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# Why does this process map every fraction to the golden ratio?
Start with any positive fraction $$\frac{a}{b}$$. First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$ Then add the (new) numerator to the denominator: $$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$ So $$\frac{2}{5} \rightarrow \frac{7}{5} \rightarrow \frac{7}{12}$$.
Repeating this process appears to map every fraction to $$\phi$$ and $$\frac{1}{\phi}$$:
$$\begin{array}{ccccccccccc} \frac{2}{5} & \frac{7}{5} & \frac{7}{12} & \frac{19}{12} & \frac{19}{31} & \frac{50}{31} & \frac{50}{81} & \frac{131}{81} & \frac{131}{212} & \frac{343}{212} & \frac{343}{555} \\ 0.4 & 1.40 & 0.583 & 1.58 & 0.613 & 1.61 & 0.617 & 1.62 & 0.618 & 1.62 & 0.618 \\ \end{array}$$ Another example: $$\begin{array}{ccccccccccc} \frac{11}{7} & \frac{18}{7} & \frac{18}{25} & \frac{43}{25} & \frac{43}{68} & \frac{111}{68} & \frac{111}{179} & \frac{290}{179} & \frac{290}{469} & \frac{759}{469} & \frac{759}{1228} \\ 1.57143 & 2.57 & 0.720 & 1.72 & 0.632 & 1.63 & 0.620 & 1.62 & 0.618 & 1.62 & 0.618 \\ \end{array}$$
Q. Why?
• Look at the Möbius transformation $z \mapsto \frac{z+1}{z+2}$. What are its fixed points, two-cycles … – Daniel Fischer Nov 18 '19 at 12:54
• Note: $\phi=\dfrac ab=\dfrac{a+b}{a+2b}\implies \phi=\dfrac{\phi+1}{\phi+2}\implies\phi(\phi+2)=\phi^2+2\phi=\phi+1\implies\phi^2+\phi-1=0$ – J. W. Tanner Nov 18 '19 at 13:39
• Isn't this what makes it golden? – Strawberry Nov 19 '19 at 9:40
• Hint : Compare the iteration's first step to the definition of the golden ratio, and the iterative process to that of the Fibonacci numbers. – Lucian Nov 20 '19 at 8:45
Instead of representing $$\frac{a}{b}$$ as a fraction, represent it as the vector $$\left( \begin{array}{c} a \\ b \end{array} \right)$$.
Then, all you are doing to generate your sequence is repeatedly multiplying by the matrix $$\left( \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right)$$. One of the eigenvectors of this matrix is $$\left( \begin{array}{c} \frac{\sqrt{5}-1}{2} \\ 1 \end{array} \right)$$, which has a slope equal to the "golden ratio".
This is a standard example of a linear discrete dynamical system, and asymptotic convergence to an eigenvector is one of the typical things that can happen. You can also guess at the long-term behavior of the system by looking at its vector field.
https://kevinmehall.net/p/equationexplorer/#%5B-100,100,-100,100%5D&v%7C(x+y)i+(x+2y)j%7C0.1
In this case you see everything that starts in the first quadrant diverges to infinity along the path of the eigenvector I mentioned before. For your sequence, you started at $$\left( \begin{array}{c} 2 \\ 5 \end{array} \right)$$, which lies in the first quadrant.
Side note: There is nothing particularly special about the golden ratio, the matrix above, or the starting point of $$\left( \begin{array}{c} 2 \\ 5 \end{array} \right)$$ for this sequence. You can change the starting point to be in the negative quadrant if you want to diverge in the opposite direction, and you can change the matrix if you want to diverge along a differently sloped eigenvector.
• @JosephO'Rourke Related: power iteration. Compare this answer. – Kamil Maciorowski Nov 19 '19 at 6:30
• @JosephO'Rourke: This is a good example of how separating the operation iteration from the initial value is valuable. In fact, this viewpoint can motivate matrices! See this explanation for more details. – user21820 Nov 19 '19 at 7:43
• Is the silver ratio (and other ratios in the family) also given by a simple matrix multiplication? – curiousdannii Nov 20 '19 at 23:14
• @curiousdannii The Pell numbers (silver ratio) use the matrix [0 1;1 2]. The Fibonacci numbers (golden ratio) use [0 1;1 1]. Any linear recurrence relation can be expressed in this way. – Brady Gilg Nov 20 '19 at 23:29
• @curiousdannii Because the OP's sequence uses that matrix. It just happens to share an eigenvector with the Fibonacci matrix. – Brady Gilg Nov 20 '19 at 23:37
Let $$f$$ be the map that takes $$a/b$$ to $$(a+b)/(a+2b)$$. We can prove inductively that the $$n$$th iteration of this process gives $$f^n(a/b) = \frac{F_{n}a + F_{n+1}b}{F_{n+1}a + F_{n+2}b},$$ where $$F_n$$ is the $$n$$th Fibonacci number. Since $$b$$ is always non-zero, asymptotically, this ratio approaches $$\lim_{n\rightarrow \infty} \frac{F_{n}a + F_{n+1}b}{F_{n+1}a + F_{n+2}b} = \lim_{n\rightarrow \infty}\frac{F_{n+1}}{F_{n+2}} = \varphi^{-1},$$ by Binet's formula. The argument for the odd convergents is basically identical.
Edit: As M. Winter points out in the comments, the last limit is a little tricky. You can follow the steps outlined in the comments, or here is an alternative. Given fractions $$a/c < b/d$$, the mediant satisfies the inequality $$\frac{a}{c} < \frac{a+b}{c+d} < \frac{b}{d}.$$ In our case, we have $$\frac{F_na}{F_{n+1}a} < \frac{F_na + F_{n+1}b}{F_{n+1}a+F_{n+2}b} < \frac{F_{n+1}b}{F_{n+2}b},$$ so the result follows by the squeeze theorem.
• I am not sure how the rearrangement in you last formula works out (it is not like e.g. $F_{n+1}$ dominates $F_n$), but here is how I would do it: $$\frac{F_{n}a + F_{n+1}b}{F_{n+1}a + F_{n+2}b} = \frac{a + F_{n+1}/F_n \cdot b}{F_{n+1}/F_n\cdot a + F_{n+2}/F_n\cdot b} \to \frac{a+\phi b}{\phi a + \phi^2b} = \frac{a+\phi b}{\phi (a+\phi b)} = \phi^{-1}.$$ – M. Winter Nov 18 '19 at 21:50
• @M.Winter Yes, you are of course correct. Thank you for the correction. – EuYu Nov 18 '19 at 22:35
• Nice to see the Fibonacci connection! – Joseph O'Rourke Nov 19 '19 at 1:39
Your numerators and denominators follow the same recursive relationship that defines the Fibonacci sequence. I.e. each time you make a new number (either a new numerator or a new denominator), the new number is equal to the sum of the two most recent previously made numbers.
Any sequence that follows this recursive relationship (the Fibonacci sequence being the most famous) has, as general term, $$x\cdot \varphi^n + y\cdot (1-\varphi)^n$$ where the exact values of $$x$$ and $$y$$ are decided by what the first two numbers are.
Now note that $$1-\varphi$$ is a number between $$-1$$ and $$0$$, so $$(1-\varphi)^n$$ becomes really small as $$n$$ grows. Which is to say, your two numbers come closer and closer to being pure powerss of the golden ratio. And since they are (close to being) pure powers of the golden ratio, with exponents one apart, the ratio between them is (close to being) the golden ratio. This conclusion is valid for any starting point that doesn't give $$x = 0$$, which apart from starting at $$\frac 00$$ is impossible to do with integers.
First consider the sequence of every second fraction: $$\frac{a_{2n+2}}{b_{2n+2}} = \frac{a_{2n}+b_{2n}}{a_{2n}+2b_{2n}} = \frac{\frac{a_{2n}}{b_{2n}} +1}{\frac{a_{2n}}{b_{2n}} + 2} = f(\frac{a_{2n}}{b_{2n}})$$ where $$f(x)$$ is defined as $$f(x) = \frac{x+1}{x+2} = 1 - \frac{1}{x+2}$$ for $$x \ge 0$$.
Use the monotony of $$f$$ to show that $$\left(\frac{a_{2n}}{b_{2n}}\right)_n$$ is a monotonic and bounded sequence, and determine its limit $$L$$ as the (unique positive) fixed point of $$f$$.
Then consider the fractions with odd indices: $$\frac{a_{2n}}{b_{2n}} \to L$$ implies $$\frac{a_{2n+1}}{b_{2n+1}} = \frac{a_{2n} + b_{2n}}{b_{2n}} \to L + 1 \, .$$
You are re-stating the Fibonacci sequence, and by the theory of linear recurrences, the terms are quasi-proportional to the powers of the largest root of the characteristic equation
$$\phi^2-\phi-1=0.$$
Hence, the ratio of successive terms quickly tends to $$\phi$$. |
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# Percent of Change
## %change=[(final amount-original amount)/original amount] x 100%
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Percent of Change
Have you ever had to go to work when it was very cold?
Jenny put on her parka and headed to work at the candy store. At noon the temperature was . At 6:00 P.M. the temperature dropped to .
What was the percent of change in the temperature?
This Concept is about using a percent of change to find a new amount. Use what you learn to figure out this Concept.
### Guidance
If we are given the percent of increase or the percent of decrease and the original amount, we can find the new amount by using the following formula.
Amount of change = percent of change original amount
Let’s apply this to a problem.
Find the new number when 75 is decreased by 40%?
First find the amount of change.
Since the original number is being decreased, we subtract the amount of change from the original number to find the new number.
When 75 is decreased by 40%, the new number is 45.
Find the new number when 28 is increased by 125%.
First find the amount of change.
Since the original number is being increased, we add the amount of change to the original number to find the new number.
When 28 is increased by 125%, the new number is 63.
Now it's your turn to try a few. Find each new amount.
#### Example A
Find the new number when 45 is increased by 10%.
Solution:
#### Example B
Find the new number when 80 is decreased by 15%.
Solution:
#### Example C
Find the new number when 50 is increased by 25%.
Solution:
Here is the original problem once again.
Jenny put on her parka and headed to work at the candy store. At noon the temperature was . At 6:00 P.M. the temperature dropped to .
What was the percent of change in the temperature?
First, figure out the amount of the change.
The amount of change is
The temperature decreased by 20%.
### Vocabulary
Here are the vocabulary words used in this Concept.
Percent of Increase
the percent that a price or cost or number has increased.
Percent of Decrease
the percent that a price or cost or number has decreased.
### Guided Practice
Here is one for you to try on your own.
The population of Westville grew from 25,000 to 27,000 in two years. What was the percent of increase for this period of time?
Answer
First, find the amount of the change by subtracting.
The amount of change is
Next, find the percent of the increase.
The population increased 8% over this period of time.
### Video Review
Here is a video for review.
### Practice
Directions: Find the percent of change and then use it to find a new amount.
1. 25 decreased by 10%
2. 30 decreased by 15%
3. 18 decreased by 10%
4. 30 decreased by 9%
5. 12 decreased by 12%
6. 90 decreased by 14%
7. 200 decreased by 80%
8. 97 decreased by 11%
9. 56 decreased by 25%
10. 15 decreased by 20%
11. 220 decreased by 5%
12. 75 decreased by 10%
13. 180 decreased by 18%
14. 1500 decreased by 12%
15. 18,000 decreased by 24%
### Vocabulary Language: English
Percent Equation
Percent Equation
The percent equation can be stated as: "Rate times Total equals Part," or "R% of Total is Part."
Percent of Decrease
Percent of Decrease
The percent of decrease is the percent that a value has decreased by.
Percent of Increase
Percent of Increase
The percent of increase is the percent that a value has increased by.
### Explore More
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# Polynomials
Polynomials are algebraic expressions that consist of variables and coefficients. Variables are also sometimes called indeterminates. We can perform arithmetic operations such as addition, subtraction, multiplication, and also positive integer exponents for polynomial expressions but not division by variable. An example of a polynomial with one variable is x2+x-12. In this example, there are three terms: x2, x and -12.
Also, Check: What is Mathematics
The word polynomial is derived from the Greek words ‘poly’ means ‘many‘ and ‘nominal’ means ‘terms‘, so altogether it is said as “many terms”. A polynomial can have any number of terms but not infinite. Let’s learn about the degrees, terms, types, properties, and polynomial functions in this article.
## What is a Polynomial?
Polynomial is made up of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is defined as an expression which is composed of variables, constants and exponents, that are combined using mathematical operations such as addition, subtraction, multiplication and division (No division operation by a variable). Based on the number of terms present in the expression, it is classified as monomial, binomial, and trinomial. Examples of constants, variables and exponents are as follows:
• Constants. Example: 1, 2, 3, etc.
• Variables. Example: g, h, x, y, etc.
• Exponents: Example: 5 in x5 etc.
## Standard Form of a Polynomial
P(x) = anxn + an-1xn-1 +an-2xn-2 + ………………. + a1x + a0
Where an, an-1, an-2, ……………………, a1, a0 are called coefficients of xn, xn-1, xn-2, ….., x and constant term respectively and it should belong to real number (⋲ R).
### Notation
The polynomial function is denoted by P(x) where x represents the variable. For example,
P(x) = x2-5x+11
If the variable is denoted by a, then the function will be P(a)
## Degree of a Polynomial
The degree of a polynomial is defined as the highest exponent of a monomial within a polynomial. Thus, a polynomial equation having one variable which has the largest exponent is called a degree of the polynomial.
Polynomial Degree Example
Zero Polynomial Not Defined 6
Constant 0 P(x) = 6
Linear Polynomial 1 P(x) = 3x+1
Quadratic Polynomial 2 P(x) = 4x2+1x+1
Cubic Polynomial 3 P(x) = 6x3+4x2+3x+1
Quartic Polynomial 4 P(x) = 6x4+3x3+3x2+2x+1
Example: Find the degree of the polynomial P(x) = 6s4+ 3x2+ 5x +19
Solution:
The degree of the polynomial is 4 as the highest power of the variable 4.
## Terms of a Polynomial
The terms of polynomials are the parts of the expression that are generally separated by “+” or “-” signs. So, each part of a polynomial in an expression is a term. For example, in a polynomial, say, 2x2 + 5 +4, the number of terms will be 3. The classification of a polynomial is done based on the number of terms in it.
Polynomial Terms Degree P(x) = x3-2x2+3x+4 x3, -2x2, 3x and 4 3
## Types of Polynomials
Depending upon the number of terms, polynomials are divided into the following categories:
• Monomial
• Binomial
• Trinomial
• Polynomial containing 4 terms (Quadronomial)
• Polynomial containing 5 terms (pentanomial ) and so on …
These polynomials can be combined using addition, subtraction, multiplication, and division but is never divided by a variable. A few examples of Non Polynomials are: 1/x+2, x-3
### Monomial
A monomial is an expression which contains only one term. For an expression to be a monomial, the single term should be a non-zero term. A few examples of monomials are:
• 5x
• 3
• 6a4
• -3xy
### Binomial
A binomial is a polynomial expression which contains exactly two terms. A binomial can be considered as a sum or difference between two or more monomials. A few examples of binomials are:
• – 5x+3,
• 6a4 + 17x
• xy2+xy
### Trinomial
A trinomial is an expression which is composed of exactly three terms. A few examples of trinomial expressions are:
• – 8a4+2x+7
• 4x2 + 9x + 7
Monomial Binomial Trinomial One Term Two terms Three terms Example: x, 3y, 29, x/2 Example: x2+x, x3-2x, y+2 Example: x2+2x+20
## Properties
Some of the important properties of polynomials along with some important polynomial theorems are as follows:
### Property 1: Division Algorithm
If a polynomial P(x) is divided by a polynomial G(x) results in quotient Q(x) with remainder R(x), then,
P(x) = G(x) Q(x) + R(x)
Where R(x)=0 or the degree of R(x) < the degree of G(x)
### Property 2: Bezout’s Theorem
Polynomial P(x) is divisible by binomial (x – a) if and only if P(a) = 0.
### Property 3: Remainder Theorem
If P(x) is divided by (x – a) with remainder r, then P(a) = r.
### Property 4: Factor Theorem
A polynomial P(x) divided by Q(x) results in R(x) with zero remainders if and only if Q(x) is a factor of P(x).
### Property 5: Intermediate Value Theorem
If P(x) is a polynomial, and P(x) ≠ P(y) for (x < y), then P(x) takes every value from P(x) to P(y) in the closed interval [x, y].
### Property 6
The addition, subtraction and multiplication of polynomials P and Q result in a polynomial where,
Degree(P ± Q) ≤ Degree(P or Q)
Degree(P × Q) = Degree(P) + Degree(Q)
### Property 7
If a polynomial P is divisible by a polynomial Q, then every zero of Q is also a zero of P.
### Property 8
If a polynomial P is divisible by two co-prime polynomials Q and R, then it is divisible by (Q • R).
### Property 9
If P(x) = a0 + a1x + a2x2 + …… + anxn is a polynomial such that deg(P) = n ≥ 0 then, P has at most “n” distinct roots.
### Property 10: Descartes’ Rule of Sign
The number of positive real zeroes in a polynomial function P(x) is the same or less than by an even number as the number of changes in the sign of the coefficients. So, if there are “K” sign changes, the number of roots will be “k” or “(k – a)”, where “a” is some even number.
### Property 11: Fundamental Theorem of Algebra
Every non-constant single-variable polynomial with complex coefficients has at least one complex zero.
### Property 12
If P(x) is a polynomial with real coefficients and has one complex zero (x = a – bi), then x = a + bi will also be a zero of P(x). Also, x2 – 2ax + a2 + b2 will be a factor of P(x).
## Polynomial Equations
Polynomial equations are those expressions which are made up of multiple constants and variables. The standard form of writing a polynomial equation is to put the highest degree first and then, at last, the constant term. An example of a polynomial equation is:
0 = a4 +3a3 -2a2 +a +1
## Polynomial Functions
A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n can be written as:
f(x) = a0xn + a1xn-1 + a2xn-2 + ….. + an-2x2 + an-1x + an
## Solving Polynomials
Any polynomial can be easily solved using basic algebra and factorization concepts. While solving the polynomial equation, the first step is to set the right-hand side as 0. The explanation of a polynomial solution is explained in two different ways:
• Solving Linear Polynomials
### Solving Linear Polynomials
Getting the solution of linear polynomials is easy and simple. First, isolate the variable term and make the equation as equal to zero. Then solve as basic algebra operation. An example of finding the solution of a linear equation is given below:
Example: Solve 3x – 9
Solution:
First, make the equation as 0. So,
3x – 9 = 0
⇒ 3x = 9
⇒ x = 9/3
Or, x = 3.
Thus, the solution of 3x-9 is x = 3.
To solve a quadratic polynomial, first, rewrite the expression in the descending order of degree. Then, equate the equation and perform polynomial factorization to get the solution of the equation. An example to find the solution of a quadratic polynomial is given below for better understanding.
Example: Solve 3x2 – 6x + x3 – 18
Solution:
First, arrange the polynomial in the descending order of degree and equate to zero.
⇒ x3 + 3x2 -6x – 18 = 0
Now, take the common terms.
x2(x+3) – 6(x+3) =0
⇒ (x2-6)(x+3)=0
So, the solutions will be x =-3 or
x2 = 6
Or, x = ±√6
More Polynomials Related Resources:
## Polynomial Operations
There are four main polynomial operations which are:
• Subtraction of Polynomials
• Multiplication of Polynomials
• Division of Polynomials
Each of the operations on polynomials is explained below using solved examples.
To add polynomials, always add the like terms, i.e. the terms having the same variable and power. The addition of polynomials always results in a polynomial of the same degree. For example,
Example: Find the sum of two polynomials: 5x3+3x2y+4xy−6y2, 3x2+7x2y−2xy+4xy2−5
Solution:
First, combine the like terms while leaving the unlike terms as they are. Hence,
(5x3+3x2y+4xy−6y2)+(3x2+7x2y−2xy+4xy2−5)
= 5x3+3x2+(3+7)x2y+(4−2)xy+4xy2−6y2−5
= 5x3+3x2+10x2y+2xy+4xy2−6y2−5
### Subtraction of Polynomials
Subtracting polynomials is similar to addition, the only difference being the type of operation. So, subtract the like terms to obtain the solution. It should be noted that subtraction of polynomials also results in a polynomial of the same degree.
Example: Find the difference of two polynomials: 5x3+3x2y+4xy−6y2, 3x2+7x2y−2xy+4xy2−5
Solution:
First, combine the like terms while leaving the unlike terms as they are. Hence,
(5x3+3x2y+4xy−6y2)-(3x2+7x2y−2xy+4xy2−5)
= 5x3-3x2+(3-7)x2y+(4+2)xy-4xy2−6y2+5
= 5x3-3x2-4x2y+6xy-4xy2−6y2+5
### Multiplication of Polynomials
Two or more polynomial when multiplied always result in a polynomial of higher degree (unless one of them is a constant polynomial). An example of multiplying polynomials is given below:
Example: Solve (6x−3y)×(2x+5y)
Solution:
⇒ 6x ×(2x+5y)–3y × (2x+5y) ———- Using distributive law of multiplication
⇒ (12x2+30xy) – (6yx+15y2) ———- Using distributive law of multiplication
⇒12x2+30xy–6xy–15y2 —————– as xy = yx
Thus, (6x−3y)×(2x+5y)=12x2+24xy−15y2
### Division of Polynomials
Division of two polynomial may or may not result in a polynomial. Let us study below the division of polynomials in detail. To divide polynomials, follow the given steps:
Polynomial Division Steps:
If a polynomial has more than one term, we use long division method for the same. Following are the steps for it.
1. Write the polynomial in descending order.
2. Check the highest power and divide the terms by the same.
3. Use the answer in step 2 as the division symbol.
4. Now subtract it and bring down the next term.
5. Repeat steps 2 to 4 until you have no more terms to carry down.
6. Note the final answer, including remainder, will be in the fraction form (last subtract term).
## Polynomial Examples
Example:
Given two polynomial 7s3+2s2+3s+9 and 5s2+2s+1.
Solve these using mathematical operation.
Solution:
Given polynomial:
7s3+2s2+3s+9 and 5s2+2s+1
= 7s3+(2s2+5s2)+(3s+2s)+(9+1)
= 7s3+7s2+5s+10
Hence, addition result in a polynomial.
Polynomial Subtraction: (7s3+2s2+3s+9) – (5s2+2s+1)
= 7s3+(2s2-5s2)+(3s-2s)+(9-1)
= 7s3-3s2+s+8
Hence addition result in a polynomial.
Polynomial Multiplication:(7s3+2s2+3s+9) × (5s2+2s+1)
= 7s3 (5s2+2s+1)+2s2 (5s2+2s+1)+3s (5s2+2s+1)+9 (5s2+2s+1))
= (35s5+14s4+7s3)+ (10s4+4s3+2s2)+ (15s3+6s2+3s)+(45s2+18s+9)
= 35s5+(14s4+10s4)+(7s3+4s3+15s3)+ (2s2+6s2+45s2)+ (3s+18s)+9
= 35s5+24s4+26s3+ 53s2+ 21s +9
Polynomial Division: (7s3+2s2+3s+9) ÷ (5s2+2s+1)
(7s3+2s2+3s+9)/(5s2+2s+1)
This cannot be simplified. Therefore, division of these polynomial do not result in a Polynomial.
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## Frequently Asked Questions – FAQs
Q1
### What is a Polynomial?
A polynomial is an expression that consists of variables (or indeterminate), terms, exponents and constants. For example, 3x2 -2x-10 is a polynomial.
Q2
### What are terms, degrees and exponents in a polynomial?
If 2x2 – 3x +19 is a polynomial, then;
Terms: 2x2,-3x & 19
Degree: 2 (the highest exponent of variable x)
Exponents: Power raised to variable x, i.e. 2 and 1.
Q3
### What is the standard form of the polynomial?
A standard polynomial is the one where the highest degree is the first term, and subsequently, the other terms come. For example, x3 – 3x2 + x -12 is a standard polynomial. So the highest degree here is 3, then comes 2 and then 1.
Q4
### What is the degree of zero and constant polynomial?
The degree of zero polynomial is not defined while the degree of a constant polynomial is 1.
Q5
### Is 8 a polynomial?
8 can be written as 8x0 or 0x2+0x+8, which represents the polynomial expression. Therefore, we can consider 8 as a polynomial.
Q6
### How to add and subtract polynomials?
To add polynomials, always add the like terms, i.e. the terms having the same variable and power. The addition of polynomials always results in a polynomial of the same degree.
For example if we add x2+3x and 2x2 + 2x + 9, then we get:
x2+3x+2x2+2x+9 = 3x2+5x+9. Subtracting polynomials is similar to addition, the only difference being the type of operation. So, subtract the like terms to obtain the solution. It should be noted that subtraction of polynomials also results in a polynomial of the same degree.
So,
x2+3x-(2x2+2x+9) = x2+3x-2x2-2x-9 = -x2+x-9 |
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# Evaluate $8.5 \times 9.5$ using a suitable standard identity.
Last updated date: 13th Jun 2024
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Hint: Here, we need to find the value of $8.5 \times 9.5$ using a suitable standard identity. We will rewrite the two numbers as a sum or difference of two numbers in such a way that the product can be found using one of the standard algebraic identities. Then, we will simplify the expression to find the required value.
Formula Used: The product of the sum of two number and the difference of two numbers can be calculated using the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
We can evaluate the given product using any of the two identities $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$ or $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
We will use the second identity to solve this problem.
First, we will rewrite the given numbers as the sum or difference of two numbers such that the identity is applicable.
We know that $8.5$ is the difference of 9 and $0.5$, and $9.5$ is the sum of 9 and $0.5$.
Therefore, substituting $8.5 = 9 - 0.5$ and $9.5 = 9 + 0.5$, we can rewrite the product as
$\Rightarrow 8.5 \times 9.5 = \left( {9 - 0.5} \right)\left( {9 + 0.5} \right)$
Now, we know that the product of the sum of two number and the difference of two numbers can be calculated using the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
Substituting $a = 9$ and $b = 0.5$ in the algebraic identity, we get
$\Rightarrow \left( {9 - 0.5} \right)\left( {9 + 0.5} \right) = {9^2} - {\left( {0.5} \right)^2}$
Simplifying the expression, we get
$\Rightarrow 8.5 \times 9.5 = 81 - 0.25$
Subtracting $0.25$ from 81, we get
$\Rightarrow 8.5 \times 9.5 = 80.75$
Therefore, the value of the product $8.5 \times 9.5$ is $80.75$.
Note: We can also solve the problem using the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$.
Substituting $8.5 = 8 + 0.5$ and $9.5 = 8 + 1.5$, we can rewrite the product as
$\Rightarrow 8.5 \times 9.5 = \left( {8 + 0.5} \right)\left( {8 + 1.5} \right)$
Now, we will use the algebraic identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$.
Substituting $x = 8$,$a = 0.5$ and $b = 1.5$ in the algebraic identity, we get
$\Rightarrow \left( {8 + 0.5} \right)\left( {8 + 1.5} \right) = {8^2} + \left( {0.5 + 1.5} \right)8 + \left( {0.5} \right)\left( {1.5} \right)$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow 8.5 \times 9.5 = 64 + 2 \times 8 + 0.75\\ \Rightarrow 8.5 \times 9.5 = 64 + 16 + 0.75\end{array}$
Adding the terms of the expression, we get
$\Rightarrow 8.5 \times 9.5 = 80.75$
$\therefore$ The value of the product $8.5 \times 9.5$ is $80.75$. |
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# 5.13: Directed Line Segments
Difficulty Level: At Grade Created by: CK-12
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While babysitting your nephew one day you have a discussion regarding math. As a second grader, he is learning to add and subtract quantities. He asks you about what you're doing in math class, and you explain to him that you have just been introduced to vectors. Once you explain to him that a vector is a mathematical quantity that has both magnitude and direction, his question is both simple and yet brilliant: "Why do you need vectors? Can't everything just be described with numbers that don't have direction?"
At the end of this Concept, you'll know how to answer your nephew.
### Guidance
A vector is represented diagrammatically by a directed line segment or arrow. A directed line segment has both magnitude and direction. Magnitude refers to the length of the directed line segment and is usually based on a scale. The vector quantity represented, such as influence of the wind or water current may be completely invisible.
A 25 mph wind is blowing from the northwest. If 1 cm=5 mph\begin{align*}1\ cm = 5\ mph\end{align*}, then the vector would look like this:
An object affected by this wind would travel in a southeast direction at 25 mph.
A vector is said to be in standard position if its initial point is at the origin. The initial point is where the vector begins and the terminal point is where it ends. The axes are arbitrary. They just give a place to draw the vector.
If we know the coordinates of a vector’s initial point and terminal point, we can use these coordinates to find the magnitude and direction of the vector.
All vectors have magnitude. This measures the total distance moved, total velocity, force or acceleration. “Distance” here applies to the magnitude of the vector even though the vector is a measure of velocity, force, or acceleration. In order to find the magnitude of a vector, we use the distance formula. A vector can have a negative magnitude. A force acting on a block pushing it at 20 lbs north can be also written as vector acting on the block from the south with a magnitude of -20 lbs. Such negative magnitudes can be confusing; making a diagram helps. The -20 lbs south can be re-written as +20 lbs north without changing the vector. Magnitude is also called the absolute value of a vector.
#### Example A
If a vector starts at the origin and has a terminal point with coordinates (3,5), find the length of the vector.
If we know the coordinates of the initial point and the terminal point, we can find the magnitude by using the distance formula. Initial point (0,0) and terminal point (3,5).
Solution: |v|=(30)2+(50)2=9+25=5.8\begin{align*}|\vec{v}| = \sqrt{(3 - 0)^2 + (5 - 0)^2} = \sqrt{9 + 25} = 5.8\end{align*} The magnitude of v\begin{align*}\vec{v}\end{align*} is 5.8.
If we don’t know the coordinates of the vector, we must use a ruler and the given scale to find the magnitude. Also notice the notation of a vector, which is usually a lower case letter (typically u,v\begin{align*}u, v\end{align*}, or w\begin{align*}w\end{align*}) in italics, with an arrow over it, which indicates direction. If a vector is in standard position, we can use trigonometric ratios such as sine, cosine and tangent to find the direction of that vector.
#### Example B
If a vector is in standard position and its terminal point has coordinates of (12, 9) what is the direction?
Solution: The horizontal distance is 12 while the vertical distance is 9. We can use the tangent function since we know the opposite and adjacent sides of our triangle.
tanθtan1912=912=36.9\begin{align*}\tan \theta & = \frac{9}{12} \\ \tan^{-1} \frac{9}{12} & = 36.9^\circ\end{align*}
So, the direction of the vector is 36.9\begin{align*}36.9^\circ\end{align*}.
If the vector isn’t in standard position and we don’t know the coordinates of the terminal point, we must a protractor to find the direction.
Two vectors are equal if they have the same magnitude and direction. Look at the figures below for a visual understanding of equal vectors.
#### Example C
Determine if the two vectors are equal.
a\begin{align*}\vec{a}\end{align*} is in standard position with terminal point (-4, 12)
b\begin{align*}\vec{b}\end{align*} has an initial point of (7, -6) and terminal point (3, 6)
Solution: You need to determine if both the magnitude and the direction are the same.
Magnitude:|a||b|Direction:ab=(0(4))2+(012)2=16+144=160=410=(73)2+(66)2=16+144=160=410tanθ=124θ=108.43tanθ=6673=124θ=108.43\begin{align*}\text{Magnitude}:\quad |\vec{a}| & = \sqrt{(0-(-4))^2 + (0 - 12)^2} = \sqrt{16+144} = \sqrt{160} = 4\sqrt{10} \\ |\vec{b}| & = \sqrt{(7-3)^2 + (-6-6)^2} = \sqrt{16+144} = \sqrt{160} = 4 \sqrt{10} \\ \text{Direction}:\quad \vec{a} & \rightarrow \tan \theta = \frac{12}{-4} \rightarrow \theta = 108.43^\circ \\ \vec{b} & \rightarrow \tan \theta = \frac{-6-6}{7-3} = \frac{-12}{4} \rightarrow \theta = 108.43^\circ\end{align*}
Because the magnitude and the direction are the same, we can conclude that the two vectors are equal.
### Vocabulary
Directed Line Segment: A directed line segment is a portion of a line that has both a magnitude and direction.
Magnitude: A magnitude is the length of a line segment or vector.
Vector: A vector is a type of mathematical quantity that has both a magnitude and a direction.
### Guided Practice
1. Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.
initial ( 2, 4) terminal (8, 6)
2. Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.
initial (5, -2) terminal (3, 1)
3. Assume a\begin{align*}\vec{a}\end{align*} is in standard position. For the terminal point (12, 18), find the magnitude and direction of thevector.
Solutions:
1. |a|=(28)2+(46)2=6.3\begin{align*}|\vec{a}| = \sqrt{(2-8)^2 + (4-6)^2} = 6.3\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{4-6}{2-8} \right ) = 18.4^\circ\end{align*}
2. \begin{align*}|\vec{a}| = \sqrt{(5-3)^2 + (-2-1)^2} = 3.6\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{-2-1}{5-3} \right ) = 123.7^\circ\end{align*}. Note that when you use your calculator to solve for \begin{align*}\tan^{-1} (\frac{-2-1}{5-3})\end{align*}, you will get \begin{align*}-56.3^\circ\end{align*}. The calculator produces this answer because the range of the calculator’s \begin{align*}y = \tan^{-1} x\end{align*} function is limited to \begin{align*}-90^\circ < y < 90^\circ\end{align*}. You need to sketch a draft of the vector to see that its direction when placed in standard position is into the second quadrant (and not the fourth quadrant), and so the correct angle is calculated by moving the angle into the second quadrant through the equation \begin{align*}-56.3^\circ + 180^\circ=123.7^\circ.\end{align*}
3. \begin{align*}|\vec{a}| = \sqrt{12^2 + 18^2} = 21.6\end{align*}, direction \begin{align*}=\tan^{-1} \left (\frac{18}{12} \right ) = 56.3^\circ \end{align*}
### Concept Problem Solution
Your nephew's thinking is quite good. Many things in the world can be described by numbers, without the use of direction. However, math needs to "line up" with reality. If something doesn't work with just numbers, there needs to be a new type of mathematical quantity to describe the behavior completely. For example, consider two cars that are both moving at 25 miles per hour. Will they collide?
You can see that there isn't enough information to answer the question. You don't know which way the cars are going. If the two cars are going in the same direction, then they won't collide. If, however, they are going directly at each other, then they will certainly collide. In order to describe the behavior of the cars completely, a quantity is needed that is not just the magnitude of the car's motion, but also the direction - which is why vectors are needed.
And that is how you should answer your nephew.
### Practice
1. What is the difference between the magnitude and direction of a vector?
2. How can you determine the magnitude of a vector if you know its initial point and terminal point?
3. How can you determine the direction of a vector if you know its initial point and terminal point?
4. How can you determine whether or not two vectors are equal?
5. If a vector starts at the origin and has a terminal point with coordinates (2, 7), find the magnitude of the vector.
6. If a vector is in standard position and its terminal point has coordinates of (3, 9), what is the direction of the vector?
7. If a vector has an initial point at (1, 6) and has a terminal point at (5, 9), find the magnitude of the vector.
8. If a vector has an initial point at (1, 4) and has a terminal point at (8, 7), what is the direction of the vector?
Given the initial and terminal coordinates below, find the magnitude and direction of the vector that results.
1. initial (4, -1); terminal (5, 3)
2. initial (2, -3); terminal (4, 5)
3. initial (3, 2); terminal (0, 3)
4. initial (-2, 5); terminal (2, 1)
Determine if the two vectors are equal.
1. \begin{align*}\vec{a}\end{align*} is in standard position with terminal point (1, 5) and \begin{align*}\vec{b}\end{align*} has an initial point (3, -2) and terminal point (4, 2).
2. \begin{align*}\vec{c}\end{align*} has an initial point (-3, 1) and terminal point (1, 2) and \begin{align*}\vec{d}\end{align*} has an initial point (3, 5) and terminal point (7, 6).
3. \begin{align*}\vec{e}\end{align*} is in standard position with terminal point (2, 3) and \begin{align*}\vec{f}\end{align*} has an initial point (1, -6) and terminal point (3, -9).
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Directed Line Segment
A directed line segment is a portion of a line that has both a magnitude and direction.
Magnitude
The magnitude of a line segment or vector is the length of the line segment or vector.
Vector
A vector is a mathematical quantity that has both a magnitude and a direction.
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Example 41 The side of a cube are increasing at the rate of 0.5 cm/sec. At what rate its surface area and volume is increasing when the side is 10 cm long?
Solution : 1) The surface area (s) of a cube of side x is s = 6x2
By Differentiating w. r. to ’ t ’ , we get
where = rate of change of surface area
and = rate of change of the side of the cube.
Given that when x = 10 cm.
= 0.5 cm /sec, then
2) The volume (v) is given by v = x3
\
\
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Example 42 The top of a ladder 20 units long is resting against a vertical wall and its foot on a level pavement, when the ladder beings to slide. When the foot of the ladder is 12 units from the wall, it is sliding away from the wall at the rate of 2 units / sec find the rate at which the top is sliding down.
Solution :
Let AB = 20 units be the length of the ladder.
We have in right angled triangle AOB,
AB2 = AO2 +OB2
\ (20)2 = y2 + x2
i.e. x2 + y2 = 256
Also given that OB = x = 12 units
\ (12)2 + y2 = 400
\ y2 = 256
\ y = 16 units
Now differentiating w. r. to ’ t ’ , x2 + y2 = 400 we get
Also. = 2 units / sec = rate of sliding away from the wall.
\ 16 = - 12(2)
\
\ The rate which the top is sliding down = 1.5 units/sec
Index
5.1 Tangent And Normal Lines
5.2 Angle Between Two Curves
5.3 Interpretation Of The Sign Of The Derivative
5.4 Locality Increasing Or Decreasing Functions 5.5 Critical Points
5.6 Turning Points
5.7 Extreme Value Theorem
5.8 The Mean-value Theorem
5.9 First Derivative Test For Local Extrema
5.10 Second Derivative Test For Local Extrema
5.11 Stationary Points
5.12 Concavity And Points Of Inflection
5.13 Rate Measure (distance, Velocity And Acceleration)
5.14 Related Rates
5.15 Differentials : Errors And Approximation |
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# Measuring Diversity
Simpson’s index of diversity
Species diversity is described as the number of species in a community. The more species, the higher the diversity. What if there are two separate communities like this:
Community #1 has 150 individuals per each of 20 different species (3000 individuals in total)
Community #2 has 10 individuals per each of 19 species, and 2990 individuals of the last species (3000 individuals in total)
It doesn’t take a complex formula to figure out that community #1 is far more diverse compared to community #2, despite them having the same number of species and individuals. The distribution of individuals to species is important in determining a community’s diversity.
The above example is easy enough, but for most purposes a formula is needed. This formula measures the index of diversity a.k.a. Simpson’s index of diversity, which is simply a measure of diversity in a community. By calculating it and obtaining a numerical value, different communities can be easily compared.
Right, here it comes…
No, don’t run away yet! Wait and see how easy it is to work out.
D = Diversity index
N = total number of all organisms
n = total number of organisms of each species
Σ = sum of
Now it’s simply a matter of replacing numbers. Look, I made it all purple so you would enjoy looking at it. Let’s work out the index of diversity for community #1 (from above).
Firstly, we need a value for N. What’s the total number of organisms? 3000. Sorted.
Next, we need a value for N – 1. No calculators! …2999, sorted.
Finally, we need a value for n and n – 1. n = 150, while n – 1 = 149.
Drawing up a table helps:
species n n – 1 n(n – 1) a 150 149 22350 b 150 149 22350 c 150 149 22350 d 150 149 22350 e 150 149 22350 f 150 149 22350 g 150 149 22350 h 150 149 22350 i 150 149 22350 j 150 149 22350 k 150 149 22350 l 150 149 22350 m 150 149 22350 n 150 149 22350 o 150 149 22350 p 150 149 22350 q 150 149 22350 r 150 149 22350 s 150 149 22350 t 150 149 22350 Total 3000 2980 447000
3000*2999 8,997,000
So, D = —————– = ————— = 20
447000 447000
20 in this case is maximum diversity (there are 20 different species). If the index was 1, then diversity would have been non-existent. An index of 10 would indicate moderate diversity.
Now work out the index of diversity for community #2 using the table above and the walk through as a guide. You should get a pretty low value. I know it’s a bit confusing that the above numbers are identical in all the columns, but if you work out community #2 then the values for 1 species should be different to the other 19.
Most of the time all species will have different values. The working of it is the same though.
Plants are difficult to count in individuals, so the percentage cover in a quadrat is usually used instead.
Ok byeeeee
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Hi! I have recently finished my first year doing biology at uni and I subscribed to you back when I was doing AQA A levels (glad I don’t live in Wales after watching this vid lol) and you and your website helped me so much. Just wanted to say it’s so nice to see you |
# How Many 4 Digit Even Numbers Can Be Formed Using 0 9?
## How many 4 digit even numbers are possible if the leftmost digit Cannot be zero?
SOLUTION: how many 4 digit odd numbers are possible if the left digit cannot be zero.
Since it cannot equal 0, there are 9 possibilities for the 1st digit.
The 2nd and 3rd digits can be any number from 0-9 so there are 10 possibilities for them..
## What is a 4 digit odd number?
Hence there are 0 4 digit numbers and 0 4 digit odd numbers. Total possible 4 digit odd numbers that cannot have the first digit as a 0 or a 3 are (5000-500-0 = 4500). Answer = 4500.
## How many numbers are there between 1000 and 9999 that contain only odd digits?
For the third digit we can likewise choose any digit from 0 to 9, with the exception of the 3 already chosen. So there are 7 possibilities for the third digit. The overall number of odd integers with 4 digits and 4 unique digits is therefore 5 * 8 * 8 * 7 = 2240.
## What is the largest odd number?
The largest two-digit odd number we could make is 97. We could not make 99 because we only have one 9. If you pulled out an odd number first this should always go in the ‘ones’ column.
## How many 4 digit numbers can be formed using the digits 1 to 9 if repetitions are not allowed?
However, the digits cannot be repeated in the 4-digit numbers and thousand’s place is already occupied with a digit. The hundred’s, ten’s, and unit’s place is to be filled by the remaining 9 digits. Hence, by FPC, total number of numbers = 9 × 9 × 8 × 7 = 4536. 3.
## What is the greatest 5 digit even number?
99998The greatest 5-digit even number is 99998.
## How many 4 digit even numbers can be formed using 0 6?
Since the first digit cannot be zero, you have five choices for the first digit, five choices for the second digit, and four choices for the third digit, giving 5⋅5⋅4⋅3=300 four digit even numbers that end in 2, 4, or 6.
## What is the largest 4 digit odd number?
99999999 is the largest 4 digit odd no because it is the largest 4 digit no which can not be divided by 2.
## How many four digit numbers are possible criteria being that all the four digits are odd?
So even number can be chosen in 4*5*5*5=500 ways. So total 625+500=1125 4-digit numbers have all their digits either even or odd.
## How many 4 digit combinations are there with 10 numbers?
10 000Finally you have 10 choices for the fourth digit and thus there are 10x10x10x10 = 10 000 possible 4 digit combinations from 0-9.
## What is the smallest odd number?
The number whose unit digit is other than 0,2,4,6,8 are odd numbers. The smallest even natural number is 2 and smallest odd natural number is 1.
## What is the sum of first 5 odd numbers?
The number series 1, 3, 5, 7, 9. Therefore, 25 is the sum of first 5 odd numbers.
## Is 0 an even number?
Zero is an even number. In other words, its parity—the quality of an integer being even or odd—is even. This can be easily verified based on the definition of “even”: it is an integer multiple of 2, specifically 0 × 2. … In this sense, 0 is the “most even” number of all.
## Are any odd numbers divisible by 4?
34,807 – the last digit is a 7. This means that the number is ODD and 2 will not divide into it evenly. There will be a remainder. … The Rule for 4 : If the last two digits of a whole number are divisible by 4, then the entire number is divisible by 4.
## How many different 4 digit even numbers can be formed?
120 waysThere are 120 ways to form 4-digit even numbers from the given digits.
## What is the greatest 4 digit even number?
The largest 4 digit number that can ever be is 9999.
## How long does it take to crack a 4 digit PIN?
As such, each PIN entry takes approximately 40 seconds, meaning that it would take up to ~111 hours to bruteforce a 4 digit PIN.
## How many 4 digit numbers are there in which all 4 digits are even?
For it to be a 4 digits integer, the first digit cannot be zero. 2nd place ( hundred place) can also be filled up by 4 number : 2, 4, 6, 8 and same logic applies to 3rd ( 10th ) and 4th( unit) . so number of ways is 4*4*4*4= 256 ways .
## What is greatest and smallest number?
We know that a four digit number has four places, i.e., thousands, hundreds, tens and ones or units from left to right as Th, H, T, O. If greatest to lowest digits are placed at these places in descending order, we get the greatest number and if placed in ascending order, we get the smallest number.
## How many 4 digit codes can be made with the numbers 0 9?
10,000There are 10,000 possible combinations that the digits 0-9 can be arranged into to form a four-digit code. Berry analyzed those to find which are the least and most predictable.
## What is a 4 digit even number?
The fourth digit must be even, and there are 5 even digits (0, 2, 4, 6, or 8). Thus, the number of 4-digit even numbers is 9 x 10 x 10 x 5 = 4500.
## What is the most common 4 digit code?
Researchers at the data analysis firm Data Genetics have found that the three most popular combinations—“1234,” “1111,” and “0000”—account for close to 20 percent of all four-digit passwords.
## What is the smallest whole number?
Which is the smallest whole number? Solution. Zero (0) is the smallest whole number. 4. How many whole numbers are there between 32 and 53? |
In this article, we"ll show you specifically how to calculate 2/3 of 6 so you can work out the portion of any type of number quickly and easily! Let"s obtain to the math!
Want to quickly learn or show students just how to convert 2/3 of 6? pat this really quick and also fun video clip now!
You probably recognize that the number over the fraction line is dubbed the numerator and also the number listed below it is called the denominator. To occupational out the fraction of any number, we very first need to transform that entirety number into a fraction as well.
You are watching: What is 2/3 of 6
Here"s a little tip for you. Any kind of number have the right to be converted to fraction if you use 1 as the denominator:
So currently that we"ve converted 6 into a fraction, to work-related out the answer, we put the portion 2/3 next by side with our brand-new fraction, 6/1 so the we deserve to multiply those two fractions.
That"s right, all you should do is transform the totality number come a portion and climate multiply the numerators and also denominators. Let"s take a look:
In this case, our new portion can in reality be streamlined down further. To do that, we need to uncover the greatest usual factor that both numbers.
You deserve to use our comfortable GCF calculator to job-related this the end yourself if you desire to. We already did that, and the GCF the 12 and also 3 is 3.
We deserve to now division both the brand-new numerator and the denominator by 3 to simplify this fraction down come its shortest terms.
12/3 = 4
3/3 = 1
When we placed that together, we have the right to see the our finish answer is:
The complete and simplified answer to the question what is 2/3 that 6 is:
4
Hopefully this tutorial has actually helped you to understand exactly how to find the portion of any whole number. You can now go give it a go with much more numbers to practice your newfound portion skills. |
# How do you integrate int 1/(4x^2-9)^(3/2) by trigonometric substitution?
Nov 26, 2016
$\frac{- x}{9 \sqrt{4 {x}^{2} - 9}} + C$
#### Explanation:
$\int \frac{\mathrm{dx}}{4 {x}^{2} - 9} ^ \left(\frac{3}{2}\right)$
Let $2 x = 3 \sec \theta$ such that $2 \mathrm{dx} = 3 \sec \theta \tan \theta d \theta$.
$= \frac{1}{2} \int \frac{2 \mathrm{dx}}{{\left(2 x\right)}^{2} - 9} ^ \left(\frac{3}{2}\right)$
$= \frac{1}{2} \int \frac{3 \sec \theta \tan \theta d \theta}{9 {\sec}^{2} \theta - 9} ^ \left(\frac{3}{2}\right)$
Factoring ${9}^{\frac{3}{2}} = 27$ from the denominator:
$= \frac{1}{2} \int \frac{3 \sec \theta \tan \theta}{27 {\left({\sec}^{2} \theta - 1\right)}^{\frac{3}{2}}} d \theta$
Since ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:
$= \frac{1}{18} \int \frac{\sec \theta \tan \theta}{\tan} ^ 3 \theta d \theta$
$= \frac{1}{18} \int \sec \frac{\theta}{\tan} ^ 2 \theta d \theta$
$= \frac{1}{18} \int \cos \frac{\theta}{\sin} ^ 2 \theta d \theta$
Letting $u = \sin \theta$ so $\mathrm{du} = \cos \theta d \theta$.
$= \frac{1}{18} \int {u}^{-} 2 \mathrm{du}$
$= - \frac{1}{18 u}$
$= - \frac{1}{18} \csc \theta$
$= - \frac{1}{18} \sec \frac{\theta}{\tan} \theta$
$= - \frac{1}{54} \frac{3 \sec \theta}{\sqrt{{\sec}^{2} \theta - 1}}$
$= - \frac{3}{54} \frac{3 \sec \theta}{\sqrt{9 {\sec}^{2} \theta - 9}}$
Using $3 \sec \theta = 2 x$:
$= - \frac{1}{18} \frac{2 x}{\sqrt{4 {x}^{2} - 9}}$
$= \frac{- x}{9 \sqrt{4 {x}^{2} - 9}} + C$ |
# A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?
Nov 9, 2015
$\text{310 K}$
#### Explanation:
If you assume that no heat is lost to the surroundings, then you can say that the heat lost by the hotter sample will be equal to the heat absorbed by the cooler sample.
Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this
$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where
$q$ - heat lost or gained
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature
Now, before doing any calculation, try to predict what will happen when the two samples are mixed.
Notice that you have more hot water than cold water, which means that you can expect the final temperature of the mixture to be closer to $\text{330 K}$ than to $\text{280 K}$.
So, you can say that
$- {q}_{\text{lost" = q_"gained}}$
Here the minus signed is used because heat lost is negative.
This is equivalent to
${m}_{\text{hot" * color(red)(cancel(color(black)(c))) * DeltaT_"hot" = m_"cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold}}$
Let's say that the final temperature is $\textcolor{b l u e}{{T}_{\text{f}}}$. This means that you have
$- 50.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * (color(blue)(T_"f") - 330)color(red)(cancel(color(black)("K"))) = 30.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 280)color(red)(cancel(color(black)("K}}}}$
$- 50.0 \cdot \textcolor{b l u e}{{T}_{\text{f") + 16500 = 30.0 * color(blue)(T_"f}}} - 8400$
$80.0 \cdot \textcolor{b l u e}{{T}_{\text{f}}} = 24900$
color(blue)(T_"f") = 24900/80.0 = "311.25 K"
Rounded to two sig figs, the number of sig figs you have for the two temperatures, the answer will be
T_"f" = color(green)("310 K") |
KS3 Maths Quiz
A train's top speed is 140 mph. Find out more in this quiz.
# Level 5-6 Numbers - Percentages - Increases and Decreases
Percentages are an important skill to master. You'll come across them in real life all the time - wage increases, inflation rates, interest rates etc. No wonder they form a key part of KS3 Maths. In this quiz we look at calculating increases and decreases in term of percentages.
You will often have to find a value when an amount increases or decreases by a percentage. There is more than one way of working it out but perhaps the easiest is to multiply by a decimal number. Say that something you want to buy for £200 has a 5% decrease in price. To work out the new price multiply £200 x 0.95 (95% of the previous price). If the price has increased by 5% then multiply £200 by 1.05 (105% of the previous price). There are other methods you can use so don't feel that you must follow my example!
Try this quiz about increases and decreases in percentages to get some practise - whatever the method you use!
Did you know...
You can play all the teacher-written quizzes on our site for just £9.95 per month. Click the button to sign up or read more.
1. To find the result of decreasing a number by 7% you multiply it by .......
Percentages are easily converted to decimals. Just divide by 100 (or move digits)
2. Harold owns a 14% stake in a company worth £250,000. How much is Harold's stake worth?
To find 14% multiply by 0.14
3. The local football team are playing badly and the number of season tickets sold reduces by 17% from last year. If 3,700 tickets were sold last year, how many were sold this year?
To find 17% multiply by 0.17
4. Thomas earns £22,500 per year. He receives a pay rise of 4.4%. How much does he now earn?
4.4% of 22,500 is 990. To work this out you could multiply 22,500 x 0.044 (4.4 hundredths)
5. In a sale, all the prices are reduced by 20%. Find the sale price of a washing machine which originally cost £299.
One way is 20 / 100 x £299 = £59.80. Then subtract £59.80 from £299 to get £239.20
6. A train's top speed is 140 mph. After a service, its top speed increases by 12%. What is the new top speed?
A 12% increase is a multiplier of 1.12. 1.12 x 140 = 156.8
7. To find the result of increasing a number by 13% you multiply it by .......
If something is increased by x% then its original value will have to be multiplied by (1 + x) to find the new value. The 1 represents 100% and the x represents the percentage increase: in this case (1 + x) = (1 + 0.13) = 1.13. If there was a decrease of x%, then you would have to multiply the original value by (1 - x)
8. To find the result of increasing a number by 66% you multiply it by .......
Calculations are much easier if a multiplier is used
9. Bob, the local car dealer, is reducing all his cars by 15%. Find the reduced price of a Nissan which originally cost £690.
Another way is to use a multiplier. A 15% decrease is a mutliplier of 100 - 15 = 85%. 85% = 0.85 (just divide 85 by 100). £690 x 0.85 = £586.50
10. Simon bought his computer for £230. A year later he sold it for 20% less than he paid for it. How much was it sold for?
To work out 20% just divide by 5
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# Difference between revisions of "Simon's Favorite Factoring Trick"
## Contents
Dr. Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization first popularized by AoPS user Simon Rubinstein-Salzedo.
## The General Statement
The general statement of SFFT is: ${xy}+{xk}+{yj}+{jk}=(x+j)(y+k)$. Two special common cases are: $xy + x + y + 1 = (x+1)(y+1)$ and $xy - x - y +1 = (x-1)(y-1)$.
The act of adding ${jk}$ to ${xy}+{xk}+{yj}$ in order to be able to factor it could be called "completing the rectangle" in analogy to the more familiar "completing the square."
## Applications
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually $x$ and $y$ are variables and $j,k$ are known constants. Also, it is typically necessary to add the $jk$ term to both sides to perform the factorization.
## Awesome Practice Problems
### Introductory
• Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }$
(Source)
### Intermediate
• $m, n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.
(Source)
• The integer $N$ is positive. There are exactly 2005 ordered pairs $(x, y)$ of positive integers satisfying:
$$\frac 1x +\frac 1y = \frac 1N$$
Prove that $N$ is a perfect square. (British Mathematical Olympiad Round 2, 2005) |
Hong Kong
Stage 1 - Stage 3
# Scales, Maps and Ratios I
Lesson
## Scales
A scale compares the distance on a map to the actual distance on the ground. Scales can be depicted in a few different ways:
## Graphic scale
A graphic scale represents a scale by using a small line with markings similar to a ruler. One side of the line represents the distance on the map, while the other side represents the true distances of objects in real life. By measuring the distance between two points on a map and then referring to the graphic scale, we can calculate the actual distance between those points.
In this picture, you can see that one centimetre on the scale represents $250$250 kilometres in real life.
## Verbal scale
A verbal scale uses words to describe the ratio between the map's scale and the real world. For example, we could say "One centimetre equals fifteen kilometres" or we could write it as $1$1cm = $15$15km. This means that one centimetre on the map is equivalent to $15$15 kilometres in the real world.
## Scale Ratio
Some maps use a representative fraction to describe the ratio between the map and the real world. If you need a refresher about ratios, see Looking at Relationships between Different Groups. We write scale ratios for maps just like other ratios with the colon in the middle. For example $1:100000$1:100000.
#### Examples
##### Question 1
Convert the following description to a proper scale ratio: $5$5cm on the map = $25$25m in real life.
Remember we need to have our two quantities in the same unit of measurement in a ratio. I'm going to convert everything to centimetres.
$25$25m $=$= $25\times100$25×100 cm $=$= $2500$2500 cm
Once we have equivalent quantities, we can write it as a scale ratio.
$5:2500$5:2500 $=$= $1:500$1:500
##### Question 2
Question: Given that the scale on a map is $1$1:$50000$50000, find the actual distance between two points that are $8$8 cm apart on the map.
Think: This means that $1$1 cm on the map represents $50000$50000cm (or $500$500m) in real life.
Do: So to work out how far $8$8cm represents, we need to multiply $8$8 by $50000$50000. Then convert to km.
$8\times50000$8×50000 $=$= $400000$400000 cm $=$= $4000$4000 m $=$= $4$4 km
Now let's look at how we can do this process in reverse.
##### Question 3
Given that the scale of a map of a garden is $1$1:$2000$2000 , how far apart should two fountains be drawn on the map if the actual distance between the fountains is $100$100 m? Express your answer in centimetres. |
Engage NY Eureka Math 3rd Grade Module 5 Lesson 15 Answer Key
Eureka Math Grade 3 Module 5 Lesson 15 Problem Set Answer Key
Question 1.
Estimate to label the given fractions on the number line. Be sure to label the fractions at 0 and 1. Write the fractions above the number line. Draw a number bond to match your number line.
Explanation :
Number line is Divided and represented with given fraction and even the number bond is drawn for every fraction as shown in above figure .
Question 2.
Draw a number line. Use a fraction strip to locate 0 and 1. Fold the strip to make 8 equal parts. Use the strip to measure and label your number line with eighths.
Count up from 0 eighths to 8 eighths on your number line. Touch each number with your finger as you count.
Explanation :
Number line is represented with count from 0 eights to 8 eighths . Each Fraction strip is also represented above the number line . Each part is $$\frac{1}{8}$$ fraction .
Question 3.
For his boat, James stretched out a rope with 5 equally spaced knots as shown.
a. Starting at the first knot and ending at the last knot, how many equal parts are formed by the 5 knots? Label each fraction at the knot.
Explanation :
With the 5 knots the given rope is divided into 4 equal parts. each part is $$\frac{1}{4}$$ .
Fraction at each knot is represented as shown in above figure .
b. What fraction of the rope is labeled at the third knot?
The fraction at 3rd knot is $$\frac{2}{4}$$ is shown in above figure .
c. What if the rope had 6 equally spaced knots along the same length? What fraction of the rope would be measured by the first 2 knots?
The fraction at the first 2 knots is $$\frac{1}{6}$$ and $$\frac{2}{6}$$.
Explanation :
if the rope had 6 equally spaced knots along the same length then each part is divided with the fraction of
$$\frac{1}{6}$$ .
Eureka Math Grade 3 Module 5 Lesson 15 Exit Ticket Answer Key
Question 1.
Estimate to label the given fraction on the number line. Be sure to label the fractions at 0 and 1. Write the fractions above the number line. Draw a number bond to match your number line.
Explanation :
The given number is marked with the given fractions.
Each part is latex]\frac{1}{5}[/latex] as the given marked fraction is latex]\frac{3}{5}[/latex] .
Number bond is
Question 2.
Partition the number line. Then, place each fraction on the number line: $$\frac{3}{6}$$, $$\frac{1}{6}$$, and $$\frac{5}{6}$$.
Explanation :
Number line is represented and the given line is divided into 6 equal parts . Each part is latex]\frac{1}{6}[/latex] .
Then the given fractions are marked on the number line .
Question 1.
Estimate to label the given fractions on the number line. Be sure to label the fractions at 0 and 1. Write the fractions above the number line. Draw a number bond to match your number line. The first one is done for you.
Explanation :
Number line is Divided and represented with given fraction and even the number bond is drawn for every fraction as shown in above figure .
Question 2.
Henry has 5 dimes. Ben has 9 dimes. Tina has 2 dimes.
a. Write the value of each person’s money as a fraction of a dollar:
Henry:
Ben:
Tina:
Henry: latex]\frac{50}{100}[/latex]
Ben: latex]\frac{90}{100}[/latex]
Tina: latex]\frac{20}{100}[/latex]
b. Estimate to place each fraction on the number line. |
If we were to toss a single die (singular for dice) 1000 times, how many 6s could we expect over the long run? What is the standard deviation?
