource allocation.
• Pattern recognition: Time series analysis allows us to identify recurring patterns and regularities in the data. These patterns could include seasonal effects cyclical variations or long-term trends. Recognizing these patterns helps in understanding the behavior of the system and can provide insights for decision-making.
• Anomaly detection: Time series analysis can identify outliers and anomalies in the data. By comparing the observed values with predicted values we can identify unexpected variations or events that deviate from the regular pattern. Anomaly detection is crucial in various domains such as fraud detection network monitoring and quality control.
• Descriptive analysis: Time series analysis helps in describing the characteristics of a time series. It involves analyzing statistical properties such as mean variance autocorrelation and stationarity which provide insights into the underlying dynamics of the data. Descriptive analysis helps in understanding the data structure and can guide the selection of appropriate modeling techniques
• Decision-making and planning: Time series analysis provides a foundation for data-driven decision-making. By analyzing historical data identifying trends and making forecasts businesses can make informed decisions about resource allocation inventory management budgeting and investment strategies. Time series analysis also assists in policy planning and formulation by providing insights into economic indicators and trends.
The time series analysis plays a vital role in understanding the dynamics of temporal data making accurate predictions detecting anomalies and supporting informed decision-making. Its applications span across various industries and domains making it an essential tool for analyzing and leveraging time-dependent information.
## JKBOSE Class 12th All Subject Notes
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## JKBOSE Class 12th Statistics Important Textual Questions
#### What is probability and how is it defined?
Probability is a fundamental concept in mathematics and statistics that quantifies the likelihood of an event occurring. It is defined as a number between 0 and 1 where 0 represents impossibility and 1 represents certainty. The probability of an event A denoted as P(A) measures the relative likelihood of A occurring.
#### What are the axioms of probability?
The axioms of probability are a set of fundamental principles that govern the behavior of probabilities. There are three axioms: non-negativity (probability of any event is a non-negative number) normalization (probability of the entire sample space is equal to 1) and additivity (probability of the union of mutually exclusive events is equal to the sum of their individual probabilities).
#### What is conditional probability and how is it calculated?
Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B) where A and B are two events. The conditional probability of A given B is calculated by dividing the probability of the intersection of A and B by the probability of event B assuming that P(B) > 0. It is defined as P(A|B) = P(A ∩ B) / P(B).
#### What is regression analysis and how does it work?
Regression analysis is a statistical concept used to analyze the relationship between variables. It aims to predict the value of a dependent variable based on one or more independent variables. In linear regression a line (regression line) is fitted to a scatterplot of data points to approximate the relationship between variables. The regression line represents the best fit to the data minimizing the distance between the line and the actual data points.
#### What are index numbers and how are they used?
Index numbers are statistical tools that measure changes in a variable over time. They provide a way to compare different observations or sets of data relative to a base period or reference point. Index numbers are commonly used in economics finance and other fields to track changes in prices quantities or other measurable factors. They help in measuring inflation analyzing stock market performance tracking economic indicators and calculating cost-of-living adjustments.
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Mathematical Functions: Definition and Properties
by | Aug 25, 2021 | Math Learning
While studying mathematics you may have noticed that functions are widely used in nearly every subject. In a matter of fact, they are one of the most basic and fundamental objects in mathematics, so in this article, we will learn about functions, their definition, and their properties.
What is a function
We call a function a given relation between elements of two sets, in a way that each element of the first set is associated with one and only one element of the second set. For example, from the set of Natural Number to the set Natural Numbers , or from the set of Integers to the set of Real Numbers .
The first set is called the Domain, it is usually denoted by and its elements by , and the second set is called the codomain noted and its elements . A function is usually denoted by a letter like: , , or …etc.
If we take an example of a function called , and is a relation between the elements of the set and elements of the set . is denoted by and it reads: of equal . The element here is called the argument or the input of the function , and we call the element the value of the function or the image of by or the output of .
For example, we note:
Evaluating a function for a value of is simply the procedure of replacing by its given value and calculate the output or the value of .
Graphical representation
A useful way to see the behavior of the function on its domain of definition is by drawing it graphicly on a plan with two axes on for the values of noted -axis and one for the or noted -axis. A function is represented by all the pairs on a plane generating the Graph of the function .
The domain of definition
As we said before, the domain of definition is the set of the possible argument (or inputs) of the function, and every element of this set has exactly one image on the codomain. For better understanding let’s see some examples of functions and their domain of definition!
Example: determine the domain of the following functions:
For , we can calculate the value of for every real number except for , because we cannot divide by i.e., , so the domain is all the real numbers except meaning we can also note .
Let’s take a look at the graph of F:
For the case of the second function, to determine the domain of definition of we need to determine and exclude the values of that make:
1. the denominator equal to zero, because we cannot divide by .
2. the square root of less than zero, because we can’t evaluate the square root of a negative number (In the space of Real Numbers ).
In other terms, we can evaluate the value of for every real number except for the ones that make the denominator less or equal to .
So, for the denominator to be equal to , the expression has to be equal to , We have:
So, the value doesn’t belong in the domain of definition .
For the square root to be less than zero, we evaluate the expression , we have then:
Therefore, the values don’t belong in the domain of definition .
So, by combining (the union) the two previous results, we get the excluded values are: .
We conclude that the domain of definition of is .
(We can also take the opposite direction and instead of looking for the values to exclude we look for the possible values of , meaning the value of that make ).
Here is the graphical representation of :
For the third function , we notice that it is in the form of a fraction, so, we need to make sure that the denominator is not equal to zero and exclude the values of for which it does. In this case, we have the denominator is a polynomial of the second degree, so we need to determine its roots, the denominator is a remarkable identity (subtraction of two squares), meaning it may be written in the factorial form . Now to determine the roots of the denominator all we need to do is to solve the equation:
and we have then: or , these two values of are excluded from the domain of definition .
Therefore, the Domain of definition of is: .
Let’s see the graph of :
The domain of values (or codomain)
Definition: The codomain or the set of destination of a function is the set containing all the output or image of i.e. the set containing all for all in the domain. In other terms, the codomain of a function is the set of all possible outputs of .
Note that the codomain can be bigger, smaller, or entirely different from the domain.
For example, A function can have its domain as Natural numbers , and the codomain as the odd natural numbers; Or it can have its domain as the natural numbers and the codomain as negative integers ; Or having the domain as the set of real numbers and the same for the codomain.
Let’s take look at an interesting example:
Is the Square root a function? If we take as the square root of a positive real number , so we have or , meaning the is defined as follows
then is not a function, because as we mentioned at the beginning every element of the domain set must have one and only one image in the codomain, but in this case, each element of the domain has two images one positive and one negative, and therefore is not a function. This can be fixed by limiting the codomain to only positive real numbers!
Also keep in mind, the Radical sign which we use when calculating the root square of a number always means the positive square root, that’s why is a function and always givesa positive real numbers as an output.
Here is a graphical representation:
Note that the codomain may contain elements that are not images of any element of the domain, in other terms an element in the codomain doesn’t necessarily mean that there is an element in the domain that if we take as input, we will get as an output. For better understanding let’s take an example: If we take as follow:
We have here the domain of is the set of real numbers , and the codomain is the set of real numbers . but the image of by (or the inputs) is not all , it is a sub-set of the codomain . In a matter of fact, The image of by is and this is what we call the Range of , Therefore the image of every element in the domain is contained in the codomain, but not necessarily the opposite.
For more clarification let’s take a look at the graph of and see its range.
Notice that the graph doesn’t go beyond on the -axis which means the range of is and the range is a sub-set of the codomain since we have: .
Even and odd functions
Even function
Definition: a function is called an even function, if and only if it verifies the following property:
Or equivalently
Note that first, the function must have and as elements of its domain which means the domain must be symmetrical in the first place.
Graphicly speaking, the graph of an even function is symmetrical with respect to the -axis.
Let’s take a look at some examples of even functions with their respective graphs:
The absolute value:
The cosine:
The function:
Odd function
Definition: a function is an odd function if and only if it verifies the following:
Or equivalently
Note that first, the function must have and as elements of its domain which means the domain must be symmetrical in the first place.
Graphically speaking, the graph of an odd function is symmetrical about the origin .
Let’s take a look at some examples of odd functions and their respective graphs:
The identity:
The sine:
The function:
Some properties of Even and odd functions
• The absolute value of an odd function is an even one (the most obvious example, the identity function is odd and the absolute value of it is even).
• The summation of two odd functions is an odd one.
• The summation of two even functions is an even one.
• The subtraction of two odd functions is an odd one.
• The subtraction of two even functions is an even one.
• The multiplication of two even functions is an even one.
• The multiplication of two odd functions is an even one.
• The multiplication of an odd function with an even function is an odd one.
• The division of two even functions is an even one.
• The division of two odd functions is an even one.
• The division of an even function and an odd function is an odd one.
You want to try out these properties! Check out the window below and see the results of the summation, subtraction, multiplication, or division of even and odd functions. Check the operation you want to display (one or multiple), and watch the respective graphs. Have fun!!!!!
Periodic functions
We call a periodic function a function that its values (outputs) repeat regularly.
Definition:
A function is called periodic with a period different than , if and only if for any , , and of its domain, it verifies the following:
Where is called the period.
Graphicly speaking, a periodic function has a part of its graph repeats regularly.
Let’s take a look at some examples of periodic functions and their respective graphs.
The sine:
The cosine:
The tangent:
Some properties of periodicity
• If is periodic with a period , then , where , is also a period for .
• To graph a periodic function with a period , all we need to do is to graph an interval of length , and then duplicate it at both ends left and right, or in other terms shift the graphed section along the -axis at a distance () left and right.
• If is periodic with a period , then for all in the domain of definition of , and for all natural number we have:
.
• If is periodic with a period , then , , and are periodic with period , where is a real number diffrent than .
Conclusion
In this article we learned a little bit about functions and some of their properties and characteristics, this may be considered as an introduction to how to study a function and detect its properties which will help for better understanding of its behavior, but keep in mind this is just an introduction and there is a lot more to learn! isn’t that amazing! So who is eager for another post to learn more about functions!!!!!
Are you looking for another interesting article! Take a look at how to solve polynomial functions of first, second, and third degree. Enjoy!!!
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```Question 88824
Given:
.
y > 3
.
If you graph y = 3, you find that the graph is a horizontal line that crosses the y-axis at
+3 as shown below:
.
{{{graph(300,300,-20,20,-3,5,3)}}}
.
This graph tells you that no matter what the value of x is, the corresponding value of
y must be 3. As examples all of the following points are on the graph: (-300,3) (-10,3), (0, 3),
(5, 3), (200,3). As stated above, no matter what the value of x is, the value of y is 3.
.
But your problem says that y is greater than three. So this time, no matter what the value
of x is the corresponding value of y must be greater than 3. As examples, the following
points satisfy this relationship: (-300,4), (-10,15), (0,5), (5, 5), (200, 3.2). The reason
these points satisfy the inequality is that in every one of those points the value of y
is greater than 3.
.
How do you show this on a graph? Go back to the above graph that shows the graph of y = 3.
Take your pencil and shade in every bit of space ABOVE the line y = 3. (That shading
would be above the line and all the way from x approaching minus infinity to x approaching
positive infinity.) Make sure your shaded area is ABOVE the line and does NOT include
the line y = 3 but does include all values of y ABOVE the line and as high as you can go
... including up to y approaching positive infinity. Any point in that shaded area will
have a value of y that is greater than 3. So the graph of y > 3 is the entire shaded area
above the line y = 3.
.
Although this is not part of the above problem, you could graph the inequality y < 3 in a similar
manner. To graph this new inequality, you would shade in the entire region BELOW the graph of
the line y = 3. Any point in this region below the line will have a value of y that is
LESS than 3, and therefore, any point in this shaded region will satisfy the inequality
y < 3. Therefore, this shaded region below the line y = 3 is the graph you need for this
new inequality of y < 3.
.
Hope this makes sense to you and gives you a feel for graphing inequalities by shading
in regions of the graph as it relates to an equation (in this problem, as it relates to
the equation y = 3).``` |
# Number Bonds = Better Understanding
[Rescued from my old blog.]
A number bond is a mental picture of the relationship between a number and the parts that combine to make it. The concept of number bonds is very basic, an important foundation for understanding how numbers work. A whole thing is made up of parts. If you know the parts, you can put them together (add) to find the whole. If you know the whole and one of the parts, you take away the part you know (subtract) to find the other part.
Number bonds let children see the inverse relationship between addition and subtraction. Subtraction is not a totally different thing from addition; they are mirror images. To subtract means to figure out how much more you would have to add to get the whole thing.
## A Picture Is Worth More than Many Words
You can draw number bonds on paper using circles or bar diagrams. Imagine each circle to be a pile of blocks or other manipulatives, and think of the bar as the blocks lined up in a row. Even a young student who does not understand math notation can clearly see the connection between these numbers: the whole (6) has been pulled apart into two piles (4 and 2), and the piles can be pushed back together to make the whole.
Math textbooks often try to communicate the same concept using four-fact families. A four-fact family looks like this:
4 + 2 = 6
2 + 4 = 6
6 – 4 = 2
6 – 2 = 4
The idea of the four-fact family is to help students realize that once they know one of the facts in the family, they know all of them. Many students never see the connection, however, and think of these equations as separate little bits of abstract information, all of which have to be memorized. This can overload their minds and make them give up on math. On the other hand, number bonds connect to the student’s understanding at a deeper level, showing all four of the fact family relationships in a single picture.
## How to Teach Number Bonds
When you start teaching number bonds, use M&Ms or popsicle sticks or whatever you have on hand to make physical piles that can be pulled apart and pushed back together, then pulled apart in another way. When I introduced number bonds to my youngest daughter, I set out six blocks on the bedspread. (We often do school work sprawled on the bed.)
“How many blocks do we have here?” I asked.
Kitten counted carefully. “1, 2, 3, 4, 5, 6.”
“I’m going to move these blocks over to make a new pile,” I said. I pushed two of the blocks to one side. “How many blocks did I move?”
She did not have to count those. “Two.”
“And how many are left?”
“1, 2, 3, 4.”
“And if I push them back together…” I did so. “Two and four are how many in all?”
“1, 2, 3, 4, 5, 6.”
“Two and four are six. You are so good at counting! Now, this time I’m going to move three blocks over here.” I moved three blocks to the side. “How many are left?”
Again, she did not need to count. “Three.”
I pushed the blocks back together. “And three and three are how many?”
I could see her lips move as she counted silently. She looked up and smiled. “Six.”
“That’s right. Three and three are six. What will happen if I move just one block over?” I did so. “How many are left?”
Kitten started to speak, but then she stopped with a puzzled look on her face. She bent over to hide the blocks with her hand so I could not see her count. “1, 2, 3, 4, 5.”
“Oh, you’re tricky!” I said. “Five blocks. And if I put them back together, one and five are how many?”
I pushed the blocks together again. She looked at the pile. “Six.”
“One and five are six. Now you try it. What piles will you make?”
Kitten pushed all six blocks to the side. She looked up guiltily, not sure that was allowed.
“Okay,” I said. “You moved six blocks. How many are left?”
She looked at the empty spot. “Zero.”
“That’s right. Zero blocks. And if you put the piles back together, six and zero make how many?”
We played this game with blocks many times, using different numbers in the original pile, throughout Kitten’s kindergarten year. The blocks give her a concrete, hands-on way to check herself, building confidence in her understanding of how numbers work. My goal was not for her to memorize specific math facts, but that she understand and be able to use the concept of taking a number apart and then putting it back together.
When teaching number bonds to an older student, I may still start with blocks, but after one or two piles of demonstration, we move quickly to the number bond pictures and games. I usually start older students with the bonds for ten:
1 + 9
2 + 8
3 + 7
4 + 6
5 + 5
Ten is the most important number in our decimal (base ten) number system, so it is vital that our children learn to recognize it in any disguise. When the student knows the bonds for ten automatically, we move on to 20, then 100, then any number we desire.
## Don’t “Drill.” Play Games!
Here are a few number bond games to enjoy with your children:
Throw two dice and tell how many more you would need to make 10.
(On the rare throws of 11 or 12, the answer is a negative number.)
Throw 3 dice and tell how many more it takes to make 20.
One player names any number 0-100, and the other tells how many more it takes to make 100.
You could also play the last game with math cards [take out the jokers and face cards, leaving just ace (1) through 10], turning up one for the tens place and one for the ones, to make a two digit number.
10’s Concentration — Turn all the math cards face down on the table. On her turn, each player turns up two cards. If they add up to 10, she gets to keep them and try again. If one of the cards is a 10, she gets to keep it and turn up another card. Whoever takes the most cards, wins.
This post is an excerpt from my book Counting & Number Bonds: Math Games for Early Learners, available now at your favorite online book dealer.
## 25 thoughts on “Number Bonds = Better Understanding”
1. I like games. These are neat. But it’s really ok to do some drill, as well. It doesn’t have to be either/or.
I give my niece and nephew the answer, and ask for a problem to go along with it. They seem willing to play that game. Adaptable by choice of number and operation and number of factors or addends required.
Jonathan
2. Yes, of course some drill is okay. Necessary, even, since any math practice page could be considered drill—even if what one needs to practice is factoring trinomials. One of my faults as a writer is that I tend to overstate what I mean.
I like your game. It reminds me of the Children’s Mathematics Calendar that Theoni Pappas used to publish. My math club had fun with that.
3. Number bonds is a good way to teach number relation. Another key thing is the learning of basic maths operations, the Addition and Subtraction. Also if the numbers are colored, the kids will like them more.
4. Hi there,
Just found your blog, I was trying to find some helpful ways to teach my daughter number bonds as we’ve just reached it in her workbook. Your post has been really helpful, and I’m gonna try with the blocks or something similar tomorrow, I think it will help her alot more to understand the concept, as she has struggled a little bit with it. Thankyou.
1. Hi, Naimah! I’m glad you dropped by, and I hope your daughter enjoyed the number bond activity.
5. Christine says:
Hi! I first encountered number bonds through my son’s textbook. He’s in K2 now. I wasn’t quite sure if this concept really works, but now that I’ve read your blog, I’m convinced! Hope to read more articles about how we can teach our kids Math in a fun and exciting way.
6. Mary says:
Just found your site. Can’t wait to try some of these ideas out on my young ones I watch in my family child care home. May make it easier when they enter school. Also help after school students with homework, one struggles with math, I feel these suggestions will help me to teach math in a fun way, not just drilling and math sheets. Some kids need to learn math in a fun environment first. I never was good in Math, this is what I need to get the kids excited about numbers/math facts
7. mathmomma says:
Well, you posted this a long time ago, but I see others are still replying. I wanted to mention a picture book I love that is great for the number pairs that add to 7. It’s called Quack and Count. I wish there were more books as fun as this one for 10 and5 and 6 and 12…
8. mathmomma says:
P.S. I hadn’t noticed I was logged in as mathmomma. This is Sue VanHattum, and I’ve started a blog at mathmamawrites.blogspot.com.
9. Tisha McBride says:
What is a good book, website, or other resource for number bonds?
10. Hi, Tisha!
If you want to read more about number bonds, you may enjoy the other blog posts I listed at the end of this article. Also, I’ve written a recent post that explains a little more about how we use the concept as the kids get older:
I like to use the Singapore Primary Math program with my homeschooled children. The books focus on teaching number bonds in the early grade levels, and I find that the program gives my kids a great foundation of mathematical understanding. I have used the 3rd Edition and U.S. Edition versions, and both are good; the new CA Standards version is probably fine, too, but I have no experience with it.
11. Sheila Jones says:
As a one-room teacher with students in grades 1-8, I encounter many levels of learning. I have found that those students who start with number bonds advance in math faster than those who depend on finger-counting or rote memorization. Sometimes a student will join my school in the later grades. They are already dependent on counting on their fingers and are resistent to change. I demonstrate the number bonds using blocks, items, pictures, etc., but could use more games that are suitable for older students. Do you have any ideas?
Sheila
12. My older students still enjoy the Concentration game. I’ve found that mental math work with big numbers helps build (or reinforce) skill with number bonds, especially when we talk about strategies.
With my kids, when they are causing trouble by poking each other or fidgeting too much, I will tell them to sit on their hands. You might try that, if your students wouldn’t be offended, as a way to enforce the mental challenge of the math.
I’ll think a bit and see if I can come up with other ideas…
13. Becky says:
I had no idea Math could be so fun! I would have had a blast learning Math using these ideas! Thank you so much for providing these resources. You will never know what a blessing this is to our family.
14. pamela December. says:
Pamela
This has been posted so long ago,but I have just found your site and I am truly greatful for the simple method you have used. I can now use same to help my second grade child who attends my family Day Care.Thank you so much.This is fun and the way it should be, so more students would gain confidence in themselves,as math is made simple.
15. sallyp@ameos.co.uk says:
How can I help my twins boys they are under the 6year, to make the games fun fun:-)!!!
16. Shalin says:
Its the best phrase that there is to say, since its easy to understand stuff from pictures than just simple text. I use creately to create infographics and other diagram for kids to teach math.
17. Stella says:
I really found your blog very helpful .i hope to use the method to teach my children
Thanks and well done
18. Kevin So says:
Hi, Do Number bond have offical Chinese Name?
19. I do not know of a special Chinese name. Sometimes number bonds might be called “Partitions.” But normally, we think of number bonds as being a pair of numbers that add up to the sum. Partitions can be any set of whole numbers that add up to the sum. So you can partition 10 into 1+2+3+4, but we wouldn’t call that a number bond.
20. Matseliso Morake says:
What is the difference between number bonds and family numbers?
1. Good question! I don’t know what family numbers are, so I can’t answer it.
I’ve heard of fact families, as I mentioned in the blog. Those are the relationships between simple numbers (often memorized as “math facts”) that are connected, like:
3 + 4 = 7
4 = 3 = 7
7 − 4 = 3
7 − 3 = 4
A number bond picture showing that three and four together make seven represents all four of these equations at once, in an easy-to-understand visual form.
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# Integrate the following with respect to x:$\int {\dfrac{1}{{2x + 3}}dx}$.
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Hint: Simple substitution of the denominator to some variable will help simplifying the integral and reducing it to a standard integral. Use this technique to evaluate this integral.
Let $I = {\text{ }}\int {\dfrac{1}{{2x + 3}}dx}$……………………… (1)
Let 2x+3 = p…………………. (2)
Now, differentiate both the sides of equation (2) we get,
$\Rightarrow 2dx = dp$………………………… (3)
Make this substitution back into the main integral$I$, substituting (3) in equation (1) we get
$\Rightarrow I = {\text{ }}\dfrac{1}{2}\int {\dfrac{1}{p}dp}$…………………. (4)
Now, we know that the standard integral of,$\int {\dfrac{1}{x}dx = \log x}$ ……………… (5)
So the value of equation (4) will be, using above equation (5) we get,
$I = \dfrac{1}{2}\log p + c$
Now, let’s substitute the value of p back into the above integral we get
$I = \dfrac{1}{2}\log \left( {2x + 3} \right) + c$ Using equation (2)
Note: Whenever we face such problems always try and simplify the integral via method of substitution. This will help simplifying the integral into a standard from. Don’t forget to substitute back the variable assumed and take the solution back to the main variable provided in question. The constant of integration is also to be taken care of in exams. |
# Class 8 Maths MCQ – Rational Numbers Between Two Rational Numbers
This set of Class 8 Maths Chapter 1 Multiple Choice Questions & Answers (MCQs) focuses on “Rational Numbers Between Two Rational Numbers”.
1. Which number lies between the numbers $$\frac{1}{4}$$ and $$\frac{1}{6}$$?
a) $$\frac{1}{5}$$
b) $$\frac{1}{6}$$
c) $$\frac{1}{3}$$
d) $$\frac{1}{2}$$
Explanation: The given fractional numbers represents $$\frac{1}{4}$$ = 0.25 and $$\frac{1}{6}$$ = 0.17
Our options are $$\frac{1}{5}$$ = 0.20
$$\frac{1}{6}$$ = 0.17
$$\frac{1}{3}$$ = 0.33
$$\frac{1}{2}$$ = 0.50
The only option between 0.17 and 0.25 is $$\frac{1}{5}$$ i.e. 0.20.
2. The rational number which is not lying between 6 and 7 is ________
a) $$\frac{36}{7}$$
b) $$\frac{45}{7}$$
c) $$\frac{46}{7}$$
d) $$\frac{13}{2}$$
Explanation: The only number which is not between 6 and 7 is $$\frac{36}{7}$$
$$\frac{36}{7}$$ = 5.14
$$\frac{45}{7}$$ = 6.42
$$\frac{46}{7}$$ = 6.57
$$\frac{13}{2}$$ = 6.5.
3. Between any consecutive two integers, there are always infinite integers.
a) True
b) False
Explanation: The set of integers is […., -1, 0, 1, ….] there cannot be any integers between two integers.
4. The maximum number of integers between two consecutive natural numbers is ________
a) zero
b) 2
c) 3
d) infinite
Explanation: There can be an infinite number of rational numbers between any two consecutive natural numbers. But there cannot be any integers between two consecutive natural numbers.
5. Which on the following number lies between $$\frac{3}{10}$$ and $$\frac{3}{10000}$$.
a) $$\frac{3}{1000}$$
b) $$\frac{3}{10000}$$
c) $$\frac{3}{100000}$$
d) $$\frac{3}{1000000}$$
Explanation: The given fractional numbers represents $$\frac{3}{10}$$ = 0.3 and $$\frac{3}{10000}$$ = 0.00003
Our options are in fractions, converting it into decimal form we get,
$$\frac{3}{1000}$$ = 0.0003
$$\frac{3}{10000}$$ = 0.00003
$$\frac{3}{100000}$$ = 0.000003
$$\frac{3}{1000000}$$ = 0.0000003.
The only option between 0.3 and 0.00003 is $$\frac{3}{1000}$$ i.e. 0.0003.
6. $$\frac{32}{9}$$ lies between which pair?
a) $$\frac{6}{5}$$ and $$\frac{7}{2}$$
b) $$\frac{3}{2}$$ and $$\frac{5}{2}$$
c) $$\frac{7}{2}$$ and $$\frac{10}{2}$$
d) $$\frac{7}{2}$$ and $$\frac{6}{5}$$
Explanation: When we place the 4 given pairs and the number on the number line we observe that the number $$\frac{32}{9}$$ lies between the pair $$\frac{7}{2}$$ and $$\frac{10}{2}$$, while other options fail to capture the given number between them. Hence the correct pair is $$\frac{7}{2}$$ and $$\frac{10}{2}$$.
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7. Which of the following is the greatest number among the given rational numbers?
a) $$\frac{139}{26}$$
b) $$\frac{72}{21}$$
c) $$\frac{99}{36}$$
d) $$\frac{101}{20}$$
Explanation: The numbers when placed on the number line, the number which is placed farthest from 0 is the greatest. Here the numbers are, $$\frac{139}{26}$$ = 5.34
$$\frac{72}{21}$$ = 3.42
$$\frac{99}{36}$$ = 2.75
$$\frac{101}{20}$$ = 5.05
So, we conclude that the greatest number is $$\frac{139}{26}$$ and hence would be placed farthest from 0.
8. [….., -1, 0, 1, …….] the given set shows which type of numbers?
a) Rational numbers
b) Integers
c) Natural numbers
d) Whole numbers
Explanation: The given set is of integers.
The sets of other numbers are,
Whole number -> [0, 1, 2, 3, 4, ……]
Natural number -> [1, 2, 3, 4, …….]
Rational number -> [……, 0, ……..] This set consists of all numbers on the number line.
9. Pick the smallest number of the following.
a) –$$\frac{139}{26}$$
b) –$$\frac{72}{21}$$
c) –$$\frac{99}{36}$$
d) –$$\frac{101}{20}$$
Explanation: The numbers when placed on the number line, the number which is placed nearest from 0 is the smallest. Here the numbers are, –$$\frac{139}{26}$$ = -5.34
–$$\frac{72}{21}$$ = -3.42
–$$\frac{99}{36}$$ = -2.75
–$$\frac{101}{20}$$ = -5.05
So, we conclude that the smallest number is –$$\frac{139}{26}$$ and hence would be placed smallest from 0. Thing to remember: When the numbers are placed on the left side of the number line, the number farthest from the center is the smallest number. This is due to the negative sign.
10. Pick the option which has the ordered pair of BIG and SMALL number.
a) $$\frac{6}{3}$$ and $$\frac{7}{3}$$
b) $$\frac{12}{11}$$ and $$\frac{11}{10}$$
c) $$\frac{13}{22}$$ and $$\frac{33}{11}$$
d) $$\frac{12}{4}$$ and $$\frac{13}{5}$$
Explanation: Here we have to find the ordered pair of numbers and the order should be big following by small. This condition is only satisfied by one pair of numbers which is $$\frac{12}{4}$$ and $$\frac{13}{5}$$.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
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# Acceleration
You are driving a steady $20\,\mathrm{m/s}$ ($72 \,\mathrm{km/hr}$) but a car behind you is rushing. You push the gas pedal and slowly speed up by gaining 2 more metres-per-second every second.
Those additional 2 metres-per-second extra per second – we could write them as $2 \,\mathrm{m/s\,/s}$ or simplify it to $2\,\mathrm{m/s^2}$ – are a change in the speed. Let’s name ‘change in speed’: acceleration.
• Speed is ‘how fast’, the change in distance, and
• acceleration is ‘how much faster’, the change in speed.
As we could move sideways and forwards simultaneously – two speeds $v_x$ and $v_y$ that combine into one velocity vector $\vec v$ – likewise accelerations in several directions, $a_x$ and $a_y$, combine into one acceleration vector $\vec a$.1
$$\vec s= (s_x,s_y)\qquad \vec v=(v_x,v_y) \qquad \vec a=(a_x,a_y)$$
Since acceleration is ‘a change in velocity’, which is ‘a change in position’, acceleration must be ‘a change in the change in position’; a double change:
$$a=v’=s^{\prime \prime } \qquad \vec a=\vec v’=\vec s^{\prime \prime }$$
‘Speeding up’ or ‘slowing down’2 as well as ‘turning’ are changes in speed or velocity, and thus require an acceleration.
Note an odd fact: Our bodies do not feel position. And also not velocity. We only feel acceleration. You can sit in a bus and drink your coffee undisturbed, unaware of whether the bus is standing still or is at full speed down the motorway. But the moment it turns, brakes or speeds up, then your coffee flies all over the place, and you feel squeezed into the seat or flung around.[2]
References:
1. Sears and Zemansky’s Univesity Physics with Modern Physics’ (book), Hugh D. Young & Roger A. Freedman, Pearson Education, 13th ed., 2012
2. Beyond velocity and acceleration: jerk, snap and higher derivatives’ (article), David Eager, Ann-marie Pendrill and others, European Journal of Physics, IOP Publishing, vol. 37, issue 6, 2016, www.iopscience.iop.org/article/10.1088/0143-0807/37/6/065008, DOI 10.1088/0143-0807/37/6/065008 |
Parallel Lines and Angles Worksheet - Page 2 | Problems & Solutions
# Parallel Lines and Angles Worksheet - Page 2
Parallel Lines and Angles Worksheet
• Page 2
11.
How should the angles 1 and 2 be classified?
a. Same side interior angles b. Alternate interior angles c. Corresponding angles d. Vertical angles
#### Solution:
Two angles are corresponding if they occupy corresponding positions.
1 and 2 occupy corresponding positions. Hence they are corresponding angles.
12.
What is the ratio of the number of pairs of same-side interior angles formed by two parallel lines and a tranversal to the number of pairs of alternate interior angles formed by two parallel lines and a transversal?
a. 1 : 2 b. 1 : 1 c. 1 : 4 d. 2 : 1
#### Solution:
The two parallel lines and a transversal is as shown.
The pairs of same side interior angles are
4 and 5
3 and 6
Hence two pairs of interior angles are formed.
The pairs of alternate interior angles are
4 and 6
3 and 5
Hence two pairs of alternate interior angles are formed.
The ratio of the number of pairs of same-side interior angles formed by two parallel lines and a tranversal is equal to the number of pairs of alternate interior angles formed by two parallel lines and a transversal = 1 : 1.
13.
A is a point on line $l$ and B is a point on line $m$. If $l$ is parallel to $m$, is $\begin{array}{c}\stackrel{↔}{\text{AB}}\end{array}$ a transversal for lines $l$ and $m$?
a. No b. Yes
#### Solution:
A is a point on line j and B is a point on line kand l and m are parallel.
[As per the question.]
Join the points A and B. AB is a transversal.
Hence AB is a transversal.
14.
Identify the relation between the lines $\begin{array}{c}\stackrel{↔}{\text{ST}}\end{array}$ and $\begin{array}{c}\stackrel{↔}{\text{ZY}}\end{array}$.
a. skew lines b. parallel lines c. oblique lines d. perpendicular lines
#### Solution:
Lines that are not in the same plane and do not intersect are called skew lines.
ST and ZY do not lie in the same plane.
The lines ST and ZY are skew lines.
15.
A is a point on line $l$ and B is a point on line $m$. If $l$ intersects $m$ (but not A or B), then which of the following is the most appropriate description for $\begin{array}{c}\stackrel{↔}{\text{AB}}\end{array}$?
a. bisector to line $l$ and $m$ b. perpendicular to line $l$ c. perpendicular to line $m$ d. transversal of lines $l$ and $m$
#### Solution:
A transversal is a line that intersects two coplanar lines at two distinct places.
A is a point on line l and B is a point on line m. l and m intersects.
[As per the question.]
AB iintersects l and m at two distinct points.
Hence AB is a transversal.
16.
Identify the parallel planes.
a. $S$$X$$Y$ and $R$$U$$T$ b. SXW and TYZ c. $S$$T$$U$ and $X$$Y$$T$ d. $R$$S$$T$ and $U$$Z$$Y$
#### Solution:
Parallel planes are planes that donot intersect.
SXY and RUT are planes that intersect. Hence they are not parallel.
RST and UZY are planes that intersect. Hence they are not parallel.
STU and XYT are planes that intersect. Hence they are not parallel.
SXW and TYZ are planes that donot intersect. Hence they are parallel.
17.
A is a point on line $l$ and B is a point on line $m$. If $l$ and $m$ are skew lines, check if $\begin{array}{c}\stackrel{↔}{\text{AB}}\end{array}$ a transversal for lines $l$ and $m$. Why?
a. Yes, because the two lines given are not parallel b. Yes, because AB intersects the two lines given c. Yes, because there are no other lines connecting the two lines given d. No, because the two lines given are skew lines
#### Solution:
Skew lines do not lie in the same plane. They are neither parallel nor intersecting.
l and m are skew lines. A is a point on line l and B is a point on line m.
A transversal is a line that intersects two coplanar lines at two distinct points.
Skew lines are non-coplanar. Hence AB is not a transversal.
18.
In the figure $\angle$4 and ________ are alternate interior angles.
a. $\angle$11 b. $\angle$10 c. $\angle$9 d. $\angle$11
#### Solution:
Two angles are alternate interior angles if they lie between the two lines on the opposite sides of the transversal.
4 and 10 lies between two lines on opposite sides of the transversal.
Hence the alternate interior angles are 4 and 10.
19.
Identify a pair of parallel planes.
a. UTYZ and WXYZ b. STYX and WXYZ c. RSXW and RSTU d. RSTU and WXYZ
#### Solution:
Planes that have no points in common are called parallel planes.
Planes UTYZ and WXYZ are not parallel as they intersect in line segment YZ.
[Points Y and Z are common to both the planes.]
Planes RSXW and RSTU are not parallel as they intersect in line segment RS.
[Points R and S are common to both the planes.]
Planes STYX and WXYZ are not parallel as they intersect in line segment XY.
[Points X and Y are common to both the planes.]
Planes RSTU and WXYZ are parallel as they do not have any point in common.
20.
Which of the following is the correct definition of skew lines?
I. Lines that are coplanar and do not intersect.
II. Lines that are coplanar and do intersect.
III. Lines that are neither parallel nor intersecting and are not coplanar.
IV. Lines that lie in the same plane and are parallel.
a. III only b. IV only c. I only d. II only
#### Solution:
Skew lines do not lie in the same plane. They are neither parallel nor intersecting.
[From Definition.]
ST and UV are skew lines. |
# NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.2
## NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.2
If you need solutions in Hindi, Click for Hindi Medium solutions of 10 Maths Exercise 11.2
Exercise 11.1
10 Maths Main Page
### Class 10 Maths Exercise 10.2 Solutions in Hindi Medium
Exercise 11.1
10 Maths Main Page
To get the solutions in English, Click for English Medium solutions.
###### Important questions from CBSE Board examination based on Circles – 10th Maths
1. Draw a line segment AB = 8 cm. Taking AB as diameter a circle is drawn with centre O. Now draw OP is perpendicular AB. Through P draw a tangent to the circle.
2. Draw a circle of radius OP = 3 cm. Draw ∆POQ = 45° such that OQ = 5 cm. Now draw two tangents from Q to given circle.
3. Draw a circle with centre O and radius 3.5 cm. Draw two tangents PA and PB from an external point P such that ∆APB = 45°. What is the value of ∆AOB + ∆APB?
4. Draw a circle of radius 4 cm. Now draw a set of tangents from an external point P such that the angle between the two tangents is half of the central angle made by joining the points of contact to the centre.
5. Draw a line segment AB = 9 cm. Taking A and B as centres draw two circles of radius 5 cm and 3 cm respectively. Now draw tangents to each circle from the centre of the other.
6. Draw a circle of radius 3.5 cm with centre O. Take point P such that OP = 6 cm. OP cuts the circle at T. Draw two tangents PQ and PR. Join Q to R. Through T draw AB parallel to QR such that A and B are point on PQ and PR.
7. Draw a circle of diameter 7 cm. Draw a pair of tangents to the circle, which are inclined to each other at an angle of 60°.
8. Draw a circle with centre O and radius 3.5 cm. Take a horizontal diameter. Extend it to both sides to point P and Q such that OP = OQ = 7 cm. Draw tangents PA and QB one above the diameter and the other below the diameter. Is PA || BQ? |
# Common Core: 1st Grade Math : Subtract Multiples of 10 From Multiples of 10: CCSS.MATH.CONTENT.1.NBT.C.6
## Example Questions
### Example Question #1 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #2 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #3 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Solve the following:
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #2 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #4 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #5 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #6 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #7 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left.
### Example Question #8 : Subtract Multiples Of 10 From Multiples Of 10: Ccss.Math.Content.1.Nbt.C.6
Explanation:
When we subtract two digit numbers, we always start with the ones place, on the right, and work our way to the left. |
# How to solve natural log equations
These sites allow users to input a Math problem and receive step-by-step instructions on How to solve natural log equations. We can solving math problem.
## How can we solve natural log equations
Do you need help with your math homework? Are you struggling to understand concepts How to solve natural log equations? Word math problems can be tricky, but there are a few tips that can help you solve them more quickly and easily. First, read the problem carefully and identify the key information. Then, determine what operation you need to perform in order to solve the problem. Next, write out the equation using numbers and symbols. Finally, solve the equation and check your work to make sure you've found the correct answer. By following these steps, you can approach word math problems with confidence and avoid making common mistakes. With a little practice, you'll be solving them like a pro in no time!
Solving an expression means to find the value of the variable(s) in the equation. In order to solve an expression, you need to use inverse operations to undo the operations that are performed on the variable(s). For example, if you have the expression 2x+3, and you want to solve for x, you would first use inverse operations to undo the addition. This would give you 2x=3. Then, you would use inverse operations to undo the multiplication, which would give you x=3/2. Solving an expression can be tricky, but with practice it can become easier. With a little bit of patience and some reverse operations, you'll be solving expressions like a pro!
How to solve partial fractions is actually not that difficult once you understand the concept. Partial fractions is the process of breaking up a fraction into simpler fractions. This is often done when dealing with rational expressions. To do this, you first need to find the greatest common factor of the numerator and denominator. Once you have found the greatest common factor, you can then divide it out of both the numerator and denominator. The next step is to take the remaining fraction and break it up into simpler fractions. This is often done by rewriting the fraction in terms of its simplest form. For example, if you have a fraction that is in the form of a/b, you can rewrite it as 1/b. In some cases, you may need to use more than one partial fraction to completely simplify a fraction. However, once you understand how to solve partial fractions, it should be a relatively straightforward process.
Composite functions can be used to model real-world situations. For example, if f(t) represents the temperature in degrees Celsius at time t, and g(t) represents the number of hours since midnight, then the composite function (fog)(t), which represents the temperature at a certain hour of the day, can be used to predict how the temperature will change over the course of 24 hours. To solve a composite function, it is important to understand the individual functions that make up the composite function and how they interact with each other. Once this is understood, solving a composite function is simply a matter of plugging in the appropriate values and performing the necessary calculations.
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# The $3 + 3 \times 3 - 3 + 3 = ?$ What is the right answer and how? A. 18B. 12C. 03D. 06
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Hint: Here in this question, we have to simplify the given terms. The question contains more than one arithmetic operation so we use the BODMAS rule. It is abbreviated as Brackets Of Division Multiplication Addition Subtraction. By applying the rule, we obtain the result.
Generally, we have 4 types of arithmetic operations namely.
Subtraction (-) which subtracts the numbers.
Multiplication ($\times$) which multiplies the numbers.
Division ($\div$) which divides the numbers.
To solve the above question, we follow the BODMAS rule and it is abbreviated as Brackets Of Division Multiplication Addition Subtraction. Here “of” signifies the operation multiplication. If we have “of” in the question we have to multiply, so any sums on simplification must be solved in that order.
