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3.7: Exercises(Concepts) - Mathematics


  1. Consider the weighted voting system ([q: 7, 3, 1])
    1. Which values of (q) result in a dictator (list all possible values)
    2. What is the smallest value for (q) that results in exactly one player with veto power but no dictators?
    3. What is the smallest value for (q) that results in exactly two players with veto power?
  2. Consider the weighted voting system ([q: 9, 4, 2])
    1. Which values of (q) result in a dictator (list all possible values)
    2. What is the smallest value for (q) that results in exactly one player with veto power?
    3. What is the smallest value for (q) that results in exactly two players with veto power?
  3. Using the Shapley-Shubik method, is it possible for a dummy to be pivotal?
  4. If a specific weighted voting system requires a unanimous vote for a motion to pass:
    1. Which player will be pivotal in any sequential coalition?
    2. How many winning coalitions will there be?
  5. Consider a weighted voting system with three players. If Player 1 is the only player with veto power, there are no dictators, and there are no dummies:
    1. Find the Banzhaf power distribution.
    2. Find the Shapley-Shubik power distribution
  6. Consider a weighted voting system with three players. If Players 1 and 2 have veto power but are not dictators, and Player 3 is a dummy:
    1. Find the Banzhaf power distribution.
    2. Find the Shapley-Shubik power distribution
  7. An executive board consists of a president (P) and three vice-presidents (left(mathrm{V}_{1}, mathrm{V}_{2}, mathrm{V}_{3} ight)). For a motion to pass it must have three yes votes, one of which must be the president's. Find a weighted voting system to represent this situation.
  8. On a college’s basketball team, the decision of whether a student is allowed to play is made by four people: the head coach and the three assistant coaches. To be allowed to play, the student needs approval from the head coach and at least one assistant coach. Find a weighted voting system to represent this situation.
  9. In a corporation, the shareholders receive 1 vote for each share of stock they hold, which is usually based on the amount of money the invested in the company. Suppose a small corporation has two people who invested $30,000 each, two people who invested $20,000 each, and one person who invested $10,000. If they receive one share of stock for each $1000 invested, and any decisions require a majority vote, set up a weighted voting system to represent this corporation’s shareholder votes.
  10. A contract negotiations group consists of 4 workers and 3 managers. For a proposal to be accepted, a majority of workers and a majority of managers must approve of it. Calculate the Banzhaf power distribution for this situation. Who has more power: a worker or a manager?
  11. The United Nations Security Council consists of 15 members, 10 of which are elected, and 5 of which are permanent members. For a resolution to pass, 9 members must support it, which must include all 5 of the permanent members. Set up a weighted voting system to represent the UN Security Council and calculate the Banzhaf power distribution.

3.7: Exercises(Concepts) - Mathematics

We have taken some pains to note that $_n$ is not a subset of $$, and in particular that $_n=<[0],[1],ldots,[n-1]>$ is not the same as $<0,1,ldots,n-1>$. The two sets certainly are closely related, however $[a]=[b]$ if and only if $a$ and $b$ have the same remainder when divided by $n$, and the numbers in $<0,1,ldots,n-1>$ are precisely all possible remainders&mdashthat's exactly why we chose them to be the "standard'' representatives when we write $_n$. This is all by way of pointing out that anything involving $_n$ can be thought of as being "about'' remainders. The principal result in this section, the Chinese Remainder Theorem, is an interesting fact about the relationship between $_n$ for different values of $n$.

NB (That's Latin for "Pay attention!''): In the next two sections $( , )$ denotes an ordered pair, not a gcd.

