Articles

Section 7.3 Answers - Mathematics


1. (y = 2 − 3x − 2x^{2} + frac{7}{2}x^{3} − frac{55}{12}x^{4} + frac{59}{8}x^{5} − frac{83}{6}x^{6} + frac{9547}{ 336}x^{7} + ldots)

2. (y = −1 + 2x − 4x^{3} + 4x^{4} + 4x^{5} − 12x^{6} + 4x^{7} + ldots )

3. (y = 1 + x^{2} − frac{2}{3}x^{3} + frac{11}{6}x^{4} − frac{9}{5} x^{5} + frac{329}{90}x^{6} − frac{1301}{315} x^{7} + dots)

4. (y = x − x^{2} − frac{7}{2} x^{3} + frac{15}{2} x^{4} + frac{45}{8} x^{5} − frac{261}{8}x^{6} + frac{207}{16} x^{7} + ldots)

5. (y = 4 + 3x − frac{15}{4} x^{2} + frac{1}{4} x^{3} + frac{11}{16} x^{4} − frac{5}{16} x^{5} + frac{1}{20} x^{6} + frac{1}{120} x^{7} + ldots)

6. (y = 7 + 3x − frac{16}{3} x^{2} + frac{13}{3} x^{3} − frac{23}{9} x^{4} + frac{10}{9} x^{5} − frac{7}{27} x^{6} − frac{1}{9} x{7} + ldots)

7. (y = 2 + 5x − frac{7}{4} x^{2} − frac{3}{16} x^{3} + frac{37}{192} x^{4} − frac{7}{192} x^{5} − frac{1}{1920} x^{6} + frac{19}{11520} x^{7} + ldots)

8. (y = 1 − (x − 1) + frac{4}{3} (x − 1)^{3} − frac{4}{3} (x − 1)^{4} − frac{4}{5} (x − 1)^{5} + frac{136}{45} (x − 1)^{6} − frac{104}{63} (x − 1)^{7} + ldots )

9. (y = 1 − (x + 1) + 4(x + 1)^{2} − frac{13}{3} (x + 1)^{3} + frac{77}{6} (x + 1)^{4} − frac{278}{15} (x + 1)^{5} + frac{1942}{45} (x + 1)^{6} − frac{23332}{315} (x + 1)^{7} + ldots)

10. (y = 2 − (x − 1) − frac{1}{2} (x − 1)^{2} + frac{5}{3} (x − 1)^{3} − frac{19}{12} (x − 1)^{4} + frac{7}{30} (x − 1)^{5} + frac{59}{45} (x − 1)^{6} − frac{1091}{630} (x − 1)^{7} + ldots)

11. (y = −2 + 3(x + 1) − frac{1}{2}(x + 1)^{2} − frac{2}{3}(x + 1)^{3} + frac{5}{8}(x + 1)^{4} − frac{11}{30} (x + 1)^{5} + frac{29}{144} (x + 1)^{6} − frac{101}{840 }(x + 1)^{7} + ldots)

12. (y = 1 − 2(x − 1) − 3(x − 1)^{2} + 8(x − 1)^{3} − 4(x − 1)^{4} − frac{42}{5}(x − 1)^{5} + 19(x − 1)^{6} − frac{604}{35} (x − 1)^{7} + ldots)

19. (y = 2 − 7x − 4x^{2} − frac{17}{6}x^{3} − frac{3}{4}x^{4} − frac{9}{40}x^{5} + ldots)

20. (y = 1 − 2(x − 1) + frac{1}{2} (x − 1)^{2} − frac{1}{6}(x − 1)^{3} + frac{5}{36}(x − 1)^{4} − frac{73}{1080}(x − 1)^{5} + ldots)

21. (y = 2 − (x + 2) −frac{7}{2}(x + 2)^{2} +frac{4}{3}(x + 2)^{3} −frac{1}{24}(x + 2)^{4} +frac{1}{60}(x + 2)^{5} + ldots)

22. (y = 2 − 2(x + 3) − (x + 3)^{2} + (x + 3)^{3} −frac{11}{12}(x + 3)^{4} +frac{67}{60}(x + 3)^{5} +ldots)

23. (y = −1 + 2x + frac{1}{3}x^{3} −frac{5}{12}x^{4} + frac{2}{5}x^{5} + ldots)

