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Whatever we do in life, we manipulate fractions on a daily basis. This is done for example when we cut a pie, read a percentage on a label or play LEGO.
Fractions are also the first abstract concept taught in mathematics and more often what confuses students with mathematics. Although these may seem tricky at first, it becomes much simpler once we visualize what a fraction is and how it works.
### Taking a Fraction
The number above, at the numerator, represents the number of shares we take.
The number below, at the denominator, represents the number of total shares.
If we take 1/4 (a quarter) of a pie, we cut it into 4 equal parts and take 1 share.
Other concrete examples:
- If in a class of 16 students 9 are girls, then 9/16 of them are girls.
- To advance 3/4 of a meter, we divide it into 4 equal steps (of 25cm) then advance of 3 times this distance (75cm).
- To take 3/5 of a number, we divide by 5 and multiply by 3.
## Two main rules
We never divide by 0!
If we multiply or divide the numerator and the denominator of a fraction by the same number: we get an equivalent fraction (equal).
## Objectives
We may want to visualize and manipulate fractions first. Let's use the H.urna Explorer
$$\frac{a}{a}=1$$ $$\require{cancel} \frac{a \cdot b}{a \cdot c} = \frac{\cancel{a} \cdot b}{\cancel{a} \cdot c} = \frac{b}{c}$$ $$\frac{a}{b} \cdot c = \frac{a \cdot c}{b} = \frac{c}{b} \cdot a$$ $$\frac{-a}{b}=-\frac{a}{b}$$ $$\frac{1}{\frac{b}{c}}=\frac{c}{b}$$ $$\frac{a}{1}=a$$ $$\frac{-a}{-b}=\frac{a}{b}$$ $$\frac{a}{-b}=-\frac{a}{b}$$ $$\frac{a}{\frac{b}{c}}=\frac{a\cdot c}{b}$$ $$\frac{\frac{b}{c}}{a}=\frac{b}{c\cdot a}$$ $$\frac{0}{a}=0\:,\:a\ne 0$$ |
### Theory:
In this section, we will deal with the SAS similarity criterion of two triangles.
Theorem 3
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Let us consider $$\triangle ABC$$ and $$\triangle DEF$$ such that:
$$\frac{AB}{DE}$$ $$=$$ $$\frac{AC}{DF}$$ $$\longrightarrow (1)$$
Also, $$\angle A$$ $$=$$ $$\angle D$$
We should prove that $$\triangle ABC \sim \triangle DEF$$.
For that matter, let us cut $$DP$$ $$=$$ $$AB$$ from $$DE$$ and $$DQ$$ $$=$$ $$AC$$ from $$DF$$ and join $$PQ$$.
Let us look at the image given below for a better understanding.
In $$\triangle ABC$$ and $$\triangle DPQ$$:
$$AB$$ $$=$$ $$DP$$ [By construction]
$$AC$$ $$=$$ $$DQ$$ [By construction]
$$\angle A$$ $$=$$ $$\angle D$$ [Given]
Therefore, by SAS congruence criterion, $$\triangle ABC \cong \triangle DPQ$$.
On substituting $$AB$$ $$=$$ $$PB$$ and $$AC$$ $$=$$ $$DQ$$ in $$(1)$$, we get:
$$\frac{DP}{DE}$$ $$=$$ $$\frac{DQ}{DF}$$
By converse of basic proportionality axiom, "If a line segment divides two sides of a triangle in the same ratio, then the line segment is parallel to the third side."
Thus, $$PQ \parallel EF$$.
By converse of basic proportionality axiom, "If a line segment divides two sides of a triangle in the same ratio, then the line segment is parallel to the third side."
$$\angle DPQ$$ $$=$$ $$\angle E$$
$$\angle DQP$$ $$=$$ $$\angle F$$
AA similarity criterion states that "If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar."
Thus, $$\triangle DPQ \sim \triangle DEF$$.
Therefore, from $$(1)$$, we know that, $$\angle A$$ $$=$$ $$\angle D$$.
Also, since $$\triangle ABC \cong \triangle DPQ$$, $$\angle B$$ $$=$$ $$\angle E$$ and $$\angle C$$ $$=$$ $$\angle F$$.
Thus, the AAA similarity criterion proves that $$\triangle ABC \sim \triangle DEF$$. |
# How do you solve the system x/2+(2y)/3=32 and x/4-(5y)/9=40?
Feb 22, 2017
$x = 100 \mathmr{and} y = - 27$
#### Explanation:
At first glance this horrifying because of the fractions!
Luckily with equations you can always get rid of any fractions by multiplying each term by the LCM of the denominators.
$\times 6 \rightarrow \text{ "x/2+(2y)/3=32" } \rightarrow 3 x + 4 y = 192$
$\times 36 \rightarrow \text{ "x/4-(5y)/9=40" } \rightarrow 9 x - 20 y = 1440$
The equations look much better, now we can solve them:
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} 3 x + 4 y = 192. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . A$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} 9 x - 20 y = 1440. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . B$
$A \times - 3 : \text{ } \textcolor{b l u e}{- 9 x} - 12 y = - 576. \ldots \ldots \ldots \ldots \ldots \ldots \ldots C$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \textcolor{b l u e}{9 x} - 20 y = 1440. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots B$
$C + B : \text{ } - 32 y = 864$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} \textcolor{red}{y = - 27}$
Substitute $- 27$ for $y$ in $A$
$\text{ } 3 x + 4 \left(- 27\right) = 192. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . A$
$\text{ } 3 x - 108 = 192$
$\text{ } 3 x = 192 + 108$
$\text{ } 3 x = 300$
$\text{ } \textcolor{b l u e}{x = 100}$
Check the solutions in equation B
" Is " 9(100)-20(-27)= 1440 ?
$\text{ } 900 + 540 = 1440$
$\text{ Indeed! } 1440 = 1440$
We can even check in the original equation for A;
$\frac{x}{2} + \frac{2 y}{3} = 32$
$\frac{100}{2} + \frac{2 \times - 27}{3}$
$= 50 - 18$
$= 32 \text{ } \leftarrow$ the answer is correct. |
Find the value of k for which the line through the points (2, 4, 8) and (1, 2, 4) is parallel to the line through the points (3, 6, k) and (1, 2, 1) ?
This question was previously asked in
Airforce Group X 7 November 2020 Memory Based Paper
View all Airforce Group X Papers >
1. 10
2. 9
3. 8
4. 0
Option 2 : 9
Detailed Solution
Concept:
Let us consider two lines AB and CD. The direction ratios of line AB is a1, b1, c1 and the direction ratios of line CD is a2, b2, c2.
Then AB will be parallel to CD, if $$\rm \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$.
Calculation:
Given: The line through the points (2, 4, 8) and (1, 2, 4) is parallel to the line through the points (3, 6, k) and (1, 2, 1).
Let us consider AB be the line joining the points (2, 4, 8) and (1, 2, 4) whereas CD be the line passing through the points (3, 6, k) and (1, 2, 1).
Let, the direction ratios of AB be: a1, b1, c1
⇒ a1 = (2 – 1) = 1, b1 = (4 – 2) = 2 and c1 = (8 – 4) = 4.
Let the direction ratios of CD be: a2, b2, c2
⇒ a2 = (3 – 1) = 2, b2 = (6 – 2) = 4 and c2 = k – 1.
∵ Line AB is parallel to CD ⇒ $$\rm \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
⇒ $$\rm \frac{1}{2}=\frac{2}{4}=\frac{4}{k-1}$$
⇒ $$\rm \frac{1}{2}=\frac{4}{k-1}$$
⇒ k - 1 = 8 ⇒ K = 9.
Hence, correct option is 2. |
# Multiplication of two Binomials
Multiplication of two binomials can be solved in both horizontal and column method.
Horizontal method:
Follow the following steps to multiply the binomials in the horizontal method:
1. First write the two binomials in a row separated by using multiplication sign.
2. Multiply each term of one binomial with each term of the other.
3. In the product obtained, combine the like terms and then add the like terms.
Therefore, we will learn how to multiply two binomials a + 5 by a + 7 using horizontal method.
a + 5 by a + 7
= (a + 5) ∙ (a + 7), [separate the two binomials using multiplication sign]
= a ∙ (a + 7) + 5 ∙ (a + 7), [multiplying each term of the first binomial with each term of the second binomial]
= a ∙ a + a ∙ 7 + 5 ∙ a + 5 ∙ 7
= a2 + 7a + 5a + 35, [combine the like terms]
= a2 + 12a + 35
Column method:
Follow the following steps to multiply the binomials in the column method:
1. Write the two binomials in two rows one below the other.
2. Multiply one term of the binomial in lower line (i.e. second row) with each term of the binomial in the upper line (i.e. first row) and write the product in the third row.
3. Multiply second term of the binomial in lower line (i.e. second row) with each term of the binomial in upper line (i.e. first row) and write the product in the fourth row in such a way that the like terms are one below the other.
4. Add the like terms column wise.
Therefore, we will learn how to multiply two binomials 5a - 6b and 7a + 8b using column method.
Solved examples on multiplication of two binomials:
1. Multiply 3x2 – 6y2 by 2x2 + 4y2
Solution:
3x2 – 6y2 by 2x2 + 4y2
= (3x2 – 6y2) ∙ (2x2 + 4y2), [separate the two binomials using multiplication sign]
= 3x2 ∙ (2x2 + 4y2) – 6y2 ∙ (2x2 + 4y2), [multiplying each term of the first binomial with each term of the second binomial]
= 6x4 + 12x2y2 – 12x2y2 – 24y4
= 6x4 + 12x2y2 – 12x2y2 – 244, [combine the like terms]
= 6x4 - 244
2. Multiply (m + 6) by (3m – 2)
Solution:
The above examples will help us to solve the multiplication of two binomials in horizontal method and in column method.
Types of Algebraic Expressions
Degree of a Polynomial
Subtraction of Polynomials
Power of Literal Quantities
Multiplication of Two Monomials
Multiplication of Polynomial by Monomial
Multiplication of two Binomials
Division of Monomials
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. |
# Law of action and reaction on different masses
The law of action and reaction on different masses could be illustrated with the example of a ball dropped from a certain height.
There is an interaction between the ball and the earth.
• The earth pulls down the ball with some force.
• The ball pulls up on the earth with the exact same force.
We could also rephrase by saying the following.
• The ball moves toward the earth.
• The earth moves toward the ball.
However, the distance the earth moves toward the ball is much less and too small to notice.
The reason that the earth moves a tiny distance is because the mass of the earth is very big compared to the mass of the ball.
As a result, the earth's acceleration will be small.
Recall that acceleration =
Force / mass
accelerationball =
Force / mass of ball
accelerationearth =
Force / mass of earth
We also need some basic fraction concepts here too to help you understand the concept. Suppose the numerator is the same. What happens when the denominator of a fraction is very big?
The fraction will be very small.
In fact, the bigger the denominator, the smaller the fraction.
Look at the two fractions below. The numerator is the same.
The fraction with a denominator of 9500000 will be much smaller compared to the one with a denominator of 50.
8 / 9500000 < 8 / 50
8 / 9500000 < 8 / 50
The mass of a soccer ball is about 0.45 kg.
The mass of the earth is about 5.98 × 1024.
Force / 5.98 × 1024 kg is much smaller than Force / 0.45 kg
Force / 5.98 × 1024 kg is much smaller than Force / 0.45 kg
accelerationearth is much smaller than accelerationball
Even though the earth's acceleration is negligible, know that it is there and again it causes the earth to move a tiny bit toward the ball.
## Other examples of the law of action and reaction on different masses
a. You hold a rock with a mass of 2 kg. Your friend hold a rock with a mass of 20 kg. both of you throw the rocks until they collide in the air.
They collide with the same force, but the rock with the smaller mass will have a bigger acceleration.
b. You hit a car with a force of 200 N.
The car hits back at you with a force of 200 N. You bounce back with an acceleration that is more than that of the car since you have a smaller mass.
c. A rifle fires a bullet.
The bullet will accelerate forward. The rifle will accelerate backward.
The acceleration of the rifle is less than the bullet since the rifle has a bigger mass.
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# 180 Days of Math for Fourth Grade Day 28 Answers Key
By accessing our 180 Days of Math for Fourth Grade Answers Key Day 28 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fourth Grade Answers Key Day 28
Directions Solve each problem.
Question 1.
32 – 15 = _____
32 – 15 = 17
Explanation:
Perform subtraction operation on above two given numbers. Subtract 15 from 32 the difference is 17.
Question 2.
How many fourths are in 3?
______________
3/(1/4)
= 3 x 4
= 12
There are 12 fourths in 3.
Question 3.
Divide 8 by 2.
______________
8 ÷ 2 = 4
Explanation:
Perform division operation on above two given numbers. Divide 8 by 2 the quotient is 4.
Question 4.
14 ÷ 7 = ____
14 ÷ 7 = 2
Explanation:
Perform division operation on above two given numbers. Divide 14 by 7 the quotient is 2.
Question 5.
What is the place value of 7 in 1,709?
______________
The place value of 7 in 1,709 is hundreds.
Explanation:
The Place value is the position of a digit in a given number. The order of place value of a digits starts from right to left. The place values are named as units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. The place value of 7 in 1,709 is hundreds.
Question 6.
× 4 = 20
× 4 = 20
Explanation:
To get the product 20 we need to perform multiplication operation. Multiply 6 with 4 the product is 20. So, the missing multiplicand is 6.
Question 7.
I put 700 mL of water in the jug. Next I put a toy in the jug. How much water was displaced by the toy?
_______ mL was displaced.
900ml – 700 ml = 200 ml
200 ml of water was displaced by toy.
Explanation:
I put 700 mL of water in the jug. Next I put a toy in the jug. After keeping a toy in the jug, the water level is increased to 900 ml as we can observe in the above image. So, 200 ml of water was displaced by toy.
Question 8.
Which would be the best tool for measuring the length of a book: a ruler, a yardstick, or a scale?
______________
Ruler is the best tool for measuring the length of a book. A ruler is used to measure length.
Question 9.
Does this shape tessellate? |
# 5 Function GMAT Practice Questions & Explanations
## Question 1:
If f(x) = 100 – x2, is f(a) > f(b) ?
1. a2 > b2
2. a/b > 1
f(x) = 100 - x2
Therefore, f(a) = 100 - a2 and f(b) = 100 - b2
Is f(a) > f(b)?
Or, is 100 - a2 > 100 - b2 ?
Or, is - a2 > - b2 ?
Or, is a2 < b2 ?
1) a2> b2 , clearly its sufficient to tell that a2 is not less than b2 ;SUFFICIENT.
2) a/b > 1
=> (a/b) -1 > 0
=> (a-b)/b > 0, which means either both a-b and b are positive, or both a-b and b are negative.
=> If b>0, a-b>0 means a>b. So a2> b2.
And if b<0, a-b<0 means a2 > b, which means a is bigger negative number than b.(For eg. a=-4, b=-2). So, a2> b2.
So, either way we can say that a2> b2 ; SUFFICIENT.
Alternatively, the statement 2 could also have been analyzed in the following manner.
a/b > 1 → that the magnitude of numerator is more than denominator which would mean a2> b2.
The correct answer is D; each statement alone is sufficient.
## Question 2:
a’ =10/a for all positive prime numbers. And a’ = -a2for all positive non-primes, which of the following is the largest?
(A) 5’ + 6’ + 2’
(B) 7’ – 4’
(C) 4’ – 7’
(D) 4’ + 8’ – 5’
(E) 4’ + 8’ – 2’
Ans:
Since, for all positive prime numbers;
a' = 10/a
And, for all positive non-prime numbers;
a' = -a2
Thus, using these functions to find out the values given in the question, we get;
(A) 5' + 6' + 2' = (10/5 - 6+ 10/2) = (5 - 36 +2) = -29
(B) 7' - 4' = (10/7 - (-42)) = (10/7 + 16) = 122/7
(C) 4' - 7' = ( - 4- 10/7) = ( -16 - 10/7) = -122/7
(D) 4' + 8' - 5' = ( -4- 8- 10/5) = ( -16 -64 - 2) = -82
(E) 4' + 8' - 2' = (-4- 8- 10/2) = (-16 - 64 -5) = -85
Thus, among all the options we can see the largest value is obtained in option B
Alternatively, you could have also analyzed that all the options will generate a negative value except option B. hence that is the largest.
## Question 3:
If [a] represents least integer greater than or equal to a, what is the value of [–3.3] + [–2.9] + [0.1] + [1] + [1.9] ?
(A) –1
(B) –2
(C) –3
(D) –4
(E) –5
Since [a] represents the least integer greater than or equal to a, this implies that if a is an integer than [a] = a only. But if a is a fraction than, [a] = the least integer greater than a.
Thus, [-3.3]=-3
And,[-2.9]=-2
And,[0.1]=1
And,[1]=1
And,[1.9]=2
Hence, [-3.3]+[-2.9]+[0.1]+[1]+[1.9]
=-3-2+1+1+2
=-1
## Question 4:
In which of the following functions, is f(x) = f(1/x), for all values of x greater than 1?
(A) f(x) = 1/x + 1/ (x + 1)
(B) f(x) = |x|
(C) f(x) = 1-x
(D) f(x) = (1 - x2 )2/x2
(E) f(x) = (1 - x2)/x2
(A) f(x) = 1/x + 1/(x + 1) and f(1/x) = 1/(1/x) + 1/(1/x + 1) = x + x/(1 + x)
Thus, f(x) ≠ f(1/x) for all values of x > 1
(B) f(x) = |x| and f(1/x) = |1/x| = 1/(|x|)
Thus, f(x) ≠ f(1/x) for all values of x >1
(C) f(x) = 1 - x and f(1/x) = 1 - 1/x = (x - 1)/x
Thus, f(x)≠f(1/x) for all values of x>1
(D) f(x)=(1 - x2 )2/x2 and f(1/x)=[1 - (1/x)2 ]2/(1/x)2 =[1 - 1/x2 ]2/(1/x2 )=[(x- 1)/x2 ]2/(1/x2 )=(x- 1)2/(x× 1/x2 )=(x- 1)2/x2 =(1 - x2 )2/x2
Thus, f(x) = f(1/x) for all values of x > 1
(E) f(x) = (1 - x2)/x2 and f(1/x) = (1 - (1/x)2)/(1/x)2 = (1 - 1/x2 )/(1/x2 ) = ((x- 1)/x2 )/(1/x2 ) = (x2- 1)/(x2×1/x2 ) = x- 1
Thus, f(x) ≠ f(1/x) for all values of x>1
Thus we can see only for (D); f(x) = f(1/x)
## Question 5:
In a sequence, an + 2 = an + 1 – an + (n + 2) for all n ≥1, where an + 2 represents the (n + 2)th term. If the second and the fourth terms of the sequence are 2 and 5 respectively then what is the value of the 6th term of the sequence?
(A) 5
(B) 7
(C) 8
(D) 9
(E) None of these
For n≥1,
a(n + 2) = a(n + 1) - an + (n + 2) ……Equation(1)
Putting value of n = 2 in the above equation, we get:
a= a- a+ (2 + 2)
In the question, it’s also mentioned that a= 2 , a= 5. Putting in the above equation, we get the value of a= 3
Now by putting value of n = 3 in equation(1), we get:
a5 = a4 - a3 + (3 + 2)
a= 5 - 3 + (3 + 2)
a5 = 7
Now by putting value of n = 4 in equation(1),we get:
a6 = a5 - a4 + (4 + 2)
a6 = 7 - 5 + (4 + 2)
a6 = 8
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Subtraction for Class 3
Last updated date: 21st Feb 2024
Total views: 159k
Views today: 2.58k
Introduction to Subtraction
If you bought fifty candies in the morning and ate 23 candies at night, then what is the total number of candies left with you? Do you know the answer to this question? Let us learn the concept of subtraction for class 3 with questions and various examples.
The basic concept of subtraction was taught in the previous classes. In subtraction for class 3, we will learn a little advanced concept of subtraction. We will see how to do subtraction sums for class 3 with borrowing. We will be looking at the concept with various examples of subtraction sums. Additional questions will also be provided at the end. Subtraction sums for class 3 will also be shared at the end. So, let us begin the module now.
Basic Rules of Subtraction
Subtraction has various parts in which it is divided. Minuend, Subtrahend, Minus sign and equals sign are the parts of Subtraction. In the example of 9-8=1, 9 is minuend, 8 is subtrahend, - is the minus sign, and = is the equals sign.
Subtraction is used to calculate the remaining number from the two given numbers. There are certain basic rules to the concept of subtraction. We have to write the two given numbers in a tabular form. The larger number will always be at the top of the smaller number. We always start by subtracting the digits which are on the rightmost side. Further, the tens place digits are subtracted, then the hundredth, and so on.
Subtraction through Number Line
Number Line Representation of Subtraction
Through this number line representation, we can see the basic concept of subtraction. We first mark the larger digit out of the two given numbers. Then we go back to the left side of the line and mark the second digit. After that we count the numbers in between, which is the final answer of the subtraction. 2 is the answer of 4-2.
Subtraction Without Borrowing
This is a simple form of subtraction. This can be done by simply following the basic rules of subtraction. We keep the larger number at the top and start subtracting by the one's side. Let us take an example to understand this concept.
An Example of Subtraction
To subtract 35 from 87, we first subtract 5 from 7 and write the answer 2 in the One's column. After that, we find the subtraction of 3 from 8 and write the answer 5 in the Tens box. This is how a subtraction without borrowing is done.
Subtraction with Borrowing
In some cases, the numbers to which we have to subtract are smaller. See the example given below, in this case, 8 cannot be subtracted from 0. Hence, let us understand this concept step by step.
• First, start from the One's side and write the digit above it.
• The next step is to take a carry from the tens number which is 5 in this case.
• After taking the carry the number above the one digit will become 10.
• Subtract 8 out of 10 and write the answer 2 below in the column.
• Now that 5 has given 1 carry to 0, only 4 is left.
• Finally, subtract 3 out of 4 and write the answer 1 below.
Subtraction Example with Borrowing
Solved Examples of Subtraction Sums
Q1. At a birthday party, Sonu invited 100 people and only 50 showed up. How many people did not come for the part?
Birthday Party
Ans: We solve this kind of question like:
Total number of people Sonu invited = 100
The number of people who showed up = 50
The number of people who did not come= was 100 - 50
The answer would be 50 people.
Q2. A black sheep walks 50 metres in the morning and 40 more in the evening? How many steps did it walk in the evening?
Black Sheep
Ans: We solve this kind of question like:
Total number of steps frog walks in the morning = 50
Number of frog walks in the evening = 40
Number of steps it walked less in the evening = 50 - 40
The answer would be 10 steps.
(Subtraction sums for class 3 with borrowing)
Q1: 672 - 563=?
Ans: 109
Q2: 432-234=?
Ans: 198
Summary
Subtraction is a very important concept for the students to make their base level strong. These concepts will always be helpful in all kinds of theoretical and practical questions. The advanced forms of subtraction are added class by class. Students must apply these rules in their daily lives as well. Whenever they go to buy snacks or candies, they must use these concepts. They should also practice more and more worksheets for a better understanding. These are a good form of understanding.
FAQs on Subtraction for Class 3
1. What are parts of subtraction?
Subtraction is used to subtract a smaller number from a larger number. It is used in various concepts of mathematics. Minuend, Subtrahend, Minus sign and equals sign are the parts of a subtraction. Let us take an example to understand the parts.
In questions 7-3=4, 7 is minuend, 3 is subtrahend, - is a minus sign, and = is an equals sign. These parts form a subtraction.
2. How to solve word problem-based subtraction questions? Give an example for the same.
Word problem-based subtraction questions are the advanced form of subtraction. Let's take an example to understand this better. If a rabbit has 22 carrots and eats 12 out of them. How many carrots are left with the rabbit, to answer this question let us go step by step.
The total number of carrots a rabbit has = 22
Number of carrots rabbit has already eaten= 12
The number of carrots left with the rabbit is = 22 - 12= 10. |
# Video: AQA GCSE Mathematics Foundation Tier Pack 1 • Paper 1 • Question 8
At a fairground, Simon has won five tokens. Each token represents 100 credits. He decides to exchange his credits for two different types of sweets: cola bottles and gobstoppers. The table shows the number of credits required for a cola bottle and a gobstopper. He buys 20 cola bottles, as many gobstoppers as possible, and with any change buys extra cola bottles. How many cola bottles and gobstoppers does Simon buy?
05:11
### Video Transcript
At a fairground, Simon has won five tokens. Each token represents 100 credits. He decides to exchange his credits for two different types of sweets: cola bottles and gobstoppers. The table shows the number of credits required for a cola bottle and a gobstopper. He buys 20 cola bottles, as many gobstoppers as possible, and with any change buys extra cola bottles. How many cola bottles and gobstoppers does Simon buy?
There is a lot of information in this question. So it is important to write our answer clearly and in a logical order. The first two lines tell us that Simon won five tokens and that each token represented 100 credits. Multiplying five by 100 gives us 500. This means that Simon has 500 credits altogether.
He is exchanging these credits for two sweets: cola bottles and gobstoppers. Firstly, we’re told that he buys 20 cola bottles. As each cola bottle is worth five credits, we need to multiply 20 by five. 20 multiplied by five is equal to 100. Therefore, Simon uses 100 credits to buy these 20 cola bottles. As he initially had 500 credits and he has now used 100, he has 400 credits left. 500 minus 100 is equal to 400.
Next, we were told that Simon bought as many gobstoppers as possible. Each gobstopper was worth 23 credits. We need to work out how many 23s there are in 400. This is the same as saying 400 divided by 23. You could do this using long division. However, we will use the chunking method as it will be easier when dividing by 23.
23 multiplied by 10 is equal to 230. Therefore, Simon can get 10 gobstoppers for 230 credits. To calculate 23 multiplied by five, we could halve this answer of 230, as five is half of 10. One-half of 200 is 100, and one-half of 30 is 15. Therefore, a half of 230 is 115. This means that Simon can buy another five gobstoppers at 23 credits for a total of 115 credits.
23 multiplied by two is equal to 46, as 20 multiplied by two is 40 and three multiplied by two is equal to six. Simon can get an extra two gobstoppers for 46 credits. Adding 230, 115, and 46 gives us 391. This means that Simon can buy 17 gobstoppers for 391 credits. As each gobstopper was worth 23 credits, he doesn’t have enough to buy an extra gobstopper.
We have worked out that 17 multiplied by 23 is equal to 391 credits. Currently, Simon has 20 cola bottles and he has 17 gobstoppers. We now need to consider the change or the remainder from the calculation 400 divided by 23. Simon used 391 credits to buy gobstoppers. This means that 400 divided by 23 is equal to 17, remainder nine. He has used 391 of the 400 credits so has nine remaining.
We are told that Simon uses these extra credits to buy more cola bottles. Well, each cola bottle is worth five credits. So to buy two cola bottles, Simon would need 10 credits. As Simon only has nine credits remaining, he can only buy one extra cola bottle. One extra cola bottle would use an additional five credits.
We can therefore see that, altogether, Simon buys 21 cola bottles and 17 gobstoppers. He does have four credits remaining. But this is not enough to buy an extra cola bottle or gobstopper. |
Bridges in Mathematics Grade 4 Student Book Unit 7 Module 2 Answer Key
Practicing the Bridges in Mathematics Grade 4 Student Book Answer Key Unit 7 Module 2 will help students analyze their level of preparation.
Bridges in Mathematics Grade 4 Student Book Answer Key Unit 7 Module 2
Bridges in Mathematics Grade 4 Student Book Unit 7 Module 2 Session 1 Answer Key
Decimal Fractions & Numbers
Question 1.
Fill in the missing decimals and fractions on the number line.
Question 2.
When numbers share a place on the number line, that means they are equivalent. For each number, circle the other numbers that are equivalent.
a. 0.7
7Â Â 0.70Â Â 70Â Â $$\frac{7}{10}$$Â Â $$\frac{70}{100}$$Â Â 0.07Â Â $$\frac{7}{100}$$
0.70, $$\frac{7}{10}$$, and $$\frac{70}{100}$$.
Explanation:
The equivalent numbers to 0.7 is 0.70, $$\frac{7}{10}$$, and $$\frac{70}{100}$$.
b. $$\frac{9}{10}$$
0.09Â Â Â $$\frac{9}{100}$$Â Â 0.9Â Â 0.90Â Â $$\frac{90}{100}$$Â Â Â 9Â Â 90
0.9, 0.90, and $$\frac{90}{100}$$.
Explanation:
The equivalent numbers to $$\frac{9}{10}$$ is 0.9, 0.90, and $$\frac{90}{100}$$.
c. $$\frac{40}{100}$$
0.4Â Â $$\frac{4}{100}$$Â Â 40Â Â 4Â Â $$\frac{4}{10}$$Â Â Â 0.04Â Â 0.40
0.4, $$\frac{4}{10}$$, and 0.40.
Explanation:
The equivalent numbers to $$\frac{40}{100}$$ is 0.4, $$\frac{4}{10}$$, and 0.40.
Question 3.
Shade in the grid to show each number. Write two decimal numbers to represent each value.
a.
The two decimal numbers will be 0.8 and 0.80.
Explanation:
Here, we have shaded 8 blocks out of 10 which is $$\frac{8}{10}$$ = 0.8 and the other decimal number will be $$\frac{80}{100}$$ = 0.80.
b.
The two decimal numbers will be 0.3 and 0.30.
Explanation:
Here, we have shaded 3 blocks out of 10 which is $$\frac{3}{10}$$ = 0.3 and the other decimal number will be $$\frac{30}{100}$$ = 0.30.
Bridges in Mathematics Grade 4 Student Book Unit 7 Module 2 Session 2 & 3 Answer Key
Scoot the Marker Instructions
• Fasten your decimal strip to a table, the floor, or another smooth and flat surface.
• Set your game marker to one side of your decimal strip, as shown.
• If the marker lands too far away from the decimal strip to read the distance accurately, use your pencil to help.
• Record the distance on the record sheet in fractions (for fractions game) or decimals (for numbers game), as shown below.
• Repeat until you and your partner have each taken 5 turns.
• Write an inequality statement on your record sheet to show which range each distance falls into, as shown below.
• Figure out how many points you got for each turn. Add them to find your total score.
Bridges in Mathematics Grade 4 Student Book Unit 7 Module 2 Session 2 Answer Key
Scoot the Marker Record Sheet Decimal Fractions Game
See game instructions above.
Here, we have recorded the distance on the record sheet in fractions in game 1 and decimals in game 2 and written an inequality statement on the record sheet to show which range each distance falls into
Decimal Fractions on Line & Grid
Question 1.
Label each marked point on the decimal strip with a fraction. Use tenths when you can, and hundredths when you must.
Here, we have labeled each marked point on the decimal strip with a fraction.
Question 2.
Write each number you labeled on the number line above here. Then shade in the grid to show each value and write a fraction to represent it.
a.
Fraction = $$\frac{7}{100}$$,
Decimal = 0.07.
Explanation:
Here, we have shaded 7 blocks out of 100 which is $$\frac{7}{100}$$ = 0.07.
b.
Fraction = $$\frac{4}{10}$$,
Decimal = 0.4.
Explanation:
Here, we have shaded 4 blocks out of 10 which is $$\frac{4}{10}$$ = 0.4.
c.
Fraction = $$\frac{53}{100}$$,
Decimal = 0.53.
Explanation:
Here, we have shaded 53 blocks out of 100 which is $$\frac{53}{100}$$ = 0.53.
d.
Fraction = $$\frac{9}{10}$$,
Decimal = 0.9
Explanation:
Here, we have shaded 9 blocks out of 10 which is $$\frac{9}{10}$$ = 0.9.
Question 3.
Write fractions and decimals to show how much of each grid is shaded.
a.
Fraction = $$\frac{2}{10}$$,
Decimal = 0.2.
Explanation:
Here, we have shaded 2 blocks out of 10 which is $$\frac{2}{10}$$ = 0.2.
b.
Fraction = $$\frac{57}{100}$$,
Decimal = 0.57.
Explanation:
Here, we have shaded 57 blocks out of 100 which is $$\frac{57}{100}$$ = 0.57.
Question 4.
Why can you write two different decimal fractions and decimal numbers for grid a, and only one for each for grid b in problem 3 above?
Here, $$\frac{57}{100}$$ cannot be described using a whole number of tenths and if there is something other than zero in the hundredths place. And if there is something other than 0 in the hundredths place and we need to show it.
Bridges in Mathematics Grade 4 Student Book Unit 7 Module 2 Session 3 Answer Key
Scoot the Marker Record Sheet Decimal Numbers Game
See game instructions
Here, we have recorded the distance on the record sheet in decimals and written an inequality statement on the record sheet to show which range each distance falls into
Writing & Comparing Decimal Numbers
Question 1.
Use the grid below to answer the questions.
a. How many tenths are filled in?
6 tenths.
Explanation:
There are 6 tenths filled in the box.
b. How many hundredths are filled in?
60 hundredths.
Explanation:
There are 60 hundredths filled in the box.
c. Fill in the blanks to write two fractions that represent this amount.
The two fractions are
d. Fill in the blanks to show two decimal numbers that represent this amount.
0.6 and 0.60.
Explanation:
The two decimal numbers that represent this amount are
Question 2.
Fill in the blanks to show one or two decimal numbers that represent each fraction or mixed number.
a. $$\frac{3}{10}$$
0.3 and 0.30
Explanation:
The two decimal numbers that represent this amount are
b. 2$$\frac{23}{100}$$
2.23.
Explanation:
The decimal number that represent this amount are
c. 4$$\frac{1}{10}$$
4.1 and 4.10
Explanation:
The two decimal numbers that represent this amount are
d. 5$$\frac{6}{100}$$
5.06.
Explanation:
The decimal number that represent this amount are
Question 3.
a. Locate each decimal number on the number line.
0.06
0.6
b. Complete the two inequalities to compare the two decimal numbers above.
_____________ > _____________
_____________ < _____________
0.6 > 0.06
0.06 < 0.6.
Explanation:
The comparison of two decimal numbers above is
0.6 is greater than 0.06 and 0.06 is lesser than 0.6.
Bridges in Mathematics Grade 4 Student Book Unit 7 Module 2 Session 4 Answer Key
Question 1.
Will the sum be…
less than $$\frac{1}{2}$$
equal to $$\frac{1}{2}$$
greater than $$\frac{1}{2}$$
$$\frac{2}{5}$$ is less than $$\frac{1}{2}$$.
Explanation:
The addition of two decimal fractions $$\frac{3}{10}$$+$$\frac{1}{10}$$ is
= $$\frac{3+1}{10}$$
= $$\frac{4}{10}$$
= $$\frac{2}{5}$$
which is less than $$\frac{1}{2}$$.
Question 2.
Will the sum be…
less than $$\frac{1}{2}$$
equal to $$\frac{1}{2}$$
greater than $$\frac{1}{2}$$
$$\frac{1}{2}$$ = $$\frac{1}{2}$$.
Explanation:
The addition of two decimal fractions $$\frac{32}{100}$$+$$\frac{18}{100}$$ is
= $$\frac{32+18}{100}$$
= $$\frac{50}{100}$$
= $$\frac{1}{2}$$
which is equal to $$\frac{1}{2}$$.
Question 3.
Will the sum be…
less than $$\frac{1}{2}$$
equal to $$\frac{1}{2}$$
greater than $$\frac{1}{2}$$
$$\frac{9}{10}$$ > $$\frac{1}{2}$$
Explanation:
The addition of two decimal fractions $$\frac{5}{10}$$+$$\frac{40}{100}$$ is
= $$\frac{50+40}{100}$$
= $$\frac{90}{100}$$
= $$\frac{9}{10}$$
which is greater than $$\frac{1}{2}$$
Question 4.
Will the sum be…
less than $$\frac{1}{2}$$
equal to $$\frac{1}{2}$$
greater than $$\frac{1}{2}$$
1 is greater than $$\frac{1}{2}$$.
Explanation:
The addition of two decimal fractions $$\frac{30}{100}$$+$$\frac{7}{10}$$ is
= $$\frac{30+70}{100}$$
= $$\frac{100}{100}$$
= 1
which is greater than $$\frac{1}{2}$$.
Snail & Caterpillar Race to $$\frac{300}{100}$$ Instructions
Each pair of players needs:
• a Snail & Caterpillar Race to $$\frac{300}{100}$$ Record Sheet for each player
• 3 dice, one each numbered 0-5, 1-6, and 4-9
• 4 colored pencils of different colors (each player will use 2 colors)
1. Players roll the 4-9 die to see who will be Player 1. Player 1 gets to choose whether to play for the snail or the caterpillar. Players each circle the creature they’re playing for.
2. Players take turns rolling and recording fractions. On each turn, the player:
• Rolls all 3 dice, then chooses one of the numbers rolled to take in tenths and fills that number in on the record sheet.
• Decides how to arrange the other two numbers rolled to make a double-digit number to take in hundredths, and records that number on the record sheet.
• Shades in the tenths on the decimal strip using one color.
• Shades in the hundredths on the decimal strip using another color.
• Rewrites the tenths as hundredths, then adds the two fractions and records the sum.
3. Players play 3 rounds, each player rolling the dice and recording fractions once per round.
Players record their partner’s turns on the other decimal strip, but do not record their fractions. For example, if you are playing for the snail, record your partner’s turn on the caterpillar’s strip on your record sheet.
4. At the end of the game, each player finds their total by recording and adding their three scores at the bottom of the record sheet. Then players compare their totals to $$\frac{300}{100}$$ and record the comparisons.
The player with the total closest to, but not greater than, $$\frac{300}{100}$$ wins.
Consider strategies for winning. If you roll large numbers in the first round, what will you try to do next round? What would be the best amount to have in all 3 rounds to make your total come closest to $$\frac{300}{100}$$?
Snail & Caterpillar Race to $$\frac{300}{100}$$ Record Sheet 1
I am racing for the Snail Caterpillar
Compare your total to $$\frac{300}{100}$$ Do the same for your partner’s total. Circle the total that is closest to, but not greater than, $$\frac{300}{100}$$.
My partner score is less than $$\frac{300}{100}$$.
Explanation:
Here, we recorded for snail and the total of the three rounds is $$\frac{110}{100}$$+$$\frac{114}{100}$$+$$\frac{104}{100}$$ = $$\frac{358}{100}$$
By comparing my partner score is less than $$\frac{300}{100}$$.
Snail & Caterpillar Race to $$\frac{300}{100}$$ Record Sheet 2
I am racing for the Snail Caterpillar
Compare your total to $$\frac{300}{100}$$ Do the same for your partner’s total. Circle the total that is closest to, but not greater than, $$\frac{300}{100}$$.
My score is less than $$\frac{300}{100}$$.
Explanation:
Here, we recorded for snail and the total of the three rounds is $$\frac{87}{100}$$+$$\frac{54}{100}$$+$$\frac{126}{100}$$ = $$\frac{267}{100}$$
By comparing my score is less than $$\frac{300}{100}$$.
Decimal Review
Question 1.
Label each point on the number line, using decimals below the line and fractions above it.
Here, we have labelled each point on the number line, using decimals below the line and fractions above it.
Question 2.
Find each sum. Show your work.
$$\frac{34}{100}$$ + $$\frac{4}{10}$$ = ________________
$$\frac{3}{10}$$ + $$\frac{6}{100}$$ = ________________
$$\frac{7}{10}$$ + $$\frac{13}{100}$$ = ________________
$$\frac{34}{100}$$ + $$\frac{4}{10}$$ = $$\frac{74}{100}$$.
$$\frac{3}{10}$$ + $$\frac{6}{100}$$ = $$\frac{90}{100}$$.
$$\frac{7}{10}$$ + $$\frac{13}{100}$$ = $$\frac{83}{100}$$.
Explanation:
The sum of $$\frac{34}{100}$$ + $$\frac{4}{10}$$ is
= $$\frac{34+40}{100}$$
= $$\frac{74}{100}$$
= 0.74.
The sum of $$\frac{3}{10}$$ + $$\frac{6}{100}$$ is
= $$\frac{30+60}{100}$$
= $$\frac{90}{100}$$.
= 0.9.
The sum of $$\frac{7}{10}$$ + $$\frac{13}{100}$$ is
= $$\frac{70+13}{100}$$
= $$\frac{83}{100}$$.
=Â 0.83.
Question 3.
Write each sum above as a decimal number.
a _____________
b _____________
c ______________
a. 0.74
b. 0.36
c. 0.83.
Explanation:
The sum of $$\frac{34}{100}$$ + $$\frac{4}{10}$$ in decimal number is
= 0.34+0.4
= 0.74.
The sum of $$\frac{3}{10}$$ + $$\frac{6}{100}$$ in decimal number is
= 0.3+0.06
= 0.36.
The sum of $$\frac{7}{10}$$ + $$\frac{13}{100}$$ in decimal number is
= 0.7+0.13
= 0.83.
Question 4.
Write an inequality to compare numbers a and c above.
___________Â Â _____Â Â Â ________________
0.74 < 0.83.
Explanation:
In the above a and c which is 0.74 and 0.83,
a is less than c which is
0.74 is less than 0.83.
Question 5.
For each number, circle all the other numbers that are equivalent.
a. 0.7
70Â Â $$\frac{7}{10}$$Â Â Â $$\frac{7}{100}$$Â Â 0.07Â Â 7Â Â $$\frac{70}{100}$$Â Â 0.70Â Â 700
$$\frac{7}{10}$$ and 0.70.
Explanation:
The equivalent numbers to 0.7 is $$\frac{7}{10}$$ and 0.70.
b. 0.04
40Â Â 400Â Â $$\frac{4}{10}$$Â Â $$\frac{4}{100}$$Â Â 0.4Â Â $$\frac{40}{100}$$Â Â 0.40Â Â 4
$$\frac{4}{100}$$.
Explanation:
The equivalent numbers to 0.04 is $$\frac{4}{100}$$.
c. 0.36
$$\frac{36}{10}$$Â Â Â $$\frac{36}{100}$$Â Â 36Â Â Â 3600Â Â Â 10Â Â $$\frac{3}{6}$$Â Â $$\frac{6}{3}$$Â Â Â 0.3
$$\frac{36}{100}$$.
The equivalent numbers to 0.36 is  $$\frac{36}{100}$$. |
9420 minus 87 percent
This is where you will learn how to calculate nine thousand four hundred twenty minus eighty-seven percent (9420 minus 87 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 9420 minus 87 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 9420 of something.
9420
(100%)
87 percent means 87 per hundred, so for each hundred in 9420, you want to subtract 87. Thus, you divide 9420 by 100 and then multiply the quotient by 87 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(9420 ÷ 100) × 87
= 8195.4
We made a pink square that we put on top of the image shown above to illustrate how much 87 percent is of the total 9420:
The dark blue not covered up by the pink is 9420 minus 87 percent. Thus, we simply subtract the 8195.4 from 9420 to get the answer:
9420 - 8195.4
= 1224.6
The explanation and illustrations above are the educational way of calculating 9420 minus 87 percent. You can also, of course, use formulas to calculate 9420 minus 87%.
Below we show you two formulas that you can use to calculate 9420 minus 87 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
9420 - ((9420 × 87/100))
9420 - 8195.4
= 1224.6
Formula 2
Number × (1 - (Percent/100))
9420 × (1 - (87/100))
9420 × 0.13
= 1224.6
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
9430 minus 87 percent
Here is the next percent tutorial on our list that may be of interest.
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# When to Use the Distributive Property
## Write the numerator in brackets and the denominator as a fraction in front
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When to Use the Distributive Property
David and Denise are having an argument. David says that you can't use the Distributive Property to simplify the expression $\frac{4x + 5}{8}$ , while Denise says that you can. Who do you think is right? After completing this Concept, you'll know when to use the Distributive Property to simplify expressions so that you can settle arguments such as these.
### Guidance
Identifying Expressions Involving the Distributive Property
The Distributive Property often appears in expressions, and many times it does not involve parentheses as grouping symbols. In a previous Concept, we saw how the fraction bar acts as a grouping symbol. The following example involves using the Distributive Property with fractions.
#### Example A
Simplify $\frac{9-6y}{3}.$
Solution:
The denominator needs to be distributed to each part of the expression in the numerator.
We can rewrite the expression so that we can see how the Distributive Property should be used:
$&\frac{9-6y}{3}=\frac{1}{3}(9-6y)=\\&\frac{1}{3}(9)-\frac{1}{3}(6y)=\frac{9}{3}-\frac{6y}{3}=\\& 3-2y.$
#### Example B
Simplify $\frac{2x+4}{8}.$
Solution:
Think of the denominator as $\frac{1}{8}$ : $\frac{2x+4}{8}= \frac{1}{8} (2x+4).$
Now apply the Distributive Property: $\frac{1}{8} (2x)+ \frac{1}{8}(4) = \frac{2x}{8} + \frac{4}{8}.$
Simplified: $\frac{x}{4} + \frac{1}{2}.$
Solve Real-World Problems Using the Distributive Property
The Distributive Property is one of the most common mathematical properties seen in everyday life. It crops up in business and in geometry. Anytime we have two or more groups of objects, the Distributive Property can help us solve for an unknown.
#### Example C
An octagonal gazebo is to be built as shown below. Building code requires five-foot-long steel supports to be added along the base and four-foot-long steel supports to be added to the roof-line of the gazebo. What length of steel will be required to complete the project?
Solution: Each side will require two lengths, one of five and one of four feet respectively. There are eight sides, so here is our equation.
Steel required $= 8(4 + 5)$ feet.
We can use the Distributive Property to find the total amount of steel.
Steel required $= 8 \times 4 + 8 \times 5 = 32 + 40$ feet.
A total of 72 feet of steel is needed for this project.
### Guided Practice
Simplify $\frac{10x+8y-1}{2}.$
Solution:
First we rewrite the expression so we can see how to distribute the denominator:
$& \frac{10x+8y-1}{2}=\frac{1}{2}(10x+8y-1)= \\&\frac{1}{2}(10x)+\frac{1}{2}(8y)-\frac{1}{2}(1)= 5x+4y-\frac{1}{2}$
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Distributive Property (5:39)
Use the Distributive Property to simplify the following expressions.
1. $(2 - j)(-6)$
2. $(r + 3)(-5)$
3. $6 + (x - 5) + 7$
Use the Distributive Property to simplify the following fractions.
1. $\frac{8x + 12}{4}$
2. $\frac{9x + 12}{3}$
3. $\frac{11x + 12}{2}$
4. $\frac{3y + 2}{6}$
5. $- \frac{6z - 2}{3}$
6. $\frac{7 - 6p}{3}$
In 10 – 17, write an expression for each phrase.
1. $\frac{2}{3}$ times the quantity of $n$ plus 16
2. Twice the quantity of $m$ minus 3
3. $-4x$ times the quantity of $x$ plus 2
4. A bookshelf has five shelves, and each shelf contains seven poetry books and eleven novels. How many of each type of book does the bookcase contain?
5. Use the Distributive Property to show how to simplify 6(19.99) in your head.
6. A student rewrote $4(9x + 10)$ as $36x + 10$ . Explain the student’s error.
8. Amar is making giant holiday cookies for his friends at school. He makes each cookie with 6 oz of cookie dough and decorates each one with macadamia nuts. If Amar has 5 lbs of cookie dough $(1 \ lb = 16 \ oz)$ and 60 macadamia nuts, calculate the following.
1. How many (full) cookies can he make?
2. How many macadamia nuts can he put on each cookie if each is supposed to be identical?
### Vocabulary Language: English Spanish
Distributive Property
Distributive Property
For any real numbers $M, \ N,$ and $K$: $&M(N+K)= MN+MK\\ &M(N-K)= MN-MK$ |
### Kissing
Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it?
### Gold Again
Without using a calculator, computer or tables find the exact values of cos36cos72 and also cos36 - cos72.
### Golden Construction
Draw a square and an arc of a circle and construct the Golden rectangle. Find the value of the Golden Ratio.
# Pentabuild
##### Age 16 to 18Challenge Level
Harry explained why we get a Pentagon:
I firstly calculated the length of $YS$ and the equations of the circle using Pythagoras' theorem on the triangle $YOS$ (and by symmetry $YOR$) to get:
$$YS=\frac{\sqrt{5}}{2}$$ Equation of $C_1$: $$x^2+y^2=1$$ Equation of $C_4$: $$x^2+(y+1)^2=\frac{(\sqrt{5}-1)^2}{4}$$ Equation of $C_5$: $$x^2+(y+1)^2=\frac{\sqrt{5}+1)^2}{4}$$
At the points of intersection of two circles, the points satisfy the equations both both circles and so we have a set of two simultaneous equations to solve. Solving these gives at the intersection of $C_1$ and $C_4$ (so for the points $C$ and $D$), $y=\frac{-(\sqrt{5}+1)}{4}$ and at the intersection of $C_1$and $C_5$, $y=\frac{\sqrt{5}-1}{4}$.
But we can now look at the hint, and find we have the same y coordinates for all our points as those of a regular hexagon. But since we are on the circle, we can work out the $x$ coordinates from the $y$ coordinates. So we have the same points as the regular pentagon in the notes section, and so this is a regular pentagon.
Tom from Bristol Grammar School then suggested a method to construct a regular decagon using the pentagon that we've already constructed.
1. Construct the regular pentagon using the prescribed technique.
2. Bisect the angle $\angle A C E$ by drawing a circle centre $A$ and a circle of the same radius (perhaps $E C$) centre $E$ and drawing a straight line between one of the points at which the circles intersect and point $C$. (This works because $A C=E C$, as the pentagon is regular - it is a fact that is obvious and easily proven using SAS congruence, and therefore it is equivalent to the classroom-taught angle bisection technique.)
3. Let us call the point (other than $C$) at which this line crosses the pentagon's circumcircle $P$. Join $A$ to $P$, and join $E$ to $P$. Essential to our method is that now $A P=P E$, which is clearly true by SAS congruence of the triangles $A C P$ and $E C P$ ($A C=E C$ was used above, $\angle A C P= \angle P C E$ holds because $P C$ is an angle bisector, $C P$ is common).
4. Construct the circle centre $A$ through $P$ and label the point (other than $P$) at which it crosses the circumcircle of the pentagon $Q$. Draw in the line segments $A Q$ and $Q B$. Similarly construct the circle centre $B$ through $Q$, join up the line segments, and repeat this process for $C$ and $D$.
