text stringlengths 22 1.01M |
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### Connecting fractions and percentages
Percentages are another way of expressing fractions with denominators that are 10, 100, 1000 and so on.
Per centum comes from the Latin meaning 'out of one hundred'. So forty-five per cent or 45% means 45 out of one hundred and we write it as a fraction $$\dfrac{45}{100}.$$
If we tallied the colours of 100 cars that passed a point on the highway and 16 were red, we could say that 16 out of one hundred or $$\dfrac{16}{100} \text{or}\ 16\%$$ of those cars were red.
We might show this by shading 16 out of 100 boxes on a grid.
There were 50 out of one hundred cars that were green. Another way of saying this is that fifty hundredths were green. This is the same as the fraction $$\dfrac{50}{100}$$ and the percentage is written as 50%.
We can show 50% on a 100-grid by colouring 50 of the 100 squares.
For a percentage such as 145%, when we write that as a fraction $$\dfrac{145}{100},$$ we notice that it is an improper fraction and can be written as a mixed number as $$1\dfrac{45}{100}\ \text{or}\ 1\dfrac{9}{20}.$$
We have shaded 100 hundredths in the first box. $$\dfrac{100}{100}$$ = 1 or 100%. We have shaded 45 hundredths in the second box. This is $$\dfrac{45}{100}$$ or 45%. |
# How do you use the quotient rule to differentiate y= (x^2+4x)/(x+2)^2?
Jun 3, 2016
$\frac{8}{x + 2} ^ 3$
#### Explanation:
differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$
$f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} \text{ then } f ' \left(x\right) = \frac{h \left(x\right) . g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$
$\text{---------------------------------------------------------------------}$
$g \left(x\right) = {x}^{2} + 4 x \Rightarrow g ' \left(x\right) = 2 x + 4$
$h \left(x\right) = {\left(x + 2\right)}^{2} \Rightarrow h ' \left(x\right) = 2 \left(x + 2\right)$
$\text{-----------------------------------------------------------------------------}$
Substitute these values into f'(x)
f'(x)=((x+2)^2(2x+4)-(x^2+4x)2(x+2))/((x+2)^4
$= \frac{\left(x + 2\right) \left[\left(x + 2\right) \left(2 x + 4\right) - 2 \left({x}^{2} + 4 x\right)\right]}{x + 2} ^ 4$
=(cancel((x+2))[2x^2+4x+4x+8-2x^2-8x])/(cancel((x+2))(x+2)^3
$= \frac{8}{x + 2} ^ 3$ |
Finding Least Common Multiple (LCM): Prime Factorisation Method
# Finding Least Common Multiple (LCM): Prime Factorisation Method Video Lecture | Mathematics (Maths) Class 6
## Mathematics (Maths) Class 6
120 videos|301 docs|39 tests
## FAQs on Finding Least Common Multiple (LCM): Prime Factorisation Method Video Lecture - Mathematics (Maths) Class 6
1. How do you find the prime factorization of a number?
Ans. To find the prime factorization of a number, you need to break it down into its prime factors. Start by dividing the number by the smallest prime number possible, such as 2 or 3. Continue dividing until you cannot divide any further, and write down the prime factors you used. Repeat this process for each prime factor until you have completely broken down the number into its prime factors.
2. What is the least common multiple (LCM) of two or more numbers?
Ans. The least common multiple (LCM) of two or more numbers is the smallest multiple that is evenly divisible by each of the numbers. It is the smallest common multiple of the given numbers.
3. How do you find the LCM using the prime factorization method?
Ans. To find the LCM using the prime factorization method, you need to find the prime factorization of each number. Then, identify the highest power of each prime factor that appears in any of the factorizations. Finally, multiply all the prime factors raised to their respective highest powers to get the LCM.
4. Can the LCM of two numbers be less than the smaller number?
Ans. No, the LCM of two numbers cannot be less than the smaller number. The LCM must always be greater than or equal to the largest number among the given numbers. This is because the LCM is a multiple of both numbers, and a multiple cannot be smaller than the number itself.
5. What is the relationship between LCM and GCD (Greatest Common Divisor)?
Ans. The relationship between LCM and GCD is that their product is equal to the product of the two given numbers. In other words, if a and b are two numbers, then LCM(a, b) * GCD(a, b) = a * b. This relationship is known as the fundamental theorem of arithmetic and holds true for any two positive integers.
## Mathematics (Maths) Class 6
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# Solve Proportions (from McDougal Littell)
```Solving Proportions
EXAMPLE 1
Solving Proportions
Solve the proportion
4
=
12
20 .
x
METHOD - Use equivalent ratios.
4
4
=
12
12
5
20
x
5
12 X 5 = 60, so x = 60.
The solution is 60.
GUIDED PRACTICE
Solve the proportion using equivalent ratios.
1.
n = 6
24
4
2.
25
=
10
5
k
n
4
25
=
10
=
6
24
n = 1
k = 2
5
k
GUIDED PRACTICE
Solve the proportion using equivalent ratios.
3.
30
= 6
11
x
4.
10
=
4
s
12
30
= 6
11
x
10
=
4
x = 55
s
12
s = 30
EXAMPLE 2
Solving Using a Verbal Model
Boating: You are on a riverboat trip. You travel 5 miles
in 3 hours. At that same rate, how long will it take you
to travel 20 miles?
SOLUTION
Use a proportion. Let t represent the total time, in hours,
of the 20 mile trip.
Write a verbal
model.
Distance traveled
Substitute
values.
Solve the proportion
=
Total distance
Time traveled
20 mi
5 mi
=
th
3h
5 mi
3h
=
20 mi
12 h
Total time
GUIDED PRACTICE
for Example 2
5. What If? In Example 2, suppose you travel 4 miles
in 2 hours. At that same rate, how long will it take
you to travel 20 miles?
4 mi
=
20 mi
th
2h
20
4
=
t
2
t = 10h
Proportions around us
SCHOOL
The ratio of boys to girls in history class is 4 to
5. How many girls are in the class if there are
12 boys in the class?
Proportions at work
• FACTORIES
• A factory produces 6 motorcycles in 9 hours.
• Write a proportion and solve it to find how
many hours it takes to produce16 motorcycles.
Proportions at home
• COOKING
• A recipe that will make 3 pies calls for 7 cups
of flour. Write a proportion and solve it to find
how many pies can be made with 28 cups of
flour.
Proportions in sports |
## About "How to determine if a point is inside or outside a circle"
How to determine if a point is inside or outside a circle ?
Here we are going to see how to determine if a point is inside or outside a circle.
Length of the tangent = √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)
To check if the given point lie on the circle, inside the circle or out side the circle, we use the formula for length of tangent.
## How to verify that points lie on a circle ?
(i) If the length is 0.Then we can say, the given point must lie on the circle.
√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) = 0
## How to verify that points lie outside the circle ?
(ii) If the length is > 0 .Then we can say, the point must lie outside the circle.
√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) > 0
## How to verify that points lie inside the circle ?
(iii) If the length is < 0 .Then we can say, the point must lie inside the circle.
√ (x₁² + y₁² + 2gx₁ +2fy₁ + c) < 0
Example 1 :
Show that the point (2, 3) lies inside the circle x2 + y2 − 6x − 8y + 12 = 0.
Solution :
The length of the tangent PT from P(x1, y1) to the circle
x2 + y2 + 2gx +2fy + c = 0 is
PT = x1+ y12 + 2gx1 + 2fy1 + c
PT = √(22 + 32 − 6.2 − 8.3 + 12)
= √(4 + 9 − 12 − 24 + 12)
= − 11 < 0
The point (2, 3) lies inside the circle
Example 2 :
Show that the point (2,-1) lies outside the circle x²+y²- 6x - 8y + 12 = 0
Solution :
Length of the tangent
= √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)
Here x₁ = 2 and y₁ = -1
= √ (2)² + (-1)² - 6(2) - 8 (-1) + 12
= √ 4 + 1 - 16 + 8 + 12
= √5 + 8 + 12 + 1 - 8
= √ 26 - 8
= √18 = 3√2 units
The length of tangent is positive. So the given point lies outside the circle.
Example 3 :
Is the point (7, − 11) lie inside or outside the circle x2 + y2 − 10x = 0 ?
Solution :
To know that where does the given point lie in the circle, we have to find the length of tangent.
Length of the tangent
= √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)
Here x₁ = 7 and y₁ = -11
= √ x2 + y2 − 10x
= √72 + (-11)2 − 10(7)
= √49 + 121 - 70
= √ 170 - 70
= √100 = 10 > 0
Hence the given point (7, -11) outside the circle.
## Related pages
After having gone through the stuff given above, we hope that the students would have understood "How to determine if a point is inside or outside a circle".
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. |
Topic 1 - Algebra
# Topic 1 - Algebra - Topic1:Algebra , ofalgebraicformulae
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Topic 1: Algebra Meaning of terms variable and function Use of function notation Rules and techniques for the manipulation, transformation and evaluation of algebraic formulae Application of algebraic techniques for substitution and factorisation Solution of linear and quadratic equations by direct methods and simultaneous equations by elimination. Introduction: Algebra is a simple language, used to create mathematical models of real world situations and to handle problems that we can't solve using just arithmetic. Rather than using words, algebra uses symbols to make statements about things. In algebra, we often use letters to represent numbers. Algebraic Expressions An algebraic expression is one or more algebraic terms in a phrase. It can include variables, constants, and operating symbols, such as plus and minus signs. It's only a phrase, not the whole sentence, so it doesn't include an equal sign. E.g.: െ 5ݕ 9ݔݕ െ 13 In an algebraic expression, terms are the elements separated by the plus or minus signs. This example has four terms: , െ5ݕ , 9ݔݕ & െ 13 . Terms may consist of variables and coefficients, or constants. Variables In algebraic expressions, letters represent variables. In this expression, the variables are ݔ and ݕ . We call these letters " var iables" because the numbers they represent can vary —that is, we can substitute one or more numbers for the letters in the expression. Coefficients Coefficients are the number part of the terms with variables. In, െ 5ݕ 9ݔݕ െ 13 the coefficient of the first term is 7. The coefficient of the second term is ( 5), and the coefficient of the third term is 9.
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If a term Constan Constan That is, value ne value. In Functio A functi depend coordin A relatio words that are We can analyzin E.g.: 1. L S c m consists nts nts are the they're th ever chang n the abov ons: ion is a rela s on the va nate) may b on is only a focus on y e the same determine ng the orde Let A = {( 1 Since each coordinate of only var e terms in t e terms w ges, since t ve expressi ation (a se alue of the be paired w a function your x coo e they mu e whether ered pairs. ,3), ( 2 ,4), ( input in bo e), we can s riables, its the algebra ithout vari there are n on the con et of ordere e other var with only O if each inp rdinate (in ust all have the given . ( 3 ,5), ( 4 , 7) old letters say that th (O coefficien aic express ables. We no variable nstant term ed pairs) w riable. How ONE outpu put is only nput). If yo e the same relation is )} is paired w he set A is a ONE TO ON t is 1. E.g.: sion that co call them es in the te m is "( 13). where the v wever, in a ut (y coordi paired wit u have 2 o output or s a function with a diffe a function. NE RELATIO : ݔ ontain onl constants erm that ca ." value of on function, inate).
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# 5.6 Minimum and maximum values
Page 1 / 6
In general context, a function may have multiple minimum and maximum values in the domain of function. These minimum and maximum values are “local” minimum and maximum, which belongs to finite sub-intervals within the domain of function. The least minimum and greatest maximum in the domain of function are “global” minimum and maximum respectively in the entire domain of the function. Clearly, least and greatest values are one of the local minimum and maximum values. The minimum and maximum, which are not global, are also known as “relative” minimum and maximum.
Note : This module contains certain concepts relating to continuity, limits and differentiation, which we have not covered in this course. The topic is dealt here because minimum, maximum, least, greatest and range are important attributes of a function and its study is required to complete the discussion on function.
## Important observations and definitions
Let us consider a very general graphic representation of a function. Following observations can easily be made by observing the graph :
1: A function may have local minimum (C, E, G, I) and maximum (B,D,F,H) at more than one point.
2: It is not possible to determine global minimum and maximum unless we know function values corresponding to all values of x in the domain of function. Note that graph above can be defined to any value beyond A.
3: Local minimum at a point (E) can be greater than local maximum at other points (B and H).
4: If function is continuous in an interval, then pair of minimum and maximum in any order occur alternatively (B,C), (C,D), (D,E), (E,F) , (F,G) , (G,H) , (H,I).
5: A function can not have minimum and maximum at points where function is not defined. Consider a rational function, which is not defined at x=1.
$f\left(x\right)=\frac{1}{x-1};\phantom{\rule{1em}{0ex}}x\ne 1$
Similarly, a function below is not defined at x=0.
| x=1; x>0 f(x) = ||x = -1; x<0
Minimum and maximum of function can not occur at points where function is not defined, because there is no function value corresponding to undefined points. We should understand that undefined points or intervals are not part of domain - thus not part of function definition. On the other hand, minimum and maximum are consideration within the domain of function and as such undefined points or intervals should not be considered in the first place. Non-occurance of minimum and maximum in this context, however, has been included here to emphasize this fact.
6: A function can have minimum and maximum at points where it is discontinuous. Consider fraction part function in the finite domain. The function is not continuous at x=1, but minimum occurs at this point (recall its graph).
7: A function can have minimum and maximum at points where it is continuous but not differentiable. In other words, maximum and minimum can occur at corners. For example, modulus function |x| has its only minimum at corner point at x=0 (recall its graph).
## Extreme value or extremum
Extreme value or extremum is either a minimum or maximum value. A function, f(x), has a extremum at x=e, if it has either a minimum or maximum value at that point.
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
What is power set
Period of sin^6 3x+ cos^6 3x
Period of sin^6 3x+ cos^6 3x |
# $\sin{(30^°)}$
## Value
$\sin{30^°}$ $\,=\,$ $\dfrac{1}{2}$
### Introduction
When angle of a right triangle is $30^°$, the value of ratio of lengths of opposite side (perpendicular) to hypotenuse is called sine of $30$ degrees and it is simply called as sin $30$ degrees.
#### Sexagesimal System
Sine of angle $30$ degrees is written as $\sin{30^°}$ mathematically in sexagesimal system.
The exact value of sin $30$ degrees in fraction is equal to $\dfrac{1}{2}$.
$\sin{30^{°}} \,=\, \dfrac{1}{2}$
The exact value of sine $30$ degrees in decimal equals to $0.5$.
$\sin{30^{°}} \,=\, 0.5$
#### Circular system
It is written as $\sin{\Big(\dfrac{\pi}{6}\Big)}$ in circular system.
$\sin{\Big(\dfrac{\pi}{6}\Big)} \,=\, \dfrac{1}{2} = 0.5$
#### Centesimal System
It is written as $\sin{\Big(33\dfrac{1}{3}\Big)}$ in centesimal system.
$\sin{\Big(\dfrac{\pi}{6}\Big)} \,=\, \dfrac{1}{2} = 0.5$
### Proof
The value of $\sin{\Big(\dfrac{\pi}{6}\Big)}$ can be derived in trigonometry by geometrical approach. It can be derived possibly in two geometrical methods.
#### Theoretical Approach
Learn how to derive the value of sin of angle $30$ degrees in theoretical geometric method by considering a special relation between lengths of opposite side and hypotenuse when angle of right triangle is $30^°$.
#### Practical Approach
Learn how to derive the value of sine $30$ degrees in geometrical approach experimentally by constructing a right triangle with an angle of $30$ degrees. |
# What Is 5/82 as a Decimal + Solution With Free Steps
The fraction 5/82 as a decimal is equal to 0.060.
A Fraction (p/q) indicates that we have to divide the number p by q. therefore p is said to be the dividend and q is said to be the divisor. The three main categories of fractions are Proper Fraction, Improper Fraction, and Mixed Fraction.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 5/82.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 5
Divisor = 82
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 5 $\div$ 82
This is when we go through the Long Division solution to our problem, as shown in figure 1.
## 5/82 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 5 and 82, we can see how 5 is Smaller than 82, and to solve this division, we require that 5be Bigger than 82.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 5, which after getting multiplied by 10 becomes 50.
We take this 50 and divide it by 82; this can be done as follows:
 50 $\div$ 82 $\approx$ 0
Where:
82 x 0 = 0
This will lead to the generation of a Remainder equal to 50 – 0 = 50. Now this means we have to repeat the process by Converting the 50 into 500 and solving for that:
 500 $\div$ 82 $\approx$ 6
Where:
82 x 6 = 492
This, therefore, produces another Remainder which is equal to 500 – 492 = 8. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 80.
 80 $\div$ 82 $\approx$ 0
Where:
82 x 0 = 0
Finally, we have a Quotient generated after combining the three pieces of it as 0.060=z, with a Remainder equal to 80.
Images/mathematical drawings are created with GeoGebra. |
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2010 | May-Jun | (P2-9709/23) | Q#5
Question
The equation of a curve y=x3e-x.
i. Show that the curve has a stationary point where x = 3.
ii. Find the equation of the tangent to the curve at the point where x = 1.
Solution
i.
We are required to show that the curve has a stationary point where x=3.
We are given equation of the curve as;
A stationary point on the curve is the point where gradient of the curve is equal to zero;
Hence, gradient of the curve at point where x=3 must be ZERO.
We can find expression for gradient of the curve and equate it with ZERO to find the x-coordinate of the point.
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Therefore;
If and are functions of , and if , then;
If , then;
Let and ;
Rule for differentiation of is:
Rule for differentiation natural exponential function , or ;
Since at stationary point the gradient of the curve must be equal to ZERO.
Hence, x-coordinate of the stationary point is 3.
ii.
We are given that curve with equation and we are required to find the equation of the tangent to the curve at point where .
To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).
We have coordinates of point on the tangent (and curve) as (1,y).
Therefore, first we need to find the y-coordinate of the curve.
Corresponding value of y coordinate can be found by substituting value of x in equation of the curve.
Hence coordinates of the point on the curve where we are required to find equation of the tangent are (1,)
Next, we need slope of the tangent to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we find gradient (slope) of the curve at point (1,) where tangent intersects the curve, then we can find slope of the tangent.
Hence, we need gradient of the curve at point (1,).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve;
We have found in (i) that;
We are looking for gradient of the curve at point (1,).
Therefore;
Therefore slope of the curve at point (1,) is;
Hence, slope of the tangent to the curve at this point is;
Now we can write equation of the tangent as follows.
Point-Slope form of the equation of the line is; |
# 5.5 Zeros of polynomial functions (Page 7/14)
Page 7 / 14
A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?
3 meters by 4 meters by 7 meters
Access these online resources for additional instruction and practice with zeros of polynomial functions.
## Key concepts
• To find $\text{\hspace{0.17em}}f\left(k\right),\text{\hspace{0.17em}}$ determine the remainder of the polynomial $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ when it is divided by $\text{\hspace{0.17em}}x-k.\text{\hspace{0.17em}}$ This is known as the Remainder Theorem. See [link] .
• According to the Factor Theorem, $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a zero of $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ if and only if $\text{\hspace{0.17em}}\left(x-k\right)\text{\hspace{0.17em}}$ is a factor of $\text{\hspace{0.17em}}f\left(x\right).$ See [link] .
• According to the Rational Zero Theorem, each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient. See [link] and [link] .
• When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
• Synthetic division can be used to find the zeros of a polynomial function. See [link] .
• According to the Fundamental Theorem, every polynomial function has at least one complex zero. See [link] .
• Every polynomial function with degree greater than 0 has at least one complex zero.
• Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form $\text{\hspace{0.17em}}\left(x-c\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a complex number. See [link] .
• The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer.
• The number of negative real zeros of a polynomial function is either the number of sign changes of $\text{\hspace{0.17em}}f\left(-x\right)\text{\hspace{0.17em}}$ or less than the number of sign changes by an even integer. See [link] .
• Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division. See [link] .
## Verbal
Describe a use for the Remainder Theorem.
The theorem can be used to evaluate a polynomial.
Explain why the Rational Zero Theorem does not guarantee finding zeros of a polynomial function.
What is the difference between rational and real zeros?
Rational zeros can be expressed as fractions whereas real zeros include irrational numbers.
If Descartes’ Rule of Signs reveals a no change of signs or one sign of changes, what specific conclusion can be drawn?
If synthetic division reveals a zero, why should we try that value again as a possible solution?
Polynomial functions can have repeated zeros, so the fact that number is a zero doesn’t preclude it being a zero again.
## Algebraic
For the following exercises, use the Remainder Theorem to find the remainder.
$\left({x}^{4}-9{x}^{2}+14\right)÷\left(x-2\right)$
$\left(3{x}^{3}-2{x}^{2}+x-4\right)÷\left(x+3\right)$
$-106$
$\left({x}^{4}+5{x}^{3}-4x-17\right)÷\left(x+1\right)$
$\left(-3{x}^{2}+6x+24\right)÷\left(x-4\right)$
$\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$
$\left(5{x}^{5}-4{x}^{4}+3{x}^{3}-2{x}^{2}+x-1\right)÷\left(x+6\right)$
the third and the seventh terms of a G.P are 81 and 16, find the first and fifth terms.
if a=3, b =4 and c=5 find the six trigonometric value sin
pls how do I factorize x⁴+x³-7x²-x+6=0
in a function the input value is called
how do I test for values on the number line
if a=4 b=4 then a+b=
a+b+2ab
Kin
commulative principle
a+b= 4+4=8
Mimi
If a=4 and b=4 then we add the value of a and b i.e a+b=4+4=8.
Tariq
what are examples of natural number
an equation for the line that goes through the point (-1,12) and has a slope of 2,3
3y=-9x+25
Ishaq
show that the set of natural numberdoes not from agroup with addition or multiplication butit forms aseni group with respect toaaddition as well as multiplication
x^20+x^15+x^10+x^5/x^2+1
evaluate each algebraic expression. 2x+×_2 if ×=5
if the ratio of the root of ax+bx+c =0, show that (m+1)^2 ac =b^2m
By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
jaks
Ryan
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching |
# What Are The Factors Of 79: Prime Factors Of 79
Here you can read all about the factors of 79. In this article, you will be able to calculate the Prime factors for 79, factors for 79, pair factors for 79, and factors 79 in math with other numbers as well. If we do not get any remainder after the division with some particular number then it simply means that number is a factor. one by one I am going to explain in a very simple format. Continue your reading to know what are all the factors of 79. Below is the list of factors of 79.
## What Are The Factors Of 79
As we know factors are the number that divides the number without leaving any remainder. So all the factors of 79 are 1 and 79. Let’s see its mathematical explanation.
• 1(79➗1=79)
• 79(79➗79=1)
1, 79 are the factors for 79.
## Prime Factors Of 79
Now Here for the calculation of Prime factors 79, you should know the prime numbers up to 79. After that, we have to divide 79 with all prime numbers. We are going to calculate the prime factors by factorization of the 79 methods.
• 79➗2 = 39.5
• 79➗3 = 26.33
• 79➗5 = 79.5
• 79➗7 = 11.28
• 79➗13 = 6.07
• 79➗17 = 4.64
• 79➗19 = 4.15
• 79➗23 = 3.43
• 79➗29 = 2.72
• 79➗31 = 2.54
Continue in the same manner…
So the prime factor for the number 79 is 79 itself.
## Factors Of 79 In Pairs
The product of the numbers whose result will be 79, that numbers are nothing but the pair factors of 120.
• 1 x 79 = 79
• 79 x 1 = 79
So the pair factors for 79 are 1 and 79.
## Factors Of -79
The negative numbers of factors of 79 are called factors of -79.
Below You can see the numbers.
• -1(-79➗-1=-79)
• -79(-79➗-79=-1)
The factors for -79 are -1, and -79.
## Factors Of 79 And Other Numbers
Here is the list of common factors of 84 with other numbers.
### The Factors For 79 And 73
The factors for 79 and 73 is 1
• 79 factors = 1 and 79
• 73 factors = 1 and 73
HCF of 79 and 73 = 1
### The Factors Of 79 And 80
The factors for 79 and 80 is 1
• 79 factors = 1 and 79
• 80 factors = 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.
• HCF of 79 and 80 = 1
### The Factors For 79 And 90
The factors for 79 and 90 are 1
• 79 factors = 1 and 79
• 90 factors = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
HCF of 79 and 90 = 1
### The Factors For 79 And 200
The factors for 79 and 200 are 1
• 79 factors = 1 and 79
• 200 factors = 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, and 200.
HCF of 79 and 200 = 1
All factors for the numbers can be seen here on Factorsweb
Click here – What Are The Factors Of 84: Prime Factors Of 84
## FAQ
### What Is The Multiple Of 79?
Solution: The first 10 multiples of 79 are 79, 158, 237, 316, 395, 474, 553, 632, 711 and 790.
### Is 79 Odd Prime Number?
If we have taken numbers from 1, 2, 3…for checking factors of 79, we found that 79 has factors 1 and the number 79 itself. Hence, we must say that 79 is a prime number. Thus, 79 is the prime number. Here, 79 is the odd prime number.
### What Numbers Can 79 Be Divided By?
When we list them out like this it’s easy to see that the numbers which 79 is divisible by are 1 and 79. You might be interested to know that all of the divisor numbers listed above are also known as the Factors of 79. Not only that, but the numbers can also be called the divisors of 79.
### Is 79 A Perfect Number?
79 is not a perfect square.
### Is 77 Or 79 A Prime Number?
What is the list of prime numbers from 1 to 100? Prime numbers from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
The answer to your question is yes. We have calculated all the numbers that 79 is evenly divisible by. The numbers that 79 is divisible by are 1 and 79.
### Why Is 79 A Prime Number?
Yes, 79 is a prime number. The number 79 is divisible only by 1 and the number itself. For a number to be classified as a prime number, it should have exactly two factors. Since 79 has exactly two factors, i.e. 1 and 79, it is a prime number.
### What Are The Sum Of Three Primes To 79?
The smallest prime that is the sum of three distinct “April fools primes” (9 + 21 + 49 = 79).
## Conclusion
Here I have given you detailed information about the factors of 79. In addition to that, I have also explained the prime factors for 79 list, factors for 79 in pairs, factors for -79, and factor numbers of 79 with another number. Now you know what are factors of 79 and also -1 and -79 are the negative factors for 79 number. So this was all about the factors of 79 and thank you for reading this article.
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What is 79 factor of 79 |
# What is 1/12 + 2/11?
Here's how you add
1 12
+
2 11
## Step 1
Of course, you can't add two fractions if the denominators (bottom numbers) don't match. To get a common denominator, multiply the denominators together. Then we fix the numerators by multiplying each one by their other term's denominator.
Now you multiply 1 by 11, and get 11, then we multiply 12 by 11 and get 132.
Do the same for the second term. We multiply 2 by 12, and get 24, then multiply 12 by 11 and get 132.
The problem now has new fractions to add:
11 132
+
24 132
## Step 2
Since our denominators match, we can add the numerators.
11 + 24 = 35
This yields the answer
35 132
## Step 3
The last step is to reduce the fraction if we can.
To find out, we try dividing it by 2...
No good. So next you try the next prime number, which is 3...
No good. So next you try the next prime number, which is 5...
No good. So next you try the next prime number, which is 7...
No good. So next you try the next prime number, which is 11...
No good. So next you try the next prime number, which is 13...
No good. So next you try the next prime number, which is 17...
No good. So next you try the next prime number, which is 19...
No good. So next you try the next prime number, which is 23...
No good. So next you try the next prime number, which is 29...
No good. So next you try the next prime number, which is 31...
No good. So next you try the next prime number, which is 37...
No good. 37 is larger than 35. So we're done reducing.
And we're done! Here's the final answer to 1/12 + 2/11
1 12
+
2 11
=
35 132
© 2014 Randy Tayler |
## Finding the Area of Rectangular Figures in Real World Mathematical Problems: Grade 3 Lesson
written by: Donna Ventura • edited by: Carly Stockwell • updated: 10/14/2013
Students will investigate the concept of area of rectangular figures in real world and mathematical problems by multiplying the two sides of the figures. They will also find the lengths of the two sides and areas of the rectangular figures.
• slide 1 of 8
Lesson Objective: The lesson is aligned to the Common Core State Standards for Mathematics – 3.MD.7b Geometric Measurement – Multiply side lengths to find areas of rectangles in context of solving real world and mathematical problems.
Materials Required: graph paper, calculator
• slide 2 of 8
### Lesson Procedure:
Finding Area of Rectangular Figures in Real World Mathematical Problems
The drawing of the front view of a tower is shown on the graph paper below. The tower has 3 levels. Each level is the shape of a rectangular figure. Each unit is equal to one meter.
1. Identify the lengths of the two sides of the rectangular figures.
2. Find the areas of the rectangular figures.
• slide 3 of 8
3. Complete the chart.
• slide 4 of 8
### Individual or Group Work:
The drawing of the front view of a house is shown on the graph paper at right. The house has a door, 2 windows, and a roof. All the shapes are rectangular figures. Each unit is equal to one meter.
1. Identify the lengths of the two sides of the rectangular figures.
2. Find the areas of the rectangular figures.
• slide 5 of 8
Complete the chart
• slide 6 of 8
3. Rhys’ bedroom is in the shape of a rectangle. The length of the bedroom is 10 feet. The width of the bedroom is 12 feet. What is the area of the bedroom, in square feet?
4. Jasper has a pond in his back yard. The pond is in the shape of a rectangle. The length of the pond is 6 feet. The width of the pond is 8 feet. What is the area of the pond, in square feet?
5. Jack planted a flower garden. The garden is in the shape of a rectangle. The length of the garden measures 4 feet. The width of the garden measures 5 feet. What is the area of the garden, in square feet?
• slide 7 of 8
The area of Rhys’ bedroom is 120 square feet.
The area of Jasper’s pond is 48 square feet.
The area of Jack’s garden is 20 square feet.
Students should be able to determine the area of rectangular figures in real world and mathematical problems by multiplying the two sides of the rectangular figures or the lengths and the widths of rectangular figures.
• slide 8 of 8 |
# 8. Integration by Trigonometric Substitution
by M. Bourne
In this section, we see how to integrate expressions like
int(dx)/((x^2+9)^(3//2))
Depending on the function we need to integrate, we substitute one of the following trigonometric expressions to simplify the integration:
For sqrt(a^2-x^2), use x =a sin theta
For sqrt(a^2+x^2), use x=a tan theta
For sqrt(x^2-a^2), use x=a sec theta
After we use these substitutions we'll get an integral that is "do-able".
Take note that we are not integrating trigonometric expressions (like we did earlier in Integration: The Basic Trigonometric Forms and Integrating Other Trigonometric Forms and Integrating Inverse Trigonometric Forms.
Rather, on this page, we substitute a sine, tangent or secant expression in order to make an integral possible.
### Example 1
int(dx)/((x^2+9)^(3//2))
We can write the question as int(dx)/((3^2+x^2)^(3//2))
It's now in the form of the second substitution suggestion given above, that is:
For sqrt(a^2+x^2), use x=a tan theta,
with a=3.
So we'll put x=3 tan theta and this gives dx=3 sec^2 theta d theta
We make the first substitution and simplify the denominator of the question before proceeding to integrate.
We'll need to use the following:
(a^2)^(3//2) = a^3.
Here's a number example demonstrating this expression:
9^(3//2) = (sqrt9)^3 = 3^3 = 27
This is a well-known trigonometric identity:
tan^2 θ + 1 = sec^2 θ
So we have:
(x^2+9)^(3//2)=((3 tan theta)^2+9)^(3//2)
=(9 tan^2 theta+9)^(3//2)
=(9[tan^2 theta+1])^(3//2)
=9^(3//2)[tan^2 theta+1]^(3//2)
=27[sec^2 theta]^(3//2)
=27[sec theta]^3
=27 sec^3 theta
Now, substituting
dx=3 sec^2 theta d theta
and
(x^2+9)^(3//2)=27 sec^3 theta
into the given integral gives us:
int(dx)/((x^2+9)^(3//2))=int(3 sec^2 theta d theta)/(27 sec^3 theta)
=1/9int(d theta)/(sec theta)
=1/9int cos theta d theta
=1/9 sin theta+K
We now need to get our answer in terms of x (since the question was in terms of x).
Since we let x = 3 tan θ, we get
tan theta=x/3
and we can draw a triangle to find the expression for sin θ in terms of x:
From the triangle we see that
sin theta=x/sqrt(x^2+9)
Therefore, we can conclude that the answer for our integral is 1/9 times this last expression.:
int(dx)/((x^2+9)^(3//2))=1/9sin theta+K
=1/9(x/(sqrt(x^2+9)))+K
=x/(9sqrt(x^2+9))+K
### Example 2
int_4^5(sqrt(x^2-16))/(x^2)dx
int_4^5(sqrt(x^2-16))/x^2 dx
This question contains a square root which is in the form of the 3rd substitution suggestion given at the top, that is:
For sqrt(x^2-a^2), use x=a sec theta
So we have a=4 and we let
x = 4 sec θ
and this gives
x^2= 16 sec^2 θ
and
dx = 4 sec θ tan θ d θ
Simplifying the square root part:
sqrt(x^2-16) =sqrt(16 sec^2 theta-16)
=sqrt(16(sec^2theta-1))
=sqrt16sqrt(tan^2theta)
=4 tan theta
Substituting
dx = 4 sec theta tan theta d theta,
x^2= 16 sec^2 theta and
sqrt(x^2-16)=4 tan theta
into the given integral gives us the following. (We take the indefinite case first and then do the substitution of upper and lower limits later, to make the writing a bit easier.)
int(sqrt(x^2-16))/x^2dx =int((4 tan theta))/(16 sec^2 theta)(4 sec theta tan theta) d theta
=int(16 tan^2 theta sec theta)/(16 sec^2 theta) d theta
=int(tan^2 theta)/(sec theta) d theta
=int(sec^2theta-1)/(sec theta) d theta
=int((sec^2 theta)/(sec theta)-1/(sec theta)) d theta
=int (sec theta-cos theta) d theta
=[ln |sec theta+tan theta|-sin theta]+K
Now, our question was a definite integral, so we need to either re-express our answer in terms of the original variable , x, or we could work it using theta.
#### Changing back to x
Earlier, we let x = 4 sec θ, so we get sec theta=x/4 (or cos theta = 4/x).
We draw an appropriate triangle like we did earlier:
sqrt(x^2-16)
Triangle to find sin theta and tan theta in terms of x.
We can see that:
tan theta=(sqrt(x^2-16))/4
and
sin theta=(sqrt(x^2-16))/x
Therefore, we can conclude that:
int_4^5(sqrt(x^2-16))/x^2 dx =[ln|sec theta+tan theta|-sin theta]_(theta=?)^(theta=?)
=[ln|x/4+(sqrt(x^2-16))/4| {:-(sqrt(x^2-16))/x]_4^5
=[ln|5/4+3/4|-3/5]-[ln|1+0|-0]
=[ln|2|-3/5]-[0]
=0.09315
(I've put theta=? as the upper and lower limits in the first line above because we don't know those limits in terms of theta, and we don't need to calculate them since we revert to x as our variable.)
#### Leaving it in terms of theta
Since sec theta=x/4, then as x ranges from 4 to 5, then sec theta will range from 1 to 1.25.
So the required upper and lower limits for theta (these are the missing question mark "theta=?" values in the above answer) will be
theta="arcsec"(1)= 0
and
theta="arcsec"(1.25)=0.6435011
Returning to our answer in theta, and substituting our upper and lower values gives:
[ln |sec theta+tan theta|-sin theta]_(theta=0)^(theta=0.6435011)
=[ln |sec 0.6435011+tan 0.6435011| {:-sin 0.6435011] -[ln |sec 0+tan 0|-sin 0]
=0.09315, which is the same as our earlier answer.
In this example, both approaches (leaving it in terms of theta or changing back to x) is about the same amount of work.
Continues below
## Exercises
Integrate each of the given functions:
1. intsqrt(16-x^2)dx
intsqrt(16-x^2) dx
This question is in the form of the first substitution suggestion in this section, that is,
For sqrt(a^2-x^2), use x =a sin theta
So we have a=4, x= 4 sin θ, and dx = 4 cos θ dθ.
Substituting and simplifying the square root part first:
sqrt(16-x^2) =sqrt(16-16 sin^2 theta)
=sqrt(16(1-sin^2 theta))
=4sqrt(cos^2 theta)
=4 cos theta
Substituting into the integral gives:
intsqrt(16-x^2) dx =int4 cos theta(4 cos theta d theta)
=int16 cos^2 theta d theta
=16int1/2(cos 2 theta+1)d theta
=8int(cos 2 theta+1)d theta
=8(1/2sin 2 theta+theta)+K
=8(sin theta cos theta + theta)+K
=8(x/4(sqrt(16-x^2))/4+arcsin {:x/4:})+K
=(xsqrt(16-x^2))/2+8 arcsin{:x/4:}+K
The second-last step comes from drawing a triangle, using sin theta = x/4 in this case, as follows:
sqrt(16-x^2)
Triangle to find theta, sin theta and cos theta in terms of x.
Quite often we can get different forms of the same final answer! That is, math software (or another human) can produce an answer which is actually correct, but in a different formto the one given here since. If this happens, don't panic! Just check your solution perhaps by substituting various values for x, or (better), drawing the graph using software.
2. int(3 dx)/(xsqrt(4-x^2))
int(3 dx)/(xsqrt(4-x^2))
This contains a sqrt(a^2-x^2) term, so we will use a substitution of x =a sin theta.
So a=2, and we let x = 2 sin θ, so dx = 2 cos θ dθ.
Substituting and simplifying the square root gives:
sqrt(4-x^2) =sqrt(4-4 sin^2 theta)
=sqrt(4(1-sin^2 theta))
=2sqrt(cos^2 theta)
=2 cos theta
This time our triangle will use sin theta = x/2, as follows:
sqrt(4-x^2)
Triangle to find csc theta and cot theta in terms of x.
Substituting everything into the integral gives:
int(3 dx)/(xsqrt(4-x^2)) = int(3(2 cos theta\ d theta))/((2 sin theta)(2 cos theta))
=3/2int(d theta)/(sin theta)
=3/2intcsc theta d theta
=3/2ln |csc theta-cot theta|+K
=3/2ln |2/x-(sqrt(4-x^2))/x|+K
=3/2ln|(2-sqrt(4-x^2))/x|+K
3. int(dx)/(sqrt(x^2+2x))
int(dx)/(sqrt(x^2+2x))
Firstly, note that
x^2+2x=(x+1)^2-1
If we put u = x + 1, then du = dx and our integral becomes:
int(dx)/(sqrt(x^2+2x))=int(du)/(sqrt(u^2-1))
Now, we use u = sec θ and so du = sec θ tan θ dθ
The square root becomes:
sqrt(u^2-1) =sqrt(sec^2theta-1) =sqrt(tan^2 theta) =tan theta
The triangle in this case starts with x+1=sec theta (that is, cos theta = 1/(x+1)), and is as follows:
sqrt(x^2+2x)
Triangle to find sec theta and tan theta in terms of x.
Returning to our integral, we have:
int(dx)/(sqrt(x^2+2x)) =int(du)/(sqrt(u^2-1))
=int(sec theta tan theta d theta)/(tan theta)
=int sec theta d theta
=ln |sec theta+tan theta|+K
=ln |x+1+sqrt(x^2+2x)|+K`
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# 4.5: Derivatives of the Trigonometric Functions
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All of the other trigonometric functions can be expressed in terms of the sine, and so their derivatives can easily be calculated using the rules we already have. For the cosine we need to use two identities,
\eqalign{ \cos x &= \sin(x+{\pi\over2}),\cr \sin x &= -\cos(x+{\pi\over2}).\cr }
Now:
\eqalign{ {d\over dx}\cos x &= {d\over dx}\sin \left(x+{\pi\over2}\right) = \cos \left(x+{\pi\over2}\right )\cdot 1 = -\sin x\cr {d\over dx}\tan x &= {d\over dx}{\sin x\over \cos x}= {\cos^2 x + \sin^2 x\over \cos^2 x}={1\over \cos^2 x}=\sec^2 x\cr {d\over dx}\sec x &= {d\over dx}(\cos x)^{-1}= -1(\cos x)^{-2}(-\sin x) = {\sin x \over \cos^2 x} = \sec x\tan x.\cr }
The derivatives of the cotangent and cosecant are similar and left as exercises. |
# Power Rule Implies Chain Rule
Having developed the Product Rule $d\left( uv \right)=u{v}'+{u}'v$ and the Power Rule $\frac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ for derivatives in your class, you can explore similar rules for the product of more than two functions and suddenly the Chain Rule will appear.
For three functions use the associative property of multiplication with the rule above:
$d\left( uvw \right)=d\left( \left( uv \right)w \right)=u\cdot v\cdot dw+w\cdot d(uv)=u\cdot v\cdot dw+w\left( udv+vdu \right)$
So expanding with a slight change in notation:
$d\left( uvw \right)=uv{w}'+u{v}'w+u'vw$
For four factors there is a similar result:
$d\left( uvwz \right)=uvw{z}'+uv{w}'z+u{v}'wz+{u}'vwz$
Exercise: Let ${{f}_{i}}$ for $i=1,2,3,...,n$ be functions. Write a general formula for the derivative of the product ${{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}}$ as above and in sigma notation
$d\left( {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} \right)={{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{{f}'}_{n}}+{{f}_{1}}{{f}_{2}}{{{f}'}_{3}}\cdots {{f}_{n}}+{{f}_{1}}{{{f}'}_{2}}{{f}_{3}}\cdots {{f}_{n}}+\cdots +{{{f}'}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}}$
$\displaystyle d\left( {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} \right)=\sum\limits_{i=1}^{n}{\frac{{{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}}}{{{f}_{i}}}{{{{f}'}}_{i}}}$
This idea may now be used to see the Chain Rule appear. Students may guess that $d{{\left( f \right)}^{4}}=4{{\left( f \right)}^{3}}$, but wait there is more to it.
Write ${{\left( f \right)}^{4}}=f\cdot f\cdot f\cdot f\text{ }$. Then from above
$d{{\left( f \right)}^{4}}=d\left( f\cdot f\cdot f\cdot f\text{ } \right)=f\cdot f\cdot f\cdot {f}'+f\cdot f\cdot {f}'\cdot f+f\cdot {f}'\cdot f\cdot f+{f}'\cdot f\cdot f\cdot f$
$d{{\left( f \right)}^{4}}=4{{\left( f \right)}^{3}}{f}'\text{ }$
Looks just like the power rule, but there’s that “extra” ${f}'$. Now you are ready to explain about the Chain Rule in the next class. |
# Sets and Subsets
The lesson introduces the important topic of sets, a simple idea that recurs throughout the study of probability and statistics.
## Set Definitions
• A set is a well-defined collection of objects.
