text
stringlengths 22
1.01M
|
|---|
# CBSE Class 10 Mathematics Real Numbers Worksheet Set D
Read and download free pdf of CBSE Class 10 Mathematics Real Numbers Worksheet Set D. Students and teachers of Class 10 Real Numbers can get free printable Worksheets for Class 10 Real Numbers in PDF format prepared as per the latest syllabus and examination pattern in your schools. Standard 10 students should practice questions and answers given here for Real Numbers in Grade 10 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 10 Real Numbers Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems provided here with solutions on daily basis to get more score in school exams and tests
## Real Numbers Worksheet for Class 10
Class 10 Real Numbers students should refer to the following printable worksheet in Pdf in standard 10. This test paper with questions and answers for Grade 10 Real Numbers will be very useful for exams and help you to score good marks
### Class 10 Real Numbers Worksheet Pdf
Real Numbers
Q.- Insert a rational and an irrational number between 2 and 3.
Sol. If a and b are two positive rational numbers such that ab is not a perfect square of a rational number, then √ab is an irrational number lying between a and b. Also, if a,b are
rational numbers, then a+b/2 is a rational number between them.
∴ A rational number between 2 and 3 is 2+3 /2 = 2.5
An irrational number between 2 and 3 is √2×3= √6
Q.- Prove that
(i) √2 is irrational number
(ii) √3 is irrational number
Similarly √5, √7, √11…... are irrational numbers.
Sol. (i) Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (≠ 0) such that .√2= r/s
Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get ,√2=a/b
where a and b are coprime.
So, b √2= a.
Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2. Now, by
Theorem it following that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is,
b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational.
(ii) Let us assume, to contrary, that √3 is rational. That is, we can find integers a and b
(≠ 0) such that √3=a/b
Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b √3= a .
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is,b2 = 3c2.
This means that b2 is divisible by 3, and so b is
also divisible by 3 (using Theorem with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.
Q.- Using Euclid’s division algorithm, find the
H.C.F. of [NCERT]
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Sol.(i) Starting with the larger number i.e., 225, we get:
225 = 135 × 1 + 90
Now taking divisor 135 and remainder 90, we
get 135 = 90 × 1 + 45
Further taking divisor 90 and remainder 45,
we get 90 = 45 × 2 + 0
∴ Required H.C.F. = 45 (Ans.)
(ii) Starting with larger number 38220, we get:
38220 = 196 × 195 + 0
Since, the remainder is 0
=> H.C.F. = 196 (Ans.)
(iii) Given number are 867 and 255
=> 867 = 255 × 3 + 102 (Step-1)
255 = 102 × 2 + 51 (Step-2)
102 = 51 × 2 + 0 (Step-3)
=> H.C.F. = 51 (Ans.)
Q.- Show that one and only one out of n; n + 2 or n + 4 is divisible by 3, where n is any positive integer.
Sol. Consider any two positive integers a and b such that a is greater than b, then according to Euclid’s division algorithm:
a = bq + r; where q and r are positive integers and 0 ≤ r < b
Let a = n and b = 3, then
a = bq + r => n = 3q + r; where 0 ≤ r < 3.
r = 0 => n = 3q + 0 = 3q
r = 1 => n = 3q + 1 and r = 2 => n = 3q + 2
If n = 3q; n is divisible by 3
If n = 3q + 1; then n + 2 = 3q + 1 + 2
= 3q + 3; which is divisible by 3
=> n + 2 is divisible by 3
If n = 3q + 2; then n + 4 = 3q + 2 + 4
= 3q + 6; which is divisible by 3
=> n + 4 is divisible by 3
Hence, if n is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3.
Hence the required result.
Q.- Consider the number 6n, where n is a natural number. Check whether there is any value of ϵ N for which 6n is divisible by 7.
Sol. Since, 6 = 2 × 3; 6n = 2n × 3n
=> The prime factorisation of given number 6n
=> 6n is not divisible by 7. (Ans)
More question-
1. If 7 x 5 x 3 x 2 + 3 is composite number? Justify your answer
2. Show that any positive odd integer is of the form 4q + 1 or 4q + 3 where q is a positive integer
3. Show that 8n cannot end with the digit zero for any natural number n
4. Prove that 3√2 is irrational 5
5. Prove that √2 + √5 is irrational
6. Prove that 5 – 2 √3 is an irrational number
7. Prove that √2 is irrational
8. Use Euclid’s Division Algorithms to find the H.C.F of a) 135 and 225 (45)
b) 4052 and 12576 (4)
c) 270, 405 and 315 (45)
9. Using Euclid’s division algorithm, check whether the pair of numbers 50 and 20 are co-prime or not.
10. Find the HCF and LCM of 26 and 91 and verify that LCM X HCF = Product of two numbers (13,182)
11. Explain why 29 is a terminating decimal expansion 23 x 53
12. 163 will have a terminating decimal expansion. State true or false .Justify your answer. 150
13. Find HCF of 96 and 404 by prime factorization method. Hence, find their LCM. (4, 9696)
14. Using prime factorization method find the HCF and LCM of 72, 126 and 168 (6, 504)
15. If HCF (6, a) = 2 and LCM (6, a) = 60 then find a (20)
16. given that LCM (77, 99) = 693, find the HCF (77, 99) (11)
17. Find the greatest number which exactly divides 280 and 1245 leaving remainder 4 and 3 (138)
18. The LCM of two numbers is 64699, their HCF is 97 and one of the numbers is 2231. Find the other (2813)
19. Two numbers are in the ratio 15: 11. If their HCF is 13 and LCM is 2145 then find the numbers (195,143)
20. Express 0.363636………… in the form a/b (4/11)
21. Write the HCF of smallest composite number and smallest prime number
22. Write whether 2√45 + 3√20 on simplification give a rational or an irrational number 2√5 (6)
23. State whether 10.064 is rational or not. If rational, express in p/q form
24. Write a rational number between √2 and √3
25. State the fundamental theorem of arithmetic
26. The decimal expansion of the rational number 74 will terminate after ………. Places
23 . 54
Please click the below link to access CBSE Class 10 Mathematics Real Numbers Worksheet Set D
## Tags:
Click for more Real Numbers Study Material
CBSE Class 10 Mathematics Real Numbers Worksheet Set E CBSE Class 10 Mathematics Real Numbers Worksheet Set F CBSE Class 10 Mathematics Real Numbers Worksheet Set G CBSE Class 10 Mathematics Real Numbers Worksheet Set K CBSE Class 10 Mathematics Real Numbers Worksheet Set H CBSE Class 10 Mathematics Real Numbers Worksheet Set I CBSE Class 10 Mathematics Real Numbers Worksheet Set C CBSE Class 10 Mathematics Real Numbers Worksheet Set J CBSE Class 10 Mathematics Real Numbers Worksheet Set A CBSE Class 10 Mathematics Real Numbers Worksheet Set B CBSE Class 10 Mathematics Real Numbers Worksheet Set D
## Latest NCERT & CBSE News
Read the latest news and announcements from NCERT and CBSE below. Important updates relating to your studies which will help you to keep yourself updated with latest happenings in school level education. Keep yourself updated with all latest news and also read articles from teachers which will help you to improve your studies, increase motivation level and promote faster learning
### Surya Namaskar Project on 75th Anniversary of Independence Day
Ministry of Education, Govt of India vide letter No. F.No. 12-5/2020-IS-4 dated 16.12.2021 has intimated that under the banner Azadi ka Amrit Mahotsav the National Yogasanasports Federation has decided to run a project of 750 million Surya Namaskar from 01 January 2022...
### Moderation of Marks Class 11 and 12 Board Exams
The portal for moderation and finalization of results for Class-12 is being opened from 16.07.2021 to 22.07.2021. As Board has to declare the result latest by 31.07.2021, schools have been requested to follow the schedule strictly and complete the moderation within...
The acquisition of 21st century competencies of communication, critical and creative thinking and the ability to locate, understand and reflect on various kinds of information has become more crucial for our learners. It is well accepted that Reading Literacy is not...
|
## Calculus (3rd Edition)
Published by W. H. Freeman
# Chapter 16 - Multiple Integration - 16.2 Double Integrals over More General Regions - Exercises - Page 859: 42
#### Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 1} \right){\rm{d}}A = 24$
#### Work Step by Step
We have $f\left( {x,y} \right) = x + 1$. The domain is not a simple region. So, we divide it into two horizontally simple regions: ${{\cal D}_1}$ and ${{\cal D}_2}$ (please see the figure attached). 1. Consider region ${{\cal D}_1}$ The lower and upper boundaries are $y=1$ and $y=3$, respectively. Whereas, the left and right boundaries are the lines: $y - 1 = 2\left( {x - 1} \right)$ and $y - 1 = \frac{1}{2}\left( {x - 1} \right)$, respectively. So, Left boundary: $y - 1 = 2\left( {x - 1} \right)$, ${\ \ \ \ }$ $y=2x-1$ $x = \frac{{y + 1}}{2}$ Right boundary: $y - 1 = \frac{1}{2}\left( {x - 1} \right)$, ${\ \ \ \ }$ $y = \frac{1}{2}x + \frac{1}{2}$ $x = 2\left( {y - \frac{1}{2}} \right) = 2y - 1$ Thus, the domain description is ${{\cal D}_1} = \left\{ {\left( {x,y} \right)|1 \le y \le 3,\frac{{y + 1}}{2} \le x \le 2y - 1} \right\}$. Evaluate the double integral of $f\left( {x,y} \right) = x + 1$ over ${{\cal D}_1}$: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 1}^3 \mathop \smallint \limits_{x = \left( {y + 1} \right)/2}^{2y - 1} \left( {x + 1} \right){\rm{d}}x{\rm{d}}y$ $= \mathop \smallint \limits_{y = 1}^3 \left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{\left( {y + 1} \right)/2}^{2y - 1}} \right){\rm{d}}y$ $= \mathop \smallint \limits_{y = 1}^3 \left( {\left( {\frac{1}{2}{{\left( {2y - 1} \right)}^2} + 2y - 1 - \frac{1}{2}{{\left( {\frac{{y + 1}}{2}} \right)}^2} - \frac{{y + 1}}{2}} \right)} \right){\rm{d}}y$ $= \mathop \smallint \limits_{y = 1}^3 \left( {\frac{{15}}{8}{y^2} - \frac{3}{4}y - \frac{9}{8}} \right){\rm{d}}y$ $= \left( {\frac{5}{8}{y^3} - \frac{3}{8}{y^2} - \frac{9}{8}y} \right)|_1^3$ $= 11$ 2. Consider region ${{\cal D}_2}$ The lower and upper boundaries are $y=3$ and $y=5$, respectively. Whereas, the left and right boundaries are the lines: $y - 1 = 2\left( {x - 1} \right)$ and $y - 3 = - 1\left( {x - 5} \right)$, respectively. Notice that the left boundary is the same with that in ${{\cal D}_1}$. So, Left boundary: $x = \frac{{y + 1}}{2}$. Right boundary: $y - 3 = - 1\left( {x - 5} \right)$, ${\ \ \ \ }$ $y=-x+8$ $x=8-y$ Thus, the domain description is ${{\cal D}_2} = \left\{ {\left( {x,y} \right)|3 \le y \le 5,\frac{{y + 1}}{2} \le x \le 8 - y} \right\}$. Evaluate the double integral of $f\left( {x,y} \right) = x + 1$ over ${{\cal D}_2}$: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 3}^5 \mathop \smallint \limits_{x = \left( {y + 1} \right)/2}^{8 - y} \left( {x + 1} \right){\rm{d}}x{\rm{d}}y$ $= \mathop \smallint \limits_{y = 3}^5 \left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{\left( {y + 1} \right)/2}^{8 - y}} \right){\rm{d}}y$ $= \mathop \smallint \limits_{y = 3}^5 \left( {\left( {\frac{1}{2}{{\left( {8 - y} \right)}^2} + 8 - y - \frac{1}{2}{{\left( {\frac{{y + 1}}{2}} \right)}^2} - \frac{{y + 1}}{2}} \right)} \right){\rm{d}}y$$= \mathop \smallint \limits_{y = 3}^5 \left( {\left( {\frac{1}{2}{{\left( {8 - y} \right)}^2} + 8 - y - \frac{1}{2}{{\left( {\frac{{y + 1}}{2}} \right)}^2} - \frac{{y + 1}}{2}} \right)} \right){\rm{d}}y$ $= \mathop \smallint \limits_{y = 3}^5 \left( {\frac{3}{8}{y^2} - \frac{{39}}{4}y + \frac{{315}}{8}} \right){\rm{d}}y$ $= \left( {\frac{1}{8}{y^3} - \frac{{39}}{8}{y^2} + \frac{{315}}{8}y} \right)|_3^5$ $= 13$ Using the linearity properties of the double integral, we have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} f\left( {x,y} \right){\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {x,y} \right){\rm{d}}A$ $= 11 + 13$ $= 24$ So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 1} \right){\rm{d}}A = 24$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
|
# Logarithmic Inequality — How do I solve this: Log base 0. 3 (x^2+8) > log base 0. 3 (9x)?
Subham asked the following question involving a logarithmic inequality:
Solve $\log_{0.3} (x^2+8) > \log_{0.3} (9x)$.
## A Brush-Up on Small-Base Logarithmic Function
When $\log(x)$ has a base that is strictly between $0$ and $1$, it behaves as a strictly decreasing function, meaning that:
$\displaystyle x_1 > x_2 \iff \log x_1 < \log x_2$
## Proof
### Preliminary fact about Small-Base Exponential Function
First, notice that while for large bases ($>1$) exponential functions are strictly increasing, for small bases (i.e., base strictly between $0$ and $1$), exponential functions are actually strictly decreasing. That is:
For all $x_1, x_2 \in \mathbb{R}$ and any small base $b$, $x_1 > x_2 \iff b^{x_1} < b^{x_2}$.
### Going Back to the Proof
Given any two positive real numbers $x_1, x_2$ and any small base $b$, we break down the proof in two parts:
1. $\log_b (x_1) < \log_b (x_2) \longrightarrow x_1 > x_2$
2. $\log_b (x_1) \ge \log_b (x_2) \longrightarrow x_1 \le x_2$
For the first part, suppose that $\log (x_1) < \log (x_2)$, then the strictly-decreasing-ness of the exponential function with base $b$ implies that:
$\displaystyle b^{\log_b (x_1)} > b^{\log_b (x_2)}$
Or equivalently,
$\displaystyle x_1 > x_2$
So the first part is done. The second part is proved similarly. $\blacksquare$
## Going Back to the Original Logarithmic Inequality
Note that the logarithmic function of base 0.3 is strictly decreasing.
Applying the strictly-decreasing-ness of $\log_{0.3} (x)$ to the original logarithmic inequality, we have:
$x^2 + 8 < 9$
or
$(x-1)(x-8) < 0$
which means that $1 < x < 8$. Pretty neat. Isn’t it? 🙂
Follow
|
# Multiply Fractions by a Whole Number
Teaching students to multiply fractions by a whole number doesn’t have to be terrible! When you throw the terms “multiply” and “fractions” in the same standard, it can sound a bit daunting. Fortunately, if students have a sound understanding of multiplication and fractions, this is somewhat natural for students. This post includes some of my favorite lessons for teaching students to multiply fractions by whole numbers.
As with most of the fraction concepts, I start with teaching the concept through Cuisenaire Rods. If the brown rod is a whole, students would first have to determine which color rod is an eighth. Then, students would use repeated addition to find the sum of three groups of one-eighth.
This progresses past unit fractions, as well as reverses the order of the factors in the number sentence. It’s far too easy for students to think there is only one way to present content or to solve problems.
## Multiply Fractions By a Whole Number in Context
Just like other math concepts, it’s important to incorporate word problems to give context to the concept. After introducing the concept of multiplying fractions by whole numbers, I have students solve multiplication word problems. If needed, students can use manipulatives. However at this point, it’s good to have students transition to pictorial representations.
## Multiplication Algorithm
One of the misconceptions that students may have is that the operation of multiplication always produces a larger product. While this is true with whole numbers, this is not true for multiplication with fractions. In this task, students use area models to begin understanding the algorithm for multiplying fractions. Students will also begin to see that the product does not always increase when multiplying a whole number by a fraction.
In the activity, students solve four multiplication of fraction problems that require students to model their work through an area model. After students show how the problem can be represented through an area model, they should give a written explanation.
## Multiply Fractions by a Whole Number – Problem Solving
The remainder of my lessons that focus on multiplying a fraction by a whole number are problem solving lessons where students apply what they’ve already learned in problem solving situations.
I do try to avoid rushing through the concrete and pictorial stages of learning. It’s easy to jump ahead to completely abstract assignments, but students need time to work with multiple representations to solidify their understanding.
Throughout these lessons, review various problem solving strategies such as making a table or working backwards, as students will more than likely use these strategies in the problem solving tasks. Even though it may be initially difficult for students to get started in these tasks, the problem solving will encourage critical thinking skills and the application and conceptual understanding of fractions.
My favorite lesson is the Playground Design. In this task, students use the multiplication of fractions to design a new playground for their school. The task is divided into three options for the new playground. In each part, students shade in the fractional parts of the 12 playground units and write the fraction of the whole and the total units. After students have completed each section, they may use the second page to design the playground using the correct fractional parts. Encourage students to be creative and to complete high quality work.
You can find all of the multiplying fractions by a whole number lessons in this post here.
## Word Problems
We can’t forget about word problems! When teaching students to multiply fractions by whole numbers, students should understand that there are two different multiplication situations. One situation is when a whole number is multiplied by a fraction. For example, Lacy buys four bags of cherries that each weigh 1/3 pound. How many pounds of cherries does Lacy buy? The second situation is when a fraction is multiplied by a whole number. For example, a timer chimes every 1/2 hour during the school day. If the school day is 8 hours long, how many times does the timer chime each school day? I created a sort that allows students to review addition and subtraction word problems, as well as discern between the two multiplication situations. You can download the sort free here.
If you’re looking for more fraction resources and ideas don’t miss this post. It’s PACKED with ideas and content!
Scroll to Top
|
Successfully reported this slideshow.
Upcoming SlideShare
×
# Different types of functions
17,751 views
Published on
by my teacher Ms. Shelamar
Published in: Education
• Full Name
Comment goes here.
Are you sure you want to Yes No
• thank you for sharing Miss Katrina
Are you sure you want to Yes No
### Different types of functions
1. 1. Different Types of Functions prepared by: Shielamar L. Labiscase
2. 2. The set of real numbers SET NOTATION A set is collection of objects. The objects in a set are elements or members of the set.
3. 3. ROSTER METHOD of writing a set encloses the elements of the set in braces, {}. Ex. The set of natural numbers less than 11. A={1,2,3,4,5,6,7,8,9,10}
4. 4. A={1,2,3,4,5,6,7,8,9,10} This set has limited number of elements and is an example of finite set. To express the fact that 10 is an element of the set, use the symbol , i.e. 10 A.
5. 5. A second way to denote a set is use to use set-builder notation, where the set is written as A = { x/x is a natural number less than 11}
6. 6. Domain of the function The domain of the function is the set of consisting of all values of x. Determine the domain of f(x) = 3/(x+7) Therefore, the dom(f) is the set of all real numbers except -7.{x/x=/ -7}
7. 7. Range of the function • The range of the functions is the set consisting of all the second components in each element of the set. – Find the of f(x) = 3/ x+7 • Range (f) = {y/y=/0}
8. 8. Constant Function Identity Function Polynomial Function Absolute Value Function Square Root Function Rational Function Greatest Integer Function Piece-wise Function
9. 9. Constant Functions The constant function C is function with the range of which is consist of a single number k for all real numbers x in its domain. In symbol C(x) = k.
10. 10. Graph f(x) = 5 x x -2 -1 0 1 2 y f(x) 5 5 5 5 5
11. 11. What is the constant function of this graph?
12. 12. Domain and Range of a Function • The domain of the constant function is all real numbers • The range is the constant k. In this function is equal to 5 • The graph is a horizontal line.
13. 13. Identity Functions The identity function I is defined by I(x) = x. The domain is the set of real numbers. The range of the identity function is also the set of all real numbers.
14. 14. Graph f(x) = x X -2 -1 0 1 2 y -2 -1 0 1 2
15. 15. Polynomial Functions A polynomial in the variable x is a function that can be written in the form, where an, an-1 , ..., a2, a1, a0 are constants. We call the term containing the highest power of x(i.e. anxn) the leading term, and we call an the leading coefficient. The degree of the polynomial is the power of x in the leading term.
16. 16. Degree of the Polynomial Name of the function 0 Constant function 1 Linear function 2 Quadratic function 3 Cubic Function 4 Quartic Function 5 Quintic Function n (where n > 5) n (where n > 5)
17. 17. Linear Function
18. 18. Domain and Range of Linear Function The domain of a linear function is the set of real numbers {x/x is a real number}. The range of the linear function is the set of real numbers {y/y is a real number}.
19. 19. Quadratic function: f(x) = x^2
20. 20. Domain and Range of Quadratic Function Domain : {xx is a real number} Range: If a > 0, {f(x)/f(x) ≥ k} If a < 0, {f(x)/f(x) ≤ k}
21. 21. Cubic Function
22. 22. Absolute Value Function It is defined by f(x) = /x/ • The domain of the absolute value function is all real numbers. • The range is all non negative numbers
23. 23. Graph of f(x) = /x/ Domain: {x/x is a real number} Range: {f(x)/f(x) ≥ 0}
24. 24. Graph in one Cartesian Coordinate Plane • y = • y = • y = /x/ + 2 • y = /x/ - 2
25. 25. Graph of y= • To graph y = , simply shift the graph of y = /x/, A units to the left. • To graph y = , simply shift the graph of y = /x/, A units to the right.
26. 26. Graph of y= • To graph y = , simply shift the graph of y = /x/, B units upward. • To graph y = , simply shift the graph of y = /x/, B units downward.
27. 27. Domain: {x/x is a real number} Range: {y/y ≥ 1}
28. 28. Absolute value function
29. 29. Domain: {x/x is a real number} Range: {y/y ≤ -2}
30. 30. Square Root Functions f (x) x Domain : Range: Graph: 0x x 0y y
31. 31. Graph and determine the domain and range • h(x) = • H(x) =
32. 32. Rational Functions x f x 1 ( ) Domain: (x)/x ≠ 0} Range: {f(x)/f(x) ≠ 0}
33. 33. Definition of Asymptote An asymptote is an imaginary line being approached but never touched or intersected by a graph as it goes through infinity
34. 34. What is the domain and range of the rational function?
35. 35. What is the domain and range of the rational function?
36. 36. Greatest Integer Function: graph
37. 37. Domain and range of Greatest Integer Function The domain of G(x) = [x] is the set of real numbers. The range is the set of integers.
38. 38. Piece-wise function • when x is less than 2, it gives x2, • when x is exactly 2 it gives 6 • when x is more than 2 and less than or equal to 6 it gives the line 10-x
39. 39. It looks like this: • a solid dot means "including", an open dot means "not including")
40. 40. Domain and Range of the Piece-wise Function • The Domain (all the values that can go into the function) is all Real Numbers up to and including 6, which we can write like this • Dom(f) = (-∞, 6] (using Interval Notation) • Dom(f) = {x | x ≤ 6} (using Set Builder Notation)
41. 41. The sign of a real number, also called sgn or signum, is for a negative number (i.e., one with a minus sign " "), 0 for the number zero, or for apositive number (i.e., one with a plus sign " ").
42. 42. • For real, this can be written
43. 43. {x/x is a set of real number} {-1, 0, 1}
44. 44. • It is defined as {x/x is the set of real numbers} {0,1}
|
# Does a perpendicular bisector bisect an angle?
Perpendicular Bisector. Putting the two meanings together, we get the concept of a perpendicular bisector, a line, ray or line segment that bisects an angle or line segment at a right angle.
An angle bisector divides an angle into two congruent angles. A perpendicular bisector splits a segment into two congruent segments and is perpendicular to that segment.
Likewise, what does the perpendicular bisector theorem state? Oh, and the perpendicular bisector theorem – the theorem states that if a point is on the perpendicular bisector of a segment, then it is equidistant from the segment’s endpoints. The converse states that if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
Herein, does an angle bisector bisect the opposite side?
The “Angle Bisector” Theorem says that an angle bisector of a triangle will divide the opposite side into two segments that are proportional to the other two sides of the triangle. An angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
Is median perpendicular?
1 Answer. Segment joining a vertex to the mid-point of opposite side is called a median. Perpendicular from a vertex to opposite side is called altitude. A Line which passes through the mid-point of a segment and is perpendicular on the segment is called the perpendicular bisector of the segment.
### Are Bisectors always perpendicular?
When it is exactly at right angles to PQ it is called the perpendicular bisector. In general, ‘to bisect’ something means to cut it into two equal parts. With a perpendicular bisector, the bisector always crosses the line segment at right angles (90°).
### What is a perpendicular line?
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). The property extends to other related geometric objects. A line is said to be perpendicular to another line if the two lines intersect at a right angle.
### What is the meaning of bisector?
A bisector is something that cuts an object into two equal parts. It is applied to angles and line segments. In verb form, we say that it bisects the other object.
### What is the symbol for perpendicular bisector?
Table of symbols in geometry: Symbol Symbol Name Meaning / definition ⊥ perpendicular perpendicular lines (90° angle) ∥ parallel parallel lines ≅ congruent to equivalence of geometric shapes and size ~ similarity same shapes, not same size
### What is a perpendicular bisector of a triangle?
The perpendicular bisector of a side of a triangle is a line perpendicular to the side and passing through its midpoint. The three perpendicular bisectors of the sides of a triangle meet in a single point, called the circumcenter . The circumcenter is equidistant from the vertices of the triangle.
### Does a bisector cut an angle in half?
That line that was used to cut the angle in half is called the angle bisector. The angle bisector theorem tells us that the angle bisector divides the triangle’s sides proportionally. When you have an angle bisector, you also have two smaller triangles.
### How do you construct an angle?
Constructing a 30º Angle Step 1: Draw the arm PQ. Step 2: Place the point of the compass at P and draw an arc that passes through Q. Step 3: Place the point of the compass at Q and draw an arc that cuts the arc drawn in Step 2 at R. Step 4: With the point of the compass still at Q, draw an arc near T as shown.
|
Education.com
# Discrete Probability Distributions Study Guide
(not rated)
## Introduction to Discrete Probability Distributions
Because some probability distributions occur frequently in practice, they have been given specific names. In this lesson, we will discuss three discrete probability distributions: the Bernoulli, the binomial, and the geometric distributions.
## Bernoulli Distribution
Suppose we flip a fair coin and observe the upper face. The sample space may be represented as S = {Head, Tail}. Suppose we define X=1 if a head is on the upper face and 0 if a tail is on the upper face. X is an example of a random variable. We have used the term random variable somewhat loosely in earlier lessons. Formally, a random variable X assigns a numerical result to each possible outcome of a random experiment. If X can assume a finite or countably infinite number of values, then X is a discrete random variable; otherwise, X is a continuous random variable. In this lesson, we will consider the distributions of some discrete random variables.
The probability function P(X = x), or p(x), assigns a probability to each possible value of X. Because these are probabilities, 0 ≤ p(x) ≤ 1 for all X = x. Further, if we sum over all possible values of X, we must get one (i.e.,
#### Bernoulli Trial
A Bernoulli trial is any random experiment that has only two possible outcomes.
A discrete probability function is any function that satisfies the following two conditions: (1) The probabilities are between 0 and 1 and (2) the probabilities sum to one. As an illustration, let X= – 1, 0, or 1 if the stock market goes down, up, or stays the same, respectively, on a given day. The probabilities associated with the particular outcomes of X change from day to day. However, suppose for a given day, they are as follows:
Each probability is between 0 and 1, and the sum of the probabilities is one. Thus, this is a valid probability function. The graph of the distribution is given in Figure 10.1.
For the moment, we are going to focus on studies in which each observation may result in one of two possible outcomes. Flipping the coin is one such study as each flip will result in either a head or a tail. In an orchard, each piece of fruit either has or has not been damaged by insects. The television set tested at the factory either works or it does not. A person has a job or does not have a job. In each case, there are only two possible outcomes; one outcome may be labeled a success and the other a failure. A Bernoulli trial is any random experiment that has only two possible outcomes. For a Bernoulli trial, let X be a random variable defined as follows:
The choice of which outcome is considered a success and which is considered a failure is arbitrary. It is only important to clearly state for which outcome X =1 and for which X = 0. The probability of success is denoted by p where 0 < p < 1. Because there are only two outcomes, the probability of a success and the probability of a failure must sum to 1. Thus, the probability of a failure is 1 – p. We can present the probability distribution of the Bernoulli random variable as shown in Table 10.2.
150 Characters allowed
### Related Questions
#### Q:
See More Questions
### Today on Education.com
#### WORKBOOKS
May Workbooks are Here!
#### WE'VE GOT A GREAT ROUND-UP OF ACTIVITIES PERFECT FOR LONG WEEKENDS, STAYCATIONS, VACATIONS ... OR JUST SOME GOOD OLD-FASHIONED FUN!
Get Outside! 10 Playful Activities
#### PARENTING
7 Parenting Tips to Take the Pressure Off
Welcome!
|
## Intermediate Algebra (12th Edition)
$p+2p^{2}$
$\bf{\text{Solution Outline:}}$ Use the Distributive Property and the laws of exponents to simplify the given expression, $p^{2/3} \left( p^{1/3}+2p^{4/3} \right) .$ $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} p^{2/3}(p^{1/3})+p^{2/3}(2p^{4/3}) .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} p^{\frac{2}{3}+\frac{1}{3}}+2p^{\frac{2}{3}+\frac{4}{3}} \\\\= p^{\frac{3}{3}}+2p^{\frac{6}{3}} \\\\= p^{1}+2p^{2} \\\\= p+2p^{2} .\end{array}
|
# Writing Equations
As you prepare for college, one of the best things that you can do for yourself, outside of studying, is to build good relationships with your teachers. Learning the proper way to ask for help from your teachers can mean the difference between finally understanding a concept and getting written off as a whiner. Read this article and think about how you can use the given advice not just in the future, but in your classes right now.
## Algebra: Writing Equations
Read the following SAT test question and then select the correct answer.
Always read each question carefully and make a note of the bottom line. Assess your options for finding the bottom line and choose the most efficient method to attack the problem. When you have an answer, loop back to verify that it matches the bottom line.
A florist buys roses at \$0.50 a piece and sells them for \$1.00 a piece. If there are no other expenses, how many roses must be sold in order to make a profit of \$300?
Bottom Line: # roses = ?
Assess your Options: You could find the profit from a single rose and then start plugging in answer choices, but that is not the fastest way to solve this problem. A better way to solve this problem is to simply write an equation. You could also solve this problem in a few seconds by using logic.
Attack the Problem: Writing an equation will not take you much time. Start by finding the profit from a single rose: \$0.50. (You know that the florist spends \$0.50 to make each dollar, so \$1.00 - \$0.50 = \$0.50.)
If each rose brings in a profit of \$0.50, then how many must you sell to get \$300? Start by writing the fifty cents, and then use x to represent the unknown number of roses. Each rose costs the same, so multiply the two numbers. Together they must all equal \$300.
\$0.50x = \$300. (Just divide 300 by .5 to isolate the variable.)
x = 600
Loop back: The x represented roses so you found your bottom line. Look down at your answer choices.
(A) 100
(B) 150
(C) 200
(D) 300
(E) 600
|
# By using the properties of determinants show that $(i) \begin{vmatrix} x+4&2x&2x \\ 2x&x+4&2x \\ 2x&2x&x+4 \end{vmatrix} = (5x+4)(4-x)^2$
Note: This is part 1 of a 2 part question, split as 2 separate questions here.
Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}x+4& 2x & 2x\\2x & x+4 & 2x\\2x &2x & x+4\end{vmatrix}$
Now let us apply $R_1\rightarrow R_1+R_2+R_3$
Let $\bigtriangleup=\begin{vmatrix}5x+4& 5x+4 & 5x+4\\2x & x+4 & 2x\\2x &2x & x+4\end{vmatrix}$
By taking (5x+4) as a common factor from first row,
Let $\bigtriangleup=\begin{vmatrix}1& 1 & 1\\2x & x+4 & 2x\\2x &2x & x+4\end{vmatrix}$
By applying $C_1\rightarrow C_1-C_2$ and $C_2\rightarrow C_2-C_3$
Let $\bigtriangleup=\begin{vmatrix}0& 0 & 1\\x-4 & x-4 & 2x\\0 &x-4 & x+4\end{vmatrix}$
Take (x-4) as a common factor from $C_2$
$\bigtriangleup=(5x+4)(x-4)\begin{vmatrix}0& 0 & 1\\x-4 & 1 & 2x\\0 &1 & x+4\end{vmatrix}$
Expanding along $R_1$ we get,
$\bigtriangleup=(5x+4)(x-4)[1(x+4)-2x(1)]$
$\bigtriangleup=(5x+4)(x-4)[1(x+4)-2x]$
$\quad=(5x+4)(x-4)(4-x)$
$\quad=(-1)(5x+4)(4-x)(4-x)$
$\quad=(5x+4)(4-x)^2$
Hence $\begin{vmatrix}x+4 & 2x & 2x\\2x & x+4 & 2x\\2x & 2x & x+4\end{vmatrix}=(5x+4)(4-x)^2$
edited Feb 24, 2013
|
# Texas Go Math Grade 2 Lesson 10.4 Answer Key 3-Digit Subtraction
Refer to our Texas Go Math Grade 2 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 2 Lesson 10.4 Answer Key 3-Digit Subtraction.
## Texas Go Math Grade 2 Lesson 10.4 Answer Key 3-Digit Subtraction
Unlock the Problem
There were 436 people at the art show. 219 people left the art show early. How many people stayed at the art show?
What information am I given?
__________ people were at the art show.
Then, ___________ people left the show.
Plan
What is my plan or strategy?
I can ___________ to solve the problem.
Solve
Show how you solve the problem.
Make a model. Then draw a quick picture of your model.
_________ people
Explanation:
There are 438 people at the art show, 111 people left the show.
I can subtract to solve the problem.
There are 327 people at the art show.
HOME CONNECTION • Your child used a model and a quick picture to represent and solve a subtraction problem.
Try Another Problem
Make a model to solve. Then draw a quick picture of your model.
Question 1.
There are 532 pieces of art at the show. 319 pieces of art are paintings. How many pieces of art are not paintings?
____________ pieces of art.
Explanation:
532 – 319 is 213.
So, 213 pieces are not paintings.
Question 2.
245 children go to the face-painting event. 114 of the children ore boys. How many of the children are girls?
___________ girls
Explanation:
245 – 114 = 131
So, 131 children are girls.
Math Talk
Mathematical Processes
Explain how you solved the first problem.
Share and Show
Make a model to solve. Then draw a quick picture of your model.
Question 3.
There were 237 books on the shelves. Miss Lane took 126 books off the shelves. How many books were still on the shelves?
__________ books
Explanation:
237 – 126 = 111
So, 111 books are still on the shelves.
Problem Solving
Solve. Write or draw to explain.
Question 4.
H.O.T. Multi-Step Maria has 127 animal cards. Ellen has twice that number of cards. How many animal cards do the girls have?
_________ animal cards
Explanation:
The sum of 127, 127 and 127 is 381
So, the girls have 381 animal cards in all.
Question 5.
H.O.T. 164 children and 31 adults saw the movie in the morning. 125 children saw the movie in the afternoon. How many fewer children saw the movie in the afternoon than in the morning?
___________ fewer children
Explanation:
The difference of 164 and 125 is 39
So, 39 fewer children saw the movie in the afternoon than in the morning.
Question 6.
Analyze In the scavenger hunt, Shawn needs to find 125 objects. He found 116 objects so far. How many more objects does he need to find?
(A) 9
(B) 19
(C) 10
(A) 9
Explanation:
In the scavenger hunt, Shawn needs to find 125 objects
He found 116 objects so far
125 – 116 is 9
So, 9 more objects he need to find.
Question 7.
Apply There are 156 game cards in the box. Then Felipe takes 138 cards out of the box. How many cards are still in the box?
(A) 8
(B) 28
(C) 18
(C) 18
Explanation:
There are 156 game cards in the box
Then Felipe takes 138 cards out of the box
156 – 138 = 18
SO, 18 cards are still in the box.
Question 8.
Dion has 175 marbles. He put 155 marbles into a box, and gave the rest to his sister. How many marbles did Dion give to his sister?
(A) 25
(B) 15
(C) 20
(C) 20
Explanation:
Dion has 175 marbles
He put 155 marbles into a box, and gave the rest to his sister
175 – 155 = 20
So, Dion gave 20 marbles to his sister.
Question 9.
TEXAS Test Prep Mr. Grant had 350 balloons. He sold 133 balloons at the park. How many balloons does he have now?
(A) 223
(B) 217
(C) 483
(B) 217
Explanation:
He sold 133 balloons at the park
350 – 133 = 217
So, he have 217 balloons now.
TAKE HOME ACTIVITY • Ask our child to choose one of the problems in this lesson and solve it in a different way.
### Texas Go Math Grade 2 Lesson 10.4 Homework and Practice Answer Key
Make a model to solve. Then draw a quick picture of your model.
Question 1.
There were 225 rubber ducks in a large box. Angela took 112 rubber ducks out of the box. How many rubber ducks were still in the box?
___________ rubber ducks
Explanation:
225 – 112 is 113
So, 113 rubber ducks were still in the box.
Problem Solving
Solve.
Question 2.
There were 112 children and 15 adults at the library on Monday. On Tuesday, 128 children and 15 adults went to the library. How many fewer children were at the library on Tuesday than on Monday?
_________ fewer children
Explanation:
The difference of 128 and 112 is 16
So, 16 fewer children were at the library on tuesday than on monday.
Question 3.
Multi-Step Paul has 172 trading cards. Eli has twice that number of cards. Then Paul gave away 21 cards. How many trading cards do the boys have now?
Explanation:
The sum of 172, 172, 172 and 21 is 537
So, the boys have 537 trading cars.
Lesson Check
Question 4.
George had 280 tickets. He sold 177 tickets at the youth club. How many tickets does he have now?
(A) 103
(B) 417
(C) 113
(A) 103
Explanation:
He sold 177 tickets at the youth club
280 – 177 = 103
So, he have 103 tickets now.
Question 5.
Laura has 182 stamps. She put 72 stamps into an envelope. The she gave the rest to her friend. How many stamps did Laura give to her friend?
(A) 100
(B) 110
(C) 120
(B) 120
Explanation:
Laura has 182 stamps. She put 72 stamps into an envelope
The she gave the rest to her friend
182 – 72 = 120
So, Laura gave 120 stamps to her friend.
Question 6.
A notebook has 144 pages. Josh gave some pages to Adam. Now he has 139 pages. How many pages did he give to Adam?
(A) 5
(B) 15
(C) 25
(A) 5
Explanation:
A notebook has 144 pages
Josh gave some pages to Adam. Now he has 139 pages
144 – 139 = 5
So, josh gave 5 pages to Adam.
Question 7.
Michelle and her group are on a school treasure hunt. They need to find 150 counters to win. They have found 121 counters. How many more counters does Michelle’s group need to find?
(A) 39
(B) 19
(C) 29
|
# How to solve $\sqrt{x+2}\geq x$?
How do you solve the inequality $$\sqrt{x+2}\geq{x}?$$
Now since $${x+2}$$ is under the radical sign, it must be greater than or equal to $${0}$$ to be defined.
So,
$${x+2}\geq{0}$$
Thus $${x}\geq{-2}$$
Now keeping this in mind, we can solve the inequality by squaring both the sides:
$${x+2}\geq{x^2}$$
So $${x^2-x-2}\leq{0}$$
Solving, $${(x-2)(x+1)}\leq{0}$$
Therefore $${x}$$ belongs to the interval $${[-1,2]}$$.
As $${x}\geq{-2}$$, the function is also defined.
Why does the answer say that $${x}$$ belongs to $${[-2,2]}$$, then?
Please feel free to point out the mistakes.
Be careful : When you square, the inequality preserves its sign direction if both sides are positive.
Note that $$\sqrt{x+2}$$ is defined for $$x \geq - 2$$, so first you need to consider $$x \geq 0$$ and work as such :
$$\sqrt{x+2} \geq x \Rightarrow x+2 \geq x^2 \Leftrightarrow x^2-x-2 \leq 0 \Leftrightarrow (x-2)(x+1) \leq 0$$
This indeed yields $$x \in [-1,2]$$ if you also consider the negative values for which the derived inequality is satisfied .
But if $$x$$ is negative $$(-2 \leq x < 0)$$, then the (positive) square root will always be bigger than the negative left-hand side. Thus, $$[-2,0)$$ will do the trick in that case.
Concluding : $$\sqrt{x+2} \geq x \implies x \in [-2,2]$$.
• But we usually take the intersection of the intervals that define the radical function (here ${[-2,∞)}$) and the intervals obtained after squaring LHS and RHS (here ${[-1,2]}$) and it works pretty well in most of the examples. Is there more specific 'rule' or method for such cases? – Parth Amritkar May 18 at 12:31
• The inequality $x^2 - x - 2 \leq 0 \implies x \in [-1, 2]$. However, since $x \geq 0$, you should have $x \in [0, 2]$. – N. F. Taussig May 19 at 9:56
Clearly $$x\geq -2$$.
• If $$x<0$$ then each $$x\in [-2,0)$$ is a solution (since negative number is always smaller than square root).
• Now if $$x\geq 0$$ then you can square it, so you get $$x^2-x-2 = (x-2)(x+1)\leq 0$$ So in this case every $$x\in[0,2]$$ is a solution.
So finally, every $$x\in [-2,2]$$ is a solution.
Once you know $$x \geqslant -2$$, consider first $$x \in [-2, 0)$$. The LHS is defined and non-negative, while the RHS is __________.
Next, consider the case $$x \geqslant 0$$, where you can freely square as you have done. Here you should get $$x \in [0, 2]$$ as the solution.
Now the solution set is the union of these cases.
Hint:
The inequation $$\;\sqrt A\ge B$$, on its domain (defined by the condition $$A\ge 0$$) is equivalent to $$A\ge B^2\quad\textbf{ or }\quad B\le 0.$$
• It is a general solution. But are you sure that $B\leq0$ rather than $B<0$? I know that it does not matter actually because of union. – Money Oriented Programmer May 21 at 11:27
• @ArtificialOdorlessArmpit: You mean it's one of the basic rules on inequalities… – Bernard May 21 at 11:29
Let $$a=\sqrt{x+2}\ge0$$ for real $$x$$
We need $$a\ge a^2-2\iff0\ge a^2-a-2=(a-2)(a+1)$$
$$\iff -1\le a\le2\ \ \ \ (1)$$
But we need to honor $$a\ge0\ \ \ \ (2)$$
Find the intersection of $$(1),(2)$$
• $$\implies0\le x+2\le2^2$$ – lab bhattacharjee May 18 at 13:18
You made a mistake squaring both sides without checking the sign first. $$1>-1$$ but $$1^2\not>(-1)^2$$.
To square both sides of an equality/inequality and obtain an equivalent statement you must be certain that both sides have the same sign (in case of inequality and both sides negative you should swich the direction of the inequality though). In your case left side is always nonnegative, therefore for nonpositive $$x$$ the inequality is aleays fulfiled as the $$LHS \ge 0 \ge RHS$$. But since the inequality doesn't make sense for $$x<-2$$ only nonpositive numbers greater than it satysfiy this inequality.
|
# JAC Class 9 Maths Solutions Chapter 11 Constructions Ex 11.2
Jharkhand Board JAC Class 9 Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.
## JAC Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.2
Page-195
Question 1.
Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Answer:
Steps of Construction:
Step 1: A line segment BC of 7 cm is drawn.
Step 2: At point B, an angle ∠XBC is constructed such that it is equal to 75°.
Step 3: A line segment BD =13 cm is cut on BX (which is equal to AB + AC).
Step 4: DC is joined.
Step 5: Draw perpendicular bisector of CD which meets BD at A.
Step 6: Join AC.
Thus, ∆ABC is the required triangle.
Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Answer:
Steps of Construction:
Step 1: A line segment BC = 8 cm is drawn and at point B, make an angle of 45° i.e. ∠XBC.
Step 2: Cut the line segment BD = 3.5 cm (equal to AB – AC) on ray BX.
Step 3: Join DC and draw the
perpendicular bisector PQ of CD.
Step 4: Let it intersect BX at point A. Join AC.
Thus, ∆ABC is the required triangle.
Question 3.
Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Answer:
Steps of Construction:
Step 1: A line segment QR = 6 cm is drawn.
Step 2: A ray QY is constructed making an angle of 60° with QR and YQ is produced backwards to form a line YY’.
Step 3: Cut off a line segment QS = 2 cm from QY’. RS is joined.
Step 4: Draw perpendicular bisector of RS intersecting QY at a point P. PR is joined. Thus, ∆PQR is the required triangle.
Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Answer:
Steps of Construction:
Step 1: A line segment PQ = il cm is drawn. (XY + YZ + ZX =11 cm)
Step 2: An angle, ZRPQ = 30° is constructed at point P and an angle ZSQP = 90° at point Q.
Step 3: ZRPQ and ZSQP are bisected. The bisectors of these angles intersect each other at point X.
Step 4: Perpendicular bisectors TU of PX and WV of QX are constructed.
Step 5: Let TU intersect PQ at Y and WV intersect PQ at Z. XY and XZ are joined.
Thus, ∆XYZ is the required triangle.
Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Answer:
Steps of Construction:
Step 1 : A line segment BC = 12 cm is drawn.
Step 2: ZCBY = 90° is constructed.
Step 3: Cut off a line segment BD = 18 cm from BY. CD is joined.
Step 4: Perpendicular bisector of CD is constructed intersecting BD at A. AC is joined.
Thus, ∆ABC is the required triangle.
|
# 2.6 BOOLEAN FUNCTIONS
Size: px
Start display at page:
Transcription
1 2.6 BOOLEAN FUNCTIONS Binary variables have two values, either 0 or 1. A Boolean function is an expression formed with binary variables, the two binary operators AND and OR, one unary operator NOT, parentheses and equal sign. The value of a function may be 0 or 1, depending on the values of variables present in the Boolean function or expression. For example, if a Boolean function is expressed algebraically as: F = AB C then the value of F will be 1, when A = 1, B = 0, and C = 1. For other values of A, B, C the value of F is 0. Boolean functions can also be represented by truth tables. A truth table is the tabular form of the values of a Boolean function according to the all possible values of its variables. For an n number of variables, 2 n combinations of 1s and 0s are listed and one column represents function values according to the different combinations. For example, for three variables the Boolean function F = AB + C truth table can be written as below in Figure 2.7. A B C F Figure SIMPLIFICATION OF BOOLEAN EXPRESSIONS When a Boolean expression is implemented with logic gates, each literal in the function is designated as input to the gate. The literal may be a primed or unprimed variable. Minimization of the number of literals and the number of terms leads to less complex circuits as well as less number of gates, which should be a designer s aim. There are several methods to minimize the Boolean function. Here, simplification or minimization of complex algebraic expressions will be shown with the help of postulates and theorems of Boolean algebra. Example 2.1. Simplify the Boolean function F=AB+ BC + B C. Solution. F = AB + BC + B C = AB + C(B + B ) = AB + C 1
2 Example 2.2. Simplify the Boolean function F= A + A B. Solution. F = A+ A B = (A + A ) (A + B) = A + B Example 2.3. Simplify the Boolean function F= A B C + A BC + AB. Solution. F = A B C + A BC + AB = A C (B +B) + AB = A C + AB Example 2.4. Simplify the Boolean function F = AB + (AC) + AB C(AB + C). Solution. F = AB + (AC) + AB C(AB + C) = AB + A + C + AB C.AB + AB C.C = AB + A + C AB C (B.B = 0 and C.C = C) = ABC + ABC + A + C + AB C (AB = AB(C + C ) = ABC + ABC ) = AC(B + B ) + C (AB + 1) + A = AC + C +A (B + B = 1 and AB + 1 = 1) = AC + (AC) = CANONICAL AND STANDARD FORMS Logical functions are generally expressed in terms of different combinations of logical variables with their true forms as well as the complement forms. Binary logic values obtained by the logical functions and logic variables are in binary form. An arbitrary logic function can be expressed in the following forms. (i) Sum of the Products (SOP) (ii) Product of the Sums (POS) Product Term. In Boolean algebra, the logical product of several variables on which a function depends is considered to be a product term. In other words, the AND function is referred to as a product term or standard product. The variables in a product term can be either in true form or in complemented form. For example, ABC is a product term. Sum Term. An OR function is referred to as a sum term. The logical sum of several variables on which a function depends is considered to be a sum term. Variables in a sum term can also be either in true form or in complemented form. For example, A + B + C is a sum term. Sum of Products (SOP). The logical sum of two or more logical product terms is referred to as a sum of products expression. It is basically an OR operation on AND operated variables. For example, Y = AB + BC + AC or Y = A B + BC + AC are sum of products expressions. Product of Sums (POS). Similarly, the logical product of two or more logical sum terms is called a product of sums expression. It is an AND operation on OR operated variables. For example, Y = (A + B + C)(A + B + C)(A + B + C ) or Y = (A + B + C)(A + B + C ) are product of sums expressions. 2
3 Standard form. The standard form of the Boolean function is when it is expressed in sum of the products or product of the sums fashion. The examples stated above, like Y =AB + BC + AC or Y = (A + B + C)(A + B + C)(A + B + C ) are the standard forms. However, Boolean functions are also sometimes expressed in nonstandard forms like F = (AB + CD)(A B + C D ), which is neither a sum of products form nor a product of sums form. However, the same expression can be converted to a standard form with help of various Boolean properties, as: F = (AB + CD)(A B + C D ) = A B CD + ABC D Minterm A product term containing all n variables of the function in either true or complemented form is called the minterm. Each minterm is obtained by an AND operation of the variables in their true form or complemented form. For a two-variable function, four different combinations are possible, such as, A B, A B, AB, and AB. These product terms are called the fundamental products or standard products or minterms. In the minterm, a variable will possess the value 1 if it is in true or uncomplemented form, whereas, it contains the value 0 if it is in complemented form. For three variables function, eight minterms are possible as listed in the following table in Figure 2.9. Figure 2-9 So, if the number of variables is n, then the possible number of minterms is 2 n. The main property of a minterm is that it possesses the value of 1 for only one combination of n input variables and the rest of the 2 n 1 combinations have the logic value of 0. This means, for the above three variables example, if A = 0, B = 1, C = 1 i.e., for input combination of 011, there is only one combination A BC that has the value 1, the rest of the seven combinations have the value 0. Canonical Sum of Product Expression. When a Boolean function is expressed as the logical sum of all the minterms from the rows of a truth table, for which the value of the function is 1, it is referred to as the canonical sum of product expression. The same can be expressed in a compact form by listing the corresponding decimal- 3
4 equivalent codes of the minterms containing a function value of 1. For example, if the canonical sum of product form of a three-variable logic function F has the minterms A BC, AB C, and ABC, this can be expressed as the sum of the decimal codes corresponding to these minterms as below. F (A,B,C) = Σ (3,5,6) = m3 + m5 + m6 = A BC + AB C + ABC where Σ (3,5,6) represents the summation of minterms corresponding to decimal codes 3, 5, and 6. The canonical sum of products form of a logic function can be obtained by using the following procedure: 1. Check each term in the given logic function. Retain if it is a minterm, continue to examine the next term in the same manner. 2. Examine for the variables that are missing in each product which is not a minterm. If the missing variable in the minterm is X, multiply that minterm with (X+X ). 3. Multiply all the products and discard the redundant terms. Here are some examples to explain the above procedure. Example 2.6. Obtain the canonical sum of product form of the following function: F (A, B) = A + B Solution. The given function contains two variables A and B. The variable B is missing from the first term of the expression and the variable A is missing from the second term of the expression. Therefore, the first term is to be multiplied by (B + B ) and the second term is to be multiplied by (A + A ) as demonstrated below. F (A, B) = A + B = A.1 + B.1 = A (B + B ) + B (A + A ) = AB + AB + AB + A B = AB + AB + A B (as AB + AB = AB) Hence the canonical sum of the product expression of the given function is F (A, B) = AB + AB + A B. Example 2.7. Obtain the canonical sum of product form of the following function. F (A, B, C) = A + BC Solution. Here neither the first term nor the second term is minterm. The given function contains three variables A, B, and C. The variables B and C are missing from the first term of the expression and the variable A is missing from the second term of the expression. Therefore, the first term is to be multiplied by (B + B ) and (C + C ). The second term is to be multiplied by (A + A ). This is demonstrated below. F (A, B, C) = A + BC = A (B + B ) (C + C ) + BC (A + A ) = (AB + AB ) (C + C ) + ABC + A BC = ABC + AB C + ABC + AB C + ABC + A BC 4
5 = ABC + AB C + ABC + AB C + A BC (as ABC + ABC = ABC) Hence the canonical sum of the product expression of the given function is F (A, B,C) = ABC + AB C + ABC + AB C + A BC Maxterm A sum term containing all n variables of the function in either true or complemented form is called the maxterm. Each maxterm is obtained by an OR operation of the variables in their true form or complemented form. Four different combinations are possible for a two-variable function, such as, A + B, A + B, A + B, and A + B. These sum terms are called the standard sums or maxterms. Note that, in the maxterm, a variable will possess the value 0, if it is in true or uncomplemented form, whereas, it contains the value 1, if it is in complemented form. Like minterms, for a threevariable function, eight maxterms are also possible as listed in the following table in Figure Figure 2-10 So, if the number of variables is n, then the possible number of maxterms is 2 n. The main property of a maxterm is that it possesses the value of 0 for only one combination of n input variables and the rest of the 2 n 1 combinations have the logic value of 1. This means, for the above three variables example, if A = 1, B = 1, C = 0 i.e., for input combination of 110, there is only one combination A + B + C that has the value 0, the rest of the seven combinations have the value 1. Canonical Product of Sum Expression. When a Boolean function is expressed as the logical product of all the maxterms from the rows of a truth table, for which the value of the function is 0, it is referred to as the canonical product of sum expression. The same can be expressed in a compact form by listing the corresponding decimal equivalent codes of the maxterms containing a function value of 0. For example, if the canonical product of sums form of a three-variable logic function F has the maxterms A + B + C, A + B + C, and A + B + C, this can be expressed as the product of the decimal codes corresponding to these maxterms as below, 5
6 F (A,B,C) = Π (0,2,5) = M0 M2 M5 = (A + B + C) (A + B + C) (A + B + C ) where Π (0,2,5) represents the product of maxterms corresponding to decimal codes 0, 2, and 5. The canonical product of sums form of a logic function can be obtained by using the following procedure. 1. Check each term in the given logic function. Retain it if it is a maxterm, continue to examine the next term in the same manner. 2. Examine for the variables that are missing in each sum term that is not a maxterm. If the missing variable in the maxterm is X, add that maxterm with (X.X ). 3. Expand the expression using the properties and postulates as described earlier and discard the redundant terms. Some examples are given here to explain the above procedure. Example 2.8. Obtain the canonical product of the sum form of the following function. F (A, B, C) = (A + B ) (B + C) (A + C ) Solution. In the above three-variable expression, C is missing from the first term, A is missing from the second term, and B is missing from the third term. Therefore, CC is to be added with first term, AA is to be added with the second, and BB is to be added with the third term. This is shown below. F (A, B, C) = (A + B ) (B + C) (A + C ) = (A + B + 0) (B + C + 0) (A + C + 0) = (A + B + CC ) (B + C + AA ) (A + C + BB ) = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) (A + B + C ) (A + B + C ) [using the distributive property, as X + YZ = (X + Y)(X + Z)] = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) (A + B + C ) [as (A + B + C ) (A + B + C ) = A + B + C ] Hence the canonical product of the sum expression for the given function is F (A, B, C) = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) (A + B + C ) Example 2.9. Obtain the canonical product of the sum form of the following function. F (A, B, C) = A + B C Solution. In the above three-variable expression, the function is given at sum of the product form. First, the function needs to be changed to product of the sum form by applying the distributive law as shown below. F (A, B, C) = A + B C = (A + B ) (A + C) Now, in the above expression, C is missing from the first term and B is missing from the second term. Hence CC is to be added with the first term and BB is to be added with the second term as shown below. F (A, B, C) = (A + B ) (A + C) 6
7 = (A + B + CC ) (A + C + BB ) = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) [using the distributive property, as X + YZ = (X + Y) (X + Z)] = (A + B + C) (A + B + C ) (A + B + C) [as (A + B + C) (A + B + C) = A + B + C] Hence the canonical product of the sum expression for the given function is F (A, B, C) = (A + B + C) (A + B + C ) (A + B + C) Deriving a Sum of Products (SOP) Expression from a Truth Table The sum of products (SOP) expression of a Boolean function can be obtained from its truth table summing or performing OR operation of the product terms corresponding to the combinations containing a function value of 1. In the product terms the input variables appear either in true (uncomplemented) form if it contains the value 1, or in complemented form if it possesses the value 0. Now, consider the following truth table in Figure 2.11, for a three-input function Y. Here the output Y value is 1 for the input conditions of 010, 100, 101, and 110, and their corresponding product terms are A BC, AB C, AB C, and ABC respectively. Figure 2-11 The final sum of products expression (SOP) for the output Y is derived by summing or performing an OR operation of the four product terms as shown below. Y = A BC + AB C + AB C + ABC In general, the procedure of deriving the output expression in SOP form from a truth table can be summarized as below. 1. Form a product term for each input combination in the table, containing an output value of Each product term consists of its input variables in either true form or complemented form. If the input variable is 0, it appears in complemented form and if the input variable is 1, it appears in true form. 7
8 3. To obtain the final SOP expression of the output, all the product terms are OR operated Deriving a Product of Sums (POS) Expression from a Truth Table As explained above, the product of sums (POS) expression of a Boolean function can also be obtained from its truth table by a similar procedure. Here, an AND operation is performed on the sum terms corresponding to the combinations containing a function value of 0. In the sum terms the input variables appear either in true (uncomplemented) form if it contains the value 0, or in complemented form if it possesses the value 1. Now, consider the same truth table as shown in Figure 2.11, for a three-input function Y. Here the output Y value is 0 for the input conditions of 000, 001, 011, and 111, and their corresponding product terms are A + B + C, A + B + C, A + B + C, and A + B + C respectively. So now, the final product of sums expression (POS) for the output Y is derived by performing an AND operation of the four sum terms as shown below. Y = (A + B + C) (A + B + C ) (A + B + C ) (A + B + C ) In general, the procedure of deriving the output expression in POS form from a truth table can be summarized as below. 1. Form a sum term for each input combination in the table, containing an output value of Each product term consists of its input variables in either true form or complemented form. If the input variable is 1, it appears in complemented form and if the input variable is 0, it appears in true form. 3. To obtain the final POS expression of the output, all the sum terms are AND operated. 8
9 2.9.5 Conversion between Canonical Forms From the above example, it may be noted that the complement of a function expressed as the sum of products (SOP) equals to the sum of products or sum of the minterms which are missing from the original function. This is because the original function is expressed by those minterms that make the function equal to 1, while its complement is 1 for those minterms whose values are 0. According to the truth table given in Figure 2.11: F (A,B,C) = Σ ( 2,4,5,6) = m2 + m4 + m5 + m6 = A BC + AB C + AB C + ABC. This has the complement that can be expressed as F (A,B,C) = (0,1,3,7) = m0 + m1 + m3 + m7 Now, if we take complement of F by DeMorgan s theorem, we obtain F as F (A,B,C) = (m0 + m1 + m3 + m7) = m0 m1 m3 m 7 = M0M1M3M7 = Π(0,1,3,7) = (A + B + C)(A + B + C ) (A + B + C ) (A + B + C ). The last conversion follows from the definition of minterms and maxterms as shown in the tables in Figures 2.9 and It can be clearly noted that the following relation holds true m j = Mj. That is, the maxterm with subscript j is a complement of the minterm with the same subscript j, and vice versa. This example demonstrates the conversion between a function expressed in sum of products (SOP) and its equivalent in product of maxterms. A similar example can show the conversion between the product of sums (POS) and its equivalent sum of minterms. In general, to convert from one canonical form to other canonical form, it is required to interchange the symbols Σ and π, and list the numbers which are missing from the original form. Note that, to find the missing terms, the total 2 n number of minterms or maxterms must be realized, where n is the number of variables in the function. 9
### CHAPTER-2 STRUCTURE OF BOOLEAN FUNCTION USING GATES, K-Map and Quine-McCluskey
CHAPTER-2 STRUCTURE OF BOOLEAN FUNCTION USING GATES, K-Map and Quine-McCluskey 2. Introduction Logic gates are connected together to produce a specified output for certain specified combinations of input
### Bawar Abid Abdalla. Assistant Lecturer Software Engineering Department Koya University
Logic Design First Stage Lecture No.6 Boolean Algebra Bawar Abid Abdalla Assistant Lecturer Software Engineering Department Koya University Outlines Boolean Operations Laws of Boolean Algebra Rules of
### Chapter 2. Boolean Expressions:
Chapter 2 Boolean Expressions: A Boolean expression or a function is an expression which consists of binary variables joined by the Boolean connectives AND and OR along with NOT operation. Any Boolean
### DKT 122/3 DIGITAL SYSTEM 1
Company LOGO DKT 122/3 DIGITAL SYSTEM 1 BOOLEAN ALGEBRA (PART 2) Boolean Algebra Contents Boolean Operations & Expression Laws & Rules of Boolean algebra DeMorgan s Theorems Boolean analysis of logic circuits
### Unit-IV Boolean Algebra
Unit-IV Boolean Algebra Boolean Algebra Chapter: 08 Truth table: Truth table is a table, which represents all the possible values of logical variables/statements along with all the possible results of
### LSN 4 Boolean Algebra & Logic Simplification. ECT 224 Digital Computer Fundamentals. Department of Engineering Technology
LSN 4 Boolean Algebra & Logic Simplification Department of Engineering Technology LSN 4 Key Terms Variable: a symbol used to represent a logic quantity Compliment: the inverse of a variable Literal: a
### Combinational Logic Circuits
Chapter 3 Combinational Logic Circuits 12 Hours 24 Marks 3.1 Standard representation for logical functions Boolean expressions / logic expressions / logical functions are expressed in terms of logical
### Chapter 2 Boolean algebra and Logic Gates
Chapter 2 Boolean algebra and Logic Gates 2. Introduction In working with logic relations in digital form, we need a set of rules for symbolic manipulation which will enable us to simplify complex expressions
### Menu. Algebraic Simplification - Boolean Algebra EEL3701 EEL3701. MSOP, MPOS, Simplification
Menu Minterms & Maxterms SOP & POS MSOP & MPOS Simplification using the theorems/laws/axioms Look into my... 1 Definitions (Review) Algebraic Simplification - Boolean Algebra Minterms (written as m i ):
### X Y Z F=X+Y+Z
This circuit is used to obtain the compliment of a value. If X = 0, then X = 1. The truth table for NOT gate is : X X 0 1 1 0 2. OR gate : The OR gate has two or more input signals but only one output
### Experiment 3: Logic Simplification
Module: Logic Design Name:... University no:.. Group no:. Lab Partner Name: Mr. Mohamed El-Saied Experiment : Logic Simplification Objective: How to implement and verify the operation of the logical functions
Get Free notes at Module-I One s Complement: Complement all the bits.i.e. makes all 1s as 0s and all 0s as 1s Two s Complement: One s complement+1 SIGNED BINARY NUMBERS Positive integers (including zero)
### IT 201 Digital System Design Module II Notes
IT 201 Digital System Design Module II Notes BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. A variable is a symbol used to represent a logical quantity.