Suppose you did this and got 200 6s. What is the z score of 200?
Jul 24, 2017
$\mu = \frac{500}{3}$, $\sigma = \frac{25}{3} \sqrt{2}$, $z = 2 \sqrt{2}$.
Explanation:
Let $X$ denote the number of 6's thrown in 1000 trials. Then $X$ has a binomial distribution as it satisfies the 4 conditions for a random variable to have a binomial distribution.
• There is a fixed number of trials. (1000)
• The probability of the event occurring is constant. (1/6)
• Each throw of the die is independent.
• There are only two outcomes. (a 6 is thrown or it is not)
In general if we had a random variable $Z$ distributed binomially we would write Z~B(n,p) where $n$ is the number of trials and $p$ is the probability of the event occurring.
So we have X~B(1000,1/6). We are asked to find the expectation of $X$ and the standard deviation.
It can be shown that the expectation of the general binomial random variable $Z$ is $n p$ and its variance is $n p \left(1 - p\right)$. See a proof here.
So,
$\text{E} \left(X\right) = 1000 \cdot \frac{1}{6}$,
$\text{E} \left(X\right) = \frac{500}{3}$,
$\text{E} \left(X\right) \approx 166.67$ to 2 decimal places.
and
$\text{Var} \left(X\right) = 1000 \cdot \frac{1}{6} \cdot \frac{5}{6}$,
$\text{Var} \left(X\right) = \frac{1250}{9}$.
As $\sigma = \sqrt{\text{Var} \left(X\right)}$,
$\sigma = \frac{25}{3} \sqrt{2}$,
$\sigma \approx 11.79$ to 2 decimal places.
The $z$ score is the number of standard deviations away from the mean. A result of 200 gives a z score of
$z = \frac{200 - \frac{500}{3}}{\frac{25}{3} \sqrt{2}}$,
$z = \left(\frac{\frac{100}{3}}{\frac{25}{3} \sqrt{2}}\right)$,
$z = \frac{4}{\sqrt{2}}$,
$z = 2 \sqrt{2}$,
$z = 2.83$ to 2 decimal places. |
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# Subtraction Methods
Reviewed by:
Last updated date: 02nd Aug 2024
Total views: 171.3k
Views today: 3.71k
## Introduction to Subtraction Methods
As we all know, subtraction is a mathematical operation that takes part away from another value. Here in this lesson, we will understand subtraction, types of subtraction models, subtraction by crossing out, and some solved examples on add and subtract for class 2. Subtraction is the series of taking away a number from another. It is a primary operation denoted by a subtraction symbol (-) and is the method of calculating the difference between two numbers. So, let’s further learn about it in this article by understanding more about it and solving some questions.
## What is Subtraction?
Subtraction is used to find the difference between numbers. When we have a group of objects, and you take away a few objects, the group becomes smaller. Subtraction methods are many ways numbers can be subtracted. The method will depend on the problem. We will be going to study in the future.
E.g. you bought 10 balls for your friend, and your friend takes 7 balls. This can be written as 10-7=3 and is read as “ ten minus seven equals three.” Here, we performed the subtraction operation on two numbers 10 and 7 to get the difference of 3.
Subtraction
## Types of Subtraction Models
Teaching all four models of subtraction-take-away, comparison, completion and whole/part/part--can better train elementary school children to think abstractly and relate their math knowledge to the real world.
• ### Take-Away Model
The takeaway model is the most commonly taught subtraction model. It teaches children that taking away some objects from a set results in fewer objects in the set. To teach children the take-away model, set a particular number (e.g. 8) of objects (apples, oranges, blocks, etc.) in front of them and ask them to take away a small number (e.g. 3) of objects. Then ask them how many objects remain. In some exercises, children learn that subtracting objects from a set results in a smaller number (e.g. 8 -3 = 5 ).
Take away model
• ### Completion Subtraction
The completion subtraction model makes children understand algebra lessons since it forces children to manipulate variables in a basic subtraction equation. For example, a completion subtraction is “7 +? = 10.” Use the same blocks or orange balls you used in the takeaway model to teach children this concept. So here, if 3 balls are added to the 7 orange balls, then it will become 10.
So to get this three, we need to do the subtraction, i.e. 10 - 7 = 3.
• ### Comparison Model
The comparison model of subtraction, which involves comparing two different numbers, is a little more challenging to visualise than the takeaway model. An example of comparison subtraction is “If Ram is 26 and Shyam is 36, how many years older is Shyan than Ram?” Since by using these types of problems, children understand concepts easily.
• ### Whole/Part Model
The part model of subtraction involves categories of objects that need to be subcategorised: “I have five mangoes for sale. Three are green mangoes. How many aren’t green apples?”. In this way, we can categorise the objects and use this model.
## How to Solve Subtraction by Crossing Out?
Crossing out is a memory game that models subtraction situations in a frame using crossing out (strike-through) and the matching equation without the difference. Subtract models subtraction situations in a frame using crossing out (strike-through) and the matching difference.
Subtraction by Crossing Out
## Solved Examples on Add and Subtract for Class 2
Q.1. Find out the number of apples in the box. If the box contains 10 apples and 6 are removed, how many apples are left in the box?
Solution: Total number of apples in the box is 10.
Out of which, 6 apples are removed.
The number of apples left is given by,
10 -6 =4
Thus the number of apples left in the box is 4.
Q2. Newton catches 25 fish in a lake. He puts 13 fish back. How many fishes does he have now?
Solution: No. of fish caught are 25 fish.
No. of fish that were put back are 13 fish.
Fishes left are given by,
25 - 13 =12.
Thus the number of fish left is 12.
Q3. There are 39 birds in a tree. A loud noise scares 16 birds away, what number of birds are left in the tree?
Solution: There are 39 birds in a tree.
Out of which, 16 birds away from a tree.
Left birds are given by,
39 - 16 =23.
Thus the number of birds left in the tree is 23.
## Practice Questions
Q.1. Mark had 10 blue marbles. Jim had 6 red marbles. What number of marbles do they have in total?
Ans: 16.
Q.2. There are 6 blue Marlins and 2 snappers in the pond. How many fish are there in total?
Ans: 4.
Q.3. Mary picked 11 cherries off the tree. Her brother Bill picked 8. How many cherries are left?
Ans: 3.
## Summary
In this chapter, we have learned that subtraction is a mathematical operation that takes part away from another value. We also learnt about subtraction, types of subtraction methods, and how to solve subtraction by crossing out. When solving subtraction problems, the 1-digit number can be subtracted more easily, but for larger numbers, we need to split the numbers into columns using their respective values. By solving subtraction questions, we develop an understanding of methods and their applications, which will help us solve such difficult questions.
## FAQs on Subtraction Methods
1. What is a subtraction fact?
Subtraction is the process of taking away a number or a thing from another number or group. The result is called the difference. For example, subtraction facts appear as numbers: 5 – 3 = 2, and 5 – 2 = 3. This can also be called a fact family, in which children learn at a young age.
2. Who invented subtraction?
Brahmagupta from India invented all the basic arithmetic operations, including subtraction.
3. When was subtraction created?
Early use of the minus sign that we use today was discovered in a German manuscript dating from 1481. After studying this manuscript, Johannes Widman published his book, Mercantile Arithmetic, using the minus sign to indicate subtraction. |
# Determinant of a Matrix
Last Updated : 06 Dec, 2023
The determinant of a Matrix is defined as a special number that is defined only for square matrices (matrices that have the same number of rows and columns). A determinant is used in many places in calculus and other matrices related to algebra, it actually represents the matrix in terms of a real number which can be used in solving a system of a linear equation and finding the inverse of a matrix.
Recommended Practice
## Determinant of 2 x 2 Matrix:
Determinant of 2 x 2 matrix
## Determinant of 3 x 3 Matrix:Â
Determinant of 3 x 3 matrix
## How to calculate?Â
The value of the determinant of a matrix can be calculated by the following procedure:Â
• For each element of the first row or first column get the cofactor of those elements.
• Then multiply the element with the determinant of the corresponding cofactor.Â
• Finally, add them with alternate signs. As a base case, the value of the determinant of a 1*1 matrix is the single value itself.Â
The cofactor of an element is a matrix that we can get by removing the row and column of that element from that matrix.
Code block
Output
```Determinant of the matrix is : 30
```
Time Complexity: O(n4)
Space Complexity: O(n2), Auxiliary space used for storing cofactors.
Note: In the above recursive approach when the size of the matrix is large it consumes more stack size.
## Determinant of a Matrix using Determinant properties:
• In this method, we are using the properties of Determinant.Â
• Converting the given matrix into an upper triangular matrix using determinant propertiesÂ
• The determinant of the upper triangular matrix is the product of all diagonal elements.Â
• Iterating every diagonal element and making all the elements down the diagonal as zero using determinant propertiesÂ
• If the diagonal element is zero then search for the next non-zero element in the same column.
There exist two cases:
• Case 1:Â If there is no non-zero element. In this case, the determinant of a matrix is zeroÂ
• Case 2:Â If there exists a non-zero element there exist two casesÂ
• Case A:Â If the index is with a respective diagonal row element. Using the determinant properties make all the column elements down to it zero
• Case B:Â Swap the row with respect to the diagonal element column and continue the Case A operation.
Below is the implementation of the above approach:
## C++
`// C++ program to find Determinant of a matrix` `#include ` `using` `namespace` `std;` `// Dimension of input square matrix` `#define N 4` `// Function to get determinant of matrix` `int` `determinantOfMatrix(``int` `mat[N][N], ``int` `n)` `{` ` ``int` `num1, num2, det = 1, index,` ` ``total = 1; ``// Initialize result` ` ``// temporary array for storing row` ` ``int` `temp[n + 1];` ` ``// loop for traversing the diagonal elements` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{` ` ``index = i; ``// initialize the index` ` ``// finding the index which has non zero value` ` ``while` `(index < n && mat[index][i] == 0) ` ` ``{` ` ``index++;` ` ``}` ` ``if` `(index == n) ``// if there is non zero element` ` ``{` ` ``// the determinant of matrix as zero` ` ``continue``;` ` ``}` ` ``if` `(index != i) ` ` ``{` ` ``// loop for swapping the diagonal element row and` ` ``// index row` ` ``for` `(``int` `j = 0; j < n; j++) ` ` ``{` ` ``swap(mat[index][j], mat[i][j]);` ` ``}` ` ``// determinant sign changes when we shift rows` ` ``// go through determinant properties` ` ``det = det * ``pow``(-1, index - i);` ` ``}` ` ``// storing the values of diagonal row elements` ` ``for` `(``int` `j = 0; j < n; j++) ` ` ``{` ` ``temp[j] = mat[i][j];` ` ``}` ` ``// traversing every row below the diagonal element` ` ``for` `(``int` `j = i + 1; j < n; j++) ` ` ``{` ` ``num1 = temp[i]; ``// value of diagonal element` ` ``num2 = mat[j][i]; ``// value of next row element` ` ``// traversing every column of row` ` ``// and multiplying to every row` ` ``for` `(``int` `k = 0; k < n; k++) ` ` ``{` ` ``// multiplying to make the diagonal` ` ``// element and next row element equal` ` ``mat[j][k]` ` ``= (num1 * mat[j][k]) - (num2 * temp[k]);` ` ``}` ` ``total = total * num1; ``// Det(kA)=kDet(A);` ` ``}` ` ``}` ` ``// multiplying the diagonal elements to get determinant` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{` ` ``det = det * mat[i][i];` ` ``}` ` ``return` `(det / total); ``// Det(kA)/k=Det(A);` `}` `// Driver code` `int` `main()` `{` ` ``/*int mat[N][N] = {{6, 1, 1},` ` ``{4, -2, 5},` ` ``{2, 8, 7}}; */` ` ``int` `mat[N][N] = { { 1, 0, 2, -1 },` ` ``{ 3, 0, 0, 5 },` ` ``{ 2, 1, 4, -3 },` ` ``{ 1, 0, 5, 0 } };` ` ``// Function call` ` ``printf``(``"Determinant of the matrix is : %d"``,` ` ``determinantOfMatrix(mat, N));` ` ``return` `0;` `}`
## Java
`// Java program to find Determinant of a matrix` `class` `GFG ` `{` ` ``// Dimension of input square matrix` ` ``static` `final` `int` `N = ``4``;` ` ``// Function to get determinant of matrix` ` ``static` `int` `determinantOfMatrix(``int` `mat[][], ``int` `n)` ` ``{` ` ``int` `num1, num2, det = ``1``, index,` ` ``total = ``1``; ``// Initialize result` ` ``// temporary array for storing row` ` ``int``[] temp = ``new` `int``[n + ``1``];` ` ``// loop for traversing the diagonal elements` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``{` ` ``index = i; ``// initialize the index` ` ``// finding the index which has non zero value` ` ``while` `(index < n && mat[index][i] == ``0``) ` ` ``{` ` ``index++;` ` ``}` ` ``if` `(index == n) ``// if there is non zero element` ` ``{` ` ``// the determinant of matrix as zero` ` ``continue``;` ` ``}` ` ``if` `(index != i)` ` ``{` ` ``// loop for swapping the diagonal element row` ` ``// and index row` ` ``for` `(``int` `j = ``0``; j < n; j++)` ` ``{` ` ``swap(mat, index, j, i, j);` ` ``}` ` ``// determinant sign changes when we shift` ` ``// rows go through determinant properties` ` ``det = (``int``)(det * Math.pow(-``1``, index - i));` ` ``}` ` ``// storing the values of diagonal row elements` ` ``for` `(``int` `j = ``0``; j < n; j++) ` ` ``{` ` ``temp[j] = mat[i][j];` ` ``}` ` ``// traversing every row below the diagonal` ` ``// element` ` ``for` `(``int` `j = i + ``1``; j < n; j++) ` ` ``{` ` ``num1 = temp[i]; ``// value of diagonal element` ` ``num2 = mat[j]` ` ``[i]; ``// value of next row element` ` ``// traversing every column of row` ` ``// and multiplying to every row` ` ``for` `(``int` `k = ``0``; k < n; k++) ` ` ``{` ` ``// multiplying to make the diagonal` ` ``// element and next row element equal` ` ``mat[j][k] = (num1 * mat[j][k])` ` ``- (num2 * temp[k]);` ` ``}` ` ``total = total * num1; ``// Det(kA)=kDet(A);` ` ``}` ` ``}` ` ``// multiplying the diagonal elements to get` ` ``// determinant` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``{` ` ``det = det * mat[i][i];` ` ``}` ` ``return` `(det / total); ``// Det(kA)/k=Det(A);` ` ``}` ` ``static` `int``[][] swap(``int``[][] arr, ``int` `i1, ``int` `j1, ``int` `i2,` ` ``int` `j2)` ` ``{` ` ``int` `temp = arr[i1][j1];` ` ``arr[i1][j1] = arr[i2][j2];` ` ``arr[i2][j2] = temp;` ` ``return` `arr;` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``/*int mat[N][N] = {{6, 1, 1},` ` ``{4, -2, 5},` ` ``{2, 8, 7}}; */` ` ``int` `mat[][] = { { ``1``, ``0``, ``2``, -``1` `},` ` ``{ ``3``, ``0``, ``0``, ``5` `},` ` ``{ ``2``, ``1``, ``4``, -``3` `},` ` ``{ ``1``, ``0``, ``5``, ``0` `} };` ` ``// Function call` ` ``System.out.printf(` ` ``"Determinant of the matrix is : %d"``,` ` ``determinantOfMatrix(mat, N));` ` ``}` `}` `// This code is contributed by Rajput-Ji`
## Python3
`# Python program to find Determinant of a matrix` `def` `determinantOfMatrix(mat, n):` ` ``temp ``=` `[``0``]``*``n ``# temporary array for storing row` ` ``total ``=` `1` ` ``det ``=` `1` `# initialize result` ` ``# loop for traversing the diagonal elements` ` ``for` `i ``in` `range``(``0``, n):` ` ``index ``=` `i ``# initialize the index` ` ``# finding the index which has non zero value` ` ``while``(index < n ``and` `mat[index][i] ``=``=` `0``):` ` ``index ``+``=` `1` ` ``if``(index ``=``=` `n): ``# if there is non zero element` ` ``# the determinant of matrix as zero` ` ``continue` ` ``if``(index !``=` `i):` ` ``# loop for swapping the diagonal element row and index row` ` ``for` `j ``in` `range``(``0``, n):` ` ``mat[index][j], mat[i][j] ``=` `mat[i][j], mat[index][j]` ` ``# determinant sign changes when we shift rows` ` ``# go through determinant properties` ` ``det ``=` `det``*``int``(``pow``(``-``1``, index``-``i))` ` ``# storing the values of diagonal row elements` ` ``for` `j ``in` `range``(``0``, n):` ` ``temp[j] ``=` `mat[i][j]` ` ``# traversing every row below the diagonal element` ` ``for` `j ``in` `range``(i``+``1``, n):` ` ``num1 ``=` `temp[i] ``# value of diagonal element` ` ``num2 ``=` `mat[j][i] ``# value of next row element` ` ``# traversing every column of row` ` ``# and multiplying to every row` ` ``for` `k ``in` `range``(``0``, n):` ` ``# multiplying to make the diagonal` ` ``# element and next row element equal` ` ``mat[j][k] ``=` `(num1``*``mat[j][k]) ``-` `(num2``*``temp[k])` ` ``total ``=` `total ``*` `num1 ``# Det(kA)=kDet(A);` ` ``# multiplying the diagonal elements to get determinant` ` ``for` `i ``in` `range``(``0``, n):` ` ``det ``=` `det``*``mat[i][i]` ` ``return` `int``(det``/``total) ``# Det(kA)/k=Det(A);` `# Drivers code` `if` `__name__ ``=``=` `"__main__"``:` ` ``# mat=[[6 1 1][4 -2 5][2 8 7]]` ` ``mat ``=` `[[``1``, ``0``, ``2``, ``-``1``], [``3``, ``0``, ``0``, ``5``], [``2``, ``1``, ``4``, ``-``3``], [``1``, ``0``, ``5``, ``0``]]` ` ``N ``=` `len``(mat)` ` ` ` ``# Function call` ` ``print``(``"Determinant of the matrix is : "``, determinantOfMatrix(mat, N))`
## C#
`// C# program to find Determinant of a matrix` `using` `System;` `class` `GFG {` ` ``// Dimension of input square matrix` ` ``static` `readonly` `int` `N = 4;` ` ``// Function to get determinant of matrix` ` ``static` `int` `determinantOfMatrix(``int``[, ] mat, ``int` `n)` ` ``{` ` ``int` `num1, num2, det = 1, index,` ` ``total = 1; ``// Initialize result` ` ``// temporary array for storing row` ` ``int``[] temp = ``new` `int``[n + 1];` ` ``// loop for traversing the diagonal elements` ` ``for` `(``int` `i = 0; i < n; i++)` ` ``{` ` ``index = i; ``// initialize the index` ` ``// finding the index which has non zero value` ` ``while``(index < n && mat[index, i] == 0) ` ` ``{` ` ``index++;` ` ``}` ` ``if` `(index == n) ``// if there is non zero element` ` ``{` ` ``// the determinant of matrix as zero` ` ``continue``;` ` ``}` ` ``if` `(index != i) ` ` ``{` ` ``// loop for swapping the diagonal element row` ` ``// and index row` ` ``for` `(``int` `j = 0; j < n; j++) ` ` ``{` ` ``swap(mat, index, j, i, j);` ` ``}` ` ``// determinant sign changes when we shift` ` ``// rows go through determinant properties` ` ``det = (``int``)(det * Math.Pow(-1, index - i));` ` ``}` ` ``// storing the values of diagonal row elements` ` ``for` `(``int` `j = 0; j < n; j++)` ` ``{` ` ``temp[j] = mat[i, j];` ` ``}` ` ``// traversing every row below the diagonal` ` ``// element` ` ``for` `(``int` `j = i + 1; j < n; j++) ` ` ``{` ` ``num1 = temp[i]; ``// value of diagonal element` ` ``num2 = mat[j,` ` ``i]; ``// value of next row element` ` ``// traversing every column of row` ` ``// and multiplying to every row` ` ``for` `(``int` `k = 0; k < n; k++)` ` ``{` ` ``// multiplying to make the diagonal` ` ``// element and next row element equal` ` ``mat[j, k] = (num1 * mat[j, k])` ` ``- (num2 * temp[k]);` ` ``}` ` ``total = total * num1; ``// Det(kA)=kDet(A);` ` ``}` ` ``}` ` ``// multiplying the diagonal elements to get` ` ``// determinant` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{` ` ``det = det * mat[i, i];` ` ``}` ` ``return` `(det / total); ``// Det(kA)/k=Det(A);` ` ``}` ` ``static` `int``[, ] swap(``int``[, ] arr, ``int` `i1, ``int` `j1, ``int` `i2,` ` ``int` `j2)` ` ``{` ` ``int` `temp = arr[i1, j1];` ` ``arr[i1, j1] = arr[i2, j2];` ` ``arr[i2, j2] = temp;` ` ``return` `arr;` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `Main(String[] args)` ` ``{` ` ``/*int mat[N,N] = {{6, 1, 1},` ` ``{4, -2, 5},` ` ``{2, 8, 7}}; */` ` ``int``[, ] mat = { { 1, 0, 2, -1 },` ` ``{ 3, 0, 0, 5 },` ` ``{ 2, 1, 4, -3 },` ` ``{ 1, 0, 5, 0 } };` ` ``// Function call` ` ``Console.Write(``"Determinant of the matrix is : {0}"``,` ` ``determinantOfMatrix(mat, N));` ` ``}` `}` `// This code is contributed by 29AjayKumar`
## Javascript
`Javascript`
Output
```Determinant of the matrix is : 30
```
Time complexity: O(n3)Â
Auxiliary Space: O(n), Space used for storing row.
Â
## Determinant of a Matrix
There is a built-in function or method in linalg module of NumPy package in python. It can be called numpy.linalg.det(mat) which returns the determinant value of the matrix mat passed in the argument.
## C++
`#include ` `// Function to calculate the determinant of a 2x2 matrix` `double` `determinant2x2(``double` `a, ``double` `b, ``double` `c, ``double` `d) {` ` ``return` `a * d - b * c;` `}` `// Function to calculate the determinant of a 3x3 matrix` `double` `determinant3x3(``double` `mat[3][3]) {` ` ``double` `det = 0;` ` ``for` `(``int` `i = 0; i < 3; i++) {` ` ``det += mat[0][i] * determinant2x2(mat[1][(i + 1) % 3], mat[1][(i + 2) % 3], mat[2][(i + 1) % 3], mat[2][(i + 2) % 3]);` ` ``}` ` ``return` `det;` `}` `// Function to calculate the determinant of a 4x4 matrix` `double` `determinant4x4(``double` `mat[4][4]) {` ` ``double` `det = 0;` ` ``for` `(``int` `i = 0; i < 4; i++) {` ` ``double` `minorMatrix[3][3];` ` ``for` `(``int` `j = 1; j < 4; j++) {` ` ``int` `col = 0;` ` ``for` `(``int` `k = 0; k < 4; k++) {` ` ``if` `(k != i) {` ` ``minorMatrix[j - 1][col] = mat[j][k];` ` ``col++;` ` ``}` ` ``}` ` ``}` ` ``double` `minorDet = determinant3x3(minorMatrix);` ` ``det += (i % 2 == 0 ? 1 : -1) * mat[0][i] * minorDet;` ` ``}` ` ``return` `det;` `}` `int` `main() {` ` ``double` `mat[4][4] = {` ` ``{1, 0, 2, -1},` ` ``{3, 0, 0, 5},` ` ``{2, 1, 4, -3},` ` ``{1, 0, 5, 0}` ` ``};` ` ``double` `det = determinant4x4(mat);` ` ``std::cout << ``"Determinant of the matrix is: "` `<< det << std::endl;` ` ``return` `0;` `}`
## Java
`public` `class` `MatrixDeterminant {` ` ``// Function to calculate the determinant of a 2x2 matrix` ` ``public` `static` `double` `determinant2x2(``double` `a, ``double` `b, ``double` `c, ``double` `d) {` ` ``return` `a * d - b * c;` ` ``}` ` ``// Function to calculate the determinant of a 3x3 matrix` ` ``public` `static` `double` `determinant3x3(``double``[][] mat) {` ` ``double` `det = ``0``;` ` ``for` `(``int` `i = ``0``; i < ``3``; i++) {` ` ``det += mat[``0``][i] * determinant2x2(mat[``1``][(i + ``1``) % ``3``], mat[``1``][(i + ``2``) % ``3``], mat[``2``][(i + ``1``) % ``3``], mat[``2``][(i + ``2``) % ``3``]);` ` ``}` ` ``return` `det;` ` ``}` ` ``// Function to calculate the determinant of a 4x4 matrix` ` ``public` `static` `double` `determinant4x4(``double``[][] mat) {` ` ``double` `det = ``0``;` ` ``for` `(``int` `i = ``0``; i < ``4``; i++) {` ` ``double``[][] minorMatrix = ``new` `double``[``3``][``3``];` ` ``for` `(``int` `j = ``1``; j < ``4``; j++) {` ` ``int` `col = ``0``;` ` ``for` `(``int` `k = ``0``; k < ``4``; k++) {` ` ``if` `(k != i) {` ` ``minorMatrix[j - ``1``][col] = mat[j][k];` ` ``col++;` ` ``}` ` ``}` ` ``}` ` ``double` `minorDet = determinant3x3(minorMatrix);` ` ``det += (i % ``2` `== ``0` `? ``1` `: -``1``) * mat[``0``][i] * minorDet;` ` ``}` ` ``return` `det;` ` ``}` ` ``public` `static` `void` `main(String[] args) {` ` ``double``[][] mat = {` ` ``{``1``, ``0``, ``2``, -``1``},` ` ``{``3``, ``0``, ``0``, ``5``},` ` ``{``2``, ``1``, ``4``, -``3``},` ` ``{``1``, ``0``, ``5``, ``0``}` ` ``};` ` ``double` `det = determinant4x4(mat);` ` ``System.out.println(``"Determinant of the matrix is: "` `+ det);` ` ``}` `}` `//This code is contributed by shivamgupta0987654321`
## Python3
`# importing the numpy package` `# as np` `import` `numpy as np` `def` `determinant(mat):` ` ` ` ``# calling the det() method` ` ``det ``=` `np.linalg.det(mat)` ` ``return` `round``(det)` `# Driver Code` `# declaring the matrix` `mat ``=` `[[``1``, ``0``, ``2``, ``-``1``],` ` ``[``3``, ``0``, ``0``, ``5``],` ` ``[``2``, ``1``, ``4``, ``-``3``],` ` ``[``1``, ``0``, ``5``, ``0``]]` `# Function call` `print``(``'Determinant of the matrix is:'``,` ` ``determinant(mat))`
## C#
`using` `System;` `class` `Program` `{` ` ``// Function to calculate the determinant of a 2x2 matrix` ` ``static` `double` `Determinant2x2(``double` `a, ``double` `b, ``double` `c, ``double` `d)` ` ``{` ` ``return` `a * d - b * c;` ` ``}` ` ``// Function to calculate the determinant of a 3x3 matrix` ` ``static` `double` `Determinant3x3(``double``[,] mat)` ` ``{` ` ``double` `det = 0;` ` ``for` `(``int` `i = 0; i < 3; i++)` ` ``{` ` ``det += mat[0, i] * Determinant2x2(mat[1, (i + 1) % 3], mat[1, (i + 2) % 3], mat[2, (i + 1) % 3], mat[2, (i + 2) % 3]);` ` ``}` ` ``return` `det;` ` ``}` ` ``// Function to calculate the determinant of a 4x4 matrix` ` ``static` `double` `Determinant4x4(``double``[,] mat)` ` ``{` ` ``double` `det = 0;` ` ``for` `(``int` `i = 0; i < 4; i++)` ` ``{` ` ``double``[,] minorMatrix = ``new` `double``[3, 3];` ` ``for` `(``int` `j = 1; j < 4; j++)` ` ``{` ` ``int` `col = 0;` ` ``for` `(``int` `k = 0; k < 4; k++)` ` ``{` ` ``if` `(k != i)` ` ``{` ` ``minorMatrix[j - 1, col] = mat[j, k];` ` ``col++;` ` ``}` ` ``}` ` ``}` ` ``double` `minorDet = Determinant3x3(minorMatrix);` ` ``det += (i % 2 == 0 ? 1 : -1) * mat[0, i] * minorDet;` ` ``}` ` ``return` `det;` ` ``}` ` ``static` `void` `Main()` ` ``{` ` ``double``[,] mat = {` ` ``{1, 0, 2, -1},` ` ``{3, 0, 0, 5},` ` ``{2, 1, 4, -3},` ` ``{1, 0, 5, 0}` ` ``};` ` ``double` `det = Determinant4x4(mat);` ` ``Console.WriteLine(``"Determinant of the matrix is: "` `+ det);` ` ``}` `}`
## Javascript
`// Function to calculate the determinant of a 2x2 matrix` `function` `determinant2x2(a, b, c, d) {` ` ``return` `a * d - b * c;` `}` `// Function to calculate the determinant of a 3x3 matrix` `function` `determinant3x3(mat) {` ` ``let det = 0;` ` ``for` `(let i = 0; i < 3; i++) {` ` ``det += mat[0][i] * determinant2x2(` ` ``mat[1][(i + 1) % 3], mat[1][(i + 2) % 3],` ` ``mat[2][(i + 1) % 3], mat[2][(i + 2) % 3]` ` ``);` ` ``}` ` ``return` `det;` `}` `// Function to calculate the determinant of a 4x4 matrix` `function` `determinant4x4(mat) {` ` ``let det = 0;` ` ``for` `(let i = 0; i < 4; i++) {` ` ``let minorMatrix = [` ` ``[0, 0, 0],` ` ``[0, 0, 0],` ` ``[0, 0, 0]` ` ``];` ` ``for` `(let j = 1; j < 4; j++) {` ` ``let col = 0;` ` ``for` `(let k = 0; k < 4; k++) {` ` ``if` `(k !== i) {` ` ``minorMatrix[j - 1][col] = mat[j][k];` ` ``col++;` ` ``}` ` ``}` ` ``}` ` ``let minorDet = determinant3x3(minorMatrix);` ` ``det += (i % 2 === 0 ? 1 : -1) * mat[0][i] * minorDet;` ` ``}` ` ``return` `det;` `}` `// Example usage` `const mat = [` ` ``[1, 0, 2, -1],` ` ``[3, 0, 0, 5],` ` ``[2, 1, 4, -3],` ` ``[1, 0, 5, 0]` `];` `const det = determinant4x4(mat);` `console.log(``"Determinant of the matrix is:"``, det);`
Output:
`Determinant of the matrix is: 30`
Time Complexity: O(n3), as the time complexity of np.linalg.det is O(n3) for an n x n order matrix.
Auxiliary Space: O(1)
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### Winning Lines
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
# Games Related to Nim
##### Stage: 1, 2, 3 and 4
Published December 2006,February 2006,December 2011,February 2011.
This small group of activities is taken from the Mathematical Games Archive on the NRICH site. They all have a related structure that can be used to develop the skills of strategic planning and reasoning as well as ideas of analogy.
Students at all levels of ability and age can access them. At the most basic level, they offer opportunities for practising arithmetical skills. At a higher level, they can be used to promote mathematical discussion by demanding detailed and reasoned explanations for a winning strategy, or an explanation of the mathematics that links the games.
#### Suggested Progression
Several of the games link into one another. One suggestion is given below. Can you see and describe the similarities between all of these games?
The following objectives are to be found in the Frameworks for teaching mathematics for Key Stages 1, 2 and 3, and apply to the activities in this theme.
• Solve mathematical problems or puzzles, recognise simple patterns and relationships, generalise and predict.
• Understand addition and subtraction mental calculation strategies.
• Use letter symbols to represent unknown numbers or variables.
• Represent problems mathematically.
• Explain and justify methods and conclusions.
• Use logical argument to establish the truth of a statement.
• Solve increasingly demanding problems and evaluate solutions.
• Present a concise, reasoned argument, using symbols, diagrams, graphs and related explanatory text.
• Suggest extensions to problems, conjecture and generalise.
#### Daisy
Play the game on line (it may take a little time to download so please be patient).
Try to work out a winning strategy.
Does one of the players have the advantage?
#### NIM
Start playing this with 7 counters.
Try to work out a winning strategy.
Does it matter who goes first?
Now play with 9 counters.
Try to work out a winning strategy.
Is it preferable to go first or second?
Can you find a generalisation about the winner for any number of counters?
#### Got It!
Play the game on line.
Start with the Got It! total of 15, using the numbers 1, 2, 3, 4 or 5.
Try to work out a winning strategy.
Does it matter who goes first?
What is the connection between NIM and Got It!?
Now change the target number, but keep the numbers 1, 2, 3, 4 and 5.
Does this alter whether it is preferable to go first or second?
Can you write a mathematical statement to link the target number to the winning player?
Try to find a generalisation for who would win for any target number, and any group of consecutive numbers from 1 to b.
Try to find a generalisation for who would win for any target number, and any group of consecutive numbers from a to b.
#### Slippery Snail
Play the game and see if you can find a winning strategy.
Does it matter who goes first?
What happens if there are more or less spots to start with?
Does it matter if the stars are in different places?
Make up your own Slippery Snail game.
What is the connection between this and NIM?
#### Pentanim
Try to work out a winning strategy by playing the game.
Does it matter who goes first?
What is the connection with NIM?
#### Last Biscuit
An interesting and engaging version of the game .
Can you play and work out a strategy to ensure you always win using some of the strategies you have already learnt?
Can you explain any patterns you see?
#### Further activities linked with these on NRICH:
An interactive version of Nim with varying numbers of piles of counters,
A whole list of NIM games: |
## What is 34.95 rounded to the nearest tenth?
35.0
34.95 rounded to the nearest tenth is 35.0.
How do you round to the tenth of a percent?
To round, we must look at the number in the thousandths place, or the second number to the right of the decimal. If the number is 5 or greater, we increase the number in the tenths place by 1. If the number is 0-4, we keep the number in the tenths place the same.
### How do you round to the nearest 10%?
Whenever you want to round a number to a particular digit, look only at the digit immediately to its right. For example, if you want to round to the nearest tenth, look to the right of the tenths place: This would be the hundredths place digit. Then, if it is 5 or higher, you get to add one to the tenths digit.
What is 0.6 rounded to the nearest 10th?
Answer and Explanation: Rounded to the nearest tenth, 0.6 remains 0.6.
#### How do you round to nearest percent?
To change any decimal to a percent, you multiply the decimal by 100 .. In order to round to the nearest tenth, you examine the value in the hundredths place to determine whether or not the number in the tenths place will be kept or rounded up.
What is 165 rounded to the nearest ten answer?
165 rounded to the nearest ten with a number line
• Determine the two consecutive multiples of 10 that bracket 165.
• 165 is between 160 and 170.
• As illustrated on the number line, 165 is the midpoint between 160 and 170.
• Therefore, 165 rounded to the nearest ten = 170.
## What is 0.5 rounded to the nearest one?
1
However, it can be converted to a whole number by rounding it off to the nearest whole number. 0.5 rounded off to the nearest whole number is 1. Since, the value after decimal is equal to 5, then the number is rounded up to the next whole number.
What is 1.5 rounded to the nearest one?
2
Both 1.5 and 2.5 are rounded to 2 . 3.5 and 4.5 are rounded to 4 .
### What is 4.78 as a percent rounded to the nearest tenth?
4.8
To the nearest tenth: 4.78 rounds to 4.8.
How do you write 0.006 as a percent?
What is Decimal 0.006 in Percent? Decimal 0.006 in Percent is 0.6%.
#### What is 0.006 as a percent?
Decimal 0.006 in Percent is 0.6%.
What is the nearest ten of 145?
Round 145 to the nearest ten. The two nearest tens to 145 are 140 and 150. Remember, the two choices of tens that we could round to is between the amount of tens we already have: 140, or the next ten along: 150.
## What is the nearest 10 of 145?
145 rounded to the nearest ten with a number line
• Determine the two consecutive multiples of 10 that bracket 145.
• 145 is between 140 and 150.
• As illustrated on the number line, 145 is the midpoint between 140 and 150.
• Therefore, 145 rounded to the nearest ten = 150.
How do you round to the nearest tenth of a percent?
To round to the nearest tenth of a percent, a person must reduce two or more decimal points to one decimal point. For instance, a percentage of 82.34 is rounded to 82.3 percent, because 82.34 is closer to 82.3 than 82.4. Rounding to the nearest tenth of a percent is a way to simplify a statistical recording of a percentage.
### What is the nearest tenth?
Nearest tenth is the first digit after the decimal point. If the digit after tenth is greater than or equal to 5, add 1 to tenth. Else remove the digit.
What is the value of 838 to the nearest tenth?
Rounding to the nearest one is 838 Rounding to the nearest tenth is 838.3 Rounding to the nearest hundredth is 838.27 Basic Rules of Rounding
#### What is the tens digit rounded to the nearest ten?
The tens digit stays the same at 2. Every digit after becomes a zero. Digits after the decimal point are dropped. 323.5 rounded to the nearest ten is 320 |
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2. CBSE
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8. NCERT Solutions class 12...
# NCERT Solutions class 12 Maths Exercise 4.3
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## NCERT Solutions class 12 Maths Determinants
1. Find the area of the triangle with vertices at the points given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii)
Ans. (i) Area of triangle = Modulus of =
= = = = sq. units
(ii) Area of triangle = Modulus of =
= =
= =
= = sq. units
(iii) Area of triangle = Modulus of =
=
=
=
=
= = 15 sq. units
### 2. Show that the points A B C are collinear.
Ans. Area of triangle ABC = Modulus of =
=
=
=
= = 0
Therefore, points A, B and C are collinear.
### 3. Find values of if area of triangle is 4 sq. units and vertices are:
(i)
(ii)
Ans. (i) Given: Area of triangle = Modulus of = 4
Modulus of = 4
Taking positive sign,
Taking negative sign,
(ii) Given: Area of triangle = Modulus of = 4
= 4
Taking positive sign,
Taking negative sign,
### 4. (i) Find the equation of the line joining (1, 2) and (3, 6) using determinants.
(ii) Find the equation of the line joining (3, 1) and (9, 3) using determinants.
Ans. (i) Let P be any point on the line joining the points (1, 2) and (3, 6).
Then, Area of triangle that could be formed by these points is zero.
Area of triangle = Modulus of = 0
Modulus of = 0
which is required line.
(ii) Let P be any point on the line joining the points (3, 1) and (9, 3).
Then, Area of triangle that could be formed by these points is zero.
Area of triangle = Modulus of = 0
Modulus of = 0
which is required line.
### 5. If area of triangle is 35 sq. units with vertices and Then is:
(A) 12
(B)
(C)
(D)
Ans. Given: Area of triangle = Modulus of = 35
Modulus of = 35
Taking positive sign,
Taking negative sign,
Therefore, option (D) is correct.
## NCERT Solutions class 12 Maths Exercise 4.3
NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 20 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide
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# Difference between revisions of "1995 AJHSME Problems/Problem 13"
## Problem
In the figure, $\angle A$, $\angle B$, and $\angle C$ are right angles. If $\angle AEB = 40^\circ$ and $\angle BED = \angle BDE$, then $\angle CDE =$
$[asy] dot((0,0)); label("E",(0,0),SW); dot(dir(85)); label("A",dir(85),NW); dot((4,0)); label("D",(4,0),SE); dot((4.05677,0.648898)); label("C",(4.05677,0.648898),NE); draw((0,0)--dir(85)--(4.05677,0.648898)--(4,0)--cycle); dot((2,2)); label("B",(2,2),N); draw((0,0)--(2,2)--(4,0)); pair [] x = intersectionpoints((0,0)--(2,2)--(4,0),dir(85)--(4.05677,0.648898)); dot(x[0]); dot(x[1]); label("F",x[0],SE); label("G",x[1],SW); [/asy]$
$\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ$
## Solution
Because $\angle BED=\angle BDE$, $\angle B=90^\circ$, and $\triangle BED$ is a triangle, we get: $$\angle B + \angle BED +\angle BDE=180$$ $$90 +\angle BED +\angle BED=180$$ $$2\angle BED=90$$ $$\angle BED=\angle BDE=45^\circ$$
So $\angle AED=\angle AEB +\angle BED=40 +45=85^\circ$ Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: $$\angle A +\angle C +\angle CDE +\angle AED=360$$ $$90 +90 +\angle CDE +85=360$$ $$\angle CDE=95^\circ \text{(E)}$$ |
# Calculating the Present Value of an Ordinary Annuity (PVOA)
In this section we will solve four exercises that calculate the present value of an ordinary annuity (PVOA). We will use PMT (“payment”) to represent the recurring identical cash payment amount.
### Exercise #1
Let’s assume we are to receive \$100 at the end of each year for two years. How do we calculate the present value of this annuity, assuming the interest rate or the required rate for discounting is 8% per year compounded annually?
The following timeline depicts the information we know, along with the unknown component (PVOA).
Because each payment is the same amount (\$100) and because there is an equal length of time between payments (one year in this example), we know this arrangement meets the definition of an annuity. Because the equal payments occur at the end of each year, we know we have an ordinary annuity.
Calculation of Exercise #1 using the PVOA Table
The equation for calculating the present value of an ordinary annuity is:
This PVOA calculation tells you that receiving \$178.30 today is equivalent to receiving \$100 at the end of each of the next two years, if the time value of money is 8% per year. If the 8% rate is a company’s required rate of return, this tells you that the company could pay up to \$178.30 for the two-year annuity.
[The \$178.30 could have been computed by using the PV of 1 Table for the two payments. In other words, receiving \$100 at the end of the first year has a present value of \$92.60 (\$100 times 0.926, the PV of 1 factor for n = 1; i = 8%). Receiving the second \$100 at the end of the second year has a present value of \$85.70 (\$100 times 0.857, the PV of 1 factor for n = 2; i = 8%). The total of those two present values (\$92.60 + \$85.70) equals \$178.30.]
The difference between the \$200 of total future payments and the present value of \$178.30 is the interest our money earns while we wait to receive the payments. This \$21.70 difference is referred to as interest, discount, or a company’s return on its investment.
### Exercise #2
Determine the present value (the value at period 0) of receiving a series of equal payments of \$200 at the end of each year for 20 years. Assume that today is June 1, 2024 and that the first payment will occur on June 1, 2025. The appropriate rate for discounting the payments is 10% per year compounded annually.
Calculation of Exercise #2 using the PVOA Table
The equation for calculating the present value of an ordinary annuity is:
This calculation tells you that receiving \$1,702.80 today is equivalent to receiving \$200 at the end of each of the next 20 years, if the time value of money is 10% per year. (Obviously, using the PVOA Table is a great time saver compared to the 20 calculations that would be required if you used the PV of 1 Table for each \$200 payment.)
Given an interest rate of 10%, the difference between the present value of \$1,702.80 and the \$4,000.00 of total payments (20 payments at \$200 each) reflects the interest earned over the years. This difference of \$2,297.20 (\$4,000 minus \$1702.80) is referred to as interest, or discount.
### Exercise #3
What is the present value of receiving a series of \$300 payments at the end of each quarter for three years, if the time value of money is 8% per year? Assume that today is June 1, 2024 and that the quarterly payments will begin on September 1, 2024.
As we can see from the timeline, this is an ordinary annuity; the payment amounts are identical, they occur at equal time intervals, and they occur at the end of each 3-month period.
Calculation of Exercise #3 using the PVOA Table
The equation for calculating the present value of an ordinary annuity is:
This calculation tells us that receiving \$3,172.50 today is equivalent to receiving \$300 at the end of each of the next 12 quarters, if the time value of money is 2% per quarter (or 8% per year).
If 8% is a firm’s targeted rate of return per year, this calculation tells us that the company can pay up to \$3,172.50 for the stream of \$300 payments. (If it pays more than \$3,172.50 it will earn less than its required rate of 8%.)
### Exercise #4
What is the maximum amount you borrow on June 1, 2024 if you are limited to a two-year loan with monthly payments of \$400? Assume the loan will have an annual interest rate of 12%. The first payment will be due on July 1, 2024.
Calculation of Exercise #4 using the PVOA Table
The equation for calculating the present value of an ordinary annuity is:
Assuming the interest rate is 12% per year (or 1% per month), \$8,497.20 is the present value amount that you could borrow today if you were to make 24 monthly payments of \$400 each starting at the end of the first month. You can see the proof of this amount by reviewing the loan amortization schedule for this scenario:
The above calculation tells us that receiving \$8,497.20 today is equivalent to receiving \$400 at the end of each of the next 24 months, if the time value of money is 1% per month (or 12% per year). It also means that a company requiring a 12% annual return compounded monthly can invest up to \$8,497.20 for this annuity of \$400 payments. |
# Find the LCM and HCF of the following integers by applying the prime factorisation method.(i) 12, 15 and 21(ii) 17, 23 and 29(iii) 8, 9 and 25.
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To find:
Here we have to find the LCM and HCF of the given integers by applying the prime factorization method.
Solution:
Calculating LCM and HCF using prime factorization method:
Writing the numbers as a product of their prime factors:
(i) Prime factorisation of 12:
• $2\ \times\ 2\ \times\ 3\ =\ 2^2\ \times\ 3^1$
Prime factorisation of 15:
• $3\ \times\ 5\ =\ 3^1\ \times\ 5^1$
Prime factorisation of 21:
• $3\ \times\ 7\ =\ 3^1\ \times\ 7^1$
Multiplying the highest power of each prime number these values together:
$2^2\ \times\ 3^1\ \times\ 5^1\ \times\ 7^1\ =\ 420$
LCM(12, 15, 21) $=$ 420
Multiplying all common prime factors:
$3^1\ =\ 3$
HCF(12, 15, 21) $=$ 3
So, the LCM and HCF of 12, 15 and 21 are 420 and 3 respectively.
(ii) Prime factorisation of 17:
• $17\ =\ 17^1$
Prime factorisation of 23:
• $23\ =\ 23^1$
Prime factorisation of 29:
• $29\ =\ 29^1$
Multiplying the highest power of each prime number these values together:
$17^1\ \times\ 23^1\ \times\ 29^1\ =\ 11339$
LCM(17, 23, 29) $=$ 11339
Multiplying all common prime factors:
There is no common prime factor. So,
HCF(17, 23, 29) $=$ 1
So, the LCM and HCF of 17, 23 and 29 are 11339 and 1 respectively.
(iii) Prime factorisation of 8:
• $2\ \times\ 2\ \times\ 2\ =\ 2^3$
Prime factorisation of 9:
• $3\ \times\ 3\ =\ 3^2$
Prime factorisation of 25:
• $5\ \times\ 5\ =\ 5^2$
Multiplying the highest power of each prime number these values together:
$2^3\ \times\ 3^2\ \times\ 5^2\ =\ 1800$
LCM(8, 9, 25) $=$ 1800
Multiplying all common prime factors:
There is no common prime factor. So,
HCF(8, 9, 25) $=$ 1
So, the LCM and HCF of 8, 9 and 25 are 1800 and 1 respectively.
Updated on 10-Oct-2022 13:19:30 |
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### Course: Digital SAT Math>Unit 13
Lesson 1: Area and volume: advanced
# Area and volume | Lesson
A guide to area and volume on the digital SAT
## What are area and volume problems?
Area and volume problems focus on using the relevant formulas for various two- and three-dimensional shapes. We'll be expected to calculate the length, area, surface area, and volume of shapes, as well as describe how changes in side length affect area and volume.
In this lesson, we'll learn to:
1. Calculate the volumes and dimensions of three-dimensional solids
2. Determine how dimension changes affect area and volume
You can learn anything. Let's do this!
## How do I calculate the volumes and dimensions of shapes?
### Volume word problem: gold ring
Volume word problem: gold ringSee video transcript
### Volume of a cone
Volume of a coneSee video transcript
### The volumes of three-dimensional solids
Good news: You do not need to remember any volume formulas for the SAT! At the beginning of each SAT math section, the following volume formulas are provided as reference.
ShapeFormula
Right rectangular prism$V=\ell wh$
Right circular cylinder$V=\pi {r}^{2}h$
Sphere$V=\frac{4}{3}\pi {r}^{3}$
Right circular cone$V=\frac{1}{3}\pi {r}^{2}h$
Rectangular pyramid$V=\frac{1}{3}\ell wh$
If the test asks for the volume of a different shape, the volume formula will be provided alongside the question.
To calculate the volume of a solid:
1. Find the volume formula for the solid.
2. Plug the dimensions into the formula.
3. Evaluate the volume.
Example: Fei Fei has a model of the Moon in the shape of a sphere. If the model has a radius of $10$ centimeters, what is the volume of the model in cubic centimeters?
Some questions will provide the volume of the solid and ask us to find a linear dimension such as length or radius.
To find an unknown dimension when the volume of a solid is given:
1. Find the volume formula for the solid.
2. Plug the volume and any known dimensions into the formula.
3. Isolate the unknown dimension.
Example: A puzzle box is shaped like a rectangular prism and has a volume of $240$ cubic inches. If the puzzle box has a length of $10$ inches and a width of $8$ inches, what is the height of the puzzle box in inches?
### Try it!
try: find the volume of a pyramid
A pyramid has a square base with a side length of $8$ centimeters. The height of the pyramid is $\frac{3}{4}$ as long as the side length of its base.
What is the height of the pyramid in centimeters?
What is the volume of the pyramid in cubic centimeters?
## How do changing dimensions affect area and volume?
### How volume changes when dimensions change
How volume changes from changing dimensionsSee video transcript
### Impact of increasing the radius
Impact of increasing the radiusSee video transcript
### The effect of changing dimensions on area and volume
When a linear dimension to the first power, e.g., the length of a rectangle or the height of a cylinder, changes by a factor, the area or volume changes by the same factor.
However, when a linear dimension to the second power, e.g., the side length of a square or the radius of a cylinder or cone, changes by a factor, the area or volume changes by the square of the factor.
### Try it!
try: compare the volumes of two cylinders
Right circular cylinder $A$ has a volume of $64\pi$ cubic feet. Which of the following right circular cylinders have the same volume as cylinder $A$ ?
practice: calculate a volume
What is the volume, in cubic meters, of a right rectangular prism that has a length of $2$ meters, a width of $0.4$ meter, and a height of $5$ meters?
practice: calculate a linear dimension
A medicine bottle is in the shape of a right circular cylinder. If the volume of the bottle is $144\pi$ cubic centimeters, what is the diameter of the base of the bottle, in centimeters?
practice: determine the effect of scaling on volume
The volume of a right circular cone $A$ is $225$ cubic inches. What is the volume, in cubic inches, of a right circular cone with twice the radius and twice the height of cone $A$?
## Want to join the conversation?
• Calculating areas and volumes are so much easier when they give you the formula. Schools in my country make students remember the formula
• In my country you have to learn it except during national exams where they give it to you.
• if there was sat art, sal would get 800
• What is the best way to remember the formulas to each shape?
• The best thing is you don't have to remember! All the formulas will be given in the SAT question paper.
• Perfect I have the formulas, no need to think!!
• who is preparing for december SAT? any tips for english section? Is Erica Meltzer's Guide worth reading?
• I'm taking it in October (1 week) and fr I've 0 tips. I will pray whoever up there to get over 1490
• can someone explain the last question pls
• The last question should be more explained. |
Review question
# A cubic has one real root—can we find an approximation to it? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R8231
## Solution
Show that the equation $x^3+3x^2 - 3 = 0$ has one positive root.
Consider the equation $y=x^3 +3x^2 - 3.$
#### Approach 1
When $x=0$, $y=-3<0$, and when $x=1$, $y=1>0$, so since the curve is continuous (we can draw it without taking our pen off the paper), it must cut the $x$-axis somewhere betwwen $x=0$ and $x=1$.
Could it cut the positive $x$-axis more than once? Well, we can see that $x^3$ and $3x^2$ are increasing functions for positive $x$.
If there were to be a second positive root, the curve would have to decrease somewhere for $x > 0$, which is not possible.
So the curve has a single positive root.
#### Approach 2
Here we find the turning points of the curve.
Differentiating gives $\frac{dy}{dx} = 3x^2 + 6x = 3x(x+2),$ and so setting $\dfrac{dy}{dx}=0$ and solving gives $x=0$ or $x=-2$. Therefore, the turning points lie at $(0,-3)$ and $(-2,1)$.
It is now clear that $(-2,1)$ is a maximum and $(0,-3)$ is a minimum, as $1>-3$.
We can now sketch the graph:
We can see that the equation has exactly one positive root.
Find to three decimal places an approximation to this root.
#### Method 1: Trial and improvement
We first find two integers that the root lies between, then use the same technique to find the root to one decimal place, and so on.
$x$ $y$ $0$ $1$ $0.5$ $0.8$ $0.9$ $0.85$ $0.86$ $0.87$ $0.88$ $0.875$ $0.876$ $0.877$ $0.878$ $0.879$ $0.8795$ -$3$ $1$ -$2.125$ -$0.568$ $0.159$ -$0.128$ -$0.145$ -$0.070$ $0.004$ -$0.033$ -$0.025$ -$0.018$ -$0.011$ -$0.003$ $0.001$
So $x$ lies between $0.879$ and $0.8795$, so to three decimal places, $x \approx 0.879$.
#### Method 2: Linear interpolation
We first find two integers that the root lies between, as in Method 1. Writing $f(x)=x^3+3x^2-3$, we find that $f(0)=-3$ and $f(1)=1$.
This suggests that the root is much closer to $1$ than to $0$, and we can use linear interpolation to improve our estimates, as shown in this diagram:
If our estimates are $x=a$ and $x=b$, giving $f(a)<0$ and $f(b)>0$, then our new estimate $c$ is given by similar triangles (as explored in the resource In-betweens). We have $\frac{c-a}{b-a}=\frac{0-f(a)}{f(b)-f(a)},$ so rearranging gives $c=a-\frac{(b-a)f(a)}{f(b)-f(a)}.$
If $f(c)$ turns out to be negative (as in the example shown), then the root lies between $c$ and $b$, so we use $c$ and $b$ for the next iteration. If $f(c)$ turns out to be positive, then the root lies between $a$ and $c$, so we use $a$ and $c$ for the next iteration.
$a$ $b$ $f(a)$ $f(b)$ $c$ $f(c)$
$0$ $1$ $-3$ $1$ $0.75$ $-0.89$
$0.75$ $1$ $-0.89$ $1$ $0.86777$ $-0.08748$
$0.86777$ $1$ $-0.08748$ $1$ $0.87841$ $-0.00743$
$0.87841$ $1$ $-0.00743$ $1$ $0.87930$ $-0.00062$
$0.87930$ $1$ $-0.00062$ $1$ $0.87938$ $-0.00005$
It looks as though we are almost there: to confirm this, we calculate $f(0.8795)=0.00087>0$, so the root lies between $0.87930$ and $0.8795$, so is $0.879$ to three decimal places.
#### Method 3: Newton-Raphson iteration
Here we start with an estimate for the root, say $x=a$, and draw a tangent to the curve from here. Our next estimate is the intersection of the tangent with the $x$-axis.
The tangent has gradient $f'(a)$, so we see that $\frac{f(a)}{a-c}=f'(a).$ Rearranging this to find $c$ gives $c=a-\frac{f(a)}{f'(a)}.$ Once we have found our $c$, we use that as our next value of $a$.
Since we therefore have an iteration here, we could write $x_0$ for our initial value of $a$, then $x_1$ for the next, and so on, and our formula can be written as $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$
In our case, $f'(x)=3x^2+6x$, so the formula is $c=a-\frac{x^3+3x^2-3}{3x^2+6x}.$
We cannot start with $a=0$, as then the denominator would be $0$ (what does this correspond to in terms of the graph?)
So we start with $a=1$:
$a$
$1$
$0.888889$
$0.879452$
$0.879385$
(We can use the ANS button on our calculator to help us here). The last two values agree to three decimal places. So checking $f(0.879)=-0.0029$ and $f(0.8795)=0.00087$ (or using $f(0.879452)=0.00051,$ which we’ve already had to calculate) shows that the root is $0.879$ to three decimal places.
Which of these methods is the simplest, do you think?
Which requires the fewest calculations? (Especially remembering that there were no electronic pocket calculators in the 1950s!)
Are there other methods which might have been even better? |
Pair Sums
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
Summing Consecutive Numbers
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?
Big Powers
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
Vectors Round a Square
Age 11 to 16
James from Wilson's School told us something he'd noticed about the second, third and fourth vectors forming a square park:
The vector journey goes: $\pmatrix{3\cr 1} \pmatrix{-1\cr 3} \pmatrix{-3\cr -1} \pmatrix{1\cr -3}$
If the first vector has both top and bottom numbers positive, say $x$ and $y$, then the four vectors will be:
$$\pmatrix{x\cr y} \pmatrix{-y\cr x} \pmatrix{-x\cr -y} \pmatrix{y\cr -x}.$$
Elliott, also from Wilson's, and ,Niharika from Leicester High School for Girls, made some observations and completed the solution:
The coordinates of these vectors must only consist of four numbers, say $x, y, -x$ and $-y$. This is always the case as all the sides must have the same length and the angle formed by them with the vertical and horizontal lines that pass through the end point must be the same.
Also after traveling along the first vector you can either move left or right. From there, you have to do the opposite of the first move, then the opposite of the second one, to get back to the original position. This is why the third vector is minus the first one and the fourth one is minus the second one. Hence, we have the following solutions for a journey:
$$\pmatrix{x\cr y} \pmatrix{-y\cr x} \pmatrix{-x\cr -y} \pmatrix{y\cr -x} \text{ or }$$
$$\pmatrix{x\cr y} \pmatrix{y\cr -x} \pmatrix{-x\cr -y} \pmatrix{-y\cr x}.$$
Josephine from The Urswick School added:
The vectors must add up to zero. This happens because you must get back to the original position. So, if we look only at the $x$ coordinates, then if you travel away the original point a distance $d$ you have to travel it back in order to get to the original point. So you must travel a distance $-d$ as well. Therefore, the sum of the $x$ components must be $0$. Similarly, for the sum of the $y$ components. Hence, the vectors add up to zero.
Beth and Susie thought about the case when you know the diagonal vector.
The sum of the first two vectors gives the diagonal one. Therefore, if the diagonal vector is $\pmatrix{m\cr n}$, then either $$\pmatrix{m\cr n}= \pmatrix{x\cr y} + \pmatrix{-y\cr x} = \pmatrix{x-y\cr x+y} \text{ or } \pmatrix{m\cr n}= \pmatrix{x\cr y} + \pmatrix{y\cr- x}= \pmatrix{x+y\cr y- x}.$$
Case 1: $\pmatrix{m\cr n}= \pmatrix{x\cr y} + \pmatrix{-y\cr x}$
Then $x-y=m$ and $x+y=n$. By adding them we get $2x=m+n$, so $x=\frac{m+n}{2}$ and $y=\frac{n-m}{2}$. Hence, the four vectors are
$$\pmatrix{\frac{m+n}{2}\cr \frac{n-m}{2}} \pmatrix{-\frac{n-m}{2}\cr \frac{m+n}{2}} \pmatrix{-\frac{m+n}{2}\cr -\frac{n-m}{2}} \pmatrix{\frac{n-m}{2}\cr -\frac{m+n}{2}}.$$
Case 2: $\pmatrix{m\cr n}= \pmatrix{x\cr y} + \pmatrix{y\cr- x}$
Then $x+y=m$ and $y-x=n$. So, $y=\frac{m+n}{2}$ and $x=\frac{m-n}{2}$. Hence, the four vectors are
$$\pmatrix{\frac{m-n}{2}\cr \frac{m+n}{2}} \pmatrix{\frac{m+n}{2}\cr -\frac{m-n}{2}} \pmatrix{-\frac{m-n}{2}\cr -\frac{m+n}{2}} \pmatrix{-\frac{m+n}{2}\cr \frac{m-n}{2}}.$$
Well done to you all. |
Home > Maths > Make a Right Angle With Knots and String
# Make a Right Angle With Knots and String
By: Suzanne Elvidge BSc (hons), MSc - Updated: 24 Oct 2012 |
A right angle is a square-shaped or L-shaped angle, and measures 90 degrees (a quarter of a circle). A square is made up of four right angles.
