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# What Are Triangles?
In Geometry, a triangle is a closed two-dimensional figure with 3 sides, 3 angles and 3 vertices. Triangles are considered as a polygon with the least number of sides. In this article, let us discuss what are triangles, its types, properties, and formulas in detail. Learn triangles in detail here.
## What are Triangles in Geometry?
In Geometry, triangles are the type of polygons, which have three sides and three vertices. This is a two-dimensional figure with three straight sides. A triangle is considered a 3-sided polygon. The sum of all the three angles of a triangle is equal to 180°. The triangle is contained in a single plane. Based on its sides and angle measurement, the triangle has six types.
### How many sides does a triangle have?
A triangle has only three sides, that are connected to each other through the vertices. Therefore, no matter what the type of triangle is, it will always have three sides.
### Types of Triangle
Based on the sides of a triangle, a triangle is classified into 3 types, namely:
1. Scalene Triangle – All the sides are of different measures.
2. Isosceles Triangle – Two sides of a triangle are of the same measure and the remaining side has a different measure.
3. Equilateral Triangle – All the 3 sides of a triangle are of the same measure.
Based on the angles of a triangle, a triangle is classified into 3 types, namely:
1. Acute Angle Triangle – All the angles of a triangle is less than 90°
2. Obtuse Angle Triangle – One of the angles of a triangle is greater than 90°
3. Right Angle Triangle – One of the angles of a triangle is equal to 90°
### Properties of Triangle
1. A triangle has three sides, three angles and three vertices.
2. Sum of all three interior angles of a triangle is 180°.
3. Sum of the length of two sides of the triangle is greater than its third side.
4. The area of a triangle is the half the product of the base and the height
5. The perimeter of a triangle is the sum of all the three sides of a triangle.
### Area of Triangle
One of the properties of a triangle is its area, which is the region covered by the three-sided polygon in a plane. The formula for any triangle’s area is given by;
Area of a Triangle = ½ × Base × Height
One more formula has been introduced by a Mathematician, which is known as Heron’s Formula. This formula is used when the sides of the triangles are known to us. It is given by;
Where ‘s’ is the semi-perimeter of the triangle and given by;
s = (a+b+c)/2
Where a, b and c are the measure of the three sides of the triangle.
### Perimeter of Triangle
The perimeter of a triangle is the length covered by the sides of the triangle in a plane. The formula of the perimeter is given by;
P = a + b + c
## Frequently Asked Questions on What Are Triangles
Q1
### Find the perimeter of a triangle, whose sides are 3 cm, 4 cm and 6 cm.
Given, sides of the triangle are 3 cm, 4 cm and 6 cm.
Let a = 3 cm, b = 4 cm and c = 6 cm
By the perimeter formula, we know;
P = a+b+c
P = 3+4+6 = 13 cm
Q2
### Find the perimeter of an isosceles triangle whose base is 9 cm and sides are equal to 12 cm.
For an isosceles triangle, two sides are equal to each other.
Therefore, the perimeter of an isosceles triangle, P = 2xa + b
P = 2 x 12 + 9
P = 24 + 9
P = 33 cm
Q3
### Does a triangle has two sides?
A triangle has three sides not two sides.
Q4
### What are the three types of triangles based on sides?
The three sides of triangles based on sides are:
Scalene triangle
Isosceles triangle
Equilateral triangle
Q5
### What are the three types of triangles based on angles?
The three types of triangles based on angles are:
Acute triangle
Obtuse triangle
Right triangle
Test your Knowledge on What Are Triangles |
SECTION 10-5 Multiplication Principle, Permutations, and Combinations
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1 10-5 Multiplication Principle, Permutations, and Combinations Can you guess what the next two rows in Pascal s triangle, shown at right, are? Compare the numbers in the triangle with the binomial coefficients obtained with the binomial formula SECTION 10-5 Multiplication Principle, Permutations, and Combinations Multiplication Principle Permutations Combinations We may expand the binomial form (a b) n in two steps: first, expand into a sum of 2 n terms, each with coefficient 1; second, group together those terms in which b appears to the same power, obtaining the sum of the n 1 terms of the binomial formula. For example, (a b) 3 (a b)(a b) 2 (a b)(aa ab ba bb) aaa aab aba abb baa bab bba bbb Step 1 a 3 3a 2 b 3ab 2 b 3 Step 2 Consider the term aba of step 1: The first a comes from the first factor of a b, the b comes from the second factor of a b, and the final a from the third factor. Therefore, 3 1 3, the coefficient of a 2 b in step 2, is the number of ways of choosing b from exactly one of the three factors of a b in (a b) 3. In the same way, 2,598,960 is the number of ways of choosing b from exactly five of the 52 factors of a b in (a b) 52. nalogously, 2,598,960 is the number of 5-card hands which can be chosen from a standard 52-card deck. In this section we study such counting techniques that are related to the sequence n 0, n 1, n 2,..., n n 52 5, and we develop important counting tools that form the foundation of probability theory. Multiplication Principle We start with an example. EXMPLE 1 Combined Outcomes Suppose we flip a coin and then throw a single die (see Fig. 1). What are the possible combined outcomes?
2 Sequences and Series Solution To solve this problem, we use a tree diagram: Heads Tails Coin Outcomes Die Outcomes Combined Outcomes Coin outcomes Die outcomes FIGURE 1 Coin and die outcomes. Start H T (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6) Thus, there are 12 possible combined outcomes two ways in which the coin can come up followed by six ways in which the die can come up. Matched Problem 1 Use a tree diagram to determine the number of possible outcomes of throwing a single die followed by flipping a coin. Now suppose you are asked, From the 26 letters in the alphabet, how many ways can 3 letters appear in a row on a license plate if no letter is repeated? To try to count the possibilities using a tree diagram would be extremely tedious, to say the least. The following multiplication principle, also called the fundamental counting principle, enables us to solve this problem easily. In addition, it forms the basis for several other counting techniques developed later in this section. Multiplication Principle 1. If two operations O 1 and O 2 are performed in order, with N 1 possible outcomes for the first operation and N 2 possible outcomes for the second operation, then there are N 1 N 2 possible combined outcomes of the first operation followed by the second. 2. In general, if n operations O 1, O 2,..., O n are performed in order, with possible number of outcomes N 1, N 2,..., N n, respectively, then there are N 1 N 2... N n possible combined outcomes of the operations performed in the given order.
3 10-5 Multiplication Principle, Permutations, and Combinations 763 In Example 1, we see that there are two possible outcomes from the first operation of flipping a coin and six possible outcomes from the second operation of throwing a die. Hence, by the multiplication principle, there are possible combined outcomes of flipping a coin followed by throwing a die. Use the multiplication principle to solve Matched Problem 1. To answer the license plate question, we reason as follows: There are 26 ways the first letter can be chosen. fter a first letter is chosen, 25 letters remain; hence there are 25 ways a second letter can be chosen. nd after 2 letters are chosen, there are 24 ways a third letter can be chosen. Hence, using the multiplication principle, there are ,600 possible ways 3 letters can be chosen from the alphabet without allowing any letter to repeat. By not allowing any letter to repeat, earlier selections affect the choice of subsequent selections. If we allow letters to repeat, then earlier selections do not affect the choice in subsequent selections, and there are 26 possible choices for each of the 3 letters. Thus, if we allow letters to repeat, there are ,576 possible ways the 3 letters can be chosen from the alphabet. EXMPLE 2 Computer-Generated Tests Many universities and colleges are now using computer-assisted testing procedures. Suppose a screening test is to consist of 5 questions, and a computer stores 5 equivalent questions for the first test question, 8 equivalent questions for the second, 6 for the third, 5 for the fourth, and 10 for the fifth. How many different 5-question tests can the computer select? Two tests are considered different if they differ in one or more questions. Solution O 1 : Select the first question N 1 : 5 ways O 2 : Select the second question N 2 : 8 ways O 3 : Select the third question N 3 : 6 ways O 4 : Select the fourth question N 4 : 5 ways O 5 : Select the fifth question N 5 : 10 ways Thus, the computer can generate ,000 different tests Matched Problem 2 Each question on a multiple-choice test has 5 choices. If there are 5 such questions on a test, how many different response sheets are possible if only 1 choice is marked for each question? EXMPLE 3 Counting Code Words How many 3-letter code words are possible using the first 8 letters of the alphabet if: () No letter can be repeated? (B) Letters can be repeated? (C) djacent letters cannot be alike?
4 Sequences and Series Solutions () No letter can be repeated. O 1 : Select first letter N 1 : 8 ways O 2 : Select second letter N 2 : 7 ways Because 1 letter has been used O 3 : Select third letter N 3 : 6 ways Because 2 letters have been used Thus, there are (B) Letters can be repeated possible code words O 1 : Select first letter N 1 : 8 ways O 2 : Select second letter N 2 : 8 ways Repeats are allowed. O 3 : Select third letter N 3 : 8 ways Repeats are allowed. Thus, there are (C) djacent letters cannot be alike possible code words O 1 : Select first letter N 1 : 8 ways O 2 : Select second letter N 2 : 7 ways Cannot be the same as the first O 3 : Select third letter N 3 : 7 ways Cannot be the same as the second, but can be the same as the first Thus, there are possible code words Matched Problem 3 How many 4-letter code words are possible using the first 10 letters of the alphabet under the three conditions stated in Example 3? EXPLORE-DISCUSS 1 The postal service of a developing country is choosing a five-character postal code consisting of letters (of the English alphabet) and digits. t least half a million postal codes must be accommodated. Which format would you recommend to make the codes easy to remember? The multiplication principle can be used to develop two additional methods for counting that are extremely useful in more complicated counting problems. Both of these methods use the factorial function, which was introduced in Section 10-4.
5 10-5 Multiplication Principle, Permutations, and Combinations 765 Permutations Suppose 4 pictures are to be arranged from left to right on one wall of an art gallery. How many arrangements are possible? Using the multiplication principle, there are 4 ways of selecting the first picture. fter the first picture is selected, there are 3 ways of selecting the second picture. fter the first 2 pictures are selected, there are 2 ways of selecting the third picture. nd after the first 3 pictures are selected, there is only 1 way to select the fourth. Thus, the number of arrangements possible for the 4 pictures is ! or 24 In general, we refer to a particular arrangement, or ordering, of n objects without repetition as a permutation of the n objects. How many permutations of n objects are there? From the reasoning above, there are n ways in which the first object can be chosen, there are n 1 ways in which the second object can be chosen, and so on. pplying the multiplication principle, we have Theorem 1: Theorem 1 Permutations of n Objects The number of permutations of n objects, denoted by P n,n, is given by P n,n n (n 1)... 1 n! Now suppose the director of the art gallery decides to use only 2 of the 4 available pictures on the wall, arranged from left to right. How many arrangements of 2 pictures can be formed from the 4? There are 4 ways the first picture can be selected. fter selecting the first picture, there are 3 ways the second picture can be selected. Thus, the number of arrangements of 2 pictures from 4 pictures, denoted by P 4,2, is given by P 4, Or, in terms of factorials, multiplying 4 3 by 1 in the form 2!/2!, we have P 4, ! 2! 4! 2! This last form gives P 4,2 in terms of factorials, which is useful in some cases. permutation of a set of n objects taken r at a time is an arrangement of the r objects in a specific order. Thus, reasoning in the same way as in the example above, we find that the number of permutations of n objects taken r at a time, 0 r n, denoted by P n,r, is given by P n,r n(n 1)(n 2)... (n r 1) Multiplying the right side of this equation by 1 in the form (n r)!/(n r)!, we obtain a factorial form for P n,r : P n,r n(n 1)(n 2)... (n r 1) (n r)! (n r)!
6 Sequences and Series But Hence, we have Theorem 2: n(n 1)(n 2)... (n r 1)(n r)! n! Theorem 2 Permutation of n Objects Taken r at a Time The number of permutations of n objects taken r at a time is given by or P n,r n(n 1)(n 2)... (n r 1) agggggggbgggggggc r factors P n,r n! (n r)! 0 r n Note that if r n, then the number of permutations of n objects taken n at a time is P n,n n! (n n)! n! 0! n! Recall, 0! 1. which agrees with Theorem 1, as it should. The permutation symbol P n,r also can be denoted by P n r, n P r, or P(n, r). Many calculators use n P r to denote the function that evaluates the permutation symbol. EXMPLE 4 Selecting Officers From a committee of 8 people, in how many ways can we choose a chair and a vicechair, assuming one person cannot hold more than one position? Solution We are actually asking for the number of permutations of 8 objects taken 2 at a time that is, P 8,2 : P 8, 2 8! (8 2)! 8! 6! 8 7 6! 56 6! Matched Problem 4 From a committee of 10 people, in how many ways can we choose a chair, vice-chair, and secretary, assuming one person cannot hold more than one position?
7 10-5 Multiplication Principle, Permutations, and Combinations 767 CUTION Remember to use the definition of factorial when simplifying fractions involving factorials. 6! 3! 2! 6! 3! ! 120 3! EXMPLE 5 Evaluating P n,r Find the number of permutations of 25 objects taken 8 at a time. Compute the answer to 4 significant digits using a calculator. Solution P 25,8 25! (25 8)! 25! ! very large number Matched Problem 5 Find the number of permutations of 30 objects taken 4 at a time. Compute the answer exactly using a calculator. Combinations Now suppose that an art museum owns 8 paintings by a given artist and another art museum wishes to borrow 3 of these paintings for a special show. How many ways can 3 paintings be selected for shipment out of the 8 available? Here, the order of the items selected doesn t matter. What we are actually interested in is how many subsets of 3 objects can be formed from a set of 8 objects. We call such a subset a combination of 8 objects taken 3 at a time. The total number of combinations is denoted by the symbol C 8,3 or 8 3 To find the number of combinations of 8 objects taken 3 at a time, C 8,3, we make use of the formula for P n,r and the multiplication principle. We know that the number of permutations of 8 objects taken 3 at a time is given by P 8,3, and we have a formula for computing this quantity. Now suppose we think of P 8,3 in terms of two operations: O 1 : N 1 : O 2 : N 2 : Select a subset of 3 objects (paintings) C 8,3 ways rrange the subset in a given order 3! ways The combined operation, O 1 followed by O 2, produces a permutation of 8 objects taken 3 at a time. Thus, P 8,3 C 8,3 3!
8 Sequences and Series To find C 8,3, we replace P 8,3 in the above equation with 8!/(8 3)! and solve for C 8,3 : 8! (8 3)! C 8,3 3! C 8,3 8! 3!(8 3)! ! ! 56 Thus, the museum can make 56 different selections of 3 paintings from the 8 available. combination of a set of n objects taken r at a time is an r-element subset of the n objects. Reasoning in the same way as in the example, the number of combinations of n objects taken r at a time, 0 r n, denoted by C n,r, can be obtained by solving for C n,r in the relationship P n,r C n,r r! C n,r P n,r r! n! r!(n r)! P n,r n! (n r)! Theorem 3 Combination of n Objects Taken r at a Time The number of combinations of n objects taken r at a time is given by C n,r n r P n,r r! n! r!(n r)! 0 r n Note that we used the combination formula in Section 10-4 to represent binomial coefficients. C(n, r). n r The combination symbols C n,r and also can be denoted by, n C r, or C n r EXMPLE 6 Selecting Subcommittees From a committee of 8 people, in how many ways can we choose a subcommittee of 2 people? Solution Notice how this example differs from Example 4, where we wanted to know how many ways a chair and a vice-chair can be chosen from a committee of 8 people. In Example 4, ordering matters. In choosing a subcommittee of 2 people, the ordering does not matter. Thus, we are actually asking for the number of combinations of 8 objects taken 2 at a time. The number is given by C 8, ! 2!(8 2)! 8 7 6! 2 1 6! 28
9 10-5 Multiplication Principle, Permutations, and Combinations 769 Matched Problem 6 How many subcommittees of 3 people can be chosen from a committee of 8 people? EXMPLE 7 Evaluating C n,r Find the number of combinations of 25 objects taken 8 at a time. Compute the answer to 4 significant digits using a calculator. Solution C 25, ! 8!(25 8)! 25! !17! Compare this result with that obtained in Example 5. Matched Problem 7 Find the number of combinations of 30 objects taken 4 at a time. Compute the answer exactly using a calculator. Remember: In a permutation, order counts. In a combination, order does not count. To determine whether a permutation or combination is needed, decide whether rearranging the collection or listing makes a difference. If so, use permutations. If not, use combinations. EXPLORE-DISCUSS 2 Each of the following is a selection without repetition. Would you consider the selection to be a combination? permutation? Discuss your reasoning. () student checks out three books from the library. (B) baseball manager names his starting lineup. (C) The newly elected President names his Cabinet members. (D) The President selects a delegation of three Cabinet members to attend the funeral of a head of state. (E) n orchestra conductor chooses three pieces of music for a symphony program. standard deck of 52 cards involves four suits, hearts, spades, diamonds, and clubs, as shown in Figure 2. Example 8, as well as other examples and exercises in this chapter, refer to this standard deck.
10 Sequences and Series FIGURE 2 standard deck of cards J Q K J Q K J Q K J Q K 2 3 EXMPLE 8 Counting Card Hands Out of a standard 52-card deck, how many 5-card hands will have 3 aces and 2 kings? Solution O 1 : Choose 3 aces out of 4 possible Order is not important. N 1 : C 4,3 O 2 : Choose 2 kings out of 4 possible Order is not important. N 2 : C 4,2 Using the multiplication principle, we have Number of hands C 4,3 C 4, Matched Problem 8 From a standard 52-card deck, how many 5-card hands will have 3 hearts and 2 spades? EXMPLE 9 Counting Serial Numbers Serial numbers for a product are to be made using 2 letters followed by 3 numbers. If the letters are to be taken from the first 8 letters of the alphabet with no repeats and the numbers from the 10 digits 0 through 9 with no repeats, how many serial numbers are possible? Solution O 1 : Choose 2 letters out of 8 available Order is important. N 1 : P 8,2 O 2 : Choose 3 numbers out of 10 available Order is important. N 2 : P 10,3
11 10-5 Multiplication Principle, Permutations, and Combinations 771 Using the multiplication principle, we have Number of serial numbers P 8,2 P 10,3 40,320 Matched Problem 9 Repeat Example 9 under the same conditions, except the serial numbers are now to have 3 letters followed by 2 digits with no repeats. nswers to Matched Problems 1. HT HT HT HT HT HT , or 3, Start 3. () ,040 (B) ,000 (C) ,290 10! 30! 8! 4. P 10, P 30,4 657, C 8,3 56 (10 3)! (30 4)! 3!(8 3)! 30! 7. C 30,4 27,405 4!(30 4)! 8. C 13,3 C 13,2 22, P 8,3 P 10,2 30,240 EXERCISE 10-5 Evaluate Problems ! 20! ! 18! 25! 9! !1! 6!3! 16! 18! P 8,5 4!(16 4)! 3!(18 3)! 10. C 8,5 11. P 52,3 12. P 13,5 13. C 13,5 14. C 13,4 15. C 52,5 16. P 20,4 32! 0!32! 7! 5!2! 17. particular new car model is available with 5 choices of color, 3 choices of transmission, 4 types of interior, and 2 types of engine. How many different variations of this model car are possible? 18. deli serves sandwiches with the following options: 3 kinds of bread, 5 kinds of meat, and lettuce or sprouts. How many different sandwiches are possible, assuming one item is used out of each category? 19. In a horse race, how many different finishes among the first 3 places are possible for a 10-horse race? Exclude ties. 20. In a long-distance foot race, how many different finishes among the first 5 places are possible for a 50-person race? Exclude ties. 21. How many ways can a subcommittee of 3 people be selected from a committee of 7 people? How many ways can a president, vice president, and secretary be chosen from a committee of 7 people? 22. Suppose 9 cards are numbered with the 9 digits from 1 to 9. 3-card hand is dealt, 1 card at a time. How many hands are possible where: () Order is taken into consideration? (B) Order is not taken into consideration? 23. There are 10 teams in a league. If each team is to play every other team exactly once, how many games must be scheduled? 24. Given 7 points, no 3 of which are on a straight line, how many lines can be drawn joining 2 points at a time? B 25. How many 4-letter code words are possible from the first 6 letters of the alphabet, with no letter repeated? llowing letters to repeat?
12 Sequences and Series 26. small combination lock on a suitcase has 3 wheels, each labeled with digits from 0 to 9. How many opening combinations of 3 numbers are possible, assuming no digit is repeated? ssuming digits can be repeated? 27. From a standard 52-card deck, how many 5-card hands will have all hearts? 28. From a standard 52-card deck, how many 5-card hands will have all face cards? ll face cards, but no kings? Consider only jacks, queens, and kings to be face cards. 29. How many different license plates are possible if each contains 3 letters followed by 3 digits? How many of these license plates contain no repeated letters and no repeated digits? 30. How many 5-digit zip codes are possible? How many of these codes contain no repeated digits? 31. From a standard 52-card deck, how many 7-card hands have exactly 5 spades and 2 hearts? 32. From a standard 52-card deck, how many 5-card hands will have 2 clubs and 3 hearts? 33. catering service offers 8 appetizers, 10 main courses, and 7 desserts. banquet chairperson is to select 3 appetizers, 4 main courses, and 2 desserts for a banquet. How many ways can this be done? 34. Three research departments have 12, 15, and 18 members, respectively. If each department is to select a delegate and an alternate to represent the department at a conference, how many ways can this be done? 35. () Use a graphing utility to display the sequences P 10,0, P 10,1,..., P 10,10 and 0!, 1!,..., 10! in table form, and show that P 10,r r! for r 0, 1,..., 10. (B) Find all values of r such that P 10,r r! (C) Explain why P n,r r! whenever 0 r n. 36. () How are the sequences and C 10,0, 0!, P 10,1 1!,..., P 10,10 10! C 10,1,..., C 10,10 related? (B) Use a graphing utility to graph each sequence and confirm the relationship of part. C P 10,0 37. sporting goods store has 12 pairs of ski gloves of 12 different brands thrown loosely in a bin. The gloves are all the same size. In how many ways can a left-hand glove and a right-hand glove be selected that do not match relative to brand? 38. sporting goods store has 6 pairs of running shoes of 6 different styles thrown loosely in a basket. The shoes are all the same size. In how many ways can a left shoe and a right shoe be selected that do not match? 39. Eight distinct points are selected on the circumference of a circle. () How many chords can be drawn by joining the points in all possible ways? (B) How many triangles can be drawn using these 8 points as vertices? (C) How many quadrilaterals can be drawn using these 8 points as vertices? 40. Five distinct points are selected on the circumference of a circle. () How many chords can be drawn by joining the points in all possible ways? (B) How many triangles can be drawn using these 5 points as vertices? 41. How many ways can 2 people be seated in a row of 5 chairs? 3 people? 4 people? 5 people? 42. Each of 2 countries sends 5 delegates to a negotiating conference. rectangular table is used with 5 chairs on each long side. If each country is assigned a long side of the table, how many seating arrangements are possible? [Hint: Operation 1 is assigning a long side of the table to each country.] 43. basketball team has 5 distinct positions. Out of 8 players, how many starting teams are possible if: () The distinct positions are taken into consideration? (B) The distinct positions are not taken into consideration? (C) The distinct positions are not taken into consideration, but either Mike or Ken, but not both, must start? 44. How many committees of 4 people are possible from a group of 9 people if: () There are no restrictions? (B) Both Juan and Mary must be on the committee? (C) Either Juan or Mary, but not both, must be on the committee? card hand is dealt from a standard 52-card deck. Which is more likely: the hand contains exactly 1 king or the hand contains no hearts? card hand is dealt from a standard 52-card deck. Which is more likely: all cards in the hand are red or the hand contains all four aces? 47. parent is placing an order for five single-dip ice cream cones. If today s flavors are vanilla, chocolate, and strawberry, how many orders are possible? Explain. (Note: This type of selection, in which repetition is allowed but order is irrelevant, is neither a combination nor a permutation.) 48. One dozen identical doughnuts are to be distributed among nine students. If each student must receive at least one doughnut, how many distributions are possible? Explain.
13 Chapter 10 Review 773 CHPTER 10 GROUP CTIVITY Sequences Specified by Recursion Formulas The recursion formula a n 5a n 1 6a n 2, together with the initial values a 1 4, a 2 14, specifies the sequence {a n } whose first several terms are 4, 14, 46, 146, 454, 1394,... The sequence {a n } is neither arithmetic nor geometric. Nevertheless, because it satisfies a simple recursion formula, it is possible to obtain an nth-term formula for {a n } that is analogous to the nth-term formulas for arithmetic and geometric sequences. Such an nth-term formula is valuable because it allows us to estimate a term of a sequence without computing all the preceding terms. If the geometric sequence {r n } satisfies the recursion formula above, then r n 5r n 1 6r n 2. Dividing by r n 2 leads to the quadratic equation r 2 5r 6 0, whose solutions are r 2 and r 3. Now it is easy to check that the geometric sequences {2 n } 2, 4, 8, 16,... and {3 n } 3, 9, 27, 81,... satisfy the recursion formula. Therefore, any sequence of the form {u2 n v3 n }, where u and v are constants, will satisfy the same recursion formula. We now find u and v so that the first two terms of {u2 n v3 n } are a 1 4, a Letting n 1 and n 2 we see that u and v must satisfy the following linear system: 2u 3v 41 4u 9v 14 Solving the system gives u 1, v 2. Therefore, an nth-term formula for the original sequence is a n ( 1)2 n (2)3 n. Note that the nth-term formula was obtained by solving a quadratic equation and a system of two linear equations in two variables. () Compute ( 1)2 n (2)3 n for n 1, 2,..., 6, and compare with the terms of {a n }. (B) Estimate the one-hundredth term of {a n }. (C) Show that any sequence of the form {u2 n v3 n }, where u and v are constants, satisfies the recursion formula a n 5a n 1 6a n 2. (D) Find an nth-term formula for the sequence {b n } that is specified by b 1 5, b 2 55, b n 3b n 1 4b n 2. (E) Find an nth-term formula for the Fibonacci sequence. (F) Find an nth-term formula for the sequence {c n } that is specified by c 1 3, c 2 15, c 3 99, c n 6c n 1 3c n 2 10c n 3. (Since the recursion formula involves the three terms which precede c n, our method will involve the solution of a cubic equation and a system of three linear equations in three variables.) Chapter 10 Review 10-1 SEQUENCES ND SERIES sequence is a function with the domain a set of successive integers. The symbol a n, called the nth term, or general term, represents the range value associated with the domain value n. Unless specified otherwise, the domain is understood to be the set of natural numbers. finite sequence has a finite domain, and an infinite sequence has an infinite domain. recursion formula defines each term of a sequence in terms of one or more of the preceding terms. For example, the Fibonacci sequence is defined by a n a n 1 a n 2 for n 3, where a 1 a 2 1. If a 1, a 2,..., a n,...is a sequence, then the expression a 1 a 2... a n... is called a series. finite sequence produces a finite series, and an infinite sequence produces an infinite series. Series can be represented using summation notation: n k m a k a m a m 1... a n where k is called the summing index. If the terms in the series are alternately positive and negative, the series is called an alternating series.
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## § 1.3 Congruences
Definition 1.3.1. Let n be a positive integer. Integers a and b are said to be congruent modulo n if they have the same remainder when divided by n. This is denoted by writing a b (mod n).
Proposition 1.3.2. Let a,b, and n > 0 be integers. Then a b (mod n) if and only if n | (a-b).
When working with congruence modulo n, the integer n is called the modulus. By the preceding proposition, a b (mod n) if and only if a-b = nq for some integer q. We can write this in the form a = b+nq, for some integer q. This observation gives a very useful method of replacing a congruence with an equation (over Z). On the other hand, Proposition 1.3.3 shows that any equation can be converted to a congruence modulo n by simply changing the = sign to . In doing so, any term congruent to 0 can simply be omitted. Thus the equation a = b+nq would be converted back to a b (mod n).
Proposition 1.3.3. Let n > 0 be an integer. Then the following conditions hold for all integers a,b,c,d:
(a) If a c (mod n) and b d (mod n), then
a ± b c ± d (mod n), and ab cd (mod n).
(b) If a+c a+d (mod n), then c d (mod n).
If ac ad (mod n) and (a,n)=1, then c d (mod n).
Proposition 1.3.4. Let a and n > 1 be integers. There exists an integer b such that ab 1 (mod n) if and only if (a,n) = 1.
Theorem 1.3.5. The congruence ax b (mod n) has a solution if and only if b is divisible by d, where d = (a,n).
If d | b, then there are d distinct solutions modulo n, and these solutions are congruent modulo n / d.
Theorem 1.3.6. [Chinese Remainder Theorem] Let n and m be positive integers, with (n,m)=1. Then the system of congruences
x a (mod n) x b (mod m)
has a solution. Moreover, any two solutions are congruent modulo mn.
## § 1.3 Solved Problems
In this section, it is important to remember that although working with congruences is almost like working with equations, it is not exactly the same.
What things are the same? You can add or subtract the same integer on both sides of a congruence, and you can multiply both sides of a congruence by the same integer. You can use substitution, and you can use the transitive property:
if a b (mod n) and b c (mod n) then a c (mod n).
You should also review Proposition 1.3.3 and the comments in the text both before and after the proof of the proposition.
What things are different? In an ordinary equation you can divide through by a nonzero number. In a congruence modulo n, you can only divide through by an integer that is relatively prime to n. This is usually expressed by saying that
if gcd (a,n) = 1 and ac ad (mod n) then c d (mod n).
Just be very careful!
One of the important techniques to understand is how to switch between congruences and ordinary equations. First, any equation involving integers can be converted into a congruence by just reducing modulo n. This works because if two integers are equal, then are certainly congruent modulo n.
The do the opposite conversion you must be more careful. If two integers are congruent modulo n, that doesn't make them equal, but only guarantees that dividing by n produces the same remainder in each case. In other words, the integers may differ by some multiple of n.
The conversion process is illustrated in Example 1.3.5 of the text, where the congruence
x 7 (mod 8)
is converted into the equation
x = 7 + 8q, for some q in Z.
Notice that converting to an equation makes it more complicated, because we have to introduce another variable. In the example, we really want a congruence modulo 5, so the next step is to rewrite the equation as
x 7 + 8q (mod 5).
Actually, we can reduce each term modulo 5, so that we finally get
x 2 + 3q (mod 5).
You should read the proofs of Theorem 1.3.5 and Theorem 1.3.6 very carefully. These proofs actually show you the necessary techniques to solve all linear congruences of the form
ax b (mod n),
and all simultaneous linear equations of the form
x a (mod n) and x b (mod n),
where the moduli n and m are relatively prime. Many of the theorems in the text should be thought of as ``shortcuts'', and you can't afford to skip over their proofs, because you might miss important algorithms or computational techniques.
26. Solve the congruence 42x 12 (mod 90). Solution
27. (a) Find all solutions to the congruence 55x 35 (mod 75).
(b) Find all solutions to the congruence 55x 36 (mod 75). Solution
28. (a) Find one particular integer solution to the equation 110x + 75y = 45.
(b) Show that if x = m; y = n is an integer solution to the equation in part (a), then so is
x = m + 15q; y = n -22q, for any integer q. Solution
29. Solve the following system of congruences.
x 2 (mod 9) x 4 (mod 10)
Solution
30. Solve the following system of congruences.
5x 14 (mod 17) 3x 2 (mod 13)
Solution
31. Solve the following system of congruences.
x 5 (mod 25) x 23 (mod 32)
Solution
32. Give integers a,b,m,n to provide an example of a system of congruences
x a (mod m) x b (mod n)
that has no solution. Solution
33. (a) Compute the last digit in the decimal expansion of 4100.
(b) Is 4100 divisible by 3? Solution
34. Find all integers n for which 13 is a factor of 4(n2 + 1). Solution
35. Prove that 10n+1 + 4 · 10n + 4 is divisible by 9, for all positive integers n. Solution
36. Prove that the fourth power of an integer can only have 0, 1, 5, or 6 as its units digit. Solution
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# Graphing Quadratic Equations Using the Axis of Symmetry
A quadratic equation is a polynomial equation of degree $2$ . The standard form of a quadratic equation is
$0=a{x}^{2}+bx+c$
where $a,b$ and $c$ are all real numbers and $a\ne 0$ .
If we replace $0$ with $y$ , then we get a quadratic function
$y=a{x}^{2}+bx+c$
whose graph will be a parabola .
The axis of symmetry of this parabola will be the line $x=-\frac{b}{2a}$ . The axis of symmetry passes through the vertex, and therefore the $x$ -coordinate of the vertex is $-\frac{b}{2a}$ . Substitute $x=-\frac{b}{2a}$ in the equation to find the $y$ -coordinate of the vertex. Substitute few more $x$ -values in the equation to get the corresponding $y$ -values and plot the points. Join them and extend the parabola.
Example 1:
Graph the parabola $y={x}^{2}-7x+2$ .
Compare the equation with $y=a{x}^{2}+bx+c$ to find the values of $a$ , $b$ , and $c$ .
Here, $a=1,b=-7$ and $c=2$ .
Use the values of the coefficients to write the equation of axis of symmetry .
The graph of a quadratic equation in the form $y=a{x}^{2}+bx+c$ has as its axis of symmetry the line $x=-\frac{b}{2a}$ . So, the equation of the axis of symmetry of the given parabola is $x=\frac{-\left(-7\right)}{2\left(1\right)}$ or $x=\frac{7}{2}$ .
Substitute $x=\frac{7}{2}$ in the equation to find the $y$ -coordinate of the vertex.
$\begin{array}{l}y={\left(\frac{7}{2}\right)}^{2}-7\left(\frac{7}{2}\right)+2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{49}{4}-\frac{49}{2}+2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{49\text{\hspace{0.17em}}-\text{\hspace{0.17em}}98\text{\hspace{0.17em}}+\text{\hspace{0.17em}}8}{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{41}{4}\end{array}$
Therefore, the coordinates of the vertex are $\left(\frac{7}{2},-\frac{41}{4}\right)$ .
Now, substitute a few more $x$ -values in the equation to get the corresponding $y$ -values.
$x$ $y={x}^{2}-7x+2$ $0$ $2$ $1$ $-4$ $2$ $-8$ $3$ $-10$ $5$ $-8$ $7$ $2$
Plot the points and join them to get the parabola.
Example 2:
Graph the parabola $y=-2{x}^{2}+5x-1$ .
Compare the equation with $y=a{x}^{2}+bx+c$ to find the values of $a$ , $b$ , and $c$ .
Here, $a=-2,b=5$ and $c=-1$ .
Use the values of the coefficients to write the equation of axis of symmetry.
The graph of a quadratic equation in the form $y=a{x}^{2}+bx+c$ has as its axis of symmetry the line $x=-\frac{b}{2a}$ . So, the equation of the axis of symmetry of the given parabola is $x=\frac{-\left(5\right)}{2\left(-2\right)}$ or $x=\frac{5}{4}$ .
Substitute $x=\frac{5}{4}$ in the equation to find the $y$ -coordinate of the vertex.
$\begin{array}{l}y=-2{\left(\frac{5}{4}\right)}^{2}+5\left(\frac{5}{4}\right)-1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-50}{16}+\frac{25}{4}-1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-50\text{\hspace{0.17em}}+\text{\hspace{0.17em}}100\text{\hspace{0.17em}}-\text{\hspace{0.17em}}16}{16}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{34}{16}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{17}{8}\end{array}$
Therefore, the coordinates of the vertex are $\left(\frac{5}{4},\frac{17}{8}\right)$ .
Now, substitute a few more $x$ -values in the equation to get the corresponding $y$ -values.
$x$ $y=-2{x}^{2}+5x-1$ $-1$ $-8$ $0$ $-1$ $1$ $2$ $2$ $1$ $3$ $-4$
Plot the points and join them to get the parabola.
Example 3:
Graph the parabola $x={y}^{2}+4y+2$ .
Here, $x$ is a function of $y$ . The parabola opens "sideways" and the axis of symmetry of the parabola is horizontal. The standard form of equation of a horizontal parabola is $x=a{y}^{2}+by+c$ where $a$ , $b$ , and $c$ are all real numbers and $a\ne 0$ and the equation of the axis of symmetry is $y=\frac{-b}{2a}$ .
Compare the equation with $x=a{y}^{2}+by+c$ to find the values of $a$ , $b$ , and $c$ .
Here, $a=1,b=4$ and $c=2$ .
Use the values of the coefficients to write the equation of axis of symmetry.
The graph of a quadratic equation in the form $x=a{y}^{2}+by+c$ has as its axis of symmetry the line $y=-\frac{b}{2a}$ . So, the equation of the axis of symmetry of the given parabola is $y=\frac{-4}{2\left(1\right)}$ or $y=-2$ .
Substitute $y=-2$ in the equation to find the $x$ -coordinate of the vertex.
$\begin{array}{l}x={\left(-2\right)}^{2}+4\left(-2\right)+2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4-8+2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-2\end{array}$
Therefore, the coordinates of the vertex are $\left(-2,-2\right)$ .
Now, substitute a few more $y$ -values in the equation to get the corresponding $x$ -values.
$y$ $x={y}^{2}+4y+2$ $-5$ $7$ $-4$ $2$ $-3$ $-1$ $-1$ $-1$ $0$ $2$ $1$ $7$
Plot the points and join them to get the parabola.
; |
Top
Rectangular to Polar Calculator
Top
The distance from x-axis and the y-axis located in the rectangular coordinate system can be represented as (x, y) coordinates. The x-coordinate and y-coordinate are the rectangular coordinate system.
The polar coordinate system is two dimensional which can be related and converted only to the other two dimensional coordinate systems alone. The coordinates used for the location of a point in the two dimensional rectangular coordinate system is (x, y). For the polar coordinate system it is (r, $\theta$).
Rectangular to Polar Calculator is used to convert Rectangular coordinates into polar coordinates.
Step by Step Calculation
Step 1 :
Observe the given rectangular coordinates(x, y).
Step 2 :
To convert rectangular coordinates into polar coordinates use the condition:
r = $\sqrt{x^{2} + y^{2}}$
$\theta$ = $tan^{-1}(\frac{y}{x})$
Example Problems
1. Convert the rectangular coordinates (4, 3) in to it's polar coordinates?
Step 1 :
Given rectangular coordinates = (4, 3) = (x, y)
Step 2 :
Polar coordinates (r , $\theta$)
r = $\sqrt{x^{2} + y^{2}}$ = $\sqrt{4^{2} + 3^{2}}$ = $\sqrt{16 + 9}$ = $\sqrt{25}$ = 5
$\theta$ = $tan^{-1}(\frac{y}{x})$ = $tan^{-1}(\frac{3}{4})$ = 36.870 degrees
(r, $\theta$) = (5, 36.870)
2. Convert the rectangular coordinates (5, 4) in to it's polar coordinates?
Step 1 :
Given rectangular coordinates = (5, 4) = (x, y)
Step 2 :
Polar coordinates (r , $\theta$)
r = $\sqrt{x^{2} + y^{2}}$ = $\sqrt{5^{2} + 4^{2}}$ = $\sqrt{25 + 16}$ = $\sqrt{41}$ = 6.40312424
$\theta$ = $tan^{-1}(\frac{4}{5})$ = $tan^{-1}(\frac{4}{5})$ = 36.66 degrees
(r, $\theta$) = (6.40312424, 36.66) |
# What are the intercepts of : 8y= - 2x - 9?
Apr 9, 2016
$x$ intercept: $\left(- \frac{9}{2} , 0\right)$
$y$ intercept: $\left(0 , - \frac{9}{8}\right)$
#### Explanation:
$x$ intercepts where $y = 0$
Therefore, substitute $y = 0$ and solve for $x$.
$0 = - 2 x - 9$
$9 = - 2 x$
$- \frac{9}{2} = x$
Therefore, $x$ intercept at $\left(- \frac{9}{2} , 0\right)$
$y$ intercepts where $x = 0$
Therefore, substitute $x = 0$ and solve for $y$.
$8 y = - 9$
$y = - \frac{9}{8}$
Therefore, $y$ intercept at $\left(0 , - \frac{9}{8}\right)$ |
# What Is a Prime Factor in Mathematics?
Prime numbers that, when multiplied, equal another number are said to be the “prime factors” of that number. For example, 2 and 7 are the prime factors of the number 14.
A prime number can only be multiplied by itself and the number 1. So, the number “8” is not prime because it’s possible to multiply “2” and “4” to produce the number 8. The number “7” is prime because no other whole numbers produce the number when multiplied together besides itself and the number 1. Determining prime factors of a number means breaking this number up into more numbers and then checking to see if those new numbers are prime. For example, the number “32” can be broken up into “8” and “4,” which when multiplied together make 32. But “8” and “4” aren’t prime numbers themselves. It’s necessary to break these numbers down further before they are prime factors. In this case, it takes four of the number “2” to make up the prime factors of number “32.” It’s also possible to write prime factors with an exponent if they include multiple of the same number. For example, the prime factors of 32 include “2” with an exponent of “4.”
Similar Articles |
## Thursday, October 11, 2012
There are many types of sequences seen in mathematics. A sequence refers to a list of numbers written in a specific order. Sequences can either be finite, which contains a set number of terms, or infinite, where the sequence continues on forever. A finite sequence is a function whose domain is the set of all natural numbers for a specified number n. An infinite sequence is a function whose domain is the set of all natural numbers. An easier way to denote terms of a sequence is to use an , which denotes the nth term of the sequence. Similarly a1 denotes the first term in the sequence, a2 denotes the second term of the sequence and so on.
The type of sequences focused on in this post is the arithmetic sequence. Suppose the sequence is 2, 6, 10, 14, 18, 22, ….... Notice that each term is 4 more than the previous term. A sequence such as this is known as an arithmetic sequence because each term is found by adding the same number to the previous term. An arithmetic sequence is a sequence in the form a1., a1 + d, a1 + 2d, a1 + 3d, a1 + 4d, …... a1 + (n – 1)d, where a1 is the first term of the sequence and d is the common difference between the terms.
The nth term of an arithmetic sequence is given by the formula "an" = a1 + (n – 1)d.
Consider the following arithmetic sequences below. Notice the common difference by subtracting the first term from the second, the second term from the third and so on.
Finite arithmetic sequence: 1, 8, 15, 22, 29
The common difference d is 7 since 8 - 1 = 7, 15 - 8 = 7, 22 - 15 = 7 and 29 – 22 = 7
Infinite arithmetic sequence: 6, 3, 0, -3, 6, -9, ….....
The common difference d is -3 since 3 - 6 = -3, -3 - 0 = -3, -6 - (-3) = -3, -9 - (-6) = -3 and so on.
In an arithmetic sequence, the first term is 8 and the common difference is -2. What are the first 6 terms of the sequence and what is the 29th term? To solve this problem we start with 8 and add the common difference to each term thereafter until we get all 6 terms.
Therefore,
a1 = 8, a2 = 8 – 2 = 6, a3 = 6 – 2 = 4, a4 = 4 – 2 = 2, a5 = 2 – 2 = 0, a6 = 0 – 2 = -2
The first 6 terms in the series are 8, 6, 4, 2, 0, -2.
To find the 29th term in the sequence, we use the formula to find the nth term in a sequence. Substitute 29 for n in the
formula "an" = a1 + (n – 1)d to get
a29 = 8 + (29 – 1)(-2)
a29 = 8 + (28)(-2)
a29 = 8 – 56
a29 = -48
Sometimes we don't know the first several terms of a sequence. Maybe we know the first term and the 50th term and we need to find the 15th term. How do we approach such a problem? First, we have to determine the common difference by substituting the value of the 50th term for an, the value of the first term for a1 and 30 for n into the formula "an" = a1 + (n – 1)d
Example: The first term of an arithmetic sequence is 6 and the 45th term is 490. What are the first 5 terms of the sequence?
To solve this problem we need to find the common difference. We use the formula an = a1 + (n – 1)d. Since the 45th term is 490, we substitute 490 for an . Since the first term is 6, we substitute 6 for a1 and 45 to get
490 = 6 + (45 – 1)d
490 = 6 + 44d
484 = 44d
11 = d
Since the common difference is 11, the first 5 terms of the sequence are 6, 17, 28, 39, 50 and are determined as follows
a1 = 6 , a2 = 6 + 11 = 17, a3 = 17 + 11 = 28, a4 = 28 + 11 = 39, a5 = 39 + 11 = 50.
Suppose we wish to find the sum of the first 10 terms of the sequence 1, 3, 5, 7, 9, 11, … We can easily solve this by adding the first 10 terms together. But what if we want the sum of the first 100 terms? Adding them would become tedious, so there is a formula we can use to solve such problems quickly.
The sum of the first n terms of an arithmetic sequence is given by
Sn = n(a1 + "an")/2
Sn = 2 , where a1 is the first term, an is the nth term and n is the number of terms.
Example: Find the sum of the first 40 terms of the sequence 7, 12, 17, 22, 27, …..
To solve this problem we need to find the common difference d. Once we find d, we can substitute the values for a1, n and d into the formula an = a1 + (n – 1)d so we can find a40. Once we find a40 we can use the formula for Sn to find the sum.
Therefore,
a40 = 7 + (40 – 1)(5)
a40 = 7 + 39(5)
a40 = 7 + 195
a40 = 202
Now we can use the formula for Sn to get
Sn = 40(7 + 202)/2
Sn = 20(209)
Sn = 4,180 |
Connect with us
The simple version of the quadratic formula was used 2000 years back by Babylonian mathematicians. The equation was almost the same as we are using today and it was written by a Hindu mathematician named Brahmagupta. This was only the quadratic equation that defined the concept of imaginary numbers and how can you show the same over a graph. This is useful for a variety of applications in science, engineering, advanced mathematics, construction, machine manufacturing and many more.
You must be surprised to know quadratic equations are a crucial part of our daily lives. Take an example of swing that is mobbing back and forth. When it is moving continuously, what type of shape will you notice? Obviously, this is a sort of arch or a part of the circle. Yes, of course!
The particular shape that is formed by swing movement can be understood better with quadratic equation when you graph it. In mathematics, quadratic equation is an expression with highest degree two and it is written in the form as given below –
$\large Quadratic\;Equation\;=ax^{2} + bx + c = 0$
Where a, b, c are three numbers that could be written as zero. This would be interesting to know that each quadratic equation has two solutions either real or imaginary. Next is the quadratic formula that is needed to compute the roots for a quadratic equation. Further, it would be interesting to know the history of quadratic formula how can you use this formula and prove it?
Every quadratic equation has two solutions that are calculated with the help of quadratic formula. In just a few simple steps, this is possible to find the solution either it is a whole number, rational number, or an imaginary number. In mathematics, the quadratic formula is given as –
$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$
### Equation Formula
For the polynomial having a degree two is called the quadratic equation that means it is squared. In mathematics, the standard quadratic equation formula is given as –
For each quadratic equation with degree two, there are two solutions that may be real or imaginary. This may be easy to solve quadratic equations with the help of quadratic formulas but to make them useful in daily application, you must have a depth understanding of the program. They are also needed to prepare yourself for the competitive exams.
### Cubic Equation Formula
The cubic equation has either one real root or it may have three-real roots. For the polynomial having a degree three is known as the cubic polynomial. In mathematics, the cubic equation formula can be given as –
$\LARGE ax^{3}+bx^{2}+cx+d=0$
Depressing the Cubic Equation
Substitute $\large x= y-\frac{b}{3a}$ in the above cubic equation, then we get,
$\large a\left ( y-\frac{b}{3a} \right )^{3}+b\left ( y-\frac{b}{3a} \right )^{2}+c\left ( y-\frac{b}{3a} \right )+d=0$
Simplifying further, we obtain the following depressed cubic equation –
$\large ay^{3}+\left ( c-\frac{b^{2}}{3a} \right )y+\left ( d+\frac{2b^{3}}{27a^{2}} +\frac{bc}{3a}\right )=0$
It must have the term in x3 or it would not be cubic ( and so a≠0), but any or all of b, c and d can be zero. For instance:
For instance: $\large x^{3}-6\times 2+11x-6=0\;or\;4x^{3}+57=0\;or\;x^{3}+9x=0$
### Exponential Equation Formula
The exponential equation is the equation where each side can be represented with the same base and it can be solved with the help of property. It can also be used to design a graph for compound interest, radioactive decay, and growth of population etc. In mathematics, the exponential equation formula can be given as –
$\large Exponential\;Equation = y=ab^{x}$
Where x and y are the constants and a, be would be the constants. |
In Each Example Given Below, Radius of Base of a Cylinder and Its Height Are Given. Then Find the Curved Surface Area and Total Surface Area. - Mathematics
Sum
In each example given below, radius of base of a cylinder and its height are given. Then find the curved surface area and total surface area.
(1) r = 7 cm, h = 10 cm
(2) r = 1.4 cm, h = 2.1 cm
(3) r = 2.5 cm, h = 7 cm
(4) r = 70 cm, h = 1.4 cm
(5) r = 4.2 cm, h = 14 cm
Solution
We know that
Curved surface area of a cylinder = 2 π rh
Total surface area = 2 πr (h + r)
(1) r = 7 cm, h = 10 cm
Curved surface area of a cylinder = 2 π rh = 2 π × 7 × 10 = 140 π = 140 × 22/7 = 440 sq cm
Total surface area = 2 π r (h + r) = 2 × 22/7 (10 + 7) = 748 sq cm
(2) r = 1.4 cm, h = 2.1 cm
Curved surface area of a cylinder = 2 π rh = 2 × 22/7 × 1.4 × 2.1 = 18.48 sq cm
Total surface area = 2 π r (h + r) = 2 × 22/7 ×1.4 (2.1 + 1.4) = 30.8 sq cm
(3) r = 2.5 cm, h = 7 cm
Curved surface area of a cylinder = 2 π rh = 2 × 22/7 × 2.5× 7 = 110 sq cm
Total surface area = 2 π r (h + r) = 2 × 22/7 × 2.5 (2.5 + 7) = 149.28 sq cm
(4) r = 70 cm, h = 1.4 cm
Curved surface area of a cylinder = 2 π rh = 2 × 22/7 × 70 × 1.4 = 616 sq cm
Total surface area = 2 π r (h + r) = 2 × 22/7 × 70 (70 + 1.4) = 31416 sq cm
(5) r = 4.2 cm, h = 14 cm
Curved surface area of a cylinder = 2 π rh = 2 × 22/7 × 4.2 × 14 = 369.6 sq cm
Total surface area = 2 π r (h + r) = 2 × 22/7 × 4.2 (4.2 + 1.4) = 480.48 sq cm
Is there an error in this question or solution?
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Balbharati Mathematics 8th Standard Maharashtra State Board
Chapter 16 Surface area and Volume
Practice Set 16.2 | Q 1 | Page 110 |
MATH
group hw 5
group hw 5 - Jessica Carlson Sarah Raubinger Christie...
• Notes
• 5
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Jessica Carlson Group Homework Sarah Raubinger 10/19/06 Christie Donahue 56. Find the lines through the origin that are tangent to the parabola. y=x 2 -2x+4 X Y -3 19 -2 12 -1 7 0 4 1 3 2 4 3 7 The derivative of the parabola is the slope of the tangent line at a given point. Using the Power Rule (d/dx(x n )=nx n-1 ), we can find the equation of y’ by bringing down the power of 2 down in front of the x in the first term of the equation making the term 2x. The second term will become a 2 since the power on the x is a 1. The last term will drop out since there is the “power on the x” is a 0. y’=2x-2 As you can see in the graph above, two lines (one on each side of the y-axis) exist that are line tangent to the graph and that pass through the point (0,0). Since we know the slope of the line is equal to the derivative of the original equation at a given point , we can use the Slope Formula (y 2 -y 1 )/(x 2 -x 1 ) to find the slope of a line that is tangent to the graph of the original function AND passes through the point (0,0). Since we don’t know the exact point on the graph that the tangent will pass through, we can use the an arbitrary values of x and y for the graph of the function (x, x 2 -2x+4). Thus, our two points for the slope formula are (x, x 2 -2x+4) and (0,0). We used these points as (x 1 ,y 1 ) and (x 2 ,y 2 ) respectively (although they could be switched to obtain the same answer). These two points in the slope formula are equal to the derivative of the equation of the graph, which we found above.
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• Spring '08
• BLAKELOCK
• Calculus, Derivative, Slope, Group Homework, Jessica Carlson Sarah Raubinger Christie Donahue, Christie Donahue, Jessica Carlson, Sarah Raubinger
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Comparing Fractions
Comparing Fractions
So far, you've learned how to easily compare fractions with like denominators and like numerators. ๐
Let's learn how to compare fractions with different denominators and different numerators.
Your first step is to find equivalent fractions that have either the same numerator or denominator.
Equivalent Fractions Review
You can find an equivalent fraction either by dividing or multiplying both numerator and denominator by the same number.
Let's try to find equivalent fractions for 6/9:
๐ First, we divide the numerator and the denominator by 3.
We get 2/3.ย
๐ The next step is to multiply the numerator and the denominator by 2.
We get 12/18.
The equivalent fractions of 6/9 are 2/3 and 12/18.
Comparing Fractions Example
Which fraction is bigger: 1/2 or 2/3?
1/2 and 2/3 do not have the same numerators, nor do they have the same denominators.
๐ So let's find some equivalent fractions for 1/2. ๐
We use multiplication to do this.
We've got 2/4, 3/6, and 4/8 as the equivalent fractions for 1/2.
Let's do the same with 2/3. ๐
We'll also use multiplication to do this.
We've got 4/6, 6/9, and 8/12.
๐ค Can you see a pair of equivalent fractions you can easily compare?
That's right!
You can compare 2/4 and 2/3 because they have the same numerators.
Since 2/4 and 2/3 have the same numerators, just compare the denominators.
Remember, when comparing fractions with the same numerators, the fraction with the smaller denominator is bigger.
๐ The denominators are 4 and 3. We know that 4 is bigger than 3.
This means 2/4 is smaller than 2/3.
Since 2/4 = 1/2, it follows that 1/2 is smaller than 2/3.ย
๐ค Is there another pair of equivalent fractions that you can easily compare?
Great!
You can also compare 3/6 and 4/6 because they have the same denominators.
Since 3/6 and 4/6 have the same denominators, just compare the numerators.
Remember, when comparing fractions with the same denominators, the fraction with the bigger numerator is greater.
๐ The numerators are 3 and 4. We know that 3 is less than 4.
This means 3/6 is smaller than 4/6.
Since 3/6 = 1/2 and 4/6 = 2/3, it follows that 1/2 is smaller than 2/3.
Secret Trick
Let's try comparing 1/2 and 2/3 using a secret trick!
๐ Multiply 1/2 by the denominator of 2/3.
1/2 = (1 x 3)/(2 x 3) = 3/6
So we see that 1/2 is equivalent to 3/6.
Now let's do the opposite. Let's multiply 2/3 with the denominator of 1/2.
2/3 = (2 x 2)/(3 x 2) = 4/6
So we see that 2/3 is equivalent to 4/6.
And now it's easy to compare 3/6 and 4/6 because they have the same denominators.
We know that:
3/6 < 4/6
When comparing fractions, multiply each fraction's denominator and numerator by the other fraction's denominator to get equivalent fractions with the same denominators. Then compare the equivalent fractions.
Great job learning how to compare fractions with different numerators and denominators!
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# Discuss the continuity of the following function at x = 0: $f(x) = \left\{ \begin{array}{l l}\large\frac{x^4+2x^3+x^2}{\tan^{-1}x}, & \quad if { x \; \neq 2 } \\ 0, & \quad if { x\; = 0 } \end{array} \right.$
Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
• $\lim\limits_{\large x\to 0}\large\frac{\tan^{-1}x}{x}$$=1 Step 1: f(x)=\left\{\begin{array}{1 1}\large\frac{x^4+2x^3+x^2}{\tan^{-1}x},&x\neq 2\\0,&x=0\end{array}\right. LHL at x\neq 2 \Rightarrow \lim\limits_{x\to 2^-}\large\frac{x^4+2x^3+x^2}{\tan^{-1}x} \Rightarrow \lim\limits_{x\to 2^-}\large\frac{x^2(x+1)^2}{\tan^{-1}x} \Rightarrow \lim\limits_{x\to 2^-}\large\frac{x(x+1)^2}{\Large\frac{\tan^{-1}x}{x}} Step 2: Since \lim\limits_{x\to 0}\large\frac{\tan^{-1}x}{x}$$=1$
$\Rightarrow \lim\limits_{x\to 2^-}\large\frac{2(2+1)^2}{1}$
$\Rightarrow 18$
Step 3:
Consider the RHL at $x=0$
$f(0)=\large\frac{0(0+1)^2}{1}$$=0$
Hence $f(0)=0$
Hence it is not continuous at $x=0$ |
## Tuesday, May 23, 2006
### Growing Post
Question 1:
A) Create a T-Chart showing how many baseball cards Jack has at the end of the next 4 months.
B) Create an algebraic formula based upon the problem above.
180 + 25T = Amount of baseball cards
T = The Month(s)
C) If Jack just turned 8 years old this month and Jack continues to buy the same amount of baseball cards each month, how many baseball cards will jack have when he turns 12?
Jack will have 1380 card by the time he's twelve.
Forty-eight months until Jack is twelve. 48 X 25 = 1200
1200 + 180= 1380
Formula: 180 + 25T= 1380
(T = 48)
Question 2:
A) If Jackie thinks that there will be four people at the party (counting herself) how many pizza’s should she order?
If Jackie has three friends she will need to have at least two pizzas so she can supply herself.
12 slices = 2 pizzas
12 slices = 4 people
B) If two more people show up at the party how many more pizza’s does she need to order?
If more people show up she will have to order three pizzas, since one pizza supports two people.
Question 3:
A) Calculate how much money Kathy will have in one year.
First multiply the months by the amount of money she recieves; 15 X 12 = 180
Add 20 to the sum; 200 + 20 = 200
B) Calculate how much money Kathy will have in one year.
This answer is quite simple. All you do is change the 20 in Kathy's formula, and change it to 25.
25 + 15n = Savings
If that was what Kathy had in her account, then her savings would be \$205.
Question Four:
A) 3N + 4N + 7 = 12 + 2N
3(1) + 4(1) + 7 = 2(1) = 12
3 + 4 + 7 = 14
N = 1
B) 8N - ( 4 + 9) = 11
8N - (4 + 9) = 11
8(3) - (4 + 9) = 11
24 - 13 = 11
N = 3
C) (8 - 3)N + 7N + 8 = 4N + 40
5N + 7N + 8 = 4N + 40
5(4) + 7(4) + 8 = 4(4) + 40
20 + 28 + 8 = 56
N = 4
At 5/23/2006 7:21 PM, Candy said...