24. (y = 2 − 3(x + 1) + frac{7}{2} (x + 1)^{2} − 5(x + 1)^{3} + frac{197}{24}(x + 1)^{4} − frac{287}{20}(x + 1)^{5} + ldots)

25. (y = −2 + 3(x + 2) − frac{9}{2}(x + 2)^{2} +frac{11}{6}(x + 2)^{3} +frac{5}{24}(x + 2)^{4} +frac{7}{20}(x + 2)^{5} + ldots)

26. (y = 2 − 4(x − 2) − frac{1}{2}(x − 2)^{2} + frac{2}{9}(x − 2)^{3} + frac{49}{432}(x − 2)^{4} + frac{23}{1080} (x − 2)^{5} + ldots)

27. (y = 1 + 2(x + 4) − frac{1}{6}(x + 4)^{2} −frac{10}{27}(x + 4)^{3} +frac{19}{648}(x + 4)^{4} +frac{13}{324}(x + 4)^{5} + ldots)

28. (y = −1 + 2(x + 1) −frac{1}{4}(x + 1)^{2} +frac{1}{2}(x + 1)^{3} −frac{65}{96}(x + 1)^{4} + frac{67}{80}(x + 1)^{5} + ldots)

31.

  1. (y=frac{c_{1}}{1+x}+frac{c_{2}}{1+2x})
  2. (y=frac{c_{1}}{1-2x}+frac{c_{2}}{1-3x})
  3. (y=frac{c_{1}}{1-2x}+frac{c_{2}x}{(1-2x)^{2}})
  4. (y=frac{c_{1}}{2+x}+frac{c_{2}x}{(2+x)^{2}})
  5. (y=frac{c_{1}}{2+x}+frac{c_{2}}{2+3x})

32. (y = 1 − 2x −frac{3}{2} x^{2} + frac{5}{3} x^{3} + frac{17}{24} x^{4} − frac{11}{20} x^{5} + ldots)

33. (y = 1 − 2x − frac{5}{2} x^{2} + frac{2}{3} x^{3} − frac{3}{8} x^{4} + frac{1}{3} x^{5} + ldots)

34. (y = 6 − 2x + 9x^{2} + frac{2}{3} x^{3} − frac{23}{4} x^{4} − frac{3}{10} x^{5} + ldots)

35. (y = 2 − 5x + 2x^{2} − frac{10}{3} x^{3} + frac{3}{2} x^{4} − frac{25}{12} x^{5} + ldots)

36. (y = 3 + 6x − 3x^{2} + x^{3} − 2x^{4} − frac{17}{20} x^{5} + ldots)

37. (y = 3 − 2x − 3x^{2} + frac{3}{2} x^{3} + frac{3}{2} x^{4} − frac{49}{80} x^{5} + ldots)

38. (y = −2 + 3x + frac{4}{3} x^{2} − x^{3} − frac{19}{54} x^{4} + frac{13}{60} x^{5} + ldots)

39. (y_{1}=sum_{m=0}^{infty}frac{(-1)^{m}x^{2m}}{m!}=e^{-x^{2}},:y_{2}=sum_{m=0}^{infty}frac{(-1)^{m}x^{2m+1}}{m!}=xe^{-x^{2}})

40. (y = −2 + 3x + x^{2} − frac{1}{6} x^{3} −frac{3}{4}x^{4} + frac{31}{120}x^{5} + ldots)

41. (y = 2 + 3x − frac{7}{2} x^{2} − frac{5}{6} x^{3} + frac{41}{24} x^{4} + frac{41}{120} x^{5} + ldots)

42. (y = −3 + 5x − 5x^{2} + frac{23}{6}x^{3} − frac{23}{12}x^{4} + frac{11}{30} x^{5} + ldots)

43. (y = −2 + 3(x − 1) + frac{3}{2}(x − 1)^{2} −frac{17}{12}(x − 1)^{3} −frac{1}{12}(x − 1)^{4} +frac{1}{8}(x − 1)^{5} + ldots)

44. (y = 2 − 3(x + 2) + frac{1}{2}(x + 2)^{2} − frac{1}{3}(x + 2)^{3} + frac{31}{24}(x + 2)^{4} −frac{53}{120}(x + 2)^{5} + ldots)