### ENGIN 112. Intro to Electrical and Computer Engineering
ENIN 2 Intro to Electrical and Computer Engineering Lecture 6 More Boolean Algebra ENIN2 L6: More Boolean Algebra September 5, 23 A B Overview Epressing Boolean functions Relationships between algebraic
### BOOLEAN ALGEBRA. 1. State & Verify Laws by using :
BOOLEAN ALGEBRA. State & Verify Laws by using :. State and algebraically verify Absorption Laws. (2) Absorption law states that (i) X + XY = X and (ii) X(X + Y) = X (i) X + XY = X LHS = X + XY = X( + Y)
### Module -7. Karnaugh Maps
1 Module -7 Karnaugh Maps 1. Introduction 2. Canonical and Standard forms 2.1 Minterms 2.2 Maxterms 2.3 Canonical Sum of Product or Sum-of-Minterms (SOM) 2.4 Canonical product of sum or Product-of-Maxterms(POM)
### Lecture 5. Chapter 2: Sections 4-7
Lecture 5 Chapter 2: Sections 4-7 Outline Boolean Functions What are Canonical Forms? Minterms and Maxterms Index Representation of Minterms and Maxterms Sum-of-Minterm (SOM) Representations Product-of-Maxterm
### Philadelphia University Faculty of Information Technology Department of Computer Science. Computer Logic Design. By Dareen Hamoudeh.
Philadelphia University Faculty of Information Technology Department of Computer Science Computer Logic Design By Dareen Hamoudeh Dareen Hamoudeh 1 Canonical Forms (Standard Forms of Expression) Minterms
### UNIT-4 BOOLEAN LOGIC. NOT Operator Operates on single variable. It gives the complement value of variable.
UNIT-4 BOOLEAN LOGIC Boolean algebra is an algebra that deals with Boolean values((true and FALSE). Everyday we have to make logic decisions: Should I carry the book or not?, Should I watch TV or not?
### Binary logic. Dr.Abu-Arqoub
Binary logic Binary logic deals with variables like (a, b, c,, x, y) that take on two discrete values (, ) and with operations that assume logic meaning ( AND, OR, NOT) Truth table is a table of all possible
### Boolean Algebra and Logic Gates
Boolean Algebra and Logic Gates Binary logic is used in all of today's digital computers and devices Cost of the circuits is an important factor Finding simpler and cheaper but equivalent circuits can
### Computer Science. Unit-4: Introduction to Boolean Algebra
Unit-4: Introduction to Boolean Algebra Learning Objective At the end of the chapter students will: Learn Fundamental concepts and basic laws of Boolean algebra. Learn about Boolean expression and will
### R.M.D. ENGINEERING COLLEGE R.S.M. Nagar, Kavaraipettai
L T P C R.M.D. ENGINEERING COLLEGE R.S.M. Nagar, Kavaraipettai- 601206 DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING EC8392 UNIT - I 3 0 0 3 OBJECTIVES: To present the Digital fundamentals, Boolean
### Gate Level Minimization
Gate Level Minimization By Dr. M. Hebaishy Digital Logic Design Ch- Simplifying Boolean Equations Example : Y = AB + AB Example 2: = B (A + A) T8 = B () T5 = B T Y = A(AB + ABC) = A (AB ( + C ) ) T8 =
### Experiment 4 Boolean Functions Implementation
Experiment 4 Boolean Functions Implementation Introduction: Generally you will find that the basic logic functions AND, OR, NAND, NOR, and NOT are not sufficient to implement complex digital logic functions.
### Assignment (3-6) Boolean Algebra and Logic Simplification - General Questions
Assignment (3-6) Boolean Algebra and Logic Simplification - General Questions 1. Convert the following SOP expression to an equivalent POS expression. 2. Determine the values of A, B, C, and D that make
### ENGINEERS ACADEMY. 7. Given Boolean theorem. (a) A B A C B C A B A C. (b) AB AC BC AB BC. (c) AB AC BC A B A C B C.
Digital Electronics Boolean Function QUESTION BANK. The Boolean equation Y = C + C + C can be simplified to (a) (c) A (B + C) (b) AC (d) C. The Boolean equation Y = (A + B) (A + B) can be simplified to
### Lecture (05) Boolean Algebra and Logic Gates
Lecture (05) Boolean Algebra and Logic Gates By: Dr. Ahmed ElShafee ١ Minterms and Maxterms consider two binary variables x and y combined with an AND operation. Since eachv ariable may appear in either
### Chapter 2 Combinational Logic Circuits
Logic and Computer Design Fundamentals Chapter 2 Combinational Logic Circuits Part 2 Circuit Optimization Overview Part Gate Circuits and Boolean Equations Binary Logic and Gates Boolean Algebra Standard
Summary Boolean Addition In Boolean algebra, a variable is a symbol used to represent an action, a condition, or data. A single variable can only have a value of or 0. The complement represents the inverse
### Boolean Analysis of Logic Circuits
Course: B.Sc. Applied Physical Science (Computer Science) Year & Sem.: IInd Year, Sem - IIIrd Subject: Computer Science Paper No.: IX Paper Title: Computer System Architecture Lecture No.: 7 Lecture Title:
### Introduction to Computer Architecture
Boolean Operators The Boolean operators AND and OR are binary infix operators (that is, they take two arguments, and the operator appears between them.) A AND B D OR E We will form Boolean Functions of
### Lecture 4: Implementation AND, OR, NOT Gates and Complement
EE210: Switching Systems Lecture 4: Implementation AND, OR, NOT Gates and Complement Prof. YingLi Tian Feb. 13, 2018 Department of Electrical Engineering The City College of New York The City University
### EEE130 Digital Electronics I Lecture #4_1
EEE130 Digital Electronics I Lecture #4_1 - Boolean Algebra and Logic Simplification - By Dr. Shahrel A. Suandi 4-6 Standard Forms of Boolean Expressions There are two standard forms: Sum-of-products form
### Chapter 2: Combinational Systems
Uchechukwu Ofoegbu Chapter 2: Combinational Systems Temple University Adapted from Alan Marcovitz s Introduction to Logic and Computer Design Riddle Four switches can be turned on or off. One is the switch
### Chapter 3 Simplification of Boolean functions
3.1 Introduction Chapter 3 Simplification of Boolean functions In this chapter, we are going to discuss several methods for simplifying the Boolean function. What is the need for simplifying the Boolean
### Specifying logic functions
CSE4: Components and Design Techniques for Digital Systems Specifying logic functions Instructor: Mohsen Imani Slides from: Prof.Tajana Simunic and Dr.Pietro Mercati We have seen various concepts: Last
### Digital Logic Design (CEN-120) (3+1)
Digital Logic Design (CEN-120) (3+1) ASSISTANT PROFESSOR Engr. Syed Rizwan Ali, MS(CAAD)UK, PDG(CS)UK, PGD(PM)IR, BS(CE)PK HEC Certified Master Trainer (MT-FPDP) PEC Certified Professional Engineer (COM/2531)
### Combinational Logic Circuits
Chapter 2 Combinational Logic Circuits J.J. Shann (Slightly trimmed by C.P. Chung) Chapter Overview 2-1 Binary Logic and Gates 2-2 Boolean Algebra 2-3 Standard Forms 2-4 Two-Level Circuit Optimization
### Code No: R Set No. 1
Code No: R059210504 Set No. 1 II B.Tech I Semester Supplementary Examinations, February 2007 DIGITAL LOGIC DESIGN ( Common to Computer Science & Engineering, Information Technology and Computer Science
### ELCT201: DIGITAL LOGIC DESIGN
ELCT201: DIGITAL LOGIC DESIGN Dr. Eng. Haitham Omran, haitham.omran@guc.edu.eg Dr. Eng. Wassim Alexan, wassim.joseph@guc.edu.eg Lecture 3 Following the slides of Dr. Ahmed H. Madian ذو الحجة 1438 ه Winter
### To write Boolean functions in their standard Min and Max terms format. To simplify Boolean expressions using Karnaugh Map.
3.1 Objectives To write Boolean functions in their standard Min and Max terms format. To simplify Boolean expressions using. 3.2 Sum of Products & Product of Sums Any Boolean expression can be simplified
### 2.1 Binary Logic and Gates
1 EED2003 Digital Design Presentation 2: Boolean Algebra Asst. Prof.Dr. Ahmet ÖZKURT Asst. Prof.Dr Hakkı T. YALAZAN Based on the Lecture Notes by Jaeyoung Choi choi@comp.ssu.ac.kr Fall 2000 2.1 Binary
### QUESTION BANK FOR TEST
CSCI 2121 Computer Organization and Assembly Language PRACTICE QUESTION BANK FOR TEST 1 Note: This represents a sample set. Please study all the topics from the lecture notes. Question 1. Multiple Choice
### ELCT201: DIGITAL LOGIC DESIGN
ELCT201: DIGITAL LOGIC DESIGN Dr. Eng. Haitham Omran, haitham.omran@guc.edu.eg Dr. Eng. Wassim Alexan, wassim.joseph@guc.edu.eg Lecture 3 Following the slides of Dr. Ahmed H. Madian محرم 1439 ه Winter
### Chapter 3. Gate-Level Minimization. Outlines
Chapter 3 Gate-Level Minimization Introduction The Map Method Four-Variable Map Five-Variable Map Outlines Product of Sums Simplification Don t-care Conditions NAND and NOR Implementation Other Two-Level
### Digital Logic Lecture 7 Gate Level Minimization
Digital Logic Lecture 7 Gate Level Minimization By Ghada Al-Mashaqbeh The Hashemite University Computer Engineering Department Outline Introduction. K-map principles. Simplification using K-maps. Don t-care
### Standard Forms of Expression. Minterms and Maxterms
Standard Forms of Expression Minterms and Maxterms Standard forms of expressions We can write expressions in many ways, but some ways are more useful than others A sum of products (SOP) expression contains:
### Gate Level Minimization Map Method
Gate Level Minimization Map Method Complexity of hardware implementation is directly related to the complexity of the algebraic expression Truth table representation of a function is unique Algebraically
### A B AB CD Objectives:
Objectives:. Four variables maps. 2. Simplification using prime implicants. 3. "on t care" conditions. 4. Summary.. Four variables Karnaugh maps Minterms A A m m m3 m2 A B C m4 C A B C m2 m8 C C m5 C m3
### CSCI 220: Computer Architecture I Instructor: Pranava K. Jha. Simplification of Boolean Functions using a Karnaugh Map
CSCI 22: Computer Architecture I Instructor: Pranava K. Jha Simplification of Boolean Functions using a Karnaugh Map Q.. Plot the following Boolean function on a Karnaugh map: f(a, b, c, d) = m(, 2, 4,
### Announcements. Chapter 2 - Part 1 1
Announcements If you haven t shown the grader your proof of prerequisite, please do so by 11:59 pm on 09/05/2018 (Wednesday). I will drop students that do not show us the prerequisite proof after this
### 1. Mark the correct statement(s)
1. Mark the correct statement(s) 1.1 A theorem in Boolean algebra: a) Can easily be proved by e.g. logic induction b) Is a logical statement that is assumed to be true, c) Can be contradicted by another
### Boolean algebra. June 17, Howard Huang 1
Boolean algebra Yesterday we talked about how analog voltages can represent the logical values true and false. We introduced the basic Boolean operations AND, OR and NOT, which can be implemented in hardware
### Code No: 07A3EC03 Set No. 1
Code No: 07A3EC03 Set No. 1 II B.Tech I Semester Regular Examinations, November 2008 SWITCHING THEORY AND LOGIC DESIGN ( Common to Electrical & Electronic Engineering, Electronics & Instrumentation Engineering,
### CS470: Computer Architecture. AMD Quad Core
CS470: Computer Architecture Yashwant K. Malaiya, Professor malaiya@cs.colostate.edu AMD Quad Core 1 Architecture Layers Building blocks Gates, flip-flops Functional bocks: Combinational, Sequential Instruction
### Points Addressed in this Lecture. Standard form of Boolean Expressions. Lecture 4: Logic Simplication & Karnaugh Map
Points Addressed in this Lecture Lecture 4: Logic Simplication & Karnaugh Map Professor Peter Cheung Department of EEE, Imperial College London Standard form of Boolean Expressions Sum-of-Products (SOP),
### Programmable Logic Devices. Programmable Read Only Memory (PROM) Example
Programmable Logic Devices Programmable Logic Devices (PLDs) are the integrated circuits. They contain an array of AND gates & another array of OR gates. There are three kinds of PLDs based on the type
### Chap-2 Boolean Algebra
Chap-2 Boolean Algebra Contents: My name Outline: My position, contact Basic information theorem and postulate of Boolean Algebra. or project description Boolean Algebra. Canonical and Standard form. Digital
### UNIT II. Circuit minimization
UNIT II Circuit minimization The complexity of the digital logic gates that implement a Boolean function is directly related to the complexity of the algebraic expression from which the function is implemented.
### Chapter 2 Combinational Logic Circuits
Logic and Computer Design Fundamentals Chapter 2 Combinational Logic Circuits Part 2 Circuit Optimization Charles Kime & Thomas Kaminski 2008 Pearson Education, Inc. (Hyperlinks are active in View Show
### DIGITAL CIRCUIT LOGIC UNIT 5: KARNAUGH MAPS (K-MAPS)
DIGITAL CIRCUIT LOGIC UNIT 5: KARNAUGH MAPS (K-MAPS) 1 Learning Objectives 1. Given a function (completely or incompletely specified) of three to five variables, plot it on a Karnaugh map. The function
### TWO-LEVEL COMBINATIONAL LOGIC
TWO-LEVEL COMBINATIONAL LOGIC OVERVIEW Canonical forms To-level simplification Boolean cubes Karnaugh maps Quine-McClusky (Tabulation) Method Don't care terms Canonical and Standard Forms Minterms and
### Code No: R Set No. 1
Code No: R059210504 Set No. 1 II B.Tech I Semester Regular Examinations, November 2007 DIGITAL LOGIC DESIGN ( Common to Computer Science & Engineering, Information Technology and Computer Science & Systems
### Code No: R Set No. 1
Code No: R059210504 Set No. 1 II B.Tech I Semester Regular Examinations, November 2006 DIGITAL LOGIC DESIGN ( Common to Computer Science & Engineering, Information Technology and Computer Science & Systems
### CS8803: Advanced Digital Design for Embedded Hardware
CS883: Advanced Digital Design for Embedded Hardware Lecture 2: Boolean Algebra, Gate Network, and Combinational Blocks Instructor: Sung Kyu Lim (limsk@ece.gatech.edu) Website: http://users.ece.gatech.edu/limsk/course/cs883
### Gate-Level Minimization. BME208 Logic Circuits Yalçın İŞLER
Gate-Level Minimization BME28 Logic Circuits Yalçın İŞLER islerya@yahoo.com http://me.islerya.com Complexity of Digital Circuits Directly related to the complexity of the algebraic expression we use to
### CprE 281: Digital Logic
CprE 28: Digital Logic Instructor: Alexander Stoytchev http://www.ece.iastate.edu/~alexs/classes/ Minimization CprE 28: Digital Logic Iowa State University, Ames, IA Copyright Alexander Stoytchev Administrative
### Bawar Abid Abdalla. Assistant Lecturer Software Engineering Department Koya University
Logic Design First Stage Lecture No.5 Boolean Algebra Bawar Abid Abdalla Assistant Lecturer Software Engineering Department Koya University Boolean Operations Laws of Boolean Algebra Rules of Boolean Algebra
### Simplification of Boolean Functions
Simplification of Boolean Functions Contents: Why simplification? The Map Method Two, Three, Four and Five variable Maps. Simplification of two, three, four and five variable Boolean function by Map method.
### Combinational Logic & Circuits
Week-I Combinational Logic & Circuits Spring' 232 - Logic Design Page Overview Binary logic operations and gates Switching algebra Algebraic Minimization Standard forms Karnaugh Map Minimization Other
### SWITCHING THEORY AND LOGIC CIRCUITS
SWITCHING THEORY AND LOGIC CIRCUITS COURSE OBJECTIVES. To understand the concepts and techniques associated with the number systems and codes 2. To understand the simplification methods (Boolean algebra
### Definitions. 03 Logic networks Boolean algebra. Boolean set: B 0,
3. Boolean algebra 3 Logic networks 3. Boolean algebra Definitions Boolean functions Properties Canonical forms Synthesis and minimization alessandro bogliolo isti information science and technology institute
### Slide Set 5. for ENEL 353 Fall Steve Norman, PhD, PEng. Electrical & Computer Engineering Schulich School of Engineering University of Calgary
Slide Set 5 for ENEL 353 Fall 207 Steve Norman, PhD, PEng Electrical & Computer Engineering Schulich School of Engineering University of Calgary Fall Term, 207 SN s ENEL 353 Fall 207 Slide Set 5 slide
### (Refer Slide Time 6:48)
Digital Circuits and Systems Prof. S. Srinivasan Department of Electrical Engineering Indian Institute of Technology Madras Lecture - 8 Karnaugh Map Minimization using Maxterms We have been taking about
### SYNERGY INSTITUTE OF ENGINEERING & TECHNOLOGY,DHENKANAL LECTURE NOTES ON DIGITAL ELECTRONICS CIRCUIT(SUBJECT CODE:PCEC4202)
Lecture No:5 Boolean Expressions and Definitions Boolean Algebra Boolean Algebra is used to analyze and simplify the digital (logic) circuits. It uses only the binary numbers i.e. 0 and 1. It is also called
### Simplification of Boolean Functions
COM111 Introduction to Computer Engineering (Fall 2006-2007) NOTES 5 -- page 1 of 5 Introduction Simplification of Boolean Functions You already know one method for simplifying Boolean expressions: Boolean
### Austin Herring Recitation 002 ECE 200 Project December 4, 2013
1. Fastest Circuit a. How Design Was Obtained The first step of creating the design was to derive the expressions for S and C out from the given truth tables. This was done using Karnaugh maps. The Karnaugh
### Variable, Complement, and Literal are terms used in Boolean Algebra.
We have met gate logic and combination of gates. Another way of representing gate logic is through Boolean algebra, a way of algebraically representing logic gates. You should have already covered the
### Chapter 2 Combinational
Computer Engineering 1 (ECE290) Chapter 2 Combinational Logic Circuits Part 2 Circuit Optimization HOANG Trang 2008 Pearson Education, Inc. Overview Part 1 Gate Circuits and Boolean Equations Binary Logic
### Date Performed: Marks Obtained: /10. Group Members (ID):. Experiment # 04. Boolean Expression Simplification and Implementation
Name: Instructor: Engr. Date Performed: Marks Obtained: /10 Group Members (ID):. Checked By: Date: Experiment # 04 Boolean Expression Simplification and Implementation OBJECTIVES: To understand the utilization
### Switching Theory And Logic Design UNIT-II GATE LEVEL MINIMIZATION
Switching Theory And Logic Design UNIT-II GATE LEVEL MINIMIZATION Two-variable k-map: A two-variable k-map can have 2 2 =4 possible combinations of the input variables A and B. Each of these combinations,
### Incompletely Specified Functions with Don t Cares 2-Level Transformation Review Boolean Cube Karnaugh-Map Representation and Methods Examples
Lecture B: Logic Minimization Incompletely Specified Functions with Don t Cares 2-Level Transformation Review Boolean Cube Karnaugh-Map Representation and Methods Examples Incompletely specified functions
### Designing Computer Systems Boolean Algebra
Designing Computer Systems Boolean Algebra 08:34:45 PM 4 June 2013 BA-1 Scott & Linda Wills Designing Computer Systems Boolean Algebra Programmable computers can exhibit amazing complexity and generality.
### 9/10/2016. The Dual Form Swaps 0/1 and AND/OR. ECE 120: Introduction to Computing. Every Boolean Expression Has a Dual Form
University of Illinois at Urbana-Champaign Dept. of Electrical and Computer Engineering ECE 120: Introduction to Computing Boolean Properties and Optimization The Dual Form Swaps 0/1 and AND/OR Boolean
### UNIT 2 BOOLEAN ALGEBRA
UNIT 2 BOOLEN LGEBR Spring 2 2 Contents Introduction Basic operations Boolean expressions and truth tables Theorems and laws Basic theorems Commutative, associative, and distributive laws Simplification
### Gate-Level Minimization
Gate-Level Minimization ( 范倫達 ), Ph. D. Department of Computer Science National Chiao Tung University Taiwan, R.O.C. Fall, 2011 ldvan@cs.nctu.edu.tw http://www.cs.nctu.edu.tw/~ldvan/ Outlines The Map Method
### Circuit analysis summary
Boolean Algebra Circuit analysis summary After finding the circuit inputs and outputs, you can come up with either an expression or a truth table to describe what the circuit does. You can easily convert
### 數位系統 Digital Systems 朝陽科技大學資工系. Speaker: Fuw-Yi Yang 楊伏夷. 伏夷非征番, 道德經察政章 (Chapter 58) 伏者潛藏也道紀章 (Chapter 14) 道無形象, 視之不可見者曰夷
數位系統 Digital Systems Department of Computer Science and Information Engineering, Chaoyang University of Technology 朝陽科技大學資工系 Speaker: Fuw-Yi Yang 楊伏夷 伏夷非征番, 道德經察政章 (Chapter 58) 伏者潛藏也道紀章 (Chapter 14) 道無形象,
Class Subject Code Subject Prepared By Lesson Plan for Time: Lesson. No 1.CONTENT LIST: Introduction to UnitI 2. SKILLS ADDRESSED: Listening I year, 02 sem CS6201 Digital Principles & System Design S.Seedhanadevi
### CMPE223/CMSE222 Digital Logic
CMPE223/CMSE222 Digital Logic Optimized Implementation of Logic Functions: Strategy for Minimization, Minimum Product-of-Sums Forms, Incompletely Specified Functions Terminology For a given term, each
### Digital logic fundamentals. Question Bank. Unit I
Digital logic fundamentals Question Bank Subject Name : Digital Logic Fundamentals Subject code: CA102T Staff Name: R.Roseline Unit I 1. What is Number system? 2. Define binary logic. 3. Show how negative
### Combinational Circuits Digital Logic (Materials taken primarily from:
Combinational Circuits Digital Logic (Materials taken primarily from: http://www.facstaff.bucknell.edu/mastascu/elessonshtml/eeindex.html http://www.cs.princeton.edu/~cos126 ) Digital Systems What is a
### Midterm Exam Review. CS 2420 :: Fall 2016 Molly O'Neil
Midterm Exam Review CS 2420 :: Fall 2016 Molly O'Neil Midterm Exam Thursday, October 20 In class, pencil & paper exam Closed book, closed notes, no cell phones or calculators, clean desk 20% of your final
### Contents. Chapter 3 Combinational Circuits Page 1 of 34
Chapter 3 Combinational Circuits Page of 34 Contents Contents... 3 Combinational Circuits... 2 3. Analysis of Combinational Circuits... 2 3.. Using a Truth Table... 2 3..2 Using a Boolean unction... 4
### EECS150 Homework 2 Solutions Fall ) CLD2 problem 2.2. Page 1 of 15
1.) CLD2 problem 2.2 We are allowed to use AND gates, OR gates, and inverters. Note that all of the Boolean expression are already conveniently expressed in terms of AND's, OR's, and inversions. Thus,
### Literal Cost F = BD + A B C + A C D F = BD + A B C + A BD + AB C F = (A + B)(A + D)(B + C + D )( B + C + D) L = 10
Circuit Optimization Goal: To obtain the simplest implementation for a given function Optimization is a more formal approach to simplification that is performed using a specific procedure or algorithm
### BOOLEAN ALGEBRA. Logic circuit: 1. From logic circuit to Boolean expression. Derive the Boolean expression for the following circuits.
COURSE / CODE DIGITAL SYSTEMS FUNDAMENTAL (ECE 421) DIGITAL ELECTRONICS FUNDAMENTAL (ECE 422) BOOLEAN ALGEBRA Boolean Logic Boolean logic is a complete system for logical operations. It is used in countless
|
# Probability Questions and Answers – Download PDF!!!
0
532
Probability Questions and Answers – Download PDF!!!. Probability theory had start in the 17th century. It is one of the branches in mathematics. The probability value is expressed from 0 to 1.Classical, Relative, Subjective are the types of probability. Here some Probability question is explained with solutions.
Download Nov 2020 Current Affairs Pd
Page Contents
## Probability – Easy
1. There are 15 boys and 10 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
A. 1/40
B. 1/ 2
C. 21/46
D. 7/ 41
E. None of these
1. Correct option is : C
Solution:
Total number of ways of selecting 3 students from 25 students = 25C3
Number of ways of selecting 1 girl and 2 boys = selecting 2 boys from 15 boys and 1 girl from 10 girls
⇒ Number of ways in which this can be done = 15C2 × 10C1
⇒ Required probability = (15C2 × 10C1)/ (25C3)
1. Two friends Harish and Kalyan appeared for an exam. Let A be the event that Harish is selected and B is the event that Kalyan is selected. The probability of A is 2/5 and that of B is 3/7. Find the probability that both of them are selected.
A. 35/36
B. 5/35
C. 5/12
D. 6/35
E. None of these
1. Correct option is : D
Solution:
Given, A be the event that Harish is selected and
B is the event that Kalyan is selected.
P(A)= 2/5
P(B)=3/7
Let C be the event that both are selected.
P(C)=P(A)×P(B) as A and B are independent events:
P(C) = 2/5*3/7
P(C) =6/35
The probability that both of them are selected is 6/35
1. A card is drawn from a well shuffled pack of 52 cards. What is the probability of getting queen or club card?
A. 17/52
B. 15/52
C. 4/13
D. 3/13
E. None of these
1. Correct option is : C
Solution:
The probability of getting queen card = 4/52
The probability of getting club card = 13/52
The club card contains already a queen card, therefore required probability is,
4/52 + 13/52 – 1/52 = 16/52 = 4/13
1. 16 persons shake hands with one another in a party. How many shake hands took place?
A. 124
B. 120
C. 165
D. 150
E. None of these
1. Correct option is : B
Solution:
Total possible ways = 16C2
= 120
1. 2 dice are thrown simultaneously. What is the probability that the sum of the numbers on the faces is divisible by either 3 or 5?
A. 7/36
B. 19/36
C. 9/36
D. 2/7
E. None of these
5. Correct option is : B
Solution:
Clearly n(s) = 6*6 = 36
Let E be the event that the sum of the numbers on the 2 faces is divisible by either 3 or 5. Then
E = {(1,2), (1,4), (1,5), (2,1), (2,3), (2,4), (3,2), (3,3), (3,6), (4,1), (4,2), (4,5), (4,6), (5,1), (5,4), (5,5), (6,3), (6,4), (6,6)}
n(E) = 19
Hence P(E) = n(E) / n(s)
= 19/ 36
Latest Government Jobs 2020
## Probability – Moderate
1. Daniel speaks truth in 2/5 cases and Sherin lies in 3/7cases. What is the percentage of cases in which both Daniel and Sherin contradict each other in stating a fact?
A. 72.6%
B. 51.4%
C. 62.3%
D. 47.5%
E. None of these
1. Correct option is : B
Solution:
Daniel and Sherin will contradict each other when one speaks truth and other speaks lies.
Probability of Daniel speak truth and Sherin lies
=2/5*3/7
=6/35
Probability of Sherin speak truth and Daniel lies
=4/7*3/5
=12/35
The two probabilities are mutually exclusive.
Hence, probabilities that Daniel and Sherin contradict each other:
=6/35 +12/35
=18/35
=18/35*100
=51.4%
1. The names of 5 students from section A, 6 students from section B and 7 students from section C were selected. The age of all the 18 students was different. Again, one name was selected from them and it was found that it was of section B. What was the probability that it was the youngest student of the section B?
A. 1/18
B. 1/15
C. 1/6
D. 1/12
E. None of these
1. Correct option is : C
Solution:
The total number of students = 18
When 1 name was selected from 18 names, the probability that he was of section B
= 6 = 1 18 3
But from the question, there are 6 students from the section B and the age of all 6 are different therefore, the probability of selecting one i.e. youngest student from 6 students will be 1/6
1. There are total 18 balls in a bag. Out of them 6 are red in colour, 4 are green in colour and 8 are blue in colour. If Vishal picks three balls randomly from the bag, then what will be the probability that all the three balls are not of the same colour?
A. 95/102
B. 19/23
C. 21/26
D. 46/51
E. None of these
1. Correct option is : D
Solution:
Number of ways in which the person can pick three balls out of 18 balls
= 18C3 = 816
Number of ways of picking 3 balls of same colour = 6C34C8C3 = (20 + 4 + 56) = 80
Probability of picking three balls of same color
= 80 = 5 816 51
Required probability = 1 – probability of picking three balls of same colour
= 1 – 5 = 46 51 51
1. Bag A contains 3 green and 7 blue balls. While bag B contains 10 green and 5 blue balls. If one ball is drawn from each bag, what is the probability that both are green?
A. 29/30
B. 1/5
C. 1/3
D. 1/30
E. None of these
4. Correct option is : B
Solution:
The required probability = 3C1/10C1 × 10C1/15C1
= 3/10 × 10/15 = 1/5
1. Ram and Shyam are playing chess together. Ram knows the two rows in which he has to put all the pieces in but he doesn’t know how to place them. What is the probability that he puts all the pieces in the right place?
A. 8!/16!
B. 8!/(2*15!)
C. 8!/15!
D. (2*8!)/16!
E. None of these
1. Correct option is : B
Solution:
Total boxes = 16
Total pieces = 16
Similar pieces = 8 pawns, 2 bishops, 2 rooks, 2 knights
Total ways of arranging these 16 pieces in 16 boxes
= 16! = 16! (8! 2! 2! 2!) (8 × 8!)
Ways of correct arrangement = 1
Probability of correct arrangement = 1 (16! / (8 × 8!)
= (8 × 8!) = 8! 16! (2 × 15!)
#### Download Probability Answers PDF
Latest Tamilnadu Jobs 2020
Latest police jobs 2020
Online Test Series Click Here To Join Whatsapp Click Here To Subscribe Youtube Click Here To Join Telegram Channel Click Here Online Classes Click Here
|
View Notes
You know 2 and 2 is 4 and you also know 5 and 5 is 10, but do you know how to add larger numbers in fraction of seconds? Math trainer addition tutorial will show you how adding larger numbers can be as simple as a kids play!
Ever imagined that every year on your birthday, when you turn one year older, how do you calculate your age? It's all about numbers, which you require to add every year!
Addition is so crucial in our lives that we cannot think of our day-to-day lives without adding numbers. So let's begin and learn about addition today!
Once you grasp the basics of addition in mathematical operations, you will understand the practice problems of addition, the importance of addition and subtraction in everyday lives. Check- below this interactive simulation to understand how we add 2-digit numbers.
                                             Â
From the above image you would know that Addition is nothing but simply putting two or more numbers together or combining them to find out the sum or the total of the numbers.
For Example,
bells when added or combined together, showing 2+4=6
Here are 2 bells, and when 4 more bells are combined they make a total of 6 bells.
mathematically, we express it as 2 + 4 = 6 and read it as Two plus four equals six (2+4=6)
### Introduction to Workout Time
You can make excellent and quick progress by having 3 sessions of 5 minutes every day. But when you only want to practice as you feel choose "1 day".
 Introduction To Cutoff Time
With just a few seconds to answer a question it makes you remember, instead of trying to count or use other slow techniques.
At first it appears tough, but with practice you get better and better. And at high speed you get ample practice.
Choose 4 seconds for excellent effect.
### The Concept of Carry Over in Addition Trainer
Wondering how to solve complicated addition questions? Note that one-digit numbers can be added simply, while larger numbers are solved by splitting them into columns of their corresponding place values, like Ones, Tens, Hundreds, Thousands, and so on.
We have to add these columns one by one:
That being said, in order to add 354 and 32, we would require writing both the numbers one below the other so that the place values are aligned and then add them.
               Â
Thus from the above example, we observe that while we add the numbers in the Ones column we obtain 6. On the other hand, when we add the numbers under the Tens Column, we obtain 12.
Here, as we retain 2 under the Tens column, we carry over 1 to the top of the Hundreds column, in such a manner that we remember to add it there.
A similar procedure is followed in big numbers whenever we get such two-digit numbers.Â
### Solve Examples on Math Trainer Addition
Example:
A football match had 4535 spectators in the 1st row of the stadium and 2339 spectators in the 2nd row. Find out the total number of spectators that were there in all?
Solution:
By adding the column of Ones, we obtain 14
While we write 4 under the One's column, we direct 1 to the top of the Tens Column going in accordance with the concept of the carry-over, in a way we remember to add it there.
Adding them all, we obtain 6874
Thus, there were 6874 stadium spectators in all.
Example:
A zoo had 1890 beers. The next day 334 new eggs of the species were hatched. Calculate the total number of beers that are there now?
Solution:
Number of beers in the zoo= 1890
Number of eggs that were hatched = 334
Hence, total number of beers in the zoo now = 1890 + 334
= 2224
Thus, 2224 beers.
### Fun Facts on Math Trainer Addition
• One of the important properties of addition states that changing the order of numbers does not change the answer. For example:Â 7 + 5 = 5 + 7, and we get 12 as their sum irrespective of the positioning.
• Terms like 'put together', ‘altogether’, 'in all', 'total' provides a hint that you need to add the given numbers.
• Begin with the larger number and add the smaller number to it. For example, adding 17 to 56 is easier than adding 56 to 17
• Break numbers as per their place values in order to make addition easier. For example, 37 + 96 can be split as 30 + 7 + 90 + 6. While this might seem difficult, it makes mental addition easier.
• Follow the 'tens' first, and then the 'ones' for easy addition of larger numbers
 1. What are Some of the Characteristics of the Math Trainer?
• Formulated for high speed such that you get lots of practice
• Shows you the appropriate answer when you get it incorrect
• Cutoff Time pushes you to remember fast, not count to obtain an answer.
• Timed Workout style just the same as the athletes use
• Recalls your performance so it provides you more practice on your weaknesses.
Answer: The main benefits of learning from trainer addition are that students get better at mental math. Apart from that they can have benefit in the form of:
• Attaining the ability to quickly perform mental calculations
• Developing better number sense and aptitude for quantifying the world around us.
• Even without applications, getting better at mental math to stimulate one’s mind.
• Strengthening the foundation for learning more advanced maths topics.
• Ability to do advanced algebra and arithmetic without having to pull out a calculator.
• Fast speed of mental calculation speed will have a direct impact on math and science test scores.
• At all grade levels, learns to know how to solve math problems when tests have a time limit on them.
• Improving mental math skills will only benefit a student in answering questions both correctly and efficiently becoming highest-scorers but enhancing the academic career.
|
# Gradients and cycling: an introduction
If you’re at all interested in cycling uphill (or even if you’re not) you would have heard people refer to a climb’s gradient (or steepness) as a percentage. A climb might have an average gradient of 3% or 5% or even 10% but just what do these numbers mean? How is gradient calculated? And how challenging are various gradients?
In the first part of this series, we answer these questions and set the scene for more detailed discussions of how gradient applies to cycling.
In cycling terms, “gradient” simply refers to the steepness of a section of road. A flat road is said to have a gradient of 0%, and a road with a higher gradient (e.g. 10%) is steeper than aย road with a lower gradient (e.g. 5%). ย A downhill road is said to have a negative gradient.
You might remember from high school maths that gradient is simply defined as rise/run — that is, the distance travelled vertically (bย in the diagram below) divided by the distanced travelled horizontally (aย in the diagram below). If we want that figure as a percentage then we multiply it by 100.
So let’s sayย we’ve got a 5km-long climb in which we climb 350 vertical metres. The average gradient is simply:
(350/5000) x 100 = 7%
Right?ย Almost.
When you climb a hill you aren’t actually riding on the bottom of the right-angled triangle. You are riding on what’s called the hypotenuse, the edge opposite the right angle. That is, the sloping part labelledย cย in the diagram below.
If we simply divide 350 by 5000, as we did in the example above, we’re dividing the rise (the elevation gain,ย b) by the hypotenuse of the triangle (c), not the run (a). We need a way to find out the horizontal length (a)ย based on the information we’ve already got.ย And to do that we need to employ a basic formula from high school maths.
You might remember the Pythagorean Theorem which shows the relationship between the lengths of the sides of a right-angled triangle. Namely:
aยฒ + bยฒย = cยฒ
Remember, cย is the length of the hypotenuse andย aย andย bย are the lengths of the other two sides.ย So, if we know two of those lengths (and we do: our hill climbs 350m in 5km) we can find out the third.
Plugging our known values into the Pythagorean Theorem we get:
aยฒ +ย 350ยฒย ย = 5000ยฒ
aยฒ +ย 122,500ย = 25,000,000
We then subtract 122,500 from both sides:
aยฒย = 25,000,000 – 122,500
aยฒย = 24,877,500
Taking the square root of either side we get:
aย = 4988
That is, in a 5km-long climb that rises 350m, we’ve travelled 4988m horizontally (imagine the road is perfectly straight).
So now that we know the horizontal distance travelled (a), as opposed to the hypotenuse (c) — the distance on the road — we can calculate the gradient accurately.
As discussed above, the percentage gradient = (rise/run) x 100.ย So, in our example:
(350/4988) x 100 = 7.02%
You’ll notice that this is extremely close to the figure we calculated before. Given the inherent inaccuracy in elevation measurements — either by a bike-mounted GPS unit, online mapping software or other method — this extra accuracy from calculating the gradient properly is virtually redundant.
Even if we’re talking about a super-steep climb — a rise of 1,500 vertical metres of 10km, say — the simple estimation of (1500/10000 ) x 100 (15% average gradient!) is more than accurate enough. Using the proper methodย really doesn’t give us any extra accuracy when we consider discrepancies in elevation measurements — an average gradient of 15.2% compared with the 15% we can calculate much faster.
So, when we’re talking about almost every road used for cycling, we can use a simple formula of (rise/hypotenuse) x 100 to estimate the road’s average gradient, even if it’s not technically accurate.
And speaking of not being technically accurate, don’t assume that a climb’s average gradient can tell us everything we need to know about how steep the climb is.
Mt. Hotham, for example, is a 30.8km long climb with an average gradient of 4.2%. From that percentage you might assume Mt. Hotham is a reasonably easy climb. But what the 4.2% doesn’t tell you is that the final third of the climb has several sections that hit 10%. By the same token, an average gradient of 4.2% doesn’t tell you that the middle third of the climb rises at less than that.
So let’s try to put all of this into a bit of context. What does climbing a road with a 5% gradient feel like? And how much easier is it than a gradient of 10%?
Of course, it really depends on how strong you are and what gearing you are running. First-time climbers might find hills with a 5% gradient challenging at first, but after a bit of training it will likely take a much higher gradient to create the same sort of challenge. That said, here’s a rough guide to how various gradients might feel:
• 1-3%: Slightly uphill but not particularly challenging. A bit like riding into the wind.
• 4-6%: A manageable gradient that can cause fatigue over long periods.
• 7-9%: Starting to become uncomfortable for seasoned riders, and very challenging for new climbers.
• 10%-15%: A painful gradient, especially if maintained for any length of time
• 16%+: Very challengingย for riders of all abilities. Maintaining this sort of incline for any length of time is very painful.
In the next part in this series, we’ll look at ways of objectively measuring how hard various gradients are and what effect a rider’s weight has on climbing speed.
## 62 Replies to “Gradients and cycling: an introduction”
A great resource, and I refer cycling newbies to the site. I lead a large number of social rides each year around Canberra (every ride is hilly) both bitumen and gravel. As part of the process, I provide a ride description and try to incorporate a consistent and objective assessment of ride difficulty, based on 1. distance, 2. surface, 3. gradient summary, and 4. the major climb.
My question:- Do you know of any software or web page that, given a .gpx route file, will provide a summary of distances at various gradient intervals over an entire ride?
I am building a spreadsheet that will do this, but not keen to re-invent the wheel. I came across something called “BikeRoll” but not very useful in this context. I plan rides in “RideWithGPS” and produce a nice profile using “GPS Visualizer”
1. Sam says:
onthegomap.com
I am an absolute beginner and use this for track elevations of commutes
2. There used to be a website called Doogle where you could click two spots on a route and get the grade percentage. Then Google disallowed the site, so I found the Veloroutes.org site with a similar tool, then it too got disallowed by Google. Is there a website where I can just click two spots and get a readout of the grade percentage between those two spots on the route?
1. Not sure to be honest, but I’ve always used RideWithGPS and found it relatively useful for such queries.
1. Thanks Matt. Yes, RWGPS is a good tool for most climbs. It shows the elevation profile and percentages along the route. Doogle gave me the same thing but almost always gave me higher percentages for each section of a route than RWGPS gave me for the same route. What RWGPS identified as a 12% grade, for example, would be a 16% climb on Doogle. My guess is that Doogle had higher resolution; that is, used more points along each section for the calculation? A 4% difference isn’t that big of a deal for grades up to 10% or so, but a 4% difference for grades higher than 10% is a big deal for me.
1. It seems to me that RideWithGPS calculates the gradient using decently long intervals, but will use shorter intervals when the control points are close together. So for climbing segments I use more control points than are needed to merely calculate the route itself. With closely spaced control points on a short climb near me, for instance, I can see that there are two sections with a grade of over 11% whereas if I just use a start and stop point at the beginning and end of the section then the RideWithGPS grade chart show a maximum of 9.3%
2. Anony. says:
Try strava.
3. William Joseph Kerley says:
Fantastic article, and series of articles. Thanks for creating these!
4. This information sure helps a lot. I live in an area that has climbs with all these gradients at any given time (mostly they all are in the 7%). Plus, they are short or long (30km).
I still use 53×17 for training purpose. T feel more comfortable using 39×15.
5. Yao says:
Nice article, but an easier way, and (I haven’t test the accuracy yet)an efficient way to calculate any incline with a phone is just simply to use the surface level meter on your phone, get the y axis angle, and the gradient for that specific spot will be tan(y angle), which is equal to the (rise/distance traveled horizontally).
Repeat this measurement multiple time, and get the average of them. The result should be fairly close to what you have been calculated above.
6. George Moncure says:
Helpful info. Thanks! While I mountain bike, it’s still relevant. I’m new in getting back into mountain biking with a fat tire electric assist bike. At 67 I made it up an 11% rocky 1/2 mile grade that I would not be able to do without e-assist. My more skilled younger friends aren’t interested in such trails and I get it now. It wasn’t easy even with e-assist but at least I got there and managed may steeper grades over shorter distances on the ride (Sheep Mountain Anticline, Wyoming). Cheers!
7. Justin Powell says:
This is such an informative post about gradient and cycling introduction. Slope style bike is a great choice for this, I think. There are many slope style bikes out there, it may be in the local shops or online. Always choose the best and quality bikes like Trek, Giant, Morpheus Bikes and others.
8. Andy Liggins says:
9. antti says:
How about degrees? 45 ^ would be 100 %.. ?
10. TG Stewart says:
DUH! I just asked a dumb question. The distance is converted to metres (meters) for the calculations. Using the same unit to measure both it won’t matter which system is used.
I should think before I ask!
11. TG Stewart says:
This example uses metric measures (kilometers and meters). Are all published gradients calculated with metric units?
1. George Moncure says:
Gradients in feet and meters are the same. Units cancel
12. Thank you so much! Wonderful
Explanation without being bogged down with science and maths. For the statistics , I used to train in Dire dawA , Ethiopia on a 5km stretch at 5 to 7% , well that is what the sign said! It was comfortable for me but best bit was
Flying back down the twisty road, wonderful.
13. Pingback: In Training
14. Michael says:
Interesting Observations throughout. Thanks for whoever put this together. I’ve always wondered how gradients of a hill were worked out.
Very simple to understand.
Back to the 2017 Paris-Nice, where the riders are to face a 20% grade on today’s time-trial. Ouch!
1. Wayne Shoudt says:
Try Lincoln Gap toughest mile in America. A four mile hill. 17% overall. The last mile does not drop below 20% and in that mile reaches 24%. It was the hardest thing I have ever done on a bike.
15. Cameron says:
Researching for my physics project. got some good info thank you!
16. Robert Black says:
Far far simpler if one sticks to something sensible such as ratios for hills. Lets just keep to 1:5 1:10 etc instaed of the ridiculous European % system.
1:5 simply means that for every 1 up there are 5 along – Simple – no calculations needed.
1. Therestoftheworld says:
Damn straight Robbie! The entire rest of the world IS ridiculous! While we’re at it we should get rid of other silly internationally standardised things like Celsius, the entire metric system, actual wire sizing (not ‘AWG’), the list is exhaustive…
2. Cian Brendan McDermott says:
Silly response Robert – % are the international standard
3. Andy Diniz says:
Ridiculous percentage system? So you’d rather represent 100 along, 7 up as 1:14.2857 than 7% ?
17. Billy Dean says:
As a beginning cyclist, I’m having trouble with anything over 7% grade in my lowest gear, which is 34/28. My thinking has been that I could revise my gearing to handle some of the 10+ grades in my area. 34/28, for example, is 1.214. So I divided 10% by 7% and got 1.428, which told me I needed to increase my current gear ratio by about 42%. Problem is, I can’t find anything on the web that explains the relationship between grade and gear. Is it linear? Or…?
1. Gordon says:
Billy, no, it is not a linear relationship. It is all about your strength/fitness level. As a grand oldie I can do 17% comfortably with 34/28 setup. 22% at a pinch. However this took me nearly two years to get there by riding 300 Kms a week and constantly challenging myself on the many steep climbs in the Dandenongs. I am afraid the only answer is hard work. Good luck with your project as it will be worth it in the end.
18. Ruby says:
This is so detailed and helpful, I’m referencing it for my maths project on the effect of the slope on caloric burn. thanks! ๐
19. Douglas Kubler says:
If you use a GPS (Garmin) then the distance recorded on a climb is “a”, the projection of your route onto a (nearly) flat earth. The gradient would be 7%.
20. chris says:
your maths is right but you have confused the sides of the triangle….
you have done aยฒ + 350ยฒ = 5000ยฒ
when you should have done 5000ยฒ + 350ยฒ = cยฒ
your words: “That is, in a 5km-long climb that rises 350m, weโve travelled 4988m horizontally (imagine the road is perfectly straight).”
you have calculated that the longest side of the triangle, the horizontal line is 18m shorter lol
good one
1. Probably worth having another close read. Especially this part:
“When you climb a hill you arenโt actually riding on the bottom of the right-angled triangle. You are riding on whatโs called the hypotenuse, the edge opposite the right angle. That is, the sloping part labelled c in the diagram below.”
21. Colin says:
So as the sportive I am about to do has climbs of 20%, I assume it is going to be painful!
Good item, well written.
1. Climbing hills on a bike is not hard. It depends how fast you want to go and your gearing. If you have the right gearing, climbing can be as comfortable as riding along the flat. The problem is psychological. Cyclists just can’t help wanting to speed up hills and making it agony for themselves. We all watch Le Tour too much!
22. Teresa says:
LOVE THIS POST!!!
Thank you so much… I and my students found this to be very useful and helpful when studying gradient percentages in geography. Thank you!
Nice to know that good ole Pythagoras is relevant to sports people!!!
23. Great article about gradients. Unfortunately here in the US, at least in my state, we don’t have any gradient % signs. I have a cyclocomputer, so I could figure out the run of the road but not sure how I could get the “rise.” You mention online software, what are some ways I could find the elevation?
My second question is about training. I’ve started cycling after being kind of sedentary for awhile and I want to start climbing hills so I can produce a lot of power in amateur races. Assuming the gradient of a hill is pretty consistent, 4%(+ or -, one percent throughout) what level of gradient do I need to be able to roll over to be competitive in beginning/intermediate amateur road races?
1. Alex says:
A simple way to find the gradient of any ride you’re thinking about doing is to map the route on google maps then paste the link into the ‘look up elevations’ section of gps visualizer . com
“ride like the wind and be home for tea”
24. Daniel says:
Hey Matt,
will you be looking at the road condition as part of this? because id say that road condition plays a BIG factor in the difficulty of a climb! for instance.
A climb that is 10% on a road that is really rough can be alot harder then a 12-15% road on a super smooth surface!
1. Sure will mate. Stay posted. Is it the Waterworks climb that we did for H10K that was super rough? I remember it being about 8% gradient but feeling like 12%. ๐
25. Manning says:
Using the Mt Hotham example as how average gradients can be misleading, I’ve often wondered if you could use a normalised gradient function (similar to normalised power) to determine the true physiological demand of a climb.
A quick calc using a similar calculation, for a 1000m long climb that averages 5% for the first 500m and 10% for the final 500m. I get the average gradient to be 7.5% but the normalised gradient to be 8.54%
i.e. the climb is as hard as one that averages 8.54% for it’s full 1000m
1. gordo says:
Manning, I get a feel for what you are asking / suggesting, however, I’m not sure what you mean by normalised power. Is it average power or RMS power or what?
The simple example of a split gradient for a 1km climb is probably in reality 2 climbs one after the other, 2 different gradients. Is it easier to do that climb in reverse for instance? 10% then 5%.
Anyhow, your input is valuable and hopefully someone can clarify it for all of us.
1. Manning says:
To quote an explanation from cyclingtips..”Normalized Power is basically an estimate of the power that you could have maintainted for the same physiological โcostโ if your power output had been constant.”
Normalised power is calculated by
1) starting at the 30 s mark, calculate a rolling 30 s average (of the preceeding time points, obviously).
2) raise all the values obtained in step #1 to the 4th power.
3) take the average of all of the values obtained in step #2.
4) take the 4th root of the value obtained in step #3.
Basically if you rode a steady power output your average and normalised power would be the same. If however you threw in some sprints (or raced a criterium) you get a larger normalised power, this reflects how taxing it was on your body.
So my thinking is using a similar formula for gradient, in the above case I took 100m increments instead of 30s increments (also should be 8.5% not 8.54%…)
Hope that explains it?
1. gordo says:
Thank you.
26. gordo says:
Good stuff Matt. Whilst I was aware of most of the technical stuff your article has certainly collated it very nicely and explains it much better than I can especially when talking to my climbing mates.
It will be interesting how you tackle the perceived hardness of a climb. I for one measure a climb by what lowest gear I need to use with a standard bike setup. For instance, I know I can do The Wall with 39/24 using my 17 kg hybrid and 8kg saddle bags. I don’t generally measure time but certainly alternate between standing and sitting constantly.
Anyhow, a good subject with lots of anecdotal information available as well as hard facts.
27. | PK says:
Is there a list of climbs in Melbourne / VIC that have “categories” i.e a category 2 climb etc….. Just curious.
1. | PK says:
|
# CBSE Class 12 Mathematics NCERT Exemplar Solutions: Chapter 1 – Relations and Functions
Rahul Tomar
In this article, we are providing you NCERT Exemplar Solutions for all the questions of Class 12 Mathematics Chapter 1 – Relations and Functions. The difficulty level of these questions is quite higher than the exercise questions usually given at the end of the chapter.
## About NCERT Exemplar Solutions Class 12 Maths:
All the questions of NCERT Exemplar Class 12 Maths book are very important for CBSE Class 12 Maths Board examination and other engineering entrance examinations.
Analysis of CBSE Class 12 previous year question paper shows that questions from NCERT Exemplar are frequently asked in CBSE board examination.
Questions from NCERT Exemplar Class 12 Mathematics are likely to be asked again in upcoming board exam as well as engineering entrance exams.
Types and number of questions in this chapter:
Types Number of questions Short answer type questions 15 Long answer type questions 12 Objective type questions 20 Fillers 5 True and False 10
Few problems along with their solutions from this chapter are given follows:
## Trending Now
Question:
Are the following set of ordered pairs functions? If so examine whether the mapping is injective or surjective.
(i) { (x, y): x is a person, y is the mother of x }.
(ii) { (a, b) : a is a person, b is an ancestor of a}.
Solution:
(i) The set of ordered pairs given here represents a function.
Here, the images of distinct elements of x under f are not distinct, so it is not injective but it is surjective.
(ii) Since, each element of domain does not have a unique image.
Therefore, the set of ordered pairs given here does not represent function.
Question:
Using the definition, prove that the function f : A b is invertible if and only if f is both one-one and onto.
Solution:
A function f :X → Y is defined to be invertible, if there exist a function g = Y → X such that gof = Ix and fog = Iy. The function is called the inverse of f and is denoted by f
We know that only bijective functions are invertible functions. A bijective function is both injective and surjective.
It means function f: X → Y is invertible iff f is a bijective function.
Question:
If the set a contains 5 elements and the set b contains 6 elements, then the number of one-one and onto mappings from a to b is
(a) 720
(b) 120
(c) 0
(d) None of these
Solution: (c)
Since, the number of elements in B is more than A.
Hence, there cannot be any one-one and onto mapping from A to B.
Question:
Every function is invertible.
Solution: False
We know that only bijective functions are invertible.
Question:
A binary operation on a set has always the identity element.
Solution: False
We have '+' is a binary operation on the set N but it has no identity element.
|
# Impulse
We discussed above the factors changing momentum which are mass and velocity. In most of the case mass is constant and for momentum change velocity changes. If velocity changes then acceleration occurs. In the first unit we said that force causes acceleration in other words change in the velocity is the result of applied net force. Change in the velocity is proportional to the applied net force. If it is big then change is also big. Another important thing is the time of applied force. How long does it act on an object? It is linearly proportional to the change in velocity. If you apply a force on an object 1 s then you see small change in the momentum. However, if you apply force on an object long period of time then you see the amount of change in momentum is bigger than the first situation. In summary, I try to say that impulse is the multiplication of applied force and time interval it applied. Impulse is also a vector quantity having both magnitude and direction. It has the same direction with applied net force.
Impulse=Force.Time Interval
Impulse and momentum are directly related to each other. Let’s find this relation now. As you can see, we found that impulse is equal to the change in momentum. In examples we will benefit from this equation.
Impulse=Change in Momentum
Example: If the time of force application is 5s find the impulse of the box given below. Impulse=Force.Time Interval
Impulse=15N.5s
Impulse=75N.s
Example: Find applied force which makes 10m/s change in the velocity of the box in 5s if the mass of the box is 4kg.
Impulse=Change in momentum
F.t=p2-p1
F.t=m. (V2-V1)
F.t=4kg.10m/s=40kg.m/s Impulse of the box is 40kgm/s
F=40kg.m/s/5s=8N Applied force
We experience the results of impulse and momentum in daily life. For example, in collisions like car crash or any other collision we can calculate the affect of force by controlling the time. Assume that, you push a box with a force of 10N for 2 seconds, the impulse is 20N.s. Then you push it with a 5N force for 4 seconds and impulse does not change. As you see increasing the time decreases the amount of force. Thus, the unwanted results of force can be eliminated by increasing the time of force application. On the contrary if you want big force then you should decrease the time and you get big force.