• Each object in a set is called an element of the set.
• Two sets are equal if they have exactly the same elements in them.
• A set that contains no elements is called a null set or an empty set.
• If every element in Set A is also in Set B, then Set A is a subset of Set B.
## Set Notation
• A set is usually denoted by a capital letter, such as A, B, or C.
• An element of a set is usually denoted by a small letter, such as x, y, or z.
• A set may be described by listing all of its elements enclosed in braces. For example, if Set A consists of the numbers 2, 4, 6, and 8, we may say: A = {2, 4, 6, 8}.
• The null set is denoted by {} or .
• Sets may also be described by stating a rule. We could describe Set A from the previous example by stating: Set A consists of all the even single-digit positive integers.
## Sets and Probability
As we learned in the previous lesson, probability is all about statistical experiments. When a researcher conducts a statistical experiment, he or she cannot know the outcome in advance. The outcome is determined by chance.
However, if the researcher can list all the possible outcomes of the experiment, it may be possible to compute the probability of a particular outcome. The list of all possible outcomes from a statistical experiment is called the sample space. And a particular outcome or collection of outcomes is called an event.
You can see that a sample space is a type of set. It is a well-defined listing of all possible outcomes from a statistical experiment. And an event in a statistical experiment is a subset of the sample space.
## Set Operations
Suppose we have a sample space S defined as follows: S = {1, 2, 3, 4, 5, 6}. Within that sample space, suppose we define two subsets as follows: X = {1, 2} and Y= {2, 3, 4}.
• The union of two sets is the set of elements that belong to one or both of the two sets. Thus, if X is {1, 2} and Y is {2, 3, 4}, the union of sets X and Y is:
X Y = {1, 2, 3, 4}
Symbolically, the union of X and Y is denoted by X Y.
• The intersection of two sets is the set of elements that are common to both sets. Thus, if X is {1, 2} and Y is {2, 3, 4}, the intersection of sets X and Y is:
X Y = {2}
Symbolically, the intersection of X and Y is denoted by X Y.
• The complement of an event is the set of all elements in the sample space but not in the event. Thus, if the sample space is {1, 2, 3, 4, 5, 6}, and Y is {2, 3, 4}, the complement of set Y is:
Y' = {1, 5, 6}
On this website, we denote the complement of set Y as Y'. In other places, you may see the complement of set Y denoted as Yc.
## Sample Problems
1. Describe the set of vowels.
If A is the set of vowels, then A could be described as A = {a, e, i, o, u}.
2. Describe the set of positive integers.
Since it would be impossible to list all of the positive integers, we need to use a rule to describe this set. We might say A consists of all integers greater than zero.
3. Set A = {1, 2, 3} and Set B = {3, 2, 1}. Is Set A equal to Set B?
Yes. Two sets are equal if they have the same elements. The order in which the elements are listed does not matter.
4. What is the set of men with four arms?
Since all men have two arms at most, the set of men with four arms contains no elements. It is the null set (or empty set).
5. Set A = {1, 2, 3} and Set B = {1, 2, 4, 5, 6}. Is Set A a subset of Set B?
Set A would be a subset of Set B if every element from Set A were also in Set B. However, this is not the case. The number 3 is in Set A, but not in Set B. Therefore, Set A is not a subset of Set B. |
# Video: Identifying Common Multiples of Numbers up to 999
Find all the common multiples of 10 and 9 up to and including 450.
04:18
### Video Transcript
Find all the common multiples of 10 and nine up to and including 450.
We need to find all the numbers that are multiples of both 10 and nine. Because they’re multiples of two numbers, we call them common multiples of 10 and nine. In other words, they must be in the 10 times table. And they must also be in the nine times table. And we’re told where we need to stop, which is 450. To begin with, let’s think about multiples of nine. What are the multiples of nine up to 10 times nine. We know that adding nine each time is the same as adding 10 and taking away one. And we can use this to spot a pattern with the nine times table. If we add one in the tens column each time and then take away one from the digit in the ones column, we can work out the next multiple of nine each time. add one 10 take away one one, add one 10 take away one one, and so on.
Five nines of 45, six nines of 54, 63, 72, 81, and then 10 times nine, we know, is 90. We can see that the number of tens increases by one each time. And we could also see that their digits in the ones place decrease by one each time. Just a quick way of working out the multiples in line. Multiples of 10 are a lot easier to spot. We know that they always end in a zero. And so the only multiple of nine between nine and 90, that’s also a most worth 10, is 90 itself. This is the same as one lot of 10 times nine. But 90 isn’t actually the first common multiple of 10 and nine. The first multiple that both numbers have in common is actually zero. And we can find that out by multiplying zero times 10 times nine. Zero is in the nine times table. It’s the same as zero times nine. And it’s also in the 10 times table. It’s the same as zero times 10.
So we can see a pattern here to help us find all the common multiples of 10 and nine up to and including 450. We just need to count in 90s. Two lots of 90 or two lots of 10 times nine equals 180. Can you see a pattern with these numbers? They look familiar, don’t they? They’re actually multiples of nine with a zero on the end. In other words, they’re multiples of nine that have been multiplied then by 10. 180 is in the nine times table, it’s the same as 20 times nine, and is also in the 10 times table, it’s the same as 18 times 10. So these numbers are all common multiples of 10 and nine. Let’s continue the pattern. We know three multiplied by nine equals 27. So three more supplied by 10 times nine or 90 equals 270. This is a number that appears in both times tables. Four nines are 36. So four times 90 or four times 10 times nine equals 360.
And we also need to include one more common multiple because we’re told to find all the common multiples of 10 and nine up to and including 450. So we need to include 452 which is the same as five times 90 or five times 10 times nine. We found that all the numbers that are both multiples of 10 and nine are actually multiples of 90. So we could just count in 90s. We knew we had to include zero because that’s a number in both the 10 and the nine times tables. We knew we also had to include 450. So all the common multiples of 10 and nine up to and including 450 are zero, 90, 180, 270, 360, and 450. |
# Five-number Summary
## Five-number Summary :
A five-number summary is a method of summarizing a dataset by providing a concise description of its key features. It consists of the minimum value, the maximum value, the median, the first quartile, and the third quartile. These five numbers are used to provide a rough sketch of the data and to identify any potential outliers.
For example, let’s say we have a dataset of 100 numbers. We can use the five-number summary to quickly summarize the data and identify any potential outliers. To do this, we would first order the numbers from least to greatest. Then, we would calculate the minimum value, the maximum value, the median, the first quartile, and the third quartile.
The minimum value is the lowest number in the dataset. For our example, the minimum value would be 1.
The maximum value is the highest number in the dataset. For our example, the maximum value would be 100.
The median is the middle number in the dataset when the numbers are ordered from least to greatest. For our example, the median would be 50.
The first quartile is the number that divides the lower half of the dataset into two parts. For our example, the first quartile would be 25.
The third quartile is the number that divides the upper half of the dataset into two parts. For our example, the third quartile would be 75.
We can use these five numbers to quickly summarize the data and identify any potential outliers. For example, if we see a number that is much lower than 1 or much higher than 100, we would know that it is an outlier.
Let’s look at another example to further understand the five-number summary.
Suppose we have a dataset of test scores for a class of 20 students. The scores are as follows:
80, 85, 90, 95, 90, 85, 80, 75, 70, 65, 60, 55, 50, 45, 40, 35, 30, 25, 20, 15
To find the five-number summary for this dataset, we would first order the numbers from least to greatest. This would give us the following:
15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 80, 85, 85, 90, 90, 95
Next, we would calculate the minimum value, the maximum value, the median, the first quartile, and the third quartile.
The minimum value is 15.
The maximum value is 95.
The median is 60.
The first quartile is 40.
The third quartile is 85.
Using these five numbers, we can quickly summarize the data and identify any potential outliers. For example, if a student scored a 100 on the test, we would know that this is an outlier because it is much higher than the maximum value of 95.
Overall, the five-number summary is a useful tool for quickly summarizing a dataset and identifying potential outliers. It provides a concise description of the key features of the data and allows us to easily visualize the distribution of the data. |
# Subtracting Mixed FractionsCalculator
## Online Subtracting Mixed Fractions Calculator
### How to Subtract Mixed Fractions With Common Denominators
• F
You have to know that mixed numbers are a combination of a whole number and a proper fraction. Even though the plus signs aren't usually showing in mixed numbers (between the whole number and the proper fraction), there are actually added together.
Example:
3
4 7
= 3 +
4 7
• 1
If both fractions have the same denominator, they can be easily subtracted by subtracting the whole numbers together, the numerators together, and the denominator remaining the same
Example:
3
4 7
-
2
1 7
3
4 7
-
2
1 7
= 3 - 2 +
4 7
-
1 7
= 1 +
4 - 1 7
=
1
3 7
• 2
Simplify the fraction if possible, in this example, the result already simplified.
### Another way to Calculate Mixed Fractions
• There is another way to subtract mixed fractions, which can be used in all cases adding, subtracting, multiplying, and dividing
• 1
Convert both mixed fractions to improper fractions, then calculated like a proper fraction and finally simplify it.
To convert a mix fraction to improper fraction (improper fraction has the numerator bigger than denominator), keep the denominator the same, then for the numerator multiply the whole number by the denominator then added to the numerator
Example how to convert mixed fraction to improper fraction:
a
b c
=
a x c + b c
example:
3
2 3
-
4
3 4
3
2 3
-
4
3 4
=
3 x 3 + 2 3
-
4 x 4 + 3 4
=
11 3
-
19 4
=
(11 x 4) - (19 x 3) 3 x 4
=
-13 12
=
-1
1 12
### How to Subtract Mixed Fractions With Different Denominators
• F
Subtracting mixed fractions with different denominators, we need first to make the fractions have the same denominators and subtract like the previous example.
Example:
4
3 4
-
3
2 3
• 1
Subtract the whole numbers together, and make the proper fractions have the same denominator. more about subtracting proper fractions click here
4
3 4
-
3
2 3
4
3 4
-
3
2 3
= 4 - 3 +
3 4
-
2 3
= 1 +
3 x 3 4 x 3
-
2 x 4 3 x 4
=
1
1 12
• 2
Simplify the fraction if possible , in this example, the result already simplified.
#### 📌 Negative signs in fractions
Positive Fractions
-1 -2
=
-
1 -2
=
-
-1 2
=
1 2
Negative Fractions
-1 2
=
1 -2
=
-
-1 -2
=
-
1 2 |
# Table of 55
In Mathematics, Table of 55 represents the repeated addition of the number 55. It means 55 is added to itself, ‘n’ number of times, where n is any natural number. For example, 55 multiplied by 3 can be expressed as:
55 x 3 ⇒ 55+ 55+ 55 = 275
Similarly, we can write the table of 55 by repeatedly adding 55 up to 20 times. Learn and memorise the table here to solve complex multiplication problems easily. [Also, Check Tables from 1 to 100].
## Multiplication Table of 55
Multiplication table of 55 is given here for students and learners, who want to prepare for exams that include mathematical computations. This table will help you to solve the multiplication problems quickly.
55 x 1 = 55 55 x 2 = 110 55 x 3 = 165 55 x 4 = 220 55 x 5 = 275 55 x 6 = 330 55 x 7 = 385 55 x 8 = 440 55 x 9 = 495 55 x 10 = 550 55 x 11 = 605 55 x 12 = 660 55 x 13 = 715 55 x 14 = 770 55 x 15 = 825 55 x 16 = 880 55 x 17 = 935 55 x 18 = 990 55 x 19 = 1045 55 x 20 = 1100
Let us see how to read the table of fifty-five in our next section.
## 55 Times Table
To memorise the table of fifty-five, we have presented here the format to read the table.
• One time fifty-five is 55
• Two times fifty-five is 110
• Three times fifty-five is 165
• Four times fifty-five is 220
• Five times fifty-five is 275
• Six times fifty-five is 330
• Seven times fifty-five is 385
• Eight times fifty-five is 440
• Nine times fifty-five is 495
• Ten times fifty-five is 550
Practise writing the 55 times table, in the above given format from 11 to 20 times to sharpen your skills.
## Solved Examples
Q.1: Find the value of 55 x 5 + 10.
Solution: 55 x 5 + 10
= (55 x 5) + 10 [BODMAS rule]
= 275 + 10
= 285
Q.2: Find the value of 55 x 11 ÷ 11.
Solution: Given, 55 x 11 ÷ 11
= 55 x (11 ÷ 11) [BODMAS rule]
= 55 x 1
= 55
Q.3: Roshan eats three apples a day. If he eats apples for 55 days, then how many apples does he eat in total?
Solution: In one day Roshan eats = 3 apples
So, in 55 days, he will eat = 55 x 3 = 165 apples
Q.4: A man runs for 4km a day. If he runs for 55 days continuously, then how many kilometers has he covered?
Solution: In one day, a man runs = 4 km
Then, in 55 days he will run = 55 x 4
= 220 kilometers
Thus, he covers 220 kms in 55 days.
### Check More Tables
For more knowledge on Multiplication Table subscribe to BYJU’S. Also, download BYJU’S-The Learning App for conceptual and interactive videos.
## Frequently Asked Questions on Table of 55
### What is table of 55?
Table of 55 is representation of repeated addition of natural number 55 to get another number.
55 x 1 = 55
55 x 2 = 55 + 55 = 110
55 x 3 = 55 + 55 + 55 = 165
55 x 4 = 55 + 55 + 55 + 55 = 220
And so on.
### What are the multiples of 55?
Multiples are the values that we get when a number is multiplied by consecutive natural numbers. The first ten multiples of 55 are 55, 110, 165, 220, 275, 330, 385, 440, 495 and 550.
### What is the sum of the first five multiples of 55?
The sum of the first five multiples of 55 is:
55+110+165+220+275 = 825
### What is the average of the first five multiples of 55?
Average is the ratio of sum of values and total number of values. Therefore,
average of first 5 multiples of 55 = (55+110+165+220+275)/5
= 825/5
= 165
### What is the value of 20 times of 55?
20 times of 55 = 20 x 55 = 1100
### Is 55 a prime number or composite number?
55 is a composite number because it has more than 2 factors. They are:
1, 5, 11 and 55. |
## Note on Locus
• Note
• Things to remember
• Videos
• Exercise
• Quiz
To every point in a plane there corresponds on ordered pair of real numbers and to every ordered pair of real numbers, there corresponds a point on a plane. If we have an algebraic relation between x and y, we get a set of points in the plane for different values of x and y satisfying the relation. If we move a set of points determined by some geometrical condition, we can represent it by an algebraic relation. This algebraic relation is called equation and the set of points in the plane is called Locus.
The locus of a point is the path traced out by the point moving under given geometrical condition( or conditions). Alternatively, the locus is the set of all those points, which satisfy the given geometrical conditions (or conditions).
For example:
1. Let a point P move such that its distance from a fixed line is always equal to d. The point P will trace out a straight line CD parallel to the fixed line. Thus, the locus of the moving point P is the straight line.
2. Consider the set of all points on the x-axis. This is the locus of a point whose ordinate is zero.
3. Let a point P move in a plane such that its distance from a fixed point, says B, is always equal to r. The point P will trace out a circle with centre B and radius r. Thus, the locus of the moving point is the circle.
4. Consider the locus of a point whose distance from the origin is 5 units. If the coordinates of any points satisfying the condition be (x,y), then (x-0)2+(y-0)2=52 is the equation of the locus.Hence, the equation of the locus of a point whose distance from the origin is equal to 5 units is x2+y2=25.The locus obviously is a circle with centre at the origin and radius 5 units.
Remarks
1. Every point which satisfies the given geometrical conditions (or conditions) lies on the locus.
2. A point which does not satisfy the given geometrical condition ( or conditions ) cannot lie on the locus.
3. Every point which lies on the locus satisfies the given geometrical condition (or conditions).
4. A point which does not lie on the locus cannot satisfy the given geometrical condition (or conditions).
5. The locus of a point moving in a plane under a given geometrical conditions is always a straight or a curved line.
6. To find the locus of moving points, plot some points satisfying the given geometrical condition, and then join these points.
To find the equation of a set of points (locus) with given geometrical conditions, the working rule is
1. Suppose (x, y) be any points in the set,
2. Find a relation between x and y satisfying the given conditions.
3. The relation is the required equation of the locus.
Notes
1. If any point belongs to the Locus, then the point satisfies the equation of the locus.
2. Any point satisfying the equation of Locus must lie on the Locus.
3. Any point out the locus does not satisfy the equation of the locus.
Methods of Finding the Equation of Locus
1. Take P(x,y) as a moving point in the coordinate plane. Represent the information in a suitable figure.
2. Find the distances according to the information provided.
3. Simplify the expressions by using algebraic operations. Make sure that there is no common number in all the terms. Also, make sure that first term is not negative. Final result or equation so obtained is the equation of the locus.
• To every point in a plane there corresponds on ordered pair of real numbers and to every ordered pair of real numbers there corresponds a point on a plane.
• The locus of a point is the path traced out by the point moving under given geometrical condition (or conditions).
.
### Very Short Questions
Soln: Let P(x,y) be a point which moves at a distance 5 units from the point(0,0)
Then, OP=5 units
or, $$\sqrt{(x-0)^2+(y-0)^2}$$=5
or, $$\sqrt{x^2+y^2}$$=5
Squaring both side,x2+y2-25=0 Ans.
0%
(2,2)
(9,2)
(5,7)
(7,8)
7
9
6
8
2
4
8
9
4
2
8
7
x-9y=8
x-5y=4
x-5y=3
x-4y=8
x-3y-1=0
x-3y-2=0
x+3y-1=0
x-3y+1=0
• ### A point moves so that its distance from the point (3,2) is always twice its distance from the point (2,1).Find the equation of a locus.
3x2+3y2-10x-4y+7=0
3x2+3y2 +10x-4y+7=0
3x2+3y2-9x-4y+7=0
4x2+3y2-10x-4y+7=0
• ### Find the equation of the locus of a point whose distance from (-1,1) is equal to the twice its distance from the X-axis.
x2-3y2+2x-2y+2=0
x2-2y2+2x-2y+2=0
x2-3y2+2x-2y+2=5
x2-3y2+2x-2y-2=0
• ### Find the equation of the locus of a point which moves so that its distance from the point (0,2) is one-third of its distance from the point (3,0).
8x2+5y2+6x+36y+27=0
8x2+8y2+6x-36y+27=0
7x2+8y2+6x-36y+7=0
8x2+8y2+6x-36y-22=0
• ### A point moves so that the ratio of its distance from the point( -a,0) and (a,0) is 2:3.Find the equation of its locus.
5x2+5y2+26ax+4a2=0
5x2+5y2+26ax+5a2=0
5x2+5y2-26ax+5a2=0
2x2+2y2+26ax+5a2=0
• ## You scored /10
Forum Time Replies Report |
+0
0
274
4
Two cubes with the faces numbered 1 through 6 are tossed and the numbers shown on the top faces are added. What is the probability that the sum is even? Express your answer as a common fraction.
Oct 22, 2018
#1
+625
+9
There are two ways that we can end up with an even number by adding two numbers.
Either both numbers have to be even, or both have to be odd.
Starting off, let's find out the probability that we will roll two cubes and they will both show an even number.
For cube A, we have $$6$$ possible outcomes. Out of those six, three numbers are even ($$2 , 4, 6$$).
Therefore, we have $$\frac{3}{6} = \frac{1}{2}$$ ways to roll one cube and have an even number show up.
This is the same for cube B. In total, we have $$\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$$ ways to roll two cubes and have both of them show even numbers.
If we do the same thing for the odd numbers, we would get the same result of $$\frac{1}{4}$$.
Adding these two possibilities together, we would get $$\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
Therefore, our answer is $$\boxed{\frac{1}{2}}$$
.
Oct 22, 2018
#2
+128
+1
Well,
even + even == even, 3#s * 3#s == 9#s
and
odd + odd == even, 3#s*3#s == 9#s
and
there are 6*6 == 36, so therefore our answer is (9+9)/36 == 18/36 == 1/2
Oct 22, 2018
#3
+1 |
Question
# A box contains 4 white balls, 6 red balls, 7 black balls and 3 blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is :A. neither white nor blackB. red or white, andC. either white or red or black or blue
Hint: Use the fundamental definition of probability that it can be measured by ratio of number of favorable cases to total cases/ sample spaces.
Here, it is given that Box contains 4 white balls, 6 red balls, 7 black balls and 3 blue balls i.e. in total 20 balls are present in the box.
Hence, Total balls in Box = 20.
A. neither white nor black
Probability of drawing neither white nor black balls means drawing red balls or blue balls.
So, total probability of drawing Red balls or Blue balls = Probability of drawing Red balls +
Probability of drawing Blue balls.
Now, we know the definition of probability of an event can be given as
P(E)=Number of favorable cases/Total cases$\ldots \ldots (1)$
Hence, Probability of drawing Red balls or Blue balls=
$P=\dfrac{6}{20}+\dfrac{3}{20}=\dfrac{9}{20}$
Therefore, Probability of drawing neither white nor black be$\dfrac{9}{20}$.
B. Red or white
Probability of drawing either Red of White balls can be given as a summation of probability
of drawing Red Balls and white balls both.
Number of Red balls= 6
Number of white balls= 4
Total number of balls= 20
Now, from equation (1) , we can give required probability as
P=Probability of drawing Red balls + Probability of drawing white balls
Hence, we get
$P=\dfrac{6}{20}+\dfrac{4}{20}=\dfrac{10}{20}=\dfrac{1}{2}$
C. either white or red or black or blue
Here, we need to involve the probability of drawing each colour as we are drawing either of
all balls of any colour.
So, here probability of either white or red or black or blue can be given as
P= Probability of drawing (Red)+(White)+(Black)+(Blue)
As we have
Number of Red balls= 6
Number of White balls= 4
Number of black balls= 7
Number of blue balls= 3
Total balls=20
Hence,
$P=\dfrac{4}{20}+\dfrac{6}{20}+\dfrac{7}{20}+\dfrac{3}{20}=\dfrac{20}{20}=1$
Therefore, answers of the problem are $\dfrac{9}{20},\dfrac{1}{2},1.$
Note: One can miss any of the colour while writing the probability of drawing that colour. He/she may involve favourable cases of any other colour ball.
First case can also be calculated by getting the probability of either white or black balls then subtract it from 1 as $1-\dfrac{1}{20}=\dfrac{19}{20}$.
One can go wrong while writing probability of red or white (2nd case) or with the last part as well. One can multiply the probability of red and white balls to get probability of red or white which is the wrong approach. So, take care of it in these kinds of questions. |
# Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2
Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2 Textbook Questions and Answers.
## BSEB Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2
Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠A.
Solution:
(i) In ∆ ABC, we have
AB = AC
⇒ ∠B = ∠C [∵ Angles opposite to equal sides are equal]
$$\frac { 1 }{ 2 }$$∠B = $$\frac { 1 }{ 2 }$$∠C
⇒ ∠OBC = ∠OCB
[∵ OB and OC bisect ∠s B and C respectively]
∴ ∠OBC = $$\frac { 1 }{ 2 }$$∠B and ∠OCB = $$\frac { 1 }{ 2 }$$∠c
⇒ OB = OC … (2)
[∵ Sides opp. to equal ∠s are equal]
(ii) Now, in ∆s ABO and ACO, we have
AB = AC [Given]
∠OBC = ∠CCB [From (1)]
∆OB = OC . [From (2)]
∴ By SAS criterion of congruence, we have
∆ABO ≅ ∆ACO
⇒ ∠BAO = ∠CAO
[∵ Corresponding parts of congruent triangles are equal]
⇒ AO bisects ∠BAC.
Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.
Solution:
In ∆s ABD and ACD, we have
DB = DC [Given]
∴ By SAS criterion of congruence, we have
∆ ABD ≅ ∆ ACD
⇒ AB = AC
[∵ Corresponding parts of congruent triangles are equal]
Hence, ∆ ABC is isosceles.
Question 3.
ABC is an isosceles triangle in Which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.
Solution:
In ∆s ABE and ACF, we have
∠AEB = ∠AFC [∵ Each = 90°]
∠BAE = ∠CAF [Common]
and, BE = CF [Given]
By AAS criterion of congruence, we have
∆ ABE ≅ ∆ ACF
⇒ AB = AC
[∵ Corresponding parts of congruent triangles are equal]
Hence, ∆ ABC is isosceles.
Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Solution:
Let BE ⊥ AC and CF ⊥ AB.
In ∆s ABE and ACF, we have
∠AEB = ∠AFC [∵ Each = 90°]
∠A = ∠A [Common]
and, AB = AC [Given]
∴ By AAS criterion of congruence,
∆ ABE ≅ ∆ ACF
⇒ BE = CF
[∵ Corresponding parts of congruent triangles are equal]
Question 5.
ABC and DBC are two isosceles triangles on tile same base BC (see figure). Show that ∠ABD = ∠ACD.
Solution:
In ∆ ABC, we have
AB = AC
⇒ ∠ABC = ∠ACB
[∵ Angles opposite to equal sides are equal]
In ∆ BCD, we have
BO = CD
⇒ ∠DBC = ∠DCB … (2)
[∵ Angles opposite to equal sides are equal]
Adding (1) and (2), we have
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD
Question 6.
∆ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.
Solution:
In ∆ ABC, we have
AB = AC
⇒ ∠ACB = ∠ABC … (1)
[∵ Angles opp. to equal sides are equal]
∴ AD = AC [∵AB = AC]
Thus, in ∆ ADC, we have
⇒ ∠ACD = ∠ADC … (2)
[∵ Angles opp. to equal sides are equal]
Adding (1) and (2), we get
∠ACB + ∠ACD = ∠ABC + ∠ADC
⇒ ∠BCD = ∠ABC + ∠BDC [∵ ∠ADC = ∠BDC]
⇒ ∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD [Adding ∠BCD on both sides]
⇒ 2∠BCD = 180°
⇒ ∠BCD = 90°
Hence, ∠BCD is a right angle.
Question 7.
ABC is a right angled triangle in which ∠A = 90 and AB = AC. Find ∠B and ∠C.
Solution: We have, ∠A = 90°
AB = AC
⇒ ∠B = ∠C
[∵ Angles opp. to equal sides of a triangle are equal]
Also, ∠A + ∠B + ∠C = 180° [Angle-sum property]
⇒ 90° + 2∠B = 180° [∵ ∠C = ∠B]
⇒ 2∠B = 180° – 90° = 90°
⇒ ∠B = $$\frac { 90° }{ 2 }$$ = 45°
∴ ∠C = ∠B = 45°
Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Let ∆ ABC be an equilateral triangle so that AB = AC = BC.
Now, ∵ AB = AC
⇒ ∠B = ∠C … (1)
[∵ Angles opp. to equal sides are equal]
Also, ∵ CB = CA
⇒ ∠A = ∠B … (2)
[∵ Angles opp. to equal sides are equal]
From (1) and (2), we have
∠A = ∠B = ∠C
Also, ∠A + ∠B + ∠C = 180° [Angle-sum property]
∴ ∠A + ∠A + ∠A = 180°
⇒ 3∠A = 180° ⇒ ∠A = 60°
∴ ∠A = ∠B = ∠C = 60°
Thus, each angle of an equilateral triangle is 60°. |
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1 9.4 Multiplying and Dividing Radicals 9.4 OBJECTIVES 1. Multiply and divide expressions involving numeric radicals 2. Multiply and divide expressions involving algebraic radicals In Section 9.2 we stated the first property for radicals: 1ab 1a 1b when a and b are any positive real numbers That property has been used to simplify radical expressions up to this point. Suppose now that we want to find a product, such as We can use our first radical rule in the opposite manner. NOTE The product of square roots is equal to the square root of the product of the radicands. 1a 1b 1ab so We may have to simplify after multiplying, as Example 1 illustrates. Example 1 Simplifying Radical Expressions Multiply then simplify each expression. (a) (b) An alternative approach would be to simplify 112 first (c) 110x 12x 220x 2 24x x x15 72
2 724 CHAPTER 9 EXPONENTS AND RADICALS CHECK YOURSELF 1 Simplify. (a) 1 16 (b) (c) 18a 1a If coefficients are involved in a product, we can use the commutative and associative properties to change the order and grouping of the factors. This is illustrated in Example 2. Example 2 Multiplying Radical Expressions NOTE In practice, it is not necessary to show the intermediate steps. (215)(16) (2 )(15 16) CHECK YOURSELF 2 Multiply ( 17)( 51). The distributive property can also be applied in multiplying radical expressions. Consider the following. Example Multiplying Radical Expressions (a) 1(12 1) The distributive property Multiply the radicals. (b) 15(216 1) The distributive property The commutative property CHECK YOURSELF (a) 15(16 15) (b) 1(215 12) The FOIL pattern we used for multiplying binomials in Section.4 can also be applied in multiplying radical expressions. This is shown in Example 4.
3 MULTIPLYING AND DIVIDING RADICALS SECTION Example 4 Multiplying Radical Expressions (a) (1 2)(1 5) Combine like terms CAUTION NOTE You can use the pattern (a b)(a b) a 2 b 2, where a 17 and b 2, for the same result and 17 2 are called conjugates of each other. Note that their product is the rational number. The product of conjugates will always be rational. Be Careful! This result cannot be further simplified: 1 and 71 are not like terms. (b) (c) (17 2)(17 2) (1 5) 2 (1 5)(1 5) CHECK YOURSELF 4 (a) (15 )(15 2) (b) (1 4)(1 4) (c) (12 ) 2 We can also use our second property for radicals in the opposite manner. NOTE The quotient of square roots is equal to the square root of the quotient of the radicands. 1a 1b a A b One use of this property to divide radical expressions is illustrated in Example 5. Example 5 Simplifying Radical Expressions Simplify. NOTE The clue to recognizing when to use this approach is in noting that 48 is divisible by. (a) (b) A A x 2 125x 2 (c) 225x 2 5x 15 A 5 There is one final quotient form that you may encounter in simplifying expressions, and it will be extremely important in our work with quadratic equations in the next chapter. This form is shown in Example 6.
4 726 CHAPTER 9 EXPONENTS AND RADICALS CHECK YOURSELF 5 Simplify. (a) (b) 281s2 19 Example 6 Simplifying Radical Expressions Simplify the expression 172 CAUTION First, we must simplify the radical in the numerator. Be Careful! Students are sometimes tempted to write This is not correct. We must divide both terms of the numerator by the common factor (1 212) Use Property 1 to simplify 172. Factor the numerator then divide by the common factor. CHECK YOURSELF 6 Simplify CHECK YOURSELF ANSWERS 1. (a) 12; (b) 16; (c) 2a (a) 10 5; (b) (a) 1 15; (b) 1; (c) (a) 5; (b) s 6. 1
5 Name 9.4 Exercises Section Date Perform the indicated multiplication. Then simplify each radical expression ANSWERS m x a b x 16x a 115a (1)(517) 22. (2 15)( 111)
6 ANSWERS (15)(2 110) (12 15) 26. (41)(16) 1(15 1) (215 1) (21 17) (1 5)(1 ) 0. (15 2)(15 1) (15 1)(15 ) 2.. (15 2)(15 2) (110 5)(110 5) 6. (12 )(12 7) (17 5)(17 5) (111 )(111 ) (1x )(1x ) 8. (1a 4)(1a 4) (1 2) (15 ) (1y 5) (1x 4) Perform the indicated division. Rationalize the denominator if necessary. Then simplify each radical expression a m
7 ANSWERS Area of a rectangle. Find the area of the rectangle shown in the figure Area of a rectangle. Find the area of the rectangle shown in the figure Complete this statement: because Explain why but When you look out over an unobstructed landscape or seascape, the distance to the visible horizon depends on your height above the ground. The equation d A 2 h is a good estimate of this, in which d distance to horizon in miles and h height of viewer above the ground. Work with a partner to make a chart of distances to the horizon given different elevations. Use the actual heights of tall buildings or 729
8 ANSWERS a. b. c. prominent landmarks in your area. The local library should have a list of these. Be sure to consider the view to the horizon you get when flying in a plane. What would your elevation have to be to see from one side of your city or town to the other? From one side of your state or county to the other? d. e. f. Getting Ready for Section 9.5 [Section 9.1] Evaluate the following. Round your answer to the nearest thousandth. (a) (c) (e) (b) (d) (f) Answers m 7. 10x x x y 101y a a. 4 b. 7 c. 11 d..464 e f
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# Solve Tangent Lines Problems in Calculus
Tangent lines problems and their solutions, using first derivatives, are presented.
## Problem 1
Find all points on the graph of y = x 3 - 3 x where the tangent line is parallel to the x axis (or horizontal tangent line).
Solution to Problem 1:
• Lines that are parallel to the x axis have slope = 0. The slope of a tangent line to the graph of y = x 3 - 3 x is given by the first derivative y '.
y ' = 3 x 2 - 3
• We now find all values of x for which y ' = 0.
3 x 2 - 3 = 0
• Solve the above equation for x to obtain the solutions.
x = - 1 and x = 1
• The above values of x are the x coordinates of the points where the tangent lines are parallel to the x axis. Find the y coordinates of these points using y = x 3 - 3 x
for x = - 1 , y = 2
for x = 1 , y = - 2
• The points at which the tangent lines are parallel to the x axis are: (-1 , 2) and (1 , -2). See the graph of y = x3 - 3 x below with the tangent lines.
## Problem 2
Find a and b so that the line y = - 3 x + 4 is tangent to the graph of y = a x3 + b x at x = 1.
Solution to Problem 2:
• We need to determine two algebraic equations in order to find a and b. Since the point of tangency is on the graph of y = a x3 + b x and y = - 3 x + 4, at x = 1 we have
a(1)3 + b(1) = - 3(1) + 4
• Simplify to write an equation in a and b
a + b = 1
• The slope of the tangent line is -3 which is also equal to the first derivative y ' of y = a x3 + b x at x = 1
y ' = 3 a x2 + x = - 3 at x = 1.
• The above gives a second equation in a and b
3 a + b = -3
• Solve the system of equations a + b = 1 and 3 a + b = - 3 to find a and b
a = - 2 and b = 3.
• See graphs of y = a x3 + b x, with a = - 2 and b = 3, and y = - 3 x + 4 below.
## Problem 3
Find conditions on a and b so that the graph of y = a e x + bx has NO tangent line parallel to the x axis (horizontal tangent).
Solution to Problem 3:
• The slope of a tangent line is given by the first derivative y ' of y = a e x + bx. Find y '
y ' = a e x + b
• To find the x coordinate of a point at which the tangent line to the graph of y is horizontal, solve y ' = 0 for x (slope of a horizontal line = 0)
a e x + b = 0
• Rewrite the above equation as follows
e x = - b/a
• The above equation has solutions for -a/b >0. Hence, the graph of y = a e x + bx has NO horizontal tangent line if -a/b <= 0
## Exercises
1) Find all points on the graph of y = x
3 - 3 x where the tangent line is parallel to the line whose equation is given by y = 9 x + 4.
2) Find a and b so that the line y = - 2 is tangent to the graph of y = a x
2 + b x at x = 1.
3) Find conditions on a, b and c so that the graph of y = a x
3 + b x 2 + c x has ONE tangent line parallel to the x axis (horizontal tangent).
### Solutions to the Above Exercises
1) (2 , 2) and (-2 , -2)
2) a = 2 and b = - 4
3) 4 b
2 - 12 a c = 0
More references on
calculus problems |
### 10-2. Angular Acceleration
Learning Objectives
• Describe uniform circular motion.
• Explain non-uniform circular motion.
• Calculate angular acceleration of an object.
• Observe the link between linear and angular acceleration.
Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity $\omega$ was defined as the time rate of change of angle $\theta$:
$ω=ΔθΔt,$
where $\theta$ is the angle of rotation as seen in Figure 1. The relationship between angular velocity $\omega$ and linear velocity $v$ was also defined in Rotation Angle and Angular Velocity as
$v = rω$
or
$ω = v r ,$
where $r$ is the radius of curvature, also seen in Figure 1. According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative
Figure 1: This figure shows uniform circular motion and some of its defined quantities.
Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer’s hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which $\omega$ changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration $\alpha$ is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:
$α=ΔωΔt,$
where $\Delta \omega$ is the change in angular velocity and $\Delta t$ is the change in time. The units of angular acceleration are $\left(\text{rad/s}\right)\text{/s}$, or ${\text{rad/s}}^{2}$. If $\omega$ increases, then $\alpha$ is positive. If $\omega$ decreases, then $\alpha$ is negative.
###### Example 1: Calculating the Angular Acceleration and Deceleration of a Bike Wheel
Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in ${\text{rad/s}}^{2}$. (b) If she now slams on the brakes, causing an angular acceleration of $–87.3\phantom{\rule{0.25em}{0ex}}{\text{rad/s}}^{2}$, how long does it take the wheel to stop?
Strategy for (a)
The angular acceleration can be found directly from its definition in $\alpha =\frac{\Delta \omega }{\Delta t}$ because the final angular velocity and time are given. We see that $\Delta \omega$ is 250 rpm and $\Delta t$ is 5.00 s.
Solution for (a)
Entering known information into the definition of angular acceleration, we get
$α = ΔωΔt = 250 rpm5.00 s.$
Because $\Delta \omega$ is in revolutions per minute (rpm) and we want the standard units of ${\text{rad/s}}^{2}$ for angular acceleration, we need to convert $\Delta \omega$ from rpm to rad/s:
$Δω = 250 rev min ⋅ 2π rad rev ⋅ 1 min 60 sec = 26.2 rad s .$
Entering this quantity into the expression for $\alpha$, we get
Strategy for (b)
In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for $\Delta t$, yielding
$Δt=Δωα .$
Solution for (b)
Here the angular velocity decreases from $\text{26.2 rad/s}$ (250 rpm) to zero, so that $\Delta \omega$ is $–\text{26.2 rad/s}$, and $\alpha$ is given to be $–\text{87.3}\phantom{\rule{0.25em}{0ex}}{\text{rad/s}}^{2}$. Thus,
$Δt = – 26.2 rad/s – 87.3 rad/s 2 = 0.300 s.$
Discussion
Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval.
If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure 2. Thus, linear acceleration is called tangential acceleration ${a}_{\text{t}}$.
Figure 2: In circular motion, linear acceleration $a$, occurs as the magnitude of the velocity changes: $a$ is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration ${a}_{\text{t}}$.
Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration, ${a}_{\text{c}}$, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in Figure 3. Thus, ${a}_{\text{t}}$ and ${a}_{\text{c}}$ are perpendicular and independent of one another. Tangential acceleration ${a}_{\text{t}}$ is directly related to the angular acceleration $\alpha$ and is linked to an increase or decrease in the velocity, but not its direction.
Figure 3: Centripetal acceleration ${a}_{\text{c}}$ occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other.
Now we can find the exact relationship between linear acceleration ${a}_{\text{t}}$ and angular acceleration $\alpha$. Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be
$at= ΔvΔt .$
For circular motion, note that $v=\mathrm{r\omega }$, so that
$at=ΔrωΔt .$
The radius $r$ is constant for circular motion, and so $\text{Δ}\left(\mathrm{r\omega }\right)=r\left(\Delta \omega \right)$. Thus,
$at= r ΔωΔt .$
By definition, $\alpha =\frac{\Delta \omega }{\Delta t}$. Thus,
$a t = rα ,$
or
$α = a t r .$
These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car’s drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration $\alpha$.
###### Example 2: Calculating the Angular Acceleration of a Motorcycle Wheel
A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels? (See Figure 4.)
Figure 4: The linear acceleration of a motorcycle is accompanied by an angular acceleration of its wheels.
Strategy
We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration ${a}_{\text{t}}$. Then, the expression $\alpha =\frac{{a}_{\text{t}}}{r}$ can be used to find the angular acceleration.
Solution
The linear acceleration is
$at = ΔvΔt = 30.0 m/s4.20 s = 7.14 m/s2.$
We also know the radius of the wheels. Entering the values for ${a}_{\text{t}}$ and $r$ into $\alpha =\frac{{a}_{\text{t}}}{r}$, we get
$α = a t r = 7.14 m/s 2 0.320 m = 22.3 rad/s 2 .$
Discussion
Units of radians are dimensionless and appear in any relationship between angular and linear quantities.
So far, we have defined three rotational quantities— , and $\alpha$. These quantities are analogous to the translational quantities , and $a$. Table 1 displays rotational quantities, the analogous translational quantities, and the relationships between them.
##### Table 1: Rotational and Translational Quantities
Rotational Translational Relationship
$\theta$ $x$ $\theta =\frac{x}{r}$
$\omega$ $v$ $\omega =\frac{v}{r}$
$\alpha$ $a$ $\alpha =\frac{{a}_{t}}{r}$
###### Making Connections: Take-Home Experiment:
Sit down with your feet on the ground on a chair that rotates. Lift one of your legs such that it is unbent (straightened out). Using the other leg, begin to rotate yourself by pushing on the ground. Stop using your leg to push the ground but allow the chair to rotate. From the origin where you began, sketch the angle, angular velocity, and angular acceleration of your leg as a function of time in the form of three separate graphs. Estimate the magnitudes of these quantities.
Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example.
Show/Hide Solution
##### Solutuon
The magnitude of angular acceleration is $\alpha$ and its most common units are ${\text{rad/s}}^{2}$. The direction of angular acceleration along a fixed axis is denoted by a + or a – sign, just as the direction of linear acceleration in one dimension is denoted by a + or a – sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis.
Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs.
#### Section Summary
• Uniform circular motion is the motion with a constant angular velocity $\omega =\frac{\Delta \theta }{\Delta t}$.
• In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is $\alpha =\frac{\Delta \omega }{\Delta t}$.
• Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as ${a}_{\text{t}}=\frac{\Delta v}{\Delta t}$.
• For circular motion, note that $v=\mathrm{r\omega }$, so that $at=ΔrωΔt.$
• The radius r is constant for circular motion, and so $\mathrm{\text{Δ}}\left(\mathrm{r\omega }\right)=r\Delta \omega$. Thus, $at=rΔωΔt.$
• By definition, $\Delta \omega /\Delta t=\alpha$. Thus, $a t = rα$
or
$α=atr.$
#### Conceptual Questions
###### Exercise 1
Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.
###### Exercise 2
Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude.
###### Exercise 3
In circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction. Explain your answer.
###### Exercise 4
Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) The plate starts to spin? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt?
#### Problems & Exercises
###### Exercise 1
At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?
Show/Hide Solution
##### Solution
$\omega =0\text{.}\text{737 rev/s}$
###### Exercise 2
Integrated Concepts
An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is its angular acceleration in ${\text{rad/s}}^{2}$? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the radial acceleration in ${\text{m/s}}^{2}$ and multiples of $g$ of this point at full rpm?
###### Exercise 3
Integrated Concepts
You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest?