GOOD JOB!!!!!!!!!!!!!!! TAYLOR BUT YOU ALSO PUT 4 INSTEAD OF 5 BUT THE ANSWER ACTUALLY IS 4 NOT 5.
At 5/24/2006 5:43 AM, Fang_the_Barbarian said...
your suppose to the brackets first
At 5/24/2006 10:56 AM, amelia_ mal said...
Hey Taylor! nice try for the post but u should mayby look over the post or edit it. it's Amalia
At 5/24/2006 11:59 AM, Mr. Reece said...
Plug 48 into the formula that you created for 1c.
For the second question 4 people include her (ie 3 + herself).
For 3a show how you plug the numbers into the formula that is already given
Good work dragon's eye
At 5/27/2006 9:40 PM, Aria said...
O_o What happened to all my text?
At 5/30/2006 9:48 AM, kimbutt said...
heey taylor.i like your growing post.
At 6/19/2006 3:00 PM, Mr. Reece said... |
1.7 Percent
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. This chapter contains many examples of arithmetic techniques that are used directly or indirectly in algebra. Since the chapter is intended as a review, the problem-solving techniques are presented without being developed. Therefore, no work space is provided, nor does the chapter contain all of the pedagogical features of the text. As a review, this chapter can be assigned at the discretion of the instructor and can also be a valuable reference tool for the student.
Overview
• The Meaning of Percent
• Converting A Fraction To A Percent
• Converting A Decimal To A Percent
• Converting A Percent To A Decimal
The meaning of percent
The word percent comes from the Latin word “per centum,” “per” meaning “for each,” and “centum” meaning “hundred.”
Percent (%)
Percent means “for each hundred” or “for every hundred.” The symbol % is used to represent the word percent.
Thus, $\begin{array}{rrrrr}\hfill 1%=\frac{1}{100}& \hfill & \hfill \text{or}& \hfill & \hfill 1%=0.01.\end{array}$
Converting a fraction to a percent
We can see how a fraction can be converted to a percent by analyzing the method that $\frac{3}{5}$ is converted to a percent. In order to convert $\frac{3}{5}$ to a percent, we need to introduce $\frac{1}{100}$ (since percent means for each hundred).
$\begin{array}{rrrrr}\hfill \frac{3}{5}& \hfill =& \frac{3}{5}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{100}{100}\hfill & \hfill & \text{Multiply\hspace{0.17em}the\hspace{0.17em}fraction\hspace{0.17em}by\hspace{0.17em}1}.\hfill \\ \hfill & \hfill =& \frac{3}{5}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill & \hfill & \text{Since \hspace{0.17em}}\frac{100}{100}=100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}.\hfill \\ \hfill & \hfill =& \frac{300}{5}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill & \hfill & \text{Divide\hspace{0.17em}}300\text{\hspace{0.17em}by\hspace{0.17em}5}.\hfill \\ \hfill & \hfill =& 60\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill & \hfill & \text{Multiply\hspace{0.17em}the\hspace{0.17em}fractions}.\hfill \\ \hfill & \hfill =& 60%\hfill & \hfill & \text{Replace\hspace{0.17em}}\frac{1}{100}\text{\hspace{0.17em}}\text{with\hspace{0.17em}the\hspace{0.17em}}%\text{\hspace{0.17em}symbol}.\hfill \end{array}$
Fraction to percent
To convert a fraction to a percent, multiply the fraction by 1 in the form $100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}$ , then replace $\frac{1}{100}$ with the % symbol.
Sample set a
Convert each fraction to a percent.
$\begin{array}{lll}\frac{1}{4}\hfill & =\hfill & \frac{1}{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \frac{100}{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 25\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 25%\hfill \end{array}$
$\begin{array}{lll}\frac{8}{5}\hfill & =\hfill & \frac{8}{5}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \frac{800}{5}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 160%\hfill \end{array}$
$\begin{array}{lll}\frac{4}{9}\hfill & =\hfill & \frac{4}{9}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \frac{400}{9}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \left(44.4...\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \left(44.\overline{4}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 44.\overline{4}%\hfill \end{array}$
Converting a decimal to a percent
We can see how a decimal is converted to a percent by analyzing the method that $0.75$ is converted to a percent. We need to introduce $\frac{1}{100}.$
$\begin{array}{lllll}0.75\hfill & =\hfill & 0.75\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill & \hfill & \text{Multiply\hspace{0.17em}the\hspace{0.17em}decimal\hspace{0.17em}by\hspace{0.17em}1}\text{.}\hfill \\ \hfill & =\hfill & 75\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill & \hfill & \hfill \\ \hfill & =\hfill & 75%\hfill & \hfill & \text{Replace\hspace{0.17em}}\frac{1}{100}\text{\hspace{0.17em}with\hspace{0.17em}the\hspace{0.17em}%\hspace{0.17em}symbol}.\hfill \end{array}$
Decimal to percent
To convert a fraction to a percent, multiply the decimal by 1 in the form $100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}$ , then replace $\frac{1}{100}$ with the % symbol. This amounts to moving the decimal point 2 places to the right.
Sample set b
Convert each decimal to a percent.
$\begin{array}{lll}0.62\hfill & =\hfill & 0.62\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 62\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 62%\hfill \end{array}$
Notice that the decimal point in the original number has been moved to the right 2 places.
$\begin{array}{lll}8.4\hfill & =\hfill & 8.4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 840\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 840%\hfill \end{array}$
Notice that the decimal point in the original number has been moved to the right 2 places.
$\begin{array}{lll}0.47623\hfill & =\hfill & 0.47623\text{\hspace{0.17em}}·\text{\hspace{0.17em}}100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 0.47623\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & 47.623%\hfill \end{array}$
Notice that the decimal point in the original number has been moved to the right 2 places.
Converting a percent to a decimal
We can see how a percent is converted to a decimal by analyzing the method that 12% is converted to a decimal. We need to introduce $\frac{1}{100}.$
$\begin{array}{lllll}12%\hfill & =\hfill & 12\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill & \hfill & \text{Replace}\text{\hspace{0.17em}}%\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}\frac{1}{100}.\hfill \\ \hfill & =\hfill & \frac{12}{100}\hfill & \hfill & \text{Multiply\hspace{0.17em}the\hspace{0.17em}fractions}.\hfill \\ \hfill & =\hfill & 0.12\hfill & \hfill & \text{Divide\hspace{0.17em}12\hspace{0.17em}by\hspace{0.17em}1}00.\hfill \end{array}$
Percent to decimal
To convert a percent to a decimal, replace the % symbol with $\frac{1}{100},$ then divide the number by 100. This amounts to moving the decimal point 2 places to the left.
Sample set c
Convert each percent to a decimal.
$\begin{array}{lll}48%\hfill & =\hfill & 48\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \frac{48}{100}\hfill \\ \hfill & =\hfill & 0.48\hfill \end{array}$
Notice that the decimal point in the original number has been moved to the left 2 places.
$\begin{array}{lll}659%\hfill & =\hfill & 659\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \frac{659}{100}\hfill \\ \hfill & =\hfill & 6.59\hfill \end{array}$
Notice that the decimal point in the original number has been moved to the left 2 places.
$\begin{array}{lll}0.4113%\hfill & =\hfill & 0.4113\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{100}\hfill \\ \hfill & =\hfill & \frac{0.4113}{100}\hfill \\ \hfill & =\hfill & 0.004113\hfill \end{array}$
Notice that the decimal point in the original number has been moved to the left 2 places.
Exercises
For the following problems, convert each fraction to a percent.
$\frac{2}{5}$
$40%$
$\frac{7}{8}$
$\frac{1}{8}$
$12.5%$
$\frac{5}{16}$
$15÷22$
$68.18%$
$\frac{2}{11}$
$\frac{2}{9}$
$22.22%$
$\frac{16}{45}$
$\frac{27}{55}$
$49.09%$
$\frac{7}{27}$
15
$1500%$
8
For the following problems, convert each decimal to a percent.
$0.36$
$36%$
$0.42$
$0.446$
$44.6%$
$0.1298$
$4.25$
$425%$
$5.875$
$86.98$
$8698%$
$21.26$
14
$1400%$
12
For the following problems, convert each percent to a decimal.
$35%$
$0.35$
$76%$
$18.6%$
$0.186$
$67.2%$
$9.0145%$
$0.090145$
$3.00156%$
$0.00005%$
$0.0000005$
$0.00034%$
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Please keep in mind that it's not allowed to promote any social groups (whatsapp, facebook, etc...), exchange phone numbers, email addresses or ask for personal information on QuizOver's platform. |
Sequences
Definition: An (infinite) sequence with rational coefficients is a function $$a:\mathbb{P}\to\mathbb{Q}$$. Normally we view it as the sequence $$a(1),a(2),\ldots$$.
Some examples:
• $$a(n)=0$$ for all $$n$$ (the zero sequence).
• $$a(n)=1/n$$ for $$n=1,2,\ldots$$
• $$a(n)=n$$ for $$n=1,2,\ldots$$
• $$a(n)=(-1)^n$$ for $$n=1,2,\ldots$$
We would like to have a way to speak about what happens to sequences as $$n$$ gets larger and larger (so for example the sequence grows, it bounces around, or it approaches a particular number.)
Limit of a sequence
Definition: Let $$a(n)$$ be a sequence. Then we say that the limit of $$a(n)$$ is $$L$$ if, for every $$\epsilon>0$$, there is an integer $$N$$, so that $$|a(n)-L|<\epsilon$$ for all $$n\ge N$$. This is written:
$\lim_{n\to\infty} a(n)=L.$
and we say that the sequence converges to $$L$$.
Examples
• Let $$a(n)$$ be the sequence defined by $$a(1)=1, a(2)=1/2, and a(n)=0$$ for $$n>2$$. Prove that the limit $$\lim_{n\to\infty} a(n)=0$$.
• Let $$a(n)$$ be the sequence $$a(n)=1/n$$. Prove that the limit of $$a(n)$$ as $$n\to\infty$$ is zero.
• Let $$a(n)=(-1)^{n}$$. Prove that the limit isn’t $$1$$. Then prove there is no limit.
• Let $$a(n)=n$$. Is there a limit?
• Let $$a(n)=(n+1)/n$$. Prove that the limit is $$1$$.
• Let $$a(n)=4+(-1/2)^{n}$$.
Non-convergence
A sequence $$a(n)$$ does not converge to a limit $$L$$ means that - there exists $$\epsilon>0$$ such that - for all $$N$$ - there exists $$n\ge N$$ such that - $$|a(n)-L|>\epsilon$$
The sequence $$a(n)=(-1)^{n-1}$$ does not converge to any limit because no matter what $$L$$ you pick and what $$N$$ you choose the distance $$|(-1)^{n-1}-L|$$ bounces back and forth between $$|1-L|$$ and $$|1+L|$$ so if you choose $$\epsilon$$ smaller than the maximum of these two you satisfy the ‘non-convergence’ requirement.
Limit rules make arguments standard
Proposition: If $$a(n)$$ converges to $$L$$ and $$b(n)$$ converges to $$M$$ then $$a(n)+b(n)$$ converges to $$L+M$$.
Proof: The estimation side calculation is that we can choose $$N$$ large enough that $$|a(n)-L|<\epsilon$$ and $$|b(n)-M|<\epsilon$$ for $$n\ge N$$. Then $$|a(n)+b(n)-L-M|<2\epsilon$$. So given $$\epsilon$$ we should choose $$N$$ large enought that $$|a(n)-L|<\epsilon/2$$ and similarly $$|b(n)-M|<\epsilon/2$$.
Proposition: Suppose that $$a(n)$$ is a sequence converging to $$L$$. Prove that there is an $$N$$ so that $$|a(n)|<2L$$ for $$n\ge N$$.
Proof: Choose $$\epsilon=L/2$$. Then there is an integer $$N$$ such that $$|a(n)-L|<L/2$$ for all $$n\ge N$$. This means that $$a(n)$$ is between $$L/2$$ and $$3L/2$$ so in particular it is less than $$2L$$.
Proposition: Suppose that $$a(n)$$ converges to $$L$$. Prove that $$a(n)^2$$ converges to $$L^2$$.
Proof: $$|a(n)^2-L^2|=|a(n)^2-a(n)L+a(n)L-L^2|\le |a(n)||a(n)-L| + |L||a(n)-L|$$. We can choose $$N$$ so that $$|a(n)|<2L$$ and $$N'$$ so that $$|a(n)-L|<\epsilon/4L$$. Then for $$n$$ bigger than both of these we have $|a(n)||a(n)-L|+|L||a(n)-L|\le (2L)\epsilon/4L+L\epsilon/4L=\epsilon/2+\epsilon/4=3\epsilon/4<\epsilon.$ Thus for $$n\ge\mathrm{max}(N,N')$$ we have $$|a(n)^2-L^2|<\epsilon.$$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.6: Inverse Trigonometric Ratios
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Use the inverse trigonometric ratios to find an angle in a right triangle.
• Solve a right triangle.
• Apply inverse trigonometric ratios to real-life situation and special right triangles.
## Review Queue
Find the lengths of the missing sides. Round your answer to the nearest hundredth.
1. Draw an isosceles right triangle with legs of length 3. What is the hypotenuse?
2. Use the triangle from #3, to find the sine, cosine, and tangent of 45\begin{align*}45^\circ\end{align*}.
3. Explain why tan45=1\begin{align*}\tan 45^\circ = 1\end{align*}.
Know What? The longest escalator in North America is at the Wheaton Metro Station in Maryland. It is 230 feet long and is 115 ft high. What is the angle of elevation, x\begin{align*}x^\circ\end{align*}, of this escalator?
## Inverse Trigonometric Ratios
The word inverse is probably familiar to you. In mathematics, once you learn how to do an operation, you also learn how to “undo” it. For example, you may remember that addition and subtraction are considered inverse operations. Multiplication and division are also inverse operations. In algebra you used inverse operations to solve equations and inequalities.
When we apply the word inverse to the trigonometric ratios, we can find the acute angle measures within a right triangle. Normally, if you are given an angle and a side of a right triangle, you can find the other two sides, using sine, cosine or tangent. With the inverse trig ratios, you can find the angle measure, given two sides.
Inverse Tangent: If you know the opposite side and adjacent side of an angle in a right triangle, you can use inverse tangent to find the measure of the angle.
Inverse tangent is also called arctangent and is labeled tan1\begin{align*}\tan^{-1}\end{align*} or arctan. The “-1” indicates inverse.
Inverse Sine: If you know the opposite side of an angle and the hypotenuse in a right triangle, you can use inverse sine to find the measure of the angle.
Inverse sine is also called arcsine and is labeled sin1\begin{align*}\sin^{-1}\end{align*} or arcsin.
Inverse Cosine: If you know the adjacent side of an angle and the hypotenuse in a right triangle, you can use inverse cosine to find the measure of the angle.
Inverse cosine is also called arccosine and is labeled cos1\begin{align*}\cos^{-1}\end{align*} or arccos.
Using the triangle below, the inverse trigonometric ratios look like this:
\begin{align*}\tan^{-1} \left ( \frac{b}{a} \right ) & = m \angle B && \tan^{-1} \left ( \frac{a}{b} \right ) = m \angle A\\ \sin^{-1} \left ( \frac{b}{c} \right ) & = m \angle B && \sin^{-1} \left ( \frac{a}{c} \right ) = m \angle A\\ \cos^{-1} \left ( \frac{a}{c} \right ) & = m \angle B && \cos^{-1} \left ( \frac{b}{c} \right ) = m \angle A\end{align*}
In order to actually find the measure of the angles, you will need you use your calculator. On most scientific and graphing calculators, the buttons look like \begin{align*}[\text{SIN}^{-1}], [\text{COS}^{-1}]\end{align*}, and \begin{align*}[\text{TAN}^{-1}]\end{align*}. Typically, you might have to hit a shift or \begin{align*}2^{nd}\end{align*} button to access these functions. For example, on the TI-83 and 84, \begin{align*}[2^{nd}][\text{SIN}]\end{align*} is \begin{align*}[\text{SIN}^{-1}]\end{align*}. Again, make sure the mode is in degrees.
When you find the inverse of a trigonometric function, you put the word arc in front of it. So, the inverse of a tangent is called the arctangent (or arctan for short). Think of the arctangent as a tool you can use like any other inverse operation when solving a problem. If tangent tells you the ratio of the lengths of the sides opposite and adjacent to an angle, then tangent inverse tells you the measure of an angle with a given ratio.
Example 1: Use the sides of the triangle and your calculator to find the value of \begin{align*}\angle A\end{align*}. Round your answer to the nearest tenth of a degree.
Solution: In reference to \begin{align*}\angle A\end{align*}, we are given the opposite leg and the adjacent leg. This means we should use the tangent ratio.
\begin{align*}\tan A = \frac{20}{25} = \frac{4}{5}\end{align*}, therefore \begin{align*}\tan^{-1} \left ( \frac{4}{5} \right ) = m \angle A\end{align*}. Use your calculator.
If you are using a TI-83 or 84, the keystrokes would be: \begin{align*}[2^{nd}][\text{TAN}]\left ( \frac{4}{5} \right )\end{align*}[ENTER] and the screen looks like:
So, \begin{align*}m \angle A = 38.7^\circ\end{align*}
Example 2: \begin{align*}\angle A\end{align*} is an acute angle in a right triangle. Use your calculator to find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.
a) \begin{align*}\sin A = 0.68\end{align*}
b) \begin{align*}\cos A = 0.85\end{align*}
c) \begin{align*}\tan A = 0.34\end{align*}
Solution:
a) \begin{align*}m \angle A = \sin^{-1} 0.68 = 42.8^\circ\end{align*}
b) \begin{align*}m \angle A = \cos^{-1} 0.85 = 31.8^\circ\end{align*}
c) \begin{align*}m \angle A = \tan^{-1} 0.34 = 18.8^\circ\end{align*}
## Solving Triangles
Now that we know how to use inverse trigonometric ratios to find the measure of the acute angles in a right triangle, we can solve right triangles. To solve a right triangle, you would need to find all sides and angles in a right triangle, using any method. When solving a right triangle, you could use sine, cosine or tangent, inverse sine, inverse cosine, or inverse tangent, or the Pythagorean Theorem. Remember when solving right triangles to only use the values that you are given.
Example 3: Solve the right triangle.
Solution: To solve this right triangle, we need to find \begin{align*}AB, m \angle C\end{align*} and \begin{align*}m \angle B\end{align*}. Use \begin{align*}AC\end{align*} and \begin{align*}CB\end{align*} to give the most accurate answers.
\begin{align*}\underline{AB}\end{align*}: Use the Pythagorean Theorem.
\begin{align*}24^2 + AB^2 & = 30^2\\ 576 + AB^2 & = 900\\ AB^2 & = 324\\ AB & = \sqrt{324} = 18\end{align*}
\begin{align*}\underline{m \angle B}\end{align*}: Use the inverse sine ratio.
\begin{align*}\sin B & = \frac{24}{30} = \frac{4}{5}\\ \sin^{-1} \left ( \frac{4}{5} \right ) & = 53.1^\circ = m \angle B\end{align*}
\begin{align*}\underline{m \angle C}\end{align*}: Use the inverse cosine ratio.
\begin{align*}\cos C & = \frac{24}{30} = \frac{4}{5}\\ \cos^{-1} \left ( \frac{4}{5} \right ) & = 36.9^\circ = m \angle C\end{align*}
Example 4: Solve the right triangle.
Solution: To solve this right triangle, we need to find \begin{align*}AB, BC\end{align*} and \begin{align*}m \angle A\end{align*}.
\begin{align*}\underline{AB}\end{align*}: Use sine ratio.
\begin{align*}\sin 62^\circ & = \frac{25}{AB}\\ AB & = \frac{25}{\sin 62^\circ}\\ AB & \approx 28.31\end{align*}
\begin{align*}\underline{BC}\end{align*}: Use tangent ratio.
\begin{align*}\tan 62^\circ & = \frac{25}{BC}\\ BC & = \frac{25}{\tan 62^\circ}\\ BC & \approx 13.30\end{align*}
\begin{align*}\underline{m \angle A}\end{align*}: Use Triangle Sum Theorem
\begin{align*}62^\circ + 90^\circ + m \angle A & = 180^\circ\\ m \angle A & = 28^\circ\end{align*}
Example 5: Solve the right triangle.
Solution: Even though, there are no angle measures given, we know that the two acute angles are congruent, making them both \begin{align*}45^\circ\end{align*}. Therefore, this is a 45-45-90 triangle. You can use the trigonometric ratios or the special right triangle ratios.
Trigonometric Ratios
\begin{align*}\tan 45^\circ & = \frac{15}{BC} && \sin 45^\circ = \frac{15}{AC}\\ BC & = \frac{15}{\tan 45^\circ} = 15 && \quad \ AC = \frac{15}{\sin 45^\circ} \approx 21.21\end{align*}
45-45-90 Triangle Ratios
\begin{align*}BC = AB = 15, AC = 15 \sqrt{2} \approx 21.21\end{align*}
## Real-Life Situations
Much like the trigonometric ratios, the inverse trig ratios can be used in several real-life situations. Here are a couple examples.
Example 6: A 25 foot tall flagpole casts a 42 feet shadow. What is the angle that the sun hits the flagpole?
Solution: First, draw a picture. The angle that the sun hits the flagpole is the acute angle at the top of the triangle, \begin{align*}x^\circ\end{align*}. From the picture, we can see that we need to use the inverse tangent ratio.
\begin{align*}\tan x & = \frac{42}{25}\\ \tan^{-1} \frac{42}{25} & \approx 59.2^\circ = x\end{align*}
Example 7: Elise is standing on the top of a 50 foot building and spots her friend, Molly across the street. If Molly is 35 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height is 4.5 feet.
Solution: Because of parallel lines, the angle of depression is equal to the angle at Molly, or \begin{align*}x^\circ\end{align*}. We can use the inverse tangent ratio.
\begin{align*}\tan^{-1} \left ( \frac{54.5}{30} \right ) = 61.2^\circ = x\end{align*}
Know What? Revisited To find the escalator’s angle of elevation, we need to use the inverse sine ratio.
\begin{align*}\sin^{-1} \left ( \frac{115}{230} \right ) = 30^\circ \qquad \text{The angle of elevation is}\ 30^\circ.\end{align*}
## Review Questions
Use your calculator to find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.
Let \begin{align*}\angle A\end{align*} be an acute angle in a right triangle. Find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.
1. \begin{align*}\sin A = 0.5684\end{align*}
2. \begin{align*}\cos A =0.1234\end{align*}
3. \begin{align*}\tan A = 2.78\end{align*}
Solving the following right triangles. Find all missing sides and angles.
1. Writing Explain when to use a trigonometric ratio to find a side length of a right triangle and when to use the Pythagorean Theorem.
Real-Life Situations Use what you know about right triangles to solve for the missing angle. If needed, draw a picture. Round all answers to the nearest tenth of a degree.
1. A 75 foot building casts an 82 foot shadow. What is the angle that the sun hits the building?
2. Over 2 miles (horizontal), a road rises 300 feet (vertical). What is the angle of elevation?
3. A boat is sailing and spots a shipwreck 650 feet below the water. A diver jumps from the boat and swims 935 feet to reach the wreck. What is the angle of depression from the boat to the shipwreck?
4. Elizabeth wants to know the angle at which the sun hits a tree in her backyard at 3 pm. She finds that the length of the tree’s shadow is 24 ft at 3 pm. At the same time of day, her shadow is 6 ft 5 inches. If Elizabeth is 4 ft 8 inches tall, find the height of the tree and hence the angle at which the sunlight hits the tree.
5. Alayna is trying to determine the angle at which to aim her sprinkler nozzle to water the top of a 5 ft bush in her yard. Assuming the water takes a straight path and the sprinkler is on the ground 4 ft from the tree, at what angle of inclination should she set it?
6. Science Connection Would the answer to number 20 be the same every day of the year? What factors would influence this answer? How about the answer to number 21? What factors might influence the path of the water?
7. Tommy was solving the triangle below and made a mistake. What did he do wrong? \begin{align*}\tan^{-1} \left ( \frac{21}{28} \right ) \approx 36.9^\circ\end{align*}
8. Tommy then continued the problem and set up the equation: \begin{align*}\cos 36.9^\circ = \frac{21}{h}\end{align*}. By solving this equation he found that the hypotenuse was 26.3 units. Did he use the correct trigonometric ratio here? Is his answer correct? Why or why not?
9. How could Tommy have found the hypotenuse in the triangle another way and avoided making his mistake?
Examining Patterns Below is a table that shows the sine, cosine, and tangent values for eight different angle measures. Answer the following questions.
\begin{align*}10^\circ\end{align*} \begin{align*}20^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}40^\circ\end{align*} \begin{align*}50^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}70^\circ\end{align*} \begin{align*}80^\circ\end{align*}
Sine 0.1736 0.3420 0.5 0.6428 0.7660 0.8660 0.9397 0.9848
Cosine 0.9848 0.9397 0.8660 0.7660 0.6428 0.5 0.3420 0.1736
Tangent 0.1763 0.3640 0.5774 0.8391 1.1918 1.7321 2.7475 5.6713
1. What value is equal to \begin{align*}\sin 40^\circ\end{align*}?
2. What value is equal to \begin{align*}\cos 70^\circ\end{align*}?
3. Describe what happens to the sine values as the angle measures increase.
4. Describe what happens to the cosine values as the angle measures increase.
5. What two numbers are the sine and cosine values between?
6. Find \begin{align*}\tan 85^\circ, \tan 89^\circ\end{align*}, and \begin{align*}\tan 89.5^\circ\end{align*} using your calculator. Now, describe what happens to the tangent values as the angle measures increase.
7. Explain why all of the sine and cosine values are less than one. (hint: think about the sides in the triangle and the relationships between their lengths)
1. \begin{align*}\sin 36^\circ = \frac{y}{7} \qquad \cos 36^\circ = \frac{x}{7}\!\\ {\;} \qquad y = 4.11 \qquad \quad \ \ x = 5.66\end{align*}
2. \begin{align*}\cos 12.7^\circ = \frac{40}{x} \qquad \ \tan 12.7^\circ = \frac{y}{40}\!\\ {\;} \qquad \ \ \ x = 41.00 \qquad \qquad \ \ y = 9.01\end{align*}
3. \begin{align*}{\;} \ \sin 45^\circ = \frac{3}{3 \sqrt{2}} = \frac{\sqrt{2}}{2}\!\\ {\;} \ \cos 45^\circ = \frac{3}{3 \sqrt{2}} = \frac{\sqrt{2}}{2}\!\\ {\;} \ \tan 45^\circ = \frac{3}{3} = 1\end{align*}
4. The tangent of \begin{align*}45^\circ\end{align*} equals one because it is the ratio of the opposite side over the adjacent side. In an isosceles right triangle, or 45-45-90 triangle, the opposite and adjacent sides are the same, making the ratio always 1.
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# Free Fall with Examples
FREE FALL
Free fall is a kind of motion that everybody can observe in daily life. We drop something accidentally or purposely and see its motion. At the beginning it has low speed and until the end it gains speed and before the crash it reaches its maximum speed. Which factors affect the speed of the object while it is in free fall? How can we calculate the distance it takes, time it takes during the free fall? We deal with these subjects in this section. First, let me begin with the source of increasing in the amount of speed during the fall. As you can guess, things fall because of the gravity. Thus, our objects gain speed approximately10m/s in a second while falling because of the gravitation. We call this acceleration in physics gravitational acceleration and show with “g”. The value of g is 9,8m/s² however, in our examples we assume it 10 m/ s² for simple calculations. Now it’s time to formulize what we said above. We talked about the increase in speed which is equal to the amount of g in a second. Thus our velocity can be found by the formula;
V=g.t where g is gravitational acceleration and t is the time.
Look at the given example below and try to understand what I tried to explain above.
Example The boy drops the ball from a roof of the house which takes 3 seconds to hit the ground. Calculate the velocity before the ball crashes to the ground. (g=10m/s²)
Velocity is;
V=g.t
V=10m/ s².3s=30m/s
We have learned how to find the velocity of the object at a given time. Now we will learn how to find the distance taken during the motion. I give some equations to calculate distance and other quantities. Galileo found an equation for distance from his experiments.
This equation is;
Using this equation we can find the height of the house in given example above. Let’s found how height the ball has been dropped? We use 10 m/s² for g.
I think the formula now a little bit clearer in your mind. We will solve more problems related to this topic. Now, think that if I throw the ball straight upward with an initial velocity. When it stops and falls back to the ground? We answer these questions now.
Picture shows the magnitudes of velocity at the bottom and at the top. As you can see the ball is thrown upward with an initial v velocity, at the top it’s velocity becomes zero and it changes it’s direction and starts to fall down which is free fall. Finally at the bottom before the crash it reaches its maximum speed which shown as V’. We have talked about the amount of increase in the velocity in free fall. It increases 9,8m/s in each second due to the gravitational acceleration. In this case, there is also g but the ball’s direction is upward; so the sign of g is negative. Thus, our velocity decreases in 9,8m/s in each second until the velocity becomes zero. At the top, because of the zero velocity, the ball changes its direction and starts to free fall. Before solving problems I want to give the graphs of free fall motion.
As you see in the graphs our velocity is linearly increases with an acceleration “g”, second graphs tells us that acceleration is constant at 9,8m/s², and finally third graphic is the representation of change in our position. At the beginning we have a positive displacement and as the time passes it decreases and finally becomes zero. Now we can solve problems using these graphs and explanations.
Example John throws the ball straight upward and after 1 second it reaches its maximum height then it does free fall motion which takes 2 seconds. Calculate the maximum height and velocity of the ball before it crashes the ground. (g=10m/s²)
Example An object does free fall motion. It hits the ground after 4 seconds. Calculate the velocity of the object after 3 seconds and before it hits the ground. What can be the height it is thrown?
Two examples given above try to show how to use free fall equations. We can find the velocity, distance and time from the given data. Now, I will give three more equations and finishes 1D Kinematics subject. The equations are;
First equation is used for finding the velocity of the object having initial velocity and acceleration. Second one is used for calculating the distance of the object having initial velocity and acceleration. Third and last equation is timeless velocity equation. If distance, initial velocity and acceleration of the object is known then you can find the final velocity of the object. Now let’s solve some problems using these equations to comprehend the subject in detail.
Example Calculate the velocity of the car which has initial velocity 24m/s and acceleration 3m/s² after 15 second.
We use the first equation to solve this question.
Example The car which is initially at rest has an acceleration 7m/s² and travels 20 seconds. Find the distance it covers during this period.
Kinematics Exams
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AP Stats Ch 6 Solutions - Chapter 6 Section 6.1 Check Your Understanding page 344 1 We are looking for the probability that the student gets either an A
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132 The Practice of Statistics for AP*, 4/e Chapter 6 Section 6.1 Check Your Understanding, page 344: 1. We are looking for the probability that the student gets either an A or a B. This probability is 0.42 0.26 0.68. + = 2. We are looking for ( ) 2 0.02 0.10 0.12. P X < = + = 3. This histogram is left skewed. This means that higher grades are more likely, but that there are a few lower grades. Check Your Understanding, page 349: 1. ( ) ( ) ( ) ( ) 0 0.3 1 0.4 2 0.2 3 0.1 1.1. X µ = + + + = The long-run average, over many Friday mornings, will be about 1.1 cars sold. 2. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 0 1.1 0.3 1 1.1 0.4 2 1.1 0.2 3 1.1 0.1 0.89. X σ = + + + = So 0.943. X σ = On average, the number of cars sold on a randomly selected Friday will differ from the mean (1.1) by 0.943 cars sold. Exercises, page 353: 6.1 (a) If you toss a coin 4 times, the sample space is {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}. All of these outcomes are equally likely and so have probability 1 0.0625. 16 = To find the probability of X taking on any specific number, count the number of outcomes with exactly this number of heads and multiply by 0.0625. For example, there are 4 ways to get exactly 3 heads so ( ) ( ) 3 4 0.0625 0.25. P X = = = This leads to the following distribution: Value 0 1 2 3 4 Probability 0.0625 0.25 0.375 0.25 0.0625
Chapter 6: Random Variables 133 (b) The histogram shows that this distribution is symmetric with a center at 2. (c) ( ) ( ) 3 1 4 1 0.0625 0.9375. P X P X = − = = − = There is a 93.75% chance that you will get three or fewer heads on 4 tosses of a fair coin. 6.2 (a) If we roll two 6-sided dice, the sample space is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. All of these are equally likely, so the probability of any one outcome is 1 . 36 To find the probability of X taking on any specific number, count the number of outcomes where the sum of the dice is exactly this number and multiply by 0.0278. For example, there are 4 ways to get a sum of 5 (the outcomes (1,4), (2,3), (3,2), and (4,1)) so ( ) 1 4 3 4 . 36 36 P X = = This leads to the following distribution: Value 2 3 4 5 6 7 8 9 10 11 12 Prob 1 36 2 36 3 36 4 36 5 36 6 36 5 36 4 36 3 36 2 36 1 36 (b) The histogram shows that the distribution is symmetric about a center of 7. (c) ( ) ( ) 1 2 3 6 1 5 5 1 4 1 1 1 . 36 36 36 36 6 6 P T P T = − = − + + = − = − = This means that about five-sixths of the time, when you roll a pair of 6-sided dice, you will have a sum of 5 or more.
134 The Practice of Statistics for AP* , 4/e 6.3 (a) “At least one nonword error” is the event { } 1 X or { } 0 . X > P(X ≥1) = 1 − P(X<1) = 1 − P(X=0) = 1 − 0.1 = 0.9. (b) The event {X 2} is “no more than two nonword errors,” or “fewer than three nonword errors.” P(X 2 ) = (X = 0 ) + P(X = 1 ) + P(X = 2 ) = 0 .1 + 0.2 + 0. 3 = 0 .6. P(X < 2) = P(X = 0 ) + P(X = 1 ) = 0 .1 + 0. 2 = 0 .3. 6.4 (a) “Plays with at most two toys” is the event { } 2 X or { } 3 . X < P(X 2 ) = (X = 0 ) + P(X = 1 ) + P(X = 2 ) = 0 .03 + 0.16 + 0.30 = 0 .49. (b) The event {X > 3} is “the child plays with more than three toys.” ( ) ( ) ( ) 3 4 5 0.17 0.11 0.28. P X P X P X > = = + = = + = ( ) ( ) 3 1 2 1 0.49 0.51. P X P X = − = − = 6.5 (a) All of the probabilities are between 0 and 1 and they sum to 1 so this is a legitimate probability distribution. (b) This is a right skewed distribution with the largest amount of probability on the digit 1.
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### Kissing Triangles
Determine the total shaded area of the 'kissing triangles'.
### Isosceles
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?
# Triangles in a Square
##### Age 11 to 14Challenge Level
We had lots of great solutions to this problem. Thank you to all of you who wrote in!
Julian from the British School Manila, Ahrus and Ben from Dixons Trinity Academy, and Kira from Wycombe High School all managed to find the area of each of the coloured triangles on the $5 \times 5$ grid. Here's what Kira did:
How to calculate the area of each triangle:
When the triangles are drawn on the dotty paper they are surrounded by $3$
right-angled triangles. The total area of all three right-angled triangles
must be subtracted from the area of the square ($25$ squares) to give the
area of the triangle.
Red triangle is $10.5$ squares
Yellow triangle is $10.5$ squares
Blue triangle is $11.5$ squares
Green triangle is $11$ squares
Purple triangle is $10.5$ squares
Therefore the blue triangle has the largest area.
Jurmana, Haidi, and Nour from the Continental School Cairo all found the largest and smallest triangles they could make on a $5 \times 5$ grid:
The largest area I found was $12.5$ cm$^2$ when the corners of the triangle are at $(0,0)$, $(0,5)$ and $(5,5)$.
The smallest area I found was $2.5$ cm$^2$.
Here's how Julian found a general formula for the area of a triangle with vertices at $(5,5)$, $(0,y)$, and $(x,0)$:
Area of the top left white triangle:
base: $5$
height: $5-y$
area: $\frac{5(5-y)}{2}$
Area of the bottom left white triangle:
base: $x$
height: $y$
area: $\frac{xy}{2}$
Area of the bottom right white triangle:
base: $5-x$
height: $5$
area: $\frac{5(5-x)}{2}$
Since the area of the colored triangle is $25$-(the area of the three
triangles covering the white space), we can add the three areas and
subtract it from $25$.
This will give us an area of $\frac{5x+5y-xy}{2}$.
Victor from Dulwich College Seoul found this formula and used it to find all the possible areas of triangles on a $5 \times 5$ grid:
The possible areas are $2.5cm^2$, $4.5cm^2$, $5cm^2$, $6.5cm^2$, $7.5cm^2$, $8cm^2$, $8.5cm^2$, $9.5cm^2$,$10cm^2$, $10.5cm^2$, $11cm^2$, $11.5cm^2$, $12cm^2$, $12.5cm^2$ which is a total of $14$ areas.
Kira thought about triangles on bigger grids:
When a triangle is drawn in a $6$ by $6$ grid the area can be represented by
$3x + 3y - 0.5xy$.
When a triangle is drawn in a $7$ by $7$ grid the area can be represented by
$3.5x + 3.5y - 0.5xy$.
This shows that there is a pattern for finding the area of the triangle. $x$
and $y$ are multiplied by half of the number of squares up or down.
Let $n$ represent size of grid eg $7$ by $7$ grid is when $n=7$:
Area of triangle in square = $0.5nx + 0.5 ny - 0.5xy$ squares. |
# Factors
Factors are the numbers we multiply together to get another number.
Example: 4 x 7 = 28
4 and 7 are factors of 28.
You might be thinking that there are other ways to multiply and make 28. You are correct!
You could also multiply 2 x 14 or 1 x 28. These are all factors of 28.
Let's list out the factors of 28: 1, 2, 4, 7, 14, 28. Each of the numbers in this list are factor pairs.
We can move from the outsides to the center and pair the numbers together to make sets of numbers that will multiply to make 28.
Take a look at another example:
List all the factors of 60.
Whenever we are asked to list out all of the factors, we can use these factor pairs to help make sure that we have not missed any. We will start on the outside and work our way in.
The first pair is 1 and 60.
Next, we will divide 60 by 2 because 60 is even. So the second pair is 2 and 30.
From here, we can continue to try the next number up in order. So the third pair is 3 and 20.
Now we will use 4 and 15.
60 ends in a zero, so we can use 5 and 12.
Then we will add 6 and 10.
Now, in between 6 and 10 is 7, 8 and 9. Once we have checked if any of these three numbers will go into 60, we know that we have found all of the factors of 60. None of them will divide evenly into 60. Therefore we have all of the factors of 60 listed above.
Factors are important because they help us solve problems. Factors can help us be able to break things into groups.
Example: There are 24 people in a room together at a party. Everyone would like to take part in games during the party. What size groups can we break the people into so that no one is left out and everyone can play?
To solve this problem, we need to know the factors of 24.
List them out: 1, 2, 3, 4, 6, 8, 12, 24
Let's see how the factor pairs can help us.
-The first pair, 1 and 24, doesn't tell us much. It just means that we could have 1 group of 24.
-The second pair tells us we could have 2 groups of 12 or 12 groups of 2.
-The third pair tells us we could have 3 groups of 8 or 8 groups of 3.
-The fourth pair tells us we could have 4 groups of 6 or 6 groups of 4.
Now we can see that there are several possibilities for grouping the party guests.
Let's check out a more geometrical example.
Example: Xin has a plot of land with an area of 36ft2. He wants to break his plot of land into different equal sized sections to plant different vegetables. How many vegetables can he plant?
Again, we will start by listing out the factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Let's again see how the factor pairs can help us.
The first pair, 1 and 36, doesn't tell us much. It just means that we could have 1 vegetable taking up all of the land or 36 vegetables taking up 1 square foot.
The second pair tells us we could have 2 veggies using 18 square feet each or 18 veggies using 2 square feet each.
The third pair tells us we could have 3 veggie using 12 square feet each or 12 veggies using 3 square feet each.
The fourth pair tells us we could have 4 veggies using 9 square feet each or 9 veggies using 4 square feet each.
The last pair tells us that we could have 6 veggies that each take up 6 square feet.
Each of these are possible solutions. And now Xin can decide how many plants to get.
Let's Review:
Factors are numbers that can be multiplied together to make another number. So 3 and 7 are factors of 21.
We can use factors to help us solve problems that ask us to break something into equal groups or equal sized sections. |
# Ex.3.1 Q4 Understanding Quadrilaterals Solution - NCERT Maths Class 8
Go back to 'Ex.3.1'
## Question
Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
FIGURES SIDE 3 4 5 6 ANGLE SUM $$180$$ \begin{align} 2 \times {180^{\rm{o}}} & = (4 - 2) \times {180^{\rm{o}}}\\ &= 360^\circ \end{align} \begin{align} 3 \times {180^{\rm{o}}}\\ & = (5 - 2) \times {180^{\rm{o}}}\\ & = 540 \end{align} \begin{align} 4 \times {180^{\rm{o}}}\\ & = (6 - 2) \times {180^{\rm{o}}}\\ & = {720^{\rm{o}}} \end{align}
What can you say about the angle sum of a convex polygon with number of sides?
(a) $$7$$
(b) $$8$$
(c) $$10$$
(d) $$n$$
Video Solution
Ex 3.1 | Question 4
## Text Solution
From the table, it can be observed that the angle sum of a convex polygon of $$n$$ sides is $$\left( {n - 2} \right){\rm{ }} \times {\rm{ }}180^\circ$$
(a) When $$n = 7$$
Then Angle sum of a polygon
\begin{align} &= (n - 2) \times {180^{\rm{\circ}}} \\&= (7 - 2) \times {180^{\rm{\circ}}} \\&= 5 \times {180^{\rm{\circ}}} \\&= {900^{\rm{\circ}}}\end{align}
(b) When $$n = 8$$
Then Angle sum of a polygon
\begin{align} &= (n - 2) \times {180^{ \rm{ \circ}}} \\&= (8 - 2) \times {180^{\rm{\circ}}} \\&= 6 \times{180^{\rm{\circ}}} \\&= {1080^{\rm{\circ}}}\end{align}
(c) When $$n = 10$$
Then Angle sum of a polygon
\begin{align} &= (n - 2) \times {180^{\rm{\circ}}} \\&= (10 - 2) \times {180^{\rm{\circ}}} \\&= 8 \times {180^{\rm{\circ}}} \\&= {1440^{\rm{\circ}}}\end{align}
(d) When $$n = n$$
Then Angle sum of a polygon $$= \left( {n - {\rm{ }}2} \right) \times 180^\circ$$
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Linear Algebra
# Application: Kirchhoff’s Laws and Circuits
Throughout this chapter, we’ve seen how linear algebra can be used to tackle systems of linear equations. This is important because of the various contexts in which systems of linear equations appear: even besides areas of math, several areas of physics rely on our ability to solve these systems.
One important area is circuits, which measures how electricity flows throughout various devices. Mathematically analyzing these circuits is extremely important for evaluating their efficacy: too little power into a device and the device won’t work; too much, and the device will overload. As we’ll see throughout this quiz, linear algebra provides the toolbox for analyzing circuits.
Let’s first define some of the basic modules present in circuits. In order to get to the actual math part, we won’t go through all of them, but we’ll touch on the important ones. Let’s start our exploration by looking at voltage (rather than current just yet).
First, the basic idea is that current travels through wires, in such a way that the voltage of each contiguous segment of wire (measured in volts, or V), known as a node, is the same. For example, the following circuit has two nodes, each with their own voltage:
Here, the boxes represent non-wire modules that we’ll look at shortly
How many nodes are in the following circuit?
The main purpose of analyzing circuits is to determine which nodes have which voltages, which in turn tells us how the voltage changes across modules, and thus tells us how much voltage the module uses.
The first of these modules is a simple battery, or voltage source. A battery with voltage $$v$$ enforces the difference between the positive node (the node connected to the positive part of the battery) and the negative node (the node connected to the negative part of the battery) is $$v$$. For example,
Here, the difference between the voltages at the green node and the red node is 10 volts.
The circuit below contains three batteries, with voltages $$a, b$$, and $$c$$. Which of the following must be true?
Before we look at our next module, we have to introduce the idea of current (measured in amperes, or A). Current is the flow of electricity through wires, which can in turn power modules such as lights. As such, current travels over modules as well as through wires.
An important property of current is Kirchoff’s law: the amount of current going into a node is the same as the amount of current going out of a node. For instance, one possible circuit could be the following:
We are now ready to introduce the second major module we’ll be working with: a resistor. Resistors relate voltage and current via Ohm’s law, which states the following: if $$v$$ is the difference in voltage across the nodes connecting to the resistor, $$i$$ is the current flowing across the resistor, and $$R$$ is the strength of the resistor $$($$measured in ohms, or $$\Omega),$$ we have $v = iR.$ The following circuit contains a battery and two resistors of strengths $$3 \Omega$$ and $$2 \Omega,$$ respectively. What is the value of the current travelling across the $$2 \Omega$$ resistor?
The last few problems illustrate a remarkable fact: using just batteries and resistors, we can induce a current across a module, which allows the module to use that current for power. This is why circuits are so important!
Unfortunately, circuits are usually much more complicated than the simple examples we’ve seen here. In the next couple problems, we’ll work through a much more complex circuit--and see how linear algebra ties into the analysis.
Over the next few problems, we’ll analyze the following complex circuit:
Here, a lightbulb is modeled as a $$50 \Omega$$ resistor. Our goal will be to find the current travelling across the bulb (remember, if the current is too little, the bulb won’t turn on; if the current is too much, the bulb will burn out).
First of all, how many nodes are in this circuit?
From the previous problem, we see there are 4 nodes. We can assume for simplicity that the pink node has $$0$$ V voltage. This means there are three nodes we don’t know the voltage of, and thus there are 3 variables we will need to solve for. As we’ve seen previously, this means we need to find 3 equations linking them.
Suppose the red node has voltage $$v_1$$ V, the green node has voltage $$v_2$$ V, and the blue node has voltage $$v_3$$ V. Analyzing the red node first, which of the following is true?
Now let’s look at a more complicated equation, by analyzing the green node. Recall that Ohm’s law allows us to write the currents in terms of $$v_1, v_2$$, and $$v_3$$. For instance, the current across the $$50 \Omega$$ resistor (from the green node to the red node) is $$\frac{v_2 - v_1}{50}$$ (make sure you see why!).
Finding the three currents and applying Kirchoff’s law to the green node, which of the following equations must be true?
In the last problem, we saw that applying Ohm’s and Kirchoff’s laws to the green node gives $\frac{v_2 - v_1}{50} + \frac{v_2 - 0}{300} + \frac{v_2 - v_3}{100} = 0,$ which can be rewritten as $-6v_1 + 10v_2 - 3v_3 = 0.$ Applying the same process to the blue node, which of the following equations is true?
We’ve now reduced to a system of three equations: $\begin{pmatrix}1&0&0\\-6&10&-3\\0&-1&3\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} = \begin{pmatrix}5\\0\\0\end{pmatrix}.$ Now we can finally find the current traveling across the bulb. What is this current?
In this chapter, we explored the mathematical analysis of circuits, which is integral to correctly setting up electrical systems. As we saw, using just the simple modules of batteries and resistors, we are able to induce a current that can power a lightbulb (or many other things, like the computer you are reading this on!).
As we saw, simple circuits can be analyzed “by hand,” but more complex circuits require a chunk of linear algebra to fully understand. Fortunately, the tools we’ve developed throughout this chapter reduce the problem to a systematic process, from which we can understand precisely what setup is necessary to--for instance--light a bulb.
× |
# Combination Formula
## Introduction to Combination:
• We have studied about arrangement i.e permutation in the previous chapter
• We learnt that arrangement of r objects taken from n objects is given by
$^{n}P_{r}=\frac {n!}{(n-r)!}$
We would be discussing the combination in this chapter
• Combination is defined as the different selection made by taking some objects out of many objects irrespective of their arrangement
So arrangement does not matter in Combination while it matters in Permutation
Example
We have three letter XYZ,
We have to select two letters and arrange it
Different arrangement would be
XY YZ ZX YX,ZY,XZ
So this is permuatation
Now if we have to just select two letter and do not worry about the arrangement then
XY, YZ ,ZX
This is combination
• Combination generally happens with committee selection problems as it does not matter how your committee is arranged. In other words, it generally does not matter whether you think about your committee as Vijay, Arun and Nita or as Nita, Vijay and Arun. In either case, the same three people are in the committee. Contrast this with permutations examples in previous where arrangement matters
## Combination Formula
A formula for the number of possible combinations of $r$ objects from a set of $n$ objects. This is written in any of the ways shown below
$^{n}C_{r} =\frac {n!}{r! (n-r)!}$
Proof:
Let $x$ be the combination of $n$ distinct objects taken $r$ at a time.
Now lets us consider one of the $x$ ways. There are $r$ objects in this one way which can be arranged in $r!$ways . Similarly all these $x$ ways can be arranged in this manner
So permutation of the n things taken r at a time will be given as
$^{n}P_{r}=x r!$
or $x=\frac { n!}{r! (n-r)!}$
$^{n}C_{r} =\frac {n!}{r! (n-r)!}$
Example:
Eleven students put their names on slips of paper inside a box. Three names are going to be taken out. How many different ways can the three names be chosen?
Solution
As the arrangement does not matter, so different ways are
$^{n}C_{r} =\frac {n!}{r! (n-r)!}$
$^{11}C_{3} =\frac {11!}{3! (11-3)!}=165$
## Properties of Combination Formula
$^{n}C_{r} =\frac {n!}{r! (n-r)!}$
1 $^{n}C_{n} = 1$ , $^{n}C_{0} = 1$ as $^{n}C_{n} =\frac {n!}{n! (n-n)!}= \frac {1}{0!} =1$ 2 for $0 \leq r <\leq n$ , we have $^{n}C_{r} = ^{n}C_{n-r}$ As $^{n}C_{r} =\frac {n!}{r! (n-r)!}$ $^{n}C_{n-r}= \frac {n!}{(n-r)! (n-n+r)!}= \frac {n!}{r! (n-r)!}$ 3 $^{n}C_{x} = ^{n}C_{y}$ Then $x=y$ or $x+y=n$ 4 $^{n}C_{r}+ ^{n}C_{r-1}= ^{n+1}C_{r}$ This property is called Pascal law’s 5 for $1 \leq r <\leq n$ , we have $^{n}C_{r} =\frac {n}{r} ^{n-1}C_{r-1}$ 6 for $1 \leq r <\leq n$, we have $n ^{n-1}C_{r-1} =(n-r+1) ^{n}C_{r-1}$
## Solved Examples
Question -1
$^{2n}C_{3} : ^{n}C_{3} =11 :1$
Find the value of n
Solution:
Writing this in factorial form
$\frac {\frac {2n!}{3! (2n-3)!}}{\frac {n!}{3!(n-3)!}} =\frac {11}{1}$
Simplifying it we get
$\frac {2n(2n-1)(2n-2)}{n(n-1)(n-2)}=\frac {11}{1}$
Solving this
$n=6$
Question-2
Over the weekend, your family is going on vacation, and your mom is letting you bring your favorite video game console as well as five of your games. How many ways can you choose the five games if you have 10 games in all?
Solution
Applying combination formula
$^{n}C_{r} =\frac {n!}{r! (n-r)!}$
$^{10}C_{5}=\frac {10!}{5! (10-5)!}=\frac {10 \times 9 \times 8 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=36$
Question-3
In how many ways can a T20 eleven can be chosen out of batch of 15 players if
1. There is no restriction on the selection
2. A particular player is always chosen
3. A particular player is never chosen
Solution:
1. If 11players to be chosen from 15 ,then $^{15}C_{11}=1365$
2. If the particular player is always choosen,then we need to choose 10 players from remaining 14 players $^{14}C_{10} =1001$
3. If the particular player is never choosen,then we need to choose 11 players from remaining 14 players $^{14}C_{11} =364$
## Related Topics
Go back to Class 12 Main Page using below links |
## Precalculus (6th Edition) Blitzer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right]$.
Consider the given matrix $A=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} \right]$. Compute the matrix of the form: $\left[ \left. A \right|I \right]$ The augmented matrix is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, using row operation we will reduce matrix in row echelon form for inverse as below: \begin{align} & {{R}_{4}}\to {{R}_{4}}-1\times {{R}_{1}}, \\ & {{R}_{3}}\to \frac{1}{3}\times {{R}_{3}}, \\ & {{R}_{2}}\to -1\times {{R}_{2}} \\ \end{align} The resulting matrix is: \begin{align} & \left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Where ${{A}^{-1}}=\left[ B \right]$ So, The inverse of matrix is: ${{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{4}}$ And, ${{A}^{-1}}A={{I}_{4}}$ Here, $A=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} \right]$ and ${{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times 1+0+0+0 & 0+0+0+0 & 0+0+0+0 & 0+0+0+0 \\ 0+0+0+0 & 0+\left( -1 \right)+0+0 & 0+0+0+0 & 0+0+0+0 \\ 0+0+0+0 & 0+0+0+0 & 0+0+3\times \frac{1}{3}+0 & 0+0+0+0 \\ 1+0+0+1\times \left( -1 \right) & 0+0+0+0 & 0+0+0+0 & 0+0+0+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{4}} \end{align} And, \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1+0+0+0 & 0+0+0+0 & 0+0+0+0 & 0+0+0+0 \\ 0+0+0+0 & 0+1+0+0 & 0+0+3+0 & 0+0+0+0 \\ 0+0+0+0 & 0+0+0+0 & 0+0+1+0 & 0+0+0+0 \\ \left( -1 \right)+0+0+1 & 0+0+0+0 & 0+0+0+0 & 0+0+0+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{4}} \end{align} Hence, $A{{A}^{-1}}={{I}_{4}}$ and ${{A}^{-1}}A={{I}_{4}}$. |
### Hellenica World
The differentiation of trigonometric functions is the mathematical process of finding the rate at which a trigonometric function changes with respect to a variable--the derivative of the trigonometric function. Commonplace trigonometric functions include sin(x), cos(x) and tan(x). For example, in differentiating f(x) = sin(x), one is calculating a function f ′(x) which computes the rate of change of sin(x) at a particular point a. The value of the rate of change at a is thus given by f ′(a). Knowledge of differentiation from first principles is required, along with competence in the use of trigonometric identities and limits. All functions involve the arbitrary variable x, with all differentiation performed with respect to x.
It turns out that once one knows the deriatives of sin(x) and cos(x), one can easily compute the derivatives of the other circular trigonometric functions because they can all be expressed in terms of sine or cosine; the quotient rule is then implemented to differentiate this expression. Proofs of the derivatives of sin(x) and cos(x) are given in the proofs section; the results are quoted in order to give proofs of the derivatives of the other circular trigonometric functions. Finding the derivatives of the inverse trigonometric functions involves using implicit differentiation and the derivatives of regular trigonometric functions also given in the proofs section.
Derivatives of trigonometric functions and their inverses
$$\left(\sin(x)\right)' = \cos(x)$$
$$\left(\cos(x)\right)' = -\sin(x)$$
$$\left(\tan(x)\right)' = \left(\frac{\sin(x)}{\cos(x)}\right)' = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} = \sec^2(x)$$
$$\left(\cot(x)\right)' = \left(\frac{\cos(x)}{\sin(x)}\right)' = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)} = -(1+\cot^2(x)) = -\csc^2(x)$$
$$\left(\sec(x)\right)' = \left(\frac{1}{\cos(x)}\right)' = \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)}.\frac{\sin(x)}{\cos(x)} = \sec(x)\tan(x)$$
$$\left(\csc(x)\right)' = \left(\frac{1}{\sin(x)}\right)' = -\frac{\cos(x)}{\sin^2(x)} = -\frac{1}{\sin(x)}.\frac{\cos(x)}{\sin(x)} = -\csc(x)\cot(x)$$
$$\left(\arcsin(x)\right)' = \frac{1}{\sqrt{1-x^2}}$$
$$\left(\arccos(x)\right)' = \frac{-1}{\sqrt{1-x^2}}$$
$$\left(\arctan(x)\right)' = \frac{1}{x^2+1}$$
Proofs of derivative of the sine and cosine functions
Limit of sin(θ)/θ as θ → 0
Limit circle
Consider the unit circle shown in the figure. Assume that the angle, say θ, made by radii OB and OC is small, e.g. less than π/2 radians, i.e. 90°. Let T1 denote the triangle with vertices O, B and C. Let S denote the sector given by the chords OB and OC (i.e. the "slice of pizza" given by cutting along the lines OB and OC). Let T2 denote the triangle with vertices O, B and D. Clearly, the area of T1 is less than the area of S, which is itself less than the area of T2, i.e. area(T1) < area(S) < area(T2).
The area of a triangle is given by one half of the length of the base multiplied by the height. Using u to denote the chosen unit of measurement, we find that the area of T1 is exactly 12 × ||OB|| × ||CA|| = 12 × 1 × sin(θ) = 12·sin(θu2. The area of the sector S is exactly 12·θ u2. Finally, the area of the triangle T2 is exactly 12 × ||OB|| × ||BD|| = 12·tan(θu2.
Since area(T1) < area(S) < area(T2) we find that, for small θ,
$$\frac{1}{2}\sin\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta \, .$$
(Recall that tan(θ) = sin(θ)/cos(θ).) If this is true then multiplication by two gives sin(θ) < θ < tan(θ). Take the reciprocals of each term will reverse the inequalities, e.g. 2 < 3 while 1⁄2 > 1⁄3. It follows that
$$\frac{1}{\sin \theta} > \frac{1}{\theta} > \frac{\cos\theta}{\sin\theta} \, .$$
Since θ is small, and so less than π/2 radians, i.e. 90°, it follows that sin(θ) > 0. We can multiply through by the positive quantity sin(θ) without changing the inequalities; whence:
$$1 > \frac{\sin\theta}{\theta} > \cos\theta \, .$$
This tells us that for very small θ, sin(θ)/θ is less than 1, but is bigger than cos(θ). However, as θ becomes smaller, cos(θ) becomes bigger and gets closer to 1 (see the cosine graph). The inequality tells us that sin(θ)/θ is always less than 1 and more than cos(θ); but as θ becomes smaller, cos(θ) gets closer to 1. So, sin(θ)/θ is sandwiched, or squeezed, between 1 and cos(θ) as θ becomes smaller and smaller. Obviously, the number 1 is fixed, and so as θ becomes smaller and smaller, cos(θ) forces sin(θ)/θ to move closer and closer to 1. This is made precise by the so-called sandwich theorem, also known as the squeeze theorem.