We have to know about the sign conventions
$+ \,\, \times \,\, + \,\, = \,\, +$
$\Rightarrow - \,\, \times \,\, - \,\, = \,\, +$
$\Rightarrow + \,\, \times \,\, - \,\, = \,\, -$
$\Rightarrow - \,\, \times \,\, + \,\, = \,\, -$
Now we consider the given question
$3 + 3 \times 3 - 3 + 3$
First, perform the multiplication operation $\left( {3 \times 3 = 9} \right)$, then we get
$\Rightarrow 3 + 9 - 3 + 3$
On addition of the number $\left( {9 + 3 + 3 = 15} \right)$, we have
$15 - 3$
On subtracting the number 3 from 15, we have
$12$
Hence, the value of $3 + 3 \times 3 - 3 + 3$ equals $12$.
Therefore, option B is the correct answer.
Note: The knowledge of arithmetic operations and tables of multiplication plays an important role in solving these types of questions. BODMAS rule is applied where the question involves more than one arithmetic operation. If we simplify randomly we don’t obtain the correct solution. |
Conversion of Units of Measurement | For Grades 1-4
# Conversion of Unit of Measurement
## Conversion of Units of Measurement - Sub Topics
• What is the Unit of Conversion?
• How to Convert Unit of Measurement?
• Conversion of Units of Weight
• Conversion of Units of Capacity
• Conversion of Units of Length
• Conversion of Units of Time
• Comparison of Measurements
• ## What is the Unit of Conversion?
The "unit of conversion" refers to the specific unit or measurement that you are using to convert from one unit to another in a conversion process. In other words, it's the unit of measurement that you are starting with and want to convert into a different unit, for example, converting kilometres to metres, milligrams to kilograms or days to hours. These units of conversion make it easier to compare and understand quantities expressed in different units.
## How to Convert Unit of Measurement?
To convert units of measurement, follow these steps:
1. Identify the starting unit and target unit.
2. Find the conversion factor relating the two units (e.g., 1 metre = 100 centimetres).
3. Use multiplication or division to apply the conversion factor to the quantity being converted.
4. Perform the calculation to obtain the converted value in the target unit.
5. Express the result with the appropriate unit and precision.
Remember that the numerical value may change, but the quantity remains the same, ensuring accurate communication and compatibility across different systems.
## Conversion of Units of Weight
1 kilogram (kg) is equal to 1000 grams (g).
To convert from kg to g: Multiply the number of kilograms by 1000.
Example: 5 kg = 5 x 1000 = 5000 g
To convert from g to kg: Divide the number of grams by 1000.
Example: 5000 g = 5000/1000 = 5 kg
## Conversion of Units of Capacity
To convert litres (L) to millilitres (mL): Multiply the number of litres by 1000.
Example: 2 litres = 2 x 1000 = 2000 millilitres
To convert millilitres (mL) to litres (L): Divide the number of millilitres by 1000.
Example: 2000 millilitres = 2000 / 1000 = 2 liters
## Conversion of Units of Length
1 kilometre (km) is equal to 1000 metres (m).
To convert kilometres(km) to metres(m): Multiply the number of kilometres by 1000.
Example: 7 km = 7 x 1000 = 7000 metres
To convert metres (m) to kilometres(km): Divide the number of metres by 1000.
Example: 7000 metres = 7000 / 1000 = 7 km
## Conversion of Units of Time
1 hour(h) is equal to 60 minutes(m).
To convert hours to minutes: Multiply the number of hours by 60.
For example, 3 hours = 3 x 60 = 180 minutes
To convert minutes to seconds: Multiply the number of minutes by 60:
For example, 180 minutes = 180 x 60 = 10,800 seconds
To convert minutes to hours: Divide the number of minutes by 60:
For example, 180 minutes ÷ 60 = 3 hours
To convert seconds to minutes: Divide the number of seconds by 60:
For example, 10,800 seconds ÷ 60 = 180 minutes
## Conversion of Day to Hours, Week to Days, Year to Days
Day to hours:
Conversion factor: 24 hours/day
Example: 3 days = 3 x 24 hours = 72 hours
Week to days:
Conversion factor: 7 days/week
Example: 2 weeks = 2 x 7 days = 14 days
Year to days:
Conversion factor: 365 days/year (or 366 days in a leap year)
Example 1: 1 year = 365 days = 365 x 1 = 365 days
Example 2: 2 years = 2 x 365 days = 730 days
## Use of >, < and = in Problems Based on Comparison of Measurements
The symbols '>', '<' and '=' are used in problems based on comparison of measurements to represent the relative size of two or more quantities.
1. Greater than: The symbol '>' is used to indicate that one quantity is greater in value than another. For example:
A = 51 cm
B = 49 cm
A > B
2. Less than: The symbol '<' is used to indicate that one quantity is smaller in value than another. For example:
X = 12 cm
Y = 13 cm
X < Y
3. Equal to: The symbol '=' is used to indicate that two or more quantities have the same value. For example:
P = 45 cm
Q = 45 cm
P = Q
Note: It is important to keep in mind that these symbols are used to compare only the values of the quantities not the units of measurement. |
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how to find a nonlinear equation from a table
Assuming you want a conic section (as implied by your "Line, Parabola, Hyperbola etc"): in general $a x^2 + b x y + c y^2 + d x + e y + f = 0$; you get five linear equations in the parameters $a,b,\ldots f$ by plugging in your given points for $(x,y)$. Enter in a value of 0.03 for f … Often, students are asked to write the equation of a line from a table of values. We will substitute $y=3x - 5$ into the equation for the circle. Answer: (2, –1) Therefore, the solution set to the given system of nonlinear equations consists of two points which are (– 3, 4) and (2, –1). In the unit on Slope, we talked about measuring the slope of a straight line.Now we will discuss how to find the slope of a point on a curve. One of the differences between the slope of a straight line and the slope of a curve is that the slope of a straight line is constant, while the slope of a curve changes from point to point.. As you should recall, to find the slope of a line you need to: nonlinear. If the nonlinear algebraic system is a polynomial equation, we could use the MATLAB routine roots to find the zeros of the polynomial. answer choices . All quizzes. Solving for $y$ gives $y=2$ and $y=1$. Suppose two people, Fermat and Sophie, go out for a jog. Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line. Identifying a possible non-linear rule for a given table of values Question 1. Yes, but because $x$ is squared in the second equation this could give us extraneous solutions for $x$. Figure 2 illustrates possible solution sets for a system of equations involving a parabola and a line. Menu. A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. To solve this kind of problem, simply chose any 2 points on the table and follow the normal steps for writing the equation of a line from 2 points. For example, suppose a problem asks you to solve the following system: Doesn’t that problem just make your skin crawl? After you solve for a variable, plug this expression into the other equation and solve for the other variable just as you did before. While this type of depreciation applies to many model… Expand the equation and set it equal to zero. We define the system LHS equations in A1:A3 using X1:X3 for variables with 1 for the initial guess as shown in Table 1. Substitute the value of the variable into the nonlinear equation. When you distribute the y, you get 4y 2 + 3y = 6. In this lesson you will learn how to write a quadratic equation by finding a pattern in a table. The line is tangent to the parabola and intersects the parabola at exactly one point. Before analyzing the solutions to the nonlinear population model, let us make a pre-liminary change of variables, and set u(t) = N(t)/N⋆, so that u represents the size of the population in proportion to the carrying capacity N⋆. Reports. Remember that you’re not allowed, ever, to divide by a variable. The relationship between two variables, x and y, is shown in the table. equation. BACK TO EDMODO. A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. There is actually a way to do that. A system of equations where at least one equation is not linear is called a nonlinear system. The following table shows the raw data for performing nonlinear regression using Polymath (refer Table E7-4.1, Elements of chemical reaction engineering, 5th edition) Pco The nonlinear equation is given by Rate=a Pco ℎ21 1+ ℎ22 To do the nonlinear regression of the above data, first open Polymath. In a nonlinear system, at least one equation has a graph that isn’t a straight line — that is, at least one of the equations has to be nonlinear. Solve the first equation for $x$ and then substitute the resulting expression into the second equation. On the other hand, Fermat is planning on running an out-and-back course, starting and ending at his house. For example, if you were to buy a car for $25,000, and it depreciates in value by$2000 per year, then the car's value can be modeled using the following function: 1. f(x) = 25000 - 2000x, where xis the number of years that have passed since purchasing the car. A linear function graphs as a straight line. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Email address. The line crosses on the inside of the parabola and intersects the parabola at two points. Notice that $-1$ is an extraneous solution. This solution set represents the intersections of the circle and the parabola given by the equations in the system. When y is 0, 9 = x2, so, Be sure to keep track of which solution goes with which variable, because you have to express these solutions as points on a coordinate pair. Sophie is planning on ending her jog at a park, so she is getting further and further from her house as she jogs. An equation containing at least one differential coefficient or derivative of an unknown variable is known as a differential equation. She is the author of several For Dummies books, including Algebra Workbook For Dummies, Algebra II For Dummies, and Algebra II Workbook For Dummies. Follow these steps to find the solutions: Solve for x2 or y2 in one of the given equations. Any equation that cannot be written in this form in nonlinear. This type of depreciation can easily be modeled using a function. This function could be written with the linear equation y = x + 2. The substitution method we used for linear systems is the same method we will use for nonlinear systems. No solution. There are several ways to solve systems of nonlinear equations: ... We can substitute this value of x into the first equation to find all possible values for y. A differential equation can be either linear or non-linear. The general form of a nonlinear equation is ax 2 + by 2 = c, where a, b, c are constants and a 0 and x and y are variables. Is the function represented by the equation linear or nonlinear? Multiple Relationships (graphs, tables, equations) 1.1k plays . Where x and y are the variables, m is the slope of the line and c is a constant value. Quiz not found! Two solutions. Because you found two solutions for y, you have to substitute them both to get two different coordinate pairs. There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line. Create a new teacher account for LearnZillion. • With nonlinear functions, the differences between the corresponding y-values are not the same. However, finding the differences between those differences produces an interesting pattern. Because this equation is quadratic, you must get 0 on one side, so subtract the 6 from both sides to get 4y 2 + 3y – 6 Recall that a linear equation can take the form $Ax+By+C=0$. Four is the limit because conic sections are all very smooth curves with no sharp corners or crazy bends, so two different conic sections can’t intersect more than four times. The line intersects the circle at $\left(2,1\right)$ and $\left(1,-2\right)$, which can be verified by substituting these $\left(x,y\right)$ values into both of the original equations. Her distance from her house can be modeled by the function y = 4x, where x is the number of hours she has been jogging for. x = 2. x=2 x = 2, solve for. Introduction In Chapter 03.03, the bisection method described as one of the simple bracketing was methods of solving a nonlinear equation of the general form . No solution. Create a new quiz. It will depend on your knowledge of how different equations tend to form graphs. Next, substitute each value for $y$ into the first equation to solve for $x$. The line is tangent to the circle and intersects the circle at exactly one point. Recall that a linear equation can take the form $Ax+By+C=0$. For data in a table or dataset array, you can use formulas represented as the variable names from the table or dataset array. Tap for more steps... Simplify each equation. Figure 4 illustrates possible solution sets for a system of equations involving a circle and a line. Create your free account Teacher Student. Recall that a linear equation can take the form $Ax+By+C=0$. To see if a table of values represents a linear function, check to see if there's a constant rate of change. Understanding the difference between linear and nonlinear equations is foremost important. Solve the nonlinear equation for the variable. Prior to using Chart Wizard, we need to select the data (table of values) we wish to graph. Find a quiz. Substitute the value of the variable into the nonlinear equation. x + y = 1. Any equation that cannot be written in this form in nonlinear. 0. Solving for one of the variables in either equation isn’t necessarily easy, but it can usually be done. You’ll use the “Outputs” table to calculate the left and right side of the Colebrook equation. This tutorial shows you how to tell if a table of values represents a linear function. And any time you can solve for one variable easily, you can substitute that expression into the other equation to solve for the other one. When you plug 3 + 4y into the second equation for x, you get (3 + 4y)y = 6. Substitute the value from Step 1 into the other equation. Let y = mx + c be the equation. Now, we factor and solve for $x$. The constant term is 1 which is the case for all the alternatives. x2 + y = 5, x2 + y2 = 7 xy + x − 4y = 11, xy − x − 4y = 4 3 − x2 = y, x + 1 = y xy = 10, 2x + y = 1 His distance from his house can be … One solution. Email confirmation. One method of finding the correct answer is to try each of the options with a value of x.If an option does not give the correct y value it cannot be a correct response to the question.. Password. Solve a = 2 - b for a. The equation becomes y … In this example, the top equation is linear. Note that the inequalities formulas are listed after the equality formula as required by the solver. In this situation, you can solve for one variable in the linear equation and substitute this expression into the nonlinear equation, because solving for a variable in a linear equation is a piece of cake! You now have y + 9 + y2 = 9 — a quadratic equation. Remember that equations and inequalities formulas are defined with respect to zero on one side, and any inequalities are interpreted as greater than zero by the solver. You may be familiar with the belief that once you buy a new car, it's already depreciated in value as soon as you've driven it off the lot. One of the equations has already been solved for $y$. When plotted on the graph we get the below curve. 9 = 0x + c. i.e. Any equation that cannot be written in this form in nonlinear. Graphically, we can think of the solution to the system as the points of intersections between the linear function. Solve the nonlinear equation for the variable. Yes. Because this equation is quadratic, you must get 0 on one side, so subtract the 6 from both sides to get 4y2 + 3y – 6 = 0. If one equation in a system is nonlinear, you can use substitution. These unique features make Virtual Nerd a viable alternative to private tutoring. Always substitute the value into the linear equation to check for extraneous solutions. Use the zero product property to solve for y = 0 and y = –1. Unless one variable is raised to the same power in both equations, elimination is out of the question. The scope of this article is to explain what is linear differential equation, what is nonlinear differential equation, and what is the difference between linear and nonlinear differential equations. Find the intersection of the given circle and the given line by substitution. Subtract 9 from both sides to get y + y2 = 0. For example, follow these steps to solve this system: Solve the linear equation for one variable. Substitute the two x-values into the original linear equation to solve for $y$. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. Two solutions. OBS – Using Excel to Graph Non-Linear Equations March 2002 Using Chart Wizard Selecting Data on the Spreadsheet Chart Wizard is a four-step process for creating graphs. Calculate the values of a and b. My quizzes. After you set up those calculations, it will be easy to use Excel to iterate through guesses to determine the value of f that causes the left side of the equation to equal the right side. In this non-linear system, users are free to take whatever path through the material best serves their needs. Put the response variable name at the left of the formula, followed by a ~, followed by a character vector representing the response formula.. y. y y. The line does not intersect the circle. The general representation of nonlinear equations is; ax2 + by2 = c. The substitution method we used for linear systems is the same method we will use for nonlinear systems. When you distribute the y, you get 4y2 + 3y = 6. Don’t break out the calamine lotion just yet, though. The solutions are $\left(1,2\right)$ and $\left(0,1\right),\text{}$ which can be verified by substituting these $\left(x,y\right)$ values into both of the original equations. c = 9. The user must create a vector of the coefficients of the polynomial, in descending order, p = [1 5 … Substitute the expression obtained in step one into the equation for the circle. Problem 4. Identifying a possible non-linear rule for a given table of values Solution (substitution) When x = 0, y = 1. You will also need to get the pairs out of the graph. Writing Equation from Table of Values. f (x Substitute the value(s) from Step 3 into either equation to solve for the other variable. Unlike linear systems, many operations may be involved in the simplification or solving of these equations. The second equation is attractive because all you have to do is add 9 to both sides to get y + 9 = x2. Examples of nonlinear equations include, but are not limited to, any conic section, polynomial of degree at least 2, rational function, exponential, or logarithm. Q. One solution. Substitute the expression obtained in step one into the parabola equation. This example shows how to create a character vector to represent the response to the reaction data that is in a dataset array. SURVEY . Your answers are. Your pre-calculus instructor will tell you that you can always write a linear equation in the form Ax + By = C (where A, B, and C are real numbers); a nonlinear system is represented by any other form. Here’s what happens when you do: Therefore, you get the solutions to the system: These solutions represent the intersection of the line x – 4y = 3 and the rational function xy = 6. If both of the equations in a system are nonlinear, well, you just have to get more creative to find the solutions. Difference Between Linear and Nonlinear Equations. Build a set of equations from the table such that q ( x) = a x + b. If you solve for x, you get x = 3 + 4y. Solve the linear equation for one of the variables. All fields are required. Mary Jane Sterling aught algebra, business calculus, geometry, and finite mathematics at Bradley University in Peoria, Illinois for more than 30 years. You have to use the quadratic formula to solve this equation for y: Substitute the solution(s) into either equation to solve for the other variable. Just remember to keep your order of operations in mind at each step of the way. If there is, you're looking at a linear function! A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Putting x = 0, y = 9 in the equation y = mx + c, we get. 1. follow the algorithm of the false-position method of solving a nonlinear equation, 2. apply the false-position method to find roots of a nonlinear equation. It forms a curve and if we increase the value of the degree, the curvature of the graph increases. 2 = a ( 1) + b 162 = a ( 9) + b 8 = a ( 2) + b 128 = a ( 8) + b 18 = a ( 3) + b. linear. When both equations in a system are conic sections, you’ll never find more than four solutions (unless the two equations describe the same conic section, in which case the system has an infinite number of solutions — and therefore is a dependent system). Tags: Question 6 . y = a x + b. There is, however, a variation in the possible outcomes. Solve the given system of equations by substitution. This gives us the same value as in the solution. Who says it is nonlinear ? $\begin{array}{l}x-y=-1\hfill \\ y={x}^{2}+1\hfill \end{array}$, $\begin{array}{llll}x-y=-1\hfill & \hfill & \hfill & \hfill \\ \text{ }x=y - 1\hfill & \hfill & \hfill & \text{Solve for }x.\hfill \\ \hfill & \hfill & \hfill & \hfill \\ \text{ }y={x}^{2}+1\hfill & \hfill & \hfill & \hfill \\ \text{ }y={\left(y - 1\right)}^{2}+1\hfill & \hfill & \hfill & \text{Substitute expression for }x.\hfill \end{array}$, $\begin{array}{l}y={\left(y - 1\right)}^{2}\hfill \\ \text{ }=\left({y}^{2}-2y+1\right)+1\hfill \\ \text{ }={y}^{2}-2y+2\hfill \\ 0={y}^{2}-3y+2\hfill \\ \text{ }=\left(y - 2\right)\left(y - 1\right)\hfill \end{array}$, $\begin{array}{l}\text{ }x-y=-1\hfill \\ x-\left(2\right)=-1\hfill \\ \text{ }x=1\hfill \\ \hfill \\ x-\left(1\right)=-1\hfill \\ \text{ }x=0\hfill \end{array}$, $\begin{array}{l}y={x}^{2}+1\hfill \\ y={x}^{2}+1\hfill \\ {x}^{2}=0\hfill \\ x=\pm \sqrt{0}=0\hfill \end{array}$, $\begin{array}{l}y={x}^{2}+1\hfill \\ 2={x}^{2}+1\hfill \\ {x}^{2}=1\hfill \\ x=\pm \sqrt{1}=\pm 1\hfill \end{array}$, $\begin{array}{l}3x-y=-2\hfill \\ 2{x}^{2}-y=0\hfill \end{array}$, $\begin{array}{l}{x}^{2}+{y}^{2}=5\hfill \\ y=3x - 5\hfill \end{array}$, $\begin{array}{c}{x}^{2}+{\left(3x - 5\right)}^{2}=5\\ {x}^{2}+9{x}^{2}-30x+25=5\\ 10{x}^{2}-30x+20=0\end{array}$, $\begin{array}{l}10\left({x}^{2}-3x+2\right)=0\hfill \\ 10\left(x - 2\right)\left(x - 1\right)=0\hfill \\ x=2\hfill \\ x=1\hfill \end{array}$, $\begin{array}{l}y=3\left(2\right)-5\hfill \\ =1\hfill \\ y=3\left(1\right)-5\hfill \\ =-2\hfill \end{array}$, $\begin{array}{l}{x}^{2}+{y}^{2}=10\hfill \\ x - 3y=-10\hfill \end{array}$, CC licensed content, Specific attribution, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. When you plug 3 + 4y into the second equation for x, you get (3 + 4y)y = 6. The line crosses the circle and intersects it at two points. The line will never intersect the parabola. This tells Chart wizard what to graph. Name. All quizzes. … The substitution method we used for linear systems is the same method we will use for nonlinear systems. Consider the same function f(x) = x3 - 5x2-x +2 that we discussed earlier. The general representation of linear equation is; y = mx +c. This example uses the equation solved for in Step 1. • A table can be used to determine whether ordered pairs describe a linear or nonlinear relationship. 30 seconds . You must factor out the greatest common factor (GCF) instead to get y(1 + y) = 0. |
# Counting and Probability for AMC-8
The notion of probability arose from games of chance like lotteries, bingos, and roulette. Though probability has now become a sophisticated field of Mathematics involving the use of Measure Theory and Analysis, its basics can be learnt using principles of counting.
if we have $x$ ways of doing something and $y$ ways of doing another thing and we can not do both at the same time, then there are $x+y$ ways to choose one of the actions.
In terms of set theory, if $S_1,S_2,\cdots S_n$ are disjoint sets, then $|\bigcup_i S_i|=\sum_i |S_i|$.
Example : On a typical saturday evening, Jill either watches a film or a play. This week, there are 4 movies on the charts and the theatre has 3 plays. In how many ways can Jill spend her evening?
The Multiplication Principle
Suppose that a task can be done in two parts. If the first part can be done in $a$ ways and the second in $b$ ways, then the task can be performed in $a\cdot b$ ways.
In terms of set theory, $|S_1\times S_2\times\cdots\times S_n|=\prod_i |S_i|$.
Example : Suppose that all roads from city A to city C pass through city B. There are 3 paths from city A to city B and 4 paths from city B to city C. How many paths are there from city A to city C?
Answer: $3\times 4=12$.
Sample Space and Events
Consider an experiment that can produce various outcomes. The set of all possible outcomes is called the sample space ($S$) of the experiment. Elements of the power set $P(S)$ of the sample space are known as events.
If event A and event B never occur simultaneously on a single performance of an experiment, then they are called mutually exclusive events.
Classical definition of probability
Given an event $A$,
$P(A):=\frac{|A|}{|S|}$.
Lesson Content
0% Complete 0/1 Steps |
## Solve Lesson 11 Page 62 SBT Math 10 – Kite>
Topic
In the coordinate plane Oxygenfor three noncollinear points USA(twelfth), WOMEN(thirty first), P(− 1 ; 2). Find the coordinates of the point Q such that quadrilateral MNPQ is a trapezoid with MN // PQ and PQ = 2MN.
From the assumption find the coordinates of the point Q satisfy $$\overrightarrow {PQ} = 2\overrightarrow {NM}$$
Detailed explanation
We have: MN // PQ so $$\overrightarrow {MN}$$ and $$\overrightarrow {PQ}$$ have the same direction
Other way, PQ = 2MN $$\Rightarrow \overrightarrow {PQ} = 2\overrightarrow {NM}$$
Call point coordinates Q is $$Q(a;b)$$. We have: $$\overrightarrow {PQ} = (a + 1;b – 2)$$ and $$\overrightarrow {NM} = ( – 2; – 3)$$
$$\Rightarrow \overrightarrow {PQ} = 2\overrightarrow {NM} \Leftrightarrow \left\{ \begin{array}{l}a + 1 = 2.( – 2)\\b – 2 = 2.( – 3)\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a + 1 = – 4\\b – 2 = – 6\end{array} \right \Leftrightarrow \left \{ \begin{array}{l}a = – 5\\b = – 4\end{array} \right.$$ . So Q(-5 ; -4) |
Changing Percents to Decimals
# Changing Percents to Decimals
Need some basic practice with decimals and percents first? Changing Decimals to Fractions and Multiplying and Dividing Decimals by Powers of Ten and Changing Decimals to Percents
Whenever computations need to be done with percents, the percents are first renamed as decimals.
Here are examples of changing percents to decimals. Notice that the $\,\%\,$ symbol is replaced with a factor of $\,\frac{1}{100}\,$:
## Examples
$3.5\% = 3.5\cdot\frac{1}{100} = \frac{3.5}{100} = 0.035$
$25\% = 25\cdot \frac{1}{100} = \frac{25}{100} = 0.25$
$50\% = 50\cdot \frac{1}{100} = \frac{50}{100} = 0.5$
$100\% = 100\cdot \frac{1}{100} = \frac{100}{100} = 1$
$250\% = 250\cdot \frac{1}{100} = \frac{250}{100} = 2.5$
As these examples illustrate, the percent symbol instructs multiplication by $\,\frac{1}{100}\,$ (or, equivalently, division by $\,100\,$), which is accomplished by moving the decimal point two places to the left. Remember that if you don't see a decimal point, then it gets inserted just to the right of the ones place.
Now, you can go from percents to decimals in one easy step,
by moving the decimal point two places to the left:
## Examples
$3\% = 0.03$
$2.37\% = 0.0237$
$0.01\% = 0.0001$
$5032\% = 50.32$
Some of my students find this memory device helpful: PuDdLe DiPpeR (Imagine a little kid, barefoot, dipping piggy-toes in puddles!)
Percent to Decimal, two places to the Left
Decimal to Percent, two places to the Right
## Practice
Here, you will practice renaming percents as decimals. For example, $\,54\%\,$ gets renamed as $\,0.54\,$. |
# Section D: Practice Problems Add and Subtract
## Section Summary
Details
In this section, we used our understanding of place value and expanded form to add and subtract large numbers using the standard algorithm.
We learned how to use the algorithm to keep track of addition of digits that results in a number greater than 9.
Whenever we have 10 in a unit, we make a new unit and record the new unit at the top of the column of numbers in the next place to the left.
When we subtract numbers it may be necessary to decompose tens, hundreds, thousands or ten-thousands before subtracting.
Finally, we learned that if the digit we are subtracting is a zero, we may need to decompose one unit of the digit in the next place to the left.
Sometimes, it is necessary to look two or more places to the left to find a unit to decompose. For example, here is one way to decompose a ten and a thousand to find .
## Problem 1 (Lesson 18)
Clare took 11,243 steps on Saturday and 12,485 steps on Sunday.
1. How many steps did Clare take altogether on Saturday and Sunday?
2. How many more steps did Clare take on Sunday than on Saturday?
## Problem 2 (Lesson 19)
1. Find the value of the sum. Explain your calculations.
2. Find the value of the difference. Explain your calculations.
## Problem 3 (Lesson 20)
Find the value of each sum and difference using the standard algorithm.
## Problem 4 (Lesson 21)
Here is how Han found
1. How can you tell by estimating that Han has made an error?
2. What error did Han make?
3. Find the value of .
## Problem 5 (Lesson 22)
In 2018 the population of Boston is estimated as 694,583 and the population of Seattle is estimated as 744,995.
1. Is the population difference between Boston and Seattle more or less than 100,000? Explain how you know.
2. Is the population difference more or less than 50,000? Explain how you know.
3. Find the difference in the populations of the two cities.
## Problem 6 (Exploration)
Han says he has a method to find the value of without any carrying: “I just write 1,000,000 as .
1. How might rewriting 1,000,000, as Han suggested, help with finding the difference of ?
2. Try Han’s method to find .
## Problem 7 (Exploration)
Use the information to determine when the airplane, telephone, printing press, and automobile were first invented.
• The airplane was invented in 1903.
• The printing press was invented 453 years before the most recent invention.
• The automobile was invented 15 years before 1900.
• It was 426 years after the invention of the printing press that the telephone was invented.
• The automobile and telephone were invented the closest together in time with only 9 years between them. |
# How do you find the domain and range of f(x)=(12x)/(x^2-36)?
Jul 16, 2018
Below
#### Explanation:
Looking at the graph, you can immediately see that there are 2 vertical asymptotes because ${x}^{2} - 36 \ne 0$ so $x = \pm 6$ are the vertical asymptotes. Therefore, the graph cannot have the points with the x-coordinates $x = 6$ and $x = - 6$
The horizontal asymptote is $y = 0$ since the degree of the numerator is less than the degree of the denominator. This is because if you imagine letting $x$ be any number, then ${x}^{2} - 36$ will be a whole lot bigger than $12 x$ and since the small number divided by a larger number, then $\frac{12 x}{{x}^{2} - 36} \to 0$
Therefore, the graph cannot have the points with the y-coordinate $y = 0$
However, what asymptotes really tell you about the graph is that the end points of the graph will be approaching the horizontal and vertical asymptotes but they will never touch the asymptotes. Basically, it tells about the shape of the graph which can help you determine the domain and range of the graph.
Intercepts
When $y = 0$, $x = 0$
When $x = 0$, $y = 0$
You will notice that the graph can pass through $\left(0 , 0\right)$ but the endpoints of the graph will be approaching $y = 0$ and not cross $y = 0$. This is because asymptotes affect the end points only.
Hence,
Domain: all reals $x$ except when $x = \pm 6$
Range: all reals $y$
Below is the graph
graph{(12x)/(x^2-36) [-10, 10, -5, 5]}
Jul 16, 2018
The domain is x in (-oo, -6)uu-6,6)uu(6,+oo). The range is $y \in \mathbb{R}$
#### Explanation:
The denominator must be $\ne 0$
Therefore,
${x}^{2} - 36 \ne 0$
$\left(x + 6\right) \left(x - 6\right) \ne 0$
$x \ne - 6$ and $x \ne 6$
The domain is x in (-oo, -6)uu-6,6)uu(6,+oo)
To find the range, let
$y = \frac{12 x}{{x}^{2} - 36}$
$y \left({x}^{2} - 36\right) = 12 x$
$y {x}^{2} - 12 x - 36 y = 0$
This is a quadratic equation in $x$ and in order to have solutions, the discriminant $\ge 0$
Therefore,
$\Delta = {\left(- 12\right)}^{2} - 4 \left(y\right) \left(- 36 y\right)$
$= 144 + 144 {y}^{2} \ge 0$
$\implies$, $144 \left(1 + {y}^{2}\right) \ge 0$
Therefore,
$\forall y \in \mathbb{R} , \Delta > 0$
The range is $y \in \mathbb{R}$
graph{12x/(x^2-36) [-32.49, 32.46, -16.24, 16.25]} |
systems of equations lab
Since there is no imaginary part, no oscillation occurs. They cost \$7.50 if purchased at the game. The inputs to solve are a vector of equations, and a vector of variables to solve the equations for. Example (Click to view) x+y=7; x+2y=11 Try it now. A system of linear equations is two or more linear equations that are being solved simultaneously. First, we will practice graphing two equations on the same set of axes, and then we will explore the different considerations you need to make when graphing two linear inequalities on the same set of axes. Before we begin, we'll introduce some terminology. She needs to make 200 milliliters of a 40% solution of sulfuric acid for a lab experiment. 522 Systems of Differential Equations Let x1(t), x2(t), x3(t) denote the amount of salt at time t in each tank. A system of equations can also be solved by graphing both equations and finding the point where they intersect. One way is to use a two-output call. Subjects. Then, solve. Therefore, the salt in all the tanks is eventually lost from the drains. Systems of Linear Equations in Two Variables: Given 2−5= 3 −2= 9 4+ 2= 12 −2−= −6 + = 3 2+ 2= 7 Solve Algebraically −2= 9 = 2+ 9 2−5() = 3 Even though we hadn’t formally introduced systems of equations yet, students looked at a variety of situations with systems of equations. One fundamental property of all linear dynamical systems (meaning most objects that move) is that they have resonances. The stability of this system is completely determined by the eigenvalues of (A + BF). These images are provided by NASA and are in the public domain. syms x y a. Note that when the matrix produced by eigvec is nonsingular, A must be diagonalizable. What are the eigenvalues and eigenvectors of the matrix, According to the mathematical definitions, is the system. This is a differential equation that describes the effect of rate of change of rudder angle on the rate of change in yaw. We suppose added to tank A water containing no salt. Here, u is the rate of change of rudder angle, and the four components of x are: A is the 4×4 matrix , and B is the 4×1 matrix . While we could illustrate this with fluids in pipes, electrical circuits, signals in the air, the effects of earthquakes on buildings, and more, we have chosen to illustrate this property with an airplane. sol = solve ( [eqn1, eqn2, eqn3], [x, y, z]); xSol = sol.x ySol = sol.y zSol = sol.z. Example A 2 F4 U L4 4 E5 U L21 High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. If all lines converge to a common point, the system is said to be consistent and has a … In this section, we will look at systems of linear equations and inequalities in two variables. The quadrant in which the solution to the following system lies: The sum of two numbers is 104. Christa Lemily. Write the general solution of the system (6) in your Word document. For a particular game, 600 tickets were sold and the receipts were \$3500. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. Students love this game and they really get into completing their work while playing it. These are associated with eigenvalues and eigenvectors of the coefficient matrix of the system. in Three Variables. What was the cost of one shirt? Write equations that represent the cost of printing digital photos at each lab. For linear systems, they combine very well with t… There is a conjugate pair of complex eigenvalues (call them. Indeed, this eigenvalue has little effect on the performance of the airplane. The amount of money each child received when Mr. Vogel left \$25,000 divided between his son and daughter, with the daughter receiving \$5000 less than the son. MAT131 Lab 4 Systems of Equations Objectives. So, the solution of the system is (6, ±2). This illustration of Dutch roll was made by Wikipedia user Picascho and is in the public domain. Or click the example. What happens to the system as. Which component is biggest? Here, up(t) represents the pilot's instructions to the rudder, and the product Fx(t) is what the plane's computer tells the rudder to do in order to damp the plane's bad resonant oscillations. solx = 0 a soly = -2*a 0. The yaw rate model is the 4×4 system. High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. This lesson covers Section 9.4: Nonlinear Systems of Equations in Two Variables. There are several ways to address the output of solve . Graph the equations 8 x ± 4y = 50 and x + 4 y = ±2. Ultimately, the goal of this exercise is not to design a real control system, but rather to demonstrate that eigenvalues and eigenvectors are associated with basic behaviors (the resonant modes) of the airplane. Lesson Notes. The cascade is modeled by the chemical balance law rate of change = input rate − output rate. Systems of Equations and Inequalities. Plan your 60-minute lesson in Math or Systems of Equations and Inequalities with helpful tips from Christa Lemily For most values of the unknowns, the equation will be false: \$y + 1 = 3\$ is a false statement for infinitely many possible choice… Set up Graph mode by selecting it from the Menu in the top left. Roll is the angle by which the wings deviate from being level, so that one wing rises up and the other drops down. Lab 5: Systems of Equations My Solutions > Heat is conducted along a metal rod positioned between two fixed temperature walls. WeBWorK. In this lab, we saw how matrices and a little bit of linear algebra can give us powerful tools for working with linear systems, even very large ones. Solve a system of nonlinear equations in two variables. We will also learn about a very useful application of systems of linear equations to economics and computer science. Question: 2.8 MATLAB: Solve Systems Of Linear Equations Revisited LAB ACTIVITY 2.8.1: MATLAB: Solve Systems Of Linear Equations Revisited Recorded A Page Refresh May Be Needed To Fill The Hanner This Tool Is Provided By A Third Party. xSol = 3 ySol = 1 zSol = -5. solve returns the solutions in a structure array. What is \$15,000 for the son and \$10,000 for the daughter. In practice, models requiring many differential equations are much more common than models using only one. Solve the following system of equations all three ways: Graphing: Elimination: Substitution: STATION E: Define the variables and write a system of equations to represent each situation. At any time t, sensors tell us the state x(t) of the plane, and (roughly speaking) we can at that time ensure that that the rate of change of rudder angle u(t) is whatever we want. Solution of a System In general, a solution of a system in two variables is an ordered pair that makes BOTH equations true. Learn how to set up a mathematical model in Excel. where F = [F1 F2 F3 F4] is a 1×4 matrix. This corresponds to the responsiveness of the airplane to the pilots' commands, which is very desirable. Solving Systems of Linear Equations UVU Math Lab. CCSS MODELING Refer to the table below. In this scenario, our design amounts to choosing the entries of F: F1, F2, F3, and F4. Press F11 Select menu option View > Enter Fullscreen for full-screen mode. More than one equation to be solved at the same time is know as this., The ratio of rise to run, The answer to an equation or system of equations., y = mx + b is more formally known as this. Large commercial airplanes require a yaw damper. Below is an example. Based on this mathematical model, engineers design and implement a control algorithm called a yaw damper that automatically moves the rudder back and forth and compensates for this phenomenon. , λn. The first system of equations is represented by intersecting lines, which shows that the system is consistent and has a unique solution, i.e., x = -1, y = 2 (see the first observation table). Solving systems of equations using substitiuon Khan Academy: Systems of Equations with Substitution Practice Interactive Systems of equations exercises MathCat: Solving Systems Using Substitution Interactive Solving systems of equations using substitution ... Online Math Lab Home: The resources in this bundle make a perfect addition to the math classroom for differentiated instruction, 13 Under what … Systems of differential equations constitute the mathematical models central to many technological and scientific applications. Exactly what is involved in designing this control system? Based on that, which type of rotation is this eigenvector most closely associated with: yaw, roll, or pitch?
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Module 4 topic 3 similar problems
1. 1. Module 4 Topic 3<br />Writing Equations of Linear Functions<br />
2. 2. Example 1<br />Similar Problem: What is the equation in slope-intercept form of the following?<br />Let's pretend that the x-intercept is 3 so the ordered pair would be (3, 0) and the other ordered pair is (1, 1).<br />1) Find the slope.<br />(3, 0) and (1, 1)<br />m = (1 - 0)/(1 - 3) = 1/-2<br />m = -1/2<br />2) Use the slope and either point to plug into point-slope form, let's use (1, 1)<br />y - y = m(x - x)<br />y - 1 = -1/2(x - 1)<br />y - 1 = -1/2 x + 1/2<br />y = -1/2 x + 3/2<br />
3. 3. Example 2<br />Similar Problem: What is the equation of the following line?<br />Anytime there is a horizontal line the equation of the line is always going to be y = y. If you were to take the ordered pairs off of the line given you would have (2, 3), (5, 3), (4, 3), (-2, 3), etc.<br />So the equation of the line is y = 3.<br />
4. 4. Example 3<br />Similar Problem: What is the equation of the following line?<br />Anytime there is a vertical line the equation of the line is always going to be x = x. If you were to take the ordered pairs off of the line given you would have (2, 0), (2, 1), (2, 2), (2, 3), etc.<br />So the equation of the line is x = 2. <br />
5. 5. Example 4<br />Similar Problem: Write an equation in slope-intercept form of the line that passes through the point (-5, 2) and has a slope of -5/4.<br />Use the point and the slope and plug into point-slope form.<br />Solve for y.<br />y - y = m(x - x)<br />y - 2 =-5/4 (x - -5)<br />y - 2 = -5/4 (x + 5)<br />y - 2 = -5/4 x + -25/4<br />y = -5/4 x - 17/4<br />
6. 6. Example 5<br /> Similar Problem: What is the equation in slope-intercept form of the line through the points (3, -1) and (-2, 9)?<br />1) Find the slope.<br />m = (9 - -1)/(-2 - 3)<br />m = 10/-5 = -2<br />2) Use the slope and either point to plug into point slope form, let's use (3, -1)<br />y - y = m(x - x)<br />y - -1 = -2(x - 3)<br />y + 1 = -2(x - 3)<br />y + 1 = -2x + 6<br />y = -2x - 5<br />
7. 7. Example 6<br />Similar Problem: Write the equation in slope-intercept form: 6x - 4y = 24.<br />6x - 4y = 24 (solve for y.)<br />Subtract 6x to both sides.<br />-4y = -6x + 24<br />Divide each side by -4.<br />y = 3/2 x - 6<br />
8. 8. Example 7<br />Similar Problem: Write the equation in slope-intercept form: 6x - 4y = 24.<br />6x - 4y = 24 (solve for y.)<br />Subtract 6x to both sides.<br />-4y = -6x + 24<br />Divide each side by -4.<br />y = 3/2 x - 6<br />
9. 9. Example 8<br />Similar Problem: What is the equation in slope-intercept form of the line through the points (3, -1) and (-2, 9)?<br />1) Find the slope.<br />m = (9 - -1)/(-2 - 3)<br />m = 10/-5 = -2<br />2) Use the slope and either point to plug into point slope form, let's use (3, -1)<br />y - y = m(x - x)<br />y - -1 = -2(x - 3)<br />y + 1 = -2(x - 3)<br />y + 1 = -2x + 6<br />y = -2x - 5<br />
10. 10. Example 9 <br />Similar Problem: What is the equation in slope-intercept form of the line through the points (3, -1) and (-2, 9)?<br />1) Find the slope.<br />m = (9 - -1)/(-2 - 3)<br />m = 10/-5 = -2<br />2) Use the slope and either point to plug into point slope form, let's use (3, -1)<br />y - y = m(x - x)<br />y - -1 = -2(x - 3)<br />y + 1 = -2(x - 3)<br />y + 1 = -2x + 6<br />y = -2x - 5<br /> |
1. ## sum of roots
x is a real number.
x+8+x+10+x+12+.........5x-2+5x=578
How many times bigger is the multiplication of x values that supplies this equation than the sum of the x values?
2. ## Re: sum of roots
sorry I can't understand this at all.
3. ## Re: sum of roots
First we will find the sum of x values that supplies this equation. After that we will find the multiplication of x values that supplies this equation and we will find the difference. Answer is -590 but I couldn't get it.
4. ## Re: sum of roots
It looks like the equation can be rewritten as $\displaystyle \sum_{k=4}^{2x} (x+2k) = 578$.