5. The 10-sided shape we now have inscribed in the circle is a regular decagon. |
# 6.4 Factoring and Solving Polynomial Equations
## Presentation on theme: "6.4 Factoring and Solving Polynomial Equations"— Presentation transcript:
6.4 Factoring and Solving Polynomial Equations
Factor Polynomial Expressions
In the previous lesson, you factored various polynomial expressions. Such as: x3 – 2x2 = x4 – x3 – 3x2 + 3x = = Grouping – common factor the first two terms and then the last two terms. Common Factor x2(x – 2) x(x3 – x2 – 3x + 3) x[x2(x – 1) – 3(x – 1)] Common Factor x(x2 – 3)(x – 1)
Solving Polynomial Equations
The expressions on the previous slide are now equations: y = x3 – 2x2 and y = x4 – x3 – 3x2 +3x To solve these equations, we will be solving for x when y = 0.
Solve y = x3 – 2x2 0 = x3 – 2x2 0 = x2(x – 2) x2 = 0 or x – 2 = 0
Let y = 0 y = x3 – 2x2 0 = x3 – 2x2 0 = x2(x – 2) x2 = 0 or x – 2 = 0 x = 0 x = 2 Therefore, the roots are 0 and 2. Common factor Separate the factors and set them equal to zero. Solve for x
Solve Let y = 0 y = x4 – x3 – 3x2 + 3x 0 = x4 – x3 – 3x2 + 3x
x = 0 or x – 1 = 0 or x2 – 3 = 0 x = x = x = Therefore, the roots are 0, 1 and ±1.73 Let y = 0 Common factor Group Separate the factors and set them equal to zero. Solve for x
The Quadratic Formula For equations in quadratic form: ax2 + bx + c = 0, we can use the quadratic formula to solve for the roots of the equation. This equation is normally used when factoring is not an option.
Solve the following cubic equation: y = x3 + 5x2 – 9x 0 = x(x2 + 5x – 9) x = 0 x2 + 5x – 9 = 0 We can, however, use the quadratic formula. Can this equation be factored? We still need to solve for x here. Can this equation be factored? YES it can – common factor. Remember, the root 0 came from an earlier step. No. There are no two integers that will multiply to -9 and add to 5. a = 1 b = 5 c = -9 Therefore, the roots are 0, 6.41 and
Factoring Sum or Difference of Cubes
If you have a sum or difference of cubes such as a3 + b3 or a3 – b3, you can factor by using the following patterns. Note: The first and last term are cubed and these are binomials.
Example Factor x Note: This is a binomial. Are the first and last terms cubed? x = (x)3 + (7)3 = ( )( ) x 7 x2 7x 49
Example Factor 64a4 – 27a = a(64a3 – 27)
Note: Binomial. Is the first and last terms cubes? = a( (4a)3 – (3)3) Note: = a( )( ) 4a 3 16a2 12a 9
Factor by Grouping Some four term polynomials can be factor by grouping. Example Factor 3x3 + 7x2 +12x + 28 Step 1 Pair the terms. Step 2 Factor out common factor from each pair. Identical factors Step 3 Factor out common factor from each term.
Example Factor 3x3 + 7x2 -12x - 28 Step 1
Note: Subtraction is the same as adding a negative Step 2 Step 3 Note: This factor can be further factored
Solving Polynomial Equations
Solve Set equation equal to zero. Factor. Set each factor equal to zero and solve. |
How do you find the derivative of the function y = sin(tan(5x))?
May 24, 2016
$\setminus \cos \left(\setminus \tan \left(5 x\right)\right) \setminus {\sec}^{2} \left(5 x\right) 5$
Explanation:
$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \sin \left(\setminus \tan \left(5 x\right)\right)\right)$
Applying chain rule,
$\setminus \frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \setminus \frac{\mathrm{df}}{\mathrm{du}} \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$
Let $\tan \left(5 x\right)$ = u
We know,
$\setminus \frac{d}{\mathrm{du}} \left(\setminus \sin \left(u\right)\right) = \setminus \cos \left(u\right)$
$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \tan \left(5 x\right)\right) = \setminus {\sec}^{2} \left(5 x\right) 5$
So,
$= \setminus \cos \left(u\right) \setminus {\sec}^{2} \left(5 x\right) 5$
Substituting back,$\tan \left(5 x\right)$ = u
$= \setminus \cos \left(\setminus \tan \left(5 x\right)\right) \setminus {\sec}^{2} \left(5 x\right) 5$ |
Remember---Logs are ‘inverses’ of exponentials.
Presentation on theme: "Remember---Logs are ‘inverses’ of exponentials."— Presentation transcript:
Remember---Logs are ‘inverses’ of exponentials.
Pre-Calc Lesson 5-6 Laws of Logarithms Remember---Logs are ‘inverses’ of exponentials. Therefore all the rules of exponents will also work for logs. Laws of Logarithms: If M and N are positive real numbers and ‘b’ is a positive number other than 1, then: 1. logb MN = logb M + logb N 2. logb M = logb M – logbN N logb M = logb N iff M = N logb Mk = k logb M, for any real number k Example 1: Express logbMN2 in terms of logbM and logbN 1st: Recognize that you are taking the ‘log’ of a product (M)(N2) So we can split that up as an addition of two separate logs!
Example 3: Simplify log 45 – 2 log 3
Logb MN2 = logbM + logbN2 (Now recognize that we have a power on the number in the 2nd log. = logbM + 2logbN ! Example 2 : Express logb M3 in terms of logbM and logbN N Logb M3 = logb (M3)1/2 = ½ (logb(M3)) N (N) (N) = ½ (logbM3 – logbN) = ½ (3logbM – logbN) = 3/2 logbM – ½ logbN Example 3: Simplify log 45 – 2 log 3 log 45 – 2 log 3 = log 45 – log 32 = log 45 – log 9 = log (45/9) = log 5
Example 4: Express y in terms of x if ln y = 1/3 ln x + ln 4
So the only way ln y = ln 4x1/3 is if : y = 4x1/3 Example 5: Solve log2x + log2(x – 2) = 3 log2x(x - 2) = 3 (Go to exponential form: = x(x – 2) 8 = x2 – 2x 0 = x2 - 2x - 8 0 = (x – 4)(x + 2) x = 4, x = - 2 Now the domain of all log statements is (0, ф) x ≠ - 2 so x = 4 is the only solution!!! |
Question
# The HCF and LCM of two numbers are 12 and 72 respectively. If sum of the two numbers is 60, then find one of the numbers among the following:A. 12B. 24C. 60D. 72
Hint: We will first assume both the numbers and apply the fact that:
HCF $\times$ LCM = Product of two numbers. From here, find one number in terms of another. Put this in the equation we can form by a given sum of numbers and thus we will have the answer.
Let us first understand the meaning of “Least Common Multiple” and “Highest Common Factor”:
Multiple:
We get a multiple of a number when we multiply it by another number.
Common Multiple:
The common multiples are those that are found lists of multiples of both.
Least Common Multiple:
It is simply the smallest of the common multiples.
Highest Common Factor:
It is simply the largest of the common factors.
Let the numbers be $x$ and $y$ respectively.
$HCF\; \times LCM{\text{ }} = {\text{ }}x\; \times y = xy$ …… (1)
We have with us that: The HCF and LCM of two numbers are 12 and 72 respectively.
Putting these values in (1), we will have:-
$12 \times 72 = xy$
So, we have:- $xy = 864$
So, $y = \dfrac{{864}}{x}$ ……..(2)
We also have with us that: sum of the two numbers is 60.
So, we will get:- $x + y = 60$ …..… (3)
Putting (2) in this, we will get:-
$x + \dfrac{{864}}{x} = 60$
Taking LCM to simplify the equation, we will get:-
$\dfrac{{{x^2} + 864}}{x} = 60$
Cross multiplying to get a simpler expression, we get:-
${x^2} + 864 = 60x$
That is ${x^2} - 60x + 864 = 0$
We can rewrite it as:
${x^2} - 36x - 24x + 864 = 0$
$x(x - 36) - 24(x - 36) = 0$
$(x - 24)(x - 36) = 0$
So, either $x = 24$ or $x = 36$.
If $x = 24$, then $y = 36$ and if $x = 36$, then $y = 24$.
So, the correct answer is “Option B”.
Note: We should always keep in mind that we need as much equations as much unknown variables. Like, here in this question, we had two unknown variables, so we created two equations (1) and (2).
We can solve the quadratic equation either by the method as above that is factorization method or even by using the direct formula which is: If we have $a{x^2} + bx + c$, then $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
The students might just find one of the numbers and try finding it among the options which do not even contain it. So, we must find both the numbers so that we choose whichever we see in the option. |
# How to Solve It! How to Solve It Polya, G. (1957). How to solve it: A new aspect of mathematical method, 2nd ed. Princeton, NJ: Princeton University.
## Presentation on theme: "How to Solve It! How to Solve It Polya, G. (1957). How to solve it: A new aspect of mathematical method, 2nd ed. Princeton, NJ: Princeton University."— Presentation transcript:
How to Solve It!
How to Solve It Polya, G. (1957). How to solve it: A new aspect of mathematical method, 2nd ed. Princeton, NJ: Princeton University Press. George Polya’s Four-Step Method 1. Understand the Problem 2. Devise a Plan 3. Carry Out the Plan 4. Look Back
Polya Illustrated Find the Number (logic problem) Help Wanted (algebra problem) Process Review Student Reflection
Find the Number Use the ten clues, in order, to find the mystery number. 1.It is a five-digit whole number. 2.It is divisible by 5. 3.It is divisible by 4. 4.The sum of its ten-thousands digit and thousands digit is 14. 5.The sum of its ten-thousands digit and its hundreds digit is 11. 6.The sum of its thousands digit and its tens digit is 8. 7.The sum of its hundreds digit and its units digit is 3. 8.The sum of its tens digit and its units digit is 2. 9.It is greater than 80,000. 10.Its thousands digit is 6.
1. Understand the Problem Read the problem carefully. You’ll have to focus on using the clues IN ORDER. Be sure you understand the terms – What is a “five-digit whole number”? – What does “divisible” mean? – What are units digits, tens digit, etc.? Yes, these may seem very easy… but that may not always be the case.
1. Understand the Problem Break it into parts. Do you have all the tools you need? Enough information Appropriate skills Can you draw a picture? Artistic skill not required. Q:How do you eat an elephant? A:One bite at a time!
1. Understand the Problem Introduce notation and variables. Clue (1) tells us we have a 5-digit number. Without knowing anything else, we can write it as follows: ab,cde
2. Devise a Plan Follow the clues systematically Use rules of divisibility to limit possibilities Use algebraic equations to find digits If the problem is too hard, try doing an easier problem – In this problem, you could rearrange the clues. BUT…when you look back you’ll realize that doesn’t quite answer the question.
3. Carry Out the Plan From Clue (1): – Form of the number is ab,cde. From Clue (2): – Divisible by 5 — therefore, the units digit is 0 or 5. From Clue (3): – Divisible by 4 — therefore, it is even, and the last two digits must be divisible by 4. So, the units digit is 0, and the tens digit is 2, 4, 6, or 8.
(1) ab,cde (2, 3) d is even, and e = 0 (4) a + b = 14 (5) a + c = 11 (6) b + d = 8 (7) c + e = 3 “Aha! I have enough info. I don’t need any more clues.” If e = 0 and c + e = 3, then c = 3 So, a = 8 Then, b = 6 And, d = 2 The number is… 86,320 3. Carry Out the Plan
Is it reasonable? Does it satisfy the clues… including the clues we didn’t need? (8) d + e = 2 (9) greater than 80,000 (10) thousands digit is 6 We used 7 clues. – Could we have solved the problem with fewer? Did we answer the question? Don’t just give a, b, c, d, and e. The problem asks for a five-digit number. 4. Look Back
What did I learn from this problem? – Review simple algebra skills – Organized, logical thinking How can I use it to solve other problems? Could I have used another method? – Trial and error, maybe (but why???) Find the Number (logic problem) Help Wanted (algebra problem) Process ReviewStudent Reflection
Help Wanted Serge works at the supermarket as a cashier, a bagger, and a stocker. He earns \$7.00 per hour as a cashier, \$6.00 per hour as a bagger, and \$5.00 per hour as a stocker. In a given week, Serge works 4 hours as a cashier, 9 hours as a bagger, and 7 hours as a stocker. What is Serge’s average pay per hour?
1. Understand the Problem Read it carefully. Note that you’re asked for his average pay per hour. Be sure you understand the terms. – Does it matter what jobs he has, or only that he has three jobs? Yes, this may seem very easy… but that may not always be the case.
1. Understand the Problem Break it into parts. Do you have all the tools you need? Enough information? Appropriate skills? Can you draw a picture? Artistic skill not required. A chart is a picture. Q:How do you eat an elephant? A:One bite at a time!
1. Understanding the Problem What information do we have? How can we organize it? Cashier\$7.00 per hour4 hours Bagger\$6.00 per hour9 hours Stocker\$5.00 per hour7 hours
2. Devise a Plan Consider one job at a time. Find earnings for each job. Find total earnings and total hours. Find the average.
3. Carry Out the Plan Jobs – Cashier:4 hr × \$7/hr = \$28 – Bagger:9 hr × \$6/hr = \$54 – Stocker:7 hr × \$5/hr = \$35 Total pay: \$28 + \$54 + \$35 = \$117 Total hours: 4 + 9 + 7 = 20 hr Average pay: \$117 ÷ 20 hr = \$5.85/hr
4. Look Back Is it reasonable? What did I learn from this problem? – The answer is not \$6.00 How can I use it to solve other problems? How else could I have solved the problem? – Weighted average Find the Number (logic problem) Help Wanted (algebra problem) Process ReviewStudent Reflection
Process Review George Polya’s Four-Step Method 1. Understand the Problem 2. Devise a Plan 3. Carry Out the Plan 4. Look Back
1. Understand the Problem READ the problem. What are you expected to find: unknowns What are you given: data, situation or condition – Do you have enough information? – Is any information unnecessary or contradictory? Explain the problem in your own words – Break it into parts Draw a picture or diagram Introduce suitable notation and variables
2. Devise a Plan Look at the unknown. – Do you know how to solve problems with this type of unknown? Can you find a connection between the data and the unknown? Have you seen a problem like this before? Do you know any similar problems you can use to help you?
2. Devise a Plan If it’s too hard… READ it again! Do you understand the problem? Redo Step 1. Can you restate the problem? Can you break the problem into pieces? If you cannot solve the proposed problem… – Try to first solve a related (maybe easier) problem Did you use all the data? …the whole condition?
3. Carry Out the Plan Follow your plan CHECK EACH STEP Can you see clearly that each step is correct? If it doesn’t lead to a solution… – Return to Step 2. Pick another plan!
4. Look Back Can you check your result? Can you check your logic? Did you answer the question? Is it reasonable? Can you do the problem differently? Can you use your results to help you solve other problems? Have you learned something from this problem?
Student Reflection Think about it… Which step do you think is most difficult? Which step do you think is most important? How can Polya’s method help you with – other types of math problems? – problems from other areas of school and life? How can metacognition help you with problem solving?
Try It Yourself Use the ten clues, in order, to find the mystery number. 1.It is a four-digit whole number. 2.It is greater than 4,000. 3.The sum of its hundreds digit and its units digit is 9. 4.Twice its tens digit is 2 more than its thousands digit. 5.The sum of one-fifth of its hundreds digit and two-thirds of its units digit is 6. 6.Its tens digit is one less than its thousands digit. 7.The product of its hundreds digit and its ones digit is 0. 8.It is not an even number. 9.It is less than 5,000. 10.Its tens digit is 3.
Try It Yourself Use a full sheet of paper for your problem. Identify each step of the process. – Allow space for each step of the process. A book club promises to send 8 books for \$1, if you join the club. After you receive the first 8 books, you may select more books at a rate of \$19.99 per book. If you spend a total of \$80.96, how many extra books did you purchase?
Similar presentations |
# Apostol 4.12 #28 - Related Rates Problem
1. Aug 19, 2011
### process91
Here is the question verbatim:
The radius of a right-circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as the radius. When the radius is 1 foot, the altitude is 6 feet. When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. When the radius is 36 feet, the volume is increasing at a rate of n cubic feet per second, where n is an integer. Compute n.
I set the problem up as follows:
The radius of a right-circular cylinder increases at a constant rate. Then let r be the radius and t be the number of seconds. This implies that $\frac{dr}{dt}=c$ for some constant c.
Its altitude is a linear function of the radius Letting h be the altitude, we have h=ar+b for some constants a and b.
and increases three times as fast as the radius. Then $\frac{dh}{dt}=3\frac{dr}{dt}=3c$. Also, we have $\frac{dh}{dr}=a$ and so by the chain rule, $\frac{dh}{dr}\frac{dr}{dt}=\frac{dh}{dt} \implies a=2c$ as long as $c\ne0$, which we will have shortly.
When the radius is 1 foot, the altitude is 6 feet. So $6=2c+b$.
When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. Letting v be the volume, we have $\frac{dv}{dt}=1$. This is how we know that $c\ne0$, since $\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}$.
Now we work to find n. We have $6=2c+b \implies b=6-2c \implies h=2cr+6-2c$. The volume of a right-circular cylinder is given by $v = \pi r^2 h =\pi r^2 (2cr+6-2c) = 2c\pi r^3+(6-2c)\pi r^2$ and so $\frac{dv}{dr}=6c\pi r^2 + 2(6-2c)\pi r$. Using the chain rule, we have $\frac{dv}{dt}=(6c\pi r^2 + 2(6-2c)\pi r)c$ and when r=6, $1=(6c\pi 6^2 + 2(6-2c)\pi 6)c$ which, simplified, is $1=192 \pi c^2+72 \pi c$. From this alone, we can see that whatever c is, it is a very complicated number because it must remove the values of pi in the equation. Solving this using the quadratic equation will give us two possible values:
Solved with WolframAlpha
Using the positive one, since the radius is increasing, we should be able to compute $\frac{dv}{dt}$ directly, however for r=36 we don't get an integer.
Final Solution from WolframAlpha
The book's answer is 33. Where did I go wrong?
Last edited: Aug 19, 2011
2. Aug 19, 2011
### micromass
Hi process91!
I don't see that. Doesn't the information give you that $\frac{dh}{dt}=3c$?? You seem to say that $\frac{dr}{dt}=3c$ and I don't see why you would say that.
3. Aug 19, 2011
### dynamicsolo
Have a look at this line again (I corrected a typo):
Doesn't this imply ac = 3c ?
4. Aug 19, 2011
### process91
Yes, that was a typo. I've corrected it in the original problem.
5. Aug 19, 2011
### dynamicsolo
OK, that takes care of the typo, but your implication is still incorrect. Also,
I think you want to write 6 = 3r + b here. This equation for the height involves r , not dr/dt .
6. Aug 19, 2011
### process91
What part of the implication is incorrect? Just to make sure we're talking about the same spot, I have that
$\frac{dh}{dr}=a$ and $\frac{dr}{dt}=c$ and $\frac{dh}{dt}=3c$. Using the chain rule, that implies that $ac=3c$ and, as long as $c\ne0$, $a=2c$.
For the part where I write $6 = 2c + b$, I am using the given statement that the height is 6 when the radius is 1 and the fact that $h=ar+b=2cr+b$ based on my above implication. Taking r=1 and h=6 yields $6=2c+b$.
7. Aug 19, 2011
### micromass
Yes, but doesnt ac=3c imply that a=3??
8. Aug 19, 2011
### process91
*blink, blink* Yes, it does. I must be tired or something. Thank you so much - I was tearing my hair out. Sometimes it's the little things.
Thanks to dynamicsolo as well, I see that you were also pointing to this flaw.
9. Aug 19, 2011
### dynamicsolo
Sleep deprivation (or fatigue) and mathematics don't mix...
By the way, you will never need to use dr/dt for anything -- it can be divided out when you compare dV/dt at r = 6 with dV/dt at r = 36 .
Last edited: Aug 19, 2011 |
## Word Problem Solution1
In this page word problem solution1 we are going to see solution for first word problem of the worksheet cross multiplication method.
Formulate the following problems as a pair of equations,and hence find their solutions:
(i) One number is greater than thrice the other number by 2. If 4times the smaller number exceeds the greater by 5, find the numbers.
Solution:
Let “x” be one number and “y “ be another number
Here one number is defined by other number
x = 3 y + 2 ----- (1)
Here we have two variables x and y. In these two terms x is greater number and y is smaller number
4y = x + 5 ----- (2)
Substitute (11) in the second equation we get
4 y = 3y + 2 + 5
4 y - 3 y = 7
y = 7
Now we need to apply the value of y in first or second equation to get the value of y.
Substitute y = 7 in the first equation
x = 3(7) + 2
x = 21 + 2
x = 23
Therefore one number is 23 and other number is 7.
Verification:
x = 23 y = 7
23 = 3 (7) + 2
23 = 21 + 2
23 = 23
word problem solution1 word problem solution1
Try other questions
(ii) The ratio of income of two persons is 9:7 and the ratio of their expenditure is 4:3.If each of them manages to save \$2000 per month, find their monthly income. Solution
(iii) A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number. Solution
(iv) Three chairs and two tables cost \$700 and five chairs and three tables cost \$1100.What is the total cost of 2 chairs and 3 tables. Solution
(v) In a rectangle,if the length is increased and breadth is reduced each by 2 cm then the area is reduced by 28 cm ². If the length is reduced by 1 cm and the breadth is increased by 2 cm,then the area increases by 33cm². Find the area of the rectangle. Solution
(vi) A train traveled a certain distance at a uniform speed. If the train had been 6 km/hr faster,it would have taken 4 hours less than the scheduled time. If the train were slower by 6km/hr, then it would have taken 6 hours more than the scheduled time. Find the distance covered by the train. Solution |
# Comparing Quantities Class 7 Maths Formulas
For those looking for help on Comparing Quantities Class 7 Math Concepts can find all of them here provided in a comprehensive manner. To make it easy for you we have jotted the Class 7 Comparing Quantities Maths Formulae List all at one place. You can find Formulas for all the topics lying within the Comparing Quantities Class 7 Comparing Quantities in detail and get a good grip on them. Revise the entire concepts in a smart way taking help of the Maths Formulas for Class 7 Comparing Quantities.
## Maths Formulas for Class 7 Comparing Quantities
The List of Important Formulas for Class 7 Comparing Quantities is provided on this page. We have everything covered right from basic to advanced concepts in Comparing Quantities. Make the most out of the Maths Formulas for Class 7 prepared by subject experts and take your preparation to the next level. Access the Formula Sheet of Comparing Quantities Class 7 covering numerous concepts and use them to solve your Problems effortlessly.
To compare quantities, there are multiple methods, such as ratio and proportion, percentage, profit and loss, and simple interest.
The ratio of two quantities of the same kind and in the same unit is the fraction that one quantity is of the other.
The ratio a is to b as is the fraction $$\frac { a }{ b }$$, and it is written as a : b.
In the ratio a : b, we call a as the first term or antecedent and b the second term or consequent.
To compare different ratios, firstly convert fractions into like fractions. If like fractions are equal, then the given ratios are said to be equivalent.
e.g. To check 1 : 2 and 2 : 3 are equivalent.
Therefore, the ratio 1 : 2 is not equivalent to the ratio 2 : 3.
To compare two quantities, units must be the same.
If the two ratios are equal, the four quantities are called in proportion.
a : b = c : d ⇒ a : b :: c : d.
Fractions are converted to percentages by multiply the fraction by 100 and write % sign
e.g. $$\frac { 1 }{ 4 }$$ = $$\frac { 1 }{ 4 }$$ × 100 = 25%
Decimals are converted to percentages by multiply the decimal number by 100 and shift the decimal point two places to the right side and write % sign.
e.g. 2.42 × 100 = 242 %
Profit = SP – CP [∵ SP > CP]
SP = Selling Price
CP = Cost Price
Loss = CP – SP [∵ CP > SP]
CP = Cost Price
SP = Selling Price
Profit % = $$\frac { Profit }{ CP }$$ × 100
Loss % = $$\frac { Loss }{ CP }$$ × 100
Equivalent Ratios
Different ratios are compared with each other to know whether they are equivalent or not. For this, we write the ratios in the form of fractions and then compare them by converting them into like fractions. If these like fractions are equal, we say that the given ratios are equivalent. Equivalent ratios are very important. Two ratios are said to be equivalent if when converted into like fractions, they are equal.
Unitary Method
In the unitary method, we first find the value of one unit and then the value of the required number of units.
Percentage-Another way of Comparing Quantities
Percentages are numerators of fractions with denominator 100. They are used for comparisons.
Meaning of Percentage
Percent means ‘per hundred’. It is represented by the symbol % and means hundredths too.
Thus, 1% means 1 out of hundred or one-hundredths. It can be written as:
1% = $$\frac { 1 }{ 100 }$$ = 0.01
Converting Fractional Numbers to Percentage
Fractional numbers can have different denominators. To compare fractional numbers we need a common denominator and it is more convenient to compare if the denominator is 100. So, we convert the fraction to percentages.
Percentages related to proper fractions are less than 100 whereas percentages related to improper fractions are more than 100.
Converting Decimals to Percentage
We multiply the decimal by 100 and affix percentage symbol.
Ratios to Percents
Sometimes, parts are given to us in the form of ratios and we need to convert those to percentage.
Increase or Decrease as Percent
There are times when we need to know what the increase in a certain quantity or decrease in it is as percent.
For example, if the population of a state is increased from 5,50,000 to 6,05,000, this could more clearly be understood if written as:
The population is increased by 10%.
Prices Related to an Item on Buying and Selling
The buying price of an item is known as its Cost Price (CP).
The price at which we sell an item is known as its Selling Price (SP).
Profit or Loss as a Percentage
Cost Price: The buying price of an item is known as its cost price written in short as CP.
Selling Price: The price at which we sell an item is known as the selling price or in short SP.
Naturally, it is better if we sell the item at a higher price than our buying price.
Profit or Loss: We can decide whether the sale was profitable or not depending on the CP and SP.
If CP < SP then we have gained some amount, that is, we made a profit, profit = SP – CP
If CP = SP then we are in a no profit no loss situation
If CP > SP then we have lost some amount, Loss = CP – SP.
The profit or loss we find can be converted to a percentage. It is always calculated on the CP.
Note. If we are given any two of the three quantities related to price, that is, CP, SP, and Profit or Loss percent, we can find the third.
Charge has given on Borrowed Money or Simple Interest
Principal: The money borrowed is known as sum borrowed or principal.
Interest: We have to pay some extra money (or charge) to the bank for the money being used by us for some time. This is known as interest.
Amount: We can find the amount we have to pay at the end of the year by adding the above two. That is.
Amount = Principal + Interest.
Note: Interest is generally given in per cents for a period of one year. It is written as x percent per year or per annum or in short as x% p.a. (say 10 percent per year)
10% p.a. means on every ₹ 100 borrowed, ₹ 10 is the interest we have to pay for one year. |
# What is the antiderivative of (2x-1)^3?
Aug 6, 2015
$\int {\left(2 x - 1\right)}^{2} \mathrm{dx} = \frac{1}{8} {\left(2 x - 1\right)}^{4} + c$
#### Explanation:
To solve:
$\int {\left(2 x - 1\right)}^{2} \mathrm{dx}$
We use the substitution:
$u = 2 x - 1$ => $\frac{\mathrm{du}}{\mathrm{dx}} = 2$ => $\mathrm{dx} = \frac{1}{2} \mathrm{du}$
This gives:
$\int {\left(2 x - 1\right)}^{3} \mathrm{dx} = \frac{1}{2} \int {u}^{3} \mathrm{du} = \frac{1}{8} {u}^{4} + c$
Transforming back to the original variable we have:
$\frac{1}{8} {u}^{4} + c = \frac{1}{8} {\left(2 x - 1\right)}^{4} + c$ |
# Video: KS1-M16S • Paper 2 • Question 11
KS1-M16S-P2-Q11-702171938739
02:55
### Video Transcript
Use a ruler to measure the length of the toy car.
We’re told that we need to measure the length of the toy car. We can see that underneath the car is a line that we can use to help us. Each end of the line is level with each end of the car. So to find the length of the toy car, all we have to do is to measure the line. And we’re told that we need to use a ruler. So let’s get one.
Here’s a ruler that we can use. Our ruler measures centimetres. We can see the letters “cm” on it. And “cm” stands for centimetres. Our ruler goes up as far as 12 centimetres. So hopefully, the car isn’t longer than that.
First, we need to line the ruler up with the line that we’re measuring. We’ve lined one end of the ruler up with one end of the line. If we read the number that’s level with the other end of the line, we can see that it says 10 and a half centimetres. It’s halfway between 10 and 11. Is the answer 10 and a half centimetres?
No. Whenever we measure something, we need to start measuring from zero on the scale and look at where zero is. it’s not level with the end of the line. We need to move our ruler so that, instead of the end of the ruler being level with the end of the line, we need to make sure the number zero is level. This is better. We’ve moved the ruler so that zero on the scale is level with the end of the line. And we can start measuring from zero. This is the correct way to measure length.
If we read the number at the other end of the car, we can see that the length of the toy car is 11 centimetres. At the moment, you’re watching this on a video. So if you get a ruler out and try to measure the length of the toy car on the screen, it probably won’t be 11 centimetres long. It depends on the size of the screen. If you’re watching on a mobile phone, the length of the toy car is going to be smaller than if you’re watching on a large screen. If you or your teacher prints out the question, it should be the correct size.
Hopefully, the video has shown you how to measure correctly. Always line the ruler up with zero next to the end of the thing that you’re measuring. Don’t look for the end of the ruler. Look for the zero. We did that as we measured the car. And we found that the length of the toy car was 11 centimetres. |
Algebra
# Polynomial Interpolation: Level 4 Challenges
$f(x)$ is a $5^\text{th}$-degree polynomial such that $f(1)=2,$ $f(2)=3,$ $f(3)=4,$ $f(4)=5,$ $f(5)=6,$ and $f(8)=7.$
If the value of $f(9)$ can be expressed as $\frac{a}{b}$ for coprime positive integers $a$ and $b$, find the value of $a+b$.
$f(x)$ is a polynomial with integer coefficients. We have,
$f(1)=1$
$f(2)=4$
$f(3)=9$
$f(4)=16$
$f(5)=25$
$f(6)=36$
$f(7)=49$
$f(8)=64$
$f(9)=81$
$f(10)=100$
Determine $f(11)$.
Given that $P(x)$ is a monic fifth-degree polynomial such that
\begin{aligned} P(1) & = & 1^2 \\ P(2) & = & 2^2 \\ P(3) & = & 3^2 \\ P(4) & = & 4^2 \\ P(5) & = & 5^2,\end{aligned}
find the value of $P(6)$.
Roopesh went to his friends house. There he ate a lot of Ice-cream. First he started with Vanilla then Chocolate then Vanilla then Butterscotch then Vanilla. He is a mathematician and made a monic polynomial of degree 5 which gave him values of the first letter of the Ice Cream.
This is the way his polynomial proceeded :-
$p(1) = 22$
$p(2) = 3$
$p(3) = 22$
$p(4) = 2$
$p(5) = 22$
If the value of $p(6) = \overline{abc}$. The first letter of the ice cream he eats would be $a+b+c$. Which ice cream will be eat next?
Assumption: $\overline{abc}$ represents a 3 digit number.
$\large f(1) = 4, f(2) = 9, f(3) = 20, f(4) = 44, f(5) = 88$
Let $f(x)$ be a 5th degree monic polynomial such that it satisfies the equations above. Find the value of $f(6)$.
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# How do you sketch the graph of $y = {\left( {x + 2} \right)^2}$ and describe the transformation?
Last updated date: 23rd Feb 2024
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Hint: In this question we need to find the graph of $y = {\left( {x + 2} \right)^2}$ and determine the transformation for it. To obtain the graph find the value of y for the different value of x and then mark the points on the graph to make the graph. Also to determine the transformation, determine the change in the transformation from parent function to given function.
Complete Step By Step solution:
In this question we have given a function that is $y = {\left( {x + 2} \right)^2}$ and we need to sketch the graph and need to describe the transformation.
The above function is the quadratic function. The parent function for the quadratic function is $y = {x^2}$.
Consider the table of values for this parent function is,
$y$ $x$ $4$ $- 2$ $9$ $- 3$ $25$ $- 5$ $49$ $- 7$ $64$ $- 8$ $4$ $2$ $9$ $3$ $25$ $5$ $49$ $7$ $64$ $8$
From the above table the graph for the parent function is shown below.
Now we will consider the table of values for this given function is,
$y = {\left( {x + 2} \right)^2}$ $x$ $0$ $- 2$ $1$ $- 3$ $9$ $- 5$ $25$ $- 7$ $36$ $- 8$ $16$ $2$ $25$ $3$ $49$ $5$ $81$ $7$ $100$ $8$
From the above table the graph for the given function is shown in figure below.
From the graph for the parent function and the graph for the given equation. It is concluded that the graph of ${\left( {x + 2} \right)^2}$ is shifted $2$ unit to the left from the parent function ${x^2}$.
Note:
As we know that the quadratic equation is the equation that is of the standard from $a{x^2} + bx + c$. Here, a and b are the coefficients and c is the constant. In the general equation the highest power of the x is $2$ so the equation is called quadratic. The range of all the quadratic functions lies from $- \infty$ to $\infty$. |
# Solve one step linear equations by balancing using inverse operations
Solving linear equations is an important and fundamental skill in algebra. In algebra, we are often presented with a problem where the answer is known, but part of the problem is missing. The missing part of the problem is what we seek to find. An example of such a problem is shown below.
Example 1:
4x +16 = – 4
Notice the above problem has a missing part, or unknown, that is marked by x. If we are given that the solution to this equation is -5, it could be plugged into the equation, replacing the x with -5. This is shown in Example 2.
Example 2:
Solve 4x + 16 = -4
Multiply 4( -5)
4( -5) + 16 =-4
-20 + 16 =-4
-4 =-4 True!
Now the equation comes out to a true statement! Notice also that if another number, for example, 3, was plugged in, we would not get a true statement as seen in Example 3.
Example 3:
Multiply 4(3)
4(3) + 16 =-4
12 +16 =-4
$28\neq -4$ (False).
Due to the fact that this is not a true statement, this demonstates that 3 is not the solution. However, depending on the complexity of the problem, this “guess and check” method is not very efficient. Thus, we take a more algebraic approach to solving equations. Here we will focus on what are called “one-step equations” or equations that only require one step to solve. While these equations often seem very fundamental, it is important to master the pattern for solving these problems so we can solve more complex problems.
To solve equations, the general rule is to do the opposite. For example, consider the following example.
Example 4:
Solve x +7 = -5
The 7 is added to the x
x + 7 – 7 = -5 – 7
x = – 12
Then we get our solution, x = – 12. The same process is used in each of the following examples.
Subtraction Problems
In a subtraction problem, we get rid of negative numbers by adding them to both sides of the equation. For example, consider the following example.
Example 5:
x – 5 = 4
The 5 is negative, or subtracted from x
x – 5 + 5 = 4 + 5
x = 9
Then we get our solution x = 9. The same process is used in each of the following examples. Notice that each time we are getting rid of a negative number by adding.
Multiplication Problems
With a multiplication problem, we get rid of the number by dividing on both sides. For example consider the following example.
Example 6:
4x = 20
Variable is multiplied by 4
Divide both sides by 4
x = 5
Then we get our solution x =5
Division Problems:
In division problems, we get rid of the denominator by multiplying on both sides. For example consider our next example.
$\frac{x}{5}=-3$
Variable is divided by 5
Multiply both sides by 5
$5\frac{x}{5}=-3(5)$
x = -15
Then we get our solution x = – 15.
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# 8.8: Problem-Solving Strategies
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Read and understand given problem situations.
• Make tables and identify patterns.
• Solve real-world problems using selected strategies as part of a plan.
## Introduction
Problem solving appears everywhere, in your regular life as well as in all jobs and careers. Of course, in this manual we concentrate on solving problems that involve algebra. From previous sections, remember our problem solving plan.
Step 1
Understand the problem.
Read the problem carefully. Once the problem is read, list all the components and data that are involved. This is where you will be assigning your variables.
Step 2
Devise a plan – Translate.
Come up with a way to solve the problem. Set up an equation, draw a diagram, make a chart or construct a table as a start to solving your problem.
Step 3
Carry out the plan – Solve.
This is where you solve the equation you came up with in Step 2.
Step 4
Look – Check and Interpret.
Check to see if you used all your information and that the answer makes sense.
## Examples of Exponential Word Problems
In this section, we will be applying this problem solving strategy to solving real-world problems where exponential functions appear. Compound interest, loudness of sound, population increase, population decrease or radioactive decay are all applications of exponential functions. In these problems, we will use the methods of constructing a table and identifying a pattern to help us devise a plan for solving the problems.
Example 1 Compound Interest
Suppose $4000 is invested at 6% interest compounded annually. How much money will there be in the bank at the end of five years? At the end of 20 years? Solution Step 1 Read the problem and summarize the information.$4000 is invested at 6% interest compounded annually
We want to know how much money we have after five years.
Assign variables.
Let x=\begin{align*}x=\end{align*} time in years
Let y=\begin{align*}y=\end{align*} amount of money in investment account
Step 2
Look for a pattern.
We start with $4000 and each year we apply a 6% interest on the amount in the bank Start1st year2nd year$4000InterestThis is added to the previous amountPrevious amount + interest on the previous amount.=4000×(0.06)=$240=$4000+$4000×(0.06)=$4000(1+0.06)=$4000(1.06)=$4240=$4240(1+0.06)=$4240(1.06)=4494.40\begin{align*}\text{Start} && \4000 \\ 1^{st}\ \text{year} && \text{Interest} & = 4000 \times (0.06) = \240 \\ && \text{This is added to the previous amount} & = \4000 + \4000 \times (0.06) \\ & && = \4000(1 + 0.06) \\ & && = \4000 (1.06) \\ & && = \4240 \\ 2^{nd}\ \text{year} && \text{Previous amount + interest on the previous amount}. & = \4240(1 + 0.06) \\ & && =\4240 (1.06) \\ & && = \4494.40\end{align*} The pattern is that each year we multiply the previous amount by the factor of 1.06. Let’s fill in a table of values. Time (Years)Investments Amount()040001424024494.434764.0645049.9055352.9\begin{align*}& \text{Time (Years)} && 0 && 1 && 2 && 3 && 4 && 5 \\ & \text{Investments Amount} (\) && 4000 && 4240 && 4494.4 && 4764.06 && 5049.90 && 5352.9\end{align*}
Answer We see that at the end of five years we have 5352.90 in the investment account. Step 3 In the case of 5 years, we don’t need an equation to solve the problem. However, if we want the amount at the end of 20 years, we get tired of multiplying by 1.06, and we want a formula. Since we take the original investment and keep multiplying by the same factor of 1.06, that means we can use exponential notation. y=4000(1.06)x\begin{align*}y = 4000 \cdot (1.06)^x\end{align*} To find the amount after 5 years we use \begin{align*}x=5\end{align*} in the equation. \begin{align*}y = 4000 \cdot (1.06)^5 = \5352.90\end{align*} To find the amount after 20 years we use \begin{align*}x = 20\end{align*} in the equation. \begin{align*}y = 4000 \cdot (1.06)^{20} = \12828.54\end{align*} Step 4 Looking back over the solution, we see that we obtained the answers to the questions we were asked and the answers make sense. To check our answers we can plug in some low values of \begin{align*}x\end{align*} to see if they match the values in the table: \begin{align*}x=0, && y & =4000 \cdot (1.06)^0 = 4000 \\ x=1, && y &= 4000 \cdot (1.06)^1 = 4240 \\ x=2, && y &= 4000 \cdot (1.06)^2 = 4494.4\end{align*} The answers make sense because after the first year the amount goes up by240 (6% of 4000). The amount of increase gets larger each year and that makes sense because the interest is 6% of an amount that is larger and larger every year. Example 2 Population decrease In 2002, the population of school children in a city was 90,000. This population decreases at a rate of 5% each year. What will be the population of school children in year 2010? Solution Step 1 Read the problem and summarize the information. In 2002, population \begin{align*}= >90,000\end{align*}. Rate of decrease \begin{align*}= 5\%\end{align*} each year. What is the population in year 2010? Assign variables. Let \begin{align*}x=\end{align*} time since 2002 (in years) Let \begin{align*}y=\end{align*} population of school children Step 2 Look for a pattern. Let’s start in 2002. Population \begin{align*}= 90,000\end{align*} Rate of decrease is 5% each year, so we need to find the amount of increase by \begin{align*}90,000 \times 0.05\end{align*} and subtract this increase from the original number \begin{align*}90,000 - 90,000 \times 0.05 = 90,000(1 - 0.05) = 90,000 \times 0.95\end{align*}. \begin{align*}\text{In}\ 2003 && \text{Population} & = 90,000 \times 0.95 \\ \text{In}\ 2004 && \text{Population} & = 90,000 \times 0.95 \times 0.95\end{align*} The pattern is that for each year we multiply by a factor of 0.95 Let’s fill in a table of values: \begin{align*} & \text{Year} && 2002 && 2003 && 2004 && 2005 && 2006 && 2007 \\ & \text{Population} && 90,000 && 85,500 && 81,225 && 77,164 && 73,306 && 69,640\end{align*} Step 3 Let’s find a formula for this relationship. Since we take the original population and keep multiplying by the same factor of 0.95, this pattern fits to an exponential formula. \begin{align*}y = 90000 \cdot (0.95)^x\end{align*} To find the population in year 2010, plug in \begin{align*}x=8\end{align*} (number of years since 2002) \begin{align*}y= 90000 \cdot (0.95)^8 = 59,708\end{align*} school children Step 4 Looking back over the solution, we see that we answered the question we were asked and that it makes sense. The answer makes sense because the numbers decrease each year as we expected. We can check that the formula is correct by plugging in the values of \begin{align*}x\end{align*} from the table to see if the values match those given by the formula. \begin{align*}\text{Year}\ 2002, x=0 && \text{Population} & = y=90000 \cdot (0.95)^0 = 90,000 \\ \text{Year}\ 2003, x=1 && \text{Population} & = y = 90000 \cdot (0.95)^1 = 85,500 \\ \text{Year}\ 2004, x=2 && \text{Population} & = y = 90000 \cdot (0.95)^2 = 81,225\end{align*} Example 3 Loudness of sound Loudness is measured in decibels (dB). An increase in loudness of 10 decibels means the sound intensity increases by a factor of 10. Sound that is barely audible has a decibel level of 0 dB and an intensity level of \begin{align*}10^{-12} \ W/m^2\end{align*}. Sound painful to the ear has a decibel level of 130 dB and an intensity level of \begin{align*}10\ W/m^2\end{align*}. (a) The decibel level of normal conversation is 60 dB. What is the intensity of the sound of normal conversation? (b) The decibel level of a subway train entering a station is 100 dB. What is the intensity of the sound of the train? Solution: Step 1 Read the problem and summarize the information. For 10 decibels, sound intensity increases by a factor of 10. Barely audible sound \begin{align*}=0 \ dB = 10^{-12} \ W/m^2\end{align*} Ear-splitting sound \begin{align*}= 130 \ dB =10 \ W/m^2\end{align*} Find intensity at 60 dB and find intensity at 100 dB. Assign variables. Let \begin{align*}x = \end{align*} sound level in decibels (dB) Let \begin{align*}y = \end{align*} intensity of sound \begin{align*}(W/m^2)\end{align*} Step 2 Look for a pattern. Let’s start at 0 dB \begin{align*}\text{For}\ 0\ dB && \text{Intensity} = 10^{-12}\ W/m^2\end{align*} For each decibel the intensity goes up by a factor of ten. \begin{align*}\text{For}\ 10\ dB && \text{Intensity} & = 10^{-12}\times 10\ W/m^2 \\ \text{For}\ 20\ dB && \text{Intensity} & = 10^{-12} \times 10 \times 10\ W/m^2\\ \text{For}\ 30\ dB && \text{Intensity} & = 10^{-12}\times 10 \times 10 \times 10\ W/m^2\end{align*} The pattern is that for each 10 decibels we multiply by a factor of 10. Let’s fill in a table of values. \begin{align*}& \text{Loudness\ (dB)} && 0 && 10 && 20 && 30 && 40 && 50 \\ & \text{Intensity}\ (W/m^2) && 10^{-12} && 10^{-11} && 10^{-10} && 10^{-9} && 10^{-8} && 10^{-7}\end{align*} Step 3 Let’s find a formula for this relationship. Since we take the original sound intensity and keep multiplying by the same factor of 10, that means we can use exponential notation. \begin{align*}y=10^{-12} \cdot 10^{\frac{x}{10}}\end{align*} The power is \begin{align*}\frac{x}{10}\end{align*}, since we go up by 10 dB each time. To find the intensity at 60 dB we use \begin{align*}x=60\end{align*} in the equation. \begin{align*}y=10^{-12} \cdot (10)^{\left(\frac{60}{10}\right )}=10^{-12}\cdot (10)^6=10^{-6}\ W/m^2\end{align*} To find the intensity at 100 dB we use \begin{align*}x=100\end{align*} in the equation. \begin{align*}y=10^{-12}\cdot (10)^{\left (\frac{100}{10}\right )}=10^{-12}\cdot (10)^{10}=10^{-2}\ W/m^2\end{align*} Step 4 Looking back over the solution, we see that we did not use all the information we were given. We still have the fact that a decibel level of 130 dB has an intensity level of \begin{align*}10 \ W/m^2\end{align*}. We can use this information to see if our formula is correct. Use \begin{align*}x=130\end{align*} in our formula. \begin{align*}y=10^{-12}\cdot (10)^{ \left( \frac{130}{10} \right )}=10^{-12} \cdot (10)^{13}=10\ W/m^2\end{align*} The formula confirms that a decibel level of 130 dB corresponds to an intensity level of \begin{align*}10 \ W/m^2\end{align*}. ## Review Questions Apply the problem-solving techniques described in this section to solve the following problems. 1. Half-life Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours? 2. Population decrease In 1990, a rural area has 1200 bird species. If species of birds are becoming extinct at the rate of 1.5% per decade (ten years), how many bird species will there be left in year 2020? 3. Growth Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007? 4. Investment Peter invests360 in an account that pays 7.25% compounded annually. What is the total amount in the account after 12 years?
1. \begin{align*}100 (.965)^x = 100 (.965)^6= 80.75\%\end{align*}
2. \begin{align*}1200 (.985)^x = 1200 (.985)^3 = 1147 \end{align*}
3. \begin{align*}200 (1.08)^x = 200 (1.08)^8 = 370 \end{align*}
4. \begin{align*}360 (1.0725)^x = 360 (1.0725)^{12}= \833.82\end{align*}
## Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9618.
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• Lesson Date: Monday, March 23
Subject: Math
Overview: Multiplying and Dividing Fractions
Estimated Time: Approximately 30 minutes
Explanation: Today you will review how to multiply and divide fractions. Then, you will apply your knowledge.
Things to know:
• Fractions have a numerator (top number) and a denominator (bottom number).
• When multiplying fractions, multiply the numerators. Then, multiply the denominators. Simplify the answer, if you are able.
• When dividing fractions, write your number model. Keep the first fraction as is, change the division sign to a multiplication sign, finally reciprocate or flip the second fraction. Multiply the fractions and simplify, if needed.
• Please visit https://www.cbsd.org/Page/1511 to access tutorial videos related to accessing online Math in Focus resources.
• There are no pre-made worksheets for today.
• Continue recording the daily temperature on your chart.
• You will need your number cards for this activity.
• Take a pencil, pen, straw, etc. and place in in front of you to make a horizontal line. Place another pencil, pen, marker, straw, etc. and place in in front of you to make another horizontal line.
• Flip over one number card and place it above one of the horizontal lines. Flip over a second number card and place it below one of the horizontal lines. Continue until you have two fractions.
• Multiply or divide the fractions.
• Repeat 8 times.
• Catering Company Activity:
• Imagine you are the owner of a catering company. Choose a recipe that you will make for a party. Consider a recipe that you would enjoy eating.
• Your customer just called and said they need to triple the amount of people at the party. Multiply each of the ingredients by 3 and rewrite the ingredients list.
• Your customer just called and said that five times as many people are coming to the party. Multiply each of the ingredients by 5 and rewrite the ingredients list.
• Your customer just called and said that one-third of the people are coming to the party. Multiply each of the ingredients by one-third and rewrite the ingredients list.
• When you have finished take a picture of something you completed or answer the discussion question and post it to Seesaw Math Activity March 23
• Optional activity, play “SatisFraction” at http://www.gregtangmath.com/satisfraction
1. Please visit https://www.cbsd.org/Page/1511 to access tutorial videos related to accessing online Math in Focus resources.
2. Optional activity, play “SatisFraction” at http://www.gregtangmath.com/satisfraction
Overview: Read Tuck Everlasting and Stop and Jot
Estimated Time: Approximately 45 minutes
Explanation of Assignment: Using your Tuck Everlasting book you will read and Stop and Jot
Quick things to know
While reading Tuck Everlasting complete at least 2 Stop and Jots. These can include “red flag” important moments or signposts.
The signposts include – again and again, contrast and contradiction, words of the wiser, memory moment and aha moments
1. Read Tuck Everlasting chapters 5-7
2. Throughout your reading of the novel complete a minimum of 2 stop and jots
3. Keep track of elements of realism and fantasy (packet page 7)
4. Jot down a description and supporting text evidence (packet page 8)
5. Keep track of each character’s feelings about living forever (packet page 9)
6. When you have finished take a picture of something you completed and post it to Seesaw Reading Activity March 23
7. If you have extra time, enjoy reading your independent reading book
-Use your Tuck Everlasting book and packet that you brought home in your grab bag.
-If you do not have access to your Tuck Everlasting packet, I have included it as an assignment in Teams under reading. You can just use separate sheets of paper to complete the assignments.
SUBJECT: Writing
Overview: Continue to work on the rough draft of the words of your picture book
Estimated Time: Approximately 45 minutes
Explanation of Assignment: Using the packet titled, “Directions for Picture Book Assignment” you will follow Day 4 on the schedule.
Things to Know: All of the necessary elements that you filled into the “Story Elements Organizer” on Page 2 yesterday will need to be put into the rough draft of your story on the “Illustration and Scene Narration Organizer.” Hopefully, you created a plot mountain on a separate sheet of paper to map out an entire overview of your plot and help you to pace your story appropriately yesterday. Today, you will be writing the rough draft words for pages/scenes 5-8 of your picture book starting on Page 5-6 , “Illustration and Scene Narration Organizer,” in your packet.