#### Make a Right Angle with Knots and String
Find a long piece of string and tie 13 knots in it. It doesn’t matter how far apart they are, just as long as they are the same distance apart. Lay the string out in a triangle with knots at the corners and three gaps between knots on one side, four on the next and five on the final side. Stretch the string taut (get people to hold the corners, or do it outside and peg it into the ground) – this makes a triangle with a right angle.
This works because, according to Pythagoras’ theorem, in a right-angled triangle, the square of the hypotenuse (the sloping side) is equal to the sum of the squares on the other two sides.
In this square, the hypotenuse is five, and the other two sides are three and four.
Just to check it…
5x5 = 3x3 + 4x425 = 9+1625=25
Another way to do this is to tie a knot in a piece of string. Using this as the beginning of the length of string, wrap the string three times round a book, and tie a second knot. Wrap four times round the book and tie a third knot. Wrap five times round a book and tie a fourth knot. To keep it accurate, try not to wind the string over itself too much. Holding the string taut, make a triangle using the knots as corners – this will be a right-angled triangle.
#### Other Ways to Make Right Angles
Fold a square of paper in half to make a rectangle, and then in half again to make a square. Unfold it – the two folds cross at right angles to each other.
Tie a weight to the end of a piece of string. Let the weight hang (it might take a moment for it to stop swinging) – once it is still, the string is at a right angle to the table or ground.
Use a protractor (it will be easier if it is a clear one). Draw a line on a piece of paper, and then draw a cross on it. Match the cross with the line at the bottom of the protractor (line it up with the centre of the protractor – there will probably be a dot or two lines crossing). Make a dot where the protractor says ’90 degrees’. Join the dot and the cross – a right angle.
Make three strips of paper or card, one 3 cm long, another 4 cm long, and the third 5 cm long. This is a similar method to the knotted string, using Pythagoras’ theorem. Make a triangle with the long sides of the three pieces of card – when all three are in place, this will be a right-angled triangle.
Draw a line. Open a pair of compasses so that the pencil and point are more than half the length of the line apart (so, for example, if the line is about 10 cm long, the compasses should be open by about 6 or 7 cm). Put the point of the compass (be careful, it’s sharp) at one end of the line and use the pencil to draw a half circle across the line.
Put the point of the compass at the other end of the line and draw another half circle – this should cross over the first half circle above and below the line (if it doesn’t, don’t worry, just go back and make the curves bigger).
Join up the two crossover points – this will make a right angle with the first line.
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Is a mixed number considered an integer?
Is a mixed number considered an integer?
A mixed number has an integer part and a fractional part.
Why is a fraction not an integer?
Step-by-step explanation: Any fraction, upon division of numerator by denominator, yields a remainder which is not equal to zero is not an integer. If the remainder obtained is zero (i.e. exactly divisible) then the fraction is an integer.
What makes a number not an integer?
The integers are the set of whole numbers and their opposites. Fractions and decimals are not included in the set of integers. For example, 2,5,0,−12,244,−15 and 8 are all integers. The numbers such as 8.5,23 and 413 are not integers.
Why a mixed number is not in the set of integers or whole numbers?
Mixed number is not in the set of integers or whole numbers, because it contains a proper fraction. Thus, mixed fraction is in the set of rational numbers.
What is improper fraction and mixed number?
A mixed number is a whole number plus a fractional part. An improper fraction is a fraction where the numerator (top number) is larger than the denominator (bottom number). You can convert between mixed numbers and improper fractions without changing the value of the figure.
Why are all integers rational numbers?
An integer can be written as a fraction by giving it a denominator of one, so any integer is a rational number. A terminating decimal can be written as a fraction by using properties of place value.
Is a mixed number always greater than a whole number?
Yes. A mixed number includes a whole number and a fraction, such as 1 is the lowest whole number, and any mixed number will have to have a whole number as part of it. So a mixed number is always greater than some whole number.
Is a mixed number a whole number?
A mixed number is a whole number plus a fractional part. An improper fraction is a fraction where the numerator (top number) is larger than the denominator (bottom number). You can convert between mixed numbers and improper fractions without changing the value of the figure.
Is a mixed number irrational?
A rational number can be written as a fraction or mixed number. An irrational number can’t. It can be written as a mixed number, so therefore it is rational.
Can mixed numbers be whole numbers?
A mixed number is a whole number and a proper fraction. Mixed numbers or mixed fractions are used to express an amount greater than a whole but less than the next whole number. Mixed numbers can be formed from improper fractions. |
# Solving y’ + a(x) y = b(x), Part 3
Yet another way to find the solution of
$y' + a(x) y = b(x)\quad\quad\quad(1)$
is to seek a function $f(x) > 0 \; \forall x$ such that the result of multiply (1) by $f(x)$, namely
$f(x)y' + f(x) a(x) y =f(x) b(x)\quad\quad\quad(2)$
can be written as
$(f(x) y)' = b(x) f(x)$,
i.e.,
$f(x) y =\int b(x) f(x)\; dx + c_{3}$
where $c_{3}$ is a constant.
Or,
$y = {1 \over f(x)} (\int b(x) f(x)\; dx + c_{3})\quad\quad\quad(3)$
since $f(x) > 0$.
Let us proceed to find such $f(x)$.
From (2) we see that if
$f'(x)=a(x) f(x)$
then the left side of (2) is certainly $(f(x) y)'$ and consequently for $f(x) > 0 \;\forall x$,
${1 \over f(x)} f'(x) = a(x)\quad\quad\quad(4)$
or,
$log(f(x)) = \int a(x) dx + c_{1}$
where $c_{1}$ is a constant.
Therefore, a solution to (4) is
$f(x) = c_{2}e^{\int a(x)\;dx}\quad\quad\quad(5)$
where $c_{2} = exp(c_{1})$. This is a positive function $\forall x$ indeed.
With (5), (3) becomes
$y = e^{-\int a(x)\;dx} (\int b(x) e^{ \int b(x) dx} dx + c)\quad\quad\quad(6)$
where $c = {c_{3} \over c_{2}}$.
In fact, for any constant c,
$e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)$
is a solution of (1):
$(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx+ c))' +$
$a(x)\dot(e^{-\int a(x)\;dx} (\int b(x) e^{\int a(x) dx} dx + c))$
$= -a(x) (e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))$
$+ e^{-\int a(x) dx} b(x) e^{\int a(x) dx}$
$+ a(x)(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))$
$=b(x)$
Therefore, the solution of (1) is
$y = e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)$
where $c$ is any constant.
Exercise: prove that $e^x$ is a positive function. |
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David Terr
Ph.D. Math, UC Berkeley
2.7. Rational Functions and Their Graphs
Thus far we have looked at graphs of polynomials as well as some simple rational functions. In this section, we will investigate graphs of rational functions in more detail.
As we know, rational functions have the general form f(x) = p(x) / q(x), where p and q are polynomials. Whenever q(x) is zero and p(x) is not, f(x) is undefined. This results in a singularity of f(x). For values of x near the singular value, |f(x)| becomes very large. Thus, when graphing rational functions, it is a good idea to first find its singularities.
Example 1: Graph the function f(x) = 1 / (1 - x2). (We already looked at this function in Section 1.6, but now we take a closer look.)
Solution: Before trying to graph f, we compute its singularities. The denominator is 1 - x2, which factors as (1 + x)(1 - x), which has zeros at x=±1. Since the numerator of f is never zero, we see that these values must be the singularities of f. We draw vertical dashed lines, known as asymptotes, at these values of x as shown below:
Next we make a table of values of f(x) versus values of x. In general, we should use several values of x near the singular values. The values we used in constructing the table in Example 3 of Section 1.6 work quite well. Below we reproduce the table and the resulting graph, along with the asymptotes.
x f(x)
-2.0
-0.33
-1.8
-0.45
-1.6
-0.64
-1.4
-1.04
-1.2
-2.27
-1.0
-
-0.8
2.78
-0.6
1.56
-0.4
1.19
-0.2
1.04
0.0
1.00
0.2
1.04
0.4
1.19
0.6
1.56
0.8
2.78
1.0
-
1.2
-2.27
1.4
-1.04
1.6
-0.64
1.8
-0.45
2.0
-0.33
Example 2: Graph the function f(x) = (x2 - 4) / (3x2 + 8x - 3).
Solution: By trial and error, we find that the denominator factors as 3x2 + 8x - 3 = (x + 3)(3x - 1), which has zeros at x = -3 and x = 1/3, while the numerator has zeros at x = ±2. Thus we compute f(x) for values of x from -4 to 4, with more values near x = -1/3 and x = 3. A table of these values is given below, followed by the resulting graph.
x f(x) x f(x)
-4.0
0.92
-0.5
0.60
-3.5
1.43
0.0
1.33
-3.4
1.69
0.1
1.84
-3.3
2.11
0.2
3.09
-3.2
2.94
0.3
11.85
-3.1
5.45
0.3
-
-3.0
-
0.4
-5.65
-2.9
-4.55
0.5
-2.14
-2.8
-2.04
1.0
-0.38
-2.7
-1.21
1.5
-0.11
-2.6
-0.78
2.0
0.00
-2.5
-0.53
2.5
0.06
-2.0
0.00
3.0
0.10
-1.5
0.21
3.5
0.13
-1.0
0.38
4.0
0.16 |
# Decimals: Dividing by 100
In this worksheet, stidemts divide numbers by 100 by shifting digits two places to the right in a grid.
Key stage: KS 2
Curriculum topic: Number: Multiplication and Division
Curriculum subtopic: Multiply/Divide Using Decimals
Difficulty level:
### QUESTION 1 of 10
When we divide numbers by 100, each digit must move two places to the right.
When we look at the columns in a Th, H, T, U grid we notice that Tens are 100 times smaller than Thousands, Units are 100 times smaller than Hundreds etc...
*Note that the U means Units, which is another word for Ones.
e.g. Work out:
54.6 × 10 = 0.546
Here is 54.6 in a Th, H, T, U grid.
Each digit moves two places to the right.
Remember to fill in that 0 in the units column.
So 54.6 ÷ 100 = 0.546
Want to understand this further and learn how this links to other topics in maths?
Why not watch this video?
Work out:
538 ÷ 100
Work out:
507 ÷ 100
Work out:
699 ÷ 100
Work out:
12.3 ÷ 100
Work out:
10.3 ÷ 100
Work out:
1.7 ÷ 100
Work out:
3 ÷ 100
Work out:
0.7 ÷ 100
Work out:
5.2 ÷ 100
Work out:
317.4 ÷ 100
• Question 1
Work out:
538 ÷ 100
5.38
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 538 becomes 5.38
• Question 2
Work out:
507 ÷ 100
5.07
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 507 becomes 5.07
• Question 3
Work out:
699 ÷ 100
6.99
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 699 becomes 6.99
• Question 4
Work out:
12.3 ÷ 100
0.123
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 12.3 becomes 0.123
• Question 5
Work out:
10.3 ÷ 100
0.103
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 10.3 becomes 0.103
• Question 6
Work out:
1.7 ÷ 100
0.017
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 1.7 becomes 0.017
• Question 7
Work out:
3 ÷ 100
0.03
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 3 becomes 0.03
• Question 8
Work out:
0.7 ÷ 100
0.007
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 0.7 becomes 0.007
• Question 9
Work out:
5.2 ÷ 100
0.052
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 5.2 becomes 0.052
• Question 10
Work out:
317.4 ÷ 100
3.174
EDDIE SAYS
When we divide numbers by 100, each digit must move two places to the right. 317.4 becomes 3.174
---- OR ----
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Data Structures & Algorithms
# Topological Sorting Algorithm
The topological sorting algorithm works with the DAG (Direct Acyclic Graph). The meaning of the topological sort is that if any node points to another node, then the node that points to another node will come after that. So, in this case, if we have a cyclic graph, we cannot predict which node comes after which node. So that is the reason the topological sort algorithm only works with the acyclic graph and not with the cyclic graph.
Every graph has more than one topological sort sequence because it depends upon the in-degree of the edges of the nodes. The first starting node that we choose with an in-degree number of nodes is 0.
Let’s understand the topological sort algorithm with an example.
Step 1: We insert those nodes whose incoming edge count is equal to 0. So those nodes are node 1 and node 4.
Step 2:
a. We begin with Node 1. We can choose any node between Node 1 and Node 4.
b. We decrease every node edge by 1, which is connected to node 1. We are decreasing the edge of the nodes (0, 2, and 3).
c. We move Node 1 from the list to the topologically sorted list, as shown below.
Step 3:
a. We now proceed from the list, which is Node 4.
b. We decrease all the outgoing edges of the nodes connected to node 4, which are nodes (0 and 3).
c. Now, node 3 has no incoming edges, so we push it into the list, and node 4 shifts to the topologically sorted list.
Step 4:
a. We now proceed from the list, which is Node 3.
b. We decrease all the outgoing edges of the nodes connected to node 3, which are nodes (0 and 2).
c. Now, nodes 0 and Node 2 have no incoming edges, so we push it into the list, and node 3 shifts to the topologically sorted list.
Step 5:
a. We now proceed from the list, which is Node 0.
b. As no outgoing edges from Node 0, so we simply add them to the topological sort list.
Step 6:
a. We now proceed from the list, which is Node 2.
b. As no outgoing edges from Node 2, so we simply add them to the topological sort list.
Step 7:
Finally, we have sorted the list here.
## Topological Sort Algorithm
The below are the steps for the topological sorting algorithm which we have to follow.
Step 0: Calculate the in-degree of each graph node.
Step 1: We first have to find a node that has incoming edges of zero.
Step 2: We remove that node from the graph and add it to the list of topological sorting orders.
Step 3: Remove those nodes that have outgoing edges.
Step 4: Reduce the in-degree by the number of related edges that were removed.
Step 5: Repeat Steps 1–4 until no nodes with 0 in-degree remain.
Step 6: Verify that all of the items are in the correct sequence.
Step 7: Now, we have sorted the order from Step 6.
Step 8: Put an end to the algorithm.
Python Code: The below is a python implementation of the above example.
fromcollectionsimportdefaultdict
classbuildGraph :
def__init__(self, nodes : int) :
self.nodes= nodes
# We are now storing the graph in adjacent list format
# It will stores the information about a particular node incoming
# edges in a graph
self.count_numbers_of_incoming_edge_of_a_node= []
# We stores the sorted nodes in topological order
self.topological_sorted_order= []
# We stores the information of all those nodes those don't
# have any incoming edges in a graph
self.nodes_have_zero_incoming_edges= []
# Now we are creating a adjacent list of all the graphs to sort
defAddGraphEdge (self, source :int, destination : int) :
self.count_numbers_of_incoming_edge_of_a_node[destination] +=1
defTopologicalSortAlgorithm (self) :
for node inrange(self.nodes) :
ifself.count_numbers_of_incoming_edge_of_a_node[node] ==0 :
self.nodes_have_zero_incoming_edges.append(node)
whileself.nodes_have_zero_incoming_edges :
self.nodes_have_zero_incoming_edges.sort()
source =self.nodes_have_zero_incoming_edges.pop(0)
self.count_numbers_of_incoming_edge_of_a_node[node] -=1
ifself.count_numbers_of_incoming_edge_of_a_node[node] ==0 :
self.nodes_have_zero_incoming_edges.append(node)
self.topological_sorted_order.append(source)
print("Topological Sorting Order : "+str(self.topological_sorted_order))
defmain() :
number_of_nodes=7
graph =buildGraph(number_of_nodes)
graph.count_numbers_of_incoming_edge_of_a_node= [0] *number_of_nodes
graph.TopologicalSortAlgorithm()
if __name__ =="__main__" :
main()
Output:
Topological Sorting Order : [0, 1, 3, 5, 6, 2, 4]
## Topological Sorting Algorithm Time Complexity:
The total time to process the algorithm is O (E + N), where E represents the number of edges and N represents the number of nodes in the graph. Then, in the following step, we must calculate each node’s in-degree, which generally takes O(E) times, and then place all of those nodes in a sorted list where their in-degree is zero, which takes O(N) times. So the total time complexity of the topological sorting algorithm is O (E+N).
But the space complexity of the topological sorting algorithm is O (N), which is the total number of nodes in the graph.
## Application :
1. Topological sort is very useful for finding the graph’s cycle.
2. Topological sort algorithm is used to determine the deadlock conditions in an operating system.
3. Topological sort algorithm is used to find the shortest path in a weighted acyclic graph.
## Conclusion:
This article has learned about one more important algorithm, topological sorting. We have seen that this algorithm only works with acyclic graphs. The topological sorting algorithm also helps determine the task compilation order. The topological sorting algorithm has many real-time advantages, like finding the shortest path. Because the topological sort is extremely useful, every programmer and student must thoroughly understand this algorithm. |
# Difference between revisions of "2014 AMC 10B Problems/Problem 3"
## Problem
Randy drove the first third of his trip on a gravel road, the next $20$ miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Randy's trip?
$\textbf {(A) } 30 \qquad \textbf {(B) } \frac{400}{11} \qquad \textbf {(C) } \frac{75}{2} \qquad \textbf {(D) } 40 \qquad \textbf {(E) } \frac{300}{7}$
## Solution 1
Let the total distance be $x$. We have $\dfrac{x}{3} + 20 + \dfrac{x}{5} = x$, or $\dfrac{8x}{15} + 20 = x$. Subtracting $\dfrac{8x}{15}$ from both sides gives us $20 = \dfrac{7x}{15}$. Multiplying by $\dfrac{15}{7}$ gives us $x = \fbox{(E)} \dfrac{300}{7}$.
## Solution 2
The first third of his distance added to the last one-fifth of his distance equals 8/15. Therefore, 7/15 of his distance is 20. Let x be his total distance, and solve for x. Therefore, x is equal to 300/7, or E. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Angle Bisectors in Triangles
## Construction and properties of bisectors, which cut angles in half.
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Practice Angle Bisectors in Triangles
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Angle Bisectors in Triangles
What if the cities of Verticville, Triopolis, and Angletown were joining their city budgets together to build a centrally located airport? There are freeways between the three cities and they want to have the freeway on the interior of these freeways. Where is the best location to put the airport so that they have to build the least amount of road? In the picture below, the blue lines are the proposed roads. After completing this Concept, you'll be able to use angle bisectors to help answer this question.
### Guidance
Recall that an angle bisector cuts an angle exactly in half. Let’s analyze this figure.
BD\begin{align*}\overrightarrow{BD}\end{align*} is the angle bisector of ABC\begin{align*}\angle ABC\end{align*}. Looking at point D\begin{align*}D\end{align*}, if we were to draw ED¯¯¯¯¯\begin{align*}\overline{ED}\end{align*} and DF¯¯¯¯¯\begin{align*}\overline{DF}\end{align*}, we would find that they are equal. Recall that the shortest distance from a point to a line is the perpendicular length between them. ED\begin{align*}ED\end{align*} and DF\begin{align*}DF\end{align*} are the shortest lengths between D\begin{align*}D\end{align*}, which is on the angle bisector, and each side of the angle.
Angle Bisector Theorem: If a point is on the bisector of an angle, then the point is equidistant from the sides of the angle.
In other words, if BD\begin{align*}\overleftrightarrow{BD}\end{align*} bisects ABC,BEED¯¯¯¯¯\begin{align*}\angle ABC, \overrightarrow{BE} \bot \overline{ED}\end{align*}, and BFDF¯¯¯¯¯\begin{align*}\overrightarrow{BF} \bot \overline{DF}\end{align*}, then ED=DF\begin{align*}ED = DF\end{align*}.
Proof of the Angle Bisector Theorem:
Given: BD\begin{align*}\overrightarrow{BD} \end{align*} bisects ABC,BAAD¯¯¯¯¯¯\begin{align*}\angle ABC, \overrightarrow{BA} \bot \overline{AD}\end{align*}, and BCDC¯¯¯¯¯\begin{align*}\overrightarrow{BC} \bot \overline{DC}\end{align*}
Prove: AD¯¯¯¯¯¯DC¯¯¯¯¯\begin{align*}\overline{AD} \cong \overline{DC}\end{align*}
Statement Reason
1. BD\begin{align*}\overrightarrow{BD}\end{align*} bisects ABC,BAAD¯¯¯¯¯¯,BCDC¯¯¯¯¯\begin{align*}\angle ABC, \overrightarrow{BA} \bot \overline{AD}, \overrightarrow{BC} \bot \overline{DC}\end{align*} Given
2. ABDDBC\begin{align*}\angle ABD \cong \angle DBC\end{align*} Definition of an angle bisector
3. DAB\begin{align*}\angle DAB\end{align*} and DCB\begin{align*}\angle DCB\end{align*} are right angles Definition of perpendicular lines
4. DABDCB\begin{align*}\angle DAB \cong \angle DCB\end{align*} All right angles are congruent
5. BD¯¯¯¯¯¯BD¯¯¯¯¯¯\begin{align*}\overline{BD} \cong \overline{BD}\end{align*} Reflexive PoC
6. ABDCBD\begin{align*}\triangle ABD \cong \triangle CBD\end{align*} AAS
7. AD¯¯¯¯¯¯DC¯¯¯¯¯\begin{align*}\overline{AD} \cong \overline{DC}\end{align*} CPCTC
The converse of this theorem is also true.
Angle Bisector Theorem Converse: If a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of the angle.
Because the Angle Bisector Theorem and its converse are both true we have a biconditional statement. We can put the two conditional statements together using if and only if. A point is on the angle bisector of an angle if and only if it is equidistant from the sides of the triangle. Like perpendicular bisectors, the point of concurrency for angle bisectors has interesting properties.
##### Investigation: Constructing Angle Bisectors in Triangles
Tools Needed: compass, ruler, pencil, paper
1. Draw a scalene triangle. Construct the angle bisector of each angle. Use Investigation 1-4 and #1 from the Review Queue to help you.
Incenter: The point of concurrency for the angle bisectors of a triangle.
2. Erase the arc marks and the angle bisectors after the incenter. Draw or construct the perpendicular lines to each side, through the incenter.
3. Erase the arc marks from #2 and the perpendicular lines beyond the sides of the triangle. Place the pointer of the compass on the incenter. Open the compass to intersect one of the three perpendicular lines drawn in #2. Draw a circle.
Notice that the circle touches all three sides of the triangle. We say that this circle is inscribed in the triangle because it touches all three sides. The incenter is on all three angle bisectors, so the incenter is equidistant from all three sides of the triangle.
Concurrency of Angle Bisectors Theorem: The angle bisectors of a triangle intersect in a point that is equidistant from the three sides of the triangle.
If AG¯¯¯¯¯,BG¯¯¯¯¯\begin{align*}\overline{AG}, \overline{BG}\end{align*}, and GC¯¯¯¯¯\begin{align*}\overline{GC}\end{align*} are the angle bisectors of the angles in the triangle, then EG=GF=GD\begin{align*}EG = GF = GD\end{align*}.
In other words, EG¯¯¯¯¯,FG¯¯¯¯¯\begin{align*}\overline{EG}, \overline{FG}\end{align*}, and DG¯¯¯¯¯¯\begin{align*}\overline{DG}\end{align*} are the radii of the inscribed circle.
#### Example A
Is Y\begin{align*}Y\end{align*} on the angle bisector of XWZ\begin{align*}\angle XWZ\end{align*}?
In order for Y\begin{align*}Y\end{align*} to be on the angle bisector XY\begin{align*}XY\end{align*} needs to be equal to YZ\begin{align*}YZ\end{align*} and they both need to be perpendicular to the sides of the angle. From the markings we know XY¯¯¯¯¯WX\begin{align*}\overline{XY} \bot \overrightarrow{WX}\end{align*} and ZY¯¯¯¯¯WZ\begin{align*}\overline{ZY} \bot \overrightarrow{WZ}\end{align*}. Second, XY=YZ=6\begin{align*}XY = YZ = 6\end{align*}. From this we can conclude that Y\begin{align*}Y\end{align*} is on the angle bisector.
#### Example B
If J,E\begin{align*}J, E\end{align*}, and G\begin{align*}G\end{align*} are midpoints and KA=AD=AH\begin{align*}KA = AD = AH\end{align*} what are points A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} called?
A\begin{align*}A\end{align*} is the incenter because KA=AD=AH\begin{align*}KA = AD = AH\end{align*}, which means that it is equidistant to the sides. B\begin{align*}B\end{align*} is the circumcenter because JB¯¯¯¯¯,BE¯¯¯¯¯\begin{align*}\overline{JB}, \overline{BE}\end{align*}, and BG¯¯¯¯¯\begin{align*}\overline{BG}\end{align*} are the perpendicular bisectors to the sides.
#### Example C
AB\begin{align*}\overrightarrow{AB}\end{align*} is the angle bisector of CAD\begin{align*}\angle CAD\end{align*}. Solve for the missing variable.
CB=BD\begin{align*}CB=BD\end{align*} by the Angle Bisector Theorem, so we can set up and solve an equation for x\begin{align*}x\end{align*}.
x+7x+715x=2(3x4)=6x8=5x=3
Watch this video for help with the Examples above.
#### Concept Problem Revisited
The airport needs to be equidistant to the three highways between the three cities. Therefore, the roads are all perpendicular to each side and congruent. The airport should be located at the incenter of the triangle.
### Vocabulary
An angle bisector cuts an angle exactly in half. Equidistant means the same distance from. A point is equidistant from two lines if it is the same distance from both lines. When we construct angle bisectors for the angles of a triangle, they meet in one point. This point is called the incenter of the triangle.
### Guided Practice
1. Is there enough information to determine if AB\begin{align*}\overrightarrow{A B}\end{align*} is the angle bisector of CAD\begin{align*}\angle CAD\end{align*}? Why or why not?
2. MO\begin{align*}\overrightarrow{MO}\end{align*} is the angle bisector of LMN\begin{align*}\angle LMN\end{align*}. Find the measure of x\begin{align*}x\end{align*}.
3. A 100\begin{align*} 100^\circ\end{align*} angle is bisected. What are the measures of the resulting angles?
1. No because B\begin{align*}B\end{align*} is not necessarily equidistant from AC¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} and AD¯¯¯¯¯¯\begin{align*}\overline{AD}\end{align*}. We do not know if the angles in the diagram are right angles.
2. LO=ON\begin{align*}LO = ON\end{align*} by the Angle Bisector Theorem.
4x54xx=23=28=7
3. We know that to bisect means to cut in half, so each of the resulting angles will be half of 100\begin{align*}100\end{align*}. The measure of each resulting angle is 50\begin{align*}50^\circ\end{align*}.
### Practice
For questions 1-6, AB\begin{align*}\overrightarrow{AB}\end{align*} is the angle bisector of CAD\begin{align*}\angle CAD\end{align*}. Solve for the missing variable.
Is there enough information to determine if AB\begin{align*}\overrightarrow{AB}\end{align*} is the angle bisector of CAD\begin{align*}\angle CAD\end{align*}? Why or why not?
1. Fill in the blanks in the Angle Bisector Theorem Converse.
Given: AD¯¯¯¯¯¯DC¯¯¯¯¯\begin{align*}\overline{AD} \cong \overline{DC}\end{align*}, such that AD\begin{align*}AD\end{align*} and DC\begin{align*}DC\end{align*} are the shortest distances to BA\begin{align*}\overrightarrow{BA}\end{align*} and BC\begin{align*}\overrightarrow{BC}\end{align*}
Prove: BD\begin{align*}\overrightarrow{BD}\end{align*} bisects ABC\begin{align*}\angle ABC\end{align*}
Statement Reason
1.
2. The shortest distance from a point to a line is perpendicular.
3. DAB\begin{align*}\angle DAB\end{align*} and DCB\begin{align*}\angle DCB\end{align*} are right angles
4. \begin{align*}\angle DAB \cong \angle DCB\end{align*}
5. \begin{align*}\overline{BD} \cong \overline{BD}\end{align*}
6. \begin{align*}\triangle ABD \cong \triangle CBD\end{align*}
7. CPCTC
8. \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC\end{align*}
Determine if the following descriptions refer to the incenter or circumcenter of the triangle.
1. A lighthouse on a triangular island is equidistant to the three coastlines.
2. A hospital is equidistant to three cities.
3. A circular walking path passes through three historical landmarks.
4. A circular walking path connects three other straight paths.
Multi- Step Problem
1. Draw \begin{align*}\angle ABC\end{align*} through \begin{align*}A(1, 3), B(3, -1)\end{align*} and \begin{align*}C(7, 1)\end{align*}.
2. Use slopes to show that \begin{align*}\angle ABC\end{align*} is a right angle.
3. Use the distance formula to find \begin{align*}AB\end{align*} and \begin{align*}BC\end{align*}.
4. Construct a line perpendicular to \begin{align*}AB\end{align*} through \begin{align*}A\end{align*}.
5. Construct a line perpendicular to \begin{align*}BC\end{align*} through \begin{align*}C\end{align*}.
6. These lines intersect in the interior of \begin{align*}\angle ABC\end{align*}. Label this point \begin{align*}D\end{align*} and draw \begin{align*}\overrightarrow{BD}\end{align*}.
7. Is \begin{align*}\overrightarrow{BD}\end{align*} the angle bisector of \begin{align*}\angle ABC\end{align*}? Justify your answer.
### Vocabulary Language: English
Angle Bisector Theorem
Angle Bisector Theorem
The angle bisector theorem states that if a point is on the bisector of an angle, then the point is equidistant from the sides of the angle.
Angle Bisector Theorem Converse
Angle Bisector Theorem Converse
The angle bisector theorem converse states that if a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of that angle.
incenter
incenter
The incenter is the point of intersection of the angle bisectors in a triangle. |
# Write Up #4
### Before reading this write up, it might be helpful to open up this GSP file to be able to follow along!!
When you first open the GSP file, you will see that our whole triangle is split into 6 smaller triangles labeled 1 through 6. These have been created by the three medians, which as defined above, create the centriod. The calculated areas, which were found simply using our measurement tool in GSP, are shown for each of the 6 triangles. Notice that these values are all equal to each other no matter where or how you drag the vertices of the whole triangle to change its shape/appearance. However, this isn't enough to prove how or why these 6 smaller triangles are equal in area, so this is what we will discuss below.
All 6 triangles share a common vertex: the centroid, which in our GSP file is the point G. The three medians create midpoints D, E, and F on each of the 3 sides of the big triangle. Now, let's start breaking the 6 smaller triangles into pairs and discuss their areas.
Let's first look at triangles 2 and 3. They share altitude FG, which we can also call their height. Their bases, AF and CF are equal in length because we know F is the midpoint of segment AC. Using the area formula for a triangle, A = 1/2bh, we can say their areas are equal because their bases and heights are equal to each other, therefore, the area formula holds true to be equal for both triangles 2 and 3. The same concept holds true for the other two pairs of triangles. Triangles 4 and 5 share height EG and midpoint E means their bases, CE and BE, are of equal length, therefore, the areas of triangles 4 and 5 are equal to each other. Triangles 1 and 6 share height DG and midpoint D means their bases, AD and BD, are of equal length, therefore, the areas of triangles 1 and 6 are equal to each other. So, we now have established 3 pairs of triangles with equal areas, but how do we know that all 6 triangles have areas equal to each other?
Going back to triangles 2 and 3 and also considering parts of the whole picture, it's fair to say:
(Δ2 + Δ3) = (Δ2 + Δ3 + Δ4) - Δ4 where(Δ2 + Δ3 + Δ4) is the right half of the big triangle
Now consider the left half of the big triangle (Δ1 + Δ6 + Δ5) and another pair of small triangles that we know have equal areas (triangles 1 and 6). We can say:
(Δ1 + Δ6) = (Δ1 + Δ6 + Δ5) - Δ5
Now that we have established equations about the right half and left half of our whole triangle, which are two other triangle shapes within our figure, we can discuss these two triangles in a way similar to the way we were discussing the three pairs of the six small triangles. These two triangles are equal in area as well. The right and left triangle share the same altitude/height of AE, one of our medians. Then, the same bases for our triangle 4 and 5 pair, CE and BE, are the bases for our right and left half triangles as well, therefore we know the bases are of equal length. So finally, we can say that their areas are equal and write it as:
(Δ2 + Δ3 + Δ4) = (Δ1 + Δ6 + Δ5)
So this shows that our six small triangles are of equal area to each other! |
# USING A GRAPH TO ESTIMATE THE SOLUTION OF A SYSTEM
## About "Using a graph to estimate the solution of a system"
Using a graph to estimate the solution of a system :
We can use a graph to estimate the solution of a system of equations before solving the system algebraically.
## Using a graph to estimate the solution of a system - Examples
Example 1 :
Estimate the solution by sketching a graph of each linear function. Then solve the system algebraically. Use your estimate to judge the reasonableness of your solution.
x - 4y = 4
2x - 3y = -3
Solution :
Step 1 :
To sketch the graph of the equations, write them in slope-intercept form.
That is,
y = mx + b
x - 4y = 4
y = (1/4)x - 1
Slope = 1/4
y-intercept = -1
2x - 3y = -3
y = (2/3)x + 1
Slope = 2/3
y-intercept = 1
Step 2 :
Find the point of intersection of the lines. The point of intersection is near to the point (-5, -2).
Let us give (-5, -2) as our estimation to the solution of the system.
Step 3 :
Solve the system algebraically.
Select one of the equation, say x - 4y = 4.
Solve for the variable x in terms of y.
Add 4y to both sides.
(x - 4y) + 4y = (4) + 4y
x = 4 + 4y
Step 4 :
Substitute the expression for x in the other equation and solve.
2x - 3y = -3
2(4 + 4y) - 3y = -3
8 + 8y - 3y = -3
Combine the like terms.
8 + 5y = -3
Subtract 8 from both sides.
5y = -11
Divide by 5 on both sides.
5y / 5 = -11 / 5
y = -11/5
Step 5 :
Substitute the value of y we got above (y = -11/5) into one of the equations and solve for the other variable, y.
x - 4y = 4
x - 4(-11/5) = 4
x + 44/5 = 4
Subtract 44/5 from both sides.
x = 4 - 44/5
x = -24/5
Hence, the solution of the system is (-24/5, -11/5).
Step 6 :
Use the estimate which has been made using the graph to judge the reasonableness of the solution.
-24/5 is close to -5 and -11/5 is close to -2.
So the solution estimated from the graph (-5, -2) seems reasonable.
Example 2 :
Estimate the solution by sketching a graph of each linear function. Then solve the system algebraically. Use your estimate to judge the reasonableness of your solution.
x + y = 4
2x - y = 6
Solution :
Step 1 :
To sketch the graph of the equations, write them in slope-intercept form.
That is,
y = mx + b
x + y = 4
y = - x + 4
Slope = - 1
y-intercept = 4
2x - y = 6
y = 2x - 6
Slope = 2
y-intercept = -6
Step 2 :
Find the point of intersection of the lines. The point of intersection is near to the point (3, 1).
Let us give (3, 1) as our estimation to the solution of the system.
Step 3 :
Solve the system algebraically.
Select one of the equation, say x + y = 4.
Solve for the variable y in terms of x.
Subtract x from both sides.
(x + y) - x = (4) - x
y = 4 - x
Step 4 :
Substitute the expression for y in the other equation and solve.
2x - y = 6
2x - (4 - x) = 6
2x - 4 + x = 6
Combine the like terms.
3x - 4 = 6
Add 4 to both sides.
3x = 10
Divide by 3 on both sides.
3x / 3 = 10 / 3
x = 10/3
Step 5 :
Substitute the value of x we got above (x = 10/3) into one of the equations and solve for the other variable, y.
x + y = 4
10/3 + y = 4
Subtract 10/3 from both sides.
y = 4 - 10/3
y = 2/3
Hence, the solution of the system is (10/3, 2/3).
Step 6 :
Use the estimate which has been made using the graph to judge the reasonableness of the solution.
10/3 is close to 3 and 2/3 is close to 1.
So the solution estimated from the graph (3, 1) seems reasonable.
After having gone through the stuff given above, we hope that the students would have understood "Using a graph to estimate the solution of a system".
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## Note on Trigonometric ratios of any angle
• Note
• Things to remember
• Videos
• Exercise
• Quiz
#### Trigonometric ratios of all quadrants
Let a line OP start from OX revolve round O in the anticlockwise direction. Then it will trace positive angle XOP. If OP is in the first quadrant, then $$\angle$$XOP lies between 0° and 90°. If OP is in the second quadrant, then $$\angle$$XOP lies between 90° and 180°. Similarly, when OP is in the third or the fourth quadrant, then $$\angle$$XOP lies between 180° and 270° or 270° and 360° respectively.If OP revolves in the clockwise direction, the angle traced is negative.
#### Rule of signs
In the above figure, lines measured along OX' or OY' are considered positive. Lines measured along OX and OY are considered negative. If PM is perpendicular to XX', then
1. Both OM and MP are positive when OP is in the first quadrant.
2. OM is negative and MP is positive when OP is in the second quadrant.
3. Both OM and MP are negative when OP is in the third quadrant.
4. OM is positive and MP is negative when OP is in the fourth quadrant.
5. The line OP is always positive.
Thus the values of trigonometric ratios in different quadrants are as follows:-
1st quadrant 2nd quadrant 3rd quadrant 4th quadrant all the ratios are (+e) positive sin and cosec (+ve) remaining (-ve) tan and cot (+ve) remaining (-ve) cos and sec (+ve) remaining (-v)
Ratios of (90° - A)
Let revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ (= OP) revolve from OX to OYn and return back to OQ through an angle YOQ = A.
Then $$\angle$$XOQ = 90° - A
Draw perpendiculars PM and QN from P and Q respectively to OX.
Now, in $$\triangle$$s ONQ and OMP, $$\angle$$OQN = $$\angle$$MOP, $$\angle$$ONQ $$\angle$$OMP and OQ = OP. So, triangles ONQ and OMP are congruent.
Corresponding sides of a congruent triangle are equal.
$$\therefore$$ OQ = OP, NQ = OM and ON = MP.
Now,
sin(90° - A) = $$\frac{NQ}{OQ}$$= $$\frac{OM}{OP}$$ = cos A
cos(90° - A)= $$\frac{ON}{OQ}$$= $$\frac{MP}{OP}$$ = sin A
tan(90° - A) = $$\frac{NQ}{ON}$$= $$\frac{OM}{MP}$$ = cot A
cosec(90° - A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{OM}$$ = sec A
sec(90° - A)= $$\frac{OQ}{ON}$$= $$\frac{OP}{MP}$$ = cosec A
cot(90° - A)= $$\frac{ON}{NQ}$$= $$\frac{MP}{OM}$$ = tan A
Ratios of (90° + A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ (=OP) revolve from OX to OY and agian to OQ through an angle YOQ = A.
Then, $$\angle$$XOQ = 90° + A
Draw perpendicular PM and QN from P and Q respectively to XOX'. In $$\triangle$$s ONQ and OMP, OQ = OP, $$\angle$$OQN = $$\angle$$MOP and $$\angle$$ONQ =$$\angle$$OMP. So, $$\triangle$$ ONQ ≅$$\triangle$$OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = MP and NQ = OM.
By the rule of signs, OQ = OP, ON = -MP and NQ = OM.
Now,
sin(90° + A) = $$\frac{NQ}{OQ}$$= $$\frac{OM}{OP}$$ = cos A
cos(90° + A)= $$\frac{ON}{OQ}$$= $$\frac{-MP}{OP}$$ = -sin A
tan(90° + A) = $$\frac{NQ}{ON}$$= $$\frac{OM}{-MP}$$ = -cot A
cosec(90° + A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{OM}$$ = sec A
sec(90° + A)= $$\frac{OQ}{ON}$$= $$\frac{OP}{-MP}$$ = -cosec A
cot(90° + A)= $$\frac{ON}{NQ}$$= $$\frac{-MP}{OM}$$ = -tan A
Ratios of (180° - A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OP (= OP) revolve from OX to OX' and return to OQ through an angle X'OQ = A.
Then, $$\angle$$XOQ = 180° - A.
Draw perpendicular PM and QN from P and Q respectively to XOX'.
In $$\triangle$$sONQ and OMP, $$\angle$$ONQ =$$\angle$$OMP,$$\angle$$QON =$$\angle$$POM and OQ = OP. So, $$\triangle$$ONQ≅ $$\triangle$$OMP.
Corresponding sides of congruent triangles are equal.
$$\therefore$$OQ = OP, ON = OM and NQ = MP.
By the rule of signs, OQ = OP, ON = -OM and NQ = MP
Now,
sin(180° - A) = $$\frac{NQ}{OQ}$$= $$\frac{MP}{OP}$$ =sin A
cos(180° - A)= $$\frac{ON}{OQ}$$= $$\frac{-OM}{OP}$$ = -cos A
tan(180° - A) = $$\frac{NQ}{ON}$$= $$\frac{MP}{-OM}$$ = -tan A
cosec(180° - A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{MP}$$ = cosec A
sec(180° - A)= $$\frac{OQ}{ON}$$= $$\frac{OM}{-OM}$$ = -sec A
cot(180° - A)= $$\frac{ON}{NQ}$$= $$\frac{-OM}{MP}$$ = -cot A
Ratios of (180° + A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ (= OP) revolve from OX to OX' and again to OQ through an angle X'OQ = A.
Then, $$\angle$$XOQ = 180° + A.
Draw perpendiculars PM and QN from P and Q respectively to XOX'.
In $$\triangle$$sONQ and OMP, $$\angle$$ONQ =$$\angle$$OMP,$$\angle$$QON =$$\angle$$POM and OQ = OP.
$$\therefore$$ $$\triangle$$ONQ = $$\triangle$$OMP.
Corresponding sides of congruent triangles are equal.
$$\therefore$$ OQ = OP, ON = OM and NQ = MP.
By the rule of signs, OQ = OP, ON = -OM and NQ = -MP.
Now,
sin(180° + A) = $$\frac{NQ}{OQ}$$= $$\frac{-OM}{OP}$$ = -sin A
cos(180° + A)= $$\frac{ON}{OQ}$$= $$\frac{-OM}{OP}$$ = -cos A
tan(180° + A) = $$\frac{NQ}{ON}$$= $$\frac{-MP}{-OM}$$ = tan A
cosec(180° + A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{-MP}$$ = -cosec A
sec(180° +A)= $$\frac{OQ}{ON}$$= $$\frac{OP}{-OM}$$ = -sec A
cot(180° +A)= $$\frac{ON}{NQ}$$= $$\frac{-OM}{-MP}$$ = cot A
Ratios of (270° - A)
Let a revolving line OP start from OX and trace out an angle XOp = A. Let another line OQ = (= OP) revolve from OX to OY' and return to OQ through an angle Y'OQ = A.
Then, $$\angle$$XOQ = 270° - A.
Draw perpendicular PM and QN from P and Q respectively to XOX'.
In $$\triangle$$sONQ and OMP, $$\angle$$ONQ =$$\angle$$OMP,$$\angle$$OQN =$$\angle$$MOP ans OQ = OP.
$$\therefore$$ $$\triangle$$ONQ ≅$$\triangle$$OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = MP, NQ = OM
By the rule of signs, OQ= OP, ON = -MP and NQ = -OM.
Now,
sin(270° - A) = $$\frac{NQ}{OQ}$$= $$\frac{-OM}{OP}$$ = -cos A
cos(270° -A)= $$\frac{ON}{OQ}$$= $$\frac{-MP}{OP}$$ = -sin A
tan(270° - A) = $$\frac{NQ}{ON}$$= $$\frac{-OM}{-MP}$$ = cot A
cosec(270° - A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{-OM}$$ = -sec A
sec(270° -A)= $$\frac{OQ}{ON}$$= $$\frac{OP}{-MP}$$ = -cosec A
cot(270° - A)= $$\frac{ON}{NQ}$$= $$\frac{-MP}{-OM}$$ = tan A
Ratios of (270° + A)
Let s revolving
line OP start from OX and trace out an angle XOP = A. Let another line OQ = (- OP) revolve from OX to OY' and return to OQ through an angle Y'OQ = A.
Then, $$\angle$$XOQ = 270° - A.
Draw perpendicular PM and QN from P and Q respectively to XOX'.
In $$\triangle$$sONQ and OMP, $$\angle$$ONQ =$$\angle$$OMP,$$\angle$$OQN =$$\angle$$MOP ans OQ = OP.
$$\therefore$$ $$\triangle$$ONQ ≅$$\triangle$$OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = MP, NQ = OM
By the rule of signs, OQ= OP, ON = -MP and NQ = -OM.
Now,
sin(270° +A) = $$\frac{NQ}{OQ}$$= $$\frac{-OM}{OP}$$ = -cos A
cos(270° +A)= $$\frac{ON}{OQ}$$= $$\frac{MP}{OP}$$ = sin A
tan(270° +A) = $$\frac{NQ}{ON}$$= $$\frac{-OM}{MP}$$ = -cot A
cosec(270° +A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{-OM}$$ = -sec A
sec(270° + A)= $$\frac{OQ}{ON}$$= $$\frac{OP}{MP}$$ = cosec A
cot(270° +A)= $$\frac{ON}{NQ}$$= $$\frac{MP}{-OM}$$ = -tan A
Ratios of (360° - A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ = OP revolve from OX to OX and return back to OQ through an angle XOQ = A.
Then, $$\angle$$XOQ = 360° - A.
Draw perpendiculars PM and QN from P and Q respectively to OX.
In $$\triangle$$sONQ and OMP, $$\angle$$ONQ = $$\angle$$OMP, $$\angle$$NOQ = $$\angle$$MOP and OQ = OP.
$$\therefore$$ $$\triangle$$ ONQ≅ $$\triangle$$ OMP.
Corresponding sides of congruent triangles are equal.
So, OQ = OP, ON = OM and NQ = MP.
By the rule of signs, OQ = OP, ON = OM and NQ = -MP.
Now,
sin(360° - A) = $$\frac{NQ}{OQ}$$= $$\frac{-MP}{OP}$$ = -sin A
cos(360° - A)= $$\frac{ON}{OQ}$$= $$\frac{OM}{OP}$$ = cos A
tan(360° - A) = $$\frac{NQ}{ON}$$= $$\frac{-MP}{OM}$$ = -tan A
cosec(360° - A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{-MP}$$ = -cosec A
sec(360° - A)= $$\frac{OQ}{ON}$$= $$\frac{OP}{OM}$$ = sec A
cot(360° - A)= $$\frac{ON}{NQ}$$= $$\frac{OM}{-MP}$$ = -cot A
Ratios of negative angle (-A)
Let a revolving line OP start from OX and trace out an angle XOP = A. Let another line OQ = (-OP) revolve from OX to OX and return back to OQ through an angle XOQ = A.
Then $$\angle$$XOQ = -A.
Draw perpendiculars PM and QN from P and Q respectively to OX. Then similarly, as above,
OQ = OP, ON = OM and NQ = -MP
Then,
sin(-A) = $$\frac{NQ}{OQ}$$= $$\frac{-MP}{OP}$$ = -sin A
cos(-A)= $$\frac{ON}{OQ}$$= $$\frac{OM}{OP}$$ = cos A
tan(-A) = $$\frac{NQ}{ON}$$= $$\frac{-MP}{OM}$$ = -tan A
cosec(-A)= $$\frac{OQ}{NQ}$$= $$\frac{OP}{-MP}$$ = -cosec A
sec(-A)= $$\frac{OQ}{ON}$$= $$\frac{OP}{OM}$$ = sec A
cot(-A)= $$\frac{ON}{NQ}$$= $$\frac{OM}{-MP}$$ = -cot A
Values of every trigonometric ratio in different quadrants are:
1st quadrant 2nd quadrant 3rd quadrant 4th quadrant all the ratios are (+e) positive sin and cosec (+ve) remaining (-ve) tan and cot (+ve) remaining (-ve) cos and sec (+ve) remaining (-v)
.
### Very Short Questions
LHS =
= sin 9° × sin 27° = sin(90° - 81°) sin(90° - 63°)
= cos 63° × cos 81° = RHS
$$\therefore$$ LHS = RHS proved.
Soln
LHS = sin1°. sin2°. sin45° .sec88° .sec89° = 1
= sin1°.sin2°.1.sec(90° - 2°).sec(90° - 1°)
= sin1°.sin2°.1.cosec2°.cosec1°
= sin1°.cosec1°.sin2°.cosec2°
= 1×1
= 1 = RHS
$$\therefore$$ LHS = RHS
LHS = sec$$\theta$$.cosec$$\theta$$(90° - $$\theta$$) - tan$$\theta$$.cot(90° - $$\theta$$) = 1
= sec$$\theta$$.sec$$\theta$$ - tan$$\theta$$.tan$$\theta$$
= sec2$$\theta$$ - tan2$$\theta$$
= 1
= 1 = RHS
$$\therefore$$ LHS = RHS proved.
Soln
LHS = $$\frac{cos(90°+θ).sec(-θ).tan(180° - θ)}{sec(360° + θ).sin(180°+θ).cot(90°-θ)}$$ = -1
= $$\frac{cos(90°+θ).sec(-θ).tan(180° - θ)}{sec(4×90° + θ).sin(2×90°+θ).cot(90°-θ)}$$
= $$\frac{-sinθ.secθ(-tanθ)}{secθ(-sinθ).tanθ}$$
= -1 = RHS
$$\therefore$$ LHS = RHS proved.
Soln
LHS = sin70°.cos20° + sin20°.cos70°
= sin70°.cos(90° - 70°) + sin(90° - 70°).cos70°
= sin70°. sin70° + cos70°.cos70°
= sin270° + cos270° = 1
0%
• ### If sin 9° × sin 27° = cos 63° × cos 81° then, what's it identity value?
cos 63° × cos 81°
tan 63° × cos 81°
sin 63° ×tan 81°
sin 63° × cos 81°
2
3
-1
1
-2
-1
1
3
1
-2
-1
2
• ### What is the numerical value of sin70°.cos20° + sin20°.cos70°?
sin270° + cos270° = -1
sin270° + cos270° = 1
sin270° + cos270° = -1
cos270° + tan270° = 1
• ### What is the identical value of sin 240°?
(frac{sqrt{3}}{2})
-(frac{sqrt{3}}{2})
(frac{1}{sqrt{2}})
-(frac{1}{sqrt{2}})
-(sqrt{3})
-(sqrt{2})
(sqrt{2})
(sqrt{3})
sin A
cos A
sec A
tan A
cot A
cosec A
sin A
cos A
cot A
sec A
tan A
cosec A
sec A
cot A
cosA
sin A
cos A
cot A
sin A
cosec A
tan A
sinA
cot A
cos A
tan A
cosec A
sin A
cos A
cos A
sin A
;i:1;s:13:
sin A
;i:1;s:11:
;i:3;s:5:
sec A
;i:2;s:5:
cot A
tan A
;i:1;s:13:
;i:2;s:7:
;i:1;s:5:
;i:3;s:5:
sin A
;i:2;s:5:
;i:3;s:5:
;i:1;s:4:
;i:2;s:5:
;i:3;s:5:
secA
sin A
cos A
cosec A
;i:1;s:11:
cot A
;i:1;s:11:
tan A
tan A
cos A
sin A
cot A
cot A
sin A
cosec A
sec A
cot A
sin A
cos A
tan A
cos A
cosec A
sin A
cot A
cos A
cosc A
sin A
tan A
sin A
tan A
cos A
cosec A
secA
cos A
cosec A
sin A
tan A
sin A
cos A
sec A
sin A
sec A
cosec A
cot A
tan A
sec A
cos A
sin A
cosec A
sec A
tan A
sin A
## DISCUSSIONS ABOUT THIS NOTE
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##### Krischal Dhungel
Find the value of x: cosec(90 a) xcosa.cos(90 a)=sin(90 a)
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##### 9999
prove that: {sin(-a)tan(180 a)sin(180 a)sec(270 a)}/{sin(360-a)cos(270-a)cosec(180-a)cot(360-a)}=-tan^2a
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# Complex Numbers Questions and Answers – Logarithm of Complex Numbers
This set of Complex Analysis Questions and Answers for Campus interviews focuses on “Logarithm of Complex Numbers”.
1. Find the value of log(-6).
a) log6+2iπ
b) log36+iπ
c) log6+2iπ
d) log6+iπ
Explanation: We know that
$$log(x-iy)=\frac{1}{2} log(x^2+y^2)+itan^{-1} (\frac{y}{x})$$
Putting x=-6 and y=0.
$$log(-6)=\frac{1}{2} log(36)+itan^{-1} (\frac{0}{-6})$$
$$log(-6)=log6+iπ$$.
2. Find the value of log2(-3).
a) $$\frac{log_3+i8\pi}{log_2}$$
b) $$\frac{log_3+3i\pi}{log_2}$$
c) $$\frac{log_3+i\pi}{log_2}$$
d) $$\frac{log_2+i\pi}{log_3}$$
Explanation: In this problem, we change the base to e
$$log_2(-3)=\frac{log_e(-3)}{loge(2)}$$
$$log_2(-3)=\frac{log_3+i\pi}{log_2}$$.
3. Represent ii in terms of e.
a) $$e^{\frac{-\pi}{3}}$$
b) $$e^{\frac{-3\pi}{2}}$$
c) $$e^{\frac{-\pi}{2}}$$
d) $$e^{\frac{-\pi}{6}}$$
Explanation: We know that
$$a^x=e^{x loga}$$
$$i^i=e^{i logi}$$
We also know from the definition of logarithm,
$$logi=\frac{i\pi}{2}$$
$$i^i=e^{i(\frac{i\pi}{2})}=e^{\frac{-\pi}{2}}$$.
Sanfoundry Global Education & Learning Series – Complex Analysis.
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# Solve: ${{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1$. $\left( a>0 \right)$.
Last updated date: 20th Jun 2024
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Hint: We have been given a sum of two logarithms. Using the theorem of ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$, we convert it into a single logarithm. Then we use theorem ${{\log }_{x}}a=b\Rightarrow {{x}^{b}}=a$ to find the quadratic equation. We solve the equation to find the solution of the problem.
We have to find the value of x for the equation ${{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1$.
We are going to use some logarithmic formula where ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$.
The base of the logarithm has to be the same.
For our given equation the left-hand side equation can be changed as
\begin{align} & {{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1 \\ & \Rightarrow {{\log }_{e}}\left[ x\left( x-a \right) \right]=1 \\ \end{align}
Also, we have ${{\log }_{x}}a=b\Rightarrow {{x}^{b}}=a$.
Now we apply the theorem on the equation ${{\log }_{e}}\left[ x\left( x-a \right) \right]=1$.
\begin{align} & {{\log }_{e}}\left[ x\left( x-a \right) \right]=1 \\ & \Rightarrow x\left( x-a \right)={{e}^{1}}=e \\ \end{align}
Now we know $\left( a>0 \right)$.
We form the quadratic equation ${{x}^{2}}-ax-e=0$.
We have the formula of quadratic equation solving as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4pc}}{2p}$ for general equation $p{{x}^{2}}+bx+c=0$. For our equation $p=1,b=-a,c=-e$.
Using the form, we get $x=\dfrac{-\left( -a \right)\pm \sqrt{{{\left( -a \right)}^{2}}-4\times 1\times \left( -e \right)}}{2\times 1}=\dfrac{a\pm \sqrt{{{a}^{2}}+4e}}{2}$.
As $\left( a>0 \right)$, we can find that ${{a}^{2}}>0$. This means ${{a}^{2}}+4e>0$. So, the roots of the term $\sqrt{{{a}^{2}}+4e}$ won’t be complex.
But when we are taking $x=\dfrac{a-\sqrt{{{a}^{2}}+4e}}{2}$, the value of $x-a=\dfrac{a-\sqrt{{{a}^{2}}+4e}}{2}-a=\dfrac{-a-\sqrt{{{a}^{2}}+4e}}{2}<0$ becomes negative. We know logarithm can’t have value for less than or equal to 0. So, the root of $x=\dfrac{a-\sqrt{{{a}^{2}}+4e}}{2}$ is not possible.
So, the roots $x=\dfrac{a+\sqrt{{{a}^{2}}+4e}}{2}$ of the equation ${{x}^{2}}-ax-e=0$ is the solution of the function ${{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1$.
Note: Both a and e are constant. We don’t need to break $x=\dfrac{a+\sqrt{{{a}^{2}}+4e}}{2}$ to find the exact answer. Also, we need to always put the value of the root to double check the answer of the equation. As $\sqrt{{{a}^{2}}+4e}>a$, that’s why the value of $x-a=\dfrac{a+\sqrt{{{a}^{2}}+4e}}{2}-a=\dfrac{\sqrt{{{a}^{2}}+4e}-a}{2}>0$ is positive. |
# Lesson 3Building Strong RootsSolidify Understanding
## Jump Start
Which One Doesn’t Belong?
Determine how each of the five equations differs from the others and be prepared to justify your answer with sound mathematical reasoning.
A.
B.
C.
D.
E.
## Learning Focus
Find roots and factors of quadratic and cubic functions.
Write quadratic and cubic equations in factored form.
Identify multiple roots of quadratic and cubic functions.
Do all polynomial functions of degree have roots?
## Open Up the Math: Launch, Explore, Discuss
When working with quadratic functions, we learned the Fundamental Theorem of Algebra:
An degree polynomial function has roots.
In this lesson, we will be exploring this idea further with other polynomial functions.
First, let’s brush up on what we learned about quadratics. The equations and graphs of four different quadratic equations are given below. Find the roots for each and identify whether the roots are real or complex/imaginary.
Roots:
Type of roots:
Roots:
Type of roots:
Roots:
Type of roots:
Roots:
Type of roots:
### 2.
Did all of the quadratic functions have roots, as predicted by the Fundamental Theorem of Algebra? Explain.
### 3.
It’s always important to keep what you’ve previously learned in your mathematical bag of tricks so that you can pull it out when you need it. What strategies did you use to find the roots of the quadratic equations?
Pause and Reflect
### 4.
Using your work from problem 1, write each of the quadratic equations in factored form. When you finish, check your answers by graphing, when possible, and make any corrections necessary.
Factored form:
Factored form:
Factored form:
Factored form:
### 5.
Based on your work in problem 1, would you say that roots are the same as -intercepts? Explain.
### 6.
Based on your work in problem 4, what is the relationship between roots and factors?
Now let’s take a closer look at cubic functions. We’ve worked with transformations of , but what we’ve seen so far is just the tip of the iceberg. For instance, consider:
### 7.
Use the graph to find the roots of the cubic function. Use the equation to verify that you are correct. Show how you have verified each root.
### 8.
Write in factored form. Verify that the factored form is equivalent to the standard form.
### 9.
Are the results you found in problem 7 consistent with the Fundamental Theorem of Algebra? Explain.
Here’s another example of a cubic function.
### 10.
Use the graph to find the roots of the cubic function.
### 11.
Write in factored form. Verify that the factored form is equivalent to the standard form. Make any corrections needed.
### 12.
Are the results you found in problem 10 consistent with the Fundamental Theorem of Algebra? Explain.
### 13.
We’ve seen the most basic cubic polynomial function, , and we know its graph looks like this:
Explain how is consistent with the Fundamental Theorem of Algebra.
### 14.
Here is one more cubic polynomial function for your consideration. You will notice that it is given to you in factored form. Use the equation and the graph to find the roots of .
### 15.
Use the equation to verify each root. Show your work.
### 16.
Are the results you found in problem 14 consistent with the Fundamental Theorem of Algebra? Explain.
### 17.
Explain how to find the factored form of a polynomial, given the roots.
### 18.
Explain how to find the roots of a polynomial, given the factored form.
Here’s a challenge: Find a cubic function in standard form with real coefficients that has three complex/imaginary roots.
## Takeaways
Verify a root:
Roots and -intercepts:
Finding factored form of a polynomial when the roots are known to be , , :
Multiple roots, or roots of multiplicity :
Finding roots of a polynomial in factored form:
## Lesson Summary
In this lesson, we found roots of cubic functions using the same methods we learned for quadratic functions. We found that cubic functions can have multiple roots, like quadratic functions. We learned to verify roots and write equivalent equations in factored and standard form. During the lesson, we applied the Fundamental Theorem of Algebra to cubic functions to consider the number and types of possible roots.
## Retrieval
Divide:
### 2.
Use the quadratic formula to find the zeros. |
Prime and Composite Numbers | Definition, Facts, and Examples
1. Math Lessons >
2. Prime and Composite Numbers
In this lesson, we'll take a look at what prime numbers and composite numbers are and will explain how to identify them. Let’s first figure out what prime numbers and composite numbers mean.