Example 3.7.1 Both $_<12>$ and $_3 imes _4$ have 12 elements. In fact, there is a natural way to associate the elements of $_<12>$ and $_3 imes _4$ given by the following: $ egin [0] &leftrightarrow&([0],[0])&& [6]&leftrightarrow&([6],[6])=([0],[2]) [1] &leftrightarrow&([1],[1])&& [7]&leftrightarrow&([7],[7])=([1],[3]) [2] &leftrightarrow&([2],[2])&& [8]&leftrightarrow&([8],[8])=([2],[0]) [3] &leftrightarrow&([3],[3])=([0],[3])&& [9]&leftrightarrow&([9],[9])=([0],[1]) [4]&leftrightarrow&([4],[4])=([1],[0])&& [10]&leftrightarrow&([10],[10])=([1],[2]) [5]&leftrightarrow&([5],[5])=([2],[1])&& [11]&leftrightarrow&([11],[11])=([2],[3]) end $ The relationship used here, $[x]leftrightarrow([x],[x])$, is about the simplest one could imagine, and this is one of those happy circumstances in which the simple, obvious choice is the one that works. Be sure you understand that the whole point of this example is to notice that every pair in $_3 imes _4$ appears exactly once. When two sets are paired up in this way, so that every element of each set appears in exactly one pair, we say that there is a one-to-one correspondence between the sets. Note also that in the expression $[x]leftrightarrow([x],[x])$, the symbol $[x]$ means three different things in the three places it appears, namely, $[x]in_<12>$, $[x]in_<3>$, and $[x]in_<4>$, respectively. This is an example of a general phenomenon. $square$

Theorem 3.7.2 (Chinese Remainder Theorem) Suppose $n=ab$, with $a$ and $b$ relatively prime. For $x=0,1,&hellip, n-1$, associate $[x]in _$ with $([x], [x])in _a imes _b$ (note that the symbol $[x]$ means different things in $_n$, $_a$ and $_b$). This gives a one-to-one correspondence between $_n$ and $_a imes _b$.

Proof. Observe that the two sets have the same number of elements. If we can show that associating $[x]$ with $([x], [x])$ does not associate any two distinct elements of $_n$ with the same ordered pair in $_a imes _b$, then every element of $_n$ will have to be associated with exactly one element of $_a imes _b$, and vice versa.

Suppose that $[x_1]$ and $[x_2]$ are assigned to the same pair in $_a imes _b$. We wish to show that $[x_1]=[x_2]$. We have $ <>equiv x_2 pmod a quad and quad <>equiv x_2 pmod b, $ in other words, both $a$ and $b$ must divide $<>-x_2$. Since $a$ and $b$ are relatively prime, their product, $n$, also divides $<>-x_2$. (See exercise 3.) That is, $<>equiv x_2pmod n$ so $[x_1]=[x_2]$.$qed$

Example 3.7.3 The theorem produces a correspondence between $_<168>$ and $_8 imes _<21>$. This correspondence, for example, takes $[97]$ to $([97], [97])=([1], [13])$. $square$

Given an element of $_n$, it is easy to find the corresponding element of $_a imes _b$: simply reduce modulo $a$ and $b$. Is there a way to reverse this? In other words, given $([y], [z])in _a imes _b$ can we find the element of $_n$ to which it corresponds? In fact, using the ubiquitous Extended Euclidean Algorithm it is easy.

Example 3.7.4 Which element of $_<168>$ corresponds to the pair $([7],[5])in _8 imes _<21>$? If we apply the Extended Euclidean Algorithm to 8 and 21, we get $1=8cdot 8 + (-3)cdot 21$. Note that $8cdot 8=64$ is congruent to 0 mod 8 and 1 mod 21, and $(-3)cdot 21=-63$ is congruent to 1 mod 8 and 0 mod 21. Therefore $ eqalign< 5cdot 64+ 7cdot (-63)equiv 5cdot 0 +7cdot 1& =7 pmod 8,cr 5cdot 64+ 7cdot (-63)equiv 5cdot 1 +7cdot 0& =5 pmod <21>,cr> $ so $5cdot 64+ 7cdot (-63)=-121equiv 47pmod <168>$ works, that is, $[47]leftrightarrow ([7],[5])$. $square$

This last example can be phrased in a somewhat different way. Given $7$ and $5$, we are asking whether the two simultaneous congruences $xequiv 7pmod 8$ and $xequiv 5pmod <21>$ can be solved, that is, is there an integer $x$ that has remainder $7$ when divided by $8$ and remainder $5$ when divided by $21$. This makes the name "Chinese Remainder Theorem'' seem a little more appropriate.