In this unit we will again benefit from the graphs. Look at the given graph below that shows the relationship of the force and time of a given system. Since impulse is equal to the multiplication of force and time then, area under this graph also gives us impulse. As shown in the graph, A1 is positive impulse and A2 is negative impulse. Total impulse gives us the change in momentum as we said before. We can also draw momentum versus time and velocity versus time graph of the system. As you can see momentum vs. time graph and velocity vs. time graphs of the system are similar because momentum is directly proportional to the velocity.
Example: The graph given below belongs to an object having mass 2kg and velocity 10m/s. It moves on a horizontal surface. If a force is applied to this object between (1-7) seconds find the velocity of the object at 7. Second. Area under the graph gives us impulse. First, we find the total impulse with the help of graph given above then total impulse gives us the momentum change. Finally, we find the final velocity of the object from the momentum change.
Impulse Momentum Exams and Solutions
|
# Line and Rotational Symmetries
Lesson
Symmetry is a pattern of self similarity, and we see symmetry often in the real world. A lot of objects are built with symmetry in mind. There is symmetry in nature, and mathematics has a lot of symmetry as well.
We can define three important types of symmetry: line, point and rotational.
Let's explore what these different symmetries look like using the applet below. (Watch this video if you would like to see this interactive in action - )
## Line Symmetry
Line symmetry occurs when a shape or image is reflected across a line. It's like you can fold the shape or image on that line and one side is an exact match of the other.
A popular art activity uses symmetry to create shapes like this butterfly.
The actual line used in line symmetry is also referred to as the axis of symmetry.
Here are some of the lines of symmetry evident in these geometric shapes. A shape could have more than one line of symmetry. For example, if you have a look at the square or the pentagon, can you see more lines of symmetry?
## Rotational Symmetry
Rotational symmetry exists when the shape can be rotated around and the original shape appears again.
See in this shape, as we turn it clockwise, the same external shape appears 3 times in one full rotation. We call this the order of the rotational symmetry. (I coloured the ends so you can see how the shape rotates around.)
## Point Symmetry
Point symmetry exists when every point on a shape has a matching point exactly the same distance from a central point.
A circle is an obvious shape with point symmetry because every point on the circumference is the same distance from the centre and is the reflection of another point also on the circumference.
Shapes or images with point symmetry, also have a rotational symmetry of order 2.
Let's have a look at some worked examples.
##### Question 1
Which of the following letters has rotational symmetry?
##### Question 2
What type of symmetry do the following shapes have?
##### Question 3
Deduce whether or not these shapes have point symmetry.
|
# Difference between revisions of "2020 AMC 10B Problems/Problem 25"
The following problem is from both the 2020 AMC 10B #25 and 2020 AMC 12B #24, so both problems redirect to this page.
## Problem
Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product$$n = f_1\cdot f_2\cdots f_k,$$where $k\ge1$, the $f_i$ are integers strictly greater than $1$, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$, $2\cdot 3$, and $3\cdot2$, so $D(6) = 3$. What is $D(96)$?
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$
## Solution 1
Note that $96 = 2^5 \cdot 3$. Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$, we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$: We see that there is $1$ way, merely $96$.
$k=2$: This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$'s leave us with $5$ ways and symmetry doubles us so we have $10$.
$k=3$: We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of $2$ in a manner that is $3-0-0$, $2-1-0$ or $1-1-1$. A $3-0-0$ split has $6 + 3 = 9$ ways of happening ($24-2-2$ and symmetry; $2-3-16$ and symmetry), a $2-1-0$ split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a $1-1-1$ split has $3$ ways of happening ($6-4-4$ and symmetry) so in this case we have $9+18+3=30$ ways.
$k=4$: We have $3, 2, 2, 2$ as our baseline, and for the two other $2$'s, we have a $2-0-0-0$ or $1-1-0-0$ split. The former grants us $4+12=16$ ways ($12-2-2-2$ and symmetry and $3-8-2-2$ and symmetry) and the latter grants us also $12+12=24$ ways ($6-4-2-2$ and symmetry and $3-4-4-2$ and symmetry) for a total of $16+24=40$ ways.
$k=5$: We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.
$k=6$: We have $3, 2, 2, 2, 2, 2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways.
Thus, adding, we have $1+10+30+40+25+6=\boxed{\textbf{(A) } 112}$.
~kevinmathz
## Solution 2
As before, note that $96=2^5\cdot3$, and we need to consider 6 different cases, one for each possible value of $k$, the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization needs to contain one factor that is itself a multiple of $3$. There are $k$ to choose from. That leaves $k-1$ slots left to fill, each of which must contain at least one factor of $2$. Once we have filled in a $2$ to each of the remaining slots, we're left with $5-(k-1)=6-k$ twos.
Consider the remaining $6-k$ factors of $2$ left to assign to the $k$ factors. Using stars and bars, the number of ways to do this is: $${{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}$$ This makes $k{5\choose{6-k}}$ possibilities for each k.
To obtain the total number of factorizations, add all possible values for k: $$\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{\textbf{(A) } \text{112}}.$$
## Solution 3
Begin by examining $f_1$. $f_1$ can take on any value that is a factor of $96$ except $1$. For each choice of $f_1$, the resulting $f_2...f_k$ must have a product of $96/f_1$. This means the number of ways the rest $f_a$, $1 can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)$, since the product of $f_2 \cdot f_3... \cdot f_k=x$ is counted as one valid product if and only if $f_1 \cdot x=96$, the product $x$ has the properties that factors are greater than $1$, and differently ordered products are counted separately.
For example, say the first factor is $2$. Then, the remaining numbers must multiply to $48$, so the number of ways the product can be written beginning with $2$ is $D(48)$. To add up all the number of solutions for every possible starting factor, $D(96/f_1)$ must be calculated and summed for all possible $f_1$, except $96$ and $1$, since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$. This means
$D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$.
Instead of calculating D for the larger factors first, reduce $D(48)$, $D(32)$, and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation. Following the recursive definition $D(n)=($sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to
$D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)$ +$(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.$
$D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$, so the sum further simplifies to
$D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5$, after combining terms. From quick casework,
$D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1$ and $D(2)=1$. Substituting these values into the expression above,
$D(96)=3 \cdot 8+4 \cdot 8+5 \cdot 4+4 \cdot 3+5 \cdot 2+4 \cdot 1+5 \cdot 1+5=\boxed{\textbf{(A) } 112}$.
~monmath a.k.a Fmirza
## Solution 4
Note that $96 = 3 \cdot 2^5$, and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$.
First, calculate $D(32)$. Since $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$, all you have to do was find the number of ways to divide the $2$'s into groups, such that each group has at least one $2$. By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$, is not needed for the remaining calculations.)
Then, to get $D(96)$, in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five.
The resulting number of possible sequences is $3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{\textbf{(A) }112}$. ~emerald_block
## Solution 5 (Minimal Casework)
Consider the arrangement of the prime factors of 96 in a line $(2,2, 2, 2, 2, 3)$. An arrangement of factors can be created by placing "dividers" to group primes. For example, $(2, 2, |, 2, 2, 2, |, 3)$ is equivalent to the arrangement $4 \cdot 8 \cdot 3$. Because there are $6$ ways to order the prime factors, and $2^5$ ways to place dividers, this gives us an initial $6 \cdot 2^5$ ways to arrange divisors.
However, through this method, we overcount cases where $3$ is combined with another factor. For example, the arrangement $4 \cdot 6 \cdot 4$ can be written as $(2, 2, |, 2, 3, |, 2, 2)$ or $(2, 2, |, 3, 2, |, 2, 2)$. Precisely, we double count any case with $6$ as a factor, triple count any case with $12$, quadruple count any case with $24$, etc.
Now, consider all cases where $3$ must be grouped with at least one $2$. This can be expressed in the same "line" format as $(2, 2, 2, 2, 6)$, where dividers can again be placed to group divisors. In this case, there are $5$ ways to order divisors, and $2^4$ ways to place dividers, so we have an $5 \cdot 2^4$ possible sequences for this case. Notice that in this format, we double count cases where $12$ is a factor, we triple count cases where $24$ is a factor, etc. Precisely, for any case counted $n$ times in the first step, it is counted $n - 1$ times in this step. Thus, if we subtract, we count each case exactly once.
So, we get:
$6 \cdot 2^5 - 5 \cdot 2^4 = \boxed{\textbf{(A) }112}$. ~hdai1122
## Solution 6 (Another Fast Way)
First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$. By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$'s, or multiply it to one of the $g_i$'s, a total of $2m+1$ ways. Hence the answer to the problem is
$$\sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{\textbf{(A) }112}.$$
~ asops
## Video Solutions
### Video Solution 1
https://youtu.be/8WrdYLw9_ns?t=1444 - Quicker Way
~ pi_is_3.14
|
# Learning Objectives
### Learning Objectives
By the end of this section, you will be able to do the following:
• Describe the right-hand rule to find the direction of angular velocity, momentum, and torque
• Explain the gyroscopic effect
• Study how Earth acts like a gigantic gyroscope
The information presented in this section supports the following AP® learning objectives and science practices:
• 4.D.3.1 The student is able to use appropriate mathematical routines to calculate values for initial or final angular momentum, or change in angular momentum of a system, or average torque or time during which the torque is exerted in analyzing a situation involving torque and angular momentum. (S.P. 2.2)
• 4.D.3.2 The student is able to plan a data collection strategy designed to test the relationship between the change in angular momentum of a system and the product of the average torque applied to the system and the time interval during which the torque is exerted. (S.P. 4.1, 4.2)
Angular momentum is a vector and, therefore, has direction as well as magnitude. Torque affects both the direction and the magnitude of angular momentum. What is the direction of the angular momentum of a rotating object like the disk in Figure 10.28? The figure shows the right-hand rule used to find the direction of both angular momentum and angular velocity. Both $LL size 12{L} {}$ and $ωω size 12{ω} {}$ are vectors—each has direction and magnitude. Both can be represented by arrows. The right-hand rule defines both to be perpendicular to the plane of rotation in the direction shown. Because angular momentum is related to angular velocity by $L=IωL=Iω size 12{L=Iω} {}$, the direction of $LL size 12{L} {}$ is the same as the direction of $ωω size 12{ω} {}$. Notice in the figure that both point along the axis of rotation.
Figure 10.28 Figure (a) shows a disk is rotating counterclockwise when viewed from above. Figure (b) shows the right-hand rule. The direction of angular velocity $ωω$ size and angular momentum $LL$ are defined to be the direction in which the thumb of your right hand points when you curl your fingers in the direction of the disk's rotation as shown.
Now, recall that torque changes angular momentum as expressed by
10.138 $net τ=ΔLΔt.net τ=ΔLΔt. size 12{"net "τ= { {ΔL} over {Δt} } } {}$
This equation means that the direction of $ΔLΔL size 12{ΔL} {}$ is the same as the direction of the torque $ττ size 12{τ} {}$ that creates it. This result is illustrated in Figure 10.29, which shows the direction of torque and the angular momentum it creates.
Let us now consider a bicycle wheel with a couple of handles attached to it, as shown in Figure 10.30. This device is popular in demonstrations among physicists, because it does unexpected things. With the wheel rotating as shown, its angular momentum is to the woman's left. Suppose the person holding the wheel tries to rotate it as in the figure. Her natural expectation is that the wheel will rotate in the direction she pushes it—but what happens is quite different. The forces exerted create a torque that is horizontal toward the person, as shown in Figure 10.30(a). This torque creates a change in angular momentum $LL size 12{L} {}$ in the same direction, perpendicular to the original angular momentum $LL size 12{L} {}$, thus changing the direction of $LL size 12{L} {}$ but not the magnitude of $LL size 12{L} {}$. Figure 10.30 shows how $ΔLΔL size 12{ΔL} {}$ and $LL size 12{L} {}$ add, giving a new angular momentum with direction that is inclined more toward the person than before. The axis of the wheel has thus moved perpendicular to the forces exerted on it, instead of in the expected direction.
Figure 10.29 In figure (a), the torque is perpendicular to the plane formed by $rr size 12{r} {}$ and $FF size 12{F} {}$ and is the direction your right thumb would point to if you curled your fingers in the direction of $FF size 12{F} {}$. Figure (b) shows that the direction of the torque is the same as that of the angular momentum it produces.
Figure 10.30 In figure (a), a person holding the spinning bike wheel lifts it with her right hand and pushes down with her left hand in an attempt to rotate the wheel. This action creates a torque directly toward her. This torque causes a change in angular momentum $ΔLΔL$ in exactly the same direction. Figure (b) shows a vector diagram depicting how $ΔLΔL$ and $LL$ add, producing a new angular momentum pointing more toward the person. The wheel moves toward the person, perpendicular to the forces she exerts on it.
### Applying Science Practices: Angular Momentum and Torque
You have seen that change in angular momentum depends on the average torque applied and the time interval during which the torque is applied. Plan an experiment similar to the one shown in Figure 10.30 to test the relationship between the change in angular momentum of a system and the product of the average torque applied to the system and the time interval during which the torque is exerted. What would you use as your test system? How could you measure applied torque? What observations could you make to help you analyze changes in angular momentum? Remember that, since angular momentum is a vector, changes can relate to its magnitude or its direction.
This same logic explains the behavior of gyroscopes. Figure 10.31 shows the two forces acting on a spinning gyroscope. The torque produced is perpendicular to the angular momentum, thus the direction of the torque is changed, but not its magnitude. The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to $LL size 12{L} {}$. If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque ($L=ΔLL=ΔL size 12{L=ΔL} {}$), and it rotates around a horizontal axis, falling over just as we would expect.
Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star. But Earth is slowly precessing (once in about 26,000 years) due to the torque of the Sun and the Moon on its nonspherical shape.
Figure 10.31 As seen in figure (a), the forces on a spinning gyroscope are its weight and the supporting force from the stand. These forces create a horizontal torque on the gyroscope, which create a change in angular momentum $ΔLΔL size 12{L} {}$ that is also horizontal. In figure (b), $ΔLΔL size 12{L} {}$ and $LL size 12{L} {}$ add to produce a new angular momentum with the same magnitude, but different direction, so that the gyroscope precesses in the direction shown instead of falling over.
|
# Chebyshev’s Theorem: How to Calculate it by Hand and in Excel
Hypothesis Testing > Chebyshev’s Theorem
Before you start, you may want to read this article on Chebyschev’s Inequality, which explains when you would want to use the theorem. Chebyshev’s Interval refers to the intervals you want to find when using Chebyshev’s theorem.
## How to Calculate Chebyshev’s Theorem
Watch the video or read the steps below:
For normal distributions, about 68% of results will fall between +1 and -1 standard deviations from the mean. About 95% will fall between +2 and -2 standard deviations. Chebyshev’s Theorem allows you to use this idea for any distribution, even if that distribution isn’t normal. The theorem states that for a population or sample, the proportion of observations is no less than (1 – (1 / k2 )), as long as the z score’s absolute value is less than or equal to k. You can only use Chebyshev’s Theorem to get results for standard deviations more than 1; It can’t be used to find results for smaller values like 0.1 or 0.9. Technically, you could use it and get some kind of a result, but those results wouldn’t be valid.
In a normal distribution, the percentages of scores you can expect to find for any standard deviations from the mean are the same.
Sample problem: a left-skewed distribution has a mean of 4.99 and a standard deviation of 3.13. Use Chebyshev’s Theorem to find the proportion of observations you would expect to find within two standard deviations from the mean:
Step 1: Square the number of standard deviations:
22 = 4.
1 / 4 = 0.25.
Step 3: Subtract Step 2 from 1:
1 – 0.25 = 0.75.
At least 75% of the observations fall between -2 and +2 standard deviations from the mean.
That’s:
mean – 2 standard deviations
4.99 – 3.13(2) = -1.27
mean + 2 standard deviations
4.99 + 3.13(2) = 11.25
Or between -1.27 and 11.25
That’s it!
Warning: As you may be able to tell, the mean of your distribution has no effect of Chebyshev’s theorem! That fact can cause some wide variations in data, and some inaccurate results.
Like the explanation? Check out the Practically Cheating Statistics Handbook, which has hundreds more step-by-step explanations, just like this one!
## How to Calculate Chebyshev’s Formula in Excel.
Microsoft Excel has a wide variety of built-in functions and formulas that can help you with statistics. However, it does not have a built-in formula for Chebyshev’s Theorem. In order to calculate Chebyshev’s theorem in Excel, you’ll need to add the formula yourself. If you want to use Chebyshev’s formula just once or twice, you can type the formula into a cell. However, if you intend on using Chebyshev’s formula several times over time, you can add a custom function (=CHEBYSHEV) to Microsoft Excel.
## How to Calculate Chebyshev’s Formula in Excel (Temporary Use).
Step 1: Type the following formula into cell A1: =1-(1/b1^2).
Step 2: Type the number of standard deviations you want to evaluate in cell B1.
Step 3: Press “Enter.” Excel will return the percentage of results you can expect to find within that number of standard deviations in cell B1.
## How to Calculate Chebyshev’s Formula in Excel (Adding a Custom Formula)
Step 1: Open the Visual Basic editor in Excel. To open the Visual Basic editor, click the “Developer” tab and then click “Visual Basic.”
Step 2: Click “Insert” and then click “New Module.”
Step 3: Type the following code into the blank window:
Function Chebyshev(stddev)
If stddev >= 0 Then
Chebyshev = (1 – (1 / stddev ^ 2))
Else: Chebyshev = 0
End If
End Function
.
Typing the code for a Chebyshev custom function in Microsoft Excel.
Step 4: Close the visual basic window and return to the worksheet. The custom function is now ready to use: type “=chebyshev(x)” into a blank cell, where “x” is the number of standard deviations. Excel will calculate Chebyshev’s theorem and return the result in the same cell.
## Where did Chebyshev’s Theorem Come From?
Pafutny Lvovich Chebyshev (1821-1894)was a Russian mathematician. His friend, mathematician and gifted linguist Irenée-Jules Bienaymé translated many of Chebyshev’s works into French.
In 1867 Chebyshev published a paper On mean values which first mentioned the inequality to give a generalized law of large numbers. however, the inequality actually first appeared fourteen years earlier in Bienaymés Considérations à l’appui de la découverte de Laplace. The editor who discovered Chebyshev’s use of Bienaymé’s inequality (without mention of the original author) said:
It is a pity that their common interest in the Inequality somehow “slipped through the cracks” in the early contacts between Bienaymé and Chebyshev. Possibly the Inequality was regarded by Bienaymé as a minor result compared with his main themes of linear least squares and Laplacian defence. Chebyshev’s recognition of its significance and its clear statement has, at any rate, always been a defensive point in his favour stressed by some historiographers. From The University of St. Andrews.
Chebyshev’s theorem is often spelled many different ways — you’ll find it spelled as Chebychev’s theorem, Chebyschev’s theorem and even Tschebyscheff’s theorem. That’s mostly due to the fact that his original name was Russian, which uses a different alphabet (cyrillic). “Chebyshev” is just the word, taken as it sounds and translated into an English approximation.
Fun fact: There is a crater on the moon named after him: Crater Chebyschev.
Chebyschev’s crater on the moon.
------------------------------------------------------------------------------
If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.
Chebyshev’s Theorem: How to Calculate it by Hand and in Excel was last modified: November 21st, 2017 by
# 20 thoughts on “Chebyshev’s Theorem: How to Calculate it by Hand and in Excel”
1. Lyon
what do you mean by “z” here in this sentence–as long as the z score’s absolute value is less than or equal to k? thank you.
2. Andale
I’m working on algebrahowto.com this year. MathHowTo.com is also in the works :)
3. Andale
K is the number of standard deviations either side of the mean. You shouldn’t need to find it — it’s usually stated in the question/problem you’re given (in the sample problem, you’re finding how many points fall within two standard deviations, so k=2).
Stephanie
4. Sammy
Use Chebyshev’s theorem to find what percent of the values will fall between 236 and 338 for a data set with a mean of 287 and standard deviation of 17.
Help
5. Nasim
dear all the above question is still not clear for me, becuase it only used the formula not that mean and standard diviation as he was exampled, please anyone clear that for me
thanks
6. rube
what if the question is a little different like avg car in the lot = 200,000 and the standard deviation is 32,000 and the deviation but the question is something like find the range in which 80% of of cars fall?? I can’t see how to fit this formula to that. Thanks a lot!
7. Shahrokh
|
Eureka Math Algebra 1 Module 3 End of Module Assessment Answer Key
Engage NY Eureka Math Algebra 1 Module 3 End of Module Assessment Answer Key
Eureka Math Algebra 1 Module 3 End of Module Assessment Task Answer Key
Question 1.
Given h(x) = |x + 2| – 3 and g(x) = – |x| + 4:
a. Describe how to obtain the graph of g from the graph of a(x) = |x| using transformations.
To obtain the graph of g, reflect the graph of a about the x-axis, and translate this graph up 4 units.
b. Describe how to obtain the graph of h from the graph of a(x) = |x| using transformations.
To obtain the graph of h, translate the graph of a 2 units to the left and 3 units down.
c. Sketch the graphs of h(x) and g(x) on the same coordinate plane.
d. Use your graphs to estimate the solutions to the equation:
|x + 2| – 3 = – |x| + 4
Solution: x ≈ 2.5 or x ≈-4.5
The solutions are the x-coordinates of the intersection points of the graphs of g and h.
e. Were your estimations in part (d) correct? If yes, explain how you know. If not, explain why not.
Let x = 2.5
Is |2.5 + 2|-3 = -|2.5| + 4 true?
Yes, 4.5 – 3 = -2.5 + 4 is true.
Let x = -4.5
Is |-4.5 + 2| – 3 = – |4.5| + 4 true?
Yes, 2.5 – 3 = -4.5 + 4 is true.
Yes, the estimates are correct. They each make the equation true.
Question 2.
Let f and g be the functions given by f(x) = x2 and g(x) = x|x|.
a. Find f($$\frac{1}{3}$$), g(4), and g( – $$\sqrt{3}$$).
f($$\frac{1}{3}$$) = $$\frac{1}{9}$$, g(4) = 16, g(-$$\sqrt{3}$$) = -3
b. What is the domain of f?
D: all real numbers
c. What is the range of g?
R: all real numbers
d. Evaluate f( – 67) + g( – 67).
(-67)2 + -67|-67| = 0
e. Compare and contrast f and g. How are they alike? How are they different?
When x is positive, both functions give the same value. But when x is negative, f gives the always-positive value of x2, whereas g gives a value that is the opposite of what f gives.
f. Is there a value of x such that f(x) + g(x) = – 100? If so, find x. If not, explain why no such value exists.
No. f and g are either both zero, giving a sum of zero; both positive, giving a positive sum; or the opposite of each other, giving a sum of zero. So, there is no way to get a negative sum.
g. Is there a value of such that (x) + g(x) = 50? If so, find x. If not, explain why no such value exists.
Yes. If x = 5, f(x) = g(x) = 25; thus, f(x) + g(x) = 50.
Question 3.
A boy bought six guppies at the beginning of the month. One month later, the number of guppies in his tank had doubled. His guppy population continued to grow in this same manner. His sister bought some tetras at the same time. The table below shows the number of tetras, t, after n months have passed since they bought the fish.
a. Create a function g to model the growth of the boy’s guppy population, where g(n) is the number of guppies at the beginning of each month and n is the number of months that have passed since he bought the six guppies. What is a reasonable domain for g in this situation?
g(n) = 6•2n Domain: n is a whole number.
b. How many guppies will there be one year after he bought the six guppies?
g(12) = 6•212 = 24576 guppies
c. Create an equation to determine how many months it will take to reach 100 guppies.
100 = 6•2n
d. Use graphs or tables to approximate a solution to the equation from part (c). Explain how you arrived at your estimate.
At the end of 4 months, there are 96 guppies which is not quite 100. So during the 5th month, the guppy population reaches 100.
e. Create a function, t, to model the growth of the sister’s tetra population, where t(n) is the number of tetras after n months have passed since she bought the tetras.
t(n) = 8(n + 1), n is a whole number.
Or, t(n) = 8n + 8, n is a whole number.
f. Compare the growth of the sister’s tetra population to the growth of the guppy population. Include a comparison of the average rate of change for the functions that model each population’s growth over time.
The guppies’ population is increasing faster than the tetras’ population. Each month, the number of guppies doubles, while the number of tetras increases by 8. The rate of change for the tetras is constant, but the rate of change for the guppies is always
g. Use graphs to estimate the number of months that will have passed when the population of guppies and tetras will be the same.
g(2) = t(2) = 24
The guppier and tetras populations will be the same, 24, after 2 months.
h. Use graphs or tables to explain why the guppy population will eventually exceed the tetra population even though there were more tetras to start with.
The guppy population’s growth is exponential, and the tetra population’s growth is linear. The graph in part (g) shows how the population of the guppies eventually overtakes the population of the tetras. The table below shows that by the end of the third month, there are more guppies than tetras.
i. Write the function g(n) in such a way that the percent increase in the number of guppies per month can be identified. Circle or underline the expression representing the percent increase in the number of guppies per month.
g(n) = 6(200%)n
Question 4.
Regard the solid dark equilateral triangle as Figure 0. Then, the first figure in this sequence is the one composed of three dark triangles, the second figure is the one composed of nine dark triangles, and so on.
a. How many dark triangles are in each figure? Make a table to show this data.
b. Given the number of dark triangles in a figure, describe in words how to determine the number of dark triangles in the next figure.
The number of triangles in a figure is 3 times the number of triangles in the previous figure.
c. Create a function that models this sequence. What is the domain of this function?
T(n) = 3n, D: n is a whole number.
d. Suppose the area of the solid dark triangle in Figure 0 is 1 square meter. The areas of one dark triangle in each figure form a sequence. Create an explicit formula that gives the area of just one of the dark triangles in the nth figure in the sequence.
A(n) = ($$\frac{1}{4}$$)n
e. The sum of the areas of all the dark triangles in Figure 0 is 1 m2; there is only one triangle in this case. The sum of the areas of all the dark triangles in Figure 1 is $$\frac{3}{4}$$ m2. What is the sum of the areas of all the dark triangles in the nth figure in the sequence? Is this total area increasing or decreasing as n increases?
I(n) = ($$\frac{3}{4}$$)n
The total area is decreasing as n increases.
f. Let P(n) be the sum of the perimeters of all the dark triangles in the nth figure in the sequence of figures. There is a real number k so that
P(n + 1) = kP(n)
is true for each positive whole number n. What is the value of k?
Let x represent the number of meters long of one side of the triangle in Figure 0.
P is a geometric sequence, and k is the ratio between any term and the previous term, so k= P(n+1)/P(n).
So, for example, for n = 0, k = $$\frac{(P(1))}{P(O)}=\frac{\frac{9}{2} x}{3 x}$$ = $$\frac{3}{2}$$ .
For n = 1, k = $$\frac{P(2)}{P(1)}$$ = $$\frac{\frac{27}{4} x}{\frac{9}{2} x}$$ = $$\frac{3}{2}$$ .
K = $$\frac{3}{2}$$
Question 5.
The graph of a piecewise function f is shown to the right. The domain of f is – 3≤x≤3.
a. Create an algebraic representation for f. Assume that the graph of f is composed of straight line segments.
b. Sketch the graph of y = 2f(x), and state the domain and range.
Domain: -3 ≤ x ≤ 3
Range: -6 ≤ y ≤ 2
c. Sketch the graph of y = f(2x), and state the domain and range.
Domain: -1.5 ≤ x ≤ 1.5
Range: -3 ≤ y ≤ 1
d. How does the range of y = f(x) compare to the range of y = kf(x), where k > 1?
y = f(x) by $$\frac{1}{k}$$, giving –$$\frac{3}{k}$$ ≤ x ≤ $$\frac{3}{k}$$.
|
Connect with StatsExamples here
TRANSCRIPT OF VIDEO:
Slide 1
The normal distribution is the most important probability distribution in statistics. Let's take a look at where it comes from, what it's used for, and why it's so important
Slide 2
the normal distribution is the most common probability distribution in statistics. It is based on the binomial distribution which is the most fundamental probability distribution.
In fact, we can think of the normal distribution as the limit of the binomial distribution as the number of trials becomes large and the probability of success of each trial is 0.5.
If you don't remember much about the binomial distribution, this channel an playlist has a video all about the binomial distribution that you can take a look at.
The figure here illustrates the binomial distribution for 10 trials and the probability of success of 0.5.
You can see that as we increase the number of trials the shape of the binomial distribution changes slightly.
And as we increase the number of trials even more, the shape of the distribution changes even more.
And here we can see how the shape of the binomial distribution with lots of trials on the left can be represented by a smooth curve as shown on the right. Even though these two figures look different, one is a histogram and the other is a curved line, they are representing the same thing - a distribution of probability values corresponding to values on the X axis.
Slide 3
As we begin to think about this normal probability distribution, it's important to remember exactly what is being represented.
On the left, the histogram is showing a particular probability corresponding to each number of observations on the X axis.
If we wanted to know the probability of some particular range or group of observations, seeing somewhere between 25 and 35 observations for example, we would add up those individual probabilities to create a sum of those probabilities.
We would essentially be calculating the sum of the Heights of a set of those individual bars.
On the right, the curve is showing the probability corresponding to each location along the X axis.
If we wanted to know the probability of some particular range on the X-axis, the proportion of the total probability corresponding to between 25 and 35 for example, we would calculate the area underneath that curve.
Note that since this is a probability distribution, we know that the total area underneath the curve is equal to 1. The value we get for the area of a region therefore represents a proportion of the overall set of probabilities
Slide 4
Looking at our histogram situation in more detail we would calculate the overall probability as the sum of all those individual probabilities for the numbers of observations we're interested in.
Slide 5
Looking at our probability curve situation in more detail we would calculate the overall probability as the integral, the area underneath the curve, along the range of values on the X axis we're interested in
Slide 6
To perform these integrals or calculate these areas we would need the equation for that curve that represents those probabilities.
The top equation to the right is the general equation for a normal distribution curve. F of X equals one over Sigma times the square root of 2π multiplied by E to the exponent negative 1/2 times , in parentheses, X minus mu divided by Sigma, end parentheses squared.
The bottom equation to the right is the equation we would get for a normal distribution corresponding to the binomial with 50 trials and a probability of success of 0.5. The mean would be 25, the variance would be 12.5, and the standard deviation would be 3.54.
that gives us F of X equals one over 3.5 four times the square root of 2π multiply by E to the exponent negative 1/2 times , in parentheses, X minus 25 divided by 3.5 four , end parentheses squared.
This equation causes the normal distribution to be centered around the mean with a width that depends on the standard deviation. And because the overall area must be equal to 1, if the standard deviation is larger, then the height of the normal distribution is lower.
Slide 7
The equation for the normal distribution can be simplified if the mean is equal to 0 and the variance and standard deviations are both equal to 1. Under these conditions we get a normal distribution called the standard normal distribution.
The standard normal distribution has the same shape as the normal distribution but you can see that the equation is much simpler, F of X equals one over the square root of 2π times E to the negative 1/2 X squared.
The standard normal distribution is centered around 0 and you can see that by the time it gets to negative two and positive 2 the individual probability values are quite small.
But keep in mind, we generally care much less about the individual probability values and much more about the area underneath the curve for certain regions along the X axis.
It's very unusual that you would ever want to calculate an individual probability value for a particular value on the X axis.
Slide 8
The standard normal distribution is useful because any set of values with a normal distribution of mean equals mu and standard deviation equals Sigma can be transformed into a new set of values that have the standard normal distribution by subtracting mu from every value and then dividing that result by Sigma.
When we do this we transform those values X in the original normally distributed data set into new values called Z values in the standard normal distribution.
Doing this transformation allows us to calculate probabilities for values in regions defined by a mean and standard deviation.
For example, if we wanted to know what proportion of the values in our original data set are in the center region bounded by 1 standard deviation above and below the mean, that corresponds to the center region of the standard normal distribution from Z equals negative one to positive one.
Instead of having to think about doing integrals and calculating areas for all the different possible normal distributions, we can transform all of those distributions of data into the standard normal distribution and calculate our areas with this to answer the questions we're interested in.
Slide 9
For example, for a data set with a mean of 23 and standard deviation of five we might be interested in how many values are between 22 and 27.
The values of 22 and 27 from the original data set can be transformed into the Z values of -0.2 and +0.6 using the equation Z equals X minus mean divided by Sigma .
When we do this, our original question is equivalent to asking what proportion of values in the standard normal distribution are between the values of Z = -0.2 and Z = 0.6.
Slide 10<
So to recap.
We have an equation for a normal distribution with mean of mu and standard deviation of Sigma and we can calculate probabilities by calculating integrals for the area under the curve between two values
If we transform these values into z values, then that procedure is equivalent to finding the area under the standard normal distribution curve between the z values corresponding to the original values.
Slide 11
So how do we calculate the area under the standard normal distribution curve? Unfortunately, it turns out that this integral cannot be solved exactly. not that it's hard, but it's impossible to get an exact equation.
This is actually one of those secrets they don't tell you about in early calculus courses, but many of the most interesting equations don't integrate easily, polynomials and trig functions do, but this equation for the normal distribution does not.
However, there are more advanced mathematical techniques that can be used to create extremely accurate numerical approximations. Instead of us learning those techniques, what we will do is borrow tables of the results from those numerical approximations.
These tables of areas are called normal distribution tables or Z tables. the most common format for these tables is to indicate the area under the curve to the left of a particular Z value. We can then calculate areas for ranges by looking up values in the tables and subtracting.
Of course if you're doing this using a computer, it has all these tables in a database somewhere and it's referencing those values without showing you.
Slide 12
As mentioned, Z tables are tabulated areas that are usually for the area to the left of a given Z value.
That's not a guarantee however, sometimes this same information is presented in a different way. For this reason it's important to understand why this procedure works and makes sense, so that if you have to use tables with different formats you can still obtain the areas you want.
We can figure out the area between A & B in our original distribution by using this table to figure out the area between the Z value corresponding to A and the Z value corresponding to B.
If the table tells us the area to the left of the Z value corresponding to A is 0.3, and the area to the left of the Z by value corresponding to B is 0.9 then the area between those Z values must be 0.6.
Slide 13
Let's look at a more specific example, what is the area between the Z scores of negative one and positive one?
We would go to our table of Z values and locate the area value in the table that corresponds to the Z score of 1.00.
In this table, which is the one provided on the StatsExamples website, the first 2 digits of the Z scores are in the column at the far left and the columns correspond to the 2nd decimal place as shown in the row at the top. The values in the table are the areas to the left of the Z score indicated.
When we look in the table and find the location for 1.00 the area listed there is 0.8413. when we look in the table and find the location for negative 1.00 the area listed there is 0.1587. subtracting 0.1587 from 0.8413 gives us 0.6826 as the area under the standard normal distribution curve between the Z value of negative one and one.
This is equivalent to the proportion of a normally distributed data set, whatever its mean and standard deviation are, having 68.26% of its values in a region 1 standard deviation above and below the mean.
This calculation we just did leads to a rule of thumb that about 2/3 or 66% of the data in a normal distribution is within one standard deviation of the mean.
Slide 14
Let's look at another example, what is the area between the Z scores of negative 2 and positive 2?
Again we would go to the table and find the areas in it that correspond to AZ score of 2.00 and AZ score of negative 2.00. Those values are 0.9772 and 0.0228. 0.9772 - 0.0228 equals 0.9544.
This calculation leads to the rule of thumb that 95% of the data in a normal distribution is within 2 standard deviations of the mean.
Slide 15
It's actually quite useful to remember these three rules of thumb for regions within the normal distribution.
About 66% of the data will be within one standard deviation of the mean.
About 95% of the data will be within 2 standard deviations of the mean.
And over 99% of the data will be within 3 standard deviations of the mean.
I've been talking about the normal distribution representing data values , which is often the case, but they also represent probabilities.
These same calculations apply to normally distributed probabilities.
For example, if we randomly choose a value from a normal distribution there is an approximately 66% probability it's within one standard deviation of the mean. If we randomly choose a value from a normal distribution there is an approximately 95% probability it's within 2 standard deviations of the mean. And if we randomly choose a value from a normal distribution there is over a 99% probability it's within 3 standard deviations of the mean.
Slide 16
OK, So what is the normal distribution used for? there are two main uses.
First, many populations exhibit a normal distribution. A normal distribution of values is a natural outcome of summing many independent factors - for example multiple genetic effects that contribute to a trait or sequential actions that lead to an outcome will tend to generate normally distributed traits or outcomes.
Therefore, the normal distribution is useful for estimating frequencies and proportions in many populations from just knowing the mean and variance of the population or estimating the population mean and variance from the sample mean and sample variance.
If we wanted to know what range of heights 95% of people are included within, instead of trying to measure everybody and create a detailed distribution we could just calculate the mean and variance and use the normal distribution.
The second reason to study normal distributions is because of something called the central limit theorem. This states that the means of samples from a population form a normal distribution, no matter what the population distribution looks like.
Therefore, the normal distribution is useful for estimating the means of populations. It allows us to calculate confidence intervals, which are regions that describe where we think the population mean probably is, based on our sample mean and variance.
Estimating the mean value of a population is the most common thing that people want to do when they do statistics. The central limit theorem allows us to use the properties of the normal distribution to do that. Details about estimating the mean of populations using confidence intervals are in this channel's video about confidence intervals.
Zoom out
The normal distribution is the most important probability distribution in all of statistics. Many statistical tests actually require that the values be normally distributed in order to work properly.
The normal distribution is the foundation of calculating confidence intervals to estimate the mean of a population and is the distribution that is the basis for performing t-tests that compare the means of different groups to one another.
End screen
The linked video shows some examples of calculations with the normal distribution and the playlist includes videos about confidence intervals and t-tests.
Like or subscribe to make sure you can find this channel again in the future.
Connect with StatsExamples here
This information is intended for the greater good; please use statistics responsibly.
|
# The Easiest solution isn’t always the best solution, even in Math Should we always believe what we are taught in the classroom?
## Presentation on theme: "The Easiest solution isn’t always the best solution, even in Math Should we always believe what we are taught in the classroom?"— Presentation transcript:
The Easiest solution isn’t always the best solution, even in Math Should we always believe what we are taught in the classroom?
Purpose Statisticians use a well selected sample to estimate an unknown value of a population. The unknown value may be the mean income, or the proportion of defective products, or proportion of “yes” responses. Estimating an unknown population proportion is the topic of interest.
Background/symbols p= Population proportion (unknown) n= # of subjects/ objects randomly selected. X= # of subjects/ objects in the sample with Yes responses. p^ = sample proportion= x/n Traditionally p^ is used as an estimate of p. Is there a better alternative?
We often provide an interval estimate of p, p^ ± error of estimation: Confidence interval A well known interval is 95% confidence. To determine error, we need to understand how p^ value varies from sample to sample.
About p^ p = fixed value of a population, while p^= varies from sample to sample, and thus it has a distribution. What we know is Under certain conditions, p^ is normally distributed with a mean value of p, and standard deviation of √p(1-p)/n A normally distributed value can be changed to a standard normal score,called a z score. A well known result is that about 95% of the z scores fall between -2 and 2.
Lets standardize p^ = # of yes /n, ( p^- mean)/ std dev = z ( standard normal) ( p^- p)/ √p(1-p)/n = z ( p^- p)/ √p(1-p)/n = ± 2 ( p is unknown). Note: We need to solve the above equation for p Easy approach for non math majors : solve for p in numerator p= p^- 2 √p(1-p)/n, p= p^+ 2 √p(1-p)/n, (p^- 2 √p^(1-p^)/n, p^+ 2 √ p^(1-p^) / n) Makes an approximate 95% confidence interval of p. ( Mathematically incorrect)
Lets try again ( p^- p)/ √p(1-p)/n = ± 2 Solve it the right way by squaring both sides, and solving the quadratic equation for p. We get two solutions of p, ( mathematically tedious) Those solutions make the 95% interval of p.This interval is very tedious, and lacks logical explanation. we take the average of those solutions, we get p =( # of yes +2)/ (n+4)= our new estimate =p~
A new interval of p Recall old interval of p: (p^- 2 √p ^ ( 1-p^ )/n, p^+ 2 √ p ^ ( 1-p^ )/ n) An alternative interval of p (p~ -2 √ p~(1- p~)/n, p~+ 2 √ p~(1- p~)/n Recall p~ = ( # of yes +2)/ n+4 p^ = (# of yes)/n
How good is the new interval Simulation results coming soon.
Similar presentations
|
# Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.1
## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.1
Question 1.
Fill in the blanks.
(i) Ratio of Rs 3 to Rs 5 = …….
(ii) Ratio of 3m to 200 cm = ……
(iii) Ratio of 5 km 400 m to 6 km = …….
(iv) Ratio of 75 paise to Rs 2 = ……..
Solution:
(i) 3 : 5
(ii) 3 : 2
(iii) 9 : 10
(iv) 3 : 8
Question 2.
Say whether the following statements are True or False.
(i) The ratio of 130 cm to 1 m is 13 : 10.
(ii) One of the terms in a ratio cannot be 1.
Solution:
(i) True
(ii) False
Question 3.
Find the simplified form of the following ratios.
(i) 15 : 20
(ii) 32 : 24
(iii) 7 : 15
(iv) 12 : 27
(v) 75 : 100
Solution:
Question 4.
Akilan walks 10 km in an hour while Selvi walks 6 km in an hour. Find the simplest ratio of the distance covered by Akilan to that of Selvi.
Solution:
Distance covered by Akilan in 1 hour = 10 km
Distance covered by Selvi in 1 hour = 6 km
$$\frac{\text { Distance covered by Akilan }}{\text { Distance covered by Selvi }}=\frac{10 \mathrm{km}}{6 \mathrm{km}}=\frac{10 \div 2}{6 \div 2}=\frac{5}{3}$$
Distance covered by Akilan : Selvi = 5 : 3.
Question 5.
The cost of parking a bicycle is ₹ 5 and the cost of parking a scooter is ₹ 15. Find the simplest ratio of the parking cost of a bicycle to that of a scooter.
Solution:
Parking cost of bicycle = ₹ 5
Parking cost of Scooter = ₹ 15
$$\frac{\text { Parking cost of bicyle }}{\text { Parking cost of scooter }}=\frac{5}{15}=\frac{1}{3}$$
Parking cost of bicycle : scooter = 1 : 3
Question 6.
Out of 50 students in class 30 are boys. Find the ratio of
(i) number of boys to the number of girls
(ii) number of girls to the total number of students
(iii) number of boys to the total number of students
Solution:
The total number of students = 50.
The number of boys = 30.
Then number of girls = 50 – 30 = 20.
Objective Type Questions
Question 7.
The ratio of Rs 1 to 20 paise is
(a) 1 : 5
(b) 1 : 2
(c) 2 : 1
(d) 5 : 1
Solution:
(d) 5 : 1
Question 8.
The ratio of 1 m to 50 cm is _____
(a) 1 : 50
(b) 50 : 1
(c) 2 : 1
(d) 1 : 2
Solution:
(c) 2 : 1
Hint: 1 m = 100 cm
Question 9.
The length and breadth of a window are 1 m and 70 cm respectively. The ratio of the length to the breadth is _____
(a) 1 : 7
(b) 7 : 1
(c) 7 : 10
(d) 10 : 7
Solution:
(d) 10 : 7
Hint: 1 m = 100 cm
Question 10.
The ratio of the number of sides of a triangle to the number of sides of a rectangle is ____
(a) 4 : 3
(b) 3 : 4
(c) 3 : 5
(d) 3 : 2
Solution:
(b) 3 : 4
Triangle has three sides and a rectangle has four sides.
Question 11.
If Azhagan is 50 years old and his son is 10 years old than the simplest ratio between the age of Azhagan to his son is ____
(a) 10 : 50
(b) 50 : 10
(c) 5 : 1
(d) 1 : 5
Solution:
(c) 5 : 1
|
Probability
Probability is often tested in the Quantitative Aptitude section of various exams like- SBI PO, IBPS PO & SO, SEBI Grade A, SSC CGL and many others. It is a scoring yet tricky chapter.
Therefore, it’s essential that you are familiar with the short tricks and tips to solve these problems quickly and accurately.
Definition of Probability
Probability deals with the uncertainty of the occurrence of an event using numbers. The chance that a particular event will or will not occur is expressed on a scale ranging from 0-1. Another way to represent probability is by way of percentage. For instance, 0% denotes an impossible event and 100% denotes a certain event.
Types of Probability
• Experimental/Empirical Probability
It is based on actual experiments and adequate recordings of the happening of events. A series of experiments are conducted to determine the occurrence of any event. Experiments with no results are called random experiments, where the outcome of such experiments is uncertain. To determine their likelihood, random experiments are repeated multiple times. The experiment is repeated a fixed number of times and each repetition is known as a trial.
The formula of experimental probability is-
P(E)= Number of times an event occurs/Total number of trials.
Let’s understand with an example-
You ask 3 friends A, B and C to toss a fair coin 15 times each in a row and the outcome of this experiment is given below-
Coins Tossed byNumber of HeadsNumber of Tails
A
6
9
B
7
8
C
8
7
Calculate the probability of occurrence of heads and tails.
The experiential probability for the occurrence of of heads and tails can be calculated as:
• Experimental Probability of Occurrence of heads = Number of times head occurs/Number of times coin is tossed.
• Experimental Probability of Occurrence of tails = Number of times tails occurs/Number of times coin is tossed.
Coins Tossed byNumber of HeadsNumber of TailsExperimental Probability of Occurrence of heads Experimental Probability of Occurrence of tails
A
6
9
6/15 = 0.4
9/15 = 0.6
B
7
8
7/15 = 0.47
8/15 = 0.53
C
8
7
8/15 = 0.53
7/15 = 0.47
It can be observed that if the number of tosses of the coin increases, the probability of occurrence of heads or tails increases to 0.5.
Geometric Probability
Geometric probability is the calculation of the likelihood that one will hit a particular area of a figure. It is calculated by dividing the desired area by the total area. In the case of Geometrical probability, there are infinite outcomes.
Let’s understand the basics of Probability in detail.
Outcome and Event
Outcome is the result of a random experiment.
For instance, when we roll a dice, getting four is an outcome.
On the other hand, Event is a set of outcomes of an experiment. It is denoted by E.
An event in probability is the subset of the corresponding sample space.
Sample space is the entire possible set of outcomes of a random experiment. It is denoted by S.
For instance, when we roll a dice, the probability of getting a number less than 5 is an event.
Please note- An Event can have a single outcome
Probability of Occurrence of an Event
Probability of occurrence of any event is the number of favourable outcomes divided by the total number of outcomes.
The formula is as follows:
P(E) = Number of Favourable Outcomes/ Total Number of Outcomes
Let’s understand with an example-
Q) Find the probability of rolling a 5 on a fair dice
To find the probability of getting 5 while rolling a dice, an experiment is not needed. We know that there are 6 possible outcomes when rolling a dice. They are 1, 2, 3, 4, 5, 6.
Therefore, the probability is,
Probability of Event P(E) = No. of. Favourable outcomes/ No. of. Possible outcomes.
P(E) = 1/6.
Hence, the probability of getting 5 while rolling a fair dice is 1/6.
Types of Events in Probability
Following are the important probability events-
• Impossible and Sure Events
Where the probability of occurrence of an event is 0, such an event is called an impossible event.
Let us understand with an example of impossible event-
Q) A bag contains only orange flavoured candies. Radha takes out one candy without looking into the bag. What is the probability that she takes out a lemon flavoured candy?
Let us take the number of candies in the bag to be 100
Number of lemon flavoured candies= 0 (since the bag contains only orange flavoured candies)
Thus, the probability that she takes out lemon flavored candies= P= Number of lemon flavoured candies/ Total number of candies= 0/100= 0
Therefore, the probability that Radha takes out a lemon flavoured candy is 0 which proves that the probability of an impossible event is 0.
If the probability of occurrence of an event is1, it will be a sure event.
Let’s look at an example of a sure event-
Q) What is the probability that a number obtained after throwing a dice is less than 7?
P(E)= P (getting a number less than 7)= 6/6= 1
• Simple/Elementary Events
Any event having only one outcome of the experiment is called a simple event.
For example- If we toss a coin ‘n’ number of times, we will get only two possible outcomes: Heads or Tails. Thus, for an individual toss, it has only one outcome i.e. Heads or Tails.
The sum of the probabilities of all simple events of an experiment is one.
For example- Let us take the coin toss example again . P(Heads) + P(Tails)= ½ + ½ = 2/2 = 1.
• Compound Events
On the contrary, if any event consists of more than one single point of the sample space, it is called a compound event.
For example-
If S= {56, 78, 96, 54, 89}
E1= {56, 54}
E2= {78,56, 89}
Then, E1 and E2 represent compound events.
• Independent and Dependent Events
If the occurrence of any event is completely unaffected by the occurrence of any other event, such events are known as independent events. The events which are affected by other events are called dependent events.
• Mutually Exclusive Events
Here, the occurrence of one event excludes the occurrence of another event. Such events are mutually exclusive events. In other words, two events do not have a common point.
For example-
S= {1, 2, 3, 4, 5, 6} and E1 and E2 are two events such that E1 consists of numbers less than 3 and E2 consists of numbers greater than 4.
Then, E1= {1,2} and E2= {5,6}
Which means, E1 and E2 are mutually exclusive.
• Exhaustive Events
If all the events consume the entire sample space, the set of events is called exhaustive.
• Complementary Events
For any event E1 there exists another event E1’ which represents the remaining elements of the sample space S.
E1= S - E1
For example-
If a dice is rolled, then the sample space S is given as S= {1, 2, 3, 4, 5, 6}.
If event E1 represents all the outcomes which is greater than 4, then E1= {5,6} and E1’ = {1, 2, 3, 4}
Thus, E1’ is the complement of the event E1.
Likewise, the complement of E1, E2, E3……….En will be represented as E1’, E2’, E3’...........En’
• Events Associated with “Or”
If two events E1 and E2 are associated with Or means that it can be either E1 or E2 or both. The union symbol (U) is used to represent Or in probability.
Therefore, the event E1 U E2 denotes E1 Or E2.
If there are mutually exhaustive events E1, E2, E3……….En associated with sample space S then,
E1 U E2 U E3 U……….En= S
• Events Associated with “And”
If two events E1 and E2 are connected with And, this means that there is an intersection of elements which is common to both the events. The intersection symbol (n) is used to denote And in probability.
Thus, the events E1 n E2 denotes E1 and E2
• Event E1 but not E2
It denotes the difference between both the events. E1 but not E2 represents all the outcomes which are present in E1 but not in E2. So, the event E1 but not E2 is denoted as E1, E2= E1 - E2
Practice Questions
Q1) It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992 . What is the probability that the
2 students have the same birthday?
It is provided to us that, probability of 2 students not having the same birthday is 0.992 .
So,
P(2 students having the same birthday)+P(2 students not having the same birthday)=1
= P(2 students having the same birthday)+0.992=1
Simplifying further,
= P(2 students having the same birthday)=0.008.
Q2) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i)red and (ii) not red?
The bag has 3 red balls and 5 black balls
(i) So, the probability of getting a red ball= number of red balls/total number of balls
P(E)= 3/ 3+5= ⅜
(ii) the probability of getting a red ball= 1- number of red balls/total number of balls
So, P(E)= 1- ⅜= ⅝
Q3) A dice is thrown once. Find the probability of getting (i) a prime number;
There are 6 results that can be obtained from a dice.
There are 3 prime numbers, 2,3,5 among those results.
Thus, the probability of getting a prime number= no of prime numbers in a dice/total numbers on dice= 3/6= ½
(ii) a number lying between 2 and 6?
There are 3 numbers between 2 and 6, 3,4,5.
Thus, the probability of getting a number between 2 and 6= no. of numbers between 2 and 6 in a dice/total numbers on a dice= 3/6= ½ no
Q4) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a king of red color?
We know there are 52 cards in the deck.
There are 2 kings of red colour in the deck.
Thus, P(E)= total number of red kings/total number of cards= 2/52= 1/26
Q5) 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
|
# Connect Numbers Activity Worksheets For Kindergarten
A Reasonable Phone numbers Worksheet might help your son or daughter become a little more knowledgeable about the concepts behind this ratio of integers. Within this worksheet, pupils are able to resolve 12 distinct issues relevant to reasonable expression. They are going to learn to flourish a couple of phone numbers, class them in sets, and figure out their items. They will also practice simplifying rational expression. As soon as they have mastered these concepts, this worksheet will certainly be a important resource for advancing their scientific studies. Connect Numbers Activity Worksheets For Kindergarten.
## Logical Phone numbers are a percentage of integers
There are 2 kinds of figures: irrational and rational. Reasonable figures are defined as total phone numbers, in contrast to irrational numbers tend not to repeat, and also have an endless amount of digits. Irrational numbers are no-zero, non-terminating decimals, and sq . origins that are not excellent squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To define a logical amount, you must understand such a rational quantity is. An integer is a complete quantity, as well as a logical number is really a percentage of two integers. The percentage of two integers will be the amount at the top split by the number on the bottom. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be manufactured into a small fraction
A logical quantity includes a denominator and numerator that are not absolutely nothing. Which means that they can be indicated like a fraction. Together with their integer numerators and denominators, realistic numbers can furthermore have a adverse worth. The unfavorable worth should be located left of and its particular absolute benefit is its range from zero. To simplify this illustration, we will state that .0333333 can be a portion that can be composed as a 1/3.
As well as adverse integers, a rational amount can even be manufactured right into a small fraction. For instance, /18,572 is really a logical quantity, when -1/ is just not. Any fraction consisting of integers is reasonable, as long as the denominator fails to consist of a and can be created being an integer. Furthermore, a decimal that leads to a point can be another rational number.
## They are perception
In spite of their name, logical figures don’t make a lot sensation. In math, they can be one organizations with a special length around the quantity line. Which means that if we count something, we are able to order the dimensions by its percentage to the unique volume. This holds correct even when you will find unlimited logical numbers among two certain figures. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To get the length of a pearl, as an example, we could add up its width. One particular pearl weighs in at ten kilograms, that is a reasonable number. Additionally, a pound’s bodyweight is equal to twenty kilograms. Thus, we should be able to break down a lb by 15, without the need of be worried about the duration of a single pearl.
## They could be expressed like a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal quantity might be created as a a number of of two integers, so four times 5 various is equal to 8. An identical dilemma requires the repeated small percentage 2/1, and either side ought to be divided by 99 to find the proper answer. But how do you have the transformation? Here are several cases.
A rational number can be designed in great shape, including fractions and a decimal. A great way to represent a reasonable number inside a decimal is to split it into its fractional equal. You will find 3 ways to separate a reasonable amount, and every one of these techniques brings its decimal equal. One of these simple approaches is usually to separate it into its fractional equal, and that’s what’s known as the terminating decimal.
|
# AP Statistics Curriculum 2007 Normal Critical
## General Advance-Placement (AP) Statistics Curriculum - Nonstandard Normal Distribution & Experiments: Finding Critical Values
### Nonstandard Normal Distribution & Experiments: Finding Scores (Critical Values)
In addition to being able to compute probability (p) values, we often need to estimate the critical values of the Normal distribution for a given p-value.
• The back and forth linear transformations converting between Standard and General Normal distributions are always useful in such analyses (Let X denotes General ($X\sim N(\mu,\sigma^2)$) and Z denotes Standard ($X\sim N(0,1)$) Normal random variables):
$Z = {X-\mu \over \sigma}$ converts general normal scores to standard (Z) values.
X = μ + Zσ converts standard scores to general normal values.
### Examples
#### Textbook prices
Suppose the amount of money college students spend each semester on textbooks is normally distributed with a mean of $195 and a standard deviation of$20. If we ask a random college students from this population how much he spent on books this semester, what is the maximum dollar amount that would guarantee she spends only as much as 30% of the population? (P(X < 184.512) = 0.3)
You can also do this problem exactly using the SOCR high-precision Nornal Distribution Calculator. If zo = − 0.5243987892920383, then P( − zo < Z < zo) = 0.4 and P(Z<z_o)=0.3. Thus, xo = μ + zoσ = 195 + ( − 0.5243987892920383) * 20 = 184.512024214159234.
#### Human Weights and Heights
Human weights and heights are known to be approximately Normally distributed. Look at the SOCR Weighs and Heights Dataset and use the SOCR Charts to validate these statements based on this sample dataset.
• Suppose the heights of college women are approximately Normally distributed with a mean of 65 inches and a standard deviation of 2 inches. If a randomly chosen college woman is at the 10th percentile (shortest 10% for women) in height for college women, then what is the largest height closest to hers?
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Voltage
The amount of potential energy between two points on a circuit.
Estimated8 minsto complete
%
Progress
Practice Voltage
Progress
Estimated8 minsto complete
%
Voltage
Like gravity, the electric force can do work and has a potential energy associated with it. But like we use fields to keep track of electromagnetic forces, we use electric potential, or voltage to keep track of electric potential energy. So instead of looking for the potential energy of specific objects, we define it in terms of properties of the space where the objects are.
The electric potential difference, or voltage difference (often just called voltage) between two points (A and B) in the presence of an electric field is defined as the work it would take to move a positive test charge of magnitude 1 from the first point to the second against the electric force provided by the field. For any other charge q\begin{align*} q \end{align*}, then, the relationship between potential difference and work will be:
ΔVAB=WABq [4] Electric Potential\begin{align*}\Delta V_{AB} &= \frac{W_{AB}}{q} \ \ \ \ \text{[4] Electric Potential}\end{align*}
Rearranging, we obtain:
WWork=ΔVABPotential Difference×qCharge\begin{align*}\underbrace{W}_{\text{Work}} &= \underbrace{\Delta V_{AB}}_{\text{Potential Difference}} \times \underbrace{q}_{\text{Charge}}\end{align*}
The potential of electric forces to do work corresponds to electric potential energy:
\begin{align*}\Delta U_{E,AB} &= q\Delta V_{AB} \ \ \text{[5] Potential energy change due to voltage change}\end{align*}
The energy that the object gains or loses when traveling through a potential difference is supplied (or absorbed) by the electric field --- there is nothing else there. Therefore, it follows that electric fields contain energy.
Key Equations
\begin{align*}E = \frac{- \Delta V}{\Delta x} \text{ }\end{align*} Electric field vs electric potential.
\begin{align*}\Delta U_E = q \Delta V \text{ }\end{align*} Change in potential energy due to travel through changing voltage.
\begin{align*}V = \frac{k q}{r} \text{ }\end{align*} Electric potential of a single charge.
To summarize: just as an electric field denotes force per unit charge, so electric potential differences represent potential energy differences per unit charge. Voltage is by definition the electric potential energy per Coulomb. So it is the electrical potential energy value divided by the charge. Thus, voltage difference is the potential value for potential energy. A 12V battery can not produce energy without charge flowing (i.e. you must connect the two ends). Electric potential is measured in units of Volts \begin{align*}(V)\end{align*} – thus electric potential is often referred to as “voltage.” Electric potential is the source of the electric potential energy. You can read the electric potential lines (that is the voltage lines) just like you would a contour map while backpacking in the mountains. Positive charges move towards lower electric potential; negative charges move toward higher electric potential. Thus, positive charges go 'downhill' and negative charges go 'uphill'.
Example
You have a negative charge of unknown value and a positive charge of magnitude \begin{align*}q_1\end{align*} and mass \begin{align*}m\end{align*}. After fixing the negative charge in place, you place the positive charge a distance \begin{align*}r_i\end{align*} away from the negative charge and then release it. If the speed of the positive charge when it is a distance \begin{align*}r_f\end{align*} away from the negative charge is \begin{align*}v\end{align*}, what was the magnitude of the negative charge in terms of the given values?