Show/Hide Solution
##### Solution
(a) $-0\text{.}{\text{26 rad/s}}^{2}$
(b) $\text{27}\phantom{\rule{0.25em}{0ex}}\text{rev}$
###### Exercise 4
Unreasonable Results
You are told that a basketball player spins the ball with an angular acceleration of . (a) What is the ball’s final angular velocity if the ball starts from rest and the acceleration lasts 2.00 s? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent? |
Investigation #1
y = asin[b(x + c)]
by
Brock F. Miller
This investigation is to examine the effect of real numbers a, b, and c on the curve of y =sinx. To begin we will look at each variable individually on the sine curve. When we finish looking individually at the variables we can easily show what happens to the sine curve when two or more of the parameters change.
To begin with let's take at a brief look at the basic sine curve:
Graph 1`
y = sinx
As you might have noticed from previous studies of the sine function the domain of sinx is all real numbers while the range is [-1,1].
Now that we have refreshed ourselves with the sine curve let's take a look at what happens with various changes in the parameter "a".
Amplitude - "a"
The amplitude of the sine function controls the range of the function. In Graph 1 the value of a was 1 (sinx = (1)sinx). Let's compare positive integers for "a".
Graph 2
y = sinx = (1)sinx
y = (2)sinx
y = (3)sinx
y = (4)sinx
When we look at the four graphs on Graph 2 you should notice that the ranges of each sine function is [-a, a].
Now let's compare y = (1)sinx to sine curves with negative integers for a
Graph 3
y = sinx
y = (-1)sinx
y = (-2)sinx
y = (-3)sinx
In Graph 3 let's look at the first two curves (y = sinx and y = (-1)sinx). You should notice that the range of both of these functions is [-1, 1] but curves mirror each other. From the positive values of "a" we said that the range was [-a, a] but you should notice for negative values of a that the range is [a, -a]. Therefore for this form of sine function, y = asin[b(x + c)] we can conclude that the range is [, ]
In conclusion about the parameter "a" we can conclude that the value of a controls the distance above and below the x-axis that the sine curve travels and stipulates either exactly what the range is for the function or will be used to calculate the range
Try These I:
1. For the function y = 6sinx:
A. What is the range of this function?
B. Graph the function when x is in the interval [-6, 6].
2. For the function y = -7sinx
A. What is the range of this function?
B. Graph the function when x is the interval [-6, 6].
Period Change - "b"
For this next section let's let a=1 and c=0. Now let's compare y=sinx to y=sin(bx) for a couple of values of b
Graph IV
y = sinx = sin(1x)
y = sin 2x
y = sin 3x
Visually, it is not as easy to see what happens when we change values of "b" but from your previous work with sine you should remember that the period is . Therefore when b = 1 (y = sin (1x) ) the period is. If you notice when we change the values of b the period of the each function changes. In fact, as "b" increases the period of the function actually decreases.
By Rule: period of sine function = .
So if we notice for y = sin(2x), b=2, the period = = . (Approx 3.142).
So the Graph does one full period during a distance on the x-axis of .
Now that we have looked at positive values of "b" it would make sense to look at negative values of "b" but this is not that exciting since sin(-x) = -sinx. Therefore if we were to look at y = sin(-2x) that should be the same as y = -sin(2x). Which tells me that a = -1 and b = 2. From the amplitude section a negative value for "a" tells us that the graph is a mirror image of the positive values and when b = 2 that tells us that the period is .
Graph V
y = sinx
y = sin(2x)
y = sin(-2x) = -sin(2x)
Notice that y = sin(2x) and y = sin(-2x) are "mirror" images of each other and notice that their respective periods are the same.
Try These II:
1. y = sin(4x)
A. What is the range of this function? What is the period of this function?
B. Graph this function over [0, 3.142]
2.
A. What is the range of this function? What is the period of this function?
B. Graph this function over [0, ].
3. y = 6sin(2x)
A. What is the range of this function? What is the period of this function?
B. Graph this function over [0, ].
Phase Shift - "c"
(Translation of y-axis)
For our investigation into what happens when we change values of "c" we will use the values of a = 1 and b = 1 for our initial look. The term "phase shift" means an actual shift of the y-axis. To see this lets look at a few examples:
Graph VI
y = sinx = sin(x + 0)
y = sin(x + 1)
y = sin(x + 2)
If we look at these changes in "c" we can see for y = sinx, c = 0 therefore the y-axis remains the same but when we look at y = sin(x + 1) and y = sin(x + 2) the values of "c" are 1 and 2 respectively. Notice that a value of c = 1 meant a translation of the y-axis 1 unit to the left and a value of c = 2 meant a translation of the y-axis two units to the left. If you look at the three different graphs you will see that there is neither a change in range for each function nor is there a change in period.
It seems as if any positive value of "c" means the sine curve will shift "c" units to the left. On the other hand it would make intuitive sense that a negative value for c would mean a shift of c units to the right for all points on the curve. Let's investigate:
Graph VII
y = sinx = sin(x + 0)
y = sin(x + (-1))
y = sin(x + (-2))
Upon investigation we see that is actually the case. We can conclude the the value of c has a "movement" feature to the sine curve.
Try These III:
1. y = sin(x + 3)
A. What is the range, period, and phase shift for this function?
B. Graph this function.
2. y = 5sin(x - 3)
A. What is the range, period, and phase shift for this function?
B. Graph this function.
Summary
Summarizing the three parameters we can conclude the following:
I. For change in "a" - The parameter "a" controls the range of this type of function. It also controls the maximum and minimum for the function.
II. For change in "b" - The parameter "b" controls the period of the function.
III. For change in "c" - The parameter "c" controls the translation of the y-axis and all points on the curve.
With this knowledge the only "graph" you need to know is that of the base sine curve and then what each parameter does and you can graph any function of the form:
y = asin[b(x + c)]
So let's look at a example:
Graph y = 7sin2(x - 4) showing at least two full periods.
a = 7: range is [-7, 7]
b = 2: period is = (Approx = 3.142)
c = -4: y-axis is shifted 4 units to the right. The graph would have one full period between x = [4, 4 + ] and another [4 - , 4]
So the graph should look like this:
Try These IV:
1. For the function: y = 3sin[0.25(x + 2)]
A. What is the range, period, and phase shift for this function?
B. Graph at least two full periods of this function.
2. For the function y = 2sin[6(x - 1)]
A. What is the range, period, and phase shift for this function?
B. Graph at least two full periods of this function.
I hope that this write-up has been helpful to you the student/reader. If there are any questions or problems with any type of sine function of this form please send me a comment at my email on my home page. Have a good one.
BFM |
# 10.03 Solving quadratic equations using square roots
Lesson
Solving basic or simple quadratics involves using algebraic manipulation. Let's look at a few examples of this process.
Remember!
We will be manipulating expressions with squared terms. Remember that when we take the square root of both sides of an equation we need to consider both the positive and negative square roots.
$x^2$x2 $=$= $25$25 $x$x $=$= $\pm\sqrt{25}$±√25 $x$x $=$= $\pm5$±5
Both $x=5$x=5 and $x=-5$x=5 are solutions because when squared they both give us $25$25.
#### Worked examples
##### Question 1
Solve the quadratic equation $x^2-9=0$x29=0.
Think: We want to end up with an expression like $x=\dots$x=, and to get there we will need to isolate $x$x
Do: Considering the order of operations, we will first move the constant term to the other side, then undo the square:
$x^2-9$x2−9 $=$= $0$0 $x^2$x2 $=$= $9$9 $x$x $=$= $\pm\sqrt{9}$±√9 $x$x $=$= $\pm3$±3
Since there are two numbers that, when squared, give $9$9, the solutions to this equation are $x=3$x=3 or $x=-3$x=3.
##### Question 2
Find solutions to $\left(2x+7\right)^2=64$(2x+7)2=64.
Think: Identify the order of operations necessary to isolate the $x$x variable. In this case we will deal with the square root first, then the addition and finally the multiplication by $2$2.
Do:
$\left(2x+7\right)^2$(2x+7)2 $=$= $64$64 $2x+7$2x+7 $=$= $\pm\sqrt{64}$±√64
We use inverse operations to remove the square from the $LHS$LHS, the opposite of a square is a square root. Next we have
$2x+7$2x+7 $=$= $\pm8$±8 (from here on, this is just like solving linear equations)
Where we have evaluate the square root. There are two solutions now, let's continue to solve both for $x$x:
$2x+7$2x+7 $=$= $8$8 or $2x+7$2x+7 $=$= $-8$−8 $2x$2x $=$= $8-7$8−7 or $2x$2x $=$= $-8-7$−8−7 $2x$2x $=$= $1$1 or $2x$2x $=$= $-15$−15 $x$x $=$= $\frac{1}{2}$12 or $x$x $=$= $\frac{-15}{2}$−152
##### Question 3
Find the solutions to $x^2-17=0$x217=0.
Think: Identify the order of operations necessary to isolate the $x$x variable. In this case we will deal with the subtraction first, then the square.
Do:
$x^2-17$x2−17 $=$= $0$0 $x^2$x2 $=$= $17$17
Here we use inverse operations to remove the $-17$17 from the $LHS$LHS, the opposite of a $-17$17 is a $+$+ $17$17. Then we have
$x$x $=$= $\sqrt{17}$√17 OR $x$x $=$= $-\sqrt{17}$−√17
Where we use inverse operations to remove the square from the $LHS$LHS, the opposite of a square is a square root.
### A special note about exact solutions
For nearly all of our work with solutions to functions and equations it is standard practice to leave our final expression in exact form. In this case, $\sqrt{17}$17 is as far as we will go, we cannot simplify the root any further.
In questions involving applications of quadratics, we may be asked to evaluate the square root at the very end and then approximate to a specific number of decimal places. For example, perhaps we want the time taken to travel a certain distance or an estimate of the area of a block of land.
### A special note about number of solutions
All of the worked examples above had two solutions, but let's consider some other possibilities.
Let's consider the simplest quadratic to solve $x^2=a$x2=a. We will have three cases for $a$a
• $a>0$a>0, that is a is positive, for example $x^2=16$x2=16
• $a=0$a=0, for example $x^2=0$x2=0
• $a<0$a<0, that is a is negative, for example $x^2=-16$x2=16
Consider how many solutions there are to each of the scenarios above. What happens when you take the square root of both sides?
• $x^2=a$x2=a, $a>0$a>0, for example $x^2=16$x2=16, will give $2$2 solutions $x=\sqrt{a}$x=a and $x=-\sqrt{a}$x=a
• $x^2=0$x2=0, for example $x^2=0$x2=0, will have $1$1 solutions $x=0$x=0
• $x^2=a$x2=a, $a<0$a<0, for example $x^2=-16$x2=16, will have no real solutions as we cannot square root a negative in the real numbers
Remember!
A quadratic equation can have $0$0, $1$1 or $2$2 solutions. We will explore this further with graphing quadratics.
#### Practice questions
##### Question 5
Solve for $p$p:
$5\left(p^2-3\right)=705$5(p23)=705
##### Question 6
Solve $\left(x-4\right)^2=10$(x4)2=10 for $x$x.
1. Write all solutions on the same line, separated by commas.
### Outcomes
#### A1:A-REI.B.4
Solve quadratic equations in one variable.
#### A1:A-REI.B.4b
Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as "no real solution." |
## Mathematics Of Finance – CS Foundation Statistics Notes
1. Interest
The amount charged, expressed as a percentage of the principal, by a lender to a borrower for the use of assets. It involves two persons
Borrower – When you borrow money, you pay interest Moneylender- When you lend money, you earn interest.
Some of the terms used in interest calculation are
• Principal or Capital – sum borrowed
• Interest – Extra money charged by the lender for use of his money
• Moneylender – a person who gives money to the borrower
• Conversion period – the period of time for which interest is calculated
• Simple interest – Simple interest is just the amount of money paid on a loan.
• Compound interest – over here Interest is charged on interest too.
2. Simple interest
Simple interest is called simple because it ignores the effects of compounding. The interest charge is always based on the original principal, so interest on interest is not included.
The formula for this is very simple:
I=PRT,
where I am interested,
P is Principal,
R is the percentage rate expressed as decimal and
T is time, which is generally expressed in years, assuming your rate is an annual rate.
Steps to calculate simple interest
Find the Principal. This is the amount of money borrowed or lent at the start of the year for which the interest will be calculated for.
Find the Rate as a decimal. This is the percentage of the Principal you will pay back each year. Divide the percentage by 100 to give the decimal value.
Specify the Time in years over which you want the interest calculating
Multiply Principle × Rate × Time to calculate the simple interest. This is the money you will pay/be paid on top of what was lent or borrowed.
Example
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The Principal sum is Rs.
Explanation:
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs. (39 × 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.
Example
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c. per annum, in 5 years. The principal sum is Rs. 8925
Principal = Rs. $\left(\frac{100 \times 4016.25}{9 \times 5}\right)$
= Rs. $\left(\frac{401625}{45}\right)$
= Rs. 8925.
3. Compound interest
Compound interest is where interest is paid on the amount already earned leading to greater and greater amounts of interest.
Formula
Total Amount = Principal + Cl (Compound Interest)
(a) Formula for Interest Compounded Annually Total Amount = P (1+(R/100))n
(b) Formula for Interest Compounded Half Yearly Total Amount = P(1+(R/200))2n
(c) Formulae for Interest Compounded Quarterly Total Amount = P( 1 +(R/400))4n
(d) Formulae for Interest Compounded Annually with fractional years (e.g. 2.5 years)
Total Amount = P(1+(R/100))a ×(1+(bR/100))
here if year is 2.5 then a =2 and b=0.5
(e) With different interest rates for different years Say x% for year 1, y% for year2, z% for year3
Total Amount = P(1+(x/100))*(1+(y/100))*(1+(z/100))
where,
CI = Compound Interest,
P = Principal or Sum of amount,
R = % Rate per annum,
” n = Time Span in years
Example
An amount of Rs 1,500.00 is deposited in a bank paying an annual interest rate of 4.3%, compounded quarterly. Find the balance after 6.years.
Solution
Use the regular compound interest formula,
A = P (1 + r/n) with P = 1500, r = 4.3/100 = 0.043,
n = 4 (not 1/4), t = 6. Therefore,
A = 1500 $\left(1+\frac{0.043}{4}\right)^{4(6)}$ = \$ 1,938.84
So, the balance after 6 years is approximately Rs. 1,938.84.
4. Annuities
An annuity is a type of investment in which regular payments are made over the course of multiple periods. Annuities help both the creditor and debtor have predictable cash flows, and it spreads payments of the investment out over time.
5. Various types of annuity
1. Contingent annuity – An annuity arrangement in which the beneficiary does not begin receiving payments until
a specified event occurs. A contingent annuity may be set up to begin sending payments to a beneficiary upon the death of another individual who wishes to ensure financial stability for the beneficiary or upon retirement or disablement of the beneficiary. ”
2. Ordinary annuity – Payments are required at the end of each period. For example, straight bonds usually pay coupon payments at the end of every six months until the bond’s maturity date.
3. Annuity due – Payments are required at the beginning of each period. Rent is an example of an annuity due. You are usually required to pay rent when you first move in at the beginning of the month, and then on the first of each month thereafter.
4. Deferred Annuity- An annuity in which the annuitant does not begin to receive payments until some future date. A deferred annuity has two phases: a savings phase and an income phase. During the savings phase, the annuitant places money into the annuity, which invests it on behalf of the annuitant. In the income phase, the annuitant receives payments. It is important to note that a deferred annuity is not taxed until the income phase begins. It also pays a death benefit to the survivor(s) of the annuitant. Nearly all retirement plans are deferred annuities
5. Immediate annuity – If at the end of the period the periodic payments are made.
6. Forborne annuity – The annuity that is left unpaid for years is called a forborne annuity.
6. Amount of annuity or future value of the annuity
The value of a group of payments at a specified date in the future. These payments are known as an annuity or a set of cash flows. The future value of annuity measures how much you would have in the future given a specified rate of return or discount rate.
The total amount of Annuity = sum of all periodic payments+ sum of all interest on periodic payments.
1. Ordinary annuity – Here the payments are made when the period ends. It is also called an immediate annuity. Most installment loans can be classified as ordinary annuities. Mortgages with the first payment due a month after the initial loan date are one of the most common examples of an ordinary annuity.
2. Annuity due – The payments are made at the beginning of each period. Any fixed payment for a service or property that occurs before a service period begins is an example of an annuity payment. Common applications include rent payments
Present Value – Present Value (PV) is a formula used that calculates the present-day value of an amount that is received at a future date. The premise of the equation is that there is a “time value of money”.
Time value of money is the concept that receiving something today is worth more than receiving the same item at a future date.
7. Present value of the annuity
The present value of the annuity formula determines the value of a series of future periodic payments at a given time. The present value of the annuity formula relies on the concept of the time value of money, in that one Rupee present day is worth more than that same Rupee at a future date. It is helpful for calculating the series of retirement payments etc.
8. Present value of ordinary annuity
The Present Value of an Ordinary Annuity is the value of a stream of expected or promised future payments that have been discounted to a single equivalent value today. It is extremely useful for comparing two separate cash flows that differ in some way.
9. Present value of an annuity due
The Present Value of an Annuity Due is identical to an ordinary annuity except that each payment occurs at the beginning of a period rather than at the end. Since each payment occurs one period earlier, we can calculate the present value of an ordinary annuity and then multiply the result by (1 + i).
PV Ordinary Annuity = C * $\left[\frac{1-(1+\mathrm{i})^{-\mathrm{n}}}{\mathrm{i}}\right]$
C = Cash flow per period
i = Interest rate
n = Number of payments
This calculates the present value of an ordinary annuity.
To calculate the present value of an annuity due, multiply the result by (1+i). (The payments start at time zero instead of one period into the future.)
Mathematics Of Finance MCQ Questions
Question 1.
Interest is
b. Money borrowed
c. Extra money paid on borrowed money
d. Borrowed run and above
c. Extra money paid on borrowed money
Question 2.
The conversion period is
a. The time for which loan is given
b. The time for which interest is calculated
c. Period of conversion of simple interest into compound interest
d. None of the above
b. The time for which interest is calculated
3. It is correct for annuity
a. It is the run paid
b. Fixed run paid at regular intervals
c. Fixed sum paid at regular interest under stared condition
d. All of the above
c. Fixed sum paid at regular interest under the stared condition
Question 4.
Annuity due is the periodic payment made
a. At the end of each period
b. At the beginning of each period
c. Start only after a specified period
d. After a fixed number of interest
b. At the beginning of each period
Question 5.
The time between two successive payment dates of an annuity is called
a. Annuity certain
b. For borne annuity
c. Payment Period
d. Terms of an annuity
c. Payment Period
6. Amount of an annuity is
a. Total amount of all provide payment and total interest on the payments
b. Total time from the beginning of the first payment to the last payment
c. The total amount paid from the time loan taken till loan and
d. None of the above
a. Total amount of all provide payments and total interest on the payments
Question 7.
If the principal amount is 99 and the amount payable is 199 at the end of the year then the rate of interest is
a. 100%
b. 99%
c. 100.5%
d. 200%
c. 100.5%
Question 8.
There are __________types of annuity
a. 7
b. 8
c. 9
d. 6
b. 8
Question 9.
An insurance company designed to pay a certain amount at a specified intervals is
a. Annuity
b. Simple interest
c. Compound interest
d. none of the center
a. Annuity
Question 10.
The rate of interest is decided by
a. Borrower
b. Lender
c. Bank
d. Both a & b
d. Both a & b
Question 11.
Compound interest is
a. The interest on principal amount
b. The interest on previously accumulated interest as well as interest on principal amount
c. The interest on principle for a number of years
d. All of the above
b. The interest on previously accumulated interest as well as interest on the principal amount
Question 12.
If an investment yields an interest rate of 10% annually compounded quarterly. What is the effective interest rate?
a. 10.4%
b. 4%
c. 10%
d. 10.2%
a. 10.4%
Question 13.
Calculate the total amount received after 5 years if yearly Rs. 200 is being invested and the rate of interest is 6% annually compounded
a. Rs. 1000
b. Rs. 1200
c. Rs. 1500
d. Rs. 1127.42
d. Rs. 1127.42
Question 14.
If after 10 years the desired balance is Rs. 5000 then how much should be invested if the rate of interest is 5% compounded quarterly
a. Rs. 3042
b. Rs. 3042.07
c. Rs. 3042.70
d. Rs. 3040
b. Rs. 3042.07
Question 15.
Historically which type of annuity did all the companies possessed?
a. Fixed annuity
b. Annuity due
c. Deferred annuity
d. None of the above
a. Fixed annuity
Question 16.
Raj borrowed Rs. 8000 for 180 days at 5% interest. On the 90th day he makes a partial payment of Rs. 2500. What will be the adjusted principal on the maturity date?
a. Rs. 2500
b. Rs. 8000
c. Rs. 5600
d. None of the above
c. Rs. 5600
Question 17.
Ajay has purchased a second-hand car by taking a loan of Rs. 5000 from the bank at a rate of interest of 8% for 4 years. Calculate the total amount he will be paying to the bank on a loan of Rs. 5000 at the end of 4 years
a. Rs. 6820
b. Rs. 6820.44
c. Rs. 6800.44
d. Rs. 6802.44
d. Rs. 6802.44
Question 18.
If it is written, “interest is compounded semi-annually”. It means.
a. In 2 years there is one conversion period
b. The conversion period is 2 per year
c. Every year interest is halved
d. None of the above
b. Conversion period is 2 per year
Question 19.
If the final amount received is 1938.84 for a deposit made for 6 years. What will be the principal amount? The annual interest rate is 4.3% compounded quarterly
a. Rs. 1200
b. Rs. 1000
c. Rs. 1600
d. Rs. 1500
d. Rs. 1500
Question 20.
Interest rate per conversion period is calculated by.
a. Rate of interest
b. Conversion period/annual interest rate
c. Annual interest rate/conversion period
d. None of the above
c. Annual interest rate/conversion period
Question 21.
Suppose you purchase a share of a company at a price of Rs. 77 a share and sold it after a month for Rs. 82 a share. You have received a dividend of Rs. 1 at the end of the month. Calculate your annual rate of return, assuming monthly compounding
a. 146%
b. 140%
c. 100%
d. 200%
a. 146%
Question 22.
If the question arises that how much the company should pay lump sum today so that a future pension plan could be purchased then it is.
a. Future value of annuity
b. Present value of annuity
c. Both a & b
d. None of the above
b. Present value of the annuity
Question 23.
What will be the total amount of interest (rounded off) received .by Ajay if he invests Rs. 10000 for 5 years at an interest rate of 7.5 % compounded quarterly.
a. Rs. 4499
b. Rs. 4490
c. Rs. 4500
d. Rs. 4994
a. Rs. 4499
Question 24.
If a firm’s debt ratio is 45% this means ________ of the firm’s assets are financed by equity financing provided debt ratio = debt/Total Assets
a. 50%
b. 55%
c. 45%
b. 55%
Question 25.
A firm has paid out Rs. 150,000 as dividends from its net income of Rs. 250,000 what is the retention ratio for the firm? (Retention ratio = net income – dividend/Net income)
a. 12%
b. 25%
c. 40%
d. 60%
c. 40%
Question 26.
When the market’s required rate of return for a particular bond is much less than its coupon rate, the bond is selling at:
b. Discount
c. Par
Question 27.
If you plan to save Rs. 5,000 with a bank at an interest rate of 8% what will be the worth of your amount after 4 years if interest is compounded annually?
a. Rs. 5,400
b. Rs. 5,900
c. Rs. 6,600
d. Rs. 6,802
d. Rs. 6,802
Question 28.
Which of the following measure reveals how much profit a company generates with the money shareholders have invested ?
a. Profit margin
b. Return on Assets
c. Return on equity
d. Debt-Equity Ratio
c. Return on equity
Question 29.
If you have Rs. 850 and you plan to save it for 4 years with an interest rate of 10%, what will be the future value of your savings?
a. Rs. 1,000
b. Rs. 1,244
c. Rs. 1,331
d. Rs. 1,464
b. Rs. 1,244
Question 30.
You need Rs. 10,000 to buy a new television. If you have Rs. 6, 000 to invest at 5 percent compounded annually, how long will you have to wait to buy the televisions?
a. 8.42 years
b. 10.51 years
c. 15.75 years
d. 18.78 years
b. 10.51 years
Question 31.
How many years will it take to pay a Rs. 11,000 loan with an Rs. 1241.08 annual payment and a 5% interest rate?
a. 6 years
b. 12 years
c. 24 years
d. 48 years
b. 12 years
Question 32.
Which one of the following terms refers to the risk that arises for bond owners from fluctuating interest rates?
a. Fluctuations Risk
b. Interest rate Risk
c. Real-Time Risk
d. Inflation Risk
b. Interest rate Risk
Question 33.
A sum of Rs.1,200 becomes Rs.1,323 in two years at compound interest compounded annually. Find the rate percent?
a. 5%
b. 6%
c. 7%
d. 8%
a. 5%
Hint
Amount=P(1+r)n
1323 = 1200(1 + r)2
r=0 .05
r = 5%
Question 34.
The difference between simple interest and compound interest compounded annually on X sum of money for 2 years at 4% per annum is Rs.1. The value of ‘X’ is
a. Rs. 100
b. Rs. 300
c. Rs. 500
d. Rs. 625
d. Rs. 625
Hint
S.I. = PRT
where I am interested,
P is Principal,
R is the percentage rate expressed as decimal and T is time x is principal
S.I. =x 2.4/100 = ,08x
Total Amount = Principal + Cl (Compound Interest)
Cl = Total amount – Principal
Amount =P(1+r)n
= X (1+.04)2
Cl = X(1+.04)2 – X = x({1 +.04}2 – 1)
= 0816x
According to question C.l. – S.l. = 1
X (.0816- .08) = 1
X = 170016 = 625
Question 35.
The time between two successive dates of an annuity is called
a. Payment interval
b. Future value of annuity
c. Contingent annuity
d. Annuity certain
a. Payment interval
Hint
The time period between two successive dates of an annuity is called payment interval.
Question 36.
Compound interest for Rs. 1 ,000 for 4 years at 5% per annum when it is compounded quarterly-
a. Rs. 215
b. Rs. 218
c. Rs. 220
d. Rs. 225
c. Rs. 220
Hint
Total amount when Interest compounded quarterly = P(1+(R/400))4n
A = 1000(1+(5/400)4×4
A = 1220
C.l. = A-P = 1220-1000 = 220
Question 37.
Amjad invested a sum of money at 8% per annum at simple interest for’t’ years. At the end of’t’ year, Amjad got back 4 times his original investment.’ The value of ‘t; 3
a. 5 Years
b. 10.5 Years
c. 25.5 Years
d. 37.5 Years
d. 37.5 Years
Hint
r = 8% , Let Principal = P
S.I. = Prt
A = S.I. + P
As per question amount = 4 P
Thus putting these value we get
A = S.I. + P
4P = S.I. + P
S.I. = 3P
Prt = 3 P
P x ,08 x t = 3P
t = 3P/Px .08 = 3/.08 =37.5 yrs
Question 38.
An annuity is a fixed sum paid at regular intervals. An annuity that continues for a number of years is called:
a. Deferred annuity
b. Immediate annuity
c. Uniform annuity
d. Perpetual annuity
d. Perpetual annuity
Hint
A perpetual annuity is an annuity in which the periodic payments begin on a fixed date and continue indefinitely.
Question 39.
In what period, the compound interest on ~ 30,000 at 7% per annum amounts to Rs.4,347
a. 2 years
b. 1.5 years
c. 3 years
d. 4 years
a. 2 years
Hint
C.I.= 4347
r = 7%
P= 30000
A = P + C.I.
A = 34347
A = P(1+(R/100))n
34347 = 30000 (1+(7/100)n
34347= 30000 (1+ .07)n
1.1449 = 1.07n
n = 2 yrs
Question 40.
What is the present value of Annuity on t1 for 2 years @ 10% p.a.?
a. 0.18
b. 2
c. 3
d. 0.67
b. 2
Hint
PVOrdinary Annuity = C *
C=cash flow per period
i = interest rate
n=number of payments
i= 10%/12 = .1/12 = .00833
n= 1 × 2 = 2
c = 1
PV = 1(1 – $\frac{.00833)^{-2}}{(.00833)}$
PV = 1 × 1.975
PV =2
Question 41.
Cl = Rs. 30,000, @7% rate is Rs. 4347. Find the number of years.
a. 4 years
b. 2 years
c. 2.5 years
d. 3 years
b. 2 years
Hint
P =A – C.l.
= 30000-4347 = 25653
A =P (1+r/100)n
30000 = 25653 (1+.07)n
1.1695 = (1.07)n
(1.07)2 = (1.07)n
2 = n
Question 42.
When the loan amount is given @ 9% then what will be the outstanding amount after 3 months?
a. 1014
b. 1000
c. 1012.5
d. 1013.6
c. 1012.5
Hint
t= 3 months = 1/4 yrs
r =9% p.a.= 2.25 per quarter
A = 1000(1+ 2.25/100)1/4×4
A = 1000 x 1.0225 = 1022.5
Question 43.
The difference, between 81 & Cl is 1, the time period is 2 years, rate 4% p.a., find the principal amount?
a. 629
b. 625
c. 700
d. 600
b. 625
Hint
S.I. = PRT
where I am interested,
P is Principal,
R is the percentage rate expressed as decimal and T is time
x is principal
5.1. = × 2 .4/100 = .08x
Total Amount = Principal + Cl (Compound Interest)
CI = Total amount – Principal
Amount =P(1+r)n
= x (1+.04)2
CI = x (1+.04)2 – X
= X({1+.04}2 – 1)
= 0816x
According to question C.I. – S.l. = 1 x(.0816 – .08) = 1 x= 1/.0016 = 625
Question 44.
On what sum will the compound compounded interest at 5% p.a. for 2 yeas annually be Rs. 1,640?
a. 10,000
b. 12,000
c. 15,525
d. 16,000
d. 16,000
Hint
Amount = P(1+r)n
1640 = P{(1 +5/100)2 -1}
1640 = P{1.1025-1}
1640 = 1025 P
P = 16000
Question 45.
A man took a loan from a bank at the rate of 12% р.a. on simple interest. After 3 years he had to pay Rs.5,400 interest only for the period. The principal amount borrowed by him was:
a. Rs. 10,000
b. Rs. 20,000
с. Rs. 15,000
d. Rs. 18,000
с. Rs. 15,000
Hint
S.l. =PRT
P = S.I./RT = 5400 × 100/12 × 3 =15000
Question 46.
With an interest rate of 5 percent, the present value of ~ 100 received one year from now is approximate:
a. Rs. 95.238
b. Rs. 105
c. Rs. 100
d. Rs. 95
b. Rs. 105
Hint
A =P(1+R/100)t
= 100(1+5/100)t = 105
Question 47.
Vidhya Signed a contract in which she will receive ~ 3 million immediately and 1 million will be paid to her every year for the next 5 years. If the interest rate is 10%, the present value of the contract is approximate:
a. Rs. 7 million
b. Rs. 6.79 million
c. Rs. 8 million
d. Rs. 8.79 million
b. Rs. 6.79 million
Hint
PV = 3 + 1 × $\frac{1}{(1+0.01)^{1}}$ + 1 × $\frac{1}{(1+0.01)^{2}}$ + 1 × $\frac{1}{(1+0.01)^{3}}$ + 1 × $\frac{1}{(1+0.01)^{4}}$ + 1 × $\frac{1}{(1+0.01)^{5}}$
= 3 + 3.79 = 6.79 million
Question 48.
A sum fetched a total Simple Interest of Rs 4016.25 at the rate of 9% p.a. in 5 years. Find the sum?
a. Rs. 8900
b. Rs. 4462.50
c. Rs. 8032.50
d. Rs. 8925 |
# Maharashtra Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.6
Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.6 Questions and Answers.
## Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.6
Question 1.
Find the value of
(a) 15C4
Solution:
(b) 80C2
Solution:
(c) 15C4 + 15C5
Solution:
(d) 20C1619C16
Solution:
Question 2.
Find n if
(a) 6P2 = n(6C2)
Solution:
(b) 2nC3 : nC2 = 52 : 3
Solution:
(c) nCn-3 = 84
Solution:
Question 3.
Find r if 14C2r : 10C2r-4 = 143 : 10.
Solution:
∴ 2r(2r – 1) (2r – 2) (2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1) (2r – 2) (2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4
Question 4.
Find n and r if,
(a) nPr = 720 and nCn-r = 120
Solution:
(b) nCr-1 : nCr : nCr+1 = 20 : 35 : 42
Solution:
Question 5.
If nPr = 1814400 and nCr = 45, find n+4Cr+3.
Solution:
Question 6.
If nCr-1 = 6435, nCr = 5005, nCr+1 = 3003, find rC5.
Solution:
Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 8 green balls, and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 8 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in 6C3 ways.
3 reen balls can be selected from 8 green balls in 8C3 ways.
3 blue balls can be selected from 7 blue balls in 7C3 ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour
Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in 6C3 ways.
2 girls can be selected from 4 girls in 4C2 ways.
∴ Number of ways the team can be selected
Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = nC2
Given 66 handshakes were exchanged.
66 = nC2
66 = $$\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}$$
66 × 2 = $$\frac{n(n-1)(n-2) !}{(n-2) !}$$
132 = n (n – 1)
n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.
Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points
Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
(iv) n = 8
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides.
∴ Through ‘n’ points we can draw nC2 lines including sides.
∴ Number of diagonals in n sided polygon = nC2 – n (n = number of sides)
Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel, they intersect at a point.
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = 20C2
= $$\frac{20 !}{2 ! 18 !}$$
= $$\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}$$
= 190
Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if (a) no three points are collinear (b) four points are collinear
Solution:
There are 10 points on a plane.
(a) When no three of them are collinear.
A line is obtained by joining 2 points.
∴ Number of lines passing through these points = 10C2
= $$\frac{10 !}{2 ! 8 !}$$
= $$\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}$$
= 5 × 9
= 45
(b) When 4 of them are collinear.
If no three points are collinear, we get a total of 10C2 = 45 lines by joining them. …..[From (i)]
Since 4 points are collinear, only one line passes through these points instead of 4C2 lines.
4C2 – 1 extra lines are included in 45 lines.
Number of lines passing through these points
= 45 – (4C2 – 1)
= 45 – $$\frac{4 !}{2 ! 2 !}$$ + 1
= 45 – $$\frac{4 \times 3 \times 2 !}{2 \times 2 !}$$ + 1
= 45 – 6 + 1
= 40
Question 14.
Find the number of triangles formed by joining 12 points if
(a) no three points are collinear
(b) four points are collinear
Solution:
There are 12 points on the plane.
(a) When no three of them are collinear.
A triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = 12C3
= $$\frac{12 !}{3 ! 9 !}$$
= $$\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}$$
= 220
(b) When 4 of these points are collinear.
If no three points are collinear, total we get 12C3 = 220 triangles by joining them. ……[From (i)]
Since 4 points are collinear, no triangle can be formed by joining these four points.
4C3 extra triangles are included in 220 triangles.
∴ Number of triangles that can be obtained from these points = 12C34C3
= 220 – $$\frac{4 !}{3 ! \times 1 !}$$
= 220 – $$\frac{4 \times 3 !}{3 !}$$
= 220 – 4
= 216
Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
There are 8 consonants and 3 vowels.
From 8 consonants, 4 can be selected in 8C4
= $$\frac{8 !}{4 ! 4 !}$$
= $$\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}$$
= 70 ways.
From 3 vowels, 2 can be selected in 3C2
= $$\frac{3 !}{2 ! 1 !}$$
= $$\frac{3 \times 2 !}{2 !}$$
= 3 ways.
Now, to form a word, these 6 ietters (i.e., 4 consonants and 2 vowels) can be arranged in 6P6 = 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.
Question 16.
Find n if,
(i) nC8 = nC12
Solution:
nC8 = nC12
If nCx = nCy, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20
(ii) 23C3n = 23C2n+3
Solution:
23C3n = 23C2n+3
If nCx = nCy, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4
(iii) 21C6n = $${ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}$$
Solution:
21C6n = $${ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}$$
If nCx = nCy, then either x = y or x = n – y
∴ 6n = n2 + 5 or 6n = 21 – (n2 + 5)
∴ n2 – 6n + 5 = 0 or 6n = 21 – n2 – 5
∴ n2 – 6n + 5 = 0 or n2 + 6n – 16 = 0
If n2 – 6n + 5 = 0, then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then
n2 + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n2 + 6n – 16 = 0, then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2
∴ n = 1 or n = 2
Check:
n = 2
∴ n2 + 5 = 22 + 5 = 9
21C6n = 21C12
and $${ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}$$ = 21C9
nCr = nCn-r
21C12 = 21C9
∴ n = 2 is a right answer.
(iv) 2nCr-1 = 2nCr+1
Solution:
2nCr-1 = 2nCr+1
If nCx = nCy, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n
Check:
2nCr-1 = 2nCn-1
and 2nCr+1 = 2nCn+1
using nCr = nCn-r, we have
2nCn+1 = 2nC2n-(n+1) = 2nCn-1
2nCr-1 = 2nCr+1
(v) nCn-2 = 15
Solution:
nCn-2 = 15
nC2 = 15 …..[∵ nCr = nCn-r]
∴ $$\frac{n !}{(n-2) ! 2 !}=15$$
∴ $$\frac{n(n-1)(n-2) !}{(n-2) ! \times 2 \times 1}=15$$
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Equating both sides, we get
∴ n = 6
Question 17.
Find x if nPr = x nCr.
Solution:
Question 18.
Find r if 11C4 + 11C5 + 12C6 + 13C7 = 14Cr.
Solution:
Question 19.
Find the value of $$\sum_{r=1}^{4}{ }^{(21-r)} \mathrm{C}_{4}$$.
Solution:
Question 20.
Find the differences between the greatest values in the following:
(a) 14Cr and 12Cr
Solution:
Greatest value of 14Cr.
Here, n = 14, which is even.
Greatest value of nCr occurs at r = $$\frac{n}{2}$$ if n is even.
∴ r = $$\frac{n}{2}$$
∴ r = $$\frac{14}{2}$$ = 7
∴ Difference between the greatest values of 14Cr and 12Cr = 14Cr12Cr
= 3432 – 924
= 2508
(b) 13Cr and 8Cr
Solution:
Greatest value of 13Cr.
Here n = 13, which is odd.
Greatest value of nCr occurs at r = $$\frac{n-1}{2}$$ if n is odd.
∴ r = $$\frac{\mathrm{n}-1}{2}$$
∴ r = $$\frac{13-1}{2}$$ = 6
∴ Difference between the greatest values of 13Cr and 8Cr = 13Cr8Cr
= 1716 – 70
= 1646
(c) 15Cr and 11Cr
Solution:
Greatest value of 15Cr.
Here n = 15, which is odd.
Greatest value of nCr occurs at r = $$\frac{n-1}{2}$$ if n is odd.
∴ r = $$\frac{n-1}{2}$$
∴ r = $$\frac{15-1}{2}$$ = 7
∴ Difference between the greatest values of 15Cr and 11Cr = 15Cr11Cr
= 6435 – 462
= 5973
Question 21.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:
∴ Number of ways a boy can invite his friends to a party so that three or more of join the party = 10 + 5 + 1 = 16
Question 22.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many of possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consist of at least 3 women.
Consider the following table:
∴ No. of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.
Question 23.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women.
Consider the following table:
∴ Number of committees with at least 5 women
= 14112 + 14700 + 6720 + 1260 + 81
= 36873
(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= 18C10 – 36873
= 18C8 – 36873
= 43758 – 36873
= 6885
Question 24.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:
∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.
Question 25.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 player is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicket keeper and at least 4 bowlers are to be included in the team.
Consider the following table:
∴ Number of ways a team of 11 players can be selected
= 45045 + 6006
= 51051
Question 26.
Five students are selected from 11. How many ways can these students be selected if
(a) two specified students are selected?
(b) two specified students are not selected?
Solution:
5 students are to be selected from 11 students.
(a) When 2 specified students are included,
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = 9C3
= $$\frac{9 !}{3 ! \times 6 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}$$
= 84
∴ Selection of students is done in 84 ways when 2 specified students are included.
(b) When 2 specified students are not included, then 5 students can be selected from the remaining (11 – 2) = 9 students.
∴ Number of ways of selecting 5 students from 9 students = 9C5
= $$\frac{9 !}{5 ! 4 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}$$
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not included. |
# differentiation of x-2
#### Understand the Problem
The question is asking for the derivative of the function x - 2 with respect to x. This can be solved by applying the basic rules of differentiation.
The derivative is $$1$$.
The derivative of the function ( x - 2 ) with respect to ( x ) is ( 1 ).
#### Steps to Solve
1. Identify the function
The given function is ( f(x) = x - 2 ).
2. Apply the differentiation rule
To differentiate the function, we apply the power rule. The power rule states that the derivative of ( x^n ) is ( nx^{n-1} ). Here, we can rewrite ( x ) as ( x^1 ).
Thus,
( \frac{d}{dx}(x) = 1 ) (since ( n = 1 ))
( \frac{d}{dx}(-2) = 0 ) (since the derivative of a constant is 0).
3. Combine the derivatives
Now combine the derivatives we found:
[ \frac{d}{dx}(x - 2) = \frac{d}{dx}(x) - \frac{d}{dx}(2) = 1 - 0 = 1 ]
The derivative of the function ( x - 2 ) with respect to ( x ) is ( 1 ). |
# Video: Finding the Union of a Set and the Complement of Another Using Venn Diagrams
Using the Venn diagram, find π βͺ π bar.
02:09
### Video Transcript
Using the Venn diagram, find π union π bar.
Letβs begin by defining some of the notation in this question. The U-shaped symbol between π and π bar means the union. In probability, this means or. We need to find the set of values that are in π or in π bar. The notation π bar means the complement of π. This is the set of all values not in π.
We can see from the Venn diagram that set π is all the numbers shown. This is the integer values from one to nine. The set of values contained in circle π are one, two, six, and nine. The set of values in circle π are one, three, and six. This means that the complement of π will be all the values except these. The numbers that are not in set π are two, four, five, seven, eight, and nine.
We recall that we need to find the values that are in set π or in set π bar. The numbers two and nine appear in both of these sets, but we do not need to write this number twice. The set of numbers contained within the union of π and π bar are one, two, four, five, six, seven, eight, and nine. In fact, the only number that is not in this union is three. This is because it appears in set π, but not in set π. As we want all the values that are either in set π or not in set π, we do not include the three. |
# To graph: The scatter plot of the data.
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 1, Problem 1P
(a)
To determine
## To graph: The scatter plot of the data.
Expert Solution
### Explanation of Solution
Graph:
Consider the femur length as the x coordinates and height as the y coordinates.
The scatter plots of the given data are shown below in Figure 1.
In Figure 1, all the points are plotted on the graph. No two points coincide each other.