Limit of [cos(θ) – 1]/θ as θ → 0
The last section enables us to calculate this new limit easily. We find that
$$\lim_{\theta \to 0} \left(\frac{\cos\theta - 1}{\theta}\right) = \lim_{\theta \to 0} \left[ \left( \frac{\cos\theta - 1}{\theta} \right) \left( \frac{\cos\theta + 1}{\cos\theta + 1} \right) \right] = \lim_{\theta \to 0} \left( \frac{\cos^2\theta - 1}{\theta(\cos\theta + 1)} \right) .$$
The well-known identity sin2θ + cos2θ = 1 tells us that cos2θ – 1 = –sin2θ. Using this, the fact that the limit of a product is the product of the limit, and the result from the last section, we find that
$$\lim_{\theta \to 0} \left(\frac{\cos\theta - 1}{\theta}\right) = \lim_{\theta \to 0} \left( \frac{-\sin^2\theta}{\theta(\cos\theta+1)} \right) = \lim_{\theta \to 0} \left( \frac{-\sin\theta}{\theta}\right) \times \lim_{\theta \to 0} \left( \frac{\sin\theta}{\cos\theta + 1} \right) = (-1) \times \frac{0}{2} = 0 \, .$$
Derivative of the sine function
To calculate the derivative of the sine function, sin(θ) we use first principles. By definition:
$$\frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\sin\theta = \lim_{\delta \to 0} \left( \frac{\sin(\theta + \delta) - \sin \theta}{\delta} \right) .$$
Using the well-known angle formula sin(α+β) = sin(α)cos(β) + sin(β)cos(α) and the two limits from this section and this section, we find that
$$\frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\sin\theta = \lim_{\delta \to 0} \left( \frac{\sin\theta\cos\delta + \sin\delta\cos\theta-\sin\theta}{\delta} \right) = \lim_{\delta \to 0} \left[ \left(\frac{\sin\delta}{\delta} \cos\theta\right) + \left(\frac{\cos\delta -1}{\delta}\sin\theta\right) \right] = (1\times\cos\theta) + (0\times\sin\theta) = \cos\theta \, .$$
Derivative of the cosine function
To calculate the derivative of the cosine function, cos(θ) we use first principals. By definition:
$$\frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\cos\theta = \lim_{\delta \to 0} \left( \frac{\cos(\theta+\delta)-\cos\theta}{\delta} \right) . Using the well-known angle formula cos(α+β) = cos(α)cos(β) – sin(α)sin(β) and the two limits from this section and this section, we find that \( \frac{\operatorname{d}}{\operatorname{d}\!\theta}\,\cos\theta = \lim_{\delta \to 0} \left( \frac{\cos\theta\cos\delta - \sin\theta\sin\delta-\cos\theta}{\delta} \right) = \lim_{\delta \to 0} \left[ \left(\frac{\cos\delta -1}{\delta}\cos\theta\right) - \left(\frac{\sin\delta}{\delta} \sin\theta\right) \right] = (0 \times \cos\theta) - (1 \times \sin\theta) = -\sin\theta \, . Proofs of derivatives of inverse trigonometric functions The following derivatives are found by setting a variable y equal to the inverse trigonometric function that we wish to take the derivative of. Using implicit differentiation and then solving for dy/dx, the derivative of the inverse function is found in terms of y. To convert dy/dx back into being in terms of x, we can draw a reference triangle on the unit circle, letting θ be y. Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x. Differentiating the inverse sine function We let \( y=\arcsin x\,\!$$
Where
$$-\frac{\pi}{2}\le y \le \frac{\pi}{2}$$
Then
$$\sin y=x\,\!$$
Using implicit differentiation and solving for dy/dx:
$${d \over dx}\sin y={d \over dx}x$$
$${dy \over dx}\cos y=1\,\!$$
Substituting $$\cos y = \sqrt{1-\sin^2 y}$$ in from above,
$${dy \over dx}\sqrt{1-\sin^2 y}=1$$
Substituting x=\sin y in from above,
$${dy \over dx}\sqrt{1-x^2}=1$$
{dy \over dx}=\frac{1}{\sqrt{1-x^2}} \)
Differentiating the inverse cosine function
We let
$$y=\arccos x\,\!$$
Where
$$0 \le y \le \pi$$
Then
$$\cos y=x\,\!$$
Using implicit differentiation and solving for dy/dx:
$${d \over dx}\cos y={d \over dx}x$$
$$-{dy \over dx}\sin y=1$$
Substituting $$\sin y = \sqrt{1-\cos^2 y}\,\!$$ in from above, we get
$$-{dy \over dx}\sqrt{1-\cos^2 y} =1$$
Substituting x=\cos y\,\! in from above, we get
$$-{dy \over dx}\sqrt{1-x^2} =1$$
$${dy \over dx} = -\frac{1}{\sqrt{1-x^2}}$$
Differentiating the inverse tangent function
We let
$$v y=\arctan x\,\!$$
Where
$$-\frac{\pi}{2} < y < \frac{\pi}{2}$$
Then
$$\tan y=x\,\!$$
Using implicit differentiation and solving for dy/dx:
$${d \over dx}\tan y={d \over dx}x$$
$${dy \over dx}\sec^2 y=1$$
Substituting $$1+\tan^2 y = \sec^2 y\,\!$$ into the above,
$${dy \over dx}(1+\tan^2 y)=1$$
Substituting $$x=\tan y\,\!$$ in from above,
$${dy \over dx}(1+x^2)=1$$
$${dy \over dx}=\frac{1}{1+x^2}$$
Trigonometry
Calculus
Derivative
Table of derivatives
References
Bibliography
Handbook of Mathematical Functions, Edited by Abramowitz and Stegun, National Bureau of Standards, Applied Mathematics Series, 55 (1964). |
AREAS OF CIRCLES AND SECTORS
About "Areas of Circles and Sectors"
Areas of Circles and Sectors :
In this section, we are going to see, how to find areas of circles and sectors using formulas.
Area of a Circle :
The area of a circle is π times the square of the radius.
A = πr2
Area of a Sector :
The sector is the region bounded by two radii of the circle and their intercepted arc. In the diagram shown below, sector APB is bounded by radii AP, BP and arc arc AB.
The ratio of the area A of a sector of a circle to the area of the circle is equal to the ratio of the measure of the intercepted arc to 360°.
A / πr2 = m∠arc AB / 360°
So, the area of the sector is
A = [m∠arc AB / 360°] ⋅ πr2
Finding the Area of a Circle
Example :
Find the area of the circle shown below.
Solution :
Formula area of a circle is given by
A = πr2
Plug r = 8.
A = π(8)2
A = 64π
Use calculator.
A ≈ 201.06
So, the area is 64π, or about 201.06, square inches.
Using the Area of a Circle
Example :
If the area of a circle is 96 square centimeters, find its diameter.
Solution :
Formula area of a circle is given by
A = πr2
Plug A = 96.
96 = πr2
Divide each side π.
96/π = πr2/π
96/π = r2
Use calculator.
30.56 r2
Take square root on each side.
5.53 r
So, the diameter of the circle is about 2(5.53), or about 11.06, centimeters.
Finding the Area of a Sector
Example 1 :
Find the area of the sector shown at the right.
Solution :
Sector CPD intercepts an arc whose measure is 80°. The radius is 4 feet.
Formula for area of a sector is given by
A = [m∠arc CD / 360°] ⋅ πr2
Plug m∠arc CD = 80° and r = 4.
A = [80° / 360°] ⋅ π(4)2
A = (2 / 9) ⋅ 16π
Use calculator.
A ≈ 11.17
So, the area of the sector is about 11.17 square feet.
Example 2 :
A and B are two points on a P with radius 9 inches and APB = 60°. Find the areas of the sectors formed by APB.
Solution :
Draw a diagram of ⊙P and APB. Shade the sectors. Label a point Q on the major arc.
Find the measures of the minor and major arcs.
Because m∠APB = 60°, we have
m∠arc AB = 60°
and
m∠AQB = 360° - 60° = 300°
Use the formula for the area of a sector.
A = [m∠arc CD / 360°] ⋅ πr2
Plug m∠arc CD = 80° and r = 4.
A = [80° / 360°] ⋅ π(4)2
A = (2 / 9) ⋅ 16π
Use calculator.
A ≈ 11.17
So, the area of the sector is about 11.17 square feet.
Area of Smaller SectorA = 60°/360° ⋅ π(9)2A = 1/6 ⋅ π ⋅ 81A ≈ 42.41 square inches Area of Larger SectorA = 300°/360° ⋅ π(9)2A = 5/6 ⋅ π ⋅ 81A ≈ 212.06 square inches
Finding the Area of a Region
We may need to divide a figure into different regions to find its area. The regions may be polygons, circles, or sectors. To find the area of the entire figure, add or subtract the areas of the separate regions as appropriate.
Example 1 :
Find the area of the shaded region shown below.
Solution :
The diagram shows a regular hexagon inscribed in a circle with radius 5 meters. The shaded region is the part of the circle that is outside of the hexagon.
Area of shaded region = Area of circle - Area of hexagon
Area of shaded region = πr2 - 1/2 ⋅ a ⋅ p
Radius of the circle is 5 and the apothem of a hexagon is
= 1/2 ⋅ side length ⋅ √3
= 1/2 ⋅ 5 ⋅ √3
5√3/2
So, the area of the shaded region is
= [π ⋅ 52] - [1/2 ⋅ (5√3/2) ⋅ (6 ⋅ 5)]
= 25π - 75√3/2
Use calculator.
≈ 13.59
So, the area of the shaded region is about 13.59 square meters.
Example 2 :
You are cutting the front face of a clock out of wood, as shown in the diagram. What is the area of the front of the case ?
Solution :
The front of the case is formed by a rectangle and a sector, with a circle removed. Note that the intercepted arc of the sector is a semicircle.
So, the required area is
= Area of rectangle + Area of sector - Area of circle
= [6 ⋅ 11/2] + [180°/360° ⋅ π ⋅ 32] - [π ⋅ (1/2 4)2]
= 33 + 9/2 ⋅ π - 4π
Use calculator.
≈ 34.57
The area of the front of the case is about 34.57 square inches.
After having gone through the stuff given above, we hope that the students would have understood, "Areas of Circles and Sectors".
Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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# 8.7: Exponential Growth
Difficulty Level: Basic Created by: CK-12
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Suppose that 1000 people visited an online auction website during its first month in existence and that the total number of visitors to the auction site is tripling every month. Could you write a function to represent this situation? How many total visitors will the auction site have had after 9 months?
### Exponential Growth
Previously, we have seen the variable only used as the base in an exponential expression. In exponential functions, the exponent is the variable and the base is a constant.
The General Form of an Exponential Function is , where initial value and .
In exponential growth situations, the growth factor must be greater than one.
#### Let's use an exponential function to solve the following problem:
A colony of bacteria has a population of 3,000 at noon on Sunday. During the next week, the colony’s population doubles every day. What is the population of the bacteria colony at noon on Saturday?
Make a table of values and calculate the population each day.
Day 0 (Sun) 1 (Mon) 2 (Tues) 3 (Wed) 4 (Thurs) 5 (Fri) 6 (Sat)
Population (thousands) 3 6 12 24 48 96 192
To get the population of bacteria for the next day we multiply the current day’s population by 2 because it doubles every day. If we define as the number of days since Sunday at noon, then we can write the following: . This is a formula that we can use to calculate the population on any day. For instance, the population on Saturday at noon will be thousand bacteria. We use , since Saturday at noon is six days after Sunday at noon.
#### Graphing Exponential Functions
Graphs of exponential growth functions show you how quickly the values of the functions get very large.
#### Let's use tables of values to complete the following problems:
1. Graph .
Make a table of values that includes both negative and positive values of . Substitute these values for to get the value for the variable.
–3
–2
–1
0 1
1 2
2 4
3 8
Plot the points on the coordinate axes to get the graph below. Exponential functions always have this basic shape: They start very small and then once they start growing, they grow faster and faster, and soon they become huge.
1. In the last problem, we produced a graph for . Compare that graph with the graph of .
–2
–1
0
1
2
3
We can see that the function is bigger than the function . In both functions, the value of doubles every time increases by one. However, starts with a value of 3, while starts with a value of 1, so it makes sense that would be bigger.
The shape of the exponential graph changes if the constants change. The curve can become steeper or shallower.
### Examples
#### Example 1
Earlier, you were told that 1000 people visited an online auction website during its first month in existence and that the total number of visitors to the auction site is tripling every month. What function represents this situation? How many total visitors will the auction site have had after 9 months?
To write the function, it is helpful to write out a table of values where is the month since the website opened and is the number of visitors
1 1000
2 3000
3 9000
4 27000
5 81000
6 243000
7 729000
8 2187000
9 6561000
Notice that the initial value is 1000 and the growth factor is 3. Therefore, you would expect the exponential function that represents this situation to be:
However, if you plug in 1 to this equation, you will get 3000 instead of 1000. The exponent value is 1 above what it should be. Therefore, you need to decrease the exponent by 1 and the exponential function that represents this situation is:
If you plug in 1 to this equation, you will get 1000 as necessary. You should test the other numbers from the table in this equation to verify that the function is correct.
To find how many total visitors that the website would have had after 9 months, add all the values of in the table from above:
In the first 9 months, the website had 9,841,000 total visitors.
#### Example 2
A population of 500 E. Coli organisms doubles every fifteen minutes. Write a function expressing the population size as a function of hours.
Since there are four 15 minute periods in an hour, this means the population will double 4 times in an hour. Doubling twice is the same thing as quadrupling since:
This means that doubling 4 times can be calculated as . So the population is 16 times as big every hour. With an initial population size of 500, the function is:
where is in hours and is the number of organisms after hours.
### Review
1. What is the general form for an exponential equation? What do the variables represent?
2. How is an exponential growth equation different from a linear equation?
3. What is true about the growth factor of an exponential equation?
4. True or false? An exponential growth function has the following form: , where and ?
5. What is the intercept of all exponential growth functions?
Graph the following exponential functions by making a table of values.
Solve the following problems involving exponential growth.
1. A chain letter is sent out to 10 people telling everyone to make 10 copies of the letter and send each one to a new person. Assume that everyone who receives the letter sends it to 10 new people and that it takes a week for each cycle. How many people receive the letter in the sixth week?
2. Nadia received \$200 for her birthday. If she saves it in a bank with a 7.5% interest rate compounded yearly, how much money will she have in the bank by her birthday?
Mixed Review
1. Suppose a letter is randomly chosen from the alphabet. What is the probability the letter chosen is , or ?
2. Evaluate when .
3. Simplify .
4. Graph .
To see the Review answers, open this PDF file and look for section 8.7.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
Exponential growth Exponential growth occurs when a quantity increases by the same proportion in each given time period.
general form of an exponential function $y=a (b)^x$, where $a=$ initial value and $b= growth \ factor$
Asymptotes An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions).
Model A model is a mathematical expression or function used to describe a physical item or situation.
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# Lesson Plan: Identifying Proportional Relationships
## Lesson Objectives
This lesson can be completed in one 50-minute class period but may require additional time depending on your class.
• Identify proportional relationships in various representations
• Understand and calculate the constant of proportionality
• Apply proportional reasoning to solve real-world problems
## TEKS Standards
• 7.4A: Represent constant rates of change in various forms
• 7.4C: Determine the constant of proportionality
• 7.4D: Solve problems involving ratios, rates, and percents
## Prerequisite Skills
• Understanding of ratios and unit rates
• Basic graphing skills
## Key Vocabulary
• Proportion
• Constant of proportionality
• Origin
• Linear relationship
As needed, refer to these grade 6 lesson plans:
## Warm-up Activity (10 minutes)
Identify equivalent ratios in a table
Example: Given the following table of values, identify pairs of numbers that form equivalent ratios:
xy
26
412
618
824
1030
Solution: All pairs form equivalent ratios (1:3), as y is always 3 times x. You can use this Desmos activity to graph the coordinates and find the line of best fit:
https://www.desmos.com/calculator/gazl6xwvfg
## Teach (20 minutes)
### Definitions
• Proportion: An equation stating that two ratios are equal
• Constant of proportionality: The constant ratio between two proportional quantities
• Origin: The point (0,0) on a coordinate plane
• Linear relationship: A relationship that forms a straight line when graphed
Use this slide show to review these other related definitions:
https://www.media4math.com/library/slideshow/definitions-proportions-and-proportional-relationships
### Example 1. Science: Hooke's Law (Force and Spring Extension)
This slide show discusses Hooke's Law in detail.
https://www.media4math.com/library/slideshow/application-linear-functions-hookes-law
Use it as background to frame the following problem solving scenario:
A physics student is investigating the relationship between the force applied to a spring and its extension. She records the following measurements:
Force (N)Extension (cm)
00
21
42
63
84
• Graph: Plot points and observe direct proportion. Use this Desmos activity:
https://www.desmos.com/calculator/lr5b4efzgq
• Constant of proportionality: Extension/Force = 0.5 cm/N
• Explain how this demonstrates a direct proportional relationship.
• Discuss real-world applications, such as in the design of suspension systems or measuring instruments.
### Example 2. Engineering: Gear Ratios
Use this slide show to demonstrate this problem solving scenario with gear ratios:
https://www.media4math.com/library/slideshow/applications-gear-ratios
Here is a summary of the scenario
An engineer is designing a gear system for a new machine. She needs to determine the relationship among the number of teeth for each of the gears:
• Gear A has 20 teeth
• Gear B has 12 teeth
• Gear C has 8 teeth
Determine the number of turns gear C has to make in order for gears A and B complete at least one turn.
• The gear ratio, in simplified form is this: 5:3:2
• The revolution ratio, in simplified form is this: 2:3:5
• Gear C must complete at least 2.5 turns for gears A and B to complete at least one turn
### Example 3. Art: Color Mixing
Word problem: An artist is creating a new shade of green by mixing yellow and blue paint. He wants to ensure he can consistently reproduce this color:
Yellow Paint (mL)Blue Paint (mL)
52
104
156
208
• Graph: Plot points and observe direct proportion
• Constant of proportionality: Blue/Yellow = 0.4
• Explain how this is used in creating consistent shades of green
• Discuss applications in graphic design, painting, and digital art
## Review (10 minutes)
• Practice identifying proportional relationships in various representations
• Determine the constant of proportionality in different contexts
• Have students work in pairs or small groups to analyze given data sets
### Example 1 (Business): Sales Commission
A real estate agent earns a 5% commission on each house sale. The following table shows the commission earned for different house prices:
House Price (\$)Commission (\$)
100,0005,000
200,00010,000
300,00015,000
400,00020,000
1. Determine if this is a proportional relationship
2. Identify the constant of proportionality
3. Calculate the commission for a \$350,000 house sale Solutions: 1. Yes, this is a proportional relationship. The ratio of commission to house price is constant (1:20 or 0.05). 2. The constant of proportionality is 0.05 or 5%. 3. Commission for \$350,000 sale: 350,000 * 0.05 = \$17,500 ### Example 2 (Sports): Running Pace A runner is training for a marathon and records her distance and time for several runs: Distance (miles)Time (minutes) 324 540 756 1080 Ask students to: 1. Determine if this is a proportional relationship 2. Identify the constant of proportionality (pace in minutes per mile) 3. Predict the time for a 13-mile run (half marathon) Solutions: 1. Yes, this is a proportional relationship. The ratio of time to distance is constant (8:1). 2. The constant of proportionality is 8 minutes per mile. 3. Time for a 13-mile run: 13 * 8 = 104 minutes or 1 hour and 44 minutes ## Assess (10 minutes) Use this 10-question quiz for assessment. ## Quiz 1. Is the relationship between x and y proportional? xy 26 412 618 824 2. What is the constant of proportionality in the relationship from question 1? 3. Does the graph of y = 2x + 1 represent a proportional relationship? 4. If a car travels 240 miles in 4 hours at a constant speed, what is the constant of proportionality? 5. In the equation y = kx, what does k represent? 6. Is the origin always included in the graph of a proportional relationship? 7. If 3 shirts cost$24, how much would 5 shirts cost in this proportional relationship?
8. What is the constant of proportionality if 8 ounces of a liquid occupy 10 cubic inches?
9. Does the table represent a proportional relationship?
xy
03
25
47
69
10. If y is proportional to x and y = 15 when x = 3, what is the constant of proportionality?
1. Yes
2. 3
3. No
4. 60 miles per hour
5. The constant of proportionality
6. Yes
7. \$40
8. 1.25 cubic inches per ounce
9. No
10. 5 |
Semi-circle perimeter
Finding the perimeter of semi-circles using the formula $\large~P\,=\,\frac{\pi~d}{2}\,+\,d$.
What is the perimeter of a semi-circle with a diameter of 8cm?
To answer this question, we need to realize that the figure is just half of a circle. This means we will still be using the formula . However, there will be a twist in what we do, because the formula gives us the circumference of a complete circle.
You will solve this problem using three steps: find the circumference of a full circle with the same diameter, divide that circumference in half, and then add the diameter.
Example 1 What is the perimeter of a semi-circle with a diameter of 8cm? Step 1: Find the circumference of the full circle. Step 2: Divide the circumference in half. This is the length of just the curvy part of the semi-circle. Step 3: Add the diameter. 12.56 + 8 = 20.56 cm This is the perimeter of the semi-circle!
Example 2 Find the perimeter of this quarter circle.
Example 3 A wire is bent to form three semicircles. Find the length of the wire in terms of .
Semi-circle circumference – CLASSWORK
Solve each of the following problems.
Find the perimeter of this shape. Find the perimeter of this shape.
Self-Check
Question 1 A rope is arranged to form two semi-circles as shown. Find the length of the rope in terms of π. [show answer]
Question 2 A garden is in the shape of a quarter circle. Find its perimeter. [show answer]
Question 3 A circle is drawn inside a square. Find the circumference of the circle. [show answer] |
# Binary Numbering System 1
## Introduction to the Binary numbering system
Simply put, binary is a base 2 numbering system. The only numbers available for use to count in binary are 1, and 0. You can think of this in many ways: True/False, Yes/No, or High/Low. Computers work by using the Binary numbering system, because computers only consist of billions of on/off switches.
If we think about our normal every day numbering system, that is called “Decimal”, or Base 10. The only reason we are used to base 10 is because we have 10 fingers to count on. When the numbering systems were developed, humans became used to Base 10. There are 10 “symbols” available for counting (0 through 9). Remember that leading 0’s do not change the value of a number. 01 is the exact same as 1. The least significant digit is on the right, because it has less impact on the overall value of the number. As our numbers get larger, we start adding digits to the right of that number. We can look at the value of 1000 and the value of 0001. In the number 1000, the value of 1 has more impact on the value than it does in the number 0001. As we add numbers to the left, those numbers become more significant.
## Decimal breakdown
Before we understand binary, let’s take a look at this decimal number of 1000. We need to first change how you think about this number. Consider the following to understand how numbers are broken down before we try this on other numbering systems.
100 = 1 (Any number to the power of 0 = 1)
101 = 10 (Any number to the power of 1 equals itself)
102 = 100 (10 x 10)
103 = 1000 (10 x 10 x 10)
Value: 103 102 101 100
Original number: 1 0 0 0
Using the Example above, we can see that we have a “1” in the thousands place, so in decimal, we would calculate this as (1 x 1000) + (0 x 100) + (0 x 10) + (0 x 1). Of course our answer is 1000.
Let’s try this with a different number, such as 2017:
Value: 103 102 101 100
Original number: 2 0 1 7
We would calculate this as (2 x 1000) + (0 x 100) + (1 x 10) + (7 x 1)… The answer is 2017.
## Binary Breakdown
Now that you understand how this works in Decimal, let’s convert a binary number. Binary is a base 2 numbering system.
20 = 1 (Any number to the power of 0 = 1)
21 = 10 (Any number to the power of 1 equals itself)
22 = 100 (2 x 2)
23 = 8(2 x 2 x 2)
Value: 23 2 2 21 20
Original number: 1 0 0 0
Here we have (1 x 8) + (0 x 4) + (0 x 2) + (0 x 1) This number equals 8. Let’s try another one.
Value: 23 2 2 21 20
Original number: 1 0 0 1
In this instance, we have (1 x 8) + (0 x 4) + (0 x 2) + (0 x 1) = 9
Let’s look at this on a ControlLogix tag:
Notice that “MyTag” is a Double Integer (DINT). A DINT consists of 32 bits (showing 16 here). We can look at this number as an Integer, or we can expand the tag to break it down into Binary. We see the value of 1001 in binary is the value of “9” in decimal. When we look at a tag, we need to ask ourselves what we want to look at. If we want to look at it as a number (such as a temperature), then we look at the number at the DINT level. If the tag is used for individual discrete (on/off) devices, then we can expand the tag to see which individual bits are turned on. |
# Formula for Mode with Examples
Here you will learn what is the formula for mode of grouped and ungrouped data and how to find mode with examples.
Let’s begin –
## What is Mode ?
Mode is the size of the variable which occurs most frequently.
## Formula for Mode :
(i) For ungrouped distribution : The value of that variate which is repeated maximum number of times.
Example : Find the mode of the following data 1, 2, 3, 1, 5, 6, 2, 8, 2, 9.
Solution : Here, 2 is repeated maximum number of times.
Hence, Mode is 2.
(ii) For ungrouped frequency distribution : The value of that variate which have maximum frequency.
Example : Find the mean of the following freq. dist.
Size of the shoes 4 5 6 7 8 Number of pairs sold 10 15 20 35 16
Solution : In the above table we notice that the size 7 has the maximum frequency i.e. 35
Therefore, 7 is the mode of distribution.
(iii) For grouped frequency distribution : First we find the class which have maximum frequency, this is model class.
$$\therefore$$ Mode = ($$l$$ + $$f_0 – f_1\over {2f_0 – f_1 – f_2}$$)$$\times$$h
where $$l$$ = lower limit of model class
$$f_0$$ = freq. of model class
$$f_1$$ = freq. of the class preceding model class
$$f_2$$ = freq. of the class succeeding model class
h = class interval of model class
Example : Find the mean of the following freq. dist.
Size of the shoes 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Number of pairs sold 2 18 30 45 35 20 6 3
Solution : Here the class 30-40 has maximum frequency, so this is the modal class
$$l$$ = 30, $$f_0$$ = 45, $$f_1$$ = 30, $$f_2$$ = 35, h = 10
$$\therefore$$ Mode = ($$l$$ + $$f_0 – f_1\over {2f_0 – f_1 – f_2}$$)$$\times$$h = $$45 – 30\over 2\times 45 – 30 – 35$$$$\times$$ 10 = 36 |
Background
You have undoubtedly noticed that in college we don’t use the two column proof method you (may) have seen in high school geometry. Here is a particularly egregious example of it.
To prove $1 = 0$ Proof $0 \times 1 = 0 \times 0$ Multiply both sides by the same thing $0 = 0$ Zero times anything is zero q.e.d.
To avoid such errors we prefer to treat a proof as an argument designed to convince the reader, not as a ritual exercise for which you fill in the details.
So, there are many possible formats for a proof. But one thing a proof must be, it must be understandable. In particular, it must consist of complete sentences: subject and predicate, just as you learned in English class.
But mathematics consists of notation that abbreviates common speech. In particular, an equation is a complete sentence. It might be true, false, or conditionally true, for some values of any variables it contains. Other such sentences involve inequalities among numbers, or set membership, $x \in S$.
Propositions
In mathematics we have a particular fondness of sentences that have an additional property, namely that it is possible to decide whether they are true of false. When pressed, we call such sentences propositions. For example $\sqrt{4}=2$ is true. But, regardless how pretty is looks
$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ (1)
this equation is neither true nor false, it all depends on the variables. For instance
$ax^2 + bx + c = 0 \implies x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ (2)
is almost a proposition. Note that $x$ occurs on both sides of the $\implies$ sign. So it’s a dummy variable. But the parameters $a,b,c$ are still "floating". And even adding something reasonable like like $(\forall a,b,c \in \mathbb{R})$ isn’t enough unless we allow complex numbers.
Quantifiers
In the previous paragraph we reminded you of quantifiers you learned about in a college course on elemntary logic, like MA347 at Illinois. In a formally quantifies proposition we prevent ambiguities by using the $\forall \exists$ statements in parentheses, at the beginning of a proposition, also in parentheses.
In practice, we’re usually less fussy, and append some conditions at the end of a sentence, or even leave it to the context of the discussion. So, you might what to do.
Here is a principle which you can apply as you evaluate your own argument:
Mathematical Notation
You will be expected to use correct mathematical notation. For this we have some rudimentary text authoring tools. However, even though we read certain symbols using common phrases, the symbols should never be used instead of these phrases, outside of a mathematical proposition.
Thus, do not use $=, \exists, \implies$ in an English sentence instead of the verb "is", or when you mean "for some", or "implies".
In the same breath, please do not use an arrow when you don’t know anything better. Even arrows have specific meanings within their contexts, as for limits in calculus.
Don't replace common English phrases with a mathematical symbol that happens to be "pronounced" the same way.
Logical argument
Of course your arguments seem logical to you. But perhaps not to the reader. So we build our arguments up in a logical order. You don’t necessarily have to obey the King’s injunction to Alice:
Start at the beginning, and go till you come to the end; then stop.
But to jump around the argument too much is bad too. Sometimes we work from the hypotheses a while, and then work backward from the conclusion a bit, until we meet in the middle, like building the transcontinental railway. But if you do this, you must say so. Such prefaces as
-To prove ...
-This follows if ...
establish the clues for where that line fits into your argument.
Gaps in the arguments
It is never possible to write out every possible reason and step in an argument. Some, or even much of it must be "left to the reader" to fill in. Sometimes, this ends up being impossible. Then the proof is said to have developed a gap. Even very famous proofs begin life with gaps that other mathematicians become famous for in filling the gaps.
However, you are just beginning to be mathematicians. So the gaps in your proof will be of a different kind. They will be unwarranted inferences for which you do not in fact know the reasons. You’re "jumping to conclusions". How do I know that? Well, experience. Is it always true? No, I’m not mind reader. So, here’s the trickiest principle of them all.
What Exactly Are You Doing?
The reader is never in a good position to guess what you’re doing. Even if the problem is clearly stated in the book, or the notes, you need to restate it, preferably in your own words. Only then can the reader tell what you are doing. Thus problem solutions like these are not acceptable
• The answer to problem 1.23.b is yes, because we did this in class.
You say what the problem asks you to prove, and then you prove it.
- |
How do you integrate intx^(2/3) *ln x from 1 to 4?
Apr 1, 2015
$\int {x}^{r} \ln x \mathrm{dx}$ is a 'standard' question.
For $r \ne - 1$ integrate by parts.
We don't (most of us) know the integral of $\ln x$, but we do know its derivative, so
Let $u = \ln x$ and $\mathrm{dv} = {x}^{r} \mathrm{dx}$ (in this question $\mathrm{dv} = {x}^{\frac{2}{3}} \mathrm{dx}$
With these choices, we get:
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $v = \int {x}^{r} \mathrm{dx} = {x}^{r + 1} / \left(r + 1\right)$ (Here: $\frac{3}{5} {x}^{\frac{5}{3}}$)
$\int u \mathrm{dv} = u v - \int v \mathrm{du} = \frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{3}{5} \int {x}^{\frac{5}{3}} \cdot \frac{1}{x} \mathrm{dx}$
$= \frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{3}{5} \int {x}^{\frac{2}{3}} \mathrm{dx}$ (You'll always get the same integral fo this kind of problem.)
$= \frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{3}{5} \cdot \frac{3}{5} {x}^{\frac{5}{3}} + C$ Or for your definite integral:
$= {\left[\frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{9}{25} {x}^{\frac{5}{3}}\right]}_{1}^{4}$
Now do the arithmetic. (remember that $\ln 1 = 0$) |
# Lesson 3
Multiply Unit Fractions
## Warm-up: Estimation Exploration: How Much is Shaded? (10 minutes)
### Narrative
The purpose of this Estimation Exploration is for students to estimate the area of a shaded region. In the synthesis, students discuss whether the product is greater or less than the expression $$\frac{1}{2} \times \frac{1}{6}$$. This allows them to connect the shaded area to their previous work with multiplication expressions (MP7).
### Launch
• Groups of 2
• Display the image.
• “What is an estimate that’s too high? Too low? About right?”
• 1 minute: quiet think time
### Activity
• 1 minute: partner discussion
• Record responses.
### Student Facing
What is the area of the shaded region? Record an estimate that is:
too low about right too high
$$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$
### Activity Synthesis
• “Is the area of the shaded region more or less than $$\frac{1}{2} \times \frac{1}{6}$$? How do you know?” (More. It looks like it is $$\frac{1}{2}$$ the length and more than $$\frac{1}{6}$$ the width.)
• “What is the value of $$\frac{1}{2} \times \frac{1}{6}$$?” ($$\frac{1}{12}$$)
## Activity 1: Notice Patterns in Expressions (20 minutes)
### Narrative
The purpose of this activity is for students to notice structure in a series of diagrams and the expressions that represent them. They investigate how these expressions vary as the number of rows and columns in the diagram change. Students see how the diagram represents the multiplication expression and also how the diagram helps to find the value of the expression (MP7). Through repeated reasoning they also begin to see how to find the value of a product of any two unit fractions (MP8).
### Launch
• Groups of 2
• Display the images from the task.
• “What is different about these diagrams?” (The number of rows increases by 1. The blue shaded piece gets smaller.)
• “What is the same?” (The size of the big square. There are always 4 columns. Only one piece is shaded.)
• 1 minute: quiet think time
• Share and record responses.
### Activity
• 1–2 minutes: independent work time to complete the first problem
• 1–2 minutes: partner discussion
• “Now, complete the rest of the problems with your partner.”
• 5 minutes: partner work time
• Monitor for students who:
• choose different diagrams to represent with multiplication expressions
• represent the same diagram with different multiplication expression, for example, $$\frac{1}{2}\times\frac{1}{4}$$ and $$\frac{1}{4}\times\frac{1}{2}$$
### Student Facing
1. Choose one of the diagrams and write a multiplication expression to represent the shaded region. How much of the whole square is shaded? Explain or show your thinking.
2. If the pattern continues, draw what you think the next diagram will look like. Be prepared to explain your thinking.
### Student Response
If students do not write correct expressions to represent the diagrams, write the correct expressions and ask, “How do the expressions represent the area of the shaded piece of the diagram?”
### Activity Synthesis
• Display the diagrams from the student workbook.
• Select previously identified students to share.
• As students explain where they see the multiplication expression in each diagram, record the expressions under the diagram for all to see.
• Refer to the diagram that shows $$\frac{1}{2}\times\frac{1}{4}$$ or $$\frac{1}{4}\times\frac {1}{2}$$.
• Display both expressions.
• “How does this diagram represent both of these expressions?” (It shows one half of one fourth shaded in and it also shows one fourth of one half shaded in.)
• Represent student explanation on the diagrams.
• “Why is the area of the shaded region getting smaller in each diagram?” (Because we are shading a smaller piece of $$\frac{1}{4}$$ each time.)
• Display:
$$\frac{1}{2} \times \frac{1}{4}= \frac{1}{2\times 4}=\frac{1}{8}$$
$$\frac{1}{3} \times \frac{1}{4}= \frac{1}{3\times 4}=\frac{1}{12}$$
$$\frac{1}{4} \times \frac{1}{4}= \frac{1}{4\times 4}=\frac{1}{16}$$
$$\frac{1}{5} \times \frac{1}{4}= \frac{1}{5\times 4}=\frac{1}{20}$$
• “These equations represent the diagrams. What patterns do you notice?” (They all have $$\frac {1}{4}$$ in them. The denominators in the first fractions go 2, 3, 4, 5. The denominators in the middle fractions are multiplication expressions, the denominators in the middle are all multiplied by 4, the denominator in the fraction that shows the value of the shaded piece goes up by 4 each time.)
• “How do the diagrams represent $$\frac{1}{2\times 4}$$, $$\frac{1}{3\times4}$$, $$\frac{1}{4\times4}$$, $$\frac{1}{5\times4}$$ ?” (The numerator tells us that there is 1 piece shaded and the denominator tells us the size of the piece. The denominator also tells us the rows and columns that the whole is divided into.)
## Activity 2: Write a Multiplication Equation (15 minutes)
### Narrative
The purpose of this activity is for students to use the structure of diagrams to calculate products of unit fractions. They also represent their work using an equation. As students become more familiar with this structure they may not need diagrams as a scaffold to find these products. Drawing their own diagrams, however, will also reinforce student understanding of how to calculate products of unit fractions.
This activity uses MLR1 Stronger and Clearer Each Time. Advances: Reading, Writing.
Engagement: Provide Access by Recruiting Interest. Invite students to share a connection between the diagram and something in their own lives that represent the fractional values.
Supports accessibility for: Attention, Conceptual Processing
• Groups of 2
### Activity
• 3–5 minutes: independent work time
• 1–2 minutes: partner discussion
### Student Facing
1. Write a multiplication equation to represent the area of the shaded piece.
2. Explain how the diagram represents the equation $$\frac{1}{5}\times\frac{1}{3}=\frac{1}{15}$$.
3. Find the value that makes each equation true. Use a diagram, if it is helpful.
1. $$\frac{1}{2} \times \frac{1}{6} = {?}$$
2. $$\frac{1}{4} \times \frac{1}{6} = {?}$$
### Student Response
If students do not refer to the rows and columns when they explain how the diagram represents the equation $$\frac{1}{5} \times \frac{1}{3} = \frac{1}{15}$$, ask “How are the rows and columns in the diagram represented in the equation?”
### Activity Synthesis
• Display: $$\frac{1}{2}\times\frac{1}{4}=\frac {1}{(2\times4)} = \frac {1}{8}$$ and the corresponding diagram.
• “How does this equation represent the diagram?” (One fourth of one half is shaded which is the same as 1 piece of the whole square that is divided into 2 columns and 4 rows so one eighth of the whole square is shaded.)
MLR1 Stronger and Clearer Each Time
• “Share your explanation about how the last diagram represents $$\frac{1}{5} \times \frac{1}{3}= \frac{1}{15}$$ with your partner. Take turns being the speaker and the listener. If you are the speaker, share your ideas and writing so far. If you are the listener, ask questions and give feedback to help your partner improve their work.”
• 3–5 minutes: structured partner discussion
• Repeat with 2–3 different partners.
• If needed, display question starters and prompts for feedback.
• “Can you give an example to help show . . . ?”
• “Can you use the word _____ in your explanation?”
• “The part that I understood best was . . . .”
• “Revise your initial draft based on the feedback you got from your partners.”
• 2–3 minutes: independent work time.
## Lesson Synthesis
### Lesson Synthesis
“Today we represented products of unit fractions with diagrams and with equations.”
“How is multiplying unit fractions the same as multiplying whole numbers? How is it different?” (We use the same multiplication facts to find the value of expressions, but the value is less than one because we are multiplying the denominators. We use diagrams that show rows and columns to multiply whole numbers and unit fractions, but the rows and columns show fractions of 1 instead of more than 1.)
“In future lessons, we are going to multiply fractions that have a numerator greater than 1. What do you wonder about that?” (Will we use the same diagrams? Will it work the same way as unit fractions?) |
# Quick Answer: What Is The Meaning Of 1 4?
## What do 1/4 means?
The fraction one-fourth, written in symbols as 1/4, means “one piece, where it takes four pieces to make a whole.” The fraction one-quarter, written in symbols as 1/4, means “one piece, where it takes 4 pieces to make a whole.”.
## Is one fourth a quarter?
A “quarter” is far more commonly used in conversation than a “fourth” but both mean “one part in four” (1/4). A “quarter” is a countable part so you can have two quarters or three quarters (such as pieces of a cake or pie).
## What is a one quarter of 24?
6Answer and Explanation: One-fourth of 24 is 6. To get to this answer, use the following steps: Divide 24 by 4. Recall the table of 4.
## How do you write 4 in words?
Learn to write numbers from 1 to 9.1 = one.2 = two.3 = three.4 = four.5 = five.6 = six.7 = seven.8 = eight.More items…
## What is every 4 months called?
The term for a four month period is quadrimester. Quad = 4 mense= month.
## What is one quarter of a fraction?
A quarter is one out of four equal parts. It can also be written as 25% or 0.25.
## What is another way to say 1 4?
We pronounce 1/2 as one-half, 1/4 as one-quarter, and 3/4 as three-quarters.
## Why is 1/4 called a quarter?
Michael divides a pizza into 4 equal parts and each one of them gets equal share. When a whole is divided into 4 equal parts, and each part is called one-quarter. One-quarter is one of four equal parts. It is written as 14.
## What is 1/4 in a whole number?
One fourth is equivalent to the fraction: 1/4. Therefore, it is a quarter of an amount. Fourths are calculated by dividing by 4.
## What is 3/4th called?
4 Answers. You call 3/4 “three fourths” or “three quarters”, and 3/5 “three fifths”.
## What is the meaning of 1 3?
1:3 (for every one boy there are 3 girls) 1/4 are boys and 3/4 are girls.
## What is 1/4 as a percentage?
25%Decimals, Fractions and PercentagesA Quarter can be written…As a fraction:1/4As a decimal:0.25As a percentage:25%1 more row |
Systems of Linear Equations
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# Systems of Linear Equations - PowerPoint PPT Presentation
Systems of Linear Equations. Lesson 2 – Solving Linear systems graphically. Todays Objectives. Students will be able to solve problems that involve systems of linear equations in two variables graphically and algebraically, including:
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### Systems of Linear Equations
Lesson 2 – Solving Linear systems graphically
Todays Objectives
• Students will be able to solve problems that involve systems of linear equations in two variables graphically and algebraically, including:
• Determine and verify the solution of a system of linear equations graphically, with and without technology
• Explain the meaning of the point of intersection of a system of linear equations
Vocabulary
• System of Linear Equations (Linear System):
• A system of linear equations is often referred to as a Linear System
• A linear system consists of two or more linear equations plotted on the same coordinate plane
Solving linear systems graphically
• The solution of a linear system can be estimated by graphing both equations on the same grid. If the two lines intersect, the coordinates (x,y) of the point of intersection are the solution of the linear system.
Solution to the system
Solving Linear Systems graphically
• Each equation of the following linear system is graphed on the grid.
3x + 2y = -12
We can use the graphs to estimate the solution of the linear system.
The set of points that satisfy both equations lie where the two graphs intersect. From the graph, the point of intersection appears to be (-2, -3).
• -2x + y = 1
• To verify the solution, we check that the coordinates (-2, -3) satisfy both equations:
• For each equation, the left side equals the right side. Since x = -2 and y = -3 satisfy each equation these numbers are the solution of the linear system
• Solve the linear system:
• Solution:
• Determine the x-intercept and y-intercept of the graph of equation (1).
• Both the x and y intercepts are 8.
• Write equation (2) in slope y-intercept form.
• The slope of the graph of equation (2) is 3/2, and its y-intercept is -7.
• Graph each line.
The point of intersection appears to be (6, 2). Verify the solution by substituting x = 6 and y = 2.
• 6 + 2 = 8; 8 = 8
• 3(6) – 2(2) = 14; 18 – 4 = 14; 14 = 14
• The left sides equal the right sides, so x = 6, and y = 2 is the solution of the linear system.
Solving a Problem by Graphing a Linear System
• One plane left Shanghai at noon to travel 1400 km to Urumqi at an average speed of 400 km/h. Another plane left Urumqi at the same time to travel to Shanghai at an average speed of 350 km/h. A linear system that models this situation is:
• where d is the distance in km’s from Shanghai to Urumqi and t is the time in hours since the planes took off.
Solving a Problem by Graphing a Linear System
• A) Graph the linear system above.
• B) Use the graph to solve this problem: When do the planes pass each other and how far are they from Urumqi?
• Solution:
• The planes pass each other when they have been travelling for the same time and they are the same distance from Urumqi.
• Solve the linear system to determine values of d and t that satisfy both equations.
• For the graph of equation (1), the slope is -400 and the vertical intercept is 1400.
• For the graph of equation (2), the slope is 350 and the vertical intercept is 0.
A) Graph the linear system
d = 1400 – 400t
d = 350t
Distance (km)
Time (h)
• The graphs appear to intersect at (1.9, 650); that is, the planes appear to pass each other after travelling for 1.9 hours and at a distance of 650 kilometers from Urumqi. Use the coordinates of the point of intersection to verify the solution.
• 400(1.9) = 760 km. So, it will be: (1400 – 760) km, or 640 km’s from Urumqi. The plane travelling from Urumqi to Shanghai travels at 350 km/h, so, in 1.9 hours, its distance from Urumqi will be 350(1.9) km = 665 km.
• These times and distances are approximate because these measures cannot be read accurately from the graph. 0.9 hours = 54 minutes.
• The planes pass each other after travelling for approximately 1 hour, 54 minutes and when they are approximately 650 km from Urumqi.
• A) Write a linear system to model this situation:
• To visit the Yu Garden in Shanghai, the ticket price is \$5 for a student and \$9 for an adult. In one hour, 32 people entered the garden and a total of \$180 in admission fees was collected.
• B) Graph the linear system then solve this problem:
• How many students and how many adults visited the garden during this time? |
### I'm Eight
Find a great variety of ways of asking questions which make 8.
### Let's Investigate Triangles
Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
### Noah
Noah saw 12 legs walk by into the Ark. How many creatures did he see?
# What Was in the Box?
##### Age 5 to 7Challenge Level
We did not have many solutions sent in showing their thoughts and recording, but here are a few. This first one came from Michael at Cloverdale Catholic School in Canada:
The number that was put into that box was 10, and out came 18. So, therefore, that wonderful big box added 8. (10 + 8 = 18)
Four other boxes were put into that bigger box, and out popped these numbers: 12, 8, 15 and 10. If the box added eight onto ten to make eighteen, then the other numbers must have been added from eight. How do you figure that out? You use the opposite operation, in which in this case it is addition, so we use subtraction. (12-8=4, so 4+8=12) and (8-8=0, so 0+8=8) and etc.
For the other box, one of the boxes that went into the bigger box was 10, and out came the numbers: 0, 19, 1 and 11. If the number 10 was put in and became the number 0, then 10 was subtracted from 10. Thus, 10-10=0, 29-10=19, 11-10=1, and 21-10=11. If 10 became 19, then 9 was added onto ten. Thus, 10+9=19, (-9)+9=0, (-8)+9=1, and 2+9=11. (A more advanced solution would have been multiplying by 1.9, in which case the numbers before the big box would be 0, 10, ≈0.526 (which means almost 0.526) and ≈5.789.)
There are two possible ways ten could have turned into one: dividing by 10, or subtracting by 9. In division's case: 10÷10=1, 0÷10=0, 190÷10=19, and 110÷10=11. In subtraction's case: 10-9=1, 9-9=0, 28-9=19, and 20-9=11. If 10 was put in that bigger box and became 11, then one was added to ten. Thus, 10+1=11, (-1)+1=0, 18+1=19, and 0+1=1. (An advanced solution is to multiply by 1.1, in which the numbers before the bigger box would have been, ≈17.27, ≈0.9 and 10.)
The next one is from Lisa at the Tokyo International School in Japan:
First you have to choose what number/numbers to minus or plus then you choose what box should start with the number. I choose to minus nine in all the boxes. The numbers that come out are 0, 19, 1, 11.
I know that if you take away 9 from ten you get 1. Then you just have to add nine to the rest of the numbers. The number that you get that's the number that you start with. So if you choose to take away the number 9 your solution is 9-9, 28-9, 10-9, 20-9.
Lastly from Louis at the International School, Seychelles:
Minette, Freddie, Ursie from Grandtully Primary in Scotland sent in the following:
We kept the number 10 in our heads, if it was a smaller number we knew it was a take away answer and if it was a bigger number we knew it was an add up.
So once we knew if it was a take away or an add up we kept the 10 in our heads and then we counted from 10 to the number it was going to end with and how many that was.
Then that was what happened in the box!
First the box took away 10 to become 0
Then the box added on 9 to become 19
Then the box took away 9 to become 10
Finally the box added on 1 to become 11
Well done those of you who sent in your ideas to solve this challenge, I wonder how many of you went on to have a look at What's in the Box?
---------------- |
# How do you differentiate f(x) =xsec4x^3 using the chain rule?
##### 1 Answer
Oct 5, 2017
$f ' \left(x\right) = 12 {x}^{3} \sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) + \sec \left(4 {x}^{3}\right)$
#### Explanation:
We seek $f ' \left(x\right)$ where:
$f \left(x\right) = x \sec \left(4 {x}^{3}\right)$
First we apply the product rule:
$f ' \left(x\right) = \left(x\right) \left(\frac{d}{\mathrm{dx}} \sec \left(4 {x}^{3}\right)\right) + \left(\frac{d}{\mathrm{dx}} x\right) \left(\sec \left(4 {x}^{3}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = x d \left(\frac{d}{\mathrm{dx}} \sec \left(4 {x}^{3}\right)\right) + \left(1\right) \left(\sec \left(4 {x}^{3}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = x d \left(\frac{d}{\mathrm{dx}} \sec \left(4 {x}^{3}\right)\right) + \sec \left(4 {x}^{3}\right)$
Then, we apply the chain rule:
$f ' \left(x\right) = x \left(\sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) \frac{d}{\mathrm{dx}} \left(4 {x}^{3}\right)\right) + \sec \left(4 {x}^{3}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = x \sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) \left(12 {x}^{2}\right) + \sec \left(4 {x}^{3}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 12 {x}^{3} \sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) + \sec \left(4 {x}^{3}\right)$ |
# Addition and Subtraction Part 3: Facts Strategies KG-3rd
by C. Elkins, OK Math and Reading Lady
This is part three in a series of strategies regarding addition and subtraction strategies. This part will focus on a variety of strategies to help toward memorization of facts, meaning automatic computation. While children are learning their number bonds (building up to 5 in KG, to 10 in first grade, and to 20 in second grade), there are other facts which cross several number bonds that students can work towards. These strategies to build mental math automaticity are highlighted below. Get some freebies in the section on doubles / near doubles.
Identity (or Zero) Property:
• The value of the number does not change when zero is added or subtracted.
• 3 + 0 = 3
• 9 – 0 = 9
Subtracting All:
• The answer is always zero when you take away / subtract all.
• 9 – 9 = 0
• 50 – 50 = 0
Adding 1 or Subtracting 1:
• Adding 1 results in the next number in the counting sequence.
• Subtracting 1 means naming the number that comes right before it in the counting sequence.
• With manipulatives, lay out an amount for student to count. Slide one more and see if he/she can name the amount without recounting.
• Do the same as above, but take one away from the group to see if he/she can name the amount without recounting.
• Show this concept using a number line.
• 6 + 1 = 7; 26 + 1 = 27
• 7 – 1 = 6; 37 – 1 = 36
• After +1 or -1 strategies are in place, then go for +2 or -2 for automatic processing.
Next-Door Neighbor Numbers:
• If subtracting two sequential numbers (ie 7 subtract 6), the answer is always one because you are taking away almost all of the original amount.
• Help students identify these types of problems: 8-7; 10-9; 98-97; 158-157
• Guide students to writing these types of problems.
• Relate these to subtracting 1 problems. If 10-1 = 9; then 10 – 9 = 1.
• Show on a number line.
Doubles (with freebies): Continue reading |
# 11A
UNIT
Full Digi-Block Activity
Equal Shares
OBJECTIVES:
• To model sharing equally (division) by dividing sets into two, three, four, or more equal parts • To express a division situation with a number sentence
summary
materials
This lesson introduces the sharing, or partitive, model of division. Students use blocks to model a division story problem. They separate the same set of blocks into 2, 3, 4, and more equal groups and explore the relationship between the growing number of subsets and the number within each subset. Students also express the division with a number sentence.
Each individual or pair of students needs: 18 – 36 single blocks 1 Dividing Blocks activity sheet paper plates or construction paper circles for organizing groups
Everyday Math Connection
In Everyday Mathematics Lesson 11.4, students model division stories with counters. In “Equal Shares” they use Digi-Blocks to explore what happens when a number is divided in to an increasing numbers of groups. They record their findings in a table, writing a number sentence to match each situation. Students conclude that as the number of groups increases, the quantity within each group decreases. The activity, “Fair Shares” in Packed With Math (Grades 2-3) further develops this idea.
Unit 11 Activity A
Equal Shares
DB-75
Class Introduction
20 min
Provide small paper plates to pairs of students. Suggest that students use them to help organize blocks as they are solving problems. Use student volunteers to play the roles of (Michelle), (Tony), and (Natalie) in the following story:
(Michelle) has a box of Digi-Blocks. She opens the box and pours all 18 onto her desk. She is looking forward to playing a game with them. How happy she is to have all 18 blocks to herself!
How many blocks in all? How many students? How many blocks does each student have?
How can we show the story with numbers?
Number Sentence
18 18
1 2
18 9
18 ÷ 1 = 18 18 ÷ 2 = 9
Explain that often a table can help organize what happens in a story, especially the one they are about to hear. Draw a table on the board. (See figure.) • Students suggest you write 18, 1, and 18 in the first three columns. They will likely not know how to write a number sentence to match the action in the story. Help them eliminate possibilities and explain why the operations with which they are familiar don’t work: 18 + 1, 18 – 1. Students may suggest 18 × 1 which does reflect the organization of the blocks. Although this is not the intended response, record this number sentence and then go back later and edit it with students. Continue the story with:
Just as (Michelle) is about to begin playing, her friend, (Tony,) asks, “May I join you?” (Michelle) says, “Of course you may, it’s more fun to play with a friend than alone!” So (Michelle) and (Tony) decide to share the blocks equally.
• Have two students show how they share the blocks, placing 9 on each plate. Have students describe why this is equal. • Refer to the table, ask: How can we show what happened with numbers? Students will suggest writing 18, 2, and 9 as they answer each question.
If we want to write a number sentence that tells “18 shared equally between 2 is 9 each,” we use a division symbol like this:
18 ÷ 2 = 9 Continue the story:
Just as (Michelle) and (Tony) are about to start their game, along comes (Natalie). (Natalie) asks, “May I play, too?” (Michelle) and (Tony) say, “Of course you may. It’s more fun to
DB-76
Unit 11 Activity A
Equal Shares
play with three people than two!”
• Have students share the blocks. They will figure out that each student will get 6 blocks. • Ask, How can we show what we did with numbers? Record 18, 3, and 6 in each column. • Have students help to write the number sentence: 18 ÷ 3 = 6. Continue the story in a similar way, having students share the 18 blocks among 4, 5, and 6 students. • When dividing 4 ways, students will discover that 2 blocks are left over. Add that in order to be fair, these blocks must be put aside because they cannot be sawed in halves – they wouldn’t be blocks anymore! Explain that we call these remainders and show how to record the remaining blocks as R2.
The authors of Everyday Mathematics suggest that the arrow symbol ( ) be used instead of the equal symbol (=) in problems that have remainders. See page 790 for further reading.
Pair / Independent Practice 15 min
Distribute Dividing Blocks activity sheets to pairs or individual students. Distribute plates and blocks and review the directions. • Name numbers for students to practice sharing. Assign the same number to at least two different students so that they can compare and discuss their work, especially if their tables look different in the end. • Suggested numbers are: 12, 20, 24, and 36 depending on students’ levels.
Assessment Observe students during the introduction and as they complete the activity sheet, do they:
Closure
5 min
Have students think about what they did with blocks and have them study their tables. If they need prompting, suggest that they look for patterns in their numbers. Ask questions, such as:
What number did you divide? Who else worked with the same number? What happened when you shared 2 (3, 4, . . .6) ways? How did you show this with a number sentence? What kinds of patterns do you see on your table? What happens to everyone’s share as the blocks are shared among more and more people? Why does this happen?
Understand equal shares? Accurately name the total number, number of equal shares, and number in each share? Express the division in a number sentence? Understand the concept of a remainder as a leftover and how to record it?
During the discussion, observe and note, do students:
Describe the action of sharing or partitive division accurately? Understand the big idea that each share gets smaller as the whole is shared more ways?
Unit 11 Activity A
Equal Shares
DB-77
Name Dividing Blocks
ACTIVITY SHEET
______________
Materials: • Your teacher will tell you how many blocks you need to get started. • You will need 6 plates. Directions: 1. Share the blocks equally, beginning with one, two, and three plates. Continue to 6 plates! 2. After each sharing write the numbers and the division number sentence in the table. Don’t forget remainders!
How many blocks in all?
How How many many blocks does children? each child have?
Blocks left over:
Number sentence
16
5
3
1
16 ÷ 5 = 3 R1
1 2 3
DB-78
Unit 11 Activity A
Equal Shares
Student Book p. 32 |
# How to Solve Simultaneous Equations With Elimination Method In WAEC & JAMB
Simultaneous equation is quite an easy mathematics to solve if you are familiar with the required steps which has been broken down below. Note that there are two methods of solving any simultaneous equations either in WAEC, NECO, JAMB or any examination or test you are tried on this topic.
Simultaneous equation is one of the most common topics in mathematics that the West African Examination Council, WAEC test candidates for every year. You will either see it as an objective question or as a theoretical question and sometimes it could appear on both sections.
When simultaneous equation appears under theory section, most likely, you would be mandated to attempt the question using Graphical method because the body wants to test your in-depth understanding on this maths topic. Not to worry, we have broken down the steps on how to solve simultaneous equation graphically, it would assist candidates or students in a long way in achieving the full score awarded to the question.
For the sake of this post, i will be putting you through the simple and complete breakdown on how to solve simultaneous equations using Elimination method.
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Ideally, there are 2 methods of solving simultaneous equations;
1. Graphical method
2. Algebraic methods (Substitution and Elimination)
I will be citing some examples on how to solve simultaneous equations using The Elimination Method.
Example 1:
2x + 3y = 12 —- (1)
5x + 2y = 8 —- (2)
Firstly we look for a way to eliminate either the x variable and coefficient (2,5) or the y variable and coefficient.) (3,2).
In order to eliminate either the x or y variable we make sure they have equal values. Now let’s eliminate the x variables to make y stand alone. The only way to make this possible is to take the coefficient of x which is 5 in equation (2) & multiply it with the whole of equation (1), and also take the coefficient of x which is 2 in equation (1) & multiply it the whole of equation (2).
5 2x + 3y = 12___(1)
X
2 5x + 2y = 8____(2)
We multiply equation (1), (2) by 5 and 2 respectively….
:. We have;
10x + 15y = 60____(3)
10x + 4y = 16____(4)
Now we can safely eliminate x variable and coefficient.