The Chinese Remainder Theorem is a useful tool in number theory (we'll use it in section 3.8), and also has proved useful in the study and development of modern cryptographic systems.


Scale Factor

The number used to multiply the lengths of a figure to stretch or shrink it into a similar image.

A scale factor larger than 1 will enlarge a figure. A scale factor between 0 and 1 will reduce a figure.

The scale factor of two similar figures is given by a ratio that compares the corresponding sides:
length of a side on the image/ length of a side on the original.

Example

If we use a scale factor of 1 /2, all lengths in the images are 1 /2 as long as the corresponding lengths in the original.

The base of the original triangle is 3 units.

The base of the image is 1.5 units.


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Concepts covered in Concise Mathematics Class 7 ICSE chapter 11 Fundamental Concepts (Including Fundamental Operations) are Fundamental Concepts, Performs Operations (Addition and Subtraction) on Algebraic Expressions with Integral Coefficients Only., Terms, Factors and Coefficients of Expression, Algebraic Expressions.

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Triangles

Most of the mathematics in Escher's work is related to geometry. One of the fundamental objects in geometry is the triangle, and in this section we will explore various ways to classify triangles.

First, we introduce two notions of "sameness" for triangles:

Congruent triangles Two triangles are congruent if they have the same side lengths and angle measures. They are the same size and shape. Similar triangles Two triangles are similar if they have the same angle measures. They are the same shape, but may be different sizes.

One simple classification comes from considering angles:

It is not hard to see that every triangle falls into exactly one of these three classes. Every triangle is either acute, obtuse, or right.

Considering side lengths leads to other classes of triangle:

Equilateral triangle An equilateral triangle is one in which all three sides have the same length. Isoceles triangle An isoceles triangle is one in which two sides have the same length. Scalene triangle A scalene triangle is one in which all three sides have different lengths.

This classification is of a different nature than the angle classification because equilateral triangles are also considered to be isoceles. That is, the class of equilateral triangles is contained in the class of isoceles triangles. In this classification, not every triangle falls into exactly one of the classes.

As a side note, a triangle with two equal angles must be isoceles, and a triangle with three equal angles (equiangular) must be equilateral.


Probability of a Complement

Time To Pay A Complement: The complement A’ of a set is simply all elements in the sample space that are NOT in set A.

(Note: I was joking above. Note the spelling is different in compliment and complement.)

1) Two dice are tossed. What is the probability that the dice will be different numbers?

A=event the two dice are different.

A’=event the two dice are not different (they must be the same)

2) Suppose two fair dice are rolled. Find the probability that the sum of the numbers rolled is greater than 3.

A= event of a sum greater than 3.

A’= event of a sum not greater than 3. <(1,1),(1,2),(2,1)>, n(A’)=3

P(A)=1-3/36= 33/36=11/12

3) The previous examples could have been solved without using the complement. This problem forces you to use the complement. Otherwise you have to do many, many calculations. Eight cards are drawn from a standard deck of cards. What is the probability of at least one face-card?

Solution: n(S)=52C8

E is the event of getting at least 1 face card

E’ is the event of not getting a face card.

P(E’) can be calculate using combinations (they come later). Once you find P(E’), then P(E) is 1-P(E’)

We can do this later.

Here is another interesting application of the complement.

Birthday Problem: Read the Birthday Problem in the text. Notice how the complement is utilized.