(b)
To determine
### To find: The linear function that represents the given data and draw the graph for it.
Expert Solution
The linear function that represents the given data is y=1.881x+82.65 .
### Explanation of Solution
By the use of online calculator, the regression line for the given data obtained is y=1.881x+82.65 , where y is the height and x is the femur length.
The above regression line is in the form of linear equation.
Therefore, the linear function that represents the given data is y=1.881x+82.65 .
The graph that represents the equation y=1.881x+82.65 is shown below in Figure 2.
From Figure 2, the graph for the linear equation y=1.881x+82.65 is a straight line.
(c)
To determine
### To find: The height of the person if the femur length is 58 cm.
Expert Solution
The height of the person if the femur length is 58 cm is 191.748 cm.
### Explanation of Solution
Consider the length of the femur is 58 cm.
From part (b), it is obtained that the linear equation that represents the given data is y=1.881x+82.65 .
Substitute x = 58 in the above equation to compute the height.
y=1.881(58)+82.65=109.098+82.65=191.748
Thus, the height of the person if the femur length is 58 cm is 191.748 cm.
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! |
## Target Profit Analysis
### Learning Outcomes
• Define target profit analysis and use it to make sales volume calculations
Minnesota Kayak has a few investors who are interested in getting a return on their investment. They have talked with your supervisor, and between them all, would like to get $30,000 a month in profit to divide between them. You have been tasked with figuring out how many kayaks need to be sold in order to get the investors their return! Target profit analysis helps us to know how much in dollar sales a company will need to reach a certain profit point. This is one of the key uses of the CVP analysis. Once the basic data is calculated, it can offer a great deal of insight and help in planning. Minnesota Kayak Company needs to sell 28 kayaks in our example to break even. The equation method or the formula method can be used with the same result. Remember the formula method is simply a shortened version of the equation method, so both ways should come to the same conclusion. With the previous information you can then figure out, the dollar sales needed to break even: $28\text{ kayaks}\times\500\text{ per kayak}=\14,000\text{ in sales}$ What if they now want to show a$30,000 a month profit?
So with that information we now have the following:
Price per kayak $500 variable costs per kayak$225 Contribution margin per kayak $275 Fixed costs/month$7,700
With this information, how many kayaks do we need to sell to show a $30,000 profit at the end of the month? It is the same exact formula we used to calculate the break-even point! Remember, we put -0- for the profit in when we were looking to break-even. We simply replace the -0- with$30,000 and now we can calculate how many kayaks we need to sell to meet our profit goal. Pretty neat huh?
Using the equation method:
$\begin{array}{rcl}\text{Profit}&=&\text{Unit CM}\times\text{Q}-\text{Fixed Expense}\\\30,000&=&\275\times\text{Q}-\7,700\\\30,000+\7,700&=&\275\times\text{Q}\\\dfrac{\37,700}{275}&=&\text{Q}\end{array}$
So we now need to sell 138 kayaks to profit $30,000! How much in sales do we need? 138 ×$500 each= $69,000 in sales. Note: Whenever you are doing break-even or target profit analysis, you should round units up. If you used your calculator to divide$37,700 by 275, you would have gotten 137.090909. Why did we round up to 138 kayaks? Well, you can’t sell only 0.09 of a kayak, so you will either sell 137 or 138. If you round down to 137 units and then plug it into the formula to calculate profit, then your profit only ends up being $29,975 – just short of our$30,000 target profit we stated. So we have to round up to 138 kayaks to meet our target profit (and in this case, slightly exceed, as selling 138 kayaks leads to a profit of $30,250). How would we get there using the formula method? $\text{Unit Sales to attain the target profit}=\dfrac{\text{Target Profit}+\text{Fixed expenses}}{\text{Unit CM}}$ So in our kayak case $\begin{array}{rcl}\text{Unit sales needed}&=&\dfrac{\30,000+\7,700}{\275}\\&=&\dfrac{\37,700}{\275}\end{array}$ So again, we need 138 kayaks sold to make a$30,000 profit!
138 kayaks × $500 selling price per kayak =$69,000 in sales.
We can now plug in any amount of desired profit and calculate how many units we need to sell! This is amazing information for business owners and managers to have available. But see the importance of good numbers for your fixed and variable costs? Keep in mind how much a small difference in costs can affect our profit!
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
# Math homework solver
In addition, Math homework solver can also help you to check your homework. We will also look at some example problems and how to approach them.
## The Best Math homework solver
This Math homework solver helps to fast and easily solve any math problems. To solve an equation using square roots, we must first isolate the square root term on one side of the equation. We can then square both sides of the equation to cancel out the square root terms, leaving us with a much simpler equation to solve. This method can be very effective for equations that are otherwise difficult to solve.
The system of equations is the mathematical representation of a set of related equations. It is an ordered list of equations with and without solutions. The solution of a system of equations is the set of values that satisfies all the given equations. To solve system of equations, first we need to identify all the variables involved in the given system. Then we need to add all unknowns and solve for them individually. Once all unknowns are known, we can add all knowns and solve for them individually. This way, we get a single solution from a set of individual solutions. We use algebra to find a solution or to solve a system of linear equations or inequalities. Algebra is used to simplify, manipulate and evaluate expressions and questions involving variables. Algebra is also used for solving more complicated problems such as quadratic equations, polynomial equations, rational expressions, exponential expressions etc. Algebra can be used to solve systems with several variables or when there are different types of questions (such as multiple choice, fill-in-the-blank). There are various methods one can use to solve system of linear equations like substitution method, elimination method and combination method etc. In this article, we will discuss several approaches on solving systems of linear equation i.e substitution method etc.
Imaginary numbers are numbers that cannot be expressed as a real number. They are often used in mathematical and scientific equations to represent an unknown or unmeasured quantity. In order to solve an equation with an imaginary number, you must first determine what the number is representing. Once you know what the number is representing, you can use algebra and other mathematical operations to solve the equation.
Once I have a good understanding of the problem, I start brainstorming potential solutions. I try to come up with as many possible solutions as I can, no matter how crazy they might sound at first. After I have a good list of potential solutions, I start to narrow them
While there are several reasons why this could occur, the main culprit is usually inaccurate body weight measurements. While it may be tempting to call this out as a potential error in your paper, remember that this is only one part of a larger investigation. Once you have corrected your data and reanalysed your results, point slope form should not be present. You will be able to find the underlying issue and correct it before publishing your paper. This type of error is hard to detect because it is so small. You can try to make sure that your patients are not underweight or overweight for many reasons: If possible, take an impression of their foot before surgery to get an exact measurement of their leg length. If they are too short, then they will have more difficulty getting into comfortable shoes after surgery. If they are too tall, then they will have more difficulty taking off their shoes when they leave the hospital. |
# Math Expressions Grade 4 Unit 3 Lesson 7 Answer Key Just-Under Quotient Digits
## Math Expressions Common Core Grade 4 Unit 3 Lesson 7 Answer Key Just-Under Quotient Digits
Math Expressions Grade 4 Unit 3 Lesson 7 Homework
Solve.
Explanation:
The division is a method of distributing a group of things into equal parts.
It is one of the four basic operations of arithmetic,
Every division equation has 4 main parts – the dividend, the divisor ,the quotient and remainder.
A dividend is a number that is divided by the divisor.
The divisor is the factor that divides the dividend.
The result obtained by the division process is called the quotient.
The number left behind after the completion of the division is called the remainder.
So, all the above divisions are done based on this.
Math Expressions Grade 4 Just-Under Quotient Digits Lesson 7 Pdf Answer Key Question 2.
Explanation:
The division is a method of distributing a group of things into equal parts.
It is one of the four basic operations of arithmetic,
Every division equation has 4 main parts – the dividend, the divisor ,the quotient and remainder.
A dividend is a number that is divided by the divisor.
The divisor is the factor that divides the dividend.
The result obtained by the division process is called the quotient.
The number left behind after the completion of the division is called the remainder
So, all the above divisions are done based on this.
Unit 3 Lesson 7 Just-Under Quotient Digits Math Expressions Question 3.
Describe how the repeating pattern in row 1 ¡s different from the pattern in row 2. Explain why.
Explanation:
The answer is dependent on Divisor
If divisor changes the quotient will vary and the answer changes respectively
In row 1 the divisor is 4 and dividend is 21, 22, 23, 24 and 25 , only 24 perfectly divisible by 4 with remainder 0
In row 2 the divisor is 6 and dividend is 21, 22, 23, 24 and 25,only 24 perfectly divisible by 6 with remainder 0
Use any method to solve.
Math Expressions Grade 4 Pdf Unit 3 Lesson 7 Question 4.
Quotient = 840, Remainder = 6
Explanation:
Math Expressions Grade 4 Just-Under Quotient Digits Pdf Question 5.
Quotient = 1322, Remainder = 5
Explanation:
Just-Under Quotient Digits Grade 4 Answer Key Unit 3 Math Expressions Question 6.
Quotient = 501, Remainder = 1
Explanation:
Question 7.
Quotient = 4018, Remainder = 1
Explanation:
Question 8.
Quotient = 376, Remainder = 1
Explanation:
Question 9.
Quotient = 469, Remainder = 2
Explanation:
Question 10.
Quotient = 1500, Remainder = 3
Explanation:
Question 11.
Quotient = 2090, Remainder = 0
Explanation:
Solve.
Question 12.
Altogether, the members of an exercise club drink 840 bottles of water each month. Each member drinks 8 bottles. How many members are there?
105 members are there in the exercise club
Explanation:
Question 13.
There are 7,623 pencils ready to be packaged in boxes at a factory. Each box holds 6 pencils. How many full boxes of pencils can be packaged?
1270 full boxes of pencils can be packaged
Explanation:
Math Expressions Grade 4 Unit 3 Lesson 7 Remembering
Question 1.
Explanation:
By subtracting smaller number 1,390 from bigger number 5,267
We get 3877
Question 2.
Explanation:
By subtracting smaller number 2482 from bigger number 9000
We get 6518
Question 3.
Explanation:
By subtracting smaller number 5,773 from bigger number 6,129
We get 356
Question 4.
Rick is selling fresh-squeezed lemonade for \$2 a serving. Rick makes each serving with 2 lemons and 4 tablespoons of sugar. If he sells 27 servings of lemonade, how much sugar does he use?
108 table spoons of sugar used for 27 servings
Explanation:
27
x4
108
Question 5.
An animal shelter receives 9 large bags of dog food every month for 14 years. Each bag weighs 55 pounds. How many pounds of dog food does the animal shelter receive each month?
693 pounds of dog food does the animal shelter receive each month
Explanation:
9 large bags, each bag weighs 55 pounds, total 495 pounds
55
x 9
495 pounds
Dog food receives for 14 years, that is 12 x 14 = 168 months
495
x168
83160 pounds of dog food received for 14 years
The question is how many pounds of dog food does the animal shelter receive each month
693 pounds of dog food does the animal shelter receive each month
Solve using any method.
Question 6.
Quotient = 150, Remainder = 2
Explanation:
Question 7.
Quotient = 65, Remainder = 7
Explanation:
Question 8. |
# Compute the products $A B$ and $B A$ whichever exists in each of the following cases:
Question:
Compute the products $A B$ and $B A$ whichever exists in each of the following cases:
(i) $A=\left[\begin{array}{rr}1 & -2 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$
(ii) $A=\left[\begin{array}{rr}3 & 2 \\ -1 & 0 \\ -1 & 1\end{array}\right]$ and $B=\left[\begin{array}{lll}4 & 5 & 6 \\ 0 & 1 & 2\end{array}\right]$
(iii) $A=\left[\begin{array}{lll}1 & -1 & 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{l}0 \\ 1 \\ 3 \\ 2\end{array}\right]$
(iv) $[a, b]\left[\begin{array}{l}c \\ d\end{array}\right]+[a, b, c, d]\left[\begin{array}{l}a \\ b \\ c \\ d\end{array}\right]$
Solution:
(i) $A B=\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right] \Rightarrow A B=\left[\begin{array}{ccc}1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3\end{array}\right] \Rightarrow A B=\left[\begin{array}{ccc}-3 & -4 & 1 \\ 8 & 13 & 9\end{array}\right]$
Since the number of columns in is greater then the number of rows in ABA does not exists.
(ii) $A B=\left[\begin{array}{cc}3 & 2 \\ -1 & 0 \\ -1 & 1\end{array}\right]\left[\begin{array}{lll}4 & 5 & 6 \\ 0 & 1 & 2\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc}12+0 & 15+2 & 18+4 \\ -4+0 & -5+0 & -6+0 \\ -4+0 & -5+1 & -6+2\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc}12 & 17 & 22 \\ -4 & -5 & -6 \\ -4 & -4 & -4\end{array}\right]$ Also, $B A=\left[\begin{array}{lll}4 & 5 & 6 \\ 0 & 1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 2 \\ -1 & 0 \\ -1 & 1\end{array}\right]$
$\Rightarrow B A=\left[\begin{array}{cc}12-5-6 & 8+0+6 \\ 0-1-2 & 0+0+2\end{array}\right]$
$\Rightarrow B A=\left[\begin{array}{cc}1 & 14 \\ -3 & 2\end{array}\right]$
(iii) $A B=\left[\begin{array}{llll}1 & -1 & 2 & 3\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 3 \\ 2\end{array}\right]$
$\Rightarrow A B=[0+(-1)+6+6]$
$\Rightarrow A B=\left[\begin{array}{l}11\end{array}\right]$ Also $B A=\left[\begin{array}{l}0 \\ 1 \\ 3 \\ 2\end{array}\right]\left[\begin{array}{llll}1 & -1 & 2 & 3\end{array}\right]$
Also,
$B A=\left[\begin{array}{l}0 \\ 1 \\ 3 \\ 2\end{array}\right]\left[\begin{array}{llll}1 & -1 & 2 & 3\end{array}\right]$
$\Rightarrow B A=\left[\begin{array}{cccc}0 & 0 & 0 & 0 \\ 1 & -1 & 2 & 3 \\ 3 & -3 & 6 & 9 \\ 2 & -2 & 4 & 6\end{array}\right]$
(iv)
$\left[\begin{array}{ll}a & b\end{array}\right]\left[\begin{array}{l}c \\ d\end{array}\right]+\left[\begin{array}{llll}a & b & c & d\end{array}\right]\left[\begin{array}{l}a \\ b \\ c \\ d\end{array}\right]$
\begin{aligned} \Rightarrow &[a c+b d]+\left[a^{2}+b^{2}+c^{2}+d^{2}\right] \\ &\left[a^{2}+b^{2}+c^{2}+d^{2}+a c+b d\right] \end{aligned} |
# Difference between revisions of "1988 IMO Problems/Problem 4"
## Problem
Show that the solution set of the inequality
$\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$
is a union of disjoint intervals, the sum of whose length is $1988$.
## Solution
Consider the graph of $f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$. On the values of $x$ between $n$ and $n+1$ for $n\in\mathbb{N}$ $1\le n\le 69$, the terms of the form $\frac{k}{x-k}$ for $k\ne n,n+1$ have a finite range. In contrast, the term $\frac{n}{x-n}$ has an infinite range, from $+\infty$ to $n$. Similarly, the term $\frac{n+1}{x-n-1}$ has infinite range from $-n-1$ to $-\infty$. Thus, since the two undefined values occur at the distinct endpoints, we can deduce that $f(x)$ takes on all values between $+\infty$ and $-\infty$ for $x\in(n,n+1)$. Thus, by the Intermediate Value Theorem, we are garunteed a $n such that $f(r_n)=\frac{5}{4}$. Additionally, we have that for $x>70$, the value of $f(x)$ goes from $+\infty$ to $0$, since as $x$ increases, all the terms go to $0$. Thus, there exists some $r_{70}>70$ such that $f(r_{70})=\frac{5}{4}$ and so $f(x)\ge\frac{5}{4}$ for $x\in(70,r_{70})$.
So, we have $70$ $r_i$ such that $f(r_i)=\frac{5}{4}$. There are obviously no other such $r_i$ since $f(x)=\frac{5}{4}$ yields a polynomial of degree $70$ when combining fractions. Thus, we have that the solution set to the inequality $f(x)\ge\frac{5}{4}$ is the union of the intervals $(n,r_n]$ (since if $f(x)<\frac{5}{4}$ for $x\in(n,r_n)$ then there would exist another solution to the equation $f(x)=\frac{5}{4}$.
Thus we have proven that the solution set is the union of disjoint intervals. Now we are to prove that the sum of their lengths is $1988$.
The sum of their lengths is $r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71$. We have that the equation $f(x)=\frac{5}{4}$ yields a polynomial with roots $r_i$. Thus, opposite of the coeficient of $x^{69}$ divided by the leading coefficient is the sum of the $r_i$. It is easy to see that the coefficient of $x^{69}$ is $-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71$. Thus, since the leading coefficient is $5$ we have $r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71$. Thus, the sum of the lengths of the intervals is $63\cdot71-35\cdot71=28\cdot71=1988$ as desired. |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Session 8, Part B:
Fractions With Cuisenaire Rods
In This Part: Representing Fractions With Rods | Other Denominators
Modeling Operations | Try It Yourself
Use the following Interactive Activity to work on the problems below, which let you use the rods to try out representations and operations with fractions. Hint: Which rod can you fit five times along the length of the orange rod? Hint: Use three identical rods that each represent one-fifth (1⁄5). Hint: Which rod can you fit 10 times along the length of the orange rod? Hint: 3⁄5 + 1⁄10 = 7⁄10 Move the rod that represents one-tenth (1⁄10) next to rods that each represent one-fifth (1⁄5). Hint: 7⁄10 – 2⁄5 = 3⁄10 Move the purple rod above your model for seven-tenths (7⁄10). Remove the red or white rods that take up the same space as one purple rod. Hint: How many yellow rods can you fit along the length of the orange rod? Hint: Choose a rod that can fit twice along the length of the rod that represents one whole. Hint: There are many fractions you can model. Try to model 1⁄3, 3⁄4, and 5⁄9. You can model a fraction by stacking two or more Cuisenaire® Rods. If the orange rod represents the number one whole, which rod would you use to represent one-fifth (1⁄5)? If the orange rod represents the number one whole, how would you model three-fifths (3⁄5)? If the orange rod represents the number one whole, how would you model one-tenth (1⁄10)? Place your models for three-fifths (3⁄5) and one-tenth (1⁄10) on the grid. How would you represent the addition of these two fractions? Using one purple rod, place a model for two-fifths (2⁄5) on the grid. How would you represent the subtraction 7⁄10 – 2⁄5? What fraction of the orange rod does the yellow rod represent? Use a red, purple, or brown rod to represent one whole. How can you represent one-half (1⁄2) of the rod you chose? What other fractions can you model using these Cuisenaire Rods? Drag the Cuisenaire Rods onto the grid Cuisenaire® is a trademark of ETA hand2mind. All rights reserved.
Video Segment In this video segment, Rhonda and Andrea use rods to model multiplication and division with thirds and fourths. First they must figure out what their model is going to be in order to do their computations. Watch this segment after you've completed Problems B2 and B3. If you are using a VCR, you can find this segment on the session video approximately 11 minutes and 42 seconds after the Annenberg Media logo.
Session 8: Index | Notes | Solutions | Video |
# McGraw Hill Math Grade 6 Lesson 16.2 Answer Key Commutative and Associative Properties
Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 16.2 Commutative and Associative Properties will engage students and is a great way of informal assessment.
## McGraw-Hill Math Grade 6 Answer Key Lesson 16.2 Commutative and Associative Properties
Exercises
IDENTIFY THE PROPERTY
Question 1.
3 × 4 × 2 = 2 × 4 × 3
3 × 4 × 2 = 2 × 4 × 3 = 24.
Associative property.
Explanation:
The associative property, in Math, states that while adding or multiplying numbers, the way in which numbers are grouped by brackets (parentheses), does not affect their sum or product.
3 × 4 × 2 = 2 × 4 × 3
= 8 × 3
= 24.
Question 2.
1 + 9 + 22 = 22 + 9 + 1
1 + 9 + 22 = 22 + 9 + 1 = 32.
Commutative property.
Explanation:
The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer.
1 + 9 + 22 = 22 + 9 + 1
= 31 + 1
= 32.
Question 3.
2 + 3 + 3 + 2 = 2 + (3 + 3) + 2
2 + 3 + 3 + 2 = 2 + (3 + 3) + 2 = 10.
Associative property.
Explanation:
The associative property, in Math, states that while adding or multiplying numbers, the way in which numbers are grouped by brackets (parentheses), does not affect their sum or product.
2 + 3 + 3 + 2 = 2 + (3 + 3) + 2
= 2 + 6 + 2
= 8 + 2
= 10.
Question 4.
4 × 2 + 2 × 3 = 2 × 4 + 3 × 2
4 × 2 + 2 × 3 = 2 × 4 + 3 × 2 = 14.
Associative property.
Explanation:
The associative property, in Math, states that while adding or multiplying numbers, the way in which numbers are grouped by brackets (parentheses), does not affect their sum or product.
4 × 2 + 2 × 3 = 2 × 4 + 3 × 2
= 8 + 6
= 14.
Question 5.
7 × 2 × 7 = 7 × (2 × 7)
7 × 2 × 7 = 7 × (2 × 7) = 98.
Associative property.
Explanation:
The associative property, in Math, states that while adding or multiplying numbers, the way in which numbers are grouped by brackets (parentheses), does not affect their sum or product.
7 × 2 × 7 = 7 × (2 × 7)
= 7 × 14
= 98.
Question 6.
7 × 7 × 8 × 7 = 8 × 7 × 7 × 7
7 × 7 × 8 × 7 = 8 × 7 × 7 × 7 = 2,744.
Associative property.
Explanation:
The associative property, in Math, states that while adding or multiplying numbers, the way in which numbers are grouped by brackets (parentheses), does not affect their sum or product.
7 × 7 × 8 × 7 = 8 × 7 × 7 × 7
= 56 × 49
= 2,744.
Question 7.
(9 + 7) + 6 = 9 + (7 + 6)
(9 + 7) + 6 = 9 + (7 + 6) = 22.
Associative property.
Explanation:
The associative property, in Math, states that while adding or multiplying numbers, the way in which numbers are grouped by brackets (parentheses), does not affect their sum or product.
(9 + 7) + 6 = 9 + (7 + 6)
= 9 + 13
= 22.
Question 8.
4 × 5 × 5 × 4 = 5 × 5 × 4 × 4
4 × 5 × 5 × 4 = 5 × 5 × 4 × 4 = 400.
Commutative property.
Explanation:
The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer.
4 × 5 × 5 × 4 = 5 × 5 × 4 × 4
= 25 × 16
= 400.
Question 9.
2 + 6 + 8 + 3 = 6 + 3 + 8 + 2
2 + 6 + 8 + 3 = 6 + 3 + 8 + 2 = 19.
Commutative property.
Explanation:
The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer.
2 + 6 + 8 + 3 = 6 + 3 + 8 + 2
= 9 + 8 + 2
= 17 + 2
= 19.
Question 10.
(6 × 4) × 2 = 6 × (4 × 2)
(6 × 4) × 2 = 6 × (4 × 2) = 48.
Associative property.
Explanation:
The associative property, in Math, states that while adding or multiplying numbers, the way in which numbers are grouped by brackets (parentheses), does not affect their sum or product.
(6 × 4) × 2 = 6 × (4 × 2)
= 6 × 8
= 48.
Question 11.
12 + 13 + 13 + 12 = 12 + 12 + 13 + 13
12 + 13 + 13 + 12 = 12 + 12 + 13 + 13 = 50.
Commutative property.
Explanation:
The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer.
12 + 13 + 13 + 12 = 12 + 12 + 13 + 13
= 24 + 13 + 13
= 37 + 13
= 50.
Question 12.
7 × 8 + 8 × 7 = 8 × 7 + 8 × 7 |
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# M28-27
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### HideShow timer Statistics
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. $$\frac{7}{8}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{5}{8}$$
E. $$\frac{3}{8}$$
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Posts: 58176
### Show Tags
16 Sep 2014, 01:29
2
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Official Solution:
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. $$\frac{7}{8}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{5}{8}$$
E. $$\frac{3}{8}$$
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is $$\frac{5}{8}$$.
The same for blue marble: the probability of drawing blue marble is $$\frac{3}{8}$$, the probability that for instance the 8th marble drawn is blue is still $$\frac{3}{8}$$. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
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Posts: 58176
### Show Tags
14 Nov 2014, 02:52
2
6
francisdarby wrote:
Hi there,
A new member here, just joined a day or two back
I'm quite confused about the solution to this problem. How is it possible that the probability remains constant even when we are removing marbles without replacement? Wouldn't the denominators in the probabilities adjust for each draw?
Any help would be greatly appreciated!
Thanks,
Francis
Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.
Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?
Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?
There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?
Similar questions to practice:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
Hope this helps.
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### Show Tags
13 Jan 2018, 07:06
2
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A simple answer using combinatorics
if the 7th slot is NOT BLUE, it is RED
now ignore this slot completely. it will not affect the other arrangements. for the 7 remaining slots, we have 7 marbles = (7! arrangements) but since 4 Red repeats and 3 blue repeats, the possible arrangements = 7!/(3!4!) = 35
total arrangements without the restriction for the 7th slot = 8!(5!3!) = 56
desired probability = 35/56 = 5/8
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### Show Tags
09 Feb 2017, 18:10
1
But if i ask the question what is P that the 2nd marble is red, wouldnt it be like that:
P(b,r)+P(r,r) = 5/8*3/7 + 3/8*2/7 = 3/8
So how can P that the 7th marble is red be 5/8? How can it be for all the same if there is NO replacement?
The reason why you got 3/8 is because you found the probability of the second draw being blue. It should be:
P(b,r)+P(r,r) = 3/8*5/7 + 5/8*4/7= 5/8
Heavily recommend drawing a probability tree (try excel) to see how the probability doesn't change up until you have no more draws left. After 8 draws, the probability then drops down to zero for either color .
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### Show Tags
17 Feb 2018, 22:58
1
I think this is a high-quality question. I have an alternate solution, 8!/(5!*3!) - Total # of permutations - 56 -----(i)
Of these fixing the 7th position as red - # of permutations is
7!/(4!*3!) - 35------(ii)
(ii)/(i) is what we want = (5/8)
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Joined: 11 Nov 2014
Posts: 1
### Show Tags
13 Nov 2014, 17:36
Hi there,
A new member here, just joined a day or two back
I'm quite confused about the solution to this problem. How is it possible that the probability remains constant even when we are removing marbles without replacement? Wouldn't the denominators in the probabilities adjust for each draw?
Any help would be greatly appreciated!
Thanks,
Francis
Intern
Joined: 24 Nov 2014
Posts: 18
### Show Tags
19 Mar 2015, 14:01
But if i ask the question what is P that the 2nd marble is red, wouldnt it be like that:
P(b,r)+P(r,r) = 5/8*3/7 + 3/8*2/7 = 3/8
So how can P that the 7th marble is red be 5/8? How can it be for all the same if there is NO replacement?
Math Expert
Joined: 02 Sep 2009
Posts: 58176
### Show Tags
20 Mar 2015, 05:58
But if i ask the question what is P that the 2nd marble is red, wouldnt it be like that:
P(b,r)+P(r,r) = 5/8*3/7 + 3/8*2/7 = 3/8
So how can P that the 7th marble is red be 5/8? How can it be for all the same if there is NO replacement?
The question asks about the probability of seventh ball being not blue, while not knowing the results of previous draws. PLEASE follow the links in my previous posts to understand better.
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### Show Tags
18 Aug 2016, 10:40
Hi Bunuel,
My understanding about this problem is that this is a case of a non-mutually exclusive event, as the execution of one event affects another, below is written what I understand from this problem.
If there are 8 marbles, and if, someone drew marbles in the manner below without replacement;
B B R R R R R R
Then the respective probabilities of drawing at succession are;
3/8 2/7 5/6 4/5 3/4 2/3 1/2 1
See that above drawing leaves only a red at the seventh draw, but to reach to the seventh draw event we should also change the sample space along with the no. of favorable cases.
Do you agree?
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### Show Tags
18 Aug 2016, 10:41
There will be more such cases to leave a non-blue at 7th draw we just need to add those cases
Math Expert
Joined: 02 Sep 2009
Posts: 58176
### Show Tags
18 Aug 2016, 10:52
subhajit1 wrote:
Hi Bunuel,
My understanding about this problem is that this is a case of a non-mutually exclusive event, as the execution of one event affects another, below is written what I understand from this problem.
If there are 8 marbles, and if, someone drew marbles in the manner below without replacement;
B B R R R R R R
Then the respective probabilities of drawing at succession are;
3/8 2/7 5/6 4/5 3/4 2/3 1/2 1
See that above drawing leaves only a red at the seventh draw, but to reach to the seventh draw event we should also change the sample space along with the no. of favorable cases.
Do you agree?
Unfortunately your understanding is wrong. I suggest you to follow the links given above.
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### Show Tags
24 Mar 2017, 11:13
Bunuel wrote:
Official Solution:
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. $$\frac{7}{8}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{5}{8}$$
E. $$\frac{3}{8}$$
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is $$\frac{5}{8}$$.
The same for blue marble: the probability of drawing blue marble is $$\frac{3}{8}$$, the probability that for instance the 8th marble drawn is blue is still $$\frac{3}{8}$$. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Hi Bunuel
the question says without replacement.
It means after each draws the no of ball reduces by 1. so how the probability of drawing 7th ball remains same as probability of drawing 1st ball red
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### Show Tags
24 Mar 2017, 11:16
daboo343 wrote:
Bunuel wrote:
Official Solution:
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. $$\frac{7}{8}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{5}{8}$$
E. $$\frac{3}{8}$$
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is $$\frac{5}{8}$$.
The same for blue marble: the probability of drawing blue marble is $$\frac{3}{8}$$, the probability that for instance the 8th marble drawn is blue is still $$\frac{3}{8}$$. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Hi Bunuel
the question says without replacement.
It means after each draws the no of ball reduces by 1. so how the probability of drawing 7th ball remains same as probability of drawing 1st ball red
I tried to explain this here: https://gmatclub.com/forum/m28-184525.html#p1442558
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31 May 2017, 09:59
This is a very good questions and hits right on the basics.
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### Show Tags
13 Jul 2017, 10:11
For those of you who are confused by Bunuel's beautiful answer, you can arrive at the same solution using permutations.
Let's find the possible permutations for the first 6 balls, and then worry about the 7th and 8th after that
5 Reb and 1 Blue - We are arranging 6 balls of two different types. To arrange 6 objects without respect for repetition of each type, the formula is simply $$6!$$. To account for repeating types, reduce $$6!$$ by the factorial of the repetition number for each type.
$$\frac{6!}{5!1!}=6$$
4 Red and 2 Blue
$$\frac{6!}{4!2!}15$$
3 Red and 3 Blue
$$\frac{6!}{3!3!}=20$$
Now multiply each by the available remaining permutations of the 7th and 8th ball:
5 Reb and 1 Blue?
Only BB is left, so $$6*1=6$$
4 Reb and 2 Blue?
RB and BR are left, so $$15*2=30$$
3 Reb and 3 Blue?
Only RR is left, so $$20*1=20$$
Thus, there are $$6+30+20=56$$ total arrangements
In how many of these last three options are the 7th ball red?
5 Red and 1 Blue: $$0$$
4 Red and 2 Blue: $$15$$
3 Red and 3 Blue: $$20$$
$$\frac{35}{56}=\frac{5}{8}$$
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10 Mar 2018, 15:49
This is a simple problem disguised as something a lot more difficult and it's easy to waste a lot of time considering permutations
Consider it like this - you draw the marbles 1 at a time, therefore you have 8 positions. In all the permutations, in any one position, the probability of that marble being red is 5/8. We are considering position 7, and the probability of the marble being red is the same as all the other positions - 5/8
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25 Mar 2018, 07:09
If there was a replacement of marbles, then we could say that all draws will have the same probability.
Not understood why every successive draw has the same probability.
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### Show Tags
06 Oct 2018, 04:17
Bunuel wrote:
Official Solution:
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. $$\frac{7}{8}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{5}{8}$$
E. $$\frac{3}{8}$$
Basically we need to find the probability that the seventh marble drawn is red (so not blue).
Now, the initial probability of drawing red marble is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is $$\frac{5}{8}$$.
The same for blue marble: the probability of drawing blue marble is $$\frac{3}{8}$$, the probability that for instance the 8th marble drawn is blue is still $$\frac{3}{8}$$. There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Bunuel From what part of the question stem are we judging that we are unaware of the previous outcomes of other draws? How do we differentiate this question from other replacement questions where each number of marbles would be reduced with each draw.
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08 Oct 2018, 02:27
I used a different method using seating arrangements.
Suppose there are 8 baskets placed in linear fashion, with each basket capable of holding only 1 ball. Suppose we place 8 balls 1-by-one in each basket
Then, total number of such arrangements - 8!/5!*3! = 56
Now, fix 7th basket with red ball placed in that. Now, we have to find number of arrangements of 7 balls with 4 red balls and 3 blue balls
==> 7!/4!*3! = 35
Probability = Number of desired outcomes/Total Outcomes = 35/56 or 5/8
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# What are we going to do? What does calculate mean? Calculate means __________. CFU Students, you already know that the relationship between quantities.
## Presentation on theme: "What are we going to do? What does calculate mean? Calculate means __________. CFU Students, you already know that the relationship between quantities."— Presentation transcript:
What are we going to do? What does calculate mean? Calculate means __________. CFU Students, you already know that the relationship between quantities is called a ratio. Now, we will use the ratio to calculate the unit rate. Make Connection 1 find the answer mathematically Vocabulary We will calculate 1 unit rates. Learning Objective Activate Prior Knowledge Write the ratio in a:b form and form. A ratio shows the relationship between one quantity and another quantity. abab 1. There were 12 buckets of water balloons and 8 kids. Write the ratio of buckets to kids. 2. Darla bought 8 granola bars for \$4. Write the ratio of granola bars to dollars. 12 : 8 8 : 4 8 8 4 CCSS 7.RP.1 - Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks ½ mile in each ¼ hour, compute the unit rate as the complex fraction ½ / ¼ miles per hour, equivalently 2 miles per hour. Name: ______________________ Period: ___ Tuesday 10.7.14
1 A unit rate is a number that shows the relationship between a quantity and one unit of another quantity. An equivalent 2 ratio table can be used to organize the information. Two unit rates can be found given one ratio. Which of the following contain a unit rate? How do you know? A 45 tree logs on 5 trucks B 4.5 pounds in every 1 box C \$1:3ft D 32 miles per hour E 7cm:\$2 In your own words, what is a unit rate? “A unit rate is ______________.” CFU 2 equal value 3 for every 1 Vocabulary Concept Development Darla bought 8 granola bars for \$4. Unit Rates 2 granola bars per 3 dollar granola bars 8484 dollars 2 0.50 \$0.50 per granola bar ÷ 4 ÷ 8 Unit Rates 1 st Unit abab 2 nd Unit 1 1 ÷ b ÷ a a b b a 1 Animated 1 st quantity abab 2 nd quantity = 1 a b b a b b a a Unit Rates
pounds (lb) dollar \$ 2 4 1 21 1 2 1. Dean bought 2 pounds (lbs) of bananas from A-Market for \$4 to make a banana smoothie after his bike ride. Calculate both unit rates: cost(\$) per pound(lb) pounds(lbs) per \$ 4 figure out 5 explain Vocabulary Skill Development/Guided Practice A unit rate shows the relationship between a quantity and one unit of another quantity. An equivalent ratio table can be used to organize the information. Two unit rates can be found given one ratio. 2. Dean just finished his bike ride. His smartphone biking app says he traveled 5 miles in 20 minutes. Calculate both unit rates: minutes per mile miles per minute Create an equivalent ratio table using the given units and quantities. Identify the unit quantity. (write 1) Determine 4 the division or multiplication which produces 1. Apply the operation to the other quantity. Interpret 5 the unit rate. Calculate unit rates. 1 2 3 5 4 \$2 per pound half-pound per dollar miles (mi) minutes (min) 5 20 1 41 1 4 4 minutes per mile Quarter-mile per minute How did I/you identify which would be the unit quantity? How did I/you determine the related quantity. CFU 2 4 ÷ 2 ÷ 4 ÷ 5 ÷ 20 ÷ 5 ÷ 20 Unit Rates
Skill Development/Guided Practice (continued) A unit rate shows the relationship between a quantity and one unit of another quantity. An equivalent ratio table can be used to organize the information. Two unit rates can be found given one ratio. 3. Mario ran 12 miles in one hour and 18 minutes (78 minutes). Find Mario’s minute-per-mile running rate. 4. Tracy ran 14 miles in one hour and 38 minutes (98 minutes). Find Tracy’s minute-per-mile running rate. Mario and Tracy are training to run a marathon (about 26.2 miles). Create an equivalent ratio table using the given units and quantities. Identify the unit quantity. (write 1) Determine the division or multiplication which produces 1. Apply the operation to the other quantity. Interpret the unit rate. Calculate unit rates. 1 2 3 5 4 How did I/you identify which would be the unit quantity? How did I/you determine the related quantity. CFU 2 4 miles (mi) minutes (min) 12 78 1 6.5 minutes per mile 6.5 miles (mi) minutes (min) 14 98 1 7 minutes per mile 7 Application Who runs a faster mile? How do you know? ÷ 12 ÷ 14 Unit Rates
5.James bought 12 pounds of almonds for \$54. Amina bought her almonds for \$4.25 per pound. Who got a better deal? How do you know? pounds (lbs) dollar (\$) 12 54 1 Amina got a better deal. She only paid \$4.25 per pound while James paid \$4.50 per pound. 4.5 Skill Development/Guided Practice (continued) How did I/you determine what the question is asking? How did I/you determine the math concept required? How did I/you determine the relevant information? How did I/you solve and interpret the problem? How did I/you check the reasonableness of the answer? CFU 2 1 3 4 5 6.Almaz bought a new car. Her new car gets 32 miles per gallon. Her old car could go 324 miles on a single tank of 12 gallons. Which car gets better gas mileage? How do you know? gallons miles (mi) 12 324 1 The new car gets better gas mileage. The old car gets 27 miles to the gallon while the new one gets 32 miles to the gallon. 27 Application Why is the better deal in question 5 lower but the better mileage in question 6 higher? ÷ 12 Unit Rates
What did you learn today about calculating unit rates? (Pair-Share) Use words from the word bank. Skill Closure Access Common Core Summary Closure A unit rate shows the relationship between a quantity and one unit of another quantity. An equivalent ratio table can be used to organize the information. Two unit rates can be found given one ratio. Create an equivalent ratio table using the given units and quantities. Identify the unit quantity. (write 1) Determine the division or multiplication which produces 1. Apply the operation to the other quantity. Interpret the unit rate. Calculate unit rates. 1 2 3 5 4 1. The school rented 6 vans to transport students to the local zoo. There will be enough seats for 54 students. Find the unit rate of students per van. Describe how you found the unit rate in the question above. Word Bank ratio unit rate one unit quantity vans students 6 54 1 9 9 students per van I divided the number of students by the number of vans to show how many students would be in a single van. ÷ 6 Unit Rates
Independent Practice A unit rate shows the relationship between a quantity and one unit of another quantity. An equivalent ratio table can be used to organize the information. Two unit rates can be found given one ratio. Create an equivalent ratio table using the given units and quantities. Identify the unit quantity. (write 1) Determine the division or multiplication which produces 1. Apply the operation to the other quantity. Interpret the unit rate. Calculate unit rates. 1 2 3 5 4 1. 10 pounds of pork meat is on sale for \$8. Calculate both unit rates: cost(\$) per pound(lb) pounds(lbs) per \$ 2. Lisa is looking for a summer job. Her friend, Harrison, told her about being a camp counselor. He said he worked for 4 weeks and made \$2240. If Lisa became a camp counselor, how much money would she make per week? How much money would she make per day? 3. A large pond has 14 geese and 84 ducks. How many ducks are there for every goose? 4.The Curry family drove 360 miles on 12 gallons. Find the unit rate of miles per gallon (mpg) of the Curry family’s car. pounds (lb) dollar \$ 10 8 1 0.8 1 \$0.80 per pound 1.25 pounds per dollar 1 4 1 weeks dollars \$ 4 2240 1 560 1 day \$560 per week 80 \$80 per day 1 week = 7 days geese ducks 14 84 1 6 ducks per goose 6 miles gallons 360 12 1 30 miles per gallon (30 mpg) 30 ÷ 10 ÷ 8 ÷ 10 ÷ 8 ÷ 4 ÷ 7 ÷ 14 ÷ 12 Unit Rates
Periodic Review 1 Access Common Core 1.Juanita left her secret recipe for apple pie on her desk. Maria peeked over and saw “10 apples” and “4 cups of sugar” in the recipe before looking away. Find both unit rates of apples per cup of sugar and cups of sugar per apple. 2. A middle school has 330 students and 15 teachers on staff. What is the unit rate of students per teacher? If Maria wants to bake apple pies using 15 apples, which unit rate above could she use to find out how many cups of sugar she should use? Explain why you chose this unit rate. How many cups of sugar will she need? apples cups of sugar 10 4 1 1 2 and a half apples for every cup of sugar 1 2 2 2 5 cup of sugar per apple 2 5 students teachers 330 15 22 22 students to every 1 teacher 1 She should use the 2/5 cup of sugar per apple rate 15 = 6 2 5 She will need 6 cups of sugar. ÷ 4 ÷ 10 ÷ 4 ÷ 10 ÷ 15 She needed to know how many cups of sugar to use for a number of apples.
Periodic Review 2 Access Common Core 1. Valerie is a softball star that set a record for homeruns in a season (48) and homeruns in a game (4). If Valerie went up to bat 240 times this season, find her attempts per homerun. 2. Geoffrey bought 15 marbles for 75 cents. Find the unit rate of marbles per quarter. Find the unit rate of marbles per nickel. Describe how you found the unit rate “marbles per quarter” in problem #2 above. homeruns attempts (at-bats) 48 240 5 1 Valerie averaged 1 homerun for every 5 times she went up to bat. marbles cents 15 75 5 25 5 1 25 cents = 1 quarter5 cents = 1 nickel 5 marbles per quarter 1 marble per nickel I found how many quarters were in 75 cents. Then I divided the number of marbles by the number of quarters. ÷ 5 ÷ 3 ÷ 5
Periodic Review 3 Access Common Core 1. Relationship specialists say that when talking to a friend, you should not talk too much more than your friend if you want to have a happy friendship. During a 22-minute phone call, Lucy talked for 18 minutes and Richard talked for 4 minutes. For every 1 minute Richard talked, how many minutes did Lucy talk? 2. Salvador ordered a metal pipe for the farm. The pipe was 5 yards long and cost \$90. Find the cost per yard, cost per foot, and cost per inch. Write a word problem with the stated unit rate answer. Hint: Fill in the table first, then write your word problem. Students Teams 6 1 1. Muffins \$ 1 3 2. Lucy’s talking minutes Richard’s talking minutes 18 4 1 4 1 2 yards dollars \$ 5 90 1 18 6 1 yard = 3 feet \$18 per yard \$6 per foot \$0.50 per inch 1 foot1 inch 1 foot = 12 inches 1 2 18 3 There are 18 students divided equally into three teams. How many students are on each team? 25 75 Julio makes large muffins for his bakery every morning. He sold 25 muffins yesterday and made \$75. How much does a muffin cost? ÷ 4 ÷ 5 ÷ 3 ÷ 12 Lucy talked for 4½ minutes for every minute that Richard talked.