45. (y = 1 − 2x + frac{3}{2} x^{2} − frac{11}{6} x^{3} + frac{15}{8} x^{4} − frac{71}{60} x^{5} + ldots)

46. (y = 2 − (x + 2) − frac{7}{2}(x + 2)^{2} − frac{43}{6}(x + 2)^{3} − frac{203}{24} (x + 2)^{4} − frac{167}{30} (x + 2)^{5} + ldots)

47. (y = 2 − x − x^{2} + frac{7}{6} x^{3} − x^{4} + frac{89}{120} x^{5} + ldots)

48. (y = 1 + frac{3}{2} (x − 1)^{2} + frac{1}{6} (x − 1)^{3} − frac{1}{8} (x − 1)^{5} + ldots)

49. (y = 1 − 2(x − 3) + frac{1}{2}(x − 3)^{2} − frac{1}{6}(x − 3)^{3} + frac{1}{4}(x − 3)^{4} − frac{1}{6}(x − 3)^{5} + ldots)


The Stacks project

Definition 7.3.1 . Let $mathcal$ be a category, and let $varphi : mathcal o mathcal$ be a map of presheaves of sets.

We say that $varphi $ is injective if for every object $U$ of $mathcal$ the map $varphi _ U : mathcal(U) o mathcal(U)$ is injective.

We say that $varphi $ is surjective if for every object $U$ of $mathcal$ the map $varphi _ U : mathcal(U) o mathcal(U)$ is surjective.

Lemma 7.3.2 . The injective (resp. surjective) maps defined above are exactly the monomorphisms (resp. epimorphisms) of $ extit(mathcal)$. A map is an isomorphism if and only if it is both injective and surjective.

Proof. We shall show that $varphi : mathcal o mathcal$ is injective if and only if it is a monomorphism of $ extit(mathcal)$. Indeed, the “only if” direction is straightforward, so let us show the “if” direction. Assume that $varphi $ is a monomorphism. Let $U in mathop> olimits (mathcal)$ we need to show that $varphi _ U$ is injective. So let $a, b in mathcal(U)$ be such that $varphi _ U (a) = varphi _ U (b)$ we need to check that $a = b$. Under the isomorphism (7.2.1.1), the elements $a$ and $b$ of $mathcal(U)$ correspond to two natural transformations $a', b' in mathop> olimits _< extit(mathcal)>(h_ U, mathcal)$. Similarly, under the analogous isomorphism $mathop> olimits _< extit(mathcal)>(h_ U, mathcal) = mathcal(U)$, the two equal elements $varphi _ U (a)$ and $varphi _ U (b)$ of $mathcal(U)$ correspond to the two natural transformations $varphi circ a', varphi circ b' in mathop> olimits _< extit(mathcal)>(h_ U, mathcal)$, which therefore must also be equal. So $varphi circ a' = varphi circ b'$, and thus $a' = b'$ (since $varphi $ is monic), whence $a = b$. This finishes (1).

We shall show that $varphi : mathcal o mathcal$ is surjective if and only if it is an epimorphism of $ extit(mathcal)$. Indeed, the “only if” direction is straightforward, so let us show the “if” direction. Assume that $varphi $ is an epimorphism.

For any two morphisms $f : A o B$ and $g : A o C$ in the category $ extit$, we let $ ext_$ and $ ext_$ denote the two canonical maps from $B$ and $C$ to $B coprod _ A C$. (Here, the pushout is evaluated in $ extit$.)

Now, we define a presheaf $mathcal$ of sets on $mathcal$ by setting $mathcal(U) = mathcal(U) coprod _(U)> mathcal(U)$ (where the pushout is evaluated in $ extit$ and induced by the map $varphi _ U : mathcal(U) o mathcal(U)$) for every $U in mathop> olimits (mathcal)$ its action on morphisms is defined in the obvious way (by the functoriality of pushout). Then, there are two natural transformations $i_1 : mathcal o mathcal$ and $i_2 : mathcal o mathcal$ whose components at an object $U in mathop> olimits (mathcal)$ are given by the maps $ ext_$ and $ ext_$, respectively. The definition of a pushout shows that $i_1 circ varphi = i_2 circ varphi $, whence $i_1 = i_2$ (since $varphi $ is an epimorphism). Thus, for every $U in mathop> olimits (mathcal)$, we have $ ext_ = ext_$. Thus, $varphi _ U$ must be surjective (since a simple combinatorial argument shows that if $f : A o B$ is a morphism in $ extit$, then $ ext_ = ext_$ if and only if $f$ is surjective). In other words, $varphi $ is surjective, and (2) is proven.

