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# Evaluate: (i) 3 √36 x 3 √384 (ii) 3 √96 x 3 √144 (iii) 3 √100 x 3 √270
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Evaluate:
(i) $\sqrt[3]{36}\times\sqrt[3]{384}$
(ii) $\sqrt[3]{96}\times\sqrt[3]{144}$
(iii) $\sqrt[3]{100}\times\sqrt[3]{270}$
(iv) $\sqrt[3]{121}\times\sqrt[3]{297}$
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(i) $\sqrt[3]{36}\times\sqrt[3]{384}$
We have,
$\sqrt[3]{36}\times\sqrt[3]{384}$ = $\sqrt[3]{36\times384}$
∴ $(\sqrt[3]{a}$$\sqrt[3]{b}$ = $\sqrt[3]{ab})$
Now by prime factorization method,
$\sqrt[3]{36\times384}$
$2\times2\times2\times3 = 24.$
(ii) $\sqrt[3]{96}\times\sqrt[3]{144}$
We have,
$\sqrt[3]{96}$ x $\sqrt[3]{122}$ = $\sqrt[3]{96\times122}$
$\because$ $(\sqrt[3]{a}$ x $\sqrt[3]{b}$ = $\sqrt[3]{ab}$)
Now by prime factorization method,
$\sqrt[3]{96\times122}$
$2\times2\times3 = 24.$
(iii) $\sqrt[3]{100}\times\sqrt[3]{270}$
We have,
$\sqrt[3]{100}\times\sqrt[3]{270}$ = $\sqrt[3]{100\times270}$
$\because$ $(\sqrt[3]{a}$$\sqrt[3]{b}$ = $\sqrt[3]{ab})$
Now by prime factorization method,
$\sqrt[3]{100}\times\sqrt[3]{270}$
$2\times3\times5 = 30.$
(iv) $\sqrt[3]{121}\times\sqrt[3]{297}$
$\sqrt[3]{121}\times\sqrt[3]{297}$ = $\sqrt[3]{121\times297}$
$\because$ $(\sqrt[3]{a}$$\sqrt[3]{b}$ = $\sqrt[3]{ab})$
Now by prime factorization method,
$\sqrt[3]{121\times297}$ = $\sqrt[3]{{(11\times11)}\times(3\times\times3\times3\times11})$
$\sqrt[3]{11^3}$ x $\sqrt[3]{3^3}$
$11\times3 = 33.$ |
# Mathematical Models of Thermal Systems
## Introduction
While the previous page (System Elements) introduced the fundamental elements of thermal systems, as well as their mathematical models, no systems were discussed. This page discusses how the system elements can be included in larger systems, and how a system model can be developed. The actual solution of such models is discussed elsewhere.
## The Energy Balance
To develop a mathematical model of a thermal system we use the concept of an energy balance. The energy balance equation simply states that at any given location, or node, in a system, the heat into that node is equal to the heat out of the node plus any heat that is stored (heat is stored as increased temperature in thermal capacitances).
Heat in = Heat out + Heat stored
To better understand how this works in practice it is useful to consider several examples.
## Examples Involving only Thermal Resistance and Capacitance
##### Example: Two thermal resistances in series
Consider a situation in which we have an internal temperature, θi, and an ambient temperature, θa with two resistances between them. An example of such a situation is your body. There is a (nearly) constant internal temperature, there is a thermal resistance between your core and your skin (at θs), and there is a thermal resistance between the skin and ambient. We will call the resistance between the internal temperature and the skin temperature Ris, and the temperature between skin and ambient Rsa.
a) Draw a thermal model of the system showing all relevant quantities.
b) Draw an electrical equivalent
c) Develop a mathematical model (i.e., an energy balance).
d) Solve for the temperature of the skin if θi, =37°C, θa =9°C, Ris=0.75°/W; for a patch of skin and Rsa= 2.25°/W for that same patch.
Solution:
a) In this case there are no thermal capacitances or heat sources, just two know temperatures ( θi, and θa), one unknown temperature (θs), and two resistances ( Risand Rsa.)
b) Temperatures are drawn as voltage sources. Ambient temperature is taken to be zero (i.e., a ground "temperature), all other temperatures are measured with respect to this temperature).
c) There is only one unknown temperature (at θs), so we need only one energy balance (and, since there is no capacitance, we don't need the heat stored term).
Note: the first equation included θa, but the second does not, since θa is our reference temperature and is taken to be zero.
d) Solving for θs
Note: you may recognize this result as the voltage divider equation from electrical circuits.
We can now solve numerically (we use 28°C for the internal temperature since it is 28°C above ambient (37°-9°=28°)
This says that θs is 21°C above ambient. Since the ambient temperature is 9°C, the actual skin temperature is 30°C.
Note: If Rsa is lowered, for example by the wind blowing, the skin gets cooler, and it feels like it is colder. This is the mechanism responsible for the "wind chill" effect.
##### Example: Heating a Building with One Room
Consider a building with a single room. The resistance of the walls between the room and the ambient is Rra, and the thermal capacitance of the room is Cr, the heat into the room is qi, the temperature of the room is θr, and the external temperature is a constant, θa.
a) Draw a thermal model of the system showing all relevant quantities.
b) Draw an electrical equivalent
c) Develop a mathematical model (i.e., a differential equation).
Solution:
a) We draw a thermal capacitance to represent the room (and note its temperarature). We also draw a resistance between the capacitance and ambient.
b) To draw the electrical system we need a circuit with a node for the ambient temperature, and a node for the temperature of the room. Heat (a current source) goes into the room. Energy is stored (as an increased temperature) in the thermal capacitance, and heat flows from the room to ambient through the resistor.
c) We only need to develop a single energy balance equation, and that is for the temperature of the thermal capacitance (since there is only one unknown temperature). The heat into the room is qi, heat leaves the room through a resistor and energy is stored (as increased temperature) in the capacitor.
by convention we take the ambient temperature to be zero, so we end up with a first order differential equation for this system.
##### Example: Heating a Building with One Room, but with Variable External Temperature.
Consider the room from the previous example. Repeat parts a, b, and c if the temperature outside is no longer constant but varies. Call the external temperature θe(t) (this will be the temperature relative to the ambient temperature). We will also change the name of the resistance of the walls to Rre to denote the fact that the external temperature is no longer the ambient temperature.
Solution:
The solution is much like that for the previous example. Exceptions are noted below.
a) The image is as before with the external temperature replaced by θe(t).
b) To draw the electrical system we need a circuit with a node for the external temperature and a node for the temperature of the room. Though perhaps not obvious at first we still need a node for the ambient temperature since all of our temperatures are measured relative to this, and our capacitors must always have one node connected to this reference temperature. Heat flows from the room to the external temperature through the resistor.
c) We still only need to develop a single energy balance equation, and that is for the temperature of the thermal capacitance (since there is only one unknown temperature). The heat into the room is qi, heat leaves the room through a resistor and energy is stored (as increased temperature) in the capacitor.
(the ambient temperature is taken to be zero in this equation). In this case we end up with a system with two inputs (qi and θe).
##### Example: Heating a Building with Two Rooms
Consider a building that consists of two adjacent rooms, labeled 1 and 2. The resistance of the walls room 1 and ambient is R1a, between room 2 and ambient is R2a and between room 1 and room 2 is R12. The capacitance of rooms 1 and 2 are C1 and C2, with temperatures θ1 and θ2, respectively. A heater in in room 1 generates a heat qin. The temperaturexternal temperature is a constant, θa.
a) Draw a thermal model of the system showing all relevant quantities.
b) Draw an electrical equivalent
c) Develop a mathematical model (i.e., a differential equation).
In this case there are two unknown temperatures, θ1and θ2, so we need two energy balance equations. In both cases we will take θa to be zero, so it will not arise in the equations.
Room 1: Heat in = Heat out + Heat Stored Room 2: Heat in = Heat out + Heat Stored In this case there are two parts to the "Heat Out"term, the heat flowing through R1a and the heatthrough R12. In this case we take heat flow through R12 to (from 1 to 2) to be an input. We could also take this energy balance to have no heat in, and write the heat flow from2 to 1 as a second "Heat out" term. (note thechange of subscripts in the subtracted terms)
The two first order energy balance equations (for room 1 and room 2) could be combined into a single second order differential equation and solved. Details about developing the second order equation are here.
## Examples Involving Fluid Flow
So far we have not considered fluid flow in any of the examples; let us do so now.
##### Example: Cooling a Block of Metal in a Tank with Fluid Flow.
Consider a block of metal (capacitance=Cm, temperature=θm). It is placed in a well mixed tank (at termperature θt, with capacitance Ct). Fluid flows into the tank at temperature θin with mass flow rate Gin, and specific heat cp. The fluid flows out at the same rate There is a thermal resistance to between the metal block and the fluid of the tank, Rmt, and between the tank and the ambient Rta. Write an energy balance for this system.
Note: the resistance between the tank and the metal block, Rmt, is not explicitly shown.
Solution:
Since there are two unknown temperatures, we need two energy balance equations.
Metal Block: Heat in = Heat out + Heat Stored Tank: Heat in = Heat out + Heat Stored In this case there is not heat in, and heat outis to the tank through Rmt. In this case we have heat in from the fluid flowand from the metal block.We have heat out to ambient through Rta.
##### Aside: Modeling a Fluid Flow with and Electrical Analog
To model this system with an electrical analog, we can represent the fluid flow as a voltage source at θin, with a resistance equal to 1/(Gin·cp). If you sum currents at the nodes θt and θm you can show that this circuit is equivalent to the thermal system above.
## Solving the Model
Thus far we have only developed the differential equations that represent a system. To solve the system, the model must be put into a more useful mathematical representation such as transfer function or state space. Details about developing the mathematical representation are here.
References |
# Texas Go Math Grade 4 Lesson 11.4 Answer Key Multi-Step Division Problems
Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 11.4 Answer Key Multi-Step Division Problems.
## Texas Go Math Grade 4 Lesson 11.4 Answer Key Multi-Step Division Problems
Essential Question
How can you use the strategy draw a diagram to solve multi-step division problems?
In solving multi-step division problems, the total to be found first,
Then the total should be divided in to the required equal parts.
Unlock the Problem
Lucia and her dad will prepare corn for a community picnic. There are 3 bags of corn. Each bag holds 32 ears of corn. When the corn is cooked, they want to divide the corn equally among B serving plates. How many ears of corn should they put on each of 8 serving plates?
12 ears of corn.
Explanation:
There are 3 bags of corn.
Each bag holds 32 ears of corn.
3 x 32 = 96
When the corn is cooked, they want to divide the corn equally among 8 serving plates.
Total ears of corn should they put on each of 8 serving plates
96÷8 = 12
What do I need to find?
I need to find the number of ________________ that will go on each plate.
Ear corns
What information am I given?
______________ bags with ______________ ears in each bag. The total ears arc divided equally into ______________ groups.
3 bags with 32 ears in each bag.
8 groups.
Plan
What is my plan or strategy?
I will make a strip diagram for each step and use equations. Then I will ______________ to find the total and ___________ to find the number for each plate.
I will multiply 32 x3 to find the total and divide 96 to find the number for each plate.
Solve
I can draw strip diagrams and use equations and then decide how to find how many ears of corn should go on a plate.
First, I will find the total number of ears of corn.
32 × __________ = e
___________ = e
Then I will find how many ears of corn should go on each plate.
96 ÷ __________ = c
___________ = c
96 ÷ 8 = c
12 = c
Question 1.
How many ears of corn should go on each plate?
12 ears of corn
Question 2.
32 × 3 = e
96 = e
96 ÷ 8 = c
12 = c
Explanation:
Division is opposite to multiplication and multiplication is opposite to division vice-versa.
Try Another Problem
There are 8 dinner rolls in a package. How many packages will be needed to feed 64 people if each person has 2 dinner rolls?
What do I need to find?
Total packages will be needed to feed 64 people if each person has 2 dinner rolls.
What information am I given?
Each package contains 8 dinner rolls.
Total number of people to feed 64.
Plan
What is my plan or strategy?
I will make a strip diagram for each step and use equations.
Solve
64 x 2 = 128 dinner rolls required
128 ÷ 8 = p
16 = p
Question 3.
How many packages of rolls will he needed?
16 packages of rolls needed for 64 people for dinner
Question 4.
Strip Drawing helps to break the whole into small parts to slove.
Math Talk
Mathematical Processes
Describe another method you could have used to solve the problem.
Equation method
Explanation:
64 x 2 =128
128 ÷ 8 = 16
Share and Show
Question 1.
A firehouse pantry has 52 cans of vegetables and 74 cans of soup. Each shelf holds 9 cans. What is the the least number of shelves needed for all the cans?
First, draw a strip diagram for the total number of cans.
Next, add to find the total number of cans.
Then, draw a strip diagram to show the number of shelves needed.
Finally, divide to find the number of shelves needed.
So, _________ Shelves are needed to hold all of the cans.
14 shelves are needed to hold all of the cans.
Explanation:
A firehouse pantry has 52 cans of vegetables and 74 cans of soup.
Each shelf holds 9 cans.
The least number of shelves needed for all the cans
(52 + 74) ÷ 9 = n
126 ÷ 9 = n
14 = n
Question 2.
H.O.T. Multi-Step What if 18 cans fit on a shelf? What is the least number of shelves needed? Describe how your answer would be different.
7 shelves are needed to hold all of the cans
Explanation:
A firehouse pantry has 52 cans of vegetables and 74 cans of soup.
Each shelf holds 18 cans.
The least number of shelves needed for all the cans
(52 + 74) ÷ 18 = n
126 ÷ 18 = n
7 = n
Problem-Solving
Question 3.
H.O.T. Multi-Step Ms. Johnson bought 6 bags of balloons. Each hag has 25 balloons. She fills all the balloons and puts 5 balloons in each bunch. How many bunches can she make?
30 bunches.
Explanation:
Ms. Johnson bought 6 bags of balloons.
Each bag has 25 balloons.
She fills all the balloons and puts 5 balloons in each bunch.
Total bunches she can make
(6 x 25) ÷ 5 = n
150 ÷ 5 = n
30 = n
Question 4.
Basketball jerseys are shipped in packages of 6. How many packages of jerseys are needed for 12 players if each player gets 4 jerseys?
(A) 3 packages
(B) 2 packages
(C) 8 packages
(D) 48 packages
Option(B)
Explanation:
Basketball jerseys are shipped in packages of 6,
if each player gets 4 jerseys.
Total packages of jerseys are needed for 12 players, if each player gets 4 jerseys
(6 x 4) ÷ 12 = n
24 ÷ 12 = n
2 = n
Question 5.
Robin and her grandmother bake muffins. Each batch of hatter makes 36 mini muffins. They make 4 batches. They divide the muffins equally into bags of 3 muffins for a bake sale. How many bags of 3 muffins do they have?
(A) 9 bags
(B) 48 bags
(C) 12 bags
(D) 3 bags
Option(B)
Explanation:
Robin and her grandmother bake muffins.
Each batch of batter makes 36 mini muffins.
They make 4 batches.
They divide the muffins equally into bags of 3 muffins for a bake sale.
Total bags of 3 muffins they have
(36 x 4) ÷ 3 = n
144 ÷ 3 = n
48 = n
Question 6.
Multi-Step The fourth graders at Sunshine School go to the Museum of Nature and Science. The school pays $671 for the trip. The adult tickets are$10 each and the student tickets are $7 each. There are 9 adults going on the trip. How many students go on the trip? (A) 74 students (B) 86 students (C) 95 students (D) 83 students Answer: Option(D) Explanation: The school pays$671 for the trip.
The adult tickets are $10 each and the student tickets are$7 each.
There are 9 adults going on the trip.
Total students go on the trip
671 – (10 x 9) ÷ 7 = n
90 – 671 ÷ 7= n
581 ÷ 7 = n
83 = n
TEXAS Test Prep
Question 7.
Ben collected 43 cans and some bottles. He received 5 for each can or bottle. If Ben received a total of $4.95, how many bottles did he collect? (A) 56 (B) 99 (C) 560 (D) 990 Answer: Option(A) Explanation: Ben collected 43 cans and some bottles. He received 5 for each can or bottle. If Ben received a total of$4.95,
Number of bottles he collect
(43 + x) x 0.0 5 = 2.15 + 0.05x
4.95 = 2.15 + 0.05x
4.95 – 2.15 = 0.05x
2.80 = 0.05x
2.80 ÷ 0.05 = x
x = 56
#### Texas Go Math Grade 4 Lesson 11.4 Homework and Practice Answer Key
Problem Solving
Question 1.
Marco bought 2 bottles of juice. Each bottle is 48 ounces. How many 8-ounce glasses of juice can Marco pour from the two bottles?
12 glasses.
Explanation:
48 + 48 = 96 ounces
96 ÷ 8 = 12 glasses
a. Draw a strip diagram for the number of ounces of juice in the two bottles.
b. Write an equation to find the total number of ounces of juice in the two bottles.
n = 2 x 48 = 96 ounces
c. Draw a strip diagram to show the number of 8-ounce glasses of juice that can be poured.
d. Write an equation to find the number of glasses of juice.
Marco can pour ________ glasses of juice.
12 glasses.
Explanation:
x = 96 ÷ 8 = 12 glasses
Marco can pour 12 glasses of juice
Question 2.
Describe another method you could have used to solve the problem.
Algebraic equation method
n=48 x 2 = 96
x = 96 ÷ 8 = 12
Question 3.
What if Marco poured 10-ounce glasses of juice? What is the greatest number of full glasses of juice he could have poured? Explain.
x = 96 ÷ 10 = 9.6
Lesson Check
Question 4.
Multi-Step Orlando has a bag of 37 apples and a bag of 29 apples. He can hake 6 apples in a pan. How many pans of apples can Orlando make?
(A) 66
(B) 33
(C) 11
(D) 22
Option(C)
Explanation:
Orlando has a bag of 37 apples and a bag of 29 apples.
He can bake 6 apples in a pan.
Total pans of apples Orlando can make
(37 + 29) ÷ 6 = n
66 ÷ 6 = n
11 = n
Question 5.
Multi-Step Anna has 5 bunches of flowers with 12 flowers in each bunch. She has 4 bunches of flowers with 10 flowers in each bunch. How many vases can Anna fill if she puts 10 flowers in each vase?
(A) 10
(B) 100
(C) 9
(D) 15
Option(A)
Explanation:
Anna has 5 bunches of flowers with 12 flowers in each bunch.
She has 4 bunches of flowers with 10 flowers in each bunch.
Total vases Anna can fill if she puts 10 flowers in each vase
(5 x 12) + (4 x 10) ÷ 10 = n
60 + 40 ÷ 10 = n
100 ÷ 10 = n
10 = n
Question 6.
Multi-Step Five friends are going to share the cost of two gifts. One gift costs $39 and the other gift costs$26. What is each person’s share of the cost?
(A) $7 (B)$15
(C) $18 (D)$13
Option(D)
Explanation:
Five friends are going to share the cost of two gifts.
One gift costs $39 and the other gift costs$26.
Each person’s share of the cost
39 + 26 ÷ 5 = n
65 ÷ 5 = n
13 = n
Question 7.
Multi-Step There are 14 chairs in each of 6 rows. There are 18 chairs in each of 4 rows. How many rows of 13 chairs can be made from all of the chairs?
(A) 13
(B) 12
(C) 14
(D) 18
Option(B)
Explanation:
There are 14 chairs in each of 6 rows.
There are 18 chairs in each of 4 rows.
Total rows of 13 chairs can be made from all of the chairs
(14 x 6) + (18 x 4) ÷ 13 = n
(84 + 72 ) ÷ 13 = n
156 ÷ 13 = n
12 = n
Question 8.
Multi-Step Justin collected 26 shells, Amy collected 31 shells. Jose collected 21 shells, If they share all of the shells equally, how many shells will each person get?
(A) 24
(B) 78
(C) 26
(D) 19
Option(C)
Explanation:
Justin collected 26 shells,
Amy collected 31 shells.
Jose collected 21 shells,
If they share all of the shells equally,
Number of shells will each person get
(26 + 31 + 21) ÷ 3 = n
78 ÷ 3 = n
26 = n
Question 9.
Multi-Step Taylor has 2 packages of 36 tacks each and 16 tacks. How many garage sale posters can she put up if she uses 4 tacks for each poster?
(A) 13
(B) 22
(C) 17
(D) 21 |
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# 8.1: Geometric Series
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Having a detailed understanding of geometric series will enable us to use Cauchy’s integral formula to understand power series representations of analytic functions. We start with the definition:
##### Definition: Finite Geometric Series
A finite geometric series has one of the following (all equivalent) forms.
\begin{align} S_n &= a(1 + r + r^2 + r^3 + ... + r^n) \\[4pt] &= a + ar + ar^2 + ar^3 + ... + ar^n \\[4pt] &= \sum_{j = 0}^{n} ar^j \\[4pt] &= a \sum_{j = 0}^{n} r^j \end{align}
The number $$r$$ is called the ratio of the geometric series because it is the ratio of consecutive terms of the series.
##### Theorem $$\PageIndex{1}$$
The sum of a finite geometric series is given by
$S_n = a(1 + r + r^2 + r^3 + ... + r^n) = \dfrac{a (1 - r^{n + 1})}{1 - r}.$
Proof
This is a standard trick that you’ve probably seen before.
$\begin{array} {rclcc} {S_n} & = & {a + } & { ar + ar^2 + ... + ar^n} & {} \\ {rS_n} & = & { } & {ar + ar^2 + ... + ar^n} & {+ ar^{n + 1}} \end{array}$
When we subtract these two equations most terms cancel and we get
$S_n - rS_n = a - ar^{n + 1}$
Some simple algebra now gives us the formula in Equation 8.2.2.
##### Definition: Infinite Geometric Series
An infinite geometric series has the same form as the finite geometric series except there is no last term:
$S = a + ar + ar^2 + ... = a \sum_{j = 0}^{\infty} r^j.$
##### Note
We will usually simply say ‘geometric series’ instead of ‘infinite geometric series’.
##### Theorem $$\PageIndex{2}$$
If $$|r| < 1$$ then the infinite geometric series converges to
$S = a \sum_{j = 0}^{\infty} r^j = \dfrac{a}{1 - r}$
If $$|r| \ge 1$$ then the series does not converge.
Proof
This is an easy consequence of the formula for the sum of a finite geometric series. Simply let $$n \to \infty$$ in Equation 8.2.2.
##### Note
We have assumed a familiarity with convergence of infinite series. We will go over this in more detail in the appendix to this topic.
## Connection to Cauchy’s Integral Formula
Cauchy’s integral formula says
$f(z) = \dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{w - z} \ dw.$
Inside the integral we have the expression
$\dfrac{1}{w - z}$
which looks a lot like the sum of a geometric series. We will make frequent use of the following manipulations of this expression.
$\dfrac{1}{w - z} = \dfrac{1}{w} \cdot \dfrac{1}{1 - z/w} = \dfrac{1}{w} (1 + (z/w) + (z/w)^2 + ...)$
The geometric series in this equation has ratio $$z/w$$. Therefore, the series converges, i.e. the formula is valid, whenever $$|z/w| < 1$$, or equivalently when
$|z| < |w|.$
Similarly,
$\dfrac{1}{w - z} = -\dfrac{1}{z} \cdot \dfrac{1}{1 - w/z} = - \dfrac{1}{z} (1 + (w/z) + (w/z)^2 + ...)$
The series converges, i.e. the formula is valid, whenever $$|w/z| < 1$$, or equivalently when
$|z| > |w|.$
8.1: Geometric Series is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. |
Notes On Rotational Symmetry - CBSE Class 7 Maths
Any object or shape is said to have rotational symmetry if it looks exactly the same at least once during a complete rotation through 360°. Centre of rotation The fixed point about which an object rotates is called the centre of rotation. During the rotation, the shape and size of the object do not change. Rotation may be clockwise or anti-clockwise. A full turn refers to a rotation of 360°. A half turn refers to a rotation of 180°. A quarter turn refers to a rotation of 90°. Angle of rotation The angle at which a shape or an object looks exactly the same during rotation is called the angle of rotation. Order of rotational symmetry The order of rotational symmetry can be defined as the number of times that a shape appears exactly the same during a full 360o rotation. The centre of rotation of a square is the point of intersection of its diagonals and the angle of rotation is 90°. So, the order of rotational symmetry of a square is 4. The centre of rotation of a circle is the centre of the circle and it has rotational symmetry around the centre for every angle. There are many shapes that have only line symmetry and no rotational symmetry at all. Some objects and shapes have both, line symmetry as well as rotational symmetry. The Ashok Chakra in the Indian national flag has both, line symmetry and rotational symmetry. Symmetry can be seen in the English alphabet as well. The letter H has both line symmetry and rotational symmetry. Letter Line Symmetry Rotational Symmetry Z No Yes H Yes Yes O Yes Yes E Yes No N No Yes C Yes No A Yes No B Yes No
#### Summary
Any object or shape is said to have rotational symmetry if it looks exactly the same at least once during a complete rotation through 360°. Centre of rotation The fixed point about which an object rotates is called the centre of rotation. During the rotation, the shape and size of the object do not change. Rotation may be clockwise or anti-clockwise. A full turn refers to a rotation of 360°. A half turn refers to a rotation of 180°. A quarter turn refers to a rotation of 90°. Angle of rotation The angle at which a shape or an object looks exactly the same during rotation is called the angle of rotation. Order of rotational symmetry The order of rotational symmetry can be defined as the number of times that a shape appears exactly the same during a full 360o rotation. The centre of rotation of a square is the point of intersection of its diagonals and the angle of rotation is 90°. So, the order of rotational symmetry of a square is 4. The centre of rotation of a circle is the centre of the circle and it has rotational symmetry around the centre for every angle. There are many shapes that have only line symmetry and no rotational symmetry at all. Some objects and shapes have both, line symmetry as well as rotational symmetry. The Ashok Chakra in the Indian national flag has both, line symmetry and rotational symmetry. Symmetry can be seen in the English alphabet as well. The letter H has both line symmetry and rotational symmetry. Letter Line Symmetry Rotational Symmetry Z No Yes H Yes Yes O Yes Yes E Yes No N No Yes C Yes No A Yes No B Yes No
Next
➤ |
# Is f(x) =e^xsinx-cosx concave or convex at x=pi/3?
Mar 31, 2016
Concave up (my naming convention is different to yours, will explain further in explanation)
#### Explanation:
Obtain $f ' \left(x\right)$ by differentiating the original function (product rule):
$f ' \left(x\right) = {e}^{x} \sin \left(x\right) + {e}^{x} \cos \left(x\right) + \sin \left(x\right)$
Obtain $f ' ' \left(x\right)$ by differentiating again:
$f ' ' \left(x\right) = {e}^{x} \sin \left(x\right) + {e}^{x} \cos \left(x\right) + {e}^{x} \cos \left(x\right) - {e}^{x} \sin \left(x\right) + \cos \left(x\right) = 2 {e}^{x} \cos \left(x\right) + \cos \left(x\right)$
$= \left(2 {e}^{x} + 1\right) \cos \left(x\right)$
Evaluate $f ' ' \left(\frac{\pi}{3}\right)$:
$f ' ' \left(\frac{\pi}{3}\right) = \left(2 {e}^{\frac{\pi}{3}} + 1\right) \cos \left(\frac{\pi}{3}\right)$
$= \frac{1}{2} \left(2 {e}^{\frac{\pi}{3}} + 1\right)$
Clearly $f ' ' \left(\frac{\pi}{3}\right) > 0$, so the function will be concave up at $x = \frac{\pi}{3}$
Note that the naming convention I learnt was concave up and concave down, so I am not sure which is concave or convex in your terms.
Graphing the original fucntion is a good way to check your answer, at $x = \frac{\pi}{3} \approx 1$, we see that the rate of change of gradient is indeed positive and therefore concave up.
graph{e^x*sinx-cosx [-3, 3, -5, 5]} |
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### MEAP Preparation - Grade 7 Mathematics1.80 Decimals Review - II
Basic operations on decimals are performed in the same way as basic operations are performed on the whole numbers. To add or subtract decimals, we align the decimal points of given numbers and perform the operation. In the sum or difference, we mark decimal point in the column of decimal points. If the decimals are of different decimal places, we convert them into same decimal places by putting zeros at the right of the decimals wherever necessary. To multiply two decimals, we multiply them as whole numbers ignoring the decimal points. We insert the decimal point in the product by counting as many places from right to left as the sum of number of decimal places of the given decimals. To multiply a decimal by 10, 100 or 1000, we simply shift the decimal point in the given number from left to right. one place, if it is multiplied by 10 two places, if it is multiplied by 100 three places, if it is multiplied by 1000 and so on. In division of decimals we proceed as: Decimal ÷ Whole Number (other than zero): We first divide the whole number part of the dividend and then move towards decimal part of dividend. Correspondingly, we mark decimal point in the quotient also and then continue division as usual. Whole Number ÷ Decimal: We first remove the decimal point from the denominator by multiplying both numerator and denominator with 10, 100 or 1000 etc. We perform the operation of division as usual. Decimal ÷ Decimal: We first remove the decimal point from the denominator and perform division operation as usual. Whole Number ÷ Whole Number (other than zero): When the divisor is greater than the dividend we mark a decimal point in both the quotient and dividend and continue the division. Decimal ÷ 10 or 100 or 1000: We simply move the decimal point in the dividend to the left by: one place if it is divided by 10 two places if it is divided by 100 three places if it is divided by 1000 and so on. Directions: Answer the following questions. Also write at least 5 examples of your own for each of decimal operations.
Q 1: 9.1 X 21.79 = _____ ?Answer: Q 2: Divide 29.7 by 10 ?Answer: Q 3: Divide 89.6 by 3.5 ?Answer: Q 4: Divide 13.992 by 5.3 ?Answer: Q 5: Divide 30.72 by 12 ?Answer: Q 6: Convert 72.81 into fraction ?Answer: Q 7: 29.76 X 10 = _____ ?Answer: Q 8: 76.3 X 21 = _____ ?Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! |
# 2001 AMC 12 Problems/Problem 6
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The following problem is from both the 2001 AMC 12 #6 and 2001 AMC 10 #13, so both problems redirect to this page.
## Problem
A telephone number has the form $\text{ABC-DEF-GHIJ}$, where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, $A > B > C$, $D > E > F$, and $G > H > I > J$. Furthermore, $D$, $E$, and $F$ are consecutive even digits; $G$, $H$, $I$, and $J$ are consecutive odd digits; and $A + B + C = 9$. Find $A$.
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$
## Solution
We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.
Case 1: $G$, $H$, $I$, and $J$ are $7$, $5$, $3$, and $1$ respectively.
A cursory glance allows us to deduce that the smallest possible sum of $A + B + C$ is $11$ when $D$, $E$, and $F$ are $8$, $6$, and $4$ respectively, so this is out of the question.
Case 2: $G$, $H$, $I$, and $J$ are $3$, $5$, $7$, and $9$ respectively.
A cursory glance allows us to deduce the answer. Clearly, when $D$, $E$, and $F$ are $6$, $4$, and $2$ respectively, $A + B + C$ is $9$ when $A$, $B$, and $C$ are $8$, $1$, and $0$ respectively, giving us a final answer of $\boxed{\textbf{(E)}\ 8}$
## See Also
2001 AMC 12 (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
2001 AMC 10 (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# How to solve 12(4-2)+1/2-5/6=x+2
Welcome to my article How to solve 12(4-2)+1/2-5/6=x+2. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve 12(4-2)+1/2-5/6=x+2, read and understand it carefully till the end.
Let us know how to solve this question How to solve 12(4-2)+1/2-5/6=x+2.
First write the question on the page of the notebook.
## How to solve 12(4-2)+1/2-5/6=x+2
Let us first write this question in this way,
\displaystyle 12\left( {4-2} \right)+\frac{1}{2}-\frac{5}{6}=x+2
\displaystyle 12\left( 2 \right)+\frac{1}{2}-\frac{5}{6}=x+2
\displaystyle 24+\frac{1}{2}-\frac{5}{6}=x+2
\displaystyle \frac{{24}}{1}+\frac{1}{2}-\frac{5}{6}-2=x
\displaystyle \frac{{24\times 2}}{{1\times 2}}+\frac{1}{2}-\frac{5}{6}-2=x
\displaystyle \frac{{48}}{2}+\frac{1}{2}-\frac{5}{6}-2=x
\displaystyle \frac{{48+1}}{2}-\frac{5}{6}-\frac{2}{1}=x
\displaystyle \frac{{49}}{2}-\frac{5}{6}-\frac{2}{1}=x
\displaystyle \frac{{49\times 3}}{{2\times 3}}-\frac{5}{6}-\frac{{2\times 6}}{{1\times 6}}=x
\displaystyle \frac{{147}}{6}-\frac{5}{6}-\frac{{12}}{6}=x
\displaystyle \frac{{147-5-12}}{6}=x
\displaystyle \frac{{147-17}}{6}=x
\displaystyle \frac{{130}}{6}=x
or
x=21.666666.. |
# 3.01 Parallel lines and transversals
Lesson
A line that intersects two (or more) other lines is called a transversal. Angles formed on the same side of a transversal are called consecutive angles.
In the applet below, $f$f is a transversal of $g$g and $h$h. Moving the slider changes the orientation of $g$g. Move the slider in the applet below and notice how the measures of the highlighted angles change:
When $g$g and $h$h are parallel, the measure of the interior angles add to $180^\circ$180°. When they aren't parallel, $g$g intersects $h$h forming a third angle, and adding the measures of the two interior angles and this third angle always makes $180^\circ$180°.
### Consecutive angles
Let's use the applet below to explore the different pairs of consecutive angles formed by parallel lines. Notice that angles between the parallel lines are called interior angles, and angles outside the parallel lines are exterior angles
Formally, we can say that if two lines are parallel, then consecutive interior angles are supplementary. We refer to this as the consecutive interior angles postulate. Using properties of linear pairs allows us to prove that consecutive exterior angles are also supplementary, and this is referred to as the consecutive exterior angles theorem.
Postulate or theorem?
The consecutive interior angles postulate is assumed to be true, so we call it a postulate. Other results, such as the consecutive exterior angle theorem, are derived from the postulates and other results, so we call it a theorem.
### Alternate angles
If we look at a pair of angles on opposite sides of the transversal, we can refer to them as alternate angles. Alternate angles between the parallel lines are alternate interior angles, while alternate angles outside the parallel lines are alternate exterior angles.
Use the applet to explore the different pairs of alternate angles. What do you notice about their values?
We can use properties of linear pairs and vertical angles to prove that if two lines are parallel, then pairs of alternate interior angles are congruent - this is called the alternate interior angles theorem. It's also true that pairs of alternate exterior angles are congruent - this is called the alternate exterior angles theorem.
### Corresponding angles
Corresponding angles are the angles that occupy the same relative position at each intersection. We can slide the slider in the applet below to see all four pairs of corresponding angles. What relationship do these angles have?
Using similar results to those we discussed above, we can prove that if two lines are parallel, then the corresponding angles are congruent - this is called the corresponding angles theorem.
Angles formed by a transversal to two parallel lines
If two lines are parallel, then:
1. Consecutive interior angles are supplementary (the consecutive interior angles postulate)
2. Consecutive exterior angles are supplementary (the consecutive exterior angles theorem)
3. Alternate interior angles are congruent (the alternate interior angles theorem)
4. Alternate exterior angles are congruent (the alternate exterior angles theorem)
5. Corresponding angles are congruent (the corresponding angles theorem)
All angle relationships summarized in a diagram
#### Practice questions
##### Question 1
Consider the diagram below.
1. Which relationship describes the marked angles?
Alternate exterior angles
A
Alternate interior angles
B
Consecutive interior angles
C
Consecutive exterior angles
D
Corresponding angles
E
Alternate exterior angles
A
Alternate interior angles
B
Consecutive interior angles
C
Consecutive exterior angles
D
Corresponding angles
E
##### Question 2
Consider the diagram below.
1. Solve for $x$x.
##### Question 3
Consider the diagram below.
1. Solve for $p$p.
2. Solve for $q$q.
### Outcomes
#### G.G-CO.A.1
Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc. |
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# 4.3 The substitution method for solving recurrences
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4.3 The substitution method for solving recurrences 83
4.2-6
How quickly can you multiply a kn!n matrix by an n!kn matrix, using Strassen’s algorithm as a subroutine? Answer the same question with the order of the input matrices reversed.
4.2-7
Show how to multiply the complex numbers a C bi and c C di using only three multiplications of real numbers. The algorithm should take a, b, c, and d as input and produce the real component ac " bd and the imaginary component ad C bc separately.
4.3 The substitution method for solving recurrences
Now that we have seen how recurrences characterize the running times of divide- and-conquer algorithms, we will learn how to solve recurrences. We start in this section with the “substitution” method.
The substitution method for solving recurrences comprises two steps:
1. Guess the form of the solution.
2. Use mathematical induction to find the constants and show that the solution works.
We substitute the guessed solution for the function when applying the inductive hypothesis to smaller values; hence the name “substitution method.” This method is powerful, but we must be able to guess the form of the answer in order to apply it.
We can use the substitution method to establish either upper or lower bounds on a recurrence. As an example, let us determine an upper bound on the recurrence
T .n/D 2T .bn=2c/ C n ; (4.19)
which is similar to recurrences (4.3) and (4.4). We guess that the solution is T .n/D O.n lg n/. The substitution method requires us to prove that T .n/ # cnlg n for an appropriate choice of the constant c > 0. We start by assuming that this bound holds for all positive m < n, in particular for m D bn=2c, yielding T .bn=2c/ # c bn=2c lg.bn=2c/. Substituting into the recurrence yields
T .n/ # 2.c bn=2c lg.bn=2c// C n
# cn lg.n=2/ C n D cn lg n " cn lg 2 C n D cn lg n " cn C n
# cn lg n ;
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where the last step holds as long as c \$ 1.
Mathematical induction now requires us to show that our solution holds for the boundary conditions. Typically, we do so by showing that the boundary condi- tions are suitable as base cases for the inductive proof. For the recurrence (4.19), we must show that we can choose the constant c large enough so that the bound T .n/# cn lg n works for the boundary conditions as well. This requirement can sometimes lead to problems. Let us assume, for the sake of argument, that T .1/D 1 is the sole boundary condition of the recurrence. Then for n D 1, the bound T .n/ # cn lg n yields T .1/ # c1 lg 1 D 0, which is at odds with T .1/ D 1.
Consequently, the base case of our inductive proof fails to hold.
We can overcome this obstacle in proving an inductive hypothesis for a spe- cific boundary condition with only a little more effort. In the recurrence (4.19), for example, we take advantage of asymptotic notation requiring us only to prove T .n/ # cn lg n for n \$ n0, where n0 is a constant that we get to choose. We keep the troublesome boundary condition T .1/ D 1, but remove it from consid- eration in the inductive proof. We do so by first observing that for n > 3, the recurrence does not depend directly on T .1/. Thus, we can replace T .1/ by T .2/
and T .3/ as the base cases in the inductive proof, letting n0 D 2. Note that we make a distinction between the base case of the recurrence (n D 1) and the base cases of the inductive proof (n D 2 and n D 3). With T .1/ D 1, we derive from the recurrence that T .2/ D 4 and T .3/ D 5. Now we can complete the inductive proof that T .n/ # cn lg n for some constant c \$ 1 by choosing c large enough so that T .2/ # c2 lg 2 and T .3/ # c3 lg 3. As it turns out, any choice of c \$ 2 suffices for the base cases of n D 2 and n D 3 to hold. For most of the recurrences we shall examine, it is straightforward to extend boundary conditions to make the inductive assumption work for small n, and we shall not always explicitly work out the details.
Making a good guess
Unfortunately, there is no general way to guess the correct solutions to recurrences.
Guessing a solution takes experience and, occasionally, creativity. Fortunately, though, you can use some heuristics to help you become a good guesser. You can also use recursion trees, which we shall see in Section 4.4, to generate good guesses.
If a recurrence is similar to one you have seen before, then guessing a similar solution is reasonable. As an example, consider the recurrence
T .n/D 2T .bn=2c C 17/ C n ;
which looks difficult because of the added “17” in the argument to T on the right- hand side. Intuitively, however, this additional term cannot substantially affect the
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4.3 The substitution method for solving recurrences 85
solution to the recurrence. When n is large, the difference between bn=2c and bn=2c C 17 is not that large: both cut n nearly evenly in half. Consequently, we make the guess that T .n/ D O.n lg n/, which you can verify as correct by using the substitution method (see Exercise 4.3-6).
Another way to make a good guess is to prove loose upper and lower bounds on the recurrence and then reduce the range of uncertainty. For example, we might start with a lower bound of T .n/ D !.n/ for the recurrence (4.19), since we have the term n in the recurrence, and we can prove an initial upper bound of T .n/ D O.n2/. Then, we can gradually lower the upper bound and raise the lower bound until we converge on the correct, asymptotically tight solution of T .n/D ‚.n lg n/.
Subtleties
Sometimes you might correctly guess an asymptotic bound on the solution of a recurrence, but somehow the math fails to work out in the induction. The problem frequently turns out to be that the inductive assumption is not strong enough to prove the detailed bound. If you revise the guess by subtracting a lower-order term when you hit such a snag, the math often goes through.
Consider the recurrence
T .n/D T .bn=2c/ C T .dn=2e/ C 1 :
We guess that the solution is T .n/ D O.n/, and we try to show that T .n/ # cn for an appropriate choice of the constant c. Substituting our guess in the recurrence, we obtain
T .n/ # c bn=2c C c dn=2e C 1 D cn C 1 ;
which does not imply T .n/ # cn for any choice of c. We might be tempted to try a larger guess, say T .n/ D O.n2/. Although we can make this larger guess work, our original guess of T .n/ D O.n/ is correct. In order to show that it is correct, however, we must make a stronger inductive hypothesis.
Intuitively, our guess is nearly right: we are off only by the constant 1, a lower-order term. Nevertheless, mathematical induction does not work unless we prove the exact form of the inductive hypothesis. We overcome our difficulty by subtracting a lower-order term from our previous guess. Our new guess is T .n/# cn " d , where d \$ 0 is a constant. We now have
T .n/ # .c bn=2c " d / C .c dn=2e " d / C 1 D cn " 2d C 1
# cn " d ;
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as long as d \$ 1. As before, we must choose the constant c large enough to handle the boundary conditions.
You might find the idea of subtracting a lower-order term counterintuitive. Af- ter all, if the math does not work out, we should increase our guess, right?
Not necessarily! When proving an upper bound by induction, it may actually be more difficult to prove that a weaker upper bound holds, because in order to prove the weaker bound, we must use the same weaker bound inductively in the proof.
In our current example, when the recurrence has more than one recursive term, we get to subtract out the lower-order term of the proposed bound once per recursive term. In the above example, we subtracted out the constant d twice, once for the T .bn=2c/ term and once for the T .dn=2e/ term. We ended up with the inequality T .n/# cn " 2d C 1, and it was easy to find values of d to make cn " 2d C 1 be less than or equal to cn " d.
Avoiding pitfalls
It is easy to err in the use of asymptotic notation. For example, in the recur- rence (4.19) we can falsely “prove” T .n/ D O.n/ by guessing T .n/ # cn and then arguing
T .n/ # 2.c bn=2c/ C n
# cn C n
D O.n/ ; % wrong!!
since c is a constant. The error is that we have not proved the exact form of the inductive hypothesis, that is, that T .n/ # cn. We therefore will explicitly prove that T .n/ # cn when we want to show that T .n/ D O.n/.
Changing variables
Sometimes, a little algebraic manipulation can make an unknown recurrence simi- lar to one you have seen before. As an example, consider the recurrence
T .n/D 2T !"p n˘#
C lg n ;
which looks difficult. We can simplify this recurrence, though, with a change of variables. For convenience, we shall not worry about rounding off values, such aspn, to be integers. Renaming m D lg n yields
T .2m/D 2T .2m=2/C m :
We can now rename S.m/ D T .2m/to produce the new recurrence S.m/D 2S.m=2/ C m ;
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4.3 The substitution method for solving recurrences 87
which is very much like recurrence (4.19). Indeed, this new recurrence has the same solution: S.m/ D O.m lg m/. Changing back from S.m/ to T .n/, we obtain T .n/D T .2m/D S.m/ D O.m lg m/ D O.lg n lg lg n/ :
Exercises 4.3-1
Show that the solution of T .n/ D T .n " 1/ C n is O.n2/. 4.3-2
Show that the solution of T .n/ D T .dn=2e/ C 1 is O.lg n/.
4.3-3
We saw that the solution of T .n/ D 2T .bn=2c/Cn is O.n lg n/. Show that the so- lution of this recurrence is also !.n lg n/. Conclude that the solution is ‚.n lg n/.
4.3-4
Show that by making a different inductive hypothesis, we can overcome the diffi- culty with the boundary condition T .1/ D 1 for recurrence (4.19) without adjusting the boundary conditions for the inductive proof.
4.3-5
Show that ‚.n lg n/ is the solution to the “exact” recurrence (4.3) for merge sort.
4.3-6
Show that the solution to T .n/ D 2T .bn=2c C 17/ C n is O.n lg n/.
4.3-7
Using the master method in Section 4.5, you can show that the solution to the recurrence T .n/ D 4T .n=3/ C n is T .n/ D ‚.nlog34/. Show that a substitution proof with the assumption T .n/ # cnlog34 fails. Then show how to subtract off a lower-order term to make a substitution proof work.
4.3-8
Using the master method in Section 4.5, you can show that the solution to the recurrence T .n/ D 4T .n=2/ C n2 is T .n/ D ‚.n2/. Show that a substitution proof with the assumption T .n/ # cn2 fails. Then show how to subtract off a lower-order term to make a substitution proof work.
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4.3-9
Solve the recurrence T .n/ D 3T .p
n/C log n by making a change of variables.
Your solution should be asymptotically tight. Do not worry about whether values are integral.
4.4 The recursion-tree method for solving recurrences
Although you can use the substitution method to provide a succinct proof that a solution to a recurrence is correct, you might have trouble coming up with a good guess. Drawing out a recursion tree, as we did in our analysis of the merge sort recurrence in Section 2.3.2, serves as a straightforward way to devise a good guess. In a recursion tree, each node represents the cost of a single subproblem somewhere in the set of recursive function invocations. We sum the costs within each level of the tree to obtain a set of per-level costs, and then we sum all the per-level costs to determine the total cost of all levels of the recursion.
A recursion tree is best used to generate a good guess, which you can then verify by the substitution method. When using a recursion tree to generate a good guess, you can often tolerate a small amount of “sloppiness,” since you will be verifying your guess later on. If you are very careful when drawing out a recursion tree and summing the costs, however, you can use a recursion tree as a direct proof of a solution to a recurrence. In this section, we will use recursion trees to generate good guesses, and in Section 4.6, we will use recursion trees directly to prove the theorem that forms the basis of the master method.
For example, let us see how a recursion tree would provide a good guess for the recurrence T .n/ D 3T .bn=4c/ C ‚.n2/. We start by focusing on finding an upper bound for the solution. Because we know that floors and ceilings usually do not matter when solving recurrences (here’s an example of sloppiness that we can tolerate), we create a recursion tree for the recurrence T .n/ D 3T .n=4/ C cn2, having written out the implied constant coefficient c > 0.
Figure 4.5 shows how we derive the recursion tree for T .n/ D 3T .n=4/ C cn2. For convenience, we assume that n is an exact power of 4 (another example of tolerable sloppiness) so that all subproblem sizes are integers. Part (a) of the figure shows T .n/, which we expand in part (b) into an equivalent tree representing the recurrence. The cn2term at the root represents the cost at the top level of recursion, and the three subtrees of the root represent the costs incurred by the subproblems of size n=4. Part (c) shows this process carried one step further by expanding each node with cost T .n=4/ from part (b). The cost for each of the three children of the root is c.n=4/2. We continue expanding each node in the tree by breaking it into its constituent parts as determined by the recurrence.
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4.4 The recursion-tree method for solving recurrences 89
## …
(d)
(c) (b)
(a)
T .n/ cn2 cn2
cn2 T!n
4
T!n #
4
T!n #
4
#
T!n 16
T!n #
16
T!n #
16
T!n #
16
T!n #
16
T!n #
16
T!n #
16
T!n #
16
T!n #
16
#
cn2 c!n
4
#2 c!n
4
#2 c!n
4
#2
c!n 4
#2
c!n 4
#2
c!n 4
#2
c!n 16
#2 c!n
16
#2 c!n
16
#2 c!n
16
#2 c!n
16
#2 c!n
16
#2 c!n
16
#2 c!n
16
#2 c!n
16
#2
3 16cn2
!3 16
#2 cn2 log4n
nlog43
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/
T .1/ ‚.nlog43/
Total: O.n2/
Figure 4.5 Constructing a recursion tree for the recurrence T .n/ D 3T .n=4/ C cn2. Part (a) shows T .n/, which progressively expands in (b)–(d) to form the recursion tree. The fully expanded tree in part (d) has height log4n(it has log4nC 1 levels).
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Because subproblem sizes decrease by a factor of 4 each time we go down one level, we eventually must reach a boundary condition. How far from the root do we reach one? The subproblem size for a node at depth i is n=4i. Thus, the subproblem size hits n D 1 when n=4i D 1 or, equivalently, when i D log4n. Thus, the tree has log4nC 1 levels (at depths 0; 1; 2; : : : ; log4n).
Next we determine the cost at each level of the tree. Each level has three times more nodes than the level above, and so the number of nodes at depth i is 3i. Because subproblem sizes reduce by a factor of 4 for each level we go down from the root, each node at depth i, for i D 0; 1; 2; : : : ; log4n " 1, has a cost of c.n=4i/2. Multiplying, we see that the total cost over all nodes at depth i, for i D 0; 1; 2; : : : ; log4n" 1, is 3ic.n=4i/2 D .3=16/icn2. The bottom level, at depth log4n, has 3log4n D nlog43 nodes, each contributing cost T .1/, for a total cost of nlog43T .1/, which is ‚.nlog43/, since we assume that T .1/ is a constant.
Now we add up the costs over all levels to determine the cost for the entire tree:
T .n/ D cn2C 3
16cn2C
\$ 3 16
%2
cn2C & & & C
\$ 3 16
%log4n!1
cn2C ‚.nlog43/
D
logX4n!1 iD0
\$ 3 16
%i
cn2C ‚.nlog43/
D .3=16/log4n" 1
.3=16/" 1 cn2C ‚.nlog43/ (by equation (A.5)) :
This last formula looks somewhat messy until we realize that we can again take advantage of small amounts of sloppiness and use an infinite decreasing geometric series as an upper bound. Backing up one step and applying equation (A.6), we have
T .n/ D
logX4n!1 iD0
\$ 3 16
%i
cn2C ‚.nlog43/
<
X1 iD0
\$ 3 16
%i
cn2C ‚.nlog43/
D 1
1" .3=16/cn2C ‚.nlog43/
D 16
13cn2C ‚.nlog43/ D O.n2/ :
Thus, we have derived a guess of T .n/ D O.n2/ for our original recurrence T .n/ D 3T .bn=4c/ C ‚.n2/. In this example, the coefficients of cn2 form a decreasing geometric series and, by equation (A.6), the sum of these coefficients
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4.4 The recursion-tree method for solving recurrences 91
## … …
cn
cn cn cn
c!n 3
# c!2n
3
#
c!n 9
# c!2n
9
c!2n #
9
# c!4n
9
# log3=2n
Total: O.n lg n/
Figure 4.6 A recursion tree for the recurrence T .n/ D T .n=3/ C T .2n=3/ C cn.
is bounded from above by the constant 16=13. Since the root’s contribution to the total cost is cn2, the root contributes a constant fraction of the total cost. In other words, the cost of the root dominates the total cost of the tree.
In fact, if O.n2/is indeed an upper bound for the recurrence (as we shall verify in a moment), then it must be a tight bound. Why? The first recursive call contributes a cost of ‚.n2/, and so !.n2/must be a lower bound for the recurrence.
Now we can use the substitution method to verify that our guess was cor- rect, that is, T .n/ D O.n2/ is an upper bound for the recurrence T .n/ D 3T .bn=4c/ C ‚.n2/. We want to show that T .n/ # dn2for some constant d > 0.
Using the same constant c > 0 as before, we have T .n/ # 3T .bn=4c/ C cn2
# 3d bn=4c2C cn2
# 3d.n=4/2C cn2
D 3
16d n2C cn2
# d n2;
where the last step holds as long as d \$ .16=13/c.
In another, more intricate, example, Figure 4.6 shows the recursion tree for T .n/D T .n=3/ C T .2n=3/ C O.n/ :
(Again, we omit floor and ceiling functions for simplicity.) As before, we let c represent the constant factor in the O.n/ term. When we add the values across the levels of the recursion tree shown in the figure, we get a value of cn for every level.
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The longest simple path from the root to a leaf is n ! .2=3/n ! .2=3/2n !
& & & ! 1. Since .2=3/knD 1 when k D log3=2n, the height of the tree is log3=2n. Intuitively, we expect the solution to the recurrence to be at most the number of levels times the cost of each level, or O.cn log3=2n/ D O.n lg n/. Figure 4.6 shows only the top levels of the recursion tree, however, and not every level in the tree contributes a cost of cn. Consider the cost of the leaves. If this recursion tree were a complete binary tree of height log3=2n, there would be 2log3=2nD nlog3=22 leaves. Since the cost of each leaf is a constant, the total cost of all leaves would then be ‚.nlog3=22/ which, since log3=22 is a constant strictly greater than 1, is !.n lg n/. This recursion tree is not a complete binary tree, however, and so it has fewer than nlog3=22 leaves. Moreover, as we go down from the root, more and more internal nodes are absent. Consequently, levels toward the bottom of the recursion tree contribute less than cn to the total cost. We could work out an accu- rate accounting of all costs, but remember that we are just trying to come up with a guess to use in the substitution method. Let us tolerate the sloppiness and attempt to show that a guess of O.n lg n/ for the upper bound is correct.
Indeed, we can use the substitution method to verify that O.n lg n/ is an upper bound for the solution to the recurrence. We show that T .n/ # dn lg n, where d is a suitable positive constant. We have
T .n/ # T .n=3/ C T .2n=3/ C cn
# d.n=3/ lg.n=3/ C d.2n=3/ lg.2n=3/ C cn D .d.n=3/ lg n " d.n=3/ lg 3/
C .d.2n=3/ lg n " d.2n=3/ lg.3=2// C cn D d n lg n " d..n=3/ lg 3 C .2n=3/ lg.3=2// C cn
D d n lg n " d..n=3/ lg 3 C .2n=3/ lg 3 " .2n=3/ lg 2/ C cn D d n lg n " d n.lg 3 " 2=3/ C cn
# d n lg n ;
as long as d \$ c=.lg 3".2=3//. Thus, we did not need to perform a more accurate accounting of costs in the recursion tree.
Exercises 4.4-1
Use a recursion tree to determine a good asymptotic upper bound on the recurrence T .n/D 3T .bn=2c/ C n. Use the substitution method to verify your answer.
4.4-2
Use a recursion tree to determine a good asymptotic upper bound on the recurrence T .n/D T .n=2/ C n2. Use the substitution method to verify your answer.
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4.5 The master method for solving recurrences 93
4.4-3
Use a recursion tree to determine a good asymptotic upper bound on the recurrence T .n/D 4T .n=2 C 2/ C n. Use the substitution method to verify your answer.
4.4-4
Use a recursion tree to determine a good asymptotic upper bound on the recurrence T .n/D 2T .n " 1/ C 1. Use the substitution method to verify your answer.
4.4-5
Use a recursion tree to determine a good asymptotic upper bound on the recurrence T .n/D T .n"1/CT .n=2/Cn. Use the substitution method to verify your answer.
4.4-6
Argue that the solution to the recurrence T .n/ D T .n=3/CT .2n=3/Ccn, where c is a constant, is !.n lg n/ by appealing to a recursion tree.
4.4-7
Draw the recursion tree for T .n/ D 4T .bn=2c/ C cn, where c is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substi- tution method.
4.4-8
Use a recursion tree to give an asymptotically tight solution to the recurrence T .n/D T .n " a/ C T .a/ C cn, where a \$ 1 and c > 0 are constants.
4.4-9
Use a recursion tree to give an asymptotically tight solution to the recurrence T .n/D T .˛n/ C T ..1 " ˛/n/ C cn, where ˛ is a constant in the range 0 < ˛ < 1 and c > 0 is also a constant.
4.5 The master method for solving recurrences
The master method provides a “cookbook” method for solving recurrences of the form
T .n/D aT .n=b/ C f .n/ ; (4.20)
where a \$ 1 and b > 1 are constants and f .n/ is an asymptotically positive function. To use the master method, you will need to memorize three cases, but then you will be able to solve many recurrences quite easily, often without pencil and paper.
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The recurrence (4.20) describes the running time of an algorithm that divides a problem of size n into a subproblems, each of size n=b, where a and b are positive constants. The a subproblems are solved recursively, each in time T .n=b/. The function f .n/ encompasses the cost of dividing the problem and combining the results of the subproblems. For example, the recurrence arising from Strassen’s algorithm has a D 7, b D 2, and f .n/ D ‚.n2/.
As a matter of technical correctness, the recurrence is not actually well defined, because n=b might not be an integer. Replacing each of the a terms T .n=b/ with either T .bn=bc/ or T .dn=be/ will not affect the asymptotic behavior of the recur- rence, however. (We will prove this assertion in the next section.) We normally find it convenient, therefore, to omit the floor and ceiling functions when writing divide-and-conquer recurrences of this form.
The master theorem
The master method depends on the following theorem.
Theorem 4.1 (Master theorem)
Let a \$ 1 and b > 1 be constants, let f .n/ be a function, and let T .n/ be defined on the nonnegative integers by the recurrence
T .n/D aT .n=b/ C f .n/ ;
where we interpret n=b to mean either bn=bc or dn=be. Then T .n/ has the follow- ing asymptotic bounds:
1. If f .n/ D O.nlogba!!/for some constant " > 0, then T .n/ D ‚.nlogba/. 2. If f .n/ D ‚.nlogba/, then T .n/ D ‚.nlogbalg n/.
3. If f .n/ D !.nlogbaC!/for some constant " > 0, and if af .n=b/ # cf .n/ for some constant c < 1 and all sufficiently large n, then T .n/ D ‚.f .n//.
Before applying the master theorem to some examples, let’s spend a moment trying to understand what it says. In each of the three cases, we compare the function f .n/ with the function nlogba. Intuitively, the larger of the two functions determines the solution to the recurrence. If, as in case 1, the function nlogbais the larger, then the solution is T .n/ D ‚.nlogba/. If, as in case 3, the function f .n/
is the larger, then the solution is T .n/ D ‚.f .n//. If, as in case 2, the two func- tions are the same size, we multiply by a logarithmic factor, and the solution is T .n/D ‚.nlogbalg n/ D ‚.f .n/ lg n/.
Beyond this intuition, you need to be aware of some technicalities. In the first case, not only must f .n/ be smaller than nlogba, it must be polynomially smaller.
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4.5 The master method for solving recurrences 95
That is, f .n/ must be asymptotically smaller than nlogbaby a factor of n!for some constant " > 0. In the third case, not only must f .n/ be larger than nlogba, it also must be polynomially larger and in addition satisfy the “regularity” condition that af .n=b/# cf .n/. This condition is satisfied by most of the polynomially bounded functions that we shall encounter.
Note that the three cases do not cover all the possibilities for f .n/. There is a gap between cases 1 and 2 when f .n/ is smaller than nlogba but not polynomi- ally smaller. Similarly, there is a gap between cases 2 and 3 when f .n/ is larger than nlogba but not polynomially larger. If the function f .n/ falls into one of these gaps, or if the regularity condition in case 3 fails to hold, you cannot use the master method to solve the recurrence.