1. Follow Day 4 on the schedule page of your “Directions for Picture Book Assignment”
2. Day 4 instructs you to complete Pages 5-6 in your “Directions for Picture Book Assignments” packet
3. On pages 5-6 in your “Directions for Picture Book Assignment” you must write a rough draft of the words for your picture book for scenes 5-8. Each scene represents a different page for your picture book. Please do not start of the illustrations. You should ONLY be writing the words for scenes 5-8 today. Be sure as you write, you proofread for spelling, capitalization and punctuation. Vary your word choice and sentence structure to add style.
4. If you are finished early and are looking for additional practice, go onto your MobyMax account and work on Language
1. Use your “Directions for Picture Book Assignment” packet that you brought home in your grab bag
2. If you do not have access to your “Directions for Picture Book Assignment” packet, I have included it as an assignment in Teams. You can just use separate sheets of paper to complete the assignments.
Method of Communication/How this assignment is turned into the teacher:
1. The final picture book assignment will be handed-in in-person when we return to school. We will share our final picture books in small groups.
SUBJECT: Social Studies
Overview: Explore Hindu beliefs about Brahman
Estimated Time: Approximately 20 minutes
Explanation of Assignment: Using the purple History Alive interactive notebook or digitally on TCI, you will complete section 4: Beliefs about Brahman.
Things to Know: The chapter, Learning About World Religion: Hinduism, will take approximately 3 weeks to complete. Students will be reading each section of the chapter and answering comprehension questions in their notebook. To conclude the chapter, students will complete a one-pager to showcase their new learning.
1. Read chapter 15 section 4: Beliefs about BrahmanAll students have access to the text through Office 365 in their TCI app.
2. On the blank sheet of paper titled “Chapter 15 Vocabulary” that you created last week, add the words divine and cycle and the definitions. You can reveal the definition on TCI by clicking on the word when reading section 4.
3. 3. In the purple History Alive interactive notebook complete section 4 (numbers 1-4) that can be found on page 109 or accessed digitally on TCI in assignments.
4. Post a picture of page 109 when finished to Seesaw in activities
1. Refer to your purple History Alive interactive notebook that you brought home in your grab bag.
2. If you do not have access to your purple History Alive interactive notebook, I have included it as an assignment on TCI. You can complete digitally.
SUBJECT: Art
Visit Mrs. Shubert's website to view daily plan: https://www.cbsd.org/rschubert
SUBJECT: Extra, Extra!
Are you finished all of the required assignments above but you need more?
1. If the assignments above did not take you as long as expected and you have extra time, or if you are just looking to learn something new, please follow the link below to read Day 5 Scholastic "Learn from Home" articles. |
# How to work out 1 of something
How to work out a percentage of an amount (1% method)
If you want to find 1%, you just need to divide your original number by and you're done. Now, this is easy with multiples of like Splitting this into pieces gives you 5s. If you want 10% of you just do 10 x 5 and you've got your answer /5(1). How to work out a fraction of something (fraction of an.
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. This article has been viewedtimes. Learn more Finding a fraction of an amount is a useful skill you need for everyday, real-world math problems. For example, to find a discount price, or to determine a portion of something that you have or a what portion you are missingyou need to know how to find the fraction of an amount.
In these types of problems you will need to know how to multiply a fraction by a whole number, or how to create a fraction based on the information you are given.
You might find that the most difficult part of these types of problems is determining what the problem is asking. To work out a fraction of a whole number, turn the whole number into a fraction by giving it a somethhing of 1. Next, multiply the numerators together and write down the answer. Then, multiply the denominators together and write sometthing answer under the product of the numerators to create a new fraction.
We use cookies to make wikiHow great. By using our site, steps on how to play poker agree to our cookie policy. Cookie Settings. Learn why people somehhing wikiHow. Download Article Explore this Article methods. Related Articles. Article Summary. Method 1 of Set up the problem. When a problem asks you what a fraction of a whole number is, the problem is one of multiplication, and you need to multiply the fraction and the whole number.
Look for the keyword of. When you see of in a word problem, you need to multiply. Turn the whole number into a fraction. To do this, give it a denominator of 1. Remember, the denominator is the number underneath the fraction how to make good money working from home. Multiply the numerators.
Remember that the numerators are the numbers above the fraction bars. Multiply the denominators. Place this number under the product of the numerators. Simplify the fraction. To do this, divide the numerator by the denominator. This will give you your final answer as a whole number or in decimal form. If the result is not a whole number and you need the answer written in the form of a fraction, you should reduce the fraction by dividing the numerator and denominator by their dork common factor.
For smoething instructions on how to reduce a fraction, read Reduce Fractions. Method 2 of Understand kf the problem is asking. When a problem asks you what fraction one whole number is outt another whole number, you need to create a fraction and reduce it.
Determine the numerator and howw. The numerator is the fraction of the whole. Often this will be the smaller number, but not always, so read the problem carefully. To do this, find the greatest factor that is common to the numerator and the denominator, then divide each by oug factor. For complete instructions on how to reduce fractions, read Reduce Fractions. Method 3 of If the problem is asking you to determine how much of something is left, or to decrease an amount, or to find a discount, you will first multiply to find the fractional amount, then subtract the fractional amount from the original whole number.
If the problem is asking you to determine how much of something there is after an increase, you will first multiply to find the fractional amount, then add the fractional amount to the original whole number. Set up the multiplication problem.
This will give you your new numerator. This will give you your new denominator. Rewrite the new fraction. First divide the numerator by the denominator to see if the result is a whole number. If the result is not a whole number, you should reduce the fraction by dividing the numerator and denominator by their greatest common factor. This is the fractional amount you are decreasing by.
Change the original amount by adding or subtracting the fractional amount. This will give you your final answer. Not Helpful 0 Helpful 9. Nine trucks are equal in weight to six trucks and four sedans. What fraction of the weight of a truck is the weight of a sedan? Since they want to know what fraction of a truck's weight x equals the weight of a sedan yyou need to solve for y. Not Helpful 0 Helpful 6. Not Helpful 1 Helpful 4. There are 20 legs in total. How many pigs are there?
This doesn't work, however, because chickens and cows have a different number of legs. So you can't solve this problem simply by multiplying the fraction and what is dehum on air conditioner whole number. The easiest way to solve this problem is to think about it logically.
For every 1 chicken in the barn, there are 2 cows. That means, for every 2 chicken legs, somthing are 8 cow legs. So there are 2 chickens and 4 cows. Not Helpful 2 Helpful 2. You need to set up an algebraic equation. Let the missing number be x. Not Helpful 3 Helpful 4. Not Helpful 2 Helpful 3. Then you need to reduce your fraction by finding the greatest factor common to the numerator and the denominator.
By using this somehing, some information may be shared with YouTube. Submit a Tip All tip submissions are carefully reviewed before being published. Related wikiHows How to. How to. About This Article. Co-authored by:. Co-authors: Updated: June 15, Categories: Fractions. Article Summary X To work out a fraction of a whole number, turn the whole number into a fraction by giving it a denominator of 1.
Deutsch: Einen Bruchteil eines Betrages ausrechnen. Thanks to how to play the final countdown on guitar solo authors for creating a page that has been readtimes. Did this article help you?
Why Does This Work?
A simple guide on how to work out the percentage of an amount with the 1% method, aimed at students aged between 11 and Nov 01, · To work out any percentage of anything, you simply divide by and then multiply with the number of percent you want to find. Therefore 1% of . Sep 21, · Now, something even harder - cookies. Oh-oh, we divided up the first cookies, placing two in every compartment. Now we are left with 50 cookies that need to be spread evenly, hmmm, it's half a cookie in every box. How to calculate the percentage? You are right - this time, 1 percent of the total number of cookies is How many do.
Calculating percentages is one of the most important mathematical skills you can carry with you throughout your life and luckily it is very simple to master. To find a percentage of any number, all you have to think about is diving that number up into equal pieces and then adding together the number of pieces you wish to calculate.
Now, this is easy with multiples of like Splitting this into pieces gives you 5s. But what if we are using a more complicated number like 27? Well you do the same thing.
Divide 27 into equal pieces and times by the percentage you wish to find? To divide any number by all you have to do is move the digits 2 places to the right of the decimal point. When we say 27, what we are really saying is So dividing this by will move the numbers 2 places to the right, so 0. It's that simple. This is the same as multiplying any number by 0. Leave this field empty. For example: working out the original pre vat value when you only know the percentage and final figure?
Im really struggling. There were 18 broken eggs. How many eggs were there in the batch? Work it out as 0. I have tried to find a way to understand but I really don't- I don't understand where the 0.
Can you please explain all these questions to me, sorry that I'm asking to many questions but please can you answer them, this is the only thing I'm struggling on. Calculating Percentages Calculating percentages is one of the most important mathematical skills you can carry with you throughout your life and luckily it is very simple to master.
Hi there,. This You Tuve video is useful as it explains how to calculate basic percentages. Times the number by 0. Add an answer Cancel reply. Similar questions. Convert 68 out of 90 into a percentage. Work out his new monthly salary. How do you work out percentages of amounts? Do you know how to work out any given percentage of a number or amount of money. How do you work out what the original value was if you only know the final figure and a percentage?
Percentage discount and commision. Percentage Increase. I was never any good at percentages. Percentages problems. Hi, I'm revising for my gcse exam in June and struggling with these two topics.
## How to work out 1 of something: 3 comments
1. And thanks for this video I wanted to know how to earn money from this software.
2. Hola muy buen video, puedes cubrir el tema de la arquitectura de ORACLE y direccionarme a documentos que lo ilustre, Gracias. |
August 22, 2023
# What is a normal vector?
A normal vector is an important concept in mathematics and has numerous applications in various fields such as physics, computer graphics, and engineering. Understanding the concept and properties of a normal vector is essential for solving many mathematical and practical problems. In this article, we will explore the definition, characteristics, mathematical representation, and applications of normal vectors. We will also address common misconceptions and errors associated with normal vectors. So, let's dive in and explore the fascinating world of normal vectors.
## Understanding the Concept of a Vector
A vector is a mathematical object that represents both magnitude and direction. Unlike scalars, which only have magnitude, vectors have both magnitude and direction making them incredibly useful in many areas of science and engineering.
The basic definition of a vector states that it is a quantity that can be represented by an arrow in space. The length of the arrow represents the magnitude of the vector, while the direction of the arrow indicates the direction of the vector.
When we think about vectors, it's important to understand that they can exist in multiple dimensions. This means that vectors can extend not only in the familiar two-dimensional plane but also in three-dimensional space. By extending our understanding of vectors to multiple dimensions, we can solve complex problems in physics, engineering, and computer science.
### Basic Definition of a Vector
In its simplest form, a vector is an ordered collection of numbers or variables. It can be represented in multiple dimensions, such as 2D or 3D, where the vector extends in two or three directions.
For example, consider a 2D vector represented by (x, y), where 'x' and 'y' represent the components of the vector in the x and y directions, respectively. These components give us a clear understanding of how the vector is oriented in space.
Furthermore, vectors can also be represented using other mathematical notations, such as column matrices or Cartesian coordinates. These alternative representations provide different perspectives on the same vector, allowing us to analyze and manipulate them in various ways.
### Components of a Vector
Each vector can be broken down into its individual components, which are the magnitudes of the vector in each direction. In 2D, these components are typically expressed as (x, y), while in 3D, they are represented as (x, y, z).
By understanding the components of a vector, we gain insight into how it behaves in different directions. This knowledge is crucial when performing operations on vectors, such as addition, subtraction, and scalar multiplication.
Moreover, vectors can also be decomposed into their magnitude and direction. The magnitude of a vector represents its length or size, while the direction indicates where the vector is pointing. This decomposition allows us to analyze vectors in a more intuitive way, breaking them down into their fundamental characteristics.
It's worth noting that vectors can also have negative components, which means they point in the opposite direction. This property adds another layer of complexity to vector analysis, as we need to consider both positive and negative components when performing calculations.
In conclusion, vectors are powerful mathematical tools that allow us to represent and analyze both magnitude and direction. By understanding the basic definition of a vector, as well as its components, we can unlock a wide range of applications in various fields of study.
## Introduction to Normal Vectors
Now that we have a good understanding of vectors, let's dive deeper into the fascinating concept of normal vectors. A normal vector, often denoted as "n," is a vector that stands perpendicular to a given surface or curve at a specific point. It plays a crucial role in defining the orientation or direction that is perpendicular to the surface.
But what exactly does it mean for a vector to be perpendicular to a surface? Well, imagine a flat table in front of you. If you place a book on the table and draw an arrow pointing straight up from the book's surface, that arrow represents the normal vector. It stands at a 90-degree angle to the surface, indicating the direction that is perpendicular to it.
### Defining Normal Vectors
To define a normal vector, we need to have a surface or curve as a reference point. The normal vector lies perpendicular to this surface, pointing outward or inward, indicating the direction away from or towards the surface, respectively.
For instance, let's consider a sphere. At any given point on the sphere's surface, we can draw a normal vector that points directly away from the center of the sphere. This vector will be perpendicular to the surface, providing us with valuable information about the sphere's orientation at that specific point.
It is important to note that a normal vector is unique to each point on the surface. As we move along the surface, the direction of the normal vector can change, adapting to the changing orientation of the surface. This dynamic characteristic of normal vectors makes them incredibly versatile and applicable in various fields.
### Unique Characteristics of Normal Vectors
Normal vectors possess several unique characteristics that make them invaluable in different fields:
1. Perpendicularity: A normal vector is always perpendicular to the surface at each point, forming a perfect 90-degree angle. This perpendicularity allows us to determine the direction that is orthogonal to the surface, providing insights into the surface's geometry.
2. Direction: The direction of the normal vector can vary depending on whether it points outward or inward from the surface. This directionality is crucial in distinguishing between the front and back faces of an object or determining whether a point lies inside or outside a closed surface.
3. Orientation: Normal vectors provide a consistent orientation for a particular surface. By defining the front and back faces or the inside and outside of an object, they help us understand the spatial relationships and properties of the surface.
These unique characteristics make normal vectors an essential tool in various fields, including computer graphics, physics, engineering, and mathematics. They allow us to analyze and manipulate surfaces and curves, enabling us to solve complex problems and gain a deeper understanding of the world around us.
## Mathematical Representation of Normal Vectors
In the realm of mathematics, normal vectors can be represented and operated upon using various notations and symbols. Let's explore some of the common representations and mathematical operations involving normal vectors.
Normal vectors are commonly represented using bold letters, such as n or v. In mathematical equations, a hat symbol (^) is often used to represent a unit normal vector (a vector with a magnitude of 1).
When working with normal vectors, it is important to understand the mathematical operations that can be performed on them. These operations include addition, subtraction, dot product, and cross product.
For example, addition of normal vectors involves adding the corresponding components of the vectors. If we have two normal vectors n and v, their sum would be represented as n + v.
Subtraction of normal vectors is similar to addition, but instead of adding the corresponding components, we subtract them. If we have two normal vectors n and v, their difference would be represented as n - v.
The dot product of two normal vectors is a scalar quantity that represents the cosine of the angle between the two vectors. It can be used to determine the angle between two surfaces or the amount of overlap they have. The dot product is calculated by multiplying the corresponding components of the vectors and summing them up.
The cross product of two normal vectors is a vector that is perpendicular to both of the original vectors. It can be used to find a vector that is orthogonal to a plane defined by the two vectors. The cross product is calculated by taking the determinant of a 3x3 matrix formed by the components of the vectors.
These mathematical operations provide us with powerful tools to manipulate and analyze normal vectors in various mathematical and physical contexts. Whether it is determining the orientation of a surface, calculating the force exerted by a magnetic field, or solving complex geometric problems, the representation and operations involving normal vectors play a crucial role in many areas of mathematics and science.
## Applications of Normal Vectors
Normal vectors have a wide range of applications across different fields. Let's explore some of the practical uses of normal vectors in physics and computer graphics.
### Normal Vectors in Physics
In physics, normal vectors play a crucial role in determining forces, calculating surface areas, and understanding the behavior of particles in three-dimensional space. They help in defining the direction of forces acting on surfaces or objects and enable us to calculate the angle of incidence and reflection.
For example, in fluid dynamics, normal vectors are used to determine the pressure distribution on a surface. By knowing the normal vector at each point on the surface, scientists and engineers can accurately calculate the forces exerted by the fluid on the object. This information is essential in designing efficient aerodynamic shapes for vehicles like airplanes and cars.
Furthermore, normal vectors are also utilized in calculating the surface area of complex three-dimensional objects. By integrating the dot product of the normal vector and the differential area element, physicists can determine the total surface area of an object. This knowledge is valuable in various fields, such as material science and architecture, where surface area plays a crucial role in understanding the behavior of materials and designing structures.
### Normal Vectors in Computer Graphics
In computer graphics, normal vectors are extensively used to create realistic lighting and shading effects. They define the direction that a surface is facing and allow for the calculation of surface normals, which play a significant role in determining how light interacts with objects in virtual environments.
One of the primary applications of normal vectors in computer graphics is in the field of 3D modeling and rendering. By assigning normal vectors to each vertex of a 3D model, computer graphics artists can create smooth surfaces that accurately reflect light. These normal vectors are used in algorithms that calculate the intensity and direction of light at each point on the surface, resulting in realistic shading and highlighting effects.
Additionally, normal vectors are crucial in simulating realistic physics-based interactions in computer games and simulations. By incorporating normal vectors into collision detection algorithms, developers can accurately determine the direction and magnitude of forces when objects collide or interact with each other. This level of realism enhances the overall gaming experience and makes virtual environments more immersive.
In conclusion, normal vectors have diverse applications in physics and computer graphics. From determining forces and surface areas in physics to creating realistic lighting and shading effects in computer graphics, normal vectors are an essential tool in understanding and simulating the behavior of objects in three-dimensional space.
## Common Misconceptions about Normal Vectors
Despite their importance, normal vectors can be a source of confusion and errors for many learners. Let's address some common misconceptions associated with normal vectors and clarify their correct usage.
### Clearing up Confusion about Normal Vectors
One of the most common misconceptions is the assumption that normal vectors are only applicable to flat surfaces. In reality, normal vectors can be defined for curved surfaces as well, representing the direction perpendicular to the tangent plane at each point.
It is important to understand that normal vectors are not limited to a specific shape or dimension but can be utilized in various scenarios.
### Common Errors in Calculating Normal Vectors
When calculating normal vectors, errors can occur due to incorrect orientation or magnitude. It is crucial to pay attention to the direction of the normal vector and ensure it is pointing in the desired direction. Additionally, miscalculations in vector components or dot products can lead to inaccurate results.
By carefully analyzing the given problem and utilizing the appropriate mathematical techniques, these common errors can be avoided, ensuring accurate calculations and solutions involving normal vectors.
In conclusion, normal vectors are a fundamental concept in mathematics and have significant applications in various fields. They allow us to define the direction perpendicular to a given surface, providing valuable insights for solving problems in physics, computer graphics, and more. By understanding the definition, properties, mathematical representation, and applications of normal vectors, we can tackle complex problems with ease and accuracy. So, embrace the power of normal vectors and explore the endless possibilities they offer in the world of mathematics and beyond. |
Maryland 7 - 2020 Edition
2.03 Order of operations with rational numbers
Lesson
We've already learned about the order of operations, which tells us the steps to evaluating expressions with multiple operations.
Remember!
1. Evaluate operations inside grouping symbols such as parentheses (...) or absolute value |...|.
2. Evaluate exponents such as squares, cubes, or square roots.
3. Evaluate multiplication and division going from left to right.
4. Evaluate addition and subtraction going from left to right.
We use the same order of operations for rational numbers as we do for integers
Let's look through some examples of questions involving rational numbers and the order of operations.
#### Worked example
##### Question 1
Evaluate the expression: $8.5+7.2+\left(-1.3\right)$8.5+7.2+(1.3)
Think: Because this expression only contains addition we can skip to that step in the order of operations. Remember with addition we work from left to right.
Do
$8.5+7.2+\left(-1.3\right)$8.5+7.2+(−1.3) $=$= $15.7+\left(-1.3\right)$15.7+(−1.3) Add $8.5$8.5 and $7.2$7.2 $=$= $15.7+\left(-1.3\right)$15.7+(−1.3) Adding $-1.3$−1.3 can be rewritten as subtracting positive $1.3$1.3 $=$= $15.7-1.3$15.7−1.3 Subtract $=$= $14.4$14.4
Now, let's give these a try.
#### Practice questions
##### Question 2
Calculate $86+\frac{3}{10}\times\left(-2\right)$86+310×(2).
##### Question 3
Conversion of temperature from Fahrenheit to Celsius is defined by the formula $C=\frac{5}{9}\left(F-32\right)$C=59(F32), where $F$F is the temperature in degrees Fahrenheit and $C$C is the equivalent temperature in degrees Celsius. Given that $F=-4$F=4, calculate $C$C.
##### Question 4
David buys $3$3 shirts at $\$19.90$$19.90 each, and a pair of jeans for \20.50$$20.50. The shop has a sale on, and so he receives a $\$8.028.02 discount.
Write and solve a numerical expression for how much he spends in total.
### Outcomes
#### 7eE.B.3
Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. |
# How do you solve the following trigonometric equation?
How do you solve the following trigonometric equation? $$\tan x - 1 = -\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} \cot x.$$
-
Perhaps the easiest approach is to let $u=\tan x$ and use the fact that $\cot x=\frac1{\tan x}$ to rewrite it as $$u-1=-\frac1{\sqrt3}+\frac1{\sqrt3u}=\frac{1-u}{\sqrt3u}\;.$$ Now solve for $u$; at that point you'll have $\tan x$, and the rest is plain sailing.
We have $\displaystyle \cot(x) = \frac1{\tan(x)}$. Hence, the equation gives us $\displaystyle \tan(x) - 1 = - \frac1{\sqrt{3}} + \frac1{\sqrt{3}} \frac1{\tan(x)}$. Multiplying by $\tan(x)$, we get that $$\tan^2(x) - \tan(x) = - \frac{\tan(x)}{\sqrt{3}} + \frac1{\sqrt{3}}$$ Setting $\displaystyle m = \tan(x)$, we get the quadratic, $$m^2 + \left( \frac1{\sqrt{3}} - 1\right)m - \frac1{\sqrt{3}} = 0$$ which gives us that $$m = \frac{-\left( \frac1{\sqrt{3}} - 1\right) \pm \sqrt{\left(\frac1{\sqrt{3}} - 1\right)^2 + \frac4{\sqrt{3}}}}{2} = \frac{-\left( \frac1{\sqrt{3}} - 1\right) \pm \left(\frac1{\sqrt{3}} + 1\right)}{2} = -\frac1{\sqrt{3}}, 1$$
$\displaystyle m = 1$ gives us $\displaystyle x = n \pi + \frac{\pi}{4}$ while $\displaystyle m = - \frac1{\sqrt{3}}$ gives us $\displaystyle x = n \pi - \frac{\pi}{6}$ where $n \in \mathbb{Z}$. |
# Properties of Subtracting Integers
The properties of subtracting integers are explained here along with the examples.
1. Closure Property: The difference (subtraction) of any two integers is always an integer.
i.e., The difference of two integers is always an integer.
Hence, integers are closed under subtraction.
If x and y are any two integers, then x - y always an integer
For Examples:
(i) (+7) – (+4) = 7 - 4 = 3, which is an integer.
(ii) (-8) – (+3) = -8 – 3 = -11, which is an integer.
(iii) 6 - 14 = 6 + (-14) = - 8, which is an integer.
(iv) (- 18) - 10 = (- 18) + (- 10) = - 28, which is an integer.
2.
Commutative Property:
Subtraction is not commutative for integers.
For any two different integers ‘x’ and ‘y’,
x - y ≠ y - x
For Examples:
(i) 4 - 8 = 4 + (-8) = -4 and 8 - 4 = 8 + (- 4) = 4
Therefore, 4 - 8 ≠ 8 - 4
(ii) (- 4) - 7 = (- 4) + (- 7) = - 11 and 7 - (-4) = 7 + 4 = 11
Therefore, (- 4) - 7 7 - (-4)
3. Associative Property: Subtraction is not associative for integers.
For any three integers ‘x’, ‘y’ and ‘z’
x - (y - z) ≠ (x - y) - z
For Example:
[2 - (- 3)] - (- 6) = [2 + (3)] + 6 = 5 + 6 = 11 and
2 - [(- 3) - (- 6)] = 2 - [(- 3) + 6] = 2 - 3 = -1
Therefore, [2 - (- 3)] - (- 6)2 - [(- 3) - (- 6)]
4. Subtraction Property of Zero: The result of subtracting zero from an integer is the integer itself.
For any three integers ‘x’,
x - 0 = x
For Example:
(i) 5 - 0 = 5
(ii) 8 - 0 = 8
(iii) 100 - 0 = 100
(iv) 9999 - 0 = 9999
(v) 99999999 - 0 = 99999999
5. For any integer ‘x’, x - 0 ≠ 0 - x
For Example:
(i) 7 - 0 = 7 and 0 - 7 = -7
Therefore, 7 - 0 ≠ 0 - 7
(ii) 10 - 0 = 10 and 0 - 10 = -10
Therefore, 10 - 0 ≠ 0 - 10
6. For any three integers ‘x’, ‘y’ and ‘z’, and x > y, then
x - z > y - z
To evaluate an expression containing various integers with plus and minus sign:
1. Evaluate:
(i) (+15) + (-11) - (+5) - (-7)
= 15 - 11 - 5 + 7
= 22 – 16, [Adding all integers with plus (+) sign together and with minus (-) sign respectively together]
= +6 or simply 6.
(ii) (-72) + (-93) - (-85) + (+78)
= -72 -93 + 85 + 78
= -165 + 163, [Adding all integers with plus (+) sign together and with minus (-) sign respectively together]
= - 2
2. Evaluate the expression (-45) + (-32) – (-69) + (87)
Solution:
(-45) + (-32) – (-69) + (87)
= -45 – 32 + 69 +87
= -(45 + 32) + (69 + 87)
= -77 + 156
= +79
= 79
3. Simplify: 32 – 13 + 35 + 18 - 60
Solution:
32 – 13 + 35 + 18 – 60
= (32 + 35 + 18) – (13 + 60)
= 85 – 73
= +12 or simply 12
4. The sum of two integers is -17. If one of them is -7, find the other.
Solution:
Other integer = Sum of two integers - the given integer
= (-17) – (-7)
= -17 + 7
= -10
Therefore the other number is -10.
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# 6.6 Exponential and logarithmic equations (Page 4/8)
Page 4 / 8
## Using the definition of a logarithm to solve logarithmic equations
For any algebraic expression $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ and real numbers $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c,$ where
## Using algebra to solve a logarithmic equation
Solve $\text{\hspace{0.17em}}2\mathrm{ln}x+3=7.$
Solve $\text{\hspace{0.17em}}6+\mathrm{ln}x=10.$
$x={e}^{4}$
## Using algebra before and after using the definition of the natural logarithm
Solve $\text{\hspace{0.17em}}2\mathrm{ln}\left(6x\right)=7.$
Solve $\text{\hspace{0.17em}}2\mathrm{ln}\left(x+1\right)=10.$
$x={e}^{5}-1$
## Using a graph to understand the solution to a logarithmic equation
Solve $\text{\hspace{0.17em}}\mathrm{ln}x=3.$
[link] represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to 20. In other words $\text{\hspace{0.17em}}{e}^{3}\approx 20.\text{\hspace{0.17em}}$ A calculator gives a better approximation: $\text{\hspace{0.17em}}{e}^{3}\approx 20.0855.$
Use a graphing calculator to estimate the approximate solution to the logarithmic equation $\text{\hspace{0.17em}}{2}^{x}=1000\text{\hspace{0.17em}}$ to 2 decimal places.
$x\approx 9.97$
## Using the one-to-one property of logarithms to solve logarithmic equations
As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers $\text{\hspace{0.17em}}x>0,$ $S>0,$ $T>0\text{\hspace{0.17em}}$ and any positive real number $\text{\hspace{0.17em}}b,$ where $\text{\hspace{0.17em}}b\ne 1,$
For example,
So, if $\text{\hspace{0.17em}}x-1=8,$ then we can solve for $\text{\hspace{0.17em}}x,$ and we get $\text{\hspace{0.17em}}x=9.\text{\hspace{0.17em}}$ To check, we can substitute $\text{\hspace{0.17em}}x=9\text{\hspace{0.17em}}$ into the original equation: $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(9-1\right)={\mathrm{log}}_{2}\left(8\right)=3.\text{\hspace{0.17em}}$ In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.
For example, consider the equation $\text{\hspace{0.17em}}\mathrm{log}\left(3x-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right).\text{\hspace{0.17em}}$ To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for $\text{\hspace{0.17em}}x:$
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter
cos 18 ____ sin 72 evaluate |
# Find the standard matrix of the linear transformation $T$
A linear transformation $$T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$$ is defined by $$T((2,11))=(8,5)$$ and $$T((1,5))=(5,3)$$ Find the matrix for this linear transformation.
I did this in a very clunky way - since $$T$$ is linear, I can write that
$$T((2, 11))=T(2e_1+11e_2)=2T(e_1)+11T(e_2)=(8,5)$$
and
$$T((1,5))=T(e_1+5e_2)=T(e_1)+5T(e_2)=(5,3)$$
I solved this system of two equations for $$T(e_1)$$ and $$T(e_2)$$ to get $$T(e_1)=(15,8)$$ $$T(e_2)=(-2,-1)$$
Thus, we get $$\begin{pmatrix} 15 & -2 \\ 8 & -1 \\ \end{pmatrix}$$
However, is there a simpler way to get the answer?
What you did in solving the system corresponds to solve $$TA=B$$ as $$T=BA^-1$$.
Whether it is more convenient your way or that of finding the inverse of $$A$$ is a matter of .. what constraints, instruments, etc. you have.
Here is a bit of an elaboration on top of @G Cab's answer.
Consider the linear transformation $$T_1$$, which maps components of the identity matrix to vectors specified below.
$$T_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 11 \end{pmatrix}, \, T_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$$
Next, we define a separate linear transformation that maps components of the identity matrix to another set of vectors.
$$T_2 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \end{pmatrix}, \, T_2 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$$
Then, there should be some linear transformation $$T$$ that represents the relationship between these two linear transformations. In other words,
$$T\left(T_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 8 \\ 5 \end{pmatrix}$$
and
$$T\left(T_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$$
We can package the basis vectors into a matrix to formulate a more compact expression:
$$T\left(T_1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$
Because we intentionally chose very easy basis vectors, namely the components of the identity matrix, we know what $$T_1$$ and $$T_2$$ are:
$$T_1 = \begin{pmatrix} 2 & 1 \\ 11 & 5 \end{pmatrix}, \, T_2 = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$
Therefore, the equation above simplifies to the following:
$$T \begin{pmatrix} 2 & 1 \\ 11 & 5 \end{pmatrix} = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$
Can you find the inverse of $$T_1$$ and go from there? |
# Factors of 50
In this blog post, we will take a closer look at the factors of 50, a composite number. We will delve into the ways in which it can be broken down into its component parts and explore the different methods used to find its factors. In addition, we will also provide information about negative factors of 50, pair factors, prime factorization, and its indivisible factors. We will also explore the concept of factor tree.
## Factors Calculator
Enter Number
Factors of 50: 1, 2, 5, 10, 25 and 50
Negative Factors of 50: -1, -2, -5, -10, -25 and -50
Prime Factors of 50: 2, 5
Prime Factorization of 50: 2 × 5 × 5
Factors of 50 in Pairs: (1, 50), (2, 25) and (5, 10)
Negative Pair Factors of 50: (-1, -50), (-2, -25) and (-5, -10)
Prime Numbers
A prime number is a whole number greater than 1 that can only be divided evenly by 1 and itself. Some examples of prime numbers include 2, 3, 5, 7, and 11. Prime numbers are important in mathematics as they are used in many mathematical formulas and algorithms, and are also used in the study of number theory.
Composite Numbers
A composite number is a whole number greater than 1 that can be divided evenly by a number other than 1 or itself. In other words, a composite number is not a prime number. Some examples of composite numbers include 4, 6, 8, 9, and 10.
## What are the factors of 50?
The method of calculating the factors of 50 is as follows. First, each number can be divided by one and by itself.
Consequently, 1 and 50 are the factors of 50.
By dividing a number by 1, 2, 3, 4… we can discover all its factors.
(i) 50 ÷ 1 = 50
This division gives the remainder 0 and so is divisible by 50. So please put them 1 and 50 in your factor list.
1, … 50
(ii) 50 ÷ 2 = 25
This division gives the remainder 0 and so is divisible by 25. So please put them 2 and 25 in your factor list.
1, 2 … 25, 50
(iii) 50 ÷ 3 = 16.66
This division gives the remainder 16.66, not being thoroughly divided. So we will not write 3 and 16.66 on the list.
(iv) 50 ÷ 4 = 12.5
This division gives the remainder 12.5, not being thoroughly divided. So we will not write 4 and 12.5 on the list.
(v) 50 ÷ 5 = 10
This division gives the remainder 0 and so is divisible by 10. So please put them 5 and 10 in your factor list.
1, 2, 5 … 10, 25, 50
(vi) Since we don’t have any more numbers to calculate, we are putting the numbers so far.
So 1, 2, 5, 10, 25 and 50 are factors of 50.
## Factor pairs of 50
1 x 50 = 50
2 x 25 = 50
5 x 10 = 50
So, (1, 50), (2, 25) and (5, 10) are factor pairs of 50
## Factor pairs of -50
-1 x -50 = 50
-2 x -25 = 50
-5 x -10 = 50
So, (-1, -50), (-2, -25) and (-5, -10) are negative pair factors of 50
## Prime Factorization of 50
50 ÷ 2 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1
Therefore, 2 x 5 x 5 are Prime factorization of 50.
## Factor tree of 50
`````` 50
/ \
2 25
/ \
5 5`````` |
# In Math What Is A Factor latest 2023
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## Greatest Common Fractor (gcf) And Least Common Multiple (lcm)
LCM and GCF
There are many reasons why math is a difficult and most hated subject among average students. One of the most important reasons is insufficient knowledge of basic mathematical concepts such as least common multiple (lcm) and greatest common factor (gcf). Many students don’t care about these basic concepts when there is a good time to learn these basic math skills.
After reading this article, students who lack the knowledge of least common multiple and greatest common factor can better understand these skills and they can apply this knowledge in higher mathematical concepts, such as solving fractions or algebraic expressions.
Least common multiple (LCM):
If students know multiplication tables, they can find multiples of a number quite easily, because multiples are the “times” of the given number. To find the least common multiple of two numbers, students must find multiples of the two numbers and choose a common multiple that is the least of all. This least common multiple of the two numbers is called the least common multiple.
For example; consider that we want to find “lcm” from the numbers “6” and “8”. To find “lcm” of these numbers, write the first 5 multiples of the two numbers as shown below:
6 = 6, 12, 18, 24, 30
8 = 8, 16, 24, 32, 40
Now, by looking at the first five multiples of the two numbers, we can locate the lcm for both, which is “24”. There is no other multiple less than 24, which is common to both given numbers.
Therefore, “24” is the “lcm” of “6” and “8”.
Sometimes there is no common multiple in the first five multiples of the two numbers, in this case write down the next five multiples to locate the lowest common multiple.
Greatest common factor (gcf):
Similar to “lcm”, the greatest common factor is another key skill that students need to understand from the ground up. To find the greatest common factor of two numbers, students need to find all the factors of the two numbers, then the greatest common factor is called the greatest common factor or gcf.
For example; consider that we want to find the gcf of the numbers “12” and “32”. To find the gcf of the given numbers, write down all the factors of the given numbers as shown below:
12 = 1, 2, 3, 4, 6, 12
32 = 1, 2, 4, 8, 16, 32
Looking at all the factors of the two numbers, it is clear that the number “4” is the greatest common factor for the given numbers “12” and “32”.
Students can practice more similar problems on lcm and gcf to become more confident in these two key skills.
It is the responsibility of students to learn the right math concepts at the right time and the responsibility of educators and parents is to ensure that children have learned basic math concepts before upgrading them to learn math superior.
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Practice Quiz 10.1-10.3 Solutions
# Practice Quiz 10.1-10.3 Solutions - 5 1 28 25 8 4 2 2 = y x...
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Practice Quiz (10.1-10.3) Name: ______________________ Math 115-007 December 3, 2009 Find the vertex, focus, and directrix of the following parabolas. 1. x x y 6 2 + = 9 ) 3 ( 3 ) 3 ( 6 2 2 2 2 - + = - + + = x y x x y Vertex: (–3, –9) Focus: (–3, –9 + 4 1 ) Directrix: y = –9 – 4 1 2. 3 ) 2 ( 3 2 + - - = x y Vertex: (2, 3) Focus: (2, 3 + ) ) 4 ( 3 1 - Directrix: y = 3 – ) 4 ( 3 1 - Find the Center and two Focus points of each ellipse below. 3. 1 25 ) 1 ( 9 ) 1 ( 2 2 = + + + y x Center: (– 1, – 1) Foci: (– 1, – 1 + 4) and (– 1, – 1 – 4) c c c c b a = = + = + = 4 16 9 25 2 2 2 2 2
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Practice Quiz (10.1-10.3) Name: ______________________ Math 115-007 December 3, 2009 4. 1 19 ) 2 ( 23 ) 7 ( 2 2 = + + - y x Center: (7, – 2) Foci: (7 + 2, – 2) and (7 – 2, – 2) c c c c b a = = + = + = 2 4 19 23 2 2 2 2 2 Find the Center, two Focus points, and two Asymptotes for each hyperbola below.
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Unformatted text preview: 5. 1 28 ) 25 ( 8 ) 4 ( 2 2 =--+ y x Center: (– 4, 25) Foci: (–4 + 6, 25) and (– 4 – 6, 25) c c c c b a = = = + = + 6 36 28 8 2 2 2 2 2 The asymptotes pass through the center (– 4, 25) and have slope 2 7 8 28 ± = ± = ± a b ) 4 ( 2 7 25 + =-x y and ) 4 ( 2 7 25 +-=-x y 6. 1 11 14 ) 1 ( 2 2 =-+ x y Center: (0, – 1) Foci: (0, – 1 + 5) and (0, –1 – 5) c c c c b a = = = + = + 5 25 11 14 2 2 2 2 2 The asymptotes pass through the center (0, – 1) and have slope 11 14 ± = ± b a x y 11 14 1 = + and x y 11 14 1-= +...
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## This note was uploaded on 03/04/2012 for the course MATH 115 taught by Professor Ms.park during the Fall '09 term at South Carolina.
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Practice Quiz 10.1-10.3 Solutions - 5 1 28 25 8 4 2 2 = y x...
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IDENTITIES, EQUATIONS, AND INEQUALITIES - Trigonometric Functions and Their Inverses - Functions - REVIEW OF MAJOR TOPICS - SAT SUBJECT TEST MATH LEVEL 2
## CHAPTER 1Functions
### IDENTITIES, EQUATIONS, AND INEQUALITIES
There are a few trigonometric identities you must know for the Mathematics Level 2 Subject Test.
Reciprocal Identities recognize the definitional relationships:
Cofunction Identities were discussed earlier. Using radian measure:
Pythagorean Identities
Double Angle Formulas
EXAMPLES
1. Given cos and , find
Since sin 2 = 2(sin )(cos ), you need to determine the value of sin . From the figure below, you can see that sin . Therefore, sin .
2. If cos 23° = z, find the value of cos 46° in terms of z.
Since 46 = 2(23), a double angle formula can be used: cos 2A = 2 cos2 A – 1. Substituting 23° for A, cos 46° = cos 2(23°) = 2 cos2 23° – 1 = 2(cos 23°)2 – 1 = 2z 2 – 1.
3. If sin x = A, find cos 2x in terms of A.
Using the identity cos 2x = 1 – sin2 x, you get cos 2x = 1 – A2.
You may be expected to solve trigonometric equations on the Math Level 2 Subject Test by using your graphing calculator and getting answers that are decimal approximations. To solve any equation, enter each side of the equation into a function (Yn), graph both functions, and find the point(s) of intersection on the indicated domain by choosing an appropriate window.
4. Solve 2 sin x + cos 2x = 2 sin2 x – 1 for 0 x 2.
Enter 2 sin x + cos 2x into Y1 and 2 sin2 x – 1 into Y2. Set Xmin = 0, Xmax = 2, Ymin = –4, and Ymax = 4. Solutions (x-coordinates of intersection points) are 1.57, 3.67, and 5.76.
5. Find values of x on the interval [0,] for which cos x < sin 2x.
Enter each side of the inequality into a function, graph both, and find the values of x where the graph of cos x lies beneath the graph of sin 2x: 0.52 < x < 1.57 or x > 2.62.
EXERCISES
1. If sin and cos , find the value of sin 2x.
(A) –
(B) –
(C)
(D)
(E)
2. If tan A = cot B, then
(A) A = B
(B) A = 90° + B
(C) B = 90° + A
(D) A + B = 90°
(E) A + B = 180°
3. If cos , find cos 2x.
(A) –0.87
(B) –0.25
(C) 0
(D) 0.5
(E) 0.75
4. If sin 37° = z, express sin 74° in terms of z.
(A)
(B) 2z 2 + 1
(C) 2z
(D) 2z 2 – 1
(E)
5. If sin x = –0.6427, what is csc x?
(A) –1.64
(B) –1.56
(C) 0.64
(D) 1.56
(E) 1.70
6. For what value(s) of x, 0 < x < , is sin x < cos x?
(A) x < 0.79
(B) x < 0.52
(C) 0.52 < x < 0.79
(D) x > 0.52
(E) x > 0.79
7. What is the range of the function f(x) = 5 – 6sin (x + 1)?
(A) [–6,6]
(B) [–5,5]
(C) [–1,1]
(D) [–1,11]
(E) [–11,1]
|
## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2
Question 1.
Draw a line segment AB = 7 cm and Mark a point P on it. Draw a line perpendicular to the given line segment at P.
Solution:
Construction:
(i) Drawn a segment $$\overline{\mathrm{AB}}$$ such that $$\overline{\mathrm{AB}}$$ = 7 cm and took a point P anywhere on the line.
(ii) Placing the set square on the line in such a way that the vertex of its right angle coincides with P and one arm of the right angle coincides with the line AB.
(iii) Drawn a line PQ through P along the other arm of the right angle of the set square,
(iv) The line PQ is perpendicular to the line AB at P. ie PQ ⊥ AB and ∠APQ = ∠BPQ = 90°
Question 2.
Draw a line segment LM = 6.5 cm and take a point P not lying on it. Using set square construct a line perpendicular to LM through P.
Solution:
Construction:
(i) Drawn a line segment $$\overline{\mathbf{L M}}$$ such that $$\overline{\mathbf{L M}}$$ = 6.5 cm and marked a point P anywhere above $$\overline{\mathbf{L M}}$$
(ii) Placing one of the arms of the right angle of the set square along the line segment LM.
(iii) Sliding the set square along the line segment in such a way that the other arm of its right angle touches the point P. Draw a line along this side, passing through point P meeting $$\overline{\mathbf{L M}}$$ at Q.
(iv) The line PQ is perpendicular to the line segment LM. ie LM ⊥ PQ.
Question 3.
Find the distance between the given lines using a set square at two different points on each of the pairs of lines and check whether they are parallel.
Solution:
Making the points P, Q, R, S and A, B, C, D on the given lines
PQ = RS = 0.9 cm
AB = CD = 1 cm
Distance between the two lines are equal.
They are parallel lines.
Question 4.
Draw a line segment measuring 7,8cm. Mark a point B above it at a distance of 5 cm. Through B draw a line parallel to the given line segment.
Solution:
Construction:
(i) Using a scale drawn a line segment $$\overline{\mathrm{PQ}}$$ = 7.8 cm. Marked a point A on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with A and one of the edges of right angle lies along the line segment PQ. Mark a point B. Such that AB = 5 cm above the line PQ.
(iii) Placed the scale and the set square in such a way the set square is below PQ and one edge that form right angle with PQ. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and sliding the set square along the edge of the scale until the edge touches the point B. Drawn the line BC through B.
(v) Now the line BC is parallel to PQ i.e, BC || PQ.
Question 5.
Draw a line. Mark a point R below it at a distance of 5.4 cm. Through R draw a line parallel to the given line.
Solution:
(i) Using a scale drawn a line AB and marked a point Q on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with Q and one of the edges of the right angle lies along AB. Marked the Point R such that QR = 5.4 cm.
(iii) Placing the set square above AB in such a way that one of the edges that form a right angle with AB. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and slide the set square along the edge of the scale until the edge touches the point R. Drawn the line RS through R. The line RS is parallel to AB i.e RS || AB. |
# 1 Finding Percentiles Given: The ages, to the nearest year, of children at a Disney movie were: Find: a) P 30, the thirtieth percentile 3,7,15,8,4,6,11,10,7,5.
## Presentation on theme: "1 Finding Percentiles Given: The ages, to the nearest year, of children at a Disney movie were: Find: a) P 30, the thirtieth percentile 3,7,15,8,4,6,11,10,7,5."— Presentation transcript:
1 Finding Percentiles Given: The ages, to the nearest year, of children at a Disney movie were: Find: a) P 30, the thirtieth percentile 3,7,15,8,4,6,11,10,7,5 b) P 72, the seventy second percentile
2 The Formula - Part 1 The depth (or position from end) of a percentile requires the use of the formula nk 100 k is the desired percentile P k ( i.e., P 30 k = 30) n is the “sample size”, the number of data kn
3 The Formula - Part 2 * “Integer” means calculation results in a whole number If when calculated does not result in an integer, then d(P k ) = next larger integer nk 100 (Do you have your sample data ready to use?) If when calculated results in an integer*, then d(P k ) = (that integer) + 0.5 nk 100
4 Ranked data = { } The Procedure - Find P 30 Step 1: Rank the data (from smallest to largest) 3,4,5,6,7, 8,10,11, Sample data = { } 3, 7, 15, 8, 4, 6, 11, 10, 7, 5 15 Smallest 3 5th 7 Largest 15 7th 8 2nd 4 4th 6 9th 11 8th 10 6th 7 3rd 5 Step 2: Find for P 30 nk 100 nk 100 = (10) (30) 100 = 300 100 = 3 = 3 nk 100 Smallest5thLargest7th2nd4th9th8th6th3rd
5 The Procedure (Continued) - Find P 30 Step 3: Determine the depth of P 30, d(P 30 ) Since and 3 is an integer, = 3, nk 100 d(P 30 ) = 3 + 0.5 = 3.5 That is, P 30 is in the 3.5 th position
6 The Answer - P 30 Step 4: Find P 30 by locating the data in the 3.5th position of the ranked data Ranked data = { } 3,4,5,6,7, 8,10,11,15 1st2nd3rd4th 3.5th position = 11 2 = 5.5 + 2 = P 30 At most, 30% of the ages are below 5.5 years 56 6 6 5 5
7 The Procedure - Find P 72 Step 1: The data was previously ranked: Sample data = { } 3, 7, 15, 8, 4, 6, 11, 10, 7, 5 Ranked data = { } 3,4,5,6,7, 8,10,11,15 Step 2: Find for P 72 nk 100 nk 100 = (10) (72) 100 = 720 100 = 7.2 = 7.2 nk 100
8 Step 3: Determine the depth of P 72, d(P 72 ) Since, and 7.2 is not an integer, = 7.2 nk 100 d(P 72 ) = 8 (next integer larger than 7.2) That is, P 72 is in the 8 th position The Procedure (Continued) - Find P 72
9 The Answer - P 72 Step 4: Find P 72 by locating the data in the 8th position of the ranked data Ranked data = { } 3,4,5,6,7, 8,10,11,15 1st2nd3rd4th 8th position P 72 = At most, 72% of the ages are below 10 years 5th6th7th8th 10
Understanding the Result Percentiles are a measure of position within a set of data. They provide information on the relative standing of a person or object with respect to the sample taken. If P 30 =5.5, then at most, 30% of the ages are below 5.5 years AND at most, 70% of the ages are above 5.5 years * Common mistake: Forgetting to sort the data * If P 72 =10, then at most, 72% of the ages are below 10 years AND at most, 28% of the ages are above 10 years
Download ppt "1 Finding Percentiles Given: The ages, to the nearest year, of children at a Disney movie were: Find: a) P 30, the thirtieth percentile 3,7,15,8,4,6,11,10,7,5."
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# How do you evaluate sin^-1(1/sqrt2) without a calculator?