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An aeroplane heads due west at 500km/h. It experiences a 15 km/h crosswind in the flowing direction S30°W. Find the speed of the aeroplane. And the direction of the aeroplane in square root form
Asked on 6/11/2023, 42 pageviews
We can use vector addition to solve this problem. Let's represent the velocity of the plane as a vector pointing due west with magnitude 500 km/h. We can call this vector P: P = 500 km/h [W] Now let's represent the velocity of the wind as a vector pointing in the direction S30°W with magnitude 15 km/h. We can call this vector W: W = 15 km/h [S30°W] To find the resulting velocity of the plane (which includes the effect of the crosswind), we need to add these vectors together: V = P + W To add these vectors, we can break them down into their components. The westward component of P is simply 500 km/h, since it is pointing due west. The southward component of W can be found using trigonometry: sin(30°) = opposite/hypotenuse sin(30°) = S/h S = h sin(30°) = 15 km/h * 0.5 = 7.5 km/h Since the wind is blowing to the south, its southward component is negative: W = -7.5 km/h [S] Now we can add the vectors: V = P + W = 500 km/h [W] + (-7.5 km/h [S]) To find the magnitude and direction of V, we can use the Pythagorean theorem and trigonometry: |V| = sqrt((500 km/h)^2 + (-7.5 km/h)^2) ≈ 500 km/h theta = tan^-1((-7.5 km/h) / 500 km/h) ≈ -0.86° (Note that we convert the southward component of W to a negative number before dividing by the westward component of P in the arctan function.) Therefore, the speed of the aeroplane including the effect of the crosswind is approximately 500 km/h, and its direction is approximately W0.86°S (note that this is almost exactly due west).
by gpt-3.5-turbo |
# Prime or Composite? Factors or Multiples?
## Presentation on theme: "Prime or Composite? Factors or Multiples?"— Presentation transcript:
Prime or Composite? Factors or Multiples?
Objective: I can differentiate between prime and composite numbers. I can differentiate between factors and multiples. Created by Chris Garrison Sycamore 5th grade
Prime Numbers? What are... Composite Numbers?
have more than two factors have “exactly two factors” 1 and itself
Factors are small Multiples are BIG!
Let’s practice what we know!
Get Ready! Let’s practice what we know!
Is this number prime or composite
Is this number prime or composite? Show all work and carefully find all the factors! 16 Composite!
Is this number prime or composite
Is this number prime or composite? Show all work and carefully find all the factors! 24 Composite!
Is this number prime or composite
Is this number prime or composite? Show all work and carefully find all the factors! 17 Prime!
Is this number prime or composite
Is this number prime or composite? Show all work and carefully find all the factors! 64 Composite!
Tip Test for success! Think smarter not harder! Example X X X X
Which of following are all prime numbers? X X a. 2, 3, 5, 7, 9, 13, b. 1, 4, 5, 7, 11, 13, c. 3, 5, 7, 13, 17, 19 X X Tip Eliminate composite numbers first. Then double check remaining numbers to make sure they are all prime.
Is this number a factor or a multiple of 9? Prove it!
27 Multiple
Is this number a factor or a multiple of 36? Prove it!
12 Factor
Is this number a factor or a multiple of 18? Prove it!
6 Factor
Is this number a factor or a multiple of 7? Prove it!
49 Multiple
Is this number a factor or a multiple of 24? Prove it!
8 Factor
Is this number a factor or a multiple of 12? Prove it!
96 Multiple
Greatest Common Factor (GCF)
The greatest factor two numbers share. 1. List the factors for each number. 2. Circle common factors. 3. Choose the greatest factor. 30: 1, 2, 3, 5, 6, 10, 15, 30 GCF = 15 15: 1, 2, 3, 5, 15
Least Common Multiple LCM: 6
Least multiple that more than one number share. 1. List the first 5 multiples of each number. 2. Choose the smallest multiple they both share. LCM: 6 2: 2, 4, 6, 8, 10 3: 3, 6, 9, 12, 15
Tips Test for success! Factor Rainbow 1, 2, 3, 4, 6, 8, 12, 24
Think smarter not harder! Example Which of following are the factors of 24? a. 1, 2, 3, 4, 6, 8, 12, 24 b. 1, 4, 6, 5, 8, 12, 24, X X c. 2, 3, 6, 8, 12, 24, 48 X Tips 1, 2, 3, 4, 6, 8, 12, 24 Double check all factors with... Eliminate any answer choice that includes a multiple. Factor Rainbow
Now tell them the difference between prime and composite number.
Closure Turn to your shoulder partner and tell them the difference between a factor and multiple. Now tell them the difference between prime and composite number. |
# Arithmetic Sequences and Sums
GCSELevel 6-7Edexcel iGCSE
## Arithmetic Sequences and Sums
An arithmetic sequence is a list of numbers, called terms. There is a common difference between consecutive terms, to for each new term the same number is added to or subtracted from the previous term. The numbers in the sequence can then be generated by a rule. We call this the $n$th term rule.
This rule allows us to find a value at the $n$th position in the sequence, which we call the $n$th term
We are also able to use the $n$th term rule to take a sum of the first $n$ terms of a sequence
## Finding the $n$th term
An arithmetic sequence is a sequence where every term has the same difference $d$, between it and the next term. We can use the $n$th term rule to generate any value in an arithmetic sequence. The rule is
The $n$th term $= a + (n-1)d$,
where $d$ is the common difference, and $a$ is the very first term in the sequence.
Example: Find the $10$th term in the following sequence,
Step 1: Find the first term ($a$)
In this sequence the first term is given as $3$.
Step 2: Find the common difference ($d$)
The common difference is the difference between two consecutive terms. In this case we can chose the first and second terms to find $d$
$d = 2\text{nd term} - 1\text{st term} = 7 - 3 = 4$.
Step 3: Use the formula
$a = 3$, $d = 4$, so the $10$th term is
$3 + (10 - 1) \times 4 = 3 + 9\times 4 = 3 + 36 = 39$
So the $10$th term in the sequence is $39$.
Level 6-7GCSEEdexcel iGCSE
## Finding the Sum of a Sequence
We can use a formula to find the sum of a sequence from the first term to the $n$th term, where $n$ is the number of terms. We denote this sum as $S_n$
There are two formulas that can be used, although they are essentially the same.
The first formula:
$S_n = \dfrac{n}{2} (2a + (n-1)d)$,
Where $a$ is the first term in the sequence, $n$ is the number of terms and $d$ is the common difference
From this first formula can find the second formula. We can deduce that $2a + (n-1)d = a + (a + (n-1)d)$ and by the $n$th term rule we know that the $n$th term $= a + (n-1)d$. We can call the $n$th term $a_n$ and the first term $a = a_1$.
So $2a + (n-1)d = a_1 + a_n$, which can be substituted back into the first formula.
The second formula:
$S_n = \dfrac{n}{2} (a_1 + a_n)$
Where $a_1$ is the first term, $a_n$ is the final term, and $n$ is the number of terms.
Level 6-7GCSEEdexcel iGCSE
## Example 1: Finding the $13$th Term
Consider the sequence:
$1, 4, 7, 10, \dots$
Find the $13$th term in the sequence.
[3 marks]
Firstly we can find the common difference:
$2$nd term $- 1$st term $= 4 - 1 = 3$
$3$nd term $- 2$st term $= 7 - 4 = 3$
$4$nd term $- 3$st term $= 10 - 7 = 3$
The common difference $(d) = 3$.
The first term $(a) =1$.
This gives us the $n$th term rule of
$a_n = 1 + (n-1)\times 3$.
For the $13$th term,
$a_{13} = 1 + (13-1) \times 3$
$a_{13} = 1 + 12 \times 3$,
$a_{13} = 1 + 36$,
$a_{13} = 37$,
The $13$th term of the sequence is $37$.
Level 6-7GCSEEdexcel iGCSE
## Example 2: Finding the sum to the $1000$th term
Consider the sequence:
$1, 2, 3, 4, \dots$
Find the sum of the first $1000$ terms in the sequence.
[3 marks]
The common difference, $d = 1$. The first term, $a = 1$
We can use the first formula for sum of an arithmetic sequence to find the sum up to $n =1000$.
$S_{1000} = \dfrac{1000}{2} (2 \times 1 + (1000-1) \times 1)$
$S_{1000} = \dfrac{1000}{2} (2 + (999))$
$S_{1000} = \dfrac{1000}{2} (1001)$
$S_{1000} = 500 \times (1001)$
$S_{1000} = 500,500$
Level 6-7GCSEEdexcel iGCSE
## Example 3: Determining if a value is a term in a sequence
Consider the sequence:
$15, 21, 27, 33, \dots$
determine whether or not the value $99$ is a term in the sequence.
[3 marks]
Firstly we must find the common difference:
$2$nd term $- 1$st term $= 21 - 15 = 6$
$3$rd term $- 2$nd term $= 27 - 21 = 6$
$4$th term $- 3$rd term $= 33 - 27 = 6$
The common difference $(d) = 6$.
Therefore the $n$th term rule is $15 + (n - 1) \times 6$
Now we can equate that to $99$ to see if it is a term,
$99 = 15 + (n-1) \times 6$,
$84 = (n-1) \times 6$,
$84 \div 6 = n - 1$,
$14 = n - 1$,
$n = 15$,
So yes, $99$ is the $15$th term in the sequence.
Level 6-7GCSEEdexcel iGCSE
## Example 4: Using the Second Summation Formula
An arithmetic sequence has two known terms. The first term is $a_1 = 7$ and the $43$rd term is $a_{43} = 343$.
Find the $n$th term rule and the sum of the sequence up to the $43$rd term.
[4 marks]
The $n$th term rule:
$a_n = a_1 + (n-1)d$
Let $n = 43$ then,
$a_{43} = 343 = 7 + (43 - 1)d$,
$343 - 7 = 42d$,
$336 = 42d$,
$336 \div 42 = d$,
$d = 8$.
The $n$th term rule then is:
$a_n = 7 + (n-1)\times 8$
The sum to the $43$rd term is:
$S_{43} = \dfrac{43}{2} \times (a_1 + a_{43})$
$S_{43} = \dfrac{43}{2} \times (7 + 343)$
$S_{43} = \dfrac{43}{2} \times 350$
$S_{43} = 7525$
Level 6-7GCSEEdexcel iGCSE
## Arithmetic Sequences and Sums Example Questions
The first term, $a = 3$. We must now find the common difference.
$2$nd term $- 1$st term $= 8 - 3 = 5$.
The common difference, $d = 5$.
So the $n$th term formula is
$3 + (n-1) \times 5=5n-2$
The first term, $a = -6$. We must now find the common difference.
$2$nd term $- 1$st term $= -4 - (-6) = 2$.
The common difference, $d = 2$.
Summation formula:
$S_n = \dfrac{n}{2} (2a + (n-1)d)$,
so,
$S_{30} = \dfrac{30}{2} (2 \times (-6) + (30 - 1) \times 2)$
$S_{30} = 15 (-12 + (29) \times 2)$
$S_{30} = 15 (-12 + 58) = 15 \times 46 = 690$
a) To generate the first $5$ terms of this sequence, we will substitute $n=1, 2, 3, 4, 5$ into the formula given.
$1$st $= 1080 + ((1)-1)(-40) = 1080$
$2$nd $= 1080 + ((2)-1)(-40) = 1040$
$3$rd $= 1080 + ((3)-1)(-40) = 1000$
$4$th $= 1080 + ((4)-1)(-40) = 960$
$5$th $= 1080 + ((5)-1)(-40) = 920$
So, the first $5$ terms are $1080$, $1040$, $1000$, $960$, and $920$
b) We can equate $-140$ to the $n$th term rule, to find it’s position in the sequence. If it’s position is a whole number, it must be in the sequence, if not it can’t be.
$-140 = 1080 + (n-1)(-40)$
$-1220 = (n-1)(-40)$
$-1220 \div -40 = (n-1)$
$-1220 \div (-40) = 30.5 = (n-1)$
$n = 31.5$
As $n$ is not an integer, $-140$ is not in the sequence.
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# 2002 AMC 12A Problems/Problem 5
The following problem is from both the 2002 AMC 12A #5 and 2002 AMC 10A #5, so both problems redirect to this page.
## Problem
Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
$[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$
$\text{(A)}\ \pi \qquad \text{(B)}\ 1.5\pi \qquad \text{(C)}\ 2\pi \qquad \text{(D)}\ 3\pi \qquad \text{(E)}\ 3.5\pi$
## Solution
The outer circle has radius $1+1+1=3$, and thus area $9\pi$. The little circles have area $\pi$ each; since there are 7, their total area is $7\pi$. Thus, our answer is $9\pi-7\pi=\boxed{2\pi\Rightarrow \text{(C)}}$. |
# Solving Inequalities
### Popular Tutorials in Solving Inequalities
#### What's an Inequality?
Inequalities come up all the time when you're working algebra problems. In this tutorial you'll learn what an inequality is, and you'll see all the common inequality symbols that you're likely to see :)
#### How Do You Solve an Inequality Using Subtraction?
Solving an inequality for a variable? Just perform the order of operations in reverse! Always make sure to follow the rules for solving an inequality!
#### How Do You Solve an Inequality Using Addition?
Solving an inequality for a variable? Just perform the order of operations in reverse! Always make sure to follow the rules for solving an inequality!
#### How Do You Solve an Inequality by Dividing by a Positive Number?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality with Positive Numbers Using Multiplication?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality with Negative Numbers Using Multiplication?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### How Do You Solve an Inequality with Negative Numbers Using Division?
Solving an inequality for a variable? Just perform the order of operations in reverse! Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.
#### What's the Subtraction Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the subtraction property of inequality, and it basically says that if you minus a number from one side of an inequality, you have to minus that same number from the other side of the inequality as well. Watch the tutorial to see how this looks in terms of algebra!
#### What's the Addition Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the addition property of inequality, and it basically says that if you add a number from one side of an inequality, you have to add that same number from the other side of the inequality as well. Watch the tutorial to see how this looks in terms of algebra!
#### What's the Division Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the division property of inequality, and it basically says that if you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number. However, you have to be very careful about the direction of the inequality! Watch the tutorial to see how this looks in terms of algebra!
#### What's the Multiplication Property of Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the multiplication property of inequality, and it basically says that if you multiply one side of an inequality by a number, you can multiply the other side of the inequality by the same number. However, you have to be very careful about the direction of the inequality! Watch the tutorial to see how this looks in terms of algebra! |
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### Course: 3rd grade (Eureka Math/EngageNY)>Unit 5
Lesson 4: Topic D: Fractions on the number line
# Representing 1 as a fraction
Let's explore various ways to represent the number 1 as a fraction, like 1/1, 2/2, and 3/3. We see how dividing a whole into equal parts and shading them in can represent 1 whole. We also use a number line to visualize these fractions, emphasizing that they all represent the same value: 1 whole. Created by Sal Khan.
## Want to join the conversation?
• would 1/1 be considered a fraction?
• Absolutely! 1/1 is a fraction, with 1's in both the numerator and the denominator.
• hi, why is a decimal considered a fraction?
• Because a decimal is part of a whole number like a fraction.
• why is 0/0 unidentified? Why can't 0/0 just equal 0.
• Good question!
If we did 6/2, for example, we would ask ourselves
“what number times 2 is 6?”. The answer would be 3.
In the same way, when we do 0/0, we ask ourselves “what number times 0 is 0?”. Any number times 0 is 0. So 0/0 can be anything and is therefore called indeterminate.
Have a blessed, wonderful day!
• Why is this easier than I thought
• i dont understand
(1 vote)
• how can a fraction be represented on a number line?
• Fractions exist between whole numbers, so the point representing a fraction would be between two points that represent whole numbers. For example, the fraction 1/2 would be between 0 and 1.
Draw a line and label a point 0 on the far left and another point 1 on the far right. Draw a point in the center of that line between 0 and 1, and write 1/2.
• I am new to fractions. So is 8/8 1 whole?
• Yes - 8/8 is the same thing as 1 - just shown in a different way.
Imagine the pizza. If it was cut into eight slices - 1 slice would be 1/8 of the pizza (1 of the 8 slices) while 8/8 (all 8 slices) would be 1 whole pizza.
• At , "Sal said "one oneth", but meant "one over one".
What is the proper way to say 1/1? One over one, one onethS?
• One over one is correct. I'm sure both can be used (correct me if I'm wrong), but one over one is generally the most universally used.
• would 1/1 be considered a fraction
• Anything counts, unless the denominator is `0`.
So yeah, `1/1` is a fraction. It equals `1` whole
(1 vote)
• is o/o a fraction
• `0/0` is indeterminate, it could be equal to any real number but there's no single agreed-upon value for it.
If `0/0 = ?` then that means `0*? = 0`. But we know that anything times `0` equals `0`, so `?` and thus `0/0` could be anything. |
Special Right Triangles Worksheets
In these worksheets, students will learn to work with special right triangles.
#### Special Right Triangles are triangles whose angles are in a particular ratio (30°, 60°, 90° and 45°, 45°, 90°). You can find the right triangle's third side by using the Pythagorean Theorem. It will become more best when you already know the two sides. There are some triangles, such as 45-45-90 and 30-60-90 triangles. People usually find their ratios without using the Pythagorean Theorem each time. With such patterns, you can solve the special triangle's sides that are missing but, you must know at least the length of one side of the triangle. Suppose you have a 45-45-90 right triangle with two triangle's sides each one is 6 inches, and one is x inches. You will find the value of x. Special right triangle There are angles of the right triangle that are 30-60-90 and 45-45-90. This way, you will have Pythagorean triple triangles. 30-60-90 angle triangles Such triangles have ratios x, x√3,2x. Solve this by using the formula of Pythagorean Theorem, X2 + (x√3)2 = (2x)2 X2 + 3x2 = 4 x 2 = 4 x2 45-45-90 angles of triangles A 45-45-90 right angle triangle side has ratios x, x, x√2 Here, we will use Pythagorean Theorem, X2 + x2 = (x√2)2 = 2x2 = 2x2 Remember that the order of side ratios should be important x, x, x√2 as each of the triangle has the same angle. The smallest sides and angles are the same. You will consider the same rule for the largest values.
There are 18 worksheets in this set, divided into three sets of 6. In the first set, students will solve problems with a geometry emphasis. In the second set, students will solve problems with a trigonometry emphasis. In the third set, students will work with right triangles in general. A sound understanding of the basic principles of geometry and trigonometry is required in order for students to be successful with these worksheets. Students will find the value of the specified side or angle of a right triangle. This set of worksheets contains lessons, step-by-step solutions to sample problems, both simple and more complex problems, reviews, and quizzes. It also includes ample worksheets for students to practice independently. Most worksheets contain between eight and ten problems. When finished with this set of worksheets, students will be able to find the value of a specified part of a right triangle. These worksheets explain how to find the value of a specified part of a right triangle. Sample problems are solved and practice problems are provided.
# Special Right Triangles Worksheets
## Special Right Triangles (Geometry Emphasis) Lesson
This worksheet explains how to find a length or an angle of a right triangle. A sample problem is solved, and two practice problems are provided.
## Worksheet
Students will find the requested measurement value of a given right triangle. Ten problems are provided.
## Practice
The goal here is to lean more on the geometry that presents itself to help you solve these exercises.
## Review and Practice
The concept of how to find the requested measurements based on the given information about a right triangle is reviewed. A sample problem is solved. Six practice problems are provided.
## Quiz
Students will demonstrate their proficiency with working with special right triangles by using a geometrically emphasis with these problems.
## Check
This sheet is really helpful to understand where your students are with this skill.
## Special Right Triangles (Trigonometry Emphasis) Lesson
This worksheet explains how to find the requested measurement value of a special right triangle. A sample problem is solved, and two practice problems are provided.
## Worksheet
Students will find all these missing values by using their knowledge of trigonometry.
## Practice
Find the exact value of the following: cosec 60° + cot 30°+ sin 30°
## Review and Practice
The concept of how to find the requested measurement value of a special right triangle is reviewed. A sample problem is solved. Six practice problems are provided.
## Quiz
Students will demonstrate their proficiency with relying on their skills to solve these types of problems. Ten problems are provided.
## Check
Students will find the requested measurement value of a special right triangle. Three problems are provided, and space is included for students to copy the correct answer when given.
## Lesson
This worksheet explains how to solve this: : In a right triangle normal is 12 and the ,hypotenuse is 14. Find the value of opposite side?
## Work with Right Triangles Worksheet
We will apply our math skills to solve real world problems like: A tree in a back yard is to be cut down. The base of the tree from the house is 23 meters away, and the angle of elevation from the house to the top of the tree is 45.5 degrees. Could the tree hit the house when it is cut down?
## Practice
Example: A fisherman drops a boat anchor and pays out 93 meters of anchor rope. The rope makes an angle of depression of 59.5 degrees with the horizontal. Assuming a level bottom, how deep in the water?
## Review and Practice
The concept of how to find the value of a specified part of a right triangle is reviewed. A sample problem is solved. Six practice problems are provided.
## Quiz
Students will demonstrate their proficiency in finding the value using the concepts with have explored in this section. Ten problems are provided.
## Check
A great way to approach solving these problems. |
# 180 Days of Math for Fourth Grade Day 4 Answers Key
By accessing our 180 Days of Math for Fourth Grade Answers Key Day 4 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fourth Grade Answers Key Day 4
Directions Solve each problem.
Question 1.
26 – 9 = ___
26 – 9 = 17
Explanation:
Perform subtraction operation on above two given numbers. Subtract 9 from 26 the difference is 17.
Question 2.
Explanation:
Perform addition operation on above two given numbers. Add $2.00 to$1.25 the sum is \$3.25.
Question 3.
Explanation:
Perform division operation on above two given numbers. Divide 40 by 8 the quotient is 5.
Question 4.
32 ÷ 4 = ___
32 ÷ 4 = 8
Explanation:
Perform division operation on above two given numbers. Divide 32 by 4 the quotient is 8.
Question 5.
6 tens + 4 ones = ___
6 tens + 4 ones = 64
Explanation:
Based on the position in the number the place value of digits is determined as ones, tens, hundreds, thousands, ten-thousands and so on. So, 6 tens + 4 ones is equal to 64.
Question 6.
7 × = 28
7 ×  = 28
Explanation:
To get the product 28 we need to perform multiplication operation. Multiply 7 with 4 the product is 28. The missing multiplicand is 4.
Question 7.
What is the term for the amount of space inside a container?
______________
Volume is the term for the amount of space inside a container.
Question 8.
____ quarts = 1 gallon
We know that
1 gallon = 4 quarts
Question 9.
Name this shape.
______________
The shape is pentagon.
Explanation:
In the above we can observe one shape. The name of the shape is pentagon. A pentagon have five sides and five angles. In a pentagon the side lengths are equal and angles are of equal measure then it called as a regular pentagon.
Question 10.
$$\frac{1}{10}$$ of 20 = 2, so $$\frac{3}{10}$$ of 20 =
______________________
Given, $$\frac{1}{10}$$ of 20 = 2
$$\frac{3}{10}$$ of 20
So $$\frac{3}{10}$$ of 20 = 6 |
# What is an example of an inconsistent equation?
## What is an example of an inconsistent equation?
Inconsistent equations is defined as two or more equations that are impossible to solve based on using one set of values for the variables. An example of a set of inconsistent equations is x+2=4 and x+2=6.
## How do you find the inconsistent equation?
To see if the pair of linear equations is consistent or inconsistent, we try to gain values for x and y. If both x and y have the same value, the system is consistent. The system becomes inconsistent when there are no x and y values that satisfy both equations.
Which equation is consistent or inconsistent?
Solutions to systems of equations: consistent vs. inconsistent. A consistent system of equations has at least one solution, and an inconsistent system has no solution.
Which equation can pair with 3x 4y 8?
Answer: Ax + By = C with 4A ≠ 3B can be paired with 3x + 4y = 8 to create a consistent and independent system. Example: 5x + 8y = 12. Let us explore more about the consistent and independent system or unique solution case of the system of linear equations.
### What makes a matrix inconsistent?
If a system of equations has no solutions, then it is inconsistent. If the last column (in an augmented matrix) is a pivot column, that is, it has a pivot, then it’s inconsistent.
### What is meant by consistent equations?
In mathematics and particularly in algebra, a linear or nonlinear system of equations is called consistent if there is at least one set of values for the unknowns that satisfies each equation in the system—that is, when substituted into each of the equations, they make each equation hold true as an identity.
How do you do substitution?
The method of substitution involves three steps:
1. Solve one equation for one of the variables.
2. Substitute (plug-in) this expression into the other equation and solve.
3. Resubstitute the value into the original equation to find the corresponding variable.
Which equation can pair 3x 4y 8 to create a consistent and independent system?
#### How many solutions exist for the system of equations in the graph?
Each shows two lines that make up a system of equations. If the graphs of the equations intersect, then there is one solution that is true for both equations. If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.
#### Can an overdetermined system be consistent?
An overdetermined system is almost always inconsistent (it has no solution) when constructed with random coefficients. However, an overdetermined system will have solutions in some cases, for example if some equation occurs several times in the system, or if some equations are linear combinations of the others.
Which system of equations is inconsistent?
Inconsistent equations of linear equations are equations that have no solutions in common. In this system, the lines will be parallel if the equations are graphed on a coordinate plane. Let’s consider an inconsistent equations as x – y = 8 and 5x – 5y = 25. They don’t have any common solutions.
What does inconsistent mean in linear algebra?
In mathematics and in particular in algebra, a linear or nonlinear system of equations is consistent if there is at least one set of values for the unknowns that satisfies every equation in the system—that is, that when substituted into each of the equations makes each equation hold true as an identity.
## What are the methods to solve linear equations?
The three methods most commonly used to solve systems of equation are substitution, elimination and augmented matrices. Substitution and elimination are simple methods that can effectively solve most systems of two equations in a few straightforward steps.
## What is consistent system of linear equations?
a linear or nonlinear system of equations is consistent if there is at least one set of values for the unknowns that satisfies every equation in the system—that is, that when substituted into each of the equations makes the equation hold true as an identity. |
# Cross-Multiplication Method to Solve System of Linear Equations | Cross Multiplication Method Examples
Cross multiplication method is used to solve a pair of simultaneous linear equations in two variables. This method helps you to find the accurate values of variables easily. It is applicable only when we have two linear equations in two variables. Get the steps to solve the system of linear equations with the cross multiplication method in the following sections. Also, check out the solved examples for a better understanding of the concept.
## Cross Multiplication Method for Solving Pair of Linear Equations
Follow the easy and simple guidelines listed below while solving Pair of Linear Equations. They are in the following fashion
• Observe the given pair of linear equations and transpose them in the form of ax + by + c = 0
• And represent them as the coefficient constant of term x, the coefficient constant of term y, coefficient of x, and coefficient of y.
• Cross multiply those constants.
• Express the cross multiplicands of x, y, and constants under x, y, and 1.
• Equate them to find the values of x, y.
Check out the derivation to get more details about the Cross-Multiplication Method.
### General Solution Using the Method of Cross-Multiplication
Let the pair of linear equations in two variables are
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
For solving these linear equations follow the below included steps
Given simultaneous linear equations in two variables are
a₁x + b₁y + c₁ = 0 ——- (i)
a₂x + b₂y + c₂ = 0 ——– (ii)
Multiply equation (i) by b₂ and equation (ii) by b₁.
b₂ (a₁x + b₁y + c₁ = 0)
b₂ (a₁x + b₁y + c₁) = b₂ x 0
a₁b₂x + b₁b₂y + b₂c₁ = 0 ——- (iii)
b₁ (a₂x + b₂y + c₂ = 0)
b₁ (a₂x + b₂y + c₂) = b₁ x 0
a₂b₁x + b₂b₁y + c₂b₁ = 0 —- (iv)
Subtract equation (iv) from equation (iii)
[a₁b₂x + b₁b₂y + b₂c₁ = 0] – [a₂b₁x + b₂b₁y + c₂b₁ = 0]
[a₁b₂x + b₁b₂y + b₂c₁] – [a₂b₁x + b₂b₁y + c₂b₁] = 0
a₁b₂x + b₁b₂y + b₂c₁ – a₂b₁x – b₂b₁y – c₂b₁ = 0
x(a₁b₂ – a₂b₁) + b₂c₁ – c₂b₁ = 0
x(a₁b₂ – a₂b₁) = c₂b₁ – b₂c₁
x = (c₂b₁ – b₂c₁) / (a₁b₂ – a₂b₁)
Substitute x = (c₂b₁ – b₂c₁) / (a₁b₂ – a₂b₁) in equation (i)
a₁(c₂b₁ – b₂c₁) / (a₁b₂ – a₂b₁) + b₁y + c₁ = 0
b₁y = -a₁(c₂b₁ – b₂c₁) / (a₁b₂ – a₂b₁) – c₁
y = (-a₁(c₂b₁ – b₂c₁) / (a₁b₂ – a₂b₁) – c₁) / b₁
y = (c₁a₂ – c₂a₁) / (b₂a₁ – b₁a₂)
The solution of the equations is
x / (c₂b₁ – b₂c₁) / (a₁b₂ – a₂b₁) = y / (c₁a₂ – c₂a₁) / (b₂a₁ – b₁a₂) = 1/(b₂a₁ – b₁a₂)
The above method is called Cross-Multiplication Methos as it uses cross-multiplication technique.
You can also represent the cross-multiplication as the following diagram.
### Cross Multiplication Method Examples
Example 1.
Solve the linear equations x + y = 5, x – y = 3 by using the cross multiplication method?
Solution:
Given system of linear equations are
x + y = 5
x – y = 3
On transposition, we get
x + y – 5 = 0
x – y – 3 = 0
Writing the co-efficient in the following way, we get:
By cross-multiplication method:
x/[(1(-3) – (-1)(-5)] = y/[(-5)1 – 1(-3)] = 1//[1(-1) – 1(1)]
x/(-3 – 5) = y/(-5 + 3) = 1/(-1 – 1)
x/(-8) = y/(-2) = 1/(-2)
x/(-8) = 1/(-2), y/(-2) = 1/(-2)
x/8 = 1/2, y/2 = 1/2
x = 8/2, y = 2/2
x = 4, y = 1
Therefore, required solution is x = 4, y = 1.
Example 2.
Solve the simultaneous linear equations 3x – 4y = 0, 9x – 8y = 12 by the method of cross-multiplication?
Solution:
Given simultaneous equations are,
3x – 4y = 0
9x – 8y = 12
On transposition, we get
3x – 4y = 0 ——– (i)
9x – 8y – 12 = 0 ———- (ii)
Multiply equation (i) by -8 and equation (ii) by -4.
-8(3x – 4y) = 0 x -8
-24x + 32y = 0 —— (iii)
-4(9x – 8y – 12) = 0 x -4
-36x + 32y + 48 = 0 ——- (iv)
Subtract equation (iv) from equation (iii)
[-24x + 32y = 0] – [-36x + 32y + 48 = 0]
-24x + 32y – (-36x + 32y + 48) = 0
-24x + 32y + 36x – 32y – 48 = 0
12x – 48 = 0
12x = 48
x = 48/12
x = 4
Substitute x = 4 in equation (ii)
9(4) – 8y – 12 = 0
36 – 8y – 12 = 0
-8y + 24 = 0
-8y = -24
y = 24/8
y = 3
Therefore, required solution is x = 4, y = 3.
Example 3.
Solve the linear equations 2x + 3y = 5, x – 4y = 8 by using the cross multiplication method?
Solution:
Given linear equations are 2x + 3y = 5, x – 4y = 8
On transposing, we get
2x + 3y – 5 = 0 ——- (i)
x – 4y – 8 = 0 ——- (ii)
Multiply equation (ii) by 2, equation (i) by 1.
From equation (i), we get
2x + 3y – 5 = 0 —– (iii)
From equation (ii), we get
2(x – 4y – 8) = 0
2x – 8y – 16 = 0 —— (iv)
Subtract equation (iii) from equation (iv).
[2x – 8y – 16] – [2x + 3y – 5] = 0
2x – 8y – 16 – 2x – 3y + 5 = 0
-11y – 11 = 0
-11y = 11
y = -11/11
y = -1
Substitute y = -1 in equation (i)
2x + 3(-1) – 5 = 0
2x – 3 – 5 = 0
2x – 8 = 0
2x = 8
x = 8/2
x = 4
Therefore, the required solution is x = 4, y = -1.
Example 4.
Solve the system of linear equations 2x + 5y = 20, x + 2y = 4?
Solution:
Given simultaneous linear equations are 2x + 5y = 20, x + 2y = 4
On transposing, we get
2x + 5y – 20 = 0 —— (i)
x + 2y – 4 = 0 —— (ii)
Multiply equation (i) by 2.
2(2x + 5y – 20) = 2 x 0
4x + 10y – 40 = 0 ——- (iii)
Multiply equation (ii) by 5.
5(x + 2y – 4) = 0 x 5
5x + 10y – 20 = 0 —– (iv)
Subtract equation (iii) from equation (iv)
[5x + 10y – 20] – [4x + 10y – 40] = 0
5x + 10y – 20 – 4x – 10y + 40 = 0
x + 20 = 0
x = -20
Put x = -20 in equation (iii)
4(-20) + 10y – 40 = 0
-80 + 10y – 40 = 0
10y – 120 = 0
10y = 120
y = 120/10
y = 12
Therefore the required solution is x = -20, y = 12. |
Using Subtraction Theorems in Proofs - dummies
# Using Subtraction Theorems in Proofs
There are four subtraction theorems you can use in geometry proofs: two are for segments and two are for angles. Each of these corresponds to one of the addition theorems.
Here are the subtraction theorems for three segments and three angles (abbreviated as segment subtraction, angle subtraction, or just subtraction):
• Segment subtraction (three total segments): If a segment is subtracted from two congruent segments, then the differences are congruent.
• Angle subtraction (three total angles): If an angle is subtracted from two congruent angles, then the differences are congruent.
Check out the above figure, which provides the visual aids for these two theorems.
Last but not least, here are the subtraction theorems for four segments and for four angles (abbreviated just like the subtraction theorems for three things):
• Segment subtraction (four total segments): If two congruent segments are subtracted from two other congruent segments, then the differences are congruent.
• Angle subtraction (four total angles): If two congruent angles are subtracted from two other congruent angles, then the differences are congruent.
The above figure illustrates these two theorems.
Before reading the formal solution of the next proof, try to think through your own game plan or commonsense argument about why the prove statement has to be true.
Statement 1:
Reason for statement 1: Given.
Statement 2:
Reason for statement 2: If an angle (angle QUR) is subtracted from two congruent angles (angles PUR and SUQ), then the differences are congruent.
Statement 3:
Reason for statement 3: Given.
Statement 4:
Reason for statement 4: If a ray bisects an angle, then it divides it into two congruent angles (definition of bisect).
Statement 5:
Reason for statement 5: If two congruent angles (the angles from statement 2) are added to two other congruent angles (the ones from statement 4), then the sums are congruent. |
How do you evaluate arcsin(sqrt 3/2)?
Jun 10, 2015
arcsin(sqrt(3)/2)=60°=pi/3
Explanation:
$\frac{\sqrt{3}}{2}$ is a known value, and the main angle $\alpha$ that has $\sin \left(\alpha\right) = \frac{\sqrt{3}}{2}$ is alpha=60°=pi/3.
Because arcsin is a function RR->[-1;1], we take only the value $\alpha = \frac{\pi}{3}$, without the periodic values.
So arcsin(sqrt(3)/2)=60°=pi/3.
Mar 6, 2018
Make a right triangle with one side = sqrt 3 and the hypotenuse = 2 and use Pythagoras to find the other leg = 1
Explanation:
If you know that the sin 30 deg = 1/2 .............
Make a right triangle with one side = sqrt 3 and the hypotenuse = 2 and use Pythagoras to find the other leg = 1
That makes the sign of the complementary angle = 1/2 which implies the angle = 30 deg, $\pi$/6, so
the angle in question = 90 - 30 = 60 degrees or $\pi$ / 3
OR
Just calculate (sqrt 3) / 2 and find the arcsin with a calculator
Mar 6, 2018
$\frac{\pi}{3} , \frac{2 \pi}{3}$
Explanation:
$\sin x = \frac{\sqrt{3}}{2}$
Trig Table gives as solution:
$x = \frac{\pi}{3}$ , or $x = {60}^{\circ}$
The unit circle gives another x that has the same sin value (sqrt3/2)
$x = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$, or $x = {120}^{\circ}$
Answers for $\left(0 , 2 \pi\right)$:
$\frac{\pi}{3} , \frac{2 \pi}{3}$
For general answer, add $2 k \pi$ |
# RD Sharma Solutions - Chapter 22 - Mensuration - III (Part - 3), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8
## Class 8: RD Sharma Solutions - Chapter 22 - Mensuration - III (Part - 3), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8
The document RD Sharma Solutions - Chapter 22 - Mensuration - III (Part - 3), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8
PAGE NO 22.26:
Question 20:
The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can be obtained from the trunk.
Circumference of the tree = 176 cm = 2πr
Length of the trunk, h = 3 m = 300 cm
So, the radius (r) can be calculated by:
Thus, the volume (V) of the timber can be calculated using the following formula:
V = πr2(h) = 22 (28 )2 (300) cm³ = 739200 cm³ = 0.74 m³
Question 21:
A well is dug 20 m deep and it has a diameter of 7 m. The earth which is so dug out is spread out on a rectangular plot 22 m long and 14 m broad. What is the height of the platform so formed?
Height of the well = h m = 20 m
Diameter of the well = d m = 7 m
Radius of the well = r m = 3.5 m
Volume of the well = πr²h = 22/7(3.5)2(20 ) m³ = 770 m³
Volume of the well = Volume of the rectangular plot
Length of the rectangular plot = 22 m
Breadth of the rectangular plot = 14 m
Volume of the rectangular plot = 770 m³ = (Length × Breadth × Height) of the rectangular plot
Height = = 2.5 m
Thus, the height of the platform is 2.5 m.
Question 22:
A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.
Diameter of the well = d m = 14 m
Height of the well = h m = 8 m
Radius of the well = r m = 7 m
Volume of the well = πrh = π(7 m)2(8 m) = 1232 m³
Volume of the well = Volume of the embankment
An embankment is a hollow cylinder with thickness. Its inner radius would be equal to the radius of the well, i.e. r = 7 m, and its outer radius is R = 7 + 21 = 28 m.
Volume of the embankment = πh(R2-r2)
To find the height (h), we use the fact that the volume of the embankment is equal to the volume of the well.
1232 = πh ((28)2-(7)2)
Hence, the height of the embankment is 0.533 m or 53.3 cm.
Question 23:
A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm × 22 cm × 14 cm. Find the rise in the level of the water when the solid is completely submerged.
Diameter of the cylindrical container = d cm = 56 cm
Radius of the cylindrical container = r cm = 28 cm
Volume of cylindrical container = Volume of the rectangular solid
Length of the rectangular solid = 32 cm
Breadth of the rectangular solid = 22 cm
Height of the rectangular solid = 14 cm
Volume of the rectangular solid = Length x Breadth x Height = 32 cm x 22 cm x 14 cm = 9856 cm³
Volume of the cylindrical container = 9856 cm³ = πr²h
9856 cm³ = 22/7 (28 cm)2h
h = 4 cm
Thus, when the solid is completely submerged, the water will rise up to 4 cm.
Question 24:
A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.
Case 1:
Height = h cm = 30 cm
Diameter = d cm = 18 cm
Radius = r cm = 9 cm
∴ Volume = (πr²h) or (π(9 )2(30)) = 2430π cm³
Case 2:
Height = h cm = 18 cm
Diameter = d cm = 30 cm,
Radius = r cm = 15 cm
∴ Volume = (πr²h) or (π(15 )2(18)) = 4050π cm³
Hence, the ratio of the volumes of the two cylinders formed is 3:5.
Question 25:
The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter. If it rains 10 cm on a day, what is the rise of water level in the tank due to it?
Length of the water on a roof = 18 m
Breadth of the water on a roof = 16.5 m
Height of the water on a roof = 10 cm = 0.1 m
Volume of the water on a roof = Length × Breadth × Height = 18 m × 16.5 m × 0.1 m = 29.7 m³
Since water is to be stored in the cylindrical tank, the volume of water on a roof is equal to the volume of a cylindrical tank.
Volume of cylindrical tank = πr²h = 29.7 m³
= 0.5906 m = 59.06 cm
Thus, the rise of water level in the tank is 59.06 cm.
Question 26:
A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm. It is drawnout into a wire of diameter 1 mm. What will be the length of the wire so formed?
Diameter of the ductile metal = 1 cm
Radius of the ductile metal = 0.5 cm
Volume of the ductile metal = πr2(length) = π(0.5 cm)2(5 cm) = 1.25π cm³
Ductile metal is drawn into a wire of diameter 1 mm.
Radius of the wire = 0.5 mm = 0.05 cm
Length of wire = = 500 cm = 5 m
Thus, the length of wire is 5 m.
Question 27:
Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.
Density of copper = Weight/Volume = 8.4 gram/1 cm³ = 8.4 gram/cm³
Volume = Weight/Density = 13.2 kg × 1000 gram/kg/8.4 gram/cm³ = 1571.43 cm³
Thus, length of 13.2 kg of copper is 125 m.
Question 28:
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Diameter of the cylindrical wire = 0.25 cm
Radius of the cylindrical wire = 0.125 cm
Volume of the brass = 2.2 dm³ = 2200 cm³
Volume of the brass = Volume of the cylindrical wire
Length of wire = = 44800 cm = 448 m
Thus, length of the wire is 448 m.
Question 29:
The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.
r = Inner radii of the tube
R = Outer radii of the tube
h = Length of the tube
2πh(R-r) = 88 ... (1)
πh(R2-r2) = 176 ... (2)
Substituting h = 14 cm in equation (1) and (2):
π(R-r) = 88/28 ... (1)
π(R-r)(R+r) = 176/14 ... (2)
Simplifying the second equation by substituting it with the first equation:
R+r = 4cm or R = (4−r)cm
Re-substituting R = 4−r into equation (1):
22/7 (4-r-r) = 88/28
4-2r = 1
r = 1.5 cm
R = 4-1.5 = 2.5 cm
Hence, the inner and the outer radii of the tube are 1.5 and 2.5 cm, respectively.
