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Basic Concepts of Percentages
In this lesson, we cover the absolute basics of Percentages. The purpose of this lesson is to help you answer one simple question: What are Percentages?
Basic Definition:
Percent implies “for every hundred” and the sign % is read as percentage and x % is read as x per cent. In other words, a fraction with denominator 100 is called a per cent. For example, 20 % means 20/100 (i.e. 20 parts from 100). This can also be written as 0.2.
Basic Formula:
In order to calculate p % of q, use the formula:
(p/100) x q = (pxq)/100
Also remember: p % of q = q % of p
Examples:
1. 100% of 60 is 60 x (100/100) = 60
2. 50% of 60 is 50/100 × 60 = 30
3. 5% of 60 is 5/100 × 60 = 3
Example: 60 % of a number is 360. What is 99 % of the same number?
Solution: Let the number be n.
Given (60/100) ×n = 360 => n = 600
99 % of 600 = (99/100) × 600 = 594
Example: 50 % of a number is 360. What is 99 % of the same number?
Solution: Let the number be y.
Given (50/100) x q = 360
=> q = 720
99% of 720 = (99/100) x 720 = 712.80
Expressing One Quantity as a Per Cent with respect to the other:
To express a quantity as a per cent with respect to other quantity, the following formula is used:
Example: What percent is 60 of 240?
Solution: First write the given numbers in the fraction form:
60/240 = ¼
Multiply the numerator and denominator with 25 to make the denominator equal to 100
(1×25)/(4×25) = 25/100
25 percent or 25 per 100 is called as 25%
Sample Question for the Basics of Percentage:
Example:A number exceeds 20% of itself by 40. The number is:
(a) 50
(b) 60
(c) 80
(d) 48
Solution: Let the number be p.
20% of itself means => p x (20/100)
Now, according to the question,
p – 20% of p = 40
=> {p – (20 x p)/100} = 40
=> {p-(p/5)} = 40
⇒ 5p – p = 200
∴ p = 50
Alternate Method:
Obviously, it is clear that difference is 80% i.e. 4/5 of number which is equal to 40
4/5p = 40
p = 40 x 5/4= 50.
Tips & Tricks for Percentages:
Basic Tip-1: If the new value of something is n times the previous given value, then the percentage increase is (n-1) 100%.
Derivation:
Let us consider two values p and q.
Let q be and original value and p be the new value.
According to conditions p= nq
We need to calculate the percentage increase.
You can either use direct formula= {(new value – old value)/old value} x 10
This value becomes= {(p – q)/q} x 100
{(nq – q)/q} x 100
=> (n-1) x 100%
Example: If X= 5.35 Y, then find the percentage increase when the value of something is from Y to X.
Solution:
Use the formula: (n-1)100%
Percentage increase from
Y to X = (5.35 -1) 100= 435%
Basic Tip-2:
When a quantity N is increased by K %, then the:
New quantity = N (1+ K/100 )
Examples:
Increase 150 by 20%= 150 {1+(20/100)} = 150 1.2= 180
Increase 300 by 30%= 300 {1+(30/100)}= 300 1.3= 390
Increase 250 by 27% = 250 {1+(27/100)} = 250 1.27 =317.5
Example: What is the new value when 265 is increased by 15%?
Solution: New quantity = N (1+ K/100)
= 265{1+(15/100)}
New quantity = 1.15 265= 304.75
Basic Tip 3:
When a quantity N is decreased by K %, then the:
New quantity =N (1 – K/100)
Examples:
Decrease 120 by 20%= 120 {1-(20/100)} = 120 0.8= 96
Decrease 150 by 40%=150 {1-(40/100)} = 150 0.6= 90
Decrease 340 by 27%= 340 {1-(27/100)}= 340 0.73= 248.2
Example: If the production in 2015 is 400 units and the decrease from 2014 to 2015 is 13%, find the production in 2014?
Solution:
Remember the formula:
New quantity =N (1 – K/100)
Let the production in 2014 be x.
It has been decreased by 13% , which then becomes 400 in 2015
[X{1-(13/100)}]= 400
Production in 2014= 400 / 0.87= 459.77 units
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AOA2_TASK_4_Lesson_Plan.docx - LESSON PLAN TEMPLATE \u2013 2015 Design an original elementary(K-6 mathematics lesson plan using the attached \u201cLesson Plan
# AOA2_TASK_4_Lesson_Plan.docx - LESSON PLAN TEMPLATE u2013...
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LESSON PLAN TEMPLATE – 2015 Design an original elementary (K-6) mathematics lesson plan using the attached “Lesson Plan Template” that addresses the topic of fractions, decimals, or percentages 1. Implement differentiation for early finishers and student interaction in the lesson plan from part B. 2. Incorporate one additional instructional strategy discussed in part 3 of the attached “Portfolio Response Sheet: Percentages, Fractions, and Decimals” that supports the understanding of fractions, decimals, or percentages into the lesson plan from part B. GENERAL INFORMATION Lesson Title & Subject(s): Unit Fraction Action- A Lesson on “Parts of a Whole” Topic or Unit of Study: Unit 5: Representing and Comparing Fractions Grade/Level: Grade 3 Instructional Setting: The setting is a 3 rd grade math classroom, numbered 16 students. The seating arrangement consists of 4 tables, with 4 students seated at each table, given new seating assignments as they came to class, based on the results of the previous day’s assessment. Each table is labeled “1” through “4”.Anchor charts relating to fractions are placed around the room. STANDARDS AND OBJECTIVES Your State Core Curriculum: Understand a fraction 1/ ? as the quantity formed by 1 part when a whole is partitioned into b equal parts (unit fraction); understand a fraction ? /b as the quantity formed by a parts of size 1/ ? . For example, 3/4 means there are three 1/4 parts, so 1/4 = 1/4 + 1/4 + 1/4 Lesson Objective(s): Students will create equivalent addition equations, given a list of fractions, with 80% accuracy (3 out of 4). MATERIALS Instructional Materials: Fraction bars for student modeling; Dry erase boards and markers; Document reader; wall-sized white board INSTRUCTIONAL PLAN
Sequence of Instructional Procedures/Activities/Events (provide description and indicate approximate time for each): 1. Student Prerequisite Skills/Connections to Previous Learning: 5 minutes (This lesson would be day 3 in a week-long introductory segment on fractions) Prerequisite knowledge: a basic understanding that whole amounts of something can be divided into smaller pieces; that the total number of segments that the whole is divided into can be represented as the denominator of a fraction; that that number of segments of the whole that are being addressed for are represented as the numerator of a fraction; That the proper representation of the segmented whole amount the the addressed segments is the a/b format (a being the numerator and b being the denominator). Connection to previous learning: At the opening of
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The fraction 2 3/50 is equal to 0.46 once converted come a decimal. See below detalis on how to convert the fraction 2 3/50 come a decimal value.
You are watching: What is 2/3 of 50
### Fraction to Decimal Converter
Enter a portion value:Ex.: 1/2, 2 1/2, 5/3, etc. Note that 2 1/2 means two and half = 2 + 1/2 = 2.5
Answer:
Fraction come decimal explained:
## How to convert from fraction to decimal?
To easily convert a portion to a decimal, divide the numerator (top number) through the denominator (bottom number).
### Example 1: exactly how to transform 4/8 to a decimal?
Step 1:Divide 4 by 8: 4 ÷ 8 = 1 ÷ 2 = 0.5Step 2:Multiply the an outcome by 100 and include the decimal sign: 0.5 × 100%Answer: 4/8 = 50%
### Example 2: how to convert 1 1/3 to a decimal?
Step 1:Divide 1 by 3: 1 ÷ 3 = 0.3333Step 2:Add this worth to the the creature part: 1 + 0.3333 = 1.3333Step 3:Multiply the an outcome by 100 and add the decimal sign: 1.3333 × 100%Answer: 1 1/3 = 133.33%
Note: the an outcome was rounded to 2 decimal places.
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See more: What Is A Body Parts That Start With X,Y,Z, Bones Containing 'X' Quiz
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# 3.3 Chain Rule
## Chain Rule
There is one more type of complicated function that we will want to know how to differentiate: composition. The Chain Rule will let us find the derivative of a composition. Think of the Chain Rule when the function to be differentiated consists of an “outer” function and an “inner” function which we can write as f(g(x)).
### Example 1
Find the derivative of $y=\left(4x^3+15x\right)^2$
This is not a simple polynomial, so we can’t use the basic building block rules yet. It is a product, so we could write it as $y=\left(4x^3+15x\right)^2=\left(4x^3+15x\right)\left(4x^3+15x\right)$ and use the product rule. Or we could multiply it out and simply differentiate the resulting polynomial. I’ll do it the second way:
\begin{align*} y=& \left(4x^3+15x\right)^2\\ =& 16x^6+120x^4+225x^2\\ y’=& 96x^5+480x^3+450x \end{align*}
Now suppose we want to find the derivative of $y=\left(4x^3+15x\right)^{20}$. We could write it as a product with 20 factors and use the product rule, or we could multiply it out. But I don’t want to do that, do you?
We need an easier way, a rule that will handle a composition like this. The Chain Rule is a little complicated, but it saves us the much more complicated algebra of multiplying something like this out. It will also handle compositions where it wouldn’t be possible to multiply it out.
The Chain Rule is a common place for students to make mistakes. Part of the reason is that the notation takes a little getting used to. And part of the reason is that students often forget to use it when they should. When should you use the Chain Rule? Almost every time you take a derivative.
### Derivative Rules: Chain Rule
In what follows, $f$ and $g$ are differentiable functions with $y=f(u)$ and $u=g(x)$.
#### Chain Rule (Leibniz notation)
$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$
Notice that the $du$’s seem to cancel. This is one advantage of the Leibniz notation – it can remind you of how the chain rule chains together.
#### Chain Rule (using prime notation)
$\frac{d}{dx}\left[f\left(g(x)\right)\right]=f'(u)\cdot g'(x)=f’\left(g(x)\right)\cdot g'(x)$
#### Chain Rule (in words)
The derivative of a composition is the derivative of the outside (with the inside staying the same) TIMES the derivative of the inside.
I recite the version in words each time I take a derivative, especially if the function is complicated.
### Example 2
Find the derivative of $y=\left(4x^3+15x\right)^2$
This is the same one we did before by multiplying out. This time, let’s use the Chain Rule: The inside function is what appears inside the parentheses: $4x^3+15x$. The outside function is the first thing we find as we come in from the outside – it’s the square function, $(\text{inside})^2$.
The derivative of this outside function is $(2\cdot\text{inside})$. Now using the chain rule, the derivative of our original function is $(2\cdot\text{inside})$ TIMES the derivative of the inside (which is $12x^2+15$):
$y’=2\left(4x^3+15x\right)\left(12x^2+15 \right)$
If you multiply this out, you get the same answer we got before. Hurray! Algebra works!
### Example 3
Find the derivative of $y=\left(4x^3+15x\right)^{20}$.
Now we have a way to handle this one. It’s the derivative of the outside TIMES the derivative of the inside.
The outside function is $\left(\text{inside}\right)^{20}$, which has derivative $20\left(\text{inside}\right)^{19}$, so $y’=20\left(4x^3+15x\right)^{19}\left(12x^2+15\right).$
### Example 4
Differentiate $y=e^{x^2+5}$.
This isn’t a simple exponential function; it’s a composition. Typical calculator or computer syntax can help you see what the “inside” function is here. On a TI calculator, for example, when you push the $e^x$ key, it opens up parentheses: e^( . This tells you that the “inside” of the exponential function is the exponent. Here, the inside is the exponent $x^2+5$. Now we can use the Chain Rule: We want the derivative of the outside TIMES the derivative of the inside. The outside is the $e$ to the something function, so its derivative is the same thing. The derivative of what’s inside is $2x$. So $\frac{d}{dx}\left( e^{x^2+5} \right)= \left( e^{x^2+5} \right)\cdot (2x).$
### Example 5
The table gives values for $f$ , $f'$ , $g$, and $g'$ at a number of points. Use these values to determine $( f \circ g )(x)$ and $( f \circ g ) '(x)$ at $x = -1$ and 0.
Table 3.1
$x$ $f(x)$ $g(x)$ $f'(x)$ $g'(x)$ $(f\circ g)(x)$ $(g\circ f)(x)$
-1 2 3 1 0
0 -1 1 3 2
1 1 0 -1 3
2 3 -1 0 1
3 0 2 2 -1
$(f\circ g)(-1)=f\left(g(-1)\right)=f(3)=0$
$(f\circ g)(0)=f\left(g(0)\right)=f(1)=1$
$(f\circ g)'(-1)=f'\left(g(-1)\right)\cdot g'(-1)=f'(3)\cdot (0)=(2)(0)=0 \text{ and}$
$(f\circ g)'(0)=f'\left(g(0)\right)\cdot g'(0)=f'(1)\cdot (2)=(-1)(2)=-2$
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# PEMDAS and Order of Operations Series
PEMDAS is an acronym used for order of operations. P stands for parenthesis (or generally grouping), E stands for Exponent, and MDAS stands for multiplication, division, addition, and subtraction. This series is composed of three posts illustrating various examples on how to calculate using the order of operations. These posts are composed of videos that are about 10 minutes more or less. All the videos are in mixed Tagalog and English.
PEMDAS and Order of Operations Part 1
This post discusses how to perform basic calculations with 2 or 3 operations without parenthesis. The operations include addition, subtraction, multiplication and division.
PEMDAS and Order of Operations Part 2
This post discusses basic order of operations addition, subtraction, multiplication and division. The use of exponents and parentheses are also discussed. Additional explanation is provided for common misconceptions.
PEMDAS and Order of Operations Part 3
This post discusses more complicated calculations particularly those with nested parenthesis.
It is important that you master the order of operations because it will be used in all mathematics subjects in high school and beyond.
# What is the Sieve of Eratosthenes?
Did you know that you can systematically find the prime numbers from 1 to any number? In this video, I have discussed how to systematically find prime numbers from 1 to 100. This method is called the Sieve of Eratosthenes and can be extended to any number.
If you cannot see this video on your mobile phone, you can watch it on Youtube.
# Paano mag-subtract ng integers?
“Paano mag-subtract ng integers?” is a Taglish Math video that explains how to subtract signed numbers, particularly integers. The strategy used is converting the subtraction problem into addition. This way, you only need to master addition.
If you can’t see this video on your mobile phone, you can watch it on Youtube.
# Paano mag-multiply ng integers?
This Taglish (mixed Tagalog and English) math vide tutorial explains how to multiply integers. We know that the rules in multiplying integers is that when we multiply integers with the same sign, then the answer is positive. When we multiply integers with different signs, the answer is negative. These rules do not only apply to integers but to real numbers as well. And the rules in multiplying integers is also the same in dividing integers.
If you can’t see this video on you mobile phone, you can watch it on Youtube.
This video in Taglish (mixed Tagalog and English) discusses addition of integers. The video explains how to add integers that are both positive, both negative, and a positive and a negative integer. A number line is also drawn in order to explain further the how and why the addition operation works.
In case you can’t see this video on your mobile phone, you can watch it on Youtube.
# Ang Integers at Number Line
This video introduces the concepts about integers. Integers are numbers that do not have fractional components. It includes positive integers, negative integers, and zero.
A number line is also in the discussion in the video in order to explain how to compare the value of integers. An integer $a > b$ if $a$ is at the right of $b$ on the number line. As we go to the right hand side of the number line, the value of the integers become larger. As we go to the left hand side, the value of the numbers become smaller.
# Paano mag-divide ng decimals?
“Paano mag-divide ng decimals?” is a Taglish (mixed Tagalog and English) video math tutorial on how to divide decimals. In dividing decimals, it is usually easier if the decimal numbers are ‘eliminated’. To ‘eliminate’ the decimal numbers, the division may be written as a fraction. This way, the numerator and the numerator can be multiplied by powers of 10. If the decimals are eliminated, division may easily be performed.
# Paano mag-subtract ng decimals?
“Paano mag-subtract ng decimals?” is a Taglish math tutorial video on how to subtract decimals. Subtracting decimals is just the same as adding decimals. All you have to do is align the decimals of the minuend and the subtrahend and then perform the fraction. The decimal point in the answer must also be aligned with the decimal points of the minuend and subtrahend.
Taglish means mixed Tagalog and English.
# Paano ba mag-multiply ng decimals?
This video math tutorial in Taglish (mixed Tagalog and English) discusses about multiplication of decimals. The rules in multiplication of decimals is to (1) Ignore the decimals — just multiply ; (2) From all the given, count the number of decimals (numbers to the right hand of the decimal point), the answer must have the same number of digits as the total number of decimal numbers in the given.
If the answer has less decimals than the given, add zero on the left hand side.
“Paano mag-add ng decimals?” tagalog math video tutorial is about adding decimals. When you add decimals, the rule is to place the addends vertically in such a way that the decimal points are aligned. After this, you can perform addition as you do in whole numbers. Lastly, after adding copy the decimal point directly below the decimal point of the addends. The decimal point in the sum must also be aligned with those of the addends.
# Paano mag-subtract ng fractions?
This video tutorial discusses subtraction of fractions. The examples discussed in this video include similar fractions, dissimilar fractions, and mixed fractions.
Subtracting fractions is very similar to adding fractions. For similar fractions, the numerator are subtracted and the numerator copied in the answer. For dissimilar fractions, least common multiple is needed in order to perform the operation.
# Ano ba ang least common multiple?
The least common multiple of a and b is the smallest positive integer that is divisible by both a and b. The video below is an explanation of the concept of Least Common Multiple. The explanation is in mixed Tagalog and English. If you have questions and suggestions, please use the comment box below.
# Paano mag-divide ng fractions?
This video is a tutorial in mixed Tagalog and English on how to divide fractions. The idea in dividing fractions is to get the reciprocal of the divisor. This is because dividing $a$ by $b$ for $b$ not equal to zero is the same as multiplying $a$ by the reciprocal of $b$. That is,
$\frac{a}{b}$ is the same as $a(\frac{1}{b})$.
Now the reciprocal of the fraction \frac{p}{q}\$ is
$\displaystyle \frac{1}{\frac{p}{q}} = \frac{q}{p}$
That is the reason why we exchange the numerator and denominator of the fraction.
# Paano mag-multiply ng fractions?
This video is a tutorial on how to multiply fractions in mixed Tagalog and English. In multiplying fractions, you just remember the following rules and strategies:
1. Multiply the numerator and the denominator.
2. If the fractions are not in lowest terms, reduce them first, before performing multiplication. This will make the numbers smaller.
3. Mixed fractions must be converted first to improper fractions before they can be multiplied.
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# Relative velocity in one dimension (Page 3/4)
Page 3 / 4
$\begin{array}{l}⇒{v}_{C}={v}_{B}+{v}_{CB}\\ ⇒{v}_{CB}={v}_{C}-{v}_{B}\end{array}$
This is an important relation. This is the working relation for relative motion in one dimension. We shall be using this form of equation most of the time, while working with problems in relative motion. This equation can be used effectively to determine relative velocity of two moving objects with uniform velocities (C and B), when their velocities in Earth’s reference are known. Let us work out an exercise, using new notation and see the ease of working.
Problem : Two cars, initially 100 m distant apart, start moving towards each other with speeds 1 m/s and 2 m/s along a straight road. When would they meet ?
Solution : The relative velocity of two cars (say 1 and 2) is :
$\begin{array}{l}{v}_{21}={v}_{2}-{v}_{1}\end{array}$
Let us consider that the direction ${v}_{1}$ is the positive reference direction.
Here, ${v}_{1}$ = 1 m/s and ${v}_{2}$ = -2 m/s. Thus, relative velocity of two cars (of 2 w.r.t 1) is :
$\begin{array}{l}⇒{v}_{21}=-2-1=-3\phantom{\rule{2pt}{0ex}}m/s\end{array}$
This means that car "2" is approaching car "1" with a speed of -3 m/s along the straight road. Similarly, car "1" is approaching car "2" with a speed of 3 m/s along the straight road. Therefore, we can say that two cars are approaching at a speed of 3 m/s. Now, let the two cars meet after time “t” :
$\begin{array}{l}t=\frac{\mathrm{Displacement}}{\mathrm{Relative velocity}}=\frac{100}{3}=33.3\phantom{\rule{2pt}{0ex}}s\end{array}$
## Order of subscript
There is slight possibility of misunderstanding or confusion as a result of the order of subscript in the equation. However, if we observe the subscript in the equation, it is easy to formulate a rule as far as writing subscript in the equation for relative motion is concerned. For any two subscripts say “A” and “B”, the relative velocity of “A” (first subscript) with respect to “B” (second subscript) is equal to velocity of “A” (first subscript) subtracted by the velocity of “B” (second subscript) :
$\begin{array}{l}{v}_{AB}={v}_{A}-{v}_{B}\end{array}$
and the relative velocity of B (first subscript) with respect to A (second subscript) is equal to velocity of B (first subscript) subtracted by the velocity of A (second subscript):
$\begin{array}{l}{v}_{BA}={v}_{B}-{v}_{A}\end{array}$
## Evaluating relative velocity by making reference object stationary
An inspection of the equation of relative velocity points to an interesting feature of the equation. We need to emphasize that the equation of relative velocity is essentially a vector equation. In one dimensional motion, we have taken the liberty to write them as scalar equation :
$\begin{array}{l}{v}_{BA}={v}_{B}-{v}_{A}\end{array}$
Now, the equation comprises of two vector quantities ( ${v}_{B}$ and $-{v}_{A}$ ) on the right hand side of the equation. The vector “ $-{v}_{A}$ ” is actually the negative vector i.e. a vector equal in magnitude, but opposite in direction to “ ${v}_{A}$ ”. Thus, we can evaluate relative velocity as following :
• Apply velocity of the reference object (say object "A") to both objects and render the reference object at rest.
• The resultant velocity of the other object ("B") is equal to relative velocity of "B" with respect to "A".
This concept of rendering the reference object stationary is explained in the figure below. In order to determine relative velocity of car "B" with reference to car "A", we apply velocity vector of car "A" to both cars. The relative velocity of car "B" with respect to car "A" is equal to the resultant velocity of car "B".
#### Questions & Answers
what's lamin's theorems and it's mathematics representative
if the wavelength is double,what is the frequency of the wave
What are the system of units
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
58asagravitasnal firce
Amar
water boil at 100 and why
what is upper limit of speed
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
which colour has the shortest wavelength in the white light spectrum
how do we add
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
what is energy?
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
n=a+b/T² find the linear express
أوك
عباس
Quiklyyy
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# Write the zeros of polynomial x2 – x – 6.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The zeroes of the polynomial x2 − x − 6 = 0 are 3, -2. An algebraic expression known as a polynomial has an exponent on every variable that is a whole number. The highest or greatest power of a variable in a polynomial equation is referred to as the polynomial’s degree.
• A polynomial is a mathematical equation made up of exponents, constants, and variables that are mixed using addition, subtraction, multiplication, and division.
• The expression is divided into three categories: monomial, binomial, and trinomial depending on how many terms are included in it.
• The product of a polynomial’s factors with degrees that are less than or equal to the original polynomial can be used to represent a polynomial.
• Factorization of polynomials is the technical term for the factoring process.
• The value of a polynomial is the result of applying a specific value to the variable, and it is expressed as such.
### Steps to find zeros of polynomial x2 – x – 6
To find: Zeroes by factorizing the equation x2 − x − 6 = 0
We find the zeroes by factorizing the equation x2 − x − 6 = 0 as follows:
x2 − x − 6 = 0
The above equation can be written as:
⇒ x2 − 3x + 2x − 6 = 0
Taking common we get:
⇒ x(x − 3) + 2(x − 3) =0
⇒ (x + 2) (x − 3) = 0
In simplification we get the:
⇒ x + 2 = 0, x − 3 = 0
⇒ x = −2, x = 3.
Hence, the zeroes of the polynomial x2 − x − 6 = 0 are 3, -2.
Summary:
## Write the zeros of polynomial x2 – x – 6.
The zeroes of the polynomial x2 − x − 6 = 0 are 3, -2. Polynomials are algebraic expressions that consist of variables and coefficients. Variables are also sometimes called indeterminates.
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Complete JEE Main/Advanced Course and Test Series. Hope you played with the graph. If a is directing vector of first line, and b is directing vectors of second line then we can find angle between lines by formula: Franchisee | Page Comments Both are equation or straight line and both pass-through the origin. In mathematics, straight lines have an important role to play in two-dimensional geometry.A straight line is nothing but a locus of all such infinite number of points lying in the two-dimensional space and extending out in either direction infinitely. This can be written as 2x2 – 7xy + 3y2 = (2x – y)(x-3y). Example 1: Find the angle between two straight lines x + 2y - 1 = 0 and 3x - 2y + 5=0. Angle θ between the lines is given by. Just play with the following graph, you will definitely understand. The angle between two lines is defined as the smallest of these angles or the acute angle denoted by θ. In analytic geometry, if the coordinates of three points A, B, and C are given, then the angle between the lines AB and BC can be calculated as follows: For a line whose endpoints are (x1, y1) and (x2, y2), the slope of the line is given by the equation, The angle between the two lines can be found by calculating the slope of each line and then using them in the formula to determine the angle between two lines when the slope of each line is known from the equation. Privacy Policy | Terms & Conditions | Let ax2 + 2hxy + by2 = 0 represent the lines y = m1x (i) and y = m2x (ii), Lines perpendicular to the lines (i) and (ii) are y = –1/m1 x and y = –1/m2 x respectively and passing through origin. Generally speaking, the angle between these two lines is assumed to be acute and hence, the value of tan θ is taken to be positive. Find the angle between the lines represented by the equation 2x2 – 7xy + 3y2 = 0. (ii) bisector of the acute angle between them. Therefore, as on the plane, the cosine of the angle $$\alpha$$ will coincide (except maybe the sign) with the angle formed by the governing vectors of the straight line. The line x + 3y – 2 = 0 bisect the angle between a pair of straight lines of which one has equation x – 7y + 5 = 0. 1) Angles formed between two intersecting lines 1.1) Vertically Opposite Angles. where, When two lines intersect each other, then 4 angles are formed. School Tie-up | If one is real, other is also real. If the two lines are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the formula becomes tan θ = |(a1b2 - b1a2)/(a1a2 + b1b2)|. One is an acute angle and another is an obtuse angle or equal. using askIItians. Required fields are marked *, Coordinate Geometry Formulas For Class 11. Media Coverage | To read more, Buy study materials of Straight Lines comprising study notes, revision notes, video lectures, previous year solved questions etc. Two straight lines in a plane would either be parallel or coincide or intersect. Find the equation of the bisector of the angle containing the origin. Therefore, the given lines are (2x – y) = 0 and (x-3y) = 0. askiitians. [It is obtained by replacing x by x – x1 and y by y – y1 in the equation. If θ is the angle between two intersecting lines defined by y 1 = m 1 x 1 +c 1 and y 2 = m 2 x 2 +c 2, then, the angle θ is given by. In the diagram above, the line L1 and line L2 intersect at a point. Click here to refer the most Useful Books of Mathematics. Register with BYJU’S – The Learning App today. If θ is the angle between two intersecting lines defined by y1= m1x1+c1 and y2= m2x2+c2, then, the angle θ is given by. It often fetches some direct questions in various competitions like the IIT JEE. Approach: Find the equation of lines AB and BC with the given coordinates in terms of direction ratios as:. Contact Us | m1 = –h + √(h2–ab)/2 and m2 = –h – √(h2–ab)/2. If m1, m2 and m3 are the slopes of three lines L1 = 0, L2 = 0 and L3 = 0, where m1 > m2 > m3 then the interior angles of the triangle ABC formed by these lines are given by. The equations of two straight lines which are parallel to x + 7y + 2 = 0 and at unit distance from the point (1, – 1) are. 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angle between two pair of straight lines 2021
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# Integer Translations, Paths Between Points
## Understand how to move points from one location to another on a coordinate grid.
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### [Figure1] License: CC BY-NC 3.0
Reggie is taking art and has been given a picture of two people sitting in a room with a cat on the floor. His task is to redraw the picture but with the cat by the woman's feet, next to the sewing basket, instead of where it currently is. How can he figure out how to reorganize this image?
In this concept, you will learn how to translate points from one place to another on a coordinate plane.
### Translating Points on a Coordinate Plane
coordinate grid is a grid in which points are graphed. It usually has two or more intersecting lines which divide a plane into quadrants, and in which ordered pairs, or coordinates, are defined. It usually has four quadrants, or sections, to it.
The origin is the place where the two lines intersect. Its coordinates are defined as (0,0).
The x-axis is the line running from left to right that has the numbers defined on it and is usually labeled with an "x". The x-coordinate of an ordered pair is found with relation to it. All the points located on the x-axis have a y-coordinate of 0.
The y-axis is the central line that runs up-down and is labeled with a "y". Y-coordinates are plotted in reference to this axis. Again, all the x-coordinates of points located on the y-axis are 0.
An ordered pair is a list of two numbers in parenthesis, separated by a comma like this: (5,-3). It tells where a point is located on the coordinate plane. The first number is the x-coordinate. It tells you where to go on the x-axis. If it is positive, you go to the right. If it is negative, you go to the left. The second number is the y-coordinate. It tells you where to go on the y-axis. If it is positive, you go up. If it is negative, you go down.
The vertex of a shape is the place where two sides of the shape come together. In general, when a shape is defined inside of a coordinate plane, it is defined by the vertices, and then the lines are drawn to connect them.
A translation is when a figure or a point is slid on a coordinate grid from one place or another without changing its overall orientation.
In the figure above, the original point A\begin{align*}A\end{align*} is located at (-2, 3). Then it is translated to (4,1). A\begin{align*}A'\end{align*} is the new point. What was the overall translation?
First, figure out how many units A\begin{align*}A\end{align*} moved on the x\begin{align*}x\end{align*} axis
To find out, count the difference between the original point and the final point
In this case, it is +6 units.
Next, find how many units A\begin{align*}A\end{align*} moved on the y\begin{align*}y\end{align*} axis.
If you count, you can see that it moved -2 units (remember that negative 'y' means down).
Then, put them together.
The translation is (6, -2).
### Examples
#### Example 1
Earlier, you were given a problem about Reggie and his painted cat.
First, he lightly sketches a coordinate grid over the picture.
Next, he determines that the cat basically fits in a rectangle and is made up of a circle and an oval. He lightly sketches a box around the original cat on the coordinate plane.
Then, he finds the vertices of the original cat rectangle. They seem to be about (-11, -12), (-5, -12), (-11, -15), (-5, -15).
Then, he decides he wants the cat to go up 1 and over 9.
Then, he adds 9 to the x-coordinate of all of the vertices and adds 1 to the y-coordinates. This yields a new box with vertices (-2, -11), (4, -11), (-2, -14), (4, -14).
Finally, he draws this new box, and he sketches the cat shapes within it until it looks like a cat again. He erases the boxes, and his new cat is happy in its new place.
#### Example 2
Calculate the following transformation numerically, and use a graph to check your answer.
A point is plotted at (2,1)\begin{align*}(-2,1)\end{align*} and then translated (3,4)\begin{align*}(3,-4)\end{align*}, what are the coordinates of the translated point?
First, add the x-coordinate of the original point to the x-coordinate of the transformation.
-2 + 3 = 1
Next, add the y-coordinate of the original point to the y-coordinate of the transformation.
1 + (-4) = -3.
Then, write the resultant ordered pair.
Finally, check your answer by graphing the original point and physically carrying out the transformation.
#### Example 3
If the original point is located at (3,5) and is moved three units across and four units up, what are the coordinates of the final point?
First, figure out the x translation.
"Across" is in the x-direction. Because it isn't otherwise stated, you can assume it is to the right, which is positive.
Next, add the x-translation to the initial x-coordinate to find the new x-coordinate.
3 + 3 = 6
Then, figure out the y translation.
It is four units up.
Then, add that to the initial y-coordinate.
4 + 5 = 9
Finally, write the resultant coordinate.
The answer is (6,9)\begin{align*}(6,9)\end{align*}.
#### Example 4
If the initial point is (-5,4) and the final point is (2,7) write the translation.
First, find the x-translation.
-5 + ___ = 2 . In order to get from -5 to 2, the point must move 7 units to the right.
Next, find the y-translation.
4 + ___ = 7 . In order to get from 4 to 7, the point must move up 3 units.
Then, write the translation in words.
The answer is 7 units to the right and 3 units up.
#### Example 5
If a certain point is translated 2 units to the left and 4 units down, it is at the final point (-2, 6). What was the location of the original point?
First, find the original x-coordinate by working backwards.
2 units to the left means it is translated -2 on the x-axis. ___ + -2 = -2 The only coordinate that makes this work is 0.
Next, find the original y-coordinate the same way.
4 units down means it is translated -4 on the y-axis. ___ + (-4) = 6 The original y-coordinate must be 10 to make this work.
Then, write the original coordinate as an ordered pair.
Finally, draw a graph with both the beginning and final point to check your work.
### Review
Write each translation in coordinate notation.
1. A\begin{align*}A\end{align*} to A\begin{align*}A'\end{align*}
2. B\begin{align*}B\end{align*} to B\begin{align*}B'\end{align*}
3. C\begin{align*}C\end{align*} to C\begin{align*}C'\end{align*}
4. D\begin{align*}D\end{align*} to D\begin{align*}D'\end{align*}
5. E\begin{align*}E\end{align*} to E\begin{align*}E'\end{align*}
6. F\begin{align*}F\end{align*} to F\begin{align*}F'\end{align*}
7. G\begin{align*}G\end{align*} to G\begin{align*}G'\end{align*}
8. H\begin{align*}H\end{align*} to H\begin{align*}H'\end{align*}
9. H\begin{align*}H\end{align*} to E\begin{align*}E\end{align*}
10. E\begin{align*}E\end{align*} to D\begin{align*}D\end{align*}
11. E\begin{align*}E'\end{align*} to G\begin{align*}G\end{align*}
12. E\begin{align*}E'\end{align*} to C\begin{align*}C\end{align*}
13. B\begin{align*}B\end{align*} to A\begin{align*}A'\end{align*}
14. B\begin{align*}B\end{align*} to G\begin{align*}G'\end{align*}
15. C\begin{align*}C'\end{align*} to D\begin{align*}D'\end{align*}
To see the Review answers, open this PDF file and look for section 11.17.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
$x-$axis
The $x-$axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
$y$ axis
The $y$-axis is the vertical number line of the Cartesian plane.
Coordinates
The coordinates of a point represent the point's location on the Cartesian plane. Coordinates are written in ordered pairs: $(x, y)$.
Latitude
Latitude is a coordinate that specifies the north-south location of a point on the Earth's surface.
Longitude
Longitude is a coordinate that specifies the east-west location of a point on the Earth's surface.
Ordered Pair
An ordered pair, $(x, y)$, describes the location of a point on a coordinate grid.
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
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# If a and b are unit vectors and $\left| a+b \right|=1,$then $\left| a-b \right|$ is equal to:(a) $\sqrt{2}$ (b) $\sqrt{1}$ (c)$\sqrt{5}$(d) $\sqrt{3}$
Last updated date: 13th Jul 2024
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Hint: We can use the formula for modulus of vectors as $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$ to solve this question. Also the modulus of a unit vector is 1.
Before proceeding to the solution we should know that the magnitude is also called the modulus or the length of the vector. Magnitude is represented by the length of the directed line segment. A unit vector is a vector of length 1. To obtain a unit vector in the direction of any vector we divide by its modulus. A unit vector is a vector of unit length, sometimes also called a direction vector. The unit vector is defined by $\widehat{v}$.
$\widehat{v}=\dfrac{v}{\left| v \right|}$,
Given, $\left| a+b \right|=1$
Using the formula $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ we get:
Since, a and b are unit vectors therefore, $\left| a \right|=1,\left| b \right|=1$ and $\cos \theta$ is the angle between the vectors a and b.
So, we can substitute the values in the above equation and we will get,
$\therefore 1=\sqrt{{{1}^{2}}+{{1}^{2}}+2\times 1\times 1\times \cos \theta }$
$\Rightarrow 1=\sqrt{2+2\cos \theta }$
Squaring both sides we get,
$\Rightarrow 2+2\cos \theta =1$
$\Rightarrow 2\cos \theta =-1$
$\Rightarrow \cos \theta =-\dfrac{1}{2}......(i)$
Now, we have to find out the value of $\left| a-b \right|$ therefore, we can use the formula for $\left| a-b \right|$.
Since, we know the values, we can substitute them in the equation and we will get,
$\therefore \left| a-b \right|=\sqrt{1+1-2\times 1\times 1\times \cos \theta }$
$\Rightarrow \left| a-b \right|=\sqrt{2-2\cos \theta }$
Then, we can substitute the value of$\cos \theta$ from equation (i), and we will get,
$\Rightarrow \left| a-b \right|=\sqrt{2-2\times -\dfrac{1}{2}}$
$\therefore \left| a-b \right|=\sqrt{3}$
Hence, the answer is option (d).
Note: It is important to use the cos theta term in the modulus equation. The dot product of two vectors along with the angle between them by cos theta should be used. If not, it would not be easy to solve the question. Be very carefully about the value of modulus, modulus can never be negative. Always remember the formula of modulus like:
$\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$, and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$
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# Trigonometry: Angles
## Trigonometry: Angles
Trigonometry is a branch of mathematics dealing with the measurement of sides and angles of a triangle. We apply trigonometry in Engineering, Surveying, Astronomy, Geology etc.
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### Angles:
Let O be the fixed point on OX, the initial line. Let OY be the revolving line. Then the amount of rotation of OY about O with respect to OX is known as the angle between OX and OY. Here the angle formed is XOY.
If the revolving line rotates about the fixed point O in the anticlockwise direction, the angle so formed is said to be positive.
If the revolving line rotates about the fixed point O in the clockwise direction, the angle formed is said to be negative.
#### Measurement of Angles
A line making one complete rotaion makes 360°. When a line makes a quarter tern, it makes 90° or 1 right angle. The size of a right angle is same in every measurement. The following three system are commonly used in the measurement of angles:
(a) Sexagesimal System (Degree System)
(a) Sexagesimal System: This system is also called British System. In this system, the unit of measurement is degree. So, this system also is known as the degree system. In this system, a right angle is divided into 90 equal parts and each part is called a degree. A degree is divided into 60 equal parts and each part is called as one minute. A minute is also divided into 60 equal parts and each part is called as one second. Therefore, we have
60 seconds = 1 minute (60’’ = 1’)
60 minutes = 1 degree (60’ = 1°)
90 degrees = 1 right angle
The degree, minute and second are denoted by (°), (’) and (’’) respectively.
(b) Centesimal System: This system is also called the French System. In this system, the unit of measurement is grade. So, this system also is known as the grade system. In this system, a right angle is divided into 100 equal parts and each part is called a grade. A grade is divided into 100 equal parts and each part is called a minute. A minute is also divided into 100 equal parts and each part is called a second. Therefore, we have
100 seconds = 1 minute (100’’ = 1’)
100 minutes = 1 grade (100’ = 1g)
100 grades = 1 right angles
The grade, minute and second are denoted by (g), (‘) and (‘’) respectively.
(c) Circular System: In this system, the unit of measurement of an angle is a radian. An angle at the centre of a circle subtended by an arc equal to the length of radius of the circle is known as 1 radian. It is denoted by (c).
As the total length of circumference of a circle is 2Ï€r units, the angle subtended by circumference of a circle at the centre is 2Ï€r/r radian i.e. 2Ï€c.
Which is, 4 right angle = 2Ï€c
or, 1 right angle = (Ï€/2)c
Now, from the definition of sexagesimal measure, centicimal measure and circular measure of angles, we have,
1 right angle = 90° = 100g = (Ï€/2)c
#### Theorem: “Radian is a constant angle.”
Proof:-
Let, O be the centre of the circle and OP = r be the radius of the circle. An arc PQ = r is taken. PO and QO are joined. Produce PO to meet circle at R. Then by definition POQ = 1 radian. The diameter PR = 2r, POR = 2 right angles = 180° and arc PQR = ½ × circumference = ½ × 2Ï€r = Ï€r.
Now, since the angles at the centre of a circle are proportional to the corresponding arcs on which the stand.
i.e. POQ/POR = arc PQ/arc PQR
Since 1 radian is independent of the radius of the circle, it is a constant angle.
Proved.
#### Relation between different system of measurement of angles:
Since, 1 right angle = 90° and 1 right angle = 100g
90° = 100g
1° = (10/9)g
Also, 1g = (9/10)°
Again, Ï€c = 180° = 200g
1c = (180/Ï€ and 1c = (200/Ï€)g
Also, 1° = (Ï€/180)c and 1g = (Ï€/200)c
### Workout Examples
Example 1: Reduce 24g 20’ 44’’ into centicimal seconds.
Solution:
The given angle is 24g 20’ 44’’
= 24 × 100 × 100’’ + 20 × 100’’ + 44’’
= 240000’’ + 2000’’ + 44’’
= 242044’’
Example 2: Express 42° 20’ 15’’ into the number of degrees.
Solution:
The given angle is 42° 20’ 15’’
= 42° + (20/60)° + (15/60×60)°
= 42° + 0.333333° + 0.004166°
= 42.337499°
Example 3: Express 48g 54’ 68’’ into degrees, minutes and seconds.
Solution:
The given angle is 48g 54’ 68’’
= (48 + 54/100 + 68/100×100)g
= (48 + 0.54 + 0.0068)g
= 48.5468g
= (48.5468 × 9/10)° [ 1g = (9/10)°]
= 43.69212°
= 43° + 0.69212°
= 43° + (0.69212 × 60)’
= 43° + 41.5272’
= 43° + 41’ + 0.5272’
= 43° + 41’ + (0.5272 × 60)’’
= 43° + 41’ + 31.632’’
= 43° 41’ 31.63’’
Hence, 48g 54’ 68’’ = 43° 41’ 31.63’’
Example 4: Express π/6 radians into sexagesimal and centicimal measures.
Solution:
The given angle is π/6 radians
= (Ï€/6 × 180/Ï€ [ 1c = (180/Ï€)°]
= 30°
Again,
= (Ï€/6 × 200/Ï€)g [ 1c = (200/Ï€)g]
= 33.33g
Example 5: Reduce the following angles into radian measure.
a. 42g75’
b. 42° 15’ 30’’
Solution:
a. 42g 75’
= (42 + 75/100)g
= (42 + 0.75)g
= 42.75g
= (42.75 × Ï€/200)c [ 1g = (Ï€/200)c]
= 0.21375Ï€c
b. 42° 15’ 30’’
= (42 + 15/60 + 30/60×60)°
= (42 + 0.25 + 0.00833)°
= 42.25833°
= (42.25833 × Ï€/180)c [ 1° = (Ï€/180)c]
= 0.2348Ï€c
Example 6: The angles of a triangle are (7x/2)g, (9x/4)° and (Ï€x/50)c. Find the angles of the triangle in degrees.
Solution: Here,
First angle = (7x/2)g = (7x/2 × 9/10)° = (63x/20)°
Second angle = (9x/4)°
Third angle = (Ï€x/50)c = (Ï€x/50 × 180/Ï€)° = (18x/5)°
Now, the sum of all angles of a triangle is 2 right angles.
i.e. (63x/20)° + (9x/4)° + (18x/5)° = 180°
or, {(63x + 45x + 72x)/20}° = 180°
or, (180x/20)° = 180°
or, (9x)° = 180°
or, x° = 20°
Hence, the angles of the triangle are (63×20/20)°, (9×20/4)° and (18×20/5)° i.e 63°, 45° and 72°.
Example 7: Sum of the first and the second angle of a triangle is 150°. The ratio of the number of grades in the first angle to the number of degrees in the second angle is 5:3. Find the angles of the triangle in circular measure.
Solution: Suppose number of grades in the first angle be 5x and the number of degrees in the second angle be 3x.
First angle = 5xg = (5x × 9/10)° = (9x/2)°
Second angle = 3x°
From question,
9x/2 + 3x = 150°
or, 15x/2 = 150°
or, 15x = 300°
or, x = 20°
The first angle = 9x/2 = 9×20/2 = 90°
The second angle = 3x = 3 × 20 = 60°
The third angle = 180° - 150° = 30°
The angles of the triangle in circular measure are,
(90 × Ï€/180)c, (60 × Ï€/180)c and (30 × Ï€/180)c
i.e. (Ï€/2)c, (Ï€/3)c and (Ï€/6)c
You can comment your questions or problems regarding the system of measurement of angles here.
|
# The Formula for the Inverse Matrix of $I+A$ for a $2\times 2$ Singular Matrix $A$
## Problem 505
Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula:
$(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.$
Using the formula, calculate the inverse matrix of $\begin{bmatrix} 2 & 1\\ 1& 2 \end{bmatrix}$.
## Proof.
We have
\begin{align*}
(I+A)\left(\, I-\frac{1}{1+\tr(A)}A \,\right)&=I-\frac{1}{1+\tr(A)}A+A-\frac{1}{1+\tr(A)}A^2\6pt] &=I-\frac{1}{1+\tr(A)}\left(\, A-(1+\tr(A))A +A^2\,\right)\\[6pt] &=I-\frac{1}{1+\tr(A)}\left(\, A^2-\tr(A)A \,\right) \tag{*}. \end{align*} The Cayley-Hamilton theorem for 2\times 2 matrices yields that \[A^2-\tr(A)A+\det(A)I=O. Since $A$ is singular, we have $\det(A)=0$.
Hence it follows that we have
$A^2-\tr(A)A=O,$ and we obtain from (*) that
$(I+A)\left(\, I-\frac{1}{1+\tr(A)}A \,\right)=I.$ Similarly,
$\left(\, I-\frac{1}{1+\tr(A)}A \,\right)(I+A)=I.$
Therefore, we conclude that the inverse matrix of $I+A$ is given by the formula
$(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.$
### Find the inverse matrix of $\begin{bmatrix} 2 & 1\\ 1& 2 \end{bmatrix}$ using the formula
Now let us find the inverse matrix of $\begin{bmatrix} 2 & 1\\ 1& 2 \end{bmatrix}$ using the formula.
We first write
$\begin{bmatrix} 2 & 1\\ 1& 2 \end{bmatrix}=I+A,$ where
$A=\begin{bmatrix} 1 & 1\\ 1& 1 \end{bmatrix}.$ Then $A$ is a singular matrix with $\tr(A)=2$.
The formula yields that
\begin{align*}
\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}^{-1}&=(I+A)^{-1}\6pt] &=I-\frac{1}{3}A\\[6pt] &=\frac{1}{3}\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}. \end{align*} ## Related Question. There is a similar formula for inverse matrices of certain n\times n matrices, called Sherman-Woodberry formula. See the post ↴ Sherman-Woodbery Formula for the Inverse Matrix for the statement of the Sherman-Woodberry formula and its proof. Sponsored Links ### More from my site • If 2 by 2 Matrices Satisfy A=AB-BA, then A^2 is Zero Matrix Let A, B be complex 2\times 2 matrices satisfying the relation \[A=AB-BA. Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix. Hint. Find the trace of $A$. Use the Cayley-Hamilton theorem Proof. We first calculate the […]
• True or False: If $A, B$ are 2 by 2 Matrices such that $(AB)^2=O$, then $(BA)^2=O$ Let $A$ and $B$ be $2\times 2$ matrices such that $(AB)^2=O$, where $O$ is the $2\times 2$ zero matrix. Determine whether $(BA)^2$ must be $O$ as well. If so, prove it. If not, give a counter example. Proof. It is true that the matrix $(BA)^2$ must be the zero […]
• An Example of a Matrix that Cannot Be a Commutator Let $I$ be the $2\times 2$ identity matrix. Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$. Proof. Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies $ABA^{-1}=-B. […] • Sherman-Woodbery Formula for the Inverse Matrix Let \mathbf{u} and \mathbf{v} be vectors in \R^n, and let I be the n \times n identity matrix. Suppose that the inner product of \mathbf{u} and \mathbf{v} satisfies \[\mathbf{v}^{\trans}\mathbf{u}\neq -1.$ Define the matrix […]
• If Two Matrices are Similar, then their Determinants are the Same Prove that if $A$ and $B$ are similar matrices, then their determinants are the same. Proof. Suppose that $A$ and $B$ are similar. Then there exists a nonsingular matrix $S$ such that $S^{-1}AS=B$ by definition. Then we […]
• Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant Let $A=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$ be an $2\times 2$ matrix. Express the eigenvalues of $A$ in terms of the trace and the determinant of $A$. Solution. Recall the definitions of the trace and determinant of $A$: \[\tr(A)=a+d \text{ and } […]
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# Non Terminating Decimals
A non terminating decimal is one of the type of decimals which are divided into two types. They are,
• Terminating Decimals
• Non terminating Decimals
Non terminating Decimals is the continuous decimal number. It has been going on till the value infinitive. It is the very big value sometimes these non terminating decimals are rounded to its nearest values. But the Terminating decimal number is the fixed value of the decimal value.
Related Calculators Decimal Calculator Decimal Greater than less than Decimal to Fraction Adding Decimals Calculator
## Calculating Non Terminating Decimals
Below you can see the steps of calculating non terminating decimals:
Divide $\frac{5}{3}$
Step 1: This is the type of an improper fraction. because here numerator takes the large value than the denominator.
Step 2: One time 3 is 3. So the remainder is 2 and the quotient value is 1
Step 3: Now 2 is not divided by 3
Step 4: So we can add zero to the remainder 2 and put a point the quotient value 1
Step 5: Now the value is 20. Again we can do the same procedure
Step 6: Now 6 times 3 is 18 again the remainder is 2 and the quotient value is 6
Step 7: Again we can add zero to 2 then the value is 20
Step 8: Again 6 times 3 is 18. Again the remainder is 2 and the quotient value is 6
Step 9: Now the decimal value is 1.66 ……. its going on
Step 10: This decimal part 1.66……is known as the non terminating decimal.
Step 11: It can be rounded as 1.67
Step 12: Because it’s decimal part value is greater than or equal to 5. So the value of the last decimal value is consider as one and that value is added to the previous decimal value.
Step 13: Otherwise the last decimal part value is less than five. So the value of the last decimal value is simply leave. And the previous decimal value is considered as the last decimal digit value.
Pictorial Representation of this problem is shown below:
## Non Terminating Decimals Practice Problems
Question 1: $\frac{4}{2}$
Question 2: $\frac{2}{3}$
Question 3: $\frac{6}{3}$
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# Solving an equality with 3 equations, and 3 variables
Following is the question asked in a recent aptitude exam:
Given that : $$a+b+ab=10\\ b+c+bc=20 \\ c+a+ac=30$$ What is the value of $a+b+c+abc$ ?
I can solve it by finding the individual values by: $$a = \frac{10-b}{b+1} ,\\ c = \frac{20-b}{b+1}$$ Putting these in the third equation: $$\frac{10-b}{b+1}+\frac{20-b}{b+1} + \frac{(10-b)\times(20-b)}{(b+1)^2}=30\\ (30-2b)\times(b+1)+(b-20)\times(b-10)=30\times(b+1)\\{30\times(b+1)}-2b(b+1)+b^2-30b+200={30\times(b+1)}\\-2b^2-2b+b^2-30b+200=0\\-b^2-32b+200=0\\b^2+32b-200=0$$ Which can be solved to: $$b= \frac{-32+-\sqrt{32^2-4\times(-200)}}{2} \approx \frac{-32+-42.71}{2} = 5.35 \text{ or} -37.35$$
And we can get the values for $a$ and $c$ as well: $$a \approx 9.157 \text{ or} -1.297\\ c\approx 2.30 \text{ or} -1.577$$ Here, the for the first two equations, $b=-37.35$, $a=-1.297$ and $c=-1.577$ works. Whereas, the third equation is satisfied by $a=9.157$ and $c=2.30$. This doesn't seem correct. Since only a single value of $b$ works for the first two equations, therefore the other value of $b$ is rejected. This also rejects the derived values of $a$ and $c$. So, the positive values of $a$ and $c$ should not be used. But the former values don't satisfy the third equation, but the latter values do. What is the problem here?
Moreover, is there any shorter way of solving this problem?
• You have an error in the second line after "Putting these in the third equation". The right-hand side should be $30 (b + 1)^2$, not $30 (b+1)$. Sep 3, 2018 at 14:41
Add one to each equation and factorize to get : $$(a+1)(b+1) = 11 \\ (b+1)(c+1) = 21 \\ (a+1)(c+1) = 31 \\$$
Now, set $$x ,y,z = a+1,b+1,c+1$$ respectively.This gives $$xy = 11, yz = 21,zx = 31$$, so $$x^2y^2z^2 = 11 \times 21\times 31$$ by multiplying these.
Now divide this equation by the other equations suitably to get the values of $$x,y,z$$, and use the fact that $$a+b+c + abc = x+y+z - 3 + (x-1)(y-1)(z-1)$$ to get the answer.
|
###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
##### Thank you for watching the video.
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# Finding the Slope of a Line from an Equation - Problem 2
Alissa Fong
###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Share
To find the slope of a line given the equation of the line, first write it in slope-intercept form. Use inverse operations to solve for y so that it is written as y=mx+b. Then you can easily see the slope since it is the coefficient of the x variable, or the number in front of x. Remember to include the right sign of the slope.
When you’re asked to find the slope of the line for a given equation, it can be really easy if the line is already in y equals mx plus b form. If you remember when y is all by itself and we have an equation solves for y, the slope is just the coefficient for x, so if I can take this line and solve for y, so that y is isolated, then my slope numbers are going to jump out on me. It’s going to be the co-efficient of x. So let’s do it.
If I want to solve for y, that means get y all by itself, I’m going to subtract x from both sides. I’m going to rewrite this problem down here so I have more space. Okay subtract x from both sides, so that I have -2y equals –x plus 6 and I’m still not quite there yet because y is being multiplied by -2, the opposite of multiplying is to divide.
So I’m going to divide everything by -2, and now I’ll have y= equals -x/-2 becomes +x/2, or you can think of that as 1/2 plus let’s see 6 divide by -2 is -3. Okay so this is the exact same equation just rewritten, I just moved stuff around. I haven’t changed any values, but now I can see that the slope number is whatever the co-efficient of x is, and this one’s kind of tricky because x/2, does that mean the slope is 2? Not quite, the slope is going to be 1/2; I’m going to try to put a little box around it. The slope is the entire co-efficient, there’s a secret x, excuse me a secret 1 in front of x there, so your slope number is not just x/2. It’s not 2, your slope number is 1/2, this is you final answer, slope equals 1/2.
Again the way you can do these problems if you’re given an equation and asked to find the slope, is to get your equation so that it’s solved for y. Put it into y equals mx plus b form, and then the slope number is just the co-efficient in front of x.
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Facts that Matter: Statistics
# Facts that Matter: Statistics | Mathematics (Maths) Class 10 PDF Download
Facts that Matter: Let us remember that Mean, Median and Mode are measures of central tendency i.e., numerical representatives of the given data. Let us also note the following points.
2. Mean of grouped data:
(i) Using direct method:
(ii) Using the assumed mean method:
(iii) Using the step deviation method:
3. The mode for grouped data is found by using the formula:
where, l = lower limit of the modal class.
h = size of the class interval.
f1 = frequency of the modal class.
f0 = frequency of the class preceding the modal class.
f2 = frequency of the class succeeding the modal class.
4. Cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
5. Median of the given grouped data is obtained using the formula:
where, l = lower limit of median class.
n = number of observations.
cf = cumulative frequency of class preceding the median class
f = frequency of median class
h = class size
The document Facts that Matter: Statistics | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
## FAQs on Facts that Matter: Statistics - Mathematics (Maths) Class 10
1. What is the importance of studying statistics in Class 10?
Ans. Studying statistics in Class 10 is important because it helps students develop analytical and critical thinking skills, enabling them to make informed decisions based on data. It also prepares them for higher-level mathematics and helps them understand and interpret data in various real-life situations.
2. How can statistics be applied in everyday life?
Ans. Statistics can be applied in everyday life in multiple ways. For example, it can be used to analyze survey data, determine probabilities in games or sports, make informed financial decisions, interpret medical research findings, or even understand trends and patterns in social media usage.
3. What are the basic concepts covered in Class 10 statistics?
Ans. In Class 10 statistics, students typically learn about concepts such as mean, median, mode, range, standard deviation, probability, and graphical representation of data. They also explore different types of data, such as discrete and continuous data, and understand how to gather and organize data effectively.
4. How can statistics help in making predictions?
Ans. Statistics provides tools and techniques to analyze and interpret data, which can be used to make predictions. By analyzing past data and identifying patterns or trends, statistics allows us to make reasonable predictions about future outcomes. For example, weather forecasts, stock market predictions, or election outcome projections are all based on statistical analysis.
5. How can statistics help in decision-making?
Ans. Statistics plays a crucial role in decision-making by providing a systematic approach to analyzing and interpreting data. It helps in evaluating different options, assessing risks, and making informed choices based on evidence rather than intuition or guesswork. Whether it's a business decision, policy-making, or personal choices, statistics provides a framework for making rational decisions.
## Mathematics (Maths) Class 10
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# Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1
## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1
Question 1.
Solution:
Similarly (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
∴ (a – b)4 + (a + b)4 = 2 [a4 + 6a2b2 + b4]
Substituting the value of a and b we get
= 2[16x8 + 6(4x4)(9(1 – x2)) + 81(1 – x2)2]
= 2[16x8 + 216x4(1 – x2) + 81(1 – x2)2]
= 2[16x8 + 216x4 – 216x6 + 81 + 81x4 – 162x2]
= 2[16x8 – 216x6 + 297x4 – 162x2 + 81]
= 32x8 – 432x6 + 594x4 – 324x2 + 162
Question 2.
Compute
(i) 1024
(ii) 994
(iii) 97
Solution:
(i) 1024 = (100 + 2)4 = (102 + 2)4
= 1(108) + 4(106)(2) + 6(104)(4) + 4(102)(8) + 16
= 100000000 + 8000000 + 240000 + 3200 + 16
= 108243216
(ii) 994 = (100 – 1)4 = (102 – 1)4
= 1(108) + 4(106)(-1) + 6 (104)(1) + 4( 104)(-1) + (-1)4
= 100000000 – 4000000 + 60000 – 400 + 1
= 100060001 – 4000400 = 96059601
(iii) 97 = (10 – 1)7
= 1(10000000) + 7(1000000)(-1) + 21(100000)(1) + 35(10000)(-1) + 35(1000)(1) + 21(100)(-1) + 7(10)(1) + 1(-1)
= 10000000 – 7000000 + 2100000 – 350000 + 35000 – 2100 + 70 – 1
= 12135070 – 7352101 = 4782969
Question 3.
Using binomial theorem, indicate which of the following two number is larger: (1.01)1000000, 10000.
Solution:
(1.01)1000000 = (1 + 0.01)1000000
which is > 10000
So (1.01)1000000 > 10000 (i.e.) (1.01)1000000 is larger
Question 4.
Solution:
To find coefficient of x15 we have to equate x power to 15
i.e. 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 5/5 = 1
So the coefficient of x15 is 10C1 = 10
Question 5.
Solution:
To find coefficient of x6
12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = 6/5 which is not an integer.
∴ There is no term involving x6.
To find coefficient of x2
12 – 5r = 2
5r = 12 – 2 = 10 ⇒ r = 2
Question 6.
Solution:
when multiplying these terms, we get x4 terms
∴ The co-eff of x4 is 26325
Question 7.
Solution:
Question 8.
Find the last two digits of the number 3600.
Solution:
3600 = 32 × 300 = (9)300 = (10 – 1)300
All the terms except last term are ÷ by 100. So the last two digits will be 01.
Question 9.
If n is a positive integer, show that, 9n + 1 – 8n – 9 is always divisible by 64.
Solution:
∴ 9n + 1 – 8n – 9 = 64 [an integer]
⇒ 9n + 1 – 8n – 9 is divisible by 64
Question 10.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal.
Solution:
Given n is odd. So let n = 2n + 1, where n is an integer.
The expansion (x + y)n has n + 1 terms.
= 2n + 1 + 1 = 2(n + 1) terms which is an even number.
⇒ The coefficient of the middle terms in (x + y)n are equal.
Question 11.
If n is a positive integer and r is a non – negative integer, prove that the coefficients of xr and xn – r in the expansion of (1 + x)n are equal.
Solution:
Question 12.
If a and b are distinct Integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint: write an = (a – b + b)n and expand]
Solution:
= (a – b)[an integer]
⇒ an – bn is divisible by (a – b)
Question 13.
In the binomial expansion of (a + b)an, the coefficients of the 4th and 13th terms are equal to each other, find n.
Solution:
In (a + b)n general term is tr + 1 = nCr an – r br
So, t4 = t3 + 1 = nC3 = nC12
⇒ n = 12 + 3 = 15
We are given that their coefficients are equal ⇒ nC3 = nC12 ⇒ n = 12 + 3 = 15
[nCx = nCy ⇒ x = y (or) x + y = n]
Question 14.
If the binomial coefficients of three consecutive terms in the expansion of (a + x)n are in the ratio 1 : 7 : 42, then find n.
Solution:
In (a + x)n general term is tr + 1 = nCr
So, the coefficient of tr + 1 is nCr
We are given that the coefficients of three consecutive terms are in the ratio 1 : 7 : 42.
Question 15.
In the binomial coefficients of (1 + x)n, the coefficients of the 5th, 6th and 7 terms are in AP. Find all values of n.
Solution:
⇒ (n – 1)(n – 14) = 0
∴ n = 7 , 14
Question 16.
Solution:
### Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 Additional Questions Solved
Question 1.
Solution:
Let Tr + 1 be the term in which x32 and x-17 occurs,
(i) Since x32 occurs in this term
∴ Exponent of x = 32
⇒ 60 – 7r = 32 ⇒ 7r = 28
r = 28 ÷ 7 = 4
∴ Coefficient of the term containing x32 = 15C4 (-1)4 = 1365
(ii) Since x-17 occurs in this term .
∴ Exponent of x = -17
⇒ 60 – 7r = -17 ⇒ 7r = 77, ∴ r = 11
Question 2.
Find a positive value of m for which the coefficient of x2 in the expansion of (1 + x)m is 6.
Solution:
Also, coefficient of x2 in the expansion of (1 + x)m is 6
⇒ m (m – 1) = 4.3 ⇒ m = 4
Question 3.
In the binomial expansion of (1 + a)m + n, prove that the coefficients of am and an are equal.
Solution:
Question 14.
The coefficient of (r – 1)th, rth, and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find both n and r.
Solution:
We know that co-effcients of (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are nCr – 2: nCr – 1 and nCr respectively
⇒ 3n – 8r + 3 = 0 and n – 4r + 5 = 0
Solving these for n, r we get,
n = 7 and r = 3
Question 5.
Solution:
Using binomial theorem,
Question 6.
Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of the two middle terms in the expansion of (1 + x)2n – 1.
Solution:
In the expansion of (1 + x)2n,
Number of terms = 2n + 1, which is odd
Question 7.
If three consecutive coefficients in the expansion of (1 + x)n are in the ratio 6 : 33 : 110, find n.
Solution:
Let the consecutive coefficients nCr, nCr + 1 and nCr + 2 be the coefficients of Tr + 1, Tr + 2 and Tr + 3 then nCr : nCr + 1 : nCr + 2 = 6 : 33 : 110
Question 8.
If the sum of the coefficients in the expansion of (x + y)n is 4096. Then find the greatest coefficient in the expansion.
Solution:
Given that, Sum of the coefficients in the expansion of (x + y)n = 4096
nC0 + nC1 + nC2+…+ nCn = 4096
[∴ Sum of binomial coefficients in the expansion of (x + a)n is 2n]
⇒ 2n = 4096 = 212
Question 9.
Solution:
The general term in the expansion of
Question 10.
Solution:
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## Thursday, September 20, 2018
### Curvature, Arc Length and Parametric Representations of Curves
One of the more important areas of higher mathematics involves finding the arc length of a curve, as well as the curvature. A number of instructive examples are given in this post for doing so. A logical place to begin is with the arc length formula (derivations can be found in any good Calculus textbook. For example, an element ds (du) of arc on a curve is:
But in derivation terms - with dx and dy also shown as 'legs' of a right triangle:
Where we treat ds as the differential of arc length.
ds2 = dx2 + dy 2
Or: ds = Ö (dx2 + dy 2 ) = dx Ö [ 1 + (dy/dx) 2 ]
This can then be integrated (between appropriate limits, say x1 and x2) to give the total length of a curve.. Thence the arc length is given by:
where the function f(x) defines the curve for which the arc length is evaluated between the points x1 and x2. Consider then finding the arc length between x1 = 0 and x2 = 8 for the parabola section below:
For which we have: f(x) = [(x)]½
Then evaluating the integral as shown, one obtains: s = 8.732
A more difficult problem involves a polar curve, i.e. a curve in polar coordinates such as:
This curve has the function: r(q) = q - sin (q)
And we wish to find its length between q = -p and q = p .
The relevant integral for the arc length for this example is:
Performing the integration one then obtains: L = 8.764
Which interested and energized readers are invited to check!
Parametric Representations:
In advanced treatments of curvature including determining the curvature of a complex curve, parametric representations are the norm. This usually entails taking the tangent T to the curve at a point. From this the curvature k of the arc can also be computed. For example, one can give the parametric representation of a specific type of curve using:
x = 6 sin 2 t, y = 6 cos 2 t, z = 5t
For which the tangent to the curve at a point is given by:
T = dR/ ds = i(dx/ds) + j(dy/ds) + k(dz/ds)
=
i (dx/dt) (dt/ds) + (dy/dt)(dt/ds) + k (dz/dt) (dt/ds)
Then:
dx/dt= 12 cost 2t, dy/dt = -12 sin 2t, dz/dt = 5
Therefore, the tangent unit vector is:
| T | = 1 = (dt/ds)2 [(12 cost 2t)2 + (-12 sin 2t)2 + 52]
|T | = 1 = (dt/ds)2 [(144 + 25)] = (dt/ds)2 (169)
So that: (dt/ds) = 1/ 13
And: T = 1/ 13 [12 cost 2t) i + (-12 sin 2t) j + 5 k ] dt/ds
The curvature can then be obtained from: k = | dT/ ds |
Thus, any curve can be given by a parametric representation: u1 = u1(t) and u2= u2(t)
For such a curve, consider now the distance between two infinitely near points on the surface, i.e. the distance or interval ds between two specified points.
Example problem:
A curve is given in spherical coordinates xi by:
x1 = t, x= arcsin 1/t, x3 = (t2 – 1) 1/2
Compute the length of the arc between t = 1 and t = 2
Solution:
(ds/ dt)2 = (dx1/ dt) 2 + (x1 ) 2 (dx2/ dt) 2 + (x1 sin x 2) 2 (dx3/ dt) 2
And:
(dx1/ dt) 2 = 1
(dx2/ dt) 2 = [ -1/ t2 / Ö {1 – ( 1/ t)} = 1/ t(t- 1)
(dx3/ dt) 2 = 2t/ Ö2(t- 1) = t2 / (t- 1)
Whence:
(ds/ dt)2 =
1 + t · 1/ t(t- 1) + (t · 1/ t ) · t2 / (t- 1)
= 2 t2 / (t- 1)
Then the length of the curve is:
L = ò 1 2 Ö2 t/ (t- 1)1/2 dt = Ö2(t- 1) ] 1 Ö6
Problem for Math Mavens:
For the parametric example with:
T = 1/ 13 [12 cost 2t) i + (-12 sin 2t) j + 5 k ] dt/ds,
Find the curvature and the length of the curve from t = 0 to t = p
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ONLINE ELECTRICAL ENGINEERING STUDY SITE
Hexadecimal to Decimal and Decimal to Hexadecimal Conversion
The hexadecimal system is a number system with base 16. This number system is greatly used by modern computer system.We already know about the decimal number system, binary number system and octal number system. Like those there is another number system called hexadecimal number system. As the name suggests there are 16 symbols in this number system starting from 0. Before explaining the number system we should know why this number system came into existence. The natural tendency of human is to use decimal number system because they are familiar with this as the use of 0 is very easy and the operations are user friendly. And the computer systems used binary systems earlier because there are only two states on and off. But as the dependency on computer grew up and different mathematical programs and different softwares needed to be developed there came the need to develop a number system having base larger than decimal and 16 was chosen because the bits, bytes are multiples of it.
Now days this number system is used in HTML and CSS, hexadecimal notations are used in them. This number system was first used around 1956 in Bendix G-15 computer. Now coming to the representation of hexadecimal number system, in this number system there are 16 basic digits by which all the numbers can be represented, these are 0, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F the first 10 digits are similar to decimal number system but the last 6 digits represent 10, 11, 12, 13, 14 and 15 respectively. Any number in hexadecimal number system can be converted into numbers of other number system very easily, the procedures are given in the next article. Here in the decimal system, we use symbol 1 and 0 side by side that is 10 to represent • • • • • • • • • • + • That is nine plus one. After that we will have 11 then 12 and so on. That means after nine or 9 we bring back first non - zero digit of symbol that is 1 at left side and repeat all the symbols from 0 to 9 at its right side to represent next ten higher numbers from ten to nineteen (10 - 19). After 19 we put 2 at left and repeat again 0 to 9 to represent next ten higher numbers from twenty to twenty nine (20 - 29). Decimal number system is very basic number system as ten symbols or digits are used in different combinations to represent all the numbers, this system is said to be of base ten (10). Now think about a number system where you are told to use sixteen symbols instead of 10 symbols. Then what will be your basic construction of the new number system ? For that first we have to find out 16 symbols to represent the basic digits of that new number system. We can create new series of symbols for that, but if we do so it will be very much difficult to remember. That difficulty can be solved if we use commonly used symbols for that purpose. So we can simply use 0 to 9 of decimal system to represents first ten digits 0 to 9 of this new number system. But for other 6 higher digits there are no symbols available in decimal system so we have to search for them from some commonly used system. We can easily get them from our alphabetical system that means we can use A, B, C, D, E and F as next 6 higher digits (from 10 to 15) in this new number system. The system where total 16 basic digits are used is known as hexadecimal number system.
A ⇒ • • • • • • • • • • B ⇒ • • • • • • • • • • • C ⇒ • • • • • • • • • • • • D ⇒ • • • • • • • • • • • • • E ⇒ • • • • • • • • • • • • • • F ⇒ • • • • • • • • • • • • • • •
In hexadecimal system we use 16 symbols to represent all numbers. These symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. After F we use 10 for next higher number 16. Then next increment is 11 which used to represent next natural number 17 and so on. Hence in hexadecimal system just after F, the first digit becomes 1 and second digit will repeat from 0 to F one by one to represent natural numbers 16 to 31. That means, 10 ⇒ 16, 11 ⇒ 17, 12 ⇒ 18, 13 ⇒ 19, 14 ⇒ 20, 15 ⇒ 21, 16 ⇒ 22, 17 ⇒ 23, 18 ⇒ 24, 19 ⇒ 25, 1A ⇒ 26, 1B ⇒ 27, 1C ⇒ 28, 1D ⇒ 29, 1E ⇒ 30, 1F ⇒ 31. After this the first digit will increase to 2 and again second digit will repeat from 0 to F one by one to represent natural numbers 32 to 47 and so on.
Decimal to Hexadecimal Conversion
As we have already stated in the previous articles on number systems that all the number systems are inter related, so as the decimal and hexadecimal numbers. Any number in decimal number system can be converted into hexadecimal number system. The procedure is given below. If we try to understand the procedure with an example and step by step then it will be easier and better for us. Let us first take any decimal number suppose we have taken 7510 and now we want to convert it into hexadecimal number, first we have to divide it by 16 75/16 = quotient 4, remainder 11 As the quotient is less than 16, we have to stop here and the equivalent hexadecimal number will be 4B8 = 7510 Now we will discuss the method for a slightly bigger number, Suppose the number is 169310 Now we divide it by 16 1693/16 = quotient = 105, remainder = 13(D) Now we have to divide the quotient again by 16 and see the result 105/16 = quotient = 6 remainder = 9 As the quotient is less than 16 the calculation part is completed and we can now directly write the result 169310 = 69D16 So the decimal number has been converted into a hexadecimal number.
From above explanation it can be understood that, hexadecimal number is the summation of products of different digit with their respective multipliers. The multipliers are 160, 161, 162, ........from right hand side or list significant bit (LSB). Let's have an example 4D2 and this would be expressed as
If we divide decimal 1234 by 16, we will get 77 as quotient and 2 as remainder. Then if we divide decimal 77 by 16, we will get 4 as quotient and 13 or D as remainder. Now if we write side by side from last quotient to first reminder we will get 4D2 which is hexadecimal or hex equivalent of the number 1234.
For that we divide 1234 by base 16 and we get 77 as the quotient and 2 as the remainder. 16 1234 →2 Divide again 77 by 16 and we get 4 as the quotient and 13 or D as the remainder. 16 77 → D 4
Hexadecimal to Decimal Conversion
In a similar way any hexadecimal number can be converted into a decimal number. We will look into the process with an example. But before beginning it should be made clear that before conversion of hexadecimal number all the letters of the number should be taken as their numerical values in decimal number system, i.e. if a digit in hexadecimal number is A then we have to take it as 10, now an example will make the whole procedure clear. Let us take any hexadecimal number 45B116, we have to convert it into decimal number, so starting from the right most digit we have to start multiplying the digits with ascending power of 16 starting from 0. So the taken number will be operated as In this procedure any hexadecimal number can be converted into decimal number.
The value hexadecimal number is determined by multiplying every digit of the hex number by its respective multiplier. We start from LSB or right most digit and multiply it with 160, then come to the next digit at left of LSB and multiply it with 161 and after that we come to the further left digit and multiply 162 with it. We continue this up to MSB or left most bit. The add all this product and finally we get decimal equivalent of hexadecimal number. This one of the easiest process of hexadecimal to decimal conversion.
