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By BYJU'S Exam Prep
Updated on: September 25th, 2023
Ratio and proportion are crucial topics in the quantitative aptitude section of competitive exams, particularly the SSC exams. Typically, 4-5 questions in the SSC exams are based on this topic. Understanding the concepts of ratio and proportion is essential because they are the foundation for grasping other mathematical concepts. In simple terms, a fraction written as a:b represents a ratio, whereas a proportion indicates that two ratios are equal. Both a and b can be any two integers.
However, some candidates often get perplexed by the concepts of ratio and proportion. If you are one of them, fret not. As this blog aims to provide candidates with a clear understanding of ratio and proportion through comprehensive explanations, formulas, ratio and proportion questions for SSC Exams and much more.
## Ratio and Proportion
Fractions are a way of expressing a portion of a whole. They are used to represent ratios and proportions in mathematics. A ratio compares two quantities by dividing one by the other, and it can be written in the form of a fraction a:b. A proportion is a statement that two ratios are equal, which means that they can be expressed as a fraction with the same value.
## Definition of Ratio
A ratio is a way of comparing two quantum measurements by employing a division of the same.
A ratio is a comparison of two quantities obtained by dividing one by the other. If a and b are two values of the same type and units, and b is not equal to zero, the quotient a/b is known as the ratio between a and b. Ratios are expressed using the colon (:) sign. This means that the a/b ratio has no unit and can be written as a: b.
## Definition of Proportion
It is an equation applied to two ratios to know whether they are equal or not.
Proportion refers to the equality of two ratios. Two proportionate ratios are always equal. The sign (::) represents proportions and helps us solve for unknown quantities. In other words, a percentage is an equation or statement that demonstrates that two ratios or fractions are equivalent. Four non-zero numbers, a, b, c, and d, are said to be in proportion if a: b = c: d. Consider the two following ratios: 3: 5 and 15: 25. In this scenario, 3:5 is represented as 3:5 = 3/5 = 0.6, and 15:25 is written as 15/25 = 3/5 = 0.6. We can argue that both ratios are proportionate since they are equal.
## Ratio and Proportion Formula
The formula for Ratio and Proportion is provided below:
Ratio
• a: b ⇒ a/b
Where a and b are any two quantities and “a” is called the antecedent (first term), and “b” is called the consequent (second term)
Proportion
• a/b = c/d or a : b :: c : d
In proportion, the two ratios are a:b & c:d. The two terms ‘b’ and ‘c’ are called ‘means or mean term,’ whereas the terms ‘a’ and ‘d’ are known as ‘extremes or extreme terms.’
## Types of Proportion
There are different types of Proportions that are listed and described below:
Direct Proportion: The term direct proportion refers to the direct link between two quantities. When there is a rise in one quantity then the other one rises as well and vice versa. As a result, a direct proportion is written as y ∝ x. For example, increasing the speed of an automobile causes it to travel more distance in a given amount of time.
Inverse Proportion: Inverse proportion explains the relationship between two quantities in which one increases while the other falls and vice versa. As a result, an inverse proportion is expressed as y ∝ 1/x. For example, when a vehicle’s speed increases, it will travel a constant distance in less time.
## Difference between Ratio and Proportion
To understand the concept of Ratio and Proportion a little better, candidates can go through the difference between these two provided below:
Proportion Ratio The proportion is used to express the relation between two ratios The ratio is used to compare the size of two things with the same unit It is expressed using the double colon (::) or equal to the symbol (=) It is expressed using a colon (:), slash (/) It is an equation It is an expression The keyword to identify proportion in a problem is “out of” The keyword to identify the ratio in a problem is “to every”
## Ratio and Proportion Questions with Solutions PDF
We have provided the most straightforward, easy-to-understand, and effective tips and tactics for solving ratio and proportion questions. Below we are going to provide you all with the Ratio and Proportion Questions with solutions PDF in order to make your SSC exam preparation more fruitful.
## Importance of SSC Ratio and Proportion Questions
SSC Ratio and Proportion Questions are of great relevance in the SSC Exams 2023 as:
• Competitive exams such as the SSC and all of its versions assess candidates’ numeric aptitude because demonstrating it allows exam makers to evaluate the taker’s fit for the profile. As a result, it is critical for students to be well-prepared in areas such as ratio and proportion.
• Ratio and proportion, unitary technique, fraction and percentage are fundamentals that serve as the foundation for more complex commercial mathematics topics covered in the SSC exam.
• Because ratio and proportion as a topic is strongly related to %, being competent in addressing its problems also helps applicants with percentage questions. This works in students’ favour because percentages are an important topic covered in the quantitative component of the SSC exam.
• Finally, studying ratio and proportion assists with preparation for other competitive government exams because this topic is covered in all of them.
## Ratio and Proportion Questions for SSC CGL
The topics of ratio and proportion hold great significance in the quantitative aptitude section of competitive exams like SSC CGL. Based on the SSC CGL Previous Year Papers, it has been observed that 4-5 questions are generally based on this topic. Therefore, it is crucial to prepare for these questions thoroughly. To help candidates prepare better, we have provided some important ratio and proportion questions that are frequently asked in SSC CGL.
1. Three numbers are in the ratio 3 : 5 : 11. If the sum of the first and the third is greater than the second by 333. The sum of three numbers is :
A. 741
B. 703
C. 1197
D. None of these
2. Rs 13450 are to be distributed between B and A such that B gets Rs 3400 less than A. The ratio of the amount received by A to that received by B is
A. 296 : 117
B. 7 : 3
C. 337 : 201
D. 199 : 105
3. The fourth proportional to 0.15, 0.27 and 16 is
A. 22.6
B. 33.7
C. 25.5
D. 28.8
4. Which of the following is the smallest fraction
A. 19/27
B. 7/13
C. 11/15
D. 12/23
5. A company employs experts, amateurs and novices in the proportion 2 : 7 : 11 and the wages of an expert, an amateur and a novice are in the ratio 7 : 4 : 2. When 33 novices are employed, the total monthly remuneration of all amounts to ₹ 1881600. Find the monthly remuneration of a person in each category of employees.
A. ₹ 63000, ₹ 36000, ₹ 18000
B. ₹ 68600, ₹ 39200, ₹ 19600
C. ₹ 66500, ₹ 38000, ₹ 19000
D. ₹ 67900, ₹ 38800, ₹ 19400
6. A person divided a certain sum between his three sons in the ratio 1 : 3 : 5. Had he divided the sum in the ratio the son, who got the least share earlier, would have got ₹1456 more. The sum (in ₹) was:
A. 2691
B. 2961
C. 2916
D. 2196
7. A man plays 100 games with the cards. He gets Rs. 5 if he wins and pays Rs. 2 if he loses. If he wins Rs. 3 on the whole in how many games did he win?
A. 25
B. 28
C. 29
D. 32
8. The marks of three students A, B and C are in the ratio 10:12:15. If the maximum marks of the paper are 90, then the maximum marks of B can be in the range:
A. 70 – 80
B. 80 – 90
C. 20 – 30
D. 40 – 50
9. 4 years back, the age of Garvit was three times that of Yuvansh, but one year back the age of Garvit was two times that of Yuvansh, what is the age difference between their present ages?
A. 7 years
B. 13 years
C. 11 years
D. 6 years
10. A purse has Rs. 80 in the form of Rs. 2, Rs. 1, 50 Paise and 10 Paise coins in the ratio of 2:3:5:5, find the total number of coins.
A. 105 coins
B. 120 coins
C. 135 coins
D. 150 coins
## Important Ratio and Proportion Questions for SSC CHSL
Ratio and Proportion questions are an important component of the SSC CHSL exam as well. There are approximately 3-4 questions asked in the tier 1 examination and now that Maths is an integral part of the SSC CHSL tier 2 exam, this chapter has an additional weightage. Important Ratio and Proportion Questions for SSC CHSL are given below:
1.If (x + y):(x – y) = 11:1, find value of (5x + 3y)/(x – 2y).
A. 45/4
B. 4/45
C. -45/4
D. -4/45
2.A bag has Rs 34.2 in the form of 1-rupee, 50-paise and 10-paise coins in the ratio of 3:5:2. Find the number of 50-paise coins.
A. 30
B. 12
C. 18
D. 6
3.P started a business by investing Rs.35000 and Q joined him after one year with an amount of Rs.21000. After two years from the starting of the business, they earned a profit of Rs.26000. What will be the P’s share (in Rs.) in the profit?
A. 15000
B. 18000
C. 20000
D. 16000
4.Mohit and Anshu invested some money in a business in the ratio 8: 15 respectively. After 9 months Mohit withdraws his capital. If they receive profits in the ratio 4 : 5, then for how many months did Anshu invest in the business?
A. 4
B. 5
C. 8
D. 6
5.The ratio of two numbers is 2 : 3. If the sum of both the numbers is 75, then what is the larger number among both the numbers?
A. 45
B. 30
C. 50
D. 48
6.If 4P = 7Q = 21R, then what is P : Q : R?
A. 5 : 3 : 4
B. 21 : 12 : 4
C. 3 : 5 : 15
D. 15 : 5 : 3
7.The ratio of two numbers is 7 : 11. If both numbers are increased by 10, the ratio becomes 9 : 13. What is the sum of the two numbers?
A. 90
B. 32
C. 48
D. 160
8.If 5A = 3B, then what is the value of (A + B)/B?
A. 8/3
B. 8/5
C. 5/8
D. 5/3
9.The ratio of two numbers is 3:5. If both numbers are increased by 8, the ratio becomes 13:19. What is the sum of the two numbers?
A. 32
B. 48
C. 40
D. 72
10.The ratio of the speed of A, B and C is 2 : 3 : 6 respectively. What is the ratio of the time taken by A, B and C respectively to cover the same distance?
A. 2 : 3 : 6
B. 6 : 3 : 2
C. 3 : 2 : 1
D. 1 : 2 : 3
## Tips and Tricks to Solve SSC Ratio & Proportion Questions
Many methods and ideas for ratio and proportion questions can be discovered in books and easily taught. These methods and suggestions assist students in solving these problems, which are a component of competitive examinations such as the SSC, in a lot more enhanced manner and achieving speedier outcomes.
• Many students are perplexed when they see a ratio and proportion separated by fractions rather than colons. Fractions, like a colon, are used to determine whether the ratios are equal or proportional.
• Practice popular mathematics strategies like visualization to better understand the fundamentals of ratio and proportion.
• Individual values have the same denominator value for proportions with more than two ratio components.
## Best Books for SSC Ratio & Proportion Questions
As noted in the preceding section, ratio and percentage are significant components of the SSC CGL syllabus and SSC CHSL syllabus. As a result, it is critical to prepare thoroughly for this section. Here are some of the best books on ratio and proportion:
Books Authors/Publishers Quantitative Aptitude for Competitive Exams RS Aggarwal Fast Track Objective Arithmetic Rajesh Verma Quantitative Aptitude for All Competitive Exams Abhijit Guha Objective Arithmetic S. Chand Publishing
In conclusion, candidates must comprehend with SSC Ratio and Proportion Questions to fetch maximum marks from this topic. If you are facing any difficulties, you should consider enrolling in SSC online coaching and attempting SSC mock tests to work on your weak areas.
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# Find the Area, in Square Metres, of the Trapezium Whose Bases and Altitude is as Under: Bases = 8 M and 60 Dm, Altitude = 40 Dm - Mathematics
Course
ConceptArea of Trapezium
#### Question
Find the area, in square metres, of the trapezium whose bases and altitude is as under:
bases = 8 m and 60 dm, altitude = 40 dm
#### Solution
Given:
Bases:
8 m
And, 60 dm $=\frac{60}{10}m = 6 m$
Altitude = 40 dm $=\frac{40}{10}m = 4 m$
Area of trapezium $=\frac{1}{2}\times(\text{ Sum of the bases })\times(\text{ Altitude })$
$= \frac{1}{2} \times (8 + 6) m \times (4) m$
$= 28 \times m\times m$
${=28 m}^2$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for Mathematics for Class 8 by R D Sharma (2019-2020 Session) (2017 to Current)
Chapter 20: Mensuration - I (Area of a Trapezium and a Polygon)
Ex. 20.2 | Q: 1.3 | Page no. 22
#### Video TutorialsVIEW ALL [1]
Solution Find the Area, in Square Metres, of the Trapezium Whose Bases and Altitude is as Under: Bases = 8 M and 60 Dm, Altitude = 40 Dm Concept: Area of Trapezium.
S
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# Indeterminate form explained
In calculus and other branches of mathematical analysis, when the limit of the sum, difference, product, quotient or power of two functions is taken, it may often be possible to simply add, subtract, multiply, divide or exponentiate the corresponding limits of these two functions respectively. However, there are occasions where it is unclear what the sum, difference, product, quotient, or power of these two limits ought to be. For example, it is unclear what the following expressions ought to evaluate to:
00,~
infty infty
,~0 x infty,~infty-infty,~00,~1infty,andinfty0.
These seven expressions are known as indeterminate forms. More specifically, such expressions are obtained by naively applying the algebraic limit theorem to evaluate the limit of the corresponding arithmetic operation of two functions, yet there are examples of pairs of functions that after being operated on converge to 0, converge to another finite value, diverge to infinity or just diverge. This inability to decide what the limit ought to be explains why these forms are regarded as indeterminate. A limit confirmed to be infinity is not indeterminate since it has been determined to have a specific value (infinity).[1] The term was originally introduced by Cauchy's student Moigno in the middle of the 19th century.
The most common example of an indeterminate form is the quotient of two functions each of which converges to zero. This indeterminate form is denoted by
0/0
. For example, as
x
approaches
0~
, the ratios
x/x3
,
x/x
, and
x2/x
go to
infty
,
1
, and
0~
respectively. In each case, if the limits of the numerator and denominator are substituted, the resulting expression is
0/0
, which is indeterminate. In this sense,
0/0
can take on the values
0~
,
1
, or
infty
, by appropriate choices of functions to put in the numerator and denominator. A pair of functions for which the limit is any particular given value may in fact be found. Even more surprising, perhaps, the quotient of the two functions may in fact diverge, and not merely diverge to infinity. For example,
x\sin(1/x)/x
.
f(x)
and
g(x)
converge to
0~
as
x
approaches some limit point
c
is insufficient to determinate the limit
An expression that arises by ways other than applying the algebraic limit theorem may have the same form of an indeterminate form. However it is not appropriate to call an expression "indeterminate form" if the expression is made outside the context of determining limits.For example,
0/0
which arises from substituting
0~
for
x
in the equation
f(x)=|x|/(|x-1|-1)
is not an indeterminate form since this expression is not made in the determination of a limit (it is in fact undefined as division by zero).Another example is the expression
00
. Whether this expression is left undefined, or is defined to equal
1
, depends on the field of application and may vary between authors. For more, see the article Zero to the power of zero. Note that
0infty
and other expressions involving infinity are not indeterminate forms.
## Some examples and non-examples
### Indeterminate form 0/0
The indeterminate form
0/0
is particularly common in calculus, because it often arises in the evaluation of derivatives using their definition in terms of limit.
As mentioned above,whileThis is enough to show that
0/0
is an indeterminate form. Other examples with this indeterminate form includeandDirect substitution of the number that
x
approaches into any of these expressions shows that these are examples correspond to the indeterminate form
0/0
, but these limits can assume many different values. Any desired value
a
can be obtained for this indeterminate form as follows:The value
infty
can also be obtained (in the sense of divergence to infinity):
### Indeterminate form 00
See main article: Zero to the power of zero.
The following limits illustrate that the expression
00
is an indeterminate form:
Thus, in general, knowing that
style\limxf(x) = 0
and
style\limxg(x) = 0
is not sufficient to evaluate the limit
If the functions
f
and
g
are analytic at
c
, and
f
is positive for
x
sufficiently close (but not equal) to
c
, then the limit of
f(x)g(x)
will be
1
.[2] Otherwise, use the transformation in the table below to evaluate the limit.
### Expressions that are not indeterminate forms
The expression
1/0
is not commonly regarded as an indeterminate form, because if the limit of
f/g
exists then there is no ambiguity as to its value, as it always diverges. Specifically, if
f
approaches
1
and
g
approaches
0~
, then
f
and
g
may be chosen so that:
f/g
approaches
+infty
f/g
approaches
-infty
1. The limit fails to exist.
In each case the absolute value
|f/g|
approaches
+infty
, and so the quotient
f/g
must diverge, in the sense of the extended real numbers (in the framework of the projectively extended real line, the limit is the unsigned infinity
infty
in all three cases[3]). Similarly, any expression of the form
a/0
with
a\ne0
(including
a=+infty
and
a=-infty
) is not an indeterminate form, since a quotient giving rise to such an expression will always diverge.
The expression
0infty
is not an indeterminate form. The expression
0+infty
obtained from considering
\limxf(x)g(x)
gives the limit
0~
, provided that
f(x)
remains nonnegative as
x
approaches
c
. The expression
0-infty
is similarly equivalent to
1/0
; if
f(x)>0
as
x
approaches
c
, the limit comes out as
+infty
.
To see why, let
L=\limxf(x)g(x),
where
\limx{f(x)}=0,
and
\limx{g(x)}=infty.
By taking the natural logarithm of both sides and using
\limxln{f(x)}=-infty,
we get that
lnL=\limx({g(x)} x ln{f(x)})=infty x {-infty}=-infty,
which means that
L={e}-infty=0.
## Evaluating indeterminate forms
The adjective indeterminate does not imply that the limit does not exist, as many of the examples above show. In many cases, algebraic elimination, L'Hôpital's rule, or other methods can be used to manipulate the expression so that the limit can be evaluated.
### Equivalent infinitesimal
When two variables
\alpha
and
\beta
converge to zero at the same limit point and
style\lim
\beta \alpha
=1
, they are called equivalent infinitesimal (equiv.
\alpha\sim\beta
).
Moreover, if variables
\alpha'
and
\beta'
are such that
\alpha\sim\alpha'
and
\beta\sim\beta'
, then:
Here is a brief proof:
Suppose there are two equivalent infinitesimals
\alpha\sim\alpha'
and
\beta\sim\beta'
.
\lim
\beta \alpha
=\lim
\beta\beta'\alpha' \beta'\alpha'\alpha
=\lim
\beta \beta'
\lim
\alpha' \alpha
\lim
\beta' \alpha'
=\lim
\beta' \alpha'
For the evaluation of the indeterminate form
0/0
, one can make use of the following facts about equivalent infinitesimals (e.g.,
x\sim\sinx
if x becomes closer to zero):[4]
For example:
\begin{align}\limx
1 \left[\left( x3
2+\cosx 3
\right)x-1\right]&=\limx
xln{ 2+\cosx 3
e
-1}{x
3}\&=\limx
1 x2
ln
2+\cosx 3
\&=\limx
1 x2
ln\left(
\cosx-1 3
+1\right)\&=\limx
\cosx-1 3x2
\&=\limx-
x2 6x2
\&=-
1 6
\end{align}
In the 2nd equality,
ey-1\simy
where
y=xln{2+\cosx\over3}
as y become closer to 0 is used, and
y\simln{(1+y)}
where
y={{\cosx-1}\over3}
is used in the 4th equality, and
1-\cosx\sim{x2\over2}
is used in the 5th equality.
### L'Hôpital's rule
See main article: L'Hôpital's rule. L'Hôpital's rule is a general method for evaluating the indeterminate forms
0/0
and
infty/infty
. This rule states that (under appropriate conditions)where
f'
and
g'
are the derivatives of
f
and
g
. (Note that this rule does not apply to expressions
infty/0
,
1/0
, and so on, as these expressions are not indeterminate forms.) These derivatives will allow one to perform algebraic simplification and eventually evaluate the limit.
L'Hôpital's rule can also be applied to other indeterminate forms, using first an appropriate algebraic transformation. For example, to evaluate the form 00:The right-hand side is of the form
infty/infty
, so L'Hôpital's rule applies to it. Note that this equation is valid (as long as the right-hand side is defined) because the natural logarithm (ln) is a continuous function; it is irrelevant how well-behaved
f
and
g
may (or may not) be as long as
f
is asymptotically positive. (the domain of logarithms is the set of all positive real numbers.)
Although L'Hôpital's rule applies to both
0/0
and
infty/infty
, one of these forms may be more useful than the other in a particular case (because of the possibility of algebraic simplification afterwards). One can change between these forms by transforming
f/g
to
(1/g)/(1/f)
.
## List of indeterminate forms
The following table lists the most common indeterminate forms and the transformations for applying l'Hôpital's rule.
Indeterminate formConditionsTransformation to
0/0
Transformation to
infty/infty
\limxf(x)=0,\limxg(x)=0
\limx
f(x) g(x)
=\limx
1/g(x) 1/f(x)
\limxf(x)=infty,\limxg(x)=infty
\limx
f(x) g(x)
=\limx
1/g(x) 1/f(x)
0 ⋅ infty
\limxf(x)=0,\limxg(x)=infty
\limxf(x)g(x)=\limx
f(x) 1/g(x)
\limxf(x)g(x)=\limx
g(x) 1/f(x)
infty-infty
\limxf(x)=infty,\limxg(x)=infty
\limx(f(x)-g(x))=\limx
1/g(x)-1/f(x) 1/(f(x)g(x))
\limx(f(x)-g(x))=ln\limx
ef(x) eg(x)
00
\limxf(x)=0+,\limxg(x)=0
\limxf(x)g(x)=\exp\limx
g(x) 1/lnf(x)
\limxf(x)g(x)=\exp\limx
lnf(x) 1/g(x)
1infty
\limxf(x)=1,\limxg(x)=infty
\limxf(x)g(x)=\exp\limx
lnf(x) 1/g(x)
\limxf(x)g(x)=\exp\limx
g(x) 1/lnf(x)
infty0
\limxf(x)=infty,\limxg(x)=0
\limxf(x)g(x)=\exp\limx
g(x) 1/lnf(x)
\limxf(x)g(x)=\exp\limx
lnf(x) 1/g(x)
## Notes and References
1. Web site: Indeterminate. Weisstein. Eric W.. mathworld.wolfram.com. en. 2019-12-02.
2. 10.2307/2689754 . Louis M. Rotando . Henry Korn . The indeterminate form 00 . Mathematics Magazine . January 1977 . 50 . 1 . 41 - 42. 2689754 .
3. Web site: Undefined vs Indeterminate in Mathematics. www.cut-the-knot.org. 2019-12-02.
4. Web site: Table of equivalent infinitesimals. Vaxa Software.
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# 4.4: Right Triangle Trigonometry
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Trigonometry is the study of triangles. If you know the angles of a triangle and one side length, you can use the properties of similar triangles and proportions to completely solve for the missing sides.
Imagine trying to measure the height of a flag pole. It would be very difficult to measure vertically because it could be several stories tall. Instead walk 10 feet away and notice that the flag pole makes a 65 degree angle with your feet. Using this information, what is the height of the flag pole?
## Trigonometric Functions
The six trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. Opp stands for the side opposite of the angle $$\theta,$$ hyp stands for hypotenuse and adj stands for side adjacent to the angle $$\theta$$.
\begin{aligned} \sin \theta &=\frac{o p p}{h y p} \\ \cos \theta &=\frac{a d j}{h y p} \\ \tan \theta &=\frac{o p p}{a d j} \\ \cot \theta &=\frac{a d j}{o p p} \\ \sec \theta &=\frac{h y p}{a d j} \\ \csc \theta &=\frac{h y p}{o p p} \end{aligned}
The reason why these trigonometric functions exist is because two triangles with the same interior angles will have side lengths that are always proportional. Trigonometric functions are used by identifying two known pieces of information on a triangle and one unknown, setting up and solving for the unknown. Calculators are important because the operations of sin, cos and tan are already programmed in. The other three (cot, sec and csc) are not usually in calculators because there is a reciprocal relationship between them and tan, cos and sec.
$$\sin \theta=\frac{o p p}{h y p}=\frac{1}{\csc \theta}$$
$$\cos \theta=\frac{a d j}{h y p}=\frac{1}{\sec \theta}$$
$$\tan \theta=\frac{o p p}{a d j}=\frac{1}{\cot \theta}$$
Keep in mind that your calculator can be in degree mode or radian mode. Be sure you can toggle back and forth so that you are always in the appropriate units for each problem.
Note that the images throughout this concept are not drawn to scale. If you were given the following triangle and asked to solve for side $$b$$, you would use sine to find $$b$$.
\begin{aligned} \sin \left(\frac{2 \pi}{7}\right) &=\frac{b}{14} \\ b &=14 \cdot \sin \left(\frac{2 \pi}{7}\right) \approx 10.9 \mathrm{in} \end{aligned}
## Examples
### Example 1
Earlier, you were asked about the height of a flagpole that you are 10 feet away from. You notice that the flag pole makes a $$65^{\circ}$$ angle with your feet.
If you are 10 feet from the base of a flagpole and assume that the flagpole makes a $$90^{\circ}$$ angle with the ground, you can use the following triangle to model the situation.
\begin{aligned} \tan 65^{\circ} &=\frac{x}{10} \\ x &=10 \tan 65^{\circ} \approx 21.4 f t \end{aligned}
### Example 2
Solve for angle $$A$$.
This problem can be solved using sin, cos or tan because the opposite, adjacent and hypotenuselengths are all given.
The argument, or input, of a sin function is always an angle. The arcsin, or $$\sin ^{-1} \theta,$$ function on the calculator has an argument that is a ratio of the triangle sides.
\begin{aligned} \sin A &=\frac{5}{13} \\ \sin ^{-1}(\sin A) &=\sin ^{-1}\left(\frac{5}{13}\right) \\ A &=\sin ^{-1}\left(\frac{5}{13}\right) \approx 0.39 \text { radian } \approx 22.6^{\circ} \end{aligned}
### Example 3
Given a right triangle with $$a=12$$ in $$, m \angle B=20^{\circ},$$ and $$m \angle C=90^{\circ}$$, find the length of the hypotenuse.
It is helpful to draw a diagram to represent the data given in a question.
\begin{aligned} \cos 20^{\circ} &=\frac{12}{c} \\ c &=\frac{12}{\cos 20^{\circ}} \approx 12.77 \mathrm{in} \end{aligned}
### Example 4
Given $$\triangle A B C$$ where $$B$$ is a right angle, $$m \angle C=18^{\circ},$$ and $$c=12 .$$ What is $$a$$ ?
Drawing out this triangle, it looks like:
\begin{aligned} \tan 18^{\circ} &=\frac{12}{a} \\ a &=\frac{12}{\tan 18^{\circ}} \approx 36.9 \end{aligned}
### Example 5
Given $$\triangle M N O$$ where $$O$$ is a right angle, $$m=12$$, and $$n=14$$. What is the measure of angle $$M$$ ?
Drawing out the triangle, it looks like:
\begin{aligned} \tan M &=\frac{12}{14} \\ M &=\tan ^{-1}\left(\frac{12}{14}\right) \approx 0.7 \text { radian } \approx 40.6^{\circ} \end{aligned}
Review
For $$1-15$$, information about the sides and/or angles of right triangle $$A B C$$ is given. Completely solve the triangle (find all missing sides and angles) to 1 decimal place.
Problem Number $$A$$ $$B$$ $$C$$ $$a$$ $$b$$ $$c$$ 1. $$90^{\circ}$$ 4 7 2. $$90^{\circ}$$ $$37^{\circ}$$ 18 3. $$90^{\circ}$$ $$15^{\circ}$$ 32 4. $$90^{\circ}$$ 6 11 5. $$90^{\circ}$$ $$12^{\circ}$$ 19 6. $$90^{\circ}$$ 17 10 7. $$90^{\circ}$$ $$10^{\circ}$$ 2 8. $$4^{\circ}$$ $$90^{\circ}$$ 0.3 9. $$\frac{\pi}{2}$$ radian 1 radian 15 10. $$\frac{\pi}{2}$$ radian 12 15 11. $$\frac{\pi}{2}$$ radian 9 14 12. $$\frac{\pi}{4}$$ radian $$\frac{\pi}{4}$$ radian 5 13. $$\frac{\pi}{2}$$ radian 26 13 14. $$\frac{\pi}{2}$$ radian 19 16 15. $$\frac{\pi}{2}$$ radian 10 $$10 \sqrt{2}$$
This page titled 4.4: Right Triangle Trigonometry is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
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# Base System Concepts and Questions for CAT
We humans use decimal system for our mathematical calculations. We know that the computers work on binary system. Some of us who are from engineering or science background are also aware of octal and hexadecimal systems. But the base is not just limited to only these few; any number can be the base. It all depends upon the number of different digits we use in that system.
In decimal system the base is 10, and we use 0 – 9 digits. Similarly in base 2 we use 0 and 1 as digits and in hexadecimal system since the base is 16, we use 0-9 numbers and A – F alphabets (A, B, C, D, E and F represent values from 10 to 15 respectively) to have 16 different digits. Similarly, for new base say 5 we use the 0-4 digits and so on.
Note: If the number say 3241 is in base x, to avoid confusion it is represented by (3241)x.
The problems in competitive exams from base systems are based on two concepts:
1. Conversion of a number from one base to another base
2. Simple mathematical operations like addition, subtraction and multiplication of numbers in base other than 10.
### Process of converting a number from base 10 into another base
When we convert a number from base 10 to any other base say x, we divide the number by x and whatever is the remainder becomes the unit digit of the number in base x and the quotient obtained is further divided and the remainder thus obtained becomes the ten’s digit in base x and so on. The process is repeated till the quotient becomes ZERO.
Example: Convert 435 into base 6.
Solution:
Dividing 435 by 6 the quotient is 72 and the remainder is 3. This remainder 3, is the unit digit of the number in base 6.
Again dividing the quotient 72 by 6, the new quotient of 12 and the remainder is 0. This remainder 0 is the ten’s place digit of the number in base 6. The graphical representation is given below for better understanding
Hence, the number 435 in base 10 = 2003 in base 6.
Or ${\left( {435} \right)_{10}} = {\left( {2003} \right)_6}$
### Process of converting a number from another base into base 10
The process is reverse of what we did in the above example. Instead of successive division, we do successive multiplication i.e.
Converting number ${\left( {abcd} \right)_p}$ into base 10 = $d \times {p^0} + c \times {p^1} + b \times {p^2} + a \times {p^3}$
Example: Convert ${\left( {3564} \right)_7}$ into base 10.
Solution:
${\left( {3564} \right)_7}$
$= {\left( {3 \times {7^3} + 5 \times {7^2} + 6 \times {7^1} + 4 \times {7^0}} \right)_{10}}$
$= {\left( {1230} \right)_{10}}$
### Process of converting a decimal / Fraction into another base (say x)
Step 1: Multiply the decimal with the base x
Step 2: Remove the whole number part after the multiplication. This is the first digit after decimal in base x.
Step 3: Repeat step 2 till all the numbers after decimal get exhausted. The whole number after 2nd multiplication is the ten’s place digit, and after third multiplication, the whole number part is hundredth digits and so on.
Let us understand the working of above steps with the help of an example:
Example: Convert 0.256 from base 10 to base 5
Solution:
Following the above steps we get:
$0.256 \times 5 = 1.28$ , The whole number part is 1.
$0.28 \times 5 = 1.4$, The whole number part is 1.
$0.4 \times 5 = 2.0$, The whole number part is 2.
Hence, ${\left( {0.256} \right)_{10}} = {\left( {0.112} \right)_5}$
### Process of converting a fraction from another base (say x) back to decimal
Say, 0.abc is in base x. The process of converting it into based 10 is as follow:
${\left( {0.abc} \right)_x} = {\left( {a \times \frac{1}{x} + b \times \frac{1}{{{x^2}}} + c \times \frac{1}{{{x^3}}}} \right)_{10}}$
For example, let us convert ${\left( {0.112} \right)_5}$ back to decimal.
${\left( {0.112} \right)_5} = {\left( {1 \times \frac{1}{5} + 1 \times \frac{1}{{{5^2}}} + 2 \times \frac{1}{{{5^3}}}} \right)_{10}}$
$= {\left( {0.2 + 0.04 + 0.016} \right)_{10}}$
$= {\left( {0.256} \right)_{10}}$
### Addition, Subtraction and Multiplication in different bases
The process of addition, subtraction and multiplication in bases other than 10 might sound difficult because we haven’t done much practice of it during our school days. But the process is exactly same as we do in decimal system.
Let us see the process of addition of two numbers in decimal system, and then we will extend the logic to add two numbers in bases other than 10.
Say, we have to add 368 to 437 i.e
We start from right and add the unit digits, i.e. 8 + 7 = 15. Then we write 5 at the unit digit of the final answer and carry over the number 1 to next column of ten’s digits. We repeat the same process till the end.
The entire logic lies on the answer of the question that why did we carry over 1 and kept 5 at the unit place? The logic is, 15 can be written as 10+5 or $1 \times 10 + 5$ (read 1 time 10 plus 5). So we kept 5 at the unit place and carried over 1.
(If the sum was say 47, it would have been written as $4 \times 10 + 7$, so we would have kept 7 at unit place and carried over the number 4).
Hence the logic is, whenever the sum is greater than the base, we divide the sum by the base, keep the remainder and carry over the quotient.
Example: Calculate (456)7 + (234)7
Solution:
The addition of unit digits (6 and 4) is 10, which is more than 7, and would be written as $1 \times 7 + 3$. The quotient is 1 and the remainder of 3. The Remainder is kept at the units place of the answer and the quotient gets carried over to the ten’s place. The process is repeated till the end.
Similarly, the process of subtraction and multiplication is same as explained above. Let us take few worked out examples.
Example: Calculate ${\left( {54} \right)_8} \times {\left( {36} \right)_8}$
Solution:
First we multiple 6 with 4 = 24
24 = 3$\times$8 + 0, so the remainder 0 will be written at the unit place of the answer and 3 will be carried over to next column (ten’s digits column)
Next, we multiply 6 with 5 and add 3 (carry) = 33. Again, 33 = 4$\times$8 + 1 = (41)8. The Process till now looks like:
Similarly, 3 multiplied by 4 = 12 = 1$\times$8 + 4, so 4 is written in units place and 1 is carried over. Finally 3 multiplied by 5 + 1 (carry) = 16 = 2$\times$8 +0 = (20)8. The process till now looks like:
The last step is to add the two rows in base 8 to get the final answer which is equal to (2450)8
Example: Calculate (753)9 – (476)9
Solution:
The unit digits of the given numbers are 3 and 6, as 3 < 6 (we can’t subtract 6 from 3), we will take 1 carry from ten’s digit of the first number i.e 5 and the new ten’s place digit of the first number will become 4. As we are working in base 9, the weight-age of 1 carry is 9. So 3 + 1 (carry) = 3 + 9 = 12.
Now, 12 – 6 = 6 is the unit digit of the answer.
Similarly moving to the next column, the new ten’s digit of first number is 4 and ten’s digit of second number is 7, again we will repeat the same concept of carry as above. The final answer is illustrated below.
Therefore, (753)9 – (476)9 = (266)9
### Converting number from base x to base y, none of x and y is equal to 10
The standard approach is to convert the number in base x to the number in base 10 and then again convert this number in base 10 to the number in base y. i.e.
(number)x ————–> (number)10 —————> (number)y
The process is repetitive and tedious if the number is large. However if the base y is some power of x, then the conversion is easy.
Example: Convert (10011110101)2 to base 4.
Solution:
Note that 22 = 4, also the digits used in base 4 are 0, 1, 2, and 3.
The conversions of these digits in base 4 to base 2 are as follow:
(0)4 = (00)2
(1)4 = (01)2
(2)4 = (10)2
(3)4 = (10)2
To convert (10011110101)2 into base 4, we write the digits in pairs starting from the right hand side. i.e.
(10011110101)2 = (01 00 11 11 01 01)2
Now substituting the above conversion from base 2 to base 4 we get
(10011110101)2 = (1 0 3 3 1 1)4
Similarly, if we have to convert (2322310)4 into base 2, it can easily be done by substitution as we did in the above problem, i.e.
(2322310)4 = (10 11 10 10 11 01 00)2
The logic can be extended to base 8 also by using the following substitution:
Binary Octal 000 0 001 1 010 2 011 3 100 4 101 5 110 6 111 7
Example: convert (100010101010100011110)2 into base 8.
Solution:
Grouping digits of the number in groups of three from the right hand side we get (100 010 101 010 100 011 110)2
Now substituting each group of digits with the corresponding equivalence in base 8 we get:
(100 010 101 010 100 011 110)2 = (4 2 5 2 4 3 6)8
### Important results related to divisibility rules in different bases
1. If the sum of all the digits of a number in base x is divisible by x-1, then the number itself is divisible by x-1.
Example: Is (343626)9 divisible by 8?
Solution:
The sum of digits = 3 + 4 + 3 + 6 + 2 + 6 = 24. As 24 is divisible by 8, the number (343626)9 is also divisible by 8.
1. If a number in base x has even number of digits, and the number is a palindrome i.e. the digits equidistant from each end are the same, then the number is divisible by x+1.
Example: (234432)7 is a six digit palindrome number and by the above result, it is divisible by 7+1 = 8.
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### One Response
1. Mohd Farhan says:
Hi, bu I think the base coversion of (3564) base-7 to base-10 is (1320) base-10 not 1230
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# Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling
## Section1.2Definition
Here is the formal definition of a differential equation.
### Definition1.Differential Equation.
A differential equation (DE) is an equation that involves one or more derivatives of an unknown function. If the function depends on a single variable, the equation is an ordinary differential equation (ODE). Otherwise, it is called a partial differential equation (PDE).
### Note2.Convention: DE means ODE.
Since this text focuses exclusively on ordinary differential equations, any use of DE will imply ODE.
According to the definition, a differential equation must include at least one derivative (e.g., $$f^\prime\text{,}$$ $$\frac{dy}{dx}$$) and an equality sign ("="). This distinction helps us identify the following expressions as differential equations:
\begin{equation*} \frac{dy}{dx} + 1 = y, \qquad f^{\prime\prime} + x^2 + 3x = 19, \qquad e^t = \tan(y^\prime) \end{equation*}
In contrast, the following are not differential equations because they either lack a derivative or an equality sign:
\begin{equation*} \frac{d^2 y}{dx^2} + 2\frac{dy}{dx}, \qquad x^2 + 3x = 19, \qquad \sin y + e^x = 0 \end{equation*}
### Note3.Derivative Notation.
We will use either prime notation or Leibniz notation to denote derivatives. For higher-order derivatives, the following conventions apply:
Derivative Order 1 2 3 4 $$...$$ n Prime $$y^\prime$$ $$y^{\prime\prime}$$ $$y^{\prime\prime\prime}$$ $$y^{(4)}$$ $$...$$ $$y^{(n)}$$ Leibniz $$\ds\frac{dy}{dx}$$ $$\ds\frac{d^2y}{dx^2}$$ $$\ds\frac{d^3y}{dx^3}$$ $$\ds\frac{d^4y}{dx^4}$$ $$...$$ $$\ds\frac{d^ny}{dx^n}$$
Be careful not to confuse $$y^{(7)}$$ with $$y$$ raised to the power of 7!
#### 1.An equation that contains an "=" sign and at least one derivative is called a derivative equation.
An equation that contains an "=" sign and at least one derivative is called a derivative equation
• True
• Incorrect, derivative equation is not a standard term in mathematics.
• False
• Correct!
#### 2.The expression $$z^{(18)}$$ is the same as $$z$$ to the power of 18.
The expression $$z^{(18)}$$ is the same as $$z$$ to the power of 18
• True
• False
• Correct!
#### 3.Identify the differential equation.
Identify the differential equation
• $$\frac{dy}{dx} + 1 = y$$
• Correct! This equation involves a derivative, making it a differential equation.
• $$x^2 + 3x = 19$$
• Incorrect. This equation does not contain any derivatives, so it is not a differential equation.
• $$\sin y + e^x = 0$$
• Incorrect. This equation does not contain any derivatives, so it is not a differential equation.
• $$y^2 + 5 = 0$$
• Incorrect. This equation does not contain any derivatives, so it is not a differential equation.
#### 4.In this textbook, what does the abbreviation "DE" stand for?
In this textbook, what does the abbreviation "DE" stand for?
• An Ordinary Differential Equation
• Correct! In this book, DE is shorthand for Differential Equation.
• An Partial Differential Equation
• Incorrect! Please review the note “Convention: DE means ODE”.
• Dependent Equation
• Incorrect. While DE could theoretically stand for Dependent Equation, in this book it always refers to Differential Equation.
• Derivative Equation
• Incorrect. While DE could theoretically stand for Derivative Equation, is not a standard term in mathematics. In this book it always refers to Differential Equation.
#### 5.What distinguishes an ordinary differential equation (ODE) from a partial differential equation (PDE)?
What distinguishes an ordinary differential equation (ODE) from a partial differential equation (PDE)?
• The number of variables the unknown function depends on.
• Correct! An ODE has derivatives with respect to a single variable, while a PDE involves multiple variables.
• The number of derivatives in the equation.
• Incorrect. Please review the definition of ODEs and PDEs.
• The number of solutions the equation has.
• Incorrect. Please review the definition of ODEs and PDEs.
• The number of hours it takes to solve the equation.
• Incorrect. Please review the definition of ODEs and PDEs.
#### 6.Which of the following is NOT required for an equation to be classified as a differential equation?
Which of the following is NOT required for an equation to be classified as a differential equation?
• An unknown function.
• Incorrect. A differential equation does include an unknown function, which we are solving for.
• An $$x$$-variable.
• Correct! An $$x$$-variable is not a requirement for a differential equation.
• A derivative.
• Incorrect. The presence of at least one derivative is essential to define a differential equation.
• An "=" sign.
• Incorrect. An equality sign is required for an equation to be classified as a differential equation.
#### 7.What notation will this textbook primarily use for derivatives?
What notation will this textbook primarily use for derivatives?
• Both prime and Leibniz notation.
• Correct! The textbook will use both prime and Leibniz notation for derivatives.
• Only prime notation.
• Incorrect. While prime notation will be used, Leibniz notation will also be utilized.
• Only Leibniz notation.
• Incorrect. The book will use both Leibniz and prime notation for derivatives.
• Subscript notation.
• Incorrect. Subscript notation is not used for derivatives in this textbook.
#### 8.Click on all the Differential Equations.
Hint.
There are only 5 Differential Equations in this set.
|
# The Charlotte Mason Way to Teach Multiplication Tables
“There is no royal road to the multiplication table; it must be learnt by heart. This is a fact which faces every teacher of elementary arithmetic, and which each must prepare for in the best way possible.”
The Teaching of Arithmetic to Young Children, p. 10
These words were spoken by Irene Stephens, the head of the mathematics department for Charlotte Mason’s school. The term “royal road” means an easy path—one free from any difficulty. So, how do we prepare the learning of multiplication tables “in the best way possible”? As always, Charlotte Mason had a plan.
Today, we’re going to take a look at Charlotte’s approach to learning the multiplication tables and when to advance to the next table or new concept. You probably won’t be surprised to hear that Charlotte’s approach to multiplication facts and written tables is distinctive. This is no “kill and drill,” rather, her “best way possible” allows a child to discover patterns, investigate how numbers behave and relate with each other, while also advancing certain ideas—such as multiplication as repeated addition or division as the complement of multiplication.
## Begin with the idea of multiplication as repeated addition
You will need a slate or gridded paper for yourself as well as a slate for each child. Your child will also be using his coin bag full of pennies and dimes.
To begin, we will reinforce multiplication as repeated addition through simple, interesting problems using coins and other concrete objects.
You will be writing on the slate until you instruct your child to do so. We begin with very simple questions where knowing the multiplication table isn’t necessary.
1. Say: John had 2¢ and a friend gave him 2¢ more. How many cents had he then? ()
2. Write on the board: 2 + 2 = 4
1. Now ask: How many times do we have 2 cents? (2 times)
3. Write on the board: 2 + 2 + 2 = 6
1. Ask: How many times do we have 2 cents? (3 times)
2. Say: Macy bought 4 candies at 4¢ apiece. How much did they cost all together? (16¢)
3. Write on the board: 4 + 4 + 4 + 4 = 16
1. Ask: How many times do we have 4 cents? (4 times
2. Say: You bought five apples at 7¢ apiece. How much did they cost all together? (35¢)
3. Write on the board: 7 + 7 + 7 + 7 + 7 = 35
1. Ask: How many times do we have 7 cents? (5 times)
2. Say: Jack saved 5¢ every day for 3 days. How much did he save in total? (15¢)
3. Write on the board: 5 + 5 + 5 = 15
1. Ask: How many times do we have 5 cents? (3 times)
## Introduce the times “x” symbol
Begin by stating: We can write this in a shorter way. Let’s take a look at a problem we’ve done before:
1. Say: You bought five gumballs at 2¢ apiece. How much did they cost all together? (10¢ or 1 dime)
2. Write on the board: 2 + 2 + 2 + 2 + 2 = 10
1. Ask: How many times do we have 2 cents? (5 times)
2. Explain: We can write it this way: 2 × 5 = 10
1. Point to or circle the “× 5.” Tell your child that the symbol means “multiplied by 5 “—that is, each of the quantities is to be taken 5 times, so that 2 × 5 means five twos.
2. Explain: The sign “×” is read multiplied by and means the first is to be multiplied by the second; so “2 × 5 = 10” shows that 2 multiplied by 5 = 10.
3. Also explain: The “ ×” sign may also be read times. 2 x 5 shows 2 is taken five times.
4. So, 4 × 3 would mean 4 multiplied by 3 = 12, or 4 times 3 or 4 taken 3 times.
Now have your child write and work a few sums on his slate using multiplication sign, then have him read the entire multiplication sentence.
1. If 4 children had 5¢ each, how much had they all together?
2. Gumballs cost 2¢. How much must you pay for 6 gumballs?
3. 3 girls have 3 ribbons each. How many ribbons have they all together?
Tip: Your child may word these in different ways. As long as the wording and answer are correct, it does not have to specifically match the wording below.
1. 5 × 4 = 20; i.e., 5¢ taken 4 times equals 20¢
2. 2 × 6 = 12; i.e., 2¢ times 6 equals 12¢
3. 3 × 3 = 9; i.e., 3 multiplied by 3 equals 9 ribbons
## Construct a multiplication table using concrete objects
### Step 1: Construct the table
Remember that concrete objects are used to explore ideas, prove facts, and build a comfort level working with a concept. Children progress naturally from working in the concrete to mental images to more abstract work, but each child will progress at her own rate.
To solidify understanding, you and your child will make a simple multiplication table using concrete objects in order to see its rationale before creating written tables.
Charlotte Mason tells us,
The child may learn the multiplication-table and do a subtraction sum without any insight into the rationale of either. He may even become a good arithmetician, applying rules aptly, without seeing the reason of them; but arithmetic becomes an elementary mathematical training only in so far as the reason why of every process is clear to the child.
Home Education, pp. 255, 256
Say: Let’s take our coins and make a multiplication table for 2. We’ll make 10 rows of coins with 2 coins in each row. Let your student read down the column of coins and answer as you guide him:
• 2 and 2 are (4),
• and 2 are (6),
• and 2 are (8),
• and 2 are (10),
• and 2 are (12),
• and 2 are (14),
• and 2 are (16),
• and 2 are (18),
• and 2 are (20).
Ask questions that cover the lines of the table out of order. Such as:
1. How many 2s are in 16? (8) Remark: So it is right to say 2 x 8 = 16.
2. If a paperclip costs 2¢ each, how much would 6 paperclips cost? (12¢)
3. 2 x 4 = (8)
4. 2 x 3 = (6)
5. 2 x 7 = (14)
6. How many 2s in 14? (7)
7. How many 2s in 6? (3)
8. How many 2s in 18? (9)
9. 2 taken 8 times? (16)
10. 2 taken 2 times? (4)
#### Written Table
Now you are going to help your child construct a written multiplication table. The way we do this will act as a tool to understanding the rationale behind the table. Again, you will want your slate and gridded math notebook. These will be used in the horizontal position. We will be taking each number here up to x 10. But you can decide if you’d like your child to learn up to x 12, which comes in handy when working with measures.
Here is a finished multiplication table for 2. A multiplication table in a Charlotte Mason math lesson is formatted like several vertical equations in a row, rather than a grid such as you may have learned in elementary school. Notice that the multiplier is written smaller than the multiplicand, in order to focus attention on the main number of the particular table.
Now, let’s look at the step-by-step process for writing a multiplication table. You will make the first table for your child to see.
Write down the numeral 2 on the (horizontal) slate or graph paper leaving room both above and below:
Write a small 1 above the 2
Write a 2 below:
Next write another 2 beside the first one.
Point to the middle line and ask, How many 2s have we? Child answers: Two 2s. Write a small 2 above the second 2:
Ask: How much is two 2s? Child answers: 4. Write 4 underneath the second 2.
Now add a third 2 to the middle line:
Say something like, Having three 2s, or 2 “three times,” we now have how many all together?
Remember, if, at any time, this is not clear, get out objects such as pennies or beans to show it in the concrete.
Continue in the same way until the whole table is written.
### Step 2: Say the table through and write it
Once the entire multiplication table for 2 is constructed, give your child a few minutes to look at it and try to visualize it in his head (it will remind you a lot of Picture Study, where a child looks at a painting until it hangs in the gallery of his mind); then have him say it through several times: One 2 is 2, two 2s are 4, three 2s are 6, four 2s are 8, etc.
Do you notice how the learning of the multiplication tables somewhat resembles picture study?
Erase a few numbers and ask your child to fill them in. Here’s an example:
Have your child say the table through again. Now erase different numbers and allow him to fill the table in again, then have him repeat the table aloud once more. For example:
Have the child write the table in his gridded math notebook, turned in a horizontal position, for ease in referral and as an aid in committing it to memory. As he writes his table, be sure he pays attention to neatness and keeping the proper place value with one number per grid square.
### Step 3: Practice with Table Work until learned
The next step is to give your child a variety of questions. Give only as many as you are able in the time allotted. Questions may be spread out over a few days. He may refer to his written table until it is learned by heart.
1. Three 2s are?
2. How much is 2 taken 5 times?
3. 2 taken 7 times is?
4. How many 2s in 12?
5. How many 2s in 16?
Note: Questions such as these serve as an introduction to division while also showing division as the complement of multiplication.
1. How much do 7 marbles cost at 2¢ each?
2. Each duck had 2 ducklings. If there are 3 ducks, how many ducklings are there?
3. Find the cost of 2 gummy bears at 8¢ each.
4. How many \$2 pizza slices am I able to buy with \$12?
5. How many 2s make 8?
6. 2 x 4 is?
7. How many 2s in 14?
Each multiplication table will be learned by heart using the same steps, with the work being steady and deliberate. In Charlotte’s schools, tables through 6 were learned the second year of lessons (our grade 2) and tables 7-10 (or 7-12) were taken the following year (our grade 3). Be sure to work with the table in a variety of ways—not merely having your child parrot it in consecutive order.
Here is a simple outline of the steps you will follow. As your student progresses you may determine if any steps may be omitted.
1. Construct the table.
2. Say the table and write it.
3. Practice with Table Work (and review previous tables) until learned.
4. Work with larger multiplication for that table (done with coins).
Keep in mind that lessons should last no longer than 20 minutes at this state, including 5 minutes given to mental math at some point in the lesson. This oral review may include include addition, subtraction, and multiplication covered so far.
At this point, your child may have moved past the need to create a table using concrete objects. If so, she might construct a table with dots or hash marks on graph paper before advancing to making the written table, or she may proceed directly to the construction of the written table. It’s going to be up to you whether to include this concrete activity as future tables are learned. Some students may need to see the concept only once; others may benefit from repeating the activity for each table studied.
Remember, the use or disuse of manipulatives should never cause worry or anxiety, so feel free to put them away or get them back out as needed. If your child needs more work with a particular table, allow plenty of practice using both the written table (including the steps laid out for visualization of the table) and mental work. The variety of questions together with the experimental work of multiplying the current number being worked on with larger numbers—before moving to the next table—can help ensure lessons remain interesting.
A few other ways to solidify multiplication facts are
• Mental Math. If you’ve read our post about mental math, you’ll remember a distinguishing characteristic of Charlotte Mason Math is the 5 minutes of scheduled mental arithmetic each day. Since arithmetic lessons are mainly oral at this stage and require mental work, this activity is somewhat set apart by giving questions in a rapid, lively fashion to fix attention while promoting a child’s promptness and concentration.
• While the child is learning her tables, multiplication can be part of the time set aside for mental math, but you’ll also want to include a variety of different questions to keep it interesting. For example, questions that involve more than one operation, such as: Charlotte picked 8 tulips. She gave 5 away, then picked 3 more, how many has she now?
## Number Sentence Cards
Before a lesson begins, pull out a card or two for additional practice of previously-learned tables. If you unexpectedly need to attend to something else for a few minutes, give your child the selected cards so she may continue working on her own without interruption. She should complete the number sentences, using manipulatives if needed, and write the answers on a dry erase board or in her math notebook. If your lesson wasn’t interrupted, feel free to use them for the “pure number” part of your mental arithmetic.
Each child is unique and no book can determine the amount of time or exact number of equations needed for your child to gain complete fluency with her multiplication facts.
In general, tables through 6 are taken in 2nd year and 7-10 (or 12) are taken in the 3rd year, but we always want to progress at a pace that ensures each step is taken on solid ground.
Your child might move a bit more quickly through certain tables (like the 2s table and the 5s table) but take longer with others. This is perfectly normal. Consistency in daily table work, mental math, and using the Number Sentence cards are keys to progress. When a child no longer needs to use her written tables, she’s ready to advance to the next table.
Your child’s progress should always be measured against herself only. We’ve all heard the quote “Comparison is the thief of joy.” Take care not to compare your child’s progress in learning her math facts to that of her sibling, neighbor, cousin, etc. or measure her against arbitrary standards that don’t take the uniqueness of every child into consideration. These types of comparisons can severely undermine a child’s confidence.
Even children who grasp math concepts quickly and seem to have an innate sense of numbers might take more time to learn their tables.
We don’t want our children to become frustrated, bored, or lose confidence in her abilities.
If you’ve followed the steps laid out here using concrete and written table work, and you are convinced your child has spent a good amount of time on a table, you may move on to the next table or concept while continuing to have daily consistent math lessons that include the important time of mental math review.
## In conclusion
This is a foundational time for your child. Give her plenty of opportunity to investigate, discover patterns, and make connections with each table. Advance to the next table when the child has mastered the table, while providing the important time of mental math and review of tables learned to help the facts become fully internalized and build speed.
Again, there is no “royal road” to learning the multiplication tables. It does take effort. Don’t give up too soon, but, if your student has worked sufficiently with a table but still hasn’t learned it by heart, continue to the next concept while providing daily mental math that includes a good amount of work with multiplication facts.
These methods prepare the child to learn the facts in the “best way possible” and provide the groundwork necessary as she progresses to more advanced arithmetic.
1. #### Melanie
I was wondering if you have a post like this for helping children learn addition and subtraction facts? My almost 8 year old is still really struggling with these facts and it feels like we are stuck. Do you have any tips or strategies to help us?
• #### Richele
Hi Melanie,
You might want to watch the podcast here on Mental Math as well as download the free samples for Book 1 and Book 2 of the Charlotte Mason Elementary Arithmetic series found in the SCM Bookstore on this site for all the details on learning those facts. The Number Sentence Cards are also quite helpful.
Richele
2. #### Julia
I’d like to hear a little about the multiplication with larger numbers sections after the introduction of the tables in books 2 and 3. I think they are there to give the student a head start on the concepts of adding multiplied numbers together, eg showing with coins 15*5= 25 units plus 5 tens = 2 tens and 5 units plus 5 tens = 7 units and 5 tens = 75, but when we first started this it felt like a huge jump. Maybe it’s important not to try to introduce vertical notation for this yet? And maybe only after the table is learnt by heart?
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# Question 1
1. $$T(n) = 2T(\frac{n}{4})+n^{0.51}$$
Using Master’s Theorem $$T(n)=aT(\frac{n}{b})+f(n)$$, we let $$a=2$$, $$b=4$$, and $$d=.51$$.
$$T(n) = \theta(n^{0.51})$$
2. $$T(n) = 16T(\frac{n}{4}) + n!$$
We can use Case 3 that was given ($$f(n) = \Omega(n^{log_{b}a+\epsilon})$$ ) to analyze the recurrence relation.
$$f(n)= n!$$ and $$n^{log_{b}a+\epsilon}$$ for $$\epsilon > 0$$
Case 3 states that if $$f(n)=\Omega(n^{log_{b}a+\epsilon})$$, then $$T(n) = \theta(f(n))$$
Since $$n!=\Omega(n^{2}+\epsilon)$$, then:
$$T(n)=\theta(n!)$$
3. $$T(n)= \sqrt{2}T(\frac{n}{2})+logn$$
For this problem, we can first use Case 1 to analyze the recurrence relation.
$$f(n)=log(n)$$ and $$n^{log_{b}a-\epsilon}=n^{\frac{1}{2}-\epsilon}$$
We know from Case 3, that if $$f(n)=O(log_{b}a-\epsilon), then T(n) = \theta(n^{log_{b}a})$$.
Since we know that a polynomial is greater than a log function, we can apply case 1, therefore:
$$T(n) = \theta(\sqrt{n})$$
4. $$T(n) = T(n-1) + lg(n)$$
$$T(n-1) = T(n-2) + lg(n-1)$$
$$T(n-2) = T(n-3) + lg(n-2)$$
$$T(1) = T(0) + lg(1)$$
$$T(n) + \sum_{i=1}^{n-1} T(i) = \sum_{i=1}^{n-1} T(i)$$ since $$T(0) = T(1) + \sum_{k=1}^{n} lg(k)$$
Then, subtract the summation terms from both sides and:
$$T(n) = \sum_{k=1}^{n} lg(k) = lg(n!) <= \theta(nlgn)$$
5. $$T(n) = 5T(\frac{n}{5})+ \frac{n}{lgn}$$ So if we are to look at the recursion tree this relation creates we can observe that at the jth level there will be $$5^j$$ nodes. Each node will only have a problem of the size $$\frac{n}{5^j}$$ so at each node the time to complete it will be $$\frac{(\frac{n}{5^j})}{log(\frac{n}{5^j})}$$
Know this if we sum over all the $$log(n)$$ levels we get:
$$T(n) = \sum_{j=0}^{log(n-1)} 5^{j}\frac{\frac{n}{5^j}}{log(\frac{n}{5^j})}$$
$$T(n) = \sum_{j=0}^{log(n-1)} \frac{n}{log(n-j)}$$
$$T(n) = n\sum_{j=1}^{log(n-1)} \frac{1}{j}= \theta(nlog(log(n)))$$
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# Manick is $12$ years old and is three times as old as his brother Rahul. How old will Manick be when he is twice as old as Rahul?(A) $14$years (B) $16$years (C) $18$years (D) $20$years
Last updated date: 13th Jul 2024
Total views: 450k
Views today: 12.50k
It is given in the question that the present age of Manick is $12$ years old and his age is thrice the age of his brother Rahul.
$\therefore$Then, the present age of Rahul $= \dfrac{{12}}{3} = 4$years old.
We have to find the age of Manick when he will be twice the age of Rahul. So, after $y$more years his age will be twice that of Rahul. Then we’ll get:
Manick’s age after $y$years$= 12 + y$,
and Rahul’s age after $y$years $= 4 + y$
$\Rightarrow 12 + y = 2 \times \left( {4 + y} \right), \\ \Rightarrow 12 + y = 8 + 2y, \\ \Rightarrow y = 4 \\$
Thus, after $4$years Manick will be twice as old as Rahul. Consequently Manick’s age after $4$ years will be $12 + 4 = 16$ years old. (B) option is correct.
Note: We can also solve the problem by basic aptitude method without forming any equation. As it is clear from the first statement that Rahul’s present age is $4$ years old. And although Manick’s present age is $12$ years old, after a few years his age will be twice as that of Rahul. Thus, his age will be an even number then also. Now, we can hit and trial for a few even numbers after $12$ to get our desired result. More complex problems of such type may lead us to form and solve simultaneous equations.
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# NCERT Solutions for Class 7 Mathematics Chapter 14 - Visualising Solid Shapes
##### Question 1:
Complete the following table:
##### Answer:
Complete table is as follow:
##### Question 2:
Here you find four nets (Figure). There are two correct nets among them to make a tetrahedron. See if you can work out which nets will make a tetrahedron.
##### Answer:
In given figures part (i) and part (iii) are two correct nets for a tetrahedron.
##### Question 3:
Identify the nets which can be used to make cubes (cut out copies of the nets and try it):
##### Answer:
Nets in (ii), (iii), (iv) and (vi) form cubes.
##### Question 5:
Can this be a net for a die? Explain your answer.
##### Answer:
Reason: Because one pair of opposite faces will have 1 and 4 on them whose total is not 7 and another pair of opposite faces will have 3 and 6 on them whose total is also not 7.
##### Question 6:
Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation).
##### Question 7:
The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.
##### Answer:
Dimensions of cuboid are 5 cm, 3 cm and 2 cm. By changing length, breadth and height, we get three different isometric sketches of cuboid are as follow:
##### Question 8:
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid.
##### Question 9:
Try to guess the number of cubes in the following arrangements (Fig.) :
##### Answer:
Number of cubes in the arrangements are as follows:
(i) 24 cubes (ii) 8 cubes
(iii) 9 cubes.
##### Answer:
Numbers on the opposite face:
(a) 5 + 6 is 2 + 1
and (b) 4 + 3 is 3 + 4.
##### Question 11:
What cross-sections do you get when you give a
(i) vertical cut
(ii) horizontal cut
to the following solids?
(a) A brick
(b) A round apple
(c) A die
(d) A circular pipe
(e) An ice cream cone.
##### Answer:
When we give a vertical cut then following cross-sections are obtained:
(a) A brick — Cuboid
(b) A round apple — Circular
(c) A die — Square
(d) A circular pipe — Circular
(e) An ice cream cone — Curved shape
When we give a horizontal cut then following cross-sections are obtained:
(a) A brick —Cuboid
(b) A round apple —Circular
(c) A die —Cuboid
(d) A circular pipe —Hemispherical
(e) An ice cream cone —Frustrum
##### Question 12:
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions.)
##### Question 13:
Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these !)
##### Answer:
Shadows of some 3-D objects, when seen under the lamp of an overhead projector.
##### Question 14:
The cube can cast a shadow in the shape of a rectangle.
1. TRUE
2. FALSE
TRUE
##### Question 15:
The cube can cast a shadow in the shape of a hexagon.
1. TRUE
2. FALSE
FALSE
##### Question 16:
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
##### Answer:
(1) Front (2) Side (3) Top
##### Question 17:
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
##### Answer:
(1) Top (2) Side (3) Front
##### Question 18:
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
##### Answer:
(1) Side (2) Front (3) Top
##### Question 19:
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
##### Answer:
(1) Side (2) Top (3) Front
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In Exponent Basics we worked with whole number exponents. The whole numbers are the set of numbers {0, 1, 2, 3, 4, ...}. The whole numbers are the positive integers, plus zero.
On this page, we will be examing exponents that are negative integers.
Negative Integer Exponents
The use of a negative integer exponent has special meaning and a rule explaining its use.
For any non-zero number x, and for any positive integer n,
Let's take a closer look at why this Rule is true:
One of the Laws of Exponents is that xm • xn = xm+n.
"When multiplying exponential expressions, if the bases are the same, add the exponents."
If we apply this law to work with a negative exponent, we get 43 • 4-3 = 43+(-3) = 40 = 1.
This application shows us that 43 • 4-3 = 1, which means that 4-3 must the multiplicative identity of 43. Therefore, 4-3 must be a fraction and it must be the reciprocal of 43.
Consequently,
.
There are three important concepts at work in this Rule:
For any non-zero number x, and for any positive integer, n:
Remember, any number (or expression) with a negative exponent ends up
on the opposite side of the fraction bar as a positive exponent.
The use of a positive exponent is an application of repeated multiplication by the base:
43 = 4 • 4 • 4 = 64.
The use of a negative exponent produces the opposite of repeated multiplication.
It can be thought of as a form of repeated division by the base:
4-3 = 1 ÷ 4 ÷ 4 ÷ 4 = 0.015625
Examples:
1 2 3 4 5 6 7
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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4
# NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4
## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4.
### Ex 9.4 Class 8 Maths Question 1.
Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq – 2q2)
(vi) (3/4 a2 + 3b2) and 4(a2 – 2/3 b2)
Solution:
(i) (2x + 5) × (4x – 3)
= 2x × (4x – 3) + 5 × (4x – 3)
= (2x × 4x) – (2x × 3) + (5 × 4x) – (5 × 3)
= 8x2 – 6x + 20x – 15
= 8x2 + 14x – 15
(ii) (y – 8) × (3y – 4)
= y × (3y – 4) – 8 × (3y – 4)
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4)
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32
(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
= 6.25l2 + 1.25ml – 1.25ml – 0.25m2
= 6.25l2 + 0 – 0.25m2
= 6.25l2 – 0.25m2
(iv) (a + 3b) × (x + 5)
= a × (x + 5) + 3b × (x + 5)
= (a × x) + (a × 5) + (3b × x) + (3b × 5)
= ax + 5a + 3bx + 15b
(v) (2pq + 3q2) × (3pq – 2q2)
= 2pq × (3pq – 2q2) + 3q2 (3pq – 2q2)
= (2pq × 3pq) – (2pq × 2q2) + (3q2 × 3pq) – (3q2 × 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4
### Ex 9.4 Class 8 Maths Question 2.
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2 + b) (a + b2)
(iv) (p2 – q2)(2p + q)
Solution:
(i) (5 – 2x) (3 + x)
= 5(3 + x) – 2x(3 + x)
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
= 15 + 5x – 6x – 2x2
= 15 – x – 2x2
(ii) (x + 7y) (7x – y)
= x(7x – y) + 7y(7x – y)
= (x × 7x) – (x × y) + (7y × 7x) – (7y × y)
= 7x2 – xy + 49xy – 7y2
= 7x2 + 48xy – 7y2
(iii) (a2 + b) (a + b2)
= a2(a + b2) + b(a + b2)
= (a2 × a) + (a2 × b2) + (b × a) + (b × b2)
= a3 + a2b2 + ab + b3
(iv) (p2 – q2) (2p + q)
= p2(2p + q) – q2(2p + q)
= (p2 × 2p) + (p2 × q) – (q2 × 2p) – (q2 × q)
= 2p3 + p2q – 2pq2 – q3
### Ex 9.4 Class 8 Maths Question 3.
Simplify.
(i) (x2 – 5) (x + 5) + 25
(ii) (a2 + 5)(b3 + 3) + 5
(iii) (t + s2) (t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y)(x2 – xy + y2)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c) (a + b – c)
Solution:
(i) (x2 – 5) (x + 5) + 25
= x2(x + 5) – 5(x + 5) + 25
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x + 0
= x3 + 5x2 – 5x
(ii) (a2 + 5) (b3 + 3) + 5
= a2(b3 + 3) + 5(b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20
(iii) (t + s2) (t2 – s)
= t(t2 – s) + s2(t2 – s)
= t3 – st + s2t2 – s3
= t3 + s2t2 – st – s3
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd
= 4ac + 0 + 0 + 0
= 4ac
(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= 2x2 + x2 + xy + 2xy – xy + 2xy + y2 – 2y2
= 3x2 + 4xy – y2
(vi) (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 – 0 + 0 + y3
= x3 + y3
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x(1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y
= 2.25x2 + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y2
= 2.25x2 + 0 + 0 + 0 – 16y2
= 2.25x2 – 16y2
(viii) (a + b + c) (a + b – c)
= a(a + b – c) + b(a + b – c) + c(a + b – c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + ab + ab – bc + bc – ac + ac + b2 – c2
= a2 + 2ab + b2 – c2 + 0 + 0
= a2 + 2ab + b2 – c2
You can also like these:
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
|
# 3.2. Random Variables#
Random numerical quantities such as “the number of heads in ten tosses of a coin” are called random variables. The terminology and notation of random variables helps reduce the amount of writing involved in phrases like “the chance that there are no more than 4 heads in ten tosses of a coin.”
Formally, suppose you have an outcome space $$\Omega$$. A random variable is a numerical function on $$\Omega$$. That is, a random variable is a function whose domain is $$\Omega$$ and whose range is the number line.
## 3.2.1. Number of Heads in Three Tosses#
As our first example, we will formally define a random count: the number of heads in three tosses of a coin.