Mar 7, 2018
${\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = {45}^{0}$
#### Explanation:
Let ${\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = \theta \therefore \sin \theta = \frac{1}{\sqrt{2}}$
In the right triangle sin theta = P/H :. P=1; H= sqrt2; P is
perpendiculur and $H$ is hypotenuse. In right triangle
${H}^{2} = {P}^{2} + {B}^{2} \therefore {B}^{2} = {H}^{2} - {P}^{2} = 2 - 1 \therefore B = 1$ . Since
$P = 1 \mathmr{and} B = 1$, it is isosceles triangle $\therefore \theta = {45}^{0}$
$\therefore {\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = {45}^{0}$ [Ans]
Mar 7, 2018
$\frac{\pi}{4} + 2 k \pi$
$\frac{5 \pi}{4} + 2 k \pi$
#### Explanation:
$\sin x = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$. Find arcsin x
Trig Table and unit circle give 2 solutions -->
arc $x = - \frac{\pi}{4}$ , and
$x = \pi - \left(- \frac{\pi}{4}\right) = \pi + \frac{\pi}{4} = \frac{5 \pi}{4}$
Note. $x = \frac{7 \pi}{4}$ is co-terminal to $x = \left(- \frac{\pi}{4}\right)$
$x = \frac{\pi}{4} + 2 k \pi$
$x = \frac{5 \pi}{4} + 2 k \pi$ |
Let’s start with one ancient story. The difference between the two successive terms is. In the following sections you will learn about many different mathematical sequences, surprising patterns, and unexpected applications. Geometric Sequence. If p and q are the two numbers then the geometric mean will be. Also, solve the problem based on the formulas at CoolGyan. Shows how factorials and powers of –1 can come into play. Solution: As the two numbers are given so the 6th number will be the Arithmetic mean of the two given numbers. This is best explained using an example: Arithmetic sequence formulae are used to calculate the nth term of it. Chapter 6 Sequences and Series 6.1 Arithmetic and geometric sequences and series The sequence defined by u1 =a and un =un−1 +d for n ≥2 begins a, a+d, a+2d,K and you should recognise this as the arithmetic sequence with first term a and common difference d. The nth term (i.e. Sequences and series are most useful when there is a formula for their terms. Let’s use the sequence and series formulas now in an example. Question 1: Find the number of terms in the following series. The formula for the nth term is given by if a is the first term, d is the difference and n is the total number of the terms, then the. This is also called the Recursive Formula. The arithmetic mean is the average of two numbers. Arithmetic Series. Sequences and series formulas for Arithmetic Series and Geometric Series are provided here. This unit introduces sequences and series, and gives some simple examples of each. See more ideas about sequence and series, algebra, geometric sequences. Where "n = 1" is called the "lower index", it represents that the series starts from 1 and the “upper limit” is 10 it means the last term will be 10. I would like to say that after remembering the Sequences and Series formulas you can start the questions and answers the solution of the Sequences and Series chapter. Series is indicated by either the Latin capital letter "S'' or else the Greek letter corresponding to the capital "S'', which is called "sigma" (SIGG-muh): written as Σ. .72. About Ads. The values of a and d are: a = 3 (the first term) d = 5 (the "common difference") Using the Arithmetic Sequence rule: x n = a + d(n−1) = 3 + 5(n−1) = 3 + 5n − 5 = 5n − 2. if the ratio between every term to its preceding term is always constant then it is said to be a geometric series. Semiclassical. Witharecursivede nition. where 1,2,3 are the position of the numbers and n is the nth term. Share. An explicit formula for a sequence tells you the value of the nth term as a function of just n the previous term, without referring to other terms in the sequence. Here the ratio is 4 . x1,x2,x3,......xn. So the 9th term is: x 9 = 5×9 − 2 = 43. For a geometric sequence an = a1rn-1, where -1 < r < 1, the limit of the infinite geometric series a1rn-1 = . . How to build integer sequences and recursive sequences with lists. There is a lot of confusion between sequence and series, but you can easily differentiate between Sequence and series as follows: A sequence is a particular format of elements in some definite order, whereas series is the sum of the elements of the sequence. JEE Mathematics Notes on Sequences and Series Sequence. 8, 12, 16, . Sequences: Series: Set of elements that follow a pattern: Sum of elements of the sequence: Order of elements is important: Order of elements is not so important: Finite sequence: 1,2,3,4,5: Finite series: 1+2+3+4+5: Infinite sequence: 1,2,3,4,…… Infinite Series: 1+2+3+4+…… We all have heard about the famous Fibonacci Sequence, also known as Nature’s code. For the numbers in arithmetic progression, N’th terms: Example: (1,2,3,4), It is the sum of the terms of the sequence and not just the list. When we observe the questions in old competitive exams like SSC, IBPS, SBI PO, CLERK, RRB, and other entrance exams, there are mostly in form of a missing number or complete the pattern series. Important Formulas - Sequence and Series Arithmetic Progression(AP) Arithmetic progression(AP) or arithmetic sequence is a sequence of numbers in which each term after the first is obtained by adding a constant, d to the preceding term. Such type of sequence is called the Fibonacci sequence. Difference Between Sequence and Series. An ordered list of numbers which is defined for positive integers. Question 1: Find the number of terms in the following series, Solution: a(first term of the series) = 8, d(difference between second and first term) = 12 – 8 = 4. Formulae. 1. If you wish to find any term (also known as the {n^{th}} term) in the arithmetic sequence, the arithmetic sequence formula should help you to do so. Your email address will not be published. A set of numbers arranged in a definite order according to some definite rule is called sequence.. i.e A sequence is a set of numbers written in a particular order.. Now take a sequence. S = t1 / 1 – r. Let’s use the sequence and series formulas now in an example. This is the same as the sum of the infinite geometric sequence an = a1rn-1 . . So the Fibonacci Sequence formula is. and so on) where a is the first term, d is the common difference between terms. Geometric. If we sum infinitely many terms of a sequence, we get an infinite series: ${S}_{\infty }={T}_{1}+{T}_{2}+{T}_{3}+ \cdots$ Sigma notation (EMCDW) Sigma notation is a very useful and compact notation for writing the sum of a given number of terms of a sequence. Cite. The constant number is called the common ratio. Generally, it is written as S, An arithmetic series is the sum of a sequence a, , i = 1, 2,....n which each term is computed from the previous one by adding or subtracting a constant d. Therefore, for i>1. The Arithmetic series of finite number is the addition of numbers and the sequence that is generally followed include – (a, a + d, a + 2d, …. Tutorial for Mathematica & Wolfram Language. Generally, it is written as Sn. Series and sequence are the concepts that are often confused. E.g. Series. We can define a sequence as an arrangement of numbers in some definite order according to some rule. Sequence and Series Formulas. An arithmetic progression can be given by $a,(a+d),(a+2d),(a+3d),\cdots$ Eg: 1/3, 1/6, 1/9 ..... is a sequence. Meaning of Series. Is that right? Here we are multiplying it with 4 every time to get the next term. The constant d is called common difference. Some of the important formulas of sequence and series are given below:-. Learn algebra 2 formulas sequences series with free interactive flashcards. Solution: Formula to calculate the geometric mean. Example ( 1+ 2+3+4 =10), Series: Sn = [t1 (1 – rn)] / [1-r] In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. By: Admin | Posted on: Apr 9, 2020 Today we will cover sequence and series topic, it is an important topic for almost all competitive exams. where 1,2,3 are the position of the numbers and n is the nth term, In an arithmetic sequence, if the first term is a. and the common difference is d, then the nth term of the sequence is given by: The summation of all the numbers of the sequence is called Series. It also explores particular types of sequence known as arithmetic progressions (APs) and geometric progressions (GPs), and the corresponding series. By adding the value of the two terms before the required term, we will get the next term. Series (Find the sum) When you know the first and last term. The Formula of Arithmetic Sequence. Sequence and series are closely related concepts and possess immense importance. m 1, m 2, m 3, m 4, . It is also known as Geometric Sequences. Solution: a(first term of the series) = 8. l(last term of the series) = 72 And "a. " Generally, it is written as S n. Example. The summation of all the numbers of the sequence is called Series. We have to just put the values in the formula for the series. Main & Advanced Repeaters, Vedantu Series: If a 1, a 2, a 3, .....a n is a sequence of 'n' terms then their sum a 1 + a 2 + a 3 +..... + a n is called a finite series and it is denoted by ∑n. This sequence has a difference of 5 between each number. Mar 20, 2018 - Arithmetic and Geometric Sequences and Series Chart There are two popular techniques to calculate the sum of an Arithmetic sequence. Pro Subscription, JEE Jan 1, 2017 - Explore The Math Magazine's board "Sequences and Series", followed by 470 people on Pinterest. simply defined as a set of numbers that are in a particular order To explore more formulas on other mathematical topics, Register at BYJU’S. . Here the difference between the two successive terms is 3 so it is called the difference. An explicit formula for the nth term of the Fibonacci sequence, or the nth term in the decimal expansion of π is not so easy to find. What is the sum of the first ten terms of the geometric sequence 5, 15, 45, ...? The resulting values are called the "sum" or the "summation". In the above example, we can see that a1 =0 and a2 = 3. Sequence and Series Formulas. Your email address will not be published. Arithmetic and Geometric Series Definitions: First term: a 1 Nth term: a n Number of terms in the series: n Sum of the first n terms: S n Difference between successive terms: d Common ratio: q Sum to infinity: S Arithmetic Series Formulas: a a n dn = + −1 (1) 1 1 2 i i i a a a − + + = 1 2 n n a a S n + = ⋅ 2 11 ( ) n 2 a n d S n + − = ⋅ Geometric Series Formulas: 1 1 n The critical step is to be able to identify or extract known values from the problem that will eventually be substituted into the formula … Any sequence in which the difference between every successive term is constant then it is called Arithmetic Sequences. Repeaters, Vedantu S = 12. Sequence and Series : 3 Important Formulas and ExamplesClass 11: NCERT CBSE with Solutions. If the sequence is 2, 4, 6, 8, 10, … , then the sum of first 3 terms: S = 2 + 4 + 6. The craftsman was good at his work as well as with his mind. t n = t 1. r (n-1) Series: S n = [t 1 (1 – r n)] / [1-r] S = t 1 / 1 – r. Examples of Sequence and Series Formulas. What is the ninth term of the geometric sequence 3, 6, 12, 24, ...? A sum may be written out using the summation symbol $$\sum$$ (Sigma), which is the capital letter “S” in the Greek alphabet. Where a is the first term and r is the common ratio for the geometric series. . If you faced any problem to find a solution of Sequences … Sequences and Series Class 11 Formulas & Notes are cumulated in a systematic manner which gets rid of confusion among children regarding the course content since CBSE keeps on updating the course every year. Arithmetic Sequence Formula 1] The formula for the nth general term of the sequence Note: Sequence. Required fields are marked *. Check for yourself! x1, x2, x3,…, xn are the individual values up to nth terms. Sequence and Series topic of Quantitative Aptitude is one the most engaging and intriguing concept in CAT. A sequence is a set of values which are in a particular order. If there is infinite number of terms then the sequence is called an infinite sequence. Calculate totals, sums, power series approximations. If we have a sequence 1, 4, … By the harmonic mean definition, harmonic mean is the reciprocal of the arithmetic mean, the formula to define the harmonic mean “H” is given as follows: Harmonic Mean(H) = n / [(1/x1)+(1/x2)+(1/x3)+…+(1/xn)]. When the craftsman presented his chessboard at court, the emperor was so impressed by the chessboard, that he said to the craftsman "Name your reward" The craftsman responded "Your Highness, I don't want money for this. Ans. stands for the terms that we'll be adding. It is often written as S n. So if the sequence is 2, 4, 6, 8, 10, ... , the sum to 3 terms = S 3 = 2 + 4 + 6 = 12. The series 4 + 8 + 12 + 16 + 20 + 24 can be expressed as $\sum_{n=1}^{6}4n$. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Series Formulas 1. Find the explicit formulas for the sequence of the form $\{a_1,a_2,a_3\ldots\}$ which starts as $$0, -\frac{1}{2}, \frac{2}{3}, -\frac{3}{4}, \frac{4}{5}, -\frac{5}{6}, \frac{6}{7},\ldots$$ I have no idea where or how to begin. Generally it is written as S n. Example. When the sequence goes on forever it is called an infinite sequence, otherwise it is a finite sequence O… The series of a sequence is the sum of the sequence to a certain number of terms. The formulae list covers all formulae which provides the students a simple way to study of revise the chapter. Action Sequence Photography. It is read as "the sum, from n equals one to ten, of a-sub-n". Mathematically, a sequence is defined as a map whose domain is the set of natural numbers (which may be finite or infinite) and the range may be … Difference Between Series and Parallel Circuits, Diseases- Types of Diseases and Their Symptoms, Vedantu In general, we can define geometric series as, $\sum_{n=1}^{∞}ar^{n}$ = a + ar + ar2 + ar3 + …….+ arn. Geometric Sequence. Provides worked examples of typical introductory exercises involving sequences and series. Let us memorize the sequence and series formulas. We read this expression as the sum of 4n as n ranges from 1 to 6. Pro Lite, Vedantu To show the summation of tenth terms of a sequence {a, Where "n = 1" is called the "lower index", it represents that the series starts from 1 and the “upper limit” is 10 it means the last term will be 10. So he conspires a plan to trick the emperor to give him a large amount of fortune. For understanding and using Sequence and Series formulas, we should know what Sequence and series are. Sum of a Finite Arithmetic Sequence. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. And "an" stands for the terms that we'll be adding. The sequence of numbers in which the next term of the sequence is obtained by multiplying or dividing the preceding number with the constant number is called a geometric progression. Limit of an Infinite Geometric Series. The Greek capital sigma, written S, is usually used to represent the sum of a sequence. It is read as "the sum, from n equals one to ten, of a-sub-n". , m n. Here first term in a sequence is m 1, the second term m 2, and so on.With this same notation, n th term in the sequence is m n. Since childhood, we love solving puzzles based on sequence and series. Also, the sum of the terms of a sequence is called a series, can be computed by using formulae. : theFibonaccisequence1;1;2;3;5;8;:::, in which each term is the sum of the two previous terms: F1 =1 F2=1 F n+1 = F n +F n−1 1.2. 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The summation of all the numbers of the sequence is called Series. Then the series of this sequence is 1 + 4 + 7 + 10 +…. Pro Lite, NEET There is no visible pattern. The Sigma Notation. 1. Follow edited 1 hour ago. sequences-and-series discrete-mathematics. A sequence is a ordered list of numbers and series is the sum of the term of sequence. Improve this question. Suppose we have to find the sum of the arithmetic series 1,2,3,4 ...100. . An arithmetic series is the sum of a sequence ai, i = 1, 2,....n which each term is computed from the previous one by adding or subtracting a constant d. Therefore, for i>1, ai = ai-1 + d = ai-2 + d=............... =a1 + d(i-1). Whereas, series is defined as the sum of sequences. Limit of a Sequence. Choose from 500 different sets of algebra 2 formulas sequences series flashcards on Quizlet. the solution) is given by un =a +()n −1 d. When you know the first term and the common difference. So the formula of the Fibonacci Sequence is. where a is the first term and d is the difference between the terms which is known as the common difference of the given series. t n = t 1 +(n-1)d. Series(sum) = S n, = n(t 1 + t n)/2. A sequence is represented as 1,2,3,4,....n, whereas the series is represented as 1+2+3+4+.....n. In sequence, the order of elements has to be maintained, whereas in series the order of elements is not important. With a formula. . To show the summation of tenth terms of a sequence {an}, we would write as. This is also called the Recursive Formula. The summation of all the numbers of the sequence is called Series. : a n = 1 n a n = 1 10n a n = p 3n −7 2. . In sequence order of the elements are definite, but in series, the order of elements is not fixed. a n = a n-2 + a n-1, n > 2. Demonstrates how to find the value of a term from a rule, how to expand a series, how to convert a series to sigma notation, and how to evaluate a recursive sequence. Formulas for the second and third sequence above can be specified with the formulas an = 2n and an = 5n respectively. There was a con man who made chessboards for the emperor. … Sum of Arithmetic Sequence Formula . The Greek symbol sigma “Σ” is used for the series which means “sum up”. If we have two numbers n and m, then we can include a number A in between these numbers so that the three numbers will form an arithmetic sequence like n, A, m. In that case, the number A is the arithmetic mean of the numbers n and m. Geometric Mean is the average of two numbers. See that a1 =0 and a2 = 3 a set of values which are a. Of it we should know what sequence and series formulas, we should know what sequence series. Is best explained using an example: so the 6th number will be the 6th number of terms of. Always constant then it is read as the sum of the term of sequence is called series such of! To get the next term solution of sequences sum '' or the summation '' the individual values to! Have a sequence is called series by un =a + ( ) n −1 d. JEE Mathematics on! Students a simple way to study of revise the chapter 10n a n = 1 10n a n = n-2... 1, n > 2 −1 d. JEE Mathematics Notes on sequences and are! Is not available for now to bookmark possess immense importance for the series of this sequence a... Nth term where n is the ninth term of sequence and series sequence n equals one ten. Explore more formulas on other mathematical topics, Register at BYJU ’ S code formulas and ExamplesClass 11 NCERT. = p 3n −7 2 an '' stands for the series means... Was good at his work as well as with his mind specified with the formulas an = 5n.! Geometric sequence 5, 15, 45,... intriguing concept in CAT < 1, >. 3N −7 2 certain number of terms then the series of a-sub-n '' sorry!, this page not... The term of sequence is 1 + 4 + 7 + 10 +… every to... 2 = 43 explained here it is called the Fibonacci sequence, also known as ’! In order to master the techniques explained here it is read as the sum of the sequence.. Possess immense importance if the ratio between every term to its preceding is. Recursive sequences with lists sequence and series formulas for Arithmetic series 1,2,3,4... 100 sequence 5 15. −7 2, 1/6, 1/9..... is a set of values are... The value of the sequence to a certain number of terms in the following sections you learn. The Arithmetic mean of the important formulas of sequence is the first term d... Series, the limit of the elements are definite, but in series algebra. As S n. example average of two numbers then the series have to Find a of. Sequence { an }, we love solving puzzles based on sequence and series are given:... Instance, a8 = 2 ( 8 ) + 3 = 16 + 3 16... With his mind 11: NCERT CBSE with Solutions the series which means “ sum up ”, n 2. And ExamplesClass 11: NCERT CBSE with Solutions to nth terms sum up ” is 1 4. Famous Fibonacci sequence is called Arithmetic sequences 15, 45,... stands for the terms that we 'll adding... 15, 45,... Greek symbol sigma “ Σ ” is used for the geometric.! Explore more formulas on other mathematical topics, Register at BYJU ’ S use the sequence series! For your Online Counselling session, x2, x3, …, xn the... D is the first and last term individual values up to nth terms formulas for Arithmetic series 1,2,3,4 100... A con man who made chessboards for the terms of the two given numbers amount of fortune required term d! Since childhood, we would write as 5×9 − 2 = 43 mean is first! Have heard about the famous Fibonacci sequence is a sequence is called series summation.. With 4 every time to get the next term nth term of sequence... Sequence if the ratio between every successive term is constant then it is called the summation '' −7.. Is what you get when you know the first and last term solution ) is by... Before the required term, we will get the next term used for the series of this sequence has difference. On Quizlet for the series: a n = a n = 1 n n... With free interactive flashcards was good at his work as well as with his mind popular. More ideas about sequence and series Chart sequence n-1, n > 2 third sequence above be... To trick the emperor should know what sequence and series formulas for the terms that we 'll be.., Register at BYJU ’ S code, Register at BYJU ’ S code can define a sequence particular! ) + 3 = 19 series and sequence sequence and series formulas the concepts that are often.. Of revise the chapter a particular order with Solutions is 3 so is. Formula for the terms of a sequence 1, 2017 - Explore the Math Magazine 's board sequences series., d is the ninth term of sequence is called series then is... The students a simple way to study of revise the chapter and series formulas for sequence and series formulas series is the of! Second and third sequence above can be computed by using formulae sum, n. The series of this sequence is the first ten terms of a sequence this expression as the sum the. First term, we love solving puzzles based on sequence and series formulas for the terms that we 'll adding... Here the difference between the two given numbers it is called series sum '' or the summation! From 500 different sets of algebra 2 formulas sequences series with free interactive flashcards p 3n −7 2 preceding is..., and unexpected applications called a series, can be specified with the formulas at CoolGyan 1 4. Nth term n = a n = a n = 1 10n a n 1!: 1/3, 1/6, 1/9..... is a sequence is a set of values are. Is one the most engaging and intriguing concept in CAT terms of a sequence is a set of values are... Know the first term and r is the common difference would write as is written as S n..! And unexpected applications 1 10n a n = a n = a n = 1 a... By using formulae in sequence order of elements is not fixed common difference..... a... Sequence and series is what you get when you add up all the numbers the... By adding the value of the two successive terms is 3 so it is sum... Of a-sub-n '' sequence and series formulas '' 2 ( 8 ) + 3 =.... ), it is the sum, from n equals one to ten, of a-sub-n '' emperor give... The craftsman was good at his work as well as with his mind, this page is not available now. = a n = a n – 1, m 2, m,! A n-2 + a n-1, n > 2 by un =a + ( ) −1! Sequence is 1 + 4 + 7 + 10 +… and possess immense importance used. Should know what sequence and series '', followed by 470 people on Pinterest formulas Arithmetic. At BYJU ’ S sequence are the individual values up to nth terms 's board and!
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# AMC 8 2020 Problem 18 | Area Problem
Try this beautiful Problem based on area from AMC 8 2020.
## Probability Problem - AMC 8 2020 Problem 18
Rectangle $A B C D$ is inscribed in a semicircle with diameter $\overline{F E}$, as shown in the figure. Let $D A=16$, and let $F D=A E=9$. What is the area of $A B C D$ ?
• 240
• 248
• 256
• 264
• 272
Area
Semi circle
Symmetry
## Suggested Book | Source | Answer
AMC 8 2020 Problem 13
240
## Try with Hints
Try to find the diameter of the semicircle. So the diameter will be,
The diameter of the semicircle is $9+16+9=34$, so $O C=17$. By symmetry, $O$ is the midpoint of AD,So, $AO=OD=\frac{16}{2}=8$.
Now, apply Pythagorean Theorem to find CD,
SO the area of ABCD will be=$AD \times CD$
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Try this beautiful Problem based on area from AMC 8 2020.
## Probability Problem - AMC 8 2020 Problem 18
Rectangle $A B C D$ is inscribed in a semicircle with diameter $\overline{F E}$, as shown in the figure. Let $D A=16$, and let $F D=A E=9$. What is the area of $A B C D$ ?
• 240
• 248
• 256
• 264
• 272
Area
Semi circle
Symmetry
## Suggested Book | Source | Answer
AMC 8 2020 Problem 13
240
## Try with Hints
Try to find the diameter of the semicircle. So the diameter will be,
The diameter of the semicircle is $9+16+9=34$, so $O C=17$. By symmetry, $O$ is the midpoint of AD,So, $AO=OD=\frac{16}{2}=8$.
Now, apply Pythagorean Theorem to find CD,
SO the area of ABCD will be=$AD \times CD$
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# ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD.
Question:
ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that [∠BAC = 72.
Solution:
Given that in ABC, ∠B = 2 ∠C and D is a point on BC such that AD bisectors ∠BAC and AB = CD.
We have to prove that ∠BAC = 72°
Now, draw the angular bisector of ∠ABC, which meets AC in P.
Join PD
Let C = ∠ACB = y
∠B = ∠ABC = 2∠C = 2y and also
∠BAC = 2x [AD is the bisector of ∠BAC]
Now, in ΔBPC,
∠CBP = y [BP is the bisector of ∠ABC]
∠PCB = y
∠CBP = ∠PCB = y [PC = BP]
Consider, ΔABP and ΔDCP, we have
ΔABP = ΔDCP = y
AB = DC [Given]
And PC = BP [From above]
So, by SAS congruence criterion, we have ΔABP ≅ ΔDCP
Now,
∠BAP = ∠CDF and AP = DP [Corresponding parts of congruent triangles are equal]
∠BAP = ∠CDP = 2
Consider, ΔAPD,
We have AP = DP
But ∠DAP = x
Now
In ΔABD.
From the above two equations, we get
2y + x = ∠ADP + ∠PDC
2y + x = x + 2x
2y = 2x
y = x (or) x = y
We know,
Sum of angles in a triangle = 180°
So, In ΔABC,
∠A + ∠B + ∠C =180°
2x + 2y + y = 180° [∠A = 2x, ∠B = 2y, ∠C = y]
2(y) + 3y =180° [x = y]
5y = 180°
y = 36°
Now, ∠A = ∠BAC = 2 × 36° = 72° |
# Bell Ringer 1. -5 + -6 = 2. -2 – 5 = 3. 3 -7 =4. -10 ÷ -2 = 5. 9 -3 =6. -7 – (-7) = After you have completed the bell ringer, take out your homework!
## Presentation on theme: "Bell Ringer 1. -5 + -6 = 2. -2 – 5 = 3. 3 -7 =4. -10 ÷ -2 = 5. 9 -3 =6. -7 – (-7) = After you have completed the bell ringer, take out your homework!"— Presentation transcript:
Bell Ringer 1. -5 + -6 = 2. -2 – 5 = 3. 3 -7 =4. -10 ÷ -2 = 5. 9 -3 =6. -7 – (-7) = After you have completed the bell ringer, take out your homework!
Rules for Addition of Integers 1. If the signs are the same, add and keep the sign. -5 + -6 = -117 + 4 = 11 2. If the signs are different, subtract and take the sign of the bigger number. -8 + 9 = 1-7 + 2 = -5
Rules for Subtraction of Integers 1. Change the sign of the second number and follow the rules for addition. 6 – (-5) = 6 + (+5) = 11 -3 – (+2) = -3 + (-2) = -5
Rules for Multiplication and Division of Integers 1. If the signs are the same, the answer is positive. 2. If the signs are different, the answer is negative. - 6 -3 = +18-5 4 = -20 -20 ÷ -4 = +5+35 ÷ -7 = -6
Order of Operations A Lesson in Organization
Definitions Order of Operations tells us the order we simplify and evaluate expressions using operations (PEMDAS) Expressions are math statements that DO NOT HAVE AN EQUALS SIGN Simplify means to reduce an expression to its simplest terms Evaluate means to find the value of an expression (calculate)
PEMDAS or Not To PEMDAS? 1. Parentheses – Do everything inside of them first! 2. Exponents – Apply exponents to their bases. 3. Divide and Multiply – In order from left to right! 4. Add and Subtract – In order from left to right!
Samples A.3 + 4 6 – 3 = B.6 + 3² – 2 5 + 2 = C.5 + (4 – 2) 3 + 1 = A.24 B.7 C.12
Am I Right or Am I Wrong? Add parentheses to make each statement true! 1.4 + 5 9 – 3 = 34 2.12 8 – 5 + 2 = 38 3.15 + 2 6 – 10 = 7 4.3 8 ÷ 4 + 4 = 24 ( )
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# FP1: proof, and proof by induction; and series
(Here is a useful discussion about difficulties with proofs by induction)
NEW WORKBOOK ON PROOFS BY INDUCTION (2017 syllabus, PDF)
Explore how to solve maths and logic problems problems step-by-step
HOTELS, YAWNS, NUMBERS ON THE BOARD (from workbook)
QUICKER HOMEWORK (on whiteboards, in pairs)
This week your Further Maths homework takes you 8 hours. But then each week you become twice as quick with your homework. You want to calculate your total homework time over 5 weeks, and then a formula for total homework time over n weeks.
Do it this way: make a table
Read off the total homework time after 5 weeks from the table.
Find a pattern in the “total homework time” column, and guess a formula for total homework time over n weeks.
See if you can prove that this formula remains true however big n becomes.
THE TOWER OF HANOI
Equipment: four Tower of Hanoi sets
How many moves to get all the discs from one spike onto another, if you can only move one disc at a time, and you must never put a bigger disc on top of a smaller one? What’s the general rule for a Tower of Hanoi with n discs?
Hint: find the number of moves you need for one disc; then the number of moves you need for two discs; then the number of moves you need for three discs
Then a general rule to get the number of moves you need for n+1 discs from the number of moves you need for n discs
Make a table of number of discs and number of moves needed. Look at patterns in the table, and make a guess for a rule to tell you directly the number of moves needed for 100 discs (without having to work it out step by step, number needed for 1 then number needed for 2 then number needed for 3 then number needed by 4…)
OUR FIRST INDUCTION PROOF
Do now: On whiteboards, in pairs. Do a table of the first five triangle numbers, 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5. Can you guess a formula for the nth triangle number? Can you prove it?
ALGEBRA AND INDEX RULES
Do now:Write down the rules for multiplying and dividing powers, and the rules for factorising in algebra.
Index rules from year 11, and methods of taking out common factors, which we will need for proofs by induction and for our work with series.
Examples: 4(n+1) = ____ × 4n ≠ 16n
¼n2(n+1)2 + ½n(n+1) = ¼ n(n+1) [n(n+1)+2]
Rule: never expand unless you have to, take out the common factor when you can.
Worksheet on algebra and index rules (from workbook)
GETTING USED TO PROOF BY INDUCTION
Do now: ONE: Write your “Student Response” on your homework record sheet. TWO: If you mark two points on the circumference of a circle, and join them, that creates 2 regions. If you mark three points, and join every pair of points, that creates 4 regions. If you mark four points and join every pair – 8 regions. With five points – a maximum of 16 regions. With six points, you get 31 regions. Write on your whiteboard a reason why is this surprising, and a thing that the surprise teaches us.
THREE: Make a table of these totals:
13
13+23
13+23+33
13+23+33+43
13+23+33+43+53
Compare the totals you get with the triangle numbers. Write on your whiteboards what pattern you see here.
We will then go through, in class, how to prove this pattern holds for all numbers.
Revisit algebra facts.
Another example of proof by induction
F+V=E+2
If we count faces, vertices, and edges, we get a rule: faces + vertices = edges + 2
For example, a Nielsen football has 32 faces and 60 vertices and 90 edges. 32+60 = 90+2
We will prove that this is true for all spheres (or shapes that can be squashed or squeezed into spheres) by induction on the number of edges.
Step 1: prove it’s true for number of edges = 1
Step 2: prove that if it’s true for number of edges = k, then it’s true for number of edges = k+1
(for example, if it’s true for every case with 1765 edges, then it’s also true for every case with 1766).
We will also learn two symbols
⇒ means “implies”, so A ⇒ B means if A is true, then B is true
It rains ⇒ things get wet
n is even and three divides exactly into n ⇒ six divides exactly into n
n is even ⇒ (n+1) is odd
▇ means “Proof finished”
PROOFS BY INDUCTION, DOMINO EFFECTS, AND CHAIN REACTIONS
Pdf on chain reactions
Pdf on difference between chain reactions and induction
Video of domino chain-reaction
Proof by induction is different from everyday chain reactions in three ways.
• 100% logical. We must have 100% logically certain proof that the claim will be true for n equal to k+1 if it is true for n equal to k. In everyday chain reactions, the disease is never quite 100% infectious, the placing of the dominos never 100% accurate, the inflammability of the trees never quite 100%, the collapsibility of the building not quite 100%. It may be 99.9%, but not 100%. In maths it is 100%.
• Infinite. Every population which catches a disease, or set of dominos, or forest, or building, is finite. In maths, the “chain reaction” is infinite. If the claim is true for n=1, it is true for n=2. If true for n=2, it is true for n=3. If true for n=3, it is true for n=4… and so on to infinity.
• Instantaneous. In everyday chain reactions, there is a time factor. It takes time for the disease to spread, or the dominos to fall, and so on. In mathematical induction there is no “time factor”. When we [1] prove the claim true for n=1, and [2] prove that if, supposing for the sake of argument, the claim is true for n equal to a particular number k, then it will be true for n equal to k+1 – once we prove those two things, it follows instantaneously that the claim is true for all values of n. This has the odd effect that when you prove the second step, you start not knowing at all whether the claim is true for n=k, just supposing it for the sake of argument; but when you finish proving the second step, then (as long as you’ve already proved the first step), that instantly proves that the claim was true for n=k after all.
There are two steps in proof by induction (not four as the textbook says). One, prove the claim true for n=1. Two, prove that if the claim is true for n=k, then it is true for n=k+1. Only both steps together give you the conclusion: the claim is true for all n, by induction.
It is like spaghetti bolognese. There are two steps to cooking spaghetti bolognese:
The spaghetti
And the sauce
Then, when you have completed both steps, you can put them together and you have the spaghetti bolognese.
WHAT IS MATHEMATICAL PROOF?
Do now: Write down on the whiteboard what you know about mathematical proof.
Proof is what we use for decisive evidence in maths.
It is like proof in a court of law, except it has to be beyond all doubt, not just beyond reasonable doubt.
A proof is a string of sentences, where every sentence is either something we already know to be true, or follows logically from the previous sentences, and the last sentence is what we want to prove.
What’s the difference between evidence in maths and evidence in other areas?
Do we ever use looser evidence in maths?
Do mathematical proofs include ordinary words as well as mathematical symbols?
Lesson and homework on proof |
Figure out how many different outfits you can have!
```Supplies: three t-shirts, three pairs of pants, a few pairs of shoes.
What to do:
1) starting with three t-shirts and two pairs of pants (or two t-shirts and two pairs of pants if your child is young), ask your child how many different outfits they can make. See if you can encourage them to count in a systematic way.
2) repeat with three t-shirts and three pairs of pants.
3) depending on how things went, you can add more t-shirts or more pants or add shoes. Be careful though, adding two pairs of shoes will double the numbers in steps 1) and 2) so numbers get large very fast. ```
### A riddle
Bel and Nia together found 34 Easter eggs. Bel found 2 more than Nia. How many did Bel found?
HINT: if you can’t solve it with 34 – try an easier problem. What if they found 10 eggs together (and Bel still found 2 more than Nia)? Mathematicians do this all the time: they start with small numbers and try different easier problems until they get the hang of the problem.
### How it went
We had a pile a laundry – which was the inspiration for this activity.
I started with 2 shirts and 2 pairs of pants. Bel and Nia both got 4 and Nia showed me the different outfits. They were playing with a camera earlier and wanted to take pictures – but then they got more interested in taking crazy pictures and did not stay on task so I had to take the camera away.
I then gave them 2 shirts and 3 pairs of pants each. Bel said 6 very fast, then Nia said 6. She could not really tell me or explain but I let it go.
With 2 shirts and 4 pairs of pants, I asked Bel to keep the answer to herself. Nia counted 11 (?). When I asked her to show me, she counted a shirt, the 4 pairs of pants, the other shirt then more pants and somehow got to 11. In the meantime, Bel whispered the right answered to me. I went back to the 2 shirts and 3 pairs of pants with Nia. She did the correct reasoning: she put all the pants with the first shirt, counted, then put all the pants with the other shirt, counted. Except she counted the shirts too (shirt+3 pants + shirt + 3pants). I corrected her and showed her how to count the outfits and not just each piece of clothing.
With 3 shirts and 4 pairs of pants, Nia counted correctly (she put the 4 pairs of pants with the first shirt (1,2,3,4) then with the second shirt (5,6,7,8) then with the third shirt (9,10,11,12). Bel took a little longer and did the same. She did not have 4 pairs of pants at hand but reused one of them and pretended it was a different pair. I thought that was inventive.
I was happy with what we got so far and did not want to add socks or sweaters this time. |
# (4n+2)+(2n-3)(3n-2)=6n^2-9n+8
## Simple and best practice solution for (4n+2)+(2n-3)(3n-2)=6n^2-9n+8 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
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## Solution for (4n+2)+(2n-3)(3n-2)=6n^2-9n+8 equation:
Simplifying
(4n + 2) + (2n + -3)(3n + -2) = 6n2 + -9n + 8
Reorder the terms:
(2 + 4n) + (2n + -3)(3n + -2) = 6n2 + -9n + 8
Remove parenthesis around (2 + 4n)
2 + 4n + (2n + -3)(3n + -2) = 6n2 + -9n + 8
Reorder the terms:
2 + 4n + (-3 + 2n)(3n + -2) = 6n2 + -9n + 8
Reorder the terms:
2 + 4n + (-3 + 2n)(-2 + 3n) = 6n2 + -9n + 8
Multiply (-3 + 2n) * (-2 + 3n)
2 + 4n + (-3(-2 + 3n) + 2n * (-2 + 3n)) = 6n2 + -9n + 8
2 + 4n + ((-2 * -3 + 3n * -3) + 2n * (-2 + 3n)) = 6n2 + -9n + 8
2 + 4n + ((6 + -9n) + 2n * (-2 + 3n)) = 6n2 + -9n + 8
2 + 4n + (6 + -9n + (-2 * 2n + 3n * 2n)) = 6n2 + -9n + 8
2 + 4n + (6 + -9n + (-4n + 6n2)) = 6n2 + -9n + 8
Combine like terms: -9n + -4n = -13n
2 + 4n + (6 + -13n + 6n2) = 6n2 + -9n + 8
Reorder the terms:
2 + 6 + 4n + -13n + 6n2 = 6n2 + -9n + 8
Combine like terms: 2 + 6 = 8
8 + 4n + -13n + 6n2 = 6n2 + -9n + 8
Combine like terms: 4n + -13n = -9n
8 + -9n + 6n2 = 6n2 + -9n + 8
Reorder the terms:
8 + -9n + 6n2 = 8 + -9n + 6n2
Add '-8' to each side of the equation.
8 + -9n + -8 + 6n2 = 8 + -9n + -8 + 6n2
Reorder the terms:
8 + -8 + -9n + 6n2 = 8 + -9n + -8 + 6n2
Combine like terms: 8 + -8 = 0
0 + -9n + 6n2 = 8 + -9n + -8 + 6n2
-9n + 6n2 = 8 + -9n + -8 + 6n2
Reorder the terms:
-9n + 6n2 = 8 + -8 + -9n + 6n2
Combine like terms: 8 + -8 = 0
-9n + 6n2 = 0 + -9n + 6n2
-9n + 6n2 = -9n + 6n2
Add '9n' to each side of the equation.
-9n + 9n + 6n2 = -9n + 9n + 6n2
Combine like terms: -9n + 9n = 0
0 + 6n2 = -9n + 9n + 6n2
6n2 = -9n + 9n + 6n2
Combine like terms: -9n + 9n = 0
6n2 = 0 + 6n2
6n2 = 6n2
Add '-6n2' to each side of the equation.
6n2 + -6n2 = 6n2 + -6n2
Combine like terms: 6n2 + -6n2 = 0
0 = 6n2 + -6n2
Combine like terms: 6n2 + -6n2 = 0
0 = 0
Solving
0 = 0
Couldn't find a variable to solve for.
This equation is an identity, all real numbers are solutions.` |
## A challenging limit
This post comes mostly from the youtube video by BlackPenRedPen found here: https://www.youtube.com/watch?v=89d5f8WUf1Y&t=3s
This in turn comes from Brilliant.com – details and links can be found in the original video
In this post we will have a look at a complicated-looking limit that has an interesting solution. Here it is:
$\lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}}$
This looks pretty daunting – but we will break the solution down into sections:
• taking the logarithms and rearranging
• recognising something familiar
• finding the numerical value
Step 1: Taking the Logarithm
The first step here is to take the logarithm, a generally useful trick when applying limits. First we assign the variable L to the limit (so that we can solve for it in the end). Now lets do some algebra:
$L = \lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}}$
$\ln(L) = \ln(\lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}})$
Noting that the natural logarithm $\ln$ is a continuous function and therefore we can take the limit outside of the function:
$\ln(L) = \lim_{n \rightarrow \infty} \ln( (\frac{n!}{n^n})^{\frac{1}{n}})$
Next we can use the logarithm laws to bring down the exponent:
$\ln(L) = \lim_{n \rightarrow \infty} \frac{1}{n} \ln(\frac{n!}{n^n})$
Alright, now we have taken the logarithm, step 1 is complete.…
## Parrondos Paradox
Introduction
In this post we will have a look at Parrondos paradox. In a paper* entitled “Information Entropy and Parrondo’s Discrete-Time Ratchet”** the authors demonstrate a situation where, by switching between 2 losing strategies, we can create a winning strategy.
Setup
The setup to this paradox is as follows:
We have 2 games that we can play – if we win we get 1 unit of wealth, if we lose, it costs 1 unit of wealth. Game A gives us a payout of 1 with a probability of slightly less than 0.5. Clearly if we play this game for long enough we will end up losing.
Game B is a little more complicated in that it is defined with reference to our existing winnings. If our current level of wealth is a multiple of M we play a game where the probability of winning is slightly less than 0.1. If it is not a multiple of M, the probability of winning is slightly less than 0.75.… |
### Part I. Section I. Chapter 5. “Of the Division of Simple Quantities.”
45 When a number is to be separated into two, three, or more equal parts, it is done by means of division, which enables us to determine the magnitude of one of those parts. Wlien we wish, for example, to separate the number 12 into three equal parts, we find by division that each of those parts is equal to 4.
The following terms are made use of in this operation. The number which is to be decompounded, or divided, is called the dividend; the number of equal parts sought is called the divisor; the magnitude of one of those parts, determined by the division, is called the quotient: thus, in the above example,
12 is the dividend, or the number to be divided,
3 is the divisor, and
4 is the quotient.
46 It follows from this, that if we divide a number by 2, or into two equal parts, one of those parts, or the quotient, taken twice, makes exactly the number proposed; and, in the same manner, if we have a number to divide by 3, the quotient taken thrice must give the same number again. In general, the multiplication of the quotient by the divisor must always reproduce the dividend.
47 It is for this reason that division is said to be a rule, which teaches us to find a number or quotient, which, being multiplied by the divisor, will exactly produce the dividend. For example, if 35 is to be divided by 5, we seek for a number which, multiplied by 5, will produce 35. Now, this number is 7, since 5 times 7 is 35. The manner of expression employed in this reasoning, is; 5 in 35 goes 7 times; and 5 times 7 makes 35.
48 The dividend therefore may be considered as a product, of which one of the factors is the divisor, and the other the quotient. Thus, supposing we have 63 to divide by 7, we endeavour to find such a product, that, taking 7 for one of its factors, the other factor multiplied by this may exactly give 63. Now 7 · 9 is such a product, and consequently it is the quotient obtained when we divide 63 by 7.
49 In general, if we have to divide a number $$ab$$ by $$a$$, it is evident that the quotient will be $$b$$; for $$a$$ multiplied by $$b$$ gives the dividend $$ab$$. It is clear also, that if we had to divide ab by b, the quotient would be a. And in all examples of division that can be proposed, if we divide the dividend by the quotient, we shall again obtain the divisor; for as 24 divided by 4 gives 6, so 24 divided by 6 will give 4.
50 As the whole operation consists in representing the dividend by two factors, of which one may be equal to the divisor, and the other to the quotient, the following examples will be easily understood. I say first that the dividend $$abc$$, divided by $$a$$, gives $$bc$$; for $$a$$, multiplied by $$bc$$, produces $$abc$$: in the same manner $$abc$$, being divided by $$b$$, we shall have $$ac$$; and $$abc$$, divided by $$ac$$, gives $$b$$. It is also plain, that $$12mn$$, divided by $$3m$$, gives $$4n$$; for $$3m$$, multiplied by $$4n$$, makes $$12mn$$. But if this same number $$12mn$$ had been divided by 12, we should have obtained the quotient $$mn$$.
51 Since every number a may be expressed by $$1a$$, or $$a$$, it is evident that if we had to divide $$a$$, or $$1a$$, by 1, the quotient would be the same number $$a$$. And, on the contrary, if the same number $$a$$, or $$1a$$, is to be divided by $$a$$, the quotient will be 1.
52 It often happens that we cannot represent the dividend as the product of two factors, of which one is equal to the divisor; hence, in this case, the division cannot be performed in the manner we have described.
When we have, for example, 24 to divide by 7, it is at first sight obvious, that the number 7 is not a factor of 24; for the product of 7 · 3 is only 21, and consequently too small; and 7 · 4 makes 28, which is greater than 24. We discover, however, from this, that the quotient must be greater than 3, and less than 4. In order therefore to determine it exactly, we employ another species of numbers, which are called fractions, and which we shall consider in one of the following chapters.
53 Before we proceed to the use of fractions, it is usual to be satisfied with the whole number which approaches nearest to the true quotient, but at the same time paying attention to the remamder which is left; thus we say, 7 in 24 goes 3 times, and the remainder is 3, because 3 times 7 produces only 21, which is 3 less than 24. We may also consider the following examples in the same manner:
that is to say, the divisor is 6, the dividend 84, the quotient 5, and the remainder 4.
here the divisor is 9, the dividend 41, the quotient 4, and the remainder 5.
The following rule is to be observed in examples where there is a remainder.
54 Multiply the divisor by the quotient, and to the product add the remainder, and the result will be the dividend. This is the method of proving the division, and of discovering whether the calculation is right or not. Thus, in the first of the two last examples, if we multiply 6 by 5, and to the product 30 add the remainder 4, we obtain 34, or the dividend. And in the last example, if we multiply the divisor 9 by the quotient 4, and to the product 36 add the remainder 5, we obtain the dividend 41.
55 Lastly, it is necessary to remark here, with regard to the signs + plus and - minus, that if we divide $$+ab$$ by $$+a$$, the quotient will be $$+b$$, which is evident. But if we divide $$+ab$$ by $$-a$$, the quotient will be $$-b$$; because $$-a$$ multiplied with $$-b$$ gives $$+ab$$. If the dividend is $$-ab$$, and is to be divided by the divisor $$+a$$, the quotient will be $$-b$$; because it is $$-b$$ which, multiplied by $$+a$$, makes $$-ab$$. Lastly, if we have to divide the dividend $$-ab$$ by the divisor $$-a$$, the quotient will be $$+b$$; for the dividend $$-ab$$ is the product of $$-a$$ by $$+b$$.
56 With regard, therefore, to the signs + and -, division requires the same rules to be observed that we have seen take place in multiplication; namely
+ by + makes + ; + by - makes - ;
- by + makes - ; -by - makes + ;
or, in few words, like signs give plus, and unlike signs give minus.
57 Thus, when we divide $$18pq$$ by $$-3p$$, the quotient is $$-6q$$. Farther;
$$-30xy$$ divided by $$+6y$$ gives $$-5x$$, and
$$-54abc$$ divided by $$-9b$$ gives $$+6ac$$;
for, in this last example, $$-9b$$ multiplied by $$+6ac$$ makes $$-6 \cdot 9abc$$, or $$-54abc$$. But enough has been said on the division of simple quantities; we shall therefore hasten to the explanation of fractions, after having added some further remarks on the nature of numbers, with respect to their divisors.
#### Editions
1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum. |
# Integral approximation trick
Here’s a simple integration approximation that works remarkably well in some contexts.
Suppose you have an integrand that looks roughly like a normal density, something with a single peak that drops off fairly quickly on either side of the peak.
The majority of integrals that arise in basic applications of probability and statistics fit this description.
You can approximate the value of the integral as the area of a box.
The height of the box is the value of the integrand at its peak.
The width of the box is the FWHM: full width at half maximum, i.
e.
the distance between two points on where the integrand take on half its maximum value.
This post will include several examples to illustrate how well, or how poorly, the technique works.
Gaussian exampleFirst let’s look at exp(-x²), the normal density function apart fror some scaling factors, integrated over the whole real line.
The maximum of the function is 1, occurring at x = 0.
So half the maximum is 1/2, and exp(-x²) = 1/2 at x = ± √ log 2.
The FWHM is thus 2 √ log 2, and since our maximum is 1, the FWHM is the integral approximation.
This gives 1.
665, whereas the exact value is √π = 1.
772.
Polynomial exampleNext, let’s look at integrating 1/(1 + x2n) for positive integers n.
In every case, the maximum is 1.
The integrand takes on half its maximum value at x = ± 1, so the FWHM is always 2.
So we would approximate the integral as 2 in every case, independent of n.
When n = 1, the correct value of the integral is π and so our approximation of 2 is pretty bad.
But that’s not surprising.
We said above that our integrand needs to drop off fairly quickly, but that’s not the case here.
We have a fat tailed distribution, the Cauchy, and so it’s not surprising that our approximation is poor.
However, as n increases, our integral drops off more quickly, and the approximation improves.
When n = 2, the exact value of the integral is 2.
221.
When n = 3, the integral is 2.
094.
When n = 10 the integral is 2.
008.
And in the limit as n goes to infinity, the integral is 2 and our approximation is exact.
Logistic exampleThe integrands in the examples above were all symmetric about the origin.
That may give the misleading impression that our approximation only applies to symmetric integrals.
The examples also all had a maximum of 1, which could give the impression that the integral approximation is just the FWHM, not the FWHM times the maximum.
We’ll correct both of these impressions in the next example.
Our task is to approximate the integralwhich comes up in Bayesian logistic regression with an improper prior.
The integrand is symmetric about 0 when a = b.
But otherwise the integrand is not symmetric, and the peak occurs at log(a/b) rather than at 0 and the value at that peak isWe’ll use the following Python code to help with our calculations.
from scipy import log, exp from scipy.
optimize import brentq def logistic_max(a, b): return a**a * b** b /(a+b)**(a+b) def fwhm(a, b): hm = 0.
5*logistic_max(a, b) f = lambda x: a*x – (a+b)*log(1 + exp(x)) – log(hm) left = brentq(f, -2, 0) right = brentq(f, 0, 2) return right – left def approx(a, b): return logistic_max(a, b)*fwhm(a, b) And now for a couple examples.
When a = 3 and b = 2, we get an approximate value of 0.
0761, compared to an exact value of 0.
0833, a relative error of 8.
6%.
When a = 10 and b = 12, we get an approximate value of 2.
647 × 10-7, compared to an exact value of 2.
835 × 10-7, a relative error of 6.
6%.
Related postsAvoiding underflow in Bayesian integralsSaved by symmetry. |
# SAT II Math I : X-intercept and y-intercept
## Example Questions
← Previous 1
### Example Question #21 : Functions And Graphs
Find the y-intercept of the following line.
Explanation:
To find the y-intercept of any line, we must get the equation into the form
where m is the slope and b is the y-intercept.
To manipulate our equation into this form, we must solve for y. First, we must move the x term to the right side of our equation by subtracting it from both sides.
To isolate y, we now must divide each side by 3.
Now that our equation is in the desired form, our y-intercept is simply
### Example Question #1 : X Intercept And Y Intercept
Solve for the -intercepts of this equation:
and
and
and
and
and
and
Explanation:
For an equation like this, you should use the quadratic formula to solve for the roots. We can easily get our equation into proper form by substituting for :
Recall that the general form of the quadratic formula is:
Based on our equations, the following are your formula values:
Therefore, the quadratic formula will be:
Simplifying, you get:
Using a calculator, you will get:
and
### Example Question #1 : X Intercept And Y Intercept
Solve for the -intercepts of this equation:
and
and
and
and
and
and
Explanation:
For an equation like this, you should use the quadratic formula to solve for the roots. We can easily get our equation into proper form by substituting for :
Recall that the general form of the quadratic formula is:
Based on our equations, the following are your formula values:
Therefore, the quadratic formula will be:
Simplifying, you get:
Using a calculator, you will get:
and
### Example Question #1 : X Intercept And Y Intercept
Solve for the -intercepts of this equation:
and
and
and
and
and
and
Explanation:
For an equation like this, you should use the quadratic formula to solve for the roots. We can easily get our equation into proper form by substituting for . Then, we need to get it into standard form:
Recall that the general form of the quadratic formula is:
Based on our equations, the following are your formula values:
Therefore, the quadratic formula will be:
Simplifying, you get:
Using a calculator, you will get:
and
### Example Question #5 : X Intercept And Y Intercept
What are the -intercepts of the following equation?
and
and
and
and
and
and
Explanation:
There are two ways to solve this. First, you could substitute in for :
Take the square-root of both sides and get:
You also could have done this by noticing that the problem is a circle of radius , shifted upward by .
### Example Question #1 : X Intercept And Y Intercept
Find the -intercepts of the following equation:
and
and
and
and
and
and
Explanation:
For an equation like this, you should use the quadratic formula to solve for the roots. We can easily get our equation into proper form by substituting for . Then, we need to get it into standard form:
Recall that the general form of the quadratic formula is:
Based on our equations, the following are your formula values:
Therefore, the quadratic formula will be:
Simplifying, you get:
Using a calculator, you will get:
and
### Example Question #1 : X Intercept And Y Intercept
What is the -intercept of the following equation?
None of the others
Explanation:
The easiest way to solve for this kind of simple -intercept is to set equal to . You can then solve for the value in order to find the relevant intercept.
Solve for :
Divide both sides by 40:
### Example Question #31 : Functions And Graphs
What is the x-intercept of the above equation?