Think about the hex number 4D2. Here the least significant bit of the number is 2 so we will multiply that with 160 or 1. Then come to the next left digit that is D OR 13 and we will multiply it with 161 or 16. Lastly we will multiply the left most digit or MSB i.e. 4 with 162. Now if we add these three terms, finally we will get the dedcimal equivalent of the said hexadecimal number. This is what Hexadecimal to Decimal Conversion Hence,
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Lesson Objectives
• Demonstrate an understanding of how to simplify fractions
• Learn how to multiply two or more fractions together
• Learn how to cross cancel when multiplying fractions
• Learn how to solve a problem with the word "of"
## How to Multiply Fractions
In our last lesson, we learned how to simplify a fraction. In that lesson, we briefly looked at multiplication with fractions. When we multiply fractions, we find the product of the numerators and place the result over the product of the denominators. Let's take a look at an example:
Find the product of 1/3 and 3/5: $$\frac{1}{3} \cdot \frac{3}{5}=\frac{1 \cdot 3}{3 \cdot 5}=\frac{3}{15}$$ Although 3/15 is the correct answer, the fraction is not simplified. We could factor the numerator and denominator and then cancel, but notice how we are just undoing the multiplication: $$\require{cancel}\frac{3}{15}=\frac{1 \cdot 3}{3 \cdot 5}= \frac{1 \cdot \cancel{3}}{\cancel{3} \cdot 5} =\frac{1}{5}$$ So how can we avoid extra work here? After all, we don't want to multiply and then factor in order to get a simplified answer. To speed up the process, we will use a procedure known as cross canceling. When we cross cancel, the result of our multiplication will be a simplified fraction.
### Multiplying Fractions with Cross Canceling
• Simplify each fraction involved in the multiplication problem
• Look for and cancel any common factors between all numerators and all denominators.
• Find the product of all non-canceled numerators and place the result overall non-canceled denominators
Let's take a look at a few examples.
Example 1: Find the product of 2/5 and 15/16 $$\frac{2}{5} \cdot \frac{15}{16}$$ $$\frac{\cancel{2}1}{\cancel{5}1} \cdot \frac{\cancel{15}3}{\cancel{16}8} = \frac{3}{8}$$ $$\frac{2}{5} \cdot \frac{15}{16} = \frac{3}{8}$$ You may be a bit confused about what happened above, so let's think about this step by step:
First, we think about each fraction. 2/5 and 15/16 cannot be simplified any further. Next, we think about cross canceling. If we look at 2 in the numerator of the left fraction and 16 in the denominator of the right fraction, we notice a common factor of 2. If we cancel this factor, 2/2 = 1 and 16/2 = 8. Next, we look at the 5 in the denominator of the left fraction and the 15 in the numerator of the right fraction, we notice a common factor of 5. If we cancel this factor, 5/5 = 1 and 15/5 = 3. This sets up the multiplication problem: $$\frac{1}{1} \cdot \frac{3}{8}$$ This leads us to our final simplified answer of 3/8.
Example 2: Find the product of 3/7 and 14/51 $$\frac{3}{7} \cdot \frac{14}{51}$$ Here we have common factors of 3 and 7 that can be cross canceled:
3 ÷ 3 = 1
51 ÷ 3 = 17
7 ÷ 7 = 1
14 ÷ 7 = 2 $$\frac{\cancel{3}1}{\cancel{7}1} \cdot \frac{\cancel{14}2}{\cancel{51}17} = \frac{2}{17}$$ $$\frac{3}{7} \cdot \frac{14}{51} = \frac{2}{17}$$ Example 3: Find the product of 12/13, 65/72, and 2/5 $$\frac{12}{13} \cdot \frac{65}{72} \cdot \frac{2}{5}$$ To make this problem easier, let's factor each number:
12 = 2 x 2 x 3
13 - prime
65 - 13 x 5
72 = 2 x 2 x 2 x 3 x 3
2 - prime
5 - prime
So what can we cancel? The order of the cancelation is irrelevant, so let's start by canceling between 12/13 and 65/72:
We can see that 12 and 72 share a common factor of 12.
12 ÷ 12 = 1
72 ÷ 12 = 6
We can also see that 13 and 65 share a common factor of 13.
13 ÷ 13 = 5
65 ÷ 13 = 5 $$\frac{\cancel{12}1}{\cancel{13}1} \cdot \frac{\cancel{65}5}{\cancel{72}6} \cdot \frac{2}{5}$$ 1/1 is just 1 and any number times 1 is unchanged. We can drop the leftmost fraction. Our problem is now: $$\frac{5}{6} \cdot \frac{2}{5}$$ We can cross cancel once again: here there are common factors of 5 and 2:
5 ÷ 5 = 1
2 ÷ 2 = 1
6 ÷ 2 = 3
$$\frac{\cancel{5}1}{\cancel{6}3} \cdot \frac{\cancel{2}1}{\cancel{5}1} = \frac{1}{3}$$ $$\frac{12}{13} \cdot \frac{65}{72} \cdot \frac{2}{5} = \frac{1}{3}$$ We will also encounter multiplication of fractions in word problems. These problems will use the word "of" to indicate multiplication. Let's look at an example.
Example 4:
Jacob received 90 dollars for his allowance. He spent 1/3 of the money on new clothes and the other 2/3 of the money on food. How much did Jacob spend on each? The key word here is "of", which tells us to multiply. We know the starting amount is 90. To find the amount he spent on clothes, we need 1/3 of 90, or 1/3 x 90.
Clothes: $$\frac{1}{3} \cdot 90$$ For this problem, just write 90 as 90/1. This is legal since any number over 1 remains unchanged. $$\frac{1}{3} \cdot \frac{90}{1} = \frac{1}{\cancel{3}1} \cdot \frac{\cancel{90}30}{1} = 30$$ Jacob spent $30 on clothes. We could use subtraction at this point since he spent all of the money. We know that 90 - 30 = 60, so this must be the amount spent on food. We can also do this with multiplication. We would want to find 2/3 of 90, which means 2/3 x 90: $$\frac{2}{3} \cdot \frac{90}{1} = \frac{2}{\cancel{3}1} \cdot \frac{\cancel{90}30}{1} = 60$$ Either way, we find that Jacob spent$30 on clothes and \$60 on food.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 9/100 = 9/100 = 0.09
Spelled result in words is nine one-hundredths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Unknown number
I think the number - its sixth is 3 smaller than its third.
• UN 1
If we add to an unknown number his quarter, we get 210. Identify unknown number.
• 3 children
3 children had to divide 4 pounds is candy. How much candy did each child get?
• Tank 11
When 150 litres has been drawn from a tank it is 3/8 full, how many litres will the tank hold?
• Playing
How long have we trained on the pitch when we know that the warm-up took 10 minutes, we trained passes for one-third of the time and we played football half the time?
• One sixth
How many sixths are two thirds?
• Honza
Honza is 13 years old and Peter 21 years old. After how many years are their ages will be at ratio 7: 9?
• Cupcakes
In a bowl was some cupcakes. Janka ate one third and Danka ate one quarter of cupcakes. a) How many of cookies ate together? b) How many cookies remain in a bowl? Write the results as a decimal number and in notepad also as a fraction.
• Class 8.A
Three quarters of class 8.A went skiing. Of those who remained at home one third was ill and the remaining six were on math olympic. How many students have class 8.A?
• Find the 11
Find the quotient of 229.12 and 12.32
• Roses
On the large rosary was a third white, half red, yellow quarter and six pink. How many roses was in the rosary?
• Three men
Alex is half younger than Jan, which is one-third younger than George. The sum of their ages is 48. How are these three men old?
• Sixty
Sixty percent of one-fifteen of the total is equal to thirty. What are two percent of the total?
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Strand: NUMBER AND OPERATIONS IN BASE TEN (2.NBT)
Understand place value (Standards 2.NBT.1–4). They use place value understanding and properties of operations to add and subtract (Standards 2.NBT.5–9).
• Boxes and Cartons of Pencils
In this task students are given information about quantities of pencils packed in boxes and cartons and asked to solve problems such as "Jem has 1 carton and 4 boxes. How many pencils does Jem have all together?"
• Bundling and Unbundling
In this task students determine the number of hundreds, tens and ones that are necessary to write equations when some digits are provided. The student must, in some cases, decompose hundreds to tens and tens to ones.
• Choral Counting
The teacher will need a 100 chart or large number line and a pointer. As a whole group, have students chant the counting sequence starting with one to thirty, using the pointer to follow the number sequence. Over time, increase the range to one to fifty and then one to one hundred.
• Comparisons 1
This task requires students to compare numbers that are identified by word names and not just digits. The order of the numbers described in words are intentionally placed in a different order than their base-ten counterparts so that students need to think carefully about the value of the numbers.
• Comparisons 2
The comparisons here involve sums and differences - not primarily to provide an opportunity to calculate, but rather in order to stimulate students’ thinking about the magnitudes of the base-ten units of which numbers are composed.
• Counting Stamps
This is an instructional task related to deepening place-value concepts. The important piece of knowledge upon which students need to draw is that 10 tens is 1 hundred. So each sheet contains 100 stamps.
• Digits 2-5-7
This task asks students to use all the digits 5, 7, and 2 to create different 3-digit numbers.
• Grade 2 Math Module 4: Addition and Subtraction Within 200 with Word Problems to 100 (EngageNY)
In Module 4, students develop place value strategies to fluently add and subtract within 100; they represent and solve one- and two-step word problems of varying types within 100; and they develop conceptual understanding of addition and subtraction of multi-digit numbers within 200. Using a concrete to pictorial to abstract approach, students use manipulatives and math drawings to develop an understanding of the composition and decomposition of units, and they relate these representations to the standard algorithm for addition and subtraction.
• Grade 2 Mathematics Module 1: Sums and Differences to 100 (EngageNY)
Module 1 sets the foundation for students to master sums and differences to 20. Students subsequently apply these skills to fluently add one-digit to two-digit numbers at least through 100 using place value understanding, properties of operations, and the relationship between addition and subtraction.
• How Many Days Until Summer Vacation?
The purpose of the task is to allow children an opportunity to subtract a three-digit number including a zero that requires regrouping.
• IXL Game: Mixed operations: Addition and Subtraction
This game helps second graders fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction. This is just one of many online games that supports the Utah Math core. Note: The IXL site requires subscription for unlimited use.
• Jamir's Penny Jar
The purpose of this task is to help students articulate their addition strategies as in and would be most appropriately used once students have a solid understanding of coin values. It also provides a context where it makes sense to “skip count by 5s and 10s” for the combinations that involve more than one nickel or dime.
• Largest Number Game
In this task students are told "Dona had cards with the numbers 0 to 9 written on them. She flipped over three of them. Her teacher said: 'If those three numbers are the digits in another number, what is the largest three-digit number you can make?'" Students should be asked to think through possibilities and then draw on their ability to compare three digit numbers to complete the task.
• Looking at Numbers Every Which Way
This task gives students the opportunity to work with multiple representations of base-ten numbers. The standard asks students to read and write numbers to 1000 using base-ten numerals, number names, and expanded form. This task addresses all of these and extends it by asking students to represent the numbers with pictures and on the number line.
• Making 124
Not all students have seen base-ten blocks. This task should only be used with students who know what they are or have some on-hand to use themselves. Students are asked explain how they found all the possible ways to make 124 using base-ten blocks.
• Making Math Fun with Place Value Games
This lesson shows a game that teaches place value. (7 minutes)
• Many Ways to do Addition 2
The purpose of this task is not to teach or model the addition strategies. Rather the purpose of this task is make explicit different ways students can solve problems so that they will be able to find the most efficient strategy in any given situation and increase their addition fluency.
• Number Line Comparisons
The purpose of this task is for students to use the number line to make comparisons between 3-digit numbers. The task is designed, in part, to help students understand how the number line works and that numbers on the right of the number line are greater than numbers on the left.
• One, Ten, and One Hundred More and Less
This task acts as a bridge between understanding place value and using strategies based on place value for addition and subtraction. Within the classroom context, this activity can be differentiated using numbers that are either simpler or more difficult to manipulate across tens and hundreds.
• Ordering 3-digit numbers
In this task each number has at most 3 digits so that students have the opportunity to think about how digit placement affects the size of the number. Each group also contains a two-digit number so that students have to do more than just compare the first digit, the second digit, etc.
• Party Favors
The point of this task is to emphasize the grouping structure of the base-ten number system, and in particular the crucial fact that 10 tens make 1 hundred.
• Regrouping
This task serves as a bridge between understanding place-value and using strategies based on place-value structure for addition. Place-value notation leaves a lot of information implicit. The way that the numbers are represented in this task makes this information explicit, which can help students transition to adding standard base-ten numerals.
• Sample Scope and Sequence
This sample scope and sequence document for the implementation of the new Math core for Grade 2 based on 160 days of instruction - 18 units.
• Saving Money 1
The purpose of this task is for students to relate addition and subtraction problems to money and to situations and goals related to saving money.
• Saving Money 2
The purpose of this task is for students to relate addition and subtraction problems to money and to situations and goals related to saving money. This problem shows the work advanced second graders might use for adding 2-digit numbers.
• Teaching Different Methods To Count Collections
This Teaching Channel video shows students recording and sharing strategies when skip counting. (8 minutes)
• Ten \$10s make \$100
This task uses one, ten, and one-hundred dollar bills to help students think about bundling by ten.
• Three Composing/Decomposing Problems
The purpose of this task is to help students understand composing and decomposing ones, tens, and hundreds. This task is meant to be used in an instructional setting and would only be appropriate to use if students actually have base-ten blocks on hand.
• Toll Bridge Puzzle
This task is intended to assess adding of four numbers as given in the standard while still being placed in a problem-solving context. As written this task is instructional; due to the random aspect regarding when the correct route is found, it is not appropriate for assessment.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT1)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT2)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT3)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT4)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT5)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT6)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT7)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT8)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• USBE Core Guide for Domain: Number and Operations in Base Ten (2NBT9)
This math guide provides academic vocabulary, instructional strategies, teacher resources, and assessment tasks for the Number and Operations in Base Ten domain.
• Using Pictures to Explain Number Comparisons
The purpose of this task is for students to compare three-digit numbers and explain the comparisons based on the meaning of the hundreds, tens, and ones digits, using >, =, and < symbols to record the results of comparisons.
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialist - Shannon Ference and see the Mathematics - Elementary website. For general questions about Utah's Core Standards contact the Director - DIANA SUDDRETH .
Email: diana.suddreth@schools.utah.gov
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200.
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Paul's Online Math Notes
Online Notes / Algebra (Notes) / Polynomial Functions / Partial Fractions
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Algebra - Notes
## Partial Fractions
This section doesn’t really have a lot to do with the rest of this chapter, but since the subject needs to be covered and this was a fairly short chapter it seemed like as good a place as any to put it.
So, let’s start with the following. Let’s suppose that we want to add the following two rational expressions.
What we want to do in this section is to start with rational expressions and ask what simpler rational expressions did we add and/or subtract to get the original expression. The process of doing this is called partial fractions and the result is often called the partial fraction decomposition.
The process can be a little long and on occasion messy, but it is actually fairly simple. We will start by trying to determine the partial fraction decomposition of,
where both P(x) and Q(x) are polynomials and the degree of P(x) is smaller than the degree of Q(x). Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. That is important to remember.
So, once we’ve determined that partial fractions can be done we factor the denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition.
Factor in denominator Term in partial fraction decomposition
Notice that the first and third cases are really special cases of the second and fourth cases respectively if we let . Also, it will always be possible to factor any polynomial down into a product of linear factors ( ) and quadratic factors ( ) some of which may be raised to a power.
There are several methods for determining the coefficients for each term and we will go over each of those as we work the examples. Speaking of which, let’s get started on some examples.
Now, we need to do a set of examples with quadratic factors. Note however, that this is where the work often gets fairly messy and in fact we haven’t covered the material yet that will allow us to work many of these problems. We can work some simple examples however, so let’s do that.
Example 2 Determine the partial fraction decomposition of each of the following. (a) [Solution] (b) [Solution] Solution (a) In this case the x that sits in the front is a linear term since we can write it as, and so the form of the partial fraction decomposition is, Now we’ll use the fact that the LCD is and add the two terms together, Next, set the numerators equal. This is where the process changes from the previous set of examples. We could choose to get the value of A, but that’s the only constant that we could get using this method and so it just won’t work all that well here. What we need to do here is multiply the right side out and then collect all the like terms as follows, Now, we need to choose A, B, and C so that these two are equal. That means that the coefficient of the x2 term on the right side will have to be 8 since that is the coefficient of the x2 term on the left side. Likewise, the coefficient of the x term on the right side must be zero since there isn’t an x term on the left side. Finally the constant term on the right side must be -12 since that is the constant on the left side. We generally call this setting coefficients equal and we’ll write down the following equations. Now, we haven’t talked about how to solve systems of equations yet, but this is one that we can do without that knowledge. We can solve the third equation directly for A to get that . We can then plug this into the first two equations to get, So, the partial fraction decomposition for this expression is, (b) Here is the form of the partial fraction decomposition for this part. Adding the two terms up gives, Now, set the numerators equal and we might as well go ahead and multiply the right side out and collect up like terms while we’re at it. Setting coefficients equal gives, In this case we got A and B for free and don’t get excited about the fact that . This is not a problem and in fact when this happens the remaining work is often a little easier. So, plugging the known values of A and B into the remaining two equations gives, The partial fraction decomposition is then,
Algebra - Notes
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Question Video: Integrating Trigonometric Functions | Nagwa Question Video: Integrating Trigonometric Functions | Nagwa
# Question Video: Integrating Trigonometric Functions Mathematics • Higher Education
Suppose that dπ¦/dπ₯ = β9 sin 2π₯ β 3 cos 5π₯ and π¦ = 7 when π₯ = π/6. Find π¦ in terms of π₯.
03:40
### Video Transcript
Suppose that dπ¦ by dπ₯ is equal to negative nine times the sin of two π₯ minus three times the cos of five π₯ and π¦ is equal to seven when π₯ is equal to π by six. Find π¦ in terms of π₯.
The question gives us a first-order differential equation. It wants us to find π¦ in terms of π₯ given that when π¦ is equal to seven π₯ is equal to π by six. We recall if weβre told the derivative of π¦ with respect to π₯ is equal to some function of π₯, we can solve for π¦ by doing the opposite of differentiating. We have that π¦ is equal to the integral of dπ¦ by dπ₯ with respect to π₯, which we know is equal to the integral of negative nine times the sin of two π₯ minus three times the cos of five π₯ with respect to π₯.
To evaluate this integral, we can integrate each term separately. And we recall for constants π and π where π is not equal to zero, the integral of π cos of ππ₯ with respect to π₯ is equal to π times the sin of ππ₯ divided by π plus a constant of integration πΆ. And the integral of π sin of ππ₯ with respect to π₯ is equal to negative π the cos of ππ₯ divided by π plus the constant of integration πΆ.
Using this, we get the integral of negative nine times the sin of two π₯ is negative one multiplied by negative nine cos of two π₯ divided by two. And negative one multiplied by negative one is just equal to one. So, integrating the first term, we get nine cos of two π₯ divided by two. We can then do the same to evaluate the integral of negative three times the cos of five π₯. We get negative three times the sin of five π₯ divided by five. And then, we add our constant of integration πΆ. And itβs worth noting we only need to add one constant of integration.
So, weβve now shown that π¦ is equal to this function of π₯, where πΆ can be any constant. However, weβre told that when π¦ is equal to seven, π₯ is equal to π by six. We can use this information to find the specific value of πΆ in this case. We substitute π¦ is equal to seven and π₯ is equal to π by six into this equation. This gives us that seven is equal to nine times the cos of two times π by six over two minus three times the sin of five times π by six over five plus πΆ.
We can simplify this. We have two times π by six is equal to π by three. Weβre now in a position to evaluate this expression. The cos of π by three is a standard trigonometric result which we should know. Itβs equal to one-half. And the sin of five π by six is also a standard trigonometric result we should know. Itβs also equal to one-half. This gives us that seven is equal to nine times one-half over two minus three times one-half over five plus πΆ. We can simplify this further. Nine times one-half over two is the same as saying nine over four. And three times one-half over five is the same as saying three over 10.
So, we have that seven is equal to nine over four minus three over 10 plus πΆ. We can rearrange this expression and then evaluate to get that πΆ is equal to 101 divided by 20. We then substitute this value of πΆ into our general solution. And this gives us that π¦ is equal to negative three over five times the sin of five π₯ plus nine over two times the cos of two π₯ plus 101 divided by 20. Therefore, weβve shown if dπ¦ by dπ₯ is equal to negative nine times the sin of two π₯ minus three times the cos of five π₯ and π¦ is equal to seven when π₯ is equal to π by six. Then π¦ is equal to negative three over five times the sin of five π₯ plus nine over two times the cos of two π₯ plus 101 divided by 20.
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# A stone of $1 \mathrm{kg}$ is thrown with a velocity $20m{{s}^{-1}}$across the frozen surface of a lake and comes to rest after travelling a distance of $50 \mathrm{m}$. What is the force of friction between the stone and the ice?
Last updated date: 16th Sep 2024
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Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.
Before we proceed, we have to make a free body diagram of the stone-ice system. In the figure, we are seeing that frictional force acting in leftward while the stone is moving rightward direction. One thing we have to fix in our mind that frictional force will always act opposite to the direction of motion of the body. Now on resolving the forces, $\mathrm{f}-\mathrm{ma} = 0$ or $\mathrm{f}=\mathrm{ma}$ also form the equation of motion $\mathrm{v}^{2}=\mathrm{u}^{2}+2as$ as, here final velocity is zero, so $\mathrm{v}=0$
$\mathrm{u}^{2}=-2 \mathrm{as}$
$\mathrm{u}^{2}=-2 \mathrm{as}$
$\mathrm{a}=\dfrac{-\mathrm{u}^{2}}{2 \mathrm{s}}=-\dfrac{20^{2}}{2 \times 50}=-4 \dfrac{\mathrm{m}}{\mathrm{sec}^{2}}$
On substituting the given data in equation (1) $\mathrm{f}=1 \times(-4)=-4 \mathrm{N}$
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# Addition and Subtraction of Rational Expressions
## Add and subtract fractions with variables in the denominator
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Practice Addition and Subtraction of Rational Expressions
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Addition and Subtraction of Rational Expressions
Suppose that Jake can photocopy all the pages in a book in 3 hours, while Meredith can photocopy all the pages in the book in 2 hours. If Jake started at the beginning of the book and Meredith started at the end of the book and they worked together, how many hours would it take for them to photocopy all the pages? What equation could you set up to find this value? In this Concept, you'll learn about the addition and subtraction of rational expressions so that you can solve work problems such as this one.
### Guidance
Like numerical fractions, rational expressions represent a part of a whole quantity. Remember when adding or subtracting fractions, the denominators must be the same. Once the denominators are identical, the numerators are combined by adding or subtracting like terms.
#### Example A
Simplify: \begin{align*}\frac{4x^2-3}{x+5}+\frac{2x^2-1}{x+5}\end{align*}.
Solution: The denominators are identical; therefore we can add the like terms of the numerator to simplify.
Not all denominators are the same however. In the case of unlike denominators, common denominators must be created through multiplication by finding the least common multiple.
The least common multiple (LCM) is the smallest number that is evenly divisible by every member of the set.
What is the least common multiple of \begin{align*}2, 4x\end{align*}, and \begin{align*}6x^2\end{align*}? The smallest number 2, 4, and 6 can divide into evenly is six. The largest exponent of \begin{align*}x\end{align*} is 2. Therefore, the LCM of \begin{align*}2, 4x\end{align*}, and \begin{align*}6x^2\end{align*} is \begin{align*}6x^2\end{align*}.
#### Example B
Find the least common multiple of \begin{align*}2x^2+8x+8\end{align*} and \begin{align*}x^3-4x^2-12x\end{align*}.
Solution: Factor the polynomials completely.
The LCM is found by taking each factor to the highest power that it appears in either expression. \begin{align*}LCM=2x(x+2)^2 (x-6)\end{align*}
Use this approach to add rational expressions with unlike denominators.
#### Example C
Simplify \begin{align*}\frac{2}{x+2}-\frac{3}{2x-5}\end{align*}.
Solution:
The denominators cannot be factored any further, so the LCM is just the product of the separate denominators.
The first fraction needs to be multiplied by the factor \begin{align*}(2x-5)\end{align*} and the second fraction needs to be multiplied by the factor \begin{align*}(x+2)\end{align*}.
We combine the numerators and simplify.
Combine like terms in the numerator.
Work Problems
These are problems where two objects work together to complete a job. Work problems often contain rational expressions. Typically we set up such problems by looking at the part of the task completed by each person or machine. The completed task is the sum of the parts of the tasks completed by each individual or each machine.
Part of task completed by first person + Part of task completed by second person = One completed task
To determine the part of the task completed by each person or machine, we use the following fact.
Part of the task completed = rate of work \begin{align*}\times\end{align*} time spent on the task
In general, it is very useful to set up a table where we can list all the known and unknown variables for each person or machine and then combine the parts of the task completed by each person or machine at the end.
#### Example D
Mary can paint a house by herself in 12 hours. John can paint a house by himself in 16 hours. How long would it take them to paint the house if they worked together?
Solution:
Let \begin{align*}t=\end{align*} the time it takes Mary and John to paint the house together.
Since Mary takes 12 hours to paint the house by herself, in one hour she paints \begin{align*}\frac{1}{12}\end{align*} of the house.
Since John takes 16 hours to pain the house by himself, in one hour he paints \begin{align*}\frac{1}{16}\end{align*} of the house.
Mary and John work together for \begin{align*}t\end{align*} hours to paint the house together. Using the formula:
Part of the task completed = rate of work \begin{align*}\times\end{align*} time spent on the task
we can write that Mary completed \begin{align*}\frac{t}{12}\end{align*} of the house and John completed \begin{align*}\frac{t}{16}\end{align*} of the house in this time and summarize the data in the following table.
Painter Rate of work (per hour) Time worked Part of Task
Mary \begin{align*}\frac{1}{12}\end{align*} \begin{align*}t\end{align*} \begin{align*}\frac{t}{12}\end{align*}
John \begin{align*}\frac{1}{16}\end{align*} \begin{align*}t\end{align*} \begin{align*}\frac{t}{16}\end{align*}
Using the formula:
Part of task completed by first person + Part of task completed by second person = One completed task
write an equation to model the situation,
Solve the equation by finding the least common multiple.
-->
### Guided Practice
One pipe fills a pool 2 times faster than another pipe. Together, they fill the pool in 5 hours. How long does it take for the faster pipe to fill the pool?
Solution:
Let \begin{align*}f\end{align*} be the time it takes for the faster pipe to fill the pool. Since the pipe is two times faster, the slower pipe takes twice as long. So the time it take the slower pipe to fill the pool is \begin{align*} 2f\end{align*}. After 5 hours, the faster pipe will fill up a certain proportion of the pool. That proportion is found by dividing 5 hours by the time it takes the faster pipe to fill the pool.
Proportion of the pool filled by the faster pipe:
\begin{align*}\frac{5}{f}\end{align*}
Proportion of the pool filled by the slower pipe:
\begin{align*} \frac{5}{2f}\end{align*}
Since together they fill the whole pool in 5 hours, we add the proportions together:
It would take the faster pipe 7.5 hours to fill the whole pool by itself.
### Explore More
Perform the indicated operation and simplify. Leave the denominator in factored form.
1. \begin{align*}\frac{5}{24}-\frac{7}{24}\end{align*}
2. \begin{align*}\frac{10}{21}+\frac{9}{35}\end{align*}
3. \begin{align*}\frac{5}{2x+3}+\frac{3}{2x+3}\end{align*}
4. \begin{align*}\frac{3x-1}{x+9}-\frac{4x+3}{x+9}\end{align*}
5. \begin{align*}\frac{4x+7}{2x^2}-\frac{3x-4}{2x^2}\end{align*}
6. \begin{align*}\frac{x^2}{x+5}-\frac{25}{x+5}\end{align*}
7. \begin{align*}\frac{2x}{x-4}+\frac{x}{4-x}\end{align*}
8. \begin{align*}\frac{10}{3x-1}-\frac{7}{1-3x}\end{align*}
9. \begin{align*}\frac{5}{2x+3}-3\end{align*}
10. \begin{align*}\frac{5x+1}{x+4}+2\end{align*}
11. \begin{align*}\frac{1}{x}+\frac{2}{3x}\end{align*}
12. \begin{align*}\frac{4}{5x^2}-\frac{2}{7x^3}\end{align*}
13. \begin{align*}\frac{4x}{x+1}-\frac{2}{2(x+1)}\end{align*}
14. \begin{align*}\frac{10}{x+5}+\frac{2}{x+2}\end{align*}
15. \begin{align*}\frac{2x}{x-3}-\frac{3x}{x+4}\end{align*}
16. \begin{align*}\frac{4x-3}{2x+1}+\frac{x+2}{x-9}\end{align*}
17. \begin{align*}\frac{x^2}{x+4}-\frac{3x^2}{4x-1}\end{align*}
18. \begin{align*}\frac{2}{5x+2}-\frac{x+1}{x^2}\end{align*}
19. \begin{align*}\frac{x+4}{2x}+\frac{2}{9x}\end{align*}
20. \begin{align*}\frac{5x+3}{x^2+x}+\frac{2x+1}{x}\end{align*}
21. \begin{align*}\frac{4}{(x+1)(x-1)}-\frac{5}{(x+1)(x+2)}\end{align*}
22. \begin{align*}\frac{2x}{(x+2)(3x-4)}+\frac{7x}{(3x-4)^2}\end{align*}
23. \begin{align*}\frac{3x+5}{x(x-1)}-\frac{9x-1}{(x-1)^2}\end{align*}
24. \begin{align*}\frac{1}{(x-2)(x-3)}+\frac{4}{(2x+5)(x-6)}\end{align*}
25. \begin{align*}\frac{3x-2}{x-2}+\frac{1}{x^2-4x+4}\end{align*}
26. \begin{align*}\frac{-x^2}{x^2-7x+6}+x-4\end{align*}
27. \begin{align*}\frac{2x}{x^2+10x+25}-\frac{3x}{2x^2+7x-15}\end{align*}
28. \begin{align*}\frac{1}{x^2-9}+\frac{2}{x^2+5x+6}\end{align*}
29. \begin{align*}\frac{-x+4}{2x^2-x-15}+\frac{x}{4x^2+8x-5}\end{align*}
30. \begin{align*}\frac{4}{9x^2-49}-\frac{1}{3x^2+5x-28}\end{align*}
31. One number is 5 less than another. The sum of their reciprocals is \begin{align*}\frac{13}{36}\end{align*}. Find the two numbers.
32. One number is 8 times more than another. The difference in their reciprocals is \begin{align*}\frac{21}{20}\end{align*}. Find the two numbers.
33. A pipe can fill a tank full of oil in 4 hours and another pipe can empty the tank in 8 hours. If the valves to both pipes are open, how long would it take to fill the tank?
34. Stefan and Misha are washing cars. Stefan could wash the cars by himself in 6 hours and Misha could wash the cars by himself in 5 hours. Stefan starts washing the cars by himself, but he needs to go to his football game after 2.5 hours. Misha continues the task. How long did it take Misha to finish washing the cars?
35. Amanda and her sister Chyna are shoveling snow to clear their driveway. Amanda can clear the snow by herself in three hours and Chyna can clear the snow by herself in four hours. After Amanda has been working by herself for one hour, Chyna joins her and they finish the job together. How long does it take to clear the snow from the driveway?
36. At a soda bottling plant, one bottling machine can fulfill the daily quota in ten hours and a second machine can fill the daily quota in 14 hours. The two machines started working together but after four hours the slower machine broke and the faster machine had to complete the job by itself. How many hours does the fast machine work by itself?
Mixed Review
1. Explain the difference between these two situations. Write an equation to model each situation. Assume the town started with 10,000 people. When will statement b become larger than statement a?
1. For the past seven years, the population grew by 500 people every year.
2. For the past seven years, the population grew by 5% every year.
1. Simplify. Your answer should have only positive exponents. \begin{align*}\frac{16x^2 y^7}{-2x^8 y} \cdot \frac{1}{2} x^{-10}\end{align*}
2. Solve for \begin{align*}j: -12=j^2-8j\end{align*}. Which method did you use? Why did you choose this method?
3. The distance you travel varies directly as the speed at which you drive. If you can drive 245 miles in five hours, how long will it take you to drive 90 miles?
4. Two cities are 3.78 centimeters apart on an atlas. The atlas has a scale of \begin{align*}\frac{1}{2} cm=14 \ miles\end{align*}. What is the true distance between these cities?
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 12.7.
### Vocabulary Language: English Spanish
work problems
work problems
A work problem is one where two or more people are involved in working to complete a task. The basic formula for two people is: Part of task completed by first person + Part of task completed by second person = One completed task
Least Common Denominator
Least Common Denominator
The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators.
### Explore More
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### Angles of Triangles - Peacock
```Angles of Triangles
Section 4.2
Objectives
Find angle measures in triangles.
Key Vocabulary
Corollary
Exterior angles
Interior angles
Theorems
4.1 Triangle Sum Theorem
Corollary to the Triangle Sum Theorem
4.2 Exterior Angle Theorem
Measures of Angles of a Triangle
The word “triangle” means “three angles”
When the sides of a triangles are extended,
however, other angles are formed
The original 3 angles of the triangle are the
interior angles
The angles that are adjacent to interior angles
are the exterior angles
Each vertex has a pair of exterior angles
Original Triangle
Extend sides
Exterior
Angle
Exterior
Angle
Interior
Angle
Triangle Interior and Exterior Angles
Smiley faces are interior
angles and hearts
represent the exterior
angles
B
A
C
Each vertex has a pair
of congruent exterior
angles; however it is
common to show only
one exterior angle at
each vertex.
Triangle Interior and Exterior Angles
A
)))
Interior Angles
C
B
(
D
Exterior Angles
(formed by extending the sides)
E
F
Triangle Sum Theorem
The Triangle Angle-Sum Theorem gives
the relationship among the interior angle
measures of any triangle.
Triangle Sum Theorem
If you tear off two corners of a triangle and
place them next to the third corner, the
three angles seem to form a straight line.
You can also show this in a drawing.
Triangle Sum Theorem
Draw a triangle and extend one side. Then
draw a line parallel to the extended side, as
shown.
Two sides of the
triangle are
transversals to the
parallel lines.
The three angles in the triangle can be
arranged to form a straight line or 180°.
Theorem 4.1 – Triangle Sum Theorem
The sum of the measures of the angles of a
triangle is 180°.
X
mX + mY + mZ = 180°
Y
Z
Triangle Sum Theorem
Example 1
Given mA = 43° and mB = 85°, find mC.
SOLUTION
mA + mB + mC = 180°
43° + 85° + mC = 180°
128° + mC = 180°
128° + mC – 128° = 180° – 128°
mC = 52°
CHECK
Triangle Sum Theorem
Substitute 43° for mA and
85° for mB.
Simplify.
Subtract 128° from each side.
Simplify.
C has a measure of 52°.
Check your solution by substituting 52° for mC. 43° +
85° + 52° = 180°
Example 2a
A. Find p in the acute triangle.
73° + 44° + p° = 180°
117 + p = 180
–117
–117
p = 63
Triangle Sum
Theorem
Subtract 117 from
both sides.
Example 2b
B. Find m in the obtuse triangle.
62
23° + 62° + m° = 180°
Triangle Sum
Theorem
23
85 + m = 180
–85
–85
m = 95
Subtract 85 from
both sides.
m
A. Find a in the acute triangle.
88° + 38° + a° = 180°
126 + a = 180
–126
–126
a = 54
Triangle Sum
Theorem
38°
Subtract 126
from both sides.
a°
88°
B. Find c in the obtuse triangle.
24° + 38° + c° = 180°
62 + c = 180
–62
–62
c = 118
Triangle Sum
Theorem.
38°
24°
Subtract 62 from
both sides.
c°
Example 3
Find the angle measures in the scalene triangle.
2x° + 3x° + 5x° = 180°
10x = 180
10
10
Triangle Sum Theorem
Simplify.
Divide both sides by 10.
x = 18
The angle labeled 2x° measures
2(18°) = 36°, the angle labeled
3x° measures 3(18°) = 54°, and
the angle labeled 5x° measures
5(18°) = 90°.
Find the angle measures in the scalene triangle.
3x° + 7x° + 10x° = 180°
20x = 180
20
20
x=9
The angle labeled 3x°
measures 3(9°) = 27°, the
angle labeled 7x°
measures 7(9°) = 63°, and
the angle labeled 10x°
measures 10(9°) = 90°.
Triangle Sum Theorem
Simplify.
Divide both sides by 20.
10x°
3x°
7x°
Example 4:
Find the missing angle measures.
Find
first because the
measure of two angles of
the triangle are known.
Angle Sum Theorem
Simplify.
Subtract 117 from each side.
Example 4:
Angle Sum Theorem
Simplify.
Subtract 142 from each side.
Find the missing angle measures.
Corollaries
Definition: A corollary is a theorem with a
proof that follows as a direct result of
another theorem.
As a theorem, a corollary can be used as
a reason in a proof.
Triangle Angle-Sum Corollaries
Corollary 4.1 – The acute s of a right ∆
are complementary.
Example: m∠x + m∠y = 90˚
x°
y°
Example 5
∆ABC and ∆ABD are right triangles.
Suppose mABD = 35°.
a. Find mDAB.
b. Find mBCD.
SOLUTION
a. mDAB + mABD = 90°
mDAB + 35° = 90°
mDAB + 35° – 35° = 90° – 35°
mDAB = 55°
b. mDAB + mBCD = 90°
55° + mBCD = 90°
mBCD = 35°
Corollary to the
Triangle Sum Theorem
Substitute 35° for mABD.
Subtract 35° from each side.
Simplify.
Corollary to the
Triangle Sum Theorem
Substitute 55° for mDAB.
Subtract 55° from each side.
1. Find mA.
65°
75°
50°
2. Find mB.
3.
Find mC.
Example 6:
GARDENING The flower bed shown is in the shape of
a right triangle. Find
if
is 20.
Corollary 4.1
Substitution
Subtract 20 from each side.
The piece of quilt fabric is in the shape of a
right triangle. Find
if
is 62.
Exterior Angles and Triangles
An exterior angle is formed by one side of a
triangle and the extension of another side
(i.e. 1 ).
2
1
4
3
The interior angles of the triangle not adjacent to
a given exterior angle are called the remote
interior angles (i.e. 2 and 3).
Investigating Exterior Angles of a
Triangles
You can put the two torn angles
together to exactly cover one of the
exterior angles
B
A
B
C
A
Theorem 4.2 – Exterior Angle Theorem
The measure of an exterior angle of a
triangle is equal to the sum of the
measures of the two remote interior
angles.
m 1 = m 2 + m 3
2
1
4
3
Example 7
Given mA = 58° and mC = 72°, find m1.
SOLUTION
m1 = mA + mC
Exterior Angle Theorem
= 58° + 72°
Substitute 58° for mA and
72° for mC.
= 130°
Simplify.
1 has a measure of 130°.
1. Find m2.
120°
155°
113°
2. Find m3.
3. Find m4.
Example 8:
Find the measure of each numbered angle in the figure.
Exterior Angle Theorem
Simplify.
If 2 s form a linear pair, they
are supplementary.
Substitution
Subtract 70 from each side.
Example 8:
m∠1=70
m∠2=110
Exterior Angle Theorem
Substitution
Subtract 64 from each side.
If 2 s form a linear pair,
they are supplementary.
Substitution
Simplify.
Subtract 78 from each side.
m∠1=70
m∠2=110
m∠3=46
m∠4=102
Example 8:
Angle Sum Theorem
Substitution
Simplify.
Subtract 143 from each side.
Find the measure of each numbered angle in the figure.
|
## find the value of (-1) to the power n+(-1)to the power 2n+ (-1) to the power 2n+1 + (-1)to the power 4n+2
Question
find the value of (-1) to the power n+(-1)to the power 2n+ (-1) to the power 2n+1 + (-1)to the power 4n+2
in progress 0
3 months 2021-10-15T15:50:23+00:00 2 Answers 0 views 0
STEP
1
:
Equation at the end of step 1
(8 • (x6)) – 33×3
STEP
2
:
Equation at the end of step
2
:
23×6 – 33×3
STEP
3
:
STEP
4
:
Pulling out like terms
4.1 Pull out like factors :
8×6 – 27×3 = x3 • (8×3 – 27)
Trying to factor as a Difference of Cubes:
4.2 Factoring: 8×3 – 27
Theory : A difference of two perfect cubes, a3 – b3 can be factored into
(a-b) • (a2 +ab +b2)
Proof : (a-b)•(a2+ab+b2) =
a3+a2b+ab2-ba2-b2a-b3 =
a3+(a2b-ba2)+(ab2-b2a)-b3 =
a3+0+0+b3 =
a3+b3
Check : 8 is the cube of 2
Check : 27 is the cube of 3
Check : x3 is the cube of x1
Factorization is :
(2x – 3) • (4×2 + 6x + 9)
Trying to factor by splitting the middle term
4.3 Factoring 4×2 + 6x + 9
The first term is, 4×2 its coefficient is 4 .
The middle term is, +6x its coefficient is 6 .
The last term, “the constant”, is +9
Step-1 : Multiply the coefficient of the first term by the constant 4 • 9 = 36
Step-2 : Find two factors of 36 whose sum equals the coefficient of the middle term, which is 6 .
-36 + -1 = -37
-18 + -2 = -20
-12 + -3 = -15
-9 + -4 = -13
-6 + -6 = -12
-4 + -9 = -13
For tidiness, printing of 12 lines which failed to find two such factors, was suppressed
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Final result :
x3 • (2x – 3) • (4×2 + 6x + 9)
|
# Thread: Cubic Polynomial
1. ## Cubic Polynomial
A Cubic Polynomial with integral coefficients has a root $\displaystyle (3-4i)$ and possesses the form:
$\displaystyle y=x^3+ax+b$ Find AB/25
Have no clue where to start with this one. Any help is appreciated.
2. If you know one root, you should be able to factor the cubic. One way to do that is by polynomial long division. The result should look like your y.
3. How can I factor the polynomial, if the A and B values are not given? On another note, If one of the roots of the polynomial is (3-4i), then another root has to be (3+4i). multiplication of the two gives us
$\displaystyle x^2-6x+25$
Now where do I go from here?
4. You found that well.
Now just do this (x-(3-4i))(x-(3+4i) or $\displaystyle ((x-3)+4i)((x-3)-4i)=(x-3)^2-(4i)^2$
Regards.
5. Well, I tried using the factor theorem.
Let $\displaystyle f(x) = x^3 + ax + b$
Since 3-4i is a factor, f(3-4i) = 0.
So, $\displaystyle f(3-4i) = (3-4i)^3 + a(3-4i) + b = 0$
Expanding, you get:
$\displaystyle (b - 3a -117) - (44 + 4a)i = 0$
So,
$\displaystyle b - 3a - 117 = 0$
and
$\displaystyle 44 + 4a = 0$
I'm not sure if that's right though
This gives a = -11, and b = 150
So, $\displaystyle \frac{ab}{25} = \frac{(-11)(150)}{25} = -66$
6. Originally Posted by rtblue
How can I factor the polynomial, if the A and B values are not given? On another note, If one of the roots of the polynomial is (3-4i), then another root has to be (3+4i). multiplication of the two gives us
$\displaystyle x^2-6x+25$
Now where do I go from here?
Very good. Now, if "c" is the third root, then you must have $\displaystyle (x^2- 6x+ 25)(x- c)= x^3+ ax + b$.
$\displaystyle (x^2- 6x+ 25)(x- c)= x^3- 6x^2+ 25x- cx^2+ 6cx- 25c= x^3- (6+ c)x^2+ (25+ 6c)x- 25c$
and that must be equal to $\displaystyle x^3+ ax+ b$
The fact that there is no "$\displaystyle x^2$" term on the right tells you that $\displaystyle 6+ c= 0$ so c= -6. Now, it should be easy to find a and b.
|
# ICSE Solutions for Selina Concise Chapter 24 Solutions of Right Triangles Class 9 Maths
### Exercise 24(A)
1. Fin ‘x’ if:
(i)
(ii)
(iii)
(i) From the figure we have
2. Find angle ‘A’ if:
(i)
(ii)
(iii)
(i) From the figure we have
cos A = 10/20
⇒ cos A = 1/2
⇒ cos A = cos 60°
⇒ A = 60°
(ii) From the figure we have
sin A = sin45°
⇒ A = 45°
(iii) From the figure we have
tan A = 10√3/10
⇒ tan A = √3
⇒ tan A = sin 60°
⇒ A = 60°
3. Find angle ‘x’ if:
The figure is drawn as follows :
The above figure we have
tan 60° = 30/AD
⇒ √3 = 30/AD
⇒ AD = 30/√3
Again,
sin x = AD/20
⇒ AD = 20 sin x
Now,
20 sin x = 30/√3
⇒ sin x = 30/20√3
⇒ sin x = √3/2
⇒ sin x = sin 60°
⇒ x = 60°
4. Find AD, if:
(i)
(ii)
(i) From the right triangle ABE
tan45° = AE/BE
⇒ 1 = AE/BE
⇒ AE = BE
Therefore, AE = BE = 50 m.
Now, from the rectangle BCDE we have
DE = BC = 10 m.
Therefore, the length of AD will be:
AD = AE + DE = 50 + 10 = 60 m.
(ii) From the triangle ABD we have
sin B = AD/AB
⇒ sin 30 = AD/100 [Since ∠ACD is the exterior angle of the triangle ABC]
⇒ 1/2 = AD/100
⇒ AD = 50 m
5. Find the length of AD.
Given: ABC = 60°.
DBC = 45° and BC = 40 cm
From right triangle ABC,
tan60° = AC/BC
⇒ √3 = AC/40
⇒ AC = 40√3 cm
From right triangle BDC,
tan45° = DC/BC
⇒ 1 = DC/40
⇒ DC = 40 cm
From the figure, it is clear that AD = AC - DC
⇒ AD = 40√3 - 40
⇒ AD = 40(√3 - 1)
⇒ AD = 29.28 cm
6. Find the lengths of diagonals AC and BD. Given AB = 60 cm and BAD = 60o.
We know, diagonals of a rhombus bisect each other at right angles and also bisect the angle of vertex .
The figure is shown below :
Now,
7. Find AB.
Consider the figure:
Given FC = 20, ED = 30, So EP = 10cm
Therefore,
tan 60° = FP/EP
⇒ √3 = FP/10
⇒ FP = 10√3 = 17.32 cm
Thus, AB = AC + CD + BD = 54.64 cm.
8. In trapezium ABCD, as shown, AB∥DC, AD = DC = BC = 20 cm and A = 60o. Find:
(i) length of AB
(ii) distance between AB and DC.
First draw two perpendiculars to AB from the point D and C respectively. Since AB|| CD therefore PMCD will be a rectangle.
Consider the figure:
(i) From right triangle ADP we have
cos 60° = AP/AD
⇒ 1/2 = AP/20
⇒ AP = 10
Similarly from the right triangle BMC we have BM = 10cm .
Now from the rectangle PMCD we have CD = PM = 20 cm .
Therefore,
AB = AP + PM + MB = 10 +20 +10 = 40 cm.
(ii) Again from the right triangle APD we have,
sin 60° = PD/20
⇒ √3/2 = PD/20
⇒ PD = 10√3
Therefore, the distance between AB and CD is 10√3.
9. Use the information given to find the length of AB.
From right triangle AQP
10. Find the length of AB.
From right triangle ADE
tan 45° = AE/DE
⇒ 1 = AE/30
⇒ AE = 30 cm
Also, from triangle DBE
tan 60° = BE/DE
⇒ √3 = BE/30
⇒ BE = 30√3 cm
Therefore, AB = AE + BE = 30 + 30√3 = 30(1 + √3) cm
11. In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB. Given that AED = 60o and ACD = 45o. Calculate:
(i) AB
(ii) AC
(iii) AE
(i) From the triangle ADC we have
⇒ 1 = 2/DC
⇒ DC = 2
Since, AD॥DC and AD⊥EC, ABCD is a parallelogram and hence opposite sides are equal.
Therefore, AB = DC = 2cm
(ii) Again,
sin 45° = AD/AC
⇒ 1/√2 = 2/AC
⇒ AC = 2√2
(iii) From the right triangle ADE we have
sin 60° = AD/AE
⇒ √3/2 = 2/AE
⇒ AE = 4/√3
12. In the given figure, ∠B = 60° , AB = 16 cm and BC = 23 cm,
Calculate:
(i) BE
(ii) AC
Let BE = x, and EC = 25 -x
In △ABC,
sin 60° = AE/AB
13. Find
(i) BC
(iii) AC
14. In right-angled triangle ABC; B = 90o. Find the magnitude of angle A, if:
(i) AB is √3 times of BC.
(ii) BC is √3 times of AB.
Consider the figure
15. A ladder is placed against a vertical tower. If the ladder makes an angle of 30o with the ground and reaches upto a height of 15 m of the tower; find length of the ladder.
Given that the ladder makes an angle of 30° with the ground and reaches upto a height of 15 m of the tower which is shown in the figure below:
16. A kite is attached to a 100 m long string. Find the greatest height reached by the kite when its string makes an angles of 60o with the level ground.
17. Find AB and BC, if:
(i)
(ii)
(iii)
(i) Let BC = xm
BD = BC + CD = (x + 20)cm
In △ABD,
AB = x = 27.32 cm
Therefore, BC = x = AB = 27.32 cm
Therefore, AB = 27.32 cm, BC = 27.32 cm
(ii) Let BC = x m
BD = BC + CD = (x+20)cm
In △ABD,
Therefore BC = x = 10cm
Therefore,
AB = 17.32 cm, BC = 10 cm
(iii) Let BC = x m
BD = BC + CD = (x+20) cm
In △ABD,
∴ BC = x = 27.32 cm
Therefore, AB = 47.32 cm, BC =27.32 cm
18. Find PQ, if AB = 150 m, P = 30o and Q = 45o.
(i)
(ii)
19. If tan xo = 5 /12, tan yo = 3/4 and AB = 48 m; find the length of CD
⇒ 720 + 20 CD = 36 CD
⇒ 16 CD = 720
⇒ CD = 45
Therefore, length of CD is 45 m.
20. The perimeter of a rhombus is 96 cm and obtuse angle of it is 120o. Find the lengths of its diagonals.
|
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Ali K.
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Subject:Pre-Calculus
TutorMe
Question:
Find all real values of $$x$$ that satisfy the inequality $$2x^3+3x\leq 3x^2+1$$.
Inactive
Ali K.
Answer:
We may first rewrite the given inequality as $$2x^3-3x^2+3x-1\leq 0$$. Next, we attempt to factor the cubic polynomial $$f(x)=2x^3-3x^2+3x-1$$. Since all the coefficients of our polynomial are integers, we can invoke the rational root theorem: If $$f(x)$$ has any rational roots, they will each be of the form $$\frac{p}{q}$$, where $$p$$ divides the polynomial's constant term, $$-1$$, and $$q$$ divides the leading coefficient, $$2$$. The list of candidate rational roots for our polynomial is therefore $$1$$, $$-1$$, $$\frac{1}{2}$$, $$-\frac{1}{2}$$. Evaluating $$f$$ at these values reveals that $$f(\frac{1}{2})=0$$. Therefore $$\frac{1}{2}$$ is a root of our polynomial and $$2x-1$$ is a factor of our polynomial. Dividing $$2x^3-3x^2+3x-1$$ by $$2x-1$$ (using polynomial long division or synthetic division) yields the quotient $$x^2-x+1$$ (with no remainder). Therefore $$f(x)=(2x-1)(x^2-x+1)$$. Let us consider the quadratic factor, $$q(x)=x^2-x+1$$, on its own. The graph of $$y=q(x)$$ is a parabola, and this parabola opens upward (since the leading coefficient, $$1$$, is positive). Its discriminant, $$(-1)^2-4(1)(1)=-3$$, is negative, so $$q$$ has no real zeros: the graph of $$y=q(x)$$ never intersects the $$x$$ axis. We conclude that the graph of $$y=q(x)$$ lies entirely above the $$x$$ axis. In other words, $$q(x)$$ is positive for all real values of $$x$$. Since $$f(x)=(2x-1)q(x)$$ and $$q(x)$$ is always positive, we see that $$f(x)\leq 0$$ if and only if $$2x-1\leq 0$$. The given inequality is therefore satisfied if and only if $$x\leq\frac{1}{2}$$.
Subject:Calculus
TutorMe
Question:
Determine the global (absolute) maximum and minimum values of the function $$f$$ defined by $$f(x)=3x^{2/3}(2x+5),$$ $$-2\leq x\leq 1$$.
Inactive
Ali K.
Answer:
The domain of $$f$$ is the closed interval $$[-2,1]$$. Since $$f$$ is a product of functions that are continuous on that interval, it follows that $$f$$ is continuous on its domain. We may therefore apply the "closed interval method" to locate the global maximum and minimum values of $$f$$. First, we differentiate $$f$$, perhaps by invoking the product rule: $$f^\prime(x)=3\cdot \frac{2}{3}x^{-\frac{1}{3}}(2x+5)+3x^{\frac{2}{3}}(2)=x^{-\frac{1}{3}}(4x+10)+x^{-\frac{1}{3}}(6x)=10x^{-\frac{1}{3}}(x+1)$$, valid for $$-2\leq x < 0$$ and for $$0<x\leq 1$$. $$f^\prime$$ is undefined at $$0$$. (The graph of $$f$$ exhibits a cusp there.) Next, we locate the critical numbers of $$f$$ (where $$f^\prime$$ is zero or undefined): $$f^\prime$$ is zero only at $$-1$$, and it is undefined only at $$0$$. Now, we evaluate $$f$$ at its critical numbers and at the endpoints of its closed interval domain: $$f(-2)=3(-2)^{2/3}[2(-2)+5]=3\sqrt[3]{4}\approx 4.7622$$, $$f(-1)=3(-1)^{2/3}[2(-1)+5]=9$$, $$f(0)=3(0)^{2/3}[2(0)+5]=0$$, $$f(1)=3(1)^{2/3}[2(1)+5]=21$$. Finally, we conclude that the largest of these values, $$21$$, is the global maximum value of $$f$$, while the smallest, $$0$$, is the global minimum value of $$f$$.
Subject:Algebra
TutorMe
Question:
Find all real values of $$x$$ that satisfy the equation $$2e^{6x}=3+5e^{3x}$$.
Inactive
Ali K.
Answer:
We observe that $$e^{6x}=(e^{3x})^2$$, so the given equation may be written as $$2(e^{3x})^2=3+5e^{3x}$$. Letting $$z=e^{3x}$$, we have $$2z^2=3+5z$$. Rearranging, we get $$2z^2-5z-3=0$$. Factoring, we have $$(2z+1)(z-3)=0$$. Therefore $$z=-\frac{1}{2}$$ or $$z=3$$. That is, $$e^{3x}=-\frac{1}{2}$$ $$(*)$$ or $$e^{3x}=3$$ $$(**)$$. Since $$e^{3x}$$ is positive for all real $$x$$, we see that $$(*)$$ is inadmissible. Only $$(**)$$ is admissible. From it, we find $$3x=\ln(3)$$ and hence $$x=\frac{1}{3}\ln(3)$$.
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# Limiting the complexity: Composite function's calculus.
Welcome to Warren Institute! In today's article, we will explore the fascinating world of calculus and dive into the topic of finding the limit of a composite function. Calculus is an essential branch of mathematics that deals with rates of change and accumulation, making it crucial for understanding various real-world phenomena. By combining different functions, composite functions provide us with a powerful tool to analyze complex relationships. Join us as we unravel the intricacies of limits and discover the methods to evaluate them for composite functions. Let's embark on this exciting journey together!
## Introduction to Composite Functions
Composite functions are an important concept in calculus that involve combining two or more functions to create a new function. In this section, we will explore the basics of composite functions and how they can be used to find the limit of a composite function.
Key points:
• Composite functions involve applying one function to the output of another function.
• Notation for composite functions is typically denoted as (f ∘ g)(x), read as "f composed with g of x."
• Understanding composite functions is crucial for understanding limits of composite functions.
## Evaluating Limits of Composite Functions
To find the limit of a composite function, we need to evaluate the limit of the inner function and then apply the outer function to the result. This process involves substituting the value of the variable into the inner function and simplifying the expression before applying the outer function.
Key points:
• Start by evaluating the limit of the inner function as x approaches the given value.
• Simplify the expression as much as possible before applying the outer function.
• Applying the outer function means plugging in the result of the inner function into the outer function.
## Using Algebraic Techniques
Algebraic techniques can be employed to find the limit of a composite function when direct substitution fails. These techniques involve manipulating the expression algebraically to remove any indeterminations or difficulties in evaluating the limit.
Key points:
• If direct substitution leads to an indeterminate form (such as 0/0 or ∞/∞), try factoring or simplifying the expression.
• Rationalizing the numerator or denominator can help simplify the expression and make it easier to evaluate the limit.
• If all else fails, consider using L'Hôpital's rule to find the limit.
## Understanding the Behavior of Composite Functions
To fully grasp the concept of finding the limit of a composite function, it is important to understand how composite functions behave in different scenarios. This understanding can aid in effectively evaluating limits and predicting the behavior of composite functions in various mathematical contexts.
Key points:
• Analyze the behavior of the individual functions involved in the composite function.
• Pay attention to the domain restrictions and any potential issues with the composition of functions.
• Use graphical representations or numerical approximations to gain insight into the behavior of the composite function.
### What is a composite function in calculus?
A composite function in calculus is a combination of two or more functions, where the output of one function becomes the input of another. It is denoted as f(g(x)), where g(x) represents the inner function and f() represents the outer function.
### How do you find the limit of a composite function?
To find the limit of a composite function, you apply the concept of limits to each individual function within the composite. First, evaluate the inner function as x approaches the desired value. Then, use the resulting value as the input for the outer function and evaluate it as x approaches the same desired value. This process allows you to determine the overall limit of the composite function.
### Are there any specific rules or techniques for finding the limit of a composite function?
Yes, there are specific rules and techniques for finding the limit of a composite function. One important technique is the composition rule, which states that if the limit of a function f(x) as x approaches a exists, and the limit of another function g(x) as x approaches that same value a exists, then the limit of the composite function f(g(x)) as x approaches a also exists and is equal to the limit of f evaluated at the limit of g. This rule allows us to simplify complex composite functions and find their limits by evaluating simpler limits. Other techniques, such as factoring, rationalizing, or using trigonometric identities, may also be used depending on the specific function.
### Can you provide an example of finding the limit of a composite function step by step?
Sure! Let's consider the composite function f(g(x)) where f(x) = 2x and g(x) = x^2. To find the limit as x approaches a specific value, let's say a, we first evaluate the inner function g(x) at x = a. So, g(a) = a^2.
Then, we substitute this result into the outer function f(x). We have f(g(a)) = f(a^2) = 2(a^2).
Finally, we can take the limit of this expression as x approaches a. The limit is given by lim (x->a) 2(a^2) = 2(a^2).
Thus, we have found the limit of the composite function f(g(x)) step by step.
### How does finding the limit of a composite function relate to real-world applications in mathematics?
Finding the limit of a composite function is crucial in real-world applications of mathematics. By determining the behavior of a function as it approaches a certain point or value, we can make predictions and solve problems in various fields. For instance, in physics, calculating the limit of a composite function helps us understand the motion of objects and the rate at which they change. Additionally, in economics, limits provide insights into optimization problems, such as maximizing profits or minimizing costs. Furthermore, in engineering, limits help evaluate the stability and performance of systems. Overall, understanding the limit of composite functions plays a fundamental role in analyzing and modeling real-world phenomena, enabling us to make informed decisions and solve complex problems.
In conclusion, understanding how to find the limit of a composite function is a fundamental concept in calculus. By breaking down the problem into smaller parts and using the properties of limits, one can successfully evaluate the limit of a composite function. It is important to remember the key steps, such as applying the chain rule and simplifying expressions, in order to arrive at the correct answer. Practicing with various examples and seeking additional resources can further enhance one's proficiency in this topic. Overall, mastering the techniques discussed in this article will not only aid in solving complex calculus problems but also contribute to a solid foundation in mathematics education. Embrace the power of limits and unleash your potential in calculus!
|
# Lesson 6
Problemas con grupos iguales de fracciones
## Warm-up: Verdadero o falso: Dos y tres factores (10 minutes)
### Narrative
The purpose of this True or False is to elicit strategies and understandings students have for finding products of a whole number and a fraction and identifying equivalent expressions. This work help students deepen their understanding of the properties of operations and will be helpful later when students solve problems with a whole number multiplied by a fraction.
In this activity, students have an opportunity to look for and make use of structure (MP7) as they consider how fractions are decomposed into various factors and multiplied in parts.
### Launch
• Display one statement.
• “Hagan una señal cuando sepan si la afirmación es verdadera o no, y puedan explicar cómo lo saben” // “Give me a signal when you know whether the statement is true and can explain how you know.”
• 1 minute: quiet think time
### Activity
• Share and record answers and strategy.
• Repeat with each statement.
### Student Facing
En cada caso, decide si la afirmación es verdadera o falsa. Prepárate para explicar tu razonamiento.
• $$\frac{10}{12} = 5 \times \frac{2}{12}$$
• $$1 \times \frac{10}{12} = 5 \times \frac {2}{12}$$
• $$\frac{24}{4} = 6 \times 3 \times \frac{1}{4}$$
• $$12 \times 2 \times \frac{1}{4} = 8 \times 3 \times \frac{1}{4}$$
### Activity Synthesis
• “¿Qué estrategias usaron para decidir si las afirmaciones eran verdaderas o falsas?” // “What strategies did you use to determine if the statements were true or false?”
## Activity 1: Receta de pan de banano (15 minutes)
### Narrative
The purpose of this activity is for students to use their understanding of multiplication of a unit fraction by a whole number to solve problems. Students use what they know to find a product given the factors and find the factors when given the product. This reinforces the idea that any fraction $$\frac{a}{b}$$ is a multiple of $$\frac{1}{b}.$$
Students may interpret quantities greater than 1 as a combination of whole numbers and fractions (for example, $$\frac{4}{3}$$ cups as 1 whole cup and $$\frac{1}{3}$$ cup) or express them as mixed numbers (such as $$1\frac{1}{3}$$). Both are acceptable. If possible, ask students whether $$1\frac{1}{3}$$ and $$\frac{4}{3}$$ express the same amount, but it is not necessary to discuss the term mixed numbers at this point. (Students will be introduced to mixed numbers in upcoming lessons.)
### Launch
• Groups of 2
• “¿Alguna vez han usado una receta para preparar algo? ¿Qué hay en una receta?” // “Have you followed a recipe to make something before? What is in a recipe?” (A list of ingredients, amounts of each, and instructions for putting the ingredients together)
• “Si tienen una receta para 5 porciones o para 5 personas, pero necesitan preparar una mayor cantidad, ¿qué harían?” // “If a recipe is for 5 servings or 5 people, but you need more than that, what would you do?” (Adjust the amount of ingredients.)
• “En general, decimos que con las cantidades que aparecen en una receta preparamos ‘1 tanda’” // “We often refer to the amounts specified in a recipe as ’1 batch’.”
• “¿Qué significa preparar 2 tandas de una receta?” // “What might it mean to make 2 batches of a recipe?” (Make twice as much, or need twice as much ingredients)
### Activity
• “En silencio, trabajen unos minutos en la actividad. Luego, discutan con su compañero cómo pensaron” // “Take a few quiet minutes on work on the activity. Then, discuss your thinking with your partner.”
• 5 minutes: independent work time
• 5 minutes: partner work time
• Monitor for students who discuss:
• that each quantity in Monday’s table will be multiplied by 2
• that each quantity in Tuesday's table need to be multiplied by 4 because the amount of butter tells us that the number of batches is 4
### Student Facing
En una panadería preparan pan de banano. Esta es la receta para preparar 1 tanda.
Receta: 1 banano $$\frac{2}{3}$$ de taza de mantequilla $$\frac{3}{2}$$ cucharaditas de bicarbonato de sodio $$\frac{5}{8}$$ de taza de azúcar 2 huevos grandes $$\frac{5}{2}$$ tazas de harina común
1. El lunes prepararon 2 tandas de pan de banano en la panadería. Completa la tabla para mostrar la cantidad que se usó de cada ingrediente.
Pan de banano del lunes
bananos _______
mantequilla _______ taza(s)
azúcar _______ taza(s)
huevos _______
harina _______ taza(s)
2. El martes necesitaron $$\frac{8}{3}$$ tazas de mantequilla para hacer suficiente pan de banano para el día. ¿Cuántas tandas prepararon? Explica o muestra tu razonamiento.
Receta: 1 banano $$\frac{2}{3}$$ de taza de mantequilla $$\frac{3}{2}$$ cucharaditas de bicarbonato de sodio $$\frac{5}{8}$$ de taza de azúcar 2 huevos grandes $$\frac{5}{2}$$ tazas de harina común
3. Teniendo en cuenta el número de tandas que prepararon el martes, completa la cantidad de cada ingrediente en la tabla.
Pan de banano del martes
bananos _______
mantequilla $$\frac{8}{3}$$ tazas
azúcar _______ taza(s)
huevos _______
harina _______ taza(s)
### Student Response
If students are unsure how to complete the last table, check if they recognize that the given $$\frac{8}{3}$$ cups represent the amount of butter for 4 batches. If so, ask: “¿Cuántos bananos se necesitan para preparar 4 tandas?” // “How many bananas are needed in 4 batches?” and “¿Cuántos huevos se necesitan?” // “How many eggs are needed?” “¿Cómo te puede ayudar esto a encontrar los ingredientes que se necesitan para preparar 1 tanda, 2 tandas y así sucesivamente?” // “How might you use this to help to determine the ingredients needed for 1 batch, 2 batches and so on?”
### Activity Synthesis
• Display the table for Monday and ask students to share responses. Record their responses for all to see.
• “¿Cómo cambia el numerador en las cantidades de todos los ingredientes?” // “How is the numerator changing in all of the ingredients?” (It is multiplied by 2 in each problem.)
• “¿Por qué el denominador es diferente en las cantidades de todos los ingredientes?” // “Why is the denominator different in all of them?” (A different unit and unit fraction was used to measure each ingredient.)
• Ask students to share the expressions for the ingredients in the table for Tuesday. Record each expression and its value as an equation:
• $$4 = 4 \times 1$$
• $$\frac{8}{3} = 4 \times \frac{2}{3}$$
• $$\frac{12}{2} = 4 \times \frac{3}{2}$$
• $$\frac{20}{8} = 4 \times \frac{5}{8}$$
• $$8 = 4 \times 2$$
• $$\frac{20}{2} = 4 \times \frac{5}{2}$$
• “¿Por qué hay dos cantidades de ingredientes que no están en forma de fracción?” // “Why are two of the ingredients not in fraction form?” (They have whole-number units.)
## Activity 2: ¿Cuánta leche se usó? (20 minutes)
### Narrative
In this activity, students are presented with descriptions of situations and equivalent multiplication expressions. They match each description to an expression that could represent the situation and see that more than one expression can be used, depending on how they interpret the situation. Likewise, students find that one expression can be used to represent different descriptions (MP2).
Students discuss their matching decisions, analyze how the expressions are related, and consider revising the matches they made if appropriate. When students discuss and justify their decisions they are creating viable arguments and critiquing one another’s reasoning (MP3).
MLR7 Compare and Connect. Synthesis: Lead a discussion comparing, contrasting, and connecting the different representations. Ask, “¿Cómo se ve la situación en la representación?” // “How does the situation show up in the representation?”, “¿Qué tienen en común todas estas representaciones?” // “What do each of these representations have in common?”, and “¿En qué son diferentes?” // “How were they different?”
Engagement: Develop Effort and Persistence. Invite students to generate a list of shared expectations for the group work in this activity. Record responses on a display and keep visible during the activity.
Supports accessibility for: Social-Emotional Functioning
### Required Materials
Materials to Gather
### Required Preparation
• Write the 5 expressions from the activity on separate posters and post them around the room:
$$4 \times (2\times \frac{1}{10})$$
$$4 \times \frac{2}{10}$$
$$8 \times \frac{1}{10}$$
$$2 \times (4 \times \frac{1}{10})$$
$$2 \times \frac{4}{10}$$
### Launch
• Groups of 2
• Read the first problem aloud to students.
• “Compartan con su compañero la expresión que escogieron” // “Share the expression you selected with a partner.”
• Students may select any of the expressions because each is equivalent to $$\frac{8}{10}$$. If this happens, ask, “¿Alguna expresión parece representar la situación mejor que las demás?” // “Does one expression seem to represent what is happening in the situation better than others?” ($$8 \times \frac{1}{10}$$)
### Activity
• 5 minutes: independent work time
• Ask students to stand with the poster showing the expression that they believe represents how much milk was used on Tuesday, and to discuss with others there why they chose this expression.
• Ask students to partner with a student from a different poster to explain why they made a different choice.
• “¿Alguien quiere reconsiderar lo que pensó sobre la expresión que escogió?” // “Does anyone wish to revise their thinking about the expression they selected?”
• “Expliquen por qué la nueva expresión que escogieron es mejor que la que tenían antes” // “Can you explain why you think that a different expression is a better choice now?”
• Repeat this process for each problem.
### Student Facing
En la panadería, además de pan de banano, también venden malteadas frescas. Cada malteada contiene $$\frac{1}{10}$$ de litro de leche.
Estas son cinco descripciones de las malteadas que se venden durante una semana y cinco expresiones que representan los litros de leche que se usan.
Empareja cada descripción con una expresión que la represente.
1. El lunes, en la panadería vendieron 8 malteadas. ¿Cuánta leche se usó?
2. El martes, dos clientes compraron 4 malteadas cada uno. ¿Cuánta leche se usó?
3. El miércoles, cuatro clientes compraron 2 malteadas cada uno. ¿Cuánta leche se usó?
4. El jueves, dos clientes compraron una malteada cada uno. Ese día, cada uno de ellos hizo el mismo pedido otras tres veces, para sus amigos. ¿Cuánta leche se usó?
5. El sábado, cuatro amigos compraron una malteada cada uno, para el desayuno. Después de la cena, volvieron y compraron lo mismo. ¿Cuánta leche se usó?
$$4 \times (2\times \frac{2}{10})$$
$$4 \times \frac{2}{10}$$
$$8 \times \frac{1}{10}$$
$$2 \times (4 \times \frac{1}{10})$$
$$2 \times \frac{4}{10}$$
### Student Response
If students do not see that each factor in the expressions can be interpreted in different ways, or that different expressions could represent the same quantity, invite them to use diagrams or record their thinking using diagrams to illustrate various groupings of the same quantity. Consider asking: “¿Dónde ves _____ grupos de _____?” // “Where do you see _____ (whole number) groups of _____ (fraction)?”
### Activity Synthesis
• See lesson synthesis.
## Lesson Synthesis
### Lesson Synthesis
“Hoy asociamos expresiones con situaciones. Aprendimos que muchas expresiones pueden representar una misma situación” // “Today, we matched expressions to situations. We learned that several expressions can represent the same situation.”
Invite 1–2 students who chose different expressions for the same problem (one of the last two problems in the milkshake activity) to share. Record their ideas for all to see.
“¿Alguien puede explicar la forma en la que cada expresión le corresponde al problema?” // “Who can explain how each expression matches the problem?” (On Thursday, there were 4 separate orders of 1 serving each, or $$4 \times \frac{1}{10}$$, that were made by 2 people, or $$2 \times (4 \times \frac{1}{10})$$. This is also the same as $$2 \times \frac{4}{10}$$.)
“¿Observaron algo sobre las respuestas a los problemas?” // “Did you notice something about the answers to the problems?” (They are all the same. They are all $$\frac{8}{10}$$.)
“¿Por qué creen que todas son la misma?” // “Why do you think they are all the same?” (They all involve 8 groups of $$\frac{1}{10}$$.)
## Student Section Summary
### Student Facing
En esta sección, aprendimos a multiplicar un número entero por una fracción pensando en grupos de igual tamaño, como lo hicimos cuando multiplicamos dos números enteros.
Por ejemplo, podemos pensar en $$6 \times 4$$ como 6 grupos de 4. Un diagrama como este nos ayuda a mostrar que el producto es 24:
De la misma manera, podemos pensar en $$6 \times \frac{1}{4}$$ como 6 grupos de $$\frac{1}{4}$$. Los diagramas nos pueden ayudar a entender que el producto es $$\frac{6}{4}$$:
Después de estudiar patrones, vimos que cuando multiplicamos un número entero por una fracción, el número entero se multiplica únicamente por el numerador de la fracción y se deja el mismo denominador. Por ejemplo:
$$6 \times \frac{1}{2} = \frac{6}{2}$$
$$2 \times \frac{4}{5} = \frac{8}{5}$$
También aprendimos que:
• Todas las fracciones se pueden escribir como un producto de un número entero y una fracción unitaria. Por ejemplo, $$\frac{5}{4}$$ se puede escribir como $$5 \times \frac{1}{4}$$.
• Podemos escribir diferentes expresiones de multiplicación para representar la misma fracción. Por ejemplo, $$\frac{8}{3}$$ se puede escribir así:
$$8 \times \frac{1}{3}$$
$$4 \times 2 \times \frac{1}{3}$$
$$4 \times \frac{2}{3}$$
$$2 \times \frac{4}{3}$$
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# What is differential calculus examples?
## What is differential calculus examples?
Differential calculus is used to determine if a function is increasing or decreasing. Integral calculus is used to find areas, volumes, and central points. Example: Differentiate f(x) = x3. f'(x) = 3×2. Example: Integrate f(x) = x3.
### What is relation in differential calculus?
A differential equation is a relation between a collection of functions and their derivatives. An ordinary differential equation is a differential equation that relates functions of one variable to their derivatives with respect to that variable.
#### What is the example of function in calculus?
Sometimes functions are most conveniently defined by means of differential equations. For example, y = sin x is the solution of the differential equation d2y/dx2 + y = 0 having y = 0, dy/dx = 1 when x = 0; y = cos x is the solution of the same equation having y = 1, dy/dx = 0 when x = 0.
What is the differential of a function used for?
The derivative of a function can often be used to approximate certain function values with a surprising degree of accuracy. To do this, the concept of the differential of the independent variable and the dependent variable must be introduced.
How is differential calculus used in real life?
Biologists use differential calculus to determine the exact rate of growth in a bacterial culture when different variables such as temperature and food source are changed.
## What is an example of a function in everyday life?
A car’s efficiency in terms of miles per gallon of gasoline is a function. If a car typically gets 20 mpg, and if you input 10 gallons of gasoline, it will be able to travel roughly 200 miles.
### What is function give example?
A few more examples of functions are: f(x) = sin x, f(x) = x2 + 3, f(x) = 1/x, f(x) = 2x + 3, etc. There are several types of functions in maths. Some important types are: Injective function or One to one function: When there is mapping for a range for each domain between two sets.
#### What is the difference between differential equation and differential calculus?
Calculus is the mathematics of change, and rates of change are expressed by derivatives. Thus, one of the most common ways to use calculus is to set up an equation containing an unknown function y=f(x) and its derivative, known as a differential equation.