What Are Prime and Composite Numbers?
A composite number is a number that can be divided by a whole number other than 1 and itself. A prime number, on the other hand, can’t be divided by any number other than 1 and itself.
Identifying If a Number Is Prime or Composite by Sharing
Imagine you have 6 donuts. How many different ways can you share them in?
We can make 2 groups with 3 donuts in each,
Or 3 groups of 2 donuts in each,
Or 6 groups with 1 donut in each.
How would you share 9 donuts?
We can, obviously, group them into 9 groups of 1 each. And also divide the donuts evenly in groups of 3 donuts each.
How would you share if there were 7 donuts?
The only way of sharing the 7 donuts would be by putting them into 7 groups of 1 each.
We can’t divide the 7 donuts into equal groups without having any left over.
The numbers 6 and 9 can be divided by other whole numbers, such as 2 and 3; but 7 can be divided by only 1 or itself.
So we understand that 6 and 9 are composite numbers, and 7 is a prime number.
Identifying If a Number Is Prime or Composite Using Its Factors
Another method of quickly checking whether a number is prime or composite is by using its factors. If you’re wondering what factors are, they are the numbers that you multiply to get another number.
A number that has only two factors, 1 and the number itself, is called a prime number.
A number that has more than two factors is called a composite number.
Let’s take a look at the first few prime numbers.
1 = 1 x 1. 1 is the only factor of 1. So, it is neither prime nor composite.
2 = 1 x 2. 2 has exactly 2 factors: 1 and 2. So, it is a prime number.
3 = 1 x 3. 3 also has exactly 2 factors: 1 and 3. So, it is a prime number.
4 = 1 x 4; 4 = 2 x 2. 4 has 3 factors: 1, 2, and 4. So, 4 is a composite number.
5 = 1 x 5. 5 has just 2 factors: 1 and 5. So, 5 is a prime number.
6 = 1 x 6; 6 = 2 x 3. Thus, 6 has 4 factors: 1, 2, 3, and 6. So, 6 is a composite number.
So with small numbers, your job of telling whether a number is prime or composite is super-easy. The quickest way of finding the factors of larger numbers, though, is by using the divisibility rules.
Here are a few examples.
Example 1
Is the number 267 prime or composite?
Listing the factors of this number would be time-consuming. So, we’ll bring the divisibility rules into play.
267 is not an even number, so it’s not divisible by 2.
Now, the divisibility test for 3 is to check if the sum of the digits is divisible by 3.
The sum of the digits of 267 is:
2 + 6 + 7 = 15
Now, 15(= 3 x 5) is a multiple of 3. So, 267 is divisible by 3.
Thus, 267 is composite.
Example 2
Is 37 a prime or composite number?
Let’s perform the divisibility tests one by one.
37 is not an even number, so it’s not divisible by 2. Subsequently, 37 is not divisible by 4, 6, or 8.
Next, the sum of the digits of 37 is: 3 + 7 = 10. 10 is not divisible by 3. So, 37 is not divisible by 3.
Now, the last digit of 37 is neither 0 nor 5, so it’s not divisible by 5 as well.
As 37 cannot have any factor greater than 6, we can stop with 6.
Thus, 37 is a prime number.
Finding Prime Numbers Using Sieve of Eratosthenes
An interesting way of finding prime numbers is by using the Sieve of Eratosthenes. This method works on the idea of sifting the composite numbers to identify the prime numbers. Let's pass the numbers up to 50 through the sieve now!
Filter out the composite numbers with a click on the chart!
Go ahead and extend this method to any number and scout out the prime and composite numbers in no time!
Lists of Prime and Composite Numbers from 1 to 100
Prime Numbers
2, 3, 5, 7,
11, 13, 17, 19,
23, 29,
31, 37,
41, 43, 47
53, 59,
61, 67,
71, 73, 79,
83, 89
97
Composite Numbers
4, 6, 8, 9, 10,
12, 14, 15, 16, 18, 20,
21, 22, 24, 25, 26, 27, 28, 30,
32, 33, 34, 35, 36, 38, 39, 40,
42, 44, 45, 46, 48, 49, 50
51, 52, 54, 55, 56, 57, 58, 60,
62, 63, 64, 65, 66, 68, 69, 70,
72, 74, 75, 76, 77, 78, 80
81, 82, 84, 85, 86, 87, 88, 90
91, 92, 93, 94, 95, 96, 98, 99, 100
Click on the circles to spot the six prime numbers
33
71
139
15
167
58
61
23
13
35
Reset
• The largest prime number found so far has 24,862,048 digits!
• Thanks to Euclid, who was instrumental in theorizing the idea, every composite number can be written as a product of prime numbers. Check it out for yourself!
• There’s an organization dedicated to research on prime numbers - Great Internet Mersenne Prime Search (GIMPS).
A Quick Recap
• A prime number has exactly two factors: 1 and the number itself.
• Composite numbers have more than two factors.
• The numbers 0 and 1 are neither prime nor composite.
• Prime numbers are also called non-composite numbers.
• We can identify whether a number is prime and composite by checking if the number can be shared into equal groups, or listing all its factors, or applying the divisibility rules, or using the sieve of Eratosthenes.
Polish up your skills with our free printable Prime and Composite Numbers worksheets!
Progress
Score
Restart Quiz |
# Prove that 1/a(a+1) = 1/a - 1/(a+1)
Asked on by kikiri
### 2 Answers |Add Yours
neela | High School Teacher | (Level 3) Valedictorian
Posted on
To prove 1/a(a+1) = 1/a -1/(a+1).
We start from the right side and show it is equal to the expression on the left.
RHS = 1/a -1/(a+1) .
The LCM of the denominators is a(a+1).
Therefore we convert each of the fractions 1/a and 1/(a+1) into equivalent fraction with LCM as the denominator.
So the first term 1/a = 1*(a+1)/a*(a+1) = (a+1)/a(a+1).
The second term 1/(a+1) = 1*a/(a+1)*a = a/(a(a+1).
Therefore 1/a -1/(a+1) = (a+1 -a)/a(a+1) = 1/a(a+1) = LHS.
Thus we proved 1/a -1/(a+1) = 1/a(a+1). Or 1/a(a+1) = 1/a -1/(a+1).
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We notice that the denominator of the left side ratio is the least common denominator of 2 irreducible ratios.
We'll suppose that the ratio 1/a(a+1) is the result of addition or subtraction of 2 elementary fractions:
1/a(a+1) = A/a + B/(a+1) (1)
We'll multiply the ratio A/a by (a+1) and we'll multiply the ratio B/(a+1) by a.
1/a(a+1) = [A(a+1) + Ba]/a(a+1)
Since the denominators of both sides are matching, we'll write the numerators, only.
1 = A(a+1) + Ba
We'll remove the brackets:
1 = Aa + A + Ba
We'll factorize by a to the right side:
1 = a(A+B) + A
If the expressions from both sides are equivalent, the correspondent coefficients are equal.
A+B = 0
A = 1
1 + B = 0
B = -1
We'll substitute A and B into the expression (1):
1/a(a+1) = 1/a - 1/(a+1)
We remark that we've obtained the request from enunciation.
The identity has been proved.
We’ve answered 317,631 questions. We can answer yours, too. |
# Identify missing fractions on a number line—Let's Practise!
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## Basics on the topicIdentify missing fractions on a number line—Let's Practise!
In this video, learn with Razzi about how to write fractions on a number line!
### TranscriptIdentify missing fractions on a number line—Let's Practise!
Razzi says get these items ready because today we're going to identify missing fractions on a number line. It's time to begin! Identify the missing fraction on the number line. Pause the video to work on the problem and press play when you are ready to see the solution! First, find the interval by identifying the pattern between the given fractions. Then, count on using the pattern starting with the first fraction. Did you also get two-fifths? Let's keep practising! Identify the missing fraction on the number line. Pause the video to work on the problem and press play when you are ready to see the solution! First, find the interval by identifying the pattern between the given fractions. Then, count backwards from seven-twelfths. Did you also get four-twelfths? Let's do one more! This time, identify all the missing fractions on the number line. Pause the video to work on the problem and press play when you are ready to see the solution! There are seven parts of one whole, so the denominator is seven then starting with one-seventh, fill in the missing numerators. Did you also get one-seventh, two-sevenths, three-sevenths, four-sevenths, five-sevenths, six-sevenths and seven-sevenths? Razzi had so much fun practising with you today! See you next time!
## Identify missing fractions on a number line—Let's Practise! exercise
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Identify missing fractions on a number line—Let's Practise!.
• ### Razzi's missing fraction.
Hints
Find the interval by identifying the pattern between the given fractions.
Count on using the pattern starting with the first fraction.
Solution
The missing fraction on the number line is $\frac{3}{6}$.
• ### More missing fractions.
Hints
How many parts of one whole are on the number line? That is your denominator.
$\frac{1}{9}$ belongs after 0. $\frac{2}{9}$ is next. Use this pattern to determine the missing fraction.
Solution
$\frac{4}{9}$ is the missing fraction.
• ### Can you complete the number line?
Hints
How many parts of one whole are on the number line? That is your denominator.
Continue counting along the number line to find the missing fractions.
Solution
The image shows the correct fractions on the number line.
• ### Find the missing fractions.
Hints
Use the pattern to find the missing fractions.
Count back from $\frac{9}{14}$ to find the missing fractions.
Solution
The missing fractions are $\frac{1}{14}$ and $\frac{5}{14}$.
• ### What is the missing fraction?
Hints
Look at the denominator of the fractions given. The missing fraction will have the same denominator.
Which fraction comes after $\frac{2}{4}$?
Which fraction comes before $\frac{4}{4}$?
Solution
The missing fraction is $\frac{3}{4}$.
• ### Measure flour to help Razzi bake biscuits.
Hints
What is the denominator?
Begin at $\frac{1}{5}$ on the number line. Add $\frac{3}{5}$ more.
Count on three jumps forwards from $\frac{1}{5}$.
Solution
• Begin at $\frac{1}{5}$ and jump three spaces forwards to add $\frac{3}{5}$.
• You land at $\frac{4}{5}$ .
• Razzi needs $\mathbf{\frac{4}{5}}$ cup of flour. |
4.12
## Precalculus
Skip to 0 minutes and 11 secondsLet's look for radical solutions, or rather solutions to equations that contain a radical. These arise rather often in applications. Here's our general equation, the nth root of f equals g. Now, we're tempted to identify the domain, because we'll see that it has some interest, and the natural domain here would be one that includes a restriction on f being positive. Why? Because the nth root is only defined for a positive number. We're also tempted to take the nth power on each side of our original equation, because that will get rid of the radical and possibly make it easier to deal with.
Skip to 0 minutes and 56 secondsNow, bear in mind that when you have positive numbers, then a and b, a and b are equal, if and only if their nth roots are equal. So that's fine, but you must remember that this is for positive numbers. So when you take the original equation, and you raise each side to the nth power, you get f equals g to the n, but in doing so, you have to remember to take account of the condition, g of x greater than or equal 0, in order to have fully equivalence of the left hand side and the right hand side.
Skip to 1 minute and 28 secondsNow remember, equivalence means if x satisfies the equation on the left, then it satisfies both conditions on the right, and vice versa, if x satisfies both conditions on the right, then it satisfies the left hand side. A common error is to assert that nth root of f equals g is the same as f equals g to the n. This is simply not true. When n is even, the new equation that you have generated will usually have more solutions, solutions for which g of x is negative. Let's look at an example to clarify this. We want to solve the equation root x plus 3 equals x plus 1. The domain of this equation?
Skip to 2 minutes and 13 secondsWell, we might say that it's the interval minus 3 to infinity, because we want x plus 3 to be positive, since it's under the square root sign. Now, equivalence would mean that we could square both sides, and we do this here provided we bear in mind the condition, as I just said, that x plus 1 will have to be positive. You'll notice that this introduces a further restriction on x. If we were very insightful, we might have included that further restriction right away in the initial definition of the domain. But it doesn't matter, we note it, and we pursue the study.
Skip to 2 minutes and 53 secondsWe look at the new equation, we work on its terms a little bit, and we see that it's a quadratic equation, which we can easily solve, the roots are 1 and minus 2. Now can we stop at this point? No, we really need to check the compatibility of these candidates with the domain conditions that we have. Now, these compatibility conditions are going to lead us to reject x equals minus 2. Why? Because minus 2 does not satisfy x greater or equal to minus 1. On the other hand, the other candidate, the 1, that's fine, it satisfies all the compatibility conditions. We check it out in the original equation. Naturally, we see that it works, and we have our answer.
Skip to 3 minutes and 42 secondsThe solution set is the singleton set 1. Now step back and look at how we solved this problem. We applied a sort of surgical approach. We dissected the equation, we cut away the irrelevant parts of the domain, and we were sensitive to the domain throughout. It turns out that there's another way to attack such problems. It's the insensitive approach, you might say, I call it the cowboy approach. We're going to ignore the domains as we work. We're going to go fast. We're not going to check anything. We're going to get some candidates, and then we'll simply come back and see if they work or not. Let's solve an example in this cowboy mode. Here's an equation we wish to solve.
Skip to 4 minutes and 27 secondsHow do we begin? Well, the first step here would be to isolate the radical on one side of the equality by itself. This will simplify the resulting arithmetic. Now we square both sides. There we go. We don't even think about the domains or equivalents. We massaged the equation to make it look simpler. We come across a quadratic equation. It factors, therefore, the roots are minus 3 and 6. And so we have this conclusion, which is what? Well, it's a necessary condition only. What it is, is the following conclusion. If x is a solution of the original equation, then it is necessary that x be 6 or minus 3, but it doesn't work the other way. It's not sufficient.
Skip to 5 minutes and 12 secondsWe need to go back and see whether they are solutions. It's quite possible that one or both are not, so we check. We plug in x equals 6 into the original equation. Does it work? Nope, doesn't work. Well, we reject 6. We put x equals minus 3 into the equation. Does that work? Well, yes it works, therefore, we keep x equals minus 3. Our solution set is the singleton set minus 3. Now that was the cowboy approach. You see how fast it was? It can be faster, and it involves less thinking. But it is true that the domain sensitive approach, the surgical approach, can also save time.
Skip to 5 minutes and 58 secondsIn fact, in mathematics, it's often the case that a little thinking at the beginning can save quite a bit of work later on. Consider this example. If we worry about the domain, we see that x should be greater than 1/2 for the square root to be defined. On the other hand, the square root of something is positive, so minus x is going to be positive, and these two conditions, we immediately see, are contradictory, so there are no solutions of this equation. So it would be unnecessary to go ahead and square and come up with a false possibility that we would then eliminate. Now, the question arises, is it better to be a surgeon or a cowboy?
Skip to 6 minutes and 38 secondsWe need not resolve this question immediately, but it will arise again pretty soon. Here's another example of solving equations. Comes up a lot in physics, engineering, chemistry, economics, and so forth. You have a formula in which parameters appear, and you wish to interchange parameters in the formula. Let's take an example. In mechanics, when you're looking at the oscillations of a pendulum in the plane, there is a formula that expresses the period of the oscillation-- that's the time it takes to make one full swing back and forth-- the period capital T, as a function of the length of the pendulum.
Skip to 7 minutes and 21 secondsAnd the formula then is T equals 2 pi square root of L over g, where g is a certain positive constant. Now, in this formula, we're thinking of L as an independent variable, which, once you have its value, determines the value of T, the period through the formula. However, it's possible to think of things in the opposite way. We could imagine that you choose the period first, and then you have to find the length that will give you that period. In other words, we'd like to express L as a dependent variable of the independent variable T. You just solve the equation for L. Here's what we do.
Skip to 8 minutes and 2 secondsFirst of all, I've switched the sides around, because I'm going to be writing at the end L equals, and I like my dependent variable to be on the left hand side of the equation. Now, all the quantities are known to be strictly positive, L, T, g, pi, and so forth. So we don't have to worry too much about equivalence, and whether we're getting extraneous solutions. We just go ahead and square both sides. And now we multiply by g and divide by 4 pi squared, and there is a formula for L. So, you know, when engineers do this kind of thing, they're not being Cowboys. Well, not usually. |
## Chapter 9 - Combinational Logic Functions
The half-adder is extremely useful until you want to add more than one binary digit quantities. The slow way to develop a two binary digit adders would be to make a truth table and reduce it. Then when you decide to make a three binary digit adder, do it again. Then when you decide to make a four digit adder, do it again. Then when ... The circuits would be fast, but development time would be slow.
Looking at a two binary digit sum shows what we need to extend addition to multiple binary digits.
11
11
11
---
110
Look at how many inputs the middle column uses. Our adder needs three inputs; a, b, and the carry from the previous sum, and we can use our two-input adder to build a three input adder.
Σ is the easy part. Normal arithmetic tells us that if Σ = a + b + Cin and Σ1 = a + b, then Σ = Σ1 + Cin.
What do we do with C1 and C2? Let’s look at three input sums and quickly calculate:
Cin + a + b = ?
0 + 0 + 0 = 0 0 + 0 + 1 = 1 0 + 1 + 0 = 1 0 + 1 + 1 = 10
1 + 0 + 0 = 1 1 + 0 + 1 = 10 1 + 1 + 0 = 10 1 + 1 + 1 = 11
If you have any concern about the low order bit, please confirm that the circuit and ladder calculate it correctly.
In order to calculate the high order bit, notice that it is 1 in both cases when a + b produces a C1. Also, the high order bit is 1 when a + b produces a Σ1 and Cin is a 1. So We will have a carry when C1 OR (Σ1 AND Cin). Our complete three input adder is:
For some designs, being able to eliminate one or more types of gates can be important, and you can replace the final OR gate with an XOR gate without changing the results.
A0 is the low order bit of A, A1 is the high order bit of A, B0 is the low order bit of B, B1 is the high order bit of B, Σ0is the low order bit of the sum, Σ1 is the high order bit of the sum, and Cout is the Carry.
A two binary digit adder would never be made this way. Instead the lowest order bits would also go through a full adder.
There are several reasons for this, one being that we can then allow a circuit to determine whether the lowest order carry should be included in the sum. This allows for the chaining of even larger sums. Consider two different ways to look at a four bit sum.
111 1<-+ 11<+-
0110 | 01 | 10
1011 | 10 | 11
----- - | ---- | ---
10001 1 +-100 +-101
If we allow the program to add a two bit number and remember the carry for later, then use that carry in the next sum the program can add any number of bits the user wants even though we have only provided a two-bit adder. Small PLCs can also be chained together for larger numbers.
These full adders can also can be expanded to any number of bits space allows. As an example, here’s how to do an 8 bit adder.
This is the same result as using the two 2-bit adders to make a 4-bit adder and then using two 4-bit adders to make an 8-bit adder or re-duplicating ladder logic and updating the numbers.
For any large combinational circuit there are generally two approaches to design: you can take simpler circuits and replicate them; or you can design the complex circuit as a complete device. Using simpler circuits to build complex circuits allows a you to spend less time designing but then requires more time for signals to propagate through the transistors.
The 8-bit adder design above has to wait for all the Cxout signals to move from A0 + B0 up to the inputs of Σ7. If a designer builds an 8-bit adder as a complete device simplified to a sum of products, then each signal just travels through one NOT gate, one AND gate and one OR gate.
A seventeen input device has a truth table with 131,072 entries, and reducing 131,072 entries to a sum of products will take some time. When designing for systems that have a maximum allowed response time to provide the final result, you can begin by using simpler circuits and then attempt to replace portions of the circuit that are too slow.
That way you spend most of your time on the portions of a circuit that matter.
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# Difference between revisions of "2005 AIME II Problems/Problem 11"
## Problem
Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$
## Solution 1
For $0 < k < m$, we have
$a_{k}a_{k+1} = a_{k-1}a_{k} - 3$.
Thus the product $a_{k}a_{k+1}$ is a monovariant: it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$, so when $k = \frac{37 \cdot 72}{3} = 888$, $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$, our answer.
Note: In order for $a_{m} = 0$ we need $a_{m-1}a_{m}=3$ simply by the recursion definition.
## Solution 2
Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$. Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$, we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$. Setting the RHS of this equation equal to $3$, we find that $m$ must be $\boxed{889}$.
~ anellipticcurveoverq
## Video solution
2005 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions |
# Permutations and Combinations - Formulas, Problems and Video Tutorials
Permutations and Combinations Solved Examples - Page 3
Permutations and Combinations Important Questions - Page 4
Permutations and Combinations Video Lecture - Page 5
Permutations and Combinations are arrangements or selections of objects out of groups. For example, selecting a team of 11 players out of 20 players.
## Permutations
The arrangements made by taking some or all elements out of a group in a particular manner are called permutations.
For example, in how many way can the letters word ENGLISH be arranged, so that vowels never come together?
The number of permutation of n thing taking r at a time is denoted by $^{n} P_{r}$ and it is defined under :
$^{n} P_{r}\: =\: \frac{n!}{(n-r)!}\: ;$ r ≤ n.
Important notations
Product of first n positive integers is called n factorial.
n! = 1, 2, 3, 4, 5…….n
n! = (n-1)! N
A special case 0 ! = 1
Various Types of Permutations
Case 1
When in a permutation of n thing taken r at a time, a particular thing always occurs.
The required numbers of permutations = $r(^{n-1}P_{r-1})$
Case 2 The number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement is $^{n-1}P_{r}$
Case 3 (Permutation of like things) The number of n things taken all at a time, given that $p_{1}$ things are $1^{st}$ alike, $p_{2}$ things are $2^{nd}$ alike, and $p_{r}$ things are $r^{th}$ alike is
$\frac{n!}{p_{1}!p_{2}!......p_{r}!}$
Case 4 (Permutation with repetitions) The number of n different things taken r at a time when each may be repeated any number of times in each arrangements is n.
Case 5 (Circular permutations) Circular permutations are the permutations of things along the circumstance of a circle. We have to consider the relative position of the different things in a circular arrangement.
For example, if there are five letters P, Q, R, S and T, two of the arrangements would be PQRST, TPQRS. These two arrangements are obviously different if the things are to be placed in a straight line. But if the arrangements are written along the circumstance of a circle, then the two arrangements PQRST and TPQRS are one and the same.
As the number of circular permutations depends on the relative position of the objects, we fix the position of one object and then arrange the remaining (n-1)! Ways. Thus the circular arrangements of five letters P, Q, R, S, T will be
(5-1)! = 4! = 4.3.2.1= 24 ways.
Some important result of permutations
1. $np_{n-1}\:=\: np_{n}$
2. $np_{n}\:=\: n$ !
3. $np_{r}\:=\:n\left (n-1 p_{r-1} \right )$
4. $np_{r}\:=\:\left ( n-r+1 \right )\times\: n p_{r-1}$
5. $np_{r}\:=\:n-1p_{r}+ r \left(n-1p _{r-1} \right )$ |
# Common Core: 4th Grade Math : Solving for Area
## Example Questions
### Example Question #111 : How To Find The Area Of A Rectangle
Annie has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #112 : How To Find The Area Of A Rectangle
Annie has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #113 : How To Find The Area Of A Rectangle
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #281 : Measurement & Data
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #282 : Measurement & Data
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #283 : Measurement & Data
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #284 : Measurement & Data
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #285 : Measurement & Data
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #286 : Measurement & Data
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper?
Explanation:
This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula
### Example Question #287 : Measurement & Data
Joe has a piece of wallpaper that is by . How much of a wall can be covered by this piece of wallpaper? |
Fsc Part 2 Mathematics (Complete Solutions)
# Q2 Find and Simplify
Find
f(a + h) - f(a) h
and simplify where,
(i) f(x) = 6x - 9 (ii) f(x) = sin x (iii) f(x) = x3 + 2x2 - 1 (iv) f(x) = cos x
Solutions
i) f(x) = 6x - 9
We have to find
f(a + h) - f(a) h
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = 6x - 9
f(a + h) = 6(a + h) - 9
f(a + h) = 6a + 6h - 9
f(a) = ?
as f(x) = 6x - 9
f(a) = 6(a) - 9
f(a) = 6a - 9
Put the value of f(a + h) and f(a) in the equation (i)
=
(6a + 6h - 9) - (6a - 9) h
=
6a + 6h - 9 - 6a - 9 h
=
6h h
= 6
ii) f(x) = sin x
We have to find
f(a + h) - f(a) h
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = sin x
f(a + h) = sin(a + h)
f(a) = ?
as f(x) = sin x
f(a) = sin a
Put the value of f(a + h) and f(a) in the equation (i)
=
sin(a + h) - sin a h
=
1 h
[ sin(a + h) - sin a ]
=
1 h
[
2 cos (
a + h + a 2
) sin (
a + h - a 2
)
] ∴ [sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 )]
=
2 h
[
cos (
2a + h 2
) sin (
h 2
)
]
=
2 h
[
cos (
2a 2
+
h 2
) sin (
h 2
)
]
=
2 h
cos ( a +
h 2
) sin (
h 2
)
iii) f(x) = x3 + 2x2 - 1
We have to find
f(a + h) - f(a) h
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = x3 + 2x2 - 1
f(a + h) = (a + h)3 + 2(a + h)2 - 1
f(a) = ?
as f(x) = x3 + 2x2 - 1
f(a) = a3 + 2a2 - 1
Put the value of f(a + h) and f(a) in the equation (i)
=
[(a + h)3 + 2(a + h)2 - 1] - (a3 + 2a2 - 1) h
=
[(a + h)3 + 2(a + h)2 - 1] - a3 - 2a2 + 1 h
=
[a3 + h3 + 3ah(a + h) + 2(a2 + h2 + 2ah) - 1] - a3 - 2a2 + 1 h
=
a3 + h3 + 3a2h + 3ah2 + 2a2 + 2h2 + 4ah - 1 - a3 - 2a2 + 1 h
=
h3 + 3a2h + 3ah2 + 2h2 + 4ah h
=
h h
[ h2 + 3a2 + 3ah + 2h + 4a ]
= h2 + 3a2 + 3ah + 2h + 4a
= h2 + 3ah + 2h + 3a2 + 4a
= h2 + (3a + 2)h + 3a2 + 4a
iv) f(x) = cos x
We have to find
f(a + h) - f(a) h
= ? ---------------------------------- (i)
f(a + h) = ?
as f(x) = cos x
f(a + h) = cos(a + h)
f(a) = ?
as f(x) = cos x
f(a) = cos a
Put the value of f(a + h) and f(a) in the equation (i)
=
cos(a + h) - cos a h
=
1 h
[ cos(a + h) - cos a ]
=
1 h
[
-2 sin (
a + h + a 2
) sin (
a + h - a 2
)
] ∴ [cos x - cos y = -2 sin( (x - y)/2 ) sin( (x + y)/2 )]
=
-2 h
[
sin (
2a + h 2
) sin (
h 2
)
]
=
-2 h
[
sin (
2a 2
+
h 2
) sin (
h 2
)
]
=
-2 h
sin ( a +
h 2
) sin (
h 2
) |
Right triangle
Right triangle
A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90-degree angle). The relation between the sides and angles of a right triangle is the basis for trigonometry.
The side opposite the right angle is called the hypotenuse (side c in the figure). The sides adjacent to the right angle are called legs (or catheti, singular: cathetus). Side a may be identified as the side adjacent to angle B and opposed to (or opposite) angle A, while side b is the side adjacent to angle A and opposed to angle B.
If the lengths of all three sides of a right triangle are integers, the triangle is said to be a Pythagorean triangle and its side lengths are collectively known as a Pythagorean triple.
Principal properties
Area
As with any triangle, the area is equal to one half the base multiplied by the corresponding height. In a right triangle, if one leg is taken as the base then the other is height, so the area of a right triangle is one half the product of the two legs. As a formula the area T is
$T=\tfrac{1}{2}ab$
where a and b are the legs of the triangle.
If the incircle is tangent to the hypotenuse AB at point P, then denoting the semi-perimeter (a + b + c) / 2 as s, we have PA = sa and PB = sb, and the area is given by
$T=\text{PA} \cdot \text{PB} = (s-a)(s-b).$
This formula only applies to right triangles.[1]
Altitudes
Altitude of a right triangle
If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangles which are both similar to the original and therefore similar to each other. From this:
• The altitude to the hypotenuse is the geometric mean (mean proportional) of the two segments of the hypotenuse.[2]:243
• Each leg of the triangle is the mean proportional of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.
In equations,
$\displaystyle f^2=de,$ (this is sometimes known as the right triangle altitude theorem)
$\displaystyle b^2=ce,$
$\displaystyle a^2=cd$
where a, b, c, d, e, f are as shown in the diagram.[3] Thus
$f=\frac{ab}{c}.$
Moreover, the altitude to the hypotenuse is related to the legs of the right triangle by[4][5]
$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{f^2}.$
For solutions of this equation in integer values of a, b, f, and c, see here.
The altitude from either leg coincides with the other leg. Since these intersect at the right-angled vertex, the right triangle's orthocenter—the intersection of its three altitudes—coincides with the right-angled vertex.
Pythagorean theorem
The Pythagorean theorem states that:
In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
This can be stated in equation form as
$\displaystyle a^2+b^2=c^2$
where c is the length of the hypotenuse, and a and b are the lengths of the remaining two sides.
Pythagorean triples are integer values of a, b, c satisfying this equation.
Illustration of the Pythagorean Theorem
The radius of the incircle of a right triangle with legs a and b and hypotenuse c is
$r = \frac{a+b-c}{2} = \frac{ab}{a+b+c}.$
The radius of the circumcircle is half the length of the hypotenuse,
$R = \frac{c}{2}.$
Thus the sum of the circumradius and the inradius is half the sum of the legs:[6]
$R+r = \frac{a+b}{2}.$
One of the legs can be expressed in terms of the inradius and the other leg as
$\displaystyle a=\frac{2r(b-r)}{b-2r}.$
Characterizations
A triangle ABC with sides $a \le b < c$, semiperimeter s, area T, altitude h opposite the longest side, circumradius R, inradius r, exradii ra, rb, rc (tangent to a, b, c respectively), and medians ma, mb, mc is a right triangle if and only if any one of the statements in the following six categories is true. All of them are of course also properties of a right triangle, since characterizations are equivalences.
Sides and semiperimeter
• $\displaystyle a^2+b^2=c^2\quad (\text{Pythagorean theorem})$
• $\displaystyle (s-a)(s-b)=s(s-c)$
• $\displaystyle s=2R+r.$[7]
• $\displaystyle a^2+b^2+c^2=8R^2.$[8]
Angles
• A and B are complementary.[9]
• $\displaystyle \cos{A}\cos{B}\cos{C}=0.$[8][10]
• $\displaystyle \sin^2{A}+\sin^2{B}+\sin^2{C}=2.$[8][10]
• $\displaystyle \cos^2{A}+\cos^2{B}+\cos^2{C}=1.$[10]
• $\displaystyle \sin{2A}=\sin{2B}=2\sin{A}\sin{B}.$
Area
• $\displaystyle T=\frac{ab}{2}$
• $\displaystyle T=r_ar_b=rr_c$
• $\displaystyle T=r(2R+r)$
• $T=PA\cdot PB,$ where P is the tangency point of the incircle at the longest side AB.[11]
• $\displaystyle r=s-c=(a+b-c)/2$
• $\displaystyle r_a=s-b=(a-b+c)/2$
• $\displaystyle r_b=s-a=(-a+b+c)/2$
• $\displaystyle r_c=s=(a+b+c)/2$
• $\displaystyle r_a+r_b+r_c+r=a+b+c$
• $\displaystyle r_a^2+r_b^2+r_c^2+r^2=a^2+b^2+c^2$
• $\displaystyle r=\frac{r_ar_b}{r_c}$
Trigonometric ratios
The trigonometric functions for acute angles can be defined as ratios of the sides of a right triangle. For a given angle, a right triangle may be constructed with this angle, and the sides labeled opposite, adjacent and hypotenuse with reference to this angle according to the definitions above. These ratios of the sides do not depend on the particular right triangle chosen, but only on the given angle, since all triangles constructed this way are similar. If, for a given angle α, the opposite side, adjacent side and hypotenuse are labeled O, A and H respectively, then the trigonometric functions are
$\sin\alpha =\frac {O}{H},\,\cos\alpha =\frac {A}{H},\,\tan\alpha =\frac {O}{A},\,\sec\alpha =\frac {H}{A},\,\cot\alpha =\frac {A}{O},\,\csc\alpha =\frac {H}{O}.$
For the expression of hyperbolic functions as ratio of the sides of a right triangle, see the hyperbolic triangle of a hyperbolic sector.
Special right triangles
The values of the trigonometric functions can be evaluated exactly for certain angles using right triangles with special angles. These include the 30-60-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/6, and the 45-45-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/4.
Thales' theorem
Median of a right angle of a triangle
Thales' theorem states that if A is any point of the circle with diameter BC (except B or C themselves) ABC is a right triangle where A is the right angle. The converse states that if a right triangle is inscribed in a circle then the hypotenuse will be a diameter of the circle. A corollary is that the length of the hypotenuse is twice the distance from the right angle vertex to the midpoint of the hypotenuse. Also, the center of the circle that circumscribes a right triangle is the midpoint of the hypotenuse and its radius is one half the length of the hypotenuse.
Medians
The following formulas hold for the medians of a right triangle:
$m_a^2 + m_b^2 = 5m_c^2 = \frac{5}{4}c^2.$
The median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles, because the median equals one-half the hypotenuse.
Relation to various means and the golden ratio
Let H, G, and A be the harmonic mean, the geometric mean, and the arithmetic mean of two positive numbers a and b with a > b. If a right triangle has legs H and G and hypotenuse A, then[13]
$\frac{A}{H} = \frac{A^{2}}{G^{2}} = \frac{G^{2}}{H^{2}} = \phi \,$
and
$\frac{a}{b} = \phi^{3}, \,$
where $\phi$ is the golden ratio $\tfrac{1+ \sqrt{5}}{2}. \,$
Euler line
In a right triangle, the Euler line contains the median on the hypotenuse—that is, it goes through both the right-angled vertex and the midpoint of the side opposite that vertex. This is because the right triangle's orthocenter, the intersection of its altitudes, falls on the right-angled vertex while its circumcenter, the intersection of its perpendicular bisectors of sides, falls on the midpoint of the hypotenuse.
Inequalities
In any right triangle the diameter of the incircle is less than half the hypotenuse, and more strongly it is less than or equal to the hypotenuse times $(\sqrt{2}-1).$[14]:p.281
In a right triangle with legs a, b and hypotenuse c,
$c \geq \frac{\sqrt{2}}{2}(a+b)$
with equality only in the isosceles case.[14]:p.282,p.358
If the altitude from the hypotenuse is denoted hc, then
$h_c \leq \frac{\sqrt {2}}{4}(a+b)$
with equality only in the isosceles case.[14]:p.282
Other properties
If segments of lengths p and q emanating from vertex C trisect the hypotenuse into segments of length c/3, then[2]:pp. 216–217
$p^2 + q^2 = 5\left(\frac{c}{3}\right)^2.$
The right triangle is the only triangle having two, rather than one or three, distinct inscribed squares.[15]
Let h and k (h > k) be the sides of the two inscribed squares in a right triangle with hypotenuse c. Then
$\frac{1}{c^2} + \frac{1}{h^2} = \frac{1}{k^2}.$
These sides and the incircle radius r are related by a similar formula:
$\displaystyle \frac{1}{r}=-{\frac{1}{c}}+\frac{1}{h}+\frac{1}{k}.$
The perimeter of a right triangle equals the sum of the radii of the incircle and the three excircles:
$a+b+c=r+r_a+r_b+r_c.$
The medians ma and mb from the legs satisfy[6]:p.136,#3110
$4c^4+9a^2b^2=16m_a^2m_b^2.$
References
1. Di Domenico, Angelo S., "A property of triangles involving area", Mathematical Gazette 87, July 2003, pp. 323-324.
2. Posamentier, Alfred S., and Salkind, Charles T. Challenging Problems in Geometry, Dover, 1996.
3. Wentworth p. 156
4. Voles, Roger, "Integer solutions of $a^{-2} + b^{-2} = d^{-2}$," Mathematical Gazette 83, July 1999, 269–271.
5. Richinick, Jennifer, "The upside-down Pythagorean Theorem," Mathematical Gazette 92, July 2008, 313–317.
6. Inequalities proposed in “Crux Mathematicorum”, [1]. Cite error: Invalid <ref> tag; name "Crux" defined multiple times with different content
7. Triangle right iff s = 2R + r, Art of problem solving, 2011
8. Andreescu, Titu and Andrica, Dorian, "Complex Numbers from A to...Z", Birkhäuser, 2006, pp. 109-110.
9. Properties of Right Triangles
10. CTK Wiki Math, A Variant of the Pythagorean Theorem, 2011, [2].
11. Darvasi, Gyula (March 2005), "Converse of a Property of Right Triangles", The Mathematical Gazette, 89 (514): 72–76<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>.
12. Bell, Amy (2006), "Hansen's Right Triangle Theorem, Its Converse and a Generalization" (PDF), Forum Geometricorum, 6: 335–342<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
13. Di Domenico, A., "The golden ratio — the right triangle — and the arithmetic, geometric, and harmonic means," Mathematical Gazette 89, July 2005, 261. Also Mitchell, Douglas W., "Feedback on 89.41", vol 90, March 2006, 153-154.
14. Posamentier, Alfred S., and Lehmann, Ingmar. The Secrets of Triangles. Prometheus Books, 2012.
15. Bailey, Herbert, and DeTemple, Duane, "Squares inscribed in angles and triangles", Mathematics Magazine 71(4), 1998, 278-284.
• Wentworth, G.A. (1895). A Text-Book of Geometry. Ginn & Co.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles> |
Rectifying Inverse Relations into Lines
Introduction
• In this lesson you will learn how to write equations of quantities which vary inversely.
• Graphs of inverse relationships will be modified to show a linear relationship.
Definitions
• Quantities vary inversely if they are related by the relationship .
• Another way to express this is .
• We also say that y varies inversely with x.
• When quantities vary inversely, the constant k is called the constant of proportionality.
• Quantities which vary inversely are also said to be inversely proportional.
Consider the following example:
Suppose that . The constant of proportionality is 3. A graph of this relationship for x > 0 is shown below.
This is in fact one branch of a hyperbola. The other branch is located in Quadrant III and is found if the values of x are negative. Features of the graph to notice are the characteristic shape sloping negatively and the fact that the graph approaches the x-axis as x gets large (end behavior).
A table of values containing x, , and is shown below.
x 0.4 2.5 7.5 0.5 2 6 0.667 1.5 4.5 1 1 3 1.5 0.667 2 2 0.5 1.5 2.5 0.4 1.2 3 0.333 1 3.5
This initial graph plots points from the table using (x, y) as coordinates.
This second graph also plots the points (w,y) or as coordinates.
Notice that these points are collinear with the slope 3.
Notice in the second graph that the points lie on a straight line. The slope of this line is 3. This is no accident. The equation can be written as . In this form, acts as the independent variable. If we let , we get y = 3w and can clearly see a linear equation with slope 3. The process of rewriting as y = 3w is known as “rectifying into a line.” The equation for the original branch of a hyperbola is rewritten so that the data can be represented linearly.
An Application
According to Boyle’s Law, the product of the pressure, P, and volume, V, of a gas under constant temperature is a constant. Calling this constant c, we get PV = c. This can also be written as where w = . In this case, a graph of would be a hyperbolic curve, whereas a graph of P = cw would be a line of slope c.
For example, if P = 4 x 105 N/m2 when V = 0.5 m3, we have PV = c = 2 x 105 Nm.
Notice that our variables need units of measurement since they represent experimental data. Pressure is the ratio of force per unit area. One atmosphere of pressure equals 1.01 x 105 N/m2 where the units N/m2 are often called a Pascal. The volume of the confined gas is measured in cubic meters, or m3.
The linear representation of this data is given by rectifying this data into a line. . This would be a line with slope 2 x 105 Nm and is easier to visualize graphically.
To see Boyle’s Law in action, visit this site. Or, if you would like to view Boyle’s original data from his experimentation in the 17th century, visit here.
Example
Suppose that y varies inversely with x and x = 0.5 when y = 0.9. What is the equation of its rectified line? What is your answer?
The table below shows the values for x, , and .
x 0.4 2.5 1.125 0.5 2 0.9 0.8 1.25 0.56 1 1 0.45 1.5 0.667 0.3 2 0.5 0.23 2.5 0.4 0.18 3 0.333 0.15 3.5 0.29 0.13
This first graph shows the hyperbolic representation . The WINDOW for this graph is X: (0, 3, 1) and Y: (-1, 6, 1)
Using the same WINDOW, this second graph plots the points where as shown in the table.
Notice that these points are collinear with the slope 0.45.
Examples
If y varies inversely as x, and y = 0.3 when x = 1.2, find the constant of proportionality. What is your answer?
Describe the graphs of both the hyperbolic and linear representations of this inverse variation. What is your answer?
M Ransom
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# Quick Answer: What Is Closure Property Of Real Numbers?
## What is an example of closure property?
The closure property means that a set is closed for some mathematical operation.
For example, the set of even natural numbers, [2, 4, 6, 8, .
.
.], is closed with respect to addition because the sum of any two of them is another even natural number, which is also a member of the set.
….
## What is properties of real numbers?
Property (a, b and c are real numbers, variables or algebraic expressions)1.Distributive Property a • (b + c) = a • b + a • c2.Commutative Property of Addition a + b = b + a3.Commutative Property of Multiplication a • b = b • a4.Associative Property of Addition a + (b + c) = (a + b) + c17 more rows
## What is the closure property in math?
In summary, the Closure Property simply states that if we add or multiply any two real numbers together, we will get only one unique answer and that answer will also be a real number. The Commutative Property states that for addition or multiplication of real numbers, the order of the numbers does not matter.
## What is the closure property of integers?
Closure property under multiplication states that the product of any two integers will be an integer i.e. if x and y are any two integers, xy will also be an integer. Example 2: 6 × 9 = 54 ; (–5) × (3) = −15, which are integers.
## How do you find a closure property?
The Closure Properties Real numbers are closed under addition, subtraction, and multiplication. That means if a and b are real numbers, then a + b is a unique real number, and a ⋅ b is a unique real number.
## What is the formula of closure property?
Closure property for addition : If a and b are two whole numbers and their sum is c, i.e. a + b = c, then c is will always a whole number. For any two whole numbers a and b, (a + b) is also a whole number. This is called the Closure-Property of Addition for the set of W. |
# Into Math Grade 3 Module 8 Lesson 2 Answer Key Find Unknown Factors and Numbers
We included HMH Into Math Grade 3 Answer Key PDF Module 8 Lesson 2 Find Unknown Factors and Numbers to make students experts in learning maths.
## HMH Into Math Grade 3 Module 8 Lesson 2 Answer Key Find Unknown Factors and Numbers
I Can use multiplication and division equations with unknown numbers to solve problems.
Jon is building a rectangle-shaped koi pond. He uses 24 one-foot square tiles to cover the bottom of the pond. He sets the same number of tiles in each row. How many rows of tiles can he set if he uses all the tiles?
Make a visual model of the problem to show the rows of tiles.
He could set 4×6 rows of tiles.
Explanation:
Given that Jon is building a rectangle-shaped koi pond and he uses 24 one-foot square tiles to cover the bottom of the pond then he sets the same number of tiles in each row. So the number of rows of tiles can he set if he uses all the tiles is 4×6 = 24.
Turn and Talk How could you use a related multiplication or division fact to solve the problem?
Build Understanding
1. Jon has 4 aquariums that hold a total of 32 gallons of water. Each aquarium holds the same amount of water. How much water is in each aquarium?
A. Show your work using a visual model.
Each aquarium can hold 8 gallons of water.
Explanation:
Given that Jon has 4 aquariums that hold a total of 32 gallons of water and each aquarium holds the same amount of water. So each aquarium can hold 32÷4 which is 8 gallons of water.
B. How many equal groups did you draw? Explain.
__________________
4 equal number of groups.
Explanation:
The number of equal groups can we draw is 4 equal numbers.
C. What is the total number of gallons shown in your visual model? __________________
32 gallons.
Explanation:
The total number of gallons shown in the visual model is 32 gallons.
D. How can you use your visual model to write a multiplication equation with an unknown factor to represent the problem? Use to represent the unknown factor. __________________
4 × s = 32,
s = 8 gallons.
Explanation:
The multiplication equation for the visual model with an unknown factor is 4 × s = 32 where s is 32÷4 = 8 gallons.
E. Find the unknown factor. How much water is in each aquarium? __________________
8 gallons.
Explanation:
The unknown factor that how much water in each aquarium is 8 gallons.
Turn and Talk How could the problem be rewritten so that 4 is the unknown factor?
2. Jon buys 35 new fish. He wants to place the fish into aquariums in groups of 7. How many aquariums does he need?
A. Show the problem using a visual model.
He needs 5 aquariums.
Explanation:
Given that Jon bought 35 new fish and he wants to place the fish into aquariums in groups of 7. So the number of aquariums does he need is 35÷7 which is 5 aquariums.
B. What is the total number of fish shown in your visual model? ______________
35 fishes.
Explanation:
The total number of fish shown in the visual model is 35 fish.
C. How many fish are in each equal group? __________
7 fish.
Explanation:
The number of fish in each equal group is 7 fish.
D. How can you use your visual model to write a division equation with an unknown to represent the problem? Use to represent the unknown factor. _______
35÷7 = ,
is 5.
Explanation:
The division equation to represent the problem is 35÷7 = where is 5.
E. What related multiplication equation with an unknown factor can you write to solve the problem? ____________
=35.
Explanation:
The related multiplication equation is 7×=35.
F. Find the unknown factor. How many aquariums does Jon need? ____________
5 aquariums.
Explanation:
The unknown factor will be 5. So Jon needs 5 aquariums.
Turn and Talk How can you tell that your answer is reasonable?
Step It Out
3. Nan buys 18 plants to place in aquariums for her betta fish. She places 3 plants in each aquarium. If Nan places all the plants, how many aquariums does she have?
A. Write the division equation with an unknown.
18 ÷ __ =
18 ÷ 3 = 6.
Explanation:
The division equation is 18 ÷ 3 = where is 6.
B. Write the related multiplication equation.
× ___ = 18
× 3 = 18.
Explanation:
The related multiplication equation is × 3 = 18
C. Solve. = ___; Nan has ___ aquariums.
6 aquariums.
Explanation:
The are 6 and Nan has 6 aquariums.
4. Nan has 8 packs of treats for her fish. She has a total of 40 treats. If each pack has the same number of treats, how many treats are in each pack?
A. Write a multiplication equation with an unknown and the related division equation to represent the problem.
___ × = 40 40 ÷ ___ =
8 × = 40 and 40 ÷ 8 = .
Explanation:
The multiplication equation with an unknown and the related division equation to represent the problem is
8 × = 40 and 40 ÷ 8 = .
B. Solve. = ___; Each pack has ___ treats.
5.
Explanation:
The is 5 and each pack has 5 treats.
Check Understanding Math Board
Question 1.
Sami has some boxes of pens. Each box has 6 pens. Sami has 54 pens. How many boxes does Sami have? Write a multiplication equation and a division equation with an unknown to represent the problem. Solve.
Sami has ___ boxes.
Sami has 9 boxes.
Explanation:
Given that Sami has some boxes of pens and each box has 6 pens and Sami has 54 pens. So the multiplication equation and a division equation with an unknown to represent the problem is 6 × = 54 and 54 ÷ 6 = . So Sami has 9 boxes.
Model with Mathematics Write a multiplication equation and a division equation with an unknown. Solve.
Question 2.
Laurence goes to the aquarium shop 5 days in a row and sees 15 new fish. He sees the same number of new fish each day. How many new fish does he see each day?
Laurence will see 3 new fish each day.
Explanation:
Given that Laurence goes to the aquarium shop 5 days in a row and sees 15 new fish. As he sees the same number of new fish each day, the new fish does he see each day is 5 × = 15 and 15 ÷ 5 = where is 3. So Laurence will see 3 new fish each day.
Question 3.
Liam is making sandwiches for a family picnic. He uses 18 slices of bread. Each sandwich uses 2 slices of bread. How many sandwiches does Liam make?
Liam makes 9 sandwiches.
Explanation:
Given that Liam uses 18 slices of bread and each sandwich uses 2 slices of bread. So the number of sandwiches does Liam make is 2 × = 18 and 18 ÷ 2 = . The unknown value is 9, Liam makes 9 sandwiches.
Question 4.
A mountain lion has a litter of cubs. Her litter has 24 paws. If each cub has 4 paws, how many cubs are in her litter?
The number of cubs in her litter is 6.
Explanation:
Given that a mountain lion has a litter of cubs as her litter has 24 paws and if each cub has 4 paws. So the number of cubs are in her litter is 4 × = 24 and 24 ÷ 4 = where the unknown value is 6. So the number of cubs in her litter is 6.
Question 5.
Tamara wants to give each of her 7 friends an equal number of friendship bracelets. She gives out 28 friendship bracelets. How many bracelets does each friend get?
The number of bracelets are 4.
Explanation:
Given that Tamara wants to give each of her 7 friends an equal number of friendship bracelets and she gives out 28 friendship bracelets. So the number of braclets does each friend get is 7 × = 28 and 28 ÷ 7 = where the unknown value is 4. So the number of bracelets are 4.
Find the unknown number.
Question 6.
12 ÷ ___ = 4
12 ÷ 3 = 4.
Explanation:
Let the missing digit be X, the equation will be 12 ÷ X = 4. So X will be
X = 12 ÷ 4
X = 3.
Question 7.
27 ÷ __ = 9
27 ÷ 3 = 9.
Explanation:
Let the missing digit be X, the equation will be 27 ÷ X = 9. So X will be
X = 27 ÷ 9
X = 3.
Question 8.
24 = 3 × ___
24 = 3 × 8.
Explanation:
Let the missing digit be X, the equation will be 24 = 3 × X. So X will be
X = 24 ÷ 3
X = 8.
Question 9.
___ ÷ 7 = 7
49 ÷ 7 = 7.
Explanation:
Let the missing digit be X, the equation will be X ÷ 7 = 7. So X will be
X ÷ 7 = 7,
X = 7 × 7
= 49.
Question 10.
___ × 9 = 18
2 × 9 = 18.
Explanation:
Let the missing digit be X, the equation will be X × 9 = 18. So X will be
X × 9 = 18,
X = 18 ÷ 9,
X = 2.
Question 11.
8 = 48 ÷ __
8 = 48 ÷ 6.
Explanation:
Let the missing digit be X, the equation will be 8 = 48 ÷ X. So X will be
8 = 48 ÷ X,
X = 48 ÷ 8,
X = 6.
Question 12.
__ ÷ 9 = 5
45 ÷ 9 = 5.
Explanation:
Let the missing digit be X, the equation will be X ÷ 9 = 5. So X will be
X ÷ 9 = 5,
X = 5 × 9
= 45.
Question 13.
9 × __ = 36
9 × 4 = 36.
Explanation:
Let the missing digit be X, the equation will be 9 × X = 36. So X will be
9 × X = 36,
X = 36 ÷ 9
= 4.
Solve.
Question 14.
Jack has $18 to spend on fish for an aquarium. He buys 6 fish. Each fish costs the same amount. How much does each fish cost? Answer: The cost of each fish is$3.
Explanation:
Given that Jack has $18 to spend on fish for an aquarium and he bought 6 fish. As each fish costs the same amount. So the cost of each fish is 18÷6 which is$3.
Question 15.
Nykeba is assigning tour guides for people going on museum tours. Each guide leads 4 people around the museum. There are 20 people going on a tour. How many tour guides does Nykeba assign?
The number of tour guides does Nykeba assigned is 5 people.
Explanation:
Given that Nykeba is assigning tour guides for people going on museum tours and each guide leads 4 people around the museum as there are 20 people going on a tour. So the number of tour guides does Nykeba assign is 20÷4 = 5.
Question 16.
Open Ended Write a word problem that could be solved using these related multiplication and
division equations.
12 ÷ = 3 and 3 × = 12 |
Just for a change, an FP3 topic. I’ve been struggling to tutor complex mappings properly (mainly because I’ve been too lazy to look them up), but have finally seen - I think - how to solve them with minimal headache.
A typical question gives you a mapping from the (complex) $z$-plane to the $w$-plane of $w = \frac{z - i}{z}$, and asks what happens to a given line in one plane or the other.
My recipe for approaching this boils down to three steps:
1. Cross-multiply to get rid of the ugly fraction
2. Multiply out the brackets to get a real equation and a complex equation
3. Use the equation of the line to eliminate something you don’t want
4. Eliminate the other variable you don’t want, leaving you with a relation between two variables
Sounds complicated? Let’s see what happens to $y = x$. First, I multiply up the $z$:
$wz = z+i$, or $(u+vi)(x + yi) = x + (y+1)i$
Then expand:
$(ux - vy) + (vx + uy)i) = x + (y+1)i$, giving two equations:
$ux - vy = x$ (1) and; $vx + uy = y+1$ (2)
Since $y=x$, substituting into (1) and dividing by $x$ gives: $u - v = 1$; the substitution gives you a straight line.
Another? How about $x + y + 1 = 0$? I have good news: we can start from (1) and (2) rather than working it all out again. We also know $y = (-1-x)$, so:
$ux + v(1+x) = x$ $vx - u(1+x) = -x$
Group the $x$s together and divide:
$x(u + v - 1) = -v$ $x(v - u + 1) = u$
$\frac{u+v-1}{v-u+1} = \frac{-v}{u}$
Cross-multiply:
$u(u+v-1) = -v(v-u+1)$ $u^2 + uv - u = -v^2 +vu - v$ $u^2 + v^2 - u + v = 0$
… so we get a circle. |
# Solve problems involving two right-angled triangles
Lesson
The next step in our trigonometric problem solving adventure is to solve 2, 3 or more step problems. What I mean is, solve problems where you have to solve other intermediate steps along the way.
The best way to learn these is to watch some examples, and then try the set of questions.
You'll need to remember these right-angled triangle rules:
Right-angled triangles
Pythagoras' theorem: $a^2+b^2=c^2$a2+b2=c2, where $c$c is the hypotenuse
$\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse = $\frac{O}{H}$OH
$\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse = $\frac{A}{H}$AH
$\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent =$\frac{O}{A}$OA
Angle of Elevation: the angle from the observer's horizontal line of sight looking UP at an object
Angle of Depression: the angle from the observer's horizontal line of sight looking DOWN at an object
Exact value triangles
#### Examples
##### Question 1
Consider the following diagram.
1. Find the length of $AD$AD, correct to 2 decimal places.
2. Find the length of $BD$BD, correct to 2 decimal places.
3. Hence, find the length of $AB$AB correct to 2 decimal places.
##### Question 2
Consider the following diagram.
1. Find $y$y, correct to two decimal places.
2. Find $w$w, correct to two decimal places.
3. Hence, find $x$x, correct to one decimal place.
### Outcomes
#### 10P.MT2.02
Determine the measures of the sides and angles in right triangles, using the primary trigonometric ratios and the Pythagorean Theorem |
# Lesson 5Give Me FivePractice Understanding
## Learning Focus
Interpret function notation to match a function with its features.
How can I visualize a graph from knowing its features?
How can function notation describe a function?
How does function notation describe relationships between two functions?
Technology guidance for today’s task:
• Add Two Functions Together Graphically:
## Open Up the Math: Launch, Explore, Discuss
### 1.
You have two sets of cards, A cards and B cards.
1. Lay out all the A cards. Each A card gives you a representation of a function.
2. Select a B card and read each statement aloud.
3. Find an A card that matches all 5 statements on the B card and justify each statement.
4. Record the match and your justification for it.
5. Keep the pairs together for the next part of the lesson.
Proceed until all the A and B cards are matched. Be sure that you can justify all 5 statements on the B card with the function that you match it with.
A card
B card
Reason
A1 matches with:
A2 matches with:
A3 matches with:
A4 matches with:
A5 matches with:
A6 matches with:
After you match the cards, complete the information below. Use both the A and the B cards in the pair for reference:
### 2.
Complete this information about the pair of cards that includes A1
Write the equation of and use it to:
Find an exact value for :
Find where :
### 3.
Complete this information about the pair of cards that includes A2
What are the intervals of increase and decrease for ?
Increasing:
Decreasing:
### 4.
Complete this information about the pair of cards that includes A3
Graph from :
### 5.
Complete this information about the pair of cards that includes A4
Write the equation and graph :
Equation:
### 6.
Complete this information about the pair of cards that includes A5
Which function is changing faster in the interval ? Why?
### 7.
Complete this information about the pair of cards that includes A6
Write the equation for :
Find :
## Ready for More?
### 1.
Explain the meaning of and .
### 2.
Find the value of .
## Takeaways
Common Phrases About Functions
We write:
We mean:
## Lesson Summary
In this lesson we interpreted function notation to match features with functions. We learned common phrases that can be used to “translate” function notation depending on the context. For instance, can be interpreted as the height of the graph at is or that when is substituted into , the output is .
## Retrieval
Find the indicated function values.
### 2.
Graph of :
#### c.
Determine whether the following statements represent a discrete or continuous relationship. Be ready to explain why.
### 3.
The amount of water in a pool that is being emptied.
### 4.
The number of runs scored at a baseball game. |
Comment
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Q)
# Find the equation of the tangent and normal at $(-2 , \large\frac{1}{4})$ to the rectangular hyperbola $2xy-2x-8y-1=0$
Comment
A)
Toolbox:
• The equation of the tangent at $(x_1,y_1)$ to a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is given by $Axx_1+\large\frac{B}{2}$$(xy_1+x_1y)+Cyy_1+D\large\frac{(x+x_1)}{2}+$$E\large\frac{(y+y_1)}{2}$$+F=0 • The normal is the perpendicular to this line that passes through (x_1,y_1). • Any line perpendicular to ax+by+c=0 is of the form bx-ay+k=0.This can be used to find the equation of the normal. Step 1: Equation of the tangent to 2xy-2x-8y-1=0 at (x_1,y_1) \large\frac{1}{2}$$\times 2(xy_1+yx_1)-2\large\frac{(x+x_1)}{2}$$-8\large\frac{(y+y_1)}{2}$$-1=0$
Here $(x_1,y_1)=(-2,\large\frac{1}{4})$
The tangent is $(\large\frac{x}{4}-$$2y)-(x-2)-4(y+\large\frac{1}{4})$$-1=0$
$x-8y-4x+8-16y-4-4=0$
$-3x-24y=0$
$3x+24y=0$
$3(x+8y)=0$
$\Rightarrow x+8y=0$
Step 2:
The normal is $\perp$ to the tangent.Its equation is of the form $8x-y+k=0$
It passes through $(-2,\large\frac{1}{4})$.
Therefore $8(-2)-\large\frac{1}{4}$$+k=0 k=16+\large\frac{1}{4}=\large\frac{65}{4} The equation of the normal is 8x-y+\large\frac{15}{4}$$=0$
$\Rightarrow 32x-4y+65=0$
Step 3:
Equation of tangent $x+8y=0$
Equation of normal $32x-4y+65=0$ |
# What is the meaning of slope in math
The slope of a line is a measure of its steepness. Mathematically, slope is calculated as “rise over run” (change in y divided by change in x).
## How do you define slope in math?
slope, Numerical measure of a line’s inclination relative to the horizontal. In analytic geometry, the slope of any line, ray, or line segment is the ratio of the vertical to the horizontal distance between any two points on it (“slope equals rise over run”).
## How do I calculate slope?
Pick two points on the line and determine their coordinates. Determine the difference in y-coordinates of these two points (rise). Determine the difference in x-coordinates for these two points (run). Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).
## What's the slope on a graph?
The slope equals the rise divided by the run: Slope =riserun Slope = rise run . You can determine the slope of a line from its graph by looking at the rise and run. One characteristic of a line is that its slope is constant all the way along it.
## How do you explain slope to a child?
The slope is defined as the “change in y” over the “change in x” of a line. If you pick two points on a line — (x1,y1) and (x2,y2) — you can calculate the slope by dividing y2 – y1 over x2 – x1.
## How do you find slope with 5 points?
1. Identify the coordinates (x₁,y₁) and (x₂,y₂) . …
2. Input the values into the formula. …
3. Subtract the values in parentheses to get 2/(-5) .
4. Simplify the fraction to get the slope of -2/5 .
5. Check your result using the slope calculator.
## What is the slope for 3?
Answer: The slope of the line y = 3 is equal to 0.