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:. 10x-10x+15y-4y=60-16
:. 15y – 4y = 60 – 16
:. 11y = 44
We divide both sides by 11 to make y stand alone
:. (11y)/11 = 44/11
:. y = 4
Example 2;
4x – 2y = 10
x + y = 4
Multiply x + y = 4 by 2 to give 2x + 2y = 8
4x – 2y = 10
2x + 2y = 8
6x = 18
x = 3
You are to add the two equations which “eliminates” the 2y (since -2y + 2y = 0) and leave 6x = 18 which, after dividing both sides by 6 leaves x = 3
Therefor; 3 + y = 4
y = 1
Then you replace x in one of the equations with the value 3. (in this example, doing this in x + y = 4 is simpler than in 4x – 2y = 10)
4x – 2y = 10
(4 x 3) – (2 x 1) = 10
12 – 2 = 10
After getting your x and y variables, always confirm the correctness of your solution by checking the values for x and y in the other equation.
Did you find this simple to understand? if you have questions on how to best solve simultaneous equation with Elimination method in WAEC or any examination, please reach us via the comment section below and we shall respond accordingly.
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10
Q:
# A vertical toy 18 cm long casts a shadow 8 cm long on the ground. At the same time a pole casts a shadow 48 m. long on the ground. Then find the height of the pole ?
A) 1080 cm B) 180 m C) 108 m D) 118 cm
Explanation:
We know the rule that,
At particular time for all object , ratio of height and shadow are same.
Let the height of the pole be 'H'
Then
$\frac{18}{8}=\frac{H}{48}$
=> H = 108 m.
Q:
Angle of Elevation means
In Heights and Distances,
The angle of elevation of an object is the angle between the horizontal and the line of sight that is the line from the object to the observer's eye.
25
Q:
A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions ?
A) 14 cm B) 13.5 cm C) 12.5 cm D) 11.4 cm
Explanation:
Let the required height of the building be x meter
More shadow length, More height(direct proportion)
Hence we can write as
(shadow length) 40.25 : 28.75 :: 17.5 : x
⇒ 40.25 × x = 28.75 × 17.5
⇒ x = 28.75 × 17.5/40.25
= 2875 × 175/40250
= 2875 × 7/1610
= 2875/230
= 575/46
= 12.5 cm
5 676
Q:
The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ?
A) 42.2 mts B) 33.45 mts C) 66.6 mts D) 58.78 mts
Explanation:
From above diagram
AC represents the hill and DE represents the tower
Given that AC = 100 m
angleXAE = angleAEC = 60° (∵ AX || CE)
Let DE = h
Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE
tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)
tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)
∵ BD = CE and Substitute the value of CE from equation 1
100/√3 = 100−h(√3) => h = 66.66 mts
The height of the tower = 66.66 mts.
6 524
Q:
An observer 1.6 m tall is $20\sqrt{3}$away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:
A) 21.6 m B) 23.2 m C) 24.72 m D) None of these
Explanation:
Draw BE // CD
Then, CE = AB = 1.6 m,
BE = AC =
Therefore, CD = CE + DE = (1.6 + 20) m = 21.6 m.
13 5777
Q:
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?
A) Data inadequate B) 8 units C) 12 units D) None of these
Explanation:
One of AB, AD and CD must have given.
9 4258
Q:
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A) 173 m B) 200 m C) 273 m D) 300 m
Explanation:
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100m,
$\frac{AB}{AC}=\mathrm{tan}30°=\frac{1}{\sqrt{3}}=>AC=AB*\sqrt{3}=100\sqrt{3m}$
$\frac{AB}{AD}=\mathrm{tan}45°=1=>AD=AB=100m$
CD=(AC+AD)=$\left(100\sqrt{3}+100\right)m=100\left(\sqrt{3}+1\right)=100*2.73=273m$
8 2649
Q:
Jack takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?
A) 6 B) 8 C) 9 D) 10
Explanation:
Average speed = total distance / total time
Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours
Average speed = 6/ 0.75 = 8 miles/hour
10 2026
Q:
If an object travels at five feet per second, how many feet does it travel in one hour?
A) 30 B) 3000 C) 18 D) 1800
Explanation:
If an object travels at 5 feet per second it covers 5x60 feet in one minute, and 5x60x60 feet in one hour.
$\inline \fn_cm \therefore$Answer = 1800 |
### Home > CC1MN > Chapter 9 Unit 9 > Lesson CC1: 9.1.1 > Problem9-14
9-14.
Express each length below in the specified unit.
It is important to remember that 12 inches are equal to one foot and 3 feet are equal to one yard.
1. $1\frac{1}{2}$ feet is _____ inches.
Knowing the information provided above, can you find how many inches are in half of a foot? It may be easiest to use addition to solve this problem
1. $6$ inches is _____ feet.
If 12 inches is one whole foot, how much of a foot is 6 inch
1. $\frac { 3 } { 4 }$feet is _____ inches.
$\text{How many inches are in }\frac{1}{4} \text{ feet?}\\ \text{ Try finding this first.}$
$\text{There are }9\text{ inches in }\frac{3}{4}\text{ feet and }\\(3\text{ inches in }\frac{1}{4}\text{ feet}).$
1. $27$ inches is _____ feet.
If 24 inches are in 2 feet, how many inches would be left in this conversion?
1. $7$ yards is _____ feet.
If each yard is 3 feet, how many feet should be in 7 yards? Try drawing a diagram.
1. $42$ inches is _____ feet.
How many groups of 12 inches fit into 42 inches?
How many inches will be left?
$\text{ There are }3\frac{1}{2}\text{ feet in }42\text{ inches}.$ |
# Chapter 1 – Sets and Functions
The following Topics and Sub-Topics are covered in this chapter and are available on MSVgo:
Introduction
You might have come across sets and functions while solving Maths problems! In this article, we have discussed sets and functions to help you get a better understanding of the subject. Sets and functions assist in performing logical and mathematical operations. We hope this article helps you grasp the concept conveniently and equips you to score well in your exams.
#### Types of sets
A collection of objects is called a set. The number of elements in every set leads to their classification into different types. Therefore, you can say a set is a collection of dissimilar elements of the same kind. The following are the types of sets:-
• Singleton Set
A set with only one element present is called a singleton set. For example, Set Y = (4) is a singleton set.
• Null Set
When the set is empty, or it doesn’t have any elements, it is called a null or void set. It is represented by {} or ϕ.
For example A = (x:x is a leap year between 2000 and 2004)
Between 2000 and 2004, we cannot find any leap year so, A = ϕ
• Proper Set
When a set consisting of some elements from the original is considered a proper subset; when a set contains original elements, along with the null set, it is called an improper subset.
• Finite Set
When in a set, the number of elements is finite, it is called a finite set. All the empty sets come under this category. In other words, a collection of no, or a constant number of elements is known as a finite set. For example:
C= ( x : x in a month in a year); Set C will have 12 elements.
D = (y : y is the zero of a polynomial (x4 – 6x2 + x + 2)); Set D will have 4 zeroes.
• Infinite Set
It is just the opposite of a finite set. When in a set, the number of elements is infinite, it is called an infinite set. For example:
D = (x : x is a natural number); There are infinite natural numbers. Thus, Set D is an infinite set.
E = (y : y is the ordinate of a point on a given line); Here, you can see there are infinite points on a line. So, E is an infinite set.
• Universal Set
Every set is formed based on the universal set and, as per the context, the Universal set is ascertained. Subsets of Universal sets are all the other sets, represented by U.
For example:-
The universal set of integers, rational numbers, and irrational numbers is the set of real numbers.
• Equal Set
Two sets C and D will be equal only when each element of set C is also the element of the set D. Even if they are subsets of each other, they will be called equal. It can be illustrated, as:
C = D
C ⊂ D and D ⊂ C ⟺ C = D
If the condition is not met, which is mentioned above, the sets will be considered unequal. It can be shown as C ≠ D.
#### Domain Co Domain and Range of a Function
In mathematics, functions are considered a fundamental concept. It has various applications around the world. The set consisting of all possible values, regarded as inputs to a function, is called the domain of the function. The digit is positive, which is present under the square root bracket. To find out the range, you have to subtract the possible x-values to find the y-values.
For example:
The domain of the function C is set B, i.e. (USA, Canada, UK, France)
#### Functions in Maths
A relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
#### FAQs
What are the sets and functions?
Answer. Sets are the well-defined collection of different objects. Different mathematicians have defined it in different ways. The relationship of one variable with another is determined by function. In other words, it is a law or expression which is used to define a relationship.
What are the basic concepts of sets?
Answer. The main concept is that a set has elements, and both the sets may be termed as equal only if each set has the elements of the other set.
What are the types of sets?
Answer. The different type of sets are:
• Empty Set
• Singleton Set
• Finite Set
• Infinite Set
• Power Set
• Sub Set
• Universal Set
What are the four types of functions?
Answer. The following are the four types of functions :
• One – One function
• Many – one function
• Onto – function
• Into – function
What are the main functions of classification?
Answer. It assists in solving mathematical problems conveniently. Classification helps to allocate various objects in groups.
Sets and functions are extremely important topics for class 12 exams. It is paramount to clear the concept behind sets and functions to excel in the exams. We will first explain sets and functions, with in-depth concept notes and explanatory video on the MSVgo app. MSVgo app has a video library that explains concepts with examples or explanatory visualizations or animation. To learn more about it, check out the MSVgo app and its official site. Stay Tuned with the MSVgo app and relish learning!
### High School Physics
• Alternating Current
• Atoms
• Communication Systems
• Current Electricity
• Dual nature of Radiation and Matter
• Electric Charges and Fields
• Electricity
• Electromagnetic Induction
• Electromagnetic Waves
• Electrons and Photons
• Electrostatic Potential and Capacitance
• Fluid Pressure
• Force and Acceleration
• Force And Laws Of Motion
• Gravitation
• Internal Energy
• Kinetic Theory
• Law of motion
• Light – Reflection And Refraction
• Magnetic Effects Of Electric Current
• Magnetism and Matter
• Management Of Natural Resources
• Mechanical properties of Fluids
• Mechanical properties of Solids
• Motion
• Motion in a plane
• Motion in a straight line
• Moving Charges and Magnetism
• Nuclear Energy
• Nuclei
• Oscillations
• Our Environment
• Paths of Heat
• Physical world
• Ray optics and optical instruments
• Semiconductor Devices
• Semiconductor Electronics: Materials, Devices and Simple Circuits
• Simple Machines
• Sound
• Sources Of Energy
• Specific and Latent Heats
• Spherical Mirrors
• Static Electricity
• Systems of Particles and Rotational motion
• Thermal properties of matter
• Thermodynamics
• Units and Measurement
• Vectors, Scalar Quantities and Elementary Calculus
• Wave Optics
• Waves
• Work, Power and Energy
### High School Chemistry
• Acids, Bases and Salts
• Alcohols, Phenols and Ethers
• Aldehydes, Ketones and Carboxylic Acids
• Aliphatic and Aromatic Hydrocarbons
• Alkyl and Aryl Halides
• Amines
• Analytical Chemistry
• Atomic Structure
• Atoms And Molecules
• Basic concepts of Chemistry
• Biomolecules
• Carbon And Its Compounds
• Carboxylic acids and Acid Derivatives
• Chemical Bonding and Molecular Structures
• Chemical Energetics
• Chemical Equilibria
• Chemical Kinetics
• Chemical Reactions And Equations
• Chemical Reactions and Their Mechanisms
• Chemistry in Everyday Life
• Chemistry of p-Block elements
• Chemistry of Transition and Inner Transition
• Classification of Elements
• Coordination Compounds
• Cyanide, Isocyanide, Nitro compounds and Amines
• Electrochemistry
• Electrolysis
• Elements, Compounds and Mixtures
• Environmental Chemistry
• Equilibrium
• Ethers and Carbonyl compounds
• Haloalkanes and Haloarenes
• Hydrocarbons
• Hydrogen
• Ideal solutions
• Introduction to Organic Chemistry
• Ionic equilibria
• Matter
• Matter Around Us
• Matter In Our Surroundings
• Metallurgy
• Metals And Non-Metals
• Mole Concept and Stoichiometry
• Natural Resources
• Organic Chemistry – Basic Principles
• Periodic Classification of Elements
• Physical and Chemical Changes
• Physical and Chemical Properties of Water
• Polymers
• Preparation, Properties and Uses of Compounds
• Principles and Processes of Isolation of Elements
• Redox Reactions
• Relative Molecular Mass and Mole
• States of Matter
• Structure Of The Atom
• Study of Compounds
• Study of Gas Laws
• Study of Representative Elements
• Surface Chemistry
• The d-block and f-block elements
• The Gaseous State
• The p-Block Elements
• The Periodic Table
• The s-Block Elements
• The Solid State
• Thermodynamics
### High School Biology
• Absorption and Movement of Water in Plants
• Anatomy of Flowering Plants
• Animal Kingdom
• Bacteria and Fungi-Friends and Foe
• Biodiversity and Conservation
• Biofertilizers
• Biological Classification
• Biomedical Engineering
• Biomolecules
• Biotechnology and its Applications
• Biotic Community
• Body Fluids and Circulation
• Breathing and Exchange of Gases
• Cell – Unit of Life
• Cell Cycle and Cell Division
• Cell Division and Structure of Chromosomes
• Cell Reproduction
• Cellular Respiration
• Chemical Coordination and Integration
• Circulation
• Control And Coordination
• Crop Improvement
• Digestion and Absorption
• Diversity In Living Organisms
• Ecosystem
• Environmental Issues
• Excretory Products and their Elimination
• Flowering Plants
• Genes and Chromosomes
• Health and Diseases
• Health and Its Significance
• Heredity And Evolution
• Heredity and Variation
• How Do Organisms Reproduce?
• Human Diseases
• Human Eye And Colourful World
• Human Health and Disease
• Human Population
• Human Reproduction
• Hygiene
• Improvement In Food Resources
• Integumentary System- Skin
• Kingdom Fungi
• Kingdom Monera
• Kingdom Protista
• Life Processes
• Locomotion and Movement
• Microbes in Human Welfare
• Mineral Nutrition
• Molecular Basis of Inheritance
• Morphology of Flowering Plants
• Neural Control And Coordination
• Nutrition in Human Beings
• Organism and Population
• Photosynthesis
• Photosynthesis in Higher Plants
• Plant Growth and Development
• Plant Kingdom
• Pollination and Fertilization
• Pollution; Sources and its effects
• Principles of Inheritance and Variation
• Reproduction and Development in Angiosperms
• Reproduction in Organisms
• Reproductive Health
• Respiration in Human Beings
• Respiration in Plants
• Respiratory System
• Sexual Reproduction in Flowering Plants
• Strategies for Enhancement in Food Production
• Structural Organisation in Animals
• Structural Organisation of the Cell
• The Endocrine System
• The Fundamental Unit Of Life
• The Living World
• The Nervous System and Sense Organs
• Tissues
• Transpiration
• Transport in Plants
### High School Math
• Algebra – Arithmatic Progressions
• Algebra – Complex Numbers and Quadratic Equations
• Algebra – Linear Inequalities
• Algebra – Pair of Linear Equations in Two Variables
• Algebra – Polynomials
• Algebra – Principle of Mathematical Induction
• Binomial Theorem
• Calculus – Applications of Derivatives
• Calculus – Applications of the Integrals
• Calculus – Continuity and Differentiability
• Calculus – Differential Equations
• Calculus – Integrals
• Geometry – Area
• Geometry – Circles
• Geometry – Conic Sections
• Geometry – Constructions
• Geometry – Introduction to Euclid’s Geometry
• Geometry – Three-dimensional Geometry
• Geometry – Lines and Angles
• Geometry – Straight Lines
• Geometry – Triangles
• Linear Programming
• Matrices and Determinants
• Mensuration – Areas
• Mensuration – Surface Areas and Volumes
• Number Systems
• Number Systems – Real Numbers
• Permutations and Combinations
• Probability
• Sequence and Series
• Sets and Functions
• Statistics
• Trignometry – Height and Distance
• Trignometry – Identities
• Trignometry – Introduction
### Middle School Science
• Acids, Bases And Salts
• Air and Its Constituents
• Basic Biology
• Body Movements
• Carbon and Its Compounds
• Cell – Structure And Functions
• Changes Around Us
• Chemical Effects Of Electric Current
• Coal And Petroleum
• Combustion And Flame
• Components Of Food
• Conservation Of Plants And Animals
• Crop Production And Management
• Electric Current And Its Effects
• Electricity And Circuits
• Elements and Compounds
• Fibre To Fabric
• Food production and management
• Force And Pressure
• Forests: Our Lifeline
• Friction
• Fun With Magnets
• Garbage In, Garbage Out
• Getting To Know Plants
• Health and Hygiene
• Heat
• Hydrogen
• Life Processes: Nutrition in Animals and Plants
• Materials: Metals And Non-Metals
• Matter and Its States
• Metals and Non-metals
• Micro Organisms: Friend And Foe
• Motion And Measurement Of Distances
• Motion And Time
• Nutrition In Animals
• Nutrition In Plants
• Organization in Living Things
• Our Environment
• Physical And Chemical Changes
• Pollution and conservation
• Pollution Of Air And Water
• Reaching The Age Of Adolescence
• Reproduction In Animals
• Reproduction In Plants
• Respiration In Organisms
• Rocks and Minerals
• Separation Of Substances
• Simple Machines
• Soil
• Some Natural Phenomena
• Sorting Materials Into Groups
• Sound
• Stars And The Solar System
• Structure of Atom
• Synthetic Fibers And Plastics
• The Living Organisms And Their Surroundings
• Transfer of Heat
• Transformation of Substances
• Transportation In Animals And Plants
• Universe
• Waste-water Story
• Water: A Precious Resource
• Weather, Climate And Adaptations Of Animals To Climate
• Winds, Storms And Cyclones
### Middle School Math
• Area and Its Boundary
• Boxes and Sketches
• Data Handling
• Fun With Numbers
• Heavy and Light
• How Many
• Long And Short
• Mapping
• Measurement
• Money
• Multiplication and Factors
• Multiply and Divide
• Numbers
• Parts and Wholes
• Pattern Recognition
• Patterns
• Play With Patterns
• Rupees And Paise
• Shapes And Angles
• Shapes And Designs
• Shapes and Space
• Similarity
• Smart Charts
• Squares
• Subtraction
• Tables And Shares
• Tenths and Hundredths
• Time |
# What Do The Stars Say About Shar Jackson? (12/04/2019)
How will Shar Jackson do on 12/04/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is not scientifically verified – do not take this too seriously. I will first find the destiny number for Shar Jackson, and then something similar to the life path number, which we will calculate for today (12/04/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology experts.
PATH NUMBER FOR 12/04/2019: We will consider the month (12), the day (04) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. This is how it’s calculated. First, for the month, we take the current month of 12 and add the digits together: 1 + 2 = 3 (super simple). Then do the day: from 04 we do 0 + 4 = 4. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 3 + 4 + 12 = 19. This still isn’t a single-digit number, so we will add its digits together again: 1 + 9 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the path number for 12/04/2019.
DESTINY NUMBER FOR Shar Jackson: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Shar Jackson we have the letters S (1), h (8), a (1), r (9), J (1), a (1), c (3), k (2), s (1), o (6) and n (5). Adding all of that up (yes, this can get tedious) gives 38. This still isn’t a single-digit number, so we will add its digits together again: 3 + 8 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for Shar Jackson.
CONCLUSION: The difference between the path number for today (1) and destiny number for Shar Jackson (2) is 1. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is of questionable accuracy. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
##### Tools
This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.
# Preview of calculus: part 2
This is a part of Calculus 1: course.
## 1 Piecewise Defined Functions
The absolute value function $$f(x) = |x|$$ is computed as follows: $$f(x) = \text{ if } x < 0 \text{ then } y = - x$$ $$f(x) = \text{ if } x \geq 0 \text{ then } y = x$$
We can re-write this algebraically as $$f(x) = \begin{cases} -x & \text{ if } x < 0 \\ x & \text{ if } x \geq 0 \end{cases}$$
This is called a piecewise definition. It's a special case of algebraic representation of functions.
Where do functions like this comes from?
Example:
Hypothetically...
The law says $$\begin{cases} \text{if income } & < 10000 \text{ then tax rate } = 0\% \\ \text{if } 10000 & \leq \text{ income } < 20000 \text{ then tax rate } = 10\% \\ \text{if income } & \geq 20000 \text{ then tax rate } = 20\% \end{cases}$$
We can express this algebraically.
Suppose $$x$$ is the income and $$y = f(x)$$ is the tax rate, then $$f(x) = \begin{cases} 0 & \text{if } x < 10000 \\ 10 & \text{if } 10000 < x < 20000 \\ 20 & \text{if } x \geq 20000 \end{cases}$$
There is a possible problem with the above. There is no definition for $$x = 10000$$! Fix it.
## 2 Characteristics of Functions
Functions: Even Odd Neither
Graphical
Algebraic $$y = x^{2}$$ $$y = x^{3}$$ $y=x+1$, just about anything...
No symmetry!
We discover Same $$y$$ for two different $$x$$'s ($$x$$ & $$-x$$). $$y$$ for $$x$$ and $$-y$$ for $$-x$$.
Definition:
$$f(x) = f(-x)$$ $$f(-x) = -f(x)$$
Examples $$(-x)^{2} = x^{2} => y = x^{2}$$ is even $$(-x)^{3} = -x^{3} => y = x^{3}$$ is odd
Other Examples $$\cos x, x^{4}, x^{6}, x^{6} + x^{4} - 17x^{2}$$. $$x, x^{3}, x^{5}, x^{7}, x + x^{5} - 17 x^{17}, \sin x$$.
Numerical
Similar
Warning: there could be more values and these observations could turn out to be false:
## 3 Growth: Increasing/Decreasing Behavior
This verbal definition is simple and the geometric meaning is very clear. However, both are imprecise. Let's work out this concept algebraically.
Function $$y=f(x)$$ is increasing on interval $$(a, b)$$ if for any given $$x_{1}, x_{2}$$ in $$(a, b)$$,
$$\begin{array}% \text{ if } x_{1} < x_{2} & \text{ then } & f(x_{1}) < f(x_{2}), \\ \end{array}$$
How do we verify these conditions? It's hard (but easy with calculus).
Example:
$$f(x) = 3x - 7$$
If $$x_{1} < x_{2}$$ then
\begin{aligned} f(x_{1}) = 3 x_{1} - 7 & \overset{?}{<} f(x_{2}) = 3 x_{2} - 7 \\ 3 x_{1} & \overset{?}{<} 3 x_{2} \\ \therefore x_{1} & < x_{2} \end{aligned}
The computation suggests that $y=f(x)$ is increasing. To finish the proof, retrace your steps.
This is even harder for quadratic, cubic, ... functions as they lead to quadratic, cubic, .... equations.
Function $$y=f(x)$$ is decreasing on interval $$(a, b)$$ if for any given $$x_{1}, x_{2}$$ in $$(a, b)$$,
$$\begin{array}% \text{ if } x_{1} < x_{2} & \text{ then } & f(x_{1}) > f(x_{2}), \\ \end{array}$$
Notation:
• $\nearrow$ for increasing, and
• $\searrow$ for decreasing.
Review exercise:
$$y = \sqrt{u} + \sqrt{4 - u}$$
Find the domain, i.e. find all $$u$$'s for which the formula makes sense.
Start
Make sure that $$\sqrt{}$$ makes sense, i.e. what's inside can not be negative.
1. $$\sqrt{u} \Rightarrow u \geq 0$$, solved already
2. $$\sqrt{4 - u} \Rightarrow 4 - u$$, solve
• Subtract 4 from both sides, $$-u \geq -4$$
• Multiply by $$(-1)$$, $$u \leq 4$$
Each $$u$$ in the domain must satisfy both inequalities, $$u \geq 0$$ AND $$u \leq 4$$. Hence
$$A: D = [0,4]$$
## 4 Word Problems
Problem. A farmer with 100 yards of fencing material wants to build as large a rectangular enclosure as possible.
He would first try to apply his common sense.
Trial and Error:
\begin{aligned} 25 \times 25 & = 625 \text{ sq. yards} \\ 24 \times 26 & = 624 \ldots \\ \vdots & = \vdots \vdots \end{aligned}
More trials lead to better results but we cannot try all possible rectangles.
Numerical analysis with Excel:
Start with the perimeter : $$2( W + D ) = 100$$, where $W,D$ are the width and the depth of the rectangle.
Solve it $$D = 50 - W$$ What about the area? \begin{aligned} A & = WD \\ & = W(50-W) \end{aligned}
Using Excel, we see this:
Excel fills gaps with straight lines. We can see that the graphical representation is better and it suggests that 25 is the best solution.
But gaps are still an issue.
Algebra:
We can restate the problem as \begin{aligned} A & = W( 50 - W ) \\ & \textrm{Find the maximum} \end{aligned}
Later, we will use the derivative. Now we use the fact that this is a parabola.
What do we know about it?
• There is one max or min.
• But which one?
$$f(x) = -x^{2} + 50 x$$ (We replaced $$W$$ with $$x$$ here.)
Because of the "-", this one opens down, with a max.
Question. Where is the maximum?
Answer. It is the midpoint between the two $$x$$-intercepts.
In our case, the $$x$$-intercepts are 0 and 50. So the "vertex" of the parabola is at
$$x = \frac{ 0 + 50 }{ 2 } = 25$$
We've solved the problem but it's clear that with even a bit more complicated functions we wouldn't be able to handle the algebra. Calculus will help...
## 5 We need more functions
We need a library of elementary functions to build others.
A parabola (curve) is the graph of a function. But what kind of function?
\begin{aligned} & x^{2} \\ & x^{2} + 1 \\ & x^{2} + x + 1 \end{aligned}
Example:
Is this a parabola?
We don't know. Probably not because it has a flat bottom.
In fact, it looks like $$y = x^{4}$$, which is not a parabola.
## 6 Linear Functions
Linear functions are simpler than quadratic functions.
Slope-intercept form: $$\begin{matrix} y = & mx & b \\ & \uparrow & \uparrow \\ & \textrm{the slope} & y\textrm{-intercept} \end{matrix}$$
Monotinicity is very simple:
• $$m > 0$$ - $$f$$ is increasing, on the whole domain.
• $$m < 0$$ - $$f$$ is decreasing, on the whole domain.
• $$m = 0$$ - $$f$$ is constant, on the whole domain (technically not "linear").
Also, there are no max/min points.
The general form is $$f(x) = ax^{2} + bx + c$$
Facts we know
$$a > 0$$
parabola opens up
$$a < 0$$
parabola opens down
$$a = 0$$
The x-coordinate of the vertex of parabola (i.e., max or min) is $$v = - \frac{b}{2 a }$$ Find y.
Domain is all reals.
Now, a quadratic function will exhibit either:
• $$\underbrace{(-\infty, v)}_{\nearrow}$$ and $$\underbrace{(v, \infty)}_{\searrow}$$ - Maximum
• $$\underbrace{(-\infty, v)}_{\searrow}$$ and $$\underbrace{(v, \infty)}_{\nearrow}$$ - Minimum
Polynomials are "made up" of powers of $$x$$ as follows.
## 8 Power Functions:
$$x^{0} = 1 , \underbrace{x}_{\text{linear}}, \underbrace{x^{2}}_{\text{quadratic}}, \underbrace{x^{3}}_{\text{cubic}}, \cdots , x^{n}, \cdots$$
The degree, odd vs even, affects the shape of the graph.
Power is even
Power is even, then it looks "like" $$x^{2}$$ .
Power is odd
Power is odd, then it and looks "like" $$x^{3}$$ .
We can think algebraically and justify the pictures.
Later, we'll use the derivative.
$$f(x) = 5x^{3} - 3x^{2} + 17x - 18$$, cubic
Fact: The domain is all real numbers (no division, etc).
## 9 Rational Functions
Called this way because they are ratios of polynomials.
$$\left. \frac{ x - 1 }{ x + 1 } \right\} \text{ polynomials }$$
Also
$$\left(\frac{1 + x}{x^{2}}\right) \frac{x^{3}}{x^{2} + 1}$$
Definition: Function $f$ is called rational if
$$f(x) = \frac{P(x)}{Q(x)}$$
where $$P$$ and $$Q$$ are polynomials.
Question:
What is the domain of a rational function?
Example:
What is the domain of $$\dfrac{x - 1}{x + 1}$$?
Answer: $x \neq -1$
I.e., the $f(x)$ makes sense for all $$x$$ but -1.
Solution:
We look at the denominator, $$x + 1 = 0$$ and solve.
$$\Longrightarrow x = -1$$
not in the domain.
If we plug $$x = -1$$, we have a division by 0.
Problem:
What is the domain for
$$f(x) = \frac{x - 1}{x^{2} - 4}$$
Set the denominator to 0 and solve
\begin{aligned} x^{2} - 4 & = 0 \\ x^{2} & = 4 \\ x & = \pm 2 \end{aligned}
So the domain consists of all real numbers except $$\pm 2$$.
## 10 Algebraic Functions
I.e., the ones "with roots".
Problem. Find the domain to the function
$$\sqrt{ x - \sqrt{x}}$$
Whatever is inside the square root can't be negative. So,
1. $$x \geq 0$$, i.e. $$x$$ is positive.
2. $$x - \sqrt{x} \geq 0$$, solve it. \begin{aligned}x & \geq \sqrt{x} \\ x^{2} & \geq x \\x & \geq 1\end{aligned}
The domain is
$$D = [1, \infty).$$
## 11 Trigonometric Functions
These three will suffice 00% of the time: $$\sin x, \cos x, \tan x = \frac{\sin x}{\cos x}$$
Fact: The domain for sin and cos are all real numbers.
Fact: These functions are periodic, with period $2\pi$.
\begin{aligned} \sin( x + 2 \pi ) & = \sin x \\ \cos(x + 2\pi) & = \cos x \\ \tan(x+2\pi) &= \tan x \end{aligned}
for all $$x$$.
Recall the definitions (comes from the Pythagorean Theorem):
Domain of $$\tan x$$?
$$\tan x = \frac{ \sin x}{\cos x}$$
To find domain, set denominator to 0 and solve. Find all $$x$$'s
\begin{aligned} \cos x & = 0 \\ x & = \ldots, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \end{aligned}
Question:
Is it increasing? It looks like it's increasing everywhere... But the correct question is: on what intervals is it increasing? In particulr, is it increasing on $$(-\infty, \infty)$$?
No. Because $$\tan$$ is undefined at $$\dfrac{\pi}{2}, \dfrac{3\pi}{2}, \ldots$$ in the first place. Plus, you can see the jumps.
Review exercise:
Plot the graph of the piece-wise defined function:
$$f(x) = \begin{cases} 3 - \frac{1}{2}x & \text{ for } x \leq 2 \\ 2x - 5 & \text{ for } x > 2 \end{cases}$$
Plot:
1. Plot $$3 - \dfrac{1}{2} x$$.
2. Plot $$3 - \dfrac{1}{2}x$$ with domain $$x \leq 2$$.
3. Plot $$2x - 5$$.
4. Plot $$2x - 5$$ with domain $$x > 2$$.
5. Combine 2) and 4).
## 12 New Functions From Old
We set up a library of elementary functions: powers, roots, exponent, sin, cos. Then we combine them algebraically with the 4 operations (+ one more). For example:
$$x^{3}, x^{5} \rightarrow x^{3} + x^{5}$$
The idea is below:
Graphical representations: How do these transformations of graphs affect the functions?
1. Shift (Two types: Horizontal and Vertical)
2. Stretch (Two types: Horizontal and Vertical)
3. Flip (Two types: Horizontal and Vertical)
Six in Total
Shift (Vertical)
Rule 1:
If the graph $$y = g(x)$$ is that of $$y = f(x)$$ shifted $$k$$ units up, then $$g(x) = f(x) + k$$ This is what happens to every point on the graph:
Fact: Same exact shape of the graph.
Example: \begin{aligned}x^{2}, x^{2} + 1, x^{2} + 10, & \textrm{ same shape} \\x^{2} - 1 , x^{2} - 10, & \textrm{ same shape}\end{aligned} What if we shift down? We have to rewrite the rule?
Rule 1 (rewritten):
If the graph $$y = g(x)$$ is that of $$y = f(x)$$ shifted $$k$$ units up/down, then $$g(x) = f(x) \pm k$$ No need for "down". Indeed, "-10 units up" mean "10 units down".
Example:
Suppose "I drive West, drive one hour at -60 m/h." Where am I? A: 60 miles east of here. \begin{aligned}\textrm{distance} & = \textrm{time} \times \textrm{velocity} \\& = (1) \times (-60) \\& = -60\end{aligned} What about the horizontal shift? Compare.
Vertical Shift: $$y = f(x) \Rightarrow y = f(x) + k$$
$$\therefore y_{1} = y_{0} + k,$$ So $$g(x) = f(x) + k$$
Horizontal Shift:
$$y_{0} = f(x_{0}) = g(x_{0} + k),$$ So $$g(x) = f(x - k)$$
Example:
$$f(x) = x^{2}$$, shift 2 units to the right and we get $$g(x) = f(x-2) = (x-2)^{2}$$ $$x$$ changes.
What is the $$x$$ -intercept of $$g$$?
Set $$f(x) = 0$$, and solve. $$\begin{array}{c | c}( x + 2) ^{2} = 0 & \textrm{opposite?} \\x - 2 = 0 & h(x) = (x + 2)^{2} \\x = 2 & x + 2 = 0 \\& x = -2\end{array}$$ $$h(x)$$ is shifted 2 units to the left.
Rule 1-2:
If the graph of $$g$$ is that of $$f$$, shifted $$k$$ units to the right, then $$g(x) = f(x-k)$$ Algorithmic interpretation of these shifts.
Vertical Shift
Horizontal Shift
Example:
Let $$h(x) = (x + 2)^{2} - 3$$ What is the graph?
A different kind of transformation.
Flip
Then $$y_{1} = - y_{0}$$, so $$g(x) = - f(x)$$
Rule 2-1:
If the graph of $$g$$ is that of $$f$$, flipped about the $$x$$-axis then $$g(x) = - f(x)$$
Example:
$$h(x) = - f(x) + 3$$ How do we get the graph of $$h$$? $$x \longmapsto f(x) \underbrace{\longmapsto}_{\textrm{vertical flip}} - f(x) \underbrace{\longmapsto}_{\textrm{vertical shift}} - f(x) + 3$$ The order matters... To know the order, read the function from inside-out.
Rule 2-2
If the graph of $$y = g(x)$$ is that of $$y = f(x)$$ flipped about the $$y$$- axis, then $$g(x) = f(-x)$$
Stretch:
The graph is not a wireframe anymore! The $$xy$$-plane is the one stretched, if you think of it as a rubber sheet, with a graph on it.
Consider the stretch by a factor of 2
Rule 3-1
If the graph of $$y = g(x)$$ is that of $$y = f(x)$$ stretched vertically by a factor of $$k > 0$$, then $$g(x) = kf(x)$$. (Factor outside of $$k$$ ).
Review exercise: Match graphs and formulas.
Rule 3-2
If graph $$y = g(x)$$ is that of $$y = f(x)$$ stretched by the factor $$k$$-horizontally, then $$g(x) = f(x/k)$$. (Factor inside of $$f$$ ).
Example: Given $$f(x) = x$$ Stretched by 2x vertically is $$g(x) = 2 x$$ On the other hand:
We can get $$y = g(x)$$ by shrinking $$y = f(x)$$ horizontally, factor 2x. $$2x = \frac{x}{\frac{1}{2}}$$ Stretch by factor $$\frac{1}{2}$$ = shrink by 2.
Example:
$$f(x) = x^{2}$$ Vertical stretch by 2 is $$g(x) = 2x^{2}$$.
Horizontal stretch by $$\sqrt{2}$$ is $$g(x) = (\sqrt{2} x )^{2}$$. \begin{aligned}(\sqrt{2} x)^{2} & = (\sqrt{2})^{2} \cdot x^{2} \\ & 2 x^{2}\end{aligned}
Example:
Verify the plot $$2 x^{2}$$ vs. $$(2 x )^{2}$$.
Not the same!
More generally we interpret these operations as compositions.
## 13 Compositions:
Consider:
More Generally:
Variables have to match!
Example:
Represent $$z = h(x) = \sqrt[3]{x^{2} + 1 }$$ as the composition of two functions. (Easy if you know how). $$x \rightarrow x^{2} + 1 \xrightarrow{y} \sqrt[3]{y} \xrightarrow{z}$$
Example:
Given Speed (60 m/h), Miles per gallon (30 m/gal) and Cost per gallon (\$5/gal), represent the expense as a function of time.
Analysis:
$$\text{time (hours)} \xrightarrow{\text{60 m/h}} \text{distance (miles)}\xrightarrow{\text{30 m/gal}} \text{gas used (gal)} \xrightarrow{\\text{5/gal}}\text{expense } (\)$$
Composition:
$$t \rightarrow \underset{f}{ 60 t } \rightarrow d \rightarrow \underset{k}{\frac{d}{30}} \rightarrow g \rightarrow \underset{h}{5 g} \rightarrow e$$
Compute:
$$d = f(t) = 60t, \; g = k(d) = \frac{d}{30} \; e = h(g) = 5g$$
Substitute \begin{aligned}g = k(d) & = k(f(t)) \\e = h(g) & = h(k(d)) \\& = h(k(f(t)))\end{aligned} Read from right to left. ( $$f(t)$$ is not $$(t)f$$.
Example: $$f(x) = x^{2}$$ and $$g(x) = \cos x$$. Find $$f(g(x))$$, replace $$x$$ in $$f$$ with $$(\cos x )$$, always with parentheses. $$x^{2} \Longrightarrow ( \cos x )^{2}$$ or $$cos^{2} x$$.
Example:
Find $$g(f(x))$$. Replace $$x$$ in $$g$$ with $$(x^{2})$$. $$\cos x \rightarrow \cos (x^{2})$$
Review Exercise.
1. Stretch vertically by 2 (about $$y$$: $$g(x) = 2 f(x)$$ ).
2. Shift horizontally - right by 2 (about $$x$$: $$h(x) = g(x-2)$$ ).Replace $$x$$ with $$( x-2 )$$.(Suggestion: use parentheses a lot).
Algebraically:
$$f(x) = \sqrt{3 x - x^{2}} \Longrightarrow ?????\sqrt{3x - x^{2} - 2}$$
$$\sqrt{3 ( x - 2 ) - ( x - 2 )^{2} }$$
## 14 Algebra of Functions
For opertions, we need to align the variables first -- but in a different way.
1. Addition -- rename the variables:
2. Composition -- rename the variables:
Example: Find the composition of these functions. $$f(x) = x + 1, \; g(x) = 2x, \; h(x) = x - 1.$$ Find $$f \circ g \circ h$$.
Diagram:
Rename the variable to make them match:
$$y = h(x) = x - 1, \; z = g(y) = 2y, \; u = f(z) = z + 1$$
Algebraically:
Algorithm:
\begin{alignat}{3}& & y = x - 1 \\& z = 2y & \\u = z + 1 & &\end{alignat}
Algebra:
\begin{aligned}u = z + 1 & = (2y) + 1 \\& = 2y + 1 \\& = 2(x - 1) + 1\end{aligned} |
CCSP03 Mathematical Ideas
Document Sample
``` Singapore Mathematical Society
Association of Mathematics Educators
Peter Pang
15/2/03
The Constant
All circles are similar.
C=circumference
The relative size of
the circumference C
to diameter D will be
exactly the same, D=diameter
that is,
C/D = constant
=
C = D = 2R
The Constant (2)
Thus provides the
connection between C=circumference
two lengths, which
are the circumference R=radius
and the diameter.
D=diameter
This same constant
connects the circle’s area
Approximating the Circle
The critical idea is to
approximate the circle C=circumference
with an inscribed
D=diameter
area
Approximating the Circle
(2)
We inscribe a regular
pentagon in a circle
Break up the
pentagon into five
triangular pieces. r
h
Each triangle has b
base b and height h.
The dotted line is
called the apothem.
Connecting Length and
Area
Area of each triangle
=(base)(height)/2
=bh/2
Area of inscribed
pentagon
=5(Area of triangle) r
h
=5(bh/2) b
=(h/2)(5b)
=(h/2)(Perimeter of
inscribed pentagon)
Connecting Length and
Area (2)
Area of inscribed
pentagon
=(h/2)(Perimeter)
What if we choose
any other regular
polygon? r
h
b
Connecting Length and
Area - Hexagon
If we inscribe a
regular hexagon, then
the hexagon will be
divided into 6 little
triangles.
Area of inscribed hexagon r
h
=6(Area of triangle) b
=6(bh/2)
=(h/2)(6b)
=(h/2)(Perimeter)
Connecting Length and
Area - Octagon
If we inscribe a
regular octagon, then
the octagon will be
divided into 8 little
triangles.
Area of inscribed octagon
=8(Area of triangle) h r
b
=8(bh/2)
=(h/2)(8b)
=(h/2)(Perimeter)
Inscribing a regular n-gon
If we inscribe a regular n-gon, we obtain
Area of polygon = (h/2)(Perimeter)
If we increase the number of sides from 10 to
10,000 to 10,000,000, what happens?
The polygons will gradually “fill up” the circle.
Area of circle
= lim (Area of regular inscribed polygon)
= lim [(h/2)(Perimeter)]
Two Questions
What happens to the apothem and to the
perimeter as the number of polygonal
sides increases indefinitely?
Clearly h will have as its limit the radius of
the circle.
The limiting value of the perimeters of the
inscribed regular n-gons will be the circle’s
circumference.
The Connection Between
Length and Area
Hence
Area of circle = lim [(h/2)(Perimeter)]
= [(r/2)(C)]
= rC/2
= r(2r)/2
= 2r2/2
= r2.
Approximating
The simplest way to approximate the ratio C/D
is to measure the circumference and diameter
of a particular circle and divide the former by
the latter.
However physical measurements introduce
inaccuracies and in any case tangible objects
like bicycle wheels and coffee cans are not
perfect, mathematical circles.
Archimedes’
Approximation
Requires only algebra and the
Pythagorean theorem.
Pythagorean theorem:
a2+b2=c2.
c
a
b
Inscribing a Square
circle r =1.
A s B
Pythagorean theorem
tells us that s2+s2=22
s 2
1
2s2=4
s =2
D C Perimeter of the
square P =4s=42
Inscribing a Square (2)
We get
A s B circumfere nce
diameter
perimeter
s 2
1 diameter
4 2
D C 2
2 2
2.828427125
A Better Approximation
Improve upon the first estimate by doubling the
number of sides of the inscribed polygon to get
an octagon and let its perimeter be our next
estimate of the circle’s circumference.
Double again to get a 16-gon, then a 32-gon,
and so on.
The major obstacle is how to determine the
relationship between the perimeter of one these
polygons to the next.
Overcoming the Obstacle
A Line segment AB of
b length a is a side of a
C regular inscribed n-
a/2 gon.
1 1-x
D Segment AC, a side of
x a/2
a regular inscribed
B
2n-gon is generated.
1
O
Overcoming the Obstacle
(2)
A Pythagorean Theorem
b tells us that
C 12 =(a/2)2+x2
a/2 =a2/4 + x2
1 1-x
D a2
x a/2 x2 1
B 4
a2
1 x 1
O 4
Overcoming the Obstacle
(2)
A Pythagorean Theorem
b tells us that
2
C a
a/2 b (1 x )
2 2
1-x 2
1 D a2
x a/2 1 2x x 2
4
B
a 2
a 2
a 2
1 2 1 1
1 4 4 4
O
a 2
22 1
4
Overcoming the Obstacle
(3)
A a2
b
b2 2 2 1
4
C a2
a/2 2 4 1
1 1-x 4
D
x a/2 2 4 a2
B
1
b 2 4 a 2 .
O
Back to Approximating -
Octagon
Apply the above
C formula to calculate
A s B
1
the side of a regular
2 inscribed octagon as
s
O b 2 4 a2
2 4 2
2
D C’
2 42
2 2
Back to Approximating -
Octagon (2)
We get
C
A s B circumfere nce
1 diameter
2
perimeter
s
O diameter
8 2 2
D C’ 2
4 2 2
3.061467459
Back to Approximating -
16-gon
Apply the formula
C again to calculate the
A s B
1
side of a regular
2 inscribed 16-gon as
s
O b 2 4 a2
2 4 2 2
2
D C’
2 4 2 2
2 2 2
Back to Approximating -
16-gon (2)
We get
C
A s B circumfere nce
1 diameter
2
perimeter
s
O diameter
16 2 2 2
D C’ 2
8 2 2 2
3.121445153
Back to Approximating -
32-gon
Another doubling and
C application of the formula
A s B
1
yields the perimeter of a
2 regular inscribed 32-gon as
s 32 2 2 2 2
O
Now we get
D C’ perimeter
diameter
16 2 2 2 2
3.136548491
Estimate of
We double seven more times, to a 64-gon, a
128-gon, a 256-gon, a 512-gon, a 1024-gon, a
2048-gon and a 4096-gon.
The 4096-gon yields an estimate of
perimeter
diameter
3.141594618
Estimate of (2)
Thus can be approximated as closely as we
wish.
This basic approach dates back 22 centuries to
Archimedes.
It however has one liability: the need to
calculate square roots of square roots.
Archimedes pushed through the calculations
with bare hands, without the benefit of the
decimal notation, up to the 96-gon.
Methodologies highlighted
Iterative process
Limiting process
Computation of √2
It is well-known that 2 is irrational, i.e., it has
an infinite non-repeating decimal expansion.
The following algorithm appeared in the
Jiuzhang Suanshu for obtaining this decimal
expansion.
An Iterative Approach
First, it is clear that the integer part of the
decimal expansion is 1, i.e., the decimal
expansion is of the form 1.abc…. We shall refer
to a as the first decimal digit, b as the second
decimal digit, etc. The strategy is to obtain one
decimal digit at a time.
Write 2 as 1 + x. Then,
2 (1 x) 1 2 x x
2 2
2 x x 1.
2
Find an x that has only 1 decimal digit, i.e., x =
0.a, such that 2x + x2 is as close to 1 as possible
but without exceeding 1 (note that this x is not a
solution to the equation above). When you carry
this out, you will find the answer to be a = 4, or x
= 0.4.
Indeed (1.4)2 is less than 2, but (1.5)2 exceeds 2.
Thus, the decimal expansion is of the form 1.4bc…,
and our next step is to find the decimal digit b.
Now, given that the integer part is 1 and the
first decimal digit is 4, we write 2 as 1.4 + x.
Then,
2 (1.4 x) 1.4 21.4x x
2 2 2
2(1.4) x x 2 2 (1.4) 2 .
We now seek a number x = 0.0b, i.e., only the
second decimal digit is possibly non-zero, such
that 2(1.4)x + x2 is as close to 2 (1.4)2 as
possible but without exceeding it.
We will get b = 1.
Squaring the Circle:
The Transcendence of
Let us start with a straightedge of unit length.
Obviously, by putting line segments drawn using
such a straightedge together, one can make line
segments of any (positive) integer length.
Now, suppose we have two such line segments
of lengths x and y (x and y being positive
whole numbers), what are all the possible
lengths of new segments we can construct
using the straightedge and compass only?
Here are five possibilities:
It is possible to make a segment length x+y by
lying the two next to each other.
Likewise, drawing segment x starting at the
right end of y and heading left, the distance
from the left endpoint of y to the left endpoint
of x is x y.
Through a more sophisticated construction, one
can make segments of length xy, x/y, and the
square root of x or y .
y x
x+ y
Subtraction:
x-y y
x
Multiplication:
xy
y
1 x
Division:
y
y/x
1
x
Square Root:
a
1 a
Thus we can add, subtract, multiply, divide, and/or
take square roots to obtain any line segment whose
length is the sum, difference, product and/or
quotient of any (positive) rational number and/or
square root of rational number.
At the same time, using a straightedge and compass,
we can construct two perpendicular lines on which a
scale of length is imposed. This of course is nothing
else but a (2-dimensional) Cartesian coordinate
system, sometimes also known as the x-y plane.
Thus, using addition, subtraction, multiplication,
division and the extraction of square roots, we can
construct geometrically points on the plane whose
coordinates are sum/difference/product/quotient of
any (positive) rational number and/or their square
roots.
Geometric Constructions
(Using Compass and Straightedge Only)
The five constructions above cover five powerful
operations on the coordinates of points in the
plane. But are any additional operations possible
by geometric construction? The answer is no. To
see this, let us scrutinize what we can do using
a straightedge or a compass.
Since the only shapes one can draw with a
straightedge or a compass are line segments
and circles (or portions thereof), the following
is an exhaustive list of techniques for
constructing new points out of old ones:
1. Draw a line segment through two points.
2. Find the intersection point of two (non-parallel)
lines.
3. Draw a circle of a given radius with a given
centre, or equivalently, draw a circle with a
given centre through another given point.
4. Find the intersection points of a circle with
another circle or line.
We have already seen that geometric
construction allows us to reach all points with
rational coordinates. Starting with points with
rational coordinates, and using 1 – 4 above,
what points can we reach?
If two points have rational coordinates (a, b)
and (c, d), i.e., a, b, c, d are rational numbers
(below, as an abbreviation, we shall simply call
such points rational points), then points on the
line segment between them satisfy the equation
x(bd) y(ac) = (bcad),
which is of the form
jx+ky = m
where j, k, m are rational numbers.
Given two non-parallel lines of this form:
jx+ky = m and nx+py = q
where j, k, m, n, p, q rational,
the lines intersect at the unique point
x = (mp-kq) / (jp-kn), y = (jq-mn) / (jp-kn),
which again has rational coordinates.
Thus 1 and 2 do not give any additional points.
Next, points on a circle with rational centre (a, b)
through a rational point (c, d) satisfy the equation
(xa)2+(yb)2 = (ca)2+(db)2
which is of the form
x2+y2+jx+ky = m
where j, k, m are rational numbers.
Given a circle of this form and the line nx+py=q
where n, p, q are rational numbers, their
intersection points will be solutions of the
n2 2 2nq nk q 2 kq
2 1 x j 2 x 2 m 0
p
p p
p
p
or
p2 2 2 pq pj q 2 jq
2 1 y k 2 y 2 m 0,
n n
n n n
which are of the form
() a b c , d e f
where a, b, c, d, e, f are rational.
Similarly, we see that coordinates of points of
intersection of two circles are again solutions of
quadratic equations and hence also of the form
().
Finally, it remains to examine points constructed
from 1 – 4 starting with points whose
coordinates are of the form ().
It is not hard to see that coordinates of such
points are again combinations of
sum/difference/product/ quotient of rational
numbers and their square roots.
After knowing exactly which numbers are
"geometrically constructible", the important
point really is this:
Since a "geometrically constructible" number is
obtainable from the rationals using only
addition, subtraction, multiplication, division,
and the taking of square roots, such a number
must be a solution to some polynomial equation
with rational coefficients.
For example, take x = 6+2. Then x 6=2, or
(x6)2 2=0. Thus, x is a solution of the
equation x2 12x + 34 = 0.
In fact, more is true: If one factors the
polynomial until one gets down to an "irreducible"
polynomial equation (one that cannot be factored
any further and still has rational coefficients), the
degree of this polynomial will always be a power
of 2.
In the previous example, we have seen that since
x contains a square root, we need to square x in
order to arrive at a polynomial equation satisfied
by x. Now, intuitively speaking, every additional
square root that appears in the number requires
you to double the degree of the polynomial
equation satisfied by that number. Thus, a
number obtained by taking a square root twice is
typically the root of an irreducible 4th degree
polynomial; a number obtained by taking a
square root thrice is typically the root of an
irreducible 8th degree polynomial, etc.
```
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# 12.2 Boundary conditions (Page 4/5)
Page 4 / 5
$\begin{array}{ccc}\hfill \frac{1}{2}\lambda +\frac{1}{2}\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \frac{2}{4}\lambda +\frac{1}{4}\lambda & =& L\hfill \\ \hfill \frac{3}{4}\lambda & =& L\hfill \\ \hfill \lambda & =& \frac{4}{3}L\hfill \end{array}$
Case 3 : In this case both ends have to be nodes. This means that the length ofthe tube is one half wavelength: So we can equate the two and solve for the wavelength:
$\begin{array}{ccc}\hfill \frac{1}{2}\lambda & =& L\hfill \\ \hfill \lambda & =& 2L\hfill \end{array}$
If you ever calculate a longer wavelength for more nodes you have made a mistake. Remember to check if your answers make sense!
## Three nodes
To see the complete pattern for all cases we need to check what the next step for case 3 is when we have an additional node. Below is the diagram for the casewhere $n=3$ .
Case 1 : Both ends are open and so they must be anti-nodes. We can have threenodes inside the tube only if we have two anti-nodes contained inside the tube and one on each end. This means we have 4 anti-nodes in thetube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between every adjacent pairof anti-nodes. We count how many gaps there are between adjacent anti-nodes to determine how many half wavelengths there are and equatethis to the length of the tube L and then solve for $\lambda$ .
$\begin{array}{ccc}\hfill 3\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda & =& \frac{2}{3}L\hfill \end{array}$
Case 2 : We want to have three nodes inside the tube. The left end must be anode and the right end must be an anti-node, so there will be two nodes between the ends of the tube. Again we can count the number ofdistances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember that the distance between the node and anadjacent anti-node is only half the distance between adjacent nodes. So starting from the left end we count 3 nodes, so 2 half wavelength intervals and then only anode to anti-node distance:
$\begin{array}{ccc}\hfill 2\left(\frac{1}{2}\lambda \right)+\frac{1}{2}\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda +\frac{1}{4}\lambda & =& L\hfill \\ \hfill \frac{5}{4}\lambda & =& L\hfill \\ \hfill \lambda & =& \frac{4}{5}L\hfill \end{array}$
Case 3 : In this case both ends have to be nodes. With one node in between there aretwo sets of adjacent nodes. This means that the length of the tube consists of two half wavelength sections:
$\begin{array}{ccc}\hfill 2\left(\frac{1}{2}\lambda \right)& =& L\hfill \\ \hfill \lambda & =& L\hfill \end{array}$
## Superposition and interference
If two waves meet interesting things can happen. Waves are basicallycollective motion of particles. So when two waves meet they both try to impose their collective motion on the particles. This can havequite different results.
If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are able to achieve the sum oftheir efforts. The resulting motion will be a peak which has a height which is the sum of the heights of the two waves. If two waves areboth trying to form a trough in the same place then a deeper trough is formed, the depth of which is the sum of the depths of the twowaves. Now in this case, the two waves have been trying to do the same thing, and so add together constructively. This is called constructive interference .
If one wave is trying to form a peak and the other is trying to form a trough, then they are competing to do different things. Inthis case, they can cancel out. The amplitude of the resulting wave will depend on the amplitudes of the two waves that are interfering. If the depth ofthe trough is the same as the height of the peak nothing will happen. If the height of the peak is bigger than the depth of thetrough, a smaller peak will appear. And if the trough is deeper then a less deep trough will appear. This is destructive interference .
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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# Factors of 101 and How to Find Them
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In mathematics, a factor is a number that can be divided evenly into another number. The factors of 101 are 1, 101. However, there are other ways to find the factors of 101. In this article, we will explore some of the other methods to find the factors of 101.
## What are the factors of 101?
When it comes to finding the factors of 101, there are a few different methods that can be used. The most common method is to simply list out all of the numbers that 101 can be evenly divided by. However, this method can be time consuming and may not always produce the most accurate results.
For a more precise method, one can use the Euclidean algorithm. This algorithm involves taking two numbers and finding the greatest common divisor between them. With this method, the first step is to determine whether 101 is a prime number. If it is, then there are no factors of 101. If not, then the next step is to find the largest number that 101 can be divided by evenly.
Once the largest number has been determined, the next step is to find the largest number that can be evenly divided into that number. This process is continued until only two numbers remain. At this point, the smaller of the two numbers is the greatest common divisor and the other number is the smallest factor.
Applying this algorithm to 101 gives us a few different results. The first result is that 101 is not a prime number, as it can be evenly divided by 3. The largest number that 101 can be divided by is 3, and the largest number that can be evenly divided into 3 is 1. Therefore, the greatest common divisor between 101 and 3 is 1 and the smallest factor of 101 is 3.
The second result we get from using the Euclidean algorithm is that the greatest common divisor between 101 and 9 is 1 and the smallest factor of 9 is 9. Therefore, we can see that 9 is also a factor of 101.
Overall, we can see that there are three factors of 101: 1, 3, and 9.
## How do you find the factors of 101?
The factors of 101 are 1, 101.
To find the factors of 101, start by identifying the smallest number that 101 is divisible by. In this case, that number is 1. Then, divide 101 by 1 to get the quotient, which is 101. The factors of 101 are therefore 1 and 101.
## What is the greatest common factor of 101?
The greatest common factor of 101 is 1.
## What are the prime factors of 101?
In mathematics, a prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example, the integer 14 is a composite number because it has the divisors 2 and 7 in addition to the divisor 1. The integer 15, on the other hand, is a prime number because it has no other positive divisors besides 1 and itself.
The first prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
As of January 2017, the largest known prime number has 24,862,048 decimal digits. It was found in 2016 by researchers at Google.
Prime numbers have been studied throughout history. They play an important role in number theory because they are used to construct some of the most fundamental objects in mathematics such as the integers, the rational numbers, and the real numbers. In addition, primes are used in cryptography and they are also used to generate pseudo-random numbers.
## What is the least common multiple of 101?
The least common multiple (LCM) of two numbers is the smallest number that is evenly divisible by both numbers. In other words, it’s the lowest number you can count up to using both numbers.
To find the LCM of 101, we need to find the prime factorization of each number.
101 = 10 × 10 + 1
10 = 2 × 5
1 = 1
Therefore, the LCM of 101 is: 2 × 5 × 10 × 10 × 1 = 100
## How do you simplify 101?
In a world where there’s an app for everything and we’re constantly inundated with choices, it can be hard to know how to simplify your life. But it’s not impossible. Here are 101 ways to help you get started:
1. Wake up earlier.
3. Eat breakfast.
4. Exercise.
5. Get enough sleep.
6. Drink plenty of water.
7. Set goals for yourself.
8. Make a to-do list.
12. Donate or sell items you no longer need or want.
14. Unsubscribe from emails you don’t read.
15. Delete apps you never use.
16. Follow the one-in, one-out rule when it comes to buying new things.
18. Hang up your clothes instead of leaving them on the floor or in a pile on a chair.
19. Put away your shoes when you come home.
20. Do laundry regularly so you don’t have a huge pile to deal with later on
## What is 101 in simplest form?
When we talk about numbers in mathematics, we often use the term “simplest form.” What this means is that we are looking at a number in its most basic form. In other words, a number in simplest form is one that has been reduced to its lowest possible denominator.