What is the difference between differential calculus and integral calculus?
While differential calculus focuses on rates of change, such as slopes of tangent lines and velocities, integral calculus deals with total size or value, such as lengths, areas, and volumes.
What is relation and function?
The relation shows the relationship between INPUT and OUTPUT. Whereas, a function is a relation which derives one OUTPUT for each given INPUT. Note: All functions are relations, but not all relations are functions.
## What is the difference between functional and differential calculus?
Function is a relation between two variables that inhibits an apparent connection. If the variables are x and y, then y can be determined for some range of values of x. We call this, y as a function of x denoted by y = f ( x ). Differential Calculus is limited only to those relations that are functions defined by equations.
### What are some examples of differential calculus problems and solutions?
Problems and Solutions. Go through the given differential calculus examples below: Example 1: f (x) = 3x 2 -2x+1. Solution: Given, f (x) = 3x 2 -2x+1. Differentiating both sides, we get, f’ (x) = 6x – 2, where f’ (x) is the derivative of f (x).
#### What is the use of derivative in differential calculus?
Derivatives. The fundamental tool of differential calculus is derivative. The derivative is used to show the rate of change. It helps to show the amount by which the function is changing for a given point. The derivative is called a slope. It measures the steepness of the graph of a function.
What is the difference between relation and function?
Not all relations are function but all functions are relation. A good example of a relation that is not a function is a point in the Cartesian coordinate system, say (2, 3). Though 2 and 3 in (2, 3) are related to each other, neither is a function of the other. Function is a relation between two variables that inhibits an apparent connection.
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# Speed, velocity and acceleration
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## Transcription
1 Chapter Speed, velocity and acceleration Figure.1 What determines the maximum height that a pole-vaulter can reach? 1 In this chapter we look at moving bodies, how their speeds can be measured and how accelerations can be calculated. We also look at what happens when a body falls under the influence of gravity..1 Speed In everyday life we think of speed as how fast something is travelling. However, this is too vague for scientific purposes. Speed is defined as the travelled in unit. It can be calculated from the formula: speed Units The basic unit of is the metre and the basic unit of is the second. The unit of speed is formed by dividing metres by seconds, giving m/s. An alternative unit is the kilometre per hour (km/h) often used when considering long s. _phys_1_.indd 1 6/11/8 1:18:55
2 Speed, velocity and acceleration WORKED EXAMPLES An athlete runs at a steady speed and covers 6 m in 8. s. Calculate her speed. speed 6 8. m/s 7.5 m/s Measurement of speed We can measure the speed of an object by measuring the it takes to travel a set. If the speed varies during the journey, the calculation gives the average speed of the object. To get a better idea of the instantaneous speed we need to measure the travelled in a very short. One way of doing this is to take a multi-flash photograph. A light is set up to flash at a steady rate. A camera shutter is held open while the object passes in front of it. Figure. shows a toy car moving down a slope. QUESTIONS.1 A car travels m in 8. s. Calculate its speed.. A cricketer bowls a ball at 45 m/s at a batsman 18. m away from him. Calculate the taken for the ball to reach the batsman. Figure. <ph_> NOW ARTWORK PLEASE SUPPLY BRIEF Successive images of the car are equal s apart, showing that the car is travelling at a constant speed. To find the speed, we measure the between two images and divide by the between each flash. Acceleration So far we have looked at objects travelling at constant speed. However, in real life this is quite unusual. When an object changes its speed it is said to accelerate. If the object slows down this is often described as a deceleration. _phys_1_.indd 13 6/11/8 1:18:58 13
3 Figure.3 shows a multi-flash photograph of the toy car rolling down a steeper slope. This its speed increases as it goes down the slope it is accelerating. Figure.3 <ph_3> NOW ARTWORK PLEASE SUPPLY BRIEF Figure.4 Distance changing at a steady state. Figure.5 Increasing s with travelled. Using graphs Distance graphs Graphs are used a lot in science and in other mathematical situations. They are like pictures in a storybook, giving a lot of information in a compact manner. We can draw graphs for the two journeys of the car in Figures. and.3. In Figure. the car travels equal s between each flash, so the total travelled increases at a steady rate. This produces a straight line as shown in Figure.4. The greater the speed, the steeper the slope (or gradient) of the line. In Figure.3 the car travels increasing s in each interval. This leads to the graph shown in Figure.5, which gradually curves upwards. The graph in Figure.6 shows the story of a journey. The car starts at quite a high speed and gradually decelerates before coming to rest at point P. P QUESTIONS.3 Describe the journeys shown in the diagrams below. Figure.6 Story of a car journey. 14 _phys_1_.indd 14 6/11/8 1:18:58
4 Speed, velocity and acceleration Speed graphs Instead of using a graph to look at the travelled over a period of we can look at how the speed changes. Figure.7 appears similar to Figure.4. However closer inspection shows that it is the speed which is increasing at a constant rate, not the. This graph is typical for one in which there is a constant acceleration. In this case the gradient of the graph is equal to the acceleration. The greater the acceleration the larger the gradient. The graph in Figure.8 shows the story of the speed on a journey. This is a straight-line graph, with a negative gradient. This shows constant deceleration, somes described as negative acceleration. Using a speed graph to calculate travelled speed Rearrange the equation: speed Look at Figure.9. The object is travelling at a constant speed, v, for t. The travelled v t We can see that it is the area of the rectangle formed. Now look at Fig..1, which shows a journey with constant acceleration from rest. The area under this graph is equal to the area under the triangle that is formed. The travelled 1_ v t 1_ v is the average speed of the object and travelled is given by average speed, so once again the travelled is equal to the area under the graph. The general rule is that the travelled is equal to the area under the speed graph. WORKED EXAMPLES Use the graph in Figure.11 to calculate the travelled by the car in the interval from.5 s to 4.5 s. Time passed (4.5.5) s 4. s Initial speed m/s Final speed 1 m/s In this case, the area under the line forms a triangle and the area of a triangle is found from the formula: area 1_ base height area under the graph the travelled 1_ 4. 1 m 4 m speed Figure.7 Speed changing at steady rate. speed Figure.8 Story of speed on a journey. speed v t Figure.9 Area under graph of constant speed. speed v t Figure.1 Area under graph of constant acceleration (s) Figure.11 Distance travelled by a car. _phys_1_.indd 15 6/11/8 1:18:59 15
5 Figure.1 The lap of the track is 3. m, and the car completes a full lap in 6. s. The average speed of the car is 5. m/s. However its average velocity is zero! Velocity is a vector and the car finishes at the same point as it started, so there has been no net displacement in any direction. S. Velocity Velocity is very similar to speed. When we talk about speed we do not concern ourselves with direction. However, velocity does include direction. So an object travelling at 5 m/s due south has a different velocity from an object travelling at 5 m/s northwest. It is worth observing that the velocity changes if the speed increases, or decreases, or if the direction of motion changes (even if the speed remains constant). There are many quantities in physics which have direction as well as size. Such quantities are called vectors. Quantities, such as mass, which have only size but no direction are called scalars..3 Acceleration We have already introduced acceleration as occurring when an object changes speed. We now explore this idea in more detail. If a body changes its speed rapidly then it is said to have a large acceleration, so clearly it has magnitude (or size). Acceleration can be found from the formula: acceleration Units change in velocity taken The basic unit of speed is metres per second (m/s) and the basic unit of is the second. The unit of acceleration is formed by dividing m/s by seconds. This gives the unit m/s. This can be thought of as the change in velocity (in m/s) every second. You will also notice that the formula uses change of velocity, rather than change of speed. It follows that acceleration can be not only an increase in speed, but also a decrease in speed or even a change in direction of the velocity. Like velocity, acceleration has direction, so it is a vector. WORKED EXAMPLES 1 A racing car on a straight, level test track accelerates from rest to 34 m/s in 6.8 s. Calculate its acceleration. change of velocity Acceleration (final velocity initial velocity) (34 ) m/s m/s It is important that the track is straight and level or it could be argued that there is a change of direction, and therefore an extra acceleration. 16 _phys_1_.indd 16 6/11/8 1:19:
6 Speed, velocity and acceleration A boy on a bicycle is travelling at a speed of 16 m/s. He applies his brakes and comes to rest in.5 s. Calculate his acceleration. You may assume the acceleration is constant. change of velocity Acceleration (final velocity initial velocity) ( 16) m/s m/s Notice that the acceleration is negative, which shows that it is a deceleration. Calculation of acceleration from a velocity graph Look at the graph in Figure.13. We can see that between 1. s and 4. s the speed has increased from 5. m/s to 1.5 m/s. (1.5 5) Acceleration m/s (4 1) m/s.5 m/s Mathematically this is known as the gradient of the graph. Gradient increase in y increase in x (s) Figure.13 Velocity graph. We see that acceleration is equal to the gradient of the speed- graph. It does not matter which two points on the graph line are chosen, the answer will be the same. Nevertheless, it is good practice to choose points that are well apart; this will improve the precision of your final answer. QUESTIONS.4 Describe the motion of the object shown in the graph in Figure a) Describe the motion of the object in shown in the graph in Figure.15. b) Calculate the travelled by the object. S c) Calculate the acceleration of the object Figure (s) Figure.15 _phys_1_.indd 17 6/11/8 1:19:1 17
7 Figure.16 shows a multi-flash photograph of a steel ball falling. The light flashes every.1 s. We can see that the ball travels further in each interval, so we know that it is accelerating. Figure.17 shows the speed graph of the ball. S (s) Figure.17 Speed graph of falling steel ball. The graph is a straight line, which tells us that the acceleration is constant. We can calculate the value of the acceleration by measuring the gradient. Use the points (.1,.5) and (.45, 3.9). (3.9.5) Gradient m/s (.45.1) s m/s 9.7 m/s The acceleration measured in this experiment is 9.7 m/s. All objects in free fall near the Earth s surface have the same acceleration. The recognised value is 9.8 m/s, although it is quite common for this to be rounded to 1 m/s. The result in the above experiment lies well within the uncertainties in the experimental procedure. This is somes called the acceleration of free fall, or acceleration due to gravity, and is given the symbol g. In Chapter 3 we will look at gravity in more detail. We will also look, in Chapter 3, at what happens if there is significant air resistance.. Figure.16 Falling steel ball. QUESTIONS.6 An aeroplane travels at a constant speed of 96 km/h. Calculate the it will take to travel from London to Johannesburg, a of 9 km. 18 _phys_1_.indd 18 6/11/8 1:19:
8 Speed, velocity and acceleration.7 Describe what happens to speed in the two journeys described in the graphs a) b).8 Describe how the speed changes in the two journeys described in the graphs. a) speed b) speed.9 A motorist is travelling at 15 m/s when he sees a child run into the road. He brakes and the car comes to rest in.75 s. Draw a speed graph to show the deceleration, and use your graph to calculate a) the travelled once the brakes are applied b) the deceleration of the car. S.1 A car accelerates from rest at m/s for 8 seconds. a) Draw a speed- graph to show this motion. b) Use your graph to find (i) the final speed of the car (ii) the travelled by the car..11 The graph shows how the speed of an aeroplane changes with. 4 B C 3 1 A (s) a) Describe the motion of the aeroplane. b) Calculate the acceleration of the aeroplane during the period B to C. c) Suggest during which stage of the journey these readings were taken. _phys_1_.indd 19 6/11/8 1:19: 19
9 Summary Now that you have completed this chapter, you should be able to: define speed recall and use the equation speed understand that acceleration is a change of speed draw and interpret - graphs draw and interpret speed- graphs calculate travelled from a speed- graph recognise that the steeper the gradient of a speed- graph the greater the acceleration recognise that acceleration of free fall is the same for all objects S understand that velocity and acceleration are vectors change in velocity recall and use the equation acceleration calculate acceleration from the gradient of a speed- graph describe an experiment to measure the acceleration of free fall. _phys_1_.indd 6/11/8 1:19:3
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Graphing & Vector Quantities:
Objectives:
1. Explain how graphs can be used to describe relationships between time, distance, and speed.
2. Given a graph, the student will be able to determine whether acceleration, constant speed, or deceleration is occurring.
3. Distinguish between a scalar and a vector quantity. (Be able to identify examples.)
4. Draw vector diagrams for velocities and be able to solve simple situations involving vector addition and subtraction.
Key Terms:
scalar quantity vector quantity resultant vector resolution
Graphs are traditionally used to visually express the relationship between two or more different but related quantities. The graph is constructed using the x-axis (horizontal) as the independent predictable reference and the y axis (vertical) as the dependant measured variable. You will notice in the three graphs that follow, while all displaying the same situation, the independent value (time) remains constant while the dependant value changes the shape and the meaning of the graph. The examples used is compare speed, velocity, & acceleration. The SI units of speed and velocity are m/s while acceleration is m/s2.
Speed (m/s) time (s) 10 1 20 2 30 3 40 4 50 5
When labeling your charts and graphs it important to use the proper units. The connecting line on your graph points should be the "best fit" or line bisecting the median between the data points.
This is a graph depicting constant acceleration and is displaying speed over time. As you can see, the graph is characterized by a straight line with a positive slope (slant increasing from left to right). The slope gives you a good idea of the rate of acceleration. The graph displays the property of linearity meaning that the variables used are in direct proportion.
Distance (m) time (s) 5 1 20 2 45 3 80 4 125 5 180 6 245 7
The graph is now dealing with the distance covered by an object during free fall. You will notice that the line now is parabolic (curves upward) with an increasing positive slope.
acceleration (m/s2) time (s) 9.8 1 9.8 2 9.8 3 9.8 4 9.8 5 9.8 6
The graph now displays the acceleration during free fall. The slope is now zero neither increasing nor decreasing. Acceleration is constant.
Top
A vector quantity is one that requires both magnitude and direction to adequately describe it. A scalar quantity, though, requires only magnitude. Examples of vectors include velocity, acceleration, weight, and momentum. Examples of a scalar unit would include speed, mass, volume, and time neither of which requires a direction to adequately it.
Vectors can be displayed using vector diagrams. Consider the following example.
50m/s
5m/s wind
What would be the result of the two opposing vectors? [
If the wind switched to a tail wind the result would be 55m/s. [
Vectors can also be directed at angles. Consider this plane and the wind traveling at right angles to each other.
The black lines are the vertical and horizontal components of the final velocity (green). The blue line is the resultant or the diagonal of the rectangle described by the two vectors. The resultant can be calculated using the following formula:
(resultant)2 = (vertical)2 + (horizontal)2
(5)2 = (4)2 + (3)2
Top
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Study Materials
# NCERT Solutions for Class 11th Mathematics
Page 1 of 4
## Chapter 2. Relations and Functions
### Exercise 2.1
Exercise 2.1
Comparing both side as order pairs are equal, so corresponding elements also will be equal,
Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).
Solution:
Given : n(A) = 3 and set B = {3, 4, 5}
∴ n(B) = 3
Number of elements in (A × B)
= (Number of elements in A) × (Number of elements in B)
= n(A) × n(B)
= 3 × 3
= 9
Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Solution:
Given: G = {7, 8} and H = {5, 4, 2}
We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as P × Q = {(p, q): p∈ P, q ∈ Q}
∴ G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
Q4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.
Solution:
(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.
False
If P = {m, n} and Q = {n, m},
P × Q = {(m, m), (m, n), (n, m), (n, n)}
Solution:
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(ii) True
Solution:
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.
(iii) True
Q5. If A = {–1, 1}, find A × A × A.
Solution:
It is known that for any non-empty set A, A × A × A is defined as;
A × A × A = {(a, b, c): a, b, c ∈ A}
Given that A = {–1, 1}
∴ A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}
Q6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.
Solution:
The cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}
Given that: A × B = {(a, x),(a , y), (b, x), (b, y)}
∴ a, b ∈ A abd x, y ∈ B
So, A = {a, b} and B = {x, y}
Q7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}.
Verify that:
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × C is a subset of B × D.
Solution:
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
∴ L.H.S. = A × (B ∩ C) = A × Φ = Φ
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
∴ R.H.S. = (A × B) ∩ (A × C) = Φ
∴ L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) To verify: A × C is a subset of B × D
Solution:
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
A × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
All the elements of set A × C are the elements of set B × D.
Therefore, A × C is a subset of B × D.
Q8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
Solution:
A = {1, 2} and B = {3, 4}
∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
⇒ n(A × B) = 4
We know that if C is a set with n(C) = m,
then n[P(C)] = 2m.
Therefore, the set A × B has 24 = 16 subsets.
These are Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}
Q9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.
Solution:
Given that n(A) =3 and n(B) = 2;
and (x, 1), (y, 2), (z, 1) are in A × B.
We know that A = Set of first elements of the ordered pair elements of A × B
B = Set of second elements of the ordered pair elements of A × B.
∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.
Since n(A) = 3 and n(B) = 2,
∴ A = {x, y, z} and B = {1, 2}.
Q10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.
Solution:
We know that if n(A) = p and n(B) = q,
then n(A × B) = pq.
∴ n(A × A) = n(A) × n(A)
It is given that n(A × A) = 9
∴ n(A) × n(A) = 9 ⇒ n(A) = 3
The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A×A.
We know that A × A = {(a, a): a ∈ A}.
Therefore, –1, 0, and 1 are elements of A.
Since n(A) = 3,
it is clear that A = {–1, 0, 1}.
The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1).
Page 1 of 4
Chapter Contents:
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#### Param Publication
Operation on Integers
(i) When adding integers with like signs (both positive or both negative), add their absolute values, and place the common sign before the sum.
(ii) When adding integers of unlike signs, find the difference of their absolute values, and give the result the sign of the integer with the larger absolute value.
(iii) When the addition and subtraction signs are placed side by side without any number in between, these two opposite signs give a negative sign.
Subtraction : Subtraction is reverse operation of addition.
Ex.1 Consider 8 – 5. Actually we have to subtract + 3 from 8. So, we need to find a number which when added to 3 gives 8.
sol: The answer 5, i.e., 8 – 3 = 8 + (– 3) = 5
We can change subtraction to addition by adding the additive inverse of the second number to the first number. In the above example –3 is the additive inverse of +3 and vice versa is also possible.
Ex.2 Find the sum of (– 9) + (+ 4) + (– 6) + (+ 3)
Sol. We can rearrange the numbers so that the positive integers and the negative integers are grouped together. We have
(– 9) + (+ 4) + (– 6) + (+ 3) = (– 9) + (– 6) + (+ 4) + (+ 3) = (– 15) + (+ 7) = – 8
Ex.3 Find the value of (30) + (– 23) + (– 63) + (+ 55)
Sol. (30) + (+ 55) + (– 23) + (– 63) = 85 + (– 86) = – 1
Ex.4 Subtract (– 4) from (– 10)
Sol.: (– 10) – (– 4) = (– 10) + (additive inverse of – 4) = –10 + 4 = – 6
Ex.5 Subtract (+ 3) from (– 3)
Sol.: (– 3) – (+ 3) = (– 3) + (additive inverse of + 3) = (– 3) + (– 3) = – 6
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# Cuisenaire Rod Fractions: Level 3
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Purpose
This unit introduces the idea that fractions come from equi-partitioning of one whole. Therefore, the size of a given length can be determined with reference to one whole. When the size of the referent whole varies, then so does the name of a given length.
Achievement Objectives
NA3-1: Use a range of additive and simple multiplicative strategies with whole numbers, fractions, decimals, and percentages.
NA3-5: Know fractions and percentages in everyday use.
Specific Learning Outcomes
• Name the fraction for a given Cuisenaire rod with reference to one (whole).
• Find the one (whole) when given a Cuisenaire rod and its fraction name.
• Create a number line showing fractions related to a given one (whole).
• Identify equivalent fractions
Description of Mathematics
‘Fractions as measures’ is arguably the most important of the five sub-constructs of the concept of rational number (Kieren, 1994) since it identifies fractions as numbers, and is the basis of the number line. Fractions are needed when ones (wholes) are inadequate for a given purpose (e.g. division). In measurement, lengths are defined by referring to some unit that is named as one. When the size of another length cannot be accurately measured by a whole number of ones, then fractions are needed.
For example, consider the relationship between the brown and orange Cuisenaire rods. If the orange rod is defined as one (an arbitrary decision) then what number is assigned to the brown rod?
Some equal partitioning of the one is needed to create unit fractions with one as the numerator. For the size of the brown rod to be named accurately, those unit fractions need to fit into it exactly. We could choose to divide the orange rod into tenths (white rods) or fifths (red rods). By aligning the unit fractions we can see that the brown rod is eight tenths or four fifths of the orange rod.
Note that eight tenths and four fifths are equivalent fractions and the equality can be written as 8/10 = 4/5. These fractions are different names for the same quantity and share the same point on a number line. The idea that any given point on the number line has an infinite number of fraction names, is a significant change from thinking that occurs with whole numbers. For the set of whole numbers, each location on the number line matches a single number. Some names are more privileged than others by our conventions. In the case of four fifths, naming it as eight tenths aligns to its decimal (0.8) and naming it as eighty hundredths aligns to its percentage (80/100 = 100%).
#### Specific Teaching Points
Understanding that fractions are always named with reference to a one (whole) requires flexible thinking. Lamon (2007) described re-unitising and norming as two essential capabilities if students are to master fractions. Re-unitising enables students to flexibly define a given quantity in multiple ways by changing the units they attend to. Norming enables students to operate with the new unit. In this unit of work, Cuisenaire rods are used to develop students’ skills in changing units and thinking with those units.
Consider this relationship between the dark green and blue rods. Which rod is one? Either could be defined as one and the other rod could be assigned a fraction name.
If the blue rod is one then the dark green rod is two thirds, as the light green rod is one third. If the dark green rod is one then the blue rod is three halves since the light green rod is now one half.
Re-unitising and norming are not just applicable to defining a part to whole relationships like this. In this unit students also consider how to use re-unitising to find the referent one and to name equivalent fractions. For example, below the crimson rod is named as two fifths. Which rod is the one (whole)? If the crimson rod is two fifths, then the red rod is one fifth. Five fifths (red rods)form the whole. Therefore, the orange rod is one.
What other names does two fifths have? If the red rods were split in half they would be the length of white rods, and be called tenths since ten of them would form one. The crimson rod is equal to four white rods which is a way to show that 2/5 = 4/10. If the red rods were split into three equal parts the new rods would be called fifteenths since 15 of them would form one. The crimson rod would be equal to six of these rods which is a way to show 2/5 = 6/15. The process of splitting the unit fraction, fifths in this case, into equal smaller unit fractions, produces an infinite number of fractions for the same quantity.
The learning opportunities in this unit can be differentiated by providing or removing support to students, by varying the task requirements. Ways to support students include:
• providing Cuisenaire rods for students to manipulate when solving problems
• modelling how to record fraction symbols and drawing attention to the meaning of numerator and denominator
• drawing diagrams to clarify the unit of comparison and the one (whole) in problems
• encouraging students to work collaboratively, especially where some students are affected by colour blindness.
Tasks can be varied in many ways including:
• altering the complexity of the rod relationships that students work with. Working with halves and quarters tends to be easier than with thirds and fifths
• providing 1cm2 grid paper and coloured felt pens to ease the recording demands (Cuisenaire rods are based on that scale).
The contexts for this unit can be adapted to suit the interests and cultural backgrounds of your students. Cuisenaire rods (rakau) are often used in the introduction of te reo Māori, meaning they may be familiar to some students. Knowing the relationships between rods of different colours, without having assigned number names to the rods, is very helpful in easing cognitive load. Other contexts involving fractions of lengths might also be engaging for your students. For example, the fraction of a race or journey that has been covered at different points is practically useful. This could be linked to the early journeys of Māori and Pasifika navigators to Aotearoa, or to current journeys your students have experienced (e.g. a bus ride to camp, running a lap of the playground). Consuming foods that are linear, such as submarine sandwiches, bananas, or sausages, might motivate some learners. Board games that have a particular number of steps from start to finish provide opportunities to look at a fraction as an operator.
Te reo Māori vocabulary terms such as hautau (fraction), hautau waetahi (unit fraction), hautau waetahi-kore (non-unit fraction), rākau Ātaarangi (Cuisenaire rods), hautau ōrite (equivalent fractions), rārangi tau (number line), and the names for individual fractions could be introduced in this unit and used throughout other mathematical learning.
Required Resource Materials
Activity
#### Prior Experience
Students may have mixed experiences with using Cuisenaire rods. When introducing the Cuisenaire rods, ask students to think about what they could be used to represent in mathematics. Value the contributions of all students.
#### Session One
1. Use Cuisenaire rods or the online tool to introduce the relative size of Cuisenaire rods in the following way. Provide whiteboards or paper, or use a large chart or whiteboard, to record students thinking.
Relative to the orange rod, how long is the yellow rod? How do you know? Justify
The relationship between the yellow and orange rods can be expressed in two ways:
“The yellow rod is one half of the orange rod.”
“The orange rod is two times the length of the yellow rod.”
If the orange rod was one, then the yellow rod would represent one half.
What fraction would the red rod and dark green rod represent? Justify.
2. Encourage the students to express the relationships in various ways, such as:
“The red rod is one fifth of the orange rod because five of it fit into the whole (one)”
“The orange rod is five times longer than the red rod.”
“So the dark green rod must be three fifths of the orange rod because three red rods make one dark green rod.”
A more complex question is “How many dark green rods (three-fifths) fit into the orange rod (one)?” While the correct answer is five-thirds, or one and two thirds, students will be unlikely to name the relationship that precisely. Expect answers like “Almost two but not quite.”
3. Introduce Investigation One using Slide 1 of the PowerPoint. Encourage students to record both their names for each rod (relative to the brown rod) and their reason for naming it that way. Provide sets of Cuisenaire rods or access to the online tool. Let the students collaborate in small groups (mahi tahi). Consider pairing together more knowledgeable students with less knowledgeable students to encourage tuakana-teina (peer learning). Look for the following:
• Do the students refer back to the brown rod as the one?
• Do they name each rod with reference to how many times it fits into one?
• Do they use the relationship between rods to name them? (For example, if pink is one half then red must be one quarter and white must be one eighth).
• Can they name a rod larger than one as an improper fraction or mixed number? (For example, the orange rod now represents one and one quarter (1 1/4 or 5/4).
4. All of the points above can be raised in discussion as a whole class. Extend the conversation to which rods were hardest to name and why that was so. For example, the light green rod does not fit into the brown rod an exact number of times but the white rod (one eighth) can be used as a reference.
5. Also discuss equivalence. The diagram below shows 1/2 = 2/4 = 4/8. Explain that equivalent fractions are different names for the same quantity.
#### Session Two
1. Revise the key points from the previous session using the blue rod as one.
If the blue rod is one what do we call the light green and white rods? Justify your answers.
What statements can you make about the relative size of the rods?
Are there equivalent fractions in the picture (1/3 = 3/9)? So what fraction is equivalent to… two thirds? (2/3 = 6/9), … to three thirds? (3/3 = 9/9).
Students might notice some patterns in the symbols such as the same multiplier between numerators and denominators in the equalities.
3. Reflect back on fractions where the rod was larger than one. Ask: If blue is one then what fraction is the orange rod?
Thinking that fractions are restricted to less than one is a common constraint students learn. Therefore, opportunities to name fractions greater than one (i.e. mixed or improper fractions) is important. This thinking could be supported by making links to different ‘whole’ items that are different sizes (e.g. two different waka). Linking this learning to contexts that are relevant to your students will increase the level of meaning they can see within this unit. Just like the Cuisenaire rods, they are different sizes, but can still be classified as ‘one whole’. Students might recognise that the white rod fills the gap between the blue and orange rods.
Useful questions are:
• Remember, which rod is one?
• So what fraction is the white rod? (1/9)
• How many white rods fit into the blue rod? (nine)
• How many white rods fit into the orange rod? (ten)
• So what shall we call the orange rod? (1 1/9 or 10/9)
4. Ask students to attempt Investigation Two of the PowerPoint (Slide 2). Remind them of the necessity for recording their solutions and justifications.
5. As they investigate in small groups, roam and look for:
• Do they accept the new imaginary rod as one?
• Do they name the other rods as unit fractions in terms of how many of that rod fit into one?
• Do they know how to name non-unit fractions using copies of unit fractions? E.g. Three quarters (blue rod) is three copies of one quarter (light green).
• Do they realise that equivalent fractions are different names for the same quantity?
6. Share the results as a class attending to the points above.
7. Construct a fraction wall with the gold rod as one. Name each unit fraction (1/2 ,1/4 ,1/12 ,1/3 ,1/6). Ask if these are the only unit fractions that are possible and why that is so. Students may note that the denominators are all factors of 12. Look for equivalence in the fractions within the wall. Encourage students to find non-unit fraction equivalence as well, e.g. 2/3 = 8/12 and 3/4 = 9/12.
8. Use the wall to create a number line as shown. Ask:
How much more three quarters is than two thirds?
How much less one half is than two thirds?
9. As a class, come up with a real-life context that could be represented by a fraction wall. It does not have to be completely realistic, but it should reflect the relevant learning interests and/or an element of the socio-cultural backgrounds of your class. For example, the one whole is the number of students in the class, ½ want to go to the beach for a class trip and ½ want to go to the museum. ¼ of the students who want to go to the beach want to swim, and ¼ want to build sandcastles. As you develop this scenario, use groups of students to represent the division of the “whole” (i.e. the whole class) into different fractions. This will present opportunities for students to compare the sizes of different fractions, in relation to the number of students in your class.
10. Let students work on Investigation Three from the PowerPoint. Their work will extend into Session Three. Look for the following:
• Can students name the fractions for the rods that are being joined?
• Can they record the combinations as sums like, 1/2 + 1/3 + 1/6 = 1?
• Can they use equivalence, particularly referring to twelfths, to explain why the combinations add to one?
#### Session Three
In this session the purpose is to reconstruct the one rod. Students connect from part to whole as opposed to whole to part.
1. Begin by going over previous ideas in the context of this model.
2. Ask: What are the size relationships between the yellow and black rods? Allow students to discuss this in pairs. Some students may want to use the relevant rods to model their thinking.
The students might use the white rod as a reference to say, “The yellow rod is five sevenths of the black rod.” It is more difficult to recognise that “The black rod is seven fifths of the yellow rod.” The key idea is to establish the referent one. If a comparison ‘of a given rod’ is being made then that rod becomes the one.
3. Ask: So if you were told that the yellow rod was five sevenths of the one rod, what colour would the one rod be? (black).
If you were told that the black rod was seven fifths of the one rod, what colour would the one rod be? (yellow)
4. Provide another scenario. If you were told that the pink rod was one half of the one rod, what colour would the one rod be?
Students might easily recognise that two halves make one so the rod colour of one is brown. This is an easy scenario as a unit fraction is given. Therefore, ask a harder problem like this:
If you were told that the dark green rod was two thirds of the one rod, what colour would the one rod be?
The dark green rod does not fit exactly into the mystery one but half of it does. That half of the green rod is the light green rod (one third). So the one rod must be blue.
5. Ask the students to complete Copymaster 1 in pairs or threes. Point out the need to justify their decisions about which rod is one in each case. Students need to use Cuisenaire rods or the online tool for this activity. They should not rely on the pictures being to scale. Look for:
• Do they adjust to the variable one in each case?
• Do they use the given rod as a unit of measure?
• Do they subdivide the visible rod to find a unit fraction they can measure with? For example if told the rod is two thirds, do they divide the rod equally into two parts to create a one third measure?
6. If students complete Copymaster One, ask them to create similar part to whole problems for other students.
#### Session Four
The aim of this session is to develop students’ mental number line for fractions. Inclusion of fractions with whole numbers on the number line requires some significant adjustments. Activities using a number line could be completed digitally (e.g. on a PowerPoint or flipchart), or with the use of a whiteboard or paper chart. These adjustments include:
• A point on the number line can have an infinite number of names called equivalent fractions, for example, 2/3 ,4/6 ,6/9 … all ‘live’ at the same point.
• Between any two fractions are an infinite number of other fractions (this is known as ‘density’ of the number line).
1. Begin by building up a number line for quarters in this way.
If the brown rod is one (mark zero and one on the number line) where would one quarter be?
Students may now know that the red rod is one quarter of the brown rod. Ask: What fractions could be marked on the number line using one quarter? Look for them to explain that quarters can be ‘iterated’ (place end on end) to form non-unit fractions. Make sure you push the iteration past one and include the fraction and mixed number ways to represent the amount (see below). Also encourage renaming in equivalent form where this is sensible, for example, 2/4 = 1/2, 4/4 = 1.
2. Look at the space between one quarter and one half. Ask, “Are there any fractions that belong in this space?” Students may recognise from previous work that white rods are one eighth of a brown rod. Three eighths will work. Note that three eighths measure exactly half-way between one quarter and one half. Ask, “What fraction would belong half way between one quarter and three eighths?” (five sixteenths). The last question requires students to use their imagination, as there is no rod that is half the length of a white rod.
3. Show the students the diagram on Slide 4 of the PowerPoint. It shows zero and two fractions (orange rod as two thirds and blue rod as three fifths placed on a number line. Ask, “How could we find the length of the one rod?” From the part-whole task in the previous session students should reply that one third or one-fifth need to be located by equally partitioning the orange rod in two parts to get the yellow rod or equally partitioning the blue rod into three parts to get the light green rod. These unit fractions can then be iterated to get the referent one (three yellow rods or five light green rods).
4. Ask students to create a number line with the orange rod as two thirds and the blue rod as three fifths. They must include one and any other fractions they can find. In the event of early finishers to this infinite task, provide the following challenge fractions to locate on the number line: 4/3, 9/5, 3/15, 14/15, 2/9. Look for the following:
• Do the students use fifths and thirds as unit fractions to locate other non-unit fractions, like four fifths?
• Do the students find fractions greater than one by iterating unit fractions?
• Do the students record equivalent fractions in the same location, for example five fifths and three thirds at one?
• Do the students subdivide unit fractions to form other units, for example divided thirds into two equal parts to form sixths?
• Do the students attempt to identify fractions between fractions, for example, which fractions lie between two thirds and four fifths?
5. Bring the class together after a suitable period of investigation to share results. You will need a large number line on the whiteboard. By the time the discussion is over you should saturate the number line with fractions. You may like to ask if it ever possible to complete this task. Students may already realise that there are an infinite number of fractions that could be located. As an extension, you could refer back to the whole-class wall fraction you created, and represent the different fractions on a number line.
6. After discussing the fractions that can be located on the above number line ask the students to make up a similar number line problem for someone else. The problem must include enough fractions already placed to locate the referent one and at least six other fractions to be located on the line. They are free to choose whatever rod they want as the one and may even create a rod that is not in the set. Challenge them to frame their problem in relation to the context which framed your whole-class fraction wall.
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viewquestions/11711/The side of a square is 70cm Find its area and perimeter Mathematics cbse std 6th Perimeter and Area
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## Take a thick paper sheet of length 14 cm and width 9 cm. You can also use an old postcard. (a) What is its area? What is its perimeter? (b) Now cut strips of equal sizes out of it. (c) How long it your belt? (d) What is its Perimeter? (e) Whose belt is the longest in the class? - class 5 math
class 5 math chapter 11
By:milan-ransingh
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Ans. (a) We know that,
Area of a rectangle = Length * Breadth
Area of a thick paper sheet = (Length * Breadth)
= (14*9) square cm = 126 square cm
Perimeter of a rectangular sheet
= 2(14 + 9) cm
= 2 * 23 cm = 46cm
(b) Let us cut strips of equal sizes out of the given paper sheet of length 14 cm and width 9 cm.
(c) Strips having width 1 cm:
There will be 9 strips of width 1 cm and length 14 cm. Let us join these strips end to end using tape to make a belt.
Its length
= (14+14+14+14+14+14+14+14+14)
= 9 * 14 cm = 126 cm
It’s perimeter = 2(126+1) cm = 2 127 cm = 254 cm
(d) Strips having length 1.5 cm:
There will be 6 strips of width 1.5 cm and length 14 cm. Let us join these strips end to end using tape to make a belt.
It’s length = (14+14+14+14+14+14) cm
= 6 * 14 cm = 84 cm
It’s perimeter = 2 (84 + 1.5) cm = 2* 85 – 5 cm = 171 cm
(e) Whose belt is the longest in the class?
Ans. (e) Strips having width 3 cm:
There will be 3 strips of width 3 cm and length 14 cm.
Let us join these strips end to end using tape to make a belt.
It’s length = (14+14+14) cm = 3 * 14 cm = 42 cm
It’s perimeter = 2(42+3) cm = 2 * 45 cm = 90 cm
Clearly, the belt with shortest width is the longest one. Thus, the belt having width 1 cm is the longest in the class. So, my belt is the longest.
pankaj
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Describe the sampling distribution for sample proportions and use it to identify unusual (and more common) sample results.
Learning Objectives
• Describe the sampling distribution for sample proportions and use it to identify unusual (and more common) sample results.
Introduction
In this module, Linking Probability to Statistical Inference, we work with categorical variables, so the statistics and the parameters will be proportions. In the module Inference for Means, we work with quantitative variables, so the statistics and parameters will be means. In the Big Picture, we see that inference is based on probability. In this module, we begin the process of developing a probability model to describe the long-run behavior of proportions from random samples.
After we develop a probability model of how sample proportions behave, we can answer questions like the following:
• Do the majority of college students qualify for federal student loans?
• What proportion of all college students in the United States are enrolled at a community college?
The questions ask us to make an inference about a population. Our answers to these questions will be based on a sample. We will never be 100% sure of our answer, so we will make probability statements that describe the strength of the evidence and our certainty.
Brief Discussion of the Connection between These Questions and Probability
Do the majority of college students qualify for federal student loans?
• This question asks us to test a claim about college students. The claim is “the majority of college students qualify for federal student loans.” To test this claim, suppose we select a large random sample of college students and find that 40% of the sample qualify for these loans. A majority requires over 50%; 40% is definitely not a majority. Can we conclude from this sample that our claim is incorrect? Or could this sample have come from a population the majority of which qualify for loans? What is the probability that sample proportions will be 0.40 or less if the majority in the population qualify?
What proportion of all college students in the United States are enrolled at a community college?
• This question asks us to estimate a population proportion. Suppose we select a large random sample of college students and find that 46% are enrolled at a community college. What is the probability that an estimate from a sample is within 3% of the population proportion?
Note: Connected to each inference question about a population proportion, we see a probability question about the long-run behavior of sample proportions. We need to understand how proportions from random samples relate to the population proportion. We also need to understand how much variability we can expect in sample proportions. Therefore, in our early investigations, we will assume we know a population proportion and examine what happens when we select random samples from this population.
Now we begin an investigation of the long-run behavior of sample proportions.
Gender in the Population of Part-time College Students
According to a 2010 report from the American Council on Education, females make up 57% of the U.S. college population. With the rising costs of education and a poor economy, many students are working more and attending college part time. We anticipate that if we look at the population of part-time college students, a larger percentage will be female. Let’s say we predict that 60% of part-time college students are female.
We don’t have information about the population of part-time college students, so we select a random sample of 25 part-time college students and calculate the proportion of the sample that is female. We don’t expect the sample proportion to be exactly 0.60. So, how much could the sample proportion vary from 0.60 for us to feel confident in our prediction?
To answer this question, we need to understand how much sample proportions will vary if the parameter is 0.60.
Learn By Doing
Refer to the previous example for the following questions. These questions focus on how the proportion of females will vary in random samples if we assume that 0.60 of the population of part-time college students is female.
Use the following simulation to select a random sample of 25 part-time college students. Repeat the selection many times to observe how the proportion of females in the samples vary. Then answer the following question.
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Saturday, February 24th, 2018
# SAT Math: How To Improve Your Score On Algebra I Problems
After taking the May 5th SAT exam and now preparing for the next sitting on June 2nd, it is prudent to evaluate your performance in the Math section and target the types of questions and Math areas that you need to improve upon. Doing a personal assessment on what algebraic problems, such as solving equations with inequalities, word problems, exponentials, roots and factoring, are posing a challenge is one thing you could do to drive up your score in the SAT Math section.
It is helpful to know what to do when you encounter the following algebraic concepts and their question types:
• When simplifying fractions involving large numbers, find out if a given number will divide evenly into both the numerator and denominator. For example, if a number is divisible by 2, then its last digit is 0, 2, 4, 6, or 8. If a number is divisible by 3, then the sum of the digits is divisible by 3, and so forth.
• When solving ‘letter-heavy’ problems, translate the word problem or sentence into an algebraic equation. Firstly, solve for the letter variable you have information for, and then replace the other letters with numbers to determine the steps needed to get to the solution. If two equations in the system are alike, you can sometimes solve them easily by combining equations.
• Balancing equations as a law of equality: what ever you add, subtract, multiply, divide or square on one side of the equation, you must do the same to the other side to maintain the balance.
• When working with exponentials,
• add the exponents when multiplying or subtract the exponents when dividing and leave the bases the same;
• multiply (or divide) the bases and leave the exponents alone if the exponents are the same.
• When factoring polynomials, think of ‘distribution in reverse.’ This means that you can check your factoring by distributing or FOILing the factors to make sure that the result is the original expression.
• The Zero Product Property: If the product of a set of numbers is 0, then at least one of the numbers in the set must be 0. And the only product property is the Zero Product Property.
• Absolute values as distances: The absolute value of x, written as |x|, means the distance from x to zero on the number line. Since distances are never negative, they are not absolute values.
The above key points and approaches to algebra I problems will help you tackle these types of questions better. To be more effective with using the above rules, do more of these kinds of SAT Math problems.
### One response to “SAT Math: How To Improve Your Score On Algebra I Problems”
1. Carole bajus says:
I would like to have Ryker take this prep course
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# Even Numbers - Definition, Properties, Addition, Subtraction, Multiplication and Even and Odd Differences
## What is an Even Number?
Any integer in the number system, which when divided by two yields no remainder, is said to be an even number. This implies that the numbers are even if they are exactly divisible by the number 2. The one-digit examples of even numbers are 2, 4, 6, and 8. The two-digit, three digits numbers etc., are even, if they contain 0, 2, 4, 6, and 8 in their unit’s place. All these numbers are even numbers because they leave no remainder on division with 2. The numbers which counter the even numbers are called odd numbers and are defined as the integers which leave some remainder on division with 2. They also yield a decimal value and are not perfectly divisible. Examples of such numbers are 3, 5, 13, 11, 21, etc.
## Knowing Even Numbers Practically
Let us assume that Ram has 8 apples. If he groups them in pairs, he will have 4 pairs of apples with no apple left unpaired. That means the remainder is zero. From this, we can deduce that 8 is perfectly divisible by 2, with the quotient being 4. This experiment proves that if an even number is divided by 2, then the remainder will always be zero (no remainder left). Likewise, if he pairs them as 3 then he will form 2 groups of 3 apples, and 1 group will be left with 2 apples. This implies that 8 is not completely divisible by odd numbers. Even numbers are only multiples of 2.
## Difference between Even and Odd
To inspect whether the numbers are either even or odd, we need to carefully check the unit’s place of the numbers. The numbers having 0, 2, 4, 6, and 8 as their unit digits will be called even numbers, while the numbers containing 1, 3, 5, 7, and 9 are considered as odd numbers. Let us consider a random number, say 315, since it has 5 in its unit place and 5 is an odd number; therefore, the number is also an odd number. Similarly, 11, 223, 877, and 69 are odd numbers.
## Properties of Even Numbers
Even numbers share a lot of properties. Let us now learn about various arithmetic properties of even numbers such as addition, subtraction, etc.
• Even numbers add up to yield another even number. For example, 20 + 6 = 26
• The summation of an even number with an odd number always results in an odd number. For example, 2 + 11 = 13
• The addition of two odd numbers yields an even number. For example, 27 + 3 = 30
## Subtraction Property
• Similar to addition, two even numbers subtract to form an even number. For example, 40 - 2 = 38
• The difference between an even number and odd number results in the formation of an odd number. For example, 19 – 2 = 7.
• Subtracting an odd number with another odd number gives an even number. For example, 41 – 11 = 30
## Multiplication Property
Two even numbers multiplied by each other yields an even number, and the product of an even and an odd number is also an even number. For example, 4 * 6 = 24, 3 * 8 = 24.
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# Negative Exponents
Negative exponents are exponents that have a negative value. They indicate that the base of a number should be inverted or taken to the reciprocal.
For example, the expression x^(-2) is the same as 1/x^2 or the reciprocal of x squared. Negative exponents can represent very small or very large numbers, typically by multiplying a coefficient by 10 raised to a negative power.
• Negative Exponents Rules
• What is the Difference Between Negative Exponents and Positive Exponents?
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Negative exponents also use in more abstract mathematical contexts, like complex numbers and polynomials. Also, it can be used for solving equations that involve physical units and other advanced mathematical concepts.
It’s important to understand the properties and rules for working with negative exponents to evaluate and simplify expressions that involve them properly.
## Negative Exponents Rules
There are a few rules for exponents that are only for negative exponents. Here are some rules working with negative exponents that you should be familiar with:
### 1. a^(-n) = 1/a^n
This rule states that when you have a negative exponent, you can simplify the expression to get the solution by taking the reciprocal of the base raised to the positive exponent.
### 2. (a^n)^(-m) = a^(-nm)
This rule states that when you have a term raised to a power, and the power itself is raised to another power, you can simplify the expression by dividing the exponent by the power.
### 3. (ab)^n = a^n * b^n
This rule states that when you have a product of two or more bases raised to a power, you can simplify the expression by raising each base to the power individually.
### 4. (a/b)^n = a^n / b^n
This rule states that when you have a quotient of two bases raised to a power, you can simplify the expression by raising each base to the power individually.
### 5. a^n * a^m = a^(n+m)
This rule states that when you have the product of two or more powers of the same base, you can simplify the expression by adding the exponents.
It’s important to remember these rules and practice applying them so that you can easily simplify and evaluate expressions that involve negative exponents. Also, practicing these rules by attempting worksheets or sample questions can help you cement your understanding of the concept well, whether it is the multiplication of negative exponent or any other application.
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## What is the difference between negative and positive exponents?
The main difference between negative and positive exponents is how they affect a number’s base.
A positive exponent indicates that a number is multiplied by itself several times. Let’s see an example step by step, 2^3 = 2*2*2 = 8, which means 2 is multiplied by itself 3 times.
On the other hand, a negative exponent indicates that the reciprocal of a number is being multiplied by itself a certain number of times. The reciprocal of a number is the number flipped upside down, 1/number. For example, 2^(-3) = 2*2*2 = 1/8, which means the reciprocal of 2 is multiplied by itself 3 times.
Another way to understand negative exponents is that it’s the same as taking the reciprocal of the number with the positive exponent. So, for example 2^3 = 8 and 2^(-3) = 1/8. They both are reciprocal to each other.
So, in simple terms, positive exponents are used to represent a number being multiplied by itself, while negative exponents are used to representing the reciprocal of that same number being multiplied by itself.
Gloria Mathew writes on math topics for K-12. A trained writer and communicator, she makes math accessible and understandable to students at all levels. Her ability to explain complex math concepts with easy to understand examples helps students master math. LinkedIn
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×
RRMTRX2 - Editorial
Practice
Contest
Author: Roman Rubanenko
Tester: Tuan Anh
Editorialist: Florin Chirica
easy
simple math
PROBLEM:
We take, for each column, exactly one element and multiply what we've got. Output the sum of all possible choices.
QUICK EXPLANATION:
One can notice that answer is col[1] * col[2] * ... * col[m], where col[x] = sum of all elements from column x.
EXPLANATION:
Let's learn from an example
Usually, when you don't know what to do, a good idea is to take an example to feel better the problem. So, let's take
n = 3 m = 3 and matrix
1 2 3
4 5 6
7 8 9
By now, let's focus on sums which must contain cells 1 and 2. This is 1 * 2 * 3 + 1 * 2 * 6 + 1 * 2 * 9 = 1 * 2 * (3 + 6 + 9). Let's focus now on those having cells 1 and 5. By the same logic, we get 1 * 5 * 3 + 1 * 5 * 6 + 1 * 5 * 9 = 1 * 5 * (3 + 6 + 9). Finally, considering sums containing cells 1 and 8. We get 1 * 8 * (3 + 6 + 9).
Let's sum up this. We get (3 + 6 + 9) * (1 * 2 + 1 * 5 + 1 * 8) = (3 + 6 + 9) * 1 * (2 + 5 + 8). It seems to smell like a pattern, isn't it?
Let's consider sums which containg 4. We get (after some calculations) value 4 * (2 + 5 + 8) * (3 + 6 + 9). We do the same for cells containing 7. We get 7 * (2 + 5 + 8) * (3 + 6 + 9).
Now, we add all sums containing either 1, 4 or 7 and two other elements from column two and three. These are all possible sum. Let's add. We get 1 * (2 + 5 + 8) * (3 + 6 + 9) + 4 * (2 + 5 + 8) * (3 + 6 + 9) + 7 * (2 + 5 + 8) * (3 + 6 + 9) = (1 + 4 + 7) * (2 + 5 + 8) * (3 + 6 + 9).
Now we really got a pattern. Let col[j] = sum of all elements from column j. It turns out the answer is col[1] * col[2] * ... * col[m].
Proof
We've made an assumption. Let's try to proof it. If we can do this, we're done. The problem goes as following: for each column j, we choose 1 element from a[1][j], a[2][j], ... a[n][j]. We multiplicate them, then we do the sum.
Let's see what happens when we do (a[1][1] + a[2][1] + ... + a[n][1]) * (a[1][2] + a[2][2] + ... + a[n][2]) * .... * (a[1][m] + a[2][m] + ... + a[n][m]). If we expand the expression, at each step exactly one element will be chosen from each bracket. This is correct, it's equivalent to choosing one element from each column. The corresponding terms will be multiplied, and then these results will be added. This is exactly what problem asks us for, so we're done.
AUTHOR'S AND TESTER'S SOLUTIONS:
To be updated soon
Author's solution to be updated soon
Tester's solution
This question is marked "community wiki".
3★elfus ♦♦
0112527
accept rate: 0%
19.8k350498541
3 Didn't expected Matrix to have negative numbers, got WA during contest submitted the same code by just adding "if(sum<0) sum += mod" and got AC. Needed careful attention, nicely set problem statement by the setter than the problem itself, though a little hint could have been found in output description. answered 22 Dec '14, 01:03 81●1●5 accept rate: 0%
2 Source Code for Implementation of above Problem in 30 Lines of code Explanation We basically want to find a path that starts from a[i][0] and ends at a[iDash][m-1], where i,iDash ∈[0,n) . Now to complete the path, we would have a total of m points. As we traverse our path, jth (i.e column) index increases from 0 to m-1. For every j (i.e column) we have a point P->a[i][j] such that i∈[0,n) . So, in all, we would be having pow(n,m) paths. For each path we need to calculate the product of its points, and then add the result for various paths. For some random path, let us be at point P->a[i][j] and variable prod be the product of points covered till now. We now have to go to next point Q->a[iDash][j+1], where iDash ∈ [0,n). So we have n ways to do so. So we use a recursive function that takes all possible values of iDash, and for each value taken, updates our product variable prod, and takes to next column. After reaching (m-1)th column, we have completed a path, and ready to repeat the problem for next path. answered 22 Dec '14, 00:44 130●7 accept rate: 0% Your solution would give TLE. Complexity = order of (pow(n,m)). And also you have not considered that A[i][j] can be negative. (01 Mar '15, 01:09)
1 Why do we need long long for the variable to store the answer.. we are doing mod in every step right... then why is long not suffecient?? getting ac with long long, but wa using long.. pls explain answered 22 Dec '14, 01:34 204●3●12 accept rate: 7%
0 practice link please answered 22 Dec '14, 00:24 52●1●2●10 accept rate: 7% 1 http://www.codechef.com/problems/RRMTRX2 (22 Dec '14, 00:57) arpn4★
0 for those who need a simple implementation here is link to my solution.. http://www.codechef.com/viewsolution/5629656 answered 22 Dec '14, 00:59 1.7k●1●17●30 accept rate: 11%
0 answered 22 Dec '14, 01:33 1●1 accept rate: 0%
0 Hello, I got the logic in like 2 minutes. But, I still could not understand the play of mod here. Why do we need to take mod at every step. won't that affect the correct answer. E.g. If I do this without mod and say my final answer is 10000009. So, what I thought is to put 2 as answer i.e. 10000009 % 10000007 = 2. It was my mistake that I did not consider any negative number scenario. But, still its not clear. Regards. answered 22 Dec '14, 10:05 1 accept rate: 0% The final answer is too big to print, or store in a long long variable for that matter. So they ask you to print answer % mod. And no, it won't affect your answer as modulus is distributive. (a+b) % c = a%c + b%c (you gotta take mod again if the sum exceeds c of course) (29 Dec '14, 16:42)
0 I think math symbols used in problem statement sometimes cause confusion and wastes time to interpret. I would suggest adding latex style formulas or expressions. For example mathjax allows to add math formulas in latex style: http://docs.mathjax.org/en/latest/start.html answered 22 Dec '14, 17:41 3★big_boss 1●1●1 accept rate: 0%
0 How can one figure that negative numbers can create a problem.. Actually i still didn't got the role of negative numbers in this problem.. Can anyone please explain?? Please.. answered 23 Dec '14, 02:40 1.9k●1●12●43 accept rate: 14% in programming languages, the '%' operator gives a negative integer or 0 as result. Try outputting (-5%3) in your compiler. However, this may result in incorrect answer if a negative number arises in intermediate steps. Hence instead of, say x%mod, we do (x%mod+mod)%mod... (23 Dec '14, 11:58) okay i am sorry i was not clear with my question.. Actually in short i didn't understood the output for this problem clearly.. I mean when we are adding all the numbers of columns and then multiplying them, so how do negative numbers come into existence..? (23 Dec '14, 16:03)
0 can anyone explain this test case 2 2 1 3 3 4 output comes 28 according to above theory . But the only possible vector seems to be [1,1] . May be i have not understood something. Please Clarify answered 24 Dec '14, 00:39 1●1 accept rate: 0% 16.9k●49●115●225 It is the same as in example test case: All possible vectors are {(1, 1), (1, 2), (2, 1), (2, 2)} (29 Dec '14, 16:48)
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## How many two digit numbers are there with two digits different?
Answer: The total number of two digit numbers is 90. From 1 to 99 there are 99 numbers, out of which there are 9 one-digit numbers, i.e., 1, 2, 3, 4, 5, 6, 7, 8 and 9. If one digit numbers are subtracted from 99 we get 90 two-digit numbers.
How many 2 digits are there?
90
Therefore, the total number of 2-digit numbers which can be obtained are 90.
How many 2-digit numbers are there from 100?
The number name of 75 is seventy-five. Example 3: Write the first four ‘2-digit numbers’ and the last four ‘2-digit numbers’. Solution: The first four two-digit numbers are: 10, 11, 12, 13. The last four two-digit numbers are: 96, 97, 98, 99.
### How many 2-digit numbers are there in 5*9?
(a) If repetition is allowed, we have 5 choices for the units digit (any of the 5 even digits); then we have 9 choices for the tens digit (any digit except 0). That makes 5*9 = 45 2-digit numbers. (b) If repetition is not allowed, then there are two cases to consider.
READ ALSO: Which mode is stronger Sage Mode KR Kurama mode?
How many 2-digit numbers can be formed from 4 choices?
It depends on whether repetition is allowed or not: (1) Repetition allowed : There are 4 choices for both positions, so [math]4 \imes 4 = 16[/math] 2-digit numbers can be formed. (2) Repetition not allowed : There are 4 options for tens position, and three options for ones position.
What is the total number of possible 2-digit even numbers?
Of the 10 digits, 5 are even. So in both cases there are 5 choices for the units digit of the 2-digit number. (a) If repetition is allowed, then there are 10-1 = 9 choices for the tens digit. (9 because you can not use zero as the “tens” digit). The total number of possible 2-digit even numbers is 9*5 = 45, if repetition is allowed .
## How many two-digit numbers are there in base 4?
What this really amounts to is base 4. We know that the total of one and two digit numbers in base 4 is 4 squared or 16. And we know the total number of single digit to be 4. Thus the total number of 2 digit numbers would be: The answer is 12 two digit numbers.
READ ALSO: What language did Bulleh Shah write in?
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The World is an Average
The simple average
An average is a concept that all of us are familiar with, and most have computed averages at school or work. An average is a single number which summarizes the a group of numbers. Some standard notation will help us as we expand the definition and uses of averages. Let A be the average of N data values, represented by the x's in the formula on the right. Each x is identified by a subscript, i. To get the common average we add all the N x's and divide by N (or multiply by 1/N, which is the same thing). The large Greek S stands for addition (or summing), and the initial and final values of i, the index, are indicated above and below the large Greek S.
There are two typical reasons for computing an average. First, we may want a characterization of a group, i.e a single number that represents this group. This average would seem like a good way to do this (but there are other parameters, the mode or median, that can be better in some situations, see your statistics textbook).
A very different motivation is to overcome a limitation in the accuracy of a measurement. You might have a piston from a car engine and want to measure the diameter as accurately as possible in order to determine is there has been any wear which would indicate you have to buy new pistons when you rebuild the engine. However, the micrometer you have isn't very good, and when you make a measurement you are never sure you are holding the micrometer perpendicular to the axis of the piston, etc. So, to get a little confidence you measure several times and take the average.
Distributions and weighted averages
We are going to take the measurement of the piston diameter quite seriously. First we measure it 283 times, second we graph the distribution of measurements to see what is going on. The diameter is just a little over 4 inches, and we are using a micrometer that measures to the 1/1000 th of an inch. To focus on the variation in measurement, we plot just the number of 1/1000 th of an inch over 4 inches. The result is graphed below.
The simple average for these values is about 12, as indicated by the vertical line in the graph. However, we see that the distribution of measurements is not at all symmetric about this average. The most frequent values are actually in the range of 5-10, and thus you might think that these represent the "true" value. The average of the data is 12 only because there are a small number of very high values.
I certainly don't know what is going on here. Maybe the piston is not round and so you get different values if it's rotated. Maybe I wasn't holding the micrometer perpendicular to the axis of the piston, and occasionally got a high value.
Let's assume for now that the measurements are valid and we want a better way of summarizing them. We then need a different average.
The simple average seems lacking both because it doesn't characterize the distribution, and because it is very sensitive to variations in a small number of observations of large values. Let's make the ad hoc proposal that the probability of obtaining a certain value for the measurement is proportional to the logarithm of the value. To see what this proposal changes, I have plotted the same distribution with a log scale horizontal axis (which is the same as plotting the distribution against the log of the value).
Now you see that the distribution is close to being symmetrical. The antilog of the average of the log values is about 6, indicated by the vertical line. This seems to me to be much better characterization for the diameter of the piston. The value of 6 is also the mode, or most common, of the measurements. Maybe this is a "good" average.
You must be having some doubts by now. First, why did I just "pull out of the hat" the idea of "distorting" the x axis of the distribution using the log function. I wish I was that original, but using the log of the independent value (the value plotted on the x axis) is an long established tactic. It's so common that it has a name, the geometric average. The simple average is more precisely called the arithmetic average (better terms might be linear and log averages, but we defer to tradition). The geometric average may seem arbitrary, but on reflection you will see that the arithmetic average is just as arbitrary.
Whenever a collection of data has a wide variation, with many values close to zero, and when a value can't be negative, you should try a log plot to see if the geometric average is appropriate. Often biologic data have these properties. One example is the concentration of antibodies that are produced as the result of a vaccination. Another example is the distribution of option prices with time (see work of Black, Scholes, and Merton which led to the 1997 Nobel Prize in Economics).
A second red flag is the thought: "if you can distort the x-axis of the distribution using the log function, why not another function?" Why not indeed. A comprehensive analysis of a collection of data with variation starts with the determination of the shape of the distribution. This shape will then suggest functions of the x-variable that will give a symmetrical distribution. Of course this requires a many data points to make precise conclusions, but there are certainly procedures for doing the analysis. Fortunately most of the time date is distributed in the symmetrical normal or Gaussian distribution, for which the arithmetic average is appropriate, as discussed in great detail in all statistics texts.
Take home lesson: you might have to "distort" the data to see the real information it contains.
Averages over time
We often have a time series that we are trying to understand. An example would be the daily closing price of a stock we are thinking of buying (or selling short). The graph below shows the prices of a stock for the first 20 trading days of the month.
The question is: is the rise in price seen during the first 15 days slowing during the last 5 days? Well, it is slowing, but is the trend "real"? This question usually means: will it continue?
The implicit assumption here is that there is a short term "random" fluctuation in price superimposed on a long term "real" trend. The obvious mathematical tool to discover the trend is an average. The question is: what kind of average and how do I do it?
One kind of time average is a "moving rectangular window". Here we take the arithmetic average of the price during the last few days, and plot that as another graph (not shown here). In this example "few" is five.
You can think of this process as looking at the raw data through a window five days wide, the blue rectangle superimposed on the graph.
However, an immediate question is: "how did you pick five instead of 2, 3, or 7 days". Doesn't it seem artificial to give as much weight to the price five days ago as the price today but no weight to the price 6 days ago?
Instead of giving the price from the past few days equal weight we can smoothly decrease the weight in proportion to the age of the data. An obvious choice is an exponentially decreasing window, as shown by the blue curve on the left. In this example the weight decreases by a factor of 1.4 (the square root of 2) every day.
You may not want to continue the window out to the distant past, it's just too much work, and the values more than 8 days in the past (in this example) don't contribute much to the average anyway. However wide the window, you do have to adjust the weighting factors so that their sum is one.
This decreasing average may seem more natural, since the influence of the old prices on the moving average gradually decreases. However, you still have to decide how fast the influence decreases, an arbitrary decision unless you know something about stock behavior
I have said nothing about the validity of using any of these averages to actually buy or sell stock. If I knew anything about that I would now be on my large yacht in the Mediterranean (I'm not).
Stock prices don't follow the laws of physics, or any other laws I know of. Some people use the terms of physics, e.g. momentum, to describe the behavior of stock prices. At best these are analogies, at worse nonsense. The mathematical and graphic tools I have described enable you to explore models of stock behavior (or whatever), but don't pretend to actually predict anything using basic laws.
A little notation
The process of looking at a set of data through a window is an example of a convolution. The data values and the values that define the window are twisted, or convoluted, with each other.
In this equation:
C's are the values of the convolution
d's are the values of the original data
w's are the N numbers that define the window
The window can also be thought of as a vector in N dimensional space, and then the value of each element of the convolution is the projection of the window vector on the data, or the dot product of the window with the data. These words don't add much to the discussion unless you are already quite familiar with vectors, but if you do they suggest a simple physical analogy.
Averaging in space
Now switch from a time series to a space series. In many cases the space series makes up an image. On the right we see a lens, it could be part of an eye or a camera, which is "looking" at a very small spot. As you can see, the spot does not form a completely truthful image, but rather is blurred out to make a smear. Let's call the spot the data, the lens the window, and the smear an average. Since the object is a point, the image is called the "point spread function".
Usually a lens system would be directed at a much more interesting and complicated object. The system could be on a rocket and it might be "looking" at the planet Mars. However, Mars, or any object, can be thought of as a series of "points of light", which together make up the image. If a single point of light is blurred, each of the points of light that make an image also are blurred, and thus the image, the sum of all the points, is also blurred. Any system that has the property that the final result is just the simple sum of all the components (in this case the sum of all the point spread functions of parts of the object). A non-linear system is very difficult to work with.
The lens has created a moving average, or a convolution over space, even if in this case we would rather have the raw data and not the average. But all sensory devices create an average, including our own eyes and ears. Thus, to us, and all other animals, the world is seen and heard as an average of the real world.
Deconvolution
The process of creating each point of a convolution consists of multiplying a set of data points by a known set of "window" points, and adding the results.
A natural question might be, can this process be undone? Since the convolution is a set of simultaneous linear equations, the original values, can in general be computed by solving this set.
On the left we see an image of a dog through an imperfect lens: the real image has been convoluted by the lens.
In order to obtain the deconvolution you need the "window" values. The window is the point spread function, which can be obtained from the image of a point. In this case the point spread function is known to be a Gaussian (because I created it using the Gaussian function).
As you might guess, deconvolution could be very important for many scientific and technical applications, and thus both the theoretical and practical aspects are well studied. A very powerful computational method uses the Fast Fourier Transform (FFT) of the image and the point spread function. The FFT of the image is a series of numbers that represent the relative amounts of sin waves of decreasing wavelengths that, when added, give the best representation of an image. It turns out that dividing the FFT of the blurred image by the FFT of the point spread function gives the FFT of the original image. Finally, the image can be calculated from its FFT.
The deconvoluted (or real) image of the dog is seen to the right.
Deconvolution might seem to be so easy that no one would pay for a good lens, or work to get the subject in focus before snapping the picture. Unfortunately there are two problems. A minor problem is that it takes a great many calculations to deconvolute an image. Computers are so powerful and cheap that this would only mean that it would typically take tens of seconds to minutes to deconvolute each image, but it would still be a bother.
A more serious problem is that even small levels of noise, i.e. random variations in the pixel intensities, causes big errors in the deconvoluted image. Deconvolution thus improves an image, but in practice not back to its best representation.
Thus it is important to get the best image possible, and then do a deconvolution if the image is very important. As an example, images taken by cameras sent to Mars are very important, and a lot of work is done on them to obtain the most detail possible.
All images of the real world are obtained by cameras or eyes, and all these devices introduce distortion defined by a point spread function.
Q.E.D., the world as we know it is an average.
|
# Area of a Circle
## Area of a Circle
The space occupied by a circle in a two-dimensional plane is called the Area of the Circle.
Let us consider a Circle as shown in the figure below. From the figure, $r$ is the Radius, $d$ is the Diameter and $C$ is the Circumference or Perimeter of the circle. The formula for finding the Area of the Circle can be represented in three different forms in terms of Radius, Diameter and Circumference as discussed below.
## Area of a Circle Formula
### Area of a Circle in terms of Radius
The Radius of a circle is the distance from the centre to any point on the circle. The formula for finding the Area of a Circle whose Radius is $r$ as shown in the above figure is given by,
$A = \pi \times {r^2}$
Where, $\pi = \frac{{22}}{7}$ or $\pi = 3.141592….$
### Area of a Circle in terms of Diameter
The Diameter of a circle is twice the Radius of the circle, therefore,
$d = 2 \times r$
$\therefore r = \frac{d}{2}$
$Area,\,\,A = \pi {r^2}$
$A = \pi {\left( {\frac{d}{2}} \right)^2}$
$A = \frac{{\pi \times {d^2}}}{4}$
### Area of a Circle in terms of Circumference
The Circumference or Perimeter of a circle is given by,
$C = 2\pi r$ or $C = \pi d$
$r = \frac{C}{{2\pi }}$ or $d = \frac{C}{\pi }$
Therefore, the Area of the Circle in terms of Circumference will be,
$A = \pi {r^2}$ or $A = \frac{{\pi \times {d^2}}}{4}$
$A = \pi \frac{{{C^2}}}{{4{\pi ^2}}}$ or $A = \frac{\pi }{4} \times \frac{{{C^2}}}{{{\pi ^2}}}$
$A = \frac{{{C^2}}}{{4\pi }}$ or $A = \frac{{{C^2}}}{{4\pi }}$
## List of Formulas for Area of a Circle
The above mentioned formulas on Area are tabulated below:
## Solved Examples on Area of a Circle
1. If the radius of a circle is 7 cm then find the area of the circle?
Solution:
Given,
Radius, $r= 7 cm$
We know that,
$A = \pi {r^2}$
$A = \frac{{22}}{7} \times {7^2}$
$A = \frac{{22}}{7} \times 7 \times 7$
$A = 22 \times 7$
$A = 154c{m^2}$
Therefore, the area of the circle is $154c{m^2}$.
2. If the diameter of a circle is 21 cm then find the area of the circle?
Solution:
Given,
$d=21cm$
Method 1: Converting diamater to radius:
$r = \frac{d}{2} = \frac{{21}}{2} = 10.5$
$A = \pi {r^2}$
$A = 3.14 \times {\left( {10.5} \right)^2}$
$A = 346.1c{m^2}$
Method 2: Calculating area from diameter:
$A = \frac{{\pi {d^2}}}{4}$
$A = \frac{{3.14 \times {{\left( {21} \right)}^2}}}{4}$
$A = \frac{{1384.74}}{4}$
$A = 346.1c{m^2}$
Therefore, the area of the circle is $346.1c{m^2}$
3. If the circumference of a circle is 40 cm then find the area of the circle?
Solution:
Given,
Circumference, $C=40 cm$
We know that,
$A = \frac{{{C^2}}}{{4\pi }}$
$A = \frac{{{{\left( {40} \right)}^2}}}{{4 \times 3.14}}$
$A = \frac{{1600}}{{12.56}}$
$A = 127.3c{m^2}$
Therefore, the area of the circle is $127.3c{m^2}$
4. If the area of a circle is $400 c{m^2}$, then find the circumference, radius and diameter of the circle?
Solution:
Given, Area, $A=400 c{m^2}$
Circumference:
We know that,
$A = \frac{{{C^2}}}{{4\pi }}$
$C = \sqrt {\frac{A}{{4\pi }}}$
$C = \sqrt {\frac{{400}}{{4 \times 3.14}}}$
$C = \sqrt {\frac{{400}}{{12.56}}}$
$C = \sqrt {31.84}$
$C = 5.64cm$
$A = \pi {r^2}$
$r = \sqrt {\frac{A}{\pi }}$
$r = \sqrt {\frac{{400}}{{3.14}}}$
$r = \sqrt {127.3}$
$r = 11.28cm$
Diameter:
$d=2\times r$
$d=2\times 11.28$
$d=22.56 cm$
5. The area of a circle is 49π. what is its circumference?
Solution: The formula for finding the area of a circle is $A=\pi\times r^2$
Given, $A=49\pi$
Therefore,
$49\pi=\pi\times r^2$
$\therefore\,\,\,\,r=7$
The circumference of the Circle will be:
$C=2\times\pi\times r$
$C=2\times\frac{22}{7}\times 7$
$C=44$
Therefore, the circumference of the circle is $44$ units.
6. The radius of a circle is increased by 1%. find how much % does its area increases?
Solution: Let the radius of the circle is $r$.
Therefore, the area of the circle will be, $A=\pi\times r^2$
If the radius of increased by $1%$, then the new radius will be,
$r_1=\frac{1}{100}\times r$
$r_1=1.01r$
The new area will be:
$A_1=\pi\times r_1^2$
$A_1=\pi\times (1.01r)^2$
$A_1=1.0201\pi r^2$
Difference between the new and the old area,
$A_1-A=1.0201\pi r^2-\pi\times r^2$
$A_1-A=.0201$
Therefore, the percentage increase in area is 2.01%
7. Find the area of a quadrant of a circle whose circumference is 22cm?
Solution: The formula for finding the area of a circle in terms of circumference is given by
$A = \frac{{{C^2}}}{{4\pi }}$
$A=\frac{{{22^2}}}{{4\pi }}$
$A=\frac{{{121}}}{{pi }}$
$A=38.51\,cm^2$
The area of a quadrant is $= \frac{1}{4}\times Total Area$
$=\frac{1}{4}\times 38.51$
$=9.62\,cm^2$
What is the formula of area of a circle?
The formula for area of a circle is $A = \pi \times {r^2}$ or $A = \frac{{\pi \times {d^2}}}{4}$.
What is the area and perimeter of a circle?
The area of a circle is the space occupied by the circle in a plane. The formula for finding area of a circle is $A = \pi \times {r^2}$. The perimeter of a circle is the length of a complete circle. The formula for finding perimeter is $2\pi r$.
How to you find the area with diameter?
The formula for finding the area of a circle when diameter is given is $A = \frac{{\pi \times {d^2}}}{4}$, where, d is the diameter.
What is the area and radius?
The area of a circle is the space occupied by the circle in a plane and the radius is the distance from the centre of a circle to any point in the circle.
What is a circle in your own words?
A circle is a collection of points which are at a fixed distance from a fixed point called Centre.
What are the types of circle?
The types of circle are: Concentric circle, Contact of circle and Orthogonal Circle.
How do you find the area of a circle using circumference?
The area of a circle can be found out using circumference by the following formula:
$A = \frac{{{C^2}}}{{4\pi }}$
|
1 Year 10 Months ago
Date: 25-09-2019
# HERE = COMES – SHE, (assume S = 8). Find values of R + H + O?
PrepInsta Main
We can make this addition problem by re-writing it as – HERE + SHE = COMES
These are the rules to solve Cryptarithmetic Questions -
1. Every Character/letter must have a unique and distinct value
2. The values of a character/letter can not be changed, and should remain same throughout
3. Starting character of number can not be zero example – 0341 should be simply 341.
4. The problem will have only and only one solution
5. Addition of two numbers is always even
6. In case of addition of two numbers, if there is carry then, the carry can only be 1
7. Once all the characters/letters are replaced with numbers, arithmetic operations must be correct
``` H E R E
+ S H E
---------
C O M E S
```
### Step 1
• C = 1 ( Rule 1, as carry is there)
Now, since –
``` H
+ _
---------
C O
```
• H + _ (nothing), can be written as H + 0 now the result of H + 0 should also be H right?
• But, the answer is coming as O, which means there must be a carry from previous step
• So, H + 1 generates a number >10 as result is C O
So, H value is 9, and O value will be H + 1 = 9 + 1 = 0 (1 carry to next step)
Rewriting the problem again
``` 9 E R E
+ S 9 E
---------
1 0 M E S
```
### Step 2
• Now, they have already given S value as 8
• Now, E + E = 8, thus, E = 4
Lets find value R
• R + 9 = E
• after substituting E value R + 9 = 4
• Thus, R = 5 and (1 carry to next step)
Lets, find the value of M
• E + S + 1 (Carry) = M
• after substituting the value 4 + 8 + 1 = 3 (1 carry to next step)
The final values are –
H = 9, E = 4, R = 5, S = 8, C = 1, O = 0, M = 3
R + H + O = 5 + 9 + 0 = 14
You can check more Cryptarithmetic Questions here - https://prepinsta.com/infosys/logical/cryptarithmetic/
Video for Cryptarithmetic -
|
# 1.8: Decimals
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##### Learning Objectives
By the end of this section, you will be able to:
• Name and write decimals
• Round decimals
• Multiply and divide decimals
• Convert decimals, fractions, and percents
##### Note
A more thorough introduction to the topics covered in this section can be found in the Prealgebrachapter, Decimals.
## Name and Write Decimals
Decimals are another way of writing fractions whose denominators are powers of 10.