## What does a slope of 1 look like?
A slope of 1 means that if the function is plotted on a graph, the line will rise 1 unit for every 1 unit farther to the right you go. Slope is equal to rise (change in y value) divided by run (change in x value), so if x and y change at the same rate, the slope is 1.
## How do you graph a slope?
1. Slope = rise/run.
2. Count the rise. Since the rise is positive 2, I counted up 2.
3. Count the run. …
4. Repeat the process to plot a third point.
5. Draw a straight line through your points.
6. Count the rise. …
7. Count the run. ( …
8. Repeat the process if you’d like to plot a 3rd point.
What is a slope in 8th grade math?
The slope of a line describes how steep a line is. Slope is the change in y values divided by the change in x values.
Article first time published on askingthelot.com/what-is-the-meaning-of-slope-in-math/
## How do you teach students to slope?
1. Day 1: Start with graphing unit rates as y=mx and introduce the different types of slopes – positive, negative, undefined and zero.
2. Day 2: Have students practice finding slope from the graph of the line using rise over run.
3. Day 2 or 3: Introduce the slope formula.
## What is the slope of 4x 2y 10?
Using the slope-intercept form, the slope is −2 .
## What is the slope of 2x 7y 6?
Using the slope-intercept form, the slope is 27 .
## What is the slope of the line y =- 1?
Precalculus Examples Using the slope-intercept form, the slope is 0 .
## How do you find a slope without a graph?
Explanation: To find the slope given two points without using a graph, we use the formula riserun , or y2−y1x2−x1 . Therefore, the slope is −72 or −3.5 . Hope this helps!
## What is the slope of 0?
The slope of a line can be thought of as ‘rise over run.’ When the ‘rise’ is zero, then the line is horizontal, or flat, and the slope of the line is zero. Put simply, a zero slope is perfectly flat in the horizontal direction.
## What is the slope of 1 0 and 3 8?
We will let (3, 8) be 1 and (1, 0) be 2 because it has a 0 and subtracting 0 is easier. Remember – whichever point contains y1, must also contain x1. 8 / 2 = 4. So the slope is 4.
## What is the slope of the line through (- 9 6 and (- 3 9?
This means the slope of the line is 0.5.
## What is the slope of the line through (- 1 8 and 3?
Explanation: The to calculate slope, use m=y2−y1x2−x1 . Plugging in the coordinates gives m=−2−(−8)3−(−1) . When simplified, the answer is 64 , which reduces to 32 .
## What is the slope of 4/5 and B 9 7?
Summary: If line AB contains points A(4, 5) and B(9, 7) then the slope of AB is 2/5.
## How do you find the slope with 3 points?
If the three points are collinear, take ANY two of these and use the formula (y2-y1)/(x2-x1) to find the required slope.
## What is the slope of 7x 5y 3?
Using the slope-intercept form, the slope is −75 . All lines that are parallel to y=−7×5−35 y = – 7 x 5 – 3 5 have the same slope of −75 .
## What is the slope of 3x y 8?
Using the slope-intercept form, the slope is −3 .
## What are AB and C in AX by C?
where A,B,andC are placeholders for constants (the x and y are variables). By normal usage A should be an integer value which is not negative. |
# How to find Surface Area and Volume of Composite Figures
Now, let’s look at composite figures in 3D. In order to calculate the volume of such composite figures, you have to separate the composite figure into smaller figures so that you can determine the volume of each, then add them up.
Example 1
Find the volume of the composite figure.
You can separate the figure into a cylinder and a cone, and find the volume of each of them.
Volume of the cylinder:
$V=\pi {r}^{2}h=\pi \cdot {6}^{2}\cdot 9\approx 1017.88\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}$
Volume of the cone:
$V=\frac{\pi {r}^{2}h}{3}=\frac{\pi \cdot {6}^{2}\cdot 5}{3}\approx 188.50\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}$
Volume of the whole figure:
$V\approx 1017.88+188.50\approx 1206.38\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}$
Example 2
Find the volume of this ice cream
Half of the scoop of ice cream is inside the wafer. The volume of the wafer is the volume of a cone:
${V}_{\text{cone}}=\frac{\pi \cdot {4}^{2}\cdot 15}{3}\approx 251.2\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}$
The volume of the entire scoop of ice cream is the volume of a sphere:
${V}_{\text{sphere}}=\frac{4\cdot \pi \cdot {4}^{3}}{3}\approx 267.95\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}$
If you now add the volume of the scoop of ice cream to the volume of the cone, you will be adding the entire ice cream scoop. But half of the scoop is inside the wafer cone, so you can’t count that volume twice! You have to divide the volume of the sphere by 2 first. Then you can add it to the volume of the cone to find the total volume of the ice cream.
The volume of the entire ice cream is
$V\approx 251.2\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}+\frac{267.95\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}}{2}\approx 385.18\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}$ |
# 1.6 Rational expressions
Page 1 / 6
In this section students will:
• Simplify rational expressions.
• Multiply rational expressions.
• Divide rational expressions.
• Add and subtract rational expressions.
• Simplify complex rational expressions.
A pastry shop has fixed costs of $\text{\hspace{0.17em}}\text{}280\text{\hspace{0.17em}}$ per week and variable costs of $\text{\hspace{0.17em}}\text{}9\text{\hspace{0.17em}}$ per box of pastries. The shop’s costs per week in terms of $\text{\hspace{0.17em}}x,$ the number of boxes made, is $\text{\hspace{0.17em}}280+9x.\text{\hspace{0.17em}}$ We can divide the costs per week by the number of boxes made to determine the cost per box of pastries.
$\frac{280+9x}{x}$
Notice that the result is a polynomial expression divided by a second polynomial expression. In this section, we will explore quotients of polynomial expressions.
## Simplifying rational expressions
The quotient of two polynomial expressions is called a rational expression . We can apply the properties of fractions to rational expressions, such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown.
$\frac{{x}^{2}+8x+16}{{x}^{2}+11x+28}$
We can factor the numerator and denominator to rewrite the expression.
$\frac{{\left(x+4\right)}^{2}}{\left(x+4\right)\left(x+7\right)}$
Then we can simplify that expression by canceling the common factor $\text{\hspace{0.17em}}\left(x+4\right).$
$\frac{x+4}{x+7}$
Given a rational expression, simplify it.
1. Factor the numerator and denominator.
2. Cancel any common factors.
## Simplifying rational expressions
Simplify $\text{\hspace{0.17em}}\frac{{x}^{2}-9}{{x}^{2}+4x+3}.$
Can the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term be cancelled in [link] ?
No. A factor is an expression that is multiplied by another expression. The $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ term is not a factor of the numerator or the denominator.
Simplify $\text{\hspace{0.17em}}\frac{x-6}{{x}^{2}-36}.$
$\frac{1}{x+6}$
## Multiplying rational expressions
Multiplication of rational expressions works the same way as multiplication of any other fractions. We multiply the numerators to find the numerator of the product, and then multiply the denominators to find the denominator of the product. Before multiplying, it is helpful to factor the numerators and denominators just as we did when simplifying rational expressions. We are often able to simplify the product of rational expressions.
Given two rational expressions, multiply them.
1. Factor the numerator and denominator.
2. Multiply the numerators.
3. Multiply the denominators.
4. Simplify.
## Multiplying rational expressions
Multiply the rational expressions and show the product in simplest form:
$\frac{\left(x+5\right)\left(x-1\right)}{3\left(x+6\right)}\cdot \frac{\left(2x-1\right)}{\left(x+5\right)}$
Multiply the rational expressions and show the product in simplest form:
$\frac{{x}^{2}+11x+30}{{x}^{2}+5x+6}\cdot \frac{{x}^{2}+7x+12}{{x}^{2}+8x+16}$
$\frac{\left(x+5\right)\left(x+6\right)}{\left(x+2\right)\left(x+4\right)}$
## Dividing rational expressions
Division of rational expressions works the same way as division of other fractions. To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. Using this approach, we would rewrite $\text{\hspace{0.17em}}\frac{1}{x}÷\frac{{x}^{2}}{3}\text{\hspace{0.17em}}$ as the product $\text{\hspace{0.17em}}\frac{1}{x}\cdot \frac{3}{{x}^{2}}.\text{\hspace{0.17em}}$ Once the division expression has been rewritten as a multiplication expression, we can multiply as we did before.
Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1 |
# Right Angled Triangle and Trigonometry
Topics: Trigonometry, Hypotenuse, Law of cosines Pages: 6 (794 words) Published: April 1, 2011
Right Triangle Trigonometry
Trigonometry is a branch of mathematics involving the study of triangles, and has applications in fields such as engineering, surveying, navigation, optics, and electronics. Also the ability to use and manipulate trigonometric functions is necessary in other branches of mathematics, including calculus, vectors and complex numbers. Right-angled Triangles In a right-angled triangle the three sides are given special names. The side opposite the right angle is called the hypotenuse (h) – this is always the longest side of the triangle. The other two sides are named in relation to another known angle (or an unknown angle under consideration).
If this angle is known or under consideration
h
θ
this side is called the opposite side because it is opposite the angle
This side is called the adjacent side because it is adjacent to or near the angle Trigonometric Ratios In a right-angled triangle the following ratios are defined sin θ = opposite side length o = hypotenuse length h cosineθ = adjacent side length a = hypotenuse length h
tangentθ =
opposite side length o = adjacent side length a
where θ is the angle as shown
These ratios are abbreviated to sinθ, cosθ, and tanθ respectively. A useful memory aid is Soh Cah Toa pronounced ‘so-car-tow-a’
Page 1 of 5
Unknown sides and angles in right angled triangles can be found using these ratios. Examples Find the value of the indicated unknown (side length or angle) in each of the following diagrams. (1) Method 1. Determine which ratio to use. 2. Write the relevant equation. 3. Substitute values from given information. 4. Solve the equation for the unknown.
b
27o
42
In this problem we have an angle, the opposite side and the adjacent side. The ratio that relates these two sides is the tangent ratio. tan θ = opposite side adjacent side
Substitute in the equation: (opposite side = b, adjacent side = 42, and θ = 27o)
b 42 b = 42 × tan 27° b = 21.4 tan 27° =
(2)
transpose to give
θ
In this triangle we know two sides and need to find the angle θ. 13.4 cm The known sides are the opposite side and the hypotenuse. The ratio that relates the opposite side and the hypotenuse is the sine ratio.
19.7 cm
opposite side hypotenuse 13.4 sin θ = 19.7 sin θ = 0.6082 sin θ =
opposite side = 13.4cm. hypotenuse = 19.7cm.
This means we need the angle whose sine is 0.6082, or sin −1 0.6082 from the calculator. ∴ θ = 42.90
Page 2 of 5
Pythagoras’ Thoerem Pythagoras’ Theorem states that in a right angled triangle the square of the length of the hypotenuse side (h), is equal to the sum of the squares of the other two sides.
h
h
2
= a
2
+ b
2
b
a
Pythagoras’ Theorem can be used to find a side length of a right angled triangle given the other two side lengths
Example 1
find the value of h h =6 +8 ∴ h 2 = 100 ∴ h = 10
2 2 2
∴ h 2 = 36 + 64
6 cm h
Pythagoras’ Theorem for this triangle
square root of 100
8 cm
Note Measurements must be in the same units and the unknown length will be in these same units so h will be 10 cm
Example 2
find the value of x
4.2 2 = 2.7 2 + x 2
x 2.7
∴17.64 = 7.29 + x 2 ∴10.35 = x 2 ∴ x = 3.22
4.2
Page 3 of 5
Exercise Find the value of the indicated unknown (side length or angle) in each of the following diagrams.
(a)
(b)
35o
4.71 mm a 62o a 14 cm
(c) 4.8 cm
(d)
θ
z
6.2 cm 20.2
6.5
α
(e) 500
(f)
b 34 27o 42
a
Page 4 of 5
Special angles and exact values
There are some special angles that enable us to obtain exact solutions for the functions and tan. sin, cos
If we take the two triangles below, and apply the basic trigonometry rules for sine, cosine and tangent –
sine =
opposite hypotenuse
cosine =
tangent =
60o
45o
2
2
1
30o 45o
1
3
1 From these two triangles, exact answers for sine,... |
# If the sum of two consecutive integers is 51, what are the numbers?
Feb 1, 2016
The two consecutive integers are 25 and 26.
#### Explanation:
Divide $51$ by $2$.
$\frac{51}{2} = 25.5$
Decrease 25.5 to 25, and increase 25.5 to 26.
$25 + 26 = 51$
Feb 1, 2016
Let the first number be $x$
Then,the second number is $x + 1$
So,
$x + \left(x + 1\right) = 51$
Remove brackets:
$\rightarrow x + x + 1 = 51$
$\rightarrow 2 x + 1 = 51$
$\rightarrow 2 x = 51 - 1$
$\rightarrow 2 x = 50$
$\rightarrow x = \frac{50}{2} = 25$
Then the second number is $x + 1 = 25 + 1 = 26$
So,the two consecutive integers are $25 \mathmr{and} 26$ |
# How do you write a number cubed
Jeremy Anderberg August 6, Last updated: Whether enjoying whiskey or soda, most people want some chill before drinking. How do you go about it, though?
Math Home Search Dr. Math Cube Root by Hand Date: Adrian Van Cauwenberghe Subject: Cube root Dear Sir, Many years ago I learned to calculate the square root of a number without calculators or computers, just pencil and paper.
## How Do I Type a "squared 2" Symbol? | benjaminpohle.com
An example will tell you how square root of Later we learned a comparable but of course more complicated way to calculate the cube root of a number.
I forgot almost completely how. What I remember is that we had to split the number up in groups of three figures instead of two example: But then I am lost. Can you help me further? Many thanks in anticipation.
Adrian Bruges - Belgium Date: Cube root To find a cube root by the "longhand" method, we proceed very much as we do to find a square root by hand. I intersperse numbered steps with an example. We will find the cube root of to two decimal places.
Draw a cube root symbol, or radical, with the number whose root you are seeking underneath. Start with the decimal point and mark off digits in both directions in groups of three.
Put a decimal point above the radical, and directly above the other decimal point. Start with the first group of 1, 2, or 3 digits. Find the largest cube of a single-digit integer less than it. Write the single digit above the radical, and its cube under the first group.
Draw a line under that cube, and subtract it from the first group. Bring down the next group below the last line drawn. This forms the current remainder. Draw a vertical line to the left of the resulting number, and to the left of that line put three hundred times the square of the number above the radical, a plus sign, thirty times the number above the radical, a multiplication sign, an underscore character, another plus sign, another underscore character, the exponent 2, an equals sign, and some blank space for the answer.
Pick the biggest digit D that would fit into both underscore places, and give a number such that D times it is less than the current remainder.The Big Secret.
The method for converting units comes right from one simple principle: Numbers with units, like meters or 32 ft/sec², are treated exactly the same as coefficients with variables, like x or 32y/z².. Once you grasp this, you see at once why the laws of units work as they do.
Relate the expanded form of a number to the number of hundreds, tens, and ones that comprise it. For example, = + 50 + 9 corresponds to 3 hundreds + 5 tens + 9 ones. Then, have the student explore other possible expanded forms such as 3 hundreds + 59 ones or 35 tens + 9 ones.
Oct 02, · Just type 2 in the cell wherever you ned it. Select the "2" in the formula bar, click Format, Cells, click Superscript. Won't affect any other text in the cell if you do it correctly.
Jan 28, · This feature is not available right now. Please try again later. To complete John's thought, there are three distinct cube roots of every non-zero number (positive real, negative real, complex), not just of the negative real numbers.
And as John points out, some of these roots are complex, so you need to know how the tools you are using behave in order to get the answer(s) you want.
As you can tell, all of the questions above deal with Algebraic expressions that deal with the addition of numbers — remember to think "addition" when you hear or read the words add, plus, increase or sum, as the resulting Algebraic expression will require the addition sign (+).
Cubes and Cube Roots |
# Daily Sudoku Puzzle (Easy)
SCORE:
## How to play
A Sudoku is a logic puzzle, there is no guessing involved, there is no maths involved, it is a pure logic puzzle.
This is an example of a starting grid for Sudoku - there is always some numbers that are given to you to start with. The more numbers you are given to start with, the easier the puzzle (normally).
There are three rules to Sudoku,
Each column must contain the numbers 1-9 once, and only once. The highlighted column already contains the numbers 5 and 7, you need to work out where the numbers 1, 2, 3, 4, 6, 8, and 9 need to go.
Each row must also contain the numbers 1-9 once, and only once. The highlighted row already contains the number 4, you need to work out where the numbers 1, 2, 3, 5, 6, 7, 8, and 9 need to go.
Each 3x3 region must contain the numbers 1-9 once, and only once. The highlighted 3x3 box already contains the numbers 2, 4, 6, and 7, you need to work out where the numbers 1, 3, 5, 8, and 9 need to go.
The easiest way to start is to look for cells where there is only one possible candidate. We know the bottom-left 3x3 region must contain the number 5 somewhere, but we don't know where. We also know that each row and column can only contain the number 5 once. This means we can eliminate all of the cells in that bottom-left square, apart from one. We can write in a '5' for the highlighted square.
The highlighted cell is another cell that we can fill right away. We can look at the numbers already in that column, row and 3x3 region, and by a process of elimination, work out what the highlighted cell should be.
That column already contains 2, 3, 4, 6 and 7. That row already contains 5, 6 and 8. That 3x3 region already contains 6 and 9. We know the highlighted cell can't be any of these numbers, because we can only have one of each number in each column, row, and 3x3 region. The highlighted cell can't be any of 2, 3, 4, 5, 6, 7, or 9. This leaves us with just the number 1!
I have highlighted three squares that you can fill in using the two techniques you have just learnt.
The rest of the puzzle can be solved using the two techniques you have just learnt.
All of our puzzles are graded easy, medium or hard. Go to the archive page and find an easy puzzle to start with!
This page will automatically load the puzzle for today. If you want to play a different puzzle, go to the archive page and choose your puzzle.
There are two ways to play a Sudoku puzzle, you can just use the mouse/touchscreen, or you can use the mouse and keyboard. You can switch between the two methods any time you like, and can use a combination of both.
### Playing with a mouse/touchscreen.
• When you have found a square where you can enter a number, click/touch that square. The square will turn light blue.
Above and below the puzzle is the number selection. Click/touch the number you want to enter in to that cell. If there is already a number in that square, it will be over-written.
• If you want to enter a pencil mark, click/touch the square you want to put in a pencil mark. It will turn light blue. Click/touch the pencil icon above or below the puzzle. This icon will turn light blue, and you are now in pencil marks mode.
Whenever you click/touch a number now, a pencil mark will be put in the square instead. To remove a number as a pencil mark, make sure you are in pencil marks mode, and click/touch the number again.
You can exit pencil mark mode by clicking/touching the pencil icon, it will turn back to normal.
• If you want to clear a particular square, make sure that square is selected and is in light blue. Click/touch the eraser icon. If there is a number in that square, it will be removed. If you click/touch it again, any pencil marks in that square will be removed.
### Playing with a mouse and keyboard.
• You will need to select a square by clicking on it with the mouse, it will turn light blue. You can change the current square by using the cursor keys on your keyboard.
• To enter a number, press that number on the keyboard. If there is already a number in that square, it will be overwritten. To remove a number, press the backspace or delete key on your keyboard.
• To enter a pencil mark, press control, shift, or alt on your keyboard at the same time as pressing a number key. Do the same thing again to remove that pencil mark.
Any mistakes you make will be hilighted in red. The website will know when you have completed a puzzle and will tell you. If you have an account and are logged in, the website will remember that you have completed that puzzle. You will also take part in out leaderboards. It's free to create an account! |
# What Are Factors: Explained for Teachers, Parents and Kids
Factors start to be mentioned in elementary school from about 3rd grade. Here you’ll find out what children will be taught and need to know about factors. A secure knowledge of multiplication facts is essential in understanding factors.
### What is a factor?
A factor is a number that fits exactly into a given number, or divides a particular number with no remainder (fraction or decimal). They can also be identified as pairs of numbers that multiply together to make another number. A factor is always a positive integer (whole number).
Note: Children often confuse factors with multiples.
### What are factors of a number?
Factors are divisors (but divisors can be positive or negative numbers) of a given whole number. A number may have many factors. Possible factors can be found using a factor tree or divisibility rules
Examples of factors in math
5 is a factor of 10, 15, 20, 25, etc. because 10 ÷ 5 = 2, 15 ÷ 3 = 3, 20 ÷ 5 = 4, 25 ÷ 5 = 5, etc.; therefore, all the numbers in the 5 times table have 5 as a factor.
When finding factors, it’s useful to look for them in pairs as two factors will multiply to make another number. The factor pairs of 12 are 1 x 12, 2 x 6 and 3 x 4, so the factors of 12 are 1, 2, 3, 4, 6 and 12.
Square numbers have an odd number of factors as one of the factors is multiplied by itself – for example, the factor pairs of 25 are 1 x 25 and 5 x 5, so the factors of 25 are 1, 5 and 25.
Prime numbers only have two factors – themselves and 1.
### When do children learn about factors in elementary school?
Students will first encounter factors in 3rd grade, where they will:
• recognize and use factor pairs and commutativity mentally and fluently
• use factors and multiples to determine unknown values in multiplication or division equations (i.e. 8 x ? = 48)
• identify multiples and factors, including finding all factor pairs of a number, and common factors of two numbers
• know and use the vocabulary of prime numbers, prime factors and composite numbers
• solve problems involving multiplication and division including using their knowledge of factors and multiples, square numbers and cube numbers
And finally in 5th grade, where children will be expected to:
• identify common factors, common multiples and prime numbers
• use common multiples to create common denominators between two or more fractions when comparing, adding, or subtracting them
• use common factors to simplify fractions
### What is the greatest common factor?
The greatest common factor is the largest factor shared between two given numbers. For example, the greatest common factor of 12 and 16 is 4 as it is the largest number that both given numbers can be divided by without remainders. Greatest common factors can be used to simplify fractions.
Read more: What is the Greatest Common Factor?
### How do factors relate to other areas of math?
Common factors are used when simplifying fractions. Factors are also useful when discussing the area of a rectangle or the volume of a rectangular prism.
### Factors in real life
Prime factors are often used in cyber security, to encrypt information.
Factors are also useful when organizing a number into groups – for example, a class of 28 children could be split into 2 groups of 14, 4 groups of 7, 7 groups of 4 or 14 groups of 2.
### 6 factor practice questions and answers
1. The factors of a number are 1, 2, 3, 5, 6, 10, 15 and 30. What is the number?
2. List the factors of 24.
3. What’s the greatest common factor of 24 and 30?
4. Write three factors of 30 that are not factors of 15.
5. Tick the numbers that are common factors of both 12 and 18: 2, 3, 6, 9, 12
6. Here are five numbers: 2 3 4 5 6. Write each number on the correct cards. The number 2 has been written on the correct cards for you.
1. 30
2. 1, 2, 3, 4, 6, 8, 12, 24
3. 6
4. 2, 6, 30
5. 2, 3, 6
6. Prime numbers: 2, 3, 5
Factors of 12: 2, 3, 4, 6
Factors of 15: 3, 5
What is a factor of 12?
The factors of 12 are 1, 2, 3, 4, 6 and 12.
What are the factors of 21?
The factors of 21 are 1, 3, and 7.
How do you explain factors to children?
An easy way to explain factors to children is that “factors are friends – they come in pairs”. Using a “factor rainbow”is a good way of visualizing these pairs.
For guides on introducing more new words in primary math, take a look at our math dictionary for parents and kids
Do you have students who need extra support in math?
Give your students more opportunities to consolidate learning and practice skills through personalized math tutoring with their own dedicated online math tutor.
Each student receives differentiated instruction designed to close their individual learning gaps, and scaffolded learning ensures every student learns at the right pace. Lessons are aligned with your state’s standards and assessments, plus you’ll receive regular reports every step of the way.
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# When to use a permutation?
Hence, Permutation is used for lists (order matters) and Combination for groups (order doesn't matter). Famous joke for the difference is: A “combination lock” should really be called a “permutation lock”. The order you put in the numbers of lock matters.
## What is the purpose of permutation?
A permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangements matters.
## How can we use permutation in real life?
An example of permutations would be the arrangement of books on a shelf. An easy one is to say there are 5 different books...how many ways can you arrange them on the shelf (in the typical upright position of libraries) ??
## How do you know if something is a permutation or not?
If order does not matter, it is a combinations problem. If order does matter, it is a permutations problem. With your example, it does not seem the order of the heads or tails occurance matters, so combinations is the way to go here.
## How do you calculate permutations?
To calculate the number of permutations, take the number of possibilities for each event and then multiply that number by itself X times, where X equals the number of events in the sequence. For example, with four-digit PINs, each digit can range from 0 to 9, giving us 10 possibilities for each digit.
## Permutations and Combinations | Counting | Don't Memorise
33 related questions found
### What does N and R stand for in permutation?
n = total items in the set; r = items taken for the permutation; "!" denotes factorial.
### How did you calculate the different permutations?
To calculate permutations, we use the equation nPr, where n is the total number of choices and r is the amount of items being selected. To solve this equation, use the equation nPr = n! / (n - r)!.
### What are the real life examples of permutations and combinations?
What are the real-life examples of permutations and combinations? Arranging people, digits, numbers, alphabets, letters, and colours are examples of permutations. Selection of menu, food, clothes, subjects, the team are examples of combinations.
### What is permutation example?
A permutation is an arrangement of objects in a definite order. ... For example, the permutation of set A={1,6} is 2, such as {1,6}, {6,1}. As you can see, there are no other ways to arrange the elements of set A.
### Is permutation with replacement?
When selecting more than one item without replacement and order is important, it is called a Permutation. When order is not important, it is called a Combination.
### What is permutation rule?
The special permutation rule states that anything permute itself is equivalent to itself factorial. Example: (3)3 = 3! (3 − 3)! ... Remark: The difference between a combination and a permutation is that order of the objects is not important for a combination.
### How do you calculate permutations and combinations?
If the order doesn't matter then we have a combination, if the order does matter then we have a permutation. One could say that a permutation is an ordered combination. The number of permutations of n objects taken r at a time is determined by the following formula: P(n,r)=n!
### What are the keywords that you should consider in permutation?
The keywords like-selection, choose, pick, and combination-indicates that it is a combination question. Keywords like-arrangement, ordered, unique- indicates that it is a permutation question.
### What is the formula of distinguishable permutation?
To find the number of distinguishable permutations, take the total number of letters factorial divide by the frequency of each letter factorial. Basically, the little n's are the frequencies of each different (distinguishable) letter. Big N is the total number of letters.
### What is the difference between combinations and permutations?
Combination is the counting of selections that we make from n objects. Whereas permutation is counting the number of arrangements from n objects. The point we need to keep in our mind is that combinations do not place an emphasis on order, placement, or arrangement but on choice.
### What do N and R stand for in combinations?
Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time.
### What is the V looking thing in math?
The Math Symbols from the Question about Union and Intersection. The “V” symbols in the reader's question are ∨ and ∧, which mean “Logical Or” and “Logical And.” The ∧ is a capital Greek Lambda. The small ^ or “caret” is available on most keyboards as “shift-6”; it symbolizes the exponentiation function.
### What is the focus of permutation?
A permutation, in contrast, focuses on the arrangement of objects with regard to the order in which they are arranged. Thus, the letters AB and BA represent two different permutations, because the order is different.
### What is an example of a combination?
A combination is a selection of all or part of a set of objects, without regard to the order in which objects are selected. For example, suppose we have a set of three letters: A, B, and C. ... Each possible selection would be an example of a combination. The complete list of possible selections would be: AB, AC, and BC. |
What is 1/333 as a decimal?
Solution and how to convert 1 / 333 into a decimal
1 / 333 = 0.003
1/333 or 0.003 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Both represent numbers between integers, in some cases defining portions of whole numbers Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
1/333 is 1 divided by 333
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. Now we divide 1 (the numerator) into 333 (the denominator) to discover how many whole parts we have. Here's 1/333 as our equation:
Numerator: 1
• Numerators are the portion of total parts, showed at the top of the fraction. Comparatively, 1 is a small number meaning you will have less parts to your equation. 1 is an odd number so it might be harder to convert without a calculator. Smaller numerators doesn't mean easier conversions. Let's take a look at the denominator of our fraction.
Denominator: 333
• Denominators represent the total parts, located at the bottom of the fraction. 333 is a large number which means you should probably use a calculator. But 333 is an odd number. Having an odd denominator like 333 could sometimes be more difficult. Ultimately, don't be afraid of double-digit denominators. So grab a pen and pencil. Let's convert 1/333 by hand.
Converting 1/333 to 0.003
Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 333 \enclose{longdiv}{ 1 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
Step 2: Extend your division problem
$$\require{enclose} 00. \\ 333 \enclose{longdiv}{ 1.0 }$$
Uh oh. 333 cannot be divided into 1. So we will have to extend our division problem. Add a decimal point to 1, your numerator, and add an additional zero. Now 333 will be able to divide into 10.
Step 3: Solve for how many whole groups you can divide 333 into 10
$$\require{enclose} 00.0 \\ 333 \enclose{longdiv}{ 1.0 }$$
Now that we've extended the equation, we can divide 333 into 10 and return our first potential solution! Multiply this number by 333, the denominator to get the first part of your answer!
Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 333 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you still have a remainder, continue to the next step.
Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. This is also true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But they all represent how numbers show us value in the real world. Here are just a few ways we use 1/333, 0.003 or 0% in our daily world:
When you should convert 1/333 into a decimal
Dining - We don't give a tip of 1/333 of the bill (technically we do, but that sounds weird doesn't it?). We give a 0% tip or 0.003 of the entire bill.
When to convert 0.003 to 1/333 as a fraction
Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.
Practice Decimal Conversion with your Classroom
• If 1/333 = 0.003 what would it be as a percentage?
• What is 1 + 1/333 in decimal form?
• What is 1 - 1/333 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.003 + 1/2? |
# AP Statistics Curriculum 2007 Limits Bin2HyperG
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===References=== ===References=== - * TBD
## General Advance-Placement (AP) Statistics Curriculum - Binomial Approximation to HyperGeometric Distribution
### Binomial Approximation to HyperGeometric Distribution
Suppose $X \sim Hypergeometric(m, N, n)$ and p = m / N.
• If N and m are large compared to n and p is not close to 0 or 1, then $P(X \le x) \approx P(Y \le x)$, where Y has a Binomial(n, p) distribution.
### Examples
An urn contains 50 marbles (35 green and 15 white). Fifteen marbles are selected without replacement. Find the probability that exactly 10 out of the 15 selected are green marbles. The answer to this question can be found using the formula: $P(X=10)=\frac{{35 \choose 10}{15 \choose 5}}{{50 \choose 15}}=0.2449.$ Using SOCR simply enter population size 50, sample size 15, and number of good objects 35, to get the figure below.
Now, select without replacement only 2 marbles. Compute the exact probability that 1 green marble is obtained. This is equal to $P(X=1)=\frac{{35 \choose 1}{15 \choose 1}}{{50 \choose 2}}=0.4286.$ This is also shown on the figure below.
We will approximate the probability of obtaining 1 green marble using Binomial as follows. Select the SOCR Binomial distribution and choose number of trials 2 and probablity of success $p=\frac{35}{50}=0.7$. Compare the figure below with the figure above. They are almost the same! Why? Using the Binomial formula we can compute the approximate probability of observing 1 green marble as $P(X=1)={2 \choose 1}0.70^10.30^1=0.42$ (very close to the exact probability, 0.4286).
### References
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# How to Calculate a Sloped Roof Area
Save
There are two ways to define a sloped roof area: the surface area of the roof itself and the area protected by the roof. These have different areas because the roof lies on the hypotenuse of a triangle, whereas the area protected by it is horizontal. When determining sloped roof area for rainfall catchment, for example, you calculate the area protected by the roof, but when replacing the tiles you calculate the surface area of the roof itself.
### Things You'll Need
• Tape measure
• Calculator
• Determine the area protected by the sloping roof by multiplying the length of the roof by the width. The result is the area covered by the sloping roof. For example, a single sloping roof covering a house 30 feet long and 20 feet wide covers an area of 600 square feet -- 30 x 20 = 600.
If the same house had a central roof-ridge dividing the roof into two sloping areas, each would cover 300 square feet or half the house.
• Determine the surface area of the roof by first measuring the distance along an imaginary horizontal line from the edge of the roof to a spot vertically beneath the highest point. This is the base of the triangle. Record the distance.
• Determine the difference in height between the lowest and the highest points on the roof. These are usually the points where the roof meets the house side walls and where it reaches the central roof ridge. Deduct the greater height from the lesser one to determine the height of the roof. This is the vertical side of the triangle.
For example, if the roof ridge is 25 feet above ground level and the lower edge of the roof is 20 feet above ground level the vertical difference is 5 feet: 25 - 20 = 5.
• Substitute the width and height determined in Step 2 and 3 into the Pythagorean Theorem which states that a^2 + b^2 = c^2, where a and b are the base and vertical sides of a triangle, and c is the hypotenuse. Solve the equation to determine the sloping width of the roof. For example, if the horizontal width is 10 feet and the height is 5 feet, the hypotenuse is 11.18 feet -- 10^2 + 5^2 = 125. The square root of 125 is 11.18.
• Multiply the hypotenuse by the length of the roof. The result is the surface area of the sloping roof. For example, if the sloping surface is 11.18 feet wide, and the roof is 30 feet long, the sloping area is 335.4 square feet -- 11.18 x 30 = 335.4.
## Tips & Warnings
• Measure the length and width of the area covered by the roof at ground level. If the walls are vertical the dimensions will be identical to those at roof level.
• Calculate the difference in height between the lowest and highest points on the roof by counting the number of rows of bricks and then measuring an equal number of rows at ground level.
• If you climb up to measure the roof take all necessary safety precautions. Check ladders and harnesses before you use them, and have another person with you in case of accidents.
## References
• Photo Credit Hemera Technologies/AbleStock.com/Getty Images
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# Polynomial components
## Understanding Polynomials
How do you tell when you’re working with polynomials? And how do you explain different components in a polynomial?
The polynomial definition is quite simple. The prefix “Poly” means “many” and polynomials are sums of variables and exponents.
You can divide up a polynomial into “terms”, separated by each part that is being added. Polynomial terms do not have square roots of variables, factional powers, nor does it have variables in the denominator of any fractions it may have. The polynomial terms can only have variables with exponents that are whole-numbers.
In general, polynomials are written with its terms being ordered in decreasing order of exponents. The term with the largest exponent goes first, followed by the term with the next highest exponent and so on till you reach a constant term.
Although polynomials can range from one to a very large amount of terms, you may hear specific names referencing to polynomials of a set number of terms. They are as follows:
- Monomial: A one-term polynomial (e.g. $3x$)
- Bionomial: A two-term polynomial (e.g. $x^4 + 3x$)
- Trinomial: A three-term polynomial (e.g. $x^4 + 2x^2 + 3x$)
If you see the above three names being used in a question, don’t worry. It’s really just another, more specific, word for polynomials.
Let’s learn how to describe the different parts of polynomials through an example.
Question:
Describe the following polynomial: $2x^5 y^2-3xy^2-7$
Solution:
Terms: $2x^5 y^2$, $-3xy^2$, $-7$
To get polynomial coefficients, take the numbers in front of each terms.
Coefficients: $2, -3, -7$
Leading coefficient: $2$
The get the degree of each terms, add the exponents in each terms to get their degrees.
Degree: $7, 3, 0$
Leading term is the term that has highest degree
Leading term: $2x^5 y^2$
Degree of polynomial is the degree of the leading term.
Degree of polynomial: 7
Constant: $- 7$
## How to find polynomial terms
Let’s look into this one-by-one. The idea of “terms” is explained in the previous section above, where we explored what is and isn’t a polynomial term. To recap, polynomial terms do not have square roots of variables, factional powers, nor does it have variables in the denominator of any fractions it may have. The polynomial terms can only have variables with exponents that are whole-numbers.
## How to find polynomial coefficients
Polynomial coefficients are the numbers that come before a term. Terms usually have a number and a variable (e.g. $2x^2$, where $2$ is the number, and $x$ is the variable). The number portion is the coefficient. So what is a leading coefficient? Simply stated, it’s the coefficient of the leading term. Since polynomials are sorted based on descending powers of exponents, the first term’s (after a polynomial’s terms are sorted properly) coefficient is the leading coefficient.
## How to find polynomial degree
The degrees are the exponents of terms. When you have multiple exponents in a term, such as in this example, you’ll have to add together their exponents to find the total exponents of that term. Keep in mind that if the variable has no exponent, it actually has an exponent of 1. It the term does not have variable, it is degree 0.
## How to find degree of polynomial
The leading term is the term that has the highest polynomial degree. In our case, since the exponents of “$5$” and “$2$” add together to get $7$, it has a higher degree than any of the other polynomial terms. Therefore, the leading term is the whole of: $2x^5 y^2$. The leading term helps us find the degree of polynomial. It is simply the degree of the leading term.
## How to find the constant
The last component is called the constant. The constant does not have a variable in it, and therefore remains constant no matter what the value of $x$ is when we are trying to evaluate the polynomial equation.
### Polynomial components
Polynomials can involve a long string of terms that are difficult to comprehend. So, before we dive into more complex polynomial concepts and calculations, we need to understand the components of polynomials and be able to identify its terms, coefficients, degree, leading term, and leading coefficient.
#### Lessons
• 1.
Describe the following polynomial:
a)
$2 {x^5}{y^2} - 3x{y^2} - 7$
Terms:
Coefficients:
Degree:
Degree of the polynomial:
b)
$4 {x^5} - {x^3} - 4x + 15$
Terms:
Coefficients:
Degree:
Degree of the polynomial: |
NCERT solution class 9 chapter 5 Introduction to Euclid’s Geometry exercise 5.1 mathematics
EXERCISE 5.1
Question 1:
Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In the following figure, if AB = PQ and PQ = XY, then AB = XY.
(i) False. Since through a single point, infinite number of lines can pass. In the following figure, it can be seen that there are infinite numbers of lines passing through a single point P.
(ii) False. Since through two distinct points, only one line can pass. In the following figure, it can be seen that there is only one single line that can pass through two distinct points P and Q.
(iii) True. A terminated line can be produced indefinitely on both the sides.
Let AB be a terminated line. It can be seen that it can be produced indefinitely on both the sides.
(iv)True. If two circles are equal, then their centre and circumference will coincide and hence, the radii will also be equal.
(v) True. It is given that AB and XY are two terminated lines and both are equal to a third line PQ. Euclid’s first axiom states that things which are equal to the same thing are equal to one another. Therefore, the lines AB and XY will be equal to each other.
Question 2:
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) parallel lines (ii) perpendicular lines (iii) line segment
(iv) radius of a circle (v) square
(i) Parallel Lines
If the perpendicular distance between two lines is always constant, then these are called parallel lines. In other words, the lines which never intersect each other are called parallel lines.
To define parallel lines, we must know about point, lines, and distance between the lines and the point of intersection.
(ii) Perpendicular lines
If two lines intersect each other at, then these are called perpendicular lines. We are required to define line and the angle before defining perpendicular lines.
(iii) Line segment
A straight line drawn from any point to any other point is called as line segment. To define a line segment, we must know about point and line segment.
It is the distance between the centres of a circle to any point lying on the circle. To define the radius of a circle, we must know about point and circle.
(v) Square
A square is a quadrilateral having all sides of equal length and all angles of same measure, i.e., To define square, we must know about quadrilateral, side, and angle.
Question 3:
Consider the two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C, which is between A and B.
(ii) There exists at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent?
Do they follow from Euclid’s postulates? Explain.
There are various undefined terms in the given postulates.
The given postulates are consistent because they refer to two different situations. Also, it is impossible to deduce any statement that contradicts any well known axiom and postulate.
These postulates do not follow from Euclid’s postulates. They follow from the axiom, “Given two distinct points, there is a unique line that passes through them”.
Question 4:
If a point C lies between two points A and B such that AC = BC, then prove that. Explain by drawing the figure.
It is given that,
AC = BC
(Equals are added on both sides) … (1)
Here, (BC + AC) coincides with AB. It is known that things which coincide with one another are equal to one another.
∴ BC + AC = AB … (2)
It is also known that things which are equal to the same thing are equal to one another. Therefore, from equations (1) and (2), we obtain
AC + AC = AB
2AC = AB
Question 5:
In the above question, point C is called a mid-point of line segment AB, prove that every line segment has one and only one mid-point.
Let there be two mid-points, C and D.
C is the mid-point of AB.
AC = CB
(Equals are added on both sides) … (1)
Here, (BC + AC) coincides with AB. It is known that things which coincide with one another are equal to one another.
∴ BC + AC = AB … (2)
It is also known that things which are equal to the same thing are equal to one another. Therefore, from equations (1) and (2), we obtain
AC + AC = AB
⇒ 2AC = AB … (3)
Similarly, by taking D as the mid-point of AB, it can be proved that
From equation (3) and (4), we obtain
2AC = 2AD (Things which are equal to the same thing are equal to one another.)
⇒ AC = AD (Things which are double of the same things are equal to one another.)
This is possible only when point C and D are representing a single point.
Hence, our assumption is wrong and there can be only one mid-point of a given line segment.
Question 6:
In the following figure, if AC = BD, then prove that AB = CD.
From the figure, it can be observed that
AC = AB + BC
BD = BC + CD
It is given that AC = BD
AB + BC = BC + CD (1)
According to Euclid’s axiom, when equals are subtracted from equals, the remainders are also equal.
Subtracting BC from equation (1), we obtain
AB + BC − BC = BC + CD − BC
AB = CD
Question 7:
Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)
Axiom 5 states that the whole is greater than the part. This axiom is known as a universal truth because it holds true in any field, and not just in the field of mathematics. Let us take two cases − one in the field of mathematics, and one other than that.
Case I
Let represent a whole quantity and only abc are parts of it.
t = a + b + c
Clearly, t will be greater than all its parts ab, and c.
Therefore, it is rightly said that the whole is greater than the part.
Case II
Let us consider the continent Asia. Then, let us consider a country India which belongs to Asia. India is a part of Asia and it can also be observed that Asia is greater than India. That is why we can say that the whole is greater than the part. This is true for anything in any part of the world and is thus a universal truth. |
Sovel Pre-Algebra Class
- x - = +
In mathematics, we can use the term 'rule', but in reality, there are no rules, only demonstrtions and proofs. Therefore, we must be able to demonstrate and prove each mathematical statement and expression before we can apply the tag 'rule' to it.
Listed below are five such demonstrations that discuss and illustrate the ideas of multiplying and dividing positive and negative values..
An Inductive Method Number Line/Opposite of Division Property of One Reciprocals Positive and Negative Exponents return to homework page
Remember, a proof can only build upon a foundation of other related proofs
An Inductive approach1 to demonstrating
that a negative multiplied by a negative equals a positive
+ · + = +
We normally work in a world of all positives. Notice the pattern of change as we decrease the multiplier
+10 +10 +10 +10 +10 +10 x 5 x +4 x +3 x +2 x +1 x 0 + 50 + 40 + 30 + 20 + 10 0
+ · - = -
If we continue this pattern, notice what happens to the product.
+10 +10 +10 +10 +10 x -1 x -2 x -3 x -4 x -5 - 10 - 20 - 30 - 40 - 50
- · + = -
We have demonstrated that a + · - = -; notice that a -· + = - is also true.
-10 -10 -10 -10 -10 -10 x 5 x +4 x +3 x +2 x +1 x 0 -50 -40 -30 -20 -10 0
- · - = +
If we continue this pattern, notice what happens to the product.
- 10 - 10 - 10 - 10 - 10 x -1 x -2 x -3 x -4 x -5 +10 +20 +30 +40 +50
1 Farrell, Margaret A. and Farmer, Walter A. Secondary Mathematics Instruction: An Integrated Approach. Janson Publications, Inc.: Providence, R.I., 1995: pages 94-96.
The Number Line and Opposites
Understanding Integer opposites can also demonstrate a negative times a negative equals a positive
an integer it's opposite it's opposite 5 -5 -(-5) + - +
The illustration above can be demonstrated visually by the tracking of these integers on a number line.
Division Property of One
Other Properties of real numbers can also demonstrate that a negative divided by a negative can equal a positive, such as the Division Property of One.
This can be stated as,
for every non-zero number a,
a ÷ 1 = a and a ÷ a = 1, such that any value divided by itself will equal a positive 1;
therefore, if a is positive, a positive divided by a positive equals a positive 1; and
if a is a negative value, a negative divided by a negative must, therefore, also equal a positive 1
Reciprocal
Certain mathematical definitions can also be used to demonstrate that a negative times a negative equals a positive. For example,
two numbers, like , whose product is 1 are called reciprocals.
The reciprocal of because = 1. Every non-zero rational number has exactly one reciprocal.
Therefore, the following negative number, and its reciprocal, also produce the product +1:
Positive and Negative Exponents
coefficient --> 4 x 2 <-- exponent variable
The exponent rests on a base. When two terms have the same base and are multiplying each other, you may add the exponents [as a shortcut to simplifying]. For example:
(4 2) (4 4) = 4 2 + 4 = 4 6 OR (4 5) (4 -3) = 4 5 + -3 = 4 2
It is important to remember that while the base is a factor, the exponent is not. The exponent simply tells us the number of times the base multiplies itself, as a factor.
The path path for converting a negative exponent to a positive exponent
x 4 = (1) (x) (x) (x) (x) = x 4 4 4 = (1) (4) (4) (4) (4) = 256 x 3 = (1) (x) (x) (x) = x 3 4 3 = (1) (4) (4) (4) = 64 x 2 = (1) (x) (x) = x 2 4 2 = (1) (4) (4) = 16 x 1 = (1) (x) = x 1 4 1 = (1) (4) = 4 x 0 = 1 = 1 4 0 = 1 = 1 x -1 = (1) (1/x 1) = 1/x 1 4 -1 = (1) (1/4 1) = 1/4 x -2 = (1) (1/x 2) = 1/x 2 4 -2 = (1) (1/4 2) = 1/16 x -3 = (1) (1/x 3) = 1/x 3 4 -3 = (1) (1/4 3 ) = 1/64
\ |
# AP Statistics Curriculum 2007 GLM MultLin
(Difference between revisions)
Revision as of 00:24, 19 February 2008 (view source)IvoDinov (Talk | contribs)← Older edit Current revision as of 23:56, 1 May 2013 (view source)IvoDinov (Talk | contribs) (→Categorical Variables in Multiple Regression) (17 intermediate revisions not shown) Line 1: Line 1: ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Multiple Linear Regression == ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Multiple Linear Regression == - In the previous sections we saw how to study the relations in bivariate designs. Now we extend that to any finite number of varaibles (mulitvariate case). + In the [[AP_Statistics_Curriculum_2007#Chapter_X:_Correlation_and_Regression | previous sections]], we saw how to study the relations in bivariate designs. Now we extend that to any finite number of variables (multivariate case). === Multiple Linear Regression === === Multiple Linear Regression === Line 10: Line 10: The coefficient $\beta_0$ is the intercept ("constant" term) and $\beta_i$s are the respective parameters of the '' p'' independent variables. There are ''p+1'' parameters to be estimated in the multilinear regression. The coefficient $\beta_0$ is the intercept ("constant" term) and $\beta_i$s are the respective parameters of the '' p'' independent variables. There are ''p+1'' parameters to be estimated in the multilinear regression. - * Multilinear vs. non-linear regression: This multilinear regression method is "linear" because the relation of the response (the dependent variable $Y$) to the independent variables is assumed to be a [http://en.wikipedia.org/wiki/Linear_function linear function] of the parameters $\beta_i$. Note that multilinear regression is a linear modeling technique '''not''' because is that the graph of $Y = \beta_{0}+\beta x$ is a straight line '''nor''' because $Y$ is a linear function of the ''X'' variables. But the "linear" terms refers to the fact that $Y$ can be considered a linear function of the parameters ( $\beta_i$), even though it is not a linear function of $X$. Thus, any model like + * Multilinear vs. non-linear regression: This multilinear regression method is "linear" because the relation of the response (the dependent variable $Y$) to the independent variables is assumed to be a [http://en.wikipedia.org/wiki/Linear_function linear function] of the parameters $\beta_i$. Note that multilinear regression is a linear modeling technique '''not''' because that the graph of $Y = \beta_{0}+\beta x$ is a straight line '''nor''' because $Y$ is a linear function of the ''X'' variables. But the "linear" term refers to the fact that $Y$ can be considered a linear function of the parameters ( $\beta_i$), even though it is not a linear function of $X$. Thus, any model like - : $Y = \beta_o + \beta_1 x + \beta_2 x^2 + \varepsilon$ + : $Y = \beta_o + \beta_1 x + \beta_2 x^2 + \varepsilon$
is still one of the '''linear''' regression, that is, linear in $x$ and $x^2$ respectively, even though the graph on $x$ by itself is not a straight line. - is still one of '''linear''' regression, that is, linear in x and $x^2$ respectively, even though the graph on $x$ by itself is not a straight line. + ===Parameter Estimation in Multilinear Regression=== + A multilinear regression with ''p'' coefficients and the regression intercept β0 and ''n'' data points (sample size), with $n\geq (p+1)$ allows construction of the following vectors and matrix with associated standard errors: - ===Approach=== + :$\begin{bmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{n} \end{bmatrix} = \begin{bmatrix} 1 & x_{11} & x_{12} & \dots & x_{1p} \\ 1 & x_{21} & x_{22} & \dots & x_{2p} \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & x_{n1} & x_{n2} & \dots & x_{np} \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{bmatrix} + \begin{bmatrix} \varepsilon_1\\ \varepsilon_2\\ \vdots\\ \varepsilon_n \end{bmatrix}$ - Models & strategies for solving the problem, data understanding & inference. + - * TBD + or, in '''vector-matrix notation''' - ===Model Validation=== + :$\ y = \mathbf{X}\cdot\beta + \varepsilon.\,$ - Checking/affirming underlying assumptions. + Each data point can be given as $(\vec x_i, y_i)$, $i=1,2,\dots,n.$. For n = p, standard errors of the parameter estimates could not be calculated. For n less than p, parameters could not be calculated. - * TBD + * '''Point Estimates''': The estimated values of the parameters $\beta_i$ are given as + :$\widehat{\beta}$$=(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T {\vec y}$ - ===Computational Resources: Internet-based SOCR Tools=== + * '''Residuals''': The residuals, representing the difference between the observations and the model's predictions, are required to analyse the regression and are given by: - * TBD + + :$\hat\vec\varepsilon = \vec y - \mathbf{X} \hat\beta\,$ + + The standard deviation, $\hat \sigma$ for the model is determined from + + :${\hat \sigma = \sqrt{ \frac {\hat\vec\varepsilon^T \hat\vec\varepsilon} {n-p-1}} = \sqrt {\frac{{ \vec y^T \vec y - \hat\vec\beta^T \mathbf{X}^T \vec y}}{{n - p - 1}}} }$ + + The variance in the errors is Chi-square distributed: + :$\frac{(n-p-1)\hat\sigma^2}{\sigma^2} \sim \chi_{n-p-1}^2$ + + * '''Interval Estimates''': The $100(1-\alpha)%$ [[AP_Statistics_Curriculum_2007_Estim_S_Mean#Interval_Estimation_of_a_Population_Mean | confidence interval]] for the parameter, $\beta_i$, is computed as follows: + + :${\widehat \beta_i \pm t_{\frac{\alpha }{2},n - p - 1} \hat \sigma \sqrt {(\mathbf{X}^T \mathbf{X})_{ii}^{ - 1} } }$, + + where ''t'' follows the [[AP_Statistics_Curriculum_2007_StudentsT | Student's t-distribution]] with $n-p-1$ degrees of freedom and $(\mathbf{X}^T \mathbf{X})_{ii}^{ - 1}$ denotes the value located in the $i^{th}$ row and column of the matrix. + + The '''regression sum of squares''' (or sum of squared residuals) ''SSR'' (also commonly called ''RSS'') is given by: + + :${\mathit{SSR} = \sum {\left( {\hat y_i - \bar y} \right)^2 } = \hat\beta^T \mathbf{X}^T \vec y - \frac{1}{n}\left( { \vec y^T \vec u \vec u^T \vec y} \right)}$, + + where $\bar y = \frac{1}{n} \sum y_i$ and $\vec u$ is an ''n'' by 1 unit vector (i.e. each element is 1). Note that the terms $y^T u$ and $u^T y$ are both equivalent to $\sum y_i$, and so the term $\frac{1}{n} y^T u u^T y$ is equivalent to $\frac{1}{n}\left(\sum y_i\right)^2$. + + The '''error''' (or explained)''' sum of squares''' (''ESS'') is given by: + + :${\mathit{ESS} = \sum {\left( {y_i - \hat y_i } \right)^2 } = \vec y^T \vec y - \hat\beta^T \mathbf{X}^T \vec y}.$ + + The '''total sum of squares''' ('''TSS''') is given by + + :${\mathit{TSS} = \sum {\left( {y_i - \bar y} \right)^2 } = \vec y^T \vec y - \frac{1}{n}\left( { \vec y^T \vec u \vec u^T \vec y} \right) = \mathit{SSR}+ \mathit{ESS}}.$ + + ===Partial Correlations=== + For a given linear model +
$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \cdots +\beta_p X_p + \varepsilon$
+ the partial correlation between $X_1$ and ''Y'' given a set of ''p-1'' controlling variables $Z = \{X_2, X_3, \cdots, X_p\}$, denoted by $\rho_{YX_1|Z}$, is the correlation between the residuals ''R''''X'' and ''R''''Y'' resulting from the linear regression of ''X'' with '''Z''' and that of ''Y'' with '''Z''', respectively. The first-order partial correlation is just the difference between a correlation and the product of the removable correlations divided by the product of the coefficients of alienation of the removable correlations. + + * Partial correlation coefficients for three variables is calculated from the pairwise simple correlations. + : If, $Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \varepsilon$, + : then the partial correlation between $Y$ and $X_2$, adjusting for $X_1$ is: + : $\rho_{YX_2|X_1} = \frac{\rho_{YX_2} - \rho_{YX_1}\times \rho_{X_2X_1}}{\sqrt{1- \rho_{YX_1}^2}\sqrt{1-\rho_{X_2X_1}^2}}$ + + * In general, the sample partial correlation is + :$\hat{\rho}_{XY\cdot\mathbf{Z}}=\frac{N\sum_{i=1}^N r_{X,i}r_{Y,i}-\sum_{i=1}^N r_{X,i}\sum r_{Y,i}} + {\sqrt{N\sum_{i=1}^N r_{X,i}^2-\left(\sum_{i=1}^N r_{X,i}\right)^2}~\sqrt{N\sum_{i=1}^N r_{Y,i}^2-\left(\sum_{i=1}^N r_{Y,i}\right)^2}},$ where the [http://en.wikipedia.org/wiki/Partial_correlation residuals] $r_{X,i}$ and $r_{X,i}$ are given by: + :: $r_{X,i} = x_i - \langle\mathbf{w}_X^*,\mathbf{z}_i \rangle$ + :: $r_{Y,i} = y_i - \langle\mathbf{w}_Y^*,\mathbf{z}_i \rangle$, + :: with $x_i$, $y_i$ and $z_i$ denoting the random (IID) samples of some joint probability distribution over X, Y and Z. + + ====Computing the partial correlations==== + The ''n''th-order partial correlation (|'''Z'''| = ''n'') can be computed from three (''n'' - 1)th-order partial correlations. The ''0''th-order partial correlation $\rho_{YX|\empty}$ is defined to be the regular [http://en.wikipedia.org/wiki/Correlation correlation coefficient] $\rho_{YX}$. + + For any $Z_0 \in \mathbf{Z}$: + :$\rho_{XY| \mathbf{Z} } = + \frac{\rho_{XY| \mathbf{Z}\setminus\{Z_0\}} - \rho_{XZ_0| \mathbf{Z}\setminus\{Z_0\}}\rho_{YZ_0 | \mathbf{Z}\setminus\{Z_0\}}} + {\sqrt{1-\rho_{XZ_0 |\mathbf{Z}\setminus\{Z_0\}}^2} \sqrt{1-\rho_{YZ_0 | \mathbf{Z}\setminus\{Z_0\}}^2}}.$ + + Implementing this computation recursively yields an exponential time [http://en.wikipedia.org/wiki/Computational_complexity_theory complexity]. + + Note in the case where Z is a single variable, this reduces to: + :$\rho_{XY | Z } = + \frac{\rho_{XY} - \rho_{XZ}\rho_{YZ}} + {\sqrt{1-\rho_{XZ}^2} \sqrt{1-\rho_{YZ}^2}}.$ + + ===Categorical Variables in Multiple Regression=== + + When using categorical variables with more than two levels in a multiple regression modeling, we need to make sure the results are correctly interpreted. For categorical variables with more than 2 levels, a number of separate, dichotomous variables need to be created – this is called “dummy coding” or “dummy variables”. + + Dichotomous categorical predictor variables, variables with two levels, may be directly entered as predictor or predicted variables in a multiple regression model. Their use in multiple regression is a direct extension of their use in simple linear regression. The interpretation of their regression weights depends upon how the variables are coded. If dichotomous variables are coded as 0 and 1, their regression weights are added or subtracted to the predicted value of the response variable, Y, depending upon whether it is positive or negative. + + If a dichotomous variable is coded as -1 and 1, then if the regression weight is positive, it is subtracted from the group coded as -1 and added to the group coded as 1. If a regression weight is negative, then addition and subtraction is reversed. Dichotomous variables are also included in hypothesis tests for R2 change like any other quantitative variable. + + Adding variables to a linear regression model will always increase the (unadjusted) $$R^2$$ value. If additional predictor variables are correlated with the predictor variables already in the model, then the combined results may be difficult to predict. In some cases, the combined result will only slightly improve the prediction, whereas in other cases, a much better prediction may be obtained by combining two correlated variables. + + If the additional predictor variables are uncorrelated (their correlation is trivial) with the predictor variables already in the model, then the result of adding additional variables to the regression model is easy to predict. Thus, the $$R^2$$ change will be equal to the correlation coefficient squared between the added variable and predicted variable. In this case it makes no difference what order the predictor variables are entered into the prediction model. For example, if $$X_1$$ and $$X_2$$ were uncorrelated $$\rho_{12} = 0$$ and $$\rho^2_{1y} = 0.4$$ and $$\rho^2_{2y} = 0.5$$, then $$R^2$$ for $$X_1$$ and $$X_2$$ would equal $$0.4 + 0.5 = 0.9$$. The value for $$R^2$$ change for $$X_2$$, given that X1 is included in the model, would be 0.5. The value for $$R^2$$ change for $$X_2$$ given no variable was in the model would be 0.5. It would make no difference at what stage $$X_2$$ was entered into the model, the value for $$R^2$$ change would always be 0.5. Similarly, the $$R^2$$ change value for $$X_1$$ would always be 0.4. Because of this relationship, uncorrelated predictor variables will be preferred, when possible. + + Look at the [[SOCR_Data_AD_BiomedBigMetadata| Modeling and Analysis of Clinical, Genetic and Imaging Data of Alzheimer’s Disease dataset]]. It is fairly clear that '''DX_Conversion''' could be directly entered into a regression model predicting '''MMSE''', because it is dichotomous. The problem is how to deal with the two categorical predictor variables with more than two levels (e.g., '''GDTOTAL'''). + + '''Dummy Coding''' refers to making many dichotomous variables out of one multilevel categorical variable. Because categorical predictor variables cannot be entered directly into a regression model and be meaningfully interpreted, some other method of dealing with information of this type must be developed. In general, a categorical variable with k levels will be transformed into k-1 variables each with two levels. For example, if a categorical variable had 4 levels, then 3 dichotomous (dummy) variables could be constructed that would contain the same information as the single categorical variable. Dichotomous variables have the advantage that they can be directly entered into the regression model. The process of creating dichotomous variables from categorical variables is called dummy coding. + + Depending upon how the dichotomous variables are constructed, additional information can be gleaned from the analysis. In addition, careful construction will result in uncorrelated dichotomous variables. These variables have the advantage of simplicity of interpretation and are preferred to correlated predictor variables. + + For example, when using dummy coding with three levels, the simplest case of dummy coding is when the categorical variable has three levels and is converted to two dichotomous variables. School in the example data has three levels, 1=Math, 2=Biology, and 3=Engineering. This variable could be dummy coded into two variables, one called Math and one called Biology. If School = 1, then Math would be coded with a 1 and Biology with a 0. If School=2, then Math would be coded with a 0 and Biology would be coded with a 1. If School=3, then both Math and Biology would be coded with a 0. The dummy coding is represented below. + +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + |+ Dummy Coded Variables + |- + | colspan=2|Subject||Math||Biology + |- + | Math||1||1||0 + |- + | Biology||2||0||1 + |- + | Engineering||3||0||0 + |}
+ + See the [[SOCR_EduMaterials_AnalysisActivities_MLR |SOCR Regression analysis]] and compare it against [[SOCR_EduMaterials_AnalysisActivities_ANOVA_1|SOCR ANOVA]]. + + Dummy Coding into Independent Variables, and the selection of an appropriate set of dummy codes, will result in new variables that are uncorrelated or independent of each other. In the case when the categorical variable has three levels this can be accomplished by creating a new variable where one level of the categorical variable is assigned the value of -2 and the other levels are assigned the value of 1. The signs are arbitrary and may be reversed, that is, values of 2 and -1 would work equally well. The second variable created as a dummy code will have the level of the categorical variable coded as -2 given the value of 0 and the other values recoded as 1 and -1. In all cases the sum of the dummy coded variable will be zero. + + We can directly interpret each of the new dummy-coded variables, called a contrast, and compare levels coded with a positive number to levels coded with a negative number. Levels coded with a zero are not included in the interpretation. + + For example, School in the example data has three levels, 1=Math, 2=Biology, and 3=Engineering. This variable could be dummy coded into two variables, one called Engineering (comparing the Engineering School with the other two Schools) and one called Math_vs_Bio (comparing Math versus Biology schools). The Engineering contrast would create a variable where all members of the Engineering Department would be given a value of -2 and all members of the other two Schools would be given a value of 1. The Math_vs_Bio contrast would assign a value of 0 to members of the Engineering Department, 1 divided by the number of members of the Math Department to member of the Math Department, and -1 divided by the number of members of the Biology Department to members of the Biology Department. The Math_vs_Bio variable could be coded as 1 and -1 for Math and Biology respectively, but the recoded variable would no longer be uncorrelated with the first dummy coded variable (Engineering). In most practical applications, it makes little difference whether the variables are correlated or not, so the simpler 1 and -1 coding is generally preferred. The contrasts are summarized in the following table. + +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + |+ Dummy Coded Variables + |- + | colspan=2|School||Engineering||Math_vs_Bio + |- + | Math||1||1||1/N1 = 1/12= 0.0833 + |- + | Biology||2||1||-1/N2 = -1/7 = -0.1429 + |- + | Engineering||3||-2||0 + |}
+ + Note that the correlation coefficient between the two contrasts is zero. The correlation between the Engineering contrast and Salary is -.585 with a squared correlation coefficient of .342. This correlation coefficient has a significance level of .001. The correlation coefficient between the Math_vs_Bio contrast and Salary is -.150 with a squared value of .023. + + Generate the corresponding [[AP_Statistics_Curriculum_2007_ANOVA_1Way#ANOVA_Table|SOCR ANOVA table]]. + + Show that the significance level is identical to the value when each contrast was entered last into the regression model. In this case the Engineering contrast was significant and the Math_vs_Bio contrast was not. The interpretation of these results would be that the Engineering Department was paid significantly more than the Math and Biology Schools, but that no significant differences in salary were found between the Math and Biology Schools. + If a categorical variable had four levels, three dummy coded contrasts would be necessary to use the categorical variable in a regression analysis. For example, suppose that a researcher at a pain center did a study with 4 groups of four patients each (N is being deliberately kept small). The dependent measure is subjective experience of pain. The 4 groups consisted of 4 different treatment conditions. + +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + ! Group||Treatment + |- + |1||None + |- + |2||Placebo + |- + |3||Psychotherapy + |- + |4||Acupuncture + |}
+ + An independent contrast is a contrast that is not a linear combination of any other set of contrasts. Any set of independent contrasts would work equally well if the end result was the simultaneous test of the five contrasts, as in an ANOVA. One of the many possible examples is presented below. + +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + |+ Dummy Coded Variables + |- + | colspan=2|Group||C1||C2||C3||C4 + |- + | None||1||0||0||0||0 + |- + | Placebo||2||1||0||0||0 + |- + | Psychotherapy||3||0||1||0||0 + |- + | Acupuncture||4||0||0||1||0 + |}
+ + Application of this dummy coding in a regression model entering all contrasts in a single block would result in an ANOVA table similar to the one obtained using Means, ANOVA, or General Linear Model. + + This solution would not be ideal, however, because there is considerable information available by setting the contrasts to test specific hypotheses. The levels of the categorical variable generally dictate the structure of the contrasts. In the example study, it makes sense to contrast the two control groups (1 and 2) with the other four experimental groups (3 and 4). Any two numbers would work, one assigned to groups 1 and 2 and the others assigned to the other four groups, but it is conventional to have the sum of the contrasts equal to zero. One contrast that meets this criterion would be (-2, -2, 1, 1). + Generally it is easiest to set up contrasts within subgroups of the first contrast. For example, a second contrast might test whether there are differences between the two control groups. This contrast would appear as (1, -1, 0, 0). As can be seen, this would be a contrast within the experimental treatment groups. Within the non-drug treatment, a contrast comparing Group 3 with Group 4 might be appropriate (0, 0, 1, -1). Combined, the contrasts are given in the following table. + +
+ {| class="wikitable" style="text-align:center; width:75%" border="1" + |+ Dummy Coded Variables + |- + | colspan=2|Group||C1||C2||C3 + |- + | None||1||-2||1||0 + |- + | Placebo||2||-2||-1||0 + |- + | Psychotherapy||3||1||0||1 + |- + | Acupuncture||4||1||0||1 + |}
+ + Equal sample sizes are important as the results are much cleaner when the sample sizes are presumed to be equal. However equal sample size are not common in real applications, even in well-designed experiments. Unequal sample size makes the effects no longer independent. This implies that it makes difference in hypothesis testing when the effects are added into the model, first, middle, or last. The same dummy coding that was applied to equal sample sizes will now be applied to the original data with unequal sample sizes. ===Examples=== ===Examples=== - Computer simulations and real observed data. - * TBD + We now demonstrate the use of [[SOCR_EduMaterials_AnalysisActivities_MLR | SOCR Multilinear regression applet]] to analyze multivariate data. - + - ===Hands-on activities=== + - Step-by-step practice problems. + - * TBD + ====Earthquake Modeling==== + This is an example where the relation between variables may not be linear or explanatory. In the [[AP_Statistics_Curriculum_2007_GLM_Regress | simple linear regression case]], we were able to compute by hand some (simple) examples. Such calculations are much more involved in the multilinear regression situations. Thus we demonstrate multilinear regression only using the [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Multiple Regression Analysis Applet]. + + Use the SOCR California Earthquake dataset to investigate whether Earthquake magnitude (dependent variable) can be predicted by knowing the longitude, latitude, distance and depth of the quake. Clearly, we do not expect these predictors to have a strong effect on the earthquake magnitude, so we expect the coefficient parameters not to be significantly distinct from zero (null hypothesis). SOCR Multilinear regression applet reports this model: + + : $Magnitude = \beta_o + \beta_1\times Close+ \beta_2\times Depth+ \beta_3\times Longitude+ \beta_4\times Latitude + \varepsilon.$ + + : $Magnitude = 2.320 + 0.001\times Close -0.003\times Depth -0.035\times Longitude -0.028\times Latitude + \varepsilon.$ + +
[[Image:SOCR_EBook_Dinov_GLM_MLR_021808_Fig1.jpg|500px]] + [[Image:SOCR_EBook_Dinov_GLM_MLR_021808_Fig2.jpg|500px]]
+ + ====Multilinear Regression on Consumer Price Index==== + Using the [[SOCR_Data_Dinov_021808_ConsumerPriceIndex | SOCR Consumer Price Index Dataset]] we can explore the relationship between the prices of various products and commodities. For example, regressing '''Gasoline''' on the following three predictor prices: '''Orange Juice''', '''Fuel''' and '''Electricity''' illustrates significant effects of all these variables as significant explanatory prices (at $\alpha=0.05$) for the cost of ''Gasoline'' between 1981 and 2006. + + : $Gasoline = 0.083 -0.190\times Orange +0.793\times Fuel +0 .013\times Electricity +$ + +
[[Image:SOCR_EBook_Dinov_GLM_MLR_021808_Fig3.jpg|500px]] + [[Image:SOCR_EBook_Dinov_GLM_MLR_021808_Fig4.jpg|500px]]
+ + ====2011 Best Jobs in the US==== + Repeat the multiliniear regression analysis using hte [[SOCR_Data_2011_US_JobsRanking| Ranking Dataset of the Best and Worst USA Jobs for 2011]].