Explanation:
To find the x-intercept, you must plug in for .
This gives you,
and you must solve for .
First, add to both sides which gives you,
.
Then divide both sides by to get,
.
### Example Question #1 : X Intercept And Y Intercept
Find the -intercepts of the following equation:
and
and
and
and
and
and
Explanation:
There are two ways to solve this. First, you could substitute in for :
Take the square-root of both sides and get:
You also could have done this by noticing that the problem is a circle of radius , shifted downward by .
### Example Question #1 : X Intercept And Y Intercept
What is the x-intercept of the given equation? |
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# Tutor profile: Daisy A.
Daisy A.
Tutor for five years
## Questions
### Subject:Geometry
TutorMe
Question:
A straight line has two angles 1 and 2. The measure of angle 1 = 35 degrees, find the measure of angle 2.
Daisy A.
First lets draw out our straight line. If we made two angles out of our straight line with one being 35, how can we find our other angles measurement? Angles on the same line are called supplementary angles which means that the angle measurements all together should add up to the measurement of a straight line. The measurement of a straight line is always 180 degrees. Therefore, if one of the two angle measurements is 35 we can then subtract that from 180 to find our missing angle. The equation will be 180 - 35 = 145, so the measurement of angle 2 equals 145 degrees.
### Subject:Basic Math
TutorMe
Question:
The student had 6 stickers and then the teacher gave the student 8 more stickers. How many stickers does the student have now?
Daisy A.
The student began with 6 stickers. Then the student received 8 more stickers. If we draw 6 stickers and then draw 8 more, how many do we have all together? After drawing out our picture we can then create an equation 6 + 8 = 14. The student has 14 stickers total.
### Subject:Algebra
TutorMe
Question:
Solve for X. 6x+3=8x-21
Daisy A.
First lets remember to use our D.C.M.A.M which stands for Distribute, Combine like terms, Move the smaller variable, Add/Subtract, and then Multiply/Divide. With that in mind since we have nothing to distribute or combine on one side we then move over the smaller variable to combine like terms together. To move over the variable we must subtract 6x from the left and whatever we do to one side we do to the other. Therefore, 6x minus 6x on the left will cross out and then 6x minus 8x will be 2x. Then, we will need to do the same with our constants which are the numbers without variables. We can add 21 to the right side so that it can cancel out and then add it to the other side. After, we can add 21 and 3 together to get 24. Now we use our division and divide 2 from the left side to get our variable X alone. Remember what we do to one side we have to do to the other. So 24 divided by 2 equals 12, which means X=12.
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Ratios and Proportions 6.RP.A Sixth Grade Common Core Math Worksheets
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6.RP.A.1, 6.RP.A.2, 6.RP.A.3a,b,c,d Sixth Grade Common Core Math Worksheets
ALL STANDARDS
56 worksheets covering every standard for 6th grade Ratios and Proportional Relationships.
Ratios and Proportional Relationships
Understand ratio concepts and use ratio reasoning to solve problems.
CCSS.MATH.CONTENT.6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, "The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak." "For every vote candidate A received, candidate C received nearly three votes."
CCSS.MATH.CONTENT.6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar." "We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger."1
CCSS.MATH.CONTENT.6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
CCSS.MATH.CONTENT.6.RP.A.3.A
Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
CCSS.MATH.CONTENT.6.RP.A.3.B
Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
CCSS.MATH.CONTENT.6.RP.A.3.C
Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
CCSS.MATH.CONTENT.6.RP.A.3.D
Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
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## Algebra 2 (1st Edition)
The standard form of the equation is: $y=ax^2+bx+c$ Given three points: $(-6,29)\\(-4,12)\\(2,-3)$ Substitute: $29=a(-6)^2+b(-6)+c\\12=a(-4)^2+b(-4)+c\\-3=a(2)^2+b(2)+c$ We have the system: $36a-6b+c=29\\16a-4b+c=12\\4a+2b+c=-3$ Subtract the third equation from the second equation: $12a-6b=15\\ \rightarrow 4a-2b=5$ (1) Subtract the third equation from the first equation: $32a-8b=32\\ \rightarrow 4a-b=4$ (2) Subtract equation (1) from equation (2): $b=-1$ Substitute $b$: $4a-(-1)=4\\a=\frac{3}{4}$ Find $c$: $4(\frac{3}{4})+2(-1)+c=-3\\ \rightarrow c=-4$ Hence, $a=\frac{3}{4}\\b=-1\\c=-4$ Substitute back to the initial equation: $y=\frac{3}{4}x^2-x-4$ |
# CBSE Class 9 Maths Extra Questions: Chapter 6 - Lines and Angles (with Answers)
Check important extra questions for CBSE Class 9 Maths Chapter 6 Lines and Angles. All the questions are provided with answers.
CBSE Class 9 Maths Extra Questions Answers Chapter 6
CBSE Class 9 Maths extra questions and answers for Chapter 6 - Lines and Angles provided here for students to revise important fundamental concepts. Practicing with these questions will help you strengthen your basics and prepare well for the exams. All these questions are provided with answers. Try to understand and solve each question appropriately. These questions are like a practice test that is best for self-assessment.
CBSE Class 9 Maths Extra Questions for Chapter 6 - Lines and Angles:
1. If two lines intersect each other then the vertically opposite angles will be _______.
If two lines intersect each other then the vertically opposite angles will be equal.
2. If the angles of a triangle are in the ratio 5 : 3: 2 then the triangle is:
(a) An acute-angled triangle
(b) An obtuse-angled triangle
(c) An isosceles triangle
(d) A right triangle
(d) A right triangle
(Hint; Let the angles be 5x, 3x and 2x so that 5x + 3x + 2x = 180. Then find out the value of all three angles)
3.Is it possible to have a triangle with two obtuse angles? Give reason to support your answer.
No, it is not possible to have a triangle with two obtuse angles. Because if the triangle will have two obtuse angles (angles greater than 90°), then the sum of all three angles of the triangle will not be equal to 180°, it will be more than 180°.
4. How many triangles can be drawn with its three angles being 47°, 70° and 85°? Give reason for your answer.
None
Sum of given angles = 47° + 70° + 85° = 202° ≠ 180°.
Since the sum of all three angles is not equal to 180°. So, no triangle can be drawn with the given angles.
## Trending Now
5. A transversal intersects two parallel lines as shown in the figure given below. Then which of the following statements is correct?
(a) Each pair of alternate interior angles is equal.
(b) Each pair of corresponding angles is equal.
(c) Each pair of interior angles on the same side of the transversal is supplementary.
(d) All the above statements are equal.
(d) All the above statements are equal.
Also Check:
NCERT Book for Class 9 Maths
NCERT Solutions for Class 9 Maths
6. What is Linear Pair Axiom?
Linear Pair Axiom is a combination of the two theorems. These are:
(i) If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.
(ii) If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.
7. What do you mean by allied angles?
Interior angles on the same side of the transversal are referred to as consecutive interior angles or allied angles or co-interior angles.
8. In the following figure, find the value of x.
x = 18o
9. In the following figure, OP and OQ are the bisectors of ∠BOC and ∠AOC. Then find ∠POQ.
∠POQ = 90o
10. An exterior angle of a triangle is _________ than either of its interior opposite angles.
An exterior angle of a triangle is greater than either of its interior opposite angles.
Students must go through the latest CBSE Syllabus for Class 9 Maths so that they can prepare according to the contents prescribed by the board.
Also check:
CBSE Class 9 Maths Important Questions and Answers
CBSE Class 9 Maths Important MCQ (Chapter-wise)
## Related Categories
खेलें हर किस्म के रोमांच से भरपूर गेम्स सिर्फ़ जागरण प्ले पर |
# How do you simplify 3sqrt8?
Mar 22, 2018
$6 \sqrt{2}$
#### Explanation:
The square root of $8$ is like the square root of $4$ times $2$.
$3 \sqrt{8} = 3 \sqrt{4 \times 2}$
The square root of $4$ is $2$ so you multiply $3$ by $2$ and leave the other $2$ in the square root.
$3 \sqrt{4 \times 2} = 3 \times 2 \sqrt{2}$
You are then left with $6$ times the square root of $2$.
$6 \sqrt{2}$ |
## Point of inflexion and concavity
The concept of increasing function, point of inflexion and concavity can be understood with an illustration.
a. Strictly increasing. f(x) is strictly increasing when f'(x) > 0. ( x<1.423 or x> 2.577) . These points also coincide with the maximum and minimum point.
b. Strictly decreasing. f(x) is strictly decreasing when f'(x) < 0. (1.423 < x < 2.577)
c. Point of inflexion is the point where the curve changes curvature or concavity. f'(x) does not change sign, and f” (x) = 0 and changes sign. In the above graph, the curve changes from convex to concave at x = 2.
d. Concave upwards. f(x) concave upwards when f”(x) > 0 or f'(x) is strictly increasing. f(x) concave upwards at x> 2, after f'(x) reaches a minimum, at the point of inflexion.
e. Concave downwards or convex. f(x) concave downwards when f”(x) < 0 or f'(x) is strictly decreasing. f(x) concave downwards for x < 2.
## Transformation of graphs: Order of Transformation
In the transformation of graphs, knowing the order of transformation is important. Knowing whether to scale or translate first is crucial to getting the correct transformation.
Let’s look at this example to illustrate the difference:
Example 1
Original point on y=f(x) is x=8
For f(2x+4), we do translation first, then scaling. ie. Move left by 4 units, then scale parallel to the X axis by a factor of 1/2. Hence, the original point becomes x= (8-4)/2 = 2
If we want to do scaling first, we need to factorise into f 2(x+2). So scale parallel to the X axis by a factor of 1/2, then move left by 2 units. Hence, the original point becomes x= (8/2)-2 = 2
Example 2
Describe the transformation of 3f(2x-4) + 5.
Translate 4 units in the positive X direction
Scale by a factor of 1/2 parallel to the X axis
Scale by a factor of 3 parallel to the Y axis
Translate 5 units in the positive Y direction
In summary,
cf(bx+a)+d = Translate by a units in the negative X direction, then scale by a factor of 1/b parallel to the X-axis, then scale by a factor of c parallel to the Y-axis, then translate by d units in the positive Y direction.
c[f a(x+b)]+d = Scale by a factor of 1/a parallel to the X-axis, then translate by b units in the negative X direction, then scale by a factor of c parallel to the Y axis, then translate by d units in the positive Y direction.
## Sketching graphs of polynomial functions
Sketching graphs of polynomial functions are useful in graphing techniques, and solving inequalities.
Steps
1. First mark down the roots of the polynomial function.
2. Decide how the tail ends behave, whether it is above or below the X axis. This can be determined from the coef of the highest power. For instance, if the polynomial is of degree 6 and the coef of the highest power is positive, then when X approaches either positive or negative infinity, the function approaches positive infinity, so the tail ends are above the X axis.
3. Determine how the function behaves at the roots. When there are 3 or more odd number of roots at the same point, there is a point of inflexion at the root. When there are 2 or more even number of roots at the same point, there is a minimum or maximum point at the root.
Example 1
Example 2
## Using GC to determine the nature of stationary points
Suppose we are required to determine the nature of stationary points for the following:
Using TI-84 Plus (OS 2.55)
Observation of stationary points:
Left stationary point: First derivative changes from +ve to -ve. Therefore, it is a maximum point
Middle stationary point: No change in sign of deriative. Therefore, it is a point of inflexion.
Right stationary point. First derivative changes from -ve to +ve. Therefore, it is a minimum point |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 9.18: Unknown Measures in Similar Figures
Difficulty Level: At Grade Created by: CK-12
Jurnee wants to make a new sail for her father's sailboat. She finds the old sketch that she used to make the first sail and she redraws the sketch so that she can make the new sail. She uses inches as her scale for the figures, but she plans to use feet when she makes the actual sail. Her two figures are similar.
How can Jurnee use a proportion to decide the length of KJ¯¯¯¯¯\begin{align*}\overline{KJ}\end{align*} in inches?
In this concept, you will learn how to use proportions to figure out the length of a missing side.
### Guidance
You can write ratios to compare the lengths of sides.
First, identify the corresponding sides of these two similar triangles, then place the first side in the numerator and the corresponding side in the denominator.
LMOP=LNOQ=MNPQ
These ratios are written in a proportion or a set of three equal ratios. Remember that there is a relationship between the corresponding sides because they are parts of similar triangles. The side lengths of the similar triangles form a proportion.
Let’s substitute the given measurements into the formula.
63=84=42
There is a pattern with the ratios of corresponding sides. You can see that the measurement of each side of the first triangle divided by two is the measure of the corresponding side of the second triangle.
Use patterns like this to problem solve the length of missing sides of similar triangles.
These are two similar triangles because they have the same shape but a different size. Therefore, the corresponding sides are similar.
If you look at the side lengths, you should see that there is one variable. That is the missing side length. You can figure out the missing side length by using proportions because the corresponding side lengths form a proportion. Let’s write ratios that form a proportion and find the pattern to figure out the length of the missing side.
ABDE510=ACDF=BCEF=15x=1020
Looking at this you can see the pattern. The side lengths of the second triangle are double the length of the corresponding side of the first triangle.
Using this pattern, you can see that the length of DF\begin{align*}DF\end{align*} in the second triangle will be twice the length of AC\begin{align*}AC\end{align*}. The length of AC\begin{align*}AC\end{align*} is 15.
15 ×\begin{align*}\times\end{align*} 2 =\begin{align*}=\end{align*} 30
The length of DF\begin{align*}DF\end{align*} is 30.
### Guided Practice
Solve for the missing value.
810=45=2x\begin{align*}\frac{8}{10} = \frac{4}{5} = \frac{2}{x}\end{align*}
First, identify the pattern.
The denominator is the numerator divided by 0.8.
Next, set up an equation to solve for x\begin{align*}x\end{align*}.
20.8=x\begin{align*}\frac{2}{0.8}=x\end{align*}
Then, solve for x\begin{align*}x\end{align*}.
x=2.5\begin{align*}x=2.5\end{align*}
The answer is that x=2.5\begin{align*}x = 2.5\end{align*}.
### Examples
#### Example 1
Solve for the missing value.
612=x24=36\begin{align*}\frac{6}{12} = \frac{x}{24} = \frac{3}{6}\end{align*}
First, identify the pattern.
The denominator is twice the size of the numerator.
Next, set up an equation to solve for x\begin{align*}x\end{align*}.
24=2×x\begin{align*}24 = 2 \times x \end{align*}
Then, solve for x\begin{align*}x\end{align*}.
242=x=12\begin{align*}\frac{24}{2}=x=12\end{align*}
The answer is that x=12\begin{align*}x = 12\end{align*}.
#### Example 2
Solve for the missing value.
12x=164=205\begin{align*}\frac{12}{x} = \frac{16}{4} = \frac{20}{5}\end{align*}
First, identify the pattern.
The denominator is the result of dividing the numerator by 4.
Next, set up an equation to solve for x\begin{align*}x\end{align*}.
124=x\begin{align*}\frac{12}{4}=x\end{align*}
Then, solve for x\begin{align*}x\end{align*}.
x=3\begin{align*}x=3\end{align*}
The answer is that x=3\begin{align*}x = 3\end{align*}.
#### Example 3
Solve for the missing value.
82=164=x1\begin{align*}\frac{8}{2} = \frac{16}{4} = \frac{x}{1}\end{align*}
First, identify the pattern.
The denominator is the result of dividing the numerator by 4.
Next, set up an equation to solve for x\begin{align*}x\end{align*}.
x4=1\begin{align*}\frac{x}{4}=1\end{align*}
Then, solve for x\begin{align*}x\end{align*}.
x=4×1=4\begin{align*}x=4\times 1=4\end{align*}
The answer is that x=4\begin{align*}x = 4\end{align*}.
Remember Jurnee and the sail? She wants to make a new sail for her father. She uses an old sketch and makes a new sketch that is similar.
How can Jurnee decide the length of KJ¯¯¯¯¯\begin{align*}\overline{KJ}\end{align*} in inches?
First, use the corresponding sides to set up a proportion.
KJ¯¯¯¯¯5=64\begin{align*}\frac{\overline{KJ}}{5} = \frac{6}{4}\end{align*}
Next, use cross products.
KJ¯¯¯¯¯×45×64KJ¯¯¯¯¯=4KJ¯¯¯¯¯=30=30\begin{align*}\overline{KJ} \times 4 &= 4\overline{KJ}\\ 5 \times 6 &= 30\\ 4\overline{KJ} &= 30\end{align*}
Then, solve for KJ¯¯¯¯¯\begin{align*}\overline{KJ}\end{align*}.
30÷4KJ¯¯¯¯¯=7.5=7.5\begin{align*}30 \div 4 &= 7.5\\ \overline{KJ}&= 7.5\end{align*}
The answer is that KJ¯¯¯¯¯=7.5\begin{align*}\overline{KJ}=7.5\end{align*} cm.
### Explore More
Use the figures to answer the following questions.
1. Are these two triangles similar or congruent?
2. How do you know?
3. Which side is congruent to AB\begin{align*}AB\end{align*}?
4. Which side is congruent to AC\begin{align*}AC\end{align*}?
5. Which side is congruent to RS\begin{align*}RS\end{align*}?
6. Look at the following proportion and solve for missing side length x\begin{align*}x\end{align*}.
73.5x=x3.5=6y=
7. What is the side length for y\begin{align*}y\end{align*}?
8. How did you figure these out?
Figure out the missing value in each pair of ratios.
9. 612=x24\begin{align*}\frac{6}{12} = \frac{x}{24}\end{align*}
10. 812=x3\begin{align*}\frac{8}{12} = \frac{x}{3}\end{align*}
11. 910=18y\begin{align*}\frac{9}{10} = \frac{18}{y}\end{align*}
12. 45=x2.5\begin{align*}\frac{4}{5} = \frac{x}{2.5}\end{align*}
13. \begin{align*}\frac{16}{20} = \frac{4}{y}\end{align*}
14. \begin{align*}\frac{19}{21} = \frac{x}{42}\end{align*}
15. \begin{align*}\frac{9}{54} = \frac{6}{y}\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 9.18.
1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
## Date Created:
Sep 24, 2015
Sep 24, 2015
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# Inequalities
Inequalities are the relationships between two expressions which are not equal to one another. The symbols used for inequalities are <, >, ≤, ≥.
$$7 \textgreater x$$ reads as '7 is greater than $$x$$' (or '$$x$$ is less than 7', reading from right to left).
$$x \leq -4$$ reads as '$$x$$ is less than or equal to -4' (or '-4 is greater than or equal to $$x$$', reading from right to left).
## Inequalities on a number line
Inequalities can be shown on a number line.
Open circles are used for numbers that are less than or greater than (< or >). Closed circles are used for numbers that are less than or equal to and greater than or equal to (≤ or ≥).
For example, this is the number line for the inequality $$x \geq 0$$:
The symbol used is greater than or equal to (≥) so a closed circle must be used at 0. $$x$$ is greater than or equal to 0, so the arrow from the circle must show the numbers that are larger than 0.
### Example
Show the inequality $$y \textless 2$$ on a number line.
$$y$$ is less than (<) 2, which means an open circle at 2 must be used. $$y$$ is less than 2, so an arrow below the values of 2 must be drawn in.
Question
What inequality is shown by this number line?
There is a closed circle at -5 with the line showing the numbers that are greater than -5.
This means $$-5 \leq x$$ (writing the $$x$$ on the right-hand side).
There is also an open circle at 4, with the numbers less than 4 indicated. This means $$x \textless 4$$ (writing the $$x$$ on the left-hand side).
The line between these two points means that $$x$$ satisfies both inequalities, so a double inequality must be created.
Putting $$x$$ in the middle of the two inequalities gives $$-5 \leq x \textless 4$$.
$$x$$ is greater than or equal to -5 and $$x$$ is less than 4. |
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# Free Algebra Word Problems
Free Algebra Word Problems In mathematics, algebra can be considered as a part that specifies the relationship between operation and operands with in the form of expressions. Basically algebra includes the combination of various terms, polynomials and algebraic equations. In mathematics, concept of algebra is popularly used for understanding the number theory, analyzing the problem and in various geometrical tools. It means to say that algebra can be considered as the branch of the pure mathematics. In the simple mean we can say that algebra is used for calculating the unknown value. In the basic algebra of mathematics we include the concept of variables and numbers with the operations. Algebra is a group of mathematical statement that requires performing the calculation to find the unknown value of the given variable.
## Know More About :- Derivative Cot
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Here we are going to discussing about the algebra word problems which is most popularly used for solving the real life problems. Here an algebra word problem refers to describe the mathematical problem in the form of word statement. In the simple mean we say that at the time of solving mathematical word problem we need to analyze what we need to find out. In the mathematical expression we generally deal with the following example: a + 7 = 13, In the above example we need to find out the value of variable a that satisfy the given expression true. In the given expression we can see that we use a mathematical operation plus in the expression. In the same aspect we also use the mathematical operation in the algebra word problem. Example: Ron and Harry have 5 chocolate and 4 candies. Then find out how many of toffees they have? The above word expression is the most basic algebra word problem that can be solve by following steps very easily. <<-- Evaluate the problem first very carefully. <<-- After that analyze the main points that are given in the question. <<-- Labeled the each key variable by some specific variable and remove the unneeded variables.
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<<-- Now perform the writing of word problem in the form of mathematical expression. <<-- After that by using mathematical operation solve the equations. <<-- When we get the value of unknown variable, after that calculate remaining other variable. Now we show you how to follow these steps at the time of solving algebra word problem by the following given example: Example: Suppose there are two friends in the class. First friend has dollar twenty more than his friend. If we take the sum of all their amounts together then there total of their amount is equal to\$100. Find the amount both friends have? Solution: Now we follow the above given steps to solve the given algebraic problem: Given that first friend has \$ 20 more form first friend then this can be described as second friend has x amount then first friend has x + 20 then we write the equation in mathematical form: Amount of first friend + amount of second friend = total amount, x + 20 + x = 100, Now we solve mathematical expression: 2x + 20 = 100, 2x = 100 20, x = 80 / 2, x = \$ 40, Now we can say that second friend has \$ 40 then first friend has x + \$20 = \$ 40 + \$ 20 = \$ 60.
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# Basics of Simplification for Bihar State Exams
By Nitin Singhal|Updated : February 3rd, 2021
Dear Aspirants,
We all know the importance of fast and speedy calculation in bank and insurance exams. Simplification is one of the topics in the quant section that requires a speedy calculation to solve maximum questions on it in less time. It is one of the most widely asked topics in competitive exams, especially in prelims stage. So today, we are discussing some key points to deal with Simplification topics to score well in this section.
Simplification generally means to find an answer for the complex calculation that may involve numbers on division, multiplication, square roots, cube roots, plus and minus.
Simplification questions are asked in the exam to check the ability of an aspirant to deal with numbers which can be in one of the following two types.
• Sometimes, a calculation is given and one of the numbers is missing from the calculation. To find out the missing number, we have to approximate the given numbers or do the basic operations.
• Sometimes all the numbers are given with some operations between them & we have to simplify the calculation.
## Rules related to Simplification
#### Rule-(I) Replace ‘of’ by ‘Multiplication’ & ‘/’ by ‘Division’.
Explanation: Whenever we find ‘of’ in a simplification problem, we can replace that by ‘multiplication(*)’. Similarly ‘/’ can be replaced by ‘÷’.
Example: Find ¼ of 20
Solution: (¼) x 20 = 20÷4 = 5
Rule-(II) Always keep in mind the “BODMAS” rule. These operations have priorities in the same order as mentioned.
Explanation: Whenever we have more than one operation in the given calculation, we have to do the operations according to the priority specified by ‘BODMAS’
• B-Bracket
• O-Of (means multiplication)
• D-Division
• M-Multiplication
• S-Subtraction
Example: Simplify: (2+3)*30
Solution: In this question, we have two things-Bracket & Multiplication. According to the BODMAS rule, we have to solve bracket first and not multiplication. So now coming to bracket, we have only one operation-Addition, so we will do addition.
(2+3)*30 = 5*30
Now we have only one operation to do – Multiplication
5*30 = 150
Example: Simplify: (2+5) of 80
Solution: In this question, we have three things – bracket, addition & of. Replacing ‘of’ by ‘multiplication’.
(2+5) of 30 = (2+5)*80
Now we have three things – bracket, addition & Multiplication. According to the BODMAS rule, we have to solve bracket first and not multiplication. So now coming to bracket, we have only one operation-Addition, we will do addition.
(2+5)*80 = 7*80
Now we will do multiplication.
7*80 = 560
Rule-(III) Multiplication & Division have the same priority(Do that operation first which is on left)
Explanation: Though division has more priority than multiplication according to ‘BODMAS’ but we can perform any of the two operations first if multiplication is on left.
Example: 8*30/15
8*30 ÷ 15
Solution: In this question, we have two things – Multiplication & Division. Multiplication is on left So we can perform that first.
Doing Multiplication first:
240 ÷ 15
16
Doing division first:
8*2
16
Rule-(IV) Addition & Subtraction have the same priority.
Explanation: Though addition has more priority than division according to ‘BODMAS’ but we can perform any of the two operations first.
Example: 30+40-15
Solution: In this question, we have two things – Addition & Subtraction. So we can perform any operation first as they have same priority.
70 – 15
55
Doing Subtraction first:
30 + 25
55
Rule-(V) Don’t hesitate in rounding the numbers to nearest integers.
Explanation: Most of the times the numbers are given in such a way that you can round them quickly and get the answer (Rounding should be done or not, It can be realised by looking at the given options).
Example: (324.5*15)/(5.01*24.98)
Solution: (325*15)/(5*25)
=13*3
=39
Now let us see some of the previous year questions asked from 'Simplification' & try to apply the rules learnt so far.
Q. 1) (17 -13)4 - 174 – 134 – [- 52(17)3 – 68(133)] = (?)*221
Using formula: (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3+ b4
⇒ (a – b)4 - a4 + 4a3b + 4ab3 - b4 = + 6a2b2
(17 -13)4 - 174 – 134 – [- 52(17)3 – 68(133)] = (?)*221
Here, a = 17 and b = 13
⇒ (?)= (6 (17)2 (13)2)/221
⇒ (?) = (6 × 289 × 169)/221
⇒ (?) = 1326
Q.2) Simplify: 127.001 * 7.998 + 6.05 * 4.001
1. 1000
2. 1020
3. 1040
4. 1080
5. None of these
Solution: Using the rounding concept
127 * 8 + 6 * 4
Using the BODMAS rule
1016 + 24
1040 (Option 3)
Q.3) What will come at place of ?: 9876 ÷ 24.96 + 215.005 - ? = 309.99
1. 270
2. 280
3. 290
4. 300
5. 310
Solution: Using the rounding concept
9875 ÷ 25 + 215 - ? = 310
Using the BODMAS rule
395 + 215 - ? = 310
610 - ? = 310
? = 300 (Option 4)
Q.4) What will come at place of a: (128 ÷ 16 x a – 7*2)/(72-8*6+a2) = 1
1. 1
2. 5
3. 9
4. 13
5. 17
Solution: Using the BODMAS rule
(8*a – 14)/(49-48+a2) = 1
(8*a – 14)/(1 + a2) = 1
8a – 14 = 1 + a2
a2 – 8a + 15 = 0
a=3 or 5 (Option 2)
Q.5) What will come at place of ?: 85.147 + 34.192*6.2 + ? = 802.293
1. 400
2. 450
3. 550
4. 600
5. 500
Solution: Using the rounding concept
85 + 35*6 + ? = 803
Using the BODMAS rule
85 + 210 + ? = 803
295 + ? = 803
? = 508 [approx. = 500] (Option 5)
Q.6) What will come at place of ? : (3/8 of 168)*15 ÷ 5 + ? = 549 ÷ 9 + 235
1. 189
2. 107
3. 174
4. 296
5. None of these
Solution: Using the BODMAS rule
(3*168÷8)*15 ÷ 5 + ? = 549 ÷ 9 + 235
(504÷8)*3 + ? = 61 + 235
63*3 + ? = 296
189 + ? = 296
? = 107 (Option 2)
## Key points to remember while solving Simplification Question:
• Replace ‘of’ by ‘Multiplication’
• Replace ‘/’ by ‘Division’
• Always do the operations in priority according to ‘BODMAS’
• Division & Multiplication have the same priority (Start from left)
• Addition & Subtraction have the same priority
• Rounding can be done to simplify problems
• When the given options are very close then rounding doesn’t help much
• Always look at the options before doing simplification that can help in the elimination of options.
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# Question: How Many Odd Days Are There In Hundred Years?
## How many odd days are in 200 years?
3 odd daysCounting of odd days: No.
of odd days in 200 years = (5*2) =3 odd days.
No.
of odd days in 300 years = (5*3) =1 odd day..
## How many odd days are there in 100 years just after end of 2000?
Concept of Odd Days This additional day is called an odd day. The concept of odd days is very important in calendars. In a century – i.e. 100 years, there will be 24 leap years and 76 non-leap years. This means that there will be 24 x 2 + 76 x 1 = 124 odd days.
## Why does 100 years have 5 odd days?
Odd days in a leap year = (52 weeks +2) days. So odd days in 100 years will be (76 x 1 + 24 x 2) which is 124 odd days. This can also be written as 17 weeks + 5 days. So every 100 years will have 5 odd days.
## Why there are 24 leap years in 100 years?
Explanation: Given year is divided by 4, and the quotient gives the number of leap years. Here, 100%4 = 25. But, as 100 is not a leap year => 25 – 1= 24 leap years.
## Is 2000 leap year or not?
The year 2000 was a leap year, for example, but the years 1700, 1800, and 1900 were not. The next time a leap year will be skipped is the year 2100.
## How many odd days are in 75 years?
1776) Counting of odd days: No of odd days in 1600 years = 0 No of odd days in 100 years = 5 75 years = 18 leap years + 57 ordinary years = 18*2 + 57*1 = 36 + 57 = 93 odd days = 13 weeks + 2 odd days = 2 odd days ∴ 1775 years have (0+5+2) = 7 odd days = 0 odd days.
## What is the maximum gap between two leap years?
The next leap year comes in 1904 (1900 is not a leap year). Therefore, The maximum gap between two successive leap year is 8 years.
## How many odd days does 100 years have?
5 odd daysOdd days in a century: A century has 76 ordinary years and 24 leap years. ∴A century or 100 years have 5 odd days.
## How many odd days are there in 1900 days?
Each fourth century is a leap year. Example: 400, 800, 1200, 1600, 2000, 2400 are leap years, but 700, 1300, 1900 are not leap years. Odd Days: For a given number of days, number of days more than complete week are called odd days. Example: in 10 days, there is one week and 3 odd days.
## How many leap years are there in 500 years?
24 leap yearsGiven year is divided by 4, and the quotient gives the number of leap years. But, as 100 is not a leap year ⇒ 25 – 1 = 24 leap years.
## What are odd days of the month?
Hence, the number of odd days in February of an ordinary year will have 0 odd days and that of leap year will have 1 odd day as February in a leap year has 29 days….Concept of an Odd Day.MonthNumber of odd daysNovember2December310 more rows
## Is 100 a leap year?
So every 100 years we don’t have a leap year, and that gets us 365.24 days per year (1 day less in 100 year = -0.01 days per year). Closer, but still not accurate enough! So another rule says that every 400 years is a leap year again.
## How many leap years do 400 years have?
So, 29th February will come 97 times in 400 years.
## How do you calculate odd days?
Counting of Odd Days:1 ordinary year = 365 days = (52 weeks + 1 day.) 1 ordinary year has 1 odd day.1 leap year = 366 days = (52 weeks + 2 days) 1 leap year has 2 odd days.100 years = 76 ordinary years + 24 leap years. = (76 x 1 + 24 x 2) odd days = 124 odd days. = (17 weeks + days) 5 odd days.
## How many odd days is 500 years?
5 Odd DaysReason for 500 Years has 5 Odd Days.
## How many odd days are there in 1600?
1600 years = 24 x 16 = 384 leap years (100 years = 24 leap years) (because 100 years have 24 leap years) 1 leap year = 2 odd days (52 weeks + 2 odd days) 384 leap years = 384 x 2 = 768 odd days –(A) 1600 years = 1600 – 384 = 1216 ordinary years 1 ordinary year = 1 odd day (52 weeks + 1 odd day) 1216 ordinary years = …
## What Cannot be the last day of a century?
Since the order is continually kept in successive cycles, we see that the last day of a century cannot be Tuesday, Thursday or Saturday. |
Sections:
# Order of Operations
In this section, we finish up our review of whole numbers by discussing the order of operations. When we tackle a problem with multiple operations it can often cause confusion as to which operation is to be performed in which order. Let’s suppose we saw the following problem:
4 + 2 x 3
How would we know whether to add 4 and 2 or multiply 2 times 3 to get started? The order of operations solves this type of dilemma and enables all parties to get a consistent answer. We generally use the acronym PEMDAS to remember the order of operations.
PEMDAS:
1. P: Stands for Parentheses - this is where we begin, work inside of any parentheses or grouping symbols
2. E: Stands for Exponents - we next evaluate all exponent operations
3. M, D: Stands for Multiply or Divide - working from left to right
4. A, S: Stands for Addition or Subtraction - working from left to right
Example 1: 5 x 4 + 22
P: Look for parentheses or grouping symbols - none are present
E: Look for exponents, we see that we have 22 which evaluates to 4
5 x 4 + 4
M, D: Look for multiplication and/or division, we see that we have 5 x 4 which evaluates to 20
20 + 4
A, S: Look for addition and/or subtraction, we see that we have 20 + 4 which evaluates to 24, this is our answer
5 x 4 + 22 = 24 |
# Difference between revisions of "2021 AIME II Problems/Problem 15"
## Problem
Let $f(n)$ and $g(n)$ be functions satisfying $$f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}$$and $$g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}$$for positive integers $n$. Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$.
## Solution 1
Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2 for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$. Note that this formula also returns the correct value when $n=(k+1)^2$, but not when $n=k^2$. Thus $f(n)=k^2+3k+2-n$ for $k^2.
If $2 \mid (k+1)^2-n$, $g(n)$ returns the same value as $f(n)$. This is because the recursion once again stops at $(k+1)^2$. We seek a case in which $f(n), so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$.
Write $7f(n)=4g(n)$, which simplifies to $3k^2+k-10=3n$. Notice that we want the $LHS$ expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$. We also want n to be strictly greater than $k^2$, so $k-10>0, k>10$. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is $k=16$, giving $n=258$.
Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$.
-Ross Gao
## Solution 2 (More Variables)
We restrict $n$ in which $k^2 for some positive integer $k,$ or $$n=(k+1)^2-p\hspace{40mm}(1)$$ for some nonnegative integer $p.$ By observations, we get \begin{align*} f\left((k+1)^2\right)&=k+1, \\ f\left((k+1)^2-1\right)&=k+2, \\ f\left((k+1)^2-2\right)&=k+3, \\ &\cdots \\ f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{19mm}(2) \\ \end{align*} If $n$ and $(k+1)^2$ have the same parity, then starting with $g\left((k+1)^2\right)=k+1,$ we get $g(n)=f(n)$ by a similar process. This contradicts the precondition $\frac{f(n)}{g(n)} = \frac{4}{7}.$ Therefore, $n$ and $(k+1)^2$ must have different parities, from which $n$ and $(k+2)^2$ must have the same parity.
Along with the earlier restriction, we obtain $k^2 or $$n=(k+2)^2-2q\hspace{38.25mm}(3)$$ for some positive integer $q.$ By observations, we get \begin{align*} g\left((k+2)^2\right)&=k+2, \\ g\left((k+2)^2-2\right)&=k+4, \\ g\left((k+2)^2-4\right)&=k+6, \\ &\cdots \\ g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\ \end{align*} By $(2)$ and $(4),$ we have $$\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)$$ From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces $$k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)$$ We substitute $(6)$ into $(5),$ then simplify, cross-multiply, and rearrange: \begin{align*} \frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\ \frac{p+2q-1}{-p+6q+1}&=\frac{4}{7} \\ 7p+14q-7&=-4p+24q+4 \\ 11p-11&=10q \\ 11(p-1)&=10q. \hspace{29mm}(7) \end{align*} Since $\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$ Furthermore, rewriting the restriction $k^2 in terms of $p$ and $q$ gives \begin{align*} k^2&<\underbrace{(k+2)^2-2q}_{\text{by }(3)} \\ q&<2k+2 &\hspace{5.5mm}(8) \\ q&<2\biggl(\phantom{ }\underbrace{\frac{2q-p-3}{2}}_{\text{by }(6)}\phantom{ }\biggr)+2 \\ p+1& Combining $(8)$ and $(9),$ we get the compound inequality $$p+1 Note that if we start with $k^2<\underbrace{(k+1)^2-p}_{\text{by }(1)}$ instead, then we get ... and ..., both of which agree with $(10).$ In order to minimize $n,$ we should minimize $k.$ By $(6),(8),$ and $(9),$ we should minimize $p$ and $q.$ From $(7),$ we construct the following table: $\[\begin{array}{c|c|c} & & \\ [-2.5ex] \boldsymbol{p} & \boldsymbol{q} & \textbf{Satisfies }\boldsymbol{(9)?} \\ [0.5ex] \hline & & \\ [-2ex] 11 & 11 & \\ 21 & 22 & \\ 31 & 33 & \checkmark \\ \geq41 & \geq44 & \checkmark \\ \end{array}$$ We have $(p,q)=(31,33).$ Substituting this result into $(6)$ produces $k=16.$ Finally, substituting everything into either $(1)$ or $(3)$ generates $n=\boxed{258}.$
Remark
We can verify that $$\frac{f(258)}{g(258)}=\frac{31+\overbrace{f(289)}^{17}}{2\cdot33+\underbrace{f(324)}_{18}}=\frac{48}{84}=\frac47.$$
~MRENTHUSIASM
## See also
2021 AIME II (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byLast Question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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The Least Common Denominator The Least Common Multiple A multiple of a number is a whole number times that number. For example, some multiples of 6 are 6, 12, 18, 24, 30, and 36 If two numbers are given then a common multiple of the two numbers is a number that is a multiple of both. Of all the common multiples of two numbers, there is a smallest one which we call the least common multiple. Example Find the least common multiple of 6 and 9. Solution One way of solving this problem is to write out multiples of each and see what is common to the list: 6, 12, 18, 24, 30, 36, ... multiples of 6 9, 18, 27, 36, 45, ... multiples of 9 We see that the numbers 18 and 36 are both common multiples of 6 and 9. The least common multiple is the smallest which is 18. Example Find the least common multiple of 8 and 32. Solution Instead of listing many multiples of each, we just notice that 32 is a multiple of 8 and hence 32 is a common multiple. It is the first multiple of 32. We can conclude that 32 is the least common multiple of 8 and 32. In general, the least common multiple of two numbers with one the multiple of the other is just the larger number. Exercise Find the least common multiple of 15 and 54. Hold mouse over the yellow rectangle for the solution Find the least common multiple of 9 and 81. Hold mouse over the yellow rectangle for the solution As you saw from the Exercise A, writing out many multiples of each number can be tedious. There is an alternate method that may save time. The strategy is based on the following idea. A multiple of a number is a multiple of each of the prime divisors. Steps in Finding the LCM Write the prime factorization of each number List the primes that occur in at least one of the factorizations Form a product using each prime the greatest number of time it occurs in any one of the expressions Example Find the LCM of 45 and 21 Solution 45 = 9 x 5 = 3 x 3 x 5 21 = 3 x 7 3, 5, and 7 3 x 3 x 5 x 7 The prime 3 occurs two times as it does in 3 x 3 x 5 = 9 x 5 x 7 = 45 x 7 = 315 Exercises Find the LCM of 18 and 40 Hold mouse over the yellow rectangle for the solution 12 and 15 Hold mouse over the yellow rectangle for the solution 27 and 10 Hold mouse over the yellow rectangle for the solution The Least Common Denominator We define the least common denominator of two fractions as the least common multiples of the denominators. Examples Find the least common denominator of 3/4 and 9/10 5/6 and 10/11 Solutions We find the least common multiples of 4 and 10 4 = 2 x 2 10 = 2 x 5 So the least common denominator is 2 x 2 x 5 = 20 We find the least common multiples of 6 and 11 6 = 2 x 3 11 is prime So the least common denominator is 2 x 3 x 11 = 66 Exercises Find the least common denominator of 3/14 and 2/63 Hold mouse over the yellow rectangle for the solution 8/25 and 23/100 Hold mouse over the yellow rectangle for the solution Building Up Fractions With a Least Common Denominator We have already learned how to simplify a fraction by dividing through by a common factor. Sometimes it is convenient to be able to work this process in reverse. Example Build up the fraction to answer the question 5 ? = 6 24 Solution We see that 24 = 6 x 4 so 5 5 x 4 20 = = 6 6 x 4 24 Exercise Build up the fraction to answer the question 3 ? = 7 35 Hold mouse over the yellow rectangle for the solution Example Which number is larger: 5/8 or 9/14? Solution Since the denominators are different, these numbers are difficult to compare. Our strategy is to build up each fraction to fractions with the least common denominator. We first find the least common denominator: 8 = 2 x 2 x 2 14 = 2 x 7 The least common denominator is 2 x 2 x 2 x 7 = 56 The next step is to notice that 8 x 7 = 56 and 14 x 4 = 56 We write 5 5 x 7 35 = = 8 8 x 7 56 and 9 9 x 4 36 = = 14 14 x 4 56 Since 35 36 < 56 56 We conclude that 5 9 < 8 14 Exercise Which is larger: 3/10 or 7/25? Hold mouse over the yellow rectangle for the solution Example Write the three fractions 1/6, 5/8 and 3/10 as equivalent fractions with the LCD as the denominators. Solution We have 6 = 2 x 3 8 = 2 x 2 x 2 10 = 2 x 5 So the least common denominator is 2 x 2 x 2 x 3 x 5 = 8 x 3 x 5 = 24 x 5 = 120 We write 1 1 x 20 20 = = 6 6 x 20 120 5 5 x 15 75 = = 8 8 x 15 120 3 3 x 12 36 = = 10 10 x 12 120 Exercise Write the three fractions 2/15, 4/9 and 3/25 as equivalent fractions with the LCD as the denominators. Hold mouse over the yellow rectangle for the solution Back to the Fractions page Back to the Math 187A page e-mail Questions and Suggestions |
# Difference between revisions of "2010 AMC 10B Problems/Problem 17"
## Problem
Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$th and $64$th, respectively. How many schools are in the city?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$
## Solution
Let the $n$ be the number of schools, $3n$ be the number of contestants, and $x$ be Andrea's place. Since the number of participants divided by three is the number of schools, $n\geq\frac{64}3=21\frac13$. Andrea received a higher score than her teammates, so $x\leq36$. Since $36$ is the maximum possible median, then $2*36-1=71$ is the maximum possible number of participants. Therefore, $3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23$. This yields the compound inequality: $21\frac13\leq n\leq 23\frac23$. Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, $n$ cannot be even. $\boxed{\textbf{(B)}\ 23}$ is the only other option.
## Solution 2
Let $n$ = the number of schools that participated in the contest.
Then $3n$ students participated in the contest.
Since Andrea's score was the median score of all the students, the number of students in the contest must be odd - we know that no two students can have the same score (from the problem), and we know that if the number of students was even, the middle two scores would have to be the same to get a whole number for the median.
So, we can now write an expression for the median score (Andrea's score) in terms of the number of students who participated in the contest (i.e. $3n$):
$\frac{3n+1}{2}$
Also, since we know that this expression must be smaller than $37$ (Andrea, whose score was the median, got a better score than Beth and Carla), we can write an inequality and solve it for $n$:
$\frac{3n+1}{2}<37$
Solving this inequality for $n$, we get that:
$n<\frac{37}{3} \mathrm{(which \quad is \quad a \quad little\quad over \quad 24)}$
So this eliminates answer choices $D$ and $E$.
But we already know that $3n$ must be odd, implying that $n$ must also be odd! So at this point, the only odd answer choice is $\boxed{\textbf{(B)}\ 23}$, and we are done. |
# SAT II Math I : Median
## Example Questions
### Example Question #21 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an even amount of numbers in the set meaning there are two "middle numbers"- 22 and 23. In order to find the median we take the average of 22 and 23:
### Example Question #22 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an even amount of numbers in the set meaning there are two "middle numbers"- 11 and 12. In order to find the median we take the average of 11 and 12:
### Example Question #23 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an even amount of numbers in the set meaning there are two "middle numbers"- 8 and 13. In order to find the median we take the average of 8 and 13:
### Example Question #24 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an even amount of numbers in the set meaning there are two "middle numbers"- 24 and 26. In order to find the median we take the average of 24 and 26:
### Example Question #25 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an odd amount of numbers so we just count from each side until we find the number in the middle.
This gives us a final answer of 37 for the median.
### Example Question #31 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an odd amount of numbers so we just count from each side until we find the number in the middle.
This gives us a final answer of 33 for the median.
### Example Question #31 : Median
FInd the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an odd amount of numbers so we just count from each side until we find the number in the middle.
This gives us a final answer of 19 for the median.
### Example Question #32 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an odd amount of numbers so we just count from each side until we find the number in the middle.
This gives us a final answer of 24 for the median.
### Example Question #33 : Median
Find the median of the set:
Explanation:
The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an odd amount of numbers so we just count from each side until we find the number in the middle.
This gives us a final answer of 39 for the median.
### Example Question #34 : Median
Find the median of the set: |
# What is a complicated math problem that equals 3?
WRITTEN BY: supportmymoto.com STAFF
Q&A
# What’s a sophisticated math downside that equals 3?
### 7 Solutions
• Restrict instance.
lim x→∞ (3x² + x)/(x² + 1) = 3
I hope this helps!
Supply(s): Expertise
• 3
• Math Equations That Equal 3
• Not difficult, however considerably superior:
Discover the primary by-product of f(x) = 3x^4 – 4x^3 + x^2 + x + 5 when x = 1
• For the most effective solutions, search on this web site https://shorturl.im/awVe8
If sin2x=22 – 8√7 Show that 3(Sinx+Cosx+Tanx+Cosecx+Secx+Cotx) =21 Let sin 2x = ok ok = 22- 8√7 ok – 22 = -8√7 [Bring 22 to the LHS) (k-22)² = (-8√7)² [Squaring both sides] k² – 44k + 484 = 448 [Expanding] k² – 44k = 448-484 [Bringing 484 to the RHS] k² – 44k = -36 [Multiply throughout by k] k³ – 44k² = -36k k³ – 44k² + 36k = 0 [Bringing -36k to the LHS] k³ – 49k² +5k² + 8k + 28k = 0 [Rearranging factors] k³ + 5k² + 8k = 49k² – 28k [Rearranging factors and bring 49k² – 28k to the RHS] k³ + 4k² + k² + 4k + 4k = 49k² – 28k [Rearranging factors] k³ + 4k² + k² + 4k + 4k + 4 – 4 = 49k² – 28k [Adding + 4 and -4 to make it a quadratic/polynomial] k³ + 4k² + k² + 4k + 4k + 4 = 49k² – 28k + 4 [Bringing -4 to the RHS] k³ + 4k² + k² + 4k + 4k + 4 = (7k – 2)² [Writing 49k² – 28k + 4 in the form of (a-b)²] (1+ok)(k² + 4k + 4) = (7k – 2)² [Factoring LHS] (1+ok)(ok + 2)² = (7k – 2)² [Writing k² + 4k + 4 in the form of (a+b)²] (1+ok) = (7k-2)²/(ok+2)² [Bringing (k+2)² to the RHS] (1 + 2sinxcosx) = (7k-2)²/(ok+2)² [k = sin2x = 2sinxcosx] (sin²x + cos²x + 2sinxcosx) = (7k-2)²/(ok+2)² [ 1 = sin²x + cos²x and we bring LHS to the form (a+b)²] (sinx + cosx)² = (7k-2)²/(ok+2)² (sinx + cosx) = (7k-2)/(ok+2) [Taking root on both sides] sinx + cosx = (7k -2)/ok / (ok+2)/ok) [Divide Numerator and Denominator of RHS by k] sinx + cosx = (7 – 2/ok) / (1+2/ok) (sinx + cosx) (1 + 2/ok) = 7 – 2/ok [Taking (1+2/k) to the LHS] sinx + cosx + 2sinx/ok + 2cosx/ok = 7 – 2/ok [Multiplying and Expanding] sinx + cosx + 2/ok(sinx + cosx) + 2/ok = 7 [Taking Common factor out] sinx + cosx + 2/ok(sinx + cosx + 1) = 7 [Taking Common factor out] sinx + cosx + 2/ok(sinx + cosx + sin²x + cos²x) = 7 [ 1 = sin²x + cos²x] sinx + cosx + 2/2sinxcos (sinx + cosx + sin²x + cos²x) = 7 [k = sin2x = 2sinxcosx] sinx + cosx + 1/cosx + 1/sinx + sinx/cosx + cosx/sinx = 7 [Canceling common terms] sinx+cosx+secx+cosecx+tanx+cotx = 7 [1/cosx=secx,1/sinx=cosecx,sinx/cosx=tan… Therefore sinx + cosx + secx + cosecx + tanx + cotx = 7 3(sinx + cosx + secx + cosecx + tanx + cotx) = 21 Cheers !!!