Breaking it down:
\displaystyle \begin{align*}\sum_{k=4}^{2x} (x+2k) & = \sum_{k=4}^{2x}x + \sum_{k=4}^{2x}2k \\ & = x\sum_{k=4}^{2x}1 + 2\sum_{k=4}^{2x}k \\ & = x(2x-3) + 2\left(\sum_{k=1}^{2x}k - \sum_{k=1}^3 k\right) \\ & = x(2x-3) + 2\left(\dfrac{2x(2x+1)}{2} - \dfrac{3(4)}{2}\right) \\ & = x(2x-3) + 2x(2x+1)-12 \\ & = 6x^2-x-12 = 578\end{align*}
So, you get a quadratic equation, and find $x = \dfrac{1\pm 119}{12}$. So, the sum of $x$-values is $10 - \dfrac{59}{6}= \dfrac{1}{6}$ and the product of them is $10 \left(-\dfrac{59}{6}\right) = -\dfrac{295}{3}$.
So, $\displaystyle -\dfrac{\dfrac{295}{3}}{\dfrac{1}{6}} = -6\dfrac{295}{3} = -590$
many thanks.
6. ## Re: sum of roots
I'm trying to understand this equation:
Originally Posted by SlipEternal
$\displaystyle \sum_{k=4}^{2x} (x+2k) = 578$.
for the case of x = -59/6. X=59/6 makes the series:
$\frac {-59} 6 + 8 +\frac {-59} 6 + 10 + \frac {-59} 6 + 12 + ....$
It never reaches 5 x -59/6 = -49.1666
I think the problem is that the formula
$\displaystyle \sum _{k=1} ^{2x} = \frac {2x(2x+1)} 2$
works only for positive integer values of 2x. So if we find x= -59/6 it doesn't hold up. I maintain that there is only one solution to the series, and that is x=10.
7. ## Re: sum of roots
Originally Posted by ebaines
I'm trying to understand this equation:
for the case of x = -59/6. X=59/6 makes the series:
$\frac {-59} 6 + 8 +\frac {-59} 6 + 10 + \frac {-59} 6 + 12 + ....$
It never reaches 5 x -59/6 = -49.1666
I think the problem is that the formula
$\displaystyle \sum _{k=1} ^{2x} = \frac {2x(2x+1)} 2$
works only for positive integer values of 2x. So if we find x= -59/6 it doesn't hold up. I maintain that there is only one solution to the series, and that is x=10.
x = 10 is the only solution to the sum equation, true, but the question asked about the underlying equation that generated the sum equation, which was the quadratic $6x^2-x-590 = 0$.
8. ## Re: sum of roots
The queston didn't ask about a quadratic - it asked about a sum of a series, and that quadratic equation only defines the sum for positive integers of x. This is like asking what values of x satisfy x+1=5 and approaching it be squaring both sides, rearranging and getting the quadratic x^2 +2x-24 = 0, and concluding that the two possible values for x are 4 and -6. Oh well, it does appear that your method matched what the person who devides the question was expecting. |
# Basics
## What is a matrix?
$$A \in \mathcal{R}^{m \times n}, A = a_{ij}$$
$$m$$ = number of rows
$$n$$ = number of columns
$$m=n \Rightarrow$$ "square matrix"
If $$A$$ and $$B$$ are the same size,
$$A, B \in \mathcal{R}^{m \times n}$$
$$\Rightarrow (A+B)_{ij} = A_{ij} + B_{ij}$$
It is associative.
$$(A+B)+C = A+(B+C)$$
It is commutative.
$$A+B = B+A$$
the matrix of all zeros (written as $$0$$).
$$0_{ij} = 0 \; \forall i, j$$
$$A+0 = 0+A = A$$
## Multiplication
If $$A$$ has the same number of columns as $$B$$ has rows,
then we can multiply them.
$$A\in \mathcal{R}^{m \times n}, B\in \mathcal{R}^{n \times p} \Rightarrow AB\in \mathcal{R}^{m \times p}$$
$$(AB)_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$$
This is the dot product
of the $$i$$th row of $$A$$
with the $$j$$th column of $$B$$.
## Properties of Multiplication
It is associative.
$$(AB)C$$ = $$A(BC)$$
It is not necessarily commutative.
$$AB \stackrel{?}{=} BA$$
The identity for multiplication is the matrix
with $$1$$s on the diagonals and zero otherwise.
It is written as $$I$$.
$$I_{ij} = 1 \mbox{ if } i=j, 0 \mbox{ otherwise}$$
$$IA = AI = A$$
## Transpose
The transpose of a matrix $$A$$ is another matrix
with the rows and columns of $$A$$ inverted.
It is written as $$A^T$$.
$$A^T_{ij} = A_{ji}$$
Note: $$(A^T)^T = A$$
# Inverse of a Matrix
## What is it?
The inverse of a square matrix $$A$$ is a matrix that
multiplies with $$A$$ to make the identity matrix.
It is written as $$A^{-1}$$.
$$AA^{-1} = A^{-1}A = I$$
BUT it does not exist for all $$A$$.
## Why is it important?
Invertibility comes up again and again in linear algebra.
We'll examine four different use cases.
# Row Operations
## Elementary Row Operations
There are only three!
1. Switch two rows: $$r_i, r_j \rightarrow r_j, r_i$$
2. Multiply a row by a constant: $$r_i \rightarrow Cr_i$$
3. Add two rows together: $$r_i \rightarrow r_i + r_j$$
You can also do 2 and 3 at the same time:
$$r_i \rightarrow C_1 r_i + C_2 r_j$$
## Can you invert a matrix? (Part 1)
$$A$$ is invertible if
it can be reduced to $$I$$
using elementary row operations.
$$A^{-1}$$ is the result of
the same operations applied to $$I$$.
The method to do this
on the augmented matrix $$[A \; \vert \; I]$$
is called Gaussian elimination.
## Intuition for inverting Part 1
By going from $$A$$ to $$I$$ with row operations,
we basically multiply by $$A^{-1}$$ (since $$AA^{-1}=I$$).
Furthermore, since $$IA^{-1} = A^{-1}$$,
we can do the same operations on $$I$$
to get $$A^{-1}$$ explicitly.
# Linear independence
## Vectors
A vector $$v$$ is a matrix with one column.
$$v \in R^n$$
## Linear independence
A set of vectors $$\{v_i\}_{i=1}^n$$ is linearly independent if
the only linear combination equal to zero
has all coefficients equal to zero.
$$\{v_i\}_{i=1}^n$$ linearly independent
$$\Longrightarrow \left( \sum_{i=1}^n c_i v_i = 0 \Leftrightarrow c_i = 0 \; \forall i \right)$$
Alternately, $$\{v_i\}_{i=1}^n$$ is linearly independent if
no vector $$v_j$$ can be created from
a linear combination of the other vectors.
## Can you invert a matrix? (Part 2)
$$A$$ is invertible if
its rows are linearly independent.
$$A$$ is also invertible if
its columns are linearly independent.
# Determinants
## What is it?
The determinant is a value
associated with square matrices.
It is written as $$\det(A)$$.
## Can you invert a matrix? (Part 3)
A square matrix $$A$$ is invertible if
$$\det(A) \neq 0$$.
## How do you calculate it?
It's best shown with examples.
## 2 x 2 Determinant
$A = \left(\begin{array}{cc} a & b\\ c & d\\ \end{array}\right)$
$$\Rightarrow \det(A) = ad-bc$$
## 3 x 3 Determinant
$A = \left( \begin{array}{ccc} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{array} \right)$
$$\Rightarrow \det(A) = a_1(b_2c_3-b_3c_2) -$$
$$a_2(b_1c_3-b_3c_1) +$$
$$a_3(b_1c_2-b_2c_1)$$
## n x n Determinant
For any row $$A_i$$, $\det(A) = \sum_{j=1}^n a_{ij} \cdot (-1)^{i+j} \cdot M_{ij},$ where $$M_{ij}$$ is the (i, j) minor matrix.
# Eigenvalues,Eigenvectors
## What are they?
For a square matrix $$A$$,
we often examine special vectors $$v$$
that do not rotate when $$A$$ is applied.
That is, for a certain scalar $$\lambda$$,
$$Av = \lambda v$$.
In this case, we call $$\lambda$$ and $$v$$
an eigenvalue and eigenvector of $$A$$.
## Why are they important?
After doing a bit of algebra, we can notice:
$$Av = \lambda v$$
$$\Rightarrow Av - \lambda Iv = 0$$
$$\Rightarrow (A - \lambda I)v = 0.$$
Now we can make an assertion...
## Can you invert a matrix? (Part 4)
#### Assertion
If $$v \neq 0$$, then $$(A-\lambda I)$$ is not invertible.
Thus, $$\det{(A-\lambda I)} = 0$$.
#### Proof
Suppose $$(A-\lambda I)^{-1}$$ exists.
$$(A-\lambda I)v = 0$$
$$\Rightarrow (A-\lambda I)^{-1}(A-\lambda I)v = 0$$
$$\Rightarrow Iv = 0$$
$$\Rightarrow v=0.$$
This contradicts our original assumption that $$v \neq 0$$.
QED
## How do I find eigenvalues?
The $$v=0$$ case is boring.
$$A\times 0 = 0 = \lambda \times 0 \Rightarrow \lambda$$ can be anything.
So, let's focus on $$v \neq 0$$, so that $$\det{(A-\lambda I)} = 0$$.
The left-hand side is an $$n^{th}$$-order polynomial in $$\lambda$$.
Thus, by the Fundamental Theorem of Algebra,
every square matrix has $$n$$ eigenvalues.
However, the eigenvalues are not necessarily distinct.
## How do I find eigenvectors?
Suppose the eigenvalues $$\{\lambda_i\}_{i=1}^n$$ are all distinct.
To get the eigenvectors $$\{v_i\}_{i=1}^n$$ , solve the original equations:
$$(A - \lambda_i)v_i = 0, i = 1,\ldots,n$$.
This produces $$n$$ simultaneous linear equations.
Then, write each $$v_i$$ in terms of one common component.
Finally, factor the common component out.
For example, if at first we obtain $$(v_1, 3v_1, 2v_1)$$,
then the actual eigenvector is simply $$(1, 3, 2)$$. |
# Arrays of charges Contents: Basic concept: Sum of forces Linear arrays Example Whiteboards 2-D Arrays Example Whiteboards.
## Presentation on theme: "Arrays of charges Contents: Basic concept: Sum of forces Linear arrays Example Whiteboards 2-D Arrays Example Whiteboards."— Presentation transcript:
Arrays of charges Contents: Basic concept: Sum of forces Linear arrays Example Whiteboards 2-D Arrays Example Whiteboards
Basic Concept TOC Three equal positive charges A, B, C The force on B is the sum of the forces of repulsion from A and C ABC F CB F AB
Basic Concept TOC Three equal positive charges A, B, C The force on B is the sum of the forces of repulsion from A and C ABC F CB F AB F AB + F CB It’s Vector Time Kiddies!!
Linear arrays of charge (Let’s start simple) TOC Find the force on charge B: 1.Calculate the forces 2.Figure out direction 3.Add forces (Vectors) ABC +45 C -15 C +12 C 67 cm 143 cm
Linear arrays of charge (Let’s start simple) TOC Find the force on charge B: 1.Calculate the forces 2.Figure out direction 3.Add forces ABC +45 C -15 C +12 C 67 cm 143 cm F CB = kq C q B = -.791 N (to the left, attract) r 2 F AB = kq A q B = 1.101N to the right (repel) r 2
Linear arrays of charge (Let’s start simple) TOC ABC +45 C -15 C +12 C 67 cm 143 cm F CB = kq C q B = -.791 N (to the left, attract) r 2 F AB = kq A q B = 1.101N to the right (repel) r 2 +1.101 N-.791 N Finally, +1.101 -.791 = +.31N (To the right)
Whiteboards: Linear Charge Arrays 11 | 2 | 323 TOC
41 N left W What is the force on C? Which direction? ABC +120 C -15 C -180 C 70 cm 170 cm F AC = kq A q C = 33.02 (to the left, attract) r 2 F BC = kq B q C = 8.40 N to the left (repel) r 2 = 33.02 left + 8.40 left = 41.42 = 41 N left
4.4 N right W What is the force on A? Which direction? ABC +11 C -140 C +520 C 110 cm 160 cm F CA = kq C q A = 11.44 (to the right, attract) r 2 F BA = kq B q A = 7.05 N to the left (repel) r 2 = 11.44 - 7.05 (left) = 4.39 = 4.4 N right
17.4 N right W What is the force on C? Which direction? ABC +213 C -14 C +213 C 40 cm 130 cm F AC and F BC cancel out F DC = kq D q C = 17.4 N (to the right, repel) r 2 D 90 cm -112 C
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# Mean, Median and Mode on the GRE
Over the last six months, I’ve gone through quite a few Revised GRE prep books. One thing I’ve noticed missing is difficult questions relating to mean, median, and mode. Sure, most books describe how to find the average, and what the difference between the mean and the median. Many already know the above, but mean, median and mode questions on the actual Revised GRE are going to require much more practice than you are going to get from simply applying the basic formulas.
Below are five questions that should hopefully give you a good mental workout. If you whip through them, answering all correctly, then you are very well prepared for the concepts you’ll find on the GRE. If you stumble, don’t despair. Remember, we have plenty more questions like these—some that are every more difficult—waiting for you at gre.magoosh.com.
1. Set S is comprised of six distinct positive integers less than 10. Which of the following must be true?
I: The median is an integer
II: The median is less than the average
III: The range of digits in Set S is less than 8
(A) I only
(B) I & II
(C) II & III
(D) III only
(E) None of the above.
2. A list is comprised of five positive integers: 4, 4, x, 7, y. What is the range of the possible values of the medians?
(A) 2
(B) 3
(C) 6
(D) 7
(E) Cannot be determined by information provided.
3. The average of five positive integers is less than 20. What is the smallest possible median of this set?
(A) 19
(B) 10
(C) 4
(D) 3
(E) 1
4. Set S is comprised of 37 integers
Column A Column B
The median of Set S The mean of the lowest and the highest term
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
1. A good idea is to choose numbers. For instance, just using 1, 2, 3, 4, 5, 6 we can see that the first condition does not need to be true. The median of this range of numbers is 3.5. For the second condition, we can also use numbers and determine that the average and the median are equal. Therefore the second condition also does not have to hold true. Finally, for the third condition, the range of digits is 9 – 1 = 8. The third condition says less than 8. Therefore the answer is (E) none of the above.
2. Here we want to find the lowest possible median and the great possible median. Then we want to subtract the least possible median from the greatest possible median to find the range of the medians. If x and y are less than or equal to 4, then the median is 4. If x and y are greater than or equal to 7, the median is 7. Therefore the range of median is 3, Answer (B).
3. Choosing numbers will help us on this one. If we choose 1, 1, 1, 1, 16, the sum is equal to 20. The median is 1, Answer (E).
4. This is a conceptual problem. Nonetheless we can still pick numbers. Imagine all 37 numbers are the number ‘1’. The mean is ‘1’ and the average of the least and greatest is 1. n this case both columns are equal (C).Now, just change the last number to a 3. That is you have the number ‘1’ 36 times and the number three. The median is still 1. But now the average of the least and greatest is greater than ‘1.’ In this case column (B) is larger. If the answer switches depending on the numbers we use, the answer is (D).
### 17 Responses to Mean, Median and Mode on the GRE
1. Aman July 19, 2018 at 7:27 am #
Hi. I had a doubt in finding mode of a sequence. What is the mode of the sequence 1,2,3,4,5,6 .
• Magoosh Test Prep Expert July 19, 2018 at 9:47 am #
Sequences actually can’t have a mode. A set of numbers has a mode only if there is a number that occurs more frequently than any other number. And in a sequence, each number occurs just once.
2. Eunice Lim March 26, 2018 at 6:07 am #
For Q1, the median has to be an integer.
3.5 is the mean.
• Magoosh Test Prep Expert March 26, 2018 at 12:22 pm #
Actually, the median only needs to be an integer if the quantity of integers in the set is odd. This is because in an odd quantity of integers, such as a string of five integers, there is one exact middle integer. So for example, in the number set 2, 3, 4, 6, 9, a single number– 4– is in the exact middle of the set, with two numbers that are less than it, and two numbers that are more than it.
On the other hand, if you have an even quantity of integers, the median will not necessarily be an integer. This is because int hat case, there will be two middle numbers, and the median is derived from adding those two middle numbers and dividing by 2, to get a quantity that’s exactly between the two middle numbers. In other words, the median of an even quantity of numbers is a mean, but it’s a mean of just the two middle numbers and not the whole set. Let’s take the example immediately above and remove the 3. We now have four numbers, as follows: 2, 4, 6, 9. Here, the median is 5, because the two middle numbers, when averaged, produce 5: (4+6)/2 = 5. Conversely, if we remove the number 6 form the original five-number set, we get: 2, 3, 4, 9. In that case, the median is the average of the two middle numbers 3 and 4. (3+4)/2 = 3.5. So in this case, we have a set of integers, but a median that isn’t an integer. We have a non integer median in Q1 for similar reasons.
3. test August 31, 2016 at 9:35 am #
Why not simple choose all 1’s for the set of 5 digits then the average is 1 and median is 1?
• Magoosh Test Prep Expert September 2, 2016 at 5:19 pm #
Hi there,
The example shown here is just one of many combinations that we could use! You can also use a set of five 1’s and get the same answer 🙂
4. sarah October 29, 2015 at 9:11 pm #
I thought that zero was considered a positive integer? Why isn’t it in these solutions?
5. saul October 21, 2015 at 2:53 am #
There are some problems here with Q2. It asks for “the range of possible medians” which can either be taken for (i) how many possible medians are there or (ii) the greatest median minus the smallest median.
In both cases, the answer would be E – it can’t be determined based on the information given. In case (i) the medians could 4, 7, x or even y (i.e. 4,4,x,y,7 or x,4,4,7,y or 4,4,7,x,y and 4,4,y,x,7). In case (ii) based on the info above in case (i) the information cannot be determined.
• Chris Lele October 21, 2015 at 1:37 pm #
Hi Saul,
Thanks for pointing out! That ambiguity is very subtle. I’m clarifying it now 🙂
6. csk August 2, 2015 at 10:26 am #
I think question 3 requires clarification.
Question says that it is “set” of numbers, accordingly, all the elements of set are distinct, so {1, 1, 1, 1, 16} is equivalent to {1, 16}.
We have to consider distinct numbers, accordingly median would be 3.
• PotatoParadox June 27, 2017 at 9:56 am #
I agree that they should be distinct, since it is a set. But… if they are distinct, the least median can only be 19, since at least they can be equidistant, 1,2,3,4… the mean in this case is equal to median. However, if any of the difference is more, the mean might become more than the median… the answer is still D.
Cheers
7. Akhilesh April 26, 2015 at 12:09 am #
I think the second question solution is not clear to me it says ” average of the 5 positive integers is less than 20 i.e the sum should be <100… why are we taking the sum of 16+1+1+1+1??? any one to clarify?
• Malek February 7, 2016 at 10:49 am #
I agree with you. Can anybody help with this question?
• Magoosh Test Prep Expert February 7, 2016 at 12:39 pm #
Hi Malek,
Sure! 🙂 The set of 1, 1, 1, 1, 16 isn’t the only set we could have used–it is just an example that fits our criteria (average less than 20, all positive integers). This set therefore proves that the lowest median we can possibly have given those criteria is 1, so the answer is (E).
8. Livia February 20, 2014 at 11:20 am #
To be fair, I think question two needs to be re-worded. “A list is comprised of five positive integers: 4, 4, x, 7, y. What is the range of possible medians?” leads one to believe that these numbers have not yet been arranged in increasing order. I selected ‘E’ because those numbers could be anything.
Also, for question five, if you randomly pick the numbers 1, 3, 5 and 6, they don’t give you the same answer. Median = 4; Mean = 3.75 (which fulfills the original requirement of median being greater than mean) yet the answer to this set would be C.
I have come to realize that my greatest enemy in the GRE is my attention to detail when reading, so perhaps I’m incorrectly reading the questions and am making my own mistakes!
• Chris Lele February 25, 2014 at 10:54 am #
Hi Livia,
Thanks for the message :).
For the first question, if x and y are less than 4, then the median has to be 4, no matter how negative x and y are. If x and y are both larger than 7, even if they are really large, the median has to be 7. And we don’t care about those instances in between (where x and/or y are between 4 and 7) because we are looking for the range–which means we are looking for the extreme numbers.
For the last question, there were some other issues with it, so I took it down. Sorry for any confusion!
Hopefully, that clear up any confusion 🙂
9. AArendy February 18, 2014 at 11:17 pm #
Hi Chris,
another set of good questions…. 🙂
i solved 1 to 4 correctly but didn’t get the meaning of the question 5 , neither i get the solution, specially the digit thing. means actually what we have to do ??
One more query , I was trying to upgrade my magoosh account , but i get the notification prompting some kind of payment issues in India. i contacted my bank to authorize the payment but they denied for any such authorization, Please help me, i need premium account asap
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# Question 7, Exercise 2.2
Solutions of Question 7 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
If $A=\left[\begin{array}{ll}x & 0 \\ y & 1\end{array}\right]$ then prove that for all positive integers $n, A^{n}=\left[\begin{array}{cc}x^{n} & 0 \\ \dfrac{y\left(x^{n}-1\right)}{x-1} & 1\end{array}\right]$.
Solution.
Given: $$A = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}.$$ We use mathematical induction to prove the given fact. For C-1, put $n = 1$ \begin{align}A^1 =\begin{bmatrix} x^1 & 0 \\ \dfrac{y(x^1 - 1)}{x - 1} & 1 \end{bmatrix} \\ = \begin{bmatrix} x & 0 \\ \dfrac{y(x - 1)}{x - 1} & 1 \end{bmatrix}\\ = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}\end{align} C-1 is satisfied.
For C-2, suppose given statement is true for $n=k$. \begin{align} A^k = \begin{bmatrix} x^k & 0 \\ \frac{y(x^k - 1)}{x - 1} & 1 \end{bmatrix}\end{align} We need to show that the formula holds for $k + 1$: \begin{align*} A^{k+1} &= A^k \cdot A\\ &= \begin{bmatrix} x^k & 0 \\ \dfrac{y(x^k - 1)}{x - 1} & 1 \end{bmatrix}\begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}\\ &= \begin{bmatrix} x^k \cdot x + 0 \cdot y & x^k \cdot 0 + 0 \cdot 1 \\ \frac{y(x^k - 1)}{x - 1} \cdot x + 1 \cdot y & \dfrac{y(x^k - 1)}{x - 1} \cdot 0 + 1 \cdot 1 \end{bmatrix}\\ &=\begin{bmatrix} x^{k+1} & 0 \\ \dfrac{y(x^{k+1} - x) + y(x-1)}{x - 1} & 1 \end{bmatrix}\\ &=\begin{bmatrix} x^{k+1} & 0 \\ \dfrac{y(x^{k+1} - x+x-1)}{x - 1} & 1 \end{bmatrix}\\ &=\begin{bmatrix} x^{k+1} & 0 \\ \dfrac{y(x^{k+1} -1)}{x - 1} & 1 \end{bmatrix} \end{align*} This is true for $n=k+1$.
Hence C-2 is satisfied, and the proof is complete.
If $A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ then prove that for all positive integers $n$, $A^{n}=\begin{bmatrix} 1+2 n & -4 n n & 1-2 n \end{bmatrix}$
Solution. we use mathematical induction. Put $n = 1$ \begin{align*} A^1 = \left[\begin{array}{cc} 1 + 2(1) & -4(1) \\ 1(1) & 1 - 2(1) \end{array}\right]\\ = \left[\begin{array}{cc} 1 + 2 & -4 \\ 1 & 1 - 2 \end{array}\right] \\ = \left[\begin{array}{cc} 3 & -4 \\ 1 & -1 \end{array}\right] \end{align*} Assume that the formula holds for some positive integer $k$, i.e., \begin{align*}A^k = \left[\begin{array}{cc} 1 + 2k & -4k \\ k & 1 - 2k \end{array}\right] \end{align*} For $n=k+1$ \begin{align*} A^{k+1} &= A^k A \\ &= \left[\begin{array}{cc} 1 + 2k & -4k \\ k & 1 - 2k \end{array}\right] \left[\begin{array}{cc} 3 & -4 \\ 1 & -1 \end{array}\right]\\ &= \left[\begin{array}{cc} (1 + 2k) \cdot 3 + (-4k) \cdot 1 & (1 + 2k) \cdot (-4) + (-4k) \cdot (-1) \\ k \cdot 3 + (1 - 2k) \cdot 1 & k \cdot (-4) + (1 - 2k) \cdot (-1) \end{array}\right] \\ &= \left[\begin{array}{cc} 3 + 6k - 4k & -4 - 8k + 4k \\ 3k + 1 - 2k & -4k - 1 + 2k\end{array}\right] \\ &= \left[\begin{array}{cc} 3 + 2k & -4k - 4 \\ k + 1 & -2k - 1\end{array}\right]\\ &= \left[\begin{array}{cc} 1 + 2k + 2 & -4k - 4 \\ k + 1 & 1 - 2k - 2 \end{array}\right] \\ &=\left[\begin{array}{cc} 1 + 2(k + 1) & -4(k + 1) \\ k + 1 & 1 - 2(k + 1) \end{array}\right] \end{align*} By mathematical induction, the formula $A^n = \left[\begin{array}{cc} 1 + 2n & -4n \\ n & 1 - 2n \end{array}\right]$ holds for all positive integers $n$. |
$$\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}}$$
# 5.4: Integration Formulas and the Net Change Theorem
In this section, we use some basic integration formulas studied previously to solve some key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite integral represents a family of functions, all of which differ by a constant. As you become more familiar with integration, you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind, think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the proper notation based on your choice.
## Basic Integration Formulas
Recall the integration formulas given in [link] and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.
Example $$\PageIndex{1}$$: Integrating a Function Using the Power Rule
Use the power rule to integrate the function $$∫^4_1\sqrt{t}(1+t)dt$$.
Solution
The first step is to rewrite the function and simplify it so we can apply the power rule:
$∫^4_1\sqrt{t}(1+t)dt=∫^4_1t^{1/2}(1+t)dt=∫^4_1(t^{1/2}+t^{3/2})dt.$
Now apply the power rule:
$∫^4_1(t^{1/2}+t^{3/2})dt=(\frac{2}{3}t^{3/2}+\frac{2}{5}t^{5/2})∣^4_1$
$=[\frac{2}{3}(4)^{3/2}+\frac{2}{5}(4)^{5/2}]−[\frac{2}{3}(1)^{3/2}+\frac{2}{5}(1)^{5/2}]=\frac{256}{15}.$
Exercise $$\PageIndex{1}$$:
Find the definite integral of $$f(x)=x^2−3x$$ over the interval $$[1,3].$$
Hint: Follow the process from Example to solve the problem.
Solution
$−\frac{10}{3}$
## The Net Change Theorem
The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.
Net Change Theorem
The new value of a changing quantity equals the initial value plus the integral of the rate of change:
$F(b)=F(a)+∫^b_aF'(x)dx$
or
$∫^b_aF'(x)dx=F(b)−F(a).$
Subtracting $$F(a)$$ from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.
We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in Figure.
Figure 1: The graph shows speed versus time for the given motion of a car.
Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by
$∫^5_2v(t)dt=∫^4_240dt+∫^5_4−30dt=80−30=50.$
Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by
$∫^5_2|v(t)|dt=∫^4_240dt+∫^5_430dt=80+30=110.$
Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.
Example $$\PageIndex{2}$$: Finding Net Displacement
Given a velocity function $$v(t)=3t−5$$ (in meters per second) for a particle in motion from time $$t=0$$ to time $$t=3,$$ find the net displacement of the particle.
Solution
Applying the net change theorem, we have
$∫^3_0(3t−5)dt=\frac{3t^2}{2}−5t∣^3_0=[\frac{3(3)^2}{2}−5(3)]−0=\frac{27}{2}−15=\frac{27}{2}−\frac{30}{2}=−\frac{3}{2}.$
The net displacement is $$−\frac{3}{2}$$ m (Figure).
Figure 2: The graph shows velocity versus time for a particle moving with a linear velocity function.
Example $$\PageIndex{3}$$: Finding the Total Distance Traveled
Use Example to find the total distance traveled by a particle according to the velocity function $$v(t)=3t−5$$ m/sec over a time interval $$[0,3].$$
Solution
The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.
To continue with the example, use two integrals to find the total distance. First, find the t-intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for t. Thus,
$$3t−5=0$$
$$3t=5$$
$$t=\frac{5}{3}.$$
The two subintervals are $$[0,\frac{5}{3}]$$ and $$[\frac{5}{3},3]$$. To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval $$[0,\frac{5}{3}]$$, we have $$|v(t)|=−v(t)$$ over that interval. Over $$[ \frac{5}{3},3]$$, the function is positive, so $$|v(t)|=v(t)$$. Thus, we have
$$∫^3_0|v(t)|dt=∫^{5/3}_0−v(t)dt+∫^3_{5/3}v(t)dt$$
$$=∫^{5/3}_05−3tdt+∫^3_{5/3}3t−5dt$$
$$=(5t−\frac{3t^2}{2})∣^{5/3}_0+(\frac{3t^2}{2}−5t)∣^3_{5/3}$$
$$=[5(\frac{5}{3})−\frac{3(5/3)^2}{2}]−0+[\frac{27}{2}−15]−[\frac{3(5/3)^2}{2}−\frac{25}{3}]$$
$$=\frac{25}{3}−\frac{25}{6}+\frac{27}{2}−15−\frac{25}{6}+\frac{25}{3}=\frac{41}{6}$$.
So, the total distance traveled is $$\frac{14}{6}$$ m.
Exercise $$\PageIndex{2}$$:
Find the net displacement and total distance traveled in meters given the velocity function $$f(t)=\frac{1}{2}e^t−2$$ over the interval $$[0,2]$$.
Hint: Follow the procedures from Example and Example. Note that $$f(t)≤0$$ for $$t≤ln4$$ and $$f(t)≥0$$ for $$t≥ln4$$.
Solution
Net displacement: $$\frac{e^2−9}{2}≈−0.8055m;$$ total distance traveled: $$4ln4−7.5+\frac{e^2}{2}≈1.740 m$$
## Applying the Net Change Theorem
The net change theorem can be applied to the flow and consumption of fluids, as shown in Example.
Example $$\PageIndex{4}$$: How Many Gallons of Gasoline Are Consumed?
If the motor on a motorboat is started at $$t=0$$ and the boat consumes gasoline at the rate of $$5−t^3$$ gal/hr, how much gasoline is used in the first 2 hours?
Solution
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval [0,2]. We have
$∫^2_0(5−t^3)dt=(5t−\frac{t^4}{4})∣^2_0=[5(2)−\frac{(2)^4}{4}]−0=10−\frac{16}{4}=6.$
Thus, the motorboat uses 6 gal of gas in 2 hours.
Example $$\PageIndex{5}$$: Chapter Opener: Iceboats
As we saw at the beginning of the chapter, top iceboat racers can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed.
Figure 3: (credit: modification of work by Carter Brown, Flickr)
Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function $$v(t)=20t+5.$$ For the second half hour of Andrew’s outing, the wind remains steady at 15 mph. In other words, the wind speed is given by
$v(t)=\begin{cases}20t+5& for 0≤t≤\frac{1}{2}\\15 & for \frac{1}{2}≤t≤1\end{cases}.$
Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?
Solution
To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then
Distance =$$∫^1_02v(t)dt.$$
Substituting the expressions we were given for $$v(t)$$, we get
$$∫^1_02v(t)dt=∫^{1/2}_02v(t)dt+∫^1_{1/2}2v(t)dt$$
$$=∫^{1/2}_02(20t+5)dt+∫^1_{1/3}2(15)dt$$
$$=∫^{1/2}_0(40t+10)dt+∫^1_{1/2}30dt$$
$$=[20t^2+10t]|^{1/2}_0+[30t]|^1_{1/2}$$
$$=(\frac{20}{4}+5)−0+(30−15)$$
$$=25.$$
Andrew is 25 mi from his starting point after 1 hour.
Exercise $$\PageIndex{4}$$
Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function $$v(t)=−10t+15.$$ In other words, the wind speed is given by
$$v(t)=\begin{cases}20t+5 & for 0≤t≤\frac{1}{2}\\−10t+15& for\frac{1}{2}≤t≤1\end{cases}$$.
Under these conditions, how far from his starting point is Andrew after 1 hour?
Hint: Don’t forget that Andrew’s iceboat moves twice as fast as the wind.
Solution
$$17.5 mi$$
## Integrating Even and Odd Functions
We saw in Functions and Graphs that an even function is a function in which $$f(−x)=f(x)$$ for all x in the domain—that is, the graph of the curve is unchanged when x is replaced with −x. The graphs of even functions are symmetric about the y-axis. An odd function is one in which $$f(−x)=−f(x)$$ for all x in the domain, and the graph of the function is symmetric about the origin.
Integrals of even functions, when the limits of integration are from −a to a, involve two equal areas, because they are symmetric about the y-axis. Integrals of odd functions, when the limits of integration are similarly $$[−a,a],$$ evaluate to zero because the areas above and below the x-axis are equal.
Rule: Integrals of Even and Odd Functions
For continuous even functions such that $$f(−x)=f(x),$$
$∫^a_{−a}f(x)dx=2∫^a_0f(x)dx.$
For continuous odd functions such that $$f(−x)=−f(x),$$
$∫^a_{−a}f(x)dx=0.$
Example $$\PageIndex{6}$$: Integrating an Even Function
Integrate the even function $$∫^2_{−2}(3x^8−2)dx$$ and verify that the integration formula for even functions holds.
Solution
The symmetry appears in the graphs in Figure. Graph (a) shows the region below the curve and above the x-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetry about the y-axis of an even function. We have
$$∫^2_{−2}(3x^8−2)dx=(\frac{x^9}{3}−2x)∣^2_{−2}$$
$$=[\frac{(2)^9}{3}−2(2)]−[\frac{(−2)^9}{3}−2(−2)]$$
$$=(\frac{512}{3}−4)−(−\frac{512}{3}+4)$$
$$=\frac{1000}{3}$$.
To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then check to make sure we get the same answer.
$∫^2_0(3x^8−2)dx=(\frac{x^9}{3}−2x)∣^2_0=\frac{512}{3}−4=\frac{500}{3}$
Since $$2⋅\frac{500}{3}=\frac{1000}{3},$$ we have verified the formula for even functions in this particular example.
Figure 4: Graph (a) shows the positive area between the curve and the x-axis, whereas graph (b) shows the negative area between the curve and the x-axis. Both views show the symmetry about the y-axis.
Example $$\PageIndex{7}$$: Integrating an Odd Function
Evaluate the definite integral of the odd function $$−5sinx$$ over the interval $$[−π,π].$$
Solution
The graph is shown in Figure. We can see the symmetry about the origin by the positive area above the x-axis over $$[−π,0]$$, and the negative area below the x-axis over $$[0,π].$$ we have
$∫^π_{−π}−5sinxdx=−5(−cosx)|^π_{−π}=5cosx|^π_{−π}=[5cosπ]−[5cos(−π)]=−5−(−5)=0.$
Figure 5: The graph shows areas between a curve and the x-axis for an odd function.
Exercise $$\PageIndex{4}$$:
Integrate the function $$∫^2_{−2}x^4dx.$$
Hint: Integrate an even function.
Solution
$$\dfrac{64}{5}$$
## Key Concepts
• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive x-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.
## Key Equations
• Net Change Theorem
$$F(b)=F(a)+∫^b_aF'(x)dx$$ or $$∫^b_aF'(x)dx=F(b)−F(a)$$
## Glossary
net change theorem
if we know the rate of change of a quantity, the net change theorem says the future quantity is equal to the initial quantity plus the integral of the rate of change of the quantity |
# Converting Weight using Metric System
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## Objective
The students will be able to convert measurements of weight within the Metric System.
#### Big Idea
Boing, boing, boing!
## Opener
15 minutes
Today’s lesson focuses on converting weights within the Metric System. Students measure the weight of various objects around the room using gram spring scales. They then convert these measurements into other units within the Metric System. During the second part of the lesson students will be practicing converting amongst all the units of Metric weight. The lesson will end with an exit slip to check for understanding that includes three real-world story problems involving converting weight in the Metric system.
To begin this lesson I provide students with an index card and then show them how to use a spring scale. I have students set up a four column chart on the lined side of the index card. I explain that they will be measuring the weight of various objects around the room.
Today we are going to be measuring the weight of objects around the room using the metric system. What units of measurement did we use in the U.S. Customary system? What unit are we going to use in the metric system? Can you think of some other units of grams?
The spring scales we are going to use today will help us measure things in grams. We won’t be able to measure things that are super heavy or super lightweight. When you choose objects keep that in mind.
On your index card you are going to write down five items that you and your partner are going to weigh. The name of the object goes in the first column. In the second column put the weight in grams. For now, leave the other two columns blank.
Most of the time when everyday objects are weighed we see the measurements in kg, g, or mg. So for our conversions today we will focus on converting our gram measurements to kilograms and milligrams.
Once students have finished measuring their five objects we move back to our seats and then I start to model conversions for students. I call on students to share their objects and then go through the process of showing how we convert from grams to kilograms and then from grams to milligrams.
After doing a few examples with students I have them finish converting the remaining objects on their cards using the blank columns. I then call on students to come up to the document camera and present their conversions for their one of their objects. For example, my penciled weighed 6 grams. 6 grams is equivalent to 6000mg which is equivalent to 0.006 kg.
## Practice
30 minutes
After modeling for the students how to do conversions of grams I now have students work in pairs to complete a skills sheet for today. I allow the students about twenty minutes to complete the conversions while I circulate the room and support students. I target students who struggled on their exit slip from yesterday.
At the end I display the correct answers on the document camera and allow students to self-correct their work. After a few minutes I ask students to share examples of what they did wrong on problems they did not get correct. My goal is that students are able to reason through the problem and identify where they got off track.
I think that if one student struggled with a certain problem, another student probably struggled with the same problem. It is good to have the student reason through their mistakes to make changes in their thought process.
We then add grams to the graphic organizer under metric weight.
## Closer
15 minutes
To wrap up this lesson I have students complete an exit ticket which includes three story problems I created involving converting weight units. I use this information to guide me in deciding if further clarification is needed in converting weight in the Metric System. I look to see where students struggled to pin point exactly what part of converting is unclear to them. |
Word Lesson: Volume and Surface Area of Spheres
In order to solve problems which require application of the volume and surface area for spheres, it is necessary to
A typical problem involving the volume or surface area of a sphere gives us various information about the size of the sphere - usually one or more values for the volume, surface area, radius, or diameter. You will then be asked to calculate the other missing values based on this given information.
Suppose the diameter of a sphere is 10 inches. What are its volume and surface area?
To get started, notice that the formulas for volume and surface area involve the radius and not the diameter. The radius is half the diameter or 5 inches.
We use r = 5 in the formulas for volume and surface area.
Volume: V = cubic inches
Surface area: S = square inches
It is difficult to inspect these answers for reasonableness. Care must be taken in arithmetic when squaring and cubing the radius.
Notice also that both formulas involve use of 4 p . If we rewrite the formula for volume, we can get a relationship between volume and surface area:
where S is the surface area. This is not often a helpful relationship, but we may refer to it when comparing volume and surface area.
For example, suppose the surface area of a sphere equals twice the numerical value of its volume. What is the sphere's radius?
r = 3/2 = 1.5
Examples
Suppose the surface area of a sphere is 64 p square feet. Find the radius and volume of this sphere. What is your answer?
The volume of a sphere is 288 p cubic centimeters. What are the radius and surface area? What is your answer?
Examples
A sphere has a diameter of 8 feet. What are its volume and surface area? V = and S = V = and S = V = and S = What is your answer?
The surface area of a sphere is 28 p . What are the radius and the volume of this sphere? What is your answer?
The formulas for surface area and volume involve use of the radius. It is important to be using the radius and not the diameter of the sphere. Finding the radius if we are given volume or surface area involves taking a square or cube root. The biggest hurdle with this type of problem is doing the arithmetic carefully. Note that the surface area formula uses r squared (r 2 ) and the volume formula uses r cubed (r 3 ). Correspondingly, remember that surface area is measured in square units and volume is measured in cubic units.
M Ransom
Show Related AlgebraLab Documents
AlgebraLAB Project Manager Catharine H. Colwell Application Programmers Jeremy R. Blawn Mark Acton Copyright © 2003-2024 All rights reserved. |
# How do you integrate int tan^-1x by integration by parts method?
Oct 31, 2016
The integral $= x \arctan x - \ln \frac{1 + {x}^{2}}{2} + C$
#### Explanation:
The integration by parts is $\int u ' v = u v - \int u v '$
Here we have $u ' = 1$$\implies$$u = x$
and $v = \arctan x$$\implies$$v ' = \frac{1}{1 + {x}^{2}}$
If $v = \arctan x$$\implies$$x = \tan v$
So by differentiating, $1 = \left(\frac{1}{\cos} ^ 2 v\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right)$
$v ' = \frac{\mathrm{dv}}{\mathrm{dx}} = {\cos}^{2} v = \frac{1}{1 + {x}^{2}}$
The integration
$\int \arctan x \mathrm{dx} = x \arctan x - \int \frac{x \mathrm{dx}}{1 + {x}^{2}}$
$\int \frac{x \mathrm{dx}}{1 + {x}^{2}} = \left(\frac{1}{2}\right) \int \frac{2 x \mathrm{dx}}{1 + {x}^{2}}$
Let $u = 1 + {x}^{2}$$\implies$$\mathrm{du} = 2 x \mathrm{dx}$
$\int \frac{2 x \mathrm{dx}}{1 + {x}^{2}} = \int \frac{\mathrm{du}}{u} = \ln u$
$\int \frac{x \mathrm{dx}}{1 + {x}^{2}} = \frac{1}{2} \ln \left(1 + {x}^{2}\right)$
And finally $\int \arctan x \mathrm{dx} = x \arctan x - \ln \frac{1 + {x}^{2}}{2} + C$ |
## Math Tools for Journalists: Part Three
The last four chapters of “Math Tools for Journalists” teaches young journalists about directional, area and volume measurements, as well as the metric system. Both are important for journalists to understand for a number of reasons.