We shall show that $varphi : mathcal o mathcal$ is both injective and surjective if and only if it is an isomorphism of $ extit(mathcal)$. This time, the “if” direction is straightforward. To prove the “only if” direction, it suffices to observe that if $varphi $ is both injective and surjective, then $varphi _ U$ is an invertible map for every $U in mathop> olimits (mathcal)$, and the inverses of these maps for all $U$ can be combined to a natural transformation $mathcal o mathcal$ which is an inverse to $varphi $. $square$

Definition 7.3.3 . We say $mathcal$ is a subpresheaf of $mathcal$ if for every object $U in mathop> olimits (mathcal)$ the set $mathcal(U)$ is a subset of $mathcal(U)$, compatibly with the restriction mappings.

In other words, the inclusion maps $mathcal(U) o mathcal(U)$ glue together to give an (injective) morphism of presheaves $mathcal o mathcal$.

Lemma 7.3.4 . Let $mathcal$ be a category. Suppose that $varphi : mathcal o mathcal$ is a morphism of presheaves of sets on $mathcal$. There exists a unique subpresheaf $mathcal' subset mathcal$ such that $varphi $ factors as $mathcal o mathcal' o mathcal$ and such that the first map is surjective.

Proof. To prove existence, just set $mathcal'(U) = varphi _ U left(mathcal(U) ight)$ for every $U in mathop> olimits (C)$ (and inherit the action on morphisms from $mathcal$), and prove that this defines a subpresheaf of $mathcal$ and that $varphi $ factors as $mathcal o mathcal' o mathcal$ with the first map being surjective. Uniqueness is straightforward. $square$

Definition 7.3.5 . Notation as in Lemma 7.3.4. We say that $mathcal'$ is the image of $varphi $.


Mental Arithmetic 3 Answers

Take a look inside.

Product description for parents

Mental Arithmetic provides rich and varied practice to develop pupils&rsquo essential maths skills and prepare them for all aspects of the Key Stage 2 national tests. It may also be used as preparation for the 11+, and with older students for consolidation and recovery.

Tailored to meet the requirements of the National Curriculum for primary mathematics, each book contains 36 one-page tests designed to build confidence and fluency and keep skills sharp. Each test is presented in a unique three-part format comprising:

  • Part A: questions where use of language is kept to a minimum
  • Part B: questions using number vocabulary
  • Part C: questions focusing on one- and two-step word problems.

Structured according to ability rather than age, the series allows children to work at their own pace, building confidence and fluency. Two Entry Tests are available in the Mental Arithmetic Teacher&rsquos Guide and on the Schofield & Sims website, enabling teachers, parents and tutors to select the appropriate book for each child. All the books can be used flexibly for individual, paired, group or whole-class maths practice, as well as for homework and one-to-one intervention.

Mental Arithmetic 3 Answers contains answers to all the questions included in Mental Arithmetic 3, as well as guidance on how to use the series and mark results. Answers are clearly laid out in the format of a correctly completed pupil book making marking quick and easy, while a card cover ensures durability.


Monday, March 29, 2021

March 29 - April 2nd

No School Thursday-Friday.

Monday: Asia Vocab. Geography, Religion (Maps are due. )

Tuesday: Asia, Government Vietnam, Korea

Wednesday: Asia Escape Room

Tuesday: Finish any Poem and start Poetry Slam

Tuesday: Finish any Poem and start Poetry Slam

Monday: P.R. 1.3 Births and Deaths (Worksheet 1.3)

Tuesday: P.R. 1.4 Jellyfish population

Wednesday: P.R. Activity oh-deer (Worksheet)

Monday: M.3 Lesson 19 (area of different shape)

HW: Problem Set #'s: 1-9 odd, Quizizz (Due Wednesday), IXL's: AA.5, AA.6 to 80%-Due Wed

Tuesday: M.3 Lesson 20 (composite areas)

Wednesday: Quiz on Circles


In geometry it is the shape made when a solid is cut through by a plane.