How many people would you randomly select in order to expect there to be a 97% chance of two or more having the same birthday? (excluding twins etc)

A= event of two or more having same birthday. A’= event noone having same birthday.

It is much easier to find P(A’) than P(A). Then we can find P(A) = 1 – P(A’)

We can do this later.


Exercises 4.3

Ex 4.3.1 Decide if the following functions from $R$ to $R$ are injections, surjections, or both.

a) $2x+1$d) $(x+1)^3$
b) $1/2^x$e) $x^3-x$
c) $sin x$f) $|x|$

a) Find an example of an injection $fcolon A o B$ and a surjection $g,colon B o C$ such that $gcirc f$ is neither injective nor surjective.

b) Find an example of a surjection $fcolon A o B$ and an injection $g,colon B o C$ such that $gcirc f$ is neither injective nor surjective.

a) Suppose $A$ and $B$ are finite sets and $fcolon A o B$ is injective. What conclusion is possible regarding the number of elements in $A$ and $B$? Justify your answer.

b) If instead of injective, we assume $f$ is surjective, what conclusion is possible? Justify your answer.

Ex 4.3.4 Suppose $A$ is a finite set. Can we construct a function $fcolon A o A$ that is injective, but not surjective? Surjective, but not injective?

a) Find a function $fcolon N o N$ that is injective, but not surjective.

b) Find a function $g,colon N o N$ that is surjective, but not injective.

Ex 4.3.6 Suppose $A$ and $B$ are non-empty sets with $m$ and $n$ elements respectively, where $mle n$. How many injective functions are there from $A$ to $B$?

Ex 4.3.7 Find an injection $fcolon N imes N o N$. (Hint: use prime factorizations.)

Ex 4.3.8 If $fcolon A o B$ is a function, $A=Xcup Y$ and $fvert_X$ and $fvert_Y$ are both injective, can we conclude that $f$ is injective?


3.2: Complements, Intersections and Unions

Basic

  1. For the sample space (S=) identify the complement of each event given.
    1. (A=)
    2. (B=)
    3. (S)
    1. (R=)
    2. (T=)
    3. (varnothing) (the &ldquoempty&rdquo set that has no elements)
    1. List the outcomes that comprise (H) and (M).
    2. List the outcomes that comprise (Hcap M), (Hcup M), and (H^c).
    3. Assuming all outcomes are equally likely, find (P(Hcap M)), (P(Hcup M)), and (P(H^c)).
    4. Determine whether or not (H^c) and (M) are mutually exclusive. Explain why or why not.
    1. List the outcomes that comprise (T) and (G).
    2. List the outcomes that comprise (Tcap G), (Tcup G), (T^c), and ((Tcup G)^c).
    3. Assuming all outcomes are equally likely, find (P(Tcap G)), (P(Tcup G)), and (P(T^c)).
    4. Determine whether or not (T) and (G) are mutually exclusive. Explain why or why not.
    1. List the outcomes that comprise (B), (R), and (N).
    2. List the outcomes that comprise (Bcap R), (Bcup R), (Bcap N), (Rcup N), (B^c), and ((Bcup R)^c).
    3. Assuming all outcomes are equally likely, find the probabilities of the events in the previous part.
    4. Determine whether or not (B) and (N) are mutually exclusive. Explain why or why not.
    1. List the outcomes that comprise (Y), (I), and (J).
    2. List the outcomes that comprise (Ycap I), (Ycup J), (Icap J), (I^c), and ((Ycup J)^c).
    3. Assuming all outcomes are equally likely, find the probabilities of the events in the previous part.
    4. Determine whether or not (I^c) and (J) are mutually exclusive. Explain why or why not.