$\begin{array} {ll} {0.1 = \frac { 1 } { 10 }} &{0.1 \text { is "one tenth" }} \\ {0.01 = \frac { 1 } { 100 }} &{0.01 \text { is "one hundredth }} \\ {0.001 = \frac { 1 } { 1,000 }} &{0.001 \text { is "one thousandth }} \\ {0.0001 = \frac { 1 } { 10,000 }} &{0.0001 \text { is "one ten-thousandth" }} \end{array}$
Notice that “ten thousand” is a number larger than one, but “one ten-thousandth” is a number smaller than one. The “th” at the end of the name tells you that the number is smaller than one.
When we name a whole number, the name corresponds to the place value based on the powers of ten. We read 10,000 as “ten thousand” and 10,000,000 as “ten million.” Likewise, the names of the decimal places correspond to their fraction values. Figure $$\PageIndex{1}$$ shows the names of the place values to the left and right of the decimal point.
##### Exercise $$\PageIndex{1}$$
Name the decimal $$4.3$$.
##### Exercise $$\PageIndex{2}$$
Name the decimal $$6.7$$.
six and seven tenths
##### Exercise $$\PageIndex{3}$$
Name the decimal $$5.8$$.
five and eight tenths
We summarize the steps needed to name a decimal below.
##### NAME A DECIMAL.
1. Name the number to the left of the decimal point.
2. Write “and” for the decimal point.
3. Name the “number” part to the right of the decimal point as if it were a whole number.
4. Name the decimal place of the last digit.
##### Exercise $$\PageIndex{4}$$
Name the decimal: $$−15.571$$.
$$−15.571$$ Name the number to the left of the decimal point. negative fifteen __________________________________ Write “and” for the decimal point. negative fifteen and ______________________________ Name the number to the right of the decimal point. negative fifteen and five hundred seventy-one __________ The $$1$$ is in the thousandths place. negative fifteen and five hundred seventy-one thousandths
##### Exercise $$\PageIndex{5}$$
Name the decimal: $$−13.461$$.
negative thirteen and four hundred sixty-one thousandths
##### Exercise $$\PageIndex{6}$$
Name the decimal: $$−2.053$$.
negative two and fifty-three thousandths
When we write a check we write both the numerals and the name of the number. Let’s see how to write the decimal from the name.
##### Exercise $$\PageIndex{7}$$: How to Write decimals
Write “fourteen and twenty-four thousandths” as a decimal.
##### Exercise $$\PageIndex{8}$$
Write as a decimal: thirteen and sixty-eight thousandths.
13.068
##### Exercise $$\PageIndex{9}$$
Write as a decimal: five and ninety-four thousandths.
5.094
We summarize the steps to writing a decimal.
##### WRITE A DECIMAL.
1. Look for the word “and”—it locates the decimal point.
• Place a decimal point under the word “and.” Translate the words before “and” into the whole number and place it to the left of the decimal point.
• If there is no “and,” write a “0” with a decimal point to its right.
2. Mark the number of decimal places needed to the right of the decimal point by noting the place value indicated by the last word.
3. Translate the words after “and” into the number to the right of the decimal point. Write the number in the spaces—putting the final digit in the last place.
4. Fill in zeros for place holders as needed.
## Round Decimals
Rounding decimals is very much like rounding whole numbers. We will round decimals with a method based on the one we used to round whole numbers.
##### Exercise $$\PageIndex{10}$$
Round 18.379 to the nearest hundredth.
##### Exercise $$\PageIndex{11}$$
Round to the nearest hundredth: 1.047.
1.05
##### Exercise $$\PageIndex{12}$$
Round to the nearest hundredth: 9.173.
9.17
We summarize the steps for rounding a decimal here.
##### ROUND DECIMALS.
1. Locate the given place value and mark it with an arrow.
2. Underline the digit to the right of the place value.
3. Is this digit greater than or equal to 5?
• Yes—add 1 to the digit in the given place value.
• No—do not change the digit in the given place value.
4. Rewrite the number, deleting all digits to the right of the rounding digit.
##### Exercise $$\PageIndex{13}$$
Round 18.379 to the nearest
1. tenth
2. whole number.
Round 18.379
1. to the nearest tenth
Locate the tenths place with an arrow. Underline the digit to the right of the given place value. Because 7 is greater than or equal to 5, add 1 to the 3. Rewrite the number, deleting all digits to the right of the rounding digit. Notice that the deleted digits were NOT replaced with zeros. So, 18.379 rounded to the nearest tenth is 18.4.
2. to the nearest whole number
Locate the ones place with an arrow. Underline the digit to the right of the given place value. Since 3 is not greater than or equal to 5, do not add 1 to the 8. Rewrite the number, deleting all digits to the right of the rounding digit. So, 18.379 rounded to the nearest whole number is 18.
##### Exercise $$\PageIndex{14}$$
Round 6.582 to the nearest
1. hundredth
2. tenth
3. whole number.
1. 6.58
2. 6.6
3. 7
##### Exercise $$\PageIndex{15}$$
Round 15.2175 to the nearest
1. thousandth
2. hundredth
3. tenth.
1. 15.218
2. 15.22
3. 15.2
To add or subtract decimals, we line up the decimal points. By lining up the decimal points this way, we can add or subtract the corresponding place values. We then add or subtract the numbers as if they were whole numbers and then place the decimal point in the sum.
1. Write the numbers so the decimal points line up vertically.
2. Use zeros as place holders, as needed.
3. Add or subtract the numbers as if they were whole numbers. Then place the decimal point in the answer under the decimal points in the given numbers.
##### Exercise $$\PageIndex{16}$$
Add: $$23.5+41.38$$.
$\text{Write the numbers so that the decimal points line up vertically.} \quad \begin{array} {r} { 23.50 } \\ { + 41.38 } \\ \hline \end{array}$
$\text{Put 0 as a placeholder after the 5 in 23.5. Remember, } \frac{5}{10} = \frac{50}{100}, \text{ so } 0.5 = 0.50 \quad \begin{array} {r} { 23.50 } \\ { + 41.38 } \\ \hline \end{array}$
$\text{Add the numbers as if they were whole numbers . Then place the decimal point in the sum.} \quad \begin{array} {r} { 23.50 } \\ { + 41.38 } \\ \hline 64.88 \end{array}$
##### Exercise $$\PageIndex{17}$$
Add: $$4.8+11.69$$.
$$16.49$$
##### Exercise $$\PageIndex{18}$$
Add: $$5.123+18.47$$.
$$23.593$$
##### Exercise $$\PageIndex{19}$$
Subtract: $$20−14.65$$.
\begin{array} {ll} {\text{Write the numbers so that the decimal points line up vertically.}} &{ \begin{align} {20 - 14.65} \\ {20.} \\ {-14.65} \\ \hline \end{align}} \\ {\text{Remember, 20 is a whole number, so place the decimal point after the 0.}} &{} \end{array}
\begin{array} {ll} {\text{Put zeros to the right as placeholders.}} &{ \begin{align} {20.00} \\ {-14.65} \\ \hline \end{align}} \end{array}
\begin{array} {ll} {\text{Write the numbers so that the decimal points line up vertically.}} &{ \begin{align} {\tiny{9} \quad \tiny{9}\qquad} \\ {\small{1} \not{\small{10}} \not{\small10}\not{\small10}}\\ {\not{2}\not{0.}\not{0}\not{0}} \\ {-14.65} \\ \hline \\{5.35} \end{align}} \end{array}
##### Exercise $$\PageIndex{20}$$
Subtract: $$10−9.58$$.
0.42
##### Exercise $$\PageIndex{21}$$
Subtract: $$50−37.42$$.
12.58
## Multiply and Divide Decimals
Multiplying decimals is very much like multiplying whole numbers—we just have to determine where to place the decimal point. The procedure for multiplying decimals will make sense if we first convert them to fractions and then multiply.
So let’s see what we would get as the product of decimals by converting them to fractions first. We will do two examples side-by-side. Look for a pattern!
Convert to fractions. Multiply. Convert to decimals.
Notice, in the first example, we multiplied two numbers that each had one digit after the decimal point and the product had two decimal places. In the second example, we multiplied a number with one decimal place by a number with two decimal places and the product had three decimal places.
We multiply the numbers just as we do whole numbers, temporarily ignoring the decimal point. We then count the number of decimal points in the factors and that sum tells us the number of decimal places in the product.
The rules for multiplying positive and negative numbers apply to decimals, too, of course!
When multiplying two numbers,
• if their signs are the same the product is positive.
• if their signs are different the product is negative.
When we multiply signed decimals, first we determine the sign of the product and then multiply as if the numbers were both positive. Finally, we write the product with the appropriate sign.
##### MULTIPLY DECIMALS.
1. Determine the sign of the product.
2. Write in vertical format, lining up the numbers on the right. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
3. Place the decimal point. The number of decimal places in the product is the sum of the number of decimal places in the factors.
4. Write the product with the appropriate sign.
##### Exercise $$\PageIndex{22}$$
Multiply: $$(−3.9)(4.075)$$.
$$(−3.9)(4.075)$$ The signs are different. The product will be negative. Write in vertical format, lining up the numbers on the right. Multiply. Add the number of decimal places in the factors $$(1 + 3)$$. Place the decimal point 4 places from the right. The signs are different, so the product is negative. $$(−3.9)(4.075) = −15.8925$$
##### Exercise $$\PageIndex{23}$$
Multiply: $$−4.5(6.107)$$.
$$−27.4815$$
##### Exercise $$\PageIndex{24}$$
Multiply: −10.79(8.12).
$$−87.6148$$
In many of your other classes, especially in the sciences, you will multiply decimals by powers of 10 (10, 100, 1000, etc.). If you multiply a few products on paper, you may notice a pattern relating the number of zeros in the power of 10 to number of decimal places we move the decimal point to the right to get the product.
##### MULTIPLY A DECIMAL BY A POWER OF TEN.
1. Move the decimal point to the right the same number of places as the number of zeros in the power of 10.
2. Add zeros at the end of the number as needed.
##### Exercise $$\PageIndex{25}$$
Multiply 5.63
1. by 10
2. by 100
3. by 1,000.
By looking at the number of zeros in the multiple of ten, we see the number of places we need to move the decimal to the right.
$$5.63(10)$$ There is 1 zero in 10, so move the decimal point 1 place to the right.
$$5.63(100)$$ There are 2 zeros in 100, so move the decimal point 2 places to the right.
There are 3 zeros in 1,000, so move the decimal point 3 places to the right. A zero must be added at the end.
Multiply 2.58
1. by 10
2. by 100
3. by 1,000.
1. 25.8
2. 258
3. 2,580
##### Exercise $$\PageIndex{27}$$
Multiply 14.2
1. by 10
2. by 100
3. by 1,000.
1. 142
2. 1,420
3. 14,200
Just as with multiplication, division of decimals is very much like dividing whole numbers. We just have to figure out where the decimal point must be placed.
To divide decimals, determine what power of 10 to multiply the denominator by to make it a whole number. Then multiply the numerator by that same power of 10. Because of the equivalent fractions property, we haven’t changed the value of the fraction! The effect is to move the decimal points in the numerator and denominator the same number of places to the right. For example:
$\begin{array} { c } { \frac { 0.8 } { 0.4 } } \\ { \frac { 0.8 ( 10 ) } { 0.4 ( 10 ) } } \\ { \frac { 8 } { 4 } } \end{array}$
We use the rules for dividing positive and negative numbers with decimals, too. When dividing signed decimals, first determine the sign of the quotient and then divide as if the numbers were both positive. Finally, write the quotient with the appropriate sign.
We review the notation and vocabulary for division:
$\begin{array} {ll} {} &{\underset{\text{quotient}}{c}} \\ {\underset{\text{dividend}}{a} \div \underset{\text{divisor}}{b} = \underset{\text{quotient}}{c}} & {\underset{\text{divisor}}{b})\overline{\underset{\text{dividend}}{a}}} \end{array}$
We’ll write the steps to take when dividing decimals, for easy reference.
##### DIVIDE DECIMALS.
1. Determine the sign of the quotient.
2. Make the divisor a whole number by “moving” the decimal point all the way to the right. “Move” the decimal point in the dividend the same number of places—adding zeros as needed.
3. Divide. Place the decimal point in the quotient above the decimal point in the dividend.
4. Write the quotient with the appropriate sign.
##### Exercise $$\PageIndex{28}$$
Divide: $$−25.65\div (−0.06)$$.
Remember, you can “move” the decimals in the divisor and dividend because of the Equivalent Fractions Property.
$$−25.65\div (−0.06)$$ The signs are the same. The quotient is positive. Make the divisor a whole number by “moving” the decimal point all the way to the right. “Move” the decimal point in the dividend the same number of places. Divide. Place the decimal point in the quotient above the decimal point in the dividend. Write the quotient with the appropriate sign. $$−25.65\div (−0.06) = 427.5$$
##### Exercise $$\PageIndex{29}$$
Divide: $$−23.492\div (−0.04)$$.
687.3
##### Exercise $$\PageIndex{30}$$
Divide: $$−4.11\div(−0.12)$$.
34.25
A common application of dividing whole numbers into decimals is when we want to find the price of one item that is sold as part of a multi-pack. For example, suppose a case of 24 water bottles costs $$3.99$$. To find the price of one water bottle, we would divide $$3.99$$ by 24. We show this division in Exercise $$\PageIndex{31}$$. In calculations with money, we will round the answer to the nearest cent (hundredth).
##### Exercise $$\PageIndex{31}$$
Divide: $$3.99\div 24$$.
.99 divided by 24 is given. A long division problem is set up with 24 dividing 3.99. A table is given with directions on the left and the mathematical steps on the right. The first step reads “Place the decimal point in the quotient above the decimal point in the dividend. Divide as usual. When do we stop? Since this division involves money, we round it to the nearest cent (hundredth). To do this, we must carry the division to the thousandths place.” To the right of this, we have a long division problem set up with 24 dividing 3.990. The quotient is given as 0.166. To show the work, below 3.990 it reads 24, solid horizontal line, 159, 144, solid horizontal line, 150, 144, solid horizontal line, and finally 6. The fifth step reads “Round to the nearest cent.” To the right of this, we have $0.166 is approximately equal to$0.17 and hence >.99 divided by 24 is \$0.17.">
$$3.99\div 24$$ Place the decimal point in the quotient above the decimal point in the dividend. Divide as usual. When do we stop? Since this division involves money, we round it to the nearest cent (hundredth.) To do this, we must carry the division to the thousandths place. Round to the nearest cent. $$0.166\approx 0.17$$ $$3.99\div 2\approx 0.17$$
##### Exercise $$\PageIndex{32}$$
Divide: $$6.99\div 36$$.
$$0.19$$
##### Exercise $$\PageIndex{33}$$
Divide: $$4.99\div 12$$.
$$0.42$$
## Convert Decimals, Fractions, and Percents
We convert decimals into fractions by identifying the place value of the last (farthest right) digit. In the decimal 0.03 the 3 is in the hundredths place, so 100 is the denominator of the fraction equivalent to 0.03.
$00.03 = \frac { 3 } { 100 }$
Notice, when the number to the left of the decimal is zero, we get a fraction whose numerator is less than its denominator. Fractions like this are called proper fractions.
The steps to take to convert a decimal to a fraction are summarized in the procedure box.
##### CONVERT A DECIMAL TO A PROPER FRACTION.
1. Determine the place value of the final digit.
2. Write the fraction.
• numerator—the “numbers” to the right of the decimal point
• denominator—the place value corresponding to the final digit
##### Exercise $$\PageIndex{34}$$
Write 0.374 as a fraction.
0.374 Determine the place value of the final digit. Write the fraction for 0.374: The numerator is 374. The denominator is 1,000. $$\dfrac{374}{1000}$$ Simplify the fraction. $$\dfrac{2\cdot 187}{2\cdot 500}$$ Divide out the common factors. $$\dfrac{187}{500}$$ so, $$0.374=\dfrac{187}{500}$$
Did you notice that the number of zeros in the denominator of $$\dfrac{374}{1000}$$ is the same as the number of decimal places in 0.374?
##### Exercise $$\PageIndex{35}$$
Write 0.234 as a fraction.
$$\dfrac{117}{500}$$
##### Exercise $$\PageIndex{36}$$
Write 0.024 as a fraction.
$$\dfrac{3}{125}$$
We’ve learned to convert decimals to fractions. Now we will do the reverse—convert fractions to decimals. Remember that the fraction bar means division. So $$\dfrac{4}{5}$$ can be written $$4\div 5$$ or $$5)\overline{4}$$. This leads to the following method for converting a fraction to a decimal.
##### CONVERT A FRACTION TO A DECIMAL.
To convert a fraction to a decimal, divide the numerator of the fraction by the denominator of the fraction.
##### Exercise $$\PageIndex{37}$$
Write $$-\dfrac{5}{8}$$ as a decimal.
Since a fraction bar means division, we begin by writing $$\dfrac{5}{8}$$ as $$8)\overline{5}$$. Now divide.
##### Exercise $$\PageIndex{38}$$
Write $$-\dfrac{7}{8}$$ as a decimal.
−0.875
##### Exercise $$\PageIndex{39}$$
Write $$-\dfrac{3}{8}$$ as a decimal.
−0.375
When we divide, we will not always get a zero remainder. Sometimes the quotient ends up with a decimal that repeats. A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly. A bar is placed over the repeating block of digits to indicate it repeats.
##### REPEATING DECIMAL
A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly.
A bar is placed over the repeating block of digits to indicate it repeats.
##### Exercise $$\PageIndex{40}$$
Write $$\dfrac{43}{22}$$ as a decimal.
##### Exercise $$\PageIndex{41}$$
Write $$\dfrac{27}{11}$$ as a decimal.
$$2.\overline{45}$$
##### Exercise $$\PageIndex{42}$$
Write $$\dfrac{51}{22}$$ as a decimal.
$$2.3\overline{18}$$
Sometimes we may have to simplify expressions with fractions and decimals together.
##### Exercise $$\PageIndex{43}$$
Simplify: $$\dfrac{7}{8}+6.4$$.
First we must change one number so both numbers are in the same form. We can change the fraction to a decimal, or change the decimal to a fraction. Usually it is easier to change the fraction to a decimal.
$$\dfrac{7}{8}+6.4$$ Change $$\dfrac{7}{8}$$ to a decimal. Add. $$0.875+6.4$$ $$7.275$$ So, $$\dfrac{7}{8}+6.4 = 7.275$$
##### Exercise $$\PageIndex{44}$$
Simplify: $$\dfrac{3}{8}+4.9$$.
$$5.275$$
##### Exercise $$\PageIndex{45}$$
Simplify: $$5.7 + \dfrac{13}{20}$$.
$$6.35$$
A percent is a ratio whose denominator is 100. Percent means per hundred. We use the percent symbol, %, to show percent.
##### PERCENT
A percent is a ratio whose denominator is 100.
Since a percent is a ratio, it can easily be expressed as a fraction. Percent means per 100, so the denominator of the fraction is 100. We then change the fraction to a decimal by dividing the numerator by the denominator.
$\begin{array} {llll} {} &{\text{6%}} &{\text{78%}} &{\text{135%}} \\ {\text { Write as a ratio with denominator } 100. } &{\dfrac{6}{100}} &{\dfrac{78}{100}} &{\dfrac{135}{100}} \\ { \text { Change the fraction to a decimal by dividing}} &{0.06} &{0.78} &{1.35}\\ {\text{the numerator by the denominator.}} &{} &{} &{} \end{array}$
Do you see the pattern? To convert a percent number to a decimal number, we move the decimal point two places to the left.
##### Exercise $$\PageIndex{46}$$
Convert each percent to a decimal:
1. 62%
2. 135%
3. 35.7%.
1. Move the decimal point two places to the left. 0.62
2. Move the decimal point two places to the left. 1.35
3. Move the decimal point two places to the left. 0.057
##### Exercise $$\PageIndex{47}$$
Convert each percent to a decimal:
1. 9%
2. 87%
3. 3.9%.
1. 0.09
2. 0.87
3. 0.039
##### Exercise $$\PageIndex{48}$$
Convert each percent to a decimal:
1. 3%
2. 91%
3. 8.3%.
1. 0.03
2. 0.91
3. 0.083
Converting a decimal to a percent makes sense if we remember the definition of percent and keep place value in mind.
To convert a decimal to a percent, remember that percent means per hundred. If we change the decimal to a fraction whose denominator is 100, it is easy to change that fraction to a percent.
$\begin{array} {llll} {} &{0.83} &{1.05} &{0.075} \\ {\text {Write as a fraction }} &{\frac{83}{100}} &{\small{1}\frac{5}{100}} &{\frac{75}{1000}} \\ { \text {The denominator is 100.}} &{} &{\frac{105}{100}} &{\frac{7.5}{100}}\\ {\text{Write the ratio as a percent.}} &{\text{83%}} &{\text{105%}} &{\text{7.5%}} \end{array}$
Recognize the pattern? To convert a decimal to a percent, we move the decimal point two places to the right and then add the percent sign.
##### Exercise $$\PageIndex{49}$$
Convert each decimal to a percent:
1. 0.51
2. 1.25
3. 0.093.
1. Move the decimal point two places to the right. $$51%$$
2. Move the decimal point two places to the right. $$125%$$
3. Move the decimal point two places to the right. $$9.3%$$
##### Exercise $$\PageIndex{50}$$
Convert each decimal to a percent:
1. 0.17
2. 1.75
3. 0.0825
1. 17%
2. 175%
3. 8.25%
##### Exercise $$\PageIndex{51}$$
Convert each decimal to a percent:
1. 0.41
2. 2.25
3. 0.0925.
1. 41%
2. 225%
3. 9.25%
## Key Concepts
• Name a Decimal
1. Name the number to the left of the decimal point.
2. Write ”and” for the decimal point.
3. Name the “number” part to the right of the decimal point as if it were a whole number.
4. Name the decimal place of the last digit.
• Write a Decimal
1. Look for the word ‘and’—it locates the decimal point. Place a decimal point under the word ‘and.’ Translate the words before ‘and’ into the whole number and place it to the left of the decimal point. If there is no “and,” write a “0” with a decimal point to its right.
2. Mark the number of decimal places needed to the right of the decimal point by noting the place value indicated by the last word.
3. Translate the words after ‘and’ into the number to the right of the decimal point. Write the number in the spaces—putting the final digit in the last place.
4. Fill in zeros for place holders as needed.
• Round a Decimal
1. Locate the given place value and mark it with an arrow.
2. Underline the digit to the right of the place value.
3. Is this digit greater than or equal to 5? Yes—add 1 to the digit in the given place value. No—do not change the digit in the given place value.
4. Rewrite the number, deleting all digits to the right of the rounding digit.
1. Write the numbers so the decimal points line up vertically.
2. Use zeros as place holders, as needed.
3. Add or subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers.
• Multiply Decimals
1. Determine the sign of the product.
2. Write in vertical format, lining up the numbers on the right. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
3. Place the decimal point. The number of decimal places in the product is the sum of the decimal places in the factors.
4. Write the product with the appropriate sign.
• Multiply a Decimal by a Power of Ten
1. Move the decimal point to the right the same number of places as the number of zeros in the power of 10.
2. Add zeros at the end of the number as needed.
• Divide Decimals
1. Determine the sign of the quotient.
2. Make the divisor a whole number by “moving” the decimal point all the way to the right. “Move” the decimal point in the dividend the same number of places - adding zeros as needed.
3. Divide. Place the decimal point in the quotient above the decimal point in the dividend.
4. Write the quotient with the appropriate sign.
• Convert a Decimal to a Proper Fraction
1. Determine the place value of the final digit.
2. Write the fraction: numerator—the ‘numbers’ to the right of the decimal point; denominator—the place value corresponding to the final digit.
• Convert a Fraction to a Decimal Divide the numerator of the fraction by the denominator.
This page titled 1.8: Decimals is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Applications of linear equations
We have learned all about linear relationships, and linear functions. From what we learned about, we know that linear relationships and functions is consisted of two variables, one dependent, and the other independent. We also know that the line is a straight line. Linear equations are nonetheless the same, adding the criteria that its variables should not be raised to an exponent greater than one, and that the variables are not used as a denominator. We are to learn more of linear equations in this chapter, especially in 10.1.
Apart from linear equations, we also need to learn about nonlinear equations. As the word suggests, these are the other equations that aren’t linear in nature, like the quadratic equations, circle equations, cubic equations, and more. We will have exercises in 10.2 that will show us the difference between these equations and that of the linear equations.
As we have learned in the previous chapters, we know that linear equations show a straight line, that would appear to fall diagonally to on the Cartesian plane, to show the linear relationship between the values found in the line. However, in some special cases, linear equations can show horizontal lines and vertical lines. In chapter 10.3 and 10.4 we will understand these special cases a bit more.
We are also going to learn how to find the parallel line and the perpendicular line of a given linear equation in chapter 10.5 and 10.6. From our previous discussion in chapter, we learned that parallel lines have the same slope, whereas perpendicular lines have the slopes that are negative reciprocals with each other.
In 10.7 and 10.8 we have more practice in looking for the parallel lines and perpendicular lines, by looking for both of them for every equation given. There are also other applications of linear equation in 10.8. To find out more about linear equations you can check out a video made by Education Alberta in Canada about “Exploring Linear Equations
### Applications of linear equations
We always encounter situations that can be expressed in linear equations. Watch this lesson and learn how to solve linear equations word problems.
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Explanation. \end{aligned} $$. A weibull distributions is the probability of a value x of a function to fall between two arbitrary points x1 and x2 along that function. You also learned about how to solve numerical problems based on Weibull distribution. There are four main steps in performing a Weibull Analysis: Collect life data for a part or product and identify the type of data you are working with (Complete, Right Censored, Interval, Left Censored) Choose a lifetime distribution that fits the data and model the life of the part or product â¢White's Formula (F(tâ±¼) = [j - 3/8]/[n + 1/4]) Example: The shear strength (in pounds) of a spot weld is a Weibull distributed random variable, X ËWEB(400;2=3). Define the Weibull variable by setting the scale (λ > 0) and the shape (k > 0) in the fields below. 3. How to Make a Weibull Analysis in 5 Steps â Part 1. Mean and variance of X,$$ \begin{aligned} E(X) &= \beta \Gamma (\dfrac{1}{\alpha}+1)\\ &=3\Gamma(\dfrac{1}{2}+1)\\ &=3\Gamma(3/2)\\ &=3\times\dfrac{1}{2}\Gamma(1/2)\\ &=\dfrac{3}{2}\times\sqrt{\pi}\\ &=\dfrac{3}{2}\times1.7725\\ &=2.6588 \end{aligned} $$,$$ \begin{aligned} V(X) &= \beta^2 \bigg[\Gamma (\dfrac{2}{\alpha}+1) -\bigg(\Gamma (\dfrac{1}{\alpha}+1) \bigg)^2\bigg]\\ &=3^2 \bigg[\Gamma (\dfrac{2}{2}+1) -\bigg(\Gamma (\dfrac{1}{2}+1) \bigg)^2\bigg]\\ &=9\bigg[\Gamma(2)-\big(\Gamma(3/2)\big)^2\bigg]\\ &=9\bigg[1-\bigg(\frac{1}{2}\Gamma(1/2)\bigg)^2\bigg]\\ &=9\bigg[1-\bigg(\frac{\sqrt{\pi}}{2}\bigg)^2\bigg]\\ &=9\bigg[1-\bigg(\frac{\sqrt{3.1416}}{2}\bigg)^2\bigg]\\ &=1.931846 \end{aligned} $$,$$ \begin{aligned} P(X\leq 6) &=F(6)\\ &= 1-e^{-(6/3)^{2}}\\ &= 1-e^{-(2)^{2}}\\ &= 1-e^{-(4)}\\ &=1-0.0183\\ &=0.9817 \end{aligned} 2. The exact value of the median rank is calculated using the formula ð¹ð¹ ð¡ð¡ðð = 1 1+ ððâðð+1 ðð ð¹ð¹0.5,2(ððâðð+1),2ðð. Percent Point Function Thus, the Weibull distribution can be used to model devices with decreasing failure rate, constant failure rate, or increasing failure rate. d.Find the 95th percentile. b.Find P(X >410 jX >390). Weibull Distribution. P (X1 0 \) The following is the plot of the Weibull cumulative distribution function with the same values of γ as the pdf plots above. 4. This tutorial will help you to understand Weibull distribution and you will learn how to derive mean, variance, distribution function, median, mode, moment and other properties of Weibull distribution. Weibull Distribution Calculator is an online probability and statistics tool for data analysis programmed to calculate precise accurate failure analysis and risk predictions with extremely small samples using a simple and useful graphical plot. â¢Mean (F(tâ±¼) = j/[n + 1]) The mean rank is sometimes recommended. The inverse cumulative distribution function is I(p) =. Here β > 0 is the shape parameter and α > 0 is the scale parameter.. Weibull Distribution in Excel (WEIBULL.DIST) Excel Weibull distribution is widely used in statistics to obtain a model for several data sets, the original formula to calculate weibull distribution is very complex but we have an inbuilt function in excel known as Weibull.Dist function which calculates Weibull distribution.. \end{aligned}, eval(ez_write_tag([[250,250],'vrcbuzz_com-leader-2','ezslot_6',120,'0','0']));a. Weibull Distribution. Male or Female ? If the set matches Weibull distribution, then the shape parameter is the slope of the straight line through the set of points with the coordinates given by numbers in Columns C and D. Calculate it using this formula: =SLOPE(D2:D101,C2:C101) (This assumes your set This versatility is one reason for the wide use of the Weibull distribution in reliability. Standard Normal Distribution Calculator Solutions are possible at the earliest stage of a problem without the requirement to crash a few more. Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student Tip: For a quick demonstration, select a test data set from the last pull-down in the Options area (#2) and click calculate. Scale (λ > 0) : Shape (k > 0) : How to Input Interpret the Output. Given that $X\sim W(\alpha = 300, \beta=0.5)$. The probability density function of $X$ is, \begin{aligned} f(x;\alpha, \beta)&=\frac{\alpha}{\beta} \big(\dfrac{x}{\beta}\big)^{\alpha-1}e^{-\big(\dfrac{x}{\beta}\big)^\alpha};\; x>0,\alpha,\beta>0. Assume that the life of a packaged magnetic disk exposed to corrosive gases has a Weibull distribution with \alpha = 300 hours and \beta = 0.5. eval(ez_write_tag([[580,400],'calculator_academy-medrectangle-3','ezslot_0',169,'0','0'])); The following formula is used to calculate a weibull distribution/probability of a function. In this tutorial, you learned about how to calculate probabilities of Weibull distribution. The pdf of two parameter Weibull distribution is given by, \begin{align*} f(x;\alpha, \beta)&= \begin{cases} \frac{\alpha}{\beta} \big(\dfrac{x}{\beta}\big)^{\alpha-1}e^{-\big(\dfrac{x}{\beta}\big)^\alpha}, & x>0, \alpha, \beta>0; \\ 0, & Otherwise. Weibull Distribution in Excel (WEIBULL.DIST) Excel Weibull distribution is widely used in statistics to obtain a model for several data sets, the original formula to calculate weibull distribution is very complex but we have an inbuilt function in excel known as Weibull.Dist function which calculates Weibull distribution.. X (required argument) â This is the value at which the function is to be calculated. Weibull Distribution | Standard | Two Parameter | Mean | Mode, Sample size calculator to test hypothesis about mean, Moment coefficient of kurtosis calculator for grouped data. Suppose that X has the Weibull distribution with shape parameter k. For a three parameter Weibull, we add the location parameter, δ. Beta (required argument) â This is the scale parameter to the Excel Weibull distribution and it must be greater than 0. The variance of Two-parameter Weibull distribution is V ( X) = β 2 ( Î ( 2 α + 1) â ( Î ( 1 α + 1)) 2). Weibull distribution parameters are estimated using âWindchill quality solution 10.1Tryoutâ software tool very easily and statistical computation & charts are presented in fig (1, 2,3,4,5, and 6) the Fig 1. Gamma (Î) distribution calculator, formulas, work with steps & solved examples to estimate the probability density function (PDF) of random variable x in statistical experiments. The probability that a disk lasts at least 600 hours, $P(X\geq 600)$, \begin{aligned} P(X\geq 600) &=1-P(X< 600)\\ &= 1-F(600)\\ &= 1-\bigg[1-e^{-(600/300)^{0.5}}\bigg]\\ &= e^{-(2)^{0.5}}\\ &=0.2431 \end{aligned}. The inverse Weibull distribution has the ability to model failures rates which are most important in the reliability and biological study areas. Choose the parameter you want to calculate and click the Calculate! Distribution Calculator Wikipedia â Weibull Distribution Wolfram Math World â Weibull Distribution⦠The old Weibull tool is available here; however, it may be slow, or non-working, depending on Google image chart availability. Calculation of Weibull distribution coefficients, from wind speed measurements. The Weibull distribution is one of the most widely used lifetime distributions in reliability engineering. This revised Weibull analysis tool makes use of JavaScript based charts. a.Find P(X >410). ... two steps, we need various tools and techniques. To learn more about other probability distributions, please refer to the following tutorial: Let me know in the comments if you have any questions on Weibull Distribution Examples and your thought on this article. It is a versatile distribution that can take on the characteristics of other types of distributions, based on the value of the shape parameter, ${\beta} \,\! Like Weibull distribution, a three-parameter inverse Weibull distribution is introduced to study the ⦠It is the shape parameter to the distribution. The scale or characteristic life value is close to the mean value of the distribution. Here I describe three different methods to estimate the coefficients (the scale factor A and the shape factor k) of the cumulative Weibull distribution function (equation 4.6). Compute the following: Let X denote the lifetime (in hundreds of hours) of vaccume tube. To improve this 'Weibull distribution Calculator', please fill in questionnaire. To read more about the step by step tutorial on Weibull distribution refer the link Weibull Distribution. It must be greater than or equal to zero. - Weibull Distribution -. Lets solve few of the Weibull distribution examples with detailed guide to compute probbility and variance for different numerical problems. Cumulative (required argu⦠Presented the Weibull probability plot with parameters are estimated & failure pattern of diesel engine. In practical situations, = min(X) >0 and X ⦠This Weibull calculator is featured to generate the work with steps for any corresponding input values to help beginners to learn how the input values are being used in such calculations. Table 4.3: Weibull Probability Distribution â ð ðº.ð¬(â) Wald CI ð ðº.ð¬( ) 0.9218841 0.176140 0.7457409, 1.0980280 17.2019258 4.951544 17.02579, 22.15347 4.3.2 Proï¬le Likelihood of Weibull Distribution Figure 4.4: Profile Log-likelihood graph of a Weibull distribution. It must be greater than 0. =WEIBULL.DIST(x,alpha,beta,cumulative) The WEIBULL.DIST function uses the following arguments: 1. How to Calculate Probabilities of Weibull distribution? Variance of Two-parameter Weibull Distribution. It should also help both in evaluating other studies using different methods of Weibull parameter estimation and in discussions on American Society for Testing and Materials Standard D5457, which appears to allow a choice for the Copyright © 2021 VRCBuzz All rights reserved. Weibull Distribution Calculator Enter alpha, beta, x1, and x2 into the calculator to determine the weibull distribution of the function. Published February 6, 2017 ... Also, some of the advantages are the flexibility of the Weibull distribution, interpretability of the parameters, and the straight relation to failure rates and bathtub curve. In this case, the formula is ð¹ð¹ ð¡ð¡ðð = ðð ðð+1. Poisson distribution calculator calculates the probability of given number of events that occurred in a fixed interval of time with respect to the known average rate of events occurred. How to Calculate Probabilities of Weibull distribution? The cumulative distribution function (cdf) is.$. eval(ez_write_tag([[250,250],'vrcbuzz_com-large-mobile-banner-1','ezslot_1',110,'0','0']));Let $X$ denote the life of a packaged magnetic disk exposed to corrosive gases in hours. The lifetime $X$ (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters $\alpha = 2$ and $\beta = 3$. Definition 1: The Weibull distribution has the probability density function (pdf). Step 5 - ⦠button to proceed. , respectively. Weibull distribution to represent a data set realize some advantages and disadvantages of some basic methods. It's an online statistics and probability tool requires an average rate of success and Poisson random variable to find values of Poisson and cumulative Poisson distribution. Calculator Academy© - All Rights Reserved 2021, Empirical Rule Calculator (68%, 95%, 99.7%), how to find alpha and beta for weibull distribution, how to find alpha and beta for weibull distribution in excel, Where P (X1 0 is the scale parameter E ^ ( -x1/B ) –... Possible at the earliest stage of a problem without the requirement to crash a few.... Point function this revised Weibull analysis tool makes use of JavaScript based charts ððâðð+1 ),2ðð ð¹ð¹0.5,2 ( ððâðð+1,2ðð... Where $\alpha =2$ and $\beta=3$ step tutorial on Weibull distribution the WEIBULL.DIST function the! Beta ( required argu⦠distribution Calculator Wikipedia â Weibull distribution X\sim W ( \alpha = 300, \beta=0.5 ).! X, alpha, beta, cumulative ) the mean value of the rank! Which the function is to be calculated Calculator Wikipedia â Weibull Distribution⦠inverse Weibull distribution the distribution crash few! The formula ð¹ð¹ ð¡ð¡ðð = 1 1+ ððâðð+1 ðð ð¹ð¹0.5,2 ( ððâðð+1 ).. The Output = ðð ðð+1, x1, and x2 into the to... X1, and x2 into the Calculator to determine the Weibull distribution has the probability (... Case, the formula ð¹ð¹ ð¡ð¡ðð = ðð ðð+1, the formula ð¹ð¹ ð¡ð¡ðð = ðð ðð+1 WEIBULL.DIST uses! Distribution function is to be calculated... two steps, we need various tools and techniques the. ; however, it may be slow, or increasing failure rate constant! Without the requirement to crash a few more ) & expected mean ( μ of. Expected mean ( μ ) of gamma distribution WEIBULL.DIST function uses the following arguments:.! Interpret the Output distribution refer the link Weibull distribution can be used to model failures rates are... Using the formula is ð¹ð¹ ð¡ð¡ðð = 1 1+ ððâðð+1 ðð ð¹ð¹0.5,2 ( ððâðð+1 ),2ðð coefficients. ) = j/ [ n + 1 ] ) the WEIBULL.DIST function uses the:. And techniques and α > 0 is the scale or characteristic life value is close the. ÐÐÂÐÐ+1 ),2ðð greater than 0, where $\alpha =2$ $! Realize some advantages and disadvantages of some basic weibull distribution calculator with steps ) ^a – e^ ( )... For different numerical problems based on Weibull distribution refer the link Weibull distribution jX > 390 )$ and \beta=3... > 0 ): Shape ( k > 0 ): Shape ( k > 0 is the value which... Distribution and it must be greater than 0 enter alpha, beta,,... Which the function is I ( P ) = to read more about the step by step tutorial on distribution. Presented the Weibull probability plot with parameters are estimated & failure pattern of diesel engine most important in the and... 300, \beta=0.5 ) $, where$ \alpha =2 $and \beta=3. > 0 ): how to calculate probabilities of Weibull distribution and must..., beta, cumulative ) the mean rank is calculated using the formula ð¹ð¹ ð¡ð¡ðð ðð... Use of the median rank is sometimes recommended weibull distribution calculator with steps x2 into the Calculator to the! Is normally approximated with a Weibull distribution has the ability to model failures rates which are most important in reliability! ( X ) step 5 - ⦠to improve this 'Weibull distribution Calculator ', please fill in.... This 'Weibull distribution Calculator Wikipedia â Weibull distribution of the distribution vaccume tube lets solve few of Weibull. Pattern of diesel engine Interpret the Output exact value of the Weibull distribution examples with detailed to. It may be slow, or non-working, depending on Google image chart.! Increasing failure rate, constant failure rate by using this Calculator, users may find probability... Of JavaScript based charts are estimated & failure pattern of diesel engine ⦠to improve this distribution. Denote the lifetime ( in hundreds of hours ) of vaccume tube pattern of diesel.. The step by step tutorial on Weibull distribution$ denote the lifetime in! Solve few of the Weibull distribution has the ability to model failures rates which are most important in reliability. ) = about how to solve numerical problems JavaScript based charts P ( X ) and V ( )... Google image chart availability required argument ) â this is the scale or characteristic life value is to! X2 into the Calculator to determine the Weibull distribution in reliability,,! Given that $X\sim W ( \alpha = 300, \beta=0.5 )$ more about the step step... Beta ( required argu⦠distribution Calculator Wikipedia â Weibull Distribution⦠inverse Weibull distribution increasing failure,! ) $to represent a data set realize some advantages and disadvantages of some basic methods step tutorial Weibull! - ⦠to improve this 'Weibull distribution Calculator ', please fill in.! ( required argument ) â this is the Shape parameter and α > 0 ) Shape... Into the Calculator to determine the Weibull probability plot with parameters are estimated & failure pattern of engine. 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# MATH PROBLEMS ON CALCULATING NUMBER OF SIBLINGS
Math Problems on Calculating Number of Siblings :
If we "b" be the number of boys, then "g" be the number of girls.
According to each boy's perspective :
He will have (b - 1) brothers and g sisters.
According to each girl's perspective :
She will have (g - 1) sisters and b brothers.
Note :
Here, we subtract 1 from the number of boys to calculate the number of brothers he has. Because each boy will not be a brother of himself.
Like that, we subtract 1 from the number of girls to calculate the number of sisters she has. Because each girl will not be a sister of herself.
## Finding Number of Siblings in Math Problems - Examples
Example 1 :
Among the children in a family, each boy has as many sisters as brothers, but each girl has only half as many sisters as brothers. How do you find the number of children in the family?
Solution :
Let "b" and "g" be the number of boys and number of girls respectively in the family.
According to each boy's perspective :
He will have (b - 1) brothers and g sisters.
According to each girl's perspective :
She will have (g - 1) sisters and b brothers.
According to the question, each boy is thinking about his brothers and he says that he has as many sisters as brothers.
We decide that :
Number of sisters = Number of brothers he has
g = b - 1 ---------(1)
Now each girl is thinking about her sisters. That is each girl is having only half as many sisters as brothers.
Number of sisters (g - 1) = b/2 ---------(2)
From (1),
b = g + 1
From (1),
b = 2(g - 1)
g + 1 = 2(g - 1)
g + 1 = 2g - 2
g - 2g = -2 - 1
g = 3
b = 3 + 1
b = 4
So, the number of boys and girls in the family are 4 and 3 respectively.
Hence the number of children in the family is 7.
Example 2 :
In a particular family each boy has as many brothers as sisters but each girl has twice as many brothers as that of sisters. How many siblings are there in the family?
Solution :
Let "b" and "g" be the number of boys and number of girls respectively in the family.
Each boy has "g" sisters and "b - 1" brothers
Each girl has "b" brothers and "g - 1" sisters
According to each boy's perspective :
Number of brothers he has = Number of sisters
b - 1 = g ----(1)
According to girl's perspective :
b = 2(g - 1) ----(2)
From (1),
b = g + 1
g + 1 = 2g - 2
g - 2g = -2 - 1
-g = -3
g = 3
By applying the value of g, we get the value of b.
b = 3 + 1
b = 4
So, the number of boys and girls in the family are 4 and 3 respectively.
Hence the number of children in the family is 7.
Example 3 :
In a family, each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many sons are there in family?
Solution :
Let "b" and "g" be the number of boys and number of girls respectively in the family.
Each boy has "g" sisters and "b - 1" brothers
Each girl has "b" brothers and "g - 1" sisters
According to girl's perspective :
g - 1 = b ---(1)
According to boy's perspective :
g = 2(b - 1) ---(2)
From (1)
g = b + 1
By applying the value of g in (2), we get
b + 1 = 2(b - 1)
b + 1 = 2b - 2
b - 2b = -2 - 1
-b = -3
b = 3
By applying the value of
g = 3 + 1
g = 4
Hence the family has 3 sons.
After having gone through the stuff given above, we hope that the students would have understood how to solve math problems involving the concept calculating number of siblings.
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# FAQ: Why Square Root Of 2 Is Irrational?
## Why is square root of 2 an irrational number?
Specifically, the Greeks discovered that the diagonal of a square whose sides are 1 unit long has a diagonal whose length cannot be rational. By the Pythagorean Theorem, the length of the diagonal equals the square root of 2. So the square root of 2 is irrational!
## How do you prove that √ 2 is irrational?
Let’s suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. A proof that the square root of 2 is irrational.
2 = (2k) 2 /b 2
b 2 = 2k 2
## Is 2 root 2 an irrational number?
Answer. To find:- 2 root 2 is an irrational number. because 8 is not a perfect square. So 2 root 2 is an irrational number.
## Is the square root of 2 a rational number?
Sal proves that the square root of 2 is an irrational number, i.e. it cannot be given as the ratio of two integers.
You might be interested: Often asked: How To Square Root Negative Numbers?
## Is 2 rational or irrational?
2 is a rational number because it can be expressed as the quotient of two integers: 2 ÷ 1.
## Is 2 a perfect square?
Answer: YES, 2 is in the list of numbers that are never perfect squares. The number 2 is NOT a perfect square and we can stop here as there is not need to complete the rest of the steps.
## Is √ 3 an irrational number?
It is denoted mathematically as √3. It is more precisely called the principal square root of 3, to distinguish it from the negative number with the same property. The square root of 3 is an irrational number.
## How do you prove 3 Root 2 is irrational?
3 +√ 2 = a/b,where a and b are integers and b is not equal to zero.. therefore, √ 2 = (3b – a)/b is rational as a, b and 3 are integers.. But this contradicts the fact that √ 2 is irrational..
## How do you prove √ 3 is irrational?
Therefore there exists no rational number r such that r2= 3. Hence the root of 3 is an irrational number. Thank you.
Which Of The Following Has Less Hydrogen Bond S H Or O H What Are Amino Acids
## Is 2/3 a rational or irrational number?
In mathematics rational means “ratio like.” So a rational number is one that can be written as the ratio of two integers. For example 3=3/1, −17, and 2/3 are rational numbers.
## Is 3 rational or irrational?
Explanation: A rational number is a number, which can be expressed as a fraction. Since 3 can be expressed as 3 = 3 1=62=124 and so on, it is a rational number.
## Is 2 a real number?
Any number that can be put on a number line is a real number. Integers like − 2, rational numbers /decimals like 0.5, and irrational numbers like √ 2 or π can all be plotted on the number line, so they are real.
You might be interested: Quick Answer: What Is The Square Root Of 441?
## Is 2/3 an irrational number?
A number that can be written as a ratio of two integers, of which denominator is non-zero, is called a rational number. As such 23 is a rational number. 23 is a rational number.
## Is Zero is a rational number?
Answer: Zero is an example of a rational number. Any fraction with non- zero denominators is a rational number. The set of rational numbers include positive, negative numbers, and zero and can be expressed as a fraction. -2 is an example of a negative rational number.
## What are 5 irrational numbers?
What are the five examples of irrational numbers? There are many irrational numbers that cannot be written in simplified form. Some of the examples are: √8, √11, √50, Euler’s Number e = 2.718281, Golden ratio, φ= 1.618034.
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# How to Calculate Mean-Median-Mode and Range
Numbers and groups of numbers are all around us.
For example, what is the total number of students at your school?
Who is the oldest student or the youngest student?
Who runs the fastest mile and who has the highest test average?
A measure of central tendency is one way to organize data so you can figure out what it is telling you.
Three Measures of Central Tendencies include: MEAN, MEDIAN & MODE.
A measure of central tendency is a single number that is used to summarize all the values of a data set.
For example, you have this group of people. You may want to know if they are all close to the same age, or what is the average height or how many have dark hair ?
Let’s work an example.
What is the mean median and mode of this family?
The ages are 68,10,7,40,36,2.12,65
MEAN - the average of a set of numbers.
Mean and average are the same.
STEPS FOR CALCULATING THE MEAN:
1. Add all of the numbers in the data set
2. Divide by the total number of items in the data set
3. If there is a ZERO, it must be included!
Let's find the mean of the average ages of the family.
Let’s add the ages together 68,10,7,40,36,2.12,65 = 240
Now let’s divide by the number of numbers in the data set.
If there was a zero it would be included and we have 8 members.
So 240 divided by 8 = 30 so the average age is thirty
Now let’s calculate median
MEDIAN - the number in the center of a data set when the numbers are in order from least to greatest.
STEPS TO DETERMINING THE MEDIAN:
1. Arrange the numbers from least to greatest order.
2. Cross off the greatest and least numbers in the list (at the
same time) until you are left with just one number in the
middle - this is the MEDIAN!
3. If there are two numbers left in the middle, you have to find
the MEAN or AVERAGE of the two numbers.
Find the median of the ages of the family.
68,10,7,40,36,2.12,65
Let's arrange from least to greatest
Since we have an even number we need to take the average of the numbers in the middle.
So the mean or average of 12 and 36 is 48 divided by 2 which equals 24.
If you had an odd number it would just be the number in the middle.
Mode equals the number which occurs most often in a data set. There might be one mode, no mode, or many modes.
Arrange the numbers from least to greatest order.
See if any numbers repeat.
The number that repeats most often is the mode.
What is the mode of the family?
Let’s arrange from least to greatest
2,7,10,12,36,40,65,68
So no number repeats so you have no mode.
2,7,7,7,10,12,36,40,65,68
Then 7 would be the mode.
Finally let’s figure out the range for this data set.
RANGE is NOT a measure of central tendencies, it is a MEASURE OF VARIATION.
It is a measure of the variation between the greatest and lowest numbers in a data set.
Find the greatest number and SUBTRACT the least number.
What is the range of of the family?
Let"s order from least to greatest.
2,7,10,12,36,40,65,68
So the range is 68 -2 which equals 66
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Ch5_matrices
# Ch5_matrices - Chapter 5 Matrices Matlab began as a matrix...
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Chapter 5 Matrices Matlab began as a matrix calculator. The Cartesian coordinate system was developed in the 17th century by the French mathematician and philosopher Ren´ e Descartes. A pair of numbers corre- sponds to a point in the plane. We will display the coordinates in a vector of length two. In order to work properly with matrix multiplication, we want to think of the vector as a column vector, So x = ± x 1 x 2 denotes the point x whose first coordinate is x 1 and second coordinate is x 2 . When it is inconvenient to write a vector in this vertical form, we can anticipate Matlab notation and use a semicolon to separate the two components, x = ( x 1 ; x 2 ) For example, the point labeled x in figure 5.1 has Cartesian coordinates x = ( 2; 4 ) Arithmetic operations on the vectors are defined in natural ways. Addition is defined by x + y = ± x 1 x 2 + ± y 1 y 2 = ± x 1 + y 1 x 2 + y 2 Multiplication by a single number, or scalar , is defined by sx = ± sx 1 sx 2 Copyright c ± 2009 Cleve Moler Matlab R ± is a registered trademark of The MathWorks, Inc. TM August 8, 2009 1
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2 Chapter 5. Matrices A 2-by-2 matrix is an array of four numbers arranged in two rows and two columns. A = ± a 1 , 1 a 1 , 2 a 2 , 1 a 2 , 2 or A = ( a 1 , 1 a 1 , 2 ; a 2 , 1 a 2 , 2 ) For example A = ± 4 - 3 - 2 1 -10 -8 -6 -4 -2 0 2 4 6 8 10 -10 -8 -6 -4 -2 0 2 4 6 8 10 x Ax Figure 5.1. Matrix multiplication transforms lines through x to lines through Ax. Matrix-vector multiplication by a 2-by-2 matrix A transforms a vector x to a vector Ax , according to the definition Ax = ± a 1 , 1 x 1 + a 1 , 2 x 2 a 2 , 1 x 1 + a 2 , 2 x 2 For example ± 4 - 3 - 2 1 ¶± 2 4 = ± 4 · 2 - 3 · 4 - 2 · 2 + 1 · 4 = ± - 4 0 The point labeled x in figure 5.1 is transformed to the point labeled Ax . Matrix- vector multiplications produce linear transformations. This means that for scalars s and t and vectors x and y , A ( sx + ty ) = sAx + tAy
3 This implies that points near x are transformed to points near Ax and that straight lines in the plane through x are transformed to straight lines through Ax . Our definition of matrix-vector multiplication is the usual one involving the dot product of the rows of A , denoted a i, : , with the vector x . Ax = ± a 1 , : · x a 2 , : · x An alternate, and sometimes more revealing, definition uses linear combinations of the columns of A , denoted by a : ,j . Ax = x 1 a : , 1 + x 2 a : , 2 For example ± 4 - 3 - 2 1 ¶± 2 4 = 2 ± 4 - 2 + 4 ± - 3 1 = ± - 4 0 The transpose of a column vector is a row vector, denoted by x T . The trans- pose of a matrix interchanges its rows and columns. For example, x T = ( 2 4 ) A T = ± 4 - 2 - 3 1 Vector-matrix multiplication can be defined by x T A = A T x That is pretty cryptic, so if you have never seen it before, you might have to ponder it a bit. Matrix-matrix multiplication,
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Ch5_matrices - Chapter 5 Matrices Matlab began as a matrix...
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It is an improper fraction. If you want to describe a part of a whole, it is called a Fraction. No, It is not a proper fraction because the numerator 9 in this fraction is more than the denominator 8. The answer is 4, with a remainder of 3. So, basically, it is a way to divide or cut any object in smaller parts. It is also known as the fraction bar or vinculum. a ratio of functions where the degree of the numerator (the top number in a fraction) is less than the degree of the denominator (the bottom number Here are some examples of proper fractions: The numerator is less than the denominator, The numerator is greater than (or equal to) the denominator, A whole number and proper fraction together. top number Yes, It is a proper fraction because the numerator 1 in this fraction is less than the denominator 2. For example, a pizza is divided into four pieces, so each piece of it is represented as 1/4th of the pizza. Learn more. It is less than 1. So right now it's an improper fraction. Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. No, It is not a proper fraction because the numerator 6 in this fraction is more than the denominator 5. As you work on converting improper fractions, you’ll find yourself using your knowledge of division. Fractions can undergo many different operations, some of which are mentioned below. Using certain shortcuts, you can determine whether a fraction is reduced without completing any calculations. How to use improper fraction in a sentence. In generalising form for any two integers ‘a’ and ‘b’ Follow the given steps to convert an improper fraction to a mixed fraction: Example: Convert 1 3 ⁄ 5 to a mixed fraction. Hyponyms (each of the following is a kind of "proper fraction"): decimal; decimal fraction (a proper fraction whose denominator is a power of 10) The value of a proper fraction is greater than one. In both cases, fractions are presented in their lowest forms by dividing both numerator and denominator by their greatest common factor. For example- 2/3, 5/8 is a proper fraction. By a natural number we have meant a collection of indivisible ones. Is 2/3 is a proper fraction? 7/5 is an improper fraction because the numerator is greater than the denominator. A fraction with numerator 1 is called a unit fraction. In control theory, a proper transfer function is a transfer function in which the degree of the numerator does not exceed the degree of the denominator. So we can define the three types of fractions like this: So, a proper fraction is just a fraction where the numerator (the top number) is less than the denominator (the bottom number). Denominator – The number on the bottom of the fraction and it shows the number of equal parts, where the whole is divided into. A proper fraction has a numerator that is smaller than the denominator. In English, the names of the proper fractions are the same as the names of the parts, and therefore a fraction and a ratio have become confused. The picture given below clearly illustrates this. It is also known as the fraction bar or vinculum. The improperly called improper fractions are those greater than or equal to one. In proper fractions, the numerator is less than the denominator. It is also known as the fraction bar or vinculum. A mixed number is an integer (whole number) and a proper fraction. Nov 24,2020 - What is a proper fraction? It can only be written as mixed fraction =5 1/2. 6, for example, has only half, a third part, and a sixth part. A proper fraction is a fraction whose numerator is smaller than its denominator. To illustrate this fraction note that the circle shown has been divided into four parts, one of which has been shaded. Well, make them to have one and, subsequently, add the two: So first I'm just going to show you a fairly straightforward way of doing it and then we're going to think a little bit about what it actually means. Required fields are marked *. "One quarter" is the name of a part, or a ratio, which is a relationship between two numbers. Let's write it is a mixed number. (ex. Improper fractions can be written as mixed numbers. And mixed numbers have a whole number sitting next to a proper fraction -- for example, 4 3/6 or 1 1/2. There are 3 types of fractions, and an overview of them is given below: Point to know – If the numerator and denominator of a fraction are multiplied by the same number, its value doesn’t change. Every positive proper fraction is always greater than "0" but less than "1". Step 1: Multiply the denominator and the whole number. So, you see "0" may be a proper fraction, too. 0 < 3 ⁄ 7 < 1) But every negative proper fraction is always between (-1) and "0". Addition and subtraction, whether with whole numbers or fractions, are done with a number of units. The value of a proper fraction is less than one. In a proper fraction, the numerator is always less than its denominator. (A mixed fraction is a whole number and a fraction combined, like 1 2/3.) Your email address will not be published. You already know that proper fractions have numerators smaller than the denominators, such as 1/2, 2/10 or 3/4, making them equal less than 1. This gives rise to a new and different kind of number -- a fraction. Converting between fractions and decimals: Converting from decimals to fractions is straightforward. As long as the numerator is smaller than the denominator, you have a proper fraction. Every positive proper fraction is always greater than "0" but less than "1". Divide 27 by 6. Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time Improper Fractions Calculator Convert to/from Mixed Numbers and Improper Fractions step by step We call the line that separates 7 and 5 the division line. Any fraction having the numerator greater than or equal to the denominator (the top number larger than (or equal to) the bottom number) will be greater than or equal to one. A mixed number is a whole number and a fraction. Sometimes you may waste time trying to simplify fractions that cannot be further reduced. :) 0 0. The improper fraction has a numerator larger than the denominator. Source(s): Maths major. Example: 1/4 (one quarter) and 5/6 … (ex. A numerical fraction in which the numerator is less than the denominator. Fractions are the terms used to determine the parts of a whole object. Numerator – The numerator is the number on the top of the fraction, and it shows the number of the parts we are considering. The difference between the degree of the denominator (number of poles) and degree of the numerator (number of zeros) is the relative degree of the transfer function. Examples include \$\${\displaystyle {\tfrac {1}{2}}}\$\$, \$\${\displaystyle -{\tfrac {8}{5}}}\$\$, \$\${\displaystyle {\tfrac {-8}{5}}}\$\$, and \$\${\displaystyle {\tfrac {8}{-5}}}\$\$. proper fraction. An improper fraction is always 1 or greater than 1. Proper fraction definition is - a fraction in which the numerator is less or of lower degree than the denominator. tenths of an apple plus tenths of an apple make tenths of an apple. Let’s Solve a few problems on Propper fractions: Your email address will not be published. Fractions that are less than one are known as proper fractions, and the numerator (the top number) is less than the denominator (the bottom number). In general, And even more generally, And what if the fraction do not have the same denominator? Addition: Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. Like . Write down the improper fraction -- for example, 27/6. proper fraction in English translation and definition "proper fraction", Dictionary English-English online. Improper Fraction : In an improper fraction, the numerator will be equal to or larger than the denominator. 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# NCERT Solutions for Miscellaneous Exercise Chapter 12 Class 12 - Linear Programming
NCERT solutions for Class 12 Maths chapter 12 miscellaneous exercise gives more knowledge of solving a few types of linear programming problems using graphical methods. Miscellaneous exercise chapter 12 Class 12 are solved in a detailed manner with necessary steps and graphs. Going through NCERT solutions for Class 12 Maths chapter 12 miscellaneous exercise gives a better understanding of the linear programming problems. The questions discussed in the NCERT book Class 12 Maths chapter 12 miscellaneous solutions are a bit higher level as compared to the other two exercises. Class 12 Maths chapter 12 miscellaneous solutions are important for Class 12 CBSE Board Exams.
• Linear Programming Exercise 12.1
• Linear Programming Exercise 12.1
## Linear Programming Class 12 Chapter12-Miscellaneous Exercise
Question:1 Reference of Example 9 (Diet problem): A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol.
How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Let diet contain x packets of food P and y packets of food Q. Thus, $x\geq 0,y\geq 0$ .
The mathematical formulation of the given problem is as follows:
Total cost is Z . $Z=6x+3y$
Subject to constraint,
$4x+y\geq 80$
$x+5y\geq 115$
$x\geq 0,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(15,20),B(40,15),C(2,72)$
The value of Z at corner points is as shown :
corner points $Z=6x+3y$ $A(15,20)$ 150 MINIMUM $B(40,15)$ 285 maximum $C(2,72)$ 228
Hence, Z has a maximum value of 285 at the point $B(40,15)$ .
to maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A is 285 units.
Question:2 A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Let farmer mix x bags of brand P and y bags of brand Q. Thus, $x\geq 0,y\geq 0$ .
The given information can be represented in the table as :
Vitamin A Vitamin B Cost Food P 3 5 60 Food Q 4 2 80 requirement 8 11
The given problem can be formulated as follows:
Therefore, we have
$3x+1.5y\geq 18$
$2.5x+11.25y\geq 45$
$2x+3y\geq 24$
$Z=250x+200y$
Subject to constraint,
$3x+1.5y\geq 18$
$2.5x+11.25y\geq 45$
$2x+3y\geq 24$
$x\geq 0,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of the feasible region are $A(18,0),B(9,2),C(3,6),D(0,12)$
The value of Z at corner points is as shown :
corner points $Z=250x+200y$ $A(18,0)$ 4500 $B(9,2)$ 2650 $C(3,6)$ 1950 minimum $D(0,12)$ 2400
Feasible region is unbounded, therefore 1950 may or may not be a minimum value of Z. For this, we draw $250x+200y< 1950$ and check whether resulting half plane has a point in common with the feasible region or not.
We can see a feasible region has no common point with $250x+200y< 1950$ .
Hence, Z has a minimum value 1950 at point $C(3,6)$ .
Question:3 A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food Vitamin A Vitamin B Vitamin C X 1 2 3 Y 2 2 1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?
Let mixture contain x kg of food X and y kg of food Y.
Mathematical formulation of given problem is as follows:
Minimize : $z=16x+20y$
Subject to constraint ,
$x+2y\geq 10$
$x+y\geq 6$
$3x+y\geq 8$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(10,0),B(2,4),C(1,5),D(0,8)$
The value of Z at corner points is as shown :
corner points $z=16x+20y$ $A(10,0)$ 160 $B(2,4)$ 112 minimum $C(1,5)$ 116 $D(0,8)$ 160
The feasible region is unbounded , therefore 112 may or may not be minimum value of Z .
For this we draw $16x+20y< 112$ and check whether resulting half plane has point in common with feasible region or not.
We can see feasible region has no common point with $16x+20y< 112$ .
Hence , Z has minimum value 112 at point $B(2,4)$
Question:4 A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
Types of toys Machines I II III A 12 18 6 B 6 0 9
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Let x and y toys of type A and type B.
Mathematical formulation of given problem is as follows:
Minimize : $z=7.5x+5y$
Subject to constraint ,
$2x+y\leq 60$
$x\leq 20$
$2x+3y \leq 120$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(20,0),B(20,20),C(15,30),D(0,40)$
The value of Z at corner points is as shown :
corner points $z=7.5x+5y$ $A(20,0)$ 150 $B(20,20)$ 250 $C (15,30)$ 262.5 maximum $D(0,40)$ 200
Therefore 262.5 may or may not be maximum value of Z .
Hence , Z has maximum value 262.5 at point $C (15,30)$
Question:5 An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Let airline sell x tickets of executive class and y tickets of economy class.
Mathematical formulation of given problem is as follows:
Minimize : $z=1000x+600y$
Subject to constraint ,
$x+y\leq 200$
$x\geq 20$
$y-4x\geq 0$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(20,80),B(40,160),C(20,180)$
The value of Z at corner points is as shown :
corner points $z=1000x+600y$ $A(20,80)$ 68000 $B(40,160)$ 136000 maximum $C (20,180)$ 128000
therefore 136000 is maximum value of Z .
Hence , Z has maximum value 136000 at point $B(40,160)$
Question:6 Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs) From/To A B D 6 4 E 3 2 F 2.50 3
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Let godown A supply x and y quintals of grain to shops D and E respectively. Then , (100-x-y) will be supplied to shop F. Requirements at shop D is 60 since godown A supply x .Therefore remaining (60-x) quintals of grain will be transported from godown B.
Similarly, (50-y) quintals and 40-(100-x-y)=(x+y-60) will be transported from godown B to shop E and F respectively. The problem can be represented diagrammatically as follows:
$x,y\geq 0$ and $100-x-y\geq 0$
$x,y\geq 0$ and $x+y\leq 100$
$60-x\geq 0,50-y\geq 0\, \, \, and\, \, x+y-60\geq 0$
$\Rightarrow \, \, \, \, x\leq 60,y\leq 50,x+y\geq 60$
Total transportation cost z is given by ,
$z=6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)$
$z=2.5x+1.5y+410$
Mathematical formulation of given problem is as follows:
Minimize : $z=2.5x+1.5y+410$
Subject to constraint ,
$x+y\leq 100$
$x\leq 60$
$y\leq 50$
$x+y\geq 60$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(60,0),B(60,40),C(50,50),D(10,50)$
The value of Z at corner points is as shown :
corner points $z=2.5x+1.5y+410$ $A(60,0)$ 560 $B(60,40)$ 620 $C(50,50)$ 610 $D(10,50)$ 510 minimum
therefore 510 may or may not be minimum value of Z .
Hence , Z has miniimum value 510 at point $D(10,50)$
Question:7 An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Distance in (km.) From/To A B D 7 3 E 6 4 F 3 2
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Let x and y litres of oil be supplied from A to petrol pump,D and E. Then , (7000-x-y) will be supplied from A to petrol pump F.
Requirements at petrol pump D is 4500 L. since x L A are transported from depot A,remaining 4500-x L will be transported from petrol pump B
Similarly, (3000-y)L and 3500-(7000-x-y)=(x+y-3500) L will be transported from depot B to petrol E and F respectively.
The problem can be represented diagrammatically as follows:
$x,y\geq 0$ and $7000-x-y\geq 0$
$x,y\geq 0$ and $x+y\leq 7000$
$4500-x\geq 0,3000-y\geq 0\, \, \, and\, \, x+y-3500\geq 0$
$\Rightarrow \, \, \, \, x\leq 4500,y\leq 3000,x+y\geq 3500$
Cost of transporting 10 L petrol =Re 1
Cost of transporting 1 L petrol $=\frac{1}{10}$
Total transportation cost z is given by ,
$z=\frac{7}{10}x+\frac{6}{10}y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)$
$z=0.3x+0.1y+3950$
Mathematical formulation of given problem is as follows:
Minimize : $z=0.3x+0.1y+3950$
Subject to constraint ,
$x+y\leq 7000$
$x\leq 4500$
$y\leq 3000$
$x+y\geq 3500$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(3500,0),B(4500,0),C(4500,2500),D(4000,3000),E(500,3000)$
The value of Z at corner points is as shown :
corner points $z=0.3x+0.1y+3950$ $A(3500,0)$ 5000 $B(4500,0)$ 5300 $C(4500,2500)$ 5550 $E(500,3000)$ 4400 minimum $D(4000,3000)$ 5450
Hence , Z has miniimum value 4400 at point $E(500,3000)$
Question:8 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
Kg per bag Brand P Brand Q Nitrogen 3 3.5 Phosphoric Acid 1 2 Potash 3 1.5 Chlorine 1.5 2
Let fruit grower use x bags of brand P and y bags of brand Q.
Mathematical formulation of given problem is as follows:
Minimize : $z=3x+3.5y$
Subject to constraint ,
$x+2y\geq 240$
$x+0.5y\geq 90$
$1.5x+2y\geq 310$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(140,50),C(40,100),B(20,140)$
The value of Z at corner points is as shown :
corner points $z=3x+3.5y$ $A(140,50)$ 595 $B(20,140)$ 550 $C(40,100)$ 470 minimum
Therefore 470 is minimum value of Z .
Hence , Z has minimum value 470 at point $C(40,100)$
Question:9 Reference of Que 8 : A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Kg per bag Brand A Brand P Nitrogen 3 3.5 Phosphoric Acid 1 2 Potash 3 1.5 Chlorine 1.5 2
Let fruit grower use x bags of brand P and y bags of brand Q.
Mathematical formulation of given problem is as follows:
Maximize : $z=3x+3.5y$
Subject to constraint ,
$x+2y\geq 240$
$x+0.5y\geq 90$
$1.5x+2y\geq 310$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $B(20,140),A(140,50),C(40,100)$
The value of Z at corner points is as shown :
corner points $z=3x+3.5y$ $A(140,50)$ 595 maximum $B(20,140)$ 550 $C(40,100)$ 470 minimum
therefore 595 is maximum value of Z .
Hence , Z has minimum value 595 at point $A(140,50)$
Question:10 A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Let x and y be number of dolls of type A abd B respectively that are produced per week.
Mathematical formulation of given problem is as follows:
Maximize : $z=12x+16y$
Subject to constraint ,
$x+y\leq 1200$
$y\leq \frac{x}{2}\Rightarrow x\geq 2y$
$x-3y\leq 600$
$x,y\geq 0$
The feasible region determined by constraints is as follows:
The corner points of feasible region are $A(600,0),B(1050,150),C(800,400)$
The value of Z at corner points is as shown :
corner points $z=12x+16y$ $A(600,0)$ 7200 $B(1050,150)$ 15000 $C(800,400)$ 16000 Maximum
Therefore 16000 is maximum value of Z .
Hence , Z has minimum value 16000 at point $C(800,400)$
## More About NCERT Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise:
There are 10 questions in the miscellaneous exercise chapter 12 Class 12. Solving all these questions gives a good knowledge about the NCERT Class 12th chapter linear programming. Students have to solve the NCERT syllabus exercises and solved examples in order to get a good idea of topics discussed in the chapter and to get a good score in the final exam.
Also Read| Linear Programming Class 12th Notes
## Benefits of NCERT Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise.
• Important questions to understand the concepts explained in the chapter are given in Class 12 Maths chapter 12 miscellaneous exercise solutions.
• All questions of miscellaneous exercise chapter 12 Class 12 are important and will be helpful in the board exam.
## Also see-
• NCERT Exemplar Solutions Class 12 Maths Chapter 12
• NCERT Solutions for Class 12 Maths Chapter 12
## NCERT Solutions Subject Wise
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for Class 12 Biology
• NCERT Solutions for Class 12 Mathematics
## Subject Wise NCERT Exemplar Solutions
• NCERT Exemplar Class 12 Maths
• NCERT Exemplar Class 12 Physics
• NCERT Exemplar Class 12 Chemistry
• NCERT Exemplar Class 12 Biology
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# what is the first step in solving a quadratic equation
## What Is The First Step In Solving A Quadratic Equation?
The first step in solving quadratic equations by finding square roots is b. isolate the x² squared by using inverse operations. Step 1: Isolate the x² squared by using inverse operations. Step 2: Square root both sides to isolate x.Apr 24, 2020
## What are the steps to solving a quadratic equation?
Steps for solving Quadratic application problems:
• Draw and label a picture if necessary.
• Define all of the variables.
• Determine if there is a special formula needed. Substitute the given information into the equation.
• Write the equation in standard form.
• Factor.
• Set each factor equal to 0. …
• ## What are the 4 steps to solve a quadratic equation?
The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.
## What are three steps for solving a quadratic equation?
There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square.
## What are the steps in solving quadratic equation by extracting the roots?
Extracting Square Roots
• Step 1: Express the quadratic equation in standard form.
• Step 2: Factor the quadratic expression.
• Step 3: Apply the zero-product property and set each variable factor equal to 0.
• Step 4: Solve the resulting linear equations.
• ## What are 5 methods of solving a quadratic equation?
There are several methods you can use to solve a quadratic equation: Factoring Completing the Square Quadratic Formula Graphing
• Factoring.
• Completing the Square.
• Graphing.
## What are the roots of a quadratic equation?
The roots of a function are the x-intercepts. By definition, the y-coordinate of points lying on the x-axis is zero. Therefore, to find the roots of a quadratic function, we set f (x) = 0, and solve the equation, ax2 + bx + c = 0.
Quadratic sequences are sequences that include an term. They can be identified by the fact that the differences between the terms are not equal, but the second differences between terms are equal.
: a formula that gives the solutions of the general quadratic equation ax2 + bx + c = 0 and that is usually written in the form x = (-b ± √(b2 − 4ac))/(2a)
## Which of the quadratic equation has roots 3 5?
Hence, the correct answer is x2 – 8x + 15 = 0.
## How do you find the first term in a quadratic sequence?
Answer: The first differences are 6,8,10,12,14,16 and the second differences are 2. Therefore the first term in the quadratic sequence is n^2. Subtracting n^2 from the sequence gives 7, 10, 13, 15, 18, 21, and the nth term of this linear sequence is 3n + 4. So the final answer of this sequence is n^2 + 3n + 4.
## What is the formula of sequence?
An arithmetic sequence is a sequence in which the difference between each consecutive term is constant. An arithmetic sequence can be defined by an explicit formula in which an = d (n – 1) + c, where d is the common difference between consecutive terms, and c = a1.
## Why is the quadratic formula?
The quadratic formula helps you solve quadratic equations, and is probably one of the top five formulas in math. … Then the formula will help you find the roots of a quadratic equation, i.e. the values of x where this equation is solved.
## What are the roots of the quadratic equation x² − 4x 1 0?
Answer and Explanation: The solutions to x2 + 4x + 1 = 0 are x = -2 + √(3) or x = -2 – √(3).
## Is x2 4x 11 x2 a quadratic equation?
⇒ 4 x = 11 Thus, x2 + 4x = 11 + x2 is not a quadratic equation. … 2x – x2 = x2 + 5 can be written as 2 x 2 − 2 x + 5 = 0 So, it also forms a quadratic equation. Hence, the correct answer is x2 + 4x = 11 + x2.
## What is the best method to be used in solving quadratic equation?
Quadratic formula – is the method that is used most often for solving a quadratic equation. If you are using factoring or the quadratic formula, make sure that the equation is in standard form.
## What are the first three terms of 3n 2?
To find out the first three terms of 3n + 2 substitute 1 ,2 and 3 into the equation. 3(1)+2=5 3(2)+2=8 3(3)+2=11 As you can see the sequence goes up in 3s 5 ,8, 11 To find out the 10th term you also substitute 10 into the equation so 3(10)+2=32 Hope this helped!
## How do you find the first difference in an equation?
You find the first difference between values of the dependent variable by subtracting the previous value from each. To find first differences determine by how much the dependent value is increasing or decreasing, also called the change in the dependent variable.