- ===References=== + - * TBD + ===[[EBook_Problems_GLM_MultLin|Problems]]===
## General Advance-Placement (AP) Statistics Curriculum - Multiple Linear Regression
In the previous sections, we saw how to study the relations in bivariate designs. Now we extend that to any finite number of variables (multivariate case).
### Multiple Linear Regression
We are interested in determining the linear regression, as a model, of the relationship between one dependent variable Y and many independent variables Xi, i = 1, ..., p. The multilinear regression model can be written as
$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \cdots +\beta_p X_p + \varepsilon$, where $\varepsilon$ is the error term.
The coefficient β0 is the intercept ("constant" term) and βis are the respective parameters of the p independent variables. There are p+1 parameters to be estimated in the multilinear regression.
• Multilinear vs. non-linear regression: This multilinear regression method is "linear" because the relation of the response (the dependent variable Y) to the independent variables is assumed to be a linear function of the parameters βi. Note that multilinear regression is a linear modeling technique not because that the graph of Y = β0 + βx is a straight line nor because Y is a linear function of the X variables. But the "linear" term refers to the fact that Y can be considered a linear function of the parameters ( βi), even though it is not a linear function of X. Thus, any model like
$Y = \beta_o + \beta_1 x + \beta_2 x^2 + \varepsilon$
is still one of the linear regression, that is, linear in x and x2 respectively, even though the graph on x by itself is not a straight line.
### Parameter Estimation in Multilinear Regression
A multilinear regression with p coefficients and the regression intercept β0 and n data points (sample size), with $n\geq (p+1)$ allows construction of the following vectors and matrix with associated standard errors:
$\begin{bmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{n} \end{bmatrix} = \begin{bmatrix} 1 & x_{11} & x_{12} & \dots & x_{1p} \\ 1 & x_{21} & x_{22} & \dots & x_{2p} \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & x_{n1} & x_{n2} & \dots & x_{np} \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{bmatrix} + \begin{bmatrix} \varepsilon_1\\ \varepsilon_2\\ \vdots\\ \varepsilon_n \end{bmatrix}$
or, in vector-matrix notation
$\ y = \mathbf{X}\cdot\beta + \varepsilon.\,$
Each data point can be given as $(\vec x_i, y_i)$, $i=1,2,\dots,n.$. For n = p, standard errors of the parameter estimates could not be calculated. For n less than p, parameters could not be calculated.
• Point Estimates: The estimated values of the parameters βi are given as
$\widehat{\beta}$$=(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T {\vec y}$
• Residuals: The residuals, representing the difference between the observations and the model's predictions, are required to analyse the regression and are given by:
$\hat\vec\varepsilon = \vec y - \mathbf{X} \hat\beta\,$
The standard deviation, $\hat \sigma$ for the model is determined from
${\hat \sigma = \sqrt{ \frac {\hat\vec\varepsilon^T \hat\vec\varepsilon} {n-p-1}} = \sqrt {\frac{{ \vec y^T \vec y - \hat\vec\beta^T \mathbf{X}^T \vec y}}{{n - p - 1}}} }$
The variance in the errors is Chi-square distributed:
$\frac{(n-p-1)\hat\sigma^2}{\sigma^2} \sim \chi_{n-p-1}^2$
• Interval Estimates: The 100(1 − α)% confidence interval for the parameter, βi, is computed as follows:
${\widehat \beta_i \pm t_{\frac{\alpha }{2},n - p - 1} \hat \sigma \sqrt {(\mathbf{X}^T \mathbf{X})_{ii}^{ - 1} } }$,
where t follows the Student's t-distribution with np − 1 degrees of freedom and $(\mathbf{X}^T \mathbf{X})_{ii}^{ - 1}$ denotes the value located in the ith row and column of the matrix.
The regression sum of squares (or sum of squared residuals) SSR (also commonly called RSS) is given by:
${\mathit{SSR} = \sum {\left( {\hat y_i - \bar y} \right)^2 } = \hat\beta^T \mathbf{X}^T \vec y - \frac{1}{n}\left( { \vec y^T \vec u \vec u^T \vec y} \right)}$,
where $\bar y = \frac{1}{n} \sum y_i$ and $\vec u$ is an n by 1 unit vector (i.e. each element is 1). Note that the terms yTu and uTy are both equivalent to $\sum y_i$, and so the term $\frac{1}{n} y^T u u^T y$ is equivalent to $\frac{1}{n}\left(\sum y_i\right)^2$.
The error (or explained) sum of squares (ESS) is given by:
${\mathit{ESS} = \sum {\left( {y_i - \hat y_i } \right)^2 } = \vec y^T \vec y - \hat\beta^T \mathbf{X}^T \vec y}.$
The total sum of squares (TSS) is given by
${\mathit{TSS} = \sum {\left( {y_i - \bar y} \right)^2 } = \vec y^T \vec y - \frac{1}{n}\left( { \vec y^T \vec u \vec u^T \vec y} \right) = \mathit{SSR}+ \mathit{ESS}}.$
### Partial Correlations
For a given linear model
$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \cdots +\beta_p X_p + \varepsilon$
the partial correlation between X1 and Y given a set of p-1 controlling variables $Z = \{X_2, X_3, \cdots, X_p\}$, denoted by $\rho_{YX_1|Z}$, is the correlation between the residuals RX and RY resulting from the linear regression of X with Z and that of Y with Z, respectively. The first-order partial correlation is just the difference between a correlation and the product of the removable correlations divided by the product of the coefficients of alienation of the removable correlations.
• Partial correlation coefficients for three variables is calculated from the pairwise simple correlations.
If, $Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \varepsilon$,
then the partial correlation between Y and X2, adjusting for X1 is:
$\rho_{YX_2|X_1} = \frac{\rho_{YX_2} - \rho_{YX_1}\times \rho_{X_2X_1}}{\sqrt{1- \rho_{YX_1}^2}\sqrt{1-\rho_{X_2X_1}^2}}$
• In general, the sample partial correlation is
$\hat{\rho}_{XY\cdot\mathbf{Z}}=\frac{N\sum_{i=1}^N r_{X,i}r_{Y,i}-\sum_{i=1}^N r_{X,i}\sum r_{Y,i}} {\sqrt{N\sum_{i=1}^N r_{X,i}^2-\left(\sum_{i=1}^N r_{X,i}\right)^2}~\sqrt{N\sum_{i=1}^N r_{Y,i}^2-\left(\sum_{i=1}^N r_{Y,i}\right)^2}},$ where the residuals rX,i and rX,i are given by:
$r_{X,i} = x_i - \langle\mathbf{w}_X^*,\mathbf{z}_i \rangle$
$r_{Y,i} = y_i - \langle\mathbf{w}_Y^*,\mathbf{z}_i \rangle$,
with xi, yi and zi denoting the random (IID) samples of some joint probability distribution over X, Y and Z.
#### Computing the partial correlations
The nth-order partial correlation (|Z| = n) can be computed from three (n - 1)th-order partial correlations. The 0th-order partial correlation $\rho_{YX|\empty}$ is defined to be the regular correlation coefficient ρYX.
For any $Z_0 \in \mathbf{Z}$:
$\rho_{XY| \mathbf{Z} } = \frac{\rho_{XY| \mathbf{Z}\setminus\{Z_0\}} - \rho_{XZ_0| \mathbf{Z}\setminus\{Z_0\}}\rho_{YZ_0 | \mathbf{Z}\setminus\{Z_0\}}} {\sqrt{1-\rho_{XZ_0 |\mathbf{Z}\setminus\{Z_0\}}^2} \sqrt{1-\rho_{YZ_0 | \mathbf{Z}\setminus\{Z_0\}}^2}}.$
Implementing this computation recursively yields an exponential time complexity.
Note in the case where Z is a single variable, this reduces to:
$\rho_{XY | Z } = \frac{\rho_{XY} - \rho_{XZ}\rho_{YZ}} {\sqrt{1-\rho_{XZ}^2} \sqrt{1-\rho_{YZ}^2}}.$
### Categorical Variables in Multiple Regression
When using categorical variables with more than two levels in a multiple regression modeling, we need to make sure the results are correctly interpreted. For categorical variables with more than 2 levels, a number of separate, dichotomous variables need to be created – this is called “dummy coding” or “dummy variables”.
Dichotomous categorical predictor variables, variables with two levels, may be directly entered as predictor or predicted variables in a multiple regression model. Their use in multiple regression is a direct extension of their use in simple linear regression. The interpretation of their regression weights depends upon how the variables are coded. If dichotomous variables are coded as 0 and 1, their regression weights are added or subtracted to the predicted value of the response variable, Y, depending upon whether it is positive or negative.
If a dichotomous variable is coded as -1 and 1, then if the regression weight is positive, it is subtracted from the group coded as -1 and added to the group coded as 1. If a regression weight is negative, then addition and subtraction is reversed. Dichotomous variables are also included in hypothesis tests for R2 change like any other quantitative variable.
Adding variables to a linear regression model will always increase the (unadjusted) $$R^2$$ value. If additional predictor variables are correlated with the predictor variables already in the model, then the combined results may be difficult to predict. In some cases, the combined result will only slightly improve the prediction, whereas in other cases, a much better prediction may be obtained by combining two correlated variables.
If the additional predictor variables are uncorrelated (their correlation is trivial) with the predictor variables already in the model, then the result of adding additional variables to the regression model is easy to predict. Thus, the $$R^2$$ change will be equal to the correlation coefficient squared between the added variable and predicted variable. In this case it makes no difference what order the predictor variables are entered into the prediction model. For example, if $$X_1$$ and $$X_2$$ were uncorrelated $$\rho_{12} = 0$$ and $$\rho^2_{1y} = 0.4$$ and $$\rho^2_{2y} = 0.5$$, then $$R^2$$ for $$X_1$$ and $$X_2$$ would equal $$0.4 + 0.5 = 0.9$$. The value for $$R^2$$ change for $$X_2$$, given that X1 is included in the model, would be 0.5. The value for $$R^2$$ change for $$X_2$$ given no variable was in the model would be 0.5. It would make no difference at what stage $$X_2$$ was entered into the model, the value for $$R^2$$ change would always be 0.5. Similarly, the $$R^2$$ change value for $$X_1$$ would always be 0.4. Because of this relationship, uncorrelated predictor variables will be preferred, when possible.
Look at the Modeling and Analysis of Clinical, Genetic and Imaging Data of Alzheimer’s Disease dataset. It is fairly clear that DX_Conversion could be directly entered into a regression model predicting MMSE, because it is dichotomous. The problem is how to deal with the two categorical predictor variables with more than two levels (e.g., GDTOTAL).
Dummy Coding refers to making many dichotomous variables out of one multilevel categorical variable. Because categorical predictor variables cannot be entered directly into a regression model and be meaningfully interpreted, some other method of dealing with information of this type must be developed. In general, a categorical variable with k levels will be transformed into k-1 variables each with two levels. For example, if a categorical variable had 4 levels, then 3 dichotomous (dummy) variables could be constructed that would contain the same information as the single categorical variable. Dichotomous variables have the advantage that they can be directly entered into the regression model. The process of creating dichotomous variables from categorical variables is called dummy coding.
Depending upon how the dichotomous variables are constructed, additional information can be gleaned from the analysis. In addition, careful construction will result in uncorrelated dichotomous variables. These variables have the advantage of simplicity of interpretation and are preferred to correlated predictor variables.
For example, when using dummy coding with three levels, the simplest case of dummy coding is when the categorical variable has three levels and is converted to two dichotomous variables. School in the example data has three levels, 1=Math, 2=Biology, and 3=Engineering. This variable could be dummy coded into two variables, one called Math and one called Biology. If School = 1, then Math would be coded with a 1 and Biology with a 0. If School=2, then Math would be coded with a 0 and Biology would be coded with a 1. If School=3, then both Math and Biology would be coded with a 0. The dummy coding is represented below.
Subject Math Biology Math 1 1 0 Biology 2 0 1 Engineering 3 0 0
See the SOCR Regression analysis and compare it against SOCR ANOVA.
Dummy Coding into Independent Variables, and the selection of an appropriate set of dummy codes, will result in new variables that are uncorrelated or independent of each other. In the case when the categorical variable has three levels this can be accomplished by creating a new variable where one level of the categorical variable is assigned the value of -2 and the other levels are assigned the value of 1. The signs are arbitrary and may be reversed, that is, values of 2 and -1 would work equally well. The second variable created as a dummy code will have the level of the categorical variable coded as -2 given the value of 0 and the other values recoded as 1 and -1. In all cases the sum of the dummy coded variable will be zero.
We can directly interpret each of the new dummy-coded variables, called a contrast, and compare levels coded with a positive number to levels coded with a negative number. Levels coded with a zero are not included in the interpretation.
For example, School in the example data has three levels, 1=Math, 2=Biology, and 3=Engineering. This variable could be dummy coded into two variables, one called Engineering (comparing the Engineering School with the other two Schools) and one called Math_vs_Bio (comparing Math versus Biology schools). The Engineering contrast would create a variable where all members of the Engineering Department would be given a value of -2 and all members of the other two Schools would be given a value of 1. The Math_vs_Bio contrast would assign a value of 0 to members of the Engineering Department, 1 divided by the number of members of the Math Department to member of the Math Department, and -1 divided by the number of members of the Biology Department to members of the Biology Department. The Math_vs_Bio variable could be coded as 1 and -1 for Math and Biology respectively, but the recoded variable would no longer be uncorrelated with the first dummy coded variable (Engineering). In most practical applications, it makes little difference whether the variables are correlated or not, so the simpler 1 and -1 coding is generally preferred. The contrasts are summarized in the following table.
School Engineering Math_vs_Bio Math 1 1 1/N1 = 1/12= 0.0833 Biology 2 1 -1/N2 = -1/7 = -0.1429 Engineering 3 -2 0
Note that the correlation coefficient between the two contrasts is zero. The correlation between the Engineering contrast and Salary is -.585 with a squared correlation coefficient of .342. This correlation coefficient has a significance level of .001. The correlation coefficient between the Math_vs_Bio contrast and Salary is -.150 with a squared value of .023.
Generate the corresponding SOCR ANOVA table.
Show that the significance level is identical to the value when each contrast was entered last into the regression model. In this case the Engineering contrast was significant and the Math_vs_Bio contrast was not. The interpretation of these results would be that the Engineering Department was paid significantly more than the Math and Biology Schools, but that no significant differences in salary were found between the Math and Biology Schools. If a categorical variable had four levels, three dummy coded contrasts would be necessary to use the categorical variable in a regression analysis. For example, suppose that a researcher at a pain center did a study with 4 groups of four patients each (N is being deliberately kept small). The dependent measure is subjective experience of pain. The 4 groups consisted of 4 different treatment conditions.
GroupTreatment
1None
2Placebo
3Psychotherapy
4Acupuncture
An independent contrast is a contrast that is not a linear combination of any other set of contrasts. Any set of independent contrasts would work equally well if the end result was the simultaneous test of the five contrasts, as in an ANOVA. One of the many possible examples is presented below.
Group C1 C2 C3 C4 None 1 0 0 0 0 Placebo 2 1 0 0 0 Psychotherapy 3 0 1 0 0 Acupuncture 4 0 0 1 0
Application of this dummy coding in a regression model entering all contrasts in a single block would result in an ANOVA table similar to the one obtained using Means, ANOVA, or General Linear Model.
This solution would not be ideal, however, because there is considerable information available by setting the contrasts to test specific hypotheses. The levels of the categorical variable generally dictate the structure of the contrasts. In the example study, it makes sense to contrast the two control groups (1 and 2) with the other four experimental groups (3 and 4). Any two numbers would work, one assigned to groups 1 and 2 and the others assigned to the other four groups, but it is conventional to have the sum of the contrasts equal to zero. One contrast that meets this criterion would be (-2, -2, 1, 1). Generally it is easiest to set up contrasts within subgroups of the first contrast. For example, a second contrast might test whether there are differences between the two control groups. This contrast would appear as (1, -1, 0, 0). As can be seen, this would be a contrast within the experimental treatment groups. Within the non-drug treatment, a contrast comparing Group 3 with Group 4 might be appropriate (0, 0, 1, -1). Combined, the contrasts are given in the following table.
Group C1 C2 C3 None 1 -2 1 0 Placebo 2 -2 -1 0 Psychotherapy 3 1 0 1 Acupuncture 4 1 0 1
Equal sample sizes are important as the results are much cleaner when the sample sizes are presumed to be equal. However equal sample size are not common in real applications, even in well-designed experiments. Unequal sample size makes the effects no longer independent. This implies that it makes difference in hypothesis testing when the effects are added into the model, first, middle, or last. The same dummy coding that was applied to equal sample sizes will now be applied to the original data with unequal sample sizes.
### Examples
We now demonstrate the use of SOCR Multilinear regression applet to analyze multivariate data.
#### Earthquake Modeling
This is an example where the relation between variables may not be linear or explanatory. In the simple linear regression case, we were able to compute by hand some (simple) examples. Such calculations are much more involved in the multilinear regression situations. Thus we demonstrate multilinear regression only using the SOCR Multiple Regression Analysis Applet.
Use the SOCR California Earthquake dataset to investigate whether Earthquake magnitude (dependent variable) can be predicted by knowing the longitude, latitude, distance and depth of the quake. Clearly, we do not expect these predictors to have a strong effect on the earthquake magnitude, so we expect the coefficient parameters not to be significantly distinct from zero (null hypothesis). SOCR Multilinear regression applet reports this model:
$Magnitude = \beta_o + \beta_1\times Close+ \beta_2\times Depth+ \beta_3\times Longitude+ \beta_4\times Latitude + \varepsilon.$
$Magnitude = 2.320 + 0.001\times Close -0.003\times Depth -0.035\times Longitude -0.028\times Latitude + \varepsilon.$
#### Multilinear Regression on Consumer Price Index
Using the SOCR Consumer Price Index Dataset we can explore the relationship between the prices of various products and commodities. For example, regressing Gasoline on the following three predictor prices: Orange Juice, Fuel and Electricity illustrates significant effects of all these variables as significant explanatory prices (at α = 0.05) for the cost of Gasoline between 1981 and 2006.
$Gasoline = 0.083 -0.190\times Orange +0.793\times Fuel +0 .013\times Electricity$
#### 2011 Best Jobs in the US
Repeat the multiliniear regression analysis using hte Ranking Dataset of the Best and Worst USA Jobs for 2011. |
### Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 8*x^2-(9-7*x)=0
## Step 1 :
Equation at the end of step 1 :
23x2 - (9 - 7x) = 0
## Step 2 :
Trying to factor by splitting the middle term2.1Factoring 8x2+7x-9 The first term is, 8x2 its coefficient is 8.The middle term is, +7x its coefficient is 7.The last term, "the constant", is -9Step-1 : Multiply the coefficient of the first term by the continuous 8•-9=-72Step-2 : Find two factors of -72 whose sum equals the coefficient of the middle term, which is 7.
-72 + 1 = -71 -36 + 2 = -34 -24 + 3 = -21 -18 + 4 = -14 -12 + 6 = -6 -9 + 8 = -1 -8 + 9 = 1 -6 + 12 = 6 -4 + 18 = 14 -3 + 24 = 21 -2 + 36 = 34 -1 + 72 = 71
Observation : No two such factors can be found no Conclusion : Trinomial can not be factored
Equation at the end of step 2 :
8x2 + 7x - 9 = 0
## Step 3 :
Parabola, Finding the Vertex:3.1Find the Vertex ofy = 8x2+7x-9Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting "y" because the coefficient of the first term,8, is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Ax2+Bx+C,the x-coordinate of the vertex is given by -B/(2A). In our case the x coordinate is -0.4375Plugging into the parabola formula -0.4375 for x we can calculate the y-coordinate:y = 8.0 * -0.44 * -0.44 + 7.0 * -0.44 - 9.0 or y = -10.531
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 8x2+7x-9 Axis of Symmetry (dashed) x=-0.44 Vertex at x,y = -0.44,-10.53 x-Intercepts (Roots) : Root 1 at x,y = -1.58, 0.00 Root 2 at x,y = 0.71, 0.00
Solve Quadratic Equation by Completing The Square
3.2Solving8x2+7x-9 = 0 by Completing The Square.Divide both sides of the equation by 8 to have 1 as the coefficient of the first term :x2+(7/8)x-(9/8) = 0Add 9/8 to both side of the equation : x2+(7/8)x = 9/8Now the clever bit: Take the coefficient of x, which is 7/8, divide by two, giving 7/16, and finally square it giving 49/256Add 49/256 to both sides of the equation :On the right hand side we have:9/8+49/256The common denominator of the two fractions is 256Adding (288/256)+(49/256) gives 337/256So adding to both sides we finally get:x2+(7/8)x+(49/256) = 337/256Adding 49/256 has completed the left hand side into a perfect square :x2+(7/8)x+(49/256)=(x+(7/16))•(x+(7/16))=(x+(7/16))2 Things which are equal to the same thing are also equal to one another.
You are watching: 8x-2=-9+7x
See more: Corgi’S Speed: Are Corgis Fast? How Fast Can A Corgi Run ? How Fast Can Pembroke Welsh Corgis Run
Sincex2+(7/8)x+(49/256) = 337/256 andx2+(7/8)x+(49/256) = (x+(7/16))2 then, according to the law of transitivity,(x+(7/16))2 = 337/256We"ll refer to this Equation together Eq. #3.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of(x+(7/16))2 is(x+(7/16))2/2=(x+(7/16))1=x+(7/16)Now, applying the Square Root Principle to Eq.#3.2.1 we get:x+(7/16)= √ 337/256 Subtract 7/16 from both sides to obtain:x = -7/16 + √ 337/256 Since a square root has two values, one positive and the other negativex2 + (7/8)x - (9/8) = 0has two solutions:x = -7/16 + √ 337/256 orx = -7/16 - √ 337/256 Note that √ 337/256 can be written as√337 / √256which is √337 / 16 |
+0
0
3
1
+17
At the office, mice vary inversely with cats, that is, mice = k/cats, for some value of k. When there are 3r-19 cats, there are 2r+1 mice, and when there are 6r-24 mice, there are r-3 cats. Find k.
Aug 10, 2024
#1
+480
0
Understanding the Problem
We're given an inverse relationship between mice and cats:
Mice = k/cats
We have two sets of data:
When cats = 3r - 19, mice = 2r + 1
When mice = 6r - 24, cats = r - 3
We need to find the value of k.
Setting up the Equations
Using the given information, we can create two equations:
2r + 1 = k / (3r - 19)
6r - 24 = k / (r - 3)
Solving for k
We can solve either equation for k and then substitute it into the other equation. Let's solve the first equation for k:
k = (2r + 1)(3r - 19)
Now, substitute this expression for k in the second equation:
6r - 24 = [(2r + 1)(3r - 19)] / (r - 3)
Simplify the equation:
(6r - 24)(r - 3) = (2r + 1)(3r - 19)
6r^2 - 42r + 72 = 6r^2 - 35r - 19
7r = 91
r = 13
Now, substitute r = 13 back into either of the original equations to find k. Let's use the first equation:
k = (213 + 1)(313 - 19)
k = (27)(20)
k = 540
Therefore, the value of k is 540.
Aug 10, 2024 |
# Reteaching 6-1
```000200010271713722_CH06_L01.qxp
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2/5/13
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Reteaching 6-1
Angle Measures
Acute angles have a measurement of less than 90. In the diagram
below, &1 and &2 are acute angles.
Obtuse angles have a measurement of more than 90. In the diagram,
&5 is an obtuse angle.
Vertical angles are formed across from each other where two lines
intersect. They always have equal measurements. In the diagram,
&1 and &4 are vertical angles. They both measure 50.
3
2
50 1
4
5
When the sum of two angles is 90, the angles are complementary.
In the diagram, &1 and &2 are complementary angles. To find the
measurement of a complementary angle, subtract from 90.
90 - m&1 = m&2
90 - 50 = m&2
40 = m&2
m&2 = 40
When the sum of two angles is 180, the angles are supplementary.
In the diagram, &4 and &5 are supplementary angles. To find the
measurement of a supplementary angle, subtract from 180.
180 - 50 = m&5
130 = m&5
m&5 = 130
Use the diagram to find the measurement of each angle.
Then classify each angle as obtuse, right, or acute.
146
A
1. &A
2. &B
B
C
D
40
3. &C
4. &D
5. Angles 1 and 2 are supplementary. If angle 1 measures (2x 2)°,
and angle 2 measures (x 5)°, what is the measure of each angle?
Course 2 Lesson 6-1
Reteaching
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
180 - m&4 = m&5
2/5/13
5:10 PM
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Class
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Reteaching 6-2
Area of a Parallelogram
You can use the area of a rectangle to find the area of
a parallelogram.
1
2
3
4 cm
Draw a perpendicular segment from one vertex to the
opposite side to form a triangle.
8 cm
Move the triangle to the right side of the parallelogram
to form a rectangle.
4 cm
Find the area of the rectangle.
A = length width = base height = bh
8 cm
A = bh
A=8?4
A = 32 cm2
The parallelogram has the same base, height, and area
as the rectangle.
000200010271713722_CH06_L02.qxd
Find the area of each figure.
1.
2.
6 cm
3.
5m
5 cm
8 ft
7m
4.
5.
4.3 in.
6.
0.7 ft
3.6 in.
2.1 in.
0.9 ft
7.2 in.
Find the area of a parallelogram with base length b and height h.
7. b 7 in., h 4 in.
Course 2 Lesson 6-2
8. b 9 m, h 1.5 m
9. b 1.25 cm, h 2 cm
Reteaching
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4 ft
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11:15 AM
Name
Page 2
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Date
Reteaching 6-3
Area of a Triangle
You can use the area of a parallelogram to find the area of
a triangle. Two identical triangles, together as shown, form
a parallelogram. Each triangle has half the area of
the parallelogram.
4 cm
Area of parallelogram: A bh
Area of triangle:
A
1
2 bh
1
2
7 cm
? 7 ? 4 14
cm2
This triangle has an area of 14 cm2.
1.
2.
3.2 ft
10 cm
6 cm
Find the area of each triangle.
3.
3.2 m
3.2 ft
5m
4.
7.8 cm
Solve.
5. Ryan took measurements of his new kite and made the drawing
shown on the right.
in.
6
.
7
4.1 in.
What is the area of Ryan’s kite?
in.
12.4
17.4 in.
10.3 in.
Course 2 Lesson 6-3
Reteaching
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
9 cm
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3/20/13
9:09 PM
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Class
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Reteaching 6-4
Areas of Other Figures
Trapezoid
Irregular Figures
Two identical trapezoids, together as shown, form
a parallelogram. The trapezoid has half the area
of the parallelogram.
Not all geometric figures are shapes with which
you are familiar. Some of them, however, can be
7 ft
divided into familiar shapes.
Find the area of the figure.
6 ft
10 ft
Use the area formulas to find
the areas of the triangle and
the rectangle.
5 12 (2)(4)
A 5 12 h(b1 1 b2)
Area of trapezoid:
5 12bh
5 12 (8)
5 12 (4)(10 1 8)
4 ft
2 ft
7 ft
5 4 ft2
⫽ 2(18) ⫽ 36 in.2
Area of a triangle
Area of parallelogram: A 5 (b1 1 b2)h
9 ft
Area of a rectangle ⫽ bh
⫽ (7)(10)
⫽ 70 ft2
10 ft
Total area ⫽ area of triangle ⫹ area of rectangle
⫽ 4 ⫹ 70
⫽ 74
The total area is 74 ft2.
Based on appearance, find the area of each figure.
1.
2.
48 ft
22 ft
4 ft
7 ft
3.
3 12 in.
9 12 in.
20 ft
6 ft
4.
5.
7m
6.
26 km
4m
17 km
11 m
Course 2 Lesson 6-4
36 yd
37 yd
11 km
46 km
13 m
20 yd
80 yd
Reteaching
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Find the total area by adding
the area of each figure.
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Reteaching 6-5
Date
Circumference and Area of a Circle
The circumference of a circle is the distance around it. To find the
circumference of a circle with radius r and diameter d, use either
the formula C = 2pr or C = pd. Use 3.14 for p.
d ⫽ 8 cm
C ⫽ pd
≈ 3.14 ? 8
⫽ 25.12 cm
6 ft
To the nearest centimeter, the
circumference is 25 cm.
To the nearest foot, the
circumference is 38 ft.
8 cm
r ⫽ 6 ft
C ⫽ 2pr
≈ 2 ? 3.14 ? 6
⫽ 37.68 ft
To find the area of a circle, use A ⫽ pr2.
The diameter of the circle is 8 cm, so the radius is 4 cm.
8 cm
A = pr2
≈ 3.14 ? 4 ? 4
⫽ 50.24 cm2
To the nearest square centimeter, the area is 50 cm2.
2.
3.
in
.
1.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Find the circumference and area of each circle. Round your
answer to the nearest whole unit.
4.
10
2
7 cm
5.
m
6.
2 cm
8y
d
3 ft
Course 2 Lesson 6-5
Reteaching
``` |
Google Classroom
GeoGebra Classroom
# Finding the Sum of the Interior Angles of Pentagons
In order to find the interior angles of a pentagon, we must first specify what exactly is being meant by interior as well as exterior and take a look back at triangles.
In the interactive diagram above, the pink shaded portion is the exterior angle as it is the angle on the exterior of the shape. As for the interior, that would represent the blue shaded angle, as it is inside of the shape. Furthermore, both sides of the exterior and interior of a triangle will always add up to 180. For this lesson, we will mainly be focusing on interior angles. Move around the top of the triangle to see that in action!
----------------------------------------------------------------------------------------------------------------------
Let's have another look at triangles again.
Try and move each triangle to your heart's content. Although each triangle looks vastly different, do you notice what amount the interior angles always equals to? Try adding them and find out!
The sum of the angles equals to?
Select all that apply
• A
• B
• C
• D
• E
Check my answer (3)
Yep, no matter how you move it, a triangle will always equal to 180. Now that you're a triangle master, let's add one more vertex point (the dots) and turn these triangles into a square and rhombus.
----------------------------------------------------------------------------------------------------------------------
For this next part, you will have two triangles and you will check to see if you can pair them up to make either a square or rhombus. (Hint: If you're having trouble, try flipping one of the triangles upside down then connecting and squeezing them together.)
(Spoilers) So in the end, the two triangles put together should like something like one of the two...
Did you come to a similar conclusion? It's pretty interesting to see how two triangles can make a square like shape, isn't it? Now that we know two triangles can be fit into making a square, remember how the sum of the interior angles of a triangle equals 180? Since it's two triangles to make a square we are dealing with, we would double the amount of the sum of the interior triangles to find the sum of the square. Let's take a look at the following example to get a better look at this.
----------------------------------------------------------------------------------------------------------------------
And so the complete shapes would simply look like...
And so with the description above and with the use of the shapes, If you were to add all of the measurements together, then for both shapes what would you get?
You would get?...
Select all that apply
• A
• B
• C
• D
• E
Check my answer (3)
So from here, what do you suppose would happen if you added yet another triangle? There's a few shapes it can turn into actually but one way turns into a familiar shape...
----------------------------------------------------------------------------------------------------------------------
Well would you look at that, it turns into an octagon! Now given what we know about the sum of the interior angles of a triangle (180) and the sum of the interior angles of a square (360) what do you think the sum of the interior angles of this shape would turn out to be? Take a wild guess.
I think it would be...
----------------------------------------------------------------------------------------------------------------------
Well, in order to find this, I will now present a formula in order to find the sum of the interior angles that will be true for any given polygon or n-gon.
. Remember that and remember it well. And as for what it stands for, The n - 2 is the number of sides that the n-gon has subtracted by 2 and 180 is the sum of the interior angles of a triangle- being 180° . So, now that you know the formula, let's get back to finding the interior angles of a pentagon and see if you were right!
----------------------------------------------------------------------------------------------------------------------
Let's start then by counting the number of sides the shape has then.
So as we can clearly see, this shape has 5 sides. So from here I'll leave you to it to plug in the amount of sides (n) into the formula to find your answer.
After using the formula, I found my answer to be?...
Select all that apply
• A
• B
• C
• D
• E
• F
• G
• H
Check my answer (3)
Once you have your answer, continue onwards.
----------------------------------------------------------------------------------------------------------------------
And there you have it! You have now found the sum of the interior angles of a pentagon! 540°.
Hopefully with this new formula that you have now discovered,you can now find what the interior angles of any pentagon and n-gon that come your way are easily. With that being said, could you kindly remind me what the formula was again? Thanks in advance!
Yes, the formula for finding the interior angles of an n-gon was... |
We have provided the solutions for not only the exercised but also for the mid-chapter checkpoint and review tests. So, the students can check the solutions after practicing from Go Math Grade Chapter 11 Answer Key. Check out the topics given below before starting your preparation. Get step by step procedure for each and every problem with suitable examples in our Go Math Answer Key for Grade 4 Chapter 11 Angles. Hence make use of the links and start practicing now.
Lesson 1:
Lesson 2:
Lesson 3: Measure and Draw Angles
Mid-Chapter Checkpoint
Lesson 4:
Common Core – New
Lesson 5:
Common Core – New
Chapter 11 Review/Test
### Common Core – New – Page No. 605
Angles and Fractional Parts of a Circle
Tell what fraction of the circle the shaded angle represents.
Question 1.
The figure shows that the $$\frac{1}{4}$$th part of the circle is shaded. So, the fraction of the shaded angle is $$\frac{1}{4}$$
Question 2.
$$\frac{□}{□}$$
Answer: $$\frac{1}{2}$$
Explanation:
Half of the circle is shaded. Thus the fraction of the shaded angle is $$\frac{1}{2}$$
Question 3.
$$\frac{□}{□}$$
Answer: $$\frac{1}{1}$$
Explanation:
From the above figure, we can observe that the complete circle is shaded. So, the fraction of the shaded angle is $$\frac{1}{1}$$ or 1.
Tell whether the angle on the circle shows a $$\frac{1}{4}, \frac{1}{2}, \frac{3}{4}$$, or 1 full turn clockwise or counterclockwise.
Question 4.
$$\frac{□}{□}$$
Answer: $$\frac{1}{2}$$ turn counter clockwise
Explanation:
From the figure, we can see that the circle is rotating in the anti-clockwise direction. And it has completed the half turn.
Thus the fraction is $$\frac{1}{2}$$ turn counter clockwise
Question 5.
$$\frac{□}{□}$$
Answer: $$\frac{3}{4}$$ turn clockwise
Explanation:
The arrow is turned in a clockwise direction. It has completed $$\frac{3}{4}$$ turn. So, the angle with direction is $$\frac{3}{4}$$ turn clockwise.
Question 6.
_________
Answer: 1 full turn counter clockwise
Explanation:
From the above picture, we can observe that the circle has completed the full turn in the counter clockwise direction.
Problem Solving
Question 7.
Shelley exercised for 15 minutes. Describe the turn the minute hand made.
Type below:
_________
Answer: The minute hand made a turn of $$\frac{1}{4}$$ clockwise.
Explanation:
Given that,
Shelley exercised for 15 minutes.
So, the fraction of the minute hand made is $$\frac{1}{4}$$.
The direction of the minute hand made is clockwise.
So, the answer is the minute hand made a turn of $$\frac{1}{4}$$ clockwise.
Question 8.
Mark took 30 minutes to finish lunch. Describe the turn the minute hand made.
Type below:
_________
Answer: The minute hand made a turn of $$\frac{1}{2}$$ clockwise.
Explanation:
Given, Mark took 30 minutes to finish lunch.
The minute hand made a turn in the clockwise direction from 12 to 6.
That means the fraction of the angle is $$\frac{1}{2}$$.
Thus the turn minute hand made is $$\frac{1}{2}$$ clockwise.
### Common Core – New – Page No. 606
Lesson Check
Question 1.
What fraction of the circle does the shaded angle represent
Options:
a. $$\frac{1}{1}$$ or 1
b. $$\frac{3}{4}$$
c. $$\frac{1}{2}$$
d. $$\frac{1}{4}$$
Answer: $$\frac{1}{4}$$
Explanation:
From the figure we can say that the fraction of the shaded angle is $$\frac{1}{4}$$.
Thus the answer is option D.
Question 2.
Which describes the turn shown below?
Options:
a. $$\frac{1}{4}$$ turn clockwise
b. $$\frac{1}{2}$$ turn clockwise
c. $$\frac{1}{4}$$ turn counterclockwise
d. $$\frac{1}{2}$$ turn counterclockwise
Answer: $$\frac{1}{2}$$ turn clockwise
Explanation:
From the figure, we can see that the circle is rotating in the clockwise direction. And it has completed the half turn.
So, the answer is $$\frac{1}{2}$$ turn clockwise.
Spiral Review
Question 3.
Which shows $$\frac{2}{3}$$ and $$\frac{3}{4}$$ written as a pair of fractions with a common denominator?
Options:
a. $$\frac{2}{3} \text { and } \frac{4}{3}$$
b. $$\frac{6}{9} \text { and } \frac{6}{8}$$
c. $$\frac{2}{12} \text { and } \frac{3}{12}$$
d. $$\frac{8}{12} \text { and } \frac{9}{12}$$
Answer: $$\frac{8}{12} \text { and } \frac{9}{12}$$
Explanation:
$$\frac{2}{3}$$ and $$\frac{3}{4}$$
The denomintors are different here. So you have to make the denominators common.
$$\frac{2}{3}$$ × $$\frac{4}{4}$$ = $$\frac{8}{12}$$
$$\frac{3}{4}$$ × $$\frac{3}{3}$$ = $$\frac{9}{12}$$
So the answer is option D.
Question 4.
Raymond bought $$\frac{3}{4}$$ of a dozen rolls. How many rolls did he buy?
Options:
a. 3
b. 6
c. 7
d. 9
Explanation:
Raymond bought $$\frac{3}{4}$$ of a dozen rolls.
Dozen = 12
$$\frac{3}{4}$$ × 12 = 9
Thus the correct answer is option D.
Question 5.
Which of the following lists all the factors of 18?
Options:
a. 1, 2, 4, 9, 18
b. 1, 2, 3, 6, 9, 18
c. 2, 3, 6, 9
d. 1, 3, 5, 9, 18
Answer: 1, 2, 3, 6, 9, 18
Explanation:
The factors of 18 are
1 × 18 = 18
2 × 9 = 18
3 × 6 = 18
6 × 3 = 18
9 × 2 = 18
18 × 1 = 18
Thus the correct answer is option B.
Question 6.
Jonathan rode 1.05 miles on Friday, 1.5 miles on Saturday, 1.25 miles on Monday, and 1.1 miles on Tuesday. On which day did he ride the shortest distance?
Options:
a. Monday
b. Tuesday
c. Friday
d. Saturday
Explanation:
Jonathan rode 1.05 miles on Friday, 1.5 miles on Saturday, 1.25 miles on Monday, and 1.1 miles on Tuesday.
The shortest among all is 1.05 miles.
Therefore the answer is option C.
### Page No. 609
Question 1.
Find the measure of the angle.
Through what fraction of a circle does the angle turn?
$$\frac{1}{3}=\frac{■}{360}$$
Think: 3 × 12 = 36, so 3 × _____ = 360.
So, the measure of the angle is _____.
_____ degrees
Explanation:
The fraction of the shaded angle is $$\frac{1}{3}$$
To measure the angle we have to multiply the fraction of the shaded angle with the total angle.
That means, $$\frac{1}{3}$$ × 360
360/3 = 120 degrees.
Thus the angle of the shaded part is 120°
Tell the measure of the angle in degrees.
Question 2.
____ °
Explanation:
The fraction of the shaded angle is $$\frac{45}{360}$$
Multiply the fraction with the complete angle
$$\frac{45}{360}$$ × 360° = 45°
Thus the angle of the above figure is 45°
Question 3.
____ degrees
Explanation:
The figure shows the fraction of the shaded angle is $$\frac{1}{12}$$
Multiply the fraction with the complete angle
$$\frac{1}{12}$$ × 360° = 30°
Therefore the measure of the shaded angle is 30°
Tell the measure of the angle in degrees.
Question 4.
____ °
Explanation:
We observe that the circle is shaded completely.
$$\frac{360}{360}$$ × 360° = 360°
Thus the above figure is the complete angle.
Question 5.
____ °
Explanation:
The fraction of the shaded angle is $$\frac{1}{10}$$
Multiply the fraction with the complete angle
$$\frac{1}{10}$$ × 360° = 36°
Therefore the measure of the shaded angle is 36°
Classify the angle. Write acute, obtuse, right, or straight.
Question 6.
_________
An obtuse angle has a measurement greater than 90 degrees but less than 180 degrees. However, A reflex angle measures more than 180 degrees but less than 360 degrees.
Question 7.
_________
A right angle is an angle of exactly 90° (degrees), corresponding to a quarter turn. If a ray is placed so that its endpoint is on a line and the adjacent angles are equal, then they are right angles.
Question 8.
_________
The acute angle is the small angle which is less than 90°.
Question 9.
_________
A straight angle is 180 degrees. A straight angle changes the direction to point the opposite way.
Question 10.
Is this an obtuse angle? Explain.
Type below:
_________
An obtuse angle has a measurement greater than 90 degrees but less than 180 degrees. However, A reflex angle measures more than 180 degrees but less than 360 degrees.
Question 11.
Alex cut a circular pizza into 8 equal slices. He removed 2 of the slices of pizza. What is the measure of the angle made by the missing slices of pizza?
____ °
Explanation:
Alex cut a circular pizza into 8 equal slices.
He removed 2 of the slices of pizza.
The fraction of the missing slices = $$\frac{2}{8}$$ = $$\frac{1}{4}$$
The fraction of the missing slices is $$\frac{1}{4}$$
To know the angle we have to multiply the fraction with complete angle i.e., 360°
$$\frac{1}{4}$$ × 360° = 90°
Thus the angle of the missing slices is 90°
### Page No. 610
Question 12.
Ava started reading at 3:30 p.m. She stopped for a snack at 4:15 p.m. During this time, through what fraction of a circle did the minute hand turn? How many degrees did the minute hand turn?
a. What are you asked to find?
Type below:
_________
Answer: I am asked to find the fraction of a circle did the minute hand turn and how many degrees did the minute hand turn
Question 12.
b. What information can you use to find the fraction of a circle through which the minute hand turned?
Type below:
_________
Answer: The fraction of a circle through which the minute hand-turned $$\frac{3}{4}$$ Clockwise.
Question 12.
c. How can you use the fraction of a circle through which the minute hand turned to find how many degrees it turned?
Type below:
_________
The figure shows that the fraction of a circle through which the minute hand turned is $$\frac{3}{4}$$ Clockwise.
Let the shaded part be x
And the nonshaded part is 90°
x + 90° = 360°
x = 360°- 90°
x = 270°
Therefore the minute hand turns 270° clockwise.
Question 12.
d. Show the steps to solve the problem.
Step 1:
$$\frac{3 × ■}{4 × ■}=\frac{?}{360}$$
Step 2:
$$\frac{3 × 90}{4 × 90}=\frac{■}{360}$$
Type below:
_________
$$\frac{3 × 90}{4 × 90}=\frac{■}{360}$$
$$\frac{270}{360} = \frac{■}{360}$$
If the denominators are equal then the numerators must be equated.
■ = 270
Question 12.
e. Complete the sentences. From 3:30 p.m. to 4:15 p.m., the minute hand made a ______ turn clockwise. The minute hand turned ______ degrees.
Type below:
_________
From 3:30 p.m. to 4:15 p.m., the minute hand made a $$\frac{3}{4}$$ turn clockwise. The minute hand turned 270 degrees.
Question 13.
An angle represents $$\frac{1}{15}$$ of a circle. Select the number to show how to find the measure of the angle in degrees.
$$\frac{1}{15}=\frac{1 × □}{15 × □}=\frac{□}{360}$$
Type below:
_________
$$\frac{1}{15} × 360° = 24° ### Common Core – New – Page No. 611 Degrees Tell the measure of the angle in degrees. Question 1. Answer: 60° Explanation: Given that the fraction of the shaded angle is [latex]\frac{60}{360}$$
$$\frac{60}{360}$$ × 360 = 60°
Thus the angle for the above figure is 60°
Question 2.
____ °
Explanation:
Half of the circle is shaded. The fraction of the shaded angle is $$\frac{1}{2}$$
$$\frac{1}{2}$$ × 360 = 360/2 = 180°
Question 3.
____ °
Explanation:
The fraction of the shaded angle is $$\frac{1}{4}$$
To find the angle we need to multiply the fraction with the total angle.
$$\frac{1}{4}$$ × 360° = 90°
Classify the angle. Write acute, obtuse, right, or straight.
Question 4.
_________
Explanation:
25° < 90°
So, the above figure is an acute angle.
Question 5.
_________
Explanation:
110° > 90°
So, the figure shown above is an obtuse angle.
Question 6.
_________
Explanation:
60° < 90°
Acute angles measure less than 90 degrees. Thus the above angle is an acute angle.
Classify the triangle. Write acute, obtuse, or right.
Question 7.
_________
Explanation:
65 + 25 = 90
The sum of two angles = 90°
Thus the above figure is a right-angled triangle.
Question 8.
_________
Explanation:
110° > 90°
So, the above triangle is an obtuse angle.
Question 9.
_________
Explanation:
50° is less than 90°
Thus the above triangle is an acute angle triangle.
Problem Solving
Ann started reading at 4:00 P.M. and finished at 4:20 P.M.
Question 10.
Through what fraction of a circle did the minute hand turn?
$$\frac{□}{□}$$
Answer: $$\frac{1}{3}$$ turn clockwise
Explanation:
The fraction of the shaded clock is $$\frac{12}{4}$$
$$\frac{12}{4}$$ = $$\frac{1}{3}$$
The minute hand turn clockwise direction.
So, the answer is $$\frac{1}{3}$$ turn clockwise
Question 11.
How many degrees did the minute hand turn?
____ °
Explanation:
The fraction of the minute hand turn is $$\frac{1}{3}$$
$$\frac{1}{3}$$ × 360° = 120°
The minute hand turn 120°
### Common Core – New – Page No. 612
Lesson Check
Question 1.
What kind of angle is shown?
Options:
a. acute
b. obtuse
c. right
d. straight
A straight angle is 180 degrees. This is a straight angle. A straight angle changes the direction to point the opposite way.
So, the answer is option D.
Question 2.
How many degrees are in an angle that turns through $$\frac{1}{4}$$ of a circle?
Options:
a. 45°
b. 90°
c. 180°
d. 270°
Explanation:
$$\frac{1}{4}$$ × 360°
$$\frac{360}{4}$$ = 90°
Thus the correct answer is option B.
Spiral Review
Question 3.
Mae bought 15 football cards and 18 baseball cards. She separated them into 3 equal groups. How many sports cards are in each group?
Options:
a. 5
b. 6
c. 11
d. 12
Explanation:
Mae bought 15 football cards and 18 baseball cards.
She separated them into 3 equal groups.
Total number of cards = 15 + 18 = 33
33/3 = 11
There are 11 sports cards in each group.
Question 4.
Each part of a race is $$\frac{1}{10}$$ mile long. Marsha finished 5 parts of the race. How far did Marsha race?
Options:
a. $$\frac{1}{10}$$ mile
b. $$\frac{5}{12}$$ mile
c. $$\frac{1}{2}$$ mile
d. 5 $$\frac{1}{10}$$ miles
Answer: $$\frac{1}{2}$$ mile
Explanation:
Each part of a race is $$\frac{1}{10}$$ mile long.
Marsha finished 5 parts of the race.
$$\frac{1}{10}$$ × 5 = 5/10 = $$\frac{1}{2}$$ mile
Thus the correct answer is option C.
Question 5.
Jeff said his city got $$\frac{11}{3}$$ inches of snow. Which shows this fraction written as a mixed number?
Options:
a. 3 $$\frac{2}{3}$$
b. 3 $$\frac{1}{3}$$
c. 2 $$\frac{2}{3}$$
d. 1 $$\frac{2}{3}$$
Answer: 3 $$\frac{2}{3}$$
Explanation:
Jeff said his city got $$\frac{11}{3}$$ inches of snow.
The mixed fraction of $$\frac{11}{3}$$ is 3 $$\frac{2}{3}$$
The correct answer is option A.
Question 6.
Amy ran $$\frac{3}{4}$$ mile. Which decimal shows how many miles she ran?
Options:
a. 0.25 mile
b. 0.34 mile
c. 0.5 mile
d. 0.75 mile
Explanation:
Amy ran $$\frac{3}{4}$$ mile.
$$\frac{3}{4}$$ = $$\frac{75}{100}$$
The decimal form of $$\frac{75}{100}$$ is 0.75
So, the answer is option D.
### Page No. 615
Question 1.
Measure ∠ABC.
Place the center of the protractor on point ____.
Align ray BC with ____ .
Read where ____ intersects the same scale.
So, m∠ABC is _____.
Type below:
_________
Use a protractor to find the angle measure.
Question 2.
m∠ONM = ____ °
Question 3.
m∠TSR = ____ °
Use a protractor to draw the angle.
Question 4.
170°
Type below:
_________
Question 5.
78°
Type below:
_________
Use a protractor to find the angle measure.
Question 6.
m∠QRS = ____ °
Question 7.
m∠XYZ = ____ °
Use a protractor to draw the angle.
Question 8.
115°
Type below:
_________
Question 9.
67°
Type below:
_________
Draw an example of each. Label the angle with its measure.
Question 10.
an acute angle
Type below:
_________
Question 11.
an obtuse angle
Type below:
_________
Question 12.
Elizabeth is making a quilt with scraps of fabric. What is the difference between m∠ABC and m∠DEF?
____ °
Question 13.
Draw an angle with a measure of 0°.
Type below:
_________
### Page No. 616
Question 14.
Hadley wants to divide this angle into three angles with equal measure. What will the measure of each angle be?
____ °
Explanation:
Given,
Hadley wants to divide this angle into three angles with equal measure.
The above figure is a right angle = 90°
If he divides into three equal angles
90/3 = 30°
So, the measure of angle will be 30°
Question 15.
Tracy measured an angle as 50° that was actually 130°. Explain her error.
Type below:
_________
Answer: She has measured the angle in the counterclockwise direction. So, that is why she got 50°.
Question 16.
Choose the word or number to complete a true statement about ∠QRS.
∠QRS is a(n) angle that has a measure of
Type below:
_________
Answer: ∠QRS is an obtuse angle that has a measure of 135°.
Earth’s Axis Earth revolves around the sun yearly. The Northern Hemisphere is the half of Earth that is north of the equator. The seasons of the year are due to the tilt of Earth’s axis.
Use the diagrams and a protractor for 17–18.
Question 17.
In the Northern Hemisphere, Earth’s axis is tilted away from the sun on the first day of winter, which is often on December 21. What is the measure of the marked angle on the first day of winter, the shortest day of the year?
____ °
Explanation:
By seeing the above figure we can say that the angle is an obtuse angle. The mark is above 90° and the marked angle is 115°.
Therefore the measure of the marked angle on the first day of winter, the shortest day of the year is 115°.
Question 18.
Earth’s axis is not tilted away from or toward the sun on the first days of spring and fall, which are often on March 20 and September 22. What is the measure of the marked angle on the first day of spring or fall?
____ °
Explanation:
The mark is exactly 90°. So, the angle on the first day of spring or fall is 90°
### Common Core – New – Page No. 617
Measure and Draw Angles
Use a protractor to find the angle measure.