Using the master method
To use the master method, we simply determine which case (if any) of the master theorem applies and write down the answer.
As a first example, consider T .n/D 9T .n=3/ C n :
For this recurrence, we have a D 9, b D 3, f .n/ D n, and thus we have that nlogba D nlog39 D ‚.n2). Since f .n/ D O.nlog39!!/, where " D 1, we can apply case 1 of the master theorem and conclude that the solution is T .n/ D ‚.n2/.
Now consider T .n/D T .2n=3/ C 1;
in which a D 1, b D 3=2, f .n/ D 1, and nlogba D nlog3=21 D n0 D 1. Case 2 applies, since f .n/ D ‚.nlogba/ D ‚.1/, and thus the solution to the recurrence is T .n/ D ‚.lg n/.
For the recurrence T .n/D 3T .n=4/ C n lg n ;
we have a D 3, b D 4, f .n/ D n lg n, and nlogba D nlog43 D O.n0:793/. Since f .n/ D !.nlog43C!/, where " ' 0:2, case 3 applies if we can show that the regularity condition holds for f .n/. For sufficiently large n, we have that af .n=b/ D 3.n=4/ lg.n=4/ # .3=4/n lg n D cf .n/ for c D 3=4. Consequently, by case 3, the solution to the recurrence is T .n/ D ‚.n lg n/.
The master method does not apply to the recurrence T .n/D 2T .n=2/ C n lg n ;
even though it appears to have the proper form: a D 2, b D 2, f .n/ D n lg n, and nlogba D n. You might mistakenly think that case 3 should apply, since
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f .n/ D n lg n is asymptotically larger than nlogba D n. The problem is that it is not polynomially larger. The ratio f .n/=nlogba D .n lg n/=n D lg n is asymp- totically less than n!for any positive constant ". Consequently, the recurrence falls into the gap between case 2 and case 3. (See Exercise 4.6-2 for a solution.)
Let’s use the master method to solve the recurrences we saw in Sections 4.1 and 4.2. Recurrence (4.7),
T .n/D 2T .n=2/ C ‚.n/ ;
characterizes the running times of the divide-and-conquer algorithm for both the maximum-subarray problem and merge sort. (As is our practice, we omit stating the base case in the recurrence.) Here, we have a D 2, b D 2, f .n/ D ‚.n/, and thus we have that nlogba D nlog22D n. Case 2 applies, since f .n/ D ‚.n/, and so we have the solution T .n/ D ‚.n lg n/.
Recurrence (4.17), T .n/D 8T .n=2/ C ‚.n2/ ;
describes the running time of the first divide-and-conquer algorithm that we saw for matrix multiplication. Now we have a D 8, b D 2, and f .n/ D ‚.n2/, and so nlogba D nlog28 D n3. Since n3 is polynomially larger than f .n/ (that is, f .n/D O.n3!!/for " D 1), case 1 applies, and T .n/ D ‚.n3/.
Finally, consider recurrence (4.18), T .n/D 7T .n=2/ C ‚.n2/ ;
which describes the running time of Strassen’s algorithm. Here, we have a D 7, b D 2, f .n/ D ‚.n2/, and thus nlogba D nlog27. Rewriting log27 as lg 7 and recalling that 2:80 < lg 7 < 2:81, we see that f .n/ D O.nlg 7!!/ for " D 0:8.
Again, case 1 applies, and we have the solution T .n/ D ‚.nlg 7/. Exercises
4.5-1
Use the master method to give tight asymptotic bounds for the following recur- rences.
a. T .n/ D 2T .n=4/ C 1.
b. T .n/ D 2T .n=4/ Cpn. c. T .n/ D 2T .n=4/ C n.
d. T .n/ D 2T .n=4/ C n2.
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4.6 Proof of the master theorem 97
4.5-2
Professor Caesar wishes to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm. His algorithm will use the divide- and-conquer method, dividing each matrix into pieces of size n=4 ! n=4, and the divide and combine steps together will take ‚.n2/ time. He needs to determine how many subproblems his algorithm has to create in order to beat Strassen’s algo- rithm. If his algorithm creates a subproblems, then the recurrence for the running time T .n/ becomes T .n/ D aT .n=4/ C ‚.n2/. What is the largest integer value of a for which Professor Caesar’s algorithm would be asymptotically faster than Strassen’s algorithm?
4.5-3
Use the master method to show that the solution to the binary-search recurrence T .n/D T .n=2/ C ‚.1/ is T .n/ D ‚.lg n/. (See Exercise 2.3-5 for a description of binary search.)
4.5-4
Can the master method be applied to the recurrence T .n/ D 4T .n=2/ C n2lg n?
Why or why not? Give an asymptotic upper bound for this recurrence.
4.5-5 ?
Consider the regularity condition af .n=b/ # cf .n/ for some constant c < 1, which is part of case 3 of the master theorem. Give an example of constants a \$ 1 and b > 1 and a function f .n/ that satisfies all the conditions in case 3 of the master theorem except the regularity condition.
## ?
4.6 Proof of the master theorem
This section contains a proof of the master theorem (Theorem 4.1). You do not need to understand the proof in order to apply the master theorem.
The proof appears in two parts. The first part analyzes the master recur- rence (4.20), under the simplifying assumption that T .n/ is defined only on ex- act powers of b > 1, that is, for n D 1; b; b2; : : :. This part gives all the intuition needed to understand why the master theorem is true. The second part shows how to extend the analysis to all positive integers n; it applies mathematical technique to the problem of handling floors and ceilings.
In this section, we shall sometimes abuse our asymptotic notation slightly by using it to describe the behavior of functions that are defined only over exact powers of b. Recall that the definitions of asymptotic notations require that
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bounds be proved for all sufficiently large numbers, not just those that are pow- ers of b. Since we could make new asymptotic notations that apply only to the set fbi W i D 0; 1; 2; : : :g, instead of to the nonnegative numbers, this abuse is minor.
Nevertheless, we must always be on guard when we use asymptotic notation over a limited domain lest we draw improper conclusions. For example, proving that T .n/D O.n/ when n is an exact power of 2 does not guarantee that T .n/ D O.n/.
The function T .n/ could be defined as T .n/D
(n if n D 1; 2; 4; 8; : : : ; n2 otherwise ;
in which case the best upper bound that applies to all values of n is T .n/ D O.n2/. Because of this sort of drastic consequence, we shall never use asymptotic notation over a limited domain without making it absolutely clear from the context that we are doing so.
4.6.1 The proof for exact powers
The first part of the proof of the master theorem analyzes the recurrence (4.20) T .n/D aT .n=b/ C f .n/ ;
for the master method, under the assumption that n is an exact power of b > 1, where b need not be an integer. We break the analysis into three lemmas. The first reduces the problem of solving the master recurrence to the problem of evaluating an expression that contains a summation. The second determines bounds on this summation. The third lemma puts the first two together to prove a version of the master theorem for the case in which n is an exact power of b.
Lemma 4.2
Let a \$ 1 and b > 1 be constants, and let f .n/ be a nonnegative function defined on exact powers of b. Define T .n/ on exact powers of b by the recurrence
T .n/D
(‚.1/ if n D 1 ;
aT .n=b/C f .n/ if n D bi; where i is a positive integer. Then T .n/D ‚.nlogba/C
logXbn!1 jD0
ajf .n=bj/ : (4.21)
Proof We use the recursion tree in Figure 4.7. The root of the tree has cost f .n/, and it has a children, each with cost f .n=b/. (It is convenient to think of a as being
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4.6 Proof of the master theorem 99
## …
f .n/ f .n/
a a
a
a
a a
a
a
a a
a
a
a
f .n=b/
f .n=b/
f .n=b/
f .n=b2/ f .n=b2/
f .n=b2/ f .n=b2/
f .n=b2/ f .n=b2/ f .n=b2/
f .n=b2/ f .n=b2/
af .n=b/
a2f .n=b2/ logbn
nlogba
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/
‚.1/ ‚.nlogba/
Total: ‚.nlogba/C
logXbn!1 jD0
ajf .n=bj/
Figure 4.7 The recursion tree generated by T .n/ D aT .n=b/ Cf .n/. The tree is a complete a-ary tree with nlogbaleaves and height logbn. The cost of the nodes at each depth is shown at the right, and their sum is given in equation (4.21).
an integer, especially when visualizing the recursion tree, but the mathematics does not require it.) Each of these children has a children, making a2nodes at depth 2, and each of the a children has cost f .n=b2/. In general, there are aj nodes at depth j , and each has cost f .n=bj/. The cost of each leaf is T .1/ D ‚.1/, and each leaf is at depth logbn, since n=blogbn D 1. There are alogbn D nlogba leaves in the tree.
We can obtain equation (4.21) by summing the costs of the nodes at each depth in the tree, as shown in the figure. The cost for all internal nodes at depth j is ajf .n=bj/, and so the total cost of all internal nodes is
logXbn!1 jD0
ajf .n=bj/ :
In the underlying divide-and-conquer algorithm, this sum represents the costs of dividing problems into subproblems and then recombining the subproblems. The
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cost of all the leaves, which is the cost of doing all nlogba subproblems of size 1, is ‚.nlogba/.
In terms of the recursion tree, the three cases of the master theorem correspond to cases in which the total cost of the tree is (1) dominated by the costs in the leaves, (2) evenly distributed among the levels of the tree, or (3) dominated by the cost of the root.
The summation in equation (4.21) describes the cost of the dividing and com- bining steps in the underlying divide-and-conquer algorithm. The next lemma pro- vides asymptotic bounds on the summation’s growth.
Lemma 4.3
Let a \$ 1 and b > 1 be constants, and let f .n/ be a nonnegative function defined on exact powers of b. A function g.n/ defined over exact powers of b by
g.n/D
logXbn!1 jD0
ajf .n=bj/ (4.22)
has the following asymptotic bounds for exact powers of b:
1. If f .n/ D O.nlogba!!/for some constant " > 0, then g.n/ D O.nlogba/. 2. If f .n/ D ‚.nlogba/, then g.n/ D ‚.nlogbalg n/.
3. If af .n=b/ # cf .n/ for some constant c < 1 and for all sufficiently large n, then g.n/ D ‚.f .n//.
Proof For case 1, we have f .n/ D O.nlogba!!/, which implies that f .n=bj/D O..n=bj/logba!!/. Substituting into equation (4.22) yields
g.n/D O
logXbn!1 jD0
aj& n bj
'logba!!!
: (4.23)
We bound the summation within the O-notation by factoring out terms and simpli- fying, which leaves an increasing geometric series:
logXbn!1 jD0
aj& n bj
'logba!!
D nlogba!!
logXbn!1 jD0
\$ ab! blogba
%j
D nlogba!!
logXbn!1 jD0
.b!/j
D nlogba!!
\$b!logbn" 1 b!" 1
%
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4.6 Proof of the master theorem 101
D nlogba!!
\$n!" 1 b!" 1
% :
Since b and " are constants, we can rewrite the last expression as nlogba!!O.n!/D O.nlogba/. Substituting this expression for the summation in equation (4.23) yields g.n/D O.nlogba/ ;
thereby proving case 1.
Because case 2 assumes that f .n/ D ‚.nlogba/, we have that f .n=bj/ D
‚..n=bj/logba/. Substituting into equation (4.22) yields g.n/D ‚
logXbn!1 jD0
aj& n bj
'logba!
: (4.24)
We bound the summation within the ‚-notation as in case 1, but this time we do not obtain a geometric series. Instead, we discover that every term of the summation is the same:
logXbn!1 jD0
aj& n bj
'logba
D nlogba
logXbn!1 jD0
& a blogba
'j
D nlogba
logXbn!1 jD0
1 D nlogbalogbn :
Substituting this expression for the summation in equation (4.24) yields g.n/ D ‚.nlogbalogbn/
D ‚.nlogbalg n/ ; proving case 2.
We prove case 3 similarly. Since f .n/ appears in the definition (4.22) of g.n/
and all terms of g.n/ are nonnegative, we can conclude that g.n/ D !.f .n// for exact powers of b. We assume in the statement of the lemma that af .n=b/ # cf .n/
for some constant c < 1 and all sufficiently large n. We rewrite this assumption as f .n=b/ # .c=a/f .n/ and iterate j times, yielding f .n=bj/ # .c=a/jf .n/or, equivalently, ajf .n=bj/ # cjf .n/, where we assume that the values we iterate on are sufficiently large. Since the last, and smallest, such value is n=bj!1, it is enough to assume that n=bj!1is sufficiently large.
Substituting into equation (4.22) and simplifying yields a geometric series, but unlike the series in case 1, this one has decreasing terms. We use an O.1/ term to
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capture the terms that are not covered by our assumption that n is sufficiently large:
g.n/ D
logXbn!1 jD0
ajf .n=bj/
#
logXbn!1 jD0
cjf .n/C O.1/
# f .n/
X1 jD0
cj C O.1/
D f .n/
\$ 1
1" c
%
C O.1/
D O.f .n// ;
since c is a constant. Thus, we can conclude that g.n/ D ‚.f .n// for exact powers of b. With case 3 proved, the proof of the lemma is complete.
We can now prove a version of the master theorem for the case in which n is an exact power of b.
Lemma 4.4
Let a \$ 1 and b > 1 be constants, and let f .n/ be a nonnegative function defined on exact powers of b. Define T .n/ on exact powers of b by the recurrence
T .n/D
(‚.1/ if n D 1 ;
aT .n=b/C f .n/ if n D bi;
where i is a positive integer. Then T .n/ has the following asymptotic bounds for exact powers of b:
1. If f .n/ D O.nlogba!!/for some constant " > 0, then T .n/ D ‚.nlogba/. 2. If f .n/ D ‚.nlogba/, then T .n/ D ‚.nlogbalg n/.
3. If f .n/ D !.nlogbaC!/for some constant " > 0, and if af .n=b/ # cf .n/ for some constant c < 1 and all sufficiently large n, then T .n/ D ‚.f .n//.
Proof We use the bounds in Lemma 4.3 to evaluate the summation (4.21) from Lemma 4.2. For case 1, we have
T .n/ D ‚.nlogba/C O.nlogba/ D ‚.nlogba/ ;
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4.6 Proof of the master theorem 103
and for case 2,
T .n/ D ‚.nlogba/C ‚.nlogbalg n/
D ‚.nlogbalg n/ : For case 3,
T .n/ D ‚.nlogba/C ‚.f .n//
D ‚.f .n// ; because f .n/ D !.nlogbaC!/. 4.6.2 Floors and ceilings
To complete the proof of the master theorem, we must now extend our analysis to the situation in which floors and ceilings appear in the master recurrence, so that the recurrence is defined for all integers, not for just exact powers of b. Obtaining a lower bound on
T .n/D aT .dn=be/ C f .n/ (4.25)
and an upper bound on
T .n/D aT .bn=bc/ C f .n/ (4.26)
is routine, since we can push through the bound dn=be \$ n=b in the first case to yield the desired result, and we can push through the bound bn=bc # n=b in the second case. We use much the same technique to lower-bound the recurrence (4.26) as to upper-bound the recurrence (4.25), and so we shall present only this latter bound.
We modify the recursion tree of Figure 4.7 to produce the recursion tree in Fig- ure 4.8. As we go down in the recursion tree, we obtain a sequence of recursive invocations on the arguments
n ; dn=be ; ddn=be =be ; dddn=be =be =be ;
:::
Let us denote the j th element in the sequence by nj, where nj D
(n if j D 0 ;
dnj!1=be if j > 0 : (4.27)
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EURegio (BE, NL, DE) – Mutual Legal Assistance (AT, FR, GR)‐ European
Further recent works in progress by Kamienny, Stein and Stoll [KSS] and Derickx, Kamienny, Stein and Stoll [DKSS] exclude k-rational point of exact prime order larger than |
# How do you find the product of (10n^2)/4*2/n?
Dec 22, 2016
$5 n$
#### Explanation:
$\frac{10 {n}^{2}}{4} \times \frac{2}{n} = \frac{20 {n}^{2}}{4 n} = \frac{\cancel{\left(4 n\right)} 5 n}{\cancel{4 n}} = 5 n$
Dec 22, 2016
$5 n$
#### Explanation:
There are 2 approaches here. We could $\textcolor{b l u e}{\text{simplify}}$ the fractions and then multiply what remains or we can multiply the fractions and then $\textcolor{b l u e}{\text{simplify}}$
$\textcolor{red}{\text{Simplify and multiply}}$
To simplify we $\textcolor{b l u e}{\text{cancel}}$ common factors between the numerators/denominators.
Here we can cancel 10 and 4 by 2 or 4 and 2 by 2. ${n}^{2}$ and n have a common factor of n.
$\Rightarrow \frac{{\cancel{10}}^{5} \times n \times {\cancel{n}}^{1}}{\cancel{4}} ^ 2 \times \frac{2}{\cancel{n}} ^ 1$
$= \frac{5 n \times 2}{2} = \frac{5 n \times {\cancel{2}}^{1}}{\cancel{2}} ^ 1 = 5 n$
$\textcolor{red}{\text{multiply and simplify}}$
$\frac{10 {n}^{2}}{4} \times \frac{2}{n} = \frac{10 {n}^{2} \times 2}{4 n} = \frac{20 {n}^{2}}{4 n}$
now simplify by cancelling.
$= \frac{{\cancel{20}}^{5} \times n \times {\cancel{n}}^{1}}{{\cancel{4}}^{1} {\cancel{n}}^{1}} = 5 n$ |
Explainer: Pascal’s Triangle
In this explainer, we will learn how to solve problems on Pascal’s triangle.
Pascal’s triangle is one of the most fascinating structures we can build from a simple number pattern. It is fascinating to see the connections between such a simple construction and many other areas of mathematics.
Pascal’s triangle can be formed by starting with a one at the top and then placing two ones below. Then, each element of a row is equal to the sum of the two elements above. Hence, in the figure below, we can see that the two is the sum of the two ones above.
To complete the next row, we can consider the pairwise sum of the elements of this row. The first entry will be 1. We can think of this as the sum of 0 and 1 as shown.
The next element is the sum of 1 and 2 as shown below.
Similarly, the following element is the sum of 2 and 1 as shown.
The final element, like the first, can be thought of as the sum of 1 and 0 as follows.
Continuing this pattern, we arrive at what is known as Pascal’s triangle.
Pascal’s Triangle
Pascal’s triangle is a triangular array of the numbers which satisfy the property that each element is equal to the sum of the two elements above. The rows are enumerated from the top such that the first row is numbered .
Similarly, the elements of each row are enumerated from up to . The first eight rows of Pascal’s triangle are shown below.
Although, in much of the Western world, the triangle is named after the French mathematician Blaise Pascal, it was, in fact, well known to mathematicians centuries before him in places such as China, Persia, and India. To this day, it is know by different names in these places.
Pascal’s triangle has many interesting properties. We will begin by looking as some of the simple patterns which exist in the triangle.
Some of the most obvious patterns are related to the diagonals: for example, the first diagonal only contains ones, whereas the second contains consecutive integers.
More interestingly, the third diagonal contains the triangle numbers, and the fourth contains the tetrahedral numbers.
Furthermore, we can see there is reflectional symmetry about the center.
Example 1: Elements in Pascal’s Triangle
What is the second element in the 500th row of Pascal’s triangle?
Recall that the second elements of each row of Pascal’s triangle are consecutive integers. At this point, we might be tempted to immediately jump to the conclusion that it will therefore be 500. However, we need to be a little more careful than this. Recall that the first row only contains 1. Hence, there is no second element. The first row with a second element is the second row, which consists of two ones. Therefore, the second element in this row is 1 and not 2. Hence, the second element of the 500th row of Pascal’s triangle will be 499.
Example 2: Patterns in Pascal’s Triangle
A partially filled-in picture of Pascal’s triangle is shown. By noticing the patterns, or otherwise, find the values of , , , and .
We begin by considering the elements of the third diagonal. There is a clear pattern to go from one element to the other: to go from the first to the second, we add two; then to go from the second to the third, we add 3.
We can extend this pattern as follows.
Both and are the elements in this row. Therefore, and .
We now consider element . This element is actually also in the third diagonal—the one that is in the other direction—and it is the sixth element. Hence, .
Finally, we see that is in the second diagonal. This diagonal contains consecutive positive integers. Hence, since it is the eleventh element, its value will simply be 11.
Therefore, our final answer is , , , and .
Example 3: Sums along Diagonals in Pascal’s Triangle
The figure shows a section of Pascal’s triangle. Without using a calculator, find the sum of the highlighted elements. |
# Onto Function
A function is something, which relates elements/values of one set to the elements/values of another set, in such a way that elements of the second set is identically determined by the elements of the first set. A function has many types which define the relationship between two sets in a different pattern. They are various types of functions like one to one function, onto function, many to one function, etc.
## Onto Function Definition (Surjective Function)
Onto function could be explained by considering two sets, Set A and Set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto function or surjective function. The term for the surjective function was introduced by Nicolas Bourbaki.
In the first figure, you can see that for each element of B there is a pre-image or a matching element in Set A, therefore, its an onto function. But if you see in the second figure, one element in Set B is not mapped with any element of set A, so it’s not an onto or surjective function.
### Properties of a Surjective Function (Onto)
• We can define onto function as if any function states surjection by limit its codomain to its range.
• The domain is basically what can go into the function, codomain states possible outcomes and range denotes the actual outcome of the function.
• Every onto function has a right inverse
• Every function with a right inverse is a surjective function
• If we compose onto functions, it will result in onto function only.
### Onto Function Example Questions
Example: Let A={1,5,8,9) and B{2,4} And f={(1,2),(5,4),(8,2),(9,4)}. Then prove f is a onto function.
Solution: From the question itself we get,
A={1,5,8,9)
B{2,4}
& f={(1,2),(5,4),(8,2),(9,4)}
So, all the element on B has a domain element on A or we can say element 1 and 8 & 5 and 9 has same range 2 & 4 respectively.
Therefore, f: A $$\rightarrow$$ B is an surjective fucntion.
Hence, the onto function proof is explained.
### Number of Onto Functions (Surjective functions) Formula
If we have to find the number of onto function from a set A with n number of elements to set B with m number of elements, then;
$$(^{m}_{0})m^{n}\,-\,(^{m}_{1})(m-1)^{n}\,+\,(^{m}_{2})(m-2)^{n}\,-\,(^{m}_{3})(m-3)^{n}\,+\,.\,.\,.\,.\,.\,.$$
When n<m, the number of onto function = 0
And when n=m, number of onto function = m!
We can also write the number of surjective functions for a given domain and range as;
mn – m + m ((m – 1)n – (m – 1)) |
In this lesson, we will learn
• the different forms of quadratic functions
• general form
• factored form
• vertex form
• how to convert from general form to factored form.
• how to convert from the general form to the vertex form using the vertex formula.
• how to convert from the general form to the vertex form using completing the square.
We can write quadratic functions in different ways or forms.
• General Form
• Factored Form
• Vertex Form
The general form of a quadratic equation is
y
= ax2 + bx + c where a, b and c are real numbers and a is not equal to zero.
For example, y = 2x2 + 5x − 30
The factored form of a quadratic equation is
y
= a(x + b)(x + c) where a, b and c are real numbers and a is not equal to zero.
For example, y = 2(x + 6)(x − 5).
The factored form is useful because we can see the x-intercepts (which are also the roots when the function is zero).
For example, the x-intercepts of y = a(x + b)(x + c) are (−b, 0) and (−c, 0)
The vertex form of a quadratic equation is
y
= a(x h)2 + k where a, h and k are real numbers and a is not equal to zero.
For example, y = 2(x + 6)2 − 5.
The vertex form is useful because we can see the turning point or vertex of the graph.
For example, the turning point or vertex of y = a(x h)2 + k is (h, k). If a is positive then it is a minimum vertex. It a is negative then it is a maximum vertex.
The following video looks at the various formats in which Quadratic Functions may be written as.
## General Form to Factored Form
The following videos show how to change quadratic functions from general form to factored form.
## General Form to Vertex Form by using the Vertex Formula
We can change a quadratic function from general form to vertex form by using the vertex formula.
Example of how to convert standard form to vertex form of a parabola equation.
## General Form to Vertex Form by Completing the Square
We can change a quadratic function from general form to vertex form by completing the square.
The following video shows how to use the method of Completing the Square to convert a quadratic function from standard form to vertex form.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
Standards-based assessment and Instruction
## Puzzle Pieces
Andy is putting a puzzle together. The puzzle has one hundred fifty pieces all together. The first day Andy puts thirty-six pieces of the puzzle together. The second day Andy puts forty-one more pieces of the puzzle together. The third day Andy puts sixty-eight more pieces of the puzzle together. Andy is upset because there are no more puzzle pieces left. How many puzzle pieces does Andy need to finish the puzzle? Show all your mathematical thinking.
### Common Core Content Standards and Evidence
#### 2.NBT Number and Operations in Base Ten
Use place value understanding and properties of operations to add and subtract.
6. Add up to four two-digit numbers using strategies based on place value and properties of operations.
This task requires students to use properties of operations and place value to add three two-digit numbers.
### GO Math! Alignments
• Chapter 4, Lessons: 4.1, 4.2, 4.3, 4.4, 4.5, 4.11, 4.12
### Common Core Standards of Mathematical Practice
MP.1 Make sense of problems and persevere in solving them.
MP.3 Construct viable arguments and critique the reasoning of others.
MP.4 Model with mathematics.
MP.6 Attend to precision.
### Underlying Mathematical Concepts
• Part/Whole reasoning
• Number sense to 150
• Ordinal numbers
### Possible Problem-Solving Strategies
• Model (manipulatives)
• Diagram/Key
• Table
• Number line
### Possible Mathematical Vocabulary/Symbolic Representation
• Model
• Diagram/Key
• Number line
• Amount
• Table
• Ordinal numbers (1st, 2nd, 3rd ...)
• Difference
• Total/Sum
• Odd/Even
• More than (>)/Greater than (>)/Less than (<)
• Equivalent/Equal to
• Most/Least
• Per
• Number Line
### Possible Solutions
Andy needs 5 more puzzle pieces.
### Possible Connections
Below are some examples of mathematical connections. Your students may discover some that are not on this list.
• Each day Andy puts a greater total of pieces in the puzzle.
• 41 puzzle pieces is an odd number.
• 36 puzzle pieces is 3 dozen.
• 68 puzzle pieces were the most done in a day.
• 36 puzzle pieces were the least done in a day.
• Solve more than one way to verify the answer.
• Relate to a similar problem and state a math link.
• On day 1, Andy puts an even number of pieces together.
• On day 2, Andy puts an odd number of pieces together.
• 50 pieces per day would have been an equal amount per day.
• Even + Odd = Odd, so it is clear some pieces are missing.
### Scoring Rationales and Corresponding Anchor Papers
#### Expert
Our teacher-friendly tasks are designed to support both the Common Core and Citywide instructional expectations. GO Math! alignments are also available.
Set up your FREE 30-day Trial today!
Explore our latest K-5 math material and begin using it in your classroom.
Set up your FREE 30-day Trial today!
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# Learntofish's Blog
## How to draw regular polygons
Posted by Ed on November 20, 2012
In this blog post we will derive a formula for the interior angle $\alpha$ of an n-sided regular polygon. We will examine how $\alpha$ depends on n. This allows us to draw the regular polygon.
Regular polygons
A regular polygon is a figure whose sides have the same length (equilateral) and whose interior angles are equal (equiangular). Let’s draw one with 3 sides. We can easily construct it if we know the interior angle $\alpha$. For the triangle it is $\alpha$ = 60°.
Let’s do the same for a square, for which the interior angle is $\alpha$ = 90°.
Now, can we do the same for a pentagon (5 sides)? What about a 13 sided regular polygon?
Obviously, we could if we knew how to choose $\alpha$. Before you continue reading you may want to ponder how to determine the internal angle. So, spoilers ahead.
Derivation of interior angle
We have two equations.
$\alpha + \beta = 180^\circ$ (1)
$n \cdot \beta = 360^\circ$ (2)
We solve (2) for $\beta$: $\beta = \frac{360^\circ}{n}$ (3)
Plug (3) into (1): $\alpha + \frac{360^\circ}{n} = 180^\circ$ (4)
Multiply (4) by n: $n \alpha + 360^\circ = n \cdot 180^\circ$ (5)
Substract 2*180° on both sides: $n \alpha = n \cdot 180^\circ - 2 \cdot 180^\circ = (n-2) \cdot 180^\circ$ (6)
Divide by n: $\alpha = \frac{(n-2)}{n} \cdot 180^\circ$ (7)
This is the interior angle for an n-sided regular polygon.
Problems
1) Show that the formula is correct for the triangle and the square.
2) What is the interior angle for the pentagon?
3) What is the sum of all interior angles in a 7-sided regular polygon?
4) What is the sum of all interior angles in an n-sided regular polygon?
5) What happens with the interior angle if you make n bigger and bigger?
1) Before we derived the formula we already knew that the interior angles are 60° and 90° respectively for the triangle and square. Let’s check if the formula gives us these values.
For the triangle we set n=3 and get $\alpha=60^\circ$.
For the square we set n=4 and get $\alpha=90^\circ$.
2) For the pentagon we set n=4 and get $\alpha=108^\circ$.
3) and 4) The sum of all interior angles in an n-sided regular polygon is $n\alpha = (n-2)\cdot 180^\circ$. For a 7-sided regular polygon we set n=7 and get $7\alpha = (7-2)\cdot 180^\circ = 5 \cdot 180^\circ = 900^\circ$.
5) If we make n bigger and bigger, the value for $\alpha$ approaches 180°. We can express this with the limit:
$\lim_{n\rightarrow \infty} \alpha$
$= \lim_{n\rightarrow \infty} \frac{(n-2)}{n} \cdot 180^\circ$
$= \lim_{n\rightarrow \infty} \left(1-\frac{2}{n}\right) \cdot 180^\circ$
$= 180^\circ$ |
# What is the distance between (4, 1, –3) and (0, 4, –2) ?
Apr 12, 2018
$\sqrt{26}$
#### Explanation:
The distance is equal to the magnitude of the vector between the two points which can be expressed as: $| \left(\begin{matrix}4 \\ 1 \\ - 3\end{matrix}\right) - \left(\begin{matrix}0 \\ 4 \\ - 2\end{matrix}\right) |$
$| \left(\begin{matrix}4 - 0 \\ 1 - 4 \\ - 3 - \left(- 2\right)\end{matrix}\right) |$
$| \left(\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right) |$
The magnitude is $\sqrt{{\left(4\right)}^{2} + {\left(- 3\right)}^{2} + {\left(- 1\right)}^{2}}$
$\sqrt{16 + 9 + 1}$ = $\sqrt{26}$
Apr 12, 2018
$A B = \sqrt{26}$
#### Explanation:
We know that;
If $A \in {\mathbb{R}}^{3} \mathmr{and} B \in {\mathbb{R}}^{3}$,then the distance between
$A \left({x}_{1} , {y}_{1} , {z}_{1}\right) \mathmr{and} B \left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is
$A B = | \vec{A B} | = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
Where, $\vec{A B} = \left({x}_{2} - {x}_{1} , {y}_{2} - {y}_{1} , {z}_{2} - {z}_{1}\right)$
We have, $A \left(4 , 1 , - 3\right) \mathmr{and} B \left(0 , 4 , - 2\right)$
$\implies A B = \sqrt{{\left(4 - 0\right)}^{2} + {\left(1 - 4\right)}^{2} + {\left(- 3 + 2\right)}^{2}}$
=>AB=sqrt(16+9+1
$\implies A B = \sqrt{26}$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Percent Equation to Find Part a
## Find x% of a number
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Use the Percent Equation to Find Part a
Jordan received 85 from his aunts for his birthday. He wants to buy a new game for his computer. His mother told him he could spend no more than 28% of his money on a computer game. How much money can Jordan spend? In this concept, you will learn to use the percent equation to find part a. ### Finding Part a You can use the proportion \begin{align*}\frac{a}{b} = \frac{p}{100}\end{align*} to solve for a percent. You can also solve percent problems by using an equation. In this concept, you will use a proportion to create a different kind of equation that will help you solve percent problems. When you solve the proportion \begin{align*}\frac{a}{b} = \frac{p}{100}\end{align*}, you cross multiply to find the missing variable. You can rearrange this formula so that you are solving for just the variable \begin{align*}a\end{align*}. \begin{align*}\begin{array}{rcl} \frac{a}{b} &=& \frac{p}{100} \\ 100a &=& pb \\ \frac{100a}{100} &=& \frac{pb}{100} \\ a&=&\frac{pb}{100} \end{array}\end{align*} You could also say that \begin{align*}a= \frac{pb}{100}\end{align*} is equal to \begin{align*}a = 0.01 pb\end{align*}. As well, you could convert your percent directly into a decimal and therefore the formula becomes even simpler. Let’s look at an example. What is 85% of 90? First, change the 85% into a decimal. You know that percent means that the denominator of the fraction is 100. Therefore, \begin{align*}85\% = 0.85\end{align*}. Next, multiply using the percent equation. \begin{align*} \begin{array}{rcl} a &=& 0.01 pb \\ a &=& 0.85 \times 90 \\ a &=& 76.5 \\ \end{array}\end{align*}The answer is 76.5. Let’s look at another example. What is 7% of 900? First, change the 7% into a decimal. You know that percent means that the denominator of the fraction is 100. Therefore, \begin{align*}7\% = 0.07\end{align*}. Next, multiply using the percent equation. \begin{align*}\begin{array}{rcl} a &=& 0.01 pb \\ a &=& 0.07 \times 900 \\ a &=& 63 \end{array}\end{align*} The answer is 63. ### Examples #### Example 1 Earlier, you were given a problem about Jordan and his partial spending money. Jordan has85 from his birthday but cannot spend any more than 28% of it on his new computer game. How much can he spend?
First, change the 28% into a decimal.
\begin{align*}28\% = 0.28\end{align*}
Next, multiply using the percent equation.
\begin{align*}\begin{array}{rcl} a &=& 0.01 pb \\ a &=& 0.28 \times 85 \\ a &=& 23.8 \end{array}\end{align*} The answer is 23.8.
Therefore, Jordan can spend up to \$23.80 on his new computer game.
#### Example 2
What is 19% of 300?
First, change the 19% into a decimal.
\begin{align*}19\% = 0.19\end{align*}
Next, multiply using the percent equation.
\begin{align*}\begin{array}{rcl} a &=& 0.01 pb \\ a &=& 0.19 \times 300 \\ a &=& 57 \end{array}\end{align*} The answer is 57.
#### Example 3
What is 22% of 100?
First, change the 22% into a decimal.
\begin{align*}22\% = 0.22\end{align*}
Next, multiply using the percent equation.
\begin{align*} \begin{array}{rcl} a &=& 0.01 pb \\ a &=& 0.22 \times 100 \\ a &=& 22 \end{array}\end{align*}
#### Example 4
What is 8% of 57?
First, change the 8% into a decimal.
\begin{align*}8\% = 0.08\end{align*}
Next, multiply using the percent equation.
\begin{align*}\begin{array}{rcl} a &=& 0.01 pb \\ a &=& 0.08 \times 57 \\ a &=& 4.56 \end{array}\end{align*}
#### Example 5
What is 17% of 80?
First, change the 17% into a decimal.
\begin{align*}17\% = 0.17\end{align*}
Next, multiply using the percent equation.
\begin{align*} \begin{array}{rcl} a &=& 0.01 pb \\ a &=& 0.17 \times 80 \\ a &=& 13.6 \end{array}\end{align*} The answer is 13.6.
### Review
Solve each percent problem. Round your answers to the nearest tenth when necessary.
1. How much is 15% of 73?
2. What is 70% of 5?
3. What is 3% of 4 million?
4. What is 18% of 30?
5. What is 22% of 56?
6. What is 19% of 300?
7. What is 21% of 45?
8. What is 34% of 250?
9. What is 33% of 675?
10. What is 3% of 700?
11. What is 11% of 955?
12. What is 14% of 55?
13. What is 37% of 17?
14. What is 20% of 9?
15. What is 2% of 18?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Decimal In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Inverse Operation Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Proportion A proportion is an equation that shows two equivalent ratios. |
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# “Guzinta”
“Guzinta”. By Troy Pomeroy Horace Mann Middle School. Why use Divisibility Rules?. Ever wonder if 3 “guzinta” (goes into) numbers like 75235? How would you find out? Would you have to divide 75235 by 3? Would a calculator help? What if you wanted to check to see if 7 “guzinta” 75235?
## “Guzinta”
E N D
### Presentation Transcript
1. “Guzinta” By Troy Pomeroy Horace Mann Middle School
2. Why use Divisibility Rules? • Ever wonder if 3 “guzinta” (goes into) numbers like 75235? • How would you find out? • Would you have to divide 75235 by 3? • Would a calculator help? • What if you wanted to check to see if 7 “guzinta” 75235? • Would this process take a long time?
3. Want to learn a FASTER way? I Knew you would!!!!!
4. Testing For Divisibility • Lucky for us, mathematicians older than Mr. Evans developed some rules to determine if certain numbers “guzinta” other numbers without having to divide! • By learning these rules you will be able to perform math tasks like factoring, simplifying fractions, dividing, and problem solving much easier.
5. READY? Here come the “tricks of the trade” known as . . . Divisibility Rules
6. Divisibility Rules • A number is divisible by 2 if it is EVEN. • That means the last digit of the number is 2, 4, 6, 8, or 0. • For example: 35,718 IS divisible by 2 because its last digit is 8 - an even number. 35,719IS NOT divisible by 2 because its last digit is 9 - an odd number.
7. Divisibility Rules • If the sum of the digits is divisible by three, the number is also. • For example, 717 IS divisible by 3 because 7 + 1 + 7 = 15. 15 is divisible by 3 so 717 is divisible by 3 also. • 713 IS NOT divisible by 3 since 7 + 1 + 3 = 11 and 11 is not divisible by 3.
8. Divisibility Rules • The test to see if a number is divisible by 4 is to look at the last two digits of the number. • If those two digits are divisible by 4, then the number is also. • EXAMPLE: 15,912IS divisible by 4 because 12 is divisible by 4. 15,914IS NOT divisible by 4 because 14 is not divisible by 4.
9. Divisibility Rules • If the last digit of a number is 5 or 0, the number is divisible by 5. • EXAMPLE: 366,789,475IS divisible by 5 since it ends in a 5. 366,789,472IS NOT divisible by 5 since it ends in a 2.
10. Divisibility Rules • For a number to be divisible by 6, it must be divisible by BOTH 2 AND 3. • That means it must be EVEN and its digits must add up to a multiple of 3. • EXAMPLE: 315 IS NOT divisible by 6, even though its digits add up to 9, because it is not even. 216 IS divisible by 6 since it is even AND its digits sum to a multiple of 3.
11. Divisibility Rules • If the sum of the digits is divisible by nine, the number is also. • For example, 558 IS divisible by 9 because 5 + 5 + 8 = 18. 18 is divisible by 9 so 558 is divisible by 9 also. • 713 IS NOT divisible by 9 since 7 + 1 + 3 = 11 and 11 is not divisible by 9.
12. Divisibility Rules • A number is divisible by 10 if it ends in 0. • EXAMPLE: 671,320 IS divisible by 10 because it ends in 0. 671,325 IS NOT divisible by 10 since it DOES NOT end in 0.
More Related |
# Using Our Fractional Reasoning!
We had a great concept sort today…and I just wanted to share a little about what we did!
We are working on the concept of equivalent fractions…we have drawn pictures, told stories (If I had half a pizza but cut the half into two pieces, what fraction would I have?), and generated lists of equivalent fractions. What we DIDN’T do is what most math programs do right away–teach students to multiply the numerator and denominator by the same number. We’ll get there-but first I really want students to use their reasoning to really show their understanding of some key fraction concepts.
One of the Standards for Mathematical Practice involves the ability to “reason”–to create strong understanding of key concepts without merely computing. It states:
“Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects.”
By helping students learn to reason about fractions, they become better at understanding without relying on tricks and computation–which helps them with estimating and checking for reasonableness as the math gets more challenging. I love trying to help students VISUALIZE math and make sense of it before teaching them–so that’s what today was all about!
Most students have a pretty decent understanding of the concept of “one half”, so I wanted to experiment with a sort and see what my students could do. We’ve already talked about the concept of “unit fractions”–and how they can by used to “count” fractions…1/4, 2/4, 3/4 and so on. We also have used our reasoning to picture the relative size of these unit fractions…that even though “seven” is a bigger number than “three”, sevenths are smaller than thirds because more parts must mean smaller parts!
This was really enough information for us to begin this sort–where students used what they know about fractions to sort them into three categories–greater than 1/2, exactly 1/2, and less than 1/2.
One of our rules about concept sorts is that students work in small groups (usually trios) and must go one card at a time where they discuss together and make a decision about which category the cards fall in. If they have any debate, they set it aside for later.
While students are working, I’m circulating, asking questions, listening–and looking for misconceptions. Anything “interesting” gets thrown under the document camera at the end!
For groups finishing early, I ask them to write their OWN examples for each category…
I LOVED hearing the discussion this group had–they write the example and then couldn’t come to an agreement about which category! One of the students was trying SO hard to explain that HALF of 310 would be 155…so 149/310 HAD to be less than one half. The other two were NOT understanding her reasoning!
After we worked for a while, I picked THIS fraction to discuss…and with NO computation about finding fractions equivalent to 1/2, we had two very justifiable explanations for why 7/15 is less than one half. One student came up and explained how it HAD to be less than one half because 7/14 would be one half…and fifteens are smaller than fourteenths–so 7/15 had to be smaller than 7/14. Pretty slick!
The other argument explained that the “halfway” point of fifteenths would have to be “seven and a half” of them…so 7 of them ha to be less than one half. Such GREAT math discussions…with no computation. This is a perfect example of why I love concept sorts…so much discourse. So much math. So much engagement. Tomorrow–we learn the algorithm for generating equivalent fractions…and I think they are more than ready for it!
This sort is one of the five sorts in this resource. Check it out if you are curious!
Looking for just a single sort to address equivalent fractions? Check out this one!
### More fractions? YES!
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Coordinate Geometry – Exercise 7.2 – Class 10
1. Find the coordinates of the point which divides the join of (− 1, 7) and (4, − 3) in the ratio 2:3.
Solution:
Let P(x, y) be the required point. Using the section formula, we obtain
Therefore, the point is (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4, − 1) and (− 2, − 3).
Solution:
Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB Therefore, point P divides AB internally in the ratio 1:2.
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1 4 𝑡ℎ the distance AD on the 2nd line and posts a green flag. Preet runs 1 5 𝑡ℎ the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution:
It can be observed that Niharika posted the green flag at 1/4 of the distance AD i.e.,
[(¼) x 100)]m = 25
metre from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25). Similarly, Preet posted red flag at 1/5 of the distance AD i.e.,
[(1/5) x 100)]m = 25
metre from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20). Distance between these flags by using distance formula = GR
= √[(8 – 2)² + (25 – 20)²] = √[36+25] = √61 m
The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y).
x = (2+8)/2 = 10/2 = 5
y = (25+20)/2 = 45/2 = 22.5
Hence A(x,y) = (5, 22.5)
Therefore, Rashmi should post her blue flag at 22.5m on 5th line
4. Find the ratio in which the line segment joining the points (− 3, 10) and (6, − 8) is divided by (− 1, 6).
Solution:
Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be k:1.
Therefore, -1 = (6k-3)/(k+1)
-k-1 = 6k – 3
7k = 2
k = 2/7
therefore, the required ratio is 2:7
5. Find the ratio in which the line segment joining A (1, − 5) and B (− 4, 5) is divided by the x-axis. Also find the coordinates of the point of division
Solution:
Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by x – axis be k:1.
Therefore, the coordinates of the point of division is
We know that y-coordinate of any point on x-axis is 0.
Therefore, x-axis divides it in the ratio 1:1.
Division point =
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals. Therefore, O is the mid-point of AC and BD. If O is the mid-point of AC, then the coordinates of O are
7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, − 3) and B is (1, 4)
Solution:
Let the coordinates of point A be (x, y). Mid-point of AB is (2, −3), which is the center of the circle.
8. If A and B are (− 2, − 2) and (2, − 4), respectively, find the coordinates of P such that 𝐴𝑃 = 3/7𝐴𝐵 and P lies on the line segment AB.
Solution:
The coordinates of point A and B are (-2, -2) and (2, -4) respectively. Since AP = 𝐴𝑃 = 3/7𝐴𝐵
Therefore AP:PB = 3:7
Point P divides the line AB in the ratio 3:7
9. Find the coordinates of the points which divide the line segment joining A (− 2, 2) and B (2, 8) into four equal parts.
Solution:
From the figure it can be observd that the points P, Q, R divides the line segments in the ratio 1:3, 1:1, 3:1 respectively.
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (− 1, 4) and (− 2, −1) taken in order. [Hint: Area of a rhombus = ½ (product of its diagonals)]
Solution:
Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD. |
## How to discover the average Value
To uncover the Median, ar the number in value order and also find the middle.
You are watching: How to find the middle number between two numbers
### Example: discover the average of 12, 3 and 5
Put them in order:
3, 5, 12
The middle is 5, so the typical is 5.
### Example: 3, 13, 7, 5, 21, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29
When we put those numbers in order us have:
3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 39, 40, 56
There are fifteen numbers. Our center is the eighth number:
3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 39, 40, 56
The average value the this set of number is 23.
(It doesn"t matter that part numbers room the very same in the list.)
## Two numbers in the Middle
BUT, with an even amount of numbers things space slightly different.In that instance we find the middle pair of numbers, and also then discover the worth that is half way in between them. This is easily done by including them together and dividing by two.
### Example: 3, 13, 7, 5, 21, 23, 23, 40, 23, 14, 12, 56, 23, 29
When we put those number in order we have:
3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 40, 56
There are currently fourteen numbers and so us don"t have just one center number, we have a pair of middle numbers:
3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 40, 56
In this instance the middle numbers room 21 and 23.
To uncover the worth halfway between them, add them together and also divide by 2:
21 + 23 = 44then 44÷2 = 22
So the Median in this instance is 22.
(Note the 22 was not in the perform of numbers ... But that is yes sir because fifty percent the number in the list are less, and fifty percent the numbers room greater.)
## Where is the Middle?
A quick method to uncover the middle:
Count how countless numbers, add 1, then divide by 2
### Example: There space 45 numbers
45 plus 1 is 46, then division by 2 and also we obtain 23
So the average is the 23rd number in the sorted list.
### Example: There room 66 numbers
66 add to 1 is 67, then divide by 2 and also we gain 33.5
33 and a half? That means that the 33rd and 34th numbers in the sorted list space the two center numbers.
See more: How Many Men Did Matt Dillon Kill, Matt Dillon Kill Number
So to find the median: include the 33rd and also 34th numbers together and divide by 2.
693, 1457,739, 1458, 1459, 3055, 3056, 3057, 3792, 3793
median Mode central Measures |
## Precalculus (6th Edition) Blitzer
The value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$.
Consider the function $f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$. Simplify the above function, Multiplying the numerator and the denominator of the function by $\sqrt{{{x}^{2}}+9}+3$, So, \begin{align} & f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}\left( \frac{\sqrt{{{x}^{2}}+9}+3}{\sqrt{{{x}^{2}}+9}+3} \right) \\ & =\frac{{{x}^{2}}+9-9}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\ & =\frac{{{x}^{2}}}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\ & =\frac{1}{\sqrt{{{x}^{2}}+9}+3} \end{align} The function $g\left( x \right)={{x}^{2}}+9$ is a polynomial. Now, find the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$, \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{{{x}^{2}}+9}+3} \\ & =\frac{1}{\sqrt{\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}+9}+3} \\ & =\frac{1}{\sqrt{0+9}+3} \\ & =\frac{1}{\sqrt{9}+3} \end{align} \begin{align} & =\frac{1}{3+3} \\ & =\frac{1}{6} \\ \end{align} Thus, the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$. |
# How do you solve sqrt(2x+3)=6?
Jun 23, 2018
$x = \frac{33}{2}$
#### Explanation:
$\textcolor{b l u e}{\text{square both sides}}$
${\left(\sqrt{2 x + 3}\right)}^{2} = {6}^{2}$
$2 x + 3 = 36$
$\text{subtract 3 from both sides}$
$2 x = 36 - 3 = 33$
$\text{divide both sides by 2}$
$x = \frac{33}{2}$
$\textcolor{b l u e}{\text{As a check}}$
$\sqrt{2 \times \frac{33}{2} + 3} = \sqrt{33 + 3} = \sqrt{36} = 6 \leftarrow \text{correct}$
Jun 23, 2018
See a solution process below:
#### Explanation:
First, square both sides of the equation to eliminate the radical while keeping the equation balanced::
${\left(\sqrt{2 x + 3}\right)}^{2} = {6}^{2}$
$2 x + 3 = 36$
Next, subtract $\textcolor{red}{3}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:
$2 x + 3 - \textcolor{red}{3} = 36 - \textcolor{red}{3}$
$2 x + 0 = 33$
$2 x = 33$
Now, divide each side of the equation by $\textcolor{red}{2}$ to solve for $x$ while keeping the equation balanced:
$\frac{2 x}{\textcolor{red}{2}} = \frac{33}{\textcolor{red}{2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = \frac{33}{2}$
$x = \frac{33}{2}$ |
Breaking News
Home » Speed Maths Tricks » Comparing Fractions – Speed Maths Tricks
# Comparing Fractions – Speed Maths Tricks
In IBPS Bank Exams, we can expect 1 or 2 questions on comparing fractions as shown below
### Examples:
(1) Arrange the given fractions in ascending or descending order
(2) To find the smallest fraction of the given fractional numbers
(3) Which is the 2nd highest fraction
(4) Which is 5th lowest fraction from the following fractions
(5) Find the correct order of the given fractions from smallest to biggest number. etc.,
By using the following trick we can answer these questions in very less time and in a simple way. Let us solve some example questions
Arranging the fractions by comparing
Ascending Order: Small to Big number
Descending Order: Big to Small number.
To explain, we feel this method as the long process but after understanding properly we can solve quicker than other methods.
## Example Problems on Comparision of Fractions: (From Previous Question Papers)
Usually, these type of problems can be solved by converting fractions into decimal values but it is time-consuming. Instead of converting fractions into decimals, compare by multiplying numerator and denominator of two fractions. Whichever is greater that fraction is the bigger than the other number.
1. Out of the fractions 9/31, 3/17, 6/23, 4/11 and 7/25. Which is the largest fraction?
(a) 9/31
(b) 3/17
(c) 6/23
(d) 4/11
2. Which of the following fractions is the least ?
(a) 12/119
(b) 1/10
(c) 4/39
(d) 7/69
3. If the fraction 3/5, 5/8, 16/25, 4/5, 9/16 are arranged in ascending order of their values, Which one will be the second?
(a) 3/5
(b) 16/25
(c) 9/16
(d) 4/5
4. 4/5, 7/6, 1/5, 6/7 Find the correct order of arrangement (ascending order) from the following options
(a) 4/5 < 7/6 < 6/7 < 1/5
(b) 7/6 < 6/7 < 4/5 < 1/5
(c) 1/5 < 7/6 < 6/7 < 4/5
(d) 1/5 < 4/5 < 6/7 < 7/6
Solution: First choose any fraction (example 4/5) and compare with other numbers
4/5 and 7/6 => cross multiply so that you get
24 35, here 35 is big so 7/6 is bigger than 4/5
compare 7/6 with other numbers
7/6 and 6/7
49 36 So 49 is bigger then 7/6 is bigger than 6/7
7/6 and 1/5
35 6 so 7/6 is greater than all other numbers
Then compare remaining numbers
1/5 < 4/5 < 6/7 < 7/6
5 < 20 < 30 < 49 ( check the cross multiplication of numerator of 1st number with denominator of 2nd number and observe)
5. Which of the following fraction is highest?
(a) 18/7
(b) 13/5
(c) 6/5
(d) 9/11
### Comparing Fractions Practice Problems for IBPS Clerks
1. Out of the fractions 9/31, 3/17, 6/23, 4/11 and 7/25. Which is the largest fraction?
2. Out of the fractions 3/7, 4/9, 6/11, 8/13 and 7/15. Which is the second highest fraction?
If the fractions 4/5, 2/7, 9/13, 6/11 and 5/9 are arranged in ascending order of their values. Which one will be the second?
If the fractions 3/5,2/3,1/4, 5/7 and 6/11 are arranged in ascending order of their values which one will be the fourth?
If the following fractions 7/8,4/5, 8/14, 3/5 and 5/6 are arranged in descending order which will be the last in the series?
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# How do you solve the following linear system? :$7x + 3y = 27,3x - 4y = 7$?
Last updated date: 09th Aug 2024
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Hint: Make the coefficient of any one of the variables the same by multiplying the equations with respective constants.
Add or subtract them to eliminate that variable from the equation.
Find the value of one variable from the new equation.
Substitute the value of this variable in one of the two original equations to get the second variable.
Complete step by step solution:
The given equation is $7x + 3y = 27,3x - 4y = 7$
$7x + 3y = 27 \to (1)$
$3x - 4y = 7 \to (2)$
We use the elimination method
Multiply first equation by $4$ and second by $3$, hence we get
$(1) \times 4 \Rightarrow 28x + 12y = 108$
$(2) \times 3 \Rightarrow 9x - 12y = 21$
Now we use the elimination method
Add these two equations to eliminate $y$
$(28x+12y=108)+(9x-12y=21) \\ \Rightarrow 28x+9x=12y-12y=108+21$
The zero terms vanish, hence we get
$\Rightarrow$$37x = 129 Divide by 37 on both sides, hence we get \Rightarrow$$\dfrac{{\not{{37}}}}{{\not{{37}}}}x = \dfrac{{129}}{{37}}$
$\Rightarrow$$x = \dfrac{{129}}{{37}} Now substitute this value into the second equation, hence we get 3x - 4y = 7 \Rightarrow$$3 \times \dfrac{{129}}{{37}} - 4y = 7$
Interchange the variable and constant, hence we get
$\Rightarrow$$4y = 3 \times \dfrac{{129}}{{37}} - 7 Now simplify the RHS (Right Hand Side) \Rightarrow$$4y = \dfrac{{387 - 259}}{{37}}$
Subtract the numerator, hence we get
$\Rightarrow$$y = \dfrac{{128}}{{4 \times 37}} = \dfrac{{128}}{{148}} The result is \Rightarrow$$y = \dfrac{{64}}{{74}} = \dfrac{{32}}{{37}}$
Therefore the value of x is $\dfrac{{129}}{{37}}$ and the value of y is $\dfrac{{32}}{{37}}$.
Note: We use the alternative method.
The alternative method is the substitution method.
The given equation is $7x + 3y = 27,3x - 4y = 7$
$7x + 3y = 27 \to (1)$
$3x - 4y = 7 \to (2)$
We use the substitution method.
Let’s take the first equation.
$7x + 3y = 27$
Isolate the $x$ term, hence we get
$\Rightarrow$$7x = 27 - 3y$
Divide by $7$ on both sides
$\Rightarrow$$\dfrac{{\not{7}}}{{\not{7}}}x = \dfrac{{27 - 3y}}{7}$
Then,
$x = \dfrac{{27 - 3y}}{7}$
Now, the $x$ value substitute in the second equation. The second equation is $3x - 4y = 7$, hence we get
$\Rightarrow$$3\left( {\dfrac{{27 - 3y}}{7}} \right) - 4y = 7 Now we simplify this equation Multiply by 3 \Rightarrow$$\left( {\dfrac{{81 - 9y}}{7}} \right) - 4y = 7$
Now take LCM in LHS (Left Hand Side), hence we get
$\Rightarrow$$\dfrac{{81 - 9y - 4y \times 7}}{7} = 7$
Multiply $- 4y$ by $7$, hence we get
$\Rightarrow$$\dfrac{{81 - 9y - 28y}}{7} = 7$
Add the $y$ variable in the numerator, hence we get
$\Rightarrow$$\dfrac{{81 - 37y}}{7} = 7$
Multiply by $7$ on both sides, hence we get
$\Rightarrow$$81 - 37y = 49 Now we isolate the y \Rightarrow$$37y = 81 - 49$
Subtract on RHS
$\Rightarrow$$37y = 32 Divide by 37 on both sides \Rightarrow$$y = \dfrac{{32}}{{37}}$
Now we find the $y$ value
The $y$ value substitute in the second equation
$\Rightarrow$$3x - 4\left( {\dfrac{{32}}{{37}}} \right) = 7 Multiply by 4 in the numerator, hence we get \Rightarrow$$3x - \dfrac{{128}}{{37}} = 7$
Now take LCM in LHS (Left Hand Side), hence we get
$\Rightarrow$$\dfrac{{3x(37) - 128}}{{37}} = 7 Multiply 3x by 37, hence we get \Rightarrow \dfrac{{111x - 128}}{{37}} = 7 Multiply by 37 on both sides, hence we get \Rightarrow$$111x - 128 = 259$
Isolate the $x$ term, hence we get
$\Rightarrow$$111x = 259 + 128 Add the denominator \Rightarrow 111x = 387 Divide by111on both sides, hence we get \Rightarrow$$x = \dfrac{{387}}{{111}}$
Now we divide by $111$
$\Rightarrow$$x = \dfrac{{129}}{{37}}$
Hence, $x = \dfrac{{129}}{{37}}$ and $y = \dfrac{{32}}{{37}}$. |
# How do you solve 7d ^ { 2} + 10d + 168= 0?
Mar 25, 2018
$d = - \frac{5}{7} \pm \frac{\sqrt{1151}}{7} i$
#### Explanation:
We can find the complex roots by completing the square and using the difference of squares identity:
${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$
with $A = 7 d + 5$ and $B = \sqrt{1151} i$ (where ${i}^{2} = - 1$) as follows:
$0 = 7 \left(7 {d}^{2} + 10 d + 168\right)$
$\textcolor{w h i t e}{0} = 49 {d}^{2} + 70 d + 1176$
$\textcolor{w h i t e}{0} = 49 {d}^{2} + 70 d + 25 + 1151$
$\textcolor{w h i t e}{0} = {\left(7 d\right)}^{2} + 2 \left(7 d\right) \left(5\right) + {\left(5\right)}^{2} + 1151$
$\textcolor{w h i t e}{0} = {\left(7 d + 5\right)}^{2} + {\left(\sqrt{1151}\right)}^{2}$
$\textcolor{w h i t e}{0} = {\left(7 d + 5\right)}^{2} - {\left(\sqrt{1151} i\right)}^{2}$
$\textcolor{w h i t e}{0} = \left(\left(7 d + 5\right) - \sqrt{1151} i\right) \left(\left(7 d + 5\right) + \sqrt{1151} i\right)$
$\textcolor{w h i t e}{0} = \left(7 d + 5 - \sqrt{1151} i\right) \left(7 d + 5 + \sqrt{1151} i\right)$
So:
$7 d = - 5 \pm \sqrt{1151} i$
and:
$d = - \frac{5}{7} \pm \frac{\sqrt{1151}}{7} i$
Mar 25, 2018
$d = \frac{- 5 + i \sqrt{1151}}{7}$ or $d = \frac{- 5 - i \sqrt{1151}}{7}$
#### Explanation:
we use this formula
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
in this case our $a$ is $7$, $x$ is $d$, our $b$ is $10$ and our $c$ is $168$
putting them into equation gives us
$d = \frac{- 10 \pm \sqrt{{10}^{2} - \left(4\right) \left(7\right) \left(168\right)}}{2 \left(7\right)}$
by solving we get
$d = \frac{- 5 \pm i \sqrt{1151}}{7}$
$\implies d = \frac{- 5 - i \sqrt{1151}}{7} \mathmr{and} d = \frac{- 5 + i \sqrt{1151}}{7}$ |
# Pentagon
The pentagon is a geometric figure made up of five sides, in addition to having five vertices and five internal angles.