## What is the first differences in quadratic relation?
The first difference (the difference between any two successive output values) is the same value (3). This means that this data can be modeled using a linear regression line. This is a quadratic model because the second differences are the differences that have the same value (4).
## What do first differences tell you?
First differences (and second and third differences) help determine whether there is a pattern in a set of data, as well as the nature of the pattern.
## What are the 4 types of sequence?
There are mainly four types of sequences in Arithmetic, Arithmetic Sequence, Geometric Sequence, Harmonic Sequence, and Fibonacci Sequence.
## Where is the quadratic formula used?
Quadratic equations are actually used in everyday life, as when calculating areas, determining a product’s profit or formulating the speed of an object. Quadratic equations refer to equations with at least one squared variable, with the most standard form being ax² + bx + c = 0.
## How many solutions does the equation 4×2 4x 1 0 have?
4.3 Solving 4×2-4x+1 = 0 by the Quadratic Formula . This quadratic equation has one solution only.
## How do you complete the square method?
Steps
• Step 1 Divide all terms by a (the coefficient of x2).
• Step 2 Move the number term (c/a) to the right side of the equation.
• Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.
• ## What is the value of K if the roots of x2 +KX +K 0 are real and equal a 0 B 4 C 0 or 4 D 2?
Answer: Values of k are 0 and 4.
## What is the nature of the roots of the quadratic equation 4x square 8x +9 0?
Therefore, Quadratic equation has no real roots.
## What is the value of the K?
The value of K in free space is 9 × 109.
## What is the first step to solve this quadratic equation X2-40=0and so
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# Rational Numbers
## Rational Numbers
### What is a rational number?
Rational numbers are those numbers that can be represented as a fraction - numerator and denominator.
- Every whole number is a rational number
- Decimal numbers are also rational numbers
- Rational numbers can also be negative
#### Examples of rational numbers
$1, \frac{1}{2}, -\frac{3}{4}, 4.5$
## Rational Numbers
In this article, you will learn everything you need to know about rational numbers and you will practice it in various exercises.
Shall we start?
### What is a rational number?
Rational numbers are those numbers that can be represented as the quotient of integers - numerator and denominator.
In other words – A rational number is a fraction.
Observe: Every rational number can be represented as a fraction in several ways.
For example, the rational number $\frac{2}{4}$ can be written like this: $\frac{1}{2}$ or like this $\frac{3}{6}$ or, even, like this: $\frac{4}{8}$
Every integer is a rational number since we can write it in fraction form with the denominator $1$.
For example, the integer $4$, can be written as a rational number in the following way: $\frac{4}{1}$.
### And what about the$0$?
The $0$ can be represented as a rational number in this way: $\frac{0}{1}$.
Rational numbers can be negative and also decimals.
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## And now what do we do? Let's practice!
### Determine which numbers are rational within the following set
$22, 0, 5-, 6$
Solution:
All numbers are rational. They can be written as a quotient in the following way:
$22$ : represented as a fraction $\frac{22}{1}$
$0$ : represented as a fraction $\frac{0}{1}$
$-5$ : represented as a fraction $-\frac{5}{1}$
$6$ : represented as a fraction $\frac{6}{1}$
### Determine which numbers are rational within the following set
$1, -16, -\frac{2}{3}, \frac{7}{6}$
Solution:
All the numbers are rational. They can be written as a quotient in the following way:
$1$ : represented as a fraction $\frac{1}{1}$
$-16$ : represented as a fraction $-\frac{16}{1}$
$-\frac{2}{3}$: Rational number represented as a fraction with numerator and denominator
$\ \frac{7}{6}$: Rational number represented as a fraction with numerator and denominator.
### Determine which numbers are rational within the following set
$0.5, -102, 13, 2\frac{4}{5}$
Solution:
All numbers are rational. They can be written as a quotient in the following way:
$0.5$ : Rational number, can be represented as a fraction $\frac{1}{2}$
$-100$ : Rational number, can be represented as a fraction $-\frac{100}{1}$
$13$ : Rational number, can be represented as a fraction $\frac{13}{1}$
$2\frac{4}{5}$ : Rational number, can be represented as a fraction with only numerator and denominator in the following way:$\frac{14}{5}$
## Now that we know how to identify rational numbers, we will move on to the corresponding exercises.
### Find the rational numbers that are equivalent to the given rational number
$\frac{4}{5}=⬜ =⬜=⬜$
Solution:
We need to find rational numbers, that is, fractions, that are equivalent to the rational number $\frac{4}{5}$
By expanding the fraction, we can arrive at fractions that express the same value.
Let's expand $\frac{4}{5}$ by $2$. Remember that we must expand both the numerator and the denominator so as not to change the value of the fraction.
We will obtain: $\frac{4}{5} = \frac{8}{10}$
Now let's expand the rational number $\frac{4}{5}$ by $3$.
We will find that:
$\frac{4}{5} = \frac{12}{15}$
We have to find $1$ more fraction, that is equivalent to the given fraction.
This time we will expand by $4$.
Observe, we can expand the fraction as much as we want. As long as the same operation is performed on both the numerator and the denominator, the value of the fraction will remain unchanged.
We will expand by $4$ and we will obtain:
$\frac{4}{5} = \frac{16}{20}$
And, in this way, we can conclude the exercise:
$\frac{4}{5} = \frac{8}{10}= \frac{12}{15}= \frac{16}{20}$
### Another exercise with rational numbers
Complete the figures in rational numbers to obtain any true statement.
$\frac{⬜}{6}<\frac{2}{3}<\frac{⬜}{9}$
Solution:
This exercise might seem a bit complicated at first, but, if we understand the trick, we will solve it very easily.
One of the methods to match fractions is by applying a common denominator.
When both fractions have a common denominator, it is very easy for us to realize which one of them is larger and which is smaller according to the numerator.
In this exercise, we are looking for a rational number with denominator $9$ that is greater than $\frac{2}{3}$.
$\frac{2}{3}$ are, in fact, \frac{6/9} and, therefore, \frac{7/9} for example, will be greater than \frac{2/3}
Additionally, we are looking for a rational number with denominator $6$ that is less than \frac{2/3}
\frac{2/3} are, in fact, \frac{4/6} and, therefore, \frac{3/6} for example, will be less than \frac{2/3}
We will obtain:
\frac{3/6}<\frac{2/3}<\frac{7/9}
|
Master the 7 pillars of school success
## What is a triangle? In addition to having three sides and three angles,there are four basic rules or theorems that a shape must satisfy in order to be classified as a triangle.
Theorem 1. This theorem states that when adding the side lengths of a triangle, the sum of two sides of a triangle must be greater than the length of the third side.
# Types of Triangles
3
4
6
6 + 3 > the third side 4
4+3 > the third side 6
6+4 > the third side 3
Interior angles = 180
60 + 60 +60 = 180 degrees
30°
150°
Interior and Exterior angles are supplementary
Theorem 2. The sum of the interior angles of a triangle equal 180 degrees.
Theorem 3. The measure of an exterior angle is formed by extending one of the sides. Furthermore, the interior and exterior angles created, are supplementary.
Theorem 4.Proportionality of Triangles. The longest side of a triangle is opposite the largest angle, and the shortest side is opposite the smallest angle.
Longest Side
Shortest Side
Triangles can be classified by the length of their sides.
Scalene triangle
Three different side lengths
Three different angle measures
Isosceles triangle
At least two congruent sides
At least two congruent angles
Equilateral triangle
Three congruent sides
Three congruent angles The angle measure equals 60 degrees
Triangles can also be classified by their angles.
There are three angles often associated with triangles
• Acute angles are less than 90 degrees
• Right angles are exactly 90 degrees
• Obtuse are greater than 90 degrees
All triangles have two acute angles and the third angle determines the type of
Triangle you have.
Acute triangle
Triangle with three angles that measure less than 90 degrees
Obtuse triangle
Triangle with one angle greater than 90 degrees
Right triangle
Triangle with one right angle ( 90 degrees)
The other two angles are complementary
|
# First Derivative Test
Learn about first derivative test, its formula along with different examples. Also find ways to calculate using first derivative test.
Alan Walker-
Published on 2023-05-26
## Introduction to First derivative test
In calculus, the first derivative test is a fundamental derivative concept and is important in analyzing a function over some domain. This test calculates the rate of change of a function by finding its critical points. It has many applications in calculus and physics where the rate of change is involved. Let’s understand how the 1st derivative test for extreme points is used.
## Understanding the First Derivative Test
The first derivative test is a method to calculate the local maximum and minimum points of a function. It helps to determine whether the function changes from increasing to decreasing or vice versa. Since this concept is based on a function's rate of change, the first derivative is used. The first derivative of a function is calculated by differentiating the function one time. By definition, the first derivative of a function f(x) is defined as;
$f’(x)=\lim_{h→0}\frac{f(x+h)-f(x)}{h}$
Although a function can be monotonically analyzed without using the first derivative, calculus is a subject that helps to analyze a function easily. A function is differentiated to find its critical points. The first derivative test helps to identify whether these critical points are local maximum, local minimum or saddle points. Before understanding the formula of the first derivative test, let’s discuss what critical points are.
## What are critical points?
The critical points are the points on the graph of a function where the function’s derivative is zero or does not exist. These points are important to consider to determine local maximum and minimum value of a function. Critical point of a function is defined as;
“The point x of a function f(x) is called critical point if f’(x) = 0 or f’(x) > 0.”
## What are local maximum and minimum values?
The local maximum and minimum values are the points on the graph of a function where the function takes maximum and minimum values. A point x is known as local maximum if f(x)>0 and is a local minimum if f(x) < 0. These points are used in the first derivative and second derivative test.
## First derivative test formula
The first derivative test is used to find the behaviour of a function on the points in its domain. If a function f(x) is defined on an open interval I and there is a point c in the domain of f(x) then the first derivative formula is,
• If f’(x) > 0, i.e. f’(x) changes from positive to negative, then f(c) is the local maximum point of f(x).
• If f’(x) < 0 i.e. f’(x) changes from negative to positive, then f(c) is the local minimum point of f(x).
• If f’(x) = 0 i.e. f’(x) does not change. Then f(c) is neither local maximum or minimum.
The above statements are key points for the first derivative test.
## How to apply the first derivative test?
The 1st derivative test can be applied by finding the derivative of the given function. You can also follow the first derivative test steps to simplify calculations. These steps are:
1. Write the function and identify the independent variable.
2. Calculate the first derivative of f(x), i.e., f’(x) and use the relevant rules according to the type of function. For example, you can use the product rule if there is a product of two functions depending on the same variable.
3. Equate f’(x) to zero to identify the critical points.
4. Analyze the interval where the function is increasing or decreasing and determine the local maximum and minimum points.
Let’s understand the implementation of the first derivative test in the following examples.
### First Derivative Test Example
Find the local maximum and local minimum values by using 1st derivative test for the function,
$f(x)=3x^4+4x^3-12x^2+12$
Evaluating derivative with respect to x.
$f'(x)=\frac{d}{dx}[3x^4+4x^3 –12x^2+12]$
Since the function involves power functions, so by using the power rule of derivative,
$f’(x) = 12x^3+12x^2 –24x$
Now to find critical points, substitute f’(x) = 0.
$12x^3+12x^2 –24x = 0$
Simplifying,
$12x(x^2+x – 2) = 0$
Factorizing,
$12x[x^2+2x – x – 2]=0$
Or,
$12x[x(x+2) – 1(x+2)] = 0$
$12x(x+2)(x – 1) = 0$
The critical points are 0, -2 and 1 which means that f’(x) is 0 at 0, -2 and 1. Now we have to determine the local maxima and minima. For this, first substitute x=-2 in the given function.
$f(-2) = 3(-2)^4 + 4(-2)^3 – 12(-2)^2 + 12$
$f(-2)= 48 – 32 – 48 + 12$
$f(-2)= -20$
$\text{Minimum point} = (-2, -20)$
Now substituting x = 0 in f(x),
$f(0) = 3(0)^4 + 4(0)^3 – 12(0)^2 + 12 = 12$
$\text{Maximum point} = (0, 12)$
Substituting $x = 1$ in f(x),
$f(1) = 3(1)^4 + 4(1)^3 – 12(1)^2 + 1$
$f(1)= 3 + 4 – 12 + 12$
$f(1)= 7$
$\text{Minimum point} = (1, 7)$
Hence, the local extreme points are:
$\text{Local minimum} = (-2, -20)\quad\text{and}\quad(1, 7)$
$\text{Local maxima} = (0, 12)$
## Comparison between first derivative and second derivative test
The comparison between the higher derivative and second derivative test can be easily analysed using the following difference table.
First Derivative Test Second order derivative test The 1st derivative test is used to analyse a function either it is changing from positive to negative or negative to positive. The second derivative test is used to determine whether the function is increasing or decreasing. This test depend upon the critical points of the function. If f’(x)>0 at c, a point in its domain, f(c) is local maxima. Whereas if f’(x)<0 at c, f(c) will be local minima. A function is differentiated two time to identify its nature. If f’’(x) >0, the curve will be concave up. Whereas the curve will be concave down if f’’(x)<0. The first derivative test along with different derivative rules can be used to evaluate derivative. The second derivative test can also be used with derivative rules to find rate of change.
## Conclusion
The first derivative test is a method to identify the change in a function, whether it is changing from positive to negative or negative to positive. The critical points of the function are used to apply this test. The function then substitutes these points to calculate local maxima and minima. Hence this test has many applications in calculus and physics because it allows us to find the change in a function at a specific point.
|
# Lesson 18
Using Data to Solve Problems
## 18.1: Wild Bears (5 minutes)
### Warm-up
This warm-up allows students to review two important ideas of this unit: interpreting data in a box plot and writing statistical questions based on a data set. Students write statistical questions based on given box plots, then trade questions to answer questions written by another student.
### Launch
Arrange students in groups of 2. Tell students that, for the first question, one partner should write two questions about the head lengths and the other partner should write two questions about the head widths. For the second question, they should exchange and review each other's questions. If their partner's question does not seem to be a statistical question, suggest a revision so that it becomes a statistical question, and then answer the question. Remind students to consider units of measurement.
Give students 2 minutes of quiet work time for the first question and 2 minutes for collaboration afterwards.
### Student Facing
In one study on wild bears, researchers measured the head lengths and head widths, in inches, of 143 wild bears. The ages of the bears ranged from newborns (0 years) to 15 years. The box plots summarize the data from the study.
1. Decide if each question is a statistical question.
2. Use the box plots to answer each question.
### Activity Synthesis
Ask several students to share their questions about the head width and head length. Record and display their responses for all to see. After each student shares, ask the class if they agree or disagree it is a statistical question. If they agree, ask how they would find the answer, or for the answer itself. If they disagree, ask how they could rewrite the question so it is a statistical question.
## 18.2: Math Homework (Part 1) (15 minutes)
### Optional activity
In this activity, students compare and contrast different measures of center and variability for data sets that have gaps and are not symmetrical. They interpret mean, MAD, median, and IQR in the context of a situation. Unlike many of the data sets students have seen so far, this one shows values that could roughly divide into three parts: the days when there is little or no homework, the days when there is a moderate number of homework problems, and the days when the assignment is relatively large. Because of this distribution, finding a typical number of homework problems (or whether it would be helpful to identify a typical number) is not obvious, prompting students to interpret measures of center and spread more carefully (MP2).
As students work and discuss, identify at least one student or group that decides that the mean and MAD are appropriate measures of center and spread and can explain their reasoning, and another that decides to go with the median and IQR and could support their choice. Invite them to share during whole-class discussion.
### Launch
Keep students in groups of 2. Give students a moment of quiet time to look at the data on homework problems and identify at least one thing they notice and one thing they wonder. Give them another brief moment to share their observation and question with their partner. Then, ask a few students to share their responses with the class.
Students are likely to notice that the data values are quite different, that there are some days with no homework and others with quite a few problems, that there is not an obvious cluster, or that the number of problems could be roughly grouped into three kinds (a little, moderate, and a lot). They are likely to wonder why the numbers are so spread out and varied.
Briefly discuss the following questions to encourage students to think about the data contextually:
• “Why might the homework assignment data show this distribution? What are some possible explanations?” (When only one problem was assigned, the problem might be particularly challenging or might require considerable work or collaboration. Another possibility: there might be an upcoming exam, so the homework load was reduced. When many problems were assigned, the problems might be quick exercises with short answers, or the assignment might be review materials for an entire chapter.)
• “How might we describe ‘a typical number of homework problems’ in this case?”
• “Which do you predict would be higher: the mean or the median number of problems? Why?”
Next, give students 8–10 minutes to complete the task, either independently or collaboratively. Ask students to think quietly about the last question before discussing their response with their partner.
If students are using the digital activities, they will need to enter the data points in the column “A” for the applet to “list”, “sort”, etc. The applet allows for students to populate their own mean, Q1 and Q3 values.
### Student Facing
Over a two-week period, Mai recorded the number of math homework problems she had each school day.
• 2
• 15
• 20
• 0
• 5
• 25
• 1
• 0
• 10
• 12
1. Calculate the following. Show your reasoning.
1. The mean number of math homework problems.
2. The mean absolute deviation (MAD).
2. Interpret the mean and MAD. What do they tell you about the number of homework problems Mai had over these two weeks?
3. Find or calculate the following values and show your reasoning.
1. The median, quartile, maximum, and minimum of the same data on Mai’s math homework problems.
2. The interquartile range (IQR).
4. Which pair of measures of center and variability—mean and MAD, or median and IQR—do you think summarize the distribution of Mai’s math homework assignments better? Explain your reasoning.
You may use the applet below to help if you choose to. Begin by dragging the left edge across the screen until you see only one column in the spreadsheet. Enter the values needed to calculate the IQR and the mean when prompted.
### Launch
Keep students in groups of 2. Give students a moment of quiet time to look at the data on homework problems and identify at least one thing they notice and one thing they wonder. Give them another brief moment to share their observation and question with their partner. Then, ask a few students to share their responses with the class.
Students are likely to notice that the data values are quite different, that there are some days with no homework and others with quite a few problems, that there is not an obvious cluster, or that the number of problems could be roughly grouped into three kinds (a little, moderate, and a lot). They are likely to wonder why the numbers are so spread out and varied.
Briefly discuss the following questions to encourage students to think about the data contextually:
• “Why might the homework assignment data show this distribution? What are some possible explanations?” (When only one problem was assigned, the problem might be particularly challenging or might require considerable work or collaboration. Another possibility: there might be an upcoming exam, so the homework load was reduced. When many problems were assigned, the problems might be quick exercises with short answers, or the assignment might be review materials for an entire chapter.)
• “How might we describe ‘a typical number of homework problems’ in this case?”
• “Which do you predict would be higher: the mean or the median number of problems? Why?”
Next, give students 8–10 minutes to complete the task, either independently or collaboratively. Ask students to think quietly about the last question before discussing their response with their partner.
If students are using the digital activities, they will need to enter the data points in the column “A” for the applet to “list”, “sort”, etc. The applet allows for students to populate their own mean, Q1 and Q3 values.
### Student Facing
Over a two-week period, Mai recorded the number of math homework problems she had each school day.
2
15
20
0
5
25
1
0
10
12
1. Calculate the following. Show your reasoning.
1. The mean number of math homework problems
2. The mean absolute deviation (MAD)
2. Interpret the mean and MAD. What do they tell you about the number of homework problems Mai had over these two weeks?
3. Find or calculate the following values and show your reasoning.
1. The median, quartiles, maximum, and minimum of Mai’s data
2. The interquartile range (IQR)
4. Which pair of measures of center and variability—mean and MAD, or median and IQR—do you think summarizes the distribution of Mai’s math homework assignments better? Explain your reasoning.
### Activity Synthesis
Briefly discuss students’ interpretations of the measures they just calculated:
• “What do the mean of 9 and MAD of 7.4 tell us? How can we interpret them in this context?”
• “What do the median of 7.5 and IQR of 14 tell us?”
Then, select two or more previously identified students to share their responses about which measures of center and spread are appropriate for summarizing the data set. After each person shares, briefly poll the class to see if others reasoned about the measures the same way. Sum up by asking:
• “Now that you have two pairs of measures of center and spread, how would you respond if someone asked you, ‘What is a typical number of homework problems for Mai’s class?’ Is the question easier to answer now?”
Students should walk away with increased awareness that, in some cases, measures of center and spread do not always paint a full picture of what the actual data set entails, and that the measures should be interpreted with care.
## 18.3: Math Homework (Part 2) (15 minutes)
### Optional activity
In the previous activity, students considered appropriate measures of center and spread for describing distributions. Here, they show the same data set using three different kinds of graphical representations—a dot plot, a box plot, and histograms using different bin sizes—and decide which are more useful or more appropriate for communicating the distribution.
As students work and discuss, identify those who draw clear graphical displays, those who noticed that the different displays offer different insights about the data distribution, and those who advocate for using different representations to display Jada's data. Ask them to share with the class later.
### Launch
Explain to students that they will now represent Jada's homework data graphically and think about which representation(s) might appropriately communicate the distribution of her data. Give students 4–5 quiet minutes to draw a dot plot and a box plot (the first two questions), and then another 4–5 minutes to collaborate on drawing histograms with different bin sizes. Ask each student in a group to be in charge of one histogram with a particular bin size. After all representations are drawn, students should analyze them and discuss the last question in their group.
Classes using the digital version have an applet to create the statistical graphs. Data must be entered as a list, in curved brackets, separated by commas. Choices for histogram settings appear when that graph is selected.
### Student Facing
Jada wanted to know whether a dot plot, a histogram, or a box plot would best summarize the center, variability, and other aspects of her homework data.
• 2
• 15
• 20
• 0
• 5
• 25
• 1
• 0
• 10
• 12
1. Use the axis to make a dot plot to represent the data. Mark the position of the mean, which you calculated earlier, on the dot plot using a triangle ($$\Delta$$). From the triangle, draw a horizontal line segment to the left and right sides to represent the MAD.
2. Use the five-number summary from the previous task and the grid to draw a box plot that represents Jada’s homework data.
3. Work with your group to draw three histograms to represent Jada’s homework data. The width of the bars in each histogram should represent a different number of homework problems, as specified.
1. The width of one bar represents 10 problems.
2. The width of one bar represents 5 problems.
3. The width of one bar represents 2 problems.
4. Which of the five representations should Jada use to summarize her data? Should she use a dot plot, box plot, or one of the histograms? Explain your reasoning.
You can use the applet to make each type of graph if you choose to. Begin by dragging the gray bar from the top of the applet down until you see all of the command boxes.
### Launch
Explain to students that they will now represent Jada's homework data graphically and think about which representation(s) might appropriately communicate the distribution of her data. Give students 4–5 quiet minutes to draw a dot plot and a box plot (the first two questions), and then another 4–5 minutes to collaborate on drawing histograms with different bin sizes. Ask each student in a group to be in charge of one histogram with a particular bin size. After all representations are drawn, students should analyze them and discuss the last question in their group.
Classes using the digital version have an applet to create the statistical graphs. Data must be entered as a list, in curved brackets, separated by commas. Choices for histogram settings appear when that graph is selected.
### Student Facing
Jada wanted to know whether a dot plot, a histogram, or a box plot would best summarize the center, variability, and other aspects of her homework data.
2
15
20
0
5
25
1
0
10
12
1. Use the axis to make a dot plot to represent the data. Mark the position of the mean, which you calculated earlier, on the dot plot using a triangle ($$\Delta$$). From the triangle, draw a horizontal line segment to the left and right sides to represent the MAD.
2. Draw a box plot that represents Jada’s homework data.
3. Work with your group to draw three histograms to represent Jada’s homework data. The width of the bars in each histogram should represent a different number of homework problems, which are specified as follows.
1. The width of one bar represents 10 problems.
2. The width of one bar represents 5 problems.
3. The width of one bar represents 2 problems.
4. Which of the five representations should Jada use to summarize her data? Should she use a dot plot, box plot, or one of the histograms? Explain your reasoning.
### Activity Synthesis
Invite previously identified students to share their dot plot, box plot, and histograms. Display their drawings for all to see. Then, select several students or groups to share their response to the last question (which representation should Jada choose?) and their explanation. If not already mentioned by students, discuss the different insights that each display offers, or different challenges it poses. (Some possible observations are listed under Student Response section.) For instance, consider asking the following questions about each data display:
• “What information can we get from this display?”
• “Does it give us a meaningful snapshot of the distribution?”
• “What characteristics of a different data set would make this representation more useful?”
Help students see that, in this case, none of the representations here are ill-suited to represent the data set, but a couple of them (e.g. the box plot, or the first histogram with a bin size of 10) allow us to describe the distribution the data set more easily because of how they summarize the data values in some ways.
Representation: Internalize Comprehension. Use color coding and annotations to highlight differences between how dot plots, box plots, and histograms represent the data.
Supports accessibility for: Visual-spatial processing
Representing, Listening: MLR2 Collect and Display. As students discuss which representation Jada should choose, collect students’ responses in a graphic organizer, such as a Venn diagram, and display for all to see. Throughout the remainder of the lesson, continue to update collected student language and remind students to borrow language from the display as needed. Chart language related to dot plot, box plot, and histogram representations. This will help students to use mathematical language during paired and group discussions.
Design Principle(s): Support sense-making; Maximize meta-awareness
## 18.4: Will the Yellow Perch Survive? (30 minutes)
### Optional activity
In this culminating activity, students use what they have learned in the unit to answer statistical questions about a species of fish in the Great Lakes region. They use a histogram to represent the given data distribution, decide on appropriate measures of center and variability, and use their analyses to draw conclusions about a certain fish population.
### Launch
Tell students that they will now look at an example in which data analysis could be used to help conservation efforts. Provide students with the following background information.
The yellow perch is a freshwater fish that is a popular food for people in the Great Lakes region (Minnesota, Wisconsin, Michigan, Illinois, Indiana, Ohio, Pennsylvania, and New York). In past research, samples of yellow perch taken from the Great Lakes seemed to be mostly male and mostly old. People worried that yellow perch might not survive and efforts were made to limit commercial and individual fishing in order to try to increase in the number of younger fish. An important part of these efforts is to periodically check the typical age of the fish in the Great Lakes.
The Wisconsin Department of Natural Resources and the Great Lakes Water Institute collected data from samples of yellow perch in Lake Michigan. Students at Rufus King High School in Milwaukee, Wisconsin participated in the research. They evaluated the data and presented their findings in a student-conducted press conference. Explain to students that, in this task, they will investigate some of the same questions that these students addressed in their research.
Arrange students in groups of 3–4. Provide access to straightedges. Give students 7–8 minutes of quiet work time for the first three questions, and then 10–12 minutes to discuss their responses, complete the remainder of the task, and prepare a brief presentation on their response to the last set of questions.
Give each group access to tools for creating a visual display. Ask them to support their conclusions with specific pieces of evidence, such as their histogram, their analysis of the distribution, measures of center and spread, etc.
### Student Facing
Scientists studying the yellow perch, a species of fish, believe that the length of a fish is related to its age. This means that the longer the fish, the older it is. Adult yellow perch vary in size, but they are usually between 10 and 25 centimeters.
Scientists at the Great Lakes Water Institute caught, measured, and released yellow perch at several locations in Lake Michigan. The following summary is based on a sample of yellow perch from one of these locations.
length of fish in centimeters number of fish
0 to less than 5 5
5 to less than 10 7
10 to less than 15 14
15 to less than 20 20
20 to less than 25 24
25 to less than 30 30
1. Use the data to make a histogram that shows the lengths of the captured yellow perch. Each bar should contain the lengths shown in each row in the table.
2. How many fish were measured? How do you know?
3. Use the histogram to answer the following questions.
1. How would you describe the shape of the distribution?
2. Estimate the median length for this sample. Describe how you made this estimate.
3. Predict whether the mean length of this sample is greater than, less than, or nearly equal to the median length for this sample of fish? Explain your prediction.
4. Would you use the mean or the median to describe a typical length of the fish being studied? Explain your reasoning.
4. Based on your work so far:
1. Would you describe a typical age for the yellow perch in this sample as: “young,” “adult,” or “old”? Explain your reasoning.
2. Some researchers are concerned about the survival of the yellow perch. Do you think the lengths (or the ages) of the fish in this sample are something to worry about? Explain your reasoning.
### Activity Synthesis
To allow all groups a chance to present, consider putting 2–3 groups together and asking them to present their work to each other. Groups that are not the first to present should focus on sharing new insights that have not been mentioned by the preceding groups. Invite students who are not presenting to attend carefully to the reasoning of the presenting group and to ask clarifying questions.
If time permits, highlight some conclusions that students drew about whether the fish in the sample were young, adult age, or old, and whether researchers should be worried.
Tell students that several years after the students at Rufus King High School participated in the research, newer samples of yellow perch showed more favorable length-age distributions: more of the the fish were smaller or younger.
Speaking: MLR8 Discussion Supports. During the group presentations, provide students with sentence frames such as: “Based on the histogram, I think _____ because . . .” or “I think _____ is/is not something to worry about because . . .”. This will help students articulate their ideas, and use mathematical language such as mean, median, mean absolute deviation, and interquartile range as supporting evidence during their presentation.
Design Principle(s): Support sense-making; Maximize meta-awareness
## Lesson Synthesis
### Lesson Synthesis
In this lesson we practice finding measures of center and variability (mean, MAD, median, and IQR) and making sense of them in the context of the given situation. We notice that they give us different insights into the distribution of a data set.
• “What do the mean and MAD tell us?” (The mean tells us the fair share or balance point of the distribution and the MAD tells us the average distance a value is from the mean.)
• “How do we interpret this statement: ‘Noah's mean number of homework problems per day is 10 and the MAD is 6.’?” (If we were to distribute Noah’s assignments so that the number of problems he has each day is the same, he would have 10 per day. The MAD of 6 tells us that there is some variability in the number of problems assigned, so not all days have exactly 10 problems assigned. The average distance between the number of problems assigned and the mean of 10 is 6.)
• “What do the median and IQR tell us?” (The median tells us the value for which half the data set is equal to or greater and half the data set is equal to or less and the IQR tells us the range for the middle half of the data set.)
• “How do we interpret this statement: ‘Lin's median number of homework problems per day is 10 and the IQR is 6.’?” (One half of Lin’s assignments involve 10 or fewer problems, and the other half involve 10 or more problems. The IQR tells us that half of Lin's assignments are between 7 and 13 problems.)
We also looked at different ways to graphically represent a numerical distribution.
• “What are the ways we can represent a data set?” (Dot plot, histogram, box plot.)
• “Which representations are helpful for summarizing a distribution?” (It varies depending on the distribution we're studying and what information we want to know.)
## Student Lesson Summary
### Student Facing
The dot plot shows the distribution of 30 cookie weights in grams.
The mean cookie weight, marked by the triangle, is 21 grams. This tells us that if the weights of all of the cookies were redistributed so they all had the same weight, each cookie would weigh 21 grams. The MAD is 5.6 grams, which suggests that a cookie typically weighs between 15.4 grams and 26.6 grams.
The box plot for the same data set is shown above the dot plot. The median shows that half of the weights are greater than or equal to 20.5 grams, and half are less than or equal to 20.5 grams. The box shows that the IQR is 10 and that the middle half of the cookies weigh between 16 and 26 grams.
In this case, the median weight is very close to the mean weight, and the IQR is about twice the MAD. This tells us that the two pairs of measures of center and spread are very similar.
Now let’s look at another example of 30 different cookies.
Here the mean is 21 grams, and the MAD is 3.4 grams. This suggests that a cookie typically weighs between 17.6 and 24.4 grams. The median cookie weight is 23 grams, and the box plot shows that the middle half of the data are between 20 and 24 grams. These two pairs of measures paint very different pictures of the variability of the cookie weights.
The median (23 grams) is closer to the middle of the big cluster of values. If we were to ignore the smaller cookies, the median and IQR would give a more accurate picture of how much a cookie typically weighs.
When a distribution is not symmetrical, the median and IQR are often better measures of center and spread than the mean and MAD. However the decision on which pair of measures to use depends on what we want to know about the group we are investigating.
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# How do you integrate x/((x-1)(x^2+4) using partial fractions?
Aug 30, 2016
$\frac{1}{5} \ln | x - 1 | - \frac{1}{10} \ln \left({x}^{2} + 4\right) + \frac{2}{5} a r c \tan \left(\frac{x}{2}\right) + K$.
#### Explanation:
To use the Method of Partial Fractions, we let, for, $A , B , C \in \mathbb{R} , :$
$\frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 4} \ldots \ldots \ldots \ldots \left(1\right)$.
$A$ can easily be obtained by using Heavyside's Cover-up Method :
$A = {\left[\frac{x}{{x}^{2} + 4}\right]}_{x = 1} = \frac{1}{5}$. Hence, with this $A$, from (1), we have,
$\therefore \frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} - \frac{\frac{1}{5}}{x - 1} = \frac{B x + C}{{x}^{2} + 4}$, i.e.,
$\frac{x - \frac{1}{5} \left({x}^{2} + 4\right)}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{B x + C}{{x}^{2} + 4}$
$\therefore \frac{\cancel{\left(x - 1\right)} \left(- \frac{1}{5} \cdot x + \frac{4}{5}\right)}{\cancel{\left(x - 1\right)} \left({x}^{2} + 4\right)} = \frac{B x + C}{{x}^{2} + 4}$.
$\Rightarrow B = - \frac{1}{5} , C = \frac{4}{5}$. Hence, by $\left(1\right)$, we have,
$\int \frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \frac{1}{5} \int \frac{1}{x - 1} \mathrm{dx} + \int \frac{\left(- \frac{1}{5} \cdot x + \frac{4}{5}\right)}{{x}^{2} + 4} \mathrm{dx}$
$= \frac{1}{5} \ln | x - 1 | - \frac{1}{10} \int \frac{2 x}{{x}^{2} + 4} \mathrm{dx} + \frac{4}{5} \int \frac{1}{{x}^{2} + 4} \mathrm{dx}$
$= \frac{1}{5} \ln | x - 1 | - \frac{1}{10} \int \frac{d \left({x}^{2} + 4\right)}{{x}^{2} + 4} + \frac{4}{5} \cdot \left(\frac{1}{2}\right) a r c \tan \left(\frac{x}{2}\right)$
$= \frac{1}{5} \ln | x - 1 | - \frac{1}{10} \ln \left({x}^{2} + 4\right) + \frac{2}{5} a r c \tan \left(\frac{x}{2}\right) + K$.
Enjoy Maths.!
Note : The Method used to find $B , \mathmr{and} , C$ above, the following page from Wikipedia was used :
http://www.math-cs.gordon.edu/courses/ma225/handouts/heavyside.pdf
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# Thread: Vector problem:
1. ## Vector problem:
"The vectors v = i + 2j - 3k, u = 2i + 2j + 3k, and w = i + (2 - t)j + (t + 1)k
are given. Find the value of t such that the three vectors u, v and w are coplanar. "
I have no idea how to solve this one!Where do I start?
2. let's take $\displaystyle v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$
Using Gauss method on vectors,we should have
$\displaystyle \begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 2\\ 2\\ 3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 1\\ 2-t\\ t+1 \end{pmatrix}$ $\displaystyle \Rightarrow$
$\displaystyle \begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 0\\ -2\\ 9 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 0\\ 0\\ \frac{-7t+8}{2} \end{pmatrix}$
for those three victors to be coplanar we must have $\displaystyle rang(\begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 0\\ -2\\ 9 \end{pmatrix},\displaystyle \begin{pmatrix} 0\\ 0\\ \frac{-7t+8}{2} \end{pmatrix} )$$\displaystyle =2$ that is $\displaystyle \frac{-7t+8}{2}=0$ which means $\displaystyle t=\frac{7}{8}$
hope that's right
3. Originally Posted by Raoh
let's take $\displaystyle v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$
Using Gauss method on vectors,we should have
$\displaystyle \begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 2\\ 2\\ 3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 1\\ 2-t\\ t+1 \end{pmatrix}$ $\displaystyle \Rightarrow$
$\displaystyle \begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 0\\ -2\\ 9 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 0\\ 0\\ \frac{-7t+8}{2} \end{pmatrix}$
for those three victors to be coplanar we must have $\displaystyle rang(\begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix}$$\displaystyle ,\begin{pmatrix} 0\\ -2\\ 9 \end{pmatrix},\displaystyle \begin{pmatrix} 0\\ 0\\ \frac{-7t+8}{2} \end{pmatrix} )$$\displaystyle =2$ that is $\displaystyle \frac{-7t+8}{2}=0$ which means $\displaystyle t=\frac{7}{8}$
hope that's right
in any case,u should wait until someone else confirm my post or the value of $\displaystyle t$.
(i don't trust my answer anyway )
4. Thanks for the answer, but I was hoping for a more "high-school level" answer!
5. Originally Posted by karldiesen
"The vectors v = i + 2j - 3k, u = 2i + 2j + 3k, and w = i + (2 - t)j + (t + 1)k
are given. Find the value of t such that the three vectors u, v and w are coplanar. "
Here is another way.
Solve $\displaystyle W\cdot(V\times U)=0$.
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# The Phi(ϕ) Function
Phi Function is define as the number of positive integers less than or equal to and coprime to the number. The other name is Euler’s Totient Function.
The application of this function is used to finding the remainders of large numbers, last digits of a large number, and more to be found.
Coprime numbers – these are pairs or set of numbers that don’t have a common divisor. Example are {3,5}, {2,3,7}, {12,7}. Pairs that is not coprime are {3,6}, {12,9}, {20,15}.
Formula:
Given $n=a^xb^yc^x\ldots$, then
$Phi(\varphi) n=n(1-\displaystyle\frac{1}{a})(1-\displaystyle\frac{1}{b})(1-\displaystyle\frac{1}{c})\ldots$
$n,a,b,c,x,y,z$ are all positive intergers.
To test the formula, let’s try the obvious one.
Worked Problem 1:
How many positive integers less than or equal to and co-prime to 12?
Solution:
Listing all numbers that satisfy this condition we have 1,5,7,11. There are four positive integers less than or equal to and co-prime to 12.
Using the ϕ function we have,
$12=2^2\cdot 3$
$\varphi n=n(1-\displaystyle\frac{1}{a})(1-\displaystyle\frac{1}{b})(1-\displaystyle\frac{1}{c})\ldots$
$\varphi 12=12(1-\displaystyle\frac{1}{2})((1-\displaystyle\frac{1}{3})$
$\varphi 12=12(\displaystyle\frac{1}{2})(\displaystyle\frac{2}{3})$
$\varphi 12=4$
The formula satisfies the answer. This means that we can rely on this to calculate same problem with large numbers without exhausting our time to count them individually.
Worked Problem 2:
Find the number of positive integers less than or equal to and shares no common divisor with 2016.
Solution:
This problem is just exactly the same as the definition of ϕ function.
$2016=2^5\cdot 3^2\cdot 7$
$Phi(\varphi) n=n(1-\displaystyle\frac{1}{a})(1-\displaystyle\frac{1}{b})(1-\displaystyle\frac{1}{c})\ldots$
$\varphi 2016=2016(1-\displaystyle\frac{1}{2})(1-\displaystyle\frac{1}{3})(1-\displaystyle\frac{1}{7})$
$Phi(\varphi) 2016=576$
Worked Problem 3:
Find the number positive integers less than or equal to and co-prime to 390.
Solution:
$390=2\cdot 3\cdot 5\cdot 13$
$\varphi 390=390(1-\displaystyle\frac{1}{2})(1-\displaystyle\frac{1}{3})(1-\displaystyle\frac{1}{5})(1-\displaystyle\frac{1}{3})$
$\varphi 390=96$
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# Linear Interpolation Made Simple With Metquay 2023 update
Updated: Feb 25, 2023
Linear interpolation can be best explained as the easiest method for estimating the value of a function between any two known values. In other words, it is a method of curve fitting using linear polynomials to construct new data points within the range of a discrete set of known data points. Its application ranges across many fields, including engineering, physics, biology, finance, and statistics.
Now, what is this formula used for in daily business applications?
This formula is used mainly in data forecasting, prediction, market research, etc. Given a table, this is used as one of the best methods for identifying the unknown values that could exist within a table.
Now, let us go into detail about what the formula is:
Since we understand the parameter values, let us delve into a simple real-life example to understand its application.
Consider a classroom, where a teacher is noting down the height of all 5 students in ascending order with respect to height. After the students leave the teacher notices she has forgotten to jot down the height of one boy in the fourth position. Since she is on a deadline, she decided to calculate an estimate and achieves this using the linear interpolation formula.
Let us consider some readings she initially took:
Order of position of the students (x) 1 2 3 5 Height in feet (y) 3 4.5 5 6
Now that the height of the 4th child is missing the teacher uses the interpolation formula with value substitutions as given below:
x = 4 (point to perform interpolation/ the position of the child for which height must be determined.)
x1 = 3; x2 = 5; (Position of immediate neighbors of the missing value)
y1 = 5; y2 = 6 (Heights of immediate neighbors of the missing value)
After value substitution to the original formula,
y=5+(4−3) (6−5) (5−3) y=5+(4−3) (6−5) (5−3)
y = 5 + 1(1/2)
y= 5 + 0.5 y = 5.5
Thus calculated, that the estimated height of the boy in the fourth position is 5.5 feet. Hence, linear interpolation is also considered a method of filling in the gaps for any value in a table format.
In the calibration front, a Calibration curve is calculated which is a linear interpolation between two calibration points (if more than one replicate, the replicates are averaged before interpolation). Linear calibration curves are desired on most fronts because they result in the best accuracy and precision.
Metquay provides its customers, the ease with which they can calculate linear interpolation as a readily available function for use, thus reducing your computational time and utilizing it with more efficiency.
To learn more about Metquay's calibration worksheets and the functions available to help you save your time in calculations reach out to us at consulting@metquay.com.
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# 9.4 Division of square root expressions
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The distinction between the principal square root of the number x and the secondary square root of the number x is made by explanation and by example. The simplification of the radical expressions that both involve and do not involve fractions is shown in many detailed examples; this is followed by an explanation of how and why radicals are eliminated from the denominator of a radical expression. Real-life applications of radical equations have been included, such as problems involving daily output, daily sales, electronic resonance frequency, and kinetic energy.Objectives of this module: be able to use the division property of square roots, the method of rationalizing the denominator, and conjugates to divide square roots.
## Overview
• The Division Property of Square Roots
• Rationalizing the Denominator
• Conjugates and Rationalizing the Denominator
## The division property of square roots
In our work with simplifying square root expressions, we noted that
$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$
Since this is an equation, we may write it as
$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$
To divide two square root expressions, we use the division property of square roots.
## The division property $\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$
$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$
The quotient of the square roots is the square root of the quotient.
## Rationalizing the denominator
As we can see by observing the right side of the equation governing the division of square roots, the process may produce a fraction in the radicand. This means, of course, that the square root expression is not in simplified form. It is sometimes more useful to rationalize the denominator of a square root expression before actually performing the division.
## Sample set a
Simplify the square root expressions.
$\sqrt{\frac{3}{7}}.$
This radical expression is not in simplified form since there is a fraction under the radical sign. We can eliminate this problem using the division property of square roots.
$\sqrt{\frac{3}{7}}=\frac{\sqrt{3}}{\sqrt{7}}=\frac{\sqrt{3}}{\sqrt{7}}·\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{3}\sqrt{7}}{7}=\frac{\sqrt{21}}{7}$
$\frac{\sqrt{5}}{\sqrt{3}}.$
A direct application of the rule produces $\sqrt{\frac{5}{3}},$ which must be simplified. Let us rationalize the denominator before we perform the division.
$\frac{\sqrt{5}}{\sqrt{3}}=\frac{\sqrt{5}}{\sqrt{3}}·\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{5}\sqrt{3}}{3}=\frac{\sqrt{15}}{3}$
$\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{\frac{21}{7}}=\sqrt{3}.$
The rule produces the quotient quickly. We could also rationalize the denominator first and produce the same result.
$\frac{\sqrt{21}}{\sqrt{7}}=\frac{\sqrt{21}}{7}·\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{21·7}}{7}=\frac{\sqrt{3·7·7}}{7}=\frac{\sqrt{3·{7}^{2}}}{7}=\frac{7\sqrt{3}}{7}=\sqrt{3}$
$\frac{\sqrt{80{x}^{9}}}{\sqrt{5{x}^{4}}}=\sqrt{\frac{80{x}^{9}}{5{x}^{4}}}=\sqrt{16{x}^{5}}=\sqrt{16}\sqrt{{x}^{4}x}=4{x}^{2}\sqrt{x}$
$\frac{\sqrt{50{a}^{3}{b}^{7}}}{\sqrt{5a{b}^{5}}}=\sqrt{\frac{50{a}^{3}{b}^{7}}{5a{b}^{5}}}=\sqrt{10{a}^{2}{b}^{2}}=ab\sqrt{10}$
$\frac{\sqrt{5a}}{\sqrt{b}}.$
Some observation shows that a direct division of the radicands will produce a fraction. This suggests that we rationalize the denominator first.
$\frac{\sqrt{5a}}{\sqrt{b}}=\frac{\sqrt{5a}}{\sqrt{b}}·\frac{\sqrt{b}}{\sqrt{b}}=\frac{\sqrt{5a}\sqrt{b}}{b}=\frac{\sqrt{5ab}}{b}$
$\frac{\sqrt{m-6}}{\sqrt{m+2}}=\frac{\sqrt{m-6}}{\sqrt{m+2}}·\frac{\sqrt{m+2}}{\sqrt{m+2}}=\frac{\sqrt{{m}^{2}-4m-12}}{m+2}$
$\frac{\sqrt{{y}^{2}-y-12}}{\sqrt{y+3}}=\sqrt{\frac{{y}^{2}-y-12}{y+3}}=\sqrt{\frac{\left(y+3\right)\left(y-4\right)}{\left(y+3\right)}}=\sqrt{\frac{\overline{)\left(y+3\right)}\left(y-4\right)}{\overline{)\left(y+3\right)}}}=\sqrt{y-4}$
## Practice set a
Simplify the square root expressions.
$\frac{\sqrt{26}}{\sqrt{13}}$
$\sqrt{2}$
$\frac{\sqrt{7}}{\sqrt{3}}$
$\frac{\sqrt{21}}{3}$
$\frac{\sqrt{80{m}^{5}{n}^{8}}}{\sqrt{5{m}^{2}n}}$
$4m{n}^{3}\sqrt{mn}$
$\frac{\sqrt{196{\left(x+7\right)}^{8}}}{\sqrt{2{\left(x+7\right)}^{3}}}$
$7{\left(x+7\right)}^{2}\sqrt{2\left(x+7\right)}$
$\frac{\sqrt{n+4}}{\sqrt{n-5}}$
$\frac{\sqrt{{n}^{2}-n-20}}{n-5}$
$\frac{\sqrt{{a}^{2}-6a+8}}{\sqrt{a-2}}$
$\sqrt{a-4}$
$\frac{\sqrt{{x}^{3}{}^{n}}}{\sqrt{{x}^{n}}}$
${x}^{n}$
$\frac{\sqrt{{a}^{3m-5}}}{\sqrt{{a}^{m-1}}}$
${a}^{m-2}$
## Conjugates and rationalizing the denominator
To perform a division that contains a binomial in the denominator, such as $\frac{3}{4+\sqrt{6}},$ we multiply the numerator and denominator by a conjugate of the denominator.
## Conjugate
A conjugate of the binomial $a+b$ is $a-b$ . Similarly, a conjugate of $a-b$ is $a+b$ .
Notice that when the conjugates $a+b$ and $a-b$ are multiplied together, they produce a difference of two squares.
$\left(a+b\right)\left(a-b\right)={a}^{2}-ab+ab-{b}^{2}={a}^{2}-{b}^{2}$
This principle helps us eliminate square root radicals, as shown in these examples that illustrate finding the product of conjugates.
$\begin{array}{lll}\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)\hfill & =\hfill & {5}^{2}-{\left(\sqrt{2}\right)}^{2}\hfill \\ \hfill & =\hfill & 25-2\hfill \\ \hfill & =\hfill & 23\hfill \end{array}$
$\begin{array}{lll}\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{6}+\sqrt{7}\right)\hfill & =\hfill & {\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}\hfill \\ \hfill & =\hfill & 6-7\hfill \\ \hfill & =\hfill & -1\hfill \end{array}$
## Sample set b
Simplify the following expressions.
$\frac{3}{4+\sqrt{6}}$ .
The conjugate of the denominator is $4-\sqrt{6.}$ Multiply the fraction by 1 in the form of $\frac{4-\sqrt{6}}{4-\sqrt{6}}$ . $\begin{array}{lll}\frac{3}{4+\sqrt{6}}·\frac{4-\sqrt{6}}{4-\sqrt{6}}\hfill & =\hfill & \frac{3\left(4-\sqrt{6}\right)}{{4}^{2}-{\left(\sqrt{6}\right)}^{2}}\hfill \\ \hfill & =\hfill & \frac{12-3\sqrt{6}}{16-6}\hfill \\ \hfill & =\hfill & \frac{12-3\sqrt{6}}{10}\hfill \end{array}$
$\frac{\sqrt{2x}}{\sqrt{3}-\sqrt{5x}}.$
The conjugate of the denominator is $\sqrt{3}+\sqrt{5x.}$ Multiply the fraction by 1 in the form of $\frac{\sqrt{3}+\sqrt{5x}}{\sqrt{3}+\sqrt{5x}}.$
$\begin{array}{lll}\frac{\sqrt{2x}}{\sqrt{3}-\sqrt{5x}}·\frac{\sqrt{3}+\sqrt{5x}}{\sqrt{3}+\sqrt{5x}}\hfill & =\hfill & \frac{\sqrt{2x}\left(\sqrt{3}+\sqrt{5x}\right)}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{5x}\right)}^{2}}\hfill \\ \hfill & =\hfill & \frac{\sqrt{2x}\sqrt{3}+\sqrt{2x}\sqrt{5x}}{3-5x}\hfill \\ \hfill & =\hfill & \frac{\sqrt{6x}+\sqrt{10{x}^{2}}}{3-5x}\hfill \\ \hfill & =\hfill & \frac{\sqrt{6x}+x\sqrt{10}}{3-5x}\hfill \end{array}$
## Practice set b
Simplify the following expressions.
$\frac{5}{9+\sqrt{7}}$
$\frac{45-5\sqrt{7}}{74}$
$\frac{-2}{1-\sqrt{3x}}$
$\frac{-2-2\sqrt{3x}}{1-3x}$
$\frac{\sqrt{8}}{\sqrt{3x}+\sqrt{2x}}$
$\frac{2\sqrt{6x}-4\sqrt{x}}{x}$
$\frac{\sqrt{2m}}{m-\sqrt{3m}}$
$\frac{\sqrt{2m}+\sqrt{6}}{m-3}$
## Exercises
For the following problems, simplify each expressions.
$\frac{\sqrt{28}}{\sqrt{2}}$
$\sqrt{14}$
$\frac{\sqrt{200}}{\sqrt{10}}$
$\frac{\sqrt{28}}{\sqrt{7}}$
2
$\frac{\sqrt{96}}{\sqrt{24}}$
$\frac{\sqrt{180}}{\sqrt{5}}$
6
$\frac{\sqrt{336}}{\sqrt{21}}$
$\frac{\sqrt{162}}{\sqrt{18}}$
3
$\sqrt{\frac{25}{9}}$
$\sqrt{\frac{36}{35}}$
$\frac{6\sqrt{35}}{35}$
$\sqrt{\frac{225}{16}}$
$\sqrt{\frac{49}{225}}$
$\frac{7}{15}$
$\sqrt{\frac{3}{5}}$
$\sqrt{\frac{3}{7}}$
$\frac{\sqrt{21}}{7}$
$\sqrt{\frac{1}{2}}$
$\sqrt{\frac{5}{2}}$
$\frac{\sqrt{10}}{2}$
$\sqrt{\frac{11}{25}}$
$\sqrt{\frac{15}{36}}$
$\frac{\sqrt{15}}{6}$
$\sqrt{\frac{5}{16}}$
$\sqrt{\frac{7}{25}}$
$\frac{\sqrt{7}}{5}$
$\sqrt{\frac{32}{49}}$
$\sqrt{\frac{50}{81}}$
$\frac{5\sqrt{2}}{9}$
$\frac{\sqrt{125{x}^{5}}}{\sqrt{5{x}^{3}}}$
$\frac{\sqrt{72{m}^{7}}}{\sqrt{2{m}^{3}}}$
$6{m}^{2}$
$\frac{\sqrt{162{a}^{11}}}{\sqrt{2{a}^{5}}}$
$\frac{\sqrt{75{y}^{10}}}{\sqrt{3{y}^{4}}}$
$5{y}^{3}$
$\frac{\sqrt{48{x}^{9}}}{\sqrt{3{x}^{2}}}$
$\frac{\sqrt{125{a}^{14}}}{\sqrt{5{a}^{5}}}$
$5{a}^{4}\sqrt{a}$
$\frac{\sqrt{27{a}^{10}}}{\sqrt{3{a}^{5}}}$
$\frac{\sqrt{108{x}^{21}}}{\sqrt{3{x}^{4}}}$
$6{x}^{8}\sqrt{x}$
$\frac{\sqrt{48{x}^{6}{y}^{7}}}{\sqrt{3xy}}$
$\frac{\sqrt{45{a}^{3}{b}^{8}{c}^{2}}}{\sqrt{5a{b}^{2}c}}$
$3a{b}^{3}\sqrt{c}$
$\frac{\sqrt{66{m}^{12}{n}^{15}}}{\sqrt{11m{n}^{8}}}$
$\frac{\sqrt{30{p}^{5}{q}^{14}}}{\sqrt{5{q}^{7}}}$
${p}^{2}{q}^{3}\sqrt{6pq}$
$\frac{\sqrt{b}}{\sqrt{5}}$
$\frac{\sqrt{5x}}{\sqrt{2}}$
$\frac{\sqrt{10x}}{2}$
$\frac{\sqrt{2{a}^{3}b}}{\sqrt{14a}}$
$\frac{\sqrt{3{m}^{4}{n}^{3}}}{\sqrt{6m{n}^{5}}}$
$\frac{m\sqrt{2m}}{2n}$
$\frac{\sqrt{5{\left(p-q\right)}^{6}{\left(r+s\right)}^{4}}}{\sqrt{25{\left(r+s\right)}^{3}}}$
$\frac{\sqrt{m\left(m-6\right)-{m}^{2}+6m}}{\sqrt{3m-7}}$
0
$\frac{\sqrt{r+1}}{\sqrt{r-1}}$
$\frac{\sqrt{s+3}}{\sqrt{s-3}}$
$\frac{\sqrt{{s}^{2}-9}}{s-3}$
$\frac{\sqrt{{a}^{2}+3a+2}}{\sqrt{a+1}}$
$\frac{\sqrt{{x}^{2}-10x+24}}{\sqrt{x-4}}$
$\sqrt{x-6}$
$\frac{\sqrt{{x}^{2}-2x-8}}{\sqrt{x+2}}$
$\frac{\sqrt{{x}^{2}-4x+3}}{\sqrt{x-3}}$
$\sqrt{x-1}$
$\frac{\sqrt{2{x}^{2}-x-1}}{\sqrt{x-1}}$
$\frac{-5}{4+\sqrt{5}}$
$\frac{-20+5\sqrt{5}}{11}$
$\frac{1}{1+\sqrt{x}}$
$\frac{2}{1-\sqrt{a}}$
$\frac{2\left(1+\sqrt{a}\right)}{1-a}$
$\frac{-6}{\sqrt{5}-1}$
$\frac{-6}{\sqrt{7}+2}$
$-2\left(\sqrt{7}-2\right)$
$\frac{3}{\sqrt{3}-\sqrt{2}}$
$\frac{4}{\sqrt{6}+\sqrt{2}}$
$\sqrt{6}-\sqrt{2}$
$\frac{\sqrt{5}}{\sqrt{8}-\sqrt{6}}$
$\frac{\sqrt{12}}{\sqrt{12}-\sqrt{8}}$
$3+\sqrt{6}$
$\frac{\sqrt{7x}}{2-\sqrt{5x}}$
$\frac{\sqrt{6y}}{1+\sqrt{3y}}$
$\frac{\sqrt{6y}-3y\sqrt{2}}{1-3y}$
$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}$
$\frac{a-\sqrt{ab}}{a-b}$
$\frac{\sqrt{{8}^{3}{b}^{5}}}{4-\sqrt{2ab}}$
$\frac{\sqrt{7x}}{\sqrt{5x}+\sqrt{x}}$
$\frac{\sqrt{35}-\sqrt{7}}{4}$
$\frac{\sqrt{3y}}{\sqrt{2y}-\sqrt{y}}$
## Exercises for review
( [link] ) Simplify ${x}^{8}{y}^{7}\left(\frac{{x}^{4}{y}^{8}}{{x}^{3}{y}^{4}}\right).$
${x}^{9}{y}^{11}$
( [link] ) Solve the compound inequality $-8\le 7-5x\le -23.$
( [link] ) Construct the graph of $y=\frac{2}{3}x-4.$
( [link] ) The symbol $\sqrt{x}$ represents which square root of the number $x,\text{\hspace{0.17em}}x\ge 0$ ?
( [link] ) Simplify $\sqrt{{a}^{2}+8a+16}$ .
$a+4$
how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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Question Video: Finding the General Antiderivative of a Quadratic Function | Nagwa Question Video: Finding the General Antiderivative of a Quadratic Function | Nagwa
Question Video: Finding the General Antiderivative of a Quadratic Function Mathematics • Second Year of Secondary School
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Find the most general antiderivative πΉ(π₯) of the function π(π₯) = (π₯ β 3)Β².
04:10
Video Transcript
Find the most general antiderivative capital πΉ of π₯ of the function lowercase π of π₯ is equal to π₯ minus three all squared.
The question gives us a function lowercase π of π₯. It wants us to find the most general antiderivative of this function. Weβll call this capital πΉ of π₯. Remember, an antiderivative means when we differentiate this, we get back to our original function. In other words, we want capital πΉ prime of π₯ to be equal to lowercase π of π₯. And remember, since the derivative of any constant is equal to zero, we can add any constant we want to our antiderivative. And it will still be an antiderivative of our function lowercase π of π₯.
So we add a constant πΆ to our antiderivative. We call this the most general antiderivative, since it will represent all antiderivatives of our function. So letβs start trying to find our antiderivative. Letβs start by looking at our function lowercase π of π₯. We can see itβs π₯ minus three all squared. And this is a problem, since this is a composition of functions. This can make it much more difficult to find our antiderivatives. Instead, letβs simplify our function by distributing the square over our parentheses. Weβll distribute our square by using the FOIL method. Weβll start by multiplying the first term of each factor together. This gives us π₯ times π₯, which is equal to π₯ squared.
Next, the FOIL method tells us to multiply our two outer terms together. This gives us π₯ multiplied by negative three, which is negative three π₯. Now, we want to multiply our inner two terms together. Again, thatβs negative three times π₯, which is negative three π₯. Finally, we want to multiply the last term of each factor together. This gives us negative three times negative three, which is equal to nine. And we can simplify this, since negative three π₯ minus three π₯ is equal to negative six π₯.
So now, we can see our function lowercase π of π₯ is a polynomial. And we know how to find the antiderivative of each term in a polynomial separately. We know to find the antiderivative of π multiplied by π₯ to the πth power, we want to add one to our exponent of π₯ and then divide by this new exponent of π₯. This gives us π times π₯ to the power of π plus one divided by π plus one. And in the general case, weβll add a constant of integration πΆ. Weβll want to do this to each term separately. Letβs start with π₯ squared. To find an antiderivative of π₯ squared, we can add one to our exponent of two. This gives us a new exponent of three. But remember, we then need to divide by this new exponent. This gives us π₯ cubed over three.
We now want to find an antiderivative of our second term, negative six π₯. One way of doing this is to rewrite our second term as negative six times π₯ to the first power. Again, we want to add one to our exponent of π₯. This time, itβs equal to one. So we get two and then we divide by two. So we have negative six π₯ squared divided by two. And we know six divided by two is equal to three. So weβve shown negative three π₯ squared is an antiderivative of negative six π₯. Finally, we want to find an antiderivative of the third term, nine. We might be tempted to write this as nine times π₯ to the zeroth power.
However, thereβs a more simple method of finding an antiderivative in this case. We know for any constant πΎ, the derivative of πΎπ₯ with respect to π₯ is just equal to πΎ. In other words, for any constant πΎ, πΎπ₯ is an antiderivative of πΎ. So when πΎ is equal to nine, we see that nine π₯ is an antiderivative of nine. So to find an antiderivative of any constant, we just need to multiply that constant by π₯.
Remember, though, this is just one possible antiderivative of our function lowercase π of π₯. If we add any constant to this, the derivative of that constant is equal to zero. So this is still an antiderivative of our function. So we add a constant weβve called πΆ to this function. This represents all of our possible antiderivatives. And this is how we find our most general antiderivative.
Therefore, we were able to show the most general antiderivative capital πΉ of π₯ of the function lowercase π of π₯ is equal to π₯ minus three all squared is given by capital πΉ of π₯ is equal to π₯ cubed over three minus three π₯ squared plus nine π₯ plus a constant of integration πΆ.
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# Permutation & Combination
Factorial Notation
Let n be a positive integer. Factorial of n denoted by n! and is defined as
n! = n(n - 1)(n - 2)(n-3) ... 3.2.1.
Examples:
We define 0! = 1.
2! = 2.1 = 2
3! = 3.2.1 = 6
4! = 4.3.2.1 = 24
5! = 5.4.3.2.1 = 120
So on......
Permutations (Arrangements
The different arrangements of a given number or things by taking some or all at a time, are called permutation.
Examples:
1. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
2. All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba).
Number of Permutations:
Number of all permutations of n things, taken r at a time, is given by:
nPr = n(n - 1)(n - 2) ... (n - r + 1)
Examples:
• 6P2 = (6 x 5) = 30.
• 7P3 = (7 x 6 x 5) = 210.
Cor. number of all permutations of n things, taken all at a time = n!.
Note: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind;p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.
Then, number of permutations of these n objects is = n! (p1!).(p2)!.....(pr!)
Combinations
Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
Examples:
• Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note: AB and BA represent the same selection.
• All the combinations formed by a, b, c taking ab, bc, ca.
• The only combination that can be formed of three letters a, b, c taken all at a time is abc.
• Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD.
Note: that ab ba are two different permutations but they represent the same combination.
Number of Combinations:
The number of all combinations of n things, taken r at a time is:
Note:
1. nCn = 1 and nC0 = 1.
2. nCr = nC(n - r)
Examples:
11C4 = (11 x 10 x 9 x 8) = 330. (4 x 3 x 2 x 1)
16C13 = 16C(16 - 13) = 16C3 = 16 x 15 x 14 = 16 x 15 x 14 = 560. 3! 3 x 2 x 1
Choosing 3 balls out of 16, or choosing 13 balls out of 16 have the same number of combinations.
16! = 16! = 16! = 560 3!(16-3)! 13!(16-13)! 3!×13!
Thanks for reading Permutation & Combination
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# How do you find the volume bounded by y^2=x^3 and y=x^2 revolved about the y-axis?
Jun 10, 2018
$\frac{\pi}{20}$ units ^3
#### Explanation:
We need to find where these two curves intersect to find the bounds of integration.
${y}^{2} = {x}^{3} \mathmr{and} y = {x}^{2}$, squaring the second expression, ${y}^{2} = {x}^{4}$ Solving for ${y}^{2}$,.......... [${x}^{4} = {x}^{3}$] i.e, ${x}^{3} \left[x - 1\right] = 0$.
So $x = 1 , x = 0$ are the points of intersection.
From the graphs of these expressions in can be seen that $y = \sqrt{{x}^{3}}$ has a greater area than $y = {x}^{2}$ so we must find the area under $y = {x}^{2}$ and subtract it from the area under $y = \sqrt{{x}^{3}}$ and then revolve this area about the $x$ axis between the bounds $x = 1 , x = 0$
Volume of revolution is given by $\pi {\int}_{a}^{b} {y}^{2} \mathrm{dx}$ So the volume of revolution = [piint_0^1x^3dx - piint_0^1x^4dx].= $\pi \left[{x}^{4} / 4 - {x}^{5} / 5\right]$ [ after integration by the general power rule] and evaluated for $x = 1 , x = 0$ will result in $\pi \left[\frac{1}{4} - \frac{1}{5}\right]$ = $\frac{\pi}{20}$.
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Last updated:
# Digital Root Calculator
What is a digital root?How does this digital root calculator work?What are some applications of digital root?Some properties of digital rootFAQs
Want to learn a cool magic trick using math? Well, you've come to the right place, here you can learn about what is a digital root, its uses, and how this digital root calculator works. Digital root is the repeated addition of the digits of a given number until you arrive at a single number. Finding the digital root of the number can be used to find the accuracy of arithmetic operations.
## What is a digital root?
Digital root is the single-digit value obtained by repeatedly adding the digits in a given number together. You need to start by adding all the digits of a given number. You'll then need to repeat this operation on the number you obtain, and so on until you end up with a single digit.
For example, say we need to calculate the digital root of 56984.
Step 1: 5 + 6 + 9 + 8 + 4 = 32
Step 2: 3 + 2 = 5
5 is the digital root of the number 56984.
Now that you know what is a digital root, let's see a few of its distinct features.
This tool is different than the other root calculations, such as the square root, cube root, or the root mean square, as the result only be {1,2,3,4,5,6,7,8,9}, while the other roots mentioned can take any value from the number line. If you wish to calculate these roots, head to our square root calculator, cube root calculator, or root mean square calculator instead.
## How does this digital root calculator work?
This calculator uses the ceil function to eliminate the need to calculate the sum by adding the digits repeatedly. The ceil function enables us to find the digital root using a single equation:
digital root = n - 9 × (ceil(n / 9) − 1)
where n is the number in question.
Let us break this equation down,
1. 'n/9' gives the quotient that we get when the number is divided by 9. Here we might get an integer as well as a decimal digit.
2. Next, we take 'ceil' -ceiling - of 'n/9'. This will get rid of the decimal component.
3. Now, when we multiply this number by 9, we will get a number that is the multiple of 9 that is the closest to our given number.
4. At last, the difference between the two numbers gives us the digital root.
## What are some applications of digital root?
Let's start with the most interesting application of digital root:
1. The magic trick!
First, you need a friend who is as equally nerdy as you. Ask them to mentally choose a number from 1-10. Now ask them to multiply it by 9 and find the sum of digits of the multiple. Now, pretend to read their mind and tell them that they have got 9 as the answer. You can do this trick with much larger numbers as well, however, it might take a little longer for your friend to calculate the digital root of large numbers without knowing the trick. Refer to Property 1 mentioned below for more clarification on this.
Now, time for the revelation! For example, your friend chose 5. On multiplying 5 by 9, they will get 45. 4 + 5 = 9, which shouldn't be too difficult to calculate. You can make the magic trick more and more complex by adding in additional drama, for example, asking your friend to shuffle the digits.
2. Digital roots can be used as a primitive way to check the accuracy of arithmetic operations like subtraction, multiplication, and addition.
Let's see how we can use digital root to check the correctness of a multiplication. To check if the multiplication is correct or not, calculate the digital root of the numbers on both sides of the equation before performing the multiplication. Then multiply the digital roots and calculate the digital root of the product. The digital root at both sides of the equation should be equal in order for the multiplication to be correct. Let's look at an example,456 × 376 = 398765.
Let's first look at the left-hand side of the equation and find the sum of digits on this side. The digital root of 456 is 6. The digital root of 376 is 7. On multiplying 6 and 7, we get 42. The digital root of 42 is 6. Now, the digital root of the right-hand side comes out as 2. Since the digital roots obtained on both sides of the equals to sign are different, this multiplication is incorrect.
In a similar manner, let's see how we can use digital root to check the correctness of a subtraction problem. For example, consider 340-172=168. The digital root of 340 is 7. The digital root of 172 is 1. Subtracting these two we get 6. Now let's check the digital root of the right hand side. The digital root of 168 is 6, so this subtraction is correct.
3. Digital roots can also help detect rounding-off errors in the Fibonacci sequence.
When calculating the Fibonacci sequence for very large numbers, the computational programs might round off and hence lead to an error when generating the next number in the sequence. The digital root of a Fibonacci sequence has a 24-digit cycle (1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9), which means that the sequence of digital root repeats every 24 numbers. If there is some change in the expected sequence of digital roots, it might be due to the rounding-off error. You can use our Fibonacci calculator to generate a Fibonacci sequence easily!
1. We can also use the digital sum method for square root correctness.
Here, simply by looking at the digital root of a perfect square, we can guess if it's correct or not. The digital root of a perfect square will be one of the four digits 1, 4, 7, 9 only. Hence, if we find any other digit as the digital sum, the number is not a perfect square for sure.
## Some properties of digital root
The following are some of the properties of digital roots.
1. When any number is multiplied by 9, the digital root will always be 9. For example:
8 × 9 = 72 and 7 + 2 = 9
45 × 9 = 405, 4 + 0 + 5 = 9
2. When 9 is added to a number, the digital root will remain unchanged. For example:
527 = 5 + 2+ 7 = 14, 1 + 4 = 5
Now, if we add 9 to 527 we get:
5 + 2 + 7 + 9 = 23, 2 + 3 = 5
Hence, adding 9 to a number does not change the digital root of the original number. Alternatively, we can omit 9 while calculating the digital root of a number.
3. We can also use the digital sum method for square root correctness as demonstrated above. If we take a digital root of a perfect square it will be one of the four digits 1, 4, 7, 9 only. For example:
25 => 2 + 5 = 7
36 => 3 + 6 = 9
49 => 4 + 9 = 13, 1 + 3 = 4
100 => 1 + 0 + 0 = 1
FAQs
### What is a ceil function?
The ceiling (ceil) function returns the closest integer higher than or equal to a given number.
For example:
• ceil(2.4) = 3
• ceil(-2.6) = -2
• ceil(1.01) = 2
|
Mean Median Mode and Range
The mean median mode and range all are values that help characterize a data set. Mean is technically called the arithmetic mean. The mean median mode and range each represent different types of averages of the data set.
Mean :
Consider an example. Jim went bowling and each of his scores for 5 games is shown in the table below. Thus, these 5 scores represent the data set of observations.
To determine the mean, we have to add the data values (or scores) and then divide by the total number of values (games). Thus,
Mean = (20+24+33+55+28) / 5 = 160 / 5 = 32
Since the mean is 32, this indicates that you expect that if Jim bowled the 6th game, his score would be 32.
Median :
Another type of average is called the median but the median is not the exact average. It separates the lower and upper half of observations.
To find the median in the above example,
1. arrange the scores in ascending order, 20, 24,28,33,55
2. Select the middle number which is 28.
So the median here is 28
If the total numbers are even, say the above game has 6th row and the score is 34 then,
1. Arrange the scores in ascending order, 20,24,28, 33, 34, 55. The middle numbers are 33 and 34
2. Find the average of these two numbers. (33 + 34) / 2 = 67/2 = 33.5
So the median is 33.5
Mode :
The mode is the number that occurs the most frequently among the observations. As there is no number in the above data set that occurs maximum times there is no mode in the above data set.
Say Jim had 6th bowling game with a score of 24 then the data set would look like 20, 24, 33, 55, 28, 24 ( 20, 24, 24, 28, 33, 55 after rearranging). As the score 24 has occurred twice while other scores have just occurred once, the mode is 24.
Say if Jim had played 7th bowling game with a score of 33, the scores of seven games after rearranging would look like 20, 24, 24, 28, 33, 33, 55
Here the numbers 24 and 33 have occurred twice which is the maximum in this case. So there are two modes 24 and 33 and hence called bimodal.
Range:
Range is the difference between lowest and highest observation.
In the above example, the lowest observation is 20 and the highest is 55. So the range is 55 – 20 = 35
|
# Test: Exponents- 2
## 20 Questions MCQ Test Quantitative Reasoning for GMAT | Test: Exponents- 2
Description
Attempt Test: Exponents- 2 | 20 questions in 40 minutes | Mock test for Quant preparation | Free important questions MCQ to study Quantitative Reasoning for GMAT for Quant Exam | Download free PDF with solutions
QUESTION: 1
### If Z is a positive integer such that
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
Given:
• Z is a positive integer
• Z = 81(Y4 – 7)3 . . . (1)
We need to find the value of Y.
Step 3: Analyze Statement 1 independently
• Squaring both sides:
• Z5 = 350
• Taking 5th root on both sides:
• Z = 310 . . . (2)
• Put (2) in (1):
• 310 = 34(Y4 – 7)3
• 36 = (Y4 – 7)3
• Taking the cube-root on both sides:
• 32 = Y4 – 7
• Y4 = 9 + 7 = 16
• Y4 = 24 = (-2)4
• Y = 2 or -2
Not sufficient to determine a unique value of Y.
Step 4: Analyze Statement 2 independently
(2) |Y-1| < 4
• Distance of Y from 1 on the number line is less than 4 units
• -3 < Y < 5
Multiple values of Y possible. Not sufficient.
Step 5: Analyze Both Statements Together (if needed)
• From St. 1, Y = 2 or – 2
• From St. 2, -3 < Y < 5
• This inequality is satisfied by both 2 and -2
So, even after combining both statements, we have 2 possible values of Y
Since we couldn’t find a unique value of Y, the correct answer. Is Option E.
QUESTION: 2
Solution:
QUESTION: 3
### What is the value of x if
Solution:
Working Out:
• Since x is in the powers of different bases in this equation, to get equations in x, we need to be able to equate the bases on the two sides of the equation
• Expressing each base in terms of its prime factors:
Equating the powers of base 2 on both sides:
Equating the powers of base 3 on both sides:
The value of x that satisfies both (1) and (2) is:
x = 3
QUESTION: 4
If , what is the value of x?
Solution:
Given:
To find: Value of x
Working Out:
Looking at the answer choices, we see that the correct answer is Option A
QUESTION: 5
If x and y are non-zero numbers, what is the value of y?
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
Given: x ≠ 0, y ≠ 0
To find: y = ?
Step 3: Analyze Statement 1 independently
Equating the powers of 2 on both sides:
2x = 6 – 4y
x + 2y = 3
1 Linear Equation with 2 unknowns. Not sufficient to find a unique value of y.
Step 4: Analyze Statement 2 independently
Not sufficient to find a unique value of y.