Question 1.
m∠ABC = 120°
By using the protractor we can measure the angle m∠ABC i.e., 120°
Question 2.
m∠MNP = ____ °
By observing the above figure we can say that the angle of m∠MNP is 90°
Question 3.
m∠RST = ____ °
By using the protractor we can measure m∠RST = 65°
Use a protractor to draw the angle.
Question 4.
40°
Question 5.
170°
Draw an example of each. Label the angle with its measure.
Question 6.
a right angle
A right angle is an angle of exactly 90°
Question 7.
an acute angle
The acute angle is the small angle which is less than 90°.
Problem Solving
The drawing shows the angles a stair tread makes with a support board along a wall. Use your protractor to measure the angles.
Question 8.
What is the measure of ∠A?
____ °
By using the protractor we can measure the angle for A = 45°
Question 9.
What is the measure of ∠B?
____ °
The same process is used to measure ∠B = 135°
### Common Core – New – Page No. 618
Lesson Check
Question 1.
What is the measure of ∠ABC?
Options:
a. 15°
b. 25°
c. 155°
d. 165°
Explanation:
Step 1: Place the center point of the protractor on the point B.
Step 2: Align the 0° mark on the scale of the protractor with ray BC.
Step 3: Find the point where AC meet. Read the angle measure on that scale.
So, the measure of ∠ABC is 15°
Thus the correct answer is option A.
Question 2.
What is the measure of ∠XYZ?
Options:
a. 20°
b. 30°
c. 150°
d. 160°
Explanation:
Step 1: Place the center point of the protractor on the point Y.
Step 2: Align the 0° mark on the scale of the protractor with ray XY.
Step 3: Find the point where YZ meet. Read the angle measure on that scale.
So, ∠XYZ = 150°
Therefore the correct answer is option C.
Spiral Review
Question 3.
Derrick earned $1,472 during the 4 weeks he had his summer job. If he earned the same amount each week, how much did he earn each week? Options: a.$360
b. $368 c.$3,680
d. $5,888 Answer:$368
Explanation:
Given that, Derrick earned $1,472 during the 4 weeks he had his summer job. Let the amount he earned per week = x x × 4 =$1,472
x = 1472 ÷ 4
x = 1472/4 = 368
So, Derrick earned $368 per week. Question 4. Arthur baked 1 $$\frac{7}{12}$$ dozen muffins. Nina baked 1 $$\frac{1}{12}$$ dozen muffins. How many dozen muffins did they bake in all? Options: a. 3 $$\frac{2}{3}$$ b. 2 $$\frac{2}{3}$$ c. 2 $$\frac{1}{2}$$ d. $$\frac{6}{12}$$ Answer: 2 $$\frac{2}{3}$$ Explanation: Given: Arthur baked 1 $$\frac{7}{12}$$ dozen muffins. Nina baked 1 $$\frac{1}{12}$$ dozen muffins. Add both the fractions Convert mixed fraction into normal fractions 1 $$\frac{7}{12}$$ = $$\frac{19}{12}$$ 1 $$\frac{1}{12}$$ = $$\frac{13}{12}$$ $$\frac{19}{12}$$ + $$\frac{13}{12}$$ = $$\frac{32}{12}$$ = $$\frac{8}{3}$$ Convert $$\frac{8}{3}$$ into mixed fraction = 2 $$\frac{2}{3}$$ So, the answer is option B. Question 5. Trisha drew the figure below. What figure did she draw? Options: a. line segment ST b. ray ST c. ray TS d. line TS Answer: ray TS A ray can be defined as a part of a line that has a fixed starting point but no endpoint. Here the point starts from T and ends at S. So, the figure Trisha drew is ray TS. The correct answer is option C. Question 6. Which describes the turn shown by the angle? Options: a. 1 full turn clockwise b. $$\frac{3}{4 }$$ turn clockwise c. $$\frac{1}{2}$$ turn clockwise d. $$\frac{1}{4}$$ turn clockwise Answer: $$\frac{1}{4}$$ turn clockwise Explanation: The figure shows that the point turned $$\frac{1}{4}$$ in clockwise direction. So, the answer is option D. ### Page No. 619 Choose the best term from the box. Question 1. The unit used to measure an angle is called a ________. ________ Answer: The unit used to measure an angle is called a degree. Question 2. ________ is the opposite of the direction in which the hands of a clock move. ________ Answer: Counterclockwise is the opposite of the direction in which the hands of a clock move. Question 3. A ________ is a tool for measuring the size of an angle. ________ Answer: A protractor is a tool for measuring the size of an angle. Tell whether the angle on the circle shows a $$\frac{1}{4}, \frac{1}{2}, \frac{3}{4}$$, or 1 full turn clockwise or counterclockwise. Question 4. $$\frac{□}{□}$$ Answer: $$\frac{1}{4}$$ turn clockwise The figure shows that the angle turn $$\frac{1}{4}$$ in the clockwise direction. Question 5. $$\frac{□}{□}$$ Answer: $$\frac{1}{2}$$ turn counterclockwise From the above figure, we can see that the angle turn $$\frac{1}{2}$$ in the counterclockwise direction. Question 6. $$\frac{□}{□}$$ Answer: $$\frac{3}{4}$$ turn clockwise The figure shows that the angle turn $$\frac{3}{4}$$ in the clockwise direction. Question 7. ____ Answer: $$\frac{1}{1}$$ or 1 turn counterclockwise From the above figure, we can see that the angle turn $$\frac{1}{1}$$ or 1 in the counterclockwise direction. Tell the measure of the angle in degrees. Question 8. ____ ° Answer: 100° $$\frac{100}{360}$$ × 360° = 100° Question 9. ____ ° Use a protractor to draw the angle. Question 10. 75° Type below: ________ Question 11. 127° Type below: ________ ### Page No. 620 Question 12. Phillip watched a beach volleyball game from 1:45 p.m. to 2:00 p.m. How many degrees did the minute hand turn? ____ ° Answer: 90° Explanation: Phillip watched a beach volleyball game from 1:45 p.m. to 2:00 p.m. The minute hand turned for 15 minutes. That means $$\frac{1}{4}$$ turn clockwise. Complete angle = 360° $$\frac{1}{4}$$ × 360° = 360°/4 = 90° Therefore the minute hand turn 90° Question 13. What angle does this piece of pie form? ____ ° Answer: 180° From the above figure, we can see that half of the pie is completed. Complete angle = 360° $$\frac{1}{2}$$ × 360° = 180° The angle for the piece of pie form is 180° Question 14. What is m∠CBT? Use a protractor to help you. ____ ° Answer: 60° By using the protractor we can say that the angle for the above figure is 60° Question 15. Matt cut a circle into 8 equal sections. He drew an angle that measures the same as the total measure of 3 of the sections in the circle. What is the measure of the angle Matt drew? ____ ° Answer: 135° Explanation: Matt cut a circle into 8 equal sections. He drew an angle that measures the same as the total measure of 3 of the sections in the circle. Complete angle = 360° Divide the total number of sections by 360° $$\frac{360}{8}$$ = 45° So, the angle for each section is 45° The angle for 3 of the sections in the circle = 3 × 45° = 135° Thus the measure of the angle Matt drew is 135° ### Page No. 623 Add to find the measure of the angle. Write an equation to record your work. Question 1. ∠PQT = ____ ° Answer: 80° To find the ∠PQT you have to add 43° and 37° ∠PQT = 43° + 37° ∠PQT = 80° Question 2. ∠JKL = ____ ° Answer: 100° Let ∠JKL = x° ∠JKL = 90° + 10° ∠JKL = 100° Question 3. ∠RHT = ____ ° Answer: Let ∠RHT = x° x = 55° + 27° + 78° x = 160° Therefore ∠RHT = 160° Use a protractor to find the measure of each angle. Label each angle with its measure. Write the sum of the angle measures as an equation. Question 4. Type below: ________ Answer: By using the protractor we can measure the angles of the above figures. m∠KLM = 160° m∠KLJ = 80° m∠LMJ = 120° Question 5. Type below: ________ Answer: By using the protractor we can measure the angles of the above figures. m∠WVZ = 90° m∠YVZ = 90° m∠WVX = 140° m∠YVX = 40° Question 6. Use Diagrams What is m∠QRT? ∠QRT = ____ ° Answer: 20° The above figure is a straight angle. ∠QRT + ∠LRD + ∠RLT = 180 ∠QRT + 75° + 85° = 180° ∠QRT + 160° = 180° ∠QRT = 180°- 160° ∠QRT = 20° Question 7. Look back at Exercise 1. Suppose you joined an angle measuring 10° to ∠PQT. Draw the new angle, showing all three parts. What type of angle is formed? Type below: ________ ### Page No. 624 Question 8. Stephanie, Kay, and Shane each ate an equal-sized piece of a pizza. The measure of the angle of each piece was 45°. When the pieces were together, what is the measure of the angle they formed? a. What are you asked to find? Type below: ________ Answer: What is the measure of the angle for the pizza leftover? Question 8. b. What information do you need to use? Type below: ________ Answer: I need the information about the angle for each piece of pizza. Question 8. c. Tell how you can use addition to solve the problem. Type below: ________ Answer: The measure of the angle of each piece was 45° There are 3 pieces of pizza = 45° + 45° + 45° = 135° Question 8. d. Complete the sentence. The three pieces of pizza formed a _________ angle. ________ Answer: Obtuse angle Question 9. What is the measure of ∠XZW? ____ ° Answer: 113° Explanation: ∠XZW = ∠XZY + ∠YZW ∠XZY = 42° ∠YZW = 71° ∠XZW = 42° + 71° ∠XZW = 113° Question 10. What is m∠PRS? Use equations to explain and check your answer. ____ ° Answer: 12° Explanation: The above figure is a straight angle. The sum of the three angles must be equal to 180° m∠PRS + m∠PRN + m∠TRN = 180° m∠PRS + 90° + 78° = 180° m∠PRS = 180° – 90° – 78° m∠PRS = 12° ### Common Core – New – Page No. 625 Join and Separate Angles Add to find the measure of the angle. Write an equation to record your work. Question 1. 50°+75° = 125° m∠ABD = 125° Explanation: m∠ABC = 50° m∠CBD = 75° To find the measure of m∠ABD we have to add m∠ABC and m∠CBD m∠ABD = 50°+75° m∠ABD = 125° Question 2. ____ ° + ____ ° = ____ ° ; m∠FGJ = ____ ° Answer: 140° + 20° = 160° m∠FGJ = 160° Explanation: m∠FGH = 140° m∠JGH = 20° To find the measure of m∠FGJ we need to add m∠FGH and m∠JGH m∠FGJ = 140° + 20° m∠FGJ = 160° Question 3. ____ ° + ____ ° = ____ ° ; m∠KLN = ____ ° Answer: 30° + 90° + 45° = 165° m∠KLN = 165° Explanation: m∠KLM = 30° m∠MLP = 90° m∠PLN = 45° To find the measure of m∠KLN we need to add m∠KLM, m∠MLP and m∠PLN m∠KLN = 30° + 90° + 45° = 165° m∠KLN = 165° Use a protractor to find the measure of each angle in the circle. Question 4. m∠ABC = ____ ° Answer: 115° Question 5. m∠DBE = ____ ° Answer: 90° Question 6. m∠CBD = ____ ° Answer: 75° Question 7. m∠EBA = ____ ° Answer: 80° Question 8. Write the sum of the angle measures as an equation. ____ ° + ____ ° + ____ ° + ____ ° = ____ ° Answer: Sum all the angles = m∠DBE + m∠ABC + m∠CBD + m∠EBA = 115° + 90° + 75° + 80° = 360° Problem Solving Question 9. Ned made the design at the right. Use a protractor. Find and write the measure of each of the 3 angles. ____ ° ; ____ ° ; ____ ° ; Answer: 50°; 60°; 70° The above figure is a straight angle. By using the protractor we can measure the angles of the above figure. The angle of above 3 shades is 50°; 60°; 70° Question 10. Write an equation to find the measure of the total angle. ____ ° + ____ ° + ____ ° = ____ ° Answer: Sum of three angles = 50° + 60° + 70° = 180° ### Common Core – New – Page No. 626 Lesson Check Question 1. What is the measure of m∠WXZ? Options: a. 32° b. 83° c. 88° d. 97° Answer: 83° Explanation: m∠WXZ = m∠WXY + m∠YXZ Let m∠WXZ be x° x° = 58° + 25° x° = 83° Thus the correct answer is option B. Question 2. Which equation can you use to find the m∠MNQ? Options: a. 148° – 24° = ■ b. 148° × 24° = ■ c. 148° ÷ 24° = ■ d. 148° + 24° = ■ Answer: 148° + 24° = ■ Explanation: To measure the unknown angle we need to add both the angles m∠MNQ = m∠MNP + m∠PNQ ■ = 148° + 24° So, the correct answer is option D. Spiral Review Question 3. Joe bought 6 packages of envelopes. Each package contains 125 envelopes. How many envelopes did he buy? Options: a. 750 b. 723 c. 720 d. 650 Answer: 750 Explanation: Joe bought 6 packages of envelopes. Each package contains 125 envelopes. To find the total number of envelopes in all 6 packages We have to multiply number of packages and number of envelopes = 125 × 6 =750 Thus Joe bought 750 envelopes. The correct answer is option A. Question 4. The Lake Trail is $$\frac{3}{10}$$ mile long and the Rock Trail is $$\frac{5}{10}$$ long. Bill hiked each trail once. How many miles did he hike in all? Options: a. $$\frac{1}{5}$$ mile b. $$\frac{4}{10}$$ mile c. $$\frac{1}{2}$$ mile d. $$\frac{8}{10}$$ mile Answer: $$\frac{8}{10}$$ mile Explanation: The Lake Trail is $$\frac{3}{10}$$ mile long and the Rock Trail is $$\frac{5}{10}$$ long. Bill hiked each trail once. We need to both the fractions $$\frac{3}{10}$$ +$$\frac{5}{10}$$ The denominators are common. So add the numerators. = $$\frac{8}{10}$$ Bill hiked $$\frac{8}{10}$$ miles in all. Thus the correct answer is option D. Question 5. Ron drew a quadrilateral with 4 right angles and 4 sides with the same length. Which best describes the figure he drew? Options: a. square b. rhombus c. trapezoid d. parallelogram Answer: square Explanation: A square has got 4 sides of equal length and 4 right angles (right angle = 90 degrees). So, the answer is option A. Question 6. How many degrees are in an angle that turns through $$\frac{3}{4}$$ of a circle? Options: a. 45° b. 90° c. 180° d. 270° Answer: 270° Explanation: Complete angle = 360° To measure the angle that turns through is $$\frac{3}{4}$$ multiply $$\frac{3}{4}$$ with 360° 360° × $$\frac{3}{4}$$ = 270° So, the answer is option D. ### Page No. 629 Question 1. Laura cuts a square out of scrap paper as shown. What is the angle measure of the piece left over? First, draw a bar model to represent the problem. Type below: _________ Question 1. Next, write the equation you need to solve. Type below: _________ Answer: m∠MNQ + m∠QNP = m∠MNP x + 90° = 115° x = 115° – 90° Question 1. Last, find the angle measure of the piece left over. m∠MNQ = So, the angle measure of the piece left over is _____. ____ ° Answer: x + 90° = 115° x = 115° – 90° x = 25° So, the angle measure of the piece left over is 25° Question 2. Jackie trimmed a piece of scrap metal to make a straight edge as shown. What is the measure of the piece she trimmed off? x = ____ ° Answer: x + 180° = 225° x = 225°- 180° x = 45° Thus the measure of the piece she trimmed off is 45° Question 3. What if Laura cut a smaller square as shown? Would m∠MNQ be different? Explain. Type below: _________ Answer: No m∠MNQ would still be 25°. Only the size of the square changed the angle will be the same. m∠PNQ and m∠MNP did not change. Question 4. The map shows Marco’s paper route. When Marco turns right onto Center Street from Main Street, what degree turn does he make? Hint: Draw a dashed line to extend Oak Street to form a 180° angle. Answer: x° + 125° + 180° = 360° x° = 360° – 125° – 180° x° = 360° – 215° x° = 145° ### Page No. 630 Question 5. Write an Equation Two angles form a straight angle. One angle measures 89°. What is the measure of the other angle? Explain. ____ ° Answer: 91° A straight angle measures 180°, so you can subtract 89° from 180° 180° – 89° = 91° Question 6. Pose a Problem Look back at Problem 5. Write a similar problem about two angles that form a right angle. ____ ° Answer: Two angles form a right angle. The measure of one angle is 25°. What is the measure of the other angle? x + 25° = 90° x °= 90° – 25° x° = 65° The measure of other angle is 65° Question 7. Sam paid$20 for two T-shirts. The price of each T-shirt was a multiple of 5. What are the possible prices of the T-shirts?
Type below:
_________
Sam paid $20 for two T-shirts. The price of each T-shirt was a multiple of 5.$20 – 2 T-shirts
x – 1 T-shirt
x = $10 The possible prices of the T-shirts are$10, $10 Another possible price of the T-shirts are$5, $15 Question 8. Zayna has 3 boxes with 15 art books in each box. She has 2 bags with 11 math books in each bag. If she gives 30 books away, how many art and math books does she have left? _____ books Answer: 37 art and math books Explanation: Zayna has 3 boxes with 15 art books in each box = 15 × 3 = 45 She has 2 bags with 11 math books in each bag = 11 × 2 = 22 Total number of books = 45 + 22 = 67 If she gives 30 books away, then we have to subtract 30 from 67 67 – 30 = 37 37 art and math books are left. Question 9. What’s the Question? It measures greater than 0° and less than 90°. Type below: _________ Answer: What is an acute angle? Question 10. Two angles, ∠A and ∠B, form a straight angle. ∠A measures 65°. For numbers 10a–10c, select True or False for the statement. a. ∠B is an acute angle. i. True ii. False Answer: False Explanation: Two angles, ∠A and ∠B, form a straight angle. ∠A measures 65°. 65° + ∠B = 180° ∠B = 180° – 65° ∠B = 115° 115° is not an acute angle. So, the above statement is false. Question 10. b. The equation 180° – 65° = x° can be used to find the measure of ∠B. i. True ii. False Answer: True Question 10. c. The measure of ∠B is 125°. i. True ii. False Answer: False 65° + ∠B = 180° ∠B = 180° – 65° ∠B = 115° So, the above statement is false. ### Common Core – New – Page No. 631 Problem Solving Unknown Angle Measures Solve each problem. Draw a diagram to help. Question 1. Question 2. An artist is cutting a piece of metal as shown. What is the angle measure of the piece left over? x = ____ ° Answer: 95° x + 130° = 225° x = 225° – 130° x = 95° Therefore the angle of the piece leftover is 95°. Question 3. Joan has a piece of material for making a costume. She needs to cut it as shown. What is the angle measure of the piece left over? x = ____ ° Answer: 50° Joan has a piece of material for making a costume. She needs to cut it as shown. By seeing the above figure we can say that it is a right angle.| The sum of two must be equal to 90° Let the unknown angle be x x + 40° = 90° x = 90° – 40° x = 50° Thue the angle measure of the piece leftover is 50° ### Common Core – New – Page No. 632 Lesson Check Question 1. Angelo cuts a triangle from a sheet of paper as shown. What is the measure of ∠x in the triangle? Options: a. 15° b. 25° c. 75° d. 105° Answer: 15° Explanation: The above figure is a right angle. So, to measure the ∠x we have to subtract 75° from 90° ∠x + 75° = 90° ∠x = 90° – 75° ∠x = 15° Thus the correct answer is option A. Question 2. Cindy cuts a piece of wood as shown. What is the angle measure of the piece left over? Options: a. 30° b. 90° c. 120° d. 150° Answer: 120° Explanation: x + 90° = 210° x = 210° – 90° x = 120° 120° is the measure of the piece leftover. So, the correct answer is option C. Spiral Review Question 3. Tyronne worked 21 days last month. He earned$79 each day. How much did Tyronne earn last month?
Options:
a. $869 b.$948
c. $1,659 d.$2,169
Answer: $1,659 Explanation: Tyronne worked 21 days last month. He earned$79 each day.
$79 × 21 = 1659 Thus Tyronne earned$1,659 last month.
So, the correct answer is option C.
Question 4.
Meg inline skated for $$\frac{7}{10}$$ mile. Which shows this distance written as a decimal?
Options:
a. 0.07 mile
b. 0.1 mile
c. 0.7 mile
d. 7.1 miles
Explanation:
Meg inline skated for $$\frac{7}{10}$$ mile.
The decimal of the fraction $$\frac{7}{10}$$ is 0.7
So, the answer is option C.
Question 5.
Kerry ran 34 mile. Sherrie ran $$\frac{1}{2}$$ mile. Marcie ran $$\frac{2}{3}$$ mile. Which list orders the friends from least to greatest distance
run?
Options:
a. Kerry, Sherrie, Marcie
b. Kerry, Marcie, Sherrie
c. Sherrie, Kerry, Marcie
d. Sherrie, Marcie, Kerry
Explanation:
Kerry ran 34 miles. Sherrie ran $$\frac{1}{2}$$ mile. Marcie ran $$\frac{2}{3}$$ mile.
The order of the above fractions is Sherrie ran $$\frac{1}{2}$$, $$\frac{2}{3}$$, 34
The distance from least to greatest is Sherrie, Marcie, Kerry.
so, the correct answer is option D.
Question 6.
What is the measure of m∠ABC?
Options:
a. 32°
b. 84°
c. 88°
d. 94°
Explanation:
m∠ABC = m∠ABD + m∠DBC
m∠ABC = 58° + 26°
m∠ABC = 84°
So, the correct answer is option B.
### Page No. 633
Question 1.
An angle represents $$\frac{1}{12}$$ of a circle. Use the numbers to show how to find the measure of the angle in degrees.
The angle measure is ____ °
$$\frac{1}{12}$$ × $$\frac{30}{30}$$ = $$\frac{30}{360}$$
Thus the angle measure is 30°
Question 2.
Match the measure of each ∠C with the measure of ∠D that forms a straight angle.
Type below:
_________
i. 122° + 58° = 180°
ii. 35° + 145° = 180°
iii. 62° + 118° = 180°
iv. 105° + 75° = 180°
Question 3.
Katie drew an obtuse angle. Which could be the measure of the angle she drew? Mark all that apply.
Options:
a. 35°
b. 157°
c. 180°
d. 92°
An obtuse angle has a measurement greater than 90 degrees but less than 180 degrees.
From the above options, B and D are more than 90°
So, the answer is option B and D.
Question 4.
Draw an angle that represents a $$\frac{1}{4}$$ turn counterclockwise on the circle.
Type below:
_________
### Page No. 634
Question 5.
Renee drew the figure shown. For 5a–5c, select Yes or No to tell whether the statement is true.
a. The measure of a straight angle is 180°.
i. yes
ii. no
By seeing the above figure we can say that the angle is a straight angle.
So, the above statement is true.
Question 5.
b. To find the measure of x, Renee can subtract 75° from 180°.
i. yes
ii. no
To know the value of x we have to subtract 75° from 180°.
x = 180° – 75°
Thus the above statement is true.
Question 5.
c. The measure of x is 115°.
i. yes
ii. no
x = 180° – 75°
x = 105°
Thus the above statement is false.
Question 6.
Trey drew this figure with a protractor.
Part A
Write an equation that can be used to find m∠KFG.
Type below:
_________
Answer: 55° + 80° + x = 180°
The figure is a straight angle.
So, the sum of the three angles must be equal to 180°
Let m∠KFG = x
55° + 80° + x = 180°
Question 6.
Part B
What is the measure of m∠KFG? Describe how you solved the equation and how you can check your answer.
____ °
Explain:
_________
Explanation:
55° + 80° + x = 180°
x = 180° – 80° – 55°
x = 45°
Question 7.
Use a protractor to find the measure of the angle.
The angle measures ____ °
By using a protractor we can measure the angle.
The angle of the above figure is 40°
### Page No. 635
Question 8.
Alex drew this angle on the circle. Which describes the angle? Mark all that apply.
Options:
a. $$\frac{1}{4}$$ turn
b. clockwise
c. $$\frac{1}{2}$$ turn
d. counterclockwise
Answer: $$\frac{1}{2}$$ turn
The above figure shows that it is straight angle. So, the fraction of the circle is $$\frac{1}{2}$$ turn.
The correct answer is option C.
Question 9.
Miles has a piece of paper that is $$\frac{1}{4}$$ of a large circle. He cuts the paper into three equal parts from the center point of the circle. What is the angle measure of each part?
The angle measure is ____ °
Explanation:
Miles has a piece of paper that is $$\frac{1}{4}$$ of a large circle. He cuts the paper into three equal parts from the center point of the circle.
$$\frac{1}{4}$$ of a large circle = 90°
Given that he cut into 3 equal parts = $$\frac{90}{3}$$ = 30°
So, the angle for each part is 30°
Question 10.
Use a protractor to find the measure of each angle. Write each angle and its measure in a box ordered by the measure of the angles from least to greatest.
Question 11.
Use the numbers and symbols to write an equation that can be used to find the measure of the unknown angle.
What is the measure of the unknown angle?
____ °
Explanation:
Let the unknown angle be x
It is a straight angle.
The sum of three angles = 180°
90° + 33° + x = 180°
x = 180° – 90° – 33°
x = 57°
### Page No. 636
Question 12.
Choose the word or number to complete a true statement about m∠JKL.
m∠JKL is a(n) ______ angle that has a measure of ____ °.
Answer: m∠JKL is an Obtuse angle that has a measure of 120°.
Question 13.
Vince began practicing piano at 5:15 p.m. He stopped at 5:35 p.m. How many degrees did the minute hand turn during Vince’s practice time?
____ °
Explain:
_________
I shaded the part of the clock that the minute hand turned from 5:15 p.m. to 5:35 p.m. anmd found that it is $$\frac{1}{3}$$ of the circle.
Next, I multiplied $$\frac{1}{3}$$ × 360° = 120°
Thus the minute hand moved 120°
Question 14.
An angle measures 125°. Through what fraction of a circle does the angle turn?
$$\frac{□}{□}$$ of a circle
Answer: $$\frac{125}{360}$$
The fraction of the circle the angle turned is $$\frac{125}{360}$$
Question 15.
Write the letter for each angle measure in the correct box.
Type below:
__________
### Page No. 637
Question 16.
For numbers 16a–16b, select the fraction that makes a true statement about the figure.
Question 16.
a. The angle in Figure 1 represents a turn.
$$\frac{□}{□}$$ turn
Answer: The angle in Figure 1 represents a $$\frac{3}{4}$$ turn
The above figure shows that $$\frac{3}{4}$$ part of the circle is shaded. So, the angle represents $$\frac{3}{4}$$ turn.
Question 16.
b. The angle in Figure 2 represents a turn.
$$\frac{□}{□}$$ turn
Answer: The angle in Figure 2 represents a $$\frac{1}{2}$$ turn.
From the second figure, we observe that half of the circle is shaded. So, The angle in Figure 2 represents a $$\frac{1}{2}$$ turn.
Question 17.
Melanie cuts a rectangle out of a piece of scrap paper as shown. She wants to calculate the angle measure of the piece that is left over.
Part A
Draw a bar model to represent the problem.
Question 17.
Part B
Write and solve an equation to find x.
The angle measures ____ °.
m∠RST = 90°
m∠RSN = 126°
m∠TSN = x°
x + 90° = 126°
x = 126° – 90°
x = 36°
m∠TSN = 36°
### Page No. 638
Question 18.
Two angles, m∠A and m∠B, form a right angle. m∠A measures 32°.
For numbers, 18a–18c, select True or False for the statement.
a. m∠B is an acute angle.
i. True
ii. False
If the sum of two angles is 90°, if one angle is acute then the other angle will be acute.
So, the above statement is true.
Question 18.
b. The equation 180° − 32° = x° can be used to find the measure of m∠B.
i. True
ii. False
Explanation:
Given that the sum of 2 angles is 90°
The sum of m∠A and m∠B = 90°
90° – 32° = x°
So, the above statement is false.
Question 18.
c. The measure of m∠B is 58°.
i. True
ii. False
Explanation:
Let m∠B = x
x° + 32° = 90°
x = 90 – 32
x = 58°.
So, the above statement is true.
Question 19.
A circle is divided into parts. Which sum could represent the angle measures that make up the circle? Mark all that apply.
Options:
a. 120° + 120° + 120° + 120°
b. 25° + 40° + 80° + 105° + 110°
c. 33° + 82° + 111° + 50° + 84°
d. 40° + 53° + 72° + 81° + 90° + 34°
Answer: 25° + 40° + 80° + 105° + 110°; 33° + 82° + 111° + 50° + 84°
Explanation:
The sum of all the angles must be equal to 360°
i. 120° + 120° + 120° + 120° = 480° ≠ 360°
ii. 25° + 40° + 80° + 105° + 110° = 360°
iii. 33° + 82° + 111° + 50° + 84° = 360°
iv. 40° + 53° + 72° + 81° + 90° + 34° = 370° ≠ 360°
So, the correct answers are option B, C.
Question 20.
Use a protractor to find the measures of the unknown angles.
What do you notice about the measures of the unknown angles? Is this what you would have expected? Explain your reasoning.
m∠x = ____ ° m∠y = ____ °
Answer: m∠x = 70°; m∠y = 110°
By using a protractor we can find the measure of m∠y
m∠y = 110°
Let m∠x = x°
Sum of supplementary angles = 180°
110° + x = 180°
x = 180° – 110°
x = 70°
Therefore m∠x = 70°
### Page No. 643
Use benchmarks to choose the metric unit you would use to measure each.
Question 1.
mass of a strawberry
__________
The metric unit used to measure the mass of a strawberry is the gram.
Question 2.
length of a cell phone
__________
The metric unit used to measure the length of a cell phone is Centimeter.
Circle the better estimate.
Question 3.
width of a teacher’s desk
10 meters or 1 meter
__________
The estimation of the width of the teacher’s desk is 1 meter.
Question 4.
the amount of liquid a punch bowl holds
2 liters or 20 liters
__________
20 liters is greater than 2 liters.
The estimation of the amount of liquid a punch bowl holds is 2 liters.
Question 5.
distance between Seattle and San Francisco
6 miles or 680 miles
__________
The distance between Seattle and San Francisco is 680 miles.
Use benchmarks to choose the customary unit you would use to measure each.
Question 6.
length of a football field
__________
The units to measure the length of a football field is Yards.
Question 7.
weight of a pumpkin
__________
The customary unit I use to measure the weight of a pumpkin is pounds.
Circle the better estimate.
Question 8.
weight of a watermelon
4 pounds or 4 ounces
__________
The estimation of the weight of watermelon is 4 pounds.
Question 9.
the amount of liquid a fish tank holds
10 cups or 10 gallons
__________
The estimation of the amount of liquid a fish tank holds is 10 gallons.
Complete the sentence. Write more or less.
Question 10.
Matthew’s large dog weighs ________ than one ton.
________
1 ton = 1000 kgs
The weight of dogs can’t be more than a ton.
So, Matthew’s large dog weighs less than one ton.
Question 11.
The amount of liquid a sink can hold is _______ than one cup of water.
________
1 cup holds very less amount of water.
So, The amount of liquid a sink can hold is more than one cup of water.
Question 12.
A paper clip has a mass of _______ than one kilogram.
________
The weight of a paper clip is about 1 gram.
So, A paper clip has a mass of less than one kilogram.
### Page No. 644
Question 13.
Cristina is making macaroni and cheese for her family. Would Cristina use 1 pound of macaroni or 1 ounce of macaroni?
__________
Answer: Cristina should use 1 pound of macaroni.
Question 14.
Which is the better estimate for the length of a kitchen table, 200 centimeters or 200 meters?
__________
Centimeters are less than meters. The length of the kitchen will be measured in centimeters.
So, the answer is 200 centimeters.
Question 15.
Jodi wants to weigh her cat and measure its standing height. Which two units should she use?
weight: ________
height: ________
The weight of the cat should be measured in Kilograms.
The height of the cat should be measured in Centimeters.
Question 16.
Evaluate Reasonableness Dalton used benchmarks to estimate that there are more cups than quarts in one gallon. Is Dalton’s estimate reasonable? Explain.
Type below:
__________
Answer: Dalton’s reasoning is correct because the measurement of cups is smaller than the measurement of quarts, therefore there would be more cups in a gallon than quarts.
Question 17.
Select the correct word to complete the sentence. Justine is thirsty after running two miles.
She should of water.
__________ |
## LESSON 14. CONE OF FRICTION
14.1 CONE OF FRICTION
Fig.14.1
Let ON represent the normal reaction offered by a surface on a body (Fig.14.1). If OX is the direction in which the body tends to move then the force of friction acts in the opposite direction i.e, along OE. If the body be in limiting equilibrium the resultant R makes an angle λ with the normal ON.
Suppose the body is at the point of sliding in other direction, it is easily seen that the resultant reaction will make the same angle λ with the normal. Hence, when limiting friction is offered the line of action of the resultant reaction should always lie on the surface of an inverted right circular cone whose semi-vertex is λ. This cone is called the cone of friction.
Example: A block is weighing 50 kg is placed on a rough surface whose coefficient of friction is 0.30 and inclined force P is applied at its top corner as shown in Fig.14.2. Determine whether the block will tip or slide and the force P required to move the block.
Fig.14.2
Solution: W = 50 × 9.81 = 490.5 N
By referring Fig.14.2
∑ Fy = 0
R1 – W + P sin 35°
R1 = 490.5 – 0.574 P
Or
Hence limiting friction force = µ R1
= 0.3 (490.5 – 0.574 P)
= 147.15 – 0.1722 P
Now F = P cos 35° using ∑ Fx = 0
So, F = 0.819 P
Let us assume that block slides before tipping,
Then 0.819 P = 147.15 – 0.1722 P
P = 148.46 N
Now check the tipping of block,
∑ M0 = 0 = (148.46 sin 35°) (0.35) + (148.46 cos 35°) (0.5) – N1 (x)
29.83 + 60.79 – R1 x = 0
90.62 = R1 x
By putting the value of R1, we get
90.62 = (490.5 – 0.574 P) x
x = 0.224 m
As x < 0.35 m, tipping will not occur.
Hence block slides with P = 148.46 N
Example: A ladder 5 m long weighing 200 N is resting against a wall at an angle 0f 60° to the horizontal ground. A man weighing 500 N climbs the ladder. At what position along the ladder from bottom does he induce slipping. The coefficient of friction for both the wall and the ground with the ladder is 0.2.
Fig.14.3
Solution: Let the ladder be at the point of sliding when the man is at a distance x metres from the foot of the ladder. See Fig.14.3.
Let F be the position of the man.
BF = x, BE = AE = 2.5m
Let the normal reactions at the floor and the wall be R and S. Friction at the floor and the wall will be 0.2R and 0.2S respectively.
Resolving the forces on the ladder horizontally and vertically,
S = 0.2R--------------(1)
R + 0.2S = 700 N-------------(2)
From equations (1) and (2), we get R = 673.08 N
S = 134.62 N
200 × 2.5 cos 60° + 500 × x cos 60° = S × 5 sin 60° + 0.2S × 5 cos 60°
250 +250x = 2.5 S + 0.5 S
250 + 250x = S (2.5 + 0.5)
250 + 250x = 134.62 (2.5 + 0.5)
250 + 250x = 650.23
x = 1.60 m
Example: A block weighing 20 N is a rectangular prism resting on a rough inclined plane as shown in Fig.14.4.The block is tied up with a horizontal string which has a tension of 5 N. Find
(a) The frictional force on block
(b) Normal reaction of the inclined plane
(c) The coefficient of friction between the surfaces of contact
Fig.14.4
Solution: Weight of the block W = 20 N
Tension in the horizontal string T = 5 N
Angle of the inclined plane = 45°
Fig.14.5
(dotted arrows indicate the components of the force along the tangential and normal to the inclined plane)
Resolving the force parallel to inclined plane, we get
F + T = W
By substituting the values, we get
F + 5 = 20
F = 10.6 N
(b) Resolving the force in the normal direction, we have
R = W + T
= 20 + 5
R = 17.68 N
By using the relation, F = µR
10.6 = µ × 17.68
µ = 0.59
Example: A body resting on a rough horizontal plane required to pull 20N inclined at 30° to the plane just to remove it. It was found that a push of 25N inclined at 30° to the plane just removed the body. Determine the weight of the body and the coefficient of friction.
Fig.14.6
Solution: given, Pull = 20N, Push = 25N and θ = 30°
Let W = weight of the body in N, R = Normal reaction and µ = coefficient of friction
1. First of all,the pull acting on the body
Resolving the forces horizontally, we get
F = 20 = 20 × 0.866 = 17.32 N
Resolving the forces vertically, we get
R = W - 20 = W – 20 × 0.5 = (W – 10) N
According to the relation,
F = µ R
17.32 = µ (W – 10)…………………………………………………………………………..(i)
2. The push acting on the body
Fig.14.7
Resolving the forces horizontally, we get
F = 25 = 25 × 0.866 = 21.65 N
Now, resolving the forces vertically, we get
R = W + 25 = W + 25 × 0.5 = (W + 12.5) N
By using the relation,
F = µ R
21.65 = µ (W + 12.5)………………………………………………………………………...(ii)
Dividing the Eq. (i) to Eq. (ii), we get
${{17.32} \over {21.65}} = {{{\rm{\mu }}\left( {{\rm{W}} - 10} \right)} \over {{\rm{\mu }}\left( {{\rm{W}} + 12.5} \right)}}$ = ${{W - 10} \over {W + 12.5}}$
By simplification, we get
W = 100 N
Now, substituting the value of W in both Eq. (i) and Eq. (ii), we get
µ = 0.192 |
# What is the antiderivative of (ln x)^2/x^2?
Mar 17, 2016
$\int {\ln}^{2} \frac{x}{x} ^ 2 \mathrm{dx} = - \frac{{\ln}^{2} \left(x\right) + 2 \ln \left(x\right) + 2}{x} + C$
#### Explanation:
First, we will use substitution
Let $t = \ln \left(x\right) \implies \mathrm{dt} = \frac{1}{x} \mathrm{dx}$ and $x = {e}^{t}$
Then
$\int {\ln}^{2} \frac{x}{x} ^ 2 \mathrm{dx} = \int {\ln}^{2} \frac{x}{x} \cdot \frac{1}{x} \mathrm{dx} = \int {t}^{2} {e}^{-} t \mathrm{dt}$
Next, we will use the integration by parts forumla
$\int u \mathrm{dv} = u v - \int v \mathrm{du}$
Integration by Parts 1:
Let $u = {t}^{2}$ and $\mathrm{dv} = {e}^{-} t \mathrm{dt}$
Then $\mathrm{du} = 2 t$ and $v = - {e}^{-} t$
Applying the formula:
$\int {t}^{2} {e}^{-} t \mathrm{dt} = - {t}^{2} {e}^{-} t + 2 \int t {e}^{-} t \mathrm{dt}$
Integration by Parts 2:
Focusing on the remaining integral...
Let $u = t$ and $\mathrm{dv} = {e}^{-} t \mathrm{dt}$
Then $\mathrm{du} = \mathrm{dt}$ and $v = - {e}^{-} t$
Applying the formula:
$\int t {e}^{-} t \mathrm{dt} = - t {e}^{-} t + \int {e}^{-} t \mathrm{dt}$
$= - t {e}^{-} t - {e}^{-} t + C$
$= - {e}^{-} t \left(t + 1\right) + C$
Substituting back, we have
$\int {t}^{2} {e}^{-} t \mathrm{dt} = - {t}^{2} {e}^{-} t + 2 \left[- {e}^{-} t \left(t + 1\right)\right] + C$
$= - {e}^{-} t \left({t}^{2} + 2 t + 2\right) + C$
Finally, substituting $x$ back in gives our final result:
$\int {\ln}^{2} \frac{x}{x} ^ 2 \mathrm{dx} = - \frac{{\ln}^{2} \left(x\right) + 2 \ln \left(x\right) + 2}{x} + C$ |
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# Algebra I Midterm Review
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### Algebra I Midterm Review
1. 1. Algebra I Midterm Review Chapters 1 - 6 Name: Ms. Psillos Chapter 1: §1.4 - §1.6 (pg. 21 - 36) §1.4 Identity and Equality Properties Identity/Property Meaning Examples Additive Identity The sum of any number and 0 is equal to the number. (0 is the additive identity) Multiplicative Identity The product of any number and 1 is equal to the number. (1 is the multiplicative identity) Additive Inverse Two numbers whose sum is 0. Multiplicative Inverse Two numbers whose product is 1. Multiplicative Property The product of any number and 0 is equal to 0. of Zero Reflexive Property Transitive Property Symmetric Property Substitution Property §1.5 The Distributive Property Using the Distributive Property Simplify or Combine Like Terms. 1 1. (9 – p)3 2. 28 y − x 3. 4 y 3 + 3 y 3 + y 4 7 §1.6 Commutative and Associative Properties: Property Meaning Examples Commutative Property The order in which you add or multiply numbers does not change their sum or product. Associative Property The way you group three or more numbers when adding or multiplying numbers does not change their sum or product.
2. 2. Chapter 2: §2.1 - §2.4 (pg. 68 - 87) §2.1 Rational Numbers on the Number Line Rational Numbers Sets Natural Numbers {1, 2, 3, 4, …} Whole Numbers {0, 1, 2, 3, 4, …} Integers {…, -3, -2, -1, 0, 1, 2, 3, …} Rational Numbers a Numbers that can be expressed in the form , where a and b are integers and b ≠ 0 b 1. Graph the set of numbers { -9, -7, -4, -1, 0, 3, 6, 8} 0 2. Identify the Coordinates on the Numbers Line. 0 Absolute Value: the absolute value of any number n is its distance from zero on a number line. 3. If a = 8 and z = - 5 evaluate the expression 3 + | z – 17 + a2| §2.2 Adding and Subtracting Rational Numbers Find the Sum or the Difference of each example. 6 2 7 3 4. -12 + (-15) 5. -38.9 + 24.2 6. − + 7. − − − 7 3 8 16 §2.3 Multiplying Rational Numbers Find the Product of each example. 4 1 5 4 8. -9(-12) 9. (3.8)(-4.1) 10. − 1 2 11. − 5 2 12 9 12. Simplify: 7m(-3n) + 3s(-4t) 13. Simplify: 6(-2x) – 14x
3. 3. §2.4 Dividing Rational Numbers Find the Quotient of each example. 1 16 4 7 h + 35 14. -78 ÷ (-4) 15. − ÷ 4 16. ÷− 17. Simplify: 3 36 12 −7 Chapter 3: §3.1 - §3.9 (pg. 119 - 87) §3.1 Writing Equations Translate verbal sentences into equations, then solve. 1. Nine times y subtracted from 95 equals 37. 2. Two times a number t decreased by eight equals seventy. 3. Half of the sum of nine and p is the same as p minus three. Translate equations into verbal sentences, then solve. 4. 2f + 6 = 19 5. 7(m + n) = 10n +17 §3.2 Solving Equations by Using Addition and Subtraction Solve each equation. 1 5 2 4 6. s – 19 = - 34 7. -25 = -150 + q 8. − +p= 9. =v+ 2 8 3 8 §3.3 Solving Equations by Using Multiplication and Division v v 2 z 2 10. –5r = 55 11. − = −9 12. =1 13. =− 7 3 7 45 5
4. 4. §3.4 Solving Multi-Step Equations c − 3 j − ( − 4) m 14. 7 + 3c = -11 15. − +5= 7 16. = 12 17. + 6 = 31 7 −6 −5 §3.5 Solving Equations with the Variable on Each Side 1 3 1 18. 5t – 9 = -3t +7 19. 5 − ( x − 6 ) = 4 20. y− y = 4+ y 21. 3(1 + d ) − 5 = 3d − 2 2 2 2 §3.6 Ratios and Proportions Ratio: is a comparison of two numbers by division expressed in the following ways: Proportion: is an equation stating that two ratios are equal. Rate: the ratio of two measurements having different units of measure. Scale: a rate or a ratio that is used when making a model or a drawing of something that is too large or too small to be drawn at actual size. Use Cross Multiplication to determine if each pairs of ratios form a proportion. 3 21 8 16 22. , 23. , 2 14 9 17 Solve each proportion. 6 x 9 16 5 6 6 7 24. = 25. = 26. = 27. = 5 15 b 7 3 x+2 14 x − 3
5. 5. Solve Word Problems Involving Rates and Scales. 28. Sam runs 15-miles every Saturday. She runs this distance in 4 hours. At this rate, how far can she run in 6 hours? 29. The scale of a map of NYC is 2 inches per every 1.8 miles. The distance between Xavier High School and Rockefeller Center on the map is 7 inches. What is the distance between these two places? §3.7 Percent of Change Find Percent of Increase or Decrease. 30. Original: 66 31. Original: 40 32. Original: 15.6 33. Original: 85 New: 30 New: 32.5 New: 11.4 New: 90 Find Amount After Sales Tax. 34. Candle: \$7.50 35. Original: \$35.00 Tax: 5.75% Tax: 7% Find Amount After Discount. 36. Watch: \$37.55 37. Shirt: \$45.00 Discount: 35% Discount: 40% §3.8 Solving Equations and Formulas Solve an equation for a specific variable. 38. v = r + at, for a 39. 9a – 2b = c + 4a, for a 3ax − n 40. 2g – m = 5 – gh, for g 41. = −4 , for x 5
6. 6. §3.9 Weighted Average Solve a Mixture Problem with Prices 42. Anthony wants to create a special blend using two coffees, one priced at \$8.40 per pound and the other at \$7.28 per pound. How many pounds of the \$7.28 coffee should he mix with 9 pounds of the \$8.40 coffee to sell the mixture for \$7.95 per pound? Solve a Mixture Problem with Percents 43. John has 35 milliliters of 30% solution of copper sulfate. How much of a 20% solution of copper sulfate should she add to obtain a 22% solution? Travel 44. A railroad switching operator has discovered that two trains are heading toward each other on the same track. Currently, the trains are 53 miles apart. One train is traveling at 75 miles per hour and the other train is traveling at 40 miles per hour. The faster train will require 5 miles to stop safely, and the slower train will require 3 miles to stop safely. About how many minutes does the operator have to warn the train engineers to stop their trains?
7. 7. Chapter 4: §4.1 - §4.6 (pg. 192 - 231) §4.1 The Coordinate Plane 1. Write the ordered pairs for points A, B, C, and D. Name the quadrant in which each point is located. §4.3 Relations Relation: a set of ordered pairs represented as a set, a table, a graph or a mapping. Domain: the set of the first numbers of the ordered pairs (set of x values). Range: the set of the second numbers of the ordered pairs (set of all y values). Inverse: switching the coordinates in each ordered pair. 1. Given the relation, {(-2, 4), (5, -1), (8, 9), (0, -5), (8, -1)}: a. Express the relation as a table and a mapping. b. State Domain: c. State Range: d. Write the Inverse of the Relation: e. Is this relation a function? Explain why. §4.4 Equations as Relations Find the solution set for y = 2x + 3, given the replacement set {(-2, -1), (-1, 3), (3, 9), (0, 4)}. x y y = 2x + 3 True or False
8. 8. Solve 6x – y = -3 if the domain is {-2, -1, 1, 3, 4}. x 6x – y = -3 y (x, y) §4.5 Graphing Linear Equations Linear Equations in Standard Form: Ax + By = C, where A is greater than or equal to 0, A and B are not both zero, and A, B, and C are integers whose greatest common factor is 1. Intercepts: the point at which is crosses a particular axis. To find x-intercepts: To find y-intercepts: Find the x and y intercepts for each equation. x 2y 3. 2x = 6 - y 4. = 10 + 5. 4x + 6y = 8 6. 3x - 2y = 15 2 3 §4.6 Functions Functions: a relation in which each element of the domain is paired with exactly one element of the range. Vertical Line Test: if a vertical line can be drawn so that it intersects the graph more than once, then the graph is not a function. By using the Vertical Line Test, determine which graphs are functions. 7. 8. 9. 10. Determine whether each relation is a function. 11. 12. 13. 14.
9. 9. Equations that are functions can be written in function notation. y = 2x + 3 f(x) = 2x + 3 If f(x) = 2x - 6 and g(x) = x – 2x2, find each value. 1 15. f − 16. g ( − 1) 17. f ( h + 9 ) 2 18. 2[ g ( b ) + 1] 19. f ( 3 y ) 20. g ( − 3) + 13 Chapter 5: §5.1 - §5.6 (pg. 256 - 297) §5.1 Slope y 2 − y1 Formula: m = x 2 − x1 Find the slope of the line that passes through the given points and identify what type of slope it is: 1. (-1, 2) & (3, 4) 2. (1, 2) & (1, 3) 3. (-1, -2) & (-4, 1) 4. (1, 2) & (-1, 2) Find the coordinates given the slope of the line: (Find the value of r) 3 4 5. (r, 6) & (10, -3), m = − 6. (-2, 7) & (r, 3), m = 7. (4, -5) & (3, r), m = 8 2 3
10. 10. §5.2 Slope and Direct Variation Direct Variation equation: In words: “y varies directly with x.” Graph the following Equations: 1 8. y = x 9. y = −3x 2 Write and Solve a Direct Variation Equation: 10. Suppose y varies directly with x, and y = 28 when x = 7. Write a direct variation equation that relates x and y. Then use the equation to find x when y = 52. 11. Suppose y varies directly with x, and y = -7 when x = -14. Write a direct variation equation that relates x and y. Then use the equation to find x when x = 20. §5.3 Slope-Intercept Form Slope-Intercept Form: Identify the slope and y-intercept in each equation. 12. -2y = 6x - 4 13. 4 x − y = −3 14. 5x - 3y = 15 15. 3x - 2y = 6
11. 11. §5.4 Writing Equations in Slope-Intercept Form Write an equation given slope and one point, in slope-intercept form. Steps: 1. Substitute the values of m, x, and y into the slope-intercept form and solve for b. 2. Write the slope-intercept form using the values of m and b. 1 2 16. (1, 5), m = 2 17. (4, -5), m = − 18. (1, -4), m = -6 19. (-3, -1), m = − 2 3 Write an equation given two points. Steps: 1. Find the slope. 2. Choose one of the points to use. 3. Substitute the values of m, x, and y into the slope-intercept form and solve for b. 4. Write the slope-intercept form using the values of m and b. 20. (-3, -1) & (6, -4) 21. (2, -2) & (3, 2) 22. (1, 1) & (7, 4) 23. (7, -2) & (-4, -2) §5.5 Writing Equations in Point-Intercept Form Point-Slope Form: Write an equation in point-slope form given the slope and one point. 2 5 24. (-6, 1), m = -4 25. (9, 1), m = 26. (-4, -3), m = 1 27. (1, -3), m = − 3 8
12. 12. Write each equation in slope-intercept form. 2 1 3 28. y + 2 = -2(x + 6) 29. y - 1 = (x + 9) 30. y + 3 = − (x + 2) 31. y - 5 = (x + 4) 3 4 2 §5.6 Geometry: Parallel and Perpendicular Lines Parallel Lines: Write the slope-intercept form of an equation of the line that passes through the given point and is parallel to the graph of each equation. 2 32. (-3, 2), y = x – 6 33. (3, 3), y = x – 1 3 1 34. (-5, -4), y = x+1 35. (2, -1), y = 2x + 2 2 Perpendicular Lines: Write the slope-intercept form of an equation of the line that passes through the given point and is perpendicular to the graph of each equation. 1 36. (1, -3), y = x + 4 37. (-3, 1), y = -3x + 7 2
13. 13. 38. (-2, 7), 2x - 5y = 3 39. (2, 4), x – 6y = 2 Chapter 6: §65.1 - §6.6 (pg. 318 - 358) §6.1 Solving Inequalities by Addition and Subtraction Page 359 #’s (9 – 15) ODD ONLY §6.2 Solving Inequalities by Multiplication and Division Page 360 #’s (19 – 25) ODD ONLY §6.3 Solving Multi-Step Inequalities Page 361 #’s (27 – 33) ODD ONLY §6.4 Solving Compound Inequalities Page 361 #’s (37 – 41) ODD ONLY §6.5 Solving Open Sentences Involving Absolute Value Page 362 #’s (43 – 49) ODD ONLY §6.6 Graphing Inequalities in Two Variables Page 362 #’s (51 – 57) ODD ONLY Check your answers in the back of the book. |
Thursday, March 24, 2011
Albert's Great Big Book of Integers
Chapter 1:
Integers could be express using a number line or integer chips.
In integers, when adding both positive and negative with the same number they will cancel each other out making the answer a zero.
eg. (+5) + (-5) or (-5) + 5
The brackets for the integers are like training wheels for making equations more understandable
but mostly we need to use standard form.
eg. (+6) + (-6) In standard form 6 -6
Chapter 2:
Multiplying Integers
The Sign Rule:
When you have an even number of negative factors, the product will be POSITIVE.
eg. (-4) x (-4) = +16
When you have an odd number of positive factors, the product will be NEGATIVE.
eg. (+5) x (-4) = -20
Ways of showing how to multiply integers:
Positive x Positive = Positive: (+2) x (+3) = +6 , (2) x (3) = 6 , (2) (3) = 6
or
2(3) = 6
or
2 groups of (+3)
Negative x Positive = Negative: (-2) x (+3), remove 2 groups of (+3)
Negative x Negative = Positive: (-2) x (-3), remove 2 groups of (-3)
Chapter 3:
Dividing Integers
The way of reading the dividing integers is:
- How many groups of __ are in __?
- How many __'s go into __?
Partitive Division - The making of groups or parts.
Quotative Division - Sharing with groups.
The quotient of the two integers with the same sign is Positive.
The quotient of the two integers with the same sign is Negative.
Chapter 4:
Order of Operations with Integers
B.E.D.M.A.S. is used to do the order of operations for integers which stands for:
Brackets
Exponents
Division
Multiplication
Subtraction
eg.
(+5) x (-3) + (-6)
÷ (+3) =
[
(+5) x (-3)] + [(-6) ÷ (+3)] =
(-15) + (-2) = -17
1 comment:
1. Awesome post Albert. I like the pictures and you have great information, very understandable. |
## Sunday, January 16, 2011
### Final Percent Post
What is a percent? A percent is a way to show a number out of 100. You can use the sign '%' to show percent. Examples: 50%, it is also equal to 50/100 as a fraction or 0.5 as a decimal.
In our recent unit on percent, we learned how to represent percents in different ways, convert them into decimals and fractions, how to calculate the percent of a number, and how to combine percents.
Representing Percents:
- To represent a percent, you can shade squares on a grid of 100 squares called a hundred grid. One completely shaded grid represents 100%.
100%
- To represent a percent greater than 100%, shade more than one grid.
-To represent a fractional percent between 0% and 1%, shade part of one square.
- To represent a fractional percent greater than 1%, shade squares from a hundred grid to show the whole number and part of one square from the grid to show the fraction.
Fractions, Decimals, Percents:
- Fractions, decimals, and percents can be used to represent numbers in various situations.
- Percents can be written as fractions and as decimals.
Percent of a Number:
- You can use mental math strategies such as halving, doubling, and dividing by ten to find the percents of some numbers.
- To calculate the percent of a number, write the percent as a decimal and then multiply by the number.
Combining Percents:
- Percents can be combined by adding to solve problems. 5% + 7% = 12% .
- To calculate the increase in a number,
• You can add the combined percent to the original number. 12% of 100 = 0.12 x 100 = 12
• You can multiply the original number by a single percent greater than 100. 112% of 100 = 1.12 x 100 = 112
- Percents of percents can be used to determine amounts that result from consecutive percent increases or decreases.
Here is my video about percents.
Here is a link to my percent scribe post:
http://spmath81610.blogspot.com/search/label/Olivia816 |
# 1. Working in pairs
The concept of ‘square numbers’ can be very abstract. Drawing squares, or making squares with counters can help pupils begin to gain a visual understanding. What numbers of counters do we need to make square shapes (i.e. those with equal numbers of counters in each row, and as many rows as columns)?
You need to plan your lessons to ensure that all pupils are participating. In Activity 1 you will ask your pupils to work in pairs.
## Case Study 1: Drawing square numbers
Mrs Baale in South Africa wanted her pupils to do some investigations in pairs with only some guidance from her. She was keen to see if the pupils could investigate square numbers for themselves.
She began the lesson by asking pupils to work in pairs. She drew a square on the board; then she drew a larger square, made up of four smaller squares (see Resource 1: Square numbers). She asked the pupils to draw as many other squares like these as they could in five minutes. She told the class that these numbers were called ‘square numbers’.
Mrs Baale asked the class if they could make more square numbers, and to note the number of little squares needed to make each big square.
By allowing the pupils to work mainly unaided, Mrs Baale felt they would gain confidence and find enjoyment in the lesson. She found that most pairs worked well together.
## Activity 1: Making square numbers with objects
Encourage each pupil to participate by asking the pairs to choose first one member to be the scribe (the one to draw) and the other the recorder, and then to swap these tasks. In this way, you can make sure each pupil is participating.
You could also give your pupils objects (seeds or small stones) as counters. Ask pairs of pupils to find ‘square numbers’ (those with equal numbers of counters in each row, and as many rows as columns).
Section 4: Seeing multiplication visually
2. Using games to explore rectangular numbers |
# How do you use Heron's formula to determine the area of a triangle with sides of that are 35, 28, and 21 units in length?
Jan 16, 2016
$294$ ${\text{units}}^{2}$
#### Explanation:
First, determine the semiperimeter $s$ of the triangle (which has sides $a , b , c$).
$s = \frac{a + b + c}{2}$
We know that $a = 35 , b = 28 , c = 21$ so
$s = \frac{35 + 28 + 21}{2} = 42$
Plug these into Heron's formula, which determines the area of a triangle:
$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
$A = \sqrt{42 \left(42 - 35\right) \left(42 - 28\right) \left(42 - 21\right)}$
$A = \sqrt{42 \times 7 \times 14 \times 21}$
$A = \sqrt{{2}^{2} \times {3}^{2} \times {7}^{4}}$
$A = 2 \times 3 \times {7}^{2}$
$A = 294$ |
# TRIGONOMETRY HELP: NAVIGATING THE BASICS
TRIGONOMETRY HELP: NAVIGATING THE BASICS
Trigonometry can be a challenging subject for many students. It involves memorizing formulas and solving equations that can be difficult to understand and apply. However, with the right approach, you can quickly learn the basics of trigonometry and begin to use it to solve problems. In this article, we'll provide an overview of the basic concepts of trigonometry and some helpful tips to make learning the subject easier. With these tools in your arsenal, you'll be able to tackle even the most difficult trigonometry problems with confidence.
Get Into The Finest Schools
1. Introduction to Trigonometry
Trigonometry is a mathematical discipline that studies the properties of angles in three dimensions. It is used in many fields of mathematics, including engineering, physics, and calculus.
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2. Understanding the Fundamentals of Trigonometry
A trigonometry problem is a problem in geometry that can be solved using trigonometry. The problem asks how to find the value of a particular angle in a certain direction, say the angle between the vertical and the horizontal.
3. Trigonometric Ratios and Their Uses
In mathematics, trigonometric ratios are relationships between certain intervals, usually measured in degrees. They are used in various areas of mathematics, including calculus, geometry, and trigonometry.
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4. Exploring the Different Types of Trigonometric Functions
Trigonometric functions are used to solve problems involving angles and angles in combination with other numbers. They can be graphed on a coordinate plane, or on an coordinate system that includes both the vertical and horizontal axes.
1
1. "A triangle is a three-dimensional figure with three sides. The length of one side is the length of the other two sides."
2
2. "A triangle has the length of the longest side (the hypotenuse) to the length of the shortest side (the hypotenuse). The sum of the lengths of the two shorter sides is the length of the triangle."
3
3. "There are three types of triangles: right triangle, a right triangle with one side missing, and a triangle with two sides missing. A right triangle has the length of the longest side (the hypotenuse) to the length of the shortest side (the hypotenuse). The sum of the lengths of the two shorter sides is the length of the triangle."
## 5. Tips and Strategies for Solving Trigonometry Problems
There are many tips and strategies for solving trigonometry problems, but here are five of the most effective. 1. Choose the right tools: A good way to solve trigonometry problems is to use a tool that is best suited for the task at hand. choosing the right calculator, ruler, and graph paper can help make the process easier. 2. Work with a partner: When working on a problem with someone else, it is often easier to keep track of what is being done and to have a understanding of the steps involved. This can make the process more efficient and less time-consuming. 3. Use a step-by-step guide: When working on a problem, it is often helpful to have a step-by-step guide to follow. This will help you to understand the steps involved and to avoid making mistakes. 4. Use online resources: Online resources can be a valuable tool when working on trigonometry problems. This includes websites that offer step-by-step guides, as well as websites that offer more general information about the subject. 5. Use a calculator: always use a calculator when working on trigonometry problems. This will help you to solve the problem more quickly and accurately.