## Chapter 9: Directional Measurements
Knowing how to calculate distance, time and rate is an important skill for young journalists. This skill can be helpful when reporting on car or plane crashes where the distance, time and rate may be important to the To find the distance, multiply the rate and the time. For the rate, divide distance by time and for time divide distance by rate.
Similarly, speed is equal to distance divided by time. Always remember though that speed and velocity are two different things and cannot be used interchangeably. Speed measures how fast something is going and velocity measures speed and direction. Acceleration is equal to the ending velocity minus the starting velocity, which is then divided by time.
Try it Out:
Example:
Q: You are a reporter in Boston covering the Boston Marathon and doing a profile on one specific runner. You are including in your story how the runner’s marathon time has improved over the years. This years, the runner ran the marathon, 26.2 miles, at an average rate of 5.7 miles per hour. How long did she take to finish the marathon?
A: 4.6 hours
26.2 miles/5.7 miles per hour = 4.6 hours
## Chapter 10: Area Measurements
Understanding area measurements such as perimeter and area is necessary for reporting on construction and physical spaces. To calculate perimeter of a four sided area, add together two times the width and two times the length. For other shapes, add together the length of all the sides. To find the area of a square or rectangle, multiply the length and width. For a triangle, multiply the base and height and divide that by two.
Smaller spaces are measured in square inches or square feet. Larger areas are measured in square feet, square yards or square rods. One square foot equals 144 square. One square yard equals nine square feet. Finally, one square rod equals 30 square yards.
For circular spaces, you must know how to calculate circumferance and area. To find circumference, multiple pi times two and then multiply that product by the radius. For area, multiply pi by the squared radius.
Try it Out:
Example:
Q: You are covering the construction of the School of Communications and need to find out the square footage of the entire new building. You know the length of Schar is 160 feet and the width is 60 feet. You also know the length of McEwen is 180 feet and the width is 60 feet. What is the area of the whole building?
A: 20,400 sq feet
(160×60)+(180×60)=20,400 sq feet
## Chapter 11: Volume Measurements
Understanding volume measurements is important for any journalist. This can come into play when discussing a new product, writing a recipe or in business reporting.
The most important part about understanding liquid measurements is knowing basic conversions:
• 2 tablespoons = 1 fluid ounce
• 1/2 pint = 8 ounces, one cup
• 1 pint = 16 ounces, two cups
• 2 pints = 1 quart
• 2 quarts = half gallon
• 4 quarts = gallon
• 1 U.S. standard barrel = 31.5 gallons
• 1 U.S. gallon = 4/5 Imperial gallon
For solid objects, a reporter can find the area by multiplying the length, height and width. Volume formulas also vary based on the shape. Remember area is measured in cubic ____ based on the unit.
A ton is a common measurement for things with a large volume. There are different types of tons: short ton (2,000 pounds), long ton (2,240 pounds) and metric ton (2,204.62 pounds).
Try it Out:
Example:
Q: What is the volume of a cargo train that is 30 feet long, 20 feet tall and 15 feet wide?
A: 9,000 cubic feet
30 x 20 x 15 = 9,000
Chapter 12: The Metric System
Understanding the metric system is arguably the most important math skill a journalist must have. Considering almost every other country in the world uses the metric system, it is important for journalists to understand it so they can convert measurements when reporting internationally.
The metric system is rather system: it is based on multiples of 10. It is important for journalists to know the prefixes, such as milli, micro, mega, giga, etc. The three units of the metric system are meter (length), gram (mass) and liter (volume).
The metric system also uses a different formula for temperature. To convert Fahrenheit to Celsius, subtract 32 from the Fahrenheit temperature and multiply it by .56. To convert Celsius to Fahrenheit, multiply 1.8 by the Celsius temperature and add 32.
There are many style and usage rules to be followed when using the metric system. For example, all units must start with a lowercase letter except at the beginning of a sentence and Celsius. The symbols should also be written in lowercase letters, except for liters and units named after a country or person. Symbols for units are never plurals. Also remember to use a space between a number and its symbol.
A reporter also does not use a period at the end of unit names. The dot or period is used as the decimal point within numbers. If a measurement is less than one, zero needs to be written before the decimal point.
Try it Out:
Example:
Q: You are a food reporter writing a recipe for scones. To make the scones, you must preheat the oven to 425 degrees Fahrenheit. What is that in Celsius?
A: 220 degrees
.56 x (425-32) = 220
## Elon alum Al Drago speaks to communications students about his work, experiences
Al Drago graduated from Elon University in 2015. Less than two years later, his photos have been published in a number of major news outlets, such as The New York Times, Rolling Stone and The Washington Post.
On Friday he visited with current Elon students to share his insights on the industry and give advice for their future careers.
“My Elon experience was based around the journalism that I did,” Drago said.
While at Elon, Drago was the Photo Editor and held internships at the Herald Sun, Burlington Times-News Raleigh News and Observer and The Baltimore Sun. Drago told stories about his experiences at Elon, specifically the major projects he worked on.
“The key is staying with a story that you’re really passionate about and you’re really engrained in.”
Drago also explained how he got his start in journalism.
“I ferociously worked … I knew what I wanted, and I wasn’t going to stop until I got it,” Drago said. Drago centered much of his talk on tips for young journalists to remember when covering stories and working toward a professional career.
Drago also advised students on how to use social media. He reminded students that future employers always see what people post on their social media. But, a presence on social media is important in this digital age. “Embrace the platform” is one of the
Most importantly, Drago encouraged students to work hard to get what they want. While it is important to have a balance between social life and professional work, Drago said that working hard and getting out of the “Elon bubble” will help you get ahead.
“I chose work every single time. I dedicated myself to my work.”
## LIVE BLOG: Spring Convocation speaker Dan Gilbert
By Olivia Zayas Ryan
On Thursday, the Elon University community gathered in Alumni Gym for the annual Spring Convocation. This year’s speaker, Dan Gilbert, is the Edgar Pierce Professor of Psychology at Harvard University.
Gilbert is also well known for his TED Talks on the science of happiness. His first TED Talk, The Surprising Science of Happiness, has been viewed more than 13 million times and remains one of the 25 most popular TED Talks of all time.
Additionally, in 2007 he published his book Stumbling on Happiness, which made it to the New York Times bestsellers list and has since been translated into more than 35 different languages.
03:31 p.m.: Procession of graduating students, students who are on the Dean’s and President’s lists and faculty and staff members begins.
03:36 p.m.: Joel Harter, associate chaplain of protestant life, and Jessica Walden, director of jewish life at Hillel, introduce the event with a prayer.
03:38 p.m.: “Convocation is an important rite of Spring here at Elon,” President Leo Lambert said.
03:39 p.m.: “Education and higher education in particular has long been seen as a gateway to opportunity in this country,” Lambert said.
03:40 p.m.: “Higher education makes possible what we could not have imagined before … Higher education lights a path for the future,” Lambert said.
03:41 p.m.: “Education and training beyond high school leads to a better and more secure life,” Lambert said.
03:43 p.m.: “As always, I thank each of you for all you do, to make Elon, this country, this community and the world, a better place,” Lambert said.
03:44 p.m.: Lambert thanks Gilbert for coming to Elon today
03:45 p.m.: India Johnson, associate professor of psychology, takes the stage to introduce Gilbert as the convocation speaker.
03:45 p.m.: Johnson shares her own story about her education as a single mother of two.
03:46 p.m.: “I thought to myself, if Dr. Gilbert could go on to become a professor of psychology, so could I.
03:48 p.m.: “So Dr. Gilbert, Daniel, thank you. Thank you for finding your way to social psychology and the academy and sharing your journey so I could find my way as well,” Johnson said.
03:48 p.m.: Gilbert thanks Lambert and Johnson for the introduction and takes the stage.
03:50 p.m.: “Happiness is what happens when you get what you want, and that’s never what happens in this lifetime on Earth,” Gilbert on ancient theories of happiness.
03:51 p.m.: “People who have everything they want aren’t any happier than the rest of us.”
03:52 p.m.: Gilbert shows advertisements from his childhood.
03:53 p.m.: “The culture was telling my mom that cigarettes, Coke and TV were the keys to happiness.”
03:54 p.m.: “We are surrounded by people who tell us where happiness should be found … None of their theories are based on science.”
03:55 p.m.: “To do science requires really only one thing: you have to be able to measure something. If you can’t measure something, you have to write poems about it.”
03:55 p.m.: Gilbert jokes that there is a mini-bar under the podium. There is a lot of laughter from the crowd. Attendees seem engaged and entertained.
03:58 p.m.: After joking about his mother and childhood, Gilbert shows a picture of his mother. He tells the three things his mother said he needed to be happy: a good marriage, money and children.
03:59 p.m.: Gilbert asks the crowd if they believe that marriage makes people happier. An overwhelming majority of attendees did not raise their hands.
04:00 p.m.: “Every single piece of data in social science states you are wrong: marriage causes happiness … Married people are happier than single people.”
04:04 p.m.: Gilbert shows graphs on the data of happiness for married people versus single people.
04:07 p.m.: “When you look at men and women as they approach divorce, what do you see?” Gilbert said, as he shows that divorce also makes people happier.
04:07 p.m.: “It’s not just ‘I do’ that makes you happier, it’s being in a marriage that’s good.”
04:08 p.m.: Gilbert addresses the question, does money make people happier?
04:09 p.m.: “The relationship between money and happiness isn’t simple … The amount of happiness money can buy levels off.”
04:11 p.m.: “My friends who are economics say ‘If money doesn’t make you happier you’re not spending it right,’ they say this in an amusing way but they are right.”
04:13 p.m.: Gilbert explains the goods and services that people should be spending money on to make them happier: experiences and things for others.
04:16 p.m.: “Research shows that when people spend money on others, doing things for others, they themselves get a happiness boost.”
04:18 p.m.: “People with children are less happy that people without them and people with children are the least happy when their children live with them.”
04:21 p.m.: “When you have a baby, life doesn’t go on,” Gilbert said. Studies show that having children does not make people happier.
04:23 p.m.: “I wanna suggest to you that the way we look when we’re living our lives and the way we look when scientists take a step back are two very different worlds … The view of human happiness I’ve shared with you is the view from outer space.”
04:24 p.m.: “There is a parenthood penalty, on average across the world, people with children are just a little bit shy of the zero point, children bring happiness down.”
04:28 p.m.: “I wanna suggest to you that what makes human beings happy is a scientific fact.”
04:29 p.m.: I really do believe that the more we learn about the true causes of happiness the more of it we can get for ourselves and our communities.”
04:29 p.m.: Gilbert concludes his speech and Lambert takes the stage. Graduates of the class of 2017 and students on the Dean’s and President’s list stand and are recognized.
04:32 p.m.: Lambert recognizes students inducted into Phi Beta Kappa, Pi Kappa Phi and Omnicron Delta Kappa.
## Elon construction causes negative environmental effects
As Elon University continues to expand its campus, trees are being cut down to make room for new buildings and facilities.
Last month, the university began construction of a new parking lot next to the McMichael building. This new parking lot will add 165 parking spots for faculty, staff and visitors, Dan Anderson, vice president of University Communications, said in February. And construction is set to begin on Sankey Hall this summer. This new building will take the place of many parking spots in the existing visitors’ lot.
To begin construction on this new parking lot, the university had to cut down a number of trees adjacent to the McMichael building. The university also recently began construction on the Schar Center, a 161,000-square-foot convocation center that will be able to seat approximately 5,500 people. This new building is being built on the 19.5-acre parcel of land next to the Hunt Softball field. To begin construction on this building, the university had to remove almost nine acres of trees.
For many, physical expansion is of great value to the university. According to Paul Moersdorf, adjunct assistant professor of physics, it is often easy to overlook the value that trees have on campus as well.
“What’s the value of a single tree? That’s hard to say,” Moersdorf said. “If you’re a human being, and it’s where you want to put a driveway, the tree has no value at all. Or if it’s where you want to put a communications building, and you’re talking about a dozen trees, the trees have no value. The communications building is more important.
“However, if you are the squirrels that rely on the nuts from those trees, or the birds that put a nest in the trees, or all of the thousands of species of life that lives in the bark, on the bark, on the leaves, in the root system of the tree — to those, the value is infinite.”
Over the past few decades, Elon has expanded its campus significantly, adding more than 100 buildings in less than 20 years. With each new building on campus, more trees are being cut down.
According to Tom Flood, assistant director of Physical Plant and director of landscaping and grounds, the university recognizes this issue and continuously adds more trees during construction.
“Construction inevitably has to remove some trees sometimes,” Flood said. “But we are very cognizant and careful to go back with a large number of trees in the planting.”
Flood offered many examples, saying that before beginning the construction of the Global Neighborhood, Physical Plant had to remove around 100 trees. But after construction was completed, 313 trees were added in the new neighborhood.
Similarly, construction for Schar Hall caused about 10 large oak trees to be cut down, but through landscaping, Physical Plant added 54 new trees — 28 percent of which were oak trees.
Though the plans have not yet been finalized, Flood said the construction team plans to add more trees in and around the new parking lot near the McMichael Building.
“We will probably have, I imagine, in the range of 40 to 45 smaller, new trees go back in on that project,” Flood said. “So we will line the street on Haggard Avenue with them and also adjacent to the elementary school. There’s a bioretention base for storm water management on the north side of the parking lot that will have trees in it and around it. And then in the landscape buffer between the parking lot and the McMichael Science Building, there will be more trees in there as well.”
Flood also added that more trees will be planted around Sankey Hall. Regardless of upcoming construction, he said that Elon is still committed to making sure that trees remain across campus.
“It’s part of the ethic of who we are — Elon means ‘oak,’ our logo is an oak leaf, and so that sets the standard for what our campus will be,” Flood said. “We have always envisioned and designed this campus as sort of a ‘southern garden,’ if you will, set in a grove of oaks — so that still guides the landscape architecture that we do today in all of our projects.”
## Elon students join in solidarity for gender equality
On Wednesday, many Elon University students, faculty and staff participated in A Day Without a Woman, a national strike to show value of women in the U.S. socio-economic system. To participate in the strike, women were encouraged to take the day off from work, avoid shopping except for at minority- or locally-owned businesses and wear red in solidarity.
A Day Without a Woman coincides with the International Women’s Day, a day to celebrate the achievements of women across the world. The organizers of A Day Without a Woman are the same activists who organized the Women’s March on Washington in January.
Members of the Elon community participated in the strike in a number of ways. Some simply wore red in support while others canceled class.
Senior Danielle Dulchinos initially wanted to participate by not going to class, but decided to attend class out of respect for those without access to an education.
“I ended up going to class because being able to go to college is such a privilege,” Dulchinos said. “It’s such a privilege as a woman for a lot of reasons and a lot of women don’t have that privilege, a lot of women don’t get to go to school at all and so for me, while I totally, 100 percent agree with the strike and agree with the point of the strike, I feel like personally I decided to go to class because this is an opportunity afforded [to] me that’s not given to many others.”
While many students participated in the event, the movement was not supported by all students. Senior Josh Weintraub said that while he supports the general cause, he does not support the organizers behind the event.
“The woman who was the main person in putting all of this together is a known Palestinian terrorist who killed multiple Israelis in a bombing and tried to plan to bomb the British Embassy in Israel in the past,” Weintraub said. “So I feel like we can support international women’s movements without allowing these people to have their views normalized by running the events.”
Weintraub said that while he does consider himself to be a supporter of women’s rights, his position as a Jewish man makes it difficult for him to support the event considering the actions of one of its organizers against Israelis.
“I felt like it puts me in an odd situation of, do I support women’s rights, and even though I do it’s frowned upon because I don’t support the events that are happening today,” Weintraub said. “So I’m put in the middle situation of not being able to identify with the people who are protesting today but identifying with what they’re trying to say at the same time.”
Regardless of personal opinion on the need or effectiveness of the event, International Women’s Day and A Day Without a Woman served as ways to spark conversations about gender inequality on campus. For Dulchinos, the event should be motivation for people to learn more about the feelings and stories of others.
“It should be a day for us to take a step back and really get to know our fellows citizens,” Dulchinos said. “I think that there is so much that can be solved if you sat down with somebody who is very different from you and just said, ‘Why do you feel that way? What makes you scared?’”
Students had the opportunity to learn more about and further discuss issues of gender inequality at Elon’s International Women’s Day Forum, an event co-sponsored by 12 different academic departments and student organizations across campus.
The forum, hosted by sophomore Lucia Jervis, included presentations from Elon faculty, students and alumni. The presenters included Rissa Trachman, associate professor of anthropology; Alexa Sykes ’13; Carmen Monico, assistant professor of human service studies; Nina Namaste, associate professor of Spanish; Sumeyye Pakdil Kesgin, adjunct assistant professor of religious studies; and sophomore Amy Belfer. The forum concluded with a performance by some members of the cast of The Vagina Monologues.
The speakers presented on a number of topics, ranging from gender issues in archaeology to global gender-based violence.
Freshman Julia Dick who attended the forum said that the speakers and their focus on women in leadership were inspirational and uplifting.
“To me, it was just extremely empowering just to hear people’s stories and hear the things that they’ve had to go through to kind of get to where they are now,” Dick said. “You know, facing adversity obviously did not stop them from pushing through and doing what they loved.”
## Judge Anna Blackburne-Rigsby speaks to students on representation, diversity
On Wednesday, Judge Anna Blackburne-Rigsby, Associate Judge on the District of Columbia Court of Appeals, spoke to Elon University students in a lecture titled “The Third Branch: How a Trusted, Diverse Judiciary is Crucial to Ensuring our Democracy,” host by Elon’s Liberal Arts Forum.
Blackburne-Rigsby was introduced by senior Emily Hayes, the president of Elon’s Liberal Arts Forum. The Liberal Arts Forum is a group of students who meet weekly to bring to campus academic speakers for each semester who embody Elon’s commitment to a liberal arts education.
Blackburne-Rigsby began her lecture by outlining how the U.S. court system works. She shared jokes and personal anecdotes as she discussed the importance of the judicial branch in a democratic society.
Above all else, the role of the judicial branch is to interpret the constitution and determine whether or not the laws we pass uphold the constitution.
After establishing a base knowledge on the court system, Blackburne-Rigsby went on to discuss the growing need for diversity in the judicial branch.
“I think diversity is a symbol of justice,” Blackburne-Rigsby said.
She stressed the importance for diversity to bring varied opinions and perspectives into
the court room. Having a diverse court system allows for greater access to justice.
Blackburne-Rigsby specifically spoke about the concept of Miranda Rights, a precedent that gives criminal defendants the right to an attorney. But, these rights only apply to criminal cases, not civil. If someone is charged with a crime in a civil suit and cannot afford an attorney, they will still not be given one by the courts. For this reason, Blackburne-Rigsby said it is important to have judges who will be able to empathize with defendants’ backgrounds.
She concluded by opening the floor up for questions from the audience.
Sophomore Jack Thorne attended the event for a class assignment, but found the lecture to be much more interesting than he expected.
“Anna Blackburne-Rigsby was a really engaging speaker so it was really easy to become interested,” Thorne said. “I learned so much about diversity in the judicial branch and about concepts I had even considered before.”
Blackburne-Rigsby’s has a special connection to Elon too, making this lecture even more special. Her son, Julian Rigsby, is a sophomore at Elon and a member of the Liberal Arts Forum. He is the one who pitched her as a speaker last year.
At first, Rigsby wasn’t sure if the Liberal Arts Forum would support the idea of bringing his mother as a speaker, but he was pleasantly surprised when the group was as excited about the idea as he was.
“I’ve looked up to my mom forever and she’s the biggest inspiration in my life,” Rigsby said. “I’ve seen her speak so many different times in a variety of venues and to countless different types of people. She is an amazing woman who works harder than anyone I know. She is so committed and has so much passion for the work that she does and I know she loves being a judge so much.”
Unfortunately, Rigsby could not attend the event because of a stomach virus, but he still views having his mother speak on campus as one of his best memories at Elon.
“Bringing her to Elon was one of the best ideas I’ve had in college,” Rigsby said. “Sadly I had a stomach virus and the day she was supposed to speak, I couldn’t attend her speech. My other members in the forum kept me updated the entire time and told me that she spoke with all of the elegance and awesomeness that I was already aware of. She honestly changes my life daily and I love her more and more every day because of how loving, passionate, and strong she is.”
Blackburne-Rigsby was appointed to the Superior Court of the District of Columbia by former president Bill Clinton and then appointed to the D.C. Court of Appeals by George W. Bush in 2006. Later this week, she will be sworn in as chief justice of the court.
## Dean of business school works to bridge gap between students and administration
Each month, Raghu Tadepalli — in addition to his many responsibilities as dean of the Martha and Spencer Love School of Business — has lunch with senior business students to discuss their praises and critiques of the business school.
Tadepalli does not simply hear complaints from students — he listens to them.
At one of these monthly lunches last year, students voiced their frustrations about the reporting portions of the business school’s internship requirement. They felt the essays and reflections required in addition to the internship were onerous and redundant.
So, Tadepalli made changes. He worked with his colleagues to modify the internship requirements and alleviate some of the work for students. This past summer, he supervised around 20 interns because he wanted to see first-hand the work students were completing to receive internship credit. Seeing that there were still redundancies in the work students were required to complete, Tadepalli helped revise the program again.
Meeting with and listening to students is not only what Tadepalli sees as the most important part of his job, but it is also his favorite part.
“I think a large number of students know my door is open, so people drop in,” Tadepalli said. “I think [the students are] really respectful and know that they’re also very busy. I’d say that they’re quite a few students who feel comfortable dropping in to chat.”
Kristin Barrier, director of operations and accreditation in the Love School of Business, sits in an office directly across from Tadepalli and witnesses these student interactions daily.
“Dean Tadepalli has a true open-door policy, and he often meets with students who are looking for help or advice,” Barrier said.
Barrier graduated from the Love School of Business in 2004 and earned her Master’s of Business Administration from Elon in 2007, so she has seen Elon’s growth under Tadepalli’s leadership from the beginning.
“The Love School of Business has grown in numbers exponentially since Dean Tadepalli arrived, and also the quality of education and the quality of students we are attracting has improved,” Barrier said. “I graduated from the Love School of Business in 2004 and the M.B.A. program in 2007, and I have seen both undergraduate and graduate programs take huge leaps in improving the curriculum and rigor in the past 5 years that he has been dean.”
Tadepalli’s move to Elon
Tadepalli began his position as dean of the Martha and Spencer Love School of Business in July of 2012. He came to the university from Babson College, where he previously served as the Murata Dean and Professor of Marketing in the F. W. Olin Graduate School of Business.
Tadepalli received a bachelor’s and master’s degree in commerce with a major in accounting from Andrha University in India. He then went onto complete his Master of Business Administration degree with a concentration in marketing from Arizona State University and earned his doctorate from Virginia Tech.
After completing receiving his doctorate, Tadepalli was given several job offers to complete marketing research, but turned down the offers because he “didn’t see much fun in it.” Tadepalli wanted to continue doing research but was more excited by the idea of teaching and interacting with students. This led him to begin working in higher education.
Tadepalli held faculty and staff positions at a few other universities before arriving at Elon almost five years ago. While he has not taught in a classroom in a while, he is still able to conduct research and will have a new study published in a few months.
No matter what position he has held, Tadepalli has always made students his biggest priority. His dedication to students is part of the reason he enjoys working at Elon because the university as a whole mirrors that same commitment.
“It’s nice to be at a university campus when there is such an undivided attention on making sure that students learn,” Tadepalli said. “I think it’s a value that permeates everything that we as, faculty or staff, that we do. Students are really at the center of what we do … here there is no mistake about it: We are about students. We are about what students are learning and how they’re learning and how we are helping them develop into leaders for tomorrow.”
His commitment to listening to students is also reflected in his leadership style. For him, listening to those he is leading is the most important aspect of leadership.
“I think listening is very important,” Tadepalli said. “When someone walks into my office, the conversation is about them. It’s not about me. So you have to pay attention to what they’re saying.”
Barrier has seen this through her interactions with the dean. Barrier is responsible for general behind the scenes work and ensuring that the Love School of Business is meeting all of the requirements for the Association to Advance Collegiate Schools of Business (AACSB) accrediting body. They work together to meet the requirements and complete other projects in the school.
“[His leadership is] very laid-back and supportive — he trusts people to do their jobs, and he makes sure they have the resources needed to accomplish what is expected in their job,” Barrier said. “He is incredibly easy to work with, values my opinions and, in my view, a great leader for the business school.”
Junior Franki Filandro had the opportunity to witness this leadership and work closely with Tadepalli when she brought the business fraternity Delta Sigma Pi to campus. Filandro wanted to provide business students with the opportunity to be involved in a professional organization without having to give up a large portion of their time.
To bring the fraternity to campus, Filandro had to first be approved by Delta Sigma Pi’s national board and then Elon. Tadepalli was supportive of the organization and helped convince the university to allow Filandro to start a chapter at Elon.
Filandro said that Tadepalli succeeds in making himself available to students, saying that he is always around to meet with students or tries to reach out to them if he hasn’t seen them in a while. Listening and staying connected to students is a part of Tadepalli’s leadership style.
His style is certainly very open and almost a backseat kind,” Filandro said. “He gives you a chance to figure it out for yourself and then guides you in the slightest of ways. I believe that the whole time he knows where you should be going, but he let’s you find it on your own.”
The future of the business school
Still, Filandro fears that the business school may be growing in quantity, not quality. According to Tadepalli, the business school has grown by around 65 percent in the past five years. In his time at Elon, Tadepalli has worked with faculty and staff to revise every major and minor within the business school, revise the MBA program and added a Master’s of Science in Analytics and Master’s of Science in Accounting.
And, with Sankey Hall, the new business school, beginning construction this summer, the fear of getting too big too quickly is on Tadepalli’s mind, too.
“We recognize that it’s not important to just be big but to be good,” Tadepalli said. “We are constantly benchmarking and asking ourselves what we can do.”
Diversity as a priority
Though many members of the Elon community believe that increasing diversity on campus needs to be made a greater priority, Tadepalli recognizes the need and works to bring diverse voices into the business school. He believes that understanding how to manage diverse groups is important to excelling in business and said that diversity is one of the business school’s core values.
He has worked to uphold that value by implementing diversity education initiatives into various class curriculums and recruiting diverse faculty and staff members. Currently, he said the business school has about 60 faculty members representing 16 different countries.
On a personal level, of the six deans at the university, Tadepalli is the only dean of color, so he sees the need for diversity first-hand. It is not uncommon for him to be the only person of color in a room.
“Yesterday morning I was in a meeting, there must have been ten people there, and I was the only non-white in that room,” “Sometimes, you know I kind of have an out of body experience and I’m thinking, ‘Wow if I were looking down upon me, if there was a picture that was taken, how would this look?’ And I think, in that respect, I came here from Babson College in Massachusetts, which is very diverse. And so, I’d say, at Elon I think the values are there and the respect for diversity is there, but we need more demonstrable programs in that regard.”
Many accomplishments to be proud of
Even with all of the growth the Love School of Business has yet to see, whether it be in infrastructure, quality or diversity, Tadepalli has plenty to be proud of. When reflecting on his proudest moments at Elon, he recalled an experience from rather early in his career here.
During his second year at Elon, the Love School of Business had an accreditation visit where three deans from other schools visited campus and assessed the business program with the goal of giving recommendations to make the program better.
At the end of the visit, the three deans sat down with Tadepalli, President Leo Lambert and Provost and Executive Vice President Steven House to discuss their findings and give recommendations for improvements over the next 5 years.
Their recommendation was simple: they had no recommendation. To the three deans, the business school was operating perfectly and there were no changes they felt Tadepalli needed to make. Even after praise like this, Tadepalli finds a way to recognize the work of others, showing once again the interesting dynamics of his leadership style.
“So I think it’s kudos to my colleagues,” Tadepalli said, “but when a group of three deans from outside come and look at you and their 21 standards, and at the end of it they say, we have no recommendation, you feel pretty good.” |
# Mention any 4 rational numbers which are less than 5
• Last Updated : 05 Aug, 2021
We use numbers in our daily lives. A numeral is a common term used to describe them. Without numbers, we can’t count items, dates, times, money, or anything else. Sometimes these numbers are used for measuring, and other times they are used for labeling. Numbers have properties that enable them to perform arithmetic operations.
Math teaches us about many sorts of numbers. Examples include natural and whole numbers, odd and even numbers, rational and irrational numbers, and so on. There are several different types of numbers; these are whole numbers, natural numbers, real numbers, integers, complex numbers, rational numbers, and irrational numbers.
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### What are Rational Numbers?
Rational numbers are one of the most prevalent types of numbers that we learn in math after integers. These numbers are of the form p/q, where p and q are integers and q ≠ 0. Because of the underlying structure of numbers, p/q form, most individuals find it difficult to distinguish between fractions and rational numbers. 3, 4, 5, and so on are some examples of rational numbers as they can be expressed in fraction form as 3/1, 4/1, and 5/1.
### How to Identify Rational Numbers?
The number in each of the examples above may be represented as a fraction of integers. As a result, each of these figures is a rational figure. To determine whether a particular number is rational, we may see if it meets any of the following criteria:
• It can be represented as a fraction of integers.
• We can determine if the number’s decimal expansion is terminating or non-terminating.
• All whole numbers are always rational numbers.
### Mention any 4 rational numbers which are less than 5
Solution:
Rational numbers are one of the most prevalent types of numbers that we learn in math after integers. A rational number is a sort of real number that has the form p/q where q≠0. All whole numbers, natural numbers, fractions of integers, integers, and terminating decimals are rational numbers.
Here, the given rational number is 5 and it is also a whole number. It can also be expressed in fraction form as 5/1. We can determine all the whole numbers less than 5 as a rational number. Hence, 1, 2, 3, and 4 are the rational numbers less than 5.
### Similar Questions
Question 1: Mention any 2 rational numbers less than 7.
Solution:
Here, the given rational number is 7 and it is also a whole number. It can also be expressed in fraction form as 7/1. We can determine all the whole numbers less than 5 as a rational number. Hence, 3, and 4 are the rational numbers less than 7.
Question 2: Mention any 3 rational numbers less than 1.
Solution:
A rational number is a sort of real number that has the form p/q where q≠0. It can be negative, positive, or zero. Here, the given rational number is 1. We can determine all the integers less than 1 as a rational number. Hence, −2, −1, and 0 are the rational numbers less than 1.
Question 3: Mention any 1 rational number less than 2.
Solution:
Here, the given rational number is 1 and it is also a whole number. It can also be expressed in fraction form as 2/1. We can determine all the whole numbers less than 2 as a rational number. Hence, 1 is the rational number less than 2.
Question 4: Mention any 5 rational numbers less than 6.
Solution:
Here, the given rational number is 6 and it is also a whole number. It can also be expressed in fraction form as 6/1. We can determine all the whole numbers less than 6 as a rational number. Hence, 1, 2, 3, 4, and 5 are rational numbers less than 6.
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# Applications of Derivatives: Optimization
Applications of Derivatives: Optimization
Example: Design a box, open at the top, to be made in one piece from a 24inch by 9inch sheet of cardboard which will have the maximum possible volume. Like this:
__________________________________________________________________________________
What is it we are trying to optimize?
We want to maximize the volume, V.
What will our final answer look like? What variable(s) are we looking for?
The overall dimensions of the box, height, length, and width. Let’s label them x,y,z.
We need a 2 dimensional sketch now to label these and help us visualize the problem.
NEED SKETCH
If we cut out the corners and fold up along the dashed lines, we’ll have a box. The dimension, x is critical, so we label it. The height then is x.
Now we need an equation. What is the general equation for the volume of a rectangular box?
V = xyz
And we’ll need to put that in terms of only one variable, say, x, the height. Look at the sketch and express the length and width in terms of x.
V = x(9-2x)(24-2x)
Now that we have an equation for volume in terms of one variable, x, we can take the derivative, set it equal to zero, and solve for x. First we multiply it out.
V = 216x – 66x2 + 4x3
dV/dx = 216 – 132x + 12x2
dV/dx = 12(18 – 11x + x2)
0 = 12(18 – 11x + x2)
0 = (9-x)(2-x)
x = 9, 2
Of course x = 9 would be too large, so we use x = 2.
Solution: Box size: 2inch x 5inch x 20inch for a volume of 200inch3 |
# 13. Simultaneous equations and row operations
## 13. Simultaneous equations and row operations
Thus far in the course, we have focused on the geometric aspects of matrices and the transformations they determine. Now we'll approach the subject from the point of view of simultaneous equations.
### Simultaneous equations
A system of simultaneous linear equations, for example x minus 1 equals minus 1, x plus y equals 3, conmprises a finite list of linear equations in some number of variables. Linear means that each term is either constant of else linear in the variables (i.e. the terms are things like 3 x or minus 4 y, not x squared or x y e to the x).
A system of simultaneous linear equations is a matrix equation in disguise. For example, the system above can be written as the 2-by-2 matrix 1, minus 1; 1, 1 times the vector x, y equals the vector minus 1, 3. Indeed, if we multiply this expression out, we get the vector x minus y, x plus y equals the vector minus 1, 3, and these two vectors are equal if and only if their components are equal, which is equivalent to the two equations we started with.
We will abbreviate such a matrix equation by writing a so-called augmented matrix: we write the matrix of coefficients, then a vertical bar, then the column of constants: 1, minus 1, bar minus 1; 1, 1, bar, 3 This is really just a shorthand for the equation the 2-by-2 matrix 1, minus 1; 1, 1 times the vector x, y equals the vector minus 1, 3.
### Solving these equations
To solve this system, we will manipulate the equations one at a time. We will see what happens to the augmented matrix as we perform these manipulations. We start with: x minus 1 equals minus 1, x plus y equals 3 or as an augmented matrix 1, minus 1, bar minus 1; 1, 1, bar, 3 We can eliminate x from the second equation by subtracting equation 1 from equation 2: x minus y equals minus 1, 2 y equals 4, or as an augmented matrix 1, minus 1, bar, minus 1; 0, 2, bar, 4 This has had the effect on the augmented matrix of subtracting the first row from the second. We will write this operation as R_2 maps to R_2 minus R_1. Next, we can divide equation 2 by 2: x minus y equals minus 1, y equals 2, or as an augmented matrix 1, minus 1, bar, minus 1; 0, 1, bar, 2 This has the effect of dividing row 2 by 2, which we write as R_2 maps to a half R_2. Finally, we add equation 2 to equation 1: x equals 1, y equals 2, or as an augmented matrix 1, 0, bar, 1; 0, 1, 2. This is the row operation R_1 maps to R_1 plus R_2. We have now solved our system of equations.
### Row operations
This process of solving simultaneous equations can therefore be understood as performing a sequence of row operations on the augmented matrix.
Definition:
• Type I row operations: Replace row i by row i plus a multiple (say lambda) of row j. We write this as R_i mapsto R_i plus lambda R_j. In terms of equations, this means we're adding/subtracting a multiple of equation j to equation i.
• Type II row operations: Replace row i by a nonzero multiple (say lambda not equal to zero) of row i. We write this as R_i maps to lambda R_i. In terms of equations, this means we're multiplying an equation by a nonzero constant.
• Type III row operations: Swap row i and row j. This corresponds to reordering your equations.
Remark:
We don't allow ourselves to multiply an equation by zero: this will change our system of equations by effectively ignoring some of them.
In our example, we "solved" the equation when we reached x=1, y=2. This meant that the augmented matrix had the identity matrix on the left of the vertical bar. So the aim of the row operations is to put the augmented matrix in the form I bar b where I is the identity matrix and b is a column vector (of "constants"). Of course, this will sometimes fail:
Example:
Consider the system x plus y equals 1 (one equation, two variables). The augmented matrix is now 1, 1, bar 1 No matter what you do, you can't put this into the form identity bar b because the matrix on the left hand side of the vertical bar isn't a square matrix (so can't be turned into the identity matrix by row operations). Nonetheless, the equation is easy to solve: we have x equals 1 minus y, so given any y we get a solution 1 minus y, y. So we don't need to get to the identity matrix so say our system is "solved". In the next few videos, we'll see the correct thing to aim for is to put the matrix on the left of the bar into reduced echelon form. |
# Mastering the Three Trigonometric Ratios: Sin Cos Tan
POSTED ON MARCH 10, 2023
The three trigonometric ratios, sin cos tan, lay down the foundations for trigonometry. Through these ratios, you can evaluate trigonometric expressions, find unknown measures, and solve problems involving right triangles. This article will be your guide to mastering what sin cos tan represents.
## What Does the Trigonometric Ratios Sin Cos Tan Represent?
Sin cos tan represents the three most important ratios in trigonometry: sine, cosine, and tangent. These three ratios are based on the ratios formed by two sides of a right triangle. They are extremely useful when it comes to finding unknown sides and angles of a right triangle.
Before getting excited, why don’t we begin with the basics? Start by understanding the key components of the ratios and identifying the specific lengths involved with finding the value of sine, cosine, and tangent of a given angle.
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In a right triangle and with a reference angle, say A, there are three crucial components:
• Opposite side: the side facing opposite to angle A.
• Hypotenuse: the longest side of any right triangle.
This means that through sin cos tan, we can now relate one angle of a triangle with its two sides.
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Let’s break down the three ratios: sine, cosine, and tangent.
• The sine of an angle, say θ, is equal to the ratio of the side opposite θ over the right triangle’s hypotenuse. A great way to remember “SOH”, where S for Sine, O for Opposite, and H for Hyptenuse.
• Similarly, the cosine of θ is equal to the ratio of the side adjacent to θ over the right triangle’s hypotenuse. Now, easily remember this relationship using “CAH”: C for Cosine, A for Adjacent, and H for Hypotenuse.
• Tangent of θ is equal to the ratio of the side opposite θ over the side adjacent to θ. Like the two other ratios, remember the relationship using “TOA”: T for Tangent, O for Opposite, and A for Adjacent.
This table summarizes the relationships between sin cos tan - plus a reminder to use the guide: SOH CAH TOA!
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#### Example 1
What are the sin cos tan values of A for the right triangle shown below?
Identify the important sides of the right triangle when finding the values of sin A, cos A, and tan A.
1. The side opposite angle A has a length of 8 cm.
2. The adjacent side relative to A has a length of 6 cm.
3. The hypotenuse (longest side of the right triangle) is 10 cm long.
Now, apply the relationships established for sin cos tan and evaluate the values that we need.
This shows that through the relationships between the angles and sides of a right triangle, it is now possible to evaluate the values of sin cos tan. In fact, for this problem, we have A= 4/5, A= 3/5, and A= 4/3.
## Using Sin Cos Tan To Find the Exact Values
Recall that we have two special right triangles in Trigonometry: the 45-45-90 and 30-60-90 triangles. These triangles’ sides have special ratios of 1:1:√2 and 1:√3:2, respectively. By using the rules for sin cos tan, it’s now possible to find the exact trigonometric values for the following angles: 30°, 45°, and 60°.
#### Example 2
Find the exact values of the following expressions:
a. sin 30°
Use the 30-60-90 triangle since you’re looking for the exact value of 30°. Identify the side facing opposite 30° and the length of the hypotenuse.
Simplify the ratio to find the exact value of 30°. This shows that 30°= 1/2.
b. cos 45°
Apply a similar approach to evaluate 45°, but this time, use the 45-45-90 triangle.
The side adjacent to 45° is 1 unit long while the hypotenuse is √2 unit long. Simplifying this leads to 45°= 1/√2.
## Final Thoughts
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# 2016 AIME I Problems/Problem 13
## Problem
Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$. A fence is located at the horizontal line $y = 0$. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where $y=0$, with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where $y < 0$. Freddy starts his search at the point $(0, 21)$ and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river.
## Solution
Clearly Freddy's $x$-coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$-coordinate. Observe that $E(24)=0$, and $$E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}$$ for all $y$ such that $1\le y\le 23$. Also note that $E(0)=1+\frac{2E(0)+E(1)}{3}$. This gives $E(0)=E(1)+3$. Plugging this into the equation for $E(1)$ gives that $$E(1)=1+\frac{E(2)+3E(1)+3}{4},$$ or $E(1)=E(2)+7$. Iteratively plugging this in gives that $E(n)=E(n+1)+4n+3$. Thus $E(23)=E(24)+95$, $E(22)=E(23)+91=E(24)+186$, and $E(21)=E(22)+87=E(24)+273=\boxed{273}$.
## Video Solution
For those who want more explanation, here is a video explaining the solution: https://www.youtube.com/watch?v=jQCGkOGMlFQ
## See also
2016 AIME I (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# ACT Math : Transformation
## Example Questions
### Example Question #1 : How To Find Transformation For An Analytic Geometry Equation
If this is a sine graph, what is the phase displacement?
π
0
2π
4π
(1/2)π
0
Explanation:
The phase displacement is the shift from the center of the graph. Since this is a sine graph and the sin(0) = 0, this is in phase.
### Example Question #1 : Transformations
If this is a cosine graph, what is the phase displacement?