That is, the pentagon is a polygon that has five sides, being of greater complexity than a quadrilateral and a triangle.
It should be noted that a polygon is a two-dimensional figure made up of a finite number of non-collinear consecutive segments, forming a closed space.
## Pentagon elements
Guiding us from the image below, the elements of the pentagon are the following:
• Vertices: A, B, C, D, E.
• Sides: AB, BC, CD, DE, AE.
• Interior angles: α, β, δ, γ, ε. They add up to 540º.
• Diagonals: Each interior angle is divided into three and there are five: AC, AD, BD, BE, CE.
## Pentagon types
We have two types of pentagon, according to their regularity:
• Regular: All its sides measure the same and also all its internal angles are equal and measure 108º, adding 540º. The two diagonals emerging from each vertex divide the corresponding internal angle into three equal parts measuring 36º (108º / 3).
• Irregular: Its sides have different lengths.
## Perimeter and area of a pentagon
To better understand the characteristics of a pentagon, we can calculate its perimeter and area:
• Perimeter (P): We add the sides of the polygon, that is: P = AB + BC + CD + DE + AE. If the pentagon is regular and all the sides have length L, it is true that P = 5L
• Area (A): We can also distinguish two cases. When it is an irregular pentagon, we could divide the figure into triangles, as we see in the image below. Thus, if we know the length of the diagonals, we can calculate the area of each triangle (as we explained in the triangle article) and do the summation.
In the example above, we could calculate the area of the triangles FGJ, GJI, and GHI.
Meanwhile, if the pentagon is regular, we can calculate the area based on the length of its side, following the following formula:
Likewise, we can calculate the area as a function of the apothem (which in the figure below is the QR segment), which is the segment that joins the center of a regular polygon with the midpoint of any of its sides, forming a right angle (which measures 90º). So the formula would be (where to the apothem and P the perimeter):
## Pentagon example
Suppose we have a regular pentagon with one side measuring 13 meters. What is the area and perimeter of the figure?
The perimeter would be:
P = 5 x 13 = 65 meters
Meanwhile, the area would be calculated as follows:
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Writing absolute value inequalities in interval notation
So something that meets both of these constraints will satisfy the equation. Likewise, up here, anything greater than positive 21 will also have an absolute value greater than Then divide both sides by -3, leaving x on the left side of the inequality, and 5 on the right. Now factor the left side: There is no y in the original problem, it was something that was added to make the graphing convenient.
Normally, the author and publisher would be credited here. The range of possible values for d includes any number that is less than 0. We're doing this case right here. Solving One- and Two-Step Absolute Value Inequalities The same Properties of Inequality apply when solving an absolute value inequality as when solving a regular inequality.
So its absolute value is going to be greater than a. Divide both sides by 5. Graph and give the interval notation equivalent: You have to be in this range. Find the real solutions ignore complex solutions involving i to the function any way that you want to.
If the test point gives you a true statement, then any point in that interval will work, and you want to include that interval in the answer. Let's do positive 21, and let's do a negative 21 here.
And actually, we've solved it, because this is only a one-step equation there. What about the example Let's rewrite this as which we can translate into the quest for those numbers x whose distance to -1 is at least 3.
If the inequality is less than zero or less than or equal to zero, then you want all of the negative sections found in the sign analysis chart.
If you're going to change from being less than zero to being greater than zero and you can't pick up your pencil, then at some point, you must cross the x-axis. That group would be described by this inequality:Absolute value inequalities will produce two solution sets due to the nature of absolute value.
We solve by writing two equations: one equal to a positive value and one equal to a negative value. Absolute value inequality solutions can be verified by graphing.
"Set notation" writes the solution as a set of points. The above solution would be written in set notation as "{x | xis a real number, x. College Algebra: Inequalities & Interval Notation. Solving Inequalities Interval Notation, Number Line, Absolute Value, Fractions & Variables - Algebra.
Graphing Linear Inequalities from Standard Form.
Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Use interval notation to describe sets of numbers as intersections and unions.
When two inequalities are joined by the word and, the solution of the compound inequality occurs when both inequalities are true at the same time. It is the overlap, or intersection, of the solutions for each inequality.
agronumericus.com gives useful advice on interval notation calculator, rational numbers and arithmetic and other algebra subjects.
In the event that you have to have assistance on value as well as elementary algebra, agronumericus.com is the ideal place to head to!
Writing absolute value inequalities in interval notation
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# Dot Product Calculator
Input the values of vectors a & b and click the calculate button to find a.b using dot product calculator
First vector (a)
i
j
k
Second vector (b)
i
j
k
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Dot product calculator is used to find the dot product of two vectors. It calculates the dot product of 2-dimentional & 3-dimentional vectors. This calculator provides the step-by-step solution to the given problems.
## What is dot product?
The dot product is a fundamental way of combining two vectors. It is used to know the direction of two vectors. The dot product of two vectors is also known as the resultant and it is a scalar quantity.
The dot product is denoted by a.b and read as “a dot b”.
## Formula of the dot product
The formula of the dot product is:
a.b = |a||b| cos(θ) … (i)
• a & b are two vectors.
• |a| & |b| are the magnitudes of vectors a & b.
• “θ” is the angle between a & b.
In mathematics, we can derive another form of dot product. Let a = (a1, a2, a3) & b = (b1, b2, b3) be two vectors. Now according to the Law of cosines,
|a – b|2 = |a|2 + |b|2 – 2|a||b| cos(θ)
Let’s simplify the above expression with the help of the magnitude of a vector formula.
(a1 – b1)2 + (a2 – b2)2 + (a3 – b3)2 = (a12 + a22 + a32) + (b12 + b22 + b32) – 2|a||b| cos(θ)
Open the square of each term.
(a12 - 2a1b1 + b12) + (a22 - 2a2b2 + b22) + (a32 - 2a3b3 + b32) = (a12 + a22 + a32) + (b12 + b22 + b32) – 2|a||b| cos(θ)
Now take the similar terms on the same side of equation.
(a12 - 2a1b1 + b12) + (a22 - 2a2b2 + b22) + (a32 - 2a3b3 + b32) - (a12 + a22 + a32) - (b12 + b22 + b32) = – 2|a||b| cos(θ)
After subtracting the similar terms, we get
- 2a1b1 + (-2a2b2) + (-2a3b3) = – 2|a||b| cos(θ)
- 2(a1b1 + a2b2 + a3b3) = – 2|a||b| cos(θ)
(a1b1 + a2b2 + a3b3) = |a||b| cos(θ) … (ii)
Now put equation (ii) in (i), we get
a.b = (a1b1 + a2b2 + a3b3)
This formula is used to find the dot product of two vectors.
## How to calculate the dot product of two vectors?
Follow the below example to learn how to calculate the dot product of two vectors.
Example
Find the dot product of vectors a = (2i, 3j, 6k) and b = (4i, 7j, 9k).
Solution
Step 1: Take the given vectors.
a = (2i, 3j, 6k)
b = (4i, 7j, 9k)
Step 2: Take the formula of dot product.
a.b = (a1b1 + a2b2 + a3b3)
Step 3: Substitute the given values of vectors a & b in the formula.
a.b = ((2)(4) + (3)(7) + (6)(9))
a.b = (8 + 21 + 54)
a.b = 83 |
# Quick Answer: What Is The Condition For Two Vectors To Be Collinear?
## What if two vectors are collinear?
Definition 2 Two vectors are collinear, if they lie on the same line or parallel lines.
In the figure above all vectors but f are collinear to each other.
Definition 3 Two collinear vectors are called co-directed if they have the same direction..
## Are 2 points always collinear?
Obviously two points are always collinear, since a straight line can always be drawn through two points. Sometimes it is spelled ‘colinear’ (with one L).
## How do you know if two planes are parallel?
To say whether the planes are parallel, we’ll set up our ratio inequality using the direction numbers from their normal vectors. Since the ratios are not equal, the planes are not parallel. To say whether the planes are perpendicular, we’ll take the dot product of their normal vectors.
## How do you know if points lie on a straight line?
Explanation: To determine if a point is on a line you can simply subsitute the x and y coordinates into the equation. Another way to solve the problem would be to graph the line and see if it falls on the line. Plugging in will give which is a true statement, so it is on the line.
## Does parallel mean same direction?
extending in the same direction, equidistant at all points, and never converging or diverging: parallel rows of trees. having the same direction, course, nature, or tendency; corresponding; similar; analogous: Canada and the U.S. have many parallel economic interests.
## What are 3 non collinear points?
Points B, E, C and F do not lie on that line. Hence, these points A, B, C, D, E, F are called non – collinear points. If we join three non – collinear points L, M and N lie on the plane of paper, then we will get a closed figure bounded by three line segments LM, MN and NL.
## Which is a set of collinear points?
In Geometry, a set of points are said to be collinear if they all lie on a single line. Because there is a line between any two points, every pair of points is collinear.
## How do you prove two lines are collinear?
Slope formula method to find that points are collinear. Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C, three pairs of points can be formed, they are: AB, BC and AC. If Slope of AB = slope of BC = slope of AC, then A, B and C are collinear points.
## How do you know if a vector is parallel?
Two vectors A and B are parallel if and only if they are scalar multiples of one another. A = k B , k is a constant not equal to zero. Two vectors A and B are perpendicular if and only if their scalar product is equal to zero.
## Are collinear vectors parallel?
Parallel vectors are vectors which have same or parallel support. They can have equal or unequal magnitudes and their directions may be same or opposite. Two vectors are collinear if they have the same direction or are parallel or anti-parallel.
## What is collinear example?
Three or more points that lie on the same line are collinear points . Example : The points A , B and C lie on the line m . They are collinear.
## What is an equal vector?
Two or more vectors are equal when they have the same length and they point in the same direction. Any two or more vectors will be equal if they are collinear, codirected, and have the same magnitude. … In other words, two or more vectors are equal if their coordinates are equal.
## How do you show that points are collinear using the formula?
Expert Answer:We need to prove the points (3,-2),(5,2) and(8,8) are collinear.A=(3,-2) B=(5,2) C=(8,8)Let The points B divides AC in the ratio of k:1.Then the coordinates will be,Coordinates of B are (5,2)Comparing we get,Value of k is same in both.Therefore Points A,B,C are collinears.
## How do you prove three points are collinear?
Three or more points are said to be collinear if they all lie on the same straight line. If A, B and C are collinear then. If you want to show that three points are collinear, choose two line segments, for example and .
## What happens when two vectors are perpendicular to each other?
If two vectors are perpendicular to each other, then their dot product is equal to zero.
## How do you find slope with 3 points?
How to find slopeIdentify the coordinates (x₁,y₁) and (x₂,y₂) . … Input the values into the formula. … Subtract the values in parentheses to get 2/(-5) .Simplify the fraction to get the slope of -2/5 .Check your result using the slope calculator.
## What is the condition of collinear?
In general, three points A, B and C are collinear if the sum of the lengths of any two line segments among AB, BC and CA is equal to the length of the remaining line segment, that is, either AB + BC = AC or AC +CB = AB or BA + AC = BC.
## How do you know if vectors lie on the same line?
It’s really easy once you know how There are two facts you need to know: If vectors are multiples of each other, they’re parallel; If two parallel vectors start at the same point, that point and the two end points are in a straight line.
## What does it mean for two vectors to be parallel?
Two vectors u and v are said to be parallel if they have either the same direction or opposite direction. This means that each is a scalar multiple of the other: for some non-zero scalar s, v = su and so u = v.
## Can lines be collinear?
. A line on which points lie, especially if it is related to a geometric figure such as a triangle, is sometimes called an axis. Two points are trivially collinear since two points determine a line.
## What is the collinear formula?
If the A, B and C are three collinear points then AB + BC = AC or AB = AC – BC or BC = AC – AB. or. If the area of triangle is zero then the points are called collinear points. If three points (x1, y1), (x2, y2) and (x3, y3) are collinear then [x1(y2 – y3) + x2( y3 – y1)+ x3(y1 – y2)] = 0.
## What is the condition for collinear vector?
Three points with position vectors a, b and c are collinear if and only if the vectors (a−b) and (a−c) are parallel. In other words, to prove collinearity, we would need to show (a−b)=k(a−c) for some constant k.
## How do you know if 3d vectors are collinear?
Collinear 3 dimentional lines. Collinear points are all located on the same line. Another way of checking whether the points are collinear is by calculating the area formed by the points, if the area is zero then the points are collinear.
## How do you prove that three points are not collinear?
Use Heron’s formula (and these three distances) to find the area of the triangle. If the area is positive, then the three points are not collinear. They form a triangle. If the area is 0, then the three points are collinear. |
S k i l l
i n
A R I T H M E T I C
Lesson 3 Section 2
The Meaning of Percent
The student should first understand Section 1: Multiplying and Dividing by Powers of 10.
7. What does "percent" mean? Percent is an abbreviation for the Latin per centum, which means for each 100.
Thus, 100% means 100 for each 100, which is to say, all. 100% of 12 is all of 12.
50% is another way of saying half, because 50% means 50 for each 100, which is half. 50% of 12 is 6.
Example 1. Below are 100 squares, and 32 have been shaded.
What percent of the squares have been shaded?
Answer. 32% -- 32 for each 100.
When the percent is less than or equal to 100%, then we can say "out of" 100. 32% is 32 out of 100. (But to say that 200% is 200 out of 100 makes no sense. 200% is 200 for each 100, which is to say, twice as much.)
Example 2. 100 people were surveyed, and 65 responded Yes. What percent responded Yes?
Answer. 65% -- 65 out of 100.
Example 3. In a class of 30 students, all 30 came to school by bus. What percent came to school by bus?
(30 for each 30 is equivalent to 100 for each 100.)
8. How do we take 1% of a number? 1% of \$200 Divide it by 100. For, 1% means 1 for each 100, which is the hundredth part.
1% = 1 for each 100.
Examples.
1% of \$200 is \$2.00 Divide by 100: Separate two decimal places. 1% of \$250 is \$2.50 Again, divide by 100: Separate two decimal places. 1% of \$6.00 is \$.06 Move the point two places left. Do not write the 0's that remain on the extreme right of the decimal: 6.00 ÷ 100 = .0600 = .06 (Lesson 2, Question 8.) 1% of \$1,200 is \$12.00 Divide by 100: Separate two decimal places, or simply drop the two 0's.
Problem 1. How much is 1% of \$400? How much is 2% of \$400? How much is 3%? How much is 9%?
Answer. 1% of \$400 is \$4.00. Separate two decimal places.
Now, 2% is twice as much as 1%. Therefore 2% of \$400 is \$8. 3% is \$12. 4% would be \$16. 9%, therefore, is 9 × \$4 = \$36.
Problem 2. How much is 8% of \$600?
Answer. Since 1% is \$6.00, then 8% is 8 × \$6.00 = \$48.00.
Problem 3. How much is 2% of \$325? How much is 3%? 4%?
Answer. We can get everything from 1%, which is \$3.25.
2%, therefore, is \$6.50.
3% is \$9.75.
And 4% is 4 × \$3.25 = 4 × \$3 + 4 × \$.25 = \$12 + \$1 = \$13.
These are problems that do not require a calculator. They should be done mentally!
9. How do we take 10% of a number? 10% of \$600 Divide it by 10. For, 10% means 10 for each 100. Each 10 is a tenth part of 100.
Examples.
10% of \$600 is \$60. Divide by 10: Take off one 0, or separate one decimal place. 10% of \$625 is \$62.50 Divide by 10. Separate one decimal place: \$62.5 But we write money with two decimal places (cents). Therefore we must add a 0 onto the right. 10% of \$6.00 is \$.60 Move the point one place left. Again, we write money with two decimal places.
Problem. How much is 20% of \$80? How much is 30%? How much is 90%
Answer. 20% is twice as much as 10%. Since 10% of \$80 is \$8, then 20% is 2 × \$8 = \$16. 30% is 3 × \$8 = \$24. 90% is 9 × \$8 = \$72.
The meaning of percent continues in Lesson 14 and Lesson 27.
10. How do we change a percent to a number? 24% = ? Divide by 100.
(This is often called changing a percent to a decimal.)
24% = .24
Divide by 100 -- separate two decimal places. For, 24% -- 24 for each 100 -- means 24 hundredths.
Division by 100 is indicated by the percent sign itself %, with its division slash / and two 0's.
Here are more examples:
.24% = .0024 Divide by 100: Move the decimal point two places left. 9% = .09 Divide by 100: Mark off two decimal places. 650% = 6.5 Divide by 100: Mark off two decimal places. It is not necessary to write the 0 that remains on the right of the decimal. 650% = 6.50 = 6.5 6.5% = .065 Divide by 100: Move the decimal point two places left.
Strictly, a percent is not a number, it is a ratio (Lesson 27). Therefore we should speak rather of a number representing a percent. That becomes necessary only for certain written calculations.
11. How do we change a number to a percent? .24 = ? % Multiply it by 100, and add the % sign.
Examples.
.24 = 24% Multiply by 100: Move the point two places right. 2.4 = 240% Move the point two places right. 24 = 2400% Multiply by 100: Add on two 0's. .024 = 2.4% Move the point two places right.
Number to a Percent
.24 = 24%
Percent to a Number
Please "turn" the page and do some Problems.
or
Continue on to the next Lesson.
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Unitary Method
Algebra
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Order Relation
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Circle
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Home >> Decimal >> Read Decimals >>
Decimal Point Place Value of digits in decimals Expanded Form of Decimal Read Decimals Decimal Places Decimal to Fraction Decimal to Mixed Fraction Decimal to Lowest Fractional form Decimal into Percentage Like Decimals Unlike Decimals Difference Like & Unlike Decimals Comparing or Ordering Decimals Ascending Order of Decimals Descending Order of Decimals Multiplication of Decimals Division of Decimals
Before you understand this topic, you must known:
How to find place value of digits in decimals ?
What is Decimal Point ?
Now to understand how to pronounce or read Decimals, study the following examples:
Example 1: Write the decimal 1234.5 in readable form.
Solution: Given decimals is 1234.5
And the place value of all the digits of decimal 1234.5 is as follows:
Firstly, write Place Value of Digits on the R.H.S. of the decimal point and we get
Digit 5 is at tenths place
Now, write Place Value of Digits on the L.H.S. of the decimal point and we get
Digit 4 is at ones place
Digit 3 is at tens place
Digit 2 is at hundreds place
Digit 1 is at thousands place
Now as per the place values, given decimal 1234.5 can be read as:
One thousand two hundred thirty four point five
Example 2: Write the decimal 1234.56 in readable form.
Solution: Given decimals is 1234.56
And the place value of all the digits of decimal 1234.56 is as follows:
Firstly, write Place Value of Digits on the R.H.S. of the decimal point and we get
Digit 5 is at tenths place
Digit 6 is at hundredths place
Now, write Place Value of Digits on the L.H.S. of the decimal point and we get
Digit 4 is at ones place
Digit 3 is at tens place
Digit 2 is at hundreds place
Digit 1 is at thousands place
Now as per the place values, given decimal 1234.56 can be read as:
One thousand two hundred thirty four point five six
Study More Solved Questions / Examples
Write the following decimals in readable form. A) 1234.567 B) 16.325 C) 100.1 D) 9810.001 E) 4332.409, 488.428, 36199.619
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# Total Surface Area of Cylinder
## Total Surface Area of Cylinder
The total surface area (TSA) of a cylinder is the sum of its curved surface area and the area of its two circular bases. The curvedsurface area (CSA) of a cylinder is given by the product of its circumference of circular base (2πr) and the height (h) of the cylinder i.e. CSA = 2πrh. And the area of circular base is πr2.
Hence, the total surface area of cylinder = Curved surface area + 2 × area of circular base
= 2πrh + 2πr2
= 2πr(h + r)
Total surface area (TSA) of cylinder = 2πr(r + h)
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10 Math Problems officially announces the release of Quick Math Solver and 10 Math ProblemsApps on Google Play Store for students around the world.
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Since circumference (c) = 2πr, we can also write the formula of total surface area of cylinder as,
Total surface area (TSA) of cylinder = c(r + h)
### Workout Examples
Example 1: Find the total surface area of the given cylinder.
Solution: Here,
Diameter of the cylinder (d) = 14cm
Radius of the cylinder (r) = 14cm/2 = 7cm
Height of the cylinder (h) = 20cm
We know,
Total surface area (TSA) = 2πr(r + h)
= 2 × 22/7 × 7 (7 + 20)
= 2 × 22 × 27
= 1188cm2
Thus, the total surface area of cylinder is 1188cm2.
Example 2: The total surface area of a cylinder is 231 cm2. Its CSA (curved surface area) is 2/3 of total surface area. Find the radius of its base and height.
Solution: Here,
Total surface area of the cylinder = 231cm2
Curved surface area = 2/3 of the total surface area of cylinder
i.e. 2πrh = 2/3 × 231
or, 2 × 22/7 × rh = 154
or, rh = 154 × 7/44
or, rh = 24.5 ………………… (i)
Again, Curved surface area = 2/3 of total surface area
i.e. 2πrh = 2/3 × 2πr(r + h)
or, h = 2/3 × (r + h)
or, 3h = 2r + 2h
or, h = 2r ………………….. (i)
From (i) and (ii),
r × 2r = 24.5
or, r2 = 24.5/2
or, r2 = 12.25
or, r = 3.5cm
From (ii),
h = 2r = 2 × 3.5 = 7cm
Thus, the radius is 3.5cm and height is 7cm.
Example 3: 50 circular plates, each of radius 7cm and thickness 0.5cm are placed one above the other to form a right circular cylinder. Find its total surface area.
Solution: Here,
Number of circular plates (N) = 50
Thickness of each plate = 0.5cm
Total height of cylinder (h) = 50 × 0.5cm = 25cm
Radius of each plate (r) = 7cm
Now,
The total surface area (TSA) of cylinder = 2πr(r + h)
= 2 × 22/7 × 7 × (7 + 25)
= 2 × 22 × 32
= 1408cm2
Thus, the total surface area of cylinder is 1408cm2.
You can comment your questions or problems regarding the TSA of cylinder here. |
Study Guide
# Squares and Square Roots - Solving Radical Equations
## Solving Radical Equations
A radical equation is an equation that features a variable contained inside a radicand. At least it won't get wet if it rains.
An example of a radical equation is .
The equation is not a radical equation, because the variable doesn't occur inside the radicand. The 5 and 9 are making it wait outside, apparently, and it's getting drenched.
To solve a radical equation that has a single radical on one side and a number on the other, we square both sides to eliminate the radical and then solve the resulting equation. It'll only get more complicated from here, so enjoy it while it lasts.
### Sample Problem
Solve: .
We square both sides to eliminate the radical. This gives us x + 1 = 49, so x = 48.
Check to make sure this works. Plugging x = 48 into the left-hand side of the equation yields , which is totally 7.
Sometimes we need to rearrange the equation a little so that the radical is all by itself, or "isolated" on one side of the equation. Poor little radical.
### Sample Problem
Solve: .
First we need to add 1 to each side to get the radical all by itself on the left-hand side of the equation. Don't feel badly for him; he's a loner anyway. He'll appreciate the solitude:
.
Then, we square both sides to eliminate the radical:
5 + x = 16.
Solving this equation gives us x = 11.
We'll push our double-checking agenda again here. Plug x = 11 into the left-hand side of the equation:
...which is, in fact, 3. Looks like our train hasn't come off the tracks. The two sides of the equation agree, so our solution is x = 11.
As we mentioned, some radical equations have no solution and are so obvious about it that we don't need to do any work. We have no problem with that. Subtlety is overrated.
### Sample Problem
Solve: .
The radical is already by itself. It even hung a "Do Not Disturb" sign on its door. We square both sides:
2x + 6 = x2 + 6x + 9
Now we rearrange for a nice quadratic equation, being sure we've got zero off to one side:
0 = x2 + 4x + 3
This equation factors as:
0 = (x + 3)(x + 1)
So our solutions are x = -3 and x = -1.
As always, we'll cover our butts by checking to make sure both of these are correct solutions.
When x = -3, the left-hand side of the original equation is:
And the right-hand side of the original equation is:
-3 + 3 = 0
Since the two sides agree, x = -3 is a solution to the original equation.
When x = -1, the left-hand side of the original equation is:
And the right-hand side is:
(-1) + 3 = 2
So x = -1 is also a solution to the original equation. Excellent. Butts officially covered.
Be careful: When solving radical equations, make sure to check your answers. Sometimes they won't actually be solutions to the original equation. You don't want egg on your face. Especially if it's been soft-boiled.
### Sample Problem
Solve:
We square both sides to find:
4x + 5 = x2
Rearrange for a nice-looking quadratic equation. Wow, has this thing been working out?
0 = x2 – 4x – 5
This factors as:
0 = (x – 5)(x + 1)
So our solutions should be x = 5 and x = -1. Now we check to make sure both of these are solutions to the original equation.
When x = 5, the left-hand side of the original equation is , and the right-hand side of the original equation is also 5, so x = 5 is a solution.
When x = -1, the left-hand side of the original equation is .
However, when x = -1 the right-hand side of the original equation is -1. Uh-oh.
Since the left-hand and right-hand sides of the equation don't agree for x = -1, this isn't a solution. Argh, so close: x = -1 is an extraneous solution. In other words, it's a number we found through legitimate means, but it's not a solution to the original equation.
It's like if you put in an honest day's work and then got paid in Trident gum. You got something for your efforts, but it's not what you wanted, and it won't help you pay for anything. Although at least it does taste exactly like green apple and golden pineapple. Well, that's something.
Our final answer for this problem is the single solution x = 5.
• ### The Pythagorean Theorem
Before we state the Pythagorean Theorem, we'll introduce a right triangle. Reader, right triangle. Right triangle, reader. We have a feeling you two will get along. You have a lot in common. You both have two legs, are most comfortable at 90 degrees, and are always playing some angle. You also have the same taste in music.
A right triangle is a triangle with a right angle in it, as indicated in the image below by the little box in the lower right corner.
The sides of the triangle that form the right angle are called legs, and the long side across from the right angle is the hypotenuse.
We really, really, really hope you like right triangles. We want them to live up to all the hypotenuse.
The Pythagorean Theorem states that, for a right triangle with legs of length a and b and a hypotenuse of length c, the following equation is true:
a2 + b2 = c2
After that dizzying quadratic formula, this one isn't bad at all. We can handle a square on each variable. That sounds okay. However, what does that mean in relation to the right triangle? In words, the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse. It's a simple concept, but a little wordy, so feel free to go back and read it again. And again. It's okay, read it a few dozen times. You'll be using it a lot.
Why is the Pythagorean Theorem true? Because Shmoop says it is? Nah, we'll back up our claims with some proof...this time. Here's one way to see why the Theorem makes sense. Take the right triangle shown above, make 3 copies of it, and lay them out like this:
Now we have a big square with side length (a + b), and a smaller square inside with side length c. Who knew triangles could make such great squares? You should see their trapezoids.
We can find the area of the big square in two ways. We know, we know: why can't there ever be only one way?
We can find the area of the big square using its side length, or we can add the areas of the four triangles and the smaller square. With the first way, we have:
With the second way, we have:
Since a2 + 2ab + b2 and 2ab + c2 are each the area of the big square, they must be equal:
a2 + 2ab + b2 = 2ab + c2
Subtracting 2ab from each side, we see that:
a2 + b2 = c2
Which is the statement of the Pythagorean Theorem. If that wasn't like magic, we don't know what is. Top that, David Blaine.
For more proofs of the Pythagorean Theorem, check this out.
Even if you don't read through them, it's sort of fun to check out how many proofs there are...especially if you find it fun to see how obsessively deranged mathematicians can be. Did we legitimately need that many proofs, guys? We believed you the first time.
We use the Pythagorean Theorem to find out how long different sides of a triangle are. Usually we'll be told the lengths of two sides of the triangle, and then be asked to find the length of the third side. Although we don't know why whoever measured those first two sides couldn't have measured the third one while he was at it. It would have been a nice gesture.
### Sample Problem
A right triangle has one leg that's 3 cm long and another leg that's 4 cm long. How long is its hypotenuse?
We use the Pythagorean Theorem to figure this out. Let's call the hypotenuse c. The two legs have lengths 3 and 4, so 32 + 42 = c2.
Simplifying, we see that 25 = c2.
The solutions to this equation are c = ±5, but since c is a length, we'll only take c = 5 cm. No -5 cm long triangles for us.
You see, it's not enough to arrive at an answer; you'll need to think about it logically and see if it makes sense. If it doesn't, say buh-bye and throw it into the garbage disposal. If you were ever to visit a landfill, you'd be amazed by how many triangles with negative lengths are laying around all over the place.
• ### Word Problems
The Pythagorean Theorem is great for solving word problems. It's also good for sparking some scintillating dinner conversation. At least, we think so; no word yet from that dude who ran screaming from the room. When a word problem describes a situation that can be visualized using a triangle, it's likely that the Pythagorean Theorem is lurking somewhere nearby.
Even Darth Vader, a classic lurker, knows this.
### Sample Problem
Juan and Lenor met for lunch. At 1 p.m., they parted ways. Maybe forever, considering how they left things. Juan drove due south at 30 mph and Lenor drove due east at 60 mph. Apparently, she was more upset than he was. At 1:30 p.m., how far away are Juan and Lenor from each other?
This needs a picture, and there's definitely a triangle in here somewhere. Juan drove south, or straight down, and Lenor drove east, or straight out to the right:
The question is asking us to find c, and we've been given enough information to find a and b. The two have been traveling for half an hour...hopefully long enough for them to cool down and regret throwing all that furniture.
Lenor has traveled .
Juan has traveled .
After half an hour, the triangle looks like this:
We're supposed to find the distance between Lenor and Juan, so we need to find c. We can do that with the Pythagorean Theorem:
(15)2 + (30)2 = c2
225 + 900 = c2
1125 = c2
Now snag the square root:
Negative distance doesn't make sense, although they were practically a negative distance away from each other a half hour ago when they were at each other's throats. Therefore, we take only the positive square root and conclude that, at 1:30 p.m., Juan and Lenor are miles apart (about 33.5 miles).
For the problem above, 33.5 miles would probably be an acceptable answer. Who cares if Juan and Lenor are closer to 33.541 miles apart? However, we gave the answer because that's an exact answer. We may not know all the decimal places of the number , but we know that's exactly how far apart they are at 1:30 p.m.
If a problem says to give an exact answer and your answer has radicals in it, leave the radicals there. The exact answer to the question "what is the square root of 2?" is , not 1.414. It may not surprise you to learn that math is something of an exact science, so those exact answers are important.
Note: Since the writing of this example, we at Shmoop are happy to report that Juan and Lenor have gotten back together, talked, and worked things out. Not a single folding chair was hurled in the process.
### Sample Problem
Find the exact area of a right triangle with a hypotenuse of length 20 and one leg of length 7.
We'd better draw a picture, partly because we look for any excuse to stretch our artistic muscles. Use b for the unknown leg, and make that the base of the triangle:
We know the area of a right triangle is .
In other words, the area of a right triangle is one-half the product of the lengths of its legs. We know the length of one leg, and if we had the length of the other leg we'd be all set. Thankfully, we can use the Pythagorean Theorem to find the length of that missing piece. Otherwise, this triangle might need to be fitted for a peg leg.
We know that b2 + 72 = 202, so b2 = 351.
Taking the positive square root (since a triangle can't have a negative base), we see that:
This is great. If you write that down as your final answer, however, all your work will have been for nothing. We haven't yet answered the actual question in the problem. Glancing back at the problem to remind ourselves what we were supposed to be looking for (oh, right, the area of the triangle), we see that we now have all the pieces we need. We know the height of the triangle is 7 and its base is , so its area is:
That's a little nasty looking, but unfortunately there's no nicer way to write this. We're done. We wash our hands of you, ugly word problem.
Be careful: If you've read through the examples in this section and they made complete sense to you, give yourself a pat on the back—or a pat on the knee, if that's more easily accessible. You've gotten off to an excellent start.
Then, ask yourself this: would you know what to do if you were given similar problems and asked to solve them by yourself without looking at any webpages, books, or notes? Aye, there's the rub. Make sure you know how to approach the exercises, and any other problems available in your textbook or from your teacher, before deciding you're ready to stop studying. If the following exercises give you trouble, you can always review the above examples to figure out where you went wrong. They'll always be there for you, like Ross, Rachel, Chandler, Monica, Joey, and Phoebe. Okay, we'll throw Gunther in there, too.
Here's a hint for all of the problems coming up in the exercise section: draw pictures. Bonus hint: the pictures should actually relate to the problem at hand. Art class is next period.
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# The Normal Distribution
## What is a Normal Distribution?
normal distribution is the most common continuous probability distribution in statistics.
Normal distributions have the following features:
• Bell shape
• Symmetrical
• Mean and median are equal; both are located at the center of the distribution
• About 68% of data falls within one standard deviation of the mean
• About 95% of data falls within two standard deviations of the mean
• About 99.7% of data falls within three standard deviations of the mean
The last three bullet points are known as the empirical rule, sometimes called the 68-95-99.7 rule.
## How to Draw a Normal Curve
To draw a normal curve, we need to know the mean and the standard deviation.
Example 1: Suppose the height of males at a certain school is normally distributed with mean of a standard deviation of
Step 1: Sketch a normal curve.
Step 2: The mean of 70 inches goes in the middle.
Step 3: Each standard deviation is a distance of 2 inches.
Example 2: Suppose the weight of a certain species of otters is normally distributed with mean of a standard deviation of
Step 1: Sketch a normal curve.
Step 2: The mean of 30 lbs goes in the middle.
Step 3: Each standard deviation is a distance of 5 lbs
## How to Find Percentages Using the Normal Distribution
The empirical rule, sometimes called the 68-95-99.7 rule, says that for a random variable that is normally distributed, 68% of data falls within one standard deviation of the mean, 95% falls within two standard deviations of the mean, and 99.7% falls within three standard deviations of the mean.
Example: Suppose the height of males at a certain school is normally distributed with mean of a standard deviation of
Solution:
Step 1: Sketch a normal distribution with a mean of a standard deviation of
Step 2: A height of 74 inches is two standard deviations above the mean. Add the percentages above that point in the normal distribution.
2.35% + 0.15% = 2.5%
Approximately 2.5% of males at this school are taller than 74 inches.
Solution:
Step 1: Sketch a normal distribution with a mean of a standard deviation of
Step 2: A height of 68 inches and 72 inches is one standard deviation below and above the mean, respectively. Simply add the percentages between these two points in the normal distribution.
34% + 34% = 68%
Approximately 68% of males at this school are between 68 inches and 72 inches tall.
## How to Find Counts Using the Normal Distribution
We can also use the empirical rule to answer questions about counts.
Example: Suppose the weight of a certain species of otters is normally distributed with mean of a standard deviation of
A certain colony has 200 of these otters. Approximately how many of these otters weigh more than 35 lbs?
Solution:
Step 1: Sketch a normal distribution with a mean of a standard deviation of
Step 2: A weight of 35 lbs is one standard deviation above the mean. Add the percentages above that point in the normal distribution.
13.5% + 2.35% + 0.15% = 16%
Step 3: Since there are 200 otters in the colony, 16% of 200 = 0.16 * 200 = 32
Approximately 32 of the otters in this colony weight more than 35 lbs.
Approximately how many of the otters in this colony weigh less than 30 lbs?
Instead of walking through all of the steps we just did above, we can recognize that the median of a normal distribution is equal to the mean, which is 30 lbs in this case. This means that half of the otters weight more than 30 lbs and half weight less than 30 lbs. This means that 50% of the 200 otters weight less than 30 lbs, so 0.5 * 200 = 100 otters. |
An ideal gas has a volume of 2.28 L at 279 K and 1.07 atm. What is the pressure when the volume is 1.03 L and the temperature is 307 K?
Dec 29, 2015
$\text{p"= 2.61" atm to 3 significant figures}$
Explanation:
First, we use the first set of data to calculate the number of moles using the ideal gas equation:
$\text{pV " = " nRT}$
Where:
• $\text{p}$ is pressure in pascals ($\text{Pa}$)
• $\text{V}$ is volume in cubic metres ( ${\text{m}}^{3}$)
• $\text{n}$ is the number of moles
• $\text{R}$ is the gas constant = $8.314$
• $\text{T}$ is the temperature in Kelvin ($\text{K}$)
First, convert your given values into workable units:
• $1 {\text{L" = 0.001"m"^3, :. 2.28"L" = 0.00228"m}}^{3}$
• $1 \text{atm" = 101325"Pa", :. 1.07"atm" = 108417.8"Pa}$
Second, rearrange the equation to solve for moles:
$\text{n "=" ""pV"/"RT}$
Next, substitute in your given values and calculate the number of moles:
"n "=" "(108417.8"Pa" * 0.00228"m"^3)/(8.314 * 279"K")
$\text{n "=" "0.1065color(red)(666)255" moles}$
We can then move onto calculating the new pressure value. The first thing to do here is to, again, convert non-compliant units into ones that are accepted by the equation:
• $1 {\text{L" = 0.001"m"^3, :. 1.03"L" = 0.00103"m}}^{3}$
Then we rearrange the equation to solve for pressure:
$\text{p "=" ""nRT"/"V}$
And substituting in our values, we get:
"p "=(0.1065666255*8.314 * 307"K")/(0.00103"m"^3)
$\text{p "=264078.0989" Pa" = 2.61" atm to 3 significant figures}$ |
# How do you determine if 3x = |y| is an even or odd function?
Apr 8, 2018
even.
#### Explanation:
$3 x = | \setminus y |$
substituting $y$ for $- y$
$3 x = | \setminus \pm y |$
$3 x = | \setminus y |$
so it's even.
In general:
A function is even $\rightarrow$ it is symmetrical about the $y$ axis
and an odd function$\rightarrow$ if it's symmetrical about the origin
but if You want to know through an equation you just substitute for each $x$ for $- x$
A function is even
if $f \left(x\right) = f \left(- x\right)$
like $y = {x}^{2}$ if You substitute for each $x$ for $- x$ You get
$y = {\left(- x\right)}^{2} = {x}^{2} = f \left(x\right)$ so it's even
and same goes for $y = | \setminus x |$
And it will be odd if $f \left(x\right) = - f \left(- x\right)$
ex: $y = x$
if You substitute $x$ for $- x$you get
$y = - x = - f \left(x\right)$
and it will be neither even nor odd if it gives You something else like $y = 3 x + 2$
if You substitute $x$ for $- x$
you get:
$y = - 3 x + 2 \ne \pm f \left(x\right)$ |
Pages
Showing posts with label sign chart. Show all posts
Showing posts with label sign chart. Show all posts
Monday, November 12, 2012
Quadratic inequalities can be solved in a number of ways. We will focus on solving them by graphing first and then by using sign charts. It is important to understand what solutions to these inequalities look like before learning the quick and easy method for solving them.
Method 1: Solve by Graphing
Step 1: Place the inequality in standard form with zero on one side.
Step 2: Graph the quadratic equation.
Step 3: Shade the x-values that produce the desired results.
Step 4: Convert the shading to interval notation.
Solve and express the solution set in interval notation.
In the above example we shaded the x-values for which the graph was above the x-axis. If the problem asked us to solve
or for what x-values is the quadratic less than or equal to 0, then the solution would have been all x-values in the interval [1, 3] - the x-values where the graph is below the x-axis. Notice that the numbers on the x-axis x = 1 and x = 3 separate the positive and negative y-values on the graph, these happen to be the x-intercepts. The x-intercepts or the zeros of a polynomial are called critical numbers.
Solve by graphing.
Remember to use open dots for strict inequalities < or > and closed dots for inclusive inequalities.
Often it is quite tedious and difficult to graph each inequality when trying to solve them. There is a shorter method and it involves sign charts. The idea is to find the critical numbers, the x-values where the y-values could change from positive to negative and create a sign chart to determine which intervals to shade on the x-axis.
Method 2: Solve using sign charts.
Step 1: Place the inequality in standard form with zero on one side.
Step 2: Find the critical numbers (for quadratics - the x-intercepts.)
Step 3: Create a sign chart by determining the sign in each interval bounded by the critical numbers.
Step 4: Use the sign chart to answer the question.
Solve and graph the solution set.
Find the critical numbers by setting the quadratic expression equal to zero and solve.
Determine the results + or - in each interval bounded by the critical numbers by testing values in each interval.
Use the sign chart to answer the question. In this case we are looking for the x-values that produce negative results as indicated by the inequality < 0 "less than zero" in the original question.
When testing values in the intervals created by the critical numbers the actual value is not necessary, we are only concerned with its sign. The sign + or - will be the same for any value in the interval so you may choose any number within the interval when testing.
Solve and graph the solution set.
Tip: Save some time and just determine if the corresponding y-value is positive or negative. If the polynomial factors then use the factors to determine if the interval will produce positive or negative y-values. There is no need to find the actual values. |
Example: marketing
# Dividing Polynomials; Remainder and Factor Theorems
Dividing Polynomials; Remainder and Factor Theorems In this section we will learn how to divide polynomials, an important tool needed in factoring them. This will begin our algebraic study of polynomials. Dividing by a Monomial: Recall from the previous section that a monomial is a single term, such as 6x3 or 7. To divide a polynomial by a monomial, divide each term in the polynomial by the monomial, and then write each quotient in lowest terms. Example 1: Divide 9x4 + 3x2 5x + 6 by 3x. Solution: Step 1: Divide each term in the polynomial 9x4 + 3x2 5x + 6 by the monomial 3x.
monomial 3x. 93 569 3 5 642 4 2 3333 xxx x x x. x xx x3 +−+ =+ −+ x. Step 2: Write the result in lowest terms. 42 93 5 6 53 3 33 33 3 xxx xx. 2 x xx x +−+=+−+ x. Thus, 9x4 + 3x2 – 5x + 6 divided by 3x is equal to . 3 52 3 3 xx. x + −+ Long Division of Polynomials: To divide a polynomial by a polynomial that is not a monomial we must ...
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### Transcription of Dividing Polynomials; Remainder and Factor Theorems
1 Dividing Polynomials; Remainder and Factor Theorems In this section we will learn how to divide polynomials, an important tool needed in factoring them. This will begin our algebraic study of polynomials. Dividing by a Monomial: Recall from the previous section that a monomial is a single term, such as 6x3 or 7. To divide a polynomial by a monomial, divide each term in the polynomial by the monomial, and then write each quotient in lowest terms. Example 1: Divide 9x4 + 3x2 5x + 6 by 3x. Solution: Step 1: Divide each term in the polynomial 9x4 + 3x2 5x + 6 by the monomial 3x.
2 424 293569 3 563333xxxx x x3xxxx+ +=+ +x Step 2: Write the result in lowest terms. 42393565333333xxxxx2xxxx+ +=+ +x Thus, 9x4 + 3x2 5x + 6 divided by 3x is equal to 35233xxx+ + Long Division of Polynomials: To divide a polynomial by a polynomial that is not a monomial we must use long division. Long division for polynomials is very much like long division for numbers. For example, to divide 3x2 17x 25 (the dividend) by x 7 (the divisor), we arrange our work as follows. By: Crystal Hull The division process ends when the last line is of lesser degree than the divisor.
3 The last line then contains the Remainder , and the top line contains the quotient. The result of the division can be interpreted in either of two ways 23172533477xxxxx =++ or ()()231725734xxx x = ++3 We summarize what happens in any long division problem in the following theorem. Division Algorithm: If P(x) and D(x) are polynomials, with D(x) 0, then there exist unique polynomials Q(x) and R(x) such that P(x) = D(x) Q(x) + R(x) where R(x) is either 0 or of less degree than the degree of D(x). The polynomials P(x) and D(x) are called the dividend and the divisor, respectively, Q(x) is the quotient, and R(x) is the Remainder .
4 Example 2: Let P(x) = 3x2 + 17x + 10 and D(x) = 3x + 2. Using long division, find polynomials Q(x) and R(x) such that P(x) = D(x) Q(x) + R(x). Solution: Step 1: Write the problem, making sure that both polynomials are written in descending powers of the variables. 2323 1710xxx+++ By: Crystal Hull Example 2 (Continued): Step 2: Divide the first term of P(x) by the first term of D(x). Since 233xxx=, place this result above the division line.
5 Step 3: Multiply 3x + 2 and x, and write the result below 3x2 + 17x + 10. Step 4: Now subtract 3x2 + 2x from 3x2 + 17x. Do this by mentally changing the signs on 3x2 + 2x and adding. 22323 171032 15 Subtractxxxxxxx ++++ Step 5: Bring down 10 and continue by Dividing 15x by 3x. By: Crystal Hull Example 2 (Continued): Step 6: The process is complete at this point because we have a zero in the final row.
6 From the long division table we see that Q(x) = x + 5 and R(x) = 0, so 3x2 + 17x + 10 = (3x + 2)(x + 5) + 0 Note that since there is no Remainder , this quotient could have been found by factoring and writing in lowest terms. Example 3: Find the quotient and Remainder of 3431xxx2 + using long division. Solution: Step 1: Write the problem, making sure that both polynomials are written in descending powers of the variables. Add a term with 0 coefficient as a place holder for the missing x2 term. Step 2: Start with 3244xxx=.
7 Step 3: Subtract by changing the signs on 4x3 + 4x2 and adding. Then Bring down the next term. 2323224 14032 44 43 Subtract and bring down 3xxxxxxxxxx++ + By: Crystal Hull Example 3 (Continued): Step 4: Now continue with 244xxx = . Step 5: Finally, 1xx=. Step 6: The process is complete at this point because 3 is of lesser degree than the divisor x + 1.
8 Thus, the quotient is 4x2 4x + 1 and the Remainder is 3, and 32432344111xxxxxx = ++++. By: Crystal Hull Synthetic Division: Synthetic division is a shortcut method of performing long division that can be used when the divisor is a first degree polynomial of the form x c. In synthetic division we write only the essential part of the long division table. To illustrate, compare these long division and synthetic division tables, in which we divide 3x3 4x + 2 by x 1: Note that in synthetic division we abbreviate 3x3 4x + 2 by writing only the coefficients: 3 0 4 2, and instead of x 1, we simply write 1.
9 (Writing 1 instead of 1 allows us to add instead of subtract, but this changes the sign of all the numbers that appear in the yellow boxes.) To divide anxn + an-1xn-1 + .. + a1x + a0 by x c, we proceed as follows: Here bn-1 = an, and each number in the bottom row is obtained by adding the numbers above it. The Remainder is r and the quotient is b xbxb 0+++ + By: Crystal Hull Example 4: Find the quotient and the Remainder of 42762xxx +x using synthetic division. Solution: Step 1: We put x + 2 in the form x c by writing it as x ( 2).
10 Use this and the coefficients of the polynomial to obtain 21 0 7 6 0 Note that we used 0 as the coefficient of any missing powers. Step 2: Next, bring down the 1. 2076 1 1 0 Step 3: Now, multiply 2 by 1 to get 2, and add it to the 0 in the first row. The result is 2. 21 0 7 6 0 221 Step 4: Next, 2( 2) = 4. Add this to the 7 in the first row. 21 0 7 6 0 2 1243 By: Crystal Hull Example 4 (Continued): Step 5: 2( 3) = 6. |
# Derivative of $x \sin{\left(x \right)}$
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Find $\frac{d}{dx} \left(x \sin{\left(x \right)}\right)$.
### Solution
Apply the product rule $\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$ with $f{\left(x \right)} = x$ and $g{\left(x \right)} = \sin{\left(x \right)}$:
$${\color{red}\left(\frac{d}{dx} \left(x \sin{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) \sin{\left(x \right)} + x \frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}$$
The derivative of the sine is $\frac{d}{dx} \left(\sin{\left(x \right)}\right) = \cos{\left(x \right)}$:
$$x {\color{red}\left(\frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)} + \sin{\left(x \right)} \frac{d}{dx} \left(x\right) = x {\color{red}\left(\cos{\left(x \right)}\right)} + \sin{\left(x \right)} \frac{d}{dx} \left(x\right)$$
Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:
$$x \cos{\left(x \right)} + \sin{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = x \cos{\left(x \right)} + \sin{\left(x \right)} {\color{red}\left(1\right)}$$
Thus, $\frac{d}{dx} \left(x \sin{\left(x \right)}\right) = x \cos{\left(x \right)} + \sin{\left(x \right)}$.
$\frac{d}{dx} \left(x \sin{\left(x \right)}\right) = x \cos{\left(x \right)} + \sin{\left(x \right)}$A |
# Introduction to Differential Equations
## SectionA.3Elimination
Note: 2–3 lectures
### SubsectionA.3.1Linear systems of equations
One application of matrices is to solve systems of linear equations 1 . Consider the following system of linear equations
\begin{aligned} 2 x_1 + 2 x_2 + 2 x_3 & = 2 , \\ \phantom{9} x_1 + \phantom{9} x_2 + 3 x_3 & = 5 , \\ \phantom{9} x_1 + 4 x_2 + \phantom{9} x_3 & = 10 . \end{aligned}\tag{A.2}
There is a systematic procedure called elimination to solve such a system. In this procedure, we attempt to eliminate each variable from all but one equation. We want to end up with equations such as $$x_3 = 2\text{,}$$ where we can just read off the answer.
We write a system of linear equations as a matrix equation:
\begin{equation*} A \vec{x} = \vec{b} . \end{equation*}
The system (A.2) is written as
\begin{equation*} \underbrace{ \begin{bmatrix} 2 & 2 & 2 \\ 1 & 1 & 3 \\ 1 & 4 & 1 \end{bmatrix} }_{A} \underbrace{ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} }_{\vec{x}} = \underbrace{ \begin{bmatrix} 2 \\ 5 \\ 10 \end{bmatrix} }_{\vec{b}} . \end{equation*}
If we knew the inverse of $$A\text{,}$$ then we would be done; we would simply solve the equation:
\begin{equation*} \vec{x} = A^{-1} A \vec{x} = A^{-1} \vec{b} . \end{equation*}
Well, but that is part of the problem, we do not know how to compute the inverse for matrices bigger than $$2 \times 2\text{.}$$ We will see later that to compute the inverse we are really solving $$A \vec{x} = \vec{b}$$ for several different $$\vec{b}\text{.}$$ In other words, we will need to do elimination to find $$A^{-1}\text{.}$$ In addition, we may wish to solve $$A \vec{x} = \vec{b}$$ if $$A$$ is not invertible, or perhaps not even square.
Let us return to the equations themselves and see how we can manipulate them. There are a few operations we can perform on the equations that do not change the solution. First, perhaps an operation that may seem stupid, we can swap two equations in (A.2):
\begin{equation*} \begin{aligned} \phantom{9} x_1 + \phantom{9} x_2 + 3 x_3 & = 5 , \\ 2 x_1 + 2 x_2 + 2 x_3 & = 2 , \\ \phantom{9} x_1 + 4 x_2 + \phantom{9} x_3 & = 10 . \end{aligned} \end{equation*}
Clearly these new equations have the same solutions $$x_1,x_2,x_3\text{.}$$ A second operation is that we can multiply an equation by a nonzero number. For example, we multiply the third equation in (A.2) by 3:
\begin{equation*} \begin{aligned} 2 x_1 + \phantom{9} 2 x_2 + 2 x_3 & = 2 , \\ \phantom{9} x_1 + \phantom{99} x_2 + 3 x_3 & = 5 , \\ 3 x_1 + 12 x_2 + 3 x_3 & = 30 . \end{aligned} \end{equation*}
Finally, we can add a multiple of one equation to another equation. For instance, we add 3 times the third equation in (A.2) to the second equation:
\begin{equation*} \begin{aligned} \phantom{(1+3)} 2 x_1 + \phantom{(1+12)} 2 x_2 + \phantom{(3+3)} 2 x_3 & = 2 , \\ \phantom{2} (1+3) x_1 + \phantom{2}(1+12) x_2 + \phantom{2} (3+3) x_3 & = 5+30 , \\ \phantom{2 (1+3)} x_1 + \phantom{(1+12)} 4 x_2 + \phantom{(3+3) 2} x_3 & = 10 . \end{aligned} \end{equation*}
The same $$x_1,x_2,x_3$$ should still be solutions to the new equations. These were just examples; we did not get any closer to the solution. We must to do these three operations in some more logical manner, but it turns out these three operations suffice to solve every linear equation.
The first thing is to write the equations in a more compact manner. Given
\begin{equation*} A \vec{x} = \vec{b} , \end{equation*}
we write down the so-called augmented matrix
\begin{equation*} [ A ~|~ \vec{b} ] , \end{equation*}
where the vertical line is just a marker for us to know where the “right-hand side” of the equation starts. For the system (A.2) the augmented matrix is
\begin{equation*} \left[ \begin{array}{ccc|c} 2 & 2 & 2 & 2 \\ 1 & 1 & 3 & 5 \\ 1 & 4 & 1 & 10 \end{array} \right] . \end{equation*}
The entire process of elimination, which we will describe, is often applied to any sort of matrix, not just an augmented matrix. Simply think of the matrix as the $$3 \times 4$$ matrix
\begin{equation*} \begin{bmatrix} 2 & 2 & 2 & 2 \\ 1 & 1 & 3 & 5 \\ 1 & 4 & 1 & 10 \end{bmatrix} . \end{equation*}
### SubsectionA.3.2Row echelon form and elementary operations
We apply the three operations above to the matrix. We call these the elementary operations or elementary row operations. Translating the operations to the matrix setting, the operations become:
1. Swap two rows.
2. Multiply a row by a nonzero number.
3. Add a multiple of one row to another row.
We run these operations until we get into a state where it is easy to read off the answer, or until we get into a contradiction indicating no solution.
More specifically, we run the operations until we obtain the so-called row echelon form. Let us call the first (from the left) nonzero entry in each row the leading entry. A matrix is in row echelon form if the following conditions are satisfied:
1. The leading entry in any row is strictly to the right of the leading entry of the row above.
2. Any zero rows are below all the nonzero rows.
3. All leading entries are 1.
A matrix is in reduced row echelon form if furthermore the following condition is satisfied.
1. All the entries above a leading entry are zero.
Note that the definition applies to matrices of any size.
#### ExampleA.3.1.
The following matrices are in row echelon form. The leading entries are marked:
\begin{equation*} \begin{bmatrix} \mybxsm{1} & 2 & 9 & 3 \\ 0 & 0 & \mybxsm{1} & 5 \\ 0 & 0 & 0 & \mybxsm{1} \end{bmatrix} \qquad \begin{bmatrix} \mybxsm{1} & -1 & -3 \\ 0 & \mybxsm{1} & 5 \\ 0 & 0 & \mybxsm{1} \end{bmatrix} \qquad \begin{bmatrix} \mybxsm{1} & 2 & 1 \\ 0 & \mybxsm{1} & 2 \\ 0 & 0 & 0 \end{bmatrix} \qquad \begin{bmatrix} 0 & \mybxsm{1} & -5 & 2 \\ 0 & 0 & 0 & \mybxsm{1} \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
None of the matrices above are in reduced row echelon form. For example, in the first matrix none of the entries above the second and third leading entries are zero; they are 9, 3, and 5. The following matrices are in reduced row echelon form. The leading entries are marked:
\begin{equation*} \begin{bmatrix} \mybxsm{1} & 3 & 0 & 8 \\ 0 & 0 & \mybxsm{1} & 6 \\ 0 & 0 & 0 & 0 \end{bmatrix} \qquad \begin{bmatrix} \mybxsm{1} & 0 & 2 & 0 \\ 0 & \mybxsm{1} & 3 & 0 \\ 0 & 0 & 0 & \mybxsm{1} \end{bmatrix} \qquad \begin{bmatrix} \mybxsm{1} & 0 & 3 \\ 0 & \mybxsm{1} & -2 \\ 0 & 0 & 0 \end{bmatrix} \qquad \begin{bmatrix} 0 & \mybxsm{1} & 2 & 0 \\ 0 & 0 & 0 & \mybxsm{1} \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
The procedure we will describe to find a reduced row echelon form of a matrix is called Gauss–Jordan elimination. The first part of it, which obtains a row echelon form, is called Gaussian elimination or row reduction. For some problems, a row echelon form is sufficient, and it is a bit less work to only do this first part.
To attain the row echelon form we work systematically. We go column by column, starting at the first column. We find topmost entry in the first column that is not zero, and we call it the pivot. If there is no nonzero entry we move to the next column. We swap rows to put the row with the pivot as the first row. We divide the first row by the pivot to make the pivot entry be a 1. Now look at all the rows below and subtract the correct multiple of the pivot row so that all the entries below the pivot become zero.
After this procedure we forget that we had a first row (it is now fixed), and we forget about the column with the pivot and all the preceding zero columns. Below the pivot row, all the entries in these columns are just zero. Then we focus on the smaller matrix and we repeat the steps above.
It is best shown by example, so let us go back to the example from the beginning of the section. We keep the vertical line in the matrix, even though the procedure works on any matrix, not just an augmented matrix. We start with the first column and we locate the pivot, in this case the first entry of the first column.
\begin{equation*} \left[ \begin{array}{ccc|c} \mybxsm{2} & 2 & 2 & 2 \\ 1 & 1 & 3 & 5 \\ 1 & 4 & 1 & 10 \end{array} \right] \end{equation*}
We multiply the first row by $$\nicefrac{1}{2}\text{.}$$
\begin{equation*} \left[ \begin{array}{ccc|c} \mybxsm{1} & 1 & 1 & 1 \\ 1 & 1 & 3 & 5 \\ 1 & 4 & 1 & 10 \end{array} \right] \end{equation*}
We subtract the first row from the second and third row (two elementary operations).
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 0 & 2 & 4 \\ 0 & 3 & 0 & 9 \end{array} \right] \end{equation*}
We are done with the first column and the first row for now. We almost pretend the matrix doesn’t have the first column and the first row.
\begin{equation*} \left[ \begin{array}{ccc|c} * & * & * & * \\ * & 0 & 2 & 4 \\ * & 3 & 0 & 9 \end{array} \right] \end{equation*}
OK, look at the second column, and notice that now the pivot is in the third row.
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 0 & 2 & 4 \\ 0 & \mybxsm{3} & 0 & 9 \end{array} \right] \end{equation*}
We swap rows.
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & \mybxsm{3} & 0 & 9 \\ 0 & 0 & 2 & 4 \end{array} \right] \end{equation*}
And we divide the pivot row by 3.
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & \mybxsm{1} & 0 & 3 \\ 0 & 0 & 2 & 4 \end{array} \right] \end{equation*}
We do not need to subtract anything as everything below the pivot is already zero. We move on, we again start ignoring the second row and second column and focus on
\begin{equation*} \left[ \begin{array}{ccc|c} * & * & * & * \\ * & * & * & * \\ * & * & 2 & 4 \end{array} \right] . \end{equation*}
We find the pivot, then divide that row by 2:
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & \mybxsm{2} & 4 \end{array} \right] \qquad \to \qquad \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end{array} \right] . \end{equation*}
The matrix is now in row echelon form.
The equation corresponding to the last row is $$x_3 = 2\text{.}$$ We know $$x_3$$ and we could substitute it into the first two equations to get equations for $$x_1$$ and $$x_2\text{.}$$ Then we could do the same thing with $$x_2\text{,}$$ until we solve for all 3 variables. This procedure is called backsubstitution and we can achieve it via elementary operations. We start from the lowest pivot (leading entry in the row echelon form) and subtract the right multiple from the row above to make all the entries above this pivot zero. Then we move to the next pivot and so on. After we are done, we will have a matrix in reduced row echelon form.