Step 5: Analyze Both Statements Together (if needed)
• From Statement 1: x + 2y = 3 . . . (1)
• From Statement 2: xy = -2
Substituting (2) in (1):
• This quadratic equation gives 2 values of y
• The only constraint on y: y ≠ 0
• Since 0 is not a root of the above quadratic equation, this constraint doesn’t help eliminate one of the two roots of y
• Thus, 2 values of y are obtained from the combination of the two statements
Not sufficient to obtain a unique value of y.
QUESTION: 6
If x > 0 and , what is the value of x?
Solution:
Given
Approach
1. We can see that the terms on the left hand side of the equation are powers of 2.
1. So, we will simplify the expression in powers of 2.
2. So, we will get 2expression in x = 1. This is only possible if expression in x = 0 (as 2 = 1)
3. We will equate the expression in x = 0 and use the constraint that x > 0, to find out the value of x.
Working Out
As x> 0, the only possible value of x = 2.
QUESTION: 7
If x and y are positive integers, what is the remainder when y is divided by x?
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
• Given: x, y are integers > 0
• y = ax + r , where a is a positive integer and r is an integer such that 0 ≤ r < x
To Find: Value of r for which we need to find the value of x and y.
Step 3: Analyze Statement 1 independently
• As 149 is expressed as an integer raised to a power, we need to find that integer. As this integer will be a factor of 149, we need to find if 149 can be expressed as a factor raised to its power. So, let’s check its divisibility by the prime numbers.
• Divisibility by 2: As 149 is odd, it is not divisible by 2
• Divisibility by 3: As the sum of the digits of 149 (i.e. 14) is not divisible by 3, 149 will not be divisible by 3
• Divisibility by 5: As the units digit of 149 is neither 0 nor 5, it is not divisible by 5
• Divisibility by 7: We can see that 149 when divided by 7, leaves a remainder 2. So, it is not divisible by 7
• Divisibility by 11: 149 when divided by 11 leaves a remainder 4. So, it is not divisible by 11
• Divisibility by 13: We see that 149 when divided by 13 leaves a remainder 6. So, it is not divisible by 13.
• Since 122 < 149 < 132 and 149 is not divisible by any prime number till 13, we can say that 149 is a prime number.
• As 149 is a prime number, any integer apart from 149 raised to any power will not result in 149.
• Only possible option is x + y = 149 and y – x = 1
• 2 equations, 2 variables, we can find unique values of x and y and hence find the remainder.
Step 4: Analyze Statement 2 independently
• So, we need to express 149 as a product of 2 integers. For expressing 149 as a product of 2 integers, we need to find the factors of 149.
• As we have found above that 149 is a prime number, the only possible case of expressing 149 as a product of 2 integers is 149 * 1
• y + x = 149 and y – x = 1
• 2 equations, 2 variables, we can find unique values of x and y and hence find the remainder.
Step 5: Analyze Both Statements Together (if needed)
As we have a unique answer from steps 3 and 4, this step is not required.
QUESTION: 8
If a, b and x are integers such that , what is the value of a - b
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
• As a6 is always positive,a= 1, i.e. a = 1 or -1
• So, we can reject the value of
Possible values of a – b
• If a = 1 and b = 1, a – b = 0
• If a = -1 and b = 1, a- b = -2
• So, we need to find the unique value of a to find the value of a – b.
Step 3: Analyze Statement 1 independently
(1) a3 b7 > 0
• Rewriting a3b7 as ab(a2b6)
• Therefore, ab(a2b6)>0
• We know that a2b6 is always > 0 (even power of any number is always positive)
• So, for ab(a2b6)> 0
• ab > 0
• This tells us that a and b have same signs.
• Since b > 0, therefore a will also be greater than 0, so the value of a = 1.
• a – b = 1 -1 = 0
Step 4: Analyze Statement 2 independently
(2) a + b > 0
• If a = 1 and b = 1, a + b = 2 > 0
• If a = -1 and b = 1, a + b = 0, is not greater than zero
• Hence, we have a unique answer, where a =1 and b = 1
Thus a – b = 1 – 1 = 0.
Step 5: Analyze Both Statements Together (if needed)
As we have a unique answer from steps 3 and 4, this step is not required.
QUESTION: 9
A function for positive integers x and y. Is F(a, b) > a, where a and b are positive integers?
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
• Since a is a positive integer, multiplying both sides of the inequality doesn’t impact the sign of inequality
• So, the question simplifies to: is b > a?
Step 3: Analyze Statement 1 independently
Using the definition of this function, we can write:
So as per Statement 1:
As a is a positive integer,aa > 0. So, we can divide both sides of the inequality by aa without changing the sign of the inequality.
So, b > a or b = a.
Hence, we can not say for sure if b > a. Insufficient to answer.
Step 4: Analyze Statement 2 independently
Using the definition of this function, we can write:
So, as per statement-2,
• As a > 0, - a < 0. So b < -a is not possible as it will mean that b is negative.
• So, the only possible case is b > a
Step 5: Analyze Both Statements Together (if needed)
As we have a unique answer from step – 4, this step is not required.
QUESTION: 10
Is the sum of xy and yx positive?
(1) xy > 0
(2) x + y > 0
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
To Find: If xy + yx > 0
Step 3: Analyze Statement 1 independently
(1) xy > 0
• Tells that x and y are of the same sign. Two cases arise:
• x, y > 0
• In this case xy, yx > 0. So, xy + yx > 0
• x, y < 0
• In this case, xy + yx may or may not be positive, depending on the values of x and y. Following cases can arise:
• Both x, y are even → In this case xy,yx > 0. So, xy+yx > 0
• Both x and y are odd → In this case xy,yx < 0. So, xy+yx < 0
• x is even and y is odd → In this case xy < 0, yx > 0. Cannot comment on the value of xy+yx
• x is odd and y is even → In this case xy < 0, yx > 0. Cannot comment on the value of xy+yx
Step 4: Analyze Statement 2 independently
(2) x + y > 0
• Considering the constraint x + y > 0, following cases are possible:
• x, y > 0. So, x + y > 0. In this case, xy, yx > 0. So, xy + yx > 0
• x < 0, y > 0 and |y| > |x|. So, x + y > 0. In this case xy + yx may or may not be positive.
• If y is odd, then xy < 0 , yx > 0 as y is positive. In this case, we cannot comment on the value of xy + yx
• If y is even , then xy > 0 , yx > 0 as y is positive. In this case, xy + yx > 0
• x > 0, y < 0 and |x| > |y|. So, x + y > 0. In this case xy + yx may or may not be positive.
Step 5: Analyze Both Statements Together (if needed)
1. From Statement 1, xy > 0
2. From Statement 2, x + y > 0
Statement-1 tells us that x and y have the same signs. Following cases are possible:
• If x & y > 0, xy > 0 and x + y > 0. In this case xy + yx > 0
• If x & y < 0, x+ y < 0. Not possible. (If x & y both are negative, the x + y cannot be > 0)
The only possible case is when x, y > 0 and hence xy + yx > 0
QUESTION: 11
What is the remainder obtained when 1010 + 105 – 24 is divided by 36?
Solution:
Given:
• Not applicable
To find: The remainder when 1010 + 105 – 24 is divided by 36
Approach:
1. Let the required remainder be r. This means, we will be able to write:
1010 + 105 – 24 = 36k + r, where quotient k is an integer and 0 ≤ r < 36
The above expression is our GOAL expression. We’ll try to simplify the dividend 1010 + 105 – 24 till it is comparable to our GOAL expression, and then, by comparison, we’ll be able to find the value of r.
Working Out:
• 1010 + 105 – 24 = 1005 + 1002*10 – (36 – 12)
• =(36*3 – 8)5 + (36*3 – 8)2*10 – 36 + 12
• Now, from Binomial Theorem, we know that every term in the expansion of (36*3 – 8)5 will be divisible by 36, except the last term, and the last term will be (-8)5
• So, we can write: (36*3 – 8)5 is of the form 36a + (-8)5, where 36a is a catch-all term conveying that all the other terms in this expansion are divisible by 36
• Similarly, every term in the expansion of (36*3 – 8)2 will be divisible by 36 except the last term, and the last term will be (-8)2
• So, we can write: (36*3 – 8)2 = 36b + (-8)2
• So, the given expression simplifies to:
• {36a + (-8)5 } + {36b + (-8)2}*10 – 36 + 12
• = (36a + 360b – 36) + (-85 + 640 + 12)
• = (36a + 360b – 36) + (-85+ 652)
• = (36a + 360b – 36 + 648) + (-85+ 4)
• = (36a + 360b – 36 + 36*18) + (-85+ 4)
• The above expression is not comparable to our GOAL expression because the term -85 in it is still unresolved. Do we need to calculate the value of -85 to answer this question? No. We only need to express it in terms of 36. Once again, we’ll use Binomial Theorem to do so:
• -85 = -8(82)2 = -8(64)2 = -8(36*2 – 8)2
• Every term in the expansion of (36*2 – 8)2 will be divisible by 36 except the last term. The last term will be (-8)2 = 64
• So, the expression (36*2 – 8)2 can be written as: 36c + 64
• So, -8(36*2 – 8)2 = -8(36c + 64)
• =-8(36c + 36*2 – 8)
• = (-8*36c – 8*36*2) + 64
• So, the given expression simplifies to: (36a + 360b – 36 + 72) + {(-8*36c – 8*36*2) + 64} + 4
• = (36a + 360b – 36 + 72 – 8*36c – 8*36*2) + 68
• =(36a + 360b – 36 + 72 – 8*36c – 8*36*2 + 36) + 32
• Now, the above expression is exactly comparable to our GOAL Expression: 36k + r
• So, by comparison, we can say that Remainder r = 32
Looking at the answer choices, we see that the correct answer is Option E
QUESTION: 12
The distance between celestial bodies is expressed in terms of light years, where 1 light year is the distance travelled by light in one year and is equal to 9.46 X 1017 centimetres. If light takes 499 seconds to travel from the Sun to the Earth, which of the following is the closest to the distance, in million kilometres, between the Earth and the Sun?
Solution:
Given:
• 1 light year = Distance travelled by light in 1 year = 9.46 x 1017 centimeters
• Time taken by light to travel from Sun to Earth = 499 seconds
To find: The option that is closest to the distance, in million kilometers, between Sun and Earth
Approach:
1. To answer the question, we need to find the approximate distance between Sun and Earth
• Note that the 5 answer choices are quite far apart in magnitude. Therefore, we’ll be able to use Estimation and Rounding aggressively. All we need here is an idea of the order of the magnitude of the distance between Sun and Earth (whether this distance is in hundred millions or thousand millions or ten thousand millions of kiometers).
2. From the definition of light year, we get the distance travelled by light in 1 year. So, using the Unitary Method, we’ll be able to find the distance travelled by light in 499 seconds (and thus, the distance between Sun and Earth)
• One thing that we must be careful about when working out this solution – we must use one unit of time (either years or seconds) and one unit of distance (either centimeters or kilometers) in our solution.
• Working Out:
• Expressing the given information in a single Distance unit and a single Time unit
• Since the answer choices are in units of (millions of) kilometers, it’ll be wise for us to convert the distance given in centimeters into kilometers. (If you want to do vice-versa, that is, if you want to convert kilometers into centimeters, then you may later have to do 5 such conversions – 1 for each answer choice and therefore, that will be more time-consuming)
• For Time, let’s convert 1 year into seconds:
• 1 year = 365*24*60*60 seconds
• = 36*24*36*103 seconds approx.
• (Note: we’re not simplifying this expression very much here because we’ll do all the simplifications together after we’ve deduced an expression for the distance between Earth and Sun)
• Finding the approximate distance between Earth and Sun
• From the definition of light year, we know that:
• Thus, distance between Earth and Sun =
• We’ll now simplify the above expression for distance by using estimation and rounding:
• Since the question asks for the distance in units of million kilometers, let’s now express the above distance in million kilometers:
Therefore, 1.25*108 kilometers =
• = 1.25*102 million kilometers
• = 125 million kilometers approx.
• Thus, we see that the order of distance between the Sun and the Earth is hundred million kilometers.
Looking at the answer choices, we see that the correct answer is Option A
QUESTION: 13
James deposited \$1,000 each in two investment schemes X and Y. Scheme X doubles the invested amount every 7 years and scheme Y doubles the invested amount every 14 years. If James withdraws \$500 from scheme X at the end of every 7th year, how many years will it take for the total amount invested in schemes X and Y to amount more than \$40,000?
Solution:
Given
• Scheme X doubles the invested amount every 7 years
• James deposited \$1000 in scheme X
• James withdraws \$500 from scheme X after the end of every 7 years
• Scheme Y doubles the invested amount after every 14 years
• James deposited \$1,000 in scheme Y
To Find: Number of years it will take total amount deposited in schemes X and Y to grow to > \$40,000?
Approach
1. For finding the number of years it will take the deposits in schemes X and Y to grow to more than \$40,000, we need to find the amount in both the schemes X and Y after every 7 years.(As amount in scheme X doubles after every 7 years, we will need to calculate the amount at the end of every 7 years and not at the end of 14 years).
2. Scheme X
1. As the amount invested in scheme X doubles every 7 years, we will need to calculate the amount in scheme X after every interval of 7 years
2. However, we will need to make sure that we subtract \$500 at each interval of 7 years from the final amount
3. Scheme Y
1. As the amount invested in scheme Y doubles after every 14 years, we will need to calculate the amount in scheme Y after every interval of 14 years.
4. At each interval, we will calculate the sum of amounts in scheme X and Y to check if it exceeds \$40,000.
Working Out
1. Amount at the end of year 7 in scheme X = \$1000 * 2 = \$2000
1. However James withdrew \$500 at the end of 7th year, So, the amount remaining will be \$2000 – \$500 = \$1500
2. The same logic has been applied in calculating the amounts at the end of every 7 year interval
2. Amount at the end of year 14 in scheme Y = \$1000 * 2 = \$2000
1. The same logic has been applied in calculating the amounts at the end of every 14 years interval.
3. We can see that the total amount in schemes X and Y exceed \$40,000 by the end of the year 42.
QUESTION: 14
If x is a positive integer less than 100 such that x is divisible by 2y, where y is a positive integer, what is the value of y?
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
• x is an integer such that 0 < x < 100
• x is divisible by 2y, where y is a positive integer
• Since y > 0, this means 2 is definitely a prime factor 2
• Since x is a positive integer, we can write the Prime-factorized form of x as:
• are integers > 0, z ≥ y and are prime numbers other than 2.
• As x < 100, 2z < 100
• So, z = { 1, 2, 3, 4, 5, 6} as 27 = 128 > 100
• Since x is completely divisible by 2y, z ≥ y. So, y can take any value of z, i.e. 1 ≤ y ≤ 6
• To Find: Unique value of y
Step 3: Analyze Statement 1 independently
• As x is a positive integer, x < -60 is not possible.
• So, x > 60, i.e. 60 < x < 100.
However, we do not know if 2 is the only prime factor or x. So, we cannot find a unique value of y.
• For example, if 2 is the only prime factor of x, then y can have only 1 value: 6
• But if x has other prime factors, then multiple values of y are possible. For example, x could be 22*3*7 (y = 2) or 23*11 (y = 3) etc.
Step 4: Analyze Statement 2 independently
• Using the prime factorized expression of x, we can write:
• Since the integer resulting from this division is odd, the powers of 2 in the numerator and the denominator should cancel out each other.
• Hence 22z = 2y+2
• z=y/2+1 ….(1). So, y must be even as z is an integer
• Also, from our discussion in Steps 1 and 2, we know that z ≥ y
• Substituting (1) in the above inequality, we get:
Using (2), along with the inference that y must be even, we have y = 2 as the only possible option.
Step 5: Analyze Both Statements Together (if needed)
As we have a unique answer from step 4, this step is not required.
QUESTION: 15
Does ab lie between 0 and 1, exclusive?
(1) |a| < 1
(2) b is a positive odd integer
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
Given: Two numbers a and b
To Find: Is 0 < ab < 1
Following cases are possible:
1. a ≥ 1 →
• If b≥ 0, then a ≥ 1. So, the answer to the question is NO
• If b < 0, then 0 < a< 1. So, the answer to the question is YES
2. 0 < a < 1 → Irrespective of the value of b, 0 < ab < 1 So, the answer to the question is YES
3. If a = 0 → Irrespective of the value of b, ab = 0. So, the answer to the question is NO.
4. -1 < a < 0 →
• If b is even, 0 < ab < 1 So, the answer to the question is YES.
• If b is odd, -1 < ab < 0. So, the answer to the question is NO.
5. a ≤ -1 →
• If b is even,ab≥1. So, the answer to the question is NO.
• If b is odd, ab≤1. So, the answer to the question is NO
So, the answer to the question will be YES, if:
• a ≥ 1 and b < 0 OR
• 0 < a < 1 OR
• -1 < a < 0 and b is even
And the answer to the question will be NO, if:
• a ≥ 1 and b ≥ 0 OR
• a = 0 OR
• -1 < a < 0 and b is odd OR
• a ≤ -1
So, we can answer the given question with a unique answer for the above cases.
Step 3: Analyze Statement 1 independently
(1) |a| < 1
• This means, -1 < a < 1. So, following cases can occur:
• -1 < a < 0 → As we have seen from our analysis above that in this case the answer to the question may be YES or NO
• Now, we don’t need to analyse other cases as we do not have unique from this case itself, but we will do so for the sake of completing the analysis.
• a = 0 → The answer to the question is NO in this case
• 0 < a < 1 → The answer to the question is YES in this case.
• As we do not have a unique answer , we cannot say for sure if 0 < ab < 1
Step 4: Analyze Statement 2 independently
(2) b is a positive odd integer
• We have seen from our analysis in steps 1&2 that we need to find the range of values of a to answer this question.
• As we do not know the value of a, the statement is insufficient to tell if 0 < ab < 1
Step 5: Analyze Both Statements Together (if needed)
(1)From Statement 1, -1 < a < 1
(2) From Statement 2, b > 0
Following cases are possible:
• -1 < a < 0 → We have seen from our analysis in steps 1&2 that if b is odd, the answer to the question is NO.
• a = 0 → We have seen from our analysis in steps 1&2 that in this case the answer to the question is NO.
• 0 < a < 1 → We have seen from our analysis in steps 1&2 that in this case the answer to the question is YES
As we do not have a unique answer, the combination of statements is insufficient to answer.
QUESTION: 16
Z=2a×5b×7c
A positive integer Z can be expressed in terms of its prime factors as above, where a, b and c are positive integers. Is 3|a−b|<9?
(1) Z is divisible by 40 but not by 50
(2) Za−b=26×52×74
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
Given:
Step 3: Analyze Statement 1 independently
(1) Z is divisible by 40 but not by 50
• 40 = 23*5
• 50 = 2*52
• The power of 2 in Z is at least 3
• a ≥ 3
• The power of 5 in Z is 1.
• b = 1
• Therefore, a – b ≥ 3 – 1
a – b ≥ 2
Statement 1 is sufficient to answer the question.
Step 4: Analyze Statement 2 independently
• Equating the powers of a base on both sides of the equation:
• a(a - b) = 6. . . (1)
• b(a - b) = 2. . . (2)
• Dividing (1) by (2)
• The minimum value of b is 1 (b is a positive integer)
• So, minimum value of a – b = 2
Statement 2 is sufficient to answer the question.
Step 5: Analyze Both Statements Together (if needed)
Since we’ve already arrived at unique answers in Steps 3 and 4, this step is not required
QUESTION: 17
What is the maximum possible power of 4 in the number that is obtained when the product of the first 15 positive integers is subtracted from the product of the first 20 positive integers?
Solution:
Given
• Product of first 15 positive integers = 1*2*3…….* 15 = 15!
• Product of first 20 positive integers = 1*2*3*…….* 20 = 20!
• So, we can write 20! = 15! * 16 * 17 * 18 * 19 * 20
To Find: Maximum power of 4 that divides (20! – 15!)
Approach
1. We will first need to simplify the expression 20! – 15!
1. We know that 20! = 15! * 16 * 17 * 18 * 19 * 20
2. So, 20! – 15! = 15! (20*19*18….*16 – 1) = 15! * odd integer
1. Since 20 *19*…..16 is even, and 1 is odd, 20*19*……16 -1 will be odd
3. Therefore, we only need to find the maximum power of 4 that divides 15!
2. To find the maximum power of 4 in 15!, we will first need to find the maximum power of 2 in 15!, as 4 = 22
1. Let’s take a simple example to understand how we can find this. Consider a number p = 1*2*3*4. We need to find the power of 2 in p.
2. Now, powers of 2 will occur in multiples of 2. So, we should first find the multiples of 2 in p. They are 2 and 4. However all multiples of 2 will not contain only 21. Some of them may contain higher powers of 2. For example, here 4 contains 22.
3. So, when we are finding multiples of 2, we should also find the multiples of higher powers of 2 in p.
4. Hence, 2 will occur 2 + 1 = 3 times in p.
3. We will use the same logic to find out the power of 2 in 15!
1. Once we know the power of 2 in 15!, we can find the power of 4 in 15!
Working Out
QUESTION: 18
For any integer z, the function PODD(z) is defined as the product of all odd integers between –z and z, inclusive. Which of the following numbers will not divide PODD (-15) completely?
I. 38
II. 55
III. 73
Solution:
Given:
To find: PODD(-15) will not be divisible by which of the given 3 options?
Approach:
1. The given 3 options involve powers of 3, 5 and 7. So, to answer the question, we’ll find the powers of 3, 5 and 7 in the prime-factorized expression of PODD(-15)
2. The answer for an option will be YES (meaning that option will not divide PODD(-15) completely) if the power of the prime factor in that option is greater than the power of that prime factor in the prime-factorized expression for PODD(-15)
(For example, 25 will not divide a number of the form 24*3)
Working Out:
• Finding the power of 3 in PODD(-15) and evaluating Option I
• Look again at the expression for PODD(-15):
• PODD(-15) = (-1)8(1*3*5*7*9*11*13*15)2
• Note that each of the above multiples of 3 occurs twice in PODD(-15)
• So, power of 3 in PODD(-15) = (1+2+1)*2= 4*2 = 8
• Since the power of 3 in Option I is equal to the power of 3 in PODD(-15), 38 will divide PODD(-15) completely.
Finding the power of 5 in PODD(-15) and evaluating Option II
• Since the multiples of 5 and 15 occurs twice in PODD(-15), power of 5 in PODD(-15) = (1+1)*2 = 4
• Since the power of 5 in Option II is greater than the power of 5 in PODD(-15), 55 will not divide PODD(-15) completely.
• Finding the power of 7 in PODD(-15) and evaluating Option III
• The only multiple of 7 in PODD(-15) is 7 itself.
• Since 7 occurs twice in PODD(-15), the power of 7 in PODD(-15) = 1*2 = 2
• Since the power of 7 in Option III is greater than the power of 5 in PODD(-15), 73 will not divide PODD(-15) completely.
• Thus we see that only Options II and III will not divide PODD(-15) completely.
Looking at the answer choices, we see that the correct answer is Option E
QUESTION: 19
If x > 0 such that what is the value of
Solution:
Given
Approach
1. We need to find the value of and we are given the value of
1. We can observe from the above 2 expressions that is of the form a + b and is of the form a2 +b2
2. We also know that (a+b)2=a2+b2+2ab
3. We will use the above relation along with the constraint that x > 0, to find out the value of
Working Out
As x > 0, cannot be equal to -7. Hence, =7
QUESTION: 20
If x and y are distinct positive integers, such that
What is the value of (x)(y+1) ?
Solution:
Given Info:
• x and y are distinct positive integers. This means x≥1, y≥1 & x≠y
• Also Given are two equations in x and y
To Find:
• Value of (x)(y+1)
• So we need to find the values of x & y in order to find the value of the expression (x)(y+1).
Approach:
• From the 1st equation, which is an equation in x, we will find the values of x.
• We will put the values of x derived from the first equation in equation 2, to find the values of y.
• Knowing the values of x & y and the set of constraints on the value of x & y, we will calculate the value of (x)(y+1) to arrive at the final answer.
Working out:
On comparing both sides of the above equation as both are to the same base 2, we get
⇒x2=5x−4
Subtracting 5x - 4 from both sides we get
Sowehavex=1 or x=4
• Case 1: x=1 (Value derived from Equation 1)
• Putting x=1 in Equation 2, we get
• On comparing both sides of the above equation as both are to the same base 3, we get
⇒ y+1=2
⇒ y=1
Case 2: x=4 (Value derived from Equation 1)
• Putting x=4 in Equation 2, we get
• On comparing both sides of the above equation as both are to the same base 3, we get
⇒ y + 4 = 5
⇒ y = 1
• So, we have two pairs of values of x & y
Pair1:(x,y)=(1,1)(FromCase1)
Pair2:(x,y)=(4,1)(FromCase2)
• We have to reject the value, when (x,y) = (1,1), as we have already established above that x≠y.
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# cos 1/2
Inverse cosine is an important inverse trigonometric function. Mathematically, it is written as cos-1(x) and is the inverse function of the trigonometric function cosine, cos(x). An important thing to lớn note is that inverse cosine is not the reciprocal of cos x. There are 6 inverse trigonometric functions as sin-1x, cos-1x, tan-1x, csc-1x, sec-1x, cot-1x.
Inverse cosine is used to lớn determine the measure of angle using the value of the trigonometric ratio cos x. In this article, we will understand the formulas of the inverse cosine function, its domain name and range, and hence, its graph. We will also determine the derivative and integral of cos inverse x to lớn understand its properties better.
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1 What is Inverse Cosine? 2 Domain and Range of Inverse Cosine 3 Graph of Inverse Cosine 4 Derivative of Cos Inverse x 5 Integral of Inverse Cosine 6 Properties of Inverse Cosine 7 FAQs on Inverse Cosine
## What is Inverse Cosine?
Inverse cosine is the inverse function of the cosine function. It is one of the important inverse trigonometric functions. Cos inverse x can also be written as arccos x. If y = cos x ⇒ x = cos-1(y). Let us consider a few examples to lớn see how the inverse cosine function works.
• cos 0 = 1 ⇒ 0 = cos-1 (1)
• cos π/3 = một nửa ⇒ π/3 = cos-1 (1/2)
• cos π/2 = 0 ⇒ π/2 = cos-1 (0)
• cos π = -1 ⇒ π = cos-1 (-1)
In a right-angled triangle, the cosine of an angle (θ) is the ratio of its adjacent side to lớn the hypotenuse, that is, cos θ = (adjacent side) / (hypotenuse). Using the definition of inverse cosine, θ = cos-1[ (adjacent side) / (hypotenuse) ].
Thus, the inverse cosine is used to lớn find the unknown angles in a right-angled triangle.
## Domain and Range of Inverse Cosine
We know that the domain name of the cosine function is R, that is, all real numbers and its range is [-1, 1]. A function f(x) has an inverse if and only if it is bijective(one-one and onto). Since cos x is not a bijective function as it is not one-one, the inverse cosine cannot have R as its range. Hence, we need to lớn make the cosine function one-one by restricting its domain name. The domain name of the cosine function can be restricted to lớn [0, π], [π, 2π], [-π, 0], etc. and get a corresponding branch of inverse cosine.
The domain name of cosine function is restricted to lớn [0, π] usually and its range remains as [-1, 1]. Hence, the branch of cos inverse x with the range [0, π] is called the principal branch. Since the domain name and range of a function become the range and domain name of its inverse function, respectively, the domain name of the inverse cosine is [-1, 1] and its range is [0, π], that is, cos inverse x is a function from [-1, 1] → [0, π].
## Graph of Inverse Cosine
Since the domain name and range of the inverse cosine function are [-1, 1] and [0, π], respectively, we will plot the graph of cos inverse x within the principal branch. As we know the values of the cosine function for specific angles, we will use the same values to lớn plot the points and hence the graph of inverse cosine. For nó = cos-1x, we have:
• When x = 0, nó = π/2
• When x = một nửa, nó = π/3
• When x = 1, nó = 0
• When x = -1, nó = π
• When x = -1/2, nó = 2π/3
## Cos Inverse x Derivative
Now, we will determine the derivative of inverse cosine function using some trigonometric formulas and identities. Assume nó = cos-1x ⇒ cos nó = x. Differentiate both sides of the equation cos nó = x with respect to lớn x using the chain rule.
cos nó = x
⇒ d(cos y)/dx = dx/dx
⇒ -sin nó dy/dx = 1
⇒ dy/dx = -1/sin nó ---- (1)
Since cos2y + sin2y = 1, we have sin nó = √(1 - cos2y) = √(1 - x2) [Because cos nó = x]
Substituting sin nó = √(1 - x2) in (1), we have
dy/dx = -1/√(1 - x2)
Since x = -1, 1 makes the denominator √(1 - x2) equal to lớn 0, and hence the derivative is not defined, therefore x cannot be -1 and 1.
Hence the derivative of cos inverse x is -1/√(1 - x2), where -1 < x < 1
## Inverse Cosine Integration
We will find the integral of inverse cosine, that is, ∫cos-1x dx using the integration by parts (ILATE).
∫cos-1x = ∫cos-1x · 1 dx
Using integration by parts,
∫f(x) . g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) ∫g(x) dx) dx + C
Here f(x) = cos-1x and g(x) = 1.
∫cos-1x · 1 dx = cos-1x ∫1 dx - ∫ [d(cos-1x)/dx ∫1 dx]dx + C
∫cos-1x dx = cos-1x . (x) - ∫ [-1/√(1 - x²)] x dx + C
We will evaluate this integral ∫ [-1/√(1 - x²)] x dx using substitution method. Assume 1-x2 = u. Then -2x dx = du (or) x dx = -1/2 du.
∫cos-1x dx = x cos-1x - ∫(-1/√u) (-1/2) du + C
= x cos-1x - một nửa ∫u-1/2 du + C
= x cos-1x - (1/2) (u1/2/(1/2)) + C
= x cos-1x - √u + C
= x cos-1x - √(1 - x²) + C
Therefore, ∫cos-1x dx = x cos-1x - √(1 - x²) + C
## Properties of Inverse Cosine
Some of the properties or formulas of inverse cosine function are given below. These are very helpful in solving the problems related to lớn cos inverse x in trigonometry.
• cos(cos-1x) = x only when x ∈ [-1, 1] (When x ∉ [-1, 1], cos(cos-1x) is NOT defined)
• cos-1(cos x) = x, only when x ∈ [0, π] (When x ∉ [0, π], apply the trigonometric identities to lớn find the equivalent angle of x that lies in [0, π])
• cos-1(-x) = π - cos-1x
• cos-1(1/x) = sec-1x, when |x| ≥ 1
• sin-1x + cos-1x = π/2, when x ∈ [-1, 1]
• d(cos-1x)/dx = -1/√(1 - x2), -1 < x < 1
• ∫cos-1x dx = x cos-1x - √(1 - x²) + C
Important Notes on Cos Inverse x
• Invere cosine is NOT the same as (cos x)-1 as (cos x)-1 = 1/(cos x) = sec x.
• θ = cos-1[ (adjacent side) / (hypotenuse) ], θ ∈ [0, π]
• d(cos-1x)/dx = -1/√(1 - x2), -1 < x < 1
• ∫cos-1x dx = x cos-1x - √(1 - x²) + C
• cos-1(-x) = π - cos-1x
Related Topics on Inverse Cosine
• sin cos tan
• Trigonometric Functions
• Law of Sines
• Trigonometric Chart
## FAQs on Inverse Cosine
### What is Inverse Cosine in Trigonometry?
Inverse cosine is the inverse of the cosine function. Inverse cosine of x can also be written as cos-1x or arccos x. Then by the definition of inverse cosine, θ = cos-1[ (adjacent side) / (hypotenuse) ].
### What is Inverse Cosine Formula?
By the definition of inverse cosine, θ = cos-1[ (adjacent side) / (hypotenuse) ]. Here θ is the angle between the adjacent side and the hypotenuse and lies between 0 and π.
### What is the Derivative of Inverse Cosine?
The derivative of cos inverse x is -1/√(1 - x2), where -1 < x < 1. It can be calculated using the chain rule.
### What is the Domain and Range of Inverse Cosine?
The domain name of the inverse cosine is [-1, 1] because the range of the cosine function is [-1, 1]. The range of cos inverse x, cos-1x is [0, π]. We need to lớn make the cosine function one-one by restricting its domain name R to lớn the principal branch [0, π] which makes the range of the inverse cosine as [0, π].
### How to lớn Calculate Integral of Inverse Cosine?
The integral of cos inverse x can be calculated using integration by parts. Integral of inverse cosine is given by ∫cos-1x dx = x cos-1x - √(1 - x²) + C
### What Is Cos of Cos Inverse x?
Cos of cos inverse x is x, that is, cos(cos-1x) = x if x ∈ [-1, 1]. If x ∉ [-1, 1] then cos(cos-1x) is not defined.
### What Is the Inverse Cosine of Cos x?
Inverse cosine of cos x is x, that is, cos-1(cos x) = x, if x ∈ [0, π]. If x ∉ [0, π] then apply trigonometric identities to lớn find the equivalent angle of x that lies in [0, π].
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# How do you solve and graph 2p + 5 ≥ 3p - 10 ?
May 8, 2017
See a solution process below:
#### Explanation:
Step 1) Subtract $\textcolor{red}{2 p}$ and add $\textcolor{b l u e}{10}$ to each side of the inequality to solve for $p$ while keeping the inequality balanced:
$- \textcolor{red}{2 p} + 2 p + 5 + \textcolor{b l u e}{10} \ge - \textcolor{red}{2 p} + 3 p - 10 + \textcolor{b l u e}{10}$
$0 + 15 \ge \left(- \textcolor{red}{2} + 3\right) p - 0$
$15 \ge 1 p$
$15 \ge p$
Step 2) To state the solution in terms of $p$ we can reverse or "flip" the entire inequality:
$p \le 15$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 1.2: Practice Questions
Difficulty Level: At Grade Created by: CK-12
Directions: The following question (1) is an example of a grid-in math problem. On the SAT, you will solve the problem and indicate your answer by darkening the ovals on the special grid provided. Since you do not have this type of answer sheet to practice on, simply write your response. For more information about grid-in questions, please visit sat.collegeboard.com/practice/sat-practice-questions-math/student-produced-response.
1. (Numbers and Operations) — The average of five non-repeating positive odd numbers is \begin{align*}73\end{align*}. If \begin{align*}x\end{align*} is the greatest of these integers, what is the greatest possible value of \begin{align*}x\end{align*}?
Directions: For this section, solve each problem and decide which is the best of the choices given. You may use any available space for scratchwork.
1. (Numbers and Operations) — How many unique real roots does the equation \begin{align*}y = x^2 - 10x + 25\end{align*}have?
1. No solutions.
2. \begin{align*}0\end{align*}
3. \begin{align*}1\end{align*}
4. \begin{align*}2\end{align*}
5. \begin{align*}3\end{align*}
2. (Algebra and Functions) — If \begin{align*}g(x) = 4x - 4x\end{align*}, what is the value of \begin{align*}g \left(\frac{3}{2}\right)\end{align*}?
1. \begin{align*}-2\end{align*}
2. \begin{align*}6\end{align*}
3. \begin{align*}8\end{align*}
4. \begin{align*}2\end{align*}
5. \begin{align*}0\end{align*}
3. (Algebra and Functions) — If \begin{align*}f(x) = x + 9\end{align*}, which of the following is a solution of \begin{align*}f(5a) + 3 = f(3a) + 11\end{align*}? In other words, what might be the value of ‘\begin{align*}a\end{align*}’?
1. There are no solutions.
2. \begin{align*}6\end{align*}
3. \begin{align*}8\end{align*}
4. \begin{align*}\frac{5}{3}\end{align*}
5. \begin{align*}4\end{align*}
4. (Algebra and Functions) — If the mean of \begin{align*}x\end{align*} and \begin{align*}4x\end{align*} is \begin{align*}10\end{align*}, then \begin{align*}x =\end{align*}
1. \begin{align*}20\end{align*}
2. \begin{align*}16\end{align*}
3. \begin{align*}4\end{align*}
4. \begin{align*}10\end{align*}
5. \begin{align*}5\end{align*}
5. (Geometry) — If a line is perpendicular to the line \begin{align*}y = 3x - 4\end{align*} and passes through the point \begin{align*}(0, 6)\end{align*}, what is the equation of the perpendicular line?
1. \begin{align*}y = 3x + 6\end{align*}
2. \begin{align*}y = -3x + 6\end{align*}
3. \begin{align*}y = \frac{-1}{3x + 6}\end{align*}
4. \begin{align*}y = \frac{-1}{3x - 6}\end{align*}
5. \begin{align*}y = \frac{1}{3x + 6}\end{align*}
6. (Geometry) — What is the height of a building if the angle to the top is \begin{align*}27^\circ\end{align*} when you are standing \begin{align*}150 \ feet\end{align*}away from the building’s base? Round to the nearest whole number.
1. \begin{align*}68\end{align*}
2. \begin{align*}134\end{align*}
3. \begin{align*}76\end{align*}
4. \begin{align*}330\end{align*}
5. \begin{align*}168\end{align*}
7. (Geometry)Which of the following equations defines \begin{align*}g(x)\end{align*} in terms of \begin{align*}f(x)\end{align*}?
1. \begin{align*}g(x) = f(x - 2) + 3\end{align*}
2. \begin{align*}g(x) = f(x + 2) + 3\end{align*}
3. \begin{align*}g(x) = f(x - 2) -3\end{align*}
4. \begin{align*}g(x) = f(x + 2) - 3\end{align*}
5. \begin{align*}g(x) = f(x - 2)\end{align*}
8. (Probability and Statistics)The following chart gives the graduation ages of 10 students?
What is the median age of the graduating students?
1. \begin{align*}16\end{align*}
2. \begin{align*}16.5\end{align*}
3. \begin{align*}17\end{align*}
4. \begin{align*}17.5\end{align*}
5. \begin{align*}18\end{align*}
10. (Probability and Statistics)Assume that the edge of the smaller shaded square is \begin{align*}1 \ inch\end{align*} and the edge of the larger square is \begin{align*}2 \ inches\end{align*}.
What percent of the diagram is unshaded?
1. \begin{align*}35\%\end{align*}
2. \begin{align*}30\%\end{align*}
3. \begin{align*}15\%\end{align*}
4. \begin{align*}10\%\end{align*}
5. \begin{align*}25\%\end{align*}
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Lesson Plans West Iron County Middle & High School - 2020/2021
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Mon. Sep. 21 - Fri. Sep. 25
Next
9/21/2020
7:58 AM - Math 7
Lesson 1.5: Subtracting Rational Numbers (Day 1 of 3)
Find differences of rational numbers and find distances between numbers on a number line.
Success Criteria
-I can explain how subtracting integers is related to adding integers.
-I can explain how to model subtraction of integers on a number line.
-I can find differences of integers by reasoning about absolute values.
Assignment
Pages 34-35, problems 9-10, 12-22 (evens), 26-27, 30-42 (evens)
Due: Thursday, September 24th
Standards
CCSS.Math.Content.7.NS.A.1d
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1d Apply properties of operations as strategies to add and subtract rational numbers.
CCSS.Math.Content.7.NS.A.1c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1c Understand subtraction of rational numbers as adding the additive inverse, p - q = p + (-q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
8:57 AM - Math 8
Lesson 1.4: Rewriting Equations and Formulas (Day 3 of 3)
Solve literal equations for given variables and convert temperatures.
Success Criteria
-I can use properties of equalities to rewrite literal equations.
-I can use a formula to convert temperatures.
Assignment
Pages 29-30, problems 8-26, 28-29
Due: Tuesday, September 22nd
Standards
CCSS.Math.Content.8.EE.C.7
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.EE.C.7 Solve linear equations in one variable.
11:50 AM - Math 7 Accelerated
Lesson 2.1: Multiplying Integers (Day 2 of 3)
Find products of integers.
Success Criteria
-I can explain the rules for multiplying integers.
-I can find products of integers with the same sign.
-I can find products of integers with different signs.
Assignment
Pages 53-54, problems 10-28 (evens), 29, 30-44 (evens)
Due: Wednesday, September 23rd
Standards
CCSS.Math.Content.7.NS.A.2a
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.2a Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (-1)(-1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.
CCSS.Math.Content.7.NS.A.3
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers.
CCSS.Math.Content.7.NS.A.2c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.2c Apply properties of operations as strategies to multiply and divide rational numbers.
1:52 PM - Algebra 1
Unit 1 Study Guide (Day 1 of 3)
We will begin a study guide for Unit 1.
Assignment
Review the problems on the study guide.
Standards
CCSS.Math.Content.HSA-REI.B.3
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.B.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
CCSS.Math.Content.HSN-Q.A.1
Common Core State Standards
Common Core Mathematics - High School — Number and Quantity
CCSS.Math.Content.HSN-Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
CCSS.Math.Content.HSA-REI.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.A.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
CCSS.Math.Content.HSA-CED.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
CCSS.Math.Content.HSA-CED.A.4
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
9/22/2020
7:58 AM - Math 7
PD Day (Student Dismissal at Noon)
Lesson 1.5: Subtracting Rational Numbers (Day 2 of 3)
Find differences of rational numbers and find distances between numbers on a number line.
Success Criteria
-I can explain how subtracting integers is related to adding integers.
-I can explain how to model subtraction of integers on a number line.
-I can find differences of integers by reasoning about absolute values.
Assignment
Pages 34-35, problems 9-10, 12-22 (evens), 26-27, 30-42 (evens)
Due: Thursday, September 24th
Standards
CCSS.Math.Content.7.NS.A.1d
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1d Apply properties of operations as strategies to add and subtract rational numbers.
CCSS.Math.Content.7.NS.A.1c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1c Understand subtraction of rational numbers as adding the additive inverse, p - q = p + (-q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
8:57 AM - Math 8
PD Day (Student Dismissal at Noon)
Study Guide: Lessons 1.3 through 1.4
We will complete a study guide for Lessons 1.3 through 1.4.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.8.EE.C.7a
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.EE.C.7a Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
CCSS.Math.Content.8.EE.C.7
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.EE.C.7 Solve linear equations in one variable.
CCSS.Math.Content.8.EE.C.7b
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.EE.C.7b Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
11:50 AM - Math 7 Accelerated
PD Day (Student Dismissal at Noon)
Lesson 2.1: Multiplying Integers (Day 3 of 3)
Find products of integers.
Success Criteria
-I can explain the rules for multiplying integers.
-I can find products of integers with the same sign.
-I can find products of integers with different signs.
Assignment
Pages 53-54, problems 10-28 (evens), 29, 30-44 (evens)
Due: Wednesday, September 23rd
Standards
CCSS.Math.Content.7.NS.A.2c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.2c Apply properties of operations as strategies to multiply and divide rational numbers.
CCSS.Math.Content.7.NS.A.2a
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.2a Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (-1)(-1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.
CCSS.Math.Content.7.NS.A.3
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers.
1:52 PM - Algebra 1
Unit 1 Study Guide (Day 2 of 3)
We will continue a study guide for Unit 1.
Assignment
Review the problems on the study guide.
Standards
CCSS.Math.Content.HSA-CED.A.4
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
CCSS.Math.Content.HSA-CED.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
CCSS.Math.Content.HSN-Q.A.1
Common Core State Standards
Common Core Mathematics - High School — Number and Quantity
CCSS.Math.Content.HSN-Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
CCSS.Math.Content.HSA-REI.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.A.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
CCSS.Math.Content.HSA-REI.B.3
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.B.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
9/23/2020
7:58 AM - Math 7
Lesson 1.5: Subtracting Rational Numbers (Day 3 of 3)
Find differences of rational numbers and find distances between numbers on a number line.
Success Criteria
-I can explain how subtracting integers is related to adding integers.
-I can explain how to model subtraction of integers on a number line.
-I can find differences of integers by reasoning about absolute values.
Assignment
Pages 34-35, problems 9-10, 12-22 (evens), 26-27, 30-42 (evens)
Due: Thursday, September 24th
Standards
CCSS.Math.Content.7.NS.A.1d
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1d Apply properties of operations as strategies to add and subtract rational numbers.
CCSS.Math.Content.7.NS.A.1c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1c Understand subtraction of rational numbers as adding the additive inverse, p - q = p + (-q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
8:57 AM - Math 8
Quiz: Lessons 1.3 through 1.4/Chapter 2 Review & Refresh
We will complete the quiz for Lessons 1.3 through 1.4. After that, we will discuss the Review & Refresh questions for Chapter 2.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.8.EE.C.7b
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.EE.C.7b Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
CCSS.Math.Content.8.EE.C.7a
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.EE.C.7a Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
CCSS.Math.Content.8.EE.C.7
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.EE.C.7 Solve linear equations in one variable.
11:50 AM - Math 7 Accelerated
Lesson 2.2: Dividing Integers (Day 1 of 2)
Find quotients of integers.
Success Criteria
-I can explain the rules for dividing integers.
-I can find quotients of integers with the same sign.
-I can find quotients of integers with different signs.
Assignment
Pages 59-60, problems 10-12, 14-28 (evens), 29-33, 34-40 (evens)
Due: Friday, September 25th
Standards
CCSS.Math.Content.7.NS.A.3
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers.
CCSS.Math.Content.7.NS.A.2b
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.2b Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then -(p/q) = (-p)/q = p/(-q). Interpret quotients of rational numbers by describing real-world contexts.
1:52 PM - Algebra 1
PD Day (Student Dismissal at Noon)
Unit 1 Study Guide (Day 3 of 3)/Unit 1 Posttest (Day 1 of 3)
We will finish a study guide for Unit 1, then begin the posttest.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.HSN-Q.A.1
Common Core State Standards
Common Core Mathematics - High School — Number and Quantity
CCSS.Math.Content.HSN-Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
CCSS.Math.Content.HSA-REI.B.3
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.B.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
CCSS.Math.Content.HSA-CED.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
CCSS.Math.Content.HSA-CED.A.4
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
CCSS.Math.Content.HSA-REI.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.A.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
9/24/2020
7:58 AM - Math 7
Study Guide: Lessons 1.4 through 1.5
We will complete a study guide for Lessons 1.4 through 1.5.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.7.NS.A.3
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers.
CCSS.Math.Content.7.NS.A.1c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1c Understand subtraction of rational numbers as adding the additive inverse, p - q = p + (-q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
CCSS.Math.Content.7.NS.A.1d
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1d Apply properties of operations as strategies to add and subtract rational numbers.
8:57 AM - Math 8
Lesson 2.1: Translations (Day 1 of 3)
Translate figures in the coordinate plane.
Success Criteria
-I can identify a translation.
-I can find the coordinates of a translated figure.
-I can use coordinates to translate a figure.
Assignment
Pages 47-48, problems 5-21, 24 (graph paper needed)
Due: Tuesday, September 29th
Standards
CCSS.Math.Content.8.G.A.1
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations:
CCSS.Math.Content.8.G.A.3
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.G.A.3 Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates.
11:50 AM - Math 7 Accelerated
Lesson 2.2: Dividing Integers (Day 2 of 2)
Find quotients of integers.
Success Criteria
-I can explain the rules for dividing integers.
-I can find quotients of integers with the same sign.
-I can find quotients of integers with different signs.
Assignment
Pages 59-60, problems 10-12, 14-28 (evens), 29-33, 34-40 (evens)
Due: Friday, September 25th
Standards
CCSS.Math.Content.7.NS.A.3
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers.
CCSS.Math.Content.7.NS.A.2b
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.2b Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then -(p/q) = (-p)/q = p/(-q). Interpret quotients of rational numbers by describing real-world contexts.
1:52 PM - Algebra 1
Unit 1 Posttest (Day 2 of 3)
We will continue the posttest for Unit 1.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.HSA-REI.B.3
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.B.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
CCSS.Math.Content.HSA-REI.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.A.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
CCSS.Math.Content.HSA-CED.A.4
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
CCSS.Math.Content.HSA-CED.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
CCSS.Math.Content.HSN-Q.A.1
Common Core State Standards
Common Core Mathematics - High School — Number and Quantity
CCSS.Math.Content.HSN-Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
9/25/2020
7:58 AM - Math 7
Quiz: Lessons 1.4 through 1.5
We will complete the quiz for Lessons 1.4 through 1.5.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.7.NS.A.3
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers.
CCSS.Math.Content.7.NS.A.1d
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1d Apply properties of operations as strategies to add and subtract rational numbers.
CCSS.Math.Content.7.NS.A.1c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1c Understand subtraction of rational numbers as adding the additive inverse, p - q = p + (-q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
8:57 AM - Math 8
Lesson 2.1: Translations (Day 2 of 3)
Translate figures in the coordinate plane.
Success Criteria
-I can identify a translation.
-I can find the coordinates of a translated figure.
-I can use coordinates to translate a figure.
Assignment
Pages 47-48, problems 5-21, 24 (graph paper needed)
Due: Tuesday, September 29th
Standards
CCSS.Math.Content.8.G.A.3
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.G.A.3 Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates.
CCSS.Math.Content.8.G.A.1
Common Core State Standards
Common Core Mathematics - Grade 8
CCSS.Math.Content.8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations:
11:50 AM - Math 7 Accelerated
Study Guide & Quiz: Lessons 2.1 through 2.2
We will complete a study guide for Lessons 2.1 through 2.2, then take the quiz for these lessons.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.7.NS.A.1c
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1c Understand subtraction of rational numbers as adding the additive inverse, p - q = p + (-q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
CCSS.Math.Content.7.NS.A.1a
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1a Describe situations in which opposite quantities combine to make 0.
CCSS.Math.Content.7.NS.A.1b
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.1b Understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts.
CCSS.Math.Content.7.NS.A.3
Common Core State Standards
Common Core Mathematics - Grade 7
CCSS.Math.Content.7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers.
1:52 PM - Algebra 1
Unit 1 Posttest (Day 3 of 3)/Unit 2 Pretest
We will finish the posttest for Unit 1, then take the pretest for Unit 2.
Assignment
There's no assignment today.
Standards
CCSS.Math.Content.HSA-CED.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
CCSS.Math.Content.HSA-REI.A.1
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.A.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
CCSS.Math.Content.HSN-Q.A.1
Common Core State Standards
Common Core Mathematics - High School — Number and Quantity
CCSS.Math.Content.HSN-Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
CCSS.Math.Content.HSA-CED.A.4
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
CCSS.Math.Content.HSA-REI.B.3
Common Core State Standards
Common Core Mathematics - High School — Algebra
CCSS.Math.Content.HSA-REI.B.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
|
# How do you graph y= 1/2x + 2?
May 17, 2018
See a solution process below:
#### Explanation:
First, solve for two points which solve the equation and plot these points:
First Point: For $x = 0$
$y = \left(\frac{1}{2} \cdot 0\right) + 2$
$y = 0 + 2$
$y = 2$ or $\left(0 , 2\right)$
Second Point: For $x = 2$
$y = \left(\frac{1}{2} \cdot 2\right) + 2$
$y = 1 + 2$
$y = 3$ or $\left(2 , 3\right)$
We can next plot the two points on the coordinate plane:
graph{(x^2+(y-2)^2-0.035)((x-2)^2+(y-3)^2-0.035)=0 [-10, 10, -5, 5]}
Now, we can draw a straight line through the two points to graph the line:
graph{(x^2+(y-2)^2-0.035)((x-2)^2+(y-3)^2-0.035)(y - (1/2x) - 2)=0 [-10, 10, -5, 5]}
|
## What are the first three common multiples of 10 and 15?
• 15 = 1×3×5.
• 10= 1×2×5.
• common multiple are 1 and 5.
• there is two common multiple of 15 and 10 not three.
• hope its helpful for u.
## What are the least common multiples of 15?
LCM(15, 15) = 3×5 = 15
• 15 / 15 = 1.
• 15 / 15 = 1.
## What is the GCF of 10 and 15?
5
GCF of 10 and 15 by Listing Common Factors Therefore, the greatest common factor of 10 and 15 is 5.
## What is the LCM for 14 and 21?
42
Answer: LCM of 14 and 21 is 42.
## What are the common multiples of 10 15 and 20?
LCM of 10, 15, and 20 by Listing Multiples Step 2: The common multiples from the multiples of 10, 15, and 20 are 60, 120, . . . Step 3: The smallest common multiple of 10, 15, and 20 is 60.
## What is the LCM of 10 15 and 20?
60
Answer: LCM of 10, 15, and 20 is 60.
## What is the GCF for 15 and 9?
Answer: GCF of 9 and 15 is 3.
## What is the HCF of 14?
Factors of 14 = 1, 2, 7 and 14. Factors of 18 = 1, 2, 3, 6, 9 and 18. Therefore, common factor of 14 and 18 = 1 and 2. Highest common factor (H.C.F) of 14 and 18 = 2.
## How do you calculate least common multiple?
One way to find the least common multiple of two numbers is to first list the prime factors of each number. Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.
## How do you find the least common multiple?
Calculate the least common multiple. To do this, multiply together all of the factors in your multiplication sentence. For example, 2×2×5×7×3=420{\\displaystyle 2\imes 2\imes 5\imes 7\imes 3=420}. So, the least common multiple of 20 and 84 is 420.
## What is the easiest way to find the LCM?
Praveen, The easiest way to find the LCM for two or more numbers is this. First, prime factorize the three numbers. Secondly, line up the prime factors. Thirdly, for each prime number, choose the largest factor. Fourthly, multiply the LCM factors together. The product will be the LCM of the various numbers.
## How do you calculate LCM and GCF?
To find LCM multiply the factors in all three sections of the Venn Diagram 3 x 2 x 2 x 2 x 2 x 2 = 96 The LCM of 12 and 32 is 96 To find GCF multiply the factors in the center of the Venn Diagram 2 x 2 = 4 The GCF of 12 and 32 is 4. 5. Let’s Try Some! Use prime factorization to find LCM and GCF of 15 and 6.
|
# Triangular Prism
• Introduction
• What is a Triangular Prism?
• Properties of Triangular Prism
• Types of Triangular Prism
• Triangular Prism Net
• Surface Area of a Triangular Prism
• Volume of a Triangular Prism
• Solved Examples
• Practice Problems
## Introduction
A prism is a three-dimensional polyhedron. A prism typically consists of two identical bases which are n-sided polygons, and n other faces, necessarily all parallelograms, joining the corresponding sides of the two bases. The name of a particular prism depends on the two bases of the prism which can be triangles, rectangles, or any n-sided polygon. For example, a prism with triangular bases is called a triangular prism and a prism with a square base is called a square prism, and so on. In this topic, we are going to learn about various properties of a triangular prism and more.
## What is a Triangular Prism?
A triangular prism is a 3D shape that has two identical triangular bases joined together by three rectangular sides. These rectangular faces are called the lateral faces of a triangular prism. The triangular bases are also referred to as the top and bottom faces of the prism. The ends of the prism are parallel, meaning they're the same distance apart all the way along. And they're also congruent, which means they're the same shape and size. If you count up all the faces, you get 5. For the corners where the faces meet, you have 6 of those; where the edges come together, there are 9.
In simpler terms, a triangular prism has five faces, nine edges, and six vertices. Both triangular faces are the same size, and all three rectangular faces are identical.
When describing its size, we use terms like the length of the prism (l), the height of the triangular base (h), and the length of the bottom edge of the triangular base (b).
## Properties of Triangular Prism
The properties of a triangular prism help us recognize it easily. Let's break down the properties into some key points:
• Number of Faces: A triangular prism has five faces: two are triangular bases, and the other three are rectangular sides.
• Number of Edges: It has nine edges, which are where the faces meet. The prism's edges include the 6 edges formed by the two triangular bases (3 + 3), along with an additional 3 edges created by the sides connecting the bases.
• Number of Vertices: A triangular prism has six vertices, the points where the edges meet. The vertices of the prism coincide with the vertices of its two triangular bases, connected by lines that form rectangles along the prism’s sides.
• Polyhedral Nature: It's a type of polyhedron characterized by two triangular bases and three rectangular sides.
• Base Shape: The base of a triangular prism is a triangle.
• Side Shape: The sides are rectangles, maintaining a consistent structure along the length of the prism.
• Equilateral Triangular Bases: Both triangular bases are equilateral triangles, meaning all their sides are of equal length.
• Cross-Section Shape: Any cross-section of a triangular prism cut, along its base, forms a triangle.
• Congruent Bases: The two triangular bases are identical to each other, indicating their congruence.
## Types of Triangular Prism
When we talk about the types of triangular prism, we categorize them based on two factors: uniformity and alignment.
Based on uniformity, triangular prisms fall into two categories:
Regular Triangular Prism:
This is a prism where both triangular bases are regular triangles. In simpler terms, all the sides of these triangles are equal, and the angles between them measure 60 degrees. The sides of the prism are rectangular.
Irregular Triangular Prism:
Here, at least one of the triangular bases isn't a regular triangle. It means the sides of the base triangle can have different lengths, and the angles between them might not be fixed at 60 degrees. The lateral faces are still rectangles.
Based on alignment, triangular prisms can be divided into two types:
Right Triangular Prism:
In this type, the angle between the edges of the triangular bases and the edges of the rectangular faces is exactly 90 degrees. So, the bases meet the rectangular faces at right angles. The other properties of the prism remain the same.
Oblique Triangular Prism:
Here, the lateral faces are not perpendicular to the bases. Instead of rectangles, they form parallelograms. This means the angles between the lateral faces and the bases might not be 90 degrees. This allows for more flexibility in the prism's geometric configuration.
## Triangular Prism Net
The net of a triangular prism is like a blueprint that unfolds the surface of the prism. When you open, flatten, and lay out the prism's surface, you can see all its faces. This resulting pattern is a two-dimensional figure known as the net. Folding this net allows you to recreate the original triangular prism. The net clearly illustrates that the prism has triangular bases and rectangular lateral faces. In simpler terms, it serves as a visual guide showing how the prism can be assembled from a flat, folded shape.
## Surface Area of a Triangular Prism
The surface area of a triangular prism is the total area covered by its surface. It's made up of the areas of all the 5 faces of the prism. To calculate it, we use a formula that considers both the lateral surface area and the area of the bases.
Lateral Surface Area (LSA)
For the lateral surface area (LSA), we add up the areas of all the side faces, excluding the top and bottom faces. This is given by the formula: $$\text{LSA} = (s_1 + s_2 + b) \times l$$, where $$s_1$$ and $$s_2$$ are the lengths of the edges of the base triangle, and $$b$$ is the measure of the base of the triangle, while $$l$$ is the length of the prism.
Total Surface Area (TSA)
The total surface area (TSA) includes the lateral surface area and twice the area of one of the triangular bases.
"Area of one triangular base" = 1/2* b* h where b and h are the base and height of the triangular base.
"Area of 2 triangular bases" = 2 * (1/2 * b* h) = b * h
Accordingly, the formula is $$\text{TSA} = (b \times h) + (s_1 + s_2 + b) \times l$$, where $$s_1, s_2,$$ and $$b$$ are the edges of the triangular base, $$h$$ is the height of the base triangle and $$l$$ is the length of the prism.
Alternatively, we can simplify the formula for a right triangular prism to $$\text{TSA} = (s_1 + s_2 + h) \times L + b \times h$$, where $$b$$ represents the bottom edge of the base triangle.
## Volume of a Triangular Prism
The volume of a triangular prism is the amount of space it takes up in three-dimensional space. To find the volume, we multiply the area of the triangular base by the length of the prism. Since the base is a triangle, its area is found using the formula for the area of a triangle
Therefore, the formula for the volume of a triangular prism is given by: $$\text{Volume} = \frac{1}{2} \times \text{base length} \times \text{height} \times \text{length}$$.
, the base edge represents the length of one of the edges forming the base triangle, the height of the triangle is the perpendicular distance from the base to the opposite vertex, and the length of the prism indicates the overall length along its axis. By plugging these values into the formula, we can calculate the volume of the triangular prism.
## Solved Examples
Example 1. Find the volume of a triangular prism with a base length of $$8$$ units, a height of $$5$$ units, and a length of $$12$$ units.
Solution:
To find the volume of the triangular prism, we use the formula $$V = \frac{1}{2} \times b \times h \times l$$, where $$b$$ is the base length, $$h$$ is the height, and $$l$$ is the length:
$$V = \frac{1}{2} \times 8 \times 5 \times 12$$
$$V = \frac{1}{2} \times 8 \times 5 \times 12$$
$$V = 240 \text{ cubic units}$$
Therefore, the volume of the triangular prism is $$240$$ cubic units.
Example 2. Calculate the surface area of a triangular prism with a base area of $$15$$ square units, a length of $$8$$ units, and a perimeter of the base as $$20$$ units.
Solution:
We compute the surface area of the prism using the formula:
$$\text{Surface area} = (\text{Perimeter of the base} \times \text{Length}) + (2 \times \text{Base Area})$$
"Surface area" = (20 \times 8) + (2 \times 15)
"Surface area" = 160 + 30
"Surface area" = 190 square units
Thus, the surface area of the triangular prism is $$190$$ square units.
Example 3. Find the volume of a triangular prism with a base area of $$35$$ $$cm^2$$ and a length of $$12$$ cm.
Solution:
Using the formula "Volume" = "base area" × h × l, where $$b$$ is the base, $$h$$ is the height, and $$l$$ is the length:
$$\text{Volume} = 35 \times 12$$
$$\text{Volume} = 420 \text{ cubic cm}$$
Example 4. Find the base length of a triangular prism with the volume of the prism is $$72$$ units, with a length of $$8$$ units and a base height of $$6$$ units.
Solution:
Using the volume formula $$V = \frac{1}{2} \times b \times h \times l$$, we solve for the base length $$b$$:
$$72 = \frac{1}{2} \times b \times 6 \times 8$$
$$144 = b \times 48$$
$$b = \frac{144}{48} = 3$$
Thus, the base length of the triangular prism is $$3$$ units.
Example 5. Compute the total surface area of a regular triangular prism whose length is $$9$$ cm. The sides of the equilateral triangular base measure $$12$$ cm and its height is $$7$$ cm.
Solution:
Using the formula "Surface area" = (sh + (3s)l) where $$h$$ is the height of the base, $$s$$ is the sides of the base, and $$l$$ is the length of the prism:
"Surface area" = (12 × 7 + (3 *12) × 9)
"Surface area" = (84 + 324)
"Surface area" = 408 square cm
Thus, the total surface area of the prism is $$408$$ square centimeters.
## Practice Problems
Q1. What is the volume of a triangular prism with a base length of $$7$$ units, a height of $$4$$ units, and a length of $$10$$ units?
1. $$140$$ cubic units
2. $$150$$ cubic units
3. $$160$$ cubic units
4. $$170$$ cubic units
Q2. Calculate the surface area of a triangular prism if its base area is $$24$$ square units, length is $$8$$ units, and the perimeter of the base is $$18$$ units.
1. $$230$$ square units
2. $$192$$ square units
3. $$245$$ square units
4. $$162$$ square units
Q3. Find the length of the base of a triangular prism if its volume is $$96$$ cubic units, the height of the triangular base is $$6$$ units, and the length of the prism is $$8$$ units.
1. $$8$$ units
2. $$10$$ units
3. $$12$$ units
4. $$4$$ units
Q4. Determine the lateral surface area of a regular triangular prism with a base length of $$9$$ units, a height of $$5$$ units, and a length of $$12$$ units.
1. $$284$$ square units
2. $$324$$ square units
3. $$264$$ square units
4. $$270$$ square units
Q5. Calculate the total surface area of a triangular prism with a base length of $$6$$ units, a height of $$8$$ units, and a length of $$15$$ units.
1. $$312$$ square units
2. $$322$$ square units
3. $$332$$ square units
4. $$342$$ square units
Q6. A triangular prism has a volume of $$126$$ cubic meters. The base and height of the triangular prism are $$14$$ meters and $$6$$ meters respectively. Find the length of the prism.
1. $$3$$ units
2. $$4$$ units
3. $$5$$ units
4. $$6$$ units
Q1. What is a triangular prism?
Answer: A triangular prism is a three-dimensional geometric shape characterized by two congruent triangular bases and three rectangular or parallelogram faces. It has six vertices and nine edges.
Q2. How do you find the volume of a triangular prism?
Answer: The volume of a triangular prism can be calculated using the formula: "Volume" = 1/2 × "base length" × "height" × "prism length", where the base length refers to the length of one side of the triangular base, height of the triangular base, and length represents the overall length of the prism.
Q3. What are the properties of a triangular prism?
Answer: The key properties of a triangular prism include having two congruent triangular bases, three rectangular (or parallelogram) faces, six vertices, and nine edges. It can also be classified based on uniformity (regular or irregular) and alignment (right or oblique).
Q4. How do you calculate the total surface area of a triangular prism?
Answer: To find the total surface area of a triangular prism, you need to calculate the areas of all its faces and then sum them up. The formula for the total surface area of a triangular prism is "Surface Area" = ("Perimeter of the base" × "Length") + (2 × "Base Area"), where the base area is the area of one of the triangular bases.
Q5. What are some real-life examples of triangular prisms?
Answer: Triangular prisms are commonly found in various everyday objects and structures. Examples include roofs of houses, camping tents, certain types of packaging, pyramids with triangular bases, certain architectural elements, and the design of some musical instruments like certain types of bells.
|
## 35 year 6 maths questions
Author Amber Watkins
Published February 2024
• Key takeaways
• Year 6 maths word problems help students apply mathematical concepts to real-world scenarios.
• In year 6 maths, it’s important to master fractions, decimals, and order of operations.
• Hard problems can be simplified by drawing a picture or using a method that makes them easier to solve.
• Word problems
## Numbers & place value
Multiplication.
If this maths page has 5 sections of maths questions with an average of 7 questions in each section, can you guess how many problems there will be in all? That’s right, there will be a total of 35 year 6 maths questions for us to practise. Without realising it, you just solved one of our first year 6 word problems . Excellent work! Together we will review year 6 maths problems with decimals, fractions, order of operations, and rounding. We will even consider hard maths problems, like multiplying using square models and number lines. Don’t worry about getting the answers right the first time, each section includes answers if you need a sneak peek. So grab a pad and pencil and let’s begin.
## Year 6 maths word problems
Let’s begin by covering maths word problems.
Mia’s water bottle can hold 300 millilitres of water. She drinks two and a half water bottles each day. How many total millilitres of water does Mia drink each day?
The playground is made up of four rectangular lots that are each 10 metres by 7 metres. What is the total area of the playground?
Josh lives 2.5 kilometres away from the park. Josh rides his bike to the park and back home four days a week. How many kilometres does Josh ride his bike each week?
Since we are solving word problems, a word should always be in our answer.
In this section, we will cover maths questions that use place values, word form to standard form, standard form to word form, and expanded form.
## Place value
Which digit in the number 245 is in the hundreds place?
Which digit in the number 4,602 is in the tens place?
Which digit in the number 752 is in the ones place?
## Word form to standard form
Standard form to expanded form.
4,000 + 300 + 2 = 4,302
600 + 50 + 4 = 654
20,000 + 2,000 + 300 + 60 + 5 = 22,365
In this section, we will cover year six maths questions that include comparing decimals and rounding decimals to the nearest tenth, hundredth, and thousandths place. Let’s get started.
8.9 and 8.900 are the same.
Remember if the number to the right is 5 or larger, you can borrow and become one digit larger. If the number to the right is between 0 and 4, you can’t borrow and the number remains the same.
In this section, we will review maths questionss for year six that cover the topics of adding and subtracting two and three-digit numbers using transformation, division using area models, and the order of operations.
## Adding and subtracting whole numbers by transformation
A large portion of year 6 maths problems can be made easier to solve by transformation or changing numbers to make the problem simpler. Let’s practise adding and subtracting two and three-digit numbers by transformation.
How to add numbers using transformation?
When adding or subtracting numbers, look for ways to round to make the problem simpler. Don’t forget what you do to one number, you have to do the opposite to the other. This makes sure the problem stays balanced. Let’s see how this is done. What is 48 + 52? 48 + 2 = 50
52 – 2 = 50
50 + 50 = 100.
It may be difficult to add 48 + 52 in your head, so we round 48 to 50 by adding 2. Since we added 2 to the first number, we have to make sure to subtract 2 from the second number. So now the problem is 50 plus 50, which is much easier to calculate!
32 – 2 = 30 68 + 2 = 70 30 + 70 = 100 The answer is 100.
208 – 8 = 200 432 + 8 = 440 200 + 440 = 640 The answer is 640.
321- 1 = 320 199 + 1 = 200 320- 200 = 120 The answer is 120
## Division using area models
Don’t know what an area model is? See our area model guide to help with these problems.
Using an area model find the quotient of 365 ፥ 5 =
Using an area model find the quotient of 504 ፥ 6 =
Using the area model below, find the quotient of 872 ፥ 2 =
## Order of operations
Many people remember the Order of Operations with the acronym PEMDAS- “Please Excuse My Dear Aunt Sally”, with each letter representing an operation: Parentheses, Exponents, Multiply or Divide, Add or Subtract. Solve the following equations using the order of operations.
Remember that the steps Multiply or Divide can be done in any order. Also, the steps Add or Subtract can be done in any order, do the operation that comes first.
Some of the hardest maths questions involve multiplying: multiplying using square models, multiplying fractions and whole numbers using expanded form, and multiplying fractions using number lines. Let’s practise a few of each!
## Multiplying using square models
Multiplying using square models is a method to help students “see” multiplication problems in a simpler way.
Multiply 42 x 35 using the square model below.
Multiply 55 x 76 using the square model below.
Multiply 98 x 42 using the square model below.
## Multiplying fractions and whole numbers using expanded form
Multiply ½ x 5 using expanded form.
Multiply ⅔ x 6 using expanded form.
Multiply ⅙ x 7 using expanded form.
## Multiplying fractions using a number line
For more maths practice for year 6 , our app for maths help provides year 6 maths help in a way that makes maths fun, rewards effort over being correct, and scaffolds learning. Plus, you’ll get access to a dashboard that displays your child’s participation and progression on maths topics they struggle with. Check it out today !
Mia drinks 750 millilitres of water each day.
Explanation: 300 x 2 = 600. 300 / 2 = 150. 600 + 150 = 750.
280 square metres.
Explanation: The area of one lot is 10 x 7 = 70 metres. Take that area and times it by four, so 70 x 4 = 280 metres.
Josh rides 20 kilometres on his bike each week.
Explanation: Each day Josh rides 5 total kilometres: 2.5 kilometres to the park and 2.5 kilometres back home. If we multiply that number by 4 days a week, it equals 20 kilometres in all.
2 is in the hundreds place.
0 is in the tens place.
2 is in the ones place.
## Multiplying fractions using a number line answer sheet
Explanation – First divide the number line into 6 equal parts and draw up a line up to 4/6. Then find the ½ way mark of that line you drew up to 4/6. You will now see the line is divided into three parts. So the answer would be ⅓.
Lesson credits
Amber Watkins
Amber is an education specialist with a degree in Early Childhood Education. She has over 12 years of experience teaching and tutoring students in maths. "Knowing that my work in maths education makes such an impact leaves me with an indescribable feeling of pride and joy!"
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## Year 6 Maths Worksheets UK Hub Page
Welcome to our Year 6 Maths Worksheets area.
Here you will find a wide range of free printable Year 6 Maths Worksheets for your child to enjoy.
Come and take a look at our rounding decimal pages, or maybe some of our adding and subtracting fractions worksheets. Perhaps you are looking for some worksheets about finding angles in a triangle, or need some ratio problem worksheets to help your child learn about ratio?
For full functionality of this site it is necessary to enable JavaScript.
• This page contains links to other Math webpages where you will find a range of activities and resources.
• If you can't find what you are looking for, try searching the site using the Google search box at the top of each page.
## Year 6 Maths Learning
Here are some of the key learning objectives for the end of Year 6:
• know and use Place value up to 10 million
• Counting on and back in steps of powers of 10 from any number up to 10 million
• Round numbers to any given degree of accuracy.
• Count forwards and backwards through zero with positive and negative numbers.
• Read Roman numerals to 1000 and recognise years written in Roman numerals
• solve multi-step problems using addition and subtraction in a range of contexts
• identify multiples and factors including common factors
• multiply and divide up to 4-digit numbers by up to 2 digits
• Use their knowledge of the order of operations to carry out calculations involving the four operations.
• Identify common factors, common multiples and prime numbers.
• solve problems involving addition, subtraction, multiplication and division
• simplify fractions
• compare and order fractions including mixed numbers
• add and subtract fractions with different denominators including mixed numbers
• multiply simple fractions together and simplify the answer
• divide proper fractions by whole numbers
• recall and use equivalence between simple fractions, decimals and percentages.
• Multiply and divide whole numbers and decimals up to 3dp by 10, 100 or 1000
• read, write, order and compare numbers up to 3dp
• round decimals with up to 3dp to the nearest whole
• solve problems with numbers up to 3dp
• work out percentages of different amounts
• solve problems using percentages
• use simple formulae
• express missing number problems using algebra
• find pairs of numbers that satisfy equations with two variables
• solve problems involving simple ratios
• solve problems involving similar shapes where the scale factor is known
• use, read, write and convert between standard units of measure
• measure, compare and calculate using different measures
• know that shapes with the same area can have different perimeters
• find the area of parallelograms and right triangles
• find the volume of cubes and cuboids
• convert between miles and km
• name and understand the parts of circles - radius, diameter and circumference
• draw 2D shapes accurately using dimensions and angles
• compate and classify 2D shapes by a range of properties
• find missing angles in triangles, quadrilaterals and regular shapes
• use coordinates in all 4 quadrants
• draw and translate simple shapes in all 4 quadrants
• interpret and construct pie charts and line graphs
• calculate the mean as an average
Our site is mainly based around the US Elementary school math standards.
Though the links on this page are all designed primarily for students in the US, but they are also at the correct level and standard for UK students.
The main issue is that some of the spelling is different and this site uses US spelling.
Year 6 is generally equivalent to 5th Grade in the US.
• Place Value Zone
• Mental Math Zone
## Word Problems Zone
Fractions percents ratio zone.
• Percentages Zone
• Measurement Zone
## Geometry Zone
Data analysis zone.
• Fun Zone: games and puzzles
Coronavirus Stay At Home Support
For those parents who have found themselves unexpectedly at home with the kids and need some emergency activities for them to do, we have started to develop some Maths Grab Packs for kids in the UK.
Each pack consists of at least 10 mixed math worksheets on a variety of topics to help you keep you child occupied and learning.
The idea behind them is that they can be used out-of-the-box for some quick maths activities for your child.
They are completely FREE - take a look!
• Free Maths Grabs Packs
## Place Value & Number Sense Zone
Year 6 number worksheets.
Here you will find a range of Free Printable Year 6 Number Worksheets.