2π
4π
π
0
(1/2)π
π
Explanation:
The phase displacement is the shift of the graph. Since cos(0) = 1, the phase shift is π because the graph is at its high point then.
### Example Question #2 : Transformation
A regular pentagon is graphed in the standard (x,y) coordinate plane. Which of the following are the coordinates for the vertex P?
Explanation:
Regular pentagons have lines of symmetry through each vertex and the center of the opposite side, meaning the y-axis forms a line of symmetry in this instance. Therefore, point P is negative b units in the x-direction, and c units in the y-direction. It is a reflection of point (b,c) across the y-axis.
### Example Question #3 : Transformation
If g(x) is a transformation of f(x) that moves the graph of f(x) four units up and three units left, what is g(x) in relation to f(x)?
Explanation:
To solve this question, you must have an understanding of standard transformations. To move a function along the x-axis in the positive direction, you must subtract the value from the operative x-value. For example, to move a function, f(x), five units to the left would be f(x+5).
To shift a function along the y-axis in the positive direction, you must add the value to the overall function. For example, to move a function, f(x), three units up would be f(x)+3.
The question asks us to move the function, f(x), left three units and up four units. f(x+3) will move the function three units to the left and f(x)+4 will move it four units up.
Together, this gives our final answer of f(x+3)+4.
### Example Question #4 : Transformation
What is the period of the function?
2π
3π
π
1
4π |
# Congruence (geometry)
In geometry, two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other.[1]
An example of congruence. The two triangles on the left are congruent, while the third is similar to them. The last triangle is neither congruent nor similar to any of the others. Congruence permits alteration of some properties, such as location and orientation, but leaves others unchanged, like distances and angles. The unchanged properties are called invariants.
More formally, two sets of points are called congruent if, and only if, one can be transformed into the other by an isometry, i.e., a combination of rigid motions, namely a translation, a rotation, and a reflection. This means that either object can be repositioned and reflected (but not resized) so as to coincide precisely with the other object. So two distinct plane figures on a piece of paper are congruent if we can cut them out and then match them up completely. Turning the paper over is permitted.
This diagram illustrates the geometric principle of angle-angle-side triangle congruence: given triangle ABC and triangle A'B'C', triangle ABC is congruent with triangle A'B'C' if and only if: angle CAB is congruent with angle C'A'B', and angle ABC is congruent with angle A'B'C', and BC is congruent with B'C'. Note hatch marks are used here to show angle and side equalities.
In elementary geometry the word congruent is often used as follows.[2] The word equal is often used in place of congruent for these objects.
• Two line segments are congruent if they have the same length.
• Two angles are congruent if they have the same measure.
• Two circles are congruent if they have the same diameter.
In this sense, two plane figures are congruent implies that their corresponding characteristics are "congruent" or "equal" including not just their corresponding sides and angles, but also their corresponding diagonals, perimeters, and areas.
The related concept of similarity applies if the objects have the same shape but do not necessarily have the same size. (Most definitions consider congruence to be a form of similarity, although a minority require that the objects have different sizes in order to qualify as similar.)
## Determining congruence of polygons
The orange and green quadrilaterals are congruent; the blue is not congruent to them. All three have the same perimeter and area. (The ordering of the sides of the blue quadrilateral is "mixed" which results in two of the interior angles and one of the diagonals not being congruent.)
For two polygons to be congruent, they must have an equal number of sides (and hence an equal numberthe same numberof vertices). Two polygons with n sides are congruent if and only if they each have numerically identical sequences (even if clockwise for one polygon and counterclockwise for the other) side-angle-side-angle-... for n sides and n angles.
Congruence of polygons can be established graphically as follows:
• First, match and label the corresponding vertices of the two figures.
• Second, draw a vector from one of the vertices of the one of the figures to the corresponding vertex of the other figure. Translate the first figure by this vector so that these two vertices match.
• Third, rotate the translated figure about the matched vertex until one pair of corresponding sides matches.
If at any time the step cannot be completed, the polygons are not congruent.
## Congruence of triangles
Two triangles are congruent if their corresponding sides are equal in length, and their corresponding angles are equal in measure.
If triangle ABC is congruent to triangle DEF, the relationship can be written mathematically as:
${\displaystyle \triangle \mathrm {ABC} \cong \triangle \mathrm {DEF} .}$
In many cases it is sufficient to establish the equality of three corresponding parts and use one of the following results to deduce the congruence of the two triangles.
The shape of a triangle is determined up to congruence by specifying two sides and the angle between them (SAS), two angles and the side between them (ASA) or two angles and a corresponding adjacent side (AAS). Specifying two sides and an adjacent angle (SSA), however, can yield two distinct possible triangles.
### Determining congruence
Sufficient evidence for congruence between two triangles in Euclidean space can be shown through the following comparisons:
• SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.
• SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent.
• ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
The ASA Postulate was contributed by Thales of Miletus (Greek). In most systems of axioms, the three criteria – SAS, SSS and ASA – are established as theorems. In the School Mathematics Study Group system SAS is taken as one (#15) of 22 postulates.
• AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. AAS is equivalent to an ASA condition, by the fact that if any two angles are given, so is the third angle, since their sum should be 180°. ASA and AAS are sometimes combined into a single condition, AAcorrS – any two angles and a corresponding side.[3]
• RHS (Right-angle-Hypotenuse-Side), also known as HL (Hypotenuse-Leg): If two right-angled triangles have their hypotenuses equal in length, and a pair of shorter sides are equal in length, then the triangles are congruent.
#### Side-side-angle
The SSA condition (side-side-angle) which specifies two sides and a non-included angle (also known as ASS, or angle-side-side) does not by itself prove congruence. In order to show congruence, additional information is required such as the measure of the corresponding angles and in some cases the lengths of the two pairs of corresponding sides. There are a few possible cases:
If two triangles satisfy the SSA condition and the length of the side opposite the angle is greater than or equal to the length of the adjacent side (SSA, or long side-short side-angle), then the two triangles are congruent. The opposite side is sometimes longer when the corresponding angles are acute, but it is always longer when the corresponding angles are right or obtuse. Where the angle is a right angle, also known as the Hypotenuse-Leg (HL) postulate or the Right-angle-Hypotenuse-Side (RHS) condition, the third side can be calculated using the Pythagorean Theorem thus allowing the SSS postulate to be applied.
If two triangles satisfy the SSA condition and the corresponding angles are acute and the length of the side opposite the angle is equal to the length of the adjacent side multiplied by the sine of the angle, then the two triangles are congruent.
If two triangles satisfy the SSA condition and the corresponding angles are acute and the length of the side opposite the angle is greater than the length of the adjacent side multiplied by the sine of the angle (but less than the length of the adjacent side), then the two triangles cannot be shown to be congruent. This is the ambiguous case and two different triangles can be formed from the given information, but further information distinguishing them can lead to a proof of congruence.
#### Angle-angle-angle
In Euclidean geometry, AAA (Angle-Angle-Angle) (or just AA, since in Euclidean geometry the angles of a triangle add up to 180°) does not provide information regarding the size of the two triangles and hence proves only similarity and not congruence in Euclidean space.
However, in spherical geometry and hyperbolic geometry (where the sum of the angles of a triangle varies with size) AAA is sufficient for congruence on a given curvature of surface.[4]
### CPCTC
This acronym stands for Corresponding Parts of Congruent Triangles are Congruent an abbreviated version of the definition of congruent triangles.[5][6]
In more detail, it is a succinct way to say that if triangles ABC and DEF are congruent, that is,
${\displaystyle \triangle ABC\cong \triangle DEF,}$
with corresponding pairs of angles at vertices A and D; B and E; and C and F, and with corresponding pairs of sides AB and DE; BC and EF; and CA and FD, then the following statements are true:
${\displaystyle {\overline {AB}}\cong {\overline {DE}}}$
${\displaystyle {\overline {BC}}\cong {\overline {EF}}}$
${\displaystyle {\overline {AC}}\cong {\overline {DF}}}$
${\displaystyle \angle BAC\cong \angle EDF}$
${\displaystyle \angle ABC\cong \angle DEF}$
${\displaystyle \angle BCA\cong \angle EFD.}$
The statement is often used as a justification in elementary geometry proofs when a conclusion of the congruence of parts of two triangles is needed after the congruence of the triangles has been established. For example, if two triangles have been shown to be congruent by the SSS criteria and a statement that corresponding angles are congruent is needed in a proof, then CPCTC may be used as a justification of this statement.
A related theorem is CPCFC, in which "triangles" is replaced with "figures" so that the theorem applies to any pair of polygons or polyhedrons that are congruent.
## Definition of congruence in analytic geometry
In a Euclidean system, congruence is fundamental; it is the counterpart of equality for numbers. In analytic geometry, congruence may be defined intuitively thus: two mappings of figures onto one Cartesian coordinate system are congruent if and only if, for any two points in the first mapping, the Euclidean distance between them is equal to the Euclidean distance between the corresponding points in the second mapping.
A more formal definition states that two subsets A and B of Euclidean space Rn are called congruent if there exists an isometry f : RnRn (an element of the Euclidean group E(n)) with f(A) = B. Congruence is an equivalence relation.
## Congruent conic sections
Two conic sections are congruent if their eccentricities and one other distinct parameter characterizing them are equal. Their eccentricities establish their shapes, equality of which is sufficient to establish similarity, and the second parameter then establishes size. Since two circles, parabolas, or rectangular hyperbolas always have the same eccentricity (specifically 0 in the case of circles, 1 in the case of parabolas, and ${\displaystyle {\sqrt {2}}}$ in the case of rectangular hyperbolas), two circles, parabolas, or rectangular hyperbolas need to have only one other common parameter value, establishing their size, for them to be congruent.
## Congruent polyhedra
For two polyhedra with the same combinatorial type (that is, the same number E of edges, the same number of faces, and the same number of sides on corresponding faces), there exists a set of E measurements that can establish whether or not the polyhedra are congruent.[7][8] The number is tight, meaning that less than E measurements are not enough if the polyhedra are generic among their combinatorial type. But less measurements can work for special cases. For example, cubes have 12 edges, but 9 measurements are enough to decide if a polyhedron of that combinatorial type is congruent to a given regular cube.
## Congruent triangles on a sphere
As with plane triangles, on a sphere two triangles sharing the same sequence of angle-side-angle (ASA) are necessarily congruent (that is, they have three identical sides and three identical angles).[9] This can be seen as follows: One can situate one of the vertices with a given angle at the south pole and run the side with given length up the prime meridian. Knowing both angles at either end of the segment of fixed length ensures that the other two sides emanate with a uniquely determined trajectory, and thus will meet each other at a uniquely determined point; thus ASA is valid.
The congruence theorems side-angle-side (SAS) and side-side-side (SSS) also hold on a sphere; in addition, if two spherical triangles have an identical angle-angle-angle (AAA) sequence, they are congruent (unlike for plane triangles).[9]
The plane-triangle congruence theorem angle-angle-side (AAS) does not hold for spherical triangles.[10] As in plane geometry, side-side-angle (SSA) does not imply congruence.
## Notation
A symbol commonly used for congruence is an equals symbol with a tilde above it, , corresponding to the Unicode character 'approximately equal to' (U+2245). In the UK, the three-bar equal sign (U+2261) is sometimes used.
## References
1. Clapham, C.; Nicholson, J. (2009). "Oxford Concise Dictionary of Mathematics, Congruent Figures" (PDF). Addison-Wesley. p. 167. Archived from the original on 29 October 2013. Retrieved 2 June 2017.CS1 maint: bot: original URL status unknown (link)
2. "Congruence". Math Open Reference. 2009. Retrieved 2 June 2017.
3. Parr, H. E. (1970). Revision Course in School mathematics. Mathematics Textbooks Second Edition. G Bell and Sons Ltd. ISBN 0-7135-1717-4.
4. Cornel, Antonio (2002). Geometry for Secondary Schools. Mathematics Textbooks Second Edition. Bookmark Inc. ISBN 971-569-441-1.
5. Jacobs, Harold R. (1974), Geometry, W.H. Freeman, p. 160, ISBN 0-7167-0456-0 Jacobs uses a slight variation of the phrase
6. "Congruent Triangles". Cliff's Notes. Retrieved 2014-02-04.
7. Borisov, Alexander; Dickinson, Mark; Hastings, Stuart (March 2010). "A Congruence Problem for Polyhedra". American Mathematical Monthly. 117: 232–249. arXiv:0811.4197. doi:10.4169/000298910X480081.
8. Creech, Alexa. "A Congruence Problem" (PDF). Archived from the original (PDF) on November 11, 2013.
9. Bolin, Michael (September 9, 2003). "Exploration of Spherical Geometry" (PDF). pp. 6–7.
10. Hollyer, L. "Slide 89 of 112". |
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Home » Data Interpretation » Data Interpretation for Bank Exams – Practice Problems
Data Interpretation for Bank Exams
Data Interpretation is the very important topic for SBI and IBPS PO Exams. We need to learn some shortcut methods in adding, subtracting, multiplication and speed maths tricks on some Quantitative aptitude topics to solve Data Interpretation questions in less time.
Directions: Study the following graphs carefully and answer the following questions given below in each graph.
I. Data Interpretation Practice Problem 1:
Number of candidates (boys and girls) appeared and qualified in an examination over the years
(1.1) What was the approximate percentage of boys qualified to appeared in 1992?
(A) 35 (B) 70 (C) 80 (D) 65
Solution 🙁 B) The approximate percentage of boys qualified to appear in 1992.
=12.5/17.5 × 100=71.43 = 70%
(1.2) The total number of girls qualified in 1991 and 1992 together was exactly equal to the number of girls appeared in which of the following years—
(A) 1994 (B) 1995 (C) 1991 (D) 1990 (E) none of these
Solution 🙁 d) The total number of girls qualified to appear in 1991 and 92.
=2500+5000= 7500
= Total number of girls involved is 1990
(1.3) What was the percentage increase in the number of boys qualified from 1993 to 1994?
(A) 50 (B) 5 (C) 100 (D) 200 (E) none of these
Solution 🙁 d) The percentage increase in the number of boys qualified from 1993 to 1994
= (15 – 5) * 1000=10000
Percentage increase = 10000/5000* 100 = 200
(1.4) What did the difference between the total number of boys and girls appeared in 1993 and the total number of boys and girls appear in 1991?
(A) 500 (B) 2500 (C) 7500 (D) 10000 (E) none of these
Solution 🙁 b) difference between the total number of boys and girls appeared is 1993 and the total number of boys and girls appeared is 1993 and the total number of boys and girls appeared in 1991
= [(12.5 + 10) – (15 + 5)] * 1000
= (22.5 – 20) * 1000= 2.5* 1000
= 2500
(1.5) In which of the following year was the percentage of girls qualified to appear the highest among the given years?
(A) 1991 (B) 1993 (C) 1994 (D) 1992 (E) none of these
Solution 🙁 b) the percentage of girls qualified to appear in 1990 = 5.0/7.5×100 = 66.66%
The percentage of girls qualified to appear in 1991 = 2.5/5×100 = 50%
The percentage of girls qualified to appear in 1992 = 5/10×100 = 50%
The percentage of girls qualified to appear in 1993 = 7.5/10×100 = 75%
The percentage of girls qualified to appear in 1995 = 7.5/12.5×100 = 60%
The percentage of girls qualified to appear in 1996 = 12.5/17.5×100 = 71.43%
DI Practice Problem II:
Percentage profit earned by two companies over the given years.
% Profit = [(Income –Expenditure)/ Expenditure]*100.
2.1 ) If the expenditure of the company B in 2008 was Rs.200 crores. What was the income?
1. 240 Crores
2. 220 Crores
3. 160 Crores
4. Cannot be determined
5. None of these.
Let the income be Rs. X Crores.
20 = [(x-200)/200] × 100
• 40 = x-200
• X=Rs. 240 crores.
2.2) If the income of Company A in 2010 was Rs.600 crores, what was its expenditure?
1. 360 crores.
2. 480 Crores.
3. 375 Crores.
4. Cannot be determined
5. None of these
The income of Company A in 2002 is Rs. 600 crores and % profit = 60
Assume expenditure is x.
60 = [(600 – x)/x] ×100
x= [(600 – x)/ 60 ] × 100
3x = 3000 – 5x
8x = 3000
X = Rs. 375 Crores
2.3) If the income of company B in 2006 was Rs.200 crores. What was its profit in 2007?
1. 21.5 Crores
2. 153 Crores
3. 46.15 Crores
4. Cannot be determined (income and expenditure not given in 2007)
5. None of these.
2.4) If the income of the two companies in 2006 were equal, what was the ratio of their expenditure?
1. 1: 2
2. 26: 27
3. 100:67
4. Can’t solve
5. None of these.
Company A, 35 = [(I – Exp1)/Exp1 ] × 100 -> (1)
• 35Exp1 = 100 I – 100 Exp1
• 135 Exp1 = 100 I
Company B, 30 = [(I – Exp2)/Exp2 ] × 100
130 Exp2 = 100 I -> (2)
From (1) and (2)
135 Exp1 = 130 Exp2
Exp1/Exp2 = 130/135
Ratio of Expenditures of A and B ,
Exp1: Exp2 = 130 : 135
26: 27.
2.5) What is the percent increase in percent profit for company B from the year 2008 to 2009?
1. 75
2. 175
3. 86
4. Cannot be determined
= [(35- 20)/ 20 ] × 100 = 75
DI Practice Problem III:
Percent Profit earned by two companies producing Electronic Goods over the years.
Profit Earned = Total Income – Total Investment in the year.
3.1) If the profit earned in 2006 by company B was Rs. 812500. What was the total income of the company in that year?
(a) Rs. 12, 50, 000
(b) Rs. 20, 62, 500
(c ) Rs. 16, 50, 000
(d ) Rs. 18, 25, 000
Sol:
% profit = [profit earned / total investment ] × 100
65 = [812500/ 1 ] × 100
I = (812500 × 100)/65
= Rs. 1250000
Total Income = Profit earned + total investment.
= Rs. (812500 + 1250000)
= Rs. 2062500
3.2) If the amount invested by two companies in 2005 was equal, what was the ratio between the total income in 2005 of the companies A and B respectively?
(a) 31: 33
(b) 33: 31
(c ) 34: 31
(d ) 14: 11
Sol:
70 = (x/ I) × 100
• 70 I = 100x —–à (1)
55 = (y/ I) × 100
• 55 I = 100y —–à (2)
From (1) & (2 )
[70 I /100 ] +I : [ 55 I/ 100] +I
170 I : 155 I = 34 : 31.
3.3) If the total income invested by two companies in 2009 was Rs. 27 lakhs while the amount invested by company B was 50% of the amount invested by company A. What was the total profit earned by the two companies together?
(a) Rs. 21.15 lakhs
(b) Rs. 20.70 lakhs
(c ) Rs. 18.70 lakhs
(d ) Rs. 20.15 lakhs
Sol.:
Investment by company A = [2/3] × 27 = Rs. 18 lakh
Investment by company B = Rs. 9 lakh.
For company A,
75 = [p /18 ] × 100.
P= (75 × 18)/ 100 = Rs. 13.5 lakh.
For Company B,
80 = [q / 9] × 100
q = 80 × 9/ 100 lakh
= Rs. 7.2 lakh
Total profit earned = 13.5 + 7.2 = Rs. 20.7 lakh
3.4) If the incomes of company A in 2007 and 2008 were equal and the amount invested in 2007 was Rs. 12 lakhs, what was the amount invested in 2008?
(a) Rs. 10, 87, 500
(b) Rs. 10, 85, 700
(c ) Rs. 12, 45, 000
(d ) Rs. 12, 85, 000
Sol:
For the year 2007,
45 = (p/12 )× 100
=> p = ( 45 × 12 )/100 = Rs. 5.4 lakh.
Total income = 12 + 5.4 lakh
=Rs. 17.4 lakh
If the amount invested in 2008 be Rs. 1 lakh, then
60 = [17.4 – I / I] × 100
• 160 I = 1740
• I = 1740 /160 = 10.875 lakh
• = Rs. 1087500.
3.5) If the amount of profit earned by company A in 2006 was Rs. 10.15 lakhs. What was the total investment?
(a) Rs. 13. 8 lakhs
(b) Rs. 14.9 lakhs
(c ) Rs. 15.4 lakhs
(d ) Rs. 18.45 lakhs
sol.:
55 = [10.15/ Investment (I) ] ×100
I = 10.15 × 100/ 55
= Rs. 18.45 lakh.
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# Cube and Dice Questions Practice Set for Competitive Exams
Cube and Dice questions are a very important component of competitive exams including TOFEL, RRB, MBA, Bank PO, and CA. A cube, sometimes known as a “regular hexahedron,” is a three-dimensional solid object bordered by six square faces or surfaces. Dice, on the other hand, are cuboidal or cubical devices with embedded numbers, symbols, or figures. Dice are used in games like ludo, craps, and other non-gambling activities. Many questions of different kinds using Cube and Dice logic are frequently asked. The main concepts of the Cube and Dice reasoning part, as well as solved examples, practice questions, and other helpful hints, will all be covered in this article.
## What are Cube and Dice Questions?
A cube is a three-dimensional figure that can only be constructed from squares. A square becomes a cube when its height is equal to one of its sides. Die/Dice is a three-dimensional figure that displays various numbers, letters, colors, and other symbols on each of its six sides or faces. It has twelve edges and eight corners. A dice has length, breadth, and height that are all equal.
As of right now, we are aware of the questions that make up the portion on cube and dice reasoning. Let’s understand each of the many question types that could appear in your exams:.
1. Single-dice problem-based reasoning: In this kind of cube and dice reasoning, questions about a single die will be asked.
2. Problem-based on Two or more Dice: Two or more dice-related problems will be asked in this kind of cube and dice reasoning.
## How to Solve Cube and Dice Questions – Tips and Tricks
Various tips and tricks for answering the Cube and Dice reasoning section questions are provided below for candidates.
## Practice Set on Cube and Dice Questions
1. Which symbol will be on the face opposite to the face with the symbol *?
A)@
B)\$
C)8
D)+
2. Two positions of dice are shown below. How many points will appear on the opposite to the face containing 5 points?
A)3
B)1
C)2
D)4
3. Which digit will appear on the face opposite to the face with the number 4?
A)3
B)5
C)6
D)2/3
4. Two positions of a dice are shown below. Which number will appear on the face opposite to the face with the number 5?
A)2/6
B)2
C)6
D)4
5. How many points will be on the face opposite to in face which contains 2 points?
A)1
B)5
C)4
D)6
6. What is the total number of faces in a standard cube?
A) 4
B) 6
C) 8
D) 12
7. How many vertices does a cube have?
A) 4
B) 6
C) 8
D) 12
8. What is the shape of a standard six-faced dice?
A) Cube
B) Cylinder
C) Sphere
D) Pyramid
9. What is the sum of numbers on opposite faces of a standard dice?
A) 6
B) 7
C) 12
D) 14
10. When a cube is unfolded, how many faces share a common edge?
A) 1
B) 2
C) 3
D) 4
11. What is the angle between any two adjacent faces of a cube?
A) 60 degrees
B) 90 degrees
C) 120 degrees
D) 180 degrees
12. How many edges does a cube have?
A) 4
B) 6
C) 8
D) 12
13. Which of the following statements is true about the faces of a cube?
A) All faces are congruent
B) Opposite faces are congruent
D) None of the above
14. When a standard dice are rolled, what is the probability of getting an odd number?
A) 1/2
B) 1/3
C) 1/6
D) 2/3
15. What is the minimum number of moves required to solve a standard Rubik’s Cube?
A) 9
B) 12
C) 15
D) 20
16. Which of the following is true about the diagonals of a cube?
A) All diagonals are equal in length
B) There are no diagonals in a cube
C) Diagonals connecting opposite vertices are equal in length
D) Diagonals connecting opposite edges are equal in length
17. What is the volume of a cube with a side length of 5 units?
A) 15 cubic units
B) 25 cubic units
C) 125 cubic units
D) 150 cubic units
18. What is the surface area of a cube with a side length 4 units?
A) 16 square units
B) 24 square units
C) 32 square units
D) 64 square units
19. If the numbers on two faces of a standard dice are added, what is the probability of getting a sum of 7?
A) 1/6
B) 1/3
C) 1/4
D) 1/2
20. How many types of diagonals does a cube have?
A) 2
B) 3
C) 4
D) 6
21. If a cube has a volume of 64 cubic units, what is the length of one side?
A) 2 units
B) 4 units
C) 6 units
D) 8 units
22. What is the shape of the faces of a standard dice?
A) Triangle
B) Rectangle
C) Square
D) Circle
23. How many faces of a cube meet at each vertex?
A) 2
B) 3
C) 4
D) 6
24. What is the maximum possible sum of the numbers on two faces of a standard dice?
A) 6
B) 7
C) 12
D) 14
25. What is the minimum possible sum of the numbers on two faces of a standard dice?
A) 1
B) 2
C) 3
D) 4
26. How many squares are there on the surface of a standard cube?
A) 4
B) 6
C) 8
D) 12
27. What is the shape of the cross-section when a cube is sliced by a plane parallel to one of its faces?
A) Square
B) Rectangle
C) Triangle
D) Circle
28. What is the total number of edges in a standard dice?
A) 6
B) 8
C) 12
D) 24
29. If a cube is painted on all its faces and then cut into smaller cubes of equal size, how many of the smaller cubes will have paint on exactly two faces?
A) 1
B) 4
C) 8
D) 16
30. Which of the following is true about the angles between adjacent faces of a cube?
A) All angles are acute
B) All angles are obtuse
C) There are both acute and obtuse angles
D) All angles are right angles
31. What is the sum of the numbers on three faces of a standard dice if they are adjacent to each other?
A) 10
B) 11
C) 12
D) 13
32. How many lines of symmetry does a cube have?
A) 1
B) 2
C) 3
D) 4
33. What is the shape of the cross-section when a cube is sliced by a plane passing through one vertex and the midpoints of two opposite edges?
A) Triangle
B) Square
C) Rectangle
D) Pentagon
34. If a cube has a surface area of 96 square units, what is the length of one side?
A) 3 units
B) 4 units
C) 6 units
D) 8 units
35. How many right angles are formed at the intersection of three faces of a cube?
A) 2
B) 3
C) 4
D) 6
36. A cube is painted blue on all faces and is then cut into 125 cubes of equal sizes.
How many cubes are not painted on any faces?
A) 8
B) 16
C) 27
D) 36
37. Three opposite different faces of a cube are painted with three different colors – black, yellow, and white. This cube is then cut into 216 smaller but identical cubes. Answer the below questions:
A) Number of small cubes that have exactly three faces painted.
B) How many cubes have two faces painted?
C) How many cubes have only one face painted?
D) What is the number of cubes that have no face painted?
38. Each of the six different faces of a cube has been coated with a different color i.e. V I B G Y and O. The Following information is given: (UPSC 2015)
1) Colours Y O and B are on adjacent faces.
2) Colours I G and Y are on adjacent faces.
3)Colours B G and Y are on adjacent faces.
4) Colours O V and B are on adjacent faces.
Which is the color of the face opposite to the face colored with “O”?
A) B
B)V
C) G
D) I
39. A cuboid has six sides of different colors. The red side is opposite to the black. The blue side is adjacent to the white. The brown side is adjacent to the blue. The red side is face down. Which one of the following would be the opposite of brown? (UPSC 2010)
A) Red
B) Black
C) White
D) Blue
40. Six faces of a cube are numbered from 1 to 6, each face carrying one different number. Further,
1) The face 2 is opposite to the face 6.
2) The face 1 is opposite to the face 5.
3) The face 3 is between the face 1 and the face 5
4) The face 4 is adjacent to the face 2.
Which one of the following is correct?
A) The face 2 is adjacent to the face 3
B) The face 6 is between the face 2 and the face 4
C) The face 1 is between the face 5 and the face 6
D) None of the above
## FAQs
How do you solve dice and cube questions?
Dice and cube questions typically involve visualizing and analyzing patterns on dice or cubes to deduce relationships or solve problems using logical reasoning.
What is cubes and dice in IQ?
Cubes and dice are types of non-verbal reasoning questions often found in IQ tests. They assess spatial intelligence and logical thinking by presenting patterns on cubes or dice for interpretation.
How many types of dice are there?
There are typically two types of dice: standard six-sided dice, commonly used in games and gambling, and non-standard dice with varying numbers of sides, shapes, or symbols for specialized purposes.
What is standard dice?
Standard dice refer to six-sided dice commonly used in games and gambling, featuring numbers 1 through 6 on each face, with opposite faces totaling 7.
This was all about the “Cube and Dice Questions ”. For more such informative blogs, check out our Study Material Section, or you can learn more about us by visiting our Indian exams page. |
Subject:
Mathematics
Topics: Problem SolvingRatios and Proportions
Common Core State Standard: 6.RP.1, 6.RP.2, 6.RP.3b, 7.RP.2c, 7.RP.3,
Concepts:
• Ratio
• Proportion
Knowledge and Skills:
• Can use proportional reasoning to solve problems.
• Can set up and compare ratios.
• Can set up and solve a proportion.
Lesson:
Procedure:
Prepare for presentation the Futures Channel movie, “The Shape of Phones.” Tell students that as they watch the movie, you want them to think about this question (which should be posted):
What qualities or characteristics would you look for in the battery for a cell phone?
At the end of the movie, ask students to work in teams of two or three for a few minutes to list answers to this question. Generate a list of answers, and be sure that it includes the characteristics of “size,” “weight,” and “how long it will last.”
Distribute the handout to the teams. Ask the teams to study the chart on the handout and talk it over for a few minutes. Then, as a class, discuss the information shown in the chart, so that it is clear to everyone what that data means.
Then have the teams begin the task of answering the questions on the handout. Circulate as they work, and ask questions to guide them to think about the ratios of “capacity to weight” and “capacity to cost” as they evaluate the batteries, and to use proportional reasoning to answer questions 2 and 3.
1) Battery E is most expensive because it has the highest “capacity to weight” ratio.
2) Battery A is the least expensive, with the highest ratio of “capacity to cost.” Using proportional thinking, you would expect that a battery of that type with twice the capacity (2400 mAh) would have about twice the weight (80 grams) and cost (\$30).
3) Battery E is the lightest, with the highest ratio of “capacity to weight.” Using proportional thinking, you would expect that a battery of that type with about three times capacity (2400 mAh) would have three times its weight (about 30 grams) and cost (\$75)
4) Answers will vary, depending on whether cost or size is considered to be the most important factor.
### Battle of the Batteries
This chart shows the characteristics of five different batteries.
Battery Capacity Weight Thickness Cost A 1200 mAh 40 grams 13 mm \$15 B 1000 mAh 35 grams 8 mm \$14 C 900 mAh 20 grams 7 mm \$20 D 830 mAh 25 grams 7 mm \$13 E 780 mAh 10 grams 5 mm \$25
Note: The capacity of a battery is the quantity of electrical current it can produce for a length of time, measured in a unit called “milliampere hours” ( abbreviated as “mAh”). This determines how long the battery lasts.
Suppose you are designing a new cell phone, and you want to have a battery with a capacity of around 2400 mAh. Answer these questions to help make a decision as to what type of battery you will use.
1) Why do you suppose battery E is the most expensive? What ratio could you use to explain this?
2) Suppose you want to your battery to have the lowest possible price. Which of the above battery types would you want to have your battery be most similar to? About how much would you expect it to weigh? To cost?
3) Suppose you want your battery to have the lowest possible weight. Which of the above battery types would you want to have your battery be most similar to? About how much would you expect it to weigh? To cost?
4) Which battery type would you choose as a model, and why? |
## Linear Algebra and Its Applications, Exercise 2.2.12
Exercise 2.2.12. What is a 2 by 3 system of equations $Ax = b$ that has the following general solution?
$x = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + w \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$
Answer: The general solution above is the sum of a particular solution and a homogeneous solution, where
$x_{particular} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$
and
$x_{homogeneous} = w \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$
Since $w$ is the only variable referenced in the homogeneous solution it must be the only free variable, with $u$ and $v$ being basic. Since $u$ is basic we must have a pivot in column 1, and since $v$ is basic we must have a second pivot in column 2. After performing elimination on $A$ the resulting echelon matrix $U$ must therefore have the form
$U = \begin{bmatrix} *&*&* \\ 0&*&* \end{bmatrix}$
To simplify solving the problem we can assume that $A$ also has this form; in other words, we assume that $A$ is already in echelon form and thus we don’t need to carry out elimination. The matrix $A$ then has the form
$A = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ 0&a_{22}&a_{23} \end{bmatrix}$
where $a_{11}$ and $a_{22}$ are nonzero (because they are pivots).
We then have
$Ax_{homogeneous} = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ 0&a_{22}&a_{23} \end{bmatrix} w \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = 0$
If we assume that $w$ is 1 and express the right-hand side in matrix form this then becomes
$\begin{bmatrix} a_{11}&a_{12}&a_{13} \\ 0&a_{22}&a_{23} \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
or (expressed as a system of equations)
$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}a_{11}&+&2a_{12}&+&a_{13}&=&0 \\ &&2a_{22}&+&a_{23}&=&0 \end{array}$
The pivot $a_{11}$ must be nonzero, and we arbitrarily assume that $a_{11} = 1$. We can then satisfy the first equation by assigning $a_{12} = 0$ and $a_{13} = -1$. The pivot $a_{22}$ must also be nonzero, and we arbitrarily assume that $a_{22} = 1$ as well. We can then satisfy the second equation by assigning $a_{23} = -2$. Our proposed value of $A$ is then
$A = \begin{bmatrix} 1&0&-1 \\ 0&1&-2 \end{bmatrix}$
so that we have
$Ax_{homogeneous} = \begin{bmatrix} 1&0&-1 \\ 0&1&-2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
as required.
We next turn to the general system $Ax = b$. We now have a value for $A$, and we were given the value of the particular solution. We can multiply the two to calculate the value of $b$:
$b = Ax_{particular} = \begin{bmatrix} 1&0&-1 \\ 0&1&-2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
This gives us the following as an example 2 by 3 system that has the general solution specified above:
$\begin{bmatrix} 1&0&-1 \\ 0&1&-2 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
or
$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&&&-&w&=&1 \\ &&v&-&2w&=&1 \end{array}$
Finally, note that the solution provided for exercise 2.2.12 at the end of the book is incorrect. The right-hand side must be a 2 by 1 matrix and not a 3 by 1 matrix, so the final value of 0 in the right-hand side should not be present.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink.
### 2 Responses to Linear Algebra and Its Applications, Exercise 2.2.12
1. Daniel says:
I think that your final note is incorrect, due to the fact that if you find the general solution for the system Ax=b that you found, you’ll have to write the solution like Strang does it in (3) page (76). There are three entries on the solution because “x” vector lenght. The general solution (in Matlab notation) is x = [u; v; w] = [1+w; 1+2w; w]= [1; 1; 0] + w*[1; 2; 1]. The general solution he proposed at the begining of the exercise
• hecker says:
My apologies for the delay in responding. Are you referring to my final sentence about the solution to exercise 2.2.12 given on page 476 in the back of the book? If so, I think I may have confused you. I am *not* saying that Strang wrote the general solution incorrectly in the statement of the exercise on page 79, or that Strang found an incorrect solution to the exercise.
Rather my point is as follows: In the statement of the solution on page 476 Strang shows as a solution the same 2 by 3 matrix that I derived above, and Strang shows that 2 by 3 matrix multiplying the vector (u, v, w) just as I do above, representing a system of two equations in three unknowns. However on the right-hand side Strang shows that 2 by 3 matrix multiplying the vector (u, v, w) to produce the vector (1, 1, 0). This cannot be: since the matrix has only two rows, that multiplication would produce a vector with only two elements, not three (as in the book). Those two elements represent the right-hand sides of the corresponding system of two equations.
So the left-hand side in the solution of 2.2.12 on page 476 is correct, but the right-hand side of the solution of 2.2.12 on page 476, namely the vector (1, 1, 0), is not. Instead the right-hand side should be the vector (1, 1) as I derived above. |
# Lesson 9Sim CitySolidify Understanding
## Learning Focus
Find an interval that is likely to contain the population proportion from a sample.
How do we know that we found the actual population proportion in a sample distribution?
## Open Up the Math: Launch, Explore, Discuss
Alyce, Javier, and Veronica continued to collect additional artifacts around the archeological site. Javier continued to look at the proportion of artifacts that are older than years. As he collected this information, he has noticed that as they look further away from the center tower, the proportion of artifacts older than years decreases. As they were looking for artifacts, Veronica identifies a bag of artifacts that were not identified by the sector they came from. Frustrated, she looks through the bag of artifacts and notices that of artifacts in the sample dated older than years.
### 1.
Which sectors do you believe it is possible the bag came from? Justify your response.
Alyce wonders if they could use some of their work from the other day to help make a prediction about which sector the bag of artifacts came from. She wonders if they could simulate taking samples of size from each sector they already know and use this to help make a prediction about which sector they came from.
To simulate this, Alyce suggests that they assign people one of the sectors and create bags with chips that represent the proportion of artifacts older than years. For example, you could put chips in a bag and make of them black and of them white. You could then reach your hand in the bag and draw a chip out, replace it, draw another chip out, and repeat this until you had a sample of artifacts. Just like the work they did the other day, they could take samples of over and over again and create a distribution of these proportions.
### 2.
Simulate taking samples of from the sector your teacher assigns you. Use this data to plot a sampling distribution for your . Conduct the simulation enough times to get a clear picture of the distribution. The tables below provide space for recording your results.
# of artifacts of $\text{1,000}$ years old Frequency Proportion $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$
# of artifacts of $\text{1,000}$ years old Frequency Proportion $10$ $11$ $12$ $13$ $14$ $15$ $16$ $17$ $18$ $19$
# of artifacts of $\text{1,000}$ years old Frequency Proportion $20$ $21$ $22$ $23$ $24$ $25$ $26$ $27$ $28$ $29$ $30$
Based on your simulation, do you think it is likely the dropped bag of artifacts came from your sector? Use your results to explain why you think it is or is not plausible that the bag came from your sector.
### 3.
Looking at the class distributions for the other sectors:
Are there sectors you feel confident the bag did not come from?
Which sectors do you believe the bag could have come from?
How do the graphs of the distributions provide evidence for your claim?
### 4.
One way of listing the most plausible population proportions for a sample is by using what we call a margin of error. We would list the plausible values as an interval. For example, if you thought it was plausible that the bag of artifacts came from sectors containing proportions of artifacts between and , you would write this as , which means it is plausible this sample of artifacts came from any population that had an actual proportion of artifacts more than years old between and . What do you think would happen to this interval of plausible values for if the bag contained a larger sample of artifacts? Justify your answer.
### 5.
Javier is thinking that using simulations to find an interval of possible values takes time and wonders if their previous work on the Central Limit Theorem and sampling distributions could be useful in finding a way to create an interval of reasonable possible values for the population proportion . Alyce found another bag of artifacts that contains artifacts that date older than years that did not have the sector they were found in recorded.
#### a.
Design a strategy that you could use to find an interval of reasonable values of for this bag.
#### b.
Sector 6 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .047 0.240.240.240.280.280.280.320.320.320.360.360.360.40.40.40.440.440.440.480.480.480.520.520.52252525505050757575100100100${\mu }_{\stackrel{^}{p}}=.36{\sigma }_{\stackrel{^}{p}}=.047$ Sector 5 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .049 0.320.320.320.360.360.360.40.40.40.440.440.440.480.480.480.520.520.520.560.560.560.60.60.60.640.640.64252525505050757575100100100${\mu }_{\stackrel{^}{p}}=.48{\sigma }_{\stackrel{^}{p}}=.049$ Sector 4 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .051 0.440.440.440.480.480.480.520.520.520.560.560.560.60.60.60.640.640.640.680.680.680.720.720.720.760.760.76252525505050757575100100100125125125${\mu }_{\stackrel{^}{p}}=.60{\sigma }_{\stackrel{^}{p}}=.051$ Sector 3 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .044 0.60.60.60.640.640.640.680.680.680.720.720.720.760.760.760.80.80.80.840.840.840.880.880.880.920.920.92303030606060909090120120120${\mu }_{\stackrel{^}{p}}=.74{\sigma }_{\stackrel{^}{p}}=.044$ Sector 2 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .04 0.720.720.720.760.760.760.80.80.80.840.840.840.880.880.880.920.920.920.960.960.96757575150150150225225225300300300${\mu }_{\stackrel{^}{p}}=.86{\sigma }_{\stackrel{^}{p}}=.04$ Sector 1 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .026 0.840.840.840.860.860.860.880.880.880.90.90.90.920.920.920.940.940.940.960.960.960.980.980.98111505050100100100150150150${\mu }_{\stackrel{^}{p}}=.92{\sigma }_{\stackrel{^}{p}}=.026$
The simulations provided show taking samples of size from each sector. Is your interval from above reasonable based on these simulations? Is your interval too wide and include unreasonable values? Is it too narrow and does not include enough reasonable values? Refine your strategy so that it is more likely to give you an interval of reasonable values based on these simulations.
### 6.
If you were to randomly sample artifacts from the sector that contained artifacts over years old:
#### a.
How likely are you to get a sample that contains or fewer artifacts more than years old?
#### b.
Based on your answer, is it likely that a bag of artifacts containing artifacts older than years came from this sector? Why or why not?