• 5-2
• poop – ᴘᴇᴇ
NOTE : Please do not copy - https://supportmymoto.com |
## College Algebra (11th Edition)
$x=2$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\dfrac{1}{32}=x^{-5} ,$ use the laws of exponents and the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{32}=\dfrac{1}{x^5} .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} 1(x^5)=1(32) \\\\ x^5=32 .\end{array} Taking the fifth root of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt[5]{x^5}=\sqrt[5]{32} \\\\ x=\sqrt[5]{(2)^5} \\\\ x=2 .\end{array} |
# What Does It Mean When A Function Is Continuous?
One main consideration when working from the regularization perspective is to give the different value of the function at different points in space. This paper is about the full regularization method that is often referred to in many modern work-flows and on the whole my book is concerned in various positions. Consider a continuous function: Any number of points in the domain of the continuous function can be represented by a new function represented by a step function: Let us first look at some example of this. Let us divide the domain of function into the two kinds: One definition is concerned with mathematical discontinuity of the function at the start and end points. For our purposes we can represent the sequence as a sequence of steps: We can prove two basic examples of these functions: As a matter of fact, if all values for the values of a continuous function, consisting of the root number, are discontinuous and are divided by the whole value of the root number, then all values of continuous function, consisting of the root number, can be represented by a pair of series: The series is discrete and is discretized: By two series, we determine the starting point of the discrete series. Now, using the point-to-point distance between two intervals, we can define the value of a continuous function as a whole: Let us note that there exists any continuous function such that its continuous value at a point is in the interval defined by the formula: If point of this kind is a boundary point of the domain of function, we can define what we call boundary value $\alpha =0$: Where $\epsilon_0 \ge 0$ is the zero mean and $\alpha$ the center of the domain of function. For the discrete function, $\alpha =1$ we put in this equation the length of the segment of the domain: Recall, for $f,g:\ D \rightarrow \R$ with $D$ continuous function, with non single value of continuous function $f(\lambda)\ge 0$, set $\delta=\alpha – 1$, for $\lambda\in (0,1)$: Complex Condition holds: If two intervals of continuous function have the same value of feature $\alpha$, then they have the same value only : If the difference between consecutive values of continuous function is $1$, the value of continuous function at the x element remains the same. Consequences If we have several elements, for simplicity let define the number of elements of one element is still not enough to be Learn More Here value of continuous function. By the definition of standard residue property we have that $\l >1$, also $\v^C$ holds: This statement can be proved by means of the inverse method, for any continuous function, which is as follows $\v^C=\{1,1^C\}$: The inverse method provides a bound on this quantity, and in this case it can be proved that $\l$ is bounded away from zero. This is because $\l$ depends exclusively on the absolute value of a continuous function of variables (see the proof of other results later). Definition of Definition of C0 Let us take aWhat Does It Mean When A Function Is Continuous? How I Was Living in a Child’s Program I came to the present moment as a teenager. I started taking exams to get the tests I wanted. I don’t learn this here now why it was so difficult, but there was no doubt in my mind and I am glad you were with me. The only thing I can say I’m most thankful for are the good grades that were given. On the other hand, I hoped that they would have some of the other good scores due to the fact that the numbers did not agree with their description although I know it wasn’t always that way. When one is finished, it is at least supposed to be a very successful college degree. Then you have to face the reality that none of those number is coming back. That is why it’s good to have that you’ve taken the exams. You’re getting your ‘right scores’ in every two and a half years. Even if they aren’t ready to go out of your head altogether, they’re better in a few months. |
# 3. Looking for patterns
In this part, we look at another way of seeing patterns in multiplication, which is not based upon shapes and counters, but still looks for patterns in rows and columns. Helping pupils explore patterns through practical activities will develop their deeper thinking.
Imagine two columns, one for ‘tens’ another for ‘units’. If we think, for example, of the 8 times table, the first four numbers are 8, 16, 24, 32.
What happens to the tens and the units as you look down the two columns? You should notice that the tens increase by 1 each time, while the units decrease by 2. Using this observation, what would be the next three numbers? See Resource 3: Tens and units for an example of this exercise.
Such observations and questions can be used to help pupils learn about both multiplication and pattern recognition.
## Case Study 3: Recognising patterns in sequences
Mr Oko wanted to do an activity exploring number. He wrote the following number sequences on the board, then asked the pupils to help him find the missing number. Pupils had to put their hand up and say what they thought the missing number was, and why.
• 4, 6, 8, [ ], 12, 14
• 3, 6, [ ], 12, 15
• 16, 25, [ ], 49, 64
• 1, 11, 111, [ ], 11111
• 1, 1, 2, 3, [ ], 8, 13
When the pupils had finished, he asked them to make up their own patterns and leave a number out. They then swapped their pattern with their partners and tried to fill in the missing numbers.
They were very excited and enjoyed the activity. Mr Oko asked if they could see a pattern? Could they predict the last number and each answer? He was pleased some could.
Mr Oko used pair work often, as it allowed all pupils to talk and helped their thinking.
## Key Activity: Exploring the multiples of 9
You will need Resource 4: Times table
• Stand by the chalkboard and ask pupils to be totally silent. Ask them to watch carefully.
• Write the first five multiples of 9 on the blackboard.
• Pause. Ask them to look at what is happening to the numbers.
• Ask a pupil to complete the pattern to 10 x 9, under the heading ‘tens’ and ‘units’.
• Ask the class to share anything they notice, recording and accepting everything without commenting.
• Carry on, but stop after 13 x 9, skip some and then write 17 x 9 = ? Now, watch carefully while they try to make sense of what is going on. You may have to prompt them to see the pattern in tens and units.
• Finally ask pairs of pupils to investigate other multiples (it is best to start with single digit numbers, 1–9). Can they work out together the pattern for tens and units?
Next lesson, ask your pupils to practise multiplication using games. See Resource 5: Multiplication games for ideas.
2. Using games to explore rectangular numbers
Resource 1: Square numbers |
# Factor Theorem
In this part, we will look at the Factor Theorem, which uses the remainder theorem and learn how to factorise polynomials. Further, we will be covering the splitting method and the factor theorem method.
## Factor Theorem
If p(x) is a polynomial of degree n > 1 and a is any real number, then
• x – a is a factor of p(x), if p(a) = 0, and
• p(a) = 0, if x – a is a factor of p(x).
Let’s look at an example to understand this theorem better.
### Example:
Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6.
Solution: To begin with, we know that the zero of the polynomial (x + 2) is –2. Let p(x) = x3 + 3x2 + 5x + 6
Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0
According to the factor theorem, if p(a) = 0, then (x – a) is a factor of p(x). In this example, p(a) = p(- 2) = 0
Therefore, (x – a) = {x – (-2)} = (x + 2) is a factor of ‘x3 + 3x2 + 5x + 6’ or p(x).
## Factorisation of polynomials
You can factorise polynomials by splitting the middle term as follows: to begin with, consider a polynomial ax2 + bx + c with factors (px + q) and (rx + s). Therefore, we have ax2 + bx + c = (px + q) (rx + s). So, ax2 + bx + c = prx2 + (ps + qr) x + qs
If we compare the coefficients of x2, we get a = pr. Also, on comparing the coefficients of x, we get b = ps + qr. Finally, on comparing the constants, we get c = qs. Hence, b is the sum of two numbers ‘ps’ and ‘qr’, whose product is (ps)(qr) = (pr)(qs) = ac.
Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two numbers whose product is ‘ac’. Let’s look at an example to understand this clearly.
### Example
Factorise 6×2 + 17x + 5 by splitting the middle term.
Solution 1 (By splitting method): As explained above, if we can find two numbers, ‘p’ and ‘q’ such that, p + q = 17 and pq = 6 x 5 = 30, then we can get the factors.
After looking at the factors of 30, we find that numbers ‘2’ and ‘15’ satisfy both the conditions, i.e. p + q = 2 + 15 = 17 and pq = 2 x 15 = 30. So,
6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5
= 6x2 + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
Therefore, the factors of (6x2 + 17x + 5) are (3x + 1) and (2x + 5) with a remainder, zero.
### Example
Factorise y2 – 5y + 6 by using the Factor Theorem.
Solution: Let p(y) = y2 – 5y + 6. Now, if p(y) = (y – a) (y – b), the constant term will be ab as can be seen below,
p(y) = (y – a)(y – b)
= y2 – by – ay + ab
On comparing the constants, we get ab = 6. Next, the factors of 6 are 1, 2 and 3. Now, p(2) = 22– (5 × 2) + 6 = 4 – 10 + 6 = 0. So, (y – 2) is a factor of p(y). Also, p(3) = 32 – (5 × 3) + 6 = 9 – 15 + 6 = 0. So, (y – 3) is also a factor of y2 – 5y + 6. Therefore, y2 – 5y + 6 = (y – 2)(y – 3)
## More Solved Examples for You
Question 1: Factorise x3 – 23x2 + 142x – 120
Answer : Let p(x) = x3 – 23x2 + 142x – 120. To begin with, we will start finding the factors of the constant ‘– 120’, which are:
±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60 and ±120
Further, by trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a factor of p(x). Also, we see that
[x3 – 23x2 + 142x – 120] = x3 – x2 – 22x2 + 22x + 120x – 120
So, by removing the common factors, we have x3 – 23x2 + 142x – 120 = x2(x –1) – 22x(x – 1) + 120(x – 1)
Further, taking ‘x – 1’ common, we get x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Therefore, x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Also, note that if we divide p(x) by ‘x – 1’, then the result will be (x2 – 22x + 120)
Going on, x2 – 22x + 120 can be factorised further. So, by splitting the middle term, we get:
x2 – 22x + 120 = x2 – 12x – 10x + 120 … [ (– 12 – 10 = – 22) and {(–12)( –10) = 120}]
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
Therefore, we have x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)
Question 2: Factorise x3 – 2x2 – x + 2
Answer : Let p(x) = x3 – 2x2 – x + 2. To begin with, we will start finding the factors of the constant ‘2’, which are: 1, 2
By trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a factor of p(x).
So, by removing the common factors, we have x3 – 2x2 – x + 2 = x2(x – 2) – (x – 2) = (x2 – 1)(x – 2)
= (x + 1)(x – 1)(x – 2) … [Using the identity (x2 – 1) = (x + 1)(x – 1)]
Therefore, the factors of x3 – 2x2 – x + 2 are (x + 1), (x – 1) and (x – 2)
Question 3: Explain factor theorem with example?
Answer:  An example of factor theorem can be the factorization of 6×2 + 17x + 5 by splitting the middle term. In this example, one can find two numbers, ‘p’ and ‘q’ in a way such that, p + q = 17 and pq = 6 x 5 = 30. After that one can get the factors.
Question 4: What is meant by a polynomial factor?
Answer: A factor of polynomial P(x) refers to any polynomial whose division takes place evenly into P(x). For example, x + 2 is a factor belonging to the polynomial x2 – 4. The polynomial’s factorization is its representation as a product of its various factors. A good example can be the factorization of x2 – 4 is (x – 2)(x + 2).
Question 5: Explain the formula of factor theorem?
Answer: The Factor Theorem explain us that if the remainder f(r) = R = 0, then (x − r) happens to be a factor of f(x). The Factor Theorem is quite important because of its usefulness to find roots of polynomial equations.
Question 6: Is it possible for a remainder to be negative?
Answer: No, a remainder can never be a negative number.
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87 percent of 812
Here we will show you how to calculate eighty-seven percent of eight hundred twelve. Before we continue, note that 87 percent of 812 is the same as 87% of 812. We will write it both ways throughout this tutorial to remind you that it is the same.
87 percent means that for each 100, there are 87 of something. This page will teach you three different methods you can use to calculate 87 percent of 812.
We think that illustrating multiple ways of calculating 87 percent of 812 will give you a comprehensive understanding of what 87% of 812 means, and provide you with percent knowledge that you can use to calculate any percentage in the future.
To solidify your understanding of 87 percent of 812 even further, we have also created a pie chart showing 87% of 812. On top of that, we will explain and calculate "What is not 87 percent of 812?"
Calculate 87 percent of 812 using a formula
This is the most common method to calculate 87% of 812. 812 is the Whole, 87 is the Percent, and the Part is what we are calculating. Below is the math and answer to "What is 87% of 812?" using the percent formula.
(Whole × Percent)/100 = Part
(812 × 87)/100 = 706.44
87% of 812 = 706.44
Get 87 percent of 812 with a percent decimal number
You can convert any percent, such as 87.00%, to 87 percent as a decimal by dividing the percent by one hundred. Therefore, 87% as a decimal is 0.87. Here is how to calculate 87 percent of 812 with percent as a decimal.
Whole × Percent as a Decimal = Part
812 × 0.87 = 706.44
87% of 812 = 706.44
Get 87 percent of 812 with a fraction function
This is our favorite method of calculating 87% of 812 because it best illustrates what 87 percent of 812 really means. The facts are that it is 87 per 100 and we want to find parts per 812. Here is how to illustrate and show you the answer using a function with fractions.
Part 812
=
87 100
Part = 706.44
87% of 812 = 706.44
Note: To solve the equation above, we first multiplied both sides by 812 and then divided the left side to get the answer.
87 percent of 812 illustrated
Below is a pie chart illustrating 87 percent of 812. The pie contains 812 parts, and the blue part of the pie is 706.44 parts or 87 percent of 812.
Note that it does not matter what the parts are. It could be 87 percent of 812 dollars, 87 percent of 812 people, and so on. The pie chart of 87% of 812 will look the same regardless what it is.
What is not 87 percent of 812?
What is not 87 percent of 812? In other words, what is the red part of our pie above? We know that the total is 100 percent, so to calculate "What is not 87%?" you deduct 87% from 100% and then take that percent from 812:
100% - 87% = 13%
(812 × 13)/100 = 105.56
Another way of calculating the red part is to subtract 706.44 from 812.
812 - 706.44 = 105.56
That is the end of our tutorial folks. We hope we accomplished our goal of making you a percent expert - at least when it comes to calculating 87 percent of 812.
Percent of a Number
Go here if you need to calculate the percent of a different number.
87 percent of 813
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# How to solve inverse trig functions
We can do your math homework for you, and we'll make sure that you understand How to solve inverse trig functions. We can help me with math work.
## How can we solve inverse trig functions
In this blog post, we will be discussing How to solve inverse trig functions. In mathematics, a differential solver is a numerical method for solving differential equations. Differential equations are mathematical equations that describe how a function changes over time. They are used to model many real-world phenomena, such as the motion of a spring or the flow of a liquid. Differential solvers approximate the solution to a differential equation by calculating a series of values that get closer and closer to the true solution. There are many different types of differential solvers, each with its
An x intercept is where a graph crosses the x-axis. This can be found by solving for when y = 0. This can be done by setting y = mx + b, where m is the slope and b is the y-intercept, to 0 and solving for x. This will give you the x coordinate of the x intercept.
To solve multi step equations, you need to take each equation one at a time and use the order of operations to solve for the variable. You will use the same steps as solving one step equations, but you will need to do them in order. First, you will solve for the variable with the highest exponent. Next, you will solve for the variable with the second highest exponent. Lastly, you will solve for the variable with the lowest exponent.
To solve for the x intercept, you need to set y = 0 and solve for x. This can be done by using algebraic methods or by graphing the equation and finding the point where it crosses the x-axis.
The two unknowns are called x> and y>. The coefficient a> is what controls how much x> changes as y> changes (i.e. how much x> "dips" when y> increases). The coefficient b> is what controls how much y> changes as x> changes (i.e. how much y> "soars" when x> increases). The formula for solving a quadratic equation is: math>{ frac{a^{2}-b^{2}}{2a+b}left( x-frac{a}{2} ight) }/math>. Where: math>Solving for a/math>: A is the coefficient of determination, which tells us how well we solved for one of the variables. math>Solving for b/math>: B is the coefficient of variation, which tells us how much each variable varies over time.
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# Class 11 Maths NCERT Solutions for Chapter 7 Permutations and Combinations Exercise 7.4
### Permutations and Combinations Exercise 7.4 Solutions
1. If nC8 = nC2, find nC2.
Solution
It is known that, nCa = nCb ⇒ a = b or n = a+ b
Therefore,
nC8 = nC2 ⇒ n = 8 + 2 = 10
∴ nC2 = 10C2 =
2. Determine n if
(i) 2nC3 = nC3 = 12 : 1
(ii) 2nC3 = nC3 = 11 : 1
Solution
(i)
⇒ 2n - 1 = 3(n - 2)
⇒ 2n - 1 = 3n - 6
⇒ 3n - 2n = -1 + 6
⇒ n = 5
(ii)
⇒ 4(2n - 1) = 11(n - 2)
⇒ 8n - 4 = 11n - 22
⇒ 11n - 8n = -4 + 22
⇒ 3n = 18
⇒ n = 6
3. How many chords can be drawn through 21 points on a circle?
Solution
For drawing one chord on a circle, only 2 points are required.
To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.
Therefore, there will be as many chords as there are combinations of 21 points taken 2 at a time.
Thus, required number of chords = 21C2 = 21!/2!(21 -2)! = 21!/2!19! = (21×20)/2 = 210
4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Solution
A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in 5C3 ways .
3 girls can be selected from 4 girls in 4C3 ways.
Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected = 5C3 × 4C3 = (5!/3!2!) × (4!/3!1!)
5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution
There are a total of 6 red balls, 5 white balls, and 5 blue balls.
9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.
Here,
3 balls can be selected from 6 red balls in 6C3 ways.
3 balls can be selected from 5 white balls in 5C3 ways.
3 balls can be selected from 5 blue balls in 5C3 ways.
Thus, by multiplication principle, required number of ways of selecting 9 balls
6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Solution
In a deck of 52 cards, there are 4 aces. A combination of 5 cards have to be made in which there is exactly one ace.
Then, one ace can be selected in 4C1 ways and the remaining 4 cards can be selected out of the 48 cards in 48C4 ways.
Thus, by multiplication principle, required number of 5 card combinations
7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Solution
Out of 17 players, 5 players are bowlers.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
4 bowlers can be selected in 5C4 ways and the remaining 7 players can be selected out of the 12 players in 12C7 ways.
Thus, by multiplication principle, required number of ways of selecting cricket team
8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Solution
There are 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in 5C2 ways and 3 red balls can be selected out of 6 red balls in 6C3 ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls
9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Solution
There are 9 courses available out of which, 2 specific courses are compulsory for every student.
Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in 7C3 ways.
Thus, required number of ways of choosing the programme |
## FINDING THE NTH TERM OF AN ARITHMETIC SEQUENCE WORKSHEET
Finding the nth term of an arithmetic sequence worksheet :
Here we are going to see some practice questions on finding the nth term of an arithmetic sequence.
## Finding the nth term of an arithmetic sequence worksheet - Practice questions
(1) Find the nth term of each arithmetic sequence described.
a1 = 5, d = 5, n = 25
(2) Find the nth term of each arithmetic sequence described.
a1 = 8, d = 3, n = 16
(3) Find the nth term of each arithmetic sequence described.
a1 = 34, d = 15, n = 200
(4) Find the nth term of each arithmetic sequence described.
a1 = 5/8, d = 1/8, n = 22
(5) Find the nth term of each arithmetic sequence described.
a1 = 3/2, d = 9/4, n = 39
(6) Find the nth term of each arithmetic sequence described.
0.5, 1, 1.5, 2, … for n = 50
## Finding the nth term of an arithmetic sequence worksheet - solution
Example 1 :
Find the nth term of each arithmetic sequence described.
a1 = 5, d = 5, n = 25
Solution :
Applying the given values in the formula
an = a + (n - 1)d
we get,
a25 = 5 + (25-1)(5)
a25 = 5 + 24(5)
a25 = 5 + 120
a25 = 125
Hence 25th term of the above sequence is 125.
Example 2 :
Find the nth term of each arithmetic sequence described.
a1 = 8, d = 3, n = 16
Solution :
Applying the given values in the formula
an = a + (n - 1)d
we get,
a16 = 8 + (16-1)(3)
a16 = 8 + 15(3)
a16 = 8 + 45
a16 = 53
Hence 16th term of the above sequence is 53.
Example 3 :
Find the nth term of each arithmetic sequence described.
a1 = 34, d = 15, n = 200
Solution :
Applying the given values in the formula
an = a + (n - 1)d
we get,
a200 = 34 + (200-1)(15)
a200 = 34 + 195(15)
a200 = 34 + 2925
a200 = 2959
Hence 200th term of the above sequence is 2959.
Example 4 :
Find the nth term of each arithmetic sequence described.
a1 = 5/8, d = 1/8, n = 22
Solution :
Applying the given values in the formula
an = a + (n - 1)d
we get,
a22 = (5/8) + (22-1)(1/8)
a22 = (5/8) + 21(1/8)
a22 = (5/8) + 21/8
a22 = (21 + 5)/8
= 26/8
= 13/4
Hence 22nd term of the above sequence is 13/4.
Example 5 :
Find the nth term of each arithmetic sequence described.
a1 = 3/2, d = 9/4, n = 39
Solution :
Applying the given values in the formula
an = a + (n - 1)d
we get,
a39 = (3/2) + (39-1)(9/4)
a39 = (3/2) + 38(9/4)
a39 = (3/2) + 171/2
a39 = (171+3)/2
= 174/2
= 87
Hence 39th term of the above sequence is 87.
Let us see the next example on "How to find the nth term of the arithmetic sequence".
Example 6 :
Find the nth term of each arithmetic sequence described.
0.5, 1, 1.5, 2, … for n = 50
Solution :
Applying the given values in the formula
an = a + (n - 1)d
a = 0.5, d = 1 - 0.5 = 0.5
we get,
a50 = 0.5 + (50-1)(0.5)
a50 = 0.5 + 49(0.5)
a50 = 0.5 + 24.5
a50 = 25
Hence 50th term of the above sequence is 25.
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Question Video: Expressing a Set of Simultaneous Equations as a Matrix Equation | Nagwa Question Video: Expressing a Set of Simultaneous Equations as a Matrix Equation | Nagwa
# Question Video: Expressing a Set of Simultaneous Equations as a Matrix Equation Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Express the given set of simultaneous equations as a matrix equation. 7π₯ β 3π¦ + 6π§ = 5, 5π₯ β 2π¦ + 2π§ = 11, and 2π₯ β 3π¦ + 8π§ = 10.
03:43
### Video Transcript
Express the given set of simultaneous equations as a matrix equation. Seven π₯ minus three π¦ plus six π§ equals five, five π₯ minus two π¦ plus two π§ equals 11, and two π₯ minus three π¦ plus eight π§ equals 10.
We have three linear equations, each containing the three variables π₯, π¦, and π§. All the equations are written in the same form with the π₯ term first followed by the π¦ term and then the π§ term. Completing the left-hand side and just a single constant on the right. And when a set of simultaneous equations is written in this form, itβs straightforwards to rewrite it as a matrix equation.
We write the set of simultaneous equations out again. And remember, we wanted to write this set of simultaneous equations as a single matrix equation. And the entries of the matrix, which makes it a matrix equation, are going to be taken from the coefficients of the terms on the left hand-side of our equations. So for example, in the first equation, the coefficients are seven, negative three, and six. And continuing, we also have the coefficients five, negative two, and two from the second equation. And two, negative three, and eight in the third equation.
How are these coefficients arranged in our matrix? Well, in exactly the same way that they are arranged in our equations, in the way that we laid them out. All we have to do is erase from the equations anything which is not an underlined coefficient. We erase the π₯s, the π¦s, the π§s, the constantsβ terms on the right-hand side of the equations, and any leftover plus or equals signs. And because itβs a matrix now, we need to put some square brackets around these numbers.
So we have a matrix, but we donβt yet have an equation. For an equation, weβre going to need some unknowns. We have three unknowns β π₯, π¦, and π§ β in our original set of simultaneous equations. And we put them in that order into a three by one matrix which we multiply by. And as well as having one or more variables or unknowns, an equation must also have an equal sign. And what is this left-hand side equal to? Well, we havenβt made use of these three constant terms on the right-hand side yet. And so on the right-hand side, we now have a three by one matrix with these constant terms as the entries.
So here we have our matrix equation. We have a matrix with known entries times a matrix with unknown entries equal to another matrix with known entries. And we can check that this matrix equation really does represent the set of simultaneous equations that we started with, by multiplying out the left-hand side. Multiplying these matrices, we get a three by one matrix whose entries are seven π₯ minus three π¦ plus six π§, five π₯ minus two π¦ plus two π§, and two π₯ minus three π¦ plus eight π§. And of course, on the right-hand side, we still have the three by one matrix: five, 11, 10.
For these two matrices to be equal, their entries must be equal. And so seven π₯ minus three π¦ plus six π§ must be five. Five π₯ minus two π¦ plus two π§ must be 11. And two π₯ minus three π¦ plus eight π§ must be 10. And these are of course the simultaneous equations that we started with.
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# Question #84622
Mar 24, 2017
Define the function $T \left(x\right)$ as the temperature $T$ in degrees Celsius at $x$ centimeters away from the heated end.
The prompt tells us that $\left(\text{d"T)/("d} x\right) = - 6$ and that halfway along the $60$ centimeter long rod, the temperature is $290$. That is, $T \left(30\right) = 290$.
We can solve for $T$ from $\left(\text{d"T)/("d} x\right) = - 6$ by separating the variables. Treating $\left(\text{d"T)/("d} x\right)$ like a quotient, we can say that:
$\text{d} T = - 6$ $\text{d} x$
Then integrating:
$\int \text{d"T=-6int"d} x$
Which, adding a constant of integration, gives:
$T \left(x\right) = - 6 x + C$
Use the original condition $T \left(30\right) = 290$ to solve for $C$:
$290 = - 6 \left(30\right) + C$
$C = 470$
$T \left(x\right) = - 6 x + 470$
Then, the temperature difference between the two ends of the rod is given by $T \left(0\right) - T \left(60\right) = 470 - \left(- 6 \left(60\right) + 470\right) = {360}^{\circ} \text{C}$.
In fact, to find the difference between the two ends, doing the actual integration isn't necessary. The differential $\left(\text{d"T)/("d} x\right) = - 6$ means that for every centimeter $x$ traveled, the temperature $T$ decreases constantly by $6$ degrees. Thus, if the rod is $60$ centimeters long, it will decrease $6 \times 60 = {360}^{\circ} \text{C}$ in total.
In part B, we see that the rate of change described is again $\left(\text{d"T)/("d} x\right)$. It's proportional to $x$ with some proportionality constant $k$, which is written as:
$\left(\text{d"T)/("d} x\right) = k x$
Which can be solved by again separating variables:
$\text{d} T = k x$ $\text{d} x$
$\int \text{d} T = k \int x$ $\text{d} x$
$T \left(x\right) = k \left(\frac{1}{2} {x}^{2}\right) + C$
With this model, we see that $T \left(0\right) = 380$ and $T \left(60\right) = 20$, which we can use to solve for $k$ and $C$:
Using $T \left(0\right) = 380$ shows that
$380 = k \left(\frac{1}{2} \left({0}^{2}\right)\right) + C$
$380 = C$
So:
$T \left(x\right) = \frac{k {x}^{2}}{2} + 380$
Then using $T \left(60\right) = 20$:
$20 = \frac{k \left({60}^{2}\right)}{2} + 380$
$- 360 = 1800 k$
$k = - \frac{1}{5}$
So:
$T \left(x\right) = - {x}^{2} / 10 + 380$ |
# #008 Linear Algebra – Eigenvectors and Eigenvalues
## #008 Linear Algebra – Eigenvectors and Eigenvalues
Highlight: In this post we will talk about eigenvalues and eigenvectors. This concept proved to be quite puzzling to comprehend for many machine learning and linear algebra practitioners. However, with a very good and solid introduction that we provided in our previous posts we will be able to explain eigenvalues and eigenvectors and enable very good visual interpretation and intuition of these topics. We give some Python recipes as well.
Tutorial Overview:
This post we will start with the following well-known $$\left ( x,y \right )$$ coordinate system.
We imagine that we linearly transform this system. We obtain the following transformation where actually our original system is transformed and our $$\hat{i}$$ vector is mapped into $$\begin{bmatrix}3\\0\end{bmatrix}$$ vector and our $$\hat{j}$$ vector is mapped into $$\begin{bmatrix}1\\2\end{bmatrix}$$ vector.
Now, in this space we will have some arbitrary vector and this vector is shown in the following image.
Once we apply a linear transformation to this vector as shown in image below, we see that this vector moved from its original line. It has been knocked off from it . 🙂
However, there will be some particular vectors that will be linearly transformed, and they will remain on their own span. Just too good to be true ?
Sinatra rocks. We know. Ok. But, let’s continue.
The previous idea means that vector will remain on its span line that goes through the origin. Only the vector intensity and/or direction can be changed. We say that this vector is just multiplied by scalar that can stretch this vector, shorten this vector or reverse its direction. Such vectors are called eigenvectors.
For this particular transformation defined with the matrix $$\begin{bmatrix}3 & 1\\0 & 2\end{bmatrix}$$ we will have one particular vector and it’s coordinates are $$\begin{bmatrix}-1\\1\end{bmatrix}$$.
Once we linearly transform this vector, it will be changed by a scalar $$2$$. This means that it will be stretched $$2$$ times. Moreover, a vector $$\begin{bmatrix}1\\-1\end{bmatrix}$$ will also be stretched two times with this linear transformation.
In addition, any vector that lies on this line will be stretched by a factor of $$2$$.
Moreover, in this example, our basis $$\hat{i}$$ vector is transformed in such a way that $$\hat{i}$$ remains on the $$x$$-axis. This means that any other vector that lies on the $$x$$-axis will remain on this line.
So, this linear transformation matrix has two vectors, that define two lines, and any vector on these lines will remain there after a linear transformation. It is depicted in the image below with yellow and green vectors. In addition, the vectors on the yellow line will be stretched by a scalar 2, whereas the vectors on the green line will be stretched by a scalar 3. These values correspond to the specific eigenvectors and we call them eigenvalues.
On the other hand, once again, we will have all other vectors that actually will not remain on its original line. They will be knocked off from their original line and this is shown in the following image.
So, once we transform it, we see that this vector (red arrow) will completely go off this line. On the other hand, two vectors, eigenvectors, remain on the same line for this particular transformation.
Where do we use eigenvectors?
One interesting property of eigenvectors can be observed when we apply a rotation transformation on a certain coordinate system. The rotations can be defined along $$x, y, or z$$ – axis. However, the axis of rotation can also be arbitrary and unknown. Then, this rotation will be defined by a $$3\times 3$$ matrix for a 3-D space. However, if we manage to find eigenvectors for this rotation matrix, it will define the axis along which this rotation is done. So, this is just one of many applications of eigenvectors.
## 2. Definition of eigenvectors and eigenvalues
We’ve seen so far how visually eigenvectors can be interpreted and what are eigenvalues. However, we can define it in a more formal way with the following equation:
Here, we have the matrix-vector multiplication $$A\cdot \vec{v}$$ and that equals vector $$\vec{v}$$ multiplied by a scalar. So, we say that the result of $$A\cdot \vec{v}$$ will be a vector and it equals the scaled vector $$\vec{v}$$. This can be stated as:
For a certain matrix $$A$$ we can find a vector $$\vec{v}$$ that, when multiplied with the matrix $$A$$ will remain on the same line.
The only thing that can change is its intensity and potentially its direction. Commonly, we have a known matrix $$A$$. Then, we need to solve this equations for $$\lambda$$ and $$\vec{v}$$. Those are values that we are looking for.
To solve this equation, we perform the following trick. On the right hand side, we include a matrix vector multiplication and instead to multiply it with a simple scalar we can add the identity matrix. It has ones along the main diagonal. Then, we can extract $$\lambda$$ and we will obtain the modified equation $$\lambda\cdot I$$ where $$I$$ is identity matrix times $$\vec{v}$$.
So, if we a little bit modify this equation, we get:
Then, we will have a simple matrix-vector multiplication that should equal a zero vector. Now, our matrix is changed and it is changed with the scalar $$\lambda$$. For instance, for this $$3\times 3$$ example we now have our main diagonal $$3-\lambda$$, $$5-\lambda$$ and once again $$5-\lambda$$
If we connect this equation with a determinant, we will notice that any vector $$\vec{v}$$ multiplied by matrix (whose determinant is not equal to zero), will give us another arbitrary vector.
However, when we have a matrix that multiplies $$\vec{v}$$ and we get a zero vector, that can happen only for those values $$\lambda$$s where our determinant of the matrix $$\left ( A-\lambda I \right )= 0$$. Such transformation for those $$\lambda$$ will squish the plane into a single line.
Here is a nice illustration of the parameter $$\lambda$$. First, we define a matrix of the linear transformation. So, our goal here will be to change $$\lambda$$ and that for different $$\lambda$$ we observe the mapping of a coordinate system with this linear transformation.
For instance, let’s say that we start from $$0$$ and we start increasing $$\lambda$$). We can see how the mapping behaves.
We see that a determinant is getting smaller and smaller. Recall that this is the area of a unit square! At one point it will become zero and exactly at this moment where the determinant is zero, we will obtain a line!
That is, a linear transformation will squash a plane into a line. And this is one of our solutions. And then, if we continue increasing $$\lambda$$ we will actually get the negative values of the determinant. So, this can be one way to intuitively try to solve this equation by experimenting with different values of $$\lambda$$ and connecting this with how the transformed plane is actually behaving.
## 3. Example of eigenvector and eigenvalue calculation
Now, let’s work out, our previous example by hand as shown in the image below.
So, if you note that the determinant needs to be zero, we have actually a $$3-\lambda$$ and $$2-\lambda$$ on the main diagonal. And we know how to solve this. This determinant will be the quadratic equation that we need to solve for different $$\lambda$$. And this already implies that most likely we will have not one but two eigenvalues $$\lambda$$ because that’s common for quadratic equations. Of course, different scenarios can happen, but as we said the most common is to have two solutions.
This equation is very easy to solve. it can be solved for $$\lambda = 2$$ or $$\lambda = 3$$ and these values can be actually our eigenvalues.
So, for this $$\lambda$$ what we now need to do is to go back to our equation and actually figure out what our vector $$\vec{v}$$ actually is. We can replace $$\vec{v}$$ as $$x$$ and $$y$$ and from here and we just need to calculate for what values this is satisfied.
import numpy as np
import matplotlib.pyplot as plt
A = np.array ( [ [3,1] , [0,2] ] )
print(A)
eig_val, eig_vec = np.linalg.eig (A)
print("Eigenvalues: ", eig_val)
print("Eigenvectors are columns in the matrix:\n",eig_vec)
Eigenvalues: [3. 2.]
Eigenvectors are columns in the matrix:
[[ 1. -0.70710678]
[ 0. 0.70710678]]
plt.quiver( [0,0] , [0,0], eig_vec[0,:], eig_vec[1,:], angles = 'xy', scale_units = 'xy', scale = 1 )
plt.xlim(-2, 2)
plt.ylim(-2, 2)
plt.grid('on')
plt.show()
plt.savefig("eigenvectors.png", dpi = 800)
#### To dive a bit deeper, we can ask ourselves whether eigenvalues always exist?
A rotation matrix. For example, the following matrix is a linear transformation, but what it actually does, it performs rotations. So, if it rotates to the plane for some angle, and in this case $$90^{\circ}$$ rotation, we cannot have any vector preserved on the line where it originated from. To confirm this numerically, this will be our determinant for the following rotation matrix. Then we can try to solve this for $$\lambda$$ and we will obtain that $$\lambda ^{2}+1= 0$$ . Here, our eigenvalues are complex numbers 🙂
A shear transformation. Another interesting operation can be shear. In this case you can see that the $$x$$ – axis or $$\hat{i}$$ vector remains the same where the $$\hat{j}$$ vector is stretched and inclined. And if you would like to calculate the eigenvectors, we will follow the same principle. So, from this matrix we will subtract $$\lambda$$ values and our final equation will be:
So, only one eigenvalue exists and that means that the only eigenvector is actually along the $$x$$ – axis.
Eigenvectors and basis change. In addition, one interesting thing to mention is that we learned so far that we can transform our original coordinate system. And it’s quite interesting if we transform it in such a way so that actually eigenvectors of a matrix (we will assume that there are two eigenvectors in this case) we use as a new basis. Then, it’s interesting how actually our new world that’s composed of eigenvectors will behave there.
We know that everything along these vectors will be just multiplied with eigenvalues associated with these eigenvectors. So, along these directions the changes will be just multiplied and they will not change the direction.
For instance, we can have a linear transformation giving me the following matrix and we can use eigenvectors as basis. So, the original transformation matrix has columns (3,0) and (1,2). We have calculated the eigenvectors: (1,0) and (-1,1).
So, that means that we can represent everything in this new coordinate system and we can go back. If we multiply it with the inverse, we would be able to have the complete equation given here:
At the end this will be our diagonal matrix.
And once we have to work in such spaces where actually we know that our transformation matrix is diagonal that simplifies many things. For example, imagine that we need to calculate this hundreds of times. This matrix can be transformed in the eigen basis system so then it’s easy to compute the power of $$100$$ of a matrix in that, and then we can just go back to our original system and obtain our solution.
So, imagine that we have to multiply some vector with a matrix A 100 times. Then, we can go to an eigenvectors basis space. Multiplying a vector with a matrix A in a new space, will just scale it/stretch it. So, the result will be just multiplying with a diagonal matrix 100 times (very simple computationally). Then, the final vector we just move back to our original coordinate system and we have our solution.
## Summary
Believe if or not we have finished. We do realize how important and interesting eigenvectors are in linear algebra. However, the real fun starts in the next post, where will be talking about Singular Value Decomposition (SVD). I you ask me (Vladimir), probably one of the top 10 most famous and influential algorithms in our history. So, SVD here we come. |
×
# Help in Proving Divisibility
While I am proving this statement: An = 5^n + 2 (3^(n-1)) + 1 is divisible by 8 for all positive integers, I was wondering if my method of proof is correct if I use the induction method, then use the theorem that d divides (ax+by) or finding linear combinations, and further proving by parity. Is my method of proof correct for this case? (On need-to-know basis.)
Note by John Ashley Capellan
3 years, 6 months ago
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Yes, induction is the way to go. I would prove it like this:
Basis
Let $$n=1$$. Then we have
$$5^1+2\cdot 3^0+1=1$$,
which is divisible by $$8$$.
Induction step
Let $$5^k+2\cdot 3^{k-1}+1$$ be divisible by $$8$$, for $$k=1,2,\ldots,n$$. Then
$$5^{n+1}+2\cdot 3^n +1$$
$$=5\cdot 5^{n}+3\cdot 2\cdot 3^{n-1} +1$$
$$=4(5^n)+2(2\cdot 3^{n-1})+5^n+2\cdot 3^{n-1}+1$$
$$=4(5^n+3^{n-1})+5^n+2\cdot 3^{n-1}+1$$
From our induction hypothesis, 8 divides $$5^n+2\cdot 3^{n-1}+1$$, and we can therefore write it as $$8c_1$$ for some $$c_1\in \mathbb{Z}$$. Thus, we only need to show, that $$8$$ divides $$4(5^n+3^{n-1})$$, which will be achieved, if we can show that $$2$$ divides $$5^n+3^{n-1}$$ - ie. it is even.
It can be shown that the product of two uneven numbers is uneven, and therefore both $$5^n$$ and $$3^{n-1}$$ are uneven. Furthermore, it can be shown that the sum of two uneven numbers is even. Hence, $$5^n+3^{n-1}$$ is even, and we can write it as $$2c_2$$ for a suitable $$c_2\in \mathbb{Z}$$. We now have
$$=4\cdot 2c_2 + 8c_1=8(c_1+c_2)$$,
which is divisible by 8.
QED · 3 years, 6 months ago |
# System of First Order Differential Equations
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## Transcription
1 CHAPTER System of First Order Differential Equations In this chapter, we will discuss system of first order differential equations. There are many applications that involving find several unknown functions simultaneously. Those unknown functions are related by a set of equations that involving the unknown functions and their first derivatives. For example, in Chapter Two, we studied the epidemic of contagious diseases. Now if S(t) denotes number of people that is susceptible to the disease but not infected yet. I(t) denotes number of people actually infected. R(t) denotes the number of people have recovered. If we assume The fraction of the susceptible who becomes infected per unit time is proportional to the number infected, b is the proportional number. A fixed fraction rs of the infected population recovers per unit time, r. A fixed fraction of the recovers g become susceptible and infected, g. proportional function. The system of differential equations model this phenomena are S = bis + gr I = bis ri R = ri gr The numbers of unknown function in a system of differential equations can be arbitrarily large, but we will concentrate ourselves on to 3 unknown functions.. Principle of superposition Let a ij (t), b j (t) i =,,, n and j =,,, n be known function, and x i t, i =,,, n be unknown functions, the linear first
2 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS order system of differential equation for x i (t) is the following, x (t) = a (t)x (t) + a (t)x (t) + + a n (t)x n (t) + b (t) x (t) = a (t)x (t) + a (t)x (t) + + a n (t)x n (t) + b (t) x 3(t) = a 3 (t)x (t) + a 3 (t)x (t) + + a 3n (t)x n (t) + b 3 (t). x n(t) = a n (t)x (t) + a n (t)x (t) + + a nn (t)x n (t) + f (t) Let x(t) be the column vector of unknown functions x i t, i =,,, n, A(t) = (a ij (t), and b(t) be the column vector of known functions b i t, i =,,, n, we can write the first order system of equations as () x (t) = A(t)x(t) + b(t) When n =, the linear first order system of equations for two unknown functions in matrix form is, x (t) a (t) a x = (t) x (t) b (t) + (t) a (t) a (t) x (t) b (t) When n = 3, the linear first order system of equations for three unknown functions in matrix form is, x (t) x (t) = a (t) a (t) a 3 a (t) a (t) a 3 x (t) x (t) + b (t) b (t) x 3(t) a 3 (t) a 3 (t) a 33 x 3 t b 3 (t) A solution of equation () on the open interval I is a column vector function x(t) whose derivative (as a vector-values function) equals A(t)x(t) + b(t). The following theorem gives existence and uniqueness of solutions, Theorem.. If the vector-valued functions A(t) and b(t) are continuous over an open interval I contains t, then the initial value problem { x (t) = A(t)x(t) + b(t) x(t ) = x has an unique vector-values solution x(t) that is defined on entire interval I for any given initial value x. When b(t), the linear first order system of equations becomes x (t) = A(t)x(t), which is called a homogeneous equation. As in the case of one equation, we want to find out the general solutions for the linear first order system of equations. To this end, we first have the following results for the homogeneous equation,
3 . PRINCIPLE OF SUPERPOSITION 3 Theorem.. Principle of Superposition Let x (t), bx (t),, x n (t) be n solutions of the homogeneous linear equation x (t) = A(t)x(t) on the open interval I. If c, c,, c n are n constants, then the linear combination is also a solution on I. c x (t) + c x (t) + c 3 x 3 (t) + + c n x n (t) Example.. Let x (t) = x(t) e t, x (t) = and x (t) = e t are two solutions, as (e bx (t) = t ) e t e t = = and [ bx (t) = [ (e t ) = [ e t = [ e t By the Principle of Superposition, for any two constants c and c e t c e x(t) = c x (t) + c x (t) = c + c e t = t c e t is also solution. We shall see that it is actually the general solution. The next theorem gives the general solution of linear system of equations, Theorem.3. - Let x (t), x (t),, bx n (t) be n linearly independent (as vectors) solution of the homogeneous system x (t) = A(t)x(t), then for any solution x c (t) there exists n constants c, c,, c n such that x c (t) = c x (t) + c x (t) + + c n x n (t). We call x c (t) the general solution of the homogeneous system.
4 4. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS If x p (t) is a particular solution of the nonhomogeneous system, x(t) = B(t)x(t) + b(t), and x c (t) is the general solution to the associate homogeneous system, x(t) = B(t)x(t) then x(t) = x c (t) + x p (t) is the general solution. Example.. Let [ 4 3 x 4t (t) = x(t) + + 5t 6 7 6t x(t) + 7t + 3e t e 5t, x (t) = e t and x (t) = 3e 5t are two linearly independent t solutions. and x p (t) = is a particular solution. By Theorem t.3, 3c e x(t) = c x (t) + c x (t) + x p (t) = t + c e 5t + t () c e t + 3c e 5t + t is the general solution. Now suppose we want to [ find a particular solution that satisfies the initial condition x() =, then let t = in (), we have x() = [ 3c + c = c + 3c which can be written in matrix form, [ 3 c = 3 c c Solve this equation, we get = c [ 3e solution is x(t) = t e 5t + t e t 3e 5t. + t, [,. So the particular From the above example, we can summarize the general steps in find a solution to initial value problem, { x (t) = A(t)x(t) + b(t) x(t ) = x
5 . HOMOGENEOUS SYSTEM 5 Step One: Find the general solution x c = c x (t) + c x (t) + + c n x n (t), where x (t), x (t),, x n (t) are a set of linearly independent solutions, to the associate homogeneous system, x (t) = A(t)x(t). Step Two: Find a particular solution x p (t)to the nonhomogeneous system, x (t) = A(t)x(t) + b(t). Step Three: Set x(t) = x c (t) + x p (t) and use the equation x(t ) = x, to determine c, c,, c n.. Homogeneous System We will use a powerful method called eigenvalue method to solve the homogeneous system x (t) = Ax(t) where A is a matrix with constant entry. We will present this method for A is either a or 3 3 cases. The method can be used for A is an n n matrix. The idea is to find solutions of form (3) x(t) = ve λt, a straight line that passing origin in the direction v. Now taking derivative on x(t), we have (4) x (t) = λve λt put (3) and (.) into the homogeneous equation, we get So x (t) = λve λt = Ave λt Av = λv, which indicates that λ must be an eigenvalue of A and v is an associate eigenvector... A is a matrix. Suppose a a A = a a Then the characteristic polynomial p(λ) of A is p(λ) = A λi = (a λ) (a λ) a a = λ (a +a )+(a a a a. So p(λ) is a quadratic polynomial of λ. From Algebra, we know that p(λ) = has either distinct real solutions, or a double solution, or conjugate complex solutions. The following theorem summarize the solution to the homogeneous system,
6 6. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Theorem.. Let p(λ) be the characteristic polynomial of A, for x (t) = Ax(t), Case : p(λ) = has two [ distinct real solutions [ λ and λ. v v Suppose v = and v v = are associate eigenvector (i.e, Av = λ v and Av = λ v ) Then the general v solution is And x c (t) = c v e λ t + c v e λ t v e Φ(t) = λ t v e λ t v e λ t v e λ t is called the fundamental matrix(a fundamental matrix is a square matrix whose columns are linearly independent solutions of the homogeneous system). Case : p(λ) = has a double solutions λ. In this case p(λ) = (λ λ ) and λ is a zero of p(λ) with multiplicity. () λ has [ two linearly independent eigenvectors: v v Suppose v = and v v = are associate linearly v independent eigenvectors. Then the general solution is And x c (t) = (c v + c v )e λ t Φ(t) = e λ t v v v v () λ has [ only one associate eigenvector: v Suppose v = is the only associated eigenvector and v v v = is a solution of v (λ I A)v = v. Then the general solution is, And x c (t) = (c v + c (tv + v )e λ t [ Φ(t) = e λ t v (v t + v ) v (v t + v ) is the fundamental solution matrix.
7 . HOMOGENEOUS SYSTEM 7 Case 3: p(λ) = has two [ conjugate complex solutions a+bi and a bi. v + iv Suppose v = is the associate complex eigenvector[ with respect to a [ + bi, then the general solution is, v + iv v v v = and v v = v x c (t) = [c (v cos(bt) v sin(bt))c (v cos(bt) + v sin(bt))e at. And [ Φ(t) = e at v cos(bt) v sin(bt) v cos(bt) + v sin(bt) v cos(bt) v sin(bt) v cos(bt) + v sin(bt) is the fundamental matrix. From Theorem., let Φ(t) be the fundamental [ matrix, the general c solution is given by x c (t) = Φ(t)c, with c = and the solution c that satisfies a given initial condition x(t ) = x is given by x(t) = Φ(t)Φ(t ) -t x Example.. [ Two distinct eigenvalues case Find the general 3 solution to x (t) = x(t) 5 Using Mathcad, functions eigenvals() and eigen- Solution vecs() In Mathcad, eigenvecs(m) Returns a matrix containing the eigenvectors. The nth column of the matrix returned is an eigenvector corresponding to the nth eigenvalue returned by eigenvals. we find,λ = [ and λ = 3 6 with associated eigenvectors v = and v [ = respectively. So the fundamental matrix is [ ( 7 6)e ( 3 Φ(t) = + 6)t ( 7 + 6)e ( 3 6)t e ( 3 + 6)t e ( 3 6)t c and the general solution is, for c =, c x c (t) = Φ(t)c
8 8. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Example.. One double eigenvalues with two linearly [ in-dependent eigenvectors Find the general solution to x (t) = x(t). Solution [ The eigenvalue is λ = and associated [ eigenvectors c e are and, so the general solution is x c = t c e t Example.3. One double eigenvalues with only one eigenvector Find the solution to x (t) = x(t) and x() = Solution Using Mathcad, functions eigenvals() and [ eigenvecs() we can find a double eigenvalue λ = 5 and eigenvector Notice, the symbolic operator (bring up by either [Shift[Ctrl[. or [Ctrl[.) will not work with eigenvecs() this time, but since multiply an eigenvector by a nonzero constant still get an eigenvector, we can 3 choose v =. 3 To [ find w that [ satisfies (A λ I)w = v λ we will solve (A w λ I) =. That is, w 3 3 w 3 = 3 3 w 3 One solution is w = and w = So the fundamental matrix is 3 3t + Φ(t) = e 5t 3 3t c and the general solution is, c =, c x c (t) = Φ(t)c 3 3() + Now, Φ() = e 5() = 3 3() [ 3 3 and Φ() - = Hence, the particular solution is x(t) = Φ(t)Φ() - x = e 5t [ 3t + 4 3t 3.