The outcome space is
$\Omega ~ = ~ \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$
Random variables are typically denoted by upper case letters, often late letters in the alphabet.
So let the random variable $$X$$ be the number of heads in three tosses. For every outcome, $$X$$ counts the number of heads in the outcome, as shown in the table below.
$$\text{outcome}$$
$$X(\text{outcome})$$
HHH
$$~~~~~~~~~~~~~~~~~~~~~~3$$
HHT
$$~~~~~~~~~~~~~~~~~~~~~~2$$
HTH
$$~~~~~~~~~~~~~~~~~~~~~~2$$
THH
$$~~~~~~~~~~~~~~~~~~~~~~2$$
HTT
$$~~~~~~~~~~~~~~~~~~~~~~1$$
THT
$$~~~~~~~~~~~~~~~~~~~~~~1$$
TTH
$$~~~~~~~~~~~~~~~~~~~~~~1$$
TTT
$$~~~~~~~~~~~~~~~~~~~~~~0$$
The possible values of $$X$$ are $$0, 1, 2, 3$$, because those are the numbers of heads you can get if you toss a coin three times.
Now let’s introduce probabilities. The coin is fair, so all outcomes are equally likely.
$$\text{outcome}$$
$$X(\text{outcome})$$
$$\text{Probability}$$
HHH
$$~~~~~~~~~~~~~~~~~~~~~~3$$
1/8
HHT
$$~~~~~~~~~~~~~~~~~~~~~~2$$
1/8
HTH
$$~~~~~~~~~~~~~~~~~~~~~~2$$
1/8
THH
$$~~~~~~~~~~~~~~~~~~~~~~2$$
1/8
HTT
$$~~~~~~~~~~~~~~~~~~~~~~1$$
1/8
THT
$$~~~~~~~~~~~~~~~~~~~~~~1$$
1/8
TTH
$$~~~~~~~~~~~~~~~~~~~~~~1$$
1/8
TTT
$$~~~~~~~~~~~~~~~~~~~~~~0$$
1/8
There is only one outcome (HHH) for which the number of heads is 3. The chance of that outcome is 1/8, so we say, “The chance of three heads is 1/8.” And we write
$P(X = 3) ~ = ~ 1/8$
There are three outcomes (HHT, HTH, THH) for which the number of heads is two. The chance of getting two heads is the total chance of these three outcomes:
$P(X = 2) ~ = ~ 3/8$
Similarly,
$P(X = 1) ~ = ~ 3/8$
and
$P(X = 0) ~ = ~ 1/8$
We can display all this information more compactly in a probability distribution table for $$X$$, known for short as a distribution table.
$$\text{Possible value } x$$
$$~~0~~$$
$$~~1~~$$
$$~~2~~$$
$$~~3~~$$
$$P(X = x)$$
$$1/8$$
$$3/8$$
$$3/8$$
$$1/8$$
Notice the lower case letter $$x$$ as notation for a generic possible value of $$X$$. Keep in mind that $$X$$ is the name of the random variable whereas $$x$$ is a number.
For the generic value of $$X$$, you don’t have to use $$x$$ as the notation. You can use any letter you like, but it should be lower case as upper case is used for events and random variables.
## 3.2.2. Distribution#
The probability distribution of a random variable, or distribution for short, is the set of all possible values of the random variables along with all of the corresponding probabilities.
When random variables have a finite number of possible values, as in the case of our example $$X$$, the distribution can be displayed in a table. The probabilities can also be written in a formula, as we will see in the next section.
The probabilities in a distribution must add up to 1.
This is clear because $$X$$ has to have one of the possible values and it can’t have two values at once. So by the addition rule, the total chance of all the values should be the chance of the set of all outcomes, which is 1.
The distribution of a random variable is sometimes called a probability mass function, abbreviated to pmf. That is because some probabilists like to visualize probabilities as masses attached to the possible values.
## 3.2.3. Probability Histogram#
The probability histogram is a natural way to visualize the distribution of a random variable. In a histogram, the areas of the bars represent probabilities. Later in the course we will see why this is important.
## 3.2.4. Probabilities#
Once you have the distribution of a random variable, you can find the chance of any event determined by the random variable. Just identify the possible values that are in the event, and add up all their chances.
For example,
\begin{split} \begin{align*} P(X > 0) ~ &= ~ P(X = 1) + P(X = 2) + P(X = 3)\\ &= ~ \frac{3}{8} + \frac{3}{8} + \frac{1}{8} ~ = ~ \frac{7}{8} \end{align*} \end{split}
This could also have been calculated as
$P(X > 0) ~ = ~ 1 - P(X = 0) ~ = ~ 1 - \frac{1}{8} ~ = ~ \frac{7}{8}$
As you can see, the only thing new in these calculations is the notation.
## 3.2.5. Equality#
There is one important new idea to keep in mind when studying random variables, which is that random variables have two different kinds of equality.
Rather than talk in generalities, let’s look at an example. Consider rolling two dice. Let $$D_1$$ be the number on the first roll and $$D_2$$ the number on the second. For example, if the outcome of the two rolls is the pair $$(4, 3)$$ then $$D_1((4, 3)) = 4$$ and $$D_2((4, 3)) = 3$$.
This shows that as functions on the outcome space, $$D_1$$ and $$D_2$$ are not equal. For the outcome $$(4, 3)$$, the value of $$D_1$$ is different from the value of $$D_2$$.
However, $$D_1$$ and $$D_2$$ clearly have something in common. They have the same distribution, shown in the table below.
$$\text{Possible value }$$
$$~~1~~$$
$$~~2~~$$
$$~~3~~$$
$$~~4~~$$
$$~~5~~$$
$$~~6~~$$
$$\text{Probability}$$
$$1/6$$
$$1/6$$
$$1/6$$
$$1/6$$
$$1/6$$
$$1/6$$
This is called the uniform distribution on $$\{1, 2, 3, 4, 5, 6 \}$$.
We say that $$D_1$$ and $$D_2$$ are equal in distribution. But remember: they are not equal.
|
Directions: Study the following questions and choose the right answer:
Important for :
1
Two light rods AB = a + b, CD = a – b symmetrically lying on a horizontal AB. There are kept intact by two strings AC and BD. The perpendicular distance between rods in a. The length of AC is given by
» Explain it
D
Since, they are symmetrically on horizontal plane.
∴ AC = BD
∴ AE = BF = x
Now, AB = (a – b) + 2x
i.e. a + b = a – b + 2x ⇒ 2b = 2x ⇒ b = x
Now in ΔACE,
x2 + a2 = AC2
AC2 = b2 + a2 ⇒ AC = b2 + a2
Hence, option D is correct.
2
If PQRS be a rectangle such PQ = 3 QR. Then, what is ∠PRS equal to?
» Explain it
C
In rectangle PQRS,
PQ || RS
∴ ∠RPQ = ∠ PRS (vertically opposite angles)
...(i)
Now in ΔPQR,
tan ∠QPR = RQ PQ
⇒ tan ∠QPR = QR
3 QR
⇒ ∠QPR = 30°
∴ ∠PRS = 30°
[From the equation (i)]
Hence, option C is correct.
3
In a trapezium, the two non-parallel sides are equal in length, each being of 5 cm. The parallel sides are at a distance of 3 cm apart. If the smaller side of the parallel sides is of length 2 cm, then the sum of the diagonals of the trapezium is
» Explain it
B
In ΔBCF,
By the pythagoras theorem,
BF= BC2 – CF2
(BF)2 = (5)2 – (3)2 ⇒ BF = 4 cm
AB = 2 + 4 + 4 = 10 cm
Now, in ΔACF,
AC2 = CF2 + FA2 ⇒ AC2 = 32 + 62
AC = 45 cm
Similarly, BD = 45 cm
∴ Sum of diagonal = 2 × 45 = 2 × 3 5 = 6 5 cm.
Hence, option B is correct.
4
The area of a rectangle lies between 40 cm2 and 45 cm2. If one of the sides is 5 cm, then its diagonal lies between
» Explain it
B
Area of rectangle lies between 40 cm2 and 45 cm2
Now, one side = 5 cm
Since, area can't be less than 40 cm2
∴ Other side can't be less than = 40 = 8 cm 5
Since, area can't be greater than 45 cm2.
∴ Other side can't be greater than = 45 = 9 cm 5
∴ Minimum value of diagonal = 82 + 52 = 89 = 9.43 cm
∴ Maximum value of diagonal = 92 + 52 = 106 = 10.3 cm
So, diagonal lies between 9 cm and 11 cm.
Hence, option B is correct.
5
Let ABCD be a parallelogram. Let P, Q, R and S be the mid-points of sides AB, BC, CD and DA respectively. Consider the following statements.
I. Area of triangle APS < Area of triangle DSR, if BD < AC.
II. Area of triangle ABC = 4 (Area of triangle BPQ).
Select the correct answer using the codes given below.
» Explain it
B
Area of ΔAPS = Area of ΔDSR
∵ AS = SD and AP = DR
∴ ar (ΔABC) = 4 ar (ΔBPQ).
Hence, option B is correct.
|
# Search by Topic
#### Resources tagged with Practical Activity similar to Making Rectangles, Making Squares:
Filter by: Content type:
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Challenge level:
### There are 63 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Practical Activity
### Making Rectangles, Making Squares
##### Stage: 3 Challenge Level:
How many differently shaped rectangles can you build using these equilateral and isosceles triangles? Can you make a square?
### Drawing Celtic Knots
##### Stage: 3 Challenge Level:
Here is a chance to create some Celtic knots and explore the mathematics behind them.
### Notes on a Triangle
##### Stage: 3 Challenge Level:
Can you describe what happens in this film?
### Making Maths: Celtic Knot Tiles
##### Stage: 2, 3 and 4 Challenge Level:
Make some celtic knot patterns using tiling techniques
### Making Maths: Snake Pits
##### Stage: 1, 2 and 3 Challenge Level:
A game to make and play based on the number line.
### Making Maths: Archimedes' Spiral
##### Stage: 2 and 3 Challenge Level:
Make a spiral mobile.
### Making Maths: Walking Through a Playing Card?
##### Stage: 2 and 3 Challenge Level:
It might seem impossible but it is possible. How can you cut a playing card to make a hole big enough to walk through?
### Back to the Practical?
##### Stage: 2 and 3
In this article for teachers, Bernard uses some problems to suggest that once a numerical pattern has been spotted from a practical starting point, going back to the practical can help explain. . . .
### Celtic Knotwork Patterns
##### Stage: 2 and 3
This article for pupils gives an introduction to Celtic knotwork patterns and a feel for how you can draw them.
### Making Maths: String and Circles
##### Stage: 2 and 3 Challenge Level:
You could use just coloured pencils and paper to create this design, but it will be more eye-catching if you can get hold of hammer, nails and string.
### Enigma
##### Stage: 2, 3 and 4 Challenge Level:
This package contains hands-on code breaking activities based on the Enigma Schools Project. Suitable for Stages 2, 3 and 4.
### Making Maths: Equilateral Triangle Folding
##### Stage: 2 and 3 Challenge Level:
Make an equilateral triangle by folding paper and use it to make patterns of your own.
### Making Maths: Make a Pendulum
##### Stage: 2 and 3 Challenge Level:
Galileo, a famous inventor who lived about 400 years ago, came up with an idea similar to this for making a time measuring instrument. Can you turn your pendulum into an accurate minute timer?
### Making Maths: Clinometer
##### Stage: 3 Challenge Level:
Make a clinometer and use it to help you estimate the heights of tall objects.
### Plaited Origami Polyhedra
##### Stage: 2, 3 and 4 Challenge Level:
These models have appeared around the Centre for Mathematical Sciences. Perhaps you would like to try to make some similar models of your own.
##### Stage: 3, 4 and 5
Logo helps us to understand gradients of lines and why Muggles Magic is not magic but mathematics. See the problem Muggles magic.
### First Forward Into Logo 12: Puzzling Sums
##### Stage: 3, 4 and 5 Challenge Level:
Can you puzzle out what sequences these Logo programs will give? Then write your own Logo programs to generate sequences.
### First Forward Into Logo 8: More about Variables
##### Stage: 3, 4 and 5 Challenge Level:
Write a Logo program, putting in variables, and see the effect when you change the variables.
### First Forward Into Logo 10: Count up - Count Down
##### Stage: 3, 4 and 5 Challenge Level:
What happens when a procedure calls itself?
### The Best Card Trick?
##### Stage: 3 and 4 Challenge Level:
Time for a little mathemagic! Choose any five cards from a pack and show four of them to your partner. How can they work out the fifth?
### Amazing Card Trick
##### Stage: 3 Challenge Level:
How is it possible to predict the card?
### Which Solids Can We Make?
##### Stage: 4 Challenge Level:
Interior angles can help us to work out which polygons will tessellate. Can we use similar ideas to predict which polygons combine to create semi-regular solids?
### First Forward Into Logo 5: Pen Up, Pen Down
##### Stage: 2, 3 and 4 Challenge Level:
Learn about Pen Up and Pen Down in Logo
### Turning the Place Over
##### Stage: 3, 4 and 5 Challenge Level:
As part of Liverpool08 European Capital of Culture there were a huge number of events and displays. One of the art installations was called "Turning the Place Over". Can you find our how it works?
### First Forward Into Logo 9: Stars
##### Stage: 3, 4 and 5 Challenge Level:
Turn through bigger angles and draw stars with Logo.
### First Forward Into Logo 6: Variables and Procedures
##### Stage: 3, 4 and 5 Challenge Level:
Learn to write procedures and build them into Logo programs. Learn to use variables.
### First Forward Into Logo 4: Circles
##### Stage: 2, 3 and 4 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### Well Balanced
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Exploring balance and centres of mass can be great fun. The resulting structures can seem impossible. Here are some images to encourage you to experiment with non-breakable objects of your own.
### Tangram Pictures
##### Stage: 1, 2 and 3 Challenge Level:
Use the tangram pieces to make our pictures, or to design some of your own!
### First Forward Into Logo 11: Sequences
##### Stage: 3, 4 and 5 Challenge Level:
This part introduces the use of Logo for number work. Learn how to use Logo to generate sequences of numbers.
### First Forward Into Logo 7: Angles of Polygons
##### Stage: 3, 4 and 5 Challenge Level:
More Logo for beginners. Learn to calculate exterior angles and draw regular polygons using procedures and variables.
### Whirling Fibonacci Squares
##### Stage: 3 and 4
Draw whirling squares and see how Fibonacci sequences and golden rectangles are connected.
### First Forward Into Logo 2: Polygons
##### Stage: 2, 3 and 4 Challenge Level:
This is the second in a twelve part introduction to Logo for beginners. In this part you learn to draw polygons.
### Paper Folding - Models of the Platonic Solids
##### Stage: 2, 3 and 4
A description of how to make the five Platonic solids out of paper.
### Sea Defences
##### Stage: 2 and 3 Challenge Level:
These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Nine Colours
##### Stage: 3 and 4 Challenge Level:
You have 27 small cubes, 3 each of nine colours. Use the small cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of every colour.
### Cool as Ice
##### Stage: 3 and 4 Challenge Level:
Design and construct a prototype intercooler which will satisfy agreed quality control constraints.
### Straw Scaffold
##### Stage: 3 Challenge Level:
Build a scaffold out of drinking-straws to support a cup of water
### More Marbles
##### Stage: 3 Challenge Level:
I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?
### Attractive Rotations
##### Stage: 3 Challenge Level:
Here is a chance to create some attractive images by rotating shapes through multiples of 90 degrees, or 30 degrees, or 72 degrees or...
### Fractions Jigsaw
##### Stage: 3 Challenge Level:
A jigsaw where pieces only go together if the fractions are equivalent.
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Marbles
##### Stage: 3 Challenge Level:
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
### Making Maths: Double-sided Magic Square
##### Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Rolling Triangle
##### Stage: 3 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
### Getting an Angle
##### Stage: 3 Challenge Level:
How can you make an angle of 60 degrees by folding a sheet of paper twice?
### Triangles to Tetrahedra
##### Stage: 3 Challenge Level:
Starting with four different triangles, imagine you have an unlimited number of each type. How many different tetrahedra can you make? Convince us you have found them all.
|
# How do you divide (-x^4-3x^3-2x^2+7x+3)/(x^2+3) ?
##### 1 Answer
Jun 25, 2018
$- {x}^{2} - 3 x + 1 + \frac{16 x}{{x}^{2} + 3}$
#### Explanation:
Using place keepers of zero value. Example: $0 {x}^{3}$
$\textcolor{w h i t e}{\text{dddddddddddddd}} + {x}^{4} - 3 {x}^{3} - 2 {x}^{2} + 7 x + 3$
$\textcolor{m a \ge n t a}{- {x}^{2}} \left({x}^{2} + 3\right) \to \textcolor{w h i t e}{\text{d") ul(-x^4+0x^3-3x^2larr" Subtract}}$
$\textcolor{w h i t e}{\text{dddddddddddddddd}} 0 - 3 {x}^{3} + {x}^{2} + 7 x + 3$
$\textcolor{m a \ge n t a}{- 3 x} \left({x}^{2} + 3\right) \to \textcolor{w h i t e}{\text{dddd")ul(-3x^3+0x^2-9xlarr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddddddddd}} 0 + {x}^{2} + 16 x + 3$
$\textcolor{m a \ge n t a}{+ 1} \left({x}^{2} + 3\right) \to \textcolor{w h i t e}{\text{dddddddddddd")ul(x^2+color(white)("..")0x+3larr" Subtract}}$
color(magenta)("Remainder "->color(white)("dddddddddddd.")0+16x+0)
$\textcolor{w h i t e}{}$
$- {x}^{2} - 3 x + 1 + \frac{16 x}{{x}^{2} + 3}$
|
AMC 8 PREPARATION TEXTBOOKS
Top AMC 8 scorers across the country have used our Introduction series of textbooks, Art of Problem Solving Volume 1, and Competition Math for Middle School to get ready for the AMC 8.
# 2018 AMC 8 Problems
## Problem 1
An amusement park has a collection of scale models, with ratio $1 : 20$, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica to the nearest whole number?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$
## Problem 2
What is the value of the product$$\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?$$
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8$
## Problem 3
Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?
$\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}$
## Problem 4
The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$?
$[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]$
$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$
## Problem 5
What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$?
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$
## Problem 6
On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
$\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100$
## Problem 7
The $5$-digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?
$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$
## Problem 8
Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.
$[asy] size(8cm); void drawbar(real x, real h) { fill((x-0.15,0.5)--(x+0.15,0.5)--(x+0.15,h)--(x-0.15,h)--cycle,gray); } draw((0.5,0.5)--(7.5,0.5)--(7.5,5)--(0.5,5)--cycle); for (real i=1; i<5; i=i+0.5) { draw((0.5,i)--(7.5,i),gray); } drawbar(1.0,1.0); drawbar(2.0,2.0); drawbar(3.0,1.5); drawbar(4.0,3.5); drawbar(5.0,4.5); drawbar(6.0,2.0); drawbar(7.0,1.5); for (int i=1; i<8; ++i) { label(""+string(i)+"",(i,0.25)); } for (int i=1; i<9; ++i) { label(""+string(i)+"",(0.5,0.5*(i+1)),W); } label("Number of Days of Exercise",(4,-0.1)); label(rotate(90)*"Number of Students",(-0.1,2.75)); [/asy]$ What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?
$\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00$
## Problem 9
Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?
$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }91\qquad\textbf{(D) }96\qquad \textbf{(E) }120$
## Problem 10
The $\emph{harmonic mean}$ of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?
$\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}$
## Problem 11
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.
$\begin{eqnarray*} \text{X}&\quad\text{X}\quad&\text{X} \\ \text{X}&\quad\text{X}\quad&\text{X} \end{eqnarray*}$
If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}$
## Problem 12
The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual tim?
$\textbf{(A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10$
## Problem 13
Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$
## Problem 14
Let $N$ be the greatest five-digit number whose digits have a product of $120$. What is the sum of the digits of $N$?
$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$
## Problem 15
In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
$[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$
## Problem 16
Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880$
## Problem 17
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?
$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$
## Problem 18
How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$
## Problem 19
In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
$[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("+",(0,0)); draw(shift(1,0)*box); label("-",(1,0)); draw(shift(2,0)*box); label("+",(2,0)); draw(shift(3,0)*box); label("-",(3,0)); draw(shift(0.5,0.4)*box); label("-",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("-",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("-",(2.5,0.4)); draw(shift(1,0.8)*box); label("+",(1,0.8)); draw(shift(2,0.8)*box); label("+",(2,0.8)); draw(shift(1.5,1.2)*box); label("+",(1.5,1.2)); [/asy]$
$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$
## Problem 20
In $\triangle ABC,$ a point $E$ is on $\overline{AB}$ with $AE=1$ and $EB=2.$ Point $D$ is on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$ and point $F$ is on $\overline{BC}$ so that $\overline{EF} \parallel \overline{AC}.$ What is the ratio of the area of $CDEF$ to the area of $\triangle ABC?$
$[asy] size(7cm); pair A,B,C,DD,EE,FF; A = (0,0); B = (3,0); C = (0.5,2.5); EE = (1,0); DD = intersectionpoint(A--C,EE--EE+(C-B)); FF = intersectionpoint(B--C,EE--EE+(C-A)); draw(A--B--C--A--DD--EE--FF,black+1bp); label("A",A,S); label("B",B,S); label("C",C,N); label("D",DD,W); label("E",EE,S); label("F",FF,NE); label("1",(A+EE)/2,S); label("2",(EE+B)/2,S); [/asy]$
$\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}$
## Problem 21
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
## Problem 22
Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
$[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("A",(0,6),NW); label("B",(6,6),NE); label("C",(6,0),SE); label("D",(0,0),SW); label("E",(3,0),S); label("F",(4,2),E); [/asy]$
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$
## Problem 23
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
$[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]$
$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$
## Problem 24
In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of edges $\overline{FB}$ and $\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$
$[asy] size(6cm); pair A,B,C,D,EE,F,G,H,I,J; C = (0,0); B = (-1,1); D = (2,0.5); A = B+D; G = (0,2); F = B+G; H = G+D; EE = G+B+D; I = (D+H)/2; J = (B+F)/2; filldraw(C--I--EE--J--cycle,lightgray,black); draw(C--D--H--EE--F--B--cycle); draw(G--F--G--C--G--H); draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J); label("A",A,E); label("B",B,W); label("C",C,S); label("D",D,E); label("E",EE,N); label("F",F,W); label("G",G,N); label("H",H,E); label("I",I,E); label("J",J,W); [/asy]$
$\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}$
## Problem 25
How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$
|
Krishna
0
Step 1: Find the circumference of the circle
NOTE: Circumference of the circle = 2 \pi r
EXAMPLE: Circumference = 2 \pi r
= 2 \pi(15)
= 30 \pi
Step 2: Identify the type of the triangle
NOTE: The triangle are congruent because they are radii of the same circle.
So, \triangle APB is isosceles, and the
angles opposite the congruent sides are congruent to each other.
EXAMPLE: \angle A = \angle B
Step 3: Add the interior angles of a triangle up to 180\degree, to find the unknown angle.
\angle A + \angle B + \angle P = 180\degree
38\degree + 38\degree + \angle P = 180\degree
\angle P = 104\degree
Step 4: Calculate the arc length.
Arc length:
Step 1: Note down the given values
Step 2: Set up the formula for arc length.
NOTE: The formula is arc length= 2 \pi (r)(\frac{\theta }{360}) ,
where {\displaystyle r} equals the radius of the circle and {\displaystyle \theta } equals the measurement of the arc’s central angle, in degrees.
or
Arc length = r * \theta
Step 3: Plug the length of the circle’s radius into the formula.
Step 4: Plug the value of the arc’s central angle into the formula.
Step 5: Simplify the equation to find the arc length
NOTE: Use multiplication and division to simplify the equation.
Area of the circle.
Step 1: Identify the known or given information.
Step 2: Set up a formula for the sector area
NOTE: A ratio will need to be constructed. Recall that a circle is composed
of 360 degrees. Therefore, the following ratio can be made,
\frac{\theta}{360} = \frac{\text{sector area} (A_C)}{\text{Total area} (A_T)}
Sector area = \frac{\theta}{360} * \pi r^2 (since area = \pi r^2)
where, \theta = Central angle.
Step 3: Plug the sector’s central angle measurement into the formula.
Step 4: Plug the sector’s radius measurement into the formula.
Step 5: Solve for the area:
EXAMPLE: Sector area = \frac{\theta}{360} * \pi r^2 (since area = \pi r^2)
Sector area = \frac{60}{360}*\left(3.14\right)\left(5\right)^2
Sector area = 13.09 cm^2
.
|
# Algebra Tips: How to Solve Word Problems About Ages
TR Smith is a product designer and former teacher who uses math in her work every day.
Joined: 5 years agoFollowers: 216Articles: 455
These kids will only reveal their ages if you solve a word problem. | Source
Word problems about ages are a popular theme for math puzzles and brain teasers because they make you think analytically; you have to figure out how to apply your math skills to find the solution. Learning how to solve these math problems can help you understand and solve more complicated mathematical problems and riddles. If you always find yourself at a loss for where to begin, the following problem-solving strategies will make these problems a breeze.
## Mathematical Notation: Variable Names
A typical age problem names the people whose ages are unknown. The first letters of their names are good choices for variables to represent the unknown ages. For example, if you are trying to determine how old Paula, Quinn, and Robert are right now, you will have an easier time setting up and solving the equations if you represent their present ages with the letters P, Q, and R.
Math word problems about ages deal with current and future ages. | Source
## Representing Future Ages
To write equations that describe mathematical relations between future ages, you must add a certain number of years to the variable. For example, if the two current age variables in question are A and B, and you want to describe what happens to their ages 8 years from now, you must replace every instance of "A" with "A + 8," and every instance of "B" with "B + 8."
Consider this simple example problem: The sum of Ben's and Anne's current ages is 48. In 8 years, Anne will be 5/3 times as old as Ben. How old are Anne and Ben now?
The first statement is easy to translate into math terms: A + B = 48. For the second statement, we need to consider the new-ish variables "A + 8" and "B + 8." Using these as placeholders for Anne's and Ben's future ages we get the second equation A + 8 = (5/3)(B + 8).
As two people grow old together, the ratio between their ages approaches 1. | Source
## The Mathematical Relations Between People's Ages and How They Change Over Time
Given two people's ages, there are several mathematical relations you can calculate using these two numbers. For instance, if the ages are represented by the variables C and D, you could compute any of the following:
• The ratio of C to D, C/D
• The ratio of D to C, D/C
• The difference between C and D, either C - D or D - C
• The sum of C and D, C + D
• The product of C and D, CD
• The arithmetic mean (average) of C and D, (C + D)/2
• Miscellaneous: Reciprocals 1/C and 1/D, Sum of squares C^2 + D^2, Greatest common factor of C and D, Least common multiple of C and D.
The numerical values of all of these quantities change over time, except for one, the age difference. If the present age difference between Carol and Dan is 5 years, then in 20, 30, or 40 years from now they will still be 5 years apart.
The sum, product, and average of two people's ages will increase as they grow older. If Carol and Dan are currently 12 years old and 16 years old, their current age sum, product, and average are 28, 192, and 14 respectively. In 10 years, their age sum, product, and average will be 48, 572, and 24. Sums of squares and other powers also increase over time.
The ratio between two ages becomes closer to 1 as the years wear on. If Carol and Dan are 12 and 16 right now, their current age ratios are C/D = 0.75 and D/C = 1.3333. But in 10 years these ratios will be (C+10)/(D+10) = 22/26 = 0.8462 and (D+10)/(C+10) = 26/22 = 1.1818.
Reciprocals become closer to 0 as ages increase.
How the greatest common factor and least common multiple change over time depends on the number theoretic properties of the ages. For example, if Carol and Dan are currently 12 and 16, then gcd = 4 and lcm = 48. But if you increase their ages by 1 year to 13 and 17 respectively, you get gcd = 1 and lcm = 221. Increase their ages by 2 years to 14 and 18 respectively and you get gcd = 2 and lcm = 126. The gcd of two people's ages can never exceed the difference between their ages (in this example, 4). The lcm is always equal to the product of the ages divided by the gcd.
## Reduce the Number of Variables Whenever Possible
If you know the difference between two people's ages and which one of them is older, you can reduce the number of variables in the problem. For example, suppose you do not know how old Tim and Ursula are, but you do know that Tim is 3 years younger than Ursula. Instead of representing their unknown ages with the pair of variables T and U, you can either replace T with U - 3, or replace U with T + 3. This substitution leaves you with one variable to consider rather than two.
For example, if you are told that the product of Tim's and Ursula's ages is a composite number between 173 and 193, there's not enough information to find a unique solution. But if you know that Ursula is three years older than Tim, you can reduce the number of unknowns and find the unique solution.
The same trick applies to ages that are consecutive numbers (or consecutive evens/odds). If you let the youngest person's age be K, then the older people's ages can be represented by K+1, K+2, K+3, and so on. In the case of consecutive even/odd ages, the sequence is K, K+2, K+4, K+6, and so on, since a pair of consecutive even/odd numbers differ by 2.
You can sometimes reduce the number of variables if you know the sum, product, or ratio of two ages, but you must be mindful that these mathematical relations change over time. For instance, if Liam is currently twice as old as Katie, the substitution L = 2K is only valid for the present. The equation (L+5) = 2(K+5) does not still hold if you are trying to describe the relation between their ages 5 years from now.
Today is the only day you'll ever be exactly four years old. Cheer up girl! | Source
## Assume the Solutions are Whole Numbers
In reality, ages are not whole numbers. If you were born on January 1, 2000 and the current date is October 1, 2009, your true age is 9.75 years. But you will probably tell people you are 9 years old until your next birthday.
For age and birthday math problems, you should assume all the ages to be whole numbers (positive integers) unless otherwise noted. If a problem is posed in such a way that there may exist mathematically consistent solutions that are not integers, the problem may specify that you should discard solutions that are not whole numbers.
Example: Ellie is older than Fred, who in turn is older than Gerta. The sum of their three ages is 43 and the product is 2640. How old are these three people?
As this problem is stated, it actually has infinitely many correct solutions, a few of which are
• E = 18.832, F = 14.5, G = 9.9668
• E = 18.444, F = 15.037, G = 9.519
• E = 19.227, F = 13.882, G = 9.891
• E = 20.05, F = 11.544, G = 11.406
However, there is only one integer solution to this word problem.
This math problem would be clearer if it explicitly stated that you should discard non-integral solutions, or if it mentioned that today was their common birthday, which would imply that their current ages are whole numbers.
## Verbal Clues: Older, Oldest, Younger, Youngest
If a problem states that certain people are older or younger than others, this information can help you reduce the number of possibilities when solving for people's ages. For example, if Marc's and Nadine's ages add up to 44 and Nadine is older, then Nadine must be at least 23 and Marc is no older than 21.
Similary, if you are trying to find a set of ages and you are given a clue about the oldest individual member, then you can rule out the possibility that a set of twins or triplets is older than everyone else. This is a key clue in the classic age problem "The Proof Is in the Pudding."
## Remember Basic Algebra and Other Math Concepts
Most birthday and age word problems are set up to be systems of linear equations. You can solve a system of n linear equations in n variables using the techniques of substitution, elimination, matrices, or by using an algebra solver.
Age puzzles that give you sums and products of ages can be solved by finding the roots of polynomials. For example, a math puzzle that gives you the sum and product of two ages can be solved with a quadratic equation.
For math puzzles about the divisibility and factors of an age, you can apply basic number theory concepts. For example, if the product of two ages is an even number, the at least one of the ages is an even number (assuming whole number ages). If the sum of two ages is even, then either both ages are even or both are odd.
## Example Age Word Problem
Bob and Billy are twins. Together with their older cousins Romeo and Tara, the sum of all their ages at present is 36. Six years ago, the sum of the twins' ages was half the sum of their cousins' ages. If the sum of the twins' and Romeo's current ages is a square number, how old is everyone?
For this problem, we can represent Bob's and Billy's current age with B, Romeo's age with R, and Tara's age with T. The first equation we can get out of the problem is 2B + R + T = 36, or equivalently 2B = 36 - (R + T).
The second statement of the problem can be translated into mathematical notation as
(B-6) + (B-6) = 0.5[(R-6) + (T-6)]
This equation can be simplified to 2B = 0.5(R + T) + 6. Before we analyze the last piece of information in the problem, let's use these two equations to find out more about the people's ages. Since we know that 2B = 36 - (R + T) and 2B = 0.5(R + T) + 6, we can eliminate the quantity 2B to make the equation
36 - (R + T) = 0.5(R + T) + 6
After some algebraic simplification, we can see that R + T = 20. While this doesn't tell us the individual ages of Romeo and Tara, it does tell us that B = 8.
Since the problem stated that both Romeo and Tara are older than Bob and Billy, there are only a few possibilities for the values of R and T:
• R = 9 and T = 11
• R = 10 and T = 10
• R = 11 and T = 9
Only the first possibility satisfies the last condition of the problem, that 2B + R equal a square number. Therefore the complete solution is
• Bob = 8
• Billy = 8
• Romeo = 9
• Tara = 11
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• vickie 4 years ago
Thanks, this is really useful. I usually get stuck when they say "in 7 years so-and-so will be half as old as such-and-such was 8 years ago..."
• ok so how do you solve... 4 years ago
the sum of Herp and Derp's ages is a square number, and in 7 years Herp will be half as old as Derp was 8 years ago?
• Author
TR Smith 4 years ago from Germany
For the second condition, if you simplify the expression H + 7 = (1/2)(D - 8) you get D - 2H = 22. And the first condition is D + H = N^2, where N is an integer. Eliminating D from both equations gives you 3H = N^2 - 22, or H = (N^2 - 22)/3. Since N^2 can be any square number, this problem actually has infinitely many (whole number) solutions if you don't put any upper limits on Herp's and Derp's ages.
But if neither of them can be older than the world-record human age (currently 122 years), then you get the following solutions:
D = 24, H = 1
D = 40, H = 9
D = 50, H = 14
D = 74, H = 26
D = 88, H = 33
D = 120, H = 49
• Tim 3 years ago
Thanks, I solved this one with trial and error, but am curious to know how you can do it with equations.
Ben is currently 2/3 of Steven's age. When Ben is as old as Steve is now, Ben will be 3/4 of Steven's age.
• Author
TR Smith 3 years ago from Germany
Hi Tim, as stated, your age problem has infinitely many solutions. For example their current ages could be (B, S) = (2, 3), (4, 6), (6, 9), (8, 12), (10, 15), etc. The first equation you get out of the problem is
B = (2/3)S
The second equation you get is when you add the difference in their ages, S-B, to get the new relation when Ben is as old as Steven. This gives you
B + (S-B) = (3/4)[S + (S-B)]
But this equation reduces to the first:
B + (S-B) = (3/4)[S + (S-B)]
S = (3/4)[2S - B]
S - (6/4)S = -B
(-2/3)S = -B
(2/3)S = B
Without two distinct equations in two variables you can't get a unique solution. So there may be some information missing in the problem.
|
Someone help me with these algebraic equations? 1. 3x + 8 = 53 2. (x + 21) + (2x + 9) = 90 3. 3(x + 8) = 12 4. 7x + 5 = 5x + 17 5. 2(3x - 4) + 10 = 5(x + 4)
2
by toriidamons
2014-09-05T21:39:59-04:00
1. 3x + 8 = 53
3x=45
x=15
2. (x + 21) + (2x + 9) = 90
x+21+2x+9=90
3x=69
x=23
3. 3(x + 8) = 12
3x+24=12
3x=-12
x=-4
4. 7x + 5 = 5x + 17
7x-5x=17-5
2x=12
x=6
5. 2(3x - 4) + 10 = 5(x + 4)
6x-8+10=5x+20
6x-5x=20+8-10
x=18
2014-09-05T22:04:17-04:00
1. 3x + 8 = 53
3x = 45
x = 15
You have to isolate the variable on one side of the equation, so, subtract the 8 from the left side, and subtract the same quantity on the right (whatever you do to one side that CHANGES THE VALUE of the equation must be done to the other side)
Now you 3x = 45. Divide 3x by 3, and 45 by 3. x = 15
2. (x + 21) + (2x + 9) = 90
x + 21 + 2x + 9 = 90
3x + 30 = 90
3x = 60
x = 20
Start by removing the parenthesis. Because the quantities are positive, all you have to do is remove the parenthesis. If there was a negative sign before it, you would change all the signs within the parenthesis. -(1 + 1) would become -1 - 1.
Now add your like terms: x + x = 2x, 21 + 9 = 30.
Now begin isolating the variable: Subtract 30 from one side, and subtract it from the other side: 3x = 60.
Divide 3x by 3, and 60 by 3: x = 20
3. 3(x + 8) = 12
3x + 24 = 12
3x = -12
x = (-4)
Use the distributive property to simplify the left side of the equation: 3x + 24. You do not have to do anything to the left side because you aren't changing the value of the equation.
Now begin isolation of the variable: 3x = -12, x = -4.
4. 7x + 5 = 5x + 17
2x + 5 = 17
2x = 12
x = 6
Begin by getting your variables on one side of the equation: Subtract 5x from the right side, and subtract it from the left side as well 2x + 5 = 17. Now divide 2 from 2x and from 12, x = 6.
5. 2(3x - 4) + 10 = 5(x + 4)
6x - 8 + 10 = 5x + 20
6x + 2 = 5x + 20
x + 2 = 20
x = 18
Begin by using the distributive property and geting your variable on one side:
6x - 8 + 10 = 5x + 20
x - 8 + 10 = 20
|
2011 AMC 12B Problems/Problem 14
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Problem
A segment through the focus $F$ of a parabola with vertex $V$ is perpendicular to $\overline{FV}$ and intersects the parabola in points $A$ and $B$. What is $\cos\left(\angle AVB\right)$?
$\textbf{(A)}\ -\frac{3\sqrt{5}}{7} \qquad \textbf{(B)}\ -\frac{2\sqrt{5}}{5} \qquad \textbf{(C)}\ -\frac{4}{5} \qquad \textbf{(D)}\ -\frac{3}{5} \qquad \textbf{(E)}\ -\frac{1}{2}$
Solution 1
Name the directrix of the parabola $l$. Define $d(X,k)$ to be the distance between a point $X$ and a line $k$.
Now we remember the geometric definition of a parabola: given any line $l$ (called the directrix) and any point $F$ (called the focus), the parabola corresponding to the given directrix and focus is the locus of the points that are equidistant from $F$ and $l$. Therefore $FV=d(V,l)$. Let this distance be $d$. Now note that $d(F,l)=2d$, so $d(A,l)=d(B,l)=2d$. Therefore $AF=BF=2d$. We now use the Pythagorean Theorem on triangle $AFV$; $AV=\sqrt{AF^2+FV^2}=d\sqrt{5}$. Similarly, $BV=d\sqrt{5}$. We now use the Law of Cosines:
$$AB^2=AV^2+VB^2-2AV\cdot VB\cos{\angle AVB}\Rightarrow 16d^2=10d^2-10d^2\cos{\angle AVB}$$
$$\Rightarrow \cos{\angle AVB}=-\frac{3}{5}$$
This shows that the answer is $\boxed{\textbf{(D)}}$.
Solution 2
WLOG we can assume that the parabola is $y=x^2$. Therefore $V = (0,0)$ and $F = (0,\frac{1}{4})$. Also $A = (-\frac{1}{2},\frac{1}{4})$ and $B = (\frac{1}{2},\frac{1}{4})$.
$AB = 1$ and $AV = VB = \sqrt{(\frac{1}{2})^2+(\frac{1}{4})^2} = \frac{\sqrt{5}}{4}$ by the pythagorean theorem.
Now using the law of cosines on $\triangle AVB$ we have:
$AB^2 = 2AV^2-2AV^2\cos(\angle AVB) = 2AV^2(1-\cos(\angle AVB))$
$1 = \frac{5}{8} (1-\cos(\angle AVB))$
Thus, $$\cos(\angle AVB) = \boxed{\textbf{(D)} -\frac{3}{5}}.$$
(solution by mihirb)
Solution 3
After assuming that the parabola is $x^2$,find the points A and B, which are +/- 1,2,1/4.Now treat them as vectors,take the dot product,then find the magnitudes and multiply them.A well known definition of the dot product says that the quotient of the two is the cosine of the angle between them.This will give you D.
Solution 4
As we know, an expression for the equation of a parabola is $y-k=\frac{1}{4a}(x-h)^2$, where $(h,k)$ is the vertex and $a$ is the distance from the focus to the vertex, here $F$ to $V$. The length of the latus rectum, or $\overline{AB}$ here, is equal to $|4a|$. This means that $AF=BF=2a$ since $F$ is the midpoint of $AB$. Then we can use right Triangle $VAF$ to figure out that $VA=\frac{a\sqrt{5}}{4}$. Now we can use the fact that $\cos{\angle AVB}=\cos{2 \angle AVF}$ and use the double angle formula. This results in $\cos{2 \angle AVF}=2{\cos}^2{\angle AVF}-1$. We can find $\cos{\angle AVF}$ from right triangle $AVF$ using the Pythagorean Theorem, which is $\frac{a}{\sqrt{5}}$. Evaulating the expression, we find that $\cos{2 \angle AVF}=\frac{-3}{5}$.
-Indefintense
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# Functions
40 %
60 %
Education
Published on March 11, 2014
Author: tylerisaacmurphy
Source: slideshare.net
## Description
Functions and how to prove 1-1 and onto Tyler Murphy March 10, 2014 Before we get into these, let’s first talk about some Definitions and notation. When we talk about a relation, R between two sets A and B, we can define R as a set of the ordered pairs in A x B. Now, IF for all ordered pairs in R, we have that for every a ∈ A, there is ONLY one b ∈ B paired with it, then we can say that R is a function from A to B. We can notate this as: R = f : A → B This simply states that R is a function that goes from the domain A to the target B. This is the vertical line test. For every x value, there is only one y associated with it. For the rest of the notes, we will refer to our function by the name f Terminology Definition 1-1 (one to one) The function passes the horizontal line test (hLT) Onto Everything in the target gets hit. That is, ran(f) = Target. Proving Onto In order to prove a function is a onto, we have to prove that: Target ⊆ Range. Dy definition we are given that Range ⊆ Target. Together these will give us that the Range = Target. Example Prove that f : R → R given the rule f(x) = x2 − x is onto. scratch work (not in finished proof): To be in the range, y must be equal to x2 − x. 1
So, y = x2 − x 0 = x2 − x − y x = 1 ± (−1)2 − (4)(1)(−y) 2(1) (by quadratic eq) x = 1 ± √ 1 + 4y 2 So this is the x value that gets mapped to any y in the target. Proof. WTS: ∀y ∈ R, ∃x ∈ R such that f(x) = y. That is, if we choose any element y of the target, we want to show that there is some element x in the domain that gets moved to that y. Let y ∈ R (target). Let x = 1 ± √ 1 + 4y 2 , which is in dom(f) = R. So, f(x) = f 1 ± √ 1 + 4y 2 = 1 ± √ 1 + 4y 2 2 − 1 ± √ 1 + 4y 2 . Now we just need to do some algebra. f( 1 ± √ 1 + 4y 2 ) = 1 ± 2 √ 1 + 4y + 1 + 4y 4 − 1 ± √ 1 + 4y 2 . = 2 + 4y ± √ 1 + 4y 4 − 2 ± 2 √ 1 + 4y 4 . = 2 + 4y ± √ 1 + 4y − 2 − ± √ 1 + 4y 4 . = 4y 4 . = y. So, since we have chosen y to be in the target and we have found some x in the domain that goes to our y, we have proven onto. Note: The key to proving onto is to take the rule of the function and work it backwards from the target to the domain. You have y = f(x) = ruled applied to x. If you solve for x, 2
it will give you the x you need to get to a y in the target. However, be careful that when you do this, the x value you get is still in the domain. Consider how this problem would have gone if f : N → R. Then when we found in the scratch work x = 1 ± √ 1 + 4y 2 , we would be outside the domain for all but a few values of y because 1 + 4y would have to be an odd perfect square or else you wouldn’t end up with a natural number for x. Proving 1-1 Proving 1-1 is much simpler than proving onto. First, think about what it means to be 1-1. It means for every value of x, there is only one associated y and for every y there is only one associated x. The first of these we get for free when we are told it’s a function because all functions pass the vertical line test. So we only need to prove that the function satisfies the horizontal line test. Example 2 Prove that f : N → Z given by the rule f(x) = −2x + 10 is one to one. Proof. Remember that we are proving an if, then statement. If f(a) = f(b), then a = b. First, declare variables. Let f(a), f(b) ∈ ran(f). Now suppose that f(a) = f(b). Now, f(a) = −2a + 10 = −2b + 10 = f(b). So −2a + 10 = −2b + 10. So −2a = −2b by adding 10 to both sides. So a = b by dividing both sides by -2. So f is 1-1. If you can prove BOTH of these for a function, then you have proved that the function is a bijection. Having a bijection is extremely valuable. If you can create a bijection from a set to the natural numbers, then that set is countable, which is handy information. From that you also know that all subsets of your set are countable. Be careful though. Just because you can create a bijection from one set to another does NOT mean that those sets are countable. Consider f : R → R defined by f(x) = x. This function is a bijection from the reals to the reals. However, the real numbers are not countable. This is because there is no possible bijection from the reals to the natural numbers. The key for countability is that the bijective function MUST go from a set to the natural numbers. 3
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# What can 30 15 be simplified?
## What can 30 15 be simplified?
Reduce 30/15 to lowest terms
• Find the GCD (or HCF) of numerator and denominator. GCD of 30 and 15 is 15.
• 30 ÷ 1515 ÷ 15.
• Reduced fraction: 21. Therefore, 30/15 simplified to lowest terms is 2/1.
### How do you simplify equivalent fractions?
Simplifying a fraction means finding an equivalent fraction where the numbers are reduced as much as possible. For example, if I was given the fraction 6/48 and asked to simplify it, I could divide both numerator and denominator by 2 to make 3/24, I could then divide both numbers by 3 to make 1/8.
#### What is 15 20 in simplest form as a fraction?
Therefore, 15/20 simplified to lowest terms is 3/4.
What is the simplest form of 5 30?
Therefore, 5/30 simplified to lowest terms is 1/6.
What is 15 as a fraction of 30?
Therefore, 15/30 simplified to lowest terms is 1/2.
## Can 96 15 be simplified?
Therefore, 96/15 simplified to lowest terms is 32/5.
### How do you solve equivalent fractions?
To find the equivalent fractions for any given fraction, multiply the numerator and the denominator by the same number. For example, to find an equivalent fraction of 3/4, multiply the numerator 3 and the denominator 4 by the same number, say, 2. Thus, 6/8 is an equivalent fraction of 3/4.
#### What is the equivalent fraction of 15 20?
2 equivalent fractions of 15/20=30/40 & 45/60.
What is the fraction in simplest form?
A fraction is said to be in simplest form if its numerator and denominator are relatively prime , that is, they have no common factors other than 1 . (Some books use “written in lowest terms” to mean the same thing.) So, 59 is in simplest form, since 5 and 9 have no common factors other than 1 .
What is 5 30 as a decimal?
5/30 as a decimal is 0.16666666666667.
## How to calculate the simplified fraction of 15 / 30?
We do this by first finding the greatest common factor of 15 and 30, which is 15. Then, we divide both 15 and 30 by the greatest common factor to get the following simplified fraction: 1/2 Therefore, this equation is true: 15/30 = 1/2 If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction.
### Which is the numerator in the fraction 15 / 30?
In the fraction 15/30, 15 is the numerator and 30 is the denominator. When you ask “What is 15/30 simplified?”, we assume you want to know how to simplify the numerator and denominator to their smallest values, while still keeping the same value of the fraction.
#### How to reduce 15 / 30 to its simplest form?
GCD of 15 and 30 is 15. Divide both the numerator and denominator by the GCD. 15 ÷ 15. /. 30 ÷ 15. Reduced fraction: 1. /. 2. Therefore, 15/30 simplified to lowest terms is 1/2.
Which is the greatest factor of 15 and 30?
We do this by first finding the greatest common factor of 15 and 30, which is 15. If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. In that case, you could convert it into a whole number or mixed number fraction.
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# How to Calculate Percentage – Change and Difference
Knowing how to compute the level of a number is a major activity for some parts of life. For instance, you might have to know how to compute rate to gauge vehicle installments or decide the initial investment for a home.
Rate estimations are additionally significant in business and different expert settings, for example, while computing assessments or worker raises. In this article, we investigate what a rate is, the way to compute various parts of a rate and the kinds of rates.
Also you read How to Calculate Percentage – Change and Difference
## What is percentage?
Rate, which may likewise be alluded to as percent, is a negligible part of a number out of 100 percent. Rate signifies “per 100” and indicates a piece of an aggregate sum. For instance, 45% addresses 45 out of 100, or 45% of the aggregate sum.
Rate may likewise be alluded to as “out of 100” or “for each 100.” For instance, you could say possibly “it snowed 20 days out of at regular intervals” or “it snowed 20% of the time.”
A rate might be written in more than one way. One way is to depict it as a decimal. For instance, 24% could likewise be composed as .24. You can track down the decimal rendition of a percent by isolating the rate by 100.
## How to calculate a percentage
The following formula is a common strategy to calculate a percentage:
### 1. Determine the total amount of what you want to find a percentage
For example, if you want to calculate the percentage of days it rained in a month, you would use the number of days in that month as the total amount. So, let’s say we are evaluating the amount of rain during the month of April’s 30 days.
### 2. Divide the number to determine the percentage
Using the example above, let’s say that it rained 15 of the 30 days in April. You would divide 15 by 30, which equals 0.5.
### 3. Multiply the value by 100
Continuing with the above example, you would multiply 0.5 by 100. This equals 50, which would give you the answer of 50%. So, in April, it rained 50% of the time.
Related: What is Aptitude?
## Types of percentage problems
There are three main types of percentage problems you might encounter in both personal and professional settings. These include:
### 1. Finding the ending number
Coming up next is an illustration of an inquiry that would expect you to utilize a rate estimation to find the completion number in an issue: “What is half of 25?” For this issue, you as of now have both the rate and the entire sum that you need to track down a level of.
Since you as of now have the rate, you will duplicate the rate by the entire number. For this situation, you would duplicate half, or 0.5, by 25. This offers you a response of 12.5. Hence, the solution to this rate issue would be “12.5 is half of 25.”
### 2. Finding the percentage
If you need to find the percentage, a question may be posed as “What percent of 5 is 2?” In this example, you will need to determine in a percentage how much of 2 is part of the whole of 5. For this type of problem, you can simply divide the number that you want to turn into a percentage by the whole. So, using this example, you would divide 2 by 5. This equation would give you 0.4. You would then multiply 0.4 by 100 to get 40, or 40%. Thus, 2 is equal to 40% of 5.
### 3. Finding the starting number
A percentage problem that asks you to find the starting number may look like “45% of what is 2?” This is typically a more difficult equation but can easily be solved using the previously mentioned formula. For this type of percentage problem, divide the whole by the percentage given. Using the example, you divide 2 by 45% or .45. This would give you 4.4, which means that 2 is 45% of 4.4.
## How to calculate percentage change
A rate change is a numerical worth that means the level of progress over the long run. It is most often utilized in money to decide the adjustment of the cost of a security over the long run. This equation can be applied to any number that is being estimated over the long run.
A rate change is equivalent to the adjustment of a given worth. You can settle a rate change by isolating the entire worth by the first worth and afterward increasing it by 100. The recipe for settling a rate change is the accompanying:
1. At a cost or rate increment:
2. [(New Price – Old Price)/Old Price] x 100
3. At a cost or rate decline:
4. [(Old Price – New Price)/Old Price] x 100
5. Here is an illustration of a cost/rate increment:
A TV cost \$100 last year however presently costs \$125. To decide the cost increment, you would take away the old cost from the new cost: 125 – 100 = 25. You would then isolate this by the old cost: 25 separated by 100 equivalents 0.25. You will then duplicate this number by 100: 0.25 x 100 = 25, or 25%. In this way, the TV cost has expanded by 25% over the course of the last year.
An illustration of a cost/rate decline:
A TV cost \$100 last year however presently costs just \$75. To decide the cost decline, you would take away the new cost from the old cost: 100 – 75 = 25. You will then isolate this number by the old cost: 25 separated by 100 equivalents 0.25. You would then duplicate this by 100: 0.25 x 100 = 25. or then again 25%. This implies the TV costs 25% short of what it did in the earlier year.
## How to calculate percentage difference
You can utilize rates to look at two changed things that are connected with one another. For instance, you might need to decide how much an item cost last year versus how much a comparable item costs this year. This estimation would give you the percent contrast between the two item costs.
Coming up next is the equation used to compute a rate contrast:
|V1 – V2|/[(V1 + V2)/2] × 100
In this equation, V1 is equivalent to the expense of one item, and V2 is equivalent to the expense of the other item.
An instance of utilizing this equation to decide the distinction between item expenses would include:
An item cost \$25 last year and a comparable item costs \$30 this year. To decide the rate contrast, you would initially deduct the expenses from one another: 30 – 25 = 5. You would then decide the normal of these two expenses (25 + 30/2 = 27.5). You will then isolate 5 by 27.5 = 0.18. You will then duplicate 0.18 by 100 = 18. This implies that the expense of the item this year is 18% more than the expense of the item from the year before.
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# Transition to Advanced Mathematics
#### Math 321
Textbook: Numbers, groups, and codes (Humphreys and Prest), 2nd edition
Times: TuTh 12-1:15
Each homework set is worth the same amount. The top 8 out of 10 homeworks count 2.5% of the course each for a total of 20% for homework. The best out of two midterms counts 40%. The final counts 40%.
HW1: 1.1: 3, 6 (Greatest common divisors); 1.2: 3, 7 (Induction)
HW2: 1.3: 6, 7, 8 (Unique factorization); 1.4: 5, 6, 7 (Congruence classes)
HW3: 1.5: 1, 4 (Solving linear congruences); 1.6: 1, 4, 7, 10, 12 (Cryptography)
HW4: 2.1: 3, 7, 8; 2.2: 7, 8, 9, 11
HW5: 2.3: 2, 3, 5, 10; 2.4: 2, 4
HW6: 3.1: 2, 4; 3.2: 1; 3.3: 3
HW7: 4.1: 2, 5; 4.2: 3, 4, 5, 8, 12
HW8: 4.3: 3, 4, 7; 4.4: 3, 4, 5, 6, 8, 14
HW9: 5.1: 1, 2, 6, 7; 5.2: 1, 3
HW10: 5.3: 2, 3, 5, 8; 5.4: 2, 3, 5
Note on 5.4: The syndrome of $w=(x|y)$ ($x$ followed by $y$) is $xA+y$ (whereas juxtaposition denotes matrix multiplication). If $z$ is a coset leader with $z=u|v$ and the same syndrome $uA+v=xA+y$, then $w+z=(x|y)+(u|v) = (x+u)|(y+v) = (x+u)|((x+u)A)$, so $w+z$ is a codeword. Moreover since the coset leaders are low weight (sparse), $w+z$ is close to $w$.
# Statistical Inference
#### Math 472
Textbook: An introduction to mathematical statistics and its applications (Larsen and Marx), 5th edition
Times: TuTh 9-10:15
Homework list
HW1: 3.8: 1, 2, 6 (Transforming and combining random variables); 3.10: 8, 12, 14 (Order statistics)
HW2: 4.6: 1, 4, 5 (Gamma distribution); 5.2: 2, 10, 14 (Method of maximum likelihood)
HW3: 5.3: 1, 13, 22 (Interval estimation); 5.4: 14, 16, 17
HW4: 5.5: 2, 3, 7; 5.6: 1, 5, 6
HW5: 5.7: 3, 4, 5; 5.8: 3, 5, 7
HW6: 6.2: 5, 7, 9; 6.3: 3, 4, 8
HW7: 6.4: 4, 6, 9; 6.5: 2, 4, 6
HW8: 7.3: 4, 6, 8; 7.4: 1, 3, 25
HW9: 9.2: 1, 5, 16
HW10: 10.2: 2, 4, 9; 10.3: 6, 8, 9
HW11: 11.2: 10, 22, 29; 11.3: 10, 18, 19
HW12: 11.4: 1, 5, 14; 12.2: 5, 6, 8
Each homework set is worth the same amount. The top 10 of 12 homeworks count 2% of the course each. The best out of two midterms counts 40%. The final counts 40%.
# Fall 2011 Teaching
#### Math 244
HW1: 14.1: 2,4,15,19,23,25
HW2: 14.2: 3,12,15, 20,23,33, 52, 54,55,58
HW3: 14.3: 8,9,14,18 and
14.4:3,12,13,25,30,32,36
HW4: 14.5: 13, 16, 18, 22, 23, 24, 30, 40, 41, 43
HW5: 14.6: 2, 8, 10, 13, 14, 17, 18, 20
HW6: 14.7: 4, 8, 12, 35, 45, 53, 56, 60, 62, 76 and
14.8: 4, 13, 15, 20, 21
HW7: 15.1: 2, 8, 18, 19, 28, 31
15.2: 2, 8, 13, 21, 26, 34, 39, 43, 45
HW8: 15.3: 7, 16, 18, 20, 25, 33, 34, 36
HW9: 15.4: 1, 3, 9, 20, 22, 30, 33, 37
HW10: 15.5: 1, 8, 12, 19, 22, 25, 28, 36, 41, 42
HW11: 15.6: 11, 13, 14, 18, 21, 24, 37, 38, 42
HW12: 15.7: 2, 4, 7, 22, 25 and
15.8: 2, 8, 9, 21, 23, 30
Each homework set is worth the same amount. The top 10 out of 12 homeworks count 1.5% of the course each for a total of 15% for homework. The two best out of three midterms count 20% each. The final counts 45%.
Math 244 Course calendar
#### Math 471
Midterm III
HW1: Chapter 1:
2,3,4,13,15,19,23,25,26,35,37,40,43,46
HW2: Chapter 1: 51, 56 and Chapter 2:
8,9,18,21,23,30,33,35,40,41,44,45,49,56,62,67,68,70,71,79,80,85
HW3: Chapter 3:
11,12,13,14,18,19,21,24,29,37,38,42,43,44,46,50,59,60,66
HW4: Chapter 5: 1,3,7,9,20,22,23
HW5: Chapter 5: 28, 33, 36, 39
HW6: Chapter 6: 2,6,8,11,12,19,21,25,28,30,31,39,42
Each homework set is worth the same amount. The top 5 homeworks count 2% of the course each. The two best out of three midterms count 25% each. The final counts 45%.
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Mathematics 29 Online
OpenStudy (anonymous):
A nursery is selling one kind of grass mixture for $0.75/lbs while another kind is going for$1.10/lbs. How much of each kind would need to be mixed to produce 50 lbs mixture of seeds that will sell for 0.90/lbs. Thanks.
OpenStudy (amistre64):
seed @ .75 = 50(.90 - 1.10)/(.75 - 1.10), i believe
OpenStudy (amistre64):
200/7 perhaps at .75 ; 28' 4/7
OpenStudy (amistre64):
we can either move the numbers about or simply determine whats left: + 3/7 and add 21 more to get 50; 21' 3/7 28' 4/7 21' 3/7 ------- 49' 7/7
OpenStudy (anonymous):
You must express what you know in terms of equations. Let $x _{1}$ be the amount of seed 1 in pounds and $x _{2}$ be the amount of seed 2 in pounds. Then the total cost of the mixture will be: $0.75x _{1}+1.10x _{2} = 0.9\left( x_{1} + x_{2} \right)$ And the total weight of the mixture will be: $x_{1} + x_{2} = 50$ Substitute equation 2 in equation 1 to get (remember, this equation is about the cost of the mix): $0.75x _{1}+1.10x _{2} = 0.9*50 = 45$ Solve equation 1 for one of the variables $x_{1} = \frac{45 - 1.10x_{2}}{0.75} = 60 - 1.466^{-} x_{2}$ Substitute this proportion back in equation 2: $60 - 1.466^{-} x_{2} + x_2 = 50$ $0.466^{-} x_{2} = 10$ $x_{2} = \frac{10}{0.466^{-}} = 21.428571428571428571428571428571$ Substitute this value back in equation 2 to find the missing value: $x_{1} + 21.428571428571428571428571428571 = 50$ $x_{1} = 28.571428571428571428571428571429$
OpenStudy (amistre64):
a +b = t ; total the amount needed ax +by = tz ; amounts * prices equals total amount * needed price then eliminate the one your not looking for. a +b = t <-- * -x ax +by = tz -ax -bx = -tx ax +by = tz ------------ b(y-x) = t(z-x) ; and solve for "b" b@y = t(z-x)/(y-x)
OpenStudy (amistre64):
seed@.75 = 50 (.90 - 1.10) / (.75 - 1.10) seed@ 1.10 = 50 (.90 - .75) / (1.10 - .75)
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dark
# Unit 10 Circles Homework 1 Answer Key
Unit 10 Circles Homework 1 Answer Key: Providing comprehensive solutions for circle-related exercises.
## Introduction
Unit 10 Circles Homework 1 is an important assignment for students studying circles in mathematics. This article aims to provide a detailed answer key to help students understand the concepts and solve the problems effectively. The comprehensive solutions presented here will guide students through the various exercise questions, allowing them to check their answers and improve their understanding of circle-related concepts.
## Unit 10 Circles Homework 1: Detailed Answer Key
1. Question: Find the circumference of a circle with a radius of 7 cm.
Solution: The formula for finding the circumference of a circle is C = 2πr, where r represents the radius of the circle. Substituting the given radius value, we get C = 2π(7) = 14π cm. Therefore, the circumference of the circle is 14π cm.
1. Question: Determine the area of a circle with a diameter of 10 units.
Solution: The formula for finding the area of a circle is A = πr², where r represents the radius of the circle. As the diameter is double the radius, we can find the radius by dividing the diameter by 2. In this case, the radius is 10/2 = 5 units. Substituting the value of the radius, we get A = π(5)² = 25π square units. Hence, the area of the circle is 25π square units.
1. Question: A circular field has a circumference of 36 meters. Find the radius of the field.
Solution: The formula for finding the circumference of a circle is C = 2πr. Rearranging the formula to solve for the radius, we have r = C/(2π). Substituting the given circumference value, we get r = 36/(2π) = 18/π meters. Therefore, the radius of the field is 18/π meters.
### Conclusion
Unit 10 Circles Homework 1 can be challenging, but with the detailed answer key provided above, students can easily comprehend the concepts related to circles and solve the problems accurately. By understanding the formulas and techniques used in finding the circumference and area of circles, students will not only excel in their homework but also develop a strong foundation in geometry. Practice and repetition are key to mastering these concepts, so students are encouraged to attempt similar exercises to reinforce their knowledge.
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## Skill Building Equations Answer Key
Skill-Building Equations: Unlocking the Answer Key Mastering the art of equations significantly helps in building mathematical competence. Not…
## Maxwell Boltzmann Distribution POGIL Answer Key
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# 2019 AMC 12A Problems/Problem 17
## Problem
Let $s_k$ denote the sum of the $\textit{k}$th powers of the roots of the polynomial $x^3-5x^2+8x-13$. In particular, $s_0=3$, $s_1=5$, and $s_2=9$. Let $a$, $b$, and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$, $3$, $....$ What is $a+b+c$?
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$
## Solution 1
Applying Newton's Sums (see this link), we have$$s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,$$so$$s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},$$we get the answer as $5+(-8)+13=10$.
## Solution 2
Let $p, q$, and $r$ be the roots of the polynomial. Then,
$p^3 - 5p^2 + 8p - 13 = 0$
$q^3 - 5q^2 + 8q - 13 = 0$
$r^3 - 5r^2 + 8r - 13 = 0$
Adding these three equations, we get
$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$
$s_3 - 5s_2 + 8s_1 = 39$
$39$ can be written as $13s_0$, giving
$s_3 = 5s_2 - 8s_1 + 13s_0$
We are given that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ is satisfied for $k = 2$, $3$, $....$, meaning it must be satisfied when $k = 2$, giving us $s_3 = a \, s_2 + b \, s_1 + c \, s_0$.
Therefore, $a = 5, b = -8$, and $c = 13$ by matching coefficients.
$5 - 8 + 13 = \boxed{\textbf{(D) } 10}$.
## Solution 3
Let $p, q$, and $r$ be the roots of the polynomial. By Vieta's Formulae, we have
$p+q+r = 5$
$pq+qr+rp = 8$
$pqr=13$.
We know $s_k = p^k + q^k + r^k$. Consider $(p+q+r)(s_k) =5s_k$.
$5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k$
Using $pqr = 13$ and $s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}$, we see $13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}$.
We have $$\begin{split} 5s_k + 13s_{k-2} &= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\ &= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\ &= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\ &= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}$$
Rearrange to get $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}$
So, $a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D) } 10}$.
## Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI
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# Fractions Game
Fraction Flip It - A fractions game for multiplying fractions.....
We think you'll enjoy teaching "how to multiply fractions" with this game as much as the kids will enjoy playing!
Fraction Flip It
Skills: Multiplying and or Dividing Fractions
Number of Players: 3 - 5
What You Need: Deck of playing cards, Fraction Flip It game sheet for each player, paper and pencil for each player,
Preparation: Remove the face cards from the deck. Print out a copy of the game sheet for each player.
Object of Game: To see who can earn the most points by correctly multiplying fractions.
How To Play:
1. Shuffle the cards and stack them face down.
2. Player 1 draws the top card and places it on the game sheet.
3. Players continue drawing cards and placing them until they have filled
all four spaces.....
Take a look at an example below....
Each player takes turns drawing one card at a time. After each card they draw, they place the card in one of the four empty rectangles on their game sheet. Each player continues drawing in turn until all players have four cards placed in all four spaces on their game sheets.
In this example player 1 has placed his cards on his game sheet in the spots pictured below. Player 1 will then write their multiplication statement on their paper and solve it. The correct answer that player 1 should write down on his paper for this round would be 8/15 (eight-fifteenths).
Each player multiplies their two fractions on their game sheets and simplifies the fraction if possible. Be sure to read through the directions for the rest of the details of playing the game.
Go to main Fraction Games page
More Fraction Fun
Fun activity that helps students understand fractions as parts of a whole.
Fraction Rectangles:
Multiply and Divide Fractions with Rectangles.
Fraction Strips:
Fraction Strips help students visualize fractions.
Fraction Fiasco:
Fraction game teaching adding, subtracting with like denominators.
Fraction Flip It:
A Fun Fraction Multiplication Game
Fun fraction puzzles to get students to work their fraction brain muscles!
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# Boolean theorems
The boolean theorems are a set of rules from boolean algebra to simplify logic expressions of combinational circuits.
Boolean algebra is a fundamental mathematical tool for digital computing, where variables and functions have only binary values of “0” or “1”. It was introduced in 1854 by the British mathematician George Boole. In this algebra, the add signal (+) represents the logical operation OR and the multiplier signal ($\times$ ou $\cdot$) represents the logical operation AND.
## Boolean theorems with one variable
Every variable is represented by a letter; this variable’s negation is usually represented by a bar above the letter. For example, $\overline{A}$ is the negation of $A$, therefore, $\overline{A}$ will always have the opposite binary value of $A$. This is the list of theorems with only one variable.
• $A+0=A$
• $A+1=1$
• $A+A=A$
• $A+\overline{A}=1$
• $A\cdot 0=0$
• $A\cdot 1=A$
• $A\cdot A=A$
• $A\cdot\overline{A}=0$
## Boolean theorems with two or three variables
• $A+B=B+A$
• $A\cdot B=B\cdot A$
• $A+(B+C)=(A+B)+C=A+B+C$
• $A(BC)=(AB)C=ABC$
• $A(B+C)=AB+AC$
• $(A+B)(C+D)=AC+AD+BC+BD$
• $A+AB=A$
• $A+\overline{A}B=A+B$
• $\overline{A}+AB=\overline{A}+B$
## De Morgan theorems
These are very useful in manipulation of logical expressions.
$\overline{(A+B)}=\overline{A}\cdot \overline{B}$
$\overline{(A\cdot B)}=\overline{A}+\overline{B}$
Proving De Morgan theorems with a truth table and considering all possibilities.