Sector 6 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .047 0.240.240.240.280.280.280.320.320.320.360.360.360.40.40.40.440.440.440.480.480.480.520.520.52252525505050757575100100100${\mu }_{\stackrel{^}{p}}=.36{\sigma }_{\stackrel{^}{p}}=.047$ Sector 5 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .049 0.320.320.320.360.360.360.40.40.40.440.440.440.480.480.480.520.520.520.560.560.560.60.60.60.640.640.64252525505050757575100100100${\mu }_{\stackrel{^}{p}}=.48{\sigma }_{\stackrel{^}{p}}=.049$ Sector 4 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .051 0.440.440.440.480.480.480.520.520.520.560.560.560.60.60.60.640.640.640.680.680.680.720.720.720.760.760.76252525505050757575100100100125125125${\mu }_{\stackrel{^}{p}}=.60{\sigma }_{\stackrel{^}{p}}=.051$ Sector 3 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .044 0.60.60.60.640.640.640.680.680.680.720.720.720.760.760.760.80.80.80.840.840.840.880.880.880.920.920.92303030606060909090120120120${\mu }_{\stackrel{^}{p}}=.74{\sigma }_{\stackrel{^}{p}}=.044$ Sector 2 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .04 0.720.720.720.760.760.760.80.80.80.840.840.840.880.880.880.920.920.920.960.960.96757575150150150225225225300300300${\mu }_{\stackrel{^}{p}}=.86{\sigma }_{\stackrel{^}{p}}=.04$ Sector 1 a histogram representing distribution of sample proportions with a mostly normal distribution. The standard deviation is .026 0.840.840.840.860.860.860.880.880.880.90.90.90.920.920.920.940.940.940.960.960.960.980.980.98111505050100100100150150150 |
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Question
# If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)−1 = f−1 og −1.
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Solution
## Injectivity of f: Let x and y be two elements of domain (Q), such that f(x) = f(y) $⇒$2x = 2y $⇒$x = y So, f is one-one. Surjectivity of f: Let y be in the co-domain (Q), such that f(x) = y. $⇒2x=y\phantom{\rule{0ex}{0ex}}⇒x=\frac{y}{2}\in Q\left(\text{domain}\right)$ $⇒$ f is onto. So, f is a bijection and, hence, it is invertible. Finding f -1: $\text{Let}{f}^{-1}\left(x\right)=y...\left(1\right)\phantom{\rule{0ex}{0ex}}⇒x=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x=2y\phantom{\rule{0ex}{0ex}}⇒y=\frac{x}{2}\phantom{\rule{0ex}{0ex}}\text{So,}{f}^{-1}\left(x\right)=\frac{x}{2}\left(\text{from}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}$ Injectivity of g: Let x and y be two elements of domain (Q), such that g(x) = g(y) $⇒$x + 2 = y + 2 $⇒$x = y So, g is one-one. Surjectivity of g: Let y be in the co domain (Q), such that g(x) = y. $⇒x+2=y\phantom{\rule{0ex}{0ex}}⇒x=2-y\in Q\left(\text{domain}\right)$ $⇒$ g is onto. So, g is a bijection and, hence, it is invertible. Finding g -1: $\text{Let}{g}^{-1}\left(x\right)=y...\left(2\right)\phantom{\rule{0ex}{0ex}}⇒x=g\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x=y+2\phantom{\rule{0ex}{0ex}}⇒y=x-2\phantom{\rule{0ex}{0ex}}\text{So,}{g}^{-1}\left(x\right)=x-2\left(\text{From}\left(2\right)\right)\phantom{\rule{0ex}{0ex}}$ Verification of (gof)−1 = f−1 og −1: $f\left(x\right)=2x;g\left(x\right)=x+2\phantom{\rule{0ex}{0ex}}\mathrm{and}{f}^{-1}\left(x\right)=\frac{x}{2};{g}^{-1}\left(x\right)=x-2\phantom{\rule{0ex}{0ex}}\mathrm{Now},\left({f}^{-1}o{g}^{-1}\right)\left(x\right)={f}^{-1}\left({g}^{-1}\left(x\right)\right)\phantom{\rule{0ex}{0ex}}⇒\left({f}^{-1}o{g}^{-1}\right)\left(x\right)={f}^{-1}\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒\left({f}^{-1}o{g}^{-1}\right)\left(x\right)=\frac{x-2}{2}...\left(3\right)\phantom{\rule{0ex}{0ex}}\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\phantom{\rule{0ex}{0ex}}=g\left(2x\right)\phantom{\rule{0ex}{0ex}}=2x+2\phantom{\rule{0ex}{0ex}}\text{Let}{\left(gof\right)}^{-1}\left(x\right)\text{=y ....}\left(4\right)\phantom{\rule{0ex}{0ex}}x=\left(gof\right)\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x=2y+2\phantom{\rule{0ex}{0ex}}⇒2y=x-2\phantom{\rule{0ex}{0ex}}⇒y=\frac{x-2}{2}\phantom{\rule{0ex}{0ex}}⇒{\left(gof\right)}^{-1}\left(x\right)=\frac{x-2}{2}\text{[from}\left(4\right)...\left(5\right)\right]\phantom{\rule{0ex}{0ex}}\text{From}\left(3\right)\text{and}\left(5\right)\text{,}\phantom{\rule{0ex}{0ex}}{\left(gof\right)}^{-1}={f}^{-1}o{g}^{-1}$
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A Dicey Situation
“tell me the chances of 6 on three dice (d6), 4 times in about 10 rolls”
EDIT: A non-ambiguous version of this question: what are the chances of getting a sum of exactly 6 on 3 six-sided dice exactly 4 times in 10 rolls?
And the answers fly! And they are all wrong.
So let’s look at an easy example. Given that you are rolling two d6, what is the probability that you will roll exactly one six?
• It’s the probability that you’ll roll a six plus the probability of not rolling a six? ( P(6) + P(~6) => 1/6 + 5/6 = 1 100% huh?
• It’s the probability that you’ll roll a six times the probability of not rolling a six? ( P(6) * P(~6) => 1/6 * 5/6 = 5/36 Nope, too low
• It’s one minus the probability of not rolling a six in the roll? ( 1 – P(~6) * P(~6) => 1 – 5/6 * 5/6 = 11/36 Nope, too high
To find this solution you have to remember the golden rule when dealing with dice probabilities, the probability of an event A happening is:
• P(A) = 1 – P(~A)
So lets apply that to the above problem. In what ways can we not roll exactly one six on 2d6? Well we could roll zero sixes or we could roll 2 sixes!
Thus our calculation becomes:
• 1 – P(~one six) = 1 – ( P(zero sixes) + P(two sixes) ) = 1 – P(zero sixes) – P(two sixes) = 1 – 5/6*5/6 – 1/6*1/6 = 1 – 25/36 – 1/36 = 10/36
Now obviously this is going to prove a little tedious when there are 10 or so dice/rolls since then you have to do 1 – P(ten sixes) – … – P(zero sixes) (excluding the one you are looking for)
So we have to come up with another way to do it.
Going back to our six example lets try another approach. We know that the probability is the # of successes out of the # of possible outcomes. So then let’s count up the # of successes and # of possible outcomes!
For successes we have (1,6); (2,6); … (5,6) and (6,1); … (6,5) which is 10 total successes.
How many possible outcomes are there? 6*6=36
Which gets us to our 10/36 answer.
There is one more way to do it which is what is the probability or rolling 6 , not six ( 1/6*5/36 = 5/36) and what is the probability of not six, six ( 5/6*1/6 = 5/36 ) now since those are the only two possibilities, add them together and get 10/36!
Let’s apply this to our big problem. How many ways are there on 3 dice to roll exactly a six? (There are 10) So the chances of rolling a six 10/216 and the probability of not rolling a six is 206/216. In ten rolls we’ll have 4 of the sixes and 6 of the not-sixes giving us P=( 10^4 * 206*6 ) / 216^10. Now we have to multiple that by the number of different ways we can arrange those 4 dice in the 10 trials (combination here 10 choose 4) = 210.
Finally 210* P = .07%
However I think the real number we’re looking for is what are the chances of rolling 6 or less, 4 or more times out of 10 times with three dice? The math on that is a bit more complicated, but boils down approximately 1%.
And that, dear readers, is how the dice rolls.
More on dice, statistics, and their place in WM/H later.
4 responses to “A Dicey Situation”
1. daawalsh says:
1 am a little rusty after leaving school twenty years ago but i get a different number
Heres my workings out :-) hopefully younger and sharper minds can check it.
So i read your calculation as trying to get the probability or rolling three dice and trying to score 6 on each dice in one roll. You roll all three dice together three times
Rolling 3 6s is an AND operation so multiply the probability
Prob of 1x 6 on a d6 is 1/6
Prob on 3x 6 on 6 dice = (1/6) x (1/6) x (1/6) = 1 / 216
Then prob ability for rolling 3×6 on three dice within 10 rolls is a OR operation so you add them. Each seperate roll is independent
1/216 + 1/216 + 1/216 etc etc
Short version gets you to 10/216 or approx 4.6%
• meatkat says:
You are correct sir, the chances of rolling three sixes and getting three sixes (or a sum of 18) is 1/216.
The statement of the original question was a bit vague, I apologize for that but what we are actually looking for in that post is a SUM of 6 on 3 dice, which is a very different beast.
• daawalsh says:
Ahh got you.
And with an a red face I shuffle off realising i managed to miss the part about 4 rolls |
# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 1 + 1/2 = 3/2 = 1 1/2 = 1.5
Spelled result in words is three halfs (or one and one half).
### How do you solve fractions step by step?
1. Add: 1 + 1/2 = 1/1 + 1/2 = 1 · 2/1 · 2 + 1/2 = 2/2 + 1/2 = 2 + 1/2 = 3/2
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(1, 2) = 2. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 1 × 2 = 2. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - one plus one half = three halfs.
#### Rules for expressions with fractions:
Fractions - simply use a forward slash between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Cakes
On the bowl were a few cakes. Jane ate one-third of them, and Dana ate a quarter of those cakes that remained. a) What part (of the original number of cakes) Dana ate? b) At least how many cakes could be (initially) on the bowl?
• Unknown number
I think the number - its sixth is 3 smaller than its third.
• Missing number
Blank +1/6 =3/2 find the missing number
• Cleaning windows
Cleaning company has to wash all the windows of the school. The first day washes one-sixth of the windows of the school, the next day three more windows than the first day and the remaining 18 windows washes on the third day. Calculate how many windows ha
• Discount sale
At Christmas Sale after a 20% discount, the cosmetic package was priced for 5 euros and 60 cents; later it went to 3 euros and 80 cents. How much percentage is the total discount?
• Brick
Isosceles scale has on one side all brick and second weight 1 kg and 1/4 of brick. The balance is in equilibrium. What is the weight of a brick?
• Mini-survey
In the mini-survey of our class about the popularity of individual subjects, it turned out that 11.1% of pupils like mathematics, 18.5% are enjoying languages, 30.4% of pupils like physical education, and the remaining 12 pupils have several popular subje
• Sales tax
A sales tax on a Php 10,800 appliance is Php.1,620. What is the rate of sales tax?
• Wood 11
Father has 12 1/5 meters long wood. Then I cut the wood into two pieces. One part is 7 3/5 meters long. Calculate the length of the other wood?
• A man
A man spends 5/9 of his money on rent, and 5/16 of the remainder on electricity. If the final balance remaining is 550 find how much was spent on rent
• The perimeter 2
The perimeter of the quadrilateral a = 1m b = 14/5m c = 2 3/10m d = 1 4/5m?
• Boys to girls
The ratio of boys to girls at a party is 3:5. If six more boys arrived and four girls left the party, the ratio of boys to girls would be 5:6. How many people were at the party initially?
• Average speed
When the bus stops at a bus stop driving average speed is 45 km/h. If it did not stop, it would drive at a speed of 54 km/h. How many minutes of every hour does it spend at stops? |
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# Fishing Rod
Extracts from this document...
Introduction
Min Hua Ma
5-1-09
IB SL MATH
Internal Assessment
Type II
IB SL Math Internal Assessment: type II
This assignment is an investigation to find different methods that model a given set of data. By using matrix methods, polynomial functions, and technology to find different equations, we can discover which equation best models the data. Leo has a fishing rod that has a length of 230cm, and the given data about his rod is:
Guide # 1 2 3 4 5 6 7 8 Distance from tip (cm) 10 23 38 55 74 96 120 149
To begin the investigation, we began by plotting the given points in a scatter plot:
The first method is using matrices to find a quadratic function. A quadratic equation is: ax2+bx+c=y, since there is only 3 variables (a, b, c) we can only create a 3 by 3 matrix and a 3 by 1 matrix. So I choose the first six data and separated them into two sets of three to make these equations to model the data.
=
To solve for for the first set of three data, we take x and get: Using that information, a quadratic equation can be created: 1x2+10x-1. The model with the actual points is shown by this graph from a graphing calculator:
This graph shows that the equation 1x2
Middle
8
23
55
96
149
Now plug each of the x’s into the cubic function (ax3+bx2+cx+d). Then put each group of data into a 4 by 4 matrix and set the y values into a 4 by 1 matrix.
For Odds:
x =
[A] [x] [B]
For evens:
x =
[A] [X] [B]
At this time, use the process x to solve for [X]. After doing so:
Odd: [X]= Even: [X]=
Then put each of the variables into a cubic function format and generate a graph for each equation.
The odd: f(x)=.0416666667x3 +.625x2 +10.9583333x-1.625
This graph shows that it touches all the data points, except when x=6. This equation models the given data by almost passing every point.
The even: f(x)= .0625x3 +.375x2 +12x -3
This graph illustrates that all, but the sixth point is touched by this function. Although this function is different from the odd cubic function, they both go through the same points and not the sixth. The differences of these two graphs are only the different numbers written for the cubic function.
Next, we need to create a polynomial function that passes through all of the particular data. To do this, I decided to create a polynomial function by using matrices.
Conclusion
See below:
1 2 3 4 5 6 7 8 9 10 23 38 55 74 96 120 149 178
By adding a ninth guide, it will affect the all the functions created. The cubic equations would not be as precise since there are an odd number of points. The polynomial equation must begin with ax8instead of ax7.
Then the assignment gave us a new set of data to compare with the quadratic model from the first part of this investigation. This time Mark has a fishing rod that is 300cm.
Guide number (from tip) 1 2 3 4 5 6 7 8 Distance from tip (cm) 10 22 34 48 64 81 102 124
The quadratic model fits this data by only going through the first two points, and misses the rest. The quadratic model from the first part of the investigation doesn’t really model this new set of data. To find how to modify the quadratic equation, I used technology to figure out what the QuadReg was and subtracted the first quadratic model by this QuadReg:
1.244047619x2 + 8.458333333x + .8392857143 Original function
- .9345238095x2 + 7.720238095x + 2.05351429 New function
.3095238095x2 + .738095243x -1.214228547
So, the quadratic has to modify by .3095238095x2 + .738095243x -1.214228547 to fit the data created by the 300cm rod. The regression of the function 9345238095x2 + 7.720238095x + 2.05351429, is .9997802355. The LinReg of this data is written y=(16.13)*x+(-11.95 ) with the regression (.99).
The limitations to this model is that there is can’t be an infinite number of guides and rod have only have certain lengths.
Ma
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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# Related International Baccalaureate Maths essays
1. ## Extended Essay- Math
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# Top 10 8th Grade PSSA Math Practice Questions
Preparing your student for the 8th Grade PSSA Math test? The best way to prepare for the PSSA Math test is to work through as many PSSA Math practice questions as possible. Here are the top 10 8th Grade PSSA Math practice questions to help your students review the most important 8th Grade PSSA Math concepts. These 8th Grade PSSA Math practice questions are designed to cover mathematics concepts and topics that are found on the actual test. The questions have been fully updated to reflect the latest 2022 8th Grade PSSA guidelines. Answers and full explanations are provided at the end of the post.
Help your students start their PSSA Math test prep journey right now with these sample PSSA Math questions.
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## 8th Grade PSSA Math Practice Questions
1- A rope weighs 600 grams per meter of length. What is the weight in kilograms of 12.2 meters of this rope? (1 kilograms = 1000 grams)
A. 0.0732
B. 0.732
C. 7.32
D. 7320
2- In a school, the ratio of the number of boys to girls is 3:7. If the number of boys is 180, what is the total number of students in the school?__________
3- In two successive years, the population of a town is increased by $$15\%$$ and $$20\%$$. What percent of its population is increased after two years?
A. 32
B. 35
C. 38
D. 68
4- Which graph shows a non–proportional linear relationship between x and y?
A.
B.
C.
D.
5- In the rectangle below if $$y>5$$ $$cm$$ and the area of the rectangle is $$50 cm^2$$ and the perimeter of the rectangle is 30 cm, what is the value of x and y respectively?
A. $$4, 11$$
B. $$5, 11$$
C. $$5, 10$$
D. $$4, 10$$
6- A football team had $40,000 to spend on supplies. The team spent$22,000 on new balls. New sports shoes cost $240 each. Which of the following inequalities represent how many new shoes the team can purchase. A. $$240x+22,000 ≤40,000$$ B. $$240x+22,000 ≥40,000$$ C. $$22,000x+240 ≤40,000$$ D. $$22,000x+240 ≥40,000$$ 7- Right triangle ABC has two legs of lengths 6 cm (AB) and 8 cm (AC). What is the length of the third side (BC)? A. 4 cm B. 6 cm C. 8 cm D. 10 cm 8- If $$3x-5=8.5$$, What is the value of $$5x+3$$? A. 13☐B. 15.5 C. 20.5 D. 25.5 9- A bank is offering $$4.5\%$$ simple interest on a savings account. If you deposit$8,000, how much interest will you earn in five years?
A. $360 B.$720
C. $1800 D.$3600
10- In a party, 10 soft drinks are required for every 12 guests. If there are 252 guests, how many soft drinks is required?
A. 21
B. 105
C. 210
D. 2510
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1- C
The weight of 12.2 meters of this rope is: $$12.2 × 600 g = 7320 g$$
$$1 kg = 1000 g$$
therefore,
$$7320 g ÷ 1000 = 7.32 kg$$
2- 600
The ratio of boys to girls is $$3:7$$.
Therefore, there are 3 boys out of 10 students. To find the answer, first, divide the number of boys by 3, then multiply the result by 10.
$$180 ÷ 3 = 60 ⇒ 60 × 10 = 600$$
3- C
the population is increased by $$15\%$$ and $$20\%$$. $$15\%$$ increase changes the population to $$115\%$$ of the original population.
For the second increase, multiply the result by $$120\%$$.
$$(1.15) × (1.20) = 1.38 = 138\%$$
38 percent of the population is increased after two years.
4- B
A linear equation is a relationship between two variables, $$x$$ and $$y$$, that can be put in the form $$y=mx + b$$.
A non-proportional linear relationship takes on the form $$y=mx + b$$, where $$b ≠ 0$$ and its graph is a line that does not cross through the origin.
5- C
The perimeter of the rectangle is: $$2x+2y=30→x+y=15→x=15-y$$
The area of the rectangle is: $$x×y=50→(15-y)(y)=50→y^2-15y+50=0$$
Solve the quadratic equation by factoring method.
$$(y-5)(y-10)=0→y=5$$
(Unacceptable, because y must be greater than 5) or $$y=10$$
If $$y=10 →x×y=50→x×10=50→x=5$$
6- A
Let $$x$$ be the number of new shoes the team can purchase. Therefore, the team can purchase $$240 x$$.
The team had $$40,000$$ and spent $$22,000$$. Now the team can spend on new shoes $$18,000$$ at most.
Now, write the inequality: $$120x+22.000 ≤40.000$$
7- D
Use Pythagorean Theorem:
$$a^2 + b^2 = c^2$$
$$6^2 + 8^2 = c^2 ⇒ 100 = c^2 ⇒ c = 10$$
8- D
$$3x-5=8.5→3x=8.5 + 5=13.5→x = \frac{13.5}{3}= 4.5$$
Then;
$$5x+3=5 (4.5)+3=22.5+3=25.5$$
9- C
Use simple interest formula:
$$I=prt$$
$$(I = interest, p = principal, r = rate, t = time)$$
$$I=(8000)(0.045)(5)=1800$$
10- C
Let $$x$$ be the number of soft drinks for $$252$$ guests. Write the proportion and solve for $$x$$.
$$\frac{10 soft drinks}{12 guests}=\frac{x}{252 guests}$$
$$x = \frac{252×10}{12}⇒x=210$$
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## Factorising Cubics
Factorising quadratic expressions is comparatively easy. If the coefficient ofis one you can often find the factors by inspection. For example, to factorisefind 2 numbers that add or take away to give 3 and multiply to give 18. By inspection we obtain 6 and 3. Then we can factorise:
When we try and factorise a cubic we can start by finding common factors. This may reduce the problem to one of factorising a quadratic:
The expression inside the brackets now factorises by inspection: find two numbers that add or take away to give -5 and multiply to give 6. We obtain -2 and -3. Hence,
If we can't reduce the problem to factorising a quadratic by inspection, then things get a little more involved. Consider how to factorise a quadratic where the coefficient ofis not one. For example,
Multiply the coefficient of2 by the constant term, 5 to get 10. Now look for the two factors of 10 that add to give the coefficient of7. The two factors are 2 and 5. Now
Example: Factorise the cubic expression
Factorise first with the common factor 3x to give
To factorise the quadratic in the brackets, multiply the coefficient of2 by the constant term, 7 to get 14, then find the factors of 14 that add to give 9. The answer is 2 and 7. Hence
hence the cubic factorises as
Example: Factorise the cubic expression
Factorise first with the common factor 2x to give
To factorise the quadratic in the brackets, multiply the coefficient of2 by the constant term, 9 to get 18, then find the factors of 18 that add to give 9. The answer is 3 and 6. Hence
hence the cubic factorises as |
Ex 4.3
Chapter 4 Class 10 Quadratic Equations
Serial order wise
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex 4.3 ,11 Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. Let side of square 1 be x metres Perimeter of square 1 = 4 × Side = 4x Now, it is given that Difference of perimeter of squares is 24 m Perimeter of square 1 – Perimeter of square 2 = 24 4x – perimeter of square 2 = 24 4x – 24 = perimeter of square 2 Perimeter of square 2 = 4x – 24 Now, Perimeter of square 2 = 4x – 24 4 × (Side of square 2 ) = 4x – 24 Side of square 2 = (4𝑥 − 24)/4 = (4(𝑥 − 6))/4 = x – 6 Hence, Side of square 1 is x & Side of square 2 is x – 6 Also, given that Sum of area of square is 468 m2 Area of square 1 + Area of square 2 = 468 (Side of square 1)2 + (Side of square 2)2 = 468 𝑥2+(𝑥−6)2=468 x2 + x2 – 2 ×𝑥×6+6^2=468 x2 + x2 – 12 x + 36 = 468 x2 + x2 – 12x + 36 – 468 = 0 2x2 – 12x – 432 = 0 Dividing both sides by 2 (2𝑥2 − 12𝑥 − 432)/2=0/2 x2 – 6x – 216 = 0 Comparing equation with ax2 + bx + c = 0, Here a = 1, b = –6, c = – 216 We know that D = b2 – 4ac D = (–6)2 – 4 × 1 × (– 216) D = 36 + 4 × 216 D = 36 + 864 D = 900 So, the roots to equation are x = (−𝑏 ± √𝐷)/2𝑎 Putting values x = (−(−6) ± √900)/(2 × 1) x = (−(−6) ± √900)/(2 × 1) x = (6 ± √900)/2 x = (6 ± √(9 × 100))/2 x = (6 ± √(3^2 × 〖10〗^2 ))/2 x = (6 ± √(3^2 ) × √(〖10〗^2 ))/2 x = (6 ± 3 × 10)/2 x = (6 ± 30)/2 Solving So, x = 18 & x = – 12 Since x is side of square and x cannot be negative So, x = 18 is the solution ∴ Side of square 1 = x = 18 m & Side of square 2 = x – 6 = 18 – 6 = 12 m |
Mathematical Proofs - Island of Sanity
# Island of Sanity
Pathetic Attempts at Humor
# Mathematical Proofs
I present here a couple of the more interesting mathematical proofs I've come across.
Note: These are, of course, gag proofs. If you haven't seen them before, try to find the flaw.
## Theorem 1: 1=2
a=b postulate a2=ab multiply both sides by a a2-b2=ab-b2 subtract b2 from both sides (a+b)(a-b)=b(a-b) factor a+b=b divide by (a-b) b+b=b as a=b, substitute 2b=b b+b=2b 2=1 divide by b
QED
## Theorem 2: All numbers are equal
Proof is by induction.
Let us consider a set of k numbers.
First, consider the case where k=1. Clearly, if there is only one number in the set, as it is equal to itself (reflexive property of equality), all the numbers in this set are equal.
Second, assume that we have proven the theorem true for some value k. (As we just did above for k=1.) Now consider a set with k+1 numbers. Let us remove any one number from the set, call it x. That leaves k numbers, and by our preceding assumption, these numbers are all equal. Now replace x and remove another number, call it y. Again, all the remaining numbers must be equal. But note that when we removed x from the set, y was still a member, and therefore equal to all the other members; and when we removed y from the set, we had replaced x, so it was a member equal to all the other members. Therefore x and y are both equal to all the other members of the set, so they must equal each other (transitive property of equality). Thus, given the assumption that the theorem is true for a set of k numbers, it must also be true for a set of k+1 number.
By the principle of induction, we have proven that the theorem is true for k=1, and that whenever the theorem is true for k, it is also true for k+1. Therefore it must be true for 2, 3, 4, ... infinity. So even in an infinite set of numbers, they must all be equal.
QED.
## Obvious
A true story:
I once attended a mathematicians' convention. The convention lasted three days. On the first day, I went to a lecture in a certain room. The speaker presented a proof of a theorem he was working on. He wrote out the proof on the blackboard (there were blackboards covering three walls of the room), and at each step he would explain the justification, "divide both sides by x", "as q must be a prime ...", etc. Then he reached step 14, said, "And of course it's obvious that ..." and wrote step 15.
Someone in the audience interrupted and said, "Wait a minute, I don't think that's obvious at all."
The speaker began, "No, you see, because ..." He got about two sentences into his explanation and then he paused. He stared at the blackboard for what seemed a long time, and then turned and walked out of the room. We all waited, but his time ran out and the next speaker came in and he hadn't returned.
On the last day of the conference, I happened to be in that same room hearing another lecture. Suddenly, the speaker from that first day walked into the room. It was obvious he hadn't shaved or changed his clothes since he'd left two days before. Completely ignoring the present speaker, he walked up to the first blackboard and began writing equations in a long and convoluted proof. He began with the equation from step 14 of the other day, filled all three blackboards, and finally on the last board he wrote step 15, proving that it did indeed follow. He triumphantly slammed down the chalk and proclaimed, "There! It's obvious!". |
# Formula of Curved Surface Area Cylinder
A cylinder is a three-dimensional solid, whose circular base and top are parallel to each other. The perpendicular distance between the top and the base is defined as the total height of the cylinder.
Area of a cylinder- The area of a cylinder is defined as the space or region occupied by the cylinder. This area can be of two types:
Curved Surface Area- The curved surface area is defined as the area of only curved surface, leaving the circular top and base.
Total Surface Area- It is the area of the curved surface as well as the bases.
Formula for calculating the Curved Surface Area (C.S.A.)-
Consider a cylinder having a height ‘h’ and base radius ‘r’.
The curved surface area of a cylinder is given as-
C.S.A.= $2\pi r h$
Example- Consider a curved surface area of a cylinder having height of 5 cm and diameter of base to be 2 cm.
Solution- Given h = 5 cm, and d = 2 cm,
thus r = 1 cm
C.S.A. = $2\pi r h = 2 \pi \times 1 \times 5 =10 \pi$
Example- Given the C.S.A. of a cylinder to be $154 cm^{2}$, and the height to be equal to twice the radius of the base. Find the radius and height of a cylinder.
Solution- Given C.S.A. = $154 cm^{2}$
$h = 2r$
$2 \pi \times r \times h = 154$
Putting the value of h=2r, we have
$\Rightarrow 2 \times \frac{22}{7} \times r \times (2r) = 154$
$\Rightarrow r^{2} = \frac{49}{4}$
$\Rightarrow r = \frac{7}{2}$
$\Rightarrow r = 3.5$cm
And h = 2r = 7cm
Thus, h = 7 cm and r = 3.5 cm |
3.2k views
In the given matrix $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$ , one of the eigenvalues is 1. The eigenvectors corresponding to the eigenvalue 1 are
1. $\left\{a\left(4,2,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
2. $\left\{a\left(-4,2,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
3. $\left\{a\left(\sqrt{2},0,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
4. $\left\{a\left(- \sqrt{2},0,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
edited | 3.2k views
+4
Answer: B. $\left \{ a\left ( -4,2,1 \right ) \bigm| a\neq0,a\in\mathbb{R} \right \}$
+15
see this
0
To find an eigenvector corresponding to the eigenvalue 1
Step 1: Plug in the eigenvector to get $A- \lambda I$
$\begin{bmatrix} 0 & -1&2 \\ 0&0 &0 \\ 1&2 & 0 \end{bmatrix}$
Step 2: Use $(A- \lambda I)x=0$
$\begin{bmatrix} 0 & -1&2 \\ 0&0 &0 \\ 1&2 & 0 \end{bmatrix}$ $\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=0$
So,
$-y+2z=0$
and
$x+2y=0$ (Only Option B satisfies this)
Now, make as many linearly dependent vectors out of it. Option B has one such.
$\begin{bmatrix} 1-1 & -1 & 2\\ 0& 1-1& 0\\ 1& 2& 1-1 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 2\\ 0& 0& 0\\ 1& 2& 0 \end{bmatrix}*\begin{bmatrix} x\\ y\\ z \end{bmatrix}$
$-y + 2z = 0$
$x + 2y = 0$
Now consider each of the triplets as the value of x, y, z and put in these equations the one which satisfies is the answer.
Why so because an eigen vector represents a vector which passes through all the points which can solve these equations.
So, we can observe that only option B is satisfying the equations.
by Active (2.3k points)
edited
0
Why does applying row/column transformation first not giving the answer? Is it not advisable to do so?
0
+1
We know that
$AX=\lambda X$
$AX-\lambda X=0$
$(A-\lambda I) X=0$
In given question $\lambda=1$
+1
Words of caution:
DO NOT use row or column transformations on the matrices, when eigen vectors and values are concerned.
Let z represents the eigenvalues.
And let the given matrix be A (square matrix of order 3 x3)
The characteristic equation for this is :
AX = zX ( X is the required eigenvector )
AX - zX = 0
[ A - z I ] [X] = 0 ( I is an identity matrix of order 3 )
put z = 1 ( because one of the eigenvalue is 1 )
[ A - 1 I ] [X] = 0
The resultant matrix is :
[ 0 -1 2 ] [x1] [0]
| 0 0 0 ] |x2] =|0|
[ 1 2 0 ] |x3] [0]
Multiplying thr above matrices and getting the equations as:
-x2 + 2x3 = 0 ----------------(1)
x1 + 2x2 = 0-----------------(2)
now let x1 = k, then x2 and x3 will be -k/2 and -k/4
respectively.
hence eigenvector X = { (k , -k/2, -k/4) } where k != 0
put k = -4c ( c is also a constant, not equal to zero ),
we get X = { ( -4c, 2c, 1c ) }, i.e. { c ( -4, 2, 1 ) }
Hence option B.
by Boss (10k points)
0
u hav copied geeksforgeeks or geeksforgeeks copied u ???
https://www.geeksforgeeks.org/gate-gate-cs-2015-set-3-question-25/
if A is a Matrix and X is a Eigen Vector then this condition follow [A] [X] = Eigen Value * [X]
so,
here Eigen value is 1
by Boss (11.7k points)
0
How did u get the X matrix ?? By using options ???
0
yes @pooja
0
ok ...(y)
+4
@puja refer this video for more clear idea
Given matrix, and it's eigen value is 1,
I write $|A-1.I|=$\begin{vmatrix} 0& -1&2 \\ 0&0 &0 \\ 1& 2& 0\end{vmatrix}$$Now I have to find an Eigen-vector(obviously non-zero) such that |A-I|.X=0 and this is \begin{vmatrix} 0 & -1 & 2\\ 0 & 0 &0 \\ 1 & 2 &0 \end{vmatrix}.\begin{bmatrix} a\\ b\\ c \end{bmatrix}=0 Now, you should think what combinations of columns 1,2 and 3 should I take to get the zero vector \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} And, you can agree if I say, -4times the first column +2 times the second column and 1 times the third column will give you the zero vector i.e. -4\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}+ 2\begin{bmatrix} -1\\ 0\\ 2 \end{bmatrix}+$$\begin{bmatrix} 2\\ 0\\ 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$so our X becomes X=$\begin{bmatrix} -4\\ 2\\ 1 \end{bmatrix}$And any linear combination of this X would also result in$|A-I|cX=c\{|A-I|.X\}=c.0=0$provided$c \not=0\$ otherwise we would get the zero vector and the zero vector is a trivial solution to AX=0.
by Boss (30.6k points) |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2017 AMC 8 Problems/Problem 19"
## Problem 19
For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
## Solution 1
Factoring out $98!$, we have $98!(10,000)$. Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. Now $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.
## Solution 2
The number of $5$'s in the factorization of $98! + 99! + 100!$ is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by $5$, until you can't divide by $5$ anymore. Factorizing $98! + 99! + 100!$, you get $98!(1+99+9900)=98!(10000)$. To find the number of trailing zeroes in 98!, we do $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22$. Now since $10000$ has 4 zeroes, we add $22 + 4$ to get $\boxed{\textbf{(D)}\ 26}$ factors of $5$. -Rekt4 |
# Area of Polygons
Published on
How to find the area of trapezoids, hexagons, pentagons, and octagons?
• Trapezoid: Area = 1/2 x (sum of the bases) x height A trapezoid is a shape with two parallel sides and two non-parallel sides. To find the area of a trapezoid, we add the lengths of the two parallel sides (called the bases), then multiply by the height and divide by 2. For example, a trapezoid with a height of 4 units, a base of 6 units, and another base of 8 units has an area of 1/2 x (6 + 8) x 4 = 28 square units.
• Hexagon: Area = 3/2 x (side length) squared x square root of 3 A hexagon is a shape with six sides of equal length. To find the area of a regular hexagon, we multiply the square of the length of one side by 3/2, then multiply by the square root of 3. For example, a hexagon with a side length of 5 units has an area of 3/2 x 5^2 x square root of 3 ≈ 64.95 square units.
• Pentagon: Area = 1/4 x (square root of [5(5 + 2(square root of 5)))]) x side length squared A pentagon is a shape with five sides of equal length. To find the area of a regular pentagon, we use the formula above. For example, a pentagon with a side length of 7 units has an area of 1/4 x (square root of [5(5 + 2(square root of 5)))]) x 7^2 ≈ 84.3 square units.
• Octagon: Area = 2 x (1 + square root of 2) x side length squared An octagon is a shape with eight sides of equal length. To find the area of a regular octagon, we multiply the square of the length of one side by 2 times the quantity 1 plus the square root of 2. For example, an octagon with a side length of 9 units has an area of 2 x (1 + square root of 2) x 9^2 ≈ 305.36 square units.
I hope that helps you understand how to find the area of trapezoids, hexagons, pentagons, and octagons! |
# How to Calculate Percentage: A Comprehensive Guide
Baca Cepat
## Introduction
Greetings, dear reader! Whether you’re a student trying to ace your math class or a professional looking to make accurate calculations, knowing how to calculate percentages is a valuable skill. Percentage is used in various fields, such as finance, science, and business, to represent a portion of a whole. In this article, we will provide you with a comprehensive guide on how to calculate percentage accurately and easily. Let’s dive in!
### Why is it Important to Know How to Calculate Percentage?
Before we dive into the nitty-gritty of calculating percentage, let’s understand why it’s an essential skill. Knowing how to calculate percentage can help you in several ways:
• It helps you make informed decisions based on statistics and data.
• It enables you to compare different values accurately.
• It helps you understand discounts, taxes, and increase/decrease in prices.
• It’s a crucial component of financial planning and budgeting.
Now that we know why it’s important let’s move on to how to calculate percentage.
## How to Calculate Percentage
### The Basic Formula
The basic formula for calculating percentage is:
Symbol Meaning
P% Percentage
A Part
B Whole
The formula is:
P% = (A/B) x 100
Where A is the part, and B is the whole. Let’s understand this with an example.
### Example
Suppose you scored 80 marks out of 100 in your math test. What percentage did you score?
P% = (80/100) x 100 = 80%
Therefore, you scored 80% in your math test.
### Calculating Percentage Increase or Decrease
Percentage increase or decrease is the percentage difference between two values. The formula to calculate percentage increase or decrease is:
Symbol Meaning
P% Percentage Increase/Decrease
A Final Value
B Initial Value
The formula is:
P% = ((A-B)/B) x 100
Where A is the final value, and B is the initial value. Let’s see an example.
### Example
Suppose the price of a product increased from \$100 to \$120. What is the percentage increase in price?
P% = ((120-100)/100) x 100 = 20%
Therefore, the price increased by 20%.
### Calculating Percentage of a Part
Suppose you want to find out what percentage of a whole a part is. The formula to calculate the percentage of a part is:
Symbol Meaning
P% Percentage of Part
A Part
B Whole
The formula is:
P% = (A/B) x 100
Where A is the part, and B is the whole. Let’s take an example.
### Example
Suppose a pizza has eight slices, and you ate two slices. What percentage of the pizza did you eat?
P% = (2/8) x 100 = 25%
Therefore, you ate 25% of the pizza.
### Calculating Percentage of a Whole
Suppose you want to find out how much a part represents in percentage of a whole. The formula to calculate the percentage of a whole is:
Symbol Meaning
P% Percentage of Whole
A Part
B Whole
The formula is:
P% = (A/B) x 100
Where A is the part, and B is the whole. Let’s illustrate this with an example.
### Example
Suppose out of 50 students, 30 students passed the exam. What percentage of students passed?
P% = (30/50) x 100 = 60%
Therefore, 60% of the students passed the exam.
### Calculating Discount
Suppose a store is offering a discount of 20%. How much will you save on an item worth \$50?
The formula to calculate discount is:
Symbol Meaning
D Discount Amount
B Original Price
P% Discount Percentage
The formula is:
D = B x (P%/100)
Therefore, the discount you’ll get on an item worth \$50 will be:
D = 50 x (20/100) = \$10
Therefore, you’ll save \$10 on an item worth \$50 with a 20% discount.
### Calculating Tax
Suppose a tax of 5% is applicable to an item worth \$100. How much will you pay after adding tax?
The formula to calculate tax is:
Symbol Meaning
T Tax Amount
B Original Price
P% Tax Percentage
The formula is:
T = B x (P%/100)
Therefore, the tax applicable to an item worth \$100 with a tax rate of 5% is:
T = 100 x (5/100) = \$5
Hence, the total amount you’ll need to pay for the item will be:
Total Amount = B + T = \$100 + \$5 = \$105
You’ll need to pay \$105 for the item after adding tax.
### 1. What is percentage?
Percentage is a fraction of a whole expressed in terms of 100.
### 2. What is the basic formula for calculating percentage?
The basic formula for calculating percentage is: P% = (A/B) x 100.
### 3. What is the percentage increase/decrease formula?
The formula to calculate percentage increase/decrease is: P% = ((A-B)/B) x 100.
### 4. How can I calculate percentage of a part?
To calculate the percentage of a part, use the formula P% = (A/B) x 100, where A is the part, and B is the whole.
### 5. How can I calculate the percentage of a whole?
To calculate the percentage of a whole, use the formula P% = (A/B) x 100, where A is the part, and B is the whole.
### 6. What is the formula for calculating discount?
The formula for calculating discount is: D = B x (P%/100), where D is the discount amount, B is the original price, and P% is the discount percentage.
### 7. What is the formula for calculating tax?
The formula for calculating tax is: T = B x (P%/100), where T is the tax amount, B is the original price, and P% is the tax percentage.
### 8. How can I convert a fraction to a percentage?
To convert a fraction to a percentage, multiply it by 100.
### 9. How can I convert a decimal to a percentage?
To convert a decimal to a percentage, multiply it by 100.
### 10. How can I convert a percentage to a fraction?
To convert a percentage to a fraction, divide it by 100.
### 11. How can I convert a percentage to a decimal?
To convert a percentage to a decimal, divide it by 100.
### 12. Can I calculate percentage on a calculator?
Yes, most calculators have the percentage function that can help you calculate percentage easily.
### 13. Is percentage used in everyday life?
Yes, percentage is used in various fields, such as finance, science, and business, to represent a portion of a whole.
## Conclusion
Congratulations! You now have a comprehensive understanding of how to calculate percentage. We hope this guide has been helpful to you. Whether you’re a student, a professional, or just someone looking to brush up on their math skills, knowing how to calculate percentage can be invaluable. Remember, percentage is used in various aspects of our lives, and being able to calculate it accurately can help you make informed decisions based on data and statistics. So go ahead, put your skills to the test, and calculate away!
### Take Action Now!
Don’t wait! Start practicing your percentage calculations and become an expert in no time! |
CAT 2023 » CAT Study Material » Data Interpretation and Logical Reasoning » A Short Note on Properties of Cube
# A Short Note on Properties of Cube
While reading this article you may grasp the concept of cube. Here we also discussed topics like cube shape and so on.
A Cube is a solid three-dimensional shape with six square faces, eight vertices, and twelve edges in mathematics or geometry. It’s also a normal hexahedron, according to legend. You’ve probably seen the 3 3 Rubik’s cube, which is the most common example in real life and can help you improve your mental power. Similarly, many real-life examples, such as 6 sided dice, will be encountered. Solid geometry is concerned with three-dimensional figures and objects with surface areas and volumes. Cuboid, cylinder, cone, and sphere are the other solid shapes. We’ll go through its concept, qualities, and importance in mathematics.