The process of finding a number in its simplest form is called “reducing” or “simplifying” a fraction. To reduce a fraction, we divide the numerator (top number) and denominator (bottom number) by the greatest common factor (GCF). The GCF is the largest number that will divide evenly into both the numerator and denominator.
For example, let’s look at the fraction ¾. To find the GCF of ¾, we would need to find the largest number that would divide evenly into both 3 (the numerator) and 4 (the denominator). The answer is 1, so when we divide both 3 and 4 by 1, we get 3/4 in its simplest form.
Now that we know what it means to reduce or simplify a fraction, let’s look at an example of how to do it. Suppose we have the fraction ¼ and we want to find it in its simplest form. We can do this by finding the GCF of ¼.
The GCF of ¼ is 2, so when we divide both the numerator (1) and denominator (4) by 2, we get ½ in its simplest form.
As another example, let’s find the simplest form of the fraction ⅚. The GCF of ⅚ is 5, so when we divide both 5 (the numerator) and 6 (the denominator) by 5, we get 1/6 in its simplest form.
We can also use the GCF to simplify mixed numbers such as 1⅓. To do this, we first convert the mixed number to an improper fraction by multiplying the whole number (1) by the denominator (3) and adding it to the numerator (1). This gives us 4/3. We can then find the GCF of 4/3 and reduce it to its simplest form.
The GCF of 4/3 is 1, so when we divide both 4 (the numerator) and 3 (the denominator) by 1, we get 4/3 in its simplest form.
Now that you know what 101 in simplest form means, try reducing some fractions on your own!
## What is the highest common factor of 101?
The highest common factor of 101 is 1.
The highest common factor (HCF) of two or more numbers is the largest number that divides evenly into all of the numbers. So, in order to find the HCF of 101, we would need to find the largest number that would divide evenly into 101.
And what do you know? The answer is 1! Any number divided by 1 will result in that same number; 101 divided by 1 equals 101. Therefore, 1 is the highest common factor of 101.
## What is the lowest common denominator of 101?
The lowest common denominator of 101 is 1.
## How do you divide 101?
It’s simple: just divide it by three.
If you’re looking for a more challenging way to divide 101, try dividing it by seven. This will give you a remainder of four, which can be tricky to work with.
Another option is to divide 101 by eleven. This gives you a remainder of two, which is also difficult to deal with.
The best way to divide 101 is by thirteen. This gives you a remainder of zero, which is much easier to work with.
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# How Do You Find The Scale Factor Of A Circle?
## How is a circle different from other shapes?
A circle is a two-dimensional shape (it has no thickness and no depth) made up of a curve that is always the same distance from a point in the center.
An oval has two foci at different positions, whereas a circle’s foci are always in the same position..
## Are all circles congruent?
All circles of the same size are congruent to one another. “Size” can refer to radius, diameter, circumference, area, etc.
## What is the scale factor of 2?
Linear scale factor The size of an enlargement/reduction is described by its scale factor. For example, a scale factor of 2 means that the new shape is twice the size of the original. … You can get the ‘big’ and ‘small’ from the corresponding sides on the figures.
## What is scale factor in math definition?
The scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure. Example: Here, XYUV=123=4 . So, the scale factor is 4 .
## What is a scale factor in math in 7th grade definition?
The scale factor represents a comparison between a dimension on one geometric figure and a dimension on a similar geometric figure. To find a scale factor between two similar figures, find two corresponding sides and write the ratio of the two sides.
## What is a scale copy?
A scale copy of a figure is a figure that is geometrically similar to the original figure. This means that the scale copy and the original figure have the same shape but possibly different sizes. … The floor plan drawing has the same shape as but is smaller than the actual floor.
## What does a scale mean?
Definition of scale (Entry 5 of 7) 1 : a graduated series of musical tones ascending or descending in order of pitch according to a specified scheme of their intervals. 2 : something graduated especially when used as a measure or rule: such as.
## How do you find the scale factor between two circles?
A scale factor exists because any two circles are similar. You can use the radii to determine the scale factor. The ratio between the radii is \begin{align*}\frac{3}{1}\end{align*} so the scale factor is \begin{align*}\frac{3}{1}=3\end{align*}.
## What is a scale factor in math in 7th grade?
● Scale Factor: The ratio of any two corresponding lengths in two similar. geometric figures.
## What does a scale factor of less than 1 mean?
When the absolute value of the scale factor is less than one, a compression occurs. When the absolute value of the scale factor is equal to one, neither an expansion nor a compression occurs.
## What is a scale factor example?
A scale factor is a number which multiplies (“scales”) a quantity. For example,the “C” in y = Cx is the scale factor for x. If the equation were y = 5x, then the factor would be 5.
## What is a scale factor of 3 2?
This gives us the ratios 12/8, 15/10, and 9/6. These all reduce to the fraction 3/2, or 1 1/2 (the second triangle is a little less than twice as big). We usually leave the scale factor as a fraction, so we would say that the scale factor is 3/2.
## What is a scale factor of a triangle?
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles. … The ratios of corresponding sides are 6/3, 8/4, 10/5. These all reduce to 2/1. It is then said that the scale factor of these two similar triangles is 2 : 1.
## What is the scale factor of 12?
Architectural ScalesDrawing ScaleScale FactorViewport Scale3/8″ = 1′-0″321/32xp1/2″ = 1′-0″241/24xp3/4″ = 1′-0″161/16xp1″ = 1′-0″121/12xp7 more rows•Aug 12, 2018
## Are all circles the same size?
Explanations (4) Similarity is a quality of scaling: two shapes are similar if you can scale one to be like the other, like these triangles ABC and DEF. Since all circles are of the same shape (they only vary by size), any circle can be scaled to form any other circle. Thus, all circles are similar! |
# Problems with patterns with whole numbers
Lesson
If numbers increase or decrease by a regular amount, we can describe this as a number pattern.
## Increasing Number Patterns
If numbers rise by a regular amount, we can describe these as increasing number patterns. We could use addition or multiplication to make numbers get bigger.
Counting up by twos is an example of an addition pattern because each number is $2$2 bigger than the last.
This is an increasing sequence, where we add $2$2 at every step
We can describe these numbers as following a rule. In this case, the rule could be describes as, "Add $2$2 each time."
## Decreasing Number Patterns
If numbers reduce by a regular amount, we can describe these as decreasing number patterns. We could use subtraction or division to make numbers get smaller.
This is a decreasing sequence, where we subtract $3$3 at every step
#### Worked Examples
##### Question 2
The numbers in the sequence represent the number of squares in the picture pattern. Write the next numbers in the pattern.
1. $4$4, $\editable{}$, $\editable{}$, $\editable{}$, $\editable{}$, $\editable{}$
##### Question 3
Complete the pattern:
1. $8$8, $12$12, $16$16, $\editable{}$, $\editable{}$, $\editable{}$
2. What is the rule for filling in the pattern?
Add $8$8.
A
Subtract $8$8.
B
Subtract $4$4.
C
Add $4$4.
D
Add $8$8.
A
Subtract $8$8.
B
Subtract $4$4.
C
Add $4$4.
D |
NCERT solution class 9 chapter 13 Surface Areas and Volumes exercise 13.2 mathematics
EXERCISE 13.2
Question 1:
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. Assume π =
Height (h) of cylinder = 14 cm
Let the diameter of the cylinder be d.
Curved surface area of cylinder = 88 cm2
⇒ 2πrh = 88 cm2 (is the radius of the base of the cylinder)
⇒ πdh = 88 cm2 (d = 2r)
⇒ = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
Question 2:
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Height (h) of cylindrical tank = 1 m
Base radius (r) of cylindrical tank
Therefore, it will require 7.48 marea of sheet.
Question 3:
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm.
(i) Inner curved surface area,
(ii) Outer curved surface area,
(iii) Total surface area.
Height (h) of cylindrical pipe = Length of cylindrical pipe = 77 cm
(i) CSA of inner surface of pipe
(ii) CSA of outer surface of pipe
(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
Question 4:
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2
It can be observed that a roller is cylindrical.
Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end of roller =
CSA of roller = 2πrh
Area of field = 500 × CSA of roller
= (500 × 31680) cm2
= 15840000 cm2
= 1584 m2
Question 5:
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m2
Height (h) cylindrical pillar = 3.5 m
Radius (r) of the circular end of pillar =
= 0.25 m
CSA of pillar = 2πrh
Cost of painting 1 marea = Rs 12.50
Cost of painting 5.5 m2 area = Rs (5.5 × 12.50)
= Rs 68.75
Therefore, the cost of painting the CSA of the pillar is Rs 68.75.
Question 6:
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Let the height of the circular cylinder be h.
Radius (r) of the base of cylinder = 0.7 m
CSA of cylinder = 4.4 m2
rh = 4.4 m2
h = 1 m
Therefore, the height of the cylinder is 1 m
Question 7:
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) Its inner curved surface area,
(ii) The cost of plastering this curved surface at the rate of Rs 40 per m2
Inner radius (r) of circular well
Depth (h) of circular well = 10 m
Inner curved surface area = 2πrh
= (44 × 0.25 × 10) m2
= 110 m2
Therefore, the inner curved surface area of the circular well is 110 m2.
Cost of plastering 1 m2 area = Rs 40
Cost of plastering 110 m2 area = Rs (110 × 40)
= Rs 4400
Therefore, the cost of plastering the CSA of this well is Rs 4400.
Question 8:
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe = = 2.5 cm = 0.025 m
CSA of cylindrical pipe = 2πrh
= 4.4 m2
The area of the radiating surface of the system is 4.4 m2
Question 9:
Find
(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) How much steel was actually used, if of the steel actually used was wasted in making the tank.
Height (h) of cylindrical tank = 4.5 m
Radius (r) of the circular end of cylindrical tank =
(i) Lateral or curved surface area of tank = 2πrh
= (44 × 0.3 × 4.5) m2
= 59.4 m2
Therefore, CSA of tank is 59.4 m2.
(ii) Total surface area of tank = 2π(r + h)
= (44 × 0.3 × 6.6) m2
= 87.12 m2
Let A m2 steel sheet be actually used in making the tank.
Therefore, 95.04 m2 steel was used in actual while making such a tank.
Question 10:
In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.
Height (h) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm
Radius (r) of the circular end of the frame of lampshade =
Cloth required for covering the lampshade = rh
2200 cm2
Hence, for covering the lampshade, 2200 cm2 cloth will be required.
Question 11:
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Radius (r) of the circular end of cylindrical penholder = 3 cm
Height (h) of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder
= 2πrh + πr2
Area of cardboard sheet used by 1 competitor
Area of cardboard sheet used by 35 competitors
= = 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be bought. |
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# What is the sum of interior angles of a polygon of 10 sides?
## What is the sum of interior angles of a polygon of 10 sides?
1,440
The sum of the measures of the interior angles of a decagon (10 sided polygon) is 1,440. We found this by using the formula (n-2)(180).
## What is the interior angle of a regular 10 Gon?
144°
Decagon/Internal angle
What is the sum of all outer angles of a regular polygon with 10 sides?
Answer: The sum of the exterior angles of a decagon is 360°. Let’s understand the solution in detail. Explanation: A 10-sided polygon is called a decagon.
What is the sum of the interior angles of a polygon if it has 15 sides?
2340 degrees
Each triangle has an angle sum of 180 degrees, so the sum of the interior angles of the 15-gon must be 13 × 180 = 2340 degrees.
### What is the interior angle sum of a polygon with 11 sides?
1620 degrees
So, the sum of the interior angles of an 11-gon is 1620 degrees.
### What will be the sum of interior angles of a polygon of 12 sides?
1800°
Dodecagon is a 12-sided polygon with 12 angles and 12 vertices. The sum of the interior angles of a dodecagon is 1800°.
How do you calculate interior angles of a polygon?
Method 1 of 2: Calculating Interior Angles in a Polygon Count the number of sides in the polygon. In order to calculate the interior angles of a polygon, you need to first determine how many sides the polygon has. Find the total measure of all of the interior angles in the polygon. Divide the total measure of all of a regular polygon’s angles by the number of its angles.
What are the exterior angles of a polygon equal to?
In a polygon, an exterior angle is formed by a side and an extension of an adjacent side. Exterior angles of a polygon have several unique properties. The sum of exterior angles in a polygon is always equal to 360 degrees.
#### What is the sum for exterior angles for any polygon?
Exterior angles of a polygon are formed when by one of its side and extending the other side. The sum of all the exterior angles in a polygon is equal to 360 degrees. You are already aware of the term polygon. A polygon is a flat figure that is made up of three or more line segments and is enclosed.
#### Are the interior angles in a polygon equal in measure?
Interior angles of a regular polygon are equal in measure . The magnitude of an angle can be determined by the number of sides of the polygon. When the measure of one interior angle of a regular polygon is determined, it can be multiplied by the number of sides of the polygon to find the sum of the interior angles of the polygon. |
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# Important Simple Interest Questions for SSC CHSL Tier – 1 2018
Dear students, you know that QUANT is a part of getting points and every chapter is important. Therefore, we are providing 15 questions of quant. Solve all these quizzes every day so that you can improve your accuracy and speed. We also provide lots of quant questions. So you can practice that chapter which takes more time to solve the questions.
प्रिय पाठकों, आप सभी जानते हैं कि संख्याताम्क अभियोग्यता का भाग बहुत ही महत्वपूर्ण है. इसलिए हम आपको संख्यात्मक अभियोग्यता कि 15 प्रश्नों कि प्रश्नोत्तरी प्रदान कर रहे हैं. इन सभी प्रश्नोत्तरी को दैनिक रूप से हल कीजिये ताकि आप अपनी गति और सटीकता में वृद्धि कर सकें. हम आपको अन्य कई संख्यात्मक अभियोग्यता के प्रश्न प्रदान करेंगे. ताकि आप पाठ्यक्रम अनुसार उन्हें हल कर पायें.
Q1. A money lender claims to lend money at the rate of 10% per annum simple interest. However, he takes the interest in advance when he lends a sum for one year. At what interest rate does he lend the money actually?
(a) 10%
(b) 101/9%
(c) 11%
(d) 111/9%
Ans.(d)
Sol. Interest Rate = 10%
Let P → 100
Rate ⇒ 10
Actual Principal = 100 – 10 = 90
Rate = 10/90 × 100 = 11(1/9)%
Q2. A certain sum doubles in 7 years at simple interest. The same sum under the same interest rate will become 4 times in how many years?
(a) 14
(b) 28
(c) 21
(d) 10
Ans.(c)
Sol.
(n-1)/t1 =(m-1)/t₂
1/7=3/t₂
t₂ = 21 years
Q3. On a certain sum the simple interest for 121/2 year is 3/4 of the sum. Then the rate of interest is?
(a) 5% per year
(b) 6% per year
(c) 7% per year
(d) 8% per year
Ans.(b)
Sol. Let Principal be x
3/4 x=(x×25×r)/(2×100)
r = 6%
Q4. A man borrows some amount at the rate of 12% per annum at simple interest. After 6 years 8 months, he paid Rs. 720 as an interest. Find the amount borrowed by him?
(a) Rs. 900
(b) Rs. 960
(c) Rs. 920
(d) Rs. 1620
Ans.(a)
Sol. Time = 6 years 8 months
=6(8/12)=20/3 years
720=(P × 20 × 12)/(3 × 100)
P = 36 × 25
= 900 Rs.
Q5. The discount on a certain sum of money, due at the end of 21/4 years at 22/3% p.a. is Rs. 78. Find the sum?
(a) Rs. 1,278
(b) Rs. 1,300
(c) Rs. 1,378
(d) Rs. 1,400
Ans.(b)
Sol. Total % discount for 9/4 years
=9/4×8/3%
= 6%
78 = P × 6/100
P = Rs. 1300
Q6. If the simple interest on Rs. 1 for 1 month is 1 paisa, then the rate percent per annum will be?
(a) 10%
(b) 8%
(c) 12%
(d) 6%
Ans.(c)
Sol. t = 1 months = 1/12 years
S.I. = 1 paisa
= 1/100 paisa
1/100=(1 × 1 × R)/(12 × 100)
R = 12%
Q7. A money lender lends Rs. 400 for 3 years to a person and lends Rs. 500 for 4 years to the other person at the same rate of simple interest. If altogether he receives Rs. 160 as interest, what is the rate of interest per annum?
(a) 5%
(b) 7%
(c) 9%
(d) 10%
Ans.(a)
Sol. ATQ,
(400 × 3 × r)/100+(500 × 4 × r)/100=160
r (12 + 20) = 160
32r = 160
r = 5%
Q8. The simple interest on a certain sum of money at the rate of 5% per annum for 8 years is Rs. 840. Rate of interest for which the same amount of interest can be received on the same sum after 5 years is?
(a) 7%
(b) 8%
(c) 9%
(d) 10%
Ans.(b)
Sol. ATQ,
840=(P × 40)/100
P = Rs. 2100
840=(2100 × R × 5)/100
R = 8%
Q9. If a sum of money doubles itself in 8 years, then the interest rate in percentage is?
(a) 81/2%
(b) 10%
(c) 101/2%
(d) 121/2%
Ans.(d)
Sol. S.I = 2P – P = P
P=(P × 8 × r)/100
r=100/8=25/2
=12(1/2)%
Q10. Alipta got some amount of money from her father. In how many years will the ratio of the money and the interest obtained from it be 10:3 at 6% simple interest per annum?
(a) 7 years
(b) 3 years
(c) 5 years
(d) 4 years
Ans.(c)
Sol. P : SI = 10 : 3
Let, P = 10x
S.I. = 3x
3x=(10x × 6 × t)/100
t = 5 years
Q11. For what sum will the simple interest at R% for 2 years will be R?
(a) 100/2R
(b) 50
(c) 100/R
(d) 200/R
Ans.(b)
Sol.
R=(P × 2 × R)/100
P = 50 Rs.
Q12. The sum of money that will yield Rs. 60 as simple interest at 6% per annum in years for 5 years is
(a) 200
(b) 225
(c) 175
(d) 300
Ans.(a)
Sol.
60=(P × 6 × 5)/100
P = 200 Rs.
Q13. The rate of simple interest per annum at which a sum of money double itself in 162/3 years is
(a) 4%
(b) 5%
(c) 6%
(d) 62/3%
Ans.(c)
Sol. ATQ
P=P ×50×r/(3 × 100)
r = 6%
Q14. A sum of 3000 yields an interest of 1080 at 12% per annum simple interest in how many years?
(a) 4 years
(b) 3 years
(c) 5 years
(d) 21/2 years
Ans.(b)
Sol.
1080=(3000 × 12 × t)/100
t = 3 years
Q15. In simple interest rate per annum a certain sum amounts to Rs. 5,182 in 2 years and Rs. 5,832 in 3 years. The principal in rupees is?
(a) Rs. 2882
(b) Rs. 5000
(c) Rs. 3882
(d) Rs. 4000
Ans.(c)
Sol. S.I for 1 year = 5832 – 5182 = 650 Rs.
S.I for 2 years = 1300 Rs.
P = 5182 – 1300
= 3882 Rs.
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# Useful Formulas
What is a problem? A problem = a fact + a judgment. That is a simple formula that tells us something about the way the world works. Maths is full of formulas, and that can intimidate some people if they don’t understand them or can’t remember the right one to use.
However, formulas should be our friends, as they help us to do sometimes complex calculations accurately and repeatably in a consistent and straightforward way. The following is a list of the most useful ones I’ve come across while teaching Maths to a variety of students at a variety of ages and at a variety of stages in their education.
## Averages
• The mean is found by adding up all the values and dividing the total by how many there are, eg the mean of the numbers 1-10 is 5.5, as 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55, and 55 ÷ 10 = 5.5.
• The mode is the most common value (or values), eg the mode of 1, 2, 2, 3, 4, 5 is 2.
• The median of an odd number of values sorted by size is the one in the middle, eg the median of the numbers 1-5 is 3. The median of an even number of values is the mean of the two numbers in the middle, eg the median of the numbers 1-10 is 5.5, as 5 and 6 are the numbers in the middle, and 11 ÷ 2 = 5.5.
• The range is the highest value minus the lowest, eg the range of the numbers 1-10 = 10 – 1 = 9.
## Geometry
• Angles around a point add up to 360º
• Angles on a straight line add up to 180º
• Opposite angles are equal, ie the two pairs of angles opposite each other when two straight lines bisect (or cross) each other
• Alternate angles are equal, ie the angles under the arms of a ‘Z’ formed by a line (or ‘transversal’) bisecting two parallel lines
• Corresponding angles are equal, ie the angles under the arms of an ‘F’ formed by a line (or ‘transversal’) bisecting two parallel lines
• Complementary angles add up to 90º
• Any straight line can be drawn using y = mx + c, where m is the gradient and c is the point where the line crosses the y-axis (the ‘y-intercept’)
• The gradient of a straight line is shown by δy/δx (ie the difference in the y-values divided by the difference in the x-values of any two points on the line, usually found by drawing a triangle underneath it)
## Polygons
• Number of diagonals in a polygon = (n-3)(n÷2) where n is the number of sides
• The sum of the internal angles of a polygon = (n-2)180º, where n is the number of sides
• Any internal angle of a regular polygon = (n-2)180º ÷ n, where n is the number of sides
## Rectangles
• Perimeter of a rectangle = 2(l + w), where l = length and w = width
Note that this is the same formula for the perimeter of an L-shape, too.
• Area of a rectangle = lw, where l = length and w = width
## Trapeziums
• Area of a trapezium = lw, where l = average length and w = width (in other words, you have to add both lengths together and divide by two in order to find the average length)
## Triangles (Trigonometry)
• Area of a triangle = ½bh, where b = base and h = height
• Angles in a triangle add up to 180º
• Pythagoras’s Theorem: in a right-angled triangle, a² + b² = c², ie the area of a square on the hypotenuse (or longest side) is equal to the sum of the areas of squares on the other two sides
Circles
• Circumference of a circle = 2πr, where r = radius
• Area of a circle = πr², where r = radius
• π = 3.14 to two decimal places and is sometimes given as 22/7
## Spheres
• Volume of a sphere = 4/3πr³, where r = radius
• Surface area of a sphere = 4πr², where r = radius
## Cuboids
• Volume of a cuboid = lwh, where l is length, w is width and h is height
• Surface area of a cuboid = 2(lw + lh + wh), where l is length, w is width and h is height
## Number Sequences
• An arithmetic sequence (with regular intervals) = xn ± k, where x is the interval (or difference) between the values, n is the value’s place in the sequence and k is a constant that is added or subtracted to make sure the sequence starts at the right number, eg the formula for 5, 8, 11, 14…etc is 3n + 2
• The sum of n consecutive numbers is n(n + 1)/2, eg the sum of the numbers 1-10 is 10(10 + 1)/2 = 110/2 = 55
## Other
• Speed = distance ÷ time
• Profit = sales – cost of goods sold
• Profit margin = profit ÷ sales
• Mark-up = profit ÷ cost of goods sold
# Shortcuts
There is always more than one way of solving a Maths problem. That can be confusing, but it can also be an opportunity – if only you can find the right trade-off between speed and accuracy.
I’ve taught a lot of QTS numeracy candidates recently, and the Maths itself isn’t particularly difficult, particularly in the mental arithmetic section.
The trick is to be familiar with all the possible short cuts and capable of using the right one at the right time. It may mean having to do more sums, but it will be much simpler and quicker in the long run. You don’t have to use all of these all the time, but it is useful to know what they are just in case you need them.
• Multiplying and dividing by 5
The most useful short cut I’ve come across is very simple. To multiply by 5, try multiplying by 10 and then dividing by 2 (or vice versa), eg
13 x 5
= 13 x 10 ÷ 2
= 130 ÷ 2
= 65
You have to do two sums rather than one, but the point is that you should be able to save time and improve the chances of getting the answer right by doing both in your head rather than having to work out a more difficult sum on paper.
You can do divide by 5 in a similar way by multiplying by 2 and dividing by 10 (or vice versa), eg
65 ÷ 5
= 65 x 2 ÷ 10
= 130 ÷ 10
= 13
You can do a similar trick with 50, 500 etc simply by multiplying or dividing by a higher power of ten.
• Chunking
If you have to multiply by a two-digit number outside your times tables, chunking is an easy way to do the sum in your head. Instead of writing it down on paper and using long multiplication (which would take a long time and is easy to get wrong!), try multiplying by the tens and the units separately and adding up the results, eg 16 x 15 = 10 x 15 + 6 x 15 = 150 + 90 = 240. The numbers might still be too tricky to do it comfortably, but it’s often worth a try.
• Rounding
To avoid sums with ‘tricky’ numbers, try rounding them up to the nearest ‘easy’ figure and adjusting at the end. This is particularly useful when working out start and end times, eg
‘The morning session in a school began at 09:25. There were three lessons of 50
minutes each and one break of 20 minutes. At what time did the morning session end? Give your answer using the 24-hour clock.’
If you assume the lessons last an hour, you can add three hours to 09:25 to get 12:25. You would normally then knock off 3 x 10 = 30 minutes, but the 20-minute break means you only need to subtract 10 minutes, which means the session ended at 12:15.
• Money problems
There is often a ‘real world’ money problem in the QTS numeracy test. That usually means multiplying three numbers together. The first thing to say is that it doesn’t matter in which order you do it – 1 x 2 x 3 is just the same as 3 x 2 x 1. The next thing to bear in mind is that you will usually have to convert from pence to pounds. You could do this at the end by simply dividing the answer by 100, but a better way is to divide one of the numbers by 100 (or two of the numbers by 10) at the beginning or turn multiplication by a fraction of a pound into a division sum, eg
‘All 30 pupils in a class took part in a sponsored spell to raise money for charity. The pupils were expected to get an average of 18 spellings correct each. The average amount of sponsorship was 20 pence for each correct spelling. How many pounds would the class expect to raise for charity?’
The basic sum is 30 x 18 x 20p, and there are a couple of ways you could do this:
1) Knock off the zeroes in two of the numbers, change the order of the numbers to make it easier and double and halve the last pair to give yourself a sum in your times tables, ie
30 x 18 x 20p
= 3 x 18 x 2
= 3 x 2 x 18
= 6 x 18
= 12 x 9
= £108
2) Convert pence into pounds, turn it into a fraction, change the order of the numbers, divide by the denominator and, again, double and halve the last pair to give yourself a sum in your times tables, ie
30 x 18 x 20p
= 30 x 18 x £0.20
= 30 x 18 x ⅕
= 30 x 18 ÷ 5
= 30 ÷ 5 x 18
= 6 x 18
= 12 x 9
= £108
• Percentages
Many students get intimidated by percentages, fractions and decimals, but they are all just different ways of showing what share you have of something. You will often by asked to add or subtract a certain percentage. The percentage will usually end in zero (eg 20%, 30% or 40%), so the easiest way is probably to find 10% first. That just means dividing by 10, which means moving the decimal point one place to the left or, if you can, knocking off a zero. Once you know what 10% is, you can simply multiply by 2, 3 or 4 etc and add or subtract that number to find the answer, eg
‘As part of the numeracy work in a lesson, pupils were asked to stretch a spring to extend its length by 40 per cent. The original length of the spring was 45 centimetres. What should be the length of the extended spring? Give your answer in centimetres.’
You need to find 40% of 45cm, so you can start by finding 10%, which is 45 ÷ 10 or 4.5cm. You can then multiply it by 4 to find 40%, which is best done by doubling twice, ie 4.5 x 2 x 2 = 9 x 2 = 18. Finally, you just add 18cm to the original length of the spring to find the answer, which is 45 + 18 = 63cm.
• Common fractions
An awful lot of questions involve converting between fractions, percentages and decimals. There is a proper technique for doing any of those, but it’s very useful if you learn the most common fractions and their decimal and percentage equivalents by heart, eg
½ = 0.5 = 50%
¼ = 0.25 = 25%
¾ = 0.75 = 75%
⅕ = 0.2 = 20%
⅖ = 0.4 = 40%
⅗ = 0.6 = 60%
⅘ = 0.8 = 80%
⅛ = 0.125 = 12.5%
⅜ = 0.375 = 37.5%
⅝ = 0.625 = 62.5%
⅞ = 0.875 = 87.5%
• Times tables
3 x 24
= (3 x 2) x (24 ÷ 2)
= 6 x 12
= 72
Alternatively, you can halve just one of the numbers and double the result, eg
24 x 9
= 12 x 9 x 2
= 108 x 2
= 216
• Multiplying by 4
If you have to multiply by 4 and the number is not in your times tables, a simple way to do it is to double it twice, eg
26 x 4
= 26 x 2 x 2
= 52 x 2
= 104
• Multiplying by a multiple of 10
If you have to multiply by a multiple of 10 such as 20 or 30, try knocking the zero off and adding it in again afterwards. That way, you don’t have to do any long multiplication and, with any luck, the sum will be in your times tables, eg
12 x 30
= 12 x 3 x 10
= 36 x 10
= 360
• Multiplying decimals
This can be a bit confusing, so the best way of doing it is probably to ignore any decimal points, multiply the numbers together and then add back the decimal point to the answer so that you end up with the same number of decimal places as you had in the beginning, eg
0.5 x 0.5
= 5 x 5 ÷ 100
= 25 ÷ 100
= 0.25
• Using the online calculator
The second section of the QTS numeracy test consists of on-screen questions that can be answered using an online calculator. This obviously makes working out the answer a lot easier, and short cuts are therefore less useful. However, just because the calculator’s there doesn’t mean you have to use it, particularly for multiple-choice questions. If you have to add up a column of cash values, for example, and compare it with a number of options, you could simply tot up the number of pence and pick the option with the right amount. Alternatively, the level of accuracy needed in the answer may give you a helping hand if it rules out all but one of the possible answers, eg 6 ÷ 21 to one decimal place is always going to be 0.3. Why? Well, it’s a bit less than 7 ÷ 21, which would be a third or 0.3 recurring. An answer of 0.4 would be more than that, and 0.2 would be a fifth, which is far too small, so it must be 0.3.
• Don’t do more than you have to!
There are several types of question that could tempt you into doing more work than you need to do. If you’re trying to work out how many tables you need at a wedding reception for a given number of guests, the answer is always going to need rounding up to the next whole number, so you don’t need to spend any time working out the exact answer to one or two decimal places. Equally, some numbers are so close to being an ‘easy’ number that you don’t need to add or subtract anything after rounding up or down to make the basic sum easier, eg
‘For a science experiment a teacher needed 95 cubic centimetres of vinegar for each pupil. There were 20 pupils in the class. Vinegar comes in 1000 cubic centimetre
bottles. How many bottles of vinegar were needed?
If you round 95cc to 100cc, the answer is 20 x 100 ÷ 1000 or 2 bottles, and the remainder consisting of 20 lots of 5cc of vinegar can safely be ignored. |
# Find The Vertex Of A Quadratic Equation Easily
A quadratic equation is any equation in the form of ax2+bx2 +c. Quadratic equations are most commonly found in the context of quadratic function
sfunctions such as ƒ(x) = x2x + 1 or ƒ(x) = 6x24x + 9. In more precise mathematical terms, a quadratic is any polynomial expression that has a degree of 2.
In the above quadratic equation, the three coefficients a, b, c are called:
• a is the leading term
• b is the linear term
• c is the constant
The graph of a quadratic function is called a parabola.
Credit: graphfree
For example, the above graph is the graph of the simple quadratic function ƒ(x) = x2.
The exact shape and orientation of the graph are determined by the values of the coefficients of the quadratic function a, b, and c.
When |a| > 1, such as 3 or 4, the graph gets “skinnier.” This is because the graph is growing at 3 times or 4 times the rate. Conversely, if |a|<1 such as 0.5 or 0.25, then the graph gets “fatter” because it is growing at half or one-fourth the rate.
The value of a also determines which way the parabola is facing. If a is positive, then the parabola faces up and opens at the top. If a is negative, then the parabola faces down and opens downward.
Credit: graphfree
The above picture is a graph of the function ƒ(x) = –x2. Because the leading term is negative (a=-1) the graph faces down. One way to remember this relationship between a and the shape of the graph is If a is positive, then the graph is also positive and makes a smiley (“positive”) face. If a is negative, then the graph makes a frowny (“negative”) face.
The b term (linear term) determines, roughly, the amount of horizontal offset of the graph. Changing the linear term moves the graph over left and right. It also slightly changes the vertical offset of the graph, though not as much as the c term. Changing the constant term determines the amount of vertical offset of the graph. When is positive, the graph is shifted up, when c is negative, the graph is shifted down.
Here is a nice link to a GeoGebra tool that lets you play around with the different coefficient values and see how changing them around changes the appearance of the graph. Take some time messing around with this app to get an intuitive feel for how quadratic functions operate. Notice that when a = 0, the graph takes on the form of a straight line. When a = 0, the quadratic can be written as a linear equation.
Parabolas are interesting because they pop up all over nature and have a lot of engineering applications. For example, Galileo discovered in the 17th century that the motion of a projectile through the air always takes the shape of a parabola and parabola-shaped curves pop in models relating to electromagnetism, population growth, and engineering.
Notice that every parabola has a point where it changes direction. This is called the vertex of the parabola and is the minimum point on a positive parabola and the maximum point on a negative parabola. The vertex is the inflection point of the graph where it starts to change direction from the negative direction to the positive or vice versa.
The different parts of a parabola. The vertex is located at the inflection point where the graph changes direction. Credit: Melikamp via WikiCommons CC-BY SA 3,0
How do you figure out where the vertex is? More specifically, how do you find the x- and y- coordinate for the vertex of any given parabola? We will look at two different methods, one involving a different form of quadratic equations and another that uses a bit of calculus to compute the vertex.
## How To Find The Vertex Of A Parabola
### Method 1. Use Vertex Form
One way to understand the vertex is to see the quadratic function expressed in vertex form. The vertex form of a quadratic function can be expressed as:
Vertex Form: ƒ(x) = a(xh)2 + k
Where the point (h, k) is the vertex. This equation makes sense if you think about it. In the case of an upright parabola, the leftmost term will always be positive so the lowest it can possibly be is 0. Therefore, the smallest possible y-value is y=k which happens when the quantity x= 0, or in other words, when xh. The same reasoning holds for downward parabolas. y=k is the highest possible y value, which obtains when h = x.
(Note: the a in the vertex form is the same a in the standard form ax2 + bx2 + c)
So the vertex form of a quadratic equation lets us “read off” the location of the vertex just by looking at the formula. But quadratics are normally not written in vertex form, so we need a way to convert between the standard presentation and vertex form. We can do this by completing the square. We will take a look at the quadratic ƒ(x) = 2x24x + 5 as a specific example:
Complete the square
Standard form y = 2x2−4x + 5 Isolate the x2 and x terms y−5 = 2x2−4x Factor out the leading coefficient y−5 = 2(x2−2x) Create the perfect square trinomial (Note: since we factored out a 2 previously from the right and have a y on the left, whatever we add on the right has to be multiplied by 2 on the left) y−5+2 = 2(x2−2x+1) Simplify and convert the right side to a difference of squares y−3 = 2(x−1)2 Isolate the y term y = 2(x−1)2 + 3 Read off the vertex point Vertex = (h, k) = (1, 3)
There is a quick and sneaky way to quickly find the right h and k values without completing the square. When given the standard form of a quadratic (ax2 + bx2 + c) you can find the h and k values as:
h = (-b/2a) and k = ƒ(h)
Just compute the h value and plug it into the function to get the k value.
Converting from vertex form back to standard form is easy. Just multiply out the squared part and simplify the entire expression.
### Method 2: Derivatives
Another way of finding the vertex is by using the tools of calculus and derivatives.
Consider the behavior of a quadratic function as it approaches its vertex. For an upward parabola, when coming from the left, the graph initially decreases rapidly. As it approaches the vertex, this rate of decrease gets slower and slower until it reaches the vertex, after which it changes directions and begins to increase. As you go further right, the graph grows quicker and shoots off.
Imagine a line that lies tangent to the parabola. When you start on the left, the tangent line is very steep. As you move to the right, the line becomes less and less steep, until it hits the vertex and becomes a flat line. As you keep moving right, the line gets steep again the further you go.
The tangent line of the vertex of a parabola. Credit: graphfree
When the tangent line hits the vertex, it is a flat line. The reason why it is flat is because at the vertex of the parabola the rate of change of the quadratic function is 0. In calculus terms, the vertex of a parabola is located at the point where its derivative is equal to 0. Drawing a line tangent to the vertex will always result in a straight line, which is an indication that the derivative of the function is 0 at that point.
So one way to find the vertex of a parabola is to find the derivate, compute the x value where the derivative is 0, and plug that back into the quadratic to find the y value of the vertex. Let’s see this in action with the function ƒ(x) = x22x + 2.
Step 1: Find the derivative of the quadratic
ƒ(x) = x2+2x 2
ƒ'(x) = 2x+2
Step 2: Solve for x when ƒ'(x) = 0
ƒ'(x) = 2x+2
0 = 2x+2
-1 = x
So -1 is the x-coordinate of the vertex.
Step 3: Plug the result from step 2 into the original quadratic
ƒ(x) = x2+2x2
ƒ(-1) = (-1)2−2(-1) 2
ƒ(-1) = -3
So the y-coordinate of the vertex is -3. Together, we get the vertex of the parabola at the point (-1, 3).
In order to tell if the vertex is a minimum or maximum point of the function, take a look at the leading term of the quadratic:
• If a > 0 then the vertex (h, k) is a minimum
• If a < 0 then the vertex (h, k) is a maximum
To sum up the major points:
• A parabola is the shape of a graph made by a quadratic function ax2 + bx2 + c
• The inflection point where the graph changes direction is called the vertex of the parabola.
• The vertex form of a quadratic is in the form ƒ(x) = a(xh)2 + k where point (h, k) is the vertex
• The vertex is the minimum of an upward parabola and the negative of a downward parabola
• The vertex of a parabola can be found by two main methods:
• Completing the square
• Differentiation
## About Alex Bolano PRO INVESTOR
When Alex isn't nerdily stalking the internet for science news, he enjoys tabletop RPGs and making really obscure TV references. Alex has a Masters's degree from the University of Missouri-St. Louis. |
# How to Complete the Square
Completing the square is an important skill in maths and you may find yourself frequently having to solve problems that involve completing the square, particularly in algebra. This guide will show you how to complete the square!
## What is Completing the Square?
Completing the square is an algebraic process that allows you to convert a quadratic function in standard form to vertex form. Completing the square involves adding a value to and subtracting a value from a quadratic polynomial so that it contains a perfect square trinomial. You can then rewrite this trinomial as the square of a binomial.
## Definition of Polynomial and Perfect Square trinomial
So lets firstly clear up some terms.
What is a polynomial?
A polynomial is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
For example:
5xy + 2x - 7
5xy is the 1st term, 2x is the 2nd term and -7 is the third term.
A polynomial can have constants ( like the numbers 5 and -11), variables (like x and y), exponents that are positive whole numbers, (like the 2 in x²). But a polynomial cannot have division by a variable, so an expression with 2/(x+2) is not a polynomial.
A trinomial by our definition of a polynomial therefore is an algebraic expression consisting of three terms.
A perfect square trinomial looks like this:
x² + 14x+49
The above trinomial is the algebraic expression we get when we expand
(x+7)² or (x+7)(x+7).
So to make a perfect square trinomial, we square the first term ( of the binomial) (x+7)².
Then we twice the product of the binomials first and last terms. Finally we square the last term of the binomial.
so we get:
x² + 14x+49
## How to Complete the Square - Step By Step
So back to how to complete the square. Let us complete the square of:
y= x² + 6x +5 (this function is in standard form.)
For the function y= x² + 6x +5 to complete the square:
Group the First two terms.
y= (x² + 6x)
Inside the brackets, add and subtract the square of half the coefficient of the second term.
y= (x² + 6x +9 -9)
Group the perfect square trinomial.
y= (x² + 6x +9) -9 + 5
Reunite the perfect square trinomial as the square of a binomial.
y= (x+3)² -9 + 5
Simplify.
y= (x+3)² -4 (This function is in vertex form.)
Remember:
y= x² +6x +5 and y= (x+3)² -4 represent the same quadratic function. You can use both forms of the function to determine what the graph will look like. However an added benefit of the vertex form is of course that you can identify the vertex of the parabola without graphing.
## How to Complete the Square with fractions
• y= 4x² -28x -23
• y= 4(x² -7x) -23
• y= 4[(x² -7x + (7/2)² -(7/2)²] -23
• y= 4(x² -7x + 49/4 -49/4) -23
• y= 4[(x² -7x + 49/4)-49/4] -23
• y= 4[(x-7/2)²-49/4] -23
• y= 4(x-7/2)²-4(49/4) -23
• y= 4(x-7/2)²-49 -23
• y= 4(x-7/2)²-72
Completing the square can be used with fractions (as shown above) and also decimals.
So that was my guide on completing the square. Be sure to review these steps as often as you need to sharpen up your skills on this algebraic process.
Thanks for joining me!
Roli Edema.
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# Surface Area of Cuboid
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Last updated date: 17th Sep 2024
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## Area of Cuboid
Imagine objects like a lunch box, television set, shoebox, carton box, bricks, book, mattresses and you would know what a cuboid is and how it looks. These shapes are cuboid. Like said, a cuboid is a 3-D geometrical object which consists of 6 rectangular faces. All angles of a cuboid are right angles and faces opposite to each other are equal. A cuboid is also known as a rectangular solid or a rectangular prism. In a cuboid, the length, width and height may be of different measurements.
### Is Cube a Cuboid?
Objects such as Rubik’s cube, ice, dice, Sudoku, sugar cubes, casseroles and milk crates etc are examples of another 3 dimensional shape called a cube. Factually, a cube is a unique form of cuboid in which all sides are similar and squares.
### Best Way to Identify a Cuboid
In a cuboid, each face is in the form of a rectangular shape and the corners or the vertices are 90-degree angles. Also, if the opposite faces are always equal to one another then it’s a cuboid. For example, a mattress is a cuboid. It consists of 6 surfaces of which each opposite pair is of similar dimensions.
### What is the Volume of Cuboids?
We can simply find the volume of a cuboid by multiplying the base area with the height. Thus,
volume of cuboid (V) = A x h
or simply
V = l × b × h
### Total Surface Area of Cuboid
If l is the length, b is the breadth and h is the height of a given cuboid, then the sum of areas of 6 rectangles of a cuboid provides the TSA of the cuboid.
### Total Surface Area of Cuboid Formula
TSA of cuboid formula = 2 (lw + wh + hl)
Where,
L = length
H = height
### Lateral Surface Area Of Cuboid
The sum of the area of 4 side faces i.e. leaving the top and the bottom face provides the LSA of a cuboid. An example of the LSA is the sum of the area of the four walls of a room.
### Lateral Surface Area of Cuboid Formula
LSA of cuboid formula = 2 (lh + wh) = 2 h (l + w)
Or simply, 2 (l+w)h
Where,
L = length
H = height
## Solved Examples on Surface Area of Cuboid
### Example 1:
The length, width and height of a cuboid are 11cm, 9cm and 15cm respectively. Calculate the total surface area of the cuboid.
### Solution:
TSA of a cuboid is given by: 2 (l*w + w*h + w*l)
Given that:
l = 11cm
w = 9cm
h = 15cm
By substituting the values in the expression we will obtain,
TSA = 2 (11*9 + 9*15 + 15*11)
TSA = 2(99 + 135 + 165)
TSA = 2 * 399
TSA = 798cm²
### Example 2:
Find out the lateral surface area of a cube having an edge of 20cm?
### Solution:
We know that the LSA of a cuboid is given by 2(l+b)h
Now, since a cube is also a cuboid in which l=b=h=a, thus LSA of a cube = 2(a+a)
Or simply,
a = 4a2
### Formula for Lateral Surface Area of Cube = 4a2
Given that a = 20 cm.
Therefore,
LSA = 4(202) = 1600 cm2
### Example 3:
Williams built a rectangular cardboard box 20 cm high. It has a square base and a volume of 2000 cm³. Then he realized that he did not require a box that elongated, so he cut short the height of the box decreasing its volume to 1,000 cm³. Find out the height of the new box and is the new box cubicle?
### Solution:
Volume of cuboid (V) = length × width × height = Base area × height.
Given that,
V = 2000 cm³,
height = 20 cm
Substituting the values in the formula, we obtain
Base area = 2000/20 = 100 cm²
We also know that the base of this box is a square.
Thus, it indicates that the length = width.
Hence, the length of square base = √100 = 10 cm
After shortening of height, new volume = 1000
= 10 ×10 x new height
Thus, new height = 1000/ 10 × 10 = 10 cm
As all the dimensions of the solid, l, w, h measure similar, the resulting solid is also a cube.
## FAQs on Surface Area of Cuboid
Q1. What is the Diagonal of a Cuboid?
Answer: The length of the longest diagonal of a cuboid is given as:
Length of diagonal of cuboid = √ (l² + b² + h²), where l= length, b= breadth, h= height
Q2. What are the Properties of a Cube Number?
• Cubes of positive numbers are positive invariably.
For example, cube of +3 is = (+3) × (+3) × (+3) = +27
• Cubes of positive numbers are invariably negative.
For example, cube of -3 is = (-3) × (-3) × (-3) = -27
• Cubes of even numbers even invariably.
• Cubes of odd numbers are odd invariably.
Q3. How Do We Measure the Volume of Water?
Answer: Generally, it is not feasible to measure the volume of water until it is stored in a container, which can be a cube, cuboid, cone, cylinder, and cone etc. And once it is inside a container we need to compute the volume of the container in order to ascertain the volume of water.
Q4. What is Meant by Nets of a Cuboid?
Answer: Another way to have a perception of the surface area of a cuboid is to take into account a net of the cuboid. The net is a 2-D geometrical shape that can be molded to create a 3-D object.
Imagine cutting along some edges of a cuboid and opening it up to create a plane figure. The plane figure is what we call the net of the cuboid. |
# Quantitative Aptitude Quiz for SBI PO/CLERK Mains: 3rd August 2018
Dear Students,
Quantitative Aptitude Quiz for SBI PO/Clerk Mains 2018
Numerical Ability or Quantitative Aptitude Section is getting complex and convoluted every year. The questions asked in this section are calculative and very time-consuming. One needs to fight tooth and nail to get a desirable score in this section. Once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions for SBI PO and SBI Clerk Mains Exam. Not only this, these quizzes will also prove propitious for the upcoming Bank of Baroda Exam. Now, pull up your socks, it’s time for Blood, sweat and tears. This quiz is according to the SBI PO/Clerk Study Mains Preparation Study Plan and with the help of this 25 Days Plan you’ll cover all important topics for Data Interpretation and Analysis section of Mains.
Q1. The simple interest on certain sum at 5% for 9 months is Rs. 10 greater than the simple interest on the same sum @ 3% for 14 months. What is the sum of interest in both the cases (i.e., total sum of interest)?
Rs. 130
Rs. 290
Rs. 120
Rs. 330
None of these
Solution:
Q2. The population of vultures in a particular locality is decreases by a certain rate of interest (compounded annually). If the current population of vultures be 29160 and the ratio of decease in population for second year and 3rd year be 10: 9. What was the population of vultures 3 years ago?
30000
35000
40000
50000
None of these
Solution:
Q3. A car runs at the speed of 40 km/h when not serviced and runs at 65 km/h, when serviced. After servicing the car covers a certain distance in 5 h. How much approximate time will the car take to cover the same distance when not serviced?
10
7
12
8
6
Solution:
Q4. A 180-metre long train crosses another 270-meter long train running in the opposite direction in 10.8 seconds. If the speed of the first train is 60 kmph., what is the speed of the second train in kmph?
80
90
150
Cannot be determined
None of these
Solution:
Q5. In a voyage of 600 km, a ship was slowed down due to bad weather and storm in Ocean. Its average speed for the trip was reduced by 200 km/hr, and the time of trip increased by 30 minutes. What would be the duration of the voyage?
1 hour
2 hour
1⅓ hour
1½ hour
1⅔ hour
Solution:
Directions (6-10): Dominos prepares Pizzas of three different types – Cheese, Onion and Chicken. The production of the three types over a period of six Months has been expressed in the bar-graph provided below. Study the graph and answer the questions based on it.
Q6. For which of the following Months the percentage of rise/fall in Order from the previous Month is the maximum for the Onion flavor?
February
March
April
May
June
Solution:
The percentage rise/fall in Order from the previous Month for Onion type during various Months are:
Q7. For which type was the average annual Order maximum in the given period?
Cheese only
Onion only
Chicken only
Cheese and Onion
Cheese and Chicken
Solution:
Q8. The total Order of Chicken type in March and April is what percentage of the total Order of Cheese type in January and February?
96.67%
102.25%
115.57%
120%
133.33%
Solution:
Q9. What is the difference between the average Order of Cheese type in January, February and March and the average Order of Onion type in April, May and June?
50,000 orders
80,000 orders
2,40,000 orders
3,30,000 orders
5,00,000 orders
Solution:
Q10. What was the approximate decline in the Order of Chicken type in June as compared to the Order in April?
50%
42%
33%
25%
22.5%
Solution:
Directions (11- 15): In the following questions, two equations numbered I and II are given. You have to solve both questions and give answer among the following options.
Q11.
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship cannot be established.
Solution:
Q12.
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship cannot be established.
Solution:
Q13.
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship cannot be established.
Solution:
Q14.
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship cannot be established.
Solution:
Q15.
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship cannot be established.
Solution: |
## What is a general solution of a differential equation?
The general solution to a differential equation is the most general form that the solution can take and doesn’t take any initial conditions into account. Example 5 y(t)=34+ct2 y ( t ) = 3 4 + c t 2 is the general solution to 2ty′+4y=3.
## How do you find the solution of a differential equation?
Here is a step-by-step method for solving them:Substitute y = uv, and. Factor the parts involving v.Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)Solve using separation of variables to find u.Substitute u back into the equation we got at step 2.
## Can Wolfram Alpha solve differential equations?
A differential equation is an equation involving a function and its derivatives. Wolfram|Alpha can solve many problems under this important branch of mathematics, including solving ODEs, finding an ODE a function satisfies and solving an ODE using a slew of numerical methods.
## What is general solution and particular solution of differential equation?
A differential equation is an equation involving a function and its derivative(s). general solution. A general solution to a linear ODE is a solution containing a number (the order of the ODE) of arbitrary variables corresponding to the constants of integration.
## What does General solution mean?
1 : a solution of an ordinary differential equation of order n that involves exactly n essential arbitrary constants. — called also complete solution, general integral. 2 : a solution of a partial differential equation that involves arbitrary functions.
## How do you solve a differential equation with two variables?
Step 1 Separate the variables by moving all the y terms to one side of the equation and all the x terms to the other side:Multiply both sides by dx:dy = (1/y) dx. Multiply both sides by y: y dy = dx.Put the integral sign in front:∫ y dy = ∫ dx. Integrate each side: (y2)/2 = x + C.Multiply both sides by 2: y2 = 2(x + C)
You might be interested: Expected value equation
## How do you find the general solution of a second order differential equation?
It is said in this case that there exists one repeated root k1 of order 2. The general solution of the differential equation has the form: y(x)=(C1x+C2)ek1x. y(x)=eαx[C1cos(βx)+C2sin(βx)].
## How do you find the roots on a calculator?
TI-85 / TI-86Press Graph.Press More and then Math (F1)Press Lower (F1). Arrow to the left of the x-intercept and press enter.Press Upper (F2). Arrow to the right of the x-intercept and press enter.Press Root (F3). The TI-85 will return a value for x and 0 (or really close to it) for y.
## How do you find the equilibrium solution?
An equilibrium solution is a solution to a DE whose derivative is zero everywhere. On a graph an equilibrium solution looks like a horizontal line. Given a slope field, you can find equilibrium solutions by finding everywhere a horizontal line fits into the slope field.
## How does Bernoulli equation solve differential?
When n = 0 the equation can be solved as a First Order Linear Differential Equation. When n = 1 the equation can be solved using Separation of Variables. and turning it into a linear differential equation (and then solve that).
## What is linear differential equation with example?
A linear equation or polynomial, with one or more terms, consisting of the derivatives of the dependent variable with respect to one or more independent variables is known as a linear differential equation. The solution of the linear differential equation produces the value of variable y. Examples: dy/dx + 2y = sin x.
### Releated
#### Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
#### Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […] |
Volume and Surface Area of 3-D Shapes
# Volume and Surface Area of 3-D Shapes
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## Volume and Surface Area of 3-D Shapes
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Volume and Surface Area of 3-D Shapes
2. VOLUME • Volume is the measure of space inside a 3-D object. • Imagine filling the classroom with boxes measuring 1 cubic foot. How much would we use? • Area of the base x height = volume of a prism or cylinder • (Area of the base x height) ÷ 3 = volume of a pyramid
3. SURFACE AREA • Surface area is the measure of all faces on a 3-D object. • Imagine covering the floor, walls, and ceiling of this room with tiles measuring 1 square foot. How many would we use? • Surface Area = sum of areas of all faces and bases
4. Mary wants to fill the box below with candy hearts and then wrap the box in Valentine’s paper. The box is 6 inches long, 6 inches wide, and 12 inches high. Each pack of candy is 3 cubic inches. How many packs of candy will she need? She has a piece of wrapping paper that is 500 square inches. Will that be enough? Will there be any wrapping paper left over? V = 6 x 6 x 12 = 432 cubic inches 432 ÷ 3 = 144 packs of candy SA = 2(6x6) + 2(6x12) + 2(6x12) = 360 square inches 500 – 360 = 140 square inches left
5. Bill is working on a farm for the summer. He is told by Farmer Brown to remove the top of the silo, fill it with grain, and then paint the outside of the silo to match the barn. The silo is 30 feet tall, and 6 feet across the top. One acre of harvested grain is 64 cubic feet. Bill estimates it will take 10 acres of harvested grain to fill the silo. Is his estimate accurate? Why/why not? One can of red paint will cover 100 square feet. How many cans of red paint will Bill need? V = ∏r2h V = 3.14 x 3 x 3 x 30 Volume = 847.8 cubic feet 64 x 10 = 640 847.8 – 640 = 207.8 cubic feet of space left over. Bill needs about 13 acres of harvested grain to fill the silo. SA = area of 2 circles + area of rectangle. ∏r2= 3.14 x 3 x 3 = 28.26 square feet The rectangle is wrapped around the circle, so its length is the circumference of the circle. C = ∏d = 3.14 x 6 = 18.84 A = lw= 18.84 x 30 = 565.2 square feet SA = 28.26 + 28.26 + 565.2 = 621.72 square feet Bill needs 7 cans of paint, because 6 cans will only cover 600 square feet.
6. A theme park is creating a new statue to honor the recently deceased Mr. Tour I. Angle. You are on the planning committee. The designs available are a triangular prism and a square pyramid. The statues will need to be painted and filled with concrete. The triangular prism would have a 12 foot-long base, stand 8 feet tall, and be 8 feet thick with a slant height of 10 feet. The square pyramid would have an equilateral base of 12 feet, and stand 8 feet tall with a slant height of 10 feet. Each face height would be 8 feet. A \$20 can of paint will cover 100 square feet. A cubic foot of concrete costs \$5. How much will each statue cost? Which is the best choice, if the park is trying to save money? Prism V = (½ bh)h = 48 x 8 =384 cubic feet 384 x\$5 = \$1920 Prism SA = area of 2 triangles + area of 3 rectangles = 48+48+96+80+80 = 352 square feet \$20 x 4 = \$80 Pyramid V = lwh ÷ 3 = 384 cubic feet 384 x\$5 = \$1920 Pyramid SA = area of 4 triangles + area of square base = 48+48+48+48+144 = 336 square feet \$20 x 4 = \$80
7. HOMEWORK ASSIGNMENT • Write your own word problem involving volume and surface area. • You may not use one of the previous examples. • If you bring a visual aid to go with your problem, you will receive extra credit. |
## Kiss those Math Headaches GOODBYE!
Attention: Dear Aunt Sally may not be fit for teaching students algebra!
A problem has been discovered in sweet Aunt Sally’s little memory trick: Please Excuse My Dear Aunt Sally.
Actually, make that two problems.
The first, revealed in my 9/9 post below, is that Aunt Sally wrongly makes students think they’re supposed to multiply before dividing. That’s because the word My (standing for MULTIPLY) comes before Dear (standing for DIVIDE).
Countless students have been deceived into thinking they’re supposed to multiply before dividing [See the 9/9 post for the full run-down on this problem.]
Today I want to point out another problem, and offer two solutions.
The second problem is that, since “Aunt” (standing for ADD) comes before “Sally” (standing for SUBTRACT), countless other students have been led to think they are supposed to ADD before they SUBTRACT.
Well, what are students supposed to do?
First of all, students need to realize that adding and subtracting are at exactly the same level of hierarchy as each other. But if that’s true, how can students ever decide which to do first.
Easy! Same solution as with multiplying and dividing. We simply look to see which of these operations is written first as we read the problem left to right.
Example: in the expression 8 + 3 – 4, the addition symbol precedes the subtraction symbol, so here we add before subtracting. And we simplify the expression like this:
8 + 3 – 4
= 11 – 4
= 7
But in the expression 8 – 3 + 4
the subtraction symbol is written before the addition symbol, so here we subtract before we add, and we simplify the expression like this:
8 – 3 + 4
= 5 + 4
= 9
It’s really that simple. Pay attention to which operation sign comes first as you read the problem from left to right. Then do the operations in the correct order based on that.
One other solution: in my book, the Algebra Survival Guide, I get away from the Please Excuse My Dear Aunt Sally approach, as I create my own memory trick, one that involves Strawberry Mousse. If you want to take a look at this approach, check out my book at this site.
From the homepage, click the link that says: View Sample Chapters of the Algebra Survival Guide, and download the chapter on Positive and Negative
Numbers.
Cover via Amazon
. |
Rd Sharma 2020 2021 Solutions for Class 9 Maths Chapter 14 Areas Of Parallelograms And Triangles are provided here with simple step-by-step explanations. These solutions for Areas Of Parallelograms And Triangles are extremely popular among Class 9 students for Maths Areas Of Parallelograms And Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 9 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.
#### Question 1:
If the given figure, ABCD is a parallelogram, AEDC and CFAD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.
Given: Here in the question it is given
(1) ABCD is a parallelogram,
(2) and
(3) , AB = 16 cm
(4) AE = 8cm
(5) CF = 10cm
Calculation: We know that formula for calculating the
Therefore,
Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )
Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)
Therefore,
Area of paralleogram ABCD = 16 × 8 ……(1)
Taking the base of Parallelogram ABCD as AD we get
Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)
Area of paralleogram ABCD = AD × 10 ……(2)
Since equation 1 and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.