There are multiple ways to do this problem, we will solve it using conservation of energy and the change in voltage to determine the magnitude of the negative charge. When working through this problem, we'll call the positive charge \begin{align*}q_1\end{align*} and the negative charge \begin{align*}q_2\end{align*}. To start we'll say that the charge had zero potential energy when it was .5m from the negative charge; this will help us as we work through the problem. Using this assertion, we will apply conservation of energy to the positive charge.
\begin{align*} \Delta U_e&=KE_f && \text{start with conservation of energy}\\ q_1\Delta V&=\frac{1}{2}mv^2 && \text{substitute the equations for each energy term}\\ \Delta V&=\frac{mv^2}{2q_1} && \text{solve for V}\\ \end{align*}
Now, since we know the voltage difference, we will express it using the equation for voltage at a certain distance from a point charge.
\begin{align*} \Delta V&=\frac{kq_2}{\Delta r} && \text{start with the equation for voltage at a certain distance}\\ \Delta V&=\frac{kq_2}{r_f}-\frac{kq_2}{r_i} && \text{express the change in radius in terms of the initial and final radius of the positive charge}\\ \Delta V&=kq_2(\frac{1}{r_f}-\frac{1}{r_i}) && \text{factor the equation}\\ \frac{mv^2}{2q_1}&=kq_2(\frac{1}{r_f}-\frac{1}{r_i}) && \text{substitute in the value from the first step}\\ q_2&=\frac{mv^2}{2q_1k(\frac{1}{r_f}-\frac{1}{r_i})} && \text{solve for }q_2\\ \end{align*}
Simulation
Charges and Fields (PhET Simulation)
Review
1. The diagram to the right shows a negatively charged electron. Order the electric potential lines from greatest to least.
1. \begin{align*}A, B, C\end{align*}
2. \begin{align*}C, B, A\end{align*}
3. \begin{align*}B, A, C\end{align*}
4. \begin{align*}B, C, A\end{align*}
5. \begin{align*}A = B = C \ldots\end{align*} they’re all at the same electric potential
2. Below are the electric potential lines for a certain arrangement of charges. Draw the direction of the electric field for all the black dots.
3. A metal sphere with a net charge of \begin{align*}+5\ \mu \mathrm{C}\end{align*} and a mass of \begin{align*}400 \;\mathrm{g}\end{align*} is placed at the origin and held fixed there.
1. Find the electric potential at the coordinate \begin{align*}(6 \;\mathrm{m}, 0)\end{align*}.
2. If another metal sphere of \begin{align*}-3 \ \mu \mathrm{C}\end{align*} charge and mass of \begin{align*}20 \;\mathrm{g}\end{align*} is placed at the coordinate \begin{align*}(6 \;\mathrm{m}, 0)\end{align*} and left free to move, what will its speed be just before it collides with the metal sphere at the origin?
4. Collisions of electrons with the surface of your television set give rise to the images you see. How are the electrons accelerated to high speed? Consider the following: two metal plates (The right hand one has small holes allow electrons to pass through to the surface of the screen.), separated by \begin{align*}30 \;\mathrm{cm}\end{align*}, have a uniform electric field between them of \begin{align*}400 \;\mathrm{N/C}\end{align*}.
1. Find the force on an electron located at a point midway between the plates
2. Find the voltage difference between the two plates
3. Find the change in electric potential energy of the electron when it travels from the back plate to the front plate
4. Find the speed of the electron just before striking the front plate (the screen of your TV)
1. a. \begin{align*}7500\mathrm{V}\end{align*} b. \begin{align*}1.5 \;\mathrm{m/s}\end{align*}
2. a. \begin{align*}6.4 \times 10^{-17}\;\mathrm{N}\end{align*} b. \begin{align*}1300\mathrm{V}\end{align*} c. \begin{align*}2.1 \times 10 ^{-16} \;\mathrm{J}\end{align*} d. \begin{align*}2.2 \times 10^7 \;\mathrm{m/s}\end{align*}
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
|
# How do you find the zeros for y=5x^2-2?
Mar 20, 2018
You set y equal 0, then solve the resulting equation for the value(s) of x.
#### Explanation:
Given: $y = 5 {x}^{2} - 2$
Set $y = 0$:
$0 = 5 {x}^{2} - 2$
Flip the equation:
$5 {x}^{2} - 2 = 0$
$5 {x}^{2} = 2$
Divide both sides by 5:
${x}^{2} = \frac{2}{5}$
When we perform the square root operation on both sides, we obtain a negative value and a positive value:
$x = - \sqrt{\frac{2}{5}}$ and $x = \sqrt{\frac{2}{5}}$
Multiply both by 1 in the form of $\frac{5}{5}$:
$x = - \sqrt{\frac{2}{5} \frac{5}{5}}$ and $x = \sqrt{\frac{2}{5} \frac{5}{5}}$
$x = - \sqrt{\frac{10}{25}}$ and $x = \sqrt{\frac{10}{25}}$
$x = - \frac{\sqrt{10}}{5}$ and $x = \frac{\sqrt{10}}{5}$
Check that both values produce $y = 0$:
$y = 5 {\left(- \frac{\sqrt{10}}{5}\right)}^{2} - 2$ and $y = 5 {\left(- \frac{\sqrt{10}}{5}\right)}^{2} - 2$
$y = \frac{10}{5} - 2$ and $y = \frac{10}{5} - 2$
$y = 2 - 2$ and $y = 2 - 2$
$y = 0$ and $y = 0 \leftarrow$ this checks
The zeros are $x = - \frac{\sqrt{10}}{5} \mathmr{and} x = \frac{\sqrt{10}}{5}$
Mar 20, 2018
To find the answer, set up the problem so $5 {x}^{2} - 2 = 0$.
#### Explanation:
To find the answer, set up the problem so $5 {x}^{2} - 2 = 0$. Then, move everything to one side of the equation, so $x$ is by itself.
$5 {x}^{2} - 2 = 0$
Add 2 to both sides to get:
$5 {x}^{2} = 2$
Then, divide both sides by 5:
${x}^{2} = \frac{2}{5}$
Then, take the square root of both sides:
$\sqrt{{x}^{2}} = \sqrt{\frac{2}{5}}$
$x = + \sqrt{\frac{2}{5}}$ and $- \sqrt{\frac{2}{5}}$
|
# How do you find the derivative of W(u)=e^(u^2)*tan(3sqrtu)?
Dec 10, 2017
$W ' \left(u\right) = \frac{3 {e}^{{u}^{2}}}{2 \sqrt[2]{u}} {\sec}^{2} \left(3 \sqrt[2]{u}\right) + 2 u {e}^{{u}^{2}} \tan \left(3 {e}^{\frac{1}{2}}\right)$
#### Explanation:
The answer is, it is very annoying.
The formula for product rule is
$f \left(x\right) = u v$
$f ' \left(x\right) = u \mathrm{dv} + v \mathrm{du}$
which seems simple enough.
Since there are clearly two terms in your function, lets set one equal to $u$ and on equal to $v$.
$W \left(u\right) = {e}^{{u}^{2}} \cdot \tan \left(3 {u}^{\frac{1}{2}}\right)$
Our first dilema is that $u$ is already used in the function, so instead I'll use the letter $z$. Then, I'll have to change the product rule formula.
$W \left(u\right) = z v$
$W ' \left(u\right) = z \mathrm{dv} + v \mathrm{dz}$
So now,
$z = {e}^{{u}^{2}}$
and
$v = \tan \left(3 {u}^{\frac{1}{2}}\right)$
Looking at the product rule, we need four things:
$z$, $v$, $\mathrm{dz}$, and $\mathrm{dv}$.
We already have $z$ and $v$ so lets find $\mathrm{dz}$ and $\mathrm{dv}$.
This is where it gets annoying.
$z = {e}^{{u}^{2}}$
so
$\mathrm{dz} = 2 u {e}^{{u}^{2}}$
$v = \tan \left(3 {u}^{\frac{1}{2}}\right)$
so
$\mathrm{dv} = \frac{3}{2} {u}^{- \frac{1}{2}} {\sec}^{2} \left(3 {u}^{\frac{1}{2}}\right)$
If you need further explanation of how I got these, let me know.
Now we just plug these variables into the formula.
$W ' \left(u\right) = z \mathrm{dv} + v \mathrm{dz}$
$W ' \left(u\right) = {e}^{{u}^{2}} \cdot \frac{3}{2} {u}^{- \frac{1}{2}} {\sec}^{2} \left(3 {u}^{\frac{1}{2}}\right) + \tan \left(3 {u}^{\frac{1}{2}}\right) \cdot 2 u {e}^{{u}^{2}}$
Simplify it to get
$W ' \left(u\right) = \frac{3 {e}^{{u}^{2}}}{2 \sqrt[2]{u}} {\sec}^{2} \left(3 \sqrt[2]{u}\right) + 2 u {e}^{{u}^{2}} \tan \left(3 {e}^{\frac{1}{2}}\right)$
See, wasn't that annoying?
|
# A triangle has sides A, B, and C. The angle between sides A and B is pi/6. If side C has a length of 25 and the angle between sides B and C is pi/12, what is the length of side A?
May 5, 2017
#### Explanation:
In general, the Law of Sine states that if a, b, and c are the lengths of the sides opposite angles $\alpha , \beta , \gamma$ (in that order), then
$\frac{a}{\sin} \left(\alpha\right) = \frac{b}{\sin} \left(\beta\right) = \frac{c}{\sin} \left(\gamma\right)$.
We only use the Law one pair of angles at a time. In this case, we know that the angle between A and B is $\gamma$, and the angle between B and C is $\alpha$. Therefore,
a = unknown
c = 25
$\alpha = \frac{\pi}{12}$
$\gamma = \frac{\pi}{6}$
Using the Law:
$\frac{a}{\sin} \left(\frac{\pi}{12}\right) = \frac{25}{\sin} \left(\frac{\pi}{6}\right)$
$\frac{\pi}{6}$ is one of the standard angles. $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.
The value of $\sin \left(\frac{\pi}{12}\right)$ may be found using the difference formula for sine:
$\sin \left(\frac{\pi}{12}\right) = \sin \left(\frac{3 \pi}{12} - \frac{2 \pi}{12}\right)$
$= \sin \left(\frac{3 \pi}{12}\right) \cos \left(\frac{2 \pi}{12}\right) - \cos \left(\frac{3 \pi}{12}\right) \sin \left(\frac{2 \pi}{12}\right)$
$= \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) - \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right)$
$= \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)$
$= \frac{\sqrt{6} - \sqrt{2}}{4}$
Solving the proportion:
$\frac{a}{\sin} \left(\frac{\pi}{12}\right) = \frac{25}{\sin} \left(\frac{\pi}{6}\right)$
$a = \frac{\left(25\right) \left(\sin \left(\frac{\pi}{12}\right)\right)}{\sin} \left(\frac{\pi}{6}\right)$
$a = \frac{\left(25\right) \frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{1}{2}}$
$a = \frac{\left(25\right) \left(\sqrt{6} - \sqrt{2}\right)}{2}$
$a = \frac{25 \left(\sqrt{6} - \sqrt{2}\right)}{2}$
NOTE: If we use the half-angle formula for sine instead of the difference formula, we obtain an answer that looks different but is equal to the above.
|
# Cone
Have you ever wondered how much ice cream is needed to fill an ice cream cone? Doesn’t the cone look like a triangle with a curved surface? So, how is the conical shape formed? And what is its volume and surface area? Let us try to find out answers to these questions.
## Cone
In geometry, the word cone refers to a pyramid-like structure with a circle-shaped base. It is a three-dimensional figure and has a circular base that tapers to a point called a vertex, apex or top. The shortest distance between the vertex and the base is called height. And, the distance from the vertex to a point on the base is the slant height.
### Types of Cone
Right Circular Cone: In this type, the axis makes a right angle from the base.
Oblique Cone: In this type, the axis is non-perpendicular to the base.
## Surface Area of a Cone
While studying cones, we generally consider a right circular cone. If you cut a paper cone straight along its side and open it, you can see the shape of paper that forms the surface of the cone. And, when you bring the sides marked P and Q together, you can see the curved portion of the shape forms a circle. Furthermore, if this shape is cut into small pieces along the lines marked in the figure, you can see that the pieces look like a triangle whose height is the slant height, l.
Now, the area of each triangle = 1/2×base of triangle×l. So, the sum of the areas of many such triangles makes up the area of the sector (fig c). Hence, the curved surface area is
= 1/2 b1l +1/2 b2l + 1/2 b3l +………. = 1/2 l (b1 + b2+ b3 + …) = 1/2 × l× (length of the whole curved boundary)
Also, the entire curved portion makes the perimeter of the base of cones. The circumference of the base = 2Ï€r, where r is the base of the radius of the base.
So, the curved surface area = 1/2 *l* 2πr = πrl or simply
Curved Surface Area of Cone = πrl
where r is the radius of the base and l is its slant height.
We already know that cones constitute the right-angled triangle. As per Pythagoras Theorem, length of the slant side, l2 = r2+ h2 Therefore,
Slant height l =√r2+ h2
Furthermore, if a circle whose area is πr2 is used to close the base. Then, the total surface area = πrl +πr2. Therefore,
 Total Surface Area of a Cone = πr (l+r)
## Volume of Cone
Cones are 3D triangles with a circle-shaped base. According to the structure, the volume of a cone is assumed to be 1/3 of a cylinder with the same radius of base and height. As the volume of a cylinder is πr2h, so,
Volume of a Cone = 1/3 πr2h
where r is the radius of the base and h is the height.
## Solved Examples For You
Question: The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28cm, calculate the height, slant height and the curved surface area.
Answer: Volume (V) = 9856 cm3
The diameter of the base = 28 cm.
Radius of the base= d/2 = 28/2= 14 cm
1/3 πr2h = 9856 cm3 = 1/3 × 22/7 × 14 × 14 × h = 9856.
Also, h= (9856 × 3 × 7) /(22 × 14 × 14) = 48 cm
h = 48 cm. l = √r2 + h2 = √ 142 + 482
l = 50 cm.
Curved Surface area = πrl = 22/7 × 14 × 50 = 2200 cm2
Question- How many edges do cones have?
Answer- Zero. Similarly, spheres and cylinders do not have any edges because they do not comprise of flat sides.
Question- What is the difference between a cone and right circular cone?
Answer- All cones may or may not have height to be perpendicular to the radius of the base. However, right circular cones are where the altitude or height is precisely perpendicular to the radius of the base.
Question- Give a few example of cones?
Answer- A funnel, a birthday cap are a few examples of cones.
Share with friends
## Customize your course in 30 seconds
##### Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Biology
Dr. Nazma Shaik
VTU
Chemistry
Gaurav Tiwari
APJAKTU
Physics
Get Started
|
# 7 1 Matrices Vectors Addition and Scalar Multiplication
• Slides: 30
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Section 7. 1 p 1
7. 1 Matrices, Vectors: Addition and Scalar Multiplication A matrix is a rectangular array of numbers or functions which we will enclose in brackets. For example, (1) are matrices. The numbers (or functions) are called entries or, less commonly, elements of the matrix. The first matrix in (1) has two rows, which are the horizontal lines of entries. Section 7. 1 p 2
7. 1 Matrices, Vectors: Addition and Scalar Multiplication (continued) Furthermore, it has three columns, which are the vertical lines of entries. The second and third matrices are square matrices, which means that each has as many rows as columns— 3 and 2, respectively. The entries of the second matrix have two indices, signifying their location within the matrix. The first index is the number of the row and the second is the number of the column, so that together the entry’s position is uniquely identified. For example, a 23 (read a two three) is in Row 2 and Column 3, etc. The notation is standard and applies to all matrices, including those that are not square. Section 7. 1 p 3
7. 1 Matrices, Vectors: Addition and Scalar Multiplication (continued) Matrices having just a single row or column are called vectors. Thus, the fourth matrix in (1) has just one row and is called a row vector. The last matrix in (1) has just one column and is called a column vector. Because the goal of the indexing of entries was to uniquely identify the position of an element within a matrix, one index suffices for vectors, whether they are row or column vectors. Thus, the third entry of the row vector in (1) is denoted by a 3. Section 7. 1 p 4
7. 1 Matrices, Vectors: Addition and Scalar Multiplication General Concepts and Notations Let us formalize what we just have discussed. We shall denote matrices by capital boldface letters A, B, C, … , or by writing the general entry in brackets; thus A = [ajk], and so on. By an m × n matrix (read m by n matrix) we mean a matrix with m rows and n columns—rows always come first! m × n is called the size of the matrix. Thus an m × n matrix is of the form (2) Section 7. 1 p 5
7. 1 Matrices, Vectors: Addition and Scalar Multiplication General Concepts and Notations (continued) The matrices in (1) are of sizes 2 × 3, 3 × 3, 2 × 2, 1 × 3, and 2 × 1, respectively. Each entry in (2) has two subscripts. The first is the row number and the second is the column number. Thus a 21 is the entry in Row 2 and Column 1. If m = n, we call A an n × n square matrix. Then its diagonal containing the entries a 11, a 22, … , ann is called the main diagonal of A. Thus the main diagonals of the two square matrices in (1) are a 11, a 22, a 33 and e−x, 4 x, respectively. Square matrices are particularly important, as we shall see. A matrix of any size m × n is called a rectangular matrix; this includes square matrices as a special case. Section 7. 1 p 6
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Vectors A vector is a matrix with only one row or column. Its entries are called the components of the vector. We shall denote vectors by lowercase boldface letters a, b, … or by its general component in brackets, a = [aj], and so on. Our special vectors in (1) suggest that a (general) row vector is of the form Section 7. 1 p 7
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Vectors (continued) A column vector is of the form Section 7. 1 p 8
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Definition Equality of Matrices Two matrices A = [ajk] and B = [bjk] are equal, written A = B, if and only if they have the same size and the corresponding entries are equal, that is, a 11 = b 11, a 12 = b 12, and so on. Matrices that are not equal are called different. Thus, matrices of different sizes are always different. Section 7. 1 p 9
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Definition Addition of Matrices The sum of two matrices A = [ajk] and B = [bjk] of the same size is written A + B and has the entries ajk + bjk obtained by adding the corresponding entries of A and B. Matrices of different sizes cannot be added. Section 7. 1 p 10
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Definition Scalar Multiplication (Multiplication by a Number) The product of any m × n matrix A = [ajk] and any scalar c (number c) is written c. A and is the m × n matrix c. A = [cajk] obtained by multiplying each entry of A by c. Section 7. 1 p 11
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Rules for Matrix Addition and Scalar Multiplication. From the familiar laws for the addition of numbers we obtain similar laws for the addition of matrices of the same size m × n, namely, (3) Here 0 denotes the zero matrix (of size m × n), that is, the m × n matrix with all entries zero. If m = 1 or n = 1, this is a vector, called a zero vector. Section 7. 1 p 12
7. 1 Matrices, Vectors: Addition and Scalar Multiplication Rules for Matrix Addition and Scalar Multiplication. (continued) Hence matrix addition is commutative and associative [by (3 a) and (3 b)]. Similarly, for scalar multiplication we obtain the rules (4) Section 7. 1 p 13
7. 2 Matrix Multiplication Section 7. 2 p 14
7. 2 Matrix Multiplication Definition Multiplication of a Matrix by a Matrix The product C = AB (in this order) of an m × n matrix A = [ajk] times an r × p matrix B = [bjk] is defined if and only if r = n and is then the m × p matrix C = [cjk] with entries (1) Section 7. 2 p 15
7. 2 Matrix Multiplication The condition r = n means that the second factor, B, must have as many rows as the first factor has columns, namely n. A diagram of sizes that shows when matrix multiplication is possible is as follows: A B = C [m × n] [n × p] = [m × p]. Section 7. 2 p 16
7. 2 Matrix Multiplication The entry cjk in (1) is obtained by multiplying each entry in the jth row of A by the corresponding entry in the kth column of B and then adding these n products. For instance, c 21 = a 21 b 11 + a 22 b 21 + … + a 2 nbn 1, and so on. One calls this briefly a multiplication of rows into columns. For n = 3, this is illustrated by where we shaded the entries that contribute to the calculation of entry c 21 just discussed. Section 7. 2 p 17
7. 2 Matrix Multiplication EXAMPLE 1 Matrix Multiplication Here c 11 = 3 · 2 + 5 · 5 + (− 1) · 9 = 22, and so on. The entry in the box is c 23 = 4 · 3 + 0 · 7 + 2 · 1 = 14. The product BA is not defined. Section 7. 2 p 18
7. 2 Matrix Multiplication EXAMPLE 4 CAUTION! Matrix Multiplication Is Not Commutative, AB ≠ BA in General This is illustrated by Examples 1 and 2, where one of the two products is not even defined, and by Example 3, where the two products have different sizes. But it also holds for square matrices. For instance, It is interesting that AB = 0 this also shows that does not necessarily imply BA = 0 or B = 0. Section 7. 2 p 19
7. 2 Matrix Multiplication Our examples show that in matrix products the order of factors must always be observed very carefully. Otherwise matrix multiplication satisfies rules similar to those for numbers, namely. (2) provided A, B, and C are such that the expressions on the left are defined; here, k is any scalar. (2 b) is called the associative law. (2 c) and (2 d) are called the distributive laws. Section 7. 2 p 20
7. 2 Matrix Multiplication Parallel processing of products on the computer is facilitated by a variant of (3) for computing C = AB, which is used by standard algorithms (such as in Lapack). In this method, A is used as given, B is taken in terms of its column vectors, and the product is computed columnwise; thus, (5) AB = A[b 1 b 2 … bp] = [Ab 1 Ab 2 … Abp]. Columns of B are then assigned to different processors (individually or several to each processor), which simultaneously compute the columns of the product matrix Ab 1, Ab 2, etc. Section 7. 2 p 21
7. 2 Matrix Multiplication Transposition We obtain the transpose of a matrix by writing its rows as columns (or equivalently its columns as rows). This also applies to the transpose of vectors. Thus, a row vector becomes a column vector and vice versa. In addition, for square matrices, we can also “reflect” the elements along the main diagonal, that is, interchange entries that are symmetrically positioned with respect to the main diagonal to obtain the transpose. Hence a 12 becomes a 21, a 31 becomes a 13, and so forth. Also note that, if A is the given matrix, then we denote its transpose by AT. Section 7. 2 p 22
7. 2 Matrix Multiplication Definition Transposition of Matrices and Vectors The transpose of an m × n matrix A = [ajk] is the n × m matrix AT (read A transpose) that has the first row of A as its first column, the second row of A as its second column, and so on. Thus the transpose of A in (2) is AT = [akj], written out (9) As a special case, transposition converts row vectors to column vectors and conversely. Section 7. 2 p 23
7. 2 Matrix Multiplication Rules for transposition are (10) CAUTION! Note that in (10 d) the transposed matrices are in reversed order. Section 7. 2 p 24
7. 2 Matrix Multiplication Special Matrices Symmetric and Skew-Symmetric Matrices. Transposition gives rise to two useful classes of matrices. Symmetric matrices are square matrices whose transpose equals the matrix itself. Skew-symmetric matrices are square matrices whose transpose equals minus the matrix. Both cases are defined in (11) and illustrated by Example 8. (11) AT = A (thus akj = ajk), Symmetric Matrix AT = −A (thus akj = −ajk), hence ajj = 0). Skew-Symmetric Matrix Section 7. 2 p 25
7. 2 Matrix Multiplication EXAMPLE 8 Symmetric and Skew-Symmetric Matrices For instance, if a company has three building supply centers C 1, C 2, C 3, then A could show costs, say, ajj for handling 1000 bags of cement at center Cj, and ajk ( j ≠ k) the cost of shipping 1000 bags from Cj to Ck. Clearly, ajk = akj if we assume shipping in the opposite direction will cost the same. Section 7. 2 p 26
7. 2 Matrix Multiplication Triangular Matrices. Upper triangular matrices are square matrices that can have nonzero entries only on and above the main diagonal, whereas any entry below the diagonal must be zero. Similarly, lower triangular matrices can have nonzero entries only on and below the main diagonal. Any entry on the main diagonal of a triangular matrix may be zero or not. Section 7. 2 p 27
7. 2 Matrix Multiplication EXAMPLE 9 Upper and Lower Triangular Matrices Upper triangular Section 7. 2 p 28 Lower triangular
7. 2 Matrix Multiplication Diagonal Matrices. These are square matrices that can have nonzero entries only on the main diagonal. Any entry above or below the main diagonal must be zero. Section 7. 2 p 29
7. 2 Matrix Multiplication If all the diagonal entries of a diagonal matrix S are equal, say, c, we call S a scalar matrix because multiplication of any square matrix A of the same size by S has the same effect as the multiplication by a scalar, that is, (12) AS = SA = c. A. In particular, a scalar matrix, whose entries on the main diagonal are all 1, is called a unit matrix (or identity matrix) and is denoted by In or simply by I. For I, formula (12) becomes (13) Section 7. 2 p 30 AI = IA = A.
|
# Statistics - Estimating Population Proportions
A population proportion is the share of a population that belongs to a particular category.
Confidence intervals are used to estimate population proportions.
## Estimating Population Proportions
A statistic from a sample is used to estimate a parameter of the population.
The most likely value for a parameter is the point estimate.
Additionally, we can calculate a lower bound and an upper bound for the estimated parameter.
The margin of error is the difference between the lower and upper bounds from the point estimate.
Together, the lower and upper bounds define a confidence interval.
## Calculating a Confidence Interval
The following steps are used to calculate a confidence interval:
1. Check the conditions
2. Find the point estimate
3. Decide the confidence level
4. Calculate the margin of error
5. Calculate the confidence interval
For example:
• Population: Nobel Prize winners
• Category: Born in the United States of America
We can take a sample and see how many of them were born in the US.
The sample data is used to make an estimation of the share of all the Nobel Prize winners born in the US.
By randomly selecting 30 Nobel Prize winners we could find that:
6 out of 30 Nobel Prize winners in the sample were born in the US
From this data we can calculate a confidence interval with the steps below.
## 1. Checking the Conditions
The conditions for calculating a confidence interval for a proportion are:
• The sample is randomly selected
• There is only two options:
• Being in the category
• Not being in the category
• The sample needs at least:
• 5 members in the category
• 5 members not in the category
In our example, we randomly selected 6 people that were born in the US.
The rest were not born in the US, so there are 24 in the other category.
The conditions are fulfilled in this case.
Note: It is possible to calculate a confidence interval without having 5 of each category. But special adjustments need to be made.
## 2. Finding the Point Estimate
The point estimate is the sample proportion ($$\hat{p}$$).
The formula for calculating the sample proportion is the number of occurrences ($$x$$) divided by the sample size ($$n$$):
$$\displaystyle \hat{p} =\frac{x}{n}$$
In our example, 6 out of 30 were born in the US: $$x$$ is 6, and $$n$$ is 30.
So the point estimate for the proportion is:
$$\displaystyle \hat{p} = \frac{x}{n} = \frac{6}{30} = \underline{0.2} = 20\%$$
So 20% of the sample were born in the US.
## 3. Deciding the Confidence Level
The confidence level is expressed with a percentage or a decimal number.
For example, if the confidence level is 95% or 0.95:
The remaining probability ($$\alpha$$) is then: 5%, or 1 - 0.95 = 0.05.
Commonly used confidence levels are:
• 90% with $$\alpha$$ = 0.1
• 95% with $$\alpha$$ = 0.05
• 99% with $$\alpha$$ = 0.01
Note: A 95% confidence level means that if we take 100 different samples and make confidence intervals for each:
The true parameter will be inside the confidence interval 95 out of those 100 times.
We use the standard normal distribution to find the margin of error for the confidence interval.
The remaining probabilities ($$\alpha$$) are divided in two so that half is in each tail area of the distribution.
The values on the z-value axis that separate the tails area from the middle are called critical z-values.
Below are graphs of the standard normal distribution showing the tail areas ($$\alpha$$) for different confidence levels.
## 4. Calculating the Margin of Error
The margin of error is the difference between the point estimate and the lower and upper bounds.
The margin of error ($$E$$) for a proportion is calculated with a critical z-value and the standard error:
$$\displaystyle E = Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
The critical z-value $$Z_{\alpha/2}$$ is calculated from the standard normal distribution and the confidence level.
The standard error $$\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ is calculated from the point estimate ($$\hat{p}$$) and sample size ($$n$$).
In our example with 6 US-born Nobel Prize winners out of a sample of 30 the standard error is:
$$\displaystyle \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.2(1-0.2)}{30}} = \sqrt{\frac{0.2 \cdot 0.8}{30}} = \sqrt{\frac{0.16}{30}} = \sqrt{0.00533..} \approx \underline{0.073}$$
If we choose 95% as the confidence level, the $$\alpha$$ is 0.05.
So we need to find the critical z-value $$Z_{0.05/2} = Z_{0.025}$$
The critical z-value can be found using a Z-table or with a programming language function:
### Example
With Python use the Scipy Stats library norm.ppf() function find the Z-value for an $$\alpha$$/2 = 0.025
import scipy.stats as stats
print(stats.norm.ppf(1-0.025))
Try it Yourself »
### Example
With R use the built-in qnorm() function to find the Z-value for an $$\alpha$$/2 = 0.025
qnorm(1-0.025)
Try it Yourself »
Using either method we can find that the critical Z-value $$Z_{\alpha/2}$$ is $$\approx \underline{1.96}$$
The standard error $$\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ was $$\approx \underline{0.073}$$
So the margin of error ($$E$$) is:
$$\displaystyle E = Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \approx 1.96 \cdot 0.073 = \underline{0.143}$$
## 5. Calculate the Confidence Interval
The lower and upper bounds of the confidence interval are found by subtracting and adding the margin of error ($$E$$) from the point estimate ($$\hat{p}$$).
In our example the point estimate was 0.2 and the margin of error was 0.143, then:
The lower bound is:
$$\hat{p} - E = 0.2 - 0.143 = \underline{0.057}$$
The upper bound is:
$$\hat{p} + E = 0.2 + 0.143 = \underline{0.343}$$
The confidence interval is:
$$[0.057, 0.343]$$ or $$[5.7 \%, 34.4 \%]$$
And we can summarize the confidence interval by stating:
The 95% confidence interval for the proportion of Nobel Prize winners born in the US is between 5.7% and 34.4%
## Calculating a Confidence Interval with Programming
A confidence interval can be calculated with many programming languages.
Using software and programming to calculate statistics is more common for bigger sets of data, as calculating manually becomes difficult.
### Example
With Python, use the scipy and math libraries to calculate the confidence interval for an estimated proportion.
Here, the sample size is 30 and the occurrences is 6.
import scipy.stats as stats
import math
# Specify sample occurrences (x), sample size (n) and confidence level
x = 6
n = 30
confidence_level = 0.95
# Calculate the point estimate, alpha, the critical z-value, the standard error, and the margin of error
point_estimate = x/n
alpha = (1-confidence_level)
critical_z = stats.norm.ppf(1-alpha/2)
standard_error = math.sqrt((point_estimate*(1-point_estimate)/n))
margin_of_error = critical_z * standard_error
# Calculate the lower and upper bound of the confidence interval
lower_bound = point_estimate - margin_of_error
upper_bound = point_estimate + margin_of_error
# Print the results
print("Point Estimate: {:.3f}".format(point_estimate))
print("Critical Z-value: {:.3f}".format(critical_z))
print("Margin of Error: {:.3f}".format(margin_of_error))
print("Confidence Interval: [{:.3f},{:.3f}]".format(lower_bound,upper_bound))
print("The {:.1%} confidence interval for the population proportion is:".format(confidence_level))
print("between {:.3f} and {:.3f}".format(lower_bound,upper_bound))
Try it Yourself »
### Example
With R, use the built-in math and statistics functions to calculate the confidence interval for an estimated proportion.
Here, the sample size is 30 and the occurrences is 6.
# Specify sample occurrences (x), sample size (n) and confidence level
x = 6
n = 30
confidence_level = 0.95
# Calculate the point estimate, alpha, the critical z-value, the standard error, and the margin of error
point_estimate = x/n
alpha = (1-confidence_level)
critical_z = qnorm(1-alpha/2)
standard_error = sqrt(point_estimate*(1-point_estimate)/n)
margin_of_error = critical_z * standard_error
# Calculate the lower and upper bound of the confidence interval
lower_bound = point_estimate - margin_of_error
upper_bound = point_estimate + margin_of_error
# Print the results
sprintf("Point Estimate: %0.3f", point_estimate)
sprintf("Critical Z-value: %0.3f", critical_z)
sprintf("Margin of Error: %0.3f", margin_of_error)
sprintf("Confidence Interval: [%0.3f,%0.3f]", lower_bound, upper_bound)
sprintf("The %0.1f%% confidence interval for the population proportion is:", confidence_level*100)
sprintf("between %0.4f and %0.4f", lower_bound, upper_bound)
Try it Yourself »
×
## Contact Sales
If you want to use W3Schools services as an educational institution, team or enterprise, send us an e-mail:
sales@w3schools.com
|
# Transformations on the coordinate plane
## Presentation on theme: "Transformations on the coordinate plane"— Presentation transcript:
Transformations on the coordinate plane
Transformations Review
Type Diagram A translation moves a figure left, right, up, or down A reflection moves a figure across its line of reflection to create its mirror image. A rotation moves a figure around a given point.
Now we will look at how each transformation looks on a coordinate plane. The transformed figure is often named with the same letters, but adding an apostrophe. The transformation of ABC is A’B’C’.
Translation Find point A and Translate ABC 6 units to the right.
Find point B and Find point C and 6 Units A A’ count 6 units to the right. Plot point A’. B’ B C C’ count 6 units to the right. Plot point B’. count 6 units to the right. Plot point C’.
Translation Rules To translate a figure a units to the right, increase the x-coordinate of each point by a amount. To translate a figure a units to the left, decrease the x-coordinate of each point by a amount. Translate point P (3, 2) 9 units to the right. Since we are going to the right, we add 9 to the x-coordinate = 12, so the new coordinates of P’ are (12, 2) Translate point P (3, 2) 6 units to the left. Since we are going up, we subtract 6 to the x-coordinate = -3, so the new coordinates of P’ are (-3, 2)
Translation Rules To translate a figure a units up, increase the y-coordinate of each point by a amount. Translate point P (3, 2) 9 units up. To translate a figure a units down, decrease the y-coordinate of each point by a amount. Since we are going up, we add 9 to the y-coordinate = 11, so the new coordinates of P’ are (3, 11) Translate point P (3, 2) 6 units down. Since we are going down, we subtract 6 to the y-coordinate = -4, so the new coordinates of P’ are (3, -4)
Translation Example 2 The coordinates of point A are (-5, 4) Since we are moving to the right we increase the x-coordinate by 6. The coordinates of point B are (-2, 3) The coordinates of point C are (-3, 1) = 1, so the new coordinates of A’ are (1, 4). = 4, so the new coordinates of B’ are (4, 3). = 3, so the new coordinates of C’ are (3, 1).
Practice Point P (5, 8). Translate 2 to the left and 6 up. P’ (3, 14)
Point Z (-3, -6). Translate 5 to the right and 9 down. Translate LMN, whose coordinates are (3, 6), (5, 9), and (7, 12), 9 units left and 14 units up. P’ (3, 14) Z’ (2, -15) L’M’N’ (-6, 20), (-4, 23), (-2, 26)
Reflection Reflect ABC across the y-axis.
Count the number of units point A is from the line of reflection. 5 Units 5 Units A A’ Count the same number of units on the other side and plot point A’. 2 Units 2 Units B B’ 3 Units 3 Units Count the number of units point B is from the line of reflection. C C’ Count the same number of units on the other side and plot point B’. Count the number of units point C is from the line of reflection. Count the same number of units on the other side and plot point C’.
Reflection Rules To reflect point (a, b) across the y-axis use the opposite of the x-coordinate and keep the y coordinate the same. Reflect point P (3, 2) across the y-axis. To reflect point (a, b) across the x-axis keep the x-coordinate the same and use the opposite of the y-coordinate Since we reflecting across the y-axis. Keep the y the same and use the opposite of the x. (-3, 2) Reflect point P (3, 2) across the x-axis. Since we reflecting across the x-axis. Keep the x the same and use the opposite of the y. (3, -2)
Practice The coordinates of ABC are: (-5, 4), (-2, 3), (-3, 1) Reflect ABC across the y-axis and then reflect it across the x-axis. To reflect it across the y-axis keep the y the same and use the opposite x. The new coordinates are: (5, 4), (2, 3), (3, 1) To reflect it across the x-axis keep the x the same and use the opposite y. The new coordinates are: (5, -4), (2, -3), (3, -1)
Rotation Rules To rotate a point 90° clockwise, switch the coordinates, and then multiply the new y-coordinate by -1. Rotate point P (3, 2) clockwise about the origin. To rotate a point 180°, just multiply each coordinate by -1. Since we are rotating it clockwise, we switch the coordinates (2, 3) and multiply the new y by -1, so the new coordinates are (2, -3) Rotate point P (3, 2) clockwise about the origin. Since we are rotating it 180°, we simply multiply the coordinates by -1, so the new coordinates are (-3, -2).
Practice Point P (5, 8). Rotate 90° clockwise about the origin.
Point Z (-3, -6). Rotate 180° about the origin. Rotate LMN, whose coordinates are (3, 6), (5, 9), and (7, 12), 90° clockwise about the origin. P’ (8, -5) Z’ (-3, -6) L’M’N’ (6, -3), (9, -5), (7, -12)
|
In 5th grade worksheet 1 let’s solve the 10 questions.
1. The smallest whole number is ................ .
2. The smallest natural number is ............... .
3. There is ...................... largest whole number because each number has its ................... .
4. Counting numbers starts from ................ .
5. Successor of 1, 64, 539 is .................. .
6. Predecessor of 700000 is .............. .
7. Put <, > or =
(a) 3674548 ........... 637454.
(b) 10001 ............ 9999.
(c) 99999 + 1 .......... 100000.
8. Write the place value of 7 in 647321.
9. Subtract the place value of 5 from its place value in the numeral 35340.
10. Smallest 5-digit number is ................ of largest 4–digit number.
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
## Recent Articles
1. ### Method of H.C.F. |Highest Common Factor|Factorization &Division Method
Apr 13, 24 05:12 PM
We will discuss here about the method of h.c.f. (highest common factor). The highest common factor or HCF of two or more numbers is the greatest number which divides exactly the given numbers. Let us…
2. ### Factors | Understand the Factors of the Product | Concept of Factors
Apr 13, 24 03:29 PM
Factors of a number are discussed here so that students can understand the factors of the product. What are factors? (i) If a dividend, when divided by a divisor, is divided completely
3. ### Methods of Prime Factorization | Division Method | Factor Tree Method
Apr 13, 24 01:27 PM
In prime factorization, we factorise the numbers into prime numbers, called prime factors. There are two methods of prime factorization: 1. Division Method 2. Factor Tree Method
4. ### Divisibility Rules | Divisibility Test|Divisibility Rules From 2 to 18
Apr 13, 24 12:41 PM
To find out factors of larger numbers quickly, we perform divisibility test. There are certain rules to check divisibility of numbers. Divisibility tests of a given number by any of the number 2, 3, 4…
|
How do you differentiate f(x) =(sin x + tan x)/(sin x-cos x) ?
Dec 21, 2015
$f ' \left(x\right) = \frac{\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)}{\sin x - \cos x} ^ 2$
Explanation:
Original equation:
$f \left(x\right) = \frac{\sin x + \tan x}{\sin x - \cos x}$
Use the quotient rule to derive.
Derive the top and times by the bottom:
$\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right)$
Derive the bottom and multiply by the top:
$\left(\cos x + \sin x\right) \left(\sin x + \tan x\right)$
Subtract the two:
$\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)$
Place it over the bottom squared:
$\frac{\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)}{\sin x - \cos x} ^ 2$
$f ' \left(x\right) = \frac{\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)}{\sin x - \cos x} ^ 2$
|
# What Is The Fraction For .67?
Have you ever found yourself trying to figure out what the fraction for .67 is? It can be a little tricky to figure out, but with a bit of understanding, you can easily get the answer. This article will provide you with an easy-to-understand explanation of the fraction for .67 and how you can figure it out.
## What is a Fraction?
A fraction is a part of a whole number. Fractions are written with a numerator (the top number) and a denominator (the bottom number). Generally, fractions are used to represent parts of a whole, such as 1/2 or 3/4. Fractions can also be used to express ratios, such as 2:3 or 4:5.
## What is .67?
.67 is a decimal number. Decimals represent numbers with a decimal point separating the integer part from the fractional part. For example, .67 represents the fraction 67/100. The fraction 67/100 can be written in its simplest form as 2/3.
## How Do You Find the Fraction for .67?
To find the fraction for .67, you first need to convert the decimal number to a fraction. To do this, you can move the decimal point two places to the right and put the resulting number over 100. In the case of .67, the fraction would be 67/100. To simplify this fraction, you can divide both the numerator and the denominator by the same number until the fraction is in its simplest form. In this case, the fraction 67/100 can be simplified to 2/3.
### Example:
Let\’s say you have the decimal number .15 and you want to find the fraction for it. To do this, you would move the decimal point two places to the right and put the resulting number over 100. In this case, the fraction would be 15/100. To simplify this fraction, you can divide both the numerator and the denominator by the same number. In this case, the fraction 15/100 can be simplified to 3/20.
## Conclusion:
To find the fraction for .67, you first need to convert the decimal number to a fraction. To do this, you can move the decimal point two places to the right and put the resulting number over 100. In the case of .67, the fraction would be 67/100. To simplify this fraction, you can divide both the numerator and the denominator by the same number until the fraction is in its simplest form. In this case, the fraction 67/100 can be simplified to 2/3.
## What Else Can You Do With Fractions?
Fractions can be used to represent parts of a whole, calculate ratios, and even solve mathematical problems. Fractions can also be used to compare quantities, calculate discounts, and find the slope. Fractions can be used in a variety of other applications, such as architecture, engineering, construction, and finance.
## Are There Other Ways to Find Fractions?
Yes, there are many other ways to find fractions. For example, you can use a calculator, a chart, or a set of conversion tables to help you figure out fractions. You can also use a fraction converter app to quickly and easily convert decimals to fractions. Additionally, there are online calculators and tutorials available to help you understand fractions and how to calculate them.
## Final Thoughts
Finding the fraction for .67 is simple once you understand how to convert decimals to fractions. With a bit of practice, you can easily figure out the fraction for any decimal number. Additionally, understanding fractions can be helpful in many other applications and can make it easier to solve mathematical problems. So if you want to be more efficient in math, it may be worth learning more about fractions.
|
Kirchhoff’s Current Law
# Kirchhoff’s Current Law - Notes | Study Network Theory (Electric Circuits) - Electrical Engineering (EE)
1 Crore+ students have signed up on EduRev. Have you?
Introduction
• Gustav Kirchhoff’s Current Law is one of the fundamental laws used for circuit analysis. His current law states that for a parallel path the total current entering a circuits junction is exactly equal to the total current leaving the same junction. This is because it has no other place to go as no charge is lost.
• In other words the algebraic sum of ALL the currents entering and leaving a junction must be equal to zero as: Σ IIN = Σ IOUT.
• This idea by Kirchhoff is commonly known as the Conservation of Charge, as the current is conserved around the junction with no loss of current. Lets look at a simple example of Kirchhoff’s current law (KCL) when applied to a single junction.
A Single Junction
• Here in this simple single junction example, the current IT leaving the junction is the algebraic sum of the two currents, I1 and I2 entering the same junction. That is IT = I1 + I2.
Note that we could also write this correctly as the algebraic sum of: IT – (I1 + I2) = 0.
So if I1 equals 3 amperes and I2 is equal to 2 amperes, then the total current, IT leaving the junction will be 3 + 2 = 5 amperes, and we can use this basic law for any number of junctions or nodes as the sum of the currents both entering and leaving will be the same.
• Also, if we reversed the directions of the currents, the resulting equations would still hold true for I1 or I2. As I1 = IT – I= 5 – 2 = 3 amps, and I2 = IT – I1 = 5 – 3 = 2 amps. Thus we can think of the currents entering the junction as being positive (+), while the ones leaving the junction as being negative (-).
• Then we can see that the mathematical sum of the currents either entering or leaving the junction and in whatever direction will always be equal to zero, and this forms the basis of Kirchhoff’s Junction Rule, more commonly known as Kirchhoff’s Current Law, or (KCL).
Resistors in Parallel
• Let’s look how we could apply Kirchhoff’s current law to resistors in parallel, whether the resistances in those branches are equal or unequal. Consider the following circuit diagram:
• In this simple parallel resistor example there are two distinct junctions for current. Junction one occurs at node B, and junction two occurs at node E. Thus we can use Kirchhoff’s Junction Rule for the electrical currents at both of these two distinct junctions, for those currents entering the junction and for those currents flowing leaving the junction.
• To start, all the current, Ileaves the 24 volt supply and arrives at point A and from there it enters node B. Node B is a junction as the current can now split into two distinct directions, with some of the current flowing downwards and through resistor R1 with the remainder continuing on through resistor R2 via node C. Note that the currents flowing into and out of a node point are commonly called branch currents.
• We can use Ohm’s Law to determine the individual branch currents through each resistor as: I = V/R, thus:
For current branch B to E through resistor R1
For current branch C to D through resistor R
• From above we know that Kirchhoff’s current law states that the sum of the currents entering a junction must equal the sum of the currents leaving the junction, and in our simple example above, there is one current, IT going into the junction at node B and two currents leaving the junction, I1 and I2.
• Since we now know from calculation that the currents leaving the junction at node B is I1 equals 3 amps and I2 equals 2 amps, the sum of the currents entering the junction at node B must equal 3 + 2 = 5 amps. Thus ΣIN = IT = 5 amperes.
• In our example, we have two distinct junctions at node B and node E, thus we can confirm this value for IT as the two currents recombine again at node E. So, for Kirchhoff’s junction rule to hold true, the sum of the currents into point F must equal the sum of the currents flowing out of the junction at node E.
• As the two currents entering junction E are 3 amps and 2 amps respectively, the sum of the currents entering point F is therefore: 3 + 2 = 5 amperes. Thus ΣIN = IT = 5 amperes and therefore Kirchhoff’s current law holds true as this is the same value as the current leaving point A.
Applying KCL to more complex circuits.
• We can use Kirchhoff’s current law to find the currents flowing around more complex circuits. We hopefully know by now that the algebraic sum of all the currents at a node (junction point) is equal to zero and with this idea in mind, it is a simple case of determining the currents entering a node and those leaving the node. Consider the circuit below.
Kirchhoff’s Current Law Example
• In this example there are four distinct junctions for current to either separate or merge together at nodes A, C, E and node F. The supply current IT separates at node A flowing through resistors R1 and R2, recombining at node C before separating again through resistors R3, R4 and R5 and finally recombining once again at node F.
• But before we can calculate the individual currents flowing through each resistor branch, we must first calculate the circuits total current, IT. Ohms law tells us that I = V/R and as we know the value of V, 132 volts, we need to calculate the circuit resistances as follows.
Circuit Resistance RAC
• Thus the equivalent circuit resistance between nodes A and C is calculated as 1 Ohm.
Circuit Resistance RCF
Thus the equivalent circuit resistance between nodes C and F is calculated as 10 Ohms. Then the total circuit current, IT is given as:
RT = R(AC) + R(CF) = 1 + 10 = 11Ω
Giving us an equivalent circuit of:
Therefore, V = 132V, RAC = 1Ω, RCF = 10Ω’s and IT = 12A.
Having established the equivalent parallel resistances and supply current, we can now calculate the individual branch currents and confirm using Kirchhoff’s junction rule as follows.
VAC = IT X RAC = 12 x 1 = 12 Volts
VCF = IT x RCF = 12 x 10 = 120 Volts
Thus, I1 = 5A, I2 = 7A, I3 = 2A, I4 = 6A, and I5 = 4A.
We can confirm that Kirchoff’s current law holds true around the circuit by using node C as our reference point to calculate the currents entering and leaving the junction as:
At node C ∑IIN = ∑IOUT
IT = I1 + I2 = I3 + I4 + I5
∴ 12 = (5 + 7) = (2 + 6 + 4)
We can also double check to see if Kirchhoff's Current Law holds true as the currents entering the junction are positive, while the ones leaving the junction are negative, thus the algebraic sum is: I1 + I2 – I3 – I4 – I5 = 0 which equals 5 + 7 – 2 – 6 – 4 = 0. So we can confirm by analysis that Kirchhoff’s current law (KCL) which states that the algebraic sum of the currents at a junction point in a circuit network is always zero is true and correct in this example.
Kirchhoff’s Current Law Example
Find the currents flowing around the following circuit using Kirchhoff’s Current Law only.
IT is the total current flowing around the circuit driven by the 12V supply voltage. At point A, I1 is equal to IT, thus there will be an I1*R voltage drop across resistor R1.
The circuit has 2 branches, 3 nodes (B, C and D) and 2 independent loops, thus the I*R voltage drops around the two loops will be:
Loop ABC ⇒ 12 = 4I1 + 6I2
Loop ABD ⇒ 12 = 4I1 + 12I3
Since Kirchhoff’s current law states that at node B, I1 = I2 + I3, we can therefore substitute current I1 for (I2 + I3) in both of the following loop equations and then simplify.
Kirchhoff’s Loop Equations
We now have two simultaneous equations that relate to the currents flowing around the circuit.
Eq. No 1 : 12 = 10I2 + 4I3
Eq. No 2 : 12 = 4I2 + 16I3
By multiplying the first equation (Loop ABC) by 4 and subtracting Loop ABD from Loop ABC, we can be reduced both equations to give us the values of I2 and I3
Eq. No 1 : 12 = 10I2 + 4I3 ( x4 ) ⇒ 48 = 40I2 + 16I3
Eq. No 2 : 12 = 4I2 + 16I3 ( x1 ) ⇒ 12 = 4I2 + 16I3
Eq. No 1 – Eq. No 2 ⇒ 36 = 36I2 + 0
Substitution of I2 in terms of Igives us the value of I2 as 1.0 Amps
Now we can do the same procedure to find the value of I3 by multiplying the first equation (Loop ABC) by 4 and the second equation (Loop ABD) by 10. Again by subtracting Loop ABC from Loop ABD, we can be reduced both equations to give us the values of I2 and I3
Eq. No 1 : 12 = 10I2 + 4I3 (x4) ⇒ 48 = 40I2 + 16I3
Eq. No 2 : 12 = 4I2 + 16I3 (x10) ⇒ 120 = 40I2 + 160I3
Eq. No 2 – Eq. No 1 ⇒ 72 = 0 + 144I3
Thus substitution of I3 in terms of I2 gives us the value of I3 as 0.5 Amps
As Kirchhoff’s junction rule states that : I= I2 + I3
The supply current flowing through resistor R1 is given as : 1.0 + 0.5 = 1.5 Amps
Thus I1 = IT = 1.5 Amps, I2 = 1.0 Amps and I= 0.5 Amps and from that information we could calculate the I*R voltage drops across the devices and at the various points (nodes) around the circuit.
The document Kirchhoff’s Current Law - Notes | Study Network Theory (Electric Circuits) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Network Theory (Electric Circuits).
All you need of Electrical Engineering (EE) at this link: Electrical Engineering (EE)
## Network Theory (Electric Circuits)
23 videos|63 docs|60 tests
Use Code STAYHOME200 and get INR 200 additional OFF
## Network Theory (Electric Circuits)
23 videos|63 docs|60 tests
### Top Courses for Electrical Engineering (EE)
Track your progress, build streaks, highlight & save important lessons and more!
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
;
|
## What is the equation of the line that is parallel to the line y − 1 = 4(x + 3) and passes through the point (4, 32)? y = –x + 3
Question
What is the equation of the line that is parallel to the line y − 1 = 4(x + 3) and passes through the point (4, 32)?
y = –x + 33
y = –x + 36
y = 4x − 16
y = 4x + 16
in progress 0
2 weeks 2021-09-13T14:32:00+00:00 2 Answers 0
y=4x+16
Step-by-step explanation:
Let us first rearrange the equation of the given line to be in the form y=mx+c. M gives the gradient of the line and c gives the y intercept.
y-1=4(x+3)
y-1=4x+12
y=4x+13
The gradient of parallel lines is equal. For the second line m=4
m=Δy/Δx
(y-32)/(x-4)=4
y-32=4(x-4)
y-32=4x-16
y= 4x+16
The equation of line 2 will be y=4x+16
2. ## Hello!
The line that is parallel to the given line and passes through the point (4,32) is:
## Why?
To solve the problem, we need to remember that parallel lines have the same slope, so, we need to find a line that has the same slope that the given line and it also pass through the point (4,32).
So, we are given the line:
So, we have that the line has a slope which is equal to “4”.
Now, we have to find which of the given lines that have a slope equal to “4”, also pass through the point (4,32), so, we need to evaluate it into the equations.
Therefore, evaluating we have:
Third line:
We can see that the equation is not satisfied, so the line does not pass through the point.
So, evaluating the fourth line, we have:
We can see that the equation is satisfied, so the line does pass through the point (4,32)
Hence, we have that the line that is parallel to the given line and passes through the point (4,32) is:
Have a nice day!
|
# What are different ways to figure out the square roots of a number?
Ashley Kannan | Certified Educator
If the question is asking how to determine the square roots of a number, the most elemental way that is dominant in today's classroom is the use of a calculator. In most settings the use of a calculator for computational issues such as finding a square root has become accepted practice. There can be a healthy debate as to whether this helps to child, but the fact of the matter is that math curriculums are embracing so many different approaches to solving problems and advocating applications which involve higher ordered thinking. Computation has become a part of the process of advanced mathematics instruction and thus using a calculator has become quite standard and accepted practice.
If one were to try to find square roots without a calculator, one of the most time honored of systems is the "guess and check" approach. In this, you simply keep plugging away until you find an answer that works. It is not the most time efficient means of solving, but it does get the job done and does help you find a square root of a number. For example, if you wanted to find the square root of 8, guess and check means that you find a range of numbers from which to work. Squaring 2 gives you 4, and squaring the next number is 3 gives you 9. With that, you have your range. We know that 2 is too small and 3 is too large. Thus, the answer exists between both numbers.
Since 9 is closer to 8 than it is to 4, we can narrow down some of our options to work with the upper half of the bracket. This means that we can start by taking 2.5 and squaring that, giving us 6.25. Close, but we need to continue progressing. Taking the next increment of 2.6, we square that and get 6.76. Continuing on to squaring 2.7, we can get to 7.29, which is closer. This process continues on in the same manner until we get to 2.9, when we go over because 2.9 squared gives 8.41. Thus, guess and check has helped us narrow that the square root of 8 exists between 2.8 and 2.9. To further narrow it to specific decimal points, the process would be continued accordingly. Guess and check is another way of being able to do and figure out square roots without the use of a calculator.
|
Arithmetic vs. Geometric Sequences
14 teachers like this lesson
Print Lesson
Objective
SWBAT identify the differences between arithmetic and geometric sequences from a table. SWBAT write an explicit or recursive formula for the function.
Big Idea
Relate geometric sequences to exponential functions and arithmetic sequences to linear functions.
Warm up
10 minutes
I expect this Warm up to take about 10 minutes for my students to complete, including time to review the task with students. The warmup enables a comparison of linear and exponential functions, which I use as a lead into the lesson.
I expect all of my students to recognize the pattern in each table, and to plot the points successfully. However, writing a formula that represents the function will be difficult for several students. When we review the warm up as a class, I will recommend that students take notes about writing explicit and recursive formulas for each function. I review the content of the notes in the video below.
Categorize Arithmetic and Geometric Sequences
15 minutes
After we review the warm up as a class, I have students add to their notes that a sequence is a list of numbers that typically has some pattern to it. If we look at the output values in each table of the warm up, each one represents a sequence, and each of these sequences forms a pattern.
The linear function increases by a common difference d. When a sequence is formed by a common difference, we can call it an arithmetic sequence.
The exponential function increases by a common factor or common ratio. When a sequence is produced by a common factor, it is called a geometric sequence.
After defining these two types of sequences, I hand a stack of seven index cards to each pair or group. Each card is numbered and has a sequence of numbers on it (see Categorize each sequence for a list of the sequences). Each student works with their assigned elbow partner. Groups of 3 students may be used if necessary.
The task: I ask my students to categorize each card as an arithmetic or a geometric sequence. On each card the student is to write the reason for their choice on their cards individually without discussing it with their partner(s). Each student works on one card at a time, sharing the work load until the cards are all categorized. When the cards are complete, the responses on the cards are shared with the partner(s). Each student in the group critiques the category and reasoning of the other students (Math Practice 3). The cards are discussed, and a consensus must be met on each card before the categories of each card are shared with the class.
To bring this activity to a close, each group posts their selections on the board by writing the number of each sequence in the chosen category. The information is discussed on any differences until the analysis of each of the 7 sequences are complete.
Write Explicit and Recursive Formulas
15 minutes
After the class successfully classified each of the 7 sequences, the students continue to work with their elbow partner. I instruct students to write both a recursive and an explicit formula for each sequence.
This task is the most difficult part of the lesson for my students. As students are working, I monitor their progress. I ask probing questions to help move the students forward, but I try to do so in a way that allows them to continue to engage in a productive struggle (MP1). Most students do well after persevering through the first couple of sequences. They learn to better interpreting the structure of the arithmetic sequences and the geometric sequences (MP7). As the students make discoveries, I encourage them to add to the notes that they began during the warm-up section of the lesson.
If necessary, I create a workshop for students that are still struggling. Students rotate in-and-out of the workshop as they make progress with writing the formulas. In the workshop, I work with a small group of students as they work on individual white boards. I review each of the seven sequences, inviting students to join the workshop if they are encountering difficulty..
Closure
10 minutes
After students complete their work, they are to complete the Exit Slip before leaving class. I use the Exit Slip as a formative assessment to check each student's understanding of the objective of this lesson. If need be, I may assign the Exit Slip as homework.
|
# Chapter 1: Right Triangles and an Introduction to Trigonometry
Difficulty Level: At Grade Created by: CK-12
Chapter Outline
## Chapter Summary
In this chapter students learned about right triangles and special right triangles. Through the special right triangles and the Pythagorean Theorem, the study of trigonometry was discovered. Sine, cosine, tangent, secant, cosecant, and cotangent are all functions of angles and the result is the ratio of the sides of a right triangle. We learned that only our special right triangles generate sine, cosine, tangent values that can be found without the use of a scientific calculator. When incorporating the trig ratios and the Pythagorean Theorem, we discovered the first of many trig identities. This concept will be explored further in Chapter 3.
## Vocabulary
A side adjacent to an angle is the side next to the angle. In a right triangle, it is the leg that is next to the angle.
Angle of depression
The angle between the horizontal line of sight, and the line of sight down to a given point.
Angle of elevation
The angle between the horizontal line of sight, and the line of sight up to a given point.
Bearings
The direction from one object to another, usually measured as an angle.
Clinometer
A device used to measure angles of elevation or depression.
Coterminal angles
Two angles in standard position are coterminal if they share the same terminal side.