$\overline{(A+B)}=\overline{A}\cdot \overline{B}$
$\overline{(A\cdot B)}=\overline{A}+\overline{B}$
## For what serves the boolean theorems?
With these theorems, combinational circuits can be simplified to have only one type of logic door such as NAND or NOR, since commercial integrated circuits with logic door uses only a type of door.
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# Show that $A$ and $A^{-1}$ have same eigenvalues?
If $A$ is a square matrix of order $2$, and determinant of $A$ is $1$, then prove that $A$ and its inverse have the same eigenvalues.
So, let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$. Since determinant is $1$, it means $\lambda_1\,\lambda_2 = 1$. What do I do after this to reach the conclusion?
• If $\lambda$ is an eigenvalue of $A$, then $Ax = \lambda x$, so... – ua11 Feb 16 '14 at 15:19
Hint: note that if $\lambda_1$ is an eigenvalue of $A$, then $1/\lambda_1$ is an eigenvalue of $A^{-1}$ (why?).
• Let $\lambda_1$ be a eigenvalue of$A$, than $Ax=\lambda_1x$. This is equivalent with $x=A^{-1}\lambda_1x$ or $\lambda_1^{-1}x=A^{-1}x$. The last equality tells us that $\lambda_1^{-1}$ is an eigenvalue for $A^{-1}$. – Emin Feb 16 '14 at 15:27
If $A$ is 2 by 2 and has determinant $1$, then its eigenvalues are $\lambda$ and $\frac{1}{\lambda}$. If you invert $A$, the $\lambda$ eigenvalue maps to $\frac{1}{\lambda}$, and the $\frac{1}{\lambda}$ eigenvalue maps to $\frac{1}{\frac{1}{\lambda}} = \lambda$. Thus, they have the same eigenvalues.
This follows from $A x = \lambda x \iff \frac{1}{\lambda} A x = x \iff \frac{1}{\lambda}x = A^{-1} x \iff A^{-1} x = \frac{1}{\lambda} x$ for invertible $A$.
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# 7.5: Systems of Linear Inequalities
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Graph linear inequalities in two variables.
• Solve systems of linear inequalities.
• Solve optimization problems.
## Introduction
In the last chapter you learned how to graph a linear inequality in two variables. To do that, you graphed the equation of the straight line on the coordinate plane. The line was solid for \begin{align*}\le\end{align*} or \begin{align*}\ge\end{align*} signs (where the equals sign is included), and the line was dashed for < or > signs (where the equals sign is not included). Then you shaded above the line (if the inequality began with y>\begin{align*}y >\end{align*} or y\begin{align*}y \ge\end{align*}) or below the line (if it began with y<\begin{align*}y < \end{align*} or y\begin{align*}y \le\end{align*}).
In this section, we’ll see how to graph two or more linear inequalities on the same coordinate plane. The inequalities are graphed separately on the same graph, and the solution for the system is the common shaded region between all the inequalities in the system. One linear inequality in two variables divides the plane into two half-planes. A system of two or more linear inequalities can divide the plane into more complex shapes.
Let’s start by solving a system of two inequalities.
## Graph a System of Two Linear Inequalities
Example 1
Solve the following system:
2x+3yx4y1812\begin{align*}2x + 3y & \le 18\\ x - 4y & \le 12\end{align*}
Solution
Solving systems of linear inequalities means graphing and finding the intersections. So we graph each inequality, and then find the intersection regions of the solution.
First, let’s rewrite each equation in slope-intercept form. (Remember that this form makes it easier to tell which region of the coordinate plane to shade.) Our system becomes
3y2x+18y23x+64yx+12yx43\begin{align*}3y \le -2x + 18 \qquad \qquad \qquad \qquad y \le - \frac{2}{3}x + 6\\ {\;} \qquad \qquad \qquad \qquad \quad \Rightarrow\!\\ -4y \le -x + 12 \qquad \qquad \qquad \qquad y \ge \frac{x}{4} - 3\end{align*}
Notice that the inequality sign in the second equation changed because we divided by a negative number!
For this first example, we’ll graph each inequality separately and then combine the results.
Here’s the graph of the first inequality:
The line is solid because the equals sign is included in the inequality. Since the inequality is less than or equal to, we shade below the line.
And here’s the graph of the second inequality:
The line is solid again because the equals sign is included in the inequality. We now shade above the line because \begin{align*}y\end{align*} is greater than or equal to.
When we combine the graphs, we see that the blue and red shaded regions overlap. The area where they overlap is the area where both inequalities are true. Thus that area (shown below in purple) is the solution of the system.
The kind of solution displayed in this example is called unbounded, because it continues forever in at least one direction (in this case, forever upward and to the left).
Example 2
There are also situations where a system of inequalities has no solution. For example, let’s solve this system.
\begin{align*}y & \le 2x - 4\\ y & > 2x + 6\end{align*}
Solution
We start by graphing the first line. The line will be solid because the equals sign is included in the inequality. We must shade downwards because \begin{align*}y\end{align*} is less than.
Next we graph the second line on the same coordinate axis. This line will be dashed because the equals sign is not included in the inequality. We must shade upward because \begin{align*}y\end{align*} is greater than.
It doesn’t look like the two shaded regions overlap at all. The two lines have the same slope, so we know they are parallel; that means that the regions indeed won’t ever overlap since the lines won’t ever cross. So this system of inequalities has no solution.
But a system of inequalities can sometimes have a solution even if the lines are parallel. For example, what happens if we swap the directions of the inequality signs in the system we just graphed?
To graph the system
\begin{align*}y \ge 2x - 4\!\\ y < 2x + 6\end{align*},
we draw the same lines we drew for the previous system, but we shade upward for the first inequality and downward for the second inequality. Here is the result:
You can see that this time the shaded regions overlap. The area between the two lines is the solution to the system.
## Graph a System of More Than Two Linear Inequalities
When we solve a system of just two linear inequalities, the solution is always an unbounded region—one that continues infinitely in at least one direction. But if we put together a system of more than two inequalities, sometimes we can get a solution that is bounded—a finite region with three or more sides.
Let’s look at a simple example.
Example 3
Find the solution to the following system of inequalities.
\begin{align*}3x - y & < 4\\ 4y + 9x & < 8\\ x & \ge 0\\ y & \ge 0\end{align*}
Solution
Let’s start by writing our inequalities in slope-intercept form.
\begin{align*}y & > 3x - 4\\ y & < - \frac{9}{4}x + 2\\ x & \ge 0\\ y & \ge 0\end{align*}
Now we can graph each line and shade appropriately. First we graph \begin{align*}y > 3x - 4\end{align*} :
Next we graph \begin{align*}y < - \frac{9}{4}x + 2\end{align*} :
Finally we graph \begin{align*}x \ge 0\end{align*} and \begin{align*}y \ge 0\end{align*}, and we’re left with the region below; this is where all four inequalities overlap.
The solution is bounded because there are lines on all sides of the solution region. In other words, the solution region is a bounded geometric figure, in this case a triangle.
Notice, too, that only three of the lines we graphed actually form the boundaries of the region. Sometimes when we graph multiple inequalities, it turns out that some of them don’t affect the overall solution; in this case, the solution would be the same even if we’d left out the inequality \begin{align*}y > 3x - 4\end{align*}. That’s because the solution region of the system formed by the other three inequalities is completely contained within the solution region of that fourth inequality; in other words, any solution to the other three inequalities is automatically a solution to that one too, so adding that inequality doesn’t narrow down the solution set at all.
But that wasn’t obvious until we actually drew the graph!
## Solve Real-World Problems Using Systems of Linear Inequalities
A lot of interesting real-world problems can be solved with systems of linear inequalities.
For example, you go to your favorite restaurant and you want to be served by your best friend who happens to work there. However, your friend only waits tables in a certain region of the restaurant. The restaurant is also known for its great views, so you want to sit in a certain area of the restaurant that offers a good view. Solving a system of linear inequalities will allow you to find the area in the restaurant where you can sit to get the best view and be served by your friend.
Often, systems of linear inequalities deal with problems where you are trying to find the best possible situation given a set of constraints. Most of these application problems fall in a category called linear programming problems.
Linear programming is the process of taking various linear inequalities relating to some situation, and finding the best possible value under those conditions. A typical example would be taking the limitations of materials and labor at a factory, then determining the best production levels for maximal profits under those conditions. These kinds of problems are used every day in the organization and allocation of resources. These real-life systems can have dozens or hundreds of variables, or more. In this section, we’ll only work with the simple two-variable linear case.
The general process is to:
• Graph the inequalities (called constraints) to form a bounded area on the coordinate plane (called the feasibility region).
• Figure out the coordinates of the corners (or vertices) of this feasibility region by solving the system of equations that applies to each of the intersection points.
• Test these corner points in the formula (called the optimization equation) for which you're trying to find the maximum or minimum value.
Example 4
If \begin{align*}z = 2x + 5y\end{align*}, find the maximum and minimum values of \begin{align*}z\end{align*} given these constraints:
\begin{align*}2x - y & \le 12\\ 4x + 3y & \ge 0\\ x - y & \le 6\end{align*}
Solution
First, we need to find the solution to this system of linear inequalities by graphing and shading appropriately. To graph the inequalities, we rewrite them in slope-intercept form:
\begin{align*}y & \ge 2x - 12\\ y & \ge - \frac{4}{3}x\\ y & \ge x - 6\end{align*}
These three linear inequalities are called the constraints, and here is their graph:
The shaded region in the graph is called the feasibility region. All possible solutions to the system occur in that region; now we must try to find the maximum and minimum values of the variable \begin{align*}z\end{align*} within that region. In other words, which values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} within the feasibility region will give us the greatest and smallest overall values for the expression \begin{align*}2x + 5y\end{align*}?
Fortunately, we don’t have to test every point in the region to find that out. It just so happens that the minimum or maximum value of the optimization equation in a linear system like this will always be found at one of the vertices (the corners) of the feasibility region; we just have to figure out which vertices. So for each vertex—each point where two of the lines on the graph cross—we need to solve the system of just those two equations, and then find the value of \begin{align*}z\end{align*} at that point.
The first system consists of the equations \begin{align*}y = 2x - 12\end{align*} and \begin{align*}y = - \frac{4}{3}x\end{align*}. We can solve this system by substitution:
\begin{align*}- \frac{4}{3}x &= 2x - 12 \Rightarrow -4x = 6x - 36 \Rightarrow -10x = -36 \Rightarrow x = 3.6\\ y &= 2x - 12 \Rightarrow y = 2(3.6) - 12 \Rightarrow y = - 4.8\end{align*}
The lines intersect at the point (3.6, -4.8).
The second system consists of the equations \begin{align*}y = 2x - 12\end{align*} and \begin{align*}y = x - 6\end{align*}. Solving this system by substitution:
\begin{align*}x - 6 &= 2x - 12 \Rightarrow 6 = x \Rightarrow x = 6\\ y &= x - 6 \Rightarrow y = 6 - 6 \Rightarrow y = 6\end{align*}
The lines intersect at the point (6, 6).
The third system consists of the equations \begin{align*}y = - \frac{4}{3}x\end{align*} and \begin{align*}y = x - 6\end{align*}. Solving this system by substitution:
\begin{align*}x - 6 &= - \frac{4}{3}x \Rightarrow 3x - 18 = -4x \Rightarrow 7x = 18 \Rightarrow x = 2.57\\ y &= x - 6 \Rightarrow y = 2.57 - 6 \Rightarrow y = -3.43\end{align*}
The lines intersect at the point (2.57, -3.43).
So now we have three different points that might give us the maximum and minimum values for \begin{align*}z\end{align*}. To find out which ones actually do give the maximum and minimum values, we can plug the points into the optimization equation \begin{align*}z = 2x + 5y\end{align*}.
When we plug in (3.6, -4.8), we get \begin{align*}z = 2(3.6) +5(-4.8) = -16.8\end{align*}.
When we plug in (6, 0), we get \begin{align*}z = 2(6) + 5(0) = 12\end{align*}.
When we plug in (2.57, -3.43), we get \begin{align*}z = 2(2.57) + 5(-3.43) = -12.01\end{align*}.
So we can see that the point (6, 0) gives us the maximum possible value for \begin{align*}z\end{align*} and the point (3.6, –4.8) gives us the minimum value.
In the previous example, we learned how to apply the method of linear programming in the abstract. In the next example, we’ll look at a real-life application.
Example 5
You have $10,000 to invest, and three different funds to choose from. The municipal bond fund has a 5% return, the local bank's CDs have a 7% return, and a high-risk account has an expected 10% return. To minimize risk, you decide not to invest any more than$1,000 in the high-risk account. For tax reasons, you need to invest at least three times as much in the municipal bonds as in the bank CDs. What’s the best way to distribute your money given these constraints?
Solution
Let’s define our variables:
\begin{align*}x\end{align*} is the amount of money invested in the municipal bond at 5% return
\begin{align*}y\end{align*} is the amount of money invested in the bank’s CD at 7% return
\begin{align*}10000 - x - y\end{align*} is the amount of money invested in the high-risk account at 10% return
\begin{align*}z\end{align*} is the total interest returned from all the investments, so \begin{align*}z = .05x + .07y + .1(10000 - x - y)\end{align*} or \begin{align*}z = 1000 - 0.05x - 0.03y\end{align*}. This is the amount that we are trying to maximize. Our goal is to find the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} that maximizes the value of \begin{align*}z\end{align*}.
Now, let’s write inequalities for the constraints:
You decide not to invest more than 1000 in the high-risk account—that means: \begin{align*}10000 - x - y \le 1000\end{align*} You need to invest at least three times as much in the municipal bonds as in the bank CDs—that means: \begin{align*}3y \le x\end{align*} Also, you can’t invest less than zero dollars in each account, so: \begin{align*}x & \ge 0\\ y & \ge 0\\ 10000 - x - y & \ge 0\end{align*} To summarize, we must maximize the expression \begin{align*}z = 1000 - .05x - .03y\end{align*} using the constraints: \begin{align*}& 10000 - x - y \le 1000 && && y \ge 9000 - x\\ & 3y \le x && && y \le \frac{x}{3}\\ & x \ge 0 && \text{Or in slope-intercept form:} && x \ge 0\\ & y \ge 0 && && y \ge 0\\ & 10000 - x - y \ge 0 && && y \le 10000 - x\end{align*} Step 1: Find the solution region to the set of inequalities by graphing each line and shading appropriately. The following figure shows the overlapping region: The purple region is the feasibility region where all the possible solutions can occur. Step 2: Next we need to find the corner points of the feasibility region. Notice that there are four corners. To find their coordinates, we must pair up the relevant equations and solve each resulting system. System 1: \begin{align*}y = \frac{x}{3}\!\\ y = 10000 - x\end{align*} Substitute the first equation into the second equation: \begin{align*}\frac{x}{3} &= 10000 - x \Rightarrow x = 30000 - 3x \Rightarrow 4x = 30000 \Rightarrow x = 7500\\ y &= \frac{x}{3} \Rightarrow y = \frac{7500}{3} \Rightarrow y = 2500\end{align*} The intersection point is (7500, 2500). System 2: \begin{align*}y = \frac{x}{3}\!\\ y = 9000 - x\end{align*} Substitute the first equation into the second equation: \begin{align*}\frac{x}{3} &= 9000 - x \Rightarrow x = 27000 - 3x \Rightarrow 4x = 27000 \Rightarrow x = 6750\\ y &= \frac{x}{3} \Rightarrow y = \frac{6750}{3} \Rightarrow y = 2250\end{align*} The intersection point is (6750, 2250). System 3: \begin{align*}y = 0\!\\ y = 10000 - x\end{align*}. The intersection point is (10000, 0). System 4: \begin{align*}y = 0\!\\ y = 9000 - x\end{align*}. The intersection point is (9000, 0). Step 3: In order to find the maximum value for \begin{align*}z\end{align*}, we need to plug all the intersection points into the equation for \begin{align*}z\end{align*} and find which one yields the largest number. (7500, 2500): \begin{align*}z = 1000 - 0.05(7500) - 0.03(2500) = 550\end{align*} (6750, 2250): \begin{align*}z = 1000 - 0.05(6750) - 0.03(2250) = 595\end{align*} (10000, 0): \begin{align*}z = 1000 - 0.05(10000) - 0.03(0) = 500\end{align*} (9000, 0): \begin{align*}z = 1000 - 0.05(9000) - 0.03(0) = 550\end{align*} The maximum return on the investment of595 occurs at the point (6750, 2250). This means that:
$6,750 is invested in the municipal bonds.$2,250 is invested in the bank CDs.
1,000 is invested in the high-risk account. Graphing calculators can be very useful for problems that involve this many inequalities. The video at http://www.youtube.com/watch?v=__wAxkYmhvY shows a real-world linear programming problem worked through in detail on a graphing calculator, although the methods used there can also be used for pencil-and paper solving. ## Review Questions 1. Consider the system \begin{align*}y < 3x - 5\!\\ y > 3x - 5\end{align*}. Is it consistent or inconsistent? Why? 2. Consider the system \begin{align*}y \le 2x + 3\!\\ y \ge 2x + 3\end{align*}. Is it consistent or inconsistent? Why? 3. Consider the system \begin{align*}y \le -x + 1\!\\ y > -x + 1\end{align*}. Is it consistent or inconsistent? Why? 4. In example 3 in this lesson, we solved a system of four inequalities and saw that one of the inequalities, \begin{align*}y > 3x - 4\end{align*}, didn’t affect the solution set of the system. 1. What would happen if we changed that inequality to \begin{align*}y < 3x - 4\end{align*}? 2. What’s another inequality that we could add to the original system without changing it? Show how by sketching a graph of that inequality along with the rest of the system. 3. What’s another inequality that we could add to the original system to make it inconsistent? Show how by sketching a graph of that inequality along with the rest of the system. 5. Recall the compound inequalities in one variable that we worked with back in chapter 6. Compound inequalities with “and” are simply systems like the ones we are working with here, except with one variable instead of two. 1. Graph the inequality \begin{align*}x > 3\end{align*} in two dimensions. What’s another inequality that could be combined with it to make an inconsistent system? 2. Graph the inequality \begin{align*}x \le 4\end{align*} on a number line. What two-dimensional system would have a graph that looks just like this one? Find the solution region of the following systems of inequalities. 1. \begin{align*}x - y < -6\!\\ 2y \ge 3x + 17\end{align*} 2. \begin{align*}4y - 5x < 8\!\\ -5x \ge 16 - 8y\end{align*} 3. \begin{align*}5x - y \ge 5\!\\ 2y - x \ge -10\end{align*} 4. \begin{align*}5x + 2y \ge -25\!\\ 3x - 2y \le 17\!\\ x - 6y \ge 27\end{align*} 5. \begin{align*}2x - 3y \le 21\!\\ x + 4y \le 6\!\\ 3x + y \ge -4\end{align*} 6. \begin{align*}12x - 7y < 120\!\\ 7x - 8y \ge 36\!\\ 5x + y \ge 12\end{align*} Solve the following linear programming problems. 1. Given the following constraints, find the maximum and minimum values of \begin{align*}z = -x + 5y\end{align*}: \begin{align*}x + 3y \le 0\!\\ x - y \ge 0\!\\ 3x - 7y \le 16\end{align*} 2. Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice elves draw a wage of five candy canes per hour worked, but can only make four trucks an hour. Senior elves can make six trucks an hour and are paid eight candy canes per hour. There’s only room for nine elves in the truck shop, and due to a candy-makers’ strike, Santa Claus can only pay out 480 candy canes for the whole 8-hour shift. 1. How many senior elves and how many apprentice elves should work this shift to maximize the number of trucks that get made? 2. How many trucks will be made? 3. Just before the shift begins, the apprentice elves demand a wage increase; they insist on being paid seven candy canes an hour. Now how many apprentice elves and how many senior elves should Santa assign to this shift? 4. How many trucks will now get made, and how many candy canes will Santa have left over? 3. In Adrian’s Furniture Shop, Adrian assembles both bookcases and TV cabinets. Each type of furniture takes her about the same time to assemble. She figures she has time to make at most 18 pieces of furniture by this Saturday. The materials for each bookcase cost her20 and the materials for each TV stand costs her $45. She has$600 to spend on materials. Adrian makes a profit of $60 on each bookcase and a profit of$100 on each TV stand.
1. Set up a system of inequalities. What \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}values do you get for the point where Adrian’s profit is maximized? Does this solution make sense in the real world?
2. What two possible real-world \begin{align*}x-\end{align*}values and what two possible real-world \begin{align*}y-\end{align*}values would be closest to the values in that solution?
3. With two choices each for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, there are four possible combinations of \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}values. Of those four combinations, which ones actually fall within the feasibility region of the problem?
4. Which one of those feasible combinations seems like it would generate the most profit? Test out each one to confirm your guess. How much profit will Adrian make with that combination?
5. Based on Adrian’s previous sales figures, she doesn’t think she can sell more than 8 TV stands. Now how many of each piece of furniture should she make, and what will her profit be?
6. Suppose Adrian is confident she can sell all the furniture she can make, but she doesn’t have room to display more than 7 bookcases in her shop. Now how many of each piece of furniture should she make, and what will her profit be?
4. Here’s a “linear programming” problem on a line instead of a plane: Given the constraints \begin{align*}x \le 5\end{align*} and \begin{align*}x \ge -2\end{align*}, maximize the value of \begin{align*}y\end{align*} where \begin{align*}y = x + 3\end{align*}.
## Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9617.
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# Ncert Solutions For Class 8 Maths Ex 11.2
Class 8 Mathematics NCERT questions provided with appropriate solutions will come in handy to revise the Class 8 Maths CBSE syllabus. The textbook is filled with a wide range of exercises including hundreds of questions based on CBSE guidelines. NCERT Solutions for Class 8 Maths Chapter 11, Exercise 11.2 is a set of questions and answers for all the respective textbook exercise questions. All solutions are prepared by BYJU’S subject experts to help students ace the exam without fear. Download free NCERT Solutions for Maths Chapter 11- Mensuration and score well in your exams.
### Access Other Exercise Solutions of Class 8 Maths Chapter 11 Mensuration
Exercise 11.1 Solutions : 5 Questions (Long answers)
Exercise 11.3 Solutions : 10 Questions (2 Short answers, 8 Long answers)
Exercise 11.4 Solutions : 8 Questions (2 Short answers, 6 Long answers)
### Access Answers to NCERT Class 8 Maths Chapter 11 Mensuration Exercise 11.2 Page number 177
1.The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
Solution: One parallel side of the trapezium (a) = 1 m
And second side (b) = 1.2 m and
height (h) = 0.8 m
Area of top surface of the table= (½)×(a+b)h
= (½)×(1+1.2)0.8
= (½)×2.2×0.8 = 0.88
Area of top surface of the table is 0.88 m2 .
2. The area of a trapezium is 34 cm2and the length of one of the parallel sides is 10 cm and its height is 4 cm Find the length of the other parallel side.
Solution: Let the length of the other parallel side be b.
Length of one parallel side, a = 10 cm
height, (h) = 4 cm and
Area of a trapezium is 34 cm2
Formula for, Area of trapezium = (1/2)×(a+b)h
34 = ½(10+b)×4
34 = 2×(10+b)
After simplifying, b = 7
Hence another required parallel side is 7 cm.
3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
Solution:
Given: BC = 48 m, CD = 17 m,
AD = 40 m and perimeter = 120 m
∵ Perimeter of trapezium ABCD
= AB+BC+CD+DA
120 = AB+48+17+40
120 = AB = 105
AB = 120–105 = 15 m
Now, Area of the field= (½)×(BC+AD)×AB
= (½)×(48 +40)×15
= (½)×88×15
= 660
Hence, area of the field ABCD is 660m2 .
4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution:
Consider, h1 = 13 m, h2 = 8 m and AC = 24 m
Area of quadrilateral ABCD = Area of triangle ABC+Area of triangle ADC
= ½( bh1)+ ½(bh2)
= ½ ×b(h1+h2)= (½)×24×(13+8)
= (½)×24×21 = 252
Hence, required area of the field is 252 m2
5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Given: d1 = 7.5 cm and d2 = 12 cm
We know that,Area of rhombus = (½ )×d1×d2
= (½)×7.5×12 = 45
Therefore, area of rhombus is 45 cm2 .
6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Solution: Since a rhombus is also a kind of a parallelogram.
Formula for Area of rhombus = Base×Altitude
Putting values, we have
Area of rhombus = 6×4 = 24
Area of rhombus is 24 cm2
Also, Formula for Area of rhombus = (½)×d1d2
After substituting the values, we get
24 = (½)×8×d2
d2 = 6
Hence, the length of the other diagonal is 6 cm.
7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2is Rs. 4.
Solution: Length of one diagonal, d1 = 45 cm and d2= 30 cm
∵ Area of one tile = (½)d1d2
= (½)×45×30 = 675
Area of one tile is 675 cm2
Area of 3000 tiles is
= 675×3000 = 2025000 cm2
= 2025000/10000
= 202.50 m2 [∵ 1m2 = 10000 cm2]
∵ Cost of polishing the floor per sq. meter = 4
Cost of polishing the floor per 202.50 sq. meter = 4×202.50 = 810
Hence the total cost of polishing the floor is Rs. 810.
8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Solution:
Perpendicular distance (h) = 100 m (Given)
Area of the trapezium shaped field = 10500 m2 (Given)
Let side along the road be ‘x’ m and side along the river = 2x m
Area of the trapezium field = (½)×(a+b)×h
10500 = (½)×(x+2x)×100
10500 = 3x×50
After simplifying, we have x = 70, which means side along the river is 70 m
Hence, the side along the river = 2x = 2( 70) = 140 m.
9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Solution:
Octagon having eight equal sides, each 5 m. (given)
Divide the octagon as show in the below figure, 2 trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and 3rd one is rectangle having length and breadth 11 m and 5 m respectively.
Now, Area of two trapeziums = 2 [(½)×(a+b)×h]
= 2×(½)×(11+5 )×4
= 4×16 = 64
Area of two trapeziums is 64 m2
Also, Area of rectangle = length×breadth
= 11×5 = 55
Area of rectangle is 55 m2
Total area of octagon = 64+55
= 119 m2
10. There is a pentagonal shaped park as shown in the figure.
For finding its area Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution:
First way: By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
= (½)(AP+BC)×CP+(1/2)×(ED+AP)×DP
= (½)(30+15)×CP+(1/2)×(15+30)×DP
= (½)×(30+15)×(CP+DP)
= (½)×45×CD
= (1/2)×45×15
=337.5 m2
Area of pentagon is 337.5 m2
Second way: By Kavita’s diagram
Here, a perpendicular AM drawn to BE.
AM = 30–15 = 15 m
Area of pentagon = Area of triangle ABE+Area of square BCDE (from above figure)
= (½)×15×15+(15×15)
= 112.5+225.0
= 337.5
Hence, total area of pentagon shaped park = 337.5 m2
11. Diagram of the adjacent picture frame has outer dimensions = 24 cm×28 cm and inner dimensions 16 cm×20 cm. Find the area of each section of the frame, if the width of each section is same.
Solution:
Divide given figure into 4 parts, as shown below:
Here two of given figures (I) and (II) are similar in dimensions.
And also figures (III) and (IV) are similar in dimensions.
Area of figure (I) = Area of trapezium
= (½)×(a+b)×h
= (½)×(28+20)×4
= (½)×48×4 = 96
Area of figure (I) = 96 cm2
Also,Area of figure (II) = 96 cm2
Now, Area of figure (III) = Area of trapezium
= (½)×(a+b)×h
= (½)×(24+16)4
= (½)×40×4 = 80
Area of figure (III) is 80 cm2
Also, Area of figure (IV) = 80 cm2
Chapter 11 Mensuration, Exercise 11.2 is about area of trapezium, area of a general quadrilateral, area of a polygon and their real-life examples. Download and practise NCERT Class 8 Maths Solutions and improve your skills.
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### Ball Packing
If a ball is rolled into the corner of a room how far is its centre from the corner?
### Three Balls
A circle has centre O and angle POR = angle QOR. Construct tangents at P and Q meeting at T. Draw a circle with diameter OT. Do P and Q lie inside, or on, or outside this circle?
### In a Spin
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
# Conical Bottle
##### Age 14 to 16 Challenge Level:
Most people start out by calculating the volume of liquid. As with many mathematical tasks some thought in advance may save a lot of work . Failing that, if you review your method you may find a neater and more efficient way to do it. Try to evaluate your own work, think about it and ask yourself questions like: "what is the key issue here?", "does my answer suggest a connection in the problem I did not use?","have I done it the best way?" . Very often the best approach leads to a really pretty bit of maths.
The quick method is to look at the scale factor.
Consider the enlargement of the conical space above the liquid to the whole cone. The scale factor is $2$ (linearly) so the volume scale factor is $2^3=8$
The space above the liquid has an eighth of the volume of the whole cone and the liquid takes up seven eighths of the volume.
When it is inverted the volume of water is still seven eighths of the volume of the cone so we use the fact again that the cube of the linear scale factor gives the volume scale factor to get: $$(\frac{h}{x})^3 =\frac{7}{8}$$
and so $$h = { \frac{\sqrt[3] 7x}{2}}$$
That is, the height of the liquid in the upturned cone is $\frac {\sqrt[3] 7} {2}$ or 0.9565 of the original height
Other method is to equate the 2 results for the volume of the liquid.
No-one had any trouble in showing that the volume of liquid was $7 \frac{\pi r^2x}{24}$ (*) where $r$ was the base radius and x the height of the cone.
All used the properties of similar triangles to find the radius of the base of liquid in the upturned cone, which is (rh/x). Hence the volume of liquid in the upturned cone is $$\pi/3 \times (\frac {rh}{x})^2 \times h (**)$$
(*) and (**) were thus equated and the height of the liquid in the upturned cone was found, by cancelling, to be:$$h = { \frac{\sqrt[3] 7x}{2}}$$
or approximately $0.9565$ of the original height.
Easy to follow solutions to this problem were received from: Sam, Jonathan and Kevin, Tom, Euan, Michael and James of Madras College and Moray and Richard of Wellingborough School
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# 4.3: Confidence intervals
We are ready now to make the first step in the world of inferential statistics and use statistical tests. They were invented to solve the main question of statistical analysis (Figure $$\PageIndex{1}$$): how to estimate anything about population using only its sample? This sounds like a magic. How to estimate the whole population if we know nothing about it? However, it is possible if we know some data law, feature which our population should follow. For example, the population could exhibit one of standard data distributions.
Let us first to calculate confidence interval. This interval predict with a given probability (usually 95%) where the particular central tendency (mean or median) is located within population. Do not mix it with the 95% quantiles, these measures have a different nature.
We start from checking the hypothesis that the population mean is equal to 0. This is our null hypothesis, H$$_0$$, that we wish to accept or reject based on the test results.
Code $$\PageIndex{1}$$ (Python):
t.test(trees\$Height)
Here we used a variant of t-test for univariate data which in turn uses the standard Student’s t-distribution. First, this test obtains a specific statistic from the original data set, so-called t-statistic. The test statistic is a single measure of some attribute of a sample; it reduces all the data to one value and with a help of standard distribution, allows to re-create the “virtual population”.
Student test comes with some price: you should assume that your population is “parametric”, “normal”, i.e. interpretable with a normal distribution (dart game distribution, see the glossary).
Second, this test estimates if the statistic derived from our data can reasonably come from the distribution defined by our original assumption. This principle lies at the heart of calculating p-value. The latter is the probability of obtaining our test statistic if the initial assumption, null hypothesis was true (in the above case, mean tree height equals 0).
What do we see in the output of the test? t-statistic equals 66.41 at 30 degrees of freedom (df $$=30$$). P-value is really low ($$2.2\times e^{-16}$$), almost zero, and definitely much lower then the “sacred” confidence level of 0.05.
Therefore, we reject the null hypothesis, or our initial assumption that mean tree height equals to 0 and consequently, go with the alternative hypothesis which is a logical opposite of our initial assumption (i.e., “height is not equal to 0”):
However, what is really important at the moment, is the confidence interval—a range into which the true, population mean should fall with given probability (95%). Here it is narrow, spanning from 73.7 to 78.3 and does not include zero. The last means again that null hypothesis is not supported.
If your data does not go well with normal distribution, you need more universal (but less powerful) Wilcoxon rank-sum test. It uses median instead of mean to calculate the test statistic V. Our null hypothesis will be that population median is equal to zero:
Code $$\PageIndex{2}$$ (Python):
salary <- c(21, 19, 27, 11, 102, 25, 21)
wilcox.test(salary, conf.int=TRUE)
(Please ignore warning messages, they simply say that our data has ties: two salaries are identical.)
Here we will also reject our null hypothesis with a high degree of certainty. Passing an argument conf.int=TRUE will return the confidence interval for population median—it is broad (because sample size is small) but does not include zero.
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# 9.4 Rare events, the sample, decision and conclusion (Page 3/6)
Page 3 / 6
## Try it
It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows:
H 0 : p = 0.50, H a : p >0.50
α = 0.01
p -value = 0.025
Interpret the results and state a conclusion in simple, non-technical terms.
Since the p -value is greater than the established alpha value (the p -value is high), we do not reject the null hypothesis. There is not enough evidence to support It’s a Boy Genetics Labs' stated claim that their procedures improve the chances of a boy being born.
## Chapter review
When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p -value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind:
1. α > p -value, reject the null hypothesis
2. α p -value, do not reject the null hypothesis
When do you reject the null hypothesis?
The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely?
The outcome of winning is very unlikely.
The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why?
It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 years. Do the data support the claim that the mean height is less than 73 inches? The p -value is almost zero. State the null and alternative hypotheses and interpret the p -value.
H 0 : μ >= 73
H a : μ <73
The p -value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5% level. The data do support the claim.
The mean age of graduate students at a University is at most 31 y ears with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1% level? The p -value is 0.0264. State the null and alternative hypotheses and interpret the p -value.
Does the shaded region represent a low or a high p -value compared to a level of significance of 1%?
The shaded region shows a low p -value.
What should you do when α > p -value?
What should you do if α = p -value?
Do not reject H 0 .
If you do not reject the null hypothesis, then it must be true. Is this statement correct? State why or why not in complete sentences.
Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal.
Is this a test of means or proportions?
means
What symbol represents the random variable for this test?
In words, define the random variable for this test.
the mean time spent in jail for 26 first time convicted burglars
Is the population standard deviation known and, if so, what is it?
Calculate the following:
1. $\overline{x}$ _______
2. σ _______
3. s x _______
4. n _______
1. 3
2. 1.5
3. 1.8
4. 26
Since both σ and ${s}_{x}$ are given, which should be used? In one to two complete sentences, explain why.
State the distribution to use for the hypothesis test.
$\overline{X}~N\left(2.5,\frac{1.5}{\sqrt{26}}\right)$
A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time on death row could likely be 15 years.
1. Is this a test of one mean or proportion?
2. State the null and alternative hypotheses.
H 0 : ____________________ H a : ____________________
3. Is this a right-tailed, left-tailed, or two-tailed test?
4. What symbol represents the random variable for this test?
5. In words, define the random variable for this test.
6. Is the population standard deviation known and, if so, what is it?
7. Calculate the following:
1. $\overline{x}$ = _____________
2. s = ____________
3. n = ____________
8. Which test should be used?
9. State the distribution to use for the hypothesis test.
10. Find the p -value.
11. At a pre-conceived α = 0.05, what is your:
1. Decision:
2. Reason for the decision:
3. Conclusion (write out in a complete sentence):
प्रायिकता सिध्दान्त पर आधारित प्रतिदर्श सिध्दान्त का विकास किसने किया?
7.The following data give thenumber of car thefts that occurred in a city in the past 12 days. 63711438726915 Calculate therange, variance, and standard deviation.
express the confidence interval 81.4% ~8.5% in interval form
a bad contain 3 red and 5 black balls another 4 red and 7 black balls, A ball is drawn from a bag selected at random, Find the probability that A is red?
The information is given as, 30% of customers shopping at SHOPNO will switch to DAILY SHOPPING every month on the other hand 40% of customers shopping at DAILY SHOPPING will switch to other every month. What is the probability that customers will switch from A to B for next two months?
Calculate correlation coefficient, where SP(xy) = 144; SS(x) = 739; SS(y) = 58. (2 Points)
The information are given from a randomly selected sample of age of COVID-19 patients who have already survived. These information are collected from 200 persons. The summarized information are as, n= 20; ∑x = 490; s^2 = 40. Calculate 95% confident interval of mean age.
Ashfat
The mode of the density of power of signal is 3.5. Find the probability that the density of a random signal will be more than 2.5.
Ashfat
The average time needed to repair a mobile phone set is 2 hours. If a customer is in queue for half an hour, what is the probability that his set will be repaired within 1.6 hours?
Ashfat
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 9 and variance 0.04. Do you think that the mean thickness of the spearmint gum it produces is 8.4
3. The following are the number of mails received in different days by different organizations: Days (x) : 23, 35, 38, 50, 34, 60, 41, 32, 53, 67. Number of mails (y) : 18, 40, 52, 45, 32, 55, 50, 48, 26, 25. i) Fit a regression line of y on x and test the significance of regression. ii) Estimate y
The number of problem creating computers of two laboratories are as follows: Number of computers: 48, 6, 10, 12, 30, 11, 49, 17, 10, 14, 38, 25, 15, 19, 40, 12. Number of computers: 12, 10, 26, 11, 42, 11, 13, 12, 18, 5, 14, 38. Are the two laboratories similar in respect of problem creating compute
Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?
null rejected
Pratik
a quality control specialist took a random sample of n=10 pieces of gum and measured their thickness and found the mean 7.6 and standered deviation 0.10. Do you think that the mean thickness of the spearmint gum it produces is 7.5?
99. A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct? a
A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct?
Niaz
what is null Hypothesis
Niaz
what is null Hypothesis
Niaz
when median is greater than mode?
hello
Amaano
is this app useful
Worthy
little bit 😭
G-
oh
Worthy
when tail is positive
Jungjoon
define hypothesis
Worthy
I'm struggling to type it's on my laptop...statistics
Yoliswa
types of averages .mean median mode quarantiles MCQ question
what a consider data?
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# Discrete Mathematics: An Active Approach to Mathematical Reasoning
## Section1.3Introduction to Relations and Functions
We are familiar with the idea of a function from previous math courses. In this class we want to understand functions as an important mathematical tool to relate objects to each other. We are used to functions from the real numbers to the real numbers in calculus. All math courses use functions in some way, often relating other mathematical objects, such as vectors in vector calculus, matrices in linear algebra, complex numbers in complex analysis, or strings of characters in computer science.
A function really describes a relationship, so we will start with the more general mathematical concept of a relation.
### Definition1.3.1.
A relation, $$R\text{,}$$ from a set $$A$$ to a set $$B$$ is a subset of $$A\times B\text{.}$$
We can describe a relation as a subset, $$(a, b)\in R$$ or in the form $$aRb$$, which means “$$a$$ is related to $$b\text{.}$$
It is important to note that a relation is just a set. As long as the first coordinate comes from $$A$$ and the second from $$B\text{,}$$ we have a relation. The notation $$aRb$$ is a convenient way to describe whether two elements are related. In this case, we are thinking of $$R$$ as a symbol representing the relationship, rather than as a set.
Let $$A=\{1, 2, 3\}, B=\{2, 3, 4\}\text{.}$$ Let $$R=\{(1, 4), (3, 2)\}\text{.}$$ Then $$R$$ is a subset of $$A\times B\text{.}$$ Thus, it is a relation.
Since $$(1, 4)\in R\text{,}$$ $$1R4$$ is true.
Since $$(2, 3)\notin R\text{,}$$ $$2R3$$ is false. Note, since $$R$$ consists of ordered pairs, the order matters.
There are lots of familiar relations in math. For example, $$\geq$$ is a relation.
Let $$A=\{1, 2, 3\}, B=\{2, 3, 4\}\text{.}$$ Find $$R$$ from $$A$$ to $$B$$ if
$$aRb$$ is given by $$a\geq b\text{.}$$
$$R=\{(2, 2), (3, 2), (3, 3)\}$$
If $$R$$ is a relation from $$A$$ to $$B\text{,}$$ then $$A$$ is the domain of the relation, and $$B$$ is the codomain of the relation.
An arrow diagram for a relation is a picture of the two sets with an arrow from $$a$$ to $$b$$ if $$aRb\text{.}$$
### Definition1.3.5.
A function, $$f\text{,}$$ from a set $$A$$ to a set $$B\text{,}$$ $$f:A \rightarrow B\text{,}$$ is a mapping such that for every $$a\in A$$ there is a unique $$b\in B$$ with $$f(a)=b\text{.}$$
Let $$A=\mathbb{R}, B=\mathbb{R}\text{.}$$ Let $$f(x)=x+2\text{.}$$ Then $$f:\mathbb{R}\rightarrow\mathbb{R}\text{.}$$ We can check that it is a function: if we choose an $$x\in A\text{,}$$ we can see that there is a corresponding real number $$x+2\text{,}$$ which always exists. We can also see that for any $$x\in A\text{,}$$ there is only one possible number as output.
A function is really a relation with some additional properties. First, for each $$a$$ in $$A\text{,}$$ a corresponding $$b$$ must exist. Second, $$b$$ must be unique. In other words, every $$a$$ must map somewhere and each $$a$$ can only map to one $$b\text{.}$$
Since a function is a relation, we can use relation notation to represent a function. Often we use $$F$$ for the relation. In particular, $$f(x)=y$$ if and only if $$(x, y)\in F\text{.}$$
Let $$A=\mathbb{R}, B=\mathbb{R}\text{.}$$ Let $$f(x)=x+2\text{.}$$ Then $$F$$ is a subset of $$\mathbb{R}\times \mathbb{R}\text{,}$$ thus it is a relation. Give 3 examples of ordered pairs in $$F\text{.}$$
There are lots of examples, $$(1, 3), (-2, 0), (5/2, 9/2)$$ are a few.
Let $$A=\mathbb{R}, B=\mathbb{R}\text{.}$$ Let $$(x, y)\in R$$ if and only if $$x^2=y^2\text{.}$$ Then $$R$$ is a subset of $$\mathbb{R}\times \mathbb{R}\text{,}$$ thus it is a relation. Give 3 examples of ordered pairs in $$R\text{.}$$
There are lots of examples, $$(2, 2), (-5, 5), (2, -2)$$ are a few.
Note, this relation is not a function since if $$x=2\text{,}$$ we can have $$y=2$$ or $$y=-2\text{.}$$ Thus $$y$$ is not unique for a given $$x\text{.}$$
### Activity1.3.1.
Let $$A=\{1, 2, 3, 4, 5\}$$ and $$B=\{0, 2, 4\}\text{.}$$ Let $$R$$ be the relation from $$A$$ to $$B$$ where $$(x, y)\in R$$ if $$x=y\text{.}$$
#### (a)
List the elements of $$R\text{.}$$
#### (b)
Draw the arrow diagram for $$R\text{.}$$
#### (c)
Is the relation $$R$$ a function? Why or why not?
### Activity1.3.2.
Let $$A=\{1, 2, 3, 4, 5\}$$ and $$B=\{0, 2, 4\}\text{.}$$ Let $$S$$ be the relation from $$A$$ to $$B$$ where $$(x, y)\in S$$ if $$x-y$$ is positive.
#### (a)
List the elements of $$S\text{.}$$
#### (b)
Draw the arrow diagram for $$S\text{.}$$
#### (c)
Is the relation $$S$$ a function? Why or why not?
### Activity1.3.3.
The successor function is given by $$g(n)=n+1\text{.}$$ Let the domain be $$A=\{1, 2, 3, 4, 5\}\text{.}$$ Give the ordered pairs for $$g\text{.}$$
### Activity1.3.4.
Let $$U$$ be the relation from $$\mathbb{R}$$ to $$\mathbb{R}$$ given by $$(x, y)\in U$$ if $$x^2+y^2=1\text{.}$$ Is $$U$$ a function? Draw the set $$U$$ in the Cartesian plane.
Hint.
Keep in mind that relations are just sets of ordered pairs, so we can graph them in the $$xy$$-plane.
#### 1.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 2), (3, 3), (4, 4)\}$$ is a relation from $$A$$ to $$B\text{.}$$
• True.
• False.
#### 2.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 2), (4, 4), (6, 6)\}$$ is a relation from $$A$$ to $$A\text{.}$$
• True.
• False.
#### 3.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 3), (4, 3), (6, 3)\}$$ is a relation from $$A$$ to $$B\text{.}$$
• True.
• False.
#### 4.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 3), (2, 4), (2, 5)\}$$ is a relation from $$A$$ to $$B\text{.}$$
• True.
• False.
#### 5.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(4, 3), (2, 5)\}$$ is a relation from $$A$$ to $$B\text{.}$$
• True.
• False.
#### 6.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 2), (3, 3), (4, 4)\}$$ is a function from $$A$$ to $$B\text{.}$$
• True.
• False.
#### 7.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 2), (4, 4), (6, 6)\}$$ is a function from $$A$$ to $$A\text{.}$$
• True.
• False.
#### 8.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 3), (4, 3), (6, 3)\}$$ is a function from $$A$$ to $$B\text{.}$$
• True.
• False.
#### 9.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(2, 3), (2, 4), (2, 5)\}$$ is a function from $$A$$ to $$B\text{.}$$
• True.
• False.
#### 10.
Let $$A=\{2, 4, 6\}, B=\{3, 4, 5\}\text{,}$$ then $$\{(4, 3), (2, 5)\}$$ is a function from $$A$$ to $$B\text{.}$$
• True.
• False.
### ExercisesExercises
#### 1.
Let $$A=\{2, 3, 4\}$$ and $$B=\{4, 6, 8\}$$ and let $$R$$ be the relation from $$A$$ to $$B$$ where
\begin{equation*} (x, y)\in R \textrm{ if } \frac{y}{x} \textrm{ is an integer.} \end{equation*}
1. Is $$4\ R\ 6$$ ? Is $$4\ R\ 8$$ ? Is $$(3, 8) \in R$$ ? Is $$(3, 6)\in R$$ ?
2. Write $$R$$ as a set of ordered pairs.
3. Write the domain and codomain of $$R\text{.}$$
4. Draw an arrow diagram for $$R\text{.}$$
#### 2.
Let $$C=\{-2, 0, 2\}$$ and $$D=\{4, 6, 8\}$$ and let $$S$$ be the relation from $$C$$ to $$D$$ where
\begin{equation*} (x, y)\in S \textrm{ if } \frac{x-y}{4} \textrm{ is an integer.} \end{equation*}
1. Is $$2\ S\ 6$$ ? Is $$2\ S\ 8$$ ? Is $$(0, 6) \in S$$ ? Is $$(2, 4)\in S$$ ?
2. Write $$S$$ as a set of ordered pairs.
3. Write the domain and codomain of $$S\text{.}$$
4. Draw an arrow diagram for $$S\text{.}$$
#### 3.
Let $$T$$ be the relation from $$\mathbb{R}$$ to $$\mathbb{R}$$ where
\begin{equation*} (x, y)\in T \textrm{ if } y=x^2. \end{equation*}
1. Is $$-3\ T\ 9$$ ? Is $$9\ T\ -3$$ ? Is $$(2, 4) \in T$$ ? Is $$(4, 2)\in T$$ ?
2. Draw the graph of $$T$$ in the Cartesian plane.
#### 4.
Let $$H$$ be the relation from $$\mathbb{R}$$ to $$\mathbb{R}$$ where
\begin{equation*} (x, y)\in H \textrm{ if } y^2-x^2=1. \end{equation*}
Is $$H$$ a function? Explain your answer.
#### 5.
Let $$A=\{-1, 0, 1\}$$ and $$B=\{w, x, y, z\}$$ and let $$F$$ be the function $$F: A\rightarrow B$$ given by the set $$F=\{(-1, y), (0, w), (1, y)\}\text{.}$$
1. Write the domain and codomain of $$F\text{.}$$
2. Explain why $$F$$ is a function.
3. Find $$F(-1), F(0),$$ and $$F(1)\text{.}$$
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# 10 Example of Quadratic Equation in Real Life
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Our daily lives involve regular use of our mathematical knowledge to solve real-life problems. Just like other mathematical concepts, we unknowingly use quadratic equations to find answers to our questions. here we are going to describe example of quadratic equation in real life.
A quadratic equation is an equation containing variables, among which at least one must be squared. It is expressed in the following form:
ax^2+bx+c= 0
Here, ‘x’ is the unknown value we need to calculate. The letters ‘a’ and ‘b’ represent the known numbers you put in while calculating. However, one must remember that ‘a’ can never be zero.
Although simply looking at the equation you may feel that it is not something you use very often. So, let us give you a few examples of how quadratic equations find their application in our daily lives.
## 10 Relatable Applications of Quadratic Equations in Real Life
I’ve discussed ten real-world quadratic equation instances in this post. These programs will aid in the development of a solid mathematical foundation in our children. I’m hoping they’ll like the examples and use them in their everyday lives.
### Building a Home & Calculating Areas
Constructors and architects take the help of quadratic equations to develop a building. For any building, you need to calculate how much land you have, how big each room will be, what the shape of the building will be, etc. So, if you are building a rectangular house, you know one side of the house will be bigger than the other. So, you can use the ratio of these two sides to calculate the materials you need.
When you buy a piece of land, you know its total size in square feet. Your constructor or builder plans the whole house for you, dividing the total available area across your home. They do so by using the quadratic equation. The size of land, number of rooms, etc., are known factors (like a,b), while the size of rooms and space allocation for stairs or corridors are unknown (the ‘x’). Now, put these values to the equation ax2+bx+c= 0, and you have all the data you need.
To calculate the profit you earn or are likely to earn from your business, you need the quadratic equation’s help. Whatever product or service you sell, you know how much it costs to produce. You are also aware of the amount of profit you aim to make. So, you might wonder what your product should be priced at. This unknown selling price is your ‘x’ of the quadratic equation. The cost price and the profit are known factors, ‘a’ and ‘b.’ Now that you have all the elements of a quadratic equation solving it step by step will give you the selling price.
### Estimating Speed with Quadratic Equations
Calculating the required speed calls for the use of a quadratic equation. You will find it even more helpful if you are on your college or university rowing team. If you are going upstream, you have to row against the stream. But you need to know how fast you should row or what should be the speed of your boat. You can easily recognize that by using the quadratic equation. Let’s first look at the known factors. You know the distance you have to row to and back, the speed of the stream, and the total time available to you. Assuming your speed is ‘x,’ you can put these known values in the ax2+bx+c= 0 equation. Now continue to solve it step-by-step; you will have the answer ready in no time.
Velocity quadratic equations are something that athletes and sports analysts use every day, every moment, especially in the case of basketball, javelin throw, shot put, etc., which involves throwing balls or spears or other such items.
In a basketball team, you will see that one player throws the ball to another, reaching them in a moment after or before they catch it. It will feel like the player knows when he should throw the ball so that it reaches the other player at the exact time. It is really a matter of quadratic calculation. Here, the known values of ‘a’ and ‘b’ are the heights, speed of the ball, loss of speed due to gravitational force, etc. The time is unknown to you, and this is the ‘x.’ Thus, you can solve it by referring to the ax2+bx+c= 0 equation. This is how your favorite basketball players score.
### Quadratic Equations in Engineering Life
Among all professions, engineers probably use quadratic equations most extensively. Here’s a list:
• Engineers in the automobile industry use quadratic equations to design vehicle structures, especially those with curved patterns.
• Automobile engineers use quadratic equations for designing and installing brake systems.
• Aerospace engineers use these equations for calculating a vertical plane projectile. Calculating the height and velocity of an object when thrown or launched in space, they use the ax2+bx+c= 0 formula to recognize how much time it will take to reach the destination.
• Using complex systems at the workplace calls for regular and frequent use of quadratic equations by electrical and chemical engineers.
• Audio engineers also use quadratic equations to design sound systems to ensure the listeners experience the best sound quality.
### Satellite Dish Setting and Signal Transmission
You need to put up a satellite dish at a particular angle to receive the most efficient signal transmission. The dish on your roof catches the signals from two or more satellites simultaneously and transmits them to your TV through a feed horn. This whole transmission process involves using quadratic equations to identify the most efficient angle.
### Defense and Military Services
We already told you how to use quadratic equations to measure height, distance, speed, etc. These measurements are also used in defense services and military activities. For example, if the military needs to throw artillery to destroy an enemy camp, they calculate the distance and the speed of launching the artillery through quadratic equations.
### Farming-Related Activities
At first, agriculture, farming, and quadratic equations might seem like two entirely different topics, but a successful agricultural venture involves quadratic equations. These equations are used to calculate the total available area and determine how the area will be divided and how crops should be allocated. Randomly building a pen anywhere on the farm will not help you achieve high yields. The use of quadratic equations ensures agricultural efficiency.
### Quadratic Equations for Effective Management
In any industry, there are various levels of management. A Production Manager supervises the product line manufacturing and other relevant activities. Again, an Engineering Manager supervises the machinery, work efficiency, etc. To estimate all these, they take the help of quadratic equations. The 0, etc., by considering the available work as the known variable. Thus, management also applies quadratic equations quite unknowingly.
### Other Real-life Applications of Quadratic Equation
In addition to the above example of quadratic equation in real life, there are other real-life instances where quadratic equations are used. These include:
• Astronomers identify and describe solar systems, planets and their orbits, and galaxies with the help of quadratic equations.
• Computer engineers ease the use of complex systems through quadratic equation implementation.
• Criminal investigators determine the trajectories of bullets by using quadratic equations.
• Insurance agents design plans and models through the computation of data. These plans, in most cases, are unique to each consumer. Such a complex process requires in-depth quadratic equation calculation.
• If there is a car accident, quadratic equations help determine car speeds.
• In Physics, different motions are described easily with quadratic equations.
• Chemical Engineers or Chemists working in fields of Chemistry use these equations to describe specific chemical reactions, identify their equilibrium, etc.
## Wrapping Up
Here, we have known the example of quadratic equation in real life. Quadratic equations helps us evaluate the relationship between variable quantities. Right from calculating area, speed, and profits to astronomy and criminal investigations, this math concept is useful in every sphere of life.
The above examples clearly show how we use quadratic equations in our daily lives. Therefore, developing a thorough understanding of the subject is necessary if we don’t wish to get stuck finding answers to our everyday queries.
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# Indeterminate Limits & Expressions
Limit of Functions > Indeterminate Limits & Expressions
Contents:
## What is an Indeterminate Expression?
An indeterminate expression seems to have more than one answer, depending on how you evaluate it.
For example, the limit
Seems to be zero if you use direct substitution: sin(0)/0. However, sin x and x are almost equal to each other as you get close to zero, which actually makes the limit 1.
This kind of conflict is why these expressions can’t be evaluated in many software programs; The system will balk if you try and enter one because it doesn’t know which way you want it evaluated. Wolfram’s The MATHEMATICA ® Book, Version 4 goes as far as calling indeterminate expressions “poison” for arithmetic computation.
## Indeterminate Expression Examples
Many of these expressions are found in mathematics:
• 00
• 0/0
• 0 x ∞
• ∞ – ∞
• ∞ / ∞
• infin;0
• 1
For example, 0 X ∞ might be 0 or ∞ (or any finite number in between). Any limit that results in an indeterminate expression when you substitute in x = a has to be evaluated carefully.
## How Do I Evaluate an Indeterminate Expression?
One workaround is to ignore the exact limit and evaluate the function close to a. In the sin(0)/0 example, you could evaluate the function at sin(0.01)/0.01 = 0.99998.
A more formal approach is L’Hospital’s Rule. The rule can only be used in special cases, but it makes some of these expressions fairly simple to evaluate. For example, all you need to do to evaluate sin(x)/x at x = 0 is:
1. Take the derivative of the top and bottom (treating each as functions). The derivative of sin(x) is cos(x) and the derivative of x is 1.
2. Make a new quotient with your answers from Step 1: cos(x)/x.
3. Find the limit of the new quotient. cos(0)/1 = 1
## Indeterminate limits
An “indeterminate” limit is one that can’t be found, at least not with the usual rules for finding limits. Indeterminate limits may not have limits at all, and if they do, they don’t indicate what those limits might be.
The most common indeterminate limits you’ll come across in calculus are {0/0} and {∞/∞}, but there are many others. Indeterminate limits include:
• Limits involving infinity: {∞ / ∞}, {∞ – ∞}.
• Exponential indeterminate forms:0, 00, 1
• Zeros types: {0/0} or {0 * ∞).
## How to Find Indeterminate Limits
Arguably, the easiest way to find these limits is to graph the function using a graphing calculator (or alternatively, look at the associated table of values). I used the free graphing calculator at Desmos.com to graph this function and find an approximate limit:
To find these limits using a more rigorous method, apply L’Hospital’s Rule.
## The Problem of Multiple Variables
The following function has the indeterminate limit 0/0:
This particular function has several variables, so you can’t use L’Hospital’s rule to find the limit (Adillon, 2015).
To find a limit, you look at what happens when the function approaches a certain point x. You might not be able to find the limit at x exactly, but you can look at what happens from either side as the function approaches x; if both sides are equal, you’ve found the limit. If both sides aren’t equal, then the limit doesn’t exist. With a function with multiple variables, like the one above, the point can be approached by infinite different paths, making it practically impossible to hone in on a limit.
## References
Adillon, R. et al. (2015). Mathematics for Economics and Business. Edicions Universitat Barcelona. Retrieved May 21, 2019 from: https://books.google.com/books?id=HwNTDAAAQBAJ
Larson, R. & Edwards, B. (2017). Calculus of a Single Variable. Cengage Learning.
Neal, D. Indeterminate forms. Retrieved January 23, 2020 from: http://people.wku.edu/david.neal/136/Unit1/Indet.pdf
Wolfram, S. (1999). The MATHEMATICA ® Book, Version 4. Cambridge University Press.
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# Algebraic Properties Don't Change Anything
Lesson Plans > Mathematics > Algebra > Expressions
## Algebraic Properties Don't Change Anything
This morning, as we were driving to church, T piped up from the back seat, "How long until S's birthday?"
Simultanously, my wife said, "One week," and I said, "Seven days."
"Seven," I said.
He thought for a little bit longer, then judiciously commented, "Then you and mama were both right!"
"That's right," I said. "We're both right, because 'one week' and 'seven days' mean exactly the same thing. They're just different ways of saying the same thing."
This idea of "two things that look different but really aren't" is a central concept of algebraic manipulation. It's also a concept that Algebra students sometimes miss out on, or easily forget. I find sometimes that students are going through motions of simplifying expressions, or solving equations, without really understanding that what they're doing is writing exactly the same thing in different ways. This leads to them doing strange things that change the value of the expression.
Students should be taught and reminded repeatedly that our properties of arithmetic and algebra are simply ways of writing the same thing in different ways.
The simplest example of this is to recognize that 1 + 3 and 4 are two expressions that look very different, but have exactly the same meaning. In the same way, algebra students will learn that like terms can be combined. What does that mean? It means that 1x + 3x and 4x are exactly the same thing, even though they look different.
What is the distributive property? It is a rule that allows you to take an unfactored expression and write it as a factored expression that means the same thing, but just looks different. If you have the expression 3x + 5x2 the distributive property lets you rewrite this as x(3 + 5x). This is not a different expression; its the same expression written in a different way. It's like saying "seven days" instead of "one week." The two phrases look and sound very different, but they have the same meaning.
And since this is an equivalence, it also works the other way: if we start with x(3 + 5x), we can rewrite it as 3x + 5x2.
The same is true for every rule of Algebraic manipulation. The additive property of equality is really saying that
x - 5 = 7 is exactly the same equation as x = 12. They look very different, but they mean exactly the same thing.
Finally, all of algebra is built (in a sense) on the concept of substitution. The substitution property tells us that if we have two expressions that mean the same thing, then we can write one in place of the other. If "one week" and "seven days" mean the same thing, I can replace one with the other in a sentence. If x and y + 2 mean the same thing, I can replace x with (y + 2) in any expression.
With this in mind, the properties we introduce should not be seen as restrictions on what we can do - they are open doors that give us permission to do something. Knowing that "one week" means the same thing as "seven days" gives us permission to replace "one week" with "seven days" or vice-versa. Knowing the destributive property gives us the freedom to replace 3x + 6y with 3(x + 2y) - and vice-versa.
This should be emphasized with every rule that crops up along the way. I'm going to start using the "one week" is the same as "seven days" illustration every time I introduce a new rule. My students will probably get sick of hearing it, but hopefully it'll help cement in their minds the concept that we're not changing the meaning of an expression by following the rules.
Lesson by Mr. Twitchell
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# Take a picture math solver
Math can be a challenging subject for many learners. But there is support available in the form of Take a picture math solver. We will also look at some example problems and how to approach them.
## The Best Take a picture math solver
This Take a picture math solver provides step-by-step instructions for solving all math problems. As the name suggests, a square calculator is used to calculate the area of a square. A square calculator is made up of four basic parts – a base, a top, a pair of sides, and an angle. The area of any four-sided figure can be calculated by using these four components in the correct order. For example, if you want to calculate the area of a square with side lengths $x$, $y$, $z$, and an angle $heta$ (in degrees), then you simply add together the values of $x$, $y$, $z$, and $heta$ in this order: egin{align*}frac{x}{y} + frac{z}{ heta} end{align*}. The above formula can also be expressed as follows: egin{align*}frac{1}{2} x + frac{y}{2} y + frac{z}{4} z = frac{ heta}{4}\end{align*} To find the area of a cube with length $L$ and width $W$, first multiply $L$ by itself twice (to get $L^2$). Next, multiply each side by $W$. Lastly, divide the result by 2 to find the area. For example: egin{align*}left(L
A negative number is not equal to any other negative number because it can never be smaller than itself. The absolute value of a complex number must be positive or zero. The absolute value is the distance from the origin to the point that represents the number. If we take the absolute value of a number, we get its magnitude (or size). For example, if we take the absolute value of 4, we get 2 because 4 is two units away from 0. If we take the absolute value of -4, we get 4 because -4 is four units away from 0. If we take the absolute value of 7, we get 0 because 7 is zero units away from 0 (it's at zero distance from 0). Now you can solve absolute value equations!> There are several different ways of solving absolute value equations. One way is to use long division by finding all possible pairs of numbers with whose product is equal to zero (this means that one plus one equals zero). Another way is to use synthetic division by finding all possible pairs of numbers whose difference is zero (this means that subtracting one from another yields zero). A third way is to use exponents, where the base and exponent are equal to
One of the most important things to consider when choosing a calculator is its accuracy. The more accurate your calculator is, the more likely you are to get the right answer. CalcXact and Casio fx-9550BCL Digital Clutch and Fx-9750BCL Digital Clutch are both highly accurate calculators that are suitable for solving word problems throughout elementary school and beyond. If you want an even more accurate calculator, then you should check out Casio's fx-9900GE or fx-9900GII. These two calculators have a high level of accuracy and can be used for a wide range of different tasks.
Math teaching apps allow teachers to create interactive lessons that engage students in learning math concepts. These apps help teachers teach both basic and complex math concepts in an engaging manner. They can be used by both new and experienced math teachers. These apps can serve as a valuable supplement to traditional classroom instruction, most importantly for those who may not have the time or resources to dedicate to teaching math (e.g., working parents, military members). They can also be used as a tool to assess student progress during class time, potentially helping teachers identify areas of weakness and strengthen their subject knowledge. Despite their potential benefits, though, it’s important to keep in mind that these apps are not a replacement for actual classroom instruction. The best way to use them is as a supplement to your regular practice sessions and assessments.
Linear inequalities can be solved using the following steps: One-Step Method The first step is to fill in the missing values. In this case, we have two set of numbers: one for x and another for y. So we will first find all the values that are missing from both sides of the inequality. Then we add each of these values to both sides of the inequality until an answer is found. Two-Step Method The second step is to get rid of any fractions. This is done by dividing both sides by something that has a whole number on it. For example, if the inequality was "6 2x + 9", then you would divide both sides by 6: 6 2(6) + 9 = 3 4 5 6 7 8 which means the inequality is true. If you wanted to find out if 2x + 9 was greater than or less than 6 then you would divide by 2: 2(2) + 9 > 6 which means 2x + 9 is greater than 6, so the solution to this inequality is "true". These two methods can be used separately or together. They both work, but they're not always as efficient as they could be since they both involve adding and subtracting numbers from each side of the equation.
Amazing idea! You can take a picture of your math problem and it will solve it for you and guide you through the process of solving it. You can choose between methods of solving if you don't like the one it uses. It also has an amazing calculator which you can also use to solve your problems.
Camilla Hall
A real lifesaver indeed for understanding math homework. Great app. More than just an ordinary calculator. The solving steps are explained clearly. It would have been better if it worked offline like its previous versions. Still, definitely recommend. ♥️✨
Annabelle Clark
Math equation solver order of operations Precalculus math help Math word problems solver Trig math solver Geometry app Solve word problems
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# Understand the problem
$a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$ Which of the following is true? (a) sup {$$a_n|n \in \mathbb{N}$$}=3 and inf {$$a_n|n \in \mathbb{N}$$}=1 (b) lim inf ($$a_n$$)=lim sup ($$a_n$$)=$$\frac{3}{2}$$ (c) sup {$$a_n|n \in \mathbb{N}$$}=2 and inf {$$a_n|n \in \mathbb{N}$$}=1 (d) lim inf ($$a_n$$)= 1 lim sup ($$a_n$$)=3
Hint 1: $a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$ Now the limit points of this set are those points which the set does not attain.So, they might be the sup and inf which are not attained by this set. Basically sup($$a_n$$)= max{ limit points, $$a_n$$ | n $$\in$$ $$\mathbb{N}$$} Limit points are $$2,1$$ and $$a_1= 2+1=3, a_3= 2- \frac{1}{3} ; a_5= 2+\frac{1}{5}$$ $$a_0= 1+1=2 , a_2= 1+ \frac{1}{4} , a_3= 1+\frac{1}{8}$$ Now you can calculate the supremum?
Hint 2: From the observation of Hint 2 we have sup $$a_n$$= max $$\{2,1,3,2\}=3$$ Similarly, inf $$a_n$$= min$$\{$$ limit points, $$a_n | n \in \mathbb{N}\}$$ Can you calculate that by yourself? Hint 3: inf $$a_n$$= min {2,1,2 -$$\frac{1}{3}$$}=1 So, option A is correct. Now there is another question regarding lim sup and lim inf. We can observe that we have mainly $$3$$ subsequences , corresponding to $$n$$ is even; $$n=2k$$ $$n$$= $$4k+1$$ $$n=4k+3$$
Can you calculate the corresponding subsequences and their limits?
Hint 4: For $$n=2k$$ we have $$a_{2k}=1+ \frac{1}{2^{ek}} \longrightarrow 1$$ ask For $$a_{4k+1}= 2+ \frac{1}{4k+1} \longrightarrow 2$$ ask $$a_{4k+3}= 2-\frac{1}{4k+3} \longrightarrow 2$$ ask So, lim sup $$a_n$$=max$$\{1,2\}=2$$ Lim inf $$a_n$$=min$$\{1,2\}=1$$ Therefore, Option C is also correct
# Connected Program at Cheenta
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
# Understand the problem
$a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$ Which of the following is true? (a) sup {$$a_n|n \in \mathbb{N}$$}=3 and inf {$$a_n|n \in \mathbb{N}$$}=1 (b) lim inf ($$a_n$$)=lim sup ($$a_n$$)=$$\frac{3}{2}$$ (c) sup {$$a_n|n \in \mathbb{N}$$}=2 and inf {$$a_n|n \in \mathbb{N}$$}=1 (d) lim inf ($$a_n$$)= 1 lim sup ($$a_n$$)=3
Hint 1: $a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$ Now the limit points of this set are those points which the set does not attain.So, they might be the sup and inf which are not attained by this set. Basically sup($$a_n$$)= max{ limit points, $$a_n$$ | n $$\in$$ $$\mathbb{N}$$} Limit points are $$2,1$$ and $$a_1= 2+1=3, a_3= 2- \frac{1}{3} ; a_5= 2+\frac{1}{5}$$ $$a_0= 1+1=2 , a_2= 1+ \frac{1}{4} , a_3= 1+\frac{1}{8}$$ Now you can calculate the supremum?