Hence fro equation (1) and (2),
This means that,
Hence we get the result as
#### Question 2:
In Q.No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, find AB.
Given: Here in the question it is given that
(1) ABCD is a parallelogram,
(2) and
(3)
(5) AE = 8cm
(6) CF = 10cm
To Find : AB =?
Calculation: We know that formula for calculating the
Area of paralleogram = base × height
Therefore,
Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )
Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)
Therefore, Area of paralleogram ABCd = 16 × 8
Area of Parallelogram ABCD = AB× 8 ……(1)
Taking the base of Parallelogram ABCD as AD we get
Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)
Area of paralleogram ABCD = 6 × 10 ……(2)
Since equation 1and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.
Hence equation 1 is equal to equation 2
Which means that,
Hence we got the measure of AB equal to
#### Question 3:
Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.
Given: Here in the question it is given that
(1) Area of paralleogram ABCD = 124 cm2
(2) E is the midpoint of AB, which means
(3) F is the midpoint of CD, which means
To Find : Area of Parallelogram AEFD
Calculation: We know that formula for calculating the
Area of Parallelogram = base × height
Therefore,
Area of paralleogram ABCD = AB × AD (Taking base as AB and Height as AD ) ……(1)
Therefore,
Area of paralleogram AEFD = AE × AD (Taking base as AB and Height as AD ) ……(2)
()
= Area of Parallelogram ABCD (from equation1)
Hence we got the result Area of Parallelogram AEFD
#### Question 4:
If ABCD is a parallelogram, then prove that
ar (ΔABD) = ar (ΔBCD) = ar (ΔABC) = ar (ΔACD) = $\frac{1}{2}$ ar (||gm ABCD)
Given: Here in the question it is given that
(1) ABCD is a Parallelogram
To Prove :
(1)
(2)
(3)
(4)
Construction: Draw
Calculation: We know that formula for calculating the
Area of Parallelogram = base × height
Area of paralleogram ABCD = BC × AE (Taking base as BC and Height as AE ……(1)
We know that formula for calculating the
Area of ΔADC = Base × Height
= Area of Parallelogram ABCD (from equation1)
Hence we get the result
Similarly we can show that
(2)
(3)
(4)
#### Question 1:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
GIVEN: Here in the question figure 1 to 6 are shown.
To Find :
(1) The figures which lie on the same base and between the same parallels.
(2) Write the common base and parallels.
As we know that ‘Two geometric figures are said to be on the same base and between the same parallels, if they have a common side(base) and the vertices (or vertex) opposite to the common base of each figure lie on a line parallel to the base.’
(1) ΔAPB and trapezium ABCD are on the same base CD and between the same parallels AB and CD.
(2) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.
(3) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC,. but they are not on the same base AD
(4) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC,. but they are not on the same base AD .
(5) ΔQRT and Parallelogram PQRS are on the same base QR and between the same parallels PS and QR.
(6) Parallelogram PQRS, AQRD, and BCQR are between the same parallels. Also Parallelogram PQRS, BPSC and APSD are between the same parallels.
#### Question 1:
In the given figure, compute the area of quadrilateral ABCD.
Given: Here from the given figure we get
(1) ABCD is a quadrilateral with base AB,
(2) ΔABD is a right angled triangle
(3) ΔBCD is a right angled triangle with base BC right angled at B
To Find: Area of quadrilateral ABCD
Calculation:
In right triangle ΔBCD, by using Pythagoreans theorem
.So
In right triangle ABD
Hence we get Area of quadrilateral ABCD =
#### Question 2:
In the given figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of Δ OTS if PQ = 8 cm.
Given: Here from the given figure we get
(1) PQRS is a square,
(2) T is the midpoint of PS which means
(3) U is the midpoint of PS which means
(4) QU = 8 cm
To find: Area of ΔOTS
Calculation:
Since it is given that PQ = 8 cm. So
Since T and U are the mid points of PS and QR respectively. So
Therefore area of triangle OTS is equals to
Hence we get the result that Area of triangle OTS is
#### Question 3:
Compute the area of trapezium PQRS in the given figure.
Given:
(1) PQRS is a trapezium in which SR||PQ..
(2) PT = 5 cm.
(3) QT = 8 cm.
(4) RQ = 17 cm.
To Calculate: Area of trapezium PQRS.
Calculation:
In triangle
.So
No area of rectangle PTRS
Therefore area of trapezium PQRS is
#### Question 4:
In the given figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.
Given: In figure:
(1) ∠AOB = 90°
(2) AC = BC,
(3) OA = 012 cm,
(4) OC = 6.5 cm.
To find: Area of ΔAOB
Calculation:
It is given that AC = BC where C is the mid point of AB
We know that the mid point of hypotenuse of right triangle is equidistant from the vertices
Therefore
CA = BC = OC
CA = BC = 6.5
AB = 2 × 6.5 = 13 cm
Now inn triangle OAB use Pythagoras Theorem
So area of triangle OAB
Hence area of triangle is
#### Question 5:
In the given figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Given: Here from the given figure we get
(1) ABCD is a trapezium
(2) AB = 7 cm,
(3) AD = BC = 5 cm,
(4) DC = x cm
(5) Distance between AB and DC is 4 cm
To find:
(a) The value of x
(b) Area of trapezium
Construction: Draw AL CD, and BM CD
Calculation:
Since AL CD, and BM CD
Since distance between AB and CD is 4 cm. So
AL = BM = 4 cm, and LM = 7 cm
In triangle ADL use Pythagoras Theorem
Similarly in right triangle BMC use Pythagoras Theorem
Now
We know that,
We get the result as
Area of trapezium is
#### Question 6:
In the given figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2$\sqrt{5}$, find the area of the rectangle.
Given: Here from the given figure we get
(1) OCDE is a rectangle inscribed in a quadrant of a circle with radius 10cm,
(2) OE = 2√5cm
To find: Area of rectangle OCDE.
Calculation:
In right triangle ΔODE use Pythagoras Theorem
We know that,
Hence we get the result as area of Rectangle OCDE =
#### Question 7:
In the given figure, ABCD is a trapezium in which AB || DC. Prove that
ar(Δ AOD) = ar(Δ BOC).
Given:
ABCD is a trapezium with AB||DC
To prove: Area of ΔAOD = Area of ΔBOC
Proof:
We know that ‘triangles between the same base and between the same parallels have equal area’
Here ΔABC and ΔABD are between the same base and between the same parallels AB and DC.
Therefore
Hence it is proved that
#### Question 8:
In the given figure, ABCD, ABFE and CDEF are parallelograms. Prove that
Given:
(1) ABCD is a parallelogram,
(2) ABFE is a parallelogram
(3) CDEF is a parallelogram
To prove: Area of ΔADE = Area of ΔBCF
Proof:
We know that,” opposite sides of a parallelogram are equal”
Therefore for
Parallelogram ABFE, AE = BF
Parallelogram CDEF, DE = CF.
Thus, in ΔADE and ΔBCF, we have
So be SSS criterion we have
This means that
Hence it is proved that
#### Question 9:
In the given figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)
Given:
(1) ABC and ABD are two triangles on the same base AB,
(2) CD bisect AB at O which means AO = OB
To Prove: Area of ΔABC = Area of ΔABD
Proof:
Here it is given that CD bisected by AB at O which means O is the midpoint of CD.
Therefore AO is the median of triangle ACD.
Since the median divides a triangle in two triangles of equal area
Therefore Area of ΔCAO = Area of ΔAOD ...... (1)
Similarly for Δ CBD, O is the midpoint of CD
Therefore BO is the median of triangle BCD.
Therefore Area of ΔCOB = Area of ΔBOD ...... (2)
Adding equation (1) and (2) we get
Area of ΔCAO + Area of ΔCOB = Area of ΔAOD + Area of ΔBOD
Area of ΔABC = Area of ΔABD
Hence it is proved that
#### Question 10:
If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(Δ BGC)= 2 ar (Δ AGC).
Given:
(1) ABC is a triangle
(2) AD is the median of ΔABC
(3) G is the midpoint of the median AD
To prove:
(b) Area of Δ BGC = 2 Area of Δ AGC
Construction: Draw a line AM perpendicular to AC
Proof: Since AD is the median of ΔABC.
Therefore BD = DC
So multiplying by AM on both sides we get
In ΔBGC, GD is the median
Since the median divides a triangle in to two triangles of equal area. So
Area of ΔBDG = Area of ΔGCD
Area of ΔBGC = 2(Area of ΔBGD)
Similarly In ΔACD, CG is the median
Area of ΔAGC = Area of ΔGCD
From the above calculation we have
Area of ΔBGD = Area of ΔAGC
But Area of ΔBGC = 2(Area of ΔBGD)
So we have
Area of ΔBGC = 2(Area of ΔAGC)
Hence it is proved that
(1)
(2)
#### Question 11:
A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar (Δ ABD) = 2 ar (Δ ADC).
Given:
(1) ABC is a triangle
(2) D is a point on BC such that BD = 2DC
To prove: Area of ΔABD = 2 Area of ΔAGC
Proof:
In ΔABC, BD = 2DC
Let E is the midpoint of BD. Then,
BE = ED = DC
Since AE and AD are the medians of ΔABD and ΔAEC respectively
and
The median divides a triangle in to two triangles of equal area. So
Hence it is proved that
#### Question 12:
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:
(i) ar (Δ ADO) = ar(Δ CDO)
(ii) ar (Δ ABP) = ar (Δ CBP).
Given: Here from the given figure we get
(1) ABCD is a parallelogram
(2) BD and CA are the diagonals intersecting at O.
(3) P is any point on BO
To prove:
(a) Area of ΔADO = Area ofΔ CDO
(b) Area of ΔAPB = Area ofΔ CBP
Proof: We know that diagonals of a parallelogram bisect each other.
O is the midpoint of AC and BD.
Since medians divide the triangle into two equal areas
In ΔACD, DO is the median
Area of ΔADO = Area ofΔ CDO
Again O is the midpoint of AC.
In ΔAPC, OP is the median
Area of ΔAOP = Area of ΔCOP …… (1)
Similarly O is the midpoint of AC.
In ΔABC, OB is the median
Area of ΔAOB = Area of ΔCOB …… (2)
Subtracting (1) from (2) we get,
Area of ΔAOB − Area of ΔAOP = Area of ΔCOB − Area of ΔCOP
Area of ΔABP = Area of ΔCBP
Hence it is proved that
(a)
(b)
#### Question 13:
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar (Δ ADF) = ar (Δ ECF)
(ii) If the area of Δ DFB = 3 cm2, find the area of ||gm ABCD.
Given: Here from the given figure we get
(1) ABCD is a parallelogram with base AB,
(2) BC is produced to E such that CE = BC
(3) AE intersects CD at F
(4) Area of ΔDFB = 3 cm
To find:
(a) Area of ΔADF = Area of ΔECF
(b) Area of parallelogram ABCD
Proof: Δ ADF and ΔECF, we can see that
ADF = ECF (Alternate angles formed by parallel sides AD and CE)
DFA = CFA (Vertically opposite angles)
(ASA condition of congruence)
As
DF = CF
Since DF = CF. So BF is a median in ΔBCD
Since median divides the triangle in to two equal triangles. So
Since .So
Hence Area of parallelogram ABCD
Hence we get the result
(a)
(b)
#### Question 14:
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar (Δ POA) = ar (Δ QOC).
Given:
(1) Diagonals AC and BD of a parallelogram ABCD intersect at point O.
(2) A line through O intersects AB at P point.
(3) A line through O intersects DC at Q point.
To find: Area of (ΔPOA) = Area of (ΔQOC)
Proof:
From ΔPOA and ΔQOC we get that
=
OA = OC
=
So, by ASA congruence criterion, we have
So
Area (ΔPOA) = Area (ΔQOC)
Hence it is proved that
#### Question 15:
In the given figure, D and E are two points on BC such that BD = DE = EC. Show that a (Δ ABD) = ar (Δ ADE) = ar (Δ AEC).
Given: In ΔABCD, D and E are two points on BC such that BD = DE = EC
To prove:
Proof: The ΔABD, ΔADE, and ΔAEC, are on the equal bases and their heights are equal
Therefore their areas are equal
So
Hence we get the result as
#### Question 16:
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) ✕ ar (Δ CPD) = ar (Δ APD) ✕ ar (Δ BPC)
Given:
(2) Diagonals AC and BD of quadrilateral ABCD intersect at P.
To prove: Area ofΔ APB ×Area of ΔCPD = Area of ΔAPD × Area of ΔBPC
Construction: Draw AL perpendicular to BD and CM perpendicular to BD
Proof:
We know that
Area of triangle = × base× height
Area of ΔAPD = × DP × AL …… (1)
Area of ΔBPC = × CM × BP …… (2)
Area of ΔAPB = × BP × AL …… (3)
Area of ΔCPD = × CM × DP …… (4)
Therefore
Hence it is proved that
#### Question 17:
If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Given: Here from the question we get
(1) ABCD is a parallelogram
(2) P is any point in the interior of parallelogram ABCD
To prove:
Construction: Draw DN perpendicular to AB and PM perpendicular AB
Proof: Area of triangle = × base× height
Area of ΔAPB = × AB × PM …… (1)
Also we know that: Area of parallelogram = base× height
Area of parallelogram ABCD = AB × DN …… (2)
Now PM < DN (Since P is a point inside the parallelogram ABCD)
Hence it is proved that
#### Question 18:
ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram AB CD.
Given:
(1) ABCD is a parallelogram.
(2) E is a point on BA such that BE = 2EA
(3) F is a point on DC such that DF = 2FC.
To find:
Area of parallelogram
Proof: We have,
BE = 2EA and DF = 2FC
AB − AE = 2AE and DC − FC = 2FC
AB = 3AE and DC = 3FC
AE = AB and FC = DC
AE = FC [since AB = DC]
Thus, AE || FC such that AE = FC
Therefore AECF is a parallelogram.
Clearly, parallelograms ABCD and AECF have the same altitude and
AE = AB.
Therefore
Hence proved that
#### Question 19:
In a Δ ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that:
(i) ar (Δ PBQ) = ar (Δ ARC)
(ii) ar (Δ PRQ) = $\frac{1}{2}$ ar (Δ ARC)
(iii) ar (Δ RQC) = $\frac{3}{8}$ ar (Δ ABC).
Given:
(1) In a triangle ABC, P is the mid-point of AB.
(2) Q is mid-point of BC.
(3) R is mid-point of AP.
To prove:
(a) Area of ΔPBQ = Area of ΔARC
(b) Area of ΔPRQ = Area of ΔARC
(c) Area of ΔRQC = Area of ΔABC
Proof: We know that each median of a triangle divides it into two triangles of equal area.
(a) Since CR is a median of ΔCAP
Therefore …… (1)
Also, CP is a median of ΔCAB.
Therefore …… (2)
From equation (1) and (2), we get
Therefore …… (3)
PQ is a median of ΔABQ
Therefore
Since
Put this value in the above equation we get
…… (4)
From equation (3) and (4), we get
Therefore …… (5)
(b)
…… (6)
…… (7)
From equation (6) and (7)
…… (8)
From equation (7) and (8)
(c)
= …… (9)
#### Question 20:
ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(i) ar (ADEG) = ar (GBCE)
(ii) ar (Δ EGB) = $\frac{1}{6}$ ar (ABCD)
(iii) ar (Δ EFC) = $\frac{1}{2}$ ar (Δ EBF)
(iv) ar (Δ EBG) = ar (Δ EFC)
(v) Find what portion of the area of parallelogram is the area of Δ EFG
Given:
ABCD is a parallelogram
G is a point such that AG = 2GB
E is a point such that CE = 2DE
F is a point such that BF = 2FC
To prove:
(i)
(ii)
(iii)
(iv)
What portion of the area of parallelogram ABCD is the area of ΔEFG
Construction: draw a parallel line to AB through point F and a perpendicular line to AB through
PROOF:
(i) Since ABCD is a parallelogram,
So AB = CD and AD = BC
Consider the two trapeziums ADEG and GBCE:
Since AB = DC, EC = 2DE, AG = 2GB
, and
, and
So, and
Since the two trapeziums ADEG and GBCE have same height and their sum of two parallel sides are equal
Since
So
Hence
(ii) Since we know from above that
. So
Hence
(iii) Since height of triangle EFC and triangle EBF are equal. So
Hence
(iv) Consider the trapezium in which
(From (iii))
Now from (ii) part we have
(v) In the figure it is given that FB = 2CF. Let CF = x and FB = 2x
Now consider the tow triangles CFI and CBH which are similar triangles
So by the property of similar triangle CI = k and IH = 2k
Now consider the triangle EGF in which
Now
(Multiply both sides by 2)
…… (2)
From (1) and (2) we have
#### Question 21:
In the given figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of ΔCBX
(ii) Prove that ar (Δ ZDE) = ar (Δ CZA)
(iii) Prove that ar (BCZY) = ar (Δ EDZ).
Given:
(1) CD||AE.
(2) CY||BA.
To find:
(i) Name a triangle equal in area of ΔCBX.
(ii) .
(iii) .
Proof:
(i) Since triangle BCY and triangle YCA are on the same base and between same parallel, so their area should be equal. Therefore
Therefore area of triangle CBX is equal to area of triangle AXY
(ii) Triangle ADE and triangle ACE are on the same base AE and between the same parallels AE and CD.
(iii) Triangle ACY and BCY are on the same base CY and between same parallels CY and BA. So we have
Now we know that
#### Question 22:
In the given figure, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar (PQE) = ar (Δ CFD).
Given:
(i) PSDA is a parallelogram.
(ii) .
(iii)
To find:
Proof:
So PQ = CD …… (1)
In ΔBED, C is the mid point of BD and CF||BE
This implies that F is the mid point of ED. So
EF = FD …… (2)
In ΔPQE and ΔCFD, we have
PE = FD
, and [Alternate angles]
PQ = CD.
So, by SAS congruence criterion, we have
ΔPQE = ΔDCF
Hence proved that
#### Question 23:
In the given figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar (trap. DCYX) = $\frac{9}{11}$ ar (trap. (XYBA)
Given: ABCD IS A trapezium in which
(a) AB||DC
(b) DC = 40 cm
(c) AB = 60 cm
(d) X is the midpoint of AD
(e) Y is the midpoint of BC
To prove:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii)
Construction: Join DY and produce it to meet AB produced at P.
Proof:
(i) In ΔBYP and ΔCYD
Y is the midpoint of BC also X is the midpoint of AD
Therefore XY||AP and
(ii) We have proved above that XY||AP
XY|| AP and AB||DC (Given in question)
XY|| DC
(iii) Since X and Y are the midpoints of AD and BC respectively.
Therefore DCYX and ABYX are of the same height say h cm.
#### Question 24:
D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar (Δ BOE) = $\frac{1}{8}$ ar (Δ ABC).
Given: In ΔABC
(1) D is the midpoint of the side BC
(2) E is the midpoint of the side BD
(3) O is the midpoint of the side AE
To prove:
Proof: We know that the median of a triangle divides the triangle into two triangles of equal area.
Since AD and AE are the medians of ΔABC and ΔABD respectively. And OB is the median of ΔABE
…… (1)
…… (2)
…… (3)
Therefore
Hence we have proved that
#### Question 25:
In the given figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (Δ ABP) = ar (Δ ACQ).
Given:
(1) X and Y are the, midpoints of AC and AB respectively.
(2) QP|| BC
(3) CYQ and BXP are straight lines.
To prove:
Proof: Since X and Y are the, midpoints of AC and AB respectively.
So XY||BC
ΔBYC and ΔBXC are on the same base BC and between the same parallels XY and BC.
Therefore
…… (1)
Similarly the quadrilaterals XYAP and XYQA are on the same base XY and between the same parallels XY and PQ. Therefore
…… (2)
Adding equation 1 and 2 we get
#### Question 26:
In the given figure, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar (Δ APE): ar (ΔPFA) = ar Δ (QFD) : ar (Δ PFD)
(iii) ar (Δ PEA) = ar (Δ QFD).
Given: ABCD and AEFD are two parallelograms
To prove:
(i) PE = FQ
(ii)
(iii)
Proof: (i) and (iii)
In ΔAPE and ΔDQF
Therefore
, and
(ii) ΔPFA and ΔPFD are on the same base PF and between the same parallels PQ and AD.
From (1) and (2) we get
#### Question 27:
In the given figure, ABCD is a ||gm. O is any point on AC. PQ || AB and LM || AD. Prove that ar (||gm DLOP) = ar (||gm BMOQ)
Given:
(1) ABCD is a parallelogram
(2) O is any point of AC.
To prove:
Calculation:
We know that the diagonal of a parallelogram divides it into two triangles of equal area
Therefore we have
Since OC and AO are diagonals of parallelogram OQCL and AMOP respectively. Therefore
Subtracting (2) and (3) from (1) we get
Hence we get the result
#### Question 28:
In a Δ ABC, if L and M are points on AB and AC respectively such that LM || BC. Prove that:
(i) ar (Δ LCM) = ar (Δ LBM)
(ii) ar (Δ LBC) = (Δ MBC)
(iii) ar (Δ ABM) = ar (Δ ACL)
(iv) ar (Δ LOB) = ar (Δ MOC)
Given:
In ΔABC, if L and M are points on AB and AC such that LM||BC
To prove:
(i)
(ii)
(iii)
(iv)
Proof: We know that triangles between the same base and between the same parallels are equal in area.
(i) Here we can see that ΔLMB and ΔLMC are on the same base BC and between the same parallels LM and BC
Therefore
…… (1)
(ii) Here we can see that ΔLBC and ΔLMC are on the same base BC and between the same parallels LM and BC
Therefore
…… (2)
(iii) From equation (1) we have,
(iv) From (2) we have,
#### Question 29:
In the given figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that
(i) ar (Δ BDE) = $\frac{1}{4}$ ar (Δ ABC)
(ii) ar (Δ BDE) = $\frac{1}{2}$ ar (Δ BAE)
(iii) ar (Δ BFE) =ar (Δ AFD)
(iv) ar (ΔABC) = 2 ar (Δ BEC)
(v) ar (Δ FED) = $\frac{1}{8}$ ar (Δ AFC)
(vi) ar (Δ BFE) = 2 ar (EFD)
Given:
(a) ΔABC and Δ BDE are two equilateral triangles
(b) D is the midpoint of BC
(c) AE intersect BC in F.
To prove:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Proof: Let AB = BC = CA = x cm.
Then BD = = DE = BE
(i) We have
(ii) We Know that ΔABC and ΔBED are equilateral triangles
BE||AC
(iii) We Know that ΔABC and ΔBED are equilateral triangles
AB || DE
(iv) Since ED is a median of Δ BEC
(v) We basically want to find out FD. Let FD = y
Since triangle BED and triangle DEA are on the same base and between same parallels ED and BE respectively. So
Since altitude of altitude of any equilateral triangle having side x is
So
…… (1)
Now
…… (2)
From (1) and (2) we get
(vi) Now we know y in terms of x. So
……. (3)
…… (4)
From (3) and (4) we get
#### Question 30:
If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AXDE meets BC at Y.
Show that
(i) Δ MBC $\cong$ Δ ABD
(ii) ar (BYXD) = 2 ar (Δ MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) Δ FCB $\cong$ Δ ACE
(v) ar (CYXE) = 2 ar (ΔFCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Given:
(1) ABCD is a right angled triangle at A
(2) BCED, ACFG and ABMN are the squares on the sides of BC, CA and AB respectively.
(3) , meets BC at Y.
To prove:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Proof:
(i)
…… (1)
(ii) Triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.
Therefore
(Using (1)) …… (2)
(iii) Since ΔMBC and square MBAN are on the same base MB and between the same parallels MB and NC.
…… (3)
From (2) and (3) we get
(iv) In triangle FCB and ACE
…… (4)
(v) Since ΔACE and rectangle CYXE are on the same base CE and between the same parallels CE and AX.
…… (5)
(vi) Since ΔFCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG
…… (6)
From (5) and (6) we get
(vii) Applying Pythagoras Theorem in ΔACB, WE get
#### Question 1:
Two parallelograms are on the same base and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 3 : 1
Given: Two parallelogram with the same base and between the same parallels.
To find: Ratio of their area of two parallelogram with the same base and between the same parallels.
Calculation: We know that “Two parallelogram with the same base and between the same parallels, are equal in area”
Hence their ratio is ,
So the correct answer is,i.e. option (c).
#### Question 2:
A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Given: A triangle and a parallelogram with the same base and between the same parallels.
To find: The ratio of the area of a triangle and a parallelogram with the same base and between the same parallels.
Calculation: We know that,” the area of a triangle is half the area of a parallelogram with the same base and between the same parallels.”
Hence the ratio of the area of a triangle and a parallelogram with the same base and between the same parallels is
Therefore the correct answer is option is (b).
#### Question 3:
Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of the sides of Δ ABC. Then the area of ΔPQR is
(a) 12 sq. units
(b) 6 sq. units
(c) 4 sq. units
(d) 3 sq. units
Given: (1) The Area of ΔABC = 24 sq units.
(2) ΔPQR is formed by joining the midpoints of ΔABC
To find: The area of ΔPQR
Calculation: In ΔABC, we have
Since Q and R are the midpoints of BC and AC respectively.
PQ || BA PQ || BP
Similarly, RQ || BP. So BQRP is a parallelogram.
Similarly APRQ and PQCR are parallelograms.
We know that diagonal of a parallelogram bisect the parallelogram into two triangles of equal area.
Now, PR is a diagonal of APQR.
∴ Area of ΔAPR = Area of ΔPQR ……(1)
Similarly,
PQ is a diagonal of PBQR
∴ Area of ΔPQR = Area of ΔPBQ ……(2)
QR is the diagonal of PQCR
∴ Area of ΔPQR = Area of ΔRCQ ……(3)
From (1), (2), (3) we have
Area of ΔAPR = Area of ΔPQR = Area of ΔPBQ = Area of ΔRCQ
But
Area of ΔAPR + Area of ΔPQR + Area of ΔPBQ + Area of ΔRCQ = Area of ΔABC
4(Area of ΔPBQ) = Area of ΔABC
∴ Area of ΔPBQ
Hence the correct answer is option (b).
#### Question 4:
The median of a triangle divides it into two
(a) congruent triangle
(b) isosceles triangles
(c) right triangles
(d) triangles of equal areas
Given: A triangle with a median.
Calculation: We know that a ,”median of a triangle divides it into two triangles of equal area.”
Hence the correct answer is option (d).
#### Question 5:
In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar (ΔABC) = 16cm2, then ar (trapezium FBCE) =
(a) 4 cm2
(b) 8 cm2
(c) 12 cm2
(d) 10 cm2
Given: In ΔABC
(1) D is the midpoint of BC
(2) E is the midpoint of CA
(3) F is the midpoint of AB
(4) Area of ΔABC = 16 cm2
To find: The area of Trapezium FBCE
Calculation: Here we can see that in the given figure,
Area of trapezium FBCE = Area of ||gm FBDE + Area of ΔCDE
Since D and E are the midpoints of BC and AC respectively.
∴ DE || BA DE || BF
Similarly, FE || BD. So BDEF is a parallelogram.
Now, DF is a diagonal of ||gm BDEF.
∴ Area of ΔBDF = Area of ΔDEF ……(1)
Similarly,
DE is a diagonal of ||gm DCEF
∴ Area of ΔDCE = Area of ΔDEF ……(2)
FE is the diagonal of ||gm AFDE
∴ Area of ΔAFE = Area of ΔDEF ……(3)
From (1), (2), (3) we have
Area of ΔBDF = Area of ΔDCF = Area of ΔAFE = Area of ΔDEF
But
Area of ΔBDF + Area of ΔDCE + Area of ΔAFE + Area of ΔDEF = Area of ΔABC
∴ 4 Area of ΔBDF = Area of ΔABC
Area of ΔBDF = Area of ΔDCE = Area of ΔAFE = Area of ΔDEF = 4 cm2 …….(4)
Now
Hence we get
Area of trapezium FBCE
There fore the correct answer is option (c).
#### Question 6:
ABCD is a parallelogram. P is any point on CD. If ar (ΔDPA) = 15 cm2 and ar (ΔAPC) = 20 cm2, then ar (ΔAPB) =
(a) 15 cm2
(b) 20 cm2
(c) 35 cm2
(d) 30 cm2
Given: (1) ABCD is a parallelogram
(2) P is any point on CD
(3) Area of ΔDPA = 15 cm2
(4) Area of ΔAPC = 20 cm2
To find: Area of ΔAPB
Calculation: We know that , “If a parallelogram and a a triangle are on the base between the same parallels, the area of triangle is equal to half the area of the parallelogram.”
Here , ΔAPB and ΔACB are on the same base and between the same parallels.
(since AC is the diagonal of parallelogram ABCD, diagonal of a parallelogram divides the parallelogram in two triangles of equal area)
Hence the correct answer is option (c).
#### Question 7:
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is
(a) 28 cm2
(b) 48 cm2
(c) 96 cm2
(d) 24 cm2
Given: Rhombus with diagonals measuring 16cm and 12 cm.
To find: Area of the figure formed by lines joining the midpoints of the adjacent sides.
Calculation: We know that, ‘Area of a rhombus is half the product of their diagonals’.
H and F are the midpoints of AD and BC respectively.
Now ABCD is a parallelogram which means
……..(1)
……(2)
From 1 and 2 we get that ABFH is a parallelogram.
Since Parallelogram FHAB and ΔFHE are on the base FH and between the same parallels HF and AB.
……(3)
Similarly ,
……(4)
Adding 3 and 4 we get,
Hence the correct answer is option (b).
#### Question 8:
A, B, C, D are mid-points of sides of parallelogram PQRS. If ar (PQRS) = 36 cm2, then ar (ABCD) =
(a) 24 cm2
(b) 18cm2
(c) 30 cm2
(d) 36 cm2
Given:
(1) PQRS is a parallelogram.
(2) A, B, C, D are the midpoints of the adjacent sides of Parallelogram PQRS.
(3)
To find:
Calculation:
A and C are the midpoints of PS and QR respectively.
Now PQRS is a parallelogram which means
AP = CQ ……..(1)
Also, PS || QR
AP || CQ ……(2)
From 1 and 2 we get that APCQ is a parallelogram.
Since Parallelogram APCQ and ΔABC are on the base AC and between the same parallels AC and PQ.
……(3)
Similarly ,
……(4)
Adding 3 and 4 we get,
Hence the correct answer is option (b).
#### Question 9:
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is
(a) a rhombus of area 24 cm2
(b) a rectangle of area 24 cm2
(c) a square of area 26 cm2
(d) a trapezium of area 14 cm2
Given: Rectangle with sides 8 cm and 6cm
To find: Area of the figure which is formed by joining the midpoints of the adjacent sides of rectangle.
Calculation: Since we know that
For rhombus EFGH, EG is the one diagonal which is equal to DA
FH is the other diagonal which is equal to AB
Hence the result is option (a).
#### Question 10:
If AD is median of ΔABC and P is a point on AC such that
ar (ΔADP) : ar (ΔABD) = 2 : 3, then ar (Δ PDC) : ar (Δ ABC)
(a) 1 : 5
(b) 1 : 5
(c) 1 : 6
(d) 3 : 5
Given: (1) AD is the Median of ΔABC
(2) P is a point on AC such that ar (ΔADP) : ar (ΔABD) =
To find: ar (ΔPDC) : ar (ΔABC)
We know that” the medians of the triangle divides the triangle in two two triangles of equal area.”
Since AD is the median of ΔABC,
ar (ΔABD) = ar (ΔADC) ……(1)
Also it is given that
ar (ΔADP) : ar (ΔABD) = ……(2)
Now,
Therefore,
Hence the correct answer is option (c).
#### Question 11:
Medians of ΔABC intersect at G. If ar (ΔABC) = 27 cm2, then ar (ΔBGC) =
(a) 6 cm2
(b) 9 cm2
(c) 12 cm2
(d) 18 cm2
Given: (1) Median of ΔABC meet at G.
(2) Area of ΔABC = 27 cm2
To find: Area of ΔBCG.
We know that the medians of the triangle divides each other in the ratio of 2:1
Hence,
Hence the correct answer is option (b).
#### Question 12:
In a ΔABC if D and E are mid-points of BC and AD respectively such that ar (ΔAEC) = 4cm2, then ar (ΔBEC) =
(a) 4 cm2
(b) 6 cm2
(c) 8 cm2
(d) 12 cm2
Given: In ΔABC
(1) D is the midpoint of BC
(2) E is the midpoint of AD
(3) ar (ΔAEC) = 4 cm2
To find: ar (ΔBEC)
Calculation: We know that”the median of the triangle divides the triangle into two triangle of equal area”
Since AD is the median of ΔABC,
ar (ΔABD) = ar (ΔADC) …… (1)
EC is the median of ΔADC,
ar (ΔAEC) = ar (ΔDEC) …… (2)
⇒ ar (ΔDEC) = 4 cm2
EC is the median of ΔBED
ar (ΔBED) = ar (ΔDEC) …… (3)
From 2 and 3 we get,
ar (ΔBED) = ar (ΔAEC) …… (4)
⇒ ar (ΔBED) = 4 cm2
Now,
Hence the correct answer is option (c).
#### Question 13:
In the given figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =
Given: (1) ABCD is a parallelogram.
(2) AB = 12 cm
(3) AE = 7.5 cm
(4) CF = 15cm
Calculation: We know that,
Area of a parallelogram = base × height
Area of a parallelogram ABCD = DC ×AE (with DC as base and AE as height) ……(1)
Area of a parallelogram ABCD = AD ×CF (with DC as base and AE as height) ……(2)
Since equation 1 and 2 both are Area of a parallelogram ABCD
Hence the correct answer is option (b).
#### Question 14:
In the given figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal Q is joined. The ratio ar (||gm XQRY) : ar (ΔQSR) =
(i) 1 : 4
(ii) 2 : 1
(iii) 1 : 2
(iv) 1 : 1
Given: (1) PQRS is a parallelogram.
(2) X is the midpoint of PQ.
(3) Y is the midpoint of SR.
(4) SQ is the diagonal.
To find: Ratio of area of ||gm XQRY : area of ΔQRS.
Calculation: We know that the triangle and parallelogram on the same base and between the same parallels are equal in area.
∴ Ar (||gm PQRS) = Ar (ΔQRS)
(since X is the mid point of PQ and Y is the midpoint of SR)
Hence the correct answer is option (d).
#### Question 15:
Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ΔAOD is
(a) ΔAOB
(b) ΔBOC
(c) ΔDOC
Given: (1) ABCD is a trapezium, with parallel sides AB and DC
(2) Diagonals AC and BD intersect at O
To find: Area of ΔAOD equals to ?
Calculation: We know that ,” two triangles with the same base and between the same parallels are equal in area.”
Therefore,
Hence the correct answer is option (b).
#### Question 16:
ABCD is a trapezium in which AB || DC. If ar (ΔABD) = 24 cm2 and AB = 8 cm, then height of ΔABC is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Given: (1) ABCD is a trapezium, with parallel sides AB and DC
(2) Area of ΔADB = 24 cm2
(3) AB = 8 cm
To find: Height of ΔABC.
Calculation: We know that ,” two triangles with the same base and between the same parallels are equal in area.”
Here we can see that, ΔADB and ΔACB are on the same base AB.
Hence,
Area of ΔACB = Area of ΔADB
Area of ΔACB = 24
Hence the correct answer is option (c).
#### Question 17:
ABCD is a trapezium with parallel sides AB =a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is
(a) a : b
(b) (a + 3b): (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)
Given: (1) ABCD is a trapezium, with parallel sides AB and DC
(2) AB = a cm
(3) DC = b cm
(4) E is the midpoint of non parallel sides AD.
(5) G is the midpoint of non parallel sides BC.
To find: Ratio of the area of the Quadrilaterals ABFE and EFCD.
Calculation: We know that, ‘Area of a trapezium is half the product of its height and the sum of the parallel sides.’
Since, E and F are mid points of AD and BC respectively, so h1 = h2
Area of trapezium ABFE
Now, Area (trap ABCD) = area (trap EFCD) + Area (ABFE)
Therefore,
Thus,
Hence, the correct option is (c)
#### Question 18:
ABCD is a rectangle with O as any point in its interior. If ar (ΔAOD) = 3 cm2 and ar (ΔABOC) = 6 cm2, then area of rectangle ABCD is
(a) 9 cm2
(b) 12 cm2
(c) 15 cm2
(d) 18 cm2
Given: A rectangle ABCD , O is a point in the interior of the rectangles such that
(1) ar (ΔAOB) = 3 cm2
(2) ar (ΔBOC) = 6 cm2
To find: ar (rect.ABCD)
Construction: Draw a line LM passing through O and parallel to AD and BC.
Calculation: We know that ,” If a triangle and a parallelogram are on the same base and between the same parallels the area of the triangle is equal to half the area of the parallelogram”
Here we can see that ΔAOD and rectangle AMLD are on the same base AD and between the same parallels AD and LM.
Hence ,
Similarly, we can see that ΔBOC and rectangle BCLM are on the same base BC and between the same parallels BC and LM
Hence,
We known that
Hence the correct answer is option (d).
#### Question 19:
In the given figure, a parallelogram ABCD and a rectangle ABEF are of equal area. Then,
(a) perimeter of ABCD = perimeter of ABEF
(b) perimeter of ABCD < perimeter of ABEF
(c) perimeter of ABCD > perimeter of ABEF
(d) perimeter of ABCD = $\frac{1}{2}$ perimeter of ABEF
We know, opposite sides of a rectangle are equal.
Therefore, in rectangle ABEF,
AB = EF ...(1)
Also, opposite sides of a parallelogram are equal.
Therefore, in parallelogram ABCD,
AB = DC ...(3)
From (1) and (3),
DC = EF
AB + EF = AB + DC ...(5)
Now, we know that, of all the line segments, perpendicular segment is the shortest.
BE < BC
⇒ AF + BE < AD + BC ...(6)
Adding (5) and (6), we get
AB + EF + AF + BE < AB + DC + AD + BC
⇒ Perimeter of rectangle < perimeter of parallelogram
Hence, the correct option is (c).
#### Question 20:
In the given figure, the area of parallelogram ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL
We know, area of the parallelogram = Base × Altitude
Therefore,
The area of parallelogram ABCD = AB × DL
= DC × DL
Hence, the correct option is (c).
#### Question 21:
ABCD is quadrilateral whose diagonal AC divided it into two parts, equal in area, then ABCD
(a) is a rectangle
(b) is always a rhombus
(c) is a parallelogram
(d) need not be any of (a), (b) or (c)
The diagonal of a parallelogram, rectangle, rhombus or a square divides them into two parts, equal in area.
Therefore, if ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD need not be a rectangle, parallelogram or a rhombus. It can be a square as well.
Hence, the correct option is (d).
#### Question 22:
The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to
(a) ar (ΔABC)
(b) $\frac{1}{2}$ar (ΔABC)
(c) $\frac{1}{3}$ar (ΔABC)
(d) $\frac{1}{4}$ar (ΔABC)
Given: (1) ABCD is a triangle.
(2) mid points of the sides of ΔABC with any of the vertices forms a parallelogram.
To find: Area of the parallelogram
Calculation: We know that: Area of a parallelogram = base × height
Hence area of ||gm DECF = EC × EG
area of ||gm DECF = EC × EG
area of ||gm DECF = (E is the midpoint of BC)
area of ||gm DECF =
area of ||gm DECF =
Hence the result is option (b).
#### Question 23:
In the given figure, ABCD and FECG are parallelograms equal in area. If ar (ΔAQE) = 12 cm2, then ar (||gm FGBQ) =
(a) 12 cm2
(b) 20 cm2
(c) 24 cm2
(d) 36 cm2
Given: (1) Area of parallelogram ABCD is equal to Area of parallelogram FECG.
(2) If Area of ΔAQE is 12cm.
To find: Area of parallelogram FGBQ
Calculation: We know that diagonal of a parallelogram divides the parallelogram into two triangles of equal area.
It is given that,
Hence the correct answer is option (c).
#### Question 1:
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is _________.
Given:
A rhombus ABCD with diagonals 12 cm and 16 cm
i.e., AC = 16 cm and BD = 12 cm
And a quadrilateral PQRS formed by joining the mid-points of the adjacent sides of ABCD.
Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.
In ∆ABC,
PQ || AC
PQ = $\frac{1}{2}$AC
⇒ PQ =
$\frac{1}{2}$(16)
⇒ PQ = 8 cm
RS || AC
RS = $\frac{1}{2}$AC
⇒ RS =
$\frac{1}{2}$(16)
⇒ RS = 8 cm
In ∆BCD,
RQ || BD
RQ $\frac{1}{2}$BD
⇒ RQ =
$\frac{1}{2}$(12)
⇒ RQ = 6 cm
SP || BD
SP $\frac{1}{2}$BD
⇒ SP =
$\frac{1}{2}$(12)
⇒ SP = 6 cm
Since, PQ = 8 cm = RS and RQ = 6 cm = SP
and Diagonals of a rhombus intersect at right angle.
angle between AC and BD is 90°
angle between PQ and QR is 90°
Therefore, PQRS is a rectangle
Thus, Area of rectangle = PQ × QR
= 8 × 6
= 48 cm2
Hence, the area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is 48 cm2.
#### Question 2:
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a ________ of area _________.
Given:
A rectangle ABCD of sides 8 cm and 6 cm
i.e., AB = 8 cm and AD = 6 cm
And a quadrilateral PQRS formed by joining the mid-points of the adjacent sides of ABCD.
We can see that QS || AB and PR || AD
Also, QS = AB = 8 cm and PR = AD = 6 cm
Thus, angle between QS and PR is 90°
Therefore, PQRS is a rhombus.
Area of rhombus = $\frac{1}{2}$× SQ × PR
=
$\frac{1}{2}$× 8 × 6
= 4 × 6
= 24 cm2
Hence, the figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a rhombus of area 24 cm2.
#### Question 3:
If P is any point on the median AD of a ΔABC, then = _________.
Given: P is any point on the median AD of a ΔABC
We know, median of a triangle divides it into two triangles of equal area.
Also, ar (ΔPDB) = ar (ΔPDC) ...(2)
Subtracting (2) from (1), we get
⇒ ar (ΔABP) = ar (ΔACP)
= 1
Hence, 1.
#### Question 4:
If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is _________.
If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Area of triangle = $\frac{1}{2}$ Area of the parallelogram
Hence, if a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is 1 : 2.
#### Question 5:
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is _________.
Parallelograms on equal bases and between the same parallels are equal in area.
Area of first parallelogram = Area of the second parallelogram
Hence, the ratio of their areas is 1 : 1.
#### Question 6:
ABCD is a parallelogram and X is the mid-point of AB. If ar(AXCD) = 24 cm2 , then ar(ΔABC) = ________.
Given:
ABCD is a parallelogram
X is the mid-point of AB
ar(AXCD) = 24 cm2
We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔABC) $\frac{1}{2}$ar(ABCD) ...(1)
Since, X is the mid-point of AB
Therefore, ar(ΔXBC$\frac{1}{2}$ar(ABC)
= $\frac{1}{2}$×$\frac{1}{2}$ar(ABCD) (From (1))
= $\frac{1}{4}$ar(ABCD) ...(2)
Thus, ar(AXCD) = ar(ABCD) − ar(ΔXBC)
⇒ 24 = ar(ABCD) − $\frac{1}{4}$ar(ABCD) (From (2))
⇒ 24 = $\frac{3}{4}$ar(ABCD)
⇒ ar(ABCD) = $24×\frac{4}{3}$
⇒ ar(ABCD) = 8 × 4
⇒ ar(ABCD) = 32 cm2
From (1)
ar(ΔABC$\frac{1}{2}$ar(ABCD)
= $\frac{1}{2}$× 32
= 16 cm2
Hence, ar(ΔABC) = 16 cm2.
#### Question 7:
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar(ΔPAS) = _______.
Given:
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm
PS = 5 cm
A is any point on PQ
QS = radius of the circle = 13 cm ...(1)
In ΔPQS
Using pythagoras theorem,
QS2 = PS2 + PQ2
⇒ 132 = 52PQ2
⇒ PQ2 = 169 − 25
PQ2 = 144
⇒ PQ = 12 cm = SR ...(2)
Thus,
ar(ΔRAS$\frac{1}{2}$× base × height
= $\frac{1}{2}$× SR × PS
= $\frac{1}{2}$× 12 × 5
= 30 cm2
Hence, ar(ΔRAS) = 30 cm2.
Disclaimer: The question is to find the area of ΔRAS instead of the area of ΔPAS.
#### Question 8:
If ABC and BDE are two equilateral triangles such that D is the mid-point of BC then ar(ΔABC) : ar(BDE) = _________.
Given:
ABC and BDE are two equilateral triangles
is the mid-point of BC
Hence, ar(ΔABC) : ar(ΔBDE) = 4 : 1.
#### Question 9:
If PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS, then ar(ΔASR) is _______ 90 cm2?
Given:
PQRS is a parallelogram with area 180 cm2
A is any point on the diagonal QS
We know, the diagonal of the parallelogram bisects it into two triangles of equal area.
Thus, ar(ΔPQS) = ar(ΔQRS) = $\frac{1}{2}$ar(PQRS)
⇒ ar(ΔPQS) = ar(ΔQRS) = 90 cm2
Therefore, ar(ΔASR) is always less than 90 cm2 unless or until the point A coincides with Q or S.
Hence, ar(ΔASR) is less than 90 cm2
#### Question 10:
The mid points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to ________.
Given:
ABC is a triangle
Let D is the mid-point of ABE is the mid-point of BC and F is the mid-point of AC.
ADEF is a parallelogram having 2 triangles of equal area i.e., ∆ADF and ∆DEF.
But the ∆ABC is divided in 4 triangles of equal area i.e., ∆ADF, ∆DEF, ∆BED and ∆CEF.
Thus, area of ∆ABC = × area of the parallelogram ADEF.
Hence, the mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to half the area of the triangle ABC.
#### Question 11:
A median of a triangle divides it into two ___________.
Let ABC be a triangle with a mid-point D on BC.
Therefore, BD = DC
Let AE be the altitude from A on BC.
Now, ar(∆ABD) = $\frac{1}{2}$× base × height
= $\frac{1}{2}$× BD × AE
Also, ar(∆ACD) = $\frac{1}{2}$× base × height
= $\frac{1}{2}$× CD × AE
=
$\frac{1}{2}$× BD × AE (∵ BD = CD)
= ar(∆ABD
Hence, a median of a triangle divides it into two triangles of equal area.
#### Question 12:
If ABCD is a rectangle, E and F are the mid-points of BC and AD respectively and G is any point on EF. Then ar(ΔGAB) : ar (rectangle ABCD) = _________.
Given:
ABCD is a rectangle
E and F are the mid-points of BC and AD respectively
is any point on EF
Since, E and F are the mid-points of BC and AD respectively.
Therefore, ar(BEFA) = ar(ECDF) = $\frac{1}{2}$× ar(ABCD) ...(1)
We know, if a triangle and a rectangle are on the same base and between the same parallels, then the area of the triangle is half the area of the rectangle.
Thus, ar(ΔGAB$\frac{1}{2}$ar(BEFA) ...(2)
From (1) and (2),
ar(ΔGAB) = $\frac{1}{2}$×$\frac{1}{2}$× ar(ABCD)
⇒ ar(ΔGAB) = $\frac{1}{4}$× ar(ABCD)
⇒
Hence, ar(ΔGAB) : ar(rectangle ABCD) = 1 : 4.
#### Question 13:
PQRS is a square of side 8 cm. T and U and respectively the mid points of PS and QR. If TU and QS intersect at 0, then ar(ΔOTS) = ________.
Given:
PQRS is a square of side 8 cm.
and are respectively the mid-points of PS and QR
TU and QS intersect at O
In ΔQOU and ΔOTS,
QOU = ∠TOS (vertically opposite angles)
OQU = ∠OST (alternate angles)
QU = TS (mid-points of sides of a square)
By AAS property,
ΔQOU ≅ ΔOTS
Thus, OU = OT (by CPCT)
⇒ OU + OT = PQ = 8 cm
⇒ OU = OT = 4 cm ...(1)
Also, TS = 4 cm (T is the mid-point of PS) ...(2)
ar(ΔOTS) = $\frac{1}{2}$× base × height
= $\frac{1}{2}$× TS × OT
= $\frac{1}{2}$× 4 × 4 (From (1) and (2))
= 8 cm2
Hence, ar(ΔOTS) = 8 cm2.
#### Question 14:
In the given figure, ABCD and EFGD are two parallelograms and G is the mid point of CD. Then, ar(ΔDPC) : ar(EFGD) = ________.
Given:
ABCD and EFGD are two parallelograms.
G is the mid point of CD
Since, G is the mid point of CD
Therefore, DG = GC
Since, ΔDPG and ΔGPC have equal base and common height,
Thus, ar(ΔDPG) = ar(ΔGPC
⇒ ar(ΔDPC) = 2 ar(ΔDPG) ...(1)
Also, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔDPG$\frac{1}{2}$ar(EFGD) ...(2)
From (1) and (2),
ar(ΔDPC) = 2 × $\frac{1}{2}$ar(EFGD)
⇒ ar(ΔDPC) = ar(EFGD)
Hence, ar(ΔDPC) : ar(EFGD) = 1 : 1.
#### Question 15:
If the given figure, PQRS and EFRS are two parallelograms, then ar(||gm PQRS) : ar(ΔMFR) = ________.
Given:
PQRS and EFRS are two parallelograms
PQRS and EFRS are two parallelograms lying on the same base SR and between the same parallels SR and PF.
We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.
Thus, ar(PQRS) = ar(EFRS) ...(1)
Also, ΔMFR and parallelogram EFRS is lying on the same base FR and between the same parallels SR and EF.
We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔMFR$\frac{1}{2}$ar(EFRS) ...(2)
From (1) and (2),
ar(ΔMFR) = $\frac{1}{2}$ar(PQRS)
⇒ ar(||gm PQRS) = 2 ar(ΔMFR)
Hence, ar(||gm PQRS) : ar(ΔMFR) = 2 : 1.
#### Question 16:
In the given figure, if ABCD is a parallelogram of area 90 cm2. Then, ar(||gm ABEF) = ________ ar(ΔABD) = _______ and ar(ΔBEF) = _________.
Given:
ABCD is a parallelogram of area 90 cm2
ABCD and ABEF are two parallelograms lying on the same base AB and between the same parallels AB and CF.
We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.
Thus, ar(ABCD) = ar(ABEF) = 90 cm2 ...(1)
Also, ΔABD and parallelogram ABEF is lying on the same base AB and between the same parallels AF and BE.
We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔABD$\frac{1}{2}$ar(ABEF)
⇒ ar(||gm ABEF) = 2 ar(ΔABD) ...(2)
From (1) and (2),
ar(||gm ABEF) = 2 ar(ΔABD) = 90 cm2
Also, diagonal of a parallelogram divides it into two triangles of equal area.
Thus, ar(ΔBEF$\frac{1}{2}$ar(ABEF)
= $\frac{1}{2}$(90)
= 45 cm2
Hence, ar(||gm ABEF) = 2 ar(ΔABD) = 90 cm2 and ar(ΔBEF) = 45 cm2.
#### Question 1:
If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar (ΔABC) : ar (ΔBDE).
Given: (1) ΔABC is equilateral triangle.
(2) ΔBDE is equilateral triangle.
(3) D is the midpoint of BC.
To find:
PROOF : Let us draw the figure as per the instruction given in the question.
We know that area of equilateral triangle = , where a is the side of the triangle.
Let us assume that length of BC is a cm.
This means that length of BD is cm, Since D is the midpoint of BC.
------(1)
------(2)
Now, ar(ΔABC) : ar(ΔBDE) = (from 1 and 2)
=
Hence we get the result ar(ΔABC) : ar(ΔBDE) =
#### Question 2:
In the given figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.
Given: (1) ABCD is a rectangle.
(2) CD = 6 cm
To find: Area of rectangle CDEF.
Calculation: We know that,
Area of parallelogram = base × height
The Area of parallelogram and a rectangle on the same base and between the same parallels are equal in area.
Here we can see that rectangle ABCD and Parallelogram CDEF are between the same base and same parallels.
Hence,
Hence we get the result as Area of Rectangle CDEF =
#### Question 3:
In the given figure, find the area of ΔGEF.
Given: (1) ABCD is a rectangle.
(2) CD = 6 cm
To find: Area of ΔGEF.
Calculation: We know that,
Area of Parallelogram = base × height
If a triangle and a parallelogram are on the same base and between the same parallels , the area of the triangle is equal to half of the parallelogram
Here we can see that Parallelogram ABCD and triangle GEF are between the same base and same parallels.
Hence,
Hence we get the result as
#### Question 4:
In the given figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ΔEFG.
Given: (1) ABCD is a rectangle.
(2) AB = 10 cm
To find: Area of ΔEGF.
Calculation: We know that,
Area of Rectangle = base × height
If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram
Here we can see that Rectangle ABCD and triangle GEF are between the same base and same parallels.
Hence,
Hence we get the result as Area of ΔGEF =
#### Question 5:
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar (ΔRAS)
Given: Here from the given figure we get
(1) PQRS is a rectangle inscribed in a quadrant of a circle with radius 10cm,
(2) PS = 5cm
To find: Area of ΔRAS.
Calculation: In right ΔPSR, (Using Pythagoras Theorem)
Hence we get the Area of ΔRAS =
#### Question 6:
In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8cm and PQ and BD intersect at O, then find area of ΔOPB.
Given: Here from the given question we get
(1) ABCD is a square,
(2) P is the midpoint of AB
(3) Q is the midpoint of CD
(4) PQ and BD intersect at O.
(5) AB = 8cm
To find : Area of ΔOPB
Calculation: Since P is the midpoint of AB,
BP = 4cm ……(1)
Hence we get the Area of ΔOBP =
#### Question 7:
ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. IF area of ΔABC is 16 cm2, find the area of ΔDEF.
Given: Here from the given question we get
(1) ABC is a triangle
(2) D is the midpoint of BC
(3) E is the midpoint of CD
(4) F is the midpoint of A
Area of ΔABC = 16 cm2
To find : Area of ΔDEF
Calculation: We know that ,
The median divides a triangle in two triangles of equal area.
For ΔABC, AD is the median
For ΔADC , AE is the median .
Similarly, For ΔAED , DF is the median .
Hence we get Area of ΔDEF =
#### Question 8:
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ΔPBQ is 17cm2, find the area of ΔASR.
Given: Here from the given figure we get
(1) PQRS is a trapezium having PS||QR
(2) A is any point on PQ
(3) B is any point on SR
(4) AB||QR
(5) Area of ΔBPQ = 17 cm2
To find : Area of ΔASR.
Calculation: We know that ‘If a triangle and a parallelogram are on the same base and the same parallels, the area of the triangle is equal to half the area of the parallelogram’
Here we can see that:
Area (ΔAPB) = Area (ΔABS) …… (1)
And, Area (ΔAQR) = Area (ΔABR) …… (2)
Therefore,
Area (ΔASR) = Area (ΔABS) + Area (ΔABR)
From equation (1) and (2), we have,
Area (ΔASR) = Area (ΔAPB) + Area (ΔAQR)
Area (ΔASR) = Area (ΔBPQ) = 17 cm2
Hence, the area of the triangle ΔASR is 17 cm2.
#### Question 9:
ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3.1. If ar (ΔPBQ) = 10cm2, find the area of parallelogram ABCD.
It is given that CQ : QP = 3: 1 and Area (PBQ) = 10 cm2
Let CQ = x and QP = 3x
We need to find area of the parallelogram ABCD.
From the figure,
Area (PBQ) =
And,
Area (BQC) =
Now, let H be the perpendicular distance between AP and CD. Therefore,
Area (PCB) = …… (1)
Thus the area of the parallelogram ABCD is,
Area (ABCD) = AB × H
Area (ABCD) = 2BP × H
From equation (1), we get
Area (ABCD) = 4 × 30 = 120 cm2
Hence, the area of the parallelogram ABCD is 120 cm2.
#### Question 10:
P is any point on base BC of ΔABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar (ΔABC) = 12 cm2, then find area of ΔEPC.
Given: Area (ABC) = 12 cm2, D is midpoint of BC and AP is parallel to ED. We need to find area of the triangle EPC.
Since, AP||ED, and we know that the area of triangles between the same parallel and on the same base are equal. So,
Area (APE) = Area (APD)
Area (APM) + Area (AME) = Area (APM) + Area (PMD)
Area (AME) = Area (PMD) …… (1)
Since, median divide triangles into two equal parts. So,
Area (ADC) = Area (ABC) = = 6 cm2
Area (ADC) = Area (MDCE) + Area (AME)
Area (ADC) = Area (MDCE) + Area (PMD) (from equation (1)) |
Question
# Farmer Jones must determine how many acres of com and wheat to plant this year. An acre of wheat yields 25 bushels of wheat and requires 10 hours of labor per week. An acre of corn yields 10 bushels of corn and requires 4 hours of labor per week. All wheat can be sold at $4 a bushel, and all com can be sold at$3 a bushel. Seven acres of land and 40 hours per week of labor are available. Government regulations requite that at least 30 bushels of corn be produced during the current year. Let x1 = number of acres of corn planted, and x2 = number of acres of wheat planted. Using these decision variables, formulate an LP whose solution will tell Farmer Jones how to maximize the total revenue from wheat and corn.
Solution
Verified
Step 1
1 of 2
$\text{\underline{\color{#4257b2}Decision variables}}$
The decision variables are already given.
$\text{\underline{\color{#4257b2}Objective function}}$
The profits need to be maximized. Since $x_1$ acres of wheat yields $25x_1$ bushels of wheat, which sell for $4$ dollars per bushel, so the revenue from $x_1$ acres of wheat is $4 \cdot 25x_1 = 100x_1$ dollars.
Similarly, the revenue from $x_2$ acres of corn is $3 \cdot 10 x_2 = 30 x_2$ dollars.
Therefore, the objective function is
$\max \ z = 100x_1 + 30x_2$
$\text{\underline{\color{#4257b2}Constraints}}$
Constraint 1:. Only 7 acres of land are available. Therefore, $x_1 + x_2 \leqslant 7$. Constraint 2:. 40 hours of labor per week are available. Since $x_1$ acres of wheat need $10x_1$ hours of labor per week, and $x_2$ acres of corn need $4x_2$ hours of labor per week, we have $10x_1 + 4x_2 \leqslant 40$. Constraint 3:. The farmer needs to produce at least 30 bushels of corn. Since each acre of corn produces $10$ bushels of corn, we have $10x_2 \geqslant 30$, which can be simplified to $x_2 \geqslant 3$.