## Definition of Cube
A cube, as previously stated, is a three-dimensional object.
a solid form with six faces In three-dimensional space, a cube is one of the most basic shapes. A cube’s six faces are all squares, which is a two-dimensional shape.
## Cube Form
A cube’s form is sometimes referred to as “cubic.” A cube can also be thought of as a block because its length, width, and height are all the same. It also contains 8 vertices and 12 edges, with three of the edges meeting at one vertex point.
Examine the image below and identify the faces, edges, and vertices. It’s also known as a right rhombohedron, an equilateral cuboid, and a square parallelepiped. The cube is one of the platonic solids, and it is a convex polyhedron with square faces on all sides.
The cube is said to have octahedral or cubical symmetry. The cube is a variant of the square prism.
Cube Surface Area and Volume Formula
The cube’s surface area and volume are addressed below:
## Cube’s Surface Area
We know that the area of any shape is defined as the plane region it occupies. Because a cube is a three-dimensional object, the area it occupies will be in a three-dimensional plane. We must compute the surface area of the cube covered by each face because a cube has six faces. The formula for calculating surface area is given below.
Assume that an is the cube’s edge.
a2 is the area of one face divided by the area of a square.
The cube has six square-shaped faces, as we know.
Area of one face minus lateral surface area (excluding top and bottom faces) = 4
LSA = 4a2
LSA + top and bottom face area = total surface area
TSA = 4a2 + a2 + a2
TSA = 6a2
## Cube’s Volume
The space contained in the cube is its volume. For example, if an object is cubical in shape and we need to submerge any material in it, such as water, the volume of water to be preserved in the object is determined. The volume formula is as follows:
a3 cubic units = cube volume.
## Cube Characteristics
The following are some of the most essential cube properties:
It has a square form to all of its faces.
All of the faces or sides are the same size.
The cube’s plane angles are all right angles.
Each face interacts with the other four.
The three faces and three edges meet at each of the vertices.
The edges on opposite sides are parallel.
## What is the difference between a square and a cube?
The most significant distinction between the square and the cube is that the square is a two-dimensional figure with only two dimensions: length and breadth, but the cube is a three-dimensional shape with three dimensions: length, breadth, and height. The cube is made from a square shape.
## Conclusion
Cube is a solid three-dimensional shape with six square faces, eight vertices, and twelve edges in mathematics or geometry.
It’s also a normal hexahedron, according to legend. You’ve probably seen the 3× 3 Rubik’s cube, which is the most common example in real life and can help you improve your mental power. Similarly, many real-life examples, such as 6 sided dice, will be encountered.
Solid geometry is concerned with three-dimensional figures and objects with surface areas and volumes. Cuboid, cylinder, cone, and sphere are the other solid shapes. We’ll go through its concept, qualities, and importance in mathematics.
Get answers to the most common queries related to the CAT Examination Preparation.
## What exactly is a cube?
A three-dimensional figure having six faces, eight vertices, and twelve edges is known as a cube. A prism is a speci...Read full
## What is the distinction between the terms cube and cuboid?
A cube is a three-dimensional version of the square, with square faces on all sides. A cuboid, on the other hand, is...Read full
## How can you figure out the volume of a cube?
Because all of a cube’s sides are the same length, its volume is calculated as a3 cubic units, where “a&...Read full
## Is it possible to call a cube a prism?
Because a cube is considered one of the Platonic solids, it is still a prism. |
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# If $APB$ and $CQD$ are two parallel lines, then bisectors of $\angle APQ$ , $\angle BPQ,\angle CQP$ and $\angle PQD$ formsA. a squareB. a rhombusC. a rectangleD. any other parallelogram
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Hint: First we draw two parallel lines $APB$ and $CQD$ , then, we draw angle bisectors of $\angle APQ$ , $\angle BPQ,\angle CQP$ and $\angle PQD$ . Assuming that these angle bisectors meet at point $M$ and $N$ . By joining points we draw a quadrilateral, then by observing the property of quadrilateral forms, we choose the correct answer.
Let us first draw a diagram by using information given that $APB$ and $CQD$ are two parallel lines and there are angle bisectors of $\angle APQ$ , $\angle BPQ,\angle CQP$ and $\angle PQD$ .
We have given that $APB$ and $CQD$ are two parallel lines. Let us assume that the bisectors of $\angle APQ$ and $\angle CQP$ meet at point $M$ . Similarly, bisectors of $\angle BPQ$ and $\angle DQP$ meet at point $N$ .
We join $PM,QM,QN,PN$ , it forms a quadrilateral.
Now, as given $APB\parallel CQD$
So, alternate interiors angles will be equal $\angle APQ=\angle PQD$
Now, as $PM$ and $QN$ are angle bisectors of $\angle APQ$ and $\angle PQD$ respectively. So,
$\angle APM=\angle MPQ$ and $\angle BPN=\angle NPQ$
$\angle MPQ=\angle NQP$
As these are alternate interior angles so the lines $PM$ and $QN$ must be parallel.
Similarly, alternate interiors angles $\angle BPQ=\angle CQP$ are also equal.
So, the lines $PN$ and $QM$ must be parallel.
As $PM\parallel QN$ and $PN\parallel QM$ .
Opposite sides are parallel to each other, it means $PMQN$ is a parallelogram.
As $CQD$ is a straight line, so the measure of $\angle CQD=180{}^\circ$
Or we can write as $\angle CQP+\angle DQP=180{}^\circ$
Or $\angle CQM+\angle PQM+\angle DQN+\angle PQN=180{}^\circ$
As $PM$ and $QN$ are angle bisectors, So, $\angle PQM+\angle PQM+\angle PQN+\angle PQN=180{}^\circ$
$\left[ \angle CQM=\angle PQM\And \angle DQN=\angle PQN \right]$
\begin{align} & 2\angle PQM+2\angle PQN=180{}^\circ \\ & 2\left( \angle PQM+\angle PQN \right)=180{}^\circ \\ & \angle PQM+\angle PQN=\dfrac{180{}^\circ }{2} \\ & \angle PQM+\angle PQN=90{}^\circ \\ \end{align}
$\angle MQN=90{}^\circ$
So, the correct answer is “Option C”.
Note: Opposite sides of a rectangle are always equal and parallel to each other. Diagonals of a rectangle bisect each other. Measure of angles of a rectangle is equal to $90{}^\circ$. It is important to apply properties to prove it a quadrilateral. Also, applying properties is important as given in the options all are quadrilateral so we have to prove that it is a rectangle. |
Polynomials
Contributor: Mason Smith. Lesson ID: 11547
To what degree do you understand polynomials? Are you a proponent of exponents? They can be easily dealt with once you factor in watching a video, taking the interactive quiz, and real-world examples!
categories
Algebra I, Algebra II, Complex Numbers and Quantities
subject
Math
learning style
Visual
personality style
Lion
Middle School (6-8), High School (9-12)
Lesson Type
Quick Query
Lesson Plan - Get It!
Audio:
Q: What happened to the quadratic polynomial when he fell asleep on the beach?
A: He got second-degree burns.
Before you can begin to calculate polynomials, you must first learn (or remember) what they are, so let's run through some definitions!
Monomials are numbers, variables, or a product of numbers and variables, with whole-number exponents.
For example, x2, 5, and –7xy are monomials, because each example has a whole number exponent (numbers have an exponent of 0 unless stated otherwise).
X-2 is not a monomial, because X-2 has an exponent of –2, and negatives are not whole number exponents. 4x - y is not a monomial because it has two variables with exponents that are not being multiplied together.
For a quick refresher, watch Algebra Basics: What Are Polynomials? - Math Antics by Math Antics. Take notes on any points you are having difficulty understanding:
When you examine monomials, you can also look at the degree of a monomial, that is simply the sum of the exponents for the variable. Remember, constants (numbers without variables) have a degree of 0. A variable without a number (x, y, or a, for example) has a degree of 1.
Let's look at three examples of degrees:
1. 2a3b4 has a degree of 7, which can be calculated by adding the exponents together: 3 + 4 = 7.
2. 4 has a degree of 0 because it is a number, and numbers have a degree of 0.
3. 8y has a degree of 1, because any variable without a number (in this case, y) is assumed to have an exponent of 1, and the number 8 has a degree of 0.
For more examples, visit Degree (of an Expression) at MathsIsFun.com.
Try out a few examples. Find the degree of each monomial.
A polynomial is a monomial, or multiple monomials being added or subtracted.
To find the degree of a polynomial, look at the degree of the term with the greatest degree. This is easily done by finding the degree of each term, then comparing them.
Try some examples:
1. 4x - 18x15 — The degree of 4x is 1. The degree of -18x15 is 15, so the degree of the entire polynomial is 15, because 15 is bigger than 1.
2. 6x4 + 9x2 - x + 3 — The degree of 6x4 is 4. The degree of 9x2 is 2. The degree of -x is 1. The degree of 3 is 0. The largest degree is 4, so the polynomial is of degree 4.
3. x2y + xy + .75 — The degree of x2y is 3 because 2 + 1 = 3. The degree of xy is 2 because 1 + 1 = 2. The degree of .75 is 0 because .75 is a constant. The degree of the polynomial is 3 because that is the largest degree.
Try a few examples. Find the degree of each polynomial.
The terms of a polynomial can technically be written in any order; however, the most efficient way to write polynomials is with only one variable in standard form, which is in descending degrees. This means if the polynomial only has the variable x, we count down from the largest exponent to the smallest, and to the constants (Remember: constants are numbers without a variable).
For example:
If we have 20x - 4x3 + 2 - x2, we would rewrite this in standard form as -4x3 - x2 + 20x + 2. This also tells us that our polynomial has a degree of 3.
The –4 in the previous example is the leading coefficient. It is also the number attached to the monomial with the greatest degree, and is the first coefficient when written in standard form.
For example, in 6y5 + y3 + 4y, the 6 would be the leading coefficient. In x2 + 3x - 6, the coefficient is 1. Remember, any variable without a number in front of it is assumed to have a coefficient of 1.
Try a few and share your answers with a teacher or parent:
Write each polynomial in standard form, then give the leading coefficient of each polynomial.
Polynomials are classified with special names based on their degree and number of terms. Sadly, there is no easy way to remember these, so make sure you practice each name! The highlighted terms are the most common:
Degree Name Number of terms Name 0 constant 1 monomial 1 linear 2 binomial 2 quadratic 3 trinomial 3 cubic 4 or more polynomial 4 quartic 5 quintic 6 or more 6th degree; 7th degree; etc.
Let's do a few examples classifying polynomials. Then, you will practice a few on your own:
Classify each polynomial below:
1. 5x - 6
• Degree: 1
• Number of terms: 2
• 5x - 6 is a linear binomial
2. y2+ y + 4
• Degree: 2
• Number of terms: 3
• y2 + y + 4 is a quadratic trinomial
3. 6x7 + 9x2- x + 3
• Degree: 7
• Number of terms: 4
• 6x7 + 9x2 - x + 3 is a 7th-degree polynomial |
# 1. Complex Numbers¶
The lecture on complex numbers consists of three parts, each with their own video:
Total video length: 38 minutes and 53 seconds
## 1.1 Definition and basic operations¶
### Complex numbers¶
Definition I
Complex numbers are numbers of the form Here $\rm i$ is the square root of -1: or equivalently:
Usual operations on numbers have their natural extension for complex numbers, as we shall see below.
Some useful definitions:
Definition II
For a complex number $z = a + b {{\rm i}}$, $a$ is called the real part, and $b$ the imaginary part.
Complex conjugate
The complex conjugate $z^*$ of $z = a + b {{\rm i}}$ is defined as
i.e., taking the complex conjugate means flipping the sign of the imaginary part.
For two complex numbers, $z_1 = a_1 + b_1 {{\rm i}}$ and $z_2 = a_2 + b_2 {{\rm i}}$, the sum $w = z_1 + z_2$ is given as
where the parentheses in the rightmost expression have been added to group the real and the imaginary part. A consequence of this definition is that the sum of a complex number and its complex conjugate is real: i.e., this results in twice the real part of $z$.
Similarly, subtracting $z^*$ from $z$ yields i.e., twice the imaginary part of $z$ (times $\rm i$).
### Multiplication¶
Multiplication
For the same two complex numbers $z_1$ and $z_2$ as above, their product is calculated as where the parentheses have again beรจn used to indicate the real and imaginary parts.
A consequence of this definition is that the product of a complex number $z = a + b {{\rm i}}$ with its conjugate is real: The square root of this number is called the norm $|z|$ of $z$:
### Division¶
The quotient $z_1/z_2$ of two complex numbers $z_1$ and $z_2$ defined above can be evaluated by multiplying the numerator and denominator by the complex conjugate of $z_2$:
Division
Try this yourself!
Example:
### Visualization: the complex plane¶
Complex numbers can be rendered on a two-dimensional (2D) plane, the complex plane. This plane is spanned by two unit vectors, one horizontal representing the real number 1 and the vertical unit vector representing ${\rm i}$.
#### Addition in the complex plane¶
Adding two numbers in the complex plane corresponds to adding their respective horizontal and vertical components:
## 1.2. Complex functions¶
Real functions can (most of the times) be written in terms of a Taylor series expanded at a point $x_{0}$: We can write something similar for complex functions by replacing the real variable $x$ with its complex counterpart $z$:
For this course, the most important function is the complex exponential function, at which we will have a closer look below.
### The complex exponential function¶
The complex exponential is used extremely often. It occurs in Fourier transforms and it is very convenient for doing calculations involving cosines and sines. It also makes many common operations on complex number a lot easier to perform.
The exponential function and Euler identity
The exponential function $f(z) = \exp(z) = e^z$ is defined as:
The last expression is called the Euler identity.
Exercise
Check that this function obeys You will need sum and difference formulas of cosine and sine.
### The polar form¶
A complex number $z$ can be represented by two real numbers, $a$ and $b$, which correspond to the real and imaginary part of the complex number. Another representation of $z$ is a vector in the complex plane with a horizontal component that corresponds to the real part of $z$ and a vertical component that corresponds to the imaginary part of $z$. It is also possible to characterize that vector by its length and direction, where the latter can be represented by the angle that the vector makes with the horizontal axis:
Polar form of complex numbers
A complex number can be represented either by its real and imaginary part corresponding to the Cartesian coordinates in the complex plane, or by its norm and its argument corresponding to polar coordinates. The norm is the length of the vector, and the argument is the angle it makes with the horizontal axis.
We can conclude that for a complex number $z = a + b {\rm i}$, its real and imaginary parts can be expressed in polar coordinates as
Inverse equations
The inverse equations are for $a>0$. In general:
It turns out that by using the magnitude $|z|$ and phase $\varphi$, we can write any complex number as By increasing $\varphi$ by $2 \pi$, we make a full circle around the origin and reach the same point on the complex plane. In other words, by adding $2 \pi$ to the argument of $z$, we get the same complex number $z$! As a result, the argument $\varphi$ is defined up to $2 \pi$, and we are free to make any choice we like, such as in the examples in the figure below:
Some useful values of the complex exponential to know by heart are:
Useful identities:
From the first expression, it also follows that As a result, $y$ is only defined up to $2\pi$.
Furthermore, we can define the sine and cosine in terms of complex exponentials:
Complex sine and cosine
Most operations on complex numbers become easier when complex numbers are converted to their polar form using the complex exponential. Some functions and operations, which are common in real analysis, can be easily derived for their complex counterparts by substituting the real variable $x$ with the complex variable $z$ in its polar form:
Examples of some complex functions stated using polar form
Use of polar form lets us notice immediately that for example, as a result of multiplication, the norm of the new number is the product of the norms of the multiplied numbers and its argument is the sum of the arguments of the multiplied numbers.
In the complex plane, this looks as follows:
Example: Find all solutions solving $z^4 = 1$.
Of course, we know that $z = \pm 1$ are two solutions, but which other solutions are possible? We take a systematic approach:
## 1.3. Differentiation and integration¶
We only consider differentiation and integration over real variables.
We can then regard the complex ${\rm i}$ as another constant, and use our usual differentiation and integration rules:
Differentiation and Integration rules
## 1.4. Bonus: the complex exponential function and trigonometry¶
Let us show some tricks in the following examples where the simple properties of the exponential function help in re-deriving trigonometric identities.
Properties of the complex exponential function I
Take $|z_1| = |z_2| = 1$, and $\arg{(z_1)} = \varphi_1$ and $\arg{(z_2)} = \varphi_2$. It is easy to see then that $z_i = \exp({\rm i} \varphi_i)$, $i=1, 2$. Then: The left hand side can be written as
Also, the right hand side can be written as Comparing the two expressions, equating their real and imaginary parts, we find Note that we used the Euler formula in order to derive the identities of trigonometric function. The point is to show you that you can use the properties of the complex exponential to quickly find the form of trigonometric formulas, which are often easily forgotten.
Properties of the complex exponential function II
In this example, let's see what we can learn from the derivative of the exponential function: Writing out the exponential in terms of cosine and sine, we see that where the prime $'$ denotes the derivative as usual. Equating real and imaginary parts leads to
## 1.5. Summary¶
1. A complex number $z$ has the form where $a$ and $b$ are both real, and $\rm i^2 = 1$. The real number $a$ is called the real part of $z$ and $b$ is the imaginary part. Two complex numbers can be added, subtracted and multiplied straightforwardly. The quotient of two complex numbers $z_1=a_1 + \rm i b_1$ and $z_2=a_2 + \rm i b_2$ is
2. Complex numbers can also be characterised by their norm $|z|=\sqrt{a^2+b^2}$ and argument $\varphi$. These parameters correspond to polar coordinates in the complex plane. For a complex number $z = a + b {\rm i}$, its real and imaginary parts can be expressed as The inverse equations are The complex number itself then becomes
3. The most important complex function for us is the complex exponential function, which simplifies many operations on complex numbers where $y$ is defined up to $2 \pi$.\ The $\sin$ and $\cos$ can be rewritten in terms of this complex exponential as Because we only consider differentiation and integration over real variables, the usual rules apply:
## 1.6. Problems¶
1. [] Given $a=1+2\rm i$ and $b=-3+4\rm i$, calculate and draw in the complex plane the numbers:
1. $a+b$,
2. $ab$,
3. $b/a$.
2. [] Evaluate:
1. $\rm i^{1/4}$,
2. $\left(1+\rm i \sqrt{3}\right)^{1/2}$,
3. $\exp(2\rm i^3)$.
3. [] Find the three 3rd roots of $1$ and ${\rm i}$.
(i.e. all possible solutions to the equations $x^3 = 1$ and $x^3 = {\rm i}$, respectively).
4. [] Quotients
1. Find the real and imaginary part of
2. Evaluate for real $a$ and $b$:
5. [] For any given complex number $z$, we can take the inverse $\frac{1}{z}$.
1. Visualize taking the inverse in the complex plane.
2. What geometric operation does taking the inverse correspond to?
(Hint: first consider what geometric operation $\frac{1}{z^*}$ corresponds to.)
6. [] Differentation and integration
1. Compute
2. Calculate the real part of ($k$, $x$, $\omega$, $t$ and $\gamma$ are real; $\gamma$ is positive).
7. [] Compute by making use of the Euler identity. |
# Advice 1: How to find the surface area of a rectangular parallelepiped
This is called a rectangular parallelepiped, all six faces of which are rectangles. The formula for calculating its surface area is very simple: S = 2(ab + bc + ac), where a, b and c – the lengths of the edges.
Instruction
1
To begin calculate the area of three different faces of the parallelepiped. For example, the length of the box (a) is 7 cm, width (b) 6 cm, and the height (C) 4 cm Then the area of the upper (lower) faces will be equal to ab, i.e. 7x6=42 cm Area of one of the side faces will be equal to bc, i.e., 6x4=24, see Finally, the area of the front (rear) face is equal to ac, i.e. 7 × 4=28 cm.
2
Now let's add up all three scores and multiply the resulting sum by two. In our it will look like the following: 42+24+28=94; 94х2=188. Thus, the surface area of this cuboid is equal to 188 cm
Note
Be careful not to confuse a rectangular parallelepiped with direct. The direct parallelepiped rectangles are only the side faces (4 of the 6 faces), and the upper and lower bases are arbitrary parallelograms.
As a particular case of a rectangular parallelepiped can be considered CC, since all its faces are equal, then for finding the surface will be necessary to erect the edge length in the square and multiply by 6.
# Advice 2 : How to find the surface area of a rectangular parallelepiped
This is called a rectangular parallelepiped, all six faces of which are rectangles. The formula for calculating its surface area is very simple: S = 2(ab + bc + ac), where a, b and c – the lengths of the edges.
Instruction
1
To begin calculate the area of three different faces of the parallelepiped. For example, the length of the box (a) is 7 cm, width (b) 6 cm, and the height (C) 4 cm Then the area of the upper (lower) faces will be equal to ab, i.e. 7x6=42 cm Area of one of the side faces will be equal to bc, i.e., 6x4=24, see Finally, the area of the front (rear) face is equal to ac, i.e. 7 × 4=28 cm.
2
Now let's add up all three scores and multiply the resulting sum by two. In our it will look like the following: 42+24+28=94; 94х2=188. Thus, the surface area of this cuboid is equal to 188 cm
Note
Be careful not to confuse a rectangular parallelepiped with direct. The direct parallelepiped rectangles are only the side faces (4 of the 6 faces), and the upper and lower bases are arbitrary parallelograms.
As a particular case of a rectangular parallelepiped can be considered CC, since all its faces are equal, then for finding the surface will be necessary to erect the edge length in the square and multiply by 6.
# Advice 3 : How to find the area of the base of the box
The base of a parallelepiped is always a parallelogram. To find the area of the base, calculate the area of this parallelogram. As a special case, it can be a rectangle or a square. You can also find the area of the base of the box, knowing its volume and height.
You will need
• Ruler, protractor, scientific calculator
Instruction
1
In the General case, the base of a parallelepiped is a parallelogram. To find its area, using a ruler, make the measurement of the lengths of its sides and a protractor to measure the angle between them. The square base of a parallelepiped is equal to the product of these sides by the sine of the angle between them S=a • b • Sin(α).
2
To determine the area of the base of the box the other way, measure one side of the base, then lower to her height from the vertex, which lies opposite this side. Measure the length of this height. To obtain the area of the base find the area of a parallelogram by multiplying the length of a side on high, which is omitted S=a • h.
3
To obtain the values of area another way to measure the lengths of its diagonals (the distance between opposite vertices), and the angle between the diagonals. The area will be equal to half the product of the diagonals by the sine of the angle between them S=0,5•d1•d2•Sin(β).
4
For a parallelepiped, which lies at the base of the diamond, it is sufficient to measure the lengths of its diagonals and to find half their work S=0,5 • d1 • d2.
5
In the case where the base cuboid is a rectangle, measure the length and width of this geometric figure, and then multiply these values S=a • b. This will be the area of its base. In the case when the base is square, measure one of its sides and lift in the second degree S=a2.
6
If you know the volume of the box, measure its height. To do this, drop a perpendicular from any vertex of the upper base onto the plane, which belongs to the lower base. Measure the length of this segment, which is the height of the box. If the parallelepiped line (its lateral edges perpendicular to the basem), it is enough to measure the length of one of these edges, which equals the height of the box. To obtain area of base, volume of a parallelepiped divide by its height S=V/h.
# Advice 4 : How to find the area of the face of the box
Spatial figure is called a cuboid has a few numerical characteristics, including the area of the surface. To define it, we need to find the area of each face of the box and fold the resulting value.
Instruction
1
Draw a box using the pencil and ruler, placing the base horizontally. This is a classic form of representation of shapes, which can be used to demonstrate all the conditions of the problem. Then to solve it will be much easier.
2
Look at the picture. Have a box of six pairs of parallel faces. Each pair represents an equal two-dimensional shapes, which in General case are parallelograms. Accordingly, their areas are also equal. Thus, the total surface is the sum of the three matching variables: square of the top or bottom of the base, the front or rear faces, right, or left face.
3
To find the square faces of a parallelepiped, we need to consider it as a separate figure with two dimensions, length and width. At the well-known formula area of a parallelogram equals the base to the height.
4
The direct box only the bases are parallelograms, all the lateral faces are rectangular in shape. The area of this figure is obtained by multiplying the length by the width, since it coincides with the height. In addition, there exists a rectangular parallelepiped, all faces of which – rectangles.
5
The cube is also a parallelepiped, which has the unique property – equality of all measurements and the numerical characteristics of the faces. The area of each of its sides are equal to the squared length of any edge, and the full surface is obtained by multiplying this value by 6.
6
The form of a parallelepiped with square corners can often be found in everyday life, for example, in the construction of houses, construction of items of furniture, household appliances, toys, stationery, etc.
7
Example: find the area of each lateral face direct of the box, if you know that the height is 3 cm, the perimeter of the base is 24 cm and base length greater than width 2 cm Solution.Write down the formula for perimeter of parallelogram P = 2•a + 2•b. The statement b = a + 2 therefore P = 4•a + 4 = 24, whence a = 5, b = 7.
8
Find the area of the lateral faces of the figure with 5 sides and 3 cm rectangle:Ѕб1 = 5•3 = 15 (cm2).The area of parallel side faces, by definition of the box, also equal to 15 cm2. It remains to determine the area of the other pairs of sides with the sides 7 and 3:Ѕб2 = 3•7 = 21 (cm2).
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714. Best Time to Buy and Sell Stock with Transaction Fee
Description
You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.
Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.
Note:
• You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
• The transaction fee is only charged once for each stock purchase and sale.
Example 1:
Input: prices = [1,3,2,8,4,9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:
- Buying at prices[0] = 1
- Selling at prices[3] = 8
- Buying at prices[4] = 4
- Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
Example 2:
Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6
Constraints:
• 1 <= prices.length <= 5 * 104
• 1 <= prices[i] < 5 * 104
• 0 <= fee < 5 * 104
Solutions
Solution 1: Memoization
We design a function $dfs(i, j)$, which represents the maximum profit that can be obtained starting from day $i$ with state $j$. Here, $j$ can take the values $0$ and $1$, representing not holding and holding a stock, respectively. The answer is $dfs(0, 0)$.
The execution logic of the function $dfs(i, j)$ is as follows:
If $i \geq n$, there are no more stocks to trade, so we return $0$.
Otherwise, we can choose not to trade, in which case $dfs(i, j) = dfs(i + 1, j)$. We can also choose to trade stocks. If $j \gt 0$, it means that we currently hold a stock and can sell it. In this case, $dfs(i, j) = prices[i] + dfs(i + 1, 0) - fee$. If $j = 0$, it means that we currently do not hold a stock and can buy one. In this case, $dfs(i, j) = -prices[i] + dfs(i + 1, 1)$. We take the maximum value as the return value of the function $dfs(i, j)$.
The answer is $dfs(0, 0)$.
To avoid redundant calculations, we use memoization to record the return value of $dfs(i, j)$ in an array $f$. If $f[i][j]$ is not equal to $-1$, it means that we have already calculated it, so we can directly return $f[i][j]$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $prices$.
Solution 2: Dynamic Programming
We define $f[i][j]$ as the maximum profit that can be obtained up to day $i$ with state $j$. Here, $j$ can take the values $0$ and $1$, representing not holding and holding a stock, respectively. We initialize $f[0][0] = 0$ and $f[0][1] = -prices[0]$.
When $i \geq 1$, if we do not hold a stock at the current day, then $f[i][0]$ can be obtained by transitioning from $f[i - 1][0]$ and $f[i - 1][1] + prices[i] - fee$, i.e., $f[i][0] = \max(f[i - 1][0], f[i - 1][1] + prices[i] - fee)$. If we hold a stock at the current day, then $f[i][1]$ can be obtained by transitioning from $f[i - 1][1]$ and $f[i - 1][0] - prices[i]$, i.e., $f[i][1] = \max(f[i - 1][1], f[i - 1][0] - prices[i])$. The final answer is $f[n - 1][0]$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $prices$.
We notice that the transition of the state $f[i][]$ only depends on $f[i - 1][]$ and $f[i - 2][]$. Therefore, we can use two variables $f_0$ and $f_1$ to replace the array $f$, reducing the space complexity to $O(1)$.
Solution 3: Greedy
Variables
• f1: Represents the maximum profit that can be achieved up to the current day with the last operation being a buy or hold after a buy. It’s initialized to -prices[0] because if we buy the stock on the first day, our profit is negative (the cost of the stock).
• f2: Represents the maximum profit that can be achieved up to the current day with the last operation being a sell or hold after a sell. It’s initialized to 0 because we haven’t made any transactions yet.
Logic
The idea is to iterate through each day’s price after the first day, updating f1 and f2 to reflect the maximum profit that can be achieved up to that day.
1. Updating f1 (Buy or Hold after Buy):
• f1 is updated to be the maximum of itself (f1) or f2 - price.
• f1 (staying in the buy state) means we are not making any new transactions; we’re just carrying forward the profit (or loss) from the previous day.
• f2 - price means we are considering a new buy on this day. We subtract the current day’s price from f2, the best profit we could have after a sell up to the previous day, to reflect the cost of this buy.
• This comparison ensures that f1 represents the best profit achievable by either continuing to hold a stock or buying anew.
2. Updating f2 (Sell or Hold after Sell):
• f2 is updated to be the maximum of itself (f2) or f1 + price - fee.
• f2 (staying in the sell state) means we are not making any new transactions; we’re just carrying forward the profit from the previous day.
• f1 + price - fee means we are considering a sell on this day. We add the current day’s price to f1, the best profit we could have with the last operation being a buy, and subtract the transaction fee to reflect the cost of this sell.
• This comparison ensures that f2 represents the best profit achievable by either continuing to hold after a sell or selling the stock.
Return Value
• After iterating through all prices, f2 will represent the maximum profit achievable by the end of the last day, with the last operation being a sell or hold after a sell. Therefore, f2 is returned as the solution.
• class Solution {
public int maxProfit(int[] prices, int fee) {
int f0 = 0, f1 = -prices[0];
for (int i = 1; i < prices.length; ++i) {
int g0 = Math.max(f0, f1 + prices[i] - fee);
f1 = Math.max(f1, f0 - prices[i]);
f0 = g0;
}
return f0;
}
}
• class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int n = prices.size();
int f[n][2];
memset(f, 0, sizeof(f));
f[0][1] = -prices[0];
for (int i = 1; i < n; ++i) {
f[i][0] = max(f[i - 1][0], f[i - 1][1] + prices[i] - fee);
f[i][1] = max(f[i - 1][1], f[i - 1][0] - prices[i]);
}
return f[n - 1][0];
}
};
• class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
f1, f2 = -prices[0], 0
for price in prices[1:]:
f1 = max(f1, f2 - price)
f2 = max(f2, f1 + price - fee)
return f2
############
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
f0, f1 = 0, -prices[0]
for x in prices[1:]:
f0, f1 = max(f0, f1 + x - fee), max(f1, f0 - x)
return f0
############
class Solution:
def maxProfit(self, prices, fee):
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""
cash = 0
hold = -prices[0]
for i in range(1, len(prices)):
cash = max(cash, hold + prices[i] - fee)
hold = max(hold, cash - prices[i])
return cash
• func maxProfit(prices []int, fee int) int {
f0, f1 := 0, -prices[0]
for _, x := range prices[1:] {
f0, f1 = max(f0, f1+x-fee), max(f1, f0-x)
}
return f0
}
• function maxProfit(prices: number[], fee: number): number {
const n = prices.length;
let [f0, f1] = [0, -prices[0]];
for (const x of prices.slice(1)) {
[f0, f1] = [Math.max(f0, f1 + x - fee), Math.max(f1, f0 - x)];
}
return f0;
} |
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# Geometry: Praxis I Exam (page 3)
By
Updated on Jul 5, 2011
### Area of a Triangle
Area is the amount of space inside a two-dimensional object. Area is measured in square units, often written as unit2. So, if the length of a triangle is measured in feet, the area of the triangle is measured in feet2.
A triangle has three sides, each of which can be considered a base of the triangle. A perpendicular line segment drawn from a vertex to the opposite base of the triangle is called the altitude, or the height. It measures how tall the triangle stands.
It is important to note that the height of a triangle is not necessarily one of the sides of the triangle. The correct height for the following triangle is 8, not 10. The height will always be associated with a line segment (called an altitude) that comes from one vertex (angle) to the opposite side and forms a right angle (signified by the box). In other words, the height must always be perpendicular to (form a right angle with) the base. Note that in an obtuse triangle, the height can be outside the triangle, and in a right triangle the height is usually one of the sides.
The formula for the area of a triangle is given by , where b is the base of the triangle, and h is the height.
Example
Determine the area of the following triangle.
#### Volume Formulas
A prism is a three-dimensional object that has matching polygons as its top and bottom. The matching top and bottom are called the bases of the prism. The prism is named for the shape of the prism's base, so a triangular prism has congruent triangles as its bases.
Note: This can be confusing. The base of the prism is the shape of the polygon that forms it; the base of a triangle is one of its sides.
Volume is the amount of space inside a three-dimensional object. Volume is measured in cubic units, often written as unit3. So, if the edge of a triangular prism is measured in feet, the volume of it is measured in feet3.
The volume of ANY prism is given by the formula V = Abh, where Ab is the area of the prism's base, and h is the height of the prism.
Example
Determine the volume of the following triangular prism:
The area of the triangular base can be found by using the formula bh, so the area of the base is (15)(20) = 150. The volume of the prism can be found by using the formula V = Abh, so the volume is V= (150)(40) = 6,000 cubic feet.
A pyramid is a three-dimensional object that has a polygon as one base, and instead of a matching polygon as the other, there is a point. Each of the sides of a pyramid is a triangle. Pyramids are also named for the shape of their (non-point) base.
The volume of a pyramid is determined by the formula Abh.
Example
Determine the volume of a pyramid whose base has an area of 20 square feet and stands 50 feet tall.
Because the area of the base is given to us, we only need to replace the appropriate values into the formula.
The pyramid has a volume of 33 cubic feet.
### Polygons
A polygon is a closed figure with three or more sides—for example, triangles, rectangles, and pentagons.
#### Terms Related to Polygons
• Vertices are corner points, also called endpoints, of a polygon. The vertices in the above polygon are A, B, C, D, E, and F, and they are always labeled with capital letters.
• A regular polygon has congruent sides and congruent angles
• An equiangular polygon has congruent angles.
### Interior Angles
To find the sum of the interior angles of any polygon, use this formula:
S = (x – 2)180°, where x = the number of sides of the polygon.
Example
Find the sum of the interior angles in this polygon:
The polygon is a pentagon that has five sides, so substitute 5 for x in the formula:
S = (5 – 2) × 180°
S = 3 × 180°
S = 540°
### Exterior Angles
Similar to the exterior angles of a triangle, the sum of the exterior angles of any polygon equals 360 degrees.
### Similar Polygons
If two polygons are similar, their corresponding angles are congruent and the ratios of the corresponding sides are in proportion.
Example
These two polygons are similar because their angles are congruent and the ratios of the corresponding sides are in proportion
A quadrilateral is a four-sided polygon. Because a quadrilateral can be divided by a diagonal into two triangles, the sum of its interior angles will equal 180 + 180 = 360°.
### Parallelograms
A parallelogram is a quadrilateral with two pairs of parallel sides.
In the figure, . Parallel lines are symbolized with matching numbers of triangles or arrows
A parallelogram has:
• opposite sides that are congruent
• opposite angles that are congruent (A = C and B = D)
• consecutive angles that are supplementary (A + B = 180°, B + C = 180°, C + D = 180°, D + A = 180° )
• diagonals (line segments joining opposite vertices) that bisect each other (divide each other in half)
#### Special Types of Parallelograms
• A rectangle is a parallelogram that has four right angles
• A rhombus is a parallelogram that has four equal sides.
• A square is a parallelogram in which all angles are equal to 90 degrees and all sides are congruent. A square is a special case of a rectangle where all the sides are congruent. A square is also a special type of rhombus where all the angles are congruent
#### Diagonals of Parallelograms
In this diagram, parallelogram ABCD has diagonals and that intersect at point E. The diagonals of a parallelogram bisect each other, which means that .
In addition, the following properties hold true:
• The diagonals of a rhombus are perpendicular.
• The diagonals of a rectangle are congruent.
• The diagonals of a square are both perpendicular and congruent.
Example
In parallelogram ABCD, the diagonal = 5x + 10 and the diagonal = 9x. Determine the value of x.
Because the diagonals of a parallelogram are congruent, the lengths are equal. We can then set up and solve the equation 5x + 10 = 9x to determine the value of x.
5x + 10 = 9x Subtract x from both sides of the equation. 10 = 4x Divide 4 from both sides of the equation. 2.5 = x
#### Area and Volume Formulas
The area of any parallelogram can be found with the formula A = bh, where b is the base of the parallelogram, and h is the height. The base and height of a parallelogram is defined the same as in a triangle.
Note: Sometimes b is called the length (l) and h is called the width (w) instead. If this is the case, the area formula is A = lw.
A rectangular prism (or rectangular solid) is a prism that has rectangles as bases. Recall that the formula for any prism is V = Abh. Because the area of the rectangular base is A = lw, we can replace lw for Ab in the formula giving us the more common, easier to remember formula, V = lwh. If a prism has a different quadrilateralshaped base, use the general prisms formula for volume.
Note: A cube is a special rectangular prism with six congruent squares as sides. This means that you can use the V = lwh formula for it, too.
### Circles
#### Terminology
A circle is formally defined as the set of points a fixed distance from a point. The more sides a polygon has, the more it looks like a circle. If you consider a polygon with 5,000 small sides, it will look like a circle, but a circle is not a polygon. A circle contains 360 degrees around a center point.
• The midpoint of a circle is called the center.
• The distance around a circle (called perimeter in polygons) is called the circumference.
• A line segment that goes through a circle, with its endpoints on the circle, is called a chord.
• A chord that goes directly through the center of a circle (the longest line segment that can be drawn) in a circle is called the diameter.
• The line from the center of a circle to a point on the circle (half of the diameter) is called the radius.
• A sector of a circle is a fraction of the circle's area.
• An arc of a circle is a fraction of the circle's circumference.
#### Circumference, Area, and Volume Formulas
The area of a circle is A = πr2, where r is the radius of the circle. The circumference (perimeter of a circle) is 2πr, or πd, where r is the radius of the circle and d is the diameter.
Example
Determine the area and circumference of this circle:
We are given the diameter of the circle, so we can use the formula C = πd to find the circumference.
C = πd
C = π(6)
C = 6π = 18.85 feet
The area formula uses the radius, so we need to divide the length of the diameter by 2 to get the length of the radius: 6 ÷ 2 = 3. Then we can just use the formula.
A = π(3)2
A = 9π = 28.27 square feet.
Note: Circumference is a measure of length, so the answer is measured in units, whereas the area is measured in square units.
#### Area of Sectors and Lengths of Arcs
The area of a sector can be determined by figuring out what the percentage of the total area the sector is, and then multiplying by the area of the circle.
The length of an arc can be determined by figuring out what the percentage of the total circumference of the arc is, and then multiplying by the circumference of the circle.
Example
Determine the area of the shaded sector and the length of the arc AB.
Because the angle in the sector is 30°, and we know that a circle contains a total of 360°, we can determine what fraction of the circle's area it is: of the circle.
The area of the entire circle is A = πr2, so A = π(4)2 = 16π.
So, the area of the sector is square inches.
We can also determine the length of the arc AB, because it is of the circle's circumference.
The circumference of the entire circle is C = 2πr, so C = 2π(4) = 8π.
This means that the length of the arc is inches.
A prism that has circles as bases is called a cylinder. Recall that the formula for the volume of any prism is V = Abh. Because the area of the circular base is A = πr2, we can replace πr2 for Ab in the formula, giving us V = πr2h, where r is the radius of the circular base, and h is the height of the cylinder.
A sphere is a three-dimensional object that has no sides. A basketball is a good example of a sphere. The volume of a sphere is given by the formula V = πr3.
Example
Determine the volume of a sphere whose radius is 1.5'.
Replace 1.5' in for r in the formula V = πr3.
V = πr3.
V = π(1.5)3
V = (3.375)π
V = 4.5π ≈ 14.14
The answer is approximately 14.14 cubic feet.
Example
An aluminum can is 6" tall and has a base with a radius of 2". Determine the volume the can holds.
Aluminum cans are cylindrical in shape, so replace 2" for r and 6" for h in the formula V = πr2h.
V = πr2h
V = π(2)2(6)
V = 24π ≈ 75.40 cubic feet |
# How do you graph the line -4x - 16y = 4?
Mar 29, 2016
See explanation
#### Explanation:
The easiest way is to manipulate the equation into the form of
$y = m x + c$
The assign different values to x from which you can then calculate the corresponding value for y. For a straight line graph you only need two points that you can join up and extend to give the graph plot. How ever, 3 points are better as it increases the chance of showing up any error. All 3 points should line up.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving your question}}$
Given;$\text{ } - 4 x - 16 h = 4$
To make the $y$ term positive multiply everything by (-1) giving
$\textcolor{b r o w n}{+ 4 x + 16 y = - 4}$
Subtract #color(blue)(4x) from both sides
$\textcolor{b r o w n}{+ 4 x \textcolor{b l u e}{- 4 x} + 16 y = - 4 \textcolor{b l u e}{- 4 x}}$
$0 + 16 y = - 4 x - 4$
Divide both sides by 16
$\frac{16}{16} \times y = - \frac{4}{16} x - \frac{4}{16}$
But $\frac{16}{16} = 1$ and $\frac{4}{16} = \frac{1}{4}$giving
$y = - \frac{1}{4} x - \frac{1}{4}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~ |
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# Recursively break a number in 3 parts to get maximum sum
Given a number n, we can divide it in only three parts n/2, n/3 and n/4 (we will consider only integer part). The task is to find the maximum sum we can make by dividing number in three parts recursively and summing up them together.