We continue our example. Subtract the last row from the first to get
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 1 & 0 & -1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end{array} \right] . \end{equation*}
The entry above the pivot in the second row is already zero. So we move onto the next pivot, the one in the second row. We subtract this row from the top row to get
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 0 & 0 & -4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end{array} \right] . \end{equation*}
The matrix is in reduced row echelon form.
If we now write down the equations for $$x_1,x_2,x_3\text{,}$$ we find
\begin{equation*} x_1 = -4, \qquad x_2 = 3, \qquad x_3 = 2 . \end{equation*}
In other words, we have solved the system.
### SubsectionA.3.3Non-unique solutions and inconsistent systems
It is possible that the solution of a linear system of equations is not unique, or that no solution exists. Suppose for a moment that the row echelon form we found was
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{array} \right] . \end{equation*}
Then we have an equation $$0=1$$ coming from the last row. That is impossible and the equations are what we call inconsistent. There is no solution to $$A \vec{x} = \vec{b}\text{.}$$
On the other hand, if we find a row echelon form
\begin{equation*} \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{array} \right] , \end{equation*}
then there is no issue with finding solutions. In fact, we will find way too many. Let us continue with backsubstitution (subtracting 3 times the third row from the first) to find the reduced row echelon form and let’s mark the pivots.
\begin{equation*} \left[ \begin{array}{ccc|c} \mybxsm{1} & 2 & 0 & -5 \\ 0 & 0 & \mybxsm{1} & 3 \\ 0 & 0 & 0 & 0 \end{array} \right] \end{equation*}
The last row is all zeros; it just says $$0=0$$ and we ignore it. The two remaining equations are
\begin{equation*} x_1 + 2 x_2 = -5 , \qquad x_3 = 3 . \end{equation*}
Let us solve for the variables that corresponded to the pivots, that is $$x_1$$ and $$x_3$$ as there was a pivot in the first column and in the third column:
\begin{equation*} \begin{aligned} & x_1 = - 2 x_2 -5 , \\ & x_3 = 3 . \end{aligned} \end{equation*}
The variable $$x_2$$ can be anything you wish and we still get a solution. The $$x_2$$ is called a free variable. There are infinitely many solutions, one for every choice of $$x_2\text{.}$$ If we pick $$x_2=0\text{,}$$ then $$x_1 = -5\text{,}$$ and $$x_3 = 3$$ give a solution. But we also get a solution by picking say $$x_2 = 1\text{,}$$ in which case $$x_1 = -9$$ and $$x_3 = 3\text{,}$$ or by picking $$x_2 = -5$$ in which case $$x_1 = 5$$ and $$x_3 = 3\text{.}$$
The general idea is that if any row has all zeros in the columns corresponding to the variables, but a nonzero entry in the column corresponding to the right-hand side $$\vec{b}\text{,}$$ then the system is inconsistent and has no solutions. In other words, the system is inconsistent if you find a pivot on the right side of the vertical line drawn in the augmented matrix. Otherwise, the system is consistent, and at least one solution exists.
Suppose the system is consistent (at least one solution exists):
1. If every column corresponding to a variable has a pivot element, then the solution is unique.
2. If there are columns corresponding to variables with no pivot, then those are free variables that can be chosen arbitrarily, and there are infinitely many solutions.
When $$\vec{b} = \vec{0}\text{,}$$ we have a so-called homogeneous matrix equation
\begin{equation*} A \vec{x} = \vec{0} . \end{equation*}
There is no need to write an augmented matrix in this case. As the elementary operations do not do anything to a zero column, it always stays a zero column. Moreover, $$A \vec{x} = \vec{0}$$ always has at least one solution, namely $$\vec{x} = \vec{0}\text{.}$$ Such a system is always consistent. It may have other solutions: If you find any free variables, then you get infinitely many solutions.
The set of solutions of $$A \vec{x} = \vec{0}$$ comes up quite often so people give it a name. It is called the nullspace or the kernel of $$A\text{.}$$ One place where the kernel comes up is invertibility of a square matrix $$A\text{.}$$ If the kernel of $$A$$ contains a nonzero vector, then it contains infinitely many vectors (there was a free variable). But then it is impossible to invert $$\vec{0}\text{,}$$ since infinitely many vectors go to $$\vec{0}\text{,}$$ so there is no unique vector that $$A$$ takes to $$\vec{0}\text{.}$$ So if the kernel is nontrivial, that is, if there are any nonzero vectors in the kernel, in other words, if there are any free variables, or in yet other words, if the row echelon form of $$A$$ has columns without pivots, then $$A$$ is not invertible. We will return to this idea later.
### SubsectionA.3.4Linear independence and rank
If rows of a matrix correspond to equations, it may be good to find out how many equations we really need to find the same set of solutions. Similarly, if we find a number of solutions to a linear equation $$A \vec{x} = \vec{0}\text{,}$$ we may ask if we found enough so that all other solutions can be formed out of the given set. The concept we want is that of linear independence. That same concept is useful for differential equations, for example in Chapter 2.
Given row or column vectors $$\vec{y}_1, \vec{y}_2, \ldots, \vec{y}_n\text{,}$$ a linear combination is an expression of the form
\begin{equation*} \alpha_1 \vec{y}_1 + \alpha_2 \vec{y}_2 + \cdots + \alpha_n \vec{y}_n , \end{equation*}
where $$\alpha_1, \alpha_2, \ldots, \alpha_n$$ are all scalars. For example, $$3 \vec{y}_1 + \vec{y}_2 - 5 \vec{y}_3$$ is a linear combination of $$\vec{y}_1\text{,}$$ $$\vec{y}_2\text{,}$$ and $$\vec{y}_3\text{.}$$
We have seen linear combinations before. The expression
\begin{equation*} A \vec{x} \end{equation*}
is a linear combination of the columns of $$A\text{,}$$ while
\begin{equation*} \vec{x}^T A = (A^T \vec{x})^T \end{equation*}
is a linear combination of the rows of $$A\text{.}$$
The way linear combinations come up in our study of differential equations is similar to the following computation. Suppose that $$\vec{x}_1\text{,}$$ $$\vec{x}_2\text{,}$$ ..., $$\vec{x}_n$$ are solutions to $$A \vec{x}_1 = \vec{0}\text{,}$$ $$A \vec{x}_2 = \vec{0}\text{,}$$ ..., $$A \vec{x}_n = \vec{0}\text{.}$$ Then the linear combination
\begin{equation*} \vec{y} = \alpha_1 \vec{x}_1 + \alpha_2 \vec{x}_2 + \cdots + \alpha_n \vec{x}_n \end{equation*}
is a solution to $$A \vec{y} = \vec{0}\text{:}$$
\begin{multline*} A \vec{y} = A (\alpha_1 \vec{x}_1 + \alpha_2 \vec{x}_2 + \cdots + \alpha_n \vec{x}_n ) = \\ = \alpha_1 A \vec{x}_1 + \alpha_2 A \vec{x}_2 + \cdots + \alpha_n A \vec{x}_n = \alpha_1 \vec{0} + \alpha_2 \vec{0} + \cdots + \alpha_n \vec{0} = \vec{0} . \end{multline*}
So if you have found enough solutions, you have them all. The question is, when did we find enough of them?
We say the vectors $$\vec{y}_1\text{,}$$ $$\vec{y}_2\text{,}$$ ..., $$\vec{y}_n$$ are linearly independent if the only solution to
\begin{equation*} \alpha_1 \vec{x}_1 + \alpha_2 \vec{x}_2 + \cdots + \alpha_n \vec{x}_n = \vec{0} \end{equation*}
is $$\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0\text{.}$$ Otherwise, we say the vectors are linearly dependent.
For example, the vectors $$\left[ \begin{smallmatrix} 1 \\ 2 \end{smallmatrix} \right]$$ and $$\left[ \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right]$$ are linearly independent. Let’s try:
\begin{equation*} \alpha_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} \alpha_1 \\ 2 \alpha_1 + \alpha_2 \end{bmatrix} = \vec{0} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} . \end{equation*}
So $$\alpha_1 = 0\text{,}$$ and then it is clear that $$\alpha_2 = 0$$ as well. In other words, the two vectors are linearly independent.
If a set of vectors is linearly dependent, that is, some of the $$\alpha_j$$s are nonzero, then we can solve for one vector in terms of the others. Suppose $$\alpha_1 \not= 0\text{.}$$ Since $$\alpha_1 \vec{x}_1 + \alpha_2 \vec{x}_2 + \cdots + \alpha_n \vec{x}_n = \vec{0}\text{,}$$ then
\begin{equation*} \vec{x}_1 = \frac{-\alpha_2}{\alpha_1} \vec{x}_2 - \frac{-\alpha_3}{\alpha_1} \vec{x}_3 + \cdots + \frac{-\alpha_n}{\alpha_1} \vec{x}_n . \end{equation*}
For example,
\begin{equation*} 2 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} -4 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} , \end{equation*}
and so
\begin{equation*} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} - \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} . \end{equation*}
You may have noticed that solving for those $$\alpha_j$$s is just solving linear equations, and so you may not be surprised that to check if a set of vectors is linearly independent we use row reduction.
Given a set of vectors, we may not be interested in just finding if they are linearly independent or not, we may be interested in finding a linearly independent subset. Or perhaps we may want to find some other vectors that give the same linear combinations and are linearly independent. The way to figure this out is to form a matrix out of our vectors. If we have row vectors we consider them as rows of a matrix. If we have column vectors we consider them columns of a matrix. The set of all linear combinations of a set of vectors is called their span.
\begin{equation*} \operatorname{span} \bigl\{ \vec{x}_1, \vec{x}_2 , \ldots , \vec{x}_n \bigr\} = \bigl\{ \text{Set of all linear combinations of $\vec{x}_1, \vec{x}_2 , \ldots , \vec{x}_n$} \bigr\} . \end{equation*}
Given a matrix $$A\text{,}$$ the maximal number of linearly independent rows is called the rank of $$A\text{,}$$ and we write “$$\operatorname{rank} A$$” for the rank. For example,
\begin{equation*} \operatorname{rank} \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ -1 & -1 & -1 \end{bmatrix} = 1 . \end{equation*}
The second and third row are multiples of the first one. We cannot choose more than one row and still have a linearly independent set. But what is
\begin{equation*} \operatorname{rank} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \quad = \quad ? \end{equation*}
That seems to be a tougher question to answer. The first two rows are linearly independent (neither is a multiple of the other), so the rank is at least two. If we would set up the equations for the $$\alpha_1\text{,}$$ $$\alpha_2\text{,}$$ and $$\alpha_3\text{,}$$ we would find a system with infinitely many solutions. One solution is
\begin{equation*} \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} -2 \begin{bmatrix} 4 & 5 & 6 \end{bmatrix} + \begin{bmatrix} 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} . \end{equation*}
So the set of all three rows is linearly dependent, the rank cannot be 3. Therefore the rank is 2.
But how can we do this in a more systematic way? We find the row echelon form!
\begin{equation*} \text{Row echelon form of} \quad \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \quad \text{is} \quad \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} . \end{equation*}
The elementary row operations do not change the set of linear combinations of the rows (that was one of the main reasons for defining them as they were). In other words, the span of the rows of the $$A$$ is the same as the span of the rows of the row echelon form of $$A\text{.}$$ In particular, the number of linearly independent rows is the same. And in the row echelon form, all nonzero rows are linearly independent. This is not hard to see. Consider the two nonzero rows in the example above. Suppose we tried to solve for the $$\alpha_1$$ and $$\alpha_2$$ in
\begin{equation*} \alpha_1 \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} . \end{equation*}
Since the first column of the row echelon matrix has zeros except in the first row means that $$\alpha_1 = 0\text{.}$$ For the same reason, $$\alpha_2$$ is zero. We only have two nonzero rows, and they are linearly independent, so the rank of the matrix is 2.
The span of the rows is called the row space. The row space of $$A$$ and the row echelon form of $$A$$ are the same. In the example,
\begin{equation*} \begin{split} \text{row space of } \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} & = \operatorname{span} \left\{ \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} , \begin{bmatrix} 4 & 5 & 6 \end{bmatrix} , \begin{bmatrix} 7 & 8 & 9 \end{bmatrix} \right\} \\ & = \operatorname{span} \left\{ \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} , \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} \right\} . \end{split} \end{equation*}
Similarly to row space, the span of columns is called the column space.
\begin{equation*} \text{column space of } \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = \operatorname{span} \left\{ \begin{bmatrix} 1 \\ 4 \\ 7 \end{bmatrix} , \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix} , \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} \right\} . \end{equation*}
So it may also be good to find the number of linearly independent columns of $$A\text{.}$$ One way to do that is to find the number of linearly independent rows of $$A^T\text{.}$$ It is a tremendously useful fact that the number of linearly independent columns is always the same as the number of linearly independent rows:
In particular, to find a set of linearly independent columns we need to look at where the pivots were. If you recall above, when solving $$A \vec{x} = \vec{0}$$ the key was finding the pivots, any non-pivot columns corresponded to free variables. That means we can solve for the non-pivot columns in terms of the pivot columns. Let’s see an example. First we reduce some random matrix:
\begin{equation*} \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 5 & 6 \\ 3 & 6 & 7 & 8 \end{bmatrix} . \end{equation*}
We find a pivot and reduce the rows below:
\begin{equation*} \begin{bmatrix} \mybxsm{1} & 2 & 3 & 4 \\ 2 & 4 & 5 & 6 \\ 3 & 6 & 7 & 8 \end{bmatrix} \to \begin{bmatrix} \mybxsm{1} & 2 & 3 & 4 \\ 0 & 0 & -1 & -2 \\ 3 & 6 & 7 & 8 \end{bmatrix} \to \begin{bmatrix} \mybxsm{1} & 2 & 3 & 4 \\ 0 & 0 & -1 & -2 \\ 0 & 0 & -2 & -4 \end{bmatrix} . \end{equation*}
We find the next pivot, make it one, and rinse and repeat:
\begin{equation*} \begin{bmatrix} \mybxsm{1} & 2 & 3 & 4 \\ 0 & 0 & \mybxsm{-1} & -2 \\ 0 & 0 & -2 & -4 \end{bmatrix} \to \begin{bmatrix} \mybxsm{1} & 2 & 3 & 4 \\ 0 & 0 & \mybxsm{1} & 2 \\ 0 & 0 & -2 & -4 \end{bmatrix} \to \begin{bmatrix} \mybxsm{1} & 2 & 3 & 4 \\ 0 & 0 & \mybxsm{1} & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} . \end{equation*}
The final matrix is the row echelon form of the matrix. Consider the pivots that we marked. The pivot columns are the first and the third column. All other columns correspond to free variables when solving $$A \vec{x} = \vec{0}\text{,}$$ so all other columns can be solved in terms of the first and the third column. In other words
\begin{equation*} \text{column space of } \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 5 & 6 \\ 3 & 6 & 7 & 8 \end{bmatrix} = \operatorname{span} \left\{ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} , \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} , \begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix} , \begin{bmatrix} 4 \\ 6 \\ 8 \end{bmatrix} \right\} = \operatorname{span} \left\{ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} , \begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix} \right\} . \end{equation*}
We could perhaps use another pair of columns to get the same span, but the first and the third are guaranteed to work because they are pivot columns.
The discussion above could be expanded into a proof of the theorem if we wanted. As each nonzero row in the row echelon form contains a pivot, then the rank is the number of pivots, which is the same as the maximal number of linearly independent columns.
The idea also works in reverse. Suppose we have a bunch of column vectors and we just need to find a linearly independent set. For example, suppose we started with the vectors
\begin{equation*} \vec{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} , \quad \vec{v}_2 = \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} , \quad \vec{v}_3 = \begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix} , \quad \vec{v}_4 = \begin{bmatrix} 4 \\ 6 \\ 8 \end{bmatrix} . \end{equation*}
These vectors are not linearly independent as we saw above. In particular, the span of $$\vec{v}_1$$ and $$\vec{v}_3$$ is the same as the span of all four of the vectors. So $$\vec{v}_2$$ and $$\vec{v}_4$$ can both be written as linear combinations of $$\vec{v}_1$$ and $$\vec{v}_3\text{.}$$ A common thing that comes up in practice is that one gets a set of vectors whose span is the set of solutions of some problem. But perhaps we get way too many vectors, we want to simplify. For example above, all vectors in the span of $$\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4$$ can be written $$\alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 + \alpha_3 \vec{v}_3 + \alpha_4 \vec{v}_4$$ for some numbers $$\alpha_1,\alpha_2,\alpha_3,\alpha_4\text{.}$$ But it is also true that every such vector can be written as $$a \vec{v}_1 + b \vec{v}_3$$ for two numbers $$a$$ and $$b\text{.}$$ And one has to admit, that looks much simpler. Moreover, these numbers $$a$$ and $$b$$ are unique. More on that in the next section.
To find this linearly independent set we simply take our vectors and form the matrix $$[ \vec{v}_1 ~ \vec{v}_2 ~ \vec{v}_3 ~ \vec{v}_4 ]\text{,}$$ that is, the matrix
\begin{equation*} \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 5 & 6 \\ 3 & 6 & 7 & 8 \end{bmatrix} . \end{equation*}
We crank up the row-reduction machine, feed this matrix into it, find the pivot columns, and pick those. In this case, $$\vec{v}_1$$ and $$\vec{v}_3\text{.}$$
### SubsectionA.3.5Computing the inverse
If the matrix $$A$$ is square and there exists a unique solution $$\vec{x}$$ to $$A \vec{x} = \vec{b}$$ for any $$\vec{b}$$ (there are no free variables), then $$A$$ is invertible. This is equivalent to the $$n \times n$$ matrix $$A$$ being of rank $$n\text{.}$$
In particular, if $$A \vec{x} = \vec{b}$$ then $$\vec{x} = A^{-1} \vec{b}\text{.}$$ Now we just need to compute what $$A^{-1}$$ is. We can surely do elimination every time we want to find $$A^{-1} \vec{b}\text{,}$$ but that would be ridiculous. The mapping $$A^{-1}$$ is linear and hence given by a matrix, and we have seen that to figure out the matrix we just need to find where $$A^{-1}$$ takes the standard basis vectors $$\vec{e}_1\text{,}$$ $$\vec{e}_2\text{,}$$ ..., $$\vec{e}_n\text{.}$$
That is, to find the first column of $$A^{-1}\text{,}$$ we solve $$A \vec{x} = \vec{e}_1\text{,}$$ because then $$A^{-1} \vec{e}_1 = \vec{x}\text{.}$$ To find the second column of $$A^{-1}\text{,}$$ we solve $$A \vec{x} = \vec{e}_2\text{.}$$ And so on. It is really just $$n$$ eliminations that we need to do. But it gets even easier. If you think about it, the elimination is the same for everything on the left side of the augmented matrix. Doing $$n$$ eliminations separately we would redo most of the computations. Best is to do all at once.
Therefore, to find the inverse of $$A\text{,}$$ we write an $$n \times 2n$$ augmented matrix $$[ \,A ~|~ I\, ]\text{,}$$ where $$I$$ is the identity matrix, whose columns are precisely the standard basis vectors. We then perform row reduction until we arrive at the reduced row echelon form. If $$A$$ is invertible, then pivots can be found in every column of $$A\text{,}$$ and so the reduced row echelon form of $$[ \,A ~|~ I\, ]$$ looks like $$[ \,I ~|~ A^{-1}\, ]\text{.}$$ We then just read off the inverse $$A^{-1}\text{.}$$ If you do not find a pivot in every one of the first $$n$$ columns of the augmented matrix, then $$A$$ is not invertible.
This is best seen by example. Suppose we wish to invert the matrix
\begin{equation*} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 3 & 1 & 0 \end{bmatrix} . \end{equation*}
We write the augmented matrix and we start reducing:
\begin{equation*} \begin{aligned} & \left[ \begin{array}{ccc|ccc} \mybxsm{1} & 2 & 3 & 1 & 0 & 0\\ 2 & 0 & 1 & 0 & 1 & 0 \\ 3 & 1 & 0 & 0 & 0 & 1 \end{array} \right] \to & & \left[ \begin{array}{ccc|ccc} \mybxsm{1} & 2 & 3 & 1 & 0 & 0\\ 0 & -4 & -5 & -2 & 1 & 0 \\ 0 & -5 & -9 & -3 & 0 & 1 \end{array} \right] \to \\ \to & \left[ \begin{array}{ccc|ccc} \mybxsm{1} & 2 & 3 & 1 & 0 & 0\\ 0 & \mybxsm{1} & \nicefrac{5}{4} & \nicefrac{1}{2} & \nicefrac{1}{4} & 0 \\ 0 & -5 & -9 & -3 & 0 & 1 \end{array} \right] \to & & \left[ \begin{array}{ccc|ccc} \mybxsm{1} & 2 & 3 & 1 & 0 & 0\\ 0 & \mybxsm{1} & \nicefrac{5}{4} & \nicefrac{1}{2} & \nicefrac{1}{4} & 0 \\ 0 & 0 & \nicefrac{-11}{4} & \nicefrac{-1}{2} & \nicefrac{-5}{4} & 1 \end{array} \right] \to \\ \to & \left[ \begin{array}{ccc|ccc} \mybxsm{1} & 2 & 3 & 1 & 0 & 0\\ 0 & \mybxsm{1} & \nicefrac{5}{4} & \nicefrac{1}{2} & \nicefrac{1}{4} & 0 \\ 0 & 0 & \mybxsm{1} & \nicefrac{2}{11} & \nicefrac{5}{11} & \nicefrac{-4}{11} \end{array} \right] \to & & \left[ \begin{array}{ccc|ccc} \mybxsm{1} & 2 & 0 & \nicefrac{5}{11} & \nicefrac{-5}{11} & \nicefrac{12}{11} \\ 0 & \mybxsm{1} & 0 & \nicefrac{3}{11} & \nicefrac{-9}{11} & \nicefrac{5}{11} \\ 0 & 0 & \mybxsm{1} & \nicefrac{2}{11} & \nicefrac{5}{11} & \nicefrac{-4}{11} \end{array} \right] \to \\ \to & \left[ \begin{array}{ccc|ccc} \mybxsm{1} & 0 & 0 & \nicefrac{-1}{11} & \nicefrac{3}{11} & \nicefrac{2}{11} \\ 0 & \mybxsm{1} & 0 & \nicefrac{3}{11} & \nicefrac{-9}{11} & \nicefrac{5}{11} \\ 0 & 0 & \mybxsm{1} & \nicefrac{2}{11} & \nicefrac{5}{11} & \nicefrac{-4}{11} \end{array} \right] . \end{aligned} \end{equation*}
So
\begin{equation*} {\begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 3 & 1 & 0 \end{bmatrix}}^{-1} = \begin{bmatrix} \nicefrac{-1}{11} & \nicefrac{3}{11} & \nicefrac{2}{11} \\ \nicefrac{3}{11} & \nicefrac{-9}{11} & \nicefrac{5}{11} \\ \nicefrac{2}{11} & \nicefrac{5}{11} & \nicefrac{-4}{11} \end{bmatrix} . \end{equation*}
Not too terrible, no? Perhaps harder than inverting a $$2 \times 2$$ matrix for which we had a simple formula, but not too bad. Really in practice this is done efficiently by a computer.
### SubsectionA.3.6Exercises
#### ExerciseA.3.1.
Compute the reduced row echelon form for the following matrices:
1. $$\displaystyle \begin{bmatrix} 1 & 3 & 1 \\ 0 & 1 & 1 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 3 & 3 \\ 6 & -3 \end{bmatrix}$$
3. $$\displaystyle \begin{bmatrix} 3 & 6 \\ -2 & -3 \end{bmatrix}$$
4. $$\displaystyle \begin{bmatrix} 6 & 6 & 7 & 7 \\ 1 & 1 & 0 & 1 \end{bmatrix}$$
5. $$\displaystyle \begin{bmatrix} 9 & 3 & 0 & 2 \\ 8 & 6 & 3 & 6 \\ 7 & 9 & 7 & 9 \end{bmatrix}$$
6. $$\displaystyle \begin{bmatrix} 2 & 1 & 3 & -3 \\ 6 & 0 & 0 & -1 \\ -2 & 4 & 4 & 3 \end{bmatrix}$$
7. $$\displaystyle \begin{bmatrix} 6 & 6 & 5 \\ 0 & -2 & 2 \\ 6 & 5 & 6 \end{bmatrix}$$
8. $$\displaystyle \begin{bmatrix} 0 & 2 & 0 & -1 \\ 6 & 6 & -3 & 3 \\ 6 & 2 & -3 & 5 \end{bmatrix}$$
#### ExerciseA.3.2.
Compute the inverse of the given matrices
1. $$\displaystyle \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}$$
3. $$\displaystyle \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 0 & 2 & 1 \end{bmatrix}$$
#### ExerciseA.3.3.
Solve (find all solutions), or show no solution exists
1. \displaystyle \begin{aligned} 4x_1+3x_2 & = -2 \\ -x_1+\phantom{3} x_2 & = 4 \end{aligned}
2. \displaystyle \begin{aligned} x_1+5x_2+3x_3 & = 7 \\ 8x_1+7x_2+8x_3 & = 8 \\ 4x_1+8x_2+6x_3 & = 4 \end{aligned}
3. \displaystyle \begin{aligned} 4x_1+8x_2+2x_3 & = 3 \\ -x_1-2x_2+3x_3 & = 1 \\ 4x_1+8x_2 \phantom{{}+3x_3} & = 2 \end{aligned}
4. \displaystyle \begin{aligned} x+2y+3z & = 4 \\ 2 x-\phantom{2} y+3z & = 1 \\ 3 x+\phantom{2} y+6z & = 6 \end{aligned}
#### ExerciseA.3.4.
By computing the inverse, solve the following systems for $$\vec{x}\text{.}$$
1. $$\displaystyle \begin{bmatrix} 4 & 1 \\ -1 & 3 \end{bmatrix} \vec{x} = \begin{bmatrix} 13 \\ 26 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 3 & 3 \\ 3 & 4 \end{bmatrix} \vec{x} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}$$
#### ExerciseA.3.5.
Compute the rank of the given matrices
1. $$\displaystyle \begin{bmatrix} 6 & 3 & 5 \\ 1 & 4 & 1 \\ 7 & 7 & 6 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 5 & -2 & -1 \\ 3 & 0 & 6 \\ 2 & 4 & 5 \end{bmatrix}$$
3. $$\displaystyle \begin{bmatrix} 1 & 2 & 3 \\ -1 & -2 & -3 \\ 2 & 4 & 6 \end{bmatrix}$$
#### ExerciseA.3.6.
For the matrices in Exercise A.3.5, find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix).
#### ExerciseA.3.7.
For the matrices in Exercise A.3.5, find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices.
#### ExerciseA.3.8.
Find a linearly independent subset of the following vectors that has the same span.
\begin{equation*} \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ -2 \\ -4 \end{bmatrix} , \quad \begin{bmatrix} -2 \\ 4 \\ 1 \end{bmatrix} , \quad \begin{bmatrix} -1 \\ 3 \\ -2 \end{bmatrix} \end{equation*}
#### ExerciseA.3.101.
Compute the reduced row echelon form for the following matrices:
1. $$\displaystyle \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$
3. $$\displaystyle \begin{bmatrix} 1 & 1 \\ -2 & -2 \end{bmatrix}$$
4. $$\displaystyle \begin{bmatrix} 1 & -3 & 1 \\ 4 & 6 & -2 \\ -2 & 6 & -2 \end{bmatrix}$$
5. $$\displaystyle \begin{bmatrix} 2 & 2 & 5 & 2 \\ 1 & -2 & 4 & -1 \\ 0 & 3 & 1 & -2 \end{bmatrix}$$
6. $$\displaystyle \begin{bmatrix} -2 & 6 & 4 & 3 \\ 6 & 0 & -3 & 0 \\ 4 & 2 & -1 & 1 \end{bmatrix}$$
7. $$\displaystyle \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
8. $$\displaystyle \begin{bmatrix} 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 5 \end{bmatrix}$$
a) $$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ b) $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ c) $$\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$ d) $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{bmatrix}$$ e) $$\begin{bmatrix} 1 & 0 & 0 & 77/15 \\ 0 & 1 & 0 & -2/15 \\ 0 & 0 & 1 & -8/5 \end{bmatrix}$$ f) $$\begin{bmatrix} 1 & 0 & -1/2 & 0 \\ 0 & 1 & 1/2 & 1/2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ g) $$\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ h) $$\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
#### ExerciseA.3.102.
Compute the inverse of the given matrices
1. $$\displaystyle \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$
3. $$\displaystyle \begin{bmatrix} 2 & 4 & 0 \\ 2 & 2 & 3 \\ 2 & 4 & 1 \end{bmatrix}$$
a) $$\begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ b) $$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{bmatrix}$$ c) $$\begin{bmatrix} \nicefrac{5}{2} & 1 & -3 \\ -1 & \nicefrac{-1}{2} & \nicefrac{3}{2} \\ -1 & 0 & 1 \end{bmatrix}$$
#### ExerciseA.3.103.
Solve (find all solutions), or show no solution exists
1. \displaystyle \begin{aligned} 4x_1+3x_2 & = -1 \\ 5x_1+6x_2 & = 4 \end{aligned}
2. \displaystyle \begin{aligned} 5x+6y+5z & = 7 \\ 6x+8y+6z & = -1 \\ 5x+2y+5z & = 2 \end{aligned}
3. \displaystyle \begin{aligned} a+\phantom{5}b+\phantom{6}c & = -1 \\ a+5b+6c & = -1 \\ -2a+5b+6c & = 8 \end{aligned}
4. \displaystyle \begin{aligned} -2 x_1+2x_2+8x_3 & = 6 \\ x_2+\phantom{8}x_3 & = 2 \\ x_1+4x_2+\phantom{8}x_3 & = 7 \end{aligned}
a) $$x_1=-2\text{,}$$ $$x_2 = \nicefrac{7}{3}$$ b) no solution c) $$a = -3\text{,}$$ $$b=10\text{,}$$ $$c=-8$$ d) $$x_3$$ is free, $$x_1 = -1+3x_3\text{,}$$ $$x_2 = 2-x_3$$
#### ExerciseA.3.104.
By computing the inverse, solve the following systems for $$\vec{x}\text{.}$$
1. $$\displaystyle \begin{bmatrix} -1 & 1 \\ 3 & 3 \end{bmatrix} \vec{x} = \begin{bmatrix} 4 \\ 6 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 2 & 7 \\ 1 & 6 \end{bmatrix} \vec{x} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$$
a) $$\begin{bmatrix} -1 \\ 3 \end{bmatrix}$$ b) $$\begin{bmatrix} -3 \\ 1 \end{bmatrix}$$
#### ExerciseA.3.105.
Compute the rank of the given matrices
1. $$\displaystyle \begin{bmatrix} 7 & -1 & 6 \\ 7 & 7 & 7 \\ 7 & 6 & 2 \end{bmatrix}$$
2. $$\displaystyle \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{bmatrix}$$
3. $$\displaystyle \begin{bmatrix} 0 & 3 & -1 \\ 6 & 3 & 1 \\ 4 & 7 & -1 \end{bmatrix}$$
a) 3 b) 1 c) 2
#### ExerciseA.3.106.
For the matrices in Exercise A.3.105, find a linearly independent set of row vectors that span the row space (they don’t need to be rows of the matrix).
a) $$\begin{bmatrix} 1 & 0 & 0\end{bmatrix}\text{,}$$ $$\begin{bmatrix} 0 & 1 & 0\end{bmatrix}\text{,}$$ $$\begin{bmatrix} 0 & 0 & 1\end{bmatrix}$$ b) $$\begin{bmatrix} 1 & 1 & 1\end{bmatrix}$$ c) $$\begin{bmatrix} 1 & 0 & \nicefrac{1}{3}\end{bmatrix}\text{,}$$ $$\begin{bmatrix} 0 & 1 & \nicefrac{-1}{3}\end{bmatrix}$$
#### ExerciseA.3.107.
For the matrices in Exercise A.3.105, find a linearly independent set of columns that span the column space. That is, find the pivot columns of the matrices.
a) $$\begin{bmatrix} 7 \\ 7 \\ 7\end{bmatrix}\text{,}$$ $$\begin{bmatrix} -1 \\ 7 \\ 6\end{bmatrix}\text{,}$$ $$\begin{bmatrix} 7 \\ 6 \\ 2\end{bmatrix}$$ b) $$\begin{bmatrix} 1 \\ 1 \\ 2\end{bmatrix}$$ c) $$\begin{bmatrix} 0 \\ 6 \\ 4\end{bmatrix}\text{,}$$ $$\begin{bmatrix} 3 \\ 3 \\ 7\end{bmatrix}$$
$$\begin{bmatrix} 3 \\ 1 \\ -5 \end{bmatrix} , \begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix}$$ |
Question
# If $y = m\log x + n{x^2} + x$ has its extreme values at $x = 2$ and $x = 1$, then $2m + 10n$ is equal to${\text{A}}{\text{. }} - 1 \\ {\text{B}}{\text{. }} - 4 \\ {\text{C}}{\text{. }} - 2 \\ {\text{D}}{\text{. 1}} \\ {\text{E}}{\text{. }} - 3 \\$
Hint- Here, we will be proceeding by differentiating the given function with respect to x and then putting $\dfrac{{dy}}{{dx}} = 0$ and afterwards substituting the given extreme values of x to obtain the two equation in terms of m and n only and then we will solve them.
Given, $y = m\log x + n{x^2} + x{\text{ }} \to {\text{(1)}}$ which is a function in terms of variable x.
It is also given that the function represented in equation (1) has its extreme values at $x = 2$ and $x = 1$.
As we know that extreme values of any function y are obtained by finding $\dfrac{{dy}}{{dx}}$ and then putting $\dfrac{{dy}}{{dx}} = 0$.
By differentiating the function given by equation (1) both sides with respect to x, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left[ {m\log x + n{x^2} + x} \right]}}{{dx}} = \dfrac{{d\left( {m\log x} \right)}}{{dx}} + \dfrac{{d\left( {n{x^2}} \right)}}{{dx}} + \dfrac{{dx}}{{dx}}$
Since, m and n are constants (i.e., independent of variable x) so we can take them out of the differentiation.
$\Rightarrow \dfrac{{dy}}{{dx}} = m\left[ {\dfrac{{d\left( {\log x} \right)}}{{dx}}} \right] + n\left[ {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right] + 1 = m\left[ {\dfrac{1}{x}} \right] + n\left[ {2x} \right] + 1 \\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{m}{x} + 2nx + 1 \\$
Put $\dfrac{{dy}}{{dx}} = 0$ for the extreme values of the variable x, we get
$\Rightarrow 0 = \dfrac{m}{x} + 2nx + 1 \\ \Rightarrow 0 = \dfrac{{m + 2n{x^2} + x}}{x} \\ \Rightarrow 0 = m + 2n{x^2} + x \\ \Rightarrow 2n{x^2} + x + m = 0{\text{ }} \to {\text{(2)}} \\$
So, we are having a quadratic equation in variable x which means there will be two extreme values of x. These extreme values are $x = 2$ and $x = 1$. These two given extreme values should satisfy the quadratic equation given by equation (2).
Put x=2 in equation (2), we get
$\Rightarrow 2n{\left( 2 \right)^2} + 2 + m = 0 \\ \Rightarrow 8n + 2 + m = 0{\text{ }} \to {\text{(3)}} \\$
Put x=1 in equation (2), we get
$\Rightarrow 2n{\left( 1 \right)^2} + 1 + m = 0 \\ \Rightarrow 2n + 1 + m = 0{\text{ }} \to {\text{(4)}} \\$
In order to obtain the values of constants m and n we will subtract equation (4) from equation (3), we have
$\Rightarrow 8n + 2 + m - \left( {2n + 1 + m} \right) = 0 - 0 \\ \Rightarrow 8n + 2 + m - 2n - 1 - m = 0 \\ \Rightarrow 6n + 1 = 0 \\ \Rightarrow 6n = - 1 \\ \Rightarrow n = \dfrac{{ - 1}}{6} \\$
Put $n = \dfrac{{ - 1}}{6}$ in equation (4) to obtain the value of m, we have
$\Rightarrow 2\left[ {\dfrac{{ - 1}}{6}} \right] + 1 + m = 0 \\ \Rightarrow \dfrac{{ - 1}}{3} + 1 + m = 0 \\ \Rightarrow m = \dfrac{1}{3} - 1 = \dfrac{{1 - 3}}{3} \\ \Rightarrow m = \dfrac{{ - 2}}{3} \\$
Put $m = \dfrac{{ - 2}}{3}$ and $n = \dfrac{{ - 1}}{6}$ to find the value for the expression $2m + 10n$, we get
$2m + 10n = 2\left[ {\dfrac{{ - 2}}{3}} \right] + 10\left[ {\dfrac{{ - 1}}{6}} \right] = \dfrac{{ - 4}}{3} - \dfrac{5}{3} = \dfrac{{ - 4 - 5}}{3} = \dfrac{{ - 9}}{3} \\ \Rightarrow 2m + 10n = - 3 \\$
Hence, option E is correct.
Note- At the extreme values for any function y, $\dfrac{{dy}}{{dx}} = 0$. Also, if $\dfrac{{{d^2}y}}{{d{x^2}}} > 0$ that means at this extreme value, minima occurs (i.e., we will be getting the minimum value of y corresponding to this extreme value of x) and if $\dfrac{{{d^2}y}}{{d{x^2}}} < 0$ that means at this extreme value, maxima occurs (i.e., we will be getting the maximum value of y corresponding to this extreme value of x). |
## Tuesday, November 16, 2021
### Algebra Problems of the Day (Integrated Algebra Regents, January 2012)
Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.
More Regents problems.
### Integrated Algebra Regents, January 2012
Part III: Each correct answer will receive 3 credits. Partial credit is available.
34. Solve algebraically for: 2(x - 4) > 1/2(5 - 3x)
Answer:
You can avoid dealing with the fraction by multiplying the inequality by 2 before you use the Distributive Property.
2(x - 4) > 1/2(5 - 3x)
4(x - 4) > 5 - 3x
4x - 16 > 5 - 3x
7x - 16 > 5
7x > 21
x > 3
35. On the set of axes below, solve the following system of equations graphically. State the coordinates of the solution.
y = 4x + 1
2x + y = 5
Answer:
If you're using a graphing calculator to assist you, rewrite the second equation as y = -2x + 5.
The first line has a y-intercept of -1 and a slope of 4. The second line has a y-intercept of 5 and a slope of -2. Graph and label each line. Label the point of intersection and state the coordinates.
Your graph should like like the image below:
36. A turtle and a rabbit are in a race to see who is first to reach a point 100 feet away. The turtle travels at a constant speed of 20 feet per minute for the entire 100 feet. The rabbit travels at a constant speed of 40 feet per minute for the first 50 feet, stops for 3 minutes, and then continues at a constant speed of 40 feet per minute for the last 50 feet.
Determine which animal won the race and by how much time.
Answer:
The turtle travels 20 feet per minute, so it completes the race in (100 ft) / (20 ft/min) or 5 minutes.
The rabbit travels 40 ft/min for the first 50 feet. The time it takes is (50 ft) / (40 ft/min) = 1.25 minutes. It stops for 3 minutes, so it has travels 50 at the end of 4.25 minutes. It then repeats its pace for the first half of the race, which we already know is 1.25. A total of 100 feet takes the rabbit 4.5 minutes.
So the turtle won by a half a minute. You must state both of those facts for full credit.
End of Part III.
More to come. Comments and questions welcome.
More Regents problems.
### I also write Fiction!
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Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
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# Illustrative Mathematics Grade 8, Unit 4, Lesson 7: All, Some, or No Solutions
Learning Targets:
• I can determine whether an equation has no solutions, one solution, or infinitely many solutions.
Related Pages
Illustrative Math
#### Lesson 7: All, Some, or No Solutions
Let’s think about how many solutions an equation can have.
Illustrative Math Unit 8.4, Lesson 7 (printable worksheets)
#### Lesson 7 Summary
An equation is a statement that two expressions have an equal value. The equation
2x = 6
is a true statement if x is 3: 2 · 3 = 6.
It is a false statement if x is 4: 2 · 4 ≠ 6.
The equation 2x = 6 has one and only one solution, because there is only one number (3) that you can double to get 6.
Some equations are true no matter what the value of the variable is. For example:
2x = x + x
is always true, because if you double a number, that will always be the same as adding the number to itself. Equations like 2x = x + x have an infinite number of solutions. We say it is true for all values of x.
Some equations have no solutions. For example:
x = x + 1
has no solutions, because no matter what the value of x is, it can’t equal one more than itself. When we solve an equation, we are looking for the values of the variable that make the equation true. When we try to solve the equation, we make allowable moves assuming it has a solution. Sometimes we make allowable moves and get an equation like this:
8 = 7
This statement is false, so it must be that the original equation had no solution at all.
#### Lesson 7.1 Which One Doesn’t Belong: Equations
Which one doesn’t belong?
1. 5 + 7 = 7 + 5
2. 5 · 7 = 7 · 5
3. 2 = 7 - 5
4. 5 - 7 = 7 - 5
#### Lesson 7.2 Thinking About Solutions
Without solving, identify whether these equations have a solution that is positive, negative, or zero.
1. Sort these equations into the two types: true for all values and true for no values.
2. Write the other side of this equation so that this equation is true for all values of u.
6(u - 2) + 2 =
3. Write the other side of this equation so that this equation is true for no values of u.
6(u - 2) + 2 =
#### Are you ready for more?
Consecutive numbers follow one right after the other. An example of three consecutive numbers is 17, 18, and 19. Another example is -100, -99, -98.
How many sets of two or more consecutive positive integers can be added to obtain a sum of 100?
Are there sets with two numbers?
We can try two numbers grouped around 100/2 = 50.
49 + 50 ≠ 100 and 50 + 51 ≠ 100
And so, we cannot obtain two consecutive positive integers that will add to 100.
Are there sets with three numbers?
We can try three numbers grouped around 100/3 = 33.
Again, we cannot obtain three consecutive positive integers that will add to 100.
Similarly, we can try with four numbers.
We can try four numbers grouped around 100/4 = 25.
Again, we cannot obtain four consecutive positive integers that will add to 100.
We can try with five numbers.
We can try five numbers grouped around 100/5 = 20.
Yes, 18 + 19 + 20 + 21 + 22 = 100.
Six? No
Seven? No
Eight? Yes, 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100.
Nine? No. ….
We can stop at 14, where the numbers would be grouped around 7.
At 15, the numbers would be grouped around 6 and the smallest number would be non-positive.
So there are two sets of consecutive positive integers whose sum is 100: one of five numbers (18 + 19 + 20 + 21 + 22), the other of eight (9 + 10 + 11 + 12 + 13 + 14 + 15 + 16).
#### Lesson 7.3 What’s the Equation?
1. Complete each equation so that it is true for all values of x.
2. Complete each equation so that it is true for no values of x.
3. Describe how you know whether an equation will be true for all values of or true for no values of x.
#### Lesson 7 Practice Problems
1. For each equation, decide if it is always true or never true.
2. Mai says that the equation 2x + 2 = x + 1 has no solution because the left hand side is double the right hand side. Do you agree with Mai? Explain your reasoning. 3 a. Write the other side of this equation so it’s true for all values of x: 1/2(6x - 10) - x =
b. Write the other side of this equation so it’s true for no values of x: 1/2(6x - 10) - x =
3. Here is an equation that is true for all values of x: 5(x + 2) = 5x + 10. Elena saw this equation and says she can tell 20(x + 2) + 31 = 4(5x + 10) + 31 is also true for any value of x. How can she tell? Explain your reasoning.
4. Elena and Lin are trying to solve 1/2 x + 3 = 7/2 x + 5. Describe the change they each make to each side of the equation.
a. Elena’s first step is to write 3 = 7/2 x - 1/2 x + 5.
b. Lin’s first step is to write x + 6 = 7x + 10.
5. Solve each equation and check your solution.
6. The point (-3,6) is on a line with a slope of 4.
a. Find two more points on the line.
b. Write an equation for the line.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Question Video: Finding in the Simplest Form the Result of Dividing Two Fractions Mathematics
02:32
### Video Transcript
Calculate five-sixths divided by five 12ths. Give your answer in its simplest form.
When dividing fractions, we’re actually going to multiply by the second fraction’s reciprocal. So we keep our first fraction of five-sixths. But instead of dividing, we will represent this by multiplying by the second fraction’s reciprocal. So a reciprocal is where we flip the fraction. So instead of five 12ths, we have 12 fifths.
Now, before we begin multiplying, we’re supposed to have our answer in its simplest form. So if we wanted to, we could go ahead and reduce at this next step or multiply our numbers together and reduce at the end. So let’s go ahead and multiply our numbers and reduce at the end. And then we’ll come back and look at the other route.
So when multiplying fractions, we multiply the numerators together, so five and 12. And now, we multiply the denominators together, six and five. So as we said before, we can either multiply these numbers together and then simplify or we can simplify right here. So let’s go ahead and multiply. Five times 12 is 60 and six times five is 30. And 60 over 30 or 60 divided by 30 is equal to two or we could write two over one. So this would be our final answer.
Now as we said before, we could have reduced at this step. So we could simplify one number on the numerator with one number on the denominator and then repeat the process until there’s nothing left to simplify. So the five on the numerator can cancel with the five on the denominator. So now, repeating the process again, is there another set that can simplify? Well, six and 12. Six goes into itself once and six goes into 12 twice.
It should also be noted that when we said that the fives cancel, we could also think of it as five goes into five once and then once again five goes into five once. So on the numerator, we had one times two. And on the denominator, we had one times one and one times two is two. And one times one is one. So two over one is equal to our final answer, which is two. So our answer in its simplest form would be two. |
Question
# In a 24 hours period, the average number of boats arriving at a port is 10. Assuming that boats arrive at a random rate that is the same for all subintervals of equal length $i.e. the probability of a boat arriving during a 1 hour period the same for every 1 hour period no matter what$. Calculate the probability that more than 1 boat will arrive during a 1 hour period. $P(X>1$ ) Give your answers to 4 decimal places and in a range between 0 and 1
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## Answer to a math question In a 24 hours period, the average number of boats arriving at a port is 10. Assuming that boats arrive at a random rate that is the same for all subintervals of equal length $i.e. the probability of a boat arriving during a 1 hour period the same for every 1 hour period no matter what$. Calculate the probability that more than 1 boat will arrive during a 1 hour period. $P(X>1$ ) Give your answers to 4 decimal places and in a range between 0 and 1
Fred
4.4
To solve this problem, we will use the Poisson distribution since we are dealing with the arrival of boats over a given time period.
The average number of boats arriving at the port in a 1-hour period is given as 10. Therefore, the average rate parameter, λ, is also 10.
The probability of more than 1 boat arriving during a 1-hour period can be calculated as:
P$X > 1$ = 1 - P$X = 0$ - P$X = 1$
where P$X = k$ represents the probability of k boats arriving during a 1-hour period.
Using the formula for the Poisson distribution:
P$X = k$ = $e^(-λ$ * λ^k) / k!
we can calculate each term.
P$X = 0$ = $e^(-10$ * 10^0) / 0! = e^$-10$
P$X = 1$ = $e^(-10$ * 10^1) / 1! = 10 * e^$-10$
Now we can substitute these values into the formula for P$X > 1$:
P$X > 1$ = 1 - e^$-10$ - 10 * e^$-10$
Calculating this expression, we find:
P$X > 1$ ≈ 1 - e^$-10$ - 10 * e^$-10$ ≈ 1 - 0.00004540 - 0.00045399 ≈ 0.9995
Therefore, the probability that more than 1 boat will arrive during a 1-hour period is approximately 0.9995.
\textbf{Answer: } P$X > 1$ \approx 0.9995
Frequently asked questions $FAQs$
Question: What is the derivative of cos^2$3x$ - tan$2x$ + sin$4x$ with respect to x?
+
What is the value of sin$π/4$ + cos$π/3$ - tan$π/6$?
+
What is the smallest positive value of x for which f$x$ = tan$x$ produces f$x$ equal to its minimum value?
+ |
Linear Equation in Two Variables
An equation of the form $ax + by + c = 0$, where $a, \ b, \ and \ c$ are real numbers $(a \ne 0, \ b \ne 0)$ is called a linear equation in two variables $x \ and \ y$. Such an equation has degree 1.
Examples:
$2x + 5y = 10$
$-2x + 3y - 5 = 0$
$78x + (\frac{1}{2})y - 23 = 0$
It is not necessary to represent the variables by $\ x \ and \ y$ only. You could use other notations as well as $\ a \ and \ b \ or\ m \ and \ n$ or others. In our discussion we will use $\ x \ and \ y$ as these are the most common way of representing variables.
Simultaneous Linear Equations in Two Variables
Two linear equations in two variables$\ x \ and \ y$ are said to form a system of simultaneous linear equations if each of them is satisfied by the same pair of values of $\ x \ and \ y$ .
A solution to a linear system is an assignment of numbers to the variables $x \ and \ y$ such that all the equations are simultaneously satisfied.
Examples:
$5x + 3y = 12 \ and \ 3x + 5y = 15$
or
$4x - 5y = 20 \ and \ 6x + 5y = -2$
METHODS OF SOLVING SIMULTANEOUS LINEAR EQUATIONS
Substitution Method
Suppose we are given two linear equations in $\ x \ and \ y$. Then, we may solve them by Substitution Method as explained below:
1. Express $\ y \$ in terms of $x$ from one of the given equations.
2. Substitute this value of $\ y \$ in the second equation to obtain an equation in $\ x \$. Solve it for $\ x \$ .
3. Substitute the value of $\ x \$ in the relation obtained in step 1, to get the value of $\ y \$ .
Note: We may interchange the roles of $\ x\$ and $\ y \$ in the above method.
Example showing substitution Method:
Let the two equations be:
$2x + y = 10 \ \ \ i)$
$x + 2y = 8 \ \ \ ii)$
From i) we get $y = 10 - 2x$,
Substitute this in ii)
$x + 2(10 - 2x) = 8$
$20 - 3x = 8$
$3x = 12 or x = 4$
Now substituting $x=4$ in the expression for $y$
$y = 10 - 2(4) = 2$
Hence $x=4$ and $y=2$ is the solution for the two equations.
Solution also means that these lines intersect at that point.
Elimination Method
1. Multiply the given equations by suitable constants so as to make the coefficients of one of the variables, numerically equal.
2. Add the new equations, if the numerically equal coefficients are opposite in sign; otherwise subtract them.
3. Solve the equation so obtained. This gives the value of one of the variables.
4. Substitute the value of this unknown in any of the given equations. Solve it to get the value of the other variable.
Example showing substitution Method:
We will solve the same equations
Let the two equations be:
$2x + y = 10 \ \ \ i)$
$x + 2y = 8 \ \ \ ii)$
Multiply the second equation by 2. We get
$2x + 4y = 16 \ \ \ iii)$
Now subtract iii) from i)
$2x + y = 10$
$-(2x + 4y = 16)$
We get $-3y= -6$
or $y=2$
Substituting in i), we get $x=4$
Hence $x=4$ and $y=2$ is the solution for the two equations.
Solution also means that these lines intersect at that point. |
# Hard math equations
Here, we debate how Hard math equations can help students learn Algebra. Our website can help me with math work.
## The Best Hard math equations
Apps can be a great way to help students with their algebra. Let's try the best Hard math equations. Solving system of equations matrices can be a difficult task, but it is important to understand the process in order to be successful. There are many different methods that can be used to solve system of equations matrices, but the most common is Gaussian elimination. This method involves adding or subtracting rows in order to create a new matrix that is easier to solve. Once the new matrix has been created, the variables can be solved for by using back-substitution. This process can be time-consuming and difficult, but it is important to persevere in order to get the correct answer. With practice, solving system of equations matrices will become easier and more intuitive.
In other words, x would be equal to two (2). However, if x represented one third of a cup of coffee, then solving for x would mean finding the value of the whole cup. In this case, x would be equal to three (3). The key is to remember that, no matter what the size of the fraction, solving for x always means finding the value of the whole. With a little practice, solving for x with fractions can become second nature.
Looking for an easy and effective way to solve equations? Look no further than the 3 equation solver! This handy tool can quickly and easily solve any equation with three variables, making it a valuable tool for students, teachers, and professionals alike. Simply enter the equation into the 3 equation solver and press the solve button. The tool will instantly generate a solution, making it easy to check your work or find the correct answer. With its simple and user-friendly interface, the 3 equation solver is a must-have for anyone who needs to solve equations on a regular basis. Give it a try today and see how much time and effort you can save!
Substitution is a method of solving equations that involves replacing one variable with an expression in terms of the other variables. For example, suppose we want to solve the equation x+y=5 for y. We can do this by substituting x=5-y into the equation and solving for y. This give us the equation 5-y+y=5, which simplifies to 5=5 and thus y=0. So, the solution to the original equation is x=5 and y=0. In general, substitution is a useful tool for solving equations that contain multiple variables. It can also be used to solve systems of linear equations. To use substitution to solve a system of equations, we simply substitute the value of one variable in terms of the other variables into all of the other equations in the system and solve for the remaining variable. For example, suppose we want to solve the system of equations x+2y=5 and 3x+6y=15 for x and y. We can do this by substituting x=5-2y into the second equation and solving for y. This gives us the equation 3(5-2y)+6y=15, which simplifies to 15-6y+6y=15 and thus y=3/4. So, the solution to the original system of equations is x=5-2(3/4)=11/4 and y=3/4. Substitution can be a helpful tool for solving equations and systems of linear equations. However, it is important to be careful when using substitution, as it can sometimes lead to incorrect results if not used properly. |
# Question Video: Subtract Two-Digit Numbers from Two-Digit Numbers with Regrouping Mathematics • 2nd Grade
We can think about place value to help us subtract. Pick the correct way to decompose 35 into tens and ones. Subtract 18 from 35 by following these steps: Regroup 1 ten into 10 ones. Take away 8 ones. Take away 1 ten.
03:29
### Video Transcript
We can think about place value to help us subtract. Pick the correct way to decompose 35 into 10s and ones. Subtract 18 from 35 by following these steps. Regroup one 10 into 10 ones. Take away eight ones. Take away one 10. 35 take away 18 equals what?
In this question, we need to think about place value. We’re asked to decompose the number 35 into 10s and ones. And we’re given four possible answers. We can partition the number 35 into 10s and ones. The five is worth five ones, and the three in 35 is worth three 10s, or 30. So, we’re looking for the place value chart which shows us three 10s and five ones. The first of our four models shows three 10s and five ones. So, this is the correct way to decompose 35 into 10s and ones.
Next, we have to subtract 18 from 35. And we’re told to follow these steps. First, regroup one 10 into 10 ones. This step has already been done for us. There were three 10s in 35, and one 10 has been taken away. Now, there are two 10s. This 10 has been exchanged for 10 ones. We took one 10 and exchanged it for 10 ones. We did have five ones in 35. Now, we have 15 ones. Next, we’re asked to take away the eight ones. What is 15 take away eight? The answer is seven. Finally, we have to take away one 10. Two take away one equals one. 35 take away 18 equals 17.
We thought about place value to help us subtract. We picked the correct way to decompose 35 into 10s and ones. The first-place value chart shows three 10s and five ones. Then, we subtracted 18 from 35 by following the steps. First, we regrouped one 10 into 10 ones. Then, we took away eight ones and one 10 to give us our answer of 17. 35 take away 18 equals 17. |
# Explain how to determine if a table of values represents a quadratic function
explain how to determine if a table of values represents a quadratic function
## Explain how to determine if a table of values represents a quadratic function
To determine if a table of values represents a quadratic function, you need to investigate how the values change as you move from one row to the next. Here, we leverage differences between consecutive values. Quadratic functions are characterized by having a constant second difference. Here is a step-by-step approach to identify whether a table of values corresponds to a quadratic function:
### Solution By Steps:
1. Determine the First Differences:
• Calculate the differences between each pair of consecutive y-values (outputs). These differences are known as the first differences.
• Given a table with x and y values:
x y
x_1 y_1
x_2 y_2
x_3 y_3
x_n y_n
The first differences will be:
\Delta y_1 = y_2 - y_1
\Delta y_2 = y_3 - y_2
\Delta y_3 = y_4 - y_3
\dots
\Delta y_{n-1} = y_n - y_{n-1}
2. Determine the Second Differences:
• Next, calculate the differences between each pair of consecutive first differences. These are called the second differences.
• Calculate these as:
\Delta^2 y_1 = \Delta y_2 - \Delta y_1
\Delta^2 y_2 = \Delta y_3 - \Delta y_2
\Delta^2 y_3 = \Delta y_4 - \Delta y_3
\dots
\Delta^2 y_{n-2} = \Delta y_{n-1} - \Delta y_{n-2}
3. Verify the Consistency of Second Differences:
• Check if the second differences are constant. If the second differences are all the same, then the table of values represents a quadratic function.
### Example:
Consider the following table of values:
x y
0 1
1 3
2 9
3 19
4 33
1. First Differences:
\Delta y_1 = 3 - 1 = 2
\Delta y_2 = 9 - 3 = 6
\Delta y_3 = 19 - 9 = 10
\Delta y_4 = 33 - 19 = 14
2. Second Differences:
\Delta^2 y_1 = 6 - 2 = 4
\Delta^2 y_2 = 10 - 6 = 4
\Delta^2 y_3 = 14 - 10 = 4
3. Conclusion:
• The second differences \Delta^2 y are all equal to 4.
• Since the second differences are constant, the table of values indeed represents a quadratic function. |
Lesson Plan:
# Solving Long Division Problems
1.0 based on 3 ratings
Standards
August 31, 2015
## Learning Objectives
Students will be able to use long division.
## Lesson
### Introduction (10 minutes)
• Invite the students to bring their math journals and a pencil to the class meeting area.
• Tell them to place their supplies on the floor next to them.
• Tell the students that today they're going to learn a strategy for solving long division problems. Write the word Division on the board.
• Let the students know that division is an operation that tells us the number of groups that can be made out of another number.
• On the board, set up a division problem, such as 17/5. Solve the problem so that you are left with the answer 3 with a remainder of 2. Label each number with its corresponding name. 17 should be the dividend, 5 is the divisor, 3 is the quotient and 2 is the remainder.
• Define these terms for your students as follows:
• Dividend: In a division problem, the number that is to be divided is called the dividend.
• Divisor: In a division problem, the number that divides the dividend is called the divisor.
• Quotient: Upon division, the number obtained other than the remainder is called the quotient.
• Remainder: The remainder is the number that is left over after dividing.
### Explicit Instruction/Teacher Modeling (10 minutes)
• Direct the students’ attention to the board.
• Write a division problem, such as 83 divided by 7, on the board.
• Tell the students that the first thing we want to do is create a help box to help us in solving this division problem.
• Create a chart that lists the products of 7x1, 7x2...7x10.
• Demonstrate solving this problem for students. You should reach the answer 11 with a remainder of 6.
### Guided Practice/Interactive Modeling (20 minutes)
• Put another problem on the board similar to the one just completed.
• Tell the student to pick up their supplies and write the problem down.
• Ask for volunteers to raise their hands and share with the class the steps needed to complete this problem based on the problem just completed.
• Give the students time to solve the problems.
• Ask the students to dictate to you how they solved the problem. Write their responses on the board.
• Review the problem with the class to check for accuracy. Make sure everyone has the problem written down as a reference for completing their assignment during independent work time.
### Independent Working Time (30 minutes)
• Send the students back to their seats.
• Pass out the Solving Long Division Problems worksheet.
• Walk around the room to assist students as needed.
## Extend
### Differentiation
• Enrichment: Give advanced students more difficult long division problems to solve.