Hint 2: From the observation of Hint 2 we have sup $$a_n$$= max $$\{2,1,3,2\}=3$$ Similarly, inf $$a_n$$= min$$\{$$ limit points, $$a_n | n \in \mathbb{N}\}$$ Can you calculate that by yourself? Hint 3: inf $$a_n$$= min {2,1,2 -$$\frac{1}{3}$$}=1 So, option A is correct. Now there is another question regarding lim sup and lim inf. We can observe that we have mainly $$3$$ subsequences , corresponding to $$n$$ is even; $$n=2k$$ $$n$$= $$4k+1$$ $$n=4k+3$$
Can you calculate the corresponding subsequences and their limits?
Hint 4: For $$n=2k$$ we have $$a_{2k}=1+ \frac{1}{2^{ek}} \longrightarrow 1$$ ask For $$a_{4k+1}= 2+ \frac{1}{4k+1} \longrightarrow 2$$ ask $$a_{4k+3}= 2-\frac{1}{4k+3} \longrightarrow 2$$ ask So, lim sup $$a_n$$=max$$\{1,2\}=2$$ Lim inf $$a_n$$=min$$\{1,2\}=1$$ Therefore, Option C is also correct
# Connected Program at Cheenta
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
# Similar Problems
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# Calculating the Birthday Problem The Other Way Around
As most of you probably know, the birthday problem is about finding the probability that any two people share the same birthday, in a room with N people. Since 23 people in the room is the number more frequently used, let's use it here as well.
On Wikipedia and on most probability books the approach to solve this is to first calculate P', the probability that no two people will share the same birthday, and then do 1 - P'. So you basically do (364/365)(364/365)...(343/365). This multiplication will give you 0.4927, 1-P' is 0.5073. So far so good.
What I want to know is how you do the calculation the other way around. That is, how you calculate P straight away, and not by finding 1-P'.
The first idea that came to me is this one:
1. Put the first person in the room. The probability of this person sharing a birthday is 0.
2. Put the second person in the room. The probability of this person sharing the birthday with the first would be 1/365.
3. Put the third person in the room. The probability of this person sharing the birthday with anyone would be 2/365.
So on and so forth, until the 23rd person, whose probability of sharing a birthday would be 22/365.
However this reasoning is flawed, as if you sum those probabilities the result would be 244/365, which is 0.6684 and not 0.5073.
What's wrong with the above reasoning, and what's the correct approach?
Update: As Thomas Andrews points out below, the problem is probably related to some cases being counted twice (i.e., 1 and 2 share and birthday and 3 and 4 share too). In this case we need to shave the individual probabilities a bit, which makes sense since the result should be lower that what we have right now. How to do it though?
Update 2: I think I found the answer. See below.
• What are you doing with those indidual probabilities? You can't add them, or you'll get a total greater than one when you have 183 people. You can't multiply them, because you'll get $0$. The problem is that you are counting some cases twice - if person $1$ and $2$ share a birthday, and $3$ and $4$ share a birthday, you've counted that case twice. Oct 9 '13 at 17:46
• @Thomas, I agree with counting some cases twice. But you said I can't add the probabilities, but I don't agree. I might not be calculated each individual probability correctly, but once I do it to get the right answer I'll certainly need to add them, right? Oct 9 '13 at 17:50
• It's step 3. It's simply not linear like that. P(person 3 shares bday with neither person 1 nor person 2) $\neq$ P(person 3 doesn't share bday with person 1) + P(person 3 doesn't share bday with person 2). Oct 9 '13 at 17:51
• So is the probability zero? No. Is the probability ever greater than $1$? It will be if you add with $184$ people in the room. So, no, it doesn't have to be one or the other. The point is, your quest to find a "direct" way of solving this is doomed. Oct 9 '13 at 17:52
• @Thomas Andrews, see my comment above. Once the individual probabilities get calculated correctly it should never go above 1. It's going above one cause the value of each probability is wrong. Once you get this right the correct way to do this is to add the individual probabilities, since it's an OR situation (or the second person share a birthday, or the third, or the fourth, etc ) Oct 9 '13 at 17:54
Well, you can certainly compute $p(n+1)/p(n)$. But it's not a trivial expression. Let $q(n)=1-p(n)$. Then we know that $$q(n+1)=q(n)\left(1-\frac{n}{365}\right)$$ So \begin{align}p(n+1) = 1-q(n+1) &= 1-q(n)\left(1-\frac{n}{365}\right)\\& = 1-(1-p(n))\left(1-\frac{n}{365}\right)\\=p(n) +\frac{(1-p(n))n}{365}\end{align}
So $\dfrac{p(n+1)}{p(n)}$ is going to be a messy expression of $p(n)$.
Similarly $p(n+1)-p(n)=\frac{(1-p(n))n}{365}$.
Essentially, this means that we get a matching birthday in $n+1$ people if we get a matching birthday in the first $n$, or if we don't in the first $n$ and the $n+1$st matches one of the $n$ previous ones.
For the example of $n=3$:
\begin{align}p(1)&=0\\p(2)&=p(1)+\frac{1(1-p(1))}{365} = \frac{1}{365}\\p(3)&=p(2)+\frac{2(1-p(2))}{365}=\frac{1}{365} + \frac{2\cdot 364}{365^2}\approx 0.0082042\end{align}
• Party time - that's my 1000th answer. :) Oct 9 '13 at 18:13
• Thanks for that, but it's still not that clear for me. Are you saying it's too complex to calculate the probability the way I want? If not, how do you use your formula to get to the 0.5073 number? Oct 9 '13 at 18:15
• You are correct. I formulated another way of explaining it below, but it was based on your idea. I'll select yours as the right answer. Could you take a look at my answer below and confirm it's the same thing (i.e., correct too)? Oct 9 '13 at 18:44
I think I found it (partially based on Thomas answer above). Here's the correct reasoning:
1. Put the first person in the room. His probability of sharing a birthday with anyone is 0.
2. Put the second person in the room. His probability of sharing a birthday with anyone is 1/365.
3. Put the third person in the room. His probability of sharing a birthday with anyone, given that the previous people didn't, is (1 - 1/365) * (2/265). Therefore (364/365)(2/365)
4. Put the fourth person in the room. His probability of sharing a birthday with anyone, given that none of the previous people did, is (1-P(3)) * (3/365).
So on and so forth.
So basically for the Nth person, the probability of him sharing a birthday with anyone in the room already, given that no one before did, P(N), is:
P(N) = (1 - P(N-1)) * ((N-1)/365)
And P(1) = 0.
If you want to calculate the probability with 23 people in the room, therefore, you just need to add all the individual P(N).
• Line (3) is wrong, that calculates the probability that the third person in the room is the first person who matches some previous person. If person 1 and person 2 match, then the probability that the third person matches someone previous is only $\frac{1}{365}$. Oct 9 '13 at 18:54
• I think the wording is not precise then, but the numbers must be, cause they are equal to yours after all. For instance, If you stop on my step 3 you'll have 1/365 + (364*2)/(365*365) = 0.0082042 too. Ain't that the case? Oct 9 '13 at 19:04
• Yeah, I was not talking about the numbers, I was talking about the words. Oct 9 '13 at 19:07
• For example, the entire sentence starting, "So, basically,..." is wrong. Oct 9 '13 at 19:08
• Gotcha. I think I fixed the wording now. Thanks for the help! Oct 9 '13 at 19:08
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Algebra is the study of unknown numbers. With basic arithmetic, individual numbers are added, subtracted, multiplied, divided, and raised to powers or other basic functions. With algebra, however, we start looking at what happens when we compare numbers even though we do not know what specific values they have yet. Some examples of standard equations from basic arithmetic are:
• $1+2=3$
• $2+3=5$
• $3\times3=9$
• $21+23=44$
• $7+8=15$
• $9-8=1$
• ${8\over4}=2$
• ${2^3} = 8$
With Algebra, however, the equations begin to look different.
• $a+b=c$
• $2x+3y=z$
• $ax^2+bx+c=0$
• $(x+y)n=z$
• $a^2+b^2=c^2$
• $e=mc^2$
• $a^z+b^z=c^z | z<=2$
These equations seem slightly strange at first, but they soon become quite natural. Each letter simply stands for a unknown number Let's examine the first algebraic equation above: $a+b=c$. If we let $a=1$ and $b=2$ we find that $c=3$, so our equation reads $1+2=3$. The power of this equation however is that we can use other values as well. If we let $a=2$ and $b=3$ then $c=5$. We do not have to always choose $a$ and $b$ however, we could choose $a$ and $c$ and find $b$. If $a=7$ and $c=15$ we will find that $b$ must be $8$ to create a true equation. When working with $b$ in this equation it might be easier sometimes to get it into a form where $b$ is easy to find. $$a+b=c$$ We will change the order of addition around first $b+a=c$ We know that $a$ does not change so we can subtract $a$ from both sides of the equation. b+a-a=c-a We know however that if we take b and add a and then subtract it again be will be back to b. So we can remove the +a-a from the left side of the equation. b=c-a Now whenever we have numbers for c and a, b will be easy to find. Much of algebra is just learning these simple techniques and applying them to more advanced problems and functions.
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## Math Hacks and Tricks: 11 Ways to Make Math Easier
For most people, math is the bane of their existence in the classroom and elsewhere. Many would even brag about their lack of proficiency in the act of making sense of two or more numbers.
It would be fine if numbers were merely a trifle, but math really does make the world go round, or at least explain why it does.
Mathematics is the language that links human understanding to the mysteries of the universe and everything else in between.
It’s also the language of money, which is perhaps why a lot of people aren’t that financially literate. In short, math is a very important part of life, whether we understand it or not.
If that has caught your attention, then perhaps you may want to actually get better with math. There isn’t a real surefire shortcut to progress and success, but there are some math hacks and tricks that can help you get better and understand the way of numbers more.
## Easier Ways to Multiply
Multiplying by 6: Multiplying the number by 3, then multiply the result by 2.
e.g. 12 x 6 = ( 12 x 3 ) x 2 = 36 x 2 = 72
Multiplying by 9: Multiply the number by 10, then subtract the original number.
e.g. 35 x 9 = ( 35 x 10 ) – 35 = 350 – 35 = 315
Multiplying by 11: Multiply the number by 10, then add the original number.
e.g. 527 x 11 = ( 527 x 10 ) + 527 = 5270 + 527 = 5797
Multiplying two-digit numbers by 11: Add the two digits together and put the result in between them (and carrying the extra digit to the higher digit if the result has two digits).
e.g. 89 x 11 → 8 (8+9) 9 → 8 (17) 9 → 8+1 (7) 9 → 979
Multiplying by 12: Multiply the number by 10, then add twice the original number.
e.g. 23 x 12 = ( 23 x 10 ) + ( 23 x 2 ) = 230 + 46 = 276
Multiplying by 13: Multiply the number by 3 and add 10 times original number.
e.g. 47 x 13 = ( 47 x 3 ) + ( 47 x 10 ) = 141 + 470 = 611
Multiplying by 14: Multiply the number by 7 and then multiply by 2.
e.g. 8 x 14 = ( 8 x 7 ) x 2 = 56 x 2 = 112
Multiplying by 15: Multiply the number by 10 and add 5 times the original number, as above.
e.g. 10 x 15 = ( 10 x 10 ) + ( 10 x 5 ) = 100 + 50 = 150
Multiplying by 16: Multiply the number by 8, then by 2.
e.g. 40 x 16 = ( 40 x 8 ) x 2 = 320 x 2 = 640
Multiplying by 17: Multiply the number by 7 and add 10 times original number.
e.g. 7 x 17 = ( 7 x 7 ) + ( 7 x 10) = 49 + 70 = 119
Multiplying by 18: Multiply the number by 20, then subtract twice the original number.
e.g. 56 x 18 = ( 56 x 20 ) – ( 56 x 2 ) = 1120 – 112 = 1008
Multiplying by 19: Multiply the number by 20, then subtract the original number.
e.g. 72 x 19 = ( 72 x 20 ) – 72 = 1440 – 72 = 1368
Multiplying by 24: Multiply the number by 8, then multiply by 3.
e.g. 32 x 24 = ( 32 x 8 ) x 3 = 256 x 3 = 768
Multiplying by 27: Multiply the number by 30, then subtract 3 times the original number.
e.g. 194 x 27 = ( 194 x 30 ) – ( 194 – 3 ) = 5820 – 582 = 5238
Multiplying by 45: Multiply the number by 50, then subtract 5 times the original number.
e.g. 67 x 45 = ( 67 x 50 ) – ( 67 x 5 ) = 3350 – 335 = 3015
Multiplying by 98: Multiply the number by 100 and subtract twice the original number.
e.g. 19 x 98 = ( 19 x 100 ) – (19 x 2 ) = 1900 – 38 = 1862
Multiplying by 99: Multiply the number by 100 and subtract the original number.
e.g. 31 x 99 = ( 31 x 100 ) – 31 = 3100 – 31 = 3069
And…
## Quick-Squaring Two-Digit Numbers Ending in 5
Multiply the first digit by itself plus 1, then put 25 at the end of it.
e.g. 552 → 5 x (5+1) → 5 x 6 → 30 → 3025
## If you’re having trouble in Trigonometry, just use SohCahToa.
This mnemonic device helps you remember how make sense of triangles. Soh means SinA=Opposite/Hypotenuse, Cah means CosA=Adjacent/Hypotenuse, and Toa means TanA=Opposite/Adjacent.
## Convert your salary to an hourly rate.
Drop the last two zeroes of your salary, then divide by two.
e.g. \$3,000 per month → \$30 / 2 = \$15 per hour
## Calculate how long it will take to triple your investment.
The Rule of 115: An odd name for a rule of thumb, but this is very useful for those who are looking to grow their money. Simply divide 115 or 110 by the growth rate, which should give you the amount of time it takes to do just that.
It’s a good rule to know how long you have with a certain investment, especially if you’re not particularly risk-averse.
e.g. 10% per year → 115 / 10 = 11.5 years
Take number X, subtract by the nearest multiple of 10 to get the difference D. You then multiply (X-D) with (X-D), then add D2 to get your answer.
It’s easier because one of the numbers you come up with is a multiple of 10, so you’re basically multiplying a double digit number with a single digit number then adding 0 at the end before you add the square of D.
e.g. 772 → D = 80 – 77 = 3 → ( 77 – 3 ) x ( 77 + 3 ) + 32 = 74 x 80 + 9 → ( 70 x 80 ) + ( 4 x 80 ) + 9 = 5600 + 320 + 9 = 5920 + 9 = 5929
It seems hard and long-winded at first, but you should be able to get the hang of it with some practice. Since you end up dealing with multiples of 10 (once you use the multiplication shortcut for multiplying a double-digit number to a multiple of 10), you should find it easier with some practice rather than just tackling 772 head on.
## Asset Allocation by Age.
Here’s another rule of thumb for your money management needs. If you don’t know how much of your assets you should put into certain investments, just remember this. Subtract your current age from 120, and the result is the percentage you should put into stocks.
e.g. 40 years old → 120 – 40 = 80% in stocks, 20% in fixed income products
Hope this list helped get you even more excited about Math. If you’re looking for more tips, you can check out more of our Math lessons.
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# Solve the equation: $2{\log _3}x + {\log _3}({x^2} - 3) = {\log _3}0.5 + {5^{{{\log }_5}({{\log }_3}8)}}$
Last updated date: 24th Jul 2024
Total views: 452.7k
Views today: 11.52k
${a^{{{\log }_a}x}} = x$
$\log {m^n} = n\log m$
$\log x{\text{ is real }}\forall x > 0$
We have given the equation $2{\log _3}x + {\log _3}({x^2} - 3) = {\log _3}0.5 + {5^{{{\log }_5}({{\log }_3}8)}}$ which can be written as ${\log _3}{x^2} + {\log _3}({x^2} - 3) = {\log _3}0.5 + {\log _3}8$ and on further solving we’ll get $2{\log _3}x + {\log _3}({x^2} - 3) = {\log _3}0.5 + {\log _3}4$ . This equation is equivalent to the system
$\left\{ {\begin{array}{*{20}{c}} {{x^2} > 0} \\ {{x^2} - 3 > 0} \\ {{x^2}({x^2} - 3) = 4} \end{array}} \right. \\ \Rightarrow \left\{ {\begin{array}{*{20}{c}} {x < 0{\text{ and }}x > 0{\text{ }}} \\ {x < \sqrt 3 {\text{ and x}} > \sqrt 3 } \\ {({x^2} - 4)({x^2} + 1) = 0} \end{array}} \right. \\ \Rightarrow {x^2} - 4 = 0 \\ \therefore x = \pm 2,{\text{ but }}x > 0 \\$
Consequently, $x = 2$ is only the root of the given equation.
Note: When you are using log properties, be careful with the base. When the question says “$\ln$”, it means base is e. On the other hand, when it says “log”, it means the base Is 10, else wise questions will always write base.
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York
2021-02-12
Evaluate the following integrals.
${\int }_{-2}^{-1}\sqrt{-4x-{x}^{2}}dx$
### Answer & Explanation
yagombyeR
Step 1: Given that
${\int }_{-2}^{-1}\sqrt{-4x-{x}^{2}}dx$
Step 2: Formula Used
$\int \sqrt{{a}^{2}-{x}^{2}}=\frac{x}{2}\sqrt{{a}^{2}-{x}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{a}\right)+C$
Step 3: Solve
We have,
${\int }_{-2}^{-1}\sqrt{-4x-{x}^{2}}dx={\int }_{-2}^{-1}\sqrt{-\left(4x+{x}^{2}\right)}dx$
$={\int }_{-2}^{-1}\sqrt{-\left(4x+{x}^{2}+4-4\right)}dx$
$={\int }_{-2}^{-1}\sqrt{-\left({\left(x+2\right)}^{2}-4\right)}dx$
$={\int }_{-2}^{-1}\sqrt{4-{\left(x+2\right)}^{2}}dx$
$={\int }_{-2}^{-1}\sqrt{{\left(2\right)}^{2}-{\left(x+2\right)}^{2}}dx$
$={\left[\frac{x+2}{2}\sqrt{4-{\left(x+2\right)}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{x+2}{2}\right)\right]}_{-2}^{-1}$
$=\left[\frac{-1+2}{2}\sqrt{4-{\left(-1+2\right)}^{2}}+2{\mathrm{sin}}^{-1}\left(\frac{-1+2}{2}\right)-\frac{-2+2}{2}\sqrt{4-{\left(-2+2\right)}^{2}}=2{\mathrm{sin}}^{-1}\left(\frac{-2+2}{2}\right)\right]$
$=\left[\frac{1}{2}\sqrt{4-1}+2{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)-0-0\right]$
$=\frac{\sqrt{3}}{2}+2{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\pi }{6}\right)$
Do you have a similar question?
Recalculate according to your conditions!
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+0
# Hello!
0
331
1
+6
Rodger wants to carpet a rectangular room with a length of 8 3/4 feet and a width of 9 1/3 feet. If he has 85 square feet of carpet, how many square feet of carpet will be left after he covers the room?
Aug 28, 2017
#1
+2339
+1
To solve this problem, we must figure out how much square feet the entire room is. We know the length and the width of the room. Since, this room is rectangular, we can use that information to figure out how much square feet this room covers.
$$8\frac{3}{4}*9\frac{1}{3}$$ First, convert both fractions to improper fractions so that we can multiply them. I will convert each one separately. $$8\frac{3}{4}=\frac{4*8+3}{4}=\frac{35}{4}$$ $$9\frac{1}{3}=\frac{3*9+1}{3}=\frac{28}{3}$$ Now, let's multiply both together. $$\frac{35}{4}*\frac{28}{3}$$ 4 and 28 have a common factor of 4. Doing this simplification makes it easier computationally. $$\frac{35}{1}*\frac{7}{3}$$ Now, do the multiplication. One-digit multiplication is simpler than doing 2-digit. $$\frac{245}{3}$$
We aren't done yet, though! The question is asking how much carpet is remaining after covering the entire floor. This requires the subtraction operator.
$$85-\frac{245}{3}$$ Make 85 a fraction in which it has a denominator of 3. $$\frac{85}{1}*\frac{3}{3}=\frac{255}{3}$$ $$\frac{255}{3}-\frac{245}{3}$$ Now, we can subtract knowing that we have common denominators. $$\frac{10}{3}ft^2$$ And of course, do not forget to include units, if applicable.
Therefore, after completely covering the $$\left(8\frac{3}{4}\right)^{'} * \left(9\frac{1}{3}\right)^{'}$$rectangular room in carpet, Rodger will have $$\frac{10}{3}ft^2=3\frac{1}{3}ft^2=3.\overline{33}ft^2$$ of carpet left.
Aug 28, 2017
#1
+2339
+1
To solve this problem, we must figure out how much square feet the entire room is. We know the length and the width of the room. Since, this room is rectangular, we can use that information to figure out how much square feet this room covers.
$$8\frac{3}{4}*9\frac{1}{3}$$ First, convert both fractions to improper fractions so that we can multiply them. I will convert each one separately. $$8\frac{3}{4}=\frac{4*8+3}{4}=\frac{35}{4}$$ $$9\frac{1}{3}=\frac{3*9+1}{3}=\frac{28}{3}$$ Now, let's multiply both together. $$\frac{35}{4}*\frac{28}{3}$$ 4 and 28 have a common factor of 4. Doing this simplification makes it easier computationally. $$\frac{35}{1}*\frac{7}{3}$$ Now, do the multiplication. One-digit multiplication is simpler than doing 2-digit. $$\frac{245}{3}$$
We aren't done yet, though! The question is asking how much carpet is remaining after covering the entire floor. This requires the subtraction operator.
$$85-\frac{245}{3}$$ Make 85 a fraction in which it has a denominator of 3. $$\frac{85}{1}*\frac{3}{3}=\frac{255}{3}$$ $$\frac{255}{3}-\frac{245}{3}$$ Now, we can subtract knowing that we have common denominators. $$\frac{10}{3}ft^2$$ And of course, do not forget to include units, if applicable.
Therefore, after completely covering the $$\left(8\frac{3}{4}\right)^{'} * \left(9\frac{1}{3}\right)^{'}$$rectangular room in carpet, Rodger will have $$\frac{10}{3}ft^2=3\frac{1}{3}ft^2=3.\overline{33}ft^2$$ of carpet left.
TheXSquaredFactor Aug 28, 2017
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Median formula class 10th
Class 10
Median formula class 10 is included in statistics, to measure of central tendency that is often used to describe the typical value of a set of observations.. The median divides the observations into two halves: half of the observations are smaller than the median, and half are larger.
The formula for calculating the median depends on whether the number of observations in the set is odd or even. In the case of an odd number of observations, the median is the middle value of the ordered set. In the case of an even number of observations, the median is the average of the two middle values of the ordered set.
The median is a useful measure of central tendency because it is not affected by extreme values or outliers in the data set. It is commonly used in fields such as economics, finance, and healthcare to describe the typical value of a set of observations.
Steps to calculate median (Grouped data)
Calculating the median for grouped data involves the following steps:
1. Arrange the set of observations in ascending or descending order.
2. Determine the number of observations, denoted by $n$.
3. If $n$ is odd, the median is the middle observation. If $n$ is even, the median is the average of the two middle observations.
4. To find the middle observation(s), divide $n$ by 2 to get the position of the middle observation(s) in the ordered set. If $n$ is odd, the middle observation is at position $\frac{n+1}{2}$. If $n$ is even, the two middle observations are at positions $\frac{n}{2}$ and $\frac{n}{2}+1$.
5. If $n$ is odd, the median is the observation at the middle position. If $n$ is even, the median is the average of the two observations at the middle positions.
For example, suppose we have the following set of observations: 7, 2, 3, 8, 5. To calculate the median, we follow these steps:
1. Arrange the observations in ascending order: 2, 3, 5, 7, 8.
2. Determine the number of observations, $n=5$.
3. Since $n$ is odd, the median is the middle observation.
4. The middle observation is at position $\frac{n+1}{2}=\frac{5+1}{2}=3$.
5. The median is the observation at the middle position, which is 5.
Therefore, the median of the set of observations 7, 2, 3, 8, 5 is 5.
Examples on Median
Example 1: Find the median of the following set of observations: 10, 15, 20, 25, 30, 35.
Solution:
1. Arrange the observations in ascending order: 10, 15, 20, 25, 30, 35.
2. Determine the number of observations, $n=6$.
3. Since $n$ is even, the median is the average of the two middle observations.
4. The two middle observations are at positions $\frac{n}{2}=\frac{6}{2}=3$ and $\frac{n}{2}+1=\frac{6}{2}+1=4$.
5. The median is the average of the two observations at the middle positions: $\frac{20+25}{2}=22.5$.
Therefore, the median of the set of observations 10, 15, 20, 25, 30, 35 is 22.5.
Example 2 : Find the median of the following set of observations: 2, 2, 3, 3, 3, 4, 5, 6.
Solution:
1. Arrange the observations in ascending order: 2, 2, 3, 3, 3, 4, 5, 6.
2. Determine the number of observations, $n=8$.
3. Since $n$ is even, the median is the average of the two middle observations.
4. The two middle observations are at positions $\frac{n}{2}=\frac{8}{2}=4$ and $\frac{n}{2}+1=\frac{8}{2}+1=5$.
5. The median is the average of the two observations at the middle positions: $\frac{3+3}{2}=3$.
Therefore, the median of the set of observations 2, 2, 3, 3, 3, 4, 5, 6 is 3.
Steps to calculate median (Ungrouped data)
The formula to calculate the median of ungrouped data is as follows:
1. Arrange the observations in ascending or descending order.
2. Determine the number of observations, denoted by $n$.
3. If $n$ is odd, the median is the middle observation. If $n$ is even, the median is the average of the two middle observations.
If $X_{(1)}, X_{(2)}, ..., X_{(n)}$ represent the ordered observations, then the median can be calculated as:
• If $n$ is odd: Median = $X_{\left(\frac{n+1}{2}\right)}$
• If $n$ is even: Median = $\frac{X_{\left(\frac{n}{2}\right)} + X_{\left(\frac{n}{2}+1\right)}}{2}$
Here, $X_{(k)}$ represents the $k$th observation in the ordered set.
For example, consider the following set of ungrouped data: 8, 5, 6, 4, 2, 7, 3, 1.
1. Arrange the observations in ascending order: 1, 2, 3, 4, 5, 6, 7, 8.
2. Determine the number of observations, $n=8$.
3. Since $n$ is even, the median is the average of the two middle observations.
4. The two middle observations are at positions $\frac{n}{2}=\frac{8}{2}=4$ and $\frac{n}{2}+1=\frac{8}{2}+1=5$.
5. The median is the average of the two observations at the middle positions: $\frac{4+5}{2}=4.5$.
Therefore, the median of the set of ungrouped data 8, 5, 6, 4, 2, 7, 3, 1 is 4.5.
Example 1: Find the median of the following set of observations: 5, 8, 3, 9, 2.
Solution:
1. Arrange the observations in ascending order: 2, 3, 5, 8, 9.
2. Determine the number of observations, $n=5$.
3. Since $n$ is odd, the median is the middle observation.
4. The middle observation is at position $\frac{n+1}{2}=\frac{5+1}{2}=3$.
5. The median is the observation at the middle position, which is 5.
Therefore, the median of the set of observations 5, 8, 3, 9, 2 is 5.
Example 2: Find the median of the following set of observations: 10, 15, 20, 25, 30, 35.
Solution:
1. Arrange the observations in ascending order: 10, 15, 20, 25, 30, 35.
2. Determine the number of observations, $n=6$.
3. Since $n$ is even, the median is the average of the two middle observations.
4. The two middle observations are at positions $\frac{n}{2}=\frac{6}{2}=3$ and $\frac{n}{2}+1=\frac{6}{2}+1=4$.
5. The median is the average of the two observations at the middle positions: $\frac{20+25}{2}=22.5$.
Therefore, the median of the set of observations 10, 15, 20, 25, 30, 35 is 22.5.
Example 3: Find the median of the following set of observations: 2, 2, 3, 3, 3, 4, 5, 6.
Solution:
1. Arrange the observations in ascending order: 2, 2, 3, 3, 3, 4, 5, 6.
2. Determine the number of observations, $n=8$.
3. Since $n$ is even, the median is the average of the two middle observations.
4. The two middle observations are at positions $\frac{n}{2}=\frac{8}{2}=4$ and $\frac{n}{2}+1=\frac{8}{2}+1=5$.
5. The median is the average of the two observations at the middle positions: $\frac{3+3}{2}=3$.
Therefore, the median of the set of observations 2, 2, 3, 3, 3, 4, 5, 6 is 3.
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# Introduction to sequences
0/2
##### Intros
###### Lessons
1. Overview:
2. Notation of Sequences
3. Definitions and theorems of Sequences
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##### Examples
###### Lessons
1. Finding the terms of a sequence
Find the first five terms of the following sequences.
1. $a_n=3(-1)^n$
2. $a_n$= $\frac{n+1}{\sqrt{n+1}}$
3. {$cos(\frac{n\pi}{2})$}
2. Finding the formula for a sequence
Find the formula for the general term $a_n$ for the following sequences
1. {$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, ...$}
2. {$\frac{1}{2}, \frac{2}{5}, \frac{3}{8}, \frac{4}{11}, ...$}
3. {-1, 4, -9, 16, ... }
3. Convergence and divergence of sequences
Evaluate the limits and determine if the following limits are converging or diverging.
1. $\lim$n →$\infty$ $\frac{(-1)^n}{n^2}$
2. $\lim$n →$\infty$ $6(\frac{1}{2})^n$
3. $\lim$n →$\infty$ $\frac{n^3+n+1}{n^2+1}$
###### Topic Notes
In this lesson, we will talk about what sequences are and how to formally write them. Then we will learn how to write the terms out of the sequences when given the general term. We will also learn how to write the general term when given a sequence. After learning the notations of sequences, we will take a look at the limits of sequences. Then we will take a look at some definitions and properties which will help us take the limits of complicating sequences. These theorems include the squeeze theorem, absolute value sequences, and geometric sequences.
## Introduction to Sequences
Welcome to our exploration of sequences, a fundamental concept in mathematics. Our journey begins with an engaging introduction video that sets the stage for understanding these fascinating mathematical patterns. Sequences are ordered lists of numbers that follow specific rules, and they play a crucial role in various mathematical applications. In this lesson, we'll delve into the notation of sequences, providing you with the tools to read and interpret them effectively. We'll also cover the formal definition of sequences, ensuring a solid foundation for future studies. You'll learn how to write individual terms and develop general formulas for sequences, skills that are essential for solving complex mathematical problems. Understanding sequences is vital as they form the basis for many advanced mathematical concepts and real-world applications. By mastering sequences, you'll enhance your problem-solving abilities and gain insights into the beautiful patterns that underlie mathematics.
## What is a Sequence?
A sequence, in mathematical terms, is a formally defined list of numbers that follow a specific pattern. This ordered collection of elements is typically represented using the notation a1, a2, ..., an-1, an, where a1 denotes the first term, a2 the second term, and so on, with an representing the nth term or the general term of the sequence. This notation allows mathematicians to describe and analyze patterns within number sets efficiently.
Two common types of sequences are arithmetic and geometric sequences. An arithmetic sequence is characterized by a constant difference between consecutive terms. For example, the sequence 2, 5, 8, 11, 14 is arithmetic, with a common difference of 3. On the other hand, a geometric sequence has a constant ratio between consecutive terms. The sequence 2, 6, 18, 54, 162 is geometric, with each term being three times the previous one.
Understanding sequences is crucial in mathematics as they form the foundation for many advanced concepts. They play a vital role in calculus, particularly in the study of series and limits. In algebra, sequences help in solving recurrence relations and exploring patterns in number theory. Moreover, sequences find practical applications in various fields beyond pure mathematics.
In computer science, sequences are fundamental to algorithm design and analysis. They are used in generating pseudorandom numbers, creating encryption keys, and developing sorting algorithms. In physics, sequences describe natural phenomena like the Fibonacci sequence, which appears in the arrangement of leaves on plants. Economics utilizes sequences in financial modeling, such as compound interest calculations and predicting market trends.
The study of sequences also enhances problem-solving skills and logical thinking. By recognizing patterns and relationships between numbers, students develop a deeper understanding of mathematical structures. This skill is transferable to many areas of life, from scientific research to everyday decision-making processes.
As we delve deeper into the world of sequences, we uncover their beauty in describing both simple and complex patterns in nature and human-made systems. Whether it's the spiral arrangement of sunflower seeds following the Fibonacci sequence or the exponential growth models in population dynamics, sequences provide a powerful tool for understanding and predicting various phenomena. Their versatility and ubiquity make them an essential concept in mathematics education and scientific research.
## Notation and Writing Sequences
Understanding the formal notation for writing sequences is crucial in mathematics, particularly in the study of series and progressions. This notation provides a concise and standardized way to represent both finite and infinite sequences. The primary components of sequence notation include curly brackets, the general term, and indicators for the sequence's range.
The basic structure of a sequence in formal notation is as follows: {an}, where an represents the general term of the sequence. The subscript 'n' denotes the position of each term in the sequence. This notation is enclosed in curly brackets {} to indicate that it represents a set of numbers in a specific order.
The starting number of a sequence is typically denoted as n = 1, although this can vary depending on the context. For instance, a sequence might be written as {an}n=1, which indicates that the sequence starts at n = 1 and continues infinitely. The infinity symbol () is used to represent infinite sequences, showing that the terms continue without end.
Let's explore some examples to illustrate how to write sequences using this notation:
1. Simple arithmetic sequence: {2n}n=1
This represents the sequence 2, 4, 6, 8, 10, ..., where each term is twice the value of n.
2. Geometric sequence: {2n}n=0
This sequence is 1, 2, 4, 8, 16, ..., starting from n = 0 and doubling each term.
3. More complex formula: {3/(2n)}n=1
This sequence would be 3/2, 3/4, 3/6, 3/8, ..., where each term is 3 divided by twice n.
The general term an is particularly important as it defines the rule for generating each term in the sequence. It can be a simple expression like 2n or a more complex function. For example, in the sequence {(-1)nn}n=1, the general term involves both n and an exponent, resulting in the alternating sequence -1, 2, -3, 4, -5, ...
When writing finite sequences, we can specify the number of terms instead of using the infinity symbol. For instance, {n2}n=110 represents the first 10 square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Understanding and using proper sequence notation is essential for clear communication in mathematics. It allows for precise description of patterns and behaviors in number sequences, which is fundamental in various areas of mathematics, including calculus, number theory, and discrete mathematics.
In conclusion, mastering the formal notation for writing sequences enhances one's ability to analyze and work with mathematical patterns. Whether dealing with simple arithmetic progressions or complex mathematical series, this standardized notation provides a powerful tool for expressing and studying sequences of numbers.
## Finding Terms and General Formulas
Understanding sequences and their general formulas is a crucial skill in mathematics. This section will demonstrate how to find specific terms in a sequence using the general formula, calculate the first few terms, and derive the general formula from a given sequence. We'll emphasize the importance of pattern recognition throughout this process.
To find specific terms in a sequence given the general formula, we simply substitute the term number into the formula. For example, consider the arithmetic sequence with the general formula an = 3n + 2, where n is the term number. To find the 5th term, we substitute n = 5:
a5 = 3(5) + 2 = 15 + 2 = 17
Similarly, for the 10th term:
a10 = 3(10) + 2 = 30 + 2 = 32
Calculating the first few terms of a sequence helps us understand its pattern. Let's use the same formula to find the first five terms:
• a1 = 3(1) + 2 = 5
• a2 = 3(2) + 2 = 8
• a3 = 3(3) + 2 = 11
• a4 = 3(4) + 2 = 14
• a5 = 3(5) + 2 = 17
Now, let's consider the reverse process: deriving the general formula when given a sequence of terms. This requires careful pattern recognition. For instance, given the sequence 2, 6, 18, 54, 162, we can observe that each term is tripled to get the next term. This suggests a geometric sequence with a common ratio of 3.
To find the general formula, we start with the first term (a1 = 2) and multiply by 3 raised to the power of (n-1), where n is the term number. Thus, the general formula is:
an = 2 * 3n-1
We can verify this by calculating a few terms:
• a1 = 2 * 31-1 = 2 * 1 = 2
• a2 = 2 * 32-1 = 2 * 3 = 6
• a3 = 2 * 33-1 = 2 * 9 = 18
Pattern recognition is crucial in working with sequences. Look for common differences (arithmetic sequences), common ratios (geometric sequences), or more complex patterns. For example, in the sequence 1, 4, 9, 16, 25, we recognize these as perfect squares. The general formula would be an = n2.
Sometimes, sequences may have more intricate patterns. Consider 1, 3, 6, 10, 15. By looking at the differences between consecutive terms (2, 3, 4, 5), we can deduce that this is a triangular number sequence. Its general formula is an = n(n+1)/2.
In conclusion, working with sequences involves a interplay between specific terms and general formulas. By practicing these skills and honing your pattern recognition abilities, you'll become proficient at analyzing and generating sequences, a valuable skill in various mathematical and real-world applications.
## Limits of Sequences
In mathematics, the concept of limits in sequences is fundamental to understanding the behavior of infinite series and forms the basis for many advanced calculus concepts. A sequence is an ordered list of numbers that follows a specific pattern or rule. The limit of a sequence represents the value that the terms of the sequence approach as we move further along the sequence, typically as we approach infinity.
The notation for limits of sequences as n approaches infinity is typically written as: lim(n) a_n = L This notation means that as n (the term number) gets arbitrarily large, the sequence a_n approaches the limit L.
Sequences can be classified into two main categories: convergent and divergent sequences. A convergent sequence is one where the terms approach a specific finite value as n approaches infinity. In other words, the limit exists. For example, the sequence 1/n converges to 0 as n approaches infinity. We can write this as: lim(n) 1/n = 0 As n gets larger, 1/n gets closer and closer to 0.
On the other hand, a divergent sequence is one that does not have a finite limit. This can occur in two ways: either the sequence grows without bound, or it oscillates without settling on a particular value. An example of a divergent sequence that grows without bound is n^2. As n increases, n^2 grows infinitely large: lim(n) n^2 = An example of an oscillating divergent sequence is (-1)^n, which alternates between 1 and -1 indefinitely, never settling on a single value.
The squeeze theorem, also known as the sandwich theorem or pinching theorem, is a powerful tool for determining the limit of a sequence when it's difficult to calculate directly. The theorem states that if we have three sequences a_n, b_n, and c_n, where: a_n b_n c_n for all n greater than some value N, and lim(n) a_n = lim(n) c_n = L Then, lim(n) b_n must also equal L.
To apply the squeeze theorem to sequences, we "sandwich" the sequence we're interested in between two other sequences whose limits we can easily determine. If these outer sequences converge to the same limit, then our middle sequence must also converge to that limit. This theorem is particularly useful when dealing with complex sequences or those involving trigonometric functions.
For example, consider the sequence (n * sin(1/n)). We know that -1 sin(x) 1 for all x. Multiplying by n (which is always positive for n > 0), we get: -n n * sin(1/n) n Dividing all parts by n: -1 sin(1/n) 1 As n approaches infinity, both -1/n and 1/n approach 0. Therefore, by the squeeze theorem: lim(n) sin(1/n) = 0
Understanding limits of sequences is crucial for many areas of mathematics, including calculus, analysis, and even applied fields like physics and engineering. By mastering the concepts of convergence, divergence, and tools like the squeeze theorem, students can tackle more complex problems and gain deeper insights into the behavior of mathematical functions and series.
## Properties and Theorems of Sequence Limits
Sequence limits play a crucial role in mathematical analysis, providing the foundation for understanding convergence and divergence in infinite series. This section explores essential properties and theorems related to sequence limits, focusing on limit rules, absolute value sequences, and exponential sequences.
One of the fundamental aspects of sequence limits is the set of limit properties that allow us to manipulate and combine sequences. These properties include the sum rule, difference rule, product rule for sequence limits, and quotient rule for sequence limits. The sum rule states that the limit of the sum of two sequences is equal to the sum of their individual limits, provided both limits exist. Mathematically, if lim(a_n) = A and lim(b_n) = B, then lim(a_n + b_n) = A + B. Similarly, the difference rule asserts that lim(a_n - b_n) = A - B.
The product rule for sequence limits states that the limit of the product of two sequences is equal to the product of their individual limits, again assuming both limits exist. This can be expressed as lim(a_n * b_n) = A * B. The quotient rule for sequence limits, which is slightly more complex, states that if lim(a_n) = A and lim(b_n) = B, with B 0, then lim(a_n / b_n) = A / B. These properties are invaluable tools for evaluating limits of complex sequences by breaking them down into simpler components.
Another important theorem in the study of sequence limits is the absolute value sequence theorem. This theorem states that if the limit of a sequence (a_n) exists and equals L, then the limit of the absolute value of the sequence |a_n| also exists and equals |L|. Mathematically, if lim(a_n) = L, then lim(|a_n|) = |L|. This theorem is particularly significant because it allows us to relate the convergence of a sequence to the convergence of its absolute value, which can be easier to analyze in some cases.
The absolute value sequence theorem has several important implications. It provides a method for proving that certain sequences do not converge by showing that their absolute values do not converge. Additionally, it helps in establishing the relationship between the convergence of a sequence and the convergence of related sequences, such as those obtained by applying continuous functions to the original sequence.
One of the most important theorems in the study of sequence limits is the theorem for limits of exponential sequences. This theorem deals with sequences of the form (x^n), where x is a real number and n is the sequence index. The theorem states that:
1. If |x| < 1, then lim(x^n) = 0 as n approaches infinity.
2. If x = 1, then lim(x^n) = 1 as n approaches infinity.
3. If x = -1, then (x^n) oscillates between 1 and -1 and does not converge.
4. If |x| > 1, then the sequence diverges (approaches infinity or negative infinity, depending on the sign of x).
This theorem has profound implications for understanding the behavior of exponential sequences and their convergence properties. It provides a clear criterion for determining whether an exponential sequence converges or diverges based solely on the base value x. This result is extensively used in various areas of mathematics, including calculus, analysis, and applied mathematics.
The exponential sequence theorem also serves as a foundation for more advanced concepts in mathematical analysis. For instance, it is crucial in the study of power series and their convergence, as well as in the analysis of exponential and logarithmic functions. The theorem's implications extend to practical applications in fields such as physics, engineering, and economics, where exponential growth or decay processes are frequently encountered.
Understanding these properties and theorems is essential for developing a strong foundation in mathematical analysis. They provide powerful tools for evaluating limits, proving convergence and divergence of sequences, and analyzing complex mathematical expressions. By mastering these concepts, students and researchers can tackle more advanced topics in calculus, real analysis, and beyond.
## Conclusion
In this lesson, we've explored the fundamental concepts of sequences, a crucial aspect of mathematical foundations. We began with the introduction video, which laid the groundwork for understanding sequences. We then delved into the definition of sequences, learning how to represent them using proper notation. The lesson covered methods for finding specific terms within a sequence and deriving general formulas to describe entire sequences. We also touched on the concept of limits, which plays a significant role in advanced sequence analysis. It's essential to practice working with sequences to solidify your understanding. We encourage you to explore various mathematical contexts where sequences appear, such as number theory, calculus, and real-world applications. By mastering these concepts, you'll be well-equipped to tackle more complex mathematical challenges. Remember, sequences are a powerful tool in mathematics, and their applications extend far beyond this introductory lesson.
### Introduction to Sequences
Overview:
Notation of Sequences
#### Step 1: Understanding What a Sequence Is
Before diving into the notation of sequences, it's essential to understand what a sequence is. A sequence is a list of numbers that follow a specific pattern. In high school, you might have encountered arithmetic sequences and geometric sequences. These are the types of sequences we will be dealing with. Formally, a sequence is a list of numbers denoted as a1, a2, ..., an-1, an, where a1 is the first term, a2 is the second term, and so on, until an, which is the nth term of the sequence.
#### Step 2: Writing Sequences in a Formal Way
Now that we know what sequences are, the next step is to understand how to write them formally. Instead of listing all the numbers, we can use a more compact notation. One common way to write a sequence is by using the notation an, where n starts from 1 and goes to infinity. This can be written as an with curly brackets, indicating that the sequence continues indefinitely. The value of an represents the value at the nth term, and there is usually a formula associated with it, such as n2 or 4n. This formula allows us to find the value of any term in the sequence by plugging in the value of n.
#### Step 3: Examples of Sequence Notation
Let's look at some examples to understand this better. Consider the sequence denoted by {2n} where n starts from 1 and goes to infinity. To find the terms of this sequence, we plug in the values of n into the formula 2n. For n=1, the first term is 2*1=2. For n=2, the second term is 2*2=4. For n=3, the third term is 2*3=6, and so on. The sequence continues as 2, 4, 6, 8, 10, and so forth, indefinitely.
#### Step 4: Working with More Complex Formulas
Sequences can also be represented using more complex formulas. For example, consider the sequence given by the formula 3/(2n). To find the terms of this sequence, we again plug in the values of n. For n=1, the first term is 3/(2*1)=3/2. For n=2, the second term is 3/(2*2)=3/4. For n=3, the third term is 3/(2*3)=3/6, and so on. The sequence continues as 3/2, 3/4, 3/6, 3/8, and so forth. Each term can be denoted as a1, a2, a3, a4, and so on, where an represents the nth term of the sequence.
#### Step 5: Infinite Sequences
Many sequences are infinite, meaning they continue indefinitely. In the notation {an}, the infinity symbol indicates that the sequence does not have an end. For example, the sequence {2n} starts at n=1 and continues forever, producing terms like 2, 4, 6, 8, 10, and so on. Similarly, the sequence {3/(2n)} also continues indefinitely, producing terms like 3/2, 3/4, 3/6, 3/8, and so forth. Understanding that sequences can be infinite is crucial for working with them in various mathematical contexts.
#### Step 6: Practical Applications
Sequences are not just theoretical constructs; they have practical applications in various fields such as computer science, finance, and engineering. For instance, in computer science, sequences are used in algorithms and data structures. In finance, sequences can represent periodic payments or interest calculations. Understanding the notation and properties of sequences is essential for applying them effectively in these fields.
#### Conclusion
In summary, sequences are lists of numbers that follow a specific pattern, and they can be represented using various notations. The notation {an} is a compact way to represent sequences, where an denotes the value at the nth term. Sequences can be finite or infinite, and they have practical applications in various fields. By understanding the notation and properties of sequences, you can work with them more effectively in both theoretical and practical contexts.
### FAQs
#### 1. What is the difference between an arithmetic and a geometric sequence?
An arithmetic sequence has a constant difference between consecutive terms, while a geometric sequence has a constant ratio between consecutive terms. For example, 2, 5, 8, 11 is arithmetic (common difference of 3), and 2, 6, 18, 54 is geometric (common ratio of 3).
#### 2. How do you find the nth term of an arithmetic sequence?
The nth term of an arithmetic sequence is given by the formula an = a1 + (n - 1)d, where a1 is the first term, n is the position of the term, and d is the common difference.
#### 3. What is a convergent sequence?
A convergent sequence is one where the terms approach a specific finite value (called the limit) as the number of terms increases indefinitely. For example, the sequence 1/n converges to 0 as n approaches infinity.
#### 4. How can the squeeze theorem be used to find sequence limits?
The squeeze theorem states that if an bn cn for all n greater than some value, and if lim(an) = lim(cn) = L, then lim(bn) must also equal L. This is useful for finding limits of complex sequences by "sandwiching" them between simpler sequences.
#### 5. What is the limit of the sequence (1 + 1/n)n as n approaches infinity?
The limit of the sequence (1 + 1/n)n as n approaches infinity is the mathematical constant e, approximately equal to 2.71828. This sequence is important in calculus and is often used to define the number e.
### Prerequisite Topics
Before diving into the fascinating world of sequences, it's crucial to have a solid foundation in certain mathematical concepts. One of the most important prerequisite topics for understanding sequences is the Squeeze theorem. This fundamental principle plays a significant role in analyzing and determining the behavior of sequences, making it an essential tool in your mathematical toolkit.
The Squeeze theorem, also known as the sandwich theorem or pinching theorem, is a powerful concept that helps us determine the limit of a sequence by comparing it to two other sequences. This theorem is particularly useful when dealing with complex sequences where direct calculation of the limit might be challenging or impossible.
Understanding the Squeeze theorem for sequences provides a strong foundation for grasping more advanced concepts in sequence analysis. It allows you to approach problems from different angles and often simplifies the process of finding limits. This theorem is not only applicable to sequences but also extends to functions and series, making it a versatile tool in calculus and mathematical analysis.
When studying sequences, you'll frequently encounter situations where the behavior of a sequence is not immediately apparent. The Squeeze theorem comes to the rescue in such scenarios, allowing you to bound the sequence between two simpler sequences whose limits are known or easier to calculate. This approach is particularly valuable when dealing with oscillating or irregularly behaving sequences.
Moreover, the Squeeze theorem helps develop critical thinking and problem-solving skills. It encourages you to think creatively about how to bound a sequence and identify appropriate comparison sequences. These skills are invaluable not only in studying sequences but also in tackling more advanced mathematical concepts and real-world applications.
As you progress in your study of sequences, you'll find that the Squeeze theorem becomes an indispensable tool. It will help you in proving convergence, finding limits, and understanding the behavior of complex sequences. The theorem's applications extend beyond just sequences, making it a fundamental concept in calculus and mathematical analysis.
By mastering the Squeeze theorem, you'll be well-prepared to tackle more advanced topics in sequence analysis. It will provide you with a solid foundation for understanding convergence, divergence, and the limiting behavior of sequences. This knowledge will prove invaluable as you delve deeper into the study of series, infinite sums, and other advanced mathematical concepts.
In conclusion, a thorough understanding of the Squeeze theorem is essential for anyone looking to master the intricacies of sequences. It serves as a bridge between basic algebraic concepts and more advanced calculus topics, making it an indispensable prerequisite for your journey into the world of sequences and beyond.
Note:
1. If a sequence has the limit $L$, then we can say that:
$\lim$n →$\infty$ $a$$n$$=L$
If the limit is finite, then it is convergent. Otherwise, it is divergent.
2. If the limit of the sequences {$a_n$} and {$b_n$} are finite and $c$ is constant, then we can say that
i) $\lim$n →$\infty$ $(a_n+b_n)=\lim$n →$\infty$ $a_n+$$\lim$n →$\infty$ $b_n$.
ii) $\lim$n →$\infty$ $(a_n-b_n)=\lim$n →$\infty$ $a_n-$$\lim$n →$\infty$ $b_n$.
iii) $\lim$n →$\infty$ $ca_n=c$ $\lim$n →$\infty$ $a_n$.
iv) $\lim$n →$\infty$$(a_nb_n)=$ $\lim$n →$\infty$$a_n*$ $\lim$n →$\infty$ $b_n$.
v) $\lim$n →$\infty$ $[a_n$$\div$$b_n]$ $=\lim$n →$\infty$$a_n$$\div$ $\lim$n →$\infty$$b_n$$,$$b_n\neq0$.
3. If $a_n\leq c_n\leq b_n$ and $\lim$n →$\infty$ $a_n=$ $\lim$n →$\infty$ $b_n=L$, then $\lim$n →$\infty$ $c_n=L$.
4.if $\lim$n →$\infty$ $|a_n|=0$, then $\lim$n →$\infty$ $a_n=0$ as well.
5. We say that:
Where the sequence {$x^n$} is convergent for -1< $x \leq$ 1, and divergent if $x$ > 1.
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# A multiple choice test contains 12 questions, each offering 4 possible answers, only 1 of which is correct. A student knows the correct answer to 7 of the questions but randomly guesses the answers to the remaining 5. What is the probability the student will get at least 10 of the 12 correct?
Since the student is guaranteed seven correct responses, she needs at least three of the remaining questions to be correct. Thus we need the probability that she gets 3,4, or 5 of the remaining questions correct.
This situation is an example of a binomial probability distribution. Each event can be described in a binary fashion (either correct or incorrect), each event is independent (the results of one does not effect the others), the probabilities do not change, and there are a finite number of events.
The formula for a binomial probability is given by:
`P(x=k)= ._(n)C_(k)(p)^(k)(1-p)^(n-k)`
where k is the number of successes we seek, n is the number of trials, and p is the probability of success. (For example, suppose we want the probability of 3 successes. The probability of the 1st success is 1/4, the next 1/4, and the 3rd 1/4. The remaining events are failures with a probability of 3/4 each. Using the multiplication principle, we get `(1/4)^3 * (3/4)^2` . However, this can occur in 10 different ways. Suppose we number the 5 questions 1–5. Then she could give correct answers to questions 123, 124, 125, 134, 135, 145, 234, 235, 245, or 345.)
Here n=5, p=1/4, and we want k=3,4, or 5.
`P(x=3)= ._5C_3(1/4)^3(3/4)^2=10(1/64)(9/16)=90/1024`
`P(x=4) = ._5C_4(1/4)^4(3/4)=5(1/256)(3/4)=15/1024 `
`P(x=5) = ._5C_5(1/4)^5(3/4)^0=1/1024`
Since these are mutually exclusive, the probability of the three events is the sum of the probabilities: P(x=3,4, or 5)=106/1024=53/512 or about 10%.
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# Decision Maths
Prim`s Algorithm
Wiltshire
## In Lesson 1 we learnt about Kruskal`s algorithm,
which was used to solve minimum connector problems. Another method that can be used is Prim`s algorithm. Step 1 Select any node Step 2 Connect it to the nearest node Step 3 Connect one node already selected to the nearest unconnected node. Step 4 Repeat 3 until all nodes are connected.
Prim`s Algorithm
Select any node you like.
Lets select F.
Wiltshire
## Consider the example we looked at last lesson.
Prim`s Algorithm
So connect E to D.
Wiltshire
## The nearest to D, F or C is E which is 2 from D.
Prim`s Algorithm
Connect it to the nearest node.
Wiltshire
## C and D are both 3 away so we can choose either.
Lets select C.
Prim`s Algorithm
only 3 away from F. So connect D to F.
Wiltshire
## The nearest node to either of F or C is D, which is
Prim`s Algorithm
5 away from F. Connect A to F.
Wiltshire
## The nearest to any of these four nodes is A which is
Prim`s Algorithm
We now need to connect the last node, B.
Wiltshire
Connect B to A.
Prim`s Algorithm
Wiltshire
Distance Table
Wiltshire
Wiltshire
## We are going to apply Prim`s algorithm to
the distance table. This demonstrates how a computer could apply the algorithm. Prim`s is more suitable than Kruskal`s as computers have a problem recognising loops. As you go through the algorithm, see if you can relate the procedure to the last example.
## Prim`s on a Distance Table
Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7
Wiltshire
Select any arbitrary node. Delete the row and loop the column that correspond to the node selected. Choose the smallest number in the loop. Delete the row that this smallest number is in. Loop the column that corresponds to the row just deleted. Choose the smallest number in any loop. Repeat steps 4, 5 and 6 until all rows have been deleted and columns looped.
## Prim`s on a Distance Table
Here I have chosen F.
Wiltshire
Loop the column.
Delete row C.
Wiltshire
Loop column C.
Delete row D.
Wiltshire
Loop column D.
Delete row E.
Wiltshire
Loop column E.
Delete row A.
Wiltshire
Loop column A.
Delete row B.
Wiltshire
Loop column B.
Wiltshire
## The algorithm is complete when all the columns have
been looped and the rows crossed out. The circles show the edges in the minimum connector.
## Prim`s on a Distance Table
In this case they are AB, DE, AF, CF, DF
Wiltshire
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# How do you find the derivative of Arccos(x/2)?
Aug 2, 2016
$\frac{d}{\mathrm{dx}} \left({\cos}^{- 1} \left(\frac{x}{2}\right)\right) = - \frac{1}{\sqrt{4 - {x}^{2}}}$
#### Explanation:
Let's go through the derivation for the general form of the derivative of $\arccos \left(x\right)$
Consider $y = {\cos}^{- 1} \left(x\right)$
$\implies x = \cos \left(y\right)$
We can construct a triangle from this. Recall that cosine is adjacent/hypotenuse. This means that adjacent to angle $y$, we have side of length $x$ and that the hypotenuse is of length $1$. By Pythagoras' the length of the opposite side is $\sqrt{1 - {x}^{2}}$.
Now, differentiate our expression:
$\frac{\mathrm{dx}}{\mathrm{dy}} = - \sin \left(y\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin \left(y\right)}$
But from our triangle we can work out sin(y)!. Sine is opposite/hypotenuse so
$\sin \left(y\right) = \sqrt{1 - {x}^{2}}$
$\therefore \frac{d}{\mathrm{dx}} \left({\cos}^{- 1} \left(x\right)\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$
That's the general form, but here we have a function of x inside the function, ie $y \left(u \left(x\right)\right)$. This calls for the chain rule.
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
$u = \frac{x}{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2}$
$y = {\cos}^{- 1} \left(u\right) \implies \frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{\sqrt{1 - {u}^{2}}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {\left(\frac{x}{2}\right)}^{2}}} \cdot \frac{1}{2} = - \frac{1}{\sqrt{4 - {x}^{2}}}$
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# Illustrative Mathematics Grade 8, Unit 2, Lesson 10: Meet Slope
Learning Targets:
• I can draw a line on a grid with a given slope.
• I can find the slope of a line on a grid.
Related Pages
Illustrative Math
#### Lesson 10: Meet Slope
Let’s learn about the slope of a line.
Illustrative Math Unit 8.2, Lesson 10 (printable worksheets)
#### Lesson 10 Summary
The following diagram shows how to find the slope of a line on a grid.
#### Lesson 10.1 Equal Quotients
Write some numbers that are equal to 15 ÷ 12.
#### Lesson 10.2 Similar Triangles on the Same Line
1. The figure shows three right triangles, each with its longest side on the same line.
Your teacher will assign you two triangles. Explain why the two triangles are similar.
2. Complete the table.
#### Lesson 10.3 Multiple Lines with the Same Slope
Open Applet
1. Draw two lines with slope 3. What do you notice about the two lines?
2. Draw two lines with slope 1/2. What do you notice about the two lines?
#### Are you ready for more?
As we learn more about lines, we will occasionally have to consider perfectly vertical lines as a special case and treat them differently. Think about applying what you have learned in the last couple of activities to the case of vertical lines. What is the same? What is different?
#### Lesson 10.4 Different Slopes of Different Lines
Here are several lines.
1. Match each line shown with a slope from this list: 1/2, 2, 1, 0.25, 3/2, 1/2.
2. One of the given slopes does not have a line to match. Draw a line with this slope on the empty grid (F).
#### Lesson 10 Practice Problems
1. Of the three lines in the graph, one has slope 1, one has slope 2, and one has slope 1/5. Label each line with its slope.
2. Draw three lines with slope 2, and three lines with slope 1/3. What do you notice?
3. The figure shows two right triangles, each with its longest side on the same line.
a. Explain how you know the two triangles are similar.
b. How long is XY?
c. For each triangle, calculate (vertical side) ÷ (horizontal side).
d. What is the slope of the line? Explain how you know.
4. Triangle A has side lengths 3, 4, and 5. Triangle B has side lengths 6, 7, and 8.
a. Explain how you know that Triangle B is not similar to Triangle A.
b. Give possible side lengths for Triangle B so that it is similar to Triangle A.
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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Sigma notation ninja tricks 2: splitting sums
[Previous posts in this series: jumping constants]
Trick 2: splitting sums
I’ve written about this before, but it’s worth spelling it out for completeness’ sake. If you have a sum of something which is itself a sum, like this:
$\displaystyle \sum (A + B)$
you can split it up into two separate sums:
$\displaystyle \sum (A + B) = \left( \sum A \right) + \left( \sum B \right)$
(You can also sort of think of this as the sigma “distributing” over the sum.) For example,
$\displaystyle \sum_{k=1}^n (k + 2) = \left( \sum_{k=1}^n k \right) + \left( \sum_{k=1}^n 2 \right).$
Why is this? Last time, the fact that we can pull constants in and out of a sigma came down to a property of addition, namely, that multiplication distributes over it. This, too, turns out to come down to some other properties of addition. As before, let’s think about writing out these sums explicitly, without sigma notation.
First, on the left-hand side, we have something like
$\displaystyle \sum_i (A_i + B_i) = (A_1 + B_1) + (A_2 + B_2) + (A_3 + B_3) + \dots$
And on the right-hand side, we have
$\displaystyle \left(\sum_i A_i\right) + \left(\sum_i B_i \right) = (A_1 + A_2 + A_3 + \dots) + (B_1 + B_2 + B_3 + \dots)$
We can see that we get all the same terms, but in a different order. But as we are all familiar with, the order in which you add things up doesn’t matter: addition is both associative and commutative, so we can freely reorder terms and still get the same sum.1
So $\sum$ “distributes over” sums! Let’s use the example from above to see how this can be useful. Suppose we want to figure out a closed-form expression for
$\displaystyle \sum_{k=1}^n (k + 2).$
If we didn’t otherwise know how to proceed we could certainly start by trying some examples and looking for a pattern. Or we could even be a bit more sophisticated and notice that this sum will be $3 + 4 + 5 + \dots + (n+2)$, so it must be $3$ less than the triangular number $1 + 2 + 3 + \dots + (n+2)$. But we don’t even need to be this clever. If we just distribute the sigma over the addition, we transform the expression into two simpler sums which are easier to deal with on their own:
$\displaystyle \sum_{k=1}^n (k + 2) = \left( \sum_{k=1}^n k \right) + \left( \sum_{k=1}^n 2 \right)$
The first sum is $1 + 2 + 3 + \dots + n$, that is, the $n$th triangular number, which is equal to $n(n+1)/2$. The second sum is just $2 + 2 + \dots + 2$ ($n$ times), so it is equal to $2n$. Thus, an expression for the entire sum is
$\displaystyle \frac{n(n+1)}{2} + 2n = \frac{n(n+1) + 4n}{2} = \frac{n(n+5)}{2}.$
As a double check, is this indeed three less than the $(n+2)$nd triangular number?
$\displaystyle \frac{(n+2)(n+3)}{2} - 3 = \frac{n^2 + 5n + 6}{2} - \frac{6}{2} = \frac{n(n+5)}{2}$
Sure enough! Math works!
1. At least, as long as the sum is finite! This still works for some infinite sums too, but we have to be careful about convergence.
Posted in algebra, arithmetic | Tagged , , , , , | 3 Comments
Book reviews: The Joy of SET and Elements of Mathematics
I have a couple of book reviews for you today! I finished both of these books recently and really enjoyed them. Though they are quite different, both gave me new ways to think about some topics I already knew, and in particular helped me make new connections between elementary and advanced concepts.
[Disclosure of Material Connection: Princeton Press kindly provided me with free review copies of these books. I was not required to write a positive review. The opinions expressed are my own.]
The Joy of SET
Liz McMahon, Gary Gordon, Hannah Gordon, and Rebecca Gordon
Princeton University Press, 2016
Most people are probably familiar with the card game SET: each card has four attributes (number, color, shading, shape) each of which can have one of three values, for a total of $3^4 = 81$ cards. The goal is to find “sets”, which consist of three cards where each attribute is either identical on all three cards, or distinct on all three cards. It’s a fun game, and because it has to do with combinations of things and pattern recognition, many people probably have the intuitive sense that it’s a “mathy” sort of game, or the sort of game that people who enjoy math would also enjoy
Well, it turns out, as the authors convincingly demonstrate, that the mathematics behind SET actually goes very deep. For example, did you know that there are exactly $3^{n-1}(3^n - 1)/2$ distinct SETs in an $n$-dimensional version of the game? (The normal game that everyone plays has $n = 4$.) How about the fact that the SET deck is a concrete model of the four-dimensional affine geometry $AG(4,3)$? Did you know that the most cards you can have without a SET is 20, and that this is intimately connected to structures called maximal caps in affine geometries—and that no one knows how many cards you could have without a SET in a $7$-dimensional (or higher) version of the game?
The authors explain all this, and much more (with a lot of humor1 along the way!), ranging through probability, modular arithmetic, combinatorics, geometry, linear algebra, and a bunch of other topics. The book begins gently, but by the end it gets into some fairly deep mathematics, and there are lots of exercises and projects at the end of each chapter. This book would make a fantastic resource for a middle school, high school, or undergraduate math club. I could even see using it as the textbook for some sort of extra/special topics class with some motivated students.
Elements of Mathematics
John Stillwell
Princeton University Press, 2016
I am a huge fan of Stillwell’s writing (almost six years ago I wrote a short review of another one of his books, Roads to Infinity) and I wasn’t disappointed. This book is definitely aimed at a more sophisticated audience than the SET book, but due to Stillwell’s lucid explanations it still manages to start out rather gently and holds many treasures even for the intrepid high school reader.
The book has two basic goals. The first is to simply lay out an overview of “elementary” mathematics, accessible in theory to anyone with a high school level mathematical background. “Elementary” mathematics refers not just to the sort of mathematics learned in grade school (arithmetic, fractions, and so on) but to the mathematics that would nowadays be viewed as “basic” by professional mathematicians—the sort of stuff that every professional mathematician is familiar with regardless of their specialty. In this respect the book is quite a tour de force, organized by areas of mathematics—arithmetic, computation, algebra, geometry, calculus, and so on—and in each area Stillwell manages to distill down the big ideas and the connections with other areas. He is a master expositor, and the text manages to be engaging and accessible without watering down the mathematics. I definitely learned new things from the book! One thing Stillwell does very well in particular is to explain not just the big ideas but the connections between them.
The other basic goal of the book is to explore the boundary between “elementary” and “advanced” mathematics. This sounds like it would be rather vague and amorphous—after all, aren’t the notions of “elementary” and “advanced” quite relative? Doesn’t it depend on how much background you have? Can’t math that is “elementary” to one person be “advanced” to someone else? This is all true, but Stillwell isn’t really talking about which areas of math are hard and which are easy. Professional mathematicians often talk about certain proofs being “elementary”, and it is often celebrated when someone finds an “elementary” proof of a theorem, even if that theorem had already been proved by “non-elementary” means, and even if the non-elementary proof was shorter. Stillwell is trying to pin down a precise meaning of this sense of “elementary”, and makes a well-reasoned case that it all comes down to infinity: something is non-elementary precisely when infinity enters into its proof in a fundamental way. This may seem rather arbitrary at first blush, but through a number of examples and surprising connections between different areas of mathematics, Stillwell makes it clear that this is an extremely “natural” place to draw a line in the sand. Not that having such a dividing line is in and of itself of any value—it’s simply fascinating to note that there is such a natural line at all, and by exploring it in depth we shed new light on the mathematics to either side of it.
1. They are extremely fond of footnotes. Reminds me of someone I know.
Posted in books, review | Tagged , , , , , , , | 1 Comment
Sigma notation ninja tricks 1: jumping constants
Almost exactly ten years ago, I wrote a page on this blog explaining big-sigma notation. Since then it’s consistently been one of the highest-traffic posts on my blog, and still gets occasional comments and questions. A few days ago, a commenter named Kevin asked,
Could you explain how to take a constant outside of a summation and bring it inside the summation?
This made me realize there’s a lot more still to be explained! In particular, understanding what sigma notation means is one thing, but becoming fluent in its use requires learning a number of “tricks”. Of course, as always, they’re not really “tricks” at all: understanding what the notation means is the necessary foundation for understanding why the tricks work!
Trick 1: jumping constants
For today, we’ll start by considering what Kevin asked about. Consider what is meant by this sigma notation:
$\displaystyle \sum_{i=1}^{4} c X_i$
It doesn’t really matter what the $X$’s are, the point is just that each $X_i$ might be different, whereas $c$ is a constant that doesn’t change. So this can be expanded as
$\displaystyle \sum_{i=1}^{4} c X_i = c X_1 + c X_2 + c X_3 + c X_4$
Since multiplication distributes over addition, we can factor out the $c$:
$c X_1 + c X_2 + c X_3 + c X_4 = c (X_1 + X_2 + X_3 + X_4)$
The right-hand side can now be written as
$\displaystyle c \left( \sum_{i=1}^4 X_i \right),$
so overall we have shown that
$\displaystyle \sum_{i=1}^4 c X_i = c \left(\sum_{i=1}^4 X_i\right).$
We usually omit the parentheses and just write
$\displaystyle c \sum_{i=1}^4 X_i.$
Our argument didn’t really depend on any of the specifics (like the fact that $i$ goes from $1$ to $4$). The general principle is that constants can “jump” back and forth across the sigma, which corresponds to multiplication distributing across addition.