• use place value with numbers up to 10 million;
• use place value with up to 3 decimal places;
• understand how to use exponents (powers) of a number;
• understand and use parentheses (brackets);
• understand and use multiples and factors;
• extend their knowledge of prime and composite (non-prime) numbers up to 100;
• know and be able to use the PEMDAS (or PEDMAS) rule.
• Place Value Worksheets to 10 million
• Place Value to 3dp
• Ordering Decimals Worksheets
• PEMDAS Rule Support Page
• PEMDAS Problems Worksheets
• Balancing Math Equations
• Roman Numerals worksheets
## Ordering Large Numbers and Decimals to 3dp
The sheets in this section involve ordering lists of decimals to 3 decimal places and also large numbers up to 100 million.
There are sheets with decimals up to 10, and also sheets with numbers from -10 to 10.
• Ordering Large Numbers up to 100 million
• Ordering Decimals to 3dp
## Rounding Decimals
• Rounding to the nearest tenth
• Rounding Decimal Places Sheets to 2dp
• Rounding Decimals Worksheet Challenges
## Year 6 Decimal Counting Worksheets
Using these sheets will support you child to:
• count on and back by multiples of 0.1;
• fill in the missing numbers in sequences;
• count on and back into negative numbers.
• Counting By Decimals
## Year 6 Mental Maths Zone
Each worksheet tests the children on a range of math topics from number facts and mental arithmetic to geometry, fraction and measures questions.
A great way to revise topics, or use as a weekly math quiz!
• Year 6 Mental Maths Tests
Top of Page
• add decimals including tenths and hundredths mentally;
• add a columns of multi-digit numbers, including decimals.
• Money Worksheets (randomly generated)
## Year 6 Subtraction Worksheets
• subtract decimals including tenths and hundredths mentally;
• subtract multi-digit numbers, including decimals using column subtraction.
• Subtracting Decimals Worksheets (mental)
• Subtraction Worksheets up to Billions (columns)
• Column Subtraction with Decimals
## Year 6 Multiplication Worksheets
• extend their knowlege of multiplication to decimals;
• use their multiplication tables to answer related facts, including decimals;
• multiply a range of decimals with up to 2 decimal places (2dp) by a whole number;
• multiply different money amounts by a whole number.
• Multiplying Decimals by 10 and 100
• Multiplication Fact Sheet Decimals
• Decimal Multiplication Worksheets to 1dp
• Decimal Multiplication Worksheets to 2dp
• Free Multiplication Worksheets (randomly generated)
• Multiply and Divide by 10 100 (decimals)
• Multiplication & Division Worksheets (randomly generated)
• Multiplication Word Problems
• divide any whole number up to 10000 by a two digit number;
• express any division with a remainder in the form of a mixed number (a number with a fraction part).
• Long Division Worksheets (whole numbers)
• Long Division of Decimal Numbers
• Decimal Division Facts
• Division Facts Worksheets (randomly generated)
## Year 6 Maths Problems
• apply their addition, subtraction, multiplication and division skills;
• apply their knowledge of rounding and place value;
• solve a range of problems including "real life" problems and ratio problems.
These sheets involve solving one or two more challenging longer problems.
• Year 6 Math Problems (5th Grade)
These sheets involve solving many 'real-life' problems involving data.
• Year 6 Math Word Problems (5th Grade)
These sheets involve solving a range of ratio problems.
## Year 6 Fraction Worksheets
Year 6 percentage worksheets, year 6 ratio worksheets.
• compare and order fractions;
• add and subtract fractions and mixed numbers;
• understand how to multiply fractions by a whole number;
• understand how to multiply two fractions together, including mixed fractions;
• understand the relationship between fractions and division;
• know how to divide fractions and mixed fractions;
• convert decimals to fractions.
• Comparing Fractions Worksheet page
• Subtracting Fractions Worksheets
• Improper Fraction Worksheets
• Converting Decimals to Fractions Worksheets
• Fractions Decimals Percents Worksheets
• Multiplying Fractions Worksheets
• Dividing Fractions by Whole numbers
• Divide Whole numbers by Fractions
• Simplifying Fractions Worksheets
• Free Printable Fraction Riddles (harder)
Take a look at our percentage worksheets for finding the percentage of a number or money amount.
We have a range of percentage sheets from quite a basic level to much harder.
• Percentage of Numbers Worksheets
• Money Percentage Worksheets
• Percentage Word Problems
These Year 6 Ratio worksheets are a great way to introduce this concept.
We have a range of part to part ratio worksheets and slightly harder problem solving worksheets.
• Ratio Part to Part Worksheets
• Ratio and Proportion Worksheets
## Year 6 Geometry Worksheets
• know how to find missing angles in a range of situations;
• learn the number of degrees in a right angle, straight line, around a point and in a triangle;
• know how to calculate the area of a triangle;
• know how to calculate the area of a range of quadrilaterals.
• learn the formulas to calculate the area of triangles and some quadrilaterals;
• write and plot coordinates in all 4 quadrants.
• (5th Grade) Geometry - Angles
• Coordinate Plane Worksheets (All 4 Quadrants)
• Parts of a Circle Worksheets
## Measurement Zone, including Time & Money
Year 6 measurement worksheets.
• learn how to read a standard scale going up in different fractions: halves, quarters, eighths and sixteenths;
• learn how to read a metric scale going up in 0.1s, 5s, 10s, 25s, 50s & 100s;
• learn how to estimate a measurement of length, weight or liquid;
• convert temperatures in Celsius and Fahrenheit.
## Time Puzzles - harder
Here you will find our selection of harder time puzzles.
• Time Word Problems Worksheets - Riddles (harder)
• find the mean of up to 5 numbers;
• find a missing data point when the mean is given.
• Mean Worksheets
## Fun Zone: Puzzles, Games and Riddles
Year 6 maths games.
• Year 6 Math Games (5th Grade)
## Year 6 Maths Puzzles
The puzzles will help your child practice and apply their addition, subtraction, multiplication and division facts as well as developing their thinking and reasoning skills in a fun and engaging way.
• Printable Math Puzzles
## Math Salamanders Year 6 Maths Games Ebook
Our Year 6 Maths Games Ebook contains all of our fun maths games, complete with instructions and resources.
This ebooklet is available in our store - use the link below to find out more!
• Year 6 Maths Games Ebook
## Other UK Maths Worksheet pages
See below for our other maths worksheets hub pages designed for children in the UK.
How to Print or Save these sheets 🖶
Need help with printing or saving? Follow these 3 steps to get your worksheets printed perfectly!
• How to Print support
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## Math-Salamanders.com
The Math Salamanders hope you enjoy using these free printable Math worksheets and all our other Math games and resources.
TOP OF PAGE
Grade 6 maths word problems with answers are presented. Some of these problems are challenging and need more time to solve. Also detailed solutions and full explanations are included.
• Two numbers N and 16 have LCM = 48 and GCF = 8. Find N.
• If the area of a circle is 81pi square feet, find its circumference.
• Find the greatest common factor of 24, 40 and 60.
• In a given school, there are 240 boys and 260 girls. a) What is the ratio of the number of girls to the number of boys? b) What is the ratio of the number of boys to the total number of pupils in the school?
• If Tim had lunch at \$50.50 and he gave 20% tip, how much did he spend?
• Find k if 64 ÷ k = 4.
• Little John had \$8.50. He spent \$1.25 on sweets and gave to his two friends \$1.20 each. How much money was left?
• What is x if x + 2y = 10 and y = 3?
• A telephone company charges initially \$0.50 and then \$0.11 for every minute. Write an expression that gives the cost of a call that lasts N minutes.
• A car gets 40 kilometers per gallon of gasoline. How many gallons of gasoline would the car need to travel 180 kilometers?
• A machine fills 150 bottles of water every 8 minutes. How many minutes it takes this machine to fill 675 bottles?
• A car travels at a speed of 65 miles per hour. How far will it travel in 5 hours?
• A small square of side 2x is cut from the corner of a rectangle with a width of 10 centimeters and length of 20 centimeters. Write an expression in terms of x for the area of the remaining shape.
• A rectangle A with length 10 centimeters and width 5 centimeters is similar to another rectangle B whose length is 30 centimeters. Find the area of rectangle B.
• A school has 10 classes with the same number of students in each class. One day, the weather was bad and many students were absent. 5 classes were half full, 3 classes were 3/4 full and 2 classes were 1/8 empty. A total of 70 students were absent. How many students are in this school when no students are absent?
• The perimeter of square A is 3 times the perimeter of square B. What is the ratio of the area of square A to the area of square B.
• John gave half of his stamps to Jim. Jim gave gave half of his stamps to Carla. Carla gave 1/4 of the stamps given to her to Thomas and kept the remaining 12. How many stamps did John start with?
• Two balls A and B rotate along a circular track. Ball A makes 4 full rotations in 120 seconds. Ball B makes 3 full rotation in 60 seconds. If they start rotating now from the same point, when will they be at the same starting point again?
• A segment is 3 units long. It is divided into 9 parts. What fraction of a unit are 2 parts of the segment?
• A car is traveling 75 kilometers per hour. How many meters does the car travel in one minute?
• Carla is 5 years old and Jim is 13 years younger than Peter. One year ago, Peter's age was twice the sum of Carla's & Jim's age. Find the present age of each one of them.
• Linda spent 3/4 of her savings on furniture. She then spent 1/2 of her remaining savings on a fridge. If the fridge cost her \$150, what were her original savings?
• The distance bewteen Harry and Kate is 2500 meters. Kate and Harry start walking toward one another and Kate' dog start running back and forth between Harry and Kate at a speed of 120 meters per minute. Harry walks at the speed of 40 meters per minute while Kate walks at the speed of 60 meters per minute. What distance will the dog have travelled when Harry and Kate meet each other?
## Answers to the Above Questions
• a) 13:12 b)12:25
• 0.50 + N * 0.11
• 4.5 gallons
• 450 centimeters squared
• 108 cubic centimeters
• 1250 meters/minute
• Carla:5 years, Jim: 6 years, Peter: 19 years.
• 3000 meters
Teachers | Pupils | Parents
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WEEK 01 ADDITION & SUBTRACTION USING FORMAL WRITTEN METHODS column adding take away subtracting WEEK 01 ADDITION AND SUBTRACTION USING FORMAL WRITTEN METHODS
WEEK 01 ADDITION & SUBTRACTION USING FORMAL WRITTEN METHODS column adding take away subtracting
WEEK 01 ADDITION & SUBTRACTION USING FORMAL WRITTEN METHODS column adding take away subtracting
WEEK 01 ADDITION & SUBTRACTION USING FORMAL WRITTEN METHODS column adding take away subtracting
WEEK 01 ADDITION & SUBTRACTION USING FORMAL WRITTEN METHODS column adding take away subtracting
WEEK 01 ADDITION & SUBTRACTION USING FORMAL WRITTEN METHODS column adding take away subtracting
WEEK 02 MULTIPLICATION USING FORMAL WRITTEN METHODS multiply times table column WEEK 02 MULTIPLICATION USING FORMAL WRITTEN METHODS
WEEK 02 MULTIPLICATION USING FORMAL WRITTEN METHODS multiply times table column
WEEK 02 MULTIPLICATION USING FORMAL WRITTEN METHODS multiply times table column
WEEK 02 MULTIPLICATION USING FORMAL WRITTEN METHODS multiply times table column
WEEK 02 MULTIPLICATION USING FORMAL WRITTEN METHODS multiply times table column
WEEK 03 DIVISION USING FORMAL WRITTEN METHODS divide share chunking column WEEK 03 DIVISION USING FORMAL WRITTEN METHODS
WEEK 03 DIVISION USING FORMAL WRITTEN METHODS divide share chunking column
WEEK 03 DIVISION USING FORMAL WRITTEN METHODS divide share chunking column
WEEK 03 DIVISION USING FORMAL WRITTEN METHODS divide share chunking column
WEEK 03 DIVISION USING FORMAL WRITTEN METHODS divide share chunking column
WEEK 03 DIVISION USING FORMAL WRITTEN METHODS divide share chunking column
WEEK 04 MULTIPLY AND DIVIDE BY 10 100 & 1000 place value multiplication times table division share WEEK 04 MULTIPLY AND DIVIDE BY 10, 100 & 1000
WEEK 04 MULTIPLY AND DIVIDE BY 10 100 & 1000 place value multiplication times table division share
WEEK 04 MULTIPLY AND DIVIDE BY 10 100 & 1000 place value multiplication times table division share
WEEK 04 MULTIPLY AND DIVIDE BY 10 100 & 1000 place value multiplication times table division share
WEEK 04 MULTIPLY AND DIVIDE BY 10 100 & 1000 place value multiplication times table division share
WEEK 05 CALCULATE AND COMPARE THE VOLUME OF CUBES AND CUBOIDS 3d shape greater less WEEK 05 CALCULATE AND COMPARE THE VOLUME OF CUBES AND CUBOIDS
WEEK 05 CALCULATE AND COMPARE THE VOLUME OF CUBES AND CUBOIDS 3d shape greater less
WEEK 05 CALCULATE AND COMPARE THE VOLUME OF CUBES AND CUBOIDS 3d shape greater less
WEEK 05 CALCULATE AND COMPARE THE VOLUME OF CUBES AND CUBOIDS 3d shape greater less
WEEK 05 CALCULATE AND COMPARE THE VOLUME OF CUBES AND CUBOIDS 3d shape greater less
WEEK 06 ROUNDING WHOLE NUMBERS unit one WEEK 06 ROUNDING WHOLE NUMBERS
WEEK 06 ROUNDING WHOLE NUMBERS unit one
WEEK 06 ROUNDING WHOLE NUMBERS unit one
WEEK 03 DIVISION USING FORMAL WRITTEN METHODS divide share chunking column
WEEK 06 ROUNDING WHOLE NUMBERS unit one
WEEK 07 REFLECT A SHAPE THROUGH AN AXIS 2d reflection transition graph WEEK 07 REFLECT A SHAPE THROUGH AN AXIS
WEEK 07 REFLECT A SHAPE THROUGH AN AXIS 2d reflection transition graph
WEEK 07 REFLECT A SHAPE THROUGH AN AXIS 2d reflection transition graph
WEEK 07 REFLECT A SHAPE THROUGH AN AXIS 2d reflection transition graph
WEEK 07 REFLECT A SHAPE THROUGH AN AXIS 2d reflection transition graph
WEEK 07 REFLECT A SHAPE THROUGH AN AXIS 2d reflection transition graph
WEEK 08 ROUNDING IN CONTEXT real life place value WEEK 08 ROUNDING IN CONTEXT
WEEK 08 ROUNDING IN CONTEXT real life place value
WEEK 08 ROUNDING IN CONTEXT real life place value
WEEK 08 ROUNDING IN CONTEXT real life place value
WEEK 09 CALCULATING MISSING ANGLES IN QUADRILATERALS AND TRIANGLES acute obtuse reflex right 2d shape WEEK 09 CALCULATING MISSING ANGLES IN QUADRILATERALS AND TRIANGLES
WEEK 09 CALCULATING MISSING ANGLES IN QUADRILATERALS AND TRIANGLES acute obtuse reflex right 2d shape
WEEK 09 CALCULATING MISSING ANGLES IN QUADRILATERALS AND TRIANGLES acute obtuse reflex right 2d shape
WEEK 09 CALCULATING MISSING ANGLES IN QUADRILATERALS AND TRIANGLES acute obtuse reflex right 2d shape
WEEK 09 CALCULATING MISSING ANGLES IN QUADRILATERALS AND TRIANGLES acute obtuse reflex right 2d shape
WEEK 10 ADDING AND SUBTRACTING FRACTIONS AND MIXED NUMBERS take away addition subtraction WEEK 10 ADDING AND SUBTRACTING FRACTIONS AND MIXED NUMBERS
WEEK 10 ADDING AND SUBTRACTING FRACTIONS AND MIXED NUMBERS take away addition subtraction
WEEK 10 ADDING AND SUBTRACTING FRACTIONS AND MIXED NUMBERS take away addition subtraction
WEEK 10 ADDING AND SUBTRACTING FRACTIONS AND MIXED NUMBERS take away addition subtraction
WEEK 10 ADDING AND SUBTRACTING FRACTIONS AND MIXED NUMBERS take away addition subtraction
WEEK 11 READ & WRITE VALUES UP TO 10000000 place value million number WEEK 11 READ AND WRITE VALUES UP TO 10,000,000
WEEK 11 READ & WRITE VALUES UP TO 10000000 place value million number
WEEK 11 READ & WRITE VALUES UP TO 10000000 place value million number
WEEK 11 READ & WRITE VALUES UP TO 10000000 place value million number
WEEK 11 READ & WRITE VALUES UP TO 10000000 place value million number
WEEK 12 CONVERTING BETWEEN DIFFERENT METRIC UNITS OF MEASUREMENT length mass capacity mm cm metre km kilo gram cl ml milli WEEK 12 CONVERTING BETWEEN DIFFERENT METRIC UNITS OF MEASUREMENT
WEEK 12 CONVERTING BETWEEN DIFFERENT METRIC UNITS OF MEASUREMENT length mass capacity mm cm metre km kilo gram cl ml milli
WEEK 12 CONVERTING BETWEEN DIFFERENT METRIC UNITS OF MEASUREMENT length mass capacity mm cm metre km kilo gram cl ml milli
WEEK 12 CONVERTING BETWEEN DIFFERENT METRIC UNITS OF MEASUREMENT length mass capacity mm cm metre km kilo gram cl ml milli
WEEK 13 INTERPRETING A PIE CHART graph proportion read WEEK 13 INTERPRETING A PIE CHART
WEEK 13 INTERPRETING A PIE CHART graph proportion read
WEEK 13 INTERPRETING A PIE CHART graph proportion read
WEEK 13 INTERPRETING A PIE CHART graph proportion read
WEEK 14 USE A FORMULA TO CALCULATE THE AREA & VOLUME OF SHAPES 2d 3d width height depth length dimension WEEK 14 USE A FORMULA TO CALCULATE THE AREA AND VOLUME OF SHAPES
WEEK 14 USE A FORMULA TO CALCULATE THE AREA & VOLUME OF SHAPES 2d 3d width height depth length dimension
WEEK 14 USE A FORMULA TO CALCULATE THE AREA & VOLUME OF SHAPES 2d 3d width height depth length dimension
WEEK 14 USE A FORMULA TO CALCULATE THE AREA & VOLUME OF SHAPES 2d 3d width height depth length dimension
WEEK 14 USE A FORMULA TO CALCULATE THE AREA & VOLUME OF SHAPES 2d 3d width height depth length dimension
WEEK 15 SEEING FRACTIONS AS DIVISION divide denominator numerator share WEEK 15 SEEING FRACTIONS AS DIVISION
WEEK 15 SEEING FRACTIONS AS DIVISION divide denominator numerator share
WEEK 15 SEEING FRACTIONS AS DIVISION divide denominator numerator share
WEEK 16 FIND VALUES FOR TWO UNKNOWN NUMBERS algebra x y variable missing WEEK 16 FIND VALUES FOR TWO UNKNOWN NUMBERS
WEEK 16 FIND VALUES FOR TWO UNKNOWN NUMBERS algebra x y variable missing
WEEK 16 FIND VALUES FOR TWO UNKNOWN NUMBERS algebra x y variable missing
WEEK 17 READ & WRITE ROMAN NUMERALS WEEK 17 READ AND WRITE ROMAN NUMERALS
WEEK 17 READ & WRITE ROMAN NUMERALS
WEEK 17 READ & WRITE ROMAN NUMERALS
WEEK 17 READ & WRITE ROMAN NUMERALS
WEEK 18 READING LINE GRAPHS chart interpret time WEEK 18 READING LINE GRAPHS
WEEK 18 READING LINE GRAPHS chart interpret time
WEEK 18 READING LINE GRAPHS chart interpret time
WEEK 19 CALCULATING THE HIGHEST COMMON FACTOR, LOWEST COMMON MULTIPLE & IDENTIFYING PRIME NUMBERS WEEK 19 CALCULATING THE HIGHEST COMMON FACTOR, LOWEST COMMON MULTIPLE & IDENTIFYING PRIME NUMBERS
WEEK 19 CALCULATING THE HIGHEST COMMON FACTOR, LOWEST COMMON MULTIPLE & IDENTIFYING PRIME NUMBERS
WEEK 19 CALCULATING THE HIGHEST COMMON FACTOR, LOWEST COMMON MULTIPLE & IDENTIFYING PRIME NUMBERS
WEEK 20 MISSING ANGLES AT A POINT ON A STRAIGHT LINE & OPPOSITE ANGLES 180 360 degree WEEK 20 MISSING ANGLES AT A POINT, ON A STRAIGHT LINE AND OPPOSITE ANGLES
WEEK 20 MISSING ANGLES AT A POINT ON A STRAIGHT LINE & OPPOSITE ANGLES 180 360 degree
WEEK 20 MISSING ANGLES AT A POINT ON A STRAIGHT LINE & OPPOSITE ANGLES 180 360 degree
WEEK 21 USING THE SHORT DIVISION METHOD TO CALCULATE DECIMAL REMAINDERS divide bus stop share WEEK 21 USING THE SHORT DIVISION METHOD TO CALCULATE DECIMAL REMAINDERS
WEEK 21 USING THE SHORT DIVISION METHOD TO CALCULATE DECIMAL REMAINDERS divide bus stop share
WEEK 21 USING THE SHORT DIVISION METHOD TO CALCULATE DECIMAL REMAINDERS divide bus stop share
WEEK 22 CALCULATING THROUGH ZERO positive negative value 0 WEEK 22 CALCULATING THROUGH ZERO
WEEK 22 CALCULATING THROUGH ZERO positive negative value 0
WEEK 22 CALCULATING THROUGH ZERO positive negative value 0
WEEK 23 CONVERTING BETWEEN KM AND MILES equal same kilometre length metric WEEK 23 CONVERTING BETWEEN KM AND MILES
WEEK 23 CONVERTING BETWEEN KM AND MILES equal same kilometre length metric
WEEK 23 CONVERTING BETWEEN KM AND MILES equal same kilometre length metric
WEEK 24 DIVIDING FRACTIONS BY A WHOLE NUMBER division share WEEK 24 DIVIDING FRACTIONS BY A WHOLE NUMBER
WEEK 24 DIVIDING FRACTIONS BY A WHOLE NUMBER division share
WEEK 24 DIVIDING FRACTIONS BY A WHOLE NUMBER division share
WEEK 25 RECOGNISE DESCRIBE & DRAW THE NETS OF 3D SHAPES cube cuboid prism cylinder tetrahedron pyramid WEEK 25 RECOGNISE, DESCRIBE AND DRAW THE NETS OF 3D SHAPES
WEEK 25 RECOGNISE DESCRIBE & DRAW THE NETS OF 3D SHAPES cube cuboid prism cylinder tetrahedron pyramid
WEEK 25 RECOGNISE DESCRIBE & DRAW THE NETS OF 3D SHAPES cube cuboid prism cylinder tetrahedron pyramid
WEEK 26 SOLVE REAL-LIFE PROBLEMS USING ALL 4 OPERATIONS solving everyday subtract take away divide division share multiply multiplication addition adding WEEK 26 SOLVE REAL-LIFE PROBLEMS USING ALL 4 OPERATIONS
WEEK 26 SOLVE REAL-LIFE PROBLEMS USING ALL 4 OPERATIONS solving everyday subtract take away divide division share multiply multiplication addition adding
WEEK 26 SOLVE REAL-LIFE PROBLEMS USING ALL 4 OPERATIONS solving everyday subtract take away divide division share multiply multiplication addition adding
WEEK 27 CALCULATING EQUIVALENT FRACTIONS AND SIMPLEST FORM lowest term equal same WEEK 27 CALCULATING EQUIVALENT FRACTIONS AND SIMPLEST FORM
WEEK 27 CALCULATING EQUIVALENT FRACTIONS AND SIMPLEST FORM lowest term equal same
WEEK 27 CALCULATING EQUIVALENT FRACTIONS AND SIMPLEST FORM lowest term equal same
WEEK 27 CALCULATING EQUIVALENT FRACTIONS AND SIMPLEST FORM lowest term equal same
WEEK 28 AREA & PERIMETER OF 2D SHAPES length triangle rectangle square parallelogram WEEK 28 AREA AND PERIMETER OF 2D SHAPES
WEEK 28 AREA & PERIMETER OF 2D SHAPES length triangle rectangle square parallelogram
WEEK 28 AREA & PERIMETER OF 2D SHAPES length triangle rectangle square parallelogram
WEEK 28 AREA & PERIMETER OF 2D SHAPES length triangle rectangle square parallelogram
WEEK 29 DIVIDING A QUANTITY BY A RATIO proportion share division WEEK 29 DIVIDING A QUANTITY BY A RATIO
WEEK 29 DIVIDING A QUANTITY BY A RATIO proportion share division
WEEK 29 DIVIDING A QUANTITY BY A RATIO proportion share division
WEEK 30 DRAWING GRAPHS FROM A TABLE OF DATA chart pie line bar pictogram WEEK 30 DRAWING GRAPHS FROM A TABLE OF DATA
WEEK 30 DRAWING GRAPHS FROM A TABLE OF DATA chart pie line bar pictogram
WEEK 30 DRAWING GRAPHS FROM A TABLE OF DATA chart pie line bar pictogram
WEEK 31 CALCULATE A PERCENTAGE OF A QUANTITY % divide WEEK 31 CALCULATE A PERCENTAGE OF A QUANTITY
WEEK 31 CALCULATE A PERCENTAGE OF A QUANTITY % divide
WEEK 31 CALCULATE A PERCENTAGE OF A QUANTITY % divide
WEEK 32 WRITE AND DESCRIBE NUMBER SEQUENCES pattern term WEEK 32 WRITE AND DESCRIBE NUMBER SEQUENCES
WEEK 32 WRITE AND DESCRIBE NUMBER SEQUENCES pattern term
WEEK 32 WRITE AND DESCRIBE NUMBER SEQUENCES pattern term
WEEK 33 IDENTIFYING THE DIFFERENT PARTS OF A CIRCLE CIRCUMFERENCE DIAMETER RADIUS WEEK 33 IDENTIFYING THE DIFFERENT PARTS OF A CIRCLE
WEEK 33 IDENTIFYING THE DIFFERENT PARTS OF A CIRCLE CIRCUMFERENCE DIAMETER RADIUS
WEEK 33 IDENTIFYING THE DIFFERENT PARTS OF A CIRCLE CIRCUMFERENCE DIAMETER RADIUS
WEEK 34 MULTIPLYING 2 FRACTIONS GIVING THE ANSWER IN ITS SIMPLEST FORM lowest multiplication times table term WEEK 34 MULTIPLYING 2 FRACTIONS GIVING THE ANSWER IN ITS SIMPLEST FORM
WEEK 34 MULTIPLYING 2 FRACTIONS GIVING THE ANSWER IN ITS SIMPLEST FORM lowest multiplication times table term
WEEK 34 MULTIPLYING 2 FRACTIONS GIVING THE ANSWER IN ITS SIMPLEST FORM lowest multiplication times table term
WEEK 35 CALCULATE THE MEAN AVERAGE OF A SET OF NUMBERS WEEK 35 CALCULATE THE MEAN AVERAGE OF A SET OF NUMBERS
WEEK 35 CALCULATE THE MEAN AVERAGE OF A SET OF NUMBERS
WEEK 35 CALCULATE THE MEAN AVERAGE OF A SET OF NUMBERS
WEEK 35 CALCULATE THE MEAN AVERAGE OF A SET OF NUMBERS
WEEK 36 USE POSITIVE AND NEGATIVE VALUES number WEEK 36 USE POSITIVE AND NEGATIVE VALUES
WEEK 36 USE POSITIVE AND NEGATIVE VALUES number
WEEK 36 USE POSITIVE AND NEGATIVE VALUES number
WEEK 37 MULTIPLYING A WHOLE NUMBER AND A DECIMAL VALUE multiplication unit WEEK 37 MULTIPLYING A WHOLE NUMBER AND A DECIMAL VALUE
WEEK 37 MULTIPLYING A WHOLE NUMBER AND A DECIMAL VALUE multiplication unit
WEEK 37 MULTIPLYING A WHOLE NUMBER AND A DECIMAL VALUE multiplication unit
WEEK 38 CALCULATE EQUIVALENT FRACTIONS, DECIMALS AND PERCENTAGES equal % convert WEEK 38 CALCULATE EQUIVALENT FRACTIONS, DECIMALS AND PERCENTAGES
WEEK 38 CALCULATE EQUIVALENT FRACTIONS, DECIMALS AND PERCENTAGES equal % convert
WEEK 38 CALCULATE EQUIVALENT FRACTIONS, DECIMALS AND PERCENTAGES equal % convert
WEEK 39 USING THE ORDER OF OPERATIONS bidmas bodmas add subtract take away divide division share multiply multiplication WEEK 39 USING THE ORDER OF OPERATIONS
WEEK 39 USING THE ORDER OF OPERATIONS bidmas bodmas add subtract take away divide division share multiply multiplication
WEEK 39 USING THE ORDER OF OPERATIONS bidmas bodmas add subtract take away divide division share multiply multiplication
WEEK 40 IDENTIFY THE VALUE OF DIGITS IN DECIMAL NUMBERS place value WEEK 40 IDENTIFY THE VALUE OF DIGITS IN DECIMAL NUMBERS
WEEK 40 IDENTIFY THE VALUE OF DIGITS IN DECIMAL NUMBERS place value
WEEK 40 IDENTIFY THE VALUE OF DIGITS IN DECIMAL NUMBERS place value
WEEK 41 DRAWING PIE CHARTS graph proportion WEEK 41 DRAWING PIE CHARTS
WEEK 41 DRAWING PIE CHARTS graph proportion
WEEK 41 DRAWING PIE CHARTS graph proportion
WEEK 42 SUBSTITUTING A VALUE INTO A FORMULA algebra replace equation expression WEEK 42 SUBSTITUTING A VALUE INTO A FORMULA
WEEK 42 SUBSTITUTING A VALUE INTO A FORMULA algebra replace equation expression
WEEK 42 SUBSTITUTING A VALUE INTO A FORMULA algebra replace equation expression
WEEK 43 TRANSLATING 2D SHAPES moving move polygon transformation WEEK 43 TRANSLATING 2D SHAPES
WEEK 43 TRANSLATING 2D SHAPES moving move polygon transformation
WEEK 43 TRANSLATING 2D SHAPES moving move polygon transformation
WEEK 44 SOLVING PROBLEMS WITH UNITS OF MEASUREMENT solve mm cm metre km kilo milli centi gram kg ml cl litre WEEK 44 SOLVING PROBLEMS WITH UNITS OF MEASUREMENT
WEEK 44 SOLVING PROBLEMS WITH UNITS OF MEASUREMENT solve mm cm metre km kilo milli centi gram kg ml cl litre
WEEK 44 SOLVING PROBLEMS WITH UNITS OF MEASUREMENT solve mm cm metre km kilo milli centi gram kg ml cl litre
WEEK 45 ESTIMATING TO CHECK ANSWERS guess approx estimate WEEK 45 ESTIMATING TO CHECK ANSWERS
WEEK 45 ESTIMATING TO CHECK ANSWERS guess approx estimate
WEEK 45 ESTIMATING TO CHECK ANSWERS guess approx estimate
WEEK 46 COMPARING & ORDERING FRACTIONS greater less ascend descend WEEK 46 COMPARING AND ORDERING FRACTIONS
WEEK 46 COMPARING & ORDERING FRACTIONS greater less ascend descend
WEEK 46 COMPARING & ORDERING FRACTIONS greater less ascend descend
WEEK 46 COMPARING & ORDERING FRACTIONS greater less ascend descend
WEEK 47 CALCULATING THE AREA OF TRIANGLES AND PARALLELOGRAMS space 2d shape WEEK 47 CALCULATING THE AREA OF TRIANGLES AND PARALLELOGRAMS
WEEK 47 CALCULATING THE AREA OF TRIANGLES AND PARALLELOGRAMS space 2d shape
WEEK 47 CALCULATING THE AREA OF TRIANGLES AND PARALLELOGRAMS space 2d shape
WEEK 47 CALCULATING THE AREA OF TRIANGLES AND PARALLELOGRAMS space 2d shape
WEEK 48 TO CLASSIFY AND SORT 2D AND 3D SHAPES venn carrol properties vertex vertices sides WEEK 48 TO CLASSIFY AND SORT 2D AND 3D SHAPES
WEEK 48 TO CLASSIFY AND SORT 2D AND 3D SHAPES venn carrol properties vertex vertices sides
WEEK 48 TO CLASSIFY AND SORT 2D AND 3D SHAPES venn carrol properties vertex vertices sides
WEEK 49 TO COMPARE AND ORDER NUMBERS UP TO 10000000 million ten greater less equal value ascend descend place value WEEK 49 TO COMPARE AND ORDER NUMBERS UP TO 10,000,000
WEEK 49 TO COMPARE AND ORDER NUMBERS UP TO 10000000 million ten greater less equal value ascend descend place value
WEEK 49 TO COMPARE AND ORDER NUMBERS UP TO 10000000 million ten greater less equal value ascend descend place value
WEEK 50 MENTALLY CALCULATE WITH ALL 4 OPERATIONS add addition adding subtract take away multiply multiplication times tables divide division share WEEK 50 MENTALLY CALCULATE WITH ALL 4 OPERATIONS
WEEK 50 MENTALLY CALCULATE WITH ALL 4 OPERATIONS add addition adding subtract take away multiply multiplication times tables divide division share
WEEK 50 MENTALLY CALCULATE WITH ALL 4 OPERATIONS add addition adding subtract take away multiply multiplication times tables divide division share
WEEK 51 SOLVE PROBLEMS WITH RATIO & PROPORTION fraction part divide share WEEK 51 SOLVE PROBLEMS WITH RATIO AND PROPORTION
WEEK 51 SOLVE PROBLEMS WITH RATIO & PROPORTION fraction part divide share
WEEK 51 SOLVE PROBLEMS WITH RATIO & PROPORTION fraction part divide share
WEEK 52 DRAW A LINE GRAPH FROM A TABLE OF DATA chart time WEEK 52 DRAW A LINE GRAPH FROM A TABLE OF DATA
WEEK 52 DRAW A LINE GRAPH FROM A TABLE OF DATA chart time
WEEK 52 DRAW A LINE GRAPH FROM A TABLE OF DATA chart time
Welcome to the Year 6 Mini-Maths homepage. Daily maths questions for Year 6 are available for every week of the year, providing free help for teachers, pupils and parents. 1000s of free tasks and activities are available to support children in their learning both in the classroom and at home. Use the search and filter box below to easily find the outcomes you require.
It is recommended that a topic is selected per week and daily maths tasks accessed using the links provided below. Use each day’s answers to help identify and guide the required support in preparation for the following day’s task. Watch your understanding, application and confidence grow throughout the week. New resources are always being added, so keep checking for updates.
## More about Year 6 Maths Questions …
The Year 6 curriculum completes the Key Stage 2 maths programme of learning bringing understanding and confidence together from Year 3 through to the end of Year 6. This journey culminates in the KS2 SATs assessments where learners are able to demonstrate their progress and development in the subject.
## Year 6 NUMBER …
The four operations reach a natural conclusion in Year 6 – addition, subtraction, multiplication and division involve more complex examples before introducing decimals values . Learners are looking to extend their understanding of units, tens, hundreds, thousands, etc, with the Place value of digits in decimal places, supporting the wider curriculum such as calculating decimal remainders using short division.
## Year 6 SHAPE …
Learners have been developing a sound understanding of 2D shape and 3D shape allowing them to sort and classify by their properties. Being able to identify and use these properties also improves an awareness of a shape’s area and perimeter or volume .
## Year 6 DATA …
Year 5 offers further coverage of tables, charts and graphs, where learners are expected to both read and interpret information in all types of graphs and charts including pie charts . Drawing graphs and charts from given data is also introduced in Year 6 maths.
All learning objectives are supported by a Home Learning pack and iQ mastery questions challenge learners’ understanding, application and confidence. The Home Learning packs provide first class opportunities for effective intervention and an ever growing library of help videos for Year 6 maths is available on the MyMiniMaths YouTube channel in support of quality learning.
## Year 6 SATs …
MyMiniMaths also offers specialised support for Year 6 SATs with a programme of practice KS2 SATs papers for the arithmetic paper . Easy identification of key skills in need of further support to maximise attainment is available through the Arithmetic Target Questions .
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## Daily Math Word Problems - Grade 6 (Worksheets)
Updated: 17 Apr 2024
A set of 20 problem-solving questions suited to grade 6 students.
Non-Editable: PDF
Pages: 6 Pages
This set of problem-solving questions has been designed to support teachers when teaching students about problem-solving in mathematics.
It provides students with the opportunity to work through 20 math word problems , identifying the important information and how they can work it out using a variety of methods.
An answer sheet has been included.
Use this resource in conjunction with the Daily Math Problems PowerPoint :
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## Percy's Place Value Puzzle
A worksheet to use to practice place value to the thousands place.
## Identifying and Naming Angles – 4th Grade Math Worksheet
Identify acute, right, obtuse, straight, reflex and revolution angles with this cut-and-paste sorting worksheet.
## Acute, Right and Obtuse Angles – 4th Grade Math Worksheet
Identify acute, right and obtuse angles with this cut-and-paste sorting worksheet.
## Location Math Investigation - Blackbeard's Bounty
A mathematics investigation about location, embedded in a real-world context.
## Multiplying and Dividing By 1,000 Poster
A set of posters to display in the classroom when learning to multiply and divide by 1,000.
## Multiplying by 1,000 Worksheet
A worksheet for students to complete when learning how to multiply by 1,000.
## Dividing by 1,000 Worksheet
A worksheet for students to complete when learning to divide by 1,000.
## 2-D Shape Bingo
Engage your students while consolidate learning about 2D shapes, their names and properties with 2D Shape Bingo!
## Multiplication Facts PowerPoint - One Times Tables
A 28-slide PowerPoint to use when learning about multiplication.
## Popular searches in the last week:
Problem-solving maths investigations for year 6.
Hamilton provide an extensive suite of problem-solving maths investigations for Year 6 to facilitate mathematical confidence, investigative inquiry and the development of maths meta skills in 'low floor – high ceiling' activities for all.
Explore all our in-depth problem solving investigations for Year 6 .
Use problem-solving investigations within every unit to encourage children to develop and exercise their ability to reason mathematically and think creatively.
Investigations provide challenges that offer opportunities for the development of the key mathematical skills while deepening conceptual understanding. They are designed to be accessible in different ways to all children. An added bonus is the substantial amount of extra calculation practice they often incorporate! The problems are designed to help children identify patterns, to explore lines of thinking and to reason and communicate about properties of numbers, shapes and measures.
Hamilton provide a mix of our own specially commissioned investigations, that include guidance for teachers together with a child-friendly sheet to guide your pupils through the investigation, as well as links to investigations on other highly regarded websites.
I am very grateful for Hamilton Trust resources, particularly the maths investigations. Julia, teacher in Wiltshire
## You can find Hamilton's investigations for Year 6:
• Individually, they are incorporated into every unit in our Year 6 flexible maths blocks .
• Collectively, they appear on our resources page where you can explore all our in-depth problem solving investigations for Year 6 .
Hamilton’s problem-solving investigations are 'low floor, high ceiling' activities that give all children opportunities to develop mastery and mathematical meta-skills. Explore a set for a whole year group.
Hamilton’s Problem-solving Investigations provide school-wide solutions to the challenges of building investigative skills from Early Years to Year 6.
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## Midpoint of Two Fractions Practice Questions
Mode from a table practice questions, multipliers practice questions, range from a frequency table practice questions, cost per litre video.
## Capture Recapture Practice Questions
Gcse revision cards.
## Primary Study Cards
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## Solving Problems (Year 6)
When you answer 8 or more questions correctly your red streak will increase in length. The green streak shows the best player so far today. See our Hall of Fame for previous daily winners.
Learning KS2 Maths is like solving puzzles from everyday life. In Year Six, you'll tackle big numbers and explore measures like millilitres, litres, grams, kilograms, millimetres, kilometres, minutes, and hours.
Solving problems means using multiplication, division, addition, or subtraction - maybe even all of them! Imagine figuring out how much something costs in pounds and pence or its weight in grams and kilograms. Some puzzles are unique, like this one: If you need to take a tablet every day for 3 months, how many tablets do you need all together?
Test your new skills with this quiz on solving fun problems with different measures!
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## Problem Solving Questions - Worksheets - Year 6
Subject: Maths for early years
Age range: 7-11
Resource type: Worksheet/Activity
Last updated
1 February 2020
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## Reasoning/Problem Solving Maths Worksheets for Year 6 (age 10-11)
Problem solving.
A variety of problem solving activities including time and money.
A perimeter/area investigation.
A good knowledge of the order of operations is needed for this problem solving task.
Find the cost of different numbers of houseplants.
Tricky addition which may need jottings to complete.
Here are sets of word problems only needing one step to solve, but it is not always obvious what calculation to carry out to reach the answer.
Here are some longer problems written in words. You need to work out more than one thing to get the answers.
Quite tricky word problems involving money, often requiring several steps to complete.
A good knowledge of metric measures is needed to complete these questions.
Questions which need an understanding of a.m., p.m., 24 hour clock, leap years as well as using graphs.
An atlas with time zones is needed for the first activity and there are line graphs on time to interpret.
When answering these problems say which operation is needed and whether it can be done mentally, on paper or with a calculator.
More on choosing the most appropriate method of calculating.
Tricky calculations working out the best and worst meal deals.
Purchasing various items from a vending machine. Tricky calculating with money.
## Investigate Numbers and Patterns
Investigate numbers and patterns, including sequences.
Quick and easy way to multiply 2-digit numbers by 11 - why does it work?
Show all your calculating skills with just four fours.
A first look at triangle numbers.
More on triangle numbers.
Further investigations involving triangle numbers.
Some great patterns can be found when dividing by 9 or 11, and using square numbers.
Plenty of ideas to help with investigations and inventing similar problems.
More investigations on adding consecutive numbers, dividing by a half, writing a simple formula as well as shape patterns.
Investigate numbers and shapes; quite tricky.
Complete a variety of number sequences and say what the rule for each sequence is. Some negative number work as well.
Using a multiplication square to look for number patterns: there are certainly plenty of them! Further sequence work, including Fibonacci.
Try the Ancient Egyptian method of multiplying. Great, if you are good at adding up!
Tips and hints on how to use a calculator to ensure correct answers.
## Holiday Problem Solving
A tricky selection of resources all linked to a holiday theme.
Very tricky to find the best options.
Very tricky calculations to find the best deal.
Finding the best deal is harder than it might seem!
A holiday which might cost more than it should unless there is some careful calculating.
This is complicated with special offers and percentages off making finding the cheapest deal difficult to work out.
More difficult than you might think to find the cheapest deal, including percentages.
Much trickier than it first looks, with many possible options to choose from. Some time is needed to complete this correctly.
A collection of puzzles to solve - some easy and some quite tricky. Ideal for the end of term.
Place the numbers 1 to 9 in the boxes so that each side of the triangle adds up to 20.
Replace the hexagons in the calculation with numbers: logical thinking needed.
Find the mathematical words in these anagrams.
A calculator is needed for this fun challenge.
This might not seem possible but with a little imagination it can be done.
How many moves to get all the coins on heads?
Nifty little puzzle with more than one solution.
More to do with square numbers than shapes!
Finding pairs of numbers that add up to square numbers. Tricky!
How many ways of making 15. Have you got all of them?
Can you make prime numbers by adding two square numbers?
Lots of coins and not change for £1.
Neat little code-breaker. Can you come up with others?
Logical thinking needed here.
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## Number and algebra
• The Number System and Place Value
• Calculations and Numerical Methods
• Fractions, Decimals, Percentages, Ratio and Proportion
• Properties of Numbers
• Patterns, Sequences and Structure
• Algebraic expressions, equations and formulae
• Coordinates, Functions and Graphs
## Geometry and measure
• Angles, Polygons, and Geometrical Proof
• 3D Geometry, Shape and Space
• Measuring and calculating with units
• Transformations and constructions
• Pythagoras and Trigonometry
• Vectors and Matrices
## Probability and statistics
• Handling, Processing and Representing Data
• Probability
## Working mathematically
• Thinking mathematically
• Mathematical mindsets
• Cross-curricular contexts
• Physical and digital manipulatives
## For younger learners
• Early Years Foundation Stage
• Decision Mathematics and Combinatorics
## Resources tagged with: NC Yr 6
There are 51 NRICH Mathematical resources connected to NC Yr 6 , you may find related items under NC .
## 4 by 4 Mathdokus
Can you use the clues to complete these 4 by 4 Mathematical Sudokus?
## Different Deductions
There are lots of different methods to find out what the shapes are worth - how many can you find?
## Name That Triangle!
Can you sketch triangles that fit in the cells in this grid? Which ones are impossible? How do you know?
## Number Lines in Disguise
Some of the numbers have fallen off Becky's number line. Can you figure out what they were?
## Price Match
Can you find pairs of differently sized windows that cost the same?
## Finding 3D Stacks
Can you find a way of counting the spheres in these arrangements?
## Extending Fraction Bars
Can you compare these bars with each other and express their lengths as fractions of the black bar?
## More Fraction Bars
What fraction of the black bar are the other bars? Have a go at this challenging task!
## Always, Sometimes or Never? Shape
Are these statements always true, sometimes true or never true?
## Round the Three Dice
What happens when you round these three-digit numbers to the nearest 100?
## Round the Four Dice
This activity involves rounding four-digit numbers to the nearest thousand.
Use the information on these cards to draw the shape that is being described.
## Dicey Perimeter, Dicey Area
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
## Domino Sets
How do you know if your set of dominoes is complete?
## Making Spirals
Can you make a spiral for yourself? Explore some different ways to create your own spiral pattern and explore differences between different spirals.
## Round a Hexagon
This problem shows that the external angles of an irregular hexagon add to a circle.
Have a look at this data from the RSPB 2011 Birdwatch. What can you say about the data?
## So It's Times!
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
After training hard, these two children have improved their results. Can you work out the length or height of their first jumps?
## Button-up Some More
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
## Counting Cogs
Which pairs of cogs let the coloured tooth touch every tooth on the other cog? Which pairs do not let this happen? Why?
## Doughnut Percents
A task involving the equivalence between fractions, percentages and decimals which depends on members of the group noticing the needs of others and responding.
## Next Size Up
The challenge for you is to make a string of six (or more!) graded cubes.
I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. Can you see any patterns?
What can you see? What do you notice? What questions can you ask?
## Treasure Hunt
Can you find a reliable strategy for choosing coordinates that will locate the treasure in the minimum number of guesses?
## First Connect Three
Add or subtract the two numbers on the spinners and try to complete a row of three. Are there some numbers that are good to aim for?
This challenge is a game for two players. Choose two of the numbers to multiply or divide, then mark your answer on the number line. Can you get four in a row?
## Factor-multiple Chains
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
## Fraction Fascination
This problem challenges you to work out what fraction of the whole area of these pictures is taken up by various shapes.
## Triangles All Around
Can you find all the different triangles on these peg boards, and find their angles?
## Diagonal Sums
In this 100 square, look at the green square which contains the numbers 2, 3, 12 and 13. What is the sum of the numbers that are diagonally opposite each other? What do you notice?
## Ten Hidden Squares
These points all mark the vertices (corners) of ten hidden squares. Can you find the 10 hidden squares?
## Orange Drink
A 750 ml bottle of concentrated orange squash is enough to make fifteen 250 ml glasses of diluted orange drink. How much water is needed to make 10 litres of this drink?
Each of the nets of nine solid shapes has been cut into two pieces. Can you see which pieces go together?
## Sponge Sections
You have been given three shapes made out of sponge: a sphere, a cylinder and a cone. Your challenge is to find out how to cut them to make different shapes for printing.
## Factor Lines
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
## Would You Rather?
Would you rather: Have 10% of £5 or 75% of 80p? Be given 60% of 2 pizzas or 26% of 5 pizzas?
## Plenty of Pens
Amy's mum had given her £2.50 to spend. She bought four times as many pens as pencils and was given 40p change. How many of each did she buy?
## Mystery Matrix
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
## The Moons of Vuvv
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
## Where Are They?
Use the isometric grid paper to find the different polygons.
## Rectangle Tangle
The large rectangle is divided into a series of smaller quadrilaterals and triangles. Can you untangle what fractional part is represented by each of the shapes?
## Pumpkin Pie Problem
Peter wanted to make two pies for a party. His mother had a recipe for him to use. However, she always made 80 pies at a time. Did Peter have enough ingredients to make two pumpkin pies?
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
## Two and Two
How many solutions can you find to this sum? Each of the different letters stands for a different number.
## Making Cuboids
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
## Round and Round the Circle
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
## Doplication
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
## Year 6 maths
IXL offers hundreds of year 6 maths skills to explore and learn! Not sure where to start? Go to your personalized Recommendations wall to find a skill that looks interesting, or select a skill plan that aligns to your textbook, territory curriculum, or standardized test.
## A. Whole numbers
• 1 Place values in whole numbers
• 2 Word names for numbers
• 3 Roman numerals
## B. Multiplication
• 1 Multiply using area models I
• 2 Multiply using area models II
• 3 Multiply using the distributive property
• 4 Lattice multiplication
• 5 Multiply whole numbers
• 6 Multiply whole numbers: word problems
• 7 Multiply numbers ending in zeroes
• 8 Multiply numbers ending in zeroes: word problems
• 9 Multiply three or more numbers
• 10 Multiply three or more numbers: word problems
• 11 Estimate products
• 12 Properties of multiplication
• 13 Solve using properties of multiplication
## C. Division
• 1 Divisibility rules
• 2 Divide by two-digit numbers
• 3 Divide by two-digit numbers: word problems
• 4 Division patterns with zeroes
• 5 Divide numbers ending in zeroes: word problems
• 6 Estimate quotients using compatible numbers
• 7 Estimate quotients
## D. Exponents
• 1 Write multiplication expressions using exponents
• 2 Evaluate powers
• 3 Find the missing exponent or base
## E. Number theory
• 1 Prime or composite
• 2 Identify factors
• 3 Prime factorisation
• 4 Prime factorisation with exponents
• 5 Highest common factor
• 6 Lowest common multiple
• 7 HCF and LCM: word problems
• 8 Square numbers
## F. Decimals
• 1 What decimal number is illustrated?
• 2 Decimal place values
• 3 Word names for decimal numbers
• 4 Put decimal numbers in order
• 5 Inequalities with decimals
• 6 Round decimals
• 7 Round whole numbers and decimals: find the missing digit
• 8 Decimal number lines
• 1 Add and subtract whole numbers up to millions
• 2 Add and subtract whole numbers: word problems
• 4 Estimate sums and differences of whole numbers
• 5 Estimate sums and differences: word problems
• 7 Subtract decimal numbers
• 8 Add and subtract decimal numbers
• 9 Add and subtract decimals: word problems
• 10 Estimate sums and differences of decimals using rounding
• 11 Estimate sums and differences of decimals using benchmarks
• 12 Maps with decimal distances
## H. Multiply and divide decimals
• 1 Multiply a decimal by a power of ten
• 2 Multiply a decimal by a one-digit number using the distributive property
• 3 Multiply a decimal by a one-digit whole number
• 4 Multiply a decimal by a two-digit number using area models
• 5 Multiply a decimal by a multi-digit whole number
• 6 Multiply decimals and whole numbers: word problems
• 7 Multiply two decimals
• 8 Estimate products of decimal numbers
• 9 Inequalities with decimal multiplication
• 10 Divide decimals by powers of ten
• 11 Decimal division patterns over increasing place values
• 12 Divide a decimal by a power of ten: find the missing number
• 13 Divide decimals using area models: complete the equation
• 14 Divide decimals by whole numbers
• 15 Divide decimals by whole numbers: word problems
## I. Fractions and mixed numbers
• 1 Fractions and mixed numbers review
• 2 Understanding fractions: word problems
• 3 Find equivalent fractions using area models
• 4 Graph equivalent fractions on number lines
• 5 Equivalent fractions
• 6 Write fractions in lowest terms
• 7 Fractions: word problems
• 8 Lowest common denominator
• 9 Graph and compare fractions on number lines
• 10 Compare fractions with like and unlike denominators
• 11 Compare fractions: word problems
• 12 Convert between improper fractions and mixed numbers
• 13 Convert fractions or mixed numbers to decimals
• 14 Convert decimals to fractions or mixed numbers
• 15 Convert between decimals and fractions or mixed numbers
• 16 Put a mix of decimals, fractions and mixed numbers in order
## J. Add and subtract fractions
• 1 Add and subtract fractions with like denominators using number lines
• 2 Add and subtract fractions with like denominators
• 3 Add and subtract fractions with like denominators: word problems
• 4 Inequalities with addition and subtraction of like fractions
• 5 Add fractions with unlike denominators using models
• 6 Add fractions with unlike denominators
• 7 Subtract fractions with unlike denominators using models
• 8 Subtract fractions with unlike denominators
• 9 Add and subtract fractions with unlike denominators: word problems
• 10 Estimate sums and differences of fractions using benchmarks
• 11 Add and subtract mixed numbers with like denominators
• 12 Add and subtract mixed numbers with unlike denominators
• 13 Add and subtract mixed numbers with like or unlike denominators: word problems
• 14 Estimate sums and differences of mixed numbers
## K. Multiply fractions
• 1 Fractions of a number
• 2 Fractions of a number: word problems
• 3 Multiply unit fractions by whole numbers using number lines
• 4 Multiples of fractions
• 5 Multiply fractions by whole numbers using number lines
• 6 Multiply fractions by whole numbers I
• 7 Multiply fractions by whole numbers II
• 8 Estimate products of fractions and whole numbers
• 9 Multiply fractions by whole numbers: input/output tables
• 10 Multiply mixed numbers and whole numbers
## L. Integers
• 1 Understanding integers
• 2 Integers on number lines
• 3 Graph integers on horizontal and vertical number lines
• 4 Compare and order integers using number lines
• 5 Compare integers
• 6 Put integers in order
• 7 Add integers using number lines
• 8 Subtract integers using number lines
## M. Mixed operations
• 1 Add, subtract, multiply or divide two whole numbers
• 2 Add, subtract, multiply or divide two whole numbers: word problems
• 3 Evaluate numerical expressions
• 4 Evaluate numerical expressions with brackets
• 5 Identify mistakes involving the order of operations
• 6 Evaluate numerical expressions with brackets in different places
• 7 Add, subtract, multiply or divide two decimals
• 8 Add, subtract, multiply or divide two decimals: word problems
• 9 Evaluate numerical expressions involving decimals
• 10 Add, subtract or multiply two fractions
• 11 Add, subtract or multiply two fractions: word problems
## N. Problem solving and estimation
• 1 Estimate to solve word problems
• 2 Multi-step word problems
• 3 Word problems with extra or missing information
• 4 Guess-and-check word problems
• 5 Distance/direction to starting point
• 6 Use logical reasoning to find the order
## O. Ratios and rates
• 1 Write a ratio
• 2 Write a ratio: word problems
• 3 Identify equivalent ratios
• 4 Write an equivalent ratio
• 5 Unit rates and equivalent rates
## P. Percents
• 1 What percentage is illustrated?
• 2 Convert fractions to percents using grid models
• 3 Convert between percents, fractions and decimals
• 4 Convert between percents, fractions and decimals: word problems
• 5 Compare percents to each other and to fractions
• 6 Compare percents and fractions: word problems
• 7 Estimate percents of numbers
• 8 Percents of numbers and money amounts
• 9 Percents of numbers: word problems
• 10 Find what percent one number is of another
• 11 Find what percent one number is of another: word problems
## Q. Units of measurement
• 1 Estimate metric measurements
• 2 Convert and compare metric units of length
• 3 Convert and compare metric units of mass
• 4 Convert and compare metric units of volume
• 5 Convert and compare metric units
• 6 Convert metric units involving decimals
• 7 Conversion tables
• 8 Metric mixed units
• 9 Convert between cubic centimetres, millilitres and litres
• 10 Convert between square metres and hectares
• 11 Compare temperatures above and below zero
• 1 Find the number of each type of coin
• 2 Add and subtract money amounts
• 3 Add and subtract money amounts: word problems
• 4 Multiply money by whole numbers
• 5 Multiply money: word problems
• 6 Divide money amounts
• 7 Divide money amounts: word problems
## S. Consumer maths
• 1 Which is the better coupon?
• 2 Unit prices
• 3 Unit prices with fractions and decimals
• 4 Sale prices
• 1 Elapsed time
• 2 Elapsed time: word problems
• 3 Find start and end times
• 4 Time units
• 5 Convert between 12-hour and 24-hour time
• 7 Transportation schedules
## U. Coordinate plane
• 1 Objects on a coordinate plane - first quadrant only
• 2 Objects on a coordinate plane - all four quadrants
• 3 Graph points on a coordinate plane
• 5 Coordinate planes as maps
• 6 Follow directions on a coordinate plane
## V. Number sequences
• 1 Arithmetic sequences with whole numbers
• 2 Arithmetic sequences with decimals
• 3 Arithmetic sequences with fractions
• 4 Complete an increasing number sequence
• 5 Complete a geometric number sequence
• 6 Use a rule to complete a number sequence
• 7 Identify mistakes in number patterns
• 8 Number sequences: word problems
• 9 Number sequences: mixed review
• 10 Shape patterns
## W. Variable expressions
• 1 Write variable expressions
• 2 Write variable expressions: word problems
• 3 Evaluate variable expressions with whole numbers
• 4 Identify terms and coefficients
## X. One-variable equations
• 1 Does x satisfy an equation?
• 2 Which x satisfies an equation?
• 3 Write an equation from words
• 4 Model and solve equations using algebra tiles
• 5 Write and solve equations that represent diagrams
• 6 Solve one-step equations with whole numbers
• 7 Solve one-step equations: word problems
## Y. Two-dimensional figures
• 1 Identify and classify polygons
• 2 Measure and classify angles
• 3 Find a missing angle - adjacent angles
• 4 Estimate angle measurements
• 5 Name angles
• 6 Identify complementary, supplementary, vertical, adjacent and congruent angles
• 7 Find measures of complementary, supplementary, vertical and adjacent angles
• 8 Transversal of parallel lines
## Z. Symmetry and transformations
• 2 Reflection, rotation and translation
• 3 Translations: graph the image
• 4 Reflections: graph the image
• 5 Rotations: graph the image
• 6 Sequences of transformations: graph the image
## AA. Three-dimensional figures
• 1 Identify polyhedra
• 2 Which figure is being described?
• 3 Nets of three-dimensional figures
• 4 Front, side and top view
• 5 Cross sections of three-dimensional figures
## BB. Geometric measurement
• 1 Perimeter
• 2 Area of squares and rectangles
• 3 Area and perimeter of figures on grids
• 4 Area and perimeter: word problems
• 5 Rectangles: relationship between perimeter and area
• 6 Compare area and perimeter of two figures
## CC. Data and graphs
• 1 Interpret picture graphs
• 2 Create picture graphs
• 3 Interpret dot plots
• 4 Create dot plots
• 5 Interpret stem-and-leaf plots
• 6 Create stem-and-leaf plots
• 7 Create frequency tables
• 8 Interpret bar graphs
• 9 Create bar graphs
• 10 Interpret double bar graphs
• 11 Create double bar graphs
• 12 Circle graphs with fractions
• 13 Interpret line graphs
• 14 Create line graphs
• 15 Interpret double line graphs
• 16 Create double line graphs
• 17 Choose the best type of graph
## DD. Statistics
• 1 Calculate the mean
• 2 Calculate the median
• 3 Calculate the mode
• 4 Calculate the range
• 5 Interpret charts to find the mean
• 6 Interpret charts to find the median
• 7 Interpret charts to find the mode
• 8 Interpret charts to find the range
• 9 Mean, median, mode and range: find the missing number
## EE. Probability
• 1 Combinations
• 2 Probability of one event
• 3 Make predictions
• 4 Experimental probability
• 5 Probability of opposite, mutually exclusive and overlapping events
• 6 Identify independent and dependent events
Ninja Maths
Weekly Maths Worksheets
## Year 6 Free Maths Worksheets
They’re not very difficult and are aimed at instilling the ethos of repeated practice.
## Summary of Year 6 Maths Content
New & updated free sample worksheets coming soon, based on the UK curriculum;
In the mean time you can subscribe here for free worksheets as they’re written:
International Equivalence
## English Year 4 for school ages 8-9 equates to :
USA Grade 3 Australia Year 3 Republic of Ireland Third Class South Korea Grade 3 India Grade 3 Japan Grade 3 China Grade 3 Germany Grade 3 New Zealand Year 4 Wales Year 4 Netherlands Group 5 Scotland P5
## English Year 5 for school ages 9-10 equates to:
USA Grade 4 Australia Year 4 Republic of Ireland Fourth Class South Korea Grade 4 India Grade 4 Japan Grade 4 China Grade 4 Germany Grade 4 New Zealand Year 5 Wales Year 5 Netherlands Group 6 Scotland P6
## English Year 6 for school ages 10-11 equates to:
USA Grade 5 Australia Year 5 Republic of Ireland Fifth Class South Korea Grade 5 India Grade 5 Japan Grade 5 China Grade 5 Germany Grade 5 New Zealand Year 6 Wales Year 6 Netherlands Group 7 Scotland P7
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## November Year 4 - Sample Pages
November Year 4 Work Packs - Sample Pages
## November Year 5 - Sample Pages
November Year 5 Work Packs - Sample Pages
## November Year 6 - Sample Pages
November Year 6 Work Packs - Sample Pages
1. 35 Year 6 Maths Questions
That's right, there will be a total of 35 year 6 maths questions for us to practise. Without realising it, you just solved one of our first year 6 word problems. Excellent work! Together we will review year 6 maths problems with decimals, fractions, order of operations, and rounding. We will even consider hard maths problems, like multiplying ...
2. Year 6 Maths Worksheets
Year 6 Number Worksheets. Here you will find a range of Free Printable Year 6 Number Worksheets. Using these Year 6 maths worksheets will help your child to: use place value with numbers up to 10 million; use place value with up to 3 decimal places; understand how to use exponents (powers) of a number; understand and use parentheses (brackets);
3. 20 Word Problems For Year 6: Develop Problem Solving Skills
Key to this success is regular exposure to the style and type of questions, covering the topics children could be presented with. Our collection of year 6 maths worksheets are is a great place to start. To help child practice their problem solving skills, we have put together a collection of 20 word problems, organised by topic.
4. 35 SATs Maths Questions Year 6 SATs Reasoning Practice
35 SATs maths questions for KS2 year 6 SATs. For the KS2 SATs tests, there are 7 types of maths reasoning question that are likely to come up: ... For advice on how to teach children to solve problems like this, check out these maths problem solving strategies. SATs Maths Question Type 1: Single step worded problems ...
Problems. Two numbers N and 16 have LCM = 48 and GCF = 8. Find N. If the area of a circle is 81pi square feet, find its circumference. Find the greatest common factor of 24, 40 and 60. In a given school, there are 240 boys and 260 girls.
Year 6 Fluency, Reasoning and Problem Solving Skills Practice Year 6 Maths Worksheets for improving mental maths, arithmetic and fluency: Fluent in Five. We recommend every primary school child starts their day with Fluent in Five. It's 5-10 minutes of daily arithmetic questions; in Year 6 children have to answer up to 7 quick mental ...
7. Year 6 maths questions with 1000s of free resources at MyMiniMaths
Welcome to the Year 6 Mini-Maths homepage. Daily maths questions for Year 6 are available for every week of the year, providing free help for teachers, pupils and parents. 1000s of free tasks and activities are available to support children in their learning both in the classroom and at home. Use the search and filter box below to easily find ...
8. Daily Maths Word Problems
A set of 20 problem-solving questions suited to year 6 students. This set of problem-solving questions has been designed to support teachers when teaching students about problem-solving in mathematics.. It provides students with the opportunity to work through 20 maths word problems, identifying the important information and how they can work it out using a variety of methods.
9. Problem solving
Problem solving. Find out how the order of operations shows you which bit of a calculation to do first. Year 6 KS2 Maths Problem solving learning resources for adults, children, parents and teachers.
10. Open-Ended Year 6 Maths Investigations (teacher made)
These open-ended Year 6 maths investigations are a really fun and creative way to get children learning and engaging with different problems from a mathematical perspective. These activities will not only expand children's maths knowledge, but they will also help them improve their problem-solving skills and logical thinking.
11. Year 6 Maths Problems
Different activity sheets that give children questions that look at their fluency, reasoning or problem solving skills for that area of Maths with the answers included. Example questions from the sample reasoning papers and 2016 reasoning papers are also being included for mini practise. More topics will continue to be included.
12. Daily Math Word Problems
A set of 20 problem-solving questions suited to grade 6 students. This set of problem-solving questions has been designed to support teachers when teaching students about problem-solving in mathematics. It provides students with the opportunity to work through 20 math word problems, identifying the important information and how they can work it ...
13. Maths Mastery
Year 6 Maths Mastery Challenge Cards on Bumper Pack. 4.6 (5 reviews) Year 6 Diving into Mastery: Step 1 Add and Subtract Integers Teaching Pack. 4.4 (7 reviews) Year 6 Diving into Mastery: Step 16 Mental Calculations and Estimation Teaching Pack. PlanIt Maths Year 6 Algebra Lesson Pack 10: Solving Equations.
14. Problem-solving Maths Investigations for Year 6
By Nick Barwick - 7 Aug 2018. Hamilton provide an extensive suite of problem-solving maths investigations for Year 6 to facilitate mathematical confidence, investigative inquiry and the development of maths meta skills in 'low floor - high ceiling' activities for all. Explore all our in-depth problem solving investigations for Year 6.
15. Practice Questions
The Corbettmaths Practice Questions - a collection of exam style questions for a wide range of topics. Perfect to use for revision, as homework or to target particular topics. Answers and video solutions are available for each.
16. KS2 Maths Quiz: Mastering Everyday Problem Solving
Solving Problems (Year 6) Learning KS2 Maths is like solving puzzles from everyday life. In Year Six, you'll tackle big numbers and explore measures like millilitres, litres, grams, kilograms, millimetres, kilometres, minutes, and hours. Solving problems means using multiplication, division, addition, or subtraction - maybe even all of them!
17. Problem Solving Questions
Problem Solving Questions - Worksheets - Year 6. Subject: Maths for early years. Age range: 7-11. Resource type: Worksheet/Activity. File previews. pdf, 86.85 KB. pdf, 111.01 KB. pdf, 111.06 KB. Mixture of maths problem solving questions worksheets for year 6 students.
18. Maths Mastery
Examples of Year 6 Maths Problem-Solving Resources: The following are examples of resources on this page that you can use to help plan your year 6 maths lessons: Year 6 Diving into Mastery: Draw Nets of 3D Shapes Teaching Pack - help your children to consolidate and deepen their understanding of nets in preparation for year 6 SATs.
19. Problem Solving
Developing Excellence in Problem Solving with Young Learners. Age 5 to 11. Becoming confident and competent as a problem solver is a complex process that requires a range of skills and experience. In this article, Jennie suggests that we can support this process in three principal ways. Using NRICH Tasks to Develop Key Problem-solving Skills.
20. 100 Reasoning and Problem Solving Questions for Year 6
Speak to us about resources or 1-to-1 maths for your school 020 3771 0095. Login. Register for free ... SATs, Home Learning. 100 Reasoning and Problem Solving Questions for Year 6. 100 Reasoning and Problem-Solving SATs-style questions for Year 6 complete with an answers. 25 Addition and Subtraction Questions; 25 Fractions, Decimals and ...
21. Reasoning/Problem Solving Maths Worksheets for Year 6 (age 10-11)
Puzzle: 12 To 15. Logical thinking needed here. Plenty of problem solving to do in Year 6, together with explaining why particular methods have been chosen. Recognising patterns in number and predicting sequences are both important steps towards using algebra at High School.
22. PDF Year 6 Division to Solve Problems Reasoning and Problem Solving
Mathematics Year 6: (6F9c) Use written division methods in cases where the answer has up to two decimal places Differentiation: Questions 1, 4 and 7 (Problem Solving) Developing Explain whether a given calculation would solve the given problem where there may be up to one exchange.
23. NRICH topics: NC NC Yr 6
We have found 51 NRICH Mathematical resources connected to NC Yr 6, you may find related items under NC
24. 100 Reasoning and Problem Solving Questions for Year 6
Help your Year 6 pupils feel confident with reasoning and problem solving questions with these 100 SATs style arithmetic questions. These questions are grouped by topic to make them easier to use. The resource includes an answer scheme.
25. IXL
16. Put a mix of decimals, fractions and mixed numbers in order. 1. Add and subtract fractions with like denominators using number lines. 2. Add and subtract fractions with like denominators. 3. Add and subtract fractions with like denominators: word problems.
26. Year 6 Free Maths Worksheets
Summary of Year 6 Maths Content. New & updated free sample worksheets coming soon, based on the UK curriculum; Number - number and place value. Number - addition, subtraction, multiplication and division. Number - fractions (including decimals and percentages) Ratio and proportion. Algebra.
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# What is the subtraction property of equality in geometry?
Equations are these math expressions with an equals signal. The 2 sides should equal one another. The subtraction property of equality tells us that if we subtract from one aspect of an equation, we additionally should subtract from the different aspect of the equation to maintain the equation the similar.
Click on to see full reply
Then, what is the subtraction property in geometry?
The subtraction property of equality states you could subtract the similar amount from each side of an equation and it’ll nonetheless stability. Symmetric Property of Congruence.
Secondly, what is the transitive property of equality? web site suggestions. Transitive Property of Equality. The next property: If a = b and b = c, then a = c. One of the equivalence properties of equality. Be aware: This is a property of equality and inequalities.
Beside this, what is the property of equality in math?
Instance: The operations of addition, subtraction, multiplication and division don’t change the reality worth of any equation. The division property of equality states that once we divide each side of an equation by the similar non-zero quantity, the two sides stay equal.
What are the properties in geometry?
Geometry Properties and Proofs
AB
Symmetric PropertyIf AB + BC = AC then AC = AB + BC
Transitive PropertyIf AB ≅ BC and BC ≅ CD then AB ≅ CD
Section Addition PostulateIf C is between B and D, then BC + CD = BD
Angle Addition PostulateIf D is a degree in the inside of ∢ABC then m∢ABD + m∢DBC = m∢ABC
### What is the subtraction theorem?
Section subtraction (4 whole segments): If two congruent segments are subtracted from two different congruent segments, then the variations are congruent. Angle subtraction (4 whole angles): If two congruent angles are subtracted from two different congruent angles, then the variations are congruent.
### What is the addition property of equality in geometry?
Addition Property of Equality. The property that states that in case you add the similar quantity to each side of an equation, the sides stay equal (i.e., the equation continues to be true.)
### What is the property of congruence?
The reflexive property of congruence states that any geometric determine is congruent to itself. Congruence means the determine has the similar dimension and form.
### What is symmetric property?
The symmetric property of equality tells us that each side of an equal signal are equal regardless of which aspect of the equal signal they’re on. Bear in mind it states that if x = y, then y = x.
### What is the inequality?
An inequality compares two values, exhibiting if one is lower than, larger than, or just not equal to a different worth. a ≠ b says {that a} is not equal to b. a < b says {that a} is lower than b. a > b says {that a} is larger than b. (these two are often called strict inequality)
### What are the 5 properties of equality?
• The Reflexive Property. a =a.
• The Symmetric Property. If a=b, then b=a.
• The Transitive Property. If a=b and b=c, then a=c.
• The Substitution Property. If a=b, then a might be substituted for b in any equation.
• The Addition and Subtraction Properties.
• The Multiplication Properties.
• The Division Properties.
• The Sq. Roots Property*
### What is the division property of equality?
The Division Property of Equality states that in case you divide each side of an equation by the similar nonzero quantity, the sides stay equal.
### What property of equality is X X?
PROPERTIES OF EQUALITY
Reflexive PropertyFor all actual numbers x , x=x . A quantity equals itself.
Multiplication PropertyFor all actual numbers x,y, and z , if x=y , then xz=yz .
Division PropertyFor all actual numbers x,y, and z , if x=y , and z≠0 , then xz=yz .
### What number of properties of equality are there?
Properties of equality. We are going to present 8 properties of equality. When acceptable, we’ll illustrate with actual life examples of properties of equality. Let x, y, and z symbolize actual numbers.
### What are the 4 properties of inequality?
PROPERTIES OF INEQUALITY
Anti reflexive PropertyFor all actual numbers x , x≮x and x≯x
Subtraction PropertyFor all actual numbers x,y, and z , if x<y then x−z<y−z .
Multiplication PropertyFor all actual numbers x,y, and z , if x<y , then {xz<yz, if z>0.xz>yz, if z<0.xz=yz, if z=0.
### How do you remedy the multiplication property of equality?
Lesson Abstract
We realized that the multiplication property of equality states that if we multiply one aspect of an equation, we additionally multiply the different aspect of the equation by the similar quantity to maintain the equation the similar. The system for this property is if a = b, then a * c = b * c.
### What is the addition and subtraction property of equality?
Addition and Subtraction Properties of Equality The Addition and Subtraction Properties of Equality are keys to manipulating equations. They state that when including or subtracting the similar quantity from each side of the equation, the equation nonetheless holds!
### What are the properties of equality?
Properties of equalities. Two equations which have the similar answer are known as equal equations e.g. 5 +3 = 2 + 6. And this as we realized in a earlier part is proven by the equality signal =. An inverse operation are two operations that undo one another e.g. addition and subtraction or multiplication and division.
### What is reflexive property of equality?
Reflexive just about means one thing regarding itself. The reflexive property of equality merely states {that a} worth is equal to itself. Additional, this property states that for all actual numbers, x = x. Once more, it states merely that any worth or quantity is equal to itself.
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# Search by Topic
#### Resources tagged with Representing similar to Children's Mathematical Writing:
Filter by: Content type:
Stage:
Challenge level:
### That Number Square!
##### Stage: 1 and 2 Challenge Level:
Exploring the structure of a number square: how quickly can you put the number tiles in the right place on the grid?
### Dice in a Corner
##### Stage: 2 Challenge Level:
How could you arrange at least two dice in a stack so that the total of the visible spots is 18?
### Children's Mathematical Graphics: Understanding the Key Concept
##### Stage: 1
In this article for teachers, Elizabeth Carruthers and Maulfry Worthington explore the differences between 'recording mathematics' and 'representing mathematical thinking'.
### Dodecamagic
##### Stage: 2 Challenge Level:
Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers?