Examples:
Input : n = 12
Output : 13
// We break n = 12 in three parts {12/2, 12/3, 12/4}
// = {6, 4, 3}, now current sum is = (6 + 4 + 3) = 13
// again we break 6 = {6/2, 6/3, 6/4} = {3, 2, 1} = 3 +
// 2 + 1 = 6 and further breaking 3, 2 and 1 we get maximum
// summation as 1, so breaking 6 in three parts produces
// maximum sum 6 only similarly breaking 4 in three
// parts we can get maximum sum 4 and same for 3 also.
// Thus maximum sum by breaking number in parts is=13
Input : n = 24
Output : 27
// We break n = 24 in three parts {24/2, 24/3, 24/4}
// = {12, 8, 6}, now current sum is = (12 + 8 + 6) = 26
// As seen in example, recursively breaking 12 would
// produce value 13. So our maximum sum is 13 + 8 + 6 = 27.
// Note that recursively breaking 8 and 6 doesn't produce
// more values, that is why they are not broken further.
Input : n = 23
Output : 23
// we break n = 23 in three parts {23/2, 23/3, 23/4} =
// {10, 7, 5}, now current sum is = (10 + 7 + 5) = 22.
// Since after further breaking we can not get maximum
// sum hence number is itself maximum i.e; answer is 23
Recommended Practice
A simple solution for this problem is to do it recursively. In each call we have to check only max((max(n/2) + max(n/3) + max(n/4)), n) and return it. Because either we can get maximum sum by breaking number in parts or number is itself maximum. Below is the implementation of recursive algorithm.
## C++
// A simple recursive C++ program to find// maximum sum by recursively breaking a// number in 3 parts.#includeusing namespace std; // Function to find the maximum sumint breakSum(int n){ // base conditions if (n==0 || n == 1) return n; // recursively break the number and return // what maximum you can get return max((breakSum(n/2) + breakSum(n/3) + breakSum(n/4)), n);} // Driver program to run the caseint main(){ int n = 120; cout << breakSum(n); return 0;}
## Java
// A simple recursive JAVA program to find// maximum sum by recursively breaking a// number in 3 parts.import java.io.*; class GFG { // Function to find the maximum sum static int breakSum(int n) { // base conditions if (n==0 || n == 1) return n; // recursively break the number and return // what maximum you can get return Math.max((breakSum(n/2) + breakSum(n/3) + breakSum(n/4)), n); } // Driver program to test the above function public static void main (String[] args) { int n = 12; System.out.println(breakSum(n)); }}// This code is contributed by Amit Kumar
## Python3
# A simple recursive Python3 program to # find maximum sum by recursively breaking # a number in 3 parts. # Function to find the maximum sum def breakSum(n) : # base conditions if (n == 0 or n == 1) : return n # recursively break the number and # return what maximum you can get return max((breakSum(n // 2) + breakSum(n // 3) + breakSum(n // 4)), n) # Driver Codeif __name__ == "__main__" : n = 12 print(breakSum(n)) # This code is contributed by Ryuga
## C#
// A simple recursive C# program to find// maximum sum by recursively breaking a// number in 3 parts.using System; class GFG { // Function to find the maximum sum static int breakSum(int n) { // base conditions if (n==0 || n == 1) return n; // recursively break the number // and return what maximum you // can get return Math.Max((breakSum(n/2) + breakSum(n/3) + breakSum(n/4)), n); } // Driver program to test the // above function public static void Main () { int n = 12; Console.WriteLine(breakSum(n)); }} // This code is contributed by anuj_67.
## Javascript
Output
144
Memoization of Recursive Solution
In the above solution, we see that there are many overlapping functions calls become so to reduce that we memoize the above recursive solution so it’s time complexity reduce exponentially to linear.
## C++
// memoization C++ program to find maximum sum by recursively breaking a number in 3 parts. #includeusing namespace std; // Function to find the maximum sumint breakSum(int n, vector&dp){ // base conditions if (n == 0 || n == 1) { return n; } // if dp[n] already calculated then directly return so no overlapping problem calls if (dp[n] != -1) { return dp[n]; } // break the number and return what maximum you can get return dp[n] = max((breakSum(n / 2, dp) + breakSum(n / 3, dp) + breakSum(n / 4, dp)), n);} // Driver program to run the caseint main() { int n = 24; // dp array todo memoization vectordp(n + 1, -1); cout << breakSum(n, dp); return 0;}
## Java
// memoization Java program to find maximum sum by recursively breaking a number in 3 parts. class Main { // Function to find the maximum sum public static int breakSum(int n, int[] dp) { // base conditions if (n == 0 || n == 1) return n; // if dp[n] already calculated then directly return so no overlapping problem calls if (dp[n] != -1) { return dp[n]; } // break the number and return what maximum you can get return dp[n] = Math.max((breakSum(n / 2, dp) + breakSum(n / 3, dp) + breakSum(n / 4, dp)), n); } // Driver program to test the above function public static void main (String[] args) { int n = 24; int[] dp = new int[n+1]; for (int i = 0; i <= n; i++) dp[i] = -1; System.out.println(breakSum(n, dp)); }} // This code is contributed by ajaymakvana.
## Python3
# Python3 code to find maximum sum by recursively breaking a number in 3 parts. # Function to find the maximum sumdef breakSum(n, dp): # base conditions if n == 0 or n == 1: return n # if dp[n] already calculated then directly return so no overlapping problem calls if dp[n] != -1: return dp[n] # break the number and return what maximum you can get dp[n] = max((breakSum(n // 2, dp) + breakSum(n // 3, dp) + breakSum(n // 4, dp)), n) return dp[n] # Driver program to run the caseif __name__ == "__main__": n = 24 # dp array todo memoization dp = [-1 for i in range(n + 1)] print(breakSum(n, dp)) # This code is contributed by lokehpotta20.
## C#
// memoization C# program to find maximum sum by // recursively breaking a number in 3 parts.using System;using System.Collections.Generic; class GFG { // Function to find the maximum sum static int breakSum(int n, List dp) { // base conditions if (n == 0 || n == 1) { return n; } // if dp[n] already calculated then directly // return so no overlapping problem calls if (dp[n] != -1) { return dp[n]; } // break the number and return what maximum you can get return dp[n] = Math.Max((breakSum(n / 2, dp) + breakSum(n / 3, dp) + breakSum(n / 4, dp)), n); } // Driver program to run the case static public void Main() { int n = 24; // dp array todo memoization Listdp = new List(); for(int i = 0; i < n + 1; i++) dp.Add(-1); Console.Write(breakSum(n, dp)); }} // This code is contributed by ratiagrawal.
## Javascript
// memoization Javascript program to // find maximum sum by recursively breaking a number in 3 parts. // Function to find the maximum sumfunction breakSum(n, dp){ // base conditions if (n == 0 || n == 1) { return n; } // if dp[n] already calculated then directly return // so no overlapping problem calls if (dp[n] != -1) { return dp[n]; } // break the number and return what maximum you can get return dp[n] = Math.max((breakSum(parseInt(n / 2), dp) + breakSum(parseInt(n / 3), dp) + breakSum(parseInt(n / 4), dp)), n); } // Driver program to run the caselet n = 24; // dp array todo memoizationlet dp=new Array(n + 1).fill(-1); console.log(breakSum(n, dp));
Output
27
Time Complexity: O(n)
Auxiliary Space: O(n) (as we are creating dp array of size n+1)
An efficient solution for this problem is to use Dynamic programming because while breaking the number in parts recursively we have to perform some overlapping problems. For example part of n = 30 will be {15,10,7} and part of 15 will be {7,5,3} so we have to repeat the process for 7 two times for further breaking. To avoid this overlapping problem we are using Dynamic programming. We store values in an array and if for any number in recursive calls we have already solution for that number currently so we directly extract it from array.
## C++
// A Dynamic programming based C++ program// to find maximum sum by recursively breaking// a number in 3 parts.#include#define MAX 1000000using namespace std; int breakSum(int n){ int dp[n+1]; // base conditions dp[0] = 0, dp[1] = 1; // Fill in bottom-up manner using recursive // formula. for (int i=2; i<=n; i++) dp[i] = max(dp[i/2] + dp[i/3] + dp[i/4], i); return dp[n];} // Driver program to run the caseint main(){ int n = 24; cout << breakSum(n); return 0;}
## Java
// A simple recursive JAVA program to find// maximum sum by recursively breaking a// number in 3 parts.import java.io.*; class GFG { final int MAX = 1000000; // Function to find the maximum sum static int breakSum(int n) { int dp[] = new int[n+1]; // base conditions dp[0] = 0; dp[1] = 1; // Fill in bottom-up manner using recursive // formula. for (int i=2; i<=n; i++) dp[i] = Math.max(dp[i/2] + dp[i/3] + dp[i/4], i); return dp[n]; } // Driver program to test the above function public static void main (String[] args) { int n = 24; System.out.println(breakSum(n)); }}// This code is contributed by Amit Kumar
## Python3
# A Dynamic programming based Python program# to find maximum sum by recursively breaking# a number in 3 parts. MAX = 1000000 def breakSum(n): dp = [0]*(n+1) # base conditions dp[0] = 0 dp[1] = 1 # Fill in bottom-up manner using recursive # formula. for i in range(2, n+1): dp[i] = max(dp[int(i/2)] + dp[int(i/3)] + dp[int(i/4)], i); return dp[n] # Driver program to run the casen = 24print(breakSum(n)) # This code is contributed by# Smitha Dinesh Semwal
## C#
// A simple recursive C# program to find// maximum sum by recursively breaking a// number in 3 parts.using System; class GFG { // Function to find the maximum sum static int breakSum(int n) { int []dp = new int[n+1]; // base conditions dp[0] = 0; dp[1] = 1; // Fill in bottom-up manner // using recursive formula. for (int i = 2; i <= n; i++) dp[i] = Math.Max(dp[i/2] + dp[i/3] + dp[i/4], i); return dp[n]; } // Driver program to test the // above function public static void Main () { int n = 24; Console.WriteLine(breakSum(n)); }} // This code is contributed by anuj_67.
## Javascript
Output
27
Time Complexity: O(n)
Auxiliary Space: O(n)
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## What is the sum of 3 consecutive multiples of 8?
Question 7 Q7) The sum of three consecutive multiples of 8 is 888.
## What are the three multiples of 8?
The multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80… and so on.
## What is consecutive multiple?
Numbers which follow each other in order, without gaps, from smallest to largest. 12, 13, 14 and 15 are consecutive numbers. 22, 24, 26, 28 and 30 are consecutive even numbers. 40, 45, 50 and 55 are consecutive multiples of 5.
## What are consecutive multiples of 888?
It is given that the sum of all the three consecutive multiples of 8 is 888. Therefore the first multiple of 8 is 8x, by substituting the value of x in the equation we get, ⇒8×36=288.
## What are three consecutive multiples?
Let us assume that the three consecutive multiples of 3 are 3x, 3(x+1) and 3(x+2). Now as all these three consecutive numbers are multiples of the number three. Thus, the largest multiple of 3 is 27.
## What are two consecutive multiples?
Answer Expert Verified Let consecutive multiple of 2 are 2x , ( 2x +2) .
## What is the difference between two consecutive multiples of any number?
For any two consecutive even numbers, the difference is 2. For example, 6 and 8 are two consecutive even numbers, their difference = 8 – 6 = 2. Any n numbers in a row will have a multiple of n. For example, for 7 consecutive numbers in a row, there will be one multiple of 7.
## What are the consecutive multiples of 9?
Ans: Let the first multiple of 9 be 9x. Therefore, the next two multiples will be 9x+9 and 9x+18. x=109. Therefore, the multiples of 9 are 981, 990 and 999.
## What number is a multiple of 5 and 3?
How to list multiples of a number?
Multiples of 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Multiples of 3 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
Multiples of 4 4, 8, 12, 16, 20, 24, 28, 32, 36, 40.
Multiples of 5 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
Multiples of 6 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
40
## What is the LCM of 9 6 and 3?
The LCM of 3,6,9 3 , 6 , 9 is the result of multiplying all prime factors the greatest number of times they occur in either number. The LCM of 3,6,9 3 , 6 , 9 is 2⋅3⋅3=18 2 ⋅ 3 ⋅ 3 = 18 .
## What is the least common multiple of 4 and 8?
Answer: LCM of 4 and 8 is 8.
## What the LCM of 3/8 and 24?
Least Common Multiple of 3 and 8 Free LCM Calculator determines the least common multiple (LCM) between 3 and 8 the smallest integer that is 24 that is divisible by both numbers. Least Common Multiple (LCM) of 3 and 8 is 24. |
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# Cards numbered from $11$ to $60$ are kept in a box, if a card is drawn at random from the box, find the probability that the number on the card drawn is$\left( i \right)$ an odd number $\left( ii \right)$ a perfect square$\left( iii \right)$ divisible by $5$ $\left( iv \right)$ a prime number less than $20$
Last updated date: 17th Jul 2024
Total views: 451.2k
Views today: 8.51k
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451.2k+ views
Hint: In each part, the number of favourable outcomes can be calculated manually. There is no need of applying the concept of permutation and combination.
Before proceeding with the question, we must know the definition of the probability. The probability of an event $X$ is defined as the ratio of the number of outcomes favourable to event $X$ and the total number of possible outcomes in the sample space. Mathematically,’
$P\left( X \right)=\dfrac{n\left( X \right)}{n\left( S \right)}...............\left( 1 \right)$
In this question, we have cards numbered from $11$ to $60$. So, the sample space $S$ is,
$S=\left\{ 11,12,13,14,................,58,59,60 \right\}$
So, total number of possible outcomes $n\left( S \right)$ which is same for each part in the question and is equal to,
$n\left( S \right)=50..............\left( 2 \right)$
(i) It is given that the card drawn is an odd number i.e. set $X$ is,
$X=\left\{ 11,13,15,.........,53,57,59 \right\}$
So, $n\left( X \right)=25.........\left( 3 \right)$
Substituting $n\left( X \right)=25$ from equation $\left( 3 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,
\begin{align} & P\left( X \right)=\dfrac{25}{50} \\ & \Rightarrow P\left( X \right)=\dfrac{1}{2} \\ \end{align}
(ii) It is given that the card drawn is a perfect square number i.e. set $X$ is,
$X=\left\{ 16,25,36,49 \right\}$
So, $n\left( X \right)=4.........\left( 4 \right)$
Substituting $n\left( X \right)=4$ from equation $\left( 4 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,
\begin{align} & P\left( X \right)=\dfrac{4}{50} \\ & \Rightarrow P\left( X \right)=\dfrac{2}{25} \\ \end{align}
(iii) It is given that the card drawn is a number which is divisible by $5$ i.e. set $X$ is,
$X=\left\{ 15,20,25,30,35,40,45,50,55,60 \right\}$
So, $n\left( X \right)=10.........\left( 5 \right)$
Substituting $n\left( X \right)=10$ from equation $\left( 5 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,
\begin{align} & P\left( X \right)=\dfrac{10}{50} \\ & \Rightarrow P\left( X \right)=\dfrac{1}{5} \\ \end{align}
(iv) It is given that the card drawn is a prime number less than $20$ i.e. set $X$ is,
$X=\left\{ 11,13,17,19 \right\}$
So, $n\left( X \right)=4.........\left( 6 \right)$
Substituting $n\left( X \right)=4$ from equation $\left( 6 \right)$ and $n\left( S \right)=50$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get,
\begin{align} & P\left( X \right)=\dfrac{4}{50} \\ & \Rightarrow P\left( X \right)=\dfrac{2}{25} \\ \end{align}
Hence, the answer of $\left( i \right)$ is $\dfrac{1}{2}$, $\left( ii \right)$ is $\dfrac{2}{25}$, $\left( iii \right)$ is $\dfrac{1}{5}$, $\left( iv \right)$ is $\dfrac{2}{25}$.
Note: There is a possibility of committing mistakes while calculating the number of favourable outcomes. Since we are calculating the number of favourable outcomes by finding out the set $X$, there is a possibility that one may find out the wrong number. |
# How to represent irrational numbers on number line
## Irrational numbers on the number line
Irrational numbers are the numbers which have a non terminating and non-repeating decimal expansion. They don’t have any fixed value so we can’t represent them directly on the number line. we have to use a few tricks for this
Before understanding the representation of irrational numbers on number line first, Our Online Maths Tutors, first explain Pythagoras’ theorem
Pythagoras’ Theorem:: Over 2000 years ago there was an amazing discovery about triangles:
When a triangle has a right angle (90°) and squares are made on each of the three sides then the biggest square has the exact same area as the other two squares put together!
pythagorus theorem
### OrIrrational Numbers using Pythagoras theorem
It can simply be written as an equation relating the lengths of the sides a, b and c, like this:
Pythagoras’s theorem
After understanding Pythagoras’ theorem we will learn how to represent an irrational number on a number line
Suppose we want to draw √10 on the number line we will do the following steps:
first of all, we need to break 10 as a sum of two numbers which are perfect squares in itself.
for example, 10 can be written as 8+2,7+3,6+4,5+5 but we will prefer it breaking as 9+1 because both 9 and 1 are complete squares. Then we do this:
Also READ this- How to solve basic problems in trigonometry?(concept-2)
(√10)²= 10
= 9+1
=(3)²+(1)²
now if we draw a right-angled triangle taking one side as 3 unit, another side as 1 unit then hypotenuse will become √10
rep. on the number line
We can easily put this value on the number line. Many irrational numbers can be shown on the number line using this method
In the next post, IB Online Tutors will discuss the numbers which can not be written as a sum of two complete square numbers |
# IBPS RRB Quantitative Aptitude Quiz Day 1
5 Steps - 3 Clicks
# IBPS RRB Quantitative Aptitude Quiz Day 1
### Introduction
What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.
A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.
The article IBPS RRB Quantitative Aptitude Quiz Day 1 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. IBPS RRB has released IBPS RRB Officer 2019 Official Notification for Officer Scale(I, II, and II). Quantitative Aptitude plays major role to qualify examination. The article IBPS RRB Quantitative Aptitude Quiz Day 1 will assist the students to know the expected questions from Quantitative Aptitude.
### Quiz
Directions(1-5): Permutation and Combinations
1. In how many ways can the letters of the word ‘APPLE’ be arranged ?
A. 720
B. 120
C. 60
D. 180
Explanation: The word ‘APPLE’ contains 5 letters, 1A, 2P, 1L.and 1E.
∴ Required number of ways = $$\frac{5!}{(1!)(2!)(1!)(1!)}$$ = 60
2. The value of $${75}_{{P}_{2}}$$ is :
A. 2775
B. 150
C. 5550
D. None of these
Explanation: $${75}_{{P}_{2}}$$ = $$\frac{75!}{75-2!}$$
= $$\frac{75!}{73!}$$
= $$\frac{75*74*(73!)}{73!}$$
= (75 * 74) = 5550.
3. In how many ways can the letters of the word ‘LEADER’ be arranged ?
A. 72
B. 144
C. 360
D. 720
Explanation: In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter.
The word ‘LEADER’ contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
∴ Required number of ways = $$\frac{6!}{(1!)(2!)(1!)(1!)(2!)}$$ = 360
4. How many words with or without meaning, can be formed by using all the letters of the word, ‘DELHI’, using each letter exactly once ?
A. 10
B. 25
C. 60
D. 120
Explanation: The word ‘DELHI’ contains 5 different letters.
Required number of words = Number of arrangements of 5 letters, taken all at a time = $${5}_{{P}_{5}}$$ = 5 !
= (5 *4 *3 *2 *1) = 120.
5. In how many different ways can the letters of the word ‘RUMOUR’ be arranged ?
A. 180
B. 90
C. 30
D. 720
Explanation: The word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1U.
∴ Required number of ways = $$\frac{6!}{(2!)(2!)(1!)(1!)}$$ = 180
Directions(1-5): Averages
1. 3 years ago, the average of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is:
A. 5 years
B. 2 years
C. 1 year
D. 4 years
Explanation: Total age of 5 members, 3 years ago = (17 x 5) years = 85 years
Total age of 5 members now = (85 + 3 x 5) years = 100 years
Total age of 6 members now = (17 x 6) years = 102 years
Age of the baby = (102 – 100) years = 2 years
2. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is:
A. 1 : 4
B. 2 : 3
C. 3 : 4
D. 4 : 2
Explanation: Let the ratio be k:1
Then, k x 16.4 + 1 x 15.4 = (k + 1) x 15.8
(16.4 – 15.8) k = (15.8 – 15.4)
k = $$\frac{2}{3}$$
3. The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these two books?
A. Rs 10 and Rs 16
B. Rs 12 and Rs 24
C. Rs 24 and Rs 18
D. Rs 28 and Rs 12
Explanation: Total price of the two books = Rs. [(12 x 10) – (11.75 x 8)] = Rs. (120 – 94) = Rs. 26
Let the price of one book be Rs.x
Then, the price of other book = Rs. (x + 60% of x ) = x + $$\frac{3}{5}$$x = ($$\frac{8}{5}$$)x
so, x + $$\frac{8}{5}$$x = 26 , x = 10
The prices of the two books are Rs. 10 and Rs. 16
4. The average age of 30 boys of a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher?
A. 50
B. 44
C. 45
D. 42
Explanation: Total ages of 30 boys = 14 x 30 = 420 years
Total age when class teacher is included = 15 x 31 = 465 years
Age of class teacher = 465 – 420 = 45 years
5. The Average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of failed candidates is 15, what is the number of candidates who passed the examination?
A. 90
B. 100
C. 108
D. 115
Explanation: Let the number of passed candidates be a
Then total marks =>120 x 35 = 39 a + (120 – a) x 15
4200 = 39 a + 1800 – 15 a
a = 100
Directions(1-5): The following series are based on a specific pattern. In these series one number is wrong, find that odd one.
1. 6, 22, 34, 78, 146, 305, 594
A. 32
B. 22
C. 305
D. 146
Explanation: 6*2 + 10 = 22
22*2 – 10 = 34
34*2 + 10 = 78
78*2 – 10 = 146
146*2 + 10 = 302
302*2 – 10 = 594
2. 4, 4, 15, 74, 524, 4724, 51974
A. 51974
B. 15
C. 74
D. 524
Explanation: 4*1 + 0 = 4
4*3 + 2 = 14
14*5 + 4 = 74
74*7 + 6 = 524
524*9 + 8 = 4724
4724*11 + 10 = 51974
3. 3, 5, 13, 39, 177, 891
A. 177
B. 13
C. 5
D. 39
Explanation: 3*1 + 2 = 5
5*2 + 3 = 13
13*3 + 4 = 43
43*4 + 5 = 177
177*5 + 6 = 891
4. 3, 7, 12, 32, 57, 93, 142
A. 7
B. 12
C. 32
D. 57
Explanation: 3 + $${2}^{2}$$ = 7
7 + $${3}^{2}$$ = 16
16 + $${4}^{2}$$ = 32
32 + $${5}^{2}$$ = 57
16 + $${4}^{2}$$ = 32
32 + $${5}^{2}$$ = 57
57 + $${6}^{2}$$ = 93
93 + $${7}^{2}$$ = 142
5. 2, 13, 38, 130, 522, 2625
A. 2625
B. 522
C. 38
D. 130
Explanation: 2*1 + 11 = 13
11*2 + 12 = 38
38*3 + 13 = 127
127*4 + 14 = 522
522*5 = 2625
### Exams
Competitive Exams – College Entrance Exams
Diploma NITC New Delhi
PG GATE 2020
Competitive Exams – Recent Job Notifications
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MOMENT OF INERTIA
# MOMENT OF INERTIA
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## MOMENT OF INERTIA
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. MOMENT OF INERTIA • Today’s Objectives: • Students will be able to: • Determine the mass moment of inertia of a rigid body or a system of rigid bodies. In-Class Activities: • Check Homework • Reading Quiz • Applications • Mass Moment of Inertia • Parallel-Axis Theorem • Composite Bodies • Concept Quiz • Group Problem Solving • Attention Quiz
2. APPLICATIONS The large flywheel in the picture is connected to a large metal cutter. The flywheel mass is used to help provide a uniform motion to the cutting blade. What property of the flywheel is most important for this use? How can we determine a value for this property? Why is most of the mass of the flywheel located near the flywheel’s circumference?
3. APPLICATIONS (continued) The crank on the oil-pump rig undergoes rotation about a fixed axis that is not at its mass center. The crank develops a kinetic energy directly related to its mass moment of inertia. As the crank rotates, its kinetic energy is converted to potential energy and vice versa. Is the mass moment of inertia of the crank about its axis of rotation smaller or larger than its moment of inertia about its center of mass?
4. MASS MOMENT OF INERTIA Consider a rigid body with a center of mass at G. It is free to rotate about the z axis, which passes through G. Now, if we apply a torque T about the z axis to the body, the body begins to rotate with an angular acceleration of . T and are related by the equation T = I . In this equation, I is the mass moment of inertia (MMI) about the z axis. The MMI of a body is a property that measures the resistance of the body to angular acceleration. The MMI is often used when analyzing rotational motion.
5. MASS MOMENT OF INERTIA (continued) Consider a rigid body and the arbitrary axis P shown in the figure. TheMMIabout the P axis is defined as I = m r2 dm, where r, the “moment arm,” is the perpendicular distance from the axis to the arbitrary element dm. The mass moment of inertia is always a positive quantity and has a unit of kg ·m2or slug · ft2.
6. MASS MOMENT OF INERTIA (continued) The figures below show the mass moment of inertia formulations for two flat plate shapes commonly used when working with three dimensional bodies. The shapes are often used as the differential element being integrated over the entire body.
7. PARALLEL-AXIS THEOREM If the mass moment of inertia of a body about an axis passing through the body’s mass center is known, then the moment of inertia about any other parallelaxis may be determined by using the parallel axis theorem, I = IG + md2 where IG = mass moment of inertia about the body’s mass center m = mass of the body d = perpendicular distance between the parallel axes
8. PLANAR KINETIC EQUATIONS OF MOTION: TRANSLATION • Today’s Objectives: • Students will be able to: • Apply the three equations of motion for a rigid body in planar motion. • Analyze problems involving translational motion. In-Class Activities: • Check Homework • Reading Quiz • Applications • FBD of Rigid Bodies • EOM for Rigid Bodies • Translational Motion • Concept Quiz • Group Problem Solving • Attention Quiz
9. APPLICATIONS The boat and trailer undergo rectilinear motion. In order to find the reactions at the trailer wheels and the acceleration of the boat, we need to draw the FBD and kinetic diagram for the boat and trailer. How many equations of motion do we need to solve this problem? What are they?
10. APPLICATIONS (continued) As the tractor raises the load, the crate will undergo curvilinear translation if the forks do not rotate. If the load is raised too quickly, will the crate slide to the left or right? How fast can we raise the load before the crate will slide?
11. PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2) • We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. • As discussed in Chapter 16, when a body is subjected to general plane motion, it undergoes a combination of translation and rotation. • First, a coordinate system with its origin at an arbitrary point P is established. The x-y axes should not rotate and can either be fixed or translate with constant velocity.
12. EQUATIONS OF TRANSLATIONAL MOTION (continued) • If a body undergoes translational motion, the equation of motion isF =m aG. This can also be written in scalar form as Fx = m(aG)xand Fy = m(aG)y • In words: the sum of all the external forces acting on the body is equal to the body’s mass times the acceleration of it’s mass center.
13. EQUATIONS OF ROTATIONAL MOTION We need to determine the effects caused by the moments of an external force system. The moment about point P can be written as: (riFi)+ Mi=r maG+IG Mp = (Mk)p where r = x i + y j and Mp is the resultant moment about P due to all the external forces. The term (Mk)pis called the kinetic moment about point P.
14. EQUATIONS OF ROTATIONAL MOTION (continued) If point P coincides with the mass center G, this equation reduces to the scalar equation of MG= IG . In words: the resultant (summation) moment about the mass center due to all the external forces is equal to the moment of inertia about G times the angular acceleration of the body. Thus, three independent scalar equations of motion may be used to describe the general planar motion of a rigid body. These equations are: Fx = m(aG)x Fy = m(aG)y and MG= IG or Mp = (Mk)p
15. Fx = m(aG)x Fy = m(aG)y MG= 0 EQUATIONS OF MOTION: TRANSLATION (Section 17.3) When a rigid body undergoes only translation, all the particles of the body have the same acceleration so aG = a and a= 0. The equations of motion become: Note that, if it makes the problem easier, the moment equation can be applied about another point instead of the mass center. For example, if point A is chosen, MA= (m aG) d .
16. EQUATIONS OF MOTION: TRANSLATION (continued) When a rigid body is subjected to curvilinear translation, it is best to use an n-t coordinate system. Then apply the equations of motion, as written below, for n-t coordinates. Fn = m(aG)n Ft = m(aG)t MG= 0 or MB= e[m(aG)t] – h[m(aG)n]
17. 4. Apply the three equations of motion (one set or the other): Fx = m(aG)x Fy = m(aG)y Fn = m(aG)n Ft = m(aG)t MG= 0 or MP = (Mk)P MG= 0 or MP = (Mk)P PROCEDURE FOR ANALYSIS Problems involving kinetics of a rigid body in only translation should be solved using the following procedure: 1. Establish an (x-y) or (n-t) inertial coordinate system and specify the sense and direction of acceleration of the mass center, aG. 2. Draw a FBD and kinetic diagram showing all external forces, couples and the inertia forces and couples. 3. Identify the unknowns. 5. Remember, friction forces always act on the body opposing the motion of the body.
18. EXAMPLE Given: A 50 kg crate rests on a horizontal surface for which the kinetic friction coefficient k = 0.2. Find: The acceleration of the crate if P = 600 N. Plan: Follow the procedure for analysis. Note that the load P can cause the crate either to slide or to tip over. Let’s assume that the crate slides. We will check this assumption later.
19. EXAMPLE (continued) Solution: The coordinate system and FBD are as shown. The weight of (50)(9.81) N is applied at the center of mass and the normal force Nc acts at O. Point O is some distance x from the crate’s center line. The unknowns are Nc, x, and aG . Applying the equations of motion: Nc = 490 N x = 0.467 m aG = 10.0 m/s2 Fx = m(aG)x: 600 – 0.2 Nc = 50 aG Fy = m(aG)y: Nc – 490.5 = 0 MG= 0: -600(0.3) + Nc(x) – 0.2 Nc(0.5) = 0
20. EXAMPLE (continued) Since x = 0.467 m < 0.5 m, the crate slides as originally assumed. If x was greater than 0.5 m, the problem would have to be reworked with the assumption that tipping occurred.
21. GROUP PROBLEM SOLVING Given:The handcart has a mass of 200 kg and center of mass at G. A force of P=50 N is applied to the handle. Neglect the mass of the wheels. Find:The normal reactions at each of the two wheels at A and B. Plan: Follow the procedure for analysis.
22. GROUP PROBLEM SOLVING (continued) Solution: The cart will move along a rectilinear path. Draw FBD and kinetic diagram. y x = Applying the equations of motion: + Fx = m(aG)x 50 cos 60 = 200 aG aG = 0.125 m/s2
23. GROUP PROBLEM SOLVING (continued) Applying the equations of motion: y x = +↑ Fy = 0 NA + NB –1962 –50 sin 60 = 0 NA + NB = 2005 N (1) • MG = 0 • -(0.3)NA+(0.2)NB+0.3(50 cos 60) – 0.6(50 sin 60) = 0 • − 0.3 NA+ 0.2 NB = 18.48 N m (2) Using Eqs. (1) and (2), solve for the reactions, NA and NB NA = 765 N, NB = 1240 N
24. End of the Lecture Let Learning Continue |
# What Is The GCF Of 15 And 9?
## What is the LCM for 15 and 10?
Least common multiple (LCM) of 10 and 15 is 30..
## What is the GCF of 7/15 and 21?
The common factors for 7,15,21 7 , 15 , 21 are 1 . The GCF (HCF) of the numerical factors 1 is 1 .
## What is the GCF of 12 and 9?
By making factor trees you can identify the largest common factors of two numbers. 9=3⋅3 and 12=2⋅2⋅3 , in this case the only common factor is 3 and therefore the greatest.
## What is the GCF of 12 and 15?
The factors of 12 are : 1, 2, 3, 4, 6, 12. The factors of 15 are : 1, 3, 5, 15. 1 and 3 are the only common factors (numbers which are factors of both 12 and 15). Therefore, the highest common factor of 12 and 15 is 3.
## What is the greatest common factor of 9?
Greatest common factor (GCF) of 9 and 12 is 3. We will now calculate the prime factors of 9 and 12, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 9 and 12.
## What is the common denominator of 15 and 9?
You need to know the least common denominator (LCD) of 9 and 15 if you want to add or subtract two fractions with 9 and 15 as denominators. The least common denominator, also called lowest common denominator (LCD), of 9 and 15 is 45.
## What is the GCF of 10 15 and 20?
We found the factors and prime factorization of 15 and 20. The biggest common factor number is the GCF number. So the greatest common factor 15 and 20 is 5.
## What is the GCF of 24 and 18?
We found the factors and prime factorization of 18 and 24. The biggest common factor number is the GCF number. So the greatest common factor 18 and 24 is 6.
## What do 15 and 20 have in common?
The factors of 20 are 20, 10, 5, 4, 2, 1. The common factors of 15 and 20 are 5, 1, intersecting the two sets above. In the intersection factors of 15 ∩ factors of 20 the greatest element is 5. Therefore, the greatest common factor of 15 and 20 is 5.
## What is the GCF of 3 and 20?
Greatest common factor (GCF) of 3 and 20 is 1.
## What is the GCF of 3 15 and 9?
We found the factors and prime factorization of 9 and 15. The biggest common factor number is the GCF number. So the greatest common factor 9 and 15 is 3.
## What is the GCF of 18 and 27?
Example: Find the GCF of 18 and 27 The factors of 18 are 1, 2, 3, 6, 9, 18. The factors of 27 are 1, 3, 9, 27. The common factors of 18 and 27 are 1, 3 and 9. The greatest common factor of 18 and 27 is 9.
## What is the GCF of 3 and 9?
Greatest common factor (GCF) of 3 and 9 is 3.
## What is the GCF of 12 and 18?
In terms of numbers, the greatest common factor (gcf) is the largest natural number that exactly divides two or more given natural numbers. Example 1: 6 is the greatest common factor of 12 and 18.
## What is the GCF of 15?
The prime factors they have in common are 3 x 5. 3 x 5 = 15. The greatest common factor is 15. When two numbers have no prime factors in common, their greatest common factor is 1.
## What is the least common multiple for 9 12 and 15?
The LCM of 9,12,15 9 , 12 , 15 is 2⋅2⋅3⋅3⋅5=180 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 5 = 180 .
## What is the GCF of 27 and 36?
Answer and Explanation: The GCF for 27 and 36 is 9. The greatest common factor (GCF) of two or more numbers is the largest positive integer in common that divides each number… |
Just For Fun
# Hailstone Numbers
In our blog we’ve already covered Kaprekar’s constant, as well as the (not-so) magic number 10. Here’s another cool little trick to sink your teeth into…
First, pick a positive integer (whole number).
Now, we are going to generate a sequence of numbers, starting with our starting number. Each term in the sequence will depend on the previous term.
If the previous term is even, we divide the previous term by 2 to get the next term.
If the previous term is odd, we multiply the previous term by 3 and add 1 to get the next term.
This can be expressed this way:
For the previous term n, the next term of the sequence will be
n ÷ 2 if n is even
3n + 1 if n is odd
Then we repeat.
So let’s do an example.
Example 1
Say I chose my starting number to be 5…
5 is odd, so our next term will be 3 × 5 + 1 = 16.
16 is even, so our next term will be 16 ÷ 2 = 8.
8 is even, so our next term will be 8 ÷ 2 = 4.
4 is even, so our next term will be 4 ÷ 2 = 2.
2 is even, so our next term will be 2 ÷ 2 = 1.
1 is odd, so our next term will be 3 × 1 + 1 = 4.
But we’ve already been here before. We know what’s going to happen next..
4 is even, so our next term will be 4 ÷ 2 = 2.
2 is even, so our next term will be 2 ÷ 2 = 1.
1 is odd, so our next term will be 3 × 1 + 1 = 4.
And the process will just go around in a circle forever..
So our final sequence we have generated is 5, 16, 8, 4, 2, 1, 4, 2, 1, … (and it will continue 4, 2, 1 forever!)
What we have discovered here is that if the sequence reaches 1, the sequence starts repeating on a 4, 2, 1 loop.
Let’s do another example.
Example 2
Say I chose the starting number to be 6…
6 is even, so our next term will be 6 ÷ 2 = 3.
3 is odd, so our next term will be 3 × 3 + 1 = 10.
10 is even, so our next term is 10 ÷ 2 = 5.
We don’t actually need to do any more calculations, because from Example 1, we know exactly what happens after 5…
So our sequence generated in Example 2 will be 6, 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, … (and it will continue 4, 2, 1 forever!). It’s the same as Example 1, but with a few extra numbers added at the start.
What this means is that we now know how the sequence will progress if any of the numbers in the above sequence appears.
This allows us to draw a diagram (I have created the example below) showing the paths that each of the numbers we have encountered so far will follow.
This group of numbers generated are known as hailstone numbers. They are called hailstone numbers because the numbers go up and down like hailstones in a cloud.
We have already found that a number of integers will end up at 1. But the question is – do all positive integers end up at 1? This is called the Collatz conjecture, and is one of the longest-running currently unsolved mathematical problems. The conjecture is named after German mathematician Lothar Collatz after he introduced the problem in 1937.
In mathematics, a conjecture is a principle which is widely expected to be true, but as of yet has not been proven or disproven. In the case of the Collatz conjecture, no positive integer has yet been found that disproves it but there is also no proof that this is true for all positive integers. You might think why can’t we prove it by showing that every positive integer ends up at 1, however that is actually impossible because there are infinitely many positive integers! So the conjecture is in a state of limbo until either someone proves it for all positive integers or somebody finds an example that doesn’t end up at 1. For years, mathematicians have been checking positive integers to try to find the elusive number that makes the Collatz conjecture fall down. As of 2017, all numbers up to 87 × 260 have been checked (that’s a very large number – try not to think about how large it is, it’ll hurt your brain). Checking numbers this large takes immense computer processing power, so is no mean feat.
Another interesting property of hailstone numbers is how some of them end up at 1 very quickly, whereas others take much longer. Take our two examples from earlier – the number 5 fittingly took 5 moves to get to 1, while the number 6 took 8 moves. Is it the case that the larger the number, the more moves it takes to get to 1? Well, no, it isn’t actually. We already know this from the diagram from earlier – the number 16 comes after the number 5 in the diagram, meaning it will take 1 move fewer. Likewise, the number 10 comes after the number 3, meaning that will take 1 move fewer.
In the table below, I have calculated how many moves each number from 1 to 30 takes to get to 1.
You probably notice the anomaly in this table… the number 27 takes 111 moves to reach 1. That’s 88 more moves than the next highest in the table (the number 25 takes 23 moves). Throughout the marathon 111-move journey it reaches a highest number of 9,232 (incidentally proving that 9,232 takes fewer moves than 27, which is bizarre in itself) before eventually making its way down to 1. In fact, 27 takes the longest of any number until 54, which takes only one more move. This should come as no surprise, as if you start at 54, the first move will be to halve it which takes you to 27. After this, you will follow the same path as 27.
This seemingly contrived and meaningless mathematical problem has puzzled and fascinated mathematicians for decades. And more importantly, it’s interesting and a good way of practicing your numerical skills!
As a bonus, find the hailstone sequence for each of the following numbers and how many moves it takes to reach 1 (don’t worry, I’m not giving you any that will take anywhere as many as 27!).
1. 64
2. 42
3. 80
4. 96
5. 92
Hint: you will be able to use shortcuts for some of them – look out for any numbers that you have seen before!
Have a go at them and put your answers in the comments or email them to sam@metatutor.co.uk and I’ll let you know if you got them right.
If you need some help with your numeracy, book a free taster session here.
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# Absolute Value Definition
In mathematics, the absolute value, also known as the modulus of real number, is defined as the non-negative number without any sign (-) used. It is written in the form of | x |. This absolute value is applied to form the zero origins to a digit of the coordinate system. We can use the absolute value on both scalars and vector quantities. A set of upward lines is used to represent absolute value, as shown in brackets (| |). For instance, the absolute value for 3 and -3 is the same |3|. For a more precise understanding, see the below diagram:
## Formula of Absolute Value
For a more precise understanding, let's assume x is a real number. Then the absolute value of x is determined as:
$\left| x \right| = \left\{ {\begin{array}{*{20}{c}} {x|if|x \ge 0}\\ { - x|if|x < 0} \end{array}} \right.$
That is,
|x| = x for a positive x
|x| = −x for a negative x (in 2nd case −x is positive), and |0| = 0.
Note:
To calculate the absolute value, you can use our Absolute Value Calculator.
### Example
For a better understanding, let's have an example below:
Suppose we have an equation having absolute value, for these, let's find the value of x (real number)
Given data
Equation : |5 - 2x| - 11 = 0
To Find
Absolute Value = ?
Solution
Let's solve this equation
|5 - 2x| = -11
Now expanding the absolute value as per the above-given formula:
5 - 2x = 11 , 5 - 2x = -11
-2x = 6 , -2x = -16
x = -3 , x = 8
Let's have a simple here, Find the absolute value for -|2| and |-2|.
As we know:
|x| = −x for a negative x (in which case −x is positive)
And,
|0| = 0.
So,
-|2| = -2 and |-2| = 2 |
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