The one remaining question is—what counts as a “constant”? The answer is, anything that doesn’t depend on the index variable. So the “constant” can even involve some variables, as long as they are other variables! For example,
$\displaystyle \sum_{i = 1}^k (n^2 + k) g(i) = (n^2 + k) \sum_{i=1}^k g(i)$
In the context of this sum, $n^2 + k$ is a “constant”, because it does not have $i$ in it. Since it doesn’t contain $i$, it is going to be exactly the same for each term of the sum, which means it can be factored out.
Posted in algebra, arithmetic | Tagged , , , , | 1 Comment
The Riemann zeta function and prime numbers
In a previous post I defined the famous Riemann zeta function,
$\displaystyle \displaystyle \zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}.$
Today I want to give you a glimpse of what it has to do with prime numbers—which is a big part of why it is so famous.
Consider the infinite product
$\displaystyle \left(\frac{1}{1-2^{-s}}\right)\left(\frac{1}{1-3^{-s}}\right)\left(\frac{1}{1-5^{-s}}\right)\left(\frac{1}{1-7^{-s}}\right) \dots \left(\frac{1}{1-p^{-s}}\right) \dots$
where each sequential factor has the next prime number raised to the $-s$ power. Using big-Pi notation, we can write this infinite product more concisely as
$\displaystyle \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}$
(The big $\Pi$ means a $\Pi$roduct just like a big $\Sigma$ means a $\Sigma$um.)
Now let’s do a bit of algebra. First, recall that the infinite geometric series $1 + r + r^2 + r^3 + \dots$ has the sum
$\displaystyle 1 + r + r^2 + r^3 + \dots = \frac{1}{1-r},$
as long as $r < 1$. (For some hints on how to derive this formula if you haven’t seen it before, see this blog post or this one.) Of course, $1/(1-p^{-s})$ is of this form, with $r = p^{-s}$. Note that $p^{-s} = 1/p^s$ which is less than $1$ as long as $s > 0$, so the geometric series formula applies, and we have
$\displaystyle \prod_p \frac{1}{1 - p^{-s}} = \prod_p (1 + p^{-s} + p^{-2s} + p^{-3s} + \dots)$
(From now on I’ll just write $\prod_p$ instead of $\prod_{p\text{ prime}}$.) That is,
$\displaystyle (1 + 2^{-s} + 2^{-2s} + \dots)(1 + 3^{-s} + 3^{-2s} + \dots)(1 + 5^{-s} + 5^{-2s} + \dots) \dots$
So this is an infinite product of infinite sums! But you’re not scared of a little infinity, are you? Good, I thought not. Now, what would happen if we “multiplied out” this infinite product of infinite sums? Note that every term in the result would come from picking one term of the form $p^{-ks}$ from each of the factors, one for each prime $p$, and multiplying them. (Though infinitely many of the choices have to be $1 = p^{-0s}$ if we are to get a finite term as a result.) For example, one way to choose terms would be
$\displaystyle 1 \cdot 3^{-2s} \cdot 5^{-s} \cdot 1 \cdot 13^{-3s} \cdot 1 \cdot 1 \cdot \dots$
which would give us $(3^2 \cdot 5 \cdot 13)^{-s} = 585^{-s}$. In fact, because of the Fundamental Theorem of Arithmetic (every positive integer has a distinct prime factorization), each choice gives us the prime factorization of a different positive integer, and conversely, every positive integer shows up exactly once. That is, after multiplying everything out, we get one term of the form $n^{-s}$ for each positive integer $n$:
$\displaystyle \prod_p \frac{1}{1 - p^{-s}} = \sum_{n \geq 1} n^{-s} = \sum_{n \geq 1} \frac{1}{n^s}$
But that’s just our old friend $\zeta(s)$! So in fact,
$\displaystyle \zeta(s) = \prod_p \frac{1}{1 - p^{-s}}$
turns out to be an equivalent way to write the Riemann zeta function.
We can now use this in a really cute proof that there are infinitely many primes. Consider $\zeta(1)$, where we substitute $1$ for $s$. In our original definition of $\zeta$, we get
$\displaystyle \zeta(1) = \sum_{n \geq 1} \frac{1}{n} = 1 + 1/2 + 1/3 + 1/4 + 1/5 + \dots$
This is known as the harmonic series, and it is a well-known fact that it diverges, that is, as you keep adding up more and more terms of the series, the sum keeps getting bigger and bigger without bound. Put another way, pick any number you like—a hundred, a million, a trillion—and eventually, if you keep adding long enough, the sum $1 + 1/2 + 1/3 + 1/4 + 1/5 + \dots$ will become bigger than your chosen number. (Though you may have to wait a very long time—the harmonic series diverges rather slowly indeed!) One way to prove this is to note that the series $1 + 1/2 + 1/3 + 1/4 + 1/5 + \dots$ is greater than
$\displaystyle 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + \dots) + \dots$
(the original series is greater than this because we only made some of its terms smaller—I changed the $1/3$ into $1/4$, and then changed $1/5$ through $1/7$ into $1/8$, and then $1/9$ through $1/15$ into $1/16$, and so on). But this new smaller series is equal to $1 + 1/2 + 1/2 + 1/2 + 1/2 + \dots$ which will clearly get arbitrarily large. So the harmonic series, which is larger, must diverge as well.
So, $\zeta(1)$ diverges. But what happens if we plug $s = 1$ into the other expression for $\zeta(s)$? We get
$\displaystyle \prod_p \frac{1}{1 - p^{-1}} = \prod_p \frac{1}{1 - 1/p} = \prod_p \frac{p}{p-1}.$
If there were only finitely many primes, this would be a finite product of some fractions and would thus have some definite, finite value—but we know it has to diverge! Thus there must be infinitely many primes.
I will note one other thing—when I was writing up some notes for this post I was initially confused by the fact that if we set, say, $s = 2$, we already know that $\zeta(2) = \pi^2/6$; but now we also know that
$\displaystyle \zeta(2) = \prod_p \frac{p^2}{p^2 - 1} = \frac{4}{3} \cdot \frac{9}{8} \cdot \frac{25}{24} \cdot \frac{49}{48} \cdot \frac{121}{120} \dots$
But this is an infinite product of fractions which are all bigger than $1$! How could it converge? …well, my intuition was just playing tricks with me. Although I have lots of practice thinking about infinite sums that converge, I am just not used to thinking about infinite products that converge. But in the end it is not really any more surprising than the fact that an infinite sum can converge even though all its terms are positive: as long as the fractions are getting smaller quickly enough, such an infinite product certainly can converge, and in fact it does. Using a computer confirms that the more terms of this product we include, the closer the product gets to $\pi^2 / 6 \approx 1.64493\dots$
Posted in number theory | Tagged , , , , | 15 Comments
Games with factorization diagram cards
Since I published a deck of factorization diagram cards last September, a few teachers have picked up copies of the cards and started using them with their students. I’ve started collecting ideas for games you can play using the cards, and want to share here a few game ideas from Alex Ford who teaches middle school in St. Paul, Minnesota.
If you want to get your own set you can buy one here! Also, if you have any other ideas for games or activities using the cards, please send them my way.
War
First, you can play a classic game of War. The twist is that while playing you should only look at the diagram side of the cards, not the side with the written-out number. So part of the game is figuring out which factorization diagram represents a bigger number. One could of course just work out what each number is and then compare, but I imagine students may also find tricks they can use to decide which is bigger without fully working out what the numbers are.
Variant 1: primes are wild, that is, primes always beat composite numbers. (If you have two primes or two composite numbers, then the higher one beats the lower one as usual.) This may actually make the game a bit easier, since when a prime is played you don’t actually need to work out the value of any composite number played in opposition to it.
Variant 2: like variant 1, except that primes only beat those composite numbers which don’t have them as a factor. For example, 5 beats 24, but 5 loses to 30: since 30 has 5 as a prime factor it is “immune” to 5.
As a fun follow-on activity to variant 2, try listing the cards in order according to which beats which!1
Set
Alex and his students came up with a fun variant on SET. Start by dealing out twelve factorization cards, diagram-side-up. Like the usual SET game, the aim is to find and claim sets of three cards. The difference is in how sets are defined. A “set” of factorization cards is any set of three cards that either
1. Share no prime factors in common (that is, any given prime occurs on at most one of the cards), or
2. Share all their prime factors in common (each prime that appears on any of the cards must appear on all three).
Here are a few examples of valid sets:
And here are a few invalid sets:
In order to claim a set you have to state the number on each card and explain why they form a set. If you are correct, remove the cards and deal three new cards. If you are incorrect, keep looking!
Alex and his students found that, just as with the classic SET game, it is possible to have a layout of twelve cards containing no set. For example, here’s the layout they found:
Just to double-check, I confirmed with a computer program that the above layout indeed contains no valid sets. As with the usual SET, if you find yourself in a situation where everyone agrees there are no sets, you can just deal out three more cards.
The natural follow-up question is: what’s the largest possible layout with no sets? So far, this is an open question!
1. 😉
Posted in arithmetic, counting, games, pattern, pictures, primes, teaching | Tagged , , , , , | 4 Comments
The MacLaurin series for sin(x)
In my previous post I said “recall the MacLaurin series for $\sin x$:”
$\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$
Since someone asked in a comment, I thought it was worth mentioning where this comes from. It would typically be covered in a second-semester calculus class, but it’s possible to understand the idea with only a very basic knowledge of derivatives.
First, recall the derivatives $\sin'(x) = \cos(x)$ and $\cos'(x) = -\sin(x)$. Continuing, this means that the third derivative of $\sin(x)$ is $-\cos(x)$, and the derivative of that is $\sin(x)$ again. So the derivatives of $\sin(x)$ repeat in a cycle of length 4.
Now, suppose that an infinite series representation for $\sin(x)$ exists (it’s not at all clear, a priori, that it should, but we’ll come back to that). That is, something of the form
$\displaystyle \sin(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots$
What could this possibly look like? We can use what we know about $\sin(x)$ and its derivatives to figure out that there is only one possible infinite series that could work.
First of all, we know that $\sin(0) = 0$. When we plug $x=0$ into the above infinite series, all the terms with $x$ in them cancel out, leaving only $a_0$: so $a_0$ must be $0$.
Now if we take the first derivative of the supposed infinite series for $\sin(x)$, we get
$\displaystyle a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + \dots$
We know the derivative of $\sin(x)$ is $\cos(x)$, and $\cos(0) = 1$: hence, using similar reasoning as before, we must have $a_1 = 1$. So far, we have
$\displaystyle \sin(x) = x + a_2x^2 + a_3x^3 + \dots$
Now, the second derivative of $\sin(x)$ is $-\sin(x)$. If we take the second derivative of this supposed series for $\sin(x)$, we get
$\displaystyle 2a_2 + (3 \cdot 2)a_3 x + (4 \cdot 3)a_4 x^2 + \dots$
Again, since this should be $-\sin(x)$, if we substitute $x = 0$ we ought to get zero, so $a_2$ must be zero.
Taking the derivative a third time yields
$\displaystyle (3 \cdot 2) a_3 + (4 \cdot 3 \cdot 2)a_4 x + (5 \cdot 4 \cdot 3) a_5 x^2 + \dots$
and this is supposed to be $-\cos(x)$, so substituting $x = 0$ ought to give us $-1$: in order for that to happen we need $(3 \cdot 2)a_3 = -1$, and hence $a_3 = -1/6$.
To sum up, so far we have discovered that
$\displaystyle \sin(x) = x - \frac{x^3}{6} + a_4x^4 + a_5x^5 + \dots$
Do you see the pattern? When we take the $n$th derivative, the constant term is going to end up being $n! \cdot a_n$ (because it started out as $a_n x^n$ and then went through $n$ successive derivative operations before the $x$ term disappeared: $a_n x^n \to n a_n x^{n-1} \to (n \cdot (n-1)) a_n x^{n-2} \to \dots \to n! \cdot a_n$). If $n$ is even, the $n$th derivative will be $\pm \sin(x)$, and so the constant term should be zero; hence all the even coefficients will be zero. If $n$ is odd, the $n$th derivative will be $\pm \cos(x)$, and so the constant term should be $\pm 1$: hence $n! \cdot a_n = \pm 1$, so $a_n = \pm 1/n!$, with the signs alternating back and forth. And this produces exactly what I claimed to be the expansion for $\sin x$:
$\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$
Using some other techniques from calculus, we can prove that this infinite series does in fact converge to $\sin x$, so even though we started with the potentially bogus assumption that such a series exists, once we have found it we can prove that it is in fact a valid representation of $\sin x$. It turns out that this same process can be performed to turn almost any function into an infinite series, which is called the Taylor series for the function (a MacLaurin series is a special case of a Taylor series). For example, you might like to try figuring out the Taylor series for $\cos x$, or for $e^x$ (using the fact that $e^x$ is its own derivative).
Posted in calculus, infinity, iteration | Tagged , , , , , , , | 4 Comments
The Basel problem
I wanted to follow up on something I mentioned in my previous post: I claimed that
$\displaystyle \displaystyle \zeta(2) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots = \frac{\pi^2}{6}.$
At the time I didn’t know how to prove this, but I did some quick research and today I’m going to explain it! It turns out that determining the value of this infinite sum was a famous open question from the mid-1600s until it was solved by Leonhard Euler in 1734. It is now known as the Basel problem (it’s not clear to me whether it was called that when Euler solved it). Since then, there have been many different proofs using all sorts of techniques, but I think Euler’s original proof is still the easiest to follow (though it turns out to implicitly rely on some not-so-obvious assumptions, so a completely formal proof is still quite tricky). I learned about this proof from some slides by Brendan Sullivan and an accompanying document.
First, recall the MacLaurin series for $\sin x$:
$\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$
This infinite sum continues forever with successive odd powers of $x$, alternating between positive and negative. (If you’ve never seen this before, you can take my word for it I suppose; if anyone asks in a comment I would be happy to write another post explaining where this comes from.)
If we substitute $\pi x$ for $x$ we get
$\displaystyle \sin(\pi x) = \pi x - \frac{(\pi x)^3}{3!} + \frac{(\pi x)^5}{5!} - \frac{(\pi x)^7}{7!} + \dots$
Note that the coefficient of $x^3$ is $-\pi^3 / 3! = -\pi^3/6$. Remember that—it will return later!
Now, recall that for finite polynomials, the Fundamental Theorem of Algebra tells us that we can always factor them into a product of linear factors, one for each root (technically, this is only true if we allow for complex roots, though we won’t need that fact here). For example, consider the polynomial
$\displaystyle 2x^3 - 3x^2 - 11x + 6.$
It turns out that this has zeros at $x = 3$, $-2$, and $1/2$, as you can verify by plugging in those values for $x$. By the Fundamental Theorem, this means it must be possible to factor this polynomial as
$\displaystyle 2(x-3)(x+2)(x-1/2).$
Note how each factor corresponds to one of the roots: when $x = 3$, then $(x-3)$ is zero, making the whole product zero; when $x = -2$, the $(x+2)$ becomes zero, and so on. We also had to put in a constant multiple of 2, to make sure the coefficient of $x^3$ is correct.
So, we can always factorize finite polynomials in this way. Can we do something similar for infinite polynomials, like the MacLaurin series for $\sin(\pi x)$? Euler guessed so. It turns out the answer is “yes, under certain conditions”, but this is not at all obvious. This is known as the Weierstrass factorization theorem, but I won’t get into the details. You can just take it on faith that it works in this case, so we can “factorize” the MacLaurin series for $\sin(\pi x)$, getting one linear factor for each root, that is, for each integer value of $x$:
$\displaystyle \displaystyle \sin(\pi x) = \pi x (1 - x)(1 + x)\left (1 - \frac{x}{2} \right) \left(1 + \frac{x}{2} \right) \left(1 - \frac{x}{3}\right) \left(1 + \frac{x}{3}\right) \dots$
For example, $x = 3$ makes the $(1 - x/3)$ term zero, and in general $x = n$ will make the $(1 - x/n)$ term zero. Note how we also included a factor of $x$, corresponding to the root at $x = 0$. We also have to include a constant factor of $\pi$: this means that the coefficient of $x^1$ in the resulting sum (obtained by multiplying the leading $\pi x$ by all the copies of $1$) will be $\pi$, as it should be.
Now, since $(a-b)(a+b) = a^2 - b^2$ we can simplify this as
$\displaystyle \sin(\pi x) = \pi x (1 - x^2) \left(1 - \frac{x^2}{4} \right) \left( 1 - \frac{x^2}{9} \right) \dots$
Let’s think about what the coefficient of $x^3$ will be once this infinite product is completely distributed out and like degrees of $x$ are collected. The only way to get an $x^3$ term is by multiplying the initial $\pi x$ by a single term of the form $-x^2/n^2$, and then a whole bunch of $1$’s. There is one way to do this for each possible $n \geq 1$. All told, then, we are going to have
$\displaystyle \sin(\pi x) = \pi x - \pi x^3 \left(1 + \frac{1}{4} + \frac{1}{9} + \dots \right) + \dots$
And now we’re almost done: recall that previously, by considering the MacLaurin series, we concluded that the coefficient of $x^3$ in $\sin(\pi x)$ is $-\pi^3 / 6$. But looking at it a different way, we have now concluded that the coefficient is $-\pi(1 + 1/4 + 1/9 + \dots)$. Setting these equal to each other, and dividing both sides by $-\pi$, we conclude that
$\displaystyle \zeta(2) = 1 + \frac 1 4 + \frac 1 9 + \dots = -\frac{\pi^3}{6} \cdot \frac{1}{-\pi} = \frac{\pi^2}{6}.$
Magic!
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# ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS
Addition and Subtraction of Rational Expressions :
Here we are going to how to add or subtract two rational expressions.
Like adding and subtracting two rational numbers, we have two methods to add or subtract two rational expressions.
Addition and Subtraction of Rational Expressions with Like Denominators :
(i) Add or Subtract the numerators
(ii) Write the sum or difference of the numerators found in step (i) over the common denominator.
(iii) Reduce the resulting rational expression into its lowest form
Addition and Subtraction of Rational Expressions with unlike Denominators :
(i) Determine the Least Common Multiple of the denominator.
(ii) Rewrite each fraction as an equivalent fraction with the LCM obtained in step (i). This is done by multiplying both the numerators and denominator of each expression by any factors needed to obtain the LCM.
(iii) Follow the same steps given for doing addition or subtraction of the rational expression with like denominators.
## Addition and Subtraction of Rational Expressions - Questions
Question 1 :
Simplify
(i) [x (x + 1)/(x - 2)] + [x(1 - x)/(x - 2)]
Solution :
Since the denominator of both fractions are same, we have to put only one denominator and add the fractions.
(ii) (x + 2)/(x + 3) + (x - 1)/(x - 2)
Solution :
(iii) [x3/(x - y)] + [y3/(y - x)]
Solution :
= [x3/(x - y)] + [y3/(y - x)]
= [x3/(x - y)] - [y3/(x - y)]
= (x3 - y3)/(x - y)
= (x - y) (x2 + xy + y2)/(x - y)
= (x2 + xy + y2)
Hence the value of the given rational expression is (x2+xy+y2).
Question 2 :
Simplify
(i) [(2x + 1)(x - 2)/(x - 4)] - [(2x2 - 5x + 2)/(x - 4)]
Solution :
= [(2x + 1)(x - 2)/(x - 4)] - [(2x2 - 5x + 2)/(x - 4)]
= [(2x2 - 4x + x - 2) - (2x2 - 5x + 2)]/(x - 4)
= [(2x2 - 3x - 2 - 2x2 + 5x - 2)]/(x - 4)
= (2x - 4)/(x - 4)
= 2(x - 2)/(x - 4)
(ii) 4x/(x2 - 1) - (x + 1)/(x - 1)
Solution :
= 4x/(x2 - 1) - (x + 1)/(x - 1)
= 4x/(x + 1)(x - 1) - (x + 1)/(x - 1)
Taking L.C.M, we get
= 4x/(x + 1)(x - 1) - (x + 1)(x + 1)/(x - 1)(x + 1)
= [4x - (x + 1)2]/(x + 1)(x - 1)
= [4x - (x2 + 2x + 1)]/(x + 1)(x - 1)
= [4x - x2 - 2x - 1]/(x + 1)(x - 1)
= (-x2 + 2x - 1)/(x + 1)(x - 1)
= -(x2 - 2x + 1)/(x + 1)(x - 1)
= -(x - 1)(x - 1)/(x + 1)(x - 1)
= -(x - 1)/(x + 1)
= (1 - x)/(1 + x)
Question 3 :
Subtract 1/(x2 + 2) from (2x3 + x2 + 3)/(x2 + 2)2
Solution :
= [(2x3 + x2 + 3)/(x2 + 2)2] - [1/(x2 + 2)]
By taking LCM, we get
= [(2x3 + x2 + 3)/(x2 + 2)2] - [(x2 + 2)/(x2 + 2)2]
= [(2x3 + x2 + 3) - (x2 + 2)/(x2 + 2)2]
= (2x3 + x2 + 3 - x2 - 2)/(x2 + 2)2
= (2x3+ 1)/(x2 + 2)2
After having gone through the stuff given above, we hope that the students would have understood, "Operations with Rational Expressions Worksheet".
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Experience Summary
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Credentials
Type Subject Issued-By Level Year
Degree Electrical Engineering Marquette University MS 1971
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Experience Summary
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Russian Peasant Multiplication:
Fun With Binary
by John C. Darrow
(1986 or earlier)
Folklore has it that, in certain Siberian villages, the peasants have never learned to multiply numbers as we do. However, they have developed a method that uses only doubling, halving, and addition of numbers to accomplish multiplication of two numbers (See Figure 1). The numbers to be multiplied are written side by side, with space below them. One of the numbers (usually the smaller) is successively halved, with fractions dropped, and the results written below it, until the number 1 is reached. The other number is successively doubled, with results written beside the other column of numbers. Any even numbers in the first column are considered "evil" and eliminated along with the corresponding numbers in the other column. The remaining numbers in the second column are added, and the result is invariably the correct answer.
Why does it work? Figure 2 illustrates. The second column contains the second number times successive powers of two (starting with 1, the zero-th power of two), while the process of halving the first number and only keeping the odd results has effectively translated it into binary (a sum of powers of two.) Adding the numbers in the second column adds the second number times each of the appropriate powers of two to give the correct answer.
Figure 1
An example: 29 * 47
A B Successively halved Successively doubled 29 47 14 94 7 188 3 376 1 752
Now, since 14 in Column A is even, it is eliminated, along with its partner 94. The remaining numbers in Column B are added.
A B Successively halved Successively doubled 29 47 Xxxxx(14) Xxxxx(94) 7 188 3 376 1 752 ------- Total 1363
The result, 1363, is the correct answer to 29 times 47.
Figure 2
Why It Works
A B C D Successively halved Successively doubled Remainder of Column B Column A / 2 equivalent 29 47 1 47 * 1 Xxxxx(14) Xxxxx(94) 0 47 * 2 7 188 1 47 * 4 3 376 1 47 * 8 1 752 1 47 * 16 ------- Total 1363
Reading Column C up yields 11101, the binary equivalent of 29 (16 + 8 + 4 + 0 + 1). When 94 is eliminated, and the other numbers from Column B added, we can use the equivalents from Column D to see we have:
(47 * 16) + (47 * 8) + (47 * 4) + (47 * 1)
Rearranging, we have:
47 * (16 + 8 + 4 + 1)
or:
47 * 29
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Sequences
Sequences - Sequences Lets start off this section with a...
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Sequences Let’s start off this section with a discussion of just what a sequence is. A sequence is nothing more than a list of numbers written in a specific order. The list may or may not have an infinite number of terms in them although we will be dealing exclusively with infinite sequences in this class. General sequence terms are denoted as follows, Because we will be dealing with infinite sequences each term in the sequence will be followed by another term as noted above. In the notation above we need to be very careful with the subscripts. The subscript of denotes the next term in the sequence and NOT one plus the n th term! In other words, so be very careful when writing subscripts to make sure that the “+1” doesn’t migrate out of the subscript! This is an easy mistake to make when you first start dealing with this kind of thing. There is a variety of ways of denoting a sequence. Each of the following are equivalent ways of denoting a sequence.
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In the second and third notations above a n is usually given by a formula. A couple of notes are now in order about these notations. First, note the difference between the second and third notations above. If the starting point is not important or is implied in some way by the problem it is often not written down as we did in the third notation. Next, we used a starting point of in the third notation only so we could write one down. There is absolutely no reason to believe that a sequence will start at . A sequence will start where ever it needs to start. Let’s take a look at a couple of sequences. Example 1 Write down the first few terms of each of the following sequences. (a) [ Solution ] (b) [ Solution ] (c) , where [ Solution ] Solution (a) To get the first few sequence terms here all we need to do is plug in values of n into the formula given and we’ll get the sequence terms.
Note the inclusion of the “…” at the end! This is an important piece of notation as it is the only thing that tells us that the sequence continues on and doesn’t terminate at the last term. [ Return to Problems ] (b) This one is similar to the first one. The main difference is that this sequence doesn’t start at . Note that the terms in this sequence alternate in signs. Sequences of this kind are sometimes called alternating sequences. [ Return to Problems ] (c) , where This sequence is different from the first two in the sense that it doesn’t have a specific formula for each term. However, it does tell us what each term should be. Each term should be the n th digit of π. So we know that The sequence is then, [ Return to Problems ] In the first two parts of the previous example note that we were really treating the formulas as functions that can only have integers plugged into them. Or,
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This is an important idea in the study of sequences (and series). Treating the sequence terms as function evaluations will allow us to do many things with sequences that couldn’t do otherwise. Before delving further into this idea however we need to get a
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This note was uploaded on 11/10/2011 for the course MATH 136 taught by Professor Prellis during the Fall '08 term at Rutgers.
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Sequences - Sequences Lets start off this section with a...
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# Find the differential equation of the family of all straight lines passing through the origin.
Last updated date: 09th Aug 2024
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Hint: First of all, write the equation of the line passing through the origin by using formula $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$. Then differentiate both sides with respect to x and finally substitute the value of m in it.
Here, we have to find the differential equation of the family of all straight lines passing through the origin. We know that to find the differential equation of the family of curves, we have to find the equation of the curve first. So, now we will find the equation of the line passing through the origin.
We know that the equation of the line of slope m and passing through points $\left( {{x}_{1}},{{y}_{1}} \right)$ is written as
$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$
So, by substituting ${{y}_{1}}=0$ and ${{x}_{1}}=0$, we get the equation of the line passing through the origin as,
$\left( y-0 \right)=m\left( x-0 \right)$
Or $y=mx....\left( i \right)$
By dividing m on both the sides, we get
$\dfrac{y}{x}=m....\left( ii \right)$
We know that according to the product rule of differentiation, $\dfrac{d}{dx}\left( f.g \right)=g\left( \dfrac{df}{dx} \right)+f.\left( \dfrac{dy}{dx} \right)$
So, by differentiating both sides of equation (i) with respect to x, we get,
$\dfrac{dy}{dx}=m\left( \dfrac{dx}{dy} \right)+x\left( \dfrac{dm}{dx} \right)$
We know that m is constant for a particular value of x and y, so $\dfrac{dm}{dx}=0$. So, we get,
$\dfrac{dy}{dx}=m\left( 1 \right)+x\left( 0 \right)$
$\Rightarrow \dfrac{dy}{dx}=m$
By substituting the value of m from equation (ii), we get,
$\dfrac{dy}{dx}=\dfrac{y}{x}$
By subtracting $\dfrac{y}{x}$ from both sides of the above equation, we get,
$\dfrac{dy}{dx}-\dfrac{y}{x}=0$
By multiplying x dx on both the sides of the above equation, we get,
$x\text{ }dy-y\text{ }dx=0$
So, we get the differential equation of the family of all straight lines passing through the origin as
$x\text{ }dy-y\text{ }dx=0$
Note:
Students must note that to find the differential equation of any curve, they must eliminate all the constants from the equation like we eliminated ‘m’ in the above solution. Also, students can verify this differential equation by substituting the value of $\dfrac{dy}{dx}$ in the differential equation and checking if the original equation of the curve is obtained or not.
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# Q1, W7: September 24th – 28th
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Teacher Felicia Taylor Intensive Math 8th 7 Focus 1: Rational and Irrational Numbers; Focus 2: Solving Multi-Step Equations
Standard(s) Taught
MAFS.8.EE.1.2
#### Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational.
MAFS.8.EE.3.7
Solve linear equations in one variable.
1. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms until an equivalent equation of the form x=a, a=a or a=b results (where a and b are different numbers).
2. Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
Learning Targets and Learning Criteria
I know I am successful when I can:
• recognize perfect squares and non-perfect squares and numbers that are perfect cubes.
• recognize taking a square root of a number as the inverse of squaring that number and taking a cube root of a number as the inverse of cubing that number.
• evaluate the square root of a perfect square and the cube root of a perfect cube.
• justify that the square root of a non-perfect square will be irrational and that the cube root of a non-perfect cube will be irrational.
• relate perfect squares and perfect cubes to geometric squares and cubes using square tiles and square cubes.
• explain the differences between one solution, no solution, and infinitely many.
• solve a linear equation with one solution, no solution, and infinitely many.
• simplify equations using distributive property, combining like terms, and inverse operations.
• solve multi-step linear equations with rational number coefficients.
Assessment Limits:
• Square roots and cube roots may be used to represent solutions to equations.
• Radicands may not include variables.
• Numbers in items will be rational numbers.
Classroom Activities
Tuesday
1. Trash-ketball Game – Solving one-step equations in teams of 2-3 people. Each team is given five (5) one-step equations to solve. For each correctly solved equation, each team gets to shoot a basket. The team with most points at the end of the game wins. We will play five (5) rounds.
2. kahoot! – Exponents
3. kahoot! – Perfect squares and cubes
Wednesday (Short Class – Early Release Period)
1. Start two-step equations.
1. Worksheet – Solving two-step equations
Friday
1. Worksheet – Solving two-step equations placemat
2. Worksheet – Solving two-step equations maze
Assignments Due
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### Course: High school geometry>Unit 9
Lesson 4: Density
# Solid geometry FAQ
## Why do we need to know the difference between 2D and 3D objects?
Understanding the difference between 2D and 3D objects is important for solid geometry. 2D shapes, like squares and circles, only have length and width. 3D shapes, like cubes and spheres, also have depth. In order to find the volume and surface area of 3D shapes, we need to be able to identify them.
## What is Cavalieri's principle?
Cavalieri's principle states that two solids with the same height and cross-sectional area at every height have the same volume. This principle is useful for determining the volume of more complex shapes by breaking them down into simpler shapes that we already know how to find the volume of.
## How do we find the volume of a prism?
To find the volume of a prism, we use the formula $V=Bh$, where:
• $V$ is the volume
• $B$ is the area of the base of the prism
• $h$ is the height of the prism
## How do we find the volume of a pyramid?
To find the volume of a pyramid, we use the formula $V=\frac{1}{3}Bh$, where:
• $V$ is the volume
• $B$ is the area of the base of the pyramid
• $h$ is the height of the pyramid
## How do we find the surface area of a prism or pyramid?
To find the surface area of a prism (or a pyramid), we add up the areas of all of its faces.
## What is density and why is it important?
Density is a measure of how compact a material is. It is defined as the mass of the material divided by its volume. In solid geometry, we can use density to compare different materials and determine which one is heavier or lighter for a given volume.
## How are these topics used in the real world?
There are countless applications of solid geometry in the real world! For example, architects and engineers use concepts like volume and surface area to design buildings, bridges, and other structures. In manufacturing, density is an important consideration when selecting materials. And in science, researchers often use dissection methods to study the internal structure of an object.
## Want to join the conversation?
• I'm confused: Is Cavalieri's principle related to density? And if so, how?
• What alex81 said. (in comments)
Cavalieri's principle has no mention of density and only works with Volume, Area, and Lengths of the 3D shapes. No Quantities.
• wait, I'm confused, what is the difference between area and volume density and how do i know which formula to use?
• Area is simply volume in 2D. To figure out the density for a 3D object, let's say a Rectangular-Based Pyramid, we need to find the volume first.
The formula is 1/3*(l*w)*h. The (l*w) is the area of the base of the Pyramid. To find the volume you would multiply the base area by the height, and then by 1/3. Furthermore, to find the density you would need the quantity and the volume you just solved for to find the density in this equation:
Quantity / Volume = Density
To answer your question, you can tell which formula to use by whether it's 3D or 2D. Density is just the next step after finding the area. (You can also find the density in a 2D object: Quantity / Area = Density)
Hope this helps!
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Derivative of a function (Differentiation) is that which tells us how much a function changes as its input variable changes. Let y = f(x) be any function then its derivative dy/dx is given by
Where n is a degree of the polynomial.
Derivative Calculator is a online tool to find the derivative of a given function. You have to enter the given function in the block provided and get the answer. This calculator can get find the derivative up to 10th order so you can get the derivatives answer directly without actually calculating each time.
## Calculus Derivative Problems
To find the derivative of any given function observe whether the function is of the form y = f(x) then find the differentiation of the given function simplify it and get the answer.
Below you could see some problems on Derivative you can go through it:
### Solved Examples
Question 1: If f(x)= 5x3 + 3 x + 1 then find f'(x).
Solution:
The given function is y = 5x3 + 3x + 1.
It derivative is given by
f'(x) = $\frac{dy}{dx}$ = $\frac{d(5x^{3})}{dx}$ + $\frac{d(3x)}{dx}$ + $\frac{d(1)}{dx}$
= 5(3x2) + 3 + 0
= 15 x2 + 3.
Question 2: If f(x)= 4x5 + 7, then Find f''(x)
Solution:
The given function is f(x)= 4x5 + 7,
f'(x) = $\frac{d(4x^{5})}{dx}$ + $\frac{d(7)}{dx}$ = 20 x4
f''(x) = $\frac{d(20 x^{4})}{dx}$ = 20 (4x3) = 80 x3
The second derivative of function f(x) = 4x5 + 7 is 80 x3.
### Differential Equation Calculator
Calculate Partial Derivatives Derivatives All Derivative Rules Anti Derivative Arctan Derivative Derivation of E
Anti Derivative Calculator Find Second Derivative Calculator Implicit Derivative Calculator Online Partial Derivative Calculator Product Rule Derivative Calculator Calculator for Calculus
dy/dx calculator second derivative calculator instantaneous rate of change calculator
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# What Is A Like Fraction
### What is like fraction with example?
In two or good-natured fractions or a cluster of fractions when the denominator is precisely the identical genuine they are above-mentioned to be resembling fractions. Or you can say that fractions own the identical countless at the bottom. Resembling violation sample – 2/4 6/4 8/4 10/4.
### What is like and unlike fraction?
Fractions having the identical denominator are named ‘Like fractions’ The five examples of resembling fractions are. 2/7 3/7 4/7 5/7 and 6/7 Unlike fractions. Fractions having particularize denominators are named ‘Unlike fractions’
### How do you find a like fraction?
In the athwart multiplicity order we athwart multiply the numerator of the leading violation by the denominator of the subordinate fraction. genuine multiply the numerator of the subordinate violation by the denominator of the leading fraction. Now multiply twain the denominators and share it as a ordinary denominator.
### What are 5 examples unlike fractions?
Answer: Fractions immediately particularize denominators (bottom numbers) are named unlike fractions. For example: (i) ²/₃ ¹⁵/₁₃ ¹²/₁₇ and ³⁵/₁₉ are unlike fractions immediately particularize denominators (bottom numbers). (ii) ¹/₆ ¹⁷/₁₄ ²⁵/₁₆ and ¹⁶/₂₃ are unlike fractionswith particularize denominators (bottom numbers).
### How many fractions are there?
The six kinds of fractions are peculiar fractions improper fractions mixed fractions resembling fractions unlike fractions and equiponderant fractions See also how to hinder poaching
### What is a mixed fraction?
More specifically a mixed violation is simply an improper violation written as the sum of a total countless and a peculiar fraction. For sample the improper violation 3/2 can be written as the equiponderant mixed violation 1-1/2 (read audibly as “one-and-a-half” or “one-and-one-half”).
### What is like fraction for Class 5?
Fractions having the identical denominator are named resembling fractions. all resembling fractions. Fractions having particularize denominators are named unlike fractions.
### What is unlike fraction answer?
Unlike fractions are fractions that own particularize denominators. The leading violation under has a denominator of two and the subordinate violation under has a denominator of three. ant: full the denominators are particularize they are unlike fractions.
### What is fraction math?
fraction In arithmetic a countless expressed as a quotient in which a numerator is divided by a denominator. In a single violation twain are integers. A intricate violation has a violation in the numerator or denominator. In a peculiar violation the numerator is pure sooner_than the denominator.
### Which groups are of like fractions?
The fractions which own the identical denominator are named resembling fractions. i.e. their denominators are equal. For sample if we own a cluster of fractions such as 1/5 2/5 3/5 4/5. ant: full the denominators of shore of the fractions are the identical i.e. 5 they are resembling fractions.
### What are the 3 types of fraction?
So we can mark_out the three types of fractions resembling this: peculiar Fractions: The numerator is pure sooner_than the denominator. Improper Fractions: The numerator is greater sooner_than (or uniform to) the denominator. Mixed Fractions:
### Which of the following pairs are like fraction?
Like fractions are fractions whose denominators (bottom numbers)are same. Examples : 1) 2/5 3/5 8/5 11/5 are like-fractions as the denominator 5 is identical in all fractions. 2) 11/15 23/15 112/15 are like-fractions as the denominator 15 is identical in all the fractions.
### What are mixed fractions give examples?
A violation represented immediately its quotient and rest is a mixed violation See also what is an irrigation well
### How do you add 2 fractions with different denominators?
How to Add Fractions immediately particularize Denominators Cross-multiply the two fractions and add the results collectively to get the numerator of the answer. presume you deficiency to add the fractions 1/3 and 2/5. … Multiply the two denominators collectively to get the denominator of the answer. … Write your reply as a fraction.
### How do you change a mixed fraction?
Answer: To change an improper violation to a mixed violation separate the numerator by the denominator write below the quotient as the total countless and the rest as the numerator on top of the identical denominator. Let us see an sample of this conversion. Explanation: Change 23/4 inter a mixed fraction.
### How do you teach adding fractions?
3 quiet steps for adding fractions All you unnecessary to do is pursue three single steps: exceed 1: meet a ordinary denominator. exceed 2: Add the numerators (and hold the denominator) exceed 3: facilitate the fraction.
### What is a basic fraction?
Fractions portray parts of a whole — that is quantities that happen between the whole numbers. … The top countless — named the numerator — tells you the countless of shaded slices. The breast countless — named the denominator — tells you the whole countless of slices.
### What is fraction Toppr?
Answer: A violation basically tells you how numerous parts of a total you have. You can identify a violation by the slash which we write between the two numbers. The top countless is the numerator and a breast countless is the denominator. For sample 3/4 is a fraction.
### How do you shade fractions?
To shadow a violation of a form pursue these steps: fear the denominator which is the countless on the breast of the fraction. separate the total form inter this numerous equally sized parts. fear the numerator which is the countless on the top of the fraction. Shadow in this numerous of the equally sized parts.
### How do multiply fractions?
There are 3 single steps to multiply fractions Multiply the top numbers (the numerators). Multiply the breast numbers (the denominators). facilitate the violation if needed.
### What is fraction in maths for Class 3?
The aloof of a total is mysterious as violation See also how numerous oceans in the united states
### What fraction of an hour is 40 minutes?
2/3rd Answer: 40 minutes is 2/3rd of an hour.
### How do you explain fractions to a child?
Fractions are abashed to portray smaller pieces (or parts) of a whole. The parts might exult up one thing or good-natured sooner_than one thing. … It’s significant to note that a total can common good-natured sooner_than one thing. … You acquire that when you narration up the numbers own good-natured value. … real total numbers (like 1 2 or 65) are simple.
### What is a fraction class 7?
Fractions are numbers representing aloof of a whole. A violation is a countless of the agree p/q such that q is not uniform to naught or one. A violation has two parts. The countless on the top is numerator and the countless under is the denominator. … For e.g. 1/5th of a pizza is a violation that is written as 1/5.
### What are the most common fractions?
There are four kinds of ordinary fractions: peculiar improper mixed and complex. A peculiar violation has a numerator smaller sooner_than the denominator such as 3/4. accordingly the overestimate of a peculiar violation is always pure sooner_than one. In improper fractions the numerator is uniform to or larger sooner_than the denominator as 4/4 or 6/5.
### What are the examples of fraction?
The interior ordinary examples of fractions engage ant: gay vitality are uniform slices of pizza production ant: [see condiment] a bar of chocolate etc.
### What are the 10 types of fractions?
Based on the numerators and denominators fractions are classified inter the following types: peculiar Fractions. … Improper Fractions. … Mixed Fractions. … resembling Fractions. … Unlike Fractions. … equiponderant Fractions. … aggregation Fractions.
### Like and unlike fraction
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# Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $$\triangle$$ ABC ~ $$\triangle$$ PQR.
## Solution :
Given : $$\triangle$$ ABC and $$\triangle$$ PQR in which AD and PM are the medians, such that
$$AB\over PQ$$ = $$BC\over QR$$ = $$AD\over PM$$
To prove : $$\triangle$$ ABC ~ $$\triangle$$ PQR
Proof : $$AB\over PQ$$ = $$BC\over QR$$ = $$AD\over PM$$
$$\implies$$ $$AB\over PQ$$ = $${1\over 2}BC\over {1\over 2}QR$$ = $$AD\over PM$$
$$\implies$$ $$AB\over PQ$$ = $$BD\over QM$$ = $$AD\over PM$$
By SSS similarity,
$$\triangle$$ ABD ~ $$\triangle$$ PQM
$$\implies$$ $$\angle$$ B = $$\angle$$ Q
Now, in triangle ABC and PQR, we have
$$AB\over PQ$$ = $$BC\over QR$$ (given)
and $$\angle$$ B = $$\angle$$ Q
So, By SAS similarity,
$$\triangle$$ ABC ~ $$\triangle$$ PQR
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# Evaluate the following:${\text{tan 3}}{{\text{5}}^ \circ }{\text{ tan 4}}{{\text{0}}^ \circ }{\text{ tan 4}}{{\text{5}}^ \circ }{\text{ tan 5}}{{\text{0}}^ \circ }{\text{ tan 5}}{{\text{5}}^ \circ }$
Last updated date: 21st Jul 2024
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453.3k+ views
Hint: Use the conversion of tangent to cotangent.
To evaluate,
${\text{tan 3}}{{\text{5}}^ \circ }{\text{ tan 4}}{{\text{0}}^ \circ }{\text{ tan 4}}{{\text{5}}^ \circ }{\text{ tan 5}}{{\text{0}}^ \circ }{\text{ tan 5}}{{\text{5}}^ \circ }$
We know that in trigonometry,
$\tan (90 - \theta ) = \cot \theta$
and $\cot \theta = \dfrac{1}{{\tan \theta }}$
So by using this identity we will get,
$tan{35^ \circ } = tan({90^ \circ } - {55^ \circ }) = cot{55^ \circ } \\ tan{40^ \circ } = tan({90^ \circ } - {50^ \circ }) = cot{50^ \circ } \\$
Now if we substitute these values in our original question we get,
$= cot{55^ \circ }cot{50^ \circ }tan{45^ \circ }tan{50^ \circ }tan{55^ \circ } \\ = \dfrac{1}{{tan{{55}^ \circ }}} \times \dfrac{1}{{tan{{50}^ \circ }}} \times tan{45^ \circ } \times tan{50^ \circ } \times tan{55^ \circ } \\$
On solving it we get,
$\tan {45^ \circ }$ = $1$
Hence, the answer is $1$
Note: In these types of problems, the conversion from one trigonometric quantity to another is crucial. Also, it's helpful to remember the trigonometric values.
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Finding the Greatest Common Factor of 2 or 3 Numbers
FINDING THE GREATEST COMMON FACTOR OF $\,2\,$ or $\,3\,$ NUMBERS
• PRACTICE (online exercises and printable worksheets)
Want more details, more exercises? Read the full text!
EXAMPLE:
Question: Find the greatest common factor of $\,27\,$ and $\,18\,$:
Answer: $\,9$
The idea:
The factors of $\,27\,$ are: $\,1\,$ $\,3\,$ $\,9\,$ $\,27$ The factors of $\,18\,$ are: $\,1\,$ $\,2\,$ $\,3\,$ $\,6\,$ $\,9\,$ $\,18\,$
The common factors of $\,27\,$ and $\,18\,$ (the numbers that appear in both lists) are $\,1\,$, $\,3\,$, and $\,9\,$.
The greatest common factor (the greatest number in the list of common factors) is $\,9\,$.
EFFICIENT ALGORITHM FOR FINDING THE GREATEST COMMON FACTOR
Here's an efficient algorithm for finding the greatest common factor,
when there aren't too many numbers, and they aren't too big.
The process is illustrated by finding the greatest common factor of $\,18\,$, $\,36\,$, and $\,90\,$:
Line up the numbers in a row. (See the purple rectangle at left.) Find ANY number that goes into everything evenly (like $\,2\,$). Do the divisions, and write the results above the original numbers. In the example: $18\,$ divided by $\,2\,$ is $\,9\,$; write the $\,9\,$ above the $\,18\,$ $36\,$ divided by $\,2\,$ is $\,18\,$; write the $\,18\,$ above the $\,36\,$, and so on. Keep repeating the process, until there isn't any number (except $\,1\,$) that goes into everything evenly. Multiply the circled numbers together. This is the greatest common factor! In the example, $\,\text{gcf}(18,36,90) = 2\cdot 3\cdot 3 = 18\,$.
WHY DOES THIS METHOD WORK?
Think about why this method works.
As you walk through each step of this discussion, keep comparing with the chart above.
Look at the prime factorizations of each number:
$18 = 2\cdot3\cdot 3$
$36 = 2\cdot 2\cdot 3\cdot 3$
$90 = 2\cdot3\cdot 3\cdot 5$
The number $\,2\,$ goes into each evenly, so separate it off:
$18 = 2\cdot (3\cdot 3)\hphantom{\cdot 3} = 2\cdot 9$
$36 = 2\cdot (2\cdot 3\cdot 3) = 2\cdot 18$
$90 = 2\cdot (3\cdot 3\cdot 5) = 2\cdot 45$
The number $\,3\,$ goes into each remaining part evenly, so separate it off:
$18 = (2\cdot 3) \cdot (3) \hphantom{\cdot 3}= (2\cdot 3)\cdot 3$
$36 = (2\cdot 3) \cdot (2\cdot 3) = (2\cdot 3)\cdot 6$
$90 = (2\cdot 3) \cdot (3\cdot 5) = (2\cdot 3)\cdot 15$
Here's the final step:
$18 = (2\cdot 3\cdot 3)\cdot 1$
$36 = (2\cdot 3\cdot 3)\cdot 2$
$90 = (2\cdot 3\cdot 3)\cdot 5$
OTHER WAYS TO APPLY THE ALGORITHM
Here are some other ways the algorithm might be applied.
Of course, you get the same answer any correct way that you do it!
You can also zip over to wolframalpha.com and type in: gcd(18,36,90)
The ‘gcd’ stands for greatest common divisor, which is another name for greatest common factor.
BONUS: AN EFFICIENT ALGORITHM FOR FINDING THE LEAST COMMON MULTIPLE
The method described here for efficiently finding the greatest common factor
also gives an efficient way to find the least common multiple (lcm)!
Here's how:
Suppose we now want the least common multiple of $\,18\,$, $\,36\,$, and $\,90\,$.
Start by doing all the same work as above—which is repeated below, for your convenience.
To be a multiple of all three numbers, we first need the shared (common) factors.
These are the numbers in parentheses on the left below, or the vertical (circled) numbers on the right.
To be a multiple of all three numbers, we also need the non-shared factors.
These are the numbers after the parentheses on the left below, or the numbers along the top on the right.
Multiply the shared and non-shared factors together to get the least common multiple!
$18 = (\color{green}{2\cdot 3\cdot 3})\cdot \color{red}{1}$ $36 = (\color{green}{2\cdot 3\cdot 3})\cdot \color{red}{2}$ $90 = (\color{green}{2\cdot 3\cdot 3})\cdot \color{red}{5}$ shared factors in parentheses: $2\cdot 3\cdot 3$ non-shared factors after parentheses: $1\cdot 2\cdot 5$ \begin{align} \text{lcm} &= (\text{shared})(\text{non-shared})\cr &= (2\cdot 3\cdot 3)\cdot (1\cdot 2\cdot 5)\cr &= 180 \end{align} shared factors are vertical (circled): $2\cdot 3\cdot 3$ non-shared factors on the top: $1\cdot 2\cdot 5$ \begin{align} \text{lcm} &= (\text{shared})(\text{non-shared})\cr &= (2\cdot 3\cdot 3)\cdot (1\cdot 2\cdot 5)\cr &= 180 \end{align}
Want an even more efficient and reliable method?
Zip to WolframAlpha and type in:
least common multiple of 18, 36, 90 or lcm(18,36,90)
Voila!
Master the ideas from this section
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# Math Assignment Help With Number Lines
## Number Lines
15.1 Introduction: A number line is a line in which real numbers can be placed, according to their value. Each point on a number line corresponds to a real number, and each real number has a unique point that corresponds to it. For example, the number 1.5 (1 1/2) corresponds with the point on a number line that is halfway between one and two.
Often the only integers are shown as specially-marked points evenly spaced on the line.
Although in this image only the integers from −9 to 9 are shown, the line includes all numbers. The arrows on both the sides of the line mean that the line continues till infinity i.e. it contains all the real numbers from minus infinite to plus infinite.
The points on a number line are called coordinates. The zero point is called the origin. The numbers to the right of the origin are positive numbers; the numbers to the left of the origin are negative numbers.
### 15.2 Drawing the number line:
The number line is always represented as a horizontal line. Positive numbers lie on the right side of zero, and negative numbers lie on the left side of zero. An arrowhead is put on either end of the line which is meant to suggest that the line continues indefinitely in the positive and negative directions, as already suggested above.
### Characteristics of a number line
1. It is a horizontal line.
2. The points on a number line are called coordinates.
3. The zero point is called the origin.
4. The numbers to the right of the origin are called positive numbers and the numbers to the left of the origin are called negative numbers.
5. It has two arrow heads on either end of the line. This represents that the line extends up to infinity.
15.3 Decimal number line: To represent decimals on number line, divide each segment into 10 parts.
Example: represent 6.5 in the number line.
Draw a number line, dividing the segment between 6 and 7 into 10 equal parts
### 5.4 Graphing Inequalities on a Number Line
A number line is a horizontal line having points which correspond to numbers. The points are spaced according to the value of the numbers and are equally spaced.
We can graph real numbers by representing them as points on the number line. For example, we can graph 3 ¼ on the number line:
We can also graph inequalities on the number line.
The following graph represents the inequality x≤ 3 ¼.
The dark line represents all the numbers that satisfy x≤ 3 ¼.
Take any number on the dark line and plug it in for x, the inequality will be true.
### Email Based Assignment Help in Number Lines
To Schedule a Number Lines tutoring session
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# Multiply Any Number With 5 And 25 In Just 3 Seconds
## Multiply Any Number With 5 And 25 In Just 3 Seconds
Multiply any number with 5 and 25 in just 3 seconds , I think you are feeling it is impossible, but it is possible,to know how just follow the article
Multiplication with 5:
Let n be the number.In order to multiply it with 5 we need to follow 2 steps:
Step 1: Divide the number by 2.
Step 2: Multiply it with 10.
n*5=>n*(10/2)=>(n/2)*10
Ex 1: 36*5.
Step 1: (36/2) =18.
Step 2: 18*10=180.
Ex 2: 76*5.
Step 1: (76/2)=38.
Step 2: 38*10=380.
Multiplication with 25:
Let n be the number.In order to multiply it with number 25 we need to follow 2 steps:
Step 1: Divide the number by 4.
Step 2: Multiply it with 100.
n*25=>n*(100/4)=>(n/4)*100
Ex 1: 36*25.
Step 1: (36/4) =9.
Step 2: 9*100=900.
Ex 2: 76*25.
Step 1: (76/4)=19.
Step 2: 19*100=1900.
So you came to know that we can multiply any number with 5 and 25 in just 3 seconds . Now we are giving 3 problems which you can multiply any number with 5 and 25 in just 3 seconds.
Problems for practice:
Multiply the following numbers with 5.
a)62 b)84 c)48
Multiply the following numbers with 25.
a)62 b)84 c)48
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Multiplication
(often denoted by the cross symbol “ד, by a point “·“, by juxtaposition, or, on computers, by an asterisk ““) is one of the four elementary, mathematical operations of arithmetic; with the others being addition, subtraction and division.
The multiplication of whole numbers may be thought as a repeated addition; that is, the multiplication of two numbers is equivalent to adding as many copies of one of them, the multiplicand, as the value of the other one, the multiplier. Normally, the multiplier is written first and multiplicand second, though this can vary, as the distinction is not very meaningful:
{\displaystyle a\times b=\underbrace {b+\cdots +b} _{a}=\underbrace {a+\cdots +a} _{b}}
For example, 4 multiplied by 3 (often written as {\displaystyle 3\times 4} and said as “3 times 4”) can be calculated by adding 3 copies of 4 together:
{\displaystyle 3\times 4=4+4+4=12}
Here 3 and 4 are the “factors” and 12 is the “product”.
One of the main properties of multiplication is the commutative property, adding 3 copies of 4 gives the same result as adding 4 copies of 3:
{\displaystyle 4\times 3=3+3+3+3=12}
The multiplication of integers (including negative numbers), rational numbers (fractions) and real numbers is defined by a systematic generalization of this basic definition.
Multiplication can also be visualized as counting objects arranged in a rectangle (for whole numbers) or as finding the area of a rectangle whose sides have given lengths. The area of a rectangle does not depend on which side is measured first, which illustrates the commutative property. The product of two measurements is a new type of measurement, for instance multiplying the lengths of the two sides of a rectangle gives its area, this is the subject of dimensional analysis.
The inverse operation of multiplication is division. For example, since 4 multiplied by 3 equals 12, then 12 divided by 3 equals 4. Multiplication by 3, followed by division by 3, yields the original number (since the division of a number other than 0 by itself equals 1).
Multiplication is also defined for other types of numbers, such as complex numbers, and more abstract constructs, like matrices. For these more abstract constructs, the order that the operands are multiplied sometimes does matter. A listing of the many different kinds of products that are used in mathematics is given in the product (mathematics) page.
Multiplication source: Wiki
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The square root of 2 can never be written exactly as a fraction. In other words,
How can we show this? Here are four proofs.
## Algebra
The usual way is as follows.
• Suppose $\sqrt{2}$ can be written as a fraction:
• $\sqrt{2}=a/b$
• Then $2=a^2/b^2$
• Then $2b^2=a^2$
• Then $a^2$ is even
• Hence $a$ is even
• (This takes a small proof of its own.)
• So we can write a=2k
• Hence $2b^2=(2k)^2=4k^2$
• So $b^2=2k^2$
• So now $b^2$ is even
• Hence $b$ is even
• So now we can reduce $\frac{a}{b}$ to a smaller fraction.
• Lather, rinse, repeat, as they say.
• Following this process we can get smaller and smaller fractions, each of which is exactly the square root of two. Clearly this is nonsense.
• The argument is sound, so the premise must be false.
• Thus we can never write $\sqrt{2}$ as $\frac{a}{b}$
## Geometry
Suppose we can draw a square that has integral sides and an integral diagonal. Using Pythagoras' theorem we can see that this is equivalent to saying that $\sqrt{2}$ is rational.
Consider the following diagram:
By Pythagoras we know that if the sides are $a$ and the diagonal is $b$ then $2a^2=b^2.$ This diagram shows that given such a pair, $a,b,$ we can find a smaller pair. (Exercise for the reader: show that $a$ and $b$ in the smaller pair are still integers.)
Repeat.
Eventually we run out of numbers and thus complete our proof by contradiction
## Algebra, again
Exercises for the diligent reader: Show that $(2a-b)^2=2(b-a)^2$ Show that b-a < a what's the relationship between this proof and the previous geometrical one?
Suppose $b^2=2a^2$ where $a,b$ are integers and as small as possible. Then a little algebra shows that $(2a-b)^2=2(b-a)^2$ so that $(b-a,2a-b)$ is a new pair of integers with the same property. But $b-a\lt~a$ and so we've got a smaller pair of integers whose ratio is $\sqrt{2},$ contrary to our original decision to take the smallest pair.
## Continued Fractions
Another exercise for the diligent reader: Show that for every number of the form $a/b,$ its continued fraction will terminate.
Every rational number has a finite continued fraction. We show now that the continued fraction for $1+\sqrt{2}$ does not terminate. That means $1+\sqrt{2}$ is irrational, and hence $\sqrt{2}$ is irrational.
To compute the continued fraction we separate the target number into its integer part and the fractional part
• $T=1+\sqrt{2}=2+\epsilon$ where $\epsilon=\sqrt{2}-1$
$\epsilon$ is between 0 and 1 as required, so we can take its reciprocal, giving $\frac{1}{\sqrt{2}-1}$ which after rationalising the denominator is $1+\sqrt{2}.$ That's what we started with, so the continued fraction for $1+\sqrt{2}$ is:
• $\LARGE{1+\sqrt{2}=2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}$
Thus the continued fraction for $1+\sqrt{2}$ is [2;2,2,2,2,...] which does not terminate. Hence $1+\sqrt{2}$ is irrational, and so $\sqrt{2}$ is irrational.
CategoryMaths
AbsoluteValue
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## You are here
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TEKS Lesson Plan/Unit Plan
• Doc File 70.00KByte
Focus Plan
Texarkana Independent School District
|Grading Period: |Refer to Scope and Sequence |Plan Code: | |
|Writer: |Barbara Fugitt |Course/subject: |Math |
|Grade(s): |Fifth grade |Time allotted for instruction: |3 – 45 minute class periods. |
[pic]
|Title: |The Improper Math |
|Lesson Topic: |Understanding how to change from an improper fraction to a mixed number and a mixed number to an |
| |improper fraction. |
|TAKS Objective: |Objective 1 |
| |The student will demonstrate an understanding of numbers, operations, and quantitative reasoning.|
|FoCUS TEKS and Student Expectation: |5.2 Number, operation, and quantitative reasoning. The student uses fractions in problem-solving |
| |situations. |
| | |
| |(B) The student is expected to generate a mixed number equivalent to a given improper fraction or|
| |generate an improper fraction equivalent to a given mixed number. |
|Supporting TEKS and Student Expectations: |5.3 Number, operation, and quantitative reasoning. The student adds, subtracts, multiplies, and |
| |divides to solve meaningful problems. |
| |(A) The student is expected to use addition and subtraction to solve problems involving whole |
| |number and decimals. |
| |(B) The student is expected to use multiplication to solve problems involving whole numbers. |
[pic]
|Concepts |Enduring Understandings/Generalizations/Principles |
| |The student will understand that |
| |Students will review that the numerator is how many parts of a fraction is shaded. |
|Numerator | |
| | |
|Denominator |Students will review that the denominator is how many parts a whole is divided into. |
|Equivalent |Students will review that equivalent means equal. |
|Simplify |Students will learn to simplify a fraction. |
|Improper Fraction |Students will learn that an improper fraction is a fraction whose numerator is equal to or larger than |
| |the denominator. |
|Mixed Number |Students will learn that a mixed number is a number that is made up of a whole number and a fraction. |
[pic]
[pic]I. Sequence of Activities (Instructional Strategies)
A. Focus/connections/anticipatory set
1. Begin the lesson by discussing the vocabulary. Review with students what a proper fraction is. A proper fraction is less than one and is part of a whole. The numerator is smaller than the denominator.
2. Write a fraction on the board, like 6/4. Ask the students to look at the fraction and ask them what they see. They should say that the numerator is larger than the denominator. Guide them towards the idea that this fraction is greater than one because of what they know about fractions.
3. Have students work in pairs or groups and find a way to make the fraction a proper fraction. Give students an ample amount of time to find a possible solution. Have students show their possible solutions. You will get one group who will get it right. Show students by drawing a picture of the fraction. They will see that the mixed number is 1 2/4 or 1 1/2.
B. Instructional activities
1. Objectives: The student will identify, read, and write mixed numbers and rename fractions to mixed numbers.
2. Procedures: The teacher will explain improper fractions and mixed numbers and demonstrate to students how to convert an improper fraction into a mixed number.
3. Modeling: The teacher will use pattern blocks and the overhead, board, or chart paper for recording answers to demonstrate improper fractions and mixed numbers.
C. Guided activity or strategy
Day 1
1. Have students continue working in groups and pass out pattern blocks. Tell each group to pull out 16 green triangles. Have the students cover as many yellow hexagons as possible. Ask students how many hexagons they covered. 2 and 4/6 is the answer. Ask students if they could reduce 4/6. They should say yes and the answer is 2/3. Ask the students what fraction is not covered. 2/6 or 1/3 Make sure you have them reduce fractions.
2. Ask students to look at their pattern blocks and use another shape and do the same thing. Have students draw their solutions. They could use a trapezoid.
3. Now that you have shown students how to find mixed numbers by using pictures. Teach students the following steps:
A. Divide the denominator into the numerator.
B. Once you have the whole number, place the remainder over the original denominator.
C. To change mixed numbers to an improper fraction multiply the whole number by the denominator of the fraction.
D. Then add the numerator to the product.
E. Take your final answer and put it over the original denominator.
4. Go over several examples with the students. Use Math book p. 273 for examples.
Day 2
1. Review improper fractions and mixed numbers by using PowerPoint over improper fractions. Have students work several of the examples together.
D. Accommodations/modifications
See student IEP for modifications.
E. Enrichment
II. STUDENT PERFORMANCE
A. Description
The students should be able to recognize a prime or composite number and be able to decipher a prime factorization of a number.
Day 1
Students will complete On My Own p. 15.2 (Activity Sheet 1) changing improper fractions to mixed numbers.
Day 2
Students will complete Activity Sheet 2.
Day 3
Students will complete the Improper Fractions assessment.
B. Accommodations/modifications
See student IEP for modifications.
C. Enrichment
III. Assessment of Activities
A. Description
The teacher will know the students have begun to master improper fractions when they:
• Complete the Day 1 and Day 2 activities with a 70% or higher.
• Complete the Improper Fraction Assessment with 70% or higher.
Complete all activities with a 70% or higher.
C. Accommodations/modifications
See student IEP for modifications.
D. Enrichment
E. Sample discussion questions
1. What is an improper fraction? An improper fraction is when a fraction has a greater numerator than a denominator.
2. What is a mixed number? A whole number and a fraction
3. How do you change an improper fraction to a mixed number? Divide the numerator by the denominator. Take the remainder and put it over the denominator.
4. How do you change a mixed number to an improper fraction? Multiply the denominator and the whole number. Take the product and add it to the numerator. Take the final answer and put it over the original denominator.
IV. TAKS Preparation
A. Transition to TAKS context
1. Students will complete a TAKS formatted Assessment.
B. Sample TAKS questions
This TEK has not been tested, but it doesn’t mean it won’t be tested in the future.
I have enclosed a TAKS formatted transparency for students to see. (TAKS Transparency)
V. Key Vocabulary
Numerator, denominator, equivalent, simplify, fraction, reduce, simplest terms, mixed number, improper fraction.
VI. Resources
A. Textbook
B. Supplementary materials/equipment
• Pattern blocks
C. Technology
1. Overhead projector: Modeling improper fractions and mixed numbers.
2. Computers: PowerPoint of improper fractions.
Have students review changing improper fractions to mixed numbers and back using morning work activities and/or homework activities.
VIII. Teacher Notes
This lesson is one that will need to be reinforced several times over. This is just an introductory. Students will need to practice on numerous occasions. Students will need to have an understanding of fractions, multiplication, division, subtraction, and addition before teaching this lesson. has some practice sheets that can be created for additional practice.
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# Model Drawing Examples
## Model Drawing Procedure
Developed in Singapore Visual representation of details and actions which assists children with problem solving
Helps children logically think using visual models to determine their computations
## Model Drawing Procedure
Teaches the importance of language within math problems
## Provides foundation for algebraic understanding
Provides for differentiated instruction
## Model Drawing Procedure
Fosters quantitative reasoning (number sense) when teachers question Empowers students to think systematically and master more difficult problems Makes multi-step and multi-concept problems easy to work
## Model Drawing does NOT
Work on every problem
Specify ONE RIGHT model Specify ONE RIGHT operation
## Areas for Use of Model Drawing Procedure in Grade 1
Whole Number Operations
## Use Model Drawing in Four-Step Process
Step 1
Main Idea of Question
Step 2
Who
What
Unit Bar
Step 3
Work the computation (s).
Step 4
Describe how the problem was solved.
1.
2. Details
Main Idea
## Four Step Process
Read the problem. Write the main idea in the question. Write who the problem is about (related to main idea.) Write what the problem is about (related to main idea.) Draw unit bars of equal length. Re-read the problem one sentence at a time adjusting the unit bars to match the story and identify the question on the model. Write the number sentence and work the computation (s). Describe how the problem was solved.
3. Strategy
4. How
Put Together
Ann has 2 toys. Jeff has 3 toys. How many toys do they have together?
Step 1 Step 2 2
Toys together
Ann toys
3
Jeff toys
Step 3 2 Step 4 Add 2 and 3.
+3 5
Andy and Henry went to the zoo. Andy saw 4 . Henry saw 5 . How many animals did the 2 boys see?
Main Idea Details 4
Animals seen
Andy
5
Henry
Strategy 4 +5 9
How
## Put together 4 and 5 to get a sum of 9.
Take Away
Subtraction
Ann has 4 toys. She gave away 1 toy. How many toys are left?
Step 1 Step 2 4
Toys left
Ann toys
]
?
Step 3 4
## Step 4 Subtract 1 from 4.
- 1 3
Compare
Subtraction
Ann has 4 toys. Jeff has 1 toy. How many more toys does Ann have than Jeff?
Step 1 Step 2 4
## More toys Ann than Jeff
Ann toys
]
1
Jeff toys
Step 3 4 Step 4 Subtract 1 from 4.
- 1 3
Anabel bought 3 at the carnival. Leo bought 4 at the carnival. How many more were bought than
Main Idea Details 3 Anabel 4 Leo
Strategy 4
## How Compared 3 to 4 to get 1.
-3 1
Ann has 2 toys. Jeff has 4 toys. How many more toys does Jeff have than Ann?
Step 1 Step 2 2
Ann toys
Jeff toys
Step 3 4
]
?
## Step 4 Subtract 2 from 4.
- 2 2
Missing Part
Subtraction
Ann has 5 balls. Three are baseballs. The rest are footballs. How many are footballs?
Step 1 Step 2 5
footballs
Ann balls
? FB
Step 3 5 Step 4 Subtract 3 from 5.
]
3 BB
- 3 2
Carlos had 11 coins in his pocket. Eight coins were quarters and the rest were dimes. How many coins were dimes?
Main Idea Details 11
Carlos coins
coins dimes
]
Q
]
?D
Strategy
11 -8 3
## Put Together and Take Away
Lisa had 8 marbles. On Monday, she gave 3 away and on Tuesday she gave away 2 more. How many marbles does she have left? Step 1 Step 2
Lisa marbles
8
X X X X X
]
?
marbles left
Step 3 3 +2 5 8 - 5 3
## Step 4 Add 3 and 2. Subtract 5 from 8.
Extra Information
Try to keep students focused on what the question is asking them to find. If a child understands that the details are usually what is needed to answer the main idea of the question, he will be less likely to include information that is not needed in the details. However, if the child includes the extra information in the drawing, placing the ? in the model will help them understand what information is needed to answer the question.
## Hold These Thoughts
In first grade, be sure all the unit bars for each variable are touching each other so comparisons are clearer. At the beginning of first grade, show one unit for each item. Modeling each part of the model drawing with unifix cubes concretely represents the picture that is being drawn. In the drawing, list the variables in the order they appear in the problem. To show take away, mark off the appropriate unit bar segments and draw a X.
## Hold These Thoughts
Unit bars need to be proportional. Make sure that the bar of \$25 is not larger than the bar of \$30.
Continuously asking students to describe how the unit bars should be adjusted, connecting the adjustment to the vocabulary in the sentence improves students number sense and reasoning. You can always adjust the size of a unit bar as you learn more information. When you lengthen the bar, it means the number is larger, and when you shorten the bar, it means the number is smaller. If there is more than one number given in a sentence, adjust the model one number at a time. Break long sentences into partsand or commas.