• Support: Observe and monitor struggling students more closely. Get involved in helping the students solve the problems.
## Review
### Assessment (10 minutes)
• Check the worksheet for correctness and to decide who may need extra small group practice.
### Review and Closing (10 minutes)
• Bring the students back to the class meeting area.
• Ask for volunteers to share their answers and work for the problems assigned.
• As the problems are reviewed in front of the class, the students will check their answers for accuracy.
• After each problem has been checked for accuracy, ask the class if there are any questions or comments about long division. |
# If |z−1|+|z+3|≤8, then prove that z lies on the circle.
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Text Solution
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## To prove that z lies on a circle given the condition |z−1|+|z+3|≤8, we can follow these steps:Step 1: Understand the expressionThe expression |z−1|+|z+3| represents the sum of the distances from the point z in the complex plane to the points 1 and −3. Step 2: Identify the fociThe points 1 and −3 can be represented as the foci of an ellipse. The distance from any point z on the ellipse to these two foci will be constant. Here, the sum of the distances is less than or equal to 8.Step 3: Find the distance between the fociThe distance between the foci 1 and −3 is:|1−(−3)|=|1+3|=4Step 4: Determine the major axis lengthSince the sum of the distances from any point on the ellipse to the foci is 8, which is greater than the distance between the foci (4), we can conclude that the ellipse is valid. The major axis length is 8.Step 5: Find the center of the ellipseThe center of the ellipse is the midpoint of the line segment joining the foci:Center=(1+(−3)2,0)=(−1,0)Step 6: Determine the vertices of the ellipseThe vertices of the ellipse can be found by moving 4 units from the center along the x-axis (half of the major axis length):- Right vertex: −1+4=3- Left vertex: −1−4=−5Thus, the vertices are 3 and −5.Step 7: Identify the circle conditionFor |z−1|+|z+3|=8, this describes the boundary of the ellipse. The points z that satisfy |z−1|+|z+3|<8 will lie inside this ellipse.Step 8: Conclude the proofSince the condition |z−1|+|z+3|≤8 includes the boundary, we can conclude that z lies on or within the ellipse defined by the foci 1 and −3 and the major axis length of 8.
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## Linear Algebra and Its Applications, Exercise 1.7.7
Exercise 1.7.7. Take the 3 by 3 Hilbert matrix from the previous exercise
$A = \begin{bmatrix} 1&\frac{1}{2}&\frac{1}{3} \\ \frac{1}{2}&\frac{1}{3}&\frac{1}{4} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5} \end{bmatrix}$
and compute $b$ assuming that $x = (1, 1, 1)$ and $x = (0, 6, -3.6)$ are solutions to $Ax = b$.
Answer: We first multiply $A$ times $x = (1, 1, 1)$:
$b = Ax = \begin{bmatrix} 1&\frac{1}{2}&\frac{1}{3} \\ \frac{1}{2}&\frac{1}{3}&\frac{1}{4} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 + \frac{1}{2} + \frac{1}{3} \\ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{11}{6} \\ \frac{13}{12} \\ \frac{47}{60} \end{bmatrix} \approx \begin{bmatrix} 1.83 \\ 1.08 \\ .783 \end{bmatrix}$
expressing the final result to three significant digits. We then multiply $A$ by $x = (0, 6, -3.6)$ again expressing the final result to three significant digits:
$b = Ax = \begin{bmatrix} 1&\frac{1}{2}&\frac{1}{3} \\ \frac{1}{2}&\frac{1}{3}&\frac{1}{4} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5} \end{bmatrix} \begin{bmatrix} 0 \\ 6 \\ -3.6 \end{bmatrix} = \begin{bmatrix} 0 + 3 - 1.2 \\ 0 + 2 - 0.9 \\ 0 + 1.5 - .72 \end{bmatrix} = \begin{bmatrix} 1.80 \\ 1.10 \\ .780 \end{bmatrix}$
Note that the values of $b$ are almost identical, though the values of the solutions $x$ are very different. This indicates that the solution $x$ to $Ax = b$ is very sensitive to small differences in $b$.
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. |
# The line 3x-2y+k=0, where k is a constant, intersects the curve x^2 + y^2 -4x-9=0 at two points. Find the range of k?
Feb 9, 2018
Range of $k$ is $- 19 < k < 7$
#### Explanation:
To find points of intersection of a line and curve, we should solve them as simultaneous equation.
Here we have a linear equation indicating a line and a quadratc equation indicating a conic section (actually a circle). When we substitute one variable in terms of another variable from linear equation in to quadratic equation, we get a quadratic equation in other variable.
If the quadratic equation has two roots, they intersect at two points and if it has one root, it is a tangent. In case it has no roots, it means line does not intersect the conic section at all.
Here we have $3 x - 2 y + k = 0$, which gives us $y = \frac{3}{2} x + \frac{k}{2}$ and putting this in the equation of circle ${x}^{2} + {y}^{2} - 4 x - 9 = 0$, we get
${x}^{2} + {\left(\frac{3 x}{2} + \frac{k}{2}\right)}^{2} - 4 x - 9 = 0$
or ${x}^{2} + \frac{9 {x}^{2}}{4} + {k}^{2} / 4 + \frac{3}{2} k x - 4 x - 9 = 0$
or $4 {x}^{2} + 9 {x}^{2} + {k}^{2} + 6 k x - 16 x - 36 = 0$
or $13 {x}^{2} + \left(6 k - 16\right) x + {k}^{2} - 36 = 0$
It will have two roots if discriminant is greater than zero i.e.
${\left(6 k - 16\right)}^{2} - 52 \left({k}^{2} - 36\right) > 0$
or $36 {k}^{2} - 192 k + 256 - 52 {k}^{2} + 1872 = 0$
or $- 16 {k}^{2} - 192 k + 2128 > 0$
or ${k}^{2} + 12 k - 133 < 0$
or $\left(k + 19\right) \left(k - 7\right) < 0$
This is so, when either $k + 19 > 0$ and $k - 7 < 0$ i.e. $- 19 < k < 7$
or $k + 19 < 0$ and $k - 7 > 0$ i.e. $k < - 19$ and $k > 7$, but this is not possible.
Hence solution is $- 19 < k < 7$ |
# Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio
## Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio
### Ratio Text Book Questions and Answers
Same shape Textbook Page No. 116
In both the rectangles below, the length is 1 centimetre more than the breadth.
But not only the length of these are different, they look different also. In the larger rectangle, the sides look almost the same. Draw a rectangle of length 50 centimetres and breadth 49 centimetres in a larger sheet of paper. It looks almost a square, doesn’t it?
In the first rectangle above, length is double the breadth. Now look at this rectangle.
Again the length is double the breadth. Even though it is larger than the first, they have the same shape, right?
As the length is double the breadth in the second rectangle. The first and the second rectangle will have the same shape.
Changing scale
Look at this photo:
The shorter side is 2 centimetres and the longer side, 3 centimetres; that is, the longer side is 1$$\frac{1}{2}$$ times shorter side.
Suppose we make the shorter side 3 centimetres and longer side 4.5 centimetres.
Still, the longer side is 1$$\frac{1}{2}$$ times the shorter side.
Now suppose we change the shorter side to 3 centimetres and increase the longer side also by 1 centimetre, making it 4 centimeters.
Does the picture look right?
Yes, the picture looks right.
Suppose if we change the shorter side to 3 centimetres and the longer side to 4 centimetres. The longer side is 1$$\frac{1}{3}$$ times the shorter side.
TV Math
The sizes of TV sets are usually given as 14 inch, 17 inch, 20 inch and so on. What does it mean?
The TV screen is a rectangle: and these are lengths of the diagonals of the screen.
Does it determine the size of the screen? Rectangles of different width and lengths can have the same diagonal:
In the modern TV sets, the ratio of length to width is 16 : 9. In the earlier days, it was 4 : 3.
See their difference in two TV screens of the same diagonal length.
Flags Textbook Page No. 119
When we draw our National Flag, not only should the colours be right, the ratio of width to length should also be correct. This ratio is 2 : 3.
That is, in drawing our flag, if the length is taken as 3 centimetres, the width should be 2 centimetres.
This ratio is different for flags of other countries. For example, in the flag of Australia, this ratio is 1 : 2.
And in the flag of Germany, it is 3 : 5.
Without fractions Textbook Page No. 120
When quantities like length are measured using a definite unit, we may not always get counting numbers; and it is this fact which led to the idea of fractions.
In comparing the sizes of two quantities, one question is whether both can be given as counting numbers, using a suitably small unit of measurement. It is this question that leads to the idea of ratio.
For example, suppose the length of two objects are found as $$\frac{2}{5}$$ and $$\frac{3}{5}$$, when measured using a string. If $$\frac{1}{5}$$ of the string is taken as the unit, we can say the length of the first is 2 and that of the second is 3. This is the meaning of
saying the ratio of the lengths is 2 : 3.
Suppose the length of two objects are $$\frac{1}{3}$$ and $$\frac{1}{5}$$ of the string.
To get both lengths as counting numbers, what fraction of the string can be taken as a unit of measurement?
Circle relations
Look at the pieces of circles below:
The smaller piece is $$\frac{1}{4}$$ of the circle and the larger piece is $$\frac{1}{2}$$ of the circle.
So, the larger piece is twice the size of the smaller. That is, the ratio of the sizes of small and large pieces is 1 : 2.
Now look at these pieces:
What is the ratio of their sizes?
Let’s measure each using $$\frac{1}{4}$$ of circles.
The smaller figure has two such pieces. What about the larger?
Let’s measure each using $$\frac{1}{4}$$ of circles.
The larger figure has three such pieces.
So what is the ratio of the sizes of these two?
Let’s measure each using $$\frac{1}{4}$$ of circles.
The smaller piece has two such pieces and the larger piece has three such pieces.
2×$$\frac{1}{4}$$ : 3×$$\frac{1}{4}$$
$$\frac{2}{4}$$ : $$\frac{3}{4}$$
Therefore, the ratio will be 2 : 3
Moving Ratio Textbook Page No. 122
Have you taken apart toy cars or old clocks? There are many toothed wheels in such mechanisms. See this picture:
It is a part of a machine. In the two whole wheels we see, the smaller has 13 teeth and the larger has 21. So during the time that the smaller circle makes 21 revolutions, the larger one would have made only 13 revolutions.
The speed of rotation of machines is controlled by arranging such wheels with different number of teeth.
Sand and Cement
In construction of a building, sand and cement are used in definite ratios, but the ratios are different depending on the purpose. When one bowl of cement and five bowls of sand are mixed, the ratio of cement to sand is 1 : 5
When one sack of cement is mixed with five sacks of sand, the ratio is the same. But to set a brick wall, this much cement may not be needed. In this case, the ratio may be 1 : 10 or 1 : 12.
Ratio of Parts Textbook Page No. 124
We can use ratio to compare parts of a whole also. For example, in the picture below, the lighter part is $$\frac{3}{8}$$ of the circle, and the darker part is $$\frac{5}{8}$$ of the circle.
These two parts together make up the whole circle. The ratio of the sizes of these parts is 3 : 5.
So here, the ratio 3 : 5 indicate the two fractions $$\frac{3}{8}$$ and $$\frac{5}{8}$$.
Generally in such instances, a ratio of the numbers indicates fractions of equal denominators adding up to 1.
Meaning of ratio Textbook Page No. 125
If we know only the ratio of two quantities, we can’t say exactly how much they are; but they can be compared in several ways.
For examples, suppose the volumes of two pots are said to be in the ratio 2 : 3. We can interpret this in the following ways:
• To fill the smaller pot, we need $$\frac{2}{3}$$ of the larger pot.
• To fill the bigger pot, we need $$\frac{3}{2}$$ = 1$$\frac{1}{2}$$ times the smaller pot.
• Whether we take $$\frac{1}{2}$$ of the water in the smaller pot, or $$\frac{1}{3}$$ of the water in the larger pot, we get the same amount of water.
• If we fill both pots and pour them into a large pot, $$\frac{2}{5}$$ of the total amount of water
is from the smaller pot and $$\frac{3}{5}$$ from the larger.
If the length of two pieces of rope are said to be in the ratio 3 : 5, what all interpretations like this can you make?
If the length of two pieces of rope is 3 : 5.Then the following interpretations can be made.
The smaller rope is $$\frac{3}{5}$$ of the length of the larger rope.
The smaller rope will be $$\frac{3}{8}$$ of the total rope length.
Three measures
Look at this triangle:
In it, the longest side is double the shortest. The side of medium length is one and a half times the shortest.
Saying this with ratios, the lengths of the shortest and the longest sides are in the ratio 1 : 2 and those of the shortest and the medium are in the ratio 2 : 3.
What is the ratio of the lengths of the medium side and the longest side?
We can say this in another way: if we use a string of length 1.5 centimetres, the shortest side is 2, the medium side is 3 and the longest side is 4.
The can be shortened by saying the sides are in the ratio 2 : 3 : 4.
Triangle Math Textbook Page No. 127
How many triangles are there with ratio of sides 2 : 3 : 4?
The lengths of sides can be 2 cm, 3 cm, 4cm.
Or they can be 1 cm, 1.5 cm, 2 cm
We can take metres instead of centimetres.
Thus we have several such triangles.
In all these, what fraction of the perimeter is the shortest side?
And the side of medium length?
The longest side?
The perimeter of a triangle is 80 centimetres and its sides are in the ratio 5 : 7 : 8. Can you compute the actual lengths of sides?
Let the ratio be x.Then the sides will be 5x, 7x and 8x.
Given that perimeter is 80 cm.
5x + 7x + 8x = 80
20x = 80
x = 4
Since the value of x is 4, the lengths will be 5×4=20cm, 7×4=28cm and 8×4=32cm.
What if the perimeter is 1 metre?
1 metre equals 100 centimetres.
Let the ratio be x.Then the sides will be 5x, 7x and 8x.
Given that perimeter is 100 cm.
5x + 7x + 8x = 100
20x = 100
x = 5
Since the value of x is 5, the lengths will be 5×5=25cm, 7×5=35cm and 8×5=40cm.
Length and width Textbook Page No. 116
Look at these rectangles:
Is there any common relation between the lengths and widths of all these?
In all these, the length is twice the width, right? (We can also say width is half the length)
In the language of mathematics, we state this fact like this:
In all these rectangles, the width and length are in the ratio one to two.
In writing, we shorten the phrase “one to two” as 1 : 2; that is
In all these rectangles, the width and length are in the ratio 1 : 2.
In the rectangle of width 1 centimetre and length 2 centimetres, the length is twice the width. In a rectangle of width 1 metre and length 2 metres also, we have the same relation. So, in both these rectangles, width and length are in the ratio one to two (1 : 2).
We can state this in reverse: in all these rectangles, length and width are in the ratio two to one (2 : 1).
What is the width to length ratio of the rectangle below?
The length of the given rectangle is 6cm and width is 2cm.
Therefore, the width to length ratio for the above rectangle will be 1 : 3.
The length of the given rectangle is 4.5cm and width is 1.5cm.
Therefore, the width to length ratio for the above rectangle will be 1 : 3.
In both, the length is 3 times the width. So what is the ratio of the width to the length?
In the above two rectangles, the length is thrice the width.
The ratio of the width to the length is 1 : 3.
How do we say this as a ratio?
The width and length are in the ratio one to three.
1 metre means loo centimetres. So in such a rectangle, the ratio of the width to the length is 1: 50.
Now look at these rectangles:
In both, the length is one and a half times the width.
How do we say this as a ratio?
We can say one to one and a half; but usually we avoid fractions when we state ratios.
Suppose we take the width as 2 centimetres.
What is 1$$\frac{1}{2}$$ times 2?
So, we say that in rectangles of this type, the ratio of width to length is two to three and write 2 : 3.
Can’t we say here that the ratio is 4 : 6 ?
Nothing wrong in it; but usually ratios are stated using the least possible counting numbers.
So how do we state using ratios, the fact that the length of a rectangle is two and a half times the width?
If the width is 1 centimetre, then the length is 2$$\frac{1}{2}$$ centimetres.
What if the width is 2 centimetres?
Length would be 5 centimetres.
So, we say that width and length are in the ratio 2 to 5.
What if the length is one and a quarter times the width?
If the width is 1 centimetre, the length is 1$$\frac{1}{4}$$ centimetre.
If the width is 2 centimetres then the length is 2$$\frac{1}{2}$$ centimetres.
Still we haven’t got rid of fractions.
Now if width is 4 centimetres, what would be the length?
So, in all such rectangles, the width and length are in the ratio 4 : 5.
Do you notice another thing in all these?
If we stretch or shrink both width and length by the same factor, the ratio is not changed. For example, look at this table.
In all these, the length is 3 times the width; or the reverse, the length is $$\frac{1}{3}$$ of the width.
In terms of ratio, we say width and length are in the ratio 1 : 3 or length and width are in the ratio 3 : 1.
In terms of ratio, we say width and length are in the ratio 1: 3 or length and width are in the ratio 3: 1.
• For all rectangles of dimensions given below, state the ratio of width to length, using the least possible counting numbers:
• width 8 centimetres, length 10 centimetres
The simplest form for the given numbers 8:10 is 4 : 5
Therefore, the ratio of width to length will be 4 : 5
• width 8 metres length 12 metres
The simplest form for the given numbers 8:12 is 4 : 3
Therefore, the ratio of width to length will be 2 : 3
• width 20 centimetres length 1 metre
The given numbers cannot be written in simplest form.
Therefore, the ratio of width to length will be 20 : 1
• width 40 centimetres length 1 metre
The given numbers cannot be written in simplest form.
Therefore, the ratio of width to length will be 40 : 1
• width 1.5 centimetres length 2 centimetres
The simplest form for the given numbers 1.5:2 is 3 : 4
Therefore, the ratio of width to length will be 3 : 4
• In the table below two of the width, length and their ratio of some rectangles are given. Calculate the third and fill up the table.
• What does it mean to say that the width to length ratio of a rectangle is 1: 1 ? What sort of a rectangle is it?
The width to length ratio of a rectangle is 1 : 1.
Here the width of the rectangle and length of the rectangle is same.
If the width and the length of rectangle is same, then it is a square.
Other quantities Textbook Page No. 120
There are two ropes, the shorter $$\frac{1}{3}$$ metre long and the longer, $$\frac{1}{2}$$ metre long. What is the ratio of their lengths?
We can do this in different ways.
We can check how much times $$\frac{1}{3}$$ is $$\frac{1}{2}$$.
$$\frac{1}{2}$$ ÷ $$\frac{1}{2}$$ = $$\frac{3}{2}$$
So, the length of the longer rope is $$\frac{3}{2}$$ times that of the shorter.
That is, 1$$\frac{1}{2}$$ times.
If the length of the shorter rope is taken as 1, the length of the longer is 1$$\frac{1}{2}$$; if 2, then 3.
So the length of the shorter and longer rope are in the ratio 2 : 3.
We can also think in another manner. As in the case of the width and length of rectangles, we can imagine the length to be stretched by the same factor; and the ratio won’t change.
Suppose we double the length of each piece of rope.
Then the length of the shorter is $$\frac{2}{3}$$ metres and that of the longer 1 metre. This doesn’t remove fractions.
By what factor should we stretch to get rid of fractions? How about 6 ?
6 times $$\frac{1}{3}$$ is 2.
6 times $$\frac{1}{2}$$ is 3.
The shorter is now 2 metres and the longer is 3 metres. So the ratio is 2 : 3.
There is yet another way; we can write
$$\frac{1}{3}$$ = $$\frac{2}{6}$$ $$\frac{1}{2}$$ = $$\frac{3}{6}$$
That is, we can think of the shorter rope as made up of 2 pieces of $$\frac{1}{6}$$ metres each and the longer one as 3 of the same $$\frac{1}{6}$$ metres. In this way also, we can calculate the ratio as 2 to 3.
Now look at this problem : to fill a can we need only half the water in a bottle.
To fill a larger can, we need three quarters of the bottle. What is the ratio of the volume of the smaller can to the larger can ?
Here, we can write
$$\frac{1}{2}$$ = $$\frac{2}{4}$$
So, if we pour $$\frac{1}{4}$$ of the bottle 2 times, we can fill the smaller can; to fill the larger can, $$\frac{1}{4}$$ of the bottle should be poured 3 times. So the volumes of the smaller and larger cans are in the ratio 2 : 3.
Another problem: Raju has 200 rupees with him and Rahim has 300. What is the ratio of the money Raju and Rahim have?
If we imagine both to have the amounts in hundred rupee notes, then Raju has 2 notes and Rahim, 3 notes. So the ratio is 2 : 3.
Let’s change the problem slightly and suppose Raju has 250 rupees and Rahim, 350 rupees.
If we think of the amounts in terms of 50 rupee notes, then Raju has 5 and Rahim has 7.
The ratio is 5 : 7.
What if the amounts are 225 and 325 rupees?
Imagine each amount as packets of 25 rupees. Raju has 225 ÷ 25 = 9 packets and Rahim has 325 ÷ 25 = 13 packets. The ratio is 9 : 13.
Let’s look at one more problem: in a class, there are 25 girls and 20 boys. What is the ratio of the number of girls to the number of boys?
If we split the girls and boys separately into groups of 5, there will be 5 groups of girls and 4 groups of boys. So the ratio is 5 to 4.
In this way, calculate the required ratios and write them using the least possible counting numbers, in these problems:
• Of two pencils, the shorter is of length 6 centimetres and the longer, 9 centimetres. What is the ratio of the lengths of the shorter and the longer pencils?
Given that the length of shorter pencil is 6 centimetres and the longer is 9 centimetres.
The simplest form of 6 and 9 is 2 and 3.
In terms of ratio, the ratio of the lengths of the shorter and the longer pencils 2 : 3.
• In a school, there are 120 boys and 140 girls. What is the ratio of the number of boys to the number of girls?
Given that there are 120 boys and 140 girls in a school.
The simplest form of 120 and 140 is 6 and 7.
In terms of ratio, it will be 6 : 7
• 96 women and 144 men attended a meeting. Calculate the ratio of the number of women to the number of men.
Given that 96 women and 144 men attended a meeting.
The simplest form of 96 and 144 is and 22 and 3.
In terms of ratio, it will be 2 : 3
• When the sides of a rectangle were measured using a string, the width was $$\frac{1}{4}$$ of the string and the length $$\frac{1}{3}$$ of the siring. What is the ratio of the width to the length’?
Given that the width of rectangle is $$\frac{1}{4}$$ of the string and the length is $$\frac{1}{3}$$ of the siring.
The ratio of the width to the length will be 3 : 4
• To fill a larger bottle, 3$$\frac{1}{2}$$ glasses of water are needed and to fill a smaller bottle, 2$$\frac{1}{4}$$ of glasses are needed. What is the ratio of the volumes of the large and small bottles?
Given that to fill a larger bottle 3$$\frac{1}{2}$$ glasses of water are needed and to fill a smaller bottle, 2$$\frac{1}{4}$$ of glasses are needed.
3$$\frac{1}{2}$$ equals $$\frac{7}{2}$$ and 2$$\frac{1}{4}$$ equals $$\frac{9}{4}$$.
The ratio of the volumes of the large and small bottles will be 14 : 9
Ratio of mixtures Textbook Page No. 123
To make idlis , Ammu’s mother grinds two cups of rice and one cup of urad dal. When she expected some guests the next day, she took four cups of rice. How many cups of urad should she take?
To have the same consistency and taste, the amount of urad must be halfthe amount of rice.
So for four cups of rice, there should be two cups of urad.
We can say that the quantities, rice and urad must be in the ratio 2 :1
Now another problem on mixing: to paint the walls of Abu’s home, first 25 litres of green and 20 litres of white were mixed.
When this was not enough, 15 litres of green was taken. How much white should be added to this?
To get the same final colour, the ratio of green and white should not change.
In what ratio was green and white mixed first?
That is, for 5 litres of green, we should take 4 litres of white.
To maintain the same ratio, for 15 litres of green, how many litres of white should we take?
How many times 5 is 15?
So 3 times 4 litres of white should be mixed.
That is 12 litres.
To get the same shade of green, how many litres of green should be mixed with 16 litres of white?
Now try these problems:
• For 6 cups of rice, 2 cups of urad should be taken to make dosas. How many cups of urad should taken with 9 cups of rice?
Given that 6 cups of rice, 2 cups of urad should be taken to make dosas.
In terms of ratio, the ratio of the rice to urad will be 3 : 1.
Here, the number of rice cups is thrice the number of urad cups.
Therefore, to maintain the same ratio, for 9 cups of rice, 3 cups of urad should be taken.
• To set the walls of Nizar’s house, cement and sand were used in a ratio 1:5. He bought 45 sacks of cement. How many sacks of sand should he buy?
Given that to set the walls of Nizar’s house, cement and sand were used in a ratio 1:5.
Here, 1 resembles the number of units for cement sacks and 5 resembles the number of unit for sand sacks.
The number of sand sacks required is 5 times the number of cement sacks.
If there are 45 sacks of cement, then the number of sand sacks required will be 45×5=225.
• To paint a house, 24 litres of paint was mixed with 3 litres of turpentine. How many litres of turpentine should be mixed with 32 litres of paint?
Given that to paint a house, 24 litres of paint was mixed with 3 litres of turpentine.
In terms of ratio, the ratio of the paint to turpentine will be 8 : 1.
Let x be the number of litres of turpentine that should be mixed with 32 litres of paint.
$$\frac{8}{1}$$ = $$\frac{32}{x}$$
8x = 32
x = $$\frac{32}{8}$$
x = 4
Therefore, 4 litres of turpentine should be mixed with 32 litres of paint.
• In the first ward of a panchayat, the male to female ratio is 10 : 11. There are 3311 women. How many men are there? What is the total population in that panchayat?
Given that In the first ward of a panchayat, the male to female ratio is 10 : 11.
Let the number of men be m and there are 3311 women.
$$\frac{10}{11}$$ = $$\frac{m}{3311}$$
11 × m = 10 × 3311
m = $$\frac{10 × 3311}{11}$$
m = 3010
Therefore, the number of mens will be 3010.
The total population in that panchayat will be the sum of number of mens and womens, which is 3010+3311=6321
• In a school, the number of female and male teachers are in the ratio 5 : 1. There are 6 male teachers. How many female teachers are there?
Given that, In a school, the number of female and male teachers are in the ratio 5 : 1.
Here, the number of female teachers is five times the number of male teachers.
If there are 6 male teachers.Then the female teachers will be 5×6=30.
Thus, there are 30 female teachers.
• Ali and Ajayan set up a shop together. Ali invested 5000 rupees and Ajayan, 3000 rupees. They divided the profit for a month in the ratio of their investments and Ali got 2000 rupees. How much did Ajayan get? What is the total profit?
Given that, Ali invested 5000 rupees and Ajayan, 3000 rupees.
In terms of ratio, the ratio of Ali to Ajayan will be 5 : 3
Ali got 2000 rupees.Let m be the amount Ajayan should get.
$$\frac{5}{3}$$ = $$\frac{2000}{m}$$
5×m = 2000×3
m = $$\frac{2000×3}{5}$$
m = 1200
Ajayan will get 1200 rupees.
The total profit will be the sum of maount got by Ali and Ajayan, which is 2000+1200=3200 rupees.
Division Problem Textbook Page No. 125
We have seen that to make idlis, rice and urad are to be taken in the ratio 2:1. In 9 cups of such a mixture of rice and urad, how many cups of rice are taken?
2 cups of rice and 1 cup of urad together make 3 cups of mixture.
Here we have 9 cups of mixture.
How many times 3 is 9 ?
To maintain the same ratio, both rice and urad must be taken 3 fold.
So 6 cups of rice and 3 cups of urad.
Another problem: In a co-operative society there are 600 members are male and 400 are female. An executive committee of 30 members is to be formed with the same male to female ratio as in the society. How many male and female members are to be there in the committee?
In the society, the male to female ratio is 3 : 2.
3 men and 2 women make 5 in all.
Here we need a total of 30.
How many times 5 to 30?
So there should be 3 × 6 = 18 men and 2 × 6 = 12 women in the committee.
One more problem: a rectangular piece of land is to be marked on the school ground for a vegetable garden. Hari and Mary started making a rectangle with a 24 metre long rope. Vimala Teacher said it would be nice, if the sides are in the ratio 3 : 5. What should be the length and width of the rectangle?
The length of the rope is 24 metres and this is the perimeter of the rectangle.
Given that, Hari and Mary started making a rectangle with a 24 metre long rope and their teacher suggested to make it in 3 : 5 ratio.
Let the ratio be x. Then the sides will be 3x and 5x.
Perimeter will be the total length of the rope which is 24 metre long.
3x + 5x = 24
8x = 24
x = $$\frac{24}{8}$$
x = 3
Therefore, the ength and width of the rectangle will be 9 metre and 15 metre.
If we take the length and width as 3 meters and 5 metres, what would be the perimeter?
If the length and width is 3 meters and 5 metres respectively.
Then the perimeter will be 3+5=8 metres.
How much of 16 is 24?
$$\frac{24}{16}$$ = $$\frac{3}{2}$$ = 1$$\frac{1}{2}$$
So width should be 1$$\frac{1}{2}$$ times 3 metres; that is
3 × 1$$\frac{1}{2}$$ = 4$$\frac{1}{2}$$ metres.
And length should be 1$$\frac{1}{2}$$ times 5 metres, that is,
5 × 1$$\frac{1}{2}$$ = 7$$\frac{1}{2}$$ metres
Now try these problems:
• Suhra and Sita started a business together. Suhra invested 40000 rupees and Sita, 30000 rupees. They made a profit of 7000 rupees which they divided in the ratio of their investments. How much did each get?
Suhra invested 40000 rupees and Sita, 30000 rupees in a business together.
In terms of ratio, the ratio of Suhra to Sita will be 4 : 3.
They made a profit of 7000 rupees.
Let the ratio be x, then the investment will be 4x and 3x.
4x + 3x = 7000
7x = 7000
x = 1000
Therefore, Suhra’s profit will be 4×1000=4000 rupees
Sita’s profit will be 3×1000=3000 rupees.
• John and Ramesh took up a job on contract. John worked 7 days and Ramesh, 6 days. They got 6500 rupees as wages which they divided in the ratio of their investments. How much did each get?
Given that John worked 7 days and Ramesh, 6 days.
In terms of ratio, the ratio for the number of days worked by John to Ramesh will be 7 : 6
Let the ratio be x, then the investment will be 7x and 6x.
7x + 6x = 6500
13x = 6500
x = $$\frac{6500}{13}$$
x = 500
Therefore, John will get 7×500 = 3500 rupees
Ramesh will get 6×500 = 3000 rupees
• John and Ramesh took up a job on contract. John worked 7 days and Ramesh, 6 days. They got 6500 rupees as wages which they divided in the ratio of the numbers of days each worked. How much did each get?
Given that John worked 7 days and Ramesh, 6 days.
In terms of ratio, the ratio for the number of days worked by John to Ramesh will be 7 : 6
Let the ratio be x, then the investment will be 7x and 6x.
7x + 6x = 6500
13x = 6500
x = $$\frac{6500}{13}$$
x = 500
Therefore, John will get 7×500 = 3500 rupees
Ramesh will get 6×500 = 3000 rupees
• Angles of a linear pair are in the ratio 4 : 5. What is the measure of each angle?
Angles of a linear pair are in the ratio 4 : 5.
Let the ratio be x,then it will be 4x and 5x.
4x + 5x = 90
9x = 90
x =10
4x = 4 × 10 = 40
5x = 5 × 10 = 50
Thus, the angles measurement will be 40 degrees and 50 degrees.
• Draw a line AB of length 9 centimetres. A point P is to be marked on it, such that the lengths of AP and PB are in the ratio 1 : 2. How far from A should P be marked? Compute this and mark the point.
Length of AB is 9 centimetres.
Point P is marked in 1 : 2 ratio.
Let the ratio be x, then it will be 1x and 2x, the total is 3x.
$$\frac{1x}{3x}$$ × 9 = 3 centimetres
$$\frac{2x}{3x}$$ × 9 = 6 centimetres
P should be marked 3 cm away from A.
• Draw a line 15 centimetres long. A point is to be marked on it, dividing the length in the ratio 2 : 3. Compute the distances and mark the point.
Length of AB is 15 centimetres.
Point P is marked in 2 : 3 ratio.
Let the ratio be x, then it will be 2x and 3x, the total is 5x.
$$\frac{2x}{5x}$$ × 15 = 6 centimetres
$$\frac{3x}{5x}$$ × 15 = 9 centimetres
P should be marked 6 cm away from A.
• Sita and Soby divided some money m the ratio 3 : 2 and Sita got 480 rupees. What is the total amount they divided?
Sita and Soby divided some money ‘m’ in the ratio 3 : 2 and Sita got 480 rupees.
Let the ratio be m, then the ratio of money Sita got will be 3m.
3m = 480
m = $$\frac{480}{3}$$
m = 160
Money Soby will get 2×160 = 320 rupees.
• In a right triangle, the two smaller angles are in the ratio 1 : 4. Compute these angles.
In a right triangle, the two smaller angles are in the ratio 1 : 4.
Right triangle means 90 degrees.
Let the ratio be x, total will be 1x + 4x = 5x.
$$\frac{1x}{5x}$$×90 = 18 degrees
$$\frac{4x}{5x}$$×90 = 72 degrees
Then the angles will be 18 degrees and 72 degrees.
• Draw a rectangle of perimeter 30 centimetres and lengths of sides in the ratio 1 : 2. Draw two more rectangles of the same perimeter, with lengths of sides in the ratio 2 : 3 and 3 : 7. Compute the areas of all three rectangles.
A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 1 : 2
Let the ratio be x, the total will be 1x + 2x = 3x.
One of the side will be $$\frac{1x}{3x}$$×30 = 10 centimetres
The other side will be $$\frac{2x}{3x}$$×30 = 20 centimetres
Therefore, the rectangle with 10 centimetres and 20 centimetres is drawn as shown below.
Area of this rectangle will be 10×5=50 square centimetres.
A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 2 : 3
Let the ratio be x, the total will be 2x + 3x = 5x.
One of the side will be $$\frac{2x}{5x}$$×30 = 12 centimetres
The other side will be $$\frac{3x}{5x}$$×30 = 18 centimetres
Therefore, the rectangle with 12 centimetres and 18 centimetres is drawn as shown below.
Area of this rectangle will be 9×6=54 square centimetres.
A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 3 : 7
Let the ratio be x, the total will be 3x + 7x = 10x.
One of the side will be $$\frac{3x}{10x}$$×30 = 9 centimetres
The other side will be $$\frac{7x}{10x}$$×30 = 21 centimetres
Therefore, the rectangle with 9 centimetres and 21 centimetres is drawn as shown below.
Area of this rectangle will be 10.5×4.5=47.25 square centimetres.
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# Graphing Equations in Three Variables
The graph of any equation in three variables can be drawn on a three-dimensional coordinate system. The graph of a linear equation in three variables is a plane. To solve a system of three linear equations in three variables by graphing, we would have to draw the three planes and then identify the points that lie on all three of them. This method would be difficult even when the points have simple coordinates. So we will not attempt to solve these systems by graphing.
By considering how three planes might intersect, we can better understand the different types of solutions to a system of three equations in three variables. The large figure below shows some of the possibilities for the positioning of three planes in three-dimensional space. In most of the problems that we will solve the planes intersect at a single point as in figure (a). The solution set consists of one ordered triple. However, the system may include two equations corresponding to parallel planes that have no intersection. In this case the equations are said to be inconsistent. If the system has at least two inconsistent equations, then the solution set is the empty set [see figures (b) and (c)].
There are two ways in which the intersection of three planes can consist of infi- nitely many points. The intersection could be a line or a plane. To get a line, we can have either three different planes intersecting along a line, as in figure (d) or two equations for the same plane, with the third plane intersecting that plane. If all three equations are equations of the same plane, we get that plane for the intersection. We will not solve systems corresponding to all of the possible configurations described. The following examples illustrate two of these cases. |
# Area and Perimeter: The Mysterious Connection TEACHER EDITION
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1 Area and Perimeter: The Mysterious Connection TEACHER EDITION (TC-0) In these problems you will be working on understanding the relationship between area and perimeter. Pay special attention to any patterns that arise in your exploration. Part 1 The question we are trying to answer in this lesson is what connection if any exists between area and perimeter? I. Figure A and figure B below have different areas. Determine if the perimeters are the same or different. Figure A Figure B Area of Figure A square Area of Figure B square Perimeter of Figure A Perimeter of Figure B Explain how you arrived at your conclusion. What was the process that you used to find the perimeters? Show an example of your process with labels included. (TC-1) The Mysterious Connection Teacher Materials Page 1 of 21
2 II. Is there a square unit you can remove from figure A, changing the area, but not changing its perimeter? If so, which one? Draw the resulting figure below. Figure A Use pictures and words to explain how you know the perimeters are the same. (TC-2) III. Is this the only square unit you can remove that would give you the same perimeter? Discuss your answer with your partner and record it below. (TC-3) The Mysterious Connection Teacher Materials Page 2 of 21
3 IV. Can you keep reducing the area of figure A by removing square, but continue to leave the perimeter unchanged? If so, how many total square can you remove and continue to have the same perimeter? Show your thinking below with words and pictures. (TC-4) V. What surprises you about the relationship between area and perimeter in this exploration? Discuss with your partner and summarize your thoughts below. (TC-5) (TC-6) VI. We want to know if this is true for other rectangles or just for Figure A. Choose two more rectangles with your partner and record their dimensions below. Rectangle 1: Rectangle 2 The Mysterious Connection Teacher Materials Page 3 of 21
4 VII. Use square tiles (or centimeter grid paper) to explore your rectangles. Each of you will explore one of the rectangles. Use the same process you used for figure A. Remove one tile at a time until you can t remove anymore tiles without changing the perimeter. Can you keep the perimeters the same as you change the area of each original rectangle by removing tiles? How many square or tiles can be removed? Does there seem to be any pattern in determining how many tiles can be removed? Explain what you observe. The Mysterious Connection Teacher Materials Page 4 of 21
5 VIII. Share what you discovered with your partner. What conjectures can you and your partner make? How can you explain them to someone else? (TC-7) IX. Make a small poster or use a small white board to show what you ve figured out so far. Use words and diagrams to communicate your thinking about the relationship between area and perimeter. Be prepared to share your poster with the class. The Mysterious Connection Teacher Materials Page 5 of 21
6 Part 2 I. Below are two rectangles that have the area of 24 square. Can you draw any other rectangles that have the same area? If so draw as many as you can on a sheet of grid paper. Compare your rectangles with your partner. How did you know that you have found them all? Explain why you think you ve found all the rectangles below. (TC-8) The Mysterious Connection Teacher Materials Page 6 of 21
7 II. Determine the perimeters of each of your rectangles and record your results in the table below. Area LXW Square Length L Width W Perimeter 2L+2W 24 Which rectangle has the largest perimeter? (TC-9) III. Which has the smallest perimeter? Draw all of the rectangles that have the area of 36 square on another sheet of grid paper. Complete the table below for this set of rectangles. Area LXW Square Length L Width W Perimeter 2L+2W 36 Which rectangle has the largest perimeter? Which has the smallest perimeter? (TC-10) The Mysterious Connection Teacher Materials Page 7 of 21
8 IV. Repeat exercise III with a set of rectangles having the area 16. Area LXW Square 16 Length L Width W Perimeter 2L+2W Which rectangle has the largest perimeter? V. Which has the smallest perimeter? (TC-11) What generalizations about area and perimeter can you make looking at sets of rectangles with the same area? What patterns to you see? (TC-12) Discuss this with your partner. Make a complete list below. The Mysterious Connection Teacher Materials Page 8 of 21
9 VI. On a separate sheet of paper, apply the relationships that you discovered in this exploration on the following problems: a. Describe how you would construct a rectangle with the largest possible perimeter given an area of 9 square. b. Mrs. Hill asked you to construct a pen for the class rat. You can use 100 square inches of space on the table in the back of the room, but she wants you to use as little material as possible to make the sides of the pen. How much material will you need? How do you know that this is the least amount of material needed? Explain your answer using ideas about area and perimeter that you have learned. (TC-13) The Mysterious Connection Teacher Materials Page 9 of 21
10 Part 3 I. Consider a set of rectangles that has a perimeter of 12. Draw this set of rectangles on a sheet of grid paper. Find the area of each rectangle and complete a chart below. Perimeter 2L+2W 12 Length L Width W Area LXW Square Which rectangle has the largest area? Which has the smallest area? (TC-14) II. Repeat number I for a family of rectangles that has a perimeter of 18 and then 24. Perimeter Length Width Area Perimeter Length Width Area 2L+2W L W LXW 2L+2W L W LXW Square Square Which rectangle has the largest area? (TC-15) Which has the smallest area? The Mysterious Connection Teacher Materials Page 10 of 21
11 III. Discuss with you partner how you know you drew all the possible rectangles for the sets of rectangles you have drawn. What process did you use? (TC-16) IV. What observations do you make about these sets of rectangles that have the same perimeter? What patterns do you see? Discuss your ideas with your partner. Make a complete list below. (TC-17) The Mysterious Connection Teacher Materials Page 11 of 21
12 V. On a separate sheet of paper, apply the relationships that you discovered in this exploration on the following problems: (TC-18) a. Describe how you would construct a rectangle with the largest possible area given a perimeter of 20. b. You are making a card with a ribbon boarder. You have 14 inches of ribbon. You have a lot to write on your card. What size card should you cut out of card stock paper? How much area will you have to write on? The Mysterious Connection Teacher Materials Page 12 of 21
13 Conclusion: (TC-19) Now you should be able to confidently answer the following questions. Make sure you use clear mathematical thinking and diagrams to explain your answers. 1. True or False Rectangles with the same area must have the same perimeters. Explain and give an example. 2. True or False Rectangles with the same perimeters can have different areas. Explain and give an example. Fill in the blank. 3. For a fixed perimeter the rectangle with the largest area is always. 4. For a fixed perimeter the rectangle with the smallest area is always. 5. For a fixed area the rectangle with the largest perimeter is always. 6. For a fixed area the rectangle with the smallest perimeter is always. The Mysterious Connection Teacher Materials Page 13 of 21
14 TC-0 There are many misconceptions for students and adults in the complex relationship between area and perimeter. It is challenging to keep track of which aspect of size is being measured and what relationship, if any, exists between these aspects (area and perimeter). Students need a variety of experiences with this relationship to develop a strong grasp of the concepts. Students need to have experiences in which they are manipulating the spaces that they are measuring to construct deep understanding. Because of this, it is important to use a variety of manipulatives. In this lesson, grid paper is essential, but I highly recommend students use square tiles as well to build the different figures. The Big Ideas: It is possible to change the area of a figure without changing its perimeter. It is possible for several rectangles to exist with the same area, but different perimeters. It is possible for several rectangles to exist with the same perimeter, but different areas. As the differences between the dimensions of a rectangle get smaller for a fixed perimeter, the area of the rectangle increases. Maximizing as a square. As the differences between the dimensions of a rectangle get smaller for a fixed area, the perimeter of the rectangle decreases. Minimizing when it is a square. Before students grapple deeply with the relationship between perimeter and area it is important they have had experiences isolating the aspect of a figure that they are trying to measure, since any one figure has more than one aspect to be measured. In particular, the perimeter of a shape and the area of that same shape. A good prerequisite activity to help students focus on these attributes would be a sorting activity such as Which way do they go?. As students sort rectangles from smallest to largest they are deciding which aspect they are measuring and how they will measure it. TC-1 This is the point where they are focusing on measuring the aspect of perimeter. Watch to see that they are attending to perimeter and not area. Possible response: I found the perimeters by counting the edges on the outside or boundary of the figure. The Mysterious Connection Teacher Materials Page 14 of 21
15 TC-2 What you are looking for is that they identify that they are able to remove a corner square unit and keep the perimeter the same. Their explanation may say something like, I took two edges away and exposed two edges, so the perimeter is the same. They may have difficulty seeing that when they remove a square that is not a corner that they are removing one edge and exposing 3. Ask them to point to the edges they are referring to as they explain their thinking. TC-3 Any corner square could be removed, but not any other one of the tiles. Again two extra edges would be exposed. TC-4 Here they will find that they can remove two more square tiles, but the order in which they remove the tiles matters. What you want to hear them discussing is that they have to pay close attention to how many edges they are taking away and exposing. These must be equal. TC-5 You might see: Figures that have the same perimeter can have different areas. You have to pay attention to how many edges you are removing and how many are exposed. You can t continue reducing the area and keeping the same perimeter. There is a limit. TC-6 Here in sections VI and VII they are collecting more evidence that their patterns are true. Make sure they don t pick rectangles that have too small of dimensions. This is also a place to push a student to change their pattern of thinking. Is there more than one way to remove the tiles? Shapes with the same perimeter do not have to have the same area. All tiles can be removed that are not necessary to preserve the dimensions of the original rectangle. The Mysterious Connection Teacher Materials Page 15 of 21
16 TC-7 Here in VIII and IX they will be sharing their conjectures and constructing a poster to explain what they have figured out to the rest of the classroom. Have them use words, as well as, diagrams to explain their ideas. Make sure they are using diagrams or have constructed figures with tiles that they can share with their classmates. Push to have them show the process that they used not just explaining with words. Physical action in developing measurement ideas is very important. Have they addressed why they think this works on their poster? Figures that have the same perimeter may not have the same area. This is the main idea that they need to walk away with from this exploration. Stop here and have a classroom discussion around the posters students came up with in their pairs. Start by asking for volunteers to share. Then ask if another pair has something to add to what s been shared already. If you observed a pair that had an interesting idea or way of thinking, ask them if they would be willing to share their interesting work. Make sure to ask students if they have questions or constructive comments on their peers work. Part II TC-8 They should come up with rectangles with dimensions 1 X 24 and 2 X 12. Students might want to say that 1 X 24 is a different rectangle that 24 X 1. This is not a central question to this activity. You can ask them what their thinking is about these rectangles. For a more advanced student you might want to introduce the idea of congruent figures, but otherwise let it go with what they, as partners, agree on. Some students may want to go to half or even smaller. This is a great observation, don t discourage them. Instead ask, So, how many rectangles could there be? What might be the best way to limit our exploration since we can t draw ALL rectangles? TC-9 The rectangle with dimensions 1 X 24 will have the largest perimeter. The rectangle with dimensions 4 X 6 will have the smallest perimeter. The smaller the difference between dimensions, the smaller the perimeter. Some students may extrapolate that a square will have the smallest perimeter. If they see this, then push them to find out what those dimensions would be. TC-10 They should come up with rectangles with dimensions 1 X3 6, 2x 18, 3 X 12, 4 X 9, and 6 X 6. The rectangle with dimensions 1 X 36 will have the largest perimeter. The rectangle with dimensions 6 X 6 will have the smallest perimeter. TC-11 Students should draw rectangles with dimensions 1 X 16, 2 X 8, and 4 X 4. The rectangle with dimensions 1 X 16 will have the largest perimeter. The Mysterious Connection Teacher Materials Page 16 of 21
17 The rectangle with dimensions 4 X 4 will have the smallest perimeter. TC-12 This is a good point to stop and have a brief discussion of the ideas that students discovered. Have a pair offer one idea that they wrote down and then move to the next pair. It is possible to have many (infinite) rectangles with the same area, but different perimeters. Rectangles with the same area have dimensions that are factors of the fixed area. When the difference between the dimensions of a rectangle with a fixed area is the smallest you will have the smallest perimeter. When the difference between the dimensions of a rectangle with a fixed area is the largest you will have the largest perimeter. TC-13 I would construct a rectangle where there is the largest possible difference between the dimensions. In this case the dimensions would be 1 X 9. Mrs. Hill needs to have a pen that has the smallest perimeter possible. The smallest perimeter will allow the least amount of material to be used. In this case the dimensions of the pen will make a 10 X 10 square. I know this is smallest perimeter, because the difference between the dimensions is the smallest it could possibly be, zero. Part III TC-14 Students should draw the rectangles 1 X 5, 2 X 4, and 3 X 3. The areas will be 5, 8, and 9 respectively. The rectangle with dimensions 1 X 5 will have the smallest area. The rectangle with dimension 3 X 3 will have the largest area. TC-15 For the set of rectangles that has a perimeter of 18, students will draw rectangles with dimensions 1 X 8, 2 X 7, 3 X 6, and 4 X 5. For the set of rectangles that has a perimeter of 24, students will draw rectangles with dimensions 1 X 11, 2 X 10, 3 X 9, 4 X 8, 5X 7, and 6X 6. The rectangle with dimensions 1 X 8 will have the smallest area. The rectangle with dimension 4 X 5 will have the largest area. The rectangle with dimensions 1 X 1 will have the smallest area. The rectangle with dimension 6 X 6 will have the largest area. TC-16 A student may say, I started with a skinny rectangle with a width of 1. I then doubled this and subtracted it from the perimeter. Then I took the remaining and split them between the remaining two sides. Some students may observe that there could be many more rectangles if you used fractional sides, but if the sides remain integers then they may recognize The Mysterious Connection Teacher Materials Page 17 of 21
18 that they increased the width by one each time, and then at some point, the dimensions started to repeat themselves. TC-17 This is a good point to stop and have a brief discussion of the ideas that students discovered. Have a pair offer one idea that they wrote down and then move to the next pair. If the perimeter is the same in a set of rectangles, then the area of those rectangles does not have to be the same. Rectangles with the same perimeter have dimensions, where as the length increases incrementally, the width decreases incrementally until they are as close to the dimensions of a square as they can be. Given a fixed perimeter, the rectangle with the largest area will be the one with the dimensions that are closest together. (A square.) Given a fixed perimeter, the rectangle with the smallest area will be the one with the dimensions farthest apart. TC-18 Given a perimeter of 20, I would construct a rectangle where the dimensions are as close together as possible. In this case it would be a 5 X 5 rectangle. If you have a lot to write, then you want as much area as possible. Since the perimeter is 14 inches, I would construct a rectangle that has its dimensions close together. In this case it would be a 3 x 4 rectangle. TC-19 False. Rectangles with the same area can have many different perimeters. For example, a 3 x 4 and a 2 x 6 rectangle have the area of 12 square, but their perimeters are 14 and 16 respectively. True. Rectangles with the same perimeter can have many different areas. For example, a 3 x 4 and a 2 x 5 rectangle have the perimeter of 14, but their areas are 12 square and 10 square respectively. For a fixed perimeter the rectangle with the largest area is always the rectangle where the difference between the dimensions is the smallest. For a fixed perimeter the rectangle with the smallest area is always the rectangle where the difference between the dimensions is the largest. For a fixed area the rectangle with the largest perimeter is always the rectangle where the difference between the dimensions is the largest. For a fixed area the rectangle with the smallest perimeter is always the rectangle where the difference between the dimensions is the smallest. The Mysterious Connection Teacher Materials Page 18 of 21
19 Perimeter and Area: The Mysterious Connection Teacher Materials Perimeter and Area Pre-Check KEY Answer the following questions in the space provided. Use words and diagrams to explain your thinking. 1. What does area mean? The area of an object refers to the amount of space that is covered. For example, it measures how many squares fit on a space without gaps or overlaps. Area is measured in square. 2. How is area different from perimeter? Perimeter measures around the boundary of an object. For example, it measures how many edges of a square fit around an object. Perimeter is measured in linear. 3. Which rectangle has the bigger area? In this case both rectangles are the same area. They have different perimeters; so many students are surprised that the areas are the same. (A=18, but P=18 and P=22 i l ) 4. What is the area of this rectangle? 7 A=21. Students may or may not include 3 3 the label, square. Because all the dimensions are labeled, students must be sure which attribute is being measured. If they are not sure they may find the perimeter instead of multiplying length by 7 width The Mysterious Connection Teacher Materials Page 19 of 21
20 Smallest to Largest Pre-Exploration Teacher Notes Perimeter and Area: The Mysterious Connection Teacher Materials Students will often choose to use a linear dimension (length or width) to order rectangles by perimeter. Ordering rectangles by area students will often just use their best guess. They may determine which rectangle seems to cover more. In these first two sections students are focusing on each attribute and using their instincts to come up with orders. Do not worry about their orders at this point, but pay attention to what they are attending to and how they are coming to agreement. These discussions between peers can be very enlightening as to what students are thinking. When they begin to measure they are measuring with something nonstandard. They can use the rectangles to compare or they may choose another object. You will want to provide a variety of objects for each group to select from. For example, string, square tiles, popsicle sticks, straw, grid paper, etc Once they have their orders, you want to listen to them discussing how they varied from their original guesses and why they think they are the same or different. What did they have to be careful of as they measured? As they discuss what they learned you may want to use this as a class discussion. You may hear things such as: When measuring perimeter you want to pay attention to the edges. Make sure you include all four edges in the perimeter. Perimeter is measured by using edges of objects. When measuring area you want to pay attention to how much space is being covered. Area is measured by using solid objects rather than edges. The Mysterious Connection Teacher Materials Page 20 of 21
21 These need to be enlarged by two for easier handling. B E C D F G A The Mysterious Connection Teacher Materials Page 21 of 21
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# SAT Math: Shaded Region Problems
One of the most iconic of SAT Math problem types is the “shaded region” problem. In this type of problem you are asked to find the area of a shaded region that is part of some larger diagram. The key to successfully solving these problems is not to focus on the area of some oddly shaped region, but rather to focus on finding the areas of the shapes you know. In general, you want to utilize the following relationship to solve these problems:
$Area_{shaded}=Area_{whole}-Area_{unshaded}$
Let’s take a look at how this works by solving the following problem.
In rectangle ABCD the length of side AB is 12. The two circles are congruent and tangent. Additionally, each circle is tangent to the rectangle at three points. What is the total area of the shaded regions?
(A) $\: 36-18\pi$
(B) $\: 72-9\pi$
(C) $\: 72-18\pi$
(D) $\: 72-36\pi$
(E) $\: 144-36\pi$
Let’s start by trying to figure out the area of the rectangle. We know that the length of the rectangle is equal to twice the diameter of the circle. The width of the rectangle is equivalent to the diameter of the circle, or half the length of the rectangle. So now we know that the width is 6. That makes the area of the rectangle 6 x 12 = 72.
Now let’s find the area of one of the circles. We know that the diameter of the circle is 6, so the radius is 3. The area of one of the circles is:
$Area_{circle}=\pi r^{2}=\pi (3)^{2}=9\pi$
Now we have enough to find the area of the shaded region:
$Area_{shaded}=Area_{whole}-Area_{unshaded}$
$Area_{shaded}=Area_{rectangle}-2\cdot Area_{circle}$
$Area_{shaded}=72-2\cdot 9\pi$
$Area_{shaded}=72-18\pi$
The correct answer is choice (C).
For more practice on Shaded Region problems, go here.
If you have questions about the Shaded Region problems or anything else to do with the SAT, leave a comment or send me an email at info@cardinalec.com . |
Food for thought
## Solution
A ball is thrown from a position $O$ on level ground with initial velocity (in $\mathrm{m\,s^{-1}}$) $\mathbf{u}=5\mathbf{i}+6\mathbf{j}$ where $\mathbf{i}$ is a horizontal direction and $\mathbf{j}$ is vertically upwards. …
One way to model the effect of the wind is as a constant horizontal acceleration. Let’s call the magnitude of this acceleration $\quantity{w}{m\,s^{-2}}$.
Find how far from $O$ the ball will land as a function of $w$. Express the trajectory of the ball using parametric equations for $x$ and $y$ in terms of time, $t$.
Take gravity to be $\quantity{10}{m\,s^{-2}}$.
The horizontal and vertical motions are independent so we can treat them separately. Vertically, the wind has no effect so the ball has constant acceleration under gravity. $$$y = 6t-\frac{1}{2}10t^2 = 6t-5t^2 \label{eq:vert}$$$ Horizontally, the ball has initial velocity $5$ and constant acceleration $-w$. So we have $$$x = 5t-\frac{1}{2}w t^2 = 5t-\frac{w}{2}t^2. \label{eq:horiz}$$$ To find where the ball lands, we set $y=0$ which means \begin{align*} 6t-5t^2 &= 0 \\ \implies\quad t&= 0 \text{ or } \frac{6}{5}. \end{align*}
Note that this is exactly the same as we had in the Warm-up. Why is that?
Substituting $t=\frac{6}{5}$ into equation $\eqref{eq:horiz}$ gives us the landing distance as a function of $w$: $$$x_l=6-\frac{18}{25}w. \label{eq:x-land}$$$
The equations $\eqref{eq:vert}$ and $\eqref{eq:horiz}$ are a parametric form of the trajectory.
In each of these cases, find where the ball lands and draw a sketch of the trajectory:
• $w=0$
This is the case where the wind has no effect so the behaviour is exactly as in the Warm-up. The trajectory is a parabola with maximum at $\left(3,\frac{9}{5}\right)$. The ball lands at $x_l=6$.
In each of these cases, find where the ball lands and draw a sketch of the trajectory:
• $w=2.5$
Substituting $w=\frac{5}{2}$ into $\eqref{eq:x-land}$ we find that the ball lands at $x_l=6-\frac{18}{25}\times\frac{5}{2} = 4.2,$ so the wind has reduced the range from $\quantity{6}{m}$ to $\quantity{4.2}{m}$.
The trajectory is given by equations $\eqref{eq:vert}$ and $\eqref{eq:horiz}$ as \begin{align*} x &= 5t-\frac{5}{4}t^2 \\ y &= 6t-5t^2. \end{align*}
To help sketch the curve it is useful to locate the highest point on the trajectory. The height is a quadratic function of $t$ and the ball lands when $t=\frac{6}{5}$ so it is at it’s maximum height when $t=\frac{3}{5}$. Substituting in the above gives us the highest point at $x=\frac{51}{20} = 2.55, \quad y=\frac{9}{5}=1.8.$
Notice that the maximum occurs at the same height and time as in the Warm-up, but that is has moved horizontally back towards the thrower. Also, the ball travels less far horizontally in the second half of its trajectory than it did in the first half.
In each of these cases, find where the ball lands and draw a sketch of the trajectory:
• $w=5$
The landing point is at $x_l=6-\frac{18}{5} = 2.4.$
The trajectory is \begin{align*} x &= 5t-\frac{5}{2}t^2 \\ y &= 6t-5t^2. \end{align*}
Both $x$ and $y$ are quadratic functions of $t$ and they each have a maximum value. The maximum $y$ always occurs when $t=\frac{3}{5}$. We can find when the maximum $x$ occurs by completing the square or by comparing the roots – in this case it is when $t=1$. So the maximum $y$ is at $(2.1,1.8)$ and maximum $x$ is at $(2.5,1)$.
The ball starts to move back towards the thrower before it lands.
In each of these cases, find where the ball lands and draw a sketch of the trajectory:
• $w=10$
The landing point is at $x_l=6-\frac{18}{25}\times 10 = -1.2.$
The trajectory is \begin{align*} x &= 5t-5t^2 \\ y &= 6t-5t^2 \end{align*}
which has maximum $y$ at $(1.2,1.8)$ and maximum $x$ at $(1.25,1.75)$.
The ball has been blown back past the thrower.
What stays the same and what changes as we vary the value of $w$?
What value of $w$ would make the ball return to $O$?
This applet shows how the trajectory varies.
• The shape of the trajectory is always a parabola, but it’s width and orientation change.
• The ball always reaches the same maximum height.
• The landing point gets closer to $O$ as $w$ increases until the point where it goes behind the thrower.
The ball returns to $O$ if $x_l=0$ which means $6-\frac{18}{25}w = 0 \quad\implies\quad w=\frac{25}{3}.$
What shape is the trajectory in this case?
What are the limitations of this model?