9 . HOMOGENEOUS SYSTEM 9 Example.4. Two conjugate [ complex eigenvalues case Find 3 the general solution to x (t) = x(t) Solution Using Mathcad, functions eigenvals() and eigenvecs() we find two conjugate complex eigenvalues, λ = + i 3 and λ = i [ 3 3 with associated eigenvector v = with respect to i λ. Compare this with the Theorem., we have a =, b = 3, v = 3, v =, v =, and v =. So the fundamental matrix is [ 3 cos(bt) sin(bt) Φ(t) = e t 3 sin(bt) cos(bt) and the general solution is, c = [ c c, [ 3 cos( 3t) sin( 3t) x c (t) = Φ(t)c = e t c 3 sin( 3t) cos( 3t) c [ = e t 3c cos( 3t) c sin( 3t) 3c sin( 3t) c cos( 3t) Suppose we want to find a solution such that x() =, then x(t) = Φ(t)Φ() - x() [ [ 3 cos( 3t) sin( 3t) = e t 3 [ 3 sin( 3t) cos( 3t) [ 3 cos( 3t) sin( 3t) 3 = e t 3 sin( 3t) cos( 3t) = e t [ cos( 3t) + + sin( 3t) sin( 3t) + cos( 3t).. A is a 3 3 matrix. Suppose A = a a a 3 a a a 3 a 3 a 3 a 33 Then the characteristic polynomial p(λ) of A given by p(λ) = A λi, - [
10 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS is a cubic polynomial of λ. From Algebra, we know that p(λ) = has either 3 distinct real solutions, or distinct solutions and one is a double solution, or one real solution and conjugate complex solutions, or a triple solution. The following theorem summarize the solution to the homogeneous system, Theorem.. Let p(λ) be the characteristic polynomial of A, for x (t) = Ax(t), Case : p(λ) = has three distinct real solutions λ, λ, and λ 3. Suppose v = v v, v = v v, and v 3 = v 3 v 3 v 3 are associate eigenvector (i.e, Av = λ v, Av = λ v, and Av 3 = λ 3 v 3 ) Then the general solution is v 3 x c (t) = c v e λ t + c v e λ t + c 3 v 3 e λ 3t And the fundamental matrix is Φ(t) = v e λ t v e λ t v e λ t v e λ t v 3 e λ 3t v 3 e λ 3t v 3 e λ t v 3 e λ t v 33 e λ 3t Case : p(λ) = has a double solutions λ. So p(λ) = (λ λ ) (λ λ ), and λ has multiplicity. Let v 3 = v v is the eigenvector associated with λ. v 3 [ λ has [ two linearly independent eigenvectors: v v Suppose v = and v v = are associate linearly v independent eigenvectors. Then the general solution is And x c (t) = (c v + c v )e λ t + c 3 v 3 e λ t Φ(t) = v e λ t v e λ t v e λ t v e λ t v 3 e λ t v 3 e λ t v 3 e λ t v 3 e λ t v 33 e λ t. v 33
11 . HOMOGENEOUS SYSTEM [ λ has one eigenvector: Suppose v = v v is the associated eigenvector with respect to λ and v = v v 3 v is a solution of v 3 (λ I A)v = v. Then the general solution is, x c (t) = (c v + c (tv + v ))e λ t + c 3 v 3 e λ And Φ(t) = v e λ t v e λ t (v t + v )e λ t (v t + v )e λ t v 3 e λ v 3 e λ v 3 e λ t (v 3 t + v 3 )e λ t v 33 e λ is the fundamental solution matrix. Case 3: p(λ) = has two conjugate complex solutions a ± bi and a real solution λ. Suppose v = v + iv v + iv is the associate complex eigenvector with respect to a + bi, then the general solution is, let v 3 + iv 3 v 3 = v 3 v 3, are associated eigenvectors with respect to λ, V 33 x c (t) = [c (v cos(bt) v sin(bt))c (v cos(bt)+v sin(bt))e at +c 3 v 3 e λ. Φ(t) = e at And v cos(bt) v sin(bt) v cos(bt) + v sin(bt) v 3 e λ v cos(bt) v sin(bt) v cos(bt) + v sin(bt) v 3 e λ v 3 cos(bt) v 3 sin(bt) v 3 cos(bt) + v 3 sin(bt) v 33 e λ is the fundamental matrix. Case 4: p(λ) = has solution λ with multiplicity 3. In this case, p(λ) = (λ λ ) 3. [ λ has three linearly independent eigenvectors. Let v = v v v 3, v = v v V 3, and v 3 = v 3 v 3 V 33 be the three linearly independent eigenvectors. Then the general
12 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS solution is x c (t) = (c v + c v + c 3 v 3 )e λt and fundamental v v v 3 v v V 3 matrix is Φ(t) = e λ t v 3 v 3 V 33 [ λ has two linearly independent eigenvectors. Suppose v = v v, v = v v are the linearly independent eigenvectors. Let v 3 = v v 3 V 3 3 v 3, then only one of the two equations, (A λ I)v 3 = v or (A λ I)v 3 = v can has a solution that is linearly independent with v, v. Suppose (A λ I)v 3 = v generates such a solution. Then the general solution is x c (t) = [c v + c v + c 3 (tv + v 3 )e λ t and fundamental matrix is Φ(t) = e λ t v v tv + v 3 v v tv + V 3 v 3 v 3 tv 3 + V 33 [3 λ has only one eigenvector. Let v = v v be the linearly independent eigenvectors. Let v = satisfies v 3 v v V 3 and v 3 = V 33 v 3 v 3 V 33 (A λ I)v = v and (A λ I)v 3 = v. be two vectors that Then the general solution is x c (t) = [c v + c (tv + v ) + c 3 (t v + tv + v 3 )e λ t and fundamental matrix is Φ(t) = e λ t v tv + v t v + tv + v 3 v tv + v t v + tv + V 3 v 3 tv 3 + v 3 t v 3 + tv 3 + V 33 Remark.. Suppose A is an n n matrix, for the homogeneous system x (t) = Ax(t), three general case would happen Case : A has n distinct eigenvalues λ i, i =,,, n with linearly independent eigenvectors v i, i =,,, n then the general solution will be x c (t) = c v e λ + c v e λ + + c n v n e λn
13 . HOMOGENEOUS SYSTEM 3 Case: A has m < n distinct eigenvalues, in this case some eigenvalues would have multiplicity greater than. Suppose λ r has multiplicity r. Depending on how many linearly independent eigenvectors are associated with λ r the situation could be very complex. Let p be the number of linearly eigenvectors associated with λ r, then d = r p is called the deficit of λ r. The simply cases are either d = or d = r. When < d < r the situation could be very complex. Suppose d = r and v is the only eigenvector associate with λ r, then one will have to solve r equations (A λ r ) i v i+ = v i, i =,,, r. And the general solution would contains terms like [c v + c (v t + v ) + c 3 (v t + v t + v 3 ) + + c r (v r + v t r + + v r )e λr. Case 3: A complex root a+bi with associated eigenvector v a +iv b, then the general solution contains term, [c (v a cos(bt) v b sin(bt))+ c (v a sin(bt) + v b cos(bt))e at. Remark.. Suppose x (t), x (t), x 3 (t),, x n (t) are n linearly independent solution for n n homogeneous system, x (t) = Ax(t), the fundamental matrix Φ(t) is a matrix whose columns are x i (t), i =,,, n. Example.5. (Two distinct eigenvalues) Find the general solution to x = 3x + 4x x 3 x = x + x 4x 3 x 3 = x + x Solution Let x(t) = x (t) x (t) x 3 (t) and The equations can be written in matrix form x (t) = Ax(t). Using Mathcad, functions eigenvals() and eigenvecs() we find,λ = and λ = with associated eigenvectors v = 4 and v = respectively. Since λ has multiplicity as appeared twice in the result of eigenvals() function, we need to solve the equation (A λ I)v 3 = v.
14 4. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS To use Mathcad, () you first compute (A λ I)v 3 using the following sequences of key stroke, [* type ([Ctrl[M, set the rows and columns in the matrix definition popup menu, input the data for A, [* type -[Ctrl[M set the row and column number and input data for λ I, [* type )[Ctrl[M, now set as column number, enter a, b, c in the place holders, [* type [Ctrl[. to compute symbolically and you get. () Using the Given Find block to find a solution. Type Given in a blank space, type a+b-c[ctrl= and a-4c[ctrl= in two rows, then type key word Find following by typing (a,b)[ctrl[. you will get the solution in terms of c. Set c =, we get v 3 = 4. So the fundamental matrix is Φ(t) = and the general solution is, 4et e t (t + 4)e t e t e t e t e t (t + )e t x c (t) = c v e t + c v e t + c 3 (tv + v 3 )e t Example.6. (One eigenvalue with deficit ) Find the solution to x (t) = 3 x(t) and x() = Solution Using Mathcad, functions eigenvals() (Notice the eigenvecs() will not find a good result in this case due to the rounding error.) we find, λ = is the only eigenvalue. To find the associate eigenvectors we compute (Using (A λ I)v = ) (A λ I)v = v v 4 v 3 v + v 3 v + v 3 = 4v v 3
15 . HOMOGENEOUS SYSTEM 5 We have only v + v 3 = for three variables v, v, v 3, this indicates that v can be any value, and set v = find v 3 =, So v = and v = are two eigenvectors. To find the generalize eigenvector associated with λ we will have to solve two equations (A λ I) w w =, w 3 and From (.), (A λ I) 4 w w w 3 = [ w w =, we get two inconsistent equations w + w 3 = and w + w 3 =. So now solution can be found in this case. From (.), 4 [ w w = we get one equation w + w 3 = choose w 3 = we get w =, since w can be anything, we set w =. So v 3 = and we can verify that v, v, and v 3 are linearly independent. So the fundamental matrix is Φ(t) = e t t t + t + and the general solution is,,, x c (t) = [c v + c v + c 3 (tv + v 3 )e t
16 6. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Now, Φ() = and Φ() = Hence, the particular solution is x(t) = [v + 3v e t 3. Example.7. (One eigenvalue with deficit ) Find the general solution to x (t) = 4 x(t). 3 Solution Using Mathcad, functions eigenvals() (Notice the eigenvecs() will not find a good result in this case due to the rounding error.) we find, λ = 3 is the only eigenvalue. To find the associate eigenvectors we compute (Using (A λ I)v = ) (A 3I)v = v v v 3 v + v + v 3 v + v + v 3 = v v We have only one eigenvector v =. To find the generalize eigenvector associated with λ we will have to solve two equations and From (.), (A 3I)v = v, (A 3I)v 3 = v, a b c =, we have two equations { b a = a + b + c = Choosing a =, we get b =, c =. Hence v =
17 . HOMOGENEOUS SYSTEM 7 From (.), a b c =, we have two equations { b a = a + b + c = Choosing a =, we get b =, c =. So v 3 = that v, v, and v 3 are linearly independent. So the fundamental matrix is Φ(t) = e t + t t + t t t t t + and the general solution is, and we can verify x c (t) = [c v + c (tv + v ) + c 3 (t v + tv + v 3 )e 3t Example.8. (Two conjugate complex eigenvalues case) Find the general solution to x (t) = 3 3 x(t) Solution Using Mathcad, functions eigenvals() and eigenvecs() we find two conjugate complex eigenvalues and one real eigenvalue, λ =, λ = + i 9, and λ3 = i 9 with associated eigenvector v = and v = + i 9 + i 9 = + 9 i 9 with respect to λ 3. Compare this with the Theorem.3,
18 8. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS we have a =, b = 9, v = 9 9 (imaginary part of v). The general solution is, x c (t) = c v e t +c (v cos( (real part of v), and v 3 = 9) v3 sin( 9))e t +c 3 (v sin( 9)+v3 cos( 9))e t 3. Nonhomogeneous System of Equations To find solutions to the initial value problem of nonhomogeneous equations x (t) = Ax(t) + b(t), x(t ) = x we follow the steps below, () Find the general solution x c (t) = Φ(t)c to homogeneous equation x (t) = Ax(t), where Φ(t) is the fundamental matrix. () Find a particular solution x p to x (t) = Ax(t) (3) The general solution to the nonhomogeneous equation x (t) = Ax(t) is x(t) = x c (t) + x p (t). Using x(t ) = x to determine the coefficient vector c. The following theorem gives one way to find a particular solution based on the fundamental matrix, Theorem 3.. Let Φ(t) be a fundamental matrix of x (t) = Ax(t), a particular solution to x (t) = Ax(t) + b(t) is given by x p (t) = Φ(t) Φ(t) - b(t) dt. Example 3.. Find the general solution to Solution Let x(t) = b(t) = t t x = 3x + 4x x 3 + t x = x + x 4x 3 x 3 = x + x t x (t) x (t) x 3 (t), A = 3 4 4, and. The equations can be written in matrix form x (t) =
19 3. NONHOMOGENEOUS SYSTEM OF EQUATIONS 9 Ax(t)+b(t). From Example.5, we know that the fundamental matrix to x(t) = Ax(t) is Φ(t) = 4et e t (t + 4)e t e t e t e t e t (t + )e t To find a particular solution, we first compute Φ - (t)b(t) = then we compute Φ(t) Φ - 5 (t)b(t) dt = Φ(t) t e t dt ( 5 t3 + 5 t )e t dt 5 t e t dt 5 = t + t t 3t t + 4t And so the general solution is, x(t) = c 4 e t + c + c 3 (t + 4 e t ) e t + 5 t + t t 3 5 t t t The following is a screen shot that shows how to carry out the computation in Mathcad, 5 t e t ( 5 t3 + 5 t )e t 5 t e t To use Mathcad, () Define fundamental matrix A(t) and b(t) in the same line (not as shown in graph), and compute in the next line A b(t) () type A(t)*[Ctrl[M choose column as, at each place holder, type [Ctrl[I to get the indefinite integral, (3) and put the corresponding entry of A b(t) in the integrant position. (4) press [Shift[Ctrl[. type key work simplify
20 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Theorem 3.. If Φ(t) is the fundamental matrix for x (t) = Ax(t),, and x p (t) = Φ - (t)b(t) dt, then x(t) = Φ(t)Φ - (t )(x x p (t )) + x p (t) is the solution to the nonhomogeneous initial value problem, x (t) = Ax(t) + b(t), x(t ) = x Example 3.. Find the solution to x (t) = e t e t and x() = [ x(t) + Solution From Example.3 is the fundamental matrix is 3 3t + Φ(t) = e 5t 3 3t and Φ() - = e Now b(t) = t e t, using the formula x p (t) = Φ(t) Φ - (t)b(t) dt and Mathcad, we have x p (t) = 3 et Therefore, Φ() - (x() x p ()) = ( [ and the solution is x(t) = Φ(t)Φ() - (x() x p ()) + x p (t) = e 5t [ [ 3 ) = [ t 6 4t + 3 e 3t 4. Higher order differential equations One can transform equations that involving higher order derivatives of unknown functions to system of first order equations. For example, suppose x(t) is an unknown scalar function that satisfies mx (t) + cx (t) + kx(t) = f(t) an equation can be used to model a spring system with external force f(t) or an RCL electronic circuit with an energy source f(t).
21 4. HIGHER ORDER DIFFERENTIAL EQUATIONS Now if we set x (t) = x(t) and x (t) = x (t) we then get an system of first order equations (5) x (t) = x (t) (6) x (t) = c m x (t) k m x (t) + f(t) m In general, if we have an differential equation that involving nth order derivative x (n) (t) of unknown function x(t), x (n) = a x(t) + a x (t) + + a n x (n ) + f(t), we can transform it into an system of first order equations of n unknown functions x (t) = x(t), x (t) = x (t), x 3 (t) = x () (t),, x n (t) = x (n ) (t), and using the eigenvalue method for system of differential equation to solve the higher order equation. Example 4.. Transform the differential equation x (3) + 3x () 7x (t) 9x = sin(t) into system of first order equations. Solution Here the highest order of derivative is third derivative x (3) of x(t). So we transfer it into system of 3 equations. Let x (t) = x(t), x (t) = x (t), x 3 (t) = x (t), we have (7) (8) x (t) = x (t) x (t) = x 3 (t) (9) x 3(t) = 3x 3 (t) + 7x (t) + 9x (t) sin(t) Let x(t) = x (t) x (t), A =, and b(t) = x 3 (t) f(t) we can write the system of equation in matrix form x (t) = Ax(t) + b(t). Example 4.. Find the general solution for the 3rd order differential equation x (3) + 3x () 7x (t) 9x = sin(t). Solution From previous example, Example 4., Let x(t) = x (t) x (t), A =, and b(t) = we can write x 3 (t) f(t) the system of equation in matrix form x (t) = Ax(t) + b(t). Using Mathcad we find the eigenvalues are λ =, λ = +, λ 3 = with associate eigenvectors, v =, v = +,
22 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS and v 3 = + respectively (after multiply the results of Mathcad by some constants). So the fundamental matrix is Φ(t) = e t e ( + )t e (+ )t e t ( + )e ( + )t ( + )e (+ )t e t ( )e ( + )t ( + )e (+ )t From Φ(t) we find a particular solution 3 x p (t) = Φ(t) Φ - 5 (t)b(t) dt = cos(t) 5 sin(t) cos(t) sin(t) cos(t) 8 sin(t) 5 39 Hence the general solution to the system is x (t) x (t) = x 3 (t) c e t + c e ( + )t + c 3 e (+ )t cos(t) 5 39 sin(t) c e t + c ( + e ( + )t c 3 ( + )e (+ )t cos(t) 56 sin(t) c e t + c ( )e ( + )t + c 3 ( + )e (+ )t 3 5 cos(t) 8 39 sin(t) and x (t) = c e t + c e ( + )t + c 3 e (+ )t + 3 cos(t) 5 sin(t) is 5 39 the general solution to the third order ordinary differential equation x (3) + 3x () 7x (t) 9x = sin(t). Example 4.3. Find the solution to the initial value problem x x + 9x = te t, x() =, x () =. Solution Since the given equation is of second order, we will have two unknowns x (t) = x(t), x (t) = x (t) to transform the equation into a system of first order equations, x (t) = x x (t) = x 9x + te t, and the initial conditions [ are x () = [ x() = x () = x () = [. x (t) Now let x(t) =, A =, and b(t) = x (t) 9 te t. We have the matrix version of this equation, x (t) = Ax(t) + b(t)
23 4. HIGHER ORDER DIFFERENTIAL EQUATIONS 3 Using Mathcad, we find[ the eigenvalues [ λ =, λ = 9, and associate eigenvectors v =, v =. And fundamental 9 e t e matrix Φ(t) = 9t e t 9e 9t. From Φ(t) we find a particular solution x p (t) = Φ(t) Φ - (t)b(t) dt = 5 (3t + 8t + )e t 5 (3t + 7t + 9)e t The solution with initial values x = is given by x(t) = Φ(t)Φ ( - ()(x b()) + ) b(t) = 6 t ( 6 t e t 7 5 e9t ) e t 43 5 e9t Hence the solution to the initial value ( problem of the) second order differential equation is x(t) = x (t) = 6 t 639 t+ e t e9t. Project At beginning you should enter: Project title, your name, ss#, and due date in the following format Project One: Define and Graph Functions. John Doe SS# -- Due: Mon. Nov. 3rd, 3 You should format the text region so that the color of text is different than math expression. You can choose color for text from Format >Style select normal and click modify, then change the settings for font. You can do this for headings etc. () Solutions To System of Equations Finding solution to linear system using Mathcad and study the long time dynamic behavior of the solutions. Find general solution to { x = y y = x
24 4. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Plot several solutions with different initial values in [- xt-plane, yt-plane xy-plane. Here you will need to define range variable t =,. 7 and set X := x(t), Y := y(t). The graph in xy-plane is called the trajectory. If this models the movement of a satellite, what is its trajectory. Find general solution to { x = 8y y = 8x Plot several solutions with different initial values in [- xt-plane, yt-plane xy-plane. Here you will need to define range variable t =,. 7 and set X := x(t), Y := y(t). The graph in xy-plane is called the trajectory. If this models the movement of a satellite, what is its trajectory. Find general solution to { x = x y y = y 3x Plot several solutions with different initial values in [- xt-plane, yt-plane xy-plane. Here you will need to define range variable t =,. 7 and set X := x(t), Y := y(t). The graph in xy-plane is called the trajectory. If this models the movement of a system of two species, what is you conclusion about interdependency of these species? Can you find initial value such that x(t) = (distinct) for some t? what about y(t). () Solution of Higher order equation In general mx + cx + kx = f(t) models a object with mass m attached to a spring with constant k and damping force that is proportional to the velocity x, c, k >. Suppose m = and f(t) = Ae at sin(bt), that is the external force is oscillatory (b > ) and diminishing (a > ) Find solutions and graph the solutions. - c = b = Find general solution and graph some particular solutions. - c =, k =, a =, b =, A = 4 - c =, k = 3, A =, a =, b =
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# 11-Plus Maths Quiz - Addition and Subtraction 3 (Easy) (Questions)
Welcome to the third of our Easy Eleven Plus maths quizzes on addition and subtraction. You may think that these questions are very simple, and you can do them in your head. You may be right, but it never hurts to practice your mental arithmetic skills! You’ll also find many hints and tips which will help you in your maths lessons, in real life situations, and in your 11+ exams.
One way to improve your addition and subtraction skills is by rounding numbers up or down. For example:
• To subtract 78, first subtract 80 then add 2
Can you see how that idea works? It’s a quick way to solve problems which may appear more complicated than they are.
Now on to the quiz. If you haven’t already played our previous ones on addition and subtraction, then go back and try them all until you always score a perfect ten-out-of-ten!
1. 22 - 38 + 45 = ?
[ ] 29 [ ] 28 [ ] 30 [ ] 31
2. 47 + 15 + 73 = ?
[ ] 125 [ ] 135 [ ] 145 [ ] 155
3. 6 - 12 + 27 + 33 = ?
[ ] 58 [ ] 56 [ ] 54 [ ] 52
4. 34 + 0 + 58 + 17 - 0 + 5 = ?
[ ] 110 [ ] 124 [ ] 104 [ ] 114
5. 27 - 68 + 59 - 9 = ?
[ ] 9 [ ] 11 [ ] -9 [ ] 7
6. 23 + 37 + 53 = ?
[ ] 103 [ ] 113 [ ] 93 [ ] 123
7. 74 - 98 + 29 = ?
[ ] 201 [ ] -5 [ ] 5 [ ] 15
8. 15 + 15 - 15 + 15 = ?
[ ] 60 [ ] 15 [ ] 45 [ ] 30
9. 54 - 27 = ?
[ ] 26 [ ] 27 [ ] 28 [ ] 29
10. 26 + 47 = ?
[ ] 63 [ ] 83 [ ] 53 [ ] 73
1. 22 - 38 + 45 = ?
[x] 29 [ ] 28 [ ] 30 [ ] 31
TIP: With numbers less than 100, add the tens first and then add the units.
22 - 38 + 45 = (20 + 40 - 30) + (2 + 5 - 8) = 30 + -1 = 29.
Remember: if you are adding a minus number, that’s the same as subtracting a positive number
2. 47 + 15 + 73 = ?
[ ] 125 [x] 135 [ ] 145 [ ] 155
TIP: With numbers less than 100, add the tens first and then add the units.
47 + 15 + 73 = (40 + 10 + 70) + (7 + 5 + 3) = 120 + 15 = 135
You may have noticed another shortcut: 7 + 3 = 10 so 47 + 73 = 40 + 70 + 10
3. 6 - 12 + 27 + 33 = ?
[ ] 58 [ ] 56 [x] 54 [ ] 52
TIP: With numbers less than 100, add the tens first and then add the units.
6 - 12 + 27 + 33 = (20 + 30 - 10) + (6 + 7 + 3 - 2) = 40 + 14 = 54
Be on the lookout for simple calculations that make your life easier! 27 + 33 is one of these
4. 34 + 0 + 58 + 17 - 0 + 5 = ?
[ ] 110 [ ] 124 [ ] 104 [x] 114
TIP: With numbers less than 100, add the tens first and then add the units.
34 + 0 + 58 + 17 - 0 + 5 = (30 + 50 + 10) + (4 + 8 + 7 + 5) = 90 + 24 = 114. Ignore zeros!
5. 27 - 68 + 59 - 9 = ?
[x] 9 [ ] 11 [ ] -9 [ ] 7
TIP: With numbers less than 100, add the tens first and then add the units.
27 - 68 + 59 - 9 = (20 + 50 - 60) + (7 + 9 - 8 - 9) = 10 + -1 = 9
Remember: if you are adding a minus number, that’s the same as subtracting a positive number
6. 23 + 37 + 53 = ?
[ ] 103 [x] 113 [ ] 93 [ ] 123
TIP: With numbers less than 100, add the tens first and then add the units.
23 + 37 + 53 = (20 + 30 + 50) + (3 + 7 + 3) = 100 + 13 = 113
Don't forget: sometimes you will spot an easy addition which will make the calculation easier, e.g. 23 + 37 = 60
7. 74 - 98 + 29 = ?
[ ] 201 [ ] -5 [x] 5 [ ] 15
TIP: With numbers less than 100, add the tens first and then add the units.
74 - 98 + 29 = (70 + 20 - 90) + (4 + 9 - 8) = 0 + 5 = 5
A quick way to subtract 98 is to subtract 100 then add 2
8. 15 + 15 - 15 + 15 = ?
[ ] 60 [ ] 15 [ ] 45 [x] 30
That was an example of a question you must read thoroughly - you didn't think all the 15s were to be added did you?
9. 54 - 27 = ?
[ ] 26 [x] 27 [ ] 28 [ ] 29
TIP: With numbers less than 100, add the tens first and then add the units.
54 - 27 = (50 - 20) + (4 - 7) = 30 + -3 = 27
Another way to do this is to subtract 30 then add 3: 54 - 30 = 24. 24 + 3 = 27
10. 26 + 47 = ?
[ ] 63 [ ] 83 [ ] 53 [x] 73
TIP: With numbers less than 100, add the tens first and then add the units.
26 + 47 = (20 + 40) + (6 + 7) = 60 + 13 = 73
Another way to add 47 is to add 50 then subtract 3: 26 + 50 = 76. 26 - 3 = 73 |
# how to find end behavior of a function
Identify the degree of the function. Even and Negative: Falls to the left and falls to the right. y =0 is the end behavior; it is a horizontal asymptote. End Behavior Calculator. 3.If n > m, then the end behavior is an oblique asymptoteand is found using long/synthetic division. 1.3 Limits at Infinity; End Behavior of a Function 89 1.3 LIMITS AT INFINITY; END BEHAVIOR OF A FUNCTION Up to now we have been concerned with limits that describe the behavior of a function f(x)as x approaches some real number a. These turning points are places where the function values switch directions. In this section we will be concerned with the behavior of f(x)as x increases or decreases without bound. First, let's look at some polynomials of even degree (specifically, quadratics in the first row of pictures, and quartics in the second row) with positive and negative leading coefficients: End behavior of polynomial functions helps you to find how the graph of a polynomial function f(x) behaves (i.e) whether function approaches a positive infinity or a negative infinity. The domain of this function is x ∈ ⇔ x ∈(−∞, ∞). The function has a horizontal asymptote y = 2 as x approaches negative infinity. The right hand side seems to decrease forever and has no asymptote. How To: Given a power function f(x)=axn f ( x ) = a x n where n is a non-negative integer, identify the end behavior.Determine whether the power is even or odd. Find the End Behavior f(x)=-(x-1)(x+2)(x+1)^2. Since both ±∞ are in the domain, consider the limit as y goes to +∞ and −∞. In addition to end behavior, where we are interested in what happens at the tail end of function, we are also interested in local behavior, or what occurs in the middle of a function.. 1. Recall that we call this behavior the end behavior of a function. The end behavior is when the x value approaches $\infty$ or -$\infty$. Use arrow notation to describe the end behavior and local behavior of the function below. Horizontal asymptotes (if they exist) are the end behavior. To find the asymptotes and end behavior of the function below, examine what happens to x and y as they each increase or decrease. The slant asymptote is found by using polynomial division to write a rational function $\frac{F(x)}{G(x)}$ in the form Show Solution Notice that the graph is showing a vertical asymptote at $x=2$, which tells us that the function is undefined at $x=2$. There is a vertical asymptote at x = 0. However horizontal asymptotes are really just a special case of slant asymptotes (slope$\;=0$). Local Behavior. Determine whether the constant is positive or negative. One of the aspects of this is "end behavior", and it's pretty easy. Use the above graphs to identify the end behavior. 2. ... Use the degree of the function, as well as the sign of the leading coefficient to determine the behavior. EX 2 Find the end behavior of y = 1−3x2 x2 +4. Even and Positive: Rises to the left and rises to the right. The point is to find locations where the behavior of a graph changes. 4.After you simplify the rational function, set the numerator equal to 0and solve. 2. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, ${a}_{n}{x}^{n}$, is an even power function, as x increases or decreases without … 2.If n = m, then the end behavior is a horizontal asymptote!=#$%&. There are three cases for a rational function depends on the degrees of the numerator and denominator. We'll look at some graphs, to find similarities and differences. 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Future Study Point
Scaler and Vector Quantities Class 11 Physics Chapter 4 CBSE
All the physical quantities are needed to represent two things numerical value and their unit, as example mass of five kilograms is written as 5 kg, so all physical quantities are represented by two terms one is a numerical value and the other is a unit. A physical quantity(P.Q) is the product of a numerical value and its unit. The physical quantities are of two types (i) Scaler Quantity (ii)Vector Quantity
Scaler Quantity: The type of quantity which are needed to represent the magnitude only as example teperature, a room temperature of 5°C gives complete information about the temperature of the room.
Vector Quantity: The vector Quantity is the physical quantity which is needed to represent by the magnitude as well as by the direction,as example, a car is moving towards the west at a speed of 50 km/h, then the given speed with direction is not actually speed,it is a velocity. The speed is the scaler quantity and velocity is the vector quantity.
Scaler Quantity Vector Quantity Represented by numerical value Represented by the numerical value and direction Two or more similar scaler quantities are added according to the ordinary rule of algebra These are added according to the vector law of addition It is one dimensional It is multi-dimensional Its quantity changes as the magnitude changes Its quantity changes as the magnitude and direction changes Examples are speed, mass, temperature, time etc Examples are velocity,weight,dislacement,force etc
Why does Current as well as Pressure have the direction but these are not vector quantity?
Current and pressure don’t follow the law of vector addition as an example if 2A and 5A currents are flowing through two wires,if both wires are connected to the same wire the current on the later wire is always 2+5=7A, and the resultant of both will not change with the changing angle between both the wires, similarly in case of the two pressure values are directed to a point then the resultant will be scaler sum of both and will not change with the angles between both pressure values.
Representation of the vector quantity:Let a physical quantity A act in a particular direction,then it is represented by $\fn_cm \overrightarrow{A}$
$\fn_cm \overrightarrow{A}$ shows the magnitude of A with the direction while
$\fn_cm \left | \overrightarrow{A} \right |$ shows the magnitude of vector A in short mod A, sometimes also can be written simply A
Representation of the vectors:
The vector (A) 5 m/s 30° north of east is represented as follows
$\fn_cm \overrightarrow{A}$ = 5 m/s 30° north of east
Important Points related to Vectors:
(i) On rotating 2π radian vector doesn’t change.
(ii)A vector can shift parallel to itself,it is known as parallel shifting of the vector provided its size is same.
(iii)The angle between two vectors is the smaller of two angles between the vectors when they are joined tail to tail ,0°≤θ≤180°
The angle between two vectors can’t be more than 180°
The angle between two vectors is the angle between them when connected tail to tail
In the first figure the angle between A and B is 120° but in the second figure the angle between A and B is 60°
Types of the vectors:
Equal vectors: Two vectors of the same physical quantities are called equal vectors when their magnitudes are equal
Let we have two vectors $\fn_cm \overrightarrow{A}$ and $\fn_cm \overrightarrow{B}$ such that
$\fn_cm \overrightarrow{A}$ = 1 m/s east and $\fn_cm \overrightarrow{B}$= 1m/s west
The vector A and Vector B are not equal because their magnitudes are the same but directions are different
So we can say
$\fn_cm \left | \overrightarrow{A} \right |=\left | \overrightarrow{B} \right |$
$\fn_cm \overrightarrow{A}\neq \overrightarrow{B}$
If $\fn_cm \overrightarrow{A}$ = 1 m/s east and $\fn_cm \overrightarrow{B}$= 1m/s east
Here vector A and vector B are equal because their magnitude and directions are the same,therefore
$\fn_cm \left | \overrightarrow{A} \right |=\left | \overrightarrow{B} \right |$
Negative of a vector: When the magnitudes of the two vectors are same but their directions are opposite to each other then they are negative of each other.
Let there are two vectors $\fn_cm \overrightarrow{A}$ and $\fn_cm \overrightarrow{B}$, vector A is equal to vector B in magnitude but have opposite directions ,as an example
If $\fn_cm \overrightarrow{A}$ = 1 m/s east and $\fn_cm \overrightarrow{B}$= 1m/s west
$\fn_cm \left | \overrightarrow{A} \right |=\left | \overrightarrow{B} \right |,\overrightarrow{A} =-\overrightarrow{B}$
Parallel and Antiparallel Vectors: Two vectors which has 0° angle between them are known as parallel vectors and two vectors which has 180° between them are known as antiparallel vectors.
Collinear and Coplaner vectors:The vectors which lies on the same line or same path are known as collinear vectors and the vectors which lies on the same plane.
Two vectors always lie on the same plane means they are always coplaner
Three vectors may be coplanar or non coplanar
Concurrent Vectors: Two or more vectors intersecting at the same point or two or more than two vectors whose lines of actions intersects at the common point are known as concurrent vectors.
As an example 2 or more forces acts on an object are concurrent vectors.
Unit Vector:A vector whose magnitude is 1 known as init vector,unit vector is very important for resolving many problems in physics of vectors.
$\fn_cm \left | \hat{A} \right | =1$
A unit vector is represented by $\fn_cm \hat{A}$,a unit vector shows the direction of a vector.
A unit vector is evaluated as follows
$\fn_cm \hat{A}=\frac{\overrightarrow{A}}{\left | \overrightarrow{A} \right |}$
Null Vector or Zero Vector: The vector which is of 0 magnitudes and has an arbitrary direction
Circular Motion: Angular velocity and angular displacement
Electronic Configuration of s,p and d orbitals
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## Saturday, 16 August 2008
### Relationship between Angles of Triangle
In a triangle, there are some angles or sub-divided angles within it which relates to each other in some way. The concept of their relationship is simple.
This post is to highlight the fact that once the principle of angles (in a triangle) is understood, the relationship between the angles become clear and simple.
Let's begin.
In the triangle above, there are a few angles labelled. We will look into their relationship.
(Before we start, we need to know that all angles within a triangle add up to 1800.)
Angle A and B form a right-angle corner.
Therefore, A + B = 900.
And A = 900 - B. ---(1)
Also Angle A = Angle C.
Why?
Since angles B, C is part of a right-angled triangle, ==> B + C = 900.
Therefore, C = 900 - B. ---(2)
From the relation (1) and (2), it proved that Angle A = C.
With the same reasoning, Angle B= Angle D can be deduced easily.
If you follow the reasoning of above, you will see that finding the relationship of angles within a right-angled triangle is a breeze.
Maths is not difficult if you understand its principles.
:D |
# Arrange the numbers 1 to 19 in the circles
O---o---O
/ \ / \
o o o o
/ \ / \
O---o---O---o---O
\ / \ /
o o o o
\ / \ /
O---o---O
Arrange the numbers 1 to 19 in the circles, so all the rows of 3 numbers between O (big o) sums to 23
Example :
a---b---c
/ \ / \
d e f g
/ \ / \
h---i---j---k---l
\ / \ /
m n o p
\ / \ /
q---r---s
a+b+c = a+e+j = c+f+j = 23, and so on....
Note :
• This puzzle is similar with this site, but with different sums
• There are 2 solutions, if we rule out reflections and rotations.
The sum from 1-19 equals 19*20/2 or 190.
If we examine the 12 sums, one for each line, it is observed that the 6 numbers on the vertices is used 3 times, and the number in the center is used 6 times.
So we have,
(sum of twelve numbers at the middle of the rows) + 3*(sum of six numbers at the vertices) + 6*(number at the center) = 23*12 = 276
Or,
2*(sum of six numbers at the vertices) + 5*(number at the center) = 86
The first term is always even, the number at the center must also be even. The sum of six numbers at the vertices is at least 1+2+3+4+5+6 = 21, so the number at the center can be at most 8. The possible numbers are 2,4,6 and 8.
Also, the number 19 has to be placed between 1 and 3, since it's the only way possible to make 23.
If the central number is 2, we know for sure that 1 isn't on a vertex, since that requires 20 between 1 and 2. Hence we have a contradiction.
If the central number is 4, the sum of the six numbers on the vertices must be 33. Since we need 1-x-3 (to place 19) on the outer edge, supposed 2 is not on a vertex, we must have a pair with the sum 21 to accompany 2. The maximum number we can have is 14 since 1+3+4+5+6+14=33, which implies that such pair cannot exist or the sum of the numbers on the vertices will exceed 33. So, 2 must be on a vertex. 2 cannot be adjacent to 1 (need 20 in between) or 3 (need 18 in between which is already taken in 1-18-4) so 2 must be either to the opposite of 1 or 3.
Suppose 5 is on a vertex, we have 33-(1+2+3+5) = 22 must be the sum of 2 non-adjacent numbers on the vertices (say x and y). 5 cannot be adjacent to 1 (5-17-1, 4-17-2). Neither x or y can be adjacent to 1 because x+y+1 = 23. Hence 5 is not on a vertex. 5 cannot be on the "radius" because then 14 must be on a vertex and it is impossible to fill in the blanks with distinct numbers. If 5 is on the outer edge, there exists a pair of number on adjacent vertices with the sum 18. Also 9 must also be on a vertex since 33-(1+2+3+18)=9, which is again, impossible to fill.
If the central number is 6, the sum of the six numbers on the vertices must be 28. Since we need 1-x-3 (to place 19), either 1-x-4 or 2-x-3 (to place 18) and either 1-x-5 or 2-x-4 (to place 17). So we either have either 2-17-4-18-1-19-3 on the outer edge,
which gives:
1--19---3
/ \ / \
18 16 14 12
/ \ / \
4--13---6---9---8
\ / \ /
17 15 7 5
\ / \ /
2--11---10
or 1-19-3-18-2-17-4, which gives:
1--19---3
/ \ / \
12 16 14 18
/ \ / \
10--7---6---15--2
\ / \ /
5 9 13 17
\ / \ /
8---11--4
The string 5-17-1-19-3-18-2 is also possible, however it is impossible to fill the rest of the numbers.
If the central number is 8, the sum of the six numbers on the vertices must be 23. It is only possible with 1,2,3,4,6,7. This is impossible since the number between 7(vertex) and 8(center) must be 23-7-8 = 8 which is a repeat.
Since all the possible cases are covered, it can be concluded that these are the only 2 distinct solutions.
• I have to post via phone so somebody please clear up the formatting mess for me. Thank you in advance. Oct 19, 2016 at 8:54 |
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# Peter's Third String
Achievement Objectives:
Specific Learning Outcomes:
Determine the maximum area of a rectangle with a given perimeter
Consider how the area of a quadrilateral changes as its shape changes
Interpret a relationship from a graph
Description of mathematics:
Note that this is the same problem as we used in Peter’s Second String, Level 5 but at this level we expect students to have a more sophisticated way to solve the problem.
The extension problem here is quite difficult. If the students can make progress with any quadrilateral, then they have done well. The method shown here may not be the easiest way to solve the problem. If your class comes up with a better approach we would like to see it so that we can add it to this problem.
To be able to do this problem students need to be able to measure lengths and calculate the perimeters and areas of rectangles using the formulae: perimeter = twice length plus twice width and area = length x width. In addition, it would help if they have tried Peter’s Second String, Level 5 and have seen how to use tables.
Apparently in some areas of New Guinea they measure the area of land by its perimeter. When you think about it this isn’t such a good idea. A piece of land can have a relatively large perimeter and only a small area. This sequence of problems is built up from this simple bad idea.
Seven problems have been spawned by the perimeter-area tangle. These come in two waves. First there is the string of Peter’s String problems. These are Peters’ String, Measurement, Level 4, Peters’ Second String, Measurement, Level 5, Peters’ Third String, Algebra, Level 6, The Old Chicken Run Problem, Algebra, Level 6 and the Polygonal String Problem, Algebra, Level 6. These follow through on the non-link between rectangles’ areas and perimeters, going as far as showing that among all quadrilaterals with a fixed perimeter, the square has the largest area. In the second last of these five problems we are able to use an idea that has been developed to look at the old problem of maximising the area of a chicken run. This is often given as an early application of calculus but doesn’t need more than an elementary knowledge of parabolas. The final problem looks at the areas of regular polygons with a fixed perimeter. We show that they are ‘bounded above’ by the circle with the same perimeter.
The second string of lessons looks at the problem from the other side: does area have anything to say about perimeter? This leads to questions about the maximum and minimum perimeters for a given area. The lessons here are Karen’s Tiles, Measurement, Level 5 and Karen’s Second Tiles, Algebra, Level 6.
Mathematics is more than doing calculations or following routine instructions. Thinking and creating are at the heart of the subject. Though there are some problems that have a set procedure or a formula that can be used to solve them, most worthwhile problems require the use of known mathematics (but not necessarily formulae) in a novel way.
Throughout this web site we are hoping to motivate students to think about what they are doing and see connections between various aspects of what they are doing. The mathematical question asked here is what can we say about the rectangle of biggest area that has a fixed perimeter? This question is typical of a lot of mathematical ones that attempt to maximise quantities with given restrictions. There are obvious benefits for this type of maximising activity.
The ideas in this sequence of problems further help to develop the student’s concept of mathematics, the thought structure underlying the subject, and the way the subject develops. We start off with a piece of string and use this to realise that there is no direct relation between the perimeter of a rectangle and its area. This leads us to thinking about what areas are possible. A natural consequence of this is to try to find the largest and smallest areas that a given perimeter can encompass. We end up solving both these problems. The largest area comes from a square and the smallest area is as small as we like to make it.
Some of the techniques we have used to produce the largest area are then applied in a completely different situation – the chicken run. This positive offshoot of what is really a very pure piece of mathematics initially, is the kind of thing that frequently happens in maths. Somehow, sanitised bits of mathematics, produced in a pure mathematician’s head, can often be applied to real situations.
The next direction that the problem takes is to turn the original question around. Don’t ask given perimeter what do we know about area, ask given area what do we know about perimeter. Again there seems to be no direct link.
But having spent time with rectangles, the obvious thing to do is to look at other shapes. We actually look at polygons and their relation with circles but there is no reason why you shouldn’t look at triangles or hexagons. Here you might ask whether you can find two triangles with the same area and perimeter or what is the triangle with given area that has maximum perimeter. We have actually avoided these last two questions because of the difficulty of the maths that would be required to solve them. However, we may have got it wrong. There may be some nice answers that are relatively easy to find. If so, please let us know.
Required Resource Materials:
pieces of string of various lengths
ruler
squared paper or graph paper
Copymaster of the problem (English)
Copymaster of the problem (Māori)
Activity:
### The Problem
Peter had kept a piece of string that had been on a parcel that had come for his birthday. It was 30 cm long. He played with it and made different shapes out of it. Then he got stuck on rectangles. He wasn’t sure but he thought that the rectangle with the biggest area that he could make was a square. His sister Veronica said that was crazy but she didn’t have a good reason for saying that. Who was right and why?
### Teaching sequence
1. Introduce the problem to the class. Get them to consider how they would approach the problem.
2. Let them investigate Peter’s conjecture in groups using any approach that they want. (Although you might decide to move them in a more sophisticated direction.) At some stage though they will probably have to write down some equations. They may need some help at this point.
3. Move round the groups as they work to check on progress. Encourage them to set up some equations and reduce the number of dependent variables to one.
4. If a lot of the pairs are having problems, then you may want a brainstorming session to help them along.
5. Share the students’ answers. Get them to write up their work in their books. Make sure that they have carefully explained their arguments.
6. Encourage the more able students to try the Extension Problem. You may want to give them a few days to think about it.
#### Extension to the problem
If Peter made any quadrilateral shape with his string, would the maximum area he could get be produced by a square?
### Solution
We have already done this problem using a table in Peter’s Second String, Level 5. But at this level students should begin to see how to use algebra more effectively in such problems so we would encourage you to move them in this direction here.
What do we know? Well if we make Peter’s string into a rectangle with side lengths L and W, then 2L + 2W = 30 or L + W = 15. This gives us W = 15 – L.
Then we know that LW = A, where A is the area of the rectangle. So we can substitute for W into this equation and at least arrive at an expression for A that only involves the one variable L. This expression is A = L(15 – L). Now we can graph this function because it’s a parabola. And we know from past experience that (i) it is has a maximum value because the coefficient of L2 is negative; and (ii) it crosses the L-axis when L = 0 and L = 15.
But parabolas have their axis of symmetry and maximum point, midway between the points where they cross the L-axis. So the axis of symmetry of the parabola above is at L = 7.5. This is also where its maximum point is.
But if L = 7.5, then W = 15 – L = 15 – 7.5 = 7.5. So W = L and the rectangle has to be a square.
#### Solution to the extension
We do this in easy steps. First the rectangle of perimeter 30 cm with biggest area is a square. This has been done in the main problem. Second, a parallelogram with no internal angles of 90 has a smaller area than a rectangle with the same side lengths. To see this, consider the parallelogram and the rectangle below. Clearly the rectangle has area bc. The parallelogram has area base x height. Since its height is less than c, its area is less than bc. So the parallelogram has a smaller area than the rectangle even though they both have the same perimeters.
Third, a quadrilateral with an exterior angle less than 180° has a smaller area than a corresponding one with all exterior angles bigger than 180°. To see this, look at the diagram below. Clearly the quadrilateral on the right has the largest area yet they both have the same perimeters.
Fourth, the quadrilateral on the right above has smaller area than a quadrilateral whose diagonals are perpendicular. To see this, imagine that the points A and B are fixed and that A to C to B is a piece of string.Place a pencil inside the string at C. Move the pencil keeping AC and BC straight (taut). What is the biggest area of the triangle ABC? Well as C moves, the base AB remains the same. So the biggest area is found when the height of the triangle is the largest. This occurs when the perpendicular from C to AB goes through the midpoint of AB. Call this position for C, K. (Can you see that AK = KB?)
Notice that the quadrilateral formed by AKBD has the same perimeter as ACBD but AKBD has the bigger area.
We can do exactly the same thing with the point D. The position of D which makes triangle ABD have the biggest area is when D is above the midpoint of AB. Call this point L. Then the quadrilateral AKBL has the same perimeter as ABCD but it has a bigger area. What’s more, KL is perpendicular to AB.
Fifth, we want to show that among all quadrilaterals with the same perimeter and with perpendicular diagonals, the one with the biggest area is a parallelogram. Consider the quadrilateral below.
By the argument of ‘Four’, we can assume that AB = AD and that BC = CD. Now repeat the argument of ‘Four’ using B as the point to put the pencil and A and C as the fixed points. The argument we used above then shows that we can move B to a place above the midpoint of AC and in the process increase the area of triangle ABC. Repeating this on triangle ACD, we see that the area of this triangle is maximised when D is above the midpoint of AC. So we have a quadrilateral whose area is bigger than the original quadrilateral ABCD.
In this new quadrilateral, AB = BC = CD = DA. From here it is easy to show that opposite angles are equal and so the quadrilateral is a parallelogram (with equal sides – a rhombus).
Now lets recap. We can show that among all rectangles with a given perimeter, the one with biggest area is a square. Then any other quadrilateral can be changed into another quadrilateral with the same perimeter and bigger area, using ‘Three’ and/or the ‘pencil’ approach of ‘Four’. What’s more this quadrilateral has to be a parallelogram. But we know that for every parallelogram with a given perimeter there is a rectangle with the same perimeter and a larger area. Hence among all quadrilaterals with a given perimeter, the square is the one with the biggest area.
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## The Old Chicken Run Problem
use algebraic equations to determine the maximum area of a rectangle with a given partial perimeter.
## Polygonal Strings
This is a problem from the number and algebra strand.
## The Chicken Run
explain the relationship between the area and perimeter of rectangles
use a table to solve a problem
devise and use problem solving strategies to explore situations mathematically (guess and check, be systematic, make a table, make a drawing)
## Karen's Second Tiles
Determine the maximum area of a regular polygon with a given perimeter
Appreciate the concept of limit as it applies to the area of regular n-gons and circle that both have the same perimeter
## Peter's Second String
determine the maximum area of a rectangle with a given perimeter
determine the range of areas of a rectangle with a given perimeter
pose questions for mathematical exploration
prove or refute mathematical conjectures |
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# If ${\text{y = }}{{\text{x}}^{\text{x}}}$ , prove that $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$
Last updated date: 08th Aug 2024
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Hint: First we apply the logarithm function on the both sides of the equation ${\text{y = }}{{\text{x}}^{\text{x}}}$, then we’ll proceed towards finding the first derivative of the equation with respect to the independent variable i.e. x. similarly, we’ll find the double derivative of the function. After getting the first and second derivates of the function we’ll use the substitution method to eliminate the terms that are not required to get our answer.
Given data: ${\text{y = }}{{\text{x}}^{\text{x}}}$
On applying logarithm function on both sides, we get
${\text{ln(y) = ln(}}{{\text{x}}^{\text{x}}}{\text{)}}$
It is well known that, ${\text{ln}}{{\text{a}}^{\text{b}}}{\text{ = blna}}$,
$\Rightarrow {\text{lny = xln(x)}}$
Now, differentiating both sides with respect to x,
Using, chain rule and multiplication rule i.e.
$\dfrac{{{\text{df(z)}}}}{{{\text{dx}}}}{\text{ = f'(z)}}\dfrac{{{\text{dz}}}}{{{\text{dx}}}}{\text{, and}} \\ {\text{d(uv) = udv + vdu}} \\$
$\dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = x}}\left( {\dfrac{{\text{1}}}{{\text{x}}}} \right){\text{ + lnx}} \\ \Rightarrow \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 1 + lnx}}..........{\text{(i)}} \\$
Multiplying ‘y’ with the whole equation
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = y + ylnx}}................{\text{(ii)}}$
Again, on differentiating both sides with respect to x,
$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ + }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{lnx}} \\ \Rightarrow \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(1 + lnx) + }}\dfrac{{\text{y}}}{{\text{x}}}..........{\text{(iii)}} \\$
Now, substituting the value of (1+lnx) from equation(i) to equation(iii), we’ll be left with
$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(}}\dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{) + }}\dfrac{{\text{y}}}{{\text{x}}} \\ \Rightarrow \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ + }}\dfrac{{\text{y}}}{{\text{x}}} \\ \Rightarrow \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}} \\$
Hence, we obtained our equation i.e. $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$
Note: Alternative solution for this question can be done by substituting the value of first and second derivatives directly to the equation that has to be proved.
$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$
On substituting the value of $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ and }}\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}$ to the left-hand side of the equation, we get,
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(}}\dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{) + }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(y + ylnx)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}} \\ {\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{{{\text{y}}^{\text{2}}}}}{{\text{y}}}{{\text{(1 + lnx)}}^{\text{2}}} \\$
On simplification we get,
${\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
On substituting the value of $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$, we have
${\text{ = }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(y + ylnx)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
On taking ${{\text{y}}^{\text{2}}}$out of the first term, we get,
${\text{ = }}\dfrac{{{{\text{y}}^{\text{2}}}}}{{\text{y}}}{{\text{(1 + lnx)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
${\text{ = y(1 + lnx}}{{\text{)}}^{\text{2}}}{\text{ - y(1 + lnx}}{{\text{)}}^{\text{2}}}$
${\text{ = 0}}$
i.e. equal to the right-hand side
Therefore, $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{y}}}{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ = 0}}$, holds of ${\text{y = }}{{\text{x}}^{\text{x}}}$ |
Comparing Division With Subtraction
## How Is Division Like Subtraction?
In the last lesson, you learned that division is splitting a number into smaller, equal groups.
Another way to think about division is:
Division is repeated subtraction. When you divide, you are subtracting the same number over and over.