Too often, students rush through a problem and answer the wrong question. Placing the question mark helps to prevent that.
## Hold These Thoughts
The computation is the differentiated part of the lesson. The model looks the same for all students, but the way they achieve success with computation is differentiated. No matter how you calculate it, the answer is the same! This is very important for students to see. Draw a dotted line between unit bars to point out segments of equal value. Dont get distracted looking for extraneous information. Focus on what the question is asking you to find. When a problem says, three times as many add one unit bar at a time. Otherwise, many students will add three MORE unit bars instead of adding just two.
__________
Date
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# 6.2: Graphs of the Other Trigonometric Functions
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##### Learning Objectives
• Analyze the graph of $$y=\tan x$$.
• Graph variations of $$y=\tan x$$.
• Analyze the graphs of $$y=\sec x$$ and $$y=\csc x$$.
• Graph variations of $$y=\sec x$$ and $$y=\csc x$$.
• Analyze the graph of $$y=\cot x$$.
• Graph variations of $$y=\cot x$$.
We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions.
## Analyzing the Graph of $$y =\tan x$$
We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that
$\tan \, x=\dfrac{\sin \, x}{\cos \, x}$
The period of the tangent function is $$\pi$$ because the graph repeats itself on intervals of $$k\pi$$ where $$k$$ is an integer. If we graph the tangent function on $$−\frac{\pi}{2}$$ to $$\frac{\pi}{2}$$, we can see the behavior of the graph on one complete cycle. If we look at any larger interval, we will see that the characteristics of the graph repeat.
We can determine whether tangent is an odd or even function by using the definition of tangent.
\begin{align*} \tan(-x)&= \dfrac{\sin(-x)}{\cos(-x)} \qquad \text{Definition of tangent}\\[4pt] &= \dfrac{-\sin \, x}{\cos \, x} \qquad \text{Sine is an odd function, cosine is even}\\[4pt] &= -\dfrac{\sin \, x}{\cos \, x} \qquad \text{The quotient of an odd and an even function is odd}\\[4pt] &= -\tan \, x \qquad \text{Definition of tangent} \end{align*}
Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in Table $$\PageIndex{1}$$.
$$x$$ $$\tan x$$ $$−\dfrac{\pi}{2}$$ $$−\dfrac{\pi}{3}$$ $$−\dfrac{\pi}{4}$$ $$−\dfrac{\pi}{6}$$ 0 $$\dfrac{\pi}{6}$$ $$\dfrac{\pi}{4}$$ $$\dfrac{\pi}{3}$$ $$\dfrac{\pi}{2}$$ undefined $$-\sqrt{3}$$ $$–1$$ $$-\dfrac{\sqrt{3}}{3}$$ 0 $$\dfrac{\sqrt{3}}{3}$$ 1 $$\sqrt{3}$$ undefined
These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values when $$\dfrac{\pi}{3}<x<\dfrac{\pi}{2}$$, we can use a table to look for a trend. Because $$\dfrac{\pi}{3}≈1.05$$ and $$\dfrac{\pi}{2}≈1.57$$, we will evaluate $$x$$ at radian measures $$1.05<x<1.57$$ as shown in Table $$\PageIndex{2}$$.
$$x$$ $$\tan x$$ 1.3 1.5 1.55 1.56 3.6 14.1 48.1 92.6
As $$x$$ approaches $$\dfrac{\pi}{2}$$, the outputs of the function get larger and larger. Because $$y=\tan \, x$$ is an odd function, we see the corresponding table of negative values in Table $$\PageIndex{3}$$.
$$x$$ $$\tan x$$ −1.3 −1.5 −1.55 −1.56 −3.6 −14.1 −48.1 −92.6
We can see that, as $$x$$ approaches $$−\dfrac{\pi}{2}$$, the outputs get smaller and smaller. Remember that there are some values of $$x$$ for which $$\cos \, x=0$$. For example, $$\cos \left (\dfrac{\pi}{2} \right)=0$$ and $$\cos \left (\dfrac{3\pi}{2} \right )=0$$. At these values, the tangent function is undefined, so the graph of $$y=\tan \, x$$ has discontinuities at $$x=\dfrac{\pi}{2}$$ and $$\dfrac{3\pi}{2}$$. At these values, the graph of the tangent has vertical asymptotes. Figure $$\PageIndex{1}$$ represents the graph of $$y=\tan \, x$$. The tangent is positive from $$0$$ to $$\dfrac{\pi}{2}$$ and from $$\pi$$ to $$\dfrac{3\pi}{2}$$, corresponding to quadrants I and III of the unit circle.
## Graphing Variations of $$y = \tan \, x$$
As with the sine and cosine functions, the tangent function can be described by a general equation.
$y=A\tan(Bx) \nonumber$
We can identify horizontal and vertical stretches and compressions using values of $$A$$ and $$B$$. The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.
Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant $$A$$.
##### FEATURES OF THE GRAPH OF $$Y = A \tan(Bx)$$
• The stretching factor is $$|A|$$.
• The period is $$P=\dfrac{\pi}{|B|}$$.
• The domain is all real numbers $$x$$,where $$x≠\dfrac{\pi}{2| B |}+\dfrac{π}{| B |}k$$ such that $$k$$ is an integer.
• The range is $$(−\infty,\infty)$$.
• The asymptotes occur at $$x=\dfrac{\pi}{2| B |}+\dfrac{π}{| B |}k$$ where $$k$$ is an integer.
• $$y=A\tan(Bx)$$ is an odd function.
### Graphing One Period of a Stretched or Compressed Tangent Function
We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the form $$\boxed{f(x)=A\tan(Bx)}$$. We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our limited domain is then the interval $$\left (−\dfrac{P}{2},\dfrac{P}{2} \right )$$ and the graph has vertical asymptotes at $$\pm \dfrac{P}{2}$$ where $$P=\dfrac{\pi}{B}$$. On $$\left (−\dfrac{\pi}{2},\dfrac{\pi}{2} \right )$$, the graph will come up from the left asymptote at $$x=−\dfrac{\pi}{2}$$, cross through the origin, and continue to increase as it approaches the right asymptote at $$x=\dfrac{\pi}{2}$$. To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use
$f \left (\dfrac{P}{4} \right )=A\tan \left (B\dfrac{P}{4} \right )=A\tan \left (B\dfrac{\pi}{4B} \right )=A \nonumber$
because $$\tan \left (\dfrac{\pi}{4} \right )=1$$.
##### Given the function $$f(x)=A \tan(Bx)$$, graph one period.
1. Identify the stretching factor, $$| A |$$.
2. Identify B and determine the period, $$P=\dfrac{\pi}{| B |}$$.
3. Draw vertical asymptotes at $$x=−\dfrac{P}{2}$$ and $$x=\dfrac{P}{2}$$.
4. For $$A>0$$, the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for $$A<0$$).
5. Plot reference points at $$\left (\dfrac{P}{4},A \right )$$, $$(0,0)$$, and $$\left (−\dfrac{P}{4},−A \right )$$, and draw the graph through these points.
##### Example $$\PageIndex{1}$$: Sketching a Compressed Tangent
Sketch a graph of one period of the function $$y=0.5\tan \left (\dfrac{\pi}{2}x \right )$$.
Solution
First, we identify $$A$$ and $$B$$.
Because $$A=0.5$$ and $$B=\dfrac{\pi}{2}$$, we can find the stretching/compressing factor and period. The period is $$\dfrac{\pi}{\dfrac{\pi}{2}}=2$$, so the asymptotes are at $$x=±1$$. At a quarter period from the origin, we have
\begin{align*} f(0.5)&= 0.5\tan \left (\dfrac{0.5\pi}{2} \right )\\[4pt] &= 0.5\tan \left (\dfrac{\pi}{4} \right )\\[4pt] &= 0.5 \end{align*}
This means the curve must pass through the points $$(0.5,0.5)$$, $$(0,0)$$,and $$(−0.5,−0.5)$$. The only inflection point is at the origin. Figure $$\PageIndex{3}$$ shows the graph of one period of the function.
##### Exercise $$\PageIndex{1}$$
Sketch a graph of $$f(x)=3\tan \left (\dfrac{\pi}{6}x \right )$$.
## Graphing One Period of a Shifted Tangent Function
Now that we can graph a tangent function that is stretched or compressed, we will add a vertical shift and/or horizontal (or phase) shift. In this case, we add $$C$$ and $$D$$ to the general form of the tangent function.
$f(x)=A\tan(Bx−C)+D \nonumber$
The graph of a transformed tangent function is different from the basic tangent function $$\tan x$$ in several ways:
##### FEATURES OF THE GRAPH OF $$Y = A\tan(Bx−C)+D$$
• The stretching factor is $$| A |$$.
• The period is $$\dfrac{\pi}{| B |}$$.
• The phase shift is $$\dfrac{C}{B}$$.
• The domain is $$x≠\dfrac{C}{B}+\dfrac{\pi}{2| B |}k$$, where $$k$$ is an odd integer.
• The range is $$(−∞,∞)$$.
• The vertical asymptotes occur at $$x=\dfrac{C}{B}+\dfrac{\pi}{2| B |}k$$, where $$k$$ is an odd integer.
• There is no amplitude.
• $$y=A \tan(Bx)$$ is and odd function because it is the quotient of odd and even functions(sin and cosine respectively).
##### Howto: Given the function $$y=A\tan(Bx−C)+D$$, sketch the graph of one period.
1. Express the function given in the form $$y=A\tan(Bx−C)+D$$.
2. Identify the stretching/compressing factor, $$| A |$$.
3. Identify $$B$$ and determine the period, $$P=\dfrac{\pi}{|B|}$$.
4. Identify $$C$$ and determine the phase shift, $$\dfrac{C}{B}$$.
5. If $$B$$ is negative, use the Negative Angle Identity which (for the tangent function) makes $$B$$ positive and changes the sign of $$A$$. Then, draw the graph of $$y=A\tan(Bx)$$ shifted to the right by $$\dfrac{C}{B}$$ and up by $$D$$.
6. Sketch the vertical asymptotes, which occur at $$x=\dfrac{C}{B}+\dfrac{\pi}{2| B |}k$$, where $$k$$ is an odd integer.
7. Plot any three reference points and draw the graph through these points.
##### Example $$\PageIndex{2}$$: Graphing One Period of a Shifted Tangent Function
Graph one period of the function $$y=−2\tan(\pi x+\pi)−1$$.
Solution
• Step 1. The function is already written in the form $$y=A\tan(Bx−C)+D$$.
• Step 2. $$A=−2$$, so the stretching factor is $$|A|=2$$.
• Step 3. $$B=\pi$$, so the period is $$P=\dfrac{\pi}{| B |}=\dfrac{\pi}{\pi}=1$$.
• Step 4. $$C=−\pi$$, so the phase shift is $$\dfrac{C}{B}=\dfrac{−\pi}{\pi}=−1$$.
• Steps 5-7. The asymptotes are at $$x=−\dfrac{3}{2}$$ and $$x=−\dfrac{1}{2}$$ and the three recommended reference points are $$(−1.25,1)$$, $$(−1,−1)$$, and $$(−0.75,−3)$$. The graph is shown in Figure $$\PageIndex{5}$$.
Analysis
Note that this is a decreasing function because $$A<0$$.
##### Exercise $$\PageIndex{2}$$
How would the graph in Example $$\PageIndex{2}$$ look different if we made $$A=2$$ instead of $$−2$$?
It would be reflected across the line $$y=−1$$, becoming an increasing function.
##### Howto: Given the graph of a tangent function, identify horizontal and vertical stretches.
1. Find the period $$P$$ from the spacing between successive vertical asymptotes or x-intercepts.
2. Write $$f(x)=A\tan \left (\dfrac{\pi}{P}x \right )$$.
3. Determine a convenient point $$(x,f(x))$$ on the given graph and use it to determine $$A$$.
##### Example $$\PageIndex{3}$$: Identifying the Graph of a Stretched Tangent
Find a formula for the function graphed in Figure $$\PageIndex{6}$$.
Solution
The graph has the shape of a tangent function.
• Step 1. One cycle extends from $$–4$$ to $$4$$, so the period is $$P=8$$. Since $$P=\dfrac{\pi}{| B |}$$, we have $$B=\dfrac{π}{P}=\dfrac{\pi}{8}$$.
• Step 2. The equation must have the form $$f(x)=A\tan \left (\dfrac{\pi}{8}x \right )$$.
• Step 3. To find the vertical stretch $$A$$, we can use the point $$(2,2)$$. \begin{align*} 2&=A\tan \left (\dfrac{\pi}{8}\cdot 2 \right )\\[4pt] &=A\tan \left (\dfrac{\pi}{4} \right ) \end{align*}
Because $$\tan \left (\dfrac{\pi}{4} \right )=1$$, $$A=2$$.
This function would have a formula $$f(x)=2\tan \left (\dfrac{\pi}{8}x \right )$$.
##### Exercise $$\PageIndex{3}$$
Find a formula for the function in Figure $$\PageIndex{7}$$. (The point $$(\dfrac{\pi}{8}, 4 )$$ is on the graph.)
$$g(x)=4\tan(2x)$$
## Analyzing the Graph of $$y = \cot x$$
The next trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity $$\cot \, x=\dfrac{1}{\tan x}$$. Notice that the function is undefined when the tangent function is $$0$$, leading to a vertical asymptote in the graph at $$0$$, $$\pi$$, etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.
We can graph $$y=\cot x$$ by observing the graph of the tangent function because these two functions are reciprocals of one another. See Figure $$\PageIndex{19}$$. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases.
The cotangent graph has vertical asymptotes at each value of $$x$$ where $$\tan x=0$$; we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, $$\cot x$$ has vertical asymptotes at all values of $$x$$ where $$\tan x=0$$, and $$\cot x=0$$ at all values of $$x$$ where $$\tan x$$ has its vertical asymptotes.
##### FEATURES OF THE GRAPH OF $$y = A \cot(Bx)$$
• The stretching factor is $$|A|$$.
• The period is $$P=\dfrac{\pi}{|B|}$$.
• The domain is $$x≠\dfrac{\pi}{|B|}k$$, where $$k$$ is an integer.
• The range is $$(−∞,∞)$$.
• The asymptotes occur at $$x=\dfrac{\pi}{| B |}k$$, where $$k$$ is an integer.
• $$y=A\cot(Bx)$$ is an odd function.
##### Given a modified cotangent function of the form $$f(x)=A\cot(Bx)$$, graph one period.
1. Express the function in the form $$f(x)=A\cot(Bx)$$.
2. Identify the stretching factor, $$|A|$$.
3. Identify the period, $$P=\dfrac{\pi}{|B|}$$.
4. If $$B$$ is negative, use the Negative Angle Identity which (for the cotangent function) makes $$B$$ positive and changes the sign of $$A$$. Then, draw the graph of $$y=A\tan(Bx)$$.
5. Plot any two reference points.
6. Use the reciprocal relationship between tangent and cotangent to draw the graph of $$y=A\cot(Bx)$$.
7. Sketch the asymptotes.
##### Example $$\PageIndex{8}$$: Graphing Variations of the Cotangent Function
Determine the stretching factor, period, and phase shift of $$y=3\cot(4x)$$, and then sketch a graph.
Solution
• Step 1. Expressing the function in the form $$f(x)=A\cot(Bx)$$ gives $$f(x)=3\cot(4x)$$.
• Step 2. The stretching factor is $$|A|=3$$.
• Step 3. The period is $$P=\dfrac{\pi}{4}$$.
• Step 4. Sketch the graph of $$y=3\tan(4x)$$.
• Step 5. Plot two reference points. Two such points are $$\left (\dfrac{\pi}{16},3 \right )$$ and $$\left (\dfrac{3\pi}{16},−3 \right )$$.
• Step 6. Use the reciprocal relationship to draw $$y=3\cot(4x)$$.
• Step 7. Sketch the asymptotes, $$x=0$$, $$x=\dfrac{\pi}{4}$$.
The orange graph in Figure $$\PageIndex{20}$$ shows $$y=3\tan(4x)$$ and the blue graph shows $$y=3\cot(4x)$$.
## Graphing Variations of $$y =\cot x$$
We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.
$y=A\cot(Bx−C)+D$
##### PROPERTIES OF THE GRAPH OF $$Y = A \cot(Bx-c)+D$$
• The stretching factor is $$| A |$$.
• The period is $$\dfrac{\pi}{|B|}$$
• The phase shift is $$\dfrac{C}{B}$$.
• The domain is $$x≠\dfrac{C}{B}+\dfrac{\pi}{| B |}k$$, where $$k$$ is an integer.
• The range is $$(−∞,∞)$$.
• The vertical asymptotes occur at $$x=\dfrac{C}{B}+\dfrac{\pi}{| B |}k$$, where $$k$$ is an integer.
• There is no amplitude.
• $$y=A\cot(Bx)$$ is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively)
##### Given a modified cotangent function of the form $$f(x)=A\cot(Bx−C)+D$$, graph one period.
1. Express the function in the form $$f(x)=A\cot(Bx−C)+D$$.
2. Identify the stretching factor, $$| A |$$.
3. Identify the period, $$P=\dfrac{\pi}{|B|}$$.
4. Identify the phase shift, $$\dfrac{C}{B}$$.
5. If $$B$$ is negative, use the Negative Angle Identity which (for the cotangent function) makes $$B$$ positive and changes the sign of $$A$$. Then, draw the graph of $$y=A\tan(Bx)$$ shifted to the right by $$\dfrac{C}{B}$$ and up by $$D$$.
6. Sketch the asymptotes $$x=\dfrac{C}{B}+\dfrac{\pi}{| B |}k$$, where $$k$$ is an integer.
7. Plot any three reference points and draw the graph through these points.
##### Example $$\PageIndex{9}$$: Graphing a Modified Cotangent
Sketch a graph of one period of the function $$f(x)=4\cot \left (\dfrac{\pi}{8}x−\dfrac{\pi}{2} \right )−2$$.
Solution
• Step 1. The function is already written in the general form $$f(x)=A\cot(Bx−C)+D$$.
• Step 2. $$A=4$$, so the stretching factor is $$4$$.
• Step 3. $$B=\dfrac{\pi}{8}$$, so the period is $$P=\dfrac{\pi}{| B |}=\dfrac{\pi}{\dfrac{\pi}{8}}=8$$.
• Step 4. $$C=\dfrac{\pi}{2}$$,so the phase shift is $$\dfrac{C}{B}=\dfrac{\dfrac{\pi}{2}}{\dfrac{\pi}{8}}=4$$.
• Step 5. We draw $$f(x)=4\tan \left (\dfrac{\pi}{8}x−\dfrac{\pi}{2} \right )−2$$.
• Step 6-7. Three points we can use to guide the graph are $$(6,2)$$, $$(8,−2)$$, and $$(10,−6)$$. We use the reciprocal relationship of tangent and cotangent to draw $$f(x)=4\cot \left (\dfrac{\pi}{8}x−\dfrac{\pi}{2} \right )−2$$.
• Step 8. The vertical asymptotes are $$x=4$$ and $$x=12$$.
The graph is shown in Figure $$\PageIndex{21}$$.
## Analyzing the Graphs of $$y = \sec x$$ and $$y = \csc x$$
The secant was defined by the reciprocal identity $$sec \, x=\dfrac{1}{\cos x}$$. Notice that the function is undefined when the cosine is $$0$$, leading to vertical asymptotes at $$\dfrac{\pi}{2}$$, $$\dfrac{3\pi}{2}$$ etc. Because the cosine is never more than $$1$$ in absolute value, the secant, being the reciprocal, will never be less than $$1$$ in absolute value.
We can graph $$y=\sec x$$ by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure $$\PageIndex{8}$$. The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.
The secant graph has vertical asymptotes at each value of $$x$$ where the cosine graph crosses the $$x$$-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.
Note that, because cosine is an even function, secant is also an even function. That is, $$\sec(−x)=\sec x$$.
As we did for the tangent function, we will again refer to the constant $$| A |$$ as the stretching factor, not the amplitude.
##### FEATURES OF THE GRAPH OF $$Y = A \sec(Bx)$$
• The stretching factor is $$| A |$$.
• The period is $$\dfrac{2\pi}{| B |}$$.
• The domain is $$x≠\dfrac{\pi}{2| B |}k$$, where $$k$$ is an odd integer.
• The range is $$(−∞,−|A|]∪[|A|,∞)$$.
• The vertical asymptotes occur at $$x=\dfrac{\pi}{2| B |}k$$, where $$k$$ is an odd integer.
• There is no amplitude.
• $$y=A\sec(Bx)$$ is an even function because cosine is an even function.
##### HOWTO: Given a function of the form $$y=A\sec(Bx)$$, graph one period
1. Express the function given in the form $$y=A\sec(Bx)$$.
2. Identify the stretching/compressing factor, $$|A|$$.
3. Identify $$B$$ and determine the period, $$P=\dfrac{2\pi}{| B |}$$.
4. Sketch the graph of $$y=A\cos(Bx)$$.
5. Use the reciprocal relationship between $$y=\cos \, x$$ and $$y=\sec \, x$$ to draw the graph of $$y=A\sec(Bx)$$.
6. Sketch the asymptotes.
7. Plot any two reference points and draw the graph through these points.
##### Example $$\PageIndex{4}$$: Graphing a Variation of the Secant Function
Graph one period of $$f(x)=2.5\sec(0.4x)$$.
Solution
• Step 1. The given function is already written in the general form, $$y=A\sec(Bx)$$.
• Step 2. $$A=2.5$$ so the stretching factor is $$2.5$$.
• Step 3. $$B=0.4$$ so $$P=\dfrac{2\pi}{0.4}=5\pi$$. The period is $$5\pi$$ units.
• Step 4. Sketch the graph of the function $$g(x)=2.5\cos(0.4x)$$.
• Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function.
• Steps 6–7. Sketch two asymptotes at $$x=1.25\pi$$ and $$x=3.75\pi$$. We can use two reference points, the local minimum at $$(0,2.5)$$ and the local maximum at $$(2.5\pi,−2.5)$$. Figure $$\PageIndex{10}$$ shows the graph.
##### Exercise $$\PageIndex{4}$$
Graph one period of $$f(x)=−2.5\sec(0.4x)$$.
This is a vertical reflection of the preceding graph because $$A$$ is negative.
Similar to the secant, the cosecant is defined by the reciprocal identity $$\csc x=\dfrac{1}{\sin x}$$. Notice that the function is undefined when the sine is $$0$$, leading to a vertical asymptote in the graph at $$0$$, $$\pi$$, etc. Since the sine is never more than $$1$$ in absolute value, the cosecant, being the reciprocal, will never be less than $$1$$ in absolute value.
We can graph $$y=\csc x$$ by observing the graph of the sine function because these two functions are reciprocals of one another. See Figure $$\PageIndex{7}$$. The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases.
The cosecant graph has vertical asymptotes at each value of $$x$$ where the sine graph crosses the $$x$$-axis; we show these in the graph below with dashed vertical lines.
Note that, since sine is an odd function, the cosecant function is also an odd function. That is, $$\csc(−x)=−\csc x$$.
The graph of cosecant, which is shown in Figure $$\PageIndex{9}$$, is similar to the graph of secant.
##### FEATURES OF THE GRAPH OF $$Y = A \csc(Bx)$$
• The stretching factor is $$| A |$$.
• The period is $$\dfrac{2\pi}{|B|}$$.
• The domain is $$x≠\dfrac{\pi}{|B|}k$$, where $$k$$ is an integer.
• The range is $$(−∞,−|A|]∪[|A|,∞)$$.
• The asymptotes occur at $$x=\dfrac{\pi}{| B |}k$$, where $$k$$ is an integer.
• $$y=A\csc(Bx)$$ is an odd function because sine is an odd function.
##### Given a function of the form $$y=A\csc(Bx)$$, graph one period.
1. Express the function given in the form $$y=A\csc(Bx)$$.
2. $$|A|$$.
3. Identify $$B$$ and determine the period, $$P=\dfrac{2\pi}{| B |}$$.
4. If $$B$$ is negative, use the Negative Angle Identity to make $$B$$ which for the $$\csc$$ function, also negates $$A$$ and $$C$$. Then, draw the graph of $$y=A\sin(Bx)$$.
5. Use the reciprocal relationship between $$y=sin \, x$$ and $$y=\csc \, x$$ to draw the graph of $$y=A\csc(Bx)$$.
6. Sketch the asymptotes.
7. Plot any two reference points and draw the graph through these points.
##### Example $$\PageIndex{6}$$: Graphing a Variation of the Cosecant Function
Graph one period of $$f(x)=−3\csc(4x)$$.
Solution
• Step 1. The given function is already written in the general form, $$y=A\csc(Bx)$$.
• Step 2. $$| A |=| −3 |=3$$,so the stretching factor is $$3$$.
• Step 3. $$B=4$$, so $$P=\dfrac{2\pi}{4}=\dfrac{\pi}{2}$$. The period is $$\dfrac{\pi}{2}$$ units.
• Step 4. Sketch the graph of the function $$g(x)=−3\sin(4x)$$.
• Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function.
• Steps 6–7. Sketch three asymptotes at $$x=0$$, $$x=\dfrac{\pi}{4}$$, and $$x=\dfrac{\pi}{2}$$. We can use two reference points, the local maximum at $$\left (\dfrac{\pi}{8},−3 \right )$$ and the local minimum at $$\left (\dfrac{3\pi}{8},3 \right )$$. Figure $$\PageIndex{14}$$ shows the graph.
##### Exercise $$\PageIndex{6}$$
Graph one period of $$f(x)=0.5\csc(2x)$$.
## Graphing Variations of $$y = \sec x$$ and $$y= \csc x$$
For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following.
$y=A\sec(Bx−C)+D \nonumber$
$y=A\csc(Bx−C)+D \nonumber$
##### FEATURES OF THE GRAPH OF $$Y = A\sec(Bx−C)+D$$
• The stretching factor is $$|A|$$.
• The period is $$\dfrac{2\pi}{|B|}$$.
• The domain is $$x≠\dfrac{C}{B}+\dfrac{\pi}{2| B |}k$$, where $$k$$ is an odd integer.
• The range is $$(−∞,−|A|]∪[|A|,∞)$$.
• The vertical asymptotes occur at $$x=\dfrac{C}{B}+\dfrac{π}{2| B |}k$$, where $$k$$ is an odd integer.
• There is no amplitude.
• $$y=A\sec(Bx)$$ is an even function because cosine is an even function.
##### Q&A: Do the vertical shift and stretch/compression affect the secant’s range?
Yes. The range of $$f(x)=A\sec(Bx−C)+D$$ is $$(−∞,−|A|+D]∪[|A|+D,∞)$$.
##### Given a function of the form $$f(x)=A\sec(Bx−C)+D$$, graph one period.
1. Express the function given in the form $$y=A \sec(Bx−C)+D$$.
2. Identify the stretching/compressing factor, $$| A |$$.
3. Identify $$B$$ and determine the period, $$\dfrac{2\pi}{|B|}$$.
4. Identify $$C$$ and determine the phase shift, $$\dfrac{C}{B}$$.
5. Draw the graph of $$y=A \sec(Bx)$$. but shift it to the right by $$\dfrac{C}{B}$$ and up by $$D$$.
6. Sketch the vertical asymptotes, which occur at $$x=\dfrac{C}{B}+\dfrac{\pi}{2| B |}k$$, where $$k$$ is an odd integer.
##### Exercise $$\PageIndex{7}$$
Given the graph of $$f(x)=2\cos \left (\dfrac{\pi}{2}x \right )+1$$ shown in Figure $$\PageIndex{17}$$, sketch the graph of $$g(x)=2\sec \left (\dfrac{\pi}{2}x \right )+1$$ on the same axes.
##### Example $$\PageIndex{5}$$: Graphing a Variation of the Secant Function
Graph one period of $$y=4\sec \left (\dfrac{\pi}{3}x−\dfrac{\pi}{2} \right )+1$$.
Solution
• Step 1. Express the function given in the form $$y=4\sec \left (\dfrac{\pi}{3}x−\dfrac{\pi}{2} \right )+1$$.
• Step 2. The stretching/compressing factor is $$| A |=4$$.
• Step 3. The period is $$\dfrac{2\pi}{|B|}= \dfrac{2\pi}{\dfrac{\pi}{3}}= 2\pi \cdot \dfrac{3}{\pi}= 6$$
• Step 4. The phase shift is $$\dfrac{C}{B}= \dfrac{\dfrac{\pi}{2}}{\dfrac{\pi}{3}}= \dfrac{\pi}{2}\cdot \dfrac{3}{\pi}= 1.5$$
• Step 5. Draw the graph of $$y=A\sec(Bx)$$, but shift it to the right by $$\dfrac{C}{B}=1.5$$ and up by $$D=1$$.
• Step 6. Sketch the vertical asymptotes, which occur at $$x=0$$, $$x=3$$, and $$x=6$$. There is a local minimum at $$(1.5,5)$$ and a local maximum at $$(4.5,−3)$$. Figure $$\PageIndex{12}$$ shows the graph.
##### Exercise $$\PageIndex{5}$$
Graph one period of $$f(x)=−6\sec(4x+2)−8$$.
##### Q&A: The domain of $$\csc \, x$$ was given to be all $$x$$ such that $$x≠k\pi$$ for any integer $$k$$. Would the domain of $$y=A\csc(Bx−C)+D$$ be $$x≠\dfrac{C+k\pi}{B}$$?
Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.
##### FEATURES OF THE GRAPH OF $$Y = A\csc(Bx−C)+D$$
1. The stretching factor is $$|A|$$.
2. The period is $$\dfrac{2\pi}{|B|}$$.
3. The domain is $$x≠\dfrac{C}{B}+\dfrac{\pi}{2| B |}k$$, where $$k$$ is an integer.
4. The range is $$(−∞,−|A|]∪[|A|,∞)$$.
5. The vertical asymptotes occur at $$x=\dfrac{C}{B}+\dfrac{\pi}{|B|}k$$, where $$k$$ is an integer.
6. There is no amplitude.
7. $$y=A\csc(Bx)$$ is an odd function because sine is an odd function.
##### Given a function of the form $$f(x)=A \csc(Bx−C)+D$$, graph one period
1. Express the function given in the form $$y=A\csc(Bx−C)+D$$.
2. Identify the stretching/compressing factor, $$|A|$$.
3. Identify $$B$$ and determine the period, $$\dfrac{2\pi}{| B |}$$.
4. Identify $$C$$ and determine the phase shift, $$\dfrac{C}{B}$$.
5. If $$B$$ is negative, use the Negative Angle Identity to make $$B$$ which for the $$\csc$$ function, also negates $$A$$ and $$C$$. Then, draw the graph of $$y=A\csc(Bx)$$ but shift it to the right by and up by $$D$$.
6. Sketch the vertical asymptotes, which occur at $$x=\dfrac{C}{B}+\dfrac{\pi}{| B |}k$$, where $$k$$ is an integer.
##### Example $$\PageIndex{7}$$: Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant
Sketch a graph of $$y=2\csc \left (\dfrac{\pi}{2}x \right )+1$$. What are the domain and range of this function?
Solution
• Step 1. Express the function given in the form $$y=2\csc \left (\dfrac{\pi}{2}x \right )+1$$.
• Step 2. Identify the stretching/compressing factor, $$| A |=2$$.
• Step 3. The period is $$\dfrac{2\pi}{| B |}=\dfrac{2\pi}{\dfrac{\pi}{2}}=2\pi⋅\dfrac{2}{\pi}=4$$.
• Step 4. The phase shift is $$\dfrac{0}{\dfrac{\pi}{2}}=0$$.
• Step 5. Draw the graph of $$y=A\csc(Bx)$$ but shift it up $$D=1$$.
• Step 6. Sketch the vertical asymptotes, which occur at $$x=0$$, $$x=2$$, $$x=4$$.
The graph for this function is shown in Figure $$\PageIndex{16}$$.
Analysis
The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of $$f(x)=2\sin \left (\frac{\pi}{2}x \right )+1$$,shown as the orange dashed wave.
## Using the Graphs of Trigonometric Functions to Solve Real-World Problems
Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function.
##### Example $$\PageIndex{10}$$: Using Trigonometric Functions to Solve Real-World Scenarios
Suppose the function $$y=5\tan(\dfrac{\pi}{4}t)$$ marks the distance in the movement of a light beam from the top of a police car across a wall where $$t$$ is the time in seconds and $$y$$ is the distance in feet from a point on the wall directly across from the police car.
1. Find and interpret the stretching factor and period.
2. Graph on the interval $$[0,5]$$.
3. Evaluate $$f(1)$$ and discuss the function’s value at that input.
Solution
1. We know from the general form of $$y=A\tan(Bt)$$ that $$| A |$$ is the stretching factor and $$\dfrac{\pi}{B}$$ is the period.
We see that the stretching factor is $$5$$. This means that the beam of light will have moved $$5$$ ft after half the period.
The period is $$\dfrac{\pi}{\tfrac{\pi}{4}}=\dfrac{\pi}{1}⋅\dfrac{4}{\pi}=4$$. This means that every $$4$$ seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.
1. To graph the function, we draw an asymptote at $$t=2$$ and use the stretching factor and period. See Figure $$\PageIndex{23}$$
1. period: $$f(1)=5\tan(\frac{\pi}{4}(1))=5(1)=5$$; after $$1$$ second, the beam of has moved $$5$$ ft from the spot across from the police car.
##### Media
Access these online resources for additional instruction and practice with graphs of other trigonometric functions.
• Graphing the Tangent
• Graphing Cosecant and Secant
• Graphing the Cotangent
## Key Equations
Shifted, compressed, and/or stretched tangent function $$y=A \tan(Bx−C)+D$$ Shifted, compressed, and/or stretched secant function $$y=A \sec(Bx−C)+D$$ Shifted, compressed, and/or stretched cosecant function $$y=A \csc(Bx−C)+D$$ Shifted, compressed, and/or stretched cotangent function $$y=A \cot(Bx−C)+D$$
## Key Concepts
• The tangent function has period $$π$$.
• $$f( x )=A\tan( Bx−C )+D$$ is a tangent with vertical and/or horizontal stretch/compression and shift. See Example $$\PageIndex{1}$$, Example $$\PageIndex{2}$$, and Example $$\PageIndex{3}$$.
• The secant and cosecant are both periodic functions with a period of $$2\pi$$. $$f( x )=A\sec( Bx−C )+D$$ gives a shifted, compressed, and/or stretched secant function graph. See Example $$\PageIndex{4}$$ and Example $$\PageIndex{5}$$.
• $$f( x )=A\csc( Bx−C )+D$$ gives a shifted, compressed, and/or stretched cosecant function graph. See Example $$\PageIndex{6}$$ and Example $$\PageIndex{7}$$.
• The cotangent function has period $$\pi$$ and vertical asymptotes at $$0,±\pi,±2\pi$$,....
• The range of cotangent is $$( −∞,∞ )$$, and the function is decreasing at each point in its range.
• The cotangent is zero at $$±\dfrac{\pi}{2},±\dfrac{3\pi}{2}$$,....
• $$f(x)=A\cot(Bx−C)+D$$ is a cotangent with vertical and/or horizontal stretch/compression and shift. See Example $$\PageIndex{8}$$ and Example $$\PageIndex{9}$$.
• Real-world scenarios can be solved using graphs of trigonometric functions. See Example $$\PageIndex{10}$$.
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# How do you simplify 4^3·4^5?
May 29, 2016
${4}^{3} \cdot {4}^{5} = {4}^{8} = 65536$
#### Explanation:
For positive integer exponents we have:
${x}^{n} = {\overbrace{x \cdot x \cdot . . \cdot x}}^{\text{n times}}$
Hence:
${x}^{m} \cdot {x}^{n} = {\overbrace{x \cdot x \cdot . . \cdot x}}^{\text{m times" * overbrace(x * x * .. * x)^"n times}}$
$= {\overbrace{x \cdot x \cdot . . \cdot x}}^{\text{m + n times}} = {x}^{m + n}$
So in our example:
${4}^{3} \cdot {4}^{5} = {4}^{3 + 5} = {4}^{8}$
$\textcolor{w h i t e}{}$
If we know our powers of $2$, then it is also helpful to use another property of exponents:
If $a , b , c > 0$ then:
${\left({a}^{b}\right)}^{c} = {a}^{b c}$
So:
${4}^{8} = {\left({2}^{2}\right)}^{8} = {2}^{2 \cdot 8} = {2}^{16} = 65536$
$\textcolor{w h i t e}{}$
Alternatively we could write:
${4}^{2} = 16$
${4}^{4} = {4}^{2} \cdot {4}^{2} = 16 \cdot 16 = 256$
${4}^{8} = {4}^{4} \cdot {4}^{4} = 256 \cdot 256 = 65536$
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# Auxiliary fraction
Auxiliary fraction
In Swami Bharati Krishna Tirtha's Vedic mathematics, the auxiliary fraction method is used to convert a fraction to its equivalent decimal representation. The "auxiliary fraction" is not a true fraction, but is simply a mnemonic aid used in the calculation. The method is essentially the long division algorithm adapted for mental calculation. It is simplest when the fraction's denominator is one less than a multiple of 10, when it uses the identity
: $a=qb+r Rightarrow 10a=q\left(10b-1\right)+\left(10r+q\right).$
Variants of the method used when the denominator is not one less than a multiple of 10 become progressively more complex but still in the realm of mental math or with one line of notation.
Dividing by powers of ten
To divide by a power of ten first move the decimal point in both the numerator and the denominator to the left the same number of places as the number of zeros at the end of the denominator. Then divide. For example, 1/800 = 0.01/8; 39/70 = 3.9/7; 3741/110000 = 0.3741/11; and 97654/90,000,000 = 0.0097654/9.
Forming the auxiliary fraction
The formation of the auxiliary fraction depends on the denominator. There are four cases:
*Type One A: denominators which end in a single nine
*Type One B: denominators which end in several nines
*Type Two: denominators which end in one
*Type Three: denominators which end in the digits, 2, 3, 4, 5, 6, 7, and 8.
If the denominator ends in zero(s), its first non-zero digit (from the right) identifies the family of the denominator.
Type One A
When the fraction's denominator ends in a single nine use the "Ekādhika Purva." First replace the denominator by its "Ekādhika" which means adding one to the denominator and dividing the result by 10. Divide the numerator by 10 also. [Pages 255-256, "Vedic Mathematics"] Hereafter, F is the actual fraction to be converted to a decimal value and A.F. is the auxiliary fraction used to remember the original dividend and the recurrent divisor.
Examples:
:When F = 1/19, then the A.F. = 1/20 = 0.1/2;
:When F = 4/29, then the A.F. = 4/30 = 0.4/3;
:When F = 8/59, then the A.F. = 8/60 = 0.8/6;
Type One B
When the denominator ends in several nines, increase the denominator by one, then divide both numerator and denominator by a power of 10 equal to the number of terminal nines in the denominator.
Examples:
*When F = 174/1299, then the A.F. = 174/1300 = 1.74/13;
*When F = 1/6999, then the A.F. = 1/7000 = 0.001/7;
If the denominator ends in 1, 3, or 7, multiply both denominator and numerator by 9, 3 or 7 respectively to convert to an equivalent fraction in which the denominator ends in 9.
Examples:
* When F = 36/121 = 324/1089, then the A.F. = 32.4/109;
* When F = 53/93 = 159/279, then the A.F. = 15.9/28;
* When F = 15/37 = 105/259, then the A.F. = 10.5/26.
Type Two
When the fraction has a denominator ending in one, form the auxiliary fraction by subtracting one from the denominator and from the numerator. Then divide both numerator and denominator by a power of 10 equal to the number of terminal zeros in the new denominator. [Pages 259-262, "Vedic Mathematics"]
Examples:
* When F = 3/61, then the A.F = 2/60 = 0.2/6;
* When F = 28/71, then the A.F. = 27/70 = 2.7/7;
* When F = 1/81, then the A.F. = 0/80 = 0.0/8;
* When F = 14/131, then the A.F. = 13/130 = 1.3/13;
* When F = 1/301, then the A.F. = 0/300 = 0.00/3;
* When F = 6163/8001, then the A.F. = 6162/8000 = 6.162/8;
If the denominator ends in 3 or 7, multiply both denominator and numerator by 7 or 3 respectively to convert to an equivalent fraction in which the denominator ends in 1.
Examples:
* When F = 2/3 = 14/21, then the A.F. = 13/20 = 1.3/2;
* When F = 10/27 = 30/81, then the A.F. = 29/80 = 2.9/8;
* When F = 5/67 = 15/201, then the A.F. = 14/200 = 0.14/2;
* When F = 4/17 = 12/51, then the A.F. = 11/50 = 1.1/5.
Type Three
For denominators that do not end in 1 or 9, use the nines family auxiliary fraction and count the number of units (above or below) that the ending is from the normal nine.
Using the auxiliary fraction - Type One
Given a fraction whose denominator ends in 9, first write the auxiliary fraction. Next divide in the auxiliary fraction to generate one (or more) quotient digit(s) at-a-time. Then write the quotient digit(s) and the remainder. The remainder at each step is prefixed to the just generated quotient digit for the next division. So if the quotient and remainder from one step are "q" and "r", the dividend for the next step is 10"r"+"q".
This algorithm meets the Vedic ideal of mental math with one line notation. ["Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas", Auxiliary Fractions, chapter XXVIII, pages 255-272.]
Word examples - Type One A
Example One
To convert the fraction F = 1/169 to a (repeating) decimal:
First, estimate the quotient as about six thousandths. Then, set up the auxiliary fraction, AF = 0.1/17. The first dividend is 0.1 and the working divisor is 17. Calculate one quotient digit at a time. Set down the remainder as a (sub-scripted) prefix to the quotient-digit just produced. Continue dividing to generate the quotient of the precision desired. Remember that the prefixed remainders are not parts of the quotient but only prefixes to the quotient-group in question and are dropped out of the answer - the lower row is a mere scaffolding and goes out. [Page 258, "Vedic Mathematics"]
As there are 168 remainders for the repeating decimal value of one one-hundred-sixty-ninth, there can be at most 168 digits before the decimal expansion repeats. If the remainder were to be zero or 169, then the fraction would terminate as an exact decimal value. There are in fact 78 digits before the decimal expansion of 1/169 repeats, and 78 is half of 156, which is the totient of 169.
17 into 0.1 goes 0 rem 1. 17 into 10 goes 0 rem 10. 17 into 100 goes 5 rem 15. 17 into 155 goes 9 rem 2. (The third dividend, 155, is formed by the remainder 15 prefixed to quotient from the previous step, 5.)
F = 1/169 = .10. 10 100 155 29 121 27 101 165 129 107 56 53 23 61 103 16 150 79 114 126 77 ...
This all the notation that is needed. The calculating is mental using the multiples of 17: 17, 34, 51, 68, 85, 102, 119, 136, 153, 170.
F = 1/169 ≈ 0.00591715976331361...
Example Two
If the denominator is 73, this is a special case because (73)(137) = 10001. F = 1/73 = 137/10001 = (137)(9,999)/(99,999,999) = 0.0136,9863 repeating.Thus we may expect an eight-digit repeating decimal with complementary halves, i.e., the digits in the first half of the repeating decimal digit set are complements of nine for the digits of the second half. [Page 227, "Vedic Mathematics"]
Example Three
When F = 3/73 = 9/219, then the A.F. = 9/22. The working divisor is 22.The dividend at each step is created by prefixing the remainder to the quotient from the previous step.
F = 0.90 24 21 210 129 195 198 09 90 24 21 210...
F = 3/73 ≈ 0.041095890410... The fraction repeats after eight decimal places. Furthermore, the eight digits have complementary halves (see Midy's theorem).
Worked example - Type One B
When F = 53/799, then the A.F. = 0.53/8
The working divisor is 8. As the denominator has two nines, the division proceeds in steps of two dividend digits at a time, generating two quotient digits at each step. The remainder is prefixed to the pair of quotient digits from the previous step to create the dividend for the next step. [Page 257, "Vedic Mathematics"]
F = 53/799 = 0.506 263 732 491 361 145 18...
F = 53/799 ≈ 0.06633291614518...
Generating a Denominator with More Terminal Nines
An additional technique is available to find another A.F. with a smaller divisor. By a judicious choice of the multiplier one can produce an equivalent fraction with more nines in the denominator. [Page 259, "Vedic Mathematics"]
When F = 1/7 = 7/49, then the A.F. = 0.7/5; We use a multiplier of 7 to produce an equivalent fraction, 7/49. Look at the denominator of the equivalent fraction, 49. Consider the 4. To build the 4 to a nine we need to add 5 in the tens place. The 5th multiple of 7 ends in 5, so we may use 5 tens or 57 as the multiplier.
F = 1/7 = 57/399, and the A.F. = 0.57/4 (dividing in bundles of two digits).
We may proceed likewise, until we have F = 1/7 = 142,857/999,999 giving a bundle of six nines and an A.F. = 0.142857/1. This means we have the repeating decimal digit set on sight because when we divide this bundle of six digits (0.142857) by one and we have no remainder and this bundle repeats!
Using the auxiliary fraction - Type Two
After forming the auxiliary fraction, divide in the first step, but create the next dividend by prefixing the remainder not to the quotient digit, but to the quotient digit's complement from nine. So if the quotient and remainder from one step are "q" and "r", the dividend for the next step is 10"r" + (9 - "q").
When the fraction has a numerator of one, the auxiliary fraction will have a numerator of zero. Having a dividend of zero is not a problem because on the second step the complement of zero is nine, an adequate dividend.
Worked example - Type Two
F = 13/31, then the A.F. = 12/30 = 1.2/3
The working divisor is 3. The lower row is the dividend.
F = 13/31 = 0. 4 1 9 3 5 4 8 3 8 7 0 9 Prefix R to complement of quotient: 05 28 10 16 14 25 11 26 21 02 29 20
F = 13/31 ≈ 0.419354838709...
Using the auxiliary fraction - Type Three
When the fraction has a denominator that does not end in 1 or 9 apply the "Ānurūpya" Sūtra, whereby, after prefixing each remainder to the quotient-digit in question, add to (or subtract from) the dividend at each step, as many times the quotient-digit as the divisor (the denominator) is below (or above) the normal nine. The process can be wholly mental with practice. [Pages 262-263, "Vedic Mathematics"]
Worked examples - Type Three
Example one
If the denominator is 68, as 68=(4)(17), we may expect two fixed digits and 16 repeating digits. When F = 15/68, the A.F. = 1.5/7 (As D ends in 8, one below the normal ending, 9, we add the quotient-digit to the dividend at each step). The working divisor is 7. Continue dividing to achieve the desired precision. F = 15/68, A.F. = 1.5/7 A prefixed rem: F = 0.12 02 40 55 48 08 22 33 15 62 19 04 11 51 Plus the Q-digit: 2 2 0 5 8 8 2 3 5 2 9 4 1 1 7 into actual dividend: 14 04 40 60 56 16 24 36 20 64 28 08 12 52 F = 15/68 ≈ 0.22058823529411...
Example two
When F = 163/275, A.F. = 16.3/28. (Since D ends in 5, four below the normal ending, 9, we add four times the Q-digit to the dividend at each step.) The working divisor is 28. Since D = 275 = (52)(11), we may expect two fixed digits, then a two-digit repeater. Remember that every factor of 2, 5, or 10 in the denominator generates one fixed decimal digit. Multiples of 28: 28, 56, 84, 112, 140, 168, 196, 224, 252, 280.
Prefix the rem: F = 0. 235 39 192 47 192 47 Plus four times the Q-digit: x 20 36 08 28 08 28 28 into the actual dividend: 163 255 75 200 75 200 75 F = 163/275 ≈ 0.592727...
References
* "Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas", by Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaja (1884-1960), Motilal Banarsidass Indological Publishers and Booksellers, Varnasi, India, 1965; reprinted in Delhi, India, 1975, 1978. 367 pages.
*Fraction (mathematics)
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## Step by action solution because that calculating 15 is 30 percent that what number
We already have our an initial value 15 and also the second value 30. Let"s i think the unknown value is Y which answer we will discover out.
You are watching: 15 is 30 of what number
As we have all the compelled values us need, currently we have the right to put castle in a simple mathematical formula as below:
STEP 115 = 30% × Y
STEP 215 = 30/100× Y
Multiplying both political parties by 100 and dividing both political parties of the equation by 30 we will certainly arrive at:
STEP 3Y = 15 × 100/30
STEP 4Y = 15 × 100 ÷ 30
STEP 5Y = 50
Finally, we have discovered the value of Y i m sorry is 50 and that is our answer.
You can quickly calculate 15 is 30 percent the what number through using any regular calculator, simply enter 15 × 100 ÷ 30 and also you will get your answer i m sorry is 50
Here is a portion Calculator to solve similar calculations such together 15 is 30 percent of what number. You deserve to solve this type of calculation through your values by beginning them into the calculator"s fields, and also click "Calculate" to acquire the result and explanation.
is
percent of what number
Calculate
## Sample questions, answers, and how to
Question: her friend has actually a bag that marbles, and also he speak you that 30 percent of the marbles space red. If there are 15 red marbles. How plenty of marbles walk he have actually altogether?
How To: In this problem, we understand that the Percent is 30, and also we are additionally told that the component of the marbles is red, therefore we understand that the component is 15.
So, that way that it should be the full that"s missing. Here is the means to figure out what the complete is:
Part/Total = Percent/100
By making use of a basic algebra we deserve to re-arrange our Percent equation choose this:
Part × 100/Percent = Total
If we take the "Part" and multiply the by 100, and also then we division that through the "Percent", us will obtain the "Total".
Let"s try it the end on ours problem around the marbles, that"s very straightforward and it"s simply two steps! We understand that the "Part" (red marbles) is 15.
So action one is to simply multiply that part by 100.
15 × 100 = 1500
In action two, us take that 1500 and divide that by the "Percent", i m sorry we space told is 30.
So, 1500 split by 30 = 50
And that way that the total number of marbles is 50.
Question: A high institution marching band has 15 flute players, If 30 percent that the tape members pat the flute, climate how countless members room in the band?
Answer: There room 50 members in the band.
How To: The smaller sized "Part" in this trouble is 15 because there room 15 flute players and we are told that they consist of 30 percent that the band, so the "Percent" is 30.
Again, it"s the "Total" that"s lacking here, and to find it, we simply need to follow our 2 action procedure as the ahead problem.
For step one, us multiply the "Part" by 100.
15 × 100 = 1500
For action two, we divide that 1500 by the "Percent", which is 30.
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1500 divided by 30 equals 50
That way that the total variety of band members is 50.
## Another action by step method
Step 1: Let"s i think the unknown value is Y
Step 2: an initial writing the as: 100% / Y = 30% / 15
Step 3: autumn the percent marks to simplify your calculations: 100 / Y = 30 / 15
Step 4: main point both sides by Y to relocate Y ~ above the best side the the equation: 100 = ( 30 / 15 ) Y
Step 5: simplifying the appropriate side, us get: 100 = 30 Y
Step 6: separating both sides of the equation by 30, we will arrive in ~ 50 = Y
This leaves us with our final answer: 15 is 30 percent the 50
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1. ## Solve the following.
1. 4/x = 5/7
2. 3a/7 = -2/5
3. x+1/6 = 4/3
4. 24/x-3 = 72/x+3
Thanks.
2. I'll only solve the 1st one, I think you can do the rest.
$\displaystyle \frac{4}{x} = \frac{5}{7}$
Multiply both sides by $\displaystyle x$,
$\displaystyle \frac{4}{x}\cdot x = \frac{5}{7}\cdot x$
$\displaystyle \frac{4}{\not x}\cdot \not x = \frac{5x}{7}$
$\displaystyle 4 = \frac{5x}{7}$
Now multiply both sides by $\displaystyle 7$,
$\displaystyle 4\cdot 7 = \frac{5x}{7}\cdot 7$
$\displaystyle 28 = \frac{5x}{\not 7}\cdot \not7$
$\displaystyle 28 = 5x$
$\displaystyle x = \frac{28}{5}$
----------
If you have an equation in the from $\displaystyle \frac{a}{b} = \frac{c}{d}$, products of the terms crosswise are equal, $\displaystyle a\cdot d = b \cdot c$
So, $\displaystyle \frac{4}{x} = \frac{5}{7}$
$\displaystyle 4\cdot 7 = 5\cdot x$
$\displaystyle 5x = 28$
$\displaystyle x = \frac{28}{5}$
3. Originally Posted by elliotfsl
1. 4/x = 5/7
Initial Equation
$\displaystyle \frac 4x = \frac 57$
You want to get x in the numerator so multiply both sides by x
$\displaystyle \frac x1 * \frac 4x = \frac 57*\frac x1$
Simplifly
$\displaystyle 4 = \frac {5x}7$
You want x to be by itself, so you need to get rid of the 7 in the denominator. You can do this by multiplying both sides by 7 since 7/7 =1
$\displaystyle \frac 71 *\frac 41 = \frac {5x}7*\frac 71$
Simplify
$\displaystyle 28 = 5x$
You want to get x by itself, so you need to get rid of the 5, do this by multiplying both sides by 1/5 since 5/5=1
$\displaystyle \frac 15*28 = 5x\frac 15$
Simplify
$\displaystyle \frac{28}5 = x$
4. Originally Posted by elliotfsl
4. 24/x-3 = 72/x+3
Initial equation
$\displaystyle \frac {24}{x-3} = \frac{72}{x+3}$
Again, you need to get x out of the denominator, so multiply both sides by x-3
$\displaystyle 24 = \frac{72(x-3)}{x+3}$
X is still in the denominator on the right side, so multiply both sides by x+3
$\displaystyle 24(x+3) = 72(x-3)$
Distribute the coefficients
$\displaystyle 24x+72 = 72x-216$
You need all your x's to be on the same side, so subtract 24x
$\displaystyle 72 = 72x-24x-216$
Simplify
$\displaystyle 72 = 48x-216$
You need all the other numbers to be on the other side of x, so add 216
$\displaystyle 72 +216= 48x$
Simplify
$\displaystyle 288= 48x$
You need x to be by itself, so multiply by 1/48
$\displaystyle \frac{288}{48}= x$
And simplify
$\displaystyle 6 = x$
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# Thread: intersection of planes
1. ## intersection of planes
1. The line of intersection of the planes 2x+y-3z=3 and x-2y+z=-1 is L. If L meets the xy plane at Point A and the z-axis at Point B, determine the length of line segment AB.
2. Determine the Cartesian equation of the plane that is parallel to the line with equation x = -2y = 3z and that contains the line of intersection of the planes with equation x-y+z=1 and 2y-z=0
Thanks, any help is greatly appreciated!
2. Hello, checkmarks!
1. The line of intersection of the planes: $\displaystyle 2x+y-3z\:=\:3\:\text{ and }\:x-2y+z\:=\:-1\:\text{ is }L.$
If $\displaystyle L$ meets the $\displaystyle xy$-plane at point $\displaystyle A$ and the $\displaystyle z$-axis at point $\displaystyle B,$
determine the length of line segment $\displaystyle AB.$
First, find the equation of the plane.
We have: .$\displaystyle \begin{array}{cccc}2x + y - 3x &=& 3 & [1] \\ x - 2y + z &=& \text{-}1 & [2] \end{array}$
$\displaystyle \begin{array}{ccccc}\text{Multiply [1] by 2:} & 4x + 2y - 6x &=& 6 \\ \text{Add [2]:} & x - 2y + z &=& \text{-}1 \end{array}$
And we have: .$\displaystyle 5x - 5z \:=\:5 \quad\Rightarrow\quad x \:=\:z+1$
Substitute into [2]: .$\displaystyle (z+1) - 2y + z \:=\:\text{-}1 \quad\Rightarrow\quad y \:=\:z+1$
We have these equations: .$\displaystyle \begin{Bmatrix}x &=& z+1 \\ y &=& z+1 \\ z &=&z \end{Bmatrix}$
On the right, replace $\displaystyle z$ with a parameter $\displaystyle t\!:\quad \begin{Bmatrix}x &=& t+1 \\ y &=& t+1 \\ z &=& t\end{Bmatrix}$
These are the parametric equations of the line of intersection.
Line $\displaystyle L$ intersects the $\displaystyle xy$-plane at $\displaystyle A$ . . . Then: .$\displaystyle z = 0.$
Then $\displaystyle t = 0 \quad\Rightarrow\quad x = 1,\;y = 1$
. . We have point $\displaystyle A(1,1,0)$
Line $\displaystyle L$ intersected the $\displaystyle z$-axis at $\displaystyle B$ . . . Then: .$\displaystyle x = 0,\:y = 0$
Then: .$\displaystyle t = \text{-1} \quad\Rightarrow\quad z \:=\:\text{-1}$
. . We have point $\displaystyle B(0,0,\text{-}1)$
Therefore, length of $\displaystyle AB \;=\;\sqrt{(0-1)^2 + (0-1)^2 + (-1-0)^2} \;=\;\sqrt{3}$
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# Circumference
Calculating the circumference of a circle helps in many ways. Suppose you were building a circular pool with a diameter of 30 feet, and you want to put a decorative border around the outside edge. You'd need to know the length of the border material required to go around the entire pool. To find this, you would need to calculate the circumference of the circle. Fortunately, calculating circumference is easy, and we can do it in just a few steps.
## Why you need pi to calculate circumference
First, you need pi. No, we're not talking about dessert -- we're talking about the ratio of a circle's circumference to its diameter. Fortunately, we don't need to actually calculate pi, since ancient cultures such as the Egyptians and Greeks already accomplished this thousands of years ago. The value of pi is approximately 3.14259 -- although the number contains an infinite number of non-repeating digits!
The formula is simple: C = 2πr
Where "C" represents the circumference, π represents pi, and "r" represents the radius.
You could also write the formula like this: C = πd
Where "d" represents the diameter of a circle.
Example 1:
A bicycle tire has a diameter of 26 inches. What is the circumference of the tire?
To solve this problem, we can use the formula for the circumference of a circle, which is C = πd, where C is the circumference, π (pi) is a mathematical constant approximately equal to 3.14, and d is the diameter of the circle.
We are given that the diameter of the bicycle tire is 26 inches, so we can substitute this value into the formula:
C = πd, C = 3.14 * 26, C = 81.7 inches
Therefore, the circumference of the bicycle tire is approximately 81.7 inches.
Example 2:
A circular swimming pool has a diameter of 30 feet. The pool has a walkway around it that is 5 feet wide. What is the total area of the walkway? What is the total circumference of the pool and the walkway?
To solve this problem, we can start by finding "r" the radius of the pool, which is half of the diameter:
r = d/2 = 30/2 = 15 feet
Next, we can find the radius of the pool and the walkway together "R", by adding the 5 feet of the walkway to the radius of the pool:
R = r + 5 = 15 + 5 = 20 feet
Finally, we can find the total circumference of the pool and the walkway together by adding the circumference of the larger circle with radius R to the circumference of the pool with radius r:
Circumference of larger circle: 2πR = 2 * 3.14 * 20 ≈ 125.7 feet
Circumference of pool: 2πr = 2 * 3.14 * 15 ≈ 94.2 feet
Total circumference of pool and walkway is equal to the Circumference of the larger circle plus the Circumference of the pool which is approximately: 125.7 + 94.2 ≈ 219.9 feet
Circle
Chord
π (Pi)
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Sequences and Series Notes Class 11th Maths
0
2582
1. Sequence: Sequence is a function whose domain is a subset of natural numbers. It represents the images of 1, 2, 3,… ,n, as f1, f2, f3, …., fn , where fn = f(n).
2. Real Sequence: A sequence whose range is a subset of R is called a real sequence.
3. Series: If a1, a2, a3 , … , an is a sequence, then the expression a1 + a2 + a3 + … + an is a series.
4. Progression: A sequence whose terms follow certain rule is called a progression.
5. Finite Series: A series having finite number of terms is called finite series.
6. Infinite Series: A series having infinite number of terms is called infinite series.
Arithmetic Progression (AP)
A sequence in which the difference of two consecutive terms is constant, is called Arithmetic Progression (AP).
Properties of Arithmetic Progression
(i) If a sequence is an AP, then its nth term is a linear expression in n, i.e., its nth term is given by An + B, where A and B are constants and A = common difference.
(ii) nth Term of an AP If a is the first term, d is the common difference and / is the last term of an AP, then
(a) nth term is given by 1= an = a + (n – 1)d
(b) nth term of an AP from the last term is a’n = l – (n – 1)d
(c) an + a’n = a + 1
i.e., nth term from the start + nth term from the end
= constant
= first term + last term
(d) Common difference of an AP
d = Tn – Tn-1, ∀ n > 1
(e) Tn = 1/2[Tn-k + Tn+k], k < n
(iii) If a constant is added or subtracted from each term of an AP, then the resulting sequence is an AP with same common difference.
(iv) If each term of an AP is multiplied or divided by a non-zero constant k, then the resulting sequence is also an AP, with common difference kd or d/k where d = common difference.
(v) If an, an+1 and an+2 are three consecutive terms of an AP, then 2an+1 = an + an+2.
(vi) (a) Any three terms of an AP can be taken as a – d, a, a + d.
(b) Any four terms of an AP can be taken as a-3d,a- d, a + d, a + 3d.
(c) Any five terms of an AP can be taken as a-2d,a – d, a, a + d, a + 2d.
(vii) Sum of n Terms of an AP
(a) Sum of n terms of AP, is given by Sn = n/2[2a + (n – 1)d] = n/2[a + l]
(b) A sequence is an AP, iff the sum of n terms is of the form An2 + Bn, where A and B are constants. Common difference in such case will be 2A.
(c) Tn = Sn – Sn-1
(viii) a2, b2 and c2 are in AP.
(ix) If a1, a2,…, an are the non-zero terms of an AP, then
(x) Arithmetic Mean
(a) If a, A and b are in AP, then A= (a + b)/2 is called the 2 arithmetic mean of a and b.
(b) If a1, a2, a3 , an are n numbers, then their AM is given by,
(c) If a, A1 , A2 , A3 ,…,An, b are in AP, then A1, A2, A3,…, An are n arithmetic mean between a and b, where
(d) Sum of n AM’s between a and b is nA
i.e., A1 + A2 + A3 + + = nA
Geometric Progression (GP)
A sequence in which the ratio of two consecutive terms is constant is called GP. The constant ratio is called common ratio (r).
i.e., an+1/an = r, ∀ n ≥ 1
Properties of Geometric Progression (GP)
(i) nth Term of a GP If a is the first term and r is the common ratio
(a) nth term of a GP from the beginning is an = arn-1
(b) nth term of a GP from the end is a’n = l/rn-1, l = last term
(c) If a is the first term and r is the common ratio of a GP, then the GP can be written as a, ar, ar2,… , arn-1, …
(d) The nth term from the end of a finite GP consisting of m terms is arm-n, where a is the first term and r is the common ratio of the GP.
(e) ana’n = al i.e., nth term from the beginning x nth term from the end = constant = first term x last term.
(ii) If all the terms of GP be multiplied or divided by same non-zero constant, then the resulting sequence is a GP with the same common ratio.
(iii) The reciprocal terms of a given GP form a GP.
(iv) If each term of a GP be raised to same power, the resulting sequence also forms a GP.
(v) If the terms of a GP are chosen at regular intervals, then the resulting sequence is also a GP.
(vi) If a1, a2, a3, … , an are non-zero, non-negative term of a GP, then
(a) GM = (a1a2a3… an )1/n
(b) log a1, log a2, log a3,…, log an are in an AP and vice-versa.
(vii) If a, b and c are three consecutive terms of a GP, then b2 = ac
(viii) (a) Three terms of a GP can be taken as a/r, a and ar.
(b) Four terms of a GP can be taken as a/r3, a/r, ar and ar3.
(c) Five terms of a GP can be taken as a/r2, a/r, ar and ar2.
(ix) Sum of n Terms of a GP
(a) Sum of n terms of a GP is given by
(x) Geometric Mean (GM)
(a) If a, G, b are in GP, then G is called the geometric mean of a and b and is given by G = √ab
(b) If a, G1, G2, G3, , Gn, b are in GP, then G1, G2, G3,… , Gn, are in GM’s between a and b, where
(c) Product of n GM’s, G1 X G2 X G3 X … X Gn = Gn
Harmonic Progression (HP)
A sequence a1, a2, a3 ,…, an of non-zero numbers is called a Harmonic Progression (HP), if the sequence 1/a1, 1/a2, 1/a3, …, 1/an is an AP.
Properties of Harmonic Progression (HP)
(i) nth term of HP, if a1, a2, a3 ,…, an are in HP, then
(a) nth term of the HP from the beginning
(b) nth term of the HP from the end
(d) an = 1/a+(n-1)d are the first term and common difference of the corresponding AP.
(ii) Sum of harmonic progression does not exist.
Harmonic Mean
(i) If a, H,b are in HP, then H is called the harmonic mean of a and b i.e., H= 2ab/(a + b)
(ii) If a, H1, H2, H3, …, Hn, b are in HP, then
H1, H2, H3, …, Hn
are n harmonic means between a and b where
(iii) Harmonic Mean (HM) between a1, a2, a3, …, an is given by
Properties of AM, GM and HM between Two Numbers
If A, G and H are arithmetic, geometric and harmonic means of two positive numbers a and b, then
(i) A=(a+b)/2, G=√ab, H=(2ab)/(a+b)
(ii) A≥G≥H
(iii) A, G, H are in GP and G2 = AH
(iv) If A,G,H are AM, GM and HM between three given numbers a, b and c, then the equation on having a, b and c as its root is
(v) If A1,A2 be two AM’s, G1, G2 be two GM’s and H1, H2 be two HM’s between two numbers a and b, then
(vi) If A,G and H be AM, GM and HM between two numbers a and b, then
Arithmetico-Geometric Progression
A sequence in which every term is a product of a term of AP and GP is known as arithmetico-geometric progression.
The series may be written as
Sum of Arithmetico-Geometric Series
Type 1 Let al + a2 + a3 + … be a given series. If a2 – al, a3 – a2, … are in AP or GP, then an and Sn can be found by the method of difference.
where T1, T2, T3 ,… are terms of new series and Sn = Σan
Type 2 It is not always necessary that the series of first order of differences i.e., a2 – a1, a3 – a2, …, an – an-1 is always either in AP or in GP in such case.
Now, the series (T2 — T1) + (T3 — T2) + …+ (Tn – Tn-1) is series of second order of differences and when it is either in AP or in GP, then an = a1 + ΣTr
Otherwise, in the similar way, we find series of higher order of differences and the nth term of the series.
Exponential Series
The sum of the series is denoted by thenumber e.
(i) e lies between 2 and 3.
(ii) e is an irrational number.
Exponential Theorem
Let a>0, then for all real value of x,
Logarithmic Series
Important Result and Useful Series
24. If number of terms in AP/GP/HP are odd, then AM/GM/HM of first and last term in middle term of progression.
25. If pth, qth and rth term of geometric progression are also in geometric progression.
26. If a,b and c are in AP and also in GP, then a=b=c
27. If a, b and c are in AP, then xa, xb and xc are in geometric progression.
Previous articleQuadratic Equations and Inequalities Notes Class 11th Maths
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# Free Playing with Numbers 02 Practice Test - 8th Grade
123 can be expressed as:
A.
10×1+10×2+10×3
B.
100×1+100×2+100×3
C.
100×1+10×2+1×3
D.
1×1+1×2+1×3
#### SOLUTION
Solution : C
Let the number be abc.
Then the general form of abc will be:
100×a+10×b+1×c
Thus, 123=(100×1)+(10×2)+(1×3)
1000×2+100×3+10×2+1×4 is expressed as:
A.
4232
B.
1111
C.
2024
D.
2324
#### SOLUTION
Solution : D
1000×2+100×3+10×2+1×4=2000+300+20+4=2324.
Aarush asked Soham to take any 2 or 3 digit number and reverse the digits, then subtract the larger number from the smaller number. Soham did the same and concluded that the result is exactly divisible by 9. What Soham concluded is:
A.
True
B.
False
#### SOLUTION
Solution : A
Let's consider a two digit number ab.Its general form will be 10×a+1×b.
If we reverse the digits, its general form will be 10×b+1×a. If we subtract both of the numbers, we get:
(10×a+1×b)(10×b+1×a)
= (10a + b) - (10b + a)
= 10a + b -10b - a
= 9a - 9b
= 9 (a-b)
Observe that the result is a multiple of 9 and hence is exactly divisible by 9.