4. "A right triangle has the length of the longest side (the hypotenuse) to the length of the shortest side (the hypotenuse). The hypotenuse is the longest side of the triangle. The sum of the lengths of the two shorter sides is the length of the triangle."
Unlock the Secrets of Trigonometry: Get Started with the Basics!
Trigonometry is a mathematics topic that can be learned in a short amount of time. By following the basic steps in this topic, anyone can learn how to work with angles and trigonometry basics. This will help them understand how to solve problems and solve trigonometric problems. |
# Circle Problems
#### Chapter 45
5 Steps - 3 Clicks
# Circle Problems
### Introduction
A circle is a set of points that equidistant from a fixed point and the point from which all the points on a circle are equidistant is called the center of the circle and represented by letter ‘O’. The distance from that point to the circle is called the radius of the circle.
### Methods
Each and every term of the circle is defined with the help of the below shown figure:
From the figure:
AB – diameter(d).
CD – chord.
O – centre.
• Chord: It is defined as any line segment whose endpoints lie on the circle.
• Sector: It is defined as a region bounded by any two radii and an arc lying between the radii.
• Arc: It is defined as any connected part of the circle.
• Segment: It is defined as a region, bounded by a chord and an arc lying between the chord’s endpoints and most importantly not including the center.
• Secant: It is defined as a straight line that intersects any curve(here circle) in 2 points.
• Tangent: It is defined as a straight line that touches the circle at a single point.
• To any point on the circle itself, the distance from the centre of the circle is known as radius.
• It is represented by the letter ‘r’.
Diameter:
• Twice the size of the radius is known as diameter.
• The length of a line segment that passes across a circle through the centre.
• It is indicated by the letter ‘d’.
Circumference:
• The circumference of a circle is the distance around the circle, which is analogous to the perimeter of a polygon.
• The ratio of the circumference (C) to the diameter (d) is equal for all the circles and is denoted by the Greek letter ‘π’ i.e.
$$\frac{C}{d}$$ = π
Here, the value of π is approximately equal to 3.14 and can also be approximated by the fraction i.e. $$\frac{22}{7}$$.
• If the radius of the circle is ‘r’, then
$$\frac{C}{d}$$ = $$\frac{C}{2r}$$ = π
So, the circumference is related to the radius by the equation i.e.
C = 2πr
Chord:
• A line segment joining any two points on the circumference of a circle is known as chord.
Congruent chords:
• If two chords are the same distance from the centre, in a circle or in congruent circles, then they are congruent.
Arcs:
• Separating the circumference of a circle by any two points into two parts is known as arc.
• Major arc is the bigger curve and minor arc is the smaller curve.
Congruent arcs:
• The arcs that the same degree measure and are in the same circle or in congruent circles is known as congruent arcs.
Theorem 1: If two minor arcs are congruent, then their central angles are congruent, for a circle or for congruent circles.
Theorem 2: If two central angles are congruent, then their arcs are congruent, for a circle or for congruent circles.
Inscribed angles:
• The angle with its vertex on a circle and with sides that contain chords of the circle is known as inscribed angle.
It is as shown in the figure below:
Arc length:
• When degree measure and measure of radius is known, length of any arc can be found by using the following formula i.e.
L = $$(\frac{n}{360})$$2r
Here,
L – length,
n – degree measure of arc, and
Area of the circle:
• The space its surface occupies is the area of a circle.
• To determine the area of a circle, use the formula i.e.
Area = π$$r^2$$
Tangent lines:
From the above figure,
• The line that intersects the circle is known as a tangent, at exactly one point called the point of tangency (P).
• If a line is tangent to a circle, then a radius drawn to the point of tangency is perpendicular to the tangent line.
• Also, if a line is perpendicular to a radius at its end point on the circle, then the line is a tangent to the circle at that end point.
Properties of circle:-
1. Tangents or secants intersecting:
According to the theorem, when two chords intersect inside a circle, the product of the measures of the two segments of one chord is equal to the product of the measures of the two segments of the other chord.
From the figure above, Chords AB and CD intersect.
BY the theorem stated above, OP x OQ = OR x OS.
2. A straight line joining the centre of a circle to the mid point of a chord is perpendicular to the chord.
3. a) Equal chords are equidistant from the centre of the circle.
b) Chords that are equidistant from centre are equal in length.
From the figure,
AB = CD
OM = ON
4.
Angle in a semicircle is always right angle i.e.
if AC is the diameter then angle B = 90°.
5. Every point in space outside the circle can extend exactly two tangent lines to the circle.
The distance from the origin of the two tangents to the points of tangency is always equal.
In the figure below, XY = XZ.
6. Angle subtended at the centre of a circle by an arc is double the angle subtended by it at any point on the remaining part of the circumference.
i.e. if O is the centre and P, Q, R are three points on a circle then $$\angle POQ$$ = $$\angle PRQ$$.
7. Angles in the same segment of a circle are equal.
Tangent properties of circle:-
1. $${OT}^2$$ = OA x OB.
2. If two circles touch then the point of contact lies on the line of the centres.
3. Common tangents:
a) If a line touches more than one circle at a time, it is said to be a common tangent to both the circles.
b) If two circles are apart from each other, four common tangents are possible.
c) If two circles touch externally, three common tangents are possible.
d) If two circles touch internally, only one common tangent is possible.
Area = $$\sqrt{(S – a)(S – b)(S – c)(S – d)}$$
Where,
a, b, c, d are four sides and S is the semi perimeter.
Ptolemy’s theorem:
If ABCD is a cyclic quadrilateral, then the product of the two diagonals is equal to the sum of the products of opposite sides.
i.e. AC x BD = (AB x CD) + (BC X AD).
### Samples
1. In the figure above, if the square inscribed in the circle has an area of 16, what is the area of the shaded region?
Solution:
It is clear from the figure that the area of the shaded region is $$\frac{1}{4}$$ of the difference between the area of the circle and the area of the square.
Given that,
Area of the square is 16.
So, each side has length 4.
If the radius is known, the area of the circle can be found.
So, draw a diagonal of the square and i.e. the diameter of the circle.
By Pythagorean theorem,
The length of the diagonal = $$\sqrt{4^2 + 4^2}$$ = 4$$\sqrt{2}$$.
Then, radius of the circle is r = $$\frac{4\sqrt{2}}{2}$$ = 2$$\sqrt{2}$$
The area of the circle is π$$r^2$$ = π$$(2\sqrt{2})^2$$ = 8π.
Therefore, area of the shaded region = $$\frac{8π – 16}{4}$$ = 2π – 4.
2. The radius of circle A is r, and the radius of circle B is $$\frac{3}{4}$$r. What is the ratio of the area of circle A to the area of circle B?
Solution:
So, its area is π$$r^2$$.
Circle B has radius $$\frac{3}{4}$$r.
So, its area is π$$(\frac{3r}{4})^2$$ = $$\frac{9πr^2}{16}$$.
Therefore, the ratio of the area of circle A to the area of circle B is π$$r^2$$ to $$\frac{9πr^2}{16}$$, which is the same as the ratio 1 to $$\frac{9}{16}$$, which is the same as the ratio 16 to 9.
3:
In the figure above, triangle RST is inscribed in a circle. The measure of angle RST is greater than 90 degrees, and the area of the circle is 25π.
Quantity A Quantity B The length of line segment RT 10
A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.
Solution:
The radius of the circle is 5 and the diameter is 10.
Since the area of the circle is 25π.
Line segment Rt is a diameter of the circle if and only if angle RST is a right angle.
Since, the measure of angle RST is greater than 90 degrees, it follows that the angle RST is not a right angle and the line segment RT is a chord but not a diameter.
Therefore, the length of line segment RT is less than 10.
Hence, option (B) is the right answer.
4.
Quantity A Quantity B The area of the circle in the figure The area of the circle with diameter 3y
A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.
olution:
The area of the circle is defined as A = π$$r^2$$.
The radius of the circle drawn is 2$$x$$.
So, area = 4$$x^2$$π.
The area of the circle with diameter 3$$y$$ would be a = π$$x^2$$
= π$$(3y)^2$$
= 9π$$y^2$$
However, since the values of $$x$$ and $$y$$ are not defined, it is impossible to evaluate whether quantity A(4$$x^2$$[/post_solution_box]π) or quantity B 9π$$y^2$$ is greater.
5. If circumference of circle is 10 cm. The measure of $$X$$ is 36 degrees.
Quantity A Quantity B The length of arc ABC 1 cm
A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.
Solution:
The length of arc ABC can be determined by
Length = $$\frac{X}{360}$$C.
Where,
X is the measure of the angle whose rays intersect the arc.
C is the circumference of the circle.
Length = $$\frac{X}{360}$$C
= $$\frac{36}{360}$$ x 10
= $$\frac{1}{10}$$ x 10
= 1 cm
Therefore, the values in both cases are equal.
Hence, correct option is C. |
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Inverse Variation Models
Identify and solve y=k/x form equations
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Inverse Variation Models
What if you were paid \$500 per week regardless of the number of hours you worked? The more hours you worked in a week (increasing quantity), the less your hourly rate (decreasing quantity) would be. How could you write and solve a function to model this situation? After completing Concept, you'll be able to write inverse variation equations and solve inverse variation applications like this one.
Watch This
Watch this video to see some more variation problems worked out, including problems involving joint variation.
Guidance
Many variables in real-world problems are related to each other by variations. A variation is an equation that relates a variable to one or more other variables by the operations of multiplication and division. There are three different kinds of variation: direct variation, inverse variation and joint variation.
Distinguish Direct and Inverse Variation
In direct variation relationships, the related variables will either increase together or decrease together at a steady rate. For instance, consider a person walking at three miles per hour. As time increases, the distance covered by the person walking also increases, at the rate of three miles each hour. The distance and time are related to each other by a direct variation:
distance=speed×time
Since the speed is a constant 3 miles per hour, we can write: d=3t\begin{align*}d = 3t\end{align*}.
The general equation for a direct variation is y=kx\begin{align*}y = kx\end{align*}, where k\begin{align*}k\end{align*} is called the constant of proportionality.
You can see from the equation that a direct variation is a linear equation with a y\begin{align*}y-\end{align*}intercept of zero. The graph of a direct variation relationship is a straight line passing through the origin whose slope is k\begin{align*}k\end{align*}, the constant of proportionality.
A second type of variation is inverse variation. When two quantities are related to each other inversely, one quantity increases as the other one decreases, and vice versa.
For instance, if we look at the formula distance=speed×time\begin{align*}distance = speed \times time\end{align*} again and solve for time, we obtain:
time=distancespeed
If we keep the distance constant, we see that as the speed of an object increases, then the time it takes to cover that distance decreases. Consider a car traveling a distance of 90 miles, then the formula relating time and speed is: t=90s\begin{align*}t = \frac{90}{s}\end{align*}.
The general equation for inverse variation is y=kx\begin{align*}y=\frac{k}{x}\end{align*}, where k\begin{align*}k\end{align*} is the constant of proportionality.
In this chapter, we’ll investigate how the graphs of these relationships behave.
Another type of variation is a joint variation. In this type of relationship, one variable may vary as a product of two or more variables.
For example, the volume of a cylinder is given by:
V=πR2h
In this example the volume varies directly as the product of the square of the radius of the base and the height of the cylinder. The constant of proportionality here is the number π\begin{align*}\pi\end{align*}.
In many application problems, the relationship between the variables is a combination of variations. For instance Newton’s Law of Gravitation states that the force of attraction between two spherical bodies varies jointly as the masses of the objects and inversely as the square of the distance between them:
F=Gm1m2d2
In this example the constant of proportionality is called the gravitational constant, and its value is given by G=6.673×1011 Nm2/kg2\begin{align*}G = 6.673 \times 10^{-11} \ N \cdot m^2 / kg^2\end{align*}.
Graph Inverse Variation Equations
We saw that the general equation for inverse variation is given by the formula y=kx\begin{align*}y = \frac{k}{x}\end{align*}, where k\begin{align*}k\end{align*} is a constant of proportionality. We will now show how the graphs of such relationships behave. We start by making a table of values. In most applications, x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} are positive, so in our table we’ll choose only positive values of x\begin{align*}x\end{align*}.
Example A
Graph an inverse variation relationship with the proportionality constant k=1\begin{align*}k = 1\end{align*}.
Solution
x\begin{align*}x\end{align*} y=1x\begin{align*}y =\frac {1}{x}\end{align*}
0 y=10=undefined\begin{align*}y=\frac {1}{0} = \text{undefined}\end{align*}
14\begin{align*}\frac {1}{4}\end{align*} y=114=4\begin{align*}y =\frac {1}{\frac{1}{4}}=4\end{align*}
12\begin{align*}\frac {1}{2}\end{align*} y=112=2\begin{align*}y=\frac {1}{\frac{1}{2}}=2\end{align*}
34\begin{align*}\frac {3}{4}\end{align*} y=134=1.33\begin{align*}y =\frac {1}{\frac{3}{4}}=1.33\end{align*}
1 y=11=1\begin{align*}y =\frac {1}{1}=1\end{align*}
32\begin{align*}\frac {3}{2}\end{align*} y=132=0.67\begin{align*}y=\frac {1}{\frac{3}{2}}=0.67\end{align*}
2 y=12=0.5\begin{align*}y =\frac {1}{2}=0.5\end{align*}
3 y=13=0.33\begin{align*}y=\frac {1}{3}=0.33\end{align*}
4 y=14=0.25\begin{align*}y =\frac {1}{4}=0.25\end{align*}
5 y=15=0.2\begin{align*}y =\frac {1}{5}=0.2\end{align*}
10 \begin{align*}y=\frac {1}{10}=0.1\end{align*}
Here is a graph showing these points connected with a smooth curve.
Both the table and the graph demonstrate the relationship between variables in an inverse variation. As one variable increases, the other variable decreases and vice versa.
Notice that when \begin{align*}x = 0\end{align*}, the value of \begin{align*}y\end{align*} is undefined. The graph shows that when the value of \begin{align*}x\end{align*} is very small, the value of \begin{align*}y\end{align*} is very big—so it approaches infinity as \begin{align*}x\end{align*} gets closer and closer to zero.
Similarly, as the value of \begin{align*}x\end{align*} gets very large, the value of \begin{align*}y\end{align*} gets smaller and smaller but never reaches zero. We will investigate this behavior in detail throughout this chapter.
Write Inverse Variation Equations
As we saw, an inverse variation fulfills the equation \begin{align*}y = \frac{k}{x}\end{align*}. In general, we need to know the value of \begin{align*}y\end{align*} at a particular value of \begin{align*}x\end{align*} in order to find the proportionality constant. Once we know the proportionality constant, we can then find the value of \begin{align*}y\end{align*} for any given value of \begin{align*}x\end{align*}.
Example B
If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*}, and if \begin{align*}y = 10\end{align*} when \begin{align*}x = 5\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x = 2\end{align*}.
Solution
Compare Graphs of Inverse Variation Equations
Inverse variation problems are the simplest example of rational functions. We saw that an inverse variation has the general equation: \begin{align*}y= \frac{k}{x}\end{align*}. In most real-world problems, \begin{align*}x\end{align*} and \begin{align*}y\end{align*} take only positive values. Below, we will show graphs of three inverse variation functions.
Example C
On the same coordinate grid, graph inverse variation relationships with the proportionality constants \begin{align*}k = 1, k = 2,\end{align*} and \begin{align*}k = \frac{1}{2}\end{align*}.
Solution
We’ll skip the table of values for this problem, and just show the graphs of the three functions on the same coordinate axes. Notice that for larger constants of proportionality, the curve decreases at a slower rate than for smaller constants of proportionality. This makes sense because the value of \begin{align*}y\end{align*} is related directly to the proportionality constants, so we should expect larger values of \begin{align*}y\end{align*} for larger values of \begin{align*}k\end{align*}.
Watch this video for help with the Examples above.
Vocabulary
• The general equation for a direct variation is \begin{align*}y = kx\end{align*}, where \begin{align*}k\end{align*} is called the constant of proportionality.
• The general equation for inverse variation is \begin{align*}y= \frac{k}{x}\end{align*}, where \begin{align*}k\end{align*} is the constant of proportionality.
Guided Practice
If \begin{align*}p\end{align*} is inversely proportional to the square of \begin{align*}q\end{align*}, and \begin{align*}p = 64\end{align*} when \begin{align*}q = 3\end{align*}, find \begin{align*}p\end{align*} when \begin{align*}q = 5\end{align*}.
Solution
Practice
For 1-4, graph the following inverse variation relationships.
1. \begin{align*}y= \frac{3}{x}\end{align*}
2. \begin{align*}y= \frac{10}{x}\end{align*}
3. \begin{align*}y= \frac{1}{4x}\end{align*}
4. \begin{align*}y= \frac{5}{6x}\end{align*}
5. If \begin{align*}z\end{align*} is inversely proportional to \begin{align*}w\end{align*} and \begin{align*}z = 81\end{align*} when \begin{align*}w = 9\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}z = 24\end{align*}.
6. If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*} and \begin{align*}y = 2\end{align*} when \begin{align*}x = 8\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x = 12\end{align*}.
7. If \begin{align*}a\end{align*} is inversely proportional to the square root of \begin{align*}b\end{align*}, and \begin{align*}a = 32\end{align*} when \begin{align*}b = 9\end{align*}, find \begin{align*}b\end{align*} when \begin{align*}a = 6\end{align*}.
8. If \begin{align*}w\end{align*} is inversely proportional to the square of \begin{align*}u\end{align*} and \begin{align*}w = 4\end{align*} when \begin{align*}u = 2\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}u = 8\end{align*}.
9. If \begin{align*}a\end{align*} is proportional to both \begin{align*}b\end{align*} and \begin{align*}c\end{align*} and \begin{align*}a = 7\end{align*} when \begin{align*}b = 2\end{align*} and \begin{align*}c = 6\end{align*}, find \begin{align*}a\end{align*} when \begin{align*}b = 4\end{align*} and \begin{align*}c = 3\end{align*}.
10. If \begin{align*}x\end{align*} is proportional to \begin{align*}y\end{align*} and inversely proportional to \begin{align*}z\end{align*}, and \begin{align*}x = 2\end{align*} when \begin{align*}y = 10\end{align*} and \begin{align*}z = 25\end{align*}, find \begin{align*}x\end{align*} when \begin{align*}y = 8\end{align*} and \begin{align*}z = 35\end{align*}.
11. If \begin{align*}a\end{align*} varies directly with \begin{align*}b\end{align*} and inversely with the square of \begin{align*}c\end{align*}, and \begin{align*}a = 10\end{align*} when \begin{align*}b = 5\end{align*} and \begin{align*}c = 2\end{align*}, find the value of \begin{align*}a\end{align*} when \begin{align*}b = 3\end{align*} and \begin{align*}c = 6\end{align*}.
12. If \begin{align*}x\end{align*} varies directly with \begin{align*}y\end{align*} and \begin{align*}z\end{align*} varies inversely with \begin{align*}x\end{align*}, and \begin{align*}z = 3\end{align*} when \begin{align*}y = 5\end{align*}, find \begin{align*}z\end{align*} when \begin{align*}y = 10\end{align*}.
Vocabulary Language: English
Constant of Proportionality
Constant of Proportionality
The constant of proportionality, commonly represented as $k$ is the constant ratio of two proportional quantities such as $x$ and $y$.
Direct Variation
Direct Variation
When the dependent variable grows large or small as the independent variable does.
Inverse Variation
Inverse Variation
Inverse variation is a relationship between two variables in which the product of the two variables is equal to a constant. As one variable increases the second variable decreases proportionally.
Joint Variation
Joint Variation
Variables exhibit joint variation if one variable varies directly as the product of two or more other variables. |
# How To: Find the area of a triangle given three points
In this tutorial, we learn how to find the area of a triangle given three points. First, you will need to plot the points on a graph. After this, find the base and the height using the graph. Next substitute into area of a triangle formula and then evaluate. When you finally find the area of the triangle, then you will write down the answer ending it with the units. This is a simple way to find the area of the triangle, you just have to make sure you count correctly and have calculated the area of the triangle out correctly.
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If the triangle is not a right triangle, how would you find the base and height. For example: point 1: (1, 5), point 2: (3, 6), point 3: (2, 7).
Use two points and create a line. that line will be the base of the triangle. Then with the third point you can join it with the two points of the base to create a triangle. Then with the third point that is not the base, create a perpendicular line to the base. now find the slope of the base. The perpendicular line will have a slope that is the negative reciprocal of the base's slope. use pont slope formula and substitute the values of the slope value ,x value and y value into the formula. I will use capital letters to show the different values. Y-y=m(X-x) . m stands for the value of the slope. You can use the distance formula to find the length of the base. for example, if your given points that make up your base is (0,3) and (4,0) you can use the distance formula to get that the length between the two points is 5. Find the base length. Then find the equation of the base. use the point slope formula mentioned before. the equation of my base can be either Y-3= -3/4 (X-0) or the equation can be Y-0= -3/4 ( X-4) since you can use any points on the base (or the line) to find the equation of the base. You also need to find the slope. you know that the perpendicular line to the base and the base intersect eachother. you set the equation of the base and the perpendicular line to y-intercept form. so my base equation becomes Y=-3/4x+3. for the equation of the perpendicular line, use the third given point. the third given point of my triangle is (0,0) so the equation is Y-0= 4/3 (X-0) so the y intercept form is Y=4/3(x). Set the two equations equal to eachother. So i take -3/4x+3 and 4/3(x) and get 4/3x=-3/4x+3. Solve for X and that will give you the x value for the intersection point. and then find the y value. then use distance formula. I did this really quick so dont trust me so much on the answer. just use method
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# Percentage
Ankur Kulhari
Percentage is very important topic for aptitude subject. Direct questions are not asked from this topic but you will find use of this topic in almost all question of Data Interpretation (DI), Interest, Profit and Loss, Partnership etc.
We will be discussing concept of percentage and shortcut tricks to solve these type of questions.
Here is the table to convert percentage into number, it will be useful while solving questions if you remember it.
1% 1/100=0.01 1/100 2% 2/100=0.02 1/50 3% 3/100=0.03 3/100 4% 4/100=0.04 1/25 5% 5/100=0.05 1/20 6% double of 3% => 0.06 3/50 7% 7/100=0.07 7/100 8% double of 4% => 0.08 2/25 9% 3 times of 3% => 0.09 9/100 10% 10 times of 1% => .10 1/10 15% 3 times of 5% => 0.15 3/20 20% 2 times of 10% => 0.20 1/5 25% 5 times of 5% => 0.25 1/4 30% 30/100=0.3 3/10 40% 40/100=0.4 2/5 50% 50/100=0.5 1/2 60% 60/100=0.6 3/5 80% 80/100=0.8 4/5
Percentage is calculated on the base of 100 (As per means each, cent means 100)
Example 1. Ram obtained 75 marks in maths in his pre-board exam out of 150. How much percent he obtained?
Explanation:
per 150 he obtained 75 marks
per cent (100) he obtained = (75/150)*100 = 50%
Percentage Change:
In such questions we first need to find out the base value (reference value) with respect to which we have to find out the change.
Example 2. Ram obtained 75 marks in pre-board exam. after re-evaluation 25 marks were increased. Find out the percentage change?
Explanation:
Initially, it was 50% {(75/150)*100}, now also we will be calculating percentage on the base value 150.
So, the change in marks =25
=> change in % = (25/150)*100 = 16.66%
We can cross verify it:
Now the total marks = 75+25 = 100
% = (100/150)*100 = 200/3 = 66.66%
Initially it was 50%,
Now, it is 66.66%
So, Change = 16.66%
A change can be of two types – an increase or a decrease.
Another type of questions:
When a value is changed from initial value to a final value,
% change = (Difference between initial and final value/initial value) X 100
Example 3. If 20 changes to 40, what is the % increase?
Explanation:
% increase = (40-20)/20 X 100 = 100%.
Note:
1. With respect to initial value we have to look for the change, so base value here is initial value.
2. If a value is doubled the percentage increase is 100.
3. If a value is tripled, the percentage change is 200 and so on.
Percentage Difference:
% Difference = (Difference between values/value compared with) X 100.
Example 4. By what percent is 40 more than 30?
Explanation:
% difference = (40-30)/30 X 100 = 33.33%
{Here 40 is compared with 30. So 30 is taken as denominator}
Exampl 5. By what % is 60 more than 30?
Explanation:
% difference = (60-30)/30 X 100 = 100%.
{Here is 60 is compared with 30.}
Hint: To calculate percentage difference the value that occurs after the word “than” in the question can directly be used as the denominator in the formula.
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# Non-linear Function in Discrete mathematics
On the basis of the name of a non-linear function, it is a function that is NOT linear. The non-linear function can be described as a function that does not create a line on the graph. The graph of a non-linear function will be anything but not a line. In this section, we will learn about the definition, graph, and examples of the non-linear function.
### What is a non-linear Function?
A non-linear function can be described as a function that does not draw a straight line on the graph. The graph of a non-linear function can be a curve, but it must not be a line.
For example: If we have a pound that has 100 fish in the starting and every week the fishes become double, in this case, we can model this situation with the help of function f(x) like f(x) = 100 (2)x, where f(x) is used to indicate the number of fishes, and x is used to indicate the number of weeks. Now we can make a table and graph of this function, which is described as follows:
x y
0 100
1 200
2 400
3 800
Now we will graph the above table, which is described as follows:
In the above graph, we can see that there is NO straight line. Hence this graph has a non-linear function. In the above graph of a non-function, there is no uniform way in the slope of this graph. We can show a non-linear function with the help of a table of values, an equation, or a graph. Now we will learn each of them. There are some examples of non-linear functions, i.e., polynomial functions, cubic functions, quadratic functions, etc.
### Table of Non-linear function
There are some steps that are used to find whether a table of values finds a linear function or not. These steps are described as follows:
Step 1: In this step, we will determine the difference between every two consecutive x values.
Step 2: In this step, we will determine the difference between every two consecutive y values.
Step 3: In this step, we will determine the corresponding ratios of differences of y and differences of x.
Step 4: The function will be linear only if the ratios calculated in step 3 are not the same.
To understand this, we will consider a table of values, which is described as follows:
x y
3 15
5 23
9 33
11 41
13 43
Now we will use the above steps and find out whether the above table shows a non-linear function.
The function is non-linear because there is not the same number of ratios between the differences of y to the differences of x.
### Equation of Non-linear function
A linear function is used to contain the form f(x) = ax+b. Since a linear function can be described as a function that will not be linear. The non-linear equation can contain any type of form, but that form must not be f(x) = ax+b. There are various examples of non-linear functions, and some of them are described as follows:
• f(x) = x2 is a non-linear and this function is also a quadratic function.
• f(x) = 2x is a non-linear function and this function is also a exponential function.
• f(x) = x3 - 3x is a non-linear function and this function is also a cubic function.
### Graphs of Non-linear Function
If there is a function that is not linear, then it will be known as the non-linear function. A non-linear function will be represented by those types of functions that do not generate a straight line in the graph. In the following graphs, we can see that there is no straight line in all the images. All the graphs show the non-linear function.
### Linear and Non-linear Functions
The linear and non-linear functions contain a lot of differences, which are described as follows:
Linear Functions Non-linear functions
If we use a graph to indicate the linear function, then that graph will be in a straight line. If we use a graph to indicate the non-linear function, then that graph will NOT be a line.
The linear function must be contained the form f(x) = ax+b. The non-linear function can contain any type of form, but that form must not be f(x) = ax+b.
The slope of linear function for any two points on the curve must be a constant. The slope of non-linear function on a graph for any two points must NOT be the same.
In case of a linear function, if we draw a table, then the ratio between differences of y and differences of x will be a constant. In the case of a non-linear function, if we draw a table, then the ratio between differences of y and differences of x will NOT be a constant.
### Important points of Non-linear function
• If there is a graph of a function that is not a line, then it will be non-linear.
• If there is an equation that does not contain the form f(x) = ax+b, then that function will be known as the non-linear function.
• With the help of an objective function z = ax+by, we can either represent the linear or non-linear function.
• There are some examples of non-linear functions such as exponential functions, polynomial functions, logarithm functions, etc.
### Examples of Non-linear
The no-linear is used to contain the various examples, and some of them are described as follows:
Example 1: In this example, there are three functions, and we have to determine which function is non-linear. These functions are described as follows:
1. f(x) = 4
2. f(x) = 3x - 7
3. f(x) = sin x
Solution:
1. The first function is f(x) = 4 and we can write it in the form f(x) = 0x+4. This function is same as the form f(x) = ax+b. Hence, we can say that this function is a linear function.
2. The second function is f(x) = 3x - 7 and this function contains the exponent function. Hence we can say that this function is a non-linear function.
3. The third function is f(x) = sin x and this function contains the trigonometric function. Hence we can say that this function is a non-linear function.
Hence, only (2) and (3) both are non-linear functions.
Example 2: In this example, we have a table with some values, and we have to show whether this table shows a linear function or not. The table is described as follows:
x y
1 5000
2 2500
3 1250
4 625
5 312.5
Solution:
As we can see, 1, 1, 1, and 1 are the differences between every two successive values of x.
Similarly, -2500, -1250, -625, and -312.5 are the differences between every two successive values of y.
The ratio between the x values and y values are -2500, -1250, -625, and -312.5, and these values are not the same.
Therefore, we can say that the function which is shown in the above table is non-linear.
Hence, this table is non-linear.
Example 3: In this example, there are four graphs, and we have to show whether they show non-linear or not.
Solution: According to the definition of a non-linear function, the graph of a non-linear function must not be a line. Graph numbers (a), (b), and (c) do not contain a straight line, whereas graph (d) contains a straight line. So we can say that graphs (a), (b), and (c) are non-linear graphs, but graph (d) is not a non-linear graph or a linear graph. |
Factorization
## Factorization
#### Basic Algebra > Factorization
What is factorization?
Consider the following example:
12 = 3 × 4
i.e., 12 is product of 3 and 4.
3 and 4 are called factors or divisors of 12
12 is also equal to 2 × 6.
Here again, numbers 2 and 6 are called factors or divisors of 12.
The process of writing number 12 into product of 3 and 4 or 2 and 6 is called factorization.
Which is the other way of factorizing 12?
It is 12 = 12 × 1
Above is an example of factorizing a monomial (here 12) into three pairs of other monomials 3, 4 and 2, 6 and 1,12.
An algebraic expression can also be factored into one or more factors.
For example, the polynomial x2 + 2x can be written as product of two binomials namely, x and (x+2)
x and (x+2) are called factors or divisors of x and (x+2) .
Also, they are polynomials, while, x is monomial and x+2 is a binomial.
So, what is Factorization?
Factorization
Expressing polynomials as product of other polynomials that cannot be further factorized is called Factorization
We will discuss all the various types of factorization in this lesson.
They are standard types found across all books on algebra. Here, we go.
### 1. First Type: Factorization into Monomials:
Remember distributive property? It is
a (b + c ) = ab + ac
Example 1: Factorize ax + bx
Solution:
In ax and bx, x is common and also a common factor.
Write the common factor x in the polynomial ax + bx outside as x(a + b)
Now, both x and a + b are factors of the polynomial
ax + bx
So, factorization of
ax + bx = x(a + b)
Here the common factor x is a monomial.
Important Note:
In 10(a + b), 10 is one of the two factors and we do not factorize 10. We need not write 10 as 1×10 or 2×5. Factors that are numbers are not further factorized.
What is Complete Factorization?
Example:
Factorize: 12x2 + 18x3
Solution:
12x2 + 18x3 can be factorized into any of the following ways:
12x2 + 18x3 = x (12x +18x2)
12x2 + 18x3 = 6x (2x + 3x2)
12x2 + 18 x3 = 6x2 (2 + 3x)
There may be more ways of expressing the polynomial 12x2 + 18x3 as product of two factors. Let us limit to the above three ways.
Which of the above three ways is accepted as standard form of factorization?
The third one.
In the third one, 6x2is the greatest common divisor or highest common factor of the two algebraic expressions (monomials) 12x2 and 18x3
Complete Factorization
In complete factorization, the greatest common factor is written as the common factor
Example 1: Factorize x2y2 + y2
Solution:
Completely factorize x2y2 + y2, i.e. write the greatest common factor of x2y2 and y2 as the common factor.
y2 is the highest common factor. To find the other factor, divide each term by the H.C.F. and add the quotients. (x2y2)/y2 = x2 and y2/y2 = 1
Sum of the quotients is x2 + 1, which is the other factor. Therefore,
x2y2 + y2 = y2(x2 + 1)
### 2. Second type of Factorization: By Grouping of Terms
1. Factorize a2 + bc + ab + ac
Solution:
let us group terms in a2 + bc + ab + ac
But, which terms to group?
Those terms which yield a common factor on grouping!
Group the terms like this:
a2 + ab + bc + ac
a is the common factor in a2 + ab and b is the common factor in ab + bc
a2 + ab = a (a + b) and bc + ac = c (b + a)
now, the common factor is the binomial (a + b).
{note that a + b is same as b + a, i.e., order of addition does not matter, since addition follows closure property which states a + b = b + a}
So, factorization of
a2 + ab + bc + ac =
a (a + b) + c (a + b) =
(a + b) (a + c)
(a + b) and (a + c)
are two binomial factors into which a2 + ab + bc + ac is factorized.
2. Factorize ax + bx + ay + by
Solution:
Group terms with similar literal coefficients and numerical coefficients.
One group is ax + bx, in which x is the same literal coefficient,
and the other is ay + by, in which the y is the same literal coefficient,
In ax + bx, the common factor is x and in ay +by, the common factor is y.
on factorization with common factors, we have:
ax + bx = x ( a + b ) and
ay + by = y (a + b)
so ax + bx + ay + by = x (a + b) + y (a + b).
(a + b) is the common binomial factor for the next step of factorization:
(a + b)(x + y)
Factorization of ax + bx + ay + by = (a + b)(x + y)
3. Factorize p2 qx + pq2y + px2 y+ qxy2
Solution:
Group p2qx and px2y.
Common factor is px
Factorization of p2qx and px2y = px (pq + xy)
Again, group pq2y and qx y2
common factor is qy
Factorization of pq2y and qxy2 = qy (pq + xy)
So we have :
p2 qx + pq2y + px2 y+ qxy2 = px (pq + xy) + qy (pq + xy)
In the next factorization step,
pq + xy is the common binomial factor. We have:
(pq + xy)(px + qy)
### Type 3: Factorization of Trinomial Perfect Squares
Trinomial Perfect Squares have three monomials, in which two terms are perfect squares and one term is the product of the square roots of the two terms which are perfect squares
Example
a2 + 2ab + b2
In this trinomial, a2 and b2 are the two perfect squares and 2ab is the product of the square roots of a2 and b2 Can you do the Factorization?
Factorization ofa2 + 2ab + b2 gives the famous formula:
(a + b) 2 = a2 + 2ab + b2
Example 1:
Factorize 9p2+ 24pq + 16q2
Solution:
9p2 = (3p)2, just like a2
16q2 = (4q)2, just like b2
24pq = 2(3p)(4q), just like 2ab
Applying, (a + b) 2 = a2 + 2ab + b2
Factorization of
9p2 + 24pq + 16q2 = (3p + 4q)2 = (3p + 4q)(3p + 4q)
Example 2:
Factorize -4x2 + 12x + 9
Solution:
In 4x2 + 12x – 9, the leading coefficient is –1.
Whenever the leading coefficient is negative, express the given polynomial as follows:
-1(4x2 – 12x + 9).
Now, use factorization of perfect trinomial squares method to factorize
4x2 – 12x + 9, which is in the form of the well-known algebraic formula:
a2 – 2 ab + b2 = (a – b) (a – b).
so, 4x2 – 12x + 9 = (2x – 3) (2x – 3)
Now, factorization of
4x2 + 12x – 9 = -1(4x2 – 12x + 9) = -1(2x – 3) (2x – 3) |
# HOW TO CONVERT A DECIMAL TO A FRACTION
How to Convert a Decimal to a Fraction :
In this section, you will learn, how to convert a decimal to a fraction.
Usually we have two kind of decimals.
Terminating decimal Non terminating and recurring decimal A decimal which ends with countable number of digits is known as terminating decimal.For example, 0.25, 0.5, ............... etc Non terminating decimal means, it will not end up with any number. For example,0.525252..............0.125125..............
## How to convert a terminating decimal to a fraction ?
To convert a terminating decimal to fraction, we have to follow the steps given below.
Step 1 :
First count the number of digits after the decimal point.
Step 2 :
By multiplying by 10, 100, 1000, ......... etc, we may get rid of the decimal point.
Fro example, if we have only two digits after the decimal point, we have to multiply both numerator and denominator by 100.
Step 3 :
If it is possible, we may simplify the numerator and denominator separately.
## How to convert a non terminating recurring decimal to a fraction ?
To know, how to convert a non-terminating recurring decimal to fraction,
## How to Convert a Decimal to a Fraction - Examples
Example 1 :
Convert the following decimal number in the form p/q, where p and q are and q ≠ 0.
0.35
Solution :
Number of digits after the decimal point is 2.
So, we have to multiply both numerator and denominator by 100.
0.35 x (100/100) = 35/100
We may simplify both numerator and denominator by 5 times table.
= 7/20
Hence the fraction form of the decimal 0.35 is 7/20.
Example 2 :
Convert the following decimal number in the form p/q, where p and q are and q ≠ 0.
2.176
Solution :
Number of digits after the decimal point is 3.
So, we have to multiply both numerator and denominator by 1000.
2.176 x (1000/1000) = 2176/1000
We may simplify both numerator and denominator
= 1088/500
= 544/250
= 272/125
Hence the fraction form of the decimal 2.176 is 272/125.
Example 3 :
Covert the given repeating decimal into fraction
0.33333.............
Solution :
Let x = 0.3333........... (1)
Number of digits in the repeating pattern is 1. That is 3.
Because there is only one digit in the repeating pattern, multiply both sides of (1) by 10.
10x = 3.333......... (2)
(2) - (1) ==>
10x - x = 3.333.........00
9x = 3
Divide each side by 9.
x = 3/9
x = 1/3
Hence the fraction form of the decimal 0.333..... is 1/3.
Example 4 :
Covert the given repeating decimal into fraction
0.6868.........
Solution :
Let x = 0.6868........... (1)
Number of digits in the repeating pattern is 2. That is 68.
Because there are two digits in the repeating pattern, multiply both sides of (1) by 100.
100x = 68.6868............... (2)
(2) - (1)==>
100x - x = 68.6868.........0
99x = 68
Divide each side by 99.
x = 68/99
Hence the fraction form of the decimal 0.6868...is 68/99.
Example 5 :
Covert the given repeating decimal into fraction
32.03256256256..........
Solution :
Let x = 32.03256256256.............
Here, the repeated pattern is 256
No. of digits between the 1st repeated pattern and decimal is 2.
So, multiply the given decimal by 100. Then, we have
100x = 3203.256256256...............(1)
No. of digits between the 2nd repeated pattern and decimal = 5
So, multiply the given decimal by 100000. Then, we have
100000x = 3203256.256256256...............(2)
Subtracting (1) from (2), we get
(2) - (1) ===>
99900x = 3200053
x = 3200053/99900
Hence the fraction form of the decimal 32.03256256256. .......is 3200053/99900.
After having gone through the stuff given above, we hope that the students would have understood, how to convert a decimal to a fraction
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WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
Areas of Rectangles and Squares
## How to Find the Area of Rectangles and Squares
Rectangles and squares are two of the most common shapes.
Can you think of any objects that are a rectangle? 🤔
Here are a few examples of rectangular objects:
Can you think objects that are square?
Here are some square objects:
In this lesson, we're going to learn to find the area of rectangles and squares.
#### Let's Review
Area is the amount of space that a shape or object covers.
Look at this shape.
How many squares does this green rectangle cover?
It covers 6 squares.
Its area is 6 square units.
We use a unit of area to know exactly how a shape is measured.
For example, if each square in the grid above is 1 square meter, then the area of the green rectangle is 6 square meters.
Tip: We can write 6 square meters as 6 m². This is read as "6 square meters"
Take note of the small number 2 written at the top right side of m. That's really important.
### Area of a Rectangle
To know the area of the rectangle, we count the number of squares it covers.
Another way is to multiply the length of the sides.
In a rectangle, opposite sides are equal.
The length of the sides that go from left to right is called width.
The length of the sides that go from top to bottom is called height or length.
To find the area of a rectangle, multiply the width by the height.
width x height = area
Let's use this formula to find the width of the green rectangle.
First, know the width and height.
The width is 3 square units, and the height is 2 square units.
Now, multiply the width by height.
3 x 2 = 6 square units
✅ If 1 square unit is 1 square meter, then 6 square units is 6 square meters (6 m²).
It's the same answer we got when we counted each square. 😃
### Area of a Square
A square has 4 equal sides.
This means that the length of the sides are the same.
To find the area of a square, multiply the length of 1 side by itself.
What's the area of this square?
The length of each side is 3.
3 x 3 = 9 square units
The area of the square is 9 square units.
✅ If 1 square unit is 1 centimeter, then 9 square units is 9 square centimeter (9 cm²).
Great job! 😺
### No Grids
If there are no grids with squares to help you, you can still figure out the area of a shape.
To find the area of a rectangle, multiply:
length of width x length of height = area
To find the area of a square, multiply:
the length of any side by itself
Let's practice with examples!
#### Find the area of this rectangle:
We see that the width is 10 inches, and the height is 6 inches.
We multiply these two numbers together.
10 x 6 = 60
✅ The area of this rectangle is 60 square inches (or 60 in²).
😃 Make sure to write the unit of area after the number.
#### What is the area of this square?
The length of one side is 4 feet.
Let's multiply that to itself.
4 ft x 4 ft = 16 square feet
✅ The area of this square is 16 square feet, or 16 ft².
Great work! 🎇
### Watch and Learn
Now try the practice exercises. 💪
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1. How much does it cost?
2. How much money do you have now?
3. How much more money do you need to save to buy it?
4. How will you earn that amount?
5. How much time will it take to save it?
## Example In Action
### Luna is a high schooler who has a part-time job babysitting every Saturday.
She earns \$60 every month from that. She has \$356 saved up in her bank account. She wants to buy a new laptop that costs \$1000, and she wants it in time for final exams in 6 months.
### Set up an equation to see if she can reach her goal:
• Part A: Will she have enough money to buy the laptop in 6 months?
• Part B: If not, how many months will it take her to save for the laptop?
## Answer the 5 questions 👉
### We know that Luna has decided to buy a laptop.
1. Find out how much it costs.
The laptop costs \$1000.
2. How much money does she currently have?
She has \$365 saved up.
3. How much more money does she need to save to buy it?
See calculation below.
4. How will she earn that amount?
She saves the earnings from her part-time job. She earns \$60 per month.
5. How much time will it take to save it?
See the calculation in the next step.
Use the formula below to create an algebraic equation to find the answer:
## Setting Up The Equation: Part A
### Will Luna will have enough money to buy the laptop in 6months?
Follow this equation to see if she will or not.
Here, the total amount she saves at the end of 6 months is \$725.
### 👉 Luna won't have enough money to buy the laptop in 6 months.
\$1000 - \$725 = \$275
## Setting Up The Equation: Part B
### How many months will it take Luna to save up for the laptop?
To answer the second part of the question, identify the known information & place it in the equation:
Here, x = 10.58, but this value isn't a whole number, so we need to round it up to 11.
### Can you figure out the solution to the problem below?
Question: Ruby broke her camera, so she'd like a new one that costs \$250. She wants to buy it in 5 weeks. How much does she need to save each week?
Hint: Place the values you know into the algebraic formula shown above and find the unknown value.
#### Quiz
How much does Ruby need to save each week?
\$60
\$55
\$50
\$45
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# Lesson 12
Volume of Right Prisms
### Lesson Narrative
In grades 5 and 6, students calculated the volume of rectangular prisms. In this lesson, students learn that they can calculate the volume of any right prism by multiplying the area of the base times the height of the prism. Students make sense of this formula by picturing the prism decomposed into identical layers 1 unit tall. These layers are composed of a number of cubic units equal to the number of square units in the area of the base. The height of the prism tells how many of these layers there are. Therefore, multiplying the number of cubic units in one layer times the number of layers gives the total number of cubic units in the prism, regardless of the shape of the base.
Given some three-dimensional figures that are prisms and some that are not, students decide whether they can apply the formula $$V = Bh$$ to calculate the volume. If so, they identify the base and measure the height, before calculating the volume. Students also apply the formula $$V = Bh$$ to find the height of a prism given its volume and the area of its base.
### Learning Goals
Teacher Facing
• Determine the volume of a right prism by counting how many unit cubes it takes to build one layer and then multiplying by the number of layers.
• Generalize (orally) the relationship between the volume of a prism, the area of its base, and its height.
• Identify whether a given figure is a prism, and if so, identify its base and height.
### Student Facing
Let’s look at volumes of prisms.
### Required Preparation
You will need the Finding Volume with Cubes blackline master for this lesson. You will only use one of the two pages. If your snap cubes measure $$\frac34$$ inch, print the first page of the blackline master, with the slightly smaller shapes. If your snap cubes measure 2 cm, print the second page of the blackline master, with the slightly larger shapes. Make sure to print the blackline master at 100% scale so the dimensions are accurate. Prepare 1 copy for every 6 students, and cut the pages in half so that each group of 3 students has one half-page.
Print, cut, and assemble the nets from the Can You Find the Volume? blackline master. Card stock paper is recommended. Make sure to print the blackline master at 100% scale so the dimensions are accurate. Prepare 1 polyhedron for every 2 students (1 copy of the entire file for every 12--18 students).
Make sure students have access to snap cubes and rulers marked in centimeters.
### Student Facing
• I can explain why the volume of a prism can be found by multiplying the area of the base and the height of the prism.
### Glossary Entries
• volume
Volume is the number of cubic units that fill a three-dimensional region, without any gaps or overlaps.
For example, the volume of this rectangular prism is 60 units3, because it is composed of 3 layers that are each 20 units3. |
# A geometric sequence is a list of terms separated by a constant ratio, the number multiplied by each consecutive term in a geometric sequence. A geometric.
## Presentation on theme: "A geometric sequence is a list of terms separated by a constant ratio, the number multiplied by each consecutive term in a geometric sequence. A geometric."— Presentation transcript:
A geometric sequence is a list of terms separated by a constant ratio, the number multiplied by each consecutive term in a geometric sequence. A geometric sequence is an exponential function with a domain of positive consecutive integers in which the ratio between any two consecutive terms is equal. The rule for a geometric sequence can be expressed either explicitly or recursively.
The explicit rule for a geometric sequence is a n = a 1 r n – 1, where a 1 is the first term in the sequence, n is the term, r is the constant ratio, and a n is the nth term in the sequence. The recursive rule for a geometric sequence is a n = a n – 1 r, where a n is the nth term in the sequence, a n – 1 is the previous term, and r is the constant ratio.
Find the constant ratio, write the explicit formula, and find the seventh term for the following geometric sequence. › 3, 1.5, 0.75, 0.375, …
Find the constant ratio by dividing two successive terms. 1.5 ÷ 3 = 0.5 Identify the first term (a 1 ). a 1 = 3 Write the explicit formula. a n = a 1 r n – 1 a n = (3)(0.5) n – 1
To find the seventh term, substitute 7 for n. a 7 = (3)(0.5) 7 – 1 a 7 = (3)(0.5) 6 a 7 = 0.046875 The seventh term in the sequence is 0.046875
~Dr. Dambreville
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How To Calculate Mortgage Payments
By: John Stone
Break Studios Contributing Writer
If you have ever bought a home, or thought of buying a home, you’ve probably wondered about how to calculate mortgage payments. These days mortgages come in all shapes and sizes, each with their own intricacies. Understanding these intricacies is necessary to properly calculate mortgage payments and to determine how much house is affordable.
1. Understand the components of a mortgage payment. All mortgage payments can be broken down into two components: interest and principal. In a traditional fixed rate mortgage, there is a set monthly payment for the life of the mortgage (typically fifteen or thirty years). Most of the monthly payment goes to pay principal in the early part of the mortgage; however, by the end of the mortgage, most of the monthly payment goes to pay off principal.
2. Calculate the monthly interest rate. The effective monthly interest rate is simply the interest rate on the mortgage divided by twelve. For example, if you had a 6.00% mortgage rate, the monthly interest rate would be 6.00% / 12 = 0.50%. In decimal form, take the monthly interest rate percentage and divide by 100. Continuing the example above, 0.50% / 100 = 0.0050.
3. Calculate the monthly payment. To calculate the mortgage payment, use the monthly interest rate as described above. That interest rate can be placed into the following equation to determine the monthly mortgage payment: Monthly Payment = P[i(1 + i)n] / [(1 + i)n-1]. In this equation, P is the total mortgage amount, i is the monthly interest rate (in decimal form), and n is the number of months of the mortgage.
4. Understand this example. For example, if you had a 30 year \$200,000 mortgage with a 5.50% interest rate, the calculation of the mortgage payments would be as follows. First, calculate the monthly interest rate: 5.50% / 12 = 0.4583% = 0.004583. Next, calculate the number of months of the mortgage: 30 years * 12 = 360 months. Finally, calculate the mortgage payment: Payment = (\$200,000) [0.004583(1+0.004583)360] / [(1+0.004583)360– 1] = \$1135.58.
5. Realize that the principal and interest is only one portion of a complete monthly payment. What was calculated above is only the principal and interest component of a monthly mortgage payment. Most monthly payments include additional bills for property taxes, homeowner’s insurance, and private mortgage insurance. Property taxes vary greatly depending on where you live; consult with a real estate agent or the previous owner to get an estimate of the annual and monthly property tax bill. Consult with the insurance company of your choice to determine monthly homeowner’s insurance payments. Finally, discuss private mortgage insurance payments with your mortgage lender—typically, private mortgage insurance is only required if your down payment is less than twenty percent of the home’s value.
6. To summarize, total monthly mortgage payments consist of PITI. PITI stands for principal, interest, taxes and insurance. Add all of these sums together to get the total monthly cost of living in a home.
Owning a home is a dream for many. Properly understanding how to calculate mortgage payments will help you avoid purchasing more (or less) home than you can afford. Use the techniques above to calculate your monthly principal and interest, and consult with real estate agents, previous owners, your mortgage company and insurance company to get the monthly amounts for taxes and insurance.
Reference:
Posted on: Apr. 30, 2010 |
# 3.2: Applied Expected Value Calculations
Difficulty Level: At Grade Created by: Bruce DeWItt
(Note - Three Separate Videos for this Section)
### Learning Objectives
• Understand the concept of a fair game
• Be able to analyze a game of chance by building a probability model and calculating expected values from scratch
Casinos have a very delicate balancing act they must manage. First of all, they want to make money. However, people just don't like to lose money. In order to make money, the casino games have to be in favor of the house and not the player. Why don't casinos tilt the games even more to the house's favor? If they did, their expected value would certainly go up. On the other hand, attendance at the casino would go down.
No matter what the odds, a casino can't make money unless they can keep people coming through the doors. Setting the games up so that there are still winners, some occasionally big, is good for attendance. You might even know someone who has made a large amount of money at a casino.
In this section, we will bring together our ideas about calculating probabilities from Ch. 2 along with the concept of expected value in order to be able to analyze a game of chance. We begin where we left off in section 3.1. A fair game is a game in which neither the player nor the house has an advantage. In other words, when all is said and done, the average player will not have made or lost any money whatsoever.
#### Example 1
A bag has 10 red marbles and 8 blue marbles in it. A player reaches into the bag pulling out 2 marbles, one after the other without replacement. If the color of the two marbles match, the player wins $10. If they don't match, the player wins nothing. The game costs$5 to play.
a) Use a tree diagram to help find the probability that the two marbles match. Use your result to build a probability model.
b) Is this game a fair game? If so, explain why. If not, give the value that the game should cost in order to be fair.
#### Solution
Begin with a tree diagram that shows what might happen when two marbles are pulled.
Summarizing these results, we get the probability model shown in Table 3.8.
Result Red,Red Red,Blue Blue,Red Blue,Blue Value $10$0 $0$10 Probability \begin{align*}\frac{90}{306}\end{align*} \begin{align*}\frac{80}{306}\end{align*} \begin{align*}\frac{80}{306}\end{align*} \begin{align*}\frac{56}{306}\end{align*}
\begin{align*}EV=\left (\10\right )\left (\frac{90}{306}\right )+\left (\0\right )\left (\frac{80}{306}\right )+\left (\0\right )\left (\frac{80}{306}\right )+\left (\10\right )\left (\frac{56}{306}\right )=\frac{\1,460}{306}\approx\4.77\end{align*}
The expected value is $4.77. Notice, however, that$4.77 is the expected amount that the house pays out each game. The expected value for the house is $0.23 because every player must pay$5 to play. At $5, the game is not fair because it favors the house by an average of 23 cents every time the game is played. To be a fair game, it should cost$4.77.
The game of GREED is a game of chance in which players try to decide when they have accumulated enough points on a turn to stop. Two 6-sided dice are rolled. The player gets to keep the total that shows on the two dice. After every roll, the player can either decide to roll again and try to add to their current total for that turn or stop and put their points in the bank. The only catch in this game is that if a total of 5 is rolled, all points accumulated on that turn are lost. For example, suppose the first roll has a total of 9 and the player decides to go again. The next roll has a total of 7. The player now has 16 points accumulated on this turn and must decide to either put those 16 points in the bank or risk them. If they decide to risk the 16 points and a total of 5 comes up next, the score for that turn will be 0.
#### Example 2
Suppose a person is playing GREED and has accumulated 26 points so far. Is it to their advantage to roll one more time?
#### Solution
We will build a probability model and calculate the expected value based upon what might happen with one more roll. (See Example 3 from Section 3.1.) For example, there is a \begin{align*}\frac{1}{36}\end{align*} chance that the total will be 2. This would mean the player would have a total of 28 points with one more roll. The highest a player could have after this turn would be 38 points if they happen to roll a total of 12. The risk is that the player will roll a total of 5 and lose their 26 points.
The one item to be careful about here is that it is impossible to get a total of 31 in the chart. Remember, if you roll a total of 5, you lose all of your points. When we perform our expected value computation for the probability model above, we get approximately 29.6. In other words, if we roll exactly one more time, our average result will be almost 30 points. This is definitely better than stopping with 26 points. It is to the advantage of the player to roll again.
#### Example 3
An investor is going to make a long-term investment in a company. If all goes well, an investment of $100 will be worth$900 in twenty years. The risk is that the company may go bankrupt within twenty years in which case the investment is worthless. Suppose there is a 25% chance that the company will go bankrupt within 20 years. What is the expected value of this investment?
Start by building a probability model as shown below that shows that there is a 25% chance of making nothing and a 75% chance of making 900. \begin{align*}EV=\left (\0 \right )\left ({0.25} \right )+\left (\900 \right )\left ({0.75} \right )=\675\end{align*}. Taking into account that this investment cost100, the investor should get an average profit of $575. ### Problem Set 3.2 (Note - Two Videos for the Following Problems) #### Exercises 1) In the carnival game Wiffle Roll, a player will roll a wiffle ball across some colored cups. Suppose that if the ball stops in a blue cup, the player wins$20. If it stops in a red cup, they win $10, and if it stops in a white cup, the player wins nothing. There are 25 white cups, 4 red cups, and 1 blue cup. Assume the chances of stopping in any cup is the same. How much should this game cost if it is to be a fair game? 2) In a simple game, you roll a single 6-sided die one time. The amount you are paid is the same as the amount rolled. For example, if you roll a one, you get paid$1. If you roll a two you get paid $2 and so on. The only exception to this is if you roll a 6 in which case you get paid$12. What should this game cost in order to be a fair game?
3) Suppose you are playing the game of GREED as described in Example 2. You have accumulated a total of 55 points on one turn so far. Is it to your advantage to roll one more time?
4) Suppose you are playing the game of GREED again. This time you have accumulated a total of 60 points in one turn so far. Is it to your advantage to roll one more time?
5) Using your results from numbers 4) and 5) and a little more investigation, for what number of points in a turn in the game of GREED does it make no difference if you roll one more time or stop? In other words, at what point total does the expected value with one more roll give the same total as if you had stopped?
6) In the Minnesota Daily 3 lottery, players are given a lottery ticket based upon 3 digits that they pick. If their 3 digits match the winning digits in the correct order, then the player wins $500. If the digits don't match, then the player loses. The game costs$1 to play. What is the expected value for a player of this lottery game.
7) A bucket contains 12 blue, 10 red, and 8 yellow marbles. For $5, a player is allowed to randomly pick two marbles out of the bucket without replacement. If the colors of the two marbles match each other, the player wins$12. Otherwise the player wins nothing. What is the expected gain or loss for the player?
8) An insurance company insures an antique stamp collection worth $20,000 for an annual premium of$300. The insurance company collects $300 every year but only pays out the$20,000 if the collection is lost, damaged or stolen. Suppose the insurance company assesses the chance of the stamp collection being lost, stolen, or destroyed at 0.002. What is the expected annual profit for the insurance company?
9) A prospector purchases a parcel of land for $50,000 hoping that it contains significant amounts of natural gas. Based upon other parcels of land in the same area, there is a 20% chance that the land will be highly productive, a 70% chance that it will be somewhat productive, and a 10% chance that it will be completely unproductive. If it is determined that the land will be highly productive, the prospector will be able to sell the land for$130,000. If it is determined that the land is moderately productive, the prospector will be able to sell the land for $90,000. However, if the land is determined to be completely unproductive, the prospector will not be able to sell the land. Based upon the idea of expected value, did the prospector make a good investment? 10) A woman who is 35 years old purchases a term life insurance policy for an annual premium of$360. Based upon US government statistics, the probability that the woman will survive the year is .999057. Find the expected profit for the insurance company for this particular policy if it pays $250,000 upon the woman's death. 11) A bucket contains 1 gold, 3 silver, and 16 red marbles. A player randomly pulls one marble out of this bag. If they pull a gold marble, they get to pick one bill at random out of a money bag containing a$100 bill, five $20 bills, and fourteen$5 bills. If they pull a silver marble out of the bag, they get to pick one bill at random out of a bag containing a $100 bill, two$20 bills, and seventeen $2 bills. If your marble is red, you automatically lose. The game costs$5 to play.
a) Build a tree diagram for this situation.
b) Build a probability model for this situation.
c) Calculate the expected gain or loss for the player.
12) A spinner has four colors on it, red, blue, green, and yellow. Half of the spinner is red and the remaining half of the spinner is split evenly among the three remaining colors. A player pays some money to spin one time. If the spinner stops on red, the player receives $2. If it stops on blue, the player receives$4. If it stops on either green or yellow, the player wins $5. What should this game cost in order to be a fair game? 13) A bag contains 1 gold, 3 silver, and 6 red marbles. A second bag contains a$20 bill, three $10 bills, and six$1 bills. A player pulls out one marble from the first bag. If it is gold, they get to pick two bills from the money bag (without replacement). If it is a silver marble, they get to pick one bill from the money bag, and if the marble is red, they lose. The game costs $3 to play. Should you play? Explain why or why not. 14) Suppose the Minnesota Daily 3 lottery adds a new prize. You still get$500 if you match all three digits in order, but you can also win $80 if you have the three correct digits but not in the right order. The game still costs$1. What is the expected value for a player of this lottery game?
#### Review Exercises
15) Consider the partially complete probability model given below.
a) What is the value of 'X'?
b) What is the expected value for this situation.
16) The student council is starting to prepare for prom and decides to name a committee of 6 members. Suppose that they decide the committee will have 2 juniors and 4 seniors on it. In how many ways can the committee be selected if there are 8 juniors and 8 seniors from which to select?
17) Three cards are dealt off the top of a well-shuffled standard deck of cards. What is the probability that all three cards will be the same color?":
18) A student does not have enough time to finish a multiple choice test so they must guess on the last two questions. List the sample space of the possible guesses for the last two questions if each question has only choices a, b, and c.
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