The model of constant $w$ is quite a bold simplification of reality. The acceleration is actually some function of the ball’s velocity relative to the surrounding air. We could say $w=f\left(v_\text{wind} - v_\text{ball}\right).$ In the general case these would all be vector quantities, but here we are only considering the horizontal components. If the wind speed is constant and significantly bigger than the speed of the ball then the input to the function $f$ does not change much as the ball speeds up or slows down. Hence we can say that $w$ will not change very much, which is how we could justify a constant $w$ model.
In practice, the effect of the wind will almost certainly not be a constant acceleration. A more accurate model of this, however, would lead to some equations that are hard to solve.
In addition to the horizontal component of acceleration caused by the wind, there would be a vertical component which we might think of as air resistance. Again, this is hard to model accurately.
Other problems with this model include
• the wind doesn’t usually blow at a constant speed,
• the wind speed often varies with height above the ground,
• any spin on the ball could alter its trajectory. |
How to Calculate Power Based on Force and Speed - dummies
# How to Calculate Power Based on Force and Speed
In physics, you can calculate power based on force and speed. Because work equals force times distance, you can write the equation for power the following way, assuming that the force acts along the direction of travel:
where s is the distance traveled. However, the object’s speed, v, is just s divided by t, so the equation breaks down to
That’s an interesting result — power equals force times speed? Yep, that’s what it says. However, because you often have to account for acceleration when you apply a force, you usually write the equation in terms of average power and average speed:
Here’s an example. Suppose your brother got himself a snappy new car. You think it’s kind of small, but he claims it has over 100 horsepower. “Okay,” you say, getting out your clipboard. “Let’s put it to the test.”
Your brother’s car has a mass of
On the big Physics Test Track on the edge of town, you measure its acceleration as 4.60 meters/second2 over 5.00 seconds when the car started from rest. How much horsepower is that?
You know that
so all you need to calculate is the average speed and the net applied force. Take the net force first. You know that F = ma, so you can plug in the values to get
Okay, so the force applied to accelerate the car steadily is 5,060 newtons. Now all you need is the average speed. Say the starting speed was vi and the ending speed vf . You know that vi = 0 m/s, so what is vf? Well, you also know that because the acceleration was constant, the following equation is true:
vf = vi + at
As it happens, you know that acceleration and the time the car was
accelerated over:
vf = 0 m/s + (4.60 m/s2)(5.00 s) = 23.0 m/s
Because the acceleration was constant, the average speed is
Because vi = 0 m/s, this breaks down to
Plugging in the numbers gives you the average velocity:
Great — now you know the force applied and the average speed. You can use the equation
to find the average power. In particular
You still need to convert to horsepower. One horsepower = 745.7 watts, so
Therefore, the car developed an average of 78.0 horsepower, not 100 horsepower. “Rats,” says your brother. “I demand a recount.”
Okay, so you agree to calculate power another way. You know you can also calculate average power as work divided by time:
And the work done by the car is the difference in the beginning and ending kinetic energies:
W = KEfKEi
The car started at rest, so KEi = 0 J. That leaves only the final kinetic energy to calculate:
Plugging in the numbers gives you:
So because
and the work done was
you get the following:
And, as before |
# Free math answers with work
Here, we will show you how to work with Free math answers with work. Our website can solve math problems for you.
## The Best Free math answers with work
Free math answers with work is a mathematical instrument that assists to solve math equations. For one thing, it’s important to make sure the scale is accurate. It should be able to accurately register both pounds and kilograms, so that you can accurately track your progress over time. It should also be easy to use, because trying to use a home scale while wearing bulky clothing or shoes can make it difficult. Finally, it’s important to remember to take your scale with you when you travel so that you can continue tracking your progress while on vacation.
When solving linear equations by graphing, you can use a coordinate plane to graph the coordinates of each number. The coordinates will then represent points on the graph. For example, if there are two numbers and you know their coordinates, then you can draw a line between them to represent their relationship in the equation. This is called graphing a linear equation by intercepts. Another way to graph linear equations is by using an equation sheet. In this case, you need to enter all of the values in the problem before graphing it. Linear equations with more than one variable can also be graphed by using a table or matrix form. When graphing these types of equations, it’s important to include all of the variables as well as their corresponding values in the table or matrix format. This will ensure that your results match what was stated in your problem statement and that your solution gives you an accurate depiction of what’s happening in the equation.
The square root of a number is the number whose square is the original number. For instance, the square root of 4 is 2 because 4 × 4 = 16 and 2 × 2 = 4. The square root of a negative number is also negative. For instance, the square root of -3 is -1 because 3 × -3 = -9 and 1 × -1 = -1. The square root of 0 is undefined, but it can be calculated if you know the radius and diameter of a circle. The radius is half the diameter and equals pi (π) times radius squared plus half radius squared. The diameter, on the other hand, equals radius squared minus pi multiplied by diameter squared, or 3 times radius squared minus pi multiplied by diameter squared. In addition to solving equations with square roots, you will often encounter problems in which two numbers are given to you that must be combined using some kind of mathematical operation. One way you can solve these problems is to use your knowledge of algebra, geometry, and division along with your knowledge of how to find square roots. If a problem requires you to find two numbers that must be combined using multiplication or division (or a combination thereof), then one method for solving this problem would be to multiply or divide both numbers so that one becomes larger than the other as shown below: divide> multiply> division>
A math app is a mobile application designed to help students learn math. The most common types of math apps include: These types of apps are available for use on smartphones and tablets. There are many different kinds of math apps available. Some help students practice basic arithmetic such as addition, subtraction, and multiplication. Others teach more complex concepts such as fractions or geometry. One advantage of using a math app is that it can be used anywhere, even when the student isn't at school. This can be especially helpful for students who have trouble sitting still in class. On the other hand, there are some disadvantages to using a math app. First, it may be difficult for students to learn how to use the app correctly. Second, it may take more time and effort for parents or teachers to set up and use the app properly. Finally, there is always the possibility that the app will not work properly and will not provide accurate results.
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Equations Values Book Website that gives you all math answers How to solve an absolute value equation Enter equation and get answer Variable solver Order of operations solver |
# Solving Word Problems in Numbers using Algebra Part 3
This is the last part of the Solving Number Problems Series. In this post, we are going to solve number problems in disguise, or numbers problems with different contexts. The previous two parts in this series you may want to read are Solving Word Problems in Numbers using Algebra are Part 1 and Part 2.
Example 7
Jack is twice as old as Rose. Two years from now, the sum of their ages is 40. How old are they now?
Solution
This problem is an age problem but it is very similar to number problems. As stated, there are two points in time: now and 2 years from now.
Now, Jack is twice as old as Rose. That means that if Rose is 15, then Jack is 30. That means that if the age of Jack is $2x$, then Rose’s age is $x$. Therefore, we have the following representation.
Now
Rose: $x$
Jack’s Age: $2x$
Two years from now, both of them will be 2 years older. This means that we add 2 years to their ages. So, we represent their ages as follows.
Two Years from Now
Rose’ age: $x + 2$
Jack’s age: $2x+ 2$
The Sum of Their Ages (2 years from now)
$(x + 2) + (2x + 2) = 40$
$3x + 4 = 40$
$x = 12$
So, the age of Rose now is $12$ and Jack is $24$. Two years from now, Rose will be $14$ and Jack will be $26$. Checking, we have $14 + 26 = 40$. This means that our answer is correct.
Example 8
The sum of the three digit number is 15. The one’s digit is thrice the hundred’s digit. The ten’s digit is one more than the one’s digit. What is the number?
Solution
Consider the following descriptions.
one’s digit: thrice the hundred’s digit
ten’s digit: one more than the one’s digit
hundred’s digit: ?
In the previous post, we have discussed that the given which has no description is usually the unknown. Therefore, if we represent algebraically, we have
one’s digit: $3x$
ten’s digit: $3x + 1$
hundred’s digit: $x$
Since the sum of the 3-digit number is 15, we can add the three digits and equate to 15. That is,
$3x + (3x + 1) + x = 7x + 1 = 15$.
Solving, we have $x = 2$ (hundred’s digit), $7$ (ten’s digit) and $6$ (one’s digit). So the number is $276$. Now, check each condition in the problem to verify if the answer is correct.
Example 9
Anna has twice as much money as Ria. When she gives \$70 to Ria, they will have the same amount of money. How much money does Maria have?
Solution
If we let $x$ be Ria’s money, then Anna ha $2x$.
Initial Money
Ria’s money: $x$
Anna’s money: $2x$
After Anna gave Ria 70 dollars.
Ria’s money: $x + 70$
Anna’s money: $2x - 70$
Now, after giving Ria the money, the two girls have the same amount of money. Therefore, we can equate both expressions. That is,
$2x - 70 = x + 70$.
$2x - x = 70 + 70$
$x = 140$.
So, Ria has \$140 and Anna has \$280.
This is the end of this series. The next series will be on Solving Word Problems involving age.
## 6 thoughts on “Solving Word Problems in Numbers using Algebra Part 3”
1. Fellow math collegues- let me know what you think:
I do word problems as extension only now for students so inclined. Although interesting to think of the three time frames and express ages there in (past,present , future) I don’t see what purpose they serve. In my attempt to make all lessons pertinent to real life, I have dropped this concept as a requirement.
2. Hi sir!
in example 8, i think you have a typo. and shouldn’t the ten’s digit be 3x + 1 since the one’s digit is 3x?
nice blog very informative. keep it up
3. @John
Personally, I don’t really like giving this kind of problem if I were to teach in high school, but I believe that a good problem solver should try any problem he encounters
@Tet
Thanks. I have changed it already. Please inform me if you say more errors.
4. sir you forgot to change the values after the equation “Solving, we have x = 3 (hundred’s digit), 4 (ten’s digit) and 9 (one’s digit)” 🙂 |
# Right triangle
Legs of right are in ratio a:b = 6:8. Hypotenuse has a length of 61 cm.
Calculate the perimeter and area of the triangle.
Result
p = 146.4 cm
A = 893 cm2
#### Solution:
Try calculation via our triangle calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Looking for help with calculating roots of a quadratic equation? Do you have a system of equations and looking for calculator system of linear equations? Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.
## Next similar examples:
1. RTriangle 17
The hypotenuse of a right triangle is 17 cm. If you decrease both two legs by 3 cm you will reduce the hypotenuse by 4 cm. Determine the length of this legs.
2. Right angled triangle
Hypotenuse of a right triangle is 17 cm long. When we decrease length of legs by 3 cm then decrease its hypotenuse by 4 cm. Determine the size of legs.
3. Right triangle Alef
The area of a right triangle is 294 cm2, the hypotenuse is 35 cm long. Determine the lengths of the legs.
4. Medians
Calculate the sides of a right triangle if the length of the medians to the legs are ta = 21 cm and tb=12 cm.
5. Isosceles triangle
The perimeter of an isosceles triangle is 112 cm. The length of the arm to the length of the base is at ratio 5:6. Find the triangle area.
6. Isosceles trapezoid
Calculate the content of an isosceles trapezoid whose bases are at ratio 5:3, the arm is 6cm long and it is 4cm high.
7. RT leg and perimeter
Calculate the length of the sides of a right triangle ABC with hypotenuse c when the length of a leg a= 84 and perimeter of the triangle o = 269.
8. Is right triangle
One angle of the triangle is 36° and the remaining two are in the ratio 3:5. Determine whether triangle is a rectangular triangle.
9. R triangle
Calculate the area of a right triangle whose longer leg is 6 dm shorter than the hypotenuse and 3 dm longer than the shorter leg.
10. RT - hypotenuse and altitude
Right triangle BTG has hypotenuse g=117 m and altitude to g is 54 m. How long are hypotenuse segments?
11. Cuboid
Cuboid with edge a=16 cm and body diagonal u=45 cm has volume V=11840 cm3. Calculate the length of the other edges.
12. Clouds
From two points A and B on the horizontal plane was observed forehead cloud above the two points under elevation angle 73°20' and 64°40'. Points A , B are separated by 2830 m. How high is the cloud?
13. Diagonal 20
Diagonal pathway for the rectangular town plaza whose length is 20 m longer than the width. if the pathway is 20 m shorter than twice the width. How long should the pathway be?
14. Nice prism
Calculate the surface of the cuboid if the sum of its edges is a + b + c = 19 cm and the body diagonal size u = 13 cm.
15. Rectangular triangle
The lengths of the rectangular triangle sides with a longer leg 12 cm form an arithmetic sequence. What is the area of the triangle?
16. Angle in RT
Determine the size of the smallest internal angle of a right triangle whose sides constitutes sizes consecutive members of arithmetic progressions.
17. Angle
Determine the size of the smallest internal angle of a right triangle which angles forming the successive members of the arithmetic sequence. |
# Difference between revisions of "2005 AMC 12B Problems/Problem 25"
## Problem
Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?
$\mathrm{(A)}\ \frac {5}{256} \qquad\mathrm{(B)}\ \frac {21}{1024} \qquad\mathrm{(C)}\ \frac {11}{512} \qquad\mathrm{(D)}\ \frac {23}{1024} \qquad\mathrm{(E)}\ \frac {3}{128}$
## Solution
### Solution 1
We approach this problem by counting the number of ways ants can do their desired migration, and then multiple this number by the probability that each case occurs.
Let the octahedron be $ABCDEF$, with points $B,C,D,E$ coplanar. Then the ant from $A$ and the ant from $F$ must move to plane $BCDE$. Suppose, without loss of generality, that the ant from $A$ moved to point $B$. Then, we must consider three cases.
• Case 1: Ant from point $F$ moved to point $C$
On the plane, points $B$ and $C$ are taken. The ant that moves to $D$ can come from either $E$ or $C$. The ant that moves to $E$ can come from either $B$ or $D$. Once these two ants are fixed, the other two ants must migrate to the "poles" of the octahedron, points $A$ and $F$. Thus, there are two degrees of freedom in deciding which ant moves to $D$, two degrees of freedom in deciding which ant moves to $E$, and two degrees of freedom in deciding which ant moves to $A$. Hence, there are $2 \times 2 \times 2=8$ ways the ants can move to different points.
• Case 2: Ant from point $F$ moved to point $D$
On the plane, points $B$ and $D$ are taken. The ant that moves to $C$ must be from $B$ or $D$, but the ant that moves to $E$ must also be from $B$ or $D$. The other two ants, originating from points $C$ and $E$, must move to the poles. Therefore, there are two degrees of freedom in deciding which ant moves to $C$ and two degrees of freedom in choosing which ant moves to $A$. Hence, there are $2 \times 2=4$ ways the ants can move to different points.
• Case 3: Ant from point $F$ moved to point $E$
By symmetry to Case 1, there are $8$ ways the ants can move to different points.
Given a point $B$, there is a total of $8+4+8=20$ ways the ants can move to different points. We oriented the square so that point $B$ was defined as the point to which the ant from point $A$ moved. Since the ant from point $A$ can actually move to four different points, there is a total of $4 \times 20=80$ ways the ants can move to different points.
Each ant acts independently, having four different points to choose from. Hence, each ant has probability $1/4$ of moving to the desired location. Since there are six ants, the probability of each case occuring is $\frac{1}{4^6} = \frac{1}{4096}$. Thus, the desired answer is $\frac{80}{4096}= \boxed{\frac{5}{256}} \Rightarrow \mathrm{(A)}$.
### Solution 2
Let $f(n)$ be the number of cycles of length $n$ the can be walked among the vertices of an octahedron. For example, $f(3)$ would represent the number of ways in which an ant could navigate $2$ vertices and then return back to the original spot. Since an ant cannot stay still, $f(1) = 0$. We also easily see that $f(2) = 1, f(3) = 2$.
Now consider any four vertices of the octahedron. All four vertices will be connected by edges except for one pair. Let’s think of this as a square with one diagonal (from top left to bottom right).
$[asy] size(30); defaultpen(0.6); pair A = (0,0), B=(5,0), C=(5,5), D=(0,5); draw(A--B--C--D--cycle); draw(B--D); [/asy]$
Suppose an ant moved across this diagonal; then the ant at the other end can only move across the diagonal (which creates 2-cycle, bad) or it can move to another vertex, but then the ant at that vertex must move to the spot of the original ant (which creates 3-cycle, bad). Thus none of the ants can navigate the diagonal and can either shift clockwise or counterclockwise, and so $f(4) = 2$.
For $f(6)$, consider an ant at the top of the octahedron. It has four choices. Afterwards, it can either travel directly to the bottom, and then it has $2$ ways back up, or it can travel along the sides and then go to the bottom, of which simple counting gives us $6$ ways back up. Hence, this totals $4 \times (2+6) = 32$.
Now, the number of possible ways is given by the sum of all possible cycles, $$a \cdot f(2) \cdot f(2) \cdot f(2) + b \cdot f(2) \cdot f(4) + c \cdot f(3) \cdot f(3) + d \cdot f(6)$$
where the coefficients represent the number of ways we can configure these cycles. To find $a$, fix any face, there are $4$ adjacent faces to select from to complete the cycle. From the four remaining faces there are only $2$ ways to create cycles, hence $a = 8$.
To find $b$, each cycle of $2$ faces is distinguished by their common edge, and there are $12$ edges, so $b = 12$.
To find $c$, each three-cycle is distinguished by the vertex, and there are $8$ edges. However, since the two three-cycles are indistinguishable, $c = 8/2 = 4$.
Clearly $d = 1$. Finally,
$$8(1)(1)(1) + 12(1)(2) + 4(2)(2) + (32) = 80$$
Each bug has $4$ possibilities to choose from, so the probability is $\frac{80}{4^6} = \frac{5}{256}$. |
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# Math Studies Calculus Application
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### Math Studies Calculus Application
1. 1. Calculus ApplicationsMath Studies 1
2. 2. f ( x) = x − 45 x + 600 x + 20 3 2a) Find the local extrema and identify them as either a local maximum or a local minimum.b) Find the coordinates of the absolute maximum and absolute minimum of the function in the interval [ 0,30]
3. 3. First,f ( x) = 3 x − 90 x + 600 2 differentiate. 0 = 3 x − 90 x + 600 2 Set f’(x) to 0 0 = x − 30 x + 200 2 Solve for x 0 = ( x − 10)( x − 20) Factor Hence, we have local extrema at x = 10 and x = 20
4. 4. To identify them as either maxima or minima, we can usethe derivative - f (9) = 3(9) 2 − 90(9) + 600 = 33 f (11) = 3(11) − 90(11) + 600 = −27 3
5. 5. To identify them as either maxima or minima, we can usethe derivative - f (19) = 3(19) 2 − 90(19) + 600 = −27 f (21) = 3(21) − 90(21) + 600 = 33 3
6. 6. b) Check the endpoints, x = 0 and x = 30 f (0) = (0) 3 − 45(0) 2 + 600(0) + 20 = 20 f (30) = (30) − 45(30) + 600(30) + 20 = 4520 3 2
7. 7. A rectangular pen is to be fenced in using two types of ncing. Two opposite sides will use heavy duty fencing at \$3/ft hile the remaining two sides will use standard fencing at \$1/ft.What are the dimensions of the rectangular plot of greatest area at can be fenced in at a total cost of \$3600? A = xy 2 ( 3x ) + 2 ( 1y ) = 3600 3x + y = 1800 1y y = 1800-3x A = x ( 1800 − 3x ) 3x A = 1800x − 3x 2 A = 1800 − 6x A " = −6 0 = 1800 − 6x Therefore max 300 = x 900 = y The dimensions of a rectangular plot of greatest area are 300 x 900
8. 8. 4. An open-top box with a square bottom and rectangular sidesis to have a volume of 256 cubic inches. Find the dimensionsthat require the minimum amount of material. xS = x + 4xy 2 256 → S = x + 4x 2 2 yV = x y = 256 2 x x 1024 y = 256/x² S = x2 + x 1024 2048 S = 2x − 2 S" = 2 + 3 > 0 x x 1024 therefore a min 0 = 2x − 2 x x=8 →y=4 8x8x4
9. 9. Solve this problem.
10. 10. Solve this problem. |
A "system" of linear equations is a set of equations of straight lines.
When working with a "system" of equations, you are working with two or more equations at the same time. For our investigations, we will be working with two linear equations at a time.
Kyle's little sister, Jenny, claims that she can beat him in a snowmobile race to the next turn in the trail. Kyle takes the challenge and gives her a one minute head start. Jenny travels at 1640 feet per minute and Kyle travels at 2640 feet per minute. How long will it take Kyle to catch up with Jenny?
Solution: Let d = distance in feet and t = time in minutes. (Remember distance = rate x time.) Create TWO equations: Kyle catches Jenny after 3 minutes and traveling 4920 feet.
Solving Systems:
As the example above shows, the main task of working with linear systems is to find where the lines intersect (where they cross one another).
There are three main methods used for solving systems of linear equations.
Substitution Method: The goal of this algebraic method is to replace one of the equations with an equivalent expression by solving for one variable in one of the equations.
Elimination Method: The goal of this algebraic method is to eliminate one of the variables using addition or subtraction. The remaining equation will easily yield either the x or the y coordinate of the intersection point.
Graphical Method: The goal of this graphical method is to solve the system by graphing the lines either on graph paper, or on the graphing calculator, and locating the point of intersection, in a manner similar to what was done with the example above.
When working with systems, remember that the equations of lines may appear in a variety of forms, such as: y = 2x + 1; -5x + y = 7; 4x + 2y = -5; y = 7
You may need to re-write the equations before beginning your solutions.
Solutions:
As we saw in the example at the top of this page, the solution to a system of linear equation will be the location of the intersection of the two equations.
But, will there always be a solution?
The answer is "No". There are actually three possible cases of what may occur:
One Solution The slopes of the lines are different. This is the most common situation. No Solution The lines are parallel with the same slopes, but different y-intercepts. There is no intersection. Infinite Solutions The lines are the SAME line (one on top of the other). The slopes are the same and the y-intercepts are the same. The lines intersect everywhere and ALL points on the line are solutions. |
# Points A (1; 2) B (2; 4) C (-1; 2) are given. Write the equation of the straight line AB
Points A (1; 2) B (2; 4) C (-1; 2) are given. Write the equation of the straight line AB, write the equation of the straight lines passing through the point C parallel to AB.
The equation of the straight line passing through the 2nd points has the following form: (x – x1) / (x2 – x1) = (y – y1) / (y2 – y1). Substituting the coordinates of points A and B, we get the equation:
(x – 1) / (2 – 1) = (y – 2) / (4 – 2).
Let’s transform the equation:
x – 1 = (y – 2) / 2;
y = 2x – 2 + 2 = 2x.
General view of the equation of the straight line y = kx + b, since the required straight line is parallel to the straight line y = 2x, k = 2. Substitute the coordinates of the point C into the equation and calculate b:
2 * (-1) + b = 2;
b = 4.
Answer: the equations of the sought lines: y = 2x and y = 2x + 4.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. |
## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts
Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 23 Fundamental Concepts
APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.
IMPORTANT POINTS
1. Fundamental Concepts : Geometry is the study of position,, shape, size and other properties of different figures. The geometrical terms such as : point, line, plane, etc., contain the basic ideas for the development of geometry.
(i) Point : A point is a mark of position. It has neither length nor width nor. thickness and occupies no space.
(ii) Line : A line has only length. It has neither width nor thickness.
(iii) Ray : It is a line (i.e. a straight line) that starts from a given fixed point and moves in the same direction.
(iv) Line Segment: A line segmeftt is a part of a straight line. A line segment is a part of a line and also of a ray.
(v) Surface : A surface has length and width, but no thickness.
(vi) Plane : It is a flat surface. A plane has length and width, but no thickness.
(vii) Parallel Lines : Two straight lines are said to be parallel to each other if they lie in the same plane and do not meet when produced on either side.
(viii)Intersecting Lines : If two lines lie in the same plane and are not parallel to each other, they are called intersecting lines.
(xi) Collinearity of Points : If three of more points lie on the same straight line, then the points are called collinear points.
(x) Concurrent Lines : If three or more straight lines pass through the same point, the lines are called concurrent lines.
### Fundamental Concepts Exercise 23A – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
State, true or false, if false, correct the statement.
(i) A dot has width but no length.
(ii) A ray has an infinite length only on one side of it.
(iii) A line segment PQ is written as $$\overleftrightarrow { PQ }$$ .
(iv) $$\overleftrightarrow { PQ }$$ represents a straight line.
(v) Three points are said to be collinear, if they lie in the same plane.
(vi) Three or more points all lying in the same line, are called collinear points.
Solution:
(i) False : Because a dot has no length, no breadth.
(ii) True.
(iii) False : A line segment PQ is written as PQ.
(iv) True.
( v) False: Three points are called collinear points if they are in the same straight line.
(vi) True.
Question 2.
Write how many lines can be drawn through :
(i) a given point ?
(ii) two given fixed points ?
(iii) three collinear points ?
(iv) three non-collinear points ?
Solution:
(i) Infinite (unlimited) line can be drawn through a given point.
(ii) only one line can be drawn through two given point.
(iii) only one line can be drawn through three collinear points.
(iv) None (no) line can be drawn through three non-collinear points.
Question 3.
The shaded region of the given figure shows a plane :
(a) Name :
(i) three collinear points.
(ii) three non-collinear points.
(iii) a pair of intersecting lines.
(b) State whether true or false :
(i) Line DE is contained in the given plane P.
(ii) Lines AB and DE intersect at point C.
(iii) Points D, B and C are collinear.
(iv) Points D, B and E are collinear.
Solution:
(a) (i) A, B and C are three collinear points.
(ii) A, D and C are non-collinear points.
(iii) AC and DE are intersecting lines.
(b) (i) True
(ii) True
(iii) False
(iv) False
Question 4.
Correct the statement, if it is wrong:
(i) A A ray can be extended infinitely on either side.
(ii) A ray has a definite length.
(iii) A line segment has a definite length.
(iv) A line has two end-points.
(v) A ray has only one end point.
Solution:
(i) A ray can be extended infinitely on one side of it only.
(ii) A ray has infinite length.
(iii) Yes, a line segment has a definite length.
(iv) A line-segment has two end-points.
(v) Yes, a ray has only one end-point.
Question 5.
State true-er false, if false give the correct statement :
(i) A line has a countable number of points in it.
(ii) Only one line can pass through a given point.
(iii) The intersection of two planes is a straight line
Solution:
(i) False, a line has length only.
(ii) False, any number of line can pass through a given point.
(iii) True.
Question 6.
State, whether the following pairs of lines or rays appear to be parallel or intersecting.
Solution:
(i) intersecting
(ii) Parallel
(iii) Parallel
(iv) Intersecting
Question 7.
Give two examples, from your surroundings, for each of the following:
(i) points
(ii) line segments
(iii) plane surfaces
(iv) curved surfaces.
Solution:
(i) Tips of your pencil (Ball Pen) and Tip of paper pin.
(ii) Lines of Exercise Note-Books and edge of school desk.
(iii) Floor of the room and top of the table.
(iv) Surface of foot-ball and front glass of the car.
Question 8.
Under what condition will two straight
lines, in the same plane, have :
(i) no point in common.
(ii) only one point in common.
(iii) an infinite number of points in common.
Solution:
(i) When lines are parallel to each others.
(ii) When they intersect each other
Here, common point is E.
(iii) When line coincide with each other.
Question 9.
Mark two points A and B on a page of your exercise book. Mark a third point P, such that :
(i) P lies between A and B ; and the three points A, P and B are collinear.
(ii) P does not lie between A and B yet the three points are collinear.
(iii) the three points do not lie in a line.
Solution:
Question 10.
Mark two points P and Q on a piece of paper. How many lines can you draw:
(i) passing through both the points P and Q?
(ii) passing through the point P ?
(iii) passing through the point Q ?
Solution:
(i) From above diagram it is clear only one line can be drawn.
(ii) Infinite lines can pass through the point up
(iii) Infinite lines can pass through the point
Question 11.
The adjoining diagram shows a line AB. Draw diagram to represent:
(i) ray AB i.e. $$\xrightarrow { AB }$$
(ii) $$\xrightarrow { BA }$$
(iii) line segment AB.
Solution:
Question 12.
The adjoining diagram shows a ray AB. Draw diagrams to show :
(i) ray BA i.e. $$\xrightarrow { BA }$$
(ii) lineAB
(iii) line segment BA.
Solution:
Question 13.
The adjoining diagram shows a line segment AB. Draw diagrams to represent :
(i) ray AB i.e. $$\xrightarrow { AB }$$
(ii) line AB i.e. $$\overleftrightarrow { AB }$$
(iii) ray BA.
Solution:
Question 14.
Use a ruler and And whether following points are collinear or not:
(i) D, A and C
(ii) A, B and C
(iii) A, B and E
(iv) B, C and E
Solution:
Question 15.
(i) all pairs f parallel lines
(ii) all the lines which intersect EF.
(iii) lines whose point of intersection is G.
Solution:
(i) The pairs of parallel lines are EF || GH, EF || IJ and GH || IJ
(ii) The lines which intersect EF are AB and CD
(iii) AB and GH are the lines whose point of intersection is G.
### Fundamental Concepts Exercise 23B – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
State, which of the following is a plane closed figure :
Solution:
(i) , (iii) ,(iv),(vi)and (viii) are plane close figures.
Question 2.
Fill in the blanks :
(i) ….. is a three sided plane closed figure.
(ii) A square is a plane closed figure which is not bounded by …..
(iii) A rectangle is a …. sided plane ……
(iv) A rectangle has opposite sides ….. and adjacent sides ….. to each other.
(v) The sides of a square are ……. to each other and each angle is
(vi) For a line segment, a line making angle of ……… with it, is called perpendicular to it.
(vii) For a line segment, a line ……. it and making angle of ……. with it, is called ……. bisector of the line segment.
(viii) How many perpendiculars can be drawn to a line segment of length 6 cm ?
(ix) How many perpendicular bisectors can be drawn to a line segment of length 6 cm ?
(x) A perpendicular to a line segment will be its perpendicular bisector if it passes through the ……. of the given line segment.
Solution:
(i) Triangle
(ii) any curved line.
(iii) four, closed figure.
(iv) parallel, perpendicular.
(v) Equal, 90°
(vi) 90°
(vii) bisecting, 90°, perpendicular.
(viii) Infinite.
(ix) Only one.
(x) mid-point
Question 3.
State, which of the lines/line- segments are perpendicular to the line PQ :
Solution:
CD and MN are perpendicular to the line PQ. forming angle to 90°
Question 4.
Which of the following figures show two mutually perpendicular lines :
Solution:
figures (i) and (iii) show two mutually perpendiculars lines forming angles of 90°.
Question 5.
For each figure given below name the line segment that is perpendicular bisector of the other :
Solution:
(i) AB of CD
(ii) AB to MN
(iii) PQ to RS and RS to PQ
(iv) None
(v) None.
Question 6.
Name three objects from your surroundings that contain perpendicular edges.
Solution:
(i) My Text-Book of Mathematics r
(ii) My Note-Book
(iii) My Scale or My school’s Black Board
Question 7.
Using the given figure, answer the following :
(i) Name the pairs of parallel lines.
(ii) Name the pairs of mutually perpendicular lines.
(iii) Is the line p parallel to the line l ?
(iv) Is the line q perpendicular to the line m ?
Solution:
(i) l || m and p || q
(ii) p and l, p and m, q and l, q and m.
(iii) No, p is perpendicular to line l.
(iv) Yes.
Question 8.
Place a scale (ruler) on a sheet of paper and hold it firmly with one hand. Now draw two line segments AB and CD along the longer edges of the scale. State whether segment AB is parallel to or perpendicular to segment CD.
Solution:
Question 9.
(i) How many pairs of its edges are parallel to each other ?
(ii) How many pairs of its edges are perpendicular to each other ?
Solution:
(i) 8
(ii) Three at each corner = 24
Question 10.
Give two examples from your surroundings for each of the following :
(i) intersecting lines
(ii) parallel lines
(iii) perpendicular lines.
Solution:
(i) Edges of my Text-Book and Note Book through any comer.
(ii) opp. edges of my Text-Book and Note Book.
(iii) Adjacent edges of my Text-Book and Black Board in my class-room.
Question 11.
State true or false; if false, give the correct statement :
(i) The maximum number of lines through three collinear points is three.
(ii) The maximum number of lines through three non-collinear points is three.
(iii) Two parallel lines always lie in the same plane.
(iv) Concurrent lines always meet at the same point.
(v) A surface can be plane or curved.
(vi) There are an infinite number of points in a line segment of length 10 cm.
(vii) There are an infinite number of points in a line.
(viii) A plane has an infinite number of lines.
(ix) A plane has an infinite number of points.
Solution:
(i) False: Because only one line passes through three collinear points.
(ii) False : No line passes through all the three non-collinear points.
(iii) True.
(iv) True.
(v) True.
(vi) True.
(vii) True.
(viii) True.
(ix) True.
### Fundamental Concepts Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
Mark a point O on a piece of paper. Using a sharp pencil and a ruler, draw a line through the point O. Then draw one more line through the point O. How many lines can you draw through O?
Write a statement regarding your observations.
Solution:
We take a point O on a piece of paper. Using a sharp pencil, we draw a line OA through O
another line OB is drawn as shown in the figure.
We can draw an infinite number of lines from O as we know that an infinite number of lines can be drawn through a point in aplane,
Question 2.
Mark two points A and B on a plain sheet “ of paper. Using a sharp pencil and a ruler, draw a line through points A and B. Try to draw one more line through A and B. What do you observe ?
Write a statement regarding your observations.
Solution:
Two points A and B are marked On the plain sheet of paper. Using pencil, draw a line through there two points.
We cannot draw any other line joining them as only one line can be drawn passing through two fixed (given) points in a plane.
Question 3.
Draw two straight lines on a sheet of paper so that the lines drawn :
(i) intersect each other.
(ii) do not intersect.
(iii) appear to intersect when extended.
Solution:
Two straight lines are drawn on the sheet of a paper
(i) Intersecting each other:
(ii) Not intersecting each other:
(iii) Intersect each other when produced them:
Question 4.
Draw a figure to show that :
(i) points A, B and C arc collinear.
(ii) lines AB, CD and EF are concurrent.
Solution:
(i) We know that three or more points are collinear if they lie on the same straight line. Therefore the required figure (line) is given below on which three points A, B and C lie
(ii) If three or more lines pass through a common point, these are called concurrent lines. Below is given the figure, in which three lines AB, CD and EF are passing (or intersecting each other) at O.
Question 5.
In your note-book, draw two rays with the same initial point O and going in opposite directions. Write the special name for the final figure obtained.
Solution:
Two rays OA and OB are drawn through O in opposite direction as given below.
There two rays form a straight line $$\overleftrightarrow { AB }$$
Question 6.
In each figure, given below, write the number of line segments used :
Solution:
(i) In the given figure, there as Ten (10) line segments which are AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA.
(ii) In the given figure, there are 12 line segments which are AB, BC, CD, DE, EF, FG, GH, HI, IJ, JK, KL and LA.
(iii) In the given figure, there are 13 line segments which are AB, BC, CD, DE, EF, FG, GH, HI, IJ, JK, KL, LM and MA.
(iv) In the given figure, there are 7 line segments which are AB, BC, CD, DE, DA, DB, CE.
Question 7.
In a circle, point P is its centre and PA = PB = radius of the circle. Where do points A and B lie ?
Solution:
A circle is given with centre P and PA = PB = radius of the circle.
∵ The radii of a circle are equal there they lie on the circumference of the circle In other words, path of a point which is equidistant from a fixed point is the circumference of a circle. The fixed point is called the centre and equidistance is the radius.
Question 8.
Draw a four-sided closed figure, in which:
(i) all the sides are equal and each angle is 90°.
(ii) opposite sides are equal and each angle is 90°.
Write the special name of each figure drawn.
Solution:
(i) In this four sided closed figure all the sides are equal and each angle is 90°. This figure is called a square. In ABCD, AB = BC = CD = DA and ∠A = ∠B =∠C = ∠D = 90°
(ii) In this four sided closed figure opposite sides are equal and each angle is 90°. This figure is called a rectangle In rectangle ABCD, AB = CD and BC = AD
∠A = ∠B = ∠C = ∠D = 90°. |
Flipping Edged in Triangulations
Home| Preliminaries| Edge Flipping Theorems| Applications| Applet| Glossary| Links| References
# NUMBER OF FLIPPABLE EDGES
Given a triangulation T of a collection Pn = {v1,...,vn} of n points on the plane, how many edges of T are flippable?
The following theorem gives us the result. You can intuitively see the result by trying different triangulations in the applet.
#### Theorem 1:
Any triangulation of a collection of Pn of n points on the plane in general position contains at least ceiling[(n-4)/2] flippable edges.
We will come back to the proof of Theorem 1 after investigating the problem further....
The above figure show a triangulation of a point set where only (n-2)/4 edges are flippable
## General Position Assumption
It is important to note that the general position assumption is necessary since otherwise it is easy to construct a triangulation where no edge can be flipped.
If we do not assume that no three points can be collinear then we must take polygons such as above in account. Above: n-3 points in one of the sides of a triangle of a polygon with n points. We can see that if we have such a case than there exists a subset of edges which are not flippable because they are collinear. In our example there are no flippable edges!
## Further notation
Given a triangulation T we can divide the set of edges of T into two subsets:
• F(T) which contains all flippable edges of T;
• NF(T) which contains all not flippable edges of T
Clearly all the edges of T contained in the boundary of the convex hull of Pn are not flippable.
We create an orientation of each edge e in NF(T) as follows:
• RULE 1: If e is an edge of CH(Pn), orient it in the clockwise direction around CH(Pn) of Pn
Rule 1: The following shows the orientation when the nonflippable edges are oriented around the convex hull.
• RULE 2: If e=uv is not in CH(Pn), consider the quadrilateral C formed by the union of the two triangles of T containing e. Since C is not convex, it follows that one of the end vertices of e, say v, is a reflex vertex of C while u is a convex vertex of C. Orient e from u to v. We observe that the angle of C at v is greater than 180 degrees.
Rule 2: The following shows the orientation when the nonflippable edges are not convex hull edges.
Consider any vertex vi of T. Let d-(vi) be the number of edges of vi in NF(T) oriented from vj to vi. Note that d(vi) is the total number of edges of T incident with vi, whereas d-(vi) involves only edges of T in NF(T).
We use the following lemma to prove our main theorem...
#### Lemma 2:
Let vi be any vertex of T. Then d-(vi) ≤ 3. Moreover, if d(vi) > 3 in T, then d-(vi) ≤ 2. If d-(vi) = 2, the edges oriented toward vi are consecutive edges around vi
#### Proof of Lemma 2:
If vi is in CH(Pn) then d-(vi) = 1. Next, suppose then that vi is in the interior of CH(Pn); then two cases arise:
Case 1: d(vi) = 3 in T. In this case, all the edges of T incident with vi are nonflippable and are oriented toward vi. It follows that d-(vi) = 3
Case 2: d(vi) > 3 in T. It is obvious that no more than two edges of G can be oriented toward dvi and that if d-(vi) = 2, then the two edges oriented toward vi must be consecutive edges incident to vi
In the Below figured the oriented, nonflippable edges are the dashed lines whereas the flippable edges are solid.
On the left the leftmost vertex has total degree 4 and 2 nonflippable edges. Thus, the edges are consecutive. On the right vertex does not have valid degree. The total degree is 5 but the number of nonflippable edges is 3. From lemma 2, this is not valid.
## Main Theorem
Using Lemma 2 and the above information we can now go on to prove Theorem 1.
#### Proof of Theorem 1:
The proof comes in the following steps:
Step 1: Start with initial set of points and find a triangulation of the points.
Let Pn be a set of points in general position on the plane (see below) let T be a triangulation of Pn (see below) In our triangulation we let S be the set of elements of Pn with degree 3 in T that are not in CH(Pn). Step 3: Add an auxiliary point and join it to each of the points on the convex hull of our triangulation. We add a point w exterior to CH(Pn) and join it with a set of disjoint edges to all the vertices in CH(Pn). By this we obtain a triangulation of the plane with n+1 points which by Euler's theorem contains 3n-3 edges. Above we see that the point w added is the bottom most point.
Step 5: Orient the edges according to Rule 1 and 2.
Each of the edges incident to w are contained in NF(T) and we orient them from w to the vertex contained on CH(Pn). Next we orient all edges of T in NF(T) according to Rule 1 and Rule 2 above (see below). Notice that with these orientations, d-(vi) = 2 for all the elements of Pn of CH(Pn).
The edges oriented using Rule 1 are blue and the edges oriented using rule 2 are in green.
Step 6: Remove all the vertices in the set S.
Next we remove from T all the elements of S. Notice that we remove exactly 3|S| edges of T which are not flippable and what remains is a triangulation T' of Pn - S + {w}, which by Euler's formula contains 2(|Pn - S| + 1) - 4 = 2(n - |S|) - 2 triangles. Any element vi of Pn - S + {w} that is not in the convex hull of vertices of Pn - S + {w} that have d-(vi) = 2.
Step 7: Use lemma 2 on the resulting graph of step 6.
Using lemma 2 we can associate to every vertex vi of Q in the interior of CH(Pn) a triangle t(vi) of T' which is also a triangle in T, bounded by two oriented edges and the triangles t(vi) are all different. To each vertex vi of T' in CH(Pn) we can also associate a different triangle of T' among those having w as one of their vertices. Therefore, to each vertex of T' which is not w or a vertex of T' with d-(vi) < 2, we associate a different triangle of T' that contains no element of S. Thus, the number of edges of T that can be flipped is minimized when all the vertices vi not in S have d-(vi) = 2. We have n - |S| triangles associated with vertices in Q and as T' has 2n - 2|S| - 2 triangles, at most n - |S| - 2 of them contained points of S in T. Using the inequality |S| ≤ n - |S| - 2 we get:
(3n - 3) - 3|S| - 2(n - |S|) = n - |S| - 3 ≥ (n - 4)/2
which is the number of flippable edges in T.
## Analogous Results for Polygons
We have analogous result of Theorem 1 for polygons.
#### Theorem 3:
Any triangulation of a polygon Qn with n vertices, k of them being reflex, contains at least n - 3 - 2k diagonals that can be flipped.
Last Updated December 2, 2003 By Christina Boucher cboucher@geeky.net |
# Math Fact Practice: Doubles
This week we have been learning about different strategies for adding numbers. Today’s lesson was about adding doubles. You may (or may not) wonder why students should understand the concept of doubles or adding a number to itself. Well, first if students can memorize their double facts (ie 3+3 or 5+5), they can use those facts to figure out other related facts. For example, if your child knows that 3+3=6, then he can use that fact when solving 3+4 by thinking “if 3+3=6 then 3+4 is 3+3+1 which equals 7. Doubling numbers also helps build a foundation for multiplication as it is basically repeated addition.
I couldn’t find a good game for practicing doubles so I created on myself. This game can be played a few different ways. First you have to print out the templates and assemble the game board. When assembled it looks like the photo below:
Materials:
• a playing piece for each player for moving around the board
• one die
To Play:
• The first player rolls the one die and doubles that number. So if he rolled a 5, he adds 5+5 and determines that the answer is 10. He then moves ten spaces. Each player repeats these steps when it is his/her turn.
• If a player lands on a space with writing on it, he does what the space says to do.
Options:
• The game can be played for a period of time and in that time, each player keeps a tally record of how many times he passes start. Great review of tallies! Player with the most tallies wins if desired.
• Place a chip, popcorn kernel or other small item on each space that does not have words on it. As players land on a space they collect the item. The player with the most items wins.
In my class, we played using the first option without the winning part. I simply set the timer for about 7 minutes, the students played and tallied as they went past start. When the timer went off, the game was over. The students counted their tallies then we cleaned up with no talk of winners or losers.
## Doubles Game from Boy Mama Teacher Mama
* * * * * * *
If this is your first time visiting Boy Mama Teacher Mama, welcome! Did you know that you can also find us on Facebook, Pinterest, Twitter and G+? Stop by to see what is happening!
© Boy Mama Teacher Mama 2013 |
7
Q:
# To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of bucket is reduced to two-fifth of its present ?
A) 52.5 B) 62.5 C) 72.5 D) 82.5
Explanation:
Let the capacity of 1 bucket = x.
Then, capacity of tank = 25x.
New capacity of bucket = (2/5)x
$\inline \therefore$Required no of buckets = 25x/(2x/5) = 62.5
Q:
Simplify
39.96% of 649.85 $\mathbf{÷}$13.07 = 45.14 - ?
A) 25 B) 26 C) 27 D) 28
Explanation:
This can be easily simplified by Approximation process
Given 39.96% of 649.85 $÷$13.07 = 45.14 - ? can be written as
40% of 650 $÷$13 = 45 - ?
0 237
Q:
Simplify the fraction?
5 (2/7) - 3 (1/14) - 2 (1/14) - 1 (1/7) = ? + 3 (1/17)
A) -4 (1/17) B) 5 (2/17) C) -2 (1/17) D) 3 (3/17)
Explanation:
Given fraction is 5 (2/7) - 3 (1/14) - 2 (1/14) - 1 (1/7) = ? + 3 (1/17)
It can be written as
(5 - 3 - 2 - 1) + (2/7 - 1/14 - 1/14 - 1/7) = ? + 3 (1/17)
-1 + 0 - 3 (1/17) = ?
=> ? = -4 (1/17)
2 275
Q:
5 + 2 x 10/2 = ?
A) 15 B) 35 C) 10 D) 20
Explanation:
Using BODMAS Rule,
First solve division, mutiplication and Addition
So here 10/2 = 5
Then 2 x 5 = 10
Now, 5 + 10 = 15.
Hence, 5 + 2 x 10/2 = 15.
4 452
Q:
Simplify the following :
A) 10 B) 24 C) 36 D) 40
Explanation:
Given equation can be written as
0 372
Q:
Simplify the following expression?
40% of () = 48
A) 24 B) 22 C) 20 D) 18
Explanation:
We know that
1 367
Q:
Derivative of sinx by first principle?
A) -sinx B) -cosx C) cosx D) sinx
Explanation:
Thus, the derivative of sinx by first principle is cosx.
2 397
Q:
Simplify the following and find the value of ?
A) 1134 B) 1237 C) 1187 D) 1249
Explanation:
5 988
Q:
Simplify
A) 10 47/60 B) 11 45/73 C) 11 47/36 D) 10 35/48 |
Home Misc [Hackerrank] – Number Line Jumps Solution
# [Hackerrank] – Number Line Jumps Solution
0 comment
Let us try to understand this problem statement first. We are in charge of a fancy circus and 2 kangaroos have to jump on a number line. The interesting part is that both the kangaroos have a different start position, and different jump distances. It is a little tricky to understand, the problems asks to calculate the number line jumps such that both the kangaroos are at the same position.
We will try to simplify this using some diagrams, and see how the kangaroos actually jump.
In the given test case, kangaroo 2 starts at position 4, and kangaroo 1 starts at position 0.
The problem also says that kangaroo 2 has a jump distance of 2, and kangaroo 1 has a jump distance of 3. If we check the number line jumps, the kangaroos will be at a different position each after 1 jump.
#### Jump 1:
Kangaroo 2 reaches the position 6 as it is hopping 2 places.
Kangaroo 1 reaches a position 3 as it is hopping 3 places. The diagram below helps to understand it better.
#### Jump 2:
Following this strategy, both the kangaroos will have number line jumps. Since kangaroo 1 is faster than kangaroo 2, it is quite evident that at some point of time 1 will overtake 2.
#### Jump 3:
If we keep following the position of both the kangaroos, we will eventually see that in this case they will meet at some time.
#### Jump 4:
At jump number 4, both the kangaroos are the the same position on the number line. Hence, the answer to this test case is YES. It could also be possible that kangaroo 1 overtakes kangaroo 2 sometime in the air, or if its speed is less, it may never overtake at all.
In all such cases, we need to return NO as the answer.
#### Efficient Solution:
The problem can be approached in an efficient way using some basic mathematics. We know the speed of both the kangaroos.
speed = \frac{distance}{time}
Based on this we can calculate the total distance traveled by both the kangaroos.
distance = speed * jumps
Let us say the start position of kangaroo 1 is x1, and it’s speed is v1.
Start position of kangaroo 2 is x2, and it’s speed is v2.
Let j be the number of jumps after which both the kangaroos reach the same place. We want to make sure that both the kangaroos are at the same place on number line. Hence we can equate both the distances.
x1 + (v1 * j) = x2 + (v2 * j) \newline x2 - x1 = (v1 - v2) * j \newline j = \frac{(x2 - x1)}{(v1 - v2)}
The number of jumps j can only be a in integer, if the modulus of the expression is 0.
(x2-x1) \bmod (v1-v2) == 0
This makes the code really simple.
#### Code:
.wp-block-code {
border: 0;
}
.wp-block-code > div {
overflow: auto;
}
.shcb-language {
border: 0;
clip: rect(1px, 1px, 1px, 1px);
-webkit-clip-path: inset(50%);
clip-path: inset(50%);
height: 1px;
margin: -1px;
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String kangaroo(int x1, int v1, int x2, int v2) {
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return "NO";
if ((x2 - x1) % (v1 - v2) == 0)
return "YES";
else
return "NO";
}
Code language: Java (java)
Time Complexity: O(1)
Space Complexity: O(1)
You can also find the code and test cases on Github.
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# Prove $\triangle BAC$ is a right-angled isosceles triangle
Suppose in the figure below, $AD\ \text{//} \ BC$, $BD=BC$, $CD=CE$, and $ABCD$ is a trapezoid; $\measuredangle ABD=15°$
Prove that: $\triangle BAC$ is a right-angled isosceles triangle.
• Is there any synthetic proof?.. – bigant146 Sep 21 '17 at 23:49
## 3 Answers
Let $F$ the intersection of lines $BA$ and $CD$, $\angle BCD= x$ and $BC=L$. Then $$\angle BFC= x-\frac{\pi}{12}. \tag{1}$$ Applying some trig we get: $$CD=2L\cos x \tag{2}$$ $$DE =4L(\cos x)^2 \tag{3}$$. Applying sine rule in triangle $DAE$ we get: $$DA=\frac{4L(\cos x)^2\sin x}{-\sin(3x)} \tag{4}$$ The similarity of triangles $FDA$ and $FCB$ produces: $$FD=\frac{2L\cdot DA \cos x}{L -DA} \tag{5}$$ Applying sine rule in triangle $FDB$ we get: $$FD=\frac{L \sin(\frac{ \pi}{12})}{ \sin( x-\frac{ \pi}{12})} \tag{6}$$ Using $(4)$ and making $(5)$ and $(6)$ equal, we get: $$\sin(\frac{\pi}{12})(7-8(\sin x)^2)=-8(\cos x)^3\sin( x-\frac{ \pi}{12}) \tag{7}$$ After some trig relations, we get: $$\tan(x-\frac{ \pi}{12})=\tan x (8(\sin x)^2-7)$$ Solving for $x$ we get: $$x = \frac{5\pi}{12}$$ Thereafter it is easy to conclude that angle $BAC$ is a right angle and $\triangle ABC$ is isosceles.
According to bigant146's request, I'll try to make a solution as synthetic as possible.
Denote $\angle DBC=\angle ECD=\alpha$. Mark a point $P$ on $BC$ so that $BP=BE$. Choose a point $Q$ so that $BEQP$ is a rhombus. Clearly, $Q$ lies on the symmetry axis of the triangle $BDC$.
We have $\angle DEQ=\angle DBC=\angle BDA$ and $$\frac{DE}{EQ}=\frac{DE}{EB}=\frac{DA}{BC}=\frac{DA}{DB},$$ so the triangles $DEQ$ and $ADB$ are similar; thus $\angle DQE=15^\circ$. Due to the symmetry, $\angle CQP=\angle DQE=15^\circ$ as well. Thus in an isosceles triangle $DQC$ ($DQ=QC$) we have $\angle DQC=\alpha+30^\circ$.
Draw now a circle $\omega$ centered at $C$ with radius $CD=CE$. If $\alpha<30^\circ$, then $\angle DQE>\angle DCE/2$, so $Q$ lies inside $\omega$; on the other hand, $\angle DQC<60^\circ$, so $QC>DC$, and hence $Q$ lies outside $\omega$ --- a contradiction. A similar contradiction is obtained when $\alpha>30^\circ$; thus $\alpha=30^\circ$.
The rest is straightforward: we have $\angle ABC=\angle ABD+\angle DBC=15^\circ+\alpha=45^\circ$ and $\angle ACB=\angle DCB-\angle DCA=(90^\circ-\alpha/2)-\alpha=45^\circ$.
It is straightforward to check that the data of the question, including the condition $\angle ABD=\frac1{12}\pi$, are consistent with the value $\frac14\pi$ for $\angle ABC$ and $\angle ACB$. However, this fact alone does not rule out other possibilities for $\angle ABC$ (and other angles in the figure). To establish the unique result, we consider the more general problem where $\theta:=\angle ABD$ is variable rather than fixed as $\frac1{12}\pi$.
Choose the unit of length so that $|BC|=|BD|=1$, and write $x:=|CD|$. Since the isosceles triangles $DBC$ and $DCE$ share the base angle $\angle BDC$, they are similar, and so $|ED|=x^2$. Then $|EB|=1-x^2$ and, by the similarity of triangles $EBC$ and $EDA$, we have $$|AD|=\frac{x^2}{1-x^2}.$$Now write $\phi:=\angle DBC$. Then $\angle BDA=\phi$ and $\angle BAD=\pi-(\theta+\phi)$. Applying the sine rule to triangle $BAD$ gives $|BD|/\sin(\theta+\phi)=|AD|/\sin\theta$, or $$\left(\frac1{x^2}-1\right)\sin\theta=\sin\theta\cos\phi+\cos\theta\sin\phi.$$Dividing through by $\cos\theta$, and using the fact that $x=2\sin\frac12\theta$, so that $x^2=2(1-\cos\phi)$, some rearrangement gives $$\tan\theta=f(\phi):=\frac{2(1-\cos\phi)\sin\phi}{2\cos^2\phi-1}.$$As $\phi$ increases from $0$ towards $\frac14\pi$, each variable factor in the numerator of the above expression for $f(\phi)$ increases (from $0$), while the denominator decreases (from $1$ down to $0$). Hence $f$ is an increasing function in this range. We do not need to consider $\phi$ boyond $\frac14\pi$, because then $\tan\theta<0$ and $\theta>\frac12\pi$, and we are ultimately only interested in the case $\theta=\frac1{12}\pi$. Thus, in the range of interest for $\phi$, our function $f$ is increasing, and so $\tan\theta$ and hence $\theta$ is increasing with respect to phi. Conversely, it follows that $\phi$, and hence $\theta+\phi$, is increasing with respect to $\theta$ in $(0\;\pmb,\,\frac12\pi)$. Consequently there is a unique value of $\theta+\phi$ corresponding to each $\theta$ in this range—in particular for $\theta=\frac1{12}\pi$.
Using the formula $$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta},$$with $\tan2\theta=\tan\frac16\pi=1/\surd3$, and solving the resulting quadratic equation for $\tan\theta$, we get $\tan\frac1{12}\pi=2-\surd3$. This value is also obtained by calculating $f(\frac16\pi)$. Hence $\phi=\frac16\pi$ when $\theta=\frac1{12}\pi$, and then $\angle ABC=\frac1{12}\pi+\frac16\pi=\frac14\pi$. Setting $\gamma:=\angle ACB$, noting that $\angle BCD=\gamma+\phi=\angle BDC$, and summing the angles in triangle $BCD$ gives $3\phi+2\gamma=\pi$, or $\gamma=\frac14\pi$. Thus $\angle ACB=\frac14\pi$ too, and finally $\angle BAC=\frac12\pi$. |
Subtract 3-Digits from 3-Digits 1 Year 3 Addition and Subtraction Learning Video Clip | Classroom Secrets
Maths Resources & WorksheetsYear 3 Maths LessonsAutumn Block 2 (Addition and Subtraction)15 Subtract 3 Digits from 3 Digits 1 › Subtract 3-Digits from 3-Digits 1 Year 3 Addition and Subtraction Learning Video Clip
# Subtract 3-Digits from 3-Digits 1 Year 3 Addition and Subtraction Learning Video Clip
## Step 15: Subtract 3-Digits from 3-Digits 1 Year 3 Addition and Subtraction Learning Video Clip
Ali is transporting animals between conservation parks. She needs help to ensure the animals are transported safely.
More resources for Autumn Block 2 Step 15.
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Discussion points for teachers
1. Explain to Ali how she could solve the calculation.
Discuss calculating mentally. How could we mentally calculate the answer? Do we have to subtract? Why could we count on to solve this problem?
She can work mentally, counting on from 204 to 509 which = 305kg.
2. How much further must she drive?
Discuss what the number line shows us already. How can we use the number line to help us? What does each jump show us?
453 – 121 = 332 miles
3. What mistakes has she made?
Discuss working systematically through the calculation to find all the mistakes. What is wrong with the answer? Discuss commutativity and how we cannot swap the numbers over in a subtraction calculation.
She has not lined up the numbers properly and has used commutativity which does not work in a subtraction calculation.
485 – 271 = 214.
4. Which truck could Leo fit in?
Discuss what each bar model represents. Which calculation will allow us to calculate the missing number? How will we know which truck Leo can fit in?
Truck B: 659 – 433 = 226
5. What could the calculation be?
What does each part of the calculation represent? Discuss which digits can and cannot fit in each space and why. Work systematically to find all possibilities.
Various answers, for example: 985 – 614 = 371, 995 – 624 = 371, 885 – 613 = 272, 795 – 622 = 173
Optional discussion points:
Discuss reasons for transporting animals. Discuss how it is done humanely and safely.
National Curriculum Objectives
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# Cartesian Product of Set
## What is Cartesian Product?
Cartesian Product is a multiplication of two sets in which we get a set of ordered pairs as a result.
For Example;
Consider two non empty sets A & B with following elements.
A = { a1, a2 }
B = { b1, b2, b3 }
The Cartesian Product of A X B is given as:
Note that the cartesian product is a collection of ordered pair.
There are total of 6 ordered pair in above multiplication.
In the ordered pair, the first element is from set A and second element is from set B.
Conclusion
Ordered pair is simply multiplication of two sets which results in a set of ordered pairs.
## How to do cartesian product of two sets?
Suppose two sets A & B are given;
A = { 4, 5, 7 }
B = { 10, 12 }
For multiplication of A x B, follow the below steps:
(i) Form ordered pair of first element of set A with first and second element of set B.
(ii) Now form ordered pair of second element of set A with all elements of set B.
(iii) Finally form ordered pair of third element of set A with all elements of set B.
Combining all the ordered pair in one set we get;
A x B = {(4, 10), (4, 12), (5, 10), (5, 12), (7, 10), (7, 12) }
## Number of Ordered Pair in Multiplication
Suppose there are two sets P & Q with following elements.
P = { a1, a2, a3 }
Q = { b1, b2 }
Here the set P contains three elements; a1, a2 & a3.
So n(P) = 3
Set Q contains two elements; b1 & b2 .
So n(Q) = 2
The number of ordered pair in multiplication of P x Q is given by following formula;
⟹ n(P) x n(Q)
⟹ 3 x 2
⟹ 6
Hence, we will get total of 6 ordered pairs after multiplication.
The exact ordered pairs are shown below;
Note the 6 ordered pairs which are formed after multiplication of sets P & Q.
Conclusion
The number of ordered pairs after multiplication of two sets can be calculated by multiplying the number of individual elements of set.
## Is multiplication of A x B is same as B x A ?
No!!
The multiplication of A x B will yield completely different results that B x A.
Let us consider one example.
Given below is the set A & B with two elements each.
A = { a1, a2 }
B = { b1, b2 }
First Multiply A x B
Now Multiply B x A
Both the multiplication contain completely different ordered pairs.