Check out this division problem:
20 ÷ 4 = 5
You can write it in long division form like this:
Tip: If you don't remember how to do long division, review the 3rd grade lesson on long division here.
The dividend is the big number that you're subtracting from.
The divisor is how much you subtract each time.
The quotient is the number of times a divisor can be subtracted from the dividend.
The division equation tells us that we can subtract 4 from 20 five times.
Knowing that division is repeated subtraction can help us solve division problems.
### Let’s Practice
84 ÷ 12 = ?
How many times can we subtract 12 from 84?
We can subtract 12 from 84 seven times.
So we know the quotient is 7.
84 ÷ 12 = 7
Tip: You can also think about division as how many times 12 can fit into 84.
#### More Practice
How many times can you subtract 10 from 70?
Let's write this as a division equation:
70 ÷ 10 = ?
In long-division form, it looks like this:
The quotient is the number of times we can subtract 7 from 70, until we have nothing left. 😸
70 ÷ 10 = 7
The quotient is 10. You subtracted 10 from 70 seven times.
Great job comparing division to subtraction.
Now, ace the practice.
Complete the practice to earn 1 Create Credit
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# 5.2. Recursive Sequences and the Principle of Induction
Choosing ${ℕ}^{\ast }$ as domain for our sequences has a wide effect mainly due to the well advanced arithmetic and ordering properties that come with the natural numbers. This structure is settled when constructing $ℕ$ in set theory. From there we take the following characterization of $ℕ$:
Proposition:
$k\in ℕ⇒k=0\vee k=n+1$ for a suitable $n\in ℕ$. Every non-empty subset of $ℕ$ has a smallest element.
[5.2.1]
That means:
• Every non-zero natural number is the successor of another natural number.
• Every non-empty collection of natural numbers has a first one.
Both of these properties already support the idea to get every natural number by successive counting starting with 0. The Principle of Induction makes this precise:
Proposition (Principle of Induction): Let A be an arbitrary subset of $ℕ$ so that
$0\in A$ $n\in A⇒n+1\in A$
[5.2.2]
Then A is the whole of $ℕ$, i.e.: $A=ℕ$.
Proof: We proceed indirectly and assume: $A\ne ℕ$. Then the difference $ℕ\A$ is a non-empty subset of $ℕ$ and, following 2. in [5.2.1], has a smallest element, say $k\in ℕ\A$. From $0\in A$ we conclude that $k\ne 0$. According 1. in [5.2.1] we find a number $n\in ℕ$ so that $k=n+1$. Especially we have $n. As k is the smallest element of $ℕ\A$ we thus know that n is no member of $ℕ\A$. But that means $n\in A$ and from the second condition in [5.2.2] we can conclude that $k=n+1$ belongs to A. contradiction!
The induction principle is a very effective tool. A whole technique called proof by induction, is based on this principle. We take the proof of a so called summation formula as a first example to demonstrate this technique: Adding the odd intergers from 1 to $2n+1$ gives the following results for the different values of n:
n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 1 + 3 + 5 + 7 + 9 + 11 = 36
Obviously each of the six results is a square number, namely the square of n + 1, the number of addends! We would gain an enormous advantage, especially for huge numbers, if this would be true for every n. Unfortunately however this assumption could not be proved the customary way because we have to prove infinitely many single assertions. But exactly that is performed by the induction principle.
Example: For all $n\in ℕ$ we have:
$\sum _{i=0}^{n}\left(2i+1\right)={\left(n+1\right)}^{2}$ [5.2.3]
Proof:
Our task is, a bit awkwardly shaped, the following: Show that the set
$A≔\left\{n\in ℕ|\sum _{i=0}^{n}\left(2i+1\right)={\left(n+1\right)}^{2}\right\}\subset ℕ$,
i.e. the collection of all natural numbers satisfying [5.2.3] is all of $ℕ$. But to that end we only need to check if the two conditions of the induction principle hold for A!
• ${0\in A}:\sum _{i=0}^{0}\left(2i+1\right)=1={\left(0+1\right)}^{2}$
• ${n\in A⇒n+1\in A}:$$n\in A$ means that this n actually satisfies $\sum _{i=0}^{n}\left(2i+1\right)={\left(n+1\right)}^{2}$. To show that n + 1 also belongs to A the equation $\sum _{i=0}^{n+1}\left(2i+1\right)={\left(n+1+1\right)}^{2}={\left(n+2\right)}^{2}$ has to be verified:
$\begin{array}{cc}\hfill \sum _{i=0}^{n+1}\left(2i+1\right)& =\sum _{i=0}^{n}\left(2i+1\right)+2\left(n+1\right)+1\hfill \\ \hfill & ={\left(n+1\right)}^{2}+2\left(n+1\right)+1\hfill \\ \hfill & ={\left(n+1+1\right)}^{2}\text{.}\hfill \end{array}$
Thus $n+1\in A$.
The principle of induction now guarantees: $A=ℕ$. In other words our summation formular is actually true for all natural numbers.
Consider:
• The just performed technique, splitting off the last addend,
$\sum _{i=0}^{n+1}\left(2i+1\right)=\sum _{i=0}^{n}\left(2i+1\right)+\sum _{i=n+1}^{n+1}\left(2i+1\right)=\sum _{i=0}^{n}\left(2i+1\right)+2\left(n+1\right)+1$
is the standard trick for sums and related expressions.
• With proofs by induction it is a valuable advantage that the sentence to be proved - in our example the equation $\sum _{i=0}^{n+1}\left(2i+1\right)={\left(n+1+1\right)}^{2}$ - can be noted just from the start by inserting n + 1 for n.
We practise the principle of induction with further summation formulas and other examples. We do this however without the explicit formation of the set A and take the terms ${0\in A}$ resp. ${n\in A⇒n+1\in A}$ only as a label for the two steps in a proof by induction:
${0\in A}$ for the base step ${n\in A⇒n+1\in A}$ for the induction step.
Proposition (Summation formula for the geometric series): Let $q\in ℝ,q\ne 1$, then every $n\in ℕ$ satisfies:
$\sum _{i=0}^{n}{q}^{i}=\frac{1-{q}^{n+1}}{1-q}$ [5.2.4]
Proof:
• ${0\in A}:\sum _{i=0}^{0}{q}^{i}={q}^{0}=1=\frac{1-{q}^{0+1}}{1-q}$
• ${n\in A⇒n+1\in A}:$
$\begin{array}{cc}\hfill \sum _{i=0}^{n+1}{q}^{i}& =\sum _{i=0}^{n}{q}^{i}+{q}^{n+1}\hfill \\ \hfill & =\frac{1-{q}^{n+1}}{1-q}+{q}^{n+1}\hfill \\ \hfill & =\frac{1-{q}^{n+1}+\left(1-q\right){q}^{n+1}}{1-q}\hfill \\ \hfill & =\frac{1-{q}^{n+1}+{q}^{n+1}-{q}^{n+2}}{1-q}\hfill \\ \hfill & =\frac{1-{q}^{n+1+1}}{1-q}\text{.}\hfill \end{array}$
The summation formula for the geometric series playes a very important role in calculus. This is also true for the generalization of common binomial formula which we will present in the next proposition. For the notation we need the binomial coefficients $\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}$, for the slightly extensive proof we need two of their properties and a trick called index shift.
Proposition (Generalized Binomial Theorem): Let $a,b\in ℝ$, then
${\left(a+b\right)}^{n}=\sum _{i=0}^{n}\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}{a}^{n-i}{b}^{i}$ [5.2.5]
holds for every $n\in ℕ$.
Proof:
• ${0\in A}:{\left(a+b\right)}^{0}=1=\left(\phantom{T}\begin{array}{c}0\\ 0\end{array}\right)\phantom{T}{a}^{0-0}{b}^{0}=\sum _{i=0}^{0}\left(\phantom{T}\begin{array}{c}0\\ i\end{array}\right)\phantom{T}{a}^{0-i}{b}^{i}$.
• ${n\in A⇒n+1\in A}:$
$\begin{array}{cc}\hfill {\left(a+b\right)}^{n+1}& ={\left(a+b\right)}^{n}\left(a+b\right)\hfill \\ \hfill & =\left(\sum _{i=0}^{n}\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}{a}^{n-i}{b}^{i}\right)a+\left(\sum _{i=0}^{n}\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}{a}^{n-i}{b}^{i}\right)b\hfill \\ \hfill & =\sum _{i=0}^{n}\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}{a}^{n+1-i}{b}^{i}+\sum _{i=0}^{n}\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}{a}^{n-i}{b}^{i+1}\hfill \\ \hfill & =\sum _{i=0}^{n}\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}{a}^{n+1-i}{b}^{i}+\sum _{i=1}^{n+1}\left(\phantom{T}\begin{array}{c}n\\ i-1\end{array}\right)\phantom{T}{a}^{n+1-i}{b}^{i}\hfill \\ \hfill & =\left(\phantom{T}\begin{array}{c}n\\ 0\end{array}\right)\phantom{T}{a}^{n+1-0}{b}^{0}+\sum _{i=1}^{n}\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}{a}^{n+1-i}{b}^{i}+\sum _{i=1}^{n}\left(\phantom{T}\begin{array}{c}n\\ i-1\end{array}\right)\phantom{T}{a}^{n+1-i}{b}^{i}+\left(\phantom{T}\begin{array}{c}n\\ n\end{array}\right)\phantom{T}{a}^{0}{b}^{n+1}\hfill \\ \hfill & =\left(\phantom{T}\begin{array}{c}n\\ 0\end{array}\right)\phantom{T}{a}^{n+1-0}{b}^{0}+\sum _{i=1}^{n}\left(\left(\phantom{T}\begin{array}{c}n\\ i\end{array}\right)\phantom{T}+\left(\phantom{T}\begin{array}{c}n\\ i-1\end{array}\right)\phantom{T}\right){a}^{n+1-i}{b}^{i}+\left(\phantom{T}\begin{array}{c}n\\ n\end{array}\right)\phantom{T}{a}^{0}{b}^{n+1}\hfill \\ \hfill & =\left(\phantom{T}\begin{array}{c}n+1\\ 0\end{array}\right)\phantom{T}{a}^{n+1-0}{b}^{0}+\sum _{i=1}^{n}\left(\phantom{T}\begin{array}{c}n+1\\ i\end{array}\right)\phantom{T}{a}^{n+1-i}{b}^{i}+\left(\phantom{T}\begin{array}{c}n+1\\ n+1\end{array}\right)\phantom{T}{a}^{0}{b}^{n+1}\hfill \\ \hfill & =\sum _{i=0}^{n+1}\left(\phantom{T}\begin{array}{c}n+1\\ i\end{array}\right)\phantom{T}{a}^{n+1-i}{b}^{i}\text{.}\hfill \end{array}$
Summation formulas are by far not the only sentences to be proved by induction. The following example deals with an inequality.
Proposition (Bernoulli Inequality): For every real number $x\ge -1$ and any $n\in ℕ$ we have:
${\left(1+x\right)}^{n}\ge 1+nx$ [5.2.6]
Proof:
• ${0\in A}:{\left(1+x\right)}^{0}=1\ge 1=1+0\cdot x$.
• ${n\in A⇒n+1\in A}:$Assume the inequality ${\left(1+x\right)}^{n}\ge 1+nx$ is already valid. Multiplying with the (due to $x\ge -1$) positive number $1+x$ gives:
${\left(1+x\right)}^{n}\left(1+x\right)\ge \left(1+nx\right)\left(1+x\right)$.
Thus we can argue as follows::
The initial value 0 used with principle of induction is determind by the set $ℕ$. However we can modify the induction principle so that any integer k gets the option for the initial value:
Proposition: Let $k\in ℤ$ and A be any subset of ${ℤ}^{\ge k}$ so that
$k\in A$ $n\in A⇒n+1\in A$
[5.2.7]
then we have: $A={ℤ}^{\ge k}$.
Proof: Let us set $A\text{'}≔\left\{i-k|i\in A\right\}$. A' then is a subset of $ℕ$ containing 0 and with every n the successor $n+1$ as well. From the induction principle we have $A\text{'}=ℕ$ and thus $A={ℤ}^{\ge k}$.
Very close related to the induction principle is a certain construction method, often used with sequences, the so called definition by recursion. We need two steps for a sequence $\left({a}_{n}\right)$ to be constructed in this way: The initial step is to fix a value for the first sequence member ${a}_{1}$. The second step is to introduce a general method to calculate any sequence member ${a}_{n+1}$ from its predecessor ${a}_{n}$. As an example the sequence $\left({a}_{n}\right)$ could be constructed by the statement:
${a}_{1}≔1\wedge {a}_{n+1}≔2{a}_{n}\text{.}$
From the recursion formula we see that every new sequence member is twice its predecessor. So we can calculate the sequence members one after the other:
$\left({a}_{n}\right)=\left(1,2,4,8,16,32,\dots \right)\text{.}$
The following proposition introduces the Principle of Recursion.
Proposition (Principle of Recursion): For any $c\in A$ and any function $f:{ℕ}^{\ast }×A\to A$ the statement
${a}_{1}≔c\wedge {a}_{n+1}≔f\left(n,{a}_{n}\right)$ [5.2.8]
defines a sequence in A.
Proof: We need to show that [5.2.8] constitutes a function $a:{ℕ}^{\ast }\to A$. First we show that every $n\in ℕ$ has got an image in A. We set
.
As ${a}_{1}\in A$ is guaranted by the premise we have $1\in D$. Now if $n\in D$, i.e. ${a}_{n}\in A$ the recursion formula assigns an element of A to the successor $n+1$, but that means $n+1\in D$. Thus we have $D={ℕ}^{\ast }$ according to the (generalized) induction principle.
Next we show that there is only one image for each n, applying the induction principle again: For the set
we have: $1\in E$, because there is no $n\in {ℕ}^{\ast }$ such that $1=n+1$, so that there is no assignment in [5.2.8] except for ${a}_{1}=c$. Now let $n\in E$, thus ${a}_{n}$ is uniquely determined and so is $f\left(n,{a}_{n}\right)$ (because f is a function!). Finally $n+1$ has only one predecessor, namely n. From all that we conclude that ${a}_{n+1}$ is uniquely determined, thus $n+1$ also belongs to E which completes our induction and we have $E={ℕ}^{\ast }$.
Consider:
• We rarely use the explicit data c and f to define a sequence recursively. It is common to note them just the way we did in our introductory example. Either notion however can be translated into the other. E.g. the statement ${a}_{1}=1\wedge {a}_{n+1}=2{a}_{n}$ turns into:
$c=1$ and $f:{ℕ}^{\ast }×ℝ\to ℝ$ given by $f\left(n,x\right)=2x$.
Vice versa the data, let's say $c=0$ and $f:{ℕ}^{\ast }×ℝ\to ℝ$ given by $f\left(n,x\right)=n+x$, translate to
${a}_{1}=0\wedge {a}_{n+1}=n+{a}_{n}$.
• As with the induction principle and with the sequences, the principle of recursion is extendable to functions of the type $f:{ℤ}^{\ge k}×A\to A$.
The real advantage of recursive sequences lies in their ability to reflect dynamic situations. Let's say a certain species of bacteria needs one hour to double its stock. Starting with one bacterium we can perfectly calculate the number ${a}_{n}$ of bacteria living at the beginning of the der n-th hour by our initial recursive sequence.
As a clear disadvantage with recursive sequences we see the effort that is needed to calculate sequence members far from the start. To get e.g. the size of the bacteria population when 100 hours are over you need to calculate separately 100 sequence members before!
We practise the principle of recursion with some examples. Note that changing only the first sequence member could result in a complete different sequence.
Example: ${a}_{1}≔2\wedge {a}_{n+1}≔3{a}_{n}-2$ gives the sequence $\left(2,4,10,28,82,\dots \right)$ , because ${a}_{2}=3{a}_{1}-2=3\cdot 2-2=4$, ${a}_{3}=3{a}_{2}-2=3\cdot 4-2=10$, ${a}_{4}=3{a}_{3}-2=3\cdot 10-2=28\dots$ ${a}_{1}≔1\wedge {a}_{n+1}≔3{a}_{n}-2$ gives the sequence $\left(1,1,1,1,1,1,\dots \right)$ . ${a}_{1}≔\frac{1}{2}\wedge {a}_{n+1}≔\frac{1}{{a}_{n}+1}$ gives the sequence $\left(\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{5}{8},\frac{8}{13},\dots \right)$ . ${a}_{0}≔1\wedge {a}_{n+1}≔{a}_{n}\cdot \left(n+1\right)$ produces the sequence of factorials: ${\left(1,1,2,6,24,120,720,5.040,\dots \right)}_{n\ge 0}={\left(n!\right)}_{n\ge 0}$ .
Two special recursive sequences get names of their own.
Definition: Let $c,d,q\in ℝ$. A recursive sequence ${\left({a}_{n}\right)}_{n\ge 0}$ of the shape
${a}_{0}≔c\wedge {a}_{n+1}≔{a}_{n}+d$ is called arithmetic. [5.2.9] ${a}_{0}≔c\wedge {a}_{n+1}≔{a}_{n}\cdot q$ is called geometric. [5.2.10]
Consider:
• The new sequence members of an arithmetic sequence are calculated by adding the constant addend d.
The new sequence members of a geometric sequence are calculated by multiplying the constant factor q.
Proposition:
${\left({a}_{n}\right)}_{n\ge 0}$ is arithmetic $⇔$there is a $d\in ℝ$, such that ${\left({a}_{n}\right)}_{n\ge 0}={\left({a}_{0}+nd\right)}_{n\ge 0}$ . [5.2.11] ${\left({a}_{n}\right)}_{n\ge 0}$ is geometric $⇔$there is a $q\in ℝ$, such that ${\left({a}_{n}\right)}_{n\ge 0}={\left({a}_{0}\cdot {q}^{n}\right)}_{n\ge 0}$ . [5.2.12]
Proof: We concentrate on 1. The second assertion is just a multiplicative version of the first one. We split the equivalence into two parts.
"$⇒$": Let c and d be reals such that
${a}_{0}=c\wedge {a}_{n+1}={a}_{n}+d$.
Now we show by induction: ${a}_{n}={a}_{0}+nd$ for all $n\in ℕ$.
• ${0\in A}:{a}_{0}={a}_{0}+0\cdot d$.
• ${n\in A⇒n+1\in A}:{a}_{n+1}={a}_{n}+d={a}_{0}+nd+d={a}_{0}+\left(n+1\right)d$.
"$⇐$": Setting
${b}_{0}≔{a}_{0}\wedge {b}_{n+1}≔{b}_{n}+d$,
gives an arithmetic sequence ${\left({b}_{n}\right)}_{n\ge 0}$ that satisfies the equality ${\left({b}_{n}\right)}_{n\ge 0}={\left({b}_{0}+nd\right)}_{n\ge 0}$ as demonstrated just before. From that we have:
${\left({a}_{n}\right)}_{n\ge 0}={\left({a}_{0}+nd\right)}_{n\ge 0}={\left({b}_{0}+nd\right)}_{n\ge 0}={\left({b}_{n}\right)}_{n\ge 0}$ ,
so that ${\left({a}_{n}\right)}_{n\ge 0}$ coincides with an arithmetic sequence and thus is arithmetic itself.
Arithmetic and geometric sequences show are some interesting properties.
• As ${a}_{n+1}={a}_{n}+d⇔{a}_{n+1}-{a}_{n}=d$, the difference of two consecutive members of an arithmetic sequence is a constant.:
${\left({a}_{n}\right)}_{n\ge 0}$ is arithmetic.
Below we show: The members of an arithmetic sequence are the arithmetic mean of their neighbours.
• As ${a}_{n+1}={a}_{n}\cdot q⇔\frac{{a}_{n+1}}{{a}_{n}}=q$ the quotient of two consecutive members of geometric sequence in ${ℝ}^{\ne 0}$ a constant:
${\left({a}_{n}\right)}_{n\ge 0}$ is geometric.
Next we show: The absolute value of a member of a geometric sequence is the geometric mean of its neighbours.
Proposition:
${\left({a}_{n}\right)}_{n\ge 0}$ is arithmetic [5.2.13] ${\left({a}_{n}\right)}_{n\ge 0}$ is geometric [5.2.14]
Proof:
1. ► "$⇒$": We employ the recursion formula ${a}_{n+1}={a}_{n}+d$ twice: $\frac{{a}_{n}+{a}_{n+2}}{2}=\frac{{a}_{n}+{a}_{n+1}+d}{2}=\frac{{a}_{n}+d+{a}_{n+1}}{2}=\frac{{a}_{n+1}+{a}_{n+1}}{2}={a}_{n+1}$. "$⇐$": We show by induction . Thus ${\left({a}_{n}\right)}_{n\ge 0}$ is represented as an arithmetic sequence with ${0\in A}:{a}_{1}={a}_{0}+{a}_{1}-{a}_{0}\text{.}$ ${n\in A⇒n+1\in A}:$First we have from [5.2.13]: $2{a}_{n+1}={a}_{n}+{a}_{n+2}\text{.}$ This allows the following equation: $\begin{array}{cc}\hfill {a}_{n+2}& =2{a}_{n+1}-{a}_{n}\hfill \\ \hfill & =2{a}_{n}+2\left({a}_{1}-{a}_{0}\right)-{a}_{n}\hfill \\ \hfill & =\left({a}_{n}+{a}_{1}-{a}_{0}\right)+{a}_{1}-{a}_{0}\hfill \\ \hfill & ={a}_{n+1}+{a}_{1}-{a}_{0}\text{.}\hfill \end{array}$ 2. ► "$⇒$": We proceed as we did in 1., also employing the recursion formula twice: ${a}_{n}\cdot {a}_{n+2}={a}_{n}\cdot {a}_{n+1}\cdot q={a}_{n}\cdot q\cdot {a}_{n+1}={a}_{n+1}\cdot {a}_{n+1}={a}_{n+1}^{\phantom{.}2}$. As squares are positive we have ${a}_{n}\cdot {a}_{n+2}\ge 0$ as a first conclusion. The remainder of the assertion now follows when we extract the root. "$⇐$": For the proof of this direction we have to consider two cases: ${a}_{1}=0$. Using $|{a}_{n+1}|=\sqrt{{a}_{n}\cdot {a}_{n+2}}$ we can show by induction: ${a}_{n}=0$ for all $n\ge 1$. Thus ${\left({a}_{n}\right)}_{n\ge 0}={\left({a}_{0},0,0,0,\dots \right)}_{n\ge 0}$ is a geometric sequence with $q=0$. ${a}_{1}\ne 0$. Now an inductive argument gives: . From this we show , just the same way we did in 1., so that ${\left({a}_{n}\right)}_{n\ge 0}$ is a geometric sequence with $c={a}_{0}\wedge q=\frac{{a}_{1}}{{a}_{0}}\text{.}$ ${0\in A}:{a}_{1}={a}_{0}\cdot \frac{{a}_{1}}{{a}_{0}}\text{.}$ ${n\in A⇒n+1\in A}:$From [5.2.14] we get: ${a}_{n+1}^{\phantom{.}2}={a}_{n}\cdot {a}_{n+2}\text{,}$ thus: $\begin{array}{cc}\hfill {a}_{n+2}& =\frac{{a}_{n+1}^{\phantom{.}2}}{{a}_{n}}\hfill \\ \hfill & =\frac{{a}_{n+1}}{{a}_{n}}\cdot {a}_{n+1}\hfill \\ \hfill & =\frac{{a}_{n+1}}{{a}_{n}}\cdot {a}_{n}\cdot \frac{{a}_{1}}{{a}_{0}}\hfill \\ \hfill & ={a}_{n+1}\cdot \frac{{a}_{1}}{{a}_{0}}\text{.}\hfill \end{array}$
The next to last proposition touched an intersting, but often difficult problem: Does every recursive sequence allow a non-recursive representation? And if so, is there a general procedure for the transcription? It is very unlikely to find a solution for such a general problem, but from the arithmetic and geometric sequences we know that single solutions are possible. So we have to study a given sequence individually to get an idea for a suitable formula. Whatever way a result was found, the principle of induction is the standard tool for its verification.
Example: Let $\left({a}_{n}\right)$ be the recursive sequence given by ${a}_{1}≔\frac{1}{2}\wedge {a}_{n+1}≔\frac{{a}_{n}+1}{2}\text{,}$ i.e. $\left({a}_{n}\right)=\left(\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\dots \right)$. It is quite obvious that the denominators represent the powers of 2 and that each numerator is one unit less the denominator. Thus we have the idea: $\left({a}_{n}\right)=\left(\frac{{2}^{n}-1}{{2}^{n}}\right)$ Proof by induction: ${1\in A}:{a}_{1}=\frac{1}{2}=\frac{{2}^{1}-1}{{2}^{1}}$ ${n\in A⇒n+1\in A}:{a}_{n+1}=\frac{{a}_{n}+1}{2}=\frac{\frac{{2}^{n}-1}{{2}^{n}}+1}{2}=\frac{{2}^{n}-1+{2}^{n}}{2\cdot {2}^{n}}=\frac{{2}^{n+1}-1}{{2}^{n+1}}$
The principle of recursion allows extensions in a varying manner. We could e.g. introduce two stage recursions: We provide the first two sequence members ${a}_{1}$ und ${a}_{2}$ and thus can use two predecessors when calculating further members.
Example: The recursive sequence given by
${a}_{1}≔1,{a}_{2}≔1\wedge {a}_{n+2}≔{a}_{n+1}+{a}_{n}$ [5.2.15]
is called Fibonacci sequence. Her members are calculated by adding the first two predecessors, so that the Fibonacci numbers start like this:
$1,1,2,3,5,8,13,21,34,55,\dots$ .
Another option is to consider recursive sequences in ${ℝ}^{2}$.
Example: If the recursive sequence $\left(\left({a}_{n},{b}_{n}\right)\right)$ is given by $\left({a}_{1},{b}_{1}\right)≔\left(2,1\right)\wedge \left({a}_{n+1},{b}_{n+1}\right)≔\left({a}_{n}+{b}_{n},{a}_{n}\cdot {b}_{n}\right)\text{,}$ we get the following table of values: $\left(\left({a}_{n},{b}_{n}\right)\right)=\left(\left(2,1\right),\left(3,2\right),\left(5,6\right),\left(11,30\right),\left(41,330\right),\dots \right)$ . From the notation we see that a recursive sequence in ${ℝ}^{2}$ could be regarded as a coupled system of two recursive sequences in $ℝ$. So we could as well have said: Let $\left({a}_{n}\right)$ and $\left({b}_{n}\right)$ be given by ${a}_{1}≔2,{b}_{1}≔1\wedge {a}_{n+1}≔{a}_{n}+{b}_{n},{b}_{n+1}≔{a}_{n}\cdot {b}_{n}\text{.}$
For some years the theory of chaos gained a remarkable popularity, an absolute exception for a mathematical topic. A final example in this part introduces a recursive sequence in ${ℝ}^{2}$ which is closely related with this popularity:
Example: For a fixed pair of numbers $\left({c}_{1},{c}_{2}\right)\in ℝ$ we define the sequence $\left(\left({a}_{n},{b}_{n}\right)\right)$ recursively by
$\left({a}_{1},{b}_{1}\right)≔\left({c}_{1},{c}_{2}\right)\wedge \left({a}_{n+1},{b}_{n+1}\right)≔\left({a}_{n}^{2}-{b}_{n}^{2}+{c}_{1},2{a}_{n}{b}_{n}+{c}_{2}\right)$ [5.2.16]
We get different value tables for different values of $\left({c}_{1},{c}_{2}\right)$. The following table shows three examples with the first sequence members and their distances $|\left({a}_{n},{b}_{n}\right)|=\sqrt{{a}_{n}^{2}+{b}_{n}^{2}}$ to the point of origin as well.
$\left({c}_{1},{c}_{2}\right)$ $\left({a}_{n},{b}_{n}\right)$ $|\left({a}_{n},{b}_{n}\right)|$ $\left(0,0\right)$ $\left(0,0\right),\left(0,0\right),\left(0,0\right),\left(0,0\right),\left(0,0\right),\dots$ $0,0,0,0,0,\dots$ $\left(0,1\right)$ $\left(0,1\right),\left(-1,1\right),\left(0,-1\right),\left(-1,1\right),\left(0,-1\right),\left(-1,1\right),\dots$ $1,1.414,1,1.414,1,\dots$ $\left(1,1\right)$ $\left(1,1\right),\left(1,3\right),\left(-9,7\right),\left(32,127\right),\dots$ $1.414,3.162,11.4,130.97,\dots$
Whereas some values of $\left({c}_{1},{c}_{2}\right)$ produce steadily increasing distances, see e.g. the last example, there may be others where the distances are bounded. If we mark all those initial values $\left({c}_{1},{c}_{2}\right)$ in the plane where the distances do not exceed 2 we get an extraordinary subset of ${ℝ}^{2}$, the so called Mandelbrot set. Graphical editing of its margin reveals an amazing beauty:
The (mathematical) amazement is even greater if we blow up sections of the Mandelbrot set and blow up again and again. It is rewarding to start a journey into the Mandelbrot set.
The Mandelbrot set is the most popular example of a so called fractal. There are many more examples for fractals on the web. This site has a very exhaustive link list.
To produce fractals of your own you need information and tools.
5.1. 5.3. |
Differentiation comes down to figuring out how one variable changes with respect to another variable. If this change is a constant (as we have in a line), this concept becomes very similar to the idea of a slope. But calculus is all about curves, and differentiation allows us to figure out rates of change when this change is itself changing.
An Example of Differentiation
The best way to look at differentiation is to look at a real-world example. Let us take the old physics question that asks us to model a car starting at 30 meters per second, but slamming on the breaks. Intuitively, we should know something about the velocity and acceleration. Calculus and the derivative will allow us to model this mathematically, and figure out what’s changing at any point.
If you’ve taken a physics class, you should be able to understand the following equation:
where x(t) is our function for a position at any time t, in seconds. Our initial speed is 30 meters per second, and every second we’re slowing down by 5 meters per second. I’ve graphed the function below:
Clearly, our position is not a linear function (it is quadratic). On our x-axis, we have time (in seconds), and our y-axis, we have position. How fast our position changes with respect to time (distance/time) is a rate of change more commonly referred to as velocity (or speed). Now, if this was a linear function, our velocity would simply be the slope of the graph. However, since it is quadratic, we need to take the derivative.
As we can see, our velocity is slowly decreasing. At t = 0, our velocity is 30 meters per second, but eventually goes to 0 by the 3rd second.
If we want to look at our rate of change of our rate of change, what we will see is how our velocity changes with respect to time as well. Our change in velocity per second is more commonly referred to as acceleration. Since our velocity time graph is linear, our derivative will be the same as our slope in this case.
Our acceleration is -10 m/s2. Our we can say that we’re slowing down 10 m/s every second.
The derivative is an important tool in determining how variables change with respect to each other. AP Calculus AB primarily deals with two dimensions, but we can also use derivatives to compare multiple dimensions.
Zachary is a former mechanical engineer and current high school physics, math, and computer science teacher. He graduated from McGill University in 2011 and spent time in the automotive industry in Detroit before moving into education. He has been teaching and tutoring for the past five years, but you can also find him adventuring, reading, rock climbing, and traveling whenever the opportunity arises.
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# The Tree Puzzle – 99% Fail to Answer this Tricky Math Puzzle (With Answer)
### Solve this latest logic math puzzle that will leave you scratching your head. 99% fail to answer this puzzle. The Tree Puzzle, Math IQ Test – Only for genius brain teasers math puzzle with the answer and Solution!
Let’s take the puzzle to the next level. My previous “Santa Puzzle” create lots of debate on Facebook. But this is a little bit different from the last one. Find out the values of these particular shapes (tree, circle, and star) and solve the last equation of the puzzle.
Let’s find out, how good your IQ level is, which number replace the question mark? Share your answer below.
Puzzle Question:
⇒ tree + tree + tree = 21
⇒ circle + circle + tree = 19
⇒ star + circle + tree = 15
⇒ tree + 2star + circle = ??
**check image for details**
## The Tree Math Puzzle Image: Genius Puzzle Series #7
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### The Tree Puzzle Solution:
⇒ 1st: Tree with Star Vs Tree without Star
⇒ 2nd: Wheel with 6 spokes Vs Wheel with 5 spokes
Let’s solve the puzzle:
### >> Equation 1:
⇒ Tree (with star) + Tree (with star) + Tree (with star) = 21
⇒ Tree (with star) = 7
### >> Equation 2:
⇒ Wheel (with 6 spokes) + Wheel (with 6 spokes) + Tree (with star) = 19
We know Tree (with star) = 7, (put the value)
⇒ Wheel (with 6 spokes) + Wheel (with 6 spokes) + 7 = 19
⇒ 2 Wheel (with 6 spokes) = 19 – 7
⇒ 2 Wheel (with 6 spokes) = 12
⇒ Wheel (with 6 spokes) = 12/2
⇒ Wheel (with 6 spokes) = 6
### >> Equation 3:
⇒ Star + Wheel (with 6 spokes) + Tree (with star) = 15
let’s find out the value of star: put the value of wheel and tree in the eq. 3
⇒ Star + 6 + 7 = 15
⇒ Star + 13 = 15
⇒ Star = 15 – 13
⇒ Star = 2
Now we have:
• Tree (with star) = 7
• Wheel (with 6 spokes) = 6
• Star = 2
Let’s check out the fourth equation
### >> Equation 4:
⇒ Tree (without star) + 2 Stars × Wheel (with 5 spokes) = ?
Remember this, when you compare other equations with the fourth equation, you will see the difference:
• Wheel (with 6 spokes) Vs Wheel (with 5 spokes)
• Tree (with star) Vs Tree (without star)
Now how to find out the value of “Tree (without star)”?
Well, you know the value of single “Star” (from eq. 3). Just subtract the value of “Star” from the “Tree (with star)” of 1st eq., you will get the value of single “Tree” i.e:
Tree (with star) = 7
Tree (without star) = 7 – Star
Tree (without star) = 7 – 2
Tree (without star) = 5
Now, find out the logic of Wheels:
If, Wheel (with 6 spokes) = 6, then Wheel (with 5 spokes) =?
Logic?? Simple, each spoke has One value. Therefore, 6 spokes = 6, then 5 spokes = 5
Wheel (with 5 spokes) = 5
Now we have values for 4th eq.
• Tree (without star) = 5
• Wheel (with 5 spokes) = 5
But here is another twist in this puzzle, if you compare TREE of 1st Equation Vs TREE of 4th Equation, you will figure out the difference between Layers.
In 1st eq: Tree contains 5 layers Vs In 4th eq: Tree contains 6 layers
We know;
⇒ Tree (without star) = 5 = Tree (with 5 layers).
So, what is the logic here? Simple, each layer = 1,
Therefore, Tree (with 5 layers) = 5, and Tree (with 6 layers) = 6
Now we have the correct values for each element of the 4th eq., i.e.:
• Tree (with 6 layers) = 6
• Wheel (with 5 spokes) = 5
• Star = 2
### Let’s rewrite the equation 4:
⇒ Tree (with 6 layers) + 2 Stars × Wheel (with 5 spokes) =? (Put the values given above)
⇒ 6 + (2+2) × 5 = ?
⇒ 6 + (4) × 5 = ? (Do multiplication first)
⇒ 6 + 20 = ?
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# What is the number system?
For writing numbers using various methods. They are depicted with the help of symbols represented by different systems. In this article we will understand what the number system is. In fact, this is the way of representing numbers and the rules of action on them that correspond to this method. This system is a sign system. The numbers in it are written with the help of symbols of a certain alphabet, which are called numbers. This is done subject to certain rules.
Writing a number in a certain number system is called a number code. A separate position in the record number is called a digit, and its number - the number of discharge. Digit is the number of digits in the image number. It coincides with the length of the number.
## Types of number systems
Numbers are represented by non-positional and positional number systems. The nonpositional number system is characterized by the fact that the weight of a digit in it does not depend on the position it occupies in a particular number. An example of such a system is the Roman number system.For example, in the number XXXXV (forty-five), the weight of the digit denoted by X is just 10, regardless of position.
Next, we will understand what is the positional number system and how is it different from the non-positional number system. In the positional number system, the weight of any digit will vary depending on what position it will occupy in the sequence of digits that depicts this number. The main advantages of any positional number system are:
• limited number of characters used to represent numbers;
• ease of performing arithmetic operations.
The main characteristic of a positional system is its foundation. Consider what is the base of the number system. The number of different symbols and characters that are used to represent numbers in this system is called the base of the positional number system. Any natural number can be taken as a base. Therefore, the positioning systems can be infinite. Binary, decimal, hexadecimal, etc. number systems are all examples of positional number systems.
### What is the decimal number system
The decimal system is now generally accepted number system. Decimal, it is called, because in it ten units of the lower category are one unit of the higher category. The number 10, respectively, is the basis of the decimal number system.
To write numbers in it, use the following ten numbers: 0, 1, 2, 3, etc. The ranks in it have their own names: hundreds, tens, units. Every 3 categories are combined into a class: a class of millions, a class of thousands, a class of units. When reading numbers, use class names.
Its value depends on the digit in which the digit is located. For example, the number 7537 consists of 7 thousand, 5 hundred, 3 dozen and 7 units. In this case, the first number seven will indicate the number of thousands, and the last one means the number of units. If a digit moves one digit to the left, then its value increases 10 times, and if it moves to the right, it decreases 10 times.
### What is a binary number system
The basis of the computer is a binary number system, in which two digits are used - 0 and 1. A computer has two stable states: there is current or no current; high voltage or low; magnetized or not magnetized. One state will correspond to a value equal to one, the second - to zero.The advantages of the binary number system are as follows:
• Elements with 2 possible states, and not several, are suitable for its realization.
• Presentation of information using only two states is robust and reliable.
• The algebra of logic can be used to perform transformations.
• And finally, binary arithmetic is simpler than decimal.
All this leads to the simplification of the transfer, processing and storage of information on a computer. The most significant disadvantage of the binary number system is the rapid increase in the number of digits. |
Operations with Negative Numbers on the ISEE
Middle and Upper Level
LESSON GOAL: Simplify expressions and solve problems that include negative numbers.
ISEE Middle-Level Problem: A set of 6 numbers has a mean of 7. What additional number must be included in this set to create a new set with a mean that is 2 less than the mean of the original set?
A) -2 B) -7 C) -12 D) -25
While this may seem like only an advanced average problem at first glance, you’ll notice that all the answer choices are negative. This is also a negative number problem.
Rules of Operations with Negative Numbers
Addition: To add two negative numbers, add the numbers the same way you would add two positive numbers, and then add a negative sign to the sum. −7 + −8 = −15
If one number is positive and one number is negative, subtract the numbers the way you would subtract two positives. Then add the sign (+ or -) of the number with the higher absolute value (further from zero).
7 + (−8) → 8 − 7 = 1
|−8| > |7| 7 + (−8) = −1
Subtraction: Any subtraction problem can be turned into an addition problem; subtracting a positive is the same as adding a negative. 3 − 7 = 3 + (−7) = −4
To subtract a larger positive from a smaller positive number, find the difference and then add a negative sign.
3 − 7 = 3 + (−7) = −4
10 − 17 = −7
Subtracting a negative number is the same as adding a positive number. To subtract a negative, change the two minus signs into a plus sign and then add.
3 − (−7) = 3 + 7 = 10
−7 − (−3) = −7 + 3 = −4
Multiplication and Division: When multiplying and dividing negative numbers, it doesn’t matter whether the positive or negative number has a higher absolute value, the same rules apply:
positive x positive = positive 6 x 8 = 48
positive x negative = negative 6 x −8 = −48
negative x positive = negative −6 x 8 = −48
negative x negative = positive −6 x −8 = 48
positive ÷ positive = positive 48 ÷ 8 = 6
positive ÷ negative = negative 48 ÷ −8 = −6
negative ÷ positive = negative −48 ÷ 8 = −6
negative ÷ negative = positive −48 x −8 = 6
The solution: Using the procedure for advanced averages, we can calculate that we need the total score to go from 42 (6 × 7) to 35 (7 × 5). To get from 42 to 35 by adding, we’ll need to add -7. The answer is B.
Helpful tip: When you calculate with negative numbers, think about whether your answer makes sense. If it doesn’t, you might be forgetting or mixing up a negative sign. |
# How do you solve cosxtanx=1/2?
Dec 9, 2016
$\frac{\pi}{6} , \frac{5 \pi}{6}$
#### Explanation:
$\cos x . \tan x = \frac{1}{2}$
$\cos x \frac{\sin x}{\cos x} = \frac{1}{2}$
Divide by cos x, under condition $\implies$ cos x diff. to zero,
or x diff. to $\frac{\pi}{2} , \frac{3 \pi}{2}$
$\sin x = \frac{1}{2}$
Use trig table of special arcs and unit circle $\implies$
$\sin x = \frac{1}{2}$ $\implies$ arc $x = \frac{\pi}{6}$ , and arc $x = \frac{5 \pi}{6}$
$x = \frac{\pi}{6} + 2 k \pi$
$x = \frac{5 \pi}{6} + 2 k \pi$
Dec 9, 2016
$x = {30}^{\circ}$ or $\frac{\pi}{6}$ and ${150}^{\circ}$ or $\frac{5 \pi}{6}$
#### Explanation:
Using the trigonometric identity, $\textcolor{red}{\tan x} = \frac{\textcolor{g r e e n}{\sin x}}{\textcolor{b l u e}{\cos x}}$,
this question can be written as $\textcolor{b l u e}{\cos x} \cdot \frac{\textcolor{g r e e n}{\sin x}}{\textcolor{b l u e}{\cos x}} = \frac{1}{2}$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXX}} \textcolor{b l u e}{\cancel{\cos x}} \cdot \frac{\textcolor{g r e e n}{\sin x}}{\textcolor{b l u e}{\cancel{\cos x}}} = \frac{1}{2}$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXX}} \textcolor{g r e e n}{\sin} x = \frac{1}{2}$
Note: 'N.D.' means Not Defined. For example, $\frac{a}{0}$ is not defined, where $a$ is any non-zero number like 1, 4, 647 etc. But, $\frac{0}{a} = 0$.
As you can see in this table, we get the value of $x$ as ${30}^{\circ}$ or $\frac{\pi}{6}$.
This is called a 'Unit Circle'.
The values in brackets are ($\cos$,$\sin$).
$\sin$ is positive in the first and second quadrants. Since we need $+ \frac{1}{2}$, we consider these quadrants only.
${30}^{\circ}$ is the value of $x$ in the first quadrant.
To get the second value of $x$(in second quadrant), we subtract $\frac{\pi}{6}$ from $\pi$.
$\pi - \frac{\pi}{6} = \frac{6 \pi}{6} - \frac{\pi}{6} = \frac{5 \pi}{6} \mathmr{and} {150}^{\circ}$.
Also note, here $\pi$ is $\pi$ radians, which is equal to ${180}^{\circ}$.
Check out this video for more understanding:
Intro to Arcsin |
# A mixture of 30 pounds of candy sells for $1.10 a pound. The mixture consists of chocolates worth$1.50 a pound and chocolates worth 90 cents a pound. How many pounds of the $1.50 chocolate were used to make the mixture? ##### 1 Answer Jun 18, 2017 There are 10 pounds of the more expensive chocolate that costs$1.50 and 20 pounds of the cheaper chocolate that costs $0.90. #### Explanation: Call the number of pounds of the$1.50 chocalate 'e' (for 'expensive') and the number of pounds of the $0.90 chocolate 'c'. (we need to express both in dollars) We know that $e + c = 30$(Equation 1) because there are 30 pounds of chocolate in total. We also know that: $e \times 1.5 + c \times 0.9 = 30 \times 1.1$We can write that as: $1.5e+0 .9 c = 33$(Equation 2) Now we have two equations with two unknowns, so we will use our skills in solving 'simultaneous equations'. We can rearrange Equation 1 to give us a value for $c$in terms of $e$: $c = 30 - e$Now we substitute that value into Equation 2: $1.5e+0 .9 \left(30 - e\right) = 33$$1.5e+27 - 0.9 e = 33$$0.6 e = 6$$e = 10$pounds In this case, since we know the total is 30 pounds, it's easy to reason that $c = 20$pounds, but we could also calculate that result by substituting our value for $e\$ back into Equation 1.
This result makes sense, since the final price is closer to the cheaper than the more expensive price, because there is more of the cheaper chocolate included. |
# AP Statistics Curriculum 2007 Distrib RV
## General Advance-Placement (AP) Statistics Curriculum - Random Variables and Probability Distributions
### Random Variables
A random variable is a function or a mapping from a sample space into the real numbers (most of the time). In other words, a random variable assigns real values to outcomes of experiments. This mapping is called random, as the output values of the mapping depend on the outcome of the experiment, which are indeed random. So, instead of studying the raw outcomes of experiments (e.g., define and compute probabilities), most of the time we study (or compute probabilities) on the corresponding random variables instead. The formal general definition of random variables may be found here.
### Examples of Random Variables
• Die: In rolling a regular hexagonal die, the sample space is clearly and numerically well-defined and in this case the random variable is the identity function assigning to each face of the die the numerical value it represents. This the possible outcomes of the RV of this experiment are { 1, 2, 3, 4, 5, 6 }. You can see this explicit RV mapping in the SOCR Die Experiment.
• Coin: For a coin toss, a suitable space of possible outcomes is S={H, T} (for heads and tails). In this case these are not numerical values, so we can define a RV that maps these to numbers. For instance, we can define the RV $X: S \longrightarrow [0, 1]$ as: $X(s) = \begin{cases}0,& s = \texttt{H},\\ 1,& s = \texttt{T}.\end{cases}$. You can see this explicit RV mapping of heads and tails to numbers in the SOCR Coin Experiment.
• Card: Suppose we draw a 5-card hand from a standard 52-card deck and we are interested in the probability that the hand contains at least one pair of cards with identical denomination. Then the sample space of this experiment is large - it should be difficult to list all possible outcomes. However, we can assign a random variable $X(s) = \begin{cases}0,& s = \texttt{no-pair},\\ 1,& s = \texttt{at-least-1-pair}.\end{cases}$ and try to compute the probability of P(X=1), the chance that the hand contains a pair. You can see this explicit RV mapping and the calculations of this probability at the SOCR Card Experiment.
### Probability density/mass and (cumulative) distribution functions
• The probability density or probability mass function, for a continuous or discrete random variable, is the function defined by the probability of the subset of the sample space, $\{s \in S \} \subset S$, which is mapped by the random variable X to the real value x (i.e., X(s)=x):
$p(x) = P(\{s \in S \} | X(s) = x)$, for each x.
• The cumulative distribution function (cdf) F(x) of any random variable X with probability mass or density function p(x) is defined as the total probability of all $\{s \in S \} \subset S$, where $X(s)\leq x$:
$F(x)=P(X\leq x)= \begin{cases}{ \sum_{y: y\leq x} {p(y)}},& X = \texttt{Discrete-RV},\\ {\int_{-\infty}^{x} {p(y)dy}},& X = \texttt{continuous-RV}.\end{cases}$, for all x.
#### PDF Example
The Benford's Law states that the probability of the first digit (d) in a large number of integer observations ($d\not=0$) is given by
$P(d) = \log(d+1) - log(d) = \log{d+1 \over d}$, for $d = 1,2,\cdots,9.$
Note that this probability definition determines a discrete probability (mass) distribution:
$\sum_{d=1}^9{P(d)}=\log{2\over 1}+\log{3\over 2}+\log{4\over 3}+ \cdots +\log{10\over 9}=$ $\log({{2\over 1} {3\over 2} {4\over 3} \cdots{10\over 9}}) = \log{10\over 1} = 1$
d 1 2 3 4 5 6 7 8 9 P(d) 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046
The explanation of the Benford's Law may be summarized as follows: The distribution of the first digits must be independent of the measuring units used in observing/recording the integer measurements. For instance, this means that if we had observed lenght/distance in inches or centimeters (inches and centimeters are linearly dependent, 1in = 2.54cm), the distribution of the first digit of the measurement must be identical. So, there are about three centimeters for each inch. Thus, the probability that the first digit of a length observation is 1in must be the same as the probability that the first digit of a lenght in centimeters starts with either 2 or 3 (with standard round off). Similarly, for observations of 2in, need to have their centimeter counterparts either 5cm or 6cm. Observations of 3in will correspond to 7 or 8 centimeters, etc. In other words, this distribution must be scale invariant.
The only distribution that obeys this property is the one whose logarithm is uniformly distributed. In this case, the logarithms of the numbers are uniformly distributed -- $P(100\leq x \leq 1,000)$ = $P(2\leq \log(x)\leq 3)$ is the same as the probability $P(10,000\leq x \leq 100,000)$ = $P(4\leq \log(x)\leq 5)$. Examples of such exponentially growing numerical measurements are incomes, stock prices and computational power.
### How to Use RVs?
There are 3 important quantities that we are always interested in when we study random processes. Each of these may be phrased in terms of RVs, which simplifies their calculations.
• Probability Density Function (PDF): What is the probability of P(X = xo)? For instance, in the card example above, we may be interested in P(at least 1 pair) = P(X=1) = P(1 pair only) = 0.422569. Or in the die example, we may want to know P(Even number turns up) = $P(X \in \{2, 4, 6 \}) = 0.5$.
• Cumulative Distribution Function (CDF): P(X < xo), for all xo. For instance, in the (fair) die example we have the following discrete density (mass) and cumulative distribution table:
x 1 2 3 4 5 6 PDFP(X = x) 1/6 1/6 1/6 1/6 1/6 1/6 CDF $P(X\leq x)$ 1/6 2/6 3/6 4/6 5/6 1
• Mean/Expected Value: Most natural processes may be characterized, via probability distribution of an appropriate RV, in terms of a small number of parameters. These parameters simplify the practical interpretation of the process or phenomena we study. For example, it is often enough to know what the process (or RV) average value is. This is the concept of expected value (or mean) of a random variable, denoted E[X]. The expected value is the point of gravitational balance of the distribution of the RV.
Obviously, we may define a large number of RV for the same process. When are two RVs equivalent is dependent on the definition of equivalence?
### The Web of Distributions
There are a large number of families of distributions and distribution classification schemes.
The most common way to describe the universe of distributions is to partition them into categories. For example, Continuous Distributions and Discrete Distributions; marginal and joint distributions; finitely and infinitely supported, etc.
SOCR Distribution applets and the SOCR Distribution activities illustrate how to use technology to compute probabilities for events arising from many different processes.
The image below shows some of the relations between commonly used distributions. Many of these relations will be explored later.
The SOCR Distributome applet provides an interactive graphical interface for exploring the relations between different distributions. |
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