$\text{\underline{\color{#4257b2}Sign restrictions}}$
Obviously, $x_1 \geqslant 0$ and $x_2 \geqslant 0$ (acres of corn and wheat must both be nonnegative).
$\text{\underline{\color{#4257b2}Conclusion}}$
Finally, we can write the LP problem as
\begin{align*} \max \ z &= 100x_1 + 30x_2 \tag{Objective function}\\ x_1 + \phantom{0}x_2 &\leqslant \phantom{0}7 \tag{Land restriction}\\ 10x_1 + 4x_2 &\leqslant 40 \tag{Labor restriction}\\ x_2 & \geqslant \phantom{0}3 \tag{Corn restriction}\\ x_1 \phantom{+0x_2} & \geqslant \phantom{0}0 \tag{Sign restriction} \\ x_2 & \geqslant \phantom{0}0 \tag{Sign restriction} \\ \end{align*}
## Recommended textbook solutions
#### Introduction to Operations Research
10th EditionFrederick S. Hillier
Verified solutions
#### Operations Research: Applications and Algorithms
4th EditionWayne L Winston
1,263 solutions
#### STAT: Behavioral Sciences
2nd EditionGary Heiman
#### A Mathematical Look at Politics
1st EditionDaniel H. Ullman, E. Arthur Robinson Jr. |
## Geometric Mean Calculator
Geometric Mean <a href="https://studysaga.in/calculator-studysaga/">Calculator</a>
# Geometric Mean Calculator
Enter two or more numbers to calculate their geometric mean:
In mathematics, there are several different types of means, including the arithmetic mean, harmonic mean, and geometric mean. In this post, we will focus on the geometric mean, which is a type of average that is often used in various fields of study, including statistics, finance, and physics.
The geometric mean is a type of average that is calculated by multiplying together a set of values and then taking the nth root of the product, where n is the number of values being multiplied. Mathematically, the formula for the geometric mean is:
GM = (x1 * x2 * x3 * … * xn)^(1/n)
where GM is the geometric mean, x1, x2, x3, …, xn are the values being multiplied, and n is the number of values being multiplied.
For example, if we have the values 2, 4, and 8, we can find the geometric mean using the formula:
GM = (2 * 4 * 8)^(1/3) = 4
This means that the geometric mean of the values 2, 4, and 8 is 4.
Formula
The formula for the geometric mean can also be expressed in terms of logarithms:
GM = 10^( (log10(x1) + log10(x2) + log10(x3) + … + log10(xn)) / n )
This formula may be useful when working with large or small numbers, as it can be easier to add the logarithms of the numbers rather than multiply them directly.
Applications
The geometric mean has many applications in various fields of study. Some examples include:
1. Finance: The geometric mean is often used in finance to calculate the average return on an investment over a period of time. This is because the geometric mean takes compounding into account, which is important when considering investment returns.
2. Statistics: The geometric mean is used in statistics to calculate the geometric standard deviation, which is a measure of the spread of a set of data.
3. Physics: The geometric mean is used in physics to calculate the root-mean-square (RMS) value of a set of values. For example, the RMS value of an alternating current is calculated using the geometric mean of the squares of the current values.
4. Biology: The geometric mean is used in biology to calculate the mean size of cells or other microscopic objects, as it takes into account the relative sizes of the objects being measured.
5. Engineering: The geometric mean is used in engineering to calculate the effective value of a quantity that varies over time, such as the voltage or current in an electrical circuit.
In all of these applications, the geometric mean provides a useful way to calculate an average value that takes into account the relative sizes of the values being averaged.
Here are some examples of the geometric mean:
1. Finding the average growth rate of an investment portfolio over multiple years.
2. Determining the average rate of change of a population over time.
3. Calculating the average interest rate earned on a set of investments with different interest rates.
4. Determining the average concentration of a solution over multiple measurements.
5. Finding the average distance traveled by a moving object in a given time interval.
6. Calculating the average size of particles in a given sample.
7. Determining the average value of a set of measurements that vary exponentially.
Here are some mathematical examples of the geometric mean:
1. Find the geometric mean of 2 and 8. Solution: Geometric mean = √(2 × 8) = √16 = 4.
2. Find the geometric mean of 3, 6, and 12. Solution: Geometric mean = ∛(3 × 6 × 12) = ∛216 = 6.
3. The altitude of a right-angled triangle is 8 cm and the base is 6 cm. What is the length of the hypotenuse? Solution: Using the Pythagorean theorem, we have: h² = 8² + 6² h² = 64 + 36 h² = 100 h = 10 The geometric mean of 8 and 6 is √(8 × 6) = √48. Thus, the length of the hypotenuse is √48 × 2 = 9.8 cm (approx).
4. If the lengths of the diagonals of a parallelogram are 8 cm and 12 cm, what is the length of each side? Solution: The diagonals of a parallelogram divide it into four triangles. Let a and b be the sides of one of these triangles, and c be the side of the parallelogram opposite to this triangle. Then, we have: c² = a² + b² (using the Pythagorean theorem) Also, we have: a × b = 2 × (area of the triangle) Using the formula for the area of a triangle, we get: a × b = 2 × (1/2) × 8 × 6 = 48 Thus, the geometric mean of 8 and 12 is √(8 × 12) = √96. Therefore, we can write: c = √(a² + b²) = √(a × b + b × a) = √(2ab) = √(2 × 48) = 6√2. So, each side of the parallelogram is 6√2 cm long.
5. Find the value of x such that the geometric mean of x and 24 is 48. Solution: The geometric mean of x and 24 is √(x × 24) = √(24x). According to the problem, this is equal to 48. Thus, we have: √(24x) = 48 Squaring both sides, we get: 24x = 48² 24x = 2304 x = 96. Therefore, the value of x is 96.
The geometric mean is a type of average that is calculated by multiplying together a set of values and then taking the nth root of the product. It has many applications in various fields of study, including finance, statistics, physics, biology, and engineering. By using the geometric mean, we can calculate an average value that takes into account the relative sizes of the values being averaged, which can be useful in a wide range of situations. |
## Where math comes alive
### How to Find the LCM for Three Numbers
Several readers have said they like my trick for finding the LCM described in the post “How to Find the LCM — FAST!” but wonder how to use the trick for finding the LCM for THREE numbers. Here is how you do that.
Essentially it involves using the same LCM trick three separate times. Here’s how it’s done.
Suppose the numbers for which you need to find the LCM are 6, 8, and 14.
Step 1) Find the LCM for the any two of those. Using 6 and 8, we find that their LCM = 24.
Step 2) Find the LCM for another pair from the three numbers. Using 8 and 14, we find that their LCM = 56.
Step 3) Find the LCM of the two LCMs, meaning that we find the LCM for 24 and 56. The LCM for those two numbers = 168.
And that, my good friends, is the LCM for the three original numbers.
So, to summarize. Find the LCM for two different pairs. Then find the LCM of the two LCMs. The answer you get is the LCM for the three numbers.
Here are a few problems that give you a chance to practice this technique.
Find the LCM for each trio of numbers.
a) 10, 25, 30
b) 16, 28, 40
c) 14, 32, 40 |
Question
# In a right angle triangle ABC, right-angled at B, if $\tan A = 1$ then verify that $2\sin A\cos A = 1$.
Hint: Here, we will verify $2\sin A\cos A = 1$by finding the values of $\sin A$ and $\cos A$ with the given $\tan A$ value.
Given,
In a right angle triangle ABC, right-angled at B i.e..,$\angle ABC = {90^0}$
And it is also given that $\tan A = 1$i.e.., $\dfrac{{BC}}{{AB}} = 1[\because \tan \theta = \dfrac{{opp}}{{Adj}}]$.
Therefore, $BC = AB$
Let, $AB = BC = k$ where ‘k’ is a positive number.
As we know that $A{C^2} = A{B^2} + B{C^2}[\because$Pythagoras Theorem]
Now let us substitute the value of AB, BC as ‘k’, we get
$\Rightarrow A{C^2} = {k^2} + {k^2} \\ \Rightarrow A{C^2} = 2{k^2} \\ \Rightarrow AC = \sqrt 2 k \\$
Now, let us find the value of $\sin A$ and $\cos A$.
$\Rightarrow \sin A = \dfrac{{opp}}{{hyp}} = \dfrac{{BC}}{{AC}} = \dfrac{k}{{\sqrt 2 k}} = \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \cos A = \dfrac{{adj}}{{hyp}} = \dfrac{{AB}}{{AC}} = \dfrac{k}{{\sqrt 2 k}} = \dfrac{1}{{\sqrt 2 }}$
Now, we need to verify $2\sin A\cos A = 1$ .Let us substitute the obtained $\sin A$ and $\cos A$ values.
$\Rightarrow 2\sin A.\cos A = 1 \\ \Rightarrow 2(\dfrac{1}{{\sqrt 2 }})(\dfrac{1}{{\sqrt 2 }}) = 1 \\ \Rightarrow \dfrac{2}{2} = 1 \\ \Rightarrow 1 = 1[\therefore L.H.S = R.H.S] \\$
Therefore, we verified that the value of $2\sin A\cos A$ is 1.
Note: The alternate approach to solve the given problem is by using the formula of double angle i.e..,$2\sin A\cos A = \sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$. |
# How do you solve -1 <= x-3 <= 5?
##### 2 Answers
Mar 16, 2017
Solve for $x$ by adding $+ 3$ to all three parts of the compound the inequality.
$2 \le x \le 8$
#### Explanation:
$- 1 \le x - 3 < - 5 \text{ }$add 3 to each part
$- 1 \textcolor{red}{+ 3} \le x - 3 \textcolor{red}{+ 3} \le 5 \textcolor{red}{+ 3} \text{ }$ Doing all of the addition results in
$+ 2 \le x \le 8 \text{ } x$ is now isolated in the middle
$2 \le x \le 8 \text{ }$ this is the final answer
Mar 16, 2017
$2 \le x \le 8$
#### Explanation:
$- 1 \le x - 3 \le 5$
You can break the inequality up into two parts, work with them separately, and combine them at the end.
$\textcolor{red}{- 1 \le x - 3} \le 5 \text{ "and" } - 1 \le \textcolor{b l u e}{x - 3 \le 5}$
$\textcolor{red}{- 1 \le x - 3} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \textcolor{b l u e}{x - 3 \le 5}$
Isolate x on the right side $\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .}$ isolate x on the left side
$\textcolor{red}{- 1 + 3 \le x} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \textcolor{b l u e}{x \le 5 + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{2 \le x} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} \textcolor{b l u e}{x \le 8}$
Now combine the two parts with one $x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \textcolor{red}{2 \le} x \textcolor{b l u e}{\le 8}$ |
Question
# A man leaves a town at 8 a.m. on his bicycle moving at 10 km/hr. Another man leaves the same town at 9 a.m. on his scooter moving at 30 km/hr. At what time does he overtake the man on the bicycle? A. $8:30am$ B. $9:00am$ C. $9:30am$ D. $10:00am$
Hint: We have to consider the fact that the distance traveled by bicycle and scooter until the point of overtaking will be the same. The formula for distance, say d is $d = s \times t$, where s is the speed and t is the time taken.
Let the time taken by the bicycle is $t$.
As mentioned in the question the scooter leaves the same town after 1 hour.
Therefore, time taken will be $t - 1$
Now, the distance is the product of speed and time i.e., $d = s \times t$ where s is the speed and t is the time taken.
For bicycle men, speed is $10km/hr$ and time is $t$.
Therefore, the distance traveled by bicycle man will be
$\Rightarrow d = 10 \times t$
Now, for scooter man, speed is $30km/hr$ and time is $(t - 1)$
Therefore, The distance traveled for scooter man will be
$\Rightarrow d = 30(t - 1)$
As we know, the distance traveled by bicycle and scooter at the point of overtaking will be the same.
$\Rightarrow d = 10t = 30(t - 1) \\ \Rightarrow 20t = 30 \\ \Rightarrow t = \dfrac{3}{2}hr \\$
The time taken by the scooter man is $t - 1 = \dfrac{3}{2} - 1 = \dfrac{1}{2}hr$.
The scooter overtakes the bicycle after half an hour. i.e., $9:30am$
So, the correct answer is “Option C”.
Note: These types of questions come under the Meeting point questions type. A different kind of question may be, If two people travel from two points A and B towards each other, and they meet at point T. The Total Distance covered by them at the meeting will be AB. The Time taken by both of them to meet will be the same. As the Time is constant, Distances AT and BT will be in the ratio of their Speed. Say that the Distance between A and B is d. If two people are walking towards each other from A and B, When they meet for the First time, they together cover a Distance “d” When they meet for the second time, they together cover a Distance “3d”. |
How do you solve the system of equations x + 8y = 26 and 2x + y = 5?
Aug 5, 2018
The solution is $\left(\frac{14}{15} , \frac{47}{15}\right)$.
Explanation:
Equation 1: $x + 8 y = 26$
Equation 2: $2 x + y = 5$
I am going to use elimination and substitution to answer this question.
Multiply Equation 1 by $- 2$.
$- 2 \left(x + 8 y\right) = 26 \times - 2$
$- 2 x - 16 y = - 52$
$- 2 x - 16 y = - 52$
$\textcolor{w h i t e}{. .}$$2 x + \textcolor{w h i t e}{. .} y = \textcolor{w h i t e}{\ldots \ldots} 5$
$- - - - - - -$
$- 15 y \textcolor{w h i t e}{\ldots \ldots \ldots} = - 47$
Divide both sides by $- 15$.
$y = \frac{- 47}{- 15}$
$y = \frac{47}{15}$
Substitute $\frac{47}{15}$ for $y$ in Equation 1 and solve for $x$.
$x + 8 \left(\frac{47}{15}\right) = 26$
$x + \frac{376}{15} = 26$
Subtract $\frac{376}{15}$ from both sides.
$x = - \frac{376}{15} + 26$
Multiply $26$ by $\frac{15}{15}$.
$x = - \frac{376}{15} + 26 \times \frac{15}{15}$
$x = - \frac{376}{15} + \frac{390}{15}$
$x = \frac{14}{15}$
Solution: $\left(\frac{14}{15} , \frac{47}{15}\right)$
graph{(x+8y-26)(2x+y-5)=0 [-10, 10, -5, 5]} |
Lesson Video: Area between Curves | Nagwa Lesson Video: Area between Curves | Nagwa
# Lesson Video: Area between Curves Mathematics
In this video, we will learn how to apply integration to find the area bounded by the curves of two or more functions.
15:16
### Video Transcript
In this video, weβll learn how to apply integration to find the area bounded by the curves of two or more functions. By this stage, you should feel confident in applying processes for integration to evaluate definite and indefinite integrals. Weβll now look at how integration can help us to find the area of the regions that lie between the graphs of two or more functions.
Consider the region that lies between the curve with equation π¦ equals π of π₯ and the π₯-axis which is bounded by the vertical lines π₯ equals π and π₯ equals π. Iβve shaded this region pink. If π is a continuous function, we know that we can evaluate the area of this region by integrating the function π of π₯ with respect to π₯ between the limits of π and π.
Now, letβs add another curve to our diagram. This time, the curve has the equation π¦ equals π of π₯, where π is continuous and π of π₯ is less than or equal to π of π₯ in the closed interval π to π. Once again, we can find the area of the region between the curve π¦ equals π of π₯ and the π₯-axis and these two vertical lines, now Iβve shaded this region in yellow, by evaluating the integral of π of π₯ between the limits of π and π.
We can now see that if we subtract the area between the curve π of π₯ and the π₯-axis from the area between the curve of π of π₯ and the π₯-axis, that weβll be left with this region π΄ three. This is the region between the two curves π¦ equals π of π₯ and π¦ equals π of π₯. We can, therefore, say that the area π΄ three of the region bounded by π¦ equals π of π₯ and π¦ equals π of π₯ is the integral of π of π₯ evaluated between π and π minus the integral of π of π₯ evaluated between π and π.
But we also know that the sum or difference of the integral of two functions is equal to the integral of the sum or difference of these functions. So, we can say that the area is equal to the integral of π of π₯ minus π of π₯ with respect to π₯ evaluated between π₯ equals π and π₯ equals π. This brings us to our first definition. The area π΄ of the region bounded by the curves π¦ equals π of π₯ and π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π, where π and π are continuous and π of π₯ is greater than or equal to π of π₯ in the closed interval π to π, is the definite integral of π of π₯ minus π of π₯ evaluated between the limits of π and π.
Notice here that π of π₯ is greater than or equal to π of π₯ for all π₯ in between and including π and π. Weβll need to watch carefully for situations where this is not the case and apply some extra logic. For now though, weβll look at the application of this formula.
Find the area of the region bounded by the curves π¦ equals three π₯ squared minus five π₯ and π¦ equals negative five π₯ squared.
Weβll recall that the area of the region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π for continuous functions π and π, such that π of π₯ is greater than or equal to π of π₯ for all π₯ in the closed interval π to π, is the definite integral between the limits of π and π of π of π₯ minus π of π₯. Weβre, therefore, going to need to define the functions π of π₯ and π of π₯ really carefully and, of course, the values for π and π, ensuring that π of π₯ is greater than π of π₯ in the closed interval π to π.
The lines π₯ equals π and π₯ equals π will mark the beginning and end of the region weβre interested in. So, what are the equations of these lines? Theyβre the π₯-coordinates at the points where the two graphs intercept. So, we can set the equations three π₯ squared minus five π₯ and negative five π₯ squared equal to each other and solve for π₯.
We begin by adding five π₯ squared to both sides. And then, we factor the expression on the left-hand side by taking out that factor of π₯. And we obtain π₯ times eight π₯ minus five to be equal to zero. We know that for this statement to be true, either π₯ itself must be equal to zero or eight π₯ minus five must be equal to zero. To solve this equation on the right, we add five and then divide through by eight. And we obtain π₯ to be equal to five-eighths.
So, we can see that the π₯-coordinates of the points of intersection of our two curves are zero and five-eighths. So, π is equal to zero and π is equal to five-eighths. Now, weβre going to need to decide which function is π of π₯ and which function is π of π₯. What we do next is sketch out the graphs of π¦ equals three π₯ squared minus five π₯ and π¦ equals negative five π₯ squared. Weβre looking to establish which of the curves is essentially on top.
We know that the graph of π¦ equals three π₯ squared minus five π₯ is a U-shaped parabola. We can even factor the expression three π₯ squared minus five π₯, set it equal to zero, and solve for π₯. And we see that it passes through the π₯-axis at zero and five-thirds. So, it will look a little something like this. The graph of π¦ equals negative five π₯ squared is an inverted parabola which passes through the origin like this. And so, we obtain the region shaded.
We can now see that in the closed interval of zero to five-eighths, the function thatβs on top, if you will, is the function defined by π¦ equals negative five π₯ squared. So, we can say that π of π₯ is equal to negative five π₯ squared. Meaning, π of π₯ is three π₯ squared minus five π₯. The area that weβre interested in must, therefore, be given by the definite integral evaluated between zero and five-eighths of negative five π₯ squared minus three π₯ squared minus five π₯ with respect to π₯.
Distributing the parentheses, and our integrand becomes negative eight π₯ squared plus five π₯. But wait a minute, we know that when we evaluate areas below the π₯-axis, we end up with a funny result. We get a negative value. You might wish to pause the video for a moment and consider what that means in this example. Did you work it out? We can see that our entire region sits below the π₯-axis and weβre just working out the difference between the areas. So, the negative results that we would obtain from integrating each function individually will simply cancel each other out. So, all thatβs left is to evaluate this integral.
The integral of negative eight π₯ squared is negative eight π₯ cubed over three. And the integral of five π₯ is five π₯ squared over two. We need to evaluate this between zero and five-eighths, which is negative eight-thirds of five-eighths cubed plus five over two times five-eighths squared minus zero. Thatβs 125 over 384 square units. This question was fairly straightforward as the curve of π¦ equals negative five π₯ squared was greater than or equal to the curve of π¦ equals three π₯ squared minus five π₯ in the interval weβre interested in.
Letβs now look at what we might do if this wasnβt the case.
The curves shown are π¦ equals one over π₯ and π¦ equals one over π₯ squared. What is the area of the shaded region? Give an exact answer.
Remember, the area of a region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π for continuous functions π of π, when π of π₯ is greater than or equal to π of π₯ for π₯ in the closed interval π to π, is given by the definite integral evaluated between π and π of π of π₯ minus π of π₯.
Now, we do have a little bit of a problem here. We can see quite clearly that the region is bounded by the vertical lines π₯ equals 0.5 and π₯ equals two. But in the closed interval π₯ from 0.5 to two, we can see that one of our functions is not always greater than or equal to the other. So, we canβt actually apply this definition. We can see, however, that if we split our region up a little more, we do achieve that requirement. Iβve added a third line at the point where the two curves intersect. This has the equation π₯ equals one.
In the closed interval 0.5 to one, the values on the red line are always greater than or equal to those on the green line. And in the closed interval π₯ between one and two, the reverse is true. So, all we do is split our region up and then add the values at the end. Letβs find the area of our first region, π
one. To do so, weβre going to need to double check which line is which. We can probably deduce that the red line is more likely to be one over π₯ squared. But letβs choose a coordinate pair and substitute these values in just to be safe.
We can see that the curve passes through the point with coordinates 0.5, 4. So, letβs substitute π₯ equals 0.5 into the equation π¦ equals one over π₯ squared. When we do, we get π¦ equals one over 0.5 squared, which is one over 0.25, which is four as required. So, the red line has the equation π¦ equals one over π₯ squared and the green line has equation π¦ equals one over π₯. And when evaluating the area of π
one, π of π₯ is therefore one over π₯ squared and π of π₯ is equal to one over π₯.
The area is, therefore, given by the definite integral between the limits of 0.5 and one of one over π₯ squared minus one over π₯. So, all thatβs left here is to evaluate this integral. This is much easier to do if we rewrite one over π₯ squared as π₯ to the power of negative two and then recall some standard results. To integrate π₯ to the power of negative two, we add one to the power and then divide by this new number. That gives us π₯ to the power of negative one over negative one, which is negative one over π₯. The integral of one over π₯, however, is the natural log of the absolute value of π₯. So, our integral is negative one over π₯ minus the natural log of the absolute value of π₯.
Weβre going to now evaluate this between π₯ equals 0.5 and π₯ equals one. Thatβs negative one over one minus the natural log of one minus negative one over 0.5 minus the natural log of 0.5. And notice, Iβve lost the symbol for the absolute value because one and 0.5 are already positive. The natural log of zero is one. Negative one over one is negative one. And negative one over 0.5 is two. Iβve also rewritten the natural log of 0.5 as the natural log of a half and distributed the parentheses. And this simplifies to one plus the natural log of one-half.
A really important skill, though, is to be able to spot when we can further simplify a logarithmic term. If we rewrite the natural log of a half as the natural log of two to the power of negative one. And then, use the fact that the natural log of π to the πth power is equal to π times the natural log of π. We see that the exact area of the first region π
one is one minus the natural log of two.
Letβs clear some space and repeat this process for region two. This time, the green line is above the red line, so weβre going to let π of π₯ be equal to one over π₯ and π of π₯ be equal to one over π₯ squared. Our area is the definite integral between one and two of one over π₯ minus one over π₯ squared which, when we integrate, gives us the natural log of the absolute value of π₯ plus one over π₯. Evaluating between the limits of one and two, and we get the natural log of two plus a half minus the natural log of one plus one, which is equal to the natural log of two minus a half.
We want to find the area of the whole region, so we add these two values. Itβs one minus the natural log of two plus the natural log of two minus one-half, which simplifies to one-half. The area of the shaded region is a half square units. In this example, we saw that the area formula can be applied to find the area between two curves where one curve is above the other for part of the integration interval and the opposite in the second part of the interval, as long as we remember to split the region up at the point where the curves intersect.
Weβll now see how we can develop this formula further to help us find the region bounded by three curves.
Find the area of the region bounded by the curves π¦ equals four minus π₯ squared, π¦ equals negative π₯, and π¦ equals the square root of π₯. Give your answer correct to one decimal place.
Remember, for continuous functions π and π, the area of the region bounded by the curves π¦ equals π of π₯, π¦ equals π of π₯, and the lines π₯ equals π and π₯ equals π, as long as π of π₯ is greater than or equal to π of π₯ in the closed interval π to π, is given by the integral evaluated between π and π of π of π₯ minus π of π₯. We are going to need to be a little bit careful here, as we have three curves. So, letβs begin by sketching this out and see what weβre dealing with.
The area enclosed between the three curves looks a little something like this. Now, if weβre really clever, we can actually use the definition we looked up before. We can split this region into the region above the π₯-axis and the region below the π₯-axis. We then can split this up a little bit further. We see that we have π
one, thatβs the region between the π₯-axis and the curve π¦ equals root π₯ between π₯ equals zero and π₯ equals π. Where π is the π₯-coordinate at the point of intersection of the curve π¦ equals root π₯ and π¦ equals four minus π₯ squared.
We then have π
two. Thatβs the region between π¦ equals four minus π₯ squared, π₯ equals π, and π₯ equals two. And the reason weβve chosen π₯ equals two as our upper limit is thatβs the π₯-value at the point at which the curve crosses the π₯-axis. We can even split our three up into two further regions to make life easier. But letβs deal first with the area of π
one and π
two. We need to work out the value of π.
We said itβs the π₯-coordinate at the point of intersection of the two curves four minus π₯ squared and root π₯. So, we set these equal to each other and solve for π₯. That gives us an π₯-value of 1.648, correct to three decimal places. So, π is equal to 1.648. We can either do this by hand or use our graphical calculators to evaluate each of these integrals. The area of π
one becomes 1.4104 and so on. And the area of π
two is 0.23326 and so on.
Letβs now consider the area of π
three. Itβs the integral between zero and two of negative π₯ evaluated with respect to π₯. We need to be a little bit careful here, since this is below the π₯-axis and therefore will yield a negative result on integration. In fact, it gives us negative two. So, we can say that the area is the absolute value of this. Itβs two. And notice, we could have actually used the formula for area of a triangle to work this area out.
Now, the area of π
four is the definite integral between two and π of four minus π₯ squared minus negative π₯. And here, π is the π₯-coordinate of the point of intersection of the lines π¦ equals negative π₯ and π¦ equals four minus π₯ squared. We can once again set four minus π₯ squared equal to negative π₯ and solve for π₯. And we find, correct to three decimal places, that they intersect at the point where π₯ equals 2.562. We type this into our calculator and we find that the area of this region is 0.59106. We find the total of these four values, which gives us 4.2347, which is 4.2 square units, correct to one decimal place.
In this video, weβve seen that we can use the formula area is equal to the definite integral between π and π of π of π₯ minus π of π₯ with respect to π₯ for continuous functions, π and π as long as π of π₯ is greater than or equal to π for all π₯ in the closed interval π to π. We also saw that for more complicated regions such as those bounded by three or more curves, those which involve regions above and below the π₯-axis, or those where π of π₯ and π of π₯ switch, we might need to split this region up a little further. |
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# How To Apply The Sine, Cosine, And Tangent Functions
26 Apr 2023
The sine, cosine, and tangent functions are important mathematical tools that are used in a variety of fields, including physics, engineering, and mathematics. These functions are part of the larger family of trigonometric functions and are defined in terms of the right-angled triangle sides. In this article, we will discuss how to apply the sine, cosine, and tangent functions to solve problems in trigonometry.
### Definition
First, let’s define the sine, cosine, and tangent functions. In a right-angled triangle, the sine function is described as the ratio of the side opposite to an angle’s length to the hypotenuse’s length. The cosine function is described as the ratio of the adjacent side’s length to the hypotenuse’s length. Finally, the tangent function is described as the ratio of the opposite side’s length to the adjacent side’s length.
### Application
To apply these functions, we first need to identify the angle we are interested in and the triangle sides that are relevant to that angle. For example, if we are interested in the angle θ, we need to identify the side opposite to θ, the side adjacent to θ, and the hypotenuse. We can then use the definitions of the sine, cosine, and tangent functions to find the values of these functions for the given angle.
Let’s consider an example problem to demonstrate how to apply these functions. Suppose we have a right-angled triangle with a hypotenuse of length ten and an angle θ of 30 degrees. We want to find the length of the side opposite to θ and the length of the side adjacent to θ.
First, one will be able to employ the definition of the sine function to find the length of the side opposite to θ. The sine of θ is described as the ratio of the opposite side’s length to the hypotenuse’s
length. Therefore, sin(θ) = opposite/hypotenuse. We know the value of the hypotenuse is 10, and the value of θ is 30 degrees, so we can substitute these values into the formula to get sin(30) = opposite/10. If we solve for the opposite side, we will get opposite = 10*sin(30) = 5.
Next, we can use the definition of the cosine function to find the length of the side adjacent to θ. The cosine of θ is described as the ratio of the adjacent side’s length to the hypotenuse’s length. Therefore, cos(θ) = adjacent/hypotenuse. We know the value of the hypotenuse is 10, and the value of θ is 30 degrees, so we can substitute these values into the formula to get cos(30) = adjacent/10. Solving for the adjacent side, we get adjacent = 10*cos(30) = 8.66.
Finally, we can use the definition of the tangent function to search for the value of the tangent of θ. The tangent of θ is described as the ratio of the length’s opposite side to the adjacent side’s length. Hence, tan(θ) = adjacent/opposite. We already know the values of the opposite and adjacent sides, so we can substitute them into the formula to get tan(30) = 5/8.66 = 0.58.
### Conclusion
To apply the sine, cosine, and tangent functions, we first need to identify the angle we are interested in and the triangle’s sides that are relevant to that angle. We can then use the definitions of these functions to search for the values of these functions for the given angle. The sine function gives the ratio of the length of the side that is opposite to the angle.
If you still find yourself scratching your head about these functions, why not attend classes at the best IB tuition centre in Singapore? We are staffed by dedicated tutors and provide a comprehensive curriculum for all students, so there’s no better place to overcome challenges O Level students face. Contact us today to learn more!
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# Unit conversion: Speed Practice examples
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## Practice examples
Remember that units for speed all look like
( !"#\$%&'( )
( !"#\$ )
. If you’re converting from one speed unit to another, say 20 km/h to m/s, write down the value you start with:
20 km h
Now look at the units you want to end up with. a) You want to convert km to m, and you know
1 km = 1000 m, or 1 =
1 km
1000 m
=
1000
1 km m
You started with km on the top, so you eliminate that by multiplying by the b) equation with km on the
bottom
. Hence you multiply your number by
!""" m
!
km
You want to convert h to s, and you know
1 h = 3600 s, or 1 =
1 h
3600 s
=
3600 s
1 h
.
You started with h on the bottom, so you eliminate that by multiplying by the equation with h on the
top
. Hence you multiply your number by
!
h
!"## s
So your final answer is:
20 km/h = 20 km h
×
!""" m
!
km
×
!
h
!"## s
=
!" × !"""
!"##
= 5 .
6 m/s
.
Have a go:
Convert each quantity to the units given:
1.
80 km/h = m/s
2.
500 km/day = m/s
3.
65 miles/h = km/h
(1 mile = 1.609 km)
4.
20 m/s = km/h
5.
90 cm/s = km/h
6.
5000 km/year = cm/s
7.
20 cm/s = km/day
8.
(
Harder
) Using a pedometer, you walk 3000 steps in 20 minutes, so your speed is
150 steps/min
. Each of your steps is
0 .
7 m
long. What is your speed?
150 steps/min = m/s = km/h
9.
(
From pre-lecture quiz
): A snail travels 0.02 km in a week. What is its average speed in metres per second?
## Answers
1.
80 km/h = 80 km h
×
!""" m
!
km
×
!
h
!"## s
=
!" × !"""
!"##
= 22 .
2 m/s
2.
500 km/day = 500 km day
×
!""" m
!
km
×
!
day
!" h
×
!
h
!"## s
=
!"" × !"""
!" × !" !!
= 5 .
8 m/s
3.
65 miles/h = 65 miles h
×
!
.
!"# km
!
mile
=
!" × !
.
!"#
!
= 105 km/h
4.
20 m/s = 20 m s
×
!
km
!""" m
×
!"## s
!
h
=
!" × !"##
!"""
= 72 km/h
5.
90 cm/s = 90 cm s
×
!
m
!"" cm
×
!
km
!""" m
×
!"## s
!
h
=
!" × !"##
!"" × !"""
= 3 .
2 km/h
Note: If you remember that 100,000 cm = 1 km, you can skip a step and write
90 cm/s = 90 cm s
×
!
km
!"" , !!!
cm
×
!"## s
!
h
=
!" × !"##
!"" , !!!
= 3 .
2 km/h
6.
5000 km/year
5000 km year
×
=
!"" , !!!
cm
!
km
×
!
year
!"# × !" × !"## s
=
!""" × !"" , !!!
!"# × !" × !"##
= 15.8 cm/s
7.
20 cm/s = 20 cm s
×
!
km
!"" , !!!
cm
×
!" × !"## s
!
day
=
!" × !" × !"##
!"" , !!!
= 17 .
3 km/day
8.
Using a pedometer, you walk 3000 steps in 20 minutes, so your speed is
150 steps/min
. Each of your steps is
0 .
7 m
long. What is your speed?
150 steps/min = 150 steps min
×
!
.
!
m
!
step
×
!
min
!" s
=
!"# × !
.
!
!"
= 1.75 m/s or
150 steps/min = 150 steps min
×
!
.
!
m
!
step
×
!
km
!""" m
×
!" min
!
h
=
!"# × !
.
!
× !"
!"""
= 6.3
km/h
9.
A snail travels 0.02 km in a week. What is its average speed in metres per second?
0 .
02 km week
=
=
0 .
02 km week
3 .
3 × 10
!
!
which we round to
3 × 10
!
!
m/s
×
1000
m/s
1
km m
×
1
7
week days
×
1 day
24 × 60 × 60 s
(since we were only given 1 significant figure in the question) = answer 1. |
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# How to Multiply Three Binomials?
TopBinomial is actually a type of polynomial. When there are two terms in a polynomial, it will be referred to as binomial. Binomials can also be obtained by summing up two monomials. Multiplication of three binomials can be easily obtained.
Binomials are those algebraic expressions which are comprised of two monomials. Binomials are considered as the simplest type of polynomials.
Monomial can be defined as expression that may consist of base, variable and exponent.
Here, we will discuss how to multiply three binomials. Given below are the steps to multiply three binomials:
1. First of all, take first two binomials and rewrite them.
2. Simple multiplication of first terms of both binomials is performed. (Multiply the bases and add the exponents)
3. Then, multiply the first term of the first binomial with the second term of the second binomial.
4. In the next step, second term of the first binomial is multiplied with the first term of the second binomial.
5. Finally, multiply both the second terms of two binomials. Whole multiplication will have result of multiplication of two binomials.
6. Now, take third binomial and multiply each term of obtained result with each of the term of the third binomial. Thus, the required result will be obtained.
Let us take an example $(a + 5) (a + 3) (a + 2)$.
Here, take first two polynomials $(a + 5) (a + 3)$ and multiply the terms as per steps given above.
$(a + 5) (a + 3) = a^2 + 8a + 15$
Now, multiply the obtained result with the third polynomial as shown below:
$(a^2 + 8a + 15) (a + 2) = (a^2) (a + 2) + (8a) (a + 2) + (15) (a + 2)$
= $a^3 + 2a^2 + 8a^2 + 16a + 15a + 30$
= $a^3 + 10a^2 + 31a + 30$ |
# Simplifying fractions
Lesson
Sometimes fractions are not written in the simplest way possible. This means that there is an equivalent fraction that uses simpler (in this case smaller) numbers.
Remember from our work in equivalent fractions, (see Is this the same as that? to refresh your memory), that some fractions have exactly the same value as another, like $\frac{1}{2}=\frac{2}{4}$12=24.
In reverse, we can could say that $\frac{2}{4}$24can be simplified to $\frac{1}{2}$12.
A simplified fraction is the equivalent fraction that has no common factors between the numerator and denominator.
We have already discussed common factors as well, if you would like a refresher go back to The Highs and Lows and Trees with Multiple Branches.
So lets look at some examples for simplifying fractions.
#### Example
##### Question 1
Simplify: $\frac{15}{25}$1525
Think: Identify common factors between 15 and 25. I can see that 15 and 25 both have a common factor of 5.
Do: I can rewrite $\frac{15}{25}$1525 as $\frac{3\times5}{5\times5}$3×55×5, and then I remove the common factor from the numerator and the denominator (top and bottom). This leaves $\frac{3}{5}$35. I check to make sure there are no more common factors, if not then this is the final answer.
Let's look at another.
##### Question 2
Simplify: $\frac{16}{40}$1640
Think: Identify common factors between 16 and 40.
Do: By using factor trees I could see that the common factors are $2\times2\times2=8$2×2×2=8.
So I can rewrite $\frac{16}{40}$1640 as $\frac{2\times8}{5\times8}$2×85×8, and then I can simplify the fraction to$\frac{2}{5}$25.
##### Question 3
Simplify $\frac{24}{42}$2442
##### Question 4
Simplify $\frac{40}{25}$4025 writing your answer as an improper fraction
### Outcomes
#### 5.NN1.06
Demonstrate and explain the concept of equivalent fractions, using concrete materials |
# Video: Pack 5 β’ Paper 3 β’ Question 11
Pack 5 β’ Paper 3 β’ Question 11
04:17
### Video Transcript
Solve five minus three plus π₯ all squared equals zero. Give your answers correct to three significant figures.
So what Iβm actually gonna do is show you a couple of methods how to solve this problem. For this first method, what Iβm gonna do is gonna rearrange and solve the equation. So the first step is to actually add three plus π₯ all squared to each side. And when we do that, we get five is equal to three plus π₯ all squared.
So now the next step is to actually square-root each side of the equation. So when I do that, I actually get plus or minus the square root of five, because we see we square-rooted five and the answer could be positive or negative, is equal to three plus π₯. And then if I actually subtract three from each side, I get plus or minus the square root of five minus three is equal to π₯. So therefore, we get the possible answers for π₯. So π₯ will be equal to root five minus three or negative root five minus three.
If we check back and look at the question what it wants, it says that it wants it to three significant figures. So therefore, we get π₯ is equal to negative 0.764 or negative 5.24. And there we have it. Weβve actually solved five minus three plus π₯ all squared equals zero to three significant figures. And it gives us the π₯ values of negative 0.764 or negative 5.24.
Okay, so thatβs the first method. And itβs the method Iβll probably use to solve this kind of problem. However, Iβm just gonna give you an alternative so you can check it as well. And Iβm gonna do that using the quadratic formula. And the quadratic formula says that if we have a quadratic in the form ππ₯ squared plus ππ₯ plus π equals zero, then π₯ is equal to negative π plus or minus the square root of π squared minus four ππ over two π.
Okay, so now, great, weβve got a quadratic formula. Letβs use it to actually check and find the answer to this equation. So weβve got five minus. And then weβve got three plus π₯ all squared equals zero. So first of all, what we need to do is actually expand the brackets. So what we have is five minus three plus π₯ multiplied by three plus π₯. So there are two brackets. And thatβs because, actually, we had three plus π₯ all squared. And this is equal to zero.
So we get five minus. And then itβs three multiplied by three, which is nine, and then three multiplied by π₯. That gives us plus three π₯. And then we have π₯ multiplied by three. So weβve got plus another three π₯. And then, finally, weβve got π₯ multiplied by π₯. That gives us π₯ squared. So weβve got plus π₯ squared.
So we now have five minus and then in brackets nine plus three π₯ plus three π₯ plus π₯ squared equals zero. So then what we have is five minus nine β because remember weβve got a minus in front of our brackets, so itβs like multiplying each term by minus one β minus six π₯, cause again we have plus three π₯ plus three π₯, which is plus six π₯, multiplied by minus one is minus six π₯ and then minus π₯ squared is equal to zero.
So then if we actually multiply each side of our equation by negative one, we get π₯ squared plus six π₯ plus four is equal to zero. Okay, great! So weβve now got it in the form ππ₯ squared plus ππ₯ plus π is equal to zero. So we can use the quadratic formula. So weβve got π is equal to one. π is equal to six. And π is equal to four.
So therefore, if we substituted into the quadratic formula, we get π₯ is equal to negative six plus or minus the square root of six squared minus four multiplied by one multiplied by four. And this is all divided by two multiplied by one. So therefore, we can say that π₯ is equal to negative six plus root 20 over two, or negative six minus root 20 over two.
So if you actually work this out on the calculator, you get π₯ is equal to negative 0.764 or π₯ is equal to negative 5.24. And both of these have been rounded to three significant figures. And if we check back our previous answer from the first way that we actually used to work this out, we get the same answers. So great! We can definitely say that the solution to five minus three plus π₯ all squared equals zero is π₯ is equal to negative 0.764 or π₯ is equal to negative 5.24. |
# We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function at a. Functions.
## Presentation on theme: " We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function at a. Functions."— Presentation transcript:
We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function at a. Functions with this property are called ‘continuous at a.’ LIMITS AND DERIVATIVES
Section 2.5 Continuity LIMITS AND DERIVATIVES In this section, we will: See that the mathematical definition of continuity corresponds closely with the meaning of the word continuity in everyday language.
A continuous process is one that takes place gradually, without interruption or abrupt change. CONTINUITY A function f is continuous at a number a if: Definition 1
Notice that Definition 1 implicitly requires three things if f is continuous at a : is defined, that is, a is in the domain of f. exists. . CONTINUITY
The definition states that f is continuous at a if f ( x ) approaches f ( a ) as x approaches a. Thus, a continuous function f has the property that a small change in x produces only a small change in f( x ). In fact, the change in f( x ) can be kept as small as we please by keeping the change in x sufficiently small. CONTINUITY
If f is defined near a, that is, f is defined on an open interval containing a, except perhaps at a, we say that f is discontinuous at a or that f has a discontinuity at a, if f is not continuous at a. CONTINUITY
Physical phenomena are usually continuous. For instance, the displacement or velocity of a vehicle varies continuously with time, as does a person’s height. However, discontinuities do occur in such situations as electric currents. See Example 6 in Section 2.2, where the Heaviside function is discontinuous at 0 because does not exist. CONTINUITY
Geometrically, you can think of a function that is continuous at every number in an interval as a function whose graph has no break in it. The graph can be drawn without removing your pen from the paper. CONTINUITY
The figure shows the graph of a function f. At which numbers is f discontinuous? Why? It looks as if there is a discontinuity when a = 1 because the graph has a break there. The official reason that f is discontinuous at 1 is that f(1) is not defined. CONTINUITY Example 1
The graph also has a break when a = 3. However, the reason for the discontinuity is different. Here, f(3) is defined, but does not exist (because the left and right limits are different). So, f is discontinuous at 3. CONTINUITY Example 1
What about a = 5? Here, f(5) is defined and exists (because the left and right limits are the same). However, So, f is discontinuous at 5. CONTINUITY Example 1
Now, let us see how to detect discontinuities when a function is defined by a formula. CONTINUITY
Where are each of the following functions discontinuous? a. b. c. CONTINUITY Example 2
Notice that f (2) is not defined. Therefore, f is discontinuous at 2. Later, we will see why f is continuous at all other numbers. CONTINUITY Example 2 a
Here, f (0) = 1 is defined. However, See Example 8 in Section 2.2. So, f is discontinuous at 0. CONTINUITY Example 2 b
Here, f (2) = 1 is defined and exists. However, Therefore, f is not continuous at 2. CONTINUITY Example 2 c
The figure shows the graphs of the functions in Example 2. In each case, the graph can not be drawn without lifting the pen from the paper, because a hole or break or jump occurs in the graph. CONTINUITY
The kind of discontinuity illustrated in parts (a) and (c) is called removable. We could remove the discontinuity by redefining f at just the single number 2. The function is continuous. CONTINUITY
The discontinuity in part (b) is called an infinite discontinuity. CONTINUITY
The discontinuities in a function like the one in the figure are called jump discontinuities. The function ‘jumps’ from one value to another. CONTINUITY
A function f is continuous from the right at a number a if and f is continuous from the left at a if CONTINUITY 2. Definition
A function f is continuous on an interval if it is continuous at every number in the interval. If f is defined only on one side of an endpoint of the interval, we understand ‘continuous at the endpoint’ to mean ‘continuous from the right’ or ‘continuous from the left.’ CONTINUITY 3. Definition
Show that the function is continuous on the interval [-1, 1]. CONTINUITY Example 4
If -1 < a < 1, then using the Limit Laws, we have: CONTINUITY Example 4
Thus, by Definition 1, f is continuous at a if -1 < a < 1. Similar calculations show that So, f is continuous from the right at -1 and continuous from the left at 1. Therefore, according to Definition 3, f is continuous on [-1, 1]. CONTINUITY Example 4
The graph of f is sketched in the figure. It is the lower half of the circle CONTINUITY Example 4
Instead of always using Definitions 1, 2, and 3 to verify the continuity of a function, as we did in Example 4, it is often convenient to use the next theorem. It shows how to build up complicated continuous functions from simple ones. CONTINUITY
If f and g are continuous at a, and c is a constant, then the following functions are also continuous at a : 1. f + g 2. f - g 3. cf 4. fg 5. f/g provided that g( a )≠ 0 CONTINUITY 4. Theorem
Each of the five parts of the theorem follows from the corresponding Limit Law in Section 2.3. For instance, we give the proof of part 1. CONTINUITY
Since f and g are continuous at a, we have: Therefore, This shows that f + g is continuous at a. Proof CONTINUITY
It follows from Theorem 4 and Definition 3 that, if f and g are continuous on an interval, then so are the functions f + g, f - g, cf, fg, and (if g is never 0) f/g. The following theorem was stated in Section 2.3 as the Direct Substitution Property. CONTINUITY
a. Any polynomial is continuous everywhere, that is, it is continuous on (- , ) b. Any rational function is continuous wherever it is defined, that is, it is continuous on its domain. CONTINUITY 5. Theorem
A polynomial is a function of the form where c 0, c 1, c 2 …., c n are constants. We know that and This equation is precisely the statement that the function f(x) = x m is a continuous function. CONTINUITY Proof a
Thus, by part 3 of Theorem 4, the function g(x) = cx m is continuous. Since P is a sum of functions of this form and a constant function, it follows from part 1 of Theorem 4 that P is continuous. CONTINUITY Proof a
A rational function is a function of the form where P and Q are polynomials. The domain of f is We know from Proof (a) that P and Q are continuous everywhere. Thus, by part 5 of Theorem 4, f is continuous at every number in D. CONTINUITY Proof b
As an illustration of Theorem 5, observe that the volume of a sphere varies continuously with its radius. This is because the formula shows that V is a polynomial function of r. CONTINUITY
Similarly, if a ball is thrown vertically into the air with a velocity of 50 ft/s, then the height of the ball in feet t seconds later is given by the formula h(t) = 50t - 16t 2. Again, this is a polynomial function. So, the height is a continuous function of the elapsed time. CONTINUITY
Knowledge of which functions are continuous enables us to evaluate some limits very quickly, as the following example shows. Compare it with Example 2(b) in Section 2.3. CONTINUITY
Find The function is rational. By Theorem 5, it is continuous on its domain, which is: Therefore, CONTINUITY Example 5
It turns out that most of the familiar functions are continuous at every number in their domains. For instance, Limit Law 10 is exactly the statement that root functions are continuous. CONTINUITY
From the appearance of the graphs of the sine and cosine functions, we would certainly guess that they are continuous. CONTINUITY
We know from the definitions of and that the coordinates of the point P in the figure are As, we see that P approaches the point (1, 0) and so and. Thus, CONTINUITY 6. Definition
Since and, the equations in Definition 6 assert that the cosine and sine functions are continuous at 0. The addition formulas for cosine and sine can then be used to deduce that these functions are continuous everywhere. CONTINUITY
It follows from part 5 of Theorem 4 that is continuous except where CONTINUITY
This happens when x is an odd integer multiple of. So, y = tan x has infinite discontinuities when and so on. CONTINUITY
The inverse function of any continuous one- to-one function is also continuous. Our geometric intuition makes it seem plausible. The graph of f -1 is obtained by reflecting the graph of f about the line y = x. So, if the graph of f has no break in it, neither does the graph of f -1. Thus, the inverse trigonometric functions are continuous. CONTINUITY
In Section 1.5, we defined the exponential function y = a x so as to fill in the holes in the graph of y = a x where x is rational. In other words, the very definition of y = a x makes it a continuous function on. Therefore, its inverse function is continuous on. CONTINUITY
The following types of functions are continuous at every number in their domains: Polynomials Rational functions Root functions Trigonometric functions Inverse trigonometric functions Exponential functions Logarithmic functions CONTINUITY 7. Theorem
Where is the function continuous? We know from Theorem 7 that the function y = ln x is continuous for x > 0 and y = tan -1 x is continuous on. Thus, by part 1 of Theorem 4, y = ln x + tan -1 x is continuous on. The denominator, y = x 2 - 1, is a polynomial—so, it is continuous everywhere. CONTINUITY Example 6
Therefore, by part 5 of Theorem 4, f is continuous at all positive numbers x except where x 2 - 1 = 0. So, f is continuous on the intervals (0, 1) and. CONTINUITY Example 6
Evaluate Theorem 7 gives us that y = sin x is continuous. The function in the denominator, y = 2 + cos x, is the sum of two continuous functions and is therefore continuous. Notice that this function is never 0 because for all x and so 2 + cos x > 0 everywhere. CONTINUITY Example 7
Thus, the ratio is continuous everywhere. Hence, by the definition of a continuous function, CONTINUITY Example 7
Another way of combining continuous functions f and g to get a new continuous function is to form the composite function This fact is a consequence of the following theorem. CONTINUITY
If f is continuous at b and, then In other words, Intuitively, Theorem 8 is reasonable. If x is close to a, then g(x) is close to b; and, since f is continuous at b, if g(x) is close to b, then f(g(x)) is close to f(b). CONTINUITY 8. Theorem
Evaluate As arcsin is a continuous function, we can apply Theorem 8: CONTINUITY Example 8
Let us now apply Theorem 8 in the special case where, with n being a positive integer. Then, and If we put these expressions into Theorem 8, we get: So, Limit Law 11 has now been proved. (We assume that the roots exist.) CONTINUITY
If g is continuous at a and f is continuous at g( a ), then the composite function given by is continuous at a. This theorem is often expressed informally by saying “a continuous function of a continuous function is a continuous function.” CONTINUITY 9. Theorem
Since g is continuous at a, we have Since f is continuous at b = g(a), we can apply Theorem 8 to obtain This is precisely the statement that the function h(x) = f(g(x)) is continuous at a—that is, is continuous at a. CONTINUITY Proof
Where are the following functions continuous? a. b. CONTINUITY Example 9
We have h(x) = f (g(x)), where and Now, g is continuous on since it is a polynomial, and f is also continuous everywhere. Thus, is continuous on by Theorem 9. CONTINUITY Example 9 a
We know from Theorem 7 that f(x) = ln x is continuous and g(x) = 1 + cos x is continuous (because both y = 1 and y = cos x are continuous). Therefore, by Theorem 9, F(x) = f(g(x)) is continuous wherever it is defined. Example 9 b CONTINUITY
Now, ln(1 + cos x) is defined when 1 + cos x > 0. So, it is undefined when cos x = -1. This happens when Thus, F has discontinuities when x is an odd multiple of and is continuous on the intervals between these values. CONTINUITY Example 9 b
An important property of continuous functions is expressed by the following theorem. Its proof is found in more advanced books on calculus. CONTINUITY
Suppose that f is continuous on the closed interval [ a, b] and let N be any number between f( a ) and f(b), where. Then, there exists a number c in ( a, b) such that f(c) = N. INTERMEDIATE VALUE THEOREM
The theorem states that a continuous function takes on every intermediate value between the function values f( a ) and f(b). INTERMEDIATE VALUE THEOREM
The theorem is illustrated by the figure. Note that the value N can be taken on once [as in (a)] or more than once [as in (b)]. INTERMEDIATE VALUE THEOREM
If we think of a continuous function as a function whose graph has no hole or break, then it is easy to believe that the theorem is true. INTERMEDIATE VALUE THEOREM
In geometric terms, it states that, if any horizontal line y = N is given between y = f( a ) and f(b) as in the figure, then the graph of f can not jump over the line. It must intersect y = N somewhere. INTERMEDIATE VALUE THEOREM
It is important that the function in the theorem be continuous. The theorem is not true in general for discontinuous functions. INTERMEDIATE VALUE THEOREM
One use of the theorem is in locating roots of equations—as in the following example. INTERMEDIATE VALUE THEOREM
Show that there is a root of the equation between 1 and 2. Let. We are looking for a solution of the given equation— that is, a number c between 1 and 2 such that f(c) = 0. Therefore, we take a = 1, b = 2, and N = 0 in the theorem. We have and INTERMEDIATE VALUE THEOREM Example 10
Thus, f(1) < 0 < f(2), that is, N = 0 is a number between f(1) and f(2). Now, f is continuous since it is a polynomial. Therefore, the theorem states that there is a number c between 1 and 2 such that f(c) = 0. In other words, the equation has at least one root in the interval (1, 2). INTERMEDIATE VALUE THEOREM Example 10
In fact, we can locate a root more precisely by using the theorem again. Since a root must lie between 1.2 and 1.3. A calculator gives, by trial and error, So, a root lies in the interval (1.22, 1.23). INTERMEDIATE VALUE THEOREM Example 10
We can use a graphing calculator or computer to illustrate the use of the theorem in Example 10. INTERMEDIATE VALUE THEOREM
The figure shows the graph of f in the viewing rectangle [-1, 3] by [-3, 3]. You can see that the graph crosses the x-axis between 1 and 2. INTERMEDIATE VALUE THEOREM
The figure shows the result of zooming in to the viewing rectangle [1.2, 1.3] by [-0.2, 0.2]. INTERMEDIATE VALUE THEOREM
In fact, the theorem plays a role in the very way these graphing devices work. A computer calculates a finite number of points on the graph and turns on the pixels that contain these calculated points. It assumes that the function is continuous and takes on all the intermediate values between two consecutive points. The computer therefore connects the pixels by turning on the intermediate pixels. INTERMEDIATE VALUE THEOREM
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# Difference of Sets using Venn Diagram
How to find the difference of sets using Venn diagram?
The difference of two subsets A and B is a subset of U, denoted by A – B and is defined by
A – B = {x : x ∈ A and x ∉ B}.
Let A and B be two sets. The difference of A and B, written as A - B, is the set of all those elements of A which do not belongs to B.
Thus A – B = {x : x ∈ A and x ∉ B} or A – B = {x ∈ A : x ∉ B}.
Clearly, x ∈ A – B
⇒ x ∈ A and x ∉ B
Similarly, the difference B – A is the set of all those elements of B that do not belongs to A.
Thus, B – A = {x : x ∈ A and x ∉ B} or A – B = {x ∈ B : x ∉ A}.
In particular, A – B = ∅ if A ⊂ B and A – B = A if A ∩ B = ∅.
The subset of A – B is also called the complement of B relative to A.
The difference A – B can be expressed in terms of the complement as A – b = A ∩ B’.
Properties of difference of sets:
1. A – (B ∩ C) = (A – B) ∪ (A – C)
2. A – (B ∪ C) = (A – B) ∩ (A – C)
Solved example to find the difference of sets using Venn diagram:
1. If A = {2, 3, 4, 5, 6, 7} and B = {3, 5, 7, 9, 11, 13}, then find (i) A – B and (ii) B – A.
Solution:
According to the given statement; A = {2, 3, 4, 5, 6, 7} and B = {3, 5, 7, 9, 11, 13}
(i) A – B
= {2, 4, 6}
(ii) B – A
= {9, 11, 13}
2. Given three sets A, B and C such that: A = {x : x is a natural number between 10 and 16}, B = {set of even numbers between 8 and 20} and C = {7, 9, 11, 14, 18, 20}.
Find the difference of sets using Venn diagram:
(i) A – B
(ii) B – C
(iii) C – A
(iv) B – A
Solution:
According to the given statement
A = {11, 12, 13, 14, 15}
B = {10, 12, 14, 16, 18}
C = {7, 9, 11, 14, 18, 20}
(i) A – B
= {Those elements of set A which are not in set B}
= {11, 13, 15}
(ii) B – C
= {Those elements of set B which are not in set C}
= {10, 12, 16}
(iii) C – A
= {Those elements of set C which are not in set A}
= {7, 9, 18, 20}
(iv) B – A
= {Those elements of set B which are not in set A}
= {10, 16, 18}
Set Theory
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A ball is thrown from the top of a building $45m$ high with a speed $20m{s}^{-1}$ above the horizontal at an angle of ${30}^{\circ }$. Find (a) The time taken by the ball to reach the ground. (b) The speed of ball just before it touches the ground.
Video Solution
Text Solution
Verified by Experts
Given $v=20m{s}^{-1},\theta ={30}^{\circ },H=45m$. (a) As the ball has been projected at an angle of ${30}^{\circ }$ above horizontally, so first of all we need to analyse the velocity horizontally and vertically. This will be useful while using distance - time relation in horizontal and vertical directions. ${v}_{xi}=v{\mathrm{cos}30}^{\circ }=20×\frac{\sqrt{3}}{2}=10\sqrt{3}m{s}^{-1}$ ${v}_{yi}=v{\mathrm{sin}30}^{\circ }=20×\frac{1}{2}=10m{s}^{-1}$ It will be easy for us to use distance - time relation in vertical as it will involve less calculation. In y - direction : $-45=10t-\frac{1}{2}×{\text{gt}}^{2}⇒{t}^{2}-2t=0$ which on solving gives $t=1+\sqrt{10}\right)s$ (positive value), (other value is $1-\sqrt{10}\right)s$, a negative value of time is not acceptable). (b) ${v}_{yf}=10-10×\left(1+\sqrt{10}\right)\right)=-10\sqrt{10}\right)m{s}^{-1}$ ${v}_{f}=\sqrt{{v}_{yf}^{2}+{v}_{xf}^{2}}=\sqrt{{\left(10\sqrt{10}\right)}^{2}+\left(10{\sqrt{3}}^{2}\right)}$ =$10\sqrt{3}\right)m{s}^{-1}$ .
|
Updated on:21/07/2023
Knowledge Check
• Question 1 - Select One
A ball is thrown horizontally with a speed of 20m/s from the top of a tower of height 100m Time taken by the ball to strike the ground is
A20s
B220s
C420s
D10s
• Question 2 - Select One
A cricket ball is thrown at a speed of 30 m s−1 in a direction 30∘ above the horizontal. The time taken by the ball to return to the same level is
A2 s
B3 s
C4 s
D5 s
• Question 3 - Select One
A ball is thrown straight upward with a speed v from a point h meter above the ground. The time taken for the ball to strike the ground is
Avg[1+1+2hgv2]
Bvg[112hgv2]
Cvg[11+2hgv2]
Dvg[2+2hgv2]
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# Limits and Continuity
## Limit
Suppose f (x) is a real valued function m is a real number. The expression
$$\lim_{x \rightarrow c} f(x)= l$$
This means that f( x ) can be taken very close to l. For example let us consider the function f( x ) in the
$$f(x) = \frac{x^2-1}{x-1}$$
So if you consider x=0.9 then f( 0.9 ) = $\frac{0.9^2 -1}{0.9-1}$ = 1.900. So, if you try some more values like
1. f ( 0.99 ) = 1.990,
2. f ( 0.999 ) = 1.999
3. f ( 1 ) = undefined.