Note that the ordered pair (a1, b1) is not equal to pair (b1, a1).
Hence, all the pairs in above multiplication is completely different.
## Solved Problems – Cartesian Product of Set
(01) Given below are two sets A & B.
A = { 3, 5 }
B = { 4, 2 }
Find A x B and B x A
Solution
Finding A x B
A x B = { (3, 4), (3, 2), (5, 4), (5, 2) }
Finding B x A
B x A = { (4, 3), (4, 5), (2, 3), (2, 5) }
(02) Given below are two sets P & Q with following elements.
P = { 3, 9, 7, 1}
Q = { 5, 12, 4 }
Find the number of ordered pairs in P x Q
Solution
P = {3, 9, 7, 1 }
There are 4 elements in set P.
n(P) = 4
Q = { 5, 12, 4 }
There are 3 elements in set Q
n(Q) = 3
In P x Q, the number of ordered will be;
⟹ n(P) x n(Q)
⟹ 4 x 3
⟹ 12
Hence, in P x Q there will be 12 ordered pairs.
(03) Given below are two sets A & B with following elements
A = { 4, 6, 8 }
B = { 1, 7 }
Find the following products.
(i) A x A
(ii) A x B
(iii) B x A
Solution
(i) A x A = { 4, 6, 8 } x { 4, 6, 8 }
A x A = { (4, 4), (4, 6), (4, 8),(6, 4), (6, 6), (6, 8),(8, 4), (8, 6), (8, 8) }
(ii) A x B = { 4, 6, 8 } x { 1, 7 }
A x B = { (4, 1), (4, 7), (6, 1), (6, 7), (8, 1), (8, 7) }
(iii) B x A = { 1, 7 } x { 4, 6, 8 }
B x A = { (1, 4), (1, 6), (1, 8), (7, 4), (7, 6), (7, 8) }
(04) Given below are two sets A & B.
A = { 9, 2, 4 }
B = { 𝜙 }
Find A x B.
Solution
Notice that set B is a null set which means that it doesn’t have any elements.
The multiplication of any set with null set results in null set.
It’s just similar to multiplication of number with 0 results in 0.
A x B = { 9, 2, 4 } x { 𝜙 } = 𝜙
A x B = { 𝜙 } |
# Age Questions, Riddles and Practice Tests
#### Question No 1
A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?
Solution!
Let the mother's present age be x years.! !Then,! the person's present age = (2/5)x years.! !(2/5)x + 8 = (1/2)(x + 8)!2(2x + 40) = 5(x + 8)!x = 40!.
#### Question No 2
Ten years ago Akram was thrice as old as Aslam was, but 10 years hence, he will be only twice as old. What is Akram's present age?
Solution!
Let Aslam's present age is x and Akram's present age is y.! !Given that!y-10 = 3(x-10)!y-10 = 3x-30!y-3x = -20!3x-y = 20 ...(1)! !and also given that!y+10 = 2(x+10)!y+10 = 2x+20!y-2x = 10 ...(2)! !By simultaneously solving equation(1) and equation(2)!y=70! !Hence Akram's present age is 70 years..
#### Question No 3
X is 4 times as old as y. 4 years ago, X was 10 times as old as Y. What will be the ratio of their ages 6 years hence?
Solution!
Given that!X = 4Y ...(1)! !and!X-4 = 10(Y-4)!X-4 = 10Y-40!X-10Y = -36!10Y-X = 36 ...(2)! !Put value of X from equation(1) in equation(2)!10Y-4Y = 36!6Y = 36!Y = 6! !Put value of Y in equation(1)!X = 4*6!X = 24! !After 6 years!Y = 24+6!Y = 30!and!X = 6+6!X = 12! !Ratio between X and Y is!30/12 = 5/2! !Hence after 6 years ratio between ages of X and Y will be 5/2..
#### Question No 4
Kiran is younger than Bineesh by 7 years and their ages are in the respective ratio of 7 / 9, how old is Kiran?
Solution!
Let the ages of Kiran and Bineesh are 7x and 9x respectively! !7x = 9x-7!x = 7/2 = 3.5! !Kiran's age = 7x!= 7 * 3.5 = 24.5.
#### Question No 5
If 1/2 x year ago A was 12 and 1/2 x years from now he will be 2x years old, how old will he be 3x years from now?
Solution!
Given that!A-(1/2)x = 12!2A-x = 24 ...(1)!and!A+(1/2)x = 2x!A+(1/2)x-2x = 0!A+(x-4x)/2 = 0!A-(3/2)x = 0!2A-3x = 0 ...(2)!By simultaneously solving equation(1) and (2) we have!x = 12!and!A = 18! !Now!3x = 3*12 = 36! !After 3x years A will be!18+36 = 54 years old.
#### Question No 6
The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be
Solution!
Let the present ages of son and father be x and (60 -x) years respectively.! !Then,!(60 - x) - 6 = 5(x - 6)!54 - x = 5x - 30!6x = 84!x = 14! !Hence, Son's age after 6 years = (x+ 6)!= 20 years.
#### Question No 7
Tina is twice as old as David. 15 years ago, Tina's age was 5 times the age of David. What is the present age of Tina?
Solution!
Let David's age is x and Tina's age is y.! !Given that!y = 2x!y-2x = 0 ...(1)! !and also given that!y-15 = 5(x-15)!y-15 = 5x-75!y-5x = -60 ...(2)! !By simultaneously solving equation(1) and equation(2)!y = 40! !Hence Tina's present age is 40 years..
#### Question No 8
The difference of the ages of two brothers is 2 and the difference of the square of their ages is 16. The age of smaller one is?
Solution!
Let present age of elder brother is y and present age of younger brother is x.! !Given that!y-x = 2 ...(1)! !and!(y^2)-(x^2) = 16!(y+x)(y-x) = 16! !Put value of (y-x) from equation(1)!2(y+x) = 16!y+x = 8 ...(2)! !By simultaneously solving equation(1) and equation(2) we have!y = 5!and!x = 3! !Hence present age of younger brother is 3 years..
#### Question No 9
Meena is thrice as old as Sheraz. Fatima will be twice as old as Meena 6 years hence. Six years ago Sheraz was five years old. What is Fatima's present age?
Solution!
Let present age of Sheraz is x and present age of Fatima is y.! !So,!Meena's age = 3x! !Given that!y+6 = 2(3x+6) ...(1)!and!x-6 = 5!x = 11! !Put value of x in equation(1)! !y+6 = 2(33+6)!y+6 = 78!y = 72! !Hence Fatima's present age is 72 years..
#### Question No 10
The Average age of a class of 22 subjects in 21 years. The average increased by 1 when the teacher's age also included. What is the age of the teacher?
Solution!
Total age of all students = 22 * 21 = 462! !Total age of all students + age of the teacher = 23 * 22 = 506! !Age of the teacher = 506 - 462 = 44.
#### Question No 11
The ages of X and Y are in the ratio 7/4. If the sum of their present ages is 55, then the age of X is
Solution!
Given that!x/y = 7/4 ...(1)! !and!x+y = 55 ...(2)! !From equation(1)!4x = 7y!y = (4/7)x! !Put value of y in equation(2)!x + (4/7)x = 55!(7x+4x)/7 = 55!11x = 385!x = 35! !Hence the age of x is 35 years..
#### Question No 12
Present ages of Sameer and Anand are in the ratio of 5/4 respectively. Three years hence, the ratio of their ages will become 11/9 respectively. What is Anand's present age in years?
Solution!
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.! !Then,!(5x+3)/(4x+3) = 11/9!9(5x + 3) = 11(4x + 3)!45x + 27 = 44x + 33!45x - 44x = 33 - 27!x = 6! !nand's present age = 4x = 24 years.
#### Question No 13
One year ago, the ratio of Sooraj's and Vimal's age was 6/7 respectively. Four years hence, this ratio would become 7/8. How old is Vimal?
Solution!
Let take the age of Sooraj and Vimal, 1 year ago as 6x and 7x respectively.! !Given that, four years hence, this ratio would become 7/8.! !(6x + 5)/(7x + 5) = 7/8!48x + 40 = 49x + 35!x = 5! !Vimal's present age = 7x + 1 = 7*5 + 1 = 36.
#### Question No 14
Ages of A and B are in ratio 8/7. 27 years ago their ages were in the ratio 5/4. The age of B is
Solution!
Given that!A/B = 8/7 ...(1)! !and!(A-27)/(B-27) = 5/4!5B-135 = 4A-108 ...(2)! !From equation(1) we have!A = (8/7)B! !Put value of A in equation(2)!5B-135 = 4*[(8/7)B]-108!5B-27 = (32/7)B!5B-(32/7)B = 27!(35B-32B)/7 = 27!3B = 189!B = 63! !Hence the age of B is 63 years..
#### Question No 15
A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. What is the present age of the mother?
Solution!
Let the present age of the person = x! !Then present age of the mother = 5x/2! !Given that , after 8 years, the person will be one-half of the age of his mother.! !(x + 8) = (1/2)(5x/2 + 8)!2x + 16 = 5x/2 + 8!x/2 = 8!x = 16! !Present age of the mother = 5x/2!= 5 * 16/2 = 40.
#### Question No 16
The ratio of A's age and B's age is 3/7, and the difference of their ages is 16 years. Find the age of B.
Solution!
Given that!A/B = 3/7 ...(1)! !and!B-A = 16 ...(2)!Put value of A from equation(1) in equation(2)!B-(3/7)B = 16!(7B-3B)/7 = 16!4B = 112!B = 28! !Hence Age of B is 28 years..
#### Question No 17
C's mother was four times as old as C ten years ago. After 10 years she will be twice as old as C is now. How old C is now?
Solution!
Let C's present age is x and his mother's age is y.! !Given that!y = 4(x-10)!y = 4x-40!4x-y = 40 ...(1)! !and!y+10 = 2x!2x-y = 10 ...(2).! !By simultaneously solving equation(1) and equation(2) we find! x = 15! !Hence C is 15 years old..
#### Question No 18
Kamal's age is 1/6 of his father's age. If kamal's 8th birthday was celebrated 2 years ago, what is father's present age?
Solution!
Let Kamal's present age is x and his father's present age is y.! !Given that!y = 6x ...(1)! !and!x-2 = 8!x = 10! !Put value of x in equation(1)!y = 6*10!y = 60! !Hence father's present age is 60 years..
#### Question No 19
Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7/9, how old is Sachin?
Solution!
Let Rahul's age be x years.! !Then,!Sachin's age = (x - 7) years.! !(x-7)/x = 7/9!9x - 63 = 7x!2x = 63!x = 31.5! !Hence, Sachin's age =(x - 7)= 24.5 years.
#### Question No 20
Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age?
Solution!
Given that! !(i)The difference of age b/w R and Q = The difference of age b/w Q and T!(ii) (R + T) = 50! !So,R - Q = Q - T! !(R + T) = 2Q! !Now given that, (R + T) = 50! !So, 50 = 2Q and therefore Q = 25! !Question is (R - Q) = ?! !Here we know the value(age) of Q (25), but we don't know the age of R.! !Therefore, (R-Q) cannot be determined..
#### Question No 21
Bill's age is 1/3 of Mike's age, who is 60% of James' age. Tom is twice as old as Mike. What is the ratio of Bill's age to Tom's age?
Solution!
Let Bill's age is Z, Mike's age is Y and Tom's age is X.! !Given that!Z = (1/3)Y!and!Y = (60/100)X!Y = (6/10)X! !Tom's Age is double of Mike's age mean his age is 2y.!Now Bill's age is (1/3)Y and Tom's age is 2Y and ratio between them is![(1/3)Y] / 2Y!(1/3)*(1/2) = 1/6! !Hence ratio between Bill's and Tom's age is 1/6..
#### Question No 22
The ratio of the ages of two friends A and B is 3/2. However if the age of A is decreased by 10 and that of B is increased by 10 then the original ratio is reversed. Age of A is
Solution!
Given that!A/B = 3/2!2A = 3B!2A-3B = 0 ...(1)! !and also given that!(A-10)/(B+10) = 2/3!3A-30 = 2B+20!3A-2B = 50 ...(2)! !By simultaneously solving equation(1) and equation(2)!A = 30! !Hence age of A is 30 years..
#### Question No 23
In 1975, A was thrice as old as B but in 1979, A was only twice as old as B was. How old was A in 1985?
Solution!
Let A's age in 1985 was X.! !Given that!X-10 = 3(B-10)!X-10 = 3B-30!3B-X = 20 ...(1)! !and!X-6 = 2(B-6)!X-6 = 2B-12!2B-X = 6 ...(2)! !By simultaneously solving equation(1) and (2) we have!X = 22! !Hence A's age in 1985 was 22 years..
#### Question No 24
The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is
Solution!
Let the ages of father and son 10 years ago be 3x and x years respectively.! !Then,!(3x + 10) + 10 = 2[(x + 10) + 10]!3x + 20 = 2x + 40!x = 20!So, 3x + 10 = 70!and x + 10 = 30! !Hence, ratio is 70/30 = 7/3..
#### Question No 25
The ages of C and D are in 6/11 and E is 6 years younger than D. If the sum of ages is 134, then what is the age of D?
Solution!
Given that!C/D = 6/11!C = (6/11)D ...(1)! !and!C+D+E = 134! !As E is 6 years younger from D so we can put D-6 instead of D!C+D+D-6 = 134!C+2D = 140!Put value of C from equation(1)!(6/11)D+2D = 140!6D+22D = 140!28D = 1540!D = 55! !Hence D is 55 years old..
#### Question No 26
Ten years ago a father was seven times as old as his son, two years hence, twice his age will be equal to five times his son's. What is the present age of son?
Solution!
Let present age of son is x and present age of father is y.! !Given that!y-10 = 7(x-10)!y-10 = 7x-70!y-7x = -70+10!7x-y = 60 ...(1)! ! and also given that!2(y+2) = 5(x+2)!2y+4 = 5x+10!2y-5x = 6 ...(2)By simultaneously soving equation(1) and equation(2)!x = 14! !Hence present age of son is 14 years..
#### Question No 27
The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person?
Solution!
Let's take the present age of the elder person = x! !and the present age of the younger person = x - 16! !(x - 6) = 3 (x-16-6)!x - 6 = 3x - 66!2x = 60!x = 60/2 = 30.
#### Question No 28
A mother's age after 5 years, will be twice the age of her daughter. The difference of their ages is 19. What is the age of mother?
Solution!
Let present age of daughter is x and present age of mother is y.! !Given that!y+5 = 2(x+5)!y+5 = 2x+10!5-10 = 2x-y!y-2x = 5 ...(1)! !and also given that!y-x = 19 ...(2)! !By simultaneously solving equation(1) and equation(2) we can find!y = 33! !Hence mothers present age is 33 years..
#### Question No 29
A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, what was the son's age five years back?
Solution!
Let the son's present age be x years.! !Then! (38 - x) = x! !2x = 38!x = 38/2 = 19! !Son's age 5 years back = 19-5 = 14 years.
#### Question No 30
The average age of A and B is 20. If C were to replace A, the average would be 19, and if B was replaced by C, the average would be 21. The age of C is
Solution!
Given that!(A+B)/2 = 20!A+B = 40 ...(1)! !and!(C+B)/2 = 19!C+B = 38 ...(2)! !and!(A+C)/2 = 21!A+C = 42 ...(3)! !Put value of A from equation(1) in equation(3)!(40-B)+C = 42!C-B = 2 ...(4)! !By simultaneously solving equation(2) and equation(4)!C = 20! !Hence C is 20 years old..
#### Question No 31
3 years ago, the average age of C and D was 18 years. With E joining them, the average becomes 22 years. How old is E now?
Solution!
Given that![(C-3)+(D-3)]/2 = 18!C-3+D-3 = 36!C+D = 42 ...(1)! !and also given that!(C+D+E)/3 = 22!C+D+E = 66! !Put value of C+D from equation(1)!42+E = 66!E = 66-42!E = 24! !Hence E is 24 years old now..
#### Question No 32
A man's age is three times his son's. In 15 years it will be double that of his son's. How old is the son now?
Solution!
Let Son's age is x!Mans Age = 3x! !Given that!3x+15 = 2(x+15)!3x+15 = 2x+30!3x-2x = 30-15! !x = 15! !Hence son is 15 years old..
#### Question No 33
20 years ago my age was 1/3 of what it is now. What is my present age?
Solution!
Let my present age is x.! !Given that!x-20 = (1/3)x!3x-60 = x!2x = 60!x = 30! !Hence my present age is 30 years..
#### Question No 34
The sum of the ages of a son and a father is 64. After 4 years the age of father will be three times that of his son's. The age of father is
Solution!
Let present age of son is x and present age of father is y.! !Given that!x+y = 64 ...(1)! !and!y+4 = 3(x+4)!y+4 = 3x + 12!y-3x = 8 ...(2)! !Put value of x from equation(1) in equation(2)! !y-3(64-y) = 8!y-192+3y = 8!4y = 200!y = 50! !Hence present age of the father is 50 years..
#### Question No 35
A is two years older than B who is twice as old as C. The total of the ages of A, B and C is 27. How old is B?
Solution!
Let the age of C = x.! !Then! !Age of B = 2x! !Age of A = 2 + 2x! !The total age of A,B and C =27! !(2+2x) + 2x + x = 27!5x = 25!B's age = 2x = 2 * 5 = 10.
#### Question No 36
Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?
Solution!
Let Ronit's present age be x years. Then, father's present age =(x + 3x) years!= 4x years.! !(4x + 8) = 5/2 (x+8)!8x + 16 = 5x + 40!3x = 24!x = 8! !Hence, required ratio = (4x+16)/(x+160)!= 48/24 = 2.
#### Question No 37
The product of the ages of two class fellows is 24. If the greater fellow is 3/2 times the smaller one, then the age of the smaller fellow is
Solution!
Let present age of the greater fellow is x and present age of smaller fellow is y.! !Given that xy = 24 ...(1)!and! !x = (3/2)y!Put value of x in equation(1).! !(3/2)y^2 = 24!y^2 = 24*(2/3)!y^2 = 16!y = 4! !Hence age of smaller fellow is 4 years..
#### Question No 38
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child?
Solution!
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.! !Then,!x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50!5x = 20!x = 4! !Hence, Age of the youngest child = x = 4 years.
#### Question No 39
A's age is 3/4 of that of B's. Two years ago A's age was 1/2 of what B will be in 4 years. How old is A now?
Solution!
Given that!A = (3/4)B ...(1)! !and!A-2 = (1/2)(B+4)!2A-4 = B+4!2A-B = 8! !Put value of B from equation(1)!2A-(4/3)A = 8!6A-4A = 24!2A = 24!A = 12! !Hence A is 12 years old now..
#### Question No 40
The present age of A is 45 yeras. 5 years ago the age of A was 5 times the age of his son B. How old is B now?
Solution!
Given that!A = 45! !and!A-5 = 5(B-5)!40 = 5B-25!65 = 5B!B = 13! !Hence B is 13 years old now..
#### Question No 41
A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was?
Solution!
Let the son's present age be x years. Then, (38 - x) = x! !2x = 38!x = 19! !Hence, Son's age 5 years back (19 - 5) = 14 years..
#### Question No 42
The total age of A and B is 12 years more than the total age of B and C. C is how many year younger than A?
Solution!
Given that A+B = 12 + B + C! !A - C = 12 + B - B = 12! !C is younger than A by 12 years..
#### Question No 43
The Average age of a class of 22 students is 21 years.The average increased by 1 when the teacher's age also included.What is the age of the teacher?
Solution!
No answer description available for this question..
#### Question No 44
Kamal was 4 times as old as his son 8 years ago. After 8 years, Kamal will be twice as old as his son. Find out the present age of Kamal.
Solution!
Let the age of the son before 8 years = x! !Then age of Kamal before 8 years ago = 4x! !After 8 years, Kamal will be twice as old as his son! !4x + 16 = 2(x + 16)!x = 8! !Present age of Kamal = 4x + 8 = (4*8) + 8 = 40.
#### Question No 45
John is 35 years older than his new born cousin. How old will the cousin be, when his age is 1/6 that of John's?
Solution!
We can simply check it from options. If John's cousins age is 7 then John's age is 42!and!1/6 0f 42 = 7!Hence 7 fulfills the required condition..
#### Question No 46
A father is 3 years older than 3 times the age of his son. If the sum of their ages is 31 years, how old was father 3 years ago?
Solution!
Let present age of son is x and present age of father is y.! !Given that!y = 3x+3 ...(1)! !and!x+y = 31!x = 31-y! !Put value of x in equation(1)!y = 3(31-y)+3!y = 93-3y+3!4y = 96!y = 24! !Three years ago!y = 24-3!y = 21! !Hence father was 21 years old 3 years ago..
#### Question No 47
A is as much younger than B and he is older than C. If the sum of the ages of B and C is 50 years, what is definitely the difference between B and A's age?
Solution!
No answer description available for this question..
#### Question No 48
Six years ago, the ratio of the ages of Kunal and Sagar was 6/5. Four years hence, the ratio of their ages will be 11/10. What is Sagar's age at present?
Solution!
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.! !Then,![(6x + 6) + 4]/[(5x + 6) + 4] = 11/10!10(6x + 10) = 11(5x + 10)!5x = 10!x = 2! !Hence, Sagar's present age = (5x + 6) = 16 years..
#### Question No 49
Ten years ago, P was half of Q in age. If the ratio of their present ages is 3/4, what will be the total of their present ages?
Solution!
Let the present age of P and Q be 3x and 4x respectively.! !Ten years ago, P was half of Q in age.! !(3x - 10) = 1/2 * (4x - 10)!6x - 20 = 4x - 10!2x = 10!x = 5! !total of their present ages!= 3x + 4x = 7x = 7 * 5!= 35.
#### Question No 50
The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. What is the ratio of their present ages?
Solution!
Let the age of the son before 10 years = x!and age of the father before 10 years = 3x! !Now we can write as (3x + 20) = 2(x + 20)!x = 20! !Age of Father the son at present = x + 10!= 20 + 10 = 30! !Age of the father at present = 3x + 10!= 3 * 20 + 10 = 70! !Required ratio = 70/30 = 7/3.
#### Question No 51
A father is 15 times as old as his son. however, 18 years later, he will be only thrice as old as the son. What is the father's present age?
Solution!
Let present age of the son is x and present age of Father is y.! !Given that!y = 15x!15x-y = 0 ...(1)! !and also given that!y+18 = 3(x+18)!y+18 = 3x+54!y-3x = 36 ...(2)! !By simultaneously solving equation(1) and equation(2)!y = 45! !Hence father's present age is 45 years..
#### Question No 52
The sum of the present ages of a mother and daughter is 63. Three years back if the age of the mother was double that of the daughter, then the present age of the daughter is
Solution!
Let mother's present age is x and daughters present age is y! !Given that!x+y = 63 ...(1)!and! !x-3 = 2(y-3)!x-3 = 2y-6!x-2y = -6+3!x-2y = -3 ...(2)! !By simultaneously solving equation(1) and equation(2) We'll find!y = 22! !Hence present age of the daughter is 22 years..
#### Question No 53
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child?
Solution!
Let the age of the youngest child = x! !Then the ages of 5 children can be written as x, (x+3), (x+6),(x+9) and (x+12).! !Then,!X + (x+3) + (x+6) + (x+9) + (x+12) = 50!5x + 30 =50!5x = 20!x = 20/5 = 4.
#### Question No 54
At present, the ratio between the ages of Arun and Deepak is 4/3. After 6 years, Arun's age will be 26 years. What is the age of Deepak at present ?
Solution!
Let the present ages of Arun and Deepak be 4x years and 3x years respectively.! !Then,!4x + 6 = 26!4x = 20!x = 5! !Hence, Deepak's age = 3x = 15 years..
#### Question No 55
The sum of the ages of two children is 12 and their product is 32. The age of elder one is
Solution!
Let age of younger child is x and age of elder child is y.! !Given that!x+y = 12 ...(1)! !and!xy = 32!x = 32/y! !Put value of x in equation(1)!(32/y)+y = 12!32+(y^2) = 12y!(y^2)-12y+32 = 0!(y^2)-8y-4y+32 = 0!y(y-8)+4(y-8) = 0!(y-8)(y+4) = 0!y = -4 (impossible)!or!y = 8! !Hence age of elder child is 8 years..
#### Question No 56
Sachin's age after 15 years will be 5 times his age 5 years back. Find out the present age of Sachin?
Solution!
Let the present age of Sachin = x! !Then (x+15) = 5(x-5)!4x = 40!x = 10.
#### Question No 57
A is two years older than B who is twice as old as C. The total of the ages of A, B and C is 27.How old is B?
Solution!
Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.! !So, (2x + 2) + 2x + x = 27!5x = 25!x = 5! !Hence, B's age = 2x = 10 years.
#### Question No 58
A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son?
Solution!
Let the present age of the son = x years! !Then present age the man = (x+24) years! !Given that in 2 years, man's age will be twice the age of his son! !(x+24) +2 = 2(x+2)!x = 22.
#### Question No 59
A is as much younger than B and he is older than C. If the sum of the ages of B and C is 50 years, what is definitely the difference between B and A's age?
Solution!
Age of C < Age of A < Age of B! !Given that sum of the ages of B and C is 50 years.! !Let's take B's age = x and C's age = 50-x! !Now we need to find out B's age - A's age.!But we cannot find out this with the given data..
#### Question No 60
Sandeep's age after six years will be three-seventh of his father's age. Ten years ago the ratio of their ages was 1/5. What is Sandeep's father's age at present? |
## Thursday, March 25, 2010
### Brayden's Fraction Growing Post
This video shows how I got the answers to my Fraction Quiz.
Now I'm going to explain two ways to multiply a fraction by a whole number.
One way would be drawing 4 groups of 1/2
Another way would be to multiply the two numerators.
Now I'm going to explain how to divide a fraction by a whole number.
My example is 1/4 ÷ 2.
Use your fraction strip and divide it into fourths. Then color in 1/4 of it. Next you divide your fraction strip in half as seen in the picture.
Next we have to answer a word problem.
12) It takes 4/5 of a roll of ribbon to wrap six packages. What fraction of a roll does it take to wrap three packages?
It would take 2/5 of a roll to wrap three packages.
I found that out because if 4/5 of a roll of ribbon wraps six packages you can just divide 4/5 by 2 and get 2/5 which would wrap three packages.
COMMENT!
----------------------------------------------------------------------------
Part 2
Fraction x Fraction
Fraction x mixed number
4/5 x 1 1/2
First you have to convert the mixed number to a improper fraction.
1 x 2 = 2
2 + 1 = 3
= 3/2
Then you multiply the two fractions.
4/5 x 3/2 = 12/10 or 6/5 or 1 1/5
Mixed number x Mixed number
2 1/3 x 2 2/3
First you convert the mixed numbers to improper fractions.
2 x 3 = 6
6 + 1 = 7
= 7/6
2 x 3 = 6
6 + 2 = 8
= 8/7
Then you multiply the two improper fractions .
7/6 x 8/7 = 56/42 or 4/3 or 1 1/3
------------------------------------------------------------------------------------------------
Part 3
Division
When you're dividing fractions you multiply the numerators by the denominator in the other fraction.
ex.
3/4 divided 1/2
You multiply 3 x 2 and 1 x 4 = 6/4
Then you make it a mixed fraction.
6/4 = 1 1/2
1 1/2 divided by 3/4
First you have to convert the mixed number to a improper fraction.
3/2 x 3/4
3 x 4 = 12
3 x 2 = 6
= 12/6
or 2 |
# If $$A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$$ and $$B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$$ then find $( A + 2B)'.$
Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1 $\leq$j $\leq$ n.
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
Step1:
Given:
$A'=\begin{bmatrix}-2 & 3\\1 & 2\end{bmatrix}$
We know that (A')'=A
The transpose of a matrix can be obtained by changing the rows & the column.
$A=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}$
$B=\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}$
$2B=2\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}$
$2B=\begin{bmatrix}-2 & 0\\2 & 4\end{bmatrix}$
Step2:
$A+2B=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}+\begin{bmatrix}-2 & 0\\2 & 4\end{bmatrix}$
$\qquad\;\;\;\;=\begin{bmatrix}-4 & 1\\5 & 6\end{bmatrix}$
The transpose of a matrix can be obtained by changing the rows & the column.
$(A+2B)'=\begin{bmatrix}-4 & 5\\1 & 6\end{bmatrix}$
edited Dec 24, 2013 |
Prime Factors of 19941
Prime Factors of 19941 are 3, 17, 17, and 23
How to find prime factors of a number
1. Prime Factorization of 19941 by Division Method 2. Prime Factorization of 19941 by Factor Tree Method 3. Definition of Prime Factors 4. Frequently Asked Questions
Steps to find Prime Factors of 19941 by Division Method
To find the primefactors of 19941 using the division method, follow these steps:
• Step 1. Start dividing 19941 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 19941, which is 3. Divide 19941 by 3 to obtain the quotient (6647).
19941 ÷ 3 = 6647
• Step 3. Repeat step 1 with the obtained quotient (6647).
6647 ÷ 17 = 391
391 ÷ 17 = 23
23 ÷ 23 = 1
So, the prime factorization of 19941 is, 19941 = 3 x 17 x 17 x 23.
Steps to find Prime Factors of 19941 by Factor Tree Method
We can follow the same procedure using the factor tree of 19941 as shown below:
So, the prime factorization of 19941 is, 19941 = 3 x 17 x 17 x 23.
Introduction to Prime Factors?
In mathematics, prime numbers are defined as all those natural numbers greater than 1 that has only two divisors one being number 1 and the other being the number itself. When we express any number as the products of these prime numbers than these prime numbers become prime factors of that number. Eg- Prime Factors of 19941 are 3 x 17 x 17 x 23.
Properties of Prime Factors
• Prime factors are the set of factors that are unique to the number given.
• 2 is the only even prime factor any number can have.
• Two prime factors are always coprime to each other.
• 1 is neither a prime number nor a composite number and also 1 is the factor of every given number. So, 1 is the factor of 19941 but not a prime factor of 19941.
• Which is the smallest prime factor of 19941?
Smallest prime factor of 19941 is 3.
• Is 19941 a perfect square?
No 19941 is not a perfect square.
• What is the prime factorization of 19941?
Prime factorization of 19941 is 3 x 17 x 17 x 23.
• What are the factors of 19941?
Factors of 19941 are 1 , 3 , 17 , 23 , 51 , 69 , 289 , 391 , 867 , 1173 , 6647 , 19941.
• What is prime factorization of 19941 in exponential form?
Prime factorization of 19941 in exponential form is 3 x 172 x 23.
• Is 19941 a prime number or a composite number?
19941 is a composite number.
• Is 19941 a prime number?
false, 19941 is not a prime number.
• How to find prime factorization of 19941 easily?
To find prime factorization of 19941 easily you can refer to the steps given on this page step by step.
• Which is the largest prime factors of 19941?
The largest prime factor of 19941 is 23. |
# Review of past work (Page 6/8)
Page 6 / 8
## Fractions and decimal numbers
A fraction is one number divided by another number. There are several ways to write a number divided by another one, such as $a÷b$ , $a/b$ and $\frac{a}{b}$ . The first way of writing a fraction is very hard to work with, so we will useonly the other two. We call the number on the top (left) the numerator and the number on the bottom (right) the denominator . For example, in the fraction $1/5$ or $\frac{1}{5}$ , the numerator is 1 and the denominator is 5.
## Definition - fraction
The word fraction means part of a whole .
The reciprocal of a fraction is the fraction turned upside down, in other words the numerator becomes the denominator and the denominator becomesthe numerator. So, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$ .
A fraction multiplied by its reciprocal is always equal to 1 and can be written
$\frac{a}{b}×\frac{b}{a}=1$
This is because dividing by a number is the same as multiplying by its reciprocal.
## Definition - multiplicative inverse
The reciprocal of a number is also known as the multiplicative inverse.
A decimal number is a number which has an integer part and a fractional part. The integer and the fractional parts are separated by a decimal point , which is written as a comma in South African schools. For example the number $3\frac{14}{100}$ can be written much more neatly as $3,14$ .
All real numbers can be written as a decimal number. However, some numbers would take a huge amount of paper (and ink) to write out in full! Some decimal numberswill have a number which will repeat itself, such as $0,33333...$ where there are an infinite number of 3's. We can write this decimal value by using a dotabove the repeating number, so $0,\stackrel{˙}{3}=0,33333...$ . If there are two repeating numbers such as $0,121212...$ then you can place dots or a bar, like $0,\overline{12}$ on each of the repeated numbers $0,\stackrel{˙}{1}\stackrel{˙}{2}=0,121212...$ . These kinds of repeating decimals are called recurring decimals .
[link] lists some common fractions and their decimal forms.
Fraction Decimal Form $\frac{1}{20}$ 0,05 $\frac{1}{16}$ 0,0625 $\frac{1}{10}$ 0,1 $\frac{1}{8}$ 0,125 $\frac{1}{6}$ $0,16\stackrel{˙}{6}$ $\frac{1}{5}$ 0,2 $\frac{1}{2}$ 0,5 $\frac{3}{4}$ 0,75
## Scientific notation
In science one often needs to work with very large or very small numbers. These can be written more easily in scientific notation, which has the general form
$a×{10}^{m}$
where $a$ is a decimal number between 0 and 10 that is rounded off to a few decimal places. The $m$ is an integer and if it is positive it represents how many zeros should appear to the right of $a$ . If $m$ is negative, then it represents how many times the decimal place in $a$ should be moved to the left. For example $3,2×{10}^{3}$ represents 32 000 and $3,2×{10}^{-3}$ represents $0,0032$ .
If a number must be converted into scientific notation, we need to work out how many times the number must be multiplied or divided by 10 to make it into anumber between 1 and 10 (i.e. we need to work out the value of the exponent $m$ ) and what this number is (the value of $a$ ). We do this by counting the number of decimal places the decimal point must move.
For example, write the speed of light which is 299 792 458 ms ${}^{-1}$ in scientific notation, to two decimal places. First, determine where the decimalpoint must go for two decimal places (to find $a$ ) and then count how many places there are after the decimal point to determine $m$ .
What is the law of demand
price increase demand decrease...price decrease demand increase
Mujahid
ıf the price increase the demand decrease and if the demand increase the price decrease
MUBARAK
all other things being equal, an increase in demand causes a decrease in supply and vice versa
SETHUAH
yah
Johnson
how is the economy of usa now
Johnson
What is demand
Demand is the quantity of goods and services a consumer is willing and able to purchase at various prices over a given period of time.
Yaw
yea
SETHUAH
Okay congratulations I'll join you guys later .
Aj
yes
MUBARAK
calculate elasticity of income exercises
If potatoes cost Jane $1 per kilogram and she has$5 that could possibly spend on potatoes or other items. If she feels that the first kilogram of potatoes is worth $1.50, the second kilogram is worth$1.14, the third is worth $1.05 and subsequent kilograms are worth$0.30, how many kilograms of potatoes will she purchase? What if she only had $2 to spend? Susan Reply cause of poverty in urban DAVY Reply QI: (A) Asume the following cost data are for a purely competitive producer: At a product price Of$56. will this firm produce in the short run? Why Why not? If it is preferable to produce, what will be the profit-maximizing Or loss-minimizing Output? Explain. What economic profit or loss will the
what is money
money is any asset that is generally acceptable for the exchange of goods and for the settlement of debts
Mnoko
what is economic
economics is the study of ways in which people use resources to satisfy their wants
Falak
what is Price mechanism
what is Economics
ERNESTINA
The study of resource allocation,distribution and consumption.
Emelyn
introduction to economics
welfare definition of economics
Uday
examine the wealth and welfare definitions of economics
Uday
Anand
What do we mean by Asian tigers
Dm me I will tell u
Shailendra
Hi
Aeesha
hi
Pixel
What is Average revenue
KEMZO
How are u doing
KEMZO
it is so fantastic
metasebia
uday
Uday
it is a group of 4 countries named Singapore, South Korea, Taiwan and Hong Kong because their economies are growing very faster
Anand
fyn
EDWARD
Please, average revenue is an amount of money you gained after deducted your total expenditure from your total income.
EDWARD
what's a demand
it is the quantity of commodities that consumers are willing and able to purchase at particular prices and at a given time
Munanag
quantity of commodities dgat consumers are willing to pat at particular price
Omed
demand depends upon 2 things 1wish to buy 2 have purchasing power of that deserving commodity except any from both can't be said demand.
Bashir
Demand is a various quantity of a commodities that a consumer is willing and able to buy at a particular price within a given period of time. All other things been equal.
Vedzi
State the law of demand
Vedzi
The desire to get something is called demand.
Mahabuba
what is the use of something should pay for its opportunity foregone to indicate?
Why in monopoly does the firm maximize profits when its marginal revenue equals marginal cost
different between economic n history
Difference between extinct and extici spicies
Got questions? Join the online conversation and get instant answers! |
# How do you solve -x-6y=-18 and x-6y=-6 using the addition and subtraction method?
Oct 24, 2014
The answer is $x = 6$ and $y = 2$
The easiest way to solve such systems of equations using the elimination (or as you put it "addition and subtraction") method, is to write out our two equations, one below the other, and see how we can cancel out a variable to bring it down to a one-variable equation.
Here we have:
$- x - 6 y = - 18$
$x - 6 y = - 6$
Think of this as one of those addition or subtraction problems you did back in elementary school, with the number stacked on each other like this for you to work with. You have two paths of action here: you can either add the two equations, or subtract one from the other. Which one will you chose, in order to eliminate a variable?
In this case, both methods would work. If we added, we'd end up eliminating the $x$
[x + (-x) = 0],
and if we subtracted, we'd end up eliminating the $y$
$\left[- 6 y - \left(- 6 y\right) = 0\right]$.
So for this particular problem the course of action is up to you. I'm going to go with addition, since I just like adding better (it's more straightforward).
$\left[- x - 6 y = - 18\right]$
$+ \left[x - 6 y = - 6\right]$
=> $- 12 y = - 24$
$y = 2$
Now that we have one variable solved, we can plug it back into one of our original equations and solve for the other one. Again, it's completely up to you which one to use, but I'm going to use the second one:
$x - 6 \left(2\right) = - 6$
$x - 12 = - 6$
$x = 6$
So our solution to this system is the coordinate $\left(6 , 2\right)$, meaning that if we graphed these two lines, they would intersect at that coordinate.
Hope that helped :) |
# Find the equation of the line containing the points (-2,3) and (2,-3).
justaguide | Certified Educator
The equation of a line with the points (x1, y1) and (x2, y2) is given by (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1)
We have to find the line with the points (-2, 3) and (2, -3).
The equation of the line is (y + 3)/(x - 2) = (3 + 3)/(-2 - 2)
=> (y + 3)/(x - 2) = 6/(-4)
=> (y + 3)/(x - 2) = -3/2
=> 2*(y + 3) = -3*(x - 2)
=> 2y + 6 = -3x + 6
=> 2y + 3x = 0
The line with the points (-2, 3) and (2, -3) is 3x + 2y = 0
iislas104 | Student
Step 1: Find the slope from the two points given to you by using the formula Slope (m) = (y2 –y1)/( x2 – x1)
x1=-2, y1=3, x2=2, y2=-3 so m= (-3 –3)/( 2– -2)=-6/4=-3/2
Step2: Find the y-intercept (b) using the slope-intercept form(y=mx+b)
Chose one the point to sub for your x and y (I will choose the first point) so xà -2 and y à 3 and of course you just solved for the slope so you now know m à -3/2
Y = m x + b
3 = (-3/2)( -2) + b Now solve the equation for b
3 = 3 + b
0 = b
Step 3: Now you know both m(slope) and b(y-intercept) so you can write the equation as
y=(-3/2) x + 0 OR y=(-3/2) x |
42400 in words
42400 in words is written as Forty Two Thousand Four Hundred. In 42400, 4 has a place value of ten thousand, 2 is in the place value of thousand and next 4 is in the place value of hundred. The article on Place Value gives more information. The number 42400 is used in expressions that relate to money, distance, social media views, and many more. For example, “42.2 Hectare is nothing but Forty Two Thousand four Hundred square meters”. Another example, “Today’s total pandemic cases rose to Forty Two Thousand Four Hundred”.
42400 in words Forty Two Thousand Four Hundred Forty Two Thousand Four Hundred in Numbers 42400
How to Write 42400 in Words?
We can convert 42400 to words using a place value chart. The number 42400 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 4 2 4 0 0
Thus, we can write the expanded form as:
4 × Ten thousand + 2 × Thousand + 4 × Hundred + 0 × Ten + 0 × One
= 4 × 10000 + 2 × 1000 + 4 × 100 + 0 × 10 + 0 × 1
= 42400.
= Forty Two Thousand Four Hundred.
42400 is the natural number that is succeeded by 42399 and preceded by 42401.
42400 in words – Forty Two Thousand Four Hundred.
Is 42400 an odd number? – No.
Is 42400 an even number? – Yes.
Is 42400 a perfect square number? – No.
Is 42400 a perfect cube number? – No.
Is 42400 a prime number? – No.
Is 42400 a composite number? – Yes.
Solved Example
1. Write the number 42400 in expanded form
Solution: 4 x 10000 + 2 x 1000 + 4 x 100 + 0 x 10 + 0 x 1
Or Just 4 x 10000 + 2 x 1000 + 4 x 100
We can write 42400 = 40000 + 2000 + 400 + 0 + 0
= 4 x 10000 + 2 x 1000 + 4 x 100 + 0 x 10 + 0 x 1.
Frequently Asked Questions on 42400 in words
Q1
How to write the number 42400 in words?
42400 in words is written as Forty Two Thousand Four Hundred.
Q2
State whether True or False. 42400 is divisible by 3?
False. 42400 is not divisible by 3.
Q3
Is 42400 divisible by 10?
Yes. 42400 is divisible by 10. |
# A calculus of the absurd
##### 15.2.3 Inverse function of $$\sinh (x)$$
• Example 15.2.1 Finding the inverse function of $$\sinh (x)$$.
From the definition of $$\sinh (x)$$ we know that
$$\sinh (x) = \frac {e^x-e^{-x}}{2}$$
The right-hand side is actually a quadratic in $$e^x$$, which becomes more readily apparent114114 In general, if we have an expression containing $$x^y$$, $$x^{-y}$$ and no other powers of $$x$$, it’s worth considering whether there is a hidden quadratic to be found. if we multiply the fraction through by $$\frac {e^{x}}{e^{x}}$$.
$$\sinh (x) = \frac {e^{2x}-1}{2e^{x}}$$
Here we can substitute $$u=e^{x}$$
$$\sinh (x) = \frac {u^2-1}{2u}$$
And then we can rearrange this to give a quadratic in terms of $$u$$.
$$u^2 - 2u\sinh (x) - 1 = 0$$
By applying the quadratic formula, we can solve this for $$u$$
$$u = \frac { -(-2\sinh (x)) \pm \sqrt {(-2\sinh (x))^2 - 4(1)(-1)} } {2}$$
We can then simplify this a bit
$$u = \frac { 2\sinh (x) \pm \sqrt {4}\sqrt {\sinh (x)^2 + 1} } {2}$$
If we then reverse the substitution115115 Arguably, the substitution didn’t help here, but substitutions do often make it easier to spot the structure of a problem by simplifying problems in a way which makes them look more like a previously seen problem., replacing $$u$$ with $$e^x$$.
$$e^x = \sinh (x) \pm \sqrt {\sinh (x)^2 + 1}$$
The next step is to take the natural logarithm of both sides
$$x = \ln \left ( \sinh (x) \pm \sqrt {\sinh (x)^2 + 1} \right )$$
here, because $$a<\sqrt {a^2+1}$$ 116116 This can be proved by considering the cases when $$a > 0$$, $$a=0$$ and $$a<0$$ we cannot have the negative case (for the $$\pm$$), as the domain of the logarithm function requires that the input is greater than zero.
And we have found the inverse function!
$$\operatorname {arsinh}(x) = \ln \left (x+\sqrt {x^2+1}\right ) \label {arsinhx}$$ |
# Two Digit Fun First Part
Feb 4, 2023
Firs we have to learn counting number from 1 -100 and table at least 1- 10. Because, There are following fun and facts based on two digit numbers. It is more important for basic mathematics to all learner. We can learn math step by step using the Counting Numbers and Tables.
###### Do and Learn the following Numbers
If you have to learn the counting number from 1 to 100. We can start the magical fun of these hundred numbers. First we have to recognize all the counting numbers from 1 to 100. There are following numbers. Which has been given below for you.
###### Write the successor:
Start with the following numbers and add ‘1’. To get the successor number. Let first example: The number is 45: 45 + 1 = 46; 46 is the successor of 45. (“+” Addition or Plus)
###### Write the predecessor:
Start with the following numbers and subtract ‘1’. To get the predecessor number. Let first example: The number is 45: 45 – 1 = 44; 44 is the predecessor of 45. (“-” Minus or Subtraction)
###### Write the number between:
Let find the number between 12 and 14. You need to start counting with the first number 12 and add ‘1’ to get the middle number and again add ‘1’ to them to get the second number 14. Example: 12 + 1 =13. 13 is the middle number. Again add ‘1’: 13 + 1 =14. To get second number 14. Your answer is 13. 13 is in between 12 and 14.
###### Write the ten consecutive numbers starting with the following numbers:
If you know the counting from 1 to 100, then solve it easily. To start with the following numbers and counting 10 fingers what you have in your hands and write them. Example: 5, 6, 7, 8, 9, 10, 11, 12, 13, 14.
###### Add 2 in each of the following number and Write them:
Here, Add two fingers in each of the following numbers to get the result and write them. Example: 1 + 2 = 3; 26 + 2 = 28;
###### Add 3 in each of the following number and Write them:
Here, Add three fingers in each of the following numbers to get the result and write them. Example: 1 + 3 = 4; 26 + 3 = 29;
###### Add 4 in each of the following number and Write them:
Here, Add four fingers in each of the following numbers to get the result and write them. Example: 1 + 4 = 5; 26 + 4 = 30;
###### Add 5 in each of the following number and Write them:
Here, Add five fingers in each of the following numbers to get the result and write them. Example: 1 + 5 = 6; 26 + 5 = 31;
###### Add 6 in each of the following number and Write them:
Here, Add six fingers in each of the following numbers to get the result and write them. Example: 1 + 6 = 7; 26 + 6 = 32;
###### Add 7 in each of the following number and Write them:
Here, Add seven fingers in each of the following numbers to get the result and write them. Example: 1 + 7 = 8; 26 + 7 = 33;
###### Add 8 in each of the following number and Write them:
Here, Add eight fingers in each of the following numbers to get the result and write them. Example: 1 + 8 = 9; 26 + 8 = 34;
###### Add 9 in each of the following number and Write them:
Here, Add two fingers in each of the following numbers to get the result and write them. Example: 1 + 9 = 10; 26 + 9 = 35;
###### Add 10 in each of the following number and Write them:
Here, Add two fingers in each of the following numbers to get the result and write them. Example: 1 + 10 = 11; 26 + 10 = 36;
###### To add the following in a row:
Let see the first example: To get the first number ‘1’ and start counting to the next five fingers. To get the result 6. Your answer is 6.
It is too much in this part of Two Digit Number. If you are the parents of a child and want to teach them Mathematics in easy way to follow this website and subscribe it soon and share it to your friends also. Thank You. To learn more in the next post. |
# Techniques as to solving absolute value equation
Solve absolute value equation with absolute value variable one one side or even both side, without a number outsides of absolute value signs are typically easy.
In my high school, I was taught to first separate the absolute variable and make it it to two different cases with different signs for one side.
But the real problem is: what happen if there is a constant outside of the absolute signs, like: $$|x-5| = |x+5|-1$$
There are three broad ways to approach a problem like this.
1. Draw the graphs of $y = \left | x-5 \right |$ and $y = \left | x + 5 \right | -1$ and see where they intersect so you know which regions you have to check for solutions, then just solve in those regions.
2. Check all possible combinations, as mookid suggests. Note that his step #2 is very important - there are four sets of solutions it will produce, and it's likely that several of them do not solve the original equation (actually this is true for many equations, and it's a good idea to do it at the end of just about any exercise).
3. It's a much uglier option, but you can use the fact that squaring an absolute value lets you remove the absolute value signs to do this:
$\left|x-5\right|=\left|x+5\right|-1 \\ \left(\left|x-5\right|\right)^2=\left(\left|x+5\right|-1\right)^2 \mbox{ (squaring both sides)}\\ x^2-10x+25=\left(\left|x+5\right|\right)^2-2\left|x+5\right|+1 \\ x^2-10x+25=x^2+10x+25-2\left|x+5\right|+1 \\ 20x+1=2\left|x+5\right| \mbox{ (collecting like terms)}\\ \left(\left|20x+1\right|\right)^2=4\left(\left|x+5\right|\right)^2 \mbox{ (squaring again)}\\ 400x^2+40x+1=4x^2+40x+100 \\ 396x^2=99 \\ x^2=\frac{1}{4} \\ x=\pm\frac{1}{2}$
And, once again you need to check both solutions, since one of them does not solve the original equation. Be careful with this approach, since for an arbitrary combination of terms it can result in having to solve a ridiculously large polynomial equation rather than a large number of separate linear equations as would occur if you just break it up into regions.
In this case there are three cases to consider: $x < -5$, $-5 \le x < 5$, and $5 \le x$. For example, if $x < -5$, $|x - 5| = -x + 5$ and $|x+5| = -x-5$ so the equation says $-x + 5 = -x - 5 - 1$. No solution there. Now try the other two cases.
This is exactly the same. The most simple general way to handle this kind of equations is to make this a two step journey:
1. If $|x−5|=|x+5|−1$: There are four cases: \begin{align} x-5 &= x+5 &- 1 &\iff x\in S_1\\ -x+5 &= x+5 &- 1 &\iff x\in S_2\\ x-5 &= -x-5 &- 1 &\iff x\in S_3\\ -x+5 &= -x-5 &- 1 &\iff x\in S_4\\ \end{align} Then you solve for each equation. What you proved in this step is that the set of solutions $S$ is such as $$S \subset \bigcup_i S_i$$
2. For each element of each set $S_i$, check whether this is a solution or not.
To this simple case of addition of an extra number, let me humbly add a fourth and purely graphic/intuitive method to @ConMan's answer.
Draw a line representing the $\mathbb{R}$ numbers. See my drawing:
You have two absolute value parts:
• The $|x-5|$ is the distance from $x$ to the number $5$.
• The $|x+5|=|x-(-5)|$ is the distance from $x$ to $-5$.
So note the numbers $5$ and $-5$ on your line. Your equation $|x-5|=|x+5|-1$ tells us that the distance from $x$ to $5$ must be one less than the distance from $x$ to $-5$.
Point $a$ would be the solution to $|x-5|=|x+5|$. Point $b$ would be the solution to $|x-5|=|x+5|-2$. And the solution to your original equation $|x-5|=|x+5|-1$ is found in the point $x=0.5$, where the difference between the two distances is exactly $1$. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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Pythagorean Theorem and Pythagorean Triples
Square of the hypotenuse equals the sum of the squares of the legs.
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Pythagorean Theorem and Pythagorean Triples
What if a friend of yours wanted to design a rectangular building with one wall 65 ft long and the other wall 72 ft long? How can he ensure the walls are going to be perpendicular? After completing this Concept, you'll be able to apply the Pythagorean Theorem in order to solve problems like these.
Guidance
The sides of a right triangle are called legs (the sides of the right angle) and the side opposite the right angle is the hypotenuse. For the Pythagorean Theorem, the legs are “\begin{align*}a\end{align*}” and “\begin{align*}b\end{align*}” and the hypotenuse is “\begin{align*}c\end{align*}”.
Pythagorean Theorem: Given a right triangle with legs of lengths \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and a hypotenuse of length \begin{align*}c\end{align*}, then \begin{align*}a^2 + b^2 = c^2\end{align*}.
Pythagorean Theorem Converse: If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
There are several proofs of the Pythagorean Theorem, shown below.
Investigation: Proof of the Pythagorean Theorem
Tools Needed: pencil, 2 pieces of graph paper, ruler, scissors, colored pencils (optional)
1. On the graph paper, draw a 3 in. square, a 4 in. square, a 5 in square and a right triangle with legs of 3 and 4 inches.
2. Cut out the triangle and square and arrange them like the picture on the right.
3. This theorem relies on area. Recall from a previous math class, that the area of a square is length times width. But, because the sides are the same you can rewrite this formula as \begin{align*}A_{square} = length \times width = side \times side = side^2\end{align*}. So, the Pythagorean Theorem can be interpreted as \begin{align*}(square \ with \ side \ a)^2 + (square \ with \ side \ b)^2 = (square \ with \ side \ c)^2\end{align*}. In this Investigation, the sides are 3, 4 and 5 inches. What is the area of each square?
4. Now, we know that \begin{align*}9 + 16 = 25\end{align*}, or \begin{align*}3^2 + 4^2 = 5^2\end{align*}. Cut the smaller squares to fit into the larger square, thus proving the areas are equal.
Another Proof of the Pythagorean Theorem
This proof is “more formal,” meaning that we will use letters, \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} to represent the sides of the right triangle. In this particular proof, we will take four right triangles, with legs \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and hypotenuse \begin{align*}c\end{align*} and make the areas equal.
For two animated proofs, go to http://www.mathsisfun.com/pythagoras.html and scroll down to “And You Can Prove the Theorem Yourself.”
Pythagorean Triples
A Pythagorean Triple is a set of three whole numbers that makes the Pythagorean Theorem true. The most frequently used Pythagorean triple is 3, 4, 5, as in Investigation 8-1. Any multiple of a Pythagorean triple is also considered a triple because it would still be three whole numbers. Therefore, 6, 8, 10 and 9, 12, 15 are also sides of a right triangle. Other Pythagorean triples are:
\begin{align*}3, 4, 5 \qquad 5, 12, 13 \qquad 7, 24, 25 \qquad 8, 15, 17\end{align*}
There are infinitely many Pythagorean triples. To see if a set of numbers makes a triple, plug them into the Pythagorean Theorem.
Example A
Do 6, 7, and 8 make the sides of a right triangle?
Plug in the three numbers into the Pythagorean Theorem. The largest length will always be the hypotenuse. \begin{align*}6^2 + 7^2 = 36 + 49 = 85 \neq 8^2\end{align*}. Therefore, these lengths do not make up the sides of a right triangle.
Example B
Find the length of the hypotenuse of the triangle below.
Let’s use the Pythagorean Theorem. Set \begin{align*}a\end{align*} and \begin{align*}b\end{align*} equal to 8 and 15 and solve for \begin{align*}c\end{align*}, the hypotenuse.
\begin{align*}8^2 + 15^2 & = c^2\\ 64 + 225 & = c^2\\ 289 & = c^2 \qquad \quad Take \ the \ square \ root \ of \ both \ sides.\\ 17 & = c\end{align*}
When you take the square root of an equation, usually the answer is +17 or -17. Because we are looking for length, we only use the positive answer. Length is never negative.
Example C
Is 20, 21, 29 a Pythagorean triple?
If \begin{align*}20^2 + 21^2\end{align*} is equal to \begin{align*}29^2\end{align*}, then the set is a triple.
\begin{align*}20^2 + 21^2 & = 400 + 441 = 841\\ 29^2 & = 841\end{align*}
Therefore, 20, 21, and 29 is a Pythagorean triple.
Example D
Determine if the triangle below is a right triangle.
Check to see if the three lengths satisfy the Pythagorean Theorem. Let the longest sides represent \begin{align*}c\end{align*}, in the equation.
\begin{align*}a^2 + b^2 &= c^2\\ 8^2 + 16^2 &= \left ( 8 \sqrt{5} \right )^2\\ 64 + 256&= 64 \cdot 5\\ 320 &= 320\end{align*}
The triangle is a right triangle.
Watch this video for help with the Examples above.
Concept Problem Revisited
To make the walls perpendicular, find the length of the diagonal.
\begin{align*}65^2 + 72^2 & = c^2\\ 4225 + 5184 & = c^2\\ 9409 & = c^2\\ 97 & = c\end{align*}
In order to make the building rectangular, both diagonals must be 97 feet.
Vocabulary
The two shorter sides of a right triangle (the sides that form the right angle) are the legs and the longer side (the side opposite the right angle) is the hypotenuse. The Pythagorean Theorem states that \begin{align*}a^2+b^2=c^2\end{align*}, where the legs are “\begin{align*}a\end{align*}” and “\begin{align*}b\end{align*}” and the hypotenuse is “\begin{align*}c\end{align*}”. A combination of three numbers that makes the Pythagorean Theorem true is called a Pythagorean triple.
Guided Practice
1. Find the missing side of the right triangle below.
2. What is the diagonal of a rectangle with sides 10 and \begin{align*}16 \sqrt{5}\end{align*}?
3. Determine if the triangle below is a right triangle.
1. Here, we are given the hypotenuse and a leg. Let’s solve for \begin{align*}b\end{align*}.
\begin{align*}7^2 + b^2 & = 14^2\\ 49 + b^2 & = 196\\ b^2 & = 147\\ b & = \sqrt{147} = \sqrt{7 \cdot 7 \cdot 3} =7 \sqrt{3}\end{align*}
2. For any square and rectangle, you can use the Pythagorean Theorem to find the length of a diagonal. Plug in the sides to find \begin{align*}d\end{align*}.
\begin{align*}10^2 + \left ( 16 \sqrt{5} \right )^2 & = d^2\\ 100 + 1280 & = d^2\\ 1380 & = d^2\\ d & = \sqrt{1380} = 2 \sqrt{345}\end{align*}
3. Check to see if the three lengths satisfy the Pythagorean Theorem. Let the longest sides represent \begin{align*}c\end{align*}, in the equation.
\begin{align*}a^2 + b^2 &= c^2\\ 22^2 + 24^2&= 26^2\\ 484 + 576 &= 676\\ 1060 &\neq 676\end{align*}
The triangle is not a right triangle.
Practice
Find the length of the missing side. Simplify all radicals.
1. If the legs of a right triangle are 10 and 24, then the hypotenuse is _____________.
2. If the sides of a rectangle are 12 and 15, then the diagonal is _____________.
3. If the legs of a right triangle are \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, then the hypotenuse is ____________.
4. If the sides of a square are 9, then the diagonal is _____________.
Determine if the following sets of numbers are Pythagorean Triples.
1. 12, 35, 37
2. 9, 17, 18
3. 10, 15, 21
4. 11, 60, 61
5. 15, 20, 25
6. 18, 73, 75
Pythagorean Theorem Proofs
The first proof below is similar to the one done earlier in this Concept. Use the picture below to answer the following questions.
1. Find the area of the square with sides \begin{align*}(a + b)\end{align*}.
2. Find the sum of the areas of the square with sides \begin{align*}c\end{align*} and the right triangles with legs \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.
3. The areas found in the previous two problems should be the same value. Set the expressions equal to each other and simplify to get the Pythagorean Theorem.
Major General James A. Garfield (and former President of the U.S) is credited with deriving this next proof of the Pythagorean Theorem using a trapezoid.
1. Find the area of the trapezoid using the trapezoid area formula: \begin{align*}A = \frac{1}{2} (b_1 + b_2)h\end{align*}
2. Find the sum of the areas of the three right triangles in the diagram.
3. The areas found in the previous two problems should be the same value. Set the expressions equal to each other and simplify to get the Pythagorean Theorem.
Vocabulary Language: English
Circle
Circle
A circle is the set of all points at a specific distance from a given point in two dimensions.
Conic
Conic
Conic sections are those curves that can be created by the intersection of a double cone and a plane. They include circles, ellipses, parabolas, and hyperbolas.
degenerate conic
degenerate conic
A degenerate conic is a conic that does not have the usual properties of a conic section. Since some of the coefficients of the general conic equation are zero, the basic shape of the conic is merely a point, a line or a pair of intersecting lines.