Distance Formula
\begin{align*}d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\end{align*}
Hypotenuse
The hypotenuse is the longest side in a right triangle, opposite the right angle.
Identity
An identity is an equation that is always true, as long as the variables and expressions involved are defined.
Included Angle
The angle inbetween two sides of a polygon.
Leg
The legs of a right triangle are the two shorter sides.
Nautical Mile
A nautical mile is a unit of length that corresponds approximately to one minute of latitude along any meridian. A nautical mile is equal to 1.852 meters.
Pythagorean Theorem
\begin{align*}a^2 + b^2 = c^2\end{align*}
Pythagorean Triple
A set whole numbers for which the Pythagorean Theorem holds true.
A quadrantal angle is an angle in standard position whose terminal side lies on an axis.
The radius of a circle is the distance from the center of the circle to the edge. The radius defines the circle.
Reference angle
The reference angle of an angle in standard position is the measure of the angle between the terminal side and the closest portion of the \begin{align*}x-\end{align*}axis.
Standard position
An angle in standard position has its initial side on the positive \begin{align*}x-\end{align*}axis, its vertex at the origin, and its terminal side anywhere in the plane. A positive angle means a counterclockwise rotation. A negative angle means a clockwise rotation.
Theodolite
A device used to measure angles of elevation or depression.
Unit Circle
The unit circle is the circle with radius 1 and center (0, 0). The equation of the unit circle is \begin{align*}x^2 + y^2 = 1\end{align*}
## Review Questions
1. One way to prove the Pythagorean Theorem is through the picture below. Determine the area of the square two different ways and set each equal to each other.
2. A flute is resting diagonally, \begin{align*}d\end{align*}, in the rectangular box (prism) below. Find the length of the flute.
3. Solve the right triangle.
4. Solve the right triangle.
5. Find the exact value of the area of the parallelogram below.
6. The modern building shown below is built with an outer wall (shown on the left) that is not at a 90-degree angle with the floor. The wall on the right is perpendicular to both the floor and ceiling. Find the length of \begin{align*}w\end{align*}.
7. Given that \begin{align*}\cos(90^\circ-x) = \frac{2}{7}\end{align*}, find the \begin{align*}\sin x\end{align*}.
8. If \begin{align*}\cos(-x) = \frac{3}{4}\end{align*} and \begin{align*}\tan x = \frac{\sqrt{7}}{3}\end{align*}, find \begin{align*}\sin(-x)\end{align*}.
9. If \begin{align*}\sin y = \frac{1}{3}\end{align*}, what is \begin{align*}\cos y\end{align*}?
10. \begin{align*}\sin \theta = \frac{1}{3}\end{align*} find the value(s) of \begin{align*}\cos \theta\end{align*}.
11. \begin{align*}\cos \theta = -\frac{2}{5}\end{align*}, and \begin{align*}\theta\end{align*} is a second quadrant angle. Find the exact values of remaining trigonometric functions.
12. (3, -4) is a point on the terminal side of \begin{align*}\theta\end{align*}. Find the exact values of the six trigonometric functions.
13. Determine reference angle and two coterminal angles for \begin{align*}165^\circ\end{align*}. Plot the angle in standard position.
14. It is very helpful to have the unit circle with all the special values on one circle. Fill out the unit circle below with all of the endpoints for each special value and quadrantal value.
## Texas Instruments Resources
In the CK-12 Texas Instruments Trigonometry FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9699.
Show Hide Details
Description
Authors:
Tags:
Subjects:
|
# Kuartil
Dina statistik déskriptif, kuartil nyaéta hiji tina tilu nilai nu ngabagi susunan data kana opat bagéan.
Mangka:
• kuartil kahiji = kuartil handap = motong data sahandapeun 25% = 25th persentil
• kuartil kadua = median = motong satengahna data = 50th persentil
• kuartil katilu = upper quartile = motong data 25% ka luhur, atawa handapeun 75% = 75th percentile
béda antara kuartil luhur jeung handap disebut interquartile range.
Ilaharna penting keur interpolasi antara nilai keur ngalengkepan ieu, siga conto di handap ieu.
i x[i]
1 102
2 105
------------- kuartil kahiji, Q1 = (105+106)/2 = 105.5
3 106
4 109
------------- kuartil kadua, Q2 = (109+110)/2 = 109.5
5 110
6 112
------------- kuartil katilu, Q3 = (112+115)/2 = 113.5
7 115
8 118
Nyokot nilai méan sisi séjén tina kuartil mangrupa kaputusan teu pasti: dina conto di luhur, nilai kuartil kudu aya dina rentang [105,106], [109,110] and [112, 115].
If the sample size is not a multiple of four, some of the quartiles may be numbers in the original data set, as in this example:
i x[i]
1 102
2 105—Q[1] = 105
3 106
------------- Q[2] = 107.5
4 109
5 110—Q[3] = 110
6 112
In both of the above cases, the first and third quartiles can be taken to be the median values of the lower and upper halves of the data, respectively. However, there is more than one school of thought on how to apply this definition when the overall median is one of the original data values. The next two examples are illustrations of some of the rules of thumb which have been adopted; neither always produces correct results, and it would be better to use a precise formulation as shown later.
One may include the median in both "halves" of the data:
i x[i]
1 102
2 105
3 106—Q1 = 106
4 109
5 110
)- Q2 = 110 (note line 5 has been duplicated
5 110 to illustrate the point)
6 112
7 115—Q3 = 115
8 118
9 120
Or not include the median in either "half":
i x[i]
1 102
2 105
------------- Q1 = 105.5
3 106
4 109
5 110—Q2 = 110
6 112
7 115
------------- Q3 = 116.5
8 118
9 120
More precise mathematical formulations are possible: the quartiles of the distribution of a random variable X can be defined as the values x such that:
${\displaystyle P(X\leq x)\geq {\frac {1}{4}}\ and\ P(X\geq x)\geq {\frac {3}{4}};}$
${\displaystyle P(X\leq x)\geq {\frac {1}{2}}\ and\ P(X\geq x)\geq {\frac {1}{2}};\ {\rm {or}}}$
${\displaystyle P(X\leq x)\geq {\frac {3}{4}}\ and\ P(X\geq x)\geq {\frac {1}{4}}.}$
With these definitions the quartiles in the last example are 106, 110 and 115:
P(X ≤ 106) = 1/3 and P(X ≥ 106) = 7/9;
P(X ≤ 110) = 5/9 and P(X ≥ 110) = 5/9; and
P(X ≤ 115) = 7/9 and P(X ≥ 115) = 1/3.
|
# How do you find the limit of (a^t-1)/t as t->0?
Nov 5, 2017
The limit is $= \ln a$
#### Explanation:
We calculate the limit as follows
${\lim}_{t \to 0} \frac{{a}^{t} - 1}{t} = \frac{{a}^{0} - 1}{0} = \frac{0}{0}$
This is an indeterminate form, so apply l'Hôspital's rule
${\lim}_{t \to 0} \frac{{a}^{t} - 1}{t} = {\lim}_{t \to 0} \frac{\left({a}^{t} - 1\right) '}{t '}$
Let $y = {a}^{t}$
Taking logarithm on both sides
$\ln y = \ln \left({a}^{t}\right) = t \ln a$
Differentiating
$\frac{\mathrm{dy}}{y} = \ln a \mathrm{dt}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = y \ln a = {a}^{t} \ln a$
Therefore,
${\lim}_{t \to 0} \frac{\left({a}^{t} - 1\right) '}{t '} = {\lim}_{t \to 0} \left(\frac{{a}^{t} ' - 1 '}{t '}\right)$
$= {\lim}_{t \to 0} \frac{\left({a}^{t} \ln a - 0\right)}{1}$
$= \ln a$
|
# ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18
## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18
Question 1.
If A is an acute angle and sin A = $$\\ \frac { 3 }{ 5 }$$ find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
Question 2.
If A is an acute angle and sec A = $$\\ \frac { 17 }{ 8 }$$, find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
Question 3.
Express the ratios cos A, tan A and sec A in terms of sin A.
Solution:
Question 4.
If tan A = $$\frac { 1 }{ \sqrt { 3 } }$$, find all other trigonometric ratios of angle A.
Solution:
Question 5.
If 12 cosec θ = 13, find the value of $$\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta }$$
Solution:
Without using trigonometric tables, evaluate the following (6 to 10):
Question 6.
(i) cos² 26° + cos 64° sin 26° + $$\frac { tan{ 36 }^{ ° } }{ { cot54 }^{ ° } }$$
(ii) $$\frac { sec{ 17 }^{ ° } }{ { cosec73 }^{ ° } } +\frac { tan68^{ ° } }{ cot22^{ ° } }$$ + cos² 44° + cos² 46°
Solution:
Question 7.
(i) $$\frac { sin65^{ ° } }{ { cos25 }^{ ° } } +\frac { cos32^{ ° } }{ sin58^{ ° } }$$ – sin 28° sec 62° + cosec² 30° (2015)
(ii) $$\frac { sin29^{ ° } }{ { cosec61 }^{ ° } }$$ + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).
Solution:
Question 8.
(i) $$\frac { { sin }35^{ ° }{ cos55 }^{ ° }+{ cos35 }^{ ° }{ sin }55^{ ° } }{ { cosec }^{ 2 }{ 10 }^{ ° }-{ tan }^{ 2 }{ 80 }^{ ° } }$$
(ii) $${ sin }^{ 2 }{ 34 }^{ ° }+{ sin }^{ 2 }{ 56 }^{ ° }+2tan{ 18 }^{ ° }{ tan72 }^{ ° }-{ cot }^{ 2 }{ 30 }^{ ° }$$
Solution:
Question 9.
(i) $${ \left( \frac { { tan25 }^{ ° } }{ { cosec }65^{ ° } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ ° } }{ { sec65 }^{ ° } } \right) }^{ 2 }+{ 2tan18 }^{ ° }{ tan }45^{ ° }{ tan72 }^{ ° }$$
(ii) $$\left( { cos }^{ 2 }25+{ cos }^{ 2 }65 \right) +cosec\theta sec\left( { 90 }^{ ° }-\theta \right) -cot\theta tan\left( { 90 }^{ ° }-\theta \right)$$
Solution:
Question 10.
(i) 2(sec² 35° – cot² 55°) – $$\frac { { cos28 }^{ ° }cosec{ 62 }^{ ° } }{ { tan18 }^{ ° }tan{ 36 }^{ ° }{ tan30 }^{ ° }{ tan54 }^{ ° }{ tan72 }^{ ° } }$$
(ii) $$\frac { { cosec }^{ 2 }(90-\theta )-{ tan }^{ 2 }\theta }{ 2({ cos }^{ 2 }{ 48 }^{ ° }+{ cos }^{ 2 }{ 42 }^{ ° }) } -\frac { { 2tan }^{ 2 }{ 30 }^{ ° }{ sec }^{ 2 }{ 52 }^{ ° }{ sin }^{ 2 }{ 38 }^{ ° } }{ { cosec }^{ 2 }{ 70 }^{ ° }-{ tan }^{ 2 }{ 20 }^{ ° } }$$
Solution:
Question 11.
Prove that following:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
(ii) $$\frac { tan\theta }{ tan({ 90 }^{ ° }-\theta ) } +\frac { sin({ 90 }^{ ° }-\theta ) }{ cos\theta } ={ sec }^{ 2 }\theta$$
(iii) $$\frac { cos({ 90 }^{ ° }-\theta )cos\theta }{ tan\theta } +{ cos }^{ 2 }({ 90 }^{ ° }-\theta )=1$$
(iv) sin (90° – θ) cos (90° – θ) = $$\frac { tan\theta }{ { 1+tan }^{ 2 }\theta }$$
Solution:
Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:
Question 12.
(i) (sec A + tan A) (1 – sin A) = cos A
(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1.
Solution:
Question 13.
(i) tan A + cot A = sec A cosec A
(ii) (1 – cos A)(1 + sec A) = tan A sin A.
Solution:
Question 14.
(i) $$\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A$$
(ii) $$\frac { 1 }{ secA+tanA } +\frac { 1 }{ secA-tanA } =2{ sec }A$$
Solution:
Question 15.
(i) $$\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA }$$
(ii) $$\frac { 1-{ tan }^{ 2 }A }{ { cot }^{ 2 }A-1 } ={ tan }^{ 2 }A$$
(iii) $$\frac { sinA }{ 1+cosA } =cosecA-cotA$$
Solution:
Question 16.
(i) $$\frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA }$$
(ii) $$\frac { { tan }^{ 2 }\theta }{ { (sec\theta -1) }^{ 2 } } =\frac { 1+cos\theta }{ 1-cos\theta }$$
(iii) $${ (1+tanA) }^{ 2 }+{ (1-tanA) }^{ 2 }=2{ sec }^{ 2 }A$$
(iv) $${ sec }^{ 2 }A+{ cosec }^{ 2 }A={ sec }^{ 2 }A{ .cosec }^{ 2 }A$$
Solution:
Question 17.
(i) $$\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA$$
(ii) $$\frac { tanA }{ secA-1 } +\frac { tanA }{ secA+1 } =2cosecA$$
Solution:
Question 18.
(i) $$\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A$$
(ii) $$cotA-tanA=\frac { { 2cos }^{ 2 }A-1 }{ sinA-cosA }$$
(iii) $$\frac { cotA-1 }{ 2-{ sec }^{ 2 }A } =\frac { cotA }{ 1+tanA }$$
Solution:
Question 19.
(i) $${ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta$$
(ii) $$\frac { cos\theta }{ 1-tan\theta } -\frac { { sin }^{ 2 }\theta }{ cos\theta -sin\theta } =cos\theta +sin\theta$$
Solution:
Question 20.
(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
(ii) 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ = cot4 θ – tan4 θ.
Solution:
Question 21.
(i) $$\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta$$
(ii) $$\frac { { tan }^{ 3 }\theta -1 }{ tan\theta -1 } ={ sec }^{ 2 }\theta +tan\theta$$
Solution:
Question 22.
(i) $$\frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA }$$
(ii) $$\sqrt { \frac { 1-cosA }{ 1+cosA } } =\frac { sinA }{ 1+cosA }$$
Solution:
Question 23.
(i) $$\sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA$$
(ii) $$\sqrt { \frac { 1-cosA }{ 1+cosA } } =cosecA-cotA$$
Solution:
Question 24.
(i) $$\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA$$
(ii) $$\frac { cotAcotA }{ 1-sinA } =1+cosecA$$
Solution:
Question 25.
(i) $$\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)$$
(ii) $${ sec }^{ 4 }A-{ tan }^{ 4 }A=1+2{ tan }^{ 2 }A$$
Solution:
Question 26.
(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
(ii) sec6 A – tan6 A = 1 + 3 tan2 A + 3 tan4 A
Solution:
Question 27.
(i) $$\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta }$$
(ii) $$\frac { sin\theta }{ cot\theta +cosec\theta } =2+\frac { sin\theta }{ cot\theta -cosec\theta }$$
Solution:
Question 28.
(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
(ii) (cosecA – sinA)(secA – cosA) sec2A = tanA
(iii) (cosecθ – sinθ)(secθ – cosθ)(tan θ + cotθ) = 1
Solution:
Question 29.
(i) $$\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2$$
(ii) $$\frac { { tan }^{ 2 }A }{ { 1+tan }^{ 2 }A } +\frac { cot^{ 2 }A }{ 1+{ cot }^{ 2 }A } =1$$
Solution:
Question 30.
(i) $$\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA }$$
(ii) $${ (sinA+secA) }^{ 2 }+{ (cosA+cosecA) }^{ 2 }={ (1+secA\quad cosecA) }^{ 2 }$$
(iii) $$\frac { tanA+sinA }{ tanA-sinA } =\frac { secA+1 }{ secA-1 }$$
Solution:
Question 31.
If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1
Solution:
Question 32.
If 7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.
Solution:
Question 33.
If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.
Solution:
Question 34.
If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 – y2 = a2 – b2.
Solution:
Question 35.
If x = h + a cos θ and y = k + a sin θ, prove that (x – h)2 + (y – k)2 = a2.
Solution:
|
# 2002 AMC 12A Problems/Problem 6
The following problem is from both the 2002 AMC 12A #6 and 2002 AMC 10A #4, so both problems redirect to this page.
## Problem
For how many positive integers $m$ does there exist at least one positive integer n such that $m \cdot n \le m + n$?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ }$ infinitely many
## Solution
### Solution 1
For any $m$ we can pick $n=1$, we get $m \cdot 1 \le m + 1$, therefore the answer is $\boxed{\text{(E) infinitely many}}$.
### Solution 2
Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.
$(m-1)(n-1) \leq 1$
Let $n=1$, then
$0 \leq 1$
This means that there are infinitely many numbers $m$ that can satisfy the inequality. So the answer is $\boxed{\text{(E) infinitely many}}$.
### Solution 3
If we subtract $n$ from both sides of the equation, we get $m \cdot n - n \le m$. Factor the left side to get $(m - 1)(n) \le m$. Divide both sides by $(m-1)$ and we get $n \le \frac {m}{m-1}$. The fraction $\frac {m}{m-1} > 1$ if $m > 1$. There is an infinite amount of integers greater than 1, therefore the answer is $\boxed{\text{(E) infinitely many}}$.
|
# NCERT Circles Class 10 Exercise 10.2 Solution
Theorem 10.2
prove that ” The Lengths of tangents drawn from an external points to a circle are equal.”
Given:
C ( O , r )
P = External point from the circle
P A and P B are tangents From External Point P
Construction:
Join External Point P , Point of contacts A and B to Center O .
Proof:
In Δ PAO and Δ PBO
P O = P O { Common Sides }
O A = O B { Radii Of Same Circle }
∠PAO = ∠PBO = 90º { Radius is always perpendicular (⊥) on Point of contact of tangent by Theorem 10.1 }
Δ PAO ≅ Δ PBO { By R.H.S. }
P A = P B { By CPCT }
Hence Proved
Key Points
• ∠OPA = ∠OPB { By CPCT }
• ∠POA = ∠POB { By CPCT }
• P O is the angle bisector of ∠APB and ∠AOB
## Exercise – 10.2
In Question 1 to 3 , Choose the correct option and give justification.
Q. 1. From a point Q , the length of the tangent to a circle is 24cm and the distance of Q
From the center is 25cm.The radius of the circle is:
(A) 7cm
(B) 12cm
(C) 15cm
(D) 24.5cm
Explanation:
In Δ QPO
∠P = 90º
By PGT
(OP)² = (OQ)² – (PQ)²
=> (OP)² = (25)² – (24)²
=> (OP)² = 625 – 576
=> OP = √49
=> OP = 7
Hence Radius if circle = 7cm
Q.2. In figure 10.11, if T P and T Q are the two tangents to a circle with center O so that ∠POQ = 110º, Then ∠PTO is equal to :
(A) 60º
(B) 70º
(C) 80º
(D) 90º
According to figure
∠TPO = ∠TQO = 90º { By theorem 10.1 radius is perpendicular on point of contact to the tangent and circle }
∴ ∠POQ + ∠PTQ = 180º
=> 110º + ∠PTQ = 180º
=> ∠PTQ = 180º – 110º
=> ∠PTQ = 70º
Q.3. If tangents P A and P B From a Point P to a circle with center O are inclined to each other at angle of 80º, then ∠POA is equal to
(A) 50º
(B) 60º
(C) 70º
(D) 80º
Explanation:
According to figure
∠APB + ∠AOB = 180º
=> 80º + ∠AOB = 180º
=> ∠AOB = 180º – 80º
=> ∠AOB = 100º
=> ∠POA = ∠POB = 100º/2= 50º { ∠AOB/2 }
Q.4. Prove that the tangents drawn at the ends if a diameter of a circle are parallel.
Given:
According to figure
Circle ( O , r )
X Y and A B are tangents
To Prove:
X Y // A B { X Y parallel to A B }
Construction:
Join points of contacts P and Q to Center O.
Proof:
In figure
∠1 =∠2 { Radii is always perpendicular on point of contacts of tangent By theorem 10.1}
Here ∠1 = ∠2 Which is alternate interior angles and P Q is transversal on X Y and A B
∴ X Y // P Q
Hence the tangents drawn at the ends of a diameter of a circle are parallel.
Q.5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.
Given:
According to figure
C (O , r)
A B = tangent which is touches the circle at point P.
∴ P = Point of contact
To Prove:
The Perpendicular at the point of contact to the tangent to a circle passes through the center
Or
O P ⊥ on A B
Construction:
Join point of contact P to center O
Proof:
According to figure
A B = Tangent on circle
P = Point of contact on the circle
We know that by theorem 10.1
“Radius is always ⊥ ( perpendicular) on point of contact of tangent
∴ O P ⊥ on AB
Hence the perpendicular at the point of contact to the tangent to a circle passes through the center.
Q.6. The length of a tangent From a point A at distance 5cm From the center of the circle is 4cm. Find the radius of the circle.
Given:
According to figure
C ( O, r )
A B = Length of Tangent from point A = 4cm
O A = 5cm
To Find:
The radius of circle OB = r=?
Solution:
In ΔABO
∠ABO = 90º
A B = 4cm
A O = 5cm
O B =?
By PGT
(OA)² = (AB)² + (OB)²
=> (5)² = (4)² + (OB)²
=> 25 = 16 + (OB)²
=> 25 – 16 = (OB)²
=> 9 = (OB)²
=>√9 = OB
=> 3 = OB
Hence Radius of circle = 3cm
Q.7. Two concentric circles are of radii 5cm and 3cm.Find the length of the chord of the larger circle which touches the smaller circle.
Given:
According to Figure
Two concentric circles with radius O P = 3cm and O A = 5cm
To Find:
Chord A B =?
Solution:
According to condition
A B = Chord for Larger circle =Tangent for smaller circle
P = Point of contact
Now
In Δ APO
∠APO = 90º
A O = 5 cm
O P = 3cm
A P =?
By PGT
(AO)² = (AP)² + (3)²
=> (5)² = (AP)² + 9
=> 25 – 9 = (AP)²
=> 16 = (AP)²
=> √16 = AP
=> 4 = AP
Radius ( OP ) is always bisect the chord from the point of contact.
A P = P B = 4cm
∴ A B = AP + P B
=> A B = 4 + 4
=> A B = 8cm
Alternate Method:
In Δ APO and ΔBPO
A O = O B { Radii of same circle }
O P = O P {Common side }
∠APO = ∠BPO = 90º
=> Δ APO ≅ Δ BPO { RHS Congruency }
∴ A P = P B { By CPCT }
Hence A P = P B = 4cm
A B = A P + P B
=> A B = 4 + 4
=> A B = 8cm
∴ Length of the chord A B = 8cm.
Q.8. A Quadrilateral ABCD id drawn to circumscribe a circle. (see in figure 10.12)
Prove that A B + C D = A D + B C .
Given:
According to figure
A B , B C , C D and A D are tangents have point of contacts P , Q , R and S respectively.
To Prove:
A B + C D = A D + B C
Proof:
We know that length of tangent from an external point are equal.
∴ A P = A S ———Equation(i)
B P = B Q ———–Equation (ii)
C R = C Q ———–Equation(iii)
D R = D S ———–Equation(iv)
Adding Equation (i) + Equation (ii) + Equation(iii) +Equation(iv)
A P + B P + C R + D R = A S + B Q + C Q + D S
=> A B + CD = A S + D S + B Q + C Q
=> A B + C D = A D + B C
Hence Proved
Q.9. In Figure 10.13, X Y and X’ Y’ are two parallel tangents to a circle with center O and another tangent A B with point of contact C intersecting X Y at A and X’ Y’ at B. Prove that ∠AOB = 90º
Given:
According to figure
diagram
tangents X Y // X’ Y’
To Prove:
∠AOB = 90º
Construction:
Join O to C
Proof:
By Theorem 10.2 :We know that “Length of tangent Drawn From an External point are equal in Length”.
P A = A C ———– Equation (i)
In ΔQBO and ΔBCO
∠BQO = ∠BCO = 90º { By Theorem 10.1 Radius is always ⊥ on point of contact }
O B = O B {common side }
O Q = O C { Radii of same circle }
ΔQBO ≅ Δ BCO { By RHS Congruence }
∠1 = ∠2 ——————–(i)
Similarly
In Δ PAO and Δ CAO
P A = C A { From Equation (i) }
A O = A O { common side }
O P = O C { Radii of same circle }
∴ Δ PAO ≅ Δ CAO { By SSS congruence }
∠ PAO = ∠ CAO { By CPCPT }
=> ∠3 = ∠5 ———————(ii)
We know that sum of angles subtended same side on a line at a point are 180º
∠1 + ∠2 + ∠3 + ∠5 = 180°
=> ∠2 + ∠2 + ∠3 + ∠3 = 180º
=> 2∠2 + 2∠3 = 180º
=> 2(∠2 + ∠3) = 180º
=> ∠2 + ∠3 = 180º/2
=> ∠2 + ∠3 = 90º
=> ∠AOB = 90º
Hence Proved
Q.10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line – segment joining the points of contact at the center .
Given:
According to figure
di
P A and P B are tangents from an external point P
To Prove:
∠APB + ∠AOB = 180
Proof:
In figure
∠PAO = ∠PBO = 90º { Radii is always ⊥ on point of contact }
∠APB + ∠PAO + ∠AOB + ∠PBO = 360º
=> ∠APB + 90º + ∠AOB + 90° = 360º
=> ∠APB + ∠AOB + 180º = 360º
=> ∠APB + ∠AOB = 360º – 180º
=> ∠APB + ∠AOB = 180º
Hence Proved
Q.11. Prove that the parallelogram circumscribing a circle is rhombus.
Given:
According to figure
A B // D C and A B =C D
A D // B C and A D = B C
Quadrilateral ABCD is circumscribing a circle with center O
To Prove:
Parallelogram ABCD is a rhombus
OR
A B = B C = C D = A D
Proof:
According to figure:
We know that By theorem 10.2 “Length of tangents drawn from an external point are equal”
B P = B Q ———–Equation(i)
A P = A S ————Equation (ii)
C R = C Q ———–Equation (iii)
D R = D S ———–Equation (iv)
Adding Equation (i) + Equation (ii) + Equation (iii) + Equation (iv)
B P + A P + C R + D R = B Q + A S + C Q + D S
=> A B + C D = B Q + C Q + A S + D S
=> A B + C D = B C + A D
=> 2 A B = 2 B C
=> A B = B C
∴ A B = B C = C D = A D
Hence parallelogram ABCD is a rhombus.
Q.12. A Triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segments B D and D C into which B C is divided by the point of contact D are of length 8cm and 6cm respectively (see fig. 10.14).Find the sides A B and A C.
Given:
In ΔABC:
ΔABC is drawn to circumscribe a circle if radius 4cm
To Find:
Length of side A B = ?
Length of side A C =?
Construction:
Join vertices A , B and C to center O
BC = 6 + 8 = 14 cm
We know that tangents drawn from an external point are equal in length.
∴ A F = A E = X (let)
C D = C F = 6cm
B D = B E = 8cm
a = A B = (x + 8 );
b = B C = 14 cm
c = A C = (6+x)
By Heron’s Formula
S = (A B + B C + A C)/2
=> S =(a + b + c)/2
=> S = ( x + 8 + 14 + 6 + x )/2
=> S = ( 2x + 28 )/2
=> S = 2(x + 14)/2
=> S = 2(x + 14)/2
=> S = x + 14
s – a = x + 14 -( x + 8)
=> s – a = x + 14 – x – 8
=> s – a = 6
s – b = x + 14 – 14
=> s – b = x
s – c = x + 14 – ( x + 6)
=> s – c = x + 14 – x – 6
=> s – c = 8
Area of Δ ABC = √s(s-a)(s-b)(s-c)
=> Area of Δ ABC = √(x+14)(x)(6)(8)
=> Area of Δ ABC = √(x+14)48x ————–Equation(i)
Area of Δ ABC = Area of Δ AOB + Area of Δ BOC + Area of Δ AOC
=> Area of Δ ABC = 1/2 × b × h + 1/2 × b × h + 1/2 × b × h
=> Area of Δ ABC = 1/2 × A B × 4 + 1/2 × B C × 4 + 1/2 × A C × 4
=> Area of Δ ABC = 1/2 ×4 ( A B + B C + A C )
=> Area of Δ ABC = 2 ( A B + B C + A C )
=> Area of Δ ABC = 2 ( 8 + x + 14 + x + 6 )
=> Area of Δ ABC = 2 (2x + 28)
=> Area of Δ ABC = 2×2(x + 14)
=> Area of Δ ABC = 4 ( x + 14) ———-Equation(ii)
Equation (i) = Equation (ii)
Area of Δ ABC = Area of Δ ABC
√(x+14)48x = 4 ( x + 14)
squaring both side
=> (x+14)48x = 16 ( x + 14)²
=> 48x = 16 ( x + 14)²/(x+14)
=> 48x = 16(x + 14)
=> 48x/16 = (x+ 14)
=> 3x = x + 14
=> 3x – x = 14
=> 2x = 14
=> x = 14/2
=> x = 7
Hence
A B = x + 8 = 7 + 8 = 15cm
B C = x + 6 = 6 + 7 = 13cm
Q.13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angle at the center of the circle.
Given:
According to Diagram:
ABCD is a quadrilateral circumscribing a circle with center O
tangents A B , B C , C D , and A D have point of contact P, Q, R and S respectively.
To Prove:
∠AOB + ∠COD = 180º
∠AOD + ∠BOC = 180º
Proof:
Angle subtend by the tangents external point on the center of circle are equal.
∠2 = ∠3 ——-Equation(i)
∠1 = ∠8 ——-Equation(ii)
∠4 = ∠5 ——-Equation(iii)
∠6 = ∠7 ——-Equation(iv)
We know that sum of all angles subtended on a point are 360°
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º
=> ∠1 + ∠2 + ∠2 + ∠5 + ∠5 + ∠6 + ∠6 + ∠1 = 360º { From Equation (i),(ii),(iii) and (iv) }
=> 2∠1 + 2∠2 + 2∠5 + 2∠6 = 360°
=> 2(∠1 + ∠2 + ∠5 + ∠6) = 360º
=> 2 ( ∠AOB + ∠COD ) = 360° { ∠1 + ∠2 = ∠AOB ; ∠5 + ∠6 = ∠COD }
=> ∠AOB + ∠COD = 360°/2
=> ∠AOB + ∠COD = 180º
Similarly
∠AOD + ∠BOC = 180º
Hence Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
|
# 180 Days of Math for Fifth Grade Day 179 Answers Key
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 179 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fifth Grade Answers Key Day 179
Directions: Solve each problem.
Question 1.
248 + 39 = ___________
The addition is the term used to describe adding two or more numbers together. The addition is denoted using the plus sign ‘+‘ such as the addition of 3 and 3 can be written as 4 + 4. Also, the plus sign (+) can be used as many times as required, such as
4 + 4 + 4 + 4.
Question 2.
93 × 54 = ___________
Explanation:
In mathematics, multiplication is a method of finding the product of two or more numbers. It is one of the basic arithmetic operations, that we use in everyday life. The major application we can see in multiplication tables.
In arithmetic, the multiplication of two numbers represents the repeated addition of one number with respect to another. These numbers can be whole numbers, natural numbers, integers, fractions, etc. If m is multiplied by n, then it means either m is added to itself ‘n’ number of times or vice versa.
The formula for multiplication:
The multiplication formula is given by:
Multiplier × Multiplicand = Product
– The multiplicand is the total number of objects in each group
– A multiplier is the number of equal groups
– Product is the result of multiplication of multiplier and multiplicand
Question 3.
The square root of a number is defined as the value, which gives the number when it is multiplied by itself. The radical symbol √ is used to indicate the square root. For example, √9 = 3. The radical symbol is also called a root symbol or surds. If a number is a perfect square, we can easily find the square root of the number. If the given number is not a perfect square number, the square root can be found using the long division method.
– Here we will show you how to calculate the square root of 891 using the long division method with one decimal place accuracy.
Step 1: Set up 891 in pairs of two digits from right to left and attach one set of 00 because we want one decimal:
step 2: Starting with the first set: the largest perfect square less than or equal to 8 is 4, and the square root of 4 is 2. Therefore, put 2 on top and 4 at the bottom like this:
step 3: Calculate 8 minus 4 and put the difference below. Then move down the next set of numbers.
step 4: Double the number on top: 2 × 2 = 4. Then, use 4 and the bottom number to make this problem:
4? × ? ≤ 491
The question marks are “blank” and the same “blank”. With trial and error, we found the largest number “blank” can be is 9. Replace the question marks in the problem with 9 to get:
49 × 9 = 441.
Now, enter 9 on top, and 441 at the bottom:
step 5: Calculate 491 minus 441 and put the difference below. Then move down the next set of numbers.
step 6: Double the number in green on top: 29 × 2 = 58. Then, use 58 and the bottom number to make this problem:
58? × ? ≤ 5000
The question marks are “blank” and the same “blank”. With trial and error, we found the largest number “blank” can be is 8. Now, enter 8 on top:
The square root of 891 with one digit decimal accuracy is 29.8. You can add decimals by simply adding more sets of 00 and repeating the last two steps over and over.
The above-given question is 84 square root of 891
The square root of 891 is 29.8
Now multiply 84 and 29.8
Thus, 84 * 29.8 = 2503.2
Question 4.
Round 138,492 to the nearest thousand.
The above-given number is 138,492
The digit at the hundredth place of 138,492 is 4
That means 4 is less than 5.
So change digits at hundredth, tenth, and one’s place to zero.
Obtained rounded number for 138492 is 138000.
Question 5.
Write 2$$\frac{3}{4}$$ as an improper fraction.
2 3/4 is a mixed number, also known as a mixed fraction. It has a whole number and a proper fraction. The numbers in the mixed fraction are defined as follow:
2 = whole number
3 = numerator
4 = denominator
To convert mixed number 2 3/4 to an improper fraction, you follow these steps:
– Multiply the whole number by the denominator
2 × 4 = 8
– Add the product from Step 1 to the numerator
8 + 3 = 11
– Write answer from Step 2 over the denominator
11/4
The mixed number 2 3/4 converted to an improper fraction is, therefore :11/4
Question 6.
30 – (10 × 2) = _________
The value of 10 × 2 = 20
30-20=10
Question 7.
– What I have done is I subtracted 67 from 192 then we get 125.
– For re-verification 125+67 then we get 192.
Question 8.
8 cups = _________ pints
8 cups are equal to 4 pints
The conversion factor from cups to pints is 0.5, which means that 1 cup is equal to 0.5 pints:
1 cup=0.5 pt
To convert 8 cups into pints we have to multiply 8 by the conversion factor in order to get the volume amount from cups to pints. We can also form a simple proportion to calculate the result:
We conclude that 8 cups is equivalent to 4 pints:
8 cups = 4 pints
Question 9.
Use a protractor to draw an acute angle.
Acute angle comes under less than 90 degrees.
An angle is formed when two rays meet at a point. In the given figure, the point of intersection of the ray OA and ray OB is O, which is called the vertex.
If the measure of the angle formed by two rays is 30 degrees, then the angle is called a 30-degree angle. The angle formed by the ray OA and OB are written as ∠AOB or ∠BOA.
∠AOB=∠BOA=30°.
constructing 30 – degree angle using protractor:
step 1: Draw a line segment OA.
step 2: Place the centre tip of the protractor at point O such that the protractor perfectly aligns with line AO.
step 3: Start from ‘A’ on the protractor in the clockwise direction and stop at 30. Mark it as point ‘D’. If point ‘A’ lies to the right of ‘O’, then start measuring anticlockwise and stop at 30.
step 4: Joinpoint ‘D’ with ‘O’. ∠AOD=30° is the required 30-degree angle.
Question 10.
What is the median of this data set?
1,425; 1,595; 1,392; 1,436
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set. To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Now arrange the given numbers in ascending order:
1392, 1425, 1436, 1595
median=1425+1436/2
median=2861/2
median=1430.5
Therefore, the median of the data set is 1430.5
Question 11.
Imagine that you write each letter of the word CALIFORNIA on individual cards. You shuffle them, turn them facedown on a table, and turn over the top card. What is the probability of turning over a number?
Probability can be defined as the ratio of the number of favourable outcomes to the total number of outcomes of an event. For an experiment having ‘n’ number of outcomes, the number of favourable outcomes can be denoted by x. The formula to calculate the probability of an event is as follows.
Probability(Event) = Favorable Outcomes/Total Outcomes = x/n
The given letter is CALIFORNIA
The total number of letters=10
Therefore, the total outcomes are 10
Turning over the top card=1
The favourable outcomes are 1.
According to the definition:
P(turning over a number)=1/10
Question 12.
How many squares of any size are there in the image?
|
Basic Probability Concepts
BASIC PROBABILITY CONCEPTS
Probability is the area of mathematics devoted to predicting the likelihood of uncertain occurrences.
For example, when you roll a die, it is uncertain whether you'll see a $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, or $\,6\,$.
However, it is possible to talk about how likely it is for each number to occur.
DEFINITIONS probability, experiment, outcome, sample space, event
A probability is a number between $\,0\,$ and $\,1\,$
that represents the likelihood of occurrence of something.
An experiment is the act of making an observation or taking a measurement.
An outcome is one of the possible things that can occur as a result of an experiment.
The set of all possible outcomes for an experiment is called its sample space,
and is frequently denoted by $\,S\,$.
Any subset of the sample space is called an event, and is frequently denoted by $\,E\,$.
The probability of an event $\,E\,$ is denoted by $\,P(E)\,$.
Given any set $\,A\,$, the notation $\,n(A)\,$ denotes the number of elements in $\,A\,$.
A die is a cube with the numbers $\,1\,$ through $\,6\,$ represented on its six faces. When it is thrown (‘rolled’), one of these six faces appears on top. For a fair die, each of the numbers $\,1\,$ through $\,6\,$ is equally likely to occur.
There are lots of experiments involving a single fair die.
Here are some of them, with their corresponding outcomes and sample spaces:
EXPERIMENT OUTCOMES SAMPLE SPACE (1) Roll once;record the number that appears on the top face. six possible outcomes: $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, $\,6\,$ $S = \{1,2,3,4,5,6\}$ (2) Roll once;record the number that appears on the bottom (hidden) face. six possible outcomes: $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, $\,6\,$ $S = \{1,2,3,4,5,6\}$ (3) Roll twice;record the numbers on each of these rolls, in order. In this experiment and those below, use the notation $\,1\star5\,$ to denote a roll of $\,1\,$, followed by a roll of $\,5\,$. $36\,$ possible outcomes: $\,1\star1\,$, $\,1\star2\,$, $\,1\star3\,$, $\,1\star4\,$, $\,1\star5\,$, $\,1\star6\,$ $\,2\star1\,$, $\,2\star2\,$, $\,2\star3\,$, $\,2\star4\,$, $\,2\star5\,$, $\,2\star6\,$ $\,3\star1\,$, $\,3\star2\,$, $\,3\star3\,$, $\,3\star4\,$, $\,3\star5\,$, $\,3\star6\,$ $\,4\star1\,$, $\,4\star2\,$, $\,4\star3\,$, $\,4\star4\,$, $\,4\star5\,$, $\,4\star6\,$ $\,5\star1\,$, $\,5\star2\,$, $\,5\star3\,$, $\,5\star4\,$, $\,5\star5\,$, $\,5\star6\,$ $\,6\star1\,$, $\,6\star2\,$, $\,6\star3\,$, $\,6\star4\,$, $\,6\star5\,$, $\,6\star6\,$ $S = \{$ $\,1\star1\,$, $\,1\star2\,$, $\,1\star3\,$, $\,1\star4\,$, $\,1\star5\,$, $\,1\star6\,$, $\,2\star1\,$, $\,2\star2\,$, $\,2\star3\,$, $\,2\star4\,$, $\,2\star5\,$, $\,2\star6\,$, $\,3\star1\,$, $\,3\star2\,$, $\,3\star3\,$, $\,3\star4\,$, $\,3\star5\,$, $\,3\star6\,$, $\,4\star1\,$, $\,4\star2\,$, $\,4\star3\,$, $\,4\star4\,$, $\,4\star5\,$, $\,4\star6\,$, $\,5\star1\,$, $\,5\star2\,$, $\,5\star3\,$, $\,5\star4\,$, $\,5\star5\,$, $\,5\star6\,$, $\,6\star1\,$, $\,6\star2\,$, $\,6\star3\,$, $\,6\star4\,$, $\,6\star5\,$, $\,6\star6\,$ $\}$ (4) Roll twice;record the sum of the numbers that appear. $\,11\,$ possible outcomes: The smallest sum you can get is $\,2\,$; this results only from rolling a $\,1\,$ followed by a $\,1\,$. The largest sum you can get is $\,12\,$; this results only from rolling a $\,6\,$ followed by a $\,6\,$. Convince yourself that every whole number between $\,2\,$ and $\,12\,$ is also a possible outcome; for example, you could get $\,5\,$ in all these ways: $\,1\star4\,$, $\,2\star3\,$, $\,3\star2\,$, $\,4\star1\,$ $S = \{2,3,4,5,6,7,8,9,10,11,12\}$ (5) Roll twice;record the greatest number that appears on the two rolls. six possible outcomes: $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, $\,6\,$ For $\,1\star1\,$, the greatest number is $\,1\,$. This is the only way to get an outcome of $\,1\,$. For $\,1\star2\,$, the greatest number is $\,2\,$. There are three ways to get the outcome $\,2\,$: $\,1\star 2\,$, $\,2\star 1\,$, $\,2\star2\,$ For $\,1\star3\,$, the greatest number is $\,3\,$. There are five ways to get the outcome $\,3\,$: $\,1\star 3\,$, $\,2\star 3\,$, $\,3\star1\,$, $\,3\star 2\,$, $\,3\star 3\,$ $S = \{1,2,3,4,5,6\}$
NOT ALL SAMPLE SPACES ARE CREATED EQUAL
Some sample spaces are much easier to work with than others.
In experiments (1), (2) and (3) above, each member of the sample space is equally likely:
• In experiments (1) and (2), you're equally likely to roll a $\,1\,$, $\,2\,$, $\,3\,$, $\,4\,$, $\,5\,$, or $\,6\,$.
• In experiment (3), each of the $\,36\,$ possible outcomes are equally likely:
for example, the chance of rolling a $\,1\,$ followed by a $\,3\,$ is (say) the same as rolling a $\,5\,$ followed by a $\,4\,$.
In experiments (4) and (5), however, the outcomes are not all equally likely:
• In experiment (4), there's only one way to get a $\,2\,$.
However, there are four ways to get a $\,5\,$.
So, the outcome $\,2\,$ is less likely than the outcome $\,5\,$.
• In experiment (5), there's only one way to get the the outcome $\,1\,$.
However, there are five ways to get the outcome $\,3\,$.
So, the outcome $\,1\,$ is less likely than the outcome $\,3\,$.
When a sample space has equally likely outcomes, then computing probabilities is as easy as counting:
EQUALLY LIKELY OUTCOMES
Let $\,S\,$ be a sample space with a finite number of outcomes.
Suppose each outcome in $\,S\,$ is equally likely to occur.
Then, the probability of an event $\,E\,$ is found by taking the number of outcomes in $\,E\,$
and dividing by the number of outcomes in $\,S\,$.
That is: $$P(E) = \frac{n(E)}{n(S)}$$
This is best illustrated by an example.
Let's consider the first (or second) experiment above—a single roll of a fair die.
The example below also clarifies the idea of an ‘event’,
and illustrates notation that is frequently used in connection with probability problems.
EXAMPLE
Consider a single roll of a fair die.
The sample space is $\,S = \{1,2,3,4,5,6\}\,$.
Note that $\,n(S) = 6\,$, since there are $\,6\,$ members in $\,S\,$.
Since this is a fair die, each member of $\,S\,$ is equally likely.
Let $\,x\,$ denote the number that is rolled.
An ‘event’ is a subset of the sample space, so there are many possible events.
Any subset $\,E\,$ of $\,S\,$ will have six or fewer members, so $\,0 \le n(E) \le 6\,$.
The table below illustrates several events, their interpretations, and some conventional language and notation:
event interpretation of event probability of event some conventional language usedto report the probability $E = \{3\}$ getting a $\,3\,$ on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac16$ $P(x = 3) = \frac16$ read as: “The probability that $\,x\,$ is $\,3\,$ is $\,\frac16\,$.” $E = \{4\}$ getting a $\,4\,$ on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac16$ $P(x = 4) = \frac16$ $E = \{3,4\}$ getting a $\,3\,$ or a $\,4\,$ on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac26 = \frac13$ $P(x = 3 \text{ or } x=4) = \frac26 = \frac 13$ $E = \{2,4,6\}$ getting an even number on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac36 = \frac12$ $P(x \text{ is even}) = \frac36 = \frac 12$ $E = \{2,3,4,5,6\}$ getting a number greater than $\,1\,$on a single roll of a fair die $\displaystyle\frac{n(E)}{n(S)} = \frac56$ $P(x \gt 1) = \frac56$
Master the ideas from this section
|
# 2001 AMC 12 Problems/Problem 17
## Problem
A point $P$ is selected at random from the interior of the pentagon with vertices $A = (0,2)$, $B = (4,0)$, $C = (2 \pi + 1, 0)$, $D = (2 \pi + 1,4)$, and $E=(0,4)$. What is the probability that $\angle APB$ is obtuse?
$\text{(A) }\frac {1}{5} \qquad \text{(B) }\frac {1}{4} \qquad \text{(C) }\frac {5}{16} \qquad \text{(D) }\frac {3}{8} \qquad \text{(E) }\frac {1}{2}$
## Solution
The angle $APB$ is obtuse if and only if $P$ lies inside the circle with diameter $AB$. (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.)
$[asy] defaultpen(0.8); real pi=3.14159265359; pair A=(0,2), B=(4,0), C=(2*pi+1, 0), D=(2*pi+1,4), E=(0,4), F=(0,0); draw(A--B--C--D--E--cycle); draw(circle((A+B)/2,length(B-A)/2)); label("A",A,W); label("B",B,SE); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("F",F,SW); draw(A--F--B,dashed); [/asy]$
The area of $AFB$ is $[AFB] = \frac {AF\cdot FB}2 = 4$, and the area of $ABCDE$ is $CD\cdot DE - [AFB] = 4\cdot (2\pi+1) - 4 = 8\pi$.
From the Pythagorean theorem the length of $AB$ is $\sqrt{2^2 + 4^2} = 2\sqrt{5}$, thus the radius of the circle is $\sqrt{5}$, and the area of the half-circle that is inside $ABCDE$ is $\frac{ 5\pi }2$.
Therefore the probability that $APB$ is obtuse is $\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\text{(C) } \frac 5{16}}$.
## Solution 1.5
Remember that an obtuse angle is any angle greater than a right angle (90 degrees). This means that if we can figure out all of the points at which $P$ creates a right angle with $A$ and $B$, any points within these bounds will yield an obtuse angle. Notice that by the Pythagorean theorem, the length of $AB$ is $2\sqrt{5}$. Also notice that when there is a right angle at $P$, $AB$ becomes the hypotenuse. Therefore, the sum of the squares of lengths $AP$ and $BP$ (or the "legs") must equal $AB^2$, or $20$. This can be written as an equation.
If Point $P$ is situated at $(x, y)$, then the distance from $P$ to $A$ is as follows: $\sqrt{|x-0|^2 + |y-2|^2}$. Accordingly, the distance from $P$ to $B$ is $\sqrt{|x-4|^2 + |y-0|^2}$. Since absolute value and squaring (because we will be squaring the lengths) both eliminate negatives, we can ignore the absolute value and solve. Setting the sum of the squares of these lengths equal to 20 yields $x^2 - 4x + y^2 - 2y = 0$. Completing the square for both $x$ and $y$, we get $(x-2)^2 + (y-1)^2 = 5$, or the equation of a circle with center $(2,1)$ and radius $\sqrt{5}$. Graphing both the pentagon and the circle, we see that the area enclosed is a semicircle. This is $\frac{2\pi*(\sqrt{5})^2}{2}$ or $\frac{5\pi}{2}$. Calculating the area of the pentagon yields $16\pi$. The probability of point P being within the bounds of the semicircle is then $\frac{5\pi}{16\pi}$ $\rightarrow$ $\frac{5}{16}$ or $\boxed{\text{(C)}}$.
## Solution 2
(Alcumus Solution)
Since $\angle APB = 90^{\circ}$ if and only if $P$ lies on the semicircle with center $(2,1)$ and radius $\sqrt{5}$, the angle is obtuse if and only if the point $P$ lies inside this semicircle. The semicircle lies entirely inside the pentagon, since the distance, 3, from $(2,1)$ to $\overline{DE}$ is greater than the radius of the circle. Thus the probability that the angle is obtuse is the ratio of the area of the semicircle to the area of the pentagon.
$[asy] pair A,B,C,D,I; A=(0,2); B=(4,0); C=(7.3,0); D=(7.3,4); I=(0,4); draw(A--B--C--D--I--cycle); label("A",A,W); label("B",B,S); label("C",C,E); label("D",D,E); label("E",I,W); draw(A--(0,0)--B,dashed); draw((3,3)..A--B..cycle,dashed); dot((2,1)); [/asy]$Let $O=(0,0)$, $A=(0,2)$, $B=(4,0)$, $C=(2\pi+1,0)$, $D=(2\pi+1,4)$, and $E=(0,4)$. Then the area of the pentagon is$$[ABCDE]=[OCDE]-[OAB] = 4\cdot(2\pi+1)-\frac{1}{2}(2\cdot4) = 8\pi,$$and the area of the semicircle is$$\frac{1}{2}\pi(\sqrt{5})^2 = \frac{5}{2}\pi.$$The probability is$$\frac{\frac{5}{2}\pi}{8\pi} = \boxed{\frac{5}{16}}.$$
|
# How do you find the vertical, horizontal and slant asymptotes of: f(x)=x/(x-1)^2?
Dec 18, 2016
The vertical asymptote is $x = 1$
No slant asymptote.
The horizontal asymptote is $y = 0$
#### Explanation:
The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{1\right\}$
As you cannot divide by $0$, $x \ne 1$
The vertical asymptote is $x = 1$
The degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.
To calculate the limits as $x \to \pm \infty$, we take the termsof highest degree in the numerator and the denominator
${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$
${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$
The horizontal asymptote is $y = 0$
graph{x/(x-1)^2 [-10, 10, -5, 5]}
|
# Optimization Problems Involving Numbers
## Solved Problems
### Example 7.
Find two positive numbers whose sum is $$12$$ so that the product of the square of one and $$4\text{th}$$ power of the other is maximum.
Solution.
The objective function is written in the form
$F\left( {x,y} \right) = {x^2}{y^4},$
where $$x$$ and $$y$$ are the two numbers.
As $$x + y = 12$$ we can write
$F = {x^2}{y^4} = {x^2}{\left( {12 - x} \right)^4} = F\left( x \right).$
Compute the derivative:
$F^\prime\left( x \right) = \left[ {{x^2}{{\left( {12 - x} \right)}^4}} \right]^\prime = 2x \cdot {\left( {12 - x} \right)^4} + {x^2} \cdot 4{\left( {12 - x} \right)^3} \cdot \left( { - 1} \right) = 6x{\left( {12 - x} \right)^3}\left( {4 - x} \right).$
Determine the critical points:
$F^\prime\left( x \right) = 0,\;\; \Rightarrow 6x{\left( {12 - x} \right)^3}\left( {4 - x} \right) = 0,\;\; \Rightarrow x = 0,\,4,\,12.$
At $$x = 0$$ and $$x = 12,$$ the objective function is equal to zero.
When $$x = 4,$$ the value of $$y$$ is
$y = 12 - x = 12 - 4 = 8.$
At this point, the objective function attains the maximum value:
${F_{\max }} = {4^2} \cdot {8^4} = {2^{16}} = 65536.$
### Example 8.
Find two positive numbers whose product is $$2$$ and the sum of one number and the square of the other is a minimum.
Solution.
Let $$x$$ and $$y$$ be the two numbers. The constraint equation is written in the form
$xy = 2,\;\; \Rightarrow y = \frac{2}{x}.$
The objective function is given by
$F = x + {y^2} = x + {\left( {\frac{2}{x}} \right)^2} = x + \frac{4}{{{x^2}}}.$
Find the derivative and determine the critical points:
$F^\prime\left( x \right) = \left( {x + \frac{4}{{{x^2}}}} \right)^\prime = 1 + 4 \cdot \left( { - \frac{2}{{{x^3}}}} \right) = 1 - \frac{8}{{{x^3}}} = \frac{{{x^3} - 8}}{{{x^3}}};$
$F^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{{x^3} - 8}}{{{x^3}}} = 0,\;\; \Rightarrow {x^3} = 8,\;\; \Rightarrow x = 2.$
Thus, the function has two critical points $$x = 0$$ and $$x = 2.$$ We should take only positive number $$x = 2.$$
Using the First Derivative Test, one can show that $$x = 2$$ is a point of minimum.
The second number is $$y = 1.$$
### Example 9.
Find two positive numbers whose sum is $$7$$ and the product of the cube of one number and the exponential function of the other is a maximum.
Solution.
Let $$x$$ and $$y$$ be the two numbers. The objective function is given by
$F = {x^3}{e^y}.$
As $$x + y = 7,$$ we substitute $$y = 7 - x$$ in the function above.
$F = {x^3}{e^y} = {x^3}{e^{7 - x}} = F\left( x \right).$
Differentiate $$F\left( x \right):$$
$F^\prime\left( x \right) = \left( {{x^3}{e^{7 - x}}} \right)^\prime = \left( {{x^3}} \right)^\prime \cdot {e^{7 - x}} + {x^3} \cdot \left( {{e^{7 - x}}} \right)^\prime = 3{x^2}{e^{7 - x}} + {x^3} \cdot {e^{7 - x}} \cdot \left( { - 1} \right) = 3{x^2}{e^{7 - x}} - {x^3}{e^{7 - x}} = {x^2}{e^{7 - x}}\left( {3 - x} \right).$
It is clear that the positive critical value is only $$x = 3.$$ Using the First Derivative Test, one can show that $$x = 3$$ is a point of local maximum.
Respectively, the other number is $$y = 4.$$
### Example 10.
The sum of two positive numbers is $$24.$$ The product of one and the square of the other is maximum. Find the numbers.
Solution.
Let the two numbers be $$x$$ and $$y.$$ The objective function is written as
$F\left( {x,y} \right) = x{y^2}.$
The constraint equation has the form
$x + y = 24,\;\; \Rightarrow y = 24 - x.$
Hence
$F = x{y^2} = x{\left( {24 - x} \right)^2}.$
Expanding $${\left( {24 - x} \right)^2},$$ we obtain:
$F\left( x \right) = x{\left( {24 - x} \right)^2} = x\left( {576 - 48x + {x^2}} \right) = 576x - 48{x^2} + {x^3}.$
Differentiate:
$F^\prime\left( x \right) = \left( {576x - 48{x^2} + {x^3}} \right)^\prime = 576 - 96x + 3{x^2} = 3\left( {192 - 32x + {x^2}} \right).$
Find the critical points:
$F^\prime\left( x \right) = 0,\;\; \Rightarrow 3\left( {192 - 32x + {x^2}} \right) = 0,\;\; \Rightarrow {x^2} - 32x + 192 = 0;$
$D = {\left( { - 32} \right)^2} - 4 \cdot 192 = 1024 - 768 = 256;$
${x_{1,2}} = \frac{{ - \left( { - 32} \right) \pm \sqrt {256} }}{2} = \frac{{32 \pm 16}}{2} = 24,\,8;$
When $$x = 24,$$ then $$y = 0,$$ so the objective function is equal to zero in this case.
Note that the second derivative is
$F^{\prime\prime}\left( x \right) = \left( {576 - 96x + 3{x^2}} \right)^\prime = 6x - 96.$
Hence, the second derivative is negative for $$x = 8,$$ that is the point $$x = 8$$ is a point of maximum of the objective function.
The other number $$y$$ is equal to
$y = 24 - x = 24 - 8 = 16.$
### Example 11.
Find two positive numbers whose sum is $$32$$ and the sum of their square roots is maximum.
Solution.
We write the objective function in the form
$F = \sqrt x + \sqrt y ,$
where $$x, y$$ are two positive numbers.
As $$x + y = 32,$$ we can plug in $$y = 32 - x$$ into the objective function.
$F = \sqrt x + \sqrt y = \sqrt x + \sqrt {32 - x} = F\left( x \right).$
Differentiate $$F\left( x \right):$$
$F^\prime\left( x \right) = \left( {\sqrt x + \sqrt {32 - x} } \right)^\prime = \frac{1}{{2\sqrt x }} + \frac{{\left( { - 1} \right)}}{{2\sqrt {32 - x} }} = \frac{{\sqrt {32 - x} - \sqrt x }}{{2\sqrt x \sqrt {32 - x} }}.$
Determine the critical points:
$F^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{\sqrt {32 - x} - \sqrt x }}{{2\sqrt x \sqrt {32 - x} }} = 0,\;\; \Rightarrow \sqrt {32 - x} - \sqrt x = 0,\;\; \Rightarrow \sqrt {32 - x} = \sqrt x ,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {32 - x = x}\\ {x \lt 32}\\ {x \gt 0} \end{array}} \right., \Rightarrow x = 16.$
There are total 3 critical points: $$x = 0, 16, 32.$$ We calculate the values of the objective function at these points:
$F\left( 0 \right) = \sqrt 0 + \sqrt {32 - 0} = \sqrt {32} = 4\sqrt 2 \approx 5.66;$
$F\left( {16} \right) = \sqrt {16} + \sqrt {32 - 16} = 4 + 4 = 8;$
$F\left( {32} \right) = \sqrt {32} + \sqrt {32 - 32} = \sqrt {32} = 4\sqrt 2 \approx 5.66$
Thus, the maximum value $${F_{\max }} = 8$$ is attained at $$x = 16,$$ $$y = 16.$$
|
# PROBLEM SOLVING TACTICS
PROBLEM SOLVING TACTICS
Tactic 1: Numbers Versus Algebra
One way to avoid rounding errors and other numerical errors is to solve problems algebraically, substituting numbers only in the final step. That is easy and that is the way experienced problem solvers operate. In these early chapters. however. we prefer to solve most problems in pans. to give you a firmer numerical grasp of what is going on. Later we shall stick to the algebra longer.
Uniform Circular Motion
A particle is in uniform circular motion if it travels around a circle or a circular arc at constant (uniform) speed. Although the speed does not vary. the particle is accelerating.That fact may be surprising because we often think of acceleration (a change in velocity) as an increase or decrease in speed. However. actually velocity is a vector. not a scalar. Then even if a velocity changes only in direction. there is still an acceleration. and that is what happens in uniform circular motion.
Proof
To find the magnitude and direction of the acceleration for uniform circular motion, we consider particle p moves at constant speed v around a circle of radius r. At the instant shown, p has coordinates x and y.
Now note that the rate changes is equal to the velocity component.Similarly we see that v. Making these substitutions in. We find
This vector and its components.Following we find that the magnitude.
CHECKPOINT 6: An object moves at constant speed along a circular path in a horizontal. problem,with the center at the origin. When the object is at x velocity. Give the object’s (a) velocity and (b) acceleration when it is at).
Sample Problem
Top gun pilots have long worried about taking a turn too tightly. As a pilot’s body under goes centripetal acceleration, with the head toward the center , the blood pressure in the brain decreases, leading to loss of brain function.