Example:

The cross section of this circular cylinder is a circle

We don't draw the rest of the object, just the shape made when you cut through.

Example:

The cross section of a rectangular pyramid is a rectangle

Cross sections are usually parallel to the base like above, but can be in any direction.

Example:

The vertical cross section through the center of this torus is two circles!


Detailed Answer Key

Find the area of the figure shown below. 

By drawing a line GD parallel to AB, we can split the given picture as two parts.

(i) GDEF is a square of side length 3 cm

(ii) ABCG is a rectangle of length 10 cm and width 4 cm

Area of rectangle  =  length x breadth

Area of square (GDEF)  =  3 2   =  9 cm 2

Area of rectangle  =  10 x 4  =  40  cm 2

Area of the given figure  =  40 + 9  =  49  cm 2

Find the area of the figure shown below. 

By drawing a line BD parallel to AE, we can split the given picture as two parts.

Area of triangle  =  (1/2) x b x h

Area of rectangle  =  length x breadth

Area of rectangle  =  10 x 5  =  50  cm 2

Area of given shape  =  17.5 + 50  =  67.5  cm 2

John bought a square plot of side 60 m. Adjacent to this David bought a  rectangular plot of dimension 70 m x 50 m. Both paid the same amount. Who is  benefited ?

To find who is benefited, we have to find the area of above shapes separately. 

(i) ABFG is a square of side length 60 m

(ii) BCDE is a rectangle of length 70 m and width 50 m.

Area of rectangle  =  length x breadth

Area of ABFG  =  60 2   =  60 x 60  =  3600 m 2

Area of rectangle  =  70 x 50  =  3500  m 2

From the above calculation, we come to know that John is having more area than David. So John is more benefited.

Find the area of the figure shown below. 

(i)  ABCD is a square of side length 15 cm

(ii)  Area of rectangle FEGH  

Area of square ( ABCD)   =  a 2   =  15 2   =  225 cm 2

Area of rectangle ( FEGH)  =  length x width 

Area of the given figure  =  225 + 140  =  365  cm 2

Daniel bought a square plot of side 50 m. Adjacent to this Richard bought a  rectangular plot of length 60 m and breadth 40 m for the same price. Find out  who is benefited and how many sq. m. are more for him?

Area of land owned by Daniel  =  a 2

Area of land owned by Richard  =  length x width

Because Daniel is having more area than Richard, we can decide that Daniel is benefited.

Daniel is having 100  m 2  more area than Richard.

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

If you have any feedback about our math content, please mail us : 

We always appreciate your feedback. 

You can also visit the following web pages on different stuff in math. 


Math Books - Grade 4

What is Common Core

The Common Core is a set of high-quality academic standards in mathematics and English language arts/literacy (ELA).

These learning goals outline what a student should know and be able to do at the end of each grade.

What Parents Should Know

Today’s students are preparing to enter a world in which colleges and businesses are demanding more than ever before.

To ensure all students are ready for success after high school, the Common Core State Standards establish clear, consistent guidelines for what every student should know and be able to do in math and English language arts from kindergarten through 12th grade.

Common core - Common mistakes

Use the repeated-addition strategy to solve 5x3.

If you answer the question with 5+5+5=15, you would be wrong.

The correct answer is 3+3+3+3+3.

Mathematically, both are correct. But under Common Core, you're supposed to read 5x3 as "five groups of three." So "three groups of five" is wrong.


Order of Operations

The order of operations is a very simple concept, and is vital to correctly understanding math. Unlike reading, where we always work left-to-right, sometimes with math we need to work one part of a problem before another, or the final answer could be incorrect! We use the term "order of operations" to describe which part of the problem needs to be worked first. Take this equation as an example:

If you were to simply solve from left to right, the answer would be incorrect. Let's do that now: 4 + 6 = 10. Divide that by 2 to get 5. Multiply 5 times 11 to get 55. Unfortunately, even though it seemed ok, this answer is wrong.