Ellipse
Ellipse
Ellipses are conic sections that look like elongated circles. An ellipse represents all locations in two dimensions that are the same distance from two specified points called foci.
hyperbola
hyperbola
A hyperbola is a conic section formed when the cutting plane intersects both sides of the cone, resulting in two infinite “U”-shaped curves.
Hypotenuse
Hypotenuse
The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Legs of a Right Triangle
Legs of a Right Triangle
The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
Parabola
Parabola
A parabola is the characteristic shape of a quadratic function graph, resembling a "U".
Pythagorean number triple
Pythagorean number triple
A Pythagorean number triple is a set of three whole numbers $a,b$ and $c$ that satisfy the Pythagorean Theorem, $a^2 + b^2 = c^2$.
Pythagorean Theorem
Pythagorean Theorem
The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
Right Triangle
Right Triangle
A right triangle is a triangle with one 90 degree angle. |
# Addition & Subtraction of Complex Numbers(With Examples)
This post is also available in: हिन्दी (Hindi)
A complex number is the combination of a real number and an imaginary number. It is of the form $a + ib$ and is usually represented by $z$. While adding two or more complex numbers, the real and imaginary parts of a complex number are added individually. Similarly, for the subtraction of complex numbers, the real and imaginary parts are subtracted separately.
While adding two or more complex numbers, you add real parts of complex numbers separately and similarly add imaginary parts of complex numbers separately.
The formula used to add two complex numbers $z_{1} = a + ib$ and $z_{2} = c + id$ is
$z_{1} + z_{2} = \left(a + ib \right) + \left(c + id \right) = \left(a + c \right) + \left(ib + id \right) = \left(a + c \right) + i \left(b + d \right)$.
Hence we have $\left(a + ib \right) + \left(c + id \right) = \left(a + c \right) + i\left(b + d \right)$.
### Steps for Adding Complex Numbers
Given below are the steps for adding complex numbers:
Step 1: Segregate the real and imaginary parts of the complex numbers
Step 2: Add the real parts of the complex numbers
Step 3: Add the imaginary parts of the complex numbers
Step 4: Write the final answer in $a + ib$ format
### Examples
Let’s consider some examples to understand the process of the addition of complex numbers.
Ex 1: Add $2 + 3i$ and $5 + 4i$.
Here, $z_{1} = 2 + 3i$ and $z_{2} = 5 + 4i$
The real parts of the two complex numbers are $2$ and $5$ and the imaginary parts of the two complex numbers are $3i$ and $i$.
On adding real parts we get $2 + 5 = 7$ and on adding the imaginary parts we get $3i + 4i = 7i$.
Therefore, $\left(2 + 3i \right) + \left(5 + 4i \right) = 7 + 7i$.
Ex 2: Add $4 + 9i$ and $2 + 3i$.
The two complex numbers are $z_{1} = a + ib = 4 + 9i$ and $z_{2} = c + id = 2 + 3i$.
So, $a = 4$, $b = 9$, $c = 2$, and $d = 3$
According to the formula, $\left(a + ib \right) + \left(c + id \right) = \left(a + c \right) + i\left(b + d \right)$
Therefore, $\left(4 + 9i \right) + \left(2 + 3i \right) = \left(4 + 2 \right) + i \left(9 + 3 \right) = 6 + 12i$.
Ex 3: Add $-2 + 7i$ and $6 – 5i$.
$\left(-2 + 7i \right) + \left(6 – 5i \right) = \left(-2 + 6 \right) + \left(7 + \left(-5 \right) \right)i = 4 + 2i$.
### Properties of Addition of Complex Numbers
Following are the properties of the addition of complex numbers:
• Closure Property: The sum of complex numbers is also a complex number. Hence, it holds the closure property.
• Commutative Property: The addition of complex numbers is commutative.
• Associative Property: The addition of complex numbers is associative.
• Additive Identity: $0$ is the additive identity of the complex numbers, i.e., for a complex number $z$, we have $z + 0 = 0 + z = z$.
• Additive Inverse: For a complex number $z$, the additive inverse in complex numbers is $-z$, i.e., $z + \left(-z \right) = 0$.
## Subtraction of Complex Numbers
While subtracting two or more complex numbers, you subtract real parts of complex numbers separately and similarly subtract imaginary parts of complex numbers separately.
The formula used to subtract two complex numbers $z_{1} = a + ib$ and $z_{2} = c + id$ is |
How To Calculate The Area of a Semicircle
How To Calculate The Area of a Semicircle
What is the area of a semicircle with a radius $8\,cm$ ?
228.9k+ views
Hint: We know that when we cut the circle diametrically into two halves, one half is known as semicircle. This means that half of a circle is a semicircle and also the area of a semicircle is half that of a circle. So, we can use the formula $\dfrac{{\pi {r^2}}}{2}$ to find the area of semicircle, where $\pi {r^2}$ is the formula of circle whose radius is r.
Formula used: Area of semicircle $= \,\dfrac{{\pi {r^2}}}{2}$ .
A semicircle is a half-circle that is formed by cutting a whole circle into two halves along a diameter line. A line segment known as the diameter of a circle cuts the circle into exactly two equal semicircles.
Hence, the area of a semicircle is half the area of a circle.
The area of a semicircle or the area of a half circle is $\dfrac{{\pi {r^2}}}{2}$, where r is the radius of the semicircle.
Therefore,
Area of semicircle $= \,\dfrac{{\pi {r^2}}}{2}$
Now, putting the value of r from the given question as $r = 8\,cm$.
Area of semicircle $= \,\dfrac{{\pi {{\left( 8 \right)}^2}}}{2}$
Now putting the value of $\pi = 3.14$
$\Rightarrow \dfrac{{3.14{{\left( 8 \right)}^2}}}{2}$
$\Rightarrow \dfrac{{3.14 \times 64}}{2}$
$\Rightarrow 3.14 \times 32$
$\therefore 100.48\,c{m^2}$
Hence, the area of the semicircle is $100.48\,c{m^2}$.
Note: The semicircle is also referred to as a half-disk. Since the semicircle is half of the circle ($360$ degrees), the arc of the semicircle always measures $180$ degrees. The perimeter of a semicircle is $\pi r + 2r$, which can also be written as $r\left( {\pi + 2} \right)$ by factoring out r.
Formula used: Area of semicircle $= \,\dfrac{{\pi {r^2}}}{2}$ .
A semicircle is a half-circle that is formed by cutting a whole circle into two halves along a diameter line. A line segment known as the diameter of a circle cuts the circle into exactly two equal semicircles.
Hence, the area of a semicircle is half the area of a circle.
The area of a semicircle or the area of a half circle is $\dfrac{{\pi {r^2}}}{2}$, where r is the radius of the semicircle.
Therefore,
Area of semicircle $= \,\dfrac{{\pi {r^2}}}{2}$
Now, putting the value of r from the given question as $r = 8\,cm$.
Area of semicircle $= \,\dfrac{{\pi {{\left( 8 \right)}^2}}}{2}$
Now putting the value of $\pi = 3.14$
$\Rightarrow \dfrac{{3.14{{\left( 8 \right)}^2}}}{2}$
$\Rightarrow \dfrac{{3.14 \times 64}}{2}$
$\Rightarrow 3.14 \times 32$
$\therefore 100.48\,c{m^2}$
Hence, the area of the semicircle is $100.48\,c{m^2}$.
Note: The semicircle is also referred to as a half-disk. Since the semicircle is half of the circle ($360$ degrees), the arc of the semicircle always measures $180$ degrees. The perimeter of a semicircle is $\pi r + 2r$, which can also be written as $r\left( {\pi + 2} \right)$ by factoring out r.
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Reduce / leveling any portion to that lowest terms by making use of our portion to the Simplest form Calculator. Uncover the prize to questions like: What is 10/15 in simplest form or what is 10/15 reduced to the most basic form?
### Fractions Simplifier
Please to fill in the two left boxes below:
Inputfraction: Integerpart: Fractionpart: As a Decimal: = = Details:Details...You are watching: What is 10/15 in simplest form
## How to mitigate a fraction
Among different ways simple a fraction, we will present the two procedure below:
### Method 1 - divide by a little Number once Possible
Start by dividing both the numerator and the denomiator the the portion by the very same number, and also repeat this till it is difficult to divide. Begin dividing by little numbers favor 2, 3, 5, 7. Because that example,
### Simplify the fraction 42/98
First division both (numerator/denominator) by 2 to gain 21/49.Dividing by 3 and 5 will certainly not work, so,Divide both numerator and denominator by 7 to acquire 3/7. Note: 21 ÷ 7 = 3 and also 49 ÷ 7 = 7
In the fraction 3/7, 3 is only divisible through itself, and also 7 is no divisible by other numbers than itself and also 1, so the fraction has been simplified as much as possible. No additional reduction is possible, therefore 42/98 is equal to 3/7 when reduced to its shortest terms. This is a PROPER fraction once the absolute value of the top number or molecule (3) is smaller sized than the absolute worth of the bottom number or denomintor (7).
### Method 2 - Greatest typical Divisor
To mitigate a portion to lowest state (also called its most basic form), just divide both the numerator and also denominator by the GCD (Greatest common Divisor).
For example, 3/4 is in shortest form, yet 6/8 is not in lowest type (the GCD that 6 and 8 is 2) and also 6/8 deserve to be composed as 3/4. You can do this because the value of a fraction will remain the same as soon as both the numerator and denominator are divided by the same number.
Note: The Greatest usual Factor (GCF) because that 6 and 8, notation gcf(6,8), is 2. Explanation:
Factors that 6 space 1,2,3,6;Factors of 8 room 1,2,4,8.
See more: What Is A Man Called When His Wife Dies ? How Does Social Security Work When A Spouse Dies
So, the is ease see that the "Greatest typical Factor" or "Divisor" is 2 since it is the greatest number i beg your pardon divides same into every one of them. |
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# 15 math words and phrases that every student should know
Updated: Dec 28, 2019
When it comes to applying mathematics concepts studied in class to the real world situations, many students feel they mostly get confused when translating math concepts and formulas into the corresponding language that we use every day but do not make immediate conscious connection to the math behind it. We just “know” it means this or that, without thinking what it would look like in terms of math formulas, expressions or equations.
Let’s take a look at the words and phrases that will improve every student’s comprehension of math concepts when applied to real life situations.
## 1. Two more than three
When one quantity is greater than the other by a certain amount, we use the word “more” and it represents addition.
3 + 2 = 5
## 2. Three less than five
When one quantity is smaller than the other by a certain amount, we use the word “less” and it represents subtraction.
5 - 3 = 2
## 3. Two times (twice) eleven
When one quantity is double the other, we use “two times” or “twice”, which represents multiplication by two.
11 x 2 = 22
## 4. Half a dozen
Half means division of a given value by two.
12 / 2 = 6
## 5. Variable
We use variables to represent any number. Variables are letters of the alphabet used in mathematical formulas, expressions and equations. Variables are also used in coding. Online math games would not exist if their creators did not know how to operate the variables.
The word “variable” is derived from the verb “to vary” which means “to change depending on the situation”.
5m + 3n
## 6. Expression
A mathematical (algebraic) expression is a combination of numbers and variables put together with the help of operations into one mathematical sentence. Algebraic expressions are often used to represent certain quantities.
Expressions are different from equations.
We do not “solve” expressions and they do not contain an equal sign.
25x + 3y
## 7. Equation
An equation is an equality of two algebraic expressions containing a variable, which is the unknown. The purpose of solving equation is to determine the unknown value by relating two quantities to each other in a meaningful way. In Canada, more attention is given to solving linear equations in grade 7 math and grade 8 math courses.
3m + 1 = 2 (m - 4)
## 8. Difference/sum of the two numbers
Difference is the result of subtracting one value from another.
Sum is the result of adding the two values.
These words are frequently used in grade 6 EQAO test, as well as online mental math practice (Prodigy Game).
3 - 2 = 1 3 + 2 = 5
## 9. Product/quotient of the two numbers
Product is the result of multiplying the two values.
Quotient is the result of dividing one value by another.
2 x 3 = 6 6 / 3 = 2
## 10. Rate
A rate is a ratio between two related quantities with different units.
Rate is frequently used when applying math to real world situations.
The most common example of rate is Speed (km/h).
Khan Academy has a number of free math videos involving rate.
## 11. To prove
Sometimes we are being asked to demonstrate mathematically (frequently by the method of substitution) that something makes sense. This means “to prove” something. When teaching middle school, it is important that teachers introduce this word regularly in math classes.
## 12. Odd/even
Odd numbers are whole numbers that start with a 1 and continue up or down the number line skipping every other number.
Even numbers are whole numbers that start with a 2 and continue up or down the number line skipping every other number.
1, 3, 5,..... 2, 4, 6, 8,......
## 13. To substitute
When applying math concepts to real life situations we often need to represent numbers as variables or one expression in terms of another expression. The process of replacing a variable with a number or replacing one part of the expression/equation with another is called “substitution”.
### If the cab traveled 20 km, what is the cost of the trip?
Substitute 20 in place of d and evaluate.
## 14. Dimensions
Length, width and height are dimensions.
When we are being asked to find the dimensions, we are being asked to determine the length, width or height depending on the shape.
## 15. Complete solution
When trying to figure out a real-life mathematical problem, it is important to record the solution in a well structured and organized manner, include all required steps and statements.
Such components as “let” and “therefore” statements, equations, formulas and diagrams make the solution complete.
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# Question Video: Applications of the Counting Principle (Product Rule) Mathematics • 12th Grade
Use the fundamental counting principle to determine the total number of outcomes of picking out an outfit from 4 shirts, 8 pairs of pants, and 2 jackets.
01:39
### Video Transcript
Use the fundamental counting principle to determine the total number of outcomes of picking out an outfit from four shirts, eight pairs of pants, and two jackets.
Now we have to remember what the fundamental counting principle says. If event ๐ has ๐ possible outcomes and event ๐ has ๐ possible outcomes, then event ๐ followed by event ๐ has ๐ times ๐ possible outcomes.
In this problem, we have three separate events. First, you need to choose your shirt, then a pair of pants, and then your jacket. That means that weโll need to multiply our choices for shirts, our choices for a pair of pants, and our jackets.
How many possible outcomes are there for the choice of what shirt to wear? Four, you have four shirts to choose from, so there are four possible outcomes.
The same goes for the pair of pants. How many possible outcomes? Eight, eight choices, eight possible outcomes. And finally our jacket choices, there are two.
Once we multiply all these things together, weโre left with 64. The total number of outcomes in picking four shirts, eight pair of pants, and two jackets is 64. |
Then draw a perpendicular from one of the vertices of the triangle to the opposite base. Here’s How to Think About It. The side opposite the 30º angle is the shortest and the length of it is usually labeled as $$x$$, The side opposite the 60º angle has a length equal to $$x\sqrt3$$, º angle has the longest length and is equal to $$2x$$, In any triangle, the angle measures add up to 180º. 6. Triangle BDC has two angle measures marked, 90º and 60º, so the third must be 30º. Therefore, side b will be 5 cm. In an equilateral triangle each side is s , and each angle is 60°. If we look at the general definition - tan x=OAwe see that there are three variables: the measure of the angle x, and the lengths of the two sides (Opposite and Adjacent).So if we have any two of them, we can find the third.In the figure above, click 'reset'. Whenever we know the ratios of the sides, we can solve the triangle by the method of similar figures. The best way to commit the 30-60-90 triangle to memory is to practice using it in problems. This implies that graph of cotangent function is the same as shifting the graph of the tangent function 90 degrees to the right. This is a triangle whose three angles are in the ratio 1 : 2 : 3 and respectively measure 30° (π / 6), 60° (π / 3), and 90° (π / 2).The sides are in the ratio 1 : √ 3 : 2. One is the 30°-60°-90° triangle. Next Topic: The Isosceles Right Triangle. Solve this equation for angle x: Problem 7. Draw the equilateral triangle ABC. For any problem involving a 30°-60°-90° triangle, the student should not use a table. We will prove that below. Therefore, each side will be multiplied by . This implies that BD is also half of AB, because AB is equal to BC. Since the right angle is always the largest angle, the hypotenuse is always the longest side using property 2. Normally, to find the cosine of an angle we’d need the side lengths to find the ratio of the adjacent leg to the hypotenuse, but we know the ratio of the side lengths for all 30-60-90 triangles. (the right angle). The side corresponding to 2 has been divided by 2. If an angle is greater than 45, then it has a tangent greater than 1. Word problems relating guy wire in trigonometry. From here, we can use the knowledge that if AB is the hypotenuse and has a length equal to $$12$$, then AD is the shortest side and is half the length of the hypotenuse, or $$6$$. Problem 2. . Answer. Imagine we didn't know the length of the side BC.We know that the tangent of A (60°) is the opposite side (26) divided by the adjacent side AB - the one we are trying to find. Sine, cosine, and tangent all represent a ratio of the sides of a triangle based on one of the angles, labeled theta or $$\theta$$. Evaluate sin 60° and tan 60°. Colleges with an Urban Studies Major, A Guide to the FAFSA for Students with Divorced Parents. Sign up to get started today. While we can use a geometric proof, it’s probably more helpful to review triangle properties, since knowing these properties will help you with other geometry and trigonometry problems. One Time Payment $10.99 USD for 2 months: Weekly Subscription$1.99 USD per week until cancelled: Monthly Subscription $4.99 USD per month until cancelled: Annual Subscription$29.99 USD per year until cancelled \$29.99 USD per year until cancelled For any angle "θ": (Sine, Cosine and Tangent are often abbreviated to sin, cos and tan.) Prove: The area A of an equilateral triangle inscribed in a circle of radius r, is. Because the ratio of the sides is the same for every 30-60-90 triangle, the sine, cosine, and tangent values are always the same, especially the following two, which are used often on standardized tests: As part of our free guidance platform, our Admissions Assessment tells you what schools you need to improve your SAT score for and by how much. THERE ARE TWO special triangles in trigonometry. If line BD intersects line AC at 90º, then the lines are perpendicular, making Triangle BDA another 30-60-90 triangle. Solve the right triangle ABC if angle A is 60°, and side c is 10 cm. From here, we can use the knowledge that if AB is the hypotenuse and has a length equal to $$12$$, then AD is the shortest side and is half the length of the hypotenuse, or $$6$$. The student should draw a similar triangle in the same orientation. Problem 1. Learn to find the sine, cosine, and tangent of 45-45-90 triangles and also 30-60-90 triangles. As for the cosine, it is the ratio of the adjacent side to the hypotenuse. The adjacent leg will always be the shortest length, or $$1$$, and the hypotenuse will always be twice as long, for a ratio of $$1$$ to $$2$$, or $$\frac{1}{2}$$. , then the lines are perpendicular, making Triangle BDA another 30-60-90 triangle. To cover the answer again, click "Refresh" ("Reload"). Since the triangle is equilateral, it is also equiangular, and therefore the the angle at B is 60°. BEGIN CONTENT Introduction From the 30^o-60^o-90^o Triangle, we can easily calculate the sine, cosine, tangent, cosecant, secant, and cotangent of 30^o and 60^o. This is often how 30-60-90 triangles appear on standardized tests—as a right triangle with an angle measure of 30º or 60º and you are left to figure out that it’s 30-60-90. First, we can evaluate the functions of 60° and 30°. Solving expressions using 45-45-90 special right triangles . Similarly for angle B and side b, angle C and side c. Example 3. How was it multiplied? Join thousands of students and parents getting exclusive high school, test prep, and college admissions information. How do we know that the side lengths of the 30-60-90 triangle are always in the ratio $$1:\sqrt3:2$$ ? The other is the isosceles right triangle. The base angle, at the lower left, is indicated by the "theta" symbol (θ, THAY-tuh), and is equa… Therefore AP is two thirds of the whole AD. If ABC is a right triangle with right angle C, and angle A = , then BC is the "opposite side", AC is the "adjacent side", and AB is the hypotenuse. The lengths of the sides of this triangle are 1, 2, √3 (with 2 being the longest side, the hypotenuse. Before we come to the next Example, here is how we relate the sides and angles of a triangle: If an angle is labeled capital A, then the side opposite will be labeled small a. 30 60 90 triangle rules and properties. (Topic 2, Problem 6.). Because the. Therefore, on inspecting the figure above, cot 30° =, Therefore the hypotenuse 2 will also be multiplied by. So that’s an important point. THERE ARE TWO special triangles in trigonometry. What is special about 30 60 90 triangles is that the sides of the 30 60 90 triangle always have the same ratio. The square drawn on the height of an equalateral triangle is three fourths of the square drawn on the side. This means that all 30-60-90 triangles are similar, and we can use this information to solve problems using the similarity. To solve a triangle means to know all three sides and all three angles. Create a right angle triangle with angles of 30, 60, and 90 degrees. Sign up for your CollegeVine account today to get a boost on your college journey. Here are a few triangle properties to be aware of: In addition, here are a few triangle properties that are specific to right triangles: Based on this information, if a problem says that we have a right triangle and we’re told that one of the angles is 30º, we can use the first property listed to know that the other angle will be 60º. ABC is an equilateral triangle whose height AD is 4 cm. Solution. Taken as a whole, Triangle ABC is thus an equilateral triangle. Therefore, side nI>a must also be multiplied by 5. Want access to expert college guidance — for free? To double check the answer use the Pythagorean Thereom: The other sides must be $$7\:\cdot\:\sqrt3$$ and $$7\:\cdot\:2$$, or $$7\sqrt3$$ and $$14$$. We could just as well call it . Normally, to find the cosine of an angle we’d need the side lengths to find the ratio of the adjacent leg to the hypotenuse, but we know the ratio of the side lengths for all 30-60-90 triangles. sin 30° is equal to cos 60°. For trigonometry problems: knowing the basic definitions of sine, cosine, and tangent make it very easy to find the value for these of any 30-60-90 triangle. Now we'll talk about the 30-60-90 triangle. And of course, when it’s exactly 45 degrees, the tangent is exactly 1. We are given a line segment to start, which will become the hypotenuse of a 30-60-90 right triangle. The student should sketch the triangle and place the ratio numbers. Theorem. 30-60-90 Right Triangles. Therefore, each side must be divided by 2. It is based on the fact that a 30°-60°-90° triangle is half of an equilateral triangle. Triangles APE, BPD are conguent, and a 60 degree angle the!, on inspecting the figure above the long leg is the ratio 1: 2: degree with! To other profile factors, such as GPA and extracurriculars are actually two. For any Problem involving a 30°-60°-90° triangle the sides are also always the... That we are given a line segment to start, which will the... Can be quickly solved with this special right triangle side nI > a must also multiplied... Theorems 3 and 9 ) draw the straight line AD bisecting the measures... The long leg is the leg opposite the 90º a freelance writer specializing in education always to... A right triangle with … 30/60/90 right triangles the ratios of whose sides do!, the side CB angles to the base all 30-60-90 triangles join thousands students... Evaluate the functions of 60° and 30° talk about the 30-60-90 triangle BDA another 30-60-90 triangle a. The answer again, click Refresh '' ( Reload '' ) Orlando, Florida is. Side corresponding to the right angle b is 9.3 cm line segment to start, which will become the.... Florida, where she majored in Philosophy 18.6 cm, click Refresh '' ( ''... Be multiplied by be divided by another similar ; that is, radians... Or simply 5 cm, and x 90 degrees Studies Major, Guide! Refers to the opposite base side using property 2 measures of 30, and a 60 degree angle, longer... That graph of the vertices of the adjacent side to the angles is the of. Side r is 1 cm tangent of 90-x should be the same as the cotangent the! We take advantage of knowing those ratios and Tan. ) ( Topic 3 ), sin 30° equal... A 60 degree angle triangle BDC has two angle measures of 30 and! Up for your CollegeVine account today to get 30‑60‑90 triangle tangent boost on your college journey and x account discover. Chances, and angle calculator displays missing sides and all three angles a freelance writer specializing in education radius... To use triangle properties like the Pythagorean theorem ( with 2 being the longest side using property 2 be! In each equation, decide which of those angles is the straight line drawn from the vertex right... Of segment BC specializing in education longer leg is the longer leg of the triangle and place ratio! Are also always in the right triangle 5 × 1, 2, √3 ( with being... To was multiplied by 5 and b = 30‑60‑90 triangle tangent in triangles is that the 1!, sin 30° is equal to BC that graph of the hypotenuse, and of course, it..., such as GPA and extracurriculars shown on the new SAT, you can see cos! Well known special right triangle with angles of 30 to get a Perfect 1600 on. As for the definition of measuring angles by degrees, '' see Topic 12 currently lives in Orlando Florida. Angle measures marked, 90º and 60º, so we can easily figure out that this is proud! Two angle measures of 30, and side c. example 3 of 30°, and C. And 90º ( the right tangent is exactly 1 with 2 being the side... Fafsa for students with Divorced parents not know. ) pass your mouse the! Equilateral, it is the side corresponding to 2 has been multiplied by is... Of AD BDC has two angle measures marked, 90º and 30‑60‑90 triangle tangent, so the third must be 90.. Abc if angle a is 60°, then AD is the University of Central Florida, where she majored Philosophy... A boost on your college journey want access to expert college guidance — for free and. 90 triangles is that the side CB, or simply 5 cm, and side C 10... A right-angled triangle divided by 30‑60‑90 triangle tangent this equation for angle x: Problem 7 Score, in radians, {! Side nI > a must also be multiplied by BP, because triangles APE, BPD are conguent, each... √3 ( with 2 being the longest side, the sides opposite the 90º of 90-x be! Get expert admissions guidance — for free the FAFSA for students with Divorced parents are special because with. Evaluate the functions of 60° and 30° can know the ratios of their sides will multiplied. An area of a right triangle PQR, angle C and side q will be multiplied by 5 BPD conguent. Its complement, 30° using property 3, we can easily figure out that this is longer! The equal angles on standardized tests, this can save you time when solving problems expert admissions guidance for! In this type of special right triangle, the longer leg is the side angle. You are actually given the 30-60-90 30‑60‑90 triangle tangent Choice college Applications, List of all U.S SAT... 30°-60°-90° triangle: the 30°-60°-90° refers to the angles are 30 and degrees... In Philosophy that is, they all have their corresponding sides in ratio this is a proud mom... Particular group of triangles and also 30-60-90 triangles are one particular group of triangles and also 30-60-90 triangles similar and., cot 30° =, therefore the hypotenuse, and get expert admissions guidance — for free:... Half its hypotenuse the remaining angle b is 60°, the sides of the sides are also always in ratio! To use triangle properties like the Pythagorean theorem to show that the side adjacent 60°... Of radius r, is and so in triangle ABC has angle measures, so we can this. Basic 30-60-90 triangle on the new SAT, you can see that 60°... Always have the same ratio are each 60. side c. example 3 Every SAT practice test + free! Their base angles sin, cos and Tan. ) 9.3 cm,! For free Perfect 1600 Score on the diagram, we can use this information to solve: we ’ only. Be divided by 2 c. example 3 Topic 6, we ’ re given.... Same orientation is a graduate of the sides of this fact is clear using trigonometry.The geometric proof:! Angle P is 30°, and angle a is 60°, and side q will be multiplied by.. Of 30º, 60º, so the third must be 30º in action, we 30‑60‑90 triangle tangent this because the angles. Start with an equilateral triangle, click Refresh '' ( Reload '' ), because AB is to. Reload '' ) θ '': ( sine, cosine and tangent of 90-x be! And angles you time when solving problems θ '': ( sine, and! You can see that the side adjacent to 60° is always half AB. Degrees to the FAFSA for students with Divorced parents use triangle properties like the Pythagorean theorem show... Good, Bad, and college admissions information: Problem 8 chancing engine takes into your... Leg of the adjacent side to the right college guidance — for free involving a 30°-60°-90°:. Degrees triangle this equation for angle x: Problem 8 mouse over the colored area the corresponding... Clear using trigonometry.The geometric proof is: a is 60°, then the lines are perpendicular, making triangle another! Triangles are one particular group of triangles and one specific kind of right triangle is half its hypotenuse the. Is called the hypotenuse, you can see that directly in the ratio numbers a... Inspecting the figure above know that we are looking at two 30-60-90 triangles are actually given 30-60-90. Most well known special right triangle side and angle calculator displays missing and. Are from the vertex at right angles to the property of cofunctions ( Topic 3,... Angle P is 30°, and the hypotenuse is 18.6 cm 30‑60‑90 triangle tangent problems an Studies... ( for the cosine, and side a will be in the ratio \ (:... To discover your chances at hundreds of different schools ABC above, what is a proud mom. Geometry, we know that one of the University of Michigan Ann Acceptance. Factors, such as GPA and extracurriculars own unique website with customizable templates theorem )... Will solve right triangles this type of special right triangle with angles of 30 multiplied 5. 90 triangles is that 30‑60‑90 triangle tangent side corresponding to 2 has been divided by 2 their corresponding sides in.. Specific kind of right triangle has a tangent greater than 1 is also half of 30‑60‑90 triangle tangent equilateral triangle splits into! Solve 30‑60‑90 triangle tangent triangle always have the same ratio to each other AD bisecting angle. Opposite base the lengths of the opposite side to the opposite base cosine right this! Shifting the graph of cotangent function is the 30-60-90 triangle are 1, 2, √3 ( 2. The length of AD angle measure, we can easily figure out that this is a triangle. To discover your chances, and side b 30‑60‑90 triangle tangent the ratio of the opposite.! Those ratios adjacent to 60° is always the largest angle, the side adjacent to 60° is always largest. A Perfect 1600 Score on the new SAT, you can see that cos 60° the fact a... Since it ’ s a right triangle has a tangent greater than 1 shifting the graph of cotangent function the! Of 30 known special right triangle side and angle a is 60°, b... Our 30-60-90 degrees triangle Score on the height of an equilateral triangle ABC is thus an equilateral.! Specializing in education three radii divide the triangle into three congruent triangles student sketch... Schools, understand your chances at hundreds of different schools then are angles!
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# Adding and Subtracting Mixed Numbers
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CCSS.MATH.CONTENT.4.NF.B.3.C CCSS.MATH.CONTENT.4.NF.B.3.D
## Mixed Numbers – Adding and Subtracting
Have you ever tried making slime at home? It is a lot of fun!
For the special slime we want to make today, the recipe is difficult to read, as it contains mixed numbers. In order to figure out how much to use of each ingredient, we need to learn and practice adding and subtracting mixed numbers. Not to worry – the following text teaches us everything we need to know.
### Steps to Adding and Subtracting Mixed Numbers
A mixed number is a combination of a whole number and a fraction. When you practice adding and subtracting mixed numbers with like denominators, you can follow these three steps:
Step # What to do
1 Add or subtract the whole numbers.
2 Add or subtract the fractions.
## Adding and Subtracting Mixed Numbers – Examples
The first ingredient on the recipe is glue. It says to add three and two-eighths plus four and five-eighths. What is that in total? Let’s start by adding three and two-eighths and four and five-eighths.
First, we need to find the sum of the whole numbers. That means, we are adding three and four first to get the sum of seven.
Next, we find the sum of the fractions.
Remember: When adding and subtracting fractions with common denominators, we simply add or subtract the numerators and the denominators stay the same. Here, this means two-eighths plus five-eighths equals seven-eighths.
Our last step is to simplify our answer if we need to. A fraction is in simplest form when the numerator and denominator have no common factors other than one. Since seven and seven-eighths is in simplest form we can skip this step.
What is the next ingredient on the slime recipe? Now, we have to add five and three-fourths minus four and one-fourth of baking soda. So let’s practice subtracting mixed numbers by calculating five and three-fourths minus four and one-fourth.
First, we need to find the difference of the whole numbers. That means, we are subtracting five and four first to get the difference of one.
Next, we find the difference of the fractions. Three-fourths minus one-fourth equals two-fourths.
Our last step is again to simplify our answer if we need to. The numerator and the denominator have the common factor of two, which means it can be simplified to one-half.
## How to Add and Subtract Mixed Numbers – Summary
Remember to follow the three steps below whenever you add and subtract mixed numbers with like denominators.
• Step 1: Find the sum or the difference of the whole numbers.
• Step 2: Find the sum or the difference of the fractions.
The last step in our recipe is to add one and one-third plus two and two-thirds of contact lense solution. Can you calculate how much that is in total?
The answer is... four! Well done. If you would like to practice some more to optimize your slime, take a look at the exercises and worksheets about adding and subtracting mixed numbers. Enjoy!
## Adding and Subtracting Mixed Numbers exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Adding and Subtracting Mixed Numbers.
• ### Can you find the mixed numbers?
Hints
Is it made up of a whole number and a fraction?
Remember, a mixed number is made up of a whole number and a fraction.
Solution
These are the mixed numbers! Remember, a mixed number is made up of a whole number and a fraction.
Not mixed numbers
5 and 13 are whole numbers.
$\frac{9}{10}$ and $\frac{5}{7}$ are fractions.
• ### Can you sequence the number sentence correctly?
Hints
This is a subtraction problem so the order is important.
Have a look at this fraction. If you divide both the numerator and the denominator by 2, what do you get?
Solution
This is how the number sentence should read.
First of all we subtract the whole numbers so 3 - 1 = 2.
We then subtract the fractions so $\frac{3}{4}$ - $\frac{1}{4}$ = $\frac{2}{4}$.
We then simplify if we need to which in this case we can. $\frac{2}{4}$ can be simplified to $\frac{1}{2}$ by dividing both the numerator and the denominator by two.
We then put the whole number and the fraction together to get our answer of 2$\frac{1}{2}$.
• ### Can you find out how much of each ingredient Axel and Tank need to make gloop?
Hints
Can it be simplified?
Add or subtract the whole numbers first.
Don't forget to double check if it is an addition or subtraction problem.
Add or subtract the fractions next.
Solution
These are the correct answers. Remember to:
• Check whether it is an addition or subtraction problem.
• Add or subtract the whole numbers first.
• Add or subtract the fractions next.
• Simplify if you can.
• ### Can you find the equivalent fractions?
Hints
Remember, you always divide the numerator and the denominator by the same number.
In this example we can simplify $\frac{3}{6}$ to $\frac{1}{2}$ by dividing both the numerator and denominator by 3.
What do you get if you half the denominator on the top row? Or divide it by 3?
Solution
In the example above both the numerator and the denominator have been divided by two to simplify the fraction.
You need to divide the denominator and the numerator by the same number to simplify the fraction. This may be by two or by three or by another number.
__________________________________________________________________
Using the information above, the answers to the other problems are:
• $5\frac{8}{10}$ = $5\frac{4}{5}$ - both divided by two.
• $5\frac{3}{9}$ = $5\frac{1}{3}$ - both divided by three.
• $5\frac{12}{16}$ = $5\frac{3}{4}$ - both divided by four.
• ### Can you find out how much shaving foam Axel and Tank need to make more slime?
Hints
Remember to add the whole numbers first.
When the denominators are the same, we only need to add the numerators.
Solution
If we add the whole numbers first we get 3.
Adding the fractions next gives us $\frac{2}{3}$; we cannot simplify this any further.
• ### Can you check Axel and Tank's alternative recipe for slime?
Hints
Remember to add or subtract the whole numbers first.
Add or subtract the fractions next.
Can the fraction be simplified?
Solution
• The first number sentence was incorrect.
The answer should be 9$\frac{3}{4}$.
3 + 6 = 9
$\frac{5}{8}$ + $\frac{1}{8}$ = $\frac{6}{8}$ which can be simplified to $\frac{3}{4}$.
We put the whole number and the fraction together to get the answer.
• The second problem was correct!
4 + 4 = 8
$\frac{7}{9}$ + $\frac{2}{9}$ = 1.
8 + 1 = 9.
• The third number sentence was also incorrect.
The answer should be 3$\frac{4}{7}$.
10 - 7 = 3.
$\frac{6}{7}$ - $\frac{2}{7}$ = $\frac{4}{7}$.
This cannot be simplified further so put the whole number and the fraction together to get the answer.
• The fourth number sentence was correct!
11 - 5 = 6.
$\frac{7}{8}$ - $\frac{3}{8}$ = $\frac{4}{8}$ which can be simplified to $\frac{1}{2}$.
Put the whole number and the fraction together to get the answer. |
College Algebra
# 9.6Binomial Theorem
College Algebra9.6 Binomial Theorem
## Learning Objectives
In this section, you will:
• Apply the Binomial Theorem.
A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find $(x+y) n (x+y) n$ without multiplying the binomial by itself $n n$ times.
## Identifying Binomial Coefficients
In Counting Principles, we studied combinations. In the shortcut to finding $(x+y) n , (x+y) n ,$ we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation $( n r ) ( n r )$ instead of $C(n,r), C(n,r),$ but it can be calculated in the same way. So
$( n r )=C(n,r)= n! r!(n−r)! ( n r )=C(n,r)= n! r!(n−r)!$
The combination $( n r ) ( n r )$ is called a binomial coefficient. An example of a binomial coefficient is $( 5 2 )=C(5,2)=10. ( 5 2 )=C(5,2)=10.$
## Binomial Coefficients
If $n n$ and $r r$ are integers greater than or equal to 0 with $n≥r, n≥r,$ then the binomial coefficient is
$( n r )=C(n,r)= n! r!(n−r)! ( n r )=C(n,r)= n! r!(n−r)!$
## Q&A
Is a binomial coefficient always a whole number?
Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.
## Example 1
### Finding Binomial Coefficients
Find each binomial coefficient.
1. $( 5 3 ) ( 5 3 )$
2. $( 9 2 ) ( 9 2 )$
3. $( 9 7 ) ( 9 7 )$
### Analysis
Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.
$( n r )=( n n−r ) ( n r )=( n n−r )$
## Try It #1
Find each binomial coefficient.
1. $( 7 3 ) ( 7 3 )$
2. $( 11 4 ) ( 11 4 )$
## Using the Binomial Theorem
When we expand $(x+y) n (x+y) n$ by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand $(x+y) 52 , (x+y) 52 ,$ we might multiply $(x+y) (x+y)$ by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.
$(x+y) 2 = x 2 +2xy+ y 2 (x+y) 3 = x 3 +3 x 2 y+3x y 2 + y 3 (x+y) 4 = x 4 +4 x 3 y+6 x 2 y 2 +4x y 3 + y 4 (x+y) 2 = x 2 +2xy+ y 2 (x+y) 3 = x 3 +3 x 2 y+3x y 2 + y 3 (x+y) 4 = x 4 +4 x 3 y+6 x 2 y 2 +4x y 3 + y 4$
First, let’s examine the exponents. With each successive term, the exponent for $x x$ decreases and the exponent for $y y$ increases. The sum of the two exponents is $n n$ for each term.
Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:
$( n 0 ),( n 1 ),( n 2 ),...,( n n ). ( n 0 ),( n 1 ),( n 2 ),...,( n n ).$
These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.
$(x+y) n = ∑ k=0 n ( n k ) x n−k y k = x n +( n 1 ) x n−1 y+( n 2 ) x n−2 y 2 +...+( n n−1 )x y n−1 + y n (x+y) n = ∑ k=0 n ( n k ) x n−k y k = x n +( n 1 ) x n−1 y+( n 2 ) x n−2 y 2 +...+( n n−1 )x y n−1 + y n$
Another way to see the coefficients is to examine the expansion of a binomial in general form, $x+y, x+y,$ to successive powers 1, 2, 3, and 4.
$(x+y) 1 =x+y (x+y) 2 = x 2 +2xy+ y 2 (x+y) 3 = x 3 +3 x 2 y+3x y 2 + y 3 (x+y) 4 = x 4 +4 x 3 y+6 x 2 y 2 +4x y 3 + y 4 (x+y) 1 =x+y (x+y) 2 = x 2 +2xy+ y 2 (x+y) 3 = x 3 +3 x 2 y+3x y 2 + y 3 (x+y) 4 = x 4 +4 x 3 y+6 x 2 y 2 +4x y 3 + y 4$
Can you guess the next expansion for the binomial $(x+y) 5 ? (x+y) 5 ?$
Figure 1
See Figure 1, which illustrates the following:
• There are $n+1 n+1$ terms in the expansion of $(x+y) n . (x+y) n .$
• The degree (or sum of the exponents) for each term is $n. n.$
• The powers on $x x$ begin with $n n$ and decrease to 0.
• The powers on $y y$ begin with 0 and increase to $n. n.$
• The coefficients are symmetric.
To determine the expansion on $(x+y) 5 , (x+y) 5 ,$ we see $n=5, n=5,$ thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of $x, x,$ the pattern is as follows:
• Introduce $x 5 , x 5 ,$ and then for each successive term reduce the exponent on $x x$ by 1 until $x 0 =1 x 0 =1$ is reached.
• Introduce $y 0 =1, y 0 =1,$ and then increase the exponent on $y y$ by 1 until $y 5 y 5$ is reached.
$x 5 , x 4 y, x 3 y 2 , x 2 y 3 ,x y 4 , y 5 x 5 , x 4 y, x 3 y 2 , x 2 y 3 ,x y 4 , y 5$
The next expansion would be
$(x+y) 5 = x 5 +5 x 4 y+10 x 3 y 2 +10 x 2 y 3 +5x y 4 + y 5 . (x+y) 5 = x 5 +5 x 4 y+10 x 3 y 2 +10 x 2 y 3 +5x y 4 + y 5 .$
But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle, shown in Figure 2.
Figure 2
To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3rd row, flank the ends of the rows with 1’s, and add $1+1 1+1$ to find the middle number, 2. In the $nth nth$ row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.
To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.
## The Binomial Theorem
The Binomial Theorem is a formula that can be used to expand any binomial.
$(x+y) n = ∑ k=0 n ( n k ) x n−k y k = x n +( n 1 ) x n−1 y+( n 2 ) x n−2 y 2 +...+( n n−1 )x y n−1 + y n (x+y) n = ∑ k=0 n ( n k ) x n−k y k = x n +( n 1 ) x n−1 y+( n 2 ) x n−2 y 2 +...+( n n−1 )x y n−1 + y n$
## How To
Given a binomial, write it in expanded form.
1. Determine the value of $n n$ according to the exponent.
2. Evaluate the $k=0 k=0$ through $k=n k=n$ using the Binomial Theorem formula.
3. Simplify.
## Example 2
### Expanding a Binomial
Write in expanded form.
1. $(x+y) 5 (x+y) 5$
2. $( 3x−y ) 4 ( 3x−y ) 4$
### Analysis
Notice the alternating signs in part b. This happens because $(−y) (−y)$ raised to odd powers is negative, but $(−y) (−y)$ raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.
## Try It #2
Write in expanded form.
1. $(x−y) 5 (x−y) 5$
2. $(2x+5y) 3 (2x+5y) 3$
## Using the Binomial Theorem to Find a Single Term
Expanding a binomial with a high exponent such as $(x+2y) 16 (x+2y) 16$ can be a lengthy process.
Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.
Note the pattern of coefficients in the expansion of $(x+y) 5 . (x+y) 5 .$
$(x+y) 5 = x 5 +( 5 1 ) x 4 y+( 5 2 ) x 3 y 2 +( 5 3 ) x 2 y 3 +( 5 4 )x y 4 + y 5 (x+y) 5 = x 5 +( 5 1 ) x 4 y+( 5 2 ) x 3 y 2 +( 5 3 ) x 2 y 3 +( 5 4 )x y 4 + y 5$
The second term is $( 5 1 ) x 4 y. ( 5 1 ) x 4 y.$ The third term is $( 5 2 ) x 3 y 2 . ( 5 2 ) x 3 y 2 .$ We can generalize this result.
$( n r ) x n−r y r ( n r ) x n−r y r$
## The (r+1)th Term of a Binomial Expansion
The $(r+1)th (r+1)th$ term of the binomial expansion of $(x+y) n (x+y) n$ is:
$( n r ) x n−r y r ( n r ) x n−r y r$
## How To
Given a binomial, write a specific term without fully expanding.
1. Determine the value of $n n$ according to the exponent.
2. Determine $(r+1). (r+1).$
3. Determine $r. r.$
4. Replace $r r$ in the formula for the $(r+1)th (r+1)th$ term of the binomial expansion.
## Example 3
### Writing a Given Term of a Binomial Expansion
Find the tenth term of $(x+2y) 16 (x+2y) 16$ without fully expanding the binomial.
## Try It #3
Find the sixth term of $(3x−y) 9 (3x−y) 9$ without fully expanding the binomial.
## Media
Access these online resources for additional instruction and practice with binomial expansion.
## 9.6 Section Exercises
### Verbal
1.
What is a binomial coefficient, and how it is calculated?
2.
What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?
3.
What is the Binomial Theorem and what is its use?
4.
When is it an advantage to use the Binomial Theorem? Explain.
### Algebraic
For the following exercises, evaluate the binomial coefficient.
5.
$( 6 2 ) ( 6 2 )$
6.
$( 5 3 ) ( 5 3 )$
7.
$( 7 4 ) ( 7 4 )$
8.
$( 9 7 ) ( 9 7 )$
9.
$( 10 9 ) ( 10 9 )$
10.
$( 25 11 ) ( 25 11 )$
11.
$( 17 6 ) ( 17 6 )$
12.
$( 200 199 ) ( 200 199 )$
For the following exercises, use the Binomial Theorem to expand each binomial.
13.
$(4a−b) 3 (4a−b) 3$
14.
$(5a+2) 3 (5a+2) 3$
15.
$(3a+2b) 3 (3a+2b) 3$
16.
$(2x+3y) 4 (2x+3y) 4$
17.
$(4x+2y) 5 (4x+2y) 5$
18.
$(3x−2y) 4 (3x−2y) 4$
19.
$(4x−3y) 5 (4x−3y) 5$
20.
$( 1 x +3y ) 5 ( 1 x +3y ) 5$
21.
$( x −1 +2 y −1 ) 4 ( x −1 +2 y −1 ) 4$
22.
$( x − y ) 5 ( x − y ) 5$
For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.
23.
$(a+b) 17 (a+b) 17$
24.
$(x−1) 18 (x−1) 18$
25.
$(a−2b) 15 (a−2b) 15$
26.
$(x−2y) 8 (x−2y) 8$
27.
$(3a+b) 20 (3a+b) 20$
28.
$(2a+4b) 7 (2a+4b) 7$
29.
$( x 3 − y ) 8 ( x 3 − y ) 8$
For the following exercises, find the indicated term of each binomial without fully expanding the binomial.
30.
The fourth term of $(2x−3y) 4 (2x−3y) 4$
31.
The fourth term of $(3x−2y) 5 (3x−2y) 5$
32.
The third term of $(6x−3y) 7 (6x−3y) 7$
33.
The eighth term of $(7+5y) 14 (7+5y) 14$
34.
The seventh term of $(a+b) 11 (a+b) 11$
35.
The fifth term of $(x−y) 7 (x−y) 7$
36.
The tenth term of $(x−1) 12 (x−1) 12$
37.
The ninth term of $(a−3 b 2 ) 11 (a−3 b 2 ) 11$
38.
The fourth term of $( x 3 − 1 2 ) 10 ( x 3 − 1 2 ) 10$
39.
The eighth term of $( y 2 + 2 x ) 9 ( y 2 + 2 x ) 9$
### Graphical
For the following exercises, use the Binomial Theorem to expand the binomial $f(x)= (x+3) 4 . f(x)= (x+3) 4 .$ Then find and graph each indicated sum on one set of axes.
40.
Find and graph $f 1 (x), f 1 (x),$ such that $f 1 (x) f 1 (x)$ is the first term of the expansion.
41.
Find and graph $f 2 (x), f 2 (x),$ such that $f 2 (x) f 2 (x)$ is the sum of the first two terms of the expansion.
42.
Find and graph $f 3 (x), f 3 (x),$ such that $f 3 (x) f 3 (x)$ is the sum of the first three terms of the expansion.
43.
Find and graph $f 4 (x), f 4 (x),$ such that $f 4 (x) f 4 (x)$ is the sum of the first four terms of the expansion.
44.
Find and graph $f 5 (x), f 5 (x),$ such that $f 5 (x) f 5 (x)$ is the sum of the first five terms of the expansion.
### Extensions
45.
In the expansion of $(5x+3y) n , (5x+3y) n ,$ each term has the form $( n k ) a n–k b k ( n k ) a n–k b k$, where $k k$ successively takes on the value $0,1,2,...,n. 0,1,2,...,n.$ If $( n k )=( 7 2 ), ( n k )=( 7 2 ),$ what is the corresponding term?
46.
In the expansion of $( a+b ) n , ( a+b ) n ,$ the coefficient of $a n−k b k a n−k b k$ is the same as the coefficient of which other term?
47.
Consider the expansion of $(x+b) 40 . (x+b) 40 .$ What is the exponent of $b b$ in the $kth kth$ term?
48.
Find $( n k−1 )+( n k ) ( n k−1 )+( n k )$ and write the answer as a binomial coefficient in the form $( n k ). ( n k ).$ Prove it. Hint: Use the fact that, for any integer $p, p,$ such that $p≥1,p!=p(p−1)!. p≥1,p!=p(p−1)!.$
49.
Which expression cannot be expanded using the Binomial Theorem? Explain.
• $( x 2 −2x+1) ( x 2 −2x+1)$
• $( a +4 a −5) 8 ( a +4 a −5) 8$
• $( x 3 +2 y 2 −z) 5 ( x 3 +2 y 2 −z) 5$
• $(3 x 2 − 2 y 3 ) 12 (3 x 2 − 2 y 3 ) 12$
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# Year 2 Maths Objectives
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1 Year 2 Maths Objectives Counting Number - number and place value Count in steps of 2, 3, and 5 from 0, and in tens from any number, forward and backward Place Value Comparing and Ordering Read and write numbers to at least 100 in numerals and in words Recognise the place value of each digit in a two-digit number (tens, ones) Partition numbers in different ways (for example, 23 = and 23 = ) Identify, represent and estimate numbers using different representations, including the number line Compare and order numbers from 0 up to 100; use <, > and = signs Find 1 or 10 more or less than a given number Rounding, approximation and estimation Round numbers to at least 100 to the nearest 10 Multiplying by powers of 10 Understand the connection between the 10 multiplication table and place value Negative numbers Sequences and patterns Describe and extend simple sequences involving counting on or back in different steps Roman numerals Solving number problems Use place value and number facts to solve problems
2 Understanding addition and subtraction Addition and subtraction facts Mental methods Written methods Number - addition and subtraction Choose an appropriate strategy to solve a calculation based upon the numbers involved (recall a known fact, calculate mentally, use a jotting) Show that addition of two numbers can be done in any order (commutative) and subtraction of one number from another cannot Understand subtraction as take away and difference (how many more, how many less/fewer) Recall and use addition and subtraction facts to 20 fluently, and derive and use related facts up to 100 Recall and use number bonds for multiples of 5 totalling 60 (to support telling time to nearest 5 minutes) Select a mental strategy appropriate for the numbers involved in the calculation Add and subtract numbers using concrete objects, pictorial representations, and mentally, including: - a two-digit number and ones - a two-digit number and tens - two two-digit numbers - adding three one-digit numbers *Written methods are informal at this stage see mental methods for expectation of calculations Estimating and checking calculations Order of operations Recognise and use the inverse relationship between addition and subtraction and use this to check calculations and solve missing number problems Solving addition and subtraction problems including those with missing numbers Solve problems with addition and subtraction including those with missing numbers: - using concrete objects and pictorial representations, including those involving numbers, quantities and measures - applying their increasing knowledge of mental and written methods
3 Understanding multiplication and division Multiplication and division facts Mental methods Written methods Estimating and checking calculations Order of operations Solving multiplication and division problems including those with missing numbers Number - multiplication and division Understand multiplication as repeated addition Understand division as sharing and grouping and that a division calculation can have a remainder Show that multiplication of two numbers can be done in any order (commutative) and division of one number by another cannot Recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers Derive and use doubles of simple two-digit numbers (numbers in which the ones total less than 10) Derive and use halves of simple two-digit even numbers (numbers in which the tens are even) Calculate mathematical statements for multiplication (using repeated addition) and division within the multiplication tables and write them using the multiplication ( ), division ( ) and equals (=) signs *Written methods are informal at this stage see mental methods for expectation of calculations Solve problems involving multiplication and division (including those with remainders), using materials, arrays, repeated addition, mental methods, and multiplication and division facts, including problems in contexts Understanding fractions Fractions of objects, shapes and quantities Counting, comparing and ordering fractions Equivalence Number - fractions (including decimals and percentages) Understand and use the terms numerator and denominator Understand that a fraction can describe part of a set Understand that the larger the denominator is, the more pieces it is split Recognise, find, name and write fractions,, and of a length, shape, set of objects or quantity Count on and back in steps of and Write simple fractions for example, of 6 = 3 and recognise the Calculating with fractions Percentages Solving problems involving fractions, decimals and percentages equivalence of and Ratio and proportion Ratio and proportion Algebra Algebra
4 Length / height Perimeter Area Mass Measurement (length/height, perimeter, area and mass/weight) measure length/height in any direction (m/cm) to the nearest appropriate unit using rulers Compare and order lengths and record the results using >, < and = measure mass (kg/g) to the nearest appropriate unit using scales Compare and order mass and record the results using >, < and = Capacity / volume Temperature Conversion Measurement (capacity, volume, temperature and conversion) measure capacity and volume (litres/ml) to the nearest appropriate unit using measuring vessels Compare and order volume/capacity and record the results using >, < and = measure temperature to the nearest degree ( C) using thermometers Time Measurement (time) Compare and sequence intervals of time Know the number of minutes in an hour and the number of hours in a day Tell and write the time to five minutes, including quarter past/to the hour and draw the hands on a clock face to show these times Money Solving problems involving money and measures Measurement (money and solving problems) Recognise and use symbols for pounds ( ) and pence (p) Combine amounts to make a particular value Find different combinations of coins that equal the same amounts of money Add and subtract money of the same unit, including giving change Solve simple problems in a practical context involving addition and subtraction of money and measures (including time)
5 Properties of shape Angles and rotation Geometry - properties of shapes Identify and describe the properties of 2-D shapes, including the number of sides and line symmetry in a vertical line Identify 2-D shapes on the surface of 3-D shapes, (for example, a circle on a cylinder and a triangle on a pyramid) Identify and describe the properties of 3-D shapes, including the number of edges, vertices and faces Use mathematical vocabulary to describe movement, including rotation as a turn Understand the link between rotation and turns in terms of right angles for quarter, half and three- quarter turns (clockwise and anti-clockwise) Geometry - position and direction Patterns Order and arrange combinations of mathematical objects in patterns and sequences Position and direction Use mathematical vocabulary to describe position, movement, including movement in a straight line Coordinates (including reflection and translation) Statistics ics Sorting and classifying Compare and sort objects, numbers and common 2-D and 3-D shapes and everyday objects Present and interpret data Interpret and construct simple pictograms, tally charts, block diagrams and simple tables Solve problems using data Averages Ask and answer simple questions by counting the number of objects in each category and sorting the categories by quantity Ask and answer questions about totalling and comparing categorical data
### Provost Williams C.E. Primary School Maths Medium Term Plan
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### Year R Maths Objectives
Year R Maths Objectives In order to meet the Early Learning Goals at the end of Year R children must be able to: Numbers Count reliably with numbers from -0, place them in order and say which number is
### Year 2 Maths Objectives
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### Maths Year 2 Step 1 Targets Number and place value count in steps of 2 and 5 from 0; forwards and backwards. use number facts to solve problems
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## Tuesday, 19 August 2008
### How To Get Gradient and Intercept from Two Points
Gradient and intercept are two key items to a straight line expression.
In maths, to obtain the equation of a line from two given co-ordinates, we inevitably think of graph plotting. This is one good way to obtain the answer by finding the gradient and intercept.
Let's take an example.
2 points: (1, 5) and (3, 11) are given. What is the straight line expression?
By plotting these two points on a graph, we can easily determine the gradient and intercept, and then the mathematical expression for the equation.
Diag: Graph with 2 points.
From the graph, to determine the gradient, we can check:
• increase of the vertical unit with reference from the 2 points as 11 - 5 = 6 units, and
• increase of the horizontal unit from the 2 points as 3 - 1 = 2 units,
Gradient = change in vertical / change in horizontal = 6 / 2 = 3
The next item is the intercept, and directly from the graph, it showed the value to be 2.
Therefore, the straight line expression comes to y = 3 x + 2.
This graphical method is OK, simple and easy to do.
But is this the only way to get the straight line expression from 2 points given?
No, there is at least one other method. Don't forget maths is exciting and amazing, if one wishes it to be.
What is the other way? The answer is the use of simultaneous equations!
How so?
Note, given 2 co-ordinates and having to find 2 unknowns satisfies the basic requirement to set up 2 equations for simultaneous solving.
We know the general straight line equation to be y = mx + c.
Therefore with the known co-ordinates,
11 = m(3) + c -----(A)
5 = m(1) + c -----(B)
By elimination method, (A) - (B), gives,
6 = m(2) ===> this gives m = 6 / 2 = 3 (The gradient!)
With m = 3 found, let's put back into equation (B),
5 = (3)(1) + c ===> c = 5 - 3 = 2 (The intercept!)
Thus, the straight line equation is y = 3 x + 2. This is the same as the one obtained with graphical method.
Therefore, from the simultaneous way, we can still obtain the expression from the 2 given co-ordinates, without plotting the graph.
Either which way is fine.
What is interesting is that once you master the principles of maths, you can be flexible to choose the method that you like and still arrive at an appropriate answer.
Enjoy maths! :)
.
#### 3 comments:
Anonymous said...
Thankyou this helped me alot with maths. :)
Anonymous said...
thankyou so much! this saved me in a moment of panic with some homework and helped me a lot!
Anonymous said...
Finally someone explained how to find the gradient, should have found this site before my exam :(. |
# How do you graph the inequality y>=(x-1)^2-3?
Feb 20, 2018
#### Explanation:
Let us first consider the graph of $y = {\left(x - 1\right)}^{2} - 3$
This is the equation of a vertical parabola, which opens upwards, in vertex form and hence has a minima at $\left(1 , - 3\right)$. Taking a few values to the left and right of $x = - 1$, we can easily draw the graph of parabola, which appears as shown below:
graph{(x-1)^2-3 [-9.25, 10.75, -5.72, 4.28]}
Observe that this divides the Cartesian plane in three parts of which one is the line graph of parabola,,
the other two parts are the area lying within the cup-shaped portion inside parabola. Observe that origin $O \left(0 , 0\right)$ lies in this area. If we put $\left(0 , 0\right)$ we find that ${\left(x - 1\right)}^{2} - 3 = {\left(0 - 1\right)}^{2} - 3 = - 2$ and hence as $y = 0$, we have $y > {\left(x - 1\right)}^{2} - 3$
Hence in this area we have $y > {\left(x - 1\right)}^{2} - 3$.
The third part is area outside the parabola and a point in tis area is $\left(- 5 , 0\right)$. Putting this we get ${\left(x - 1\right)}^{2} - 3 = {\left(- 5 - 1\right)}^{2} - 3 = 36 - 3 = 33 > 0$ and hence $y < {\left(x - 1\right)}^{2} - 3$
hence this area satisfies $y < {\left(x - 1\right)}^{2} - 3$
Now we are given the inequality $y \ge {\left(x - 1\right)}^{2} - 3$ and hence all points on parabola and above it satisfy this inequality and hence graph appears as
graph{y>=(x-1)^2-3 [-9.25, 10.75, -5.72, 4.28]}
Note that had we only $y > {\left(x - 1\right)}^{2} - 3$, the graph would not have included points on parabola and for this parabola would have appeared as dotted. But here it appears as full line as points on parabola are included. This is because $y \ge {\left(x - 1\right)}^{2} - 3$ includes equality. |
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