Question 30:
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?
Radius of the circular pipe = 0.01 m
Length of the water column in 1 sec = 6 m
Volume of the water flowing in 1 s = πr2 h = π(0.01)2(6) m³
Volume of the water flowing in 30 mins = π(0.01)2(6) × 30 × 60 m³
Let hm be the rise in the level of water in the cylindrical tank.
Volume of the cylindrical tank in which water is being flown = π(0.6)2 × h
Volume of water flowing in 30 mins = Volume of the cylindrical tank in which water is being flown
π(0.01)2(6) × 30 × 60 = π(0.6)2 × h
h = 3 m
Question 31:
A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.
Here, r = Inner radius = 5.2 cm
t = Thickness = 0.8 cm
h = Length = 25 cm
R = r + t = 5.2 cm + 0.8 cm = 6 cm
Volume of the metal = π h (R2 - r2)
= 22/7 × (25) × ((6 )2 - (5.2 )2)
= 704 cm³
Thus, the volume of the metal is 704 cm³.
Question 32:
From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.
Radius of the water tap = 0.75 cm = 0.0075 m
Length of the water flowing in 1 s = 7 m = 700 cm
Volume of water flowing in 1s = π(0.0075)2 × 700
Volume of the water flowing in 1 hour = π(0.0075)2 × 700 × 60 × 60
Volume of the water flowing in 1 hour = 22/7× (0.0075)2 × 7 × 60 × 60 = 4.455 m³ = 4455l (1000l = 1 m³)
Question 33:
A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled?
Radius of the cylindrical tank = 0.7 m
Height of the cylindrical tank = 2.1 m
Volume of the cylindrical tank = π(0.7)2(2.1) m³
Length of the water column flown from the pipe in 1 s = 2 m
Let the time taken to completely fill the water tank be x sec.
Length of the water column flown from the pipe in x sec = 2 x m
Radius of the pipe = 1.75 cm = 0.0175 m
Volume of the water column flown from the pipe in x sec = π(0.0175)2 (2 x) m³
Volume of the cylindrical tank = Volume of the water column flown from the pipe
π(0.7)2(2.1) = π(0.0175)2 (2 x)
= 1680 sec = 28 min
Thus, the time required to fill the water tank is 28 min.
Question 34:
A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.
Let h cm be the length of the paper and r cm be the radius of the paper.
We know that the rectangular sheet of paper 30 cm x 18 cm can be transformed into two types of cylinder.
Type 1:
Length = 30 cm
Diameter = 18 cm
Volume = (πr²h) = (π(9cm)(30cm)) = 2430π cm³
Type 2:
Length = 18 cm
Diameter = 30 cm
Volume = (πr²h) or (π(15cm)(18cm)) = 4050π cm³
Hence, the ratio of the volumes of the two cylinders formed is 3:5.
Question 35:
How many litres of water flow out of a pipe having an area of cross-section of 5 cm² in one minute, if the speed of water in the pipe is 30 cm/sec?
We know:
Area of cross section = 5 cm²
Rate = 30 cm/s and
Time = 1 min
So, the volume of water flow is:
Volume = Volumetric rate × Time = (30 cm/s)(5 cm²)(60 s/min) = 9000 cm³ = 9 litres
Thus, 9 litres of water flows out of the pipe.
Question 36:
A solid cylinder has a total surface area of 231 cm². Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.
We know that the total surface area of the cylinder is 231 cm² and the curved surface area is 2/3 of the total surface area.
So, the curved surface area is:
2/3 × (231 cm²) = 154 cm²
Then, the radius of the cylinder can be calculated in the following manner:
Curved surface area = 2πrh
154 cm² = 2πrh ... (1)
Here, r cm is the radius of the cylinder and h cm is the length of the cylinder.
2πr2 = (231-154) cm² = 77 cm²
77 cm² = 2πr2
From here, the radius (r) can be calculated in the following manner:
r = 3.5 cm
Substituting this result into equation (1):
154 cm² = 2π(3.5 cm)h
h = 154 cm² / (2x 22/7 x (3.5cm))
h = 7 cm
∴ V = πr2h = 22/7 x (3.5 cm)2 x (7 cm) = 269.5 cm³
Hence, the volume of the cylinder is 269.5 cm³.
The document RD Sharma Solutions - Chapter 22 - Mensuration - III (Part - 3), Class 8, Maths Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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Greatest Common Factor and Least Common Multiple
GCF and LCM
# Motivation Problem for GCF
How are factors involved in the following scenario?
Gary has 20 table tennis balls and 16 paddles. He wants to sell common sized packages containing both paddles and balls. What is the greatest number of packages he can sell with no left over balls or paddles?
We use factors to search for solutions to the first problem.
1 · 20 = 20 and 1 · 16 = 16 One package with 20 balls and 16 paddles. 2 · 10 = 20 and 2 · 8 = 16 Two packages each with 10 balls and 8 paddles. 4 · 5 = 20 and 4 · 4 = 16 Four packages each with 5 balls and 4 paddles.
The question can be answered by finding the greatest common factor. If Gary wants to divide the balls and paddles into packages with each package containing the same number of balls then we are looking for a number that is a factor of both. This is what we would call a common factor. The above illustration shows that Gary has three options for packaging the balls and paddles with one, two, or four packages. If we want to create as many packages as possible then we are looking for the greatest common factor.
Example:
Set of factors of 20 is {1, 2, 4 , 5, 10, 20}
Set of factors of 16 is {1, 2, 4 , 8, 16}
From these two lists we see that the greatest common factor of 20 and 16 is 4, so Gary would be able to sell four packages each containing four paddles and five balls.
This problem motivates a need for being able to determine the greatest common factor for two or more values.
# Greatest Common Factor (GCF)
Definition of Greatest Common Factor (GCF) :
The greatest common factor (GCF) of two natural numbers n and m is the greatest natural number k that is a factor of both n and m. We symbolize this as GCF(n, m) = k.
Example:
Relating this definition to the problem we solved on the previous page, we see that the two numbers n and m are the numbers 20 and 16. We may write the solution symbolically as GCF(20, 16) = 4. This says "the greatest common factor of 20 and 16 is 4."
The number k in the definition corresponds to the 4 in the problem. It is the factor of the greatest value shared by both n and m. The number k is the Greatest Common Factor of 20 and 16.
One method for finding the greatest common factor is to choose the value that is the greatest value in the intersection of the sets of common factors.
Example: Find the GCF(20, 16).
The set of factors of 20 is {1, 2, 4, 5, 10, 20} and the set of factors of 16 is {1, 2, 4, 8, 16}.
{1, 2, 4, 5, 10, 20} ∩ {1, 2, 4, 8, 16} = {1, 2, 4} which has greatest value 4.
So, GCF(20, 16) = 4.
Notice that an intermediate step in this method applied to the problem at the beginning of the session gives all possibilities for packaging all the balls and paddles: one, two, or four packages.
Example: Find GCF(18, 24).
The set of factors of 18 is {1, 2, 3, 6, 9, 18}
and set of factors of 24 is {1, 2, 3, 4, 6, 8, 12, 24}.
{1, 2, 3, 6, 9, 18} ∩ {1, 2, 3, 4, 6, 8, 12, 24} = {1, 2, 3, 6} which has greatest value 6.
The GCF (18, 24) = 6.
# Short Cut Methods for GCF
As the number of factors becomes larger, it is not practical to list every factor and search for the greatest. Instead, we can use what we know about prime factorization to help us find the greatest common factor.
Example: Find GCF (308, 1176).
First, we find the prime factorizations of the numbers.
308 = 22 · 7 · 11 and 1176 = 23 · 3 · 72
Next, we examine how the factors correspond. For now, we will expand the exponent notation and line up the factors that match, leaving any other factors at the end of the factorization.
We see that 2 · 2 · 7 are the factors that are in both factorizations.
Therefore the GCF (308, 1176) = 2 · 2 · 7 = 28.
Further, notice that when the factors are matched up, there are three factors of 2 in 1176 but there are only two factors of 2 in 308. We were only able to match up the lesser number of 2's. This corresponds to the least exponent of the factor 2.
So to get the GCF of two numbers, we only need to take the common prime factors from the prime factorizations and compare their exponents. In each case, we take the exponential expression that has the least exponent and multiply those together.
Example: We reconsider the previous example to illustrate the use of the exponents.
Find GCF (308, 1176).
First, we find the prime factorizations of the numbers.
308 = 22 · 7 · 11
1176 = 23 · 3 · 72
For the common prime factors, the least exponent for 2 is 2 and the least exponent for 7 is 1. So, in this case we have GCF (308, 1176) = 22 · 71 = 4 · 7 = 28.
Example: Find GCF(4950, 7020)
4950 = 2 · 32 · 52 · 11
7020 = 22 · 33 · 5 · 13
The common prime factors are 2, 3, and 5. The least exponent for 2 and 5 is one (21 = 2 and 51 = 5) and for 3 is two.
So the GCF (4950, 7020) = 2 · 32 · 5 = 90.
Example: Find GCF(7920, 92664)
7920 = 24 · 32 · 5 · 11
92664 = 23 · 34 · 11 · 13
The common prime factors are 2, 3, and 11. The least exponent for 2 is 3 (23), for 3 is 2 (32), and for 11 is 1 (111).
So the GCF (7920, 92664) = 23 · 32 · 11 = 792.
# Motivation Problem for LCM
How are multiples involved in the following scenario?
A machine contains two gears. One gear has 12 teeth and the other has 20 teeth. The gears are aligned by a mark drawn from the center of the first gear to the center of the second gear. What is the smallest number of revolutions of the first gear necessary to realign the mark?
We begin by listing, for each gear, the number of teeth that would pass the starting mark after the first revolution, second revolution, third revolution, … .
First gear: 12, 24, 36, 48, 60 , 72, 84, 96, 108, 120 , 132, 144, 156, 168, 180 , 192, …
Second gear: 20, 40, 60 , 80, 100, 120 , 140, 160, 180 , 200, 220, 240 , 260, 280, 300 ,…
Notice that we are listing the multiples of 12 and 20. The first gear will be aligned with the second gear after five revolutions. (Also, note the second gear will have made three revolutions.)
This question was answered using the least common multiple. Notice that the mark on the first gear returns to its original position after each full revolution. This will happen when it has gone through 12 teeth, 24 teeth, 36 teeth, etc. The mark on the second gear returns to its original position after it has made any number of full revolutions. This will happen when it has gone through 20 teeth, 40 teeth, 60 teeth, 80 teeth, etc.
Notice that 12, 24, 36, … are multiples of 12 and 20, 40, 60, … are multiples of 20.
The mark will be realigned only when a value that is both a multiple 12 and of 20 is reached. The mark will be realigned for the first time when the first such multiple is reached. Therefore, we are looking for the Least Common Multiple of 12 and 20.
# Least Common Multiple (LCM)
Definition of Least Common Multiple (LCM) :
The least common multiple (LCM) of two natural numbers n and m is the smallest valued number h so that h is a multiple of both n and m. We symbolize this as LCM(n, m) = h.
Example:
Relating this definition to the problem we solved on the previous page, we see that the two numbers n and m are the numbers 12 and 20. We may write the solution symbolically as LCM(12, 20) = 60. This says "the least common multiple of 12 and 20 is 60."
The number h in the definition corresponds to the 60 in the problem. It has the least value if the multiples shared by both n and m. The number h is the Least Common Multiple of 12 and 20.
One method for finding the least common multiple is to choose the value that is the minimum value in the intersection of the sets of multiples.
Example: Find the LCM(12, 20).
We find the set of their natural number multiples and then find the least value in the intersection of the two sets.
Set of multiples of 12 is A = {12, 24, 36, 48, 60, 72, 84, 96, 108, 120, …}
Set of multiples of 20 is B = {20, 40, 60, 80, 100, 120, 140, 160, 180, …}
AB = {60, 120, 180, …} which has the least value 60.
The LCM(12, 20) = 60.
Notice that the intermediate step in the problem gives all the common multiples.
Since we have found the least common multiple, 60, for the gear problem, we can now answer the question to the gear problem. The question asked how many revolutions are needed to realign the mark. Since we know the first gear must pass 60 teeth and after every 12 it has made one revolution, we know it must make 5 revolutions before the marks are aligned for the first time (60 ÷ 12 = 5). (Also, note that the second gear must make three revolutions for 60 teeth to pass the mark, 60 ÷ 20 = 3.)
As with the GCF, finding the LCM of larger numbers by listing multiples is not very practical. Also, just like the GCF, we can use prime factorization to help.
Example: Find LCM(308,1176).
Again we list the prime factors of the values.
308 = 22 · 7 · 11
1176 = 23 · 3 · 72
Since we are looking for multiples, every multiple of 308 must contain 22 · 7 · 11 as factors and every multiple of 1176 must contain 23 · 3 · 72 as factors. This means that 2, 3, 7, and 11 must be prime factors in any common multiple. For 2 and 7, which appear in both lists, we need the highest power of that prime as a factor to create the LCM, since 23 is automatically a multiple of 22. So using the highest power of each prime, LCM (308,1176) = 23 · 3 · 72 · 11 = 12,936.
### Joke or Quote
Entities should not be multiplied unnecessarily.
William of Occam (1300-1439) |
Quadrilaterals are a special kind of polygon. Similar to triangles and also other polygons, quadrilaterals have actually special properties and can be classified by qualities of their angles and sides. Expertise the properties of different quadrilaterals can assist you in solving difficulties that show off this form of polygon.
You are watching: A parallelogram with four congruent sides and four right angles
Picking personally the name “quadrilateral” helps you understand what that refers to. The prefix “quad-” way “four,” and “lateral” is obtained from the Latin word for “side.” for this reason a square is a four-sided polygon.
Since the is a polygon, you know that that is a two-dimensional figure made up of right sides. A quadrilateral additionally has 4 angles formed by its 4 sides. Below are some examples of quadrilaterals. Notification that each figure has four straight sides and four angles.
The amount of the inner angles of any kind of quadrilateral is 360°. Think about the two examples below.
You could draw many quadrilaterals such as these and carefully measure up the 4 angles. Friend would uncover that because that every quadrilateral, the amount of the internal angles will constantly be 360°.
You can likewise use your understanding of triangles as a method to understand why the sum of the inner angles of any quadrilateral is 360°. Any type of quadrilateral can be split into two triangles as presented in the images below.
In the first image, the quadrilaterals have actually each been split into two triangles. The angle measurements of one triangle are displayed for each.
These measurements include up come 180º. Currently look at the dimensions for the various other triangles—they also add up to 180º!
Since the sum of the inner angles of any kind of triangle is 180° and also there are two triangles in a quadrilateral, the sum of the angle for each quadrilateral is 360°.
Specific types of Quadrilaterals
Let’s begin by evaluating the team of quadrilaterals that have two pairs of parallel sides. This quadrilaterals are dubbed parallelograms They take a selection of shapes, yet one classic example is shown below.
Imagine extending the bag of the opposite sides. Lock would never intersect since they are parallel. Notice, also, the the opposite angles of a parallelogram space congruent, as are the the opposite sides. (Remember the “congruent” means “the same size.”) The geometric symbol because that congruent is , for this reason you deserve to write
and
. The parallel sides are additionally the same length:
and
. These relationships room true for every parallelograms.
There room two special situations of parallelograms that will certainly be acquainted to friend from her earliest experiences through geometric shapes. The an initial special situation is dubbed a rectangle. By definition, a rectangle is a parallelogram due to the fact that its pairs of the contrary sides are parallel. A rectangle likewise has the one-of-a-kind characteristic that all of its angle are appropriate angles; all four of that is angles space congruent.
The other special instance of a parallelogram is a special form of rectangle, a square. A square is just one of the most simple geometric shapes. The is a special instance of a parallelogram that has four congruent sides and also four ideal angles.
A square is likewise a rectangle because it has two sets of parallel sides and four best angles. A square is also a parallelogram due to the fact that its the opposite sides are parallel. So, a square can be divide in any of these 3 ways, with “parallelogram” being the least details description and also “square,” the many descriptive.
Another quadrilateral that you can see is referred to as a rhombus. All 4 sides of a rhombus room congruent. Its properties encompass that each pair the opposite sides is parallel, likewise making that a parallelogram.
In summary, all squares space rectangles, yet not every rectangles space squares. Every rectangles space parallelograms, yet not every parallelograms are rectangles. And every one of these forms are quadrilaterals.
The diagram below illustrates the relationship between the different species of quadrilaterals.
You can use the properties of parallelograms to settle problems. Consider the instance that follows.
Example Problem Determine the procedures of and . is the contrary is opposite Identify the opposite angles. A building of parallelograms is that opposite angles are congruent. = 60°, for this reason = 60° = 120°, therefore = 120° Use the provided angle measurements to recognize measures of the opposite angles. Answer = 60° and = 120°
There is an additional special type of quadrilateral. This quadrilateral has actually the building of having actually only one pair the opposite sides that are parallel. Here is one example of a trapezoid.
Notice the
, and that and are no parallel. Girlfriend can easily imagine that if you extended sides and , they would intersect over the figure.
If the non-parallel sides of a trapezoid are congruent, the trapezoid is dubbed an isosceles trapezoid. Like the likewise named triangle that has two political parties of same length, the isosceles trapezoid has actually a pair that opposite sides of same length. The other pair of opposite political parties is parallel. Below is an example of one isosceles trapezoid.
In this trapezoid ABCD,
and
. |
## Choice variance
Let the vector $$\mathbf x=(x_1,…,x_n)\in \mathbb R^n$$ represents $$n$$ observations, i.e. choices. The variance is defined as $$\displaystyle \sigma^2=\sum_{i=1}^n (x_i-\bar x)^2$$, where $$\displaystyle \bar x=\frac 1n \sum_{i=1}^n x_i$$ is the average.
Show that $$\sigma^2$$ is a quadratic form on $$\mathbb R^n$$ and determine the matrix of this form.
• #### Resolution
We use the well known fact
$$\displaystyle \sigma^2=\sum_{i=1}^n (x_i-\bar x)^2= \sum_{i=1}^n (x_i^2-\bar x^2)$$,
hence
$$\displaystyle \sum_{i=1}^n (x_i^2-\bar x^2)= \sum_{i=1}^n x_i^2- n\bar x^2= \sum_{i=1}^n x_i^2- n\left(\frac 1n \sum_{i=1}^n x_i \right)^2= \sum_{i=1}^n x_i^2- \frac 1n \sum_{i=1}^n\sum_{j=1}^n x_ix_j=$$
$$\mathbf x^T\mathbf x -\frac1n \mathbf x^T\mathbf J_n\mathbf x= \mathbf x^T\left(\mathbf I_n-\frac 1n \mathbf J_n\right)\mathbf x$$,
where the matrix $$\mathbf J_n$$ contains only ones.
Note that the above relation can also be derived as:
$$\sum_{i=1}^n (x_i-\bar x)^2=\sum_{i=1}^n (x_i^2-2x_i\bar x+ \bar x^2)= \sum_{i=1}^n x_i^2 -2\bar x \sum_{i=1}^n x_i+ n\bar x^2=$$
$$\sum_{i=1}^n x_i^2 -n\bar x^2= \sum_{i=1}^n (x_i^2-\bar x^2)$$.
• #### Result
The form can be expressed as the product $$\sigma^2=\mathbf x^T\left(\mathbf I_n-\frac 1n \mathbf J_n\right)\mathbf x$$.
The matrix of the form is $$\begin{pmatrix} \frac{n-1}{n} & -\frac1n & … &-\frac1n \\ -\frac1n & \frac{n-1}{n} & \ddots & \vdots \\ \vdots &\ddots &\ddots &-\frac1n\\[3pt] -\frac1n & … & -\frac1n & \frac{n-1}{n} \\ \end{pmatrix}$$ |
# What is the probability of rolling a 3 twice on a six sided dice?
Contents
Probability of getting an odd number 1/2 ( 1, 3 or 5 so three good outcomes out of 6 possible outcomes; and 3/6 reduces to 1/2). Now multiply the two events: 1/6 x 1/2 = 1/12 or approximately 8.3%. (1/6)×(3/6)=1/12 Answer. Assuming that the die is a fair six sided die with faces numbered individually 1 through 6 .
## What is the probability of rolling a 3 on a die twice?
1/6 x 1/6 is 1/36. So the probability of rolling a 3 and a 2 is 1/36.
## What is the probability of rolling a multiple of 3 on a 6 sided dice?
If it is a standard fair 6 sided die, the probability of rolling a multiple of 3 is (1/3). This is because all six numbers (1,2,3,4,5,6) are equally likely and two of the six 3 and 6 are multiples of three.
## What is the total probability of rolling a die twice and getting a 3 on the first roll and a number greater than 3 on the second roll?
For the second die, you want a number greater than 3. There are 3 possibilities out of 6, so the probability is 3/6 = 1/2. The probability that both of these happened is 1/6 x 1/2 = 1/12.
## What is the probability of rolling a 1 on a 6-sided die?
Two (6-sided) dice roll probability table. Single die roll probability tables.
Probability of rolling more than a certain number (e.g. roll more than a 5).
Roll more than a… Probability
1 5/6(83.33%)
2 4/6 (66.67%)
3 3/6 (50%)
4 4/6 (66.667%)
## What’s the probability that your second roll is a 6 given that first roll is a 6 already?
1 Answer. As other people have pointed out in comments, the correct answer to the question “what is the probability of rolling another 6 given that I have rolled a 6 prior to it?” is indeed 16. This is because the die rolls are assumed (very reasonably so) to be independent of each other.
## What is the probability of getting a multiple of 3 on dice?
The probability of getting a multiple of 3 is therefore = n(E)/n(S) = 2/6 =1/3.
## What is the probability of rolling a 3 or an odd number?
In this situation, rolling a six-sided has 6 outcomes, each of which is equally likely, so we can define the probability of an event (such as rolling a 3 or rolling an odd number) as the ratio of favorable outcomes to possible outcomes: the probability of rolling a 3 is 16 and the probability of rolling an odd number …
## When rolling a die 100 times what is the probability of rolling a 6 exactly 20 times?
Assuming that each roll is independent, you have a 1/6 chance of rolling a 6. You want to find the probability of it rolling a 6 twenty times out of 100, so we have (1/6)^20. Here the order doesn’t matter so multiply this by (100/20). For eighty times, you will have (5/6)^80 chance in rolling a die that is not a 6.
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## What are the odds of rolling a 6 with 2 dice?
When you roll two dice, you have a 30.5 % chance at least one 6 will appear. This figure can also be figured out mathematically, without the use of the graphic. |
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# Three men paint a house in 20 days. How many days 30 men take to do the same?(a) 3(b) 2(c) 60(d) 12
Last updated date: 17th Jul 2024
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Hint: We first try to form the proportionality equation for the variables. We take an arbitrary constant. We use the given values of the variables to find the value of the constant. Finally, we put the constant’s value to find the equation.
Complete step-by-step solution:
We have been given the relation between two variables where we assume number of men as r and number of days as t.
The inversely proportional number is actually directly proportional to the inverse of the given number. The relation between r and t is inverse relation.
It’s given r varies inversely as t which gives $r\propto \dfrac{1}{t}$.
To get rid of the proportionality we use the proportionality constant which gives $r=\dfrac{k}{t} \Rightarrow rt=k$.
Here, the number k is the proportionality constant. It’s given $r=3$ when $t=20$.
We put the values in the equation $rt=k$ to find the value of k.
So, $3\times 20=k$. Simplifying we get $3\times 20=k=60$.
Therefore, the equation becomes with the value of k as $rt=60$.
Now we simplify the equation to get the value of t for number of men being 30
\begin{align} & 30t=60 \\ & \Rightarrow t=\dfrac{60}{30}=2 \\ \end{align}
Therefore, the number of days required to complete the work is 2.
Note: In a direct proportion, the ratio between matching quantities stays the same if they are divided. They form equivalent fractions. In an indirect (or inverse) proportion, as one quantity increases, the other decreases. In an inverse proportion, the product of the matching quantities stays the same. |
# How do you find the intervals of increasing and decreasing linear equations?
## How do you find the intervals of increasing and decreasing linear equations?
We can determine if a linear function is increasing or decreasing by determining the sign of the slope. If the slope is negative or less than zero then the function is decreasing.
## How do you find the interval of increase in a linear function?
It’s always climbing as I look at it moving from left to right. So therefore the answer is that in question a the first equation. My function is always increasing.
What are the intervals where the function is increasing and decreasing?
Value 6 is the graph. Increasing is the f of x increasing. Or decreasing or constant increasing so what we would write is the graph is increasing. From the x. Value 3 to 6. The most common mistake is
### Do linear functions have increasing and decreasing intervals?
A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 2.1.
### How do you find the intervals of increase and decrease on a graph?
To determine the intervals where a graph is increasing and decreasing: break graph into intervals in terms of , using only round parenthesis and determine if the graph is getting higher or lower in the interval. (getting higher) or decreasing (getting lower) in each interval.
How do you find the increase and decrease of a graph?
Finding where a graph is increasing or decreasing – YouTube
## How do you find the interval?
The class interval is the difference between the upper class limit and the lower class limit. For example, the size of the class interval for the first class is 30 – 21 = 9. Similarly, the size of the class interval for the second class is 40 – 31 = 9.
## How do you find the intervals of increasing and decreasing class 12?
The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain. (i) increasing on I if x1 < x2 in I => f(x1) ≤ f(x2) for all x1, x2 Є I. (ii) strictly increasing on I if x1 < x2 in I => f(x1) < f(x2) for all x1, x2 Є I.
How do you find the intervals of a function?
To find the increasing intervals of a given function, one must determine the intervals where the function has a positive first derivative. To find these intervals, first find the critical values, or the points at which the first derivative of the function is equal to zero.
### What is the rate of change of a linear function?
The rate of change of a linear function is also known as the slope. An equation in slope-intercept form of a line includes the slope and the initial value of the function. The initial value, or y-intercept, is the output value when the input of a linear function is zero.
### How do you solve a linear function?
To solve a linear function, you would be given the value of f(x) and be asked to find x.
Solving Linear Functions
1. Substitute the value of f(x) into the problem. In this case:
2. Isolate the variable.
3. Continue to isolate the variable.
4. Simplify.
How do you find increasing and decreasing?
How can we tell if a function is increasing or decreasing?
1. If f′(x)>0 on an open interval, then f is increasing on the interval.
2. If f′(x)<0 on an open interval, then f is decreasing on the interval.
## How do you find the interval of increase on a graph?
Determine the intervals that a graph is increasing and decreasing – YouTube
## How do you determine the intervals on which a function is decreasing?
Explanation: To find when a function is decreasing, you must first take the derivative, then set it equal to 0, and then find between which zero values the function is negative. Now test values on all sides of these to find when the function is negative, and therefore decreasing.
How do you find increase and decrease?
### What is increasing and decreasing functions Class 12?
For a given function, y = F(x), if the value of y is increasing on increasing the value of x, then the function is known as an increasing function and if the value of y is decreasing on increasing the value of x, then the function is known as a decreasing function.
### How do you find the interval of decrease?
To find when a function is decreasing, you must first take the derivative, then set it equal to 0, and then find between which zero values the function is negative. Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of 0, 2, and 10.
How do you find the rate of change over an interval?
Average Rate of Change of a Function Over an Interval – YouTube
## How do you find the rate of change on a linear function graph?
For a linear function, the rate of change is represented by the parameter in the slope-intercept form for a line: y = m x + b , and is visible in a table or on a graph.
## What are the 5 examples of linear equation?
Some of the examples of linear equations are 2x – 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, 3x – y + z = 3. In this article, we are going to discuss the definition of linear equations, standard form for linear equation in one variable, two variables, three variables and their examples with complete explanation.
What is an example of a linear function?
A linear function is a function that represents a straight line on the coordinate plane. For example, y = 3x – 2 represents a straight line on a coordinate plane and hence it represents a linear function. Since y can be replaced with f(x), this function can be written as f(x) = 3x – 2.
### How do you find intervals of increase and decrease on a graph?
How do you calculate increase and decrease?
## How do you find the average rate of change over an interval on a graph?
Ex: Find the Average Rate of Change from a Graph – YouTube |
### Solve equations of two unknowns, similar figures
Question Sample Titled 'Solve equations of two unknowns, similar figures'
Let ${A}$ $\text{seconds}$ be the required time of coating metal onto a suitcase of surface area ${P}$ $\text{m}^{{2}}$ . It is given that ${A}$ is the sum of two parts, one part is a constant and the other part varies as ${P}$ . When ${P}={8}$ , ${A}={1250}$ ; when ${P}={14}$ , ${A}={1730}$ .
(a) Find the required time of coating metal onto a suitcase of surface area of ${18}$ $\text{m}^{{2}}$ . (4 marks) (b) There is a larger suitcase which is similar to the suitcase described in (a). If the volume of the larger suitcase is ${64}$ times that of the suitcase described in (a), find the required time of coating metal onto the larger suitcase. (2 marks)
(a) Let ${A}={c}+{g}{P}$ , where ${c}$ and ${g}$ are non-zero constants. 1A So, we have ${c}+{8}{g}={1250}$ and ${c}+{14}{g}={1730}$ . 1M Solving, we have ${c}={610}$ and ${g}={80}$ . 1A for both correct The required cost$={610}+{80}{\left({18}\right)}$ $={2050}$ $\text{seconds}$ 1A u-1 for missing unit (b) By the concept of similar figures, the surface area of the larger suitcase is ${64}^{{\tfrac{{2}}{{3}}}}$ times that of the suitcase decribed in (a) since the volume of the larger suitcase is ${64}$ times that of the suitcase described in (a). 1M The required cost $={610}+{80}{\left({18}\right)}\cdot{64}^{{\tfrac{{2}}{{3}}}}$ $={23650}$ $\text{seconds}$ 1A u-1 for missing unit
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# How do you find the height of a parallelogram?
## How do you find the height of a parallelogram?
Formula of Height of a Parallelogram
1. Question. What will be the height of the parallelogram if the area is 30 Sq inch and the base of a parallelogram is 6 inch, then find the height of a parallelogram ?
2. Solution.
3. Height=\, \frac{Area}{Base}
4. a= 30, b = 6.
5. H = \frac{30}{6} = 5cm.
6. Hence, the height is 5 cm.
How do you know if a quadrilateral is a cyclic?
In a cyclic quadrilateral, the sum of each pair of opposite angles is 180 degrees. If a quadrilateral has one pair of opposite angles that add to 180, then you know it is cyclic. A trapezoid is cyclic if, and only if, it is isosceles.
### Which is not a rhombus?
Properties of a Rhombus If you have a quadrilateral with only one pair of parallel sides, you definitely do not have a rhombus (because two of its sides cannot be the same length). You have a trapezoid.
How do you find the diagonal measurement of a quadrilateral?
Correct answer: The four sides of a rhombus have equal measure, so each side has measure one fourth of the perimeter of 48, which is 12. \displaystyle RX = HX \cdot \sqrt{3} = 6 \sqrt{3}. Since the diagonals of a rhombus bisect each other, \displaystyle RO = 2 \cdot RX = 2 \cdot 6 \sqrt{3} = 12\sqrt{3}.
## Is every parallelogram a rhombus?
Not every parallelogram is a rhombus, though any parallelogram with perpendicular diagonals (the second property) is a rhombus. In general, any quadrilateral with perpendicular diagonals, one of which is a line of symmetry, is a kite.
Is this a parallelogram Why?
A parallelogram is a quadrilateral with two pairs of parallel sides. The opposite sides of a parallelogram are equal in length, and the opposite angles are equal in measure. Rhombus: If all the sides of a parallelogram are congruent or equal to each other, then it is a rhombus.
### Does a rhombus have 4 90 degree angles?
No, because a rhombus does not have to have 4 right angles. A square has got 4 sides of equal length and 4 right angles (right angle = 90 degrees). A Rhombus has got 4 sides of equal length and opposite sides are parallel and angles are equal.
What is a real life example of a parallelogram?
Real life examples of parallelograms include tables, desks, arrangements of streets on a map, boxes, building blocks, paper and the Dockland office building in Hamburg, Germany.
## Are all angles of a rhombus 90 degrees?
On the other hand, as the basic property of square states that all its interior angles are right angles, a rhombus is not considered as square, unless all the interior angles measure 90°. Any rhombus includes four angles, out of which the opposite ones are equal to each other.
What do you need to prove a quadrilateral is a parallelogram?
To prove a quadrilateral is a parallelogram, you must use one of these five ways.
• Prove that both pairs of opposite sides are parallel.
• Prove that both pairs of opposite sides are congruent.
• Prove that one pair of opposite sides is both congruent and parallel.
• Prove that the diagonals bisect each other.
### What defines a parallelogram?
A parallelogram is a special type of quadrilateral that has equal and parallel opposite sides. We also see a lot of parallelogram like shapes and objects around us. Properties of parallelogram. The opposite sides of a parallelogram are parallel to each other.
Which of the quadrilaterals are parallelograms Why?
A square is also a rectangle because it has two sets of parallel sides and four right angles. A square is also a parallelogram because its opposite sides are parallel. All rectangles are parallelograms, but not all parallelograms are rectangles. And all of these shapes are quadrilaterals.
## Is a parallelogram a rhombus?
So opposite sides are congruent and quadrilateral MNOP is a parallelogram. Also, adjacent sides are congruent, so parallelogram MNOP is a rhombus. 1….Geometry.
Statements Reasons
9. Parallelogram ABCD is a rhombus Definition of rhombus
Why is a kite not a rhombus?
Kites are a special type of quadrilateral with two distinct pairs of consecutive sides the same length. Because rhombi and squares also have sides the same length, they are also kites, but the reverse is not true. Every kite is not a rhombus, because all sides of a kite are not equal.
### Does a rhombus have a 90 angle?
In any rhombus, the diagonals (lines linking opposite corners) bisect each other at right angles (90°). That is, each diagonal cuts the other into two equal parts, and the angle where they cross is always 90 degrees.
How do you find the diagonal?
Explanation: You can find the diagonal of a rectangle if you have the width and the height. The diagonal equals the square root of the width squared plus the height squared.
## What does a diagonal of any quadrilateral connect?
The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices. The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides.
How do you find the diagonal of a cyclic quadrilateral?
### Does a rhombus have 4 right angles?
A square has two pairs of parallel sides, four right angles, and all four sides are equal. It is also a rectangle and a parallelogram. A rhombus is defined as a parallelogram with four equal sides. No, because a rhombus does not have to have 4 right angles.
Is the quadrilateral a parallelogram If yes state how you know?
Is the quadrilateral a parallelogram? If yes, state how you know. Both pairs of opposite sides are parallel. Both pairs of opposite angles are congruent.
## Is this quadrilateral a parallelogram?
Let’s write that as a theorem and lay it to rest. Theorem 16.2: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram….Two Pairs of Congruent Sides.
Statements Reasons
11. Quadrilateral ABCD is a parallelogram Definition of parallelogram
What is the formula of diagonal of parallelogram?
Find the diagonal of a parallelogram with sides 3 cm, 5 cm and angle 45 degrees ? Diagonal of parallelogram = 3.576 cm….Solution: |
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How do you simplify ${\left( {x - 3} \right)^3}$ ?
Last updated date: 20th Jun 2024
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Hint: In this question we need to find the simplified form of ${\left( {x - 3} \right)^3}$. Here, $\left( {x - 3} \right)$being multiplied $3$ times that is $\left( {x - 3} \right)\left( {x - 3} \right)\left( {x - 3} \right)$, the output of it will yield an equation that has ${x^3}$ as the variable with largest power, thus the equation will be cubic equation.
Complete step by step solution:
As we know that the equation of the form $a{x^3} + {x^2} + cx + d$ is known as the cubic equation. Here, $x$ is the variable, $a$, $b$, and $c$ are the coefficients and$d$ is the constant. The largest power of the variable in the equation is three and it determines that the equation is cubic.
In this question, we have given a term that is and we need to simplify it.
Here, is multiplied three times.
Here, we will consider the algebraic identity ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a + b} \right)$
Now we will apply this algebraic identity to the given equation as,
$\Rightarrow {\left( {x - 3} \right)^3} = {x^3} - {3^3} - 3x\left( 3 \right)\left( {x + 3} \right)$
Now, we will simplify the above expression as,
$\Rightarrow {\left( {x - 3} \right)^3} = {x^3} - 9 - 9x\left( {x + 3} \right)$
Now, we will multiply $9x$ in the bracket terms as,
$\Rightarrow {\left( {x - 3} \right)^3} = {x^3} - 9 - 9{x^2} - 27x$
After simplification we will get,
$\therefore {\left( {x - 3} \right)^3} = {x^3} - 9{x^2} - 27x - 9$
From above, the expanded form of the ${\left( {x - 3} \right)^3} = {x^3} - 9{x^2} - 27x - 9$.
Note:
The exponent is the number of times a number can be multiplied by itself. For example consider a variable $a$ as $a \times a = {a^2}$ , then ${a^n}$ represents $a$ multiplied by itself n number of times. The exponent form${a^n}$ is pronounced as a raise to the power n. Where,
${a^0} = 1$ And${a^1} = a$. The other properties are${\left( {ab} \right)^n} = {a^n}{b^n}$, also ${a^n}{a^m} = {a^{m + n}}$. |
# A plane flies 45km in a direction 35 degree south of east, then turns and flies in a direction 75 degree north of east for 65km. Find the final position of the plane. Solve this problem using the parallelogram method of vector addition. Be sure to give the final position in terms of the distance and direction of the plane from its starting point. This is a Physics problem (vectors).
To solve using parallelogram method, draw the two vectors on the same initial point. (See Fig.1 in the attachment.)
Then, draw another two lines in such a way that a parallelogram is formed. (See Fig.2)
The angle between the two vectors is:
`35^o +75^o=110^o`
Applying the property of a parallelogram --- consecutive angles are supplementary, then the other angle is:
`180^o - 110^o=70^o`
And draw a diagonal starting from the initial position. This represents the resultant vector. (See Fig.3)
To solve for the magnitude of the resultant vector, apply Cosine Law.
`c^2=a^2+b^2-2abcosC`
`R^2=45^2+65^2-2(45)(65)cos(70^o)`
`R^2=4249.18`
`R=65.19 km`
To solve for theta, apply Cosine Law again.
`b^2=a^2+c^2-2ac cosB`
`65^2=45^2+65.19^2-2*45*65.19cos(theta)`
`cos(theta)= (45^2+65.19^2-65^2)/(-2*45*65.19)`
`theta = cos^(-1)((45^2+65.19^2-65^2)/(-2*45*65.19))`
`theta=69.55^o`
Using the East axis as the reference for its direction, then the direction of the resultant vector is:
`theta_f = 69.55^o - 35^o`
`theta_f =34.55^o` (See Fig.4)
Therefore, the final position of the plane is 65.19km away from its starting point, and its direction is 34.55 degree North of East.
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# Normal Distribution
## From discrete to continuous
The probability that a person picked at random has a height between 150 and 160 cm can be calculated from the table:
$p=\frac{41}{15+41+67+72+65+34+3}$
You could also find the probability by looking at the areas of the bars in bar chart. The probability is then found by dividing the area of the red bar by the total area.
$p=\frac{41\cdot 10}{15\cdot 10+41\cdot 10+67\cdot 10+72\cdot 10+65\cdot 10+34\cdot 10+3\cdot 10}$
### Probability and area
One way of illustrating the probability that an outcome is in a certain interval, is to let the area of the interval be the probability. The total area must in this case be one. A new bar chart having the same appearance but the total area of one can be constructed.
#### Exercise 1
Add another column in the spreadsheet representing the relative frequency.
You can add a number of values in cells by writing:
Sum(B2:B8)
If you make a bar chart using the relative frequency instead of the absolute frequency, the total area will still not be one. Add another column where you normalize the relative frequencies in order to get the total area one. How do you construct this column? The column for the data should not be changed. Draw the bar chart!
#### Exercise 2
You can approximate a discrete distribution of data values with a continuous function. One such approximating function is the normal probability distribution function (normal pdf). The graph of the normal pdf is a completely symmetric bell-shaped curve whose appearance depends only on two values; the mean and the standard deviation.
If $$\mu$$ is the mean value and $$\sigma$$ the standard deviation, the normal pdf is:
$f(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{\left( - \dfrac{(x-\mu)^2}{2\sigma^2}\right)}$
Use the mean 173.2492 and the standard deviation 14.0333 to plot the graph of the normal distribution.
## Standard normal distribution
The pdf for the standard normal distribution $$\phi$$, has mean $$\mu =0$$ and standard deviation $$\sigma=1$$.
$\phi (x)=\frac{1}{\sqrt{2\pi }} e^{-\frac{1}{2}x^2}$
A short form for writing that a random variable $$X$$ is distributed normally with mean $$\mu$$ and standard deviation $$\sigma$$, is to write: $$X\ \sim \mathcal{N}(\mu,\sigma^2)$$. A variable that is distributed by the standard normal distribution is denoted $$Z$$. The correspondence between $$Z$$ and $$X$$ is
$Z=\frac{X-\mu}{\sigma}$
The probability is given by the area under the curve. The total area is therefore always one. If the random variable is denoted $$X$$, then the probability $$P$$ is given by
$P(a\lt X \lt b)=\int_a^bf(x)dx$
where $$f(x)$$ is the normal pdf.
## Normal distribution in GeoGebra
By using the tool Probability Calculator, you can find the probability between boundaries for a number of distribution functions. The tool is found in the menu belonging to the spreadsheet.
## The inverse normal problem
If you know the mean and the standard deviation, you can find the probability between given boundaries by using the "Probability Calculator".
The inverse problem is that you know the probability for given boundaries and want to find the mean and the standard deviation. Since there are two unknowns to be found, you must know two probabilities. The problem is solved by using the correspondence with the standard normal pdf.
Suppose that you know that $$P(X\lt 4)=0.4$$ and $$P(X\lt 3)=0.2$$. The boundaries 4 and 3, can be compared to the corresponding boundaries for the standard normal pdf. The correspondence between these boundaries is $$z=\frac{x-\mu}{\sigma}$$, where $$z$$ is the boundary for the standard normal pdf, and $$x$$ is the boundary for the unknown normal pdf.
To find the boundaries corresponding to 4 and 3, you can use GeoGebra. You must find $$z_1$$ from $$P(Z\lt z_1)=0.4$$ and $$z_2$$ from $$P(Z\lt z_2)=0.2$$.
Input 0.4 in the box shown in the picture below and press Enter. You get $$z_1=-0.2533$$.