Similarly, for 3 digit numbers, the difference of the number and number obtained by reversing the digits (if abc is the number) will be given by:
(100×a+10×b+c)(100×c+10×b+a)
=100a + 10b + c - (100c + 10b + a)
= 100a + 10b + c - 100c - 10b - a
= 99a - 99c
= 99 (a-c)
= 9×11(ac)
which is also divisible by 9.
Note that this happens with any 2 or 3 number.
When Rashi ordered pizza and coke, the bill had the digits A and B not visible clearly.
Pizza A12Coke + 31B 525
Using the information visible on her bill, find the digits A and B respectively.
A.
A = 2, B = 1
B.
A = 3, B = 5
C.
A = 1, B = 3
D.
A = 2, B = 3
#### SOLUTION
Solution : D
In the tens place, the numbers being added are 1 and 1. The sum given in the tens place is 2, which means there is no carry over throughout the summation.
A + 3 = 5 i.e A = 5 - 3 = 2 and
2 + B = 5 i.e B = 5 - 2 = 3
Pizza 212Coke + 313 525
Soham borrowed money from his mother for buying pens. He bought 7 pens and the price of each pen is a two digit number, 1A. If the total amount spent by him is 10A, then the value of A is
___
#### SOLUTION
Solution :
Price of one Pen =1×10+A
Number of Pens = 7
Total price =(10+A)×7
(10+A)×7=(100+A)
70 + 7A = 100 + A
6A = 30
A = 5
By which of the following numbers is 392 divisible?
A.
9
B. 3
C. 2
D. 4
#### SOLUTION
Solution : C and D
A number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9 respectively.
Here, sum of digits = 3 + 9 + 2 = 14 which is not divisible by either 3 or 9. Hence, the number is not divisible by 3 and 9.
Being an even number, the given number is divisible by 2.
A number is divisible by 4 if the number formed by the last two digits are divisible by 4.
In 392, 92 which is divisible by 4. Hence, the number is also divisible by 4.
198 is divisible by:
A.
11
B. 5
C. 8
D. 6
#### SOLUTION
Solution : A and D
A number is divisible by 11 if the difference between the sum of digits at its odd places and that of the digits at the even places is divisible by 11.
(1 + 8) - 9 = 0 which is divisible by 11.
198 is even number and thus, divisible by 2.
The sum of digits 1 + 9 + 8 = 18 which is divisible by 3.
So, 198 is also divisible by 3.
Since, the number is divisible by both 2 and 3, it is also divisible by 6.
A number to be divisible by 5 must end in 0 or 5. Hence, 198 is not divisible by 5.
A number is divisible by 8 if the number formed by the last 3 digits of the number are divisible by 8, hence 198 is not divisible by 8.
Soham took a 3 digit number and formed two other 3 digit numbers using the digits of the original number. He then added the three resulting numbers and divided their sum by 37 . He concluded that the result was exactly divisible by 37, no matter what number he chose. Is Soham's statement true or false?
A.
True
B.
False
#### SOLUTION
Solution : A
Let the number be abc.
abc = 100a + 10b + c.
By rearranging the digits, let us say the two other numbers formed are cab and bca.
Expressing these two numbers in the expanded form,
cab = 100c + 10a + b
bca = 100b + 10c + a
abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 100a + 10a + a + 10b + b + 100b + c + 100c + 10c
= 111a + 111b + 111c
= 111(a + b + c)
= 37 × 3 × (a + b + c), since 111 is the product of 37 and 3.
Hence the sum is divisible by 37.
Soham got a call from Manish who was not very clear with the concepts of divisibility. Manish then asked, "Is 244 divisible by 2?” . Soham replied yes. Is Soham's reply true or false?
A.
True
B.
False
#### SOLUTION
Solution : A
If the ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
For the number 244 in the ones place we find 4 which is divisible by 2. Therefore we can say that the number 244 is divisible by 2.
3 1 Q+ 1 Q 3 5 0 1
A.
1
B.
4
C.
5
D.
8
#### SOLUTION
Solution : D
Study the addition in the ones column:
Q + 3 = 1, a number whose ones digit is 1.
If Q = 8, then 8 + 3 = 11.
So, the puzzle can be solved as shown below:
3 1 8+ 1 8 3 5 0 1
Hence, Q = 8
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By M. Bourne
We can draw a circle that closely fits nearby points on a local section of a curve, as follows.
### Application of Radius of Curvature
When engineers design train tracks, they need to ensure the curvature of the track will be safe and provide a comfortable ride for the given speed of the trains. [Image source].
We say the curve and the circle osculate (which means "to kiss"), since the 2 curves have the same tangent and curvature at the point where they meet.
The radius of curvature of the curve at a particular point is defined as the radius of the approximating circle. This radius changes as we move along the curve.
How do we find this changing radius of curvature?
The formula for the radius of curvature at any point x for the curve y = f(x) is given by:
text(Radius of curvature)=([1+((dy)/(dx))^2]^(3//2))/(|(d^2y)/(dx^2)|)
### Need Graph Paper?
Find the radius of curvature for the cubic
y = 2x3 x + 3
at the point x = 1.
## Exploration
In the following interactive graph you can explore what "changing radius of curvature" means.
Slowly drag the point "P" around the curve to see the changing radius of curvature (segment CP). It works best if you use a left-right motion - don't worry about following the up-down of the graph.
You'll notice at the point of inflexion there is interesting behavior. The circle changes from being below the curve to above (when moving left to right). When we are right on the point of inflexion, what does the circle become?
Continues below
### Example 2
[This example was supplied by a reader.]
We have a curve which is defined by data points and we don't know the function for this data. How can we find the radius of curvature?
We take any 3 data points to illustrate ways of solving this. I chose the points (1, 1), (2, 3) and (3,8).
We'll do this in 3 different ways, just for fun (and for learning about how different math approaches can be used)!
Method 1: Approximation Using a Parabolic Fit and Calculus Methods
Method 2: Using Linear Approximations and Calculus Methods
Method 3: Finding the Radius of the Circle through our 3 Points
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# Приближенное вычисление определенных интегралов. Методы прямоугольников и метод Симпсона
### Содержание работы
кафедра 304
Домашнее задание
по предмету «Численные методы»
## по теме: «Приближенное вычисление определенных интегралов. Методы прямоугольников и метод Симпсона»
Выполнила студентка 325 гр.
Старцева А. В.
Проверила ст. преподаватель каф. 304
Яровая О. В.
______________________
NUMERICAL INTEGRATION
Theoretical information
Integral of the form in many cases it is impossible to find. For example:
· function can not be integrated
· function is given by the table
· function is given by a graph
Definite integral numerically equal to the area of the figure bounded by the x-axis, direct x = a and x = b and the graph of the function f (x).
f(a)=y0 f(b)=y1 h=
n- predetermined number of divisions
Drawing 1.1 - Geometric interpretation of the definition Definite integral
Numerical integration is a primary tool used by engineers and scientists to obtain approximate answers for definite integrals that cannot be solved analytically.
We approach the subject of numerical integration. The goal is to approximate the definite integral of f(x) over the interval [a, b] by evaluating f(x) at a finite number of sample points.
Newton-Cotes methods are based on polynomial approximation of the integrand.
In the process of numerical integration is necessary to calculate the approximate value of the integral and estimate the error, regardless of the method chosen.
Error will decrease with an increasing number of partitions of n of the integration interval [a, b], due to more accurate approximation of the integrand, but it will increase accuracy by summing partial integrals, and the last error with some value n0 becomes dominant.
Numerical integration formulas dimensional case called quadrature.
The idea of building the quadrature formula is to replace the integrand polynomial chosen degree. We obtain the form of a quadrature formula, depending on the degree of the polynomial.
1. Replacement of a polynomial of degree zero
1) the method left rectangles
Drawing 1.2 - Geometric interpretation of the method left rectangles
2) the method right rectangles
Drawing 1.3 - Geometric interpretation of the method right rectangles
3) the method of average rectangles
Drawing 1.4 - Geometric interpretation of the method of average rectangles
2. Replacement of first-degree polynomial
the trapezoidal rule
Drawing 1.4 - Geometric interpretation of the trapezoidal rule
3. Replacement of a second degree polynomial
the Simpson's method
Drawing 1.5 - Geometric interpretation of the Simpson's method
Unit interval is 2h; n=2m
The formula works for an even number of intervals and odd number of points.
Formula Newton-Cotes
≈
polynomial for constant grid
, ,
n- degree of the polynomial; i- number approximating polynomial
General view of the quadrature formula
=
explicit form for the weighted quadrature coefficients
Lagrange polynomial for the uniform grid
,
, – coefficients Cotes i=(0,..n)
, , i=(0,..,n)
Priori error
the methods left and right rectangles
the method of average rectangles
the trapezoidal rule
the Simpson's method
ξ є [a; b]
power h determines the order of the method
Posteriori error
General view
p-order method
A - coefficient depending on the value of the derivative and integration method
The first formula Runge
w - the exact value, which should come numerical method
- results of numerical calculations
w = ++O()
calculate increments kh, k – constant, which k>1 or k<1. A- remains unchanged, because we do not change the method. Necessary to carry out a priori error, that order of the method to determine.
w = ++O((k), equate:
+=+
h, 2h
p=1 the methods left and right rectangles
p=2 the method of average rectangles, the trapezoidal rule
p=4 the Simpson's method
Input
Tabulation of functions on a given interval [a,b]
Calculation of the definite integral using the built-in
Program block that implements the calculation of the approximate value of the definite integral.
the Simpson's method
the trapezoidal rule
the method of average rectangles
the method right rectangles
the method left rectangles
Calculating a posteriori error computation
Calculating a priori calculation errors
Calculate in Matlab
function res = Integral(func, str, a , b, n);
%str - methods name, a,b - distance, n - number of points
h = (b-a)/n;
Sum = 0;
switch (str)
case 'Simpson'
q1=0; q2=0;
for i = 1:n-1
if (mod(i,2)==1) q1 = q1+myfunc(func,a + i*h);
else q2 = q2+myfunc(func,a + i*h);
end;
end;
Sum = (h/3)*( myfunc(func,a) + 4*q1 + 2*q2 + myfunc(func,b) );
case 'Trapezium'
for i = 1:n-1
Sum = Sum + myfunc(func,a + i*h);
end;
Sum = (h/2)*( myfunc(func,a) + 2*Sum + myfunc(func,b) );
case 'LeftRectangles'
for i=1:n-1
Sum = Sum + h*myfunc(func,a + i*h);
end;
Sum = h*myfunc(func,a) + Sum;
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Given that x5 + ax3 + bx2 − 3 = (x2 − 1) Q(x) x − 2 where Q(x) is a polynomial. State the degree of Q(x) and find the value of a and b. Find also the remainder when Q(x) is divided by x + 2.
Solution
Since $\displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3$ is divided by (x2 − 1), Q(x) is a polynomial of degree 3.
$\displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$
$\displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=(x-1)(x+1)Q(x)-x-2$
When x = − 1, $\displaystyle -1-a+b-3=(-1-1)(-1+1)Q(x)-(-1)-2$
$\displaystyle \therefore -a+b=3$ --------------(1)
When x = 1, $\displaystyle 1+a+b-3=(1-1)(1+1)Q(x)-1-2$
$\displaystyle \therefore a+b=-1$ --------------(2)
$\displaystyle (2)+(1)\Rightarrow 2b=2\Rightarrow b=1$
$\displaystyle (2)-(1)\Rightarrow 2a=-4\Rightarrow a=-2$
$\displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$
$\displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1=({{x}^{2}}-1)Q(x)$
$\displaystyle \therefore Q(x)=\frac{{{{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1}}{{{{x}^{2}}-1}}$
When Q(x) is divided by x + 2,
the remainder $\displaystyle =Q(-2)$
$\displaystyle =\frac{{{{{(-2)}}^{5}}-2{{{(-2)}}^{3}}+{{{(-2)}}^{2}}+(-2)-1}}{{{{{(-2)}}^{2}}-1}}$
$\displaystyle =\frac{{-32+16+4-2-1}}{3}$
$\displaystyle =-\frac{{15}}{3}$
$\displaystyle =-5$
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# Polar form – General Form, Conversion Rules, and Examples
Polar forms are one of the many ways we can visualize a complex number. They are a way for us to visualize complex numbers on a complex plane as vectors.
A complex number in the polar form will contain a magnitude and an angle to guide us with the complex number’s orientation.
The polar forms of complex numbers help us visualize and treat the complex numbers as quantities that have distance and direction – through the polar coordinate system.
When discussing polar forms, we’re expecting to make use of past and new concepts. In this article, we’ll tackle the following topics:
• A refresher on polar coordinates and how we can convert a rectangular coordinate to a polar coordinate.
• A guide on how we can represent complex numbers in polar form and coordinate system.
• Learning how to plot complex numbers in the context of their polar form’s elements.
First, let’s recall what we know about polar and rectangular forms. Make sure to keep a handy notepad or note-taking app since this will be a thorough discussion of polar forms and complex numbers in polar form.
## What is a polar form?
The polar form of numbers, in general, makes use of the number’s modulus (or distance from the origin) and the angle it forms concerning the $x$-axis.
Some examples of coordinates in polar form are: $(2, 45^{\circ})$, $\left(3, \dfrac{\pi}{4}\right)$, and $(5, 1.2)$.
• The first coordinates, $\{2, 3, 5\}$, represent the distance of the coordinate from the origin.
• The second coordinates, $\left\{ 45^{\circ}, \dfrac{\pi}{4}, 1.2 \text{ radians}\right \}$, represent the angle formed along the $x$-axis.
Here’s a helpful diagram to better understand how a rectangular coordinate may differ when represented by its polar form. Let’s go ahead and inspect its elements.
• $r$ represents the distance of the point from the origin
• $\theta$ represents the angle formed by $r$ with the $x$-axis
• $(x, y)$ is the rectangular form of a point while $(r, \theta)$ represents its polar form
We can even relate $x$, $y$, $r$, and $\theta$, so we can better understand how rectangular and polar forms differ from each other.
\begin{aligned}x&= r\cos \theta \\ y &= r \sin \theta\\x^2 + y^2 &= r^2 \\ \tan \theta &= \dfrac{y}{x}\end{aligned}
Here’s an example on how we can convert a rectangular coordinate, $(1, \sqrt{3})$, to its polar form. We have $x = 1$ and $y = \sqrt{3}$. Using these values, we can now find the value of $r$ and $\theta$ as shown below.
$\boldsymbol{x^2 + y^2 = r^2}$ $\boldsymbol{\tan \theta = \dfrac{y}{x}}$ \begin{aligned}r &= \sqrt{x^2 + y^2}\\ &= \sqrt{(1)^2 + (\sqrt{3})^2}\\&= \sqrt{4}\\&= 2\end{aligned} \begin{aligned}\tan \theta &= \dfrac{\sqrt{3}}{1}\\\tan \theta &= \sqrt{3}\\\theta &= \tan^{-1} \sqrt{3}\\&=60^{\circ}, 240^{\circ}\end{aligned}Since the point lies on the first quadrant, $\theta = 60^{\circ}$.
This means that the rectangular coordinate, $(1, \sqrt{3})$, is equal to $(2, 60^{\circ})$ in polar form.
The graph above will help us understand how the rectangular and polar coordinates can coincide in one graph. These introductory concepts will also help us define complex numbers in terms of $r$ and $\theta$ as well.
### How to write a complex number in polar form?
When learning about complex numbers, we have learned how $a + bi$ can be graphed on a complex plane where $a$ replaces the $x$-coordinate $b$ replaces the $y$-coordinate.
Since the complex plane has similar dynamics with the rectangular coordinate. With the real axis representing the $x$-axis and the imaginary axis representing the $y$-axis.
Let’s rewrite the polar form we have from the earlier section and see how this applies for complex numbers.
\begin{aligned}x&= r\cos \theta \\ y &= r \sin \theta\\x^2 + y^2 &= r^2 \\ \tan \theta &= \dfrac{y}{x}\end{aligned} \begin{aligned}a&= r\cos \theta \\ a &= r \sin \theta\\a^2 + b^2 &= r^2 \\ \tan \theta &= \dfrac{b}{a}\end{aligned}
This also means that we can write $a + bi$ as $r \cos \theta + r\sin \theta i= r(\cos \theta + \sin \theta i)$. The expression $r(\cos \theta + \sin \theta i)$ is the polar form of the complex number.
We can also see that $r$ is equal to the complex number’s modulus or absolute value from the table.
This means that we can write complex numbers in terms of $r$ and $\theta$. In fact, a well-known abbreviation of a complex number’s polar is $r\text{cis} \theta$.
## How to convert to polar form?
We can convert numbers in polar to rectangular form. Similarly, we can convert numbers in rectangular to polar form.
The graphs above show how the two forms (rectangular and polar) reflect the same complex number and a point in the same position.
It helps to always go back to this group of equations whenever in doubt with a step or two.
\begin{aligned}x&= r\cos \theta \\ y &= r \sin \theta\\x^2 + y^2 &= r^2 \\ \tan \theta &= \dfrac{y}{x}\end{aligned}
We’ve summarized helpful steps to remember in converting complex numbers in the rectangular form to polar form. Don’t worry. We’ve also prepared steps for doing the reverse.
### How to convert rectangular form to polar form?
We can find the complex number’s polar form by following the steps below.
1. Find the complex number’s absolute value, $|z| = r = \sqrt{a^2 + b^2}$.
2. Find the value of $\theta$ using $\tan \theta = \dfrac{b}{a}$.
3. Substitute the values of $r$ and $\theta$ into the expression, $r(\cos \theta + \sin \theta i)$.
Whenever stuck with writing complex numbers in polar form, always go back to these three steps.
Why don’t we apply this to write $1 + \sqrt{3}i$ in polar form?
Let’s find the value of $r$ by finding the absolute value of $2 + i$.
\begin{aligned}r &= \sqrt{(1)^2 + (\sqrt{3})^2}\\&= \sqrt{1 + 3}\\&= 2\end{aligned}
Now, we use $a = 1$ and $b = \sqrt{3}$ to find the value of $\theta$. Since both $a$ and $b$ are positive, $\theta$ lies on the first quadrant.
\begin{aligned}\tan \theta &= \dfrac{1}{\sqrt{3}}\\\theta &= \tan^{-1} \dfrac{1}{\sqrt{3}}\\&=30\circ\end{aligned}
Hence, $r = 2$ and $\theta = 30^{\circ}$.
Using these values, we can now write $1 + \sqrt{3}i$ as in polar form, $r(\cos \theta + \sin \theta i)$.
\begin{aligned}r(\cos \theta + \sin \theta i)&= 2(\cos 30^{\circ} + \sin 30^{\circ}i)\\&=2\left(\dfrac{\sqrt{3}}{2}+ \dfrac{1}{2}i \right )\\&=2 \cdot \dfrac{\sqrt{3}}{2} + 2 \cdot \dfrac{1}{2}i\\&= \sqrt{3} + i\end{aligned}
A common abbreviation used for complex numbers’ polar forms is $r \text{cis} \theta$, so $1 + \sqrt{3}i$ will also be equal to $2 \text{cis} 30^{\circ}$.
### How to convert polar form to rectangular form?
Converting polar forms to rectangular forms is more straightforward. We can evaluate the values of $r\cos \theta$ and $r \sin \theta$ to find the rectangular form of the coordinate.
Given $(r\cos \theta, r\sin \theta)$, perform the following steps to convert polar coordinate to rectangular form:
1. Find the values of $r\cos \theta$ and $r\sin \theta$.
2. Write the results in order as $(x, y)$.
If given a complex number in polar form, we can simply write $(x, y)$ as $(a, b)$ instead and return $a + bi$.
For example, if we want to convert the complex number, $4\left(\cos \dfrac{\pi}{4} + \sin \dfrac{\pi}{4}\right)$, we simply evaluate $\cos \dfrac{\pi}{4}$ and $\sin \dfrac{\pi}{4}$.
\begin{aligned}\cos \dfrac{\pi}{4} &= \dfrac{\sqrt{2}}{2}\\\sin \dfrac{\pi}{4} &= \dfrac{\sqrt{2}}{2}\\4\left(\cos \dfrac{\pi}{4} + \sin \dfrac{\pi}{4}i\right)&= 4 \left(\dfrac{\sqrt{2}}{2} +\dfrac{\sqrt{2}}{2}i\right)\\&=4 \cdot \dfrac{\sqrt{2}}{2} + 4 \cdot \dfrac{\sqrt{2}}{2}i\\&=2\sqrt{2} + 2\sqrt{2}i\end{aligned}
This means that $2 \text{cis} \dfrac{\pi}{4}$ is equal to $2\sqrt{2} + 2\sqrt{2}i$.
We can apply a similar process when converting any polar form to a rectangular form.
## How to multiply and divide complex numbers in polar form?
We’ve learned how to multiply and divide complex numbers in the past, but what if we’re given two complex numbers in polar forms?
Let’s say we have $z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)$.
To find their product, we can multiply the two moduli, $r_1$ and $r_2$, and find the cosine and sine of the sum of $\theta_1$ and $\theta_2$.
$z_1 z_2 = r_1 r_2 [\cos( \theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]$
We apply a similar process to find the quotient of two complex numbers in polar form. Divide $r_1$ by $r_2$ and find the cosine and sine of the difference between $r_1$ and $r_2$.
$\dfrac{z_1}{z_2}= \dfrac{r_1}{r_2} [\cos(\theta_1 – \theta_2) + i\sin (\theta_1 – \theta_2)]$
This means that we can immediately multiply and divide them when given two complex numbers in polar form without converting them to rectangular form first.
We’ve learned everything we need so far for the polar and rectangular forms of complex numbers and numbers in general. It’s time that we test our knowledge by trying out these sample problems below.
### Example 1
Review our knowledge of polar coordinates by converting the following rectangular coordinates into polar form.
a. $(-\sqrt{3}, 1)$
b. $(-2, -2\sqrt{3})$
c. $(8, -8)$
Solution
Given a rectangular coordinate, $(x, y)$, we find, $r$, the distance between $(x, y)$ from the origin using the property, $r = \sqrt{x^2 + y^2}$.
The value of $\theta$ can be determined by taking the tangent inverse of $\dfrac{y}{x}$.
Applying this to $(-\sqrt{2}, 1)$, we have the following values of $r$ and $\theta$.
$\boldsymbol{r}$ $\boldsymbol{\theta}$ \begin{aligned}r &= \sqrt{(-\sqrt{3})^2 + (1)^2}\\&=\sqrt{3 + 1}\\&=\sqrt{4}\\&= 2 \end{aligned} \begin{aligned}\tan \theta &= \dfrac{1}{-\sqrt{3}}\\\theta &= \tan^{-1} \dfrac{1}{-\sqrt{3}}\\&=150^{\circ}, 330^{\circ}\end{aligned}
Since $x$ is negative and $y$ is positive, $\theta$ lies on the second quadrant, so $\theta = 150^{\circ}$. Use $r$ and $\theta$ to write the polar form of the coordinate.
This means that $(-\sqrt{2}, 1)$ is equal to $(2, 150^{\circ})$.
We’ll apply the same process for the two other coordinates to find their polar forms.
$\boldsymbol{(x, y)}$ $\boldsymbol{r}$ $\boldsymbol{\theta}$ $\boldsymbol{(r, \theta))}$ $(-2, -2\sqrt{3})$ \begin{aligned}r &= \sqrt{(-2)^2 + (-2 \sqrt{3})^2}\\&=\sqrt{4 + 12}\\&=\sqrt{16}\\&= 4 \end{aligned} \begin{aligned}\tan \theta &= \dfrac{-2\sqrt{3}}{-2}\\\theta &= \tan^{-1} \sqrt{3}\\&=240^{\circ}\end{aligned} $(4, 240^{\circ})$. $(8, -8)$ \begin{aligned}r &= \sqrt{(8)^2 + (-8)^2}\\&=\sqrt{64 + 64}\\&=\sqrt{128}\\&= 8\sqrt{2} \end{aligned} \begin{aligned}\tan \theta &= \dfrac{-8}{8}\\\theta &= \tan^{-1} -1\\&=315^{\circ}\end{aligned} $(8\sqrt{2}, 315^{\circ})$
Hence, we have the following:
a. $(-\sqrt{3}, 1) =(2, 150^{\circ})$
b. $(-2, -2\sqrt{3}) = (4, 240^{\circ})$
c. $(8, -8) = (8\sqrt{2}, 315^{\circ})$
### Example 2
Graph the following complex numbers on one complex plane, then convert them to polar form.
a. $-3+ 3i$
b. $4 + 4\sqrt{3}i$
c.$2\sqrt{3} – 2i$
Solution
When graphing complex numbers of the form, $a + bi$, the point must be $a$ units from the right or left of the vertical axis and $b$ units above or below the horizontal axis. It can also help keep in mind that it’s actually the same as graphing the point $(a, b)$.
• The number $-3 + 3i$ is $3$ units from the left of the vertical axis and $3$ units above the horizontal axis.
• Similarly, $4 + 4\sqrt{3}i$ is $4$ units from the right of the vertical axis and $4\sqrt{3}$ units above the horizontal axis.
• Lastly, we can graph $2\sqrt{3} – 2i$ by plotting a point that is $2\sqrt{3}$ units from the right and $2$ units below the horizontal axis.
Let’s go ahead and graph these values on the complex plane.
This graph will also guide us in checking whether we have the right value for $r$ and $\theta$ once we convert the three numbers to polar forms.
To convert a complex number, $a + bi$, we have to find its absolute value, $r = \sqrt{a^2 + b^2}$ and angle, $\theta = \tan^{-1} \dfrac{y}{x}$.
Once we have $r$ and $\theta$, we can now write the complex number in its polar form, $r(\cos \theta + i\sin \theta)$ or $r\text{cis} \theta$.
We can first try converting $-3 + 3i$ to polar form by first finding $r$ and $\theta$.
$\boldsymbol{r}$ $\boldsymbol{\theta}$ \begin{aligned}r&= \sqrt{(-3)^2 + (3)^2}\\&=\sqrt{9 + 9}\\&= \sqrt{18}\\&=3\sqrt{2}\end{aligned} \begin{aligned}\tan \theta &= \dfrac{3}{-3}\\\theta&=\tan^{-1} (-1)\\&=135^{\circ}, 315^{\circ} \end{aligned}
Now that we have $r= 3\sqrt{2}$ and $\theta = 135^{\circ}$, let’s convert $-3 + 3i$ to polar form.
\begin{aligned}-3 + 3i &= 3\sqrt{2}(\cos 135^{\circ} + i\sin 135^{\circ})\\&=3\sqrt{2} \text{cis}135^{\circ}\end{aligned}
Let’s apply the same process for the two remaining complex numbers.
Complex Numbers $\boldsymbol{r}$ $\boldsymbol{\theta}$ $\boldsymbol{r(\cos \theta + i\sin \theta)}$ $4 + 4\sqrt{3}i$ \begin{aligned}r&= \sqrt{(4)^2 + (4\sqrt{3})^2}\\&=\sqrt{16 + 48}\\&= \sqrt{64}\\&=8\end{aligned} \begin{aligned}\tan \theta &= \dfrac{4\sqrt{3}}{4}\\\theta&=\tan^{-1} (\sqrt{3})\\&=60^{\circ}\end{aligned} $8(\cos 60^{\circ} + i\sin 60^{\circ})$ $2\sqrt{3} – 2i$ \begin{aligned}r&= \sqrt{(2\sqrt{3})^2 + (2)^2}\\&=\sqrt{12 + 4}\\&= \sqrt{16}\\&=4\end{aligned} \begin{aligned}\tan \theta &= \dfrac{-2}{2\sqrt{3}}\\\theta&=\tan^{-1} \left(-\dfrac{1}{\sqrt{3}}\right)\\&=330^{\circ}\end{aligned} $4(\cos 330^{\circ} + i\sin 330^{\circ})$
Let’s summarize the results for the three complex numbers:
a. $-3+ 3i = 3\sqrt{2}(\cos 135^{\circ} + i\sin 135^{\circ})$
b. $4 + 4\sqrt{3}I = 8(\cos 60^{\circ} + i\sin 60^{\circ})$
c.$2\sqrt{3} – 2i = 4(\cos 330^{\circ} + i\sin 330^{\circ})$
### Example 3
The complex numbers shown below are in their polar forms. Rewrite them so that they are of the form $a + bi$.
a. $4(\cos 30^{\circ} + i\sin 30^{\circ})$
b. $-2\left(\cos \dfrac{3\pi}{4}+ i\sin \dfrac{3\pi}{4}\right)$
c. $-5 \text{cis } 45^{\circ}$
d. $\dfrac{1}{2} \text{cis } \dfrac{3\pi}{2}$
Solution
Rewriting complex numbers from polar forms to their rectangular forms is straightforward.
• Evaluate the trigonometric values based on $\theta$ inside the parenthesis.
• Distribute the absolute value, $r$, to each of the resulting values.
Why don’t we apply this with the first expression? We begin by evaluating $\cos 30^{\circ}$ and $\sin 30^{\circ}$.
\begin{aligned}4(\cos 30^{\circ} + i\sin 30^{\circ}) &= 4\left( \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right)\end{aligned}
Distribute $4$ to each of the trigonometric values to simplify the complex number.
\begin{aligned}4\left( \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right) &= 4 \cdot \dfrac{\sqrt{3}}{2} + 4\cdot \dfrac{1}{2}i\\&=2\sqrt{3} + 2i\end{aligned}
Hence, $4(\cos 30^{\circ} + i\sin 30^{\circ}$ is equivalent to $2\sqrt{3} + 2i$ in rectangular form.
We apply the same process with b, but we’re working with angles in radians this time. Make sure to review your special angles and trigonometric table of values when working with polar forms of complex numbers.
We have the following steps to obtain the polar form of $-2\left(\cos \dfrac{3\pi}{4}+ i\sin \dfrac{3\pi}{4}\right)$.
\begin{aligned}-2\left(\cos \dfrac{3\pi}{4}+ i\sin \dfrac{3\pi}{4}\right) &= -2\left(-\dfrac{\sqrt{2}}{2}+ \dfrac{\sqrt{2}}{2}i\right)\\&= -2 \cdot -\dfrac{\sqrt{2}}{2} + -2\cdot \dfrac{\sqrt{2}}{2}i\\&=\sqrt{2}-\sqrt{2}i\end{aligned}
For c, recall that $r\text{cis } \theta$ is just a shorter version of $r(\cos \theta + i\sin \theta)$.
Let’s rewrite $-5 \text{cis } 45^{\circ}$ as $-5 ( \cos 45^{\circ} + i\sin 45^{\circ})$. We then apply similar process to rewrite it in the form $a + bi$: evaluate the cosine and sine of $45^{\circ}$ then distribute $-5$.
\begin{aligned}-5 \text{cis }45^{\circ} &= -5(\cos 45^{\circ} + i\sin 45^{\circ})\\&= -5\left(\dfrac{\sqrt{2}}{2} + i\dfrac{\sqrt{2}}{2} \right )\\&= -5 \cdot \dfrac{\sqrt{2}}{2} + -5 \cdot \dfrac{\sqrt{2}}{2}i\\&=-\dfrac{5\sqrt{2}}{2} -\dfrac{5\sqrt{2}}{2}i\end{aligned}
Let’s now work on $\dfrac{1}{2} \text{cis } \dfrac{3\pi}{2}$ by first rewriting this in its expanded form. Evaluate the resulting expression to find the rectangular form of the complex number.
\begin{aligned}\dfrac{1}{2} \text{cis } \dfrac{3\pi}{2} &= \dfrac{1}{2} \left(\cos \dfrac{3\pi}{2} + i\sin \dfrac{3\pi}{2} \right )\\&= \dfrac{1}{2} [0 + i(-1)]\\&= \dfrac{1}{2}(-i)\\&= -\dfrac{1}{2}i\end{aligned}
This means that we have the following expressions when the four are converted to rectangular form.
a. $4(\cos 30^{\circ} + i\sin 30^{\circ}) = 2\sqrt{3} + 2i$
b. $-2\left(\cos \dfrac{3\pi}{4}+ i\sin \dfrac{3\pi}{4}\right) = \sqrt{2}-\sqrt{2}i$
c. $-5 \text{cis } 45^{\circ} =-\dfrac{5\sqrt{2}}{2} -\dfrac{5\sqrt{2}}{2}i$
d. $\dfrac{1}{2} \text{cis } \dfrac{3\pi}{2} = -\dfrac{1}{2}i$
### Example 4
Find the product of the following pairs of complex numbers. Express your final answer in polar form.
a. $z_1 = 3(\cos 50^{\circ} + i\sin 50^{\circ})$, $z_2 = -6(\cos 20^{\circ} + i\sin 20^{\circ})$
b. $z_1 = -4\left(\cos \dfrac{\pi}{12} + i\sin \dfrac{\pi}{12}\right)$, $z_2 =2\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right)$
c. $z_1 = 5(\cos 12^{\circ} + i\sin 12^{\circ})$, $z_2 = -2\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right)$,and $z_3 = -3(\cos 15^{\circ} + i\sin 15^{\circ})$
Solution
Recall that given two complex numbers, we can find their product by multiplying the absolute values and adding the two angles.
If we have $z_1 = r_1(\cos \theta_1 + i\sin \theta_1)$ and $z_2 = r_2(\cos \theta_2 + i\sin \theta_2)$, their product can be expressed as $z_1z_2 = r_1r_2[\cos(\theta_1 + \theta_2)+ \sin(\theta_1 + \theta_2)]. This means that the product of$z_1 = 3(\cos 50^{\circ} + i\sin 50^{\circ})$and$z_2 = -6(\cos 20^{\circ} + i\sin 20^{\circ})$can be determined using the formula shown above.$\begin{aligned}z_1z_2&= [3(\cos 50^{\circ} + i\sin 50^{\circ})][-6(\cos 20^{\circ} + i\sin 20^{\circ})]\\&= (3 \cdot -6)[\cos (50^{\circ} + 20^{\circ}) + i\sin(50^{\circ} + 20^{\circ})]\\&=-18(\cos 70^{\circ} + i\sin 70^{\circ})\end{aligned}$We apply the same process to find the product of the two complex numbers in b. The only difference is that we’re working with radians.$\begin{aligned}z_1z_2&= \left[-4\left(\cos \dfrac{\pi}{12} + i\sin \dfrac{\pi}{12}\right)\right]\left[2\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right)\right]\\&=(-4 \cdot 2) \left[\cos\left(\dfrac{\pi}{12} + \dfrac{\pi}{6}\right) + i\sin\left(\dfrac{\pi}{12} + \dfrac{\pi}{6}\right) \right ]\\&=-8 \left[\cos\left(\dfrac{\pi + 2\pi}{12}\right ) + i\sin\left(\dfrac{\pi + 2\pi}{12}\right ) \right]\\&=-8\left(\cos \dfrac{\pi}{4} + i\sin \dfrac{\pi}{4} \right )\end{aligned}$For the third item, we must make sure all angles are either in degrees or in radians. Since we already have two angles in degrees, it’s faster if we change$\dfrac{\pi}{6}$to degrees. Hence, we have$z_2 = -2(\cos 30^{\circ} + i\sin 30^{\circ})$. We will still perform the same process, but we’re working with three pairs of absolute value numbers and angles this time.$\begin{aligned}z_1z_2z_3 &= [5(\cos 12^{\circ} + i\sin 12^{\circ})]\left[-2\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right)\right][-3(\cos 15^{\circ} + i\sin 15^{\circ})]\\&=[5(\cos 12^{\circ} + i\sin 12^{\circ})]\left[-2(\cos 30^{\circ} + i\sin 30^{\circ})\right][-3(\cos 15^{\circ} + i\sin 15^{\circ})]\\&= (5 \cdot -2 \cdot -3)[\cos (12^{\circ} + 30^{\circ} + 15^{\circ}) + i\sin(12^{\circ} + 30^{\circ} + 15^{\circ})]\\&= 30(\cos 57^{\circ} + i\sin 57^{\circ})\end{aligned}$Hence, we have the following products for a, b, and c. a.$z_1z_2 = -18(\cos 70^{\circ} + i\sin 70^{\circ})$b.$z_1z_2 = -8\left(\cos \dfrac{\pi}{4} + i\sin \dfrac{\pi}{4} \right )$c.$z_1z_2z_3 = 30(\cos 57^{\circ} + i\sin 57^{\circ})$### Example 5 Find$\dfrac{z_1}{z_2}$for each of the given pairs of complex numbers. a.$z_1 = 8(\cos 60^{\circ} + i\sin 60^{\circ})$,$z_2 = -4(\cos 20^{\circ} + i\sin 20^{\circ})$b.$z_1 = 12\left(\cos \dfrac{7\pi}{12} + i\sin \dfrac{7\pi}{12}\right)$,$z_2 = 6\left(\cos \dfrac{\pi}{4} + i\sin \dfrac{\pi}{4}\right)$c.$ z_1 = -16(\cos 120^{\circ} + i\sin 120^{\circ})$,$z_2 = -4\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right)$Solution Recall that given two complex numbers, we can find their quotient by dividing the complex number’s absolute values and finding the difference between the two angles. If we have$z_1 = r_1(\cos \theta_1 + i\sin \theta_1)$and$z_2 = r_2(\cos \theta_2 + i\sin \theta_2)$, their quotient can be expressed as$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}[\cos(\theta_1 – \theta_2)+ \sin(\theta_1 – \theta_2)]$. Using the formula shown above, we can find the quotient of$z_1 = 8(\cos 60^{\circ} + i\sin 60^{\circ})$and$z_2 = -4(\cos 20^{\circ} + i\sin 20^{\circ})$.$\begin{aligned}\dfrac{z_1}{z_2}&= \dfrac{8(\cos 60^{\circ} + i\sin 60^{\circ})}{ -4(\cos 20^{\circ} + i\sin 20^{\circ})}\\&=-\dfrac{8}{4}[\cos(60^{\circ} – 20^{\circ}) + i\sin(60^{\circ} – 20^{\circ})]\\&=-2 (\cos 40^{\circ} + i\sin 40^{\circ})\end{aligned}$This time, let’s work with radians and subtract the two angles. We’ll apply a similar process of subtracting regular fractions when subtracting the two angles in radians.$\begin{aligned}\dfrac{z_1}{z_2}&= \dfrac{12\left(\cos \dfrac{7\pi}{12} + i\sin \dfrac{7\pi}{12}\right)}{ 6\left(\cos \dfrac{\pi}{4} + i\sin \dfrac{\pi}{4}\right)}\\&= \dfrac{12}{6} \left[\cos \left( \dfrac{7\pi}{12} – \dfrac{\pi}{4} \right )+ i\sin \left( \dfrac{7\pi}{12} – \dfrac{\pi}{4} \right ) \right ]\\&=2 \left[\cos\left( \dfrac{7\pi – 3\pi}{12} \right ) + i\sin\left( \dfrac{7\pi – 3\pi}{12} \right ) \right ]\\&= 2\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3} \right)\end{aligned}$For the third item, we can either write all the angles in degrees or radians. Since it’s easier to subtract angles in degrees, change$\dfrac{\pi}{6}$to$30^{\circ}$.$\begin{aligned}\dfrac{z_1}{z_2}&= \dfrac{-16(\cos 120^{\circ} + i\sin 120^{\circ})}{ = -4\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right)}\\&= \dfrac{-16(\cos 120^{\circ} + i\sin 120^{\circ})}{-4(\cos 30^{\circ} + i\sin 30^{\circ})}\\&= \dfrac{-16}{-4} [\cos(120^{\circ} – 30^{\circ}) + i\sin(120^{\circ} – 30^{\circ})]\\&= 4 (\cos 90^{\circ} + i\sin 90^{\circ})\end{aligned}$Let’s go ahead and list down the resulting quotients of the three items. a.$\dfrac{z_1}{z_2} = -2 (\cos 40^{\circ} + i\sin 40^{\circ}) $b.$\dfrac{z_1}{z_2} = 2\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3} \right) $c.$\dfrac{z_1}{z_2} = 4 (\cos 90^{\circ} + i\sin 90^{\circ})$### Example 6 Given that$z_ 1 = 4\text{cis } 60^{\circ}$,$z_2 = -6(\cos 30^{\circ} + i\sin 30^{\circ})$, and$z_3 = -12 \left(\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}\right)$, find the value of the following expressions: a.$z_1 \cdot z_2$(Write your answer in the rectangular form) b.$\dfrac{z_1}{z_3} $c.$ \dfrac{z_1 \cdot z_ 3}{z_2}$Solution The first item requires us to multiply$z_1$and$z_2$. Let’s first rewrite$z_1$as$4(\cos 60^{\circ} + i\sin 60^{\circ})$. Find their product by adding$4$and$-6$the finding the cosine and sine of$(60^{\circ} + 30^{\circ})$.$\begin{aligned}z_1 \cdot z_2 &= [4(\cos 60^{\circ} + i\sin 60^{\circ} )] \cdot [-6(\cos 30^{\circ} + i\sin 30^{\circ})]\\&= (4 \cdot -6)[\cos ( 60^{\circ} + 30^{\circ}) + i\sin ( 60^{\circ} + 30^{\circ})]\\&=-24 (\cos 90^{\circ} + i\sin 90^{\circ})\end{aligned}$The product is still in polar form, so if we want a rectangular form as our final answer, let’s evaluate$\cos 90^{\circ}$and$\sin 90^{\circ}$then distribute$-24$to each of the values.$\begin{aligned}z_1 \cdot z_2 &= -24 (\cos 90^{\circ} + i\sin 90^{\circ})\\&= -24(0 + i)\\&= -24i\end{aligned}$Next, we want to find the quotient of$z_1$and$z_3$. To do so, we can divide$4$by$-12$, then subtract the two angles, then find the cosine and sine of the difference. But before we do, let’s convert$\dfrac{\pi}{4}$from$z_3$to degrees since it’s easier to find the difference of whole numbers.$\begin{aligned}\dfrac{z_1}{z_3}&= \dfrac{4(\cos 60^{\circ} + i\sin 60^{\circ} )}{-12(\cos 45^{\circ} + i\sin 45^{\circ})}\\&= \dfrac{4}{-12} [\cos(60^{\circ} – 45^{\circ}) + i\sin (60^{\circ} – 45^{\circ})]\\&= -\dfrac{1}{3}(\cos 15^{\circ} + i\sin 15^{\circ})\end{aligned}$To find the value of$\dfrac{z_1 \cdot z_ 3}{z_2}$, let’s begin by simplifying the numerator. Again, since it’s easier to add angles in degrees, we’ll use$z_3 =-12(\cos 45^{\circ} + i\sin 45^{\circ})$.$\begin{aligned}\dfrac{z_1 \cdot z_3}{z_2}&= \dfrac{(4\text{cis } 60^{\circ}) \cdot [-12(\cos 45^{\circ} + i\sin 45^{\circ})]}{-6(\cos 30^{\circ} + i\sin 30^{\circ})}\\\\z_1 \cdot z_3 &= [4 \cos 60^{\circ} + i\sin60^{\circ}][-12(\cos 45^{\circ} + i\sin 45^{\circ})]\\&= (4 \cdot -12)[\cos(60^{\circ} + 45^{\circ}) + i\sin(60^{\circ} + 45^{\circ})]\\&= -48 (\cos 105^{\circ} + i \sin 105^{\circ}) \end{aligned}$Replace the numerator with$-48 (\cos 105^{\circ} + i \sin 105^{\circ})$then divide this by$z_2 = -6(\cos 30^{\circ} + i\sin 30^{\circ})$. To divide the two values, we divide$-48$by$-8$and find the cosine and sine of$105^{\circ} + 30^{\circ}$.$\begin{aligned}\dfrac{z_1 \cdot z_3}{z_2}&= \dfrac{-48 (\cos 105^{\circ} + i \sin 105^{\circ})}{-6(\cos 30^{\circ} + i\sin 30^{\circ})}\\&=\dfrac{-48}{-6} [\cos(105^{\circ} – 30^{\circ}) + i\sin (105^{\circ} – 30^{\circ})]\\&= 8 (\cos 75 + i \sin 75^{\circ} ) \end{aligned}$We can list down the values of$z_1 \cdot z_2$,$\dfrac{z_1}{z_3}$, and$ \dfrac{z_1 \cdot z_ 3}{z_2}$. a.$z_1 \cdot z_2 = -24i $b.$\dfrac{z_1}{z_3} = -\dfrac{1}{3}(\cos 15^{\circ} + i\sin 15^{\circ})$c.$ \dfrac{z_1 \cdot z_ 3}{z_2} = 8 (\cos 75 + i \sin 75^{\circ} )$### Example 7 Given that$z_ 1 = -2 – 2i$,$z_2 = 4\sqrt{3} + 4i$, and$z_3 = 8 – 8\sqrt{3}i$, find the value of the following expressions: a.$z_1 \cdot z_2$b.$(z_3)^2$c.$ \dfrac{z_1 \cdot z_ 2}{(z_3)^2}$Write your answers in polar form. Solution Since we want the final answers to be in polar form, it is much easier to rewrite$z_1$,$z_2$, and$z_3$in the polar form first. • For each complex number, we first find the absolute value of the complex number using$r = \sqrt{a^2 + b^2}$, where$a$represents the real number parts and$b$represents the imaginary number parts. • We can then find the angle or argument of the complex number using$\tan^{-1} \dfrac{b}{a}$. • Write each complex number in the form of$r(\cos \theta + i\sin \theta)$. Let’s summarize our calculations in a table. $\boldsymbol{a + bi}\boldsymbol{r}\boldsymbol{\theta}\boldsymbol{ r(\cos \theta + i\sin \theta)}z_ 1 = -2 – 2i\begin{aligned} r&= \sqrt{(-2)^2 + (-2)^2}\\&= 2\sqrt{2} \end{aligned}\begin{aligned} \theta &= \tan^{-1} \dfrac{-2}{-2}\\&= \tan^{-1} 1\\&= 225^{\circ} \end{aligned}2\sqrt{2}(\cos 225^{\circ} + i\sin 225^{\circ})z_2 = 4\sqrt{3} + 4i\begin{aligned} r&= \sqrt{(4\sqrt{3})^2 + (4)^2}\\&= \sqrt{48 + 16}\\&=\sqrt{64}\\&= 8\end{aligned}\begin{aligned} \theta &= \tan^{-1} \dfrac{4}{4\sqrt{3}}\\&= \tan^{-1} \dfrac{1}{\sqrt{3}}\\&= 30^{\circ} \end{aligned}8(\cos 30^{\circ} + i\sin 30^{\circ})z_3 = 8 – 8\sqrt{3}i\begin{aligned} r&= \sqrt{(8\sqrt{3})^2 + (8)^2}\\&= \sqrt{192 + 64}\\&=\sqrt{256}\\&= 16\end{aligned}\begin{aligned} \theta &= \tan^{-1} \dfrac{-8\sqrt{3}}{8}\\&= \tan^{-1} -\sqrt{3}\\&= 300^{\circ} \end{aligned}16(\cos 300^{\circ} + i\sin 300^{\circ})$Let’s begin by finding$z_1 \cdot z_2$by multiplying$2\sqrt{2}$and$8$then finding the cosine and sine of$225^{\circ} + 30^{\circ}$.$\begin{aligned}z_1 \cdot z_2 &= [2\sqrt{2}(\cos 225^{\circ} + \sin 225^{\circ})]\cdot[8(\cos 30^{\circ} + \sin 30^{\circ})]\\&=(2\sqrt{2} \cdot 8) [\cos (225^{\circ} + 30^{\circ}) + i\sin (225^{\circ} + 30^{\circ})] \\&= 16\sqrt{2}(\cos 255^{\circ} + i\sin 255^{\circ})\end{aligned}$We can express$(z_3)^2$as$z_3 \cdot z_3$then find the product of$16(\cos 300^{\circ} + \sin 300^{\circ})$and itself.$\begin{aligned}z_3 \cdot z_3 &= [16(\cos 300^{\circ} + i\sin 300^{\circ})][16(\cos 300^{\circ} + i\sin 300^{\circ})]\\&= (16 \cdot 16)[\cos(300^{\circ} + 300^{\circ}) + i\sin (300^{\circ} + 300^{\circ})]\\&= 256(\cos 600^{\circ} + i\sin 600^{\circ})\end{aligned}$We can also rewrite$\cos 600^{\circ}$and$\sin 600^{\circ}$as$\cos (600^{\circ} – 360^{\circ}) = \cos 240^{\circ}$and$\sin (600^{\circ} – 360^{\circ}) = \sin 240^{\circ}$. Hence, we have$256 (\cos 240^{\circ}+ \sin 240^{\circ})$. We can use our results from part a and b to find$ \dfrac{z_1 \cdot z_ 2}{(z_3)^2}$. Once we replace the numerator and denominator with$16\sqrt{2}(\cos 255^{\circ} + i\sin 255^{\circ})$and$256 (\cos 240^{\circ}+ \sin 240^{\circ})$, respectively, we can then simplify the expression as shown below.$\begin{aligned}\dfrac{z_1 \cdot z_ 2}{(z_3)^2} &= \dfrac{16\sqrt{2}(\cos 255^{\circ} + i\sin 255^{\circ})}{256 (\cos 240^{\circ}+ \sin 240^{\circ})}\\&= \dfrac{16\sqrt{2}}{256}[\cos(255^{\circ} – 240^{\circ}) + i\sin(255^{\circ} – 240^{\circ})]\\&= \dfrac{\sqrt{2}}{16} (\cos 15^{\circ} + i\sin 15^{\circ}) \end{aligned}$Hence, we have the values of$z_1 \cdot z_2$,$(z_3)^2$, and$ \dfrac{z_1 \cdot z_ 2}{(z_3)^2}$in polar form as shown below: a.$z_1 \cdot z_2 = 16\sqrt{2}(\cos 255^{\circ} + i\sin 255^{\circ})$b.$(z_3)^2 = 256 (\cos 240^{\circ}+ \sin 240^{\circ})$c.$ \dfrac{z_1 \cdot z_ 2}{(z_3)^2} = \dfrac{\sqrt{2}}{16} (\cos 15^{\circ} + i\sin 15^{\circ}) $### Example 8 Let’s say, we have$z = r(\cos \theta + i \sin \theta)$. Find the expressions of the following in terms of$r$and$\theta$. a.$z^2$b.$z^3$c.$z^4$d.$z^n$Solution We can rewrite$z^2$as$z \cdot z = [r(\cos \theta + i \sin \theta)] \cdot [r(\cos \theta + i \sin \theta)]$. This means that we simply multiply$r$to itself and find the cosine and sine values of$\theta + \theta$.$\begin{aligned}z^2 &= [r (\cos \theta + i\sin \theta)][r (\cos \theta + i\sin \theta)]\\&=(r \cdot r)[\cos(\theta + \theta) + i\sin(\theta + \theta)]\\&=r^2(\cos 2\theta + i\sin 2\theta) \end{aligned}$To find$z^3$, we multiply$z$to the result of$z^2$. Hence, we have the calculations as shown below.$\begin{aligned}z^3 &= z \cdot z^2\\&= [r(\cos \theta + i\sin \theta)][r^2(\cos 2\theta + i\sin 2\theta) ]\\&= (r \cdot r^2)[\cos (\theta + 2\theta) + i\sin(\theta + 2\theta)]\\&=r^3(\cos 3\theta + i\sin 3\theta)\end{aligned}$Notice a pattern yet? Let’s work on$z^4$so you one more item to observe. Again, we simply multiply$z = r(\cos \theta + i \sin \theta)$to the$z^3 = r^3(\cos 3\theta + i\sin 3\theta)$.$ \begin{aligned}z^4 &= z \cdot z^3\\&= [r(\cos \theta + i\sin \theta)][r^3(\cos 3\theta + i\sin 3\theta) ]\\&= (r \cdot r^3)[\cos (\theta + 3\theta) + i\sin(\theta + 3\theta)]\\&=r^4(\cos 4\theta + i\sin 4\theta)\end{aligned}$Why don’t we write down the four results first and find a pattern?$ \begin{aligned}z^1 &=r^1(\cos 1\theta + i\sin 1\theta)\\z^2 &=r^2(\cos 2\theta + i\sin 2\theta)\\z^3 &=r^3(\cos 3\theta + i\sin 3\theta)\\z^4 &=r^4(\cos 4\theta + i\sin 4\theta)\end{aligned}$We can see that when$z$is raised a power say,$n$, we can multiply$r$by$n$then multiply the$\theta$by$n$. This means that$z^n$can be expressed as$r^n (\cos n\theta + i\sin n\theta)$. This is actually a property of complex numbers, and we’ll divide further into this topic when we learn about DeMoivre’s Theorem. For now, let’s write down the results we have for the four items: a.$z^2 = r^2 (\cos 2\theta + i\sin 2\theta) $b.$z^3 = r^3 (\cos 3\theta + i\sin 3\theta) $c.$z^4 = r^4 (\cos 4\theta + i\sin 4\theta) $d.$z^n = r^n (\cos n\theta + i\sin n\theta) $### Practice Questions 1. Review our knowledge of polar coordinates by converting the following rectangular coordinates into their polar form. a.$(-3\sqrt{3}, -3)$b.$(-4, -4\sqrt{3})$c.$(6, -6)$2. Graph the following complex numbers on one complex plane, then convert them to polar form. a.$-4 + 4i$b.$6 + 6\sqrt{3}i$c.$3\sqrt{3} – 3i$3. The complex numbers shown below are in their polar forms. Rewrite them so that they are of the form$a + bi$. a.$6(\cos 60^{\circ} + i\sin 60^{\circ})$b.$-3\left(\cos \dfrac{2\pi}{3}+ i\sin \dfrac{2\pi}{3}\right)$c.$8 \text{cis } 30^{\circ}$d.$\dfrac{3}{4} \text{cis } \dfrac{5\pi}{6}$4. Find the product of the following pairs of complex numbers. Express your final answer in polar form. a.$z_1 = 2(\cos 40^{\circ} + i\sin 40^{\circ})$,$z_2 = -3(\cos 30^{\circ} + i\sin 30^{\circ})$b.$z_1 = -4\left(\cos \dfrac{\pi}{12} + i\sin \dfrac{\pi}{12}\right)$,$z_2 =2\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right)$c.$z_1 = 6(\cos 22^{\circ} + i\sin 22^{\circ})$,$z_2 = -3\left(\cos \dfrac{\pi}{4} + i\sin \dfrac{\pi}{4}\right)$,and$z_3 = -4(\cos 25^{\circ} + i\sin 25^{\circ})$5. Find$\dfrac{z_1}{z_2}$for each of the given pairs of complex numbers. a.$z_1 = 24(\cos 40^{\circ} + i\sin 40^{\circ})$,$z_2 = -12(\cos 15^{\circ} + i\sin 15^{\circ})$b.$z_1 = 10\left(\cos \dfrac{5\pi}{6} + i\sin \dfrac{5\pi}{6}\right)$,$z_2 = -5\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}\right)$c.$z_1 = -6(\cos 80^{\circ} + i\sin 80^{\circ})$,$z_2 = 12\left(\cos \dfrac{\pi}{4} + i\sin \dfrac{\pi}{4}\right)$6. Given that$z_ 1 = 5\text{cis } 120^{\circ}$,$z_2 = -15(\cos 45^{\circ} + i\sin 45^{\circ})$, and$z_3 = 10 \left(\cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6}\right)$, find the value of the following expressions: a.$z_1 \cdot z_2$(Write your answer in the rectangular form) b.$\dfrac{z_1}{z_3}$c.$ \dfrac{z_1 \cdot z_ 3}{z_2}$7. Given that$z_ 1 = -4 + 4i$,$z_2 = 3\sqrt{3} + 3i$, and$z_3 = -6 – 6\sqrt{3}i$, find the value of the following expressions: a.$z_1 \cdot z_3$b.$(z_2)^2$c.$ \dfrac{z_1 \cdot z_ 3}{(z_2)^2}$Write your answers in polar form. ### Answer Key 1. a.$(6, 150^{\circ})$b.$(8, 240^{\circ})$c.$(6\sqrt{2},315^{\circ})$2. a.$(4sqrt{2},135^{\circ})$b.$(12, 60^{\circ})$c.$(6, 330^{\circ})$3. a.$3 + \dfrac{3\sqrt{3}}{2}i$b.$\dfrac{3}{2} – \dfrac{3\sqrt{3}}{2}i$c.$4\sqrt{3} + 4i$d.$-\dfrac{3\sqrt{3}}{8} +\dfrac{3}{8}i$4. a.$-6(\cos 70^{\circ}+ \sin 70^{\circ})$b.$-8\left(\cos \dfrac{\pi}{4}+ i\sin \dfrac{\pi}{4}\right)$c.$72(\cos 92^{\circ}+ \sin 92^{\circ})$or$72\left(\cos \dfrac{23\pi}{45}+ i\sin \dfrac{23\pi}{45}\right)$5. a.$-2(\cos 25^{\circ}+ \sin 25^{\circ})$b.$-2\left(\cos \dfrac{\pi}{2}+ i\sin \dfrac{\pi}{2}\right)$c.$-\dfrac{1}{2}(\cos 35^{\circ}+ \sin 35^{\circ})$or$-\dfrac{1}{2}\left(\cos \dfrac{7\pi}{36}+ i\sin \dfrac{7\pi}{36}\right)$6. a.$-75(\cos 165^{\circ}+ \sin 165^{\circ})$b.$\dfrac{1}{2}(\cos 90^{\circ}+ \sin 90^{\circ})$or$\dfrac{1}{2}\left(\cos \dfrac{\pi}{2}+ i\sin \dfrac{\pi}{2}\right)$c.$-\dfrac{10}{3}(\cos 105^{\circ}+ \sin 105^{\circ})$or$-\dfrac{10}{3}\left(\cos \dfrac{7\pi}{12}+ i\sin \dfrac{7\pi}{12}\right)$7. a.$48\sqrt{2}(\cos 15^{\circ}+ \sin 15^{\circ})$b.$36(\cos 60^{\circ}+ \sin 60^{\circ})$c.$\dfrac{4\sqrt{2}}{3}(\cos 315^{\circ}+ \sin 315^{\circ})\$
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## Section5.4Indeterminate Form & L'Hôpital's Rule
### Subsection5.4.1Indeterminate Forms
Before we embark on introducing one more limit rule, we need to recall a concept from algebra. In your work with functions (see Chapter 2) and limits (see Chapter 4) we sometimes encountered expressions that were undefined, because they either lead to a contradiction or to numbers that are not in the set of numbers we started out with. Let us look at an example for either scenario to investigate the concept “undefined” more deeply.
Suppose that
\begin{equation*} f(x)=\frac{1}{x}\text{.} \end{equation*}
What happens when $x=0\text{?}$ Then $f(0)=1/0\text{,}$ but $1/0$ is undefined. Why is that? Let's assume this value is defined. This means that $1/0$ is equal to some number, call it $n\text{.}$ Then
\begin{equation*} \begin{split} \frac{1}{0} \amp = n \\ 1 \div 0 \amp = n \\ 1 \amp = n \times 0 \\ 1 \amp = 0 \end{split} \end{equation*}
Clearly, 1 is not equal to 0, and so this statement is a contradiction. In fact, if we analyze the satament
\begin{equation*} 1 = n \times 0\text{,} \end{equation*}
we notice that there is no number for $n$ that will satisfy this equation. Therefore, $1/0$ could not have been a number, and hence we say $1/0$ is undefined. This is the reason why we write that the domain of $f$ is given by
\begin{equation*} \mathcal{D}_{f} =\left\{ x \in \mathbb{R}\big\rvert x \neq 0 \right\}\text{.} \end{equation*}
###### Example5.31. Different Number Set.
Suppose that $f(x)=\sqrt{x-1}$ and that we are working over the real numbers. What happens when $x=0\text{?}$ Then
\begin{equation*} f(0) = \sqrt{-1}\text{,} \end{equation*}
but $\sqrt{-1}$ is undefined over the real numbers. Why is that? Let's assume this value is defined. Then by the definition of square root, there is a real number $n$ such that $-1 = n^2\text{.}$ Clearly, the square of a real number cannot produce a negative real number because positive × positive and negative × negative are both positive real numbers. In fact, $\sqrt{-1}$ is the imaginary number $i\text{,}$ which belongs to the set of complex numbers.
When we work out limit problems algebraically, we will often get as an initial answer something that is undefined. This is because the places where a function is undefined are the “interesting” places to look for limits. For example, if
\begin{equation*} g(x)=\frac{x^{2}-9}{x-3}\text{,} \end{equation*}
then
\begin{equation*} g(3) = \frac{3^2-9}{3-3} = \frac{0}{0}\text{,} \end{equation*}
but
\begin{equation*} \lim_{x\to 3} g(x) = \lim_{x\to 3} \frac{x^2-9}{x-3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3} \left(x+3\right) = 6\text{.} \end{equation*}
The function $g$ is a line with a hole at $x=3$ and the limit showed us that this hole can be removed with the $y$-value 6 at $x=3$ (see Fig 5.11).
However, we must remember that when we are calculating the limit of $f(x)$ as $x \to a$ we are not interested in the behavior of $f(x)$ at $a\text{,}$ but we want to know the behavior of $f(x)$ around $a\text{.}$ It is therefore important for us to identify an undefined value a of a function, and furthermore, to investigate whether the type of undefined value can tell us something about the behavior of the function around $a\text{.}$
Before we continue, we need to draw attention to a notation that we have been using when calculating limits. When we write $f(x) \to 0$ as $x \to a\text{,}$ we actually mean that $f(x)$ gets arbitrarily close to zero as $x$ gets closer and closer to $a\text{.}$ However, the function value never reaches zero. Similarly, when we write $f(x) \to \infty$ as $x \to a\text{,}$ we actually mean that $f(x)$ grows ever larger, without bound as $x$ gets closer and closer to $a\text{.}$ However, the function value never reaches infinity, since infinity is not even a number.
###### Limit Behaviour.
When calculating limits,
1. 0 represents a number arbitrarily close to zero;
2. $+\infty$ represents an arbitrarily large positive number; and
3. $-\infty$ represents an arbitrarily large negative number.
Therefore, $f(x) \to \frac{0}{0}$ as $x \to a$ means that $f(x)$ is a fraction for which both the numerator and the denominator get arbitrarily close to zero as $x$ gets closer and closer to $a\text{,}$ and $f(x) \to \frac{\infty}{\infty}$ as $x \to a$ means that $f(x)$ is a fraction for which both the numerator and the denominator grow ever larger, without bound as $x$ gets closer and closer to $a\text{.}$ We also know from experience that some limits that demonstrate $\frac{0}{0}$ or $\frac{\infty}{\infty}$ behaviour work out to be real numbers, i.e. the limit exists, while others do not, as the following four examples remind us:
###### Example5.32. Limit exists when 0/0.
\begin{equation*} \lim_{x\to 3} \frac{x^{2}-9}{x-3} \stackrel{\frac{0}{0}}{=} \lim_{x\to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3} (x+3) = 6 \end{equation*}
###### Example5.33. Limit does not exist when 0/0.
\begin{equation*} \begin{split} \lim_{x\to 0^{+}} \frac{\sqrt{x+1}-1}{x^{2}} \stackrel{\frac{0}{0}}{=} \amp \lim_{x\to 0^{+}} \frac{\sqrt{x+1}-1}{x^{2}} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} \\ \amp = \lim_{x\to 0^{+}} \frac{1}{x\left(\sqrt{x+1}+1\right)} \stackrel{\frac{1}{0^{+}}}{=} \infty \end{split} \end{equation*}
###### Example5.34. Limit exists when $\infty$/$\infty$.
\begin{equation*} \lim_{x\to \infty} \frac{1-x}{2x} \stackrel{\frac{-\infty}{\infty}}{=} \lim_{x\to \infty} \frac{\frac{1}{x}-1}{2} = -\frac{1}{2} \end{equation*}
###### Example5.35. Limit does not exist when $\infty$/$\infty$.
\begin{equation*} \lim_{x\to \infty} \frac{1-x^{2}}{2x} \stackrel{\frac{-\infty}{\infty}}{=} \lim_{x\to \infty} \frac{\frac{1}{x}-x}{2} = -\infty \end{equation*}
Upon closer inspection of the undefined expressions 0/0 and $\infty$/$\infty\text{,}$ we should realize that both terms are based on the division operation and ask ourselves whether there are other undefined expressions that we may encounter when taking limits. We therefore investigate arithmetic ($a+b\text{,}$$a-b\text{,}$$ab\text{,}$$a/b$) and exponentiating ($a^b$) operations where $a$ and $b$ are values that approach 0, 1, some arbitrary number $n \neq 0,1$ or $\infty\text{.}$ We leave it up to the reader to perform an exhaustive listing of all combinations, and instead limit ourselves to the combinations that are of interest as shown in Table 5.12.
We now encourage the reader to investigate each one of the terms shown in Table 5.12 and decide whether the undefined expression resolves to give a single number value or infinity (determinate form), or whether this cannot be determined (indeterminate form), all the while keeping in mind our earlier discussion on limit behaviour around $x=a\text{.}$ We formally define this new terminology before we explore some terms together.
###### Definition5.36. Determinate and Indeterminate Forms.
An undefined expression involving some operation between two quantities is called a determinate form if it evaluates to a single number value or infinity.
An undefined expression involving some operation between two quantities is called an indeterminate form if it does not evaluate to a single number value or infinity.
We will inspect multiplication more closely. Consider $0 \times 0\text{.}$ Clearly, a number that is getting arbitrarily close to zero that is multiplied by another number that is getting arbitrarily close to zero gets even closer to zero, i.e. $0 \times 0 \to 0\text{.}$ Now consider $\infty \times \infty\text{.}$ Here, multiplying two values that are growing large without bound simply means that their product grows large without bound, i.e. $\infty \times \infty \to \infty\text{.}$ Similarly, $\left(-\infty\right)\times\infty$ means that the magnitude of the product grows large without bound and that $\left(-\infty\right)\times\infty \to -\infty\text{.}$ What about $n\times\infty\text{,}$ when $n \neq 0\text{?}$ Here we need to differentiate between negative and positive values of $n\text{:}$ If $n>0\text{,}$ then $n\times \infty \to \infty\text{,}$ and if $n\lt 0\text{,}$ then $n\times\infty \to -\infty\text{.}$ So far, we have only encountered determinate forms involving multiplication. Lastly, consider $0\times\infty\text{.}$ Here, we have a number that is getting arbitrarily close to zero being multiplied with a value that is growing large without bounds. This is like two ends of a rope being tugged and we do not know which side is going to win. Therefore, $0\times\infty$ is an expression that cannot be determined.
We leave the remaining terms up to the reader to investigate and simply present the determinate and indeterminate forms of the expressions from Table 5.12 in Table 5.13.
### Subsection5.4.2L'Hôpital's Rule for Finding Limits
We are now in a position to introduce one more technique for trying to evaluate a limit.
###### Definition5.37. Limits of the Indeterminate Forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$.
A limit of a quotient $\lim\limits_{x\rightarrow a}\frac{f\left( x\right) }{g\left( x\right) }$ is said to be an indeterminate form of the type$\frac{0}{0}$ if both $f\left( x\right) \rightarrow 0$ and $g\left( x\right) \rightarrow 0$ as $x\rightarrow a\text{.}$ Likewise, it is said to be an indeterminate form of the type $\frac{\infty }{\infty }$ if both $f\left( x\right) \rightarrow \pm \infty$ and $g\left( x\right) \rightarrow \pm \infty$ as $x\rightarrow a$ (Here, the two $\pm$ signs are independent of each other).
This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities, so we will not prove it here.
Note:
1. There may be instances where we would need to apply L'Hôpital's Rule multiple times, but we must confirm that $\lim\limits_{x\to a}\dfrac{f'(x)}{g'(x)}$ is still indeterminate before we attempt to apply L'Hôpital's Rule again.
2. L'Hôpital's Rule is also valid for one-sided limits and limits at infinity.
###### Notation when Applying L'Hôpital's Rule.
We use the symbol $\Heq$ to denote we are using l'Hôpital's Rule in that step.