4. f( 1.001 ) = 2.001
5. f( 1.01 ) = 2.010.
In the above situation, if we take the limit from lower values of x and the other from higher values of x then the function reaches the same value which is approximately equal to 2. so here we can say that the left hand limit is equal to right hand limit and also we say that the limit exists.
1. so f(x) = l = RHS lim exists x- c
2. f(x) = l =LHS lim exits,
Therefore, it is clear that the limit will exists only when right hand side limit is equal to left hand side limit.
## Continuity
Function y = f(x) is continuous at point x=b if the following three conditions are satisfied:
1. f(b) is defined ,
2. lim f( x ) exists
Lim f( x ) exists = f(b) is defined ,
Function f is said to be continuous on an interval I if f is continuous at each point x in I. Here is a list of some well-known facts related to continuity:
1. The sum is continuous
2. The difference is continuous
3. The PRODUCT of 2 continuous functions are continuous
4. The quotient of continuous functions is continuous at all points, given that denominator is not zero.
5. The composition of continuous functions is continuous at all points.
6. Any polynomial is continuous for all values of x.
7. Function ex, sinx, cosx are all continuous for all values of x.
The graph below shows the function which is continuous.
The graph of a discontinuous function
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+0
# Algebra
-1
2
1
+109
What are the coordinates of the points where the graphs of f(x)=x^3 + x^2 - 3x + 5 and g(x) = x^3 + 2x^2 intersect?
Give your answer as a list of points separated by commas, with the points ordered such that their -coordinates are in increasing order. (So "(1,-3), (2,3), (5,-7)" - without the quotes - is a valid answer format.)
May 3, 2024
#1
+180
+1
Alright! I geuss I can try to solve this question!
First, let's note that the only points where the lines intersect is when we have $$f(x) = g(x)$$
This means that we only have to solve the equation $$x^3 + x^2 -3x + 5 = x^3 + 2x^2$$ to find the points where the lines intersect.
Combining like terms and moving all the terms to one side, we get the equation $$x^2 + 3x - 5 = 0$$.
We can't factor this polynomial with integers, so the next best thing to do is to complete the square and find x that way.
We set up the equation by moving 5 to the other side, getting us $$x^2 + 3x = 5$$. Now we complete the square by adding 9/4 to both sides.
$$x^2 + 3x + 9/4 = 5 + 9/4$$
$$(x + 3/2)^2 = 29/4$$
Now we just have to square root both sides to get us $$x+3/2= \pm \sqrt{29/4}$$
This means that $$x = \frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}$$.
Subsituting these x values back into the g(x) and f(x) functions, we get $$y = \frac{-\sqrt{29}-3} 2, \frac{\sqrt{29}-3}{2}$$
So for the final answer, we have $$(\frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}), (-\frac{\sqrt{29}-3}{2}, \frac{\sqrt{29}-3}{2})$$
Thanks! :) :) :)
May 3, 2024 |
# Is a graph a function?
## Is a graph a function?
The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value.
## How do you tell if a graph represents a function?
Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.
## How do you write numbers in a business?
In business writing, a generally accepted rule is to spell out numbers from one to nine and use numerals for 10 and above. Some organisations change at 11 rather than 10. Numerals are usually used in scientific and technical writing, and increasingly I am seeing some organisations use numerals in all documents.
## Why is x2 y2 r2?
Thus, using the theorem of Pythagoras, x2 + y2 = r2 , and this is the equation of a circle of radius r whose centre is the origin O(0, 0). The equation of a circle of radius r and centre the origin is x2 + y2 = r2 .
## What are the rules of standard form?
The Standard Form for a linear equation in two variables, x and y, is usually given as Ax + By = C where, if at all possible, A, B, and C are integers, and A is non-negative, and, A, B, and C have no common factors other than 1.
## What is a standard form number?
Any number that we can write as a decimal number, between 1.0 and 10.0, multiplied by a power of 10, is said to be in standard form. 1.98 ✕ 10¹³; 0.76 ✕ 10¹³ are examples of numbers in standard form.
## What is the standard form of an equation of a circle?
First you need to know that the equation for a circle is (x-a)^2 + (y-b)^2 = r^2 where the center is at point (a,b) and the radius is r. so for instance (x-2)^2 + (y-3)^2 = 4 would have the center at (2,3) and have a radius of 2 since 4 = 2^2.
## How do you find standard form without a calculator?
Calculate ( 2 × 10 7 ) ÷ ( 8 × 10 2 ) .
1. Divide the first numbers: 2 ÷ 8 = 0.25.
2. Apply the Law of Indices on the powers.
3. 10 7 ÷ 10 2 = 10 7 − 2 = 10 5.
4. So: ( 2 × 10 7 ) ÷ ( 8 × 10 2 ) = 0.25 × 10 5.
## What is standard form formula?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y).
## Is a circle a function?
If you are looking at a function that describes a set of points in Cartesian space by mapping each x-coordinate to a y-coordinate, then a circle cannot be described by a function because it fails what is known in High School as the vertical line test. A function, by definition, has a unique output for every input
## How do you find standard form?
When adding or subtracting numbers written in standard form the key is first write the numbers as ordinary number. Then add or subtract them using a standard column method. To multiply or divide numbers in standard form rearrange the sum so the numbers and powers of 10 are dealt with separately.
## What is the rule about numbers in writing?
It is generally best to write out numbers from zero to one hundred in nontechnical writing. In scientific and technical writing, the prevailing style is to write out numbers under ten. While there are exceptions to these rules, your predominant concern should be expressing numbers consistently.
## What does Standard Form tell you about a graph?
An equation written in standard form is yet another equation that forms a parabola when graphed. Each letter in the standard form equation tells us a piece of information about the parabola, just like the letters from the vertex form equation had. “a”, can also tell us the width of a parabola.
## Is every line a function?
1 Answer. No, every straight line is not a graph of a function. Nearly all linear equations are functions because they pass the vertical line test. The exceptions are relations that fail the vertical line test
## Which line is not a function?
The vertical line test is used to determine if a graph of a relationship is a function or not. if you can draw any vertical line that intersects more than one point on the relationship, then it is not a function.
## What is the standard form of 200000?
200,000? 200,000 = 2 x 10^5.
## How do you graph standard form?
First, find the intercepts by setting y and then x equal to zero. This is pretty straightforward since the line is already in standard form. Plot the x and y-intercepts, which in this case is (9, 0) and (0, 6) and draw the line on the graph paper!
## Whats a function on a graph?
The graph of the function is the set of all points (x,y) in the plane that satisfies the equation y=f(x) y = f ( x ) . If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that x value has more than one output.
## What is the rule of a function?
A function rule describes how to convert an input value (x) into an output value (y) for a given function. An example of a function rule is f(x) = x^2 + 3
## What does Standard Form mean in math?
more A general term meaning “written down in the way most commonly accepted” It depends on the subject: • For numbers: in Britain it means “Scientific Notation”, in other countries it means “Expanded Form” (such as 125 = 100+20+5)
## Can a function be a straight line?
A linear function is a function whose graph is a straight line. The line can’t be vertical, since then we wouldn’t have a function, but any other sort of straight line is fine.
## How do you write numbers in professional writing?
A simple rule for using numbers in writing is that small numbers ranging from one to ten (or one to nine, depending on the style guide) should generally be spelled out. Larger numbers (i.e., above ten) are written as numerals.
## What does standard form look like?
An equation in standard form looks like ax + by = c; in other words, the x and y terms are on the left side of the equation and the constant is on the right side.
## How do you find the vertex in standard form?
To find the vertex, you need to find the x- and y-coordinates. To find the y-coordinate of the vertex, substitute the value for x into the equation and solve for y . Example: Find the vertex of y=x2+4x−9 , where: a=1 , b=4 , and c=−9 .
## How can you identify a function?
Relations can be written as ordered pairs of numbers or as numbers in a table of values. By examining the inputs (x-coordinates) and outputs (y-coordinates), you can determine whether or not the relation is a function. Remember, in a function each input has only one output. A couple of examples follow.
## What is not a function?
A function is a relation in which each input has only one output. In the relation , y is a function of x, because for each input x (1, 2, 3, or 0), there is only one output y. x is not a function of y, because the input y = 3 has multiple outputs: x = 1 and x = 2. |
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Problem 830
# Find the surface area of a regular tetrahedron when each edge is of length a) $$1 ;$$ b) 2 .
Expert verified
a) The surface area of a regular tetrahedron with edge length 1 is $$\sqrt{3}$$ square units. b) The surface area of a regular tetrahedron with edge length 2 is $$4\sqrt{3}$$ square units.
See the step by step solution
## Step 1: Recall the formula for the area of an equilateral triangle
For an equilateral triangle with side length $$s$$, the area (A) can be computed using the formula: $$A = \frac{\sqrt{3}}{4}s^2$$.
## Step 2: Calculate the area of one equilateral triangular face for each case
a) When the side length ($$a$$) is 1, the area of one equilateral triangle is: $$A_1 = \frac{\sqrt{3}}{4}(1)^2 = \frac{\sqrt{3}}{4}$$ b) When the side length ($$a$$) is 2, the area of one equilateral triangle is: $$A_2 = \frac{\sqrt{3}}{4}(2)^2 = \frac{4\sqrt{3}}{4} = \sqrt{3}$$
## Step 3: Find the surface area of the regular tetrahedron for each case
Since there are four equilateral triangular faces in a tetrahedron, we need to multiply the area of one face by 4. a) Surface area of the tetrahedron with side length 1: $$SA_1 = 4A_1 = 4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}$$ b) Surface area of the tetrahedron with side length 2: $$SA_2 = 4A_2 = 4(\sqrt{3}) = 4\sqrt{3}$$
## Step 4: Results
a) The surface area of a regular tetrahedron with edge length 1 is $$\sqrt{3}$$ square units. b) The surface area of a regular tetrahedron with edge length 2 is $$4\sqrt{3}$$ square units.
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# How to solve cube roots
We will explore How to solve cube roots can help students understand and learn algebra. We will also look at some example problems and how to approach them.
## How can we solve cube roots
Solving for an exponent can be tricky, but there are a few tips that can help. First, make sure to identify the base and the exponent. The base is the number that is being multiplied, and the exponent is the number of times that it is being multiplied. For example, in the equation 8 2, the base is 8 and the exponent is 2. Once you have identified the base and exponent, you can begin to solve for the exponent. To do this, take the logarithm of both sides of the equation. This will allow you to move the exponent from one side of the equation to the other. For example, if you take the logarithm of both sides of 8 2 = 64, you getlog(8 2) = log(64). Solving this equation for x gives you x = 2log(8), which means that 8 2 = 64. In other words, when solving for an exponent, you can take the logarithm of both sides of the equation to simplify it.
Factoring is the process of breaking down an equation into smaller pieces in order to solve it. For example, the equation x^2+5x+6 can be factored as (x+3)(x+2). While this may seem like a simple task, factoring can be quite challenging, particularly when equations are complex. However, with practice, most students can develop the skills necessary to factor equations successfully. As with any skill, mastering factoring can take time and effort, but the rewards are well worth it.
Solving for x with fractions can be tricky, but there are a few steps that can make the process simpler. First, it is important to understand that when solving for x, the goal is to find the value of x that will make the equation true. In other words, whatever value is plugged into the equation in place of x should result in a correct answer. With this in mind, the next step is to create an equation using only fractions that has the same answer no matter what value is plugged in for x. This can be done by cross-multiplying the fractions and setting the two sides of the equation equal to each other. Once this is done, the final step is to solve for x by isolating it on one side of the equation. By following these steps, solving for x with fractions can be much less daunting. |
# What is a Two-Step Equation? An equation written in the form Ax + B = C.
## Presentation on theme: "What is a Two-Step Equation? An equation written in the form Ax + B = C."— Presentation transcript:
What is a Two-Step Equation? An equation written in the form Ax + B = C
a) 3x – 5 = 16 b) y/4 + 3 = 12 c) 5n + 4 = 6 d) n/2 – 6 = 4
1. Solve for any Addition or Subtraction on the variable side of equation by “undoing” the operation from both sides of the equation. 2. Solve any Multiplication or Division from variable side of equation by “undoing” the operation from both sides of the equation. PEMDAS in Reverse…
Addition Subtraction Multiplication Division
Identify what operations are on the variable side. (Add, Sub, Mult, Div) “Undo” each operation by using the opposite operation. Whatever you do to one side, you must do to the other side to keep the equation balanced.
4x – 5 = 11 +5 = +5(Add 5 to both sides) 4x = 16(Simplify) 4 = 4(Divide both sides by 4) x = 4(Simplify)
2x – 5 = 17 3y + 7 = 25 5n – 2 = 38 12b + 4 = 28
x = 11 y = 6 n = 8 b = 2
x 3 + 4 = 9 - 4 = - 4 (Subtract 4 from both sides) x 3 = 5 (Simplify) (x 3) 3 = 5 3 (Multiply by 3 on both sides) x = 15(Simplify)
x 5 – 3 = 8 c 7 + 4 = 9 r 3 – 6 = 2 d 9 + 4 = 5
x = 55 c = 35 r = 24 d = 9
1. Make sure your equation is in the form Ax + B = C 2. Keep the equation balanced. 3. Use opposite operations to “undo” Follow the rules (PEMDAS in Reverse): 1.Undo Addition or Subtraction 2.Undo Multiplication or Division
What is a Multi-Step Equation? A multi-step equations takes more than two steps to solve. The same rules for two-step equations apply, but now you have equations that require you to do additional work BEFORE you begin to isolate the variable.
a) 8x - 3x - 10 = 20 b) 7x + 2(x + 6) = 39 c) (3x + 5) = -24
1. Start by simplifying one or both sides of the equation. This may require: a.Combining like terms b.Using the distributive property c.Multiplying by a reciprocal 2. Use inverse operations (PEMDAS in reverse) to isolate the variable.
8x – 3x – 10 = 20Original equation 5x – 10 = 20Combine like terms +10 = +10Add 10 to both sides 5x = 30Simplify 5 = 5Divide both sides by 5 x = 6Simplify
7x + 2(x + 6) = 39Original equation 7x + 2x + 12 = 39 Distributive property 9x + 12 = 39Combine like terms -12 = -12Subtract 12 from both sides 9x = 27Simplify 9 = 9Divide both sides by 9 x = 3Simplify
(3/2)(3x + 5) = -24Original equation · (2/3) = · (2/3)Multiply by reciprocal 3x + 5 = -16Simplify (-24/1 * 2/3 = -48/3 = -16) -5 = -5Subtract 5 from both sides 3x = -21Simplify 3 = 3Divide both sides by 3 x = -7Simplify
12v + 14 + 10v = 80 3 + 4(z + 5) = 31 5h + 2(11 – h) = -5 (3/2)(x – 5) = -6
v = 3 z = 2 h = -9 x = 1
TB pp. 144-145 #1, 3, 7, 11, 15, 19, 21, 22, 25, 27, 31, 35, 40 TB pp. 150-151 #1, 5, 11, 13, 17, 18, 19, 23, 27, 31
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# 3880 in words
3880 in words is written as Three Thousand Eight Hundred and Eighty. 3880 represents the count or value. The article on Counting Numbers can give you an idea about count or counting. The number 3880 is used in expressions that relate to money, distance, length, year and others. Let us consider an example for 3880. ”A remote village has Three Thousand Eight Hundred and Eighty houses”.
3880 in words Three Thousand Eight Hundred and Eighty Three Thousand Eight Hundred and Eighty in Numbers 3880
## How to Write 3880 in Words?
We can convert 3880 to words using a place value chart. The number 3880 has 4 digits, so let’s make a chart that shows the place value up to 4 digits.
Thousands Hundreds Tens Ones 3 8 8 0
Thus, we can write the expanded form as:
3 × Thousand + 8 × Hundred + 8 × Ten + 0 × One
= 3 × 1000 + 8 × 100 + 8 × 10 + 0 × 1
= 3880
= Three Thousand Eight Hundred and Eighty.
3880 is the natural number that is succeeded by 3879 and preceded by 3881.
3880 in words – Three Thousand Eight Hundred and Eighty.
Is 3880 an odd number? – No.
Is 3880 an even number? – Yes.
Is 3880 a perfect square number? – No.
Is 3880 a perfect cube number? – No.
Is 3880 a prime number? – No.
Is 3880 a composite number? – Yes.
## Solved Example
1. Write the number 3880 in expanded form
Solution: 3 × 1000 + 8 × 100 + 8 × 10 + 0 × 1
We can write 3880 = 3000 + 800 + 80 + 0
= 3 × 1000 + 8 × 100 + 8 × 10 + 0 × 1
## Frequently Asked Questions on 3880 in words
Q1
### How to write 3880 in words?
3880 in words is written as Three Thousand Eight Hundred and Eighty.
Q2
### Is 3880 a perfect square number?
No. 3880 is not a perfect square number.
Q3
### Is 3880 a prime number?
No. 3880 is not a prime number. |
## Step-by-step explanation:
We are offered a straight equation in terms of variable x and also y as follows:
Now, we space asked to find the worth of x because that the given value that y.
You are watching: What is the value of x in the equation x – y = 30, when y = 15?
i.e. We need to evaluate the offered equation for the given value of y.
Now we room given: y= 15
so by putting this worth in equation (1) we have:
On including 15 ~ above both the sides of the equation us have:
A)12 months in a year52 weeks in a year\$10 x 52 main = \$520 because that lottery tickets\$400 x 12 month = (\$400 x 10) + (400 x 2) = 4000 + 800 = \$4800 because that rentadding increase the yearly complete for both \$520+\$4800 = \$5320b) 4800 split by 100 = 48 = 1% of rental 520 separated by 48 = 10.8% (11% if rounding)
ou occupational in the packaging room of a company that renders soup. The old soup deserve to is 4 inch high and has a diameter that 3 inch
katifund.org:
Step-by-step explanation:
Remark
This question is a little ambiguous. You can not it is in sure just how to act the Chiefs. Encompass their 10 or not. Transforms out friend do.
So the total score was 45
But this deserve to be resolved without making use of that information. The vital is that the total variety of plays made to be 13
Givens
Let the touchdowns = x
Let the allude after = y
Let the field goals = z
The variety of point afters = the touchdowns. X = y
The number of touchdowns is 6 time the field goals: z = x / 6
Equation.
x + y + z = 13 substitute x for y
Solution
x + x + z = 13 incorporate the left: instead of x/6 because that z
2x + x/6 = 13 main point by 6 to eliminate the fraction
6*2x + x = 13 * 6 combine the left
12x + x = 13*6 integrate the left
13x = 13 * 6 division by 13 ~ above both sides.
13x/13 = 6
x = 6
====================
katifund.org
x = 6
y = 6
z = 1
Check
It"s time to see what happened.
See more: A Colonist Who Could Be Ready To Fight In A Minute’S Notic, Battles Of Lexington And Concord (Article)
6 touchdowns = 6 * 6 = 36
6 solitary points = 6 * 1 = 6
1 ar goal = 3
Total 45
Apparently we obtained the right katifund.org. Friend made one little error in the last equation. Z must equal x/6
Very amazing problem. Sports stat males are going come love this question. Thanks for posting. |
# The Law of Cosines
We will discuss here about the law of cosines or the cosine rule which is required for solving the problems on triangle.
In any triangle ABC, Prove that,
(i) b$$^{2}$$ = c$$^{2}$$ + a$$^{2}$$ - 2ca. cos B or, cos B = $$\frac{c^{2} + a^{2} - b^{2}}{2ca}$$
(ii) a$$^{2}$$ = b$$^{2}$$ + c$$^{2}$$ - 2ab. cos A or, cos A = $$\frac{b^{2} + c^{2} - a^{2}}{2bc}$$
(iii) c$$^{2}$$ = a$$^{2}$$ + b$$^{2}$$ - 2ab. cos C or, cos C = $$\frac{a^{2} + b^{2} - c^{2}}{2ab}$$
Proof of the law of cosines:
Let ABC is a triangle. Then the following three cases arise:
Case I: When the triangle ABC is acute-angled:
Now form the triangle ABD, we have,
cos B = BD/BC
⇒ cos B = BD/c
⇒ BD = c cos B ……………………………………. (1)
Again from the triangle ACD, we have
cos C = CD/CA
⇒ cos C = CD/b
⇒ CD = b cos C
By using the Pythagoras theorem on the triangle ACD, we get
AC$$^{2}$$ = AD$$^{2}$$ + CD$$^{2}$$
⇒ AC$$^{2}$$ = AD$$^{2}$$ + (BC - BD)$$^{2}$$
⇒ AC$$^{2}$$ = AD$$^{2}$$ + BC$$^{2}$$ + BD$$^{2}$$ - 2 BC ∙ BD
⇒ AC$$^{2}$$ = BC$$^{2}$$ + (AD$$^{2}$$ + BD$$^{2}$$) - 2 BC ∙ BD
⇒ AC$$^{2}$$ = BC$$^{2}$$ + AB$$^{2}$$ - 2 BC ∙ BD, [Since From triangle, we get, AD$$^{2}$$ + BD$$^{2}$$ = AB$$^{2}$$]
⇒ b$$^{2}$$ = a$$^{2}$$ + c$$^{2}$$ - 2a ∙ c cos B, [From (1)]
⇒ b$$^{2}$$ = c$$^{2}$$ + a$$^{2}$$ - 2ca cos B or, cos B = $$\frac{c^{2} + a^{2} - b^{2}}{2ca}$$
Case II: When the triangle ABC is obtuse-angled:
The triangle ABC is obtuse angled.
Now, draw AD from A which is perpendicular to produced BC. Clearly, D lies on produced BC.
Now from the triangle ABD, we have,
cos (180° - B) = BD/AB
⇒- cos B = BD/AB, [Since, cos (180° - B) = - cos B]
⇒ BD = -AB cos B
⇒ BD = -c cos B ……………………………………. (2)
By using the Pythagoras theorem on the triangle ACD, we get
AC$$^{2}$$ = AD$$^{2}$$ + CD$$^{2}$$
⇒ AC$$^{2}$$ = AD$$^{2}$$ + (BC + BD)$$^{2}$$
⇒ AC$$^{2}$$ = AD$$^{2}$$ + BC$$^{2}$$ + BD$$^{2}$$ + 2 BC ∙ BD
⇒ AC$$^{2}$$= BC$$^{2}$$+ (AD^2 + BD^2) + 2 BC ∙ BD
⇒ AC$$^{2}$$ = BC$$^{2}$$ + AB$$^{2}$$ + 2 BC ∙ BD, [Since From triangle, we get, AD$$^{2}$$ + BD$$^{2}$$ = AB$$^{2}$$]
⇒ b$$^{2}$$ = a$$^{2}$$ + c$$^{2}$$ + 2a ∙ (-c - cos B), [From (2)]
⇒ b$$^{2}$$ = c$$^{2}$$ + a$$^{2}$$ - 2ca cos B or, cos B = $$\frac{c^{2} + a^{2} - b^{2}}{2ca}$$
Case III: Right angled triangle (one angle is right angle): The triangle ABC is right angled. The angle B is a right angle.
Now by using the Pythagoras theorem we get,
b$$^{2}$$ = AC$$^{2}$$ = BC$$^{2}$$ + BA$$^{2}$$ = a$$^{2}$$ + c$$^{2}$$
⇒ b$$^{2}$$ = a$$^{2}$$ + c$$^{2}$$
⇒ b$$^{2}$$ = a$$^{2}$$ + c$$^{2}$$ - 2ac cos B, [We know that cos 90° = 0 and B = 90°. Therefore, cos B = 0] or, cos B = $$\frac{c^{2} + a^{2} - b^{2}}{2ca}$$
Therefore, in all three cases, we get,
b$$^{2}$$ = a$$^{2}$$ + c$$^{2}$$ - 2ac cos B or, cos B = $$\frac{c^{2} + a^{2} - b^{2}}{2ca}$$
Similarly, we can prove that the formulae (ii) a$$^{2}$$ = b$$^{2}$$ + c$$^{2}$$ - 2ab. cos A or, cos A = $$\frac{b^{2} + c^{2} - a^{2}}{2bc}$$ and (iii) c$$^{2}$$ = a$$^{2}$$ + b$$^{2}$$ - 2ab. cos C or, cos C = $$\frac{a^{2} + b^{2} - c^{2}}{2ab}$$.
Solved problem using the law of Cosines:
In the triangle ABC, if a = 5, b = 7 and c = 3; find the angle B and the circum-radius R.
Solution:
Using the formula, cos B = $$\frac{c^{2} + a^{2} - b^{2}}{2ca}$$ we get,
cos B = $$\frac{3^{2} + 5^{2} - 7^{2}}{2 ∙ 3 ∙ 5}$$
cos B = $$\frac{9 + 25 - 49}{30}$$
cos B = - 1/2
cos B = cos 120°
Therefore, B = 120°
Again, if R be the required circum-radius then,
b/sin B = 2R
⇒ 2R = 7/sin 120°
⇒ 2R = 7 ∙ 2/√3
Therefore, R = 7/√3 = (7√3)/3 units. |
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## Day 104 - Lesson 7.3
##### Learning Targets
• Check the Random and Large Counts conditions for constructing a confidence interval for a population proportion.
• Determine the critical value for calculating a C% confidence interval for a population proportion using Table A or technology.
• Calculate a C% confidence interval for a population proportion.
##### Activity:
For today’s lesson, you will need A LOT of Hershey’s kisses. Each group needs 50 kisses. If you’re teaching this lesson after February 14, we suggest hitting up stores to buy some clearance Valentine’s Day kisses on February 15. We were able to get them half off! Groups will be tossing the kisses and keeping track of the proportion of kisses that land flat on the bottom, as opposed to on their side. From there they will create a 95% confidence interval for the true proportion of Hershey kisses that land flat.
When students are working through the activity, draw a number line on the board going from to 1, with tick marks every 0.1. After the groups have calculated their confidence interval, have one person from each group come up to the front and draw their interval under the scale. After going over the activity, reveal that the true proportion (as best as we can guess) is p = 0.35. Draw a vertical line through the drawn intervals showing this. Approximately 95% of the class intervals should have captured the true proportion. This is a great time to review the interpretation of a confidence level from lesson 7.2.
The structure and questioning in this activity are scaffolded really nicely. Students have all the information they need since they’ve already learned about margin of error in chapter 3 and about the standard deviation of a sampling distribution of a sample proportion in chapter 6. However, make sure you point out important information to students. When going over questions #4 and 5, add to the margin that these are the RANDOM and NORMAL conditions which must be met in order to create a confidence interval. After question #6, identify that 2 standard deviations is called a CRITICAL VALUE or z*. We add the critical values for 90%, 95% and 99% in the margin.
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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2
More Exercises
Question 1.
A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks.
Solution:
Arranging in the ascending order, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8
Here, n = 11 i.e. odd,
The middle term = $$\\ \frac { n+1 }{ 2 }$$ = $$\\ \frac { 11+1 }{ 2 }$$ = $$\\ \frac { 12 }{ 2 }$$ = 6th term
Median = 5
Question 2.
(a) Find the median of the following set of numbers : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 (1990)
(b)For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.
Solution:
(a) Arranging in ascending order :
0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9
Here, n = 12 which is even
Question 3.
Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5
Solution:
Writing in ascending order 0, 1, 1, 2, 2, 3, 3, 3, 4, 5
Here, n = 10 which is even
Question 4.
The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Observation are :
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
n = 9
Median = $$\\ \frac { 9+1 }{ 2 }$$ th term
i.e, 5th term = x + 4
Question 5.
The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m – 1 and median q. Find
(i) p
(ii) q
(iii) the mean of p and q.
Solution:
(i) Mean of 1, 7, 5, 3, 4, 4 is m.
Here n = 6
Question 6.
Find the median for the following distribution:
Solution:
Writing the distribution in cumulative frequency table:
Question 7.
Find the median for the following distribution.
Solution:
Writing the distribution in cumulative frequency table :
Question 8.
Marks obtained by 70 students are given below :
Calculate the median marks.
Solution:
Arranging the variates in ascending order and in c.f. table.
Question 9.
Calculate the mean and the median for the following distribution :
Solution:
Writing the distribution in c.f. table :
Question 10.
The daily wages (in rupees) of 19 workers are
41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35.
Find
(i) the median
(ii) lower quartile
(iii) upper quartile range,
(iv) interquartile range.
Solution:
Arranging the observations in ascending order
21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53
Here n = 19 which is odd.
(i) Median = $$\\ \frac { n+1 }{ 2 }$$ th term = $$\\ \frac { 19+1 }{ 2 }$$ = $$\\ \frac { 20 }{ 2 }$$ = 10th term = 31
Question 11.
From the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range
Solution:
Writing frequency distribution in c.f. table :
Question 12.
For the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
Solution:
Writing the distribution in cumulative frequency (c.f.) table :
We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2, drop a comment below and we will get back to you at the earliest. |
# Fractions Too?
3 teachers like this lesson
Print Lesson
## Objective
SWBAT Find percent increase and decrease using fraction multipliers.
#### Big Idea
What do fractions have to do with percents?
## Launch
5 minutes
• POD
As students enter the room, they will have a seat, take out their Problem of the Day (POD) sheet and begin to work on the question on the SMARTboard. The POD allows students to use MP 3 continually based on the discussions we have about the problem each day.
• Learning Target
The target for the day is also on the SMARTboard each day when students enter the room. The target for today’s lesson is for students to simplify expressions they are given.
To start class today I want to touch base with students and their understanding of percents and the decimal multipliers. I want them to recognize the relationship that exists between percents and decimals.
Model the answer to this question using a percent multiplier and a decimal multiplier.
A mobile phone is reduced by 60% in a sale. Give me an example of what the phone could have originally cost and what it costs now.
## Explore
30 minutes
In this lesson students are going to apply the strategies they used in prior lessons to place fraction multipliers between amounts to show the increase and decrease. They will place the fraction multipliers and operations between the amounts, similar to the previous lesson. Working with a partner, students will make decisions about what operations to place between the amounts. Each pair will receive a copy of Card Set A and be instructed on how to place the cards using the document camera. Using the operations on Card Set D (after they have been cut apart) students will determine how to make the connections between the increase and decrease between the cards and place the operations in the right locations. There should be arrows between all of the cards, horizontally and vertically, as well as diagonally. As pairs work, I will circulate to answer and ask questions to promote student thinking. I want to know, how do they know what to do? If students are struggling to get started I may be able to help them jump start their thinking by reminding them that they can find the increase or decrease then find the two numbers it applies to. Or they can determine the change between two numbers then identify the arrow needed.
## Landing
5 minutes
Now that we have used percent, decimals, and fractions to indicate increase and decrease, I want to see if students can apply them all. We will be completing the post assessment in the next lesson so I want to see where students are with the increase and decrease in case there are questions I can answer for them before we complete the post assessment.
Create two cards with amounts and show the percent increase (or decrease) along with the decimal and fraction multipliers between the amounts. |
# Percent Decrease: Formula & Calculation
Instructor: Norair Sarkissian
Norair holds master's degrees in electrical engineering and mathematics
Many values we come across on a regular basis change frequently. In this lesson we will look at examples when a quantity decreases in value, and how such decreases can be represented using percentages.
## Introduction
A brand new car depreciates in value immediately after its purchase. If a \$17,000 car depreciates by 15%, what does that actually mean? Later in the lesson, we will answer this question. But first, let us examine a couple of key concepts, which may be familiar to us already: percentage and absolute change.
## Review: Percentage
One way of looking at a percentage involves converting a ratio or fraction to a denominator of 100. We perform this conversion by multiplying the ratio by 100%.
For instance, if we have 23 students in a classroom and 10 of them are boys, the ratio of boys in the classroom is 10 out of 23. The percentage of boys in the classroom is determined by performing the multiplication
As for calculating a specific percentage of a given quantity, we first express the percentage either as a fraction or a decimal number, and then multiply it by the given number.
For example, to find 30% of 120, we first convert 30% into a decimal number. 30% actually means '30 out of a hundred', or '30 divided by 100'. Therefore, 30% of 120 is determined as follows:
## Review: Absolute Change
If the value of a quantity changes over time, then absolute change is defined as the difference between the final and initial values of that quantity. Thus, if a number A changes to a new value Z, then the absolute change, C, is given by
If C is a positive number, it means the quantity has increased in value. If C is negative, the quantity has decreased in value. Finally, if C = 0, then Initial Value = Final Value and the quantity has remained constant.
For example, if the price of a laptop computer increases from \$700 to \$750, the absolute change is the difference, or \$50. On the other hand, if the price of the computer drops from \$700 to \$650, the absolute change will be -\$50, where the negative value indicates that the price has decreased.
## Percentage Change
Combining these two key concepts, the percentage change represents the absolute change C as a percentage of the initial value A. Thus, the percentage change in a quantity, p, is defined as
Please note that we use the absolute value of A (or |A|) in the expression, as it allows us to handle the case when the initial value A is negative. In this lesson, we will only consider problems where A is positive, and quite often we will discard the absolute value symbol for that reason.
When we are given the initial and final values, A and Z respectively, the formula we use for evaluating the percentage change directly is
If the quantity has decreased in value, i.e., if Z < A, then p will be a negative percentage, and p is called the percent decrease. The formula for percent decrease is the same as that of percentage change.
## Examples
A gallon of gas was selling for \$3.67 last month. This week, the price of gas has dropped to \$3.57 a gallon. What is the percent decrease in price?
We are given the initial value A = 3.67 and the final value Z = 3.57. The percent decrease is
Therefore, the percent decrease from \$3.67 to \$3.57 is 2.72%.
Here's another example:
Cindy goes to her favorite shoe store and sees a sign that says that every item in the store is on sale, at a 20% discount. The shoes she had planned on buying had an original price of \$120. How much money will she save if she buys the shoes on sale?
We need to determine the amount of (absolute) change in the price.
Using the percent decrease formula, we have
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# Decimal to Binary
## Use our tool to change a regular number to a binary number. It will show you the steps and rules used to make this conversion.
The decimal and binary number systems are fundamental concepts in computer science and mathematics. Converting decimal numbers to binary is a crucial skill, especially for those working in fields related to computing and digital systems. This article will guide you through the process of converting decimal to binary, explain the importance of binary numbers, and provide practical examples to solidify your understanding.
## Understanding Number Systems
Before diving into the conversion process, it’s essential to understand what decimal and binary number systems are.
### The Decimal Number System
The decimal number system, also known as the base-10 system, is the most commonly used number system in everyday life. It consists of ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The value of a number in the decimal system is determined by its digits and their positions. Each position represents a power of 10, increasing from right to left.
### The Binary Number System
The binary number system, or base-2 system, is used extensively in computing and digital electronics. It consists of only two digits: 0 and 1. Each digit in a binary number is called a bit. Similar to the decimal system, the value of a binary number is determined by its digits and their positions, but each position represents a power of 2.
## Why Convert Decimal to Binary?
Converting decimal numbers to binary is essential in various fields, particularly in computer science and digital electronics. Computers operate using binary logic, so understanding binary numbers and their operations is crucial for programming, network communication, and designing digital circuits.
## Conversion Methods
There are several methods to convert decimal numbers to binary. We will explore two common methods: the division-remainder method and the subtraction method.
### Division-Remainder Method
The division-remainder method is a straightforward technique for converting a decimal number to binary. Follow these steps:
1. Divide the Decimal Number by 2: Start with the decimal number you want to convert.
2. Record the Remainder: Write down the remainder (0 or 1) after dividing by 2.
3. Update the Quotient: Update the decimal number to the quotient obtained from the division.
4. Repeat: Repeat the process with the updated quotient until it becomes 0.
5. Reverse the Remainders: The binary representation is obtained by reading the remainders in reverse order.
#### Example: Convert 23 to Binary
1. 23 ÷ 2 = 11 (remainder 1)
2. 11 ÷ 2 = 5 (remainder 1)
3. 5 ÷ 2 = 2 (remainder 1)
4. 2 ÷ 2 = 1 (remainder 0)
5. 1 ÷ 2 = 0 (remainder 1)
Reading the remainders in reverse order, we get 23 in binary as 10111.
### Subtraction Method
The subtraction method involves subtracting the largest possible power of 2 from the decimal number until you reach zero. The steps are as follows:
1. Identify the Largest Power of 2: Find the largest power of 2 less than or equal to the decimal number.
2. Subtract: Subtract this value from the decimal number.
3. Record the Binary Digit: Write a 1 in the binary place value corresponding to the power of 2 used.
4. Repeat: Repeat the process with the new value obtained from the subtraction until it becomes 0.
#### Example: Convert 23 to Binary
1. The largest power of 2 ≤ 23 is 16 (2^4).
2. 23 - 16 = 7. Record a 1 in the 2^4 place.
3. The largest power of 2 ≤ 7 is 4 (2^2).
4. 7 - 4 = 3. Record a 1 in the 2^2 place.
5. The largest power of 2 ≤ 3 is 2 (2^1).
6. 3 - 2 = 1. Record a 1 in the 2^1 place.
7. The largest power of 2 ≤ 1 is 1 (2^0).
8. 1 - 1 = 0. Record a 1 in the 2^0 place.
The binary representation is obtained by placing 0s in all other places. Thus, 23 in binary is 10111.
## Binary Conversion in Programming
In programming, converting decimal to binary is often required for various applications, such as data encoding, cryptography, and algorithm design. Most programming languages provide built-in functions to facilitate this conversion.
### Python Example
Here’s how you can convert a decimal number to binary in Python:
`decimal_number = 23`
`binary_number = bin(decimal_number)`
`print(binary_number)`
The output will be:
`0b10111`
### JavaScript Example
In JavaScript, you can use the `toString()` method:
`let decimalNumber = 23;`
`let binaryNumber = decimalNumber.toString(2);`
`console.log(binaryNumber);`
The output will be:
`10111`
## Practical Applications of Binary Numbers
Binary numbers are foundational in digital systems. Here are some practical applications:
### Computer Architecture
Binary numbers are used in computer architecture to design and implement processors, memory, and other digital components. Each instruction and piece of data is represented in binary.
### Networking
In networking, binary numbers are used to represent IP addresses and subnet masks. For instance, an IPv4 address like 192.168.0.1 is represented in binary as 11000000.10101000.00000000.00000001.
### Data Encoding
Binary encoding is used to store and transmit data efficiently. File formats, such as images, videos, and text files, are encoded in binary.
### Logic Gates and Circuits
Logic gates, the building blocks of digital circuits, operate on binary numbers. Boolean algebra, which underlies logic gates, uses binary values to perform logical operations.
## Common Pitfalls and Tips
### Accuracy in Manual Conversion
When converting manually, ensure you correctly identify and handle powers of 2. Small errors can lead to incorrect results. Double-check each step to avoid mistakes.
### Understanding Binary Arithmetic
Familiarize yourself with binary arithmetic, including addition, subtraction, multiplication, and division. These operations are essential for various computing tasks and can help you better understand binary numbers.
### Use of Tools
Don’t hesitate to use online tools and calculators for conversion, especially when dealing with large numbers. They can save time and reduce errors.
## Conclusion
Understanding how to convert decimal numbers to binary is a fundamental skill in computer science and digital electronics. Whether you use the division-remainder method, the subtraction method, or built-in programming functions, mastering this process is essential for various applications. Binary numbers are integral to computer architecture, networking, data encoding, and digital circuits, making this knowledge indispensable in the modern technological landscape.
By following the steps outlined in this article and practicing with different examples, you can enhance your proficiency in converting decimal numbers to binary, paving the way for a deeper understanding of digital systems.
### Azahar Ahmed
CEO / Co-Founder
I am Azahar Ahmed, a youthful Engineer, Entrepreneur, Digital Marketer, and Motivational speaker native to Nagaon, Assam, India. Originating from a middle-class background, I am the sole son. My accomplishments are indebted to my father, a Teacher, and my mother, formerly a Teacher but now devoted to our well-being. My mother has been my closest ally, and unitedly, my parents have fostered and realized all my aspirations, epitomizing the perfect parents. |
# Surface Area of a Cylinder – Formulas and Examples
The surface area of a cylinder is equal to the area occupied by its surface in three-dimensional space. A cylinder is a three-dimensional geometric figure that has two circular bases that are parallel to each other. We know that a cylinder is composed of two circular bases and a surface that covers the two bases. Therefore, the surface area of the cylinder equals the area of the two circular bases plus the area of the curved surface. The surface area is represented in square units, for example, m².
Here, we will learn about the formula that can be used to calculate the surface area of cylinders. In addition, we will look at some examples in which we will apply this formula to find the answer.
##### GEOMETRY
Relevant for
Learning about the surface area of a cylinder with examples.
See examples
##### GEOMETRY
Relevant for
Learning about the surface area of a cylinder with examples.
See examples
## Formula to find the surface area of a cylinder
The total area of a cylinder is made up of the following parts:
• Area of bases
• Area of curved surface
### Area of bases
The base of a cylinder is a circular figure. Therefore, we can use the formula for the area of a circle. Since we have two circular bases, the total area of the bases is:
$latex \text{Area of bases}=2\pi{{r}^2}$
### Area of curved surface
The area of the curved surface is given by the area covered between the bases of a cylinder that has a radius “r” and a height “h“. This area is also known as the lateral surface area. We can find this area with the following formula:
$latex \text{Lateral area}=2\pi r h$
### Total surface area of cylinder
The total surface area of a cylinder is equal to the sum of the area of the two bases plus the area of the curved surface. Therefore, we have:
where $latex A_{s}$ represents the surface area of the cylinder, r represents the length of the radius, and h represents the length of the height.
## Surface area of a cylinder – Examples with answers
The following examples are solved using the formula for the surface area of cubes. It is recommended that you try to solve the problems yourself before looking at the answer.
### EXAMPLE 1
What is the surface area of a cylinder that has a radius of 5 m and a height of 8 m?
We have the following values:
• Radius, $latex r=5$
• Height, $latex h=8$
We use the formula for surface area with these values:
$latex A_{S}=2\pi r(r+h)$
$latex A_{S}=2\pi (5)(5+8)$
$latex A_{S}=2\pi (5)(13)$
$latex A_{S}=408.4$
The surface area is 408.4 m².
### EXAMPLE 2
A cylinder has a height of 7 m and a radius of 6 m. What is its surface area?
From the question, we can get the following:
• Radius, $latex r=6$
• Height, $latex h=7$
We plug these values into the formula for surface area:
$latex A_{S}=2\pi r(r+h)$
$latex A_{S}=2\pi (6)(6+7)$
$latex A_{S}=2\pi (6)(13)$
$latex A_{S}=490.1$
The surface area is 490.1 m².
### EXAMPLE 3
If a cylinder has a height of 12 m and a radius of 8 m, what is its surface area?
We have the following values:
• Radius, $latex r=8$
• Height, $latex h=12$
We use these values in the formula for surface area:
$latex A_{S}=2\pi r(r+h)$
$latex A_{S}=2\pi (8)(8+12)$
$latex A_{S}=2\pi (8)(20)$
$latex A_{S}=1005.3$
The surface area is 1005.3 m².
### EXAMPLE 4
If a cylinder has a diameter of 6 m and a height of 7 m, what is its surface area?
In this case, we have the diameter instead of the radius. However, we can get the radius simply by dividing the diameter by 2. Therefore, we have:
• Radius, $latex r=3$
• Height, $latex h=7$
Using the formula with these values, we have:
$latex A_{S}=2\pi r(r+h)$
$latex A_{S}=2\pi (3)(3+7)$
$latex A_{S}=2\pi (3)(10)$
$latex A_{S}=188.5$
The surface area is 188.5 m².
### EXAMPLE 5
What is the surface area of a cube that has a diameter of 12 m and a height of 13 m?
Similar to the previous exercise, we divide the diameter by 2 to obtain the radius. Therefore, we have:
• Radius, $latex r=6$
• Height, $latex h=13$
Using these values in the formula for surface area, we have:
$latex A_{S}=2\pi r(r+h)$
$latex A_{S}=2\pi (6)(6+13)$
$latex A_{S}=2\pi (6)(19)$
$latex A_{S}=716.3$
The surface area is 716.3 m².
## Surface area of a cube – Practice problems
Use the following problems to practice using the formula for the surface area of cubes. If you need help with this, you can look at the solved examples above. |
Smartick is an advanced online program that teaches kids math and coding in only 15 min. a day
Nov05
# What Is the Difference between a Sequence and a Pattern?
In today’s post, we are going to look at the difference between a sequence and a pattern, join us!
Like we have seen in an earlier post, a sequence is a string of organized objects following criteria, which can be:
• Ordered (increasing or decreasing).
• Established by a pattern.
Today we are going to concentrate on the sequences established by a pattern, defined by one or more attributes.
What is a pattern? According to Merriam-Webster, it is a form or model proposed for imitation. In the case of sequences, their patterns are models that serve to construct them.
To practice these sequences established by a pattern, Smartick has exercises where the sequence is made up of pictures.
These sequences are always defined by two attributes: shape and color. Knowing the shape pattern and the color pattern, we can create them.
We will look at some examples to understand this better.
#### Example of a sequence with one color and various shapes
In the following exercise, let’s see if we can identify if there is a pattern in the sequence.
In this case, the sequence is the series of pictures that decorate the body of the caterpillar:
Since all of the pictures are the color blue, the color pattern is:
and since the shapes alternate between planet, moon, planet, moon, planet, moon… the shape pattern is:
#### Example of a sequence with various colors and shapes
Now we are going to see an example that is a bit more complicated.
Now the sequence is this series of pictures:
Since the pictures are the colors blue, purple, green, blue, purple, green…the color pattern is:
And since the shapes appear in the order: car, car, airplane, car, car, airplane, ca, car, airplane…the shape pattern is:
Have you seen how easy it is to find patterns in a sequence? Now we are going to practice with another exercise.
#### Example of completing a sequence
Now we are going to identify the patterns in the following sequence to finish it.
Look closely at the colors of the pictures. They are: orange, red, purple, orange, red, purple…and continue this way. Therefore, the color pattern is:
“orange, red, purple”
Now, look at what the shapes are. They are: a basketball net, a racket, a rollerblade, a rollerblade, a basketball net, a racket, a rollerblade, a rollerblade…and continue this way. We can see that the shape pattern is:
Now that you have the patterns of color and shape, you can finish the sequence!
The following picture, since it comes after the color orange, will be red and after the picture of a basketball net, there will be a racket. Therefore, the next picture will be a red racket.
On to the next! Since purple comes after red and a rollerblade comes after a racket, we know that the next picture will be a purple rollerblade.
You’ve almost completed it!
Are you motivated to complete it?
You know what to do, think of the color and shape of the next picture and I’m sure you’ll get it!
I hope that these examples have helped you understand the difference between sequence and pattern well. Don’t forget to share if you liked this post!
And you know, if you want to practice more exercises, go to Smartick and try it free. |
# 2.8 Fractions - 04
Page 1 / 1
## Memorandum
8. a) a) $\frac{2}{6}$
b) $\frac{1}{3}$
c) $\frac{3}{9}$
d) $\frac{4}{\text{12}}$
b) Every one eats the same amount of food. / Fractions are the same.
9. $\frac{2}{3}$ = $\frac{6}{9}$
$\frac{4}{5}$ = $\frac{\text{80}}{\text{100}}$
$\frac{\text{18}}{\text{20}}$ = $\frac{9}{\text{10}}$
$\frac{\text{75}}{\text{100}}$ = $\frac{3}{4}$
$\frac{2}{3}$ = $\frac{\text{16}}{\text{24}}$
$\frac{\text{25}}{\text{30}}$ = $\frac{5}{6}$
## Activity: fractions [lo 1.4.1, lo 1.11, lo 2.1.5]
8. Four learners have been rewarded with a chocolate for their good work. They don’t eat it up immediately, but only the section that has been coloured in.
i)
ii)
iii)
iv)
1. What fraction does each one eat?
i) Carli: ___________________
ii) Peter-John: ______________
iii) Kayla: __________________
iv) Vusi: ___________________
b) What do you notice?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
8.2 Do you still remember?
In our example $\frac{2}{6}=\frac{1}{3}=\frac{3}{9}=\frac{4}{\text{12}}$
We call these fractions equivalent fractions.
Equivalent fractions are thus the same quantity or equal to each other
8.3 TAKE NOTE:
To form an equivalent fraction, you must multiply or divide the numerator AND denominator by THE SAME NUMBER.
e.g.
2 × 4 5 × 4
=
8 20
;
18 6 24 6
=
3 4
9. Join the fraction in column A with its equivalent in column B:
A B $\frac{2}{3}$ $\frac{3}{4}$ $\frac{4}{5}$ $\frac{6}{9}$ $\frac{\text{18}}{\text{20}}$ $\frac{9}{\text{10}}$ $\frac{\text{75}}{\text{100}}$ $\frac{5}{6}$ $\frac{2}{3}$ $\frac{\text{16}}{\text{24}}$ $\frac{\text{25}}{\text{30}}$ $\frac{\text{80}}{\text{100}}$
## Assessment
Learning Outcome 1: The learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems.
Assessment Standard 1.4: We know this when the learner recognises and uses equivalent forms of the rational numbers listed above, including:
1.4.1: common fractions;
1.11: recognises, describes and uses:
Learning Outcome 2: The learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills.
Assessment Standard 2.1: We know this when the learner investigates and extends numeric and geometric patterns looking for a relationship or rules, including patterns:
2.1.5: represented in tables.
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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# 9.3 Modeling Linear Relationships
If you know a person’s pinky (smallest) finger length, do you think you could predict that person’s height? Imagine collecting data on this and constructing a scatterplot of the points on graph paper. Then draw a line that appears to “fit” the data. For your line, pick two convenient points and use them to find the slope of the line. Find the y-intercept of the line by extending your line so it crosses the y-axis. Using the slopes and the y-intercepts, write your equation of “best fit.” According to your equation, what is the predicted height for a pinky length of 2.5 inches? You have just started the process of linear regression.
# Linear Regression
Data rarely perfectly fit a straight line, but we can be satisfied with rough predictions. Typically, you have a set of data whose scatter plot appears to “fit” a straight line. This is called a Line of Best Fit or Least-Squares Line. This process of fitting the best-fit line is called linear regression
The equation of the regression line is ŷ =a+bx
The ŷ is read “y hat” and is the estimated value of y. It is the value of y obtained using the regression line. It may or may not be equal to values of y observed from the data.
The sample means of the x values and the y values are and , respectively. The best fit line always passes through the point .
The slope, b can be written as where sy = the standard deviation of the y values and sx = the standard deviation of the x values. r is the correlation coefficient, which is discussed in the next section.
The y-intercept, a, can then be calculated by using the slope, and means of x and y.
Example
Recall our example:
A random sample of 11 statistics students produced the following data, where x is the third exam score out of 80, and y is the final exam score out of 200. Can you predict the final exam score of a random student if you know the third exam score?
Figure 9.7 (repeat): Third and Final Exam Scores Data
x (third exam score) y (final exam score)
65 175
67 133
71 185
71 163
66 126
75 198
67 153
70 163
71 159
69 151
69 159
Consider the following diagram. Each point of data is of the the form (x, y) and each point of the line of best fit using least-squares linear regression has the form (x, ŷ).
SCUBA divers have maximum dive times they cannot exceed when going to different depths. The data in the figure below show different depths with the maximum dive times in minutes. Use your calculator to find the least squares regression line and predict the maximum dive time for 110 feet.
Figure 9.11: SCUBA Diver Stats
X (depth in feet) Y (maximum dive time)
50 80
60 55
70 45
80 35
90 25
100 22
# Understanding Slope
The slope of the line, b, describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English.
INTERPRETATION: The slope of the best-fit line tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average.
Example
[Previous Example Continued]
The slope of the line is b = 4.83.
Interpretation: For a one-point increase in the score on the third exam, the final exam score increases by 4.83 points, on average.
# Understanding the Y-Intercept
The y-intercept of the line, a, can tell us what we would predict the value of y to be when x is 0. This may make sense in some cases, but in many it may not make sense for x to be equal to 0, therefore the y intercept may not be useful.
Example
[Previous Example Continued]
The y-intercept of the line is –173.51
Interpretation: In this context it does not really make sense for x to be 0 (unless a student did not take the exam or try at all). Therefore our y intercept does not make sense.
# Prediction
The next, and most useful step in regression is to actually use that equation to predict future values of y.
Recall in our example we have examined the scatterplot and found the correlation coefficient and coefficient of determination. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction.
Example
[Previous Example Continued]
Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores (x-values) range from 65 to 75. Since 73 is between the x-values 65 and 75, substitute x = 73 into the equation. Then:
What would you predict the final exam score to be for a student who scored a 66 on the third exam?
What would you predict the final exam score to be for a student who scored a 90 on the third exam?
Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows:
ŷ = 72.5 + 2.8x
What would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week?
### Image References
Figure 9.10: Kindred Grey via Virginia Tech (2020). “Figure 9.10” CC BY-SA 4.0. Retrieved from https://commons.wikimedia.org/wiki/File:Figure_9.10.png . Adaptation of Figure 12.11 from OpenStax Introductory Statistics (2013) (CC BY 4.0). Retrieved from https://openstax.org/books/introductory-statistics/pages/12-3-the-regression-equation
definition |
Current math problems
Looking for Current math problems? Look no further! We can solve math word problems.
The Best Current math problems
Here, we will be discussing about Current math problems. Algebra can be a helpful tool for solving real-world problems. In many cases, algebraic equations can be used to model real-world situations. Once these equations are set up, they can be solved to find a solution that meets the given constraints. This process can be particularly useful when solving word problems. By taking the time to carefully read the problem and identify the relevant information, it is often possible to set up an equation that can be solved to find the desired answer. In some cases, multiple equations may need to be written and solved simultaneously. However, with a little practice, solving word problems using algebra can be a straightforward process.
Solving natural log equations requires algebraic skills as well as a strong understanding of exponential growth and decay. The key is to remember that the natural log function is the inverse of the exponential function. This means that if you have an equation that can be written in exponential form, you can solve it by taking the natural log of both sides. For example, suppose you want to solve for x in the equation 3^x = 9. Taking the natural log of both sides gives us: ln(3^x) = ln(9). Since ln(a^b) = b*ln(a), this reduces to x*ln(3) = ln(9). Solving for x, we get x = ln(9)/ln(3), or about 1.62. Natural log equations can be tricky, but with a little practice, you'll be able to solve them like a pro!
How to solve perfect square trinomial? This is a algebraic equation that can be written in the form of ax2 + bx + c = 0 . If the coefficient of x2 is one then we can use the factoring method to solve it. We will take two factors of c such that their product is equal to b2 - 4ac and their sum is equal to b. How to find such numbers? We will use the quadratic formula for this. Now we can factorize the expression as (x - r1)(x - r2) = 0, where r1 and r2 are the roots of the equation. To find the value of x we will take one root at a time and then solve it. We will get two values of x, one corresponding to each root. These two values will be the solutions of the equation.
Natural log equations can be tricky to solve, but there are a few tried-and-true methods that can help. . This formula allows you to rewrite a natural log equation in terms of a different logarithmic base. For example, if you're trying to solve for x in the equation ln(x) = 2, you can use the change of base formula to rewrite it as log2(x) = 2. Once you've rewriting the equation in this form, it's often easier to solve. Another approach is to use substitution. This involves solving for one variable in terms of the other and then plugging that value back into the original equation. For instance, if you're trying to solve the equation ln(x+1) - ln(x-1) = 2, you could start by solving for ln(x+1) in terms of ln(x-1). Once you've done that, you can plug that new value back into the original equation and solve for x. With a little practice, solving natural log equations can be a breeze. |
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# Difference between revisions of "2010 AMC 12B Problems/Problem 20"
## Problem
A geometric sequence $(a_n)$ has $a_1=\sin x$, $a_2=\cos x$, and $a_3= \tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\cos x$?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
## Solution
By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$. Since $\tan x=\frac{\sin x}{\cos x}$, we can rewrite this as $\cos^3x=\sin^2x$.
The common ratio of the sequence is $\frac{\cos x}{\sin x}$, so we can write
$$a_1= \sin x$$ $$a_2= \cos x$$ $$a_3= \frac{\cos^2x}{\sin x}$$ $$a_4=\frac{\cos^3x}{\sin^2x}=1$$ $$a_5=\frac{\cos x}{\sin x}$$ $$a_6=\frac{\cos^2x}{\sin^2x}$$ $$a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}$$ $$a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}$$
Since $\cos^3x=\sin^2x=1-\cos^2x$, we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$, which is $a_8$ , making our answer $8 \Rightarrow \boxed{E}$.
## Solution 2
Notice that the common ratio is $r=\frac{\cos(x)}{\sin(x)}$; multiplying it to $\tan(x)=\frac{\sin(x)}{\cos(x)}$ gives $a_4=1$. Then, working backwards we have $a_3=\frac{1}{r}$, $a_2=\frac{1}{r^2}$ and $a_1=\frac{1}{r^3}$. Now notice that since $a_1=\sin(x)$ and $\a_2=cos(x)$ (Error compiling LaTeX. ! LaTeX Error: Command \_ unavailable in encoding OT1.), we need $a_1^2+a_2^2=1$, so $\frac{1}{r^6}+\frac{1}{r^4}=\frac{r^2+1}{r^6}=1\implies r^2+1=r^6$. Dividing both sides by $r^2$ gives $1+\frac{1}{r^2}=r^4$, which the left side is equal to $1+\cos(x)$; we see as well that the right hand side is equal to $a_8$ given $a_4=1$, so the answer is $\boxed{E}$. - mathleticguyyy
2010 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions |
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