In a similar way you get $$z_2=-0.8416$$. To find $$\mu$$ and $$\sigma$$, solve the simultaneous equations:
\left\{ \begin{align*} -0.2533&=&\frac{4-\mu}{\sigma} \\ -0.8416&=&\frac{3-\mu}{\sigma} \end{align*} \right.
A graphical solution is shown below.
The solution is $$\mu=4.4306$$ and $$\sigma=1.6998$$. You can check the solution by using the "Probability Calculator".
by Malin Christersson under a Creative Commons Attribution-Noncommercial-Share Alike 2.5 Sweden License
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# What is the linear function that passes through the points (2,1) and (1,1)?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
A linear function is one where the largest power of x is equal to 1. This condition ensures that the graph of the function is a straight line, hence the term linear function.
Let the linear function be f(x) = y = mx + n
As this passes through (2,1) and (1,1)
y = mx + n
=> 1 = 2m + n
=> n = 1-2m
Also, 1 = m + n
substitute n = 1-2m in this
=> 1 = m + 1 - 2m
=> -2m = 0
=> m = 0
n = 1 - 0 = 1
Therefore the required linear function is y = 1 or f(x) = 1.
neela | High School Teacher | (Level 3) Valedictorian
Posted on
To find the linear function that passes through the points (2,1) and (1,1).
We know that the linear function or the line that passes through (x1,y1) and (x2,y2) is given by:
y-y1 = {(y2-y1)/(x2-x1)}(x-x1).
So the line that passes through (2,1) and (1,1) is given by:
y - 1= {(1-1)/(1-2)}(x-2)
y-1 = (0/-1)(x-2)
y-1 = 0.
So y = 1 is the line that passes through (2,1) and (1,1).
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll write the form of a linear function:
f(x) = ax + b
A linear function is determined when it's coefficients are determined. So, we'll have to determine the coefficients a and b.
Since the function is determined by the points (2,1) and (1,1), that means that if we'll substitute the coordinates of the points into the expression of the function, we'll get the relations:
f(2) = 1
f(2) = a*2 + b
2a + b = 1 (1)
f(1) = 1
f(1) = a+b
a + b = 1 (2)
We'll put (1) = (2):
2a + b = a + b
We'll combine and eliminate like terms:
2a - a = b - b
a = 0
We'll substitute a in (2):
b = 1
Since the expression of the function is: f(x) = 1, the function is not linear, but constant. |
## A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average weight of turke
Question
A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average weight of turkeys sold in Charlotte. According to the chain’s researcher, a random sample of 20 turkeys sold at the chain’s stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds. And a random sample of 24 turkeys sold at the chain’s stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds. Use a 5% level of significance to determine whether there is a difference in the mean weight of turkeys sold in these two cities. Assume the population variances are approximately the same and use the pooled t-test
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Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom
The null hypothesis is accepted .
Assume the population variances are approximately the same
Step-by-step explanation:
Explanation:-
Given data a random sample of 20 turkeys sold at the chain’s stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds
The first sample size ‘n₁’= 20
mean of the first sample ‘x₁⁻’= 17.53 pounds
standard deviation of first sample S₁ = 3.2 pounds
Given data a random sample of 24 turkeys sold at the chain’s stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds
The second sample size n₂ = 24
mean of the second sample “x₂⁻”= 14.89 pounds
standard deviation of second sample S₂ = 2.7 pounds
Null hypothesis:-H₀: The Population Variance are approximately same
Alternatively hypothesis: H₁:The Population Variance are approximately same
Level of significance ∝ =0.05
Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42
Test statistic :-
where
substitute values and we get S² = 40.988
t = 1.3622
Calculated value t = 1.3622
Tabulated value ‘t’ = 2.081
Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom
Conclusion:-
The null hypothesis is accepted
Assume the population variances are approximately the same. |
Math holidays are few and far between, so when Pi Day rolls around, our math classes celebrate! It takes place on March 14 (3-14). Pi is the ratio of a circle’s circumference to its diameter. The quantity pi is not some strange number that appears in math formulas; it is a naturally occurring and universal ratio and it holds true for any circle. We spend time learning about circles and their parts (diameter, radius, circumference). Drawing compasses are put out for students to practice perfecting their circles. We read the book Sir Cumference and the Dragon of Pi and we learn pi trivia e.g., which famous mathematician was born on Pi Day? Albert Einstein, who was born in 1879. Each student receives a pi pencil, as they love to try to memorize as many digits as possible. The part of pi day that students are the most excited about is eating pie. Parents help us by sending in a wonderful variety of pies for us to share. Before we cut the pies, we relate the pieces to fractions: if I cut the pie into 8 equal pieces, what is the name for 2 pieces? (2/8ths or 1/4th) What is half of an eighth? (1/16th). We find ways to get a tremendous amount of math out of pies. We finish our circle unit by measuring a variety of round objects, so that students can see how/where pi occurs. Students measure each object’s diameter and circumference. They multiply the diameter by pi (3.14) and the answer should match the circumference they measured: Diameter x Pi = Circumference (of all circles). An alternative application is to measure the circumference and the diameter. When the circumference is divided by the diameter, the result should be pi/3.14. This is the ratio of a circle’s circumference to its diameter and where the power of pi lies. This activity gives them the opportunity to use pi and to experience how this special number relates to circles in the real world: to see that the distance around a circle is a little more than three times as long as its diameter. In the 4th and 5th grades, students are beginning to understand the idea of a constant number such as pi. The fact that it is never-ending and that no discernible pattern or sequence of numbers within it has been found is hard for anyone to grasp, but is also what makes this number so fascinating and mysterious to us all. All in all, we had a wonderful Pi Day! |
# Finding Perpendicular Bisector
• Nov 11th 2006, 10:27 AM
Finding Perpendicular Bisector
Question: Find the perpendicular bisector of the line segment with endpoints (5,2) and (-3,6).
I don't remember the procedure to do this.
• Nov 11th 2006, 10:50 AM
topsquark
Quote:
Question: Find the perpendicular bisector of the line segment with endpoints (5,2) and (-3,6).
I don't remember the procedure to do this.
There is more than one bisector, so there is no unique solution to this. The key is that slope of the perpendicular bisector (m2) is the negative inverse of the slope of the line (m1), ie. $\displaystyle m_2 = -\frac{1}{m_1}$.
For example, the slope of your line is:
$\displaystyle m = \frac{6 - 2}{-3 -5} = \frac{3}{-8} = -\frac{3}{8}$
So the slope of the perpendicular bisector is
$\displaystyle m' = - \frac{1}{m} = - \frac{1}{- \frac{3}{8} } = \frac{8}{3}$
-Dan
• Nov 11th 2006, 12:35 PM
earboth
Quote:
Question: Find the perpendicular bisector of the line segment with endpoints (5,2) and (-3,6).
I don't remember the procedure to do this.
I'm not quite sure if I found the right explanation:
Perpendicular bisector passes through the midpoint of the line and has the perpendicular direction:
M((5+(-3))/2,(2+6)/2). So M(1,4)
The slope of the perpendicular bisector has been calculated by topsquark.
Now use point-slope-formula of a straight line:
$\displaystyle \frac{y-4}{x-1}=\frac{8}{3}$. Solve for y:
$\displaystyle y = \frac{8}{3} \cdot x+\frac{4}{3}$
EB
• Nov 11th 2006, 02:01 PM
topsquark
:o Perpendicular bisector. The key word was in my face the whole time! Oops! Thank you, earboth, for finishing that.
-Dan |
# What is 3/7 + 6/7?
This is how you add
3 7
+
6 7
## Step 1
Since our denominators match, we can add the numerators.
3 + 6 = 9
That gives us the sum, which is
9 7
## Step 2
Now, do we need to simplify this fraction?
First, we attempt to divide it by 2...
No good. So next you try the next prime number, which is 3...
No good. So next you try the next prime number, which is 5...
No good. So next you try the next prime number, which is 7...
No good. So next you try the next prime number, which is 11...
No good. 11 is larger than 9. So we're done reducing.
And we're done! Here's the final answer to 3/7 + 6/7
3 7
+
6 7
=
9 7
© 2014 Randy Tayler |
# 2011 July: Dough Doctor
I’ve heard that there is a way to calculate the amount of dough needed to make any size of pizza. Can you explain how this is done? A: What you are referring to is the use of our old friend “pi” to calculate the surface area of a circle, and then using that number to develop a dough density number. It may sound confusing, but it really isn’t. Here is the way it’s done.
Let’s say you want to make 12-, 14-, and 16-inch diameter pizzas, and you need to know what the correct dough weight will be for each size. The first thing to do is to pick a size you want to work with (any size at all will work). We’ll assume we opted to work with the 12-inch size. The first thing to do is to make our dough, then scale and ball some dough balls using different scaling weights for the dough balls. The idea here is to make pizzas from the different dough ball weights, and then, based on the characteristics of the finished pizza, select the dough ball weight that gives us the pizza that we want with regard to crust appearance, texture and thickness. Make a note of that weight. For this example, we will say that 11 ounces of dough gives us what we were looking for.
We’re now going to find the dough density number that is all-important in determining the dough weights for the other sizes. Begin by calculating the surface area of the size of pizza you elected to find the dough weight for. In this case, it is a 12-inch pizza. The formula for finding the surface area of a circle is pi x R squared. Pi equals 3.14, and R is half of the diameter. To square it we simply multiply it times itself.
Here is what the math looks like:
3.14 x 6 x 6 (or 36) = 113.04 square inches
To calculate the dough density number, we will need to divide the dough weight by the number of square inches. So, now we have 11 ounces divided by 113.04 = 0.0973106 ounces of dough per square inch of surface area on our 12-inch pizza. This number is referred to as the “dough density number.”
Our next step is to calculate the number of square inches of surface area in each of the other sizes we want to make. In this case we want to make 14- and 16-inch pizzas in addition to the 12-inch pizza.
The surface area of a 14-inch pizza is 3.14 x 49 (7 x 7 = 49) = 153.86 square inches of surface area. All we need to do now is to multiply the surface area of the 14-inch pizza by the dough density number (0.0973106) to find the dough scaling weight for the 14-inch pizza — 153.86 x 0.0973106 = 14.972208 ounces of dough. Round that off to 15 ounces of dough needed to make the 14-inch pizza crust.
For the 16-inch pizza we multiply 3.14 X 64 (8 x 8 = 64) = 200.96 square inches of surface area. Multiply this times the dough density factor to get the dough weight required to make our 16-inch crusts. 200.96 X 0.0973106 = 19.555538 ounces of dough. Round that off to 19.5 ounces of dough needed to make the 16-inch pizza crust.
In summary, the following dough weights will be needed to make our 12-, 14-, and 16-inch pizza crusts: 12-inch (11-ounces); 14-inch (15-ounces): and 16-inch (19.5-ounces).
In addition to being used to calculate dough weights for different size pizzas, this same calculation can be used to find the weights for both sauce and cheese, too.
In these applications, all you need to do is to substitute the dough weight with the sauce or cheese weight found to make the best pizza for you. This will provide you with a specific sauce or cheese weight, which can then be used in exactly the same manner to calculate the amount of sauce or cheese required for any other size pizza you wish to make. As an example, going back to that 12-inch pizza, let’s say we really like the pizza when it has 5 ounces of sauce on it. We already know that a 12-inch pizza has a surface area of 113.04 square inches, so we divide five-ounces by 113.04 = 0.0442321 ounces of sauce per square inch of surface area. Our sauce density number is 0.0442321. We know that the 14-inch pizza has a surface area of 153.86 square inches. So all we need to do is to multiply 153.86 times the sauce density number to find the correct amount of sauce to use on our 14-inch pizza. 153.86 x 0.0442321 = 6.80-ounces of sauce should be used on our 14-inch pizza.
For the 16-inch pizza, we know that it has 200.96 square inches of surface area. So all we need to do is multiply this times the sauce density factor — 200.96 x 0.0442321 = 8.88 ounces of sauce should be used on our 16-inch pizza.
To calculate the amount of cheese to use, again, we will use the 12-inch pizza and experiment with applying different amounts of cheese until we find the amount that works best for us. Then divide this amount by the surface area of our test pizza (a 12-inch, which has 113.04-inches of surface area). Lets say that we found six ounces of cheese to work well in our application. six-ounces divided by 113.04 = 0.0530785-ounce of cheese per square inch of surface area. Our cheese density number is 0.0530785.
A 14-inch pizza has 153.86 square inches of surface area. Multiply this times the cheese density number to find the amount of cheese to add on our 14-inch pizza — 153.86 x 0.0530785 = 8.16-ounces of cheese should be used on our 14-inch pizza.
A 16-inch pizza has 200.96 square inches of surface area. Multiply this times the cheese density number to find the amount of cheese to add on our 16-inch pizza — 200.96 x 0.0530785 = 10.66-ounces of cheese should be used on our 16-inch pizza.
By calculating your dough, sauce and cheese weights for each of your pizza sizes, you will find that your pizzas will bake in a more similar manner, regardless of size, this is especially true if you are baking in any of the conveyor ovens, in which the baking time is fixed, and you want to be able to bake all of your pizza sizes at similar baking times. Typically, this allows us to bake pizzas with one to three toppings on one conveyor, regardless of size, and those pizzas with four or more toppings on another conveyor, again, regardless
of size.
If you use a deck or conveyor oven, you will find that your pizzas will bake with greater predictability, and your cost control over your different size ranges will be enhanced, and that can’t hurt in today’s economy.
Tom Lehmann is a director at the American Institute of Baking in Manhattan, Kansas. |
Four 33¢ stamps and three 15¢ stamps
equals \$1.77 in postage.
Complete Solutions:
Several ways to do the problem:
Solution 1.
Using the hint on the first page, divide 177 by 33. With five 33¢ stamps, you have \$1.65 worth of postage and need 12 more cents. With four 33¢ stamps, you have \$1.32 worth of postage and need 45 more cents. Since three 15¢ stamps make 45¢, the solution is four 33¢ and three 15¢ stamps.
Number of 33¢ stamps Number of 15¢ stamps Value 6 0 \$1.98 5 0 \$1.65 5 1 \$1.80 4 1 \$1.47 4 2 \$1.62 4 3 \$1.77
Solution 2.
Another strategy is to make a table (shown at left) listing all of the possible combinations of 33¢ and 15¢ stamps that can be used, without exceeding \$1.77. Then work your way up, adding 15¢ stamps.
Solution 3.
Make a combination chart
, (shown at right) and fill it out until you either reach or pass \$1.77 in every row and column. The first row has the values of an increasing number of 33¢ stamps and the first column the values of increasing the number of 15¢ stamps. Remaining cells are the combinations of the two.
No. of 33¢ stamps 1 2 3 4 5 6 No. of 15¢ stamps \$.33 \$.66 \$.99 \$1.32 \$1.65 \$1.98 1 \$.15 \$.48 \$.81 \$1.14 \$1.47 \$1.80 2 \$.30 \$.63 \$.96 \$1.29 \$1.62 \$1.95 3 \$.45 \$.78 \$1.11 \$1.44 \$1.77 4 \$.60 \$.93 \$1.26 \$1.59 \$1.92 5 \$.75 \$1.08 \$1.41 \$1.74 \$2.07 6 \$.90 \$1.23 \$1.56 \$1.89 7 \$1.05 \$1.38 \$1.71 \$2.04 8 \$1.20 \$1.53 \$1.86 9 \$1.35 \$1.68 \$2.01 10 \$1.50 \$1.83 11 \$1.65 \$1.98 12 \$1.80
Solution 4. Another way to approach the problem is to realize that using only 15¢ stamps makes the total end in 0 or 5. Next look at the number of 33¢ stamps that can be used to make a total that ends in either 2 or 7.
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How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
# Area of Ellipse Problem – Duke Math Meet 2009: Problem 7
Try this problem from Duke Math Meet 2009 Problem 7 based on Area of Ellipse. This problem was asked in the individual round.
Let $R_I , R_{II} , R_{III} , R_{IV}$ be areas of the elliptical region $\frac{(x-10)^2}{10} + \frac {(y-31)^2}{31} \le 2009$ that lie in the first, second, third, and fourth quadrants, respectively. Find $R_I - R_{II} + R_{III} - R_{IV}$
Discussion:
Special Note: The answer to this problem is given as 1240. This is the wrong answer. It approximates the region $R_I - R_{II} + R_{III} - R_{IV}$ as a rectangle. However, we provide a solution using symmetry using that assumption and that works fine. Computing area of the ellipse is a tricky business.
First, we draw an approximate picture of the ellipse. The centre is at (10, 31). Here is a figure of it.
ellipse with centre at (10, 31)
To find $R_{I} - R_{II}$ reflect region $R_{II}$ about y-axis. Look at the figure. The shaded region is $R_{I} - R_{II}$. Its width along the line through centre (y=31) is 20 by symmetry (as the centre is 10 unit away from x-axis so $R_I$ is 20 unit 'thicker' than $R_{II}$ . You may convince yourself about this by solving for x setting y = 31 ).
reflection of R(II) about the y-axis
Now to find $R_{III} - R_{IV}$ we reflect $R_{III}$ about y-axis again. The strip (shaded in blue) is negative of $R_{III} - R_{IV}$
-(R3 - R4) shaded in blue
Note that the blue region 'begins' 31 unit 'below' the minor axis of the ellipse. So if we go 31 unit 'above' the minor axis and take the portion of the red strip, by symmetry it will be equal to the blue strip. We have shaded it in black and red.
-(R3 - R4) in the red strip
So if we want to find $R_{I} - R_{II} - { - (R_{III} -R_{IV}) }$ we remove the black stripe from the red strip and get the final region whose area is $R_I - R_{II} + R_{III} - R_{IV}$.
R1- R2 +R3 - R4
Here is where we apply approximation. Width of the strip is 20 and its height is (31+31) = 62. Hence if we approximate the area as a rectangle, then answer is 1240 (62*20).
But note that the strip is not ACTUALLY a rectangle. So this is only an approximate answer. |
# Area of a regular polygon
We will show you how to derive 3 formulas that you can use to get the area of a regular polygon also called n-gon:
It is not easy to draw an n-gon, so let's represent the regular polygon or n-gon with a pentagon
Our strategy before we derive the formula of the area of a regular polygon will be to show you how to get the area of a pentagon and then generalize the approach for an n-gon
To get the area of the pentagon above, follow the steps below:
1. Break the pentagon into 5 congruent triangles
2. Get the area for one triangle
3. Multiply by 5 to get the area for all 5 triangles or the whole pentagon
Notice that the pentagon has 5 sides and you can make 5 triangles. Similarly, the n-gon has n sides and you can make n triangles.
The formula to get the area of a triangle is A =
base × height / 2
Looking at the triangle above, you can see that the base is s and s is also the length of one side of the pentagon. The red line is the height and it is called apothem in an n-gon
Say that s = 4 and apothem = 8
A =
4 × 8 / 2
Since there are 5 triangles in a pentagon A = 5 ×
4 × 8 / 2
First generalization of the area of a regular polygon
base = s and height = apothem. An n-gon has n triangles
Since there are n triangles in an n-gon, A = n ×
s × apothem / 2
Sometimes, s is not given, but you know the apothem and the number of sides. So you need to find s in this case. You will need some basic trigonometric identities
You need to know also how to get the central angle. Take a look at the figure again, the central angle is the one in black
Find the area when n = 5 and apothem = 8
Call the angle in orange x and use trigonometric identity.
tan (x) =
s / 2 × 8
tan (x) =
s / 16
If 4 =
8 / 2
then, 8 = 4 × 2
Similarly, if tan(x) =
s / 16
then, s = tan(x) × 16
Now, we have to find x. x is half the angle in black. The angle in black is the central angle.
The central angle can be found by using the formula: Central angle =
360 degrees / n
n is the number of sides.
Central angle =
360 degrees / 5
= 72 degrees
The angle in orange or x is equal to 72 divided by 2 or 36 degrees. Thus, s = tan(36°) × 16
Putting it all together, A = 5 ×
tan(36°) × 16 × 8 / 2
This math can be done fairly easily, but we will not do it so you can better see how we can generalize and come up with a formula for an n-gon
Second generalization of the area of a regular polygon
base = s , height = apothem and the n-gon has n sides
A = n ×
s × apothem / 2
Using tan(x) =
s / 2 × apothem
, we get s = tan(x) × 2 × apothem
Find x for an n-gon.
Central angle =
360 degrees / n
Recall though that x is the orange angle, so
360 degrees / n
must be divided by 2
This gives x =
180 degrees / n
s = tan(
180 degrees / n
)× 2 × apothem
Common pitfall: Thinking that the apothem needs to be written only once! Take a look again. The apothem appears in s and also appears in A
### Derivation of the area of a regular polygon when s and n are given, but the apothem is not known
Since the apothem is missing we can use the formula s = tan(x) × 2 × apothem and solve for apothem.
Replace the apothem into the formula for the area and simplify. You will end up with an equation in terms of s and the angle
s = 2 × tan(x) × apothem
apothem =
s / 2 tan(x)
A =
s / 2 tan(x)
If we rewrite A = n ×
s × apothem / 2
it will be easier to manage
A =
n × s / 2
× apothem
A = n × s / 2 × s / 2 × tan(x)
A =
n × s2 / 4 × tan(x)
x is still the same angle, so x =
180 degrees / n
### Derivation of the area of a regular polygon when the radius is given, but the apothem and s are not known
The radius is the blue line or the hypotenuse. If x is the angle in orange and let the radius be r we get:
sin(x) =
s / 2 × r
s = 2 × r sin(x)
cos(x) =
apothem / r
apothem = cos(x) × r
Replace the value for the apothem and s into the formula A = n ×
s × apothem / 2
A = n ×
2 × r sin(x) × cos(x) × r / 2
A = n ×
2 × sin(x) × cos(x) × r2 / 2
Useful trigonometric formula: sin(2x) = 2 sin(x)cos(x)
A = n ×
sin(2x) × r2 / 2
x =
180 degrees / 2
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### Using Combinatorics to ‘count’ Divisors on the GMAT
Are you ready for a challenge? Try to solve the following question in under two minutes:
How many different positive divisors does the number 147,000 have?
If you feel like two minutes are not nearly enough to solve the problem, you’re not alone. Even the most seasoned GMAT veterans might find the problem challenging, as it requires a deep level of understanding of two mathematical concepts: Divisibility and Combinatorics (just a fancy word for ˜counting’).
If I replaced the number 147,000 with the number 24, many more people would be able to come up with an answer:
You could just pair up the divisors (factors) and count them. Start with the extremes (1×24) and work your way in:
1×24
2×12
3×8
4×6
A quick count will show the number 24 has exactly 8 different positive divisors.
The number 147,000 will have many more positive divisors “ too many to count This is a strong indication that we will need to use combinatorics.
Divisibility: Any positive integer in the universe can be expressed as the product of prime numbers.
A few examples:
The number 24 is comprised of three factors of 2 and one factor of 3.
We should be able to ˜build’ any divisor of 24 using some combination of those prime factors; for example, we could choose to use none of them. What do you get if you use none of those prime factors?
Is 1 a divisor of 24? Absolutely!
What if we choose to use all of the prime factors that we have at our disposal?
Is 24 a divisor of 24? Of course!
I can use zero, one, two, or three factors of 2, and for each of those options, I can choose to use either zero or one factor of 3.
Combinatorics: When faced with several choices that are independent of each other, the total number of combinations will be the product of the number of choices in each decision point. For example, if you can choose among 3 appetizers, 2 mains, and 4 desserts when ˜building’ your 3-course meal, the number of different meals you could construct is 3 * 2 * 4 = 24
Intuitively, for each appetizer you could choose (out of 3), there are still 2 options for your main, and for each of those there are still 4 options for your dessert. As a rule of thumb, if you’re using the word each you should multiply the options.
Back to the number 24 allow me to repeat my last point here for your convenience: I can use zero, one, two, or three factors of 2 (four options total), and for each of those options, I can choose to use either zero or one factor of 3 (two options total).
Using combinatorics, we can now say there are exactly 4 * 2 =8 different positive divisors for the number 24.
Try to solve the original question again at this point, before you continue to read my solution.
.
.
.
There are four options for the number of 2 we will use (zero, one, two, or three), two options for the number of 3, four options for the number of 5, and three options for the number of 7.
In total, the answer to the original question is:
Is it just me, or is there a pattern here? You can just add 1 to each of the exponents of the prime numbers and then multiply them all together to get the total number of different positive divisors of any number!
This is a cool rule/formula to memorize, but as you guys know by now “ memorizing little tricks and formulae will only get you so far on the GMAT. In order to crush it you really have to understand why these formulae work, and that was my goal with this blog post Let me know in the comments whether you feel like you have a deeper level of understanding now!
### Takeaways:
1. No matter how complex the problem you’re facing is, you can always simplify it by picking a nicer number (e.g. 24 rather than 147,000).
2. Any positive integer in the universe can be expressed as the product of prime numbers.
3. When faced with several choices that are independent of each other, the total number of combinations will be the product of the number of choices in each decision point.
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8. Avi Gutman October 30, 2012 at 11:18 am
Amar, in the mean time, check out this post for more on probability…
http://www.manhattangmat.com/blog/index.php/2012/10/12/and-vs-or-probability-the-1-x-trick-why-the-orioles-are-201-odds-to-win-the-world-series/
9. Avi Gutman October 27, 2012 at 8:51 pm
Hey folks, thanks for the compliments – I’m glad you found this article useful!
Amar, we will learn about probability this Wednesday in class, but I might write an article about it as well in the near future 😉
Cheers,
Avi
10. Amar October 27, 2012 at 1:56 am
This is really an important concept and you have explained it well…Can you please write something on Probability especially for those problems where we need to decide whether to add or multiply the probabilities of individual occurances.
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### Pythagorean Theorem proof with similar right triangles
```Pythagorean Theorem proof
from similar right triangles
A geometric realization of a proof in
H. Wu’s “Teaching Geometry According
to the Common Core Standards”
Given a right triangle ABC with legs a and b
and hypotenuse c.
Prove a2 + b2 = c2
A
c
b
C
a
B
In other words,
prove that the sum of the
areas of the squares
on the legs a and b
A
c2
c
is equal to
the area of
the square on
the hypotenuse c.
b2
b
C
a
a2
B
But how can we possibly prove that
is
equal
to
the sum of the
areas of these
the area of this?
c2
a2
b2
?
We don’t have to prove that the figures are equal—
they aren’t! All we have to do is prove that the area of
the two little squares equals the area of the big square.
Maybe if we could partition the big square
into 2 rectangles, one equal in area to a2,
and the other to b2…
c2
a2
b2
Hang onto that dream as we proceed…
We’ll partition ABC into 3 similar right
triangles and use the equal ratios of their
corresponding parts in our proof.
Draw a perpendicular
to the hypotenuse AB
through the vertex
of the right angle at C.
That perpendicular
intersects AB at a
point we’ll call D.
A
c
D
b
C
a
B
Now we have 3 similar right triangles:
ABC, the big one we started with,
containing A and B,
A
ACD, the little one
that contains A,
D
b
C
and BCD,
the other little one
that contains B.
c
a
B
We know all 3 triangles are similar by AA:
two pairs of equal corresponding angles make
similar triangles.
c
A
Each triangle has a
right angle, and all
right angles are equal.
D
b
C
And each triangle contains
either A or B or both.
a
B
Reflection, translation, and rotation enable a
nice arrangement of our 3 similar triangles.
A
D
A
a
b
b
D
C
C
C
a
B
D
B
A reflection across AC
A reflection across DB
A translation
A translation
Then a rotation
Then a rotation
Similar triangles means that the ratios
of their corresponding parts will be equal.
c
A
A
C
D
b
D
C
ABC
a
b
C
a
B
CBD
=
D
|| ||
=
|| ||
B
By the cross-multiplication algorithm…
c
A
A
D
C
D
a
b
b
C
a
C
B
D
|| ||
=
|| ||
||2 = ∙ ||
2 = ∙ ||
B
Now we have
= ∙ ||
∙
A
2
is the area
of the square
on leg .
c
D
C
a
a2
∙ is the area
of the rectangle
with sides and .
B
Looking at the left triangles, similarity gives us
c
A
A
D
C
D
a
b
b
C
ACD
C
a
B
ABC
=
D
|| ||
=
|| ||
B
By the cross-multiplication algorithm, we have
c
A
A
D
C
D
a
b
b
C
a
C
B
D
|| ||
=
|| ||
||2 = ∙ ||
2 = ∙ ||
B
And now we have
Could this be
the partition
dreaming of?
c
= ∙
A
D
b2
b
C
a
B
2 is the area of the square on leg .
∙ is the area of the rectangle
with sides and .
Putting it together…
c
2 = ∙ ||
A
2 = ∙ ||
D
b2
2
+
2
c
c
b
C
= ∙ + ∙ ||
a
a2
B
Since 2 + 2 = ∙ + ∙ ||
We distribute out and get
2 + 2 = ∙ ( + )
A
We notice that
= +
c2
b2
so we substitute for
+ , giving us
2 + 2 = ∙ ()
and finally 2 + 2 = 2
cc
D
b
C
a
a2
B
``` |
Q:
# Dice throw
Problem statement:
Given n dice each with m faces, numbered from 1 to m, find the number of ways to get sum XX is the summation of values on each face when all the dice are thrown.
``` Input:
n=3
m=3
X=6
Output:
Total number of ways are: 7
```
Explanation:
``` Total number of dices: 3 say x1,x2,x3
Number of faces on each dice: 3 (1 to 3)
Total sum to be achieved: 6
We will write as xi(j)which means face value of dice xi is j
So sum 6 can be achieved in following ways:
6=x1(1)+x2(2)+x3(3)
6=x1(1)+x2(3)+x3(2)
6=x1(2)+x2(2)+x3(2)
6=x1(2)+x2(3)+x3(1)
6=x1(2)+x2(1)+x3(3)
6=x1(3)+x2(2)+x3(3)
6=x1(3)+x2(3)+x3(1)
This are total 7 ways to achieve the sum.```
If it was only 1 dice, then if X<=m, the answer would be 1 else 0. Since there is only one way to achieve the sum if possible as there is only one dice.
Now when n, number of dice>1, then the problem becomes a recursive one
We can think of the recursive function as f(n,X) where n is number of dice and X is desired sum.
A single dice has m choices, which means the face can have values ranging 1 to m
So,
Recursively we can write,
That means summation of all choices for this particular dice to have face value 1 to minimum(X, m)
For our example case, n=3, m=3, X=6
So, we need to find f(3,6)
``` f(3,6)=f(2,5)+f(2,4)+f(2,3)
```
f(2,5), f(2,4), f(2,3) all are sub problems themselves which are needed to be solved further. This would generate a recursion tree.
Of course, we have base cases for single dice which is f(1,i)=1 for i=1 to m
But this recursion will generate many overlapping sub problems, hence, we need to convert it to dynamic programing.
``` 1) Declare dp[n+1][x+1] similar to f(n,x). Initialize it to 0.
2) Implement the base case f(1,i)
for i=1 to i minimum(m ,x)
dp[1][i]=1;
3) Fill the other values as per recursion relation
for i=2 to n //iterate for number of dices
for j=1 to x //iterate for sums
for k=1 to minimum(m ,j)
//iterate for face values up to minimum(m,j),j be the subsum
dp[i][j]+=dp[i-1][j-k];
end for
end for
end for
```
C++ Implementation:
``````#include <bits/stdc++.h>
using namespace std;
long long int dicethrow(int m, int n, int x) {
if (m * n < x)
return 0;
long long dp[n + 1][x + 1];
memset(dp, 0, sizeof(dp));
//base case
for (int i = 1; i <= m && i <= x; i++)
dp[1][i] = 1;
for (int i = 2; i <= n; i++) { //iterate for number of dices
for (int j = 1; j <= x; j++) { //iterate for sums
//iterate for face values up to minimum(m,j),j be the subsum
for (int k = 1; k <= m & k < j; k++) {
dp[i][j] += dp[i - 1][j - k];
}
}
}
return dp[n][x];
}
int main() {
int n, m, x;
cout << "Enter number of dices, n:\n";
cin >> n;
cout << "Enter number of faces on a dice, m:\n";
cin >> m;
cout << "Enter sum, X:\n";
cin >> x;
cout << "Number of ways to achieve sum: " << dicethrow(m, n, x) << endl;
return 0;
}
``````
Output
```RUN 1:
Enter number of dices, n:
3
Enter number of faces on a dice, m:
3
Enter sum, X:
6
Number of ways to achieve sum: 7
RUN 2:
Enter number of dices, n:
3
Enter number of faces on a dice, m:
3
Enter sum, X:
12
Number of ways to achieve sum: 0
```
In the second output there is no way to acquire the sum which can be verified as m*n<X. It's better practise to keep such base case to optimize your code :) |
Divisibility Rule For 12-The divisibility rule for 12 states that a number is divisible by 12 if it is divisible by both 3 and 4.
To use the rule, first, check if the number is divisible by 3. If it is, then check if the number is also divisible by 4. If the number is divisible by both 3 and 4, then it is also divisible by 12.
What is the divisibility rule for 12?
In mathematics, the divisibility rule for 12 is a guideline that can be used to determine whether or not a number is evenly divisible by 12. The rule tells that if a number is divisible by both 3 and 4, then it is also divisible by 12. For example, the number 36 is divisible by both 3 and 4, so it is also divisible by 12. The divisibility rule for 12 can be useful when trying to simplify fractions or solve division problems.
Divisibility rule for twelve with examples are given here for making this concept strong and understandable.
The divisibility rule for twelve states that if a number is divisible by twelve, then it is also divisible by three and four. For example, if a number is divisible by twelve, then it is also divisible by three and four. However, if a number is not divisible by twelve, then it is not necessarily true that it is also not divisible by three or four.
PDF
## Divisibility Rule For 12
When trying to determine if a number is divisible by 12, we can use the divisibility rule for 12. This states that a number is divisible by 12 if it is divisible by both 3 and 4. Therefore, to check if a number is divisible by 12, we first need to check if it is divisible by 3 and 4.
For example, let’s look at the number 36. It is seen that it is divisible by 3 because the sum of its digits (3+6) is 9 and 9 is divisible by 3. We can also see that it is divisible by 4 because the last two digits (6) form a two-digit number that is divisible by 4 (24). Therefore, we can conclude that 36 is also divisible by 12.
PDF
### Examples of Divisibility Rule of 12
There are several examples of the divisibility rule of 12. One example is that 12 is divisible by 2 because it is an even number. Another example is that 12 is divisible by 3 because it has 3 as a factor. Lastly, 12 is divisible by 4 because it has 4 as a factor.
To use the divisibility rules 12, simply add up all of the digits in the number you are trying to divide. If the sum of those digits is evenly divisible by 12, then so is the original number.
These rules of divisibility for 12 can be useful for students who are learning division. This can help them quickly determine whether or not a particular number can be divided by 12. It can also be helpful in other situations, such as when trying to simplify fractions or calculate remainders.
PDF |
Calculating Derivatives by Definition
This page on calculating derivatives by definition is a follow-up to the page An Intuitive Introduction to the Derivative. On that page, we arrived at the limit definition of the derivative through two routes: one using geometric intuition and the other using physical intuition.
Let's restate that definition. Given a function f, the derivative of that function at a point x equals:
The definition of the derivative
On that introductory page, I used Δx instead of h. Now I will be using h because it is easier to manipulate it algebraically, and we'll be doing a lot of manipulations. Let's examples of how to apply this definition to actually calculate the derivative of a function.
Example 1: Derivative of a Constant
Let's calculate the derivative by definition of the simplest of functions, a constant. That is, we consider
where c is a constant. There is only one thing to do, apply the definition. As the function is constant, we have that
This implies that the difference in the numerator of the definition equals zero
This make the calculation really easy
The derivative of a constant is zero. This is a very important fact that you will use all the time. And it makes intuitive sense also, both geometrically and physically.
The derivative gives the rate of change of the function. As the constant doesn't change, its rate of change equals zero. Geometrically, the graph of a constant function equals a straight horizontal line. Hence, its slope equals zero.
Example 2: Derivative of f(x)=x
Now, let's calculate, using the definition, the derivative of
After the constant function, this is the simplest function I can think of. In this case the calculation of the limit is also easy, because
Then, the derivative is
The derivative of x equals 1.
Example 3: Derivative of f(x)=x²
This example will make us work just a little bit more than the others. We consider the function:
In this case we have that
Replacing this expression into the limit, we get
This limit can be solved by factoring h
The derivative of x² is 2x.
Example 4: Derivative of a Sum
Let's now use the definition to deduce a simple rule for derivatives. Let's say we have a function F that is the sum of two other functions, f and g:
We'll try to calculate the derivative of F in terms of the derivatives of f and g. As one might expect, the derivative of F equals the sum of the derivatives of f and g. To prove that, we do the only thing that we can do, use the definition
We have that:
Let's replace these two expressions in the definition of the derivative
Now let's reorder the terms in the numerator to make evident where the derivatives of f and g will appear
And in fact we can break down this expression into two limits
What we proved is the following
That is, the derivative of a sum is the sum of the derivatives of each term. For example, let's consider the function:
We know the derivative of each term, so the derivative of the sum is:
The same rule applies for differences, because we can think of a difference as a sum in disguise.
Example 5: Derivative of a Constant Times a Function
Let's consider a function of the type
where c is a constant. Let's calculate the derivative of F in terms of the derivative of f. We have that
We can factor the constant c and take it out of the limit sign
That is, we can take constants out, as with limits, when calculating derivatives. For example,
Example 6: Derivative of f(x)=xn (the power rule)
This example is what is called the power rule. We'll calculate the derivative of x raised to the n-th power
This rule is very important because it allows us to calculate the derivative of any polynomial function. Let's get down to it. As always, we apply the definition
Now we have a problem, how do we expand the following
There is a theorem in algebra that solves our problem. It is the binomial theorem. To learn about it, or just to recall if you already know it, I recommend you KhanAcademy's video on the binomial theorem. The binomial theorem says the following:
The binomial theorem
Replacing a=x and b=h in the preceding formula, we get that
Applying the binomial theorem to calculate the derivative
If we subtract x to the n-th power to both sides we get
We obtain an expression for the difference in the numerator of the definition of the derivative
Now, if we divide both sides by h, we get
We obtain an expression for the quotient, dividing both sides by h. A bunch of things cancel out
If we take the limit as h approaches infinity of both sides, in the left side we get the derivative of f, and in the right side a bunch of things approach zero. Notice that the first term in the right does not have an h, so we can take it out of the limit sign:
We take the limit to obtain the derivative. The first factor comes out of the limit sign, because it doesn't have an h
Pay attention to the factors highlighted in pink. These are powers of h, that will approach zero as h approaches zero. The etc represents a bunch of terms that have higher powers of h in the numerator.
So, all the terms, starting from the second, will approach zero when h approaches zero
When h approaches 0, all terms except the first approach zero
Using Leibniz notation
The power rule in Leibniz notation
This is the hardest proof we have done so far, but it is very important. This rule includes the special case of f(x) = x² that we calculated earlier. Let's try to understand how to apply this rule.
This rule is somewhat similar to the rule for logarithms of powers: you take the power "down", and then lower the power of x by 1.
The Power Rule: Take down the exponent and then subtract one to it
As I said earlier, this important rule allows us to calculate the derivative of any polynomial. For example:
Example 7: The Derivative of f(x) = 1/(1+x)
Let's do one more. This one can't be calculated using the power rule.
Notice that all the derivatives we calculated thus far were of polynomial functions. This means that we could have calculated all of them using the power rule. This one isn't a polynomial. So, we must use the definition. We have that
Now, we can add the fractions in the numerator
This fraction can be simplified a bit more
And now, we only need to make h equal to zero to obtain the value of the limit
Exercises
Now, here are some problems for you to solve. First there are some problems where you'll need to apply the power rule. Then there are some problems where you'll need to use the definition to find the derivative of some functions.
Power Rule Exercises
Find the derivative of the following polynomial functions using the power rule.
Derivatives by Definition Exercises
Find the derivative of the following functions using the definition. |
# Video: Pack 3 β’ Paper 2 β’ Question 12
Pack 3 β’ Paper 2 β’ Question 12
03:29
### Video Transcript
π₯ to the power of seven divided by π₯ to the power of π¦ is equal to π₯ to the power of 10. Find the value of π¦.
To answer this question, we need to recall what happens to the powers when weβre dividing two terms with the same base. We subtract the powers. π₯ to the power of π divided by π₯ to the power of π is equal to π₯ to the power of π minus π. This means that the left-hand side of the equation can be written as π₯ to the power of seven minus π¦.
As the bases of the two sides of the equation are the same β theyβre both π₯ β we can equate the powers. This gives seven minus π¦ is equal to 10, which is a linear equation that we can solve in order to find the value of π¦.
As the coefficient of π¦ is negative, we can begin by adding π¦ to both sides of the equation, giving seven is equal to π¦ plus 10. Next, we subtract 10 from both sides to solve for π¦, giving negative three is equal to π¦.
The second part of the question says three squared to the power of π is equal to three to the power of 12. Find the value of π.
To answer this part of the question, we need to recall what happens when you raise a number or a letter to a power and then another power. We multiply the powers together. π₯ to the power of π to the power of π is π₯ to the power of ππ. This means that the left-hand side of the equation can be written as three to the power of two π. So we have three to the power of two π is equal to three to the power of 12.
As the bases on both sides of the equation are the same, we can again equate the powers, giving two π is equal to 12. Dividing both sides by two solves the equation and gives the value of π. π is equal to six.
The third part of the question says eight to the power of π multiplied by 16 to the power of π can be written in the form two to the power of π. Show that π is equal to three π plus four π.
Weβve been asked to express this product in the form two to the power of π, that is, as a power of two. To do so, we first recall that both eight and 16 are powers of two. Eight is equal to two cubed. And 16 is equal to two to the power of four. We can replace eight and 16 with two cubed and two to the power of four, giving two cubed to the power of π multiplied by two to the power of four to the power of π.
If we recall the result that we used in the second part of this question, we know how to simplify the brackets. We can multiply together the powers inside and outside each bracket. This gives two to the power of three π multiplied by two to the power of four π.
Finally, we need to recall what happens to the powers when we multiply together two terms with the same base. We add the powers. π₯ to the power of π multiplied by π₯ to the power of π is equal to π₯ to the power of π plus π. Adding together the powers gives two to the power of three π plus four π. And if we look back at the question, this is what we wanted to show. Weβve expressed the product in the form two to the power of π, where π is equal to three π plus four π. |
# Activities/Turtle Art/Tutorials/Fractions
This is the outline that will be fleshed out in Turtle Art.
• Cut a pie in pieces, and color some of the pieces, as Tony did. That gives the basic idea of a fraction. Point out that when you cut a pie in, say, 5 pieces, you are doing 1 divided by 1/5.
• Cut more than one pie in the same number of pieces each. This lets us talk about "improper" fractions and mixed fractions (integer plus fraction), and converting between them. We can also introduce rational numbers at some stage of child development.