There are several warning signs to signal a pilot to ease up when the centripetal acceleration is 2 or 3. the pilot feels heavy. At about 4, the pilot’s vision switches to black and white and narrows to “tunnel vision.” If that acceleration is sustained.
SOLUTION: The Key idea here is that although the pilot’s speed is constant. the circular path requites a (centripetal) acceleration with.If an unwary pilot caught in a dogfight puts the aircraft into such a tight turn. the pilot goes almost immediately. with no warningsigns to signal the danger
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Expression Evaluation with Powers and Grouping Symbols
## Evaluate expressions with parentheses and exponents.
%
Progress
MEMORY METER
This indicates how strong in your memory this concept is
Progress
%
Expression Evaluation with Powers and Grouping Symbols
Have you ever had a part time job?
Lydia and Bart both work at a bookstore. Lydia makes amount each hour; Bart has more experience, so he makes 1.5 times more than Lydia makes each hour. If Lydia and Bart both work 4 hours a day, how much will they make together in seven days if Lydia makes $8 an hour. Write an expression; then solve it. Do you know how to do this? This Concept will show you exactly how to do this. ### Guidance Previously we worked on the basic order of operations. Well, now we can expand our rules to include evaluation of more complicated expressions. In intricate expressions parentheses are used as grouping symbols. Parentheses indicate which operations should be done first. In the order of operations, operations within the parentheses are always first. In the Evaluate Powers with Variable Bases Concept, we learned how to simplify and evaluate exponents. . Exponential notation is another factor we must account for in the order of operations. After completing the operations in parentheses, we then evaluate the exponents. Then we complete the multiplication and division from left to right; finally, we complete the addition and subtraction from left to right. The chart below shows the complete order of operations. If you keep the order of operations in the forefront of your mind and are careful to take each step one at a time and show your work, evaluating complex expressions using the order of operations will become second nature. Let’s look at the order of operation once again. Evaluate the expression: if First, we substitute 4 in for . Next, we add the terms in parentheses. Since parentheses is the first order of operations. Now we can evaluate the exponent. Multiplication is next and there is no division. Finally, we complete with subtraction since there is no addition. 21 Our answer is 21. Sometimes you will also evaluate expressions with more than one variable. Just keep track and follow the order of operations and you will be all set. Now it's time for you to practice. #### Example A Evaluate if is 2 Solution: 15 #### Example B Evaluate if is 3 and is 4 Solution: 30 #### Example C Evaluate if is 5 Solution: 129 Now back to the dilemma at the bookstore. Here is the original problem once again. Lydia and Bart both work at a bookstore. Lydia makes amount each hour; Bart has more experience, so he makes 1.5 times more than Lydia makes each hour. If Lydia and Bart both work 4 hours a day, how much will they make together in seven days if Lydia makes$8 an hour.
Write an expression; then solve it.
In this problem, stands for the amount Lydia makes per hour. Because Bart makes 1.5 times the amount that Lydia makes per hour, then describes how much Bart makes per hour. The total Lydia and Bart make in 1 hour is therefore . Now the problem tells us that Lydia and Bart each work 4 hours per day. So, the amount they both make in 1 day is . We want to find out how much they make in 7 days, so our expression is . The problem gives us the value of . Lydia makes $8 an hour. We substitute 8 for the variable in our expression and solve using the order of operations. Remember: parenthesis, exponents, multiplication, division, addition, subtraction. The answer is$560.00.
### Vocabulary
Numerical Expression
an expression that uses numbers and operations.
Variable Expression
an expression that uses numbers, variables and operations.
Parentheses
grouping symbols, the first step of the order of operations.
Exponent
the little number that tells how many times to multiply the base times itself.
Order of Operations
the order that you perform each operation when evaluating an expression.
### Guided Practice
Here is one for you to try on your own.
Evaluate if is 4
First, substitute 4 into every place that an appears.
Now evaluate according to the order of operations.
### Practice
Directions: Evaluate the following variable expressions is
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Directions: Use what you have learned to answer the following questions true or false.
13. Parentheses are a grouping symbol.
14. Exponents can’t be evaluated unless the exponent is equal to 3.
15. If there is multiplication and division in a problem you always do the division first.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Evaluate
To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Numerical expression
A numerical expression is a group of numbers and operations used to represent a quantity.
Order of Operations
The order of operations specifies the order in which to perform each of multiple operations in an expression or equation. The order of operations is: P - parentheses, E - exponents, M/D - multiplication and division in order from left to right, A/S - addition and subtraction in order from left to right.
Parentheses
Parentheses "(" and ")" are used in algebraic expressions as grouping symbols.
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
|
Graphing the polynomial functions
The source or original polynomial function
Translating (parallel shifting) of the polynomial function
Coordinates of translations and their role in the polynomial expression
Translated power function
Translated sextic function example
Graphing the polynomial functions
Translating (parallel shifting) of the polynomial function
Thus, to obtain the graph of a given polynomial function f (x) we translate (parallel shift) the graph of its source function in the direction of the x-axis by x0 and in the direction of the y-axis by y0.
Inversely, to put a given graph of the polynomial function beck to the origin, we translate it in the opposite direction, by taking the values of the coordinates of translations with opposite sign.
Coordinates of translations and their role in the polynomial expression
The coordinates of translations we calculate using the formulas,
Hence, by plugging the coordinates of translations into the source polynomial function fs(x), i.e.,
y - y0 = an(x - x0)n + an-2(x - x0)n-2 + . . . + a2(x - x0)2 + a1(x - x0)
and by expanding above expression we get the polynomial function in the general form
f (x) = yanxn + an-1xn-1 + an-2xn-2 + . . . + a2x2 + a1x + a0.
Inversely, by plugging the coordinates of translations into the given polynomial f (x) expressed in the general form, i.e.,
y + y0 = an(x + x0)n + an-1(x + x0)n-1 + . . . + a1(x + x0) + a0
and after expanding and reducing above expression we get its source polynomial function.
Note that in the above expression the signs of the coordinates of translations are already changed.
Translated monomial (or power) function
If we set all coefficients, an-2 to a1, in the above expanded form of the polynomial to zero, we get
y - y0 = an(x - x0)n, x0- an-1/( n · an) and y0 = f (x0).
the translated power (or monomial) function, the exponent of which is an odd or an even positive integer.
When the exponent is even, i.e., of the form n = 2m, m Î N, the graph of the source power function
y = anxn is symmetric about the y-axis, that is f (-x) = f (x).
When the exponent is odd, i.e., of the form n = 2m + 1, m Î N, the graph of the source power function
y = anxn is symmetric about the origin, that is f (-x) = -f (x).
Example: Given is sextic y = (1/4)x6 - 6x5 + 60x4 - 320x3 + 960x2 - 1536x + 1008, find its source or original function and calculate the coordinates of translations, the zero points and the turning point.
Draw graphs of the source and the given sextic.
Solution: 1) Calculate the coordinates of translations
y0 = f (4) = (1/4) · 46 - 6 · 45 + 60 · 44 - 320 · 43 + 960 · 42 - 1536 · 4 + 1008, y0 = - 16.
2) To get the source sextic, plug the coordinates of translations into the general form of the given sextic, to draw its graph back to the origin,
y + y0 = a6(x + x0)6 + a5(x + x0)5 + a4(x + x0)4 + a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0 or
y - 16 = (1/4)(x - 4)6 - 6(x - 4)5 + 60(x - 4)4 - 320(x - 4)3 + 960(x - 4)2 - 1536(x - 4) + 1008,
after expanding and reducing above expression obtained is
y = (1/4) x6 - the source sextic.
Since all the coefficients, a4, a3, a2 and a1, of the source sextic y = a6x6 + a4x4 + a3x3 + a2x2 + a1x, are zero then, the only turning point is T(x0, y0) or T( 4, - 16).
3) Inversely, by plugging the coordinates of translations into the source sextic
y - y0 = a6(x - x0)6 or y - 16 = (1/4)(x - 4)6
what after expanding yields
y = (1/4) x6 - 6x5 + 60x4 - 320x3 + 960x2 - 1536x + 1008
the given sextic. Therefore, the given sextic is translated monomial or power function.
As the translated monomial or power function has zeros if a6 · y0 < 0 then.
Calculus contents A
|
# Fourteen Magic Number Tricks
“Fight Against Stupidity And Bureaucracy”
.
In previous posts we have had ‘Beautiful Numbers’, ‘Big Numbers’, ‘Unusual Numbers’ and lots of what I called ‘Significant Numbers’.
Today, for a bit of a change, it is the turn of ‘Magic Numbers’, or magic number tricks.
I call them magic numbers because the results of some of them are predictable and on occasions magicians have incorporated them into their magic routines, where, for example, they need their ‘stooge’ to pick a certain card or a certain page in a book and want to give the audience the illusion of a random choice.
Try some of these out. Use them to do a bit of magic yourselves, or to win friends and influence people. Or just to entertain people you like or bore people you don’t like, whichever you think is appropriate.
Anyway I hope you enjoy this selection. You’ll need a calculator if you want to check them out.
.
.
First Magic Number Trick.
Step 1: Pick a number,
Step 2: add 2,
Step 3: multiply by 3,
Step 4: subtract 6,
Step 5: divide by 3.
You should get the number you started with.
This works for other, larger numbers. This example started with add 2 and multiply 3. But any two numbers work, just multiply them together to get the next number that you subtract.
.
.
Second Magic Number Trick.
Step 1: Pick a number,
Step 2: square it (probably need a calculator for big numbers),
Step 3: add twice the original number,
Step 4: add one,
Step 5: take the square root (rounding it to the nearest whole number, 7.999… becomes 8),
Step 6: subtract 1,
You should get the number you started with.
.
Third Magic Number Trick.
Step 1: Pick a number,
Step 2: square it,
Step 3: add ten times the original number,
Step 4: add 25,
Step 5: take the square root (rounding to the nearest whole number),
Step 6: subtract your original number.
The answer should always be 5.
.
Third Magic Number Trick.
Here is a slightly more complicated one.
Step 1: Pick a number between 1 and 100,
Step 2: add 28,
Step 3: multiply by 6,
Step 4: subtract 3,
Step 5: divide by 3,
Step 6: subtract the original number plus 3,
Step 7: add 8,
Step 8: subtract the original number minus 1,
Step 9: multiply by 7.
Your answer should be 427.
.
Fourth Magic Number Trick.
Step 1: Pick a number 1 through 9,
Step 2: multiply by 12345679 (notice there is no 8 there),
Step 3: multiply by 9.
Do you see your original number?
.
Fifth Magic Number Trick.
Step 1: Pick a 3-digit number in which the first and last digits differ by more than one,
Step 2: reverse this number (for example, 531 becomes 135) and subtract the smaller from the larger,
Step 3: add this number to the reverse of itself.
Your answer is 1089.
.
Sixth Magic Number Trick.
Step 1: Think of a 3 digit number.
Step 2: Multiply it by 7, then by 11, and then by 13.
Your answer should be your original number twice,
for example, if you chose the number 456, your answer would be 456456
.
Seventh Magic Number Trick.
Step 1: Think of a 2 digit number.
Step 2: Multiply it by 3, then by 7, then by 13, and then by 37.
You should see your original number repeated three times.
For example, if your number was 45, the answer would be 454545
.
Eighth Magic Number Trick.
Step 1: Think of a 5 digit number.
Step 2: Multiply it by 11.
Step 3: Multiply it by 9091.
For example, if the number is 12345, the answer should be 1234512345
.
Ninth Magic Number Trick.
If you multiply 1089 by 9 you get 9801. The number has reversed itself!
This also works with 10989 or 109989 or 1099989 and so on.
.
.
Tenth Magic Number Trick.
19 = 1 x 9 + 1 + 9 and 29 = 2 x 9 + 2 + 9.
This also works for 39, 49, 59, 69, 79, 89 and 99.
.
Eleventh Magic Number Trick.
2 is the only number that gives the same result added to itself as it does times by itself.
In other words 2 + 2 = 4 = 2 x 2, or, (2+2) – (2 x 2) = 0
.
Twelfth Magic Number Trick.
If you multiply 21978 by 4 it turns backwards
.
Thirteenth Magic Number Trick.
153, 370, 371 and 407 are all the sum of the cubes of their digits.
In other words 153 = 13+53+33, 370 = 33+73+03, 371 = 33+73+13, 153 = 43+03+73,
.
Fourteenth Magic Number Trick.
1 divided by 37 = 0•027027027
and
1 divided by 27 = 0•037037037
.
.
===========================
.
|
# calculate the amount and compound interest on rs.10800 for 3 year’s at 12.5% per annum compound anually
calculate the amount and compound interest on rs.10800 for 3 year’s at 12.5% per annum compound anually
About the author
1. ## Question:-
• Calculate the amount and compound interest on rs.10800 for 3 year’s at 12.5% per annum compound anually.
## To Find:-
• Find the amount and the compound interest.
## Solution:-
$$\tt\implies \: A = 10800 \times { ( \dfrac { 1 + 12.5 } { 100 } ) }^{ 3 }$$
$$\tt\implies \: A = 10800 \times {1.125 }^{ 3 }$$
$$\tt\implies \: A = 15377$$
Now ,
We have to find the compound Interest:-
$$\tt\implies \: Interest = 15377 – 10800$$
$$\tt\implies \: Interest = 4577$$
Hence ,
• Compound interest is 4577
Reply
2. Answer:
1. Amount = Rs 15377
2. Compound interest = Rs 4577
Step-by-step explanation:
Given that:
• Time = 3 years
• Principal = Rs 10800
• Rate of interest = 12.5% per annum
To Find:
1. Amount
2. Compound interest
Formula used:
1. A = P(1 + R/100)ᵀ
2. CI = A – P
Where,
• P = Principal
• R = Rate of interest
• T = Time
• A = amount
• CI = Compound interest
Finding the amount:
⟶ A = 10800(1 + 12.5/100)³
⟶ A = 10800(1 + 0.125)³
⟶ A = 10800(1.125)³
⟶ A = 10800 × 1.125 × 1.125 × 1.125
⟶ A ≈ 15377
∴ Amount in compound interest = Rs 15377
Finding the compound interest:
⟶ CI = 15377 – 10800
⟶ CI = 4577
∴ Compound interest = Rs 4577
Reply
|
# about slide So let us launch
## Found at: sdf.org:70/users/melton/phlog/slide-rule-revisited
```The Slide Rule and the neighborhood
```
```about slide rules. So let us launch into another problem involving
```
```than punching the numbers in on the calculator.
```
```Our man is standing atop a 300 ft. ocean front cliff and observes a
```
```We'll refer back to our handy table for solutions to right
```
```triangles. There are a couple of approaches we can take, but in
```
```this case since we are given the angle of depression, we'll subtract
```
```that from 90° to give us our remaining angle which is 78°.
```
```To determine the distance from the base of the cliff to the boat
```
```(we'll call it 'a'), we multiply 300 * (TAN 78°) = a. So first
```
```let us find the tangent of 78° on the slide rule. Hmm...the
```
```tangent scale only goes from 5.7° to 45°. Let us see what
```
```one of my slide rules (Sterling Acumath 400) provides a solution:
```
```So if we follow this path then 90°-78°=12°. Back to
```
```Now we need the reciprocal of .212. Here we go to the 'CI' and 'D'
```
```To determine the decimal point when finding reciprocals one of the
```
```To find y = 1/x:
```
```- Convert x to scientific notation and read it's coefficient c and
```
`t's exponent, p`
```- if the coefficient is 1 or -1 exactly, y exponent is -p
```
```- otherwise y's exponent is -p-1
```
```To get our decimal in the correct place, we convert .212 to
```
```Acumath slide rule the 'CI' scale is on the slide. We place the
```
```cursor over 2.12 on the CI scale (remembering the 'CI' scale
```
`ncreases from right to left) and read down to the 'D' scale which`
```neither 1 or -1, the exponent of our answer is -(-1)-1 = 0 so our
```
`t out. Now we multiply the height of the cliff (300 or 3E2) times`
```the tangent of 78° which we just figured out is 4.71E0. On the
```
```the 'D' scale:
```
```Then we move the cursor over to 4.71E0 on the 'C' scale and read our
```
```answer directly below on the 'D' scale, in this case, 1.42.
```
```Now we add the exponents 0 + 2 = 2, but since we have to adjust for
```
```the change of magnitude (3 * 4.71 > 10), We add one to the exponent
```
```(0 + 2 + 1 = 3). Our answer is 1.42E3 or 1420. What does the
```
```calculator tell us?
```
```and calibration. The additional factor is the user's ability to set
```
`t and read the values. For most applications, three significant`
```figures is good enough.
```
```Now that we have gone through this seemingly tedious exercise, I am
```
```the mathematical equivalent to stopping and smelling the roses. In
```
```our rush for an instant answer or instant gratification, we miss out
```
```on the intimate details of the journey. In a similar way, it is
```
```like the difference between hopping in our car and driving to the
```
```local market for our groceries, or slowing down and taking the
```
```bicycle or even walking and have more intimate contact with the
```
```neighborhood and surroundings. As I have mentioned in a previous
```
```a simple tool like the scythe allows one not only to mow grass and
```
`ntimate with the subtleties of the land (and wildlife) that would`
```otherwise be lost when sitting in a tractor noisily masticating the
```
```on a journey to an answer and on the trip, we become more familiar
```
```the Luddites only destroyed machines that did not support or foster
```
```the "commonality." Sometimes I find myself pausing and reflecting
```
`f a particular technology I am using meets that standard, or have I`
```become a slave to that particular device and all of its ancillary
```
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# A luminous object is placed 20 cm from the surface of a convex mirror and a plane mirror is set so that the virtual images formed in the two mirrors coincide. If the plane mirror is at a distance of 12 cm from the object, then the focal length of the convex mirror is:A. -10 cm.B. -5 cm.C. -20 cm.D. -40 cm.
Last updated date: 20th Jun 2024
Total views: 404.4k
Views today: 12.04k
Verified
404.4k+ views
Hint:-The virtual images of both the mirrors should coincide. The position of the image formed by the plane mirror should be equal to the position at which the image formed by the convex mirror also the focal length of the plane mirror is infinite.
Formula used: The formula of the mirror formula is given by,
$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$
Where u is the object distance from the mirror v is the image distance from the mirror and f is the focal length of the mirror.
Complete step-by-step solution
It is given in the problem that a plane mirror is placed in between the convex mirror and the object and we need to find out the value of the focal length of the convex mirror such that the images formed by the plane mirror and the convex mirror coincide with each other.
The object is placed at 20 cm in front of the convex mirror and the plane mirror is 12 cm from the plane mirror.
The focal length of the plane mirror is infinite therefore,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{\infty }$
$\Rightarrow \dfrac{1}{f} = 0$
The mirror formula is,
$\Rightarrow \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$
$\Rightarrow \dfrac{1}{u} + \dfrac{1}{v} = 0$
$\Rightarrow \dfrac{1}{u} = - \dfrac{1}{v}$
$\Rightarrow v = - u$
The image of the object formed will be virtual.
Therefore the image formed will be at position$v = - u = - 12$.
The distance between the object and the image is equal to,
$\Rightarrow 12 - \left( { - 12} \right) = 24cm$
Now for the convex mirror the formula will be,
$\dfrac{1}{{u'}} + \dfrac{1}{{v'}} = \dfrac{1}{{f'}}$
Where u’ is the object distance v’ is the image distance and f’ is the focal distance.
The distance of the object is given by $u' = 20cm$ and image distance will be,
$\Rightarrow v' = - \left( {24 - 20} \right)$
$\Rightarrow v' = - 4cm$
Now let us calculate the focal length of the convex mirror,
$\Rightarrow \dfrac{1}{{u'}} + \dfrac{1}{{v'}} = \dfrac{1}{{f'}}$
$\Rightarrow \dfrac{1}{{20}} - \dfrac{1}{4} = \dfrac{1}{{f'}}$
$\Rightarrow \dfrac{1}{{f'}} = \dfrac{1}{{20}} - \dfrac{1}{4}$
$\Rightarrow \dfrac{1}{{f'}} = \dfrac{{1 - 5}}{{20}}$
$\Rightarrow \dfrac{1}{{f'}} = \dfrac{{ - 4}}{{20}}$
$\Rightarrow \dfrac{1}{{f'}} = \dfrac{{ - 1}}{5}$
$\Rightarrow f' = - 5cm$
The focal length of the convex mirror is equal to$f' = - 5cm$. The correct answer for this problem is option B.
Note:- The focal length of the plane mirror is infinite which means that the image will be formed at a point which is exactly the same distance as the object is from the plane mirror but on the same side of the object.
|
# 10 Fast Maths Tricks and Shortcuts
## List of Cool Maths Tricks
### Squaring
In this simple trick we need to modify the equation and make the units digit zero. After all it is easy to multiply when units digit is zero.
For example - Find square of 43
= (43+3) × (43-3) + (3×3)
=(46×40) + 9
= (460×4) + 9
= 1840 + 9 = 1849
### Multiplication
Vedic maths gave us the easiest method to do complex multiplications quickly. This method can be quickly explained with an example.
Multiply 62 with 32
Step 1
Step 2
Step 3
### Multiplication with 5
Simply multiply the number by 10 and then divide it by 2.
For example 99×5= 990/2= 495
Multiplication with 4
Multiplication with 9
Multiplication with 6
### Multiplication with 99
Multiply the number with 100 and then minus same number from the result. Let's take an example
Multiply 32×99 = 3200 - 32 = 3168
### Mixture
Questions related to mixtures can be easily solved by alligation method. By using alligation we can wide arrange of maths questions. Let me explain this with a simple example
Example
Price of wine of \$60 per liter. If Samuel is adding water with and selling the mixture for \$40 per liter. Profit margin remains same. What is the ratio of water and wine in the mixture.
### Square root
Best method to find square root of large numbers is by dividing the number into parts. Let's take an example
Find square √ 1936
√ 1936 = √ 4 × √ 484 = √ 4 × √ 4 × √ 121 = 2×2×11 = 44
### Time and Work
Every question in in time and work chapter can be solved easily by finding efficiency of workers or subject (such as pipes).
For example - A takes 10 days to complete a job. B takes 20 days to complete the same job. In how many days they will complete the job if they work together ?
A's efficiency = 100/10 = 10% per day
B's efficiency = 100/20 = 5% per days
A and B can do 15% of the work in a day if they work together. So they can do the whole job in 100/15 = 6.66 days or 6 days and 18 hours.
### Profit and Loss
In case of profit
25% of Cost Price (1/4 of CP) = 20% of Selling Price (1/5 of SP)
Similarly 1/3 of CP = 1/2 of SP
In case of loss
25% of Selling Price (1/4 of SP) = 20% of Cost Price (1/5 of CP)
Similarly 1/3 of SP = 1/2 of CP
### Estimation
That's the most important technique. This is not a secret that every successful candidate is using this technique during exams.
Example - 112 × 92
Simply 112 × 9 = 1008
|
Express each of the following product as a monomials and verify the result in each case for x = 1:
Question:
Express each of the following product as a monomials and verify the result in each case for x = 1:
$\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$
Solution:
We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$.
We have:
$\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$
$=\left\{4 \times(-3) \times \frac{4}{5}\right\} \times\left(x^{2} \times x \times x^{3}\right)$
$=\left\{4 \times(-3) \times \frac{4}{5}\right\} \times\left(x^{2+1+3}\right)$
$=-\frac{48}{5} x^{6}$
$\therefore\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)=-\frac{48}{5} x^{6}$
Substituting x = 1 in LHS, we get:
$\mathrm{LHS}=\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$
$=\left(4 \times 1^{2}\right) \times(-3 \times 1) \times\left(\frac{4}{5} \times 1^{3}\right)$
$=4 \times(-3) \times \frac{4}{5}$
$=-\frac{48}{5}$
Putting x = 1 in RHS, we get:
RHS $=-\frac{48}{5} x^{6}$
$=-\frac{48}{5} \times 1^{6}$
$=-\frac{48}{5}$
$\because L H S=\mathrm{RHS}$ for $x=1$; therefore, the result is correct
Thus, the answer is $-\frac{48}{5} x^{6}$.
|
How do you describe the relationship between two quantities?
How do you describe the relationship between two quantities?
Two quantities have a proportional relationship if they can be expressed in the general form y = kx, where k is the constant of proportionality. In other words, these quantities always maintain the same ratio. That is, when you divide any pair of the two values, you always get the same number k.
What expresses the relationship between two or more concepts?
A principle expresses the relationship between two or more concepts or constructs. Concepts and principles serve two important functions: 1) They help us to understand or explain what is going on around us.
How does a function model the relationship between two quantities?
A linear relationship between two quantities will produce a graph of a straight line. The line represents every possible solution for the range of the function. In order to create the line, we use the function equation and evaluate the range, or output, values based upon several of the domain, or input, values.
What are the components of theory?
This definition suggests three things:
• First, theory is logically composed of concepts, definitions, assumptions, and generalizations.
• Second, the major function of theory is to describe and explain – in fact, theory is a general explanation, which often leads to basic principles.
How do you find the linear relationship between two variables?
A linear relationship can also be found in the equation distance = rate x time. Because distance is a positive number (in most cases), this linear relationship would be expressed on the top right quadrant of a graph with an X and Y-axis.
What is an example of a linear relationship?
Linear relationships such as y = 2 and y = x all graph out as straight lines. When graphing y = 2, you get a line going horizontally at the 2 mark on the y-axis. When graphing y = x, you get a diagonal line crossing the origin.
What gives the mathematical relationship of the variables?
What is a Mathmatical Relation? A mathematical relation is, a relationship between sets of numbers or sets of elements. Often you can see relationships between variables by simply examining a mathematical equation. Many physical relationships in electrostatics, electrodynamics, thermodynamics, etc.
What expresses the same relationship between two quantities?
Equivalent Ratio – ratios the express the same relationship between two quantities.
How do you describe your relationship?
Relation“) We often use the following adjectives to describe relationships: friendly: She is generally confident, well-spoken and professional, and easily establishes friendly relationships with co-workers. happy: I’m in a happy relationship, with a growing family. brief: He had brief relationships with several women.
What shows the relationship between two or more variables?
Regression. Regression analysis is used to determine if a relationship exists between two variables.
What is the relationship between 1 and 2?
Enneagram Ones and Twos are a complementary couple since both offer the other the example of their own qualities. Both types are highly dutiful and are attracted to service roles and occupations: both may be teachers, ministers, or health care workers who have long hours and many responsibilities.
How do you find the functional relationship between two variables?
A function exists when each x-value (input, independent variable) is paired with exactly one y-value (output, dependent variable). This pairing is also referred to as a functional relationship.
How do you include a relationship between two use cases?
Include Relationship Between Two Use Cases. Include relationship show that the behavior of the included use case is part of the including (base) use case. The main reason for this is to reuse common actions across multiple use cases. In some situations, this is done to simplify complex behaviors.
What is extend relationship in use case diagram?
Extend relationship in use case diagrams Although extending use case is optional most of the time it is not a must. An extending use case can have non-optional behavior as well. This mostly happens when your modeling complex behaviors.
How to classify relationships according to relationship-types?
Just loke entities, we can classify relationships according to relationship-types: A lecturer is giving a lecture. A weak entity is a type of entity which doesn’t have its key attribute. It can be identified uniquely by considering the primary key of another entity. For that, weak entity sets need to have participation.
What is included use case relationship in Salesforce?
Include Relationship Between Two Use Cases Include relationship show that the behavior of the included use case is part of the including (base) use case. The main reason for this is to reuse common actions across multiple use cases. In some situations, this is done to simplify complex behaviors.
|
# The Breadth of a Room is Twice Its Height, One Half of Its Length and the Volume of the Room is 512 Cu. Dm. Find Its Dimensions. - Mathematics
The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.
#### Solution
$\text { Suppose that the breadth of the room = x dm }$
$\text { Since breadth is twice the height, breadth }= 2 \times \text { height }$
$\text { So, height of the room = } \frac{\text { breadth }}{2}=\frac{x}{2}$
$\text { Also, it is given that the breadth is half the length .}$
$\text { So, breadth }= \frac{1}{2} \times \text { length }$
$\text { i . e . , length }= 2 \times \text { breadth } = 2 \times x$
$\text { Since volume of the room = 512 cu dm, we have }$
$\text { Volume of a cuboid = length } \times\text { breadth } \times \text { height }$
$\Rightarrow 512 = 2 \times x \times x \times \frac{x}{2}$
$\Rightarrow 512 = x^3$
$\Rightarrow x = \sqrt[3]{512} = 8 dm$
$\text { Hence, length of the room }= 2 \times x = 2 \times 8 = 16 dm$
$\text { Breadth of the room = x = 8 dm }$
$\text { Height of the the room } = \frac{x}{2}=\frac{8}{2} = 4 dm$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 21 Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube)
Exercise 21.4 | Q 6 | Page 30
|
# 2.6: Multiplication of Rational Numbers
Difficulty Level: Basic Created by: CK-12
Estimated8 minsto complete
%
Progress
Practice Multiplication of Rational Numbers
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated8 minsto complete
%
Estimated8 minsto complete
%
MEMORY METER
This indicates how strong in your memory this concept is
Suppose you wrote a computer program that multiplies by a random number. What if the random number were –1? What if it were 0? What if it were 1? In fact, the random number doesn't even have to be an integer. What if it were ? What if it were ? In this Concept, you'll learn about the Multiplication Property of –1, the Multiplicative Identity Property, and the Zero Property of Multiplication so that you can answer these questions.
### Guidance
When you began learning how to multiply whole numbers, you replaced repeated addition with the multiplication sign . For example:
Multiplying rational numbers is performed the same way. We will start with the Multiplication Property of –1.
The Multiplication Property of –1: For any real number .
This can be summarized by saying, "A number times a negative is the opposite of the number."
#### Example A
Evaluate .
Solution:
Using the Multiplication Property of : .
This property can also be used when the values are negative, as shown in Example B.
#### Example B
Evaluate .
Solution:
Using the Multiplication Property of : .
A basic algebraic property is the Multiplicative Identity. Similar to the Additive Identity, this property states that any value multiplied by 1 will result in the original value.
The Multiplicative Identity Property: For any real number .
A third property of multiplication is the Multiplication Property of Zero. This property states that any value multiplied by zero will result in zero.
The Zero Property of Multiplication: For any real number .
Multiplication of fractions can also be shown visually, as you can see in the example below.
#### Example C
Find , drawing one model to represent the first fraction and a second model to represent the second fraction.
Solution:
By placing one model (divided in thirds horizontally) on top of the other (divided in fifths vertically), you divide one whole rectangle into smaller parts.
The product of the two fractions is the
### Guided Practice
Simplify
Solution: By drawing visual representations, you can see that
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Multiplication of Rational Numbers (8:56)
Multiply the following rational numbers.
Multiply the following by negative one.
1. 79.5
2. 25
3. –105
#### Quick Quiz
1. Order from least to greatest: .
2. Simplify
3. Simplify .
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
Multiplication Property of –1 For any real number $a, (-1) \times a = -a$.
multiplicative identity property The product of any number and one is the number itself.
Zero Property of Multiplication For any real number $a, \ (0) \times a = 0$.
Associative Property The associative property states that you can change the groupings of numbers being added or multiplied without changing the sum. For example: (2+3) + 4 = 2 + (3+4), and (2 X 3) X 4 = 2 X (3 X 4).
Commutative Property The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example $a+b=b+a \text{ and\,} (a)(b)=(b)(a)$.
distributive property The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
Integer The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Mixed Number A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
Show Hide Details
Description
Difficulty Level:
Basic
Tags:
Subjects:
|
# Some Random Tips for your Quant Section
Hello Students,
Here we are providing you some tricks which can be helpful to solve Quant questions. These can be applied for solving the questions.
### To find greater between fractions without calculating their decimal values.
Here is a short example:
Consider 4/5 and 6/7
Cross multiply the numerator and denominator of both the fractions.
4 × 7 = 28 6 × 5 = 30
30 > 28 so the fraction carrying 6 as numerator is greater, i.e. 6/7 > 4/5
* You can use this method when you are asked to find out the greatest or smallest among given fractions.
### Method to solve mixed fractions.
1. Here is an example:
Consider finding the value for 2(1/3) + 4(2/7) + 5(1/6).
This can be calculated as:
(2 + 4 + 5) + [(1/3) + (2/7) + (1/6)] = 11 + (14 + 12 + 7)/42
= 11 + 33/42
= 495/42
Now similarly 2(1/3) + 4(2/7) – 5(1/6)
= (2 + 4 – 5) + [(1/3) + (2/7) – (1/6)] = 1 + (14 + 12 – 7)/42
= 1 + 19/42
= 61/42
1. When x^2 = 25, then x = 5 and -5 because 5 × 5 = 25 and also (-5) × (-5) = 25
But when x = √25 or 25^1/2, then x = 5 only because when radical symbol is used, we can take only positive (or zero) values.
### Squares of numbers ending with 5.
1. Consider examples:
(25)^2 = 2 × 3 | 5 × 5 = 625
(85)^2 = 8 × 9 | 5 × 5 = 7225
(105)^2 = 10 × 11 | 5 × 5 = 11025
### Multiplication with number 15.
1. 120 × 15 = 120 + (120/2) = 120 + 60 = 1800
162 × 15 = 162 + (162/2) = 162 + 81 = 2430
185 × 15 = 185 + (185/2) = 185 + 92.5 = 2775
# India's Leading Institution for Competitive Exams in Chandigarh BANK PO / SSC / NDA / CDS / CAT / UGC / SSB/ CLAT / HM / RRB NTPC
Call :- 9316488822 / 88821....
|
# Solving InequalitiesPage 1
#### WATCH ALL SLIDES
Slide 1
1.6 Solving Inequalities
Slide 2
## Solving Inequalities
Solving inequalities follows the same procedures as solving equations.
There are a few special things to
consider with inequalities:
We need to look carefully at the inequality sign.
We also need to graph the solution set.
Slide 3
## Review of Inequality Signs
> greater than
< less than
greater than or equal
less than or equal
Slide 4
## How to graph the solutions
> Graph any number greater than. . .
open circle, line to the right
< Graph any number less than. . .
open circle, line to the left
Graph any number greater than or equal to. . .
closed circle, line to the right
Graph any number less than or equal to. . .
closed circle, line to the left
Slide 5
## Solve the inequality:
x + 4 < 7
-4 -4
x < 3
Subtract 4 from each side.
Keep the same inequality sign.
Graph the solution.
Open circle, line to the left.
Slide 6
## There is one special case.
Sometimes you may have to reverse the direction of the inequality sign!!
That only happens when you
multiply or divide both sides of the inequality by a negative number.
Slide 7
Example:
Solve: -3y + 5 >23
-5 -5
-3y > 18
-3 -3
y < -6
Subtract 5 from each side.
Divide each side by negative 3.
Reverse the inequality sign.
Graph the solution.
Open circle, line to the left.
Slide 8
Try these:
Solve 2x+3>x+5
Solve - c - 11>23
Solve 3(r-2)<2r+4
|
# Do Now Try to extend the following patterns. What would be next? 1.January, March, May …. 2.7, 14, 21, 28, …. 3.1, 4, 9, 16, …. 4.1, 6, 4, 9, 7, 12, 10,
## Presentation on theme: "Do Now Try to extend the following patterns. What would be next? 1.January, March, May …. 2.7, 14, 21, 28, …. 3.1, 4, 9, 16, …. 4.1, 6, 4, 9, 7, 12, 10,"— Presentation transcript:
Do Now Try to extend the following patterns. What would be next? 1.January, March, May …. 2.7, 14, 21, 28, …. 3.1, 4, 9, 16, …. 4.1, 6, 4, 9, 7, 12, 10, … July, September(Every other month) 35, 42(Multiples of 7) 25, 36(Perfect Squares. i.e. 1 2, 2 2, 3 2, 4 2 …) 15, 13, 18, 16(Add 5 then subtract 2)
2.1 Using Patterns and Inductive Reasoning Target Use inductive reasoning to identify patterns and make conjectures.
inductive reasoning conjecture Vocabulary
Find the next item in the pattern. Example 3: Identifying a Pattern In this pattern, the figure rotates 90° counter- clockwise each time. The next figure is.
Check It Out! Example 4 Find the next item in the pattern 0.4, 0.04, 0.004, … When reading the pattern from left to right, the next item in the pattern has one more zero after the decimal point. The next item would have 3 zeros after the decimal point, or 0.0004.
When several examples form a pattern and you assume the pattern will continue, you are applying inductive reasoning. Inductive reasoning is the process of reasoning that a rule or statement is true because specific cases are true. You may use inductive reasoning to draw a conclusion from a pattern. A statement you believe to be true based on inductive reasoning is called a conjecture.
Complete the conjecture. Example 5: Making a Conjecture The sum of two positive numbers is ?. The sum of two positive numbers is positive. List some examples and look for a pattern. 1 + 1 = 23.14 + 0.01 = 3.15 3,900 + 1,000,017 = 1,003,917
Check It Out! Example 6 The product of two odd numbers is ?. Complete the conjecture. The product of two odd numbers is odd. List some examples and look for a pattern. 1 1 = 1 3 3 = 9 5 7 = 35
Example 7: Biology Application The cloud of water leaving a whale’s blowhole when it exhales is called its blow. A biologist observed blue-whale blows of 25 ft, 29 ft, 27 ft, and 24 ft. Another biologist recorded humpback- whale blows of 8 ft, 7 ft, 8 ft, and 9 ft. Make a conjecture based on the data. Heights of Whale Blows Height of Blue-whale Blows25292724 Height of Humpback-whale Blows 8789
Example 7: What conjectures can we make? Potential conjectures: The height of a blue-whale’s blow is greater than a humpback whale’s blow. -or- The height of a blue whale’s blow is about three times greater than a humpback whale’s blow. Heights of Whale Blows Height of Blue-whale Blows25292724 Height of Humpback-whale Blows 8789
Check It Out! Example 8 Make a conjecture about the lengths of male and female whales based on the data. In 5 of the 6 pairs of numbers above the female is longer. Conjecture: Female whales are longer than male whales. Average Whale Lengths Length of Female (ft) 495150485147 Length of Male (ft) 4745444648
To show that a conjecture is false, you have to find only one example in which the conjecture is not true. This case is called a counterexample. To show that a conjecture is always true, you must prove it. A counterexample can be a drawing, a statement, or a number.
Inductive Reasoning 1. Look for a pattern. 2. Make a conjecture. (use complete sentences) 3. Prove the conjecture or find a counterexample.
Lesson Quiz Find the next item in each pattern. 1. 0.7, 0.07, 0.007, … 2. 0.0007 Determine if each conjecture is true. If false, give a counterexample. 3. The quotient of two negative numbers is a positive number. 4. Every prime number is odd. 5. Two supplementary angles are not congruent. 6. The square of an odd integer is odd. false; 2 true false; 90° and 90° true
Adjustment to Homework Policy Homework that consists of answers only will receive a maximum of 5 points. Write given information/sketch picture for every problem. Show work on any problem requiring a calculation.
Assignment #11 Pages 85-88 Foundation: 7-19 odd, 22, 23, 27-31 odd (on 7-19, describe the pattern in words) Core: 36-43, 51 Challenge:54
Download ppt "Do Now Try to extend the following patterns. What would be next? 1.January, March, May …. 2.7, 14, 21, 28, …. 3.1, 4, 9, 16, …. 4.1, 6, 4, 9, 7, 12, 10,"
Similar presentations
|
# Sampling Variability
Video solutions to help Grade 7 students understand the term “sampling variability” in the context of estimating a population mean.
## New York State Common Core Math Grade 7, Module 5, Lesson 17
### Lesson 17 Student Outcomes
• Students use data from a random sample to estimate a population mean.
• Students understand the term “sampling variability” in the context of estimating a population mean.
### Lesson 17 Summary
A population characteristic is estimated by taking a random sample from the population and calculating the value of a statistic for the sample. For example, a population mean is estimated by selecting a random sample from the population and calculating the sample mean.
The value of the sample statistic (e.g., the sample mean) will vary based on the random sample that is selected. This variation from sample to sample in the values of the sample statistic is called sampling variability.
Lesson 17 Classwork
Example 1: Estimating a Population Mean
The owners of a gym have been keeping track of how long each person spends at the gym. Eight hundred of these times (in minutes) are shown in the population tables located at the end of the lesson. These 8000 times will form the population that you will investigate in this lesson.
Look at the values in the population. Can you find the longest time spent in the gym in the population? Can you find the shortest?
On average, roughly how long do you think people spend at the gym? In other words, by just looking at the numbers in the two tables, make an estimate of the population mean.
You could find the population mean by typing all numbers into a calculator or a computer, adding them up, and dividing by . This would be extremely time-consuming, and usually it is not possible to measure every value in a population.
Instead of doing a calculation using every value in the population, we will use a random sample to find the mean of the sample. The sample mean will then be used as an estimate of the population mean.
Example 2: Selecting a Sample Using a Table of Random Digits
The table of random digits provided with this lesson will be used to select items from a population to produce a random sample from the population. The list of digits are determined by a computer program that simulates a random selection of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Imagine that each of these digits is written on a slip of paper and placed in a bag. After thoroughly mixing the bag, one slip is drawn and its digit is recorded in this list of random digits. The slip is then returned to the bag and another slip is selected. The digit on this slip is recorded, and then returned to the bag. The process is repeated over and over. The resulting list of digits is called a random-number table.
How could you use a table of random digits to take a random sample?
Step 1: Place the table of random digits in front of you. Without looking at the page, place the eraser end of your pencil somewhere on the table. Start using the table of random digits at the number closest to where your eraser touched the paper. This digit and the following two specify which observation from the population tables will be the first observation in your sample.
For example, suppose the eraser end of your pencil lands on the twelfth number in row of the random digit table. This number is 5 and the two following numbers are 1 and 4. This means that the first observation in your sample is observation number 514 from the population. Find observation number 514 in the population table. Do this by going to Row and moving across to the column heading “4” This observation is 53, so the first observation in your sample is 53. If the number from the random-number table is any number 800 or greater, you will ignore this number and use the next three digits in the table.
Step 2: Continue using the table of random digits from the point you reached, and select the other four observations in your sample like you did above.
Exercises 1–4
Initially, you will select just five values from the population to form your sample. This is a very small sample size, but it is a good place to start to understand the ideas of this lesson.
The values found in Example 2 are used to illustrate the answers to the following questions, but each student should select their own random sample of five.
1. Use the table of random number to select five values from the population of times. What are the five observations in your sample?
2. For the sample that you selected, calculate the sample mean.
3. You selected a random sample and calculated the sample mean in order to estimate the population mean. Do you think that the mean of these five observations is exactly correct for the population mean? Could the population mean be greater than the number you calculated? Could the population mean be less than the number you calculated?
4. In practice, you only take one sample in order to estimate a population characteristic. But, for the purposes of this lesson, suppose you were to take another random sample from the same population of times at the gym. Could the new sample mean be closer to the population mean than the mean of these five observations? Could it be further from the population mean?
Exercises 5–7
As a class, you will now investigate sampling variability by taking several samples from the same population. Each sample will have a different sample mean. This variation provides an example of sampling variability.
5. Place the table of random digits in front of you and, without looking at the page, place the eraser end of your pencil somewhere on the table of random numbers. Start using the table of random digits at the number closest to where your eraser touches the paper. This digit and the following two specify which observation from the population tables will be the first observation in your sample. Write this three-digit number and the corresponding data value from the population in the space below.
6. Continue moving to the right in the table of random digits from the place ended in Exercise 5. Use three digits at a time. Each set of three digits specifies which observation in the population is the next number in your sample. Continue until you have four more observations, and write these four values in the space below.
7. Calculate the mean of the five values that form your sample. Round your answer to the nearest tenth. Show your work and your sample mean in the space below.
Exercises 8–11
You will now use the sample means from Exercise 7 from the entire class to make a dot plot.
8. Write the sample means for everyone in the class in the space below.
9. Use all the sample means to make a dot plot using the axis given below. (Remember, if you have repeated or close values, stack the dots one above the other.)
10. What do you see in the dot plot that demonstrates sampling variability?
11. Remember that in practice you only take one sample. (In this lesson, many samples were taken in order to demonstrate the concept of sampling variability.) Suppose that a statistician plans to take a random sample of size 5 from the population of times spent at the gym and that he or she will use the sample mean as an estimate of the population mean. Approximately how far can the statistician expect the sample mean to be from the population mean?
|
# Tag Info
## New answers tagged induction
0
$3^n\ge n^2, n\ge2$ $3^{n+1}=3^n+3^n+3^n\ge n^2+n^2+n^2\ge n^2+2n+1=(n+1)^2$ This uses the fact that $n^2\ge 2n,n\ge 2$ which can be shown via induction as well and obviously $n^2\ge1$ for all $n\ge 2$. The case for $1$ can be shown and then use $2$ as the base case.
1
Essentially, you want to show that $$3n^2 > (n+1)^2$$ which is not so hard since $$3n^2 - (n^2 + 2n + 1) > 0 \iff 2n^2 - 2n-1 > 0$$ But $2n^2 - 2n - 1 = 2(n^2 -n) - 1 = (n^2-2) + (n^2 - 2n+1) =(n^2 -2) + (n-1)^2$, so that we have for all $n \geq 2$ that $2n^2 - 2n - 1 \geq 0$ since $(n-1)^2$ is always $\geq 0$ and $n^2 - 2$ is $\geq 0$ when $n\geq ... 0 Hint:$3n^2-(n+1)^2=2n^2-2n-1>0$for all$n>1$. 2 It's true for$n=1$. Assume it holds for$n$, i.e:$3^{3n} + 1 = k(3^n +1)$then consider $$3^{3(n+1)} +1 = 27 \cdot 3^{3n} + 1 = 27 (3^{3n} +1) - 26 = 27k(3^n +1) - 26$$ Let's get it into a more amenable form $$$$3^{3(n+1)} +1 = 9k3^{n+1} + 27k - 26 = 9k(3^{n+1}+1) + 18k - 26$$ \tag{1}$$ So we want to show that$3^{n+1} + 1$... 1 In my answer I'm assuming that you also have$0 \in A$and that$A$is a set of natural numbers. Otherwise as celtschk said in a comment your proof is invalid since it could be that$A$contains things besides natural numbers. The logical flaws in your proof are:$\def\nn{\mathbb{N}}$You claimed that$0 \in B$. This would require$0 \in A$by definition, ... 1 Since the meat of your question seems to be about how to prove this claim with induction (even though the other answers provide much shorter and easier proofs), that is how I will choose to approach. Checking base case: Here, we check that the statement is true for some starting value(s) of$n$. The statement's truth will depend only upon$n$as$l$is ... 0 Yeah if you use binomial theorem ..then $$\sum_{l=0}^{n} nC_l=^nC_0+^nC_1+^nC_2+......$$ Which is equal to $$1+ ^nC_1+^nC_2+....+^nC_n$$ That is$(1+1)^n$...taking x=1and a=1 in the expansion of$(x+a)^n$1 For a prove by induction: You can start the induction at$n=0$, which is trivial. So let$n$be greater than$0\begin{align} \sum_{k=0}^{n} \binom{n}{k} &= 1+ \sum_{k=1}^{n-1} \binom{n}{k} + 1 \\ &= 1+ \sum_{k=1}^{n-1} \left[ \binom{n-1}{k-1} + \binom{n-1}{k} \right] + 1 \\ &= 1+ \sum_{k=1}^{n-1} \binom{n-1}{k-1} + \sum_{k=1}^{n-1}\binom{... 0 The other approaches are much faster, but here is how you can proceed if you'd like to use induction: Forn=1$we get$\sum_{l=0}^n{n\choose l}={1\choose 0}+{1\choose 1}=1+1=2^1$For the induction step, use the identity $${n+1\choose l}={n\choose l}+{n\choose l-1}$$ for$1\leq l\leq n$to obtain $$\sum_{l=0}^{n+1}{n+1\choose l}={n+1\choose 0}+{n+1\... 1 If S is a set with n elements, then \binom{n}{l} is the number of subsets of S with l elements. \sum_{l=0}^{n}\binom{n}{l} is then the number of all subsets of S and this is equal to 2^{n} because the subsets of S can be put into bijection with the set of n-bit strings. 0 The binomial coefficient n\choose i is by definition, the coefficient of x^i in the expanded binomial (1+x)^n, i.e., (1+x)^n=\sum_{i=0}^n{n\choose i}x^i. Use this definition with x=1. 1 Hint \ To prove \,f_n = {\rm rhs} - {\rm lhs} > 0\, for all \,n\ge 2\, note \,f_2 > 0\, (base) and note$$\ \color{#c00}{f_{n+1}-f_n} =\, \frac{1}{n(n+1)}-\frac{1}{(n+1)^2}\, =\, \frac{1}{n(n+1)^2} \color{#c00}{> 0}$$thus \, f_n > 0 \,\Rightarrow\, \color{#c00}{f_{n+1} > f_n} > 0\ (induction step) Remark \ The ... 2 Let$$f(n) = \frac{1}{2^2} + \cdots + \frac{1}{n^2}.$$Now we want to show that$$f(n)<\frac{n-1}{n}\tag{1}$$for all integers greater than 1. A proof by induction consists of two equally important steps. In the base case we show that (1) indeed holds for n=2. In the inductive step we assume that (1) is true for some number n and use that to ... 0 We want to prove:$$ H_{n}^{(2)}\leq 2-\frac{1}{n}\tag{1} $$by induction. (1) holds for n=1, hence it is enough to prove that for every n\geq 1$$ H_{n+1}^{(2)}-H_{n}^{(2)}=\frac{1}{(n+1)^2}\leq \frac{1}{n}-\frac{1}{n+1} = \frac{1}{n(n+1)}\tag{2} $$holds, but that is trivial. In a similar way we me prove the improved inequality:$$\boxed{\forall n\... 1 The coefficients given by the formula are correct; for instance,${2}\choose{1}$$=2, so it gives you x^2+2xy+y^2. In general, {n}\choose{k}$$=$$$\frac{n!}{k!(n-k)!}$$ e.g.${2}\choose{1}$$=$$\frac{2!}{1!\cdot1!} = \frac{2}{1} = 2$$2 Assume the arithmetic progression, starting with \frac1{m_1}, contains k positive terms. Now the (uniform) term difference is$$ d = \frac{1}{m_1} - \frac{1}{m_2} \\ m_2 \geq m_1+1 \implies d \geq \frac{1}{m_1} - \frac{1}{m_1+1}=\frac{1}{m_1(m_1+1)} $$The entire sequence, which ends in a positive value \frac{1}{m_k}, consists of k-1 steps of size ... 2 It is important in this problem to understand that the maximun number k is determined by the choice of the two first integers m_1,m_2. Let m_1 and m_2=m_1+h (where h\ge 1); the common difference of the a. p. is d=\frac{-h}{m_1(m_1+h)} so we have$$u_1=\frac{1}{m_1}\\u_2=\frac{1}{m_1+h}\\u_3=\frac{1}{m_1+h}+\frac{-h}{m_1(m_1+h)}=\frac{m_1-h}{m_1(...
2
Since they are in arithmetic progression let us call $1/m_n-1/m_{n+1}=L>0$. We then have that \begin{align} \sum_{n=1}^{k-1}\left(\frac{1}{m_n}-\frac{1}{m_{n+1}}\right) &= \sum_{n=1}^{k-1}L \\ &= (k-1)L \\ &= \frac{1}{m_1}-\frac{1}{m_k} \end{align} Which means we just have to show k=\frac{\frac{1}{m_1}-\frac{1}{m_k}}{L}+1<m_1+2\qquad (... 0 Let S be the set of all permutations f of {1,2,3,..,, n+1} with at least one non-fixed point (i.e., a value k with f(k)≠k). Then |S| = (n+1)!-1. Now count the permutations with the highest non-fixed point first, then those with the 2nd highest non-fixed point (which are not already counted), then those with the 3rd highest non-fixed point (which ... 0 Here is how I would write up the main part of the induction proof (DeepSea and Bill handle the base case easily), in the event that you may find it useful: \begin{align} a^{4k+5}-a&= a^4(a^{4k+1}-a)+a^5-a\tag{rearrange}\\[1em] &= a^4(30\eta)+a^5-a\tag{by ind. hyp.; \eta\in\mathbb{Z}}\\[1em] &= a^4(30\eta)+30\ell\tag{by base case; \ell\in\... 0 The induction hypothesis is that a^{4n+1}-a=30k, for some integer k. Therefore a^{4(n+1)+1}-a=a^4\cdot a^{4n+1}-a=a^4(30k+a)-a=30ka^4+a^5-a $$and the proof is reduced to showing that a^5-a is divisible by 30, which is the base step. 0 Hint \ See this answer for various proofs of the base case \,30\mid \color{#0a0}{a^{\large 5}-a}.\, The inductive step is {\rm mod}\ 30\!:\,\ \color{#c00}{a^{\large 4n+1}\equiv a}\,\Rightarrow\,a^{\large 4(n+1)+1}\!\equiv a^{\large 4} \color{#c00}{a^{\large 4n+1}}\!\equiv a^{\large 4}\color{#c00}a\equiv \color{#0a0}{a^{\large 5}\equiv a}.\ \ QED ... 0 Don't factor entirely. Just factor enough to realize a^{4n+1} - a = a(a^{4n} - 1)=a(a - 1)(a^{4n-1} + a^{4n-2} + .... + a + 1). Assume for n = k that a^{4k + 1}-a= a(a^{4n} - 1)=a(a - 1)(a^{4k-1} + a^{4k-2} + .... + a + 1)= 30M then a^{4(k+1) + 1} - a = a(a-1)(a^{4k + 3} + ...)= a(a-1)(a^{4k + 3} + a^{4k + 2} + a^{4k+1} + a^{4k}) +a(a - 1)(a^{... 0 a(a^{4n}-1)=a(a^{2n}-1)(a^{2n}+1)=a(a^{n}-1)(a^{n}+1)(a^{2n}+1). Assume 30 | a(a^{4n}-1). 30 | a(a^{4(n+1)}-1) if and only if 30 | a(a^{4(n+1)}-1)-a(a^{4n}-1). Try factoring: a(a^{4(n+1)}-1)-a(a^{4n}-1). Think about how regardless of a, you can show that one of the terms of the factorization must be divisible by 2, another divisible by 3, and ... 1 We start with the base case: n = 1 \implies P(1): a^5 - a = a(a-1)(a+1)(a^2+1). The product 6 = 3! \mid a(a-1)(a+1) . If a = 0, 1, 4 \pmod 5 \implies 5 \mid a^5 - a \implies 30 \mid a^5 - a since \text{gcd}(5,6) = 1. If a = 2, 3 \pmod 5 \implies 5 \mid a^2 + 1 \implies 30 \mid a^5 - a. Thus P(1) is true. Assume P(n): 30 \mid a^{4n+1} - a is ... 9 As an interesting alternative, note that x^3 +1 = (x+1)(x^2 - x + 1), so setting x = 3^n gives 3^{3n} + 1 = (3^n + 1)(3^{2n} - 3^n + 1). 2 Divide both sides by 2, and you have the fact that the sum up to n+1 of \left( \begin{array}{c} k \\ 2 \end{array} \right) is \left( \begin{array}{c} n+2 \\ 3 \end{array} \right) . This is the reason that the total number of gifts given in The Twelve Days of Christmas song is \left( \begin{array}{c} 14 \\ 3 \end{array} \right) =364 . In ... 2 You have proved the statement via induction. Your job is done. When you write that$$\sum_\limits{k}^{n+1}k(k+1)=\frac{1}{3}(n+1)(n+2)(n+3)$$What you actually mean is$$\sum_\limits{k}^{n+1}k(k+1)=\frac{1}{3}(n+1)\{(n+1)+1\}\{(n+2)+1\}$$So if I represent the mathematical statement \sum_\limits{k}^{n}k(k+1)=\frac{1}{3}n(n+1)(n+2) as P(n), then you ... 0 You have actually proved the statement, but it is a little hard to see, because in your inductive step you went from n to n+1. Instead, go from n-1 to n: Suppose the statement holds for n-1, so$$ S_{n-1} := \sum_{k=1}^{n-1} k (k+1) = {1 \over 3} (n-1) (n) (n+1). $$Then$$ S_{n-1} + n(n+1) = n (n+1) \left[ 1 + {1 \over 3} (n-1) \right] = n (n+...
2
You already have the first step, that it is true for $n=1$. In the second step, you are assuming it is true for $n$ and you want to prove it is true for $n+1$. So by assumption your $\sum_{k=1}^{n} k(k+1)$ is $\frac{1}{3} n (n+1)(n+2)$, and you want to show that $$(n+1)(n+2) + \frac{1}{3} n (n+1)(n+2) = \frac{1}{3} (n+1)(n+2)(n+3)$$ which is a ...
0
You have proven that it is true for the case that n = 1 (i.e. that 2 = 2 - this is called the base case). You have then shown that if the relationship is true for n, it is also true for n + 1 (this is the inductive case); this is what the last line shows. Therefore, since it is true for n = 1, it is true for n + 1 = 2; since it is true for n = 2, it is true ...
1
Hint : For $k^2 \le p < (k+1)^2$, we have $k=\sqrt{k^2} \le \sqrt{p} < \sqrt{(k+1)^2}= k+1$. So using the definition of the floor function, what could be the value of $\lfloor \sqrt{p} \rfloor$?
0
Show the contrapositive: If we have more tan $2^{n-1}$ subsets, then we find two among them that are disjoint. This can be done by the pigeon-hole principle: There are exactly $2^{n-1}$ subsets $A$ with $n\notin A$. If declare $A$ and $\{1,2,,\ldots,n\}\setminus A$ as a "hole", then we have $2^{n-1}$ holes. Henec if we have more than $2^{n-1}$ subsets - or "...
2
Note that there are exactly $2^n$ subsets in total to choose from. It is easy to see how to pick $2^{n-1}$ pairwise intersecting sets because we can just take all subsets of $[n-1]$ and then add $n$ to them. To see that we cannot have more than $2^{n-1}$ pairwise intersecting sets, begin with a set of $2^{n-1}$ pairwise intersecting sets. For each of those ...