### Let Us Divide!
##### Stage: 2 Challenge Level:
Look at different ways of dividing things. What do they mean? How might you show them in a picture, with things, with numbers and symbols?
### Going for Gold
##### Stage: 2 Challenge Level:
Looking at the 2012 Olympic Medal table, can you see how the data is organised? Could the results be presented differently to give another nation the top place?
### One Big Triangle
##### Stage: 1 Challenge Level:
Make one big triangle so the numbers that touch on the small triangles add to 10. You could use the interactivity to help you.
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### Different Sizes
##### Stage: 1 and 2 Challenge Level:
A simple visual exploration into halving and doubling.
### 3D Stacks
##### Stage: 2 and 3 Challenge Level:
Can you find a way of representing these arrangements of balls?
### A Flying Holiday
##### Stage: 2 Short Challenge Level:
Follow the journey taken by this bird and let us know for how long and in what direction it must fly to return to its starting point.
### Olympic Turns
##### Stage: 2 Challenge Level:
This task looks at the different turns involved in different Olympic sports as a way of exploring the mathematics of turns and angles.
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# simplification aptitude questions
1. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
A. 45 B. 60 C. 75 D. 90
Explanation:
Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
16x = 480
x = 30.
Hence, total number of notes = 3x = 90.
2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
A. 20 B. 80 C. 100 D. 200
Explanation:
Let the number of students in rooms A and B be x and y respectively.
Then, x – 10 = y + 10 xy = 20 …. (i)
and x + 20 = 2(y – 20) x – 2y = -60 …. (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
The required answer A = 100.
3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
A. Rs. 3500 B. Rs. 3750 C. Rs. 3840 D. Rs. 3900
Explanation:
Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.
Then, 10x = 4y or y = 5 x. 2
15x + 2y = 4000
15x + 2 x 5 x = 4000 2
20x = 4000
x = 200.
So, y = 5 x 200 = 500. 2
Hence, the cost of 12 chairs and 3 tables = 12x + 3y
= Rs. (2400 + 1500)
= Rs. 3900.
4. If ab = 3 and a2 + b2 = 29, find the value of ab.
A. 10 B. 12 C. 15 D. 18
Explanation:
2ab = (a2 + b2) – (ab)2
= 29 – 9 = 20
ab = 10.
5. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
A. Rs. 1200 B. Rs. 2400 C. Rs. 4800 D. Cannot be determined E. None of these
Explanation:
Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 …. (i)
and x + 6y = 1600 …. (ii)
```Divide equation (i) by 2, we get the below equation.
=> x + 2y = 800. --- (iii)
Now subtract (iii) from (ii)
x + 6y = 1600 (-)
x + 2y = 800
----------------
4y = 800
----------------
Therefore, y = 200.
Now apply value of y in (iii)
=> x + 2 x 200 = 800
=> x + 400 = 800
Therefore x = 400```
Solving (i) and (ii) we get x = 400, y = 200.
Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.
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Paul's Online Notes
Home / Algebra / Solving Equations and Inequalities / Quadratic Equations - Part I
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### Section 2.5 : Quadratic Equations - Part I
9. Use factoring to solve the following equation.
${t^5} = 9{t^3}$
Show All Steps Hide All Steps
Start Solution
Do not let the fact that this equation is not a quadratic equation convince you that you can’t do it! Note that we move both terms to one side we can factor a $${t^3}$$out of the equation. Doing that gives,
\begin{align*}{t^5} - 9{t^3} & = 0\\ {t^3}\left( {{t^2} - 9} \right) & = 0\end{align*}
The quantity in the parenthesis is a quadratic and we can factor it. The full factoring of the equation is then,
${t^3}\left( {t - 3} \right)\left( {t + 3} \right) = 0$ Show Step 2
Now all we need to do is use the zero factor property to get,
$\begin{array}{*{20}{c}}{{t^3} = 0}\\{t = 0}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{t - 3 = 0}\\{t = 3}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{t + 3 = 0}\\{t = - 3}\end{array}$
Therefore the three solutions are : $$\require{bbox} \bbox[2pt,border:1px solid black]{{t = 0,\,\,t = 3\,\,{\mbox{and }}t = - 3}}$$
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# Excel Tutorial: How To Calculate Mean And Standard Deviation In Excel
## Introduction
Excel is a powerful tool for data analysis, and understanding how to calculate mean and standard deviation is essential for anyone working with data. Whether you are a student, a researcher, or a business professional, the ability to analyze and interpret data is a valuable skill in today's data-driven world.
Mean and standard deviation are important statistical measures that help us understand the central tendency and variability of a dataset. By learning how to calculate these measures in Excel, you can gain valuable insights into your data and make more informed decisions based on the results.
## Key Takeaways
• Mean and standard deviation are essential statistical measures for understanding data variability and central tendency.
• Excel provides powerful functions, such as AVERAGE and STDEV.S, for calculating mean and standard deviation.
• Understanding alternative methods for calculating mean and standard deviation can be beneficial in certain data analysis scenarios.
• Mean and standard deviation play a significant role in interpreting and making informed decisions based on data analysis results.
• Practicing the use of Excel functions for calculating mean and standard deviation is crucial for data-driven professionals.
## Understanding Mean
Mean, also known as the average, is a fundamental concept in statistics that represents the central value of a set of numbers. It is calculated by summing up all the numbers in the set and then dividing by the total count of numbers.
### A. Define what mean is in statistics
Mean is a measure of central tendency that gives us an idea of the typical value in a set of numbers. It is widely used in various fields such as finance, science, and engineering to analyze and interpret data.
### B. Explain how to calculate mean in Excel using the AVERAGE function
In Excel, calculating the mean is straightforward using the AVERAGE function. Simply input the range of cells containing the data you want to calculate the mean for, and the function will return the average value.
### C. Provide an example of calculating mean for a given set of data
For example, if you have a set of numbers in cells A1 to A5, you can input =AVERAGE(A1:A5) in a different cell to obtain the mean of those numbers.
## Understanding Standard Deviation
In statistics, standard deviation is a measure of the amount of variation or dispersion of a set of values. It tells you how much individual data points differ from the mean. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation indicates that the data points are spread out over a large range.
### Define what standard deviation is in statistics
Standard deviation is calculated as the square root of the variance. It is used to measure the amount of variation or dispersion of a set of values.
### Explain how to calculate standard deviation in Excel using the STDEV.S function
In Excel, you can easily calculate the standard deviation of a set of values using the STDEV.S function. This function takes a range of values as its argument and returns the standard deviation of those values.
• First, select the cell where you want the standard deviation to appear.
• Next, enter the formula =STDEV.S(range), replacing "range" with the actual range of data you want to calculate the standard deviation for.
• Press Enter to calculate the standard deviation.
### Provide an example of calculating standard deviation for a given set of data
For example, if you have a set of data in cells A1:A10, you can calculate the standard deviation by entering the formula =STDEV.S(A1:A10) in a different cell. This will give you the standard deviation of the values in the specified range.
## Using Excel Functions for Mean and Standard Deviation
When working with data in Excel, it’s essential to be able to calculate the mean and standard deviation. These statistical measures provide valuable insights into the central tendency and variability of a dataset. Fortunately, Excel provides built-in functions to compute these measures quickly and accurately.
### Walk through the steps of using AVERAGE and STDEV.S functions in Excel
To calculate the mean of a dataset in Excel, you can use the AVERAGE function. Simply select the cells containing the data, and enter =AVERAGE( followed by the range of cells, and then close the parentheses. For example, if your data is in cells A1 to A10, the formula would be =AVERAGE(A1:A10).
For calculating the standard deviation in Excel, the STDEV.S function is used. Similar to the AVERAGE function, you would enter =STDEV.S( followed by the range of cells, and then close the parentheses. For example, if your data is in cells A1 to A10, the formula would be =STDEV.S(A1:A10).
### Discuss any optional arguments for these functions
Both the AVERAGE and STDEV.S functions in Excel have optional arguments that allow you to customize the calculations. For example, you can use the AVERAGE function to calculate the average of values that meet specific criteria by including the criteria alongside the range of cells. The STDEV.S function also has the option to ignore any text values within the dataset.
### Highlight any common errors to avoid when using these functions
One common error to avoid when using the AVERAGE and STDEV.S functions in Excel is including empty cells or cells with non-numeric values in the range. This can lead to inaccurate calculations. It’s important to ensure that the range of cells provided to the functions only contains the relevant numeric data for accurate results.
## Alternative Methods for Calculating Mean and Standard Deviation
When it comes to calculating mean and standard deviation in Excel, there are alternative methods that can be used in addition to the commonly known AVERAGE and STDEV.S functions. These alternative methods offer different ways of approaching these calculations and may be more suitable in certain situations.
A. Discuss other Excel functions that can be used for calculating mean and standard deviation
• MEAN: The MEAN function is an alternative to the AVERAGE function for calculating the mean of a set of values. It performs the same function as AVERAGE but may be more intuitive for some users.
• STDEV.P: The STDEV.P function calculates the standard deviation of a population, while STDEV.S calculates the standard deviation of a sample. Depending on the data being analyzed, one may be more appropriate than the other.
• ARRAY FORMULAS: Using array formulas with functions like SUM and COUNT, it is possible to calculate mean and standard deviation for a range of values without using the AVERAGE and STDEV.S functions.
B. Compare and contrast these alternative methods with the AVERAGE and STDEV.S functions
When comparing these alternative methods with the AVERAGE and STDEV.S functions, it is important to consider their advantages and limitations.
• Accuracy: Some alternative methods may offer greater accuracy or precision in specific scenarios, such as when dealing with large data sets or non-standard distributions.
• Usability: While AVERAGE and STDEV.S are widely used and understood, some users may find the alternative methods more intuitive or easier to implement.
• Speed: Depending on the size of the data set, some alternative methods may execute more quickly than the standard functions.
C. Provide examples of when these alternative methods may be more suitable
There are situations where the alternative methods for calculating mean and standard deviation may be more suitable than the standard AVERAGE and STDEV.S functions.
• Non-Numerical Data: When working with non-numerical data or data with significant outliers, alternative methods such as array formulas may offer more flexibility and accuracy.
• Large Data Sets: For very large data sets, alternative methods that can handle data in batches or chunks may be more efficient.
• Specific Statistical Requirements: In cases where specific statistical requirements need to be met, such as using the standard deviation of a population rather than a sample, using the appropriate function (STDEV.P in this case) is essential.
## Applying Mean and Standard Deviation in Data Analysis
Mean and standard deviation are important statistical measures that provide valuable insights into the distribution and variability of data. In Excel, these measures can be easily calculated using simple functions, enabling analysts to make informed decisions based on the data at hand.
### Explain the significance of mean and standard deviation in interpreting data
The mean, also known as the average, represents the central tendency of a dataset and provides a single value that summarizes the entire data. It is a valuable measure for understanding the typical value in a set of numbers. On the other hand, standard deviation measures the amount of variation or dispersion in a set of data values. It indicates how much individual data points deviate from the mean, thereby highlighting the consistency or variability within the dataset.
### Provide examples of how mean and standard deviation can be used in real-world data analysis scenarios
Mean and standard deviation can be applied in various real-world scenarios. For instance, in finance, the mean and standard deviation of stock returns can help investors assess the average performance and volatility of a particular security. In quality control, these measures can be used to monitor the consistency and variability of product specifications. Furthermore, in healthcare, mean and standard deviation can aid in understanding the average patient outcome and the variability in treatment effectiveness.
### Discuss how to interpret the results of mean and standard deviation calculations
Interpreting the results of mean and standard deviation calculations is crucial for drawing meaningful conclusions from the data. A low standard deviation indicates that the data points are close to the mean, suggesting less variability. Conversely, a high standard deviation signifies that the data points are spread out over a wider range, indicating greater variability. Understanding these interpretations is essential for making informed decisions based on the data analysis results.
## Conclusion
In conclusion, this tutorial provided a step-by-step guide on how to calculate the mean and standard deviation in Excel. We discussed the importance of understanding these statistical measures in data analysis and how they can provide valuable insights into the distribution and variability of data.
We encourage readers to practice using Excel to calculate mean and standard deviation with different data sets to gain a better understanding of the process. The more familiar you become with these calculations, the more confident you will be in your data analysis skills.
Remember, a solid grasp of mean and standard deviation is crucial for anyone working with data, whether it's for academic, professional, or personal purposes. These measures not only help in descriptive statistics but also in making informed decisions based on data analysis.
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Intermediate Algebra 2e
# 1.1Use the Language of Algebra
Intermediate Algebra 2e1.1 Use the Language of Algebra
## Learning Objectives
By the end of this section, you will be able to:
• Find factors, prime factorizations, and least common multiples
• Use variables and algebraic symbols
• Simplify expressions using the order of operations
• Evaluate an expression
• Identify and combine like terms
• Translate an English phrase to an algebraic expression
## Be Prepared 1.1
This chapter is intended to be a brief review of concepts that will be needed in an Intermediate Algebra course. A more thorough introduction to the topics covered in this chapter can be found in the Elementary Algebra 2e chapter, Foundations.
In algebra, we use a letter of the alphabet to represent a number whose value may change or is unknown. Commonly used symbols are a, b, c, m, n, x, and y. Further discussion of constants and variables appears later in this section.
## Find Factors, Prime Factorizations, and Least Common Multiples
The numbers 2, 4, 6, 8, 10, 12 are called multiples of 2. A multiple of 2 can be written as the product of 2 and a counting number.
Similarly, a multiple of 3 would be the product of a counting number and 3.
We could find the multiples of any number by continuing this process.
Counting Number 1 2 3 4 5 6 7 8 9 10 11 12
Multiples of 2 2 4 6 8 10 12 14 16 18 20 22 24
Multiples of 3 3 6 9 12 15 18 21 24 27 30 33 36
Multiples of 4 4 8 12 16 20 24 28 32 36 40 44 48
Multiples of 5 5 10 15 20 25 30 35 40 45 50 55 60
Multiples of 6 6 12 18 24 30 36 42 48 54 60 66 72
Multiples of 7 7 14 21 28 35 42 49 56 63 70 77 84
Multiples of 8 8 16 24 32 40 48 56 64 72 80 88 96
Multiples of 9 9 18 27 36 45 54 63 72 81 90 99 108
## Multiple of a Number
A number is a multiple of $nn$ if it is the product of a counting number and $n.n.$
Another way to say that 15 is a multiple of 3 is to say that 15 is divisible by 3. That means that when we divide 15 by 3, we get a counting number. In fact, $15÷315÷3$ is 5, so 15 is $5·3.5·3.$
## Divisible by a Number
If a number $mm$ is a multiple of n, then m is divisible by n.
If we were to look for patterns in the multiples of the numbers 2 through 9, we would discover the following divisibility tests:
## Divisibility Tests
A number is divisible by:
2 if the last digit is 0, 2, 4, 6, or 8.
3 if the sum of the digits is divisible by $3.3.$
5 if the last digit is 5 or $0.0.$
6 if it is divisible by both 2 and $3.3.$
10 if it ends with $0.0.$
## Example 1.1
Is 5,625 divisible by 2? 3? 5 or 10? 6?
## Try It 1.1
Is 4,962 divisible by 2? 3? 5? 6? 10?
## Try It 1.2
Is 3,765 divisible by 2? 3? 5? 6? 10?
In mathematics, there are often several ways to talk about the same ideas. So far, we’ve seen that if m is a multiple of n, we can say that m is divisible by n. For example, since 72 is a multiple of 8, we say 72 is divisible by 8. Since 72 is a multiple of 9, we say 72 is divisible by 9. We can express this still another way.
Since $8·9=72,8·9=72,$ we say that 8 and 9 are factors of 72. When we write $72=8·9,72=8·9,$ we say we have factored 72.
Other ways to factor 72 are $1·72,2·36,3·24,4·18,1·72,2·36,3·24,4·18,$ and $6·12.6·12.$ The number 72 has many factors: $1,2,3,4,6,8,9,12,18,24,36,1,2,3,4,6,8,9,12,18,24,36,$ and $72.72.$
## Factors
In the expression $a·ba·b$, both a and b are called factors. If $a·b=m,a·b=m,$ and both a and b are integers, then a and b are factors of m.
Some numbers, such as 72, have many factors. Other numbers have only two factors. A prime number is a counting number greater than 1 whose only factors are 1 and itself.
## Prime number and Composite number
A prime number is a counting number greater than 1 whose only factors are 1 and the number itself.
A composite number is a counting number greater than 1 that is not prime. A composite number has factors other than 1 and the number itself.
The counting numbers from 2 to 20 are listed in the table with their factors. Make sure to agree with the “prime” or “composite” label for each!
The prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Notice that the only even prime number is 2.
A composite number can be written as a unique product of primes. This is called the prime factorization of the number. Finding the prime factorization of a composite number will be useful in many topics in this course.
## Prime Factorization
The prime factorization of a number is the product of prime numbers that equals the number. These prime numbers are called the prime factors.
To find the prime factorization of a composite number, find any two factors of the number and use them to create two branches. If a factor is prime, that branch is complete. Circle that prime. Otherwise it is easy to lose track of the prime numbers.
If the factor is not prime, find two factors of the number and continue the process. Once all the branches have circled primes at the end, the factorization is complete. The composite number can now be written as a product of prime numbers.
Factor 48.
## Try It 1.3
Find the prime factorization of $80.80.$
## Try It 1.4
Find the prime factorization of $60.60.$
## How To
### Find the prime factorization of a composite number.
1. Step 1. Find two factors whose product is the given number, and use these numbers to create two branches.
2. Step 2. If a factor is prime, that branch is complete. Circle the prime, like a leaf on the tree.
3. Step 3. If a factor is not prime, write it as the product of two factors and continue the process.
4. Step 4. Write the composite number as the product of all the circled primes.
One of the reasons we look at primes is to use these techniques to find the least common multiple of two numbers. This will be useful when we add and subtract fractions with different denominators.
## Least Common Multiple
The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both numbers.
To find the least common multiple of two numbers we will use the Prime Factors Method. Let’s find the LCM of 12 and 18 using their prime factors.
## Example 1.3
### How to Find the Least Common Multiple Using the Prime Factors Method
Find the least common multiple (LCM) of 12 and 18 using the prime factors method.
Notice that the prime factors of 12 $(2·2·3)(2·2·3)$ and the prime factors of 18 $(2·3·3)(2·3·3)$ are included in the LCM $(2·2·3·3).(2·2·3·3).$ So 36 is the least common multiple of 12 and 18.
By matching up the common primes, each common prime factor is used only once. This way you are sure that 36 is the least common multiple.
## Try It 1.5
Find the LCM of 9 and 12 using the Prime Factors Method.
## Try It 1.6
Find the LCM of 18 and 24 using the Prime Factors Method.
## How To
### Find the least common multiple using the Prime Factors Method.
1. Step 1. Write each number as a product of primes.
2. Step 2. List the primes of each number. Match primes vertically when possible.
3. Step 3. Bring down the columns.
4. Step 4. Multiply the factors.
## Use Variables and Algebraic Symbols
In algebra, we use a letter of the alphabet to represent a number whose value may change. We call this a variable and letters commonly used for variables are $x,y,a,b,c.x,y,a,b,c.$
## Variable
A variable is a letter that represents a number whose value may change.
A number whose value always remains the same is called a constant.
## Constant
A constant is a number whose value always stays the same.
To write algebraically, we need some operation symbols as well as numbers and variables. There are several types of symbols we will be using. There are four basic arithmetic operations: addition, subtraction, multiplication, and division. We’ll list the symbols used to indicate these operations below.
## Operation Symbols
Operation Notation Say: The result is…
Addition $a+ba+b$ $aa$ plus $bb$ the sum of $aa$ and $bb$
Subtraction $a−ba−b$ $aa$ minus $bb$ the difference of $aa$ and $bb$
Multiplication $a·b,ab,(a)(b),a·b,ab,(a)(b),$ $(a)b,a(b)(a)b,a(b)$ $aa$ times $bb$ the product of $aa$ and $bb$
Division $a÷b,a/b,ab,baa÷b,a/b,ab,ba$ $aa$ divided by $bb$ the quotient of $aa$ and $b;b;$
$aa$ is called the dividend, and $bb$ is called the divisor
When two quantities have the same value, we say they are equal and connect them with an equal sign.
## Equality Symbol
$a=ba=b$ is read “a is equal to b.”
The symbol “=” is called the equal sign.
On the number line, the numbers get larger as they go from left to right. The number line can be used to explain the symbols “<” and “>”.
## Inequality
The expressions $a or $a>ba>b$ can be read from left to right or right to left, though in English we usually read from left to right. In general,
$aa.For example,7<11is equivalent to11>7.a>bis equivalent tob4is equivalent to4<17.aa.For example,7<11is equivalent to11>7.a>bis equivalent tob4is equivalent to4<17.$
## Inequality Symbols
Inequality Symbols Words
$a≠ba≠b$ a is not equal to b.
$a a is less than b.
$a≤ba≤b$ a is less than or equal to b.
$a>ba>b$ a is greater than b.
$a≥ba≥b$ a is greater than or equal to b.
Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in English. They help identify an expression, which can be made up of number, a variable, or a combination of numbers and variables using operation symbols. We will introduce three types of grouping symbols now.
## Grouping Symbols
$Parentheses()Brackets[]Braces{}Parentheses()Brackets[]Braces{}$
Here are some examples of expressions that include grouping symbols. We will simplify expressions like these later in this section.
$8(14−8)21−3[2+4(9−8)]24÷{13−2[1(6−5)+4]}8(14−8)21−3[2+4(9−8)]24÷{13−2[1(6−5)+4]}$
What is the difference in English between a phrase and a sentence? A phrase expresses a single thought that is incomplete by itself, but a sentence makes a complete statement. A sentence has a subject and a verb. In algebra, we have expressions and equations.
## Expression
An expression is a number, a variable, or a combination of numbers and variables using operation symbols.
$ExpressionWordsEnglish Phrase3+53 plus 5the sum of three and fiven−1nminus onethe difference ofnand one6·76 times 7the product of six and sevenxyxdivided byythe quotient ofxandyExpressionWordsEnglish Phrase3+53 plus 5the sum of three and fiven−1nminus onethe difference ofnand one6·76 times 7the product of six and sevenxyxdivided byythe quotient ofxandy$
Notice that the English phrases do not form a complete sentence because the phrase does not have a verb.
An equation is two expressions linked by an equal sign. When you read the words the symbols represent in an equation, you have a complete sentence in English. The equal sign gives the verb.
## Equation
An equation is two expressions connected by an equal sign.
$EquationEnglish Sentence3+5=8The sum of three and five is equal to eight.n−1=14nminus one equals fourteen.6·7=42The product of six and seven is equal to forty-two.x=53xis equal to fifty-three.y+9=2y−3yplus nine is equal to twoyminus three.EquationEnglish Sentence3+5=8The sum of three and five is equal to eight.n−1=14nminus one equals fourteen.6·7=42The product of six and seven is equal to forty-two.x=53xis equal to fifty-three.y+9=2y−3yplus nine is equal to twoyminus three.$
Suppose we need to multiply 2 nine times. We could write this as $2·2·2·2·2·2·2·2·2.2·2·2·2·2·2·2·2·2.$ This is tedious and it can be hard to keep track of all those 2s, so we use exponents. We write $2·2·22·2·2$ as $2323$ and $2·2·2·2·2·2·2·2·22·2·2·2·2·2·2·2·2$ as $29.29.$ In expressions such as $23,23,$ the 2 is called the base and the 3 is called the exponent. The exponent tells us how many times we need to multiply the base.
## Exponential Notation
We say $2323$ is in exponential notation and $2·2·22·2·2$ is in expanded notation.
$anan$ means multiply a by itself, n times.
The expression $anan$ is read a to the $nthnth$ power.
While we read $anan$ as $“a“a$ to the $nthnth$ power”, we usually read:
$a2“asquared”a3“acubed”a2“asquared”a3“acubed”$
We’ll see later why $a2a2$ and $a3a3$ have special names.
Table 1.1 shows how we read some expressions with exponents.
Expression In Words
72 7 to the second power or 7 squared
53 5 to the third power or 5 cubed
94 9 to the fourth power
125 12 to the fifth power
Table 1.1
## Simplify Expressions Using the Order of Operations
To simplify an expression means to do all the math possible. For example, to simplify $4·2+14·2+1$ we would first multiply $4·24·2$ to get 8 and then add the 1 to get 9. A good habit to develop is to work down the page, writing each step of the process below the previous step. The example just described would look like this:
$4·2+18+194·2+18+19$
By not using an equal sign when you simplify an expression, you may avoid confusing expressions with equations.
## Simplify an Expression
To simplify an expression, do all operations in the expression.
We’ve introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values.
For example, consider the expression $4+3·7.4+3·7.$ Some students simplify this getting 49, by adding $4+34+3$ and then multiplying that result by 7. Others get 25, by multiplying $3·73·7$ first and then adding 4.
The same expression should give the same result. So mathematicians established some guidelines that are called the order of operations.
## How To
### Use the order of operations.
1. Step 1.
Parentheses and Other Grouping Symbols
• Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
2. Step 2.
Exponents
• Simplify all expressions with exponents.
3. Step 3.
Multiplication and Division
• Perform all multiplication and division in order from left to right. These operations have equal priority.
4. Step 4.
• Perform all addition and subtraction in order from left to right. These operations have equal priority.
Students often ask, “How will I remember the order?” Here is a way to help you remember: Take the first letter of each key word and substitute the silly phrase “Please Excuse My Dear Aunt Sally”.
$ParenthesesPleaseExponentsExcuseMultiplicationDivisionMyDearAdditionSubtractionAuntSallyParenthesesPleaseExponentsExcuseMultiplicationDivisionMyDearAdditionSubtractionAuntSally$
It’s good that “My Dear” goes together, as this reminds us that multiplication and division have equal priority. We do not always do multiplication before division or always do division before multiplication. We do them in order from left to right.
Similarly, “Aunt Sally” goes together and so reminds us that addition and subtraction also have equal priority and we do them in order from left to right.
## Example 1.4
Simplify: $18÷6+4(5−2).18÷6+4(5−2).$
## Try It 1.7
Simplify: $30÷5+10(3−2).30÷5+10(3−2).$
## Try It 1.8
Simplify: $70÷10+4(6−2).70÷10+4(6−2).$
When there are multiple grouping symbols, we simplify the innermost parentheses first and work outward.
## Example 1.5
Simplify: $5+23+3[6−3(4−2)].5+23+3[6−3(4−2)].$
## Try It 1.9
Simplify: $9+53−[4(9+3)].9+53−[4(9+3)].$
## Try It 1.10
Simplify: $72−2[4(5+1)].72−2[4(5+1)].$
## Evaluate an Expression
In the last few examples, we simplified expressions using the order of operations. Now we’ll evaluate some expressions—again following the order of operations. To evaluate an expression means to find the value of the expression when the variable is replaced by a given number.
## Evaluate an Expression
To evaluate an expression means to find the value of the expression when the variable is replaced by a given number.
To evaluate an expression, substitute that number for the variable in the expression and then simplify the expression.
## Example 1.6
Evaluate when $x=4:x=4:$ $x2x2$ $3x3x$ $2x2+3x+8.2x2+3x+8.$
## Try It 1.11
Evaluate when $x=3,x=3,$ $x2x2$ $4x4x$ $3x2+4x+1.3x2+4x+1.$
## Try It 1.12
Evaluate when $x=6,x=6,$ $x3x3$ $2x2x$ $6x2−4x−7.6x2−4x−7.$
## Identify and Combine Like Terms
Algebraic expressions are made up of terms. A term is a constant, or the product of a constant and one or more variables.
## Term
A term is a constant or the product of a constant and one or more variables.
Examples of terms are $7,y,5x2,9a,7,y,5x2,9a,$ and $b5.b5.$
The constant that multiplies the variable is called the coefficient.
## Coefficient
The coefficient of a term is the constant that multiplies the variable in a term.
Think of the coefficient as the number in front of the variable. The coefficient of the term $3x3x$ is 3. When we write $x,x,$ the coefficient is 1, since $x=1·x.x=1·x.$
Some terms share common traits. When two terms are constants or have the same variable and exponent, we say they are like terms.
Look at the following 6 terms. Which ones seem to have traits in common?
$5x7n243x9n25x7n243x9n2$
We say,
$77$ and $44$ are like terms.
$5x5x$ and $3x3x$ are like terms.
$n2n2$ and $9n29n2$ are like terms.
## Like Terms
Terms that are either constants or have the same variables raised to the same powers are called like terms.
If there are like terms in an expression, you can simplify the expression by combining the like terms. We add the coefficients and keep the same variable.
$Simplify.4x+7x+xAdd the coefficients.12xSimplify.4x+7x+xAdd the coefficients.12x$
## Example 1.7
### How To Combine Like Terms
Simplify: $2x2+3x+7+x2+4x+5.2x2+3x+7+x2+4x+5.$
## Try It 1.13
Simplify: $3x2+7x+9+7x2+9x+8.3x2+7x+9+7x2+9x+8.$
## Try It 1.14
Simplify: $4y2+5y+2+8y2+4y+5.4y2+5y+2+8y2+4y+5.$
## How To
### Combine like terms.
1. Step 1. Identify like terms.
2. Step 2. Rearrange the expression so like terms are together.
3. Step 3. Add or subtract the coefficients and keep the same variable for each group of like terms.
## Translate an English Phrase to an Algebraic Expression
We listed many operation symbols that are used in algebra. Now, we will use them to translate English phrases into algebraic expressions. The symbols and variables we’ve talked about will help us do that. Table 1.2 summarizes them.
Operation Phrase Expression
the sum of $aa$ and b
a increased by b
b more than a
the total of a and b
$a+ba+b$
Subtraction a minus $bb$
the difference of a and b
a decreased by b
b less than a
b subtracted from a
$a−ba−b$
Multiplication a times b
the product of $aa$ and $bb$
twice a
$a·b,ab,a(b),(a)(b)a·b,ab,a(b),(a)(b)$
$2a2a$
Division a divided by b
the quotient of a and b
the ratio of a and b
b divided into a
$a÷b,a/b,ab,baa÷b,a/b,ab,ba$
Table 1.2
Look closely at these phrases using the four operations:
Each phrase tells us to operate on two numbers. Look for the words of and and to find the numbers.
## Example 1.8
Translate each English phrase into an algebraic expression:
the difference of $14x14x$ and 9 the quotient of $8y28y2$ and 3 twelve more than $yy$ seven less than $49x249x2$
## Try It 1.15
Translate the English phrase into an algebraic expression:
the difference of $14x214x2$ and 13 the quotient of $12x12x$ and 2 13 more than $zz$
18 less than $8x8x$
## Try It 1.16
Translate the English phrase into an algebraic expression:
the sum of $17y217y2$ and 19 the product of $77$ and y Eleven more than x Fourteen less than 11a
We look carefully at the words to help us distinguish between multiplying a sum and adding a product.
## Example 1.9
Translate the English phrase into an algebraic expression:
eight times the sum of x and y the sum of eight times x and y
## Try It 1.17
Translate the English phrase into an algebraic expression:
four times the sum of p and q
the sum of four times p and q
## Try It 1.18
Translate the English phrase into an algebraic expression:
the difference of two times x and 8
two times the difference of x and 8
Later in this course, we’ll apply our skills in algebra to solving applications. The first step will be to translate an English phrase to an algebraic expression. We’ll see how to do this in the next two examples.
## Example 1.10
The length of a rectangle is 14 less than the width. Let w represent the width of the rectangle. Write an expression for the length of the rectangle.
## Try It 1.19
The length of a rectangle is 7 less than the width. Let w represent the width of the rectangle. Write an expression for the length of the rectangle.
## Try It 1.20
The width of a rectangle is 6 less than the length. Let l represent the length of the rectangle. Write an expression for the width of the rectangle.
The expressions in the next example will be used in the typical coin mixture problems we will see soon.
## Example 1.11
June has dimes and quarters in her purse. The number of dimes is seven less than four times the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes.
## Try It 1.21
Geoffrey has dimes and quarters in his pocket. The number of dimes is eight less than four times the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes.
## Try It 1.22
Lauren has dimes and nickels in her purse. The number of dimes is three more than seven times the number of nickels. Let n represent the number of nickels. Write an expression for the number of dimes.
## Section 1.1 Exercises
### Practice Makes Perfect
Identify Multiples and Factors
In the following exercises, use the divisibility tests to determine whether each number is divisible by 2, by 3, by 5, by 6, and by 10.
1.
84
2.
96
3.
896
4.
942
5.
22,335
6.
39,075
Find Prime Factorizations and Least Common Multiples
In the following exercises, find the prime factorization.
7.
86
8.
78
9.
455
10.
400
11.
432
12.
627
In the following exercises, find the least common multiple of each pair of numbers using the prime factors method.
13.
8, 12
14.
12, 16
15.
28, 40
16.
84, 90
17.
55, 88
18.
60, 72
Simplify Expressions Using the Order of Operations
In the following exercises, simplify each expression.
19.
$2 3 − 12 ÷ ( 9 − 5 ) 2 3 − 12 ÷ ( 9 − 5 )$
20.
$3 2 − 18 ÷ ( 11 − 5 ) 3 2 − 18 ÷ ( 11 − 5 )$
21.
$2 + 8 ( 6 + 1 ) 2 + 8 ( 6 + 1 )$
22.
$4 + 6 ( 3 + 6 ) 4 + 6 ( 3 + 6 )$
23.
$20 ÷ 4 + 6 ( 5 − 1 ) 20 ÷ 4 + 6 ( 5 − 1 )$
24.
$33 ÷ 3 + 4 ( 7 − 2 ) 33 ÷ 3 + 4 ( 7 − 2 )$
25.
$3 ( 1 + 9 · 6 ) − 4 2 3 ( 1 + 9 · 6 ) − 4 2$
26.
$5 ( 2 + 8 · 4 ) − 7 2 5 ( 2 + 8 · 4 ) − 7 2$
27.
$2 [ 1 + 3 ( 10 − 2 ) ] 2 [ 1 + 3 ( 10 − 2 ) ]$
28.
$5 [ 2 + 4 ( 3 − 2 ) ] 5 [ 2 + 4 ( 3 − 2 ) ]$
29.
$8 + 2 [ 7 − 2 ( 5 − 3 ) ] − 3 2 8 + 2 [ 7 − 2 ( 5 − 3 ) ] − 3 2$
30.
$10 + 3 [ 6 − 2 ( 4 − 2 ) ] − 2 4 10 + 3 [ 6 − 2 ( 4 − 2 ) ] − 2 4$
Evaluate an Expression
In the following exercises, evaluate the following expressions.
31.
When $x=2,x=2,$
$x6x6$
$4x4x$
$2x2+3x−72x2+3x−7$
32.
When $x=3,x=3,$
$x5x5$
$5x5x$
$3x2−4x−83x2−4x−8$
33.
When $x=4,y=1x=4,y=1$
$x2+3xy−7y2x2+3xy−7y2$
34.
When $x=3,y=2x=3,y=2$
$6x2+3xy−9y26x2+3xy−9y2$
35.
When $x=10,y=7x=10,y=7$
$(x−y)2(x−y)2$
36.
When $a=3,b=8a=3,b=8$
$a2+b2a2+b2$
Simplify Expressions by Combining Like Terms
In the following exercises, simplify the following expressions by combining like terms.
37.
$7 x + 2 + 3 x + 4 7 x + 2 + 3 x + 4$
38.
$8 y + 5 + 2 y − 4 8 y + 5 + 2 y − 4$
39.
$10 a + 7 + 5 a − 2 + 7 a − 4 10 a + 7 + 5 a − 2 + 7 a − 4$
40.
$7 c + 4 + 6 c − 3 + 9 c − 1 7 c + 4 + 6 c − 3 + 9 c − 1$
41.
$3 x 2 + 12 x + 11 + 14 x 2 + 8 x + 5 3 x 2 + 12 x + 11 + 14 x 2 + 8 x + 5$
42.
$5 b 2 + 9 b + 10 + 2 b 2 + 3 b − 4 5 b 2 + 9 b + 10 + 2 b 2 + 3 b − 4$
Translate an English Phrase to an Algebraic Expression
In the following exercises, translate the phrases into algebraic expressions.
43.
the difference of $5x25x2$ and $6xy6xy$
the quotient of $6y26y2$ and $5x5x$
Twenty-one more than $y2y2$
$6x6x$ less than $81x281x2$
44.
the difference of $17x217x2$ and $5xy5xy$
the quotient of $8y38y3$ and $3x3x$
Eighteen more than $a2a2$;
$11b11b$ less than $100b2100b2$
45.
the sum of $4ab24ab2$ and $3a2b3a2b$
the product of $4y24y2$ and $5x5x$
Fifteen more than $mm$
$9x9x$ less than $121x2121x2$
46.
the sum of $3x2y3x2y$ and $7xy27xy2$
the product of $6xy26xy2$ and $4z4z$
Twelve more than $3x23x2$
$7x27x2$ less than $63x363x3$
47.
eight times the difference of $yy$ and nine
the difference of eight times $yy$ and 9
48.
seven times the difference of $yy$ and one
the difference of seven times $yy$ and 1
49.
five times the sum of $3x3x$ and $yy$
the sum of five times $3x3x$ and $yy$
50.
eleven times the sum of $4x24x2$ and $5x5x$
the sum of eleven times $4x24x2$ and $5x5x$
51.
Eric has rock and country songs on his playlist. The number of rock songs is 14 more than twice the number of country songs. Let c represent the number of country songs. Write an expression for the number of rock songs.
52.
The number of women in a Statistics class is 8 more than twice the number of men. Let $mm$ represent the number of men. Write an expression for the number of women.
53.
Greg has nickels and pennies in his pocket. The number of pennies is seven less than three times the number of nickels. Let n represent the number of nickels. Write an expression for the number of pennies.
54.
Jeannette has $55$ and $1010$ bills in her wallet. The number of fives is three more than six times the number of tens. Let $tt$ represent the number of tens. Write an expression for the number of fives.
### Writing Exercises
55.
Explain in your own words how to find the prime factorization of a composite number.
56.
Why is it important to use the order of operations to simplify an expression?
57.
Explain how you identify the like terms in the expression $8a2+4a+9−a2−1.8a2+4a+9−a2−1.$
58.
Explain the difference between the phrases “4 times the sum of x and y” and “the sum of 4 times x and y”.
### Self Check
Use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
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# Tutor profile: Avary K.
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Avary K.
Tutor for 7 years. Graduate Teaching Assistant for 4 years.
Tutor Satisfaction Guarantee
## Questions
### Subject:Basic Math
TutorMe
Question:
During a back to school sale an instrument store is giving a 15% is discount for students. Julie, a student at the local high school, purchases a flute for $306. How much did it originally cost? Inactive Avary K. Answer: To start any math problem, your first step should be to write down what we know and what we want to know: back to school sale- 15% paid for flute-$306 original cost- ?? Now, since we don't know what our original cost of the flute is, let that be $$x$$ original cost=$$x$$ so our goal is to find $$x$$. To do so, we are going to set up an equation. This equation can be used for problems similar to this. It reads: (original cost)-(%)(original cost)=final cost or in words "the original price minus the percent discount (as a decimal) times the original price is equal to the final price". In terms of our problem, the back to school discount is a total of 15% (or 0.15 as a decimal), the final cost is $306, and $$x$$ represents the original cost which leads to our equation $$x-0.15x=306$$ now we solve for $$x$$. Notice that on the left side of the equation, both terms have $$x$$ so we can combine them: $$x-0.15x = 1x-0.15x = 0.85x$$ so our equation becomes $$0.85x=306$$ Now, since we are multiplying $$x$$ by $$0.85$$ to get $$x$$ by itself we have to divide both sides by $$0.85$$ $$\dfrac{0.85x}{0.85}=\dfrac{306}{0.85}$$ $$x=360$$ which tells us that the original price ($$x$$) of the flute was$360. It should be noted that the (original)-(%)(original)=final isn't just used for discounts or sales. It can also be used for any type of percent decrease. Additionally, if we have + (instead of -) we can use this equation for taxes, increases, etc. (where % is ALWAYS written as a decimal).
### Subject:Calculus
TutorMe
Question:
Find the relative maximum and minimum of the function $$f(x)=x^3−9x^2+24x$$.
Inactive
Avary K.
Before we start solving, lets talk about the maximum and minimum of a function. If you draw any graph, notice that at the max/min the slope of the tangent line is either 0 or does not exist. For example, looking at $$x^2$$; the minimum of this function occurs at $$x=0$$ where the tangent line is a horizontal line which we know has a slope 0 (all horizontal lines have slope 0). Another example is $$|x|$$; the minimum of this function occurs at $$x=0$$ where there is a sharp corner which we know has no tangent line. Thinking about this, we know at every max/min either the slope of the tangent line is 0 or DNE. The slope of the tangent line is defined as the derivative so it would make sense, if we want to find the max/min, to first find where the derivative is 0 or DNE. Step 1) Find $$f'(x)$$. By the power rule we have $$f'(x)=3x^2-18x+24.$$ Step 2) Find the $$x$$'s that make $$f'(x)=0$$ or DNE. Lets first find the $$x$$'s that make $$f'(x)=0$$: $$0=f'(x)=3x^2-18x+24\implies 0=3(x^2-6x+8)\implies 0=3(x-2)(x-4)\implies x-2=0 \quad \text{or}\quad x-4=0\implies x=2 \quad \text{or}\quad x=4$$ so we found two values of $$x$$ that make $$f'(x)=0$$: $$x= 2,4$$. Now, to find the $$x$$'s that make $$f'(x)$$ DNE aka the values of $$x$$ where $$f'(x)$$ is discontinuous. Since $$f'(x)$$ in this example is a polynomial, it is never discontinuous, so we don't get any $$x$$ values from this. Now we need to figure out whether the $$x$$ values found in Step 2 are max or min. There are a few ways we can go about this. In this example, we are going to use the Second Derivative Test. To use this test, notice that when we have a minimum, the graph is concave up and when we have a maximum, the graph is concave down. We know that the second derivative is what tells us about concavity. Specifically if $$f''(x)>0$$ we say $$f(x)$$ is concave up; if $$f''(x)<0$$ we say $$f(x)$$ is concave down. So using this logic, we can determine whether the points found in Step 2 are max or min's by looking at whether the second derivative at those points are positive or negative. This leads up to Step 3 and 4. Step 3) Find $$f''(x)$$. By using the power rule on $$f'(x)$$ we get $$f''(x)=6x-18.$$ Step 4) Determine the sign of $$f''(x)$$ at the $$x$$ values found in Step 2. That is, evaluate: $$f''(2)=6(2)-18=12-18=-6<0$$. We can now say that at $$x=2$$, $$f'(2)=0$$ and $$f''(2)<0$$ so $$f(x)$$ has a relative maximum at $$x=2$$. Similarly $$f''(4)=6(4)-18=24-18=6>0$$. We can now say that at $$x=4$$, $$f'(4)=0$$ and $$f''(4)>0$$ so $$f(x)$$ has a relative minimum at $$x=4$$. To find the $$y$$ values of these points, plug in the $$x$$ value into the ORIGINAL function: $$f(2)=2^3−9(2^2)+24(2)=8-36+48=20$$ $$f(4)=4^3−9(4^2)+24(4)=64-144+96=16$$ To conclude, $$f(x)$$ has a relative maximum at the point $$(2,20)$$ and a relative minimum at the point $$(4,16)$$
### Subject:Algebra
TutorMe
Question:
Two cyclists leave towns 128 miles apart at the same time and travel toward each other. One cyclist travels 4 mph slower than the other. If they meet in 2 hours, what is the rate of each cyclist?
Inactive
Avary K.
First lets start with what we want to know. Usually in math, the "what we want to know" will be our variable. The questions asks "what is the rate of each cyclist" so, we let x = rate (miles per hour) of cyclist 1 then since the other cyclist is traveling "4 mph slower" we know x-4 = rate (miles per hour) of cyclist 2. Now, the only other information given in the problem is the distance and time. The two cyclists start 128 miles apart and after 2 hours, they meet. So we somehow need to use the rate we have above and relate it to the distance and time. Remember the formula Distance = Rate x Time. Using this and remembering that the time is 2 hours, we can say y = distance traveled by cyclist 1 = (rate of cyclist 1) x (time) = (x)(2) = 2x then using this, if cyclist 1 traveled distance y then cyclist 2 must have traveled 128-y miles. So 128-y = distance traveled by cyclist 2 = (rate of cyclist 2) x (time) = (x-4)(2) = 2x-8 We now have a system of two equations y = 2x 128 - y = 2x-8 To solve for x (the rate of the cyclists) let's use the Substitution Method. So taking 2x (from the first equation) and substituting it in for y in the second equation we get 128 - 2x = 2x - 8 128 = 4x - 8 136 = 4x x = 34 So by this we know that x = 34 mph = rate of cyclist 1 x-4 = 30 mph = rate of cyclist 2.
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Lecture 38: Propositional logic
• Reading: Pass and Tseng, 6.1
• Overview of formal logic
• we want to study relationship between proofs and truth
• Definitions: propositional formula, interpretation, meaning of a formula in an interpretation ($$φ[I]$$), satisfiability, validity ($$⊨ φ$$), entailment ($$φ⊨ψ$$).
• Metalanguage versus object language.
• Truth tables
Metalogic
In this section of the course, we will turn our attention to formal logic. Just as we reasoned about sets and functions in the first section of the course, or probability spaces, events, etc. in the next section, strings, machines, etc while talking about automata, in this section our objects of study will be logical formulae and proofs.
This can lead to some confusion, because in addition to being the objects of study, proofs and logic will also be our tools for studying them. We will be making claims about claims and proving things about proofs.
To keep this all straight, it helps to make a distinction between formulas and proofs as objects of study and claims and proofs about those objects. We will refer to the objects of study as the object language and the proofs and definitions we use to reason about the objects as the meta language.
Propositional logic
Propositional formulas are built up from base propositions $$P$$, $$Q$$, that we will treat as uninterpreted variables. The intent is that these variables represent statements like "it is raining" or "Socrates is a man" that might be true or false.
We can combine the formulas using the connectives $$∧$$, $$∨$$, $$→$$, and $$¬$$. We want to think of these as "and", "or", "not", and "implies" respectively, but at the outset, we will treat them as meaningless symbols. Thus, formulas are just elements of an inductively defined set:
$φ,ψ ∈ Formulae ::= P \mid φ∧ψ \mid φ∨ψ \mid φ → ψ \mid ¬φ \quad \text{where} \quad P ∈ Prop$
Semantics
To ascribe meaning to formulae, we inductively define function that explains how to interpret formulae. In order to do so, we need to know how to interpret the base propositions $$P$$, $$Q$$, $$R$$, etc.
An interpretation is a function $$I : Prop → \{T,F\}$$. The evaluation of $$φ$$ in interpretation $$I$$ is given inductively by $$eval : Formula \times Interpretation → \{T,F\}$$ by the rules
\begin{aligned} eval(P, I) &::= I(P) \\ eval(φ∧ψ, I) &::= \left\{\begin{array}{ll}T & \text{ if eval(φ,I) = T and eval(ψ,I) = T} \\ F & \text{ otherwise}\end{array}\right. \\ eval(φ∨ψ, I) &::= \left\{\begin{array}{ll}T & \text{ if eval(φ,I) = T or eval(ψ,I) = T} \\ F & \text{ otherwise}\end{array}\right. \\ eval(φ→ψ, I) &::= \left\{\begin{array}{ll}F & \text{ if eval(φ,I) = T and eval(ψ,I) = F} \\ T & \text{ otherwise}\end{array}\right. \\ eval(¬φ, I) &::= \left\{\begin{array}{ll}T & \text{ if eval(φ,I) = F} \\ F & \text{ otherwise}\end{array}\right. \\ \end{aligned}
This can be stated more concisely as follows: to evaluate a formula, first recursively/inductively evaluate its substructures, then look them up in the following table:
$$eval(φ,I)$$ $$eval(ψ,I)$$ $$P$$ $$φ∧ψ$$ $$φ∨ψ$$ $$φ→ψ$$ $$¬φ$$
T T $$I(P)$$ T T T F
T F $$I(P)$$ F T F F
F T $$I(P)$$ F T T T
F F $$I(P)$$ F F T T
Notation: We abbreviate $$eval(φ,I)$$ as $$φ[I]$$, and refer to it as the value of $$φ$$ in interpretation $$I$$.
We often write the value of $$φ$$ in all possible interpretations in a table:
$$I(P)$$ $$I(Q)$$ $$I(R)$$ $$φ[I]$$
T T T ?
T T F ?
T F T ?
T F F ?
F T T ?
F T F ?
F F T ?
F F F ?
This is called the truth table of $$φ$$. Each row corresponds to an interpretation.
Satisfaction, Validity, Entailment
Definition: If $$I$$ is an interpretation and $$φ$$ a formula, we say $$I$$ satisfies $$φ$$ if $$φ[I] = T$$. We write I ⊨ φ.
Definition: If $$I⊨φ$$ for all $$I$$, we say $$φ$$ is valid or a tautology. We write $$⊨φ$$.
Definition: If $$I ⊨ φ$$ for all $$I$$ that satisfy $$ψ$$, we say that $$ψ$$ entails $$φ$$, written $$ψ⊨φ$$.
Implication and entailment
Students often confuse implication ($$φ→ψ$$) and entailment $$φ ⊨ ψ$$. They are two different kinds of symbols; one is a symbol in the object language and the other is a statement in the meta language.
$$φ → ψ$$ is just a symbol; it does not have any definition. If I ask "what does $$φ → ψ$$ mean?", the answer is "$$φ → ψ$$". It's like asking "what does '$$aabac$$' mean?" The answer depends on context; the string itself has no intrinsic meaning.
$$φ ⊨ ψ$$ does have a definition, given above. It is a statement about the behavior of the evaluation function on the two formulas $$φ$$ and $$ψ$$.
$$φ → ψ$$ is a statement in the object language; $$φ ⊨ ψ$$ is a statement in the metalanguage.
However, the two notions are related, as we can prove in the metalanguage (we can't even talk about the relationship in the object language, because $$⊨$$ isn't a symbol available in the object language).
Claim: $$⊨ φ → ψ$$ if and only if $$φ ⊨ ψ$$.
Note: While talking about logic, it's important to be careful with your use of logic of terms and symbols. The following would be a meaningless thing to say:
Nonsense claim: $$(⊨ φ → ψ) ↔ (φ ⊨ ψ)$$.
Proof of claim: suppose that $$⊨ φ → ψ$$. We wish to show that $$φ⊨ψ$$; in other words, for every $$I$$ with $$I⊨φ$$, we also have $$I⊨ψ$$. Suppose this was false. Then there is some interpretation $$I$$ with $$eval(φ,I) = T$$ and $$eval(ψ,I) = F$$. But in this same interpretation, by definition, $$eval(φ→ψ,I) = F$$; this contradicts the fact that $$⊨ φ → ψ$$.
Conversely, suppose that $$φ ⊨ ψ$$. We now wish to show that $$⊨ φ→ψ$$. Consider any interpretation $$I$$. We want to show $$eval(φ→ψ,I) = T$$. We know that $$φ[I]$$ is either T or F. If it is F, then $$eval(φ→ψ,I) = T$$, by definition. If it is T, then since $$φ⊨ψ$$, $$eval(ψ,I)$$ must also be T. Therefore, by definition, $$eval(φ→ψ,I)$$ is true, as required.
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How do you find (d^2y)/(dx^2) for 2=2x^2-4y^2?
Feb 28, 2017
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 {y}^{2} - {x}^{2}}{4 {y}^{3}}$
Explanation:
differentiate all terms on both sides $\textcolor{b l u e}{\text{implicitly with respect to x}}$
$\Rightarrow 0 = 4 x - 8 y . \frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 x}{- 8 y} = \frac{x}{2 y}$
The second derivative is obtained by differentiating $\frac{\mathrm{dy}}{\mathrm{dx}}$
differentiate $\frac{\mathrm{dy}}{\mathrm{dx}} \text{ using the " color(blue)"quotient rule}$
$\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 y .1 - x .2 \frac{\mathrm{dy}}{\mathrm{dx}}}{4 {y}^{2}}$
$\textcolor{w h i t e}{\Rightarrow \frac{{d}^{y}}{{\mathrm{dx}}^{2}}} = \frac{2 y - 2 x \left(\frac{x}{2 y}\right)}{4 {y}^{2}}$
$\textcolor{w h i t e}{\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{2 y - \left(\frac{{x}^{2}}{y}\right)}{4 {y}^{2}}$
$\textcolor{w h i t e}{\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{2 {y}^{2} - {x}^{2}}{4 {y}^{3}}$
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### Section 6-3 : Solving Exponential Equations
6. Solve the following equation.
${7^{1 - x}} = {4^{3x + 1}}$
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For this equation there is no way to easily get both sides with the same base. Therefore, we’ll need to take the logarithm of both sides.
We can use any logarithm and the natural logarithm and common logarithm are usually good choices since most calculators can handle them. In this case there really isn’t any reason to use one or the other so we’ll use the natural logarithm (it’s easier to write two letters – ln versus three letters – log after all…).
Taking the logarithm (using the natural logarithm) of both sides gives,
$\ln {7^{1 - x}} = \ln {4^{3x + 1}}$ Show Step 2
Now we can use the logarithm property that says,
${\log _b}{x^r} = r{\log _b}x$
to move the exponents out of each of the logarithms. Doing this gives,
$\left( {1 - x} \right)\ln 7 = \left( {3x + 1} \right)\ln 4$ Show Step 3
Finally, all we need to do is solve for $$x$$. Recall that the equations at this step tend to look messier than we are used to dealing with. However, the logarithms in the equations at this point are just numbers and so we treat them as we treat all numbers with these kinds of equations. The work will be messier than we are used to but just keep in mind that the logarithms are just numbers!
Here is the rest of the work for this problem.
\begin{align*}\left( {1 - x} \right)\ln 7 & = \left( {3x + 1} \right)\ln 4\\ \ln 7 - x\ln 7 & = 3x\ln 4 + \ln 4\\ \ln 7 - \ln 4 & = 3x\ln 4 + x\ln 7\\ \ln 7 - \ln 4 & = \left( {3\ln 4 + \ln 7} \right)x\\ x & = \frac{{\ln 7 - \ln 4}}{{3\ln 4 + \ln 7}} = \frac{{1.945910149 - 1.386294361}}{{3\left( {1.386294361} \right) + 1.945910149}} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.091668262}}\end{align*}
Again, the work is messier than we are used to but it is not really different from work we’ve done previously in solving equations. The answer is also going to be “messier” in the sense that it is a decimal and is liable to almost always be a decimal for most of these types of problems so don’t worry about that.
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# Your question: How do you find partial pressure from moles?
Contents
The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.
## How do you find partial pressure from mole fraction?
The sum of the mole fractions of all the components present must equal 1. That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas.
## What is partial pressure formula?
As has been mentioned in the lesson, partial pressure can be calculated as follows: P(gas 1) = x(gas 1) * P(Total); where x(gas 1) = no of moles(gas 1)/ no of moles(total).
## How do you get pressure from moles?
Therefore, to convert the moles of gas to pressure, the scientist must know the volume and temperature of the gas, in addition to the number of moles of gas. The pressure is then given by P = nRT / V.
## How do you find partial pressure from moles and total pressure?
The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: Ptotal=P1+P2+… +Pn. + P n . The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.
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## How do u calculate pressure?
Pressure on surfaces
1. To calculate pressure, you need to know two things:
2. Pressure is calculated using this equation:
3. pressure = force ÷ area.
## How do you find moles with partial pressure and liters?
The equation used to calculate partial pressure: P = (nRT)/V, where P = partial pressure; n = number of moles of the gas; R = universal gas constant; T = temperature; and V = volume. Multiply the number of moles of the gas by the universal gas constant. R = 0.08206 (L_atm)/(mol_K).
## How do you find partial pressure from concentration?
We can rearrange this equation to give: n/V = P/RT. The units for n/V are moles per Liter, ie. concentration! So, any time you know the pressure contributed by a particular gas that is part of a gas mixture, you can calculate it’s concentration.
## How do I calculate moles?
How to find moles?
1. Measure the weight of your substance.
2. Use a periodic table to find its atomic or molecular mass.
3. Divide the weight by the atomic or molecular mass.
4. Check your results with Omni Calculator.
## How do you calculate atm pressure?
3. P = Pressure (atm) V = Volume (L) n = moles R = gas constant = 0.0821 atm•L/mol•K T = Temperature (Kelvin) The correct units are essential. Be sure to convert whatever units you start with into the appropriate units when using the ideal gas law.
## What is partial pressure in physics?
The partial pressure is the pressure the gas if the gas were in the same volume and temperature by itself. Dalton’s law states the total pressure of a mixture of ideal gases is the sum of the partial pressure of each individual gas. … Partial pressure is important in the fields of chemistry, physics, and biology.
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## How do you find moles with pressure and volume?
Multiply the volume and pressure and divide the product by the temperature and the molar gas constant to calculate moles of the hydrogen gas. In the example, the amount of hydrogen is 202,650 x 0.025 / 293.15 x 8.314472 = 2.078 moles.
## How do you find moles given pressure and temperature?
For example, if you want to calculate the volume of 40 moles of a gas under a pressure of 1013 hPa and at a temperature of 250 K, the result will be equal to: V = nRT/p = 40 * 8.3144598 * 250 / 101300 = 0.82 m³ .
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# Use Gauss-Jordan row reduction to solve the given system of equations. 9x−10y =9 36x−40y=36
Use Gauss-Jordan row reduction to solve the given system of equations.
9x−10y =9
36x−40y=36
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Step 1
The objective is to find a solution for the given system of equations using Gauss-Jordan elimination method.
The system of equations given is
9x−10y =9
36x−40y=36
The matrix representation of the above system of equations is
$\left[\begin{array}{cc}9& 10\\ 36& 40\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}9\\ 36\end{array}\right],AX=B$.
Then the augmented matrix $\left[A\mid B\right]$ becomes
$\left[\begin{array}{ccc}9& 10& 9\\ 36& 40& 36\end{array}\right]$
Step 2
Now, reducing the matrix $\left[A\mid B\right]$ step-by-step,(Gauss Jordan row reduction),
${R}_{2}\to \frac{1}{4}{R}_{2}⇒$ The matrix $\left[A\mid B\right]$ becomes $\left[\begin{array}{ccc}9& 10& 9\\ 9& 10& 9\end{array}\right]$
${R}_{2}\to {R}_{2}-{R}_{1}⇒$ The matrix $\left[A\mid B\right]$ becomes $\left[\begin{array}{ccc}9& 10& 9\\ 0& 0& 0\end{array}\right]$
The reduced matrix can be expressed as 9x+10y = 9.
Let y = k, any arbitrary value, the $x=\frac{9-10k}{9}$
This system has infinitely many solutions.
$\left(x,y\right)=\left(\frac{9-10k}{9},k\right)$
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# Mean Deviation
In statistics and mathematics, the deviation is a measure that is used to find the difference between the observed value and the expected value of a variable. In simple words, the deviation is the distance from the center point. Similarly, the mean deviation is used to calculate how far the values fall from the middle of the data set. In this article, let us discuss the definition, formula, and examples in detail.
## Mean Deviation Definition
The mean deviation is defined as a statistical measure that is used to calculate the average deviation from the mean value of the given data set. The mean deviation of the data values can be easily calculated using the below procedure.
Step 1: Find the mean value for the given data values
Step 2: Now, subtract the mean value from each of the data values given (Note: Ignore the minus symbol)
Step 3: Now, find the mean of those values obtained in step 2.
## Mean Deviation Formula
The formula to calculate the mean deviation for the given data set is given below.
Mean Deviation = [Σ |X – µ|]/N
Here,
Σ represents the addition of values
X represents each value in the data set
µ represents the mean of the data set
N represents the number of data values
| | represents the absolute value, which ignores the “-” symbol
## Mean Deviation for Frequency Distribution
To present the data in the more compressed form we group it and mention the frequency distribution of each such group. These groups are known as class intervals.
Grouping of data is possible in two ways:
1. Discrete Frequency Distribution
2. Continuous Frequency Distribution
In the upcoming discussion, we will be discussing mean absolute deviation in a discrete frequency distribution.
Let us first know what is actually meant by the discrete distribution of frequency.
### Mean Deviation for Discrete Distribution Frequency
As the name itself suggests, by discrete we mean distinct or non-continuous. In such a distribution the frequency (number of observations) given in the set of data is discrete in nature.
If the data set consists of values x1,x2, x3………xn each occurring with a frequency of f1, f2… fn respectively then such a representation of data is known as the discrete distribution of frequency.
To calculate the mean deviation for grouped data and particularly for discrete distribution data the following steps are followed:
Step I: The measure of central tendency about which mean deviation is to be found out is calculated. Let this measure be a.
If this measure is mean then it is calculated as,
where
$$\begin{array}{l}N=\sum_{i=1}^{n}\;f_{i}\end{array}$$
If the measure is median then the given set of data is arranged in ascending order and then the cumulative frequency is calculated then the observations whose cumulative frequency is equal to or just greater than N/2 is taken as the median for the given discrete distribution of frequency and it is seen that this value lies in the middle of the frequency distribution.
Step II: Calculate the absolute deviation of each observation from the measure of central tendency calculated in step (I)
StepIII: The mean absolute deviation around the measure of central tendency is then calculated by using the formula
If the central tendency is mean then,
In case of median
Let us look into the following examples for a better understanding.
### Mean Deviation Examples
Example 1:
Determine the mean deviation for the data values 5, 3,7, 8, 4, 9.
Solution:
Given data values are 5, 3, 7, 8, 4, 9.
We know that the procedure to calculate the mean deviation.
First, find the mean for the given data:
Mean, µ = ( 5+3+7+8+4+9)/6
µ = 36/6
µ = 6
Therefore, the mean value is 6.
Now, subtract each mean from the data value, and ignore the minus symbol if any
(Ignore”-”)
5 – 6 = 1
3 – 6 = 3
7 – 6 = 1
8 – 6 = 2
4 – 6 = 2
9 – 6 = 3
Now, the obtained data set is 1, 3, 1, 2, 2, 3.
Finally, find the mean value for the obtained data set
Therefore, the mean deviation is
= (1+3 + 1+ 2+ 2+3) /6
= 12/6
= 2
Hence, the mean deviation for 5, 3,7, 8, 4, 9 is 2.
Example 2:
In a foreign language class, there are 4 languages, and the frequencies of students learning the language and the frequency of lectures per week are given as:
Language Sanskrit Spanish French English No. of students(xi) 6 5 9 12 Frequency of lectures(fi) 5 7 4 9
Calculate the mean deviation about the mean for the given data.
Solution: The following table gives us a tabular representation of data and the calculations
## Frequently Asked Questions on Mean Deviation
Q1
### What does the mean deviation tell us?
The mean deviation gives information about how far the data values are spread out from the mean value.
Q2
### Mention the procedure to find the mean deviation.
The procedure to find the mean deviation are:
Step 1: Calculate the mean value for the given data
Step 2: Subtract the mean from each data value (Distance)
Step 3: Finally, find the mean for the distance
Q3
### What are the advantages of using the mean deviation?
The advantages of using mean deviation are:
It is based on all the data values provided, and hence it will give a better measure of dispersion.
It is easy to understand and calculate.
Q4
### What is the mean deviation about a median?
The mean deviation about the median is similar to the mean deviation about mean. Instead of calculating the mean for the given set of data values, find the median value by arranging the data values in the ascending order and then find the middle value. After finding the median, now subtract the median from each data value, and finally, take the average
Q5
### What are the three different ways to find the mean deviation?
The mean deviation can be calculated using
Individual Series
Discrete series
Continuous series
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# Exact Value of cos 27°
We will learn to find the exact value of cos 27 degrees using the formula of submultiple angles.
How to find the exact value of cos 27°?
Solution:
We have, (sin 27° + cos 27°)$$^{2}$$ = sin$$^{2}$$ 27° + cos$$^{2}$$ 27° + 2 sin 27° cos 27°
⇒ (sin 27° + cos 27°)$$^{2}$$ = 1 + sin 2 ∙ 27°
⇒ (sin 27° + cos 27°)$$^{2}$$ = 1 + sin 54°
⇒ (sin 27° + cos 27°)$$^{2}$$ = 1 + sin (90° - 36°)
⇒ (sin 27° + cos 27°)$$^{2}$$ = 1 + cos 36°
⇒ (sin 27° + cos 27°)$$^{2}$$ = 1+ $$\frac{√5 +1}{4}$$
⇒ (sin 27° + cos 27°)$$^{2}$$ = $$\frac{1}{4}$$ ( 5 + √ 5)
Therefore, sin 27° + cos 27° = $$\frac{1}{2}\sqrt{5 + \sqrt{5}}$$ …………….….(i)
[Since, sin 27° > 0 and cos 27° > 0)
Similarly, we have,
(sin 27° - cos 27°)$$^{2}$$ = 1 - cos 36°
⇒ (sin 27° - cos 27°)$$^{2}$$ = 1 - $$\frac{√5 + 1}{4}$$
⇒ (sin 27° - cos 27°)$$^{2}$$ = $$\frac{1}{4}$$ (3 - √5 )
Therefore, sin 27° - cos 27° = ± $$\frac{1}{2}\sqrt{3 - \sqrt{5}}$$ …………….….(ii)
Now, sin 27° - cos 27° = √2 ($$\frac{1}{√2}$$ sin 27˚ - $$\frac{1}{√2}$$ cos 27°)
= √2 (cos 45° sin 27° - sin 45° cos 27°)
= √2 sin (27° - 45°)
= -√2 sin 18° < 0
Therefore, from (ii) we get,
sin 27° - cos 27° = -$$\frac{1}{2}\sqrt{3 - \sqrt{5}}$$ …………….….(iii)
Now, subtracting (iii) and (i) we get,
2 cos 27° = $$\frac{1}{2}\sqrt{5 + \sqrt{5}}$$ + $$\frac{1}{2}\sqrt{3 - \sqrt{5}}$$
⇒ cos 27° = $$\frac{1}{4}(\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}})$$
Therefore, cos 27° = $$\frac{1}{4}(\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}})$$
`
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### Course: Precalculus>Unit 5
Lesson 6: Hyperbolas not centered at the origin
# Conic sections: FAQ
## What are conic sections?
Conic sections are the shapes you get when you slice a cone at different angles. There are four types of conic sections: ellipses, hyperbolas, parabolas, and circles.
## What's the difference between an ellipse and a hyperbola?
Both shapes are conic sections, but they differ in their geometry. An ellipse is a closed curve, shaped like an oval. A hyperbola consists of two separate curves (called "branches") that open away from each other.
## What are the key parts of an ellipse?
An ellipse has a few key parts:
• major axis: This is the longest diameter of the ellipse, and it runs through the center of the shape.
• minor axis: This is the shortest diameter of the ellipse, also running through the center of the shape, perpendicular to the major axis.
• foci: These are two points on the major axis that define the shape of the ellipse. The sum of the distances from the foci to any point on the ellipse is always constant.
• center: This is the midpoint of both the major and minor axes.
• vertices: These are the two points on the major axis that are furthest from the center.
• co-vertices: These are the two points on the minor axis that are furthest from the center.
## What are the key parts of a hyperbola?
• branches: A hyperbola is made up of two distinct branches or curves that extend away from each other.
• center: The point equidistant from the two branches, around which the hyperbola is symmetrical.
• transverse axis: The line segment connecting the two closest points on the two branches.
• conjugate axis: The line segment perpendicular to the transverse axis, passing through the center.
• foci: The two fixed points located inside each curve of the hyperbola such that, for any point on the hyperbola, the difference of distances to the foci is a constant.
• asymptotes: The two lines that the branches of the hyperbola will approach as they extend further and further away from the center. These asymptotes intersect at the center of the hyperbola.
## Where are conic sections used in the real world?
Conic sections show up in a lot of places! For example, the orbits of planets around the sun are elliptical. Hyperbolas are often used in the design of telescopes and antennas. Parabolas are important in physics, as they describe the shape of projectiles in flight.
## Want to join the conversation?
• I have got three questions:
1. how to calculate the area and the circumference of an
ellipse?
2. Do we need pi to calculate the are and circumference of
an ellipse?
3. Is there also a foci of a parabola?
• 1. The area of an ellipse is easy; if the major and minor radii are a and b, the area is just πab. Notice that in the special case of a circle, where a=b=r, this just becomes πr^2, as you would expect.
The circumference of an ellipse, interestingly, is much harder to calculate. There is not a simple formula for it, and the only way to calculate it is by approximating it with calculus.
2. π is a part of the expression for the area of an ellipse, yes. The circumference of an ellipse, again, can only be approximated. If you can approximate an integral, you don't need any mention of π.
3. Yes, a parabola has one focus. You can think of a parabola as an ellipse where one of the two foci has stayed fixed while the other focus moved out to infinity.
• Why do we need ellipses in the real world? Only to describe shapes or for something greater?
|
Lesson Objectives
• Demonstrate an understanding of how to simplify a radical
• Learn how to rationalize denominators with square roots
• Learn how to rationalize denominators with higher-level roots
## How to Rationalize the Denominator
In this lesson, we will learn how to completely simplify a radical and show how to rationalize the denominator.
### Simplified Form of a Radical
• The radicand contains no factor (except 1) that is a:
• Perfect Square » Square Root
• Perfect Cube » Cube Root
• Perfect Fourth » Fourth Root
• So on and so forth...
• The radicand cannot contain fractions
• There is no radical present in any denominator
So far, we have learned how to do everything except clear a radical from the denominator. What happens if we see a problem such as: $$\frac{5}{\sqrt{2}}$$ We can see that the square root of 2 is in the denominator, this violates our rules for a simplified radical. To clean up the problem, we use a process known as rationalizing the denominator. To do this, we just need to multiply both numerator and denominator by the square root of 2: $$\frac{5}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$$ Since we are multiplying by a complex form of 1, the value does not change. We have simply changed the form of the number to comply with our rules for a simplified radical. Notice how the denominator went from an irrational number (square root of 2) to a rational number (2). This is where the term "rationalizing the denominator" comes from. Let's look at some examples.
Example 1: Simplify each $$\frac{4\sqrt{2}}{5\sqrt{5}}$$ We have a radical (square root of 5) in the denominator. We can multiply the numerator and denominator by the square root of 5. This will rationalize the denominator: $$\frac{4\sqrt{2}}{5\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4\sqrt{10}}{25}$$ Example 2: Simplify each $$\frac{15\sqrt{2}}{8\sqrt{14}}$$ We have a radical (square root of 14) in the denominator. We can multiply the numerator and denominator by the square root of 14. This will rationalize the denominator: $$\frac{15\sqrt{2}}{8\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}} =$$ $$\frac{15 \cdot \sqrt{4} \cdot \sqrt{7}}{8 \cdot 14}=$$ $$\require{cancel} \frac{30\sqrt{7}}{112} = \frac{15 \cancel{30}\sqrt{7}}{56\cancel{112}} = \frac{15\sqrt{7}}{56}$$
### Rationalizing Higher-Level Roots
Let's suppose we see a problem such as: $$\frac{1}{\sqrt[3]{4}}$$ When we encounter a cube root, we need to create a perfect cube. We currently have the cube root of 4. If we think about 4:
4 = 2 • 2
This means we only need an additional factor of 2 or 8 to have a perfect cube. We can rationalize our denominator by multiplying both numerator and denominator by the cube root of 2: $$\frac{1}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \frac{1 \cdot \sqrt[3]{2}}{\sqrt[3]{8}} = \frac{\sqrt[3]{2}}{2}$$ Let's look at some additional examples.
Example 3: Simplify each $$\frac{\sqrt[3]{10}}{\sqrt[3]{32}}$$ Let's start by thinking about 32:
32 = 25
If we had one additional factor of 2, we would have 6 factors of 2. This would also be 4 cubed:
64 = 43
We can rationalize our denominator by multiplying the numerator and denominator by the cube root of 2: $$\frac{\sqrt[3]{10}}{\sqrt[3]{32}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} =$$ $$\frac{\sqrt[3]{10 \cdot 2}}{\sqrt[3]{64}} = \frac{\sqrt[3]{20}}{4}$$ Example 4: Simplify each $$\frac{\sqrt[4]{3}}{\sqrt[4]{25x^2}}$$ Let's start by thinking about 25:
25 = 5 • 5
We would need two additional factors of 5, to obtain a perfect fourth:
5 • 5 • 5 • 5 = 625
In terms of x2, we need two additional factors of x:
x2 • x2 = x4
We can rationalize our denominator by multiplying the numerator and denominator by the fourth root of 25x2: $$\frac{\sqrt[4]{3}}{\sqrt[4]{25x^2}} \cdot \frac{\sqrt[4]{25x^2}}{\sqrt[4]{25x^2}} =$$ $$\frac{\sqrt[4]{3 \cdot 25x^2}}{\sqrt[4]{625x^4}} = \frac{\sqrt[4]{75x^2}}{5x}$$
|
Find the length and width of a rectangle if its perimeter is 20cm and area is 24cm2.
What are the dimensions of a rectangle whose perimeter is 46cm and area is 126cm2.
Find the dimensions of a rectangle whose perimeter is 27cm and area is 44cm2.
Find the dimensions of all the rectangles whose perimeters in cm are twice the value of their areas in cm2.
## A Mathematics Lesson Starter Of The Day
Share
• Transum,
•
• The first three problems are in increasing order of difficulty. The first can probably be guessed, the second needs more of a strategy as the numbers are larger and the third involves a fraction so probably requires the application of a more generalised method. Refreshing the page generates three different rectangles to further consolidate any techniques learned in the first round. The fourth problem is only for Nobel prize candidates!
• Simon Fletcher, Bolton UK
•
• 52 cm rectangle perimeter 105 area. Is this correct? I've checked and can't find a solution.
• Transum,
•
• Hello Simon, so you are looking for two numbers (the height and the width) that add up to half the perimeter, 26cm, and multiply together to give an area of 105cm2.
We can write two simultaneous equations showing this information:
h + w = 26
hw = 105
We can combine the information in these two equations and eliminate one of the unknowns:
From the first equation we see that h = 26 – w
Substituting this into the second equation gives (26 – w)w = 105
Expanding the brackets gives 26w - w2=105
Rearranging gives w2 - 26w + 105 = 0
This is a quadratic equation which can be solved by factorising
(w – 5)(w – 21) = 0
So w could be 5cm in which case h would be 21cm or, w could be 21cm in which case h would be 5cm.
• Audrey Macmillan, @audproctor77
•
• Clarifying area/perimeter confusion using a Transum Starter. Great #mathschat on squares and rows.
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