• Cut a pie in pieces, and cut the pieces into smaller pieces (multiplication of the simplest fractions, such as 1/2 times 1/3). Some fractions can be described using the bigger pieces, and some require the smaller pieces. Talk about reducing fractions to lowest terms. (You will need other materials in order to talk about Greatest Common Divisors. I'll do something on that.) Take some time on multiplying fractions. Then notice that, for example, if you divide a pie into sixths, three of the pieces make a half. 3 times 1/6 is 1/2, so 1/2 divided by 3 is 1/6, and 1/2 (= 3/6) divided by 1/6 is 3. (Assuming prior understanding that if the product of, say, 2 and 3 is 6, then 6/3 = 2 and 6/2 = 3.)
• Cut several pies. For example, cut two pies into three pieces each, and then color pairs of pieces. How many groups of two pieces make two pies? Congratulations, you have just divided 2 by 2/3.
• Work other examples, dividing whole numbers by fractions, then fractions by other fractions, choosing cases that come out even to start with.
• Now look at examples where one fraction does not go evenly into the other. What do you have to do to make sense of the remainder? Say you have a pizza cut into 8 pieces, and you have hungry pizza eaters who want three slices each. How many can you accommodate? Well, two, with two slices left over. Two slices is 2/3 of three slices, so that comes to 2 2/3 portions.
None of this requires Turtle Art. People have been learning fractions for thousands of years. You can cut pies or cakes or plots of land or the floor of the classroom or a tabletop or pieces of construction paper to do all of this. Oh, yes. How many pieces do the local pizza parlors cut pizzas of various sizes into? What fractions can you make from those pieces? Can you find pictures of pizzas from directly above, so that they appear as circles and you can print them and cut them up? (Yes.) What else? Craters on the moon? The whole moon? Circular swimming pools, fountains, ponds, coins, cups?
It remains an open question whether the children can discover the invert-and-multiply rule for dividing fractions by themselves in a sufficently rich environment, whether they will need broad hints, or whether they will have to be told. It would be interesting to me to hear how they would explain these ideas to each other. I will be interested to hear your results. |
# Hypothesis Testing Assignment Help
Hypothesis testing is a statistical process that is helpful to draw inferences about the population of interested in the given sample.
In the other word, it is a statistical method that helps to evaluate hypothesis about a population by using a sample data set. The main objective of this test is to either accept or reject the null hypothesis (Black, 2009). In this case, hypothesis testing is also useful to determine whether a claim about body fat parameter is accurate or not. Here, boss of Silver’s Gym claims that the average body fat in men attending gym is 20%, while statistical analyst makes different claim. In this situation, the set-up of null and alternative hypothesis can be useful to determine best possible outcome. The procedure of hypothesis testing is as follows: Hypotheses: Null hypotheses (H0): µ = 20 Alternative hypotheses (Ha): µ ? 20 Alpha level (?) = 0.05 Level of Significance: Alpha level (?) = 0.05 Suitable Test: For this problem, z-test will be used to conduct hypothesis testing regarding average body fat of men. The main reason behind the selection of z-test is sample size of body fat data. It is known that if sample size is greater than 30 (n > 30) then z-test is preferable to use (Jackson, 2009). In this case, sample size is 252 that greater than 30, so z-test is better to conduct hypothesis testing. Under this method, the critical value of z for a two-tailed test is +1.96 or -1.96 (At alpha level = 0.05). Testing: The following formula can be useful to determine Z-score. (Osborn, 2006). Where, Xbar = 18.9 (Mean of body fat data) µ = 20 ? = 7.8 (Standard deviation of body fat data) n = 252 Then, the score of z will be – = (-1.1) ÷ (7.8/15.87) = (-1.1) ÷ (0.4915) Z = -2.24 The outcome shows that the calculated value of Z is lower from the critical value of z at alpha level 0.05 (-2.24 > -1.96). Decision Making: On the basis of outcome, we reject the null hypotheses (Since -2.24 < -1.96). It means the average body fat in men who attending gym cannot be equal to 20%. Get 100% original and complete Hypothesis Testing Assignment Help and statistical analysis assignment help with assignmenthelpexperts.com. From our experts you will get complete and original assignment help on time. If you are first time using our assignment help services then you can first check our quality of work then you can transfer the money in our account. |
# Math Lesson: Solving A Basic Problem
Updated on January 19, 2014
## Step 1: Write Down the Problem
The most basic principle of math is the order of operations (more commonly known as PEMDAS). When I was in school they gave us a sentence for remembering the order. “Please excuse my dear aunt Sallie.”
Here is a step by step on how it works.
Remember:
1. Bring your work down once you’ve finished a step or piece of the step
2. In this example, the piece that needs to be worked is underlined
3. Also in this example, what the underlined piece turns into is highlighted in red
## Step 2: Parenthesis
No matter what operations this kind of problem requires, always do the parenthesis first. Once the inside of the parenthesis is mathematically combined, the parenthesis disappear and you will only have the answer. This is shown, highlighted in red, in the picture to the right. The only thing left from the parenthesis is the answer of six.
If there is more than one set of "Parenthesis", go from left to right and do the first set you come to, then the next and the next and so forth.
## Step 3: Exponents
Exponents are always solved immediately after inside the parenthesis are solved.
If there is more than one piece with "Exponents", go from left to right and do the first one you come to, then the next and the next and so forth.
## Step 4: Multiply or Divide
Just remember that this is where it may start to take multiple steps to work out the problem. Some problems have consecutive multiply and divide signs as in the problem to the right (6/3x7), and then you have another set further down the equation (0x4). The six has to be divided by the three ( to equal two) and the zero must be multiplied by the four (to equal zero). This will cause the two and the zero to be brought down. Then you simply repeat the multiply/divide step until all of the multiplication and division signs have been mathematically removed.
If there is more than one "Multiply or Divide" set duo, go from left to right and do the first set you come to, then the next and the next and so forth.
## Step 5: Add or Subtract
The rules for "Add or Subtract" are very similar to "Multiply or Divide". This is also where it will take multiple steps to work out the problem. If the problem has consecutive add and subtract signs as in the problem to the right, you simply go from left to right, adding or subtracting the first set (-5+14) then the next set (36+0). Then you bring those answers down and you simply repeat the "Add or Subtract" step until all of the signs have been mathematically removed.
As with the others, if there is more than one "Add or Subtract" set, go from left to right and do the first set you come to, then the next and the next and so forth.
10
202
23 |
# Core 2 AS level maths formulae OCR
Flashcards by Tess W, updated more than 1 year ago
Created by Tess W almost 5 years ago
148
8
### Description
All the formulae and notes needed for Core 2, including those in and out of the formula booklet given. For AS level Maths for OCR
## Resource summary
Question Answer Summary of Sine, Cosine, and Tangent angles and values 017c748b-f395-4c63-a9e1-a5f1ae254511.png (image/png) The properties of the function cosx Periodic property: cos(x-360) = cosx Even/odd property: cos(-x) = cosx Translation property: cos(x-180) = -cosx Supplementary property: cos(180-x) = -cosx The properties of the function of sinx Periodic property: sin(x-360) = sinx Even/odd property: sin(-x) = -sinx Translation property: sin(x-180) = -sinx Supplementary property: sin(180-x) = sinx The properties of the function tanx Periodic property: tan(x-180) = tanx Odd property: tan(-x) = -tanx Supplementary property: tan(180-x) = -tanx Solving the equation cosx = k Step 1: Find cos-1(k) Step 2: Use the symmetry property cos(-x) = cosx to find another root Step 3: Add or subtract multiples of 360 to find the roots in the required interval Solving the equation sinx = k Step 1: Find sin-1(k) Step 2: Use the symmetry property sin(180-x) = sinx to find another root Step 3: Add or subtract multiples of 360 to find the roots in the required interval Solving the equation tanx = k Step 1: Find tan-1k Step 2: Add or subtract multiples of 180 to find the roots in the required interval Relations between the trigonometric function. Sometimes called Pythagoras' theorem in trigonometry. For all values of x: tanx = sinx/cosx, provided that cosx does not equal 0; cos^2(x) + sin^2(x) = 1
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## Definite Integration
In the previous post, our IB Maths Tutors discussed indefinite integration. Now we shall discuss definite integration
► Definite Integration- We already know that this c here is an integral constant. we are not sure about its value. This c is the reason we call this process indefinite integration. But suppose we do our integration between certain limits like:-
here a lower limit while b higher limit
=g(b)-g(a)
You can clearly see that this function is independent of ‘c’. Means we can be sure about its value so this type of integration is called Definite Integration.
►Definite Integration of a function f(x) is possible in [a,b] if f(x) is continuous in the given interval
►If f(x), the integrand, is not continuous for a given value of x then it doesn’t mean that g(x), the integral, is also discontinuous for that value of x.
► Definite integration of a function between given limits like Algebraic sum of areas bounded by the given curve f(x) and given lines x=a and x=b. That’s why the answer for definite integration problems is a single number.
► If that shows a few things:-
(i) The lines between which area is bounded are co-incident(a=b)
(ii) Area covered above the x-axis=Area covered below the x-axis that means positive part of area and negative part of area is equal
(iii) there must be at least one solution/root to f(x) between x=a and x=b(this is something we study in ROLE’S THEOREM in detail)
► If given function f(x) is not continuous at x=c then we should write
► If given function f(x) > or <0 in any given interval (a,b) then >0 or <0 in given interval (a,b)
► If given function f(x) g(x) in the given interval (a,b) then
in the given interval
► If we integrate the given function f(x) in the given interval (a,b) then
Some More Properties of Definite Integration:-
1. We can interchange the limits on any definite integral, all that we need to do is tack a minus sign onto the integral when we do.
2. . If the upper and lower limits are the equal then integration of function will be zero
3. , where c is any constant/any real number
4. that means definite integration is a distributive process
5. here c is a number lying somewhere between a and b
6. If we don’t change the integrand and the limits, then change in the variable will not affect the answer
7.(a) If f(x) is an odd function i.e. f(x) = – f(-x) then
(b) If f(x) is an even function i.e. f(x) = f(-x) then
8. in particular
9.
10. where f(a) is periodic with period ‘a’.
Walli’s Formula:
Where K = if both m and n are even (m, n N)
= 1 in case the function is odd
Here
Leibnitz’s Rule- If f(x) is a continuous function and u(x) & v(x) are differentiable in the interval [a,b] then,
This rule is used when at least one of the limits is a function.
Here is a very detailed Pdf for definite integration download it solves the questions
Here are the links to articles written on calculus so far, you should read them for better understanding of calculus |
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# 5 Common O Level Maths Trigonometry Mistakes and How to Avoid Them
Trigonometry is a fundamental branch of Mathematics that often challenges O Level students. As parents, it’s essential to understand the common pitfalls students might encounter when working on O Level Maths trigonometry questions. In this article, we’ll delve into five common mistakes that students often make and provide helpful tips to assist your child
in overcoming these challenges.
## 1. Misinterpreting Inverse Trigonometric Functions:
Inverse trigonometric functions, such as tan-1, sin-1, and cos-1, play a vital role in solving trigonometric equations. However, students might misinterpret their use, leading to incorrect solutions. For example, consider the equation tan-1(x) = 0.5. Students might mistakenly treat tan-1(x) as 1/tan(x), leading to a wrong interpretation. To avoid this mistake, remind your child that tan-1(x) is an inverse trigonometric function and should be treated as such. The correct interpretation of the example equation is x = tan(0.5).
Example: Solve for x in the equation tan-1(x)=0.5.
## 2. Misinterpreting Powers of Trigonometric Functions
Another common mistake arises from misinterpreting powers of trigonometric functions. For instance, if your child encounters the equation sin x2 = 0.5, they might incorrectly treat sin x2 as (sin x)2. To steer clear of this error, encourage your child to recognise that sin x2 is not the same as sin2 x. In this scenario, the correct approach is to solve for sin x2 directly, as shown in the example sin x2 = 0.5.
Example: Solve for x in the equation sin x2=0.5.
Quadrant considerations are crucial in trigonometry, especially when dealing with inverse trigonometric functions. Students may overlook this aspect, resulting in incomplete solutions. Imagine the equation sin(x) = 0.5, where 0° ≤ x ≤ 360°. If your child only identifies one solution, they might miss the fact that trigonometric equations often have multiple solutions due to the periodic nature of trigonometric functions. To avoid this error, teach your child to consider the entire range of angles within the given constraints, as demonstrated in the example.
Example: Solve for x in the equation sin x=0.5, where 0°≤x≤360°
## 4. Ignoring the Given Range of Angles
Sometimes, students neglect the range of angles given in the question, leading to solutions that are incomplete or incorrect. For instance, when solving sin(2x) = 0.5 with 0° ≤ x ≤ 360°, students might solve for angles in quadrants 1 and 2, assuming they’ve covered the entire range. However, they might overlook angles in other quadrants, resulting in an incomplete solution. To help your child avoid this error, ensure they understand the range of angles specified and apply the correct calculations accordingly, as illustrated in the example.
Example: Solve for x in the equation sin 2x=0.5, where 0°≤x≤360°
## 5. Forgetting to Check the Validity of Solutions
It’s common for students to forget to check the validity of solutions, especially when dealing with complex equations involving multiple terms. In equations like (2sin x-1)(sin x+2)=0, students might fail to realise that certain values make the equation invalid due to the limited range of the sine rule function. To prevent this oversight, encourage your child to double-check their solutions against the possible range of the involved trigonometric functions. This practice ensures that the solutions are mathematically sound, as highlighted in the example.
Example: Solve for x in the equation (2sin x-1)(sin x+2)=0, where 0°≤x≤360°.
By understanding and addressing these common mistakes in O Level trigonometry, you can provide valuable guidance to your child as they navigate through their Mathematical studies. Keep in mind that trigonometry is a skill that improves with practice and a deep understanding of its principles.
Encourage your child to approach each problem with patience, pay attention to the specific context of the question, and double-check their solutions for accuracy and validity. By helping them recognise and rectify these common errors, you’ll contribute to their success in Trigonometry O Level and lay a strong foundation for their future mathematical endeavours.
Seeking excellence in O Level Math for your child? Choose AGrader Learning Centre! Our High-Quality Mathematics Curriculum uses a Step-by-Step Approach to tackle challenging math problems efficiently. With our unique methods, your child can solve 90% of mind-boggling math problems quickly.
AGrader Secondary Math tutors employ heuristics to develop problem-solving skills, empowering students to conquer the toughest problem sums with ease. Trigonometry, a crucial O Level topic, can be daunting – but not with AGrader. We address the 5 common trigonometry Mistakes, providing solutions for your child’s success.
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45 45 90 special right triangle formula
# 45 45 90 special right triangle formula
45- 45- 90 Triangle.As you see from the figure on the right, two 45-45-90 triangles together make a square, so the area of one of them is half the area of the square. As a formula. where S is the length of either short side. A special right triangle is a right triangle with some regular feature that makes calculations on the triangle easier, or for which simple formulas exist. For example, a right triangle may have angles that form simple relationships, such as 454590. This is called an "angle-based" right triangle. A special right triangle is a right triangle with some regular feature that makes calculations on the triangle easier, or for which simple formulas exist. For example, a right triangle may have angles that form simple relationships, such as 454590. This is called an "angle-based" right triangle. If you forget the formula for a 45-45-90 triangle, use the Pythagorean Theorem.Lesson 8-2 Special Right Triangles 425. Special Needs L1 For Example 3, have students check the answer by cutting out a 60-mm by 60-mm square. Remember the 45- 45- 90 triangle as the root two triangle. Using that tidbit, do one of the following: Look back at the lengths of the sides in triangle BAT and triangle BOY. In triangle BAT, the hypotenuse is the only side that contains a radical. 45-45-90 and 30-60-90 Triangles. Students learn that in a 45-45-90 triangle, the legs are congruent, and the length of the hypotenuse is equal to root 2 times the length of a leg. The unknown sides in a right triangle can be calculated using the Pythagoras formula.Question 2: Find the angles of the right triangle ABC if B is 90o and the length of the sides of the triangle are AB 12 cm, BC 9 cm and CA 15 cm ?Special Right Triangle Worksheet. 45 45 90 Special Right Triangle. Some Sepcific Examples.
Practice Using Special Right Triangles. Problem 1. What is the value of z in the triangle below? Each diagonal divides a square into two congruent 45 - 45 - 90 triangles. 30 - 60 - 90 Special Right TriangleSpecial Right Triangles Formulas. Given below are some of the theorems on properties of triangles. Special right triangles 45 45 90. Special triangles are a way to get exact values for trigonometric equations. Most trig questions youve done up till now have required that you round answers in the end.Solving expressions using 45-45-90 special right triangles. |
Bisectors and isosceles triangles
Prove that in a general triangle ABC if two bisectors are equal, this triangle must be isosceles.
(Proposed by Ali Abouzar/FB/Math)
The statement is known as the ”Steiner-Lehmus theorem”. There is much material on it on the web, including proofs. In Coxeter’s book ”Geometry revisited” there is a proof that is elementary as well as easy.
____________________________________________________
Proof by geometry (my own though certainly many people have constructed it or something similar before. After all the problem is a theorem that has attracted interest for well over a century).
_________________________
First we derive a formula for the length of a bisector (I got the idea for this from my friend Tomas Carnstam, Lund University, Sweden). – Let a, b, c be the sides of a triangle and L the length of the bisector from C. Then two expressions for the triangle area gives
bL sin(C/2) + aL sin(C/2) = ba sin C.
This gives together with the formula sin(2v) = 2 sin v cos v, L = 2ab/(a + b)·cos(C/2).
________________________
Using this formula, if the bisectors from C and B are equally long,
2ab/(a + b)·cos(C/2) = 2ac/(a + c)·cos(B/2). Square this to get
b²(a + c)²·cos²(C/2) = c²(a + b)²·cos²(B/2)
b²(a + c)²·(1 + cos C) = c²(a + b)²·(1 + cos B). Then insert (by the cosine theorem)
cos C = (a² + b² – c²)/(2ab) etc, to get
b²(a + c)²·(1 + (a² + b² – c²)/(2ab)) = c²(a + b)²·(1 + (a² + c² – b²)/(2ac))
b(a + c)²·(2ab + a² + b² – c²) = c(a + b)²·(2ac + a² + c² – b²)
b(a + c)²(a + b – c) = c(a + b)²(a + c – b)
Then it is straightforward to expand and then factorize to get
(b – c)(a³ + b²c + bc² + a²b + a²c + 3abc) = 0.
Here the second factor is obviously positive so b – c = 0, i e b = c.
Annonser
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# Ex 13.9 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev
## Class 9 : Ex 13.9 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev
The document Ex 13.9 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by VP Classes.
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Question 1. A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see the figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
Solution:
Here, Height = 110 cm
Depth = 25 cm
Thickness of the plank = 5 cm
Cost of polishing
∵ Area to be polished = [(110 x 85)] + [(85 x 25) x 2] + [(110 x 25) x 2] + [(5 x 110) x 2] + [(75 x 5) x 4] cm2
= [9350] + [4250] + [5500] + [1100] + [1500] cm2
= 21700 cm2
∴ Cost of polishing at the rate 20 paise per cm2
Cost of painting
∵ Area to be painted = [(75 x 20) x 6] + [(90 x 20) x 2] + [90 x 75] cm
= [9000 + 3600 + 6750] cm2 = 19350 cm2
∴ Cost of painting at the rate of 10 paise per cm2
Total Expenses
Total expenses = (Cost of polishing) + (Cost of painting)
Thus, the total required expense = ₹ 6275.
Question 2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
Solution:
For spherical part
Surface area of a sphere = 4pr2
∴ Surface area of a wooden sphere [∵ Diameter = 21 cm]
= 22 x 3 x 21 cm= 1386 cm
Surface area of the tip of a pillar = π (Radius)2
= 7.071 cm2
∴ Surface area of a wooden sphere to be painted silver = [1386 – 7.071] cm
= 1378.93 cm2
⇒ Surface area of 8 wood spheres to be painted = 8 x 1378.93 cm
= 11031.44 cm
Now, the cost of silver painting at the rate of 25 paise per cm2
For cylindrical part Radius of the base of the cylindrical part (R) = 1.5 cm
Height of the cylindrical part (h) = 7 cm
∴ Curved surface area of the cylindrical part (pillar) = 2πRh
⇒ Total curve surface area of 8 pillars = 8 x 66 cm2 = 528 cm2
∴ Cost of painting black for 8 pillars at the rate of 5 paise per cm2
Total cost of painting Total cost of painting = [Cost of silver painting] + [Cost of black painting]
Question 3. The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let the original diameter = d Case I
⇒ Original surface area of the sphere =
Case II Decreased diameter = d – (25% of d) = d –
⇒ Decreased surface area =
Per cent decrease in surface area Decrease in curved surface area = [Original surface area] – [Decreased surface area]
Now, % decrease in surface area
Thus, the required per cent decrease in curved surface area is 43.75%
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# Illustrative Mathematics Unit 6.2, Lesson 1: Introducing Ratios and Ratio Language
Learn more about ratios and how to describe the relationship between two quantities in words. After trying the questions, click on the buttons to view answers and explanations in text or video.
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Illustrative Math
Grade 6
### Introducing Ratios And Ratio Language
Let’s describe two quantities at the same time.
Illustrative Math Unit 6.2, Lesson 1 (printable worksheets)
#### 1.1 - What Kind and How Many?
1. If you sorted this set by color (and pattern), how many groups would you have?
2. If you sorted this set by area, how many groups would you have?
3. Think of a third way you could sort these figures. What categories would you use? How many groups would you have?
• See Possible Answers
1. There would be 4 groups: white (solid), green (cross-hatches), yellow (stripes) and blue (dots).
2. There would be 4 groups: 2-squares, 3-squares, 4-squares, and 5-squares.
3. One possible third way to sort these figures would be by shape. There are 7 distinct shapes (counting different rotations of a certain shape as 1 shape).
• See Video 1 for Whole Lesson
• See Video 2 for Whole Lesson
#### 1.2 - The Teacher’s Collection
Your teacher will show you a collection of objects. Alternatively, consider the following collection:
(1) Think of a way to sort your teacher’s collection into two or three categories. Record your categories in the top row of the table and the amounts in the second row.
category name
category amount
(2) Write at least two sentences that describe ratios in the collection. Remember, there are many ways to write a ratio as a sentence:
• The ratio of one category to another category is ________ to ________.
• The ratio of one category to another category is ________ : ________.
• There are _______ of one category for every _______ of another category.
(3) Make a visual display of your items that clearly shows one of your statements. Be prepared to share your display with the class.
• Definition Of Ratio
A ratio is an association between two or more quantities. We can use this to compare quantities of objects between categories.
• See Possible Answers
(1)
category name small medium large
category amount 6 6 3
(2) The ratio of small to large clips is 6 : 3.
There are 6 medium clips for every 3 large clips
(1) Use two colors to shade the rectangle so there are 2 square units of one color for every 1 square unit of the other color.
(2) The rectangle you just colored has an area of 24 square units.
Draw a different shape that does not have an area of 24 square units, but that can also be shaded with two colors in a 2:1 ratio. Shade your new shape using two colors.
• See Possible Answers
(1)
There are 16 red squares for every 8 blue squares, which is the same as 2 red squares for every 1 blue square.
(2)
There are 8 red squares for every 4 blue squares, which is the same as 2 red squares for every 1 blue square.
#### Lesson 1 Summary
A ratio is an association between two or more quantities. There are many ways to describe a situation in terms of ratios. For example, look at this collection:
Here are some correct ways to describe the collection:
• The ratio of squares to circles is 6 : 3.
• The ratio of circles to squares is 3 to 6.
Notice that the shapes can be arranged in equal groups, which allow us to describe the shapes using other numbers.
There are 2 squares for every 1 circle.
There is 1 circle for every 2 squares.
#### Glossary Terms
ratio: a ratio is an association between two or more quantities.
For example, the ratio 3:2 could describe a recipe that uses 3 cups of flour for every 2 eggs, or a boat that moves 3 meters every 2 seconds. One way to represent the ratio 3:2 is with a diagram that has 3 blue squares for every 2 green squares.
#### Practice Problems
(1) In a fruit basket there are 9 bananas, 4 apples, and 3 plums.
1. The ratio of bananas to apples is ________ : ________.
2. The ratio of plums to apples is ________ to ________.
3. For every ________ apples, there are ________ plums.
4. For every 3 bananas there is one ________.
• Answers
1. The ratio of bananas to apples is 9 : 4.
2. The ratio of plums to apples is 3 to 4.
3. For every 4 apples, there are 3 plums.
4. For every 3 bananas there is one plum.
(2) Complete the sentences to describe this picture.
1. The ratio of dogs to cats is _______.
2. For every _____ dogs, there are _____ cats.
• Answers
1. The ratio of dogs to cats is 3:4.
2. For every 3 dogs, there are 4 cats.
(3) Write two different sentences that use ratios to describe the number of eyes and legs in this picture.
• Answers
The ratio of eyes to legs is 4:8.
For every 1 eye, there are 2 legs.
(4) Choose an appropriate unit of measurement for each quantity: cm, cm2, or cm3.
1. area of a rectangle
2. volume of a prism
3. side of a square
4. area of a square
5. volume of a cube
• Answers
1. area of a rectangle: cm2 (2-dimensional)
2. volume of a prism: cm3 (3-dimensional)
3. side of a square: cm (1-dimensional)
4. area of a square: cm2
5. volume of a cube: cm3
(5) Find the volume and surface area of each prism.
a. Prism A: 3 cm by 3 cm by 3 cm
b. Prism B: 5 cm by 5 cm by 1 cm
c. Compare the volumes of the prisms and then their surface areas. Does the prism with the greater volume also have the greater surface area?
• Answers
a. Volume = 33 cm3 = 27 cm3
Surface area = 6(32) = 54 cm2
b. Volume = 5 × 5 × 1 = 25 cm3
Surface area = 2(5 × 5) + 4(5) = 70 cm2
c. No. Prism A has a greater volume, but Prism B has a greater surface area.
(6) Which figure is a triangular prism? Select all that apply.
• Answers
A, C, and D are triangular prisms.
Recall that prisms are polyhedra which consist of two congruent bases connected by rectangular faces, and that prisms are named after the shape of their bases. B is a pentagonal prism. E is a rectangular pyramid.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Math Snap
## Aesignment $y=\frac{1}{2} x-1$ Identify the slope of the graphed line: $\square$ Identify the $y$-intercept of the graphed line: $\square$ Identify the slope of the line given by the equation: $\square$ Identify the $y$-intercept of the line given by the equation: $\square$
#### STEP 1
Assumptions 1. The equation of the line is given as $y = \frac{1}{2} x - 1$. 2. We need to identify the slope of the graphed line. 3. We need to identify the $y$-intercept of the graphed line. 4. The equation is already in the slope-intercept form $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept.
#### STEP 2
Identify the slope of the graphed line. In the equation $y = \frac{1}{2} x - 1$, the coefficient of $x$ is the slope. $\text{Slope} = \frac{1}{2}$
#### STEP 3
Identify the $y$-intercept of the graphed line. In the equation $y = \frac{1}{2} x - 1$, the constant term is the $y$-intercept. $\text{y-intercept} = -1$
#### STEP 4
Summarize the identified values. - The slope of the graphed line is $\frac{1}{2}$. - The $y$-intercept of the graphed line is $-1$.
##### SOLUTION
Since the equation given is already in the slope-intercept form $y = mx + b$, the slope and $y$-intercept have already been identified. $\text{Slope of the line given by the equation} = \frac{1}{2}$ $\text{y-intercept of the line given by the equation} = -1$ |
# Divisibility Rules Worksheets | Divisibility Tests from 2 to 10
1. Math >
2. Number Sense >
3. Division >
4. Divisibility Rules
Explore our comprehensive divisibility rules worksheets with exercises designed to help children test divisibility for numbers 2 through 10 by mastering rules like checking evenness for 2, digit sums for 3, and specific endings for 5 and 10. This PDF resource offers essential practice and encourages children to apply rules methodically to determine if numbers can be evenly divided without any remainder. Use our answer keys for quick evaluation. Make a great start with our free worksheets!
Divisibility Rule for 2
Numbers ending in 0, 2, 4, 6, or 8 (even) are divisible by 2; numbers ending in 1, 3, 5, 7, or 9 (odd) are not. Remember this as you test the numbers in these divisibility rules for 2 worksheets.
Divisibility Rule for 3
A number is divisible by 3 if the sum of its digits is a multiple of 3. For instance, 924: 9 + 2 + 4 = 15, and 1 + 5 = 6, so 924 is divisible by 3. Packed with 18 numbers to test, each worksheet on divisibility rules for 3 is a must-have.
Divisibility Rule for 4
Kids in grade 4 and grade 5 observe the digits in the ten’s and one’s places. If the number formed by the last two digits can be divided by 4, the entire number is divisible by 4.
Divisibility Rule for 5
Is 250 divisible by 5? Here's how to find out easily: if the last digit is 5 or 0, the number is divisible by 5. Use this handy trick to complete these divisibility tests for 5 worksheets.
Divisibility Rule for 6
Watch kids classify numbers in these divisibility rules for 6 worksheets as divisible by 6 or not by ensuring two conditions: it must be even, and the sum of its digits should be a multiple of 3.
Divisibility Rule for 7
Repeatedly remove the last digit, double it, and subtract from the truncated number until a single digit remains. If this digit is 7 or 0, the original number is divisible by 7.
Divisibility Rule for 8
Children in 5th grade and 6th grade distinguish numbers divisible by 8 from those that aren't by checking if the last 3 digits of a number are divisible by 8, then the number is divisible by 8.
Divisibility Rule for 9
The rule for divisibility by 9 states that if the sum of the digits in a number can be divided by 9, the original number is divisible by 9. Use this rule to determine if the numbers are evenly divided by 9 without leaving a remainder.
Divisibility Rule for 10
Let kids observe the last digit if it is 0 then the number is divisible by 10 as they complete these practice sheets on the divisibility rule for 10.
Divisibility Rules for 2, 5, and 10
This compilation serves as an assessment of divisibility rules for 2, 5, and 10. Practice using the divisibility rules to check if the given number is divisible by 2, 5, or 10, as instructed.
Divisibility Rules for 3, 6, and 9
Brush up your knowledge of divisibility rules for 3, 6, and 9 with these printables, and use the appropriate divisibility rules to complete the exercises.
Divisibility Rules for 4 and 8
Charge up practice with these handouts where the learners circle the numbers divisible by 4 in the first part and then the numbers divisible by 8 in the next part.
Divisibility Rules for 2, 3, 4, 5, 6, 8, 9, and 10
Take your skills up a notch and check if the number is divisible by the specified divisor. This mixed bag of exercise is a must-try!
Divisibility Rules for 2 to 10
These divisibility test worksheets serve as a tool to test knowledge. Check if the number on the left is divisible by any number from 2 to 10 and check the appropriate box.
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Printable Worksheets
• 20,000+ Worksheets Across All Subjects |
# Tagged: Stanford.LA
## Top 10 Popular Math Problems in 2016-2017
It’s been a year since I started this math blog!!
More than 500 problems were posted during a year (July 19th 2016-July 19th 2017).
I made a list of the 10 math problems on this blog that have the most views.
Can you solve all of them?
The level of difficulty among the top 10 problems.
【★★★】 Difficult (Final Exam Level)
【★★☆】 Standard(Midterm Exam Level)
【★☆☆】 Easy (Homework Level)
## Problem 438
Determine whether each of the following statements is True or False.
(a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$.
(b) If the characteristic polynomial of an $n \times n$ matrix $A$ is
$p(\lambda)=(\lambda-1)^n+2,$ then $A$ is invertible.
(c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible.
(d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$.
(e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.
(Stanford University, Linear Algebra Exam Problem)
## Problem 216
Let
$A=\begin{bmatrix} 1 & 3 & 3 \\ -3 &-5 &-3 \\ 3 & 3 & 1 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 4 & 3 \\ -4 &-6 &-3 \\ 3 & 3 & 1 \end{bmatrix}.$ For this problem, you may use the fact that both matrices have the same characteristic polynomial:
$p_A(\lambda)=p_B(\lambda)=-(\lambda-1)(\lambda+2)^2.$
(a) Find all eigenvectors of $A$.
(b) Find all eigenvectors of $B$.
(c) Which matrix $A$ or $B$ is diagonalizable?
(d) Diagonalize the matrix stated in (c), i.e., find an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$ or $B=PDP^{-1}$.
(Stanford University Linear Algebra Final Exam Problem)
## Problem 181
Suppose that $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is an eigenvector of a matrix $A$ corresponding to the eigenvalue $3$ and that $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ is an eigenvector of $A$ corresponding to the eigenvalue $-2$.
Compute $A^2\begin{bmatrix} 4 \\ 3 \end{bmatrix}$.
(Stanford University Linear Algebra Exam Problem)
## Problem 70
Suppose that $A$ is an $n \times n$ matrix with eigenvalue $\lambda$ and corresponding eigenvector $\mathbf{v}$.
(a) If $A$ is invertible, is $\mathbf{v}$ an eigenvector of $A^{-1}$? If so, what is the corresponding eigenvalue? If not, explain why not.
(b) Is $3\mathbf{v}$ an eigenvector of $A$? If so, what is the corresponding eigenvalue? If not, explain why not.
(Stanford University, Linear Algebra Exam)
## Problem 39
Suppose that $A$ is a diagonalizable matrix with characteristic polynomial
$f_A(\lambda)=\lambda^2(\lambda-3)(\lambda+2)^3(\lambda-4)^3.$
(a) Find the size of the matrix $A$.
(b) Find the dimension of $E_4$, the eigenspace corresponding to the eigenvalue $\lambda=4$.
(c) Find the dimension of the kernel(nullspace) of $A$.
(Stanford University Linear Algebra Exam)
## Problem 38
Let $A$ be an $m \times n$ real matrix.
Then the kernel of $A$ is defined as $\ker(A)=\{ x\in \R^n \mid Ax=0 \}$.
The kernel is also called the null space of $A$.
Suppose that $A$ is an $m \times n$ real matrix such that $\ker(A)=0$. Prove that $A^{\trans}A$ is invertible.
(Stanford University Linear Algebra Exam)
## Problem 37
Suppose that $A$ is a diagonalizable $n\times n$ matrix and has only $1$ and $-1$ as eigenvalues.
Show that $A^2=I_n$, where $I_n$ is the $n\times n$ identity matrix.
(Stanford University Linear Algebra Exam)
See below for a generalized problem.
## Problem 32
Let
$A=\begin{bmatrix} 2 & 0 & 10 \\ 0 &7+x &-3 \\ 0 & 4 & x \end{bmatrix}.$ Find all values of $x$ such that $A$ is invertible.
(Stanford University Linear Algebra Exam) |
# Ratio Practice: Real Life Math for 6th Grade
Looking for ways to help your 6th graders see and make use of ratios in real life? This set of camping ratio practice problems is a great way to explore and apply ratios!
I recently spent an extended season living in and traveling in an RV with my family. It was a unique and fun way to explore and see parts of the country we’ve never seen! It was also a fun way to meet other campers and explore the great outdoors. Although we saw and did lots of fun things on our journey, the one request my kids made over and over and over (especially when they made friends with other kids at the campsites!) was to make s’mores. Apparently that never gets old! And so we have the inspiration for this set of camping-themed ratio practice problems. I hope you enjoy it!
## Using Ratios in Real Life:
Often, kids are introduced to ratios and are given simplistic problems that don’t actually challenge them or show them the usefulness of ratios.
Obviously, they need to understand how to write and read ratios, and simplify ratios, but beyond that? Can you present problems using ratios to push your kids further?
I think so!
This set of camping-themed problems present a situation and then challenge students to answer related questions using ratios and proportional reasoning.
Even if you haven’t explicitly taught your students how to solve these problems, I would encourage you to assign them anyway and see what kinds of strategies your kids use!
Giving students a context will help them to begin to make sense of ratios and proportional reasoning, laying an important foundation for Algebra.
It will also help them see how changing the ratio can change the problem.
## Ratio Practice Problems Included:
This download includes 3 pages of ratio problems that build on each other. They are all related (the Frazier family is planning their annual camping trip and needs your help!), so you’ll want to start with the first page and let students solve them in order.
Problems include things like:
• If the capacity of their tents has a ratio of 3 kids: 2 adults and there are 9 kids, how many adults can go camping?
• They want the s’mores ratio to be 2 marshmallows : 3 pieces of chocolate : 1 graham cracker. If they have 24 marshmallows, how many s’mores can they make?
• Mrs. Frazier brought 96 marshmallows, 160 pieces of chocolate, and 64 graham crackers. What is the ratio for each s’more?
And more!
## Discussing the Problems as a Group:
In order for this set of ratio practice to have the most impact, it is important that you allow students to discuss their strategies for thinking about and solving these problems.
Did they draw a picture? Did they multiply and divide? How? Why?
Allowing all students to see different ways to approach and think about these problems will help them to better understand and expose them to mathematical reasoning they may not have considered on their own.
One idea is to put students in groups of 2 or 3 to begin. Allow them to discuss and solve the problems together first, then open it up to the class.
Let each group share their solutions and their reasoning. It’s important that they share how they arrived at their solution, not just the final answer.
This will push students to learn how to articulate their strategies (this takes some work!) but will also help other students who thought about it differently (or did not get the right answer).
Remember, there is always more than one way to solve a math problem, so it’s good for students to see that!
## Seeing Ratios in Real Life
Although this may be challenging for your students at first (especially if they are not used to applying ratios to real life problems), it will be worth it to help them build up their number sense, and flexibility in problem solving.
Seeing how proportions can be used and applied is a math skill that will carry over into many other math concepts, such as scale models, indirect measurement, percent problems, and linear equations in Algebra.
I hope this fun challenge gives you an easy way to weave in meaningful math problem solving. You might even want to make s’mores together as a class when they finish it up!
Ready to grab this freebie? Click the link below to go to my shop and grab it! |
# Area of Ellipse a-axis 23 in by b-axis 61 cm Calculator
Choose
Ellipse Axis A
Ellipse Axis B
Area of Ellipse a-axis 23 in by b-axis 61 cm is 1735.2970422 in2 or 3563.62 cm2.
Area of Ellipse a-axis 23 in by b-axis 61 cm in other units
Value unit
0.0440765 km²
0.027388 mi²
44.0765449
144.6080869 ft²
1735.2970422 in²
48.2026956 yd²
4407.6544872 cm²
44076.5448719 mm²
## Area of Ellipse a-axis 23 in by b-axis 61 cm
Steps:
Given that Ellipse a-axis(x) = 23 in and b-axis(y) = 61 cm
Area = π (a*b)
Unit Conversion of Width 61 cm = 24.015748 in
To convert Centimeter to Inches
we know that, 1 Centimeter = 0.393705 inches or
1 Centimeter = 1/2.54 inches
To convert Centimeters to inches, divide the centimeter value by 2.54.
Result in Centimeters: 61 × cm/2.54 × in/cm
Cancel The Comman factor of cm
Result in Inches: 61/2.54 in
Divide the 61 by 2.54
Result in Inches: 24.015748 inches
∴ 61 Centimeters = 24.015748 inches
Put the values of a, b and π(Pi) in above Area Formula
A = 3.14 (23 in * 24.015748 in)
Simplify the above equation
Area = 1735.2970422 in²
∴Area of an Ellipse a-axis 23 in by b-axis 61 cm is 1735.2970422 cm²
or
Unit Conversion of Length 23 in = 58.42 cm
To convert Inches to Centimeters
we know that, 1 Inche = 2.54 Centimeters
To convert Inches to Centimeters, multiply the inche value by 2.54
Result in Centimeters: 23 in × 2.54 × cm/in
Cancel The Comman factor of in
Result in Centimeters: (23 x 2.54 cm)
Multiply 23 into 2.54
Result in Centimeters: 58.42 Centimeters
∴ 23 Inches = 58.42 Centimeters
A = 3.14 (58.42 cm * 61 cm)
Simplify the above equation
Area = 3563.62 cm²
∴Area of an Ellipse a-axis 23 in by b-axis 61 cm is 3563.62 cm²
### Area of a Ellipse Calculations
Here are examples of Area of Ellipse calculations. |
# PROVING BASIC PROPORTIONALITY THEOREM IN GIVEN TRIANGLE
Basic Proportionality Theorem :
If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Converse of Basic Proportionality Theorem Examples :
If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Given :
In triangle ABC and a line intersecting AB in D and AC in E, such that AD/DB = AE/EC.
Example 1 :
ABCD is a quadrilateral with AB parallel to CD. A line drawn parallel to AB meets AD at P and BC at Q. Prove that AP/PD = BQ/QC.
Solution :
Join BD by intersecting the line PQ at the point Q.
In triangle DAB, PE and AB are parallel, by using “Thales theorem”
AP/PD = BE/ED ----(1)
In triangle BCD EQ and DC are parallel, by using “Thales theorem”
BE/ED = BQ/QC ----(2)
(1) = (2)
AP/PD = BQ/QC
Example 2 :
In t he figure, PC and QK are parallel BC and HK are parallel, if AQ = 6 cm, QH = 4 cm, HP = 5 cm, KC = 18 cm, then find AK and PB.
Solution :
In triangle APC, the sides PC and QK are parallel.
By using “Thales theorem” we get
AQ/QP = AK/KC
QP = QH + HP
= 4 + 5
= 9 cm
6/9 = AK/18
AK = (6 x 18)/9
AK = 12 cm
In triangle ABC, the sides BC and HK are parallel,
By using “Thales theorem” we get
AH/HB = AK/KC
AH = AQ + QH
= 6 + 4
= 10
10/HB = 12/18
(10 x 18)/12 = HB
HB = 15 cm
Now we need to find the length of PB,
PB = HB – HP
= 15 – 5
= 10 cm
Example 3 :
In the figure DE is parallel to AQ and DF is parallel to AR prove that EF is parallel to QR.
Solution :
In triangle PQA, the sides DE is parallel to the side AQ.
By using “Thales theorem” we get
PE/EQ = PD/DA ----(1)
In triangle PAR, the sides DF is parallel to the side AR
By using “Thales theorem” we get
PD/DA = PF/FR ----(2)
(1) = (2)
PE/EQ = PF/FR
From this we can decide EF is parallel to QR in the given triangle PQR.
Example 4 :
In the figure the sides DE and AB are parallel and DF and AC are parallel. Prove that EF and BC are parallel.
Solution :
In triangle APB, the sides DE and AB are parallel.
PD/DA = PE/EB ----(1)
In triangle PAC, the sides DF and AC are parallel.
PD/DA = PF/FC ----(2)
(1) = (2)
PE/EB = PF/FC
Hence the sides EF and BC are parallel.
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