    1. (P(A)).
    2. (P(B)).
    3. (P(A^c)). Two ways: (i) by finding the outcomes in (A^c) and adding their probabilities, and (ii) using the Probability Rule for Complements.
    4. (P(Acap B)).
    5. (P(Acup B)) Two ways: (i) by finding the outcomes in (Acup B) and adding their probabilities, and (ii) using the Additive Rule of Probability.
    1. The Venn diagram provided shows a sample space and two events (A) and (B). Suppose (P(a)=0.32, P(b)=0.17, P(c)=0.28, ext P(d)=0.23). Confirm that the probabilities of the outcomes add up to (1), then compute the following probabilities.

    1. (P(A)).
    2. (P(B)).
    3. (P(A^c)). Two ways: (i) by finding the outcomes in (A^c) and adding their probabilities, and (ii) using the Probability Rule for Complements.
    4. (P(Acap B)).
    5. (P(Acup B)) Two ways: (i) by finding the outcomes in (Acup B) and adding their probabilities, and (ii) using the Additive Rule of Probability.
    1. Confirm that the probabilities in the two-way contingency table add up to (1), then use it to find the probabilities of the events indicated.
    1. (P(A), P(B), P(Acap B)).
    2. (P(U), P(W), P(Ucap W)).
    3. (P(Ucup W)).
    4. (P(V^c)).
    5. Determine whether or not the events (A) and (U) are mutually exclusive the events (A) and (V).
    1. Confirm that the probabilities in the two-way contingency table add up to (1), then use it to find the probabilities of the events indicated.
    1. (P(R), P(S), P(Rcap S)).
    2. (P(M), P(N), P(Mcap N)).
    3. (P(Rcup S)).
    4. (P(R^c)).
    5. Determine whether or not the events (N) and (S) are mutually exclusive the events (N) and (T).

    Applications

    1. Make a statement in ordinary English that describes the complement of each event (do not simply insert the word &ldquonot&rdquo).
      1. In the roll of a die: &ldquofive or more.&rdquo
      2. In a roll of a die: &ldquoan even number.&rdquo
      3. In two tosses of a coin: &ldquoat least one heads.&rdquo
      4. In the random selection of a college student: &ldquoNot a freshman.&rdquo
      1. In the roll of a die: &ldquotwo or less.&rdquo
      2. In the roll of a die: &ldquoone, three, or four.&rdquo
      3. In two tosses of a coin: &ldquoat most one heads.&rdquo
      4. In the random selection of a college student: &ldquoNeither a freshman nor a senior.&rdquo
      1. At least one child is a girl.
      2. At most one child is a girl.
      3. All of the children are girls.
      4. Exactly two of the children are girls.
      5. The first born is a girl.
      1. The person is male.
      2. The person is not in favor.
      3. The person is either male or in favor.
      4. The person is female and neutral.

      The record of a part is selected at random. Find the probability of each of the following events.

      1. The part was defective.
      2. The part was either of high quality or was at least usable, in two ways: (i) by adding numbers in the table, and (ii) using the answer to (a) and the Probability Rule for Complements.
      3. The part was defective and came from supplier (B).
      4. The part was defective or came from supplier (B), in two ways: by finding the cells in the table that correspond to this event and adding their probabilities, and (ii) using the Additive Rule of Probability.
      1. Individuals with a particular medical condition were classified according to the presence ((T)) or absence ((N)) of a potential toxin in their blood and the onset of the condition (( ext)). The breakdown according to this classification is shown in the two-way contingency table.

      One of these individuals is selected at random. Find the probability of each of the following events.

      1. The person experienced early onset of the condition.
      2. The onset of the condition was either midrange or late, in two ways: (i) by adding numbers in the table, and (ii) using the answer to (a) and the Probability Rule for Complements.
      3. The toxin is present in the person&rsquos blood.
      4. The person experienced early onset of the condition and the toxin is present in the person&rsquos blood.
      5. The person experienced early onset of the condition or the toxin is present in the person&rsquos blood, in two ways: (i) by finding the cells in the table that correspond to this event and adding their probabilities, and (ii) using the Additive Rule of Probability.
      1. The breakdown of the students enrolled in a university course by class (( ext)) and academic major (( ext)) is shown in the two-way classification table.