The correct order of operations

The order of operations will allow you to solve this problem the right way. The order is this: Parenthesis, Exponents, Multiplication and Division, and finally Addition and Subtraction. Always perform the operations inside a parenthesis first, then do exponents. After that, do all the multiplication and division from left to right, and lastly do all the addition and subtraction from left to right.

A popular way of remembering the order is the acronym PEMDAS. Parenthesis, Exponents, Multiply and Divide, Add and Subtract. You can also create a little phrase to go along with this, like "Please Excuse My Dear Aunt Sally." Whatever you choose, make sure that you know all six steps of the order of operations very well.

Let's try solving that equation again, this time using PEMDAS.

Step 1) Parenthesis. There aren't any. Move on.

Step 2) Exponents. None. Keep going.

Step 3) Multiplication and Division. Go from left to right performing all the multiplication and division as you come across it, so divide 6 by 2 to get 3, and multiply that by 11 to get 33.

Step 4) Addition and Subtraction. From left to right, 4 + 33 = 37.

$ 4+6 div 2 * 11 $ $ 4+3*11 $ $ 4+33 $ $ 37 $

The whole idea is just to follow the rule: PEMDAS. Now we can try to solve one with parenthesis and exponents.

What if you are given an equation like the one below? Just simplify it in small steps, using the order of operations at all times.

Remember, the first step is Parenthesis. Look inside the parenthesis, where we have (6*2). That comes out to 12. We can then compute (12^2) and get 144, making this equation easy to finish:

$ 5+(6*2)^2 div 5 $ $ 5+(12)^2 div 5 $ $ 5+144 div 5 $ $ 5+28.8 $ $ 33.8 $


Math Worksheets with Answers - Common Core State Standards (CCSS)

Grade based K-12 math worksheets with answers for common core state standards is available online for free in printable & downloadable (PDF) format to teach, practice or learn 1st, 2nd, 3rd, 4th, 5th & 6th grade mathematics effectively. The Common Core State Standards for Mathematics (CCSSM) was developed by Common Core State Standards Initiative (CCSSI) to enhance K-12 students all the knowledge and skills required to fully participate in the 21st century global economy. All the common core math worksheets provided over RankUpturn with reference to the code standards developed by CCSSI.

Refer the code details such as grade level, domain, clusters, standard number and content standards to select the worksheet of activities on corresponding CCSS codes. The below example shows how to decode CCSS code for mathematics. In 6.OA.A.1,
6 is the grade level.
OA is the Domain represents the major branch of mathematics "Operations and Algebraic Thinking".
A is the Cluster represents the topics covered in the domain ordered by A, B, C and so on.
1 is the standard number.

The grade based common core math worksheets for kindergarten (KG), grade-1, grade-2, grade-3, grade-4, grade-5 and grade-6 increase the student’s ability to apply mathematics in real world problems, conceptual understanding, procedural fluency, problem solving skills, critically evaluate the reasoning or prepare the students to learn the mathematics in the subsequent grade levels. The grade based CCSSM worksheets with answers includes the math domains such as counting and cardinality, number operations in base ten, number and operations of fractions, ratios and proportional relationships, number system, number and quantity, operations and algebraic thinking, expressions and equations, functions, algebra and functions, geometry, measurement and data, statistics and probability. The details of domain, cluster, cluster headings and content standards help you to understand what kind of activities are covered in the worksheets.


Here's The Answer To That Fruit Math Puzzle That's Driving Everyone Coconuts

It's one of those Facebook posts that's all over the place, and -- like "the dress" or the hidden panda -- everyone seems to have a different take on it.

What is half a coconut plus one apple plus three bananas?

On the surface, it would seem the apple has a value of 10, the bananas a value of four and the coconut a value of two. And if that were the end of the story, the answer would be 2 + 10 + 4 = 16.

But in the final frame, the fruit are slightly different.

There is half a coconut, instead of two halves in the earlier example equation that establishes a value of two. And instead of the four bananas representing the value of four, there are just three bananas.

If the numbers are reduced along with the portions of fruit, you get a totally different answer: 1 + 10 + 3 = 14.

Which is the correct answer? Like the dress, the answer is in how you see it.


Watch the video: Q 1 - Ex - Fractions - Chapter 7 - Maths Class 6th - NCERT (November 2021).