###### Example5.39. L'Hôpital's Rule and Indeterminate Form 0/0.
Compute $\ds\lim_{x\to \pi}\frac{x^2-\pi^2}{\sin x}\text{.}$
Solution
We use L'Hôpital's Rule: Since the numerator and denominator both approach zero,
\begin{equation*} \lim_{x\to \pi}\frac{x^2-\pi^2}{\sin x}\Heq \lim_{x\to \pi}\frac{2x}{\cos x}\text{,} \end{equation*}
provided the latter exists. But in fact this is an easy limit, since the denominator now approaches $-1\text{,}$ so
\begin{equation*} \lim_{x\to \pi}\frac{x^2-\pi^2}{\sin x}=\frac{2\pi}{-1} = -2\pi\text{.} \end{equation*}
###### Example5.40. L'Hôpital's Rule and Indeterminate Form $\infty$/$\infty$.
Compute $\ds\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}\text{.}$
Solution
As $x$ goes to infinity, both the numerator and denominator go to infinity, so we may apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to \infty}\frac{2x^2-3x+7}{x^2+47x+1}\Heq \lim_{x\to \infty}\frac{4x-3}{2x+47}\text{.} \end{equation*}
In the second quotient, it is still the case that the numerator and denominator both go to infinity, so we are allowed to use L'Hôpital's Rule again:
\begin{equation*} \lim_{x\to \infty}\frac{4x-3}{2x+47}\Heq\lim_{x\to \infty}\frac{4}{2}=2\text{.} \end{equation*}
So the original limit is 2 as well.
###### Example5.41. L'Hôpital's Rule and Indeterminate Form 0/0.
Compute $\ds\lim_{x\to 0}\frac{\sec x - 1}{\sin x}\text{.}$
Solution
Both the numerator and denominator approach zero, so applying L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0}\frac{\sec x - 1}{\sin x}\Heq \lim_{x\to 0}\frac{\sec x\tan x}{\cos x}=\frac{1\cdot 0}{1}=0\text{.} \end{equation*}
###### Example5.42. L'Hôpital's Rule and Indeterminate Form $\infty$/$\infty$.
Compute $\ds\lim_{x\to 0^{+}} \dfrac{\frac{1}{x^2}}{\ln x}\text{.}$
Solution
As $x$ aproaches zero from the right, the numerator approaches $+\infty$ and the denominator approaches $-\infty\text{.}$ We may therefore apply L'Hôpital's Rule:
\begin{equation*} \begin{split} \lim_{x\to 0^{+}} \dfrac{\frac{1}{x^2}}{\ln x} \amp \Heq \lim_{x \to 0^{+}} \frac{\frac{-2}{x^3}}{\frac{1}{x}} \\ \amp \Heq \lim_{x\to 0^{+}} \frac{-2}{x^{2}} \\ \amp = -\infty \end{split} \end{equation*}
Note: In order to decide which of two functions $f$ and $g$ grows faster as the independent variable, say $x\text{,}$ becomes larger, we can apply the limit as $x$ goes to infinity to the ratio $f/g$ of these two functions. If the function $f$ in the numerator grows faster, then the limit approaches infinity. If the function $g$ in the denominator grows faster, then the limit approaches zero. If the functions have similar growth rates, then then the limit approaches a constant. This type of limit is readily computed using L'Hôpital's Rule, and so L'Hôpital's Rule is a useful tool to know.
We now exemplify this idea of growth rate. Let us have a closer look at the two functions $f(x)=5x^3$ and $g(x)=x^3\text{.}$ Then the function $f(x)=5x^3$ grows exactly five times as fast as the function $g(x)=x^3\text{.}$ However, the ratio of the two functions
\begin{equation*} \frac{f(x)}{g(x)} = \frac{5x^3}{x^3} = 5, x\neq 0\text{,} \end{equation*}
is a constant, and so both functions have fundamentally the same growth rate.
### Subsection5.4.3Informally Extending L'Hôpital's Rule
L'Hôpital's Rule concerns limits of a quotient that are indeterminate forms. But not all functions are given in the form of a quotient. But all the same, nothing prevents us from re-writing a given function in the form of a quotient. Indeed, some functions whose given form involve either a product $f\left( x\right) g\left( x\right)$ or a power $f\left( x\right) ^{g\left( x\right) }$ carry indeterminacies such as $0\cdot \left(\pm \infty\right)$ or $1^{\pm \infty }\text{.}$ Something small times something numerically large (positive or negative) could be anything. It depends on how small and how large each piece turns out to be. A number close to 1 raised to a numerically large (positive or negative) power could be anything. It depends on how close to 1 the base is, whether the base is larger than or smaller than 1, and how large the exponent is (and its sign). We can use suitable algebraic manipulations to relate them to indeterminate quotients. We will illustrate with three examples, a product, a power and a difference.
###### Example5.43. L'Hôpital's Rule and Indeterminate Form $0\times\infty$.
Compute $\ds\lim_{x\to 0^+} x\ln x\text{.}$
Solution
This doesn't appear to be suitable for L'Hôpital's Rule, but it also is not “obvious”. As $x$ approaches zero, $\ln x$ goes to $-\infty\text{,}$ so the product looks like:
\begin{equation*} (\hbox{something very small})\cdot(\hbox{something very large and negative})\text{.} \end{equation*}
This could be anything: it depends on how small and how large each piece of the function turns out to be. As defined earlier, this is a type of $\pm0\cdot\infty\text{,}$ which is indeterminate. So we can in fact apply L'Hôpital's Rule after re-writing it in the form $\frac{\infty }{\infty }\text{:}$
\begin{equation*} x\ln x = \frac{\ln x}{1/x}=\frac{\ln x}{x^{-1}}\text{.} \end{equation*}
Now as $x$ approaches zero, both the numerator and denominator approach infinity (one $-\infty$ and one $+\infty\text{,}$ but only the size is important). Using L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0^+} {\ln x\over x^{-1}}\Heq \lim_{x\to 0^+} {1/x\over -x^{-2}} =\lim_{x\to 0^+} {1\over x}(-x^2)= \lim_{x\to 0^+} -x = 0\text{.} \end{equation*}
One way to interpret this is that since $\ds\lim_{x\to 0^+}x\ln x = 0\text{,}$ the $x$ approaches zero much faster than the $\ln x$ approaches $-\infty\text{.}$
Finally, we illustrate how a limit of the type $1^\infty$ can be indeterminate.
###### Example5.44. L'Hôpital's Rule and Indeterminate Form $1^{\infty}$.
Compute $\ds{\lim_{x\to 1^+}x^{1/(x-1)}}\text{.}$
Solution
Plugging in $x=1$ (from the right) gives a limit of the type $1^\infty\text{.}$ To deal with this type of limit we will use logarithms. Let
\begin{equation*} L=\lim_{x\to 1^+}x^{1/(x-1)}\text{.} \end{equation*}
Now, take the natural log of both sides:
\begin{equation*} \ln L=\lim_{x\to 1^+}\ln\left(x^{1/(x-1)}\right)\text{.} \end{equation*}
Using log properties we have:
\begin{equation*} \ln L=\lim_{x\to 1^+}\frac{\ln x}{x-1}\text{.} \end{equation*}
The right side limit is now of the type $0/0\text{,}$ therefore, we can apply L'Hôpital's Rule:
\begin{equation*} \ln L=\lim_{x\to 1^+}\frac{\ln x}{x-1}\Heq\lim_{x\to 1^+}\frac{1/x}{1}=1 \end{equation*}
Thus, $\ln L=1$ and hence, our original limit (denoted by $L$) is: $L=e^1=e\text{.}$ That is,
\begin{equation*} L=\lim_{x\to 1^+}x^{1/(x-1)}=e\text{.} \end{equation*}
In this case, even though our limit had a type of $1^\infty\text{,}$ it actually had a value of $e\text{.}$
###### Example5.45. L'Hôpital's Rule and Indeterminate Form $\infty-\infty$.
Compute $\lim\limits_{x\to 0^{+}} \left(\dfrac{1}{\sin x} - \dfrac{1}{x}\right)\text{.}$
Solution
As $x$ approaches zero from the right,
\begin{equation*} \frac{1}{\sin x} - \frac{1}{x} \to \infty - \infty\text{.} \end{equation*}
This is not a form on which we know we can use L'Hôpital's Rule, however, if we combine the fractions, the problem becomes
\begin{equation*} \lim_{x\to 0^{+}} \frac{x-\sin x}{x\sin x}\text{,} \end{equation*}
which gives us the indeterminate form $0/0\text{.}$ We can now apply L'Hôpital's Rule twice:
\begin{equation*} \begin{split} \lim_{x\to 0^{+}} \frac{x-\sin x}{x\sin x} \amp \Heq \lim_{x\to 0^{+}} \frac{1-\cos x}{\sin x + x\cos x} \\ \amp\Heq \lim_{x\to 0^{+}} \frac{\sin x}{2\cos x - x\sin x}\\ \amp= \frac{0}{2-0} = 0.\end{split} \end{equation*}
##### Exercises for Section 5.4.
Compute the following limits.
1. $\ds\lim_{x\to 0} \frac{\cos x -1}{\sin x}$ Answer
0
Solution
We see that the limit
\begin{equation*} \lim_{x\to 0} \frac{\cos x -1}{\sin x} \end{equation*}
is of the indeterminate form $0/0\text{.}$ We can therefore apply L'Hôpital's Rule once to find,
\begin{equation*} \lim_{x\to 0} \frac{\cos x -1}{\sin x} \Heq \lim_{x\to 0}\frac{-\sin x}{\cos x} = \frac{0}{1} = 0\text{.} \end{equation*}
2. $\ds\lim_{x\to \infty} \frac{e^x}{x^3}$ Answer
$\infty$
Solution
We now see the indeterminate form $\infty/\infty$ and apply L'Hôpital's Rule until the form becomes determinate.
\begin{equation*} \lim_{x\to \infty} \frac{e^x}{x^3} \Heq \lim_{x \to \infty} \frac{e^x}{3x^2} \Heq \dots \Heq \lim_{x\to \infty} \frac{e^x}{6} = \infty\text{.} \end{equation*}
3. $\ds\lim_{x\to \infty} \frac{\ln x}{x}$ Answer
0
Solution
The limit is of the indeterminate form $\infty/\infty\text{,}$ and so we apply L'Hôpital's Rule until the form becomes determinate.
\begin{equation*} \lim_{x\to \infty} \frac{\ln x}{x} \Heq \lim_{x\to\infty} \frac{1/x}{\frac{1}{2}x^{-1/2}} = \lim_{x\to\infty} \frac{2}{\sqrt{x}} = 0\text{.} \end{equation*}
4. $\ds\lim\limits_{x\to\infty} \dfrac{\ln x}{x}$ Answer
0
Solution
To evaluate $\lim\limits_{x\to\infty} \dfrac{\ln x}{x}\text{,}$ we see that this is of the indeterminate form $\dfrac{\infty}{\infty}$ and so we can apply L'Hôpital's Rule:
\begin{equation*} \lim\limits_{x\to\infty} \dfrac{\ln x}{x} \Heq \lim_{x\to\infty} \frac{\diff{}{x}\ln x}{\diff{}{x} x}= \lim_{x\to\infty} \frac{\frac{1}{x}}{1} = \lim_{x\to\infty} \frac{1}{x} = 0\text{.} \end{equation*}
5. $\ds\lim\limits_{x\to 0} \frac{\sqrt{9+x}-3}{x}$ Answer
1/6
Solution
The limit is of the indeterminate form $0/0$ and so we apply L'Hôpital's Rule:
\begin{equation*} \lim\limits_{x\to 0} \frac{\sqrt{9+x}-3}{x} \Heq \lim\limits{x\to 0} \frac{\frac{1}{2\sqrt{9+x}}}{1} = \frac{1}{2(3)} = \frac{1}{6}\text{.} \end{equation*}
6. $\ds\lim_{x\to 2} \frac{2-\sqrt{x+2}}{4-x^2}$ Answer
1/16
Solution
We see that the limit is of the indeterminate form $0/0\text{,}$ and so we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 2} \frac{2-\sqrt{x+2}}{4-x^2} \Heq \lim_{x\to 2} \frac{-\frac{-1}{2\sqrt{x+2}}}{-2x} = \frac{1}{16}\text{.} \end{equation*}
7. $\ds\lim_{x\to 1} \frac{\sqrt{x} - 1}{\sqrt[3]{x}-1}$ Answer
3/2
Solution
The limit is of the indeterminate form $0/0\text{,}$ and so we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 1} \frac{\sqrt{x} - 1}{\sqrt[3]{x}-1} \Heq \lim_{x\to 1} \frac{\frac{x^{-1/2}}{2}}{\frac{x^{-2/3}}{3}} = \frac{3}{2}\text{.} \end{equation*}
8. $\ds\lim_{x\to 0} \frac{x^2}{\sqrt{2x+1}-1}$ Answer
0
Solution
The limit is of the interderminate form $0/0\text{,}$ and so we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0} \frac{x^2}{\sqrt{2x+1}-1} \Heq \lim_{x\to 0} \frac{2x}{(2x+1)^{-1/2}} = 0\text{.} \end{equation*}
9. $\ds \lim_{u to 1} \frac{(u-1)^3}{(1/u)-u^2+(3/u)-3}$ Solution
The limit is of the indeterminate form $0/0\text{,}$ and so we apply L'Hôpital's Rule:
\begin{equation*} \lim_{u to 1} \frac{(u-1)^3}{(1/u)-u^2+(3/u)-3} \Heq \lim_{u \to 1} \frac{3(u-1)^2}{-u^{-2}-2u - 3u^{-2}} = 0\text{.} \end{equation*}
10. $\ds\lim_{x \to 0} \frac{2+(1/x)}{3-(2/x)}$ Answer
-1/2
Solution
The limit is of the indeterminate form $\infty/\infty\text{,}$ and so we could apply L'Hôpital's Rule. However, notice that
\begin{equation*} \lim_{x \to 0} \frac{2+(1/x)}{3-(2/x)} \Heq \lim_{x \to 0} \frac{2-x^{-2}}{3+2x^{-2}} = ..\text{.} \end{equation*}
and so this procedure will not lead to a determinate form. Instead, let $t= 1/x\text{.}$ Then
\begin{equation*} \lim_{x\to 0^+} \frac{2+(1/x)}{3-(2/x)} = \lim_{t\to \infty} \frac{2+t}{3-2t} \Heq \lim_{t\to \infty}\frac{1}{-2} = -\frac{1}{2}\text{,} \end{equation*}
and similarly,
\begin{equation*} \lim_{x\to 0^-} \frac{2+(1/x)}{3-(2/x)} = \lim_{t\to -\infty} \frac{2+t}{3-2t} \Heq \lim_{t\to \infty}\frac{1}{-2} = -\frac{1}{2}\text{.} \end{equation*}
Hence,
\begin{equation*} \lim_{x\to 0} \frac{2+(1/x)}{3-(2/x)} = -\frac{1}{2}\text{.} \end{equation*}
5
11. $\ds\lim_{x\to 0^+} \frac{1+5/\sqrt{x}}{2+1/\sqrt{x}}$ Solution
The limit is of the indeterminate form $\infty/\infty\text{:}$
\begin{equation*} \lim_{x\to 0^+} \frac{1+5/\sqrt{x}}{2+1/\sqrt{x}} = \lim_{x\to 0^{+}} \frac{\sqrt{x} + 5}{2\sqrt{x}+1} = 5\text{.} \end{equation*}
12. $\ds \lim_{x\to \pi/2} \frac{\cos x}{(\pi/2)-x}$ Answer
1
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to \pi/2} \frac{\cos x}{(\pi/2)-x} \Heq \lim_{x\to \pi/2} = \frac{-\sin x}{-1} = 1\text{.} \end{equation*}
13. $\ds\lim_{x\to 0} \frac{x^2}{e^x-x-1}$ Answer
2
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule twice:
\begin{equation*} \lim_{x\to 0} \frac{x^2}{e^x-x-1} \Heq \lim_{x\to 0} \frac{2x}{e^x-1} \Heq \lim_{x\to 0} \frac{2}{e^x} = 2\text{.} \end{equation*}
14. $\ds \lim_{x\to 1} \frac{\ln x}{x-1}$ Answer
1
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 1} \frac{\ln x}{x-1} \Heq \lim_{x\to 1} \frac{1/x}{1} = 1\text{.} \end{equation*}
15. $\ds\lim_{x\to 0} \frac{\ln(x^2+1)}{x}$ Answer
0
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0} \frac{\ln(x^2+1)}{x} \Heq \frac{\frac{2x}{x^2+1}}{1} = 0\text{.} \end{equation*}
16. $\ds\lim_{x\to 1} \frac{x\ln x}{x^2-1}$ Answer
1/2
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 1} \frac{x\ln x}{x^2-1} \Heq \lim_{x\to 1} \frac{\ln x + 1}{2x}= \frac{1}{2}\text{.} \end{equation*}
17. $\ds\lim_{x\to 0} \frac{\sin(2x)}{\ln(x+1)}$ Answer
2
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0} \frac{\sin(2x)}{\ln(x+1)} \Heq \frac{2\cos x}{1/(x+1)} = 2\text{.} \end{equation*}
18. $\ds\lim_{x\to 1} \frac{\sqrt{x}-1}{x-1}$ Answer
1/2
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 1} \frac{\sqrt{x}-1}{x-1} \Heq \frac{\frac{1}{2\sqrt{x}}}{1} = \frac{1}{2}\text{.} \end{equation*}
19. $\ds \lim_{x\to 0} \frac{\sqrt{x+1}-1}{\sqrt{x+4}-2}$ Answer
2
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0} \frac{\sqrt{x+1}-1}{\sqrt{x+4}-2} \Heq \lim_{x\to 0} \frac{\sqrt{x+4}}{\sqrt{x+1}} = 2\text{.} \end{equation*}
20. $\ds \lim_{x\to 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x+1}-1}$ Answer
0
Solution
The limit is of the indeterminate form $0/0\text{.}$ Therefore, we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x+1}-1} \Heq \lim_{x\to 0} \frac{2x\sqrt{x+1}}{\sqrt{x^2+1}} = 0\text{.} \end{equation*}
1. $\ds \lim_{x\to 0^+} \sqrt{x}\ln x$ Answer
0
Solution
The limit
\begin{equation*} \lim_{x\to 0^+} \sqrt{x}\ln x \end{equation*}
is of the form $0 \cdot (-\infty)\text{,}$ therefore we need to transform the limit in order to use L'Hôpital's Rule. Let $t=1/x\text{.}$ Then,
\begin{equation*} \lim_{x\to 0^+} \sqrt{x}\ln x = \lim_{t \to +\infty} \frac{\ln\left(\frac{1}{t}\right)}{\sqrt{t}} \end{equation*}
which is of the form $-\infty/\infty\text{.}$ We apply L'Hôpital's Rule until the form becomes determinate.
\begin{equation*} \begin{split} \lim_{t \to \infty} \frac{\ln\left(\frac{1}{t}\right)}{\sqrt{t}} \amp = \lim_{t \to \infty} \frac{\diff{}{t}\left(\ln(1)-\ln(t)\right)}{\frac{1}{2}t^{-1/2}} \\ \amp = \lim_{t \to \infty} \frac{-1/t}{\frac{1}{2}t^{-1/2}}\\ \amp = \lim_{t \to \infty} \frac{-2}{\sqrt{t}} = 0. \end{split} \end{equation*}
2. $\ds\lim_{x\to 0} \frac{(1-x)^{1/4} -1}{x}$ Answer
-1/4
Solution
The limit is of the indeterminate form $0/0\text{,}$ and so we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0} \frac{(1-x)^{1/4} -1}{x} \Heq \lim_{x\to 0}\frac{-\frac{1}{4} (1-x)^{-3/4}}{1} = -\frac{1}{4}\text{.} \end{equation*}
3. $\ds \lim_{t\to 0} \left(t+\frac{1}{t}\right)\left((4-t)^{3/2} - 8\right)$ Answer
-3
Solution
The limit is of the form $\infty \cdot 0\text{,}$ and so we need to transform the limit in order to apply L'Hôpital's Rule. Let
\begin{equation*} \lim_{t\to 0} \left(t+\frac{1}{t}\right)\left((4-t)^{3/2} - 8\right) = \lim_{t\to 0} t\left((4-t)^{3/2} - 8\right) + \lim_{t\to 0} \frac{1}{t}\left((4-t)^{3/2} - 8\right)\text{.} \end{equation*}
The first limit we can readily evaluate as
\begin{equation*} \lim_{t\to 0} t\left((4-t)^{3/2} - 8\right) = 0\text{.} \end{equation*}
For the second limit, we can now apply L'Hôpital's Rule:
\begin{equation*} \lim_{t\to 0} \frac{(4-t)^{3/2} - 8}{t} \Heq \lim_{t\to 0} \frac{-3/2 \sqrt{4-t}}{1} = -3\text{.} \end{equation*}
4. $\ds\lim_{t\to 0^+} \left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)\left(\sqrt{t+1}-1\right)$ Answer
1/2
Solution
The limit is of the form $\infty \cdot 0\text{,}$ and so we must transform the limit in order to apply L'Hôpital's Rule. Let
\begin{equation*} \lim_{t\to 0^+} \left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)\left(\sqrt{t+1}-1\right) = \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{t} + \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{\sqrt{t}}\text{.} \end{equation*}
Both limits are of the form $0/0$ and so we can apply L'Hôpital's Rule to each:
\begin{equation*} \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{t} \Heq \lim_{t\to 0^+} \frac{1}{2\sqrt{t+1}} = \frac{1}{2}\text{.} \end{equation*}
We now apply L'Hôpital's Rule to the second limit:
\begin{equation*} \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{\sqrt{t}} \Heq \lim_{t\to 0^+} \frac{\sqrt{t}}{\sqrt{t+1}} = 0\text{.} \end{equation*}
Therefore,
\begin{equation*} \lim_{t\to 0^+} \left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)\left(\sqrt{t+1}-1\right) = \frac{1}{2}\text{.} \end{equation*}
5. $\ds \lim_{x\to 0} \frac{e^x-1}{x}$ Answer
1
Solution
This limit is of the indeterminate form $\dfrac{0}{0}\text{,}$ and so we apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to 0} \frac{e^x-1}{x} \Heq \lim_{x\to 0} \frac{\diff{}{x} \left(e^x-1\right)}{\diff{}{x} x} = \lim_{x\to 0} \frac{e^x}{1} = e^0 = 1\text{.} \end{equation*}
6. $\ds\lim\limits_{x\to 1} \dfrac{x^{1/4}-1}{x}$ Answer
0
Solution
We evaluate:
\begin{equation*} \lim\limits_{x\to 1} \dfrac{x^{1/4}-1}{x} = \dfrac{1-1}{1} = 0\text{.} \end{equation*}
Notice that $\dfrac{0}{1}$ is not an indeterminate form and we do not apply L'Hôpital's Rule.
Discuss what happens if we try to use L'Hôpital's Rule to find $\ds \lim_{x\to \infty} \frac{x+\sin x}{x+1}\text{.}$ Solution
The limit
\begin{equation*} \lim_{x\to\infty} \frac{x+\sin x}{x+1} \end{equation*}
is of the indeterminate form $\infty/\infty\text{.}$ And so we could try to apply L'Hôpital's Rule:
\begin{equation*} \lim_{x\to\infty} \frac{x+\sin x}{x+1} \Heq \frac{1+\cos x}{1} = 1 +\lim_{x\to\infty} \cos x = DNE\text{.} \end{equation*}
\begin{equation*} \lim_{x\to\infty} \frac{x}{x+1} + \lim_{x\to \infty} \frac{\sin x}{x+1} = 1 + \lim_{x\to\infty}\text{.} \end{equation*}
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# Solving Equations with Variables On Both Sides Worksheet
The Solving Equations with Variables On both sides worksheet can help you and your students become masters of algebra. The worksheet will help your students develop their ability to use the inverse operations and solve equations with variables on both sides. In addition, the exercises will provide ample challenge and encourage creativity. The examples and solutions to these equations will make solving equations with variables on both sides easier.
Solving Equations with Variables Both Sides Worksheet from solving equations with variables on both sides worksheet , source:fadeintofantasy.net
An equation is an expression containing two variables on both sides of the equals sign. The variables are often expressed as integers. A solution is an expression expressing the unknowns in terms of the variables. A general equation can be written as 3x+5y=25x+8y=3 and has a unique solution of x = -1, y = 0.
When solving equations with variables on both sides, it is important to note that variables on both sides of the equal sign can have extraneous solutions, which are the solutions to the initial equation. These are called exponents. The values of the exponents are known as coefficients. The variable that is on both sides of the equals sign must be moved to the opposite side of the equals sign to get the solution.
Variables Both Sides Worksheet – Croefit from solving equations with variables on both sides worksheet , source:croefit.com
If an equation contains a variable on both sides of the equals sign, it must be solved by combining like terms. In addition to removing variables and modifying numbers, it must also be simplified by multiplying both sides by the common denominator. The solution may be as simple as rearranging the parentheses, or it can be as complex as putting a variable on both sides of the equation.
An equation consists of two expressions that are connected by an equals sign. The left-hand side contains the left-hand expression, while the right-hand side contains the right-hand expression. These two sides are called the right-hand and the left-hand sides. The equation with variables on both sides is the one with the lowest value. The variable term on the left-hand side is the one that is on the other-hand side. The answer must be the same for both sides.
Solving Equations with Variables Both Sides Worksheet from solving equations with variables on both sides worksheet , source:resumevalet.info
This worksheet will produce equations with variables on both sides. These worksheets will contain integers and will have problems with variables on both sides. They will also contain a number of parentheses. You can solve these equations with the inverse and additive inverse. This will simplify the problems with variables on both sides. Then, you can use the calculator to solve them.
When an equation has a variable on both sides, it is difficult to solve it without a calculator. You must write down the variables in order to simplify them. The calculator should allow you to change the values of the variables. You can move the variable term to the other side, which will give you the answer. In addition, you can add or subtract one variable to the other. Ax2+Bx+C-y=0. Using the inverse and the additive inverse will make it easier to solve such complicated equations.
Solving Equations with Variables Both Sides Worksheet from solving equations with variables on both sides worksheet , source:resumevalet.info
When solving equations with variables on both sides, you must combine like terms with each other. In other words, you must move one variable to the other side to find the new value. Alternatively, you can also use the inverse and add a term. This will simplify the equation and make it easier for you. There are two types of inverse-commutative-commuting methods: Using the inverse and the additive inverse.
In this worksheet, you should solve equations that contain two variables. Using the inverse of a variable in an equation requires you to add the variable to each side. A general equation is a one-variable equation. In reverse-commutative-commuting the two terms is a more effective method for solving equations with variables on both sides. The inverse operation will help you solve the problem with a variable in both sides.
60 best Algebra images on Pinterest from solving equations with variables on both sides worksheet , source:pinterest.com
Solving Equations with Variables Both Sides Worksheet from solving equations with variables on both sides worksheet , source:resumevalet.info
Pythagorean Theorem Worksheets from solving equations with variables on both sides worksheet , source:math-aids.com
Solving Equations with Variables Both Sides Worksheet from solving equations with variables on both sides worksheet , source:resumevalet.info
Solving Equations with Variables Both Sides Worksheet New Fancy from solving equations with variables on both sides worksheet , source:transatlantic-news.com
Variables Both Sides Worksheet – Croefit from solving equations with variables on both sides worksheet , source:croefit.com
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Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$$(cosec\; θ – cot \;θ)^2 =\dfrac{(1-cos\; θ)}{(1+cos\; θ)}$$
Asked by Abhisek | 1 year ago | 89
##### Solution :-
To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)
L.H.S. = (cosec θ – cot θ)2
The above equation is in the form of (a-b)2, and expand it
Since (a-b)2 = a2 + b2 – 2ab
Here a = cosec θ and b = cot θ
= (cosec2θ + cot2θ – 2cosec θ cot θ)
Now, apply the corresponding inverse functions and equivalent ratios to simplify
$$\dfrac{1}{sin^2θ}+ \dfrac{cos^2θ}{sin^2θ}- \dfrac{2cos θ}{sin^2θ}$$
$$\dfrac{(1 + cos^2θ – 2cos θ)}{(1 – cos^2θ)}$$
$$\dfrac{(1-cos \;θ)^2}{(1 – cos\;θ)(1+cos\; θ)}$$
$$\dfrac{ (1-cos\; θ)}{(1+cos\; θ)}$$ = R.H.S.
Therefore, (cosec θ – cot θ)
$$\dfrac{ (1-cos\; θ)}{(1+cos\; θ)}$$
Hence proved.
Answered by Pragya Singh | 1 year ago
### Related Questions
#### Prove the sinθ sin (90° – θ) – cos θ cos (90° – θ) = 0
Prove the sinθ sin (90° – θ) – cos θ cos (90° – θ) = 0
#### Prove that sin 48° sec 48° + cos 48° cosec 42° = 2
Prove that sin 48° sec 48° + cos 48° cosec 42° = 2
#### Prove that tan 20° tan 35° tan 45° tan 55° tan 70° = 1
Prove that tan 20° tan 35° tan 45° tan 55° tan 70° = 1
#### Evaluate the 2sin 3x = \sqrt{3}
Evaluate the $$2sin 3x = \sqrt{3}$$
#### Evalute the following: sin 20°/ cos 70°
Evalute the following:
(i) $$\dfrac{sin 20°}{ cos 70°}$$
(ii) $$\dfrac{ cos 19°}{ sin 71°}$$
(iii) $$\dfrac{sin 21°}{ cos 69°}$$
(iv) $$\dfrac{tan 10°}{ cot 80°}$$
(v) $$\dfrac{sec 11°}{ cosec 79°}$$
|
UPSC > Calculate Cube Roots
# Calculate Cube Roots - CSAT Preparation - UPSC
## Introduction
• We have heard many people saying that there is actually no shortcut to find the cube root of a number. This is not true. In fact, there are ways to quickly, easily and accurately calculate the cube root of a given number.
• These ways when mastered can help you to get the cube root of a number in less than 5 seconds.
• We are now going to have a look at the simplest & fastest way of calculating the cube root of a number in this article.
Calculating the Cube Root of a number quickly and accurately
• The first and the most important step is to memorize the cubes of 1 to 9. These would form an important part of your toolkit in solving the cube roots. Here is a table for your convenience.
1 –> 1
2 –> 8
3 –> 27
4 –> 64
5 –> 125
6 –> 216
7 –> 343
8 –> 512
9 –> 729
• Once you memorize this list, the next step is to remember the last digit (unit digit) of each of these cubes. Here is the list again, only with unit digits this time.
1 –> 1
2 –> 8
3 –> 7
4 –> 4
5 –> 5
6 –> 6
7 –> 3
8 –> 2
9 –> 9
• That’s it. Now that you have memorized the cubes of first 9 natural numbers and their unit digits, you are all set to amaze your friends by calculating cube roots within 5 seconds (given that they have not read this article).
Let us see how to calculate cube roots in less than 5 seconds.
Q.1. Find out the cube root of 50653.
Here is how to solve this question:
• The first step is to divide the number into 2 parts by separating the last 3 digits. So, we get 50 & 563 as the two parts of the number.
• Now, take the first part and find the largest cube contained in the first part i.e. in 50 = 27 (which is the cube of 3). The next cube i.e. 64 (cube of 4) is larger than 50. Now, as 27 is the cube of 3, your ten’s part of cube root would be 3. [This is why we memorized the cubes]
• The next step is to take the last digit of the number, which in this case is 3. Which number’s cube had 3 as the unit digit? 7… right?? (7*7*7=343) Hence, 7 is the unit digit of your solution. [This is why we memorized the endings]
Let us solve another example to show how the answer can be achieved in less than 5 seconds.
Q.2. Calculate the cube root of 941192.
By Step 1: We get two parts i.e. 941 and 192.
By Step 2: The largest cube less than 941 is 729 (cube of 9). So, ten’s digit is 9.
By Step 3: The ending digit is 2. Hence, unit’s digit is 8. That’s it. 98 is the answer.
Practice it a bit and you would be able to solve this in even less than 5 seconds.
The document Calculate Cube Roots | CSAT Preparation - UPSC is a part of the UPSC Course CSAT Preparation.
All you need of UPSC at this link: UPSC
## CSAT Preparation
197 videos|151 docs|200 tests
## FAQs on Calculate Cube Roots - CSAT Preparation - UPSC
1. How do you calculate cube roots?
Ans. To calculate the cube root of a number, you can use the following steps: 1. Identify the number whose cube root you want to find. 2. Make an estimate of the cube root that is close to the actual value. 3. Divide the number by the estimate and obtain a new estimate of the cube root. 4. Repeat step 3 until the estimate becomes accurate enough.
2. Can I use a calculator to calculate cube roots?
Ans. Yes, most calculators have a cube root function that allows you to directly calculate the cube root of a number. Simply enter the number and press the cube root button (∛) to obtain the result.
3. What is the cube root of a negative number?
Ans. The cube root of a negative number is also a negative number. For example, the cube root of -8 is -2, as (-2) x (-2) x (-2) equals -8. In general, the cube root of a negative number will always have the same sign as the original number.
4. Can you calculate the cube root of a fraction?
Ans. Yes, you can calculate the cube root of a fraction. Simply calculate the cube root of the numerator and the cube root of the denominator separately. For example, to find the cube root of 1/8, you would calculate the cube root of 1 (which is 1) and the cube root of 8 (which is 2). Therefore, the cube root of 1/8 is 1/2.
5. Are there any shortcuts or formulas for calculating cube roots?
Ans. Yes, there is a shortcut method called the "prime factorization method" that can be used to calculate cube roots. This method involves finding the prime factorization of the number and then grouping the factors in sets of three. The cube root is then obtained by taking one factor from each group. However, this method is more suitable for larger numbers and may not be practical for smaller numbers.
## CSAT Preparation
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Time remaining:
How wide is the alley?
Mathematics Tutor: None Selected Time limit: 1 Day
Say you have a building on each side of a level alley. You place a 10 foot ladder at the base of one building and lean it against the other building. Then you place a 14 foot ladder at the base of the other building and lean it across to the opposite building. The ladders cross each other at a vertical distance of 5 feet from the surface of the alley. Calculate the width of the alley.
Apr 27th, 2015
Let x1 = the angle the 10 ft ladder makes with the ground
Let x2 = the angle the 14 ft ladder makes with the ground
Then cosx1 = width of the alley / 14
cosx2 = width of the alley / 10
therefore,
14cosx1 = 10cosx2
Where the ladders cross is 5 ft off the ground. Draw a line directly from that point to the ground.
Let y = where the line divides the alley to the left
then, w - y = where the line divides the line to the right
therefore,
tanx1 = 5 / y
y = 5 / tanx1
tanx2 = 5 / w - y
wtanx2 - ytanx2 = 5
ytanx2 = wtanx2 - 5
y = wtanx2 / tanx2 - 5 / tanx2
y = w - 5 / tanx2
therefore,
5/ tanx1 = w - 5 / tanx2
w = 5 / tanx1 + 5 / tanx2
I'm going to need more time to figure this one out, if you'll let me.
Apr 27th, 2015
OK Vincent, take more time to figure it out.
Apr 27th, 2015
Ok, thanks.
Apr 27th, 2015
Fortunately, it turns out the relationship from the beginning
14cosx1 = 10cosx2 is easily solved as
14cos(60) = 10cos(45)
therefore, x1 = 60 degrees
and x2 = 45 degrees
therefore,
w = 5 / tan(60) + 5 / tan(45)
= 5 / 1.732 + 5 / 1
= 2.887 + 5 = 7.887 ft
Apr 27th, 2015
How did you determine that the angles were 45 degrees and 60 degrees?
Apr 27th, 2015
If you draw a diagram of the two ladders leaning up against building, you'll see that you can establish two relationships right away.
Let x1 = the angle the 10 ft ladder makes with the ground
Let x2 = the angle the 14 ft ladder makes with the ground
I hope you can see that, since I can't draw it here. If you're having trouble seeing that then I can draw it and attach it here.
Then,
cosx1 = width of the alley / 14
cosx2 = width of the alley / 10
Therefore,
width of the alley = 14cosx1
and, width of the alley = 10cosx2
that means,
14cosx1 = 10cosx2
By trial and error I found very quickly that x1 = 60 and x2 = 45, because
14cos(60) = 10cos(45) check it out
once you determine those two angles the rest is easier to solve.
Apr 27th, 2015
Is there a way of calculating those two angles with out using trial and error? Using the trial and error method, I find that the two angles do not come out to be exactly 60 and 45 degrees.
Apr 27th, 2015
But they are within 7/1000. That's close enough. Really, this problem is much harder than it seems using other methods. You wind up getting 4th degree polynomials. It's a lot more work.
Apr 27th, 2015
I still do not see where the angles would be 60 and 45 degrees. Suppose the ladders (diagonals) were 21.5 feet and 14 feet long, how would you determine the angles?
May 14th, 2015
The angles depend on the ratio of the ladders. If you change the lengths, you change the ratios. Then the angle calculation changes too. Then it would be 21.5cosx1 = 14cosx2. You get a different set of angles.
May 14th, 2015
...
Apr 27th, 2015
...
Apr 27th, 2015
Dec 5th, 2016
check_circle
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# How do you find the center and radius of the circle x^2-12x+y^2+4y+15=0?
Dec 18, 2016
The center is $\left(6 , - 2\right)$ and the radius $= 5$
#### Explanation:
We complete the squares and rearrange the equation
${x}^{2} - 12 x + {y}^{2} + 4 y = - 15$
${x}^{2} - 12 x + 36 + {y}^{2} + 4 y + 4 = - 15 + 36 + 4$
And now we factorise
${\left(x - 6\right)}^{2} + {\left(y + 2\right)}^{2} = 25 = {5}^{2}$
We compare this equation to the standard equation of a circle
${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
The centre is $\left(a , b\right)$ and the radius $= r$
In our case,
The center is $\left(6 , - 2\right)$ and the radius $= 5$
graph{x^2-12x+y^2+4y+15=0 [-6.22, 16.275, -6.95, 4.3]}
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DISCOVER
# An easy way to learn times tables
Updated April 17, 2017
Multiplication times tables present a challenge to some people, as those who are less mathematically inclined rack their brains trying to find ways to memorise answers. If you have trouble with times tables, the following tricks will help you memorise or solve the answers for basic times tables for the numbers 1 to 12.
## Basics
When learning times tables, it does not matter in which order you multiply. Whether it's 2 times 4 or 4 times 2, the answer does not change.
If you are great at addition, perhaps using that function will help you become more comfortable with times tables. Rather than thinking of an equation as 8 times 6, think of it as either 6 groupings of 8 or 8 groupings of 6, and add them together. When first trying to master times tables, this may take less time than trying to memorise every equation in a times table.
If you want to take the memorisation route, you can learn times tables in chunks. You should use a chart/worksheet that lays all the answers for you for numbers 1 through 12 (numbers 1 to 12 running vertically on the page and also horizontally, so you can cross-reference the information). You can skip memorising 1 times tables, as when you multiply any number by 1, the number does not change (1 times 2 is 2, 1 times 3 is 3, 1 times 4 is 4, and so forth). Start with the 2 times table and work your way up to the 10 times table.
## Patterns For Memorization
To help you memorise times tables, you should learn some patterns. When you multiply a number by 2, you are doubling that number. So, when you multiply 2 times 7, you are adding 7 and 7 together to get 14. Adding 9 to itself is the same as multiplying 9 times 2. Starting with 1, the pattern is that with every number you multiply by 2, your answers are every other number--2, 4, 6, 8, 10 and so on.
When you multiply a number by 5, the answers alternate between ending with a zero and 5. So 5 times 2 equals 10, 5 times 3 equals 15, 5 times 4 equals 20 and so forth.
Eight and nine times tables also have a pattern. Beginning with 8 times 1, you start with 8, and your answers as you move up the times table are 16, 24, 32, 40, 48, 56, 64, 72 and 80. Notice that the "units" place (the second number) moves like this: 8-6-4-2-0 and then repeats. It's similar with 9 times tables--with each equation's answer, the "units" place decreases 1 number at a time, while the "tens" place (the first number) increases 1 number at a time. Consider: 9 (with an invisible zero in front of it), 18, 27, 36, 45, 54, 63, 72, 81, 90.
Ten times tables should present the least challenge. To get your answer, you simply add a zero after the number you are multiplying by 10. So 5 times 10 equals 50, 7 times 10 equals 70 and so on.
## Times Tables For 11 and 12
Times tables for 11 should not be too difficult to remember either. When multiplying 11 by numbers 2 through 9, you simply put together the two digits of the number you're multiplying. For instance, 11 times 2 equals 22, 11 times 5 equals 55 and 11 times 7 equals 77, up through 11 times 9 equals 99. Once you reach 11 times 10, just remember that the "units" place increases one number at a time--11 times 10 equals 110, 11 times 11 equals 121, 11 times 12 equals 132 and so on.
Times tables for 12 represent the highest number in basic times tables. You already have patterns to memorise 2, 5 and 10 times tables, so that leaves you to memorise 12 times the numbers 3, 4, 6, 7, 8, 9 and 11. There's no consistent pattern for you to memorise these, so you either have to put your multiplication skills to use or think of the answers in terms of addition (if the problem is 12 times 3, you add 3 groups of 12 together to get 36).
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Division of Decimals by Decimals
## Making the divisor a whole number
0%
Progress
Practice Division of Decimals by Decimals
Progress
0%
Division of Decimals by Decimals
Do you enjoy projects?
Most students love to participate in hands-on projects, and the students in Mrs. Andersen’s class aren’t any exception. At the science museum there is a whole section that is a Discovery Center. In the Discovery Center, students can use real objects to work on experiments. Mrs. Andersen has asked her students to bring a notebook and a pencil into the Discovery Center. The students need to keep track of the experiments that they work on. They will each have an opportunity to share their discoveries when they return to the classroom. When Miles enters the Discovery Center he is immediately overwhelmed with all of the options. After looking around, he finally decides to work on an experiment that involves an hour glass. To complete the experiment, Miles needs to figure out how long it takes 1.25 pounds of sand to go through the hour glass. There is bucket of sand that is 6.25 pounds in front of Miles. He has a scale and another bucket to hold the sand he needs for his experiment. Miles needs to complete the experiment as many times as he can with the 6.25 pound bucket of sand. Miles picks up the scoop and begins to sort out the sand. Remember he needs 1.25 pounds of sand each time he does the experiment.
If Miles needs 1.25 pounds of sand, how many times can he complete the experiment if he has a 6.25 pound bucket?
Pretend you are Miles. If you were completing this experiment, how many times could you do it given the amount of sand you have been given and the amount of sand that you need?
In this Concept, you will learn how to work through this experiment to find the solution.
### Guidance
Remember Miles? In the experiment, he is working on dividing up sand. If you were going to complete this problem yourself, you would need to know how to divide decimals by decimals.
How can we divide a decimal by a decimal?
To divide a decimal by a decimal, we have to rewrite the divisor. Remember that the divisor is the number that is outside of the division box. The dividend is the number that is inside the division box.
In this problem, 2.6 is our divisor and 10.4 is our dividend. We have a decimal being divided into a decimal. Whew! This seems pretty complicated. We can make our work simpler by rewriting the divisor as a whole number.
How can we do this?
Think back to the work we did in the last section when we multiplied by a power of ten. When we multiply a decimal by a power of ten we move the decimal point one place to the right.
We can do the same thing with our divisor. We can multiply 2.6 times 10 and make it a whole number. It will be a lot easier to divide by a whole number.
2.6 \begin{align*}\times\end{align*} 10 \begin{align*}=\end{align*} 26
Because we multiplied the divisor by 10, we also need to multiply the dividend by 10. This is the only way that it works to rewrite a divisor.
10.4 \begin{align*}\times\end{align*} 10 \begin{align*}=\end{align*} 104
Now we have a new problem to work with.
What about if we have two decimal places in the divisor?
Now, we want to make our divisor .45 into a whole number by multiplying it by a power of ten. We can multiply it by 100 to make it a whole number. Then we can do the same thing to the dividend.
Here is our new problem and quotient.
Now it is time for you to practice a few. Rewrite each divisor and dividend by multiplying them by a power of ten. Then find the quotient.
#### Example A
\begin{align*}1.2 \overline{)4.8 \;}\end{align*}
Solution: 4
#### Example B
\begin{align*}5.67 \overline{)11.34 \;}\end{align*}
Solution: 2
#### Example C
\begin{align*}6.98 \overline{)13.96 \;}\end{align*}
Solution: 2
Congratulations you have finished the Concept! Now you are ready for the experiment. Here is the original problem once again.
Most students love to participate in hands-on projects, and the students in Mrs. Andersen’s class aren’t any exception. At the science museum there is a whole section that is a Discovery Center. In the Discovery Center, students can use real objects to work on experiments. Mrs. Andersen has asked her students to bring a notebook and a pencil into the Discovery Center. The students need to keep track of the experiments that they work on. They will each have an opportunity to share their discoveries when they return to the classroom. When Miles enters the Discovery Center he is immediately overwhelmed with all of the options. After looking around, he finally decides to work on an experiment that involves an hour glass. To complete the experiment, Miles needs to figure out how long it takes 1.25 pounds of sand to go through the hour glass. There is bucket of sand that is 6.25 pounds in front of Miles. He has a scale and another bucket to put the sand he needs for his experiment.
Miles needs to complete the experiment as many times as he can with the 6.25 pound bucket of sand. Miles picks up the scoop and begins to sort out the sand. Remember he needs 1.25 pounds of sand each time he does the experiment.
If Miles needs 1.25 pounds of sand, how many times can he complete the experiment if he has a 6.25 pound bucket?
Write a division problem.
You can start by multiplying the divisor by a power of ten to rewrite it as a whole number. Do this to the dividend too. Since there are two places in the divisor, we can multiply it by 100 to make it a power of ten.
Next, we divide. Our answer will tell us how many times Miles can complete the hourglass experiment.
Miles can complete the experiment 5 times using 1.25 pounds of sand from his 6.25 pound bucket.
### Vocabulary
Divisor
the number doing the dividing, it is found outside of the division box.
Dividend
the number being divided. It is found inside the division box.
Quotient
the answer in a division problem
### Guided Practice
Here is one for you to try on your own.
\begin{align*}3.45 \overline{)7.245 \;}\end{align*}
The first thing to do is to make 3.45 a whole number. We can do this by moving the decimal point two places to the right. If we do this in the divisor, we also have to do this in the dividend.
Next, we divide.
The answer is \begin{align*}2.1\end{align*}.
### Practice
Directions: Divide the following decimals.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
### Vocabulary Language: English
Dividend
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Quotient
Quotient
The quotient is the result after two amounts have been divided.
|
### Percentage: Formulas, Tricks, Examples
Percentage is a way of expressing a number, especially a ratio, as a fraction of 100. The word is derived from the Latin per centum meaning ‘by the hundred’. It is often denoted using the percent sign, ‘%’, or the abbreviation ‘pct.’ For example, 35% (read as ‘thirty-five percent’) is equal to 35/100, or 0.35.
### Quicker Methods to Solve the Problems
For converting a fraction or a decimal to a Percentage, multiply it by hundred
Example 1: Convert the fraction 3/5 into percent fraction.
Solution: % = 60%
Example 2: Convert the fraction 3.5/100 into percent fraction
Solution: % = 3.5%
For converting a percentage to a fraction or decimal, divide by hundred.
Example 3: What is the fraction of 60%?
Solution:
If price of a commodity is increased by x%, the consumption should be reduced, so that the expense remains the same, by
× 100%.
Example 4: If the price of sugar is increased by 25%, find how much percent a family must reduce their consumption of sugar so as not to increase the expenditure of the family?
Solution: Reduction in consumption of sugar
% = 20%
If A is x % greater than B, then B will be % lesser than A.
Example 5: If the price of Kerosene oil falls by 10%, find how much perecent can a householder increase its consumption, so as not to decrease expenditure on this item?
Solution: Increase in consumption of Kerosene oil
% = 11.11%
If price of a commodity is decreased by x %, the consumption can be increased, so that the expense remains the same, by
%
Example 6: If Ravi’s salary is 50% more than that of Gopal’s, then how much percent is Gopal’s salary less than that of Ravi’s salary?
Solution: Gopal’s salary is less than that of Ravi’s by
% = %
If A is x % greater than B, then B will be % lesser than A.
Example 7: If income of Rekha is 30% less than that of Vina, then how much percent is Vina’s income more than that of Rekha?
Solution: Vina’s income is more than that of Rekha by
%
%
### Population Formula
If the original population of a town is P, and the annual increase is r %, then the population after n years is and population before n years =
If the annual decrease be r %, then the population after n years is and population before n years =
Example 8: The population of a certain town increased at a certain rate per cent per annum. Now it is 456976. Four years ago, it was 390625. What will it be 2 years hence?
Solution: Suppose the population increases at r% per annum.
Then, = 456976
∴
Population 2 years hence = 456976
= 456976 × = 494265 approximately.
### First Increase and then decrease
If the value is first increased by x % and then decreased by y% then there is increase or decrease, according to the +ve or –ve sign respectively.
Example 9: A number is increased by 10%. and then it is decreased by 10%. Find the net increase or decrease per cent.
Solution: % change =
i.e., 1% decrease.
Average percentage rate of change over a period.
× % [where n = period]
The percentage error = × 100%
### Successive Increase or decrease
If the value is increased successively by x % and y % then the final increase is given by
If the value is decreased successively by x % and y % then the final decrease is given by
Example 10: The price of a car is decreased by 10 % and 20 % in two successive years. What per cent of price of a car is decreased after two years?
Solution: Put x = – 10 and y = – 20, then
= – 28%
∴ The price of the car decreases by 28%.
### Student and Marks
The percentage of passing marks in an examination is x%. If a candidate who scores y marks fails by z marks, then the maximum marks
M =
A candidate scoring x % in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more than the minimum required passing marks. Then the maximum marks
.
Example 11: Vishal requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum he could have got?
Solution: If Vishal has 15 marks more, he could have scored 40% marks.
Now, 15 marks more than 185 is 185 + 15 = 200
Let the maximum marks be x, then 40% of x = 200
or
or
Thus, maximum marks = 500
Quicker method:
Maximum marks
=
|
# Proposition: 1.25: Angles and Sides in a Triangle IV
### (Proposition 25 from Book 1 of Euclid's “Elements”)
If two triangles have two sides equal to two sides, respectively, but (one) has a base greater than the base (of the other), then (the former triangle) will also have the angle encompassed by the equal straight lines greater than the (corresponding) angle (in the latter). * Let $ABC$ and $DEF$ be two triangles having the two sides $AB$ and $AC$ equal to the two sides $DE$ and $DF$, respectively (That is), $AB$ (equal) to $DE$, and $AC$ to $DF$. * And let the base $BC$ be greater than the base $EF$. * I say that angle $BAC$ is also greater than $EDF$.
### Modern Formulation
If two triangles ($$\triangle{ABC}$$, $$\triangle{DEF}$$) have two sides of one triangle ($$\overline{AB}$$, $$\overline{AC}$$) respectively equal to two sides of the other ($$\overline{DE}$$, $$\overline{DF}$$), where the base of one $$(\overline{BC})$$ is greater than the base of the other $$(\overline{EF})$$, then the angle $$(\angle{BAC})$$ contained by the sides of the triangle with the longer base $$(\triangle{ABC})$$ is greater in measure than the angle $$(\angle{EDF})$$ contained by the sides of the other triangle $$(\triangle{DEF})$$.
In shorter words, if $$\overline{AB}=\overline{DE}$$, $$\overline{AC}=\overline{DF}$$, and $$\overline{BC} > \overline{EF}$$, then $$\angle{BAC} > \angle{EDF}$$.1
Proofs: 1 Corollaries: 1
Proofs: 1 2 3
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@Calahan
@Casey
@Fitzpatrick
### References
#### Adapted from CC BY-SA 3.0 Sources:
1. Callahan, Daniel: "Euclid’s 'Elements' Redux" 2014
|
1.
$$\sqrt{{x}^{2}}$$
Solución
2.
$$\sqrt{162}$$
Solución
3.
$$\sqrt{720}$$
Solución
4.
$$\frac{\sqrt{20}}{2}$$
Solución
5.
$$\frac{\sqrt{1.44}}{2}$$
Solución
6.
$$\sqrt{{(2\times 3)}^{2}-4\times {3}^{2}}$$
Solución
7.
$${25}^{0.5}$$
Solución
8.
$$\sqrt[3]{64}$$
Solución
9.
$$\sqrt[3]{125}$$
Solución
10.
$$\sqrt[3]{4}\sqrt[3]{4}\sqrt[3]{4}$$
Solución
# Square and Cube Roots - Introduction
Square and cube roots — also known as radicals — are powerful concepts in mathematics. It is crucial in the concept of standard deviation in the field of probability theory and statistics. It is also critical in the formula for roots of a quadratic equation.
Taking the square or cube root is the opposite of squaring or cubing, which occurs when numbers are multiplied by themselves (squared), or multiplied by themselves twice (cubed).
# How to Find Roots
For example, if you see the square root
$$\sqrt{100}$$
, you are looking for the number that yields
$$100$$
when multiplied by itself. The answer is
$$10$$
, because
$$10\times 10=100$$
. This is called a perfect square because
$$10$$
is a whole number rather than a decimal. The same applies to cube roots: The root of
$$\sqrt[3]{27}$$
is
$$3$$
because
$$3\times 3\times 3=27$$
. Similarly, this is a perfect cube, because the answer is a whole number.
Yet, what happens if you are dealing with a square or cube root that does not yield a whole number? Let’s look at how we can simplify the root in these cases.
# How to Simplify Roots
Let's use a method that involves prime factors.
Example: Simplify the square root
$$\sqrt{162}$$
.
Solution: Start by rewriting the radical as its prime factors, which are
$$\sqrt{2\times 3\times 3\times 3\times 3}$$
. Then, group the same prime factors into pairs, then rewrite them as squares to give
$$\sqrt{{3}^{2}\times {3}^{2}\times 2}$$
. Next, use
$$\sqrt{{x}^{2}}=x$$
to simplify the root and give
$$(3\times 3)\sqrt{2}$$
. Finally, simplify again to produce
$$9\sqrt{2}$$
.
For more detail, see the full solution here.
# What's Next
We hope you now understand the basic concept of roots, as well as the basic techniques to find roots and to simplify them. You can also try other practice problems to hone your skills and on other math topics. Have a difficult homework problem about square and cute roots? Try Cymath’s square and cube root calculator - you can simply enter your number and the app will provide detailed solution steps.
At Cymath, we have designed our app to help students succeed with clear, easy-to-understand solutions, so that they can get the explanations and skills they need to improve overall math ability and confidence. You can also get more help with Cymath Plus.
|
# 11.1 Systems of linear equations: two variables (Page 7/20)
Page 7 / 20
Meal tickets at the circus cost $\text{\hspace{0.17em}}\text{}4.00\text{\hspace{0.17em}}$ for children and $\text{\hspace{0.17em}}\text{}12.00\text{\hspace{0.17em}}$ for adults. If $\text{\hspace{0.17em}}1,650\text{\hspace{0.17em}}$ meal tickets were bought for a total of $\text{\hspace{0.17em}}\text{}14,200,$ how many children and how many adults bought meal tickets?
Access these online resources for additional instruction and practice with systems of linear equations.
## Key concepts
• A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
• The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See [link] .
• Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
• One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See [link] .
• Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See [link] .
• A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See [link] .
• It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See [link] , [link] , and [link] .
• Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See [link] .
• The solution to a system of dependent equations will always be true because both equations describe the same line. See [link] .
• Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See [link] and [link] .
## Verbal
Can a system of linear equations have exactly two solutions? Explain why or why not.
No, you can either have zero, one, or infinitely many. Examine graphs.
If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins.
If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company?
This means there is no realistic break-even point. By the time the company produces one unit they are already making profit.
If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point?
Given a system of equations, explain at least two different methods of solving that system.
You can solve by substitution (isolating $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ ), graphically, or by addition.
## Algebraic
For the following exercises, determine whether the given ordered pair is a solution to the system of equations.
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
|
# How do you get a vector from the magnitude? My teacher said that you should know how to change a magnitude to a vector, but I have no idea how to do it! Please give me an example with explanation!
krishna-agrawala | Student
I do not know in what context your teacher was talking about changing magnitude to a vector. So I will describe here some fundamental concepts associated with vectors.
All quantities are classified in two groups - scalars and vectors. Scalar quantities have only magnitude, for example, mass, length, time, volume, and speed.
Vector quantities have magnitude as well as direction. Some example of vector quantities are force, velocity and acceleration.
Perhaps the best way to understand the difference between scalar and vector quantities is to understand the difference between speed and velocity. When we are talking about the fastest ever train, it is enough to say that the speed of the train is 500 kilometer per hour. It is not necessary for us to know the direction of train's movement. However if we wanted to study the path taken by a cannon ball launched fired from a gun, it becomes important to know the angle and direction in which the cannon was fired.
When we want to add or subtract scalars quantity there is no concept of direction. If weight of a truck is 1 ton and it is carrying a consignment of 10 tons. the combined weight of the truck and the consignment is the addition of the two weights. This comes to 11 tons. But if we we were to calculate the rate of movement of an ant moving within the truck. We will need to know the direction of the ant's movement in relation to the direction of movement of the truck. If both are moving in the same direction, the net rate of movement or the magnitude of velocity will be addition of the two velocities of the truck and the ant. If they are moving in opposite direction the net velocity will be obtained by subtraction. If their movements are in some other directions then the net velocity cannot be obtained by simple addition or subtraction. We will need to use the method of vector addition.
A vector is often represented by a straight line. The length of the line represents the magnitude of the vector, and and the direction of line, marked by an arrow represents the direction of the vector.
neela | Student
A vector has two characteristics: (i) Magnitude and (ii) direction.
A force of 3 Newton in eastern direction and another force of 4newton in northern direction has resultant of magnitude in the direction of the diagonal of the rectangle of the sides 3 and 4 and is equal to 5.
We say a force of magnitude F in a particular direction has an effect of F*cosx along a direction of angle x with the force F.
|
Question Video: Determining the Limit of a Function from Its Graph at a Point of Jump Discontinuity | Nagwa Question Video: Determining the Limit of a Function from Its Graph at a Point of Jump Discontinuity | Nagwa
# Question Video: Determining the Limit of a Function from Its Graph at a Point of Jump Discontinuity Mathematics • Second Year of Secondary School
## Join Nagwa Classes
Determine the limit of the function as π₯ βΆ 2.
03:10
### Video Transcript
Determine the limit of the function as π₯ approaches two.
Weβre given a graph of a function π of π₯. We need to determine the limit of this function as π₯ is approaching two. If we were to try and do this directly from our graph, we would arrive at a problem. If we take values of π₯ close to two but less than two, we can see that our outputs are less than zero. However, if we take our values of π₯ close to two but greater than two, we can see that our outputs are bigger than three. And we know that the outputs of this function canβt be approaching both zero and three.
So to answer this question, weβre going to need to recall the relationship between left-hand and right-hand limits and the regular limits. We recall if the limit as π₯ approaches two from the left of π of π₯ is equal to some finite value of πΏ and the limit as π₯ approaches two from the right of π of π₯ is also equal to πΏ, then we must have the limit as π₯ approaches two of π of π₯ is also equal to πΏ. In fact, this works in reverse. If we know the limit as π₯ approaches two of π of π₯ is equal to some finite value of πΏ, then both the limit as π₯ approaches two from the left of π of π₯ and the limit as π₯ approaches two from the right of π of π₯ must also be equal to πΏ.
But we need to remember what happens if these conditions are not met. If the left-hand limit and the right-hand limit are not equal, then we say the limit as π₯ approaches two of π of π₯ does not exist. So to answer this question, we can look at the left- and right-hand limit individually. Letβs start with the limit as π₯ approaches two from the left of π of π₯. Our values of π₯ are approaching two from the left, so π₯ will be less than two. We want to see what happens to our outputs of π of π₯ as π₯ gets closer and closer to two from the left.
From our graph, we can see that π of negative three is equal to negative five. Moving closer to two, we can see when π₯ is equal to negative one, our output π of negative one is equal to negative three. Moving closer still, when π₯ is equal to one, we can see our function outputs negative one. And in fact, from this sketch, we can see as π₯ approaches two from the left, our output values are getting closer and closer to zero. So as π₯ approached to two from the left, our outputs π of π₯ go closer and closer to zero. This is the same as saying the limit as π₯ approaches two from the left of π of π₯ is equal to zero.
We can do exactly the same as π₯ approaches two from the right. Letβs start with π₯ is equal to seven. We can see from our graph that π of seven is equal to eight. From our graph, we can also see that π of five is equal to six. When π₯ is equal to three, we can see our function outputs four. And as we get closer and closer to two from the right, we can see that our outputs are getting closer and closer to three.
So from our graph, we can conclude the limit as π₯ approaches two from the right of π of π₯ is equal to three. But this means weβve shown that our limit as π₯ approached to two from the left was not equal to our limit as π₯ approached to two from the right. So this means that the limit as π₯ approaches two of π of π₯ must not exist.
Therefore, by looking at the graph of this function, we were able to determine the limit as π₯ approached to two from the left of π of π₯ was equal to zero and the limit as π₯ approached two from the right of π of π₯ was equal to three. And then, because these two values were not equal, we were able to conclude that the limit as π₯ approaches two of π of π₯ does not exist.
## Join Nagwa Classes
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# PROPERTIES OF PARALLELOGRAMS WORKSHEET
Problem 1 :
In the parallelogram given below, find ∠B, ∠C and ∠D.
Problem 2 :
In the parallelogram ABCD given below, find ∠A, ∠B, ∠C and ∠D.
Problem 3 :
In the parallelogram given below, find the measures of ∠ABO and ∠ACB.
Problem 4 :
The perimeter of the parallelogram ABCD is 30 units and the length of the side AB is 9 units, find the length of other sides of the parallelogram.
Problem 5 :
In the parallelogram given below, find the value of x, measures of ∠A and ∠C.
Problem 6 :
In the parallelogram given below,
AO = x + 40
OC = 2x + 18
Find the length of AO and OC.
Problem 7 :
In two adjacent angles of a parallelogram, if one angle is four times of the other, then find the measures of the two angles.
Problem 8 :
In the parallelogram given above, find the lengths of the sides GJ and HI (in cm).
Problem 9 :
In the parallelogram given below, find the values of x and y.
Problem 10 :
In the parallelogram given below, find the values of x and y.
Problem 1 :
In the parallelogram given below, find ∠B, ∠C and ∠D.
Solution :
In a parallelogram, adjacent angles are supplementary.
In the above parallelogram, ∠A and ∠B are adjacent angles.
So, we have
∠A + ∠B = 180°
65° + ∠B = 180°
∠B = 180° - 65°
∠B = 115°
Because opposite angles are congruent, we have
∠C = ∠A∠C = 65° ∠D = ∠B∠D = 115°
Hence, the measures of ∠B, ∠C and ∠D are 115°, 65° and 115° respectively.
Problem 2 :
In the parallelogram ABCD given below, find ∠A, ∠B, ∠C and ∠D.
Solution :
In a parallelogram, adjacent angles are supplementary.
In the above parallelogram, ∠A and ∠B are adjacent angles.
So, we have
x + 2x = 180°
3x = 180°
x = 60°
The measure of angle ∠A is
= x
= 60°
The measure of angle ∠B is
= 2x
= 2 ⋅ 60°
= 120°
According to the properties of parallelogram, the opposite angles are congruent.
So, we have
∠C = ∠A∠C = 60° ∠D = ∠B∠D = 120°
Hence, the measures of ∠A, ∠B, ∠C and ∠D are 60°, 120°, 60° and 120° respectively.
Problem 3 :
In the parallelogram given below, find the measures of ∠ABO and ∠ACB.
Solution :
In the parallelogram given above ∠AOB and ∠COD are vertically opposite angles.
Because vertically opposite angles are equal, we have
∠AOB = ∠COD
∠AOB = 105°
In triangle ABO, we have
∠OAB + ∠AOB + ∠ABO = 180°
Plug ∠OAB = 30° and ∠AOB = 105°.
30° + 105° + ∠ABO = 180°
135° + ∠ABO = 180°
∠ABO = 45°
In the parallelogram given above, AD||BC, AC is transversal and ∠OCB and ∠OAD are alternate interior angles.
If two parallel lines are cut by a transversal, alternate interior angles are equal.
So, we have
In the parallelogram given above, ∠OAD = 45°.
So, we have
∠OCB = 45°
Because ∠OCB ≅ ∠ACB, we have
∠ACB = 45°
Hence, the measures of ∠ABO and ∠ACB are 45° each.
Problem 4 :
The perimeter of the parallelogram ABCD is 30 units and the length of the side AB is 9 units, find the length of other sides of the parallelogram.
Solution :
Given : Perimeter of the parallelogram is 30 units.
That is,
AB + BC + CD + AD = 30 -----> (1)
Because it is parallelogram, length of opposite sides must be equal.
So, we have
AB = CD
Because AB = 9 units and AB = CD, we can have
Then, we have
(1)-----> 9 + BC + 9 + AD = 30
18 + BC + AD = 30
Subtract 18 from both sides.
Because AD = BC, we can have
Divide both sides by 2.
Then, the length of CD is also 6 units.
Hence, the length of CD is 9 units, AD and BC are 6 units each.
Problem 5 :
In the parallelogram given below, find the value of x, measures of ∠A and ∠C.
Solution :
According to the properties of parallelogram, opposite angles are equal.
So, we have
∠B = ∠D
(x + 29)° = 87°
x + 29 = 87
x = 58
In a parallelogram, adjacent angles are supplementary.
So, we have
∠D + ∠C = 180°
87° + ∠C = 180°
∠C = 93°
In a parallelogram, opposite angles are equal
So, we have
∠A = ∠C
∠A = 93°
Hence, the measures of ∠A and ∠C are 93° each.
Problem 6 :
In the parallelogram given below,
AO = x + 40
OC = 2x + 18
Find the length of AO and OC.
Solution :
According to the properties of parallelogram, the diagonals bisect each other.
So, we have
AO = OC
x + 40 = 2x + 18
2x - x = 40 - 18
x = 22
The length of AO is
AO = x + 40
AO = 22 + 40
AO = 62
The length of OC is
OC = 2x + 18
OC = 2⋅ 22 + 18
OC = 44 + 18
OC = 62
Hence, the lengths of AO and OC is 62 units each.
Problem 7 :
In two adjacent angles of a parallelogram, if one angle is four times of the other, then find the measures of the two angles.
Solution :
Let "x" be one of the angles.
Then, the adjacent angle of x is 4x.
In a parallelogram, adjacent angles are supplementary.
So, we have
x + 4 x = 180°
5x = 180°
Divide both sides by 5.
x = 36°
Then, the measure of the adjacent angle is
= 4x
= 4 ⋅ 36°
= 144°
Hence, the measures of the two adjacent angles are 36° and 144°.
Problem 8 :
In the parallelogram given below, find the lengths of the sides GJ and HI (in cm).
Solution :
According to the properties of parallelogram, the length of opposite sides are equal.
Length of GJ = Length of HI
x + 44 = 5x
44 = 4x
11 = x
The length of HI is
= 5x
= 5 ⋅ 11
= 55
Because opposite sides are equal, the length of GJ is also 55 units.
Hence, the lengths of GJ and HI is 55 units each.
Problem 9 :
In the parallelogram given below, find the values of x and y.
Solution :
According to the properties of parallelogram, the diagonals of a parallelogram bisect each other.
From the one of the diagonals, we have
x + y = 2y - 2
x = y - 2 -----> (1)
From the other diagonal, we have
3x = 2y -----> (2)
Plug x = y - 2 in (2).
(2)-----> 3(y - 2) = 2 y
3y - 6 = 2y
y = 6
Plug y = 6 in (1).
(1)-----> x = 6 - 2
x = 4
Hence, the values of x is 4 and y is 6.
Problem 10 :
In the parallelogram given below, find the values of x and y.
Solution :
In the parallelogram given above, the measure of angle Y is
∠Y = 45° + 70°
∠Y = 115°
In a parallelogram, adjacent angles are supplementary.
Because ∠F and ∠Y are supplementary, we have
∠F + ∠Y = 180°
Plug ∠F = 7x - 5 and ∠Y = 115°
7x - 5 + 115 = 180
7x + 110 = 180
7x = 70
x = 10
The measure of angle ∠F is
= (7x - 5)°
= (7 ⋅ 10 - 5)°
= (70 - 5)°
= 65°
In a parallelogram, opposite angles are equal.
So, we have
∠D = ∠F
(5y)° = 65°
5y = 65
y = 13
Hence, the value of x is 10 and y is 13.
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