      A student enrolled in the course is selected at random. Adjoin the row and column totals to the table and use the expanded table to find the probability of each of the following events.

      1. The student is a freshman.
      2. The student is a liberal arts major.
      3. The student is a freshman liberal arts major.
      4. The student is either a freshman or a liberal arts major.
      5. The student is not a liberal arts major.
      1. The table relates the response to a fund-raising appeal by a college to its alumni to the number of years since graduation.

      An alumnus is selected at random. Adjoin the row and column totals to the table and use the expanded table to find the probability of each of the following events.

      1. The alumnus responded.
      2. The alumnus did not respond.
      3. The alumnus graduated at least (21) years ago.
      4. The alumnus graduated at least (21) years ago and responded.

      Additional Exercises

      1. The sample space for tossing three coins is (S=)
        1. List the outcomes that correspond to the statement &ldquoAll the coins are heads.&rdquo
        2. List the outcomes that correspond to the statement &ldquoNot all the coins are heads.&rdquo
        3. List the outcomes that correspond to the statement &ldquoAll the coins are not heads.&rdquo

        Answers

          1. ()
          2. ()
          3. (varnothing)
          1. (H=, M=)
          2. (Hcap M=, Hcup M=H, H^c=)
          3. (P(Hcap M)=4/8, P(Hcup M)=7/8, P(H^c)=1/8)
          4. Mutually exclusive because they have no elements in common.
          1. (B=, R=, N=)
          2. (Bcap R=varnothing , Bcup R=, Bcap N=, Rcup N=, B^c=, (Bcup R)^c=)
          3. (P(Bcap R)=0, P(Bcup R)=8/16, P(Bcap N)=2/16, P(Rcup N)=10/16, P(B^c)=12/16, P((Bcup R)^c)=8/16)
          4. Not mutually exclusive because they have an element in common.
          1. (0.36)
          2. (0.78)
          3. (0.64)
          4. (0.27)
          5. (0.87)
          1. (P(A)=0.38, P(B)=0.62, P(Acap B)=0)
          2. (P(U)=0.37, P(W)=0.33, P(Ucap W)=0)
          3. (0.7)
          4. (0.7)
          5. (A) and (U) are not mutually exclusive because (P(Acap U)) is the nonzero number (0.15). (A) and (V) are mutually exclusive because (P(Acap V)=0).
          1. &ldquofour or less&rdquo
          2. &ldquoan odd number&rdquo
          3. &ldquono heads&rdquo or &ldquoall tails&rdquo
          4. &ldquoa freshman&rdquo
          1. &ldquoAll the children are boys.&rdquo Event: (), Complement: ()
          2. &ldquoAt least two of the children are girls&rdquo or &ldquoThere are two or three girls.&rdquo Event: (), Complement: ()
          3. &ldquoAt least one child is a boy.&rdquo Event: (), Complement: ()
          4. &ldquoThere are either no girls, exactly one girl, or three girls.&rdquo Event: (), Complement: ()
          5. &ldquoThe first born is a boy.&rdquo Event: (), Complement: ()
          1. (0.0023)
          2. (0.9977)
          3. (0.0009)
          4. (0.3014)
          1. (920/1671)
          2. (668/1671)
          3. (368/1671)
          4. (1220/1671)
          5. (1003/1671)
          1. ()
          2. ()
          3. ()

          Outdoor Math Ideas

          Counting, Number Recognition, One-to-One Correspondence –

          Count the pine cones

          Under our pine trees we have thousands of tiny little pine cones. So following an idea from Peaceful Parenting, I drew boxes with the numbers 1-20 on the driveway. Then my just-turned Four worked at filling up the numbers through 10 and my Five counted pine cones for the larger numbers.