2
An inductive proof is not the way to go. HINT: Let $[n]=\{1,\ldots,n\}$. If $A\subseteq[n]$, can such a collection contain both $A$ and $[n]\setminus A$? What I think you’re missing in your attempt to prove the result by induction is that there are many such collections of subsets of $[n]$. For instance, if $n=3$ one such collection is $$\big\{\{1\},\{... 1 Note: An inductive proof is given as an answer by David W.Farlow. Explanations of some steps and flaws in your inductive argument: We get (1) from the inductive hypothesis (which is assumed to be true.)$$(1+kh) \leq (1+h)^k \Rightarrow (1+h)(1+kh) \leq (1+h)^k(1+h)=(1+h)^{k+1}$$as h >-1 \Rightarrow h+1>0 and so multiplication of both sides ... 0 First of all, why have you been reading a bunch of solutions? I suggest you make a serious attempt at the problem yourself and only then look to other solutions to bridge whatever gap that exists in your reasoning (looking up a solution should be the very last resort). To that end, note that the inequality is trivial for when n=0,1. Thus, consider the ... 1 The induction step breaks down to: Let X = \cup_{i=0}^n A_i; a union of n finite sets that is, by assumption finite. Let Y = A_{n+1}; a finite set. So \cup_{i=0}^{n+1} A_i = X \cup Y; the union of two finite sets. Suffices to show the union of two finite sets is finite. Which is so absolutely obvious we just know it's going to be a bitch to ... 1 By definition,$$\bigcup\limits_{i=1}^{k+1} A_i = [\bigcup\limits_{i=1}^k A_i] \cup A_{k+1}$$Since p(k) is true, \bigcup\limits_{i=1}^k A_i is finite. Since p(2) is true (i.e. the union of two finite sets is finite), so is p(k+1). I want to point out that if A_1, ... , A_n are sets, \bigcup\limits_{i=1}^n A_i doesn't necessarily make sense. ... 1 I'm not sure what you mean by multiplying or adding anything. You need to show that$$\bigcup_{i=1}^{k+1}A_i$$is finite. In particular$$\bigcup_{i=1}^{k+1}A_i=\left(\bigcup_{i=1}^{k}A_i\right)\bigcup A_{k+1}$$and thus you have the union of two finite sets. Depending on your definition of finite, this may be all you have to say (i.e. the union of two ... 0 We give some hints for your question: 1) S_{n+1}-S_n=x^n 2) \displaystyle\frac{x^{n+1}-1}{x-1}-\displaystyle\frac{x^{n}-1}{x-1}=x^n. 0 Let S(n) be the statement: \displaystyle\sum_{k=0}^{n-1}x^{k}=\dfrac{x^{n}-1}{x-1} Basis step: S(1): LHS: \displaystyle\sum_{k=0}^{(1)-1}x^{k}=x^{0} \hspace{23 mm}=1 RHS: \dfrac{x^{(1)}-1}{x-1}=\dfrac{x-1}{x-1} \hspace{27 mm}=1 \hspace{73.5 mm} LHS = RHS \hspace{1 mm} (verified.) Inductive step: Assume S(m) is true, i.e. assume ... 0 Another approach is to use congruences: First note that if k is odd, then$$ 4^k\equiv 4\ (\text{mod}\,10). $$Now, if n is odd, then n+2 is odd, so$$ 1^{n+2}+2^{n+2}+3^{n+2}+4^{n+2}\equiv 1+2^{n+2}+3^{n+2}+4=5+2^{n+2}+3^{n+2}\ (\text{mod}\,10). $$It suffices to prove that 5\mid 2^{n+2}+3^{n+2}. Indeed, since n+2 is odd we get$$ 2^{n+2}+3^{n+2}\...
3
Induction: Consider the difference, $$\left(1 + 2^{n+2} + 3^{n+2} + 4^{n+2}\right)-\left(1+2^n+3^n+4^n\right).$$ Is it divisible by $10$?
2
You made lots of little mistakes but your biggest mistake is that you inductions step, doesn't actually do any inducing. A proper induction step always goes like this: ======= We assume that something is true for a specific $k = n$ $\implies$ Then something is true for $k = n+1$. ====== Thus, we start with showing it is true for $k=$ base case; thus ...
0
We can also use the Abel's summation and get $$S=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\sum_{k=1}^{n}1\cdot\frac{1}{\sqrt{k}}=\sqrt{n}+\frac{1}{2}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t^{3/2}}dt$$ where $\left\lfloor t\right\rfloor$ is the floor function. Since $t-1\leq\left\lfloor t\right\rfloor \leq t$ we have 2\sqrt{n}-2+\frac{1}{\sqrt{...
2
Hint Equivalently we seek to prove that $\,f(n) = 2^n/(4n) > 1\,$ for all $\,n\ge 5.$ Note $\,f(5)>1\,$ and $\,f(n\!+\!1)/f(n) = 2n/(n\!+\!1) \ge 1\,$ for $\,n\ge 5\,$ so $\,\color{#c00}{f(n\!+\!1) \ge f(n)}\,$ Hence the induction reduces to a trivial one: an $\rm\color{#c00}{increasing}$ sequence stays $\,\ge\,$ its initial value. From ...
3
You got the basis step correct by checking that $4n<2^n$ for $n=5$. Next, you must prove that $4N<2^N \implies 4(N+1) < 2^{N+1}$ for $N\geq5$ When trying to solve this problem, simplify the right-hand-side of this implication to a form that is easily comparable to the left-hand-side. Notice that $4(N+1) < 2^{N+1}$ holds if and only if 4N+4 &... 4 Consider the following (see if you can determine how one step relates to another): \begin{align} 4(k+1)&=4k+4\\[1em] &< 2^k+4\tag{why?}\\[1em] &< 2^k+2^k\tag{why?}\\[1em] &= 2\cdot2^k\\[1em] &=2^{k+1}. \end{align} 1 Your choice ofn$should only be for the basis of the induction, not for the inductive step. What you need to do is to show that if$4n<2^n$then$4(n+1)<2^{n+1}$. One way of doing that is to say that$4(n+1)=4n+4<2^n+4$and then argue that$x+4<2x\$ for the relevant values.
Top 50 recent answers are included
|
# Video: KS2-M16 • Paper 2 • Question 19
Miss Mills is making jam to sell at the school fair. The cost of 1 kg of strawberries is £7.50. The cost of 1 kg of sugar is 79 p. The cost of 10 glass jars is £6.90. She uses 12 kg of strawberries and 10 kg of sugar to make 20 jars full of jam. Calculate the cost to make 20 jars full of jam.
04:03
### Video Transcript
Miss Mills is making jam to sell at the school fair. The cost of one kilogram of strawberries is seven pounds 50. The cost of one kilogram of sugar is 79 p. The cost of 10 glass jars is six pounds 90. She uses 12 kilograms of strawberries and 10 kilograms of sugar to make 20 jars full of jam. Calculate the cost to make 20 jars full of jam.
Miss Mills has three costs: the cost of strawberries, the cost of sugar, and the cost of the glass jars. We know that strawberries cost seven pounds 50 per kilogram. And Miss Mills uses 12 kilograms of strawberries. To work out the cost of 12 kilograms of strawberries, we need to multiply 12 by seven pounds 50. 10 times seven pounds 50 is 75 pounds. And two times seven pounds 50 is 15 pounds. 75 pounds plus 15 pounds equals 90 pounds. So the cost of the strawberries is 90 pounds.
Now we need to work out the total cost of the sugar Miss Mills used. We know that one kilogram of sugar is 79 p. And the question tells us that Miss Mills used 10 kilograms of sugar. 79 p multiplied by 10 equals 790 pence. We can also write this as seven pounds 90.
The final cost for Miss Mills is the cost of the glass jars. We know that 10 glass jars costs six pounds 90. And the question tells us that Miss Mills needs 20 jars. So to find the cost of 20 jars, we need to double the amount it costs for 10 jars. Double six pounds is 12 pounds. And 90 pence times two is one pounds 80. 12 and one pound 80 equals 13 pounds 80. 12 pounds plus one pound 80 equals 13 pounds 80.
Now that we found the cost of the strawberries, the sugar, and the jars, we can add these three amounts together to give us the total amount of Miss Mills costs. zero plus zero plus zero equals zero. Nine plus eight is 17. We can write the seven and carry the one. Seven plus three is 10 plus the one we carried, which makes 11. Again, we can write the one and we need to carry the other one. Nine plus one is 10, plus the one we carried, again, makes 11. So the total cost for Miss Mills to make 20 jars of jam is 111 pounds and 70 pence.
We calculated the cost of the strawberries, the cost of the sugar, the cost of the jars, and then added these three amounts together to give us our total of 111 pounds 70.
|
Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Class 10
Important Questions for Exam - Class 10
Β
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v) (7π₯ β 2π¦)/π₯π¦ = 5 (8π₯ + 7π¦)/π₯π¦ = 15 Given (7π₯ β 2π¦)/π₯π¦ = 5 (7π₯ )/π₯π¦ β (2π¦ )/π₯π¦ = 5 (7 )/π¦ β(2 )/π₯ = 5 (βπ )/π +(π )/π = 5 (8π₯ + 7π¦)/π₯π¦ = 15 (8π₯ )/π₯π¦ + (7π¦ )/π₯π¦ = 15 (8 )/π¦ +(7 )/π₯ = 15 (π )/π +(π )/π = 15 Our equations are (β2 )/π₯ +(7 )/π¦ = 5 β¦(1) (7 )/π₯ +(8 )/π¦ = 15 ...(2) So, our equations become β2u + 7v = 5 7u + 8v = 15 Hence, we solve β2u + 7v = 5 β¦(3) 7u + 8v = 15 β¦(4) From (3) β2u + 7v = 5 7v = 5 + 2u v = (5 + 2π’)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2π’)/7) = 15 Multiplying 7 both sides 7 Γ 7u + 7 Γ 8 ((5 + 2π’)/7) = 7 Γ 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 β 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) β2u + 7v = 5 β2(1) + 7v = 5 β2 + 7v = 5 7v = 5 + 2 7v = 7 v = 7/7 v = 1 Hence, u = 1, v = 1 But we have to find x & y We know that u = π/π 1 = 1/π₯ x = 1 v = π/π 1 = 1/π¦ y = 1 Hence, x = 1 , y = 1 is the solution of the given equation Question 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vi) 6x + 3y = 6xy 2x + 4y = 5xy Given 6x + 3y = 6xy Diving whole equation by xy (6π₯ + 3π¦)/π₯π¦ = 6π₯π¦/π₯π¦ 6π₯/π₯π¦ +3π¦/π₯π¦ = 6 π/π +π/π = 6 2x + 4y = 5xy Diving whole equation by xy (2π₯ + 4π¦)/π₯π¦ = 5π₯π¦/π₯π¦ 2π₯/π₯π¦ +4π¦/π₯π¦ = 5 π/π +π/π = 5 Hence, our equations are 6/π¦ +3/π₯ = 6 β¦(1) 2/π¦ +4/π₯ = 5 β¦(2) So, our equations become 6v + 3u = 6 2v + 4u = 5 Now, we solve 6v + 3u = 6 β¦(3) 2v + 4u = 5 β¦(4) From (3) 6v + 3u = 6 6v = 6 β 3u v = (6 β 3π’)/6 Putting value of v in (4) 2v + 4u = 5 2((6 β 3π’)/6) + 4u = 5 ((6 β 3π’)/3) + 4u = 5 Multiplying both sides by 3 3 Γ ((6 β3π’)/3) + 3 Γ 4u = 3 Γ 5 (6 β 3u) + 12u = 15 β3u + 12u = 15 β 6 9u = 9 u = 9/9 u = 1 Putting u = 1 in (3) 6v + 3u = 6 6v + 3(1) = 6 6v + 3 = 6 6v = 6 β 3 6v = 3 v = 3/6 v = π/π Hence, u = 1 , v = 1/2 But we have to find x & y Now, u = π/π 1 = 1/π₯ x = 1 v = π/π 1/2 = 1/π¦ y = 2 Hence, x = 1 , y = 2 is the solution of the given equation
|
# What is 6/12 + 5/10?
6 12
+
5 10
## Step 1
We still have different denominators (bottom numbers), though, so we need to get a common denominator. This will make the bottom numbers match. Multiply the denominators together first. Now, multiply each numerator by the other term's denominator.
Now we multiply 6 by 10, and get 60, then we multiply 12 by 10 and get 120.
Now for the second term. You multiply 5 by 12, and get 60, then multiply 12 by 10 and get 120.
This gives us a new problem that looks like so:
60 120
+
60 120
## Step 2
Since our denominators match, we can add the numerators.
60 + 60 = 120
That gives us an answer of
120 120
## Step 3
Can this fraction be reduced?
First, we attempt to divide it by 2...
Are both the numerator and the denominator evenly divisible by 2? Yes! So we reduce it:
120 120
÷ 2 =
60 60
Let's try dividing by that again...
Are both the numerator and the denominator evenly divisible by 2? Yes! So we reduce it:
60 60
÷ 2 =
30 30
Let's try dividing by that again...
Are both the numerator and the denominator evenly divisible by 2? Yes! So we reduce it:
30 30
÷ 2 =
15 15
Let's try dividing by that again...
Nope! So now we try the next greatest prime number, 3...
Are both the numerator and the denominator evenly divisible by 3? Yes! So we reduce it:
15 15
÷ 3 =
5 5
Let's try dividing by that again...
Nope! So now we try the next greatest prime number, 5...
Are both the numerator and the denominator evenly divisible by 5? Yes! So we reduce it:
5 5
÷ 5 =
1 1
Let's try dividing by that again...
No good. 5 is larger than 1. So we're done reducing.
There you have it! Here's the final answer to 6/12 + 5/10
6 12
+
5 10
=
1 1
= 1
|
```Question 140987
{{{- 8x + 5y = 40}}} Rearrange the terms
So we now have the system of equations:
{{{5x-8y=-40}}}
{{{-8x+5y=40}}}
In order to graph these equations, we need to solve for y for each equation.
So let's solve for y on the first equation
{{{-8y=-40-5x}}} Subtract {{{5 x}}} from both sides
{{{-8y=-5x-40}}} Rearrange the equation
{{{y=(-5x-40)/(-8)}}} Divide both sides by {{{-8}}}
{{{y=(-5/-8)x+(-40)/(-8)}}} Break up the fraction
{{{y=(5/8)x+5}}} Reduce
Now lets graph {{{y=(5/8)x+5}}} (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver</a>)
{{{ graph( 600, 600, -10, 10, -10, 10, (5/8)x+5) }}} Graph of {{{y=(5/8)x+5}}}
So let's solve for y on the second equation
{{{5y=40+8x}}} Add {{{8 x}}} to both sides
{{{5y=+8x+40}}} Rearrange the equation
{{{y=(+8x+40)/(5)}}} Divide both sides by {{{5}}}
{{{y=(+8/5)x+(40)/(5)}}} Break up the fraction
{{{y=(8/5)x+8}}} Reduce
Now lets add the graph of {{{y=(8/5)x+8}}} to our first plot to get:
{{{ graph( 600, 600, -10, 10, -10, 10, (5/8)x+5,(8/5)x+8) }}} Graph of {{{y=(5/8)x+5}}}(red) and {{{y=(8/5)x+8}}}(green)
With help from a graphing calculator, we can see that the two lines intersect at the point ({{{-40/13}}},{{{40/13}}})
This means that system is independent and consistent. So you are correct.```
|
# How To Find The Area Of A Shape
Size: px
Start display at page:
Transcription
1 9 Areas and Perimeters This is is our next key Geometry unit. In it we will recap some of the concepts we have met before. We will also begin to develop a more algebraic approach to finding areas and perimeters. 9.1 Area The easiest method to find an area of a shape, particularly if it is a simple shape made up of straight lines, is to count the squares inside it. Example 1 Find the area of each of these shapes in terms of the square shown. One square Solution This can be divided into 5 of the squares, so its area is 5 square units. 5a 5b We have 4 squares (labelled 1, 2, 3 and 4), and the two triangles (labelled 5a and 5b) can be joined together to form another square. So, in total, we have an area of 5 square units. 121
2 9.1 MEP Y7 Practice Book A Example 2 Estimate the area of this shape Solution There are 4 complete squares (labelled 1, 2, 3 and 4). Region 5 and 6 together make up about squares, as do regions 7 and 8. So we have another 3 squares giving a total of 7 square units (plus a little bit more!). Exercises 1. Draw around your hand on squared paper and find its area. Who has the largest hand in your class? 2. Find the area of this circle by counting squares. 122
3 3. Find the areas of these shapes by counting squares. (d) 123
4 9.2 Area and Perimeter of a Square We now bring in standard units for measuring area and perimeter. You should always put units in your answers. The area of a square can be found by counting squares or multiplying the length of the sides. The area of a square with sides 1 cm is 1 cm 2. Area = 1 cm Area = 1 2 Area = 4 4 = 1 2 The perimeter of a square is the total length of the four sides. 1 cm 1 cm 1 cm 1 cm Perimeter = Perimeter = = = 1 124
5 Note Note also that: So that, for example, 1 m = 100 cm 1 cm = 10 mm 25 mm = 2. 8 mm = cm = 2.61 m 3 = 0.32 m = 0.06 m Exercises 1. Find the area and perimeter of each of these squares. 125
6 9.2 MEP Y7 Practice Book A 2. Find the area of squares with sides of length: 10 cm 1 (d) 9 cm (e) 1 (f) 20 cm 3. Find the perimeter of squares with sides of length: 1 1 (d) 19 cm (e) 9 cm (f) 1 4. Copy and complete each of these statements. 3. = mm 10. = mm 28 mm = cm (d) 216 mm = cm (e) 15 = m (f) 8 = m (g) 1.62 m = cm (h) 1.7 m = cm (i) 0.82 m = cm (j) 0.07 m = cm 5. A square has sides of length 20 mm. Find the area of the square in: mm 2 cm 2 6. The perimeter of a square is 40 cm. How long are its sides? 126
7 7. The area of a square is 3 2. How long are its sides? 8. The perimeter of a square is 4. What is its area? 9. The area of a square is What is its perimeter? 10. For a square the perimeter is and the area is 2. The perimeter is twice the area. What are the lengths of the sides of a square for which the perimeter is equal to the area; half of the area? 9.3 The Area and Perimeter of a Rectangle For a rectangle, say by, we can proceed either by counting squares or multiplying the lengths. So for example, the area of this rectangle is 10 cm 2 from counting squares or, alternatively; Area = 5 2 Note also that 1 cm is the same as 10 mm, = 10 cm 2 so that a 1 cm square has an area of 1 cm 2 and this can also be written as 1 cm 1 cm = 10 mm 10 mm i.e 1 cm 2 = 100 mm 2 127
8 9.3 MEP Y7 Practice Book A Example What is 1 m 2 in terms of cm 2? Solution 1 m 1 m = 100 cm 100 cm i.e. 1 m 2 = cm 2 Exercises 1. Find the area of these rectangles in cm
9 (d) 2. Find the perimeter of the rectangles in question Find the area of these rectangles in suitable units. The diagrams have not been drawn accurately. (d) 1 mm 11 mm 9 cm (e) 7 cm (f) 9 mm 3 mm 129
10 9.3 MEP Y7 Practice Book A 4. Find the perimeter of the rectangles in question Find the area and perimeter of these rectangles mm (d) 1.5 m 4.2 mm 1.4 m (e) 3. (f) 7.4 mm 6.1 cm 8 mm 6. Find the area and perimeter of this rectangle in cm 2 and cm in m 2 and m. 1 m 30 cm 7. Find the area of this rectangle in mm 2 and cm
11 8. Find the perimeter and area of this rectangle making clear which units you have decided to use mm 9. A rectangle has an area of 4 2. The length of one side is. Find the perimeter of the rectangle. 10. A rectangle has a perimeter of 2 and an area of 3 2. What are the lengths of the sides of the rectangle? 9.4 Area of Compound Shapes We illustrate this method with an example. Example Find the area of the shape shown below. 7 cm 131
12 9.4 MEP Y7 Practice Book A Solution Divide the shape into rectangles; one way is shown below. B A C 7 cm Area of A = 3 7 = 21 cm 2 Area of B = 3 2 = 2 Area of C = 3 6 = 1 2 Total Area = = 4 2 Exercises 1. Find the area of these shapes. 132
13 2. Find the area of each of these shapes. The diagrams have not been drawn accurately. 25 mm (d) 20 mm 1 cm 40 mm 35 mm (e) (f) 7 cm 133
14 9.4 MEP Y7 Practice Book A 3. Find the perimeter of the shapes in question Find the area and perimeter of these shapes. 5 mm 1 cm 2 mm 1 mm 5 mm 2 mm 2 mm (d) 10 cm 5. Find the shaded area in each of the diagrams below. 1 cm 1 cm 1 cm 1 cm 6. (d) 0.7 cm cm 9 cm
15 6. Find the area and perimeter of these shapes. 3.1 cm m 2.1 m 3.5 m m 6.5 m 3.7 cm 13.2 m (d) A lawn is 3 m by 5 m. A path 1 m wide is laid around the lawn. Find the area of the path. Path Lawn Path 8. Ben's dad buys a carpet that is 3 m wide and 4 m long. 3.6 m 2.6 m 1.4 m This is the shape of Ben's room 1.2 m 3.8 m How much carpet is wasted? 135
16 9.4 MEP Y7 Practice Book A 9. Wendy is going to paint the front of the house for her mum and dad. 2 m 1 m Find the area that needs to be painted. 1.5 m 1.5 m 1 m 2 m 1.5 m 2 m 5 m Wendy gets 50p for every 1 m 2 that she paints. How much money does Wendy get? 6 m 10. This shape can be cut out of card and folded to form a box. How much card is wasted if this shape is cut out of a sheet 1 by 20 cm? 9.5 The Area of a Triangle For a triangle, Area = 1 2 base perpendicular height Height Base 136
17 Example 1 Find the area of the triangle shown. 7 cm Solution Area = = 21 cm 2 Example 2 Find the area of the triangle shown. 7 cm Solution Area = = 1 2 Exercises 1. Find the area of each of these triangles. 5 m 7 m (d) 1 40 mm 30 mm 137
18 9.5 MEP Y7 Practice Book A (e) (f) (g) 7. (h) 5.2 mm mm 2. The diagrams show two triangles drawn inside a rectangle. Explain why the two shaded triangles have the same area. 3. Find the area of the shaded triangle. What is the area of the parallelogram? 138
19 4. Find the area of this parallelogram. 5. Find the area of each of these shapes. They have not been drawn accurately. 7 cm 7 cm (d) 10 cm 1 cm 139
20 9.5 MEP Y7 Practice Book A 6. A pyramid can be made by folding up the shape below. What is the area of this shape? Find the shaded areas
21 8. Draw this triangle. Find its area to the nearest 0.1 cm Find the area of this triangle to the nearest 0.1 cm 2. 7 cm 10. Find the area of an equilateral triangle with sides of length, giving your answer correct to 1 decimal place. 11. What is the area of the biggest triangle that can be drawn inside a parallelogram with sides of length 10 cm and 1? 141
### Perimeter is the length of the boundary of a two dimensional figure.
Section 2.2: Perimeter and Area Perimeter is the length of the boundary of a two dimensional figure. The perimeter of a circle is called the circumference. The perimeter of any two dimensional figure whose
### 9 Area, Perimeter and Volume
9 Area, Perimeter and Volume 9.1 2-D Shapes The following table gives the names of some 2-D shapes. In this section we will consider the properties of some of these shapes. Rectangle All angles are right
### Unit 8 Angles, 2D and 3D shapes, perimeter and area
Unit 8 Angles, 2D and 3D shapes, perimeter and area Five daily lessons Year 6 Spring term Recognise and estimate angles. Use a protractor to measure and draw acute and obtuse angles to Page 111 the nearest
### Student Outcomes. Lesson Notes. Classwork. Exercises 1 3 (4 minutes)
Student Outcomes Students give an informal derivation of the relationship between the circumference and area of a circle. Students know the formula for the area of a circle and use it to solve problems.
### Geometry of 2D Shapes
Name: Geometry of 2D Shapes Answer these questions in your class workbook: 1. Give the definitions of each of the following shapes and draw an example of each one: a) equilateral triangle b) isosceles
### Assessment For The California Mathematics Standards Grade 4
Introduction: Summary of Goals GRADE FOUR By the end of grade four, students understand large numbers and addition, subtraction, multiplication, and division of whole numbers. They describe and compare
### Calculating Area, Perimeter and Volume
Calculating Area, Perimeter and Volume You will be given a formula table to complete your math assessment; however, we strongly recommend that you memorize the following formulae which will be used regularly
### 16 Circles and Cylinders
16 Circles and Cylinders 16.1 Introduction to Circles In this section we consider the circle, looking at drawing circles and at the lines that split circles into different parts. A chord joins any two
### Dŵr y Felin Comprehensive School. Perimeter, Area and Volume Methodology Booklet
Dŵr y Felin Comprehensive School Perimeter, Area and Volume Methodology Booklet Perimeter, Area & Volume Perimeters, Area & Volume are key concepts within the Shape & Space aspect of Mathematics. Pupils
### GAP CLOSING. 2D Measurement. Intermediate / Senior Student Book
GAP CLOSING 2D Measurement Intermediate / Senior Student Book 2-D Measurement Diagnostic...3 Areas of Parallelograms, Triangles, and Trapezoids...6 Areas of Composite Shapes...14 Circumferences and Areas
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Santa Monica College COMPASS Geometry Sample Test MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the area of the shaded region. 1) 5 yd 6 yd
### Geometry - Calculating Area and Perimeter
Geometry - Calculating Area and Perimeter In order to complete any of mechanical trades assessments, you will need to memorize certain formulas. These are listed below: (The formulas for circle geometry
### Area and Perimeter. Name: Class: Date: Short Answer
Name: Class: Date: ID: A Area and Perimeter Short Answer 1. The squares on this grid are 1 centimeter long and 1 centimeter wide. Outline two different figures with an area of 12 square centimeters and
### Area of a triangle: The area of a triangle can be found with the following formula: You can see why this works with the following diagrams:
Area Review Area of a triangle: The area of a triangle can be found with the following formula: 1 A 2 bh or A bh 2 You can see why this works with the following diagrams: h h b b Solve: Find the area of
### Area and Circumference
4.4 Area and Circumference 4.4 OBJECTIVES 1. Use p to find the circumference of a circle 2. Use p to find the area of a circle 3. Find the area of a parallelogram 4. Find the area of a triangle 5. Convert
### Finding Volume of Rectangular Prisms
MA.FL.7.G.2.1 Justify and apply formulas for surface area and volume of pyramids, prisms, cylinders, and cones. MA.7.G.2.2 Use formulas to find surface areas and volume of three-dimensional composite shapes.
### Key Stage 2 / 35. Mathematics Paper 2: reasoning. National curriculum tests. Total Marks. Reasoning: Paper 2, Test 1: GAPPS EDUCATION.
National curriculum tests Key Stage 2 Mathematics Paper 2: reasoning MA First name Middle name Last name Date of birth Day Month Year School name Total Marks / 35 Instructions You may not use a calculator
### Area of a triangle: The area of a triangle can be found with the following formula: 1. 2. 3. 12in
Area Review Area of a triangle: The area of a triangle can be found with the following formula: 1 A 2 bh or A bh 2 Solve: Find the area of each triangle. 1. 2. 3. 5in4in 11in 12in 9in 21in 14in 19in 13in
### SGS4.3 Stage 4 Space & Geometry Part A Activity 2-4
SGS4.3 Stage 4 Space & Geometry Part A Activity 2-4 Exploring triangles Resources required: Each pair students will need: 1 container (eg. a rectangular plastic takeaway container) 5 long pipe cleaners
### Which two rectangles fit together, without overlapping, to make a square?
SHAPE level 4 questions 1. Here are six rectangles on a grid. A B C D E F Which two rectangles fit together, without overlapping, to make a square?... and... International School of Madrid 1 2. Emily has
### Geometry Final Exam Review Worksheet
Geometry Final xam Review Worksheet (1) Find the area of an equilateral triangle if each side is 8. (2) Given the figure to the right, is tangent at, sides as marked, find the values of x, y, and z please.
### PERIMETER AND AREA. In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures.
PERIMETER AND AREA In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. Perimeter Perimeter The perimeter of a polygon, denoted by P, is the
### What You ll Learn. Why It s Important
These students are setting up a tent. How do the students know how to set up the tent? How is the shape of the tent created? How could students find the amount of material needed to make the tent? Why
### Characteristics of the Four Main Geometrical Figures
Math 40 9.7 & 9.8: The Big Four Square, Rectangle, Triangle, Circle Pre Algebra We will be focusing our attention on the formulas for the area and perimeter of a square, rectangle, triangle, and a circle.
### Grade 3 Core Standard III Assessment
Grade 3 Core Standard III Assessment Geometry and Measurement Name: Date: 3.3.1 Identify right angles in two-dimensional shapes and determine if angles are greater than or less than a right angle (obtuse
### Show that when a circle is inscribed inside a square the diameter of the circle is the same length as the side of the square.
Week & Day Week 6 Day 1 Concept/Skill Perimeter of a square when given the radius of an inscribed circle Standard 7.MG:2.1 Use formulas routinely for finding the perimeter and area of basic twodimensional
### Area is a measure of how much space is occupied by a figure. 1cm 1cm
Area Area is a measure of how much space is occupied by a figure. Area is measured in square units. For example, one square centimeter (cm ) is 1cm wide and 1cm tall. 1cm 1cm A figure s area is the number
### MATHEMATICS TEST. Paper 1 calculator not allowed LEVEL 6 TESTS ANSWER BOOKLET. First name. Middle name. Last name. Date of birth Day Month Year
LEVEL 6 TESTS ANSWER BOOKLET Ma MATHEMATICS TEST LEVEL 6 TESTS Paper 1 calculator not allowed First name Middle name Last name Date of birth Day Month Year Please circle one Boy Girl Year group School
### Lesson 22. Circumference and Area of a Circle. Circumference. Chapter 2: Perimeter, Area & Volume. Radius and Diameter. Name of Lecturer: Mr. J.
Lesson 22 Chapter 2: Perimeter, Area & Volume Circumference and Area of a Circle Circumference The distance around the edge of a circle (or any curvy shape). It is a kind of perimeter. Radius and Diameter
### Geometry Unit 6 Areas and Perimeters
Geometry Unit 6 Areas and Perimeters Name Lesson 8.1: Areas of Rectangle (and Square) and Parallelograms How do we measure areas? Area is measured in square units. The type of the square unit you choose
### Possible Stage Two Mathematics Test Topics
Possible Stage Two Mathematics Test Topics The Stage Two Mathematics Test questions are designed to be answerable by a good problem-solver with a strong mathematics background. It is based mainly on material
### Hiker. A hiker sets off at 10am and walks at a steady speed for 2 hours due north, then turns and walks for a further 5 hours due west.
Hiker A hiker sets off at 10am and walks at a steady speed for hours due north, then turns and walks for a further 5 hours due west. If he continues at the same speed, what s the earliest time he could
### Area of Parallelograms (pages 546 549)
A Area of Parallelograms (pages 546 549) A parallelogram is a quadrilateral with two pairs of parallel sides. The base is any one of the sides and the height is the shortest distance (the length of a perpendicular
### The GED math test gives you a page of math formulas that
Math Smart 643 The GED Math Formulas The GED math test gives you a page of math formulas that you can use on the test, but just seeing the formulas doesn t do you any good. The important thing is understanding
### Algebra Geometry Glossary. 90 angle
lgebra Geometry Glossary 1) acute angle an angle less than 90 acute angle 90 angle 2) acute triangle a triangle where all angles are less than 90 3) adjacent angles angles that share a common leg Example:
### CIRCUMFERENCE AND AREA OF A CIRCLE
CIRCUMFERENCE AND AREA OF A CIRCLE 1. AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take = 3.14) 2. In the given
### 10-3 Area of Parallelograms
0-3 Area of Parallelograms MAIN IDEA Find the areas of parallelograms. NYS Core Curriculum 6.A.6 Evaluate formulas for given input values (circumference, area, volume, distance, temperature, interest,
### AUTUMN UNIT 3. first half. Perimeter. Centimetres and millimetres. Metres and centimetres. Area. 3D shapes PART 3 MEASURES AND PROPERTIES OF SHAPES
PART AUTUMN first half MEASURES AND PROPERTIES OF SHAPES SECTION Perimeter SECTION Centimetres and millimetres SECTION Metres and centimetres SECTION Key Stage National Strategy CROWN COPYRIGHT 00 Area
### Estimating Angle Measures
1 Estimating Angle Measures Compare and estimate angle measures. You will need a protractor. 1. Estimate the size of each angle. a) c) You can estimate the size of an angle by comparing it to an angle
### Convert between units of area and determine the scale factor of two similar figures.
CHAPTER 5 Units of Area c GOAL Convert between units of area and determine the scale factor of two. You will need a ruler centimetre grid paper a protractor a calculator Learn about the Math The area of
### 2006 Geometry Form A Page 1
2006 Geometry Form Page 1 1. he hypotenuse of a right triangle is 12" long, and one of the acute angles measures 30 degrees. he length of the shorter leg must be: () 4 3 inches () 6 3 inches () 5 inches
### VOLUME of Rectangular Prisms Volume is the measure of occupied by a solid region.
Math 6 NOTES 7.5 Name VOLUME of Rectangular Prisms Volume is the measure of occupied by a solid region. **The formula for the volume of a rectangular prism is:** l = length w = width h = height Study Tip:
### Area. Area Overview. Define: Area:
Define: Area: Area Overview Kite: Parallelogram: Rectangle: Rhombus: Square: Trapezoid: Postulates/Theorems: Every closed region has an area. If closed figures are congruent, then their areas are equal.
### Chapter 8 Geometry We will discuss following concepts in this chapter.
Mat College Mathematics Updated on Nov 5, 009 Chapter 8 Geometry We will discuss following concepts in this chapter. Two Dimensional Geometry: Straight lines (parallel and perpendicular), Rays, Angles
### SA B 1 p where is the slant height of the pyramid. V 1 3 Bh. 3D Solids Pyramids and Cones. Surface Area and Volume of a Pyramid
Accelerated AAG 3D Solids Pyramids and Cones Name & Date Surface Area and Volume of a Pyramid The surface area of a regular pyramid is given by the formula SA B 1 p where is the slant height of the pyramid.
### Geometry and Measurement
The student will be able to: Geometry and Measurement 1. Demonstrate an understanding of the principles of geometry and measurement and operations using measurements Use the US system of measurement for
### Area of Parallelograms, Triangles, and Trapezoids (pages 314 318)
Area of Parallelograms, Triangles, and Trapezoids (pages 34 38) Any side of a parallelogram or triangle can be used as a base. The altitude of a parallelogram is a line segment perpendicular to the base
### Wednesday 15 January 2014 Morning Time: 2 hours
Write your name here Surname Other names Pearson Edexcel Certificate Pearson Edexcel International GCSE Mathematics A Paper 4H Centre Number Wednesday 15 January 2014 Morning Time: 2 hours Candidate Number
### CAMI Education linked to CAPS: Mathematics
- 1 - TOPIC 1.1 Whole numbers _CAPS curriculum TERM 1 CONTENT Mental calculations Revise: Multiplication of whole numbers to at least 12 12 Ordering and comparing whole numbers Revise prime numbers to
### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, June 17, 2010 1:15 to 4:15 p.m., only Student Name: School Name: Print your name and the name of your
### Teacher Page Key. Geometry / Day # 13 Composite Figures 45 Min.
Teacher Page Key Geometry / Day # 13 Composite Figures 45 Min. 9-1.G.1. Find the area and perimeter of a geometric figure composed of a combination of two or more rectangles, triangles, and/or semicircles
### Paper 1. Calculator not allowed. Mathematics test. First name. Last name. School. Remember KEY STAGE 3 TIER 6 8
Ma KEY STAGE 3 Mathematics test TIER 6 8 Paper 1 Calculator not allowed First name Last name School 2009 Remember The test is 1 hour long. You must not use a calculator for any question in this test. You
### GAP CLOSING. Volume and Surface Area. Intermediate / Senior Student Book
GAP CLOSING Volume and Surface Area Intermediate / Senior Student Book Volume and Surface Area Diagnostic...3 Volumes of Prisms...6 Volumes of Cylinders...13 Surface Areas of Prisms and Cylinders...18
### Charlesworth School Year Group Maths Targets
Charlesworth School Year Group Maths Targets Year One Maths Target Sheet Key Statement KS1 Maths Targets (Expected) These skills must be secure to move beyond expected. I can compare, describe and solve
### Assessment For The California Mathematics Standards Grade 3
Introduction: Summary of Goals GRADE THREE By the end of grade three, students deepen their understanding of place value and their understanding of and skill with addition, subtraction, multiplication,
### EDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES. Maths Level 2. Chapter 5. Shape and space
Shape and space 5 EDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES Maths Level 2 Chapter 5 Shape and space SECTION H 1 Perimeter 2 Area 3 Volume 4 2-D Representations of 3-D Objects 5 Remember what you
### Grade 8 Mathematics Geometry: Lesson 2
Grade 8 Mathematics Geometry: Lesson 2 Read aloud to the students the material that is printed in boldface type inside the boxes. Information in regular type inside the boxes and all information outside
### Volume of Right Prisms Objective To provide experiences with using a formula for the volume of right prisms.
Volume of Right Prisms Objective To provide experiences with using a formula for the volume of right prisms. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game
### GAP CLOSING. 2D Measurement GAP CLOSING. Intermeditate / Senior Facilitator s Guide. 2D Measurement
GAP CLOSING 2D Measurement GAP CLOSING 2D Measurement Intermeditate / Senior Facilitator s Guide 2-D Measurement Diagnostic...4 Administer the diagnostic...4 Using diagnostic results to personalize interventions...4
### Paper 1. Calculator not allowed. Mathematics test. First name. Last name. School. Remember KEY STAGE 3 TIER 5 7
Ma KEY STAGE 3 Mathematics test TIER 5 7 Paper 1 Calculator not allowed First name Last name School 2009 Remember The test is 1 hour long. You must not use a calculator for any question in this test. You
### Glencoe. correlated to SOUTH CAROLINA MATH CURRICULUM STANDARDS GRADE 6 3-3, 5-8 8-4, 8-7 1-6, 4-9
Glencoe correlated to SOUTH CAROLINA MATH CURRICULUM STANDARDS GRADE 6 STANDARDS 6-8 Number and Operations (NO) Standard I. Understand numbers, ways of representing numbers, relationships among numbers,
### SURFACE AREA AND VOLUME
SURFACE AREA AND VOLUME In this unit, we will learn to find the surface area and volume of the following threedimensional solids:. Prisms. Pyramids 3. Cylinders 4. Cones It is assumed that the reader has
### Calculating the Surface Area of a Cylinder
Calculating the Measurement Calculating The Surface Area of a Cylinder PRESENTED BY CANADA GOOSE Mathematics, Grade 8 Introduction Welcome to today s topic Parts of Presentation, questions, Q&A Housekeeping
### Circumference Pi Regular polygon. Dates, assignments, and quizzes subject to change without advance notice.
Name: Period GPreAP UNIT 14: PERIMETER AND AREA I can define, identify and illustrate the following terms: Perimeter Area Base Height Diameter Radius Circumference Pi Regular polygon Apothem Composite
### How do you compare numbers? On a number line, larger numbers are to the right and smaller numbers are to the left.
The verbal answers to all of the following questions should be memorized before completion of pre-algebra. Answers that are not memorized will hinder your ability to succeed in algebra 1. Number Basics
### 1. Kyle stacks 30 sheets of paper as shown to the right. Each sheet weighs about 5 g. How can you find the weight of the whole stack?
Prisms and Cylinders Answer Key Vocabulary: cylinder, height (of a cylinder or prism), prism, volume Prior Knowledge Questions (Do these BEFORE using the Gizmo.) [Note: The purpose of these questions is
### Pizza! Pizza! Assessment
Pizza! Pizza! Assessment 1. A local pizza restaurant sends pizzas to the high school twelve to a carton. If the pizzas are one inch thick, what is the volume of the cylindrical shipping carton for the
### Surface Area Quick Review: CH 5
I hope you had an exceptional Christmas Break.. Now it's time to learn some more math!! :) Surface Area Quick Review: CH 5 Find the surface area of each of these shapes: 8 cm 12 cm 4cm 11 cm 7 cm Find
### CSU Fresno Problem Solving Session. Geometry, 17 March 2012
CSU Fresno Problem Solving Session Problem Solving Sessions website: http://zimmer.csufresno.edu/ mnogin/mfd-prep.html Math Field Day date: Saturday, April 21, 2012 Math Field Day website: http://www.csufresno.edu/math/news
### Mensuration. The shapes covered are 2-dimensional square circle sector 3-dimensional cube cylinder sphere
Mensuration This a mixed selection of worksheets on a standard mathematical topic. A glance at each will be sufficient to determine its purpose and usefulness in any given situation. These notes are intended
### Chapter 7 Quiz. (1.) Which type of unit can be used to measure the area of a region centimeter, square centimeter, or cubic centimeter?
Chapter Quiz Section.1 Area and Initial Postulates (1.) Which type of unit can be used to measure the area of a region centimeter, square centimeter, or cubic centimeter? (.) TRUE or FALSE: If two plane
### Unit 7 Circles. Vocabulary and Formulas for Circles:
ccelerated G Unit 7 ircles Name & ate Vocabulary and Formulas for ircles: irections: onsider 1) Find the circumference of the circle. to answer the following questions. Exact: pproximate: 2) Find the area
### Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers.
Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used
### How to fold simple shapes from A4 paper
How to fold simple shapes from 4 paper ndrew Jobbings www.arbelos.co.uk 18 February 2012 ontents Introduction 1 Square 2 Equilateral triangle 3 Rhombus 5 Regular hexagon 6 Kite 7 Why do the methods work?
### YOU MUST BE ABLE TO DO THE FOLLOWING PROBLEMS WITHOUT A CALCULATOR!
DETAILED SOLUTIONS AND CONCEPTS - SIMPLE GEOMETRIC FIGURES Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to ingrid.stewart@csn.edu. Thank you! YOU MUST
### Mathematics standards
Mathematics standards Grade 6 Summary of students performance by the end of Grade 6 Reasoning and problem solving Students represent and interpret routine and non-routine mathematical problems in a range
### 10-4-10 Year 9 mathematics: holiday revision. 2 How many nines are there in fifty-four?
DAY 1 Mental questions 1 Multiply seven by seven. 49 2 How many nines are there in fifty-four? 54 9 = 6 6 3 What number should you add to negative three to get the answer five? 8 4 Add two point five to
### 2nd Semester Geometry Final Exam Review
Class: Date: 2nd Semester Geometry Final Exam Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The owner of an amusement park created a circular
### CK-12 Geometry: Parts of Circles and Tangent Lines
CK-12 Geometry: Parts of Circles and Tangent Lines Learning Objectives Define circle, center, radius, diameter, chord, tangent, and secant of a circle. Explore the properties of tangent lines and circles.
### EVERY DAY COUNTS CALENDAR MATH 2005 correlated to
EVERY DAY COUNTS CALENDAR MATH 2005 correlated to Illinois Mathematics Assessment Framework Grades 3-5 E D U C A T I O N G R O U P A Houghton Mifflin Company YOUR ILLINOIS GREAT SOURCE REPRESENTATIVES:
### Areas of Polygons. Goal. At-Home Help. 1. A hockey team chose this logo for their uniforms.
-NEM-WBAns-CH // : PM Page Areas of Polygons Estimate and measure the area of polygons.. A hockey team chose this logo for their uniforms. A grid is like an area ruler. Each full square on the grid has
Geometry Progress Ladder Maths Makes Sense Foundation End-of-year objectives page 2 Maths Makes Sense 1 2 End-of-block objectives page 3 Maths Makes Sense 3 4 End-of-block objectives page 4 Maths Makes
### Geometry Notes PERIMETER AND AREA
Perimeter and Area Page 1 of 57 PERIMETER AND AREA Objectives: After completing this section, you should be able to do the following: Calculate the area of given geometric figures. Calculate the perimeter
### Circumference and Area of a Circle
Overview Math Concepts Materials Students explore how to derive pi (π) as a ratio. Students also study the circumference and area of a circle using formulas. numbers and operations TI-30XS MultiView two-dimensional
### Volume of Pyramids and Cones
Volume of Pyramids and Cones Objective To provide experiences with investigating the relationships between the volumes of geometric solids. www.everydaymathonline.com epresentations etoolkit Algorithms
### Level 1 - Maths Targets TARGETS. With support, I can show my work using objects or pictures 12. I can order numbers to 10 3
Ma Data Hling: Interpreting Processing representing Ma Shape, space measures: position shape Written Mental method s Operations relationship s between them Fractio ns Number s the Ma1 Using Str Levels
### GCSE Exam Questions on Volume Question 1. (AQA June 2003 Intermediate Paper 2 Calculator OK) A large carton contains 4 litres of orange juice.
Question 1. (AQA June 2003 Intermediate Paper 2 Calculator OK) A large carton contains 4 litres of orange juice. Cylindrical glasses of height 10 cm and radius 3 cm are to be filled from the carton. How
### Lateral and Surface Area of Right Prisms
CHAPTER A Lateral and Surface Area of Right Prisms c GOAL Calculate lateral area and surface area of right prisms. You will need a ruler a calculator Learn about the Math A prism is a polyhedron (solid
### Exercise Worksheets. Copyright. 2002 Susan D. Phillips
Exercise Worksheets Copyright 00 Susan D. Phillips Contents WHOLE NUMBERS. Adding. Subtracting. Multiplying. Dividing. Order of Operations FRACTIONS. Mixed Numbers. Prime Factorization. Least Common Multiple.
### Perimeter. 14ft. 5ft. 11ft.
Perimeter The perimeter of a geometric figure is the distance around the figure. The perimeter could be thought of as walking around the figure while keeping track of the distance traveled. To determine
### Calculating Perimeter
Calculating Perimeter and Area Formulas are equations used to make specific calculations. Common formulas (equations) include: P = 2l + 2w perimeter of a rectangle A = l + w area of a square or rectangle
### Open-Ended Problem-Solving Projections
MATHEMATICS Open-Ended Problem-Solving Projections Organized by TEKS Categories TEKSING TOWARD STAAR 2014 GRADE 7 PROJECTION MASTERS for PROBLEM-SOLVING OVERVIEW The Projection Masters for Problem-Solving
### Numeracy Targets. I can count at least 20 objects
Targets 1c I can read numbers up to 10 I can count up to 10 objects I can say the number names in order up to 20 I can write at least 4 numbers up to 10. When someone gives me a small number of objects
### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2015 8:30 to 11:30 a.m., only.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 13, 2015 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications
### Area & Volume. 1. Surface Area to Volume Ratio
1 1. Surface Area to Volume Ratio Area & Volume For most cells, passage of all materials gases, food molecules, water, waste products, etc. in and out of the cell must occur through the plasma membrane.
### Solids. Objective A: Volume of a Solids
Solids Math00 Objective A: Volume of a Solids Geometric solids are figures in space. Five common geometric solids are the rectangular solid, the sphere, the cylinder, the cone and the pyramid. A rectangular
### Target To know the properties of a rectangle
Target To know the properties of a rectangle (1) A rectangle is a 3-D shape. (2) A rectangle is the same as an oblong. (3) A rectangle is a quadrilateral. (4) Rectangles have four equal sides. (5) Rectangles
### Geometry Notes VOLUME AND SURFACE AREA
Volume and Surface Area Page 1 of 19 VOLUME AND SURFACE AREA Objectives: After completing this section, you should be able to do the following: Calculate the volume of given geometric figures. Calculate
|
5.02 Fractions of a quantity
Lesson
Let's try this problem to review how to identify a fraction of a group of objects.
Examples
Example 1
Which of the following shows that \dfrac{3}{4} of these ice creams have been selected?
A
B
C
Worked Solution
Create a strategy
Find what \dfrac{1}{4} of the ice creams would look like and choose the option that shows 3 of the 4 equal groups.
Apply the idea
Here is 1 quarter of the ice creams. We can see that 1 quarter of the ice creams means 3 ice creams.\dfrac{1}{4}\text{ of the ice creams }=3
So 3 quarter means 3 of the 4 groups. 3 lots of 3 is 9.\begin{aligned} \dfrac{3}{4} \text{ of the ice creams } &= 3 \times 3 \\ &= 9 \end{aligned}
So the correct answer is Option B because it shows 9 ice creams selected from 12 ice creams.
Idea summary
The denominator tells us the number of equal parts that we need to divide the collection of objects into.
The numerator tells us the number of equal parts that we need to select to represent our fraction.
To find a fraction of a collection, we can first find the unit fraction of the collection, then multiply it so that we get the desired fraction.
Unit fraction of a quantity
This video looks at finding a unit fraction of a quantity.
Examples
Example 2
What is \dfrac15 of 20?
Worked Solution
Create a strategy
Divide the whole number by the denominator.
Apply the idea
\dfrac15 of 20 is 4.
Idea summary
We can quickly find the unit fraction of a whole number by dividing the whole number by the denominator.
Fraction of a quantity
This video looks at finding a non-unit fraction of a quantity.
Examples
Example 3
What is \dfrac{6}{11} of \$55? Worked Solution Create a strategy We can find the fraction of the amount by dividing the amount by the denominator, then multiplying the answer by the numerator. Apply the idea \dfrac{6}{11} of \$55 is \\$30.
Idea summary
You can find a fraction of an amount by:
• Dividing the total quantity by the denominator, then multiplying the answer by the numerator, or
• Multiplying the total quantity by the numerator, then dividing the answer by the denominator.
Outcomes
VCMNA213
Find a simple fraction of a quantity where the result is a whole number, with and without digital technologies
|
We've updated our
TEXT
# Use a graph to determine where a function is increasing, decreasing, or constant
As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function.
Figure 3. The function $f\left(x\right)={x}^{3}-12x$ is increasing on $\left(-\infty \text{,}-\text{2}\right){{\cup }^{\text{ }}}^{\text{ }}\left(2,\infty \right)$ and is decreasing on $\left(-2\text{,}2\right)$.
This video further explains how to find where a function is increasing or decreasing. https://www.youtube.com/watch?v=78b4HOMVcKM
While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is "local minima." Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is "extremum.") Often, the term local is replaced by the term relative. In this text, we will use the term local.
Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain.
For the function in Figure 4, the local maximum is 16, and it occurs at $x=-2$. The local minimum is $-16$ and it occurs at $x=2$.
Figure 4
To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 5 illustrates these ideas for a local maximum.
Figure 5. Definition of a local maximum.
These observations lead us to a formal definition of local extrema.
### A General Note: Local Minima and Local Maxima
A function $f$ is an increasing function on an open interval if $f\left(b\right)>f\left(a\right)$ for any two input values $a$ and $b$ in the given interval where $b>a$.
A function $f$ is a decreasing function on an open interval if $f\left(b\right)<f\left(a\right)$ for any two input values $a$ and $b$ in the given interval where $b>a$.
A function $f$ has a local maximum at $x=b$ if there exists an interval $\left(a,c\right)$ with $a<b<c$ such that, for any $x$ in the interval $\left(a,c\right)$, $f\left(x\right)\le f\left(b\right)$. Likewise, $f$ has a local minimum at $x=b$ if there exists an interval $\left(a,c\right)$ with $a<b<c$ such that, for any $x$ in the interval $\left(a,c\right)$, $f\left(x\right)\ge f\left(b\right)$.
### Example 7: Finding Increasing and Decreasing Intervals on a Graph
Given the function $p\left(t\right)$ in the graph below, identify the intervals on which the function appears to be increasing.
Figure 6
### Solution
We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from $t=1$ to $t=3$ and from $t=4$ on.
In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval $\left(4,\infty \right)$.
### Analysis of the Solution
Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at $t=1$ , $t=3$ , and $t=4$ . These points are the local extrema (two minima and a maximum).
### Example 8: Finding Local Extrema from a Graph
Graph the function $f\left(x\right)=\frac{2}{x}+\frac{x}{3}$. Then use the graph to estimate the local extrema of the function and to determine the intervals on which the function is increasing.
### Solution
Using technology, we find that the graph of the function looks like that in Figure 7. It appears there is a low point, or local minimum, between $x=2$ and $x=3$, and a mirror-image high point, or local maximum, somewhere between $x=-3$ and $x=-2$.
Figure 7
### Analysis of the Solution
Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 7 provides screen images from two different technologies, showing the estimate for the local maximum and minimum.
Figure 8
Based on these estimates, the function is increasing on the interval $\left(-\infty ,-{2.449}\right)\\$ and $\left(2.449\text{,}\infty \right)\\$. Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at $\pm \sqrt{6}$, but determining this requires calculus.)
### Try It 4
Graph the function $f\left(x\right)={x}^{3}-6{x}^{2}-15x+20\\$ to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.
Solution
### Example 9: Finding Local Maxima and Minima from a Graph
For the function $f$ whose graph is shown in Figure 9, find all local maxima and minima.
Figure 9
### Solution
Observe the graph of $f$. The graph attains a local maximum at $x=1$ because it is the highest point in an open interval around $x=1$. The local maximum is the $y$ -coordinate at $x=1$, which is $2$.
The graph attains a local minimum at $\text{ }x=-1\text{ }$ because it is the lowest point in an open interval around $x=-1$. The local minimum is the y-coordinate at $x=-1$, which is $-2$.
# Analyzing the Toolkit Functions for Increasing or Decreasing Intervals
We will now return to our toolkit functions and discuss their graphical behavior in the table below.
Function Increasing/Decreasing Example
Constant Function $f\left(x\right)={c}$ Neither increasing nor decreasing
Identity Function $f\left(x\right)={x}$ Increasing
Quadratic Function $f\left(x\right)={x}^{2}$ Increasing on $\left(0,\infty\right)$ Decreasing on $\left(-\infty,0\right)$ Minimum at $x=0$
Cubic Function $f\left(x\right)={x}^{3}$ Increasing
Reciprocal $f\left(x\right)=\frac{1}{x}$ Decreasing $\left(-\infty,0\right)\cup\left(0,\infty\right)$
Reciprocal Squared $f\left(x\right)=\frac{1}{{x}^{2}}$ Increasing on $\left(-\infty,0\right)$ Decreasing on $\left(0,\infty\right)$
Cube Root [latex-display]f\left(x\right)=\sqrt[3]{x}[/latex-display] Increasing
Square Root $f\left(x\right)=\sqrt{x}$ Increasing on $\left(0,\infty\right)$
Absolute Value $f\left(x\right)=|x|$ Increasing on $\left(0,\infty\right)$ Decreasing on $\left(-\infty,0\right)$
|
#### Everything You Need in One Place
Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.
#### Learn and Practice With Ease
Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.
#### Instant and Unlimited Help
Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!
0/10
##### Examples
###### Lessons
1. Determining the Characteristics of a Quadratic Function Using Various Methods
Determine the following characteristics of the quadratic function $y = -2x^2 + 4x + 6$:
• Opening of the graph
$y-$intercept
$x-$intercept(s)
• Vertex
• Axis of symmetry
• Domain
• Range
• Minimum/Maximum value
1. Using factoring
3. Using completing the square
4. Using the vertex formula
2. From the graph of the parabola, determine the:
• vertex
• axis of symmetry
• y-intercept
• x-intercepts
• domain
• range
• minimum/maximum value
3. Identifying Characteristics of Quadratic function in General Form: $y = ax^2 + bx+c$
$y = 2{x^2} - 12x + 10$ is a quadratic function in general form.
i) Determine:
• y-intercept
• x-intercepts
• vertex
ii) Sketch the graph.
1. Identifying Characteristics of Quadratic Functions in Vertex Form: $y = a(x-p)^2 + q$
$y = 2{\left( {x - 3} \right)^2} - 8$ is a quadratic function in vertex form.
i) Determine:
• y-intercept
• x-intercepts
• vertex
ii) Sketch the graph.
###### Topic Notes
Three properties that are universal to all quadratic functions: 1) The graph of a quadratic function is always a parabola that either opens upward or downward (end behavior); 2) The domain of a quadratic function is all real numbers; and 3) The vertex is the lowest point when the parabola opens upwards; while the vertex is the highest point when the parabola opens downward.
|
# How to calculate mean from frequency table
how to calculate mean from frequency table
How to calculate mean from a frequency table?
Answer: Calculating the mean from a frequency table involves several systematic steps. Here’s a detailed guide on how to do it:
### 1. Understanding the Frequency Table:
A frequency table typically consists of two columns:
• Data values (or class intervals in the case of grouped data)
• Frequencies (the number of times each data value or class interval occurs)
### 2. Steps to Calculate the Mean:
#### For Ungrouped Data:
1. List the Data Values and Their Frequencies:
Suppose you have a frequency table like this:
Data Value (x) Frequency (f)
1 4
2 6
3 8
4 2
5 1
2. Multiply Each Data Value by Its Frequency:
Calculate the product of each data value and its corresponding frequency:
\begin{array}{c|c|c} \text{Data Value (x)} & \text{Frequency (f)} & \text{Product (x \cdot f)} \\ \hline 1 & 4 & 1 \cdot 4 = 4 \\ 2 & 6 & 2 \cdot 6 = 12 \\ 3 & 8 & 3 \cdot 8 = 24 \\ 4 & 2 & 4 \cdot 2 = 8 \\ 5 & 1 & 5 \cdot 1 = 5 \\ \end{array}
3. Sum the Products:
\sum (x \cdot f) = 4 + 12 + 24 + 8 + 5 = 53
4. Sum the Frequencies:
\sum f = 4 + 6 + 8 + 2 + 1 = 21
5. Calculate the Mean:
Divide the sum of the products by the sum of the frequencies:
\text{Mean} = \frac{\sum (x \cdot f)}{\sum f} = \frac{53}{21} \approx 2.52
#### For Grouped Data:
When dealing with grouped data, the process is slightly different because data values are grouped into intervals.
1. Determine the Midpoint of Each Class Interval:
Suppose you have a frequency table like this:
Class Interval Frequency (f)
0-10 5
10-20 8
20-30 12
30-40 7
40-50 3
Calculate the midpoint (m) of each class interval:
\text{Midpoint (m)} = \frac{\text{Lower limit} + \text{Upper limit}}{2}
For example:
\text{Midpoint of } 0-10 = \frac{0 + 10}{2} = 5
Repeat this for all class intervals.
2. Multiply Each Midpoint by Its Frequency:
Calculate the product of each midpoint and its corresponding frequency:
\begin{array}{c|c|c|c} \text{Class Interval} & \text{Frequency (f)} & \text{Midpoint (m)} & \text{Product (m \cdot f)} \\ \hline 0-10 & 5 & 5 & 5 \cdot 5 = 25 \\ 10-20 & 8 & 15 & 15 \cdot 8 = 120 \\ 20-30 & 12 & 25 & 25 \cdot 12 = 300 \\ 30-40 & 7 & 35 & 35 \cdot 7 = 245 \\ 40-50 & 3 & 45 & 45 \cdot 3 = 135 \\ \end{array}
3. Sum the Products:
\sum (m \cdot f) = 25 + 120 + 300 + 245 + 135 = 825
4. Sum the Frequencies:
|
+0
# Probability
0
72
1
Right triangle XYZ has legs of length XY = 12 and YZ = 6. Point D is chosen at random within the triangle XYZ. What is the probability that the area of triangle XYD is at most 20?
Mar 7, 2023
#1
+195
0
We can start by using the Pythagorean theorem to find the length of the hypotenuse XZ of triangle XYZ:
XZ^2 = XY^2 + YZ^2 = 12^2 + 6^2 = 144 + 36 = 180
Taking the square root of both sides, we get XZ = 6√5.
The area of triangle XYZ is (1/2) * XY * YZ = 36, and the area of rectangle XYX'Z, where X' is the midpoint of XZ, is XY * X'Z = 36√5.
Since triangle XYD is a right triangle with legs of length 6 and a and hypotenuse XY, its area is (1/2) * 6 * a = 3a, where a is the length of the altitude from D to XY.
If we choose D randomly within triangle XYZ, the probability that it lies within a certain region of the triangle is proportional to the area of that region. Therefore, we can find the probability that the area of triangle XYD is at most 20 by finding the ratio of the area of the region where this is true to the area of triangle XYZ.
Let H be the foot of the altitude from X to YZ, and let E be the point where the line parallel to YZ through D intersects XY. Then, triangle XYD has area at most 20 if and only if DE is at most 10.