          Sorting, Measurement – Sort sticks by size

          Despite my nervousness around little boys and very long sticks, my Four and Five love to collect the big sticks they find in our yard and the little wooded area in the back. They rang the doorbell to show me all the sticks they had collected… and sorted by size. You could also have your child put the sticks in order from shortest to longest. You could do the same thing with leaves or wildflowers.

          Measurement – Very Big & Very Small Hunt

          Last spring, I printed this set of action cards and had my kids (ages 2,4, and 5 at the time) hunt for objects in the yard. They loved this very active scavenger hunt, and it was great for teaching my almost-three-year-old about size. I think we’ll have to pull it out again this year! Learn more and get your free printable action cards in this post.

          Patterns – Create nature patterns

          Addition, Subtraction, Number Recognition –

          Water Balloon Math

          This game can be modified in so many different ways. Prepare some water balloons and write either a single number or an addition or subtraction fact on each one. Then write matching numbers (or answers to the addition and subtraction problems) on your driveway or sidewalk. As your child chooses a balloon he throws it on the matching number or answer.

          My kids loved learning and staying cool at the same time!

          Fractions – Giant Fractions on the Driveway

          Fractions can be a difficult concept, but this short lesson seemed to do the trick for my Five. I drew a large rectangle on the driveway. We talked about how this was one whole rectangle. Then I divided it in two equal parts (okay, they weren’t exactly equal, but the kids didn’t notice). I had my Five stand in one part to show “one half.” Then his sister stood in the other half to show “one half.” We talked about how two halves make a whole.

          Then we worked together to show different fractions. In the above picture we made 𔄛/4.”

          The kids liked finding ways to make fractions all by themselves. Here my Seven is showing 𔄜/4.” This also gave us the chance to talk about equivalent fractions. 𔄜/4” is the same thing as “one whole.”

          My Five likes to find any opportunity he can to demonstrate standing on his head. This worked well for 𔄙/4.”

          Number Recognition, Counting –

          Mr. Wolf, What Time Is It?

          This game was designed for my just-turned-Four, but it wouldn’t have been fun without his older siblings playing along. I had written the numbers 1-12 on pieces of construction paper. I was the Wolf and stood at the end of the yard. The kids stood facing me at the other end of the yard.

          They called, “Mr. Wolf, what time is it?” Then I showed a number, and they took turns reading it. When I showed the 𔄚,” for example, one of them would say 𔄚:00!” Then they walked toward me that number of steps.

          Periodically I would put down the pages and call out “Lunch time!” At that point the kids would try to reach the tree behind me or run back to start, where they were “safe.” I ran after them to try to catch them. If they reached the tree behind me safely, they could be the next Wolf.

          Hint: wear shoes you can run in. This is excellent exercise for the mom.

          DO YOU HAVE OUR PRESCHOOL MATH CURRICULUM?

          This curriculum contains a wide selection of no-prep/no-worry math activities for our youngest learners!


          Calculus III

          Here are my online notes for my Calculus III course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn Calculus III or needing a refresher in some of the topics from the class.

          These notes do assume that the reader has a good working knowledge of Calculus I topics including limits, derivatives and integration. It also assumes that the reader has a good knowledge of several Calculus II topics including some integration techniques, parametric equations, vectors, and knowledge of three dimensional space.

          Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

            Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus III have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn’t covered in class.

          Here is a listing (and brief description) of the material that is in this set of notes.

          3-Dimensional Space - In this chapter we will start looking at three dimensional space. This chapter is generally prep work for Calculus III and so we will cover the standard 3D coordinate system as well as a couple of alternative coordinate systems. We will also discuss how to find the equations of lines and planes in three dimensional space. We will look at some standard 3D surfaces and their equations. In addition we will introduce vector functions and some of their applications (tangent and normal vectors, arc length, curvature and velocity and acceleration).


          Watch the video: NCERT Solutions Class 6 MATHS Chapter 3: Playing With Numbers Ex. (December 2021).