The area of region XYZE is (1/2) * YZ * XE = 3XE, and the area of region XHY is (1/2) * XY * YH = 36/5. Therefore, the area of the region where the area of triangle XYD is at most 20 is:
36 - 3XE - (36/5) = 144/5 - 3XE
We want this area to be proportional to the area of triangle XYZ, which is 36, so we want:
(144/5 - 3XE)/36 = P
where P is the probability we are looking for.
Solving for XE, we get:
XE = (144/5 - 36P)/3
We need XE to be at most 6 in order for DE to be at most 10. Therefore, we have:
(144/5 - 36P)/3 ≤ 6
Multiplying both sides by 3, we get:
144/5 - 36P ≤ 18
Subtracting 144/5 from both sides, we get:
-36P ≤ -126/5
Dividing both sides by -36 and reversing the inequality, we get:
P ≥ 7/60
Therefore, the probability that the area of triangle XYD is at most 20 is at least 7/60.
Mar 7, 2023
|
# Show What You Know: Factors and Multiples + Introduction to Exponents
47 teachers like this lesson
Print Lesson
## Objective
SWBAT: • Show what they know about factors, multiples, GCF, and LCM. • Define exponent and base • Simplify exponents
#### Big Idea
What do students understand? What gaps do they have in their understanding? What's greater, 2 to the third power or 3 to the second power? Students take a quiz and then work to understand that repeated multiplication is represented with exponents.
## Do Now
10 minutes
See my Do Now in my Strategy folder that explains my beginning of class routines.
Often, I create do nows that have problems that connect to the task that students will be working on that day. Here, I want students to do a brief reflection about what they have learned related to multiples and factors. As students are working, I walk around and observe connections and questions. I have 2 students come to the front and share their maps. I ask students to share out questions. Based on the questions, I may create a review question for the students. They can use the next page to show their work.
I do my best to keep the review brief. It is easy for this review to stretch and take over the period. I set a timer and hold myself to this.
## Quiz
15 minutes
I give students the Quiz. If students do not finish in the allotted time, they set up a time (preferably that day) to come in and complete it. I use this data to inform my instruction. If students struggle with a concept, I will spiral it into do nows and homework assignments. I may also add a few problems on that topic to the next quiz.
## Repeated Operations & Notes
10 minutes
We go through problems 1 and 2 together. I want students to make the connection between being able to represent repeated addition with multiplication and repeated multiplication with exponents. It is okay if students don’t make this connection on their own. If that is the case, I will introduce that 3 to the fourth power is equivalent to 3x3x3x3. I also use this time to quickly introduce/review the parentheses and dot as additional ways to represent multiplication.
Students take notes on the vocabulary words on the next pages. Students may struggle at first that any number to the first power is itself. If we think about exponents as indicating repeated multiplication, than a number raised to the power of one is just itself. I ask students what pattern they see when they have a base of 10 (MP8: Look for and express regularity in repeated reasoning).
A common mistake is that students simply multiply the base by the exponent (3 squared as 3 x 2). When I see this I stress that exponents indicate repeated multiplication. Using the 3 squared example, I say that 3 ^2 tells me to multiply 3 by itself, which is not the same thing as multiplying 3 by 2. Then I draw models to reinforce my point. I draw a square that is 3 by 3 units and a rectangle that is 3 by 2 units. The area of these shapes is different.
## Show me the money!
15 minutes
To hype up my students I have them do a drum roll as I announce the new game show they are going to participate in. I have found that most sixth grade students have fun when anything is presented as a game or competition. I read through the rules and ask for a volunteer to go through the example round with me. As the volunteer and I go through the example, I record what I expect students to record on the paper. Even with this easy example, I have my volunteer help me at the end with the calculator.
For a game like this, I let students pick their own partners. If there is an odd number of students, there can be one group of three and they can adjust their worksheet. As students play, I walk around and monitor that students are following the rules (and making and recording a guess in 5 seconds). I am looking for what students are guessing and how they are changing their choices with the more rounds they go through.
For students who need an extra challenge have them use the 1-10 spinner with a paperclip, instead of the dice. The other player still uses the calculator to find the end product.
If students finish the game, they can work on the challenge questions.
## Closure and Ticket to Go
10 minutes
For Closure put this on the board: option a)3^4 and option b) 4^3. I give students 5 seconds to make a decision and then ask them who chooses a) and who chooses b)? We represent both options as repeated multiplication and then simplify them. I ask students what they noticed while they played with their partner. Did anything surprise you? Some students may share a round where they were surprised once they simplified both options. Other students may share their experience with rolling a one, and the pattern they noticed when you compare a number raised to the first power vs. one raised to that power. Students are engaging in MP8: Look for and express regularity in repeated reasoning.
Now that students are familiar with these types of comparisons, I can use this to fill up a couple minutes of down time throughout lessons or the day. Students understanding and number sense with exponents will continue to improve!
With the last few minutes of class I give a Ticket to Go for students to independently complete. I am looking to see if students have grasped the connection between exponents and repeated multiplication.
I pass out the HW Introduction to Exponents at the end of class.
|
# Selina Solutions Concise Mathematics Class 6 Chapter 7: Number Line
Selina Solutions is a key source of study material for students. The solutions are well structured by experts with the purpose of helping students to solve each question with ease. Practicing the Selina Solutions Concise Mathematics on a regular basis increases students’ understanding capacity and problem solving skills. The diligent practice of Selina Solutions makes them understand basic concepts thoroughly. Selina Solutions are the best reliable resource among students to grasp the concepts easily as the answers provided are accurate with illustrations
A straight line with numbers placed at equal intervals along its length is called the Number Line. Our BYJU’S experts in the respective subject, designed the solutions to boost exam preparation among students. Accurate answers from Selina enables students to solve the complex problems easily. Download the PDF of Selina Solutions Concise Mathematics Class 6 Chapter 7 Number Line, from the links given below.
## Selina Solutions Concise Mathematics Class 6 Chapter 7: Number Line Download PDF
### Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 7: Number Line
Exercise 7(A) Solutions
Exercise 7(B) Solutions
## Access Selina Solutions Concise Mathematics Class 6 Chapter 7: Number Line
#### Exercise 7(A) page no: 55
1. Fill in the blanks, using the following number line:
(i) An integer, on the given number line, is ………. than every number on its left.
(ii) An integer on the given number line is greater than every number to its ………..
(iii) 2 is greater than -4 implies 2 is to the ……….. of -4.
(iv) -3 is …………. than 2 and 3 is ………… than -2.
(v) -4 is ……….. than -8 and 4 is ……….. than 8.
(vi) 5 is ………….. than 2 and -5 is ………… than -2.
(vii) -6 is …………. than 3 and the opposite of -6 is …………. than opposite of 3.
(viii) 8 is ………….. than -5 and -8 is ……………. than 5.
Solution:
(i) An integer, on the given number line, is greater than every number on its left.
(ii) An integer, on the given number line, is greater than every number to its left.
(iii) 2 is greater than -4 implies 2 is on the right of -4
(iv) -3 is less than 2 and 3 is greater than -2
(v) -4 is greater than -8 and 4 is lessthan 8
(vi) 5 is greater than 2 and -5 is less than -2
(vii) -6 is less than 3 and the opposite of -6 is greater than opposite of 3
(viii) 8 is greater than -5 and -8 is less than 5
2. In each of the following pairs, state which integer is greater :
(i) – 15, – 23
(ii) – 12, 15
(iii) 0, 8
(iv) 0, – 3
Solution:
(i) -15, -23
-15 lies on the right side of -23 on the number line. Therefore, -15 is greater than -23
(ii) -12, 15
15 lies on the right side of -12 on the number line. Hence, 15 is greater than -12
(iii) 0, 8
8 lies on the right side of 0 on the number line. Therefore, 8 is greater than 0
(iv) 0, -3
0 lies on the right side of -3 on the number line. Hence, 0 is greater than -3
3. In each of the following pairs, state which integer is smaller :
(i) 0, – 6
(ii) 2, – 3
(iii) 15, – 51
(iv) 13, 0
Solution:
(i) -6 lies on the left side of 0 on the number line. Hence, -6 is smaller than 0
(ii) -3 lies on the left side of 2 on the number line. Therefore, -3 is smaller than 2
(iii) -51 lies on the left side of 15 on the number line. Hence, -51 is smaller than 15
(iv) 0 lies on the left side of 13 on the number line. Therefore, 0 is smaller than 13
4. In each of the following pairs, replace * with < or > to make the statement true:
(i) 3 * 0
(ii) 0 * – 8
(iii) – 9 * – 3
(iv) – 3 * 3
(v) 5 * – 1
(vi) – 13 * 0
(vii) – 8 * – 18
Solution:
(i) 3 lies on the right side of 0 on the number line ⇒ 3 > 0
Hence, 3 is greater than 0
(ii) 0 lies on the right side of -8 on the number line ⇒ 0 > -8
Therefore, 0 is greater than -8
(iii) -9 lies on the left side of -3 on the number line ⇒ -9 < -3
Hence, -9 is smaller than -3
(iv) -3 lies on the left side of 3 on the number line ⇒ -3 < 3
Therefore, -3 is smaller than 3
(v) 5 lies right side of -1 on the number line ⇒ 5 > -1
Hence, 5 is greater than -1
(vi) -13 lies left side of 0 on the number line ⇒ -13 < 0
Therefore, -13 is smaller than 0
(vii) -8 lies on the right side of -18 on the number line ⇒ -8 > -18
Hence, -8 is greater than -18
5. In each case, arrange the given integers in ascending order, using a number line:
(i) -8, 0, -5, 5, 4, -1
(ii) 3, -3, 4, -7, 0, -6, 2
Solution:
(i) -8, 0, -5, 5, 4, -1
Draw a number line and mark the given integers in ascending order as -8, -5, -1, 0, 4, 5 on the number line as shown below
(ii) 3, -3, 4, -7, 0, -6, 2
Draw a number line and mark the given integers in ascending order as -7, -6, -3, 0, 2, 3, 4 on the number line as shown below
6. In each case, arrange the given integers in descending order, using a number line:
(i) -5, -3, 8, 15, 0, -2
(ii) 12, 23, -1, 0, 7, 6
Solution:
(i) -5, -3, 8, 15, 0, -2
Draw a number line and mark the given integers on it. Arranging these integers in descending order 15, 8, 0, -2, -3, -5 as shown below on the number line
(ii) 12, 23, -1, 0, 7, 6
Draw a number line and mark the given integers on it. Arranging these integers in descending order 23, 12, 7, 6, 0, -1 as shown below on the number line
7. For each of the statements given below, state whether it is true or false:
(i) The smallest integer is 0.
(ii) The opposite of -17 is 17
(iii) The opposite of zero is zero
(iv) Every negative integer is smaller than 0
(v) 0 is greater than every positive integer
(vi) Since zero is neither negative nor positive, it is not an integer
Solution:
(i) False
(ii) True
(iii) True
(iv) True
(v) False
(vi) False
#### Exercise 7(B) page no: 59
1. Use a number line to evaluate each of the following:
1. (i) (+7) + (+4) (ii) 0 + (+6) (iii) (+5) + 0
2. (i) (-4) + (+5) (ii) 0 + (-2) (iii) (-1) + (+4)
3. (i) (+4) + (-2) (ii) (+3) + (-6) (iii) 3 + (-7)
4. (i) (-1) + (-2) (ii) (-3) + (-4) (iii) (-2) + (-5)
5, (i) (+10) – (+2) (ii) (+8) – (-5) (iii) (-6) – (+2)
(iv) (-7) – (+5) (v) (+4) – (-2) (vi) (-8) – (-4)
Solution:
1. (i) (+7) + (+4)
Move 7 units to the right of zero for (+7) and move 4 units to the right of +7 for (+4)
Hence, (+7) + (+4) = +11
(ii) 0 + (+6)
No movement for 0 and move 6 units to the right of zero for (+6)
Hence, 0 + (+6) = +6
(iii) (+5) + 0
Move 5 units to the right of 0 for (+5) and for 0, no movement
Hence, (+5) + 0 = +5
2. (i) (-4) + (+5)
Move 4 units to the left of 0 for (-4) and move 5 units to the right of -4 for (+5)
Hence, (-4) + (+5) = +1
(ii) 0 + (-2)
No movement for 0 and move 2 units to the left of 0 for -2
Hence, 0 + (-2) = -2
(iii) (-1) + (+4)
Move 1 unit to the left of 0 for (-1) and move 4 units to the right of -1 for (+4)
Hence, (-1) + (+4) = +3
3. (i) (+4) + (-2)
Move 4 units to the right of 0 for (+4) and move 2 units to the left of +4 for (-2)
Hence, (+4) + (-2) = +2
(ii) (+3) + (-6)
Move 3 units to the right of 0 for (+3) and move 6 units to the left of 3 for (-6)
Hence, (+3) + (-6) = -3
(iii) 3 + (-7)
Move 3 units to the right of 0 for (3) and move 7 units to the left of 3 for (-7)
Hence, 3 + (-7) = -4
4. (i) (-1) + (-2)
From 0, move 1 unit to the left for (-1) and move 2 units to the left of -1 for (-2)
Hence, (-1) + (-2) = -3
(ii) (-3) + (-4)
From 0, move 3 units to the left for (-3) and move 4 units to the left of -3 for (-4)
Hence, (-3) + (-4) = -7
(iii) (-2) + (-5)
Move 2 units to the left from 0 for (-2) and move 5 units to the left of -2 for (-5)
Hence, (-2) + (-5) = -7
5. (i) (+10) – (+2)
From +2, it took 8 steps to reach the position of number +10 to the right
Hence, (+10) – (+2) = +8
(ii) (+8) – (-5)
Starting from the position of -5, it took 13 steps towards right to reach +8
Hence, (+8) – (-5) = + 13
(iii) (-6) – (+2)
From +2, it took 8 steps towards left to reach -6
Hence, (-6) – (+2) = -8
(iv) (-7) – (+5)
From +5, it took 12 steps towards left to reach -7
Hence, (-7) – (+5) = -12
(v) (+4) – (-2)
From +4, it took 6 steps towards left to reach the number -2
Hence, (+4) – (-2) = +6
(vi) (-8) – (-4)
From -8, it took 4 steps towards right to reach -4
Hence, -8 – (-4) = -4
6. Using a number line, find the integer which is:
(i) 3 more than -1
(ii) 5 less than 2
(iii) 5 more than -9
(iv) 4 less than -4
(v) 7 more than 0
(vi) 7 less than -8
Solution:
(i) 3 more than -1
To get 3 more than -1, from -1 move 3 units to the right of -1 to get 2
Therefore, 3 more than -1 = 2
(ii) 5 less than 2
To get 5 less than 2, from 2 move 5 units to the left of 2 to get -3
Therefore, 5 less than 2 = -3
(iii) 5 more than -9
To get 5 more than -9, from -9 move 5 units to the right of -9 to get -4
Therefore, 5 more than -9 = -4
(iv) 4 less than -4
To get 4 less than -4, from -4 move 4 units to the left of -4 to get -8
Therefore, 4 less than -4 = -8
(v) 7 more than 0
To get 7 more than 0, from 0 move 7 units to the right of 0 to get 7
Therefore, 7 more than 0 = 7
(vi) 7 less than -8
To get 7 less than -8, from -8 move 7 units to the left of -8 to get -15
Therefore, 7 less than -8 = -15
|
# In a Wheatstone’s bridge, three resistance P,Q and R connected in the three arms and the fourth arm is formed by two resistance S1 and S2 connected in parallel. The condition for bridge to be balanced will be Option 1) $\frac{P}{Q}= \frac{R}{S_{1}+S_{2}}$ Option 2) $\frac{P}{Q}= \frac{2R}{S_{1}+S_{2}}$ Option 3) $\frac{P}{Q}= \frac{R\left ( S_{1}+S_{2} \right )}{S_{1}S_{2}}$ Option 4) $\frac{P}{Q}= \frac{R\left ( S_{1}+S_{2} \right )}{2S_{1}S_{2}}$
As we learnt in
Wheat stone Bridge -
It is an arrangement of four resistances which can be used to measure one of them in terms of rest
- wherein
We know that $\frac{P}{Q}=\frac{R}{S}$
$\because S =\frac{S_{1} \ S_{2}}{S_{1} + S_{2}}$
$\therefore \frac{P}{Q} =\frac{R\left ( S_{1}+S_{2} \right )}{S_{1} \ S_{2}}$
Option 1)
$\frac{P}{Q}= \frac{R}{S_{1}+S_{2}}$
This is incorrect.
Option 2)
$\frac{P}{Q}= \frac{2R}{S_{1}+S_{2}}$
This is incorrect.
Option 3)
$\frac{P}{Q}= \frac{R\left ( S_{1}+S_{2} \right )}{S_{1}S_{2}}$
This is correct.
Option 4)
$\frac{P}{Q}= \frac{R\left ( S_{1}+S_{2} \right )}{2S_{1}S_{2}}$
This is incorrect.
### Preparation Products
##### JEE Main Rank Booster 2021
This course will help student to be better prepared and study in the right direction for JEE Main..
₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)
An exhaustive E-learning program for the complete preparation of JEE Main..
₹ 6999/- ₹ 5/-
##### Knockout JEE Main April 2021
An exhaustive E-learning program for the complete preparation of JEE Main..
₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022
An exhaustive E-learning program for the complete preparation of JEE Main..
₹ 34999/- ₹ 24999/-
|
### Roots and Coefficients
If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?
### Target Six
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
### 8 Methods for Three by One
This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?
# Strolling Along
##### Age 14 to 18Challenge Level
Thank you very much to Pablo and Sergio from King's College Alicante in Spain, Z. Hakim from Annie Gale Jr High in Canada and Michele from Hockerill Anglo-European College in Italy, who all sent in good work on this problem. The solution below draws on their observations and explanations.
Multiplying by a real number
Here are some examples of multiplying a complex number by a positive real number:
$(3+2i)\times(3+0i)=9+6i$
$(3+2i)\times(0.5+0i)=1.5+i$
Multiplying $z_1$ by a real number $z_2$ enlarges $z_1$ by a scale factor of $z_2$, centred at $0+0i$.
The same happens if $z_2$ is negative, but this time the scale factor is negative, so $z_3$ is on the other side of $0+0i$:
$(3+2i)\times(-2+0i)=-6-4i$
Multiplying by $i$
Some examples are shown below. To help see the effect, the line segments joining $z_1$ and $z_3$ to $0+0i$ are also shown, and the right-angled triangles that these make with the axes are shaded.
$(0.9+5.8i) i=-5.8+0.9i$
$(-4.7+1.8i) i=-1.8-4.7i$
$(1.3-3.3i) i=3.3+1.3i$
Since in each case, the vertical side of one green triangle is equal to the horizontal side of the other and vice versa, the triangles must be congruent. So the distance from $0+0i$ to $z_3$ is the same as the distance from $0+0i$ to $z_1$. Also, the angle at $0+0i$ is the same in each pair of triangles, so the angles between their hypotensues must be $90^\circ$, the same as the angle between the real and imaginary axes.
So in each case, multiplication by $i$ represents a rotation of $90^\text o$ anticlockwise about $0+0i$.
We can use Pythagoras' Theorem to prove that the distance from $0+0i$ does not change.
e.g. For the first pair:
The distance of both $z_1$ and $z_3$ from $0+0i$ is $\sqrt{0.9^2+5.8^2}$.
We can use the gradients of the lines joining $z_1$ and $z_3$ to $0+0i$ to prove that they are perpendicular.
e.g. For the first pair:
The lines have gradients $\frac{5.8}{0.9}$ and $\frac{0.9}{-5.8}$
$\frac{5.8}{0.9}\times\frac{0.9}{-5.8}=-1$ so the lines are perpendicular.
We can also do this for a general complex number $z_1 = a+bi$:
$$z_3 = (a+bi)\times i=ai+bi^2=-b+ai$$
The distance of both $z_1$ and $z_3$ from $0+0i$ is $\sqrt{a^2+b^2}$
The gradients of the line segments joining $z_1$ and $z_3$ to $0+0i$ are $\frac ba$ and $\frac{a}{-b}$
$\frac ba\times\frac a{-b}=-1$, so they are perpendicular.
Multiplying by a multiple of $i$
Some examples are shown below.
$(-4-3i)\times(0+2i)=6-8i$
$(-4-3i)\times(0+0.5i)=1.5-2i$
$(-4-3i)\times(-3i)=12+9i$
The angle between the lines joining $z_1$ and $z_3$ to $0+0i$ are still perpendicular, so there is still a rotation of $90^\text o$ about $0+0i$. Notice that this rotation appears to be anticlockwise when $z_2$ is $2i$ or $0.5i$, but clockwise when $z_2=-3i$. So if $b$ is negative, then the direction of the rotation is reversed.
Also notice that $z_3$ is either closer to, or further from, $0+0i$ than $z_1$ is. So as well as a rotation, there is an enlargement centred at the origin. The scale factor of the enlargement is the size of $b$ - so when $b=2$ the scale factor is $2$, and when $b=-3$, the scale factor is $3$.
For positive values of $b$, this agrees with what we had seen so far, since multiplying by $bi$ is the same as multiplying by $i$ (rotation) and by $b$ (enlargement).
For negative values of $b$, we have said that the rotation is clockwise instead of anticlockise. We could explain this by saying that the enlargement has a negative scale factor, so everything is moved to the other side of the origin, which has the same effect as changing the direction of the rotation.
|
# A charge of 2 C is at the origin. How much energy would be applied to or released from a 3 C charge if it is moved from ( -4, 6 ) to ( 7 , -4 ) ?
Jun 5, 2018
$= - 0.7895$ $G J$
#### Explanation:
The potential energy can be given by:
$V \left(r\right) = \frac{{Q}_{1} {Q}_{2}}{4 \pi {\epsilon}_{0} r}$
Where $Q$ are the magnitudes of the charges and $r$ is the distance between the charges.
We already know that the charges (which are constant) are:
${Q}_{1} = 2 C$ and ${Q}_{2} = 3 C$ so we can plug these in to the top (ignoring units for now) to get:
$V \left(r\right) = \frac{6}{4 \pi {\epsilon}_{0} r} = \frac{3}{2 \pi {\epsilon}_{0} r}$.
We wish to see the energy difference in moving the $3 C$ charge from $\left(- 4 , 6\right)$ to $\left(7 , - 4\right)$.
So we need to find $r$ for both these cases (the scalar distance from the origin) which can be done by simple Pythagoras.
${r}_{1} = \sqrt{{\left(- 4\right)}^{2} + {6}^{2}} \approx 7.2111 m$
${r}_{2} = \sqrt{{7}^{2} + {\left(- 4\right)}^{2}} \approx 8.0623 m$
assuming the coordinates given are in metres.
As it is a repulsive force (same sign charges) and we are moving further away then energy is being released.
To calculate how much we simply need:
$\Delta V = V \left({r}_{2}\right) - V \left({r}_{1}\right)$
$= \frac{3}{2 \pi {\epsilon}_{0} \times 8.0623} - \frac{3}{2 \pi {\epsilon}_{0} \times 7.2111}$
$= 6.6887 \times {10}^{9} - 7.4783 \times {10}^{9}$
$= - 7.895 \times {10}^{8} J$
$= - 0.7895$ $G J$
|
Maths
# What is the age of both father and son? | Algebra | Maths | Tutorial
QUESTION
The age of a father is twice the age of his son. Twenty years ago, his age was 6 times his son’s age. What is the age of both father and son?
SOLUTION
Step 1: Define
1. Let X = Age of Son
2. Therefore the age of the father is 2X (as he is twice the age of his son)
Step 2: Their Age 20 years ago
Their age 20 years ago
Father = 2X – 20
Son = X – 20
Step 3: Father is Six Times Sons Age
As the age of the father is 6 times the age of the son twenty years ago.
We illustrate this as:
6 * (Age of Son) = 6 * (X – 20) = Fathers age 20 years ago
Step 4: Solving for X (Sons Age)
As we know that 20 years ago the fathers age is 2X – 20 and also 6 * (X – 20) we can solve this simultaneous equation.
2X – 20 = 6 * (X – 20)
2X – 20 = 6X – 120
-20 + 120 = 6X – 2X
100 = 4X
X = 25 (The sons age is therefore 25)
Step 5: Calculating Fathers Age
The fathers age is 2X = 2 * 25 = 50
Step 6: Verification
To make sure that this is correct we can check to see what their ages were 20 years ago and confirm if the father really was 6 times older than the son.
Father = (2X – 20) = 2(25) – 20 = 30
Son = (X – 20) = 25 – 20 = 5
5 * 6 = 30 (Confirmation that the solution is correct!)
## 3 thoughts on “What is the age of both father and son? | Algebra | Maths | Tutorial”
1. Lorin Black says:
Check your formulas in the confirmation. 2(50)-20 does not =30.
Liked by 1 person
1. Simplified By Me says:
Thank you, corrected typo!
Like
This site uses Akismet to reduce spam. Learn how your comment data is processed.
|
# How to solve by completing the square
In this blog post, we will take a look at How to solve by completing the square. Our website can solve math word problems.
## How can we solve by completing the square
In this blog post, we will be discussing How to solve by completing the square. Need help with math homework? There are lots of good math homework apps out there, but not all of them are equal. To make sure you’re getting the best math homework app for your needs, consider these things: There are various types of math homework apps available online. Some are designed for individual use, while others can be used in a group setting. It’s important to choose one that suits your needs. Math is a challenging subject, so it’s important to find an app that suits your learning style. Math homework apps can help you stay organized and keep track of assignments, which is a huge plus!
Quadratic equations can be tricky to solve. Luckily there are several ways to tackle them. Here are a few: One way is to use the quadratic formula . This method is easiest for equations that have only two terms. The formula looks like this: \$largefrac{a}{b} = frac{large c}{large b}\$ where \$a\$ and \$b\$ are the coefficients of \$x^2 + y^2 = c\$, and \$c\$ is the solution. If we plug in values for \$x\$ and \$y\$, we can find out what \$c\$ is. Another way to solve quadratic equations is by factoring them (if they're in the form of an expression, like an equation or a fraction). This means finding out which numbers can be divided into both sides of the equation without changing the value of the whole thing. When you factor an expression like this, you're reducing all the terms on both sides of the equals sign to a single number. Then you multiply that number by both sides, cancel one term on each side, and solve for the other variable. This process works best with two-term equations. And finally, there are properties of quadratics that can help you find solutions. For example, quadratics that are similar to each other usually have similar solutions. And
i need help in math is the best app for students to learn math.This app has a lot of different math problems that you can choose from, and each problem has a step-by-step solution. It’s also easy to use, and it’s easy to navigate through. There are a lot of charts, graphs, and tables that make it even easier for you to understand how to do math. You can also take notes if you want. This app is great for everyone who wants to learn math.
Solving rational expressions calculator is a simple online tool which helps to solve rational expressions. It can be used in place of standard calculators. In order to use this tool, enter the expression you want to solve, choose between log and trigonometric functions, and click ‘Calculate’. The result will be displayed on the screen. You can also select among several options for converting and simplify rational expressions using the drop-down menu. While solving rational expressions using this calculator can take some time depending on the complexity of the expression, it is still a useful tool for learning or practicing basic math skills.
This app is literally so good it helps me how to do math sums that I don't know how to solve but I just to say that the person who made this app has to add ratios in it too Whenever I ask a ratio question, they didn't get it so kindly add ratios in it too.
Hellen Sanchez
I don't really rate apps, but this one literally saved my life with the step-by-step process on how to solve so much. I don't have smart friends, but I do have a smart phone with the app installed on it and that works for me! ;)
Isobel Miller
Polynomial long division solver Intergral solver Step by step algebra solutions Indefinite integral solver How to solve equations step-by step Problem solving or problem solving
|
# Selina solutions for Concise Mathematics Class 9 ICSE chapter 17 - Circle [Latest edition]
## Chapter 17: Circle
Exercise 17 (A)Exercise 17 (B)Exercise 17 (C)Exercise 17 (D)
Exercise 17 (A) [Pages 210 - 211]
### Selina solutions for Concise Mathematics Class 9 ICSE Chapter 17 Circle Exercise 17 (A) [Pages 210 - 211]
Exercise 17 (A) | Q 1 | Page 210
A chord of length 6 cm is drawn in a circle of radius 5 cm.
Calculate its distance from the center of the circle.
Exercise 17 (A) | Q 2 | Page 210
A chord of length 8 cm is drawn at a distance of 3 cm from the center of the circle.
Calculate the radius of the circle.
Exercise 17 (A) | Q 3 | Page 210
The radius of a circle is 17.0 cm and the length of the perpendicular drawn from its center to a chord is 8.0 cm.
Calculate the length of the chord.
Exercise 17 (A) | Q 4 | Page 210
A chord of length 24 cm is at a distance of 5 cm from the center of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the center.
Exercise 17 (A) | Q 5 | Page 210
In the following figure, AD is a straight line, OP ⊥ AD and O is the centre of both circles. If OA = 34cm, OB = 20 cm and OP = 16 cm;
find the length of AB.
Exercise 17 (A) | Q 6 | Page 210
In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. Find the distance between the chords,
if both the chords are:
(i) on the opposite sides of the centre;
(ii) on the same side of the centre.
Exercise 17 (A) | Q 7 | Page 210
Two parallel chords are drawn in a circle of diameter 30.0 cm. The length of one chord is 24.0 cm and the distance between the two chords is 21.0 cm;
find the length of another chord.
Exercise 17 (A) | Q 8 | Page 210
A chord CD of a circle whose center is O is bisected at P by a diameter AB. Given OA = OB = 15 cm and OP = 9 cm.
Calculate the lengths of: (i) CD ; (ii) AD ; (iii) CB.
Exercise 17 (A) | Q 9 | Page 211
The figure given below shows a circle with center O in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm,
find the radius of the circle.
Exercise 17 (A) | Q 10 | Page 211
In the given figure, O is the center of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm,
Find the :
(i) the radius of the circle
(ii) length of chord CD.
Exercise 17 (B) [Page 217]
### Selina solutions for Concise Mathematics Class 9 ICSE Chapter 17 Circle Exercise 17 (B) [Page 217]
Exercise 17 (B) | Q 1 | Page 217
The figure shows two concentric circles and AD is a chord of a larger circle.
Prove that: AB = CD.
Exercise 17 (B) | Q 2 | Page 217
A straight line is drawn cutting two equal circles and passing through the mid-point M of the line joining their centers O and O'. Prove that the chords AB and CD, which are intercepted by the two circles, are equal.
Exercise 17 (B) | Q 3 | Page 217
M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O.
Prove that: (i) ∠BMN = ∠DNM
(ii) ∠AMN = ∠CNM
Exercise 17 (B) | Q 4 | Page 217
In the following figure; P and Q are the points of intersection of two circles with centers O and O'. If straight lines APB and CQD are parallel to OO';
prove that: (i) OO' = 1/2AB ; (ii) AB = CD
Exercise 17 (B) | Q 5 | Page 217
Two equal chords AB and CD of a circle with center O, intersect each other at point P inside the circle.
Prove that: (i) AP = CP ; (ii) BP = DP
Exercise 17 (B) | Q 6 | Page 217
In the following figure, OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q.
Prove that:
( i ) ΔOPA ≅ ΔOQC
( ii ) ΔBPC ≅ ΔBQA
Exercise 17 (B) | Q 7 | Page 217
The length of the common chord of two intersecting circles is 30 cm. If the diameters of these two circles are 50 cm and 34 cm, calculate the distance between their centers.
Exercise 17 (B) | Q 8 | Page 217
The line joining the midpoints of two chords of a circle passes through its center.
Prove that the chords are parallel.
Exercise 17 (B) | Q 9 | Page 217
In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centers of the circles.
Show that:
(i) AB = CD ;
(ii) AC = BD.
Exercise 17 (B) | Q 10 | Page 217
AB and CD are two equal chords of a circle with center O which intersect each other at a right angle at point P.
If OM ⊥ AB and ON ⊥ CD;
show that OMPN is a square.
Exercise 17 (C) [Pages 220 - 221]
### Selina solutions for Concise Mathematics Class 9 ICSE Chapter 17 Circle Exercise 17 (C) [Pages 220 - 221]
Exercise 17 (C) | Q 1 | Page 220
In the given figure, an equilateral triangle ABC is inscribed in a circle with center O.
Find: (i) ∠BOC
(ii) ∠OBC
Exercise 17 (C) | Q 2 | Page 220
In the given figure, a square is inscribed in a circle with center O.
Find:
(i) ∠BOC
(ii) ∠OCB
(iii) ∠COD
(iv) ∠BOD
Is BD a diameter of the circle?
Exercise 17 (C) | Q 3 | Page 220
In the given figure, AB is a side of regular pentagon and BC is a side of regular hexagon.
(i) ∠AOB
(ii) ∠BOC
(iii) ∠AOC
(iv) ∠OBA
(v) ∠OBC
(vi) ∠ABC
Exercise 17 (C) | Q 4 | Page 220
In the given figure, arc AB and arc BC are equal in length. If ∠AOB = 48°, find:
(i) ∠BOC
(ii) ∠OBC
(iii) ∠AOC
(iv) ∠OAC
Exercise 17 (C) | Q 5 | Page 220
In the given figure, the lengths of arcs AB and BC are in the ratio 3:2. If ∠AOB = 96°,
find: (i) ∠BOC (ii) ∠ABC
Exercise 17 (C) | Q 6 | Page 220
In the given figure, AB = BC = DC and ∠AOB = 50°.
(i) ∠AOC
(ii) ∠AOD
(iii) ∠BOD
(iv) ∠OAC
(v) ∠ODA
Exercise 17 (C) | Q 7 | Page 221
In the given figure, AB is a side of a regular hexagon and AC is a side of a regular eight-sided polygon.
Find:
(i) ∠AOB
(ii) ∠AOC
(iii) ∠BOC
(iv) ∠OBC
Exercise 17 (C) | Q 8 | Page 221
In the given figure, O is the center of the circle and the length of arc AB is twice the length of arc BC. If ∠AOB = 100°,
find: (i) ∠BOC (ii) ∠OAC
Exercise 17 (D) [Page 221]
### Selina solutions for Concise Mathematics Class 9 ICSE Chapter 17 Circle Exercise 17 (D) [Page 221]
Exercise 17 (D) | Q 1 | Page 221
The radius of a circle is 13 cm and the length of one of its chords is 24 cm.
Find the distance of the chord from the center.
Exercise 17 (D) | Q 2 | Page 221
Prove that equal chords of congruent circles subtend equal angles at their center.
Exercise 17 (D) | Q 3 | Page 221
Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points?
Exercise 17 (D) | Q 4 | Page 221
Suppose you are given a circle. Describe a method by which you can find the center of this circle.
Exercise 17 (D) | Q 5 | Page 221
Given two equal chords AB and CD of a circle with center O, intersecting each other at point P.
Prove that:
(i) AP = CP
(ii) BP = DP
Exercise 17 (D) | Q 6 | Page 221
In a circle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively.
Calculate the distance between the chords, if they are on:
(i) the same side of the center.
(ii) the opposite sides of the center.
Exercise 17 (D) | Q 7 | Page 221
In the given figure, O is the center of the circle with radius 20 cm and OD is perpendicular to AB. If AB = 32 cm,
find the length of CD.
Exercise 17 (D) | Q 8 | Page 221
In the given figure, AB and CD are two equal chords of a circle, with centre O. If P is the mid-point of chord AB, Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.
Exercise 17 (D) | Q 9 | Page 221
In the given figure, AOC is the diameter of the circle, with centre O. If arc AXB is half of arc BYC, find ∠BOC.
Exercise 17 (D) | Q 10 | Page 221
The circumference of a circle, with center O, is divided into three arcs APB, BQC, and CRA such that:
"arc APB"/2 = "arc BQC"/3 = "arc CRA"/4
Find ∠BOC.
## Chapter 17: Circle
Exercise 17 (A)Exercise 17 (B)Exercise 17 (C)Exercise 17 (D)
## Selina solutions for Concise Mathematics Class 9 ICSE chapter 17 - Circle
Selina solutions for Concise Mathematics Class 9 ICSE chapter 17 (Circle) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Concise Mathematics Class 9 ICSE solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Concise Mathematics Class 9 ICSE chapter 17 Circle are Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles, Arc, Segment, Sector, Chord Properties - a Straight Line Drawn from the Center of a Circle to Bisect a Chord Which is Not a Diameter is at Right Angles to the Chord, Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles, Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof), Theorem: Equal chords of a circle are equidistant from the centre., Converse: The chords of a circle which are equidistant from the centre are equal., Chord Properties - There is One and Only One Circle that Passes Through Three Given Points Not in a Straight Line, Arc and Chord Properties - If Two Arcs Subtend Equal Angles at the Center, They Are Equal, and Its Converse, Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles.
Using Selina Class 9 solutions Circle exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 9 prefer Selina Textbook Solutions to score more in exam.
Get the free view of chapter 17 Circle Class 9 extra questions for Concise Mathematics Class 9 ICSE and can use Shaalaa.com to keep it handy for your exam preparation
Share
|
# How do you find a standard form equation for the line with (7,-4) and perpendicular to the line whose equation is x-7y-4=0?
Oct 5, 2017
$7 x + y = 45$
#### Explanation:
$\text{the equation of a line in "color(blue)"standard form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A x + B y = C} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where A is a positive integer and B, C are integers.
• " given a line with slope m then the slope of a line"
$\text{perpendicular to it is}$
•color(white)(x)m_(color(red)"perpendicular")=-1/m
$\text{rearrange "x-7y-4=0" into "color(blue)"slope-intercept form}$
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\Rightarrow y = \frac{1}{7} x - \frac{4}{7} \Rightarrow m = \frac{1}{7}$
$\Rightarrow {m}_{\textcolor{red}{\text{perpendicular}}} = - \frac{1}{\frac{1}{7}} = - 7$
$\Rightarrow y = - 7 x + b \leftarrow \text{ partial equation}$
$\text{to find b substitute "(7,-4)" into the partial equation}$
$- 4 = - 49 + b \Rightarrow b = 45$
$\Rightarrow y = - 7 x + 45 \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$
$\Rightarrow 7 x + y = 45 \leftarrow \textcolor{red}{\text{ in standard form}}$
Oct 5, 2017
$7 x + y = 45$
#### Explanation:
The given equation is in standard form which is $a x + b y = c$
but we need to know its slope, so change it into the form $y = m x + c$
$x - 7 y - 4 = 0 \text{ "rarr x-4 =7y" } \rightarrow 7 y = x - 4$
$y = \textcolor{b l u e}{\frac{1}{7}} x - \frac{4}{7}$
$\rightarrow m = \frac{1}{7}$
If lines are perpendicular then ${m}_{1} \times {m}_{2} = - 1$
(One slope is the negative reciprocal of the other - flip and change the sign.)
The slope perpendicular to $\frac{1}{7}$ is $- 7$
Now we have a point $\left(7 , - 4\right)$ and $m = - 7$ so substitute into the point-slope formula for a straight line.
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - \left(- 4\right) = - 7 \left(x - 7\right)$
$y + 4 = - 7 x + 49 \text{ } \leftarrow$ change to standard form:
$7 x + y = 49 - 4$
$7 x + y = 45$
|
Mathematic
# Problems on Arithmetic Progression
A sequence of numbers is known as an arithmetic progression (A.P.) if the difference between the term and the preceding term is always same or constant. An arithmetic progression is a sequence of numbers such that the difference between any two successive members is constant.
For example,
• the sequence 1, 2, 3, 4, … is an arithmetic progression with common difference 1.
• the sequence 3, 5, 7, 9, 11,… is an arithmetic progression with common difference 2.
• the sequence 20, 10, 0, -10, -20, -30, … is an arithmetic progression with common difference -10.
Problems on Arithmetic Progression
Here we will learn how to solve different types of problems in the arithmetic progression.
1. Show that the sequence 7, 11, 15, 19, 23, ……… is an Arithmetic Progression. Find its 27th term and the general term.
Solution:
First term of the given sequence = 7
Second term of the given sequence = 11
Third term of the given sequence = 15
Fourth term of the given sequence = 19
Fifth term of the given sequence = 23
Now, Second term – First term = 11 – 7 = 4
Third term – Second term = 15 – 11 = 4
Fourth term – Third term = 19 – 15 = 4
Fifth term – Fourth term = 23 – 19 = 4
Therefore, the given sequence is an Arithmetic Progress with the common difference 4.
We know that nth term of an Arithmetic Progress, whose first term is a and the common difference is d is tn = a + (n – 1) × d.
Therefore, 27th term of the Arithmetic Progress = t27 = 7 + (27 – 1) × 4 = 7 + 26 × 4 = 7 + 104 = 111.
General term = nth term = an = a + (n – 1)d = 7 + (n – 1) × 4 = 7 + 4n – 4 = 4n + 3
1. The 5th term of an Arithmetic Progression is 16 and 13th term of an Arithmetic Progression is 28. Find the first term and common difference of the Arithmetic Progression.
Solution:
Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the required Arithmetic Progression.
According to the problem,
5th term of an Arithmetic Progression is 16
i.e., 5th term = 16
⇒ a + (5 – 1)d = 16
⇒ a + 4d = 16 ………………. (i)
and 13th term of an Arithmetic Progression is 28
i.e., 13th term = 28
⇒ a + (13 – 1)d = 28
⇒ a + 12d = 28 ……………….. (ii)
Now, subtract the equation (i) from (ii) we get,
8d = 12
⇒ d = 12/8
⇒ d = 3/2
Substitute the value of d = 3232 in the equation (i) we get,
⇒ a + 4 × 3/2 = 16
⇒ a + 6 = 16
⇒ a = 16 – 6
⇒ a = 10
Therefore, the first term of the Arithmetic Progression is 10 and common difference of the Arithmetic Progression is 3/2.
Information Source:
|
## Solving Problems Involving Tickets and Stamps
### Learning Outcomes
• Solve ticket and stamp word problems
The strategies we used for coin problems can be easily applied to some other kinds of problems. Problems involving tickets or stamps are very similar to coin problems, for example. Like coins, tickets and stamps have different values, so we can organize the information in tables much like we did for coin problems.
### Example
At a school concert, the total value of tickets sold was $\text{\1,506}$. Student tickets sold for $\text{\6}$ each and adult tickets sold for $\text{\9}$ each. The number of adult tickets sold was $5$ less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?
Solution:
• Determine the types of tickets involved.
There are student tickets and adult tickets.
• Create a table to organize the information.
Type $\text{Number}$ $\text{Value (\)}$ $\text{Total Value (\)}$
Student $6$
Adult $9$
$1,506$
Step 2. Identify what you are looking for.
We are looking for the number of student and adult tickets.
Step 3. Name. Represent the number of each type of ticket using variables.
We know the number of adult tickets sold was $5$ less than three times the number of student tickets sold.
Let $s$ be the number of student tickets.
Then $3s - 5$ is the number of adult tickets.
Multiply the number times the value to get the total value of each type of ticket.
Type $\text{Number}$ $\text{Value (\)}$ $\text{Total Value (\)}$
Student $s$ $6$ $6s$
Adult $3s - 5$ $9$ $9\left(3s - 5\right)$
$1,506$
Step 4. Translate: Write the equation by adding the total values of each type of ticket.
$6s+9\left(3s - 5\right)=1506$
Step 5. Solve the equation.
$\begin{array}{c}6s + 27s - 45 = 1506\hfill \\ 33s - 45 = 1506\hfill \\ 33s = 1551\hfill \\ s=47\text{ students}\hfill \\\text{Substitute to find the number of adults.}\hfill \\ 3s - 5 =\text{ number of adults}\hfill \\ 3(\color{red}{47}) - 5 = 136\text{ adults}\hfill \end{array}$
Step 6. Check. There were $47$ student tickets at $\text{\6}$ each and $136$ adult tickets at $\text{\9}$ each. Is the total value $\text{\1506}?$ We find the total value of each type of ticket by multiplying the number of tickets times its value; we then add to get the total value of all the tickets sold.
$\begin{array}{ccc}\hfill 47\cdot 6& =\hfill & 282\hfill \\ \hfill 136\cdot 9& =\hfill & \underset{\text{_____}}{1224}\hfill \\ & & 1506\quad\checkmark \hfill \end{array}$
Step 7. Answer the question. They sold $47$ student tickets and $136$ adult tickets.
### try it
Now we’ll do one where we fill in the table all at once.
### Example
Monica paid $\text{\10.44}$ for stamps she needed to mail the invitations to her sister’s baby shower. The number of $\text{49-cent}$ stamps was four more than twice the number of $\text{8-cent}$ stamps. How many $\text{49-cent}$ stamps and how many $\text{8-cent}$ stamps did Monica buy?
Watch the following video to see another example of how to find the number of tickets sold, given the total sales for two different ticket values.
|
Jump to a New ChapterAnatomy of SAT Numbers & OperationsEssential ConceptsEssential StrategiesTest-Taking StrategiesThe 8 Most Common MistakesConclusionSet 1: Multiple ChoiceSet 2: Grid-InsPosttest
Number Terms Order of Operations Odd and Even Numbers Positive/Negative/Undecided Divisibility and Remainders Fearless Factors Multiples, Multiples, Multiples Fractions
Decimals Percents Ratios Exponents, or The Powers That Be Roots and Radicals Sequences Sets
Fractions
Fractions are ubiquitous on the SAT Math section (there’s an SAT word for you to chew on). A fraction shows the relationship between a part of something and the whole of something. Here’s what a fraction looks like: .
A fraction is composed of two numbers, a numerator, and a denominator. The numerator (the part) is above the fraction bar, and the denominator (the whole) is below it. So in our example, 2 is the numerator, and 3 is the denominator.
What if we multiply both the numerator and the denominator of by 4? We would end up with:
Because we multiplied both parts by the same numbers, the two fractions ( and ) have the same part-to-whole relationship. These fractions are equivalent fractions, which means they equal each other.
Suppose we decide that is too bulky. We know that 8 and 12 have an LCM of 4—remember that term?—so we can whittle that down to a slimmer fraction:
Taking a fat fraction and dividing out similar terms from both the numerator and denominator is called reducing a fraction.
Reducing fractions and making equivalent fractions are necessary skills, because you often have to perform these tasks before adding, subtracting, multiplying, or dividing fractions. Don’t simply take our word for it. Review the following sections.
We’ll start simple. Fractions with the same denominators are the easiest. You need only add or subtract the numerators:
No rocket science there. Fractions with different denominators are one degree trickier. Before adding or subtracting them, you first need to make sure that the denominators are the same.
For instance, if you need to perform the operation , first you have to equalize the denominators of these two fractions. You do this by finding the lowest common multiple of denominators 3 and 15. This number is the lowest common denominator, or LCD.
Because 15 is a multiple of 3, the LCD is 15. So all you need to do is multiply 3 and 5 to get 15. However, if you increase the denominator by 5, you need to increase the numerator by 5 too. Remember: all fractions are the parts of a whole:
Now you can get your subtraction on:
Suppose the LCD is difficult to find, such as when you have 12 and 18 in the denominator. You could simply multiply 12 and 18 and use the product as a common denominator. This gives you some big fractions that you have to reduce. Working with big numbers increases the possibility of errors, so it’s best to find the LCD before proceeding.
Multiplying and Dividing Fractions
Compared to adding and subtracting with different denominators, multiplying fractions is easy. You just find the product of the fractions’ denominators and numerators:
Sometimes you can cross-cancel numbers and make multiplying even simpler. In the example below, the 5 in the numerator takes out the 5 in the denominator.
Division of fractions is just like multiplication of fractions with a little twist. When dividing, you flip the second fraction, then multiply:
One key concept to remember is: when you multiply a fraction by itself, the resulting fraction is less than the original fraction. For example:
The fraction is smaller than .
Weird-Looking Fractions
Not every fraction on the SAT is in the button-down conservative style. Many items are designed to freak you out. To do this, the SAT uses mixed numbers and complex fractions. We’ll cover complex fractions first.
For complex fractions, you have to calculate a fraction of a fraction:
See that bar between the 3 and the 1? It’s a fraction bar, but you can also think of it as a division sign. That bar holds the key to solving complex fractions. Take the fraction above the bar and divide it by the fraction below the bar. You just learned how to divide fractions, so this shouldn’t be tough:
Simple enough. On to mixed numbers.
A mixed number, or improper fraction, includes both an integer and a fraction. For example, is a mixed number. Mixed numbers are impenetrable to simple addition and subtraction, so you need to do one of two things before you work with them:
1. Convert them into proper fractions.
2. Use your calculator and change them to decimals.
Converting improper fractions is pretty simple. Start with a mixed number, such as . Multiply the integer 4 by the denominator 7, then add that product to the numerator 2 and place the new number over the old denominator:
Remember that on the grid-in section, you cannot write a mixed number as your answer because the computer reads as . That’s why converting mixed numbers becomes so important. Of course, you can also just convert the whole mess into a decimal.
Jump to a New ChapterAnatomy of SAT Numbers & OperationsEssential ConceptsEssential StrategiesTest-Taking StrategiesThe 8 Most Common MistakesConclusionSet 1: Multiple ChoiceSet 2: Grid-InsPosttest
Test Prep Books
Test Prep Centers
New SAT Test Center Mini SAT
SAT Vocab Novels
Rave New World S.C.A.M. Sacked Busted Head Over Heels
SAT Power Tactics
Algebra Data Analysis, Statistics & Probability Geometry Reading Passages Sentence Completions Writing Multiple-Choice Questions The Essay Test-Taking Strategies Vocabulary Builder
SparkCollege
College Admissions Financial Aid College Life
|
Block matrix
In the mathematical discipline of matrix theory, a block matrix or a partitioned matrix is a partition of a matrix into rectangular smaller matrices called blocks. Looking at it another way, the matrix is written in terms of smaller matrices written side-by-side. A block matrix must conform to a consistent way of splitting up the rows, and the columns: we group the rows into some adjacent 'bunches', and the columns likewise. The partition is into the rectangles described by one bunch of adjacent rows crossing one bunch of adjacent columns. In other words, the matrix is split up by some horizontal and vertical lines that go all the way across.
Example
The matrix
[itex]P = \begin{bmatrix}
1 & 1 & 2 & 2\\ 1 & 1 & 2 & 2\\ 3 & 3 & 4 & 4\\ 3 & 3 & 4 & 4\end{bmatrix}[itex]
can be partitioned into 4 2×2 blocks
[itex]P_{11} = \begin{bmatrix}
1 & 1 \\ 1 & 1 \end{bmatrix}, P_{12} = \begin{bmatrix} 2 & 2\\ 2 & 2\end{bmatrix}, P_{21} = \begin{bmatrix} 3 & 3 \\ 3 & 3 \end{bmatrix}, P_{22} = \begin{bmatrix} 4 & 4\\ 4 & 4\end{bmatrix}.[itex]
The partitioned matrix can then be written as
[itex]P_{\mathrm{partitioned}} = \begin{bmatrix}
P_{11} & P_{12}\\ P_{21} & P_{22}\end{bmatrix}.[itex]
Block diagonal matrices
A block diagonal matrix is a block matrix which is a square matrix, and having main diagonal blocks square matrices, such that the off-diagonal blocks are zero matrices. A block diagonal matrix A has the form
[itex]
\mathbf{A} = \begin{bmatrix} A_{1} & 0 & \cdots & 0 \\ 0 & A_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & A_{n} \end{bmatrix} [itex]
where Ak is an square matrix. Any square matrix can be trivially considered as a block diagonal matrix, taking n=1.
Application
In linear algebra terms, the use of a block matrix corresponds to having a linear mapping thought of in terms of corresponding 'bunches' of basis vectors. That again matches the idea of having distinguished direct sum decompositions of the domain and range. It is always particularly significant if a block is the zero matrix; that carries the information that a summand maps into a sub-sum.
Given the interpretation via linear mappings and direct sums, there is a special type of block matrix that occurs for square matrices (the case m = n). For those we can assume an interpretation as an endomorphism of an n-dimensional space V; the block structure in which the bunching of rows and columns is the same is of importance because it corresponds to having a single direct sum decomposition on V (rather than two). In that case, for example, the diagonal blocks in the obvious sense are all square. This type of structure is required to describe the Jordan normal form.
This technique is used to cut down calculations of matrices, column-row expansions, and many computer science applications, including VLSI chip design. An example is the Strassen algorithm for fast matrix multiplication.
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy
|
# Complex Numbers
Section 1.3 Complex Numbers
How can we solve the equation x 2= −1? x = ± It turns out that this is not a real number ! So we have to "make up " some new numbers! We define the new number i as follows i =
Complex Numbers Let i2=-1 then i= i is the square root of -1
Examples
Complex Numbers A complex number has the form a + bi a is called the real part of the complex number b is called the imaginary part of the complex number
The Complex Plane -
Adding Complex Numbers (a + bi ) + (c + di ) = (a + c) + (b + d ) i To add two complex numbers, add the real parts and add the imaginary parts, the result is a complex number.
Examples (2 + 3i ) + (−5 + 4i ) = (5 + 2i ) − (2 − 7i ) =
Multiplying Complex Numbers (a+bi ) (c+di ) = ac + adi + bci + bdi2 = (ac −bd ) + (ad + bc) i This follows from the distributive property and the fact that t2 = −1
Examples (2 + 3i ) (−5 + 4i ) = (5 + 2i ) (2 − 7i ) = (a + bi ) (a − bi ) = ( a + bi) and (a − bi ) are called complex conjugates
If z=a + bi is a complex number, then = a -bi is the conjugate of z. The product of a complex number and its conjugate is a real number.
Multiply the complex conjugates (2 + 3i ) (2 − 3i ) = (5 − 2i ) (5 + 2i ) = (c + di ) (c − di ) =
Dividing Complex Numbers
Magnitude of Complex Numbers If c = a + bi is a complex number, The absolute value (magnitude) of is the distance away from 0 0 in the complex plane. This follows from the distance formula!
The Complex Plane
Examples Find c = 2 + 3i c = 5 − 4i c = −4 + 3i c = −4 − 3i
The Quadratic Formula If ax2 + bx + c = 0 then
Use the quadratic formula to solve x2 + 4 = 0 3y2 + 2y +1= 0 2t2− t = −2
Analytical Solutions Given a quadratic equation ax 2 + bx + c = 0 Use the quadratic formula If b 2− 4ac > 0 there are two real solutions If b2 − 4ac = 0 there is one real solution If b2 − 4ac< 0there are no real solutions, but two complex conjugate solutions.
Prev Next
|
# How do you simplify sqrt80 + sqrt125 - sqrt45 + 2 sqrt20?
May 27, 2016
$10 \sqrt{5}$
#### Explanation:
Note that each of the values under the square root sign has 5 as a factor. In addition to this, in each case the other factor is a perfect square.
$\sqrt{16 \times 5} + \sqrt{25 \times 5} - \sqrt{9 \times 5} + 2 \sqrt{4 \times 5}$
$= 4 \sqrt{5} + 5 \sqrt{5} - 3 \sqrt{5} + 2 \times 2 \sqrt{5}$
Each term is now given in terms of $\sqrt{5}$ and can be simplified to give
$10 \sqrt{5}$
|
# What number comes next in this sequence?
What comes next in this sequence?
4, 3, 9, 5, 19, 9, 39, 17, 79, 33, ?
It didn't immediately jump out at me, but ended up not too challenging and thought some may enjoy it.
It is:
4, 3, 9, 5, 19, 9, 39, 17, 79, 33, 159, 65, 319, 129
Because:
The differences between every other number are:
5, 2, 10, 4, 20, 8, 40, 16
So the odd entries begin with $4$ and add $5\times2^n$ to each.
The even entries begin with $3$ and add $2^n$ to each.
In appropriate math notation, thanks to f'':
$2x\,\text{th}$ term is $2^x+1$ and the $2x+1\,\text{th}$ term is $5⋅2^x−1$.
• The closed form is that the $2x$th term is $2^x+1$ and the $2x+1$th term is $5\cdot2^x-1$. – f'' May 10 '16 at 2:21
• Ah, ninja'd. This tablet types angle brackets too slowly :) – paste May 10 '16 at 2:24
The next numbers in sequence are
159, 65, 319, ...
The formula of the sequence is:
$(2^{\frac{n}{2}}) \cdot (2^{\frac{3}{2}}+2^{-\frac{1}{2}})^{nmod2}+(-1)^{n}$, where $n$ is the nth term and $nmod2$ is n modulo 2
The formula is deduced using the following logic:
To get the odd terms (where n = 1, 3, 5, ...), the following formula is used:
$2^{(\frac{n+1}{2}+1)} + 2^{(\frac{n+1}{2}-1)} - 1$
$=(2^{\frac{n}{2}})(2^{\frac{3}{2}}+2^{-\frac{1}{2}}) - 1$
To get the even terms (where n = 2, 4, 6, ...). the following formula is used:
$2^{\frac{n}{2}}+1$
There are only two differences in the above two formulas, one is $(2^{\frac{3}{2}}+2^{-\frac{1}{2}})$ and the other is the last constant $1$. To combine those two formulas, these two differences need to be addressed
1. As can be deduced from the above two formulas, $(2^{\frac{3}{2}}+2^{-\frac{1}{2}})$ is needed to obtain the odd terms while it is not needed to obtain the even terms. Hence a power of $nmod2$ is added so that $(2^{\frac{3}{2}}+2^{-\frac{1}{2}})^{nmod2} = 1$ when $n$ is even
2. As for the last constant, using $(-1)^n$ should suffice
After combination, the final formula is obtained: $(2^{\frac{n}{2}}) \cdot (2^{\frac{3}{2}}+2^{-\frac{1}{2}})^{nmod2}+(-1)^{n}$
I hope that my logic of deduction is easy enough to understand. Do comment on unclear parts =)
|
Credit:Judith HauslerCulturaGetty Images
Q:
# What are the factors of 30?
A:
The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. A number is considered the factor of another number when it can be evenly divided into that number.
Know More
## Keep Learning
If the only factors for a number are 1 and the number itself, it is considered a prime number. Common factors between two numbers are any factors shared by both of those numbers. The greatest common factor is the biggest of those common factors. There are multiple ways to find the greatest common factor, one of which is to list all the factors of both numbers. Another method is finding the shared prime number factors and multiplying them together. The greatest common factor is useful for simplifying fractions, because dividing the numerator and the denominator by the greatest common factor simplifies the fraction as low as possible.
Numbers can be broken down into prime number factors through prime factorization, also known as integer factorization. This is done by continually breaking down a number into smaller factors until the only factors left are prime numbers. Prime factorization can be very difficult and complex for larger numbers, and no method is known to complete long factorizations quickly as of 2014.
## Related Questions
• A:
The factors of 21 are 1, 3, 7 and 21. A factor is a number that can divide evenly into a given number. Because 21 can be divided by numbers other than itself and 1, it is called a composite number.
Filed Under:
• A:
The factors of 49 are 1, 7 and 49. The factors of a given number are those numbers that divide into it evenly leaving no remainder. One and the number itself, 49 in this case, are always factors.
Filed Under:
• A:
The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72. Factors are the numbers that can be multiplied together to get 72.
Filed Under:
• A:
The factors of 26 are one, two, 13 and 26. One and the number itself, in this case 26, are always factors. Since 26 is an even number, which means that it ends in zero, two, four, six or eight, it's evenly divisible by two. Two times